- #1

Hertz

- 180

- 8

Sorry, the title doesn't match up 100% with the content of the topic, but that's because I've decided to be a little bit more explicit about my question.

I am trying to walk through the proof of Euler's Equation from Calculus of Variations, and I'm a little bit confused by the final step.

Right now I have this:

[itex]\int^{x_{2}}_{x_{1}}{(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'})\eta(x) dx}=0[/itex]

and then they proceed to say that this implies:

[itex]\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}=0[/itex]

Could someone please explain why this is a correct generalization? Isn't the left side of Euler's Equation really just a function of x? And since eta is also a function of x, couldn't their product technically be an odd function and couldn't the interval technically be a symmetric interval? Wouldn't this mean that the left side of Euler's Equation does not necessarily have to equal zero in very rare circumstances?

[Original Post]

If

∫F(x)G(x)dx=0

and F(x) is arbitrary

Does it imply that

G(x)=0

?

I am trying to walk through the proof of Euler's Equation from Calculus of Variations, and I'm a little bit confused by the final step.

Right now I have this:

[itex]\int^{x_{2}}_{x_{1}}{(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'})\eta(x) dx}=0[/itex]

and then they proceed to say that this implies:

[itex]\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}=0[/itex]

Could someone please explain why this is a correct generalization? Isn't the left side of Euler's Equation really just a function of x? And since eta is also a function of x, couldn't their product technically be an odd function and couldn't the interval technically be a symmetric interval? Wouldn't this mean that the left side of Euler's Equation does not necessarily have to equal zero in very rare circumstances?

[Original Post]

If

∫F(x)G(x)dx=0

and F(x) is arbitrary

Does it imply that

G(x)=0

?

Last edited: