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Differential and Integral Equation




                    Notes          Hence (ix) with the coefficient (x) is the solution of Laplace s equation (i), which satisfy all the
                                   given boundary conditions.
                                   Case II: Let there be a thin rectangular metallic plate bounded by the lines  x  0, x  , a y  0 and
                                   y   , b  the edges  x  0, x  , a y  0 are kept at temperature zero while the edge  y  b  is kept at
                                   temperature  ( ).f x
                                   Here the boundary conditions are given by

                                                             y
                                                          V (0, ) = 0                                      ...(xi)
                                                             y
                                                          V ( , ) = 0                                     ...(xii)
                                                            a
                                                            x
                                                          V ( ,0) = 0                                     ...(xiii)
                                                          V ( , ) =  f ( )                                ...(xiv)
                                                              b
                                                                     x
                                                            x
                                   Proceeding as in Case I and using (xi) and (xii), we get
                                                                    Figure 22.6















                                                                           n
                                                              A = 0 and                (n  1, 2, 3, ....)
                                                                            a
                                   Therefore for each value of n, we have

                                                                                     n
                                                                                 /a
                                                                               n y
                                                                      n y
                                                                        /a
                                                         V  ( , ) = C e    D e     sin  x ... (n  1, 2, 3,...)
                                                             y
                                                            x
                                                          n          n       n
                                                                                      a
                                   Hence for different values of n, the solution of (i) is
                                                                                         n
                                                            x
                                                          V ( , ) =   C e n y  /a  D e  n y  /a  sin  x
                                                             y
                                                                        n
                                                                                n
                                                                   n  1                   a
                                   In this result using (xiii), we get
                                                             D n =  C n .
                                   Therefore
                                                                                       n
                                                                         n y
                                                                                  /a
                                                                                n y
                                                                           /a
                                                             y
                                                            x
                                                          V ( , ) =   C e      e     sin  x
                                                                       n
                                                                   n  1                 a
                                   or
                                                                            n  y   n  x
                                                          V ( , ) =   C n  sinh  sin    where C  n  2 C             ...(xv)
                                                             y
                                                            x
                                                                                                   n
                                                                   n  1      a       a
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