### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Topics - Jingxuan Zhang

Pages: [1]
1
##### 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 $$\tilde{u}(\rho,\theta,\phi)=\rho^3\cos^2\phi\sin\phi\cos\theta\sin\theta,0\leq\rho\leq1,0\leq\theta\leq 2\pi, 0\leq\phi\leq\pi.$$

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

3
##### 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}).

4
##### Technical Questions / Tag each equation in an IVP
« on: February 10, 2018, 04:14:10 PM »
I am trying to typeset in latex a system such as
$$\left\{\begin{array}{cc}f(x)&=k_1\\g(x)&=k_2\end{array}\right..$$
Is there anyway that allows me to label each individual equation such as with \tag? True, with align I can label but i need also the brace. True, with array I can brace but then cannot label. Thanks.

5
##### Quiz-1 / Q1-T0101-P1,2
« on: January 25, 2018, 01:10:48 PM »
1. General solution of
$$u_{xy}=e^{x+y}\implies u_{x}=e^{x+y}+\varphi_{x}(x)\implies e^{x+y}+\varphi(x)+\psi(y)$$
2. General solution of
$$u_{t}+(x^2+1) u_{x}=0 \implies C=\arctan(x)-t \implies u=\varphi(\arctan (x)-t)$$

Pages: [1]