The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions
The solution of the Cauchy problem for differential-difference double-infinite Toda lattice by means of inverse spectral problem for semi-infinite block Jacobi matrix is given. Namely, we construct a simple linear system of three differential equations of first order whose solution gives the spectra...
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| Цитувати: | The integration of double-infinite Toda lattice by means of inverse spectral problem and related questions / Yu. Berezansky // Methods of Functional Analysis and Topology. — 2009. — Т. 15, № 2. — С. 101-136. — Бібліогр.: 48 назв. — англ. |
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irk-123456789-57112010-02-03T12:01:15Z The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions Berezansky, Yu. The solution of the Cauchy problem for differential-difference double-infinite Toda lattice by means of inverse spectral problem for semi-infinite block Jacobi matrix is given. Namely, we construct a simple linear system of three differential equations of first order whose solution gives the spectral matrix measure of the aforementioned Jacobi matrix. The solution of the Cauchy problem for the Toda lattice is given by the procedure of orthogonalization w.r.t. this spectral measure, i.e. by the solution of the inverse spectral problem for this Jacobi matrix. 2009 Article The integration of double-infinite Toda lattice by means of inverse spectral problem and related questions / Yu. Berezansky // Methods of Functional Analysis and Topology. — 2009. — Т. 15, № 2. — С. 101-136. — Бібліогр.: 48 назв. — англ. 1029-3531 http://dspace.nbuv.gov.ua/handle/123456789/5711 en Інститут математики НАН України |
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The solution of the Cauchy problem for differential-difference double-infinite Toda lattice by means of inverse spectral problem for semi-infinite block Jacobi matrix is given. Namely, we construct a simple linear system of three differential equations of first order whose solution gives the spectral matrix measure of the aforementioned Jacobi matrix. The solution of the Cauchy problem for the Toda lattice is given by the procedure of orthogonalization w.r.t. this spectral measure, i.e. by the solution of the inverse spectral problem for this Jacobi matrix. |
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Article |
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Berezansky, Yu. |
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Berezansky, Yu. The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions |
| author_facet |
Berezansky, Yu. |
| author_sort |
Berezansky, Yu. |
| title |
The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions |
| title_short |
The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions |
| title_full |
The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions |
| title_fullStr |
The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions |
| title_full_unstemmed |
The Integration of Double-Infinite Toda Lattice by Means of Inverse Spectral Problem and Related Quetions |
| title_sort |
integration of double-infinite toda lattice by means of inverse spectral problem and related quetions |
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Інститут математики НАН України |
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2009 |
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http://dspace.nbuv.gov.ua/handle/123456789/5711 |
| citation_txt |
The integration of double-infinite Toda lattice by means of inverse spectral problem and related questions / Yu. Berezansky // Methods of Functional Analysis and Topology. — 2009. — Т. 15, № 2. — С. 101-136. — Бібліогр.: 48 назв. — англ. |
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2025-07-02T08:47:25Z |
| last_indexed |
2025-07-02T08:47:25Z |
| _version_ |
1836524288824311808 |
| fulltext |
Methods of Functional Analysis and Topology
Vol. 15 (2009), no. 2, pp. 101–136
THE INTEGRATION OF DOUBLE-INFINITE TODA LATTICE BY
MEANS OF INVERSE SPECTRAL PROBLEM AND RELATED
QUESTIONS
YURIJ BEREZANSKY
Dedicated to the fond memory of A. Ya. Povzner (1915–2008)
Abstract. The solution of the Cauchy problem for differential-difference double-
infinite Toda lattice by means of inverse spectral problem for semi-infinite block
Jacobi matrix is given. Namely, we construct a simple linear system of three differ-
ential equations of first order whose solution gives the spectral matrix measure of
the aforementioned Jacobi matrix. The solution of the Cauchy problem for the Toda
lattice is given by the procedure of orthogonalization w.r.t. this spectral measure, i.e.
by the solution of the inverse spectral problem for this Jacobi matrix.
1. Introduction
The classical method of integration of the Cauchy problem for the KdV equation on
(x, t) ∈ [0,∞) × [0,∞) by means of the inverse spectral problem for Sturm-Liouville
equation on x ∈ [0,∞) can be adapted to the semi-infinite Toda lattice
α̇n(t) =
1
2
αn(t)(βn+1(t) − βn(t)),
β̇n(t) = α2
n(t) − α2
n−1(t), n = 0, 1, . . . =: N0, t ∈ [0,∞); α−1(t) = 0
(1)
(results on the inverse problem can be found in the book [33]). Corresponding results
were published in [2, 3, 5]. There instead of the Sturm-Liouville equation we used semi-
infinite Jacobi matrices; the inverse spectral problem for them is classical and simple
(see, e.g., [1]). Important results for the finite Toda lattice were obtained in [28, 37].
But the analogous Cauchy problem with given initial data
(αn(0), βn(0))∞n=−∞
for the double-infinite Toda lattice when in (1) n = . . . ,−1, 0, 1, . . . =: Z (α−1(t) is
arbitrary) is a difficult problem. In this case an application of the spectral theory for
double-infinite Jacobi matrix encounters some serious difficulties. For example, if the
duplication method [1] is used and the consideration is reduced to a difference equation
(1) on the half-line for 2× 2-matrix αn(t), βn(t) then one can not integrate the equation
for the corresponding matrix spectral measure with respect to t [11]. A direct approach
with application of Toda lattices on positive and negative half-axes gives a Riccati type
equation for this measure [47].
There are quite a few papers (in both the scalar and the non-Abelian matrix case) in
which one manages to overcome this difficulties in some special situations and for some
special initial data by not restricting oneself to the spectral approach alone. For example,
in the periodic case the equation was integrated in [32] in terms of theta functions (also
2000 Mathematics Subject Classification. Primary 39A13; Secondary 47A75.
Key words and phrases. Toda lattice, Cauchy problem, Jacobi matrix, direct and inverse spectral
problems, generalized eigenvectors expansion.
101
102 YURIJ BEREZANSKY
see previous work [27]), in [34, 21, 22] by the inverse scattering problem method (for a
difference problem), and for initial data tending to zero as |n| → ∞, in [45, 46, 47] (the
latter method is similar to the one used in [2, 3, 5]). New classes of solutions were found
in [24]. See also [16, 17, 19, 18]. Along the indicated lines of investigation, there is also
a number of papers of more physical nature. For some of them, see the collections of
papers of the conference mentioned in [6]; [29, 30], etc. The original book of M. Toda [42]
also is of a more physical nature. We note that a recently published book [41] contains a
mathematically rigorous presentation of some questions related to the integration of the
Cauchy problem for nonlinear differential-difference equations. New book [48] contains
a number of recent results in this direction.
Let us dwell in some more detail on the integration method in [2, 3, 5] for the Cauchy
problem in the case of the Toda lattice (1) on the half-line n ∈ N0 using the spectral
theory for classical Jacobi matrices. The essence of the method is as follows. To the
solution u(t) = (αn(t), βn(t))∞n=−∞, t ∈ [0,∞), one assigns a selfadjoint Jacobi matrix
J(t) with βn(t) on the main diagonal and αn(t) on the two neighboring diagonals. We
assume that the solution u(t) is bounded on every [0, T ) ∋ t, therefore ∀t ∈ [0,∞) J(t) is
a bounded selfadjoint operator on the space ℓ2. Let dρ(λ; t) be its spectral measure. It is
possible to prove that this measure changes in time t in a simple way: roughly speaking
(2) dρ(λ; t) = eλtdρ(λ; 0), λ ∈ R, t ∈ [0,∞).
Therefore, the spectral measure dρ(λ; 0) of the matrix J(0) corresponding to the initial
value u(0) of the solution u(t), makes it possible to use (2) to find the measure dρ(λ; t)
for t > 0. Then we can use the classical inverse spectral problem for Jacobi matrices in
terms of dρ(λ; t) to recover J(t), i.e. to find the solution u(t) of our Cauchy problem
(note that the case of an unbounded solution u(t) was considered in [40]).
As it was mentioned earlier, in the article [11] M. Gekhtman and I tried to apply
such approach to the case of the Toda lattice (1) on n ∈ Z. There we have double-
infinite Jacobi matrices J(t) for which the spectral measure dρ(λ; t) is a 2 × 2-matrix
measure. It is possible to understand this measure as a spectral matrix of semi-infinite
Jacobi matrix but with 2 × 2-matrix elements (duplication method [1]). This leads to
some matrix differential equation in t for dρ(λ; t) which is impossible to solve due to
noncommutativity of its coefficients (in the case of semi-axis n ∈ N0 such an equation is
very simple and its solution is (2)).
In this article we propose some way of overcoming the last difficulty. Namely, in the
standard duplication method we construct by sequence (. . . , ξ−1, ξ0, ξ1, . . .) of complex
numbers ξn the semi-infinite sequence (x0, x1, . . .) of vectors xn = (ξn, ξ−n−1) ∈ C2. As
a result, our double-infinite Jacobi matrix J(t) transforms into a semi-infinite Jacobi
matrix whose elements are 2 × 2-matrices. Such matrix acts on the space
C
2 ⊕ C
2 ⊕ C
2 ⊕ . . . ;
corresponding spectral theory is known [31, 1]. But its application to our situation
gives the noncommutativity of coefficients of the differential equation for dρ(λ; t) and
impossibility to solve it.
In this paper we go in a different direction: we use the duplication of the form:
(3) (. . . , ξ−1, ξ0, ξ1, . . .) �→ x0 = ξ0 ∈ C
1, x1 = (ξ1, ξ−1) ∈ C
2, x2 = (ξ2, ξ−2) ∈ C
2, . . . .
Corresponding Jacobi matrix J(t) acts on the space
C
1 ⊕ C
2 ⊕ C
2 ⊕ . . .
and its spectral theory is not standard. The main difficulty is that the first matrix
elements on side diagonals are not invertible since these elements act from C1 into C2 or
vice versa.
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 103
Such a situation presents, at first, essential difficulties for the finding, with the help of
recurrence relations, the corresponding orthogonal polynomials. This can be overcome
by means of a certain point of view on the construction of spectral matrix which was
used in [7, 26]. Note also, that the spectral theory of unitary and normal operator Jacobi
matrices with elements acting between different Hilbert spaces was developed in [8, 9].
In [7] another point of view was suggested: there, we considered the matrix operator
with some ”boundary conditions” at the point n = 0.
It is necessary to say that transfer (3) really means that we choose the spectral matrix
dρ(λ; t) of our difference operator on Z in a nonstandard manner. But this choice allows
one to write a simple differential equation for dρ(λ; t) w.r.t. t ∈ [0,∞) (recall, that a
choice of a spectral matrix in the case of a difference equation on Z is not unique: it
depends on two chosen fundamental solutions of our difference equations).
An approach sketched above gives a possibility to find a system of three first order
linear differential equations w.r.t. t ∈ [0,∞) for the functions ρα,β(λ; t), α, β = 0, 1
(ρ0,1(λ; t) = ρ1,0(λ; t)).
The coefficients of these equations depend in a simple manner on
(4) β0(t), t ∈ [0,∞), α0(0).
In general, it is impossible to give exact formulas for the solution of this simple lin-
ear system (if the functions (4) are constant, it is easy to do). But we get a certain
”linearization” of our Cauchy problem for (1), n ∈ Z, because the construction of its re-
quired solution u(t) = (αn(t), βn(t))∞n=−∞, i.e. the matrix J(t), is an relatively standard
procedure: it is achieved via a certain orthogonalization procedure.
Of course, we can get exact formulas for the solution of the Cauchy problem for (1),
n ∈ Z, only in some special cases, but in the general case the method of obtaining the
solution is sufficiently simple and ”computable”.
It is worth to stress that our Cauchy problem has an essential peculiarity: to find
its solution it is necessary to give the standard initial data (αn(0), βn(0))∞n=−∞ and, in
addition, the function (4). This situation reminds the so-called shock problem for Toda
lattice ([38, 43, 44], see also [4, 5]); it will be investigated in another paper. Note also,
that in the case of semi-infinite Toda lattice the data (4) is also implicitly present: α0(t)
as a coefficient at α̇n(t), β̇n(t) and β0(t) via the procedure of normalization of the spectral
measure (see [5], pp. 27–28, 37) is connected with the constant of normalization of the
spectral measure to a probability measure.
The above mentioned results concerning the double-infinite Toda lattices are contained
in Sections 6–9 of the paper. Sections 2, 3 contain preliminary constructions devoted to
the spectral theory of difference expressions on N0 with operator coefficients [1, 11] and
Toda lattices [42, 11]. These constructions help to understand the results in the next
Sections.
Sections 4, 5 contain a result about integration of matrix Toda lattice on the half-axis
n ∈ N0, the coefficients of which are commutative matrices. These results are connected
with Section 6–9 and are useful. Note that the articles [11, 23, 20] are linked with
Sections 4 and 5.
Above we have used only the spectral theory of selfadjoint Jacobi matrices, with scalar
or matrix elements. But in [12] an analogous theory was developed for a solution of the
Cauchy problem for some types of matrix nonlinear equations connected with a normal
or unitary Jacobi matrix with operator elements; article [25] gave an impetus to such an
investigation.
Note also that the described above way of integration of some nonlinear differential-
difference equations (with scalar or matrix coefficients) can be transferred to so-called
non-isospectral equations for which the spectrum of the corresponding matrix J(t), i.e.,
104 YURIJ BEREZANSKY
the support ∀t of dρ(λ; t), changes in t (from (2) we see that the spectrum of J(t) for (1) is
fixed, an analogous picture takes place in other cases mentioned above: our problems are
isospectral). For the case of nonisospectral equations see articles [10, 39, 14, 15, 6, 35, 36].
2. Preliminaries: some facts about the spectral theory of difference
expressions with operator coefficients
Let H be a Hilbert space with a scalar product (x, y) and a norm ‖x‖ ; x, y ∈ H.
Denote by L(H) the set of all bounded operators in H. Define a difference expression J
whose coefficients belong to L(H) and are collectively bounded in the norm and which
act on sequences f = (fn)∞n=0 of vectors fn ∈ H as follows :
(Jf)n =an−1fn−1 + bnfn + anfn+1, an, bn ∈ L(H), an > 0, b∗n = bn;
n ∈ N0 = {0, 1, . . .}; a−1 ∈ L(H) is fixed.
(5)
When calculating (Jf)0 we always assume that f−1 = 0; this assumption serves as a
boundary condition, therefore the value of a−1 is not essential in Section 2.
The expression (5) induces a bounded selfadjoint operator J on the space l2 :=
ℓ2(H) = H ⊕ H ⊕ . . . . Namely, for finite sequences f ∈ lfin ⊂ l2 we put f �→ Jf ∈ l2;
we denote by J the closure of this map (or sometimes shortly by J). This operator is
generated by the Jacobi Hermitian matrix with operator elements:
(6) J =
b0 c0 0 0 0 . . .
a0 b1 c1 0 0 . . .
0 a1 b2 c2 0 . . .
...
...
...
...
...
. . .
, an, bn, cn ∈ L(H), cn = an > 0, bn = b∗n, n ∈ N0.
The uniform boundedness of elements an, bn ensures that J is a bounded selfadjoint
operator on l2.
Let λ ∈ R be fixed, consider the sequence of operators (Pn(λ))
∞
n=0 =: P (λ) where
Pn(λ) are solutions of the operator equation
(JP (λ))n :=an−1Pn−1(λ) + bnPn(λ) + anPn+1(λ) = λPn(λ), n ∈ N0,
P−1(λ) = 0, P0(λ) = 1
(7)
(1 is the identity operator on H). The equation (7) is solvable: one can find P1(λ),
P2(λ), . . . recursively; it is possible because ∀n ∈ N0 a−1
n exists. In particular
(8) P1(λ) = a−1
0 (λ1 − b0), λ ∈ R.
It is easy to see that every Pn(λ) is an operator-valued polynomial w.r.t. λ ∈ R (i.e.
a polynomial with coefficients from L(H)) of degree n ∈ N0; its top coefficient is an
invertible operator on H.
It is possible to prove that every generalized eigenvector of the operator J has the form
ϕ(λ) = (ϕn(λ))
∞
n=0 , ϕn(λ) = Pn(λ)ϕ0 where ϕ0 ∈ H (initial value of ϕ(λ)) is arbitrary;
here λ belongs to the spectrum of J . In our case, the spectral measure dρ(λ) of the
operator J is a nonnegative operator-valued Borel measure B(R) ∋△ �→ ρ(△) ∈ L(H),
ρ(△) ≥ 0, which is probabilistic, ρ(R) = 1.
For an account of necessary facts of spectral theory of the operator generated by (5),
(6), it is convenient to introduce the notion of a corresponding pseudo-Hilbert space. We
will do this not in a general situation but only in two cases which are necessary for us.
1). Consider the set l2(L(H)) of all sequences U = (Un)∞n=0 of operators Un ∈ L(H)
for which the series
∞∑
n=0
U∗
nUn is weakly convergent on H. It is easy to check that such a
set is a module w.r.t. the multiplication on the right by operators Λ ∈ L(H) (”pseudo-
scalars”). In l2(L(H)) it is possible to introduce the corresponding pseudo-scalar product
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 105
{U, V } =
∞∑
n=0
U∗
nVn ∈ L(H) (this series is also weakly convergent). As a result we have
the following definition and properties of a pseudo-Hilbert space l2(L(H)) :
l2(L(H)) =
{
U = (Un)∞n=0, Un ∈ L(H)
∣∣
∞∑
n=0
U∗
nUn < ∞, i.e. weakly converges
}
,
{U, V } =
∞∑
n=0
U∗
nVn ∈ L(H);
∀U, V ∈ l2(L(H)), ∀Λ ∈ L(H) U + V := (Un + Vn)∞n=0, UΛ := (UnΛ)∞n=0.
Then: (U + V )Λ = UΛ + V Λ, {U, V }∗ = {V, U}, {U, V Λ} = {U, V }Λ,
{UΛ, V } = Λ∗{U, V }, {U + V, W} = {U, W}+ {V, W}.
(9)
Denote by δn the sequence (0, . . . , 0︸ ︷︷ ︸
n
, 1, 0, 0, . . .) =: δn ∈ l2(L(H)) and consider a
matrix A = (ajk)∞j,k=0 of type (6) with arbitrary ajk ∈ L(H) (the matrix can be full).
The pseudo-vectors δn play a role of an orthonormal basis in l2(L(H)). Therefore if we
introduce a ”pseudo-linear” operator A on l2(L(H)) in a natural way: ∀j ∈ N0
(10) (AU)j =
∞∑
k=0
ajkUk, U = (Un)∞n=0 ∈ lfin(L(H));
then it is easy to calculate the following formula for ajk :
(11) ajk = {δj,Aδk}, j, k ∈ N0.
Conversely, every pseudo-linear operator A : l2(L(H)) → l2(L(H)) which is continu-
ous in some natural topology in l2(L(H)) has a matrix representation (10), (11).
Let U = (Un)∞n=0 ∈ l2(L(H)) and x ∈ H, then Ux := (Unx)∞n=0 is a vector from the
space l2 and ∀x, y ∈ H
(12) ({U, V }x, y) = (V x, Uy)l2 , U, V ∈ l2(L(H)).
Every linear continuous operator A : l2 → l2 gives rise to some pseudo-linear operator
A : l2(L(H)) → l2(L(H)). Namely, ∀U ∈ l2(L(H)), ∀x ∈ H we put: (AU)x = A(Ux) (it
is clear that A cannot be always reconstructed by A). The matrix representation (10),
(11) for A gives the corresponding matrix representation (ajk)∞j,k=0, ajk ∈ L(H), for A :
∀j ∈ N0
(Af)j =
∞∑
k=0
ajkfk, f = (fn)∞n=0 ∈ lfin;
ajk = {δj,Aδk}, (Aδk)x = A(δkx), δkx = (0, . . . , 0︸ ︷︷ ︸
k
, x, 0, 0, . . .), x ∈ H.
(13)
The multiplication of operators on l2 and their matrices are connected in a usual way.
2). The second example of a pseudo-Hilbert space. Consider some nonnegative
operator-valued Borel probability measure dρ(λ) on R introduced above. Let R ∋ λ �→
f(λ) ∈ H be a continuous bounded vector-function, g(λ) is analogical one. In an ordi-
nary way (starting from a linear combination of vector-valued characteristic functions)
it is possible to introduce the integral
(14)
∫
R
(dρ(λ)f(λ), g(λ)) =: (f, g)L2 .
106 YURIJ BEREZANSKY
This integral defines a (quasi) scalar product f and g. Let L2 = L2(H, R, dρ(λ)) be a
completion of the set of such functions f. This Hilbert space is similar in some sense to
the space l2.
Now we can introduce a corresponding pseudo-Hilbert space,
L2(L(H)) = L2(L(H), R, dρ(λ)).
For an operator-valued continuous bounded function R ∋ λ �→ U(λ) ∈ L(H) and a
similar function V (λ) define a pseudo-scalar product by the formula
(15) {U, V } =
∫
R
U∗(λ)dρ(λ)V (λ)
(here the integral is introduced similarly to (14), starting from operator-valued char-
acteristic functions). Such set of functions U(λ) and pseudo-scalars Λ ∈ L(H) form a
module w.r.t. ordinary sum of functions and multiplication on Λ on the right. Expres-
sion (15) determines a pseudo-scalar product on this module. As a result, we get some
pseudo-Hilbert space L2(L(H)) with properties analogous to (9), (12).
Let’s return to the spectral theory of the operator J generated by Jacobi matrix (6)
on the space l2. As was said above, every generalized eigenvector ϕ(λ) of this operator
has the form
(16) ϕ(λ) = (ϕn(λ))∞n=0, ϕn(λ) = Pn(λ)ϕ0, ϕ0 ∈ H, λ ∈ spectrum J ,
where Pn(λ) are operator-valued polynomials which are solutions of equation (7).
The corresponding Fourier transform ̂ is
(17) l2 ∋ f = (fn)∞n=0 �→ f̂(λ) =
∞∑
n=0
P ∗
n(λ)fn ∈ L2
(L2 = L2(H, R, dρ(λ)) is constructed from the spectral measure dρ(λ) of J). The unitary
mapping (17) between the spaces l2 and L2 at first is defined on f ∈ lfin ⊂ l2 and then
is extended by continuity to the whole l2.
The Fourier transform̂ (17) can be rewritten in another convenient form:
lfin(L(H)) ∋ U = (Un)∞n=0 �→ Û(λ) =
∞∑
n=0
P ∗
n(λ)U(λ) ∈ L2(L(H));
∀x ∈ H (Ux)̂(λ) = (Û(λ))x.
(18)
As it was mentioned, the mapping (17), (18) is unitary between l2 and L2. The corre-
sponding Parseval equality has the form (see (9), (15))
∀f, g ∈ l2 (f, g)l2 = (f̂ , ĝ)L2 =
∫
R
(f̂(λ)dρ(λ), ĝ(λ)) or
∀U, V ∈ l2(L(H)) {U, V } = {Û , V̂ } =
∫
R
(Û(λ))∗dρ(λ)V̂ (λ).
(19)
From this Parseval equality and (17), (18) it follows that the polynomials Pn(λ) are
orthonormal in the following sense:
∀x, y ∈ H
∫
R
(P ∗
j (λ)xdρ(λ), P ∗
k (λ)y) = δj,k(x, y), j, k ∈ N0,
or
∫
R
Pj(λ)dρ(λ)P ∗
k (λ) = δj,k1, j, k ∈ N0.
(20)
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 107
Using the second equality in (20) and (7) we conclude that the following representa-
tions take place:
(21) an =
∫
R
λPn(λ)dρ(λ)P ∗
n+1(λ), bn =
∫
R
λPn(λ)dρ(λ)P ∗
n (λ), n ∈ N0.
The second equality in (20) is more convenient. It gives a possibility to introduce a
generalization of the classical Schmidt procedure of orthogonalization of a sequence of
vectors in a Hilbert space to the case of pseudo-Hilbert space. We will introduce such a
procedure in the essential for us case of operator-valued polynomials P (λ) on λ ∈ R.
So, we assume that our measure dρ(λ) with higher invertible coefficient has the prop-
erty that the following integral exists and is a positive invertible operator in H :
(22) {P, P} =
∫
R
P ∗(λ)dρ(λ)P (λ).
Consider the following sequence of polynomials from L2(L(H)) = L2(L(H), R, dρ(λ)) :
(23) 1, λ1, λ21, . . .
Put P0(λ) = 1 and consider the polynomial of the form P
′
1(λ) = λ1 − P0(λ)K1
where K1 ∈ L(H). Pick out K1 in such a manner that {P0(λ), P
′
1(λ)} = 0, for this
it is necessary to put (see (9)) K1 = {P0(λ), λ1}. Using our assumption we can assert
that {P
′
1(λ), P
′
1(λ)} is a positive invertible operator. Therefore the operator P1(λ) =
P
′
1(λ){P
′
1(λ), P
′
1(λ)}−1/2 is such that
(24) {P0(λ), P1(λ)} = 0, {P1(λ), P1(λ)} = 1.
Assume that we construct the polynomials P0(λ), . . . , Pn(λ) (∀j = 0, . . . , n Pj(λ) as a
linear combination with operator coefficients of 1, . . . , λj1) with the properties (24):
(25) {Pj(λ), Pk(λ)} = δj,k1, j, k = 0, . . . , n.
Construct Pn+1(λ). For this we put
P
′
n+1(λ) = λn+11 − P0(λ)K1 − · · · − Pn(λ)Kn
with some coefficients K1, . . . , Kn ∈ L(H) of the form Kj = {Pj(λ), λn+11}. Using (9)
we get ∀j = 0, . . . , n
{Pj(λ), P
′
n+1(λ)} = {Pj(λ), λn+11} − {Pj(λ), P0(λ)K1} − · · · − {Pj(λ), Pn(λ)Kn}
= {Pj(λ), λn+11} − {Pj(λ), Pj(λ)}Kj = 0.
Now it is necessary only to normalize P
′
n+1(λ) : we put Pn+1(λ) = P
′
n+1(λ){P
′
n+1(λ),
P
′
n+1(λ)}−1/2.
So, we construct the system P0(λ), . . . , Pn+1(λ) with the property (25) for j, k =
0, . . . , n + 1. The procedure of orthogonalization of sequence (24) is complete.
Such a construction allows us to solve the inverse spectral problem for Jacobi matrix
(6): assume we know some nonnegative operator-valued Borel probability measure dρ(λ)
for which all operator-valued polynomials are integrable and condition of invertibility of
operator (22) is fulfilled. Then such a measure is a spectral measure of the operator J
generated by the matrix (6) and elements of this matrix are reconstructed by formulas
(21) where Pn(λ) are obtained via the procedure of orthogonalization of sequence (23). If
we start from a spectral measure dρ(λ) of the operator J generated by (6) we get as a
result the initial matrix (6).
It is possible to give some conditions on the measure dρ(λ) which guarantees in-
vertibility of the operator (22). We consider here only the simplest case when H is
108 YURIJ BEREZANSKY
two-dimensional: H = C2 (such a case will be useful in what follows). An analogous
situation takes place for an arbitrary finite-dimensional H, the proof is similar.
Lemma 1. Let H = C2 and a nonnegative operator-valued Borel probability measure
dρ(λ) be such that all the integrals
(26)
∫
R
λn1dρ(λ)λn1, n ∈ N0,
exist. If the support dρ(λ) has an infinite set of points, then every operator (22) is
invertible.
Proof. Since H is finite-dimensional the invertibility of the operator {P, P} (23) is equiv-
alent to the property: if for x ∈ C2 {P, P}x = 0, then x = 0, i.e.,
∥∥{P, P}1/2x
∥∥2
= 0 ⇒
x = 0. Here P (λ) is an operator-valued polynomial of degree n ∈ N0.
For every Borel set △⊂ R, the matrix ρ(△) = (ρα,β)1α,β=0 is nonnegative. Therefore
|ρ0,1(△)|2 = |ρ1,0(△)|2 ≤ ρ0,0(△)ρ1,1(△) and the complex-valued measures ρα,β(λ), α, β =
0, 1, are absolutely continuous w.r.t. the nonnegative measure △ �→ σ(λ) := ρ0,0(△) +
ρ1,1(λ). Using the Radon-Nikodym theorem we can write that dρ(λ) = C(λ)dσ(λ), λ ∈ R,
where C(λ) = (Cα,β(λ))1α,β=0 is a nonnegative matrix with summable functions Cα,β(λ).
This matrix is positive on the set of full measure dσ(λ).
Such a representation and formula (22) gives that ∀x ∈ H
∥∥∥{P, P}1/2x
∥∥∥
2
= ({P, P}x, x) =
(( ∫
R
P ∗(λ)dρ(λ)P (λ)
)
x, x
)
=
(( ∫
R
P ∗(λ)C(λ)P (λ)dσ(λ)
)
x, x
)
=
∫
R
∥∥∥C1/2(λ)P (λ)x
∥∥∥ dσ(λ).
(27)
If the left-hand side in (27) is equal to 0, then
∥∥C1/2(λ)P (λ)x
∥∥ = 0 for σ-almost all λ ∈ R.
The possibility of the matrix C(λ) gives the same property also for ‖P (λ)x‖2
. But the
last expression is an ordinary polynomial of degree 2n with highest coefficient ‖Anx‖
2
,
where An is the invertible highest coefficient of P (λ). The equality ‖P (λ)x‖
2
= 0 means
that this ordinary polynomial is equal to zero. Therefore x = 0. �
3. Preliminaries: the double-infinite Toda lattice and its known
reduction to semi-infinite 2 × 2-matrix Toda lattice
The double-infinite Toda lattice has the form
(28) α̇n =
1
2
αn(βn+1 − βn), β̇n = α2
n − α2
n−1, n ∈ Z = {. . . ,−1, 0, 1, . . .},
where αn = αn(t), βn = βn(t) are real continuously differentiable functions of t ∈
[0, T ), T ≤ ∞; · = d
dt . Expression (28) is a differential-difference nonlinear equation and
for (28) it is possible to consider the following Cauchy problem: given the initial data
αn(0), βn(0),
n ∈ Z, find a solution αn(t), βn(t), n ∈ Z, for t ∈ (0, T ).
The equation (28) can be rewritten in the form of a semi-infinite equation, but for
the 2 × 2-matrices an(t), bn(t), n ∈ N0, introduced by αn(t), βn(t), n ∈ Z; t ∈ [0, T ) (see,
e.g. [11, 1]). We mention here a simpler procedure. Introduce the diagonal matrices:
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 109
∀t ∈ [0, T )
an(t) =
[
αn(t) 0
0 α−n−1(t)
]
, bn(t) =
[
βn(t) 0
0 −β−n(t)
]
, n ∈ N0;
a−1(t) =
[
α−1(t) 0
0 α0(t)
]
.
(29)
Construct for these 2 × 2-matrices an(t) = an, bn(t) = bn, n ∈ N0, a−1(t) = a−1; t ∈
[0, T ), the differential-difference equation similar to (28):
(30) ȧn =
1
2
an(bn+1 − bn), ḃn = a
2
n − a
2
n−1, n ∈ N0; t ∈ [0, T ).
Lemma 2. The equation (28) can be written in an equivalent form (30) where the
matrices an, bn are of the form (29).
Proof. The equalities (30) with matrices (29) have the form:
[
α̇n 0
0 α̇−n−1
]
=
1
2
[
αn 0
0 α−n−1
] ([
βn+1 0
0 −β−n−1
]
−
[
βn 0
0 −β−n
])
=
[
1
2αn(βn+1 − βn) 0
0 1
2α−n−1(β−n − β−n−1)
]
, n ∈ N0;
[
β̇n 0
0 −β̇−n
]
=
[
αn 0
0 α−n−1
]2
−
[
αn−1 0
0 α−n
]2
=
[
α2
n − α2
n−1 0
0 α2
−n−1 − α2
−n
]
, n ∈ N.
(31)
Comparing the elements in equalities (31) and corresponding for ḃ0 = a0 − a−1 we get:
α̇n =
1
2
αn(βn+1 − βn), β̇n = α2
n − α2
n−1, n ∈ N0;
α̇−n−1 =
1
2
α−n−1(β−n − β−n−1), −β̇−n = α2
−n−1 − α2
−n, n ∈ N0.
(32)
The first line in (32) gives the equality (28) for n = 0, 1, . . . Its second line gives the
equality (28) for n = −1,−2, . . . and equality −β̇0 = α2
−1 −α0, which we had in the first
line. �
The equation (30) with matrices (29) is not applicable to integration of lattice (28)
by means of inverse spectral problem for (6) since now f−1 = 0 and the value of a−1 is
not essential. But a certain modification of (30), (29) is applicable: see Sections 6–9.
4. Semi-infinite operator Toda lattice and the corresponding Lax
equation
Return now to the general situation of matrices of type (6) with elements from L(H).
Denote by J(t), t ∈ [0, T ), the matrix of the form (6) with elements a−1(t), an(t), bn(t) =
b∗n(t), cn(t) = an(t), n ∈ N0, from L(H) for every t. We assume that they depend on t in a
continuously differentiable way (in the operator norm sense) and their norms are bounded
uniformly on n ∈ N0 and t ∈ [0, T ). Let J(t), t ∈ [0, T ), be the operator constructed in
Section 2 on the space l2.
Consider another matrix A(t), t ∈ [0, T ), of type (6); its L(H)-valued entries will be
denoted by ãn = ãn(t), b̃n = b̃n(t), c̃n = c̃n(t). They are ∀t ∈ [0, T ) arbitrary operators in
L(H) with above said properties of differentiability and boundedness. Our Lax equation
has the form:
(33) J̇(t) = [J(t), A(t)] := J(t)A(t) − A(t)J(t), t ∈ [0, T ).
110 YURIJ BEREZANSKY
After simple calculations (see [5, 11, 12]) it is possible to conclude that the equality
(33) is equivalent to the following system of equalities: ∀n ∈ N0 and t ∈ [0, T )
0 = anc̃n+1 − c̃nan+1,
ȧn = bnc̃n + anb̃n+1 − b̃nan − c̃nbn+1,
ḃn = an−1c̃n−1 + bnb̃n + anãn − ãn−1an−1 − b̃nbn − c̃nan,
ȧn = anb̃n + bn+1ãn − ãnbn − b̃n+1an,
0 = an+1ãn − ãn+1an.
(34)
Above it is necessary to take into account that a−1 = ã−1 = c̃−1 = 0. Because our
operator J(t) does not depend on a−1 (by the construction), we can also assume that
a−1 = 0.
From the first and the last equalities in (34) we conclude:
(35) c̃n = a−1
n−1 . . . a−1
0 c̃0a1 . . . an, ãn = an . . . a1ã0a
−1
0 . . . a−1
n−1, n ∈ N.
From the second and the fourth equations in (34) we conclude:
(36) anb̃n+1 + b̃n+1an = anb̃n + b̃nan + c̃nbn+1 + bn+1ãn − bnc̃n − ãnbn, n ∈ N0.
Consecutively substituting n = 0, 1, . . . into (36) and using (35), we find b̃1, b̃2, . . . Here
we use the fact that the equation for ξ ∈ L(H) aξ + ξa = b, where a, b ∈ L(H), a > 0, is
uniquely solvable.
As a result, we can assert that the Lax equation (33) is equivalent to the second and
the third equations in (34) and equalities (35) and (36).
In a scalar case when H = C and the elements an(t), bn(t) of the matrix J(t) are
real-valued and ãn(t), b̃n(t), c̃n(t) are complex-valued functions, the Lax equation (33) is
equivalent to the following simple generalization of Toda lattice (see [5]): ∀t ∈ [0, T )
ȧn =
1
2
fan(bn+1 − bn), ḃn = f(a2
n − a2
n−1), n ∈ N0, a−1 = 0;
f = f(t) = a−1
0 (t)(ã0(t) − c̃0(t)).
(37)
If dimH > 1 the situation is more complicated [11, 23]. We consider here only an
interesting case when all operators an(t), bm(t), ãj(t), c̃k(t) are commuting and b̃l(t) =
0, t ∈ [0, T ); n, m, j, k, l ∈ N0.
In this case it follows from (35):
(38) c̃n(t) = c̃0(t)a
−1
0 (t)an(t), ãn(t) = ã0(t)a
−1
0 (t)an(t), n ∈ N, t ∈ [0, T ).
With the help of (38) and the assumption b̃l(t) = 0, t ∈ [0, T ), l ∈ N0, the condition
(36) gives:
0 = c̃nbn+1 + bn+1ãn − bnc̃n − ãnbn = (c̃n + ãn)(bn+1 − bn)
= (c̃0 + ã0)a
−1
0 an(bn+1 − bn), n ∈ N0, t ∈ [0, T ).
(39)
Assume in addition that
(40) ã0(t) =
1
2
a0(t), c̃0(t) = −
1
2
a0(t), t ∈ [0, T ).
Then from (39) it follows that the condition (36) is fulfilled.
Our assumption about commutativity and equality to zero of all b̃n, n ∈ N0, and
(38), (40) gives now that the second and the third equation from (34) have the form:
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 111
∀n ∈ N0, t ∈ [0, T )
ȧn = bnc̃n − c̃nbn+1 = c̃n(bn − bn+1) = c̃0a
−1
0 an(bn − bn+1) =
1
2
an(bn+1 − bn),
ḃn = an−1c̃n−1 + anãn − ãn−1an−1 − c̃nan = an(ãn − c̃n) + an−1(c̃n−1 − ãn−1)
= a2
n − a2
n−1.
So, we have proved the following statement.
Lemma 3. For t ∈ [0, T ), consider the operator Hermitian Jacobi matrix J(t) of type
(6) and an analogous matrix A(t) with elements ãn(t), b̃n(t), c̃n(t) ∈ L(H) (conditions for
smoothness and boundedness of all these operators are formulated above, in the beginning
of Section 4). Assume that all elements of these matrices ∀t are commuting operators,
b̃n(t) = 0, n ∈ N0, t ∈ [0, T ), and the elements ã0(t), c̃0(t) have the form (40).
Then the corresponding Lax equation (33) can be written in the equivalent form:
ȧn(t) =
1
2
an(t)(bn+1(t) − bn(t)),
ḃn(t) = a2
n(t) − a2
n−1(t), n ∈ N0, t ∈ [0, T ); a−1(t) = 0.
(41)
In one case, this result is applicable to the situation in Section 3: assume that the
solution of Toda equation (28) partially consist of positive functions: αn(t) > 0 and
uniformly bounded, i.e. ∃c ∈ [0,∞) : αn(t), |βn(t)| ≤ c, n ∈ N0, t ∈ [0, T ). Then all
unknowns an(t), bn(t) ∈ L(C2) of equations (30) are uniformly bounded and a
−1
n (t) exist.
Using Lemma 3 we can assert that the equations (30) with the condition a−1(t) = 0, t ∈
[0, T ), are equivalent to the Lax equation (33) with the corresponding matrix A(t). But
this condition is very essential: we cannot consider the general solution of Toda lattice
(28). We will overcome this difficulty in Sections 6–9. But now our immediate goal is
to construct in Section 5 a solution of the Cauchy problem for (41) in the case when all
unknowns are commuting operators.
We will now make an observation connected with Section 2 and 3.
Remark 1. With the Cauchy problem for double-infinite Toda lattices (28) one can
connect in a more natural way the operator semi-infinite Jacobi matrix acting in the
same space l2 = ℓ2(C
2).
Namely, the ordinary space ℓ2 on the whole Z, ℓ2(C, Z) = {ξ = (ξn)∞n=−∞, ξn
∈ C|
∞∑
n=−∞
|ξn|
2 < ∞}, and the space ℓ2(C
2) on N0 introduced above are unitary isomor-
phic. This isomorphism I can be constructed, for example, as follows :
ℓ2(C, Z) ∋ ξ = (ξn)∞n=−∞ �→Iξ = ((Iξ)n)∞n=0 ∈ ℓ2(C
2), (Iξ)n = (ξ−n−1, ξn) ∈ C
2.(42)
The difference expression L on sequences ξ = (ξn)∞n=−∞, ξn ∈ C, connected with (28)
is natural to present in the form of a double-infinite Toda scalar matrix of the form (6),
[45, 46, 47], i.e.
(43) (Lξ)n = αn−1ξn−1 + βnξn + αnξn+1, n ∈ Z.
The isomorphism (42) transforms the expression (43) into the expression ILI−1 generated
by a matrix of type (6) (with L(C2)-valued elements) on vector-valued sequences f =
(fn)∞n=0, fn ∈ C2 (for more details see [1, 11]).
The matrix of expression ILI−1 is an operator Jacobi matrix with elements from
L(C2) but these elements are somewhat different from (29). In this way it is possible to
develop a spectral theory of double-infinite Hermitian Jacobi matrices [1]. But potential
applications to the integration of the double-infinite Toda lattice meet with essential
difficulties: the corresponding differential equations w.r.t. t ∈ [0, T ) for the spectral
measure are not possible to integrate and find a simple solution [47, 11].
112 YURIJ BEREZANSKY
5. Evolution of the spectral measure of the semi-infinite operator Toda
lattice in the commutative case. Solution of the corresponding
Cauchy problem
Consider operator matrices J(t) and (in a general situation) A(t), t ∈ [0, T ), introduced
in Section 4. Let J(t) be the operator on l2 constructed according to the rule (5).
Denote by Rz(t) the resolvent of J(t) : Rz(t) = (J(t) − z1)−1, z ∈ C \ R, t ∈ [0, T )
and consider its derivative Ṙz(t) (which, of course, exists as a derivative in L(H)).
Assume that J(t) satisfies the Lax equation (33), then from (33) ∀t ∈ [0, T ) we get:
(44) (J(t)−z1)· = [J(t)−z1, A(t)]; Ṙz(t) = −Rz(t)(J(t)−z1)·Rz(t) = [Rz(t), A(t)]
(the equation (44) can be understood as matrix equalities on f ∈ lfin).
According to (13), (12) we can introduce the matrix (Rz;jk(t))∞j,k=0 of the operator
Rz(t) : l2 → l2 : ∀z ∈ C \ R, t ∈ [0, T ), j, k ∈ N0
Rz;jk(t) = {δj,Rz(t)δk} ∈ L(H);
(Rz;jk(z)x, y) = ({δj ,Rz(t)δk}x, y) = (Rz(t)(δkx), δjy)l2 .
Therefore,
(45) Rz;jk(t) = δ∗j Rz(t)δk, j, k ∈ N0.
Here we understand δk as an operator from H into l2:
H ∋ x �→ δkx := (0, . . . , 0︸ ︷︷ ︸
k
, x, 0, 0, . . .) ∈ l2; such an understanding of δk will be of use
also in what follows.
Introduce the (operator) Weyl function m(z) for a given operator J(t) : ∀z ∈ C \ R
m(z; t) = Rz;0,0(t) = δ∗0Rz(t)δ0 =
∫
R
1
λ − z
dρ(λ; t) ∈ L(H),
B(R) ∋△ �→ ρ(△; t) = δ∗0E(△; t)δ0 ∈ L(H).
(46)
Here dρ(λ; t) is the spectral measure of operator J(t) on the space l2; t ∈ [0, T ) is fixed.
From the second equality in (44) we can find how m(z; t) evolves in t. Namely, using
(46) and (44), (45), (13), (11) we get:
ṁ(z; t) = δ∗0Ṙz(t)δ0 = δ∗0 [Rz(t), A(t)]δ0 = δ∗0Rz(t)A(t)δ0 − δ∗0A(t)Rz(t)δ0
= {δ0,Rz(t)A(t)δ0} − {δ0, A(t)Rz(t)δ0} = Rz;00(t)̃b0(t) + Rz;01(t)ã0(t)
− b̃0(t)Rz;00(t) − c̃0(t)Rz;10(t), z ∈ C \ R, t ∈ [0, T ).
(47)
Let us pass now to the essential for us case when all the operators
an(t), bn(t), ãn(t), c̃n(t) ∈ L(H)
are commuting and b̃n(t) = 0, n ∈ N0, t ∈ [0, T ). We will use the following simple fact.
Lemma 4. Let A : l2 → l2 be a linear continuous operator, (ajk)∞j,k=0 its matrix (13).
Assume that the bounded inverse operator A−1 exists, denote by (bjk)∞j,k=0 its matrix.
We assert that if some operator c ∈ L(H) commutes with every ajk then it also commutes
with every bjk.
Proof. Consider the operator C = c1 where 1 is the identity operator on l2. Since ∀j, k ∈
N0 ajkc = cajk, the operator C commutes with A. Therefore it commutes also with
A−1, i.e. bjkc = cbjk, j, k ∈ N0. �
We can now deduce the difference equation for the Weyl function m(z; t).
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 113
Lemma 5. Assume as before that all the operators an(t), bm(t), ãj(t), c̃k(t) are commut-
ing and b̃l(t) = 0, n, m, j, k, l ∈ N0, t ∈ [0, T ). Then m(z; t) satisfies the following operator
differential equation (all operators below are commuting):
(48) ṁ(z; t) = a−1
0 (t)(ã0(t) − c̃0(t))(1 + (z1 − b0(t))m(z; t)), z ∈ C \ R, t ∈ [0, T ).
Proof. Since ∀z ∈ C \ R, t ∈ [0, T ), 1 = (J(t) − z1)Rz(t), therefore using (45), (11) and
the calculations as in (47) we get:
1 = δ∗0(J(t) − z1)Rz(t)δ0 = {δ0, (J(t) − z1)Rz(t)δ0}
= Rz;00(t)(b0(t) − z1) + Rz;01(t)a0(t).
From this equality we find:
(49) Rz;01(t) = (1 − m(z; t)(b0(t) − z1))a−1
0 (t), z ∈ C \ R, t ∈ [0, T ).
Using (45), (11) and (49) we get for the same z, t :
Rz;10(t) = δ∗1Rz(t)δ0 = (δ∗0R
∗
z(t)δ1)
∗ = (δ∗0Rz(t)δ1)
∗
= (Rz;01(t))
∗ = a−1
0 (t)(1 − (b0(t) − z1)m(z; t)).
(50)
Note that here we took advantage of the equality following from (46):
(51) m∗(z; t) = m(z; t), z ∈ C \ R, t ∈ [0, T ).
The representation (47), (49) and (50) was obtained without our assumption about
commutativity. With this assumptions Lemma 4 and formulas above give (48):
ṁ(z; t) = Rz;01(t)ã0(t) − c̃0(t)Rz;10(t) = a−1
0 (t)(ã0(t) − c̃0(t))(1 − (b0(t) − z1)m(z; t)).
�
As in the scalar case, the operator-valued Weyl function
(52) m(z) =
∫
R
1
λ − z
dρ(λ) ∈ L(H), z ∈ C \ R
(B(R) ∋△ �→ ρ(△) ∈ L(H) is a finite nonnegative operator-valued Borel measure) defines
uniquely the measure dρ(λ). For the proof it suffices to apply this equality to ∀x ∈ H
and then scalar multiply it by ∀y ∈ H ; thus the equation is reduced to the scalar case.
From this fact it follows that the spectral measure dρ(λ; t), as well as m(z; t), commutes
with an(t), bn(t), n ∈ N0, t ∈ [0, T ).
Therefore the equation (48) is reduced to the equivalent equation for the measure
dρ(λ; t). Namely, take ã0(t), c̃0(t) of the form (40). Then (48) has the form:
(53) ṁ(z; t) = 1 + (z1 − b0(t))m(z; t), z ∈ C \ R, t ∈ [0, T ).
Transform this equality in the following manner:
∫
R
1
λ − z
dρ̇(λ; t) =
( ∫
R
1
λ − z
dρ(λ; t)
)·
= ṁ(z; t) = 1 + (z1 − b0(t))m(z; t)
=
∫
R
λ − b0(t)
λ − z
dρ(λ; t).
This equality is equivalent to the following one for the measure dρ(λ; t) :
(54) dρ̇(λ; t) = (λ − b0(t))dρ(λ; t), λ ∈ R, t ∈ [0, T ).
For every t, the operators from L(H), ρ(△; t) and b0(t), commute, therefore a solution
of equation (54) is easy to write:
(55) dρ(λ; t) = c(t)eλtdρ(λ; 0), λ ∈ R, t ∈ [0, T ).
114 YURIJ BEREZANSKY
Here the operator-valued function c(t) is equal to
(56) c(t) = e
−
t�
0
b0(τ)dτ
∈ L(H), t ∈ [0, T ).
To find this function it is not necessary to know b0(t) : the measure (55) is probability,
i.e. ∀t ρ(R; t) = 1. Therefore from (55) we get:
(57) c(t) =
( ∫
R
eλtdρ(λ; 0)
)−1
, t ∈ [0, T ).
As a result, we can formulate the following intermediate theorem whose proof follows
from above stated facts (see also the proof of the Theorem 5).
Theorem 1. Consider the semi-infinite operator Toda lattice (41) with continuously dif-
ferentiable w.r.t. t commuting operators an(t), bn(t) ∈ L(H), n ∈ N0, t ∈ [0, T ). For (41),
consider the following Cauchy problem: for initial data (a0(0), a1(0), . . . ; b0(0), b1(0), . . .)
find the solution (a0(t), a1(t), . . . ; b0(t), b1(t), . . .), t ∈ [0, T ).
Assume that such a solution exists with the additional properties: an(t) > 0, bn(t) =
b∗n(t), ∃C ∈ [0,∞) : ‖an(t)‖ , ‖bn(t)‖ ≤ C, n ∈ N0, t ∈ [0, T ).
Construct from this solution a Jacobi matrix J(t) (6), let J(t) be the bounded selfad-
joint operator on l2 generated by J(t). Denote by dρ(λ; t) the operator spectral measure
of J(t). Then this measure is recovered from the initial measure dρ(λ; 0) with the help of
formulas (55), (57).
A solution of our Cauchy problem is given by formulas (21) in applying them to the
spectral measure dρ(λ; t) and the corresponding orthogonal polynomials Pn(λ; t) that can
be obtained via the orthogonalization procedure.
Remark 2. It is possible to investigate, in a similar fashion, a lattice that is a little
more general than (41):
ȧn(t) =
1
2
f(t)an(t)(bn+1(t) − bn(t)),
ḃn(t) = f(t)(a2
n(t) − a2
n−1(t)), n ∈ N0, t ∈ [0, T ); a−1(t) = 0,
(58)
where [0, T ) ∋ t �→ f(t) ∈ L(H) is a given operator-valued function whose values
commute with an(t), bn(t), n ∈ N0. Note that to study (58) it is necessary to take
ã0(t) = 1
2a0(t)f(t), c̃0(t) = − 1
2a0(t)f(t) instead of (40).
Remark 3. It is possible to check by direct calculations that the formulas for solutions
of the Cauchy problem for (41), (58), given in Theorem 1 are in fact solutions of these
equations. It is possible to do this in a way similar to that in the works [5, 15, 12].
6. Another reduction of the double-infinite Toda lattice to semi-infinite
matrix Toda lattice
In Section 3 one method of reduction was given, but in this case the matrix a−1 in
system (30) is not equal to zero. Therefore the constructions of Sections 4, 5 are not
applicable to an investigation of the Toda lattice in a general situation. In this Section
we will propose another approach to such a reduction.
Instead of the space l2 = ℓ2(C
2) = C
2 ⊕ C
2 ⊕ · · · , we consider its subspace,
(59) l2,0 = C
1⊕C
2⊕C
2⊕· · · = H0⊕H1⊕H2⊕· · · , H0 = C
1, H1 = H2 = · · · = C
2,
with elements x = (xn)∞n=0, where x0 ∈ C and xn = (xn;0, xn;1) ∈ C2, n ∈ N. As in
Remark 1 we consider the space ℓ2(C, Z) of sequences ξ = (ξn)∞n=−∞ and introduce
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 115
(like in (42)) a mapping K :
ℓ2(C, Z) ∋ ξ = (ξn)∞n=−∞ �→ Kξ = ((Kξ)n)∞n=0 ∈ l2,0,
(Kξ)0 = ξ0 ∈ C, (Kξ)n = (ξn, ξ−n) ∈ C
2, n ∈ N.
(60)
This mapping is an unitary isomorphism between the spaces ℓ2(C, Z) and l2,0.
The inverse mapping K−1 has the form
l2,0 ∋ x = (xn)∞n=0 �→ K−1x = ((K−1x)n)∞n=−∞ ∈ ℓ2(C, Z),
(K−1x)n = xn;0, (K
−1x)0 = x0, (K
−1x)−n = xn;1, n ∈ N.
(61)
Consider the difference expression L of the three-diagonal form, more general than
(43), and acting on sequences ξ = (ξn)∞n=−∞, ξn ∈ C :
(62) (Lξ)n = αn−1ξn−1 + βnξn + γnξn+1, n ∈ Z,
where αn, βn, γn are real coefficients. Calculate its image under isomorphism (60), namely
L = KLK−1. This image act on sequences x = (xn)∞n=0, x0 ∈ C, xn = (xn;0, xn;1) ∈ C2,
n ∈ N, and has the following form.
Let n = 2, 3, . . . . Then using (60), (62), (61) we get:
(63)
(Lx)n = (KLK−1x)n = ((LK−1x)n, (LK−1x)−n)
= (αn−1(K
−1x)n−1 + βn(K−1x)n + γn(K−1x)n+1,
α−n−1(K
−1x)−n−1 + β−n(K−1x)−n + γ−n(K−1x)−n+1)
= (αn−1xn−1;0 + βnxn;0 + γnxn+1;0, α−n−1xn+1;1 + β−nxn;1 + γ−nxn−1;1),
i.e.
(64) (Lx)n = an−1xn−1 + bnxn + cnxn+1, n = 2, 3, . . . ;
(65) an =
[
αn 0
0 γ−n−1
]
, bn =
[
βn 0
0 β−n
]
, cn =
[
γn 0
0 α−n−1
]
, n = 2, 3, . . . .
For the case n = 1 we get from (63) and (61) :
(Lx)1 = (KLK−1x)1 = ((LK−1x)1, (LK−1x)−1)
= (α0x0 + β1x1;0 + γ1x2;0, α−2x2;1 + β−1x1;1 + γ−1x0),
i.e. we have the expression (64) for n = 1 with b1, c1, given by (65) for n = 1 and a0 of
the form a0 =
[
α0
γ−1
]
.
Analogously for n = 0 we have:
(Lx)0 = (KLK−1x)0 = (LK−1x)0 = α−1(K
−1x)−1 + β0(K
−1x)0 + γ0(K
−1x)1
= α−1x1;1 + β0x0 + γ0x1;0.
In this case expression (64) also holds with matrices:
a−1 = 0, b0 = [β0], c0 = [γ0 α−1].
These simple calculations show that the following lemma is valid.
Lemma 6. The expression L = KLK−1 where L has the form (62) acts on sequences
x = (xn)∞n=0, x0 ∈ C, xn = (xn;0, xn;1), n ∈ N, by the rule:
(66) (Lx)n = an−1xn−1 + bnxn + cnxn+1, where
116 YURIJ BEREZANSKY
an =
[
αn 0
0 γ−n−1
]
, bn =
[
βn 0
0 β−n
]
, cn =
[
γn 0
0 α−n−1
]
, n ∈ N;
a0 =
[
α0
γ−1
]
, b0 = [β0], c0 = [γ0 α−1]; a−1 = 0.
Note, that in the Hermitian case, when L has the form (43) (i.e. γn = αn), the
expression L is also Hermitian:
(67) cn = an, n ∈ N; a0 =
[
α0
α−1
]
, c0 = [α0 α−1] = a∗
0, a−1 = 0.
Return now to the double-infinite Toda lattice (28) and write it in the form of a Lax
equation (of course, this fact is well known).To this end, introduce the double-infinite
Jacobi type matrices with real elements acting on sequences ξ = (ξn)∞n=−∞, ξ ∈ C :
J =
. . .
. . .
. . .
. . . 0 α−2 β−1 α−1 0 . . .
. . . 0 α−1 β0 α0 0 . . .
. . . 0 α0 β1 α1 0 . . .
. . .
. . .
. . .
,
A =
. . .
. . .
. . .
. . . 0 α̃−2 β̃−1 γ̃−1 0 . . .
. . . 0 α̃−1 β̃0 γ̃0 0 . . .
. . . 0 α̃0 β̃1 γ̃1 0 . . .
. . .
. . .
. . .
.
(68)
Assuming that the elements of matrices (68) depend smoothly on t ∈ [0, T ), T ≤ ∞
(as often before, we will not designate this dependence) introduce the Lax equation:
(69) J̇ (t) = [J (t),A(t)] := J (t)A(t) −A(t)J (t), t ∈ [0, T ).
This equality in the ”coordinate” form can be rewritten as follows (compare with (33),
(34) for operator elements; see also [5, 12]):
0 = αnγ̃n+1 − γ̃nαn+1,
α̇n = βnγ̃n + αnβ̃n+1 − β̃nαn − γ̃nβn+1,
β̇n = αn−1γ̃n−1 + αnα̃n − α̃n−1αn−1 − γ̃nαn,
α̇n = αnβ̃n + βn+1α̃n − α̃nβn − β̃n+1αn,
0 = αn+1α̃n − α̃n+1αn, n ∈ Z, t ∈ [0, T ).
(70)
We will assume that ∀t ∈ [0, T ) and ∀n ∈ Z αn(t) > 0. Then from (70) it is easy to
calculate that
(71)
α̃n = αnα−1
0 α̃0, γ̃n = αnα−1
0 γ̃0, β̃n = β̃0 + (2α0)
−1(α̃0 + γ̃0)(βn − β0),
n ∈ Z, t ∈ [0, T ).
It is possible also to give a simple formula for calculation of β̃n by α̃0, γ̃0 and α0, β0, βn.
From the second and third equalities in (70) and (71) we conclude that ∀t ∈ [0, T )
(72) α̇n =
1
2
f(t)αn(βn+1 − βn), β̇n = f(t)(α2
n − α2
n−1)
where f(t) = α−1
0 (α̃0 − γ̃0), n ∈ Z.
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 117
Let, similarly to (40),
(73) α̃0(t) =
1
2
α0(t), γ̃0(t) = −
1
2
α0(t), t ∈ [0, T ),
then the function f(t) = 1, t ∈ [0, T ), and equalities (72) have the form of the Toda
lattice (28). In this case it is possible also to put: β̃n = 0, n ∈ Z (see a similar calculation
in [5, 12]).
Conversely, it is easy to see that Toda lattice (28) gives the Lax equation (69). There-
fore Toda lattice (28) and the Lax equation (69), (70) (with conditions (73)) are equivalent
(under the assumption: αn(t) > 0, t ∈ [0, T ), n ∈ Z).
Our nearest aim is to rewrite Toda lattice (28) in terms of matrices (66), (67) (similar
to (41)). For this we need to apply to our objects the mappings K, K−1 (60), (61).
Let J (t) = L(t) have the form (43), i.e. (68), and A(t) be given by (68) with conditions
(73), t ∈ [0, T ). Denote ∀t ∈ [0, T )
(74) J(t) = KJ (t)K−1; A(t) = KA(t)K−1 : lfin → lfin.
The Lax equality (69) gives the corresponding Lax equation for J(t), A(t) :
(75)
J̇(t) = KJ̇ (t)K−1 = K(J (t)A(t) −A(t)J (t))K−1 = J(t)A(t) − A(t)J(t),
t ∈ [0, T ).
According to Lemma 6, (66), (67) the matrix elements an, bn and ãn, b̃n, c̃n, n ∈ N0,
of the three-diagonal matrices J(t) and A(t) are:
an =
[
αn 0
0 α−n−1
]
, bn =
[
βn 0
0 β−n
]
,
ãn =
[
α̃n 0
0 γ̃−n−1
]
, b̃n =
[
β̃n 0
0 β̃−n
]
, c̃n =
[
γ̃n 0
0 α̃−n−1
]
, n ∈ N;
a0 =
[
α0
α−1
]
, b0 = [β0], c0 = [α0 α−1] = a∗
0,
ã0 =
[
α̃0
γ̃−1
]
, b̃0 = [β̃0], c̃0 = [γ̃0 α̃−1].
(76)
Therefore the Lax equality (75) has the form (33) but the matrices J(t), A(t) act on the
space l2,0 (59) distinct from the space l2 = ℓ2(H) : in (59) the first term is not equal to
the other.
Nevertheless we can claim that equality (75) is equivalent to a system of type (34) with
special elements.
This fact follows from a simple general calculation in [12], §2, (9), (23), (25). Namely,
we get for the matrices (76) using their commutativity:
0 = anc̃n+1 − c̃nan+1,
ȧn = bnc̃n + anb̃n+1 − b̃nan − c̃nbn+1,
ḃn = an−1c̃n−1 + anãn − ãn−1an−1 − c̃nan,
ȧn = anb̃n + bn+1ãn − ãnbn − b̃n+1an,
0 = an+1ãn − ãn+1an, n = 2, 3, . . .
(77)
For n = 1 we also have equalities (77) but instead of the third one of them we get:
(78) ḃ1 = a0c̃0 + a1ã1 − ã0a
∗
0 − c̃1a1.
118 YURIJ BEREZANSKY
For n = 0 the corresponding equalities have the form:
0 = a∗
0c̃1 − c̃0a1,
ȧ∗
0 = b0c̃0 + a∗
0b̃1 − b̃0a
∗
0 − c̃0b1,
ḃ0 = a∗
0ã0 − c̃0a0,
ȧ0 = a0b̃0 + b1ã0 − ã0b0 − b̃1a0,
0 = a1ã0 − ã1a0.
(79)
The system (77) is a partial case of the general system (34) with H = C2 and
n = 2, 3, . . . instead of n = 0, 1, . . . ; ã1, c̃1 are not necessarily equal to zero. All the op-
erators an(t), bm(t), ãj(t), c̃k(t), n, m, j, k = 2, 3, . . . , t ∈ [0, T ), are commuting, see (76);
the operators an(t) are invertible. Also we can set β̃n = 0, n ∈ Z, then b̃n = 0, n ∈ N0.
We will rewrite this system in a more simple form at first for the case n = 2, 3, . . . .
Its coefficients ãn, c̃n have the the form of (76) with the elements α̃n, γ̃n satisfying the
conditions (71) and (73) (with an index 2 instead of 0). So, we have:
(80) α̃m = αmα−1
2 α̃2 =
1
2
αm, γ̃m = αmα−1
2 γ̃2 = −
1
2
αm, m ∈ Z;
ãn =
[
α̃n 0
0 γ̃−n−1
]
=
[
1
2αn 0
0 − 1
2α−n−1
]
= −
1
2
qan, q :=
[
−1 0
0 1
]
;
c̃n =
[
γ̃n 0
0 α̃−n−1
]
=
[
− 1
2αn 0
0 1
2α−n−1
]
=
1
2
qan, n ∈ N.
(81)
Since b̃n = 0, n ∈ N0, we get conditions equivalent to (77): ∀n = 2, 3, . . .
0 = an(
1
2
qan+1) − (
1
2
qan)an+1,
ȧn = bn(
1
2
qan) − (
1
2
qan)bn+1,
ḃn = an−1(
1
2
qan−1) − an(
1
2
qan) + (
1
2
qan−1)an−1 − (
1
2
qan)an,
ȧn = −bn+1(
1
2
qan) + (
1
2
qan)an,
0 = −an+1(
1
2
qan) + (
1
2
qan+1)an.
(82)
The first and the last conditions in (82) are obviously true; the other three equalities
in (82) are equivalent to the following:
(83) ȧn =
1
2
anq(bn − bn+1), ḃn = q(a2
n−1 − a2
n), n = 2, 3, . . .
Consider the case n = 1. As earlier, we conclude that the first and last equalities in
(82) are true and the second and the forth equalities are equivalent to the first equality
in (83). For the next calculations it is necessary to find ã0, c̃0. Using (80) and (76) we
conclude:
(84) ã0 =
1
2
[
α0
−α−1
]
= −
1
2
qa0, c̃0 =
1
2
[−α0 α−1] =
1
2
a∗
0q.
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 119
Now the equality (78) with the help of (84), (81) and (76) gives:
ḃ1 =a0(
1
2
a∗
0q) − a1(
1
2
qa1) + (
1
2
qa0)a
∗
0 − (
1
2
qa1)a1
= − qa2
1 +
1
2
[
−α2
0 α0α−1
−α−1α0 α2
−1
]
−
1
2
[
α2
0 α0α−1
−α−1α0 −α2
−1
]
= − qa2
1 +
[
−α2
0 0
0 α2
−1
]
, i.e. ḃ1 = −qa2
1 +
1
2
(a0a
∗
0q + qa0a
∗
0).
(85)
Consider the case n = 0. Now the equalities of type (77) have the form (79); using the
formulas (76), (81), (84) we get:
0 = a∗
0(
1
2
qa1) − (
1
2
a∗
0q)a1,
ȧ∗
0 = b0(
1
2
a∗
0q) − (
1
2
a∗
0q)b1,
ḃ0 = −a∗
0(
1
2
qa0) − (
1
2
a∗
0q)a0,
ȧ0 = −b1(
1
2
qa0) + (
1
2
qa0)b0,
0 = −a1(
1
2
qa0) + (
1
2
qa1)a0.
(86)
The first and the last conditions in (86) are true; the other three equalities in (86) are
equivalent to the following:
(87) ȧ∗
0 =
1
2
(b0a
∗
0 − a∗
0b1)q, ḃ0 = −a∗
0qa0, ȧ0 =
1
2
q(a0b0 − b1a0).
As a result, we can state that the Lax equation (75) is equivalent to the system (83)
for n ∈ N (where for n = 1 it is necessary to replace the second equality by (85)) and
system (87) for n = 0. Also it is clear that in (87) the first equality is equivalent to the
third one.
Now we formulate the final result of this Section.
Theorem 2. The Toda lattice (28) with an assumption αn(t) > 0 and uniform bound-
edness of αn(t), βn(t), t ∈ [0, T ), T ≤ ∞, n ∈ Z, is equivalent to the Lax equation
(88) J̇(t) = J(t)A(t) − A(t)J(t), t ∈ [0, T ),
with bounded operators J(t), A(t) acting on the space l2,0 (59) and having the form of a
Jacobi matrix J(t) of type (6) with elements respectively
(89)
an = cn =
[
αn 0
0 α−n−1
]
, bn =
[
βn 0
0 β−n
]
, n ∈ N,
a0 =
[
α0
α−1
]
, b0 = [β0], c0 = a∗
0;
ãn = −
1
2
qan, b̃n = 0, n ∈ N0, c̃n =
1
2
qan, n ∈ N, c̃0 =
1
2
a∗
0q; q =
[
−1 0
0 1
]
.
This Toda lattice is also equivalent to the system:
ȧn =
1
2
anq(bn − bn+1), n ∈ N, ḃn = q(a2
n−1 − a2
n), n = 2, 3, . . . ,
ḃ1 =
1
2
(a0a
∗
0q + qa0a
∗
0) − qa2
1, ȧ0 =
1
2
q(a0b0 − b1a0), ḃ0 = −a∗
0qa0.
(90)
The proof follows from the calculations above. See formulas (76), (81), (84) and (83),
(85), (87). Of course it is also easy to check directly that lattice (28) is equivalent to (90).
120 YURIJ BEREZANSKY
7. Direct and inverse spectral problems for a single-valued degenerate
Jacobi matrix
Our aim is to get a solution of the Cauchy problem for a double-infinite Toda lattice
by using Theorem 2 and scheme of Section 5. But now our Jacobi matrix (6) acts on the
space l2,0 and is single-valued degenerate: its element a0 is not invertible. The aim of
this Section is to develop the spectral theory for such matrices.
Thus we consider the operator Jacobi matrix on the space l2,0 (59)
(91) J =
b0 a∗
0 0 0 0 . . .
a0 b1 a1 0 0 . . .
0 a1 b2 a2 0 . . .
...
...
...
...
...
. . .
,
a0 =
[
α0
α−1
]
: C1 → C2,
a∗
0 = [α0 α−1] : C2 → C1,
b0 = [β0] : C
1 → C
1;
an, bn ∈ L(C2), an > 0, bn = b∗n, n ∈ N.
All the matrices an, bn, n ∈ N0, are real. For simplicity we will assume that the matrices
an =
[
an;0 0
0 an;1
]
, bn =
[
bn;0 0
0 bn;1
]
, n ∈ N, are diagonal and α−1 > 0 (only such
matrices will be used in what follows, see Theorem 2).
We assume that elements an, bn are uniformly bounded therefore J gives rise to a
bounded selfadjoint operator J on the space l2,0. To construct its eigenfunction expansion
consider the difference equations: let ϕ(λ) = (ϕn(λ))∞n=0, ϕ0(λ) =: ϕ0,0(λ) ∈ R1, ϕn(λ) =
(ϕn,0(λ), ϕn,1(λ)) ∈ R2, n ∈ N, λ ∈ R,
(92)
(Jϕ(λ))0 = b0ϕ0(λ) + a∗
0ϕ1(λ) = λϕ0(λ), i.e.
(Jϕ(λ))0,0 = β0ϕ0,0(λ) + α0ϕ1,0(λ) + α−1ϕ1,1(λ) = λϕ0,0(λ),
(Jϕ(λ))1 = a0ϕ0(λ) + b1ϕ1(λ) + a1ϕ2(λ) = λϕ1(λ), i.e.
((Jϕ(λ))1,0, (Jϕ(λ))1,1) = (α0ϕ0,0(λ) + b1,0ϕ1,0(λ) + a1,0ϕ2,0(λ),
α−1ϕ0,0(λ) + b1,1ϕ1,1(λ) + a1,1ϕ2,1(λ)) = (λϕ1,0(λ), λϕ1,1(λ)),
(Jϕ(λ))n = an−1ϕn−1(λ) + bnϕn(λ) + anϕn+1(λ) = λϕn(λ), n = 2, 3, . . .
We put ϕ0,0(λ) = c0,0, ϕ1,0(λ) = c1,0 where c0,0, c1,0 ∈ R are some fixed constants.
Then from recurrence relations (92) we can find recursively
(93) ϕ1,1(λ), ϕ2,0(λ), ϕ2,1(λ), ϕ3,0(λ), ϕ3,1(λ), . . .
because α−1 > 0 and all the matrices an, n ∈ N, are invertible.
As a result, we find a solution with the initial data c0,0, c1,0 :
(94)
ϕ(λ) = (ϕ0(λ) = ϕ0,0(λ), ϕn = (ϕn,0(λ), ϕn,1(λ)))∞n=1 = (ϕn,νn
(λ))∞νn=0,1;n=0,
ϕn,νn
(λ)νn=0,1 := ϕn,0(λ), ϕn,1(λ); ϕ0,0(λ) = c0,0, ϕ1,0(λ) = c1,0; λ ∈ R.
Note that in (94) we assumed that for n = 0 we have only one index ν0 = 0; for
n = 1, 2, . . . we have two indexes νn = 0, 1. Such an agreement will be used in what
follows.
The functions ϕn,νn
(λ) from (93), (94) are polynomials w.r.t. λ with real coefficients.
The form of these polynomials will be essential for the inverse problem, we will establish
it a little later.
Now we only want to note that the recursion (92) is linear and it is easy to see
that every solution (94) is a linear combination of two solutions P (0; λ), P (1; λ) with a
corresponding initial data: ∀α = 0, 1
P (α; λ) = (Pn,νn
(α; λ))∞νn=0,1;n=0, where
P0,0(0; λ) = 1, P1,0(0; λ) = 0 and P0,0(1; λ) = 0, P1,0(1; λ) = 1.
(95)
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 121
Thus we have:
(96) ϕ(λ) = c0,0P (0; λ) + c1,0P (1; λ), c0,0 = ϕ0,0, c1,0 = ϕ1,0.
This formula gives a solution of the difference equation (92) with the initial conditions
ϕ0,0 and ϕ1,0.
Remark 4. Instead of basic polynomial solutions P (0; λ), P (1; λ) of system (92) it is
possible to use two solutions of this system that satisfy initial data that differ from (95).
Note that we need two such solutions since every solution of (92) has two free parameters.
The choice of initial data (95) is the most convenient and is reminiscent of the initial
data for classical Jacobi matrices.
To construct the eigenfunctions expansion of the operator J we apply the projection
spectral theorem (see, e.g. [13], Ch. 15). Denote by l2,0(p) the weighted space of vectors
f = (fn)∞n=0, fn ∈ Hn (recall that H0 = C1,H1 = H2 = . . . = C2) for which
‖f‖
2
l2,0(p) =
∞∑
n=0
‖fn‖
2
Hn
pn < ∞, (f, g)l2,0(p) =
∞∑
n=0
(fn, gn)Hn
pn.
Here p = (pn)∞n=0, pn > 0, is a given sequence of weights. In what follows, pn ≥ 1 and
∞∑
n=0
p−1
n < ∞, therefore the embedding of the positive space l2,0(p) ⊂ l2,0 is quasinuclear.
The corresponding negative space is l2,0(p
−1), p−1 := (p−1
n )∞n=0. As a result, we construct
the quasinuclear rigging
(97) l0 = (l0,fin)
′
⊃ l2,0(p
−1) ⊃ l2,0 ⊃ l2,0(p) ⊃ l0,fin
(l0 denotes the space of all sequences f = (fn)∞n=0, fn ∈ Hn; l0,fin is the corresponding
space of finite sequences).
Now we will use the result from [1], Ch. 5, and [13], Ch. 15, on the generalized
eigenvectors expansion for a bounded selfadjoint operator connected with the chain (97)
in a standard way. For the operator J we have the representation
(98) Jf =
∫
R
λΦ(λ)dσ(λ)f, f ∈ l2,0(p),
where Φ(λ) : l2,0(p) → l2,0(p
−1) is a generalized projection operator and dσ(λ) is a
spectral measure (with a bounded support). Since the matrix J is real the operators J
and Φ(λ) are also real.
For all f, g ∈ l0,fin we have the Parseval equality
(99) (f, g)l2,0
=
∫
R
(Φ(λ)f, g)l2,0
dσ(λ).
Extending by continuity, the equality takes place for all f, g ∈ l2,0.
Let us denote by πn,νn
, n ∈ N, the operator of orthogonal projection in l2,0 onto the
one-dimensional subspace Hn,νn
of Hn consisting of the fn,νn
∈ C
1; π0 = π0,0 is an
analogous projection on H0 = C1. Hence for all f = (fn)∞n=0 ∈ l2,0 we have fn,νn
=
πn,νn
f, f0 = π0f. This operator acts analogously in the spaces l2,0(p) and l2,0(p
−1) but
possibly with a norm which is not equal to one.
Let us consider the scalar matrix (Φj,νj ;k;νk
(λ))∞νj ,νk=0,1;j,k=0 where
(100) Φj,νj ;k;νk
(λ) = πj,νj
Φ(λ)πk,νk
: l2,0 → Hj,νj
⊂ l2,0.
As we have arranged in (94), here and further we assume that the index n, νn for n = 0
is equal to 0, 0 (one time).
The generalized projection operator Φ(λ) is real, therefore the elements of its matrix
(100) are also real.
122 YURIJ BEREZANSKY
The Parseval equality (99) can be rewritten as follows: ∀f, g ∈ l2,0
(f, g)l2,0
=
∞∑
j,k=0;νj ,νk=0,1
∫
R
(Φ(λ)πk,νk
f, πj,νj
g)l2,0
dσ(λ)
=
∞∑
j,k=0;νj ,νk=0,1
∫
R
(πj,νj
Φ(λ)πk,νk
f, g)l2,0
dσ(λ)
=
∞∑
j,k=0;νj ,νk=0,1
∫
R
Φj,νj ;k,νk
(λ)fk,νk
gj,νj
dσ(λ).
(101)
It is essential to give a representation of elements of matrix (100) via polynomials (95).
Lemma 7. For every fixed j, k ∈ N0 and νj , νk = 0, 1, the matrix (100) has the following
representation: ∀λ ∈ R
Φj,νj ;k,νk
(λ) = Φ0,0;0,0(λ)Pj,νj
(0; λ)Pk,νk
(0; λ) + Φ0,0;1,0(λ)Pj,νj
(0; λ)Pk,νk
(1; λ)
+ Φ1,0;0,0(λ)Pj,νj
(1; λ)Pk,νk
(0; λ) + Φ1,0;1,0(λ)Pj,νj
(1; λ)Pk,νk
(1; λ).
(102)
Proof. For fixed k ∈ N0 and νk ∈ {0, 1}, the vector ϕ(λ) = (ϕj,νj
(λ))∞νj=0,1;j=0, where
(103) ϕj,νj
(λ) = Φj,νj ;k,νk
(λ) = (Φ(λ)πk,νk
f)j,νj
and fixed f ∈ l0,fin with fk,νk
= 1 is a generalized solution, in l0 = (l0,fin)
′
, of the
equation Jϕ(λ) = λϕ(λ), since Φ(λ) is a projector onto a generalized eigenvector of the
selfadjoint operator J with the corresponding generalized eigenvalue λ. Therefore for
all g ∈ l0,fin we have (ϕ(λ), Jg)l2,0
= λ(ϕ(λ), g)l2,0
. Transforming the finite difference
Hermitian expression J to ϕ(λ) we get (Jϕ(λ), g)l2,0
= λ(ϕ(λ), g)l2,0
. Hence, it follows
that ϕ(λ) ∈ l2,0(p
−1) exists as a usual real solution of the difference equation Jϕ = λϕ,
i.e. (92), with the initial conditions ϕ0,0(λ) = Φ0,0;k,νk
(λ) and ϕ1,0(λ) = Φ1,0;k,νk
(λ).
Using the representation (96) of solutions of equation (92) by its initial conditions we
can write:
(104)
Φj,νj ;k,νk
(λ) = Φ0,0;k,νk
(λ)Pj,νj
(0; λ) + Φ1,0;k,νk
(λ)Pj,νj
(1; λ),
j ∈ N0, νj ∈ {0, 1}.
Operator Φ(λ) is formally selfadjoint therefore its matrix (100) is selfadjoint w.r.t.
indexes j, νj and k, νk and, similarly to (103) , is a solution of equation (92) w.r.t. k, νk
with arbitrary fixed j, νj . So, using (96) we get the representation similar to (104) for
Φ0,0;k,νk
(λ) and Φ1,0;k,νk
(λ) :
Φ0,0;k,νk
(λ) = Φ0,0;0,0(λ)Pk,νk
(0; λ) + Φ0,0;1,0(λ)Pk,νk
(1; λ),
Φ1,0;k,νk
(λ) = Φ1,0;0,0(λ)Pk,νk
(0; λ) + Φ1,0;1,0(λ)Pk,νk
(1; λ), k ∈ N0, νk ∈ {0, 1}.
(105)
Substituting (105) into (104), we get the representation (102). �
Introduce the Fourier transform̂ (w.r.t. our eigenfunction expansion), at first for
finite f = (fn)∞n=0 = (fn,νn
)∞νn=0,1;n=0 ∈ l0,fin. It is a vector-valued function f̂(λ) =
(f̂0(λ), f̂1(λ)) ∈ C2 of λ ∈ R with the values
(106) f̂α(λ) =
∞∑
n=0;νn=0,1
Pn,νn
(α; λ)fn,νn
, α = 0, 1
(recall that here the index n, νn for n = 0 is equal to 0, 0;).
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 123
Substitute the representation (102) into (101). For f, g ∈ l0,fin using notation (106)
we get:
(f, g)l2,0
=
∫
R
(Φ0,0;0,0(λ)ĝ0(λ)f̂0(λ) + Φ0,0;1,0(λ)ĝ0(λ)f̂1(λ)
+ Φ1,0;0,0(λ)ĝ1(λ)f̂0(λ) + Φ1,0;1,0(λ)ĝ1(λ)f̂1(λ))dσ(λ).
(107)
Introduce the 2 × 2-matrix-valued real measure by the formula
(108) dρ(λ) =
[
Φ0,0;0,0(λ) Φ0,0;1,0(λ)
Φ1,0;0,0(λ) Φ1,0;1,0(λ)
]
dσ(λ) =: (dρα,β(λ))1α,β=0, λ ∈ R.
Then the equality (107) can be rewritten in the form:
(109)
(f, g)l2,0
=
∫
R
([
Φ0,0;0,0(λ) Φ0,0;1,0(λ)
Φ1,0;0,0(λ) Φ1,0;1,0(λ)
]
f̂(λ), ĝ(λ)
)
C2
dσ(λ) =
∫
R
(dρ(λ)f̂ (λ), ĝ(λ))C2 .
Lemma 8. The matrix in (108) is nonnegative and real for almost all λ ∈ R w.r.t. the
measure dσ(λ).
Proof. This fact follows from nonnegativity of the operator Φ(λ) as an operator from
l2,0(p) into l2,0(p
−1). Indeed, definition (100) gives, for fixed λ, arbitrary ξ = (ξ0, ξ1) ∈ C2
and the corresponding vector f = (f0, f1, 0, . . .) = (ξ0, (ξ1, 0), (0, 0), (0, 0), . . .) ∈ l0,fin, the
following:
([
Φ0,0;0,0(λ) Φ0,0;1,0(λ)
Φ1,0;0,0(λ) Φ1,0;1,0(λ)
]
ξ, ξ
)
C2
=
([
π0Φ(λ)π0 π0Φ(λ)π1,0
π1,0Φ(λ)π0 π1,0Φ(λ)π1,0
]
ξ, ξ
)
C2
= π0Φ(λ)π0ξ0ξ0 + π0Φ(λ)π1,0ξ1ξ0 + π1,0Φ(λ)π0ξ0ξ1 + π1,0Φ(λ)π1,0ξ1ξ1
= (Φ(λ)(π0 + π1,0)f, (π0 + π1,0)f)l2,0
≥ 0.
�
As a result, the measure (108) is a finite nonnegative 2 × 2-matrix-valued real Borel
measure. This measure we will call the matrix spectral measure of our operator J . We can
introduce, as in Section 1, the corresponding L2-type Hilbert space L2 = L2(C2, R, dρ(λ))
with the product (14).
The Parseval equality (109) now has the form: ∀f, g ∈ l0,fin
(110) (f, g)l2,0
=
∫
R
(dρ(λ)f̂ (λ), ĝ(λ))C2 = (f̂(·), ĝ(·))L2 .
After extending by continuity, this equality also will be true for arbitrary f, g ∈ l2,0. Note
that the Fourier transform of f ∈ l2,0 is defined as a corresponding limit in the space
L2 of the Fourier transform (106) of vectors from l0,fin approximating f. In this case in
(110) we have an extension of the integral appearing there.
As in the classical theory of Jacobi matrices the polynomials Pn,νn
(α; λ) are orthogonal
in some sense w.r.t. matrix spectral measure dρ(λ). The conditions of orthogonality are
very easy to deduce from the Parseval equality (110) and formula (106).
So, according to (106) the polynomial Pn,νn
(α; λ) for n ∈ N, νn = 0, 1, is equal to the α-
coordinate f̂α(λ) of the Fourier transform f̂(λ) = (f̂0(λ), f̂1(λ)) of the finite sequence f =
(0, . . . , 0, 1, 0, 0, . . .) =: εn,νn
with 1 at the n, νn place; for P0,0(α; λ) f = (1, 0, 0, . . .) =:
124 YURIJ BEREZANSKY
ε0,0. Therefore the Parseval equality (110) gives: ∀n, m ∈ N0, νn, νm ∈ {0, 1}
∫
R
1∑
α,β=0
dρα,β(λ)Pn,νn
(β; λ)Pm,νm
(α; λ) =
∫
R
(dρ(λ)ε̂n,νn
(λ), ε̂m,νm
(λ))C2
= (εn,νn
, εm,νm
)l2,0
= δn,mδνn,νm
.
(111)
It is convenient to present the Fourier transform (106) and orthogonality (111) in a
form which is a partial case of general definitions (17), (20) in the case H = C2. For this,
introduce ∀n ∈ N0 by Pn,νn
(α; λ) the matrices
Pn(λ) =(Pn;α,β(λ))1α,β=0, where Pn;α,β(λ) = Pn,α(β; λ), n ∈ N;
P0(λ) =[P0;0,0(λ) P0;0,1(λ)], where P0;0,0(λ) = P0,0(0; λ) = 1,
P0;0,1(λ) = P0,0(1; λ) = 0, i.e. P0(λ) = [1 0].
(112)
Then according to (106) ∀f ∈ l0,fin we have:
(113)
f̂(λ) = (f̂0(λ), f̂1(λ))
=
( ∞∑
n=0;νn=0,1
Pn,νn
(0; λ)fn,νn
,
∞∑
n=0;νn=0,1
Pn,νn
(1; λ)fn,νn
)
=
∞∑
n=0
P ∗
n(λ)fn.
Therefore, we get the formula (17) but only changing P ∗
0 (λ) from 1 to
[
1
0
]
.
The conditions (111) of orthogonality also have the form (20) for Pn(λ) of the kind
(112). Indeed, according to (112) and (111) we get ∀j, k ∈ N :
( ∫
R
Pj(λ)dρ(λ)P ∗
k (λ)
)
α,β
=
1∑
γ,ν=0
∫
R
Pj;α,γ(λ)dργ,ν(λ)Pk;β,ν (λ)
=
1∑
γ,ν=0
∫
R
Pj,α(γ; λ)dργ,ν(λ)Pk,β(ν; λ) = (εk,β , εj,α)l2,0
= δk,jδβ,α.
Therefore ∀j, k ∈ N
(114)
∫
R
Pj(λ)dρ(λ)P ∗
k (λ) = δj,k1.
Analogous calculation gives (114) for j = 0, k ∈ N and j ∈ N, k = 0. For j = k = 0
integral (114) is equal to
∫
R
dρ0,0(λ) = 1.
We can conclude that the Fourier transform (106) has the standard form (17) where
H = C2 and the operator-valued polynomials Pn(λ), n ∈ N0, are defined by (112). They
are orthonormal in the sense of (114) (for j = k = 0 unity 1 is the usual number).
Formulas of type (21) are also true in our case. Namely, we have for elements of the
matrix J (91):
(115)
a∗
n =
∫
R
λPn(λ)dρ(λ)P ∗
n+1(λ), bn =
∫
R
λPn(λ)dρ(λ)P ∗
n (λ),
n ∈ N0; an = a∗
n, n ∈ N.
Indeed, the Fourier transform of the operator J generated by J on the space l2,0
is the operator of multiplication by λ, this follows from (98). Using the orthogonality
conditions of (114) for j, k ∈ N0 we find the formulas (115) easily.
In the general case of the space l2 = ℓ2(H) = H ⊕H⊕ . . ., in Section 2 we constructed
operator-valued polynomials Pn(λ) by procedure of orthogonalization of sequence (23).
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 125
In our case of the space l2,0 = C1⊕C2⊕C2 . . . the situation is more complicated because
the first space C1 is different from the others C2 and the operator a0 is not invertible.
Now, unlike Sections 2, 5, it is more convenient to use the ordinary orthogonalization
in the Hilbert space L2 = L2(C2, R, dρ(λ)). Consider the corresponding procedure.
First, it is necessary to establish the structure of the polynomials Pn(λ), n ∈ N0,
constructed as solutions Pn,α(β; λ) of difference equations (92) with initial data (95).
So, from (92) for n = 0, 1 (the second and the fourth equalities), (95) and definition
(112) we find:
P0(λ) = [1 0],
P1(λ) =
[
P1,0(0; λ) P1,0(1; λ)
P1,1(0; λ) P1,1(1; λ)
]
=
[
0 1
1
α−1
(λ − β0) − α0
α−1
]
,
P2(λ) =
[
P2,0(0; λ) P2,0(1; λ)
P2,1(0; λ) P2,1(1; λ)
]
=
[
− α0
a1,0
1
a1,0
(−b1,0 + λ)
1
a1,1α−1
(−α2
−1 + (−b1,1 + λ)(−β0 + λ)) α0
a1,1α−1
(b1,1 − λ)
]
.
(116)
For n = 2, 3, . . . the last equality in (92) gives: ∀β = 0, 1
(an−1,0Pn−1,0(β; λ) + bn,0Pn,0(β; λ) + an,0Pn+1,0(β, λ), an−1,1Pn−1,1(β; λ)
+ bn,1Pn,1(β; λ) + an,1Pn+1,1(β; λ)) = (λPn,0(β; λ), λPn,1(β; λ)).
From this equality we conclude that
Pn+1(λ) =
[
Pn+1,0(0; λ) Pn+1,0(1; λ)
Pn+1,1(0; λ) Pn+1,1(1; λ)
]
=
[
1
an,0
(−an−1,0Pn−1,0(0; λ) − (bn,0 − λ)Pn,0(0; λ))
1
an,1
(−an−1,1Pn−1,1(0; λ) − (bn,1 − λ)Pn,1(0; λ))
1
an,0
(−an−1,0Pn−1,0(1; λ) − (bn,0 − λ)Pn,0(1; λ))
1
an,1
(−an−1,1Pn−1,1(1; λ) − (bn,1 − λ)Pn,1(1; λ))
]
, n = 2, 3, . . . .
(117)
Denote by en,0 = (1, 0), en,1 = (0, 1) the standard basis in the space Hn = C2, n ∈
N; e0,0 = e0 = 1 ∈ C1 = H0. The vectors
(118)
ε0,0 = ε0 = (e0, 0, . . .), εn,νn
= (0, . . . , 0, en,νn︸ ︷︷ ︸
n,νn place
, 0, 0, . . .) ∈ l2,0,
n ∈ N, νn ∈ {0, 1},
comprise an orthonormal basis in the space l2,0 (these vectors are the same as in (111)).
Note that the projectors πn,νn
: l2,0 → Hn used earlier are the projections onto εn,νn
.
We will use this basis here and in what follows.
The equalities (113) and (112) give a sequence of vectors (C2-valued functions of
λ ∈ R) from the Hilbert space L2 :
(ε̂0,0)(λ) = P ∗
0 (λ)e0,0 = (P0,0(0; λ), P0,0(1; λ)) = (1, 0),
(ε̂n,0)(λ) = P ∗
n(λ)en,0 = (Pn,0(0; λ), Pn,0(1; λ)),
(ε̂n,1)(λ) = P ∗
n(λ)en,1 = (Pn,1(0; λ), Pn,1(1; λ)), λ ∈ R, n ∈ N.
(119)
126 YURIJ BEREZANSKY
These vectors are orthonormal in L2. Indeed, using (114), (14) we get: ∀ej,α, ek,β where
j, k ∈ N0, α, β ∈ {0, 1}
δj,kδα,β = (δj,k1ek,β, ej,α)C2 =
( ∫
R
Pj(λ)dρ(λ)P ∗
k (λ)ek,β , ej,α
)
C2
=
∫
R
(
dρ(λ)P ∗
k (λ)ek,β , P ∗
j (λ)ej,α
)
C2
= (P ∗
k (λ)ek,β , P ∗
j (λ)ej,α)L2 .
It is easy to see from the formulas (116), (117) that the C2-valued functions (119) are
linear combinations of the following initial system of C2-valued functions:
(120) (1, 0); (0, 1), (λ, 0); (0, λ), (λ2, 0); . . . ; (0, λn−1), (λn, 0); . . . , λ ∈ R, n ∈ N.
More precisely, ∀n ∈ N the vector-valued function R ∋ λ �→ P ∗
n(λ)en,0 (R ∈ λ �→
P ∗
n(λ)en,1) is a linear combination of the first 2n (2n + 1) functions from the sequence
(120).
It is easy to understand that, similarly to Lemma 1, we have the following result.
Lemma 9. Let the support of the measure dρ(λ) have an infinite set of points, then the
functions (120) are linearly independent in the space L2.
Proof. As in the proof of Lemma 1 we construct the scalar measure dσ(λ) = dρ0,0(λ)+
dρ1,1(λ) and a nonnegative matrix C(λ) such that dρ(λ) = C(λ)dσ(λ), λ ∈ R; this matrix
is positive on the set of full measure dσ(λ), this measure also has infinite support.
Assume that some finite linear combination f(λ) of C2-valued functions (120) is equal
to zero in the space L2, i.e.
0 = ‖f(·)‖2
L2 =
∫
R
(dρ(λ)f(λ), f(λ))C2 =
∫
R
(C(λ)f(λ), f(λ))C2dσ(λ).
Then (C(λ)f(λ), f(λ))C2 = 0 for σ-almost all λ ∈ R. But C(λ) is invertible on the set
of full measure dσ(λ), therefore also f(λ) = (f0(λ), f1(λ)) = 0 for σ-almost all λ. The
functions f0(λ), f1(λ) are some ordinary polynomials, therefore their equality to zero
means that all their coefficients also equal to zero (since dσ(λ) has infinite support). In
other words, the functions (120) are linearly independent. �
This lemma allows us to apply the procedure of orthogonalization to the sequence
(120). The result is unique and gives (119).
Results obtained in this Section can be formulated as the following theorem.
Theorem 3. Consider a Jacobi matrix J of the form (91) with conditions on its entries
formulated at the beginning of Section 7. This matrix gives rise to a selfadjoint bounded
operator J on the space l2,0 (59).
The direct spectral problem for J consists of the following. The generalized eigenvectors
of operator J are equal to solutions P (α; λ) = (Pn,νn
(α; λ))∞νn=0,1;n=0, α = 0, 1, (95) of
difference equations (92) with the initial data
(121) P0,0(0; λ) = 1, P1,0(0; λ) = 0; P0,0(1; λ) = 0, P1,0(1; λ) = 1.
These eigenvectors compose a full system: if we introduce ∀f ∈ l0,fin the Fourier
transform (106)
(122) f̂(λ) = (f̂0(λ), f̂1(λ)) ∈ C
2, f̂α(λ) =
∞∑
n=0;νn=0,1
Pn,νn
(α; λ)fn,νn
, α = 0, 1,
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 127
then we have the Parseval equality (110)
(123) (f, g)l2,0
=
∫
R
(dρ(λ)f̂ (λ), ĝ(λ))C2 .
Here dρ(λ) = (dρα,β(λ))1α,β=0 is a 2 × 2-matrix nonnegative spectral measure of the
operator J ; it is a probability measure: ρ(R) = 1. The mapping l2,0 ∋ f �→ f̂(λ) after its
extension by closure is a unitary operator between the spaces l2,0 and L2(C2, R, dρ(λ)).
The inverse spectral problem for J consists of the following. Assume we are given
some 2 × 2-matrix nonnegative Borel probability measure dρ(λ) on R for which all the
functions R ∋ λ �→ (λn, λn) ∈ C2, n ∈ N, are integrable and support of dρ(λ) has an
infinite set of points.
Then such a measure is the spectral measure of some operator J generated on the
space l2,0 by Jacobi matrix J (91) of the indicated kind.
The elements of this matrix J are constructed by the following procedure. Consider in
the space L2(C2, R, dρ(λ)) a sequence of vectors (i.e. C2-valued functions)
(124) R ∋ λ �→ (1, 0), (0, λn−1), (λn, 0), n ∈ N.
Apply to these vectors the classical Gram-Schmidt procedure of orthonormalization in the
order stated in (124).
As a result, we get the sequence of C2-valued polynomials
R ∋ λ �→ (1, 0) =(P0,0(0; λ), P0,0(1; λ)), (Pn,0(0; λ), Pn,0(1; λ)),
(Pn,1(0; λ), Pn,1(1; λ)), n ∈ N.
(125)
Using these polynomials we construct the 2 × 2-matrix valued polynomials
(126) P0(λ) = [1 0], Pn(λ) =
[
Pn,0(0; λ) Pn,0(1; λ)
Pn,1(0; λ) Pn,1(1; λ)
]
, n ∈ N.
Then the elements of the matrix J are reconstructed by formulas (115).
If we started from the Jacobi matrix J (91) and the 2 × 2-matrix spectral measure
dρ(λ) of the corresponding operator J then the procedure above gives entries of our initial
matrix J.
We finish this Section with some simple facts.
Let B(R) ∋△ �→ E(△) be a resolution of identity of the operator J , then it is possible
to write
(127) E(△) =
∫
△
Φ(λ)dσ(λ)
(understanding E(△) as an operator from l2,0(p) into l2,0(p
−1)). Using (118), (127) and
(100) we conclude: ∀ △∈ B(R)
(128)
(E(△)εk,νk
, εj,νj
)l2,0
= (E(△)πk,νk
εk,νk
, πj,νj
εj,νj
)l2,0
= (πj,νj
E(△)πk,νk
εk,νk
, εj,νj
)l2,0
=
( ∫
△
Φj,νj ;k,νk
(λ)dσ(λ)εk,νk
, εj,νj
)
l2,0
=
∫
△
Φj,νj ;k,νk
(λ)dσ(λ), j, k ∈ N0; νj , νk ∈ {0, 1}.
In particular, from (128) for k, νk and j, νj = 0, 0; 1, 0 and (108) we get: ∀ △∈ B(R)
(E(△)εα,0, εβ,0)l2,0
=
∫
△
Φβ,0;α,0(λ)dσ(λ) = ρβ,α(△), α, β = 0, 1.
128 YURIJ BEREZANSKY
But (E(△)εα,0, εβ,0)l2,0
= (E(△)εβ,0, εα,0)l2,0
and the function Φj,νj ;k,νk
(λ) is real. There-
fore we get an essential identity for the nonnegative and real spectral measure (108):
(129) ρα,β(△) = (E(△)εα,0, εβ,0)l2,0
= ρβ,α(△), △∈ B(R), α, β = 0, 1; ρ(R) = 1.
Analogously to (52), we introduce for our problem the matrix Weyl function
(130) m(z) =
∫
R
1
λ − z
dρ(λ) =
[
(Rzε0,0, ε0,0)l2,0
(Rzε0,0, ε1,0)l2,0
(Rzε1,0, ε0,0)l2,0
(Rzε1,0, ε1,0)l2,0
]
= (mα,β(z))1α,β=0,
where Rz is the resolvent of operator J . The function m(z) defines the measure dρ(λ)
uniquely; from (129) it follows that the matrix m(z) is symmetric.
8. The equations for the matrix Weyl function and the spectral matrix
Every bounded operator A : l2,0 → l2,0 can be written as a matrix in the basis (118),
namely ∀f ∈ l2,0 we have:
f = (fn)∞n=0 = (fn,νn
)∞n=0,νn=0,1, fn,νn
= (f, εn,νn
)l2,0
;
∀j ∈ N0, νj ∈ {0, 1}, (Af)j,νj
=
∞∑
k=0,νk=0,1
aj,νj ;k,νk
fk,νk
,
aj,νj ;k,νk
= (Aεk,νk
, εj,νj
)l2,0
, here 0, ν0 = 0, 0.
(131)
The matrices aj;k = aj,νj ;k,νk
from (131) act in the following way:
aj;k =
[
aj,0;k,0 aj,0;k,1
aj,1;k,0 aj,1;k,1
]
: C
2 → C
2, a0;k =
[
a0,0;k,0
a0,1;k,0
]
: C
1 → C
2,
aj;0 =
[
aj,0;0,0 aj,0;0,1
]
: C
2 → C
1, j, k ∈ N; a0;0 = [a0,0;0,0] : C
1 → C
1.
(132)
Using these matrices, we can rewrite (131) in the form:
(133) (Af)j =
∞∑
k=0
aj;kfk, j ∈ N0.
The general Jacobi matrix B of type (6), connected with the space l2,0 instead of
ℓ2(H), acts on f = (fn)∞n=0, fn ∈ Hn, by
(Bf)n = an−1fn−1 + bnfn + cnfn+1, n ∈ N; (Bf)0 = b0f0 + c0f1;
an, bn, cn : C
2 → C
2, n ∈ N; a0 : C
1 → C
2, b0 : C
1 → C
1, c0 : C
2 → C
1.
(134)
Using (134) it is easy to calculate that for vectors (118) we have:
Bεn,νn
= (0, . . . , 0︸ ︷︷ ︸
n−1
, cn−1en,νn
, bnen,νn
, anen,νn
0, 0, . . .), n ∈ N, νn ∈ {0, 1},
Bε0,0 = (b0e0,0, a0e0,0, 0, 0, . . .).
(135)
Consider the Lax equation (88). As in (44) for resolvent Rz(t) of operator J(t) we
get
(136) Ṙz(t) = Rz(t)A(t) − A(t)Rz(t), z ∈ C \ R, t ∈ [0, T ).
Introduce, in agreement with (131), (132), (133), a matrix of the operator Rz(t) in the
basis (118): its entries are equal to
Rz;j,νj ;k,νk
(t) = (Rz(t)εk,νk
, εj,νj
)l2,0
, j, k ∈ N0, νj , νk ∈ {0, 1};
(Rz(t)f)j =
∞∑
k=0
Rz;j;k(t)fk, j ∈ N0.
(137)
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 129
Using these entries, we construct, according to (130), the matrix Weyl function m(z; t) =
(mα,β(z; t))1α,β=0, z ∈ C \ R, t ∈ [0, T ). Our next goal is to deduce a differential equa-
tion for m(z; t) when the resolvent Rz(t) satisfies the Lax equality (136). The proof is
analogous to the case (53) but more complicated since our operators act on the space
l2,0 = C1 ⊕ C2 ⊕ C2 ⊕ . . . instead of H ⊕ H ⊕ . . . and the matrix a0 : C1 → C2 is not
invertible.
Elements of the matrix m(z; t) have the form (see (130))
m0,0(z; t) = (Rz(t)ε0,0, ε0,0)l2,0
= (Rz̄(t)ε0,0, ε0,0)l2,0
= m0,0(z̄; t),
m0,1(z; t) = (Rz(t)ε0,0, ε1,0)l2,0
= (Rz̄(t)ε1,0, ε0,0)l2,0
= m1,0(z̄; t) = m1,0(z; t),
m1,1(z; t) = (Rz(t)ε1,0, ε1,0)l2,0
= (Rz̄(t)ε1,0, ε1,0)l2,0
= m1,1(z̄; t),
z ∈ C \ R, t ∈ [0, T ).
(138)
Consider m0,0(z; t). Using (136) we get:
(139)
ṁ0,0(z; t) = (Ṙz(t)ε0,0, ε0,0)l2,0
= (Rz(t)A(t)ε0,0, ε0,0)l2,0
− (A(t)Rz(t)ε0,0, ε0,0)l2,0
.
The second formula in (135), (137) and (89) gives (below an, bn, cn, where n ∈ N0, depend
on t):
A(t)ε0,0 = (0, ã01, 0, 0, . . .) = (0,
1
2
(α0,−α−1), 0, 0, . . .),
(Rz(t)A(t)ε0,0)
∞
j=0 = (Rz;j;1(t)
1
2
(α0,−α−1))
∞
j=0; in particular
(Rz(t)A(t)ε0,0, ε0,0)l2,0
= (Rz(t)A(t)ε0,0)0,0 =
1
2
Rz;0,0;1,0(t)α0
−
1
2
Rz;0,0;1,1(t)α−1 =
1
2
α0(Rz(t)ε1,0, ε0,0)l2,0
−
1
2
α−1(Rz(t)ε1,1, ε0,0)l2,0
.
(140)
Analogously using the first formula from (135) for n = 1 and (89) we get:
Rz(t)ε0,0 = ((Rz(t)ε0,0)0, (Rz(t)ε0,0)1, . . .) ,
A(t)Rz(t)ε0,0 = (c̃0(Rz(t)ε0,0)1, ã0(Rz(t)ε0,0)0 + c̃1(Rz(t)ε0,0)2, . . .)
=
(
1
2
[−α0 α−1](Rz(t)ε0,0)1, ã0(Rz(t)ε0,0)0 + c̃1(Rz(t)ε0,0)2, . . .
)
;
(A(t)Rz(t)ε0,0, ε0,0)l2,0
= (A(t)Rz(t)ε0,0)0,0 =
1
2
[−α0 α−1](Rz(t)ε0,0)1
= −
1
2
α0(Rz(t)ε0,0)1,0 +
1
2
α−1(Rz(t)ε0,0)1,1 = −
1
2
α0(Rz(t)ε0,0, ε1,0)l2,0
+
1
2
α−1(Rz(t)ε0,0, ε1,1)l2,0
.
(141)
From (139), (140), (141) and (138) we conclude that
ṁ0,0(z; t) =α0m0,1(z; t) −
1
2
α−1(Rz(t)ε0,0, ε1,1)l2,0
−
1
2
α−1(Rz(t)ε1,1, ε0,0)l2,0
.(142)
Consider m0,1(z; t). Analogously to (139)–(141) we have:
ṁ0,1(z; t) = (Ṙz(t)ε0,0, ε1,0)l2,0
= (Rz(t)A(t)ε0,0, ε1,0)l2,0
− (A(t)Rz(t)ε0,0, ε1,0)l2,0
,
(Rz(t)A(t)ε0,0, ε1,0)l2,0
= (Rz(t)A(t)ε0,0)1,0 =
1
2
Rz;1,0;1,0(t)α0
−
1
2
Rz;1,0;1,1(t)α−1 =
1
2
α0(Rz(t)ε1,0, ε1,0)l2,0
−
1
2
α−1(Rz(t)ε1,1, ε1,0)l2,0
.
(143)
130 YURIJ BEREZANSKY
Calculate
(144)
(A(t)Rz(t)ε0,0, ε1,0)l2,0
= (A(t)Rz(t)ε0,0)1,0 = π1,0(ã0(Rz(t)ε0,0)0 + c̃1(Rz(t)ε0,0)2).
But according to (89) ã0 = − 1
2qa0 = 1
2
[
α0
−α−1
]
, c̃1 = 1
2qa1 = 1
2
[
−α1 0
0 α−2
]
, therefore
π1,0ã0 = 1
2
[
α0
0
]
, π1,0c̃1 = 1
2
[
−α1
0
]
and (144) gives
(145) (A(t)Rz(t)ε0,0, ε1,0)l2,0
=
1
2
α0(Rz(t)ε0,0, ε0,0)l2,0
−
1
2
α1(Rz(t)ε0,0, ε2,0)l2,0
.
Using (143), (144), the last equality, and (138) we find:
ṁ0,1(z; t) = −
1
2
α0m0,0(z; t) +
1
2
α0m1,1(z; t) −
1
2
α−1(Rz(t)ε1,1, ε1,0)l2,0
+
1
2
α1(Rz(t)ε0,0, ε2,0)l2,0
.
(146)
Consider m1,1(z; t). We have as before
(147)
ṁ1,1(z; t) = (Ṙz(t)ε1,0, ε1,0)l2,0
= (Rz(t)A(t)ε1,0, ε1,0)l2,0
− (A(t)Rz(t)ε1,0, ε1,0)l2,0
.
Now according to the first formula in (135) for n = 1 and (89) we have
A(t)ε1,0 = (c̃0e1,0, 0, ã1e1,0, 0, 0, . . .)
=
(
1
2
[−α0 α−1]e1,0, 0,
1
2
[
α1 0
0 −α−2
]
e1,0, 0, 0, . . .
)
=
(
−
1
2
α0, 0,
1
2
(α1, 0), 0, 0, . . .
)
.
Analogously to (140) we get:
(Rz(t)A(t)ε1,0)
∞
j=0 = (Rz;j;0(t)(−
1
2
α0) + Rz;j;2(t)
1
2
(α1, 0))∞j=0;
(Rz(t)A(t)ε1,0, ε1,0)l2,0
= Rz;1,0;0,0(t)(−
1
2
α0) + Rz;1,0;2,0(t)
1
2
α1
= −
1
2
α0(Rz(t)ε0,0, ε1,0)l2,0
+
1
2
α1(Rz(t)ε2,0, ε1,0)l2,0
.
(148)
For the second term in (147) we get analogously to (141):
Rz(t)ε1,0 = ((Rz(t)ε1,0)0, (Rz(t)ε1,0)1, . . .),
A(t)Rz(t)ε1,0 = (c̃0(Rz(t)ε1,0)1, ã0(Rz(t)ε1,0)0 + c̃1(Rz(t)ε1,0)2, . . .)
= (
1
2
[−α0 α−1](Rz(t)ε1,0)1, ã0(Rz(t)ε1,0)0 + c̃1(Rz(t)ε1,0)2, . . .).
Calculate
(A(t)Rz(t)ε1,0, ε1,0)l2,0
= (A(t)Rz(t)ε1,0)1,0 = π1,0(ã0(Rz(t)ε1,0)0 + c̃1(Rz(t)ε1,0)2).
As is the case (144), (145) we see that the last term is equal to 1
2α0(Rz(t)ε1,0, ε0,0)l2,0
−
1
2α1(Rz(t)ε1,0, ε2,0)l2,0
. Thus, we calculated (A(t)Rz(t)ε1,0, ε1,0)l2,0
. Using (147), (148),
this equality and (138) we have:
ṁ1,1(z; t) = − α0m0,1(z; t) +
1
2
α1(Rzε1,0, ε2,0)l2,0
+
1
2
α1(Rzε2,0, ε1,0)l2,0
.(149)
Every equality (142), (146) and (149), where z ∈ C \ R, t ∈ [0, T ), contains two last
summands which do not contain the two functions from (138). We can find a system of
differential equation for functions (138) in this case if we can express these summands by
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 131
means of functions (138). Such expression can be obtained similarly to (49), using the
obvious equality
(150) 1 + zRz(t) = Rz(t)J(t), z ∈ C \ R, t ∈ [0, T ).
So, from (150), the second formula in (135), (89) and (137)
1 + zm0,0(z; t) = ((1 + zRz(t))ε0,0, ε0,0)l2,0
= (Rz(t)J(t)ε0,0, ε0,0)l2,0
,
J(t)ε0,0 = (b0e0,0, a0e0,0, 0, 0, . . .);
(Rz(t)J(t)ε0,0)
∞
j=0 = (Rz;j;0(t)b01 + Rz;j;1(t)(α0, α−1))
∞
j=0,
(Rz(t)J(t)ε0,0, ε0,0)l2,0
= (Rz(t)J(t)ε0,0)0,0 = Rz;0,0;0,0(t)β0 + Rz;0,0;1,0(t)α0
+ Rz;0,0;1,1(t)α−1 = β0(Rz(t)ε0,0, ε0,0)l2,0
+ α0(Rz(t)ε1,0, ε0,0)l2,0
+ α−1(Rz(t)ε1,1, ε0,0)l2,0
= β0m0,0(z; t) + α0m1,0(z; t) + α−1(Rz(t)ε1,1, ε0,0)l2,0
.
(151)
From (151) and (138) we conclude: ∀z ∈ C \ R, t ∈ [0, T )
(152) α−1(Rz(t)ε1,1, ε0,0)l2,0
= 1 + zm0,0(z; t) − β0m0,0(z; t) − α0m0,1(z; t).
Analogously to (151) for m0,1(z; t) :
zm0,1(z; t) = ((1 + zRz(t))ε0,0, ε1,0)l2,0
= (Rz(t)J(t)ε0,0, ε1,0)l2,0
,
(Rz(t)J(t)ε0,0, ε1,0)l2,0
= (Rz(t)J(t)ε0,0)1,0
= Rz;1,0;0,0(t)β0 + Rz;1,0;1,0(t)α0 + Rz;1,0;1,1(t)α−1
= β0(Rz(t)ε0,0, ε1,0)l2,0
+ α0(Rz(t)ε1,0, ε1,0)l2,0
+ α−1(Rz(t)ε1,1, ε1,0)l2,0
= β0m0,1(z; t) + α0m1,1(z; t) + α−1(Rz(t)ε1,1, ε1,0)l2,0
.
As a result, we get ∀z ∈ C \ R, t ∈ [0, T )
(153) α−1(Rz(t)ε1,1, ε1,0)l2,0
= zm0,1(z; t) − β0m0,1(z; t) − α0m1,1(z; t).
It is convenient to perform similar calculations for m1,0(z; t) instead of the third equa-
lity in (138). So, we have
(154) zm1,0(z; t) = ((1 + zRz(t))ε1,0, ε0,0)l2,0
= (Rz(t)J(t)ε1,0, ε0,0)l2,0
.
Analogously to calculations in (151), (152) we get:
J(t)ε1,0 = (c0e1,0, b1e1,0, a1e1,0, 0, 0, . . .) = ([α0 α−1]e1,0, b1e1,0, α1e1,0)
= (α0, (β1, 0), (α1, 0), 0, 0, . . .);
(Rz(t)J(t)ε1,0)
∞
j=0 = (Rz;j;0(t)α0 + Rz;j;1(t)(β1, 0) + Rz;j;2(t)(α1, 0))∞j=0;
(Rz(t)J(t)ε1,0, ε0,0)l2,0
= Rz;0,0;0,0(t)α0 + Rz;0,0;1,0(t)β1 + Rz;0,0;2,0(t)α1
= α0(Rz(t)ε0,0, ε0,0)l2,0
+ β1(Rz(t)ε1,0, ε0,0)l2,0
+ α1(Rz(t)ε2,0, ε0,0)l2,0
= α0m0,0(z; t) + β1m1,0(z; t) + α1(Rz(t)ε2,0, ε0,0)l2,0
.
These calculations and (154), (138) give: ∀z ∈ C \ R, t ∈ [0, T )
(155) α1(Rz(t)ε2,0, ε0,0)l2,0
= zm0,1(z; t) − α0m0,0(z; t) − β1m0,1(z; t).
Finally, for the function m1,1(z; t) we have analogously to (154), (155)
1 + zm1,1(z; t) = ((1 + zRz(t))ε1,0, ε1,0)l2,0
= (Rz(t)J(t)ε1,0, ε1,0)l2,0
,
(Rz(t)J(t)ε1,0, ε1,0)l2,0
= Rz;1,0;0,0(t)α0 + Rz;1,0;1,0(t)β1 + Rz;1,0;2,0(t)α1
= α0(Rz(t)ε0,0, ε1,0)l2,0
+ β1(Rz(t)ε1,0, ε1,0)l2,0
+ α1(Rz(t)ε2,0, ε1,0)l2,0
= α0m0,1(z; t) + β1m1,1(z; t) + α1(Rz(t)ε2,0, ε1,0)l2,0
.
132 YURIJ BEREZANSKY
So, we get ∀z ∈ C \ R, t ∈ [0, T )
(156) α1(Rz(t)ε2,0, ε1,0)l2,0
= 1 + zm1,1(z; t) − α0m0,1(z; t) − β1m1,1(z; t).
It is necessary also to add to the equalities (152), (153), (155), (156) their simple
modifications (see (138)):
α−1(Rz(t)ε0,0, ε1,1)l2,0
= α−1(Rz̄(t)ε1,1, ε0,0)l2,0
= 1 + zm0,0(z; t) − α0m0,1(z; t) − β0m0,0(z; t),
α1(Rz(t)ε0,0, ε2,0)l2,0
= α1(Rz̄(t)ε2,0, ε0,0)l2,0
= zm0,1(z; t) − α0m0,0(z; t) − β1m0,1(z; t),
α1(Rz(t)ε1,0, ε2,0)l2,0
= α1(Rz̄(t)ε2,0, ε1,0)l2,0
= 1 + zm1,1(z; t) − α0m0,1(z; t) − β1m1,1(z; t), z ∈ C \ R, t ∈ [0, T ).
(157)
Substituting (152), (153), (156), (157) into (142), (146), (149) we get:
ṁ0,0(z; t) = (β0(t) − z)m0,0(z; t) + 2α0(t)m0,1(z; t) − 1,
ṁ0,1(z; t) = −α0(t)m0,0(z; t) +
1
2
(β0(t) − β1(t))m0,1(z; t) + α0(t)m1,1(z; t),
ṁ1,1(z; t) = −2α0(t)m0,1(z; t) − (β1(t) − z)m1,1(z; t) + 1,
m1,0(z; t) = m0,1(z; t), z ∈ C \ R, t ∈ [0, T ).
(158)
As a result, we can now formulate the following assertion.
Lemma 10. The entries (138) of the matrix Weyl function (mα,β(z; t))1α,β=0 with respect
to t ∈ [0, T ) satisfy the linear system of differential equations (158).
Rewrite the equations (158) in the form of equations for elements of the corresponding
spectral matrix (dρα,β(λ; t))1α,β=0 of the operator J(t), t ∈ [0, T ). Recall that it is given
by formulas (108), (129) and connected with the matrix Weyl function by (130).
Similar to [5, 12] and Section 5 we can write using (130), (138), (158): for a fixed
t ∈ [0, T ) and z ∈ C \ R
(159)
∫
R
1
λ − z
dρ̇0,0(λ; t) = ṁ0,0(z; t)
= (β0(t) − z)m0,0(z; t) + 2α0(t)m0,1(z; t) − 1
=
∫
R
β0(t) − λ
λ − z
dρ0,0(λ; t) +
∫
R
2α0(t)
λ − z
dρ0,1(λ; t).
Here we have used that ρ(R; t) = 1 (see (129)) and therefore ρ0,0(R; t) = 1.
From (159) we conclude due to arbitrariness of z that
(160) ρ̇0,0(λ; t) = (β0(t) − λ)ρ0,0(λ; t) + 2α0(t)ρ0,1(λ; t)
for almost all λ ∈ R w.r.t. the scalar spectral measure dσ(λ; t) of the operator J(t) and
the functions ρ0,0(λ; t), ρ0,1(λ; t) are continuously differentiable in t.
Analogously to (159), (160) we can write all other equations for ρα,β(α; t) which are
equivalent to the differential equations (158). As a result, we have the following system
for three unknowns: ρ0,0(λ; t), ρ0,1(λ; t), ρ1,1(λ; t) :
ρ̇0,0(λ; t) = (β0(t) − λ)ρ0,0(λ; t) + 2α0(t)ρ0,1(λ; t),
ρ̇0,1(λ; t) = −α0(t)ρ0,0(λ; t) +
1
2
(β0(t) − β1(t))ρ0,1(λ; t) + α0(t)ρ1,1(λ; t),
ρ̇1,1(λ; t) = −2α0(t)ρ0,1(λ; t) − (β1(t) − λ)ρ1,1(λ; t);
ρ1,0(λ; t) = ρ0,1(λ; t), λ ∈ R, t ∈ [0, T ).
(161)
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 133
It is evident that the system (161) gives (158). So, we get the following result.
Lemma 11. The Lax equation (88) leads to the system (161) for elements of the spectral
matrix (ρα,β(λ; t))1α,β=0 of the operator J(t).
According to Theorem 2, the Lax equation (88) is equivalent to the Toda lattice
(28). Therefore from the first equality in (28) we get:
(162)
1
2
(β0(t) − β1(t)) = −
α̇0(t)
α0(t)
, β1(t) = β0(t) + 2
α̇0(t)
α0(t)
, t ∈ [0, T ).
The system (161) now has the form:
ρ̇0,0(λ; t) = (β0(t) − λ)ρ0,0(λ; t) + 2α0(t)ρ0,1(λ; t),
ρ̇0,1(λ; t) = −α0(t)ρ0,0(λ; t) −
α̇0(t)
α0(t)
ρ0,1(λ; t) + α0(t)ρ1,1(λ; t),
ρ̇1,1(λ; t) = −2α0(t)ρ0,1(λ; t) −
(
β0(t) − λ + 2
α̇0(t)
α0(t)
)
ρ1,1(λ; t);
ρ1,0(λ; t) = ρ0,1(λ; t), λ ∈ R, t ∈ [0, T ).
(163)
As a result, we have proved the following fact.
Theorem 4. Consider the Toda lattice (28) under the assumption that αn(t) > 0 and
uniformly boundedness of αn(t), βn(t), t ∈ [0, T ), T ≤ ∞, n ∈ Z. This lattice is equivalent
to the Lax equation (88) where J(t), A(t) are given by matrices of the form (6) with
elements (89) acting on the space l2,0 (59).
Operator J(t) is a bounded selfadjoint operator on the space l2,0. Its 2 × 2-matrix
valued nonnegative real and the probability spectral measure dρ(λ; t) = (dρα,β(λ; t))1α,β=0
with continuously differentiable in t entries ρα,β(λ; t) changes in t according to the system
(163) of differential equations.
For system (163) it is possible to consider the Cauchy problem: given the initial data
dρ(λ; 0), to find the solution dρ(λ; t), t ∈ [0, T ). This problem is solvable.
9. The investigation of system of differential equations for the spectral
matrix and the main theorem
Our first goal is to obtain some properties of solutions of system (163). Unknowns of
this system form a real matrix. Also important is the determinant of this matrix:
ρ(λ; t) =
[
ρ0,0(λ; t) ρ0,1(λ; t)
ρ0,1(λ; t) ρ1,1(λ; t)
]
, λ ∈ R, t ∈ [0, T );
r(λ; t) = Det ρ(λ; t).
(164)
The initial data ρ(λ; 0) in (163) is generated by the spectral measure dρ(λ; 0) of
the bounded selfadjoint operator J(0) : its increment on △∈ B(R) is equal to ρ(△
; 0), ρ(−∞; 0) = 0. Therefore this matrix-valued function R ∋ λ �→ ρ(α; 0) is equal to
zero for λ ∈ (−∞, a] and 1 for λ ∈ (b,∞) where [a, b] is the minimal closed interval
containing the spectrum of J(0). Its values on [a, b) are nonnegative real matrices, the
function [a, b) ∋ λ �→ ρ(λ; 0) is nondecreasing continuous from the left. The correspond-
ing scalar function r(λ; t) has analogous properties and changes from 0 to 1.
It is possible to simplify the system (163), namely the following result takes place:
Lemma 12. For the determinant r(λ; t) introduced above the following relations holds:
(165) r(λ; t) =
α2
0(0)
α2
0(t)
r(λ; 0), λ ∈ R, t ∈ [0, T ).
134 YURIJ BEREZANSKY
Proof. Using the equation (163) we get ∀λ ∈ R and ∀t ∈ [0, T )
ṙ(λ; t) = ρ̇0,0(λ; t)ρ1,1(λ; t) + ρ0,0(λ; t)ρ̇1,1(λ; t) − 2ρ0,1(λ; t)ρ̇0,1(λ; t)
= −2
α̇0(t)
α0(t)
r(λ; t).
As a result, we have a simple differential equation for r(λ; t) . Its integration gives:
log r(λ; t) = log r(λ; 0) − 2
t∫
0
α̇0(τ)
α0(τ)
dτ = log
(
r(λ; 0)
α2
0(0)
α2
0(t)
)
.
This equality is equivalent to (165). �
Let ρ(λ; 0) be the spectral measure of the operator J(0), then ρ(λ; 0) = 1 for λ > b,
therefore the corresponding r(λ; 0) also is equal to 1. For an arbitrary fixed t > 0 and
for a sufficiently large λ ρ(λ; t) = 1 (the operator J(t) is bounded) and r(λ; t) = 1 also.
For such λ the equality (165) gives that α0(t) = α0(0). Hence we have proved that
(166) α0(t) = α0(0), t ∈ [0, T ).
Now we can rewrite the system (163) in the form:
ρ̇0,0(λ; t) = (β0(t) − λ)ρ0,0(λ; t) + 2κρ0,1(λ; t),
ρ̇0,1(λ; t) = −κρ0,0(λ; t) + κρ1,1(λ; t),
ρ̇1,1(λ; t) = −2κρ0,1(λ; t) − (β0(t) − λ)(ρ1,1(λ; t)),
κ = α0(0), λ ∈ R, t ∈ [0, T ).
(167)
So, we can now formulate the following main result.
Theorem 5. Consider the double-infinite Toda lattice (28) and pose the following prob-
lem: find a solution αn(t), βn(t), t ∈ [0, T ), T ≤ ∞, n ∈ Z, of (28) which satisfies the
following conditions:
(168) αn(0), βn(0), n ∈ Z, and β0(t), t ∈ [0, T ),
are given. Here αn(t) > 0, βn(t) are real continuously differentiable uniformly (w.r.t. n)
bounded functions. So, we have the Cauchy problem for (28) with prescribed values of
β0(t).
This solution can be found via the following procedure:
1) Consider the Jacobi matrix J(0) of type (91) with elements an(0), bn(0), n ∈ N0,
constructed according to (89). Find its 2 × 2-matrix spectral measure dρ(λ; 0).
2) Find the evolution dρ(λ; t) = (dρα,β(λ; t))1α,β=0 of this measure. For this it is
necessary to find a solution of the Cauchy problem for system (167) with the initial data
dρ(λ; 0).
3) We know dρ(λ; t) for every fixed t ∈ [0, T ). Consider the Hilbert space
L2(C2, R, dρ(λ; t))
and in this space the sequence of C2-valued functions (124). Apply to these vectors the
classical procedure of orthonormalization, as a result we get C2-valued polynomials (125)
(that depend on t). Using these polynomials we form the 2×2-matrix valued polynomials
(126) Pn(λ; t), n ∈ N0.
4) The required solution αn(t), βn(t), t ∈ [0, T ), of our initial problem is given by
formulas (115) with the obtained dρ(λ; t), Pn(λ; t); transferring from the calculated a∗
n(t),
INTEGRATION OF DOUBLE-INFINITE TODA LATTICE 135
bn(t) to αn(t), βn(t) is given by the formulas (89):
[
αn(t) 0
0 α−n−1(t)
]
= an(t) = a∗
n(t),
[
βn(t) 0
0 β−n(t)
]
= bn(t), n ∈ N;
[
α0(t)
α−1(t)
]
= a0(t), [β0(t)] = b0(t), t ∈ [0, T ).
The proof of this theorem follows from the results of Sections 6–8. It should be
emphasized that we assumed that solutions of our problem (28), (168), exists.
Acknowledgments. The author is very grateful to M. I. Gekhtman and I. Ya. Ivasiuk
for essential remarks and help.
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E-mail address: berezan@mathber.carrier.kiev.ua
Received 01/10/2008
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