Author Topic: HA9 Problem 4  (Read 1474 times)

Victor Ivrii

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HA9 Problem 4
« on: March 26, 2015, 02:56:43 PM »
Apply method of descent but now looking for stationary solution of $-\Delta u=f(x_1,x_2,x_3)$ instead of non-stationary solution of
\begin{align*}
& u_{tt}-\Delta u=f(x_1,x_2,x_3),\\
& u|_{t=0}=g(x_1,x_2,x_3),\\
& u_t|_{t=0}=h(x_1,x_2,x_3)
\end{align*}
and derive for $n=3$ with $G(x,y)$ equal to
$$-\frac{1}{4\pi} \bigl((x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2\bigr)^{-\frac{1}{2}}.$$

Chaojie Li

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Re: HA9 Problem 4
« Reply #1 on: March 26, 2015, 08:08:21 PM »
By Kirch formula ,$c=1$
$$u(x,t)= \frac{\partial\ }{\partial t} \Bigl(\frac{1}{4\pi t} \iint _{S(x,|t|)} g(y)\,d\sigma\Bigr)+ \frac{1}{4\pi t} \iint _{S(x,|t|)} h(y)\,d\sigma + \int_0^t \frac{1}{4\pi (t-\tau)} \iint_{S(x,|t-\tau|) } f(y,\tau)\,d\sigma d\tau$$
$$-\Delta u = f(x_1,x_2,x_3), u_t|_{t=0}=0, u_y|_{t=0}=h(x_1,x_2,x_3)$$
so we have ::$h(x_1,x_2,x_3)=0$, u is independent of $t,g$ and h are independent of t.\\
$$\frac{\partial}{\partial 4 \pi t}\iint _{S(x,|t|)} g(y)d\sigma =0$$
$$u|_{t=0}=g(x_1,x_2,x_3)$$
$$u=\int_0^t\frac{1}{4\pi (t-\tau)}\iint_{S(x,|t-\tau|)}f(y,\tau)d\sigma dr=\int_0^{x_3}\frac{1}{4\pi(x_3-y_3)}\iint_{S(x,|x_3-y_3|)}f(y,y_3)d\sigma dy_3$$
$$\frac{1}{4\pi x_3}\iint_{S(x,|x_3|)}f(y)d\sigma = \pm \frac{1}{2\pi}\iint_{S(x,|x_3|)}\frac{f(y)}{\sqrt{|x|^2+|x-y|^2}}$$
$$u=\frac{1}{4\pi}\int_0^{x_3}\iint_{S(x,|x-x_3|)}\frac{f(y,y_3)}{\sqrt{(x_3-y_3)^2+|x-y|^2}}dydy_3=\frac{1}{4\pi}\iiint\frac{f(y_1,y_2,y_3)}{\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2}}dy$$
$$G=-\frac{1}{4\pi \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}}$$
and
$$-\Delta u=f(x_1,x_2,x_3)$$
so $$u=\iiint -G(x,y)f(y_1,y_2,y_3)dy =\iiint G(x,y)\Delta(x)dy$$
« Last Edit: March 26, 2015, 08:37:06 PM by Chaojie Li »