Answer to Question #174809 in Differential Equations for Anoop

Question

solve d²u/dt² =9d²u/dx² with 0《x《1

u(0,t)=0 , u(1,t)=0 , d u(x,0)/dt =0

u(x,0) = x 0《x《0.25

0.25 0.25《x《0.75

1-x 0.75《x《1

Answer 1

The wave equation:

$\frac{d^2u}{dt^2}=c^2\frac{d^2u}{dx^2}$

The general solution:

$u(x,t)=\sum sin(n\pi x/l)(A_ncos(cn\pi t/l)+B_nsin(cn\pi t/l))$

$f(x)=u(x,0)=\sum A_nsin(n\pi x/l)$

$g(x)=u_t(x,0)=\sum B_n(cn\pi/l)sin(n\pi x/l)$

$A_n=\frac{2}{l}\displaystyle\intop ^l_0 f(x)sin(n\pi x/l)dx$

$B_n=\frac{2}{cn\pi}\displaystyle\intop ^l_0 g(x)sin(n\pi x/l)dx$

Then:

$B_n=0$

$A_n=2(\displaystyle\intop ^{0.25}_0 xsin(n\pi x)dx+0.25\displaystyle\intop ^{0.75}_{0.25} sin(n\pi x)dx+\displaystyle\intop ^{1}_{0.75} (1-x)sin(n\pi x)dx)$

$\displaystyle\intop ^{0.25}_0 xsin(n\pi x)dx=\frac{sin(n\pi x)-n\pi xcos(n\pi x)}{n^2\pi^2}|^{0.25}_0=$

$=\frac{sin(n\pi /4)-0.25n\pi cos(n\pi /4)}{n^2\pi^2}$

$\displaystyle\intop ^{0.75}_{0.25} sin(n\pi x)dx=-\frac{cos(n\pi x)}{n\pi}|^{0.75}_{0.25}=-\frac{cos(3n\pi/4)-cos(n\pi/4)}{n\pi}$

$\displaystyle\intop ^{1}_{0.75} (1-x)sin(n\pi x)dx=(-\frac{cos(n\pi x)}{n\pi}-\frac{sin(n\pi x)-n\pi xcos(n\pi x)}{n^2\pi^2})|^{1}_{0.75}=$

$=-\frac{cos(n\pi)-cos(3n\pi/4)}{n\pi}-\frac{-n\pi cos(n\pi )-sin(3n\pi /4)+0.75n\pi cos(3n\pi/4)}{n^2\pi^2}$

$u(x,t)=\sum A_nsin(n\pi x)cos(3n\pi t)$