Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 / 3 / a / i / Solution

For this Neumann Poisson problem, interpret the forcing in $L^2(U)$, as is automatic if it is smooth up to the boundary. Literal interior smoothness alone does not ensure the integrals or bounded functionals required in this question: for example, $f(x)=x^{-2}$ on $(0,1)$ is interior smooth but even $\int f\cdot 1$ diverges. Classical regularity in the converse is likewise understood up to the boundary.

Suppose the weak solution is smooth on $\overline{U}$. Testing against compactly supported test functions gives $-\Delta u=f$ in distributions and hence pointwise. Now the weak identity and Green's first identity implyfor every smooth $v$ on $\overline{U}$. Every smooth boundary function has such an extension, so ${\partial }_{n}u=0$ on $\partial U$. This proves both the interior equation and the boundary condition.

Conversely, for $u\in C^2(\overline{U})$ satisfying the equation and zero normal derivative, Green's first identity gives the weak identity for all smooth $v$ on $\overline{U}$. The density of smooth functions in a Sobolev space and the Cauchy-Schwarz inequality extend it continuously to every $v\in H^1(U)$. Thus the classical solution is a weak solution, and the smooth weak solution is classical.