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Messages - Jingxuan Zhang

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16
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 $$\tilde{u}(\rho,\theta,\phi)=\rho^3\cos^2\phi\sin\phi\cos\theta\sin\theta,0\leq\rho\leq1,0\leq\theta\leq 2\pi, 0\leq\phi\leq\pi.$$

17
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.

18
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
$$u=g-\frac{3}{5}(\rho^4-1)=\frac{2}{5}(x^4+y^4+z^4)-\frac{6}{5}(x^2y^2+x^2z^2+y^2z^2)+\frac{3}{5}.$$

19
Term Test 2 / Re: TT2--P1
« on: March 25, 2018, 09:56:55 AM »
Done. To the posterity: my mistake was on \eqref{error}.

20
Term Test 2 / Re: TT2--P5
« on: March 25, 2018, 07:46:58 AM »
\begin{split}
\hat{f}(\omega)&=\frac{1}{2\pi}\int_{-\infty}^\infty (\frac{1}{4}e^{2ix}+\frac{1}{4}e^{-2ix}+\frac{1}{2})e^{-|x|}e^{-i\omega x}\,dx\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{4}e^{-|x|}e^{-i(\omega-2) x}\,dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{4}e^{-|x|}e^{-i(\omega+2) x}\,dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{2}e^{-|x|}e^{-i\omega x}\,dx\\
&=\frac{1}{4\pi(1+(x-2)^2)}+\frac{1}{4\pi(1+(x+2)^2)}+\frac{1}{2\pi(1+x^2)}
\end{split}

21
Term Test 2 / Re: TT2--P1N
« on: March 25, 2018, 07:36:09 AM »
Again I don't quite see the issue of sign, please inform me the mistake.
Found it. To the posterity: my mistake was on \eqref{error}.

By the way, how can I draw with tikz a picture like this? or did you use latex at all?

22
Term Test 2 / Re: TT2--P1
« on: March 25, 2018, 07:33:02 AM »
That subscript is really awkward but I don't see sign problem at $\lambda_n$?

23
Term Test 2 / Re: TT2--P1N
« on: March 24, 2018, 08:46:48 AM »
The associated eigenproblem is
$$\left\{\begin{split}&X''=\lambda X,\\&X'|_{x=0}=(X'-\alpha X)|_{x=0}=0.\end{split}\right.\label{1-4}$$
If $\alpha=0$ the we know the solution are integer $\cos$'s
$$\label{error}\lambda_n=-n^2, X_n(x)=\cos nx, n=0,1,....$$
If $\alpha\neq 0,\lambda>0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cosh \gamma x + B\sinh \gamma x, \gamma>0.$$
Plugging in boundary condition we find $B=0$ and
$$\gamma A\sinh \gamma\pi+\alpha A\cosh \gamma\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=\gamma_n^2$ where $\gamma_n$ is a nonzero root of
$$\gamma\tanh \gamma\pi+\alpha=0.$$
If $\alpha\neq 0,\lambda<0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cos \omega x + B\sin \omega x, \omega>0.$$
Plugging in boundary condition we find $B=0$ and
$$-\omega A\sin\omega\pi+\alpha A\cos \omega\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=-\omega_n^2$ where $\omega_n$ is a nonzero root of
$$\omega\tan \omega\pi-\alpha=0.$$
If $\lambda=0$ then we have only trivial solution.

24
Term Test 2 / Re: TT2--P1
« on: March 24, 2018, 08:33:49 AM »
The associated eigenproblem is
$$\left\{\begin{split}&X''=\lambda X,\\&X|_{x=0}=(X'-\alpha X)|_{x=0}=0.\end{split}\right.\label{1-4}$$
If $\alpha=0$ the we know the solution are half-integer $\sin$'s
$$\label{error}\lambda_n=-\Bigl(n+\frac{1}{2}\Bigr)^2, X_n(x)=\sin \Bigl(n+\frac{1}{2}\Bigr)x,n=0,1,....$$
If $\alpha\neq 0,\lambda>0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cosh \gamma x + B\sinh \gamma x, \gamma>0.$$
Plugging in boundary condition we find $A=0$ and
$$\gamma B\cosh \gamma\pi+\alpha B\sinh \gamma\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=\gamma_n^2$ where $\gamma_n$ is a nonzero root of
$$\gamma=-\alpha\tanh \gamma\pi.$$
If $\alpha\neq 0,\lambda<0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cos \omega x + B\sin \omega x, \omega>0.$$
Plugging in boundary condition we find $A=0$ and
$$\omega B\cos \omega\pi+\alpha B\sin \omega\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=-\omega_n^2$ where $\omega_n$ is a nonzero root of
$$\omega=-\alpha\tan \omega\pi.$$
If $\lambda=0$ then we have only trivial solution.

25
Term Test 2 / Re: TT2--P2
« on: March 24, 2018, 08:16:29 AM »
PFT$x\mapsto \omega$ \eqref{2-1} becomes
$$\label{2-4}\hat{u}_{yy}-\omega^2\hat{u}=0.$$
Due to \eqref{2-3} general solution for \eqref{2-4} is
$$\hat{u}(y)=Ae^{-|\omega|y}.$$
Plugging in \eqref{2-2}
$$\hat{g}(\omega)=-A(\omega)|\omega|+A(\omega)\alpha\implies A(\omega)=\frac{\hat{g}(\omega)}{\alpha-|\omega|}$$
and so
$$\label{2-5}\hat{u}(\omega, y)=\hat{g}(\omega)\frac{e^{-|\omega|y}}{\alpha-|\omega|}.$$
Regularity of \eqref{2-5} is guaranteed whenever $\alpha\neq 0$. Now taking IFT we have
$$u(x,y)=\int_{-\infty}^\infty \frac{\sin\omega}{\pi\omega} \frac{e^{-|\omega|y+i\omega x}}{\alpha-|\omega|}\,d\omega.$$

26
Term Test 2 / Re: TT2--P5
« on: March 23, 2018, 08:00:54 PM »
I suggest somewhat different numerical factor. Not much different.

27
Term Test 2 / Re: TT2--P4N
« on: March 23, 2018, 07:57:27 AM »
General solution for \eqref{4-1} is, due to boundary condition and consideration of regularity at zero,
\label{4-4}
u=\frac{1}{2} A_0+
\sum_n r^n\Bigl( A_n\cos n\theta \Bigr).
Plugging in \eqref{4-3} and using even continuation
\begin{split}
9^n A_n&=\frac{2}{\pi}\int_0^\pi (\pi-\theta)\cos n\theta\,d\theta
=\frac{2}{\pi}\int_0^\pi\theta'\cos n(\theta'-\pi)\,d\theta'\\
&=\frac{2}{n\pi}\int_0^\pi\sin n(\theta'-\pi)\,d\theta'\\
&=\left\{\begin{split}&0&& \text{ n even},\\ &\frac{4}{\pi n^2}&& \text{ n odd.}\end{split}\right.
\end{split}\label{4-5}
Combining \eqref{4-4}, \eqref{4-5}:
$$u=\frac{4}{\pi}\sum_k \Bigl(\frac{r}{9}\Bigr)^{2k+1} \frac{\cos(2k+1)\theta}{(2k+1)^2}.$$

28
Quiz-6 / Re: Quiz 6 T5102
« on: March 16, 2018, 11:43:33 AM »
The general bounded solution of the DE is
$$u=\frac{a_0}{2}+\sum_n r^{-n}(a_n \cos n\theta + b_0\sin n\theta)\label{a}$$
Now $f$ is odd so $a_n\equiv0$ and
$$b_n=\frac{2a^n}{\pi}\int_0^\pi \sin n\theta\,d\theta=\left\{\begin{array} &\frac{4a^n}{n\pi}&\text{n odd}\\0&\text{n even}\end{array}\right.\label{b}$$

Combining $(1),(2)$ we have the final solutoin
$$\frac{4}{\pi}\sum_k (\frac{a}{r})^{2k+1} \frac{\sin(2k+1)\theta}{2k+1}\label{c}$$

29
Term Test 1 / Re: P1 Night
« on: March 11, 2018, 02:10:44 PM »
A) Clearly
$$C=xe^{-t^2/2}$$
B)
$$du=e^{t^2/2}dx=C/xdx\implies u=x\ln x e^{-t^2/2}+\varphi(xe^{-t^2/2})$$
C)
$$x=x\ln x + \varphi(x)\implies\varphi(x)=x - x\ln x\implies u = xe^{-t^2/2}+\frac{xt^2}{2}e^{-t^2/2}$$

I am fortunate enough that I didn't write night exam...this I spent more than half an hour.

30
Quiz-5 / Re: Quiz5 tut 0201
« on: March 08, 2018, 04:40:53 PM »
I just want to remind everyone who sees this so that he won't make the same mistake as I did:
$$F(\Re f)\neq\Re F(f)\text{ and } F(\Im f )\neq\Im F(f)$$

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