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					                                  Introduction to Partial Differential Equations

                                  1       Preliminaries
                                  A partial differential equation is an equation involving a function u of several vari-
                                  ables, say, u = u(x, y, z, t) and the partial derivatives of u. For example, in the study
                                  of thermal effects in a solid body the temperature u may vary from point to point in the
                                  solid as well as from time to time, and, the derivatives
                                                                             ∂u ∂u ∂u ∂u
                                                                               , , ,
                                                                             ∂x ∂y ∂z ∂t
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                                  will, in general, be nonzero. Further in some other practical problems like wave prop-
                                  agation, potential flow, steady-state heat conduction, higher derivatives of the type
                                                                    ∂2u ∂ 2u ∂ 2u ∂3u
                                                                       ,    ,    ,      etc.
                                                                    ∂x2 ∂x∂y ∂y 2 ∂t∂x2
                                  may occur.
                                      When the laws of physics are applied to this kind of problems, we obtain some
                                  relations between the derivatives of the kind
                                                                          ∂u       ∂ 2u     ∂ 2u
                                                           F   x, t, u,      ,··· , 2,··· ,      ,···   =0                   (1)
                                                                          ∂x       ∂x       ∂x∂t
                                  Such an equation relating partial derivatives is called a partial differential equation
                                  (pde).
                                      As in the case of ODEs, we define the order of a pde to be the order of the derivative
                                  of the highest order occurring in the equations.
                                  Examples:

                                          ∂u    ∂ 2u
                                      •      = k 2 , k ∈ R+ : 2nd order equation in two variables x and t.
                                          ∂t    ∂x
                                                 3
                                            ∂u           ∂u
                                      •              +      = 0: 1st order third degree equation in two variables x and t.
                                            ∂x           ∂t
                                            ∂u    ∂u ∂u
                                      • x      +y    +    = 0: 1st order equation in three variables x, y and t.
                                            ∂x    ∂y   ∂t
                                          ∂3u    ∂ 3u ∂u
                                      •      3
                                               + 2 −     = f (x, y, t): 3rd order equation in three variables x, y and
                                          ∂x    ∂x ∂y ∂t
                                          t.
                                          ∂2u ∂ 2u ∂ 2u
                                      •      +    +     = 0: 2nd order equation in three variables x, y and z.
                                          ∂x2 ∂y 2 ∂z 2

                                                                                    1
                                  2     First order PDE in two independent variables
                                  A first order partial differential equation in two independent variables in its most gen-
                                  eral form is given by
                                                                    F (x, y, u, ux , uy ) = 0                          (2)
                                  where F is a known function of its arguments. When the function F is not a linear
                                  expression in ux and uy , equation (2) is said to be a nonlinear equation which will
                                  look like:
                                                     a(x, y, u)(ux )2 + b(x, y, u)(uy )2 = c(x, y, u)              (3)
                                  When F is a linear expression in ux and uy , but not necessarily in u, equation (2) is
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                                                            a(x, y, u)ux + b(x, y, u)uy = c(x, y, u)                     (4)

                                  where a and b depend on u also. This equation is called a quasilinear equation. A first
                                  order semilinear equation is an equation of the form

                                                              a(x, y)ux + b(x, y)uy = c(x, y, u)                         (5)

                                  where the coefficients of ux and uy do not depend on u and the nonlinearity in the
                                  equation is present only in the inhomogenous term on the right hand side of (5). A first
                                  order linear equation is of the form

                                                         a(x, y)ux + b(x, y)uy = c1 (x, y)u + c2 (x, y)                  (6)

                                  where the dependent variable u and its partial derivatives ux , uy all appear linearly with
                                  a, b, c1 and c2 as functions of x and y only.
                                      The solution u = u(x, y) represents a surface in (x, y, u) space. This surface is
                                  called an integral surface of the partial differential equation.

                                  Examples:

                                      1. Nonlinear equation
                                                                   3                      2            2
                                                              ∂u           ∂u        ∂u           ∂u
                                                                       +      = 0,            +            =1
                                                              ∂x           ∂t        ∂x           ∂y

                                      2. Quasilinear equation
                                                         ∂u    ∂u                   ∂u      ∂u
                                                     u      +k    = u2 , (x2 + u2 )    − xy    = u3 x + y 2
                                                         ∂x    ∂y                   ∂x      ∂y

                                                                                 2
                                     3. Semilinear equation
                                                                 ∂u    ∂u          ∂u    ∂u
                                                             x      +y    = u2 , x    −y    = xu2
                                                                 ∂x    ∂y          ∂x    ∂y

                                     4. Linear equation
                                                           ∂u    ∂u             ∂u    ∂u
                                                       x      +y    = xu + y, y    −x    = xyu + x
                                                           ∂x    ∂y             ∂x    ∂y

                                  2.1   Origin of first order PDEs
                                  Consider the equation
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                                                                   x2 + y 2 + (z − c)2 = a2                            (7)
                                  where a and c are arbitrary constants. Equation (7) represents the set of all spheres
                                  whose centres lie along z-axis. Differentiating (7) with respect to x and y separately,
                                  then we get

                                                                    x + p(z − c) = 0,
                                                                     y + q(z − c) = 0.
                                                                  ∂z            ∂z
                                  where we use the notations p =      and q =      . Eliminating the arbitrary constant c
                                                                  ∂x            ∂y
                                  from these two equations, we obtain the partial differential equation

                                                                         yp − xq = 0                                   (8)

                                  which is of first order. In some sense, the set of all spheres with centres on the z-axis
                                  is characterized by the PDE (8). Now consider the equation

                                                                  x2 + y 2 = (z − c)2 tan2 α,                          (9)

                                  where c and α are arbitrary constants, which represents the set of all right circular
                                  cones whose axes coincide with the line Oz. Differentiating (9) with respect to x and
                                  y separately, we obtain

                                                       p(z − c) tan2 α = x, q(z − c) tan2 α = y.                     (10)

                                  After eliminating the constants c and α, we get

                                                                        yp − xq = 0,                                 (11)

                                  which is same as equation (8). It is evident what the spheres and the cones have in
                                  common is that they are surfaces of revolution which have the line Oz as the axis

                                                                              3
                                  of symmetry. All surfaces of revolution with this property are characterized by an
                                  equation of the form
                                                                  z = f (x2 + y 2 ),                            (12)
                                  where f is an arbitrary function.
                                     Let u = x2 + y 2 and differentiate (12) w.r.t. x and y respectively to get
                                                          p = zx = 2xf (u), q = zy = 2yf (u),
                                  where f (u) = df /dx from which we obtain (11) after eliminating the arbitrary func-
                                  tion f (u). Thus, we see that the function defined by each of (7), (9) and (12) is, in
                                  some sense, a solution of the pde (11).
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                                      Let us generalize this argument as:
                                      The relations (7) and (9) are both of the type
                                                                     F (x, y, z, a, b) = 0,                          (13)
                                  where a and b are arbitrary constants. Differentiating (13) w.r.t. x and y respectively,
                                  we get
                                                            ∂F       ∂F       ∂F       ∂F
                                                                 +p      = 0,      +q      = 0.                      (14)
                                                            ∂x       ∂z        ∂y      ∂z
                                  (13) and (14) together constitute three equations involving two arbitrary constants a
                                  and b, and in the general case, it will be possible to eliminate a and b from these
                                  equations to obtain a relation of the kind
                                                                     f (x, y, z, p, q) = 0,                          (15)
                                  showing that the system of surfaces (7) or (9) or (12) gives rise to a pde (15) of first-
                                  order.
                                     The obvious generalization of (12) is a relation between x, y and z of the type
                                                                        F (u, v) = 0,                                (16)
                                  where u and v are known functions of x, y and z, and F is an arbitrary function of u
                                  and v. Differentiating (16) w.r.t. x and y respectively, we get
                                                     ∂F    ∂u    ∂u         ∂F      ∂v    ∂v
                                                              +p          +            +p         = 0,
                                                     ∂u    ∂x    ∂z         ∂v      ∂x    ∂z
                                                     ∂F    ∂u    ∂u         ∂F      ∂v    ∂v
                                                              +q          +            +q         = 0.
                                                     ∂u    ∂y    ∂z         ∂v      ∂y    ∂z
                                  If we, now, eliminate ∂F/∂u and ∂F/∂v from these equations, we obtain the equation
                                                                 ∂(u, v)    ∂(u, v)   ∂(u, v)
                                                             p           +q         =         ,                      (17)
                                                                 ∂(y, z)    ∂(z, x)   ∂(x, y)
                                  which is a pde of the type (15). Equation (17) is a linear first-order pde.

                                                                               4
                                  2.2    Cauchy’s problem for first-order partial differential equations
                                  The existence and uniqueness of solutions to first-order pdes are discussed below in
                                  the form of Cauchy’s Problem.
                                  Cauchy’s Problem.
                                  If
                                  (a) x0 (µ), y0 (µ) and z0 (µ) are functions which, together with their first derivatives, are
                                  continuous in the interval M defined by µ1 < µ < µ2 ; and
                                  (b) if F (x, y, z, p, q) is a continuous function of x, y, z, p and q in a certain region U of
                                  the xyzpq space, then it is required to establish the existence of a function φ(x, y) with
                                  the following properties:
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                                  (1) φ(x, y) and its partial derivatives with respect to x and y are continuous functions
                                  of x and y in a region R of the xy space.
                                  (2) For all values of x and y lying in R, the point {x, y, φ(x, y), φx (x, y), φy (x, y)} lies
                                  in U and
                                                              F [x, y, φ(x, y), φx (x, y), φy (x, y)] = 0.
                                  (3) For all µ belonging to the interval M , the point {x0 (µ), y0 (µ)} belongs to the region
                                  R, and
                                                                    φ{x0 (µ), y0 (µ)} = z0 .
                                  From the point of view of geometry, what we want to establish is that there exists a
                                  surface z = φ(x, y) which passes through the curve Γ whose parametric equations are

                                                             x = x0 (µ), y = y0 (µ), z = z0 (µ)                            (18)

                                  and at every point of which the direction (p, q, −1) of the normal is such that

                                                                      F (x, y, z, p, q) = 0.                               (19)

                                  This is one of the various forms of Cauchy’s problem.
                                     To prove the existence of a solution of (19) passing through a curve with equations
                                  (18) it is necessary to make some further assumptions about the form of F and the
                                  nature of the curve Γ.

                                  Theorem 2.1 If g(y) and all its derivatives are continuous for |y − y0 | < δ, if x0 is
                                  a given number, and z0 = g(y0 ), z0 = g (y0 ), and if f (x, y, z, q) and all its partial
                                  derivatives are continuous in a region S defined by

                                                          |x − x0 | < δ, |y − y0 | < δ, |q − q0 | < δ,


                                                                                5
                                  then there exists a unique function φ(x, y) such that
                                  (a) φ(x, y) and all its partial derivatives are continuous in a region R defined by |x −
                                  x0 | < δ1 , |y − y0 | < δ1 ;
                                  (b) For all (x, y) ∈ R, z = φ(x, y) is a solution of the equation
                                                                    ∂z                    ∂z
                                                                       =f      x, y, z,        ;
                                                                    ∂x                    ∂y
                                  (c) For all values of y in the interval |y − y0 | < δ1 , φ(x0 , y) = g(y).

                                  The above theorem is due to Sonia Kowalewski and is also known as Cauchy-Kowalewski
                                  theorem.
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                                     We have seen earlier that relations of the type

                                                                      F (x, y, z, a, b) = 0                          (20)

                                  led to pdes of first-order. Any such relations which contains two arbitrary constants
                                  a and b is a solution of a pde of first-order, and is said to be a complete solution or a
                                  complete integral of that equation.
                                      On the other hand, any relation of the type

                                                                          F (u, v) = 0                               (21)

                                  involving an arbitrary function F connecting a solution of a first-order pde is called a
                                  general solution or a general integral of that equation.
                                      It is obvious that in some sense a general integral provides a much broader set of
                                  solutions of the pde than does a complete integral.


                                  2.3    Linear equations of the first order
                                  Let us consider the linear pde
                                                                        P p + Qq = R,                                (22)
                                  where P, Q and R are given functions of x, y and z ( which do not involve any deriva-
                                  tives), p and q respectively denote ∂z/∂x and ∂z/∂y and we are interested in finding a
                                  relation between x, y and z involving an arbitrary function. The first systematic theory
                                  of this type was given by Lagrange and for this reason (22) is frequently referred to as
                                  Lagrange’s equation.

                                  Theorem 2.2 The general solution of the linear pde

                                                                        P p + Qq = R                                 (23)

                                                                                6
                                  is
                                                                             F (u, v) = 0,                              (24)
                                  where F is an arbitrary function and u(x, y, z) = c1 and v(x, y, z) = c2 form a
                                  solution of the equation
                                                                   dx   dy     dz
                                                                      =     = .                               (25)
                                                                   P    Q       R


                                  Proof: If equations
                                                               u(x, y, z) = c1 , v(x, y, z) = c2                        (26)
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                                  satisfy equations (25), then the equations

                                                                    ux dx + uy dy + uz dz = 0

                                  and
                                                                           dx   dy   dz
                                                                              =    =
                                                                           P    Q    R
                                  must be compatible, i.e., we must have

                                                                     P ux + Quy + Ruz = 0.                              (27)

                                  Similarly we have
                                                                     P vx + Qvy + Rvz = 0.                              (28)
                                  Solving equations (27) and (28) for P, Q and R, we have
                                                            P                 Q                 R
                                                                     =                 =                 .              (29)
                                                     ∂(u, v)/∂(y, z)   ∂(u, v)/∂(z, x)   ∂(u, v)/∂(x, y)

                                  Previously we saw that the relation

                                                                             F (u, v) = 0

                                  leads to the pde
                                                                   ∂(u, v)    ∂(u, v)   ∂(u, v)
                                                               p           +q         =         .                       (30)
                                                                   ∂(y, z)    ∂(z, x)   ∂(x, y)
                                  Substituting (29) into (30), we see that (24) is a solution of (23) if u and v are given by
                                  (26).
                                  Example: Find the general solution of
                                                                         ∂z      ∂z
                                                                    x2      + y2    = (x + y)z.
                                                                         ∂x      ∂y

                                                                                  7
                                  Solution: The integral surfaces of this equation are generated by the integral curves of
                                  the equations
                                                                  dx     dy        dz
                                                                    2
                                                                      = 2 =              .                            (31)
                                                                  x      y      (x + y)z
                                  Taking the first two equations
                                                                              dx   dy
                                                                                2
                                                                                  = 2,
                                                                              x    y
                                  we get
                                                                           1/x − 1/y = c1 ,                                       (32)
                                  and it follows from the equations that
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                                                                        dx − dy       dz
                                                                          2 − y2
                                                                                 =
                                                                        x          (x + y)z
                                  giving us
                                                                       x−y
                                                                              = c2 .                              (33)
                                                                         z
                                  Combining (32) and (33), we see that the integrals curves of equation (31) are given
                                  by (33) and the equation
                                                                         xy
                                                                             = c2                                 (34)
                                                                          z
                                  and that the curves given by these equations generate the surface

                                                                            xy x − y
                                                                       F       ,           = 0,                                   (35)
                                                                             z   z
                                  where F is an arbitrary function. It should be observed that this surface can be ex-
                                  pressed by equations such as

                                                        z = xy f (x − y/z), or z = xy g((x − y)/xy).

                                  We can generalize theorem 2.2 as follows:

                                  Theorem 2.3 If ui (x1 , x2 , x3 , . . . , xn , z) = ci , i = 1, 2, 3, . . . , n are independent solu-
                                  tions of the equations
                                                               dx1         dx2               dxn     dz
                                                                       =          = ··· =        = ,
                                                                P1          P2               Pn      R
                                  then the relation φ(u1 , u2 , . . . , un ) = 0, in which the function φ is arbitrary, is a
                                  general solution of the linear pde
                                                                  ∂z       ∂z               ∂z
                                                             P1       + P2     + · · · + Pn     = R.
                                                                  ∂x1      ∂x2              ∂xn



                                                                                   8
                                  2.4   Integral surfaces passing through a given curve
                                  Here we will discuss how a general solution may be used to determine the integral
                                  surface which passes through a given curve. Suppose we have found two solutions

                                                                u(x, y, z) = c1 , v(x, y, z) = c2                  (36)

                                  of the auxiliary equations
                                                                    dx     dy    dz
                                                                       =      = .
                                                                    P      Q      R
                                  Then any solution of the corresponding linear equation is of the form
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                                                                           F (u, v) = 0                            (37)

                                  arising from a relation
                                                                           F (c1 , c2 ) = 0                        (38)
                                  between the constants c1 and c2 . The problem we have to consider is that of deter-
                                  mining the function F in special circumstances. If we want to find the integral surface
                                  which passes through the curve C whose parametric equations are

                                                                   x = x(t), y = y(t), z = z(t),

                                  where t is a parameter, then the particular solution (36) must be such that

                                                    u{x(t), y(t), z(t)} = c1 , v{x(t), y(t), z(t)} = c2 .

                                  Therefore, we have two equations from which we may eliminate the single variable t
                                  to obtain a relation of the type (38). The solution we seek is then given by (37).
                                  Example: Find the integral surface of the linear pde

                                                            x(y 2 + z)p − y(x2 + z)q = (x2 − y 2 )z

                                  which contains the straight line x + y = 0, z = 1.
                                  Solution: The auxiliary equations are
                                                                dx          dy           dz
                                                                2 + z)
                                                                       =      2 + z)
                                                                                     = 2          .
                                                            x(y          −y(x         (x − y 2 )z
                                  Taking
                                                                y dx + x dy            dz
                                                                                   = 2
                                                            xy 3
                                                              + xyz − yx  3 − yxz   (x − y 2 )z
                                  which will ultimately give us
                                                                     d(xy)       dz
                                                                             =− ,
                                                                       xy         z

                                                                                  9
                                  whose solution is
                                                                            xyz = c1 .                                 (39)
                                  Similarly taking
                                                                   x dx + y dy                dz
                                                                                          = 2          ,
                                                         x2 y 2   +x 2 z − y 2 x2 − y 2 z  (x − y 2 )z
                                  gives us
                                                                       x2 + y 2 − 2z = c2 .                            (40)
                                  For the curve x + y = 0, z = 1, we have the equations x = t, y = −t, z = 1.
                                  Substituting these values in (39) and (40), we have
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                                                                     −t2 = c1 , 2t2 − 2 = c2 ,

                                  and eliminating t from them, we obtain the relation

                                                                        2c1 + c2 + 2 = 0,

                                  showing that the desired integral surface is

                                                                  x2 + y 2 + 2xyz − 2z + 2 = 0.


                                  2.5    Surfaces orthogonal to a given system of surfaces
                                  An interesting application of the theory of linear pdes of first-order is to determine
                                  the systems of surfaces. Suppose we are given a one-parameter family of surfaces
                                  characterized by the equation
                                                                     f (x, y, z) = c1                              (41)
                                  and we want to find a system of surfaces which cut each of these given surfaces at right
                                  angles. The normal at the point (x, y, z) to the surface of the system (41) which passes
                                  through that point is the direction given by the direction ratios
                                                                                  ∂f ∂f ∂f
                                                                  (P, Q, R) =       ,  ,         .                     (42)
                                                                                  ∂x ∂y ∂z
                                  If the surface with equation


                                                                           z = φ(x, y)                                 (43)
                                  cuts each surface of the given system orthogonally, then its normal at the point (x, y, z)
                                  which is in the direction
                                                                        ∂z ∂z
                                                                           , , −1
                                                                        ∂x ∂y

                                                                                10
                                  is perpendicular to the direction (P, Q, R) of the normal to the surface of the set (41)
                                  at that point. Therefore, we have the linear pde
                                                                         ∂z    ∂z
                                                                     P      +Q    =R                                 (44)
                                                                         ∂x    ∂y
                                  for the determination of the surfaces (43). Substituting from (42), we see that this
                                  equation is equivalent to
                                                                   ∂f ∂z ∂f ∂z          ∂f
                                                                           +         =     .
                                                                   ∂x ∂x ∂y ∂y          ∂z
                                  Conversely, any solution of the linear pde (44) is orthogonal to every surface of the
                                  system characterized by (41), for (44) simply states that the normal to any solution of
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                                  (44) is perpendicular to the normal to that member of the system (41) which passes
                                  through the same point.
                                      The linear pde (44) is therefore the general pde determining the surfaces orthogonal
                                  to members of the system (41); i.e., the surfaces orthogonal to the system (41) are the
                                  surfaces generated by the integral curves of the equations
                                                                 dx       dy       dz
                                                                      =        =        .                            (45)
                                                               ∂f /∂x   ∂f /∂y   ∂f /∂z
                                  Example: Find the surface which intersects the surfaces of the system z(x + y) =
                                  c(3z + 1) orthogonally and which passes through the circle x2 + y 2 = 1, z = 1.
                                                      z(x + y)
                                  Solution: Here f =           and hence we have
                                                       3z + 1
                                                     ∂f      z    ∂f      z    ∂f     x+y
                                                        =       ,    =       ,    =           .
                                                     ∂x   3z + 1 ∂y    3z + 1 ∂z    (3z + 1)2
                                  The integral curves are given by
                                                         dx           dy              dz
                                                                =            =
                                                     z/(3z + 1)   z/(3z + 1)   (x + y)/(3z + 1)2
                                  giving us
                                                               dx          dy        dz
                                                                      =           =     .
                                                            z(3z + 1)   z(3z + 1)   x+y
                                  Taking the first two, we get
                                                                      x − y = c1
                                  and taking
                                                         (x + y)dx + (x + y)dy = z(3z + 1)dz
                                  gives us
                                                                x2 + y 2 − 2z 3 − z 2 = c2 .

                                                                            11
                                  Thus any surface which is orthogonal to the given surfaces has equations of the form

                                                              x2 + y 2 − 2z 3 − z 2 = f (x − y).

                                  For the particular surface passing through the circle x2 + y 2 = 1, z = 1, we have
                                  f = −2.
                                     Thus the required surface is

                                                                    x2 + y 2 = 2z 3 + z 2 − 2.


                                  2.6    Compatible system of first-order equations
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                                                                               n
                                  Definition 2.1 An equation of the form             Fi dxi = 0 is called a Pfaffian equation.
                                                                              i=1


                                  Definition 2.2 A necessary condition that Pfaffian differential equation X · dr = 0
                                  should be integrable is that X · curl X = 0

                                  If every solution of the first-order pde

                                                                       f (x, y, z, p, q) = 0                              (46)

                                  is also a solution of the pde
                                                                       g(x, y, z, p, q) = 0                               (47)
                                  then the equations are said to be compatible. We need to find the conditions in order
                                  that (46) and (47) are compatible.
                                      If
                                                                         ∂(f, g)
                                                                     J≡          = 0,                             (48)
                                                                         ∂(p, q)
                                  we can solve equations (46) and (47) to obtain the explicit expressions

                                                                  p = φ(x, y, z), q = ψ(x, y, z)                          (49)

                                  for p and q. The condition that the pair of equations (46) and (47) should be compatible
                                  reduces then to the conditions that the system of equations (49) should be completely
                                  integrable, i.e., the equation

                                                                     φ dx + ψ dy − dz = 0                                 (50)

                                  should be integrable. We already know that the condition that (50) is integrable is

                                                             φ(−ψz ) + ψ(φz ) − (ψx − φy ) = 0

                                                                                12
                                  which is equivalent to
                                                                   ψx + φψz = φy + ψφz .                             (51)
                                  Substituting from equations (49) into (46) and differentiating w.r.t. x and z respec-
                                  tively, we get

                                                                  fx + fp φx + fq ψx = 0,
                                                                  fz + fp φz + fq ψz = 0,

                                  from which it is deduced that

                                                     fx + φfz + fp (φx + φφz ) + fq (ψx + φψz ) = 0.
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                                  Similarly we can deduce from (47) that

                                                     gx + φgz + gp (φx + φφz ) + gq (ψx + φψz ) = 0.

                                  Solving these two equations, we find that
                                                                        1    ∂(f, g)    ∂(f, g)
                                                           ψx + φψz =                +φ                              (52)
                                                                        J    ∂(x, p)    ∂(z, p)
                                  where J is as defined as in (48).
                                     If we differentiate the given pair of equations w.r.t. y and z respectively we will
                                  obtain
                                                                        1 ∂(f, g)        ∂(f, g)
                                                         φy + ψφz = −               +ψ                              (53)
                                                                        J ∂(y, q)        ∂(z, q)
                                  so that, substituting from (52) and (53) and replacing φ, ψ by p, q respectively, we see
                                  that the condition that the two equations should be compatible is

                                                                         [f, g] = 0,                                 (54)

                                  where [f, g] is
                                                               ∂(f, g)    ∂(f, g) ∂(f, g)    ∂(f, g)
                                                    [f, g] =           +p        +        +q         .               (55)
                                                               ∂(x, p)    ∂(z, p) ∂(y, q)    ∂(z, q)
                                  Example: Show that the equations

                                                                 xp = yq, z(xp + yq) = 2xy

                                  are compatible and solve them.
                                  Solution: Here
                                                         f = xp − yq, g = z(xp + yq) − 2xy

                                                                             13
                                  so that
                                    ∂(f, g)        ∂(f, g)                ∂(f, g)         ∂(f, g)
                                            = 2xy,         = −x2 p − xyq,         = −2xy,         = xyp + y 2 q
                                    ∂(x, p)        ∂(z, p)                ∂(y, q)         ∂(z, q)

                                  which imply that
                                                                 [f, g] = y 2 q 2 − p2 x2 = 0.
                                  Hence the given equations are compatible.
                                     From the first equations, we see that p = (y/x)q, and hence we can take p =
                                  y/z, q = x/z. Then
                                                                  z dz = y dx + x dy
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                                  giving us
                                                                      z 2 = 2xy + c1 .


                                  2.7       Charpit’s Method
                                  Let us consider solving the pde

                                                                    f (x, y, z, p, q) = 0                        (56)

                                  In Charpit’s method the fundamental idea is to introduce a second first-order partial
                                  differential equation
                                                                g(x, y, z, p, q, a) = 0                          (57)
                                  containing an arbitrary constant a and which is such that:


                                     • Equations (56) and (57) can be solved to give

                                                               p = p(x, y, z, a), q = q(x, y, z, a)

                                     • The equation
                                                              dz = p(x, y, z, a)dx + q(x, y, z, a)dy             (58)
                                        is integrable.

                                  When such a function g can be found, the solution of equation (58)

                                                                    F (x, y, z, a, b) = 0                        (59)

                                  containing two arbitrary constants a and b, will be a solution of (56).

                                                                              14
                                      The main problem is then the determination of the second equation (57) which is
                                  not much difficult as we need an equation g = 0 compatible with the given equation
                                  f = 0. That is, determining g from the following equation
                                            ∂g      ∂g               ∂g            ∂g             ∂g
                                       fp      + fq    + (pfp + qfq ) − (fx + pfz ) − (fy + qfz )    = 0.             (60)
                                            ∂x      ∂y               ∂z            ∂p             ∂q
                                  Now the problem is to find a solution of this equation, and it is equivalent to finding an
                                  integral of the subsidiary equations
                                                  dx   dy      dz           dp             dq
                                                     =    =           =              =              .                 (61)
                                                  fp   fq   pfp + qfq   −(fx + pfz )   −(fy + qfz )
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                                  These equations are known as Charpit’s equations.
                                      Once an integral g(x, y, z, p, q, a) of this kind has been found, the problem reduces
                                  to solving for p, q and then integrating the equation (58).
                                  Example: Find a complete integral of the equation

                                                                          p2 x + q 2 y = z.                           (62)

                                  Solution: The auxiliary equations are:
                                                     dx    dy         dz            dp       dq
                                                         =     =                =        =
                                                     2px   2qy   2(p2 x + q 2 y   p − p2   q − q2

                                  from which it follows that
                                                               p2 dx + 2pxdp   q 2 dy + 2qydq
                                                                             =
                                                                     p2 x             q2y

                                                               ⇒ p2 x = aq 2 y, a is a constant.                      (63)
                                  Solving (62) and (63) for p and q, we get
                                                                                1/2                           1/2
                                                                  az                               z
                                                       p=                             , q=
                                                               (1 + a)x)                       (1 + a)y)

                                  So, equation (58) in this case becomes
                                                                   1/2                                  1/2
                                                          1+a                     a    1/2          1
                                                                         dz =                dx +             dy
                                                           z                      x                 y

                                                          ⇒ {(1 + a)z}1/2 = (ax)1/2 + y 1/2 + b
                                  which is the required complete integral of (62).


                                                                                  15
                                  2.8       Special types of first-order PDEs
                                  2.8.1 Equations involving only p and q:

                                  For equations of the type
                                                                          f (p, q) = 0                              (64)
                                  Charpit’s equations reduce to
                                                              dx   dy      dz       dp   dq
                                                                 =    =           =    =                            (65)
                                                              fp   fq   pfp + qfq    0   0

                                  An obvious solution of this equation is
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                                                                            p=a                                     (66)

                                  the corresponding value of q being obtained from (64) as

                                                                          f (a, q) = 0                              (67)

                                  so that
                                                                    q = Q(a), a a constant.
                                  The solution of the equation is

                                                                     z = ax + Q(a)y + b                             (68)

                                  where b is a constant. It is to be noted that we took dp = 0 here to provide our second
                                  equation. In some problems, the amount of work involved is considerably reduced if
                                  we take dq = 0, which leads to q = a.
                                  Example: Find the complete integral of pq = 1.
                                  Solution: We know that dp = 0 ⇒ p = a, and then

                                                                      aq = 1 ⇒ q = 1/a.

                                  The complete integral is
                                                                             y
                                                                               +b
                                                                        z = ax +
                                                                             a
                                                 ⇒ a2 x + y − az = c, where a, c are arbitrary constants.




                                                                              16
                                  2.8.2 Equations not involving the independent variable:

                                  Let the pde be of the form
                                                                         f (z, p, q) = 0                            (69)
                                  Then the Charpit’s equations take the form
                                                         dx   dy      dz        dp     dq
                                                            =    =           =      =      .                        (70)
                                                         fp   fq   pfp + qfq   −pfz   −qfz
                                  which leads to the relation
                                                                             p = aq                                 (71)
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                                  Solving (69) and (71), we obtain expressions for p, q from which a complete integral
                                  follows immediately.
                                  Example: Find a complete integral of p2 z 2 + q 2 = 1.
                                  Solution: Putting p = aq, we have

                                                                       q 2 (1 + a2 z 2 ) = 1

                                                       ⇒ q = (1 + a2 z 2 )−1/2 , p = a(1 + a2 z 2 )−1/2
                                  Hence
                                                                        dz = pdx + qdy
                                                                ⇒ (1 + a2 z 2 )1/2 dz = adx + dy.
                                  which leads to the complete integral

                                              az(1 + a2 z)1/2 − log[az + (1 + a2 z 2 )1/2 ] = 2a(ax + y + b).

                                  2.8.3   Separable equations:

                                  We say that a first-order pde is separable, if it can be written in the form

                                                                       f (x, p) = g(y, q)                           (72)

                                  For such equations, Charpit’s equations become
                                                         dx   dy       dz       dp    dq
                                                            =     =           =     =     .                         (73)
                                                         fp   −gq   pfp − qgq   −fx   −gy
                                  so that we have an ODE
                                                                         dp fx
                                                                           +   = 0.
                                                                         dx fp
                                  in x and p which may be solved to give p as a function of x, and an arbitrary constant.

                                                                               17
                                     Writing this equation in the form fp dp + fx dx = 0, we see that its solution is

                                                                         f (x, p) = a.

                                  Hence we determine p and q from the relations

                                                                  f (x, p) = a; g(y, q) = a                             (74)

                                  and then proceed as in the general theory.
                                  Example: Determine the complete integral of

                                                                     p2 y(1 + x2 ) = qx2 .
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                                  Solution: Given that
                                                                     p2 y(1 + x2 ) = qx2 .
                                                                        p2 (1 + x2 )  q
                                                                    ⇒          2
                                                                                     = ,
                                                                             x        y
                                               ax
                                  so that p = √       , q = a2 y.
                                              1+x   2
                                  Hence a complete integral is
                                                                      √        1
                                                                 z = a 1 + x2 + a2 y 2 + b,
                                                                               2
                                  where a and b are constants.

                                  2.8.4   Clairaut’s Equation

                                  A first-order PDE is said to be of Clairaut type if it can be written in the form

                                                                   z = px + qy + g(p, q)                                (75)

                                  The corresponding Charpit’s equations are
                                                     dx       dy             dz            dp   dq
                                                          =        =                     =    =
                                                   x + gp   y + gq   px + qy + pgp + qgq    0   0
                                                                      ⇒ p = a, q = b.
                                  Using these values in (75), we get the complete integral

                                                                   z = ax + by + g(a, b)                                (76)

                                  as is readily verified by direct differentiation.
                                  Example: Find a complete integral of

                                                                 (p + q)(z − xp − yq) = 1.

                                                                              18
                                  Solution: The given equation can be written in the form
                                                                                   1
                                                                  z = xp + yq +       ,
                                                                                  p+q
                                  we see that the complete integral is
                                                                                   1
                                                                  z = ax + by +       .
                                                                                  a+b
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