Sharp eigenvalue asymptotics of fourth-order differential operators

Abstract

Two-term self-adjoint fourth-order differential operator with summable potential on the unit interval is considered. High energy eigenvalue asymptotics and the trace formula for this operator are obtained.

Keywords

Spectrum eigenvalue asymptotics fourth-order differential operator trace formula

1. Introduction and main results

Consider self-adjoint fourth-order operators $H = H (q)$ acting in the Hilbert space $L^{2} (0, 1)$ and defined by $\begin{matrix} (1.1) & H y = y^{(4)} + q y, y^{'} (0) = y^{‴} (0) = y^{'} (1) = y^{‴} (1) = 0, \end{matrix}$ where q is real potential and $q \in L^{1} (0, 1)$ . The operator is determined on the domain $\begin{array}{l} D (H) = & {y \in L^{2} (0, 1) : y^{'}, y^{″}, y^{‴} \in L^{1} (0, 1), y^{(4)} + q y \in L^{2} (0, 1), \\ y^{'} (0) = y^{‴} (0) = y^{'} (1) = y^{‴} (1) = 0} . \end{array}$

A great number of papers are devoted to the eigenvalue asymptotics for second-order operators. We mention only the books of Marchenko [17] and Levitan and Sargsyan [16], the review Fulton and Pruess [12], and see also the references therein.

The spectral properties of fourth-order differential operators with various boundary conditions are studied in many papers. Firstly, we discuss the asymptotic properties of general fourth-order differential operators on the unit interval. Caudill, Perry, and Schueller [10] described iso-spectral potentials for these operators. McLaughlin [18] investigated inverse spectral problems for fourth-order operators with the Neumann type boundary conditions. Polyakov [25] obtained the eigenvalue asymptotics and the estimates of spectral projections for these operators. Badanin and Korotyaev determined eigenvalue asymptotics and trace formulas for self-adjoint fourth-order operators on the circle [5] and on the unit interval [7]. The fourth-order operators with periodic coefficients on the line were considered by Badanin and Korotyaev [3,6]. Moreover, these authors [2] are studied the periodic operator H. Gunes, Kerimov, and Kaya [15] obtained the eigenvalue asymptotics for general fourth-order differential operators with periodic (antiperiodic) boundary conditions and proved that the system of eigenfunctions and associated functions of this operator form a basis in the space $L_{p} (0, 1)$ , $1 < p < \infty$ . Also these operators were considered by Polyakov [26,28]. The spectral properties of the periodic Euler-Bernoulli equation were studied by Papanicolaou [22,24].

Spectral asymptotics for higher-order operators are much less investigated. Numerous results about the regular boundary value problems for these operators are expounded by Naimark [20]. High energy asymptotics of the eigenvalues for even-order operator are determined by Akhmerova [1], Badanin and Korotyaev [4], Mikhailets and Molyboga [19], Polyakov [27].

Secondly, we describe results about trace formulas for fourth and higher-order operators. We mention only several results that are closely related to our ones. Sadovnichii and Podolskii [30] represented the survey about trace formulas. Shevchenko [31] determined the trace formula of two-term arbitrary order differential operators for general Birkhoff regular boundary conditions with smooth potentials. The trace formulas for the operator H were obtained by Sadovnichii [29] (under the Dirichlet type boundary conditions) and Bayramov, Oer, Öztürk Uslu, and Kizilbudak Caliskan [9] (under the periodic boundary conditions). Papanicolaou [23] deduced the trace formula for higher-order Schrödinger operators with Neumann or Robin boundary conditions. Nazarov, Stolyarov and Zatitskiy [21] determined the first-order trace formula for a regular differential operators on the segment perturbated by a multiplication operator. Gül [13] proved the first regularized trace formula for the self-adjoint operator H with bounded operator coefficient. Gül and Ceyhan [14] obtained the second regularized trace formula for this operator.

The main goal of the present paper is to determine eigenvalue asymptotics and the trace formula for the operator H.

The operator H has pure discrete spectrum (see [20, Ch. I]). Introduce the fundamental solutions $φ_{j}$ , $j = 1, 2, 3, 4$ , of equation $\begin{matrix} (1.2) & y^{(4)} + q y = λ y, λ \in C, \end{matrix}$ satisfying the conditions $φ_{j}^{(k - 1)} (0, λ) = δ_{j k}$ , j, $k = 1, 2, 3, 4$ , where $δ_{j k}$ is the Kronecker symbol. Each $φ_{j} (x, \cdot)$ , $x \in R$ , is an entire function.

The spectrum $σ (H)$ consists of real or non-real eigenvalues and satisfies $σ (H) = {λ \in C : D (λ) = 0}$ , where D is an entire function given by $\begin{matrix} (1.3) & D (λ) = - det (\begin{matrix} φ_{1}^{'} (1, λ) & φ_{3}^{'} (1, λ) \\ φ_{1}^{‴} (1, λ) & φ_{3}^{‴} (1, λ) \end{matrix}), λ \in C . \end{matrix}$

Consider the unperturbed operator $H_{0} y = H (0) y = y^{(4)}$ under the boundary conditions (1.1). In this case the function D has the form $\begin{matrix} D_{0} = - i z^{2} sin z sin i z, \end{matrix}$ see also (2.12). Here and below we use the spectral parameter $\begin{matrix} z = λ^{1 / 4}, arg z \in (- \frac{π}{4}, \frac{π}{4}] . \end{matrix}$ The eigenvalues of $H_{0}$ are the zeros of $D_{0}$ and have the form $μ_{n}^{0} = {(π n)}^{4}$ , $n \in Z_{+} = N \cup {0}$ . These eigenvalues $μ_{n}^{0}$ are simple.

We denote the eigenvalues of the perturbed problem by $μ_{n}$ , $n \in Z_{+}$ , (see details in Section 3.1). Our first main results are devoted to the high energy asymptotics of these eigenvalues in the case $q \in L^{1} (0, 1)$ . We give these asymptotics in terms of Fourier coefficients of q. In order to determine the trace formula for the operator H we need an additional condition $q^{'} \in L^{1} (0, 1)$ . We also derive the eigenvalue asymptotics in this case.

Introduce the Fourier coefficients $\begin{matrix} f_{0} = \int_{0}^{1} f (x) d x, {\hat{f}}_{n} = \int_{0}^{1} f (x) e^{- i 2 π n x} d x, {\hat{f}}_{c n} = \int_{0}^{1} f (x) cos 2 π n x d x, n \in Z . \end{matrix}$ Now we can formulate the main results of our paper.

Theorem 1.
Let $q \in L^{1} (0, 1)$ and let $n \in N$ be large enough. Then the eigenvalues $μ_{n}$ are real and simple. Moreover, they satisfy $\begin{matrix} (1.4) & μ_{n} = {(π n)}^{4} + q_{0} + {\hat{q}}_{c n} + O (n^{- 1}), as n \to + \infty . \end{matrix}$ If, in addition, $q^{'} \in L^{1} (0, 1)$ , then $\begin{matrix} (1.5) & μ_{n} = {(π n)}^{4} + q_{0} + {\hat{q}}_{c n} + O (n^{- 2}), as n \to + \infty . \end{matrix}$
Remark 1.

The proof of Theorem 1 is based on the matrix form of the Birkhoff method. This method was used in [6] and [8].

We use asymptotics (1.5) in order to obtain the trace formula for the operator H (see Theorem 2 below).

Our second results are devoted to the trace formula for the operator H. Consider the shifted operator $H_{t} = H (q_{t})$ , given by (1.1), where $t \in T$ , $T = R ∖ Z$ , and $q_{t} = q (\cdot + t)$ . We denote by $μ_{n} (t)$ the eigenvalues of the operator $H_{t}$ . Theorem 2.
Let $q^{″} \in L^{1} (T)$ . Then there exists $N = N (q) \in N$ such that each function $\sum_{n = 1}^{N} μ_{n} (t)$ , $μ_{n} (t)$ , $n > N$ , belongs to the space $C^{2} (T)$ . Moreover, the following trace formula holds: $\begin{matrix} (1.6) & \sum_{n = 1}^{\infty} (μ_{n} (t) - μ_{n} (0)) = \frac{q (t) - q (0)}{2}, \end{matrix}$ the series converges absolutely and uniformly on $T$ .
Remark 2.
(i) The asymptotics (1.5) show that if $q^{″} \in L^{1} (T)$ , then the series (1.6) converges absolutely and uniformly on $T$ . The proof of other results is based on the asymptotic analysis of difference of resolvents of the perturbed operator H and the unperturbed operator $H_{0}$ .

(ii) Nazarov, Stolyarov and Zatitskiy [21] determined “formal” the first-order trace formula for a regular differential operators, but they did not obtain conditions under which these series converge. Our result coincides with this trace formula, but we prove that corresponding series are convergence.

(iii) Papanicolaou [23] obtained the trace formula for higher-order Schrödinger operators with Neumann or Robin boundary conditions. Our result also coincides with this trace formula, but we use another method for the proof.

2. Fundamental solutions

2.1. Fundamental matrix

It will be convenient for us, instead of the fundamental solutions $φ_{j}$ , $j = 1, 2, 3, 4$ , introduced in §1, to use other fundamental solutions $ϕ_{j}$ , whose asymptotics are well controlled. In order to describe these fundamental solutions, we introduce additional notation.

Recall that $z = λ^{1 / 4}$ , $z \in Z$ , $λ \in C ∖ (- \infty, 0)$ , where $\begin{matrix} Z = {z \in C : \arg z \in (- \frac{π}{4}, \frac{π}{4})} . \end{matrix}$ If $λ \in C_{+}$ , then $z \in Z_{+}$ , where $\begin{matrix} Z_{+} = {z \in C : \arg z \in (0, \frac{π}{4})} . \end{matrix}$ Introduce the numbers $ω_{1} = - ω_{4} = i$ , $ω_{2} = - ω_{3} = 1$ . Then the following estimates hold: $\begin{matrix} Re (i ω_{1} z) ⩽ Re (i ω_{2} z) ⩽ Re (i ω_{3} z) ⩽ Re (i ω_{4} z), \forall z \in Z_{+} . \end{matrix}$

Note that unperturbed equation (1.2) with $q = 0$ has the fundamental solutions $\begin{matrix} (2.1) & ϕ_{j}^{0} (x, z) = e^{i z x ω_{j}}, j = 1, 2, 3, 4, (x, z) \in R \times Z_{+} . \end{matrix}$ Consider the perturbed equation (1.2). Let $r > 0$ be large enough and let $z \in Z_{+} (r)$ , where $\begin{matrix} Z_{+} (r) = {z \in Z_{+} : | z | > r}, r > 0 . \end{matrix}$ Then equation (1.2) has the fundamental solutions $ϕ_{j} (x, z)$ , $j = 1, 2, 3, 4$ , $x \in R$ , satisfying the asymptotics $\begin{matrix} (2.2) & \begin{matrix} ϕ_{1}^{'} (x, z) = - z e^{- z x} (1 + O (z^{- 1})), ϕ_{1}^{‴} (x, z) = - z^{3} e^{- z x} (1 + O (z^{- 1})), \\ ϕ_{2}^{'} (x, z) = i z e^{i z x} (1 + O (z^{- 1})), ϕ_{2}^{‴} (x, z) = - i z^{3} e^{i z x} (1 + O (z^{- 1})), \\ ϕ_{3}^{'} (x, z) = - i z e^{- i z x} (1 + O (z^{- 1})), ϕ_{3}^{‴} (x, z) = i z^{3} e^{- i z x} (1 + O (z^{- 1})), \\ ϕ_{4}^{'} (x, z) = z e^{z x} (1 + O (z^{- 1})), ϕ_{4}^{‴} (x, z) = z^{3} e^{z x} (1 + O (z^{- 1})), \end{matrix} \end{matrix}$ as $| z | \to \infty$ , $z \in Z_{+}$ , uniformly on any bounded subset of $R$ (see [20]).

Introduce the fundamental matrix $A (x, z)$ , $(x, z) \in R \times Z_{+} (r)$ , of equation (1.2) by $\begin{matrix} A = (\begin{matrix} ϕ_{1} & ϕ_{2} & ϕ_{3} & ϕ_{4} \\ ϕ_{1}^{'} & ϕ_{2}^{'} & ϕ_{3}^{'} & ϕ_{4}^{'} \\ ϕ_{1}^{″} & ϕ_{2}^{″} & ϕ_{3}^{″} & ϕ_{4}^{″} \\ ϕ_{1}^{‴} & ϕ_{2}^{‴} & ϕ_{3}^{‴} & ϕ_{4}^{‴} \end{matrix}) . \end{matrix}$ The matrix-valued function A satisfies the equation $\begin{matrix} (2.3) & A^{'} = P A, where P = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ λ - q & 0 & 0 & 0 \end{matrix}) . \end{matrix}$ Rewrite the determinant D given by (1.3) in terms of fundamental solutions ϕ. Introduce the function $\begin{matrix} (2.4) & ξ (z) = det ϕ (z), \end{matrix}$ where the matrix-valued function ϕ has the form $\begin{matrix} (2.5) & ϕ (z) = (\begin{matrix} ϕ_{1}^{'} (0, z) & ϕ_{2}^{'} (0, z) & ϕ_{3}^{'} (0, z) & ϕ_{4}^{'} (0, z) \\ ϕ_{1}^{‴} (0, z) & ϕ_{2}^{‴} (0, z) & ϕ_{3}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \\ ϕ_{1}^{'} (1, z) & ϕ_{2}^{'} (1, z) & ϕ_{3}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{2}^{‴} (1, z) & ϕ_{3}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}) . \end{matrix}$ The function ξ, like D, is the characteristic function of the spectrum of the operator H, but, unlike D, it is not an entire function of the variable λ. However, the function ξ has a well-controlled asymptotic behavior at high energy. The function ξ is analytic in $Z_{+} (r)$ and the function D given by (1.3) satisfies $\begin{matrix} (2.6) & D (λ) = \frac{ξ (z)}{det A (0, z)}, λ = z^{4} . \end{matrix}$ The proof of this equation repeats the arguments from [8, Lemma 3.2]. The function D is entire, then the identity (2.6) can be extended analytically from $Z_{+} (r)$ onto $C$ .

Consider the unperturbed case $q = 0$ . Let $z \in Z_{+}$ . In this case the fundamental matrix $A = A^{0}$ has the form $\begin{matrix} (2.7) & A^{0} = (\begin{matrix} ϕ_{1}^{0} & ϕ_{2}^{0} & ϕ_{3}^{0} & ϕ_{4}^{0} \\ {(ϕ_{1}^{0})}^{'} & {(ϕ_{2}^{0})}^{'} & {(ϕ_{3}^{0})}^{'} & {(ϕ_{4}^{0})}^{'} \\ {(ϕ_{1}^{0})}^{″} & {(ϕ_{2}^{0})}^{″} & {(ϕ_{3}^{0})}^{″} & {(ϕ_{4}^{0})}^{″} \\ {(ϕ_{1}^{0})}^{‴} & {(ϕ_{2}^{0})}^{‴} & {(ϕ_{3}^{0})}^{‴} & {(ϕ_{4}^{0})}^{‴} \end{matrix}) = Λ Y_{0}, Y_{0} = diag (ϕ_{1}^{0}, ϕ_{2}^{0}, ϕ_{3}^{0}, ϕ_{4}^{0}), \end{matrix}$ where $\begin{matrix} (2.8) & Λ = (\begin{matrix} 1 & 1 & 1 & 1 \\ - z & i z & - i z & z \\ z^{2} & - z^{2} & - z^{2} & z^{2} \\ - z^{3} & - i z^{3} & i z^{3} & z^{3} \end{matrix}), \end{matrix}$ and $ϕ_{j}^{0}$ , $j = 1, 2, 3, 4$ , are given by (2.1). Using the identity $ϕ_{1}^{0} ϕ_{2}^{0} ϕ_{3}^{0} ϕ_{4}^{0} = 1$ , we obtain $\begin{matrix} (2.9) & det A^{0} (0, z) = det Λ = - 16 i z^{6} . \end{matrix}$ The matrix-valued function $ϕ = ϕ^{0}$ has the form $\begin{matrix} (2.10) & \begin{matrix} ϕ^{0} (z) & = (\begin{matrix} {(ϕ_{1}^{0})}^{'} (0, z) & {(ϕ_{2}^{0})}^{'} (0, z) & {(ϕ_{3}^{0})}^{'} (0, z) & {(ϕ_{4}^{0})}^{'} (0, z) \\ {(ϕ_{1}^{0})}^{‴} (0, z) & {(ϕ_{2}^{0})}^{‴} (0, z) & {(ϕ_{3}^{0})}^{‴} (0, z) & {(ϕ_{4}^{0})}^{‴} (0, z) \\ {(ϕ_{1}^{0})}^{'} (1, z) & {(ϕ_{2}^{0})}^{'} (1, z) & {(ϕ_{3}^{0})}^{'} (1, z) & {(ϕ_{4}^{0})}^{'} (1, z) \\ {(ϕ_{1}^{0})}^{‴} (1, z) & {(ϕ_{2}^{0})}^{‴} (1, z) & {(ϕ_{3}^{0})}^{‴} (1, z) & {(ϕ_{4}^{0})}^{‴} (1, z) \end{matrix}) \\ = (\begin{matrix} - z & i z & - i z & z \\ - z^{3} & - i z^{3} & i z^{3} & z^{3} \\ - z e^{- z} & i z e^{i z} & - i z e^{- i z} & z e^{z} \\ - z^{3} e^{- z} & - i z^{3} e^{i z} & i z^{3} e^{- i z} & z^{3} e^{z} \end{matrix}) . \end{matrix} \end{matrix}$ Then the function $ξ^{0} = det ϕ^{0} (z)$ satisfies $\begin{matrix} (2.11) & ξ^{0} (z) = - 16 z^{8} sin z sin i z = - 16 i z^{6} D_{0} (λ), \end{matrix}$ where the entire function $D_{0}$ has the form $\begin{matrix} (2.12) & D_{0} (λ) = - i z^{2} sin z sin i z . \end{matrix}$

We analyse equation (1.2) by using the matrix form of the Birkhoff method for asymptotic analysis of higher-order equations (see [4,6,8]). In essence, the Birkhoff method is a peculiar inversion of the operator on the left-hand side of equation (2.3). This inversion is carried out in such a way that the result is an integral equation with a contracting kernel. This integral equation is solved by iterations. One problem is that the matrix coefficient on the right-hand side of equation (2.3) grows in the λ-variable. Therefore, we preliminarily transform equation (2.3) into equation (2.15) having a diagonal coefficient on the left-hand side and a decreasing matrix coefficient on the right-hand side.

Define the matrix-valued function $Y_{1} (x, z)$ by $\begin{matrix} (2.13) & A (x, z) = Λ (z) Y_{1} (x, z), z \in Z_{+} (r) . \end{matrix}$ In the unperturbed case $Y_{1} = Y_{0}$ , where $Y_{0}$ is defined by (2.7), satisfies $Y_{0}^{'} - i z T Y_{0} = 0$ , where $\begin{matrix} (2.14) & T = diag (ω_{1}, ω_{2}, ω_{3}, ω_{4}) = diag (i, 1, - 1, - i) . \end{matrix}$ We prove the following result.

Lemma 1.
Let $q \in L^{1} (0, 1)$ and let $z \in Z_{+} (r)$ , where $r > 0$ is large enough. Then the matrix-valued function $Y_{1} (x, z)$ given by ( 2.13 ) satisfies the equation $\begin{matrix} (2.15) & Y_{1}^{'} - i z T Y_{1} = \frac{1}{z^{3}} Φ_{1} Y_{1}, \end{matrix}$ where the diagonal matrix T and the matrix $Φ_{1}$ have the form $\begin{matrix} (2.16) & T = T - \frac{q}{4 z^{4}} T, Φ_{1} = \frac{q}{4} (Q + i T), \end{matrix}$ with $\begin{matrix} (2.17) & Q = (\begin{matrix} 1 & 1 & 1 & 1 \\ - i & - i & - i & - i \\ i & i & i & i \\ - 1 & - 1 & - 1 & - 1 \end{matrix}) . \end{matrix}$
Proof.
Substituting the definition (2.13) into equation (2.3) and using the identity $\begin{matrix} Λ^{- 1} P Λ = i z T + \frac{q}{4 z^{3}} Q, \end{matrix}$ where $P$ is defined in (2.3), we obtain the following equation for the matrix-valued function $Y_{1}$ : $\begin{matrix} Y_{1}^{'} - i z T Y_{1} = \frac{q}{4 z^{3}} Q Y_{1} . \end{matrix}$ Adding the term $i q T Y_{1} / (4 z^{3})$ to both sides of this equation we obtain (2.15). □

Consider the smooth potential $q^{'} \in L^{1} (0, 1)$ . In this case, we may reduce equation (2.15) to equation (2.22), where the matrix coefficient on the right-hand side decreases as $z^{- 4}$ . In order to obtain these results we use the method from [11, Ch. V.1.3]. We introduce a new unknown matrix-valued function $Y_{2} (x, z)$ by $\begin{matrix} (2.18) & Y_{1} (x, z) = U (x, z) Y_{2} (x, z), (x, z) \in [0, 1] \times Z_{+} (r), \end{matrix}$ where $Y_{1}$ is the solution of equation (2.15).

We search the matrix-valued function U in the form $\begin{matrix} (2.19) & U (x, z) = I_{4} - \frac{q (x)}{z^{4}} W, x \in [0, 1] . \end{matrix}$ We choose the matrix U so that the coefficient on the right-hand side of equation (2.22) decreases as $z^{- 4}$ . It turns out that the matrix W must satisfies the following identity (see the proof of Lemma 2 below): $\begin{matrix} (2.20) & Q + 4 i [W, T] = - i T, \end{matrix}$ where $[W, T] = W T - T W$ . Direct calculations give $\begin{matrix} (2.21) & W = \frac{1}{8} (\begin{matrix} 0 & - 1 + i & - 1 - i & - 1 \\ - 1 - i & 0 & - 1 & - 1 + i \\ - 1 + i & - 1 & 0 & - 1 - i \\ - 1 & - 1 - i & - 1 + i & 0 \end{matrix}) . \end{matrix}$ Thus, we obtain the following results. Lemma 2.
Let $q \in L^{1} (0, 1)$ and let $z \in Z_{+} (r)$ , where $r > 0$ is large enough. Then the matrix-valued function $Y_{1} (x, z)$ , $x \in [0, 1]$ , given by ( 2.18 ), satisfies the equation $\begin{matrix} (2.22) & Y_{2}^{'} - i z T Y_{2} = \frac{1}{z^{4}} Φ_{2} Y_{2}, \end{matrix}$ where T is defined by ( 2.16 ) and $\begin{matrix} (2.23) & Φ_{2} (x, z) = q^{'} (x) W + O (z^{- 1}), \end{matrix}$ as $| z | \to \infty$ uniformly on $x \in [0, 1]$ and the matrix W given by ( 2.21 ).
Proof.
Let $z \in Z_{+} (r)$ . Substituting (2.18) into (2.15), we obtain $\begin{matrix} (2.24) & Y_{2}^{'} = U^{- 1} (A_{1} U - U^{'}) Y_{2}, A_{1} = i z T + \frac{1}{z^{3}} Φ_{1} = i z T + \frac{q}{4 z^{3}} Q . \end{matrix}$ The identity (2.19) yields the asymptotics $\begin{matrix} (2.25) & U^{- 1} (x, z) = I_{4} + \frac{q (x)}{z^{4}} W + O (z^{- 8}) \end{matrix}$ uniformly on $x \in [0, 1]$ . Using (2.19) and (2.25), we get $\begin{array}{l} U^{- 1} A_{1} U = (I_{4} + \frac{q}{z^{4}} W + O (z^{- 8})) A_{1} (I_{4} - \frac{q}{z^{4}} W) = A_{1} + \frac{q}{z^{4}} [W, A_{1}] + O (z^{- 8}), \\ U^{- 1} U^{'} = - (I_{4} + \frac{q}{z^{4}} W + O (z^{- 8})) \frac{q^{'}}{z^{4}} W = - \frac{q^{'}}{z^{4}} W + O (z^{- 8}), \end{array}$ uniformly on $x \in [0, 1]$ . These asymptotics give $\begin{matrix} (2.26) & U^{- 1} (A_{1} U - U^{'}) = i z T + \frac{q}{4 z^{3}} (Q + 4 i [W, T]) + \frac{q^{'}}{z^{4}} W + O (z^{- 5}) \end{matrix}$ uniformly on $x \in [0, 1]$ . Substituting (2.20) into (2.26), we obtain $\begin{matrix} U^{- 1} (A_{1} U - U^{'}) = i z (T - \frac{q}{4 z^{4}} T) + \frac{1}{z^{4}} (q^{'} W + O (z^{- 1})) = i z T + \frac{1}{z^{4}} Φ_{2}, \end{matrix}$ where T and $Φ_{2}$ are defined by (2.16) and (2.23), respectively. Substituting the last identity into equation (2.24), we have (2.22). □

2.2. The Birkhoff method

The fundamental matrix $Y_{1}$ contains both exponentially increasing, bounded, and decreasing entries at high energy. Therefore, asymptotic analysis of this matrix is rather difficult. In various combinations of entries of the fundamental matrix, the contribution of bounded and decreasing entries completely disappears against the background of the contribution of increasing ones. Birkhoff’s method gives a factorization of the fundamental matrix, where the exponential entries are separated into a separate diagonal matrix. This allows good control over their contribution when calculating the asymptotics.

Here we write out the basic formulas for the matrix form of the Birkhoff method (see [4,6,8]). Consider the differential equation $\begin{matrix} (2.27) & Y^{'} - i z Θ Y = \frac{1}{z^{m}} Φ Y \end{matrix}$ on the interval $[0, 1]$ with the unknown $4 \times 4$ -matrix-valued function $Y$ , where $z \in Z_{+} (r)$ , $r > 0$ is large enough, $m \in N$ , and the $4 \times 4$ -matrix-valued functions $Θ (x, z)$ and $Φ (x, z)$ satisfy the following conditions:

$Θ = diag (Θ_{1}, Θ_{2}, Θ_{3}, Θ_{4})$ is diagonal;

$Φ (\cdot, z) \in L^{1} (0, 1)$ and $Θ (\cdot, z) \in L^{1} (0, 1)$ for all $z \in Z_{+} (r)$ ;

for a. e. $x \in [0, 1]$ the functions $Φ (x, z)$ and $Θ (x, z)$ are analytic in $Z_{+} (r)$ and $\begin{matrix} (2.28) & Φ (x, z) = F (x) + O (z^{- 1}), Θ (x, z) = T + O (z^{- 1}), \end{matrix}$ as $| z | \to \infty$ , $z \in Z_{+}$ , uniformly in $x \in [0, 1]$ , where the matrix $T$ is given by (2.14) and the $4 \times 4$ -matrix-valued function $F \in L^{1} (0, 1)$ ;

F is off-diagonal, that is, $F_{j j} = 0$ , $j = 1, 2, 3, 4$ .

Note that equations (2.15) and (2.22) have the form (2.27). In fact, using the Birkhoff method we rewrite the differential equation (2.27) in the form of a specific Fredholm integral equation with a small kernel at high energy.

Let K be an integral operator in the space $C [0, 1]$ of $4 \times 4$ matrix-valued functions given by $\begin{array}{l} {(K X)}_{l j} (x, z) \\ (2.29) & = \int_{0}^{1} K_{l j} (x, s, z) {(Φ X)}_{l j} (s, z) d s, l, j = 1, 2, 3, 4, (x, z) \in [0, 1] \times Z_{+} (r), \end{array}$ for all $X \in C [0, 1]$ , where $r > 0$ is large enough and $\begin{matrix} K_{l j} (x, s, z) = \{\begin{matrix} e^{i z \int_{s}^{x} (Θ_{l} (u, z) - Θ_{j} (u, z)) d u} χ (x - s), & l < j, \\ - e^{i z \int_{s}^{x} (Θ_{l} (u, z) - Θ_{j} (u, z)) d u} χ (s - x), & l ⩾ j, \end{matrix} χ (s) = \{\begin{matrix} 1, & s ⩾ 0, \\ 0, & s < 0 . \end{matrix} \end{matrix}$ Then for $z \in Z_{+} (r)$ the integral operator K is a contraction. Therefore, the matrix-valued integral equation $\begin{matrix} (2.30) & X = I_{4} + \frac{1}{z^{m}} K X, m \in N, \end{matrix}$ has a unique solution $X (\cdot, z) \in C [0, 1]$ . Moreover, $X (\cdot, z) \in C^{1} [0, 1]$ and each matrix-valued function $X (x, \cdot)$ , $x \in [0, 1]$ , is analytic on $Z_{+} (r)$ and satisfies the asymptotics $\begin{matrix} (2.31) & X (x, z) = I_{4} + \frac{(K I_{4}) (x, z)}{z^{m}} + O (z^{- 2 m}), as | z | \to \infty . \end{matrix}$

The integral equation (2.30) and the differential equation (2.27) are equivalent in the following sense.

Lemma 3.
Let $x \in [0, 1]$ , $z \in Z_{+} (r)$ , where $r > 0$ is large enough. Then the matrix-valued function $Y$ given by the identity $\begin{matrix} (2.32) & Y (x, z) = X (x, z) e^{i z \int_{0}^{x} Θ (s, z) d s} \end{matrix}$ satisfies the differential equation ( 2.27 ) if and only if $X$ is a solution of the integral equation ( 2.30 ).

The proof repeats the arguments from [8, Theorem 4.5].
2.3. Representation of the fundamental matrix

Using (2.32), in the following lemma we obtain the factorization of the fundamental matrix A of equation (1.2). We represent the matrix A as a product of the bounded matrix $X$ , the simple matrix Λ, and the diagonal matrix $e^{i z \int_{0}^{x} Θ (s, z) d s}$ . Thus, all exponentially increasing terms are removed from A into this diagonal matrix. In fact, we obtain two factorizations: the factorization (2.38) for nonsmooth potential and the factorization (2.41) for the smooth one. Despite the fact that the factorization (2.38) is also true in the case of smooth potential, it is inconvenient for calculating sharp eigenvalues asymptotics, since the remainder term in $X$ has the order $O (z^{- 4})$ . In the asymptotics (2.41), the remainder term in $X$ has the order $O (z^{- 5})$ , therefore, it is much more convenient for obtaining sharp eigenvalues asymptotics.

Introduce the matrix-valued function $B_{σ} (x, z)$ , $(x, z) \in [0, 1] \times Z_{+}$ , $σ = 1, 2$ , by the formulas $\begin{array}{l} (2.33) & B_{σ, j j} (x, z) = 0, j = 1, 2, 3, 4, \\ (2.34) & B_{σ, l j} (x, z) = - \int_{x}^{1} e^{- i z (s - x) (ω_{l} - ω_{j})} F_{σ, l j} (s) d s, 1 ⩽ j < l ⩽ 4, \\ (2.35) & B_{σ, l j} (x, z) = \int_{0}^{x} e^{i z (x - s) (ω_{l} - ω_{j})} F_{σ, l j} (s) d s, 1 ⩽ l < j ⩽ 4, \end{array}$ where the matrix-valued functions $F_{σ}$ , $σ = 1, 2$ , defined by $\begin{matrix} (2.36) & F_{1} = \frac{q}{4} (Q + i T), F_{2} = q^{'} W, \end{matrix}$ where Q and W are defined by (2.17) and (2.21), respectively. Note that each matrix-valued function $B_{σ} (x, \cdot)$ , $x \in [0, 1]$ , is analytic and bounded in $Z_{+}$ .

Introduce the functions $ζ_{σ, k j} (x, z)$ , $σ = 1, 2$ , k, $j = 1, 2, 3, 4$ , $(x, z) \in [0, 1] \times Z_{+}$ , by $\begin{matrix} (2.37) & ζ_{1, k j} = \frac{B_{1, k j}}{z^{3}}, ζ_{2, k j} = \frac{1}{z^{4}} (B_{2, k j} - q W_{k j}), \end{matrix}$ where W is defined by (2.21). Note that the functions $ζ_{σ, k j}$ , $σ = 1, 2$ , are analytic and bounded in $Z_{+}$ .

We prove the following result.

Lemma 4.
Let $q \in L^{1} (0, 1)$ and let $(x, z) \in [0, 1] \times Z_{+} (r)$ for some $r > 0$ large enough. Then
The fundamental matrix A of equation ( 1.2 ) satisfies the asymptotics $\begin{matrix} (2.38) & A (x, z) = Λ (z) (I_{4} + \frac{B_{1} (x, z)}{z^{3}} + O (z^{- 4})) e^{i z \int_{0}^{x} T (s, z) d s}, \end{matrix}$
where Λ is given by ( 2.8 ), the diagonal $4 \times 4$ matrix-valued function T has the form ( 2.16 ), and the matrix-valued function $B_{1}$ defined by ( 2.33 )–( 2.35 ). Moreover, the fundamental solutions $ϕ_{j}^{'}$ , $ϕ_{j}^{‴}$ , $j = 1, 2, 3, 4$ , given by ( 2.2 ) have the form $\begin{matrix} (2.39) & \begin{matrix} ϕ_{1}^{'} (x, z) = - z (1 + (ζ_{σ, 11} - i ζ_{σ, 21} + i ζ_{σ, 31} - ζ_{σ, 41}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{1} (s, z) d s}, \\ ϕ_{2}^{'} (x, z) = i z (1 + (i ζ_{σ, 12} + ζ_{σ, 22} - ζ_{σ, 32} - i ζ_{σ, 42}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{2} (s, z) d s}, \\ ϕ_{3}^{'} (x, z) = - i z (1 + (- i ζ_{σ, 13} - ζ_{σ, 23} + ζ_{σ, 33} + i ζ_{σ, 43}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{3} (s, z) d s}, \\ ϕ_{4}^{'} (x, z) = z (1 + (- ζ_{σ, 14} + i ζ_{σ, 24} - i ζ_{σ, 34} + ζ_{σ, 44}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{4} (s, z) d s}, \end{matrix} \end{matrix}$ and $\begin{matrix} (2.40) & \begin{matrix} ϕ_{1}^{‴} (x, z) = - z^{3} (1 + (ζ_{σ, 11} + i ζ_{σ, 21} - i ζ_{σ, 31} - ζ_{σ, 41}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{1} (s, z) d s}, \\ ϕ_{2}^{‴} (x, z) = - i z^{3} (1 + (- i ζ_{σ, 12} + ζ_{σ, 22} - ζ_{σ, 32} + i ζ_{σ, 42}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{2} (s, z) d s}, \\ ϕ_{3}^{‴} (x, z) = i z^{3} (1 + (i ζ_{σ, 13} - ζ_{σ, 23} + ζ_{σ, 33} - i ζ_{σ, 43}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{3} (s, z) d s}, \\ ϕ_{4}^{‴} (x, z) = z^{3} (1 + (- ζ_{σ, 14} - i ζ_{σ, 24} + i ζ_{σ, 34} + ζ_{σ, 44}) (x, z) + O (z^{- σ - 3})) e^{i z \int_{0}^{x} T_{4} (s, z) d s}, \end{matrix} \end{matrix}$ as $| z | \to \infty$ , $z \in Z_{+}$ , for $σ = 1$ , uniformly on $x \in [0, 1]$ , where $ζ_{1, k j}$ , $k, j = 1, 2, 3, 4$ , are defined by ( 2.37 ).
Let $q^{'} \in L^{1} (0, 1)$ . Then $\begin{matrix} (2.41) & A (x, z) = Λ (z) U (x, z) (I_{4} + \frac{B_{2} (x, z)}{z^{4}} + O (z^{- 5})) e^{i z \int_{0}^{x} T (s, z) d s}, \end{matrix}$
where T is defined by ( 2.16 ), U is given by ( 2.19 ), and the matrix-valued function $B_{2}$ defined by ( 2.33 )–( 2.35 ). Moreover, the fundamental solutions $ϕ_{j}^{'}$ , $ϕ_{j}^{‴}$ , $j = 1, 2, 3, 4$ , have the form ( 2.39 ) and ( 2.40 ) for $σ = 2$ , where $ζ_{2, k j}$ , $k, j = 1, 2, 3, 4$ , are defined by ( 2.37 ).
Proof.
(i) It follows from (2.13) that $A (x, z) = Λ (z) Y_{1} (x, z)$ , where $Y_{1}$ is the solution of equation (2.15). Comparing (2.15) and (2.27), we see that $m = 3$ , $Θ = T$ , $Φ = Φ_{1}$ , where T and $Φ_{1}$ are defined by (2.16). Using (2.32), we get $\begin{matrix} (2.42) & A (x, z) = Λ (z) X (x, z) e^{i z \int_{0}^{x} Θ (s, z) d s}, \end{matrix}$ where $X$ is the solution of the integral equation (2.30) with $m = 3$ , $Θ = T$ , $Φ = Φ_{1}$ . Substituting (2.29) into (2.31) and using the second identity in (2.16), we obtain $\begin{array}{rcl} X (x, z) & = & I_{4} - \frac{1}{4 z^{3}} \int_{0}^{1} K (x, s, z) q (s) (Q + i T) d s + O (z^{- 4}) \\ (2.43) & = & I_{4} + \frac{B_{1} (x, z)}{z^{3}} + O (z^{- 4}), \end{array}$ where $B_{1}$ is given by (2.33)–(2.35), as $| z | \to \infty$ , $z \in Z_{+}$ , uniformly in $x \in [0, 1]$ . Substituting (2.43) into (2.42), we get (2.38). The identities $ϕ_{j}^{'} = A_{2 j}$ , $ϕ_{j}^{‴} = A_{4 j}$ , $j = 1, 2, 3, 4$ , imply (2.39) and (2.40) for $σ = 1$ , where $ζ_{1, k j}$ , $k, j = 1, 2, 3, 4$ , have the form (2.37).

(ii) It follows from (2.13) and (2.18) that $A (x, z) = Λ (z) Y_{1} (x, z) = Λ (z) U (x, z) Y_{2} (x, z)$ , where $Y_{2}$ is the solution of equation (2.22). Comparing (2.22) and (2.27), we obtain $m = 4$ , $Θ = T$ , $Φ = Φ_{2}$ , where T and $Φ_{2}$ are defined by (2.16) and (2.23), respectively. Using (2.32) and (2.18), we get $\begin{matrix} (2.44) & A (x, z) = Λ (z) U (x, z) X (x, z) e^{i z \int_{0}^{x} Θ (s, z) d s}, \end{matrix}$ where $X$ is the solution of the integral equation (2.30) with $m = 4$ , $Θ = T$ , $Φ = Φ_{2}$ . The asymptotics (2.31) yields $\begin{matrix} (2.45) & X (x, z) = I_{4} + \frac{(K I_{4}) (x, z)}{z^{4}} + O (z^{- 8}) \end{matrix}$ as $| z | \to \infty$ , $z \in Z_{+}$ , uniformly in $x \in [0, 1]$ . Substituting the asymptotics (2.23) into (2.29), we get $\begin{matrix} {(K I_{4})}_{l j} (x, z) = \int_{0}^{1} K_{l j} (x, s, z) {(q^{'} (x) W)}_{l j} d s + O (z^{- 1}) = B_{2, l j} (x, z) + O (z^{- 1}), \end{matrix}$ where l, $j = 1, 2, 3, 4$ , and $B_{2}$ is given by (2.33)–(2.35). Substituting these relations into (2.45), we have $\begin{matrix} X (x, z) = I_{4} + \frac{B_{2} (x, z)}{z^{4}} + O (z^{- 5}) . \end{matrix}$ Substituting this asymptotics into (2.44), we get (2.41). The equations (2.41) and (2.19) imply $\begin{matrix} A (x, z) = Λ (z) (I_{4} - \frac{q (x)}{z^{4}}) (I_{4} + \frac{B_{2} (x, z)}{z^{4}} + O (z^{- 5})) e^{i z \int_{0}^{x} Θ (s, z) d s}, \end{matrix}$ where W has the form (2.21). The identities $ϕ_{j}^{'} = A_{2 j}$ , $ϕ_{j}^{‴} = A_{4 j}$ , $j = 1, 2, 3, 4$ , imply (2.39) and (2.40) for $σ = 2$ , where $ζ_{2, k j}$ , $k, j = 1, 2, 3, 4$ , are defined by (2.37). □

2.4. Counting lemma

Now we prove Counting Lemma for the zeros of the function D defined by (2.6). Introduce the domains $D_{n}$ , $n \in Z_{+}$ , and the contours $C_{a} (r)$ by $\begin{matrix} D_{n} = {λ \in C : | z - π n | < \frac{π}{4}}, C_{a} (r) = {λ \in C : | z - a | = r}, a \in C, r > 0 . \end{matrix}$

Always below the asymptotics for $| z | \to \infty$ are uniform with respect to $\arg z$ in the corresponding sector.

Lemma 5.
Let $q \in L^{1} (0, 1)$ . Then
The function $det A (0, z)$ satisfies $\begin{matrix} (2.46) & det A (0, z) = - 16 i z^{6} (1 + O (z^{- 1})), \end{matrix}$ as $| z | \to \infty$ , $z \in Z_{+}$ . Moreover, $\begin{matrix} (2.47) & D (λ) = D_{0} (λ) (1 + O (z^{- 1})) \end{matrix}$ as $| λ | \to \infty$ , $λ \in C ∖ ⋃_{n \in Z_{+}} D_{n}$ .

For each integer $N ⩾ 0$ large enough the function D has (counting with multiplicities) $N + 1$ zeros in the disk ${| λ | < π^{4} {(N + 1 / 2)}^{4}}$ and for each $n > N$ it has exactly one simple real zero in the domain $D_{n}$ .

Proof.
(i) It follows from (2.38) that $det A (0, z) = det Λ (z) (1 + O (z^{- 1}))$ . The definition (2.8) yields $det Λ = - 16 i z^{6}$ . This gives (2.46).

Let $z \in Z_{+}$ and $| z | \to \infty$ . Substituting (2.2) into (2.5) we obtain $\begin{matrix} (2.48) & \begin{matrix} ϕ (z) \\ = (\begin{matrix} - z (1 + O (z^{- 1})) & i z (1 + O (z^{- 1})) & - i z (1 + O (z^{- 1})) & z (1 + O (z^{- 1})) \\ - z^{3} (1 + O (z^{- 1})) & - i z^{3} (1 + O (z^{- 1})) & i z^{3} (1 + O (z^{- 1})) & z^{3} (1 + O (z^{- 1})) \\ - z e^{- z} (1 + O (z^{- 1})) & i z e^{i z} (1 + O (z^{- 1})) & - i z e^{- i z} (1 + O (z^{- 1})) & z e^{z} (1 + O (z^{- 1})) \\ - z^{3} e^{- z} (1 + O (z^{- 1})) & - i z^{3} e^{i z} (1 + O (z^{- 1})) & i z^{3} e^{- i z} (1 + O (z^{- 1})) & z^{3} e^{z} (1 + O (z^{- 1})) \end{matrix}) . \end{matrix} \end{matrix}$ Let, in addition, $λ \in C ∖ ⋃_{n \in Z_{+}} D_{n}$ . Substituting (2.48) into (2.4) and using (2.11), we get $\begin{matrix} (2.49) & ξ (z) = ξ^{0} (z) (1 + O (z^{- 1})) = - 16 i z^{6} D_{0} (λ) (1 + O (z^{- 1})) . \end{matrix}$ Substituting (2.46) and (2.49) into the identity (2.6), we obtain (2.47).

(ii) Let $N ⩾ 0$ be integer and large enough and let $N^{'} > N$ be another integer. Let λ belong to the contours $C_{0} (π (N + 1 / 2))$ , $C_{0} (π (N^{'} + 1 / 2))$ , and $\partial D_{n}$ for all $n > N$ . Using (2.47), we get $\begin{matrix} | D (λ) - D_{0} (λ) | ⩽ O (z^{- 1}) | D_{0} (λ) | < | D_{0} (λ) | \end{matrix}$ on all contours. Hence, by Rouche’s theorem, D has the same number of zeros as the function $D_{0}$ in each of the bounded domains and the remaining unbounded domain. The function $D_{0}$ has exactly one simple zero in $λ_{n} = {(π n)}^{4}$ , $n \in Z_{+}$ . Therefore, the function D has $N + 1$ zeros in the disk ${| λ | < π^{4} {(N + 1 / 2)}^{4}}$ and for each $n > N$ exactly one simple zero in the domain $D_{n}$ . Since $N^{'} > N$ can be chosen arbitrary large, the statement of Lemma follows.

We prove that the zero of D in the domain $D_{n}$ , $n > N$ , is real. The facts above shows that the function D has one zero μ in $D_{n}$ . If $μ \notin R$ , then there is another zero $\overline{μ} \in D_{n}$ . Therefore, there are two zeros of D in $D_{n}$ . We get a contradiction with the previous statement. Hence μ is real. □

2.5. Asymptotics of the determinant of fundamental matrix

Lemma 5 shows that the eigenvalues at high energy lie in the vicinities $D_{n}$ of the unperturbed eigenvalues. Our next goal is to obtain high energy asymptotics for the eigenvalues. For this we need some preliminary results.

Introduce the sector $Z_{+}^{+} = {z \in C : \arg z \in [0, π / 8]}$ .

The matrix ϕ, given by (2.5), has the form (2.10) in the unperturbed case. This shows that the first column of this matrix contains exponentially decreasing terms as $| z | \to \infty$ in $Z_{+}^{+}$ , and the fourth column contains exponentially increasing ones. Using this fact, we can determine the main term in the asymptotics of the characteristic determinant ξ, given by (2.4). This term is expressed in terms of linear combinations for product of second-order determinants. Thus, we obtain the following results.

Lemma 6.
Let $q \in L^{1} (0, 1)$ and let $| z | \to \infty$ . Then $ξ (z) = e^{\sqrt{2} Re z} O (z^{8})$ , $z \in Z_{+}^{+}$ . Moreover, $\begin{matrix} (2.50) & ξ (z) = ξ_{1} (z) ξ_{2} (z) + ξ_{3} (z) ξ_{4} (z) + O (z^{6}), z \in Z_{+}^{+}, \end{matrix}$ where $\begin{array}{l} (2.51) & ξ_{1} (z) = det (\begin{matrix} ϕ_{3}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{3}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}), ξ_{2} (z) = det (\begin{matrix} ϕ_{1}^{'} (0, z) & ϕ_{2}^{'} (0, z) \\ ϕ_{1}^{‴} (0, z) & ϕ_{2}^{‴} (0, z) \end{matrix}), \\ (2.52) & ξ_{3} (z) = det (\begin{matrix} ϕ_{2}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{2}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}), ξ_{4} (z) = det (\begin{matrix} ϕ_{1}^{‴} (0, z) & ϕ_{3}^{‴} (0, z) \\ ϕ_{1}^{'} (0, z) & ϕ_{3}^{'} (0, z) \end{matrix}) . \end{array}$
Proof.
Let $z \in Z_{+}^{+}$ and $| z | \to \infty$ . Then $0 ⩽ Im z ⩽ (\sqrt{2} - 1) Re z$ . Consider the identity (2.4). Direct calculations yield $\begin{array}{l} ξ (z) = & |\begin{matrix} ϕ_{3}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{3}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{1}^{'} (0, z) & ϕ_{2}^{'} (0, z) \\ ϕ_{1}^{‴} (0, z) & ϕ_{2}^{‴} (0, z) \end{matrix}| \\ + |\begin{matrix} ϕ_{2}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{2}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{1}^{‴} (0, z) & ϕ_{3}^{‴} (0, z) \\ ϕ_{1}^{'} (0, z) & ϕ_{3}^{'} (0, z) \end{matrix}| \\ + |\begin{matrix} ϕ_{2}^{'} (1, z) & ϕ_{3}^{'} (1, z) \\ ϕ_{2}^{‴} (1, z) & ϕ_{3}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{1}^{'} (0, z) & ϕ_{4}^{'} (0, z) \\ ϕ_{1}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \end{matrix}| \\ + |\begin{matrix} ϕ_{1}^{'} (1, z) & ϕ_{3}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{3}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{2}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \\ ϕ_{2}^{'} (0, z) & ϕ_{4}^{'} (0, z) \end{matrix}| \\ + |\begin{matrix} ϕ_{1}^{'} (1, z) & ϕ_{2}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{2}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{3}^{'} (0, z) & ϕ_{4}^{'} (0, z) \\ ϕ_{3}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \end{matrix}| \\ (2.53) & + |\begin{matrix} ϕ_{1}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{2}^{'} (0, z) & ϕ_{3}^{'} (0, z) \\ ϕ_{2}^{‴} (0, z) & ϕ_{3}^{‴} (0, z) \end{matrix}| . \end{array}$ The asymptotics (2.2) imply $\begin{array}{l} |\begin{matrix} ϕ_{2}^{'} (1, z) & ϕ_{3}^{'} (1, z) \\ ϕ_{2}^{‴} (1, z) & ϕ_{3}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{1}^{'} (0, z) & ϕ_{4}^{'} (0, z) \\ ϕ_{1}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \end{matrix}| \\ = |\begin{matrix} i z e^{i z} (1 + O (z^{- 1})) & - i z e^{- i z} (1 + O (z^{- 1})) \\ - i z^{3} e^{i z} (1 + O (z^{- 1})) & i z^{3} e^{- i z} (1 + O (z^{- 1})) \end{matrix}| |\begin{matrix} - z (1 + O (z^{- 1})) & z (1 + O (z^{- 1})) \\ - z^{3} (1 + O (z^{- 1})) & z^{3} (1 + O (z^{- 1})) \end{matrix}| \\ = z^{3} O (1) \cdot z^{3} O (1) = O (z^{6}) \end{array}$ and $\begin{array}{l} |\begin{matrix} ϕ_{1}^{'} (1, z) & ϕ_{4}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{4}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{2}^{'} (0, z) & ϕ_{3}^{'} (0, z) \\ ϕ_{2}^{‴} (0, z) & ϕ_{3}^{‴} (0, z) \end{matrix}| \\ = |\begin{matrix} - z e^{- z} (1 + O (z^{- 1})) & z e^{z} (1 + O (z^{- 1})) \\ - z^{3} e^{- z} (1 + O (z^{- 1})) & z^{3} e^{z} (1 + O (z^{- 1})) \end{matrix}| |\begin{matrix} i z (1 + O (z^{- 1})) & - i z (1 + O (z^{- 1})) \\ - i z^{3} (1 + O (z^{- 1})) & i z^{3} (1 + O (z^{- 1})) \end{matrix}| \\ = z^{3} O (1) \cdot z^{3} O (1) = O (z^{6}) . \end{array}$ The estimates $0 ⩽ Im z ⩽ (\sqrt{2} - 1) Re z$ and the asymptotics (2.2) give $\begin{array}{l} |\begin{matrix} ϕ_{1}^{'} (1, z) & ϕ_{3}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{3}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{2}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \\ ϕ_{2}^{'} (0, z) & ϕ_{4}^{'} (0, z) \end{matrix}| \\ = |\begin{matrix} - z e^{- z} O (1) & - i z e^{- i z} O (1) \\ - z^{3} e^{- z} O (1) & i z^{3} e^{- i z} O (1) \end{matrix}| |\begin{matrix} - i z^{3} O (1) & z^{3} O (1) \\ i z O (1) & z O (1) \end{matrix}| \\ = e^{- z - i z} O (z^{8}) = e^{- Re z + Im z} O (z^{8}) = e^{(\sqrt{2} - 2) Re z} O (z^{8}) \end{array}$ and $\begin{array}{l} |\begin{matrix} ϕ_{1}^{'} (1, z) & ϕ_{2}^{'} (1, z) \\ ϕ_{1}^{‴} (1, z) & ϕ_{2}^{‴} (1, z) \end{matrix}| |\begin{matrix} ϕ_{3}^{'} (0, z) & ϕ_{4}^{'} (0, z) \\ ϕ_{3}^{‴} (0, z) & ϕ_{4}^{‴} (0, z) \end{matrix}| \\ = |\begin{matrix} - z e^{- z} O (1) & i z e^{i z} O (1) \\ - z^{3} e^{- z} O (1) & - i z^{3} e^{i z} O (1) \end{matrix}| |\begin{matrix} - i z O (1) & z O (1) \\ i z^{3} O (1) & z^{3} O (1) \end{matrix}| \\ = e^{- z + i z} O (z^{8}) \\ = e^{- Re z - Im z} O (z^{8}) \\ = e^{- \sqrt{2} Re z} O (z^{8}) . \end{array}$ Substituting these asymptotics into (2.53), we obtain (2.50). Moreover, using again the asymptotics (2.2) and the estimates $0 ⩽ Im z ⩽ (\sqrt{2} - 1) Re z$ , we get $\begin{array}{l} ξ_{1} (z) ξ_{2} (z) + ξ_{3} (z) ξ_{4} (z) \\ = |\begin{matrix} - i z e^{- i z} O (1) & z e^{z} O (1) \\ i z^{3} e^{- i z} O (1) & z^{3} e^{z} O (1) \end{matrix}| |\begin{matrix} - z O (1) & i z O (1) \\ - z^{3} O (1) & - i z^{3} O (1) \end{matrix}| \\ + |\begin{matrix} i z e^{i z} O (1) & z e^{z} O (1) \\ - i z^{3} e^{i z} O (1) & z^{3} e^{z} O (1) \end{matrix}| |\begin{matrix} - z^{3} O (1) & i z^{3} O (1) \\ - z O (1) & - i z O (1) \end{matrix}| \\ = e^{z - i z} O (z^{8}) + e^{z + i z} O (z^{8}) \\ = e^{z - i z} (O (z^{8}) + e^{i 2 z} O (z^{8})) \\ = e^{Re z + Im z} O (z^{8}) = e^{\sqrt{2} Re z} O (z^{8}) . \end{array}$ This gives the statement of the lemma. □

3. Eigenvalues for $q \in L^{1} (0, 1)$

3.1. Rough eigenvalue asymptotics

Recall that the eigenvalues of the operator H are zeros of the entire function D given by (1.3).

We index the eigenvalues $μ_{n}$ of the function D by the following way. It follows from Lemma 5 that the function D has (counting with multiplicities) $N + 1$ zeros in the disk ${| λ | < π^{4} {(N + 1 / 2)}^{4}}$ for each integer $N \in N$ large enough. The eigenvalues $μ_{n}$ inside this domain are labeled by $\begin{matrix} Re μ_{0} ⩽ Re μ_{1} ⩽ Re μ_{2} ⩽ \dots ⩽ Re μ_{N} . \end{matrix}$ Each domain $D_{n}$ , $n > N$ , contains one real eigenvalue that we denote by $μ_{n}$ .

We determine the rough asymptotics of the large eigenvalues $μ_{n}$ . The identity (2.6) and the asymptotics (2.46) show that the large eigenvalues are zeros of the functions $ξ (z)$ defined by (2.4). The function $ξ (z)$ is analytic in $Z_{+} (r)$ , where $r > 0$ is large enough.

Lemma 7.
The eigenvalues $μ_{n}$ satisfy $\begin{matrix} (3.1) & μ_{n} = {(π n)}^{4} + O (n^{2}), as n \to + \infty . \end{matrix}$
Proof.
Let $λ = z^{4} = μ_{n}$ , $n \to + \infty$ . Lemma 5(ii) yields $z = π n + δ_{n}$ , where $δ_{n} = O (1)$ . Substituting the asymptotics (2.2) into (2.51) and (2.52), we obtain $\begin{array}{l} ξ_{1} (z) ξ_{2} (z) = & |\begin{matrix} - i z e^{- i z} (1 + O (n^{- 1})) & z e^{z} (1 + O (n^{- 1})) \\ i z^{3} e^{- i z} (1 + O (n^{- 1})) & z^{3} e^{z} (1 + O (n^{- 1})) \end{matrix}| \\ \times |\begin{matrix} - z (1 + O (n^{- 1})) & i z (1 + O (n^{- 1})) \\ - z^{3} (1 + O (n^{- 1})) & - i z^{3} (1 + O (n^{- 1})) \end{matrix}| \\ = & 4 z^{8} e^{z - i z} (1 + O (n^{- 1})) \end{array}$ and $\begin{array}{l} ξ_{3} (z) ξ_{4} (z) = & |\begin{matrix} i z e^{i z} (1 + O (n^{- 1})) & z e^{z} (1 + O (n^{- 1})) \\ - i z^{3} e^{i z} (1 + O (n^{- 1})) & z^{3} e^{z} (1 + O (n^{- 1})) \end{matrix}| \\ \times |\begin{matrix} - z^{3} (1 + O (n^{- 1})) & i z^{3} (1 + O (n^{- 1})) \\ - z (1 + O (n^{- 1})) & - i z (1 + O (n^{- 1})) \end{matrix}| \\ = & - 4 z^{8} e^{z + i z} (1 + O (n^{- 1})) . \end{array}$ Substitute these asymptotics into (2.50). Then $\begin{array}{l} ξ (z) & = ξ_{1} (z) ξ_{2} (z) + ξ_{3} (z) ξ_{4} (z) + O (z^{6}) \\ = 4 z^{8} e^{z} (e^{- i z} - e^{i z} + O (n^{- 1})) \\ = - 8 i z^{8} e^{z} (sin z + O (n^{- 1})) \\ = - 8 i z^{8} e^{z} ({(- 1)}^{n} sin δ_{n} + O (n^{- 1})) . \end{array}$ The equation $ξ (z) = 0$ gives $δ_{n} = O (n^{- 1})$ . Therefore, $z = π n + O (n^{- 1})$ . This yields (3.1). □

3.2. Sharp eigenvalue asymptotics

Let $q \in L^{1} (0, 1)$ . Then the fundamental matrix A satisfies (2.38). In order to clarify the eigenvalue asymptotics we need to improve the asymptotics of the function ξ. Introduce the functions α and β by the formulas $\begin{matrix} (3.2) & \begin{matrix} α (z) = \int_{0}^{1} T_{2} (s, z) d s = - \int_{0}^{1} T_{3} (s, z) d s = 1 - \frac{q_{0}}{4 z^{4}}, z \in Z_{+}, \\ β (z) = \int_{0}^{1} T_{1} (s, z) d s = - \int_{0}^{1} T_{4} (s, z) d s = i - \frac{i q_{0}}{4 z^{4}}, z \in Z_{+} . \end{matrix} \end{matrix}$ where $T_{j}$ are elements of the matrix $T = {(T_{j})}_{j = 1}^{4}$ and T is defined by (2.16). The functions α and β are real-analytic functions on the domain $Z_{+}$ . The asymptotics (2.28) and the definition (2.14) give $α (z) = 1 + o (1)$ , $β (z) = 1 + o (1)$ , $z \in Z_{+}$ , $| z | \to \infty$ .

We prove the following results.

Lemma 8.
Let $q \in L^{1} (0, 1)$ and let $| z | \to \infty$ . Then the functions ξ defined by ( 2.50 ) have the asymptotics $\begin{matrix} (3.3) & \begin{matrix} ξ (z) & = - 4 z^{8} e^{- i β z} (2 i sin α z - e^{- i α z} γ_{σ, 1} (z) + e^{i α z} γ_{σ, 2} (z)), z \in Z_{+}, \end{matrix} \end{matrix}$ where $α = α (z)$ and $β = β (z)$ , and $γ_{σ, 1} (z)$ , $γ_{σ, 2} (z)$ , $σ = 1, 2$ , $z \in Z_{+}$ , satisfy the asymptotics $\begin{matrix} (3.4) & \begin{matrix} γ_{σ, 1} (z) = (ζ_{σ, 33} - ζ_{σ, 23} - ζ_{σ, 14} + ζ_{σ, 44}) (1, z) \\ + (ζ_{σ, 11} - ζ_{σ, 41} + ζ_{σ, 22} - ζ_{σ, 32}) (0, z) + O (z^{- σ - 3}), \\ γ_{σ, 2} (z) = (ζ_{σ, 22} - ζ_{σ, 32} - ζ_{σ, 14} + ζ_{σ, 44}) (1, z) \\ + (ζ_{σ, 33} - ζ_{σ, 23} + ζ_{σ, 11} - ζ_{σ, 41}) (0, z) + O (z^{- σ - 3}) . \end{matrix} \end{matrix}$ Here $ζ_{σ, k j}$ , $σ = 1, 2, k$ , $j = 1, 2, 3, 4$ , are defined by ( 2.37 ).
Proof.
Let $z \in Z_{+}$ , $| z | \to \infty$ . We prove the statement of this lemma in the case $σ = 1$ . The proof in the case $σ = 2$ is similar. The formula (2.37) gives $\begin{matrix} (3.5) & ζ_{1, k j} (x, z) ζ_{1, k j} (y, z) = O (z^{- 6}), x, y \in [0, 1], z \in Z_{+} . \end{matrix}$ Substituting (2.39) and (2.40) into (2.51) and using (3.2), (3.5), we get $\begin{matrix} (3.6) & \begin{matrix} ξ_{1} (z) = & - i z^{4} e^{- i α z - i β z} (1 + (- i ζ_{1, 13} - ζ_{1, 23} + ζ_{1, 33} + i ζ_{1, 43}) (1, z) + O (z^{- 4})) \\ \times (1 + (- ζ_{1, 14} - i ζ_{1, 24} + i ζ_{1, 34} + ζ_{1, 44}) (1, z) + O (z^{- 4})) \\ - i z^{4} e^{- i α z - i β z} (1 + (i ζ_{1, 13} - ζ_{1, 23} + ζ_{1, 33} - i ζ_{1, 43}) (1, z) + O (z^{- 4})) \\ \times (1 + (- ζ_{1, 14} + i ζ_{1, 24} - i ζ_{1, 34} + ζ_{1, 44}) (1, z) + O (z^{- 4})) \\ = & - 2 i z^{4} e^{- i α z - i β z} (1 + (- ζ_{1, 23} + ζ_{1, 33} - ζ_{1, 14} + ζ_{1, 44}) (1, z) + O (z^{- 4})) \end{matrix} \end{matrix}$ and $\begin{matrix} (3.7) & \begin{matrix} ξ_{2} (z) = & i z^{4} (1 + (ζ_{1, 11} - i ζ_{1, 21} + i ζ_{1, 31} - ζ_{1, 41}) (0, z) + O (z^{- 4})) \\ \times (1 + (- i ζ_{1, 12} + ζ_{1, 22} - ζ_{1, 32} + i ζ_{1, 42}) (0, z) + O (z^{- 4})) \\ + i z^{4} (1 + (i ζ_{1, 12} + ζ_{1, 22} - ζ_{1, 32} - i ζ_{1, 42}) (0, z) + O (z^{- 4})) \\ \times (1 + (ζ_{1, 11} + i ζ_{1, 21} - i ζ_{1, 31} - ζ_{1, 41}) (0, z) + O (z^{- 4})) \\ = & 2 i z^{4} (1 + (ζ_{1, 11} - ζ_{1, 41} + ζ_{1, 22} - ζ_{1, 32}) (0, z) + O (z^{- 4})) . \end{matrix} \end{matrix}$ Recall that the functions $ζ_{1, k j}$ , k, $j = 1, 2, 3, 4$ , are bounded (see §2.3). Substituting (2.39) and (2.40) into (2.52) and using again (3.2) and (3.5), we obtain $\begin{matrix} (3.8) & \begin{matrix} ξ_{3} (z) = & i z^{4} e^{i α z - i β z} (1 + (i ζ_{1, 12} + ζ_{1, 22} - ζ_{1, 32} - i ζ_{1, 42}) (1, z) + O (z^{- 4})) \\ \times (1 + (- ζ_{1, 14} - i ζ_{1, 24} + i ζ_{1, 34} + ζ_{1, 44}) (1, z) + O (z^{- 4})) \\ + i z^{4} e^{i α z - i β z} (1 + (- ζ_{1, 14} + i ζ_{1, 24} - i ζ_{1, 34} + ζ_{1, 44}) (1, z) + O (z^{- 4})) \\ \times (1 + (- i ζ_{1, 12} + ζ_{1, 22} - ζ_{1, 32} + i ζ_{σ, 42}) (1, z) + O (z^{- 4})) \\ = & 2 i z^{4} e^{i α z - i β z} (1 + (ζ_{1, 22} - ζ_{1, 32} - ζ_{1, 14} + ζ_{1, 44}) (1, z) + O (z^{- 4})), \end{matrix} \end{matrix}$ and $\begin{matrix} (3.9) & \begin{matrix} ξ_{4} (z) = & i z^{4} (1 + (- i ζ_{1, 13} - ζ_{1, 23} + ζ_{1, 33} + i ζ_{1, 43}) (0, z) + O (z^{- 4})) \\ \times (1 + (ζ_{1, 11} + i ζ_{1, 21} - i ζ_{1, 31} - ζ_{1, 41}) (0, z) + O (z^{- 4})) \\ + i z^{4} (1 + (ζ_{1, 11} - i ζ_{1, 21} + i ζ_{1, 31} - ζ_{1, 41}) (0, z) + O (z^{- 4})) \\ \times (1 + (i ζ_{1, 13} - ζ_{1, 23} + ζ_{1, 33} - i ζ_{1, 43}) (0, z) + O (z^{- 4})) \\ = & 2 i z^{4} (1 + (ζ_{1, 11} - ζ_{1, 41} - ζ_{1, 23} + ζ_{1, 33}) (0, z) + O (z^{- 4})) . \end{matrix} \end{matrix}$ Substituting (3.6)–(3.9) into (2.50), we obtain (3.3). □

Now we determine sharp eigenvalues asymptotics for the case $q \in L^{1} (0, 1)$ . First, we obtain asymptotics of the functions $γ_{1, 1}$ and $γ_{1, 2}$ defined by (3.4). Introduce the sequence $\begin{matrix} (3.10) & ϰ_{1, n} = \frac{1}{4} \int_{0}^{1} e^{- 2 π n s} (q (s) + q (1 - s)) d s, n \in N . \end{matrix}$ Note that $ϰ_{1, n} = O (1)$ , as $n \to + \infty$ .
Lemma 9.
Let $q \in L^{1} (0, 1)$ . Let $z \in Z_{+}$ and let $z = \sqrt[4]{λ} = π n + O (n^{- 1})$ , $n \to + \infty$ . Then $\begin{matrix} (3.11) & γ_{1, 1} (z) = \frac{i {\hat{q}}_{c n}}{2 z^{3}} - \frac{ϰ_{1, n}}{z^{3}} + O (n^{- 4}), γ_{1, 2} (z) = - \frac{ϰ_{1, n}}{z^{3}} + O (n^{- 4}), \end{matrix}$ where $ϰ_{1, n}$ has the form ( 3.10 ).
Proof.
Let $z = π n + O (n^{- 1})$ . The first definition (3.4) and the first identity (2.37) yield $\begin{matrix} (3.12) & \begin{matrix} γ_{1, 1} (z) = & \frac{1}{z^{3}} (- B_{1, 23} (1, z) + B_{1, 33} (1, z) - B_{1, 14} (1, z) + B_{1, 44} (1, z) \\ + B_{1, 11} (0, z) - B_{1, 41} (0, z) + B_{1, 22} (0, z) - B_{1, 32} (0, z)) + O (z^{- 4}) . \end{matrix} \end{matrix}$ Substituting (2.33)–(2.35) into (3.12) and using the first identity in (2.36), we have $\begin{array}{l} γ_{1, 1} (z) = & \frac{1}{z^{3}} (- \int_{0}^{1} e^{i 2 π n (1 - s)} F_{1, 23} (s) d s - \int_{0}^{1} e^{- 2 π n (1 - s)} F_{1, 14} (s) d s \\ - \int_{0}^{1} e^{- 2 π n s} F_{1, 41} (s) d s - \int_{0}^{1} e^{i 2 π n s} F_{1, 32} (s) d s) + O (n^{- 4}) \\ = & - \frac{1}{4 z^{3}} \int_{0}^{1} q (s) (e^{- 2 π n s} + e^{- 2 π n (1 - s)}) d s + \frac{i ({\overline{\hat{q}}}_{n} + {\hat{q}}_{n})}{4 z^{3}} + O (n^{- 4}) . \end{array}$ The identity ${\overline{\hat{q}}}_{n} + {\hat{q}}_{n} = 2 {\hat{q}}_{c n}$ gives the first asymptotics in (3.11).

Similarly, the second definition (3.4) and the first identity (2.37) imply $\begin{matrix} (3.13) & \begin{matrix} γ_{1, 2} (z) = & \frac{1}{z^{3}} (B_{1, 22} (1, z) - B_{1, 32} (1, z) - B_{1, 14} (1, z) + B_{1, 44} (1, z) \\ + B_{1, 33} (0, z) - B_{1, 23} (0, z) + B_{1, 11} (0, z) - B_{1, 41} (0, z)) + O (z^{- 4}) . \end{matrix} \end{matrix}$ Substituting (2.33)–(2.35) into (3.13) and using the first identity in (2.36), we obtain $\begin{array}{l} γ_{1, 2} (z) & = \frac{1}{z^{3}} (- \int_{0}^{1} e^{- 2 π n (1 - s)} F_{1, 14} (s) d s - \int_{0}^{1} e^{- 2 π n s} F_{1, 41} (s) d s) + O (n^{- 4}) \\ = - \frac{1}{4 z^{3}} \int_{0}^{1} q (s) (e^{- 2 π n s} + e^{- 2 π n (1 - s)}) d s + O (n^{- 4}) . \end{array}$ This yields the second asymptotics in (3.11). □

Second, we determine asymptotics of the eigenvalues $μ_{n}$ . Lemma 10.
Let $q \in L^{1} (0, 1)$ . Then the eigenvalues $μ_{n}$ satisfy the asymptotics ( 1.4 ).
Proof.
Let $λ = z^{4} = μ_{n}$ , $n \to + \infty$ . Lemma 7 shows $z = π n + δ_{n}$ , $δ_{n} = O (n^{- 1})$ . Substituting the asymptotics (3.11) into (3.3), we get $\begin{matrix} (3.14) & \begin{matrix} ξ (z) = & - 4 z^{8} e^{- i β z} (2 i sin α z - e^{- i α z} (\frac{i {\hat{q}}_{c n}}{2 z^{3}} - \frac{ϰ_{1, n}}{z^{3}} + O (n^{- 4})) \\ + e^{i α z} (- \frac{ϰ_{1, n}}{z^{3}} + O (n^{- 4}))) \\ = & - 4 z^{8} e^{z - \frac{q_{0}}{4 z^{3}}} (2 i sin α z - \frac{2 i ϰ_{1, n}}{π^{3} n^{3}} sin α z - \frac{i {\hat{q}}_{c n}}{2 π^{3} n^{3}} e^{- i α z} + O (n^{- 4})), \end{matrix} \end{matrix}$ where α, β, and $ϰ_{1, n}$ have the form (3.2) and (3.10), respectively. Substituting the asymptotics $\begin{array}{l} sin α z & = sin (z - \frac{q_{0}}{4 z^{3}}) \\ = {(- 1)}^{n} sin (δ_{n} - \frac{q_{0}}{4 π^{3} n^{3}} + O (n^{- 5})) \\ = {(- 1)}^{n} (δ_{n} - \frac{q_{0}}{4 π^{3} n^{3}} + O (n^{- 5})) \end{array}$ and $\begin{matrix} e^{- i α z} = e^{- i z + \frac{i q_{0}}{4 z^{3}}} = {(- 1)}^{n} e^{- i δ_{n} + \frac{i q_{0}}{4 z^{3}}} = {(- 1)}^{n} (1 + O (n^{- 1})) \end{matrix}$ into (3.14) and using the asymptotics $ϰ_{1, n} = O (1)$ , we obtain $\begin{matrix} ξ (z) = - 4 z^{8} e^{π n} {(- 1)}^{n} (2 i δ_{n} - \frac{i q_{0}}{2 π^{3} n^{3}} - \frac{i {\hat{q}}_{c n}}{2 π^{3} n^{3}} + O (n^{- 4})) . \end{matrix}$ The equation $ξ (z) = 0$ gives $\begin{matrix} δ_{n} = \frac{q_{0} + {\hat{q}}_{c n}}{4 π^{3} n^{3}} + O (n^{- 4}) . \end{matrix}$ Then identity $z = π n + δ_{n}$ implies $\begin{matrix} z = π n + \frac{q_{0} + {\hat{q}}_{c n}}{4 π^{3} n^{3}} + O (n^{- 4}), \end{matrix}$ which yields the asymptotics (1.4). □

4. The case $q^{'} \in L^{1} (0, 1)$

Now we determine sharp eigenvalues asymptotics for the case $q^{'} \in L^{1} (0, 1)$ . First, we determine asymptotics of the functions $γ_{2, 1}$ and $γ_{2, 2}$ defined by (3.4). Introduce the sequence $\begin{matrix} (4.1) & ϰ_{2, n} = \frac{1}{8} \int_{0}^{1} e^{- 2 π n s} (q^{'} (1 - s) - q^{'} (s)) d s, n \in N . \end{matrix}$ Note that $ϰ_{2, n} = O (1)$ , as $n \to + \infty$ .

Lemma 11.
Let $q^{'} \in L^{1} (0, 1)$ . Let $z \in Z_{+}$ and let $z = π n + O (n^{- 1})$ . Then $\begin{matrix} (4.2) & \begin{matrix} γ_{2, 1} (z) = - \frac{q (0) + q (1)}{4 z^{4}} - \frac{i {\hat{q}}_{s n}^{'}}{4 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (n^{- 5}), \\ γ_{2, 2} (z) = - \frac{q (0) + q (1)}{4 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (n^{- 5}), \end{matrix} \end{matrix}$ where $ϰ_{2, n}$ has the form ( 4.1 ) and $\begin{matrix} (4.3) & {\hat{q}}_{s n}^{'} = \int_{0}^{1} q^{'} (x) sin 2 π n x d x . \end{matrix}$
Proof.
Let $z = π n + O (n^{- 1})$ . The first definition (3.4) and the second formula (2.37) yield $\begin{array}{l} γ_{2, 1} (z) = & \frac{1}{z^{4}} (- B_{2, 23} (1, z) + B_{2, 33} (1, z) - B_{2, 14} (1, z) + B_{2, 44} (1, z) + B_{2, 11} (0, z) - B_{2, 41} (0, z) \\ + B_{2, 22} (0, z) - B_{2, 32} (0, z) + q (1) (W_{23} - W_{33} + W_{14} - W_{44}) \\ + q (0) (- W_{11} + W_{41} - W_{22} + W_{32})) + O (z^{- 5}) . \end{array}$ The identity (2.21) gives $\begin{array}{l} γ_{2, 1} (z) = & - \frac{q (0) + q (1)}{4 z^{4}} + \frac{1}{z^{4}} (- B_{2, 23} (1, z) + B_{2, 33} (1, z) - B_{2, 14} (1, z) \\ + B_{2, 44} (1, z) + B_{2, 11} (0, z) - B_{2, 41} (0, z) + B_{2, 22} (0, z) - B_{2, 32} (0, z)) + O (z^{- 5}) \end{array}$ Substituting (2.33)–(2.35) into this representation and using the second identity in (2.36), we get $\begin{array}{l} γ_{2, 1} (z) = & - \frac{q (0) + q (1)}{4 z^{4}} + \frac{1}{z^{4}} (- \int_{0}^{1} e^{i 2 π n (1 - s)} F_{2, 23} (s) d s - \int_{0}^{1} e^{- 2 π n (1 - s)} F_{2, 14} (s) d s \\ + \int_{0}^{1} e^{- 2 π n s} F_{2, 41} (s) d s + \int_{0}^{1} e^{i 2 π n s} F_{2, 32} (s) d s) + O (n^{- 5}) \\ = & - \frac{q (0) + q (1)}{4 z^{4}} + \frac{{\hat{q}}_{n}^{'} - {\overline{\hat{q}}}_{n}^{'}}{8 z^{4}} + \frac{1}{8 z^{4}} \int_{0}^{1} q^{'} (s) (- e^{- 2 π n s} + e^{- 2 π n (1 - s)}) d s + O (n^{- 5}) . \end{array}$ The identity ${\hat{q}}_{n}^{'} - {\overline{\hat{q}}}_{n}^{'} = - 2 i {\hat{q}}_{s n}^{'}$ implies the first asymptotics in (4.2).

Similarly, the second definition (3.4) and the second formula (2.37) give $\begin{array}{l} γ_{2, 2} (z) = & \frac{1}{z^{4}} (B_{2, 22} (1, z) - B_{2, 32} (1, z) - B_{2, 14} (1, z) + B_{2, 44} (1, z) + B_{2, 33} (0, z) - B_{2, 23} (0, z) \\ + B_{2, 11} (0, z) - B_{2, 41} (0, z) + q (1) (- W_{22} + W_{32} + W_{14} - W_{44}) \\ + q (0) (- W_{33} + W_{23} - W_{11} + W_{41})) + O (z^{- 5}) . \end{array}$ The identity (2.21) implies $\begin{array}{l} γ_{2, 2} (z) = & - \frac{q (0) + q (1)}{4 z^{4}} + \frac{1}{z^{4}} (B_{2, 22} (1, z) - B_{2, 32} (1, z) - B_{2, 14} (1, z) + B_{2, 44} (1, z) \\ + B_{2, 33} (0, z) - B_{2, 23} (0, z) + B_{2, 11} (0, z) - B_{2, 41} (0, z)) + O (z^{- 5}) . \end{array}$ Substituting (2.33)–(2.35) into these asymptotics and using the second identity in (2.36), we get $\begin{array}{l} γ_{2, 2} (z) & = - \frac{q (0) + q (1)}{4 z^{4}} - \frac{1}{z^{4}} \int_{0}^{1} e^{- 2 π n (1 - s)} F_{2, 14} (s) d s - \frac{1}{z^{4}} \int_{0}^{1} e^{- 2 π n s} F_{2, 41} (s) d s + O (n^{- 5}) \\ = - \frac{q (0) + q (1)}{4 z^{4}} + \frac{1}{8 z^{4}} \int_{0}^{1} q^{'} (s) (- e^{- 2 π n s} + e^{- 2 π n (1 - s)}) d s + O (n^{- 5}) . \end{array}$ This implies the second asymptotics in (4.2). □

Second, we determine asymptotics of the eigenvalues $μ_{n}$ . Lemma 12.
Let $q^{'} \in L^{1} (0, 1)$ . Then the eigenvalues $μ_{n}$ satisfy the asymptotics $\begin{matrix} (4.4) & μ_{n} = {(π n)}^{4} + q_{0} - \frac{{\hat{q}}_{s n}^{'}}{2 π n} + O (n^{- 2}), as n \to + \infty, \end{matrix}$ where ${\hat{q}}_{s n}^{'}$ is defined by ( 4.3 ).
Proof.
Let $λ = z^{4} = μ_{n}$ , $n \to + \infty$ . Lemma 10 shows $\begin{matrix} (4.5) & z = π n + \frac{q_{0}}{4 π^{3} n^{3}} + δ_{n}, δ_{n} = O (n^{- 4}) . \end{matrix}$ Substituting the asymptotics (4.2) into (3.3), we get $\begin{array}{l} ξ (z) = & - 4 z^{8} e^{- i β z} (2 i sin α z - e^{- i α z} (- \frac{q (0) + q (1)}{4 z^{4}} - \frac{i {\hat{q}}_{s n}^{'}}{4 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (n^{- 5})) \\ + e^{i α z} (- \frac{q (0) + q (1)}{4 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (n^{- 5}))) \\ = & - 4 z^{8} e^{z - \frac{q_{0}}{4 z^{3}}} (2 i sin α z + (- \frac{i (q (0) + q (1))}{2 π^{4} n^{4}} + \frac{2 i ϰ_{2, n}}{π^{4} n^{4}}) sin α z \\ (4.6) & + \frac{i {\hat{q}}_{s n}^{'}}{4 π^{4} n^{4}} e^{- i α z} + O (n^{- 5})), \end{array}$ where α, β, and $ϰ_{2, n}$ have the form (3.2) and (4.1), respectively. Using again (3.2) and (4.5), we get $\begin{matrix} α z = z - \frac{q_{0}}{4 z^{3}} = π n + δ_{n} + \frac{q_{0}}{4 π^{3} n^{3}} - \frac{q_{0}}{4 z^{3}} = π n + δ_{n} + O (n^{- 5}) . \end{matrix}$ Then $\begin{matrix} sin α z = {(- 1)}^{n} (δ_{n} + O (n^{- 5})), e^{- i α z} = {(- 1)}^{n} (1 + O (n^{- 4})) . \end{matrix}$ Substituting these asymptotics into (4.6) and using the asymptotics $ϰ_{2, n} = O (1)$ , we obtain $\begin{matrix} ξ (z) = - 4 z^{8} e^{π n} {(- 1)}^{n} (2 i δ_{n} + \frac{i {\hat{q}}_{s n}^{'}}{4 π^{4} n^{4}} + O (n^{- 5})) . \end{matrix}$ The equation $ξ (z) = 0$ gives $\begin{matrix} δ_{n} = - \frac{{\hat{q}}_{s n}^{'}}{8 π^{4} n^{4}} + O (n^{- 5}) . \end{matrix}$ Then identity (4.5) implies $\begin{matrix} z = π n + \frac{q_{0}}{4 π^{3} n^{3}} - \frac{{\hat{q}}_{s n}^{'}}{8 π^{4} n^{4}} + O (n^{- 5}), \end{matrix}$ which yields the asymptotics (4.4). □
Proof of Theorem 1.
The formula (1.4) is proved in Lemma 10. Substituting the identity ${\hat{q}}_{s n}^{'} = - 2 π n {\hat{q}}_{c n}$ into asymptotics (4.4), we get (1.5). □

5. Trace formula

5.1. Asymptotics of the characteristic function

Now we determine asymptotics of the functions $γ_{2, 1}$ and $γ_{2, 2}$ defined by (3.4) in the case $q^{″} \in L^{1} (0, 1)$ . Moreover, we assume that q is 1-periodic function, i. e. $q^{″} \in L^{1} (T)$ .

Lemma 13.
Let $q^{″} \in L^{1} (T)$ and let $z \in Z_{+}^{+}$ , $| z | \to \infty$ . Then $\begin{matrix} (5.1) & γ_{2, j} (z) = - \frac{q (0)}{2 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (z^{- 5}), j = 1, 2, \end{matrix}$ where $ϰ_{2, n}$ has the form ( 4.1 ).
Proof.
Let $q^{″} \in L^{1} (0, 1)$ . Then it follows from the second identity in (2.36) that $F_{2}^{'} \in L^{1} (0, 1)$ . Now we can integrate by parts in the integrals in the asymptotics (2.34) and (2.35). Therefore, $B_{2, k j} (x, z) = O (z^{- 1})$ , $k, j = 1, 2, 3, 4$ , $k \neq j$ , as $| z | \to \infty$ . Substituting these asymptotics into (2.37), we get $\begin{matrix} ζ_{2, k j} (x, z) = - q (x) W_{k j} + O (z^{- 5}), k, j = 1, 2, 3, 4, x \in [0, 1], z \in Z_{+}^{+}, \end{matrix}$ where W is defined by (2.21). Substituting these asymptotics into (3.4), we obtain (5.1). □

In order to determine the asymptotics of the characterstic function D we use the representation (2.6). Lemma 14.
Let $q^{″} \in L^{1} (T)$ . Then $\begin{matrix} (5.2) & D (λ) = \frac{z^{2} e^{- i β z} sin α z}{2} (1 - \frac{q (0)}{2 λ} + \frac{ϰ_{2, n}}{λ} + O (z^{- 5})) \end{matrix}$ as $| z | \to \infty$ for $z \in Z_{+}^{+}$ . Here the functions α and β given by ( 3.2 ) and $λ = z^{4}$ .
Proof.
Let $z \in Z_{+}^{+}$ and $| z | \to \infty$ . Substituting (5.1) into (3.3), we obtain $\begin{matrix} (5.3) & \begin{matrix} ξ (z) = & - 4 z^{8} e^{- i β z} (2 i sin α z - e^{- i α z} (- \frac{q (0)}{2 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (z^{- 5})) \\ + e^{i α z} (- \frac{q (0)}{2 z^{4}} + \frac{ϰ_{2, n}}{z^{4}} + O (z^{- 5}))) + O (z^{6}) \\ = & - 8 i z^{8} e^{- i β z} sin α z (1 - \frac{q (0)}{2 λ} + \frac{ϰ_{2, n}}{λ} + O (z^{- 5})), \end{matrix} \end{matrix}$ where α and β have the form (3.2).

The formulas (2.21) and (2.19) imply $\begin{array}{l} det U (0, z) & = exp (tr ln (I_{4} - \frac{q (0)}{z^{4}} W)) = exp (- \frac{q (0)}{z^{4}} tr W + O (z^{- 5})) = 1 + O (z^{- 5}) . \end{array}$ Substituting these asymptotics and (2.9) into (2.41) and using (2.33), we get $\begin{array}{l} det A (0, z) & = det Λ det U (0, z) (1 + \frac{1}{z^{4}} tr B (0, z) + O (z^{- 5})) \\ = - 16 i z^{6} (1 + O (z^{- 5})) (1 + O (z^{- 5})) \\ = - 16 i z^{6} (1 + O (z^{- 5})) . \end{array}$ Substituting the last asymptotics and (5.3) into (2.6), we obtain the asymptotics (5.2). □

5.2. Trace formula

In this Section we prove the trace formula for the shifted operators $H_{t} = H (q_{t})$ given by (1.1), where $q_{t} = q (\cdot + t)$ , $t \in R$ .

Proof of Theorem 2.

Let $q^{″} \in L^{1} (T)$ . Introduce the function $F (t) = \sum_{n = 1}^{N} μ_{n} (t)$ , $t \in T$ , where N is given in Lemma 5(ii), and the resolvents $R_{t} (λ) = {(H_{t} - λ)}^{- 1}$ . Then we have $\begin{matrix} F (t_{1}) - F (t_{2}) = \frac{1}{2 π i} \oint_{Γ_{N}} λ Tr (R_{t_{1}} (λ) - R_{t_{2}} (λ)) d λ, \end{matrix}$ where the contours $Γ_{N}$ are given by $\begin{matrix} Γ_{N} = {λ \in C : | λ | = π^{4} {(N + \frac{1}{2})}^{4}} . \end{matrix}$ Using the identities $R_{t_{1}} (λ) - R_{t_{2}} (λ) = R_{t_{1}} (λ) (H_{t_{2}} - H_{t_{1}}) R_{t_{2}} (λ) = R_{t_{1}} (λ) (q_{t_{2}} - q_{t_{1}}) R_{t_{2}} (λ)$ , we get $| F^{(k)} (t_{1}) - F^{(k)} (t_{2}) | ⩽ C ‖ q_{t_{2}}^{(k)} - q_{t_{1}}^{(k)} ‖_{\infty}$ for some constant $C > 0$ and $k = 0, 1, 2$ , where $f^{(0)} = f$ , $f^{(k)} = d^{k} f / (d t^{k})$ . These estimates yield $F \in C^{2} (T)$ .

The asymptotics (1.5) show that the series (1.6) converges absolutely and uniformly in $t \in T$ .

Let $D (λ, t) = D (λ, q_{t})$ and $D (λ) = D (λ, 0)$ . The asymptotics (2.47) shows that $log (D (λ, t) / D (λ))$ is well defined on the contours $Γ_{N}$ for large $N \in N$ by the condition $log 1 = 0$ . For $N \in N$ large enough we have $\begin{array}{l} \frac{1}{2 π i} \oint_{Γ_{N}} λ d log \frac{D (λ, t)}{D (λ)} & = \frac{1}{2 π i} \oint_{Γ_{N}} \sum_{n = 1}^{\infty} (\frac{λ}{λ - μ_{n} (t)} - \frac{λ}{λ - μ_{n} (0)}) d λ \\ (5.4) & = \sum_{n = 1}^{N} (μ_{n} (t) - μ_{n} (0)) . \end{array}$

Let $N \to \infty$ . The asymptotics (5.2) give $\begin{matrix} \frac{D (λ, t)}{D (λ)} = 1 + \frac{q (0) - q (t)}{2 λ} + O (z^{- 5}) \end{matrix}$ for $| λ | \to \infty$ uniformly in $t \in T$ . These formula yields $\begin{matrix} log \frac{D (λ, t)}{D (λ)} = \frac{q (0) - q (t)}{2 λ} + O (z^{- 5}) . \end{matrix}$ Integrating by parts we have $\begin{array}{l} lim_{N \to \infty} \frac{1}{2 π i} \oint_{Γ_{N}} λ d log \frac{D (λ, t)}{D (λ)} & = - lim_{N \to \infty} \frac{1}{2 π i} \oint_{Γ_{N}} log \frac{D (λ, t)}{D (λ)} d λ \\ = - lim_{N \to \infty} \frac{1}{2 π i} \oint_{Γ_{N}} (\frac{q (0) - q (t)}{2 λ} + O (z^{- 5})) d λ \\ = \frac{q (t) - q (0)}{2} . \end{array}$ Equation (5.4) gives the identity (1.6). □

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