Past exam of the calculus of variations course of the University of Oxford / 2010 / 3

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