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Given the Poisson's Equation

$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=x,\, r\lt 3$

I want to solve the homogeneous part (Laplace's Equation) first and rewrite it in polar coordinate and as far as i get is

$\dfrac{\partial^2 u}{\partial r^2}+\dfrac{1}{r} \dfrac{\partial u}{\partial r}+\dfrac{1}{r^2} \dfrac{\partial^2 u}{\partial \theta^2}=0$

But when i see an example in a textbook, one of them assume that the solution is radially symmetric and only depend on $r$, $u=u(r)$. So it changed to

$\dfrac{\partial^2 u}{\partial r^2}+\dfrac{1}{r} \dfrac{\partial u}{\partial r}=0$

And on another reference says that the Laplace's Equation in polar coordinate is

$\dfrac{1}{r} \dfrac{\partial}{\partial r}\left(r \dfrac{\partial u}{\partial r}\right)+\dfrac{1}{r^2} \dfrac{\partial^2 u}{\partial \theta^2}=0$

Actually it's make me confused. I just learned a few days, so I don't really understand. So, how do i solve the homogeneous part?

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    $\begingroup$ It depends whether your problem is defined on a bounded set or not. If $\Omega=\mathbb{R}^n$ yo can assume the solution is radially symmetric (in your case $n=2$). But if $\Omega$ is a bounded set, you have take into account the angular part too. $\endgroup$ – CarlIO May 11 '19 at 13:43
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Note that $$\partial_r (r\partial_r u)=\partial_r u+r \partial^2_r u$$ So the 2 forms are equivalent. And you can assume that the solution has the form of $$u(r, \theta)=R(r)\Theta(\theta)$$ Which will separate your PDE into 2 ODE. After that, the general solution will be the linear combination of all possible solutions.

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