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I am not mathematician but an engineer. At some stage of derivation for a formulation I've just been stuck with the following integration which might be super easy and stupid question for all of you as a professional mathematician but your help is so much valuable for me.

$$ \delta J = \int \frac{du}{dx} \frac{d (\delta u)}{dx} dx $$

where $$ \delta $$ is variational operator which functions as

$$ \frac{d(\delta u)}{dx} = \delta \frac{du}{dx} $$

and

$$ \int \delta dx = \delta \int u dx $$

for differential and integral operators.

The term $$ \delta J$$ is supposed to yield to

$$ \int \frac{du}{dx} \frac{d (\delta u)}{dx} dx = \delta \int \frac{1}{2} (\frac{du}{dx})^2 dx $$

To be able to get this what I did was to take the variational operator out of the integral as the commutative rules are applied as instructed above.

$$ \int \frac{du}{dx} \delta \frac{du}{dx} dx = \delta \int \frac{du}{dx} \frac{du}{dx} dx = \delta \int (\frac{du}{dx})^2 dx $$

where 1/2 doesn't appear.

As a result I am not able to understand where the 1/2 coefficient is coming from.

  • 3
    I think you're missing a rule... you need a product rule, something like $\delta (uv) = u \delta v + v \delta u$, as well. Nothing in what you wrote allows you to move $du/dx$ "inside" the $\delta$, i.e., to go from $(du/dx) \delta (du/dx)$ to $\delta (du/dx)^2$. But using the product rule, you can write $\delta (du/dx)^2 = 2(du/dx)\delta(du/dx)$, which is where the $1/2$ comes from. – mjqxxxx Nov 12 '21 at 00:22

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