Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 / B / 5

Use $\Vert w{\Vert }_{W^{1,\infty }}=\Vert w{\Vert }_{\infty }+\Vert w^{\prime }{\Vert }_{\infty }$. Fix $u$. For $w$ small, all triples $(x,u+tw,u^{\prime }+t{w}^{\prime })$, $0\le t\le 1$, lie in a fixed compact set up to null sets. The continuous derivatives $f_u,f_v$ are uniformly continuous there, with a modulus $\omega (s)\to 0$ as $s\to 0$. The integral form of the mean-value expansion gives a remainder bounded byThe proposed linear derivative is bounded because its coefficient functions are bounded on that compact set. The remainder is little-oh of the norm, proving Fréchet differentiability andThe derivative $u^{\prime }$ only needs to exist almost everywhere; the uniform compact-range bound handles its measurable representative.

Put $w=t\varphi$ into the differentiability expansion. The remainder is $o(|t|\Vert \varphi \Vert )$, so division by $t$ and passage to zero gives $g^{\prime }(0)=DI(u)\varphi$. For $\varphi =0$ this is immediate as well.

For a zero-trace $\varphi$, the curve $u+t\varphi$ remains admissible. When $|t|$ is small it is within the weak-minimum neighborhood in $W^{1,\infty }$, so its scalar energy has a minimum at zero. The preceding derivative therefore vanishes. Thus $DI(u)\varphi =0$ for every zero-trace variation.

The first variation isThe endpoint terms vanish. The fundamental lemma of the calculus of variations gives $xu=0$ almost everywhere, hence $u=0$ on $(0,1)$. A Lipschitz representative is then identically zero and cannot have terminal value one. Therefore there are no admissible solutions of the stationarity equation.

Stationarity says $\int (2u^{\prime }{\varphi }^{\prime }+\varphi )=0$, or $2u^{\prime \prime }=1$ in distributions. Integrating gives $u=x^2/4+Cx+D$. The endpoint values force $D=0$ and $C=3/4$. Conversely this polynomial satisfies the first-variation identity for every zero-trace variation. Thus the only solution is $u(x)=(x^2+3x)/4$.