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### Messages - Jingxuan Zhang

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16
##### 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
$$(\varphi(-\varepsilon)-\varphi(\varepsilon))\ln\varepsilon\to 0 \text{ as } \varepsilon\downarrow0.$$

17
##### 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.

18
##### 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 $$\sin(x)\delta'''(x)=\delta(x)-3\delta''(x).$$

19
##### 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
$$\frac{ru_r}{\sqrt{1+u_r^2}}=A\implies u_r^2(r^2-A^2)=A^2\implies u_r=\frac{A}{\sqrt{r^2-A^2}}\implies u=A\cosh^{-1}(r/A)+B.$$

20
##### APM346––Home Assignments / problem in problem
« on: March 30, 2018, 06:02:18 PM »
I assume problem 2.1 of http://www.math.toronto.edu/ivrii/PDE-textbook/Chapter10/S10.P.html intends to ask "Minimizing T" subject to no constrain?

21
##### APM346--Lectures / Re: consequence of EL
« on: March 30, 2018, 02:19:45 PM »
Do you mean

\frac{d\ }{dt} L_{q'_j}(q,q',t)- L_{q_j}=0
\tag{A1}

and

\frac{d\ }{dt} \Bigl(\sum_j  q'_j L_{q'_j} -L\Bigr)=0?
\label{C1}

and do you mean $q=(q_j)?$

22
##### APM346--Lectures / consequence of EL
« on: March 30, 2018, 09:37:25 AM »
From (what I consider to be Euler-Lagrange)
$$\label{1}L_u=(L_{u'})_t$$
how can I derive
$$\label{2}L=u'L_{u'}+C?$$

Or is \eqref{2} even right? are they derived independently? if I integrate both sides of \eqref{1}, what exactly will be on the left?

23
##### 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 $$\tilde{u}(\rho,\theta,\phi)=\rho^3\cos^2\phi\sin\phi\cos\theta\sin\theta,0\leq\rho\leq1,0\leq\theta\leq 2\pi, 0\leq\phi\leq\pi.$$

24
##### 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.

25
##### 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
$$u=g-\frac{3}{5}(\rho^4-1)=\frac{2}{5}(x^4+y^4+z^4)-\frac{6}{5}(x^2y^2+x^2z^2+y^2z^2)+\frac{3}{5}.$$

26
##### APM346––Home Assignments / Re: C9 question 1 typo?
« on: March 29, 2018, 08:53:22 AM »
yes.

27
##### Term Test 2 / Re: TT2--P1
« on: March 25, 2018, 09:56:55 AM »
Done. To the posterity: my mistake was on \eqref{error}.

28
##### 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}

29
##### 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?

30
##### 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 $\lambda_n$?

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