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An Indefinite Convection-Diffusion Operator With Real Spectrum

We confirm rigorously the conjecture, based on numerical and asymptotic evidence, that all the eigenvalues of a certain non-self-adjoint operator are real.


Introduction
For 0 < ε < 2, we consider the operator (Hf )(θ) := ε ∂ ∂θ sin(θ) ∂f ∂θ + ∂f ∂θ (1) initially defined on all C 2 periodic functions f ∈ L 2 (−π, π); the exact domain is given by taking the closure of the operator defined on the above functions. That such a closure exists will be shown later. In a recent paper [1] Benilov, O'Brien and Sazonov showed that the equation approximates the evolution of a liquid film inside a rotating horizontal cylinder. We shall show that the eigenvalue problem has only real eigenvalues, which were shown to exist by Davies in [3]. In the same paper, he showed that the spectrum of −iH is equal to the set of its eigenvalues, so this implies that the spectrum is real. This was conjectured in [1], and Chugunova and Pelinovsky proved in [2] that all but finitely many eigenvalues are real, and gave numerical evidence that all eigenvalues are real. Our approach is to show that it is sufficient to consider H acting on the Hardy space H 2 (−π, π) and analytically continue any solution of (3) to the unit disc, where the corresponding ODE (8) is now self-adjoint on [0, 1] with regular singularities at the end points. In order to do this we make use of a bound on the Fourier coefficients proved by Davies in [3]. As in [3], by expanding f ∈ L 2 (−π, π) in the form one may rewrite the eigenvalue problem in the form Av = λv, where A = −iH is given by Here we have identified l 2 (Z) and L 2 (−π, π) using the Fourier transform F : l 2 (Z) → L 2 (−π, π). We have where A − acts in l 2 (Z − ), the central 0 acts in C and A + acts in l 2 (Z + ). We assume that A + has its natural maximal domain Davies has shown in [3] that A + is closed and that D is the closure under the graph norm of A + of the subspace consisting of those v ∈ l 2 (Z + ) that have finite support. Let τ be the natural identification between l 2 (Z + ) and l 2 (Z − ); then Dom (A) = τ (D) ⊕ C ⊕ D and τ induces a unitary equivalence between A + and A − . Therefore, in order to prove that all eigenvalues of A are real, we only need to prove that all eigenvalues of A + are real. The Fourier transform identifies , which is clearly a subspace of Dom (H). We now show that H is the closure of H 0 . Proof It follows from Davies' result on the domain of A + that the trigonometric polynomials are dense in Dom (H) with respect to the graph norm. Since the trigonometric polynomials are contained in C 2 per ([−π, π]), this space is also dense in Dom (H), which is closed in graph norm since Dom (A) is.

Reality Of The Eigenvalues
If A + v = λv, then v is a solution of the recurrence relation satisfying the initial condition εv 2 = (1 −λ)v 1 . We shall study the generating function, ∞ k=1 v k z k of (v k ).
Then the function u(z) := ∞ k=1 v k z k , defined for |z| < 1, satisfies the differential equation Proof The constant term in is clearly 0, and the coefficient of z is The coefficient of z n is From here on we assume that λ ∈ C is an eigenvalue of the operator A + and v is a corresponding non-zero eigenvector. In [3], Davies proved that there exist constants b, m such that where c = 1 + 1/ε and ||w|| ∞,c := sup{|w n |n c : 1 ≤ n < ∞}. Since ε > 0, this implies that v ∈ l 1 (Z + ), and hence that ∞ k=1 v k z k is absolutely convergent for |z| ≤ 1. The equation (5) can be written in the form of Heun's equation with α = 1, β = 0, γ = 0, δ = 1 + 1/ε, ǫ = 1 − 1/ε, a = −1 and µ = 2λ/ε. This is a Fuchsian equation with four regular singular points, at 0,  [4].
Lemma 2.2 Suppose that 0 < ε < 2, 1/ε / ∈ Z, λ ∈ C is an eigenvalue of A + and v ∈ l 2 (Z + ) is a corresponding non-zero eigenvector. Then there exists a solution u of (7) which is analytic in an open set containing [0, 1] and such that u(z) = ∞ k=1 v k z k for all z such that |z| ≤ 1.
Proof Define u(z) = ∞ k=1 v k z k on {z ∈ C : |z| < 1}. Let u 1 be the solution of (7) with exponent 0 about 1 and u 2 be the solution with exponent −1/ε about 1. Let U be the intersection of the open discs of unit radius about 0 and 1. The space of solutions of (7) in U is two-dimensional, and u, u 1 and u 2 lie in this space. Hence there exist constants a, b such that u = au 1 + bu 2 in U. Since v ∈ l 1 (Z + ), u(z) converges to a finite limit as z → 1 in U. Also u 1 (1) is finite, but u 2 (z) → ∞ as z → 1 in U. Therefore we must have b = 0 and u = au 1 in U. Let W be the union of the open discs of unit radius about 0 and 1. We now extend u to all of W by u(z) = au 1 (z) on the open disc of unit radius about 1. Now u is an analytic solution of (7) on W , which is an open set containing [0, 1], such that u(z) = ∞ k=1 v k z k for all z such that |z| ≤ 1.