## The Euler Lagrange Equation

Suppose we want to minimise the integral
$I= \int^b_a f(x,y, y') dx$
for a known function
$f(,x,y,y')$
. This is done by optimising the curve
$(x,y(x))$
along which the integral is taken.
Replace
$y(x)$
by
$y(x)+ \epsilon y_1(x)$
then the integral becomes
\begin{aligned} &{} \int^b_a f(x. y+\epsilon y_1, y'+ \epsilon y'_1)dx \\ &= \int^b_a f(x,y,y')+ \epsilon ( \frac{\partial f}{\partial y}y_1 + \frac{\partial f}{\partial y'}y'_1 )dx +O( \epsilon^2 )\end{aligned}

The change in the value of the integral is
$dI= \epsilon \int^b_a \frac{\partial f}{\partial y}y_1 + \frac{\partial f}{\partial y'}y'_1dx$

We require
$dI=0$
so
$\int^b_a (\frac{\partial f}{\partial y}y_1 + \frac{\partial f}{\partial y'}y'_1)dx=0$
.
Integrating the second term inside the integral by parts gives
$[ \frac{\partial f}{\partial y} ]^b_a + \int^b_a \frac{\partial f}{\partial y}-\frac{d}{dx} (\frac{\partial f}{\partial y'})y_1dx=0$

If the endpoints
$(a,y(a)), \; (b,y(b))$
are fixed then
$y_1(a)=y_1(b)=0$
so
$\frac{\partial f}{\partial y}-\frac{d}{dx} (\frac{\partial f}{\partial y'})$

This is the Euler - Lagrange equation.