Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 107 / 1

Assume real bounded coefficients, with a principal part defining a uniformly elliptic operator, say $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ for some $\lambda >0$, and assume $c\le 0$. No smoothness of the boundary is needed for this classical weak maximum principle for elliptic operators. Its conclusion isIf $c=0$, the zero may be omitted. Only the symmetric part of the principal coefficients contributes to the Hessian matrix contraction.

For a strict inequality $Lu>0$, a positive interior maximum is impossible: there $Du=0$, $D^{2}u$ is negative semidefinite, and $cu\le 0$, so $Lu\le 0$. To reduce the non-strict case to this one, let $q(x)=e^{\kappa x_1}$. Boundedness of $b^1,c$ permits choosing $\kappa$ so large thatThus $L(u+\epsilon q)>0$. The preceding maximum argument and continuity on the compact closure giveLet $\epsilon \downarrow 0$; the exponential is bounded on the bounded domain. This is the exponential perturbation proof of the weak maximum principle with drift, with the zeroth-order term included. For $c=0$, any interior maximum, of either sign, is impossible for the perturbed function, giving the sharper boundary bound. The sign restriction on $c$ is essential: a positive first Dirichlet Laplacian eigenfunction violates the conclusion for $L=\Delta +{\lambda }_1$.

Apply the weak maximum principle for elliptic operators to $q=|Dw{|}^2$. The product rule and the equation for the harmonic function giveThe function $q$ is twice differentiable inside and continuous on the closure, by the prescribed regularity. Hence ${\sup }_{\Omega }q\le {\sup }_{\partial \Omega }q$. Conversely, each boundary value of $q$ is a limit of values at interior points, so the reverse inequality holds for the suprema. Taking square roots proves the gradient maximum principle for harmonic functions:This argument controls the length of the entire gradient, without comparing its separate components.

By continuity on the compact closure, $v$ has a maximum and minimum. If its maximum were greater than one, the boundary hypothesis would force that maximum to occur at an interior point $x_0$. There $\Delta v(x_0)\le 0$, whereasa contradiction. If its minimum were less than minus one, it would again be interior, where $\Delta v\ge 0$ but $v^3-v<0$. ThereforeThis is a direct maximum/minimum argument for a nonlinear elliptic boundary value problem; treating its nonlinearity as a zeroth-order coefficient of fixed nonpositive sign would not be justified.