12-267/Derivation of Euler-Lagrange

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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