Use signs to bound elliptic solutions

Try

A function with positive second derivative on an interval cannot have a strict interior maximum. In several dimensions, the trace of the Hessian replaces that second derivative. At an interior maximum of a C² function, the Hessian is negative semidefinite, so its trace is nonpositive.

Use this observation to predict where a function satisfying Δu>0 can achieve its maximum on a bounded domain closure. The non-strict inequality Δu≥0 needs a small additional argument.

Predict

For −u″=2 on [0,1] with zero endpoints, predict whether u is positive or negative inside. Track the minus sign before appealing to a maximum principle. A wrong sign would claim that a concave arch lies below its endpoints.

Experiment

Direct integration gives u=x(1−x). Its second derivative is −2, and its maximum is 1/4 at the midpoint.

Statement	Correct consequence
Δu≥0	Maximum bounded by boundary maximum
Δu≤0	Minimum bounded by boundary minimum
−Δu≥0 with zero boundary	u≥0

Use u and −u to test each row in one dimension. These statements require a classical function inside and continuity on a bounded domain’s closure; the extrema are then attained on that compact closure.

Understand

Suppose Δu≥0. For ε positive, define v=u+ε|x|². Its Laplacian is strictly positive, so it cannot attain its maximum at an interior point. Hence u(x)+ε|x|² is at most the maximum of u+ε|x|² on the boundary. Boundedness controls the perturbation uniformly, and letting ε decrease to zero gives the weak maximum principle. Applying it to −u gives the minimum version.

For two functions with −Δu≤−Δv and u≤v on the boundary, set w=u−v. Then Δw≥0 and w≤0 on the boundary, so w≤0 everywhere. This comparison argument proves uniqueness of a classical Dirichlet Poisson solution by applying it in both directions to solutions with identical source and boundary values.

It also constructs useful bounds. If −Δu=f on a ball of radius R with zero boundary and 0≤f≤M, compare u with v=M(R²−|x|²)/(2d), where d is the dimension. Since −Δv=M and v vanishes on the sphere, 0≤u≤v. The bound follows from a calculated comparison function, not from pretending the forcing is zero.

On unbounded domains, a maximum may escape to infinity, so boundary information alone can be insufficient. Additional growth or boundedness conditions are needed. Likewise, more general elliptic operators require ellipticity and appropriate lower-order coefficient signs; the Laplacian proof does not automatically cover every operator bearing second derivatives.

Apply

Construct the quadratic comparison bound for a disk of radius 2 with source between 0 and 3. Calculate its center value and verify the Laplacian. Then explain why a numerical solution exceeding that bound would indicate a modeling, boundary or discretization issue under these assumptions. A numerical violation does not invalidate the theorem unless its hypotheses actually hold for the computed problem.

Continue

Separation of variables provides explicit harmonic solutions on rectangles and disks. Use the maximum principle as an independent check on their signs and magnitudes.