Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 12 / 1

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.