C定解达朗贝尔公式 确定待定函数的形式 无限长,即无边界条件 设初始条件为41=0=0(x)和ul=0=v(x) -0<x<∞) f(x)+f(x)=(x) f(x)-f2(x) (5)d5+f1(x0)-f2(x0) aff'(x)-af2'(x)=(x) 0)=9+wM+1(x)(x)()=0(w(M2-2U(x)( f(x+a)=(x+a)+v(5)45+,U(x)-(x 4(r-a=lo(x-a)-Ij vl()ds-1U,(26 )-52(x)I f1(x+a)+f2(x-a) (x,1)=[(x+an)+(x-am)+|v(5)d5 2a x-atC.定解 达朗贝尔公式 确定待定函数的形式 无限长，即无边界条件。 设初始条件为 ( ) 0 u x t= = 和 ( ) 0 u x x t= = (−  x  ) ( ) ( ) ( ) 1 2 f x + f x = x '( ) '( ) ( ) 1 2 af x −af x = x ( ) ( ) ( ) 1 ( ) ( ) 1 2 1 0 2 0 0 d f x f x a f x f x x x − = + −     [ ( ) ( )] 2 1 ( ) 1 ( ) 2 1 ( ) 1 1 0 2 0 0 d f x f x a f x x x x = + + −      [ ( ) ( )] 2 1 ( ) 1 ( ) 2 1 ( ) 2 1 0 2 0 0 d f x f x a f x x x x = − − −          d a u x t x at x at x at x at ( ) 2 1 [ ( ) ( )] 2 1 ( , )  + − = + + − + [ ( ) ( )] 2 1 ( ) 1 ( ) 2 1 ( ) 2 1 0 2 0 0 d f x f x a f x at x at x x − = − − − −      ( ) ( ) u = f 1 x +at + f 2 x −at [ ( ) ( )] 2 1 ( ) 2 1 ( ) 2 1 ( ) 1 1 0 2 0 0 d f x f x a f x at x at x x + = + + + −     