$\newcommand{\const}{\mathrm{const}}$ $\newcommand{\erf}{\operatorname{erf}}$

### APM 346 (2012) Home Assignment 7

This assignment is based on Lecture 22, Lecture 23 and Lecture 24 coming shortly.

#### Problem 1

1. Find the solutions that depend only on $r$ of the equation \begin{equation*} \Delta u:=u_{xx}+u_{yy}+u_{zz}=k^2u, \end{equation*} where $k$ is a positive constant. (Hint: Substitute $u=v/r$)

2. Find the solutions that depend only on $r$ of the equation \begin{equation*} \Delta u:=u_{xx}+u_{yy}+u_{zz}=-k^2u, \end{equation*} where $k$ is a positive constant. (Hint: Substitute $u=v/r$)

#### Problem 2

1. Try to find the solutions that depend only on $r$ of the equation \begin{equation*} \Delta u:=u_{xx}+u_{yy}=k^2u, \end{equation*} where $k$ is a positive constant. What ODE should satisfy $u(r)$?

2. Try to find the solutions that depend only on $r$ of the equation \begin{equation*} \Delta u:=u_{xx}+u_{yy}=-k^2u, \end{equation*} where $k$ is a positive constant. What ODE should satisftfy $u(r)$?

#### Problem 3

1. Solve \begin{align*} & \Delta :=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt] & u|_{r=a}=f(\theta). \end{align*} where we use polar coordinates $(r,\theta)$ and f(\theta)=\left\{\begin{aligned} &1 &&0<\theta<\pi\\ -&1 &&\pi<\theta<2\pi. \end{aligned}\right.

2. Solve \begin{align*} & \Delta :=u_{xx}+u_{yy}=0&& \text{in } r>a\\[3pt] & u|_{r=a}=f(\theta),\\[3pt] & \max |u| <\infty. \end{align*} where we use polar coordinates $(r,\theta)$ and f(\theta)=\left\{\begin{aligned} &1 &&0<\theta<\pi\\ -&1 &&\pi<\theta<2\pi. \end{aligned}\right.

#### Problem 4

1. Solve \begin{align*} & \Delta :=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt] & u_r|_{r=a}=f(\theta). \end{align*} where we use polar coordinates $(r,\theta)$ and f(\theta)=\left\{\begin{aligned} &1 &&0<\theta<\pi\\ -&1 &&\pi<\theta<2\pi. \end{aligned}\right.

2. Solve \begin{align*} & \Delta :=u_{xx}+u_{yy}=0&& \text{in } r>a\\[3pt] & u_r|_{r=a}=f(\theta),\\[3pt] & \max |u| <\infty. \end{align*} where we use polar coordinates $(r,\theta)$ and f(\theta)=\left\{\begin{aligned} &1 &&0<\theta<\pi\\ -&1 &&\pi<\theta<2\pi. \end{aligned}\right.