Hill-type formula and Krein-type trace formula for $S$-periodic solutions in ODEs

The present paper is devoted to studying the Hill-type formula and Krein-type trace formula for ODE, which is a continuous work of our previous work for Hamiltonian systems \cite{HOW}. Hill-type formula and Krein-type trace formula are given by Hill at 1877 and Krein in 1950's separately. Recently, we find that there is a closed relationship between them \cite{HOW}. In this paper, we will obtain the Hill-type formula for the $S$-periodic orbits of the first order ODEs. Such a kind of orbits is considered naturally to study the symmetric periodic and quasi-periodic solutions. By some similar idea in \cite{HOW}, based on the Hill-type formula, we will build up the Krein-type trace formula for the first order ODEs, which can be seen as a non-self-adjoint version of the case of Hamiltonian system.


Introduction
In the present paper, we will study the Hill-type formula and Krein-type trace formula for Speriodic solutions for the first order ODE. Hill-type formula was introduced by Hill [2] when he considered the motion of lunar perigee at 1877, and the Krein-type trace formula was built up by Krein [6,7] in 1950's when he studied the stability of Hamiltonian systems. Although they appeared separately, there is a closed relationship between them. In fact, the Krein-type trace formula could be derived by the Hill-type formula for S-periodic orbits in Hamiltonian systems. Moreover, motivated by the Krein's original work, the Krein-type trace formula was used to study the stability problem in n-body problem, details could be found in [5,4]. In the case of Hamiltonian D is a bounded operator acting on L 2 (0, T ; R n ) defined by (Dz)(t) = D(t)z(t). In this paper, we will prove the following Hill-type formula. is not a trace class operator, but a Hilbert-Schmidt operator. Hence, the infinite determinant is not the classical Fredholm determinant. In fact, we can define the determinant in the following way. LetP N be the orthogonal projections onto And the conditional Fredholm determinant 2) If S = I n , then the boundary condition problem (1.2-1.3) is the canonical periodic boundary condition problem. In this case C(S) = T −n . When the period T = 1, the Hill-type formula was obtained by Denk [1]. Based on the Hill-tpye formula, Denk developed some efficient numerical method for the ODEs.
Similar to [4], the Hill-type formula (1.4) is the starting point of the Krein-type trace formula. To get the trace formula, for D 0 (t), D(t) ∈ B(n), we consider the eigenvalue probleṁ z(t) = (D 0 (t) + αD(t))z(t), (1.5) z(0) = Sz(T ), (1.6) that is, to find the α ∈ C such that the system (1.5-1.6) has a nontrivial solution. As above, let γ α be the fundamental solution of (1.5). To state the trace formula, we need some notations. Write and for any positive integer m ≥ 2, (1). For m = 1, F is not a trace class operator but a Hilbert-Schmidt operator. And hence T r(F ) is not the usual trace but a kind of conditional trace [5]. That is For m ≥ 2, F m are trace class operators. Obviously, for ν = 0, λ i is the eigenvalues of (1.5)-(1.6), if and only if 1 λ j is an eigenvalue of F .
where the sum takes for eigenvalues counting algebraic multiplicity.
(2). The trace formula (1.7) for ODE is different from that for Hamiltonian system, and the reason is that the differential operator here is not self-adjoint any more.
Remark 1.5. The idea and the techniques are similar to those in [4], however, there are at least two reasons to write this paper. For the first, it is not like that Hamiltonian system comes from mechanic system mostly, the ODE systems come from many areas, then the Hill-type trace formula and Krein-type trace formula for ODE will be more convenient to be used. For the second, since the differential operators for ODEs are not self-adjoint, we should do some spectral analysis for ODE carefully. Although the formulas for ODE are similar to that for Hamiltonian system in [4], however, it is difficult to deduce them.
The trace formula for Hamiltonian systems is a useful tool in study the stability of Hamiltonian systems, by using the trace formula and Maslov-type index theory [8], some applications to the n-body problem is given in [4,3]. In Section 5, as an application, we will give a generalization of Krein's work on second order systems [6].
This paper is organized as follows, in section 2, we review the basic properties of conditional Fredholm determinant and conditional trace. In section 3, we derive the Hill-type formula for the S-periodic orbits in ODE. In section 4, we get the trace formula from the Hill-type formula. Finally, as an example, we will reformulate Krein's trace formula from our viewpoint.

Preliminaries
In this section, we will mainly recall some fundamental properties of conditional Fredholm determinant, which was developed in [5]. In the classical settings, the Fredholm determinant det(id + F ) is defined for a trace class operator F , details could be found in [10]. However, when we study the ODEs, the operators we encountered are of the form (id + F ) with that F is not a trace class operator, but a Hilbert-Schmidt operator. Thus, the Fredholm determinant det(id + F ) could not be defined, and the conditional Fredholm determinant will be used instead. To define the conditional Fredholm determinant , the trace finite condition plays an important role.
Let {P k } be a sequence of finite rank projections, such that the following conditions are satisfied, (1) for k ≤ m, Range(P k ) ⊆ Range(P m ), (2) P k converges to id in the strong operator topology.
A Hilbert-Schmidt operator F is called to have the trace finite condition with respect to P k , if the limit lim k→∞ T r(P k F P k ) exists, and the limit is finite. Clearly, a trace class operator has the trace finite condition. Obviously, all the Hilbert-Schmidt operators with trace finite conditions consists a linear space, that is, if F 1 and F 2 are Hilbert-Schmidt operators with trace finite condition, then α 1 F 1 + α 2 F 2 has the trace finite condition. By [10], if F is a Hilbert-Schmidt operator, then the regularized Fredholm determinant is defined by (2.1) As been pointed in [5], if F is a Hilbert-Schmidt operator with trace finite condition, then the conditional Fredholm determinant can be defined by where, P k F P k are finite rank operators, and hence det(id + P k F P k ) is well defined. As we proved in [5], many fundamental properties of conditional Fredholm determinant are similar to that of the usual Fredholm determinant.
2) Let E = E 1 ⊕ E 2 , and F i be Hilbert-Schmidt operators on E i with trace finite condition with respect to P (i) then F has the trace finite condition with respect to P Similar to that we have given in [5], it is not hard to show that det(id + αF ) is analytic on α, for a Hilbert-Schmidt operator F with trace finite condition. For reader's convenience, we will give the proof of it.
Lemma 2.2. Let F be a Hilbert-Schmidt operator with trace finite condition with respect to {P k }, then det(id + αF ) is an entire function.
Proof. Write then, by the definition, f k converges to f (α) = det(id + αP k F P k ) point-wisely. Moreover, since P k are finite rank projections, P k F P k are finite rank operators, and hence they are trace class.
It follows that f k (α) are entire functions. By Montel's Theorem, it suffices to show that, {f k } is locally bounded, that is, for any compact set K ⊆ C, there is a constant C > 0 depending only on K such that sup |f k (α)| α ∈ K < C.
Firstly, by [10, Theorem 9.2], where || · || 2 is the Hilbert-Schmidt norm. Secondly, since lim Therefore, sup |f k (α)| α ∈ K < C for some constant C which depends only on K, that is, {f k } is locally bounded. The proof is complete.

Remark 2.3.
In the proof of Lemma 2.2, we show that {det(id+αP k F P k )} is a normal family, and hence there is a subsequence of {det(id + αP k F P k )}, say {det(id + αP k j F P k j )}, which is convergent uniformly to det(id + αF ) on any compact subset of C. Denote by Following [10], we call a function f : X → Y between Banach spaces, finitely analytic if and only if, for all A 1 , ..., A n ∈ X, f (z 1 A 1 + ... + z n A n ) is an entire function of z 1 , ..., z n from C n to Y . This concept is very useful in studying the conditional Fredholm determinant. An important property is the following: Next, we will consider the Taylor expansion of det(id + αF ). Write By [10, Theorem 5.4], we have the following lemma.
Then the Taylor expansion near 0 for g k (α) is The following reasoning is similar to that in [4, Section 2], and we write here again for reader's convenience. Now, let g(α) = m am m! α m be the Taylor expansion of g(α). Since g k j (α) converges to g(α) on any compact subset of C, the coefficients a k j ,m → a m . Notice that F is a Hilbert-Schmidt operator with trace finite condition, then T rF k → T rF . Therefore Note that for α small, by [10, p.47,(5.12)], we have that (2.5) By taking limit, we have the following theorem.
Theorem 2.6. Let g(α) = det(id + αF ). Then for α small, At last of this section, we will brief review the Hill-type formula for Hamiltonian systems. AssumeS be a symplectic orthogonal matrix, forS-periodic orbits in Hamiltonian system, the Hill-type formula was proved in [5], which is listed as follows.
where λ = e νT and C(S) > 0 is a constant depending only onS. Precisely, In fact, if we denoteW = ker(S − I 2n ) ⊥ , then Obviously, C(S) > 0. IfS = I 2n , i.e. for the periodic boundary conditions, C(S) = T −2n , and if ker(S − I 2n ) = 0, then For a n × n orthogonal matrix S, let W = ker(S − I n ) ⊥ , and we also denote Then in the special caseS = S 0 n 0 n S , where S is a n × n orthogonal matrix, (2.8)

Hill-type formula for S-periodic solutions of ODEs
In this section, we will mainly derive the Hill-type formula for the first order ODE.

Hill-type formula for the first order ODEs
To derive the Hill-type formula, we will consider the conditional Fredholm determinant of ODEs. Firstly, we consider the first order ODE with the S-periodic boundary condition, where S is an orthogonal matrix on R n and D ∈ B(n). To continue, we will consider the spectrum of d dt with the S-boundary condition. Denote the domain of d dt by where 0 = θ 1 = · · · = θ k 0 < θ k 0 +1 ≤ · · · ≤ θ n < 2π. It is easy to check that ν ∈ σ(A) if and only if ν = (θ j + 2kπ)/T , for some 1 ≤ j ≤ n; furthermore, (U * SU e √ −1ν j InT − I n )ξ j = 0 if and only if ξ j = (0, 0, · · · , 0 j−1 , 1, 0, · · · , 0). We have the following lemma.
Lemma 3.1. The spectrum of d dt (with the S-boundary condition) is periodic with the period of 2π/T . Precisely, let ν j = θ j /T , then In fact ν j = θ j /T . Moreover, let − √ −1(ν j + 2kπ/T ) ∈ σ( d dt ), then the corresponding eigenvector is e (ν j +2kπ/T ) It is worth being pointed out that where k 0 = dim ker(S − I n ).
Remark 3.2. Since S is an orthogonal matrix, e iθ is an eigenvalue of S if and only if e −iθ is an eigenvalue of S. By the argument before Lemma 3.1 we know that, under the S-boundary condition, the spectrum

4)
which will be used later.
In fact, we have the following lemma. For reader's convenience, we will give the proof of it.
Lemma 3.3. Under the above assumption, we have that The above calculations imply the desired result. Now, we will consider the conditional Fredholm determinant det ( d dt − D)( d dt +P 0 ) −1 with respect toP N . As we have done before, write Hence, the conditional Fredholm determinant det id − (D +P 0 )( d dt +P 0 ) −1 is well-defined. Please note that for any D 1 ∈ B(n) such that d dt + D 1 is invertible, the operator D( d dt + D 1 ) −1 is a Hilbert-Schmidt operator with the trace finite condition with respect toP N . Therefore the infinite In this remaining part of this subsection, we will deduce the Hill-type formula for first order ODE with S-periodic boundary condition, where S is an orthogonal matrix. Consider the following equationu To obtain the Hill-type formula of the above system, we will consider the following Hamiltonian system,ż Changing the basis by V , we have Let γ(t) be the fundamental solution of (3.7). Under the new basis, it is obvious that where γ D (t) satisfiedγ D (t) = D(t)γ D (t) with γ D (0) = I n .
By the Hill-type formula (2.7) for Hamiltonian system, we have By Proposition 2.1(2), we can rewrite the above equation as , it suffices to calculate the de- with T -periodic boundary condition and det −i d dt − iνI n−k 0 −i d dt −1 with S 1 boundary condition respectively. Firstly, notice that the spectrum of i d dt with T -periodic boundary condition Moreover, by direct computation We have where σ can be ±1, to be decided. Please note that, when we take ν = 0, the left hand side of (3.11) equals to 1 and the right hand side equals to σ det (S 1 − I n−k 0 ) C(S) 1 2 . Since C(S) > 0 and σ is a square root of 1, we have σ = sign(det (S 1 − I n−k 0 )).
Proposition 3.5. Under the above condition, we have Notice that where the second equality holds true because of Proposition 2.1(1). Therefore, we have the following theorem which is just Theorem 1.1.
Theorem 3.6. Under the above assumption, (3.14) Please note that (−1) k 0 C(S) = (−1) n |C(S)|, for some monotone function G. By the above analysis, we have where d(ν) is a function depending only on ν and B(t) = V iD 0 n V −1 . Thus we have that is a finitely analytic function from B(n) to C, and satisfied is an analytic function on Ω.
By the multiplicative property of conditional Fredholm determinant, Proposition 2.1(1) , we have the following corollary.

Trace formula for 1st order ODE
In this section, we will derive the trace formula from the Hill-type formula. Firstly, we will study the Taylor expansion for linear parameterized Monodromy matrices, which is similar to the case of Hamiltonian systems [4].
In what follows, to simplify the notation, set Set M = Sγ 0 (T ), then Sγ α (T ) = M M (α). For λ ∈ C, which is not an eigenvalue of M , by some easy computations, we have that and f (α) = det(I n + · · · + α k G k + · · · ), which is an analytic function on C. Next, we will compute the Taylor expansion for f (α). Let α k−1 G k , then for α small enough, by Theorem 2.6, we have Since f (α) vanishes nowhere near 0, we can write f (α) = e g(α) , then by (4.6), some direct computation shows that For α small enough, let g(α) be the function satisfying det(Sγ α (T ) − λI n ) = det(M − λI n ) · exp(g(α)), (4.8) then the coefficients g (k) (0)/k! could be determined by (4.7). And we have the following theorem, which is the main result in this subsection.
where G k = (M − λI n ) −1 M M k , and M = Sγ 0 (T ), We only list the first 4 terms By the definition of G k , Generally, and T r(G j 1 · · · G j k ) could be given similarly.
In the case of Hamiltonian, there are some symmetry property of M k , please refer [4] for the detail. T r(G j 1 · · · G j k ) . (t 2 ) · · · t k−1

0D
(t k )dt k · · · dt 2 dt 1 , k ∈ N. For large m, the right hand side of (4.20) is a little complicated. However, for m = 1, 2, we can write it down more precisely.

Examples
Krein considered second order system in [6] y ′′ + λR(t)y = 0, y(0) + y(T ) = y ′ (0) + y ′ (T ) = 0, (5.1) where R(t) is continuous path of real symmetric matrices on R n . Set R ave = 1 where C is a constant matrix which is chosen such that X ave = 0. Let λ j be the eigenvalues of (5.1), Krein get In this section, we will give a generalization of Krein's trace formula from our viewpoint. To simplify the notation, let A(ν) = −( d dt + ν) 2 and denote by which is a constant matrix. From (4.23), we have T r(RA(ν) −1 ) = −ωT 2 · T r(R ave · S(S − ω) −2 ), (5.5) By taking derivative with respect to ν on both sides of (5.5), we get