Discrete Euler-Lagrange equations

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• JD_PM
In summary: L}}{\partial(\partial_x_2...\partial_x_n})+\frac{\partial \mathcal{L}}{\partial(\partial_x_1...\partial_x_n}+\delta_{i'}^{i} \delta_{j'}^{j} \delta_{k'}^{k}\right)##.
JD_PM
TL;DR Summary
I want to understand discretization of 3D space (lattice) and the summation-by-parts method. To do so I am deriving the discrete Euler-Lagrange equations. I am basically stuck in how to proceed with summation by parts.
I want to derive the discrete EL equations

$$\frac{d}{dt} \frac{\partial L}{\partial \dot \phi_a^{(i j k)}} - \frac{\partial L}{\partial \phi_a^{(i j k)}} = 0$$

We deal with a Lagrange density which only depends on the fields themselves and their first order derivatives.

We discretize space, so that any spatial vector can be written as

$$\vec x = il \hat e_1 + jl \hat e_2 + kl \hat e_3$$

Where $l$ is the distance between consecutive lattice points.

The Lagrangian is no longer ##L= \int d^3 \vec x \mathscr{L}## but ##L= \sum_{(i j k)} l^3 \mathscr{L}^{(i j k)}##. When we take the limit ##l \rightarrow 0## we recover ##L= \int d^3 \vec x \mathscr{L}##.

The fields are no longer ##\phi_a (\vec x, t)## but ##\phi_a^{(i j k)} (t)##. When we take the limit ##l \rightarrow 0## we recover ##\phi_a (\vec x, t)##.

The idea I have is that we will not have to integrate by parts but sum by parts.

As we know, the action is defined as follows

$$S= \int dt L; \ \text{where} \ L=l^3\sum_{(i j k)} \mathscr{L}^{(i j k)}$$

We extremize the action (i.e. ##\delta S =0##)

$$\delta S = \int dt \ l^3 \sum_{(i j k)} \delta \mathscr{L}^{(i j k)}=0$$

Let's work out the term

$$\sum_{(i j k)} \delta \mathscr{L}^{(i j k)} \tag{*}$$

I know that, for the fields ##\phi_a## with spacetime coordinate dependence, we have

$$\delta \phi_a = \frac{\partial \phi_a}{\partial x^{\mu}} \delta x^{\mu}$$

So I would naively proceed as follows

$$\delta \mathscr{L}^{(i j k)} = \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \phi_a^{(i j k)}} \delta \phi_a^{(i j k)}+ \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \dot \phi_a^{(i j k)}} \delta \dot \phi_a^{(i j k)} + \sum_b^3 \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_b \phi_a^{(i j k)}} \delta_b \phi_a^{(i j k)} \tag{**}$$

Where ##b=x,y,z##

I am stuck in the following.

The idea is to perform summation by parts; i.e.

$$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_n g_{n+1} - f_m g_m) - \sum_{k=m+1}^n g_k (f_k -f_{k-1})$$

To the terms ##\frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \dot \phi_a^{(i j k)}} \delta \dot \phi_a^{(i j k)}## and ##\sum_b^3 \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_b \phi_a^{(i j k)}} \delta_b \phi_a^{(i j k)}##

I am a bit lost here. As an example

$$\sum_{(i j k)} \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_x \phi_a^{(i j k)}} \delta_x \phi_a^{(i j k)}=\sum_{(i j k)} \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_x \phi_a^{(i j k)}} \Big(\frac{\phi_a^{(i+1, j, k)}-\phi_a^{(i, j, k)}}{l} \Big) \tag{***}$$

Where I have used the definition of derivative on the infinitesimal term. I am confused, as I've got two derivatives before performing summation by parts, instead of 1; could you please shed some light on how to perform the summation by parts ##(***)##? Once that is understood, I should be able to derive the (discrete EL equations).

Thank you

etotheipi
I would suggest following the argument in Goldstein except replace the continuous path variation, ##\eta(x)##, with a discrete one.

In the second edition, page 32 equation 2-4 would become, $$\phi_{ijk}(\alpha)=\phi_{ijk}(0)+\alpha\eta_{ijk}$$

Have you tried as a starting point , $$\frac{\partial \phi}{\partial x} \approx (\phi_{i+1jk} - \phi_{i-1jk})/2l,$$ as an approximation? With this and the corresponding expressions for y and z results in the Lagrange equation you're trying to prove being straight forward.

Hi Paul Colby.

It turns out I was overcomplicating. A colleague and I came up with an alternative approach, which basically was using the chain rule in one of the spatial components

$$\frac{\partial L}{\partial \dot \phi_a^{(i,j ,k)}} = \frac{\partial}{\partial \phi_a^{(i,j,k)}} \sum_{(i',j',k')} \Big[ l^3 \mathcal{L}^{(i',j',k')}(\phi^{(i',j',k')}- \phi^{(i'-1,j',k')}) \Big]$$
$$= l^3 \sum_{(i',j',k')} \frac{\partial \mathcal{L}^{(i',j',k')}}{\partial(\phi^{(i',j',k')}- \phi^{(i'-1,j',k')})}\frac{\partial(\phi^{(i',j',k')}- \phi^{(i'-1,j',k')})}{\partial \phi_a^{(i,j,k)}}$$

Where the term ##\frac{\partial(\phi^{(i',j',k')}- \phi^{(i'-1,j',k')})}{\partial \phi_a^{(i,j,k)}}## yields

$$\frac{\partial(\phi^{(i',j',k')}- \phi^{(i'-1,j',k')})}{\partial \phi_a^{(i,j,k)}} = \delta_{i'}^{i} \delta_{j'}^{j} \delta_{k'}^{k}-\delta_{i'-1}^{i} \delta_{j'}^{j} \delta_{k'}^{k}$$

Plugging it into and taking the limit ##l \rightarrow 0## leads to the desired term, i.e. ##l^3 \partial_x \left(\frac{\partial \mathcal{L}}{\partial(\partial_x \phi)}\right)##

PhDeezNutz and Paul Colby

1. What are Discrete Euler-Lagrange equations?

Discrete Euler-Lagrange equations are a set of mathematical equations used to solve optimization problems in discrete systems. They are derived from the continuous Euler-Lagrange equations, which are used in classical mechanics to find the equations of motion for a system.

2. How are Discrete Euler-Lagrange equations different from continuous Euler-Lagrange equations?

The main difference between the two is that discrete Euler-Lagrange equations are used for discrete systems, while continuous Euler-Lagrange equations are used for continuous systems. In discrete systems, the variables are only defined at specific points, while in continuous systems, the variables are defined over a continuous range.

3. What is the significance of Discrete Euler-Lagrange equations in scientific research?

Discrete Euler-Lagrange equations are widely used in various fields of science, such as physics, engineering, and computer science. They provide a powerful tool for solving optimization problems and can be applied to a wide range of systems, making them an essential tool for scientific research.

4. What are some common applications of Discrete Euler-Lagrange equations?

Discrete Euler-Lagrange equations have many applications, including finding the optimal path for a robot to move from one point to another, optimizing control systems in engineering, and solving problems in computer vision and image processing. They are also used in the study of fluid dynamics and in the design of efficient algorithms for data analysis.

5. Can Discrete Euler-Lagrange equations be solved analytically?

In most cases, Discrete Euler-Lagrange equations cannot be solved analytically, and numerical methods must be used to find a solution. However, for simple systems with few variables, it is possible to find an analytical solution. In general, numerical methods are more efficient and accurate in solving Discrete Euler-Lagrange equations.

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