12-267/Derivation of Euler-Lagrange

Disclamer: This is a student prepared note based on the lecure of Tuesday October 2nd.

For a function $y(x)$ defined on $[a,b]$ to be an extremum of $J(y)=\int _{a}^{b}F(x,y,y')dx$, it must be that for any function $h(x)$ defined on $[a,b]$ that preserves the endpoints of $y$ (that is, $h(a)=0$ and $h(b)=0$), we have ${\frac {d}{d\epsilon }}J(y+\epsilon h)$$|_{\epsilon =0}=0$.

${\frac {d}{d\epsilon }}J(y+\epsilon h)$$|_{\epsilon =0}$

$={\frac {d}{d\epsilon }}\int _{a}^{b}F(x,y+\epsilon h,y'+\epsilon h')dx|_{\epsilon =0}$

Let $F_{n}$ signify F differentiated with respect to its nth variable.

$=\int _{a}^{b}(F_{1}\cdot 0+F_{2}\cdot h+F_{3}\cdot h')dx|_{\epsilon =0}$

$=\int _{a}^{b}(F_{2}(x,y,y')\cdot h+F_{3}(x,y,y')\cdot h')dx$

$=\int _{a}^{b}(F_{2}\cdot h-[{\frac {d}{dx}}F_{3}]\cdot h)dx+F_{3}\cdot h|_{a}^{b}$ (integrating by parts)

Due to the constraints of $h(a)=0$ and $h(b)=0$, $F_{3}\cdot h|_{a}^{b}=0$.

$=\int _{a}^{b}(F_{2}-{\frac {d}{dx}}F_{3})\cdot hdx$

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that $F_{2}={\frac {d}{dx}}F_{3}$, or in other terms, $F_{y}-{\frac {d}{dx}}F_{y}'=0$.