Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 63 / 3

Work first in a real Hilbert space. A bounded linear operator $L$ is self-adjoint whenFor the variational interpretation, elliptic means uniformly coercive: some $\gamma >0$ satisfiesThis is ellipticity of a bounded Hilbert-space operator, expressed through its quadratic form, rather than the principal-symbol definition for an elliptic differential operator. In the nonsymmetric case this condition concerns the symmetric part of $L$ and does not itself imply self-adjointness.

A positive-definite operator is self-adjoint and strictly positive:Some variational conventions use “positive definite” for the stronger combination of self-adjointness and ellipticity; we distinguish strict from uniform positivity explicitly. A uniformly positive definite symmetric operator is coercive, whereas strict positivity alone in infinite dimension need not give a bounded inverse or solvability for every forcing. If positivity is instead defined only by the displayed real quadratic inequality, symmetry must be added separately for the next part. In a complex Hilbert space, use a self-adjoint operator, the real quadratic form, and $\mathrm{Re}⟨f,v⟩$ in the linear term.

Assume the self-adjoint interpretation of positive-definite operator from the preceding part. For $v,w\in H$ and real $\epsilon$, expand the quadratic functional:The first variation therefore vanishes in every direction precisely when $⟨Lv-f,w⟩=0$ for every $w$, or $Lv=f$. This is its Euler-Lagrange equation and its weak formulation. For any weak solution $u$, write $v=u+w$. Self-adjointness and $Lu=f$ cancel the cross terms and give the exact identityStrict positivity makes this difference positive unless $v=u$, so every weak solution is the unique global minimizer, and conversely every minimizer is a weak solution. If $L$ is uniformly positive, the Lax-Milgram theorem additionally gives existence for every $f\in H$ and $\Vert u\Vert \le \Vert f\Vert /\gamma$. The identity above then gives the quantitative gap $I(v)-I(u)\ge \gamma \Vert v-u{\Vert }^2$.

The hypotheses matter. On ${\mathbb{R}}^2$, letThen $⟨Lv,v⟩=\Vert v{\Vert }^2>0$ for nonzero $v$, but $I$ is minimized at $v=f$, whereas $L^{-1}f=(1/2,-1/2{)}^T$. Thus real quadratic positivity without symmetry does not imply the requested variational assertion: the symmetric part determines a real quadratic functional.

Also, strict self-adjoint positivity does not imply existence for arbitrary $f$. On $H={\ell }^2$, take $(Lv{)}_j=v_j/j$ and $f_j=1/j$. The operator is bounded, self-adjoint and strictly positive, and $f\in {\ell }^2$; a solution would have $u_j=1$, which is not in ${\ell }^2$. In fact the trial vectors with their first $N$ coordinates equal to one give $I(v)=-{\sum }_{j=1}^{N}1/j\to -\infty$. The printed conclusion about a weak solution is valid whenever that solution exists; an unconditional existence assertion needs uniform positivity.

The appropriate bounded-operator formulation uses the zero-boundary Sobolev spaceThe Poincare inequality makes this a Hilbert space norm equivalent to the usual $H^1$ norm, and the zero endpoint traces encode the Dirichlet boundary conditions. Under the usual regular-coefficient assumptions, for example continuous $p,q$ on the closed interval with $p>0$ and $q\ge 0$, setIf $p_{\ast }=\min p>0$, then Cauchy-Schwarz inequality and the Poincare inequality giveBy the Riesz representation theorem, a unique bounded linear operator $T:H\to H$ satisfies $(Tu,v{)}_H=a(u,v)$. Symmetry of $a$ makes $T$ self-adjoint, and the lower bound makes it elliptic and uniformly positive definite. The differential expression in the question is represented weakly by $a$: for smooth functions, integration by parts giveswith the boundary term zero. A forcing $f\in L^2$ becomes the bounded functional $v\mapsto \int fv$, or its Riesz representative in $H$.

Equivalently, for sufficiently smooth coefficients one may realize the differential operator itself on $L^2(-1,1)$ with domain $H^2\cap H_0^1$. Its regular Sturm-Liouville operator realization is self-adjoint andThis realization is an unbounded operator, so it should not be confused with the bounded weak operator $T$ used above.

The PDF states sign conditions without coefficient regularity. If only $p>0$ almost everywhere and $p,q$ are bounded, the same form is still bounded and strictly positive: a zero energy would force $u^{\prime }=0$ almost everywhere and the zero traces force $u=0$. Uniform ellipticity, however, requires a positive lower bound. For instance $p(x)=x^2$ for $x\ne 0$, $p(0)=1$, and $q=0$ satisfy literal pointwise positivity. Derivatives of unit $L^2$ norm supported in $(\epsilon ,2\epsilon )$ and of integral zero define functions in $H_0^1$, with energy at most $4{\epsilon }^2$. Thus pointwise positivity without regularity does not imply coercivity in this $H$ norm. These distinctions supply the precise regularity and operator domain behind the intended positive-definiteness proof.