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### Messages - Yiqi Shi

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##### Final Exam / Re: FE-1
« on: December 18, 2015, 10:37:02 PM »

u_{tt}-\frac{1}{2}u_{xx}=0

u(x,t)=\phi(x+\frac{1}{2}t)+\psi(x-\frac{1}{2}t)

by(2) and (3)

\phi(x)+\psi(x)=f(x)=2xe^{-2x^2}\\
\phi(x)-\psi(x)=g(x)=0

Then

\phi(x)=xe^{-2x^2}\\
\psi(x)=xe^{-2x^2}

Thus for $x>\frac{1}{2}t$

\mu(x,t)=(x+\frac{1}{2}t)e^{-2(x+\frac{1}{2}t)^2}+(x-\frac{1}{2}t)e^{-2(x-\frac{1}{2}t)^2}

Now consider  $x<\frac{1}{2}t$
by (4)

\phi(\frac{1}{2}t)+\psi(-\frac{1}{2}t)=te^{-t^2/2}

let $x=-\frac{1}{2}t$, then $t=-2x$

\phi(-x)+\psi(x)=-2xe^{-(-2x)^2/2}\\
\psi(x)=-2xe^{-2x^2}-\phi(-x)\\
\psi(x)=-2xe^{-2x^2}+xe^{-2x^2}\\
\psi(x)=-xe^{-2x^2}

Thus for $x<\frac{1}{2}t$

\mu(x,t)=(x+\frac{1}{2}t)e^{-2(x+\frac{1}{2}t)^2}-(x-\frac{1}{2}t)e^{-2(x-\frac{1}{2}t)^2}

2
##### Final Exam / Re: FE-7
« on: December 18, 2015, 10:27:19 PM »
Since g is a polynomial of degree 3, the The degree of $P$ should be 1. Also, notice that $g$ is odd about $z$ and $P$ has the same rotational symmetry as $g$, so we can assume $P=kz$.

$$\Delta \mu=2z+2z-2kz-2kz-6kz=0 \implies k=\frac{2}{5}$$\\
Thus,

\mu=z(x^2+y^2)-\frac{2}{5}z(x^2+y^2+z^2-1)

3
##### Test 1 / Re: TT1-P4
« on: October 21, 2015, 10:18:33 PM »

\frac{dE(t)}{dt}= \frac{1}{2}\int_0^L \bigl( 2u_tu_{tt}+ 2u_{x}c^2u_{xt} + \omega^22 uu_t)\,dx +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)

do integral by part on $\int_0^L \bigl(2u_{x}c^2u_{xt})\,dx$

\int_0^L \bigl(2u_{x}c^2u_{xt})\,dx=2c^2[(u_{x}u_{t})\Big|_{0}^L - \int _0^L u_{xx}u_{t}\,dx)]

plugging these into $\frac{dE(t)}{dt}$

\frac{dE(t)}{dt}=c^2\int_0^L \bigl( u_tu_{tt}- u_{xx}u_{t} + \omega^2 uu_t)\,dx +c^2(u_{x}u_{t})\Big|_{0}^L  +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)

By condition $u_{tt}-c^2u_{xx} + \omega^2 u =0$

c^2\int_0^L \bigl( u_tu_{tt}- u_{xx}u_{t} + \omega^2 uu_t)\,dx=0

\frac{dE(t)}{dt}=c^2(u_{x}u_{t})\Big|_{0}^L  +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)

by condition $(u_x -\alpha u)|_{x=0}=0$ and $(u_x +\beta u)|_{x=L}=0$, we have

\begin{split}
&\alpha u(0,t)=u_{x}(0,t)\\
&\beta u(L,t)=-u_{x}(L,t)
\end{split}

plugging (6) into  $\frac{dE(t)}{dt}$, we have:

\begin{split}
\frac{dE(t)}{dt}&=c^2[(u_{x}u_{t})\Big|_{0}^L+u_{x}(0,t)u_t(0,t)-u_{x}(L,t)u_t(L,t)]\\
&=c^2[u_{x}(L,t)u_t(L,t)-u_{x}(0,t)u_t(0,t)+u_{x}(0,t)u_t(0,t)-u_{x}(L,t)u_t(L,t)]\\
&=c^2[0+0]\\
&=0
\end{split}

Since $\frac{dE(t)}{dt}$=0, then E(t) does not depend on t.

4
##### Quiz 1 / Quiz 1-P2
« on: October 01, 2015, 11:44:08 PM »
Find the general solutions to

u_{xxyy}=\sin(x)\sin(y);
\label{eq1}

Solution:

u_{xxy}=-\sin(x)\cos(y)+\phi(x);
\label{eq2}

u_{xx}=-\sin(x)\sin(y)+y\phi(x)+\psi(x);
\label{eq3}

u_{x}=\cos(x)\sin(y)+y\int \phi(x)\,dx+\int\psi(x)\,dx+h(y)=\cos(x)\sin(y)+yf(x)+g(x)+h(y);
\label{eq4}

u=\sin(x)\sin(y)+y\int f(x)\,dx+\int g(x)\,dx+xh(y)+\eta(y)=\sin(x)\sin(y)+yF(x)+G(x)+xh(y)+\eta(yï¼‰;
\label{eq5}

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##### Web Bonus = Sept / Re: Web bonus problem : Week 3 (#1)
« on: September 27, 2015, 02:45:49 PM »
Firstly, Let's use characteristic coordinates \left\{\begin{aligned}&\xi=x+ct,\\&\eta=x-ct.\end{aligned}\right.\label{eq1} When using the characteristic coordinates, the problem become: \begin{gather}u_{\xi\eta}=0\qquad \text{as }\xi>0,\eta>0,\\[3pt]u|_{\xi=0}=g(t)\qquad \text{as }t<0 ,\\[3pt]u|_{\eta=0}=h(t)\qquad \text{as } t>0.\end{gather} Then by 2.4.1 equation(4), we have $$u=\phi(\xi)+\psi(\eta)\label{5}$$ is the general solution to (2). Then $$u|_{\xi=0}=\phi(0)+\psi(\eta)=g(t)$$ $$u|_{\eta=0}=\phi(\xi)+\psi(0)=h(t)$$ Thus by (5)(6)(7) $$u=g(t)+h(t)-\phi(0)-\psi(0)$$ Since we have $$u(0,0)=\phi(0)+\psi(0)$$ $$u(0,0)=h(0)=g(0)$$ Then $$u=g(t)+h(t)-g(0)$$ solves Goursat problem.

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