= Solution
Let $\phi_1,\phi_2$ be two sufficiently regular solutions with the same source and boundary data, and set $w=\phi_1-\phi_2$. Then $\nabla^2w=0$ and $\alpha\partial_nw+w=0$. <Green's first identity> gives
$$
\int_V|\nabla w|^2\,dV
=\int_{\partial V}w\,\partial_nw\,dS
=-\int_{\partial V}\alpha(\partial_nw)^2\,dS\le0.
$$
The left side is nonnegative, so it vanishes. Thus $w$ is constant on each connected component. Its <normal derivative> is zero, and the <boundary condition> then forces that constant to be zero. This proves \b[at most one solution]. The <Poisson uniqueness with a nonnegative Robin normal coefficient> proof never divides by $\alpha$, so it includes the points or boundary portions where $\alpha=0$.
For the exponential-sine modes, direct differentiation gives
$$
\partial_x^2(e^{\pm\ell x}\sin\ell y)=\ell^2e^{\pm\ell x}\sin\ell y,
\qquad
\partial_y^2(e^{\pm\ell x}\sin\ell y)=-\ell^2e^{\pm\ell x}\sin\ell y.
$$
Their sum is zero, so both modes are <harmonic functions> for every real $\ell$.
For the square, the horizontal Dirichlet data suggest <separation of variables> in the form $\phi(x,y)=F(x)\sin(ky)$, where the positive integer $k$ makes the sine vanish at both horizontal edges. The <Laplace equation> gives $F''-k^2F=0$, hence $F=A\cosh(kx)+B\sinh(kx)$. At $x=0$, the outward normal points in the negative $x$ direction. The <Robin boundary condition> is therefore $F(0)-F'(0)=0$, or $A=kB$. The right edge requires $F(\pi)=1$, fixing the remaining coefficient. The resulting <separated Laplace mode with a Robin edge> is
$$
\boxed{\phi(x,y)=\frac{k\cosh(kx)+\sinh(kx)}{k\cosh(k\pi)+\sinh(k\pi)}\sin(ky).}
$$
The denominator is positive for $k>0$. The function satisfies the <Laplace equation> and all four <boundary conditions> directly; for example its left-edge value and $x$ derivative agree, giving the correct outward-normal Robin sign. It extends regularly to the corners.
On the left edge the uniqueness coefficient is $\alpha=1$; on the other three edges it is $\alpha=0$ with the prescribed Dirichlet value. The same <Green's first identity> argument applies to the square's piecewise smooth boundary, whose corners have zero boundary measure. Thus \b[this is the unique regular solution], or equivalently the unique finite-energy solution with these boundary traces. Unbounded corner singularities are outside that boundary-value class.
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