Euler-Lagrange equation

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The Euler-Lagrange equation, developed by Leonhard Euler and Joseph-Louis Lagrange in the 1750s, is the major formula of the calculus of variations. It provides a way to solve for functions which extremize a given cost functional. It is widely used to solve optimization problems, and in conjunction with the action principle to calculate trajectories. It is analogous to the result from calculus that a function attains its extreme values when its derivative vanishes.

1 Statement
2 Examples
3 Multidimensional variations
4 History
5 Proof
6 External links

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Statement

Formally, given a functional F(x, f(x), f'(x)) with continuous first partial derivatives, any function f which extremizes the cost functional

must also satisfy the ordinary differential equation

<math> \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} = 0. </math>

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Examples

A standard example is finding the shortest path between two points in the plane. Assume that the points to be connected are (a,c) and (b,d). The length of a path y=f(x) between these two points is:

<math> L = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx </math>

The Euler-Lagrange equation yields the differential equation:

<math> \frac{d}{dx} \frac{\partial}{\partial y'}\sqrt{1+\left(\frac{dy}{dx}\right)^2} = 0 \Rightarrow \frac{dy}{dx} = C </math>

In other words, a straight line.

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Multidimensional variations

There are also various multi-dimensional versions of the Euler-Lagrange equations. If q is a path in n-dimensional space, then it extremizes the cost functional

only if it satisfies

<math> \frac{d}{dt} \frac{dL}{dq'_k} - \frac{dL}{dq_k} = 0 </math> <math> \forall k = 1, 2, \dots n </math>

This formulation is particularly useful in physics when L is taken to be the Lagrangian.

Another multi-dimensional generalization comes from considering a function on n variables. If <math>\Omega</math> is some surface, then

<math> J = \int_{\Omega} L(f, x_1, \dots , x_n, f_{x_1}, \dots , f_{x_n}) d\Omega </math>

is extremized only if f satisfies the partial differential equation

<math> \frac{\partial L}{\partial f} - \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \frac{\partial L}{\partial f_{x_i}} = 0 </math>

When n = 2 and L is the energy functional, this leads to the soap-film minimal surface problem.

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History

The Euler-Lagrange equation was developed in the 1750s by Euler and Lagrange in connection with their studies of the tautochrone problem. This is the problem of determining a curve on which a weighted particle will fall to a fixed point in a fixed amount of time, independent of the starting point.

Lagrange solved this problem in 1755 and sent the solution to Euler. The two further developed Lagrange's method and applied it to mechanics, which led to the formulation of Lagrangian mechanics. Their correspondence ultimately led to the Calculus of variations, a term coined by Euler himself in 1766.

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Proof

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 <math>f</math> which satisfies the boundary conditions <math>f(a)=c</math>, <math>f(b)=d</math>, and which extremizes the cost functional

We assume that <math>F</math> has continuous first partial derivatives. A weaker assumption can be used, but the proof becomes more difficult.

If <math>f</math> extremizes the cost functional subject to the boundary conditions, then any slight perturbation of <math>f</math> that preserves the boundary values must either increase <math>J</math> (if <math>f</math> is a minimizer) or decrease <math>J</math> (if <math>f</math> is a maximizer).

Let <math>\eta(x)</math> be a differentiable function satisfying <math>\eta(a)=\eta(b)=0</math>. Then define

<math> J(\epsilon) = \int_a^b F(x,f(x) + \epsilon \eta(x), f'(x) + \epsilon \eta'(x))\, dx </math>

Since <math>J(0)</math> is the cost of <math>f</math>, an extreme value, it follows that <math>J'(0)=0</math>, i.e.

<math> J'(0) = \int_a^b \left[ \eta(x) \frac{\partial F}{\partial f} + \eta'(x) \frac{\partial F}{\partial f'} \right]\,dx = 0</math>

(Note that we have to be careful here about passing the derivative through the integral.)

The crucial next step is to use integration by parts on the second term, yielding

<math> 0 = \int_a^b \left[ \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} \right] \eta(x)\,dx + \left[ \eta(x) \frac{\partial F}{\partial f'} \right]_a^b </math>

Using the boundary conditions on <math>\eta</math>, we get that

<math> 0 = \int_a^b \left[ \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} \right] \eta(x)\,dx </math>

Applying the fundamental lemma of Calculus of variations now yields the Euler-Lagrange equation:

<math> 0 = \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} </math>

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