Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2007 / A / 4 / Solution

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.