(2)的解为: w(x, t) f(a,r)da dt 2a JoLJx-a(t-t 所以,原定解问题的解为: (x0=[0(x+a)+0(x-an)+,∫ 20x(5)d2 x+a f(a, rda dr 2a JoLJx-a(t-T)0.8 1 0.6 0.4 0.2 0 x t 0 0.5 1 1.5 2 −1 −0.5 0 0.5 1 n 16 所以,原定解问题的解为: ( ) 0 ( ) 1 ( , ) ( , ) 2 t x a t x a t W x t f d d a + − − − = ( ) ( ) ( ) . . 1 1 ( , ) 2 2 x at x at u x t x at x at d a + − = + + − + (2)的解为: ( ) 0 ( ) 1 ( , ) 2 t x a t x a t f d d a + − − − +