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

Original paper: public parent PDF, pages 310–312, module C4.6, formal paper code A12201W1. The cover is page 310; questions 1–2 are on page 311 and question 3 is on page 312.

The Banach fixed-point theorem states that a contraction mapping $T:X\to X$ of a nonempty complete metric space $(X,d)$ has a unique fixed point, and iterations of a map $x_{n+1}=T(x_n)$ converge to it from every $x_0\in X$. Precisely, assume a single constant $q\in [0,1)$ satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y\in X$.

Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and the geometric series therefore give, for $m>n$,Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x_{\ast }\in X$. The contraction mapping is a continuous map, so $T{x}_{\ast }={\lim }_{n}T{x}_n={\lim }_n{x}_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing $x_{\ast }=y_{\ast }$. Letting $m\to \infty$ in the estimate also gives the contraction iteration error bound:If $q=0$, $T$ is constant and the sequence reaches its unique fixed point after the first step; this also removes any ambiguity about $0^0$ in the displayed estimate at $n=0$.

The native paper prints $n\ge 1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\ge 0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.

For the sine fixed-point iteration, after one step $x_1\in [-1,1]$. On this interval, the sine has the same sign as its argument, andFor example, $t-\sin t={\int }_0^t(1-\cos s)ds>0$, since the integrand is positive for $0<s\le 1$. By oddness of the sine, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\le a_n$ for $n\ge 1$. This monotone sequence is bounded below by zero, so it has a limit $a\in [0,1]$. The continuity of the sine gives $a=\sin a$, and the strict inequality above forces $a=0$.

Hence $x_n\to 0$, and $0$ is a fixed point of the sine. It is the only real fixed point: every such point has absolute value at most one, where the preceding strict inequality applies. ThusThe Banach fixed-point theorem does not directly apply to the sine on $[-1,1]$: the ratios $|\sin t-\sin 0|/|t-0|$ tend to one as $t\to 0$, so no uniform contraction mapping constant smaller than one exists there.

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.

The retraction of a Hilbert ball onto its sphere exists. In the real Hilbert space ${\ell }^2$, let $S(x_0,x_1,\dots )=(0,x_0,x_1,\dots )$ be the unilateral shift operator, and let $e_0=(1,0,\dots )$. For $x$ in the closed unit ball $\mathbb{B}$, defineThe orthogonality of $e_0$ and $Sx$ gives $\Vert F(x){\Vert }^2=1$. Thus $F$ is a continuous map into the unit sphere. It has no fixed point: $F(x)=x$ would imply $x_{j+1}=x_j$ for every $j\ge 0$. A constant sequence in ${\ell }^2$ must be zero, whereas $F(0)=e_0\ne 0$.

Put $d=x-F(x)$, which is nonzero everywhere, and defineThis is a continuous map on the whole closed unit ball, including zero. To check its range, observe that $⟨F(x),d⟩=⟨F(x),x⟩-1$, soIf $\Vert x\Vert =1$, then $\Vert x-F(x){\Vert }^2=2(1-⟨F(x),x⟩)$, hence $\alpha (x)=1$ and $r(x)=x$. Therefore $r:\mathbb{B}\to \partial \mathbb{B}$ is a retraction.

Geometrically this is the second intersection of the line from $F(x)$ through $x$ with the unit sphere. In factThe construction illustrates why the finite-dimensional hypothesis in the no-retraction theorem matters. The closed unit ball of ${\ell }^2$ is not a compact set in the norm topology.

Let $S_{+}^{n-1}$ be the nonnegative portion of the unit sphere specified in the paper. If $Ax=0$ for some $x\in S_{+}^{n-1}$, then this nonzero vector is an eigenvector with eigenvalue zero, which already proves the assertion.

Otherwise $Ax\ne 0$ for every $x\in S_{+}^{n-1}$, andis a continuous map from $S_{+}^{n-1}$ to itself. Indeed, every entry of $Ax$ is nonnegative because $A$ is a nonnegative matrix and every entry of $x$ is nonnegative. By the supplied homeomorphism $h:{\mathbb{B}}^{n-1}\to S_{+}^{n-1}$, the continuous map $h^{-1}Th$ maps the finite-dimensional closed unit ball into itself. The Brouwer fixed-point theorem supplies a fixed point, hence $T(x_{\ast })=x_{\ast }$ for some $x_{\ast }\in S_{+}^{n-1}$. ThereforeTogether with the zero case this proves that $A$ has a nonnegative eigenvalue, with a nonnegative eigenvector. For $n=1$, the auxiliary closed unit ball is the one-point space ${\mathbb{B}}^0$ and the argument remains valid. No claim of strict positivity of every entry of the eigenvector is needed.

For Banach spaces $X,Y$, a nonlinear compact operator $T:X\to Y$ is a continuous map that sends every bounded set to a relatively compact subset of $Y$: the closure of its image is a compact set. For a continuous map, the image condition is equivalent to requiring that every bounded sequence $(x_j)$ has a subsequence along which $T(x_j)$ converges in the norm topology of $Y$. For a linear operator, this is the usual compact operator definition; the continuity requirement is included explicitly here because the subsequent applications are nonlinear.

The compact-domain version of the Schauder fixed-point theorem saysA useful equivalent version replaces $K$ by a nonempty closed bounded convex set $C$ and assumes $T:C\to C$ is a continuous map with relatively compact image. This latter form will be used for the partial differential equation. Both forms assert existence of a fixed point, without asserting uniqueness.

Let $E=\mathrm{aff}K$ be the affine hull and put $k=\dim E\le n$. If $k=0$, $K$ is a singleton and is homeomorphic to ${\mathbb{B}}^0$. Suppose $k>0$. Choose $k+1$ affinely independent points of $K$ spanning $E$. Their simplex is contained in $K$ because $K$ is a convex set, and its barycentre $a$ has a neighbourhood in $E$ contained in that simplex. Identifying $E-a$ with ${\mathbb{R}}^k$, put $D=K-a$. For some $0<r\le R<\infty$,Here the inner inclusion follows from that neighbourhood, and the outer one follows from the compactness of $K$.

Use the Minkowski functionalThe inclusions imply $|z|/R\le p(z)\le |z|/r$. The fact that $D$ is a convex set implies $p(z+w)\le p(z)+p(w)$: if $z\in sD$ and $w\in tD$, then $(z+w)/(s+t)$ is a convex combination of points of $D$. Also $p(tz)=tp(z)$ for $t\ge 0$. ConsequentlyThus $p$ is a continuous function. Since $D$ is closed and contains zero, $D=\{p\le 1\}$: the forward inclusion is immediate, and for the reverse inclusion one takes a limit of $z/t_j\in D$ with $t_j\to p(z)\le 1$, using positive homogeneity and the fact that $D$ is a convex set. For $z\ne 0$, $p(z)>0$, so this limiting argument has no division by zero.

Define a radial continuous map byEach ray meets $D$ in exactly $\{s\theta :0\le s\le 1/p(\theta )\}$. Hence $h$ is a bijection from ${\mathbb{B}}^k$ to $K$. Away from zero its continuity follows from that of $p$, and at zero $|h(t\theta )-a|\le Rt$. A continuous map that is a bijection from a compact space to a Hausdorff space is a homeomorphism. ThereforeUsing the affine hull is essential: $K$ can have empty interior in the ambient ${\mathbb{R}}^n$.

We first prove the compact-domain Schauder fixed-point theorem. Let $K$ be a nonempty compact convex set in a Banach space $X$, and let $T:K\to K$ be a continuous map. The Brouwer fixed-point theorem states that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point. By the homeomorphism proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact convex set, including a singleton.

For $\epsilon >0$, use the supplied Schauder approximation lemma to obtain a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with $\Vert Tx-T_{\epsilon }x\Vert \le \epsilon$. The set $K_{\epsilon }=K\cap L_{\epsilon }$ is nonempty because it contains $T_{\epsilon }(K)$. It is a compact convex set, since a finite-dimensional linear subspace is closed. Restricting $T_{\epsilon }$ to $K_{\epsilon }$ and applying the Brouwer fixed-point theorem gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, and thusTake ${\epsilon }_j=1/j$. The compactness of $K$ yields a convergent subsequence $x_{{\epsilon }_{j_m}}\to x_{\ast }\in K$. The continuity of $T$ and the preceding estimate giveFor completeness, this also proves the version on a nonempty closed bounded convex set $C$ when $T(C)$ is relatively compact. Let $K=\overline{\mathrm{conv}}T(C)\subset C$. The closed convex hull of a compact set in a Banach space is compact. To verify the fact needed here, cover $T(C)$ by finitely many balls of radius $\delta$. Replacing every term of a convex combination by a centre changes it by at most $\delta$ in norm. The convex hull of those finitely many centres is compact in a finite-dimensional linear subspace, so it has a finite net of radius $\delta$. Thus $\mathrm{conv}T(C)$ is totally bounded. Its closure is complete and totally bounded, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a fixed point in $C$.

Use the real zero-trace Sobolev space $H_0^1(\Omega )$ with norm $\Vert \nabla u{\Vert }_2$. The Poincaré inequality on the bounded smooth domain gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$, making this an equivalent Sobolev norm. The weak formulation is to find $u\in H_0^1(\Omega )$ with $u\ge 0$ almost everywhere andThe zero Dirichlet boundary condition is encoded in the Sobolev trace. Formally, the identity follows by multiplying the partial differential equation by a test function and using integration by parts; it then extends to all $v\in H_0^1(\Omega )$ by density.

To check all terms, boundedness of $\Omega$ gives finite Lebesgue measure $m=|\Omega |$. The Cauchy-Schwarz inequality gives $u\in L^1(\Omega )$ andThus the right side is a bounded linear functional of $v$, since its absolute value is at most $\Vert 1+\sqrt{u}{\Vert }_2{C}_P\Vert \nabla v{\Vert }_2$. The left side is finite by the Cauchy-Schwarz inequality. These estimates work for every stated $n\ge 2$.

Let $P=\{w\in L^2(\Omega ):w\ge 0\text{ almost everywhere}\}$. For $g\in L^2(\Omega )$, define $Gg=z\in H_0^1(\Omega )$ byThe Lax-Milgram theorem guarantees a unique $z$: the bilinear form on the left is bounded and coercive in the norm $\Vert \nabla \cdot {\Vert }_2$, and the right side is a bounded linear functional by the Poincaré inequality. Testing with $z$ givesDefine $T:P\to L^2(\Omega )$ by $Tw=G(1+\sqrt{w})$. This actually maps into $P$. Indeed the weak maximum principle says that a weak solution $z\in H_0^1(\Omega )$ of $-\Delta z=g\ge 0$ is nonnegative almost everywhere. Here $1+\sqrt{w}\ge 0$ and belongs to $L^2(\Omega )$. Directly, taking the negative part $z^{-}=\max \{-z,0\}$ as a test function gives $-\Vert \nabla z^{-}{\Vert }_2^2={\int }_{\Omega }g{z}^{-}dx\ge 0$, so $z^{-}=0$ by the Poincaré inequality.

The required fixed point formulation isEvery such fixed point belongs automatically to $H_0^1(\Omega )$ and satisfies part (b)(i). Conversely, uniqueness in the Lax-Milgram theorem makes every nonnegative weak solution a fixed point of $T$.

We apply the closed-bounded-domain version of the Schauder fixed-point theorem in the Banach space $L^2(\Omega )$. Write $m=|\Omega |$ and use $G,T,C_P$ from part (b)(ii). ChooseThe set $C$ is nonempty, bounded, closed and a convex set. For closedness of its positivity condition, if $w_j\ge 0$ and $w_j\to w$ in $L^2$, then pointwise $w^{-}\le |w-w_j|$, whence $\Vert w^{-}{\Vert }_2\to 0$.

The weak maximum principle already gives $Tw\ge 0$. Our estimates give, for $w\in C$,Thus $T(C)\subset C$.

To prove continuity, use the pointwise square root estimate $|\sqrt{s}-\sqrt{t}{|}^2\le |s-t|$ for $s,t\ge 0$. The Cauchy-Schwarz inequality then givesHence $T$ is a continuous map on $C$, even though the square root need not have a bounded derivative near zero.

Finally,The Rellich-Kondrachov theorem says that $H_0^1(\Omega )$ embeds compactly into $L^2(\Omega )$ on a bounded smooth domain. For $n>2$, $2<2n/(n-2)$, and for $n=2$ every finite target exponent is available, so the required compact embedding holds in every dimension in the question. Thus $T(C)$ is relatively compact in $L^2(\Omega )$.

All hypotheses of the Schauder fixed-point theorem are met. It gives $u=Tu\in C$, and part (b)(ii) converts this fixed point into the required weak solution:Only existence is needed. The construction does not assume that $\Omega$ is small or that $T$ is a contraction mapping.

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).