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### Messages - Bruce Wu

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1
##### Final Exam / Re: FE-6
« on: December 19, 2015, 11:12:28 AM »
Vivian I think you forgot the factor of 2 when computing $A_n$ and so it is missing from your answer. It should be:
$$u(r,t)=\sum^{\infty}_{n=1}\frac{2(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

2
##### Final Exam / Re: FE-3
« on: December 18, 2015, 11:51:00 PM »
Great work everyone!

3
##### Final Exam / Re: FE-6
« on: December 18, 2015, 11:48:40 PM »
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.

4
##### Test 2 / Re: TT2-P3
« on: November 19, 2015, 02:02:19 AM »
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 $\lambda_n=(\frac{n\pi}{a})^{2}+(\frac{n\pi}{b})^{2}$?

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##### Test 2 / Re: TT2-P3
« on: November 19, 2015, 01:58:00 AM »
I did it a different way. But by Catch's method shouldn't there be the additional constraint that $\lambda=\lambda_1 +\lambda_2$?

6
##### Test 2 / Re: TT2-P2
« on: November 19, 2015, 01:42:47 AM »
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.

7
##### Test 2 / Re: TT2-P1
« on: November 19, 2015, 01:26:59 AM »
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 $$u(x,t)=\frac{\pi^2}{12}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(2nx)\cos(2nt)$$

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##### Test 2 / Re: TT2-P2
« on: November 19, 2015, 01:18:37 AM »
The final answer cannot be right if it does not satisfy the boundary conditions.

9
##### Test 2 / Re: TT2-P2
« on: November 19, 2015, 01:13:48 AM »
Xi Yue's solution makes no sense. By her answer $u(x,0)=0$, which clearly does not satisfy the boundary condition.

10
##### Test 2 / Re: TT2-P4
« on: November 19, 2015, 12:33:42 AM »
We can go one step further by noting that all $a_n$ for even $n$ are zero. Then $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^n\sin(n\theta)}{n(n+1)}$$

11
##### Test 2 / Re: TT2-P5
« on: November 18, 2015, 11:20:20 PM »
I agree with Emily. I got $$\frac{\cos(\pi k/2)}{\pi (1-k^2)}$$

12
##### Textbook errors / Re: Section 5.2 Theorem 3.c)
« on: November 18, 2015, 02:07:37 PM »
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.

13
##### Textbook errors / Section 5.2 Theorem 3.c)
« on: November 18, 2015, 02:06:20 PM »
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 $\hat{p}=-i\hbar\frac{d}{dx}$

14
##### Web Bonus = Oct / Re: Web bonus problem : Week 4 (#6)
« on: November 18, 2015, 12:08:49 PM »
Can you please clarify what you mean by $c_{1/2}$ in the line above equation (6)?

15
##### Textbook errors / Section 7.2 Equation 3
« on: November 15, 2015, 02:27:36 PM »
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

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