Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 / B / 6

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.