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

A continuous embedding $X\hookrightarrow Y$ is an injective continuous linear inclusion, so $\Vert x{\Vert }_Y\le C\Vert x{\Vert }_X$ for a fixed constant. On a finite-measure domain, Holder inequality giveswith $1/\infty =0$. For finite $p>q$ apply Holder inequality to $|u{|}^q\cdot 1$ with exponents $p/q$ and $p/(p-q)$; equality of exponents is immediate, and $p=\infty$ follows by integrating the supremum bound.

Weak convergence $u_j\rightharpoonup u$ means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. The Riemann-Lebesgue lemma says that the Fourier transform of an integrable function tends to zero at high frequency. Since $L^2(0,1)\subset L^1(0,1)$, it makes $\cos (j\pi x)$ weakly null in $L^2$. Its absolute value is at most one, so $u_j{v}_j\in L^2$. For a test $\varphi \in L^2$,while $\int u_{j}v\varphi \to 0$ by the Riemann-Lebesgue lemma applied to $v\varphi \in L^1$. Hence $u_j{v}_j\rightharpoonup 0$. Taking $w_j=u_j=\cos (j\pi x)$ instead gives $u_j{w}_j=(1+\cos (2j\pi x))/2\rightharpoonup 1/2$, so two weakly null factors need not have a weakly null product.

For ${\bar{z}}_j=j^{1/p}{1}_{(0,1/j)}$ and $p^{\prime }=p/(p-1)$, Holder inequality givesThus ${\bar{z}}_j\rightharpoonup 0$ in $L^p$. In contrast, if $z_j$ had a weak $L^1$ limit $z$, testing on every interval strictly away from zero would give integral zero for $z$, hence $z=0$ almost everywhere by the permitted fact. Testing against the constant function one would give $\int z=1$, a contradiction. Therefore $z_j$ has no weak $L^1$ limit.

Choose a smooth compactly supported mollifier of integral one. On every compact subset of $\Omega$, and for sufficiently small $\epsilon$, each translated mollifier is an allowed test, so the assumption gives $w\ast {\rho }_{\epsilon }=0$. The standard mollifier approximation theorem gives $w\ast {\rho }_{\epsilon }\to w$ in local $L^1$. Thus $w=0$ almost everywhere.

For a multi-index $\alpha$, the weak derivative $v=D^{\alpha }u$ is the locally integrable function satisfyingTwo such functions have zero pairing difference against every test, so the first result proves uniqueness. A compactly supported smooth test on an open subset $E$ extends by zero to one on $\Omega$, proving restriction of the derivative to $E$.

Put $w=D^{\beta }u$. For every test $\varphi$ the defining identity applied to $D^{\alpha }\varphi$ givesThus these two distributions are identical; one is represented by a locally integrable function if and only if the other is. Their weak derivatives then agree, proving $D^{\alpha }(D^{\beta }u)=D^{\alpha +\beta }u$ with exactly the asserted existence equivalence.

On $(0,2)$ the displayed piecewise affine function is continuous at one. Integration by parts separately on the two intervals shows its first weak derivative is one on $(0,1)$ and minus one on $(1,2)$; the first boundary terms cancel by continuity. The derivative of that step function is $-2{\delta }_1$. This distribution is not represented by a locally integrable function: tests of height one supported in shrinking neighborhoods of one have pairing $-2$ with it, whereas their integrals against any fixed local $L^1$ function tend to zero. Hence there is a first weak derivative but no locally integrable second weak derivative.

The weak formulation isThe boundary condition is encoded by $u\in H_0^1$. The Sobolev embedding in dimension three gives $H_0^1\hookrightarrow L^6$, hence also $L^4,L^3,L^2$ on the bounded domain. All energy terms and all the displayed pairings are therefore finite.

Pointwise, $t^4/4-t^3/3\ge t^4/8-C$ for a constant $C$, since the difference is a quartic with positive leading coefficient. The Poincaré inequality and Young inequality give $\int fv\ge -\frac{1}{4}\Vert Dv{\Vert }_2^2-C_f$. ConsequentlyThus the infimum is finite and a minimizing sequence is bounded in $H_0^1$. Extract a weakly convergent subsequence there. The Rellich-Kondrachov theorem states that on a bounded $C^1$ domain in dimension three the embedding $H^1\hookrightarrow L^q$ is compact for $q<6$. A subsequence therefore converges strongly in $L^2,L^3,L^4$ to the same limit $u$. The potential terms and forcing converge, while the gradient norm is weakly lower semicontinuous. The weak limit retains zero trace. Hence $I(u)\le \mathrm{liminf}I(v_j)$ and $u$ attains the absolute minimum.

For any $\varphi \in H_0^1$, the function $t\mapsto I(u+t\varphi )$ has a minimum at zero. Differentiating its polynomial terms gives $\int [Du\cdot D\varphi +(u^3-u^2+f)\varphi ]=0$. These differentiations are justified by the Sobolev integrability already noted. This is the weak formulation, proving every energy minimizer is a weak solution.

The Lax-Milgram theorem gives, for a bounded coercive bilinear form on a real Hilbert space and every bounded linear functional, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The Fredholm alternative for a compact operator $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker (I+\mu K{)}^{\ast }$ in the Hilbert-space case.

Uniform ellipticity gives $\sum a_{ij}{\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ for some $\theta >0$. Bounded coefficients make $B_{\lambda }$ bounded on $H^1$, and for $\lambda <-\gamma$,The forcing $v\mapsto \int fv$ is bounded on $H^1$. The Lax-Milgram theorem therefore gives a unique weak solution for every $\lambda <-\gamma$.

When $c=0$, choose the shift ${\lambda }_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int (Kf)v=\int fv$. Thus $K=(L+I{)}^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, orConversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves compactness of $K$ as an $L^2$ operator.

By the supplied kernel assumption, $\ker (I-K)$ consists of constants. Also $K1=1$. Since $K$ is a self-adjoint operator, the Fredholm alternative condition is $Kf\perp 1$, andTherefore a weak solution exists exactly when the mean of $f$ is zero, and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.

For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the weak formulation explicitly defined in the paper, so does not require that stronger boundary identification.