例1 u -afu=Acos-sin ot n(x.D=0=0n(x2)2==0 4=。=0(x)anl-o=v(x) 解v- -a2ν=0 v_(x,t v,(,tre=0 r+0 0 Ier+0=Acos sin uf(x,t) 由边界条件:(x;)=∑r0r)cos2n 带入泛定方程:∑n”+(D7女=0 n元 n T=0 nra(t-T nra(t 70(2r)=4+B(t-z) T,=A(T)cOS +B(T)sIn v(x, t; t)=Ao+Br+2(A, (T)cos a(t- +B(rsin例 1 t l x utt a uxx A   cos sin 2 − = ux (x,t) x=0 = 0 ux (x,t) x=l = 0 ( ) 0 u x t= = ( ). 0 u x t t= = 解 0 2 vtt − a vxx = vx (x,t) x=0 = 0 vx (x,t) x=l = 0 v t= +0 = 0 cos sin ( , ) 0 f x t l x vt t A   = + = 由边界条件：   = = 0 ( , ; ) ( , ) cos n n l n x v x t T t    带入泛定方程： [ '' ( ) ]cos 0 0 2  + =  n= n l n x T l n a T   '' ( ) 0 2 + T = l n a Tn  ( ; ) ( ) 0 0 0 T t  = A + B t − l n a t B l n a t Tn An n ( ) ( )sin ( ) ( ) cos       − + − = l n x l n a t B l n a t v x t A B A n n n          ) cos ( ) ( )sin ( ) ( , ; ) ( ( ) cos 1 0 0   = − + − = + +