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Elliptic Fredholm solvability can fail at an eigenvalue

Statement refuted

For the Dirichlet problem for $- Δ$ on a bounded domain and every real datum $f \in L^{2}$ , the equation $- Δ u - λ u = f$ is uniquely solvable.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; the interval $Ω = (0, π)$ ; the Dirichlet Laplacian with form $a (u, v) = \int_{0}^{π} u^{'} v^{'}$ ; an integer $k \geq 1$ and the eigenvalue $λ_{k} = k^{2}$ ; and $f \in L^{2} (0, π)$ .

[F1]

The eigenfunction: $u_{k} (x) = \sin (k x)$ lies in $H_{0}^{1} (0, π)$ and satisfies $a (u_{k}, v) = k^{2} (u_{k}, v)_{L^{2}}$ for every $v \in H_{0}^{1} (0, π)$ (Dirichlet Laplacian eigenpairs on an interval, Symmetric elliptic weak eigenpairs).

[F2]

One-dimensional representatives: every $H^{1} (0, π)$ class has an absolutely continuous representative $w^{*}$ on $[0, π]$ satisfying $w^{*} (x) - w^{*} (y) = \int_{y}^{x} w^{'}$ (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives). Averaging $w^{*} (x) = w^{*} (y) + \int_{y}^{x} w^{'}$ in $y$ and applying Cauchy--Schwarz (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs) gives $∣ w^{*} (x) ∣ \leq π^{- 1 / 2} ∥ w ∥_{2} + π^{1 / 2} ∥ w^{'} ∥_{2}$ uniformly on $[0, π]$ . Thus $H^{1}$ convergence implies uniform convergence of these representatives, and approximation by $C_{c}^{\infty}$ shows that every $H_{0}^{1}$ representative has zero endpoints. For continuously differentiable representatives the second fundamental theorem is The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ .

[F3]

Resonant form and Fredholm alternative: put $q_{k} (u, v) : = a (u, v) - k^{2} (u, v)_{L^{2}} = \int_{0}^{π} (u^{'} v^{'} - k^{2} u v) d x$ . This is the symmetric uniformly elliptic form with principal coefficient $1$ , zero drift and bounded constant potential $- k^{2}$ , so the Fredholm alternative applies on $(0, π)$ . Define $N : = {u \in H_{0}^{1} (0, π) : q_{k} (u, v) = 0 \forall v \in H_{0}^{1} (0, π)}$ and let $N^{*}$ be the homogeneous space for its adjoint form. Since $q_{k}^{*} = q_{k}$ , one has $N^{*} = N$ ; the weak problem $q_{k} (u, v) = (f, v)_{L^{2}}$ is solvable if and only if $(f, v)_{L^{2}} = 0$ for every $v \in N$ , and when $N \neq {0}$ uniqueness fails (The Fredholm alternative for weak elliptic Dirichlet problems, The formal adjoint and the adjoint weak Dirichlet problem, The $L^{2}$ operator associated with a symmetric elliptic form).

[F4]

Orthogonality: $(u_{k}, u_{k})_{L^{2}} = \int_{0}^{π} \sin^{2} (k x) d x = π / 2 \neq 0$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , Dirichlet Laplacian eigenpairs on an interval).

Counterexample

1.1F1F2F3givenalgebra

The homogeneous space. By [F1] and the definition of $q_{k}$ in [F3], $u_{k} = \sin (k x)$ is a nonzero homogeneous solution. Conversely, if $u \in H_{0}^{1}$ is a weak homogeneous solution, compactly supported tests give $D (u^{'}) = - k^{2} u \in L^{2}$ , so both $u$ and $u^{'}$ have absolutely continuous representatives by [F2]. Their integral identities imply that $u$ is $C^{1}$ with derivative that representative of $u^{'}$ , and that derivative is $C^{1}$ with derivative $- k^{2} u$ , since the latter is continuous. Hence $u$ is $C^{2}$ and $u^{''} = - k^{2} u$ on $[0, π]$ , with $u (0) = u (π) = 0$ by [F2]. Set $z = u - (u^{'} (0) / k) \sin (k x)$ . It satisfies $z (0) = z^{'} (0) = 0$ and $z^{''} = - k^{2} z$ . The derivative of $∣ z^{'} ∣^{2} + k^{2} ∣ z ∣^{2}$ is zero, so [F2] makes that energy identically zero; thus $z = 0$ . This proves $N = span {\sin (k x)}$ , exactly one-dimensional.

2.1F1F3F4step 1.1givenalgebra

Solvability fails on a nonzero datum. Since $q_{k}$ is symmetric, [F3] gives $N^{*} = N = span {\sin (k x)}$ , so the weak problem $- u^{''} - k^{2} u = f$ with $u \in H_{0}^{1} (0, π)$ is solvable if and only if $\int_{0}^{π} f (x) \sin (k x) d x = 0$ ; for $f = \sin (k x)$ this integral equals $π / 2 \neq 0$ by [F4], so this datum admits no weak solution. Uniqueness also fails whenever a solution exists, because adding any multiple of the nonzero homogeneous solution $\sin (k x)$ produces another solution.

3.1F3step 1.1step 2.1given∎

Conclusion. On $Ω = (0, π)$ with $λ = k^{2}$ the equation $- Δ u - λ u = f$ is neither uniquely solvable for every $f \in L^{2}$ (uniqueness fails at the eigenvalue) nor solvable for the particular datum $f = \sin (k x)$ ; hence the refuted statement fails, and the failure is exactly the one-dimensional orthogonality condition predicted by the Fredholm alternative.

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Sources

John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page notes) (standard reference, not scraped)
Richard S. Laugesen, Spectral Theory of Partial Differential Equations (University of Illinois lecture notes, arXiv:1203.2344, complete 120 pages) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)