Variational Operator Homework: Part 2

[Niles]

Homework Statement

Hi

In my book they have the following functional (δ is the variational opertator):

<br /> \delta J = \int_0^1 {\left( {\frac{{du}}{{dx}}\frac{{d(\delta u)}}{{dx}} + u\delta u - x\delta u} \right)} = \delta \int_0^1 {\left( {\frac{1}{2}\left( {\frac{{du}}{{dx}}} \right)^2 + \frac{1}{2}u^2 - xu} \right)} <br />

I don't understand the second equality. They say that the integration-operator and the variational operator commute, which I agree with, but where does the factor ½ come from?


Niles.

 

[hunt_mat]

Integration by parts is the answer you're looking for.

 

[Dick]

Integration by parts is the answer you're looking for.

No, it's not integration by parts. Write u+δu as the function plus variation. Then the variation of u^2, δ(u^2) is (u+δu)^2-u^2. Which is u^2+2δu*u+(δu)^2-u^2=2δu*u+(δu)^2. Ignore the (δu)^2 since δu is a 'small variation'. So the quadratic is really small. So δ(u^2)=2δu*u. The variation acts a lot like a differentiation operator. That's where the 1/2 factors come from.

 

[Niles]

No, it's not integration by parts. Write u+δu as the function plus variation. Then the variation of u^2, δ(u^2) is (u+δu)^2-u^2. Which is u^2+2δu*u+(δu)^2-u^2=2δu*u+(δu)^2. Ignore the (δu)^2 since δu is a 'small variation'. So the quadratic is really small. So δ(u^2)=2δu*u. The variation acts a lot like a differentiation operator. That's where the 1/2 factors come from.

Thanks, but if the variational operator commutes with the integration- and differentiation operator, then why can't we just pull the δ to the left to begin with and drop the above calculation

 

[Dick]

Thanks, but if the variational operator commutes with the integration- and differentiation operator, then why can't we just pull the δ to the left to begin with and drop the above calculation

How can you just 'pull the δ to the left'? There's not just one δ. It's mixed up in products.

 

[dongo]

Sorry if this one is old, but as far as I know, variation uses expanding the "deformed function", i.e u+\delta u, in a Taylor series and ignoring all terms of order higher than one (so to speak, we perform some linearization trick.). As Dick pointed out, we have to subtract the original integral from the linearized Taylor expansion in order to obtain the variation. Hope this aids at clarification!