u=u +u be l(x,)x=0=0u(x,)x== n(x,)-0=0 (x)x=0 L 1|==0(x) a3x=0v(x)2==0v(x1)=0 be t=T u(x,t)=v(x,t; rdr 由边界条件: v(x,tr)=∑ Ck expl nIa(t-T nA sIn 26 1丌 e sin ds=2bnt (n)2+(k)2I II u = u + u 0 2 − xx = I t I u a u ( , ) x=0 = 0 I u x t ( , ) x=l = 0 I u x t ( ) 0 u x t I = = ( , ) x=0 = 0 II u x t ( , ) x=l = 0 II u x t t=0 = 0 II u kx xx II t II u a u be− − = 2 0 2 vtt − a vxx = v(x,t) x=0 = 0 v(x,t) x=l = 0 kx v t be− = +0 =  = t II u x t v x t d 0 ( , ) ( , ; )  ]sin ; ( ) ( , ; ) exp[ 2 2 2 2 1 l n x l n a t v x t C n k     − = −  = 由边界条件：    d l n e l b C l kx k sin 2 0  − = 2 2 ( ) ( ) 1 ( 1) 2 n lk e bn kl n + − − = −  