Page 582 - Modelling in Transport Phenomena A Conceptual Approach
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562         APPENDIX B.  SOLUTIONS OF DIFFERENTIAL EQUATIONS

            the solutions, i.e.,  ClO, + C202, also satisfies the partial differential equation and
            the boundary conditions. Therefore, the complete solution is
                                        00
                                   e =  c,,         sin( nxc)               (B.3-52)
                                       n=l
            The unknown coefficients Cn can be determined by using the initial condition. The
            use of  Eq.  (B.3-36) results in

                                           00
                                       1 =    C, sin(nx0                    (B.3-53)
                                          n=l
            Since the eigenfunctions are simply orthogonal, multiplication of  Eq.  (B.3-53) by
            sinmxt and integration from  = 0 to  = 1 gives


                                                                            (B.3-54)

            Note that the integral on the right side of  Eq.  (B.3-54) is zero when m # n and
            nonzero when m = n.  Therefore, when m = n the summation drops out and Q.
            (B.3-54) reduces to the form
                                                  I'
                               Jd' sin(n?rt) d(   = Cn   sin2(nrJ) d(       (B.3-55)

            Evaluation of the integrals show that

                                            2
                                      cn = - [l - (-  l)n]                  (B.3-56)
                                            xn
            The coefficients C,, take the following values depending on the value of  n:
                                          0    n = 2,4,6, ...
                                   Cn={ - n = 1,3,5, ...                    (B .3- 57)
                                           4
                                           xn
            Therefore, the solution becomes


                                                                            (B.3-58)


            Replacing n by 2k + 1 gives


                                                                            (B .3- 59)
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