= Solution
{c}
For <separation of variables>, set $u=X(x)Y(y)$, with $X''/X=-Y''/Y=-\lambda$. The side <Neumann boundary conditions> give $X'(0)=X'(\pi)=0$. As in the standard <Sturm-Liouville problem>, <integration by parts> shows $\lambda\geq0$. The zero <eigenvalue> gives constant $X$, and $Y''=0$ together with $Y'(\pi)=0$ makes $Y$ constant too.
For positive <eigenvalues>, the side conditions give $\lambda=n^2$, $X_n=\cos(nx)$, $n\geq1$. The top condition chooses the <hyperbolic cosine> factor $\cosh(n(\pi-y))$. Normalize it to one at $y=0$, obtaining
$$
u(x,y)=A_0+\sum_{n=1}^\infty a_n\cos(nx)
\frac{\cosh(n(\pi-y))}{\cosh(n\pi)}.
$$
The <orthogonality of integer Fourier modes> and the permitted interchange of <summation> and <integration> give
$$
A_0=\frac1\pi\int_0^\pi x(\pi-x)\,dx=\frac{\pi^2}{6},
\qquad
a_n=\frac2\pi\int_0^\pi x(\pi-x)\cos(nx)\,dx.
$$
For the remaining <Fourier coefficients>, integrate by parts twice. The <sine> boundary term vanishes, and with $f=x(\pi-x)$,
$$
\begin{aligned}
\int_0^\pi f(x)\cos(nx)\,dx
&=-\frac1n\int_0^\pi f'(x)\sin(nx)\,dx\\
&=\frac{[f'(x)\cos(nx)]_0^\pi}{n^2}
-\frac1{n^2}\int_0^\pi f''(x)\cos(nx)\,dx\\
&=-\frac{\pi(1+(-1)^n)}{n^2}.
\end{aligned}
$$
The last <integral> vanishes because $f''=-2$ and $n$ is a positive <integer>. Thus $a_n=-2(1+(-1)^n)/n^2$: only even <Fourier modes> survive, as expected from <reflection symmetry> about $x=\pi/2$. Writing $n=2m$ gives
$$
\boxed{u(x,y)=\frac{\pi^2}{6}
-\sum_{m=1}^\infty\frac{\cos(2mx)}{m^2}
\frac{\cosh(2m(\pi-y))}{\cosh(2m\pi)}.}
$$
The factor involving <hyperbolic cosine> lies between zero and one on the closed <rectangle>. Comparison with $\sum m^{-2}$ proves <uniform convergence>, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\geq\varepsilon>0$, the <Fourier coefficients> have additional <exponential decay>, allowing differentiation and verifying <Laplace's equation> and all three <Neumann boundary conditions>.
To justify the endpoint calculation in <b>, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:
$$
u_y(x,0)=2\sum_{m=1}^\infty\frac{\tanh(2m\pi)}m\cos(2mx).
$$
Subtracting $1$ from the <hyperbolic tangent> leaves an absolutely convergent <series>. For the remaining part, the <power series> $-\log(1-z)=\sum_{m\geq1}z^m/m$ gives, by taking real parts at $z=re^{2ix}$ and then $r\uparrow1$,
$$
\sum_{m=1}^\infty\frac{\cos(2mx)}m=-\log(2\sin x),\qquad0<x<\pi.
$$
The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary <normal derivative> used in <b>. It also explains why the failure of the <corner compatibility condition for mixed boundary data> affects first <derivatives> at the bottom corners without spoiling continuity of $u$ itself.
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