Inverse Sturm-Liouville problem with analytical functions in the boundary condition

Inverse Problem 1.1

Let the entire functions f j ( λ ) , j = 1 , 2 , be known a priori. Given the eigenvalues { λ n } n = 1 ∞ and the number ω : = 1 2 ∫ 0 π q ( x ) d x , find the potential q.

Inverse Problem 2.1

Given the Cauchy data { K , N } and the number ω, find the potential q.

Theorem 2.2

Let { λ n } n = 1 ∞ and { λ ˜ n } n = 1 ∞ be subspectra of the problems L ( q ) and L ( q ˜ ) , respectively. Suppose that L ( q ) and { λ n } n = 0 ∞ satisfy the condition (Complete), and let λ n = λ ˜ n , n ≥ 1 , ω = ω ˜ . Then, q = q ˜ in L 2 ( 0 , π ) .

Theorem 2.3

Let { λ n } n = 1 ∞ be a subspectrum of the problem L ( q ) . Suppose that the sequence { v n } n = 0 ∞ is incomplete in ℋ . Then, there exists a complex-valued function q ˜ ∈ L 2 ( 0 , π ) , q ˜ ≠ q such that ω = ω ˜ and { λ n } n = 1 ∞ is a subspectrum of L ( q ˜ ) .

Algorithm 2.4

Let eigenvalues { λ n } n = 1 ∞ and the number ω be given. We need to find the potential q.

1. Using f j ( λ ) , j = 1 , 2 , { λ n } n = 0 ∞ and ω , construct the vector functions { v n } n = 0 ∞ and the numbers { w n } n = 0 ∞ via (2.6), (2.7), (2.10) and (2.11).

2. For the basis { v n } n = 0 ∞ , find the biorthonormal basis { v n ⁎ } n = 0 ∞ , i.e., ( v n , v k ⁎ ) ℋ = δ n k , n , k ≥ 0 , where δ n k is the Kronecker delta.

3. Construct the element u ∈ ℋ , satisfying (2.12), by the formula

   u = ∑ n = 0 ∞ w n ¯ v n ⁎ .

4. Using the components of u ( t ) = [ N ( t ) ¯ , K ( t ) ¯ ] , solve Inverse Problem 2.1 and find q.

In certain applications, it can be difficult to check the conditions (Complete) and (Basis). Therefore, we introduce some other conditions, sufficient for uniqueness and for constructive solution of Inverse Problem 1.1.

(Complete2) The sequence { c 〈 ν 〉 ( t , λ n ) } n ∈ I , ν = 0 , m n − 1 ¯ is complete in L 2 ( 0 , 2 π ) .

(Basis2) The sequence { c 〈 ν 〉 ( t , λ n ) } n ∈ I , ν = 0 , m n − 1 ¯ is a Riesz basis in L 2 ( 0 , 2 π ) .

(Separation) For every n ≥ 0 , we have f 1 ( λ n ) ≠ 0 or f 2 ( λ n ) ≠ 0 .

(Simple) There exists an integer n 0 such that m n = 1 and λ n ≠ 0 for n ≥ n 0 .

(Asymptotics) Im ρ n = O ( 1 ) , n → ∞ and { ρ n − 1 } n ≥ n 0 ∈ l 2 , where ρ n ≔ λ n , arg ρ n ∈ [ − π 2 , π 2 ) .

These conditions are natural for applications, such as the Hochstadt-Lieberman problem (see Section 4), transmission inverse eigenvalue problem, inverse problems for quantum graphs (see [37, Section 4]), etc.

Theorem 2.5

(i) (Separation) and (Complete2) together imply (Complete); (ii) (Separation), (Simple), (Asymptotics) and (Basis2) together imply (Basis).

Proof of Theorem 2.2

Suppose that the problems L ( q ) , L ( q ˜ ) and their subspectra { λ n } n = 1 ∞ , { λ ˜ n } n = 1 ∞ satisfy the conditions of Theorem 2.2. By virtue of definitions (2.6), (2.7), (2.10) and (2.11), we have v n = v ˜ n in ℋ and w n = w ˜ n for all n ≥ 0 . Hence, relation (2.12) for L ˜ has the form

Subtracting (3.1) from (2.12) and using the completeness of the sequence { v n } n = 0 ∞ , we get u = u ˜ in ℋ , i.e., K = K ˜ , N = N ˜ in L 2 ( 0 , π ) . Using the uniqueness of Inverse Problem 2.1 solution, we conclude that q = q ˜ in L 2 ( 0 , π ) .□

Proof of Theorem 2.3

Let the problem L ( q ) and the subspectrum { λ n } n = 1 ∞ be such that the sequence { v n } n = 0 ∞ is incomplete in ℋ . Then, there exists u ˆ ∈ ℋ , u ˆ ≠ 0 , such that

Since relation (3.2) is linear by u ˆ , one can choose u ˆ satisfying the estimate ∥ u ˆ ∥ ℋ ≤ ε for ε from Theorem 6.1. Set u ≔ [ N ( t ) ¯ , K ( t ) ¯ ] , u ˜ ≔ u + u ˆ = [ N ˜ ( t ) ¯ , K ˜ ( t ) ¯ ] and u ˜ ≠ u . By Theorem 6.1, there exists q ˜ ∈ L 2 ( 0 , π ) such that ω = ω ˜ and { K ˜ , N ˜ } are the Cauchy data of q ˜ . Define the functions

Clearly, Δ ˜ ( λ ) is the characteristic function of L ( q ˜ ) . Relations (2.12) and (3.2) yield (3.1). Consequently, the function λ Δ ˜ ( λ ) has zeros { λ n } n ∈ I of the corresponding multiplicities { m n } n ∈ I . Thus, { λ n } n = 1 ∞ is a subspectrum of L ( q ˜ ) , q ˜ ≠ q .□

Lemma 3.1

Suppose that (Separation) is fulfilled. Then, there exist coefficients { C n , k } such that the following relations hold:

for n ∈ I \ { 0 } , ν = 0 , m n − 1 ¯ and for n = 0 , ν = 0 , m 0 − 2 ¯ .

Proof

Fix n ∈ I \ { 0 } . Relation (2.1) can be rewritten in the form

The condition (Separation) and the relation Δ ( λ n ) = 0 imply that

where C n , 0 is a nonzero constant, i.e., relation (3.3) holds for ν = 0 .

Let us prove (3.4) for ν = 1 , m n − 1 ¯ by induction. Assume that (3.3) is already proved for η j 〈 k 〉 ( λ n ) , k = 0 , ν − 1 ¯ , j = 1 , 2 . Using (3.4) and the relation Δ 〈 ν 〉 ( λ n ) = 0 , we get

Differentiation of the products yields

Here and below the arguments ( λ n ) are omitted for brevity. Using (3.3) for η j 〈 k 〉 , k = 0 , ν − 1 ¯ , we obtain

Calculations show that

Hence,

In view of (Separation), f 1 ≠ 0 or f 2 ≠ 0 . Consequently, there exists the constant C n , ν such that

Thus, relation (3.3) is proved for n ∈ I \ { 0 } , ν = 0 , m n − 1 ¯ . Obviously, the arguments above are also valid for n = 0 , ν = 0 , m 0 − 2 ¯ .□

Lemma 3.2

The sequence { v n } n = 0 ∞ is complete in ℋ if and only if so does { u n } n = 0 ∞ .

Proof

Let an element h = [ h 1 ¯ , h 2 ¯ ] ∈ ℋ be such that

Definitions (2.6), (2.10) and (2.11) imply that

where

Obviously,

Consider the function

Let us show that

Using Lemma 3.1, (3.7) and (3.8), we derive

for all n ∈ I , ν = 0 , m n − 1 ¯ , except for ( n , ν ) = ( 0 , m 0 − 1 ) . Let us consider the case ( n , ν ) = ( 0 , m 0 − 1 ) separately. Note that

since s ( t , 0 ) = 0 , c ( t , 0 ) = 1 , v 0 = [ 0 , 1 ] and (3.6) holds for n = 0 . Combining (3.7), (3.8) and (3.11), we obtain

By using analogous ideas and Lemma 3.1, we derive

In particular, (3.10) holds for n = 0 , ν = m 0 − 1 .

Relations (3.9) and (3.10) yield

Thus, we have shown that (3.12) follows from (3.6). It can be proved similarly that (3.6) follows from (3.12). Therefore, the completeness of the sequence { v n } n = 0 ∞ is equivalent to the completeness of the sequence { g n } n = 0 ∞ .□

Proposition 3.3

Let { θ n } n = 0 ∞ be a sequence of complex numbers, satisfying the asymptotic formula

Define

Then, the sequence { c 〈 ν 〉 ( t , θ n ) } n ∈ J , ν = 0 , μ n − 1 ¯ is a Riesz basis in L 2 ( 0 , a ) .

Proposition 3.4

Let G ( λ ) be an entire function, satisfying the conditions:

for some positive constants C and a. Let { λ n } n = 0 ∞ be arbitrary complex numbers, and let the set I and the multiplicities { m n } n = 0 ∞ be defined by (2.2). Suppose that

and the sequence { c 〈 ν 〉 ( t , λ n ) } n ∈ I , ν = 0 , m n − 1 ¯ is complete in L 2 (0, a ) . Then, G ( λ ) ≡ 0 .

Proof

By the Paley-Wiener Theorem, the function G can be represented in the form

Differentiating (3.15), we get

Since the sequence { c 〈 ν 〉 ( t , λ n ) } n ∈ I , ν = 0 , m n − 1 ¯ is complete in L 2 (0, a ) , we have r = 0 in L 2 (0, a ) , so G ( λ ) ≡ 0 .□

Proof of Theorem 2.5(i)

Suppose that the conditions (Separation) and (Complete2) are fulfilled. Let us show that these two conditions imply the completeness of the sequence { g n } n = 0 ∞ . Let h ∈ ℋ be such that (3.12) is valid. We have to show that h = 0 . Consider the function G ( λ ) defined by (3.9). Relation (3.12) implies (3.10). In view of (2.3), (2.4), (3.9), (3.10) and (Complete2), the conditions of Proposition 3.4 are fulfilled for a = 2 π . Therefore, G ( λ ) ≡ 0 , i.e.,

□