Past exam of the calculus of variations course of the University of Oxford / 2009 / 3

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ makes $I(v)\ge I(\bar{u})$ whenever $v\in \mathcal{A}$ and $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$. Both the function and its derivative are required to be close.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which the same conclusion holds whenever $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. Every strong local minimizer is weak, because Sobolev closeness implies uniform closeness. A global minimizer is both, and the converse implication from weak to strong is not generally valid.

If the integrand is independent of $u$, a weak local minimizer has vanishing first variation:The bounded function $h(x)=f_v(x,{\bar{u}}^{\prime }(x))$ therefore has zero distributional derivative, so is constant almost everywhere. For an elementary test-function argument, any smooth compactly supported $\psi$ of integral zero is the derivative of a compactly supported smooth $\phi$. Hence $\int h\psi =0$; subtracting a fixed unit-integral test function from arbitrary tests shows that $h$ agrees with one constant against every test. The fundamental lemma then givesThis is the first integral for a variational integrand independent of the function value.

A point of convexity $v_0$ of a differentiable real function $g$ has a globally supporting tangent: $g(v)\ge g(v_0)+g^{\prime }(v_0)(v-v_0)$ for every $v$. The Weierstrass necessary condition for a strong minimum says that ${\bar{u}}^{\prime }(x)$ is such a point for $v\mapsto f(x,\bar{u}(x),v)$ for almost every $x$. Equivalently, the Weierstrass excess function is nonnegative at every trial slope.

Zero is neither a weak nor a strong local minimizer. Take $u_{\epsilon }(x)=-\epsilon \sin (\pi x)$. It tends to zero in $W^{1,\infty }$ and has zero endpoint values. Since $\sqrt{1+t}\le 1+t/2$ for $t\ge 0$,for sufficiently small positive $\epsilon$. Thus even weakly close admissible competitors decrease the energy.

Zero is neither a weak nor a strong local minimizer. For any nonzero smooth zero-trace $\phi$, Taylor expansion givesThe mixed integral $\int \phi {\phi }^{\prime }=0$ by the endpoint values. Writing $D=\int {\phi }^{\prime 2}$ and $U=\int {\phi }^2$, the quadratic coefficient is $-3D/8+U/2$. The elementary estimate $U\le D/2$ on $[0,1]$ makes this at most $-D/8<0$. Thus arbitrarily small variations decrease the energy. The separate coefficient one on $u^2$ is retained in this calculation.

Zero is a strict weak local minimizer, but is not a strong local minimizer. The derivative integrand has expansion $\cosh v\cos (v^2)=1+v^2/2+O(v^4)$, so for small $|v|$ it is at least $1+3v^2/8$. Concavity gives $(1+u^2{)}^{1/4}\le 1+u^2/4$ for every real $u$. Therefore, when $\Vert u^{\prime }{\Vert }_{\infty }$ is small,using $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. This proves strict weak minimality.

For the strong claim, let $v_N(x)=\sqrt{\pi }\mathrm{dist}(x,N^{-1}\mathbb{Z})$. Its amplitude tends to zero and its derivative is $\pm \sqrt{\pi }$ almost everywhere, so the derivative integrand is the constant $-\cosh \sqrt{\pi }$. The other integrand term is at most $-2$. HenceThis disproves strong minimality. The same obstruction is detected by the Weierstrass necessary condition, since the supporting-tangent inequality at slope zero fails at $v=\sqrt{\pi }$.