Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2008 / B / 7

With coefficients evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$, put $p=f_{vv}$, $r=f_{uv}$ and $s=f_{uu}$. The second variation isThe Jacobi equation in divergence form is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$; if $r$ is differentiable it is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$. A conjugate point $c>a$ admits a nonzero Jacobi solution vanishing at $a,c$.

There are two positivity conventions worth distinguishing. Strict pointwise positivity means $q(\varphi )>0$ for each nonzero fixed-endpoint variation. Coercive positivity means $q(\varphi )\ge c\int {\varphi }^{\prime 2}$ with one $c>0$ for all such variations; the endpoint Poincaré estimate makes this equivalent to a positive lower bound in the full $H^1$ norm. Under the regular Legendre hypothesis $p>0$, the necessary and sufficient Jacobi criterion is no conjugate point in $(a,b]$; in that regular case it gives both strict and coercive positivity. Equivalently the Jacobi equation has a solution positive on $[a,b]$, and the square identitygives positivity and a uniform derivative lower bound by the positive minimum of $z$ and boundedness of $z,z^{\prime }$. Conversely a conjugate Jacobi field, extended by zero beyond its second zero, is a nonzero Lipschitz variation with $q=0$, contradicting either positivity convention.

Coercivity itself forces $p>0$: local high-frequency variations make the leading coefficient at least the coercivity constant. Mere strict positivity need not force it, as $q(\varphi )=\int {\varphi }^2$ shows. If that weaker convention is intended without regularity, a complete criterion is $p\ge 0$, no conjugate point for every regularized form $q_{\eta }=q+\eta \int {\varphi }^{\prime 2}$ with $\eta >0$, and no nonzero zero-endpoint Lipschitz weak solution of the original Jacobi equation. The regularized criteria imply $q_{\eta }\ge 0$, hence $q\ge 0$ by passage to the limit. For a nonnegative quadratic form, $q(\varphi )=0$ forces all mixed terms with test variations to vanish, giving exactly the weak Jacobi kernel; excluding it gives strict positivity. Necessity follows in reverse. This specifies the degenerate case without imposing an unstated strict Legendre condition.

For weak local minimality, sufficient conditions are the Euler-Lagrange equation and a coercively positive second variation with the stated smoothness. Taylor's remainder is bounded by $\epsilon \Vert \varphi {\Vert }_{H^1}^2$ for small $W^{1,\infty }$ perturbations, so the positive quadratic term dominates it. These are the classical weak minimum sufficient conditions.

At zero the quadratic form isFor $\lambda \ge 0$ it dominates $\int u^{\prime 2}$. For $-{\pi }^2<\lambda <0$ set $\mu =\sqrt{-\lambda }<\pi$ and $z(x)=\cos (\mu (x-1/2))$, positive on the closed interval and satisfying $z^{\prime \prime }+{\mu }^{2}z=0$. The Jacobi identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2$. To obtain a uniform lower bound, let $m=\min z>0$, $K=\max |z^{\prime }|$ and $v=u/z$. Since $\max z=1$ and $\int v^2\le \frac{1}{2}\int v^{\prime 2}$,so $Q_{\lambda }(u)\ge m^2\int u^{\prime 2}/(K^2+2)$.

Finally $I(u)=Q_{\lambda }(u)+O(\int u^4)$ near zero, because ${\sin }^{2}u=u^2+O(u^4)$. The remainder is at most a constant times $\Vert u{\Vert }_{\infty }^2\int u^2\le \frac{1}{2}\Vert u{\Vert }_{\infty }^2\int u^{\prime 2}$. It is dominated by the coercive bound for small perturbations. Thus the second variation is positive and zero is a strict weak local minimizer for every $\lambda >-{\pi }^2$. This argument has not assumed the sharp Poincaré inequality that is to be deduced next.

For a fixed nonzero Lipschitz zero-trace $\varphi$, positivity for every $\lambda >-{\pi }^2$ gives $\int {\varphi }^{\prime 2}+\lambda \int {\varphi }^2>0$. Let $\lambda \downarrow -{\pi }^2$ to obtainFor general $\varphi \in W_0^{1,2}$, approximate by smooth compactly supported functions in the $W^{1,2}$ norm. Both squared norms in the inequality converge, proving it for the full required space. Equality at $\sin (\pi x)$ verifies sharpness.

For a fixed nonzero $u_0\in \mathcal{A}$,The quartic coefficient is positive. Therefore the energy is unbounded below and zero is not a global minimizer, for every value of $\lambda$.