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
\[\begin{equation}\begin{aligned} &{} \int^b_a f(x. y+\epsilon y_1, y'+ \epsilon y'_1)dx \\ &= \int^b_a f(x,y,y')+ \epsilon ( \frac{\partial f}{\partial y}y_1 + \frac{\partial f}{\partial y'}y'_1 )dx +O( \epsilon^2 )\end{aligned} \end{equation}\]

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.

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