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Messages - Jingxuan Zhang

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APM346--Misc / perturbed NLS
« on: February 19, 2019, 06:57:24 AM »
Consider the perturbed NLS
\begin{equation}\partial_t q=-\partial_x^2 q-2|q|^2q+\epsilon R\qquad R=R(q,-q^*).\label{1}\end{equation}
It is suggested that we consider also the conjugated equation at the same time
\begin{equation}\partial_t r=\partial_x^2 r+2|r|^2r-\epsilon R^*\qquad r=-q^*.\label{2}\end{equation}
It seems to me that whenever $q$ solves \eqref{1}, $r$ solves \eqref{2}.Then how does \eqref{2} help? What puzzles me more is that when we do the 1st order perturbation theory and collect coefficients according to $\epsilon^1$, it is suggested to write a coupled system of PDE involving $q,r$. Why is this better than just considering one equation, say \eqref{1} alone?

Afterwards we also consider the spectrum of that coupled linear operator, but is that why?

MAT334--Lectures & Home Assignments / Bounding gamma function
« on: January 27, 2019, 08:50:05 PM »
I have almost forgotten how to due with
\int_{|z|=r,-\pi/2\le\arg z\le 0\} e^-zz^{\alpha-1}\,dz,\quad r\to0+
Any hint?

APM346--Misc / analytic extension
« on: January 26, 2019, 10:29:57 AM »
Suppose I have a absolutely convergent series solution of an ODE on the real line and I want to extend this to the whole plane. Naïvely I will just write the same formula, replaced with a complex variable. But to what extent is this justifiable?

MAT334--Lectures & Home Assignments / Re: What can I say about f'(0)?
« on: November 21, 2018, 08:17:44 PM »
O.k., but is there anything I can say about $f'(0)$?

If there is really nothing to say, then please consider the following situation: $f:\mathbb{R}\to\mathbb{C}$ is continuous, and $\lim_{t\to 0} t^{-1}(f(t)+f(t)^{-1}-2)$ exists. What can say about this limit? (In particular I would love it to be 0).

MAT334--Lectures & Home Assignments / Re: 2.3 Q7
« on: November 21, 2018, 07:51:46 PM »
Ye: $\pm$ is \pm.

MAT334--Lectures & Home Assignments / What can I say about f'(0)?
« on: November 21, 2018, 04:45:22 PM »
Suppose $f:D\to\mathbb{C}$ is analytic near 0, such that $\|x\|=1\implies |f(x)|=1$. Does it follow that $f'(0)$ is purely imaginary?

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:
\begin{equation}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}\end{equation}
So then
\begin{equation}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}\end{equation}

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

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

I think you should have the minus sign.

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

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

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

Heed your trig! Beside there should be cases.

APM346––Home Assignments / Re: Question 9 from 4.2P
« on: April 09, 2018, 08:34:28 AM »

Two things: first, shouldn't it rather be $C\cos c\sqrt{\lambda}t + D\sin  c\sqrt{\lambda}t$? second, how do you thereof derive the $\tan$ equation?

Edit: never mind. I think I understand you now. I think your problem is that your supposed form of $T$ and the condition for $T$ doesn't match. Indeed if $\frac{T'}{T}$ is some constant then $T$ should be rather of exponential form.

From your tangent equatoin after some calculation I found, if $u\neq\text{const}$,
$$T=-\frac{\tan\omega l}{c\alpha}e^{ic\omega t},\frac{\omega l}{\pi}\notin\mathbb{Z}.$$

EDIT: Above are wrong. But not very wrong: it is right to consider $T$ as complex exponential.

Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6
« on: April 08, 2018, 01:00:33 PM »
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...

APM346––Home Assignments / Re: Clever Guesses with Heat Equation
« on: April 07, 2018, 08:37:24 PM »
So what if IC is not as in that question? $x^2e^{-x^3}$, say? Can we still use this or similar result?

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

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

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