Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 / B

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.