Poisson equation inside a ball $B(0, 1)$ Find the solution of the Poisson equation inside a ball $B(0, 1)$, i.e.:
$$\begin{cases} \Delta u=x^{2} \ \ \ in \ B(0,1)\\u=3 \ \ \ on \ \partial B(0,1)\end{cases}$$
($x^2=x\cdot x$, $x\in \mathbb{R}^{2}$)
I have the following formula for the solution of Poisson equation:
$$u(x)=-\int_{\partial\Omega}\frac{\partial G}{\partial n}(x,y) u(y) \ dS(y)-\int_{\Omega}G(x,y)\Delta u(y) \ dy$$
I know the formula for the Green function $G$ in this case. 
I tried to evaluate these integrals with polar coordinates, but they turned out quite difficult...
 A: HINT: Try solving the equation in polar coordinates instead of that Green function integration. The heuristic is that the right hand side is unchanged with respect to the polar angle, i.e., spherically symmetric, the solution should inherit this property (This can be proved by the Green function integration formula you gave). Now in polar coordinates:
$$
\Delta u = \frac{1}{r} \frac{\partial u}{\partial r} 
+ \frac{\partial^2 u}{\partial r^2}
+ \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2},
$$
where $r = \sqrt{x^2+y^2}$. Now 
$$
\frac{\partial^2 u}{\partial \theta^2} = 0
$$
because $u$ is constant on any circle $B(0,r)$, let $w(r) = w(\sqrt{x^2+y^2}) = u(x,y)$ then equation becomes:
$$
w'' + \frac{1}{r}w' -r^2 = 0,
$$
the boundary condition is:
$$
w(1) = 3.
$$
Let $v= w'$, and then try solving the first order ODE.  
A: Some suggestions along the same lines as Shuhao's, but perhaps on the lazier side.


*

*For any rotation $T:\mathbb R^2\to \mathbb R^2$ the function $u\circ T$ solves the same boundary problem, because $\Delta(u\circ T)=(\Delta u)\circ T$. 

*By an appropriate uniqueness theorem (maximum principle for harmonic functions, really), $u=u\circ T$. That is, $u$ is a function of $r$ only. 

*Since the Laplacian of $u$ is homogeneous of degree $2$, and is the second derivative of $u$, the function $u$ itself should be homogeneous of degree $4$. Up to a harmonic (constant) term, that is. All that remains is to find two real constants.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Note that $\ds{\pars{~A\ \mbox{is a constant}~}}$
\begin{align}
\Delta\pars{Ar^4}&=A\nabla\cdot\pars{4r^{2}\vec{r}}
=4A\bracks{\pars{\nabla r^{2}}\cdot\vec{r} + r^{2}\nabla\cdot\vec{r}}
=4A\bracks{2r^{2}+ 3r^{2}}=20Ar^{2}
\\[3mm]&\mbox{such that}\quad\Delta\pars{r^{2} \over 20} = r^{2}
\quad\mbox{with}\quad A = {1 \over 20}.
\end{align}

It means that
  $\ds{\Phi\pars{\vec{r}} \equiv {\rm u}\pars{\vec{r}} - {r^{2} \over 20}\quad}$
  satisfies
  $\quad\ds{\left\{%
\begin{array}{l}
\Delta\Phi\pars{\vec{r}} & = & 0
\\[2mm]
\left.\Phi\pars{\vec{r}}\right\vert_{r\ =\ 1} & = &{59 \over 20}
\end{array}\right.}$

Obviously,
$\ds{\Phi\pars{\vec{r}} = {59 \over 20}\,,\forall\ r\ \mid\
0 \leq r \leq 1\quad\imp\quad
\color{#66f}{\large{\rm u}\pars{\vec{r}} = {r^{2} + 59 \over 20}\,,\quad
0 \leq r \leq 1}}$
