Partial differential equations in $ \infty $

Question

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) $=?

Answer 1

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

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

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

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

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.