Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2008 (C5.1, DMAC 2465, DMPC 2465)

A continuous embedding $X\hookrightarrow Y$ is an injective continuous linear inclusion, so $\Vert x{\Vert }_Y\le C\Vert x{\Vert }_X$ for a fixed constant. On a finite-measure domain, Holder inequality giveswith $1/\infty =0$. For finite $p>q$ apply Holder inequality to $|u{|}^q\cdot 1$ with exponents $p/q$ and $p/(p-q)$; equality of exponents is immediate, and $p=\infty$ follows by integrating the supremum bound.

Weak convergence $u_j\rightharpoonup u$ means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. The Riemann-Lebesgue lemma says that the Fourier transform of an integrable function tends to zero at high frequency. Since $L^2(0,1)\subset L^1(0,1)$, it makes $\cos (j\pi x)$ weakly null in $L^2$. Its absolute value is at most one, so $u_j{v}_j\in L^2$. For a test $\varphi \in L^2$,while $\int u_{j}v\varphi \to 0$ by the Riemann-Lebesgue lemma applied to $v\varphi \in L^1$. Hence $u_j{v}_j\rightharpoonup 0$. Taking $w_j=u_j=\cos (j\pi x)$ instead gives $u_j{w}_j=(1+\cos (2j\pi x))/2\rightharpoonup 1/2$, so two weakly null factors need not have a weakly null product.

For ${\bar{z}}_j=j^{1/p}{1}_{(0,1/j)}$ and $p^{\prime }=p/(p-1)$, Holder inequality givesThus ${\bar{z}}_j\rightharpoonup 0$ in $L^p$. In contrast, if $z_j$ had a weak $L^1$ limit $z$, testing on every interval strictly away from zero would give integral zero for $z$, hence $z=0$ almost everywhere by the permitted fact. Testing against the constant function one would give $\int z=1$, a contradiction. Therefore $z_j$ has no weak $L^1$ limit.

Choose a smooth compactly supported mollifier of integral one. On every compact subset of $\Omega$, and for sufficiently small $\epsilon$, each translated mollifier is an allowed test, so the assumption gives $w\ast {\rho }_{\epsilon }=0$. The standard mollifier approximation theorem gives $w\ast {\rho }_{\epsilon }\to w$ in local $L^1$. Thus $w=0$ almost everywhere.

For a multi-index $\alpha$, the weak derivative $v=D^{\alpha }u$ is the locally integrable function satisfyingTwo such functions have zero pairing difference against every test, so the first result proves uniqueness. A compactly supported smooth test on an open subset $E$ extends by zero to one on $\Omega$, proving restriction of the derivative to $E$.

Put $w=D^{\beta }u$. For every test $\varphi$ the defining identity applied to $D^{\alpha }\varphi$ givesThus these two distributions are identical; one is represented by a locally integrable function if and only if the other is. Their weak derivatives then agree, proving $D^{\alpha }(D^{\beta }u)=D^{\alpha +\beta }u$ with exactly the asserted existence equivalence.

On $(0,2)$ the displayed piecewise affine function is continuous at one. Integration by parts separately on the two intervals shows its first weak derivative is one on $(0,1)$ and minus one on $(1,2)$; the first boundary terms cancel by continuity. The derivative of that step function is $-2{\delta }_1$. This distribution is not represented by a locally integrable function: tests of height one supported in shrinking neighborhoods of one have pairing $-2$ with it, whereas their integrals against any fixed local $L^1$ function tend to zero. Hence there is a first weak derivative but no locally integrable second weak derivative.

The weak formulation isThe boundary condition is encoded by $u\in H_0^1$. The Sobolev embedding in dimension three gives $H_0^1\hookrightarrow L^6$, hence also $L^4,L^3,L^2$ on the bounded domain. All energy terms and all the displayed pairings are therefore finite.

Pointwise, $t^4/4-t^3/3\ge t^4/8-C$ for a constant $C$, since the difference is a quartic with positive leading coefficient. The Poincaré inequality and Young inequality give $\int fv\ge -\frac{1}{4}\Vert Dv{\Vert }_2^2-C_f$. ConsequentlyThus the infimum is finite and a minimizing sequence is bounded in $H_0^1$. Extract a weakly convergent subsequence there. The Rellich-Kondrachov theorem states that on a bounded $C^1$ domain in dimension three the embedding $H^1\hookrightarrow L^q$ is compact for $q<6$. A subsequence therefore converges strongly in $L^2,L^3,L^4$ to the same limit $u$. The potential terms and forcing converge, while the gradient norm is weakly lower semicontinuous. The weak limit retains zero trace. Hence $I(u)\le \mathrm{liminf}I(v_j)$ and $u$ attains the absolute minimum.

For any $\varphi \in H_0^1$, the function $t\mapsto I(u+t\varphi )$ has a minimum at zero. Differentiating its polynomial terms gives $\int [Du\cdot D\varphi +(u^3-u^2+f)\varphi ]=0$. These differentiations are justified by the Sobolev integrability already noted. This is the weak formulation, proving every energy minimizer is a weak solution.

The Lax-Milgram theorem gives, for a bounded coercive bilinear form on a real Hilbert space and every bounded linear functional, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The Fredholm alternative for a compact operator $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker (I+\mu K{)}^{\ast }$ in the Hilbert-space case.

Uniform ellipticity gives $\sum a_{ij}{\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ for some $\theta >0$. Bounded coefficients make $B_{\lambda }$ bounded on $H^1$, and for $\lambda <-\gamma$,The forcing $v\mapsto \int fv$ is bounded on $H^1$. The Lax-Milgram theorem therefore gives a unique weak solution for every $\lambda <-\gamma$.

When $c=0$, choose the shift ${\lambda }_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int (Kf)v=\int fv$. Thus $K=(L+I{)}^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, orConversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves compactness of $K$ as an $L^2$ operator.

By the supplied kernel assumption, $\ker (I-K)$ consists of constants. Also $K1=1$. Since $K$ is a self-adjoint operator, the Fredholm alternative condition is $Kf\perp 1$, andTherefore a weak solution exists exactly when the mean of $f$ is zero, and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.

For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the weak formulation explicitly defined in the paper, so does not require that stronger boundary identification.

A convex function satisfies $F((1-t)x+ty)\le (1-t)F(x)+tF(y)$ for every $0\le t\le 1$ and every $x,y$.

A strictly convex function satisfies that inequality strictly whenever $x\ne y$ and $0<t<1$.

For necessity of the strict supporting-tangent property, convexity gives a nonnegative gap along a segment from $z_0$ to $z$. If the gap at its other endpoint were zero, convexity of the gap and its zero initial value would force it to be zero all along the segment. That would make $F$ affine on this nontrivial segment, contrary to strict convexity. Conversely let $z=(1-t)x+ty$ with distinct endpoints. Apply the strict supporting inequality at $z$ to both endpoints, multiply by $1-t$ and $t$, and add. The linear terms cancel, yielding strict convexity. ThusFor strict Jensen's inequality, put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting gap $F(u)-F(m)-DF(m)\cdot (u-m)$ is nonnegative everywhere and strictly positive where $u\ne m$. That set has positive measure because $u$ is not almost everywhere constant. Its integral is therefore strictly positive. The linear term integrates to zero, giving the requested strict integral Jensen inequality.

For the unlabelled Euler-Lagrange assertion, use its integrated form $f^{\prime }(u^{\prime })=c$ almost everywhere. This follows from the first variation $\int f^{\prime }(u^{\prime }){\varphi }^{\prime }=0$ for all compactly supported tests. If $u\in C^1$, the left side is continuous and so equals $c$ everywhere. Its derivative values lie in the finite fiber $(f^{\prime }{)}^{-1}(c)$, while the continuous image of the interval under $u^{\prime }$ is connected. Thus $u^{\prime }$ is constant. Endpoint values fix it to $m=(\beta -\alpha )/(b-a)$, giving exactly the affine $\bar{u}$. Conversely that affine function has constant $f^{\prime }({\bar{u}}^{\prime })$, so is a solution. This proves uniqueness among $C^1$ solutions; the scalar integrand is the one-dimensional case of the notation in the source.

If $f$ is strictly convex, Jensen's inequality applied to $u^{\prime }$ givesEquality in the strict inequality can occur only when $u^{\prime }$ is almost everywhere constant. Absolute continuity and the endpoints then give $u=\bar{u}$. Hence the affine function is the unique global minimizer.

In this special case the endpoints are both zero, overriding the previous strict ordering. The affine function is zero and $I(0)=1$. Since $0<f(v)\le 1$, every admissible energy is at most one, proving zero is a global maximizer.

For $u_N(x)=N\min \{x,1-x\}$, the endpoint values vanish and its slopes are $\pm N$ almost everywhere. HenceEvery admissible $u$ has essentially bounded derivative, so its integrand is bounded below by the positive constant $1/(1+\Vert u^{\prime }{\Vert }_{\infty }^{2008})$. Thus no admissible function has zero energy and there is no global minimizer.

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ gives $I(v)\ge I(\bar{u})$ for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$.

A strong local minimizer of a variational integral obeys the same inequality for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. The derivative of a competitor need not be close in this definition.

A global minimizer has $I(\bar{u})\le I(v)$ for every admissible $v$. Global minimality implies strong local minimality, which implies weak local minimality. The reverse implications are not valid in general.

Where $f_v$ exists, the strengthened Weierstrass condition says that for some $\delta >0$, the excessis nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ with $|y-\bar{u}(x)|<\delta$, $|p-{\bar{u}}^{\prime }(x)|<\delta$. This is a neighborhood condition in height and reference slope, not just an inequality on the reference curve.

The classical Weierstrass sufficient condition, with the usual smooth integrand regularity in addition to continuity, states that an admissible $C^1$ curve satisfying the Euler-Lagrange equation, a coercively positive second variation, and this strengthened condition is a strong local minimizer. A strict excess away from the reference slope gives the corresponding strict conclusion. The differentiability assumptions are part of that classical sufficient theorem; continuity alone does not define its first and second variations.

Zero is a strict weak local minimizer and is not a strong local minimizer. Write $D={\int }_0^1{u}^{\prime 2}$. For zero endpoint values, Cauchy-Schwarz inequality applied togives $|u(x){|}^2\le x(1-x)D$, hence ${\int }_0^1{x}^2{u}^2\le D/20$. For sufficiently small slopes, Taylor expansion gives $\frac{1}{2}[\sin (1+u^{\prime 2})-\sin 1]\ge (\frac{1}{2}\cos 1-\epsilon )u^{\prime 2}$. Thereforefor a suitably small fixed $\epsilon$ and every nonzero sufficiently weakly close competitor.

For strong minimality take $a=\sqrt{3\pi /2-1}$ and $u_N(x)=a\mathrm{dist}(x,N^{-1}\mathbb{Z})$. Their amplitudes tend to zero but slopes are $\pm a$, so $\sin (1+u_N^{\prime 2})=-1$. Thus $I(u_N)\le -1/2<I(0)=\sin 1/2$. Uniform closeness therefore does not prevent a decrease in energy.

The inequality $\cos t\ge 1-t^2/2$ givesThe elementary endpoint estimate $|u(x){|}^2\le x{\int }_0^1{u}^{\prime 2}$ implies $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. ConsequentlyThus zero is the unique global minimizer, and in particular a strict strong and strict weak local minimizer.

With coefficients evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$, put $p=f_{vv}$, $r=f_{uv}$ and $s=f_{uu}$. The second variation isThe Jacobi equation in divergence form is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$; if $r$ is differentiable it is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$. A conjugate point $c>a$ admits a nonzero Jacobi solution vanishing at $a,c$.

There are two positivity conventions worth distinguishing. Strict pointwise positivity means $q(\varphi )>0$ for each nonzero fixed-endpoint variation. Coercive positivity means $q(\varphi )\ge c\int {\varphi }^{\prime 2}$ with one $c>0$ for all such variations; the endpoint Poincaré estimate makes this equivalent to a positive lower bound in the full $H^1$ norm. Under the regular Legendre hypothesis $p>0$, the necessary and sufficient Jacobi criterion is no conjugate point in $(a,b]$; in that regular case it gives both strict and coercive positivity. Equivalently the Jacobi equation has a solution positive on $[a,b]$, and the square identitygives positivity and a uniform derivative lower bound by the positive minimum of $z$ and boundedness of $z,z^{\prime }$. Conversely a conjugate Jacobi field, extended by zero beyond its second zero, is a nonzero Lipschitz variation with $q=0$, contradicting either positivity convention.

Coercivity itself forces $p>0$: local high-frequency variations make the leading coefficient at least the coercivity constant. Mere strict positivity need not force it, as $q(\varphi )=\int {\varphi }^2$ shows. If that weaker convention is intended without regularity, a complete criterion is $p\ge 0$, no conjugate point for every regularized form $q_{\eta }=q+\eta \int {\varphi }^{\prime 2}$ with $\eta >0$, and no nonzero zero-endpoint Lipschitz weak solution of the original Jacobi equation. The regularized criteria imply $q_{\eta }\ge 0$, hence $q\ge 0$ by passage to the limit. For a nonnegative quadratic form, $q(\varphi )=0$ forces all mixed terms with test variations to vanish, giving exactly the weak Jacobi kernel; excluding it gives strict positivity. Necessity follows in reverse. This specifies the degenerate case without imposing an unstated strict Legendre condition.

For weak local minimality, sufficient conditions are the Euler-Lagrange equation and a coercively positive second variation with the stated smoothness. Taylor's remainder is bounded by $\epsilon \Vert \varphi {\Vert }_{H^1}^2$ for small $W^{1,\infty }$ perturbations, so the positive quadratic term dominates it. These are the classical weak minimum sufficient conditions.

At zero the quadratic form isFor $\lambda \ge 0$ it dominates $\int u^{\prime 2}$. For $-{\pi }^2<\lambda <0$ set $\mu =\sqrt{-\lambda }<\pi$ and $z(x)=\cos (\mu (x-1/2))$, positive on the closed interval and satisfying $z^{\prime \prime }+{\mu }^{2}z=0$. The Jacobi identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2$. To obtain a uniform lower bound, let $m=\min z>0$, $K=\max |z^{\prime }|$ and $v=u/z$. Since $\max z=1$ and $\int v^2\le \frac{1}{2}\int v^{\prime 2}$,so $Q_{\lambda }(u)\ge m^2\int u^{\prime 2}/(K^2+2)$.

Finally $I(u)=Q_{\lambda }(u)+O(\int u^4)$ near zero, because ${\sin }^{2}u=u^2+O(u^4)$. The remainder is at most a constant times $\Vert u{\Vert }_{\infty }^2\int u^2\le \frac{1}{2}\Vert u{\Vert }_{\infty }^2\int u^{\prime 2}$. It is dominated by the coercive bound for small perturbations. Thus the second variation is positive and zero is a strict weak local minimizer for every $\lambda >-{\pi }^2$. This argument has not assumed the sharp Poincaré inequality that is to be deduced next.

For a fixed nonzero Lipschitz zero-trace $\varphi$, positivity for every $\lambda >-{\pi }^2$ gives $\int {\varphi }^{\prime 2}+\lambda \int {\varphi }^2>0$. Let $\lambda \downarrow -{\pi }^2$ to obtainFor general $\varphi \in W_0^{1,2}$, approximate by smooth compactly supported functions in the $W^{1,2}$ norm. Both squared norms in the inequality converge, proving it for the full required space. Equality at $\sin (\pi x)$ verifies sharpness.

For a fixed nonzero $u_0\in \mathcal{A}$,The quartic coefficient is positive. Therefore the energy is unbounded below and zero is not a global minimizer, for every value of $\lambda$.

The continuous superlinear function $\Phi$ tends to positive infinity at both ends, so it has a finite global minimum $M$. Hence $I(u)\ge M(b-a)$ for every admissible function. An affine function with the prescribed endpoints is Lipschitz and has finite energy by continuity of the integrand on its compact range. Thuswhere the strict left inequality simply records finiteness of $M$.

A minimizing sequence has an energy upper bound $C$. Set $H(t)=\Phi (t)-M\ge 0$. Then $\int H(u_j^{\prime })\le C-M(b-a)=C_1$. Since $H(t)/|t|\to \infty$, for every $A>0$ there is $R_A$ such that $H(t)\ge A|t|$ for $|t|>R_A$. Consequently, for every measurable $E$,Taking first $A$ large and then $|E|$ small proves uniform integrability; taking $E=(a,b)$ also gives a uniform $L^1$ bound. The Dunford-Pettis theorem on a finite-measure space identifies bounded uniformly integrable subsets of $L^1$ with relatively weakly compact ones. Thus a subsequence satisfies

The compactness conclusion justified by the hypotheses is $u\in W^{1,1}$ with the prescribed traces; the printed claim $u\in W^{1,\infty }$ needs additional regularity. DefineThen $u\in W^{1,1}$ and $u^{\prime }=v$. Weak $L^1$ convergence tested against $1_{(a,x)}$ gives $u_{j_k}(x)\to u(x)$ for every $x$, including the endpoint trace at $b$. Uniform integrability of the derivatives makes the functions equicontinuous: their increment over an interval is bounded by the integral of $|u_j^{\prime }|$ on that interval, uniformly small for short intervals. Their fixed initial trace and common derivative $L^1$ bound also make them uniformly bounded. The Arzelà-Ascoli theorem therefore upgrades the convergence to uniform convergence, and in particularUniform integrability does not bound the derivative in $L^{\infty }$. The natural compact class here is absolutely continuous functions, rather than Lipschitz functions. This distinction can affect attainment: smooth strictly convex superlinear integrands can have non-Lipschitz minimizing limits, so it is not repaired merely by adding convexity in the last variable. If one additionally has a uniform Lipschitz bound on the minimizing sequence, weak testing on measurable subsets passes that bound to $v$ and proves the printed $u\in \mathcal{A}$. Alternatively one can formulate the direct method from the outset in the fixed-trace $W^{1,1}$ class. The proof above gives all the asserted convergence and trace conclusions without assuming the missing bound.

Provided the limit belongs to the original admissible class, the definition of the infimum and the assumed inequality giveThe needed fundamental property is sequential lower semicontinuity for the convergence in the compactness argument: uniform, or strong $L^1$, convergence of functions together with weak $L^1$ convergence of derivatives. The extra integrand hypothesis in the classical Tonelli existence theorem is convexity in the derivative variable $v$ for every fixed $(x,u)$, together with the stated coercivity and continuity. In one dimension the resulting integral is lower semicontinuous under the uniform/weak-derivative convergence above.

This lower-semicontinuity assertion does not itself supply the missing Lipschitz membership in (iii). With the natural fixed-trace $W^{1,1}$ admissible class and its own infimum, it completes Tonelli's direct existence argument. Retaining the printed Lipschitz class requires the additional membership estimate; changing admissible classes can otherwise encounter the Lavrentiev phenomenon.

The affine competitor $1-x$ has finite energy and the energy is nonnegative. A minimizing sequence therefore has bounded $\int u_j^{\prime 2}$. Its fixed value at zero gives $|u_j(x)|\le 1+\Vert u_j^{\prime }{\Vert }_2$, so it is also bounded in $L^2$ and hence in $W^{1,2}$. The estimategives uniform equicontinuity. Extract a weakly convergent subsequence in $W^{1,2}$ and, by the Arzelà-Ascoli theorem, a uniformly convergent subsequence to the same limit $u$. The endpoint values are retained. Uniform convergence on a common bounded range gives $\int u_j^{2008}\to \int u^{2008}$, while the derivative norm is weakly lower semicontinuous. Consequentlyso the global minimum is attained in the stated $W^{1,2}$ class. In fact the gradient term is strictly convex on this fixed-endpoint affine class and $t^{2008}$ is convex, so the minimizer is unique as well.