Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 9 / 2

For a bounded domain, the nondivergence-form elliptic operator obeys the weak maximum principle: $Lu\ge 0$ and $c\le 0$ implyHere $u^{+}=\max \{u,0\}$. Boundedness of $\Omega$ is needed for this formulation: it is not explicitly imposed in this question. On an unbounded domain one needs an appropriate condition at infinity instead. Without either condition the claim is false, as $u(x)=x_n$ and $L=\Delta$ on the half-space $\{x_n>0\}$ show.

For the bounded-domain proof choose $\gamma >0$ so large that $\lambda {\gamma }^2-\Vert b^1{\Vert }_{\infty }\gamma -\Vert c{\Vert }_{\infty }>0$. The positive exponential $h(x)=e^{\gamma x_1}$ then satisfiesLet $u_{\epsilon }=u+\epsilon h$, $M={\sup }_{\partial \Omega }{u}^{+}$ and $H={\sup }_{\overline{\Omega }}h<\infty$. If $u_{\epsilon }$ had a value greater than $M+\epsilon H$, it would attain a positive maximum at an interior point. There its gradient vanishes and its Hessian matrix is negative semidefinite. The symmetric part of $(a^{ij})$ is positive-definite by uniform ellipticity, so $a^{ij}{D}_{ij}{u}_{\epsilon }\le 0$; also $c{u}_{\epsilon }\le 0$. Hence $L{u}_{\epsilon }\le 0$, contradicting $L{u}_{\epsilon }=Lu+\epsilon Lh>0$. Thus $u_{\epsilon }\le M+\epsilon H$. Letting $\epsilon \downarrow 0$ proves the weak maximum principle, without requiring continuity of the coefficients.

For example, the unbounded-domain version follows by exhausting the domain with bounded truncations if ${\mathrm{limsup}}_{|x|\to \infty ,x\in \Omega }u(x)\le M$: the added spherical boundaries then have supremum at most $M+o(1)$.

In the bounded-domain setting of the weak maximum principle, let $u_1,u_2$ be two classical solutions with the same forcing and boundary values. Their difference $z=u_1-u_2$ satisfies $Lz=0$ and $z=0$ on the boundary. Apply the weak maximum principle first to $z$ and then to $-z$:Thus $u_1=u_2$, proving uniqueness of the Dirichlet problem. The same proof works on an unbounded domain if the difference and its negative satisfy the required condition at infinity. As literally printed without either restriction, uniqueness need not hold: on $\{x_n>0\}$ the two harmonic functions $0$ and $x_n$ have the same zero boundary values.

On the unit ball take the uniformly elliptic operator $L=\Delta -2n$ and the smooth function $u(x)=-1-|x{|}^2$. The zeroth-order coefficient is negative andNeverthelessThis is consistent with the weak maximum principle, whose upper bound here is ${\sup }_{\partial B_1}{u}^{+}=0$. At a negative interior maximum the term $cu$ can be positive; this is why a negative boundary maximum cannot replace its positive part. If $c=0$, subtracting the boundary maximum preserves the differential inequality and gives the stronger bound. That subtraction is unavailable for general $c\le 0$.