### Author Topic: HA9-P1  (Read 2694 times)

#### Emily Deibert

• Elder Member
• Posts: 100
• Karma: 0
##### HA9-P1
« on: November 13, 2015, 07:10:47 AM »
« Last Edit: November 14, 2015, 11:38:42 AM by Victor Ivrii »

#### Emily Deibert

• Elder Member
• Posts: 100
• Karma: 0
##### Re: HA9-P1
« Reply #1 on: November 13, 2015, 07:36:12 AM »
(a) As in the previous home assignment, we're asked to find solutions of the 2-D Laplace equation that rely only on $r$. Recall from home assignment 8 that for solutions depending only on $r$, our problem in polar coordinates will reduce to:

u_{rr} + \frac{1}{r}u_r = 0 \longrightarrow ru_{rr} + u_r = 0

Notice that this is the derivative of $ru_r$, so our equation becomes:

\left( r u_r \right)' = 0

This is now a simple problem; the solution is as follows:

ru_r = c_1 \longrightarrow u_r = \frac{c_1}{r} \longrightarrow u = c_1\ln{r} + c_2

Thus the solution is:

u(r) = c_1\ln{r} + c_2

(b) This is a similar problem, but now we are dealing with a 3-D Laplace equation. Recall that when we switch to spherical coordinates, the Laplacian becomes:

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

Now, can we get rid of any of the terms in the Laplacian here, as we did in part one? Indeed, since we are looking for solutions that depend only on $\rho$ and $\Lambda$ by definition has no $\rho$-dependence. We proceed to reduce our problem to:

u_{\rho \rho} + \frac{2}{\rho}u_{\rho} = 0

Proceeding as in the previous problem:

u_{\rho \rho} + \frac{2}{\rho}u_{\rho} = 0 \longrightarrow \rho u_{\rho \rho} + 2u_{\rho} = 0

This is a Euler-Cauchy equation, so we assume a solution $\rho^m$. Plugging this in yields:

\rho(m)(m-1)\rho^{m-2} + 2m\rho^{m-1} = 0 \longrightarrow (m)(m-1)\rho^{m-1} + 2m\rho^{m-1} = 0

We look for the roots of the characteristic equation:

m(m-1) + 2m = 0 \longrightarrow m^2 - m + 2m = 0 \longrightarrow m^2 + m = 0 \longrightarrow m(m+1) = 0 \longrightarrow m = 0, -1

The roots are $m=0$ and $m= -1$. Plugging these into our solution $\rho^m$ and forming the general solution as a linear combination of these two solutions yields the final answer:

u(\rho) = \frac{c_1}{\rho} + c_2
« Last Edit: November 13, 2015, 07:38:33 AM by Emily Deibert »

#### Xi Yue Wang

• Full Member
• Posts: 30
• Karma: 0
##### Re: HA9-P1
« Reply #2 on: November 16, 2015, 03:40:59 AM »
For part (c),
Given $r=(x_1^2+...+x_n^2)^\frac{1}{2}$, $$\frac{\partial r}{\partial x_i} = \frac{x_i}{(x_1^2+...+x_n^2)^\frac{1}{2}} = \frac{x_i}{r}\\\frac{\partial u}{\partial x_i} = u_r \frac{x_i}{r}\\\frac{\partial^2 u}{\partial x_i^2} = \frac{\partial u_r}{\partial x_i}\frac{x_i}{r} + u_r[\frac{1}{r} + \frac{x_i\partial r^-1}{\partial x_i}]\\=u_{rr}(\frac{x_i}{r})^2 + \frac{u_r}{r} - \frac{u_r x_i^2}{r^3}\\\Delta u = \sum_{i=1}^{n} \frac{\partial^2 u}{\partial x_i^2} = \frac{u_{rr}}{r^2}\sum_{i=1}^{n} x_i^2+\frac{nu_r}{r}-\frac{u_r}{r^3}\sum_{i=1}^{n} x_i^2 = 0\\= u_{rr} +\frac{nu_r}{r} - \frac{u_r}{r} = 0\\=u_{rr} +\frac{n-1}{r}u_r = 0$$

For part (d),
$\Longleftarrow$ Given $x\neq 0$, suppose $u(r) = Ar^{2-n}+B$, then $$u_r = (2-n)Ar^{1-n}\\u_{rr} = (2-n)(1-n)Ar^{-n}$$
From part (c), we get $$u_{rr} + \frac{n-1}{r}u_r\\=(2-n)(1-n)Ar^{-n}+\frac{n-1}{r}(2-n)Ar^{1-n}\\=(2-n)(1-n)Ar^{-n} - (2-n)(1-n)Ar^{-n} = 0 =\Delta u$$
$\Longrightarrow$ Given $n\neq 2, x\neq 0$, suppose $$\Delta u = u_{rr} + \frac{n-1}{r}u_r = 0$$
Multiply $r^{n-1}$ on both side,$$r^{n-1}u_{rr} + (n-1)r^{n-2}u_r = 0$$ which is equal to $$(r^{n-1}u_r)' = 0$$ We let $$r^{n-1}u_r = (2-n)A\\u_r = (2-n)Ar^{1-n}\\u=Ar^{2-n}+B$$ which satisfies Laplace equation.