Euler-Lagrange differential equation (elementary)

Let $q(t)$ be a twice differentiable function from $ℝ$ to $ℝ$ and let $L$ be a twice differentiable function from ${ℝ}^3$ to $ℝ$. Let $\dot{q}$ denote $\frac{d}{dt}q$.

Define the functional $I$ as follows:

$$I(q)={\int }_a^{b}L(t,q(t),\dot{q}(t))𝑑t$$

Suppose we regard the function $L$ and the limits of integratiuon $a$ and $b$ as fixed and allow $q$ to vary. Then we could ask for which functions $q$ (if any) this integral attains an extremal (minimum or maximum) value. (Note: especially in Physics literature, the function $L$ is known as the Lagrangian.)

Suppose that a differentiable function  $q_0:[a,b]\to ℝ$  is an extremum of $I$. Then, for every differentiable function  $f:[-1,+1]\times [a,b]\to ℝ$  such that  $f(0,x)=q_0(x)$, the function  $g:[-1,+1]\to ℝ$,  defined as

$$g(\lambda )={\int }_a^{b}L(t,f(\lambda ,t),\frac{\partial f}{\partial t}(\lambda ,t))𝑑t$$

will have an extremum at $\lambda =0$. If this function is differentiable, then  $dg/d\lambda =0$  when  $\lambda =0$.

By studying the condition $dg/d\lambda =0$ (see the addendum to this entry for details), one sees that, if a function $q$ is to be an extremum of the integral $I$, then $q$ must satisfy the following equation:

$$\frac{\partial }{\partial q}L-\frac{d}{dt}\left(\frac{\partial }{\partial \dot{q}}L\right)=0.$$ (1)

This equation is known as the Euler–Lagrange differential equation or the Euler-Lagrange condition. A few comments on notation might be in . The notations $\frac{\partial }{\partial q}L$ and $\frac{\partial }{\partial \dot{q}}L$ denote the partial derivatives of the function $L$ with respect to its second and third arguments, respectively. The notation $\frac{d}{dt}$ means that one is to first make the argument a function of $t$ by replacing the second argument with $q(t)$ and the third argument with $\dot{q}(t)$ and secondly, differentiate the resulting function with respect to $t$. Using the chain rule, the Euler-Lagrange equation can be written as follows:

$$\frac{\partial }{\partial q}L-\frac{{\partial }^2}{\partial t\partial \dot{q}}L-\dot{q}\frac{{\partial }^2}{\partial q\partial \dot{q}}L-\ddot{q}\frac{{\partial }^2}{\partial {\dot{q}}^2}L=0$$ (2)

This equation plays an important role in the calculus of variations. In using this equation, it must be remembered that it is only a necessary condition and, hence, given a solution of this equation, one cannot to the conclusion that this solution is a local extremum of the functional $F$. More work is needed to determine whether the solution of the Euler-Lagrange equation is an extremum of the integral $I$ or not.

In the special case $\frac{\partial }{\partial t}L=0$, the Euler-Lagrange equation can be replaced by the Beltrami identity.

Title	Euler-Lagrange differential equation (elementary)
Canonical name	EulerLagrangeDifferentialEquationelementary
Date of creation	2013-03-22 12:21:49
Last modified on	2013-03-22 12:21:49
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	34
Author	rspuzio (6075)
Entry type	Definition
Classification	msc 47A60
Classification	msc 70H03
Classification	msc 49K05
Synonym	Euler-Lagrange condition
Related topic	CalculusOfVariations
Related topic	BeltramiIdentity
Related topic	VersionOfTheFundamentalLemmaOfCalculusOfVariations
Defines	Euler-Lagrange differential equation
Defines	Lagrangian