Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 (C5.1, DMAC 2465, DMSU 2465)

Weak convergence means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. Norm convergence implies this by $|\ell (u_j-u)|\le \Vert \ell \Vert \Vert u_j-u\Vert$. The converse fails for coordinate unit vectors in ${\ell }^2$: their norms are one, but pairing with any square-summable sequence tends to zero, so they are weakly null.

For the subsequence assertion apply Gram-Schmidt process to a countable dense subset to choose a countable orthonormal basis $(e_m)$ of the separable Hilbert space and a bound $M$ for the sequence. Repeated subsequence selection and diagonalization make every coefficient $⟨u_{j_k},e_m⟩$ converge to a scalar $a_m$. Bessel inequality gives ${\sum }_{m=1}^N|a_m{|}^2\le M^2$ for every $N$, so $u={\sum }_m{a}_m{e}_m$ exists in $H$ with norm at most $M$. Pairings with finite linear combinations of the basis converge. Approximating any other test vector by such combinations and using the common norm bound proves $u_{j_k}\rightharpoonup u$. The finite-dimensional case uses the same argument with a finite basis.

Apply this to every subsequence of the bounded $L^2$ sequence. Any weakly convergent further subsequence has limit $v$ with ${\int }_a^{b}v=0$, since interval indicators are $L^2$ tests. The permitted fact gives $v=0$ almost everywhere. If the whole sequence failed to converge weakly to zero, some fixed test functional and subsequence would stay a positive distance from zero. Its weakly convergent further subsequence would then contradict the preceding conclusion. Thus the entire sequence converges weakly to zero in $L^2(0,1)$.

The first weak derivative $D_{j}u=v$ is characterized by ${\int }_{\Omega }u{\partial }_j\varphi =-{\int }_{\Omega }v\varphi$ for every $\varphi \in C_c^{\infty }(\Omega )$. The fundamental lemma of the calculus of variations in the locally integrable form says that a function pairing to zero with every such test is zero almost everywhere. Applied to the difference of two candidate derivatives, it proves uniqueness. The Sobolev space consists of $u\in L^p$ with all first weak derivatives in $L^p$, with normFor the one-dimensional tent, integration by parts on each side of one has cancelling junction terms because the function is continuous. Thus $u^{\prime }=1$ on $(0,1)$ and $u^{\prime }=-1$ on $(1,2)$, almost everywhere. Both it and its derivative are bounded, so it belongs to every stated Sobolev space; its displayed norm to the power $p$ is $2/(p+1)+2$.

For $u=x_1/r$ in the punctured ball, ordinary differentiation givesThe function is bounded and $|D_{j}u|\le 2/r$, so polar integration gives finite derivative $L^p$ norms when ${\int }_0^1{r}^{n-1-p}dr<\infty$, namely $p<n$. To verify these are weak derivatives across the origin, integrate by parts outside a ball of radius $\epsilon$ against a compactly supported smooth test. The inner boundary term is at most a fixed constant times ${\epsilon }^{n-1}$, since $u$ is bounded. It tends to zero for $n>1$. The volume terms converge by local integrability of the displayed derivative. Thus $u\in W^{1,p}(B)$ for $1\le p<n$ with exactly these derivatives. Its value at the origin can be chosen arbitrarily.

Bounded coefficients and Cauchy-Schwarz give $|B[u,v]|\le C\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Uniform ellipticity gives $B[u,u]\ge \theta \Vert Du{\Vert }_2^2$ because $c\ge 0$. Poincaré inequality $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$ holds on $H_0^1$ of a bounded domain, soA weak solution is $u\in H_0^1$ satisfying $B[u,\varphi ]=\int f\varphi$ for every $\varphi \in H_0^1$. The energy is finite by boundedness of $B$ and the forcing functional, andIt is bounded below and a minimizing sequence is bounded in the Hilbert space $H_0^1$. Its closed linear span is separable, so part 1 supplies a weakly convergent subsequence with limit $u$. Symmetry and nonnegativity of $B$ showThe forcing term converges by weak convergence. Thus $I(u)\le \mathrm{liminf}I(v_j)$, proving attainment. First variation in every zero-trace direction gives the weak equation. If $u,w$ are two solutions, test their difference equation with $u-w$ and use coercivity to get $u=w$. Hence there is exactly one weak solution, and it is the unique energy minimizer. Indeed $I(v)-I(u)=\frac{1}{2}B[v-u,v-u]$ directly verifies the minimizing property.

The spectral theorem for compact self-adjoint operators used here concerns a compact self-adjoint operator: its nonzero eigenvalues are real, have finite multiplicity and can accumulate only at zero; the orthogonal complement of its kernel is the closed span of its orthonormal eigenvectors. Self-adjointness is essential for this eigenbasis statement.

Let $Kf$ be the unique zero-parameter weak solution. Uniqueness gives linearity. Testing with $Kf$ and using coercivity gives $\Vert Kf{\Vert }_{H^1}\le {\nu }^{-1}\Vert f{\Vert }_2$. The Rellich-Kondrachov theorem makes the embedding $H_0^1(\Omega )\hookrightarrow L^2(\Omega )$ compact on this bounded $C^1$ domain, so $K$ is compact on $L^2$. Symmetry givesIt is self-adjoint and positive. It is injective: if $Kf=0$, the weak equation pairs $f$ to zero with all $H_0^1$ tests, dense in $L^2$, so $f=0$. Moreover $⟨Kf,f⟩=B[Kf,Kf]>0$ for $f\ne 0$.

The parameter equation is equivalent to $u=K(f+\lambda u)$, that isA solution of this operator equation in $L^2$ lies in $H_0^1$, because its right-side representation is $K(f+\lambda u)$, and the equivalence follows in both directions. For $\lambda \le 0$, the form $B[u,v]-\lambda ⟨u,v⟩$ remains coercive; the preceding minimization argument gives a unique weak solution.

The spectral theorem now supplies a complete orthonormal basis $({\varphi }_j)$ of $L^2$ with $K{\varphi }_j={\kappa }_j{\varphi }_j$, ${\kappa }_j>0$, and ${\kappa }_j\to 0$. The elliptic eigenvalues are ${\lambda }_j=1/{\kappa }_j>0$, with finite multiplicity and tending to infinity, and $L{\varphi }_j={\lambda }_j{\varphi }_j$ weakly. If $\lambda$ is not one of them, thenThis converges even in $H_0^1$: its squared energy norm is ${\sum }_j{\lambda }_j|⟨f,{\varphi }_j⟩{|}^2/|{\lambda }_j-\lambda {|}^2$, finite because the multiplier is bounded over $j$. Partial sums solve the weak equation with the corresponding $L^2$ projections of $f$. Passing to the limit proves the equation for every test; the same coefficients force uniqueness. Thus the expansion represents the unique weak solution away from the elliptic eigenvalues.

Use $\Vert w{\Vert }_{W^{1,\infty }}=\Vert w{\Vert }_{\infty }+\Vert w^{\prime }{\Vert }_{\infty }$. Fix $u$. For $w$ small, all triples $(x,u+tw,u^{\prime }+t{w}^{\prime })$, $0\le t\le 1$, lie in a fixed compact set up to null sets. The continuous derivatives $f_u,f_v$ are uniformly continuous there, with a modulus $\omega (s)\to 0$ as $s\to 0$. The integral form of the mean-value expansion gives a remainder bounded byThe proposed linear derivative is bounded because its coefficient functions are bounded on that compact set. The remainder is little-oh of the norm, proving Fréchet differentiability andThe derivative $u^{\prime }$ only needs to exist almost everywhere; the uniform compact-range bound handles its measurable representative.

Put $w=t\varphi$ into the differentiability expansion. The remainder is $o(|t|\Vert \varphi \Vert )$, so division by $t$ and passage to zero gives $g^{\prime }(0)=DI(u)\varphi$. For $\varphi =0$ this is immediate as well.

For a zero-trace $\varphi$, the curve $u+t\varphi$ remains admissible. When $|t|$ is small it is within the weak-minimum neighborhood in $W^{1,\infty }$, so its scalar energy has a minimum at zero. The preceding derivative therefore vanishes. Thus $DI(u)\varphi =0$ for every zero-trace variation.

The first variation isThe endpoint terms vanish. The fundamental lemma of the calculus of variations gives $xu=0$ almost everywhere, hence $u=0$ on $(0,1)$. A Lipschitz representative is then identically zero and cannot have terminal value one. Therefore there are no admissible solutions of the stationarity equation.

Stationarity says $\int (2u^{\prime }{\varphi }^{\prime }+\varphi )=0$, or $2u^{\prime \prime }=1$ in distributions. Integrating gives $u=x^2/4+Cx+D$. The endpoint values force $D=0$ and $C=3/4$. Conversely this polynomial satisfies the first-variation identity for every zero-trace variation. Thus the only solution is $u(x)=(x^2+3x)/4$.

Along the reference curve put $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$. The second variation isStrict positivity means $q(\varphi )>0$ for every nonzero zero-trace variation. Under the given regular Legendre condition $p>0$, the no-conjugate-point criterion also supplies a uniform coercive lower bound. The Jacobi equation is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$. A point $c>a$ is conjugate to $a$ when a nonzero solution vanishes at both $a$ and $c$. In this regular setting, absence of conjugate points in $(a,b]$ is equivalent to existence of a Jacobi solution positive throughout $[a,b]$.

Choose such a positive solution $z$. The given square identity extends from smooth compactly supported variations to $H_0^1$ by density and boundedness of its coefficients. With $v=\varphi /z$ it reads $q(\varphi )=\int p{z}^2{v}^{\prime 2}$. The positive minima of $p,z$ bound this below by $c\Vert v^{\prime }{\Vert }_2^2$. A zero-trace function obeys $\Vert v{\Vert }_2\le C\Vert v^{\prime }{\Vert }_2$, while ${\varphi }^{\prime }=z{v}^{\prime }+z^{\prime }v$ and $\varphi =zv$ give $\Vert \varphi {\Vert }_{H^1}\le C_1\Vert v^{\prime }{\Vert }_2$. ConsequentlyIn particular the required second variation is positive for all nonzero Lipschitz zero-trace variations.

The Jacobi equation is $z^{\prime \prime }+{\lambda }^{2}z=0$. For $|\lambda |<\pi$ the function $z=\cos (\lambda (x-1/2))$ is positive on the closed interval. The preceding square identity giveswith $z=1$ if $\lambda =0$. As in part (a), its positive minimum and bounded derivative provide a constant $\gamma >0$ with $I(u)\ge \gamma \Vert u{\Vert }_{H^1}^2$. The second variation at zero is $2I$, so is positive, and zero is the unique global minimizer.

For a fixed zero-trace $u$, positivity of $\int u^{\prime 2}-{\lambda }^2\int u^2$ for all ${\lambda }^2<{\pi }^2$ gives, by letting ${\lambda }^2\uparrow {\pi }^2$,This derivation did not assume the sharp inequality in advance. For equality, the nonnegative quadratic form $Q(u)=\int (u^{\prime 2}-{\pi }^2{u}^2)$ has $Q(u)=0$. Expanding $Q(u+t\varphi )\ge 0$ for both signs of $t$ forces $\int u^{\prime }{\varphi }^{\prime }-{\pi }^{2}u\varphi =0$ for every test. Thus $u^{\prime \prime }+{\pi }^{2}u=0$, and the zero endpoint values give exactly $u(x)=C\sin (\pi x)$, including $C=0$. These functions do give equality.

Take $u_t(x)=t\sin (\pi x)$. It is admissible andwhen ${\lambda }^2>{\pi }^2$. Hence the infimum is minus infinity. The roman label (iii) is retained exactly as printed.

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

A strong local minimizer of a variational integral has the same property for all admissible $v$ with $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$; its derivative need not be close. Since the supremum norm is bounded by the Sobolev norm, this neighborhood contains a sufficiently small weak neighborhood. Thus every strong local minimizer is a weak local minimizer of a variational integral.

A global minimizer satisfies $I(\bar{u})\le I(v)$ for every admissible $v$.

A point of convexity $s$ of a differentiable scalar function is one where its tangent supports the entire graph: $g(q)\ge g(s)+g^{\prime }(s)(q-s)$ for every $q\in \mathbb{R}$. The Weierstrass necessary condition for strong minimality, under its differentiability assumptions in the slope variable, isfor almost every $x$. In other words the reference slope is a point of convexity of the slope integrand. The later example is smooth, so these derivative-based conditions apply there.

Write $g(v)=-\cos (\pi v/2)$. For the stipulated $0<\delta <1$, the displayed tent is continuous at its junction, has zero endpoint values, and its slopes are $\delta$ and $\delta -4$. At both slopes $g^{\prime }$ has the same value $c=(\pi /2)\sin (\pi \delta /2)>0$, and $g^{\prime \prime }=({\pi }^2/4)\cos (\pi v/2)$ is strictly positive on common sufficiently small neighborhoods of them. Taylor's theorem gives, for small $\Vert {\varphi }^{\prime }{\Vert }_{\infty }$,Thus $u^{\delta }$ is a strict weak local minimizer of a variational integral. It is not a strong local minimizer of a variational integral: the global supporting-tangent requirement at either slope has nonzero slope $c$, whereas $g$ is bounded. Letting the trial slope tend to positive infinity violates that requirement. The Weierstrass necessary condition therefore fails.

The opening parameter range $\delta \in (0,1)$ is essential. The subsequent phrase “for each $\delta >0$” must be read within it; taken literally beyond that range it is false. At $\delta =2$ the tent slopes are $2,-2$, where $g=1$ and $g^{\prime \prime }<0$, so this is not a weak minimum.

If $u$ is a strong local minimizer of a variational integral, the Weierstrass condition makes $u^{\prime }(x)$ a point of convexity of the bounded function $g$ for almost every $x$. A global supporting line to a bounded function must have zero slope, and its contact value must be a global minimum. Here that value is $-1$. Therefore $g(u^{\prime })=-1$ almost everywhere and $I(u)=-1$, while every admissible energy is at least $-1$. Hence every strong local minimizer is global.

For every positive integer $n$, set $u_n(x)=4n\min \{x,1-x\}$. Its slopes are $4n,-4n$ almost everywhere, both giving $g=-1$. Hence $I(u_n)=-1$, the universal lower bound, and these tents are all distinct. Thus there are infinitely many global minimizers. Zero is another one.

Superlinearity and continuity make $\Phi$ bounded below by a finite constant $M$. Consequently $I(u)\ge M(b-a)$ for every admissible function. The affine endpoint interpolant has finite energy by continuity on its compact range, so the infimum is finite. Some other admissible energies can be infinite; this does not affect the bounded-below or minimizing-sequence argument.

The De la Vallée-Poussin criterion characterizes uniform integrability by existence of a nonnegative increasing convex function $\Psi$ with $\Psi (s)/s\to \infty$ and a uniform bound on $\int \Psi (|v_j|)$. On a finite interval this gives a weakly convergent subsequence in $L^1$ by the Dunford-Pettis theorem.

After discarding a finite prefix of the minimizing sequence if needed, put $H(t)=\Phi (t)-M\ge 0$. Its derivative sequence has $\int H(u_j^{\prime })\le C$. The superlinear bound directly gives uniform integrability: for $H(t)\ge A|t|$ outside $[-R_A,R_A]$,If the convex version of the criterion is used, choose increasing $R_m\to \infty$ so $H(t)\ge m^2|t|$ for $|t|\ge R_m$, and set $\Psi (s)={\sum }_{m\ge 1}(s-R_m{)}_{+}$. This is locally a finite sum, convex and increasing. It is superlinear because for fixed $N$ and $s\ge 2R_N$, $\Psi (s)\ge Ns/2$. If exactly $m$ terms are active, $\Psi (s)\le ms\le H(t)$ for $s=|t|$, so its integrals have the same bound. The criterion therefore supplies $u_{j_k}^{\prime }\rightharpoonup v$ in $L^1$.

Define $u(x)=\alpha +{\int }_a^{x}v$. Interval-indicator tests give $u_{j_k}(x)\to u(x)$ for every $x$, including $b$, so its endpoint values are correct and $u\in W^{1,1}$. The common derivative $L^1$ bound makes these functions uniformly bounded. Uniform integrability also gives common equicontinuity, since an increment is bounded by the derivative integral on the intervening interval. The Arzelà-Ascoli theorem yields uniform convergence to the same identified limit, and in particularHere the printed admissible space is $W^{1,1}$, so this compactness argument gives precisely the needed membership.

Write $l={\inf }_{\mathcal{A}}I$. Since the limit is admissible, the assumed lower-semicontinuity inequality gives $l\le I(u)\le \mathrm{liminf}I(u_{j_k})=l$. Thus $I(u)=l$ and $u$ is a global minimizer.

The extra integrand hypothesis in the Tonelli existence theorem for one dimension is convexity in the derivative variable for each fixed $(x,u)$. Together with the stated continuity and coercive lower bound it gives the lower-semicontinuity property along the uniform-function/weak-derivative convergence just proved.

The assumed weak-star limit is the constant $m=\theta a+(1-\theta )b$. Each integer rescaling has exactly $j$ full periods on $(0,1)$, so a change of variables on those periods givesApply the asserted lower semicontinuity to this sequence. The integral of $f$ of its constant limit is $f(m)$, givingThis holds for every real $a,b$ and $0<\theta <1$; the endpoint values of $\theta$ give equality. Therefore $f$ is convex.