# Variational operator

1. Jul 26, 2011

### Niles

1. The problem statement, all variables and given/known data
Hi

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

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

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?

Best,
Niles.

2. Jul 26, 2011

### hunt_mat

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

3. Jul 26, 2011

### Dick

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.

4. Jul 27, 2011

### Niles

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

5. Jul 27, 2011

### Dick

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

6. Jul 28, 2011

### 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!