2.1.1. Positions

Let us call *t*, the time, the independent variable, *x*(*t*) a function and *y* some linear functional that depends on *x* over the whole range of times through a relation ye<sup>t</sup> to be defined:

$$y(t) = G[t, x].\tag{43}$$

We consider a (Lagrangian) function of these, *<sup>L</sup>*(*<sup>t</sup>*, *x*, *y*), and we look for the corresponding E-L equation for the action *L* using the ideas of the calculus of variation. For instance, following Agrawall [17], we consider *x*<sup>∗</sup>(*t*) and *y*<sup>∗</sup>(*t*) the functions that make the action stationary and write *x*(*t*) = *x*<sup>∗</sup>(*t*) + *εη*(*t*). This implies, due to the assumed linearity of *G* that

$$y(t) = y^\*(t) + \varepsilon G[t, \eta(s)].\tag{44}$$

With this, the action becomes a function of *ε* with derivative:

$$\frac{d}{d\varepsilon} \int\_{0}^{T} L(t, x, y) \, dt = \int\_{0}^{T} \left( \frac{\partial L(t, x, y)}{\partial x} \eta(t) + \frac{\partial L(t, x, y)}{\partial y} G[t, \eta(s)] \right) dt = 0. \tag{45}$$

The next step is to write the integrand in terms of *η*(*t*) and declare each factor to be zero. For this, we need to suppose a specific form to the functional *G*.

Let us suppose a nonlocal dependence (in time) of *y* on *x* of the form

$$y(t) = G[t, \mathbf{x}(\mathbf{s})] := \int\_0^T K(t, \mathbf{s}) \, \mathbf{x}(\mathbf{s}) \, d\mathbf{s} \, \tag{46}$$

where *<sup>K</sup>*(*<sup>t</sup>*,*<sup>s</sup>*) is some kernel, independent of both *x* and *y*. With this we have from Equation (45)

$$\begin{split} &\int\_{0}^{T} \left( \frac{\partial L(t,x,y)}{\partial x} \eta(t) + \frac{\partial L(t,x,y)}{\partial y} \int\_{0}^{T} K(t,s) \eta(s) \, ds \right) dt = 0 \\ &\iff \quad \int\_{0}^{T} \int\_{0}^{T} \left( \frac{\partial L(t,x,y)}{\partial x} \eta(s) \delta(t-s) + \frac{\partial L(t,x,y)}{\partial y} K(t,s) \eta(s) \right) ds \, dt = 0 \\ &\iff \quad \int\_{0}^{T} \int\_{0}^{T} \left( \frac{\partial L(t,x,y)}{\partial x} \delta(t-s) + \frac{\partial L(t,x,y)}{\partial y} K(t,s) \right) dt \, \eta(s) ds = 0 \\ &\implies \quad \int\_{0}^{T} \left( \frac{\partial L(t,x,y)}{\partial x} \delta(t-s) + \frac{\partial L(t,x,y)}{\partial y} K(t,s) \right) dt = 0 \\ &\iff \quad \frac{\partial L(s,x,y)}{\partial x} + \int\_{0}^{T} \frac{\partial L(t,x,y)}{\partial y} K(t,s) \, dt = 0. \end{split} \tag{47}$$

The use of Dirac's delta is justified if we understand that *L* is zero for *t* outside [0, *<sup>T</sup>*]. Besides, for all this manipulation to make sense, we need the kernel *K* and both partials of *L* to be integrable

and *∂L*/*∂x* to have a finite L2-norm. If we choose a singular kernel, as below, we will see that some other conditions might be necessary.

Equation (47) is the corresponding E-L equation, we may exchange in it the name of the variables *t* and *s* and we finally have as necessary condition for the stationary action:

$$\frac{\partial L(t, \mathbf{x}, \mathbf{y})}{\partial \mathbf{x}} + \int\_0^T \frac{\partial L(\mathbf{s}, \mathbf{x}, \mathbf{y})}{\partial \mathbf{y}} K(\mathbf{s}, t) \, d\mathbf{s} = \mathbf{0}. \tag{48}$$

If we compare the integral with Equation (46) we see that the variables in the kernel are interchanged.

This is an important feature and it implies that the equation at a given time *t* involves values of *x* from both the past and the future. We will see this more clearly below when we deal with fractional derivatives.
