Author Topic: HA9-P1  (Read 1554 times)

Emily Deibert

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« Last Edit: November 14, 2015, 11:38:42 AM by Victor Ivrii »

Emily Deibert

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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:

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

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

\begin{equation}
\left( r u_r \right)' = 0
\end{equation}

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

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

Thus the solution is:

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

(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:

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

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:

\begin{equation}
u_{\rho \rho} + \frac{2}{\rho}u_{\rho} = 0
\end{equation}

Proceeding as in the previous problem:

\begin{equation}
u_{\rho \rho} + \frac{2}{\rho}u_{\rho} = 0 \longrightarrow \rho u_{\rho \rho} + 2u_{\rho} = 0
\end{equation}

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

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

We look for the roots of the characteristic equation:

\begin{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
\end{equation}

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:

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

Xi Yue Wang

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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.