Euler-Lagrange Integral

Given that the Euler-Lagrange equation (not the usual physics form) is

\[\frac { d }{ dx } \left( \frac { \partial F }{ \partial y' } \right) -\frac { \partial F }{\partial y } =0\]

show that if \(F\) does not explicitly depend on \(x\), then the Euler equation can be integrated as

\[F-y' \frac{\partial F}{\partial y'} = C.\]

Solution

Part 1.

We must first show that \[\frac { d }{ dx } \left( F-y' \frac { \partial F }{ \partial y' } \right) -\frac { \partial F }{\partial x} =0. \]

We observe that \[\frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx} + \frac{\partial F}{\partial y'} \frac{dy'}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} y'+ \frac{\partial F}{\partial y'} y''. \]

Since \[\frac{d}{dx} \left(y' \frac{\partial F}{\partial y'} \right)=y' \frac{d}{dx} \left(\frac{\partial F}{\partial y'} \right)+\frac{\partial F}{\partial y'} y''.\]

By subtraction \[\frac{d}{dx} \left(F-y' \frac{\partial F}{\partial y'} \right) = \frac{\partial F}{\partial x} + y' \left[\frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) \right]. \]

Since \(\frac { d }{ dx } \left( \frac { \partial F }{ \partial y' } \right) -\frac { \partial F }{\partial y } =0\), we find that \[\frac { d }{ dx } \left( F-y' \frac { \partial F }{ \partial y' } \right) -\frac { \partial F }{\partial x} =0. \]

Part 2.

Since \(F\) does not explicitly depend on \(x\), then \(\frac { \partial F }{\partial x} =0 \).

By integration \[F-y' \frac{\partial F}{\partial y'} = C \] for some arbitrary constant \(C\).

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
3 years, 7 months ago

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