Derive the open Toda equations

How do exponential interactions produce the Toda equations? Starting with a finite open chain, you will derive every force, check the two ends, and convert the result into the variables used by a Lax matrix. The three-particle calculation is small enough to do by hand, but it exposes the signs and boundary terms needed for any finite chain.

Required background. Differentiate functions of several variables while holding the other variables fixed. The entry check and repair supplies the Hamiltonian rule used here.

Helpful background. The differential-equation bridge explains first-order systems and initial data. The course introduction explains how this calculation leads to a test of integrability. The Toda model fixes the system throughout the sequence.

Forces in the open Toda chain

Take a fixed integer $N\geq2$, real coordinates $q_i$, and real canonical momenta $p_i$, with $1\leq i\leq N$. All quantities and time are dimensionless; the masses and exponential coupling have been scaled to one. There is no constraint ordering the $q_i$. The Hamiltonian is

$$H(q,p)=\frac12\sum_{i=1}^{N}p_i^2 +\sum_{i=1}^{N-1}e^{q_i-q_{i+1}}.$$

“Open” means that the interaction sum stops at the bond between sites $N-1$ and $N$. There is no bond joining $N$ to $1$, and neither endpoint is held fixed. This is the finite nonperiodic system in Moser 1975, § 1, p. 467, equations (1.1)–(1.3), PDF.

Use the canonical bracket

$$\{f,g\}=\sum_{i=1}^{N} \left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} -\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right).$$

For a function without explicit time dependence, $\dot f=\{f,H\}$. In particular,

$$\dot q_i=\frac{\partial H}{\partial p_i}, \qquad \dot p_i=-\frac{\partial H}{\partial q_i}.$$

The task is to compute these derivatives, rather than guess a force from the appearance of the exponential.

Entry check and repair

For $V(q_1,q_2)=e^{q_1-q_2}$, find $\partial V/\partial q_1$ and $\partial V/\partial q_2$. Then use the bracket definition to find $\{q_i,H\}$ and $\{p_i,H\}$. Try this before reading the repair.

Repair. The chain rule gives $\partial_{q_1}V=V$ and $\partial_{q_2}V=-V$. Also $\partial_{q_j}q_i=\delta_{ij}$ and $\partial_{p_j}q_i=0$, so only one term survives in $\{q_i,H\}$: it is $\partial_{p_i}H=p_i$. Reversing the roles of $q_i$ and $p_i$ gives $\{p_i,H\}=-\partial_{q_i}H$. The minus sign belongs to Hamilton’s equation as well as to any derivative of a negative exponent.

As a quick retry, the Hamiltonian $p^2/2+e^{-q}$ gives $\dot q=p$ and $\dot p=e^{-q}$.

Differentiate the three-particle Hamiltonian

For $N=3$, introduce the positive bond strengths

$$u=e^{q_1-q_2},\qquad v=e^{q_2-q_3}, \qquad H=\frac12(p_1^2+p_2^2+p_3^2)+u+v.$$

The middle coordinate occurs in two bonds, with opposite signs:

$$\frac{\partial H}{\partial q_2}=-u+v.$$

The end coordinates occur in only one bond each. Thus the complete equations are

$$\begin{aligned} \dot q_1&=p_1,&\dot p_1&=-u,\\ \dot q_2&=p_2,&\dot p_2&=u-v,\\ \dot q_3&=p_3,&\dot p_3&=v. \end{aligned}$$

Each bond pushes its two particles in opposite directions. Consequently the total momentum $P=p_1+p_2+p_3$ obeys

$$\dot P=-u+(u-v)+v=0.$$

This cancellation detects an unbalanced extra force, but it does not determine the boundary condition. An erroneous closing bond $w=e^{q_3-q_1}$ would add $+w$ to $\dot p_1$ and $-w$ to $\dot p_3$, leaving $\dot P=0$. Verify each endpoint equation against the open Hamiltonian as well as checking total momentum.

For general $N$, write $f_i=e^{q_i-q_{i+1}}$ for the actual bonds $1\leq i\leq N-1$, and set the missing-bond symbols $f_0=f_N=0$. Then

$$\dot q_i=p_i,\qquad \dot p_i=f_{i-1}-f_i.$$

The values $f_0=f_N=0$ do not introduce extra finite coordinates $q_0$ or $q_{N+1}$.

Exponential coordinates and a worked check

Define

$$a_i=e^{(q_i-q_{i+1})/2}\gt0,\qquad b_i=p_i, \qquad a_0=a_N=0.$$

The chain rule, including its factor of one half, gives

$$\begin{aligned} \dot a_i &=\frac12e^{(q_i-q_{i+1})/2}(\dot q_i-\dot q_{i+1})\\ &=\frac12a_i(b_i-b_{i+1}),\qquad 1\leq i\leq N-1,\\ \dot b_i&=a_{i-1}^2-a_i^2,\qquad 1\leq i\leq N. \end{aligned}$$

These variables encode the momenta and coordinate differences. They do not retain the common displacement of all particles, and they are not new canonical coordinate–momentum pairs.

For the initial state used throughout the course,

$$q(0)=(0,0,0),\qquad p(0)=(1,0,-1),$$

we have $u=v=1$ and $a_1=a_2=1$. Therefore

$$\begin{aligned} \dot q(0)&=(1,0,-1),&\dot p(0)&=(-1,0,1),\\ \dot a(0)&=(1/2,1/2),&P(0)&=0,\qquad H(0)=3. \end{aligned}$$

There is a useful independent energy check. The kinetic energy derivative is $\sum_i p_i\dot p_i=-2$, while

$$\dot u+\dot v=u(p_1-p_2)+v(p_2-p_3)=2.$$

The two contributions cancel. The particles can have nonzero accelerations while their total energy stays constant.

Exercises

Guided practice: the missing endpoint terms

For an interior site, complete

$$\partial_{q_i}f_{i-1}=\underline{\phantom{-f_{i-1}}}, \qquad \partial_{q_i}f_i=\underline{\phantom{f_i}}.$$

Use the answers to derive $\dot p_i$. Write the two endpoint equations separately, then show by telescoping that $\sum_i\dot p_i=0$. Explain why this is consistent with invariance under $q_i\mapsto q_i+c$ for a constant $c$.

Hint · Full solution

Independent practice: unequal bonds

At one instant take

$$q=(\log4,0,-\log9),\qquad p=(2,-1,0).$$

Compute $a$, $\dot q$, $\dot p$, $\dot a$, $P$, and $H$. Verify directly that the kinetic and potential energy derivatives cancel at that instant. This state tests both a nonzero middle force and unequal bond strengths.

Hint · Full solution

Changed setting: add a pinning potential

For $N=3$, replace $H$ by

$$H_\kappa=H+\frac{\kappa}{2}\sum_{i=1}^{3}q_i^2, \qquad \kappa\gt0.$$

Keep the canonical bracket. Derive the new equations for $q_i$, $p_i$, and $a_i$. Is $P$ still conserved for all initial states? Is the original $H$ conserved? Identify the energy that is conserved, and state which symmetry has changed. Do not infer whether the modified system is integrable from this calculation alone.

Hint · Full solution

Hints

Endpoint terms. $q_i$ has coefficient $-1$ in the exponent of $f_{i-1}$ and $+1$ in that of $f_i$. In the sum of forces, each actual bond appears twice.

Unequal bonds. First compute $u=4$ and $v=9$; their square roots, rather than their values, are the $a_i$. For the energy check, use $\dot u=u(p_1-p_2)$ and $\dot v=v(p_2-p_3)$.

Pinning. Differentiate the added term with respect to each $q_i$. Its derivative with respect to $p_i$ is zero. Check the two energy derivatives separately before combining them.

Solutions and checks

Endpoint terms

The derivatives are $-f_{i-1}$ and $f_i$, so the additional minus sign in Hamilton’s equation gives $\dot p_i=f_{i-1}-f_i$. At the left end, $\dot p_1=-f_1$; at the right end, $\dot p_N=f_{N-1}$. Thus

$$\dot P=\sum_{i=1}^{N}(f_{i-1}-f_i)=f_0-f_N=0.$$

A common translation changes no difference $q_i-q_{i+1}$, so it changes neither the potential nor $H$. Equivalently, differentiating $H(q+c\mathbf1,p)$ with respect to $c$ gives $\sum_i\partial_{q_i}H=0$, exactly the identity used in $\dot P=-\sum_i\partial_{q_i}H$.

Unequal bonds

Substitution gives

$$\begin{aligned} a&=(2,3),&\dot q&=(2,-1,0),\\ \dot p&=(-4,-5,9),&\dot a&=(3,-3/2),\\ P&=1,&H&=\frac{31}{2}. \end{aligned}$$

The middle force is $4-9=-5$. The kinetic energy derivative is $2(-4)+(-1)(-5)=-3$. The potential energy derivative is $4(2-(-1))+9(-1-0)=3$. Their sum is zero, as is the total force $-4-5+9$.

Pinning

The new equations are

$$\begin{aligned} \dot q_i&=p_i,\\ \dot p_i&=f_{i-1}-f_i-\kappa q_i,\\ \dot a_i&=\tfrac12a_i(p_i-p_{i+1}). \end{aligned}$$

The equation for $a_i$ retains its form because $\dot q_i$ does. However,

$$\dot P=-\kappa\sum_iq_i,\qquad \dot H=-\kappa\sum_iq_ip_i.$$

Neither quantity is conserved for arbitrary initial data. The added potential has derivative $\kappa\sum_iq_ip_i$, so $\dot H_\kappa=0$. Pinning singles out an origin and removes common-translation invariance. A derivative that happens to vanish at one instant is not a conservation law valid along every trajectory.

Check your understanding

You are ready to build the Lax pair when you can recover the endpoint signs and the factor of one half in $\dot a_i$ without looking them up. Keep these distinctions clear:

• Open endpoints remove bonds; they do not fix endpoint positions.
• $a_i^2$ is the bond strength, while $a_i$ is its positive square root.
• Deriving equations and checking energy does not yet establish Liouville integrability.

References

• Moser, Jürgen. “Finitely many mass points on the line under the influence of an exponential potential—an integrable system.” In Dynamical Systems, Theory and Applications, edited by Jürgen Moser, Lecture Notes in Physics 38, pp. 467–497. Springer, 1975. DOI. Open PDF.