# Spectrum and Generation of Solutions of the Toda Lattice

Authors: D. Barrios Rolanía; J. R. Gascón Márquez
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

Sufficient conditions for constructing a set of solutions of the Toda lattice are
analyzed. First, under certain conditions the invariance of the spectrum ofJ(t) is
established in the complex case. Second, given the tri-diagonal matrix J(t) defining a Toda lattice solution, the dynamic behavior of zeros of polynomials associated to J(t) is analyzed. Finally, it is shown by means of an example how to apply our results to
generate complex solutions of the Toda lattice starting with a given solution.


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## Body

## 1. Introduction

We consider the Toda lattice given by the following equations:(1.1)α̇n(t)=λn+12(t)-λn2(t),λ̇n+1(t)=12λn+1(t)(αn+1(t)-αn(t)),n∈ℕ(λ1≡0),
where λn(t),αn(t) are complex and differentiable functions of one real variable, α̇n(t),λ̇n(t) denote its derivatives, and we assume λn(t)≠0 for each t∈ℝ and n≥2. It is well known (see [1, page 705]) that (1.1) can be expressed in the Lax pair form as(1.2)J̇(t)=[J(t),K(t)],
where [A,B]=AB-BA is the commutator of the operators A and B, and J(t),K(t) are the operators for  which matrix representation is given, respectively, by(1.3)J(t)=(α1(t)λ2(t)λ2(t)α2(t)λ3(t)λ3(t)α3(t)⋱⋱⋱),K(t)=12(0-λ2(t)λ2(t)0-λ3(t)λ3(t)0⋱⋱⋱)
with respect to the canonical basis {ei},i≥0. (In what follows, we identify an operator and its matrix representation respect to the canonical basis.)In the particular case thatλn(t),andαn(t) are real functions, under certain conditions J(t) is a self-adjoint operator. This property has several consequences in the study of system (1.1). For instance, in this situation the unitary equivalence between operators J(t),t∈ℝ, was established in [2]. In that paper, the existence of unitary operators U(t) such that(1.4)J(t)=U(t)-1J(0)U(t)
for each t∈ℝ was proved. As it is well known that  under these conditions the spectrum σ(J(t)) of each operator J(t) verifies(1.5)σ(J(t))=σ(J(0)),t∈ℝ.
In other words, the spectrum does not depend on t∈ℝ. This fact permits to use the self-adjoint operator theory to analyze the integrability of system (1.1) (see [3, 4]). These tools can be used, also, in more general systems (see [5–7]). Then, due to some properties of the real Toda lattice (see, e.g., [1]), the associated Cauchy problem can be solved, recovering the solution J(t) from the initial values defined by J(0).Ifλn(t)andαn(t) are complex functions, then the operator J(t) given in (1.3) is not any longer a self-adjoint operator. Therefore, some of the assumptions of [2] are not verified. To the best of our knowledge there is no proof of the invariance of the spectrum of J(t) in the complex case, so we would like to establish that result in the more general possible situation, that is, when J(t) is not necessarily a bounded operator. However, we think that some advance, in this sense, is a relevant contribution in the study of solutions of the Toda lattice. This is related with our first result.Theorem 1.1.
Let{αn(t),λn+1(t)},n∈ℕ be a solution of (1.1) such that the sequence {λn+1(t)},n∈ℕ, is bounded for each t∈ℝ. Then one has
(1.6)σ(J(t))=σ(J(t0)),∀t,t0∈ℝ;
that is, the spectrum σ(J(t)) of J(t) is invariant on t∈ℝ.System (1.1) is a particular case of the generalized Toda lattice of order p;(1.7)J̇nn(t)=Jn,n+1(t)Jn,n+1p(t)-Jn-1,n(t)Jn-1,np(t),J̇n,n+1(t)=12Jn,n+1(t)[Jn+1,n+1p(t)-Jn,np(t)],n=0,1,…,
where we denote by Jij(t) (resp., Jijp(t)) the entry of J(t) (resp., Jp(t)) corresponding to the row i and the column j (see [8, 9]). The sequence {Pn(t,z)} of polynomials defined by the three-term recurrence relation(1.8)Pn+1(t,z)=(z-αn+1(t))Pn(t,z)-λn+12(t)Pn-1(t,z),n≥0P-1(t,z)≡0,P0(t,z)≡1
is an important tool in the study of complex solutions of (1.7) (see [10, 11]). From {Pn(t,z)} we can define the sequence {p̂n(t,z)} by(1.9)p̂n(t,z)=Pn(t,z)λ2(t)⋯λn+1(t),n∈ℕ.
(Obviously, the zeros of p̂n(t,z) and Pn(t,z) are the same.) Beside some other results, the bases of a method for obtaining new solutions of (1.7) from a given solution were established in [11]. In that paper, the location of zeros of the sequence {p̂n(t,z)} plays an important role, and the relevance of finding a point C∈ℂ which is not a root of any polynomial p̂n(t,z),n∈ℕ,t∈ℝ, was showed. Hence, our interest is in knowing the dynamic behaviour of p̂n(t,z) and, also, some bound for its zeros.Denote byJn(t) the finite-dimensional matrix of order n defined by the first n rows and columns of J(t) (see [12]). From (1.8), it can be easily established that(1.10)Pn(t,z)=det(zIn-Jn(t))
(see, i.e., [13]). Thus, for any t∈ℝ and n∈ℕ, the set of zeros of Pn(t,z) coincides with the spectrum σ(Jn(t)) of Jn(t). When J(t) is a self-adjoint operator, then the spectrum σ(Jn(t)) of each main section Jn(t) is contained in σ(J(t)). So, for this kind of operators, using (1.5) and the relationship between σ(Jn(t)) and σ(J(t)) it is possible to deduce some bound for the set of zeros of Pn(t,z) in terms of σ(J(0)).IfJ(t) is not a self-adjoint, to get some knowledge about the behaviour of solutions of (1.7) as well as (1.1) is very difficult. This is due to the lack of a general result about the relationship between σ(Jn(t)) and σ(J(t)) (see [11, 14]). For general banded matrices, the relation between the spectrum σ(A) of a band infinite matrix A and the spectrum σ(An) of its main sections was analyzed, under certain conditions, in [15]. More precisely, the representation A=ℜA+iℑA was used, assuming ℜA self-adjoint and ℑA bounded. In our case, if we suppose that J(t) verifies this restriction, then we have(1.11)J(t)=ℜJ(t)+iℑJ(t),t∈ℝ,
where ℜJ(t) is a self-adjoint operator and ℑJ(t) is bounded. For C∈ℝ verifying(1.12)d(C,σ(ℜJn(t)))>∥ℑJn(t)∥,
from [15, Lemmas  1, 2] we know Pn(t,C)≠0 or, what is the same, C∉σ(Jn(t)). Moreover, taking into account that ∥ℑJ(t)∥≥∥ℑJn(t)∥, from these results we can deduce Pn(t,C)≠0 for any n∈ℕ when C∈ℂ is such that d(C,σ(ℜJ(t)))>∥ℑJ(t)∥. In this way, the zeros of each sequence of polynomials {Pn(t,z)},n∈ℕ, are located in the neighborhood of σ(ℜJ(t)) given by(1.13){z:d(z,σ(ℜJ(t)))≤∥ℑJn(t)∥}.
Besides the above comments on Theorem 1.1 importance, this fact justifies our interest in obtaining relationship between σ(J(t)) for different values of t∈ℝ, because bounding the zeros in a certain region of the complex plane permits to work with the method given in [11] in the complement of the region zeros free.On the other hand, in conditions under which there are not any information about the dynamic behaviour of the spectrum, our following result gives complementary information about the knowledge and the dynamic behavior of zeros ofPn(t,z).Theorem 1.2.
Letzn1(t),zn2(t),…,znn(t) be the roots of Pn(t,z), nonnecessary distinct. Then one has
(1.14)żnk(t)=(p̂n-1(t,znk(t)))2∑j=0n-1(p̂j(t,znk(t)))2,
understanding that żnk(t)=∞ when the multiplicity of znk(t) as a zero of Pn(t,z) would be m(znk(t))>1.We stress that, in Theorem1.2, we do not need additional conditions about the operator J(t). Theorems 1.1 and 1.2 are complementary results, in the sense that both can be used for determining conditions to obtain some new solutions of (1.1) and (1.7).Section2 is devoted to prove Theorems 1.1 and 1.2. After the existence of C∈ℂ such that Pn(t,C)≠0 for any n∈ℕ,t∈ℝ, can be guaranteed, we will show, in Section 3, how to construct a new solution of (1.1) from a given solution.


## 2. Invariance of Spectrum versus Variation of Zeros of Polynomials

### 2.1. Proof of Theorem1.1

We define the antilinear operator𝒞 such that 𝒞ei=ei for each vector ei,i=0,1,…, in the canonical base. Thus, for any x∈ℓ2 we have(2.1)x=∑i≥0xiei,𝒞x=∑i≥0xi¯ei.
In [16], antilinear operators were introduced in order to study symmetric complex operators. In our case, we have the following auxiliaryy result for 𝒞, which justifies the definition of transpose operator (see [16, page 2]). We recall that we identify an operator with its matrix representation.Lemma 2.1.
(a) LetA be a linear operator and let A* be the adjoint operator of A. Then, the matrix representation of 𝒞A*𝒞 is AT, that is, AT=𝒞A*𝒞.
(b)J(t) is a symmetric complex operator, that is, J(t)=𝒞J(t)*𝒞 for each t∈ℝ.
(c)K(t) is an antisymmetric operator, that is, K(t)=-𝒞K(t)*𝒞 for each t∈ℝ.Proof.
Given a linear operatorB, it is obvious that 𝒞B𝒞 is also a linear operator. So, it is sufficient to prove the enunciated equalities for each basic vector ei. In (a), for A=(aks)k,s=0∞, the column i of A* is given by
(2.2)A*ei=∑k≥0aik¯ek,
and therefore,
(2.3)𝒞A*𝒞ei=∑k≥0aikek
is the i column of the transpose matrix AT. For proving (b) and (c), it is sufficient to take in account the following expressions,
(2.4)J(t)ei=λi+1(t)ei-1+αi+1(t)ei+λi+2(t)ei+1,2K(t)ei=-λi+1(t)ei-1+λi+2(t)ei+1,i=0,1,…,
where we understand e-1=0. In other words, (b) and (c) can be obtained directly as a consequence of the structure of the matrices J(t) and K(t).Now, we consider the following matrix initial value problem:(2.5)Q̇(t)=Q(t)K(t)Q(0)=I.
Under the restrictions of Theorem 1.1, the operator K(t) given by (1.3) is bounded. Hence, we assume that K(t) is a bounded operator in the rest of the section. Moreover, assuming continuous solutions for the Toda lattice, the operator K(t) is a continuous function on t∈ℝ. It is known that we can consider different kinds of continuity for a operator-value function t↦A(t). In our case, the function t↦K(t) of a real variable is continuous in norm (see [17, page 152]). Therefore, the existence of a solution Q(t) of (2.5) can be guaranteed (see [18, page 123]).We have the following auxiliary result.Lemma 2.2.
LetQ(t) be a solution of (2.5). Then
(2.6)Q(t)Q(t)T=Q(t)TQ(t)=I;
that is, Q(t) is an invertible matrix and Q(t)-1=Q(t)TProof.
Transposing the equations (2.5), since Lemma 2.1 we arrive to
(2.7)Q̇(t)T=-K(t)Q(t)TQ(0)T=I.
In other words, Q(t)T is a solution of differential equation R(t)=-K(t)R(t), verifying the same initial condition given by (2.5). To see this one has the following.     (1)
First of all, we showQ(t)Q(t)T=I. Using (2.5) and (2.7) we obtain(2.8)ddt(Q(t)Q(t)T)=Q̇(t)Q(t)T+Q(t)Q̇(t)T=0,
then Q(t)Q(t)T is independent on t∈ℝ. From this fact and Q(0)=I, we deduce Q(0)Q(0)T=I and the first part of (2.6) is proved.     (2)
Following [18, pages 123-124], we can write(2.9)QT(t)=I-∫0tK(τ)dτ+∑n≥2(-1)n∫0t∫0τ1⋯∫0τn-1K(τ1)K(τ2)⋯K(τn)dτn⋯dτ2dτ1.
Due to the continuity in norm of K(t), the series given in the right-hand side of (2.9) converges in norm. Even more,
(2.10)∥QT(t)∥≤etmax[0,t]∥K(τ)∥.
In a similar way, the series given in the right-hand side of
(2.11)R(t)=I+∫0tK(τ)dτ+∑n≥2(-1)n∫t0∫tτ1⋯∫tτn-1K(τ1)K(τ2)⋯K(τn)dτn⋯dτ2dτ1
converges in norm. Then, the above series defines the bounded operator R(t) for each t∈ℝ. From straightforward computations, we obtain
(2.12)QT(t)R(t)=R(t)QT(t)=I,t∈ℝ.
Then, from Q(t)QT(t)=I we get that R(t)=Q(t) and, finally, QT(t)Q(t)=I.Now, we will finish the proof of Theorem1.1. For this purpose, take the solution Q(t) of (2.5). Using (1.2), (2.5), and (2.7), we immediately arrive to(2.13)ddt(Q(t)J(t)QT(t))=Q̇(t)J(t)Q(t)T+Q(t)J̇(t)Q(t)T+Q(t)J(t)Q̇(t)T=0.
Then, taking into account the initial condition in (2.5) and (2.7),(2.14)Q(t)J(t)Q(t)T=Q(0)J(0)Q(0)T=J(0).
From this and Lemma 2.2,(2.15)J(t)=Q(t)TJ(0)Q(t).
Therefore, J(0) and J(t) are equivalent operators, and we have, as a consequence,(2.16)σ(J(t))=σ(J(0))
for each t∈ℝ. So, σ(J(t)) is independent on t∈ℝ, as we wanted to prove.


### 2.2. Proof of Theorem1.2

Takingp=1 in (2.6) of [11, Theorem  2], we obtain(2.17)Ṗn(t,z)=-λn+12(t)Pn-1(t,z)
for each n∈ℕ and all z∈ℂ. Then, writing(2.18)Pn(t,z)=∏i=1n(z-zni(t))
and taking derivatives with respect to t, we have(2.19)Ṗn(t,z)=-∑i=1nżni(t)∏j≠i(z-znj(t)).
With the notation established in Section 1, for each fixed zero z=znk(t) of Pn(t,z) the right-hand side of (2.17) is not zero. As a matter of fact, we have λn+1(t)≠0, and, if we suppose Pn(t,znk(t))=Pn-1(t,znk(t))=0, then using the recurrence relation (1.8) we will arrive to Pn-2(t,znk(t))=0 and, iterating, to P0(t,znk(t))=0, which is not possible being P0≡1.Comparing (2.17) and (2.19) for z=znk(t), we see(2.20)∑i=1nżni(t)∏j≠i(znk(t)-znj(t))=λn+12(t)Pn-1(t,znk(t)),k=1,…,n.
Moreover, ∏j≠i(znk(t)-znj(t))=0 when i≠k. Therefore, from (2.20) we have(2.21)żnk(t)∏j≠k(znk(t)-znj(t))=λn+12(t)Pn-1(t,znk(t)),k=1,…,n,
and, consequently,(2.22)żnk(t)∏j≠k(znk(t)-znj(t))≠0,k=1,…,n.
We will take in consideration the-two possible cases following.(i)
If the multiplicity ofznk(t) as a zero of Pn(t,z) is mnk(t)>1, then the factor znk(t)-znk(t) is in the left-hand side of (2.22), so żnk(t)=∞.(ii)
Ifznk(t) is a simple zero of Pn(t,z), then, from (2.23), we obtain(2.23)żnk(t)=λn+12(t)Pn-1(t,znk(t))∏j≠k(znk(t)-znj(t)).On the other hand, writing(2.24)Pn(t,z(t))=∏i=1n(z-zni(t))
and taking derivatives with respect to z,(2.25)Pn′(t,z)=∑i=1n∏j≠i(z-znj(t)).
So,(2.26)Pn′(t,znk(t))=∏j≠k(znk(t)-znj(t)).
Moreover, the following formula is well known:(2.27)∑j=0n-1(p̂j(t,znk(t)))2=Pn′(t,znk(t))Pn-1(t,znk(t))(λ2(t)⋯λn+1(t))2
(see [13, page 24]).Finally, from (2.23), (2.26), and (2.27) we arrive to (1.14).We point out that (2.27) also holds in the case (i) when znk(t) is not a simple zero and the denominator in (1.14) is zero.Remark 2.3.
(i) It follows, from Theorem1.2, that the zeros of each polynomial Pn(t,z) depend on t∈ℝ because its derivatives are not zero. Moreover, in the case of real Toda lattices, that is, when the coefficients αn(t),λn(t) in (1.8) are real functions, we have żnk(t)>0,k=1,2,…,n. Then, in this case znk(t),k=1,2,…,n, are monotonically increasing functions of t∈ℝ. For each fixed n,znk(t),k=1,2,…,n, are simple zeros of Pn(t,z). Then, znk(t)≠znk′(t) for k≠k′,k,k′=1,…,n, and, therefore, the curves {znk(t):k=1,2,…,n} have no points in common.
(ii)  LetJ(t) be a bounded operator. It is a consequence of Theorem 1.1 that ∥J(t)∥ is independent on t∈ℝ. Then, for each n∈ℕ,
(2.28)|znk(t)|≤∥J(t)∥≤M,k=1,…,n,n∈ℕ.
From this fact and (i), we deduce
(2.29)limt→∞znk(t)=mk∈ℝ,k=1,…,n;
that is, each curve znk(t),t∈ℝ, has an asymptotic line z=mk in the (t,z)—plane.
(iii) When the entries ofJ(t) are not real functions, then we do not know the multiplicity of znk(t) as a zero of Pn(t,z). Therefore, in the complex case it is possible that (i) and (ii) are not longer true.


## 2.1. Proof of Theorem1.1

We define the antilinear operator𝒞 such that 𝒞ei=ei for each vector ei,i=0,1,…, in the canonical base. Thus, for any x∈ℓ2 we have(2.1)x=∑i≥0xiei,𝒞x=∑i≥0xi¯ei.
In [16], antilinear operators were introduced in order to study symmetric complex operators. In our case, we have the following auxiliaryy result for 𝒞, which justifies the definition of transpose operator (see [16, page 2]). We recall that we identify an operator with its matrix representation.Lemma 2.1.
(a) LetA be a linear operator and let A* be the adjoint operator of A. Then, the matrix representation of 𝒞A*𝒞 is AT, that is, AT=𝒞A*𝒞.
(b)J(t) is a symmetric complex operator, that is, J(t)=𝒞J(t)*𝒞 for each t∈ℝ.
(c)K(t) is an antisymmetric operator, that is, K(t)=-𝒞K(t)*𝒞 for each t∈ℝ.Proof.
Given a linear operatorB, it is obvious that 𝒞B𝒞 is also a linear operator. So, it is sufficient to prove the enunciated equalities for each basic vector ei. In (a), for A=(aks)k,s=0∞, the column i of A* is given by
(2.2)A*ei=∑k≥0aik¯ek,
and therefore,
(2.3)𝒞A*𝒞ei=∑k≥0aikek
is the i column of the transpose matrix AT. For proving (b) and (c), it is sufficient to take in account the following expressions,
(2.4)J(t)ei=λi+1(t)ei-1+αi+1(t)ei+λi+2(t)ei+1,2K(t)ei=-λi+1(t)ei-1+λi+2(t)ei+1,i=0,1,…,
where we understand e-1=0. In other words, (b) and (c) can be obtained directly as a consequence of the structure of the matrices J(t) and K(t).Now, we consider the following matrix initial value problem:(2.5)Q̇(t)=Q(t)K(t)Q(0)=I.
Under the restrictions of Theorem 1.1, the operator K(t) given by (1.3) is bounded. Hence, we assume that K(t) is a bounded operator in the rest of the section. Moreover, assuming continuous solutions for the Toda lattice, the operator K(t) is a continuous function on t∈ℝ. It is known that we can consider different kinds of continuity for a operator-value function t↦A(t). In our case, the function t↦K(t) of a real variable is continuous in norm (see [17, page 152]). Therefore, the existence of a solution Q(t) of (2.5) can be guaranteed (see [18, page 123]).We have the following auxiliary result.Lemma 2.2.
LetQ(t) be a solution of (2.5). Then
(2.6)Q(t)Q(t)T=Q(t)TQ(t)=I;
that is, Q(t) is an invertible matrix and Q(t)-1=Q(t)TProof.
Transposing the equations (2.5), since Lemma 2.1 we arrive to
(2.7)Q̇(t)T=-K(t)Q(t)TQ(0)T=I.
In other words, Q(t)T is a solution of differential equation R(t)=-K(t)R(t), verifying the same initial condition given by (2.5). To see this one has the following.     (1)
First of all, we showQ(t)Q(t)T=I. Using (2.5) and (2.7) we obtain(2.8)ddt(Q(t)Q(t)T)=Q̇(t)Q(t)T+Q(t)Q̇(t)T=0,
then Q(t)Q(t)T is independent on t∈ℝ. From this fact and Q(0)=I, we deduce Q(0)Q(0)T=I and the first part of (2.6) is proved.     (2)
Following [18, pages 123-124], we can write(2.9)QT(t)=I-∫0tK(τ)dτ+∑n≥2(-1)n∫0t∫0τ1⋯∫0τn-1K(τ1)K(τ2)⋯K(τn)dτn⋯dτ2dτ1.
Due to the continuity in norm of K(t), the series given in the right-hand side of (2.9) converges in norm. Even more,
(2.10)∥QT(t)∥≤etmax[0,t]∥K(τ)∥.
In a similar way, the series given in the right-hand side of
(2.11)R(t)=I+∫0tK(τ)dτ+∑n≥2(-1)n∫t0∫tτ1⋯∫tτn-1K(τ1)K(τ2)⋯K(τn)dτn⋯dτ2dτ1
converges in norm. Then, the above series defines the bounded operator R(t) for each t∈ℝ. From straightforward computations, we obtain
(2.12)QT(t)R(t)=R(t)QT(t)=I,t∈ℝ.
Then, from Q(t)QT(t)=I we get that R(t)=Q(t) and, finally, QT(t)Q(t)=I.Now, we will finish the proof of Theorem1.1. For this purpose, take the solution Q(t) of (2.5). Using (1.2), (2.5), and (2.7), we immediately arrive to(2.13)ddt(Q(t)J(t)QT(t))=Q̇(t)J(t)Q(t)T+Q(t)J̇(t)Q(t)T+Q(t)J(t)Q̇(t)T=0.
Then, taking into account the initial condition in (2.5) and (2.7),(2.14)Q(t)J(t)Q(t)T=Q(0)J(0)Q(0)T=J(0).
From this and Lemma 2.2,(2.15)J(t)=Q(t)TJ(0)Q(t).
Therefore, J(0) and J(t) are equivalent operators, and we have, as a consequence,(2.16)σ(J(t))=σ(J(0))
for each t∈ℝ. So, σ(J(t)) is independent on t∈ℝ, as we wanted to prove.


## 2.2. Proof of Theorem1.2

Takingp=1 in (2.6) of [11, Theorem  2], we obtain(2.17)Ṗn(t,z)=-λn+12(t)Pn-1(t,z)
for each n∈ℕ and all z∈ℂ. Then, writing(2.18)Pn(t,z)=∏i=1n(z-zni(t))
and taking derivatives with respect to t, we have(2.19)Ṗn(t,z)=-∑i=1nżni(t)∏j≠i(z-znj(t)).
With the notation established in Section 1, for each fixed zero z=znk(t) of Pn(t,z) the right-hand side of (2.17) is not zero. As a matter of fact, we have λn+1(t)≠0, and, if we suppose Pn(t,znk(t))=Pn-1(t,znk(t))=0, then using the recurrence relation (1.8) we will arrive to Pn-2(t,znk(t))=0 and, iterating, to P0(t,znk(t))=0, which is not possible being P0≡1.Comparing (2.17) and (2.19) for z=znk(t), we see(2.20)∑i=1nżni(t)∏j≠i(znk(t)-znj(t))=λn+12(t)Pn-1(t,znk(t)),k=1,…,n.
Moreover, ∏j≠i(znk(t)-znj(t))=0 when i≠k. Therefore, from (2.20) we have(2.21)żnk(t)∏j≠k(znk(t)-znj(t))=λn+12(t)Pn-1(t,znk(t)),k=1,…,n,
and, consequently,(2.22)żnk(t)∏j≠k(znk(t)-znj(t))≠0,k=1,…,n.
We will take in consideration the-two possible cases following.(i)
If the multiplicity ofznk(t) as a zero of Pn(t,z) is mnk(t)>1, then the factor znk(t)-znk(t) is in the left-hand side of (2.22), so żnk(t)=∞.(ii)
Ifznk(t) is a simple zero of Pn(t,z), then, from (2.23), we obtain(2.23)żnk(t)=λn+12(t)Pn-1(t,znk(t))∏j≠k(znk(t)-znj(t)).On the other hand, writing(2.24)Pn(t,z(t))=∏i=1n(z-zni(t))
and taking derivatives with respect to z,(2.25)Pn′(t,z)=∑i=1n∏j≠i(z-znj(t)).
So,(2.26)Pn′(t,znk(t))=∏j≠k(znk(t)-znj(t)).
Moreover, the following formula is well known:(2.27)∑j=0n-1(p̂j(t,znk(t)))2=Pn′(t,znk(t))Pn-1(t,znk(t))(λ2(t)⋯λn+1(t))2
(see [13, page 24]).Finally, from (2.23), (2.26), and (2.27) we arrive to (1.14).We point out that (2.27) also holds in the case (i) when znk(t) is not a simple zero and the denominator in (1.14) is zero.Remark 2.3.
(i) It follows, from Theorem1.2, that the zeros of each polynomial Pn(t,z) depend on t∈ℝ because its derivatives are not zero. Moreover, in the case of real Toda lattices, that is, when the coefficients αn(t),λn(t) in (1.8) are real functions, we have żnk(t)>0,k=1,2,…,n. Then, in this case znk(t),k=1,2,…,n, are monotonically increasing functions of t∈ℝ. For each fixed n,znk(t),k=1,2,…,n, are simple zeros of Pn(t,z). Then, znk(t)≠znk′(t) for k≠k′,k,k′=1,…,n, and, therefore, the curves {znk(t):k=1,2,…,n} have no points in common.
(ii)  LetJ(t) be a bounded operator. It is a consequence of Theorem 1.1 that ∥J(t)∥ is independent on t∈ℝ. Then, for each n∈ℕ,
(2.28)|znk(t)|≤∥J(t)∥≤M,k=1,…,n,n∈ℕ.
From this fact and (i), we deduce
(2.29)limt→∞znk(t)=mk∈ℝ,k=1,…,n;
that is, each curve znk(t),t∈ℝ, has an asymptotic line z=mk in the (t,z)—plane.
(iii) When the entries ofJ(t) are not real functions, then we do not know the multiplicity of znk(t) as a zero of Pn(t,z). Therefore, in the complex case it is possible that (i) and (ii) are not longer true.


## 3. Obtaining Some New Solutions of Toda Lattice

Consider the following solution of (1.1):(3.1)αn(t)=et+n-1,λn(t)=(n-1)et,n∈ℕ.
With the notation employed in the above sections, we have(3.2)J(t)=(etetetet+12et2etet+2⋱⋱⋱).
Since ∑n≥1(1/net) is a divergent series, the Carleman condition ([19, page 59]) indicates that J(t) is a self-adjoint operator. Moreover, it is easy to see that det(Jn(t))=ent,n∈ℕ, and therefore, J(t) is a positive-definite operator. From both issues, we can get that(3.3)σ(Jn(t))⊂[0,+∞).
Then, (3.1) is an example of solution of (1.1) for which the associated polynomials {Pn(t,z)} have all their zeros in [0,+∞). The dynamic behavior of these zeros was determined in Theorem 1.2 and Remark 2.3.From (3.1), it is possible to obtain some complex solutions of (1.1). For this purpose we take C∈ℂ∖[0,+∞) and we apply the method given in [11]. Here, we explain and illustrate that method. Let(3.4)J(1)(t):=(etet1et+12et1et+2⋱⋱⋱).
Due to the fact that Pn(t,C)≠0, we have(3.5)det(Jn(t)-CIn)=det(Jn(1)(t)-CIn)≠0.
Thus, J(1)(t)-CI admits the formal representation given by(3.6)J(1)(t)-CI=L(t)U(t),
([20, Theorem  1, page 35]), where(3.7)L(t):=(l11(t)l21(t)l22(t)l32(t)l33(t)⋱⋱),U(t):=(1u12(t)1u23(t)1u34(t)⋱⋱).
(Despite the fact that the entries in both matrices depend on C, in order to simplify our notation we do not write down explicitly this dependence.) More precisely, for each m∈ℕ we obtain(3.8)lmm(t)=et+m-1-C-(m-1)etet+m-2-C-(m-2)et/⋱-et/(et-C),lm+1,m(t)=1,um,m+1(t)=metlmm(t).
Then, the new obtained solution, generated from J(t) and C, is given as {α̃n(t),λ̃n+1(t)},n∈ℕ, being(3.9)U(t)L(t):=(α̃1(t)-C(λ̃2(t))21α̃2(t)-C(λ̃3(t))21α̃3(t)-C⋱⋱⋱).
In other words, for each C∈ℂ∖[0,+∞) a new solution of the Toda lattice can be generated from the product U(t)L(t). In this way, a sequence of solutions can be obtained iterating this process. In our example, the new complex solution is given by(3.10)α̃1(t)=et+etet-C,α̃2(t)=et+1-etet-C+2etet+1-C-et/(et-C),⋯λ̃2(t)=etet-C(et+1-C-etet-C),λ̃3(t)=2etet+1-C-et/(et-C)(et+2-C2etet+1-C-et/(et-C)),⋯.
Because our initial solution {αn(t),λn+1(t)},n∈ℕ, is a real solution, we know σ(J(t))=σ(J(t0)) for any t,t0∈ℝ. Moreover,(3.11)σ(J̃(t))∖{C}=σ(J(t))∖{C}
(see [21, Proposition  3.6, page 225]). Thus(3.12)σ(J̃(t))∖{C}=σ(J(t0))∖{C}.
Because {λn(t)},n∈ℕ, is not a bounded sequence, we cannot apply Theorem 1.1. However, from (3.12) we conjecture that Theorem 1.1 could be extended to a more general situation.



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*Source: 237487-2009-10-29.xml*