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# 11.3: Derivation of the Euler-Lagrange Equation

In this section, we use the Principle of Least Action to derive a differential relationship for the path, and the result is the Euler-Lagrange equation. This derivation closely follows [163, p. 23-33], so see that reference for a more rigorous derivation. Assume that we know the Lagrangian which describes the difference between two forms of energy, and we know the action. We want to find a differential relationship for the path $$y(t)$$ which minimizes the action. This path has the smallest integral over $$t$$ of the difference between the two forms of energy.

Suppose that the path $$y(t)$$ minimizes the action and is the path found in nature. Consider a path $$\tilde y(t)$$ which is very close to the path $$y(t)$$. Path $$\tilde y(t)$$ is equal to path $$y(t)$$ plus a small difference.

$\tilde y = y + \varepsilon \eta \label{11.3.1}$

In Equation \ref{11.3.1}, $$\varepsilon$$ is a small parameter, and $$\eta = \eta(t)$$ is a function of $$t$$. We can evaluate the Lagrangian at this nearby path.

$\mathcal{L}\left(t, \tilde{y}, \frac{d \tilde{y}}{d t}\right)=\mathcal{L}\left(t, y+\varepsilon \eta, \dot{y}+\varepsilon \frac{d \eta}{d t}\right)$

The Lagrangian of the nearby path $$\tilde y(t)$$ can be related to the Lagrangian of the path $$y(t)$$.

$\mathcal{L}\left(t, \tilde{y}, \frac{d \tilde{y}}{d t}\right)=\mathcal{L}(t, y, \dot{y})+\varepsilon\left(\eta \frac{\partial \mathcal{L}}{\partial y}+\frac{d \eta}{d t} \frac{\partial \mathcal{L}}{\partial \dot{y}}\right)+O\left(\varepsilon^{2}\right) \label{11.3.3}$

Equation \ref{11.3.3} is written as an expansion in the small parameter $$\varepsilon$$. The lowest order terms are shown, and $$O(\varepsilon^2)$$ indicates that all additional terms are multiplied by $$\varepsilon^2$$ or higher powers of this small parameter.

We can also express the difference in the action for paths $$\tilde y$$ and $$y$$ as an expansion in the small parameter $$\varepsilon$$.

$\mathbb{S}(\hat{y})- \mathbb{S}(y)= \varepsilon\left[\int_{t_{0}}^{t_{1}} \eta \frac{\partial \mathcal{L}}{\partial y}+\frac{d \eta}{d t} \frac{\partial \mathcal{L}}{\partial \dot{y}} d t\right]+O\left(\varepsilon^{2}\right)$

The term in brackets is called the first variation of the action, and it is denoted by the symbol $$\delta$$.

$\delta \mathbb{S}(\eta, y)=\int_{t_{0}}^{t_{1}} \eta \frac{\partial \mathcal{L}}{\partial y}+\frac{d \eta}{d t} \frac{\partial \mathcal{L}}{\partial \dot{y}} d t$

Path $$y$$ has the least action, and all nearby paths $$\tilde y(t)$$ have larger action. Therefore, the small difference $$\mathbb{S}(\tilde y)−\mathbb{S}(y)$$ is positive for all possible choices of $$\eta(t)$$. The only way this can occur is if the first variation is zero.

$\delta \mathbb{S}(\eta, y)=0$

$\int_{t_{0}}^{t_{1}} \eta \frac{\partial \mathcal{L}}{\partial y}+\frac{d \eta}{d t} \frac{\partial \mathcal{L}}{\partial \dot{y}} d t=0 \label{11.3.7}$

If the action is a minimum for path $$y$$, then Equation \ref{11.3.7} is true. However, if path $$y$$ satisfies Equation \ref{11.3.7}, the action may or may not be at a minimum.

Use integration by parts on the second term to put Equation \ref{11.3.7} in a more familiar form.

$u = \frac{\partial \mathcal{L}}{\partial \dot{y}} \nonumber$

$du = \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{y}}dt \nonumber$

$v = \eta \nonumber$

$dv = \frac{d \eta}{dt}dt \nonumber$

$\int_{t_{0}}^{t_{1}} \frac{d \eta}{d t} \frac{\partial \mathcal{L}}{\partial \dot{y}} d t =\left[\eta \frac{\partial \mathcal{L}}{\partial \dot{y}}\right]_{t_{0}}^{t_{1}}-\int_{t_{0}}^{t_{1}} \eta \frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) d t$

Assume the endpoints of path $$y$$ and $$\tilde y$$ align.

$\eta (t_0) = \eta (t_1) = 0.$

$\int_{t_{0}}^{t_{1}} \frac{d \eta}{d t} \frac{\partial \mathcal{L}}{\partial \dot{y}} d t = -\int_{t_{0}}^{t_{1}} \eta \frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) d t \label{11.3.10}$

Combine Equation \ref{11.3.10} with Equation \ref{11.3.7}.

$0 =\int_{t_{0}}^{t_{1}} \eta \frac{\partial \mathcal{L}}{\partial y} -\eta \frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) d t$

$0 =\int_{t_{0}}^{t_{1}} \eta \frac{\partial \mathcal{L}}{\partial y} - \frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) d t \label{11.3.12}$

For Equation \ref{11.3.12} to be true for all functions $$\eta$$, the term in brackets must be zero, and the result is the Euler-Lagrange equation.

$\frac{\partial \mathcal{L}}{\partial y} -\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) =0$

We have completed the derivation. Using the Principle of Least Action, we have derived the Euler-Lagrange equation. If we know the Lagrangian for an energy conversion process, we can use the Euler-Lagrange equation to find the path describing how the system evolves as it goes from having energy in the first form to the energy in the second form.

The Euler-Lagrange equation is a second order differential equation. The relationship can be written instead as a pair of first order differential equations,

$\frac{d\mathbb{M}}{dt} = \frac{\partial \mathcal{L}}{\partial y}$

and

$\mathbb{M} = \frac{\partial \mathcal{L}}{\partial \dot{y}}.$

The Hamiltonian can be expressed as a function of the generalized momentum, [167, ch. 3].

$H(t, y, \mathbb{M}) =|\mathbb{M} \dot{y} -\mathcal{L}|$

Using the Hamiltonian, the Euler-Lagrange equation can be written as [167]

$\frac{d\mathbb{M}}{dt} = -\frac{\partial H}{\partial y}$

and

$\frac{dy}{dt} = \frac{\partial H}{\partial \mathbb{M}}.$

This pair of first order differential equations is called Hamilton's equations, and they contain the same information as the second order Euler-Lagrange equation. They can be used to solve the same types of problems as the Euler-Lagrange equation, for example finding the path from the Lagrangian.