Solving Poisson's equation on $B_1(0)\subset \mathbb{R}^2$

Question

I am trying to solve a specific Poisson equation on the following set $B_1 =\left \{ (x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1 \right \}$

\begin{cases} \Delta u = y & \text{in}\quad B_1\\ u = 1 & \text{on}\quad \partial B_1 \end{cases}

I have studies Green's functions but I don't understand them very well - I don't know how to apply them to a specific case. I have tried changing coordinates to polar, but I don't know how to handle the $y$ term. Any help would really help me understand these problems better. Hints welcome too!

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

---

\begin{align} \on{u} & = {1 \over 6}\,y^{3} + \pars{\substack{\mbox{General Solution in}\ \ds{2}D \\[1mm] \mbox{Polar Coordinates}}} \\[5mm] & = {1 \over 6}\,r^{3}\sin^{3}\pars{\theta} +\pars{\substack{\mbox{General Solution in}\ \ds{2}D \\[1mm] \mbox{Polar Coordinates}}} \\[5mm] & = {1 \over 8}\,r^{3}\sin\pars{\theta} - {1 \over 24}\,r^{3}\sin\pars{3\theta} +\pars{\substack{\mbox{General Solution in}\ \ds{2}D \\[1mm] \mbox{Polar Coordinates}}} \\[5mm] & = {1 \over 8}\,r^{3}\sin\pars{\theta} - {1 \over 24}\,r^{3}\sin\pars{3\theta} \\[2mm] & + \bracks{1 + a_{1}r\sin\pars{\theta} + a_{3}r^{3}\sin\pars{3\theta}} \\[5mm] & = 1 + \pars{{1 \over 8}r^{3} - a_{1}r}\,\sin\pars{\theta} + \pars{-{1 \over 24}\,r^{3} + a_{3}r^{3}}\sin\pars{3\theta} \end{align} Since it must be $\ds{\theta}$-independent in $\ds{\partial B_{1}}$: \begin{align} \on{u} & = 1 + {1 \over 8}\pars{r^{3} - r}\,\sin\pars{\theta} \\[5mm] & = 1 + {1 \over 8}\pars{x^{2} + y^{2} -1}\,y \end{align}

Answer 2

General method

The standard result to use here asserts that the solution $w$ to \begin{cases} \Delta w = 0 & \quad \text{in}\quad B_1(0)\\ w=P_m(x,y) & \quad \text{on} \quad \partial B_1(0) \end{cases} where $P_m(x)$ is a polynomial of $\mathbb{R}^2$ restricted to $\partial B_1(0)$, is another polynomial of degree $P_{m-2}$ and it has the form $$w(x,y)=(1-(x^2+y^2))q(x,y)+P_m(x,y),$$ where $q$ has degree $m-2$. For example, you can find the proof in Theorem 5.1 in Chapter 5 in the book "Harmonic Function Theory" of Axcler, Bourdon, Ramey.

Application

We note that $\Delta \frac{y^3}{6}=y$ and we reduce to the previous case defining $$w(x,y):=u(x,y)-\frac{y^3}{6}$$ In our case $P_3(x,y)=1-\frac{y^3}{6}$ and hence we search for $$q(x,y)=a+bx+cy.$$ Imposing $\Delta w = 0$ we compute $a=b=0$ and $c=-1/8$. Hence we obtain $$ u(x,y)=\frac{x^2+y^2}{8}-\frac{y}{8}+1. $$