# Formulating a symplectic integrator for a non-local Hamiltonian

I recently asked two questions, Q. [1] and Q. [2], regarding reformulating non-local Lagrangians as Hamiltonians.

In these questions, the Hamiltonian is formulated as an integral because of it's non-local nature. Additionally, all of the partial derivatives must be replaced by functional derivatives, for the same reason.

My question is, how does one formulate a symplectic integrator for such a Hamiltonian?

In all symplectic integrator derivations I've seen, the Hamiltonian function is used, not the integral. Is there a more generalized approach one can take in this case?

Take for example the case where:

$$\mathbb{H}=\frac{1}{2}\int^t_0 \left(p(\tau)p(t-\tau)+q(\tau)q(t-\tau)\right)\,\text{d}\tau \tag{1}$$

This Hamiltonian has the associated Hamilton's equations of (as per Q. [2]) :

$$\dot{q}(\tau)=p(\tau),\,\dot{p}(\tau)=q(\tau) \tag{2}$$

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[1] This question deals with the Legendre transform for non-local Lagrangian formulations.

[2] This question (and answer) deals the derivation of the Euler-Lagrange equations and Hamilton's equations for non-local Lagrangians.

• I had an answer which I realized was irrelevant; the problem appears to be that you have a field theory and do not realize it. (Hence why functional derivatives are used, the Hamiltonian is an integral expression, etc.) Mar 31, 2015 at 20:41
• @AlexNelson: Can you explain what you mean? I'm fairly new to all this and any input (or a source I can look at) would be helpful.
– Ron
Mar 31, 2015 at 21:15

The Hamilton's eqs. $$\dot{z}^I(t)~\approx~ \{z^I(t), \mathbb{H}\}_{PB},\tag{1}$$ for non-local Hamiltonian theories have the same form as for local theories. Here $\mathbb{H}$ is a (possibly non-local) Hamiltonian functional, cf. my Phys.SE answer here. It follows that the symplectic integrator program carries over essentially un-modified, if the Hamiltonian functional
$$\mathbb{H}[q,p]~=~\mathbb{T}[p]+\mathbb{V}[q]\tag{2}$$