Toronto Math Forum
APM3462015F => APM346Home Assignments => HA9 => Topic started by: Emily Deibert on November 13, 2015, 07:10:47 AM

Problem 1
http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter7/S7.P.html#problem7.P.1 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter7/S7.P.html#problem7.P.1)

(a) As in the previous home assignment, we're asked to find solutions of the 2D 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 3D 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 EulerCauchy equation, so we assume a solution $\rho^m$. Plugging this in yields:
\begin{equation}
\rho(m)(m1)\rho^{m2} + 2m\rho^{m1} = 0 \longrightarrow (m)(m1)\rho^{m1} + 2m\rho^{m1} = 0
\end{equation}
We look for the roots of the characteristic equation:
\begin{equation}
m(m1) + 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}

For part (c),
Given $r=(x_1^2+...+x_n^2)^\frac{1}{2}$, $$\frac{\partial r}{\partial x_i} = \frac{x_i}{(x_1^2+...+x_n^2)^\frac{1}{2}} = \frac{x_i}{r}\\\frac{\partial u}{\partial x_i} = u_r \frac{x_i}{r}\\\frac{\partial^2 u}{\partial x_i^2} = \frac{\partial u_r}{\partial x_i}\frac{x_i}{r} + u_r[\frac{1}{r} + \frac{x_i\partial r^1}{\partial x_i}]\\=u_{rr}(\frac{x_i}{r})^2 + \frac{u_r}{r}  \frac{u_r x_i^2}{r^3}\\\Delta u = \sum_{i=1}^{n} \frac{\partial^2 u}{\partial x_i^2} = \frac{u_{rr}}{r^2}\sum_{i=1}^{n} x_i^2+\frac{nu_r}{r}\frac{u_r}{r^3}\sum_{i=1}^{n} x_i^2 = 0\\= u_{rr} +\frac{nu_r}{r}  \frac{u_r}{r} = 0\\=u_{rr} +\frac{n1}{r}u_r = 0 $$
For part (d),
$\Longleftarrow$ Given $x\neq 0$, suppose $u(r) = Ar^{2n}+B$, then $$u_r = (2n)Ar^{1n}\\u_{rr} = (2n)(1n)Ar^{n}$$
From part (c), we get $$u_{rr} + \frac{n1}{r}u_r\\=(2n)(1n)Ar^{n}+\frac{n1}{r}(2n)Ar^{1n}\\=(2n)(1n)Ar^{n}  (2n)(1n)Ar^{n} = 0 =\Delta u$$
$\Longrightarrow$ Given $n\neq 2, x\neq 0$, suppose $$\Delta u = u_{rr} + \frac{n1}{r}u_r = 0$$
Multiply $r^{n1}$ on both side,$$r^{n1}u_{rr} + (n1)r^{n2}u_r = 0$$ which is equal to $$(r^{n1}u_r)' = 0$$ We let $$r^{n1}u_r = (2n)A\\u_r = (2n)Ar^{1n}\\u=Ar^{2n}+B$$ which satisfies Laplace equation.