# Euler-Lagrange Equation (1)

### From TorontoMathWiki

# Theorem of the Euler-Lagrange equation for the case of one dependent variable

Let be a functional of the form

,

where has continuous partial derivatives of second order and .

Let ,

where and are given real numbers. If is a local extremum for , then

for all .

# The Proof for the case of one dependent variable

Here we consider a particular class of problem called the **fixed endpoint variational problem** and work in .

Let be a functional of the form

,

where is a function assumed to have at least second-order continuous partial derivatives.

Given two values , the fixed endpoint variational problem consists of determining the functions such that , and has a local extremum in at , where

,

and

.

We want to derive necessary conditions for to have a local extremum in at .

Suppose that has a local extremum in at .

Then there is an such that does not change sign for all such that .

For any there is an such that , and for small Taylor's theorem implies that

.

Please note that the partial derivatives in the above expression are all evaluated at the point .

Now,

.

The quantity

is called the **first variation** of .

It is clear that if then , and . For small, the sign of is determined by the sign of the first variation, unless for all .

Since has a local extremum in at , does not change sign for all such that . Hence, for all .

Using integration by parts,

,

since .

Hence, for all .

Note that ,

and given that has at least two continuous derivatives, we see that for any the function defined by

is continuous on the interval . Note again that for a given function the partial derivatives defining are evaluated at the point .

Hence, for all .

By the **Fundamental Lemma of Calculus of Variations**, .

Hence,

and the equation is called the **Euler-Lagrange equation**.

# Theorem of the Euler-Lagrange equation for the case of several dependent variables

Let be a functional of the form

,

where , and has continuous second order partial derivatives.

Let ,

where are given vectors.

If is a local extremum for in then

for .

# The proof for case of several dependent variables

Let denote the set of functions such that for every we have for .

Consider a functional of the form

,

where denotes differentiation with respect to , and is a function having continuous partial derivatives of second order.

Given two vectors , the fixed endpoint problem consists of determining the local extrema for subject to the conditions and .

Again, we have

and

.

For small Taylor's theorem implies that

,

where .

Hence,

.

Therefore the first variation for this functional is

.

If has a local extremum at then similar arguments imply that for all .

Define . For any the above condition reduces to

.

From the one variable case, we know this condition leads to the Euler-Lagrange equation

.

Similarly, we have

for .

# Special cases

## Case 1

Suppose does not depend on .

Then, the Euler-Lagrange equation simplifies to

and hence

where is some constant.

Solving for , we get

for some function . This can be solved by a quadrature.

## Case 2

We will use a more compact notation to improve readability.

Suppose the integrand does not depend on , in which case .

First, we notice the following

again by the chain rule. The second and third terms cancel, and we obtain

- (*)

Now, from the Euler-Lagrange equation, we get

by the chain rule.

However, the right side of this equation (which is zero) can be directly substituted into the right side of (*). Hence (*) is also zero. That is,

or

where is a constant.

## Case 3

Suppose does not depend on .

Then, the Euler-Lagrange equation takes the form

# Examples

## Straight Lines

**Result**: The shortest line between two points and in is straight.

**Proof**: Consider a continuously differentiable function that parametrizes a curve with the required endpoints.

We will let denote the ith component of the function for .

We have the arc length function

where is the integrand.

Now, for every

and

Substituting into the Euler-Lagrange equation, we get for , which describes a straight line.

We have only shown that the straight line with the required endpoints an extremum. Nonetheless, in 3 (or less) dimensions, it is intuitively clear that such a line has the minimum length.

## Geodesic

See Geodesic.

## Catenary

See Catenary.

## Cycloid

For a discussion on the brachistochrone property of the cycloid, see Cycloid.

## Hypocycloid

**Result**: The frictionless tunnel connecting two points and on the surface of a (spherical, homogeneous) planet taking the shape of a hypocycloid is the one for which a ball dropped (initially from rest) at one point moves to the other the fastest.

**Proof**:
We will study the system partly in polar co-ordinates with the planet centered at the origin and partly in Cartesian co-ordinates , whereby we assume that the problem is 2-dimensional.

Out problem is to minimize the total time

where

- denotes the speed of the ball at time
- denotes the total distance traveled by time

We now find by first finding the potential energy , where we set .

Consider a spherical surface of radius concentric to the planet such that , where is the radius of the planet. Then, applying Gauss' law on this surface, we obtain

where

- is the gravitational force (positive outward) at radius
- is the planet's density
- is the mass of the ball
- is the gravitational constant

The left side is the surface flux of the gravitational force and the right side is the total mass contained in the surface.

Substituting

we obtain

Hence,

To find , we apply the conservation of mechanical energy, so that

where is the mass of the ball, and we remind that and .

Letting , we get

Furthermore,

Hence, we conclude that

where

- denotes the time derivative

and we let denote the integrand and minimized the integral using the Euler-Lagrange equation.

Since is independent of , therefore our equation reduces to (case 2 from above)

where and are constants.

Differentiating then substituting, we obtain

where

Adding both equations then squaring both sides, we get

where

The following cycloid with outer radius and inner radius satisfies the required equation

This completes the proof.

# References

- The Calculus of Variations by Bruce van Brunt
- http://www.physics.unlv.edu/~maxham/gravitytrain.pdf
- Calculus of Variations by Gelfand and Fomin