= Solution
{c}
Let $u_1,u_2$ be two solutions with the same data, and write $w=u_1-u_2$ on the rectangle $\Omega=(0,\pi)\times(-1,1)$. By <linearity>,
$$
\Delta w=0,\qquad
w(x,-1)=w(x,1)=0,\qquad
w_x(0,y)=w_x(\pi,y)=0.
$$
The <normal derivative> is $-w_x$ on the left edge and $w_x$ on the right edge, so it vanishes on both. The <energy method> applies <Green's first identity>, which follows by integrating
$$
\nabla\cdot(w\nabla w)=|\nabla w|^2+w\Delta w.
$$
It gives
$$
\iint_\Omega|\nabla w|^2\,dA
=\int_{\partial\Omega}w\,\partial_nw\,ds
-\iint_\Omega w\Delta w\,dA=0.
$$
The boundary term vanishes on each horizontal edge because $w=0$, and on each vertical edge because $\partial_nw=0$. The interior term vanishes because $w$ is a <harmonic function>. Since the continuous integrand $|\nabla w|^2$ is nonnegative, a zero <integral> forces $\nabla w=\mathbf0$. The rectangle is <connected>, so $w$ is constant; its zero value on either horizontal edge makes that constant zero. Hence
$$
\boxed{u_1=u_2}.
$$
This proof assumes enough boundary regularity for <Green's first identity>. It also works in the finite <Dirichlet energy> class using the weak identity with test function $w$, so it remains applicable when corner derivatives are not continuous. It proves uniqueness conditional on existence and does not impose a compatibility condition for arbitrary $f,g$.
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