Separation of variables

Posted October 27th, 2007 by vic777

Algebra questions

What would be the solution for:
$u_t_t-4u_x_x=0, 0<x<1, t>0,$
$u(x,0)=\sin\frac{\pi x}{2}, u_t(x,0)=1+x, 0\le x\le1,$
$u(0,t)=0,u_x(1,t)=0, t\ge0$

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Answer

On October 28th, 2007 Structure says:

Look for a nonzero solution of the form u(x,t)=X(x)T(t).
We get T"(t)X(x)-4T(t)X"(x)=0 or
$\frac{T"(t)}{4T(t)}=\frac{X"(x)}{X(x)}=-\lambda^2$
(*)We get $T"(t)+4\lambda^2T(t)=0$
(**) $X"(x)+\lambda^2X(x)=0\: X(0)=X'(1)=0$
Second equation has a solution of the form
$X(x)=a\cos(\lambda x)+b\sin(\lambda x)$
$X(0)=a=0\: X'(1)=b\lambda\cos\lambda =0$
$\lambda_n=\frac{(2n+1)\pi}{2}$
So $X_n(x)=\sin (\frac{(2n+1)\pi}{2}x)$
Equation (*) has solution
$T_n(t)=c_n\cos 2\lambda_n t+d_n\sin 2\lambda_n t$
or $T_n(t)=c_n\cos (2n+1) t+d_n\sin (2n+1) t$
Look now for a solution of our equation
$u(x,t)=\sum_{n=0}^{+\infty}T_n(t)X_n(x)=\sum_{n=0}^{+\infty}c_n\cos (2n+1) t+d_n\sin (2n+1) t\sin (\frac{(2n+1)\pi}{2}x)$
From $u(x,0)=\sum_{n=0}^{+\infty}c_n\sin \frac{(2n+1)\pi}{2}x=\sin (\frac{\pi}{2}x)$ we have
$c_n=2\int_{0}^{1}\sin (\frac{\pi}{2}x)\sin(n\pi x)dx=\delta_{0,n}$
$\frac{\partial u}{\partial t}(x,0)=\sum_{n=0}^{+\infty}d_n(2n+1)\pi \sin \frac{(2n+1)\pi}{2}x=1+x$
$\int_0^1\sum_{n=0}^{+\infty}d_n(2n+1)\pi \sin \frac{(2n+1)\pi}{2}x\sin \frac{(2m+1)\pi}{2}x dx=\int_0^1(1+x)\sin \frac{(2m+1)\pi}{2}xdx$
$\sum_{n=0}^{+\infty}d_n(2n+1)\pi \int_0^1\sin \frac{(2n+1)\pi}{2}x\sin \frac{(2m+1)\pi}{2}x dx=\frac{2}{(2m+1)\pi}+(-1)^m\frac{4}{(2m+1)^2\pi^2}$
$\sum_{n=0}^{+\infty}d_n(2n+1)\pi \frac{1}{2}\delta_{m,n}=\frac{2}{(2m+1)\pi}+(-1)^m\frac{4}{(2m+1)^2\pi^2}$
$\frac{1}{2}(2m+1)\pi d_m=\frac{2}{(2m+1)\pi}+(-1)^m\frac{4}{(2m+1)^2\pi^2}$
$d_m=\frac{4}{(2m+1)^2\pi^2}+(-1)^m\frac{8}{(2m+1)^3\pi^3}$
So finally solution can be written
$u(x,t)=\cos t \sin \frac{\pi}{2}x+\sum_{n=0}^{+\infty}(\frac{4}{(2n+1)^2\pi^2}+(-1)^n\frac{8}{(2n+1)^3\pi^3})\sin(2n+1)t\sin \frac{(2m+1)\pi}{2}x$