Question

偏微分方程里的问题

A uniform string of length πl is fastened at its ends x=0,x=πl.The point x=1/3πl is drawn aside a small distance b ,and the string released from rest at time t=0.Show that,if Tis the tension and ρ the density per unit length of string,the subsequent displacement of the string is ?

Answer 1

u_tt=a^2u_xx,a^2=T/ρ
u(0,t)=u(πl,0)=0
u(x,0)=f(x)=f_1(x)=3bx/(πl),0<x<1/3πl
= f_2(x)=3/2b(πl-x)/(πl),1/3πl<x<πl
u_t(x,0)=0
特征函数法设u(x,t)=∑T(t)sin(nπx/L),n from 1 to n ,L=πl
代入范定方程有
∑[T"(t)+(nπa/L)^2T(t)]sin(nπx/L)=0
得T"(t)+(nπa/L)^2T(t)=0
T(t)=Acos(nπa/L)t+Bsin(nπa/L)t
u(x,t)=∑[Acos(nπa/L)t+Bsin(nπa/L)t]sin(nπx/L)
系数A,B的确定
由u_t(x,0)=0可得B=0
u(x,t)=∑Acos(nπat/L)sin(nπx/L)
由u(x,0)=f(x),得
∑Asin(nπx/L)=f(x)由正交性有
A=2/L∫(0~L)sin(nπx/L)f(x)dx
=2/L∫(0~πL/3)sin(nπx/L)f_1(x)dx+2/L∫(πL/3~πL)sin(nπx/L)f_2(x)dx=A(n)
定解为
u(x,t)=∑A(n)cos(nπat/L)sin(nπx/L),
A(n)积分你分部算一算