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Differential and Integral Equation




                    Notes          In this case there is no boundary condition and the initial condition is
                                        V x ,0  f x                            x                          ...(xx)
                                   Again we assume the solution of equation (xi) as

                                        V  , x t  X . .
                                                T
                                   Proceeding as in the last two cases, we get the three solutions and here we find that (i) and (ii) do
                                   not constitute the solution. Hence we take here the third solution (iii), i.e.

                                                                     p t
                                        X  A cospx B sin px  and T  C e  c  2 2
                                                                 0
                                                                  2
                                   Here we have taken the constant as  p  instead of   2 .
                                                                       p t
                                                     C
                                   Hence ( , , )V x t p  XT  ( cospx D sin px )e  c  2 2                  ...(xxi)
                                   Since f(x) is not periodic here, therefore we will use Fourier integrals and not Fourier series.
                                   Also, we may consider C and D as functions of p

                                             p
                                                     p
                                   write C  C ( ), D D ( ).
                                   Now since the heat equation is linear and homogeneous, we have
                                           t
                                          x
                                                      p
                                                    t
                                                   x
                                        V ( , )  V ( , , )dp
                                               0
                                                                       2 2
                                                                        p t
                                                                       c
                                           t
                                          x
                                   or   V ( , )  C p  cospx D p  sin px e  dp                            ...(xxii)
                                               0
                                   (xxiii) is the solution of (i) provided this integral exists and can be differentiated w.r.t.  ,x,  and
                                   w.r.t. ,t,.
                                   Using the initial condition (xx), we get
                                        V ( ,0)   C p  cospx D p  sin px dp  X x
                                          x
                                                0
                                             1
                                        C p       f  sin p d
                                              1
                                   and  D p       f  sin p d  ;


                                               1                     c 2 2
                                                                       p t
                                        V  , x t      f   cos px  p  e   d  dp
                                                 0
                                              1           c  2 2
                                                            p t
                                                    f    e    cos x    . p dp d
                                                 0     0
                                   The change of the order of integration is justified, since inner integral exists and after changing
                                   the order of integration resulting integral also exists.
                                   Solving the inner integral by using the substitution  cp t  s  and using the well known integral



                                         e  s  2  cos2bsds  e  b  2
                                                       2
                                        0



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