what we means by radially symmetric slution in pde and the relation with polar coordinate we consider the equation 
$$\Delta u- q(x) u=0$$
in the $2$ D circle $B=\{|x| \leq 1\}$ with the boundary condition 
$$
|x|=1: u=1
$$
the function $q$ is taken in the form
$$
q(x)
=
\begin{cases}
w, & |x| < \epsilon,\\
0, & |x| \geq \epsilon
\end{cases}
$$
where $w$ and $\epsilon$ are some positive constants.
In the case of radially symmetric solutions, we can write the problem in the polar coordinate as
$$
u''(r)+\dfrac{1}{r} u'(r)- q(r)u=0
$$
$$
u'(0)=0, u(1)=1.
$$
My question are, please what we means by radially symmetric solutions? What's the relation between radially symetric solution avec polar coordinate? And how we re-write the problem in the polar coordinate?
 A: Radially symmetric solution means that $u(r,\theta)=u(r)$, so is, the solution doesn't depend on the polar angle $\theta$ or, autrament dit, the value of $u$ it's the same no mind the value of $\theta$. It is said too symmetric under rotation.
The laplacian $\Delta$ in polar coordinates is,
$\Delta=\dfrac{\partial^2}{\partial r}+\dfrac{1}{r}\dfrac{\partial}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2}{\partial\theta}$
$\Delta u=\dfrac{\partial^2u}{\partial r}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2u}{\partial\theta}$
The last term will vanish if $u$ symmetric. Now, $q(x)=q(r)$, radially symmetric, because the condition $\vert x\vert\lt\epsilon$ in its definition. We suppose now the solution has to be symmetric too. So is,
$\Delta u- q(x) u=0\to\dfrac{\partial^2u}{\partial r}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}-q(r)u=0$
As $u$ depends only on $r$, the partials become totals:
$u''(r)+\dfrac 1ru'(r)-q(r)u(r)=0$ We have to rewrite too the definition of $q$,
$q(r)=
\begin{cases}
  w, & r < \epsilon,\\
  0, & r \geq \epsilon
\end{cases}$
with $u=1$ if $r=1$ and $u'=0$ if $r=0$
