The Lucas-Uzawa Models: Closed-Form Solutions

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					                            The Lucas-Uzawa Models:
                             Closed-Form Solutions
                                         (--Preliminary Version--)
                                             9th of May 2011


                                        Solomon M. Antoniou

                                             SKEMSYS
                                  Scientific Knowledge Engineering
                                     and Management Systems
                             37 oliatsou Street, Corinthos 20100, Greece
                                             solomon@otenet.gr
                                        solomon_antoniou@yahoo.com


                                                 Abstract
We present explicit solutions in a number of Lucas-Uzawa Models. The models
are solved without dimensional reduction, using two different methods. The first
method uses a procedure similar to the dimensional reduction. However in our
method we do not consider solutions along the balanced growth path. The second
method has recently appeared in the literature. However, in our own calculations
the intermediate steps appear explicitly in an easy to follow algorithm. Some of
the results we derive exhibit differences compared to the results already found.
The solution procedure of models with externalities uses quite different techniques
to those known so far. The closed-form solutions of the models with externalities
appear for the first time in the literature.
Keywords: Economic Dynamics, Lucas-Uzawa model, Closed-Form Solutions,
Special Functions, Hypergeometric Functions.
-------------------------------------
The paper is available from: www.docstoc.com/profile/solomonantoniou




                                                                                1
1. Introduction
Modern Economic Growth Theory has received considerable attention after its
formulation by Romer [26] and [27], and Lucas [17]. A central role is played by
the Uzawa-Lucas Model (Uzawa [33] and Lucas [17]). The reader may consult
any of the well-known text books on the subject, like Acemoglu [2], Aghion and
Howitt [3], Barro and Sala-i-Martin [5], Greiner, Semmler and Gong [13] and
Romer [28]. We should also mention Greiner and Semmler [12], Uzawa [34] and
Xepapadeas [36] who consider a number of environmental issues examined within
the Lucas-Uzawa models.
There is a number of papers which claim either they have found closed-form
solutions or they have simplified considerably the equations of motion, like
Benhabib and Perli [6], Bethmann [7], Bucekkine and Ruiz-Tamarit [8], Caballè
and Santos [9], Hiraguchi [14], Mattana [19] and [20], Moro [21] and Mulligan
and Sala-i-Martin [23] among others. Most of the solution methods first use
dimensional reduction and then the model is solved along the balanced growth
path (usually known as BGP). The only exception is the paper by Bucekkine and
Ruiz-Tamarit [8]. In this paper the authors use a method which does not use
dimensional reduction.
The mathematical tool in analysing the dynamics of the Lucas-Uzawa models is
the Pontryagin optimization method. The reader may consult any of the known
references of this technique, like Malliaris and Brock [18], Pontryagin et. al. [24]
or Seierstad and Sydsaeter [30].
We use two different methods in obtaining closed-form solution to the models we
consider. The first method uses a procedure similar to the dimensional reduction.
However in our method we do not consider solutions along the balanced growth
path. The second method uses the procedure introduced by Bucekkine and Ruiz-




                                                                                 2
Tamarit [8]. However our line of reasoning is more transparent and will offer a
simple method of solving this kind of models.
The solution procedure of models with externalities uses quite different techniques
to those known so far. The closed-form solutions of the models with externalities
appear for the first time in the literature.
The closed-form solutions of the models considered is based on the evaluation of
the two integrals
              t                                    t
                   ωs        µs        ν                ϕt
              ∫e        (e        − C 0 ) ds and   ∫e        2 F ( − ν,
                                                                1         η ;1 + η ; C 0 e − µ t ) dt
              0                                    0

These two integrals are evaluated in Appendices A and B respectively, expressed
in terms of hypergeometric functions. The first one is expressed in terms of the
hypergeometric function 2 F1 () (Appendix A, equation (A.16)), while the second is

expressed in terms of the generalized hypergeometric function 3 F2 [] (Appendix
B, equation (B.11)). The evaluation of the first integral is based on the integral
representation of the hypergeometric function, while the second on an integral
formula of the generalized hypergeometric function, appearing in the existing
literature.
We should stress the fact that the paper does not examine in general growth
models. It only serves to introduce some standard methods in solving these
models. Especially the relation between BGP and exact solutions will be examined
in an expanded version of the paper (under preparation).
The paper is organized as follows:
In Section 2 we consider a simple model which is solved explicitly following two
methods. The first method of solution uses a type of dimensional reduction at the
beginning and then introducing two auxiliary functions, we find that these
functions satisfy a system of ordinary differential equations. After solving the
system, we can evaluate the function of the physical capital, the control variables
and then the human capital function. The second method uses the same steps


                                                                                                        3
followed by Bucekkine and Ruiz-Tamarit. However all the intermediate steps are
made explicit in an easy to follow way. There is however a difference to the
expression for the human capital. According to our own calculations, the human
capital is expressed in terms of the generalized hypergeometric function 3 F2 [] .
In Section 3 we consider a simple model with externalities. The solution
procedures we follow are quite different compared to the methods considered in
the model of Section 2 and appear for the first time in the literature.
In Section 4 we solve the model considered by Bucekkine and Ruiz-Tamarit which
is again solved using two methods.
In section 5 we consider the model of Bucekkine and Ruiz-Tamarit equipped with
externalities. The methods used in Section 3 are used again in this model. We thus
conclude that the methods introduced in Section 3 can be applied to more
complicated models.
In Section 6 we consider and solve explicitly a model introduced by Ruiz-Tamarit
and Sánchez-Moreno [29], on optimal regulation in a natural-resource-based
economy.
Finally in section 7 we consider some issues on optimal fiscal policy along the
lines of reasoning by Gómez [11].
A word of caution: The various parameters introduced in the text, like µ, ν, ϕ etc.,
are defined in different ways in the various Sections (or even in the various
subsections).


2. A simple Lucas-Uzawa model
2.0. The model.
We work with a Cobb-Douglas production function

                Y = A K α (u H)1−α
where


                                                                                     4
        A: represents the technological level of the economy, considered to be
             constant
        K: represents the stock of the physical capital
        H: represents the stock of the human capital
        u: is the portion of the human capital devoted to the production of output Y
The portion of the human capital devoted to the production of more human capital
will then be 1− u . This leads to the equation of motion for the human capital:
                  H( t ) = γ (1 − u ) H
where γ is a constant coefficient.
The equation of motion of the physical capital is derived by the equation
K = Y − C where C is the consumption, or using the expression for Y,

                  K = A K α (u H)1−α − C
We consider the dynamic optimization problem
                        ∞
                             C( t )1−σ − 1 − ρ t
                  max ∫                   e      dt                              (2.1)
                         0
                                1− σ

subject to

        K ( t ) = K ( t ) α ( u ( t ) H( t ))1−α − C( t )                        (2.2)

        H( t ) = γ (1 − u ( t )) H( t )                                          (2.3)
with initial conditions
        K (0) = K 0 , H(0) = H 0                                                 (2.4)
where
        C( t ) ≥ 0 , u ( t ) ∈ [ 0, 1 ] , K ( t ) ≥ 0 , H( t ) ≥ 0               (2.5)
In equation (2.1) we work with an isoelastic utility function and maximize the
integral, where σ is the instantaneous elasticity of substitution and ρ is the
instantaneous discount rate. In equation (2.2) we have considered the case A = 1
for simplicity.



                                                                                     5
2.1. The equations of motion.
The current value Hamiltonian is given by

       H c (C, u , H, K ) =

                  C( t )1−σ − 1
                =               + λ K [ K ( t ) α (u ( t ) H( t ))1−α − C( t ) ] +
                     1− σ
                  + λ H [ γ (1 − u ( t )) H( t ) ]                                   (2.6)

where λ K and λ H are the costate variables corresponding to K and H
respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method. The reader may consult any of the known references of this
technique, like Malliaris and Brock [18], Pontryagin et. al. [24] or Seierstad and
Sydsaeter [30].
The first order conditions read

       ∂H c
            = 0 ⇔ C − σ e −ρ t − λ K = 0                                             (2.7)
        ∂C

       ∂H c
            = 0 ⇔ −λ H γ H + λ K [ (1 − α) K α (u H) − α H ] = 0                     (2.8)
        ∂u
We also have the two Euler equations

       ∂H c
            + λ K = 0 ⇔ α λ K K α −1 (u H)1− α + λ K = 0                             (2.9)
       ∂K

       ∂H c
            + λH = 0 ⇔
       ∂H

       ⇔ λ H γ (1 − u ) + λ K (1 − α) K α ( u H)1− α H − 1 + λ H = 0                 (2.10)
the dynamic constraints

       K = K α ( u H)1−α − C , K (0) = K 0                                           (2.11)



                                                                                         6
       H = γ (1 − u ) H , H(0) = H 0                                         (2.12)
and the transversality conditions

       lim λ K K e − ρ t = 0                                                 (2.13)
       t →∞

       lim λ H H e − ρ t = 0                                                 (2.14)
       t →∞

2.2. First method of solution.
This method uses a procedure similar to the dimensional reduction. The derived
system however is not solved along the balanced growth path (BGP). Considering
two auxiliary functions, we get a system of decoupled first order ordinary
differential equations, which, when solved, determine the physical capital, the
control variable u and finally the human capital.
2.2.1. Simplification of the dynamical equations.
Taking logarithms and differentiation with respect to time, we obtain from
equation (2.7) that
              C     λ
       −σ       −ρ = K                                                       (2.15)
              C     λK
Similarly, taking logarithms and differentiation with respect to time, we obtain
from equation (2.8) that
       λK   K  u  H λ
          +α −α −α = H                                                       (2.16)
       λK   K  u  H λH
From equation (2.9) we obtain
       λK
          = − α K α −1 (u H)1−α                                              (2.17)
       λK
From equation (2.10) we obtain
       λH                                       λ    1
          = − γ (1 − u ) − (1 − α) K α (u H)1− α K ⋅                         (2.18)
       λH                                       λH H
From equation (2.11) we obtain



                                                                                   7
       K                       C
         = K α −1 ( u H)1− α −                                               (2.19)
       K                       K
From equation (2.12) we obtain
       H
         = γ (1 − u )                                                        (2.20)
       H
We introduce a function named Y defined by

       Y = K α ( u H )1− α                                                   (2.21)
Equations (2.17)-(2.19) can be expressed in terms of the function Y as follows:
       λK      Y
          = −α                                                               (2.22)
       λK      K

       λH                         λ Y
          = − γ (1 − u ) − (1 − α) K ⋅                                       (2.23)
       λH                         λH H

       K Y C
        = −                                                                  (2.24)
       K K K
From equation (2.8) we find that
       λK Y  γ
         ⋅ =     u                                                           (2.25)
       λH H 1− α
Because of the previous relation, we can simplify further equation (2.23):
       λH
          = −γ                                                               (2.26)
       λH
Combining equations (2.15) and (2.22) we find an expression for the ratio
       C α Y ρ
        = ⋅ −                                                                (2.27)
       C σ K σ

                                                         λK K H    λ
Equation (2.16), upon substituting the expressions for     , ,  and H given
                                                         λK K H    λH
by (2.22), (2.24), (2.20) and (2.26) respectively, we obtain the equation
       u γ (1 − α)      C
         =         +γu−                                                      (2.28)
       u     α          K



                                                                                 8
2.2.2. Auxiliary Functions and their Differential Equations.
We now introduce two more functions U and Z defined by
            Y
       U=                                                                    (2.29)
            K
and
            C
       Z=                                                                    (2.30)
            K
respectively.
We shall establish a system of two ordinary differential equations satisfied by
these two functions.
Taking logarithms and differentiation of the defining equations (2.29) and (2.30),
we obtain the equations
       U Y K
        = −                                                                  (2.31)
       U Y K
and
       Z C K
        = −                                                                  (2.32)
       Z C K
respectively.
We can express the right hand sides of the two previous equations in terms of the
functions U and Z. From equation (2.24) we obtain
       K
         =U−Z                                                                (2.33)
       K
From equation (2.27) we obtain
       C α    ρ
        = ⋅U−                                                                (2.34)
       C σ    σ
                                   Y
We now have to express the ratio     in terms of U and Z. Taking logarithms and
                                   Y
differentiation of (2.21), we obtain the equation




                                                                                  9
        Y    K         u          H
          = α + (1 − α) + (1 − α)                                               (2.35)
        Y    K         u          H
                          K u    H
Substituting the ratios    , and   given by (2.33), (2.28) and (2.20)
                          K u    H
respectively into the previous equation, we obtain
        Y        γ (1 − α)
          =αU−Z+                                                                (2.36)
        Y            α
Therefore we obtain from (2.31) and (2.32), using (2.36), (2.33) and (2.34), the
system of two equations
        U γ (1 − α)
          =         − (1 − α) U                                                 (2.37)
        U     α
and
        Z      α−σ    ρ
          = Z+     U−                                                  (2.38)
        Z       σ     σ
Equation (2.37) can be solved by separation of variables. The solution is given by

             γ    eν t
        U=     ⋅ νt                                                             (2.39)
             α e − C0

where
             γ (1 − α)                 γ
        ν=             and C 0 = 1 −                                            (2.40)
                 α                   α U(0)
Using the expression (2.39) for the function U, equation (2.38) is converted into
the equation
            α−σ
                        ν eν t    ρ
        Z−            ⋅ νt        −  Z = Z2                                    (2.41)
            σ (1 − α) e − C       σ
                               0   
This is a Bernoulli differential equation, solved under the substitution

        X = Z −1                                                                (2.42)
Equation (2.41) is converted in terms of X into the linear differential equation




                                                                                    10
            α−σ        ν eν t    ρ
        X+           ⋅ νt       −  X = −1                                                (2.43)
            σ (1 − α) e − C      σ
                              0   
The integrating factor of the above equation is

        I( t ) = e ω t (e ν t − C 0 ) ζ                                                    (2.44)
where
                  ρ                   α−σ
        ω=−             and ζ =                                                            (2.45)
                  σ                  σ (1 − α)
Multiplying (2.43) by the integrating factor, we obtain the equation
        d
           ( I( t ) X ) = − e ω t ( e ν t − C 0 ) ζ
        dt
which, upon integration in the interval [ 0, t ] , gives
                                          t
        I( t ) X − I(0) X(0) = − ∫ e ω s (e ν s − C 0 ) ζ ds                               (2.46)
                                          0

The integral on the right hand side of the previous equation is calculated in
Appendix A. We have, using the notation of this section
        t
             ωs                                 1
        ∫e        (e ν s − C 0 ) ζ ds = −         Ω( t )                                   (2.47)
        0
                                               ην

where

        Ω ( t ) = e − ην t F( − ζ , η ;1 + η ; C 0 e − ν t ) − F( − ζ , η ;1 + η ; C 0 )   (2.48)
and
                  ω+ νζ
        η=−                                                                                (2.49)
                    ν
Therefore we obtain from (2.46) the following expression for the function X:
                                           1
              (1 − C 0 ) ζ X(0) +            Ω( t )
                                          ην
        X=                                                                                 (2.50)
                      e ω t (e ν t − C 0 ) ζ
We thus have that the function Z related to X by (2.42) is given by



                                                                                               11
                e ω t (e ν t − C 0 ) ζ
        Z=                                                                            (2.51)
                                  1
           (1 − C 0 ) ζ X(0) +         Ω( t )
                                 ην
Since from (2.47) we have upon differentiation with respect to t the relation
                                       d 1           
        − e ω t (e ν t − C 0 ) ζ =         ην Ω( t ) 
                                                                                    (2.52)
                                       dt            
equation (2.51) is written as
                W
        Z=−                                                                           (2.53)
                W
where
                                         1
        W = (1 − C 0 ) ζ X(0) +            Ω( t )                                     (2.54)
                                        ην
On the other hand, equation (2.39) can be expressed as
              1 V
        U=       ⋅                                                                    (2.55)
             1− α V
where

        V = eν t − C0                                                                 (2.56)

2.2.3. Equation for the physical capital.
Equation (2.33) can be written, because of (2.55) and (2.53), as
        K   1 V W
          =   ⋅ +                                                                     (2.57)
        K 1− α V W
The above equation can be integrated once, providing us with the following
expression for the physical capital
                     1
        K = D0    V 1−α     W
or
                                          1
                           νt                             ζ             1        
        K ( t ) = D 0 (e        − C0   ) 1−α (1 − C
                                                     0)       X(0) +      Ω( t )    (2.58)
                                                                       ην        



                                                                                          12
where
                        K0                                         K0
        D0 =         1
                                       =                 1
                                                                                                      (2.59)
                    1− α                   (1 − C 0     1− α (1 − C        ζ
                                                                                X ( 0) +
                                                                                            1     
                                                                                              Ω(0)
               V(0)           W ( 0)                  )               0)
                                                                                          ην     
2.2.4. Equation for the control variable u.
From equation (2.28), using (2.30) and (2.53), we derive the equation
        u γ (1 − α)      W
          =         +γu+
        u     α          W
which is equivalent to the equation
            γ (1 − α) W 
        u −
            α        +  u = γ u2                                                                    (2.60)
                      W
This is a Bernoulli differential equation, which under the substitution

        y = u −1                                                                                      (2.61)
becomes a linear first order differential equation
           γ (1 − α) W 
        y+
           α        + y = −γ                                                                        (2.62)
                     W
The previous equation admits the integrating factor

        J(t) = e ν t W                                                                                (2.63)
Multiplying equation (2.62) by the integrating factor, we obtain the equation
        d
           ( J ( t ) y) = − γ e ν t W
        dt
which, upon integration in the interval [0, t ] , gives us
                                           t
        J ( t ) y − J (0) y(0) = − γ ∫ e ν t W (s) ds                                                 (2.64)
                                           0

Let
                   t
                        νt
        X( t ) =   ∫e        W (s) ds                                                                 (2.65)
                   0




                                                                                                          13
Using the expression (2.54), we find that
                    t
                         νs
        X( t ) =    ∫e        W (s) ds =
                    0

                                                                         t
                               1                       
        = (1 − C 0 ) ζ X(0) −    F(−ζ, η ;1 + η ; C 0 ) ∫ e ν s ds +
                              ην                       0
                              t
                       1           ν (1− η) s
                    +         ∫e                F(−ζ, η ;1 + η ; C 0 e −ν s ) ds       (2.66)
                      ην      0

The last integral is evaluated in Appendix B. We get from (B.16), using the
notation of this section,
         t
               ϕt
         ∫e         F (− ζ, η ;1 + η ; C 0 e − ν t ) dt =
         0

              1  ϕt     p, − ζ , η                       p, − ζ , η     
      =−        e 3 F2              C 0 e − ν t  − 3 F2              C0         (2.67)
             νp        p + 1, η + 1                     p + 1, η + 1    
where
                                                  ϕ
        ϕ = ν(1 − η) and p = −                                                        (2.68)
                                                  ν
Therefore we have the following expression for the function X( t ) :
                    t
                         νs
        X( t ) =    ∫e        W (s) ds =
                    0

                               1                         eν t − 1 
        = (1 − C 0 ) ζ X(0) −    F(−ζ, η ;1 + η ; C 0 )           −
                              ην                        ν 
                                                                    

               1     ϕt     p, − ζ , η                       p, − ζ , η     
        −           e 3 F2              C 0 e − ν t  − 3 F2              C0     (2.69)
             ην 2 p        p + 1, η + 1                     p + 1, η + 1    
We then obtain from (2.64)
               J(0) y(0) − γ X( t )
        y=                                                                             (2.70)
                      J(t )
and then


                                                                                           14
                          J( t )
          u(t) =                                                            (2.71a)
                   J(0) y(0) − γ X( t )
or, comparing (2.63) and (2.65),
                          X( t )
          u(t) =                                                            (2.71b)
                   J(0) y(0) − γ X( t )

2.2.5. Equation for the human capital.
Equation (2.20) for the human capital, because of the previous expression (2.71),
takes on the form
          H          − γ X( t )
            =γ+
          H     J(0) y(0) − γ X( t )
or even
          H     S
            =γ+                                                              (2.72)
          H     S
where
        S( t ) = J (0) y(0) − γ X( t )                                       (2.73)
Equation (2.72) gives upon integration
                   H0
        H( t ) =        S( t ) e γ t                                         (2.74)
                   S(0)

2.3. Second method of solution.
The second method of solution follows the procedure introduced by Bucekkine
and Ruiz-Tamarit [8]. The intermediate steps however are more transparent and
the overall method provides a standard algorithmic procedure. On the other hand
there is a difference in the human capital evaluation. In our calculation the human
capital is expressed in terms of generalized hypergeometric functions.
2.3.1. Equations for the control variables.
Solving equation (2.7) with respect to C, we obtain




                                                                                 15
                                                   1
                ρ          −
       C = exp  − t  (λ K ) σ                                                                                   (2.75)
                σ 
Solving equation (2.8) with respect to u, we obtain
                               1                   1
                           −
           γ                 α    λK           α   K
       u =                       
                                   λ          
                                                                                                                 (2.76)
          1− α                    H                H

We also get from the previous relation that
                                               1− α                1− α
                                           −
                         γ                    α       λK        α
       (u H)1−α        =                             
                                                       λ      
                                                                         K1−α                                    (2.77)
                        1− α                          H     
we shall need later on.
2.3.2. Equations for the costate variables.
Equation (2.10), because of (2.76) and (2.77), takes on the form
       λH + γ λH = 0                                                                                              (2.78)
Equation (2.78) admits the general solution

       λ H = λ H ( 0) × e − γ t                                                                                   (2.79)
Equation (2.9), because of (2.77), becomes
                                       1− α                 1− α
                                   −
               γ                      α       λK         α
       λK + α                                
                                               λ       
                                                                  λK = 0                                         (2.80)
              1− α                            H      
This equation, written as (because of (2.79))
                                           1− α                        1− α                    1− α
                                       −                           −
                                                                                                              1
                 γ                        α                           α          γ (1 − α)  α
       λK   = −α                                 (λ H (0))                  exp            t      (λ K   )α
                1− α                                                             α          
is an equation with separable variables with general solution
                    1− α                                           1− α
                −                                             −
                     α      α γ                                   α          γ (1 − α) 
       (λ K )              =       λ H (0)                              exp           t + C                   (2.81)
                            γ 1− α                                           α         
where C is a constant to be determined from the initial conditions. For t = 0 we
obtain from the above equation


                                                                                                                      16
                          1− α                                       1− α
                      −                                          −
                           α         α γ                            α
       (λ K (0))                 =          λ H ( 0)                      +C
                                     γ 1− α          
from which we determine the constant C. Therefore equation (2.81) takes the final
form
                    1− α                                       1− α
                −                                          −
                     α           α γ                          α          γ (1 − α) 
       (λ K )              =            λ H ( 0)                    exp           t +
                                 γ 1− α                                  α         
                                                    1− α                                     1− α
                                                −                                        −
                                                     α      α γ                             α
                                  + (λ K (0))              −       λ H (0)                        (2.82)
                                                            γ 1− α         
The previous equation can also be written as
                    1− α                                       1− α
                −                                          −
                     α           α γ                          α
       (λ K )              =            λ H ( 0)                    ×
                                 γ 1− α          
                                                              1− α    
                                                             −
                           γ (1 − α)  γ  1 − α λ K (0)  α         
                    × exp                 γ ⋅ λ (0) 
                                      t +                        − 1
                           α          α          H                
                                                                      
or
                    1− α                                       1− α
                −                                          −
                                 α γ                          α
       (λ K )        α     =            λ H ( 0)                    × (e µ t − C 0 )              (2.83)
                                 γ 1− α          
where we have introduced the notation
            γ (1 − α)
       µ=                                                                                           (2.84)
                α
and
                                                         1− α
                                                     −
                      γ  1 − α λ K ( 0)                 α
       C0 = 1 −               ⋅                                                                   (2.85)
                      α  γ λ H ( 0) 
                                        
We thus obtain from (2.83) the following expression for the costate variable λ K :




                                                                                                        17
                                    α                                        α
               γ γ               1−α                                  −
                                                          µt                1− α
       λK   =                          λ H ( 0) × ( e        − C0 )                     (2.86)
              1− α  α 

2.3.3. Equation for the physical capital.
We now consider equation (2.11). This equation, because of (2.77) and (2.75),
takes the form
                                1− α          1− α
                            −                                           1
          γ                    α      λK  α                     −
                                                                        σ        ρ 
       K=                            
                                       λ          K − (λ K )             exp  − t    (2.87)
         1− α                         H                                      σ 
In order to simplify the above expression, we need to find the ratio of the costate
variables. Using the expressions (2.86) and (2.79), we obtain
                   1− α                      1− α
         λK       α      γ γ              α            eµ t
        
        λ     
                         =                       ×                                     (2.88)
         H               α 1− α                     eµ t − C0
Equation (2.87), taking into account (2.88) and (2.86), takes on the form
                                                                             α
            1    µ eµ t                                        ρ 
       K−                 K = − A (e µ t − C 0 ) σ (1−α ) exp  − t                      (2.89)
          1 − α eµ t − C0                                      σ 

where A is a constant defined by
                             α                                  1
                     −                                      −
         γ              σ (1− α )      γ                    σ
       A=                            ⋅     λ H (0)                                    (2.90)
         α                            1− α         
We come now to the solution of (2.89), which is a linear first order differential
equation, with integrating factor
                                                                                   1
                       1      µ eµ t                        −
       I( t ) ≡ exp  −                  dt  = (e µ t − C 0 ) 1−α
                     1 − α ∫ eµ t − C
                                                                                          (2.91)
                                            
                                      0    
Multiplying equation (2.89) by the integrating factor I( t ) , we find
                                                               α        1
                                                     −
       d                           µt       σ (1− α ) 1− α      ρ 
          (K ( t ) I( t )) = − A (e − C 0 )                exp  − t 
       dt                                                       σ 




                                                                                              18
and integrating in the interval [ 0, t ] , we arrive at the equation
                                                                                                   α       1
                                                       t                                   −
                                                                ρ  µs           σ (1− α ) 1− α
            K ( t ) I ( t ) − K 0 I ( 0) = − A         ∫   exp  − s  (e − C 0 )
                                                                σ 
                                                                                                 ds                  (2.92)
                                                       0

Introducing the notation
                        α       1                                     ρ
            ν=               −                      and ω = −                                                        (2.93)
                    σ (1 − α) 1 − α                                   σ
equation (2.92) becomes
                                      1                             1                t
                                 −                             −
                   µt                                                                     ωs
      K ( t ) (e        − C0 )       1− α   − K 0 (1 − C 0 )       1− α   = −A       ∫e        (e µ s − C 0 ) ν ds   (2.94)
                                                                                     0

The integral appearing in (2.94) is not elementary. It can be evaluated using the
hypergeometric function, known from the theory of differential equations in the
complex domain. All the relevant calculations appear in Appendix A. We have
found (see equation (A.16))
  t
         ωs
  ∫e          (e µ s − C 0 ) ν ds =
  0

           1
  =−         [ e −µη t F (− ν, η ;1 + η ; C 0 e − µ t ) − F (− ν, η ;1 + η ; C 0 e − µ t ) ]                         (2.95)
          ηµ
We are now in a position to write down the final expression for the function K ( t )
using (2.94), (2.91) and (2.95):
                                             1                             1
                                        −                             −              A
                         µt                 1− α                          1− α
            K ( t ) (e        − C0 )               − K 0 (1 − C 0 )              =      Ψ(t)                         (2.96)
                                                                                     µη
where

            Ψ ( t ) = e −µη t F (− ν, η ;1 + η ; C 0 e − µ t ) − F (− ν, η ;1 + η ; C 0 )                            (2.97)
From equation (2.96) we obtain the final expression for the physical capital:
                                  1                               1
                    K (1 − C ) − 1−α + A Ψ ( t )  × (e µ t − C ) 1−α
            K(t) =  0                                                                                               (2.98)
                             0
                                       µη                      0
                                                 


                                                                                                                         19
2.3.4. Equation for the human capital.
We come now in finding an explicit solution of the differential equation (2.12) for
the human capital. Substituting into (2.12) the expression for u given by (2.76),
we obtain the following differential equation
                                           1               1
                                       −
                   γ                     α    λK   α
        H−γH = −γ                            
                                               λ      K
                                                                                                 (2.99)
                  1− α                        H    
From (2.88) we obtain
                    1            1               1                                     1
         λK       α    γ    1−α    γ       α µ                     −
        
        λ      
                       =                 exp     t  × (e µ t − C 0 ) 1−α
         H             α          1− α      1− α 
Using the previous expression and also (2.98) for K, we get the following
differential equation for H:
        H−γH =
                          1
                                                 −
                                                     1               
              γ  1−α      γ                    1− α + A Ψ ( t ) 
        = −γ         exp  t   K 0 (1 − C 0 )                                               (2.100)
             α           α                            µη
                                                                    
This is a linear first order differential equation with integration factor

        J(t ) = e − γ t
Therefore it can be written as
                                  1
                                                               −
                                                                   1               
   d −γ t           γ  1−α      γ                          1− α + A Ψ ( t )  (2.101)
      (e  H) = − γ         exp   − γ  t   K 0 (1 − C 0 )                    
   dt              α            α                                 µη
                                                                                  
Integrating the previous formula in the interval [ 0, t ] , we obtain
                                           1
                                                t                         1              
                                γ  1−α
    e − γ t H( t ) − H 0 = − γ                      µt    K (1 − C ) − 1−α + A Ψ ( t )  dt
                               α
                                                ∫e          0       0
                                                                                µη              (2.102)
                                                0                                        
Using the expression for the function Ψ ( t ) , given in (2.97), equation (2.102) can
be transformed into the equation



                                                                                                      20
        e − γ t H( t ) = H 0 −
                          1
                                              −
                                                   1                                          t
            γ  1−α                            1− α − A F( − ν, η ;1 + η ; C )                  µt
        − γ 
           α
                               K 0 (1 − C 0 )
                                                        µη
                                                                               0              ∫e        dt −
                              
                                                                                             0


                                       A
                                          t
                                             µ (1− η) t
                                                                                           
                                                                                           
                                   +      ∫e            F(− ν, η ;1 + η ; C 0 e − µ t ) dt                     (2.103)
                                       µη 0                                                
                                                                                           
The last integral is not elementary and needs to be evaluated using some results
from the theory of Generalized Hypergeometric Functions. We put the integral
into the form
                      t
                              ϕt
                      ∫e           F( − ν, η ;1 + η ; C 0 e − µ t ) dt                                          (2.104)
                      0

where
                     ϕ = µ (1 − η)                                                                              (2.105)
The integral has been evaluated in Appendix B. We have found (see equation
(B.11))
          t
               ϕt
          ∫e        F (− ν, η ;1 + η ; C 0 e − µ t ) dt =
          0

                1 ϕt      p, − ν , η             1        p, − ν , η     
        =−        e 3 F2              C0 e− µ t  +  3 F2              C0                                    (2.106)
               µp        p + 1, η + 1            µp      p + 1, η + 1    
                                                                 ϕ
where the parameter p is defined by p = −                          .
                                                                 µ
Therefore from (2.103), using (2.104) and (2.106), we arrive at the following final
expression for the function H( t ) :

                                                1        
                                γt        γ  1−α       
                     H ( t ) = e H 0 − γ         X( t )                                                     (2.107)
                                         α             
                                                         
where X( t ) is given by



                                                                                                                     21
                                1                                                           t
                  K (1 − C ) − 1−α − A F (− ν, η ;1 + η ; C )                                     µt
        X( t ) ≡  0       0
                                     µη
                                                            0                                ∫e         dt +
                                                                                            0

                                       t
                              A     µ (1− η) t
                            +    ∫e            F (− ν, η ;1 + η ; C 0 e − µ t ) dt =
                              µη 0

                        1                              µt
          K (1 − C ) − 1−α − A F (− ν, η ;1 + η ; C )   e − 1  −
        = 0                                                    
                                                    0 
                                                            µ 
                   0
                             µη
                                                              

              A   ϕt     p, − ν , η                       p, − ν , η     
        −        e 3 F2              C 0 e − µ t  − 3 F2              C0                                      (2.108)
            µ ηp        p + 1, η + 1                      p + 1, η + 1
             2
                                                                            

2.3.5. Expressions for the control variables.
It is now easy to find the expressions of the control variables u ( t ) and C( t )
using the functions we have already found.
The control variable C( t ) is given by (2.75). Substituting into this expression the
λ K given by (2.86), we obtain
                                   1                α                                α
                               −             −
             γ                   σ   γ       σ (1− α )          µt            σ (1− α )          ρ 
   C( t ) =      λ H ( 0)                                × (e        − C0 )               × exp  − t         (2.109)
            1− α                     α                                                           σ 
                                                                                                           λK
Using (2.76) for the control variable u ( t ) , we first find the ratio                                       from (2.88)
                                                                                                           λH
and then substitute the expressions for K and H from (2.98) and (2.107)
respectively. We have
                                                                                    1
                                                                              −                   A
                     1                                       K 0 (1 − C 0 )        1− α   +          Ψ(t)
           γ      1−α          µ                                                            µη
    u(t) =              × exp      − γ t×                                                                    (2.110)
           α                  1− α                                                1
                                                                           γ         1− α
                                                                    H0 − γ                      X( t )
                                                                           α
where all the functions and parameters are defined in the text.




                                                                                                                        22
3. A Lucas-Uzawa model with externalities.
3.0. The model.
We work with a Cobb-Douglas production function

               Y = A K α (u H L)1−α
where
        A: represents the technological level of the economy, considered to be
           constant
        K: represents the stock of the physical capital
        L: represents the labour, considered to be constant
        H: represents the stock of the human capital
        u: is the portion of the human capital devoted to the production of output Y
Therefore uHL represents the amount of labour used to produce output.
Externality is introduced to the model in the form of average schooling of the
human capital. We thus introduce the production function

               Y = A K α (u H L)1−α H β
                                      a

where H a is the measure of the externality.
The physical capital, ignoring depreciation, is expressed by

               K = A K α (u HL)1−α H β − C
                                     a

where C is the consumption. Considering L constant, and dividing through by L,
we can write the previous equation in per capita terms (retaining the notation)

               K = A K α (u H)1−α H β − C
                                    a

In equilibrium condition, we have H a = H .
The portion of the human capital devoted to the production of more human capital
will then be 1− u . This leads to the equation of motion for the human capital:
               H( t ) = γ (1 − u ) H




                                                                                  23
where γ is a constant coefficient.
We consider the dynamic optimization problem
                         ∞
                             C( t )1−σ − 1 −ρ t
                  max ∫                   e     dt                                        (3.1)
                         0
                                1− σ

subject to

        K ( t ) = A K ( t ) α ( u ( t ) H ( t ))1−α H ( t ) β − C( t )                    (3.2)

        H( t ) = γ (1 − u ( t )) H( t )                                                   (3.3)
with initial conditions
        K (0) = K 0 , H(0) = H 0                                                          (3.4)
where
        C( t ) ≥ 0 , u ( t ) ∈ [ 0, 1 ] , K ( t ) ≥ 0 , H( t ) ≥ 0                        (3.5)
In equation (3.1) we work again with an isoelastic utility function and maximize
the integral, where σ is the instantaneous elasticity of substitution and ρ is the
instantaneous discount rate.

3.1. The equations of motion.
The current value Hamiltonian is given by

 H c (C, u , H, K ) =

            C( t )1−σ − 1
        =                 + λ K [ A K ( t ) α (u ( t ) H( t ))1−α H( t ) β − C( t ) ] +
               1− σ
                             + λ H [ γ (1 − u ( t )) H( t ) ]                             (3.6)

where λ K and λ H are the costate variables corresponding to K and H
respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method. The reader may consult any of the known references of this
technique, like Malliaris and Brock [18], Pontryagin et. al. [24] or Seierstad and
Sydsaeter [30].



                                                                                              24
The first order conditions read

       ∂H c
            = 0 ⇔ C− σ − λ K = 0                                                       (3.7)
        ∂C

       ∂H c
            = 0 ⇔ λ K [ (1 − α) A K α (u H) − α H1+β ] − λ H γ H = 0                   (3.8)
        ∂u
We also have the two Euler equations

       ∂H c
            + λ K = ρ λ K ⇔ α λ K AK α −1 (u H)1− α H β + λ K = ρ λ K                  (3.9)
       ∂K

       ∂H c
            + λH = ρλH ⇔
       ∂H

       ⇔ λ K (1 − α + β) A K α ( u H )1− α H β−1 + λ H γ (1 − u ) + λ H = ρ λ H        (3.10)
the dynamic constraints

       K = A K α ( u H )1−α H β − C , K (0) = K 0                                      (3.11)

       H = γ (1 − u ) H , H(0) = H 0                                          (3.12)
and the transversality conditions

       lim λ K K e − ρ t = 0                                                           (3.13)
       t →∞

       lim λ H H e − ρ t = 0                                                           (3.14)
       t →∞

3.2. First method of solution to the model.
This method uses a procedure similar to the dimensional reduction. The derived
system however is not solved along the balanced growth path (BGP). Considering
two auxiliary functions, we get a system of decoupled ordinary first order
differential equations, which, when solved, determine the physical capital, the
control variable u and finally the human capital.
3.2.1. Simplification of the dynamical equations.




                                                                                           25
Taking logarithms and differentiation with respect to time, we obtain from
equation (3.7) that
            C λK
       −σ    =                                                               (3.15)
            C λK
Taking logarithms and differentiation with respect to time, we obtain from
equation (3.8) that
       λK    K   u         H λ
          + α − α + (β − α) = H                                              (3.16)
       λK    K   u         H λH
From equation (3.9) we obtain
       λK
          = ρ − α AK α −1 (u H)1−α H β                                       (3.17)
       λK
From equation (3.10) we obtain
       λH                                                    λ
          = ρ − γ (1 − u ) − (1 − α + β) AK α (u H)1− α H β−1 K              (3.18)
       λH                                                    λH
From equation (3.11) we obtain
       K                           C
         = AK α −1 (u H)1− α H β −                                           (3.19)
       K                           K
From equation (3.12) we obtain
       H
         = γ (1 − u )                                                        (3.20)
       H
We introduce a function named Y defined by

       Y = AK α ( u H )1− α H β                                              (3.21)
Equations (3.17)-(3.19) can be expressed in terms of the function Y as follows:
       λK        Y
          = ρ −α                                                             (3.22)
       λK        K

       λH                               λ Y
          = ρ − γ (1 − u ) − (1 − α + β) K ⋅                                 (3.23)
       λH                               λH H




                                                                                 26
       K Y C
        = −                                                                  (3.24)
       K K K
From equation (3.8) we find that
       λK Y  γ
         ⋅ =     u                                                           (3.25)
       λH H 1− α
Because of the previous relation, we can simplify further equation (3.23):
       λH                    γ (1 − α + β)
          = ρ − γ (1 − u ) −               ⋅u                                (3.26)
       λH                        1− α
Combining equations (3.15) and (3.22) we find an expression for the ratio
       C α Y ρ
        = ⋅ −                                                                (3.27)
       C σ K σ

                                                         λK K H    λ
Equation (3.16), upon substituting the expressions for     , ,  and H given
                                                         λK K H    λH
by (3.22), (3.24), (3.20) and (3.26) respectively, we obtain the equation
       u γ (1 − α + β) γ (1 − α + β)    C
         =            +              u−                                      (3.28)
       u       α           1− α         K


3.2.2. Auxiliary functions and their differential equations.
We now introduce two more functions U and Z defined by
            Y
       U=                                                                    (3.29)
            K
and
            C
       Z=                                                                    (3.30)
            K
respectively.
We shall establish a system of two ordinary differential equations satisfied by
these two functions.
Taking logarithms and differentiation of the defining equations (3.29) and (3.30),
we obtain the equations


                                                                                  27
       U Y K
        = −                                                                 (3.31)
       U Y K
and
       Z C K
        = −                                                                 (3.32)
       Z C K
respectively.
We can express the right hand sides of the two previous equations in terms of the
functions U and Z. From equation (3.24) we obtain
       K
         =U−Z                                                               (3.33)
       K
From equation (3.27) we obtain
       C α    ρ
        = ⋅U−                                                               (3.34)
       C σ    σ
                                     Y
We now have to express the ratio       in terms of U and Z. Taking logarithms and
                                     Y
differentiation of (3.21), we obtain the equation
       Y    K         u              H
         = α + (1 − α) + (1 − α + β)                                        (3.35)
       Y    K         u              H
                          K u    H
Substituting the ratios    , and   given by (3.33), (3.28) and (3.20)
                          K u    H
respectively into the previous equation, we obtain
       Y        γ (1 − α + β)
         =αU−Z+                                                             (3.36)
       Y              α
Therefore we obtain from (3.31) and (3.32), using (3.36), (3.33) and (3.34), the
system of two equations
       U γ (1 − α + β)
         =             − (1 − α) U                                          (3.37)
       U       α
and




                                                                                   28
        Z      α−σ    ρ
          = Z+     U−                                                  (3.38)
        Z       σ     σ
Equation (3.37) can be solved by separation of variables. The solution is given by

                1    ν eν t
        U=         ⋅ νt                                                         (3.39)
              1 − α e − C0

where
             γ (1 − α + β)                γ (1 − α + β)
        ν=                 and C 0 = 1 −                                        (3.40)
                   α                     α(1 − α) U(0)
Using the expression (3.39) for the function U, equation (3.38) is converted into
the equation
          ρ   α−σ        ν eν t 
        Z+ −           ⋅ νt      Z = Z2                                       (3.41)
           σ σ (1 − α ) e − C 
                               0

This is a Bernoulli differential equation, solved under the substitution

        X = Z −1                                                                (3.42)
Equation (3.41) is converted in terms of X into the linear differential equation
           α−σ        ν eν t    ρ
        X+          ⋅ νt       −  X = −1                                      (3.43)
           σ (1 − α) e − C      σ
                             0   
The integrating factor of the above equation is

        I( t ) = e ω t (e ν t − C 0 ) ζ                                         (3.44)
where
                ρ                     α−σ
        ω=−            and ζ =                                                  (3.45)
                σ                    σ (1 − α)
Multiplying (3.43) by the integrating factor, we obtain the equation
        d
           ( I( t ) X ) = − e ω t ( e ν t − C 0 ) ζ
        dt
which, upon integration in the interval [ 0, t ] , gives




                                                                                    29
                                       t
        I( t ) X − I(0) X(0) = − ∫ e ω s (e ν s − C 0 ) ζ ds                               (3.46)
                                       0

The integral on the right hand side of the previous equation is calculated in
Appendix A. We have, using the notation of this section
        t
             ωs                                 1
        ∫e        (e ν s − C 0 ) ζ ds = −         Ω( t )                                   (3.47)
        0
                                               ην

where

        Ω ( t ) = e − ην t F( − ζ , η ;1 + η ; C 0 e − ν t ) − F( − ζ , η ;1 + η ; C 0 )   (3.48)
and
                  ω+ νζ
        η=−                                                                                (3.49)
                    ν
Therefore we obtain from (3.46) the following expression for the function X:
                                        1
              (1 − C 0 ) ζ X(0) +         Ω( t )
                                       ην
        X=                                                                                 (3.50)
                      e ω t (e ν t − C 0 ) ζ
We thus have that the function Z, related to X by (3.42), is given by

                  e ω t (e ν t − C 0 ) ζ
        Z=                                                                                 (3.51)
                                    1
             (1 − C 0 ) ζ X(0) +         Ω( t )
                                   ην
Since from (3.47) we have upon differentiation with respect to t the relation
                                      d 1           
        − e ω t (e ν t − C 0 ) ζ =        ην Ω( t ) 
                                                                                         (3.52)
                                      dt            
equation (3.51) is written as
                  W
        Z=−                                                                                (3.53)
                  W
where
                                        1
        W = (1 − C 0 ) ζ X(0) +           Ω( t )                                           (3.54)
                                       ην



                                                                                               30
On the other hand, equation (3.39) can be expressed as
              1 V
        U=       ⋅                                                                                    (3.55)
             1− α V
where

        V = eν t − C0                                                                                 (3.56)

3.2.3. Equation for the physical capital.
Equation (3.33) can be written, because of (3.55) and (3.53), as
        K   1 V W
          =   ⋅ +                                                                                     (3.57)
        K 1− α V W
The above equation can be integrated once, providing us with the following
expression for the physical capital
                     1
        K = D0    V 1− α    W
or
                                          1
                           νt                               ζ             1        
        K ( t ) = D 0 (e        − C0   ) 1−α (1 − C
                                                       0)       X(0) +      Ω( t )                  (3.58)
                                                                         ην        
where
                      K0                                             K0
        D0 =         1
                                       =                 1
                                                                                                      (3.59)
                    1− α                   (1 − C 0     1− α (1 − C        ζ
                                                                                X ( 0) +
                                                                                            1     
                                                                                              Ω(0)
               V(0)         W ( 0)                    )               0)
                                                                                          ην     
3.2.4. Equation for the control variable u.
From equation (3.28), using (3.30) and (3.53), we derive the equation
        u γ (1 − α + β) γ (1 − α + β)    W
          =            +              u+
        u       α           1− α         W
which is equivalent to the equation
            γ (1 − α + β) W    γ (1 − α + β) 2
        u −
                         + u =
                                             u                                                       (3.60)
                  α       W        1− α
This is a Bernoulli differential equation, which under the substitution


                                                                                                          31
       y = u −1                                                                           (3.61)
becomes a linear first order differential equation
          γ (1 − α + β) W      γ (1 − α + β)
       y+
                       + y = −
                                                                                (3.62)
                α       W          1− α
The previous equation admits the integrating factor

       J(t) = e ν t W                                                                     (3.63)
Multiplying equation (3.62) by the integrating factor, we obtain the equation
        d                   γ (1 − α + β) ν t
           ( J ( t ) y) = −              e W
        dt                      1− α
which, upon integration in the interval [0, t ] , gives us
                                                           t
                                  γ (1 − α + β)                 νt
       J ( t ) y − J (0) y(0) = −
                                      1− α                 ∫e        W (s) ds             (3.64)
                                                           0

Let
                  t
                       νt
       X( t ) =   ∫e        W (s) ds                                                      (3.65)
                  0

Using the expression (3.54), we find that
                  t
                       νs
       X( t ) =   ∫e        W (s) ds =
                  0

                                                                         t
                              1                       
       = (1 − C 0 ) ζ X(0) −    F(−ζ, η ;1 + η ; C 0 ) ∫ e ν s ds +
                             ην                       0
                            t
                     1           ν (1− η) s
                  +         ∫e                F(−ζ, η ;1 + η ; C 0 e −ν s ) ds            (3.66)
                    ην      0

The last integral is evaluated in Appendix B. We get from (B.16), using the
notation of this section,
         t
             ϕt
        ∫e        F (− ζ, η ;1 + η ; C 0 e − ν t ) dt =
        0




                                                                                              32
             1  ϕt     p, − ζ , η                       p, − ζ , η     
      =−       e 3 F2              C 0 e − ν t  − 3 F2              C0         (3.67)
            νp        p + 1, η + 1                     p + 1, η + 1    
where
                                           ϕ
        ϕ = ν(1 − η) and p = −                                                       (3.68)
                                           ν
Therefore we have the following expression for the function X( t ) :
                   t
                        νs
        X( t ) =   ∫e        W (s) ds =
                   0

                    ζ        1                         eν t − 1 
        = (1 − C 0 ) X(0) −    F(−ζ, η ;1 + η ; C 0 )           −
                            ην                        ν 
                                                                   

              1     ϕt     p, − ζ , η                       p, − ζ , η     
        −          e 3 F2              C 0 e − ν t  − 3 F2              C0     (3.69)
            ην 2 p        p + 1, η + 1                     p + 1, η + 1    
We then obtain from (3.64)
              J ( 0) y ( 0) − µ X ( t )
        y=                                                                            (3.70)
                         J(t )
and then
                           J( t )
        u(t) =                                                                        (3.71)
                   J (0) y(0) − µ X( t )
where we have put
              γ (1 − α + β)
        µ=                                                                            (3.72)
                  1− α
3.2.5. Equation for the human capital.
Equation (3.71) can also be written, comparing (3.63) and (3.65), as
                      X( t )
        u=                                                                            (3.73)
              J (0) y(0) − µ X( t )
Equation (3.20) for the human capital, because of the previous expression, takes
on the form




                                                                                          33
          H          − γ X( t )
            =γ+
          H     J(0) y(0) − µ X( t )
or even
          H    γ S
            =γ+ ⋅                                                                    (3.74)
          H    µ S
where
        S = J ( 0) y ( 0) − µ X ( t )                                                (3.75)
Equation (3.74) gives upon integration
                              γ
                H0
        H=            γ
                          S eγ t
                              µ

                      µ
              S(0)
or
                                                1− α
                          H0
        H( t ) =           1− α
                                      S( t )   1− α +β
                                                         eγ t                        (3.76)
                          1−α +β
                   S(0)

3.3. Second method of solution to the model.
3.3.1. Equations for the control variables.
Solving equation (3.7) with respect to C, we obtain
                          1
                      −
        C = (λ K )        σ                                                          (3.77)
Solving equation (3.8) with respect to u, we obtain
                                  1                1
                                                            β−α
            (1 − α) A           α    λK         
                                                   α
                                                             α
        u =
                      
                                     
                                      λ            KH
                                                                                    (3.78)
                 γ                   H          
We also get from the previous relation that
                                       1− α                    1− α
                                                                         β (1− α )
               1− α        (1 − α) A  α               λ K  α 1−α
        (u H)         =
                                              
                                                      
                                                       λ         K   H α          (3.79)
                                 γ                    H



                                                                                         34
we shall need later on.
3.3.2. Equations for the costate variables.
Equation (3.9), because of (3.79), takes on the form
                                           1− α                     1
                                                                          β
                               (1 − α) A  α           λK        α
                                                                          α
       λK − ρλK = − α A 
                                            
                                                      
                                                       λ       
                                                                       H       λH            (3.80)
                                   γ                  H      
Equation (3.10), because of (3.78) and (3.79), becomes
                                                1− α                        1
                                                                                 β−α
                                    (1 − α) A  α             λK         α
       λ H + ( γ − ρ) λ H = − β A 
                                                      
                                                             
                                                              λ        
                                                                               H α    K λH   (3.81)
                                            γ                H       

3.3.3. Equation for the physical capital.
We now consider equation (3.11). This equation, because of (3.79), takes the form
                           1− α          1− α
                                                     β                          1
               (1 − α) A  α      λK  α           α
                                                                            −
                                                                                σ
       K =A
                         
                                 
                                  λ             H       K − (λ K )                         (3.82)
                   γ             H

3.3.4. Equation for the human capital.
Substituting into (3.12) the expression for u given by (3.78), we obtain the
following differential equation
                                        1              1
                                                            β
                  (1 − α) A           α    λK      α
       H−γH = −γ 
                            
                                           
                                            λ     
                                                          Hα   K                             (3.83)
                       γ                   H    

3.3.5. The solution strategy.
It is now evident the solution strategy we have to follow to solve the system of the
four previous equations (3.80)-(3.83).
3.3.5.1. Relation between H and λ H
Dividing equation (3.81) by equation (3.83), we obtain
               λ H + ( γ − ρ) λ H    β λH
                                  =     ⋅                                                     (3.84)
                   H−γH             1− α H
The previous equation can be written as


                                                                                                  35
               λ H + ( γ − ρ) λ H    β H−γH
                                  −      ⋅   =0
                      λH            1− α   H
or equivalently
               λH   β H
                  −   ⋅ =µ                                                               (3.85)
               λH 1− α H
where
                       βγ
               µ=−         +ρ−γ                                                          (3.86)
                      1− α
Integrating equation (3.85), we get
                                         β
                                        1− α
               λ H ( t ) = C 0 H( t )          eµ t                                      (3.87)
where
                        λ H (0)
               C0 =         β
                                                                                         (3.88)
                      H(0) 1−α
3.3.5.2. Expression for the costate variable λ K .
Using the above expression for λ H into the equation (3.80) for λ K , we get the
equation
                                                                  1
                                               µ (1 − α) 
               λK − ρλK           = − B 0 exp  −          (λ K ) α                     (3.89)
                                                   α 
where
                                        1− α
                            (1 − α) A  α
               B0 = α A 
                                              
                                                                                        (3.90)
                                  γ C0        
Equation (3.89) is a Bernoulli differential equation. Multiplying both members of
                              1                                                          1
                          −                                                         1−
this equation by (λ K )       α    and introducing a new function λ by λ = (λ K )        α,   we
obtain the linear first order differential equation




                                                                                              36
                          ρ (1 − α)   1− α           µ (1 − α) 
                  λ+                λ=     B 0  exp  −                                  (3.91)
                              α        α                 α 
The above differential equation is found to admit a solution given by

         B0                                                
  λ=         exp  (ρ − µ) (1 − α) t  − 1 − (ρ − µ) λ (0)   × exp  − ρ (1 − α) t 
                                                                                   
        ρ−µ             α                                               α       
                                                B0          
Therefore

                 1−
                      1                                                     −
                                                                                1−α  
                             B0       (ρ − µ) (1 − α)   (ρ − µ)                   
        (λ K )        α   =      exp                 t  − 1 −    (λ K (0)) α   ×
                                                                                    
                            ρ−µ             α                  B0
                                                                                  

                                    ρ (1 − α) 
                             × exp  −        t
                                        α     
from which we obtain the following expression for the costate variable λ K :
                                  α                             α
                             −                             −
              B                1− α        νt
        λK =  0 
             ρ−µ                      (e        − D0 )       1−α   × eρ t                 (3.92)
                
where
             (ρ − µ) (1 − α)
        ν=                                                                                  (3.93)
                   α
and
                                                    1− α
                 (ρ − µ)           −
        D0 = 1 −         (λ K (0))                   α                                      (3.94)
                   B0


3.3.5.3. Equation for the physical capital.
Equation (3.82) for the physical capital, because of the (3.87), gets simplified into
the equation
                           1−α
                                                    1− α                               1
               (1 − α) A  α                                   µ (1 − α)            −
      K =A
                                
                                        (λ K      ) α     exp  −          K − (λ K ) σ
                     γ C0                                         α 

and because of (3.92), into



                                                                                                37
                           1− α
               (1 − α) A  α  ρ − µ           eν t
       K − A
                              
                                     
                                       B  νt
                                                  K=
                 γ C0                0  e − D0
                       α                     α
           B  σ(1−α ) ν t                    ρ 
       = − 0 
          ρ−µ        (e − D 0 ) σ (1−α ) exp − t                            (3.95)
                                             σ 
We obtain that
                         1− α
          (1 − α) A     α ρ−µ         ρ−µ    ν
        A
          γC        
                                     =
                                    B       =                                 (3.96)
                0                 0    α    1− α

We also introduce the notation
                           α
             B  σ (1−α )
       F0 =  0 
            ρ−µ                                                               (3.97)
               
and
             ρ
       ω=−                                                                      (3.98)
             σ
Equation (3.95) then becomes
                                                            α
            1    ν eν t
       K−      ⋅ νt     K = − F0 (e ν t − D 0 ) σ (1−α ) e ω t                  (3.99)
          1 − α e − D0

Equation (3.99) is a linear first order differential equation with integrating factor
given by
                                                                1
                       1       ν eν t                       −
       I( t ) ≡ exp  −                  dt  = (e ν t − D 0 ) 1−α
                     1 − α ∫ eν t − D
                                                                               (3.100)
                                            
                                      0    
Therefore equation (3.99), after multiplying by the integrating factor, becomes
                                                  α     1
                                                    −
       d                       ωt ν t      σ (1− α ) 1− α
          ( I( t ) K ) ≡ − F0 e (e − D 0 )
       dt
Integrating the above equation we arrive at




                                                                                    38
                                          t
        I( t ) K − I(0) K 0 ≡ − F0 ∫ e ω t (e ν t − D 0 ) ζ dt                                          (3.101)
                                          0

where
                            α       1    α−σ
                   ζ=            −     =                                                                (3.102)
                        σ (1 − α) 1 − α σ (1 − α)
The integral appearing in (3.101) is evaluated to be (Appendix A)
        t
             ωt
        ∫e        (e ν t − D 0 ) ζ dt =
        0

            1        −νη t 1 η−1
                                            −ν t
                                                           1                         
                                                                                     
        =−          e      ∫ x   (1 − D 0 e         ζ
                                                  x ) dx − ∫ u η−1 (1 − D 0 u ) ζ du 
           ην                                                                       
                           0                              0                         
              1
        =−      Ψ(t)                                                                                    (3.103)
             ην
where

        Ψ ( t ) = e −νη t F ( − ζ , η ;1 + η ; D 0 e − ν t ) − F ( − ζ , η ;1 + η ; D 0 )               (3.104)
                  ω+ νζ
        η=−                                                                                             (3.105)
                    ν
Using (3.101), (3.103) and the explicit expression for the integrating factor I( t )
given by (3.100), we obtain the following expression for the human capital:
                            1                                1
                (1 − D ) − 1−α K + F0 Ψ ( t )  (e ν t − D ) 1−α
        K(t) =                                                                                         (3.106)
                       0         0
                                    ην                    0
                                              
3.3.5.4. Equation for the human capital.
We now turn to the evaluation of the human capital using equation (3.83). This
equation can be written, using (3.87) and (3.92), as
                                              1                 1
                                                           −                                  1
                   (1 − α) A                α    B0        1− α        νt
                                                                                         −
                                                                                             1− α
                   γC 
        H−γH = −γ                               ρ−µ
                                                                    (e        − D0 )              ×
                         0                          




                                                                                                            39
                                                      β
                                ρ − µ  − 1−α
                         × exp       tH      K                            (3.107)
                                α 
The previous equation, using the expression (3.106) for the physical capital, takes
the form
                                ν 
         H − γ H = − G 0 × exp     t×
                               1− α 

                                  1                         β
                      (1 − D ) − 1−α K + F0 Ψ ( t )  × H − 1−α
                    ×                                                      (3.108)
                             0         0
                                          ην         
                                                    
where
                                   1                 1
                                                −
                 (1 − α) A       α    B0        1− α
         G0 = γ 
                 γC        
                                      ρ−µ
                                                                          (3.109)
                       0                 
Equation (3.108) is a Bernoulli differential equation. Multiplying through by
   β
H 1−α   , we obtain the equation
              β              β
                       1+                         ν 
         H H 1−α   −γH      1− α   = − G 0 × exp     t×
                                                 1− α 

                                  1                 
                      (1 − D ) − 1−α K + F0 Ψ ( t ) 
                    ×                                                      (3.110)
                             0         0
                                          ην         
                                                    
Introducing a new function U by
                            β
                      1+
                U =H       1− α                                             (3.111)
we obtain from (3.110) the equation
              γ (1 − α + β)     1− α + β            ν 
         U−                 U=−          G 0 × exp     t×
                  1− α           1− α              1− α 

                                                     1                 
                                         (1 − D ) − 1−α K + F0 Ψ ( t ) 
                                       ×                                   (3.112)
                                                0         0
                                                             ην         
                                                                       



                                                                                 40
The integrating factor is given by
                       γ (1 − α + β) 
        J ( t ) = exp  −            t
                           1− α      
Therefore multiplying (3.112) by the integrating factor, we derive the equation

                           1− α + β                   −
                                                          1                  
                                         ϕt             1− α K + F0 Ψ ( t ) 
        d
           (J( t ) U ) = −          G 0 e  (1 − D 0 )         0                             (3.113)
        dt                  1− α                                  ην
                                                                            
where
              ν     γ (1 − α + β)
        ϕ=        −                                                                           (3.114)
             1− α       1− α
Integrating (3.113), we obtain

                      1− α + β
                                   t               −
                                                       1                  
                                      ϕt             1− α K + F0 Ψ ( t )  dt
   J( t ) U − U 0 = −          G 0 ∫ e  (1 − D 0 )         0                                (3.115)
                       1− α                                    ην
                                   0                                     
We have further
        t                      1                 
             ϕt    (1 − D ) − 1−α K + F0 Ψ ( t )  dt =
        ∫e               0         0
                                       ην         
        0                                        
                      1                                                   t
          (1 − D ) − 1−α K − F0 F (−ζ, η ; η + 1; D )                          ϕt
        =       0         0
                              ην
                                                    0                      ∫e        dt +
                                                                          0

                                      t
                                F          ( ϕ − ν η) t
                               + 0    ∫e                  F (−ζ, η ; η + 1; D 0 e −ν t ) dt   (3.116)
                                ην    0

We have the following formula (Appendix B)
        t
             ( ϕ − ν η) t
        ∫e                  F (−ζ, η ; η + 1; D 0 e −ν t ) dt =
        0

              1 θt      p, − ζ , η             1        p, − ζ , η     
        =−      e 3 F2              D0 e− ν t  +  3 F2              D0                    (3.117)
             νp        p + 1, η + 1            νp      p + 1, η + 1    
where




                                                                                                   41
                                                    θ
                  θ = ϕ−νη           and      p=−                                    (3.118)
                                                    ν
Therefore (3.115), because of (3.116) and (3.117), takes the form
                  1− α + β                    γ (1 − α + β) 
        U =  U0 −          G 0 X( t )  × exp               t                     (3.119)
                   1− α                           1− α      
where
                  t                      1                 
        X( t ) ≡ ∫ e   ϕt    (1 − D ) − 1−α K + F0 Ψ ( t )  dt =
                                   0         0
                                                 ην         
                  0                                        
                      1                                ϕt
          (1 − D ) − 1−α K − F0 F (−ζ, η ; η + 1; D )  e − 1 −
        =       0         0                        0 
                              ην                           ϕ
                                                      

             F0  θ t       p, − ζ , η                       p, − ζ , η     
        −          e 3 F2              D 0 e − ν t  − 3 F2              D0     (3.120)
            ην 2 p        p + 1, η + 1                     p + 1, η + 1    
We thus obtain the following formula for the human capital
                                                           1− α
               1− α +β                                  1−α +β
                                   1− α + β            
        H = H(0) 1−α
                               −            G 0 X( t )           × eγt             (3.121)
                                    1− α               
                                                       
taking into account the relation (3.111) between H and U.
3.3.5.5. Expression for the costate variable λ H .
Using equations (3.87) and (3.121), we are able to find the following expression
for the costate variable λ H :
                                                                      1
                        1− α +β                      β (1−α +β)
                  
        λ H = C 0  H(0)  1− α − 1 − α + β G X ( t )             e (µ + γ ) t
                                            0                                       (3.122)
                                  1− α              
                                                    
where C 0 is given by (3.88).



4. Application I.

                                                                                          42
(The model considered by Bucekkine and Ruiz-Tamarit).
In this Section we solve again the model considered by Bucekkine and Ruiz-
Tamarit [8]. This model is solved using the two methods we already used in
solving the simple model considered in Section 2. Despite the fact that in the
second method we use the same technique to that introduced in reference [8], we
present explicitly all the intermediate steps, in easy to follow calculations. Apart
from that, we have found that the human capital function is expressed in terms of
the generalized hypergeometric function 3 F2 [] .

4.0. The model.
We consider the dynamic optimization problem
                         ∞
                              c( t )1−σ − 1
                  max ∫                     N ( t ) e −ρ t                                         (4.1)
                          0
                                  1− σ

subject to

        K ( t ) = A K ( t ) β ( u ( t ) N ( t ) H ( t ))1−β − π K ( t ) − c( t ) N ( t )           (4.2)

        H( t ) = δ (1 − u ( t )) H( t ) − ϑ H( t )                                                 (4.3)
with initial conditions
        K (0) = K 0 , H(0) = H 0 , N(0) = N 0                                                      (4.4)
where
        c( t ) ≥ 0 , u ( t ) ∈ [ 0, 1 ] , K ( t ) ≥ 0 , H( t ) ≥ 0                         (4.5)
The notation in this model is as in previous models. In this model however has
been considered depreciation of physical capital ( π is the coefficient of
depreciation) and depreciation of human capital ( ϑ is the coefficient of
depreciation). The function N( t ) represents the population growth.

4.1. The dynamical equations.
The current value Hamiltonian is given by

 H c (c, u , H, N, K ) =


                                                                                                       43
   c( t )1−σ − 1
 =               N( t ) + λ K [ A K ( t ) β (u ( t ) N( t ) H( t ))1−β − π K ( t ) − c( t ) N( t ) ] +
       1− σ
    + λ H [ δ (1 − u ( t )) H( t ) − ϑ H( t ) ]                                                          (4.6)

where λ K and λ H are the costate variables corresponding to K and H
respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method.
The first order conditions read

         ∂H c
              = 0 ⇔ c−σ = λK                                                                             (4.7)
          ∂c

         ∂H c
              = 0 ⇔ λ K [ (1 − β) A K β ( u N H ) − β N ] − λ H δ = 0                                    (4.8)
          ∂u

We also have the two Euler equations

         ∂H c
              + λK = ρ λK ⇔
         ∂K

        ⇔ λ K = (ρ + π) λ K − λ K β A K β−1 ( u N H )1− β                                                (4.9)

         ∂H c
              + λH = ρλH ⇔
         ∂H

        λ H = (ρ + ϑ) λ H − λ K (1 − β) A K β ( u N )1− β H −β − λ H δ(1 − u )                           (4.10)
the dynamic constraints

        K = A K β ( u N H )1−β − π K − c N , K (0) = K 0                                                 (4.11)

        H = δ (1 − u ) H − ϑ H , H(0) = H 0                                                              (4.12)
and the transversality conditions

         lim λ K K e − ρ t = 0                                                                           (4.13)
        t →∞

         lim λ H H e − ρ t = 0                                                                           (4.14)
        t →∞




                                                                                                             44
We also suppose that we have the exponential population growth (exogenous)
given by

       N(t) = N 0 e n t                                                      (4.15)

4.2. First method of solution.
4.2.1. Simplification of the dynamical equations.
Taking logarithms and differentiation with respect to time, we obtain from
equation (4.7) that
            C λK
       −σ    =                                                               (4.16)
            C λK
Taking logarithms and differentiation with respect to time, we obtain from
equation (4.8) that
       λK    K   u         N   H λ
          + β − β + (1 − β) − β = H                                          (4.17)
       λK    K   u         N   H λH
From equation (4.9) we obtain
       λK
          = ρ + π − βA K β−1 (u N H)1−β                                      (4.18)
       λK
From equation (4.10) we obtain
       λH                                                λ    1
          = ρ + ϑ − δ (1 − u ) − (1 − β) AK β (u N H)1− β K ⋅                (4.19)
       λH                                                λH H
From equation (4.11) we obtain
       K                               C
         = A K β−1 (u N H)1− β − π − N                                       (4.20)
       K                               K
From equation (4.12) we obtain
       H
         = δ (1 − u ) − ϑ                                                    (4.21)
       H
From equation (4.15) we obtain
       N
         =n                                                                  (4.22)
       N


                                                                                 45
We introduce a function named Y defined by

       Y = A K β ( u NH )1− β                                                (4.23)
Equations (4.18)-(4.20) can be expressed in terms of the function Y as follows:
       λK            Y
          = ρ + π −β                                                         (4.24)
       λK            K

       λH                               λ Y
          = ρ + ϑ − δ (1 − u ) − (1 − β) K ⋅                                 (4.25)
       λH                               λH H

       K Y     C
        = −π−N                                                               (4.26)
       K K     K
From equation (4.8) we find that
       λK Y  δ
         ⋅ =     u                                                           (4.27)
       λH H 1− β
Because of the previous relation, we can simplify further equation (4.25):
       λH
          = ρ + ϑ−δ                                                          (4.28)
       λH
Combining equations (4.16) and (4.24), we find an expression for the ratio
       C β Y ρ+π
        = ⋅ −                                                                (4.29)
       C σ K  σ
From equation (4.17), upon substituting the expressions given by (4.24), (4.26),
(4.22) and (4.21), we obtain the equation
       u (1 − β) (π + δ − ϑ + n )         C
         =                        + δu −N                                    (4.30)
       u            β                     K


4.2.2. Auxiliary functions and their differential equations.
We now introduce two more functions U and Z defined by
            Y
       U=                                                                    (4.31)
            K
and



                                                                                   46
            NC
       Z=                                                                    (4.32)
            K
respectively.
We shall establish a system of two ordinary differential equations satisfied by
these two functions.
Taking logarithms and differentiation of the defining equations (4.31) and (4.32),
we obtain the equations
       U Y K
        = −                                                                  (4.33)
       U Y K
and
       Z N C K
        = + −                                                                (4.34)
       Z N C K
respectively.
We can express the right hand sides of the two previous equations in terms of the
functions U and Z. From equation (4.26) we obtain
       K
         = −π + U − Z                                                        (4.35)
       K
From equation (4.29) we obtain
       C β    ρ+π
        = ⋅U−                                                                (4.36)
       C σ     σ
                                    Y
We now have to express the ratio      in terms of U and Z. Taking logarithms and
                                    Y
differentiation of (4.23), we obtain the equation
       Y    K         u         N          H
         = β + (1 − β) + (1 − β) + (1 − β)                                   (4.37)
       Y    K         u         N          H
                          K u    H
Substituting the ratios    , and   given by (4.26), (4.30), (4.22) and (4.21)
                          K u    H
respectively, into the previous equation, we obtain




                                                                                  47
        Y                                 (1 − β) (π + δ − ϑ + n ) 
          = β U − Z + (1 − β) n + δ − ϑ +                           − βπ   (4.38)
        Y                                            β             
Therefore we obtain from (4.33) and (4.34), using (4.24), (4.21) and (4.22), the
system of two equations
        U
          = − (1 − β) U + (1 − β) d                                          (4.39)
        U
and
        Z      β−σ         ρ+π
          = Z+     U+n +π−                                                   (4.40)
        Z       σ           σ
where
                            π + δ − ϑ + (1 − β) n
        d =π+ n +δ−ϑ+                                                        (4.41)
                                     β
Equation (4.39) can be solved by separation of variables. The solution is given by

             1     ν eν t
        U=      ⋅                                                            (4.42)
           1 − β eν t − C0

where
                                        d
        ν = (1 − β) d and C 0 = 1 −                                          (4.43)
                                       U(0)
Using the expression (4.42) for the function U, equation (4.40) is converted into
the equation
           β−σ        ν eν t         ρ+π
        Z−          ⋅ νt       +n+π−     Z = Z2                            (4.44)
           σ (1 − β) e − C            σ 
                             0          
This is a Bernoulli differential equation, solved under the substitution

        X = Z −1                                                             (4.45)
Equation (4.44) is converted in terms of X into the linear differential equation
            β−σ        ν eν t          ρ+ π
        X+           ⋅ νt       +n +π−      X = −1                         (4.46)
            σ (1 − β) e − C             σ 
                              0            
The integrating factor of the above equation is


                                                                                   48
        I( t ) = e ω t (e ν t − C 0 ) ζ                                                    (4.47)
where
                         ρ+π                           β−σ
        ω= n +π−                    and ζ =                                                (4.48)
                          σ                           σ (1 − β)
Multiplying (4.46) by the integrating factor, we obtain the equation
        d
           ( I( t ) X ) = − e ω t ( e ν t − C 0 ) ζ
        dt
which, upon integration in the interval [ 0, t ] , gives
                                          t
        I( t ) X − I(0) X(0) = − ∫ e ω s (e ν s − C 0 ) ζ ds                               (4.49)
                                          0

The integral on the right hand side of the previous equation is calculated in
Appendix A. We have, using the notation of this section
        t
             ωs                                 1
        ∫e        (e ν s − C 0 ) ζ ds = −         Ω( t )                                   (4.50)
        0
                                               ην

where

        Ω ( t ) = e − ην t F( − ζ , η ;1 + η ; C 0 e − ν t ) − F( − ζ , η ;1 + η ; C 0 )   (4.51)
and
                  ω+ νζ
        η=−                                                                                (4.52)
                    ν
Therefore we obtain from (4.49) the following expression for the function X:
                                           1
              (1 − C 0 ) ζ X(0) +            Ω( t )
                                          ην
        X=                                                                                 (4.53)
                      e ω t (e ν t − C 0 ) ζ
We thus have that the function Z, related to X by (4.45), is given by

                   e ω t (e ν t − C 0 ) ζ
        Z=                                                                                 (4.54)
                                     1
              (1 − C 0 ) ζ X(0) +         Ω( t )
                                    ην
Since from (4.50), we have upon differentiation with respect to t the relation


                                                                                               49
                                     d 1           
        − e ω t (e ν t − C 0 ) ζ =       ην Ω( t ) 
                                                                                   (4.55)
                                     dt            
equation (4.54) is written as
                W
        Z=−                                                                          (4.56)
                W
where
                                        1
        W = (1 − C 0 ) ζ X(0) +           Ω( t )                                     (4.57)
                                       ην
On the other hand, equation (4.42) can be expressed as
              1 V
        U=       ⋅                                                                   (4.58)
             1− β V
where

        V = eν t − C0                                                               (4.59)

4.2.3. Equation for the physical capital.
Equation (4.26) can be written, because of (4.58) and (4.56), as
        K   1 V W
          =   ⋅ +  −π                                                               (4.60)
        K 1− β V W
The above equation can be integrated once, providing us with the following
expression for the physical capital
                      1
        K = D0 V     1−β
                           W e− π t
or
                                        1
                                       1−β                   1        
        K ( t ) = D 0 (e ν t − C 0 )                 ζ
                                          (1 − C 0 ) X(0) +    Ω( t )  e − π t   (4.61)
                                                            ην        
where




                                                                                            50
                      K0                                      K0
        D0 =         1
                                    =                 1
                                                                                                 (4.62)
                                                                      ζ               1     
               V(0) 1−α    W ( 0)       (1 − C 0   ) 1−α (1 − C
                                                                 0)       X ( 0) +      Ω(0)
                                                                                     ην     
4.2.4. Equation for the control variable u.
From equation (4.30), using (4.32) and (4.56), we derive the equation
        u (1 − β) (π + δ − ϑ + n )        W
          =                        + δu +
        u            β                    W
which is equivalent to the equation
            (1 − β) (π + δ − ϑ + n ) W 
        u −
                                    +  u = δ u2                                                (4.63)
                       β             W
This is a Bernoulli differential equation, which under the substitution

        y = u −1                                                                                 (4.64)
becomes a linear first order differential equation
           (1 − β) (π + δ − ϑ + n ) W 
        y+
                                   +  y = −δ                                                   (4.65)
                      β             W
The previous equation admits the integrating factor

        J(t) = eµ t W                                                                            (4.66)
where
             (1 − β) (π + δ − ϑ + n )
        µ=                                                                                       (4.67)
                        β
Multiplying equation (4.65) by the integrating factor, we obtain the equation
        d
           ( J ( t ) y) = − δ e µ t W
        dt
which, upon integration in the interval [0, t ] , gives us
                                        t
        J ( t ) y − J (0) y(0) = − δ ∫ e µ t W (s) ds                                            (4.68)
                                        0

Let



                                                                                                      51
                    t
                         µt
        X( t ) =    ∫e        W (s) ds                                                  (4.69)
                    0

Using the expression (4.57), with Ω( t ) given by (4.51), we find that
                    t
                         µs
        X( t ) =    ∫e        W (s) ds =
                    0

                                                                           t
                               1                       
        = (1 − C 0 ) ζ X(0) −    F(−ζ, η ;1 + η ; C 0 ) ∫ e µ s ds +
                              ην                       0
                              t
                       1           (µ − η ν ) s
                    +         ∫e                  F(−ζ, η ;1 + η ; C 0 e −ν s ) ds    (4.70)
                      ην      0

The last integral is evaluated in Appendix B. We get from (B.16), using the
notation of this section,
         t
               ϕt
         ∫e         F (− ζ, η ;1 + η ; C 0 e − ν t ) dt =
         0

              1  ϕt     p, − ζ , η                       p, − ζ , η     
      =−        e 3 F2              C 0 e − ν t  − 3 F2              C0           (4.71)
             νp        p + 1, η + 1                     p + 1, η + 1    
where
                                                  ϕ
        ϕ = µ − ην and p = −                                                            (4.72)
                                                  ν
Therefore we have the following expression for the function X( t ) :
                    t
                         µs
        X( t ) =    ∫e        W (s) ds =
                    0

                    ζ        1                         eµ t − 1 
        = (1 − C 0 ) X(0) −    F(−ζ, η ;1 + η ; C 0 )           −
                            ην                        µ 
                                                                   

               1     ϕt     p, − ζ , η                       p, − ζ , η     
        −           e 3 F2              C 0 e − ν t  − 3 F2              C0       (4.73)
             ην 2 p        p + 1, η + 1                     p + 1, η + 1    
We then obtain from (4.68)



                                                                                               52
               J (0) y(0) − δ X( t )
          y=                                                              (4.74)
                       J(t )
and then
                       J(t )
          u=                                                              (4.75a)
               J (0) y(0) − δ X( t )
or, comparing (4.66) and (4.69),
                       X( t )
          u=                                                              (4.75b)
               J (0) y(0) − δ X( t )

4.2.5. Equation for the human capital.
Equation (4.21) for the human capital, because of the previous expression (4.75b),
takes on the form
          H             − δ X( t )
            = δ−ϑ+
          H        J(0) y(0) − δ X( t )
or even
          H        S
            = δ−ϑ+                                                        (4.76)
          H        S
where
        S( t ) = J (0) y(0) − δ X( t )                                    (4.77)
Equation (4.76) gives upon integration
                   H0
        H( t ) =        S( t ) e (δ−ϑ) t                                  (4.78)
                   S(0)

4.3. Second method of solution.
4.3.1. Equations for the control variables.
Solving equation (4.7) with respect to c, we obtain
                         1
                     −
        c = (λ K )       σ                                                 (4.79)
Solving equation (4.8) with respect to u, we obtain




                                                                               53
                         1             1             1− β
             (1 − β)A  β       λK β
                                     K                β
       u =                       
                                    HN                                                                    (4.80)
                   δ         λH 
We also get from the previous relation the two relations
                                      1−β                 1−β                1− β
                           (1 − β)A  β            λ K  β 1−β
       (uNH)1−β =                                
                                                   λ    K             N    β
                                                                                                           (4.81)
                                δ                 H
and
                                    1−β                  1−β             1− β
                         (1 − β)A  β             λ K  β 1−β
       (u N)1−β =                               
                                                  λ    K          N    β
                                                                                H β−1                      (4.82)
                              δ                  H
we shall need later on.
4.3.2. Equations for the costate variables.
Equation (4.9), because of (4.81), becomes
                                                     1−β          1− β                  1−β            1
                                                                                    −
                            (1 − β)A                β            β                     β             β
       λ K = (ρ + π) λ K −                               βA N          (λ H )               (λ K )       (4.83)
                                δ 
Equation (4.10), because of (4.82) and (4.80), takes on the form
       λ H = (ρ + ϑ) λ H − λ H δ −
                                                  1−β             1−β      1− β
                                       (1 − β)A  β         λK  β
                 − λ k (1 − β) A                          
                                                            λ   K H −1 N β                          −
                                              δ            H
                                               1            1       1− β
                        (1 − β) A   λ K     β           β K      β
                 + λH δ                                 
                                                            HN
                            δ   λH                     
which is equivalent to the equation
       λ H = (ρ + ϑ − δ) λ H                                                                               (4.84)
since, as we can easily check, the other two terms cancel.
Equation (4.84) admits the general solution

       λ H ( t ) = λ H ( 0 ) × e ( ρ + ϑ− δ ) t                                                            (4.85)



                                                                                                               54
Equation (4.83), using (4.15) and (4.85), takes the simplified form
      λ K = (ρ + π) λ K −
                                       1−β                   1−β                                         1
                                                         −
       (1 − β) AN 0                   β                     β         (1 − β)(n + δ − ρ − ϑ) 
   −βA                                     (λ H (0))             exp 
                                                                                              t  (λ K ) β
                                                                                                             (4.86)
             δ                                                                  β             
Introducing the notation
                                     1−β
                       (1 − β)AN 0  β
             ε = βA                                                                                         (4.87)
                                  δ          
and
                   (1 − β)(n + δ − ρ − ϑ)
             µ=                                                                                               (4.88)
                             β
equation (4.86) can also be written as
                                                                  1−β              1
                                                              −
                                                                   β µt            β
             λ K − (ρ + π) λ K = −ε (λ H (0))                        e    (λ K )                              (4.89)
Equation (4.89) is a Bernoulli differential equation. Multiplying through by
             1
         −
             β
(λ K )           , this equation becomes
                              1                          1                         1−β
                          −                         1−                         −
                              β                          β                          β µt
             λ K (λ K )           − (ρ + π) (λ K )            = −ε (λ H (0))           e                      (4.90)
Introducing a new function λ by
                               1
                          1−
                               β
             λ = (λ K )                                                                                       (4.91)
equation (4.90) takes on the form
                                                                          1−β
                                                      −
                (1 − β)(ρ + π)    ε (1 − β)                                β µt
             λ+                λ=           (λ H (0))                         e                               (4.92)
                       β              β

The previous equation admits the function e α t as an integrating factor, where
                   (1 − β)(ρ + π)
             α=                                                                                               (4.93)
                          β



                                                                                                                  55
Therefore the general solution of (4.92) is given by
                                                   1−β
                                            −
                αt      ε (1 − β)                   β (µ + α ) t
        λe            =           (λ H (0))            e           +C                               (4.94)
                        β (µ + α)
For t = 0 , we obtain from the previous equation the relation
                                                            1−β
                                                       −
                                   ε (1 − β)                 β
                          λ ( 0) =           (λ H (0))            +C                                (4.95)
                                   β (µ + α)
from which we determine the constant C.
Therefore equation (4.94) takes the form
                                           1−β                                        1−β 
                            −                                                        −
        ε (1 − β)                           β µt                 ε (1 − β)                
     λ=           (λ H (0))                    e     +  λ ( 0) −           (λ H (0)) β  e − α t
        β (µ + α)                                                β (µ + α)                
                                                                                          
The above equation, taking into account (4.91), can also be written as
                          1−β                              1−β
                      −                               −
                           β      ε (1 − β)                 β −α t
        (λ K )                  =           (λ H (0))         e      ×
                                  β (µ + α)

                                                                          1−β 
                                                                         −
                                                  β (µ + α)  λ K (0)  β 
                          × − 1 + e ( µ + α ) t +                           
                                                  ε (1 − β)  λ H (0) 
                                                                             
                                                                              
or, in equivalent form,
                1−β                                     1−β
            −                               −
                 β          ε                            β         (1 − β) (ρ + π) 
   (λ K )             =           (λ H (0))                   exp  −
                                                                                  t×
                                                                                    
                        n −ϑ+δ+ π                                         β        
                                                                           1−β 
                                                                          −
                    (1 − β) (n − ϑ + δ + π)  n − ϑ + δ + π  λ K (0)  β 
       × − 1 + exp 
                                β
                                             t +
                                                     ε
                                                              
                                                               λ (0)         
                                                            H              
                                                                               
From the previous equation we get the following expression for the costate
variable λ K :




                                                                                                        56
                                      β
                                 −
            ε                      1−β
  λK =                                   λ H (0) e (ρ+ π) t ×
        n −ϑ+δ+ π
                                                                                                                  β
                                                                        1−β                                 −
                                                                     −                                          1−β
               (1 − β) ( n − ϑ + δ + π)  n − ϑ + δ + π  λ K (0)  β 
  × − 1 + exp 
                            β
                                         t +
                                                 ε        λ (0) 
                                                                                                                    (4.96)
                                                        H              
                                                                           
This is an explicit expression for the costate variable λ K .
In subsequent calculations, we shall need a compact form of this equation. For this
purpose we introduce the notation
                          (1 − β) (n − ϑ + δ + π)
                    ν=                                                                                                 (4.97)
                                     β
and
                                                                          1−β
                                                                      −
                                 n − ϑ + δ + π  λ K (0)                  β
                    C0 = 1 −
                                       ε
                                               
                                                λ (0)                                                               (4.98)
                                                H 
Therefore equation (4.96) takes on the compact form
                                                β                                                       β
                                           −                                                       −
                     ε                       1−β                (ρ + π) t        νt                  1−β
         λK    =                                   λ H ( 0) e               (e        − C0 )                         (4.99)
                 n −ϑ+δ+ π
4.3.3. Equation for the physical capital.
We now consider equation (4.11). This equation, because of (4.81) and (4.79),
takes the form
                            1−β             1−β           1− β                                 1
                (1 − β) A  β        λK  β              β
                                                                                           −
                                                                                               σ
      K=A                          
                                     λ             N          K − π K − (λ K )                  N                   (4.100)
                     δ              H
In the above expression, we need to find the ratio of the costate variables.
Using the expressions (4.99) and (4.85), we get
             1−β
   λK       β      n −ϑ+δ+ π        (1 − β) (π + δ − ϑ) 
  
  λ     
                  =           × exp 
                                                           t  × (e ν t − C 0 ) − 1
                                                                                                                      (4.101)
   H                   ε                     β           



                                                                                                                           57
Equation (4.100), taking into account (4.101) and (4.99), and also (4.15) for the
population, takes the form
                                1−β
                 (1 − β)A N 0  β           n −ϑ+δ+ π
        K=A                                         ×
                       δ                       ε    

               (1 − β) (n + π + δ − ϑ) 
        × exp 
                                      t  × (e ν t − C 0 ) − 1 K − π K −
                                         
                          β             
                                                  β
                                           −                            −
                                                                            1
                      ε                      σ (1−β)                      σN               ρ+π 
                −                                      (λ H (0))               0   exp n −     t×
                  n −ϑ+δ+ π                                                                 σ  
                                                            β
                                            νt            σ(1−β)
                                  × (e           − C0 )                                                   (4.102)
We can simplify the above expression considerably using the previously defined
quantities. We thus find, using the expression (4.87) which defines ε and (4.97)
which defines ν , that
                            1−β
          (1 − β)A N 0     β     n −ϑ+δ+ π n −ϑ+δ+ π    ν
        A                                 =          =
                δ                    ε         β       1− β
and
             (1 − β) (n + π + δ − ϑ)  ν t
        exp 
                                    t = e
                                      
                        β            
Therefore equation (4.102) takes on the form
                                                                            β
                  νt 
  K+ π − 1 ⋅ νe       K = − D 0 (e ν t − C 0 ) σ(1−β) exp n − ρ + π  t  (4.103)
                                                                       
        1 − β e − C0 
                νt
                                                                  σ  
                     
where
                                             β
                                      −                         −
                                                                    1
                  ε                     σ (1−β)                   σ
        D0 =                                      (λ H (0))           N0                                (4.104)
              n −ϑ+δ+ π
Equation (4.103) admits the integrating factor




                                                                                                              58
                                                                                        1
                                           νt                       −
                     
        I( t ) = exp ∫      π − 1 ⋅ νe          dt  = (e ν t − C 0 ) 1−β × e π t
                                                                                           (4.105)
                               1 − β eν t − C0  
                                                
Multiplying equation (4.103) by the integrating factor I( t ) , we find
                                                               β      1
                                                         −
        d                                                                    ρ+π 
           (K ( t ) I( t )) = − D 0 (e ν t − C 0 ) σ(1−β) 1−β × exp  π + n −     t
        dt                                                                    σ  
and integrating in the interval [ 0, t ] , we arrive at the equation
                                                       t
        K ( t ) I( t ) − K 0 I(0) = − D 0 ∫ e ω s (e ν s − C 0 ) m ds                       (4.106)
                                                       0

where
                           ρ+π
        ω=π+n−                                                                              (4.107)
                            σ
and
                  β       1
        m=             −                                                                    (4.108)
              σ (1 − β) 1 − β
The integral which appears in (4.106) has been evaluated in Appendix A. We
have, using (A.16) and the notation of this section
         t
             ωs
        ∫e        (e ν s − C 0 ) m ds =
        0

              1
        =−      [ e −νη t F (− m, η ;1 + η ; C 0 e − ν t ) − F (− m, η ;1 + η ; C 0 ) ]     (4.109)
             νη
where
                ω+νm
        η=−                                                                                 (4.110)
                  ν
We are now in a position to find the final expression for the function K ( t ) using
(4.106), (4.105) and (4.109):
                                        1                                      1
                                   −                                      −
                     νt                1−β        πt                          1−β
        K ( t ) (e        − C0 )             ×e        − K 0 (1 − C 0 )             =



                                                                                                59
             D 0 −νη t
        =       [e     F ( − m, η ;1 + η ; C 0 e − ν t ) − F ( − m, η ;1 + η ; C 0 ) ]
             νη
or
                                  1                          −
                                                                 1              
                                                                     D0        
        K ( t ) = (e ν t − R )   1−β
                                       ×e −πt
                                              K 0 (1 − C 0 )
                                                                1−β
                                                                    +    Ψ ( t )            (4.111)
                                                                     νη        
                                                                               
where

        Ψ ( t ) = e − νη t F ( − m , η ;1 + η ; C 0 e − ν t ) − F ( − m , η ;1 + η ; C 0 )   (4.112)

4.3.4. Equation for the human capital.
We come now in finding an explicit solution of the differential equation (4.12) for
the human capital. Substituting into (4.12) the expression for u given by (4.80),
we obtain the following differential equation
                                              1           1 1− β
                                  (1 − β)A  β     λK  β β
        H = (δ − ϑ) H − δ                         
                                                       N K
                                                                                            (4.113)
                                        δ         λH 
From (4.99) we obtain
                 1                           1                                          1
       λK                                                                 −
                 β     n − ϑ + δ + π  1−β       π+δ−ϑ 
      
      λ     
                    =                    × exp 
                                                       t  × (e ν t − C 0 ) 1−β
                                                          
       H                   ε                    β    
Using the previous expression and also (4.15) for N, we get the following
differential equation for H:
                                                             1
                                                        −
        H = (δ − ϑ) H − Q × e ζ t × (e ν t − C 0 )          1−β
                                                                  K                          (4.114)
where
                                 1                      1         1−β
              (1 − β)A  β  n − ϑ + δ + π  1−β
        Q = δ                                 × N 0β                                     (4.115)
                  δ              ε       
and
                 π + δ − ϑ (1 − β) n
        ζ=                +                                                                  (4.116)
                     β         β


                                                                                                  60
Using now the expression for K ( t ) given by (4.111), we obtain the equation

        H + ( ϑ − δ) H =

                                                             −
                                                                  1                    −
                                                                                           1            
                           ( ζ − π) t          νt                1−β                     1−β   D0      
        = −Q × e                        × (e        − C0 )              K 0 (1 − C 0 )       +    Ψ(t)    (4.117)
                                                                                               νη      
                                                                                                       
This is a linear first order differential equation with integration factor

        e ( ϑ− δ ) t
Therefore it can be written as
                                                                   −
                                                                       1            
        d ( ϑ− δ ) t            ( ζ − π + ϑ− δ ) t                   1−β   D0      
           (e        H ) = −Q e                     K 0 (1 − C 0 )       +    Ψ(t)                        (4.118)
        dt                                                                 νη      
                                                                                   
Integrating the previous formula in the interval [ 0, t ] , we obtain

                                              t       −
                                                          1               
      ( ϑ− δ ) t                     ωt                 1−β   D0         
    e            H( t ) = H 0 − Q ∫ e  K 0 (1 − C 0 )       +    Ψ ( t )  dt                              (4.119)
                                                              νη         
                                  0                                      
where
        ω= ζ −π+ϑ−δ                                                                                         (4.120)
The integral in the previous equation can be written as
                       
                       t              −
                                         1               
                    ωt                 1−β   D0         
        X( t ) ≡ ∫ e  K 0 (1 − C 0 )       +    Ψ ( t )  dt =
                                             νη         
                 0                                      

                          −
                              1                                 t
                                  D0                           
        =  K 0 (1 − C 0 )   1−β
                                 −    F ( − m, η ;1 + η ; C 0 )  ∫ e ω t dt +
                                  νη                           0
                                                               

          D0 t φ t                       −ν t
        +    ∫ e F (− m, η ;1 + η ; C 0 e ) dt
          νη 0
                                                                                                            (4.121)

where
        φ = ω−νη                                                                                            (4.122)



                                                                                                                61
The last integral is not elementary and needs to be evaluated using some results
from the theory of Generalized Hypergeometric Functions. We have evaluated this
integral in Appendix B. We have, using (B.11) and the notation of this section
        t
             φt
        ∫e        F ( − m, η ;1 + η ; C 0 e − ν t ) dt =
        0

               1  φt     p, − m, η                        p, − m, η      
        =−       e 3 F2              C 0 e − ν t  − 3 F2              C0       (4.123)
              νp        p + 1, η + 1                     p + 1, η + 1    


where
                  φ
        p=−                                                                          (4.124)
                  ν
We then obtain from (4.121)

                      t              −
                                         1               
                    ωt                 1−β   D0         
        X( t ) ≡ ∫ e  K 0 (1 − C 0 )       +    Ψ ( t )  dt =
                                             νη         
                 0                                      

                             1                                   ωt
                          −
                             1−β   D0                            e −1
        =  K 0 (1 − C 0 )       −    F ( − m, η ;1 + η ; C 0 )      −
                                  νη                            ω 
                                                                     

              1     φt     p, − m, η                        p, − m, η      
        −          e 3 F2              C 0 e − ν t  − 3 F2              C0     (4.125)
            ν 2η p        p + 1, η + 1                     p + 1, η + 1    
From (4.119) we arrive at the following final expression for the function H( t ) :

                   H ( t ) = { H 0 − Q ⋅ X ( t )} e (δ−ϑ) t                          (4.126)

where X( t ) is given by (4.125).


5. Application II.
(A more complicated Lucas-Uzawa model with externalities).
In this section we find a closed-form solution of the model considered by
Bucekkine and Ruiz-Tamarit (solved in Section 4) with externalities.


                                                                                          62
5.0. The model.
We consider the dynamic optimization problem
                         ∞
                             C( t )1−σ − 1
                  max ∫                    N ( t ) e −ρ t                                              (5.1)
                         0
                                1− σ

subject to

        K ( t ) = A K ( t ) α ( u ( t ) N ( t ) H ( t ))1−α H ( t ) β − π K ( t ) − C( t ) N ( t )     (5.2)

        H( t ) = γ (1 − u ( t )) H( t ) − ϑ H( t )                                                     (5.3)
with initial conditions
        K (0) = K 0 , H(0) = H 0 , N(0) = N 0                                                          (5.4)
where
        C( t ) ≥ 0 , u ( t ) ∈ [ 0, 1 ] , K ( t ) ≥ 0 , H( t ) ≥ 0                                     (5.5)
This is the model considered in previous section augmented with externalities.

5.1. The equations of motion.
The current value Hamiltonian is given by

                             C( t )1−σ − 1
        H (C, u, H, N, K ) =
          c
                                           N( t ) +
                                1− σ

        + λ K [ A K ( t ) α ( u ( t ) N ( t ) H ( t ))1−α H ( t ) β − π K ( t ) − C( t ) N ( t ) ] +

        + λ H [ γ (1 − u ( t )) H( t ) − ϑ H( t ) ]                                                    (5.6)

where λ K and λ H are the costate variables corresponding to K and H
respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method.
The first order conditions read

        ∂H c
             = 0 ⇔ C− σ N − λ K N = 0                                                                  (5.7)
         ∂C




                                                                                                           63
       ∂H c
            = 0 ⇔ λ K [ (1 − α) A K α u −α ( N H)1− α H β ] + λ H (− γ H) = 0   (5.8)
        ∂u
We also have the two Euler equations

       ∂H c
            + λK = ρλK ⇔
       ∂K

       ⇔ λ K − (ρ + π) λ K = − λ K α A K α −1 ( u N H )1− α H β                 (5.9)

       ∂H c
            + λH = ρ λH ⇔
       ∂H

       λ H − (ρ + ϑ) λ H = − λ K (1 − α + β) A K α ( u N )1− α H β−α −

                               − λ H γ (1 − u )                                 (5.10)
the dynamic constraints

       K = A K α ( u N H )1−α H β − π K − C N ,      K (0) = K 0                (5.11)

       H = γ (1 − u ) H − ϑ H , H(0) = H 0                                      (5.12)
and the transversality conditions

       lim λ K K e − ρ t = 0                                                    (5.13)
       t →∞

       lim λ H H e − ρ t = 0                                                    (5.14)
       t →∞

We also suppose that we have the exponential (exogenous) population growth
given by

       N(t) = N 0 e n t                                                         (5.15)

5.2. First method of solution to the model.
5.2.1. Simplification of the dynamical equations.
Taking logarithms and differentiation with respect to time, we obtain from
equation (5.7) that




                                                                                    64
            C λK
       −σ    =                                                               (5.16)
            C λK
Taking logarithms and differentiation with respect to time, we obtain from
equation (5.8) that
       λK    K   u         N         H λ
          + α − α + (1 − α) + (β − α) = H                                    (5.17)
       λK    K   u         N         H λH
From equation (5.9) we obtain
       λK
          = ρ + π − α AK α −1 (u N H)1−α H β                                 (5.18)
       λK
From equation (5.10) we obtain
       λH                                                        λ
          = ρ + ϑ − γ (1 − u ) − (1 − α + β) AK α (u N)1− α H β−α K          (5.19)
       λH                                                        λH
From equation (5.11) we obtain
       K                              CN
         = AK α −1 ( u N H)1− α H β −    −π                                  (5.20)
       K                               K
From equation (5.12) we obtain
       H
         = γ (1 − u ) − ϑ                                                    (5.21)
       H
From equation (5.15) we obtain
       N
         =n                                                                  (5.22)
       N
We introduce a function named Y defined by

       Y = AK α ( u N H )1− α H β                                            (5.23)
Equations (5.18)-(5.20) can be expressed in terms of the function Y as follows:
       λK            Y
          = ρ + π −α                                                         (5.24)
       λK            K

       λH                                   λ Y
          = ρ + ϑ − γ (1 − u ) − (1 − α + β) K ⋅                             (5.25)
       λH                                   λH H



                                                                                 65
        K Y CN
         = −   −π                                                            (5.26)
        K K K
From equation (5.8) we find that
        λK Y  γ
          ⋅ =     u                                                          (5.27)
        λH H 1− α
Because of the previous relation, we can simplify further equation (5.25):
        λH          βγ
           =ρ+ϑ−γ−      ⋅u                                                   (5.28)
        λH         1− α
Combining equations (5.16) and (5.24) we find an expression for the ratio
        C α Y ρ+π
         = ⋅ −                                                               (5.29)
        C σ K  σ
Equation (5.17), upon substituting the expressions given by (5.24), (5.26), (5.22)
and (5.28), we obtain the equation
        u     γ (1 − α + β)    NC
          =δ+               u−                                               (5.30)
        u         1− α         K
where
             γ + π − ϑ + (1 − α) n + β ( γ − ϑ)
        δ=                                      +ϑ−π− γ                      (5.31)
                             α
5.2.2. Auxiliary functions and their differential equations.
We now introduce two more functions U and Z defined by
             Y
        U=                                                                   (5.32)
             K
and
             NC
        Z=                                                                   (5.33)
             K
respectively.
We shall establish a system of two ordinary differential equations satisfied by
these two functions.




                                                                                  66
Taking logarithms and differentiation of the defining equations (5.32) and (5.33),
we obtain the equations
        U Y K
         = −                                                                (5.34)
        U Y K
and
        Z N C K
         = + −                                                              (5.35)
        Z N C K
respectively.
We can express the right hand sides of the two previous equations in terms of the
functions U and Z. From equation (5.26) we obtain
        K
          =U−Z−π                                                            (5.36)
        K
From equation (5.29) we obtain
        C α    ρ+π
         = ⋅U−                                                              (5.37)
        C σ     σ
                                       Y
We now have to express the ratio         in terms of U and Z. Taking logarithms and
                                       Y
differentiation of (5.23), we obtain the equation
        Y    K         u         N              H
          = α + (1 − α) + (1 − α) + (1 − α + β)                             (5.38)
        Y    K         u         N              H
                          K u    H
Substituting the ratios    , and   given by (5.26), (5.30), (5.22) and (5.21)
                          K u    H
respectively into the previous equation, we obtain
        Y
          =αU−Z+ε                                                           (5.39)
        Y
where
        ε = (1 − α) δ − α π + (1 − α) n + (1 − α + β) ( γ − ϑ)              (5.40)
Therefore we obtain from (5.34) and (5.35), using (5.36), (5.37) and (5.39), the
system of two equations



                                                                                   67
        U
          = ε + π − (1 − α) U                                          (5.41)
        U
and
        Z      α−σ         ρ+π
          = Z+     U+n +π−                                                      (5.42)
        Z       σ           σ
Equation (5.41) can be solved by separation of variables. The solution is given by

                1    ν eν t
        U=         ⋅ νt                                                         (5.43)
              1 − α e − C0

where
                                              π+ε
        ν = π + ε and C 0 = 1 −                                                 (5.44)
                                          (1 − α) U(0)
Using the expression (5.43) for the function U, equation (5.42) is converted into
the equation
            α−σ
                        ν eν t         ρ+π
                                            Z = Z2
        Z−             ⋅ νt       +n+π−                                         (5.45)
            σ (1 − α ) e − C            σ 
                               0          
This is a Bernoulli differential equation, solved under the substitution

        X = Z −1                                                                (5.46)


Equation (5.45) is converted in terms of X into the linear differential equation
           α−σ        ν eν t         ρ+ π
        X+          ⋅ νt       +n+π−      X = −1                          (5.47)
           σ (1 − α) e − C            σ 
                             0           
The integrating factor of the above equation is

        I( t ) = e ω t (e ν t − C 0 ) ζ                                     (5.48)
where
                        ρ+π                    α−σ
        ω= n +π−                    and ζ =                                 (5.49)
                         σ                    σ (1 − α)
Multiplying (5.47) by the integrating factor, we obtain the equation



                                                                                    68
        d
           ( I( t ) X ) = − e ω t ( e ν t − C 0 ) ζ
        dt
which, upon integration in the interval [ 0, t ] , gives
                                       t
        I( t ) X − I(0) X(0) = − ∫ e ω s (e ν s − C 0 ) ζ ds                               (5.50)
                                       0

The integral on the right hand side of the previous equation is calculated in
Appendix A. We have, using the notation of this section
        t
             ωs                                 1
        ∫e        (e ν s − C 0 ) ζ ds = −         Ω( t )                                   (5.51)
        0
                                               ην

where

        Ω ( t ) = e − ην t F( − ζ , η ;1 + η ; C 0 e − ν t ) − F( − ζ , η ;1 + η ; C 0 )   (5.52)
and
                  ω+ νζ
        η=−                                                                                (5.53)
                    ν
Therefore we obtain from (5.50) the following expression for the function X:
                                        1
              (1 − C 0 ) ζ X(0) +         Ω( t )
                                       ην
        X=                                                                                 (5.54)
                      e ω t (e ν t − C 0 ) ζ
We thus have that the function Z, related to X by (5.46), is given by

                e ω t (e ν t − C 0 ) ζ
        Z=                                                                                 (5.55)
                                  1
           (1 − C 0 ) ζ X(0) +         Ω( t )
                                 ην
Since from (5.51) we have upon differentiation with respect to t, the relation
                                      d 1           
        − e ω t (e ν t − C 0 ) ζ =        ην Ω( t ) 
                                                                                         (5.56)
                                      dt            
equation (5.55) is written as
                  W
        Z=−                                                                                (5.57)
                  W



                                                                                                69
where
                                         1
        W = (1 − C 0 ) ζ X(0) +            Ω( t )                                                     (5.58)
                                        ην
On the other hand, equation (5.43) can be expressed as
              1 V
        U=       ⋅                                                                                    (5.59)
             1− α V
where

        V = eν t − C0                                                                                 (5.60)

5.2.3. Equation for the physical capital.
Equation (5.36) can be written, because of (5.59) and (5.57), as
        K   1 V W
          =   ⋅ +  −π                                                                                 (5.61)
        K 1− α V W
The above equation can be integrated once, providing us with the following
expression for the physical capital
                     1
        K = D0    V 1−α     W e −πt
or
                                          1
                           νt                               ζ               1        
        K ( t ) = D 0 (e        − C0   ) 1−α (1 − C
                                                       0)       X ( 0) +      Ω( t )  e − π t        (5.62)
                                                                           ην        
where
                      K0                                              K0
        D0 =         1
                                       =                 1
                                                                                                       (5.63)
                                                                           ζ               1     
               V(0) 1−α     W ( 0)         (1 − C 0   ) 1−α (1 − C
                                                                      0)       X ( 0) +      Ω(0)
                                                                                          ην     
5.2.4. Equation for the control variable u.
From equation (5.30), using (5.33) and (5.57), we derive the equation
        u     γ (1 − α + β)    W
          =δ+               u+
        u         1− α         W
which is equivalent to the equation


                                                                                                           70
              W     γ (1 − α + β) 2
       u − δ +  u =
                                  u                                               (5.64)
              W
                         1− α
This is a Bernoulli differential equation, which under the substitution

       y = u −1                                                                    (5.65)
becomes a linear first order differential equation
              W       γ (1 − α + β)
       y + δ +  y = −
                                                                                 (5.66)
              W           1− α
The previous equation admits the integrating factor

       J(t) = e δ t W                                                              (5.67)
Multiplying equation (5.66) by the integrating factor, we obtain the equation
        d                   γ (1 − α + β) δ t
           ( J ( t ) y) = −              e W
        dt                      1− α
which, upon integration in the interval [0, t ] , gives us
                                                            t
                                  γ (1 − α + β)                  δt
       J ( t ) y − J (0) y(0) = −
                                      1− α                  ∫e        W (s) ds     (5.68)
                                                            0

Let
                  t
                       δt
       X( t ) =   ∫e        W (s) ds                                               (5.69)
                  0

Using the expression (5.52), we find that
                  t
                       δs
       X( t ) =   ∫e        W (s) ds =
                  0

                              1                       t
       = (1 − C 0 ) ζ X(0) −    F(−ζ, η ;1 + η ; C 0 ) ∫ e δ s ds +
                             ην                       0
                            t
                     1           ( δ − ν η) s
                  +         ∫e                  F(−ζ, η ;1 + η ; C 0 e −ν s ) ds   (5.70)
                    ην      0




                                                                                       71
The last integral is evaluated in Appendix B. We get from (B.16), using the
notation of this section,
         t
               ϕt
         ∫e         F (− ζ, η ;1 + η ; C 0 e − ν t ) dt =
         0

              1  ϕt     p, − ζ , η                       p, − ζ , η     
      =−        e 3 F2              C 0 e − ν t  − 3 F2              C0         (5.71)
             νp        p + 1, η + 1                     p + 1, η + 1    
where
                                           ϕ
        ϕ = δ − ν η and p = −                                                         (5.72)
                                           ν
Therefore we have the following expression for the function X( t ) :
                    t
                         δs
        X( t ) =    ∫e        W (s) ds =
                    0

                    ζ        1                          eδ t − 1 
        = (1 − C 0 ) X(0) −    F(−ζ, η ;1 + η ; C 0 )            −
                            ην                         δ 
                                                                    

               1     ϕt     p, − ζ , η                       p, − ζ , η     
        −           e 3 F2              C 0 e − ν t  − 3 F2              C0     (5.73)
             ην 2 p        p + 1, η + 1                     p + 1, η + 1    
We then obtain from (5.69)
               J ( 0) y ( 0) − µ X ( t )
        y=                                                                             (5.74)
                          J(t )
and then
                            J( t )
        u(t) =                                                                         (5.75)
                    J (0) y(0) − µ X( t )
where we have put
               γ (1 − α + β)
        µ=                                                                             (5.76)
                   1− α
5.2.5. Equation for the human capital.
Equation (5.75) can also be written as (comparing (5.67) and (5.69))



                                                                                           72
                           X( t )
          u(t) =                                                           (5.77)
                   J (0) y(0) − µ X( t )
Equation (5.21) for the human capital, because of the previous expression, takes
on the form
          H              − γ X( t )
            = γ −ϑ+
          H         J(0) y(0) − µ X( t )
or even
          H    γ S
            =γ+ ⋅                                                          (5.78)
          H    µ S
where
        S = J ( 0) y ( 0) − µ X ( t )                                      (5.79)
Equation (5.78) gives upon integration
                             γ
                H0
        H=           γ
                         S eγ t
                             µ

                     µ
              S(0)
or
                                             1− α
                         H0
        H( t ) =           1− α
                                   S( t )   1− α +β
                                                      eγ t                 (5.80)
                          1−α +β
                   S(0)


5.3. Second method of solution of the model.
5.3.1. Equations for the control variables.
From equation (5.7), solving with respect to C, we obtain
                         1
                     −
        C = (λ K )       σ                                                  (5.81)
From equation (5.8), solving with respect to u, we obtain




                                                                                   73
                         1                        1
                                 1− α                β
            (1 − α) A  α                   λK α α             K
       u =
                       
                               N α         
                                            λ   H                                                     (5.82)
                γ                          H                  H

We also get from the previous relation the two relations
                                    1− α                             1− α
                                                       1− α                β (1− α )
                        (1 − α) A  α                         λK  α
       (uNH)1−α = 
                                           
                                                     N α     
                                                              λ        H α K1−α                       (5.83)
                               γ                             H
and
                                 1− α                              1− α
                                                     1− α                 (1− α )(β − α )
                      (1 − α)A  α                          λK  α
       (uN)1−α = 
                                       
                                                   N α     
                                                            λ        H       α         K1−α           (5.84)
                           γ                               H
we shall need later on.
5.3.2. Equations for the costate variables.
Equation (5.9), because of (5.83), becomes
                                                            1− α                       1− α
                                                                       1− α                     β
                                                 (1 − α)A  α                   λK  α
       λ K − (ρ + π) λ K = − α A 
                                                             
                                                                     N α       
                                                                                λ           Hα   λK   (5.85)
                                                       γ                       H
Equation (5.10), because of (5.84) and (5.82), takes on the form
       λ H − (ρ + ϑ − γ ) λ H =
                                     1−α                                 1− α
                                                         1− α                    β−α
                          (1 − α)A  α                            λK  α
              = −β A 
                                               
                                                       N α       
                                                                  λ          H α    λK K              (5.86)
                                   γ                             H

5.3.3. Equation for the physical capital.
Equation (5.11), using (5.83) and (5.81), takes the simplified form
       K + πK =
                       1− α                                1− α
                                     1− α                         β                        1
            (1 − α)A  α                            λK  α α                         −
       = A
                       
                                   N α             
                                                    λ        H     K − N (λ K )         σ             (5.87)
                γ                                  H

5.3.4. Equation for the human capital.
Equation (5.12), because of (5.82) takes the form


                                                                                                             74
                                            1                    1
                                                    1− α            β
                                (1 − α)A  α               λK α α
        H − ( γ − ϑ) H = − γ 
                                              
                                                  N α     
                                                           λ   H     K        (5.88)
                                          γ               H

5.3.5. The solution strategy.
Equations (5.85)-(5.88) constitute a system of four equations which determine
completely the dynamics of the model. The solution strategy we follow is quite
different compared to the one we used without the externalities.
The first observation is that equation (5.86) cannot be solved as it is, since its right
hand side contains unknown quantities. Therefore we have to modify the solution
strategy in the presence of externalities.
Dividing (5.86) by (5.88) we obtain the equation
        λ H − (ρ + ϑ − γ ) λ H    β λH
                               =
           H − ( γ − ϑ) H        1− α H
which is equivalent to the equation
        λ H − (ρ + ϑ − γ ) λ H    β H − ( γ − ϑ) H
                               −      ⋅            =0                            (5.89)
                 λH              1− α     H
The previous equation can be written as
        λH   β H
           −   ⋅ =µ                                                              (5.90)
        λH 1− α H
where
                              β
        µ = (ρ + ϑ − γ ) +        (ϑ − γ )                                       (5.91)
                             1− α
Equation (5.90) admits the solution
                               β
        λ H (t) = C0   H( t ) 1−α   eµ t                                         (5.92)
where
                λ H (0)
        C0 =         β
                                                                                 (5.93)
               H(0) 1−α



                                                                                     75
Equation (5.92) expresses a relation between λ H and H. However this relation
simplifies considerably the other equations of motion. We then can obtain closed-
form solutions.
5.3.5.1. Solution of equation for the costate variable λ K .
Using this equation into the equation (5.85) for λ K , we obtain the equation
                                                 1− α
                                                          1− α                               1
                                      (1 − α)A  α                    µ (1 − α) 
  λ K − (ρ + π) λ K = − α A 
                             γC                 
                                                        N α      exp  −        t  (λ K ) α     (5.94)
                                0                                        α      
which can be solved, since it is a Bernoulli differential equation.
                                                                        1
                                                                    −
In fact multiplying this equation through by (λ K )                     α,   and taking into account
equation (5.15) for the population, we obtain the equation
                         1                       1
                     −                      1−
        λ K (λ K )       α   − (ρ + π) (λ K )    α   = − ε eν t                                   (5.95)
where
                                 1− α
                  (1 − α)AN 0  α
        ε = αA                                                                                (5.96)
                             γ C0     
and
             (n − µ) (1 − α)
        ν=                                                                                        (5.97)
                   α
Introducing a new function λ by
                          1
                     1−
        λ = (λ K )        α                                                                       (5.98)
equation (5.95) takes on the form
             (1 − α) (ρ + π)    ε (1 − α) ν t
        λ+                   λ=          e                                                        (5.99)
                   α                α

The previous equation admits the function e ξ t as an integrating factor, where




                                                                                                       76
                 (1 − α) (ρ + π)
       ξ=                                                                                        (5.100)
                       α
Therefore the general solution is given by
                            ε (1 − α) (ν + ξ) t
        λ eξ t =                       e        +C                                               (5.101)
                            α ( ν + ξ)
For t = 0 , we obtain from the previous equation the relation
                           ε (1 − α)
        λ (0) =                       +C                                                         (5.102)
                           α ( ν + ξ)
from which we determine the constant C.
Therefore we obtain from equation (5.101) the following expression for the
function λ :
            ε (1 − α) ν t          ε (1 − α)                  −ξ t
     λ=                e +  λ(0) −
                                                             e
                                                              
            α ( ν + ξ)             α ( ν + ξ)                
The above equation, taking into account (5.98), can be expressed in terms of the
function λ K as
                           1− α                            1− α
                       −
                            α       ε (1 − α) ν t 
                                                   λ (0) − α − ε (1 − α)
                                                                                      
                                                                                       e −ξ t
        (λ K )                    =            e + K                                 
                                    α ( ν + ξ)                  α ( ν + ξ)
                                                                                     
The previous equation can also be written as
                       1− α
                   −                  ε (1 − α) − ξ t
        (λ K )          α         =              e    ×
                                      α ( ν + ξ)

                                                                                  1− α 
                                                                                 −
                                                ( ν + ξ ) t α ( ν + ξ)                 
                                      × − 1 + e            +           (λ K (0)) α 
                                                             ε (1 − α)                 
                                                                                       
or, in equivalent form,
                1− α
            −                   ε             (1 − α) (ρ + π) 
   (λ K )        α     =               × exp  −              t×
                            ρ+ π+ n −µ              α         




                                                                                                     77
                                                                     1− α 
                 (1 − α) (ρ + π + n − µ)  ρ + π + n − µ          −      
    × − 1 + exp                         t +             (λ K (0)) α                        (5.103)
                            α                   ε                       
                                                                          
taking into account the expression for ξ and that
       α ( ν + ξ)
                  =ρ+ π+ n −µ
        1− α
From equation (5.103) we get the following expression for the costate variable
λK :
                                      α                               α
                                 −                               −
                  ε                1− α          ζt
        λK   =
              ρ+ π+ n −µ
                                           × (e        − D0 )       1− α   e (ρ + π ) t       (5.104)
                        
where we have introduced the notation
             (1 − α) (ρ + π + n − µ)
       ζ=                                                                                      (5.105)
                        α
                                                   1− α
                ρ+ π+ n −µ           −
       D0 = 1 −            (λ K (0))                α                                          (5.106)
                    ε
Equation (5.104) is an explicit expression for the costate variable λ K .

5.3.5.2. Solution of equation for the physical capital.
We now consider equation (5.87). This equation, because of (5.92), takes the form
    K + πK =
                      1− α
                              1− α           1− α                                          1
        (1 − α)A     α                                 µ (1 − α)               −
    =A 
        γC                N α     (λ K   ) α     exp  −        t  K − N (λ K ) σ          (5.107)
              0                                            α      
Using the expressions (5.104) for λ K and (5.15) for the population, we get from
the previous equation
       K + πK =
                         1− α
              (1 − α)A  α     ρ+ π+ n −µ     ζt
       =A 
           γC          
                                          × (e − D 0 ) ×
              0                   ε     



                                                                                                    78
                         n (1 − α) (ρ + π) (1 − α) µ (1 − α)  
                  × exp           +               −           t K −
                         α               α             α  
                                            α                                α
                 ε     σ (1−α )                                      ρ+π 
         − N0 
              ρ+π+n −µ
                                 × (e ζ t − D 0 ) σ (1−α ) × exp  n −     t  (5.108)
                                                                      σ  
The previous equation can be simplified considerably. Using (5.96) in place of ε
we find
                         1− α
              (1 − α)A  α        ρ+ π+ n −µ ρ+ π+ n −µ    ζ
         A
           γC             
                                            =           =                                        (5.109)
              0                      ε          α        1− α

where we also have used expression (5.105) for ζ .
We also have that
             n (1 − α) (ρ + π) (1 − α) µ (1 − α)   ζ t
         exp          +               −           t = e                                         (5.110)
             α               α             α  
using (5.105) again.
Therefore (5.108) takes on the form
                    ζt 
  K+ π − 1 ⋅ ζe         K = −
        1 − α eζ t − D0 
                        
                                    α                               α
          ε                    σ (1−α )          ζt            σ (1− α )              ρ+π 
  − N0 
       ρ+π+n −µ
                                           × (e        − D0 )               × exp  n −     t   (5.111)
                                                                                       σ  
The above differential equation is a linear first order differential equation with
integrating factor

                      
                                1     ζ eζ t   π t
                                                              ζt
                                                                       −
                                                                         1
         I( t ) = exp ∫   π −      ⋅           dt  = e × (e − D 0 ) 1−α                         (5.112)
                              1 − α eζ t − D0  
                                               
Therefore multiplying (5.111) by the integrating factor, we obtain the equation
d
   (K ( t ) I( t )) = −
dt




                                                                                                        79
                              α                               α       1
                                                                    −
         ε               σ (1−α )          ζt            σ (1− α ) 1− α                 ρ+π 
− N0 
     ρ+ π+ n −µ
                                     × (e        − D0 )                    × exp π + n −     t
                                                                                         σ  
and integrating in the interval [ 0, t ] , we arrive at the equation
                                             t
        K ( t ) I( t ) − K 0 I(0) = − F0 ∫ e ω t (e ζ t − D 0 ) η ds                              (5.113)
                                             0

where
                                            α
                     ε                 σ (1− α )
        F0 = N 0 
                 ρ+ π+ n −µ
                                                                                                 (5.114)
                           
                        ρ+π
        ω=π+n−                                                                                    (5.115)
                         σ
                 α       1
        η=            −                                                                           (5.116)
             σ (1 − α) 1 − α
The integral appearing in the right-hand-side of (5.113) has been evaluated in
Appendix A. We have, using the notation of this section:
        t
             ωs                                  1
        ∫e        (e ζ s − D 0 ) η ds = −           Ψ(t)                                          (5.117)
        0
                                                 υζ

where
               ω + ηζ
        υ=−                                                                                       (5.118)
                  ζ
and

        Ψ ( t ) = e − υζ t F( − η, υ ; υ + 1; D 0 e − ζ t ) − F( − η, υ ; υ + 1; D 0 )            (5.119)
We thus get the following expression for the physical capital K:
                                −
                                   1                             1
                                       F0             ζt
        K ( t ) = K 0 (1 − D 0 ) 1−α +    Ψ ( t ) × (e − D 0 ) 1−α                              (5.120)
                                       υζ        
                                                 
5.3.5.3. Solution of equation for the human capital




                                                                                                      80
We come now in finding an explicit solution of the differential equation (5.88) for
the human capital. Substituting into (5.88) the expression (5.92) for λ H , we obtain
the following differential equation
                                            1
                                                 1− α           1              β
                         (1 − α)A  α            α
                                                                          −                 µ 
                         γC  N
   H + (ϑ − γ ) H = − γ                               (λ K   )α     H       1− α   K exp  − t      (5.121)
                               0                                                          α 
Using the expressions for K and λ K given by (5.120) and (5.104) respectively,
and also the equation (5.15) for the human capital, the previous equation becomes
                                                 1                      β
                                              −                        −
        H + (ϑ − γ ) H = − G 0 K 0 (1 − D 0 )   1− α + F0 ~ ( t )  × H 1− α ×
                                                           F 
                                                       υζ         
                                                                  
                              n (1 − α) ρ + π µ  
                        × exp          +     −  t                                                  (5.122)
                              α           α   α 
where
                             1                                              1
                                   1− α                                −
                 (1 − α)A  α                      ε                    1− α
        G0 = γ 
                γC         
                                N0 α           ρ+ π+ n −µ
                                                                                               (5.123)
                   0                                    
Equation (5.122) is a Bernoulli differential equation. Multiplying this equation
                β
through by   H 1−α   , we obtain the equation
             β                        β                               1                
                                1+                                 −
        H H 1−α   + (ϑ − γ ) H       1− α   = − G 0 K 0 (1 − D 0 )   1− α + F0 Ψ ( t )  ×
                                                                                        
                                                                            υζ         
                                                                                       
                              n (1 − α) ρ + π µ  
                        × exp          +     −  t                                                  (5.124)
                              α           α   α 
                                                              β
                                                        1+
Introducing a new function U by U = H                        1− α ,   the above equation is converted to
the linear first order differential equation




                                                                                                           81
      (1 − α + β)(ϑ − γ )     (1 − α + β)                    −
                                                                 1                
                                                               1− α + F0 Ψ ( t )  ×
   U+                     U=−             G 0 K 0 (1 − D 0 )                     
             1− α                1− α                                 υζ         
                                                                                 
                                  n (1 − α) ρ + π µ  
                            × exp          +     −  t                               (5.125)
                                  α           α   α 
Multiplying the above equation by the integrating factor
                       (1 − α + β)(ϑ − γ ) 
        J ( t ) = exp                     t                                           (5.126)
                              1− α         
becomes equivalent to

                          (1 − α + β)                    −
                                                             1                
        d                                                  1− α + F0 Ψ ( t )  e φ t
           (J( t ) U) = −             G 0 K 0 (1 − D 0 )                              (5.127)
        dt                   1− α                                 υζ         
                                                                             
where
             (1 − α + β)(ϑ − γ ) n (1 − α) ρ + π µ
        φ=                      +         +     −                                       (5.128)
                    1− α             α       α    α
Integrating equation (5.127) in the interval [ 0, t ] , we obtain

                           1− α + β
                                        t                   −
                                                                1                
                                           φt                 1− α + F0 Ψ ( t )  dt
    J ( t ) U − U ( 0) = −          G 0 ∫ e  K 0 (1 − D 0 )                           (5.129)
                            1− α                                      υζ
                                        0                                       
The integral in the previous equation can be written as
                  t                        1               
        X( t ) ≡ ∫ e   φt    K (1 − D ) − 1−α + F0 Ψ ( t )  dt =
                             0       0
                                                 υζ         
                  0                                        
                        1                             t
          K (1 − D ) − 1−α − F0 F(− η, υ ;1 + υ ; D )  e φ t dt +
        = 0       0                                0 ∫
                              υζ
                                                      0

            F0 t (φ− υ ζ ) t
        +      ∫e            F(− η, υ ;1 + υ ; D 0 e − ζ t ) dt                         (5.130)
            υζ 0




                                                                                             82
The last integral is not elementary and needs to be evaluated using some results
from the theory of Generalized Hypergeometric Functions. We have, using
Appendix B
         t
         ∫e
               at
                    F(− η, υ ;1 + υ ; D 0 e − ζ t ) dt =
         0

                1     at     p, − η, υ                        p, − η, υ      
        =−           e 3 F2              D 0 e − ζ t  − 3 F2              D0        (5.131)
               ζp           p + 1, υ + 1                     p + 1, υ + 1    
where
                                                 a
        a = φ − υζ            and p = −                                                   (5.132)
                                                 ζ
We then obtain from (5.130)
                     t                       1               
        X( t ) ≡ ∫ e     φt    K (1 − D ) − 1−α + F0 ~ ( t )  dt =
                                                      F 
                               0       0
                                                   υζ
                     0                                       
                        1                              φt
          K (1 − D ) − 1−α − F0 F(− η, υ ;1 + υ ; D )   e − 1  −
        = 0                                                    
                                                    0 
                                                             φ 
                   0
                              υζ
                                                              

              F0     at     p, − η, υ                        p, − η, υ      
        −           e 3 F2              D 0 e − ζ t  − 3 F2              D0         (5.133)
             υ ζ 2p        p + 1, υ + 1                     p + 1, υ + 1    
From (5.129) we arrive at the following expression for the function U:
                    1− α + β                    (1 − α + β)(ϑ − γ ) 
        U =  U(0) −          G 0 X( t )  × exp  −                  t                  (5.134)
                     1− α                             1− α          
where X( t ) is given by (5.133).
                                         β
                                   1+
Using the fact that U = H               1− α ,   we find the following expression for the human
capital H:
                                                               1− α
                       1− α +β                    1−α +β
                  ( H ) 1−α − 1 − α + β G X( t ) 
        H( t ) =  0                                       × e ( γ − ϑ) t                 (5.135)
                                 1− α
                                          0       
                                                 


                                                                                                  83
6. Application III.
(The model considered by Ruiz-Tamarit and Sánchez-Moreno)
We now find the closed-form solution of the model introduced by Ruiz-Tamarit
and Sánchez-Moreno [29] on optimal regulation in a natural resource based
economy.

6.0. The Model.
We consider the dynamic optimization problem
                         ∞
                             C( t )1−σ − 1
                  max ∫                    N e −ρ t dt                                       (6.1)
                         0
                                1− σ

subject to

        K ( t ) = AK ( t ) β ( u ( t ) Q ( t ))1−β − N C( t )                                (6.2)

        Q( t ) = δ (1 − u ( t )) Q( t ) − u ( t ) Q( t )                                     (6.3)
with initial conditions
        K (0) = K 0 > 0 , Q(0) = Q 0 > 0                                                     (6.4)
where
        C( t ) ≥ 0 , u ( t ) ∈ [ 0, 1 ] , K ( t ) ≥ 0 , Q( t ) ≥ 0                           (6.5)
In this model Q( t ) is a renewable natural resource and u ( t ) is the aggregate
extraction rate, while N is the population (considered constant).

6.1. The equations of motion.
The current value Hamiltonian is given by

        H c (C, u , H, K ) =

                      C( t )1−σ − 1
                  =                 N + λ K [ A K ( t )β (u ( t ) H( t ))1−β − NC( t ) ] +
                         1− σ
                    + λ Q [ (δ − (1 + δ) u ( t )) Q( t ) ]                                   (6.6)




                                                                                                 84
where λ K and λ Q are the costate variables corresponding to K and Q

respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method.
The first order conditions read

       ∂H c
            = 0 ⇔ C− σ − λ K = 0                                                (6.7)
        ∂C

       ∂H c
            = 0 ⇔ −λ Q (1 + δ) + λ K [ (1 − β) AK β (u Q) − β ] = 0             (6.8)
        ∂u
We also have the two Euler equations

       ∂H c
            + λ K = ρ λ K ⇔ λ K = ρ λ K − λ K βAK β−1 (u Q)1− β                 (6.9)
       ∂K

       ∂H c
            + λQ = ρ λQ ⇔
        ∂Q

       ⇔ λ Q = ρ λ Q − λ K (1 − β) K β (u Q)1− β Q − 1 − λ Q (δ − (1 + δ) u )   (6.10)

the dynamic constraints

       K ( t ) = AK ( t ) β ( u ( t ) Q ( t ))1−β − N C( t ) , K (0) = K 0      (6.11)

       Q( t ) = δ (1 − u ( t )) Q( t ) − u ( t ) Q( t ) , Q(0) = Q 0            (6.12)
and the transversality conditions

       lim λ K K e − ρ t = 0                                                    (6.13)
       t →∞

       lim λ Q Q e − ρ t = 0                                                    (6.14)
       t →∞

6.2. The method of solution.
6.2.1. Equations for the control variables.
Solving equation (6.7) with respect to C, we obtain




                                                                                    85
                        1
                    −
       C = (λ K )       σ                                                                              (6.15)
Solving equation (6.8) with respect to u, we obtain
                         1                1
                  λ 
              (1 − β)A  β                β   K
       u =        K                                                                   (6.16)
           1+ δ   λQ 
                                            Q

We also get from the previous relation that
                                    1−β             1−β
                         (1 − β)A  β         λK  β
       (u Q)1−β =                                     K1−β                                         (6.17)
                   1+ δ                      λQ 
                                                  
we shall need later on.
6.2.2. Equations for the costate variables.
Equation (6.10), because of (6.16) and (6.17), takes on the form
       λ Q + ( δ − ρ) λ Q = 0                                                                          (6.18)

Equation (6.18) admits the general solution

       λ Q = λ Q ( 0) × e ( ρ − δ ) t                                                                  (6.19)

Equation (6.9), because of (6.17), becomes
                                             1−β                 1−β
                                  (1 − β)A  β     λK          β
       λ K = ρ λ K − βA                                            λK                              (6.20)
                         1+ δ                     λQ      
                                                            
This equation can be written as (because of (6.19))
                                                       1−β                          1−β
                                                   −                                               1
                        1+ δ                          β         (1 − β)(ρ − δ)   β             β
  λK − ρλK              (1 − β)A λ Q (0) 
                = − βA                                     exp  −
                                                                                 t
                                                                                         (λ K )
                                                                      β         
                                                                                                             1
                                                                                                         −
                                                                                                             β
is a Bernoulli differential equation. Multiplying this equation through by (λ K )                                ,
we derive the equation
                        1                 1
                    −                1−
                        β                 β
       λ K (λ K )           − ρ (λ K )        =


                                                                                                             86
                                            1−β                              1−β
                                        −
                  1+ δ                     β         (1 − β)(ρ − δ)       β
                  (1 − β)A λ Q (0) 
          = − βA                                exp  −
                                                                      t
                                                                                         (6.21)
                                                           β         
Introducing the function λ by
                          1
                     1−
                          β
          λ = (λ K )                                                                               (6.22)
equation (6.21) is converted to the equation
               ρ (1 − β)
          λ+             λ=
                   β
                                  1                1−β
                                           −
                     (1 − β)A   β                 β          (1 − β)(δ − ρ) 
          = (1 + δ)            (λ Q (0))                exp 
                                                                             t
                                                                                                  (6.23)
                     1+ δ                                          β        
which is a linear first order differential equation.
Since the integrating factor is
                        ρ (1 − β) 
          I( t ) = exp 
                        β t      
                                  
equation (6.23) can be written by multiplying through by I( t )
                                                  1             1−β
                                                         −
          d                        (1 − β)A  β                 β         δ (1 − β) 
             (I( t ) λ) = (1 + δ)            (λ Q (0))                   β t
                                                                      exp            
          dt                       1+ δ                                            
and by integration in the interval [0, t ] ,
                                                      1           1−β t
                                                             −
                                     (1 − β)A  β                 β          δ(1 − β) 
     I( t ) λ − I(0) λ(0) = (1 + δ) 
                                     1+ δ 
                                                  (λ Q (0))           ∫  β s  ds
                                                                         exp 
                                                                             
                                                                                        
                                                                                        
                                                                       0

or
                               1−β                        
                              −        
                                            δ(1 − β)          ρ (1 − β) 
       λ = λ(0) + ε (λ Q (0)) β             β t  − 1   × exp  − β t 
                                       exp                                                    (6.24)
                                                                       
                                                          
where




                                                                                                       87
                                                   1
             (1 + δ) β  (1 − β)A                 β
        ε=                                                                                             (6.25)
             δ (1 − β)  1 + δ 

Equation (6.24), written in terms of λ K , is given by
                     1−β                 1−β                1−β                                     
                 −                      −                  −                     
                      β                   β                                           δ(1 − β)   
        (λ K )             = (λ K (0))       + ε (λ Q (0)) β                          β t  −1   ×
                                                                                 exp           
                                                                                               
                                                                                                    

                                       ρ (1 − β) 
                                × exp  −
                                                t
                                                  
                                           β     
or
                     1−β                           1−β
                 −                             −
                                                                               ρ (1 − β) 
        (λ K )        β
                           = ε (λ Q (0))            β
                                                         (e ν t − C 0 ) × exp  −
                                                                                        t
                                                                                                        (6.26)
                                                                                   β     
where
             δ (1 − β)
        ν=                                                                                               (6.27)
                 β
and
                                             1−β
                                         −
                1  λ (0)                    β
        C0 = 1 −  K                                                                                    (6.28)
                ε  λ Q ( 0) 
                            
We thus derive from (6.26) the following expression for the costate variable λ K :
                          β                                        β
                     −                                        −
        λK = ε           1−β
                               λ Q (0) (e ν t − C 0 )             1−β
                                                                        × eρ t                           (6.29)

6.2.3. Equation for the physical capital.
We now consider equation (6.11). This equation, because of (6.17) and (6.15),
takes the form
                                         1−β                  1−β
                                     −                                                 1
              1+ δ                      β     λK           β                   −
        K = A                                                                       σ
              (1 − β)A 
                                               λQ       
                                                                    K − N (λ K )                         (6.30)
                                                       




                                                                                                             88
In order to simplify the above expression, we need to find the ratio of the costate
variables. Using the expressions (6.29) and (6.19), we obtain
               1−β
         λK
            
               β          −1           eν t
                     =ε         ×                                                         (6.31)
         λQ                       eν t − C0
            
Equation (6.30), taking into account (6.31) and (6.29), takes on the form
                                     1−β
                                 −
              1+ δ                  β −1              eν t
              (1 − β)A 
        K − A                          ε        ×                K=
                                                    eν t − C0
                    β                       1                        β
                                        −                                        ρ 
                 σ (1−β)
        = −Nε              (λ Q (0))        σ    (e ν t − C 0 )   σ (1−β)
                                                                            exp  − t    (6.32)
                                                                                 σ 
The previous equation can be simplified considerably using the definition of
parameters introduced earlier.
We find, using the definition (6.25) of ε and (6.27) of ν :
                               1−β
                           −
          1+ δ                β −1         δ   ν
          (1 − β)A 
        A                         ε    =     =
                                           β 1− β
Therefore (6.32) takes the form
                                                                            β
             1    ν eν t                                       ρ 
        K−      ⋅ νt     K = − D 0 (e ν t − C 0 ) σ (1−β) exp  − t                      (6.33)
           1 − β e − C0                                        σ 

where
                      β                          1
                                             −
                   σ (1−β)                       σ
        D0 = N ε               (λ Q (0))                                                  (6.34)

Equation (6.33), which is a linear first order differential equation, admits the
integrating factor
                                                                                 1
                      1         ν eν t                       −
        J( t ) ≡ exp  −                  dt  = (e ν t − C 0 ) 1−β
                      1 − β ∫ eν t − C
                                                                                          (6.35)
                                             
                                       0    
Multiplying equation (6.33) through by the integrating factor I( t ) , we find



                                                                                              89
                                                     β       1
                                                               −
            d                                                            ρ 
               (K ( t ) J ( t )) = − D 0 (e ν t − C 0 ) σ (1−β) 1−β exp  − t 
            dt                                                           σ 
and integrating in the interval [ 0, t ] , we arrive at the equation
                                            t
        K ( t ) J ( t ) − K 0 J (0) = − D 0 ∫ e ωs (e ν s − C 0 ) µ ds                         (6.36)
                                            0

where we have introduced the notation
                    α       1                         ρ
        ν=               −             and ω = −                                               (6.37)
                σ (1 − α) 1 − α                       σ
The integral appearing in (6.36) is not elementary. It can be evaluated using the
hypergeometric function, known from the theory of differential equations in the
complex domain. All the relevant calculations appear in Appendix A. We have
found using (A.16) and the notation of this section
  t
       ωs
  ∫e        (e ν s − C 0 ) µ ds =
  0

        1
  =−      [ e −νη t F (− µ, η ;1 + η ; C 0 e − ν t ) − F (− µ, η ;1 + η ; C 0 ) ]              (6.38)
       ην
where
                  ω + µν
        η=−                                                                                    (6.39)
                    ν
We are now in a position to write down the final expression for the function K ( t )

                                 −
                                    1                                 1
                                        D0        
        K ( t ) =  K 0 (1 − C 0 ) 1−β +    Ω( t )  × (e ν t − C 0 ) 1−β                      (6.40)
                                        νη        
                                                  
where

        Ω ( t ) = e − νη t F ( − µ , η ; 1 + η ; C 0 e − ν t ) − F ( − µ , η ; 1 + η ; C 0 )   (6.41)

6.2.4. Equation for the renewable natural resource Q.




                                                                                                   90
We come now in finding an explicit solution of the differential equation (6.12) for
the renewable natural resource Q. Substituting into (6.12) the expression for u
given by (6.16), we obtain the following differential equation
                                                                     1          1
                                                                 −
                               1+ δ                                β    λK   
                                                                                β
          Q − δ Q = − (1 + δ)                                                  K
                               (1 − β)A 
                                                                         λQ   
                                                                                               (6.42)
                                                                             
From (6.31) we obtain
                   1                                                1
                                  1
            λK   β         −           e       νt              1−β
                =ε             1−β
                                       × νt  
            λQ                        e −C 
                                          0

Using the above expression and also (6.40) in (6.42), we obtain the equation
                                          −
                                              1              
                                            1−β   D0                ν 
          Q − δ Q = − E 0  K 0 (1 − C 0 )       +    Ω( t )  × exp 
                                                                     1− β t 
                                                                                              (6.43)
                                                  νη                      
                                                            
where
                                                     1
                                                 −           −
                                                                  1
                         1+ δ                      β           1−β
                         (1 − β) A 
          E 0 = (1 + δ)                                ε                                     (6.44)
                                   
Equation (6.43) is a linear first order differential equation with integration factor

          I( t ) = e − δ t
Therefore it can be written as
                                    −
                                        1              
   d −δt                              1−β   D0                      ν                  
      (e Q) = − E 0  K 0 (1 − C 0 )       +    Ω( t )  × exp − δ +
                                                                                          t  (6.45)
                                                                                          
   dt                                       νη                     1− β                
                                                      
Integrating the previous formula in the interval [ 0, t ] , we obtain

                                       t                        −
                                                                    1              
        −δ t                               νt                     1−β   D0        
    e          Q − Q0 = − E 0 ∫ e                K 0 (1 − C 0 )       +    Ω( t )  dt        (6.46)
                                                                        ην        
                                       0                                          




                                                                                                   91
Using the expression for the function Ω( t ) , given in (6.41), equation (6.46) can be
transformed into the equation

        e −δ t Q = Q 0 −

                               −
                                    1                                          t
                                 1−β D 0                                          νt
        − E 0   K 0 (1 − C 0 )      −
                                        ην
                                           F(− µ, η ;1 + η ; C 0 )             ∫e        dt +
                                                                 
                                                                             0


                                D     t                                                  
                                                                                         
                               + 0    ∫   e ν (1−η) t F(− µ, η ;1 + η ; C 0 e − ν t ) dt        (6.47)
                                ην    0                                                  
                                                                                         
The last integral is not elementary and needs to be evaluated using some results
from the theory of Generalized Hypergeometric Functions. We put the integral
into the form
                     t
                         φt
                    ∫e        F(− µ, η ;1 + η ; C 0 e − ν t ) dt                                 (6.48)
                    0

where
                   φ = ν (1 − η)                                                                 (6.49)
The integral has been evaluated in Appendix B. Using (B.11) and the notation of
this section, we have
        t
             φt
        ∫e        F (− µ, η ;1 + η ; C 0 e − ν t ) dt =
        0

              1  φt     p, − µ , η                       p, − µ, η      
        =−      e 3 F2              C 0 e − ν t  − 3 F2              C0                    (6.50)
             νp        p + 1, η + 1                     p + 1, η + 1    
where the parameter p has defined by: − ν p = ϕ .
Therefore from (6.47), using (6.50), we arrive at the following final expression for
the function Q( t ) :

                   Q ( t ) = e δ t [ Q 0 − E 0 X ( t )]                                          (6.51)

where X( t ) is given by



                                                                                                     92
                   t                −
                                        1              
                   νt                 1−β   D0        
       X( t ) ≡ ∫ e  K 0 (1 − C 0 )       +    Ω( t )  dt =
                                            ην        
                0                                     

                         −
                             1                                t
                                 D0                          
       =  K 0 (1 − C 0 )   1−β
                                −    F( − µ, η ;1 + η ; C 0 )  ∫ e ν t dt +
                                 ην                          0
                                                             
                             t
                     D            ν (1− η) t
                    + 0      ∫e                F(− µ, η ;1 + η ; C 0 e − ν t ) dt =
                     ην      0

                            1                                   νt
                         −
                            1−β   D0                             e −1
       =  K 0 (1 − C 0 )       −    F ( − µ, η ; 1 + η ; C 0 )     −
                                 ην                             ν 
                                                                   

             A     φt     p, − µ, η                        p, − µ, η      
       −          e 3 F2              C 0 e − ν t  − 3 F2              C0            (6.52)
           ν 2η p        p + 1, η + 1                     p + 1, η + 1    

6.2.5. Expressions for the control variables.
It is now easy to find the expressions of the control variables u ( t ) and C( t )
using the functions we have already calculated.
The control variable C( t ) is given by (6.15). Substituting into this expression the
λ K given by (6.29), we obtain
                   β                      1                         β
                                      −                                           ρ 
                σ (1−β)                   σ        νt            σ (1−β)
   C( t ) = ε             (λ Q (0))           (e        − C0 )             × exp  − t    (6.53)
                                                                                  σ 
Using (6.16) for the control variable u ( t ) (the aggregate extraction rate), we first
                   λK
find the ratio        from (6.31) and then substitute the expressions for K and Q
                   λH
from (6.40) and (6.51) respectively. We obtain




                                                                                               93
                            1                β
                                        −              
                (1 − β)A  β               1−β                  νβ  
       u(t) =        ×ε                         × exp − δ +
                                                                     t ×
               1+ δ                                         1− β  
                                                                     
                                         1
                                    −     D0
                                        1−β
                   K 0 (1 − C 0 )            Ω( t )
                                              +
                                          νη
               ×                                    × (e ν t − C 0 )          (6.54)
                            Q 0 − E 0 X( t )
where all the functions and parameters are defined in the text.


7. Application IV. Optimal Fiscal Policy.
In this section we examine the issue of optimal fiscal policy, along the lines of the
paper by Gómez [11].

7.1. The model for the decentralized economy.
We consider the dynamic optimization problem
                      ∞
                          C( t )1−σ − 1 −ρ t
               max ∫                   e     dt                               (7.1)
                      0
                             1− σ

subject to
       H = γ (1 − u ) H , γ > 0                                               (7.2)
and
       K = (1 − τ k ) r K + (1 − τ h ) w u H − C + s h w H                    (7.3)
Equation (7.2) is the human capital evolution equation and (7.3) is the
corresponding evolution equation for the physical capital (the household budget
constraint).
In (7.3) we denote by r the rate of return on physical capital, and w the wage
rate. They are given by
                       Y                                   Y
               r=α            and w = (1 − α)                                 (7.4)
                       K                                  uH




                                                                                   94
The variable τ k is the rate government taxes the physical capital, and τ h is the

rate government taxes the labor income, while s h is the rate government
subsidizes investment in education.
Introducing the notation
       τh = τh + s h
       ˆ                                                                                   (7.5)
equation (7.3) can be written as
       K = (1 − τ k ) r K + (1 − τ h ) w u H − C + s h w H
                                 ˆ                                                         (7.6)
We further assume that the government runs a balanced-budget and has no access
to lump-sum taxation, which means that τ k r K + τ h w u H = s h w (1 − u ) H , or using
equation (7.5),
       τk r K + τh w u H = sh w H
                ˆ                                                                          (7.7)
The production function is given by

       Y = A K α ( u H)1−α H β , A > 0 , 0 < α < 1 , β > 0
                             a                                                             (7.8)
where H a is the measure of the externality. In equilibrium condition we have

Ha = H .

7.1.1. The dynamical equations.
The current value Hamiltonian of the optimization problem is given by

       H c (C, u , H, K ) =

                    C1−σ − 1
                =            + λ K [ (1 − τ k ) r K + (1 − τ h ) w u H − C + s h w H ] +
                                                           ˆ
                     1− σ
                  + λ H [ γ (1 − u ) H ]                                                   (7.9)

where λ K and λ H are the costate variables corresponding to K and H
respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method.
The first order conditions read


                                                                                               95
       ∂H c
            = 0 ⇔ C− σ − λ K = 0                                                    (7.10)
        ∂C

       ∂H c
            = 0 ⇔ λ K [ (1 − τ h ) w H ] − λ H γ H = 0
                             ˆ                                                      (7.11)
        ∂u
We also have the two Euler equations

       ∂H c
            + λ K = ρ λ K ⇔ (1 − τ k ) r λ K + λ K = ρ λ K                          (7.12)
       ∂K

       ∂H c
            + λH = ρλH ⇔
       ∂H

       ⇔ λ K [ (1 − τ h ) w u + s h w ] + λ H γ (1 − u ) + λ H = ρ λ H
                    ˆ                                                               (7.13)
the dynamic constraints
       K = (1 − τ k ) r K + (1 − τ h ) w u H − C + s h w H , K (0) = K 0
                                 ˆ                                                  (7.14)

       H = γ (1 − u ) H , H(0) = H 0                                       (7.15)
the government’s budget constraint
       τk r K + τh w u H = sh w H
                ˆ                                                                   (7.16)
and the transversality conditions

          lim λ K K e − ρ t = 0                                                     (7.17)
       t →∞

          lim λ H H e − ρ t = 0                                                     (7.18)
       t →∞

7.1.2. Transformation of the equations.
From equation (7.10), taking logarithms and differentiation with respect to time,
we find
                      C λK
                 −σ    =                                                            (7.19)
                      C λK
From (7.12) we obtain the equation




                                                                                        96
              λK
                 = ρ − (1 − τ k ) r
              λK
and using the expression for r (the first equation in (7.4)) we find
              λK                    Y
                 = ρ − α (1 − τ k )                                         (7.20)
              λK                    K
where Y is given by (7.8) in equilibrium ( H a = H ).
After combining (7.19) with (7.20), we get the following expression
              C α (1 − τ k ) Y ρ
                =             −                                             (7.21)
              C     σ        K σ
From equation (7.14), taking into account (7.16) and the expressions (7.4) for r
and w, we obtain the equation
              K Y C
               = −                                                          (7.22)
              K K K
From equation (7.15) we obtain
              H
                = γ (1 − u )                                                (7.23)
              H
From equation (7.11) we can obtain the following expression for the ratio of the
costate variables
              λK       γ
                 =                                                          (7.24)
              λ H (1 − τ h ) w
                       ˆ

From equation (7.13) we obtain the equation
              λH                                              λ
                 = ρ − γ (1 − u ) − [ (1 − τ h ) w u + s h w ] K
                                           ˆ
              λH                                              λH

which, because of (7.24), takes on the form
              λH             s         
                 = ρ − γ 1 + h
                          1− τ         
                                                                           (7.25)
              λH               ˆh      
We now come again in (7.11). Using the expression for w, we obtain the
following equation



                                                                                   97
                  (1 − α) (1 − τ h )A λ K K α u − α H β−α = γ λ H
                               ˆ                                               (7.26)
where we also have used the expression for the production function given by (7.8)
in equilibrium. Taking logarithms and differentiating with respect to time equation
(7.26), we arrive at the equation
               τh
               ˆ    λ     K   u         H λ
         −         + K + α − α + (β − α) = H                                   (7.27)
             1 − τh λ K
                 ˆ        K   u         H λH

                                                           λK K H     λ
In the previous equation we substitute the ratios            , ,  and H given by
                                                           λK K H     λH
(7.20), (7.22), (7.23) and (7.25) respectively, and we obtain the equation
   u     Y C γ (1 − α + β) γ (α − β)         γ                τh
                                                              ˆ
     = τk − +             +          u+              sh −                      (7.28)
   u     K K       α           α        α (1 − τ h )
                                               ˆ          α (1 − τ h )
                                                                 ˆ

The system of equations (7.21), (7.22), (7.23) and (7.28) determine the dynamics
of the decentralized economy.
Introducing a new function Z by
              C
         Z=                                                                    (7.29)
              K
we get
         Z C K
          = −                                                                  (7.30)
         Z C K
                                   C     K
Upon substituting the ratios         and   given by (7.21) and (7.22) respectively
                                   C     K
into (7.30), we derive the equation
         Z  α (1 − τ k )  Y C ρ
          =             − 1 + −                                              (7.31)
         Z      σ          K K σ

7.2. The centrally-planned economy.
We consider the dynamic optimization problem




                                                                                     98
                         ∞
                             C( t )1−σ − 1 −ρ t
                  max ∫                   e     dt                                        (7.32)
                         0
                                1− σ

subject to

        K ( t ) = A K ( t ) α ( u ( t ) H ( t ))1−α H ( t ) β − C( t )                    (7.33)

        H( t ) = γ (1 − u ( t )) H( t )                                                   (7.34)
with initial conditions
        K (0) = K 0 , H(0) = H 0                                                          (7.35)
where
        C( t ) ≥ 0 , u ( t ) ∈ [ 0, 1 ] , K ( t ) ≥ 0 , H( t ) ≥ 0                        (7.36)

7.2.1. The dynamical equations.
The current value Hamiltonian is given by

 H c (C, u , H, K ) =

            C( t )1−σ − 1
        =                 + λ K [ A K ( t ) α (u ( t ) H( t ))1−α H( t ) β − C( t ) ] +
               1− σ
                             + λ H [ γ (1 − u ( t )) H( t ) ]                             (7.37)

where λ K and λ H are the costate variables corresponding to K and H
respectively.
We can write down the dynamic equations of the model, using Pontryagin’s
optimization method.
The first order conditions read

        ∂H c
             = 0 ⇔ C− σ − λ K = 0                                                         (7.38)
         ∂C

        ∂H c
             = 0 ⇔ λ K [ (1 − α) A K α (u H) − α H1+β ] − λ H γ H = 0                     (7.39)
         ∂u
We also have the two Euler equations




                                                                                              99
       ∂H c
            + λ K = ρ λ K ⇔ α λ K AK α −1 (u H)1− α H β + λ K = ρ λ K                  (7.40)
       ∂K

       ∂H c
            + λH = ρλH ⇔
       ∂H

       ⇔ λ K (1 − α + β) A K α ( u H )1− α H β−1 + λ H γ (1 − u ) + λ H = ρ λ H        (7.41)
the dynamic constraints

       K = A K α ( u H )1−α H β − C , K (0) = K 0                                      (7.42)

       H = γ (1 − u ) H , H(0) = H 0                                          (7.43)
and the usual transversality conditions.
The system of equations (7.38)-(7.43) has been solved in Section 3.
We also have from (3.38)
       Z α−σ Y C ρ
         =    + −                                                                      (7.44)
       Z   σ K K σ
and from (3.28)
       u γ (1 − α + β) γ (1 − α + β)    C
         =            +              u−                                                (7.45)
       u       α           1− α         K
where u has been evaluated, its explicit expression being given by (3.71).


7.3. The optimal fiscal policy.
The question we now pose is what fiscal policy makes the decentralized economy
replicate the first-best optimum attainable by a central planner.
The first observation is that comparing (7.31) to (7.44) we see that the two
expressions coincide if τ k = 0 , which means that physical capital income should
not be taxed.
Comparing then (7.28) to (7.45), after putting τ k = 0 , we get

       γ (α − β)         γ                τh
                                          ˆ          γ (1 − α + β)
                 u+              sh −              =               u
           α        α (1 − τ h )
                           ˆ          α (1 − τ h )
                                             ˆ           1− α


                                                                                           100
from which we derive the expression for s h :

              β (1 − τ h )
                     ˆ       τ
                             ˆ
       sh =                u+ h                                                 (7.46)
               (1 − α)        γ
(i) Suppose that s h = 0 . If τ h were constant, equation (7.28) would become
                              ˆ

       u   C γ (1 − α + β) γ (α − β)
         =− +             +          u                                          (7.47)
       u   K       α           α
which is different to its corresponding equation (7.44) in a centrally planned
economy. Therefore τ h ≠ 0 , for each optimal path. But then, we get from (7.46)
                   ˆ

the equation
                β γ (1 − τ h )
                         ˆ
       τh = −
       ˆ                       u                                                (7.48)
                  (1 − α)
which is not feasible, since when the economy reaches the steady state
u = u* ∈ (0, 1) , the government’s share will grow indefinitely.

(ii) Since τ h = 0 in the steady state, the optimal tax should be constant. We thus
           ˆ

have from equation (7.46) that
              β (1 − τ h )
                     ˆ
       sh =                u                                                    (7.49)
               (1 − α)
Solving the system of equations (7.49) and (7.7), taking into account τ k = 0 , we
obtain the expression
                 β
       τh =
       ˆ                                                                        (7.50)
              1− α + β
which is the (constant) optimal tax rate on human capital income.
Substituting the expression (7.50) into (7.49), we derive the optimal subsidy rate
sh :
                 β
       sh =            u                                                        (7.51)
              1− α + β
The government size, measured as the subsidy (or taxes) share of output ϕ is
constant in time, since it is given by


                                                                                    101
            s h wH β (1 − α)
       ϕ=         =                                                           (7.52)
               Y    1− α + β
Using equations (7.50) and (7.51) we are able to calculate the tax rate τ h on labor
income calculated to be
                               β
       τh = τh − s h =
            ˆ                        (1 − u )                                 (7.53)
                            1− α + β
The government size can be calculated to be
            s h w (1 − u ) H β (1 − α)
       ϕ=                   =          (1 − u )                               (7.54)
                   Y          1− α + β
which is a time-varying quantity.
The fiscal policy reasoning is along the lines introduced by Gómez [11]. Our
difference however lies to the fact that the time-varying quantities enter through
the control variable u, calculated explicitly in Section 3.2 and given explicitly by
equation (3.71), written down here for convenience
                        J( t )
       u(t) =                                                                 (7.55)
                J (0) y(0) − µ X( t )
The various functions appearing in the previous formula have been defined or
calculated in Section 3.2.
We thus arrive at the following Proposition (Gómez [11], Proposition 2):
Proposition. The decentralized economy can attain the first-best equilibrium
solution if physical capital income is not taxed and investment in human capital is
subsidized at a rate
                               β
                s h (t) =            u(t)
                            1− α + β
The subsidy can be financed by taxing human capital income at a rate
                               β
                τh (t) =             (1 − u ( t ))
                            1− α + β




                                                                                     102
Lump-sum taxation is not required to balance the government budget either in the
steady state or in the transitional phase. The quantity u ( t ) which appears in the
two previous expressions is calculated explicitly and is given by (7.55), where the
various functions and parameters are defined in Section

Appendix A. Evaluation of the integral
                          t
                               ωs
                          ∫e         (e µ s − C 0 ) ν ds                        (A.1)
                          0

In this Appendix we shall evaluate the integral (A.1) using a two-step procedure.
In the first step we shall convert the integral into a combination of integrals each
one having limits 0 and 1.
Using the simple identity

               e ω s (e µ s − C 0 ) ν = e (ω+µν) s (1 − C 0 e − µ s ) ν         (A.2)
the integral (A.1) becomes
                t
                     ( ω+ µν ) s
                ∫e                 (1 − C 0 e − µ s ) ν ds                      (A.3)
                0

We make use of the substitution

               u = e− µ s                                                       (A.4)
Under this substitution, we get
                        1 du
               ds = −        and e (ω+µν) s = u η
                        µ u
where
                       ω+µν
               η=−                                                              (A.5)
                        µ
Taking into account that the new limits of the integral are
                s = 0 ⇒ u =1 
                                 
              s = t ⇒ u = e− µ t 
the integral in (A.3) takes on the form


                                                                                       103
                            e− µ t
                       1
                     −          ∫    u η−1 (1 − C 0 u ) ν du                                          (A.6)
                       µ        1

The above expression can be written as
            e− µ t                      1                         
          1                                                       
         −  ∫ u η−1 (1 − C 0 u ) ν du − ∫ u η−1 (1 − C 0 u ) ν du                                   (A.7)
          µ 0                                                     
                                        0                         
The first integral in the above expression can be converted into an integral with
new limits 0 and 1. This is done using the substitution:

                    u = e −µ t x                                                                      (A.8)
In fact, since
                      u =0⇒ x =0 
                                       
                    u = e− µ t ⇒ x = 1 
we get
         e− µ t                                                1
          ∫       u η−1 (1 − C 0 u ) ν du = e −µη t            ∫x
                                                                    η−1
                                                                          (1 − C 0 e − µ t x ) ν dx   (A.9)
          0                                                    0

Therefore the expression (A.7) takes the form

           1       −µη t
                  
                            1
                                    η−1                −µ t
                                                                           1                     
                                                                                                 
         −        e        ∫   x         (1 − C 0 e            ν
                                                              x ) dx − ∫ u η−1 (1 − C 0 u ) ν du     (A.10)
           µ                                                                                    
                           0                                          0                         
In the second step, we shall express each one of the integrals in terms of the
hypergeometric function                       F(a , b ; c ; z) . This is done using the integral
representation of the hypergeometric function, known from the theory of
differential equations in the complex domain, Ince [15], Morse and Feshbach [22],
Smirnov [32], Whittaker and Watson [35]. It is known that the hypergeometric
function admits the following integral representation (see any of the previously
noted books or any standard monograph on special functions, like Abramowitz
and Stegun [1], Andrews, Askey and Roy [4], Lebedev [16], Rainville [25], or
Slater [31])


                                                                                                         104
                                              1
                                 Γ (c)          b −1      c − b −1
       F(a , b ; c ; z) =                    ∫ t (1 − t )          (1 − z t ) −a dt       (A.11)
                          Γ ( b ) Γ (c − b ) 0

when Re(c) > Re(b) > 0 .
When c − b − 1 = 0 , i.e. c = 1 + b , the above integral representation can be written
as
                                              1
                               Γ(1 + b)     b −1       −a
       F(a , b ;1 + b ; z) =             ∫ t (1 − z t ) dt
                               Γ(b) Γ(1) 0
                                                                                          (A.12)


and since Γ(1 + b) = b Γ(b) and Γ(1) = 1 , we have
                                1
       F(a , b ;1 + b ; z) = b ∫ t b −1 (1 − z t ) −a dt                                  (A.13)
                                0

The above formula is going to be used for expressing (A.10) in terms of the
hypergeometric function. We find, because of (A.13)
       1
            η−1                                   1
       ∫x         (1 − C 0 e − µ t x ) ν dx =       F ( − ν , η ;1 + η ; C 0 e − µ t )    (A.14)
       0
                                                  η

and
       1
            η−1                           1
       ∫x         (1 − C 0 x ) ν dx =       F ( − ν , η ;1 + η ; C 0 )                    (A.15)
       0
                                          η

Therefore
        t
             ωs
        ∫e        (e µ s − C 0 ) ν ds =
        0

              1
       =−       [ e −µη t F (− ν, η ;1 + η ; C 0 e − µ t ) − F (− ν, η ;1 + η ; C 0 ) ]   (A.16)
             µη
where η is given by (A.5).
Note. Formula (A.16) is valid within a range of the parameters involved. The
series expansion of the hypergeometric function F(a , b ; c ; z) is given by




                                                                                             105
                                            ∞    (a ) n ( b ) n n
              F(a , b ; c ; z) = 1 + ∑                         z
                                            n =1   (c) n n!

where
              (α) n = α(α + 1)               (α + n − 1) , n ≥ 1

              (α ) 0 = 1
is the usual Pochhammer symbol.
The above series converges for | z | < 1 . For different values of z, an analytic
continuation needs to be considered.

Appendix B. Evaluation of the integral
                      t
                               ϕt
                      ∫e            F(− ν, η ;1 + η ; C 0 e − µ t ) dt          (B.1)
                      0

We shall evaluate this integral in two steps:
In the first step we shall convert the above integral into a combination of integrals
having each one of them limits 0 and 1.
Under the transformation
                                                                        ϕ
                                                 −
                    −µs         1 du          ϕs
              u=e      , ds = −       and e = u µ                               (B.2)
                                µ u
and taking into account that the new limits are
                s = 0 ⇒ u =1 
                                 
              s = t ⇒ u = e− µ t 
the integral in (B.1) transforms into
                       e− µ t           ϕ
                                    −
                1                     µ                               du
               − 
                µ
                 
                           ∫    u           F( −ν, η ; η + 1; C 0 u )
                                                                         u
                           1

which can also be written as
       e− µ t                            1                                
 1          p −1                          p −1                          
 −   ∫ u F(−ν, η ; η + 1; C 0 u ) du − ∫ u F( −ν, η ; η + 1; C 0 u ) du  (B.3)
 µ
    0                                                                   
                                         0                                


                                                                                    106
where
                  ϕ
         p=−                                                                                                 (B.4)
                  µ
Using the substitution

         u = e− µ t x                                                                                        (B.5)
we get
         e− µ t
          ∫       u p −1 F( −ν, η ; η + 1; C 0 u ) du =
          0

                                  1
                         −µp t            p −1
                    =e            ∫x             F(−ν, η ; η + 1; C 0 e − µ t x ) dx                         (B.6)
                                  0

Collecting the above, we have
          t
               ϕt
          ∫e        F(− ν, η ;1 + η ; C 0 e − µ t ) dt =
          0

                                           1
                       1
                    = − e− µ p t           ∫x
                                                  p −1
                                                         F(−ν, η ; η + 1; C 0 e − µ t x ) dx +
                       µ                   0

                            1
                        1         p −1
                      +
                        µ   ∫u            F(−ν, η ; η + 1; C 0 u ) du                                        (B.7)
                            0

In the second step we shall express each one of the integrals appearing in (B.7) in
terms of the generalized hypergeometric function 3 F2 [ ] .
We have the following general formula (see for example Rainville [25], Section
49, Theorem 28):

               Γ( b1 )                1
                                           a1 −1                                  a2,     , ap     
                                      ∫t           (1 − t ) b1 −a1 −1   p −1Fq −1              z t  dt =
         Γ(a1 ) Γ(b1 − a1 )           0                                          b 2 ,   , bq      

                                        a1 , a 2 ,         , ap 
                                = p Fq 
                                                            , bq 
                                                                 z                                           (B.8)
                                       b1 , b 2 ,                 
From the above formula we obtain


                                                                                                                107
           Γ(a1 + 1)          1
                                               a , a    
                              ∫   t a1 −1 2 F1  2 3 z t  dt =
           Γ(a1 ) Γ(1)        0                 b2      

                                                     a , a , a   
                                              = 3 F2  1 2 3 z                                 (B.9)
                                                     a1 + 1, b 2 
Therefore

           Γ(p + 1)
                          1
                                  p −1                                        p, − ν , η 
           Γ( p) Γ(1)     ∫t             2 F ( −ν, η ; η + 1; z t ) dt = 3 F2 
                                            1                                              z
                                                                             p + 1, η + 1 
                                                                                                (B.10)
                          0

where as usual, Γ(p + 1) = p Γ(p) .
We thus arrive at
            t
                 ϕt
            ∫e        F (− ν, η ;1 + η ; C 0 e − µ t ) dt =
            0

                 1  ϕt     p, − ν , η                       p, − ν , η     
           =−      e 3 F2              C 0 e − µ t  − 3 F2              C0                (B.11)
                µp        p + 1, η + 1                     p + 1, η + 1    
where p is given by (B.4).
For the generalized hypergeometric functions the reader may consult Andrews,
Askey and Roy [4], Erdélyi [10], Rainville [25], or Slater [31] among a plethora of
monographs.
Note. The generalized hypergeometric function is defined by
                                                             p

                                             ∞ ∏ i n
                                                      (a )
                 a1 , a 2 , , a p                          zn
           p Fq                   z = 1 + ∑ q  i =1      ⋅
                b1 , b 2 , , b q          n =1             n!
                                                      (b )  ∏      j n
                                                            j=1

If p ≤ q the series converges for all finite z. If p = q + 1 , the series converges for
| z |<1.


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DOCUMENT INFO
Description: Abstract We present explicit solutions in a number of Lucas-Uzawa Models. The models are solved without dimensional reduction, using two different methods. The first method uses a procedure similar to the dimensional reduction. However in our method we do not consider solutions along the balanced growth path. The second method has recently appeared in the literature. However, in our own calculations the intermediate steps appear explicitly in an easy to follow algorithm. Some of the results we derive exhibit differences compared to the results already found. The solution procedure of models with externalities uses quite different techniques to those known so far. The closed-form solutions of the models with externalities appear for the first time in the literature. Keywords: Economic Dynamics, Lucas-Uzawa model, Closed-Form Solutions, Special Functions, Hypergeometric Functions.