Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2023 / 3

A useful form of Minty's lemma is the following. Let $X$ be a real Banach space, $K\subset X$ a nonempty convex set, and $A:K\to X^{\ast }$ a monotone operator that is hemicontinuous. Here operator monotonicity means $⟨Ax-Ay,x-y⟩\ge 0$ for $x,y\in K$, and hemicontinuity means that $t\mapsto ⟨A(x+t(y-x)),z⟩$ is continuous on $[0,1]$ for $x,y\in K$ and $z\in X$. For $u\in K$,The forward implication follows from operator monotonicity:For the reverse implication, fix $w\in K$ and take $v_t=u+t(w-u)\in K$ for $0<t\le 1$. The assumed inequality givesDivide by $t$ and send $t\downarrow 0$. Hemicontinuity gives $⟨Au,w-u⟩\ge 0$, as required.

In the whole-space version, for a monotone operator $A:X\to X^{\ast }$ that is hemicontinuous,Apply the just-proved result to $A-f$ and $K=X$; the resulting inequality $⟨Au-f,v-u⟩\ge 0$ for all $v$ forces $Au=f$ by choosing $v=u\pm z$. This is the variational inequality form of Minty's lemma.

Choose the real zero-trace Sobolev spaceThe Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is $4^{\ast }=5\cdot 4/(5-4)=20$. The Sobolev embedding theorem gives $X\hookrightarrow L^{20}(\Omega )$ continuously. Boundedness of $\Omega$ and Lp inclusion on a finite measure space then give $X\hookrightarrow L^8(\Omega )$ continuously, say $\Vert u{\Vert }_8\le C_8\Vert u{\Vert }_X$.

For $f\in Y$, the weak formulation isThe zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:Thus $|\nabla u{|}^2\nabla u\in L^{4/3}(\Omega ;{\mathbb{R}}^5)$, while $u^7\in L^{8/7}(\Omega )$, and all three pairings are finite bounded linear functionals in $v$. The choice $Y=L^{8/7}(\Omega )$ is sufficient; it is not intended to be the largest possible space of data.

For the spaces in part (b)(i) and fixed $f\in Y$, define $A:X\to X^{\ast }$ byFor each $u$, this is a linear functional in $v$, and the estimates in part (b)(i) showHence $A(u)$ belongs to the continuous dual space $X^{\ast }$, and $A$ sends bounded sets to bounded sets. The weak formulation is exactlyThe forcing term is included in $A$ so that its right side is zero, as requested.

For $u,w\in X$, the forcing terms cancel, givingThe scalar function $s\mapsto s^7$ is strictly increasing on all of $\mathbb{R}$, so the second integrand is nonnegative. For the first one, use the quartic gradient monotonicity inequality. Put $d=\xi -\eta$ and $m=(\xi +\eta )/2$ for real vectors $\xi ,\eta$. Expanding givesConsequentlyThe last strict inequality uses the Poincaré inequality: if the weak gradient of $u-w\in W_0^{1,4}(\Omega )$ vanishes, then $u=w$. This proves that $A$ is a strictly monotone operator.

We prove existence by the direct method in the calculus of variations, making the step with weakly compact sets explicit. For $X=W_0^{1,4}(\Omega )$ and $f\in L^{8/7}(\Omega )$, introduce the functionalIt is finite by part (b)(i). Hölder's inequality and the Sobolev embedding theorem giveThus $E$ is a coercive function, bounded below, and every minimizing sequence is bounded in $X$.

The space $X$ is a reflexive Banach space: it is isomorphic to a closed linear subspace of the finite product of the reflexive spaces $L^4(\Omega )$ recording the function and its weak derivatives. A bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Hence, after passing to a subsequence, a minimizing sequence satisfies $u_j\rightharpoonup u_{\ast }$ in $X$. The continuous linear maps $u\mapsto \nabla u$ into $L^4$ and $u\mapsto u$ into $L^8$ preserve weak convergence, soThe norm in any Banach space has weak lower semicontinuity, since it is the supremum of absolute values of bounded linear functionals of dual norm at most one, by the Hahn-Banach theorem. Therefore the two nonnegative norm powers in $E$ have weak lower semicontinuity. Also $u\mapsto {\int }_{\Omega }fudx$ belongs to $X^{\ast }$ and is continuous for weak convergence. It follows thatFor any $v\in X$, the real function $t\mapsto E(u_{\ast }+tv)$ has a minimum at $t=0$. Its derivative there isThis differentiation is justified directly by finite polynomial expansions: the fourth power of the Euclidean norm of $\nabla u_{\ast }+t\nabla v$ and the eighth power of $u_{\ast }+tv$ have integrable coefficients by Hölder's inequality. Thus $A(u_{\ast })=0$, giving existence of a weak solution.

If $u,w$ are two weak solutions, testing the difference by $u-w$ and using part (b)(iii) givesHenceNo sign assumption on $f$ or on the weak solution is required.

Take the functionalThe factor $1/8$ is essential: the derivative of $|s{|}^8/8=s^8/8$ is $s^7$, while the gradient with respect to $\xi$ of $|\xi {|}^4/4$ is $|\xi {|}^2\xi$. As in part (c), finite polynomial expansions and Hölder's inequality justify differentiation along every direction $v\in X$, giving $DJ(u)[v]=⟨A(u),v⟩$.

We prove the global and strict minimum directly. The strictly convex functions $\Phi (\xi )=|\xi {|}^4/4$ and $\Psi (s)=|s{|}^8/8$ satisfy their tangent inequalitieswith strict inequality when the respective arguments differ. For example, strict convexity of $\Phi$ follows from strict operator monotonicity of its gradient, proved by the midpoint identity in part (b)(iii); strict convexity of $\Psi$ follows because its derivative $s^7$ is strictly increasing.

Apply these inequalities pointwise to $\nabla u_{\ast },\nabla v$ and $u_{\ast },v$, then integrate. The weak formulation at $u_{\ast }$ with test function $v-u_{\ast }$ givesIf $v\ne u_{\ast }$ in $X$, the Poincaré inequality implies $\nabla v\ne \nabla u_{\ast }$ on a set of positive Lebesgue measure. The first tangent inequality is strict on that set and the second is nonnegative everywhere. Their integrable remainders therefore have strictly positive total integral. HenceThus $u_{\ast }$ is the unique global minimizer of $J$. Conversely, every minimizer has zero directional derivative in every direction and so satisfies $A(u)=0$, which also identifies it with the unique weak solution from part (c).