Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 2 / c / Solution

If two solutions exist, their difference $w$ satisfies ${\nabla }^{2}w-w=0$ inside $B$ and $w=0$ on the boundary. Suppose $w$ has a positive value. Continuity on the closed ball gives a positive maximum at an interior point, where the Hessian matrix is negative semidefinite and hence ${\nabla }^{2}w\le 0$. But the equation gives ${\nabla }^{2}w=w>0$, a contradiction. A negative minimum similarly contradicts ${\nabla }^{2}w=w<0$.

Thus $\boxed{w=0}$, proving uniqueness by the maximum principle for the modified Helmholtz equation. Here a classical solution is understood to be $C^2$ inside and continuous up to the boundary; mere first differentiability would not define the displayed Laplacian classically. With enough boundary regularity, the alternative Green's first identity proof is $0={\int }_{B}w({\nabla }^{2}w-w)=-{\int }_B(|\nabla w{|}^2+w^2)$.