Past exam of the calculus of variations course of the University of Oxford / 2010 (C5.2b, DMAC 2B65)

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\rho >0$ such thatThe Sobolev norm controls both the function and its first weak derivative. The boundary values remain fixed because both functions are admissible.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ satisfiesOnly the function values must be close; admissible derivatives may be large. Every strong local minimizer is therefore also a weak local minimizer of a variational integral.

A point of convexity of $g\in C^1(\mathbb{R})$ is a point $v_0$ where its tangent line supports the entire graph from below:The Weierstrass necessary condition says that, at almost every $x$, ${\bar{u}}^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,\bar{u}(x),v)$. In terms of the Weierstrass excess function,

Now use the stated derivative coercivity. If $\bar{u}$ were not a local minimizer in $L^1$, there would be $v_n\in \mathcal{A}$ with $\Vert v_n-\bar{u}{\Vert }_1\to 0$ and $I(v_n)<I(\bar{u})$. The energy bound givesThe common value $v_n(a)=\alpha$ controls their function norms as well, by $|v_n(x)|\le |\alpha |+{\int }_a^b|v_n^{\prime }|$. Thus $(v_n)$ is bounded in $W^{1,2010}(a,b)$. The one-dimensional compact Sobolev embedding into $C([a,b])$ supplies a uniformly convergent subsequence. Its limit must be $\bar{u}$, since the same subsequence converges to $\bar{u}$ in $L^1$. Eventually $\Vert v_n-\bar{u}{\Vert }_{\infty }<\rho$, contradicting strong local minimality. Hence $\bar{u}$ is also a local minimizer in $L^1$. This is coercivity transfers strong minimality to L1 minimality.

Zero is a strict weak local minimizer, but is not a strong local minimizer. Put $D={\int }_0^1|u^{\prime }{|}^{2}dx$. Since $u(0)=0$, the Cauchy-Schwarz inequality gives $|u(x){|}^2\le xD$, and thereforeFor sufficiently small $\Vert u^{\prime }{\Vert }_{\infty }$, $\sin (u^{\prime 2})\ge \frac{1}{2}{u}^{\prime 2}$ pointwise. Hencewith equality only when $u=0$. Thus zero is a strict weak local minimizer of a variational integral.

For the strong claim, define a Lipschitz triangular waveIt has zero endpoint values, $\Vert u_N{\Vert }_{\infty }=\sqrt{\pi }/(2N)\to 0$, and $u_N^{\prime }=\pm \sqrt{\pi }$ almost everywhere. Thus $\sin (u_N^{\prime 2})=\sin \pi =0$, while $u_N$ is nonzero on a set of positive measure. ConsequentlyThese uniformly small competitors disprove the strong local minimizer of a variational integral property.

Zero is a strict strong local minimizer, and therefore also a strict weak local minimizer. Here $I(0)=-2/2010$, andFor $|u|\le 1$, Taylor's theorem gives $\cosh u-1\le \frac{1}{2}\cosh (1)u^2$. Also $u(0)=0$ and the Cauchy-Schwarz inequality give ${\int }_0^1{u}^{2}dx\le \frac{1}{2}{\int }_0^1{u}^{\prime 2}dx$. ThereforeThe coefficient is positive. Thus every nonzero admissible $u$ with $\Vert u{\Vert }_{\infty }\le 1$ has strictly greater energy, proving that zero is a strict strong local minimizer of a variational integral.

For a fixed-endpoint variation $\phi \in W_0^{1,\infty }(a,b)$, the second variation iswhere all coefficients are evaluated at $(x,\bar{u}(x),{\bar{u}}^{\prime }(x))$. The notation ${\delta }^{2}I(\bar{u})>0$ means this is strictly positive for every nonzero admissible variation. This is pointwise positive definiteness of the quadratic form, not an assertion of a prescribed uniform lower bound.

PutAfter integration by parts, the form is $q(\phi )={\int }_a^b(P{\phi }^{\prime 2}+Q{\phi }^2)dx$. Its Jacobi equation isIn the regular case $P>0$ on $[a,b]$, a conjugate point $c\in (a,b]$ to $a$ is a point for which a nonzero solution satisfies $w(a)=w(c)=0$. Equivalently it is a zero of the normalized solution with $w(a)=0$, $P(a)w^{\prime }(a)=1$.

Under this regular Strengthened Legendre condition, the Jacobi positivity criterion isThe endpoint $b$ must be excluded as a conjugate point too. To prove necessity, suppose a Jacobi solution vanishes at $a$ and $c$. Define $\phi =w$ on $[a,c]$ and zero on $[c,b]$. It is a nonzero Lipschitz admissible variation, anda contradiction. For completeness, the converse follows from the Jacobi identity for the second variation: when the normalized $w$ has no further zero,The boundary contribution $[P(w^{\prime }/w){\phi }^2{]}_a^b$ vanishes, including the limit at $a$ since $w$ has a simple zero and $\phi =O(x-a)$. Equality forces $\phi =Cw$, and $\phi (b)=0$ with $w(b)\ne 0$ forces $C=0$.

Independently of regularity, positivity necessarily implies the ordinary Legendre necessary condition $P\ge 0$. If $P<0$ on an interval, localized rapidly oscillating test functions have a negative derivative contribution of order frequency squared, dominating the bounded zero-order term. This contradicts positivity.

The assumptions stated here do not themselves impose $P>0$. That strengthened condition is a regularity hypothesis for the familiar Jacobi criterion, and is not necessary for strict positivity: the integrand $f=u^2$ gives $q(\phi )=2\int {\phi }^2>0$ with $P=0$. For a criterion covering the full possibly degenerate class, use Regularized Jacobi criterion for a degenerate second variation. The following conditions are necessary and sufficient:

These conditions are necessary because $q>0$ gives $q_{\eta }(\phi )=q(\phi )+\eta \int {\phi }^{\prime 2}>0$, so the regular criterion just proved applies; a weak Jacobi solution tested against itself would give $q=0$. They are sufficient as well: the regular criterion gives $q_{\eta }>0$ for every $\eta >0$, hence $q\ge 0$ by letting $\eta \downarrow 0$. If $q(\phi )=0$, nonnegativity of $q(\phi +t\psi )$ for every real $t$ forces its mixed bilinear term to vanish for every test $\psi$. This is precisely the weak Jacobi equation. The third condition then forces $\phi =0$. Thus the criterion treats degeneracy without silently assuming the strengthened condition.

Here $P=2$, $Q=-2{\lambda }^2$ and the Jacobi equation is $w^{\prime \prime }+{\lambda }^{2}w=0$. With $w(0)=0$ and $w^{\prime }(0)=1$, its solution is $w(x)=\sin (|\lambda |x)/|\lambda |$ for $\lambda \ne 0$, and $w(x)=x$ for $\lambda =0$. Its first positive zero is $\pi /|\lambda |$ when $\lambda \ne 0$.

If ${\lambda }^2<{\pi }^2/b^2$, no conjugate point occurs in $(0,b]$. The Jacobi positivity criterion givesSince $I_{\lambda }$ itself is this quadratic form divided by two, $I_{\lambda }(u)>0=I_{\lambda }(0)$ for every nonzero admissible $u$. Thus zero is the unique global minimizer. This argument establishes the result before using the sharp Poincaré inequality.

Fix an admissible $u$. The preceding result gives ${\int }_0^b{u}^{\prime 2}dx\ge {\lambda }^2{\int }_0^b{u}^{2}dx$ for every ${\lambda }^2<{\pi }^2/b^2$. Letting ${\lambda }^2$ increase to ${\pi }^2/b^2$ provesEquality is attained byIndeed the derivative and function squared integrals are $A^2{\pi }^2/(2b)$ and $A^{2}b/2$. These are all equality cases: an equality function minimizes the nonnegative threshold quadratic form, so its first variation vanishes. The resulting weak equation $u^{\prime \prime }+(\pi /b{)}^{2}u=0$ makes $u$ smooth and, with the endpoint conditions, yields precisely these multiples. This is the Wirtinger inequality.

Use $u_A(x)=A\sin (\pi x/b)$, which is admissible for every real $A$. Direct integration givesThe coefficient is strictly negative, so the energy tends to minus infinity as $|A|\to \infty$. Therefore

The function $u_{\ast }(x)=x^{1/4}$ has weak derivative $u_{\ast }^{\prime }(x)=\frac{1}{4}{x}^{-3/4}\in L^1(0,1)$ and the required endpoint values, so it belongs to the admissible Sobolev space. Its fourth power is $x$, and henceSince the integrand $(u^4-x{)}^2$ is nonnegative, zero is a global lower bound. Thus $u_{\ast }$ is a global minimizer. Its derivative does not belong to $L^{4/3}$, since its $4/3$ power is proportional to $x^{-1}$; positive regularization therefore needs different competitors.

A convenient form of the Tonelli existence theorem is: on a finite interval with prescribed endpoint values, if a continuous integrand $F(x,z,v)$ is convex in $v$ and satisfies $F(x,z,v)\ge c|v{|}^p-C$ for some $p>1$, then, provided a finite-energy admissible competitor exists, its integral attains a minimum in the corresponding $W^{1,p}$ class. The lower bound may more generally subtract an integrable function of $x$. Compact one-dimensional Sobolev embedding controls $z$, while convexity gives weak lower semicontinuity in the derivative.

Here $p=4/3$ and $F_{\epsilon }=(z^4-x{)}^2+\epsilon |v{|}^{4/3}$ is continuous, convex in $v$, and bounded below by $\epsilon |v{|}^{4/3}$. The competitor $u(x)=x$ has finite energy. Thus for every $\epsilon >0$ the Tonelli existence theorem gives an admissible minimizer in $W^{1,4/3}(0,1)$.

Any finite-energy member of the larger stated $W^{1,1}$ class necessarily has derivative in $L^{4/3}$ and bounded function values by absolute continuity; it therefore belongs to $W^{1,4/3}$. All other competitors have infinite energy. Hence the same function $u_{\epsilon }$ is a global minimizer in the full stated class.

The finite-energy comparison $I_{\epsilon }(u_{\epsilon })\le I_{\epsilon }(x)<\infty$ gives $\epsilon \int |u_{\epsilon }^{\prime }{|}^{4/3}<\infty$. Together with the endpoint value and absolute continuity, this proves $u_{\epsilon }\in W^{1,4/3}(0,1)$.

For $\alpha >1/4$, the power competitor $u_{\alpha }=x^{\alpha }$ is admissible and has finite derivative energy. Direct integration givesChoose $\delta ={\epsilon }^{1/3}$ and $\alpha =1/4+\delta$. The first three terms tend to zero as $\alpha \downarrow 1/4$, while the last equals $\frac{3}{4}{\epsilon }^{2/3}(1/4+{\epsilon }^{1/3}{)}^{4/3}\to 0$. By minimality and nonnegativity,One also obtains the quantitative bound $I_{\epsilon }(u_{\epsilon })=O({\epsilon }^{2/3})$: for $\alpha =1/4+\delta$, the mean value theorem gives $|x^{4\alpha }-x|\le 4\delta x|\log x|$, and ${\int }_0^1{x}^2{\log }^{2}xdx=2/27$. Thus the squared mismatch is at most $32{\delta }^2/27$, matching the order of the derivative penalty with this choice. This is a power-law comparison for vanishing derivative regularization.

Use the direct method in the calculus of variations. The affine function $1-x$ is admissible with finite energy, and the energy is nonnegative since $2010$ is even. Thus $m=\inf I$ is finite. Choose a minimizing sequence $u_n$ with uniformly bounded energy. ThenThe fixed value $u_n(0)=1$ and the Cauchy-Schwarz inequality give $\Vert u_n{\Vert }_{\infty }\le 1+\Vert u_n^{\prime }{\Vert }_2$ and $|u_n(x)-u_n(y)|\le \Vert u_n^{\prime }{\Vert }_2|x-y{|}^{1/2}$. Hence the sequence is bounded in $W^{1,2}$ and uniformly bounded and equicontinuous.

By reflexivity and the Arzelà-Ascoli theorem, pass to a subsequence with $u_n\rightharpoonup u$ in $W^{1,2}$ and $u_n\to u$ uniformly. The two limits coincide; uniform convergence also preserves $u(0)=1$, $u(1)=0$. Since all functions stay in a common bounded interval, uniform convergence gives $u_n^{2010}\to u^{2010}$ uniformly, and therefore convergence of the potential integrals. The derivative term is lower semicontinuous for weak convergence, soAdmissibility gives $I(u)\ge m$. Thus