# 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 $h(b) = 0$, $F_3 \cdot h |_a^b = 0$.

$= \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx$

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, $F_y - \frac{d}{dx} F_y' = 0$.

Special cases (without derivations):

In the case that F does not depend on y', we have $F_y = 0$

In the case that F does not depend on y, we have $F_{y'} = c$

In the case that F does not depend on x, we have $F - y'F_{y'} = c$