How Do You Solve the Toda Lattice Differential Equation for a Single Soliton?

[alejandrito29]

toda is a chain of particles of displacement $$q(n,t)$$ acoplated by a spring

the differential equation are

$$\frac{d^2q(n,t)}{dt^2}=e^{-(q(n,t)-q(n-1,t))}-e^{-(q(n+1,t)-q(n,t))}$$

the solution for one soliton is:

$$q= Cte+ log (\frac{1+cte2 e^{-2cte3 n + 2 sinh t}}{1+cte2 e^{-2cte3 (n+1) + 2 sinh t}} )$$

i tried various ways , for example take $$q(n-1,t), q(n+1,t)=cte$$, but, i don't obtain the solution.

one link about this is http://www.mat.univie.ac.at/~gerald/ftp/book-jac/toda.html

help please

 

[Nino MMP]

. The solution for one soliton can be found using the method of inverse scattering. In this method, one first needs to find a suitable ansatz for the solution, which is a function of the independent variables n and t. This ansatz is then substituted into the differential equation and the resulting equation is solved for the unknown coefficients in the ansatz. The final solution is then obtained by substituting these coefficients back into the ansatz. For the Toda lattice, the ansatz is:q(n,t) = A + Bn + C log(1 + D e^{-2E n + 2F sinh t})where A, B, C, D, E and F are constants to be determined. Substituting this ansatz into the differential equation yields a system of linear equations in A, B, C, D, E and F. One can then solve this system of equations to obtain the constants. Finally, the solution is obtained by substituting these constants back into the ansatz.