# What is the Euler-Lagrange equation

Definition/Summary

Also known as the Euler equation. It is the solution to finding an extrema of a functional in the form of

$$v(y)=\int_{x_{1}}^{x_{2}} F(x,y,y') dx \ .$$

The solution usually simplifies to a second order differential equation.

Equations

$$F_{y}-D_{x}F_{y'} \ = \ 0$$

or

$$\frac{\partial F}{\partial y} \ - \ \frac{\mathrm{d} }{\mathrm{d} x} \ \frac{\partial F}{\partial y'} \ = \ 0$$

Extended explanation

PROOF

Let us find the extrema of the functional

$$v(y)=\int_{x_{1}}^{x_{2}}F(x,y,y')dx \ .$$

Such a functional could be arc length, for example. For the variation of v,

$$\delta v = \frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} \ ,$$

let Δy be an arbitrary differentiable function such that Δy(x1)y(x2)=0.

Now, to find the extrema, the variation must be zero. i.e.

$$\delta v=\frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} = 0$$

or

$$\frac{\partial }{\partial a}\int_{x_{1}}^{x_{2}} F(x,y+a\Delta y,y'+a\Delta y')dx|_{a=0}=0 \ .$$

Using the chain rule of multiple variables, this simplifies to

$$\int_{x_{1}}^{x_{2}} (\frac{\partial F}{\partial y}\frac{\mathrm{d} (y+a\Delta y)}{\mathrm{d} a}+\frac{\partial F}{\partial y'}\frac{\mathrm{d}(y'+a\Delta y' ) }{\mathrm{d} a}) dx \ .$$

We then split d(y+aΔy) and d(y'+aΔy') into dy+Δyda and dy'+Δy'da respectively. Remember that y and y' is independent of a, and da/da=1. We therefore get (using different notation: $F_{y}=\frac{\partial F}{\partial y}$)

$$\int_{x_{1}}^{x_{2}} (\Delta y F_{y}+\Delta y'F_{y'})dx$$

Using integration by parts on the right side with "u"=Fy' and "dv"=Δy'dx:

$$\int_{x_{1}}^{x_{2}}\Delta yF_{y} dx \ + \ [\Delta yF_{y'}]_{x_{1}}^{x_{2}} \ - \ \int_{x_{1}}^{x_{2}}\Delta y\frac{\mathrm{d} F_{y'}}{\mathrm{d} x} \ dx \ .$$

However, Δy(x1)=Δy(x2)=0. Thus the middle term is zero, so:

$$\int_{x_{1}}^{x_{2}}\Delta y(F_{y}-\frac{\partial }{\partial x}F_{y'})dx=0 \ .$$

Applying the Fundamental Lemma of Calculus of Variations, we find

$$F_{y}-\frac{\partial}{\partial x}F_{y'}=0 \ .$$

Or, more compactly,

$$F_{y}-D_{x}F_{y'} = 0 \ ,$$

where Dx is the differential operator with respect to x.
This is a second order differential equation which, when solved, gives the desired extrema of the functional.

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