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# Euler-Lagrange differential equation (elementary)

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

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

$g(\lambda)=\int_{a}^{b}L\left(t,f(\lambda,t),{\partial f\over\partial t}(% \lambda,t)\right)\,dt$ |

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:

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

This equation is known as the *Euler–Lagrange differential equation* or the Euler-Lagrange condition. A few comments on notation might be in order. 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 $\frac{d}{d{t}}$ 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:

$\frac{\partial}{\partial{q}}L-{\partial^{2}\over\partial t\partial\dot{q}}L-% \dot{q}{\partial^{2}\over\partial q\partial\dot{q}}L-\ddot{q}{\partial^{2}% \over\partial\dot{q}^{2}}L=0$ | (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 jump 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.

## Mathematics Subject Classification

47A60*no label found*70H03

*no label found*49K05

*no label found*

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