Terminology — Question 10

Question 10 – Solution

Strategy. Separate existence compatibility from the constant ambiguity left by Neumann data.

Step 1: Integrate the equation before solving it. These are Neumann conditions. Since the square has area one and each edge has length one, the divergence theorem gives $$\boxed{s=\int_\Omega\Delta u\,dA =\int_{\partial\Omega}\partial_nu\,ds=g_L+g_R+g_B+g_T.}$$ In coordinates the derivatives are $-u_x$ on the left, $u_x$ on the right, $-u_y$ on the bottom and $u_y$ on the top. Replacing all by positive coordinate derivatives would give the wrong compatibility relation.

Step 2: Construct every compatible polynomial. For the proposed polynomial the four traces give $$c=-g_L,\quad 2a+c=g_R,\quad d=-g_B,\quad 2b+d=g_T.$$ Thus $$\boxed{u=\frac{g_R+g_L}{2}x^2+\frac{g_T+g_B}{2}y^2-g_Lx-g_By+C.}$$ Its Laplacian is the sum of the four edge values, exactly $s$ when compatible. Every boundary derivative has the prescribed value. This explicit construction proves sufficiency for constant source and constant edge data on this square; it is not a general sufficiency theorem for arbitrary domains and functions.

Step 3: Prove the exact remaining ambiguity. The difference $w$ of two solutions obeys $\Delta w=0$, $\partial_nw=0$. Green’s identity yields $$\int_\Omega|\nabla w|^2\,dA =\int_{\partial\Omega}w\partial_nw\,ds-\int_\Omega w\Delta w\,dA=0.$$ Continuity gives $\nabla w=0$ everywhere, and connectedness of the square makes $w$ constant. Conversely any added constant leaves all equations and data unchanged. A value or mean normalization is needed for actual uniqueness.

Step 4: Normalize the concrete field. Here $s=6$ and $u=x^2+2y^2+C$. Its integral is $1/3+2/3+C=1+C$. The zero-mean condition fixes $C=-1$, so $\boxed{u=x^2+2y^2-1}$ is unique among solutions with that normalization. For $s=0$, the integrated compatibility condition would require $0=6$, so no solution exists with these edge data.