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Unit 23: Wave and Diffusion Equations by Separation of Variable




               show that the function                                                           Notes
                       1      x 2
               U  , x t  exp
                       t     4xt

               is also the solution of heat equation.

          23.2.2 Heat Flow in Two Dimensional Rectangular System

          To illustrate the solution of the two dimensional diffusion equation, let us consider the following
          problem.
                                            Figure  23.5

                                    Y
                                          t = 0

                                                  b
                                    O                      X
                                            a
          A thin rectangular plate whose surface is impervious  to heat flow has at  t = 0 an arbitrary
          distribution of temperature. Its four edges are kept at zero temperature. It is required to determine
          the subsequent temperature of the plate as t increases.
          Let  the plate  extend from  x = 0 to  x =  a and from  y = 0 to  y  =  b. Expressing the  problem
          Mathematically, we must solve the equation

                   2 V  2 V   V
               c  2             .                                                  ...(i)
                   x  2  y  2  t
          Subject to the boundary conditions

                    y
                V  0, ,t  0
                    y
                V a , ,t  0
                              for all t.                                           ...(ii)
                V x ,0,t  0
                    b
                V x , ,t  0
          The initial conditions are
               V x , ,0  F  , x y   for 0  x  a ,0  y  b
                   y
               V x , ,   0                                                        ...(iii)
                   y
          To solve equation (i) assume a solution of the form

                   y
               V x , ,t  e  t X x Y y   e  t XY say .                             ...(iv)
          where X is a function of x only and Y is function of y only. Substituting (iv) in (i) we get

                  2
                         2
                1 d X  1 d Y
               X dx 2  Y dy 2  c 2
                  2
                               2
                1 d X  1    1 d Y   2
          or       2   2        2    .                                             ...(v)
               X dx   c    Y dy

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