Let be a real dynamical system with degrees of freedom. Here is the configuration space and the Lagrangian, i.e. a smooth real-valued function such that and is an -dimensional "vector of speed". (For those familiar with differential geometry, is a smooth manifold, and where is the tangent bundle of
Let be the set of smooth paths for which and
The action functional is defined via
A path is a stationary point of if and only if
Here, is the time derivative of When we say stationary point, we mean a stationary point of with respect to any small perturbation in . See proofs below for more rigorous detail.
Derivation of the one-dimensional Euler–Lagrange equation
The derivation of the one-dimensional Euler–Lagrange equation is one of the classic proofs in mathematics. It relies on the fundamental lemma of calculus of variations.
We wish to find a function which satisfies the boundary conditions , , and which extremizes the functional
We assume that is twice continuously differentiable.[4] A weaker assumption can be used, but the proof becomes more difficult.[citation needed]
If extremizes the functional subject to the boundary conditions, then any slight perturbation of that preserves the boundary values must either increase (if is a minimizer) or decrease (if is a maximizer).
Let be the result of such a perturbation of , where is small and is a differentiable function satisfying . Then define
We now wish to calculate the total derivative of with respect to ε.
The third line follows from the fact that does not depend on , i.e. .
When , has an extremum value, so that
The next step is to use integration by parts on the second term of the integrand, yielding
Using the boundary conditions ,
Applying the fundamental lemma of calculus of variations now yields the Euler–Lagrange equation
Alternative derivation of the one-dimensional Euler–Lagrange equation
Given a functional
on with the boundary conditions and , we proceed by approximating the extremal curve by a polygonal line with segments and passing to the limit as the number of segments grows arbitrarily large.
Divide the interval into equal segments with endpoints and let . Rather than a smooth function we consider the polygonal line with vertices , where and . Accordingly, our functional becomes a real function of variables given by
Extremals of this new functional defined on the discrete points correspond to points where
Note that change of affects L not only at m but also at m-1 for the derivative of the 3rd argument.
Evaluating the partial derivative gives
Dividing the above equation by gives
and taking the limit as of the right-hand side of this expression yields
The left hand side of the previous equation is the functional derivative of the functional . A necessary condition for a differentiable functional to have an extremum on some function is that its functional derivative at that function vanishes, which is granted by the last equation.
A standard example[citation needed] is finding the real-valued function y(x) on the interval [a, b], such that y(a) = c and y(b) = d, for which the path length along the curve traced by y is as short as possible.
the integrand function being .
The partial derivatives of L are:
By substituting these into the Euler–Lagrange equation, we obtain
that is, the function must have a constant first derivative, and thus its graph is a straight line.
Single function of single variable with higher derivatives
The stationary values of the functional
can be obtained from the Euler–Lagrange equation[5]
under fixed boundary conditions for the function itself as well as for the first derivatives (i.e. for all ). The endpoint values of the highest derivative remain flexible.
Several functions of single variable with single derivative
If the problem involves finding several functions () of a single independent variable () that define an extremum of the functional
then the corresponding Euler–Lagrange equations are[6]
Single function of several variables with single derivative
A multi-dimensional generalization comes from considering a function on n variables. If is some surface, then
is extremized only if f satisfies the partial differential equation
When n = 2 and functional is the energy functional, this leads to the soap-film minimal surface problem.
Several functions of several variables with single derivative
If there are several unknown functions to be determined and several variables such that
the system of Euler–Lagrange equations is[5]
Single function of two variables with higher derivatives
If there is a single unknown function f to be determined that is dependent on two variables x1 and x2 and if the functional depends on higher derivatives of f up to n-th order such that
then the Euler–Lagrange equation is[5]
which can be represented shortly as:
wherein are indices that span the number of variables, that is, here they go from 1 to 2. Here summation over the indices is only over in order to avoid counting the same partial derivative multiple times, for example appears only once in the previous equation.
Several functions of several variables with higher derivatives
If there are p unknown functions fi to be determined that are dependent on m variables x1 ... xm and if the functional depends on higher derivatives of the fi up to n-th order such that
where are indices that span the number of variables, that is they go from 1 to m. Then the Euler–Lagrange equation is
where the summation over the is avoiding counting the same derivative several times, just as in the previous subsection. This can be expressed more compactly as
Let be a smooth manifold, and let denote the space of smooth functions . Then, for functionals of the form
where is the Lagrangian, the statement is equivalent to the statement that, for all , each coordinate frame trivialization of a neighborhood of yields the following equations:
Euler-Lagrange equations can also be written in a coordinate-free form as [7]
where is the canonical momenta 1-form corresponding to the Lagrangian . The vector field generating time translations is denoted by and the Lie derivative is denoted by . One can use local charts in which and :={\frac {d}{dt}}={\dot {q}}^{\alpha }{\frac {\partial }{\partial q^{\alpha }}}+{\ddot {q}}^{\alpha }{\frac {\partial }{\partial {\dot {q}}^{\alpha }}}}
and use coordinate expressions for the Lie derivative to see equivalence with coordinate expressions of the Euler Lagrange equation. The coordinate free form is particularly suitable for geometrical interpretation of the Euler Lagrange equations.