Wave equations

Posted October 7th, 2007 by vic777

Algebra questions

What would be a solution of the initial boundary-value problem?
$\frac{\partial^2u}{\partial t^2}=\frac{\partial^2u}{\partial x^2}$
$0<x<2, t>0$
$u(x,0)=sin(\frac{\pi x}{2}), 0\le x\le2$
$\frac{\partial u}{\partial t}(x,0)=0, 0\le x\le2$
$u(0,t)=0$
$u(2,t)=0, t\ge0$

Login or register to post comments
Printer-friendly version

Comments

Answer

On October 7th, 2007 Isoscel says:

We may look for a nonzero solution using separable function method of Fourier.
Take $u(x,t)=X(x)T(t)$
Then equation becomes
$T"(t)X(x)=T(t)X"(x)$ or
$\frac{T"(t)}{T(t)}=\frac{X"(x)}{X(x)}$
As a function of variable t is equal to a function of variable x their common value has to be a constant.
$\frac{T"(t)}{T(t)}=\frac{X"(x)}{X(x)}=-c^2$
To determine the set of admissible set of constant we look for
$(*)X"(x)+c^2X(x)=0$
As
$u(0,t)=X(0)T(t)=0$
$u(2,t)=X(2)T(t)=0$
impose X(0)=X(2)=0 otherwise T(t)=0 so u(x,t) would be zero everywhere
Equation (*) has a general solution
$X(x)=a\cos (cx)+b\sin (cx)$
As $X(0)=0=a\cos 0+b\sin 0=a$ we have a=0.
Then $X(2)=b\sin (2c)=0$
If b=0 the solution would be zero so
$=\sin (2c)=0$ and $2c=n\pi$ So the set of constant is
$c_n=\frac{n\pi}{2}$
So we have a set of function
$X_n(x)=\sin \frac{n\pi x}{2}$
Second equation in the other variable t
$(**)T"(t)+c_{n}^2T(t)=0$
has also a solution
$T_n(t)=a_n\cos (c_nt)+b_n\sin (c_nt)=a_n\cos (\frac{n\pi}{2}t)+b_n\sin (\frac{n\pi}{2}t)$
Now we look for a solution of the form
$u(x,t)=\sum_{n=1}^{+\infty}X_n(x)T_n(t)=\sum_{n=1}^{+\infty}\sin \frac{n\pi x}{2}(a_n\cos (\frac{n\pi}{2}t)+b_n\sin (\frac{n\pi}{2}t))$
We look for constants $a_n,b_n$
From
$u(x,0)=\sum_{n=1}^{+\infty}X_n(x)T_n(0)=\sum_{n=1}^{+\infty}\sin \frac{\pi x}{2}(a_n\cos (\frac{n\pi}{2}0)+b_n\sin (\frac{n\pi}{2}0)=\sum_{n=1}^{+\infty}a_n \sin \frac{n\pi x}{2}=\sin \frac{\pi x}{2}$
As the coefficients are unique we observe that
$a_1=1$
$a_n=0\: n\geq 2$
From
$\frac{\partial u}{\partial t}(x,0)=\sum_{n=1}^{+\infty}\sin \frac{\pi x}{2}\frac{n\pi}{2}b_n\cos (\frac{n\pi}{2}0)=\sum_{n=1}^{+\infty}\frac{n\pi}{2}b_n\sin \frac{\pi x}{2}=0$
we get $b_n=0 \: for n\geq 1$
Finally solution is
$u(x,t)=\cos (\frac{\pi}{2}t)\sin (\frac{\pi}{2}x)$