Inverse Problems for the Quadratic Pencil of the Sturm-Liouville Equations with Impulse

Theorem 1. If $$, then the solution f_ν(x, λ) has the form

where and the function A_ν(x, t) satisfies the inequality with

Theorem 2. Let $$. Then there are the functions A(x, t),   B(x, t) whose first order partial derivatives are summable on [0, π] for each x ∈ [0, π] such that the representation

is satisfied, where Moreover, the relations are held.

Definition 3. A complex number λ₀ is called an eigenvalue of the boundary value problem L(α, a) if (1) with λ = λ₀ has a nontrivial solution y₀(x) satisfying the boundary conditions (2) and the jump conditions (3). In this case y₀(x) is called the eigenfunction of the problem L(α, a) corresponding to the eigenvalue λ₀. The number of linearly independent solutions of the problem L(α, a) for a given eigenvalue λ₀ is called the multiplicity of λ₀.

The following lemmas can be proved analogously to the corresponding assertions in [11].

Lemma 4. The eigenvalues of the boundary value problem L(α, a) are real, nonzero, and simple.

Proof. We define a linear operator L₀ in the Hilbert space L₂[0, π] as follows. The domain D (L₀) consists of all functions $$ satisfying the boundary conditions (2) and the jump conditions (3). For y ∈ D(L₀), we set L₀y = −y^′′ + q(x)y. Integration by part yields

Since condition (17) holds, it follows that (L₀y, y) > 0.

Let λ₀ be an eigenvalue of the boundary value problem L(α, a) and y₀(x) an eigenfunction corresponding to this eigenvalue and normalized by the condition (y₀, y₀) = 1. By taking the inner product of both sides of the relation $$ by y₀(x), we obtain $$ and hence

The desired assertion follows from the last relation by virtue of (L₀y₀, y₀) > 0 with regard to the fact that p(x) is real.

Let us show that λ₀ is a simple eigenvalue. Assume that this is not true. Suppose that y₁(x) and y₂(x) are linearly independent eigenfunctions corresponding to the eigenvalue λ₀. Then for a given value of λ₀, each solution y₀(x) of (1) will be given as linear combination of solutions y₁(x) and y₂(x). Moreover it will satisfy boundary conditions (2) and discontinuity conditions (3). However, it is impossible.

Lemma 5. The problem (1)–(3) does not have associated functions.

Proof. Let y₀(x) be an eigenfunction corresponding to eigenvalue λ₀ and normalized by the condition (y₀, y₀) = 1 of the problem (1)–(3). Suppose that y₁(x) is an associated function of eigenfunction y₀(x), that is, the following equalities hold:

If these equations are multiplied by y₁(x) and y₀(x), respectively, as inner product, subtracting them side by side and taking into our account that operator L₀ is symmetric, the function p(x) and λ₀ are real, we get λ₀ = (py₀, y₀). Due to the condition (6), λ₀ = (py₀, y₀) does not agree with (19′). Therefore, the assertion is not true.

Lemma 6. Eigenfunctions corresponding to different eigenvalues of the problem L(α, a) are orthogonal in the sense of the equality

where  (·, ·) denotes the inner product in L₂[0, π].

Lemma 7. Let y(x, λ) be a solution of (1) satisfying the condition (18) and the jump conditions (3). Then λ is real and nonzero and

Moreover, the sign of the left-hand side of (22) is similar to the sign of λ.

Lemma 8. The zeros {λ_n} of the characteristic function Δ(λ) coincide with the eigenvalues of the boundary value problem L(α, a). The functions φ(x, λ_n) and ψ(x, λ_n) are eigenfunctions corresponding to the eigenvalue λ_n, and there exists a sequence {β_n} such that

Proof. Let Δ(λ₀) = 0. Then by virtue of 〈ψ(x, λ₀), φ(x, λ₀)〉 = 0, φ(x, λ₀) = Cψ(x, λ₀) for some constant C. Hence λ₀ is an eigenvalue and φ(x, λ₀), ψ(x, λ₀) are eigenfunctions related to λ₀.

Conversely, let λ₀ be an eigenvalue of L(α, a), show that Δ(λ₀) = 0. Assuming the converse suppose that Δ(λ₀) ≠ 0. In this case the functions φ(x, λ₀) and ψ(x, λ₀) are linearly independent. Then y (x, λ₀) = c₁φ(x, λ₀) + c₂ψ(x, λ₀) is a general solution of the problem L(α, a). If c₁ ≠ 0, we can write

Then we have which is a contradiction.

Lemma 9. The equality $$ holds. Here $$.

Proof. If we differentiate the equalities

with respect to λ, we get

By virtue of these equalities we have

If the last equations are integrated from x to π and from 0 to x, respectively, by the discontinuity conditions we obtain

If we add the last equalities side by side, we get or For λ → λ_n, this yields The lemma is proved.

Lemma 10. The roots of the characteristic equation Δ₀(λ) = 0 are separate, that is

Proof. Let λπ − β⁺(π) = x. Then, λ(2a − π) + β⁻(π) = kx + b, where k = (2a − π)/π,   b = β⁺(π)((2a − π)/π) + β⁻(π). Since a ∈ (π/2, π), then k ∈ (0,1). Using these notations we can rewrite the equation Δ₀(λ) = 0 in the following form:

Here A = −(α⁺/α⁻) which implies that |A| > 1. Preliminarily show that there are no multiple roots of (35). Assuming the converse we suppose x₀ to be a multiple root of (35). Then holds. Now (35) and (36) imply that A² = 1 − (1 − k²)sin²(kx₀ + b) ≤ 1 which is a contradiction. Therefore, (35) has no multiple roots.

Further assuming (34) not to be true let $$ and $$ be increasing sequences of roots of (35) such that $$ and

If we assume that $$, where n ∈ ℕ  and$$ is a bounded sequence $$, then from (37) we find that $$, where $$ is a bounded sequence such that $$. It is obvious that $$, where $$ and $$. Here [·] and {·} denote the integer and fractional parts of a real number, respectively. Since sequences $$ and $$ will be bounded, without loss of generality we can assume that these sequences are convergent. Then let Therefore, we can write the equality $$ as Then by virtue of (38) and (39), from (40) we get Similarly we can obtain Further, from (40) and (42), we have Let us write this equality as Now dividing both sides of equality (44) by $$ and taking limit as p → ∞, by virtue of (3) and (39), we obtain Finally, from (41) and (45), we conclude that A² = 1 − (1 − k²)sin²(y₀ + b) ≤ 1 which is a contradiction. Hence roots of equation Δ₀(λ) = 0 are separate. The lemma is proved.

Lemma 11. For sufficiently large values of n, one has

Proof. As it is shown in [38], |Δ_o(λ)| ≥ C_δe^|Im λ|π for all $$, where C_δ > 0 is some constant. On the other hand, since

for sufficiently large values of n (see [1]) we get (47). The lemma is proved.

Lemma 12. The problem L(α, a) has countable set of eigenvalues. If one denotes by λ₁, λ₂, …  the positive eigenvalues arranged in increasing order and by λ₋₁, λ₋₂, …  the negative eigenvalues arranged in decreasing order, then eigenvalues of the problem L(α, a) have the asymptotic behavior

where δ_n ∈ l₂ and d_n is a bounded sequence, $$.

Proof. According to Lemma 11, if n is a sufficiently large natural number and λ ∈ Γ_n, we have |Δ_o(λ)| ≥ C_δe^|Im λ|π > (C_δ/2)e^|Im λ|π > |Δ(λ) − Δ_o(λ)|. Applying Rouche′s theorem we conclude that for sufficiently large n inside the contour Γ_n the functions Δ_o(λ) and Δ_o(λ)+{Δ(λ) − Δ_o(λ)} = Δ(λ) have the same number of zeros counting their multiplicities. That is, there are exactly (n + 1) zeros λ₀, λ₁, …, λ_n. in Γ_n. Analogously, it is shown by Rouche′s theorem that, for sufficiently large values of n, the function Δ(λ) has a unique zero inside each circle $$. Since δ > 0 is arbitrary, it follows that $$, where lim _n→∞ ɛ_n = 0. Further according to Δ(λ_n) = 0, we have

On the other hand, since (50) takes the form of It is easy to see that the function Δ₀(λ) is type of “Sine” [39], so there exists γ_δ > 0 such that $$ is satisfied for all n. We also have where sup _n |h_n| ≤ M for some constant M > 0 [40] (see also [41]). Further, substituting (53) into (52) after certain transformations [1, page 67], we reach ɛ_n ∈ l₂. We can obtain more precisely

Since $$, we have

where δ_n ∈ l₂. Hence we obtain where is a bounded sequence. The proof is completed.

Lemma 13. Normalizing numbers α_n of the problem L(α, a) are positive and the formula

holds, where d₁₁ = −(απ/2)p(0),   d_1n ∈ l₂.

Proof. The formula (58) can be easily obtained from the equalities

by using the asymptotic formula (49) for λ_n.

Lemma 14. If $$, then $$, that is, the sequence {λ_n} uniquely determines β^±(π).

Lemma 15. If $$, then $$, that is, the sequence {λ_n} uniquely determines numbers a and α.

Theorem 16. If $$, then $$. Thus, the boundary value problem L(α, a) is uniquely defined by the Weyl function.

Proof. Since

it is easy to observe that Let us define the matrix P(x, λ) = [P_jk(x, λ)] _j,k=1,2, where Then we have According to (60) and (65), for each fixed x, the functions P_jk(x, λ) are meromorphic in λ with poles at points λ_n and $$. Denote$$. By virtue of (65), (66), and we get It follows from (60) and (66) that if $$, then for each fixed x the functions P_1k(x, λ) are entire in k. Together with (69) this yields P₁₂(x, λ) ≡ 0,   P₁₂(x, λ) ≡ A(x). Now using (67), we obtain Therefore, for |λ| → ∞,  argλ ∈ [ɛ, π − ɛ]  (ɛ > 0), we have where b = 1 for x < a and b = α⁺ for x > a. Similarly, one can calculate Finally, taking into account the relations 〈Φ(x, λ), φ(x, λ)〉 ≡ 1 and (65), we have $$, that is, $$ for all x and λ. Consequently, $$. The theorem is proved.

Theorem 17. If $$, then $$.

Proof. It is obvious that characteristic functions Δ(λ) and ψ(0, λ) are uniquely determined by the sequences $$ and $$, respectively. If $$, then $$. It follows from (60) that $$. Therefore, applying Theorem 16, we conclude that $$. The proof is completed.

Theorem 18. If $$, then $$, that is, spectral data {λ_n, α_n} uniquely determines the problem L(α, a).

Proof. It is obvious that the Weyl function M(λ) is meromorphic with simple poles at points $$. Using the expression

and equalities $$, we have Since the Weyl function M(λ) is regular for λ ∈ Γ_n, applying the Rouche theorem, we conclude that Taking (60) and (63) into account, we arrive at |M(λ)| ≤ C_δ|λ|⁻¹,  λ ∈ G_δ. Therefore where $$. Hence, by the residue theorem, we have Finally, from the equality $$, applying Theorem 16, we conclude that $$. The theorem is proved.