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### Messages - Sheng Gao

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##### Web Bonus Problems / Re: Exam Week
« on: April 13, 2018, 11:48:40 PM »
By Laplace equation
$u_{rr} +\frac{1}{r}u_r +\frac{1}{r^2}u_{\theta\theta}=0, r<1, 0<\theta<\pi$
$u(r,0)=r^2$
$u(r,\pi)=r^2$
$u|_{r=1}=1$

let $v=x^2-y^2$, then
$\Delta v=v_{xx}+v_{yy}=2-2=0$
$v|_{y=0}=x^2$  $v|_{x^2+y^2=1}=1-2y^2=1-2r^2sin\theta=1-2sin\theta, since$  $r=1$

And
Let $w=u-v, then \Delta w=0$ $w|_{y=0}=0$, $w|_{x^2+y^2=1}=2sin^2\theta$

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##### Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 12, 2018, 07:47:21 PM »
This is my solution, not sure true or not. If it is correct, I will post more details. $u(x,t)=\sqrt2[(x+t)^\frac{3}{2}+(x-t)^\frac{3}{2}]+\frac{\sqrt2}{3}[(x+t)^\frac{1}{2}-(x-t)^\frac{1}{2}]$

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##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 08, 2018, 11:24:26 PM »
$u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$ where $\phi$, $\psi$ so far are unknown and will be chosen later. Such $u$ obviously exist for any $\phi$, $\psi$.

What about $v:=u_{yy}-g=0$? Obviously $v_{xx}=0$ and we need to satisfy $v(0,y)=0$, $v_x(0,y)=0$, which are, in fact, $\phi '' -g(0,y)=0$, $\psi ''- g_x(0,y)=0$, and we can satisfy these two equations.  Four arbitrary constants $C_1=\phi(0)$, $C_2=\phi'(0)$, $C_3=\psi(0)$ and $C_4=\psi'(0)$ appear.
May I ask ask about how we get $u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$? Since I cannot follow up starting here.....

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