Page 406 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION
P. 406

Unit 23: Wave and Diffusion Equations by Separation of Variable




                                                                                                Notes
                          1              2  2
          we get  V x ,t         f  e  x  /4c td
                       2  6  t
                   x
          Putting     w , so that dx  2c tdw , we have
                 2c t

                       1              w dw
                                        2
               V  , x t    f x  2cw t e                                         ...(xxiii)
          which is the required solution.
          Case IV: Let there be a bar of length L which is perfectly insulated. Both ends i.e. x = 0 and x = L
          are also perfectly insulated and the initial temperature of the bar is
               V x ,0  f x

                                                               V
          The flux of heat across the faces  x  0 and x  L is proportional t 0  at the end, since these ends
                                                               x
          are insulated. In this case the boundary conditions are


                 V  0,t  0                                                      ...(xxiv)
                x

                 V   , L t  0                                                    ...(xxv)
                x

          and the initial condition is
               V x ,0  f x           0  x  L                                    ...(xxvi)

          Proceeding as in Case I, here also we get three solutions. Solution (ii) is inadmissible as in this
          V = XT increases indefinitely with time. The solution (iii) by itself is inadequate since in this case
          the temperature will tend to zero as t tends to infinity. Therefore general solution will consist of
          the solution of (i) and (iii).
          Using boundary condition (xxiv) in solution (i), i.e.
               X   Ax B  and T  C

          or   V    ' A x B '
          we get  A ' 0.

          Therefore V  ' B  is one of the solution of (i). Considering solution (iii) i.e.

                                        2 2
                                         c t
               X   A cos x B sin x , T  Ce
                                         2 2
          or   V  , x t  C 'cos x D 'sin x e  c t
          Using boundary condition (xxiv) and (xxv), we get

               D ' 0
                  n
          and                             n  1,2,3,.....
                   L





                                           LOVELY PROFESSIONAL UNIVERSITY                                   399
   401   402   403   404   405   406   407   408   409   410   411