Euler-Lagrange equations


context $s\in\mathbb N $
context $L:C^2(\mathbb R^s\times\mathbb R^s\times\mathbb R,\mathbb R) $
definiendum $q\in$ it
inclusion $q:C(\mathbb R,\mathbb R^s) $
let $\diamond\ q(t)$
let $\diamond\ L(x^1,\dots,x^s,v^1,\dots,v^s,t)$
range $j\in\{1,\dots,s\}$
postulate $\left(\dfrac{\mathrm d}{\mathrm dt}\dfrac{\partial L}{\partial v^j}\right)(q,q',t) - \dfrac{\partial L}{\partial x^j}(q,q',t) = 0$


In this form, the equations of motion in classical mechanics are second order in $t$.

The quantities $\dfrac{\partial L}{\partial v^j}$ and $\dfrac{\partial L}{\partial x^j}$ are also called generalized momenta, resp. generalized forces.

We often encounter situations with


Moreover, often


where $M(q)$ is a matrix and the brackets are the standard inner product.


Least action principle

A curve $q_\text{sol}$ being a solition to the Euler-Lagrange equations is implied by the following: For any sufficiently small time interval $(t_i,t_{i+1})$, the functional

$S(t_i,t_{i+1})[q]=\int_{t_i}^{t_{i+1}}L(q,q',t)\,{\mathrm dt}.$

is minimal at $q_\text{sol}$, with respect to “small and smooth” variations $q\mapsto q+\delta q$ that leave the endpoints fixed ($\delta q(t_1)=\delta q(t_2)=0$). To remove the quotes here, one needs a little of variational calculus.

$\bullet$ Observe that the solution space modulate with different data and the action data is in some sense more finer: Starting with the Euler-Lagrange equations, we must fix the initial data $q(t_0)$ at $t_0$ as well as $q'(t_0)$, i.e. the value of $\lim_{t_1\to t_0}\frac{q(t_1)-q(t_0)}{t_1-t_0}$. Starting with the least action postulate, for computing the initial segment, we must fix $q(t_0)$ at $t_0$ as well as a $t_1$ together with a value for $q(t_1)$.

$\bullet$ Sidenote: For a given $L$ and two fixed points, there isn't necessarily any solution path connecting them. E.g. the points $\langle 3,2\rangle$ and $\langle 3,-2\rangle$ in ${\mathbb R}^2\setminus\{\langle 3,0\rangle\}$ can't be connected in a way that makes $L\propto q'^2$ minimal (solutions to that $L$ would be straight lines).

$\bullet$ A nice aspect about this correspondence is that the action requirement makes the following apparent: If we re express $q$ in terms of other coordinates ${\hat q}$, then $L(q,q',t)$ rewrites as ${\hat L}({\hat q},{\hat q}',t)$ with some new ${\hat L}$. Now optimizing $q$ w.r.t. to the action for $L$ is literally the same as optimizing the ${\hat q}$ w.r.t. to the action for ${\hat L}$. The derivation of the Euler-Lagrange equations doesn't care for details about the Lagrangian, though, so both systems lead to a problem of the form “$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial v} = \frac{\partial L}{\partial x}$”, merely for different Lagrangians. So it turns that when re-expressing coordinates, we don't actually have to think about how the differential equation transformes, we just have switch to the new induced Lagrangian in terms of the new coordinates and automatically know what we have to solve is the new Euler-Lagrange equations.




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