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### Messages - Chaojie Li

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1
##### HA10 / Re: HA10 Problem 3
« on: April 02, 2015, 08:32:49 PM »

2
##### HA10 / Re: HA10 Problem 4
« on: April 02, 2015, 08:32:20 PM »

3
##### HA9 / Re: HA9 Problem 1
« on: March 28, 2015, 01:29:47 PM »
most part code is copied from Chen

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda

with

\Lambda:=
\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$:
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

which could be rewritten as

\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+
\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0

and since the first term depends only on $\rho$ and the second only on$\phi, \theta$ we conclude that both are constant:
\begin{align}
\rho^2 P'' +2\rho P' = \lambda P,\\
\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).
\end{align}
The first equation is of Euler type and it has solutions $P:=\rho^l$ iff
$\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$

Then suppose that
\begin{align}
u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2)  + c(1-x^2-y^2-z^2) ^2\\\\
=&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2

\end{align}
\begin{align}
\Delta u &= u_{xx}+u_{yy}+u_{zz}\\
&=12 z^2  -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2\\\text{Suppose $1-\rho^2-z^2=0$}\\
&=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\
&=12z^2-6a-12bz^2-6c+20c-20cz^2\\
&=(12-12b-20c)z^2-6a-6c=0
\end{align}
Then we have
\begin{align}
12-12b-20c=0\\
-6a-6c=0
\end{align}
Then $a = -c$, $b=1-\frac{5}{3}c$, and c is arbitrary.
[/quote]

4
##### HA9 / Re: HA9 Problem 1
« on: March 28, 2015, 01:25:04 PM »

\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda

with

\Lambda:=
\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.

Plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$:
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0

which could be rewritten as

\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+
\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0

and since the first term depends only on $\rho$ and the second only on$\phi, \theta$ we conclude that both are constant:
\begin{align}
\rho^2 P'' +2\rho P' = \lambda P,\\
\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).
\end{align}
The first equation is of Euler type and it has solutions $P:=\rho^l$ iff
$\lambda= l(l+1)$. However if we are considering ball, solution must be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.Which in this case, u depends on $x^2+y^2+z^2$ and $z$

Then suppose that
\begin{align}
u= &z^4 + a (1-x^2-y^2-z^2) + bz^2 (1-x^2-y^2-z^2)  + c(1-x^2-y^2-z^2) ^2\\\\
=&z^4 + a (1-x^2-y^2-z^2) + b (z^2-z^2x^2-z^2y^2-z^4) + c(1-x^2-y^2-z^2) ^2

\end{align}
\begin{align}
\Delta u &= u_{\xx}+u_{\yy}+u_{zz}\\
&=12 z^2  -6a + 2b( 1 -\rho^2 -7 z^2) -6c +20c\rho^2\\\text{Suppose $1-\rho^2-z^2=0$}\\
&=12z^2-6a+2b(-6z^2)-6c+20c(1-z^2)\\
&=12z^2-6a-12bz^2-6c+20c-20cz^2\\
&=(12-12b-20c)z^2-6a-6c=0
\end{align}
Then we have
\begin{align}
12-12b-20c=0\\
-6a-6c=0
\end{align}
Then $a = -c$, $b=1-\frac{5}{3}c$, and c is arbitrary.
[/quote]

5
##### HA4 / Re: HA4 problem 2
« on: March 26, 2015, 08:51:16 PM »
First separate variables. Let: $\Big {[}u(x,t) = X(x)T(t)\Big {]}$ in $\Big {[}u_{tt}(x,t) + K u_{xxxx} = 0, K > 0\Big {]}$

Then:$$\Big {[}u_{tt}(x,t) = X(x)T(t), u_{xxxx}(x,t) = X(x)T(t)\Big {]}$$
$$\Big {[}u_{tt}(x,t) + K u_{xxxx} = X(x)T(t) + K X(x)T(t) = 0\Big {]}$$
$$\Big {[}\frac{X(x)}{X(x)} = \frac{-T(t)}{K T(t)} = \lambda\Big {]}$$

For some constant, as both sides are independant of the other respective variable. Now, we have by assumptions that $\Big {[}\lambda = c^4 > 0, K = k^2 > 0\Big {]}$

So we are left with two ODEs in the form: $$\Big {[}X(x) = c^4 X(x), T(t) = -c^4 k^2 T(t)\Big {]}$$

Which yield solutions: $$\Big {[}X(x) = A \cosh(c x) + B \sinh(c x) + C \cos(c x) + D \sin(c x)\Big {]}$$
$$\Big {[}T(t) = A \cos(c^2 k t) + B \sin(c^2 k t)\Big {]}$$

Using the two boundary conditions: $\Big {[}u(0,t) = 0 = u_x(0,t)\Big {]}$

$$\Big {[}X(0) = A \cosh(c 0) + B \sinh(c 0) + C \cos(c 0) + D \sin(c 0) = A + C = 0, A = -C\Big {]}$$
$$\Big {[}X(0) = A c \sinh(c 0) + B c \cosh(c 0) - C c \sin(c 0) + D c \cos(c 0) = B c + D c, D = -B\Big {]}$$

As we disregard the case where: $$\Big {[}X(0) \neq 0, X(x) \neq 0 \implies T(t)=0\Big {]}$$

We plug into the 3rd and 4th boundary conditions.

$$\Big {[}u_{xx}(l,t) = 0 = u_{xxx}(l,t)\Big {]}$$

$$X(l) = A c^2 \cosh(c l) + B c^2 \sinh(c l) - C c^2 \cos(c l) - D c^2 \sin(c l)$$
$$=A c^2 \cosh(c l) + B c^2 \sinh(c l) + A c^2 \cos(c l) + B c^2 \sin(c l)$$$$= A c^2 (\cosh(c l) + \cos(c l)) + B c^2 (\sinh(c l) + \sin(c l)) = 0$$

$$\Big {[}X(l) = A c^3 (\sinh(c l) - \sin(c l)) + B c^3 (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
Yielding:
$$\Big {[}A (\cosh(c l) + \cos(c l)) + B (\sinh(c l) + \sin(c l)) = 0\Big {]}$$
$$\Big {[}A (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
$$\Big {[}A = -B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))}\Big {]}$$
$$\Big {[}-B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
$$\Big {[}-\frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l) + (\cosh(c l) + \cos(c l)))\Big {]}$$
$$\Big {[}-\sinh(c l)^2 + \sin(c l)^2 + \cosh(c l)^2 + 2 \cosh(c l) \cos(c l) + \cos(c l)^2 =0\Big {]}$$

Where we used the facts that $$\Big {[}B = 0 \implies X(x) = 0, (\cosh(c l) + \cos(c l)) = 0 \implies B = A = 0\Big {]}$$
Which we disregard as were not so interested in the trivial case.
$$\Big {[} \cosh(c l)^2 - \sinh(c l)^2 + \sin(c l)^2+ \cos(c l)^2 + 2 \cosh(c l) \cos(c l) = 1 + 1 + 2 \cosh(c l) \cos(c l) = 0\Big {]}$$

As:$\Big {[}\cosh^2 - \sinh^2 = 1, \cos^2 + \sin^2 = 1\Big {]}$

So our eigenvalues are those c which satisfy$\Big {[}\cosh(c l) \cos(c l) = -1 \blacksquare$

Next consider: $$\Big {[}\int_0^l X_n(x)X_m(x) dx, n \neq m, \lambda_n \neq \lambda_m\Big {]}$$

Then $$\Big {[}(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) - \lambda_m X_n(x)X_m(x) dx\Big {]}$$
$$\Big {[}\int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx = (X_n(x)X_m(x) - X_n(x)X_m(x))_{x=(0,l)} - \int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx\Big {]}$$
$$\Big {[}= 0 - (X_n(x)X_m(x) - X_n(x)X_m(x))_{x=(0,l)} + \int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx = 0 + 0 = 0\Big {]}$$

Using integration by parts, and the fact that our boundary conditions vanish at:

$$\Big {[}X(0) = 0, X(0) = 0, X(l) = 0, X(l) = 0\Big {]}$$

Then we have $$\Big {[}(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0, \lambda_n \neq \lambda_m \neq 0 \implies \int_0^l X_n(x)X_m(x) dx = 0\Big {]}$$

And our eigenfunctions are orthogonal.

Let our differential operator be$$\Big {[}\mathcal{I}, st. \mathcal{I} X = \lambda X, \mathcal{I} Y = \lambda Y\Big {]}$$

We show that: $$\Big {[}\langle \mathcal{I}X,Y\rangle = \langle X,\mathcal{I}^*Y\rangle\Big {]}$$

$$\Big {[}\langle \mathcal{I}X,Y\rangle = \int_0^l \mathcal{I}X(x)Y(x)^* dx = \int_0^l \lambda X(x)Y(x)^* dx = \int_0^l X(x)\lambda^* Y(x)^* dx = \int_0^l X(x)\mathcal{I}^*Y(x)^* dx = \langle X,\mathcal{I}^*Y\rangle\Big {]}$$

As we had by assumption our eigenvalues were real. Then our operator is hermetian, so the eigenfuncitons of different eigenvalues are linearly independant, and the eigenfunctions of the same eigenvalue are linearly dependant.

6
##### HA9 / Re: HA9 Problem 4
« 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 $$u=\iiint -G(x,y)f(y_1,y_2,y_3)dy =\iiint G(x,y)\Delta(x)dy$$

7
##### HA9 / Re: HA9 Problem 3
« on: March 26, 2015, 08:07:44 PM »
(b)
As in part(a) we can do similar by reflection for Neumann problem for Laplace equation in brief use even continuation $f(-y)=f(y),x=(x_1,x_2),y=(y_1,y_2),-y=(y_1,-y_2)$\\\
Upper half plane, $y>0$
$$u(x,y)=\int_0^\infty f(x,y)G(x-y)dy+\int_0^\infty f(x,-y)G(x-y)dy$$
$$=\int_0^\infty f(x,y)G(x-y)dy +\int _0^\infty f(x,y)G(x+y)dy$$
$$=\int _0 ^\infty f(x,y)[G(x-y)+G(x+y)]dy$$
$$G(x-y)-G(x+y)=\frac{1}{2\pi}\log [(x_1-y_1)^2+(x_2-y_2^2)]^{\frac{1}{2}}+\frac{1}{2\pi}\log[(x_1-y_1)^2+(x_2+y_2)^2]^{\frac{1}{2}}$$
$$=\frac{1}{4\pi}\log [(x_1-y_1)^2+(x_2-y_2)^2][(x_1-y_1)^2+(x_2+y_)^2]$$
For upper half plane $y>0$, green function:
$$G(x,y)=\frac{1}{4\pi}\log [(x_1-y_1)^2+(x_2-y_2)^2][(x_1-y_1)^2+(x_2+y_2)^2]$$
Similarly, we consider the upper half space $z>0$,$x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$,$-y=(y_1,y_2,-y_3)$\\
$$G(x,y)=G(x-y)+G(x+y)=-\frac{1}{4\pi}\{[(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2]^{-\frac{1}{2}}+[(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2]^{-\frac{1}{2\pi}}\}$$
for $z>0$ upper half space , Green Function
$$G(x,y)=-\frac{1}{4\pi \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}}-\frac{1}{4\pi \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2}}$$

8
##### HA9 / Re: HA9 Problem 3
« on: March 26, 2015, 08:06:59 PM »
Q3(a)
Dir Problem in the hall plane for Laplace equation :
In whole place $G(x,y)=\frac{1}{2\pi}\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ where $x=(x_1,x_2), y=(y_1,y_2)$
consider the upper half plane $y>0$ yes odd continuation
$$-y=(y_1,-y_2)\implies f(-y)=-f(y)$$
so
$$u(x,y)=\int_0^\infty f(x,y)G(x-y)dy+\int_{-\infty}^0f(x-y)G(x-y)dy$$
$$=\int_0^\infty f(x,y)G(x-y)dy+\int_0^\infty-f(x,y)G(x+y)dy$$
$$=\int_0^\infty f(x,y)[G(x-y)-G(x+y)]dy$$
$$G(x-y)-G(x+y)=\frac{1}{2\pi}\log \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}-\frac{1}{2\pi}$$
$$\frac{1}{2\pi} \log \sqrt{\frac{(x_1-y_1)^2+(x_2-y_2)^2}{(x_1-y_1)^2+(x_2+y_2)^2}}$$
$$=\frac{1}{4\pi}\log \frac{(x_1-y_1)^2+(x_2-y_2)^2}{(x_1-y_1)^2+(x_2+y_2)^2}$$
For upper half plane $y>0$, Green function $G(x,y)=\frac{1}{4\pi}\log \frac{(x_1-y_1)^2+(x_2-y_2)^2}{(x_1-y_1)^2+(x_2+y_2)^2}$
similarly Dirichlet problem in half space for Laplace equation
$$G=-\frac{1}{4\pi}[(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2]^{1\frac{1}{2}}$$
where $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$,$-y=(y_1,y_2,-y_3)$
for $z>0$ upper half space:
$$G(x,y)=G(x-y)-G(x+y)=-\frac{1}{4\pi}\{[(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_2)^2]^{-\frac{1}{2}}-[(x_1-y_1)^2+(x_2-y_2)^2+(x_3+y_3)^2]^{-\frac{1}{2}}\}$$
Green function. $\blacksquare$

9
##### HA9 / Re: HA9 Problem 1
« on: March 26, 2015, 08:06:07 PM »
This is Q1

10
##### HA9 / Re: HA9 Problem 2
« on: March 26, 2015, 08:05:36 PM »
$$u(x,y,z)=\int U_3(x-x',y-y',z-z')f(x',y',z')dx'dy'dz'$$
Since $z-z'$ is a constant so that u doesn't depend on $z'$
$$u(x,y)=\int U_3(x-x',y-y',z-z')dZdx'dy' =\int U_2 (x-x',y-y') dx'dy'$$
so $U_2=\int U_3(x-x',y-y',z-z')dZ$\\
and we have :
$$U_3=-\frac{1}{4\pi}(x^2+y^2+z^2)^{\frac{1}{2}}$$
$$\text{and}$$
$$U_3(1,0,z)=-\frac{1}{4\pi(1+z^2)^{\frac{1}{2}}}$$
$$U_2=2[\int_0 ^N U_3(x,y,z)-\int_0 ^N U_3(1,0,z)]dz$$
$$\implies$$
$$U_2=-\frac{1}{2\pi}[\int_0 ^N\frac{1}{\sqrt{x^2+y^2+z^2}}+\int _0 ^N \frac{1}{\sqrt{1+z^2}}]$$
$$=-\frac{1}{2\pi}[\log (z+\sqrt{x^2+y^2+z^2})\Big {|}_0 ^N-\log (z+\sqrt{1+z^2})\Big{|}_0 ^N]$$
$$U_2=-\frac{1}{2\pi}[\log \frac{N+\sqrt{x^2+y^2+z^2}}{N+\sqrt{1+N^2}}]-\log \sqrt{x^2+y^2}$$
$$\text{As} N\implies \infty$$
$$\log \frac{N+\sqrt{x^2+y^2+z^2}}{N+\sqrt{1+N^2}}]\implies 0$$
$$\text{Therefore we have}$$
$$U_2=\frac{1}{2\pi} \log \sqrt{x^2+y^2} \blacksquare$$

11
##### HA8 / Re: question 1 (a)-(b)
« on: March 19, 2015, 09:19:32 PM »
Part c.
$$\text{Let: } r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right).$$
$$\implies \Delta u = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(r\right) = \left(\sum_{i=1}^{n} \partial_{x_i}^2\right) u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right) = 0$$
$$\implies \sum_{i=1}^{n} [ \partial_{x_i}^2 u\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) ] = \sum_{i=1}^{n} [ \partial_{x_i}\frac{ x_i u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} ]$$
$$= \sum_{i=1}^{n} [ \frac{ u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} - \frac{ x_{i}^2 u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} + \frac{ x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)}]$$
$$= \sum_{i=1}^{n} [ \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} + \frac{x_{i}^2 u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)} ]$$
$$= \sum_{i=1}^{n} [ \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{ \left(\sum_{j=1}^{n} x_j^2\right)^{\frac{3}{2}}} ] + \frac{\left(\sum_{j=1}^{n} x_{j}^2\right) u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\sum_{j=1}^{n} x_{j}^2}$$
$$= u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) + \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}} \sum_{i=1}^{n} \left( \frac{ \left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2 }{ \sum_{j=1}^{n} x_j^2} \right)$$
$$\text{Notice that: } \sum_{i=1}^{n} [ \frac{ \left(\left(\sum_{j = 1}^{n} x_j^2\right) - x_{i}^2\right) }{ \left(\sum_{j=1}^{n} x_j^2\right)} ] = \sum_{i=1}^{n} \left( \frac{ \sum_{j = 1}^{n} x_j^2 }{ \sum_{j=1}^{n} x_j^2} \right) - \sum_{i=1}^{n} \left( \frac{ x_{i}^2}{ \sum_{j=1}^{n} x_j^2} \right)$$
$$= n - \frac{ \sum_{i=1}^{n} x_{i}^2}{ \sum_{j=1}^{n} x_j^2} = n -1 \text{ so we have:}$$
$$\implies \Delta u = u_{rr}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right) + \left(n-1\right) \frac{u_{r}\left(\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}} \right)}{\left(\sum_{j=1}^{n} x_j^2\right)^{\frac{1}{2}}}$$
$$= u_{rr} + \frac{n-1}{r}u_r = 0 \text{, as needed. } \blacksquare$$

Part d.
$$\text{Let: } n \in \mathbb{N} \setminus 2, \phantom{O} \{ x_1 \dots x_n \} \in \mathbb{R}^n, \phantom{O} r = +\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}, \phantom{O} u\left(r\right) = u\left(\left(\sum_{i=1}^{n} x_i^2\right)^{\frac{1}{2}}\right).$$
By part c. we have that the Laplacian of $u\left(r\right)$ satisfies part c equation named (*):
$$\Delta u = u_{rr} + \frac{n-1}{r}u_r = 0$$
If $r \ne 0$, $u\left(r\right) = A r^{2-n} + B$, $u_{r} = A\left(2-n\right) r^{1-n}$, $u_{rr} = A\left(1-n\right)\left(2-n\right) r^{-n}$ and clearly:
$$u_{rr} + \frac{n-1}{r}u_r = A \left(1-n\right) \left(2 - n\right) r^{-n} + \frac{n-1}{r} A \left(2 - n\right) r^{1 - n}$$
$$= A \left(1 - n\right) \left(2 - n\right) r^{-n} - A \left(1 - n\right) \left(2 - n\right) r^{-n} = 0 \phantom{O} \square$$
Thus $u$ satisfies Laplace's equation in $r$. Conversely, if $u\left(r\right)$ satisfies Laplace's equation in r(*) for $r\ne 0$, then:
$$u_{rr} + \frac{n-1}{r}u_r = 0 \implies r^{n-1} u_{rr} + \left(n-1\right)r^{n-2}u_r = 0 \implies \partial_r\left(r^{n-1} u_{r}\right)= 0$$
$$\implies r^{n-1} u_{r}= \left(2-n\right) A \implies u_{r}= \left(2-n\right)\frac{A}{r^{n-1}}$$
$$\implies u\left(r\right) = A r^{2-n} + B \phantom{O} \square$$
Thus we have $u = u\left(r\right)$ satisfies Laplace's equation in $r \ne 0$, $n \in \mathbb{N} \setminus 2$,
$$\Delta u\left(r\right) = 0 \iff u\left(r\right) = A r^{2-n} + B, \phantom{O} \{ A, B \} \in \mathbb{R} \phantom{O} \blacksquare$$

12
##### HA8 / Re: Solution of Question 4
« on: March 19, 2015, 09:14:32 PM »

13
##### HA8 / Solution of Question 4
« on: March 19, 2015, 09:13:07 PM »
Using Newton shell theorem (see [Section 6 of Lecture 25]) prove that if Earth was a homogeneous solid ball then the gravity pull inside of it would be proportional to the distance to the center.

14
##### HA8 / Re: Solution of Question 3
« on: March 19, 2015, 09:10:17 PM »
This is my solution of Q3 b

15
##### HA8 / Solution of Question 3
« on: March 19, 2015, 09:08:28 PM »
1.  Using the proof of maximum principle prove the maximum principle for
subharmonic functions and minimum principle for superharmonic
functions.

2.  Show that minimum principle for subharmonic functions and maximum
principle for superharmonic functions do not hold (*Hint*: construct
counterexamples with $f=f(r)$).

3.  Prove that if $u,v,w$ are respectively harmonic, subharmonic and
superharmonic functions in the bounded domain $\Omega$,
coinciding on its boundary ($u|_\Sigma=v|_\Sigma=w|_\Sigma$)
then in $w\ge u \ge v$ in $\Omega$.

SO i just post what i write on paper

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