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Suppose that we have $$\begin{cases} u_{tt}=u_{xx} \text{ for }\ 0 < x < \pi ,\ t > 0\\ u(x,0)=8\sin x \\ u_{t}(x,0)=0\\ u(0,t)= u(\pi,t)=0\end{cases}$$

find

$ \lim_{t \rightarrow \infty }u( \frac{\pi}{4},t) $=?

behnam
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  • I think this would be a reasonable question if you gave the separation of variables solution and then asked for the limit of that expression. As it stands you have not supplied enough of your own thoughts. – Ian Apr 18 '15 at 16:31
  • @Ian : There could be a way to calculate this limit without knowing the exact solution (Even if I'm not sure it's possible here) – Tryss Apr 18 '15 at 16:59

1 Answers1

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$$u(x,t)=X(x)T(t)$$

$$u_{tt}(x,t)=X(x)T''(t)$$

$$u_{xx}(x,t)=X''(x)T(t)$$

$$u_{tt}=u_{xx} \Rightarrow X(x)T''(t)=X''(x)T(t)$$

We are looking for a non-trivial solution, so we assume that $X(x),T(t) \neq 0$ and thus we can divide the above relation by the product $X(x)T(t)$.

So we have:

$$\frac{T''(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$

Now use the initial and boundary conditions and solve the two problems that you will get.

evinda
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