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

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 $q_0 : [a,b] \to \reals$ is an extremum of $I$ . Then, for every differentiable function $f \colon [-1,+1] \times [a,b] \to \reals$ such that $f(0,x) = q_0 (x)$ , the function $g \colon [-1, +1] \to \reals$ , 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: \begin{equation} \mpderiv{q} L - \mderiv{t} \left( \mpderiv{\dot{q}} L \right) = 0. \end{equation}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 $\mpderiv{q} L$ and $\mpderiv{\dot{q}} L$ denote the partial derivatives of the function $L$ with respect to its second and third arguments, respectively. The notation $\mderiv{t}$ 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: \begin{equation} \mpderiv{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. \end{equation} 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 extrmum of the integral $I$ or not.

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