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### Messages - Jingxuan Zhang

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16
##### 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}

17
##### 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?

18
##### 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 $\lambda_n$?

19
##### 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.

20
##### 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.

21
##### 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
$$u(x,y)=\int_{-\infty}^\infty \frac{\sin\omega}{\pi\omega} \frac{e^{-|\omega|y+i\omega x}}{\alpha-|\omega|}\,d\omega.$$

22
##### Term Test 2 / Re: TT2--P5
« on: March 23, 2018, 08:00:54 PM »
I suggest somewhat different numerical factor. Not much different.

23
##### 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}:
$$u=\frac{4}{\pi}\sum_k \Bigl(\frac{r}{9}\Bigr)^{2k+1} \frac{\cos(2k+1)\theta}{(2k+1)^2}.$$

24
##### 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
$$\frac{4}{\pi}\sum_k (\frac{a}{r})^{2k+1} \frac{\sin(2k+1)\theta}{2k+1}\label{c}$$

25
##### 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.

26
##### 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:
$$F(\Re f)\neq\Re F(f)\text{ and } F(\Im f )\neq\Im F(f)$$

27
##### Quiz-5 / Re: Quiz 5, T5102
« on: March 08, 2018, 07:23:50 AM »
George:
The reason that the absolute value is missing is that the contour you choose must depend on the sign of $k$ If $k<0$ then your contour should instead be upper semicircle and if $k>0$ then you are right. This is done to satisfy the hypothesis of Jordan's lemma, which you implicitly used to control the integral over the arc as it increasese.

Also what property do you refer to? I think $\hat{xf}=i\hat{f}'$?

Beside: don't we always use $\frac{1}{2\pi}$ for the scaling? and this for one thing at least better suit Residual Theorem.
\hat{f}(\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{(x-ai)(x+ai)}=\left\{\begin{align*}\left.i\frac{e^{-i\omega x}}{x+ai}\right|_{x=ai}&&\omega<0\\ \left.-i\frac{e^{-i\omega x}}{x-ai}\right|_{x=-ai}&&\omega>0\end{align*}\right.=\frac{e^{-|\omega|a}}{2a}
And so immediately
$$\hat{xf}(\omega)=i\hat{f}'(\omega)=-i \text{sgn}(\omega)\frac{e^{-|\omega|a}}{2}$$

28
##### Quiz-4 / Quiz 4 -- both sections
« on: March 01, 2018, 10:07:07 PM »
Note: Since problems for both sections are very similar I suggest to discuss them together, so certain parts of the solutions could be used without repetition
(V.I.)

The only difference that in Wed section condition on the right end are $u_{xx}|_{x=l}=u_{xxx}|_{x=l}=0$,

This is problem 3 part 1,2 from http://www.math.toronto.edu/courses/apm346h1/20181/PDE-textbook/Chapter4/S4.2.P.html

The associated eigenvalue problem is
\begin{align}X^{iv}-\omega^4 X=0\label{1}\\X(0)=X'(0)=0\label{2}\\X(l)=X'(l)=0
\label{3}\end{align}
From (\ref{1}) we write
$$X=A\cosh (\omega x) + B\sinh(\omega x)+C\cos(\omega x)+D\sin(\omega x)\label{4}$$
whence (\ref{2}) implies, if $\omega\neq 0$,
$$A+C=B+D=0\label{5}$$
and so (\ref{4}) becomes
$$X=A(\cosh (\omega x) -\cos(\omega x)) + B(\sinh(\omega x)-\sin(\omega x))\label{6}$$

Now the algebraic system in variable of $A,B$ obtained from (\ref{3}) has nontrivial solution if and only if the coefficient matrix is singular, that is:
$$\left|\begin{array}{cc}\cosh (\omega l) -\cos(\omega l)&\sinh(\omega l)-\sin(\omega l)\\ \sinh (\omega l) +\sin(\omega l)&\cosh(\omega l)-\cos(\omega l)\end{array}\right|=2-2\cosh (\omega l)\cos(\omega l)=0\iff\cosh (\omega l)\cos(\omega l)=1\label{equation}.$$
The null space of this system is
$$(A,B)'=t(-\sinh(\omega l)+\sin(\omega l),\cosh (\omega l) -\cos(\omega l))',t\in\mathbb{R}\label{8}$$
and so (\ref{6}) becomes
$$X=(-\sinh(\omega l)+\sin(\omega l))(\cosh (\omega x) -\cos(\omega x))+(\cosh (\omega l) -\cos(\omega l))(\sinh(\omega x)-\sin(\omega x))\label{9}$$
up to a scalar multiple. This is the eigenspace.

The graphs are those of $1/\cosh$ and $\cos$, imagined to be in variable of $\omega l$. Their (infinitely many) intersections suffice (\ref{equation}).

29
##### Web Bonus Problems / Re: Web Bonus problem -- Reading Week
« on: February 17, 2018, 03:26:11 PM »
Part (b): normalize the integrand we have
$$I=\hat{u}(0)=\int e^{-\frac{1}{2}a x^2}\,dx=\sqrt{\frac{2\pi}{a}}\cdot\underbrace{\frac{1}{\sqrt{2\pi}}\int e^{-z^2/2}\,dz}_{1}=\sqrt{\frac{2\pi}{a}}$$
But here really $\sqrt{a}$ is ill-defined: $a\in\mathbb{C}$ might result in this root having multiple value. The sign therefore depends on the branch chosen for $a$. In particular the principal value of $I$ will have a positive sign if $\Im(a)\geq0$ and a negative sign otherwise.

EDIT: I consulted Ahlfors' text on complex variable, which I have completely forgotten now. Write $a=\alpha+i\beta$ then
$$\sqrt{\frac{2\pi}{\alpha}}=\pm\frac{\sqrt{2\pi}}{\alpha^2+\beta^2}\left(\sqrt{\frac{\alpha+\sqrt{\alpha^2+\beta^2}}{2}}-i\frac{\beta}{|\beta|}\sqrt{\frac{-\alpha+\sqrt{\alpha^2+\beta^2}}{2}}\right)$$

If $\alpha>0,\beta=0$ then of course this is just the simple formula on LS.

30
##### Web Bonus Problems / Re: Web Bonus problem -- Reading Week
« on: February 17, 2018, 03:10:50 PM »
I find the second part perplexing: what are we to compare $I$ with? and shouldn't $I=\hat{u}(0)$ be immediate once the formula for $\hat{u}$ is determined by Adam?

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