The Toda lattice is a prototypical example of a
completely integrable system. To see this one uses
Flaschka's variables : a(n,t) = \frac{1}{2} {\rm e}^{-(q(n+1,t) - q(n,t))/2}, \qquad b(n,t) = -\frac{1}{2} p(n,t) such that the Toda lattice reads : \begin{align} \dot{a}(n,t) &= a(n,t) \Big(b(n+1,t)-b(n,t)\Big), \\ \dot{b}(n,t) &= 2 \Big(a(n,t)^2-a(n-1,t)^2\Big). \end{align} To show that the system is completely integrable, it suffices to find a
Lax pair, that is, two operators
L(t) and
P(t) in the
Hilbert space of square summable sequences \ell^2(\mathbb{Z}) such that the Lax equation :\frac{d}{dt} L(t) = [P(t), L(t)] (where [
L,
P] =
LP -
PL is the
Lie commutator of the two operators) is equivalent to the time derivative of Flaschka's variables. The choice : \begin{align} L(t) f(n) &= a(n,t) f(n+1) + a(n-1,t) f(n-1) + b(n,t) f(n), \\ P(t) f(n) &= a(n,t) f(n+1) - a(n-1,t) f(n-1). \end{align} where
f(n+1) and
f(n-1) are the shift operators, implies that the operators
L(t) for different
t are unitarily equivalent. The matrix L(t) has the property that its eigenvalues are invariant in time. These eigenvalues constitute independent integrals of motion, therefore the Toda lattice is completely integrable. In particular, the Toda lattice can be solved by virtue of the
inverse scattering transform for the
Jacobi operator L. The main result implies that arbitrary (sufficiently fast) decaying initial conditions asymptotically for large
t split into a sum of solitons and a decaying
dispersive part. == See also ==