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A nondivergence-form elliptic operator applies its principal coefficients directly to the Hessian matrix: $Lu=a^{ij}{D}_{ij}u+b^i{D}_{i}u+cu$. Only the symmetric part of $a$ contributes to the second-derivative term. Uniform ellipticity bounds its quadratic form below by $\lambda |\xi {|}^2$ with a constant $\lambda >0$ independent of position.

Let $u,v$ be two critical points with the same classical boundary values, and put $w=u-v$. Since $F$ is independent of $z$, the Euler-Lagrange equation for each is $D_i{F}_{p_i}(x,Du)=0$. By the fundamental theorem of calculus along the segment between the gradients,Subtracting the equations gives $D_i(A_{ij}{D}_{j}w)=0$. The uniformly convex variational integrand makes $A$ symmetric and uniformly elliptic with the same positive lower bound. Since $F$ is smooth and $u,v\in C^2(\overline{\Omega })$, $A$ is continuously differentiable locally in $\Omega$, sois a classical nondivergence-form elliptic operator with no zeroth-order term.

To avoid assuming that derivatives of $F$ stay bounded all the way to the boundary, apply the proved weak maximum principle on inner open sets ${\Omega }_{\delta }=\{x\in \Omega :\mathrm{dist}(x,\partial \Omega )>\delta \}$. Their closures lie compactly in $\Omega$, so all linearized coefficients there are bounded. The maximum principle applies on each component and yieldsEvery point of the inner boundary is at distance $\delta$ from $\partial \Omega$. Since $w$ is continuous on the compact closure and equals zero on that boundary, the right-hand side tends uniformly to zero as $\delta \downarrow 0$. Every fixed interior point eventually belongs to ${\Omega }_{\delta }$, provingThis establishes uniqueness for uniformly convex gradient Dirichlet problems using precisely the boundary continuity and ellipticity needed for the comparison.

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 $M+o(1)$.

A second-order nondivergence-form elliptic operator on a domain $D\subseteq {\mathbb{R}}^d$ has the formOnly the symmetric part of $a$ affects the Hessian matrix term, so one may take $a(x)$ symmetric. The operator is a uniformly elliptic operator if there is a constant $\kappa >0$, independent of $x$, such thatA common bounded-coefficient convention additionally requires an upper bound $\sum a_{jk}(x){\xi }_j{\xi }_k\le \Lambda |\xi {|}^2$ with $\Lambda <\infty$. The lower bound says that the principal quadratic form cannot degenerate in any direction or at any location.

For an explicit stochastic example, standard $d$-dimensional Brownian motion has infinitesimal generatorIts principal quadratic form is $|\xi {|}^2/2$, so it is uniformly elliptic with $\kappa =\Lambda =1/2$. More generally, a sufficiently regular stochastic differential equation $d{X}_t=b(X_t)dt+\Sigma (X_t)d{B}_t$ has diffusion generator $b\cdot \nabla +\frac{1}{2}(\Sigma {\Sigma }^{\mathsf{T}}):D^2$; it is uniformly elliptic when $\Sigma {\Sigma }^{\mathsf{T}}$ has a uniformly positive lower bound. A conservative diffusion process has no zeroth-order term, since its infinitesimal generator annihilates constants.