§9.2无界杆的导热问题 4,-a24x=0, 4。=p(x): (-o<x<+o,t>0) 令：(x,)=X(x)T(t) 代入泛定方程 →x(x)T'()-a2x"(x)T()=0 →X"+2X=0;T+a2T=0. 令：N几=o,(o>0）X+m2x=0→X=G,coswx+4,:sin@x T'+o2a2T=0→T=c,eoa =4(x,t)=X(x)T(t)=[A(@)cosox+B(@)sinox]e-ia 4(x,）=j心[4A(w)cos@x+-B(o )小sin@x]edo 6
6 §9.2 无界杆的导热问题 ( ) 2 0 0, . t xx t u u u x = − = = (− + x t , 0) 令：u x t X x T t ( , ) = ( ) ( ) 代入泛定方程 令： = , 0 ( ) = - '' 2 X X + = 0 ( ) ( ) ( ) ( ) ' 2 '' X x T t X x T t − = 0 '' ' 2 X X T T + = + = 0; 0. '' ' 2 X T X T = ' 2 2 0 T T + = 1 2 X c x c x = + cos sin 2 2 3 t T c e− = ( ) ( ) ( ) ( ) ( ) 2 2 , cos sin t u x t X x T t A x B x e − = = + ( ) ( ) ( ) 2 2 0 , cos sin t u x t A x B x e d − = +
代入初始条件p(x)=[A（(o)cos@x+-B(o)sin@x]do 将p(x)展成三角形式的傅立叶积分， p(x=∫rC(o)coswxdw-+∫D(o)sinoxd@ C(o)-Ip(5)cos@5d5.D(@)-(5)sinofdg. 4(o）=C(o)=p5)cos®sa5 B(o)=D(0)=1-9(5)sino5d5. u(x.tp()cosogdscosox+()sin@id6sin@xedo (5)(cnso5cosox+sinmgsin@x)ed -(5)ecoso(x-5)d is
7 代入初始条件 ( ) ( ) ( ) 0 x A x B x d cos sin = + 将 ( x) 展成三角形式的傅立叶积分， ( ) ( ) ( ) 0 0 x C xd D xd cos sin = + ( ) ( ) ( ) ( ) 1 1 C d D d cos , sin . − − = = ( ) ( ) ( ) 1 A C d cos , − = = ( ) ( ) ( ) 1 B sin . D d − = = ( ) ( ) ( ) 2 2 0 1 , cos cos sin sin t u x t d x d x e d − − − = + ( ) ( ) 2 2 0 1 cos cos sin sin t x x e d d − − = + ( ) ( ) 2 2 0 1 cos t e x d d − − = −