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					                                  Second-order Partial Differential Equations

                                  1    Classification and Characteristics
                                  The general second-order partial differential equation in two independent variables is
                                  of the form:

                                  A(x, y)uxx +2B(x, y)uxy +C(x, y)uyy +D(x, y)ux +E(x, y)uy +F (x, y)u = f (x, y),

                                  which may conveniently be written as

                                               A(x, y)uxx + 2B(x, y)uxy + C(x, y)uyy = f (x, y, u, ux , uy ),
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                                                                                                                          (1)

                                  where u(x, y) is a dependent variable and x, y are independent variables. The coeffi-
                                  cients are, in general, functions of x and y. For simplicity let us consider the possibility
                                  of obtaining solutions of the form

                                                                       u = g(x + λy),                                     (2)

                                  where (x + λy) is the argument of g, for the homogeneous equation corresponding to
                                  equation (1) which is
                                                             Auxx + 2Buxy + Cuyy = 0,                            (3)
                                  where A, B and C are assumed constants. If (2) is a solution of (3), then substituting,
                                  we obtain
                                               Ag (x + λy) + 2Bλg (x + λy) + Cλ2 g (x + λy) = 0,
                                  which will be an identity if and only if

                                                                    Cλ2 + 2Bλ + A = 0.                                    (4)

                                  This is a quadratic equation in λ, and solving for λ, we have
                                                                              √
                                                                       −B ± B 2 − CA
                                                                  λ=                      .                               (5)
                                                                               C
                                  Thus,
                                  (i) λ has two distinct roots if B 2 − CA > 0;
                                  (ii) λ has two equal roots if B 2 − CA = 0; and
                                  (iii) λ has two complex roots if B 2 − CA < 0.
                                  Hence, there are two, one or no real values of λ for which solutions of the form g(x +
                                  λy) exist, according to the sign of the discriminant B 2 − AC. Equation (1) is said to
                                  be

                                                                               1
                                     1. Hyperbolic if B 2 − AC > 0,

                                     2. Parabolic if B 2 − AC = 0, and

                                     3. Elliptic if B 2 − AC < 0.

                                  Now in each case, the loci
                                                                        x + λ1 y = C1
                                                                                           ,                            (6)
                                                                        x + λ2 y = C2
                                  where λ1 and λ2 are the roots of equation (4), are called the characteristic curves
                                  or simply the characteristics of equation (3). These two characteristics define two
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                                  families of parallel lines in the xy-plane.
                                       Please note that for the classification of the PDEs, though we consider equation (3),
                                  i.e., only three terms of the equation (1), the classification applies to the general equa-
                                  tion. The above classification does not hold for PDEs in three independent variables.
                                       Consider the characteristic in its simplest form:

                                                                          x + λy = c.

                                  Then,
                                                                    dy      1
                                                                       =− .
                                                                    dx      λ
                                  Now eliminating λ from equation (4) by the slope dy/dx, we obtain the following
                                  differential equation
                                                                 2
                                                             dy            dy
                                                         A         − 2B         + C = 0.                      (7)
                                                             dx            dx
                                  This differential equation must be satisfied by any characteristics of the form x+λy = c
                                  and thus characteristics are found by the solution of (7), and hence this equation is
                                  known as the characteristic differential equation.
                                      Even if A, B and C are not constants but functions of x and y, still the classification
                                  as done above is valid and the characteristic equation in that case can be written as
                                                                    2
                                                               dy                     dy
                                                    A(x, y)             − 2B(x, y)         + C(x, y) = 0.               (8)
                                                               dx                     dx

                                  Solving (7) or (8) for dy/dx, we get
                                                                               √
                                                                    dy   B±        B 2 − AC
                                                                       =                    .                           (9)
                                                                    dx             A



                                                                               2
                                  Equation (9) represents a pair of ordinary differential equations the solutions of which
                                  exist in the form
                                                                   φ(x, y) = C1 = ξ
                                                                                           .                          (10)
                                                                   ψ(x, y) = C2 = η
                                  These two are said to be the characteristics of the partial differential equation (1).
                                  The simplest, but very useful, examples of hyperbolic, parabolic and elliptic partial
                                  differential equation are, respectively,

                                     • The one-dimensional wave equation: utt = c2 uxx .

                                     • The one-dimensional heat conduction or diffusion equation: ut = αuxx .
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                                     • The two-dimensional Laplace’s equation: uxx + uyy = 0.

                                  Using the concepts of characteristics, a general second-order partial differential equa-
                                  tion can be reduced to one of the above forms.


                                  1.1    Hyperbolic Class
                                  Equation (1) can be classified as hyperbolic type if B 2 − CA > 0. The characteristic
                                  will be given by the solution of the following pair of ordinary differential equations
                                                                              √
                                                                  dy     B ± B 2 − AC
                                                                      =                   ,
                                                                  dx            A
                                  which will give distinct and real roots. Integrating the pair we will get

                                                                     φ(x, y) = C1
                                                                                           .
                                                                     ψ(x, y) = C2

                                  Thus the characteristics are defined as

                                                                      ξ = φ(x, y)
                                                                                       .                             (11)
                                                                      η = ψ(x, y)

                                  Under this transformation, the original equation (1) can be reduced to the standard
                                  form
                                                                uξη = h(ξ, η, u, uξ , uη ).                      (12)




                                                                              3
                                  1.2   Parabolic Class
                                  Equation (1) can be classified as parabolic type if B 2 − CA = 0. As the roots will be
                                  real and equal, the characteristics will be given by the solution of the ODE
                                                                           dy  B
                                                                              = .
                                                                           dx  A
                                  Integrating we get
                                                                        φ(x, y) = c1 .
                                  Note that we will get only one characteristic from the solution of the above equation,
                                  say, η = φ(x, y). The other characteristic may be defined as
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                                                                            ξ = x.

                                  The selection of ξ, i.e., the second characteristic, is based on the fact that it should
                                  not be parallel to the other characteristic η. In other words, we choose ξ such that the
                                  Jacobian of the transformation
                                                                            ∂(ξ, η)
                                                                      J=            = 0.
                                                                            ∂(x, y)
                                  Therefore under the transformation
                                                                       ξ =x
                                                                                          ,                            (13)
                                                                       η = φ(x, y)
                                  equation (1) can be written as

                                                                   uξξ = h(ξ, η, u, uξ , uη ).                         (14)


                                  1.3   Elliptic Class
                                  Equation (1) can be classified as elliptic type if B 2 − CA < 0. The characteristic will
                                  be given by the solution of the following pair of ordinary differential equations
                                                                              √
                                                                 dy     B ± i CA − B 2
                                                                      =                    .
                                                                 dx             A
                                  Taking positive sign and integrating, we get

                                                                         φ(x, y) = c,

                                  where φ(x, y) is a complex function. The characteristics in this case are given by

                                                                    ξ = Re{φ(x, y)}
                                                                                              .                        (15)
                                                                    η = Im{φ(x, y)}

                                                                               4
                                  Under this transformation, equation (1) reduces to the standard form

                                                                  uξξ + uηη = h(ξ, η, u, uξ , uη ).                  (16)

                                  Alternatively, considering both positive and negative signs and then solving the differ-
                                  ential equations, we obtain

                                                                   φ(x, y) = c1 , ψ(x, y) = c2 .

                                  These two functions are complex conjugates. The characteristics are then given by

                                                                   ξ = φ(x, y)
                                                                                                    .                (17)
                                                                                 ¯
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                                                                   η = ψ(x, y) = φ(x, y)

                                  Under this transformation, equation (1) reduces to the standard form

                                                                     uξη = h(ξ, η, u, uξ , uη ).                     (18)

                                  Example: Find the characteristics of the following equation and reduce it to the ap-
                                  propriate standard form and then obtain the general solution:

                                                                   3uxx + 10uxy + 3uyy = 0.

                                  Solution: Here we have B 2 − CA = 16 > 0 and hence the equation is of hyperbolic
                                  type. The characteristics are given by the solutions of
                                                                          2
                                                                    dy                   dy
                                                              A               − 2B            + C = 0,
                                                                    dx                   dx
                                  i.e.,
                                                                          2
                                                                     dy                  dy
                                                              3               − 10            + 3 = 0,
                                                                     dx                  dx
                                  giving
                                                                       dy   5±4     1
                                                                          =     = 3, .
                                                                       dx    3      3
                                  Integrating these, we get

                                                                          3x − y = c1 ,
                                                                          x − 3y = c2 .

                                  Hence the characteristics are defined by

                                                                          ξ = 3x − y,
                                                                          η = x − 3y.

                                                                                     5
                                  Here we have ξx = 3, ηx = 1, ξy = −1, ηy = −3. Therefore, under this transformation,
                                  we have

                                                           ux = uξ ξx + uη ηx = 3uξ + uη ,
                                                           uy = uξ ξy + uη ηy = −uξ − 3uη .

                                  Now the expressions for the second order derivatives are

                                                            uxx = 9uξξ + 6uξη + uηη ,
                                                            uxy = −3uξξ − 10uξη − 3uηη ,
                                                             uyy = uξξ + 6uξη + 9uηη .
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                                  Substituting these values in the given equation we get

                                                                            uξη = 0.

                                  Now integrating with respect to ξ partially

                                                                           uη = F (η).

                                  Again integrating with respect to η partially, we obtain

                                                                      u = F (η) + G(ξ).

                                  Hence the solution of the original equation is

                                                           u(x, y) = F (x − 3y) + G(3x − y),

                                  where F and G are arbitrary functions.
                                  Example: Find the characteristics of the following equation and reduce it to the ap-
                                  propriate standard form and then obtain the general solution:

                                                                 uxx + 4uxy + 4uyy = 0.

                                  Solution: Here we have B 2 − CA = 0 and hence the equation is of parabolic type.
                                  The characteristics are given by the solutions of
                                                                       2
                                                                 dy                   dy
                                                            A              − 2B             + C = 0,
                                                                 dx                   dx
                                  i.e.,
                                                                       2
                                                                 dy               dy
                                                                           −4              + 4 = 0,
                                                                 dx               dx

                                                                                  6
                                  giving
                                                                          dy
                                                                             = 2,
                                                                          dx
                                  which, on integration, gives
                                                                        y − 2x = c1 .
                                  Therefore the characteristics are given by.

                                                                       ξ = x,
                                                                       η = y − 2x.

                                  Here we have ξx = 1, ηx = −2, ξy = 0, ηy = 1. Therefore, under this transformation,
Prepared by Swaroop Nandan Bora




                                  we have

                                                            ux = uξ ξx + uη ηx = uξ − 2uη ,
                                                            uy = uξ ξy + uη ηy = uη .

                                  Now the expressions for the second order derivatives are

                                                                 uxx = uξξ − 4uξη + 4uηη ,
                                                                 uxy = uξη − 2uηη ,
                                                                 uyy = uηη .

                                  Substituting these values in the given equation we get

                                                                          uξξ = 0.

                                  Now integrating with respect to ξ partially

                                                                        uξ = f (η).

                                  Again integrating with respect to ξ partially, we obtain

                                                                     u = ξf (η) + g(η).

                                  Hence the solution of the original equation is

                                                           u(x, y) = xf (y − 2x) + g(y − 2x),

                                  where f and g are arbitrary functions.
                                  Example: Find the characteristics of the following equation and reduce it to the ap-
                                  propriate standard form and then obtain the general solution:

                                                                   uxx + 4uxy + 5uyy = 0.

                                                                                7
                                  Solution: Here we have B 2 − CA = −1 < 0 and hence the equation is of elliptic type.
                                  The characteristics are given by the solutions of
                                                                         2
                                                                  dy                    dy
                                                              A              − 2B             + C = 0,
                                                                  dx                    dx

                                  i.e.,
                                                                         2
                                                                  dy                dy
                                                                             −4              + 5 = 0,
                                                                  dx                dx
                                  giving
                                                                        dy
                                                                            = 2 ± i.
                                                                        dx
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                                  Taking the positive sign and integrating,

                                                                       y − (2 + i)x = c.

                                  which can be rewritten as
                                                                    (2 − i)y − 5x = K1 .
                                  or,
                                                                     2y − 5x − iy = K1 .
                                  The characteristics in this case are

                                                                    ξ = 2y − 5x − iy,
                                                                    η = 2y − 5x + iy.

                                  Therefore, under this transformation, we will ultimately have the canonical form of the
                                  given equation as

                                                                              uξη = 0.
                                  Integrating this with respect to η partially,

                                                                             uξ = φ (ξ).

                                  Again integrating this with respect to ξ partially

                                                                       u = φ(ξ) + ψ(η).




                                                                                    8
                                     Now let us consider the solution for a nonhomogeneous equation.
                                  Example: Find the characteristics of the following equation and reduce it to the ap-
                                  propriate standard form and then obtain the general solution:

                                                            uxx − 4uxy + 4uyy = cos(2x + y).

                                  Solution: It can readily be seen that the given equation is a parabolic one and the
                                  characteristics are given by the solution of
                                                                                    2
                                                                        dy
                                                                           +2           = 0.
                                                                        dx
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                                  giving us
                                                                        y + 2x = c1 .
                                  Hence the characteristics are given by

                                                                        ξ = x,
                                                                        η = y + 2x.

                                  The canonical form will be
                                                                         uξξ = cos η.
                                  Integrating partially with respect to ξ, we get

                                                                    uξ = ξ cos η + f (η).

                                  Again integrating w.r.t. ξ, we get

                                                                        ξ2
                                                            u(ξ, η) =      cos η + ξf (η) + g(η).
                                                                        2
                                  Writing back in the variables x and y, we have the solutions as:

                                                                                               x2
                                                  u(x, y) = xf (y + 2x) + g(y + 2x) +             cos(y + 2x),
                                                                                               2
                                  where f and g are arbitrary functions.


                                  1.4 The D’Alembert solution of the wave equation
                                  In applied sciences and engineering, method of characteristics is not of much use for
                                  parabolic and elliptic equations. On the other hand, the two families of characteristics
                                  of hyperbolic equations, being real and distinct, are of considerable practical value.

                                                                              9
                                  Especially, in one-dimensional progressive wave propagation, consideration of char-
                                  acteristics can give us a good deal of information about the propagation of the wave
                                  fronts. This solution of one-dimensional wave equation, known as D’Alembert’s solu-
                                  tion, was discovered by a French mathematician named Jean Le Rond D’Alembert.
                                      Consider the one-dimensional wave equation

                                                                         utt = c2 uxx .                          (19)

                                  Here B 2 − AC = c2 > 0. The characteristics are given by the solution of the ODE
                                                                             dx
                                                                                = ±c.                            (20)
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                                                                             dt
                                  Integrating we get
                                                                       x − ct = c1
                                                                                          .                      (21)
                                                                       x + ct = c2
                                  Hence the characteristics are given by

                                                                   ξ = x − ct, η = x + ct.

                                  Under this transformation

                                                        ux = uξ ξx + uη ηx = uξ + uη ,
                                                        ut = uξ ξt + uη ηt = −cuξ + cuη ,
                                                       uxx = (uξξ ξx + uξη ηx ) + (uηξ ξx + uηη ηx )
                                                            = uξξ + 2uξη + uηη ,
                                                       utt = −c(uξξ ξt + uξη ηt ) + c(uηξ ξt + uηη ηt ),
                                                            = c2 (uξξ − 2uξη + uηη ).

                                  Substituting these into the given equation, we obtain,

                                                                             uξη = 0.                            (22)

                                  Integrating partially with respect to η,

                                                                         uξ = F (ξ).

                                  Integrating partially w.r.t. ξ, we obtain,

                                                                      u = F (ξ) + G(η).


                                                                               10
                                     Hence the solution in physical variables is

                                                             u(x, t) = F (x − ct) + G(x + ct)                          (23)

                                  where F and G are arbitrary functions.
                                        The physical interpretation of these functions is quite interesting. The functions F
                                  and G represent two progressive waves travelling in opposite directions with the speed
                                  c. To see this let us first consider the solution u = F (x − ct). At t = 0, it defines the
                                  curve u = F (x), and after time t = t1 , it defines the curve u = F (x − ct1 ). But these
                                  curves are identical except that the latter is translated to the right a distance equal to
                                  ct1 .
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                                        Thus the entire configuration moves along the positive direction of x-axis a distance
                                  of ct1 in time t1 . The velocity with which the wave is propagated is, therefore,
                                                                               ct1
                                                                          v=       =c
                                                                               t1
                                      Similarly the function G(x+ct) defines a wave progressing in the negative direction
                                  of x-axis with constant velocity c. The total solution is, therefore, the algebraic sum of
                                  these two travelling waves.
                                      Solution (23) is very convenient representation for progressive waves which travel
                                  large distances through a uniform medium.
                                      Let us consider the following two initial conditions for a uniform medium −∞ <
                                  x < ∞.

                                                            Displacement: u(x, 0) = φ(x),                              (24)
                                                                  Velocity: ut (x, 0) = ψ(x),                          (25)

                                  that is, we consider the vibration of a thin string of infinite length. Then from the
                                  solution (23) we find that

                                                                    F (x) + G(x) = φ(x),                               (26)
                                                              −cF (x) + c G (x) = ψ(x),                                (27)

                                  for all values of x. Integrating the second equation with respect to x, we get
                                                                                      x
                                                                             1
                                                             −F (x) + G(x) =               ψ(s)ds.                     (28)
                                                                             c       x0

                                  From (26) and (28), we obtain
                                                                                          x
                                                                 1        1
                                                          F (x) =  φ(x) −                     ψ(s)ds ,                 (29)
                                                                 2        c           x0
                                                                                       x
                                                                 1        1
                                                          G(x) =   φ(x) +                     ψ(s)ds ,                 (30)
                                                                 2        c           x0


                                                                             11
                                  where A is an integration constant and s is a dummy variable.
                                     Substituting these expressions into (23), we obtain
                                                                                               x+ct
                                                         1                           1
                                                    u=     [φ(x − ct) + φ(x + ct)] +                  ψ(s)ds.           (31)
                                                         2                           2c     x−ct

                                  This is D’Alembert’s solution of wave equation. Thus for a given initial displacement
                                  and velocity in the vertical direction, the wave equation is completely solved and this
                                  solution is usually called the progressive wave solution. It is to be noted that the use of
                                  string problem to demonstrate the solution of the wave problem is a matter of conve-
                                  nience. However, any variables satisfying the wave equation possess the mathematical
Prepared by Swaroop Nandan Bora




                                  properties developed for the string.
                                       It is clear that the wave equation can be handled very easily by introducing the
                                  characteristic variables (ξ, η). The relationship between the physical plane and the
                                  characteristic plane for this particular example can be demonstrated graphically.
                                       Equation (23) represents the solution as the sum of two progressive waves;one
                                  going to the right and the other to the left. The wave celerity is c. For each of the two
                                  progressive waves, we can also follow the wave motion by observing that in the xt-
                                          1                                                                       1
                                  plane 2 F (x−ct) is constant along each line x−ct = constant and similarly 2 G(x+ct)
                                  is constant along each line x + ct = constant. Thus there are two families of parallel
                                  lines called the characteristics along which the waves are propagated.
                                       Furthermore, along x-axis the values of u(x, 0) and ut (x, 0) are given as initial
                                  conditions of displacement and velocity and they just suffice to determine the constant
                                  values of F and G along the individual characteristic. The characteristics, therefore,
                                  represent the paths in xt-plane along which disturbances in the medium propagated.
                                       Finally, since the solution of wave equation is u = F (x − ct) + G(x + ct) the value
                                  of u at any point in the xt-plane is the sum of the values of F and G on the respective
                                  characteristics which pass through that point.
                                       Suppose we need to find the wave function at P (x, t), then draw two character-
                                  istics passing through that point. Measure the x-intercepts subtended by these two
                                  characteristics. Then using the formula (31), we obtain
                                                                                             β
                                                                   1                 1
                                                       u(x, t) =     [φ(α) + φ(β)] +             ψ(s)ds,
                                                                   2                 2c    α

                                  where α = x − ct and β = x + ct, which is as required.




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