Poisson equation in polar coordinates

Question

I am having trouble finding a solution to the poisson equation: $\frac{∂^2u}{∂r^2}+\frac{1}{r}\frac{∂u}{dr}+\frac{1}{r^2}\frac{∂^2u}{∂θ^2}=-r^2\sin2\theta$ with $a<r<b$, $-π<θ<π$ and $\frac{∂u}{∂r}(a,θ) = \frac{∂u}{∂r}(b,θ) = 0$. I have tried assuming that the solution is $(kr^4+lr^3+mr^2+nr+p)\sin2\theta$ and replacing in the equation but this solution doesn't satisfy the boundary conditions.

Answer 1

Your assumption $\quad (kr^4+lr^3+mr^2+nr+p)\sin2\theta\quad $ is too restrictive. Assume : $$u=f(r)\sin(2\theta)$$ Putting it into $\quad \frac{∂^2u}{∂r^2}+\frac{1}{r}\frac{∂u}{dr}+\frac{1}{r^2}\frac{∂^2u}{∂θ^2}=-r^2\sin2\theta\quad $ leads to : $$f''+\frac{1}{r}f'-\frac{4}{r^2}f=-r^2$$ The solution of this linear ODE is : $$f(r)= -\frac{1}{12}r^4+c_1r^2+c_2r^{-2}$$ $$u(r,\theta)=\left(-\frac{1}{12}r^4+c_1r^2+c_2r^{-2}\right)\sin(2\theta)$$ $$\frac{∂u}{dr}(r,\theta)=\left(-\frac{1}{3}r^3+2c_1r-2c_2r^{-3}\right)\sin(2\theta)$$ The conditions $\quad\frac{∂u}{∂r}(a,θ) = \frac{∂u}{∂r}(b,θ) = 0\quad$ implies : $$\begin{cases} -\frac{1}{3}a^3+2c_1a-2c_2a^{-3}=0 \\ -\frac{1}{3}b^3+2c_1b-2c_2b^{-3}=0 \end{cases}$$ To be solved for $c_1$ and $c_2$.