1

**APM346--Misc / Re: Thank You!**

« **on:**December 22, 2015, 06:56:18 PM »

Indeed, it was an excellent course. Enjoy the break everyone!

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

1

Indeed, it was an excellent course. Enjoy the break everyone!

2

Vivian I think you forgot the factor of 2 when computing $A_n$ and so it is missing from your answer. It should be:

$$u(r,t)=\sum^{\infty}_{n=1}\frac{2(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

$$u(r,t)=\sum^{\infty}_{n=1}\frac{2(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

4

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.

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.

5

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

Actually I think Catch did mention that, near the top right of the page.

You're right, but then in the end shouldn't the final eigenvalues be $\lambda_n=(\frac{n\pi}{a})^{2}+(\frac{n\pi}{b})^{2}$?

6

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

7

Hopefully if one actually evaluates the integral we will get the same thing. Even wolfram alpha pro couldn't handle it, I tried. However as it stands right now I agree with Emily's answer.

8

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 $$u(x,t)=\frac{\pi^2}{12}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(2nx)\cos(2nt)$$

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 $$u(x,t)=\frac{\pi^2}{12}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(2nx)\cos(2nt)$$

9

The final answer cannot be right if it does not satisfy the boundary conditions.

10

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

11

We can go one step further by noting that all $a_n$ for even $n$ are zero. Then $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^n\sin(n\theta)}{n(n+1)}$$

12

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

13

Actually, never mind. I found the error, it's in the proof. Integration by parts was used but the negative sign in front of the integral was neglected. The rule stands corrected.

14

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 $\hat{p}=-i\hbar\frac{d}{dx}$

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 $\hat{p}=-i\hbar\frac{d}{dx}$

15

Can you please clarify what you mean by $c_{1/2}$ in the line above equation (6)?