# 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
4 years, 11 months ago

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