Let be two sufficiently regular solutions with the same source and boundary data, and set . Then and . Green's first identity gives
The left side is nonnegative, so it vanishes. Thus is constant on each connected component. Its normal derivative is zero, and the boundary condition then forces that constant to be zero. This proves at most one solution. The Poisson uniqueness with a nonnegative Robin normal coefficient proof never divides by , so it includes the points or boundary portions where .
For the exponential-sine modes, direct differentiation gives
Their sum is zero, so both modes are harmonic functions for every real .
For the square, the horizontal Dirichlet data suggest separation of variables in the form , where the positive integer makes the sine vanish at both horizontal edges. The Laplace equation gives , hence . At , the outward normal points in the negative direction. The Robin boundary condition is therefore , or . The right edge requires , fixing the remaining coefficient. The resulting separated Laplace mode with a Robin edge is
The denominator is positive for . The function satisfies the Laplace equation and all four boundary conditions directly; for example its left-edge value and derivative agree, giving the correct outward-normal Robin sign. It extends regularly to the corners.
On the left edge the uniqueness coefficient is ; on the other three edges it is 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 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.