Three-dimensional Laplace equation solutions

Question

I want to find all the solutions $u=f(r)$ to the three-dimensional Laplace equation $$u_{xx}+u_{yy}+u_{zz}=0$$ where $r=\sqrt{x^2+y^2+z^2}$

I started by finding $u_{xx}$, $u_{yy}$ and $u_{zz}$ using the chain rule:

$u_{xx}=u_{rr}*\frac{x^2}{x^2+y^2+z^2}+u_{r}*\frac{y^2+z^2}{(x^2+y^2+z^2)^{3/2}}$

Is this correct? I was confused when computing $\frac{\partial}{\partial x}(u_{r}*\frac{x}{\sqrt{x^2+y^2+z^2}})$.

If it is correct, then I went on to plug in my results (finding $u_{yy}$ and $u_{zz}$ by comparing them to $u_{xx}$).

I get $r^2u_{rr}+2ru_{r}=0$ which is a Cauchy-Euler ordinary differential equation. By solving, I find $$u(r)=c_{1}+c_{2}r$$ Since $r$ itself is a function of $x$, $y$ and $z$, should $c_{1}$ and $c_{2}$ also be functions of $x$,$y$ and $z$ as opposed to constants? (In a normal Cauchy-Euler ODE, $r$ wouldn't depend on other variables)

Answer 1

Your result $\quad r^2u_{rr}+2ru_r=0\quad$ is correct.

This can be directly obtained from the Laplace's equation in spherical coordinates : https://en.wikipedia.org/wiki/Laplace%27s_equation

In the case of $u$ function of $r$ only : $$\Delta u(r,\varphi,\theta)=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r} \right)=0 \quad \text{if}\quad \frac{\partial u}{\partial \varphi} = \frac{\partial u}{\partial \theta}=0$$ $$\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r} \right)=\quad r^2u_{rr}+2ru_r=0$$ Since there is only one variable $r$, the PDE is reduced to an ODE $$r\frac{d^2u}{dr^2}+2\frac{du}{dr}=0$$ The solution is not $\quad u(r)=c_1+c_2r\quad$ (mistake in your last equation), but is : $$u(r)=c_1+\frac{c_2}{r}$$