Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 5 / b

Use the classical regularity setting $u\in C^2(D)\cap C(\overline{D})$, with $D$ bounded. Continuity on the compact set $\overline{D}$ guarantees existence of a minimum. The inequality $\Delta u=f\le 0$ means that $u$ is superharmonic function.

A strict inequality $\Delta u<0$ would immediately exclude an interior minimum, because its Hessian matrix there would be positive semidefinite and have nonnegative trace. To handle equality, introduce the quadratic perturbation proof of the minimum principle:Its minimum must lie on $\partial D$. Choose $R$ so that $x^2+y^2\le R^2$ on $\overline{D}$. For any $p\in D$,Letting $\epsilon \downarrow 0$ gives $u(p)\ge {\min }_{\partial D}u$. ThereforeThis asserts that a minimum is attained on the boundary, not that it cannot also be attained inside; a constant function attains it everywhere.

For a harmonic function, $\Delta u=0$ and $\Delta (-u)=0$. Apply the preceding minimum argument to $u$ and to $-u$:These are the weak minimum and maximum principle for harmonic functions.

The difference $w=v-u$ is a harmonic function, because the Laplacian is a linear differential operator:The boundary inequality says $w\ge 0$ on $\partial D$. The minimum principle from part (ii) givesThis is the comparison principle for harmonic functions. Applying it in both directions also shows why fixed Dirichlet boundary conditions determine at most one harmonic function.