Toronto Math Forum

Vivian I think you forgot the factor of 2 when computing $A_n$ and so it is missing from your answer. It should be:
$$u(r,t)=\sum^{\infty}_{n=1}\frac{2(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

Actually $$\lim_{r\rightarrow 0}\frac{\sinh(r)}{r}=1$$
So that does not blow up at the origin either. However the problem with that function (corresponding to positive eigenvalues) is that it can never be $0$ at $r\neq 0$, so the boundary conditions can never be satisfied.

I did it a different way. But by Catch's method shouldn't there be the additional constraint that $\lambda=\lambda_1 +\lambda_2$?

I am not sure if my solution agrees with Rong Wei's. I do not think there was a need to shift coordinates. Btw there was a hint saying to only consider eigenfunctions which are even with respect to x. I will proceed with that knowledge.

By separation of variables, we have $$\frac{X''}{X}=\frac{T''}{T}=-\lambda$$
One can check that there are only positive eigenvalues, so let $\lambda=\omega^2$. Solving the $X$ equation, and only keeping the even term, we have
$$X(x)=A\cos(\omega x)$$
The boundary conditions in $x$ imply that $$X'\left(\pm\frac{\pi}{2}\right)=\mp \omega A\sin\left(\omega\frac{\pi}{2}\right)=0\Rightarrow \omega\frac{\pi}{2}=n\pi\Rightarrow\omega=2n\Rightarrow\lambda=4n^2$$
So we have $$X_{n}(x)=A_n \cos(2nx)$$
Now solving the $T$ equation, we get $$T(t)=B\cos(2nt)+C\sin(2nt)$$
The $T'(0)=0$ boundary condition implies that $C=0$. So
$$T_n(t)=B_n \cos(2nt)$$
The general solution is, after absorbing some constants $$u(x,t)=\frac{1}{2}A_0+\sum_{n=1}^{\infty}A_n\cos(2nx)\cos(2nt)$$
$u(x,0)=x^2$ implies $$A_n=\frac{2}{\pi}\int_{-\pi/2}^{\pi/2}x^2\cos(2nx)dx=\frac{(-1)^n}{n^2}$$
This is not defined for $n=0$, so we have to calculate that term separately
$$A_0=\frac{2}{\pi}\int_{-\pi/2}^{\pi/2}x^2dx=\frac{\pi^2}{6}$$
Therefore the final solution is $$u(x,t)=\frac{\pi^2}{12}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(2nx)\cos(2nt)$$

Xi Yue's solution makes no sense. By her answer $u(x,0)=0$, which clearly does not satisfy the boundary condition.

I agree with Emily. I got $$\frac{\cos(\pi k/2)}{\pi (1-k^2)}$$

In the proof it says $$\int\left(e^{-ikx}\right)'f(x)dx=ik\hat{f}(k)$$
However, $\left(e^{-ikx}\right)'=-ike^{-ikx}$, so shouldn't it be $-ik\hat{f}(k)$? So is it the rule that is wrong or is it the proof that is wrong? Recall that the rule is $$g(x)=f'(x)\Rightarrow \hat{g}(k)=ik\hat{f}(k)$$
Should it be instead $$\hat{g}(k)=-ik\hat{f}(k)$$
Which makes sense since in quantum mechanics $\hat{p}=-i\hbar\frac{d}{dx}$

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter7/S7.2.html#mjx-eqn-eq-7.2.3

In the line above the equation it should be $\textbf{U}=w\nabla u$

Indeed