 Solution

This is Laplace's equation. Zero is not unique: with $k=\pi /2$,is nonzero and harmonic. It vanishes on $y=0$ and $x=0$, while $u_x(1,y)=k\cos (k)\sinh (ky)=0$.

A sufficient additional condition is $u(x,1)=0$ on the top edge. Green's first identity then givesOn the bottom, left and top edges $u=0$; on the right edge the normal derivative is zero. Thus the gradient vanishes and $u$ is constant; the bottom value makes that constant zero. For $C^2(D)\cap C^1(\overline{D})$, apply the identity on inset rectangles and pass to the boundary using continuity of the function and its gradient.