12-267/Derivation of Euler-Lagrange: Difference between revisions

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

Special cases (without derivations):

In the case that F does not depend on y', we have <math>F_y = 0</math>

In the case that F does not depend on y, we have <math>F_{y'} = c</math>

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

Latest revision as of 17:18, 24 October 2012

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

For a function defined on to be an extremum of , it must be that for any function defined on that preserves the endpoints of (that is, and ), we have .

Let signify F differentiated with respect to its nth variable.

(integrating by parts)

Due to the constraints of and , .

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that , or in other terms, .

Special cases (without derivations):

In the case that F does not depend on y', we have

In the case that F does not depend on y, we have

In the case that F does not depend on x, we have