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This question is based on a previous question I asked, Q. [1]

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

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Ron
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2 Answers2

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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

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

The corresponding Hamilton's eqs. read

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

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).

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Qmechanic
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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.

Ron
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