A Second Regularized Trace Formula for a Fourth Order Differential Operator

: In applications, many states given for a system can be expressed by orthonormal elements, called “state elements”, taken in a separable Hilbert space (called “state space”). The exact nature of the Hilbert space depends on the system; for example, the state space for position and momentum states is the space of square-integrable functions. The symmetries of a quantum system can be represented by a class of unitary operators that act in the Hilbert space. The operators called ladder operators have the effect of lowering or raising the energy of the state. In this paper, we study the spectral properties of a self-adjoint, fourth-order differential operator with a bounded operator coefﬁcient and establish a second regularized trace formula for this operator.


Introduction
The trace formulae of a differential operator may be seen as a generalization of the traces of matrices or trace-class operators. These formulae are, in general, referred as regularized trace formulae for operators, and they can be used to solve inverse problems [1] and can be applied to index theory [2]. The regularized trace formula of a scalar differential operator was first introduced by I. M. Gelfand and B. M. Levitan [3]. Then, several works on the regularized traces of scalar differential operators appeared (see [4][5][6][7][8][9]). The trace formulae for differential operators with operator coefficients were studied in many works [10][11][12][13][14][15][16][17][18]. Recently a second regularized trace formula was obtained in [19] for the Sturm-Liouville operator with the antiperiodic boundary conditions.
To explain our motivation here, let H be a separable Hilbert space. On Hilbert space H 1 = L 2 (H; [0, π]), we consider two differential operators L 0 and L given by the differential statements: l 0 (u) = u iv (t) and l(u) = u iv (t) + Q(t)u(t) with the identical symmetric boundary conditions u (0) = u (π) = u (0) = u (π) = 0. Our aim was to find a trace formula called the second regularized trace for the operator L by taking advantage from spectral properties of the unperturbed operator L 0 . Here we refer to [20] for the first regularized trace formula of the same operator.
Let σ 1 (H) denote the space of trace-class operators from H to H [21]. Moreover, the norms in H and H 1 are denoted by . H and . and the inner products are denoted by ·, · H and ·, · , respectively. Denote by trA the sum of the eigenvalues of a trace-class operator A [22] and the notation "·" stands for the product.
Each point in the spectrum {m 4 } ∞ m=0 of L 0 gives an eigenvalue of L 0 with its infinite multiplicity. We can easily check that the orthonormal eigenvectors corresponding to these eigenvalues are given by the system ψ mn (t) = d m cos mt · ϕ n (m = 0, 1, 2, . . . ; n = 1, 2, . . . ), where This system constitutes an orthonormal basis of the Hilbert space H 1 , and we will often refer to this fact through the paper.
At the end, we will obtain a formula for the sum of where the sequences {λ mn } ∞ n=1 represent the eigenvalues of L for m = 0, 1, 2, . . . belonging to the interval [m 4 − 1/2, m 4 + 1/2], and C is a constant depending on Q(t). This formula is said to be a second regularized trace formula of the operator L .

Some Relations about Eigenvalues and Resolvents
Let R 0 λ and R λ be resolvents of L 0 and L , respectively. Since the operator function Q(t) satisfies condition (iii) and the system (1) is an orthonormal basis of H 1 , the operator QR 0 λ : H 1 → H 1 is a trace-class operator for every λ / ∈ {m 4 } ∞ m=0 [20]. Moreover, since Q(t) satisfies also conditions (ii) and (iii), then the spectrum of L is a subset of the union of pairwise disjoint intervals I m = m 4 − Q , m 4 + Q (m = 0, 1, 2, . . . ) on the real line. Furthermore, we have: (a) Each point of the spectrum of L which is not the same as m 4 in I m is an isolated eigenvalue of finite multiplicity. (b) m 4 is the possible eigenvalue of L of any multiplicity. (c) lim n→∞ λ mn = m 4 such that {λ mn } ∞ n=1 are the eigenvalues of L in I m . Let ρ(L) be the resolvent set of L . Since QR 0 λ ∈ σ 1 (H 1 ) for every λ ∈ ρ(L), the equation gives R λ − R 0 λ ∈ σ 1 (H 1 ). On the other hand, since the series are absolutely convergent, we have: for every λ ∈ ρ(L). Multiply by λ 2 /2πi both sides of this equality and integrate it over the circle |λ| = b p = (p 2 + p + 1) 2 − 1/2 (p = 1, 2, . . . ). We obtain By using the relation (2) we find, for any positive integer N: If we use this expression in (3), we obtain where The fact that Q(t) satisfies the condition (iii) implies QR 0 λ ∈ σ 1 (H 1 ) for every λ = m 4 (m = 0, 1, . . . ), and the operator function QR 0 λ in the domain C − {m 4 } ∞ m=0 is analytic with respect to the norm in σ 1 (H 1 ). By [8], we have: Proof. According to (7) we get Since the system in (1) is an orthonormal basis of H 1 and QR 0 λ is a trace-class operator for every λ ∈ ρ(L 0 ), we have Replacing this expression in (9), we get: By (1) and the fact that Together with the condition (iii) on Q(t) and last inequality, the series ∞ ∑ n=1 QR 0 λ ψ mn , ψ mn (m = 0, 1, 2, . . . ) and QR 0 λ ψ mn , ψ mn are absolutely and uniformly convergent on the circle |λ| = b p . Therefore, using Cauchy integral formula and the system (1), the equality (10) becomes By applying partial integration four times successively to the second integral in (11), we get (8).

Theorem 2.
With the same hypothesis as in Theorem 1, we have the following equality: Proof. Using (7) we have: Inserting this expression into (13), we get: By separating the series according to m and r into four series and applying the Cauchy integral formula, we obtain: Qψ mn , ψ rq 2 . Let Hence, we get: For m = 0, p and r = p + 1, ∞ we have: Hence, (14) becomes with For any positive integers p and i, let Consider the case i ≤ p (m ≥ 1). First we get: Similarly, for i > p with m ≥ 1 we get: where O(p) depends on p and i and From (20)- (22) we get Clearly we get: These relations give: By (18) and the last inequality, we find: Using (21)- (23) we get ∑ m,r∈E . Therefore, we can show that: By the last two inequalities we find: The inequality Moreover, the first part of (15) becomes: Thus, using (15), (16), (25)-(27) and this last equation, we obtain (12).

The Second Regularized Trace Formula
In this section, we first compute D p3 and we show lim p→∞ D pj = 0 (j ≥ 4) and lim for |λ| = (p 2 + p + 1) 2 − 1 2 . Then we give the second regularized trace formula. Let us first recall that we have: Qψ m s n s , ψ m g(s) n g(s) where the sign * indicates the existence of the numbers greater or less than b p between m 4 1 , m 4 2 , . . . , m 4 j , and Hence, we get: which is absolutely convergent. Taking we rewrite D p3 as: Moreover, the inner product in H 1 says that F(m, r, s) is given by: We have: Using the Cauchy integral formulae, (28) becomes: F(m, r, s).

Let us denote by
and Then we get replaced in A 1 give: Similarly, by (33) we first get and we obtain: Thus, D p3 takes the form For any integers p, i and j such that i > j and p ≥ j, let Let us consider Then we rewrite A as the following: It is clear that where |O(1)| < const and, O(1) depends on p, i and j. Hence, Put: then (39) can be written shortly as By the fact and by (30) and (41), we write (44) as: Here E 2 expresses a set for any integers p, i and j, providing i < j ≤ p, that is, This gives: By (47) and (48), we get: Since γ ij = γ ji and, by (37), (40), (42), (43), (46) and (49), we obtain: Moreover, if Q(t) has a continuous derivative of second order with respect to the norm in σ 1 (H) on the interval [0, π], then |γ ij | ≤ const i 2 j 2 . This implies that Hence, D p3 becomes Now, we will show that By (29)-(31), and using the partial integration we can show that for m = r = s, we have: Moreover, from (30), (35) and (41), we obtain: for m = s and r = s.
Denoting by Example where H is a separable Hilbert space. Consider the operator function Q(t) = (2π) −1 tT (t ∈ [0, π]) where, for every u ∈ H, T : H −→ H is given by with the orthonormal basis {φ i } i≥1 in H. Here ·, · H is the inner product on H. We first show that, for every t ∈ [0, π] the operator function Q(t) is a trace-class (kernel) operator on H. To understand this, it is enough to see that T is a trace-class operator: For every u, v ∈ H we have Since T has eigenvalues λ i = s i = i −2 (i = 1, 2, 3, · · · ), called s-numbers, T is also a trace-class operator. It is also easy to see that Q (t) = (2π) −1 T and Q (i) (t) = 0 for i ≥ 2 with respect to the norm of σ 1 (H). This implies the self-adjointness of Q (i) (t) for i = 0, 1, 2, · · · ; that is, we have [Q (i) (t)] * = Q (i) (t) ; (i = 0, 1, 2, · · · ).
Here, we also notice that Q is a self-adjoint, trace-class operator from H 1 to H 1 .

Conclusions
We introduced and computed a new second regularized trace formula for a fourthorder differential operator with a bounded operator coefficient defined on a separable Hilbert space. This formula can be generalized to an even-order differential operator through the techniques we used here. On the other hand, the regularized trace can be also computed on a separable Banach space, which is a continuous dense embedding in a separable Hilbert space [23]. The trace formulae of these operators are used in many branches of mathematics, mathematical physics and quantum mechanics. For example, the resonant frequencies of the rotating turbine blade can be determined using fourth-order differential operators with the operator coefficients. The quantum mechanics of particles in the wave mechanical formulation cannot be completely represented by a wave-like structure. For example, electron spin degrees of freedom do not imply the action of a gradient operator. Therefore, it is useful to reformulate quantum mechanics in a framework that only includes differential operators.
Funding: This research received no external funding.