partial differential equation with polar coordinate i have difficultises to resolve the following problem. Thank you for the help.
We consider the heat equation
$$
\dfrac{\partial u}{\partial t}= c^2 (\dfrac{\partial^2 u}{\partial x^2}+ \dfrac{\partial^2 u}{\partial y^2})
$$
1. Write this equation using the polar coordinate.
2. We put $u(r,\theta,t)= R(r) \Theta(\theta) T(t)$. Gives the differential equations satisfied by $R, \Theta, T$.
 A: The first question boils down to express the laplacian in polar coordinates:
$$\triangle=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}.$$
In order to establish that simply recall that the change of variable is the following:
$$(r,\theta)\mapsto (r\cos(\theta),r\sin(\theta)).$$
Hence, one has:
$$\begin{align}\frac{\partial}{\partial r}&=\cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}\\\frac{\partial}{\partial\theta}&=r\cos(\theta)\frac{\partial}{\partial y}-r\sin(\theta)\frac{\partial}{\partial y}\end{align}.$$
From there express $\displaystyle\frac{\partial}{\partial x}$ and $\displaystyle\frac{\partial}{\partial y}$, then take again the partial derivatives.
The second question simply consists in taking the partial derivatives of this special form of $u$ to get:
$$\frac{\dot{T}(t)}{T(t)}=\frac{r^2R''(r)\Theta(\theta)+rR'(r)\Theta(\theta)+R(r)\Theta''(\theta)}{r^2R(r)\Theta(\theta)}.$$
I just isolated $t$ from the other variables and multiply by $r^2$ the right term. One has a function of $t$ which is equal to a function of $(r,\theta)$, hence they are equal and constant. Therefore, there exists $k$ such that:
$$\begin{align}\dot{T}(t)&=kT(t)\\r^2R''(r)\Theta(\theta)+rR'(r)\Theta(\theta)+R(r)\Theta''(\theta)&=kr^2R(r)\Theta(\theta)\end{align}.$$
Once again isolates the variables in the last equation to get:
$$\frac{\Theta''(\theta)}{\Theta(\theta)}=\frac{kr^2R(r)-r^2R''(r)-rR'(r)}{R(r)}.$$
One has a function of $r$ is equal to a function of $\theta$, hence they are equal and constant. Therefore, there exists $\ell$ such that:
$$\begin{align}\Theta''(\theta)&=\ell\Theta(\theta)\\r^2R''(r)+rR'(r)+(\ell-kr^2)R(r)&=0\end{align}.$$
Finally, the differential equations satisfied by $R,\Theta,T$ are:
$$\begin{align}\dot{T}(t)&=kT(t)\\\Theta''(\theta)&=\ell\Theta(\theta)\\r^2R''(r)+rR'(r)+(\ell-kr^2)R(r)&=0\end{align}.$$
Only the last equation requires some tricks to be solved, it is almost a Bessel's equation.
Remark. Some signs can be wrong, it is your duty to be careful and check my computations. I have also set $c=1$, they were already enough constant!
