Euler-Lagrange Integral

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

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

show that if FF does not explicitly depend on xx, then the Euler equation can be integrated as

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

Solution

Part 1.

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

We observe that dFdx=Fx+Fydydx+Fydydx=Fx+Fyy+Fyy.\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 ddx(yFy)=yddx(Fy)+Fyy.\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 ddx(FyFy)=Fx+y[Fyddx(Fy)].\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 ddx(Fy)Fy=0\frac { d }{ dx } \left( \frac { \partial F }{ \partial y' } \right) -\frac { \partial F }{\partial y } =0, we find that ddx(FyFy)Fx=0.\frac { d }{ dx } \left( F-y' \frac { \partial F }{ \partial y' } \right) -\frac { \partial F }{\partial x} =0.

Part 2.

Since FF does not explicitly depend on xx, then Fx=0\frac { \partial F }{\partial x} =0 .

By integration FyFy=CF-y' \frac{\partial F}{\partial y'} = C for some arbitrary constant CC.

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

#Calculus #CalculusofVariation

Note by Steven Zheng
6 years, 6 months ago

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