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Messages - Jingxuan Zhang

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31
Term Test 2 / Re: TT2--P1N
« on: March 24, 2018, 08:46:48 AM »
The associated eigenproblem is
\begin{equation}\left\{\begin{split}&X''=\lambda X,\\&X'|_{x=0}=(X'-\alpha X)|_{x=0}=0.\end{split}\right.\label{1-4}\end{equation}
If $\alpha=0$ the we know the solution are integer $\cos$'s
\begin{equation}\label{error}\lambda_n=-n^2, X_n(x)=\cos nx, n=0,1,....\end{equation}
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.

32
Term Test 2 / Re: TT2--P1
« on: March 24, 2018, 08:33:49 AM »
The associated eigenproblem is
\begin{equation}\left\{\begin{split}&X''=\lambda X,\\&X|_{x=0}=(X'-\alpha X)|_{x=0}=0.\end{split}\right.\label{1-4}\end{equation}
If $\alpha=0$ the we know the solution are half-integer $\sin$'s
\begin{equation}\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,....\end{equation}
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.

33
Term Test 2 / Re: TT2--P2
« on: March 24, 2018, 08:16:29 AM »
PFT$x\mapsto \omega$ \eqref{2-1} becomes
\begin{equation}\label{2-4}\hat{u}_{yy}-\omega^2\hat{u}=0.\end{equation}
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
\begin{equation}\label{2-5}\hat{u}(\omega, y)=\hat{g}(\omega)\frac{e^{-|\omega|y}}{\alpha-|\omega|}.\end{equation}
Regularity of \eqref{2-5} is guaranteed whenever $\alpha\neq 0$. Now taking IFT we have
$$u(x,y)=\int_{-\infty}^\infty \frac{\sin\omega}{\pi\omega} \frac{e^{-|\omega|y+i\omega x}}{\alpha-|\omega|}\,d\omega.$$

34
Term Test 2 / Re: TT2--P5
« on: March 23, 2018, 08:00:54 PM »
I suggest somewhat different numerical factor. Not much different.

35
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,
\begin{equation}\label{4-4}
u=\frac{1}{2} A_0+
\sum_n r^n\Bigl( A_n\cos n\theta \Bigr).\end{equation}
Plugging in \eqref{4-3} and using even continuation
\begin{equation}\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}\end{equation}
Combining \eqref{4-4}, \eqref{4-5}:
$$u=\frac{4}{\pi}\sum_k \Bigl(\frac{r}{9}\Bigr)^{2k+1} \frac{\cos(2k+1)\theta}{(2k+1)^2}.$$


36
APM346--Lectures / Re: Question 1 from TT2, 2015S
« on: March 21, 2018, 09:55:01 PM »
Tristan this is the trick used in 244 to solve Euler's equation:
$$\partial_t=x\partial_x;\,\partial^2_t=\partial_x+x^2\partial^2_x,\, x=e^t.$$

37
APM346––Home Assignments / Re: An ODE
« on: March 21, 2018, 08:28:35 AM »
But in the hint you say the solution should be in the form of trig polyn? and the ODE should have one variable? how am I supposed to interpret this $(x\pm iy)^{|m|}$?

38
APM346––Home Assignments / Re: An ODE
« on: March 20, 2018, 01:12:50 PM »
But what then is meant by $(x+iy)^m$? are these the solutions? apparently it doesn't seem to be.

39
APM346––Home Assignments / An ODE
« on: March 20, 2018, 08:24:05 AM »
I am referring to Q2.3 on
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter8/S8.P.html

So how do I actually solve
$$\sin^2\phi \Phi''+\sin\phi\cos\phi \Phi' -(l(l+1)\sin^2\phi-m^2)\Phi=0$$
which, suppose it's correctly derived, has cost me an entire afternoon? I remember I certain remark
in lecture that there should be constrain on $m$ and some thing like $(x+iy)^m$, but that part of my note
is very much blurred.

I heuristically plugged in $\sin,\cos,\sin^2,\cos^2,\sin\cos$ but there does not seem to be a good cancellation.


40
APM346--Lectures / excersice 1, chap 7.3
« on: March 17, 2018, 05:34:56 PM »
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter7/S7.3.html#sect-7.3.1

Exercise 1 appears quite strange. if $f=0$ on $\Omega=\{\|x\|\geq R\}$ and $u(y)=\int_\Omega G(x,y)f(x)\,dx$, then shouldn't this integral vanish instead of giving that curious form?

41
Quiz-6 / Re: Quiz 6 T5102
« on: March 16, 2018, 11:43:33 AM »
The general bounded solution of the DE is
\begin{equation}u=\frac{a_0}{2}+\sum_n r^{-n}(a_n \cos n\theta + b_0\sin n\theta)\label{a}\end{equation}
Now $f$ is odd so $a_n\equiv0$ and
\begin{equation}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}\end{equation}

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

42
APM346--Lectures / technical typos
« on: March 15, 2018, 04:40:02 PM »
1. in chap 7.3 between eq $(12),(13)$ the codes are there.
2. in chap 8.1 between eq $(15),(16)$ methinks the last term in the enumerator of the first selected fraction should have a factor $\rho^2$.

43
APM346––Home Assignments / Re: To drag
« on: March 13, 2018, 10:08:27 PM »
But I think it's quite apparent that $G(x,y)$ depends on the variable of integration, both $dV$ and $dS$?

44
APM346––Home Assignments / To drag
« on: March 13, 2018, 04:39:00 PM »
I am referring to the mean value theorem proof in
http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter7/S7.2.html

So twice we dragged the kernel $G$ out of the integral. How can we actually do that? if $\Delta u \gtrless$ can we still drag? can we in general drag?

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