Let $U:\mathbb{R}^2\to\mathbb{R}$ be two times continuous differentiable. The Laplac operator $\Delta$ is defined as follows:
$\Delta := D_1D_1+ D_2D_2$
Laplace's equation reads $\Delta U=0$, therefor:
$\frac{\partial^2 U}{\partial x_1^2}+\frac{\partial^2 U}{\partial x_2^2}=0$
We observe now $U$ in polar coordinates $r,\varphi$ in $\mathbb{R}^2$, therefor we observe
$u:[0,\infty)\times\mathbb{R}\to\mathbb{R}$ with $u(r,\varphi):=U(r\cos(\varphi),r\sin(\varphi))$
Show, that Laplace's equation takes the following form for $U$:
$\frac{\partial^2 u}{\partial r^2}+\frac1r\cdot\frac{\partial u}{\partial r}+\frac{1}{r^2}\cdot\frac{\partial^2 u}{\partial\varphi^2}=0$
I am aware, that this question was asked here befor, but they did NOT solve my questions on this problem. So I would like some help, to go over this problem.
Obviously we need to calculate $\frac{\partial u}{\partial r}, \frac{\partial^2 u}{\partial r^2}, \frac{\partial u}{\partial\varphi^2}$ using the chain rule.
As a hint it is given, that $\frac{\partial u}{\partial r}=\cos(\varphi)\frac{\partial U}{\partial\color{red} {x_1}}+\sin(\varphi)\frac{\partial U}{\partial\color{red}{ x_2}}$
Which I can not follow...
EDIT:
With the help of David Jaramillo I was able to resolve my problem.
Now I want to give:
$\frac{\partial^2 u(r,\varphi)}{\partial r^2}$
Therefor I have to give: $\frac{\partial}{\partial r}\left(\frac{\partial u(r,\varphi)}{\partial r}\right)=\frac{\partial}{\partial r}\left(\cos(\varphi)\frac{\partial U}{\partial x_1}+\sin(\varphi)\frac{\partial U}{\partial x_2}\right)$
Step by step, I want to give
$\frac{\partial}{\partial r} \left(\cos(\varphi)\frac{\partial U(x_1, x_2)}{\partial x_1}\right)$
and
$\frac{\partial}{\partial r} \left(\sin(\varphi)\frac{\partial U(x_1, x_2)}{\partial x_2}\right)$
[I made a mistake, which I edit now.]
I think you have a problem with the chain rule, to solve this kind of problems one usually does each component independently, and not using the matrix form
$$ \frac{\partial}{\partial r}=\frac{\partial x}{\partial r}\frac{\partial }{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}\\ \frac{\partial}{\partial \varphi}=\frac{\partial x}{\partial \varphi}\frac{\partial }{\partial x}+\frac{\partial y}{\partial \varphi}\frac{\partial}{\partial y} $$
from where you get the result.
In matric for the chain rule reads
$$ \left(\begin{array}{c} \frac{\partial}{\partial r}\\ \frac{\partial}{\partial \varphi}\end{array}\right)=\left(\begin{array}{cc} \frac{\partial x}{\partial r}& \frac{\partial y}{\partial r}\\ \frac{\partial x}{\partial \varphi} & \frac{\partial y}{\partial \varphi}\end{array}\right)\left(\begin{array}{c} \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\end{array}\right) $$
Which yields exactly the same result.