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

Weak convergence means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. Norm convergence implies this by $|\ell (u_j-u)|\le \Vert \ell \Vert \Vert u_j-u\Vert$. The converse fails for coordinate unit vectors in ${\ell }^2$: their norms are one, but pairing with any square-summable sequence tends to zero, so they are weakly null.

For the subsequence assertion apply Gram-Schmidt process to a countable dense subset to choose a countable orthonormal basis $(e_m)$ of the separable Hilbert space and a bound $M$ for the sequence. Repeated subsequence selection and diagonalization make every coefficient $⟨u_{j_k},e_m⟩$ converge to a scalar $a_m$. Bessel inequality gives ${\sum }_{m=1}^N|a_m{|}^2\le M^2$ for every $N$, so $u={\sum }_m{a}_m{e}_m$ exists in $H$ with norm at most $M$. Pairings with finite linear combinations of the basis converge. Approximating any other test vector by such combinations and using the common norm bound proves $u_{j_k}\rightharpoonup u$. The finite-dimensional case uses the same argument with a finite basis.

Apply this to every subsequence of the bounded $L^2$ sequence. Any weakly convergent further subsequence has limit $v$ with ${\int }_a^{b}v=0$, since interval indicators are $L^2$ tests. The permitted fact gives $v=0$ almost everywhere. If the whole sequence failed to converge weakly to zero, some fixed test functional and subsequence would stay a positive distance from zero. Its weakly convergent further subsequence would then contradict the preceding conclusion. Thus the entire sequence converges weakly to zero in $L^2(0,1)$.

The first weak derivative $D_{j}u=v$ is characterized by ${\int }_{\Omega }u{\partial }_j\varphi =-{\int }_{\Omega }v\varphi$ for every $\varphi \in C_c^{\infty }(\Omega )$. The fundamental lemma of the calculus of variations in the locally integrable form says that a function pairing to zero with every such test is zero almost everywhere. Applied to the difference of two candidate derivatives, it proves uniqueness. The Sobolev space consists of $u\in L^p$ with all first weak derivatives in $L^p$, with normFor the one-dimensional tent, integration by parts on each side of one has cancelling junction terms because the function is continuous. Thus $u^{\prime }=1$ on $(0,1)$ and $u^{\prime }=-1$ on $(1,2)$, almost everywhere. Both it and its derivative are bounded, so it belongs to every stated Sobolev space; its displayed norm to the power $p$ is $2/(p+1)+2$.

For $u=x_1/r$ in the punctured ball, ordinary differentiation givesThe function is bounded and $|D_{j}u|\le 2/r$, so polar integration gives finite derivative $L^p$ norms when ${\int }_0^1{r}^{n-1-p}dr<\infty$, namely $p<n$. To verify these are weak derivatives across the origin, integrate by parts outside a ball of radius $\epsilon$ against a compactly supported smooth test. The inner boundary term is at most a fixed constant times ${\epsilon }^{n-1}$, since $u$ is bounded. It tends to zero for $n>1$. The volume terms converge by local integrability of the displayed derivative. Thus $u\in W^{1,p}(B)$ for $1\le p<n$ with exactly these derivatives. Its value at the origin can be chosen arbitrarily.

Bounded coefficients and Cauchy-Schwarz give $|B[u,v]|\le C\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Uniform ellipticity gives $B[u,u]\ge \theta \Vert Du{\Vert }_2^2$ because $c\ge 0$. Poincaré inequality $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$ holds on $H_0^1$ of a bounded domain, soA weak solution is $u\in H_0^1$ satisfying $B[u,\varphi ]=\int f\varphi$ for every $\varphi \in H_0^1$. The energy is finite by boundedness of $B$ and the forcing functional, andIt is bounded below and a minimizing sequence is bounded in the Hilbert space $H_0^1$. Its closed linear span is separable, so part 1 supplies a weakly convergent subsequence with limit $u$. Symmetry and nonnegativity of $B$ showThe forcing term converges by weak convergence. Thus $I(u)\le \mathrm{liminf}I(v_j)$, proving attainment. First variation in every zero-trace direction gives the weak equation. If $u,w$ are two solutions, test their difference equation with $u-w$ and use coercivity to get $u=w$. Hence there is exactly one weak solution, and it is the unique energy minimizer. Indeed $I(v)-I(u)=\frac{1}{2}B[v-u,v-u]$ directly verifies the minimizing property.

The spectral theorem for compact self-adjoint operators used here concerns a compact self-adjoint operator: its nonzero eigenvalues are real, have finite multiplicity and can accumulate only at zero; the orthogonal complement of its kernel is the closed span of its orthonormal eigenvectors. Self-adjointness is essential for this eigenbasis statement.

Let $Kf$ be the unique zero-parameter weak solution. Uniqueness gives linearity. Testing with $Kf$ and using coercivity gives $\Vert Kf{\Vert }_{H^1}\le {\nu }^{-1}\Vert f{\Vert }_2$. The Rellich-Kondrachov theorem makes the embedding $H_0^1(\Omega )\hookrightarrow L^2(\Omega )$ compact on this bounded $C^1$ domain, so $K$ is compact on $L^2$. Symmetry givesIt is self-adjoint and positive. It is injective: if $Kf=0$, the weak equation pairs $f$ to zero with all $H_0^1$ tests, dense in $L^2$, so $f=0$. Moreover $⟨Kf,f⟩=B[Kf,Kf]>0$ for $f\ne 0$.

The parameter equation is equivalent to $u=K(f+\lambda u)$, that isA solution of this operator equation in $L^2$ lies in $H_0^1$, because its right-side representation is $K(f+\lambda u)$, and the equivalence follows in both directions. For $\lambda \le 0$, the form $B[u,v]-\lambda ⟨u,v⟩$ remains coercive; the preceding minimization argument gives a unique weak solution.

The spectral theorem now supplies a complete orthonormal basis $({\varphi }_j)$ of $L^2$ with $K{\varphi }_j={\kappa }_j{\varphi }_j$, ${\kappa }_j>0$, and ${\kappa }_j\to 0$. The elliptic eigenvalues are ${\lambda }_j=1/{\kappa }_j>0$, with finite multiplicity and tending to infinity, and $L{\varphi }_j={\lambda }_j{\varphi }_j$ weakly. If $\lambda$ is not one of them, thenThis converges even in $H_0^1$: its squared energy norm is ${\sum }_j{\lambda }_j|⟨f,{\varphi }_j⟩{|}^2/|{\lambda }_j-\lambda {|}^2$, finite because the multiplier is bounded over $j$. Partial sums solve the weak equation with the corresponding $L^2$ projections of $f$. Passing to the limit proves the equation for every test; the same coefficients force uniqueness. Thus the expansion represents the unique weak solution away from the elliptic eigenvalues.