### Show Posts

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.

### Messages - Jingxuan Zhang

Pages: [1] 2 3 4
1
##### Final Exam / Re: FE-P6
« on: April 14, 2018, 07:14:10 AM »
Alternatively, professor, you can consider it a bonus, which is rewarding for those who has worked out one of the previous year's final where this situation happened in almost exactly the same manner.

Observe indeed $u$ must be spherical symmetric as is the boundary. Let $v=ru$, then \eqref{6-1}-\eqref{6-2}become, once identity \eqref{6-3} is known,
\begin{align} &v_{tt}-v_{rr}  =0, \label{6-1'}\\ &v|_{t=0}=0, &&v_t|_{t=0}= \left\{\begin{aligned} &\sin(r) &&r<\pi,\\ &0 &&r\ge \pi,\end{aligned}\right.\qquad \label{6-2'} \end{align}
which is easily solved with a combined use of even continuation and D'Alembert's:
v=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\sin r\sin t&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-4}
So then
u=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\frac{\sin r\sin t}{r}&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2r}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-5}

2
##### Final Exam / Re: FE-P3
« on: April 11, 2018, 07:23:56 PM »
I agree with George. Tristan: but \eqref{3-4}.

3
##### Final Exam / Re: FE-P7
« on: April 11, 2018, 07:21:38 PM »
To All,

I think you should have the minus sign.

4
##### Final Exam / Re: FE-P1
« on: April 11, 2018, 07:20:42 PM »
Tristan,

I think
$$(e^{-x} \psi(x))' = e^{x} + 2e^{x} + e^{x},$$
not minus.

5
##### Final Exam / Re: FE-P6
« on: April 11, 2018, 07:09:45 PM »
Andrew,

Heed your trig! Beside there should be cases.

6
##### Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6
« on: April 08, 2018, 01:00:33 PM »
Zhongnan,
I think c) is alright as the way it stands. After all \eqref{eq-11.1.12} is deduced from \eqref{eq-11.1.13} so that is also fine...

7
##### Term Test 2 / Re: TT2--P3N
« on: April 06, 2018, 04:54:41 PM »
Andrew,

Your square is misplaced at the place where it is pointed out.

8
##### Web Bonus Problems / Re: Week 13 -- BP2
« on: April 05, 2018, 04:51:30 PM »
Adam: no you cannot and don't need to assume $\varphi$ is even, but rather you would need to use $\varphi$ is smooth. The gap in your argument is smoothed by
$$(\varphi(-\varepsilon)-\varphi(\varepsilon))\ln\varepsilon\to 0 \text{ as } \varepsilon\downarrow0.$$

9
##### Term Test 2 / Re: tt2-Q2 problem
« on: April 05, 2018, 04:36:45 PM »
Same question. The TA says $\alpha<0$ or $\alpha =n\pi$ which I don't really quite understand.

10
##### Quiz-B / Re: Quiz-B P2
« on: April 02, 2018, 09:46:14 PM »
Observe for nice $\varphi$
$$(\sin(x) \varphi(x))'''=-\cos(x) \varphi(x)-3\sin(x)\varphi'(x)+3\cos(x)\varphi''(x)+\sin(x)\varphi'''(x).$$
Therefore
$$\langle\sin(x)\delta'''(x),\varphi(x)\rangle=\langle\delta'''(x),\sin(x) \varphi(x)\rangle=-\langle\delta(x),(\sin(x) \varphi(x))'''\rangle=\varphi(0)-3\varphi''(0).$$
So $$\sin(x)\delta'''(x)=\delta(x)-3\delta''(x).$$

11
##### Quiz-B / Re: Quiz-B P1
« on: April 02, 2018, 09:33:04 PM »
Define
$$L(r,u_r)=r\sqrt{1+u_r^2}.$$
Then E.-L. $$(L_{u_r})_r=L_u$$ gives
$$\label{1-1}\Bigl(\frac{ru_r}{\sqrt{1+u_r^2}}\Bigr)_r=0,$$
from which we derive
$$\frac{ru_r}{\sqrt{1+u_r^2}}=A\implies u_r^2(r^2-A^2)=A^2\implies u_r=\frac{A}{\sqrt{r^2-A^2}}\implies u=A\cosh^{-1}(r/A)+B.$$

12
##### Quiz-7 / Thursday's quiz
« on: March 29, 2018, 03:21:42 PM »
It was question 3.3 as of
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter8/S8.P.html

Since $g$ is already harmonic, it's harmonic extension in unit ball has the same formula! Moreover $g(kx)=k^3g(x)$ so this formula is in fact a sum of homogeneous harmonic polynomial consisting of one term! So
$$u(x,y,z)=xyz,x^2+y^2+z^2\leq 1$$ or $$\tilde{u}(\rho,\theta,\phi)=\rho^3\cos^2\phi\sin\phi\cos\theta\sin\theta,0\leq\rho\leq1,0\leq\theta\leq 2\pi, 0\leq\phi\leq\pi.$$

13
##### Quiz-7 / Re: Wednesday's quiz
« on: March 29, 2018, 03:14:47 PM »
Of course I wasn't aware of that. But $\Delta g =12\rho^2$ is really there. I should probably have displayed it.

14
##### Quiz-7 / Wednesday's quiz
« on: March 29, 2018, 09:20:21 AM »
I was not there but I heard from my friend that they were asked to find harmonic extension in $B_1(0)$ of $g(x,y,z)=x^4+y^4+z^4$ given on $C_1(0)$.

Solution $u$ is sought in the form
$$\label{1}u=g-P(x,y,z)(\rho^2-1),\,\rho=\sqrt{x^2+y^2+z^2}.$$
Where $P$ is a polynomial even and symmetric in $x,y,z$, as does $g$, and $\deg(P)=\deg(g)-2=2$ . Therefore $P=P(\rho)=a\rho^2+b$ for some constant $a,b$, and so \eqref{1} becomes
$$\label{2}u=g-(a\rho^4+(b-a)\rho^2-b).$$
Observe $$\Delta\rho^2=6,\,\Delta\rho^4=20\rho^2,\,\Delta g=12\rho^2.$$ Now set $\Delta u=0$ and \eqref{2} gives
$$\label{3}0=12\rho^2-(20a\rho^2+6(b-a)).$$
Equating both sides of \eqref{3} term by term we find $a=b=\frac{3}{5}$ and so \eqref{2} becomes
$$u=g-\frac{3}{5}(\rho^4-1)=\frac{2}{5}(x^4+y^4+z^4)-\frac{6}{5}(x^2y^2+x^2z^2+y^2z^2)+\frac{3}{5}.$$

15
##### Term Test 2 / Re: TT2--P1
« on: March 25, 2018, 09:56:55 AM »
Done. To the posterity: my mistake was on \eqref{error}.

Pages: [1] 2 3 4