As a vector calculus exercise, I want to solve following PDE $$\nabla^2 u = r^2$$ on a 2D domain $D = \{(r,\theta) : 1 \leq r \leq 2\}$ plus a condition that $u=1$ whenever $r=1$ and $r=2$.
There was a hint: consider the case $u(r,\theta) = u(r)$ first. So I followed it, using the polar laplacian formula: $$u_{xx} + u_{yy} = u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2}u_{\theta\theta}$$ and obtained the solution $$u(r) = \frac{r^4}{12} + A \log r + B$$ for some appropriate constant $A,B$. However, I have no idea how I can "use" this $\theta$-independent solution to get a general one. The course I am taking does not presuppose advanced tools such as delta function, Green's function, or Fourier transform. I am to solve this only by elementary means of vector calculus, but I am seriously stuck.
Thanks in advance for any form of help, hint, or solution.
If by solving a PDE, you mean finding a function $u$ defined on the annulus $D=\{ 1\le r\le 2\}$, then you are more or less done: by choosing appropriate $A, B$, one of the function
$$\tag{1} u(r, \theta) = u(r) = \frac{r^4}{12} + A \log r + B$$
satisfies the PDE and also the boundary condition.
If by solving the PDE, you mean finding ALL the solutions, then you need one more step. In this case it's the uniqueness: if $v(r, \theta)$ is another solution with the same boundary condition, then $w = v-u$ (with $u$ given in (1)) satisfies $$ \nabla^2 w = 0$$ and $w$ is zero on the boundary $\{ r=1\} \cup \{ r=2\}$. The maximum principle for harmonic function implies that $w$ is identically zero. Hence $u=v$ and thus the radially symmetric function $u$ you found is the only solution.