I'm reading Farlow's Partial Differential Equations for Scientists and Engineers and am trying to solve the following problem:
The answer is:
I don't understand how he comes up to this answer. We have the heat equation, so I guess $u_{tt}(x,0)=-\pi ^2 \sin (\pi x)$ and then $u_t (x,0)=\alpha^2(- \pi ^2 \sin (\pi x))$, I think this means that $u(x,t)$ is decreasing as $t\to \infty$ but what confuses me is that we only have that information for a small neighborhood $(x,0)$. How do we know that $u(x,t)\to 0$ as $t\to \infty$? Even the function he gives as an example doesn't seems to approach zero as $t\to \infty$.
EDIT: I think there is a typo in the answer, it should be $e^{-\alpha^2 t} \sin (\pi x)$.
In this specific case, $u_{xx}(x,0)=-\pi^2 \sin(\pi x)$. So now after a little time evolution, you have $u(x,0)$ plus a multiple of $\sin(\pi x)$ which is again a multiple of $\sin(\pi x)$. Therefore $u_{xx}(x,t)=-\pi^2 u(x,t)$ for all $t$. Then the PDE tells you $u_t(x,t)=-\alpha^2 \pi^2 u(x,t)$, which means that for each fixed $x$ you have exponential decay.
That "after a little time evolution" phrase is not rigorous, but it can be sorted out rigorously by substituting $u(x,t)=f(t) \sin(\pi x)$ into the PDE and then solving the resulting ODE for $f$.
I am not really even sure what the goal was with that second hint.