= Solution
The <Fourier coefficients> satisfy $a_0=a_n=0$ and
$$
b_n=\frac1\pi\left(\int_0^\pi\sin(n\theta)\,d\theta-\int_\pi^{2\pi}\sin(n\theta)\,d\theta\right)
=\frac{2[1-(-1)^n]}{\pi n}.
$$
Thus the <square wave> has <Fourier series>
$$
\boxed{f(\theta)\sim\frac4\pi\sum_{\substack{n\ge1\\n\text{ odd}}}\frac{\sin(n\theta)}n}.
$$
It equals $f$ away from its <jump discontinuities>, and converges to $0$ at the jumps. The assigned endpoint values do not affect its <Fourier coefficients>.
For the <Poisson equation>, use <separation of variables> $\phi=\sum_{n\text{ odd}}b_nR_n(r)\sin(n\theta)$. The radial equation is $r^2R_n''+rR_n'-n^2R_n=r^2$. Its bounded solution with zero <Dirichlet boundary condition> at $r=1$ is
$$
R_n(r)=\frac{r^n-r^2}{n^2-4}.
$$
Indeed the negative power is excluded at the centre, and the boundary fixes the coefficient of $r^n$. No resonance at $n=2$ occurs because only odd modes are forced. Therefore
$$
\boxed{\phi(r,\theta)=\frac4\pi\sum_{\substack{n\ge1\\n\text{ odd}}}
\frac{r^n-r^2}{n(n^2-4)}\sin(n\theta)}.
$$
This is a <Poisson equation on a disk with angular forcing>. The series converges uniformly on the closed disk: for $n\ge3$ each term is bounded by a constant times $n^{-3}$. Its first spatial derivatives also converge uniformly. For instance the radial derivative and $r^{-1}$ times the angular derivative of each term are bounded by constants times $n^{-2}$; the $n=1$ harmonic term is smooth in Cartesian coordinates. It defines a function with zero boundary data and the required <weak formulation>. Applying the <Laplacian> to finite partial sums gives the <Fourier partial sums> of $f$, which converge in $L^2$; passing to the limit against smooth compactly supported test functions verifies the <Poisson equation>. It holds classically off the jump rays and the centre, where the forcing is locally constant. It cannot be interpreted as a global classical equation with continuous second derivatives for discontinuous forcing. Uniqueness follows from the zero-boundary <Dirichlet energy> identity for the difference of two solutions.
Finally <orthogonality> gives $\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta=\pi b_n$. The radial integral is
$$
\int_0^1R_n(r)r\,dr
=\frac{1}{n^2-4}\left(\frac1{n+2}-\frac14\right)
=-\frac1{4(n+2)^2}.
$$
Uniform convergence of $\phi$ and boundedness of $f$ justify integration term by term, so
$$
\boxed{\int_{r\le1}f\phi\,dA
=-\frac4\pi\sum_{\substack{n\ge1\\n\text{ odd}}}\frac1{n^2(n+2)^2}}.
$$
The negative sign also agrees with $\int\phi\Delta\phi=-\int|\nabla\phi|^2$. The original PDF's radial denominator is $n^2-4$; the converted TeX's $n^2-1$ is a transcription error.
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