# 11.3: Derivation of the Euler-Lagrange Equation

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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.