DIFFERENTIAL GEOMETRY AND APPLICATIONS Proc. Conf., Aug. 28 – Sept. 1, 1995, Brno, Czech Republic Masaryk University, Brno 1996, 455–462

ON AN EXTENSION OF THE

3–DIMENSIONAL TODA LATTICE

Mircea Puta

Abstract. An extension of the 3-dimensional nonperiodic Toda lattice is proposed and some of its geometrical and dynamical properties are pointed out.

After Flaschka transformation the dynamics of the 3-dimensional nonperiodic Toda lattice is given by the following system of differential equations:

(1.1)  \{ \begin{matrix}ȧ_{1} & = 2b_{1}^{2} \\ ȧ_{2} & = 2(b_{2}^{2} − b_{1}^{2}) \\ ȧ_{3} & = −2b_{2}^{2} \\ ḃ_{1} & = b_{1}(a_{2} − a_{1}) \\ ḃ_{2} & = b_{2}(a_{3} − a_{2}) \\ b_{1} & > 0 \\ b_{2} & > 0 \end{matrix}

If we drop the last two conditions we obtain a new system of differential equations which is a slight extension of the 3-dimensional nonperiodic Toda lattice, namely:

(1.2)  \{ \begin{matrix}ȧ_{1} & = 2b_{1}^{2} \\ ȧ_{2} & = 2(b_{2}^{2} − b_{1}^{2}) \\ ȧ_{3} & = −2b_{2}^{2} \\ ḃ_{1} & = b_{1}(a_{2} − a_{1}) \\ ḃ_{2} & = b_{2}(a_{3} − a_{2}) \end{matrix}

The goal of this paper is to study some of its geometrical and dynamical properties.

1991 Mathematics Subject Classification. 58F05.

Key words and phrases. Toda lattice, Poisson manifolds, Lyapunov stability, energy-Casimir method, Lax representation.

This paper is in final form and no version of it will be submitted for publication elsewhere.

2. Lax formulation

Let us take now

L = [ \begin{matrix}a_{1} & b_{1} & 0 \\ b_{1} & a_{2} & b_{2} \\ 0 & b_{2} & a_{3} \end{matrix} ] and B = [ \begin{matrix}0 & −b_{1} & 0 \\ b_{1} & 0 & −b_{2} \\ 0 & b_{2} & 0 \end{matrix} ]

Then we can prove: Theorem 2.1. The equations (1.2) have a Lax formulation, i.e. they are equivalent to

(2.1) L̇ = [L, B]

where

[L, B] = LB − BL.

The proof is an easy computation and we shall not give any detail. Remark 2.1. The equation (3.1) always has some constants of motion. Indeed, L̇ = [L, B] implies (L^{2})˙ = [L^{2}, B] and similarly for L^{k}. Since the trace of a matrix commutator [B, N] is always zero,

(trace L^{k})˙ = trace(L^{k})˙ = trace [L^{k}, B] = 0

Hence, the trace(L^{k}) are constant along the trajectories of (2.1).

Further, we can prove: Theorem 2.2 (Flaschka). The flow (1.2) is isospectral, i.e. the eigenvalues of L are independent of t. Proof. Let V (t) be the solution of the matrix equation:

(2.2)  \{ \begin{matrix}\frac{dV (t)}{dt} = −B(t) · V (t) \\ V (0) = I_{3} \end{matrix}

Then one easily check that

L(t) = V (t) · L(0) · V (t)^{−1}.

Indeed, we have successively:

\frac{d}{dt}[V^{−1}(t)L(t)V (t)] = \frac{dV^{−1}(t)}{dt}L(t)V (t) + V^{−1}(t)\frac{d}{dt}[L(t)V (t)]

= V^{−1}(t)B(t)L(t)V (t) + V^{−1}(t)\frac{dL(t)}{dt}V (t)

+ V^{−1}(t)L(t)\frac{dV (t)}{dt}

= V^{−1}(t)B(t)L(t)V (t) + V^{−1}(t)[L(t), B(t)]V (t)

+ V^{−1}(t)L(t)[−B(t) · V (t)]

= V^{−1}(t)B(t)L(t)V (t) + V^{−1}(t)L(t)B(t)V (t)

− V^{−1}(t)B(t)L(t)V (t) − V^{−1}(t)L(t)B(t)V (t)

= 0

Hence

V^{−1}(t)L(t)V (t) = L(0)

or equivalent

L(t) = V (t)L(0)V (t)^{−1}

as required. Now it is clear that the eigenvalues of L are independent on t, so the flow (1.2) is isospectral.

3. Brockett formulation

Let us take now:

L = [ \begin{matrix}a_{1} & b_{1} & 0 \\ b_{1} & a_{2} & b_{2} \\ 0 & b_{2} & a_{3} \end{matrix} ] and N = [ \begin{matrix}−1 & 0 & 0 \\ 0 & −2 & 0 \\ 0 & 0 & −3 \end{matrix} ]

Then we can prove:

Theorem 3.1. The equation (2.2) have a Brockett formulation, i.e. they are equivalent to:

L̇ = [L, [L, N]].

The proof is an easy computation and we shall omit any other details.

Remark 3.1. As a consequence of the above theorem we can conclude that (1.2) defines in a natural way a gradient system.

4. The Lie-Poisson structure

Let ê = {e_{1} = (1, 0, 0, 0, 0), e_{2} = (0, 1, 0, 0, 0), e_{3} = (0, 0, 1, 0, 0), e_{4} = (0, 0, 0, 1, 0), e_{5} = (0, 0, 0, 0, 1)} be the canonical basis of R^{5}. If we define:

(4.1)  \begin{matrix}[e_{1}, e_{4}] = −e_{4} ; [e_{2}, e_{4}] = e_{4} \\ [e_{2}, e_{5}] = −e_{5} ; [e_{3}, e_{5}] = e_{5} \end{matrix}

and all the other brackets zero, then (R^{5}, [·, ·]) is a Lie-algebra. It follows that the minus-Lie-Poisson structure on (R^{5})^{∗} ≃ R^{5} is given by the matrix:

(4.2)  ∏ = [ \begin{matrix}0 & 0 & 0 & b_{1} & 0 \\ 0 & 0 & 0 & −b_{1} & b_{2} \\ 0 & 0 & 0 & 0 & −b_{2} \\ −b_{1} & b_{1} & 0 & 0 & 0 \\ 0 & −b_{2} & b_{2} & 0 & 0 \end{matrix} ]

Now we can easily prove:

Theorem 4.1. The system (1.2) may be realized as a Hamilton-Poisson mechanical system with the phase space P = R^{5}, the Hamiltonian H given by

(4.3)  H = \frac{1}{2}[a_{1}^{2} + a_{2}^{2} + a_{3}^{2}] + b_{1}^{2} + b_{2}^{2}

and the Poisson structure (4.2).

Remark 4.1. It is easy to see that the function C given by

(4.4)  C = a_{1} + a_{2} + a_{3}

is a Casimir of our configuration, i.e.

\{C, f\}\begin{matrix}− \\ LP \end{matrix} = 0,

for each f ∈ C^{∞}(R^{5}, R).

5. Stability and stabilization

It is clear that equilibrium states of the system (1.2) are:

(5.1)  (α, β, γ, 0, 0); α, β, γ ∈ R.

Consider first the system (1.2) linearized about the equilibrium (5.1). Its eigenvalues are given by the solutions of

(5.2)  λ^{3}(λ + α − β)(λ + β − γ) = 0.

If α, β, γ ∈ R, α < β or α, β, γ ∈ R, β < γ, or α, β, γ ∈ R, α < β < γ at least one root of the equation (5.2) is positive and so the corresponding equilibrium states are unstable.

If α, β, γ ∈ R, α ≥ β ≥ γ then the equilibrium states (5.1) are spectrally stable. Are they nonlinear stable? Using the energy-Casimir method we can prove the following statement:

Theorem 5.1. The equilibrium states

(5.3)  (α, α, α, 0, 0), α ∈ R

are nonlinear stable. Proof. Recall that the energy-Casimir method (see e.g. Holm, Marsden, Raţiu and Weinstein [2] or Puta [3]) requires finding a constant of motion for the system, usually the energy H, and a family C of constants of motion such that for some C ∈ C, H + C has a critical point at the equilibrium of interest. Often the C’s are taken to be Casimir-functions that commute with all other functions under the Poisson bracket. Then in the finite dimensional case, the definiteness of δ^{2}(H + C) at the critical point is sufficient to prove stability.

Now let us take the modified energy-Casimir function:

(5.4)  \begin{aligned}H + φ(C) & = \frac{1}{2}[a_{1}^{2} + a_{2}^{2} + a_{3}^{2}] + b_{1}^{2} + b_{2}^{2} \\ & + φ(a_{1} + a_{2} + a_{3}), \end{aligned}

where φ is a smooth arbitrary real valued function, i.e. φ ∈ C^{∞}(R, R). Then we have:

\begin{aligned}δ(H + φ(C)) & = a_{1}δa_{1} + a_{2}δa_{2} + a_{3}δa_{3} + 2b_{1}δb_{1} + 2b_{2}δb_{2} \\ & + φ̇(δa_{1} + δa_{2} + δa_{3}). \end{aligned}

At the equilibrium of interest (5.3) the first variation vanishes if and only if

(5.5)  φ̇(3α) = −α

Then

\begin{aligned}δ^{2}(H + φ(C)) & = (δa_{1})^{2} + (δa_{2})^{2} + (δa_{3})^{2} + 2(δb_{1})^{2} + 2(δb_{2})^{2} \\ & + φ̈(δa_{1} + δa_{2} + δa_{3})^{2}. \end{aligned}

and at the equilibrium of interest (5.3) we have

\begin{matrix}δ^{2}(H + φ(C))(α, α, α, 0, 0) = (δa_{1})^{2} + (δa_{2})^{2} + (δa_{3})^{2} \\ +2(δb_{1})^{2} + 2(δb_{2})^{2} + φ̈(3α)(δa_{1} + δa_{2} + δa_{3})^{2}. \end{matrix}

If we choose φ such that

φ̈(3α) ≥ 0,

then the second variation at the equilibrium of interest is positive definite and we have nonlinear stability.

Remark 5.1. It is an open problem to decide the stability or the instability of the equilibrium states which are not mention in our previous considerations.

6. Stabilization by one control

In this section we shall prove that the equilibrium states:

(6.1) (α, α, 0, 0, 0); α ∈ R, α > 0 (6.2) (0, 0, α, 0, 0); α ∈ R, α < 0

(6.3) (0, α, α, 0, 0), α ∈ R, α < 0

of the system (1.2) may be nonlinear stabilized by a particular linear control applied to the axis Ob_{2} [resp. Ob_{2}, resp. Ob_{1}].

The system (1.2) with one control about the axis Ob_{2} can be writen in the following form:

(6.4)  \{ \begin{matrix}ȧ_{1} & = 2b_{1}^{2} \\ ȧ_{2} & = 2(b_{2}^{2} − b_{1}^{2}) \\ ȧ_{3} & = −2b_{2}^{2} \\ ḃ_{1} & = b_{1}(a_{2} − a_{1}) \\ ḃ_{2} & = b_{2}(a_{3} − a_{2}) + u, \end{matrix}

where u ∈ C^{∞}(R^{5}, R).

In all that follows we shall employ the feedback:

(6.5)  u = kb_{2},

where k ∈ R is the control parameter.

Theorem 6.1. The controlled system (6.4), (6.5) is a Hamilton-Poisson mechanical system with the phase space P = R^{5}, the Poisson structure (4.2) and the Hamiltonian H given by:

(6.6)  H = \frac{1}{2}[a_{1}^{2} + a_{2}^{2} + a_{3}^{2}] + b_{1}^{2} + b_{2}^{2} + ka_{3}.

Using now again the energy-Casimir method we can prove:

Theorem 6.2. The controlled system (6.4), (6.5) may be nonlinear stabilized about the equilibrium states (6.1) for k = α.

Let us employ now the feedback

(6.7)  u = −kb_{2}

then we can prove in a similar manner the following results:

Theorem 6.3. The controlled system (6.4), (6.7) is a Hamilton-Poisson mechanical system with the phase space P = R^{5}, the Poisson structure (4.2) and the Hamiltonian H given by:

(6.8)  H = \frac{1}{2}[a_{1}^{2} + a_{2}^{2} + a_{3}^{2}] + b_{1}^{2} + b_{2}^{2} + ka_{1} + ka_{2}.

Theorem 6.4. The controlled system (6.4), (6.7) may be nonlinear stabilized about the equilibrium states (6.3) for k = α.

The system (1.2) with one control about the axis Ob_{1} can be written in the following form:

(6.9)  \{ \begin{matrix}ȧ_{1} & = 2b_{1}^{2} \\ ȧ_{2} & = 2(b_{2}^{2} − b_{1}^{2}) \\ ȧ_{3} & = −2b_{2}^{2} \\ ḃ_{1} & = b_{1}(a_{2} − a_{1}) + v \\ ḃ_{2} & = b_{2}(a_{3} − a_{2}) \end{matrix}

Let us employ now the feedback:

(6.10)  v = −kb_{1}

where k ∈ R is the control parameter. Then we can prove:

Theorem 6.5. The controlled system (6.9), (6.10) is a Hamilton-Poisson mechanical system with the phase space P = R^{5}, the Poisson structure (4.2) and the Hamiltonian H given by:

(6.11)  H = \frac{1}{2}[a_{1}^{2} + a_{2}^{2} + a_{3}^{2}] + b_{1}^{2} + b_{2}^{2} + ka_{1}.

Theorem 6.6. The controlled system (6.9), (6.10) may be nonlinear stabilized about the equilibrium states (6.2) for k = α.

Remark 6.1. It is easy to see that the equilibrium states:

(6.12)  (α, 0, α, 0, 0), α ∈ R, α ≠ 0

are all unstable.

7. Stabilization by two controls

In this section we shall prove that the equilibrium states

(7.1)  (α, 0, 0, 0, 0), α ∈ R, α > 0

of the system (1.2) may be nonlinear stabilized by two controls about the axes Ob_{1} and Ob_{2}.

The system (1.2) with two controls about the axes Ob_{1} and Ob_{2} can be written in the following form:

(7.2)  \{ \begin{matrix}ȧ_{1} & = 2b_{1}^{2} \\ ȧ_{2} & = 2(b_{2}^{2} − b_{1}^{2}) \\ ȧ_{3} & = −2b_{2}^{2} \\ ḃ_{1} & = b_{1}(a_{2} − a_{1}) + u \\ ḃ_{2} & = b_{2}(a_{3} − a_{2}) + v \end{matrix}

Let us employ now the feedbacks:

(7.3)  \{ \begin{matrix}u = kb_{1} \\ v = −kb_{2} \end{matrix}

where k ∈ R is the control parameter.

Then we have:

Theorem 7.1. The controlled system (7.2), (7.3) is a Hamilton-Poisson mechanical system with the phase space P = R^{5}, the Poisson structure (4.2) and the Hamiltonian H given by:

(7.4)  H = \frac{1}{2}[a_{1}^{2} + a_{2}^{2} + a_{3}^{2}] + b_{1}^{2} + b_{2}^{2} + ka_{2} + ka_{3}.

Theorem 7.2. The controlled system (7.2), (7.3) may be nonlinear stabilized about the equilibrium states (7.1) for k = α.

Remark 7.1. It is easy to see that the equilibrium states:

(7.5)  (0, α, 0, 0, 0), α ∈ R, α ≠ 0

are all unstable.

ACKNOWLEDGEMENT. This work was finalized during the author’s stay at the Erwin Schrödinger International Institute for Mathematical Physics in Vienna. I want to express here all my gratitude to the Institute and to Professor Peter Michor, in particular, for their support and hospitality.

References

H. Flaschka The Toda lattice, I, Phys. Rev. B (1974), 1924-1925. D. Holm, J. Marsden, T. Raţiu and A. Weinstein Nonlinear stability of fluid and

plasma equilibria, Phys. Rep. 123 (1985), 1-116.

M. Puta Hamiltonian mechanics and geometric quantization, Kluwer Academic Publishers, vol. 270 (1993).

Department of Mathematics, West University of Timisoara Bvd. V. Pârvan 4, Timisoara 1900, ROMANIA