Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 12 (Elliptic Partial Differential Equations)

Take a compactly supported smooth variation $\phi \in C_c^{\infty }(\Omega )$ and set $u_t=u+t\phi$. Its support is away from the boundary, so it preserves any prescribed boundary values. Differentiating the functional at $t=0$ gives the first variationBy integration by parts, there is no boundary term. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationExpanding the total derivative, an equivalent classical form isRepeated indices are summed. In the last term involving $x_i$, the derivative holds $z,p$ fixed; the other terms account for their dependence on $u(x),Du(x)$. Compactly supported variations derive the interior equation without imposing a natural boundary condition.

The weak maximum principle for elliptic operators gives the clipped boundary boundsIf $c=0$, constants solve the equation, and applying these bounds after subtracting the boundary maximum or minimum gives the sharper conclusion that both extrema occur on the boundary. For $c<0$, the clipping by zero is necessary in the general statement.

To prove the upper bound, write $B=\Vert b_1{\Vert }_{\infty }$ and $C_0=\Vert c{\Vert }_{\infty }$, and choose $\alpha >0$ large enough that $\lambda {\alpha }^2-B\alpha -C_0>0$. For $\varphi (x)=e^{\alpha x_1}$, uniform ellipticity givesPut $u_{\epsilon }=u+\epsilon \varphi$. Then $L{u}_{\epsilon }>0$. If $u_{\epsilon }$ had a positive maximum at an interior point, its gradient would vanish there and its Hessian matrix would be negative semidefinite. Only the symmetric part of $a_{ij}$ contracts with this Hessian, and its positive definiteness implies $a_{ij}{D}_{ij}{u}_{\epsilon }\le 0$. Also $c{u}_{\epsilon }\le 0$. These facts would give $L{u}_{\epsilon }\le 0$, a contradiction.

Since $\overline{\Omega }$ is compact and $u_{\epsilon }$ is continuous on it, its maximum exists. It is therefore either nonpositive or attained on the boundary, proving ${\sup }_{\Omega }{u}_{\epsilon }\le \max \{0,{\sup }_{\partial \Omega }{u}_{\epsilon }\}$. The exponential is bounded on the bounded domain; letting $\epsilon \downarrow 0$ proves the upper estimate. Apply the same argument to $-u$ for the lower estimate. This is the exponential perturbation proof of the weak maximum principle with drift and requires no continuity of the coefficient functions.

On the bounded interval $\Omega =(0,\pi )$, take $L=d^2/d{x}^2+1$ and $u(x)=\sin x$. Its principal coefficient is $1$, so the operator is uniformly elliptic, andThusThe positive zeroth-order coefficient $c=1$ permits a zero-boundary eigenfunction, demonstrating failure of the weak maximum principle with a positive zeroth-order coefficient. The function $-\sin x$ similarly violates the lower bound.

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.

No: the conclusion need not survive without an additional restriction at infinity. A counterexample that keeps the functional finite is $\Omega ={\mathbb{R}}^n$ and $F(p)=\frac{1}{2}|p{|}^2$. The distinct functions $u\equiv 0$ and $v\equiv 1$ both lie in $C^2(\overline{\Omega })$ and have energy zero. Their boundary data agree because $\partial \Omega$ is empty. For any variation with finite gradient energy,so both are genuine critical points, not merely formal solutions with infinite functional value.

With nonempty boundary, the classical Euler-Lagrange equation also illustrates the missing condition: in dimension $n\ge 3$, on $\Omega =\{|x|>1\}$, both $0$ and $1-|x{|}^{2-n}$ are harmonic functions with zero boundary values. The latter has finite Dirichlet energy, because its gradient-energy integral is proportional to ${\int }_1^{\infty }{r}^{1-n}dr$, and is stationary under compactly supported variations. It approaches $1$ at infinity while the zero solution approaches $0$.

This is nonuniqueness of Dirichlet data without control at infinity. For a global variational formulation one must specify the admissible variations: the exterior example is not critical against every finite-energy zero-boundary variation, since varying it in its own direction changes its energy to first order. Prescribing an appropriate limit at infinity, or a variational space in which the difference is an admissible zero-boundary test function and constants are excluded, can restore uniqueness.

Write $H^1=W^{1,2}$ and $H_0^1=W_0^{1,2}$. The Dirichlet boundary condition in the Sobolev space sense means $u-\psi \in H_0^1(\Omega )$; this definition remains meaningful even without regularity assumptions on the boundary. For bounded full coefficients, a weak solution is a function satisfyingwhereThe sign follows by integration by parts in $D_i(a_{ij}{D}_{j}u)+qu=f$. No classical derivatives of the measurable coefficients are taken. Equivalently one may first test against $C_c^{\infty }(\Omega )$ and extend by density, provided this bilinear form is bounded on $H^1\times H_0^1$. All integrals in this definition must exist: as explained below, the printed quadratic bounds alone do not guarantee this when $A=(a_{ij})$ is nonsymmetric.

There is a qualification in the printed hypotheses: quadratic ellipticity does not bound a nonsymmetric coefficient matrix. The intended standard existence proof needs $a_{ij}\in L^{\infty }(\Omega )$, or more generally a bounded principal bilinear form. To see why this does not follow from the displayed inequalities, take $\Omega =(-1,1{)}^2$ andThen ${\zeta }^{T}A(x)\zeta =|\zeta {|}^2$ for every $x,\zeta$. However, choose $u_0,{\phi }_0\in C_c^{\infty }(\Omega )$ equal to $x_2,x_1$, respectively, on a small rectangle about the origin. There $(AD{u}_0)\cdot D{\phi }_0=k(x_1)$, whose positive part has infinite integral. Thus the standard principal pairing is not a finite Lebesgue integral even for these smooth test functions. This demonstrates the missing form hypothesis, rather than assuming that all unbounded skew coefficients prevent solvability.

Under the intended bounded-coefficient hypothesis the proof is as follows. On $H=H_0^1(\Omega )$ use the Hilbert space norm $\Vert Dw{\Vert }_2$, equivalent to the $H^1$ norm by the Poincaré inequality on a bounded domain. The form from part (i) is bounded, sinceIt is a coercive bilinear form even without symmetry:Here the nonpositive sign of $q$ is essential. With $u=\psi +w$, the required equation becomesThe Cauchy-Schwarz inequality, bounded coefficients and the Poincaré inequality show that $\ell$ is a bounded linear functional on $H$. The Lax-Milgram theorem states that a bounded coercive bilinear form on a real Hilbert space represents each bounded linear functional in this way with a unique $w\in H$. It therefore yields $u=\psi +w$ with the required boundary condition and weak equation. For completeness, if $u_1,u_2$ are solutions, $z=u_1-u_2\in H_0^1$ satisfies $B(z,z)=0$; coercivity and the Poincaré inequality imply $z=0$.

With the missing bounded-form hypothesis, the weak solution exists and is unique. For nonsymmetric matrices the printed quadratic inequalities omit the hypothesis needed for this standard formulation and proof. No maximum principle is used. In part (iii), symmetry makes the quadratic upper bound a bound on the whole matrix, so no such qualification is needed there.

Symmetry implies that all eigenvalues of $A(x)$ lie between $\lambda$ and $\Lambda$, so $\Vert A(x){\Vert }_{\mathrm{op}}\le \Lambda$. In particular the bilinear form $B$ from part (i) is bounded and symmetric under the printed assumptions. The symmetric elliptic Dirichlet energy with a nonpositive potential isAll its terms are finite. For an admissible variation $\phi \in H_0^1$, its first variation is $B(u,\phi )+\int f\phi$. Thus stationarity is exactly the weak formulation of $Lu=f$.

Here is the direct method in the calculus of variations. Put $u=\psi +w$. Expanding the symmetric quadratic form givesThe Poincaré inequality controls $\Vert w{\Vert }_2$ by $\Vert Dw{\Vert }_2$. Therefore $J$ is bounded below on the admissible class, and a minimizing sequence is bounded in $H^1$. By weak compactness in this reflexive Banach space, a subsequence converges weakly to $u$. The affine closed subspace $\psi +H_0^1$ is weakly closed, so $u$ remains admissible.

The symmetric square root $A^{1/2}(x)$ is measurable and bounded. Since $-q\ge 0$, the two bounded linear mapstake weak $H^1$ convergence to weak $L^2$ convergence. The weak lower semicontinuity of the Hilbert norm shows that $B(u,u)$ is weakly lower semicontinuous. The term $\int fu$ is weakly continuous. Hence $J(u)\le \mathrm{liminf}J(u_k)$ and $u$ attains the minimum. Varying in either sign yields the weak equation.

The sign $q\le 0$ was used both to make the potential energy nonnegative, hence weakly lower semicontinuous by this argument, and to obtain coercivity. Finally $J(u+z)-J(u)=B(z,z)/2\ge \lambda \Vert Dz{\Vert }_2^2/2$ for $z\in H_0^1$, since the first variation at $u$ vanishes. Therefore

Take $\Omega =(0,\pi )$, $a_{11}=1$, $q=1$, $\psi =0$ and $f(x)=\sin x$. The coefficients are bounded and the operator is uniformly elliptic. If a weak solution $u\in H_0^1(0,\pi )$ existed, testing its weak formulation with $\phi =\sin x\in H_0^1(0,\pi )$ would giveBy integration by parts for $H_0^1$ functions, the first integral on the left is ${\int }_0^{\pi }u\sin x$, so the left side is zero. The right side equals $-\pi /2$, a contradiction. ThusThe forcing is not orthogonal to the homogeneous zero-boundary eigenfunction $\sin x$. This is the obstruction described by the Fredholm alternative for an elliptic Dirichlet problem, proved here directly without assuming that theorem.

Let $d=\mathrm{dist}({\Omega }^{\prime },\partial \Omega )>0$. Choose a smooth cutoff function $\eta \in C_c^{\infty }(\Omega )$ with $0\le \eta \le 1$, $\eta =1$ on ${\Omega }^{\prime }$ and $|D\eta |\le C(n)/d$. Such cutoffs can be constructed by mollifying the indicator of a fixed small neighbourhood of $\overline{{\Omega }^{\prime }}$. Write $M=\Vert A{\Vert }_{\infty ,\mathrm{op}}$, $B=\Vert b{\Vert }_{\infty }$ and $C_0=\Vert c{\Vert }_{\infty }$; these are controlled by the individual coefficient bounds and dimension.

The weak formulation, extended to compactly supported $H^1$ tests by density, isWe may use $\phi ={\eta }^{2}u$, since $u\in H_{\mathrm{loc}}^1$. Expanding its gradient and using uniform ellipticity givesApply Young inequality to the first two terms, allocating at most $\lambda /4$ of the gradient integral to each. Bound $|fu|\le (f^2+u^2)/2$ in the last term. Absorbing those gradient contributions yields the Caccioppoli inequality with bounded lower-order termsConsequently $\Vert Du{\Vert }_{L^2({\Omega }^{\prime })}\le C(\Vert u{\Vert }_{L^2(\Omega )}+\Vert f{\Vert }_{L^2(\Omega )})$. Adding the $L^2$ norm of $u$ proveswhere $C$ has precisely the stated dependence on dimension, ellipticity, coefficient bounds and $d$. The proof uses $|c|$, so no sign condition on the zeroth-order term is required.

First use weak compactness of bounded sets in the Hilbert space $L^2(\Omega )$ to pass to a subsequence with $u_k\rightharpoonup u$ globally in $L^2$. The weak lower semicontinuity of the Hilbert norm gives $\Vert u{\Vert }_{L^2(\Omega )}\le K$. This global step also applies if $\Omega$ is unbounded.

Choose a nested exhaustion $U_j⋐U_{j+1}⋐\Omega$ by bounded smooth open sets. Part (i) bounds $u_k$ in $H^1(U_j)$ for every $j$. By the Rellich-Kondrachov compactness theorem, $H^1(U_j)$ embeds compactly in $L^2(U_j)$. Weak compactness in $H^1(U_j)$ and a diagonal subsequence argument therefore give, for the same subsequence,The local limit agrees with the global weak $L^2$ limit. Using smooth exhaustion sets avoids imposing any unprinted boundary regularity on a general ${\Omega }^{\prime }$.

For a smooth compactly supported test function, the weak equation passes to the limit: the fixed bounded coefficients multiply fixed test functions, and local weak gradient convergence handles both the principal and drift terms. The zeroth-order term passes by local $L^2$ convergence. Thus $Lu=f$ weakly, $u\in H_{\mathrm{loc}}^1(\Omega )$, and the global bound already proved gives $u\in S_K$.

To improve convergence of the gradients, put $w_k=u_k-u$. Crucially the forcing is the same for all $k$, so $L{w}_k=0$. Given ${\Omega }^{\prime }⋐\Omega$, choose $j$ with $\overline{{\Omega }^{\prime }}\subset U_j$. Apply part (i) on $U_j$ to the homogeneous equation for $w_k$:This establishes strong local Sobolev compactness for a fixed elliptic equation and proves

We prove the Interior second-derivative estimate for the Poisson equation using difference quotients, rather than differentiating the $L^2$ forcing. Fix a coordinate $e_k$ and write ${\delta }_{h}v(x)=(v(x+h{e}_k)-v(x))/h$. Choose a smooth cutoff function $\eta$ equal to one on ${\Omega }^{\prime }$ with support compactly contained in $\Omega$, leaving room for all translations with sufficiently small $|h|$. We can arrange $|D\eta |\le C(n)/d$, where $d=\mathrm{dist}({\Omega }^{\prime },\partial \Omega )$.

The standard difference quotient lemmas give discrete integration by parts, $\int a{\delta }_{-h}b=-\int ({\delta }_{h}a)b$, and $\Vert {\delta }_{h}v{\Vert }_2\le \Vert D_{k}v{\Vert }_2$ on the relevant enlarged set. In the weak formulation $\int Du\cdot D\phi =-\int f\phi$, use the admissible test function $\phi =-{\delta }_{-h}({\eta }^2{\delta }_{h}u)$. Moving a difference quotient onto $Du$ givesSet $X=\Vert \eta D{\delta }_{h}u{\Vert }_2$ and $Y=\Vert D\eta {\delta }_{h}u{\Vert }_2$. The cross term has absolute value at most $2XY$. Since $0\le \eta \le 1$, the difference quotient lemma on the right givesThe Young inequality now implies $X^2\le C(Y^2+\Vert f{\Vert }_2^2)$. Moreover $Y\le C(n)d^{-1}\Vert Du{\Vert }_{L^2(\Omega )}$, independently of $h$. Hence, for every coordinate $k$,By the Sobolev characterization by bounded difference quotients, applied to each first weak derivative on smaller interior sets, these uniform bounds give all second weak derivatives. Their bounds pass to the weak limit as $h\to 0$. Thus $u\in W_{\mathrm{loc}}^{2,2}(\Omega )$ andNo derivative of $f$ is used, and no boundary condition is needed for this interior elliptic regularity argument.

First perform homogenization of Dirichlet boundary data: put $v=u-\psi \in H_0^1(B_1^{+})$ and $g=f-\Delta \psi \in L^2(B_1^{+})$. Then $\Delta v=g$ weakly. Extend both functions oddly across the flat face:The zero Sobolev trace of $v$ makes this odd reflection an $H^1(B_1)$ function. This can also be seen by approximating $v$ by smooth compactly supported functions in $B_1^{+}$ and reflecting the approximants. Its $H^1$ norm is $\sqrt{2}$ times that of $v$, and $\Vert \tilde{g}{\Vert }_{L^2(B_1)}=\sqrt{2}\Vert g{\Vert }_{L^2(B_1^{+})}$.

For $\varphi \in C_c^{\infty }(B_1)$, the function $\varphi (x^{\prime },t)-\varphi (x^{\prime },-t)$ on the upper half-ball has zero trace on the flat face and vanishes near the curved face, hence is an admissible $H_0^1$ test. Splitting the integrals over the two half-balls and changing $t$ to $-t$ showsThus $\Delta \tilde{v}=\tilde{g}$ weakly throughout the full ball; there is no extra distribution on the reflecting face.

Apply part (i) to $B_{1/2}⋐B_1$, restrict to the upper half-ball, and then add $\psi$. Since $\Vert \Delta \psi {\Vert }_2\le C(n)\Vert \psi {\Vert }_{H^2}$, this gives $u\in W^{2,2}(B_{1/2}^{+})$ and the second-derivative estimate at a flat Dirichlet boundaryThe fixed radii leave distance $1/2$ for the interior estimate, so the constant depends only on dimension.

Let $B_r(x_0)$ have closure contained in the domain of a harmonic function $u$. Put ${\omega }_n=|B_1(0)|$, so $|B_r|={\omega }_n{r}^n$ and $|\partial B_r|=n{\omega }_n{r}^{n-1}$. The sphere and ball mean value properties for harmonic functions areTo prove the sphere formula, define its average using a fixed unit sphere:Differentiating under the integral and applying the divergence theorem givesContinuity gives $M(r)\to u(x_0)$ as $r\downarrow 0$, proving the spherical average identity. Integrating that identity over radii with the polar-coordinate weight yieldswhich proves the ball identity. For $n=1$, the sphere average is the average of the two endpoints, and the same formulas hold with the counting surface measure.

A local form of the Harnack inequality for harmonic functions is: if $u\ge 0$ is harmonic on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, the triangle inequality gives $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$. The ball mean value property for harmonic functions and nonnegativity therefore implyTaking the supremum over $x$ and infimum over $z$ proves the inequality, including the case where the infimum is zero. No strict positivity assumption was used.

On a connected domain $\Omega$, the equivalent compact-set statement is that for each nonempty compact $K⋐\Omega$ there exists $C(K,\Omega ,n)$ withHere is the passage from balls to this statement. Cover $K$ by finitely many small balls $B_{r_j}(y_j)$ whose quadrupled closed balls lie in $\Omega$. Join their centres to a fixed centre by paths in $\Omega$. Each of the finitely many paths has compact image and positive distance from the boundary, so subdividing it gives a finite chain of overlapping small balls whose quadrupled closures also lie in $\Omega$. The local inequality compares any two points in each ball; using a point in each successive overlap propagates comparisons along the chain. The number of comparisons is bounded uniformly over this finite collection of chains and covering balls. Multiplying the factors $3^n$ gives one finite constant for all pairs $x,z\in K$, proving the compact-set inequality. In particular a nonnegative harmonic function vanishing at one interior point vanishes throughout a connected domain.

Let $m={\inf }_{{\mathbb{R}}^n}u$, which is finite because $u$ is bounded, and put $v=u-m$. Then $v\ge 0$ is harmonic. Choose points $y_j$ with $v(y_j)\to 0$. For any fixed $x$, choose $r_j$ large enough that $x,y_j\in B_{r_j}(0)$. Since $v$ is harmonic on all of ${\mathbb{R}}^n$, the Harnack inequality for harmonic functions on $B_{4r_j}(0)$ givesThe comparison factor is independent of the growing radius, which is essential. Thus $v(x)=0$ for every $x$ andThis proves the Liouville theorem for harmonic functions in every dimension. In fact the argument only needs an entire harmonic function to be bounded below; changing its sign gives the corresponding bounded-above result.

Set $X=Du/\sqrt{1+|Du{|}^2}$. Since $u\in C^2({\mathbb{R}}^n)$, the field is $C^1$, and $|X|\le 1$ everywhere, even when the gradient of $u$ is unbounded. The divergence theorem on $B_R(0)$ givesLetting $R\to \infty$ proves $\kappa =0$. This is the fact that an entire bounded vector field cannot have nonzero constant divergence; it holds in every dimension.

The equation is now exactly the minimal surface equation for a graph. Apply the supplied Bernstein theorem for minimal graphs in dimensions $1\le n\le 7$ to concludeThus no nonzero constant right-hand side is possible for an entire graph, and in the stated dimensions every possible graph is an affine plane.

Let $M$ bound the operator norm of $D^{2}F$; the bound on individual second derivatives supplies such an $M$ depending also on dimension. The Taylor theorem with integral remainder givesThus quadratic growth from a bounded Hessian givesSince $\Omega$ is bounded it has finite measure, and a function in the Sobolev space $H^1(\Omega )$ has $Du\in L^2(\Omega )$. The given lower bound also implies that the integrand is nonnegative. ConsequentlyFor later use the same Hessian bound yields $|DF(p)|\le |DF(0)|+M|p|$, hence $DF(Du)\in L^2(\Omega )$.

Suppose $u_k\rightharpoonup u$ in $H^1(\Omega )$. Then $D{u}_k\rightharpoonup Du$ in $L^2(\Omega ;{\mathbb{R}}^n)$. Since $F$ is a differentiable convex function, its supporting inequality isTake $p=D{u}_k(x)$, $q=Du(x)$ and integrate. Part (i) shows that every energy is finite and that the fixed field $DF(Du)$ belongs to $L^2$, soThe last integral tends to zero by the definition of weak convergence in $L^2$. ThereforeThis is weak lower semicontinuity of convex gradient energies. It uses a fixed supporting gradient at the limit; weak convergence alone would not justify passing pointwise through the nonlinear integrand.

The affine class $C_{\psi }=\psi +H_0^1(\Omega )$ is nonempty, and part (i) gives a finite comparison energy $\mathcal{F}(\psi )$. Since $F(p)\ge \alpha |p{|}^2$, its energy infimum $m$ satisfies $0\le m\le \mathcal{F}(\psi )$. Choose a minimizing sequence $u_k\in C_{\psi }$ with $\mathcal{F}(u_k)\le \mathcal{F}(\psi )+1$. ThenThe gradient bounds also control the full Sobolev norm. Indeed $w_k=u_k-\psi \in H_0^1$, so the Poincaré inequality givesThus $u_k$ is bounded in the Hilbert space $H^1(\Omega )$. Its weak compactness yields a weakly convergent subsequence $u_k\rightharpoonup u$. The closed linear subspace $H_0^1$ is weakly closed, hence $u-\psi \in H_0^1$ and $u\in C_{\psi }$. By part (ii),The direct method in the calculus of variations therefore provesNeither a smooth boundary nor a classical trace theorem is needed: the boundary condition is encoded by the closed zero-boundary Sobolev space.

No additional hypothesis on $F$ is needed here for scalar interior $C^{1,\beta }$ regularity. In the elliptic sense of uniform convexity, the assumptions already give constants $0<\lambda \le M<\infty$ with $\lambda I\le D^{2}F(p)\le MI$ for all $p$. The following argument gives interior gradient regularity for uniformly convex autonomous energies.

The first variation at the minimizer exists because $|DF(p)|\le C(1+|p|)$, and its weak Euler-Lagrange equation isFor a coordinate difference quotient ${\delta }_{h}u$, subtract the translated equation and the original equation. The averaged linearization of a nonlinear divergence-form equation givesThese measurable symmetric matrices satisfy $\lambda I\le A_h\le MI$ independently of $h$. The Caccioppoli inequality and the standard bound $\Vert {\delta }_{h}u{\Vert }_2\le \Vert D_{k}u{\Vert }_2$ bound $D{\delta }_{h}u$ on every smaller interior set. Quoting the Sobolev characterization by bounded difference quotients yields $u\in H_{\mathrm{loc}}^2$.

Now $DF$ is globally Lipschitz continuous, so the Sobolev chain rule and commutation of distributional derivatives allow differentiation of the weak equation. Each $v_k=D_{k}u\in H_{\mathrm{loc}}^1$ satisfiesThe De Giorgi-Nash-Moser theorem states that weak solutions of this scalar divergence-form equation with bounded measurable uniformly elliptic coefficients are locally Hölder continuous. It gives a common $\beta \in (0,1)$ depending only on $n$ and the ellipticity ratio $M/\lambda$, with $D_{k}u\in C_{\mathrm{loc}}^{0,\beta }$ for every $k$. A Sobolev function whose weak gradient is continuous has a $C^1$ representative, as follows by local mollification and integration of its gradient. HenceThe higher-regularity conclusion comes from differentiating an autonomous scalar equation; it does not rely on boundary regularity or a higher derivative such as $D^{3}F$.