12-267/Derivation of Euler-Lagrange

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

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

[math]\displaystyle{ \frac{d}{d \epsilon } J(y + \epsilon h) }[/math][math]\displaystyle{ |_{\epsilon = 0} }[/math]

[math]\displaystyle{ = \frac{d}{d \epsilon } \int_a^b F(x, y + \epsilon h, y' + \epsilon h') dx |_{\epsilon = 0} }[/math]

Let [math]\displaystyle{ F_n }[/math] signify F differentiated with respect to its nth variable.

[math]\displaystyle{ = \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h') dx |_{\epsilon = 0} }[/math]

[math]\displaystyle{ = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx }[/math]

[math]\displaystyle{ = \int_a^b (F_2 \cdot h + [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b }[/math] (integrating by parts)

Due to the constraints of [math]\displaystyle{ h(a) = 0 }[/math] and [math]\displaystyle{ h(b) = 0 }[/math], [math]\displaystyle{ F_3 \cdot h |_a^b = 0 }[/math].

[math]\displaystyle{ = \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx }[/math]

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that [math]\displaystyle{ F_2 = \frac{d}{dx} F_3 }[/math], or in other terms, [math]\displaystyle{ F_y - \frac{d}{dx} F_y' = 0 }[/math].