# An inconsistency in Hamiltonian formulation for non-local Lagrangian: what am I doing wrong?

This question is based on a previous question I asked, Q. 

In this question, I proposed an example of a non-local Lagrangian (functional), I'm revisiting it here: $$\mathbb{L}=\frac{1}{2}\int^t_0 \left(\dot{q}(\tau)\dot{q}(t-\tau)-q(\tau)q(t-\tau)\right)\,\text{d}\tau \tag{1}$$

Taking the first variation of Eq. (1) with respect to $q$, I get that the functional is stationary with respect to (neglecting boundary terms): $$\ddot{q}(\tau)-q(\tau)=0 \tag{2}$$

Now, using the approach detailed in the answer to Q.  , I formulate a Hamiltonian integral as:

$$\mathbb{H}=\frac{1}{2}\int^t_0 \left(p(\tau)p(t-\tau)+q(\tau)q(t-\tau)\right)\,\text{d}\tau \tag{3}$$

Now, taking the functional derivatives of (3), we have:

$$\frac{\delta \mathbb{H}}{\delta p}=p(\tau),\,\frac{\delta \mathbb{H}}{\delta q}=q(\tau) \tag{4}$$

Now, as detailed in the answer to Q.  , (4) implies that:

$$\dot{q}(\tau)-p(\tau)=0,\,\dot{p}(\tau)+q(\tau)=0 \tag{5}$$

Substituting the first equation in (5) into the second yields:

$$\ddot{q}(\tau)+q(\tau)=0 \tag{6}$$

Eq. (6) contradicts Eq. (2), why?

It seems that the non-local nature of the Lagrangian leads to a different set of Hamilton's equations, namely:

$$\frac{\delta \mathbb{H}}{\delta p}=\dot{q},\,\frac{\delta \mathbb{H}}{\delta q}=\dot{p} \tag{7}$$

I just assumed this naively (since it would correct the contradiction), is this true, or am I making some mistake in my work?

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

• Your conclusion, ' the non-local nature of the Lagrangian leads to a different set of Hamilton's equations', cannot be correct as local Lagrangians are a special case of nonlocal ones, and the sign cannot suddenly change by going to a special case. Qmechanic's answer below does not have that defect. – Arnold Neumaier Apr 1 '15 at 9:20
• @ArnoldNeumaier: I see what you mean, the difference arises when he takes the functional derivative at (D) in his answer. He takes it at the local time, whereas I'm not (which is not consistent with the local case). What I'm curious about is what it means to do one over the other, in general. – Ron Apr 1 '15 at 12:10
• To take a functional derivative you need to replace $q(s)$ by $q(s)+\delta q(s)$, plug this into all occurrences of $q$ and its derivative, multiply out, simplify by keeping only the first order terms, use integration by parts to have every first order term to have a factor $\delta q(\tau)$, and set the coefficient of $\delta q(\tau)$ to zero. This gives a unique result, which is that of Qmechanic. – Arnold Neumaier Apr 1 '15 at 13:22
• @ArnoldNeumaier: What's the significance of taking a "functional derivative" (if that's what it would be at that point) as $\frac{\delta \mathbb{H}}{\delta f(t-\tau)}$ or collecting with respect to $\delta f(t-\tau)$ (rather than the usual $\delta f(\tau)$)? – Ron Apr 1 '15 at 14:36
• $\tau$ is only a dummy variable, which means when you vary $q(t-\tau)$ you need to replace it by $q(t-\tau)+\delta q(t-\tau)$, and similar for all other terms. After expansion, you need to make a variable transformation to bring it to a form where you can collect all required terms. Just as when you derive the equation of motion. I suggest you replace your current answer by a full derivation of the equation of motion; then it is easier to comment on mistakes and on the relation to the Hamiltonian version. – Arnold Neumaier Apr 1 '15 at 16:55

In this answer we apply the general non-local theory developed in my Phys.SE answer here to OP's non-local example. Let us for simplicity assume that time belongs to the unit interval $[t_i,t_f]=[0,1]$. OP's non-local Lagrangian action functional reads (modulo some sign conventions$^1$)

$$\left. S[q,v]\right|_{v=\dot{q}}, \tag{A}$$

where

$$S[q,v]~:=~\frac{1}{2}\int_{[0,1]^2}\! dt~du~\delta(1\!-\!t\!-\!u)\left\{ v(t)v(u) -q(t)q(u)\right\} .\tag{B}$$

The corresponding Lagrangian eq. of motion reads

$$\ddot{q}~\approx~q,\tag{C}$$

i.e., exponentially increasing/decreasing solutions. The Lagrangian momentum is

$$p(t)~:=~\frac{\delta S[q,v]}{\delta v(t)}~=~v(1\!-\!t) .\tag{D}$$

The Hamiltonian functional becomes

$$\mathbb{H}[q,p]~=~ \frac{1}{2}\int_{[0,1]^2}\! dt~du~\delta(1\!-\!t\!-\!u)\left\{ p(t)p(u) +q(t)q(u)\right\} . \tag{E}$$

$$\dot{q}(t)~\approx~ \frac{\delta \mathbb{H}}{\delta p(t)}~=~p(1\!-\!t),\qquad -\dot{p}(t)~\approx~ \frac{\delta \mathbb{H}}{\delta q(t)}~=~q(1\!-\!t). \tag{F}$$

Note that Hamilton's eqs. (F) imply the Lagrangian eq. of motion (C), as they should.

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$^1$ We choose sign conventions to match OP's Lagrangian eq. of motion (C).

• The functional $S$ over here is different from expression I used in (1) in my question, what is the difference between the two? Why use one over the other? I see you're using a double integration to effectively evaluate $v$ and $q$ at $1-t$, why would one formulate the expression this way? – Ron Mar 31 '15 at 21:54
• The chosen sign convention is to get the Lagrangian eq. of motion (C). – Qmechanic Mar 31 '15 at 22:14
• I see where I confused myself. In my derivation I defined: $p(t)=\frac{\delta S}{\delta v(1-t)}$ so that I can collect the $\delta v(1-t)$ terms to get the functional derivative. Why define the momenta as in (D) instead of this? – Ron Mar 31 '15 at 22:35
• E.g. because Def. (D) is compatible with the usual local case. – Qmechanic Mar 31 '15 at 22:52
• I see how the equations in (F) here would yield the solution for $q$ (one can differentiate one of them and reduce the system to one second order local differential equation), so the solution to the system would be the same. What's the implication a system such as (F)? I might just ask a new question about that specifically. – Ron Apr 1 '15 at 14:03

I've found the inconsistency. What follows is a derivation of the Hamiltonian for convolutional Lagrangians.

Starting with the total variation of a convolutional Lagrangian, we have: $$\delta \mathbb{L}=\int^t_0 \left(\frac{\delta \mathbb{L}}{\delta \dot{q}}\delta\dot{q}(t-\tau)+\frac{\delta \mathbb{L}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau \tag{1}$$ Using integration by parts on Eq. (1), we have: $$\delta \mathbb{L}=\int^t_0 \left(\frac{\text{d}}{\text{d}t}\frac{\delta \mathbb{L}}{\delta \dot{q}}+\frac{\delta \mathbb{L}}{\delta q}\right)\delta q(t-\tau)\,\text{d}\tau+\delta q(t) \left(\left.\frac{\delta \mathbb{L}}{\delta \dot{q}}\right|_{\tau=0}\right)-\delta q(0) \left(\left.\frac{\delta \mathbb{L}}{\delta \dot{q}}\right|_{\tau=t}\right)\tag{2}$$ This implies that for the boundary conditions: $$\left.\frac{\delta \mathbb{L}}{\delta \dot{q}}\right|_{\tau=0}=0,\delta q(0)=0 \tag{3}$$ And the relationship: $$\frac{\text{d}}{\text{d}t}\frac{\delta \mathbb{L}}{\delta \dot{q}}+\frac{\delta \mathbb{L}}{\delta q}=0 \tag{4}$$ The functional is zero. Notice, the relationship in Eq. (4) has a sign change from the typical, Euler-Lagrange equation. This is where the contradiction arises.

Moving on, we define: $$\mathbb{H}=\sup_{v} \left(\int^t_0 p(\tau)v(t-\tau)\,\text{d}\tau-\left. \mathbb{L}\right|_{\dot{q}=v}\right) \tag{5}$$ To find the $v$ in, we find: $$\frac{\delta}{\delta v} \left(\int^t_0 p(\tau)v(t-\tau)\,\text{d}\tau-\mathbb{L}\right)=0 \tag{6}$$ Eq. (6) is true for: $$p(\tau)=\frac{\delta \mathbb{L}}{\delta v} \tag{7}$$ If Eq. (7) is true, we can use Eq. (4) to get: $$\dot{p}=-\frac{\delta \mathbb{L}}{\delta q} \tag{8}$$ Back to Eq. (1), we have: $$\left.\delta \mathbb{L}\right|_{\dot{q}=v}=\int^t_0 \left(p(\tau)\delta\dot{q}(t-\tau)+\frac{\delta \mathbb{L}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau \tag{9}$$ In Eq. (7), we can make the substitution: $$\int^t_0 p(\tau)\delta\dot{q}(t-\tau)\,\text{d}\tau =\delta \!\left(\int^t_0 p(\tau)\dot{q}(t-\tau)\,\text{d}\tau\right)-\int^t_0 \dot{q}(\tau)\delta p(t-\tau)\,\text{d}\tau\tag{10}$$ This allows use to rearrange Eq. (7) using Eq. (8), as: $$\delta \!\left(\int^t_0 p(\tau)v(t-\tau)\,\text{d}\tau-\left. \mathbb{L}\right|_{\dot{q}=v}\right)=\int^t_0 \left(\dot{q}(\tau)\delta p(t-\tau)-\frac{\delta \mathbb{L}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau \tag{11}$$ The quantity on the left-hand side of Eq. (11) is nothing but $\delta \mathbb{H}$, which is:

$$\delta \mathbb{H}=\int^t_0 \left(\frac{\delta \mathbb{H}}{\delta p}\delta p(t-\tau)+\frac{\delta \mathbb{H}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau \tag{12}$$

Comparing the right-hand side of (11) and (12) yields: $$\dot{q}=\frac{\delta \mathbb{H}}{\delta p},\,\frac{\delta \mathbb{H}}{\delta q}=-\frac{\delta \mathbb{L}}{\delta q}\tag{13}$$ Using Eq. (8), we then have: $$\dot{q}=\frac{\delta \mathbb{H}}{\delta p},\,\dot{p}=\frac{\delta \mathbb{H}}{\delta q} \tag{14}$$ Which are Hamilton's equations for this type of convolutional non-local Lagrangian.