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According to page 7 of the PDF document

$$ \frac{\delta}{\delta f} \int \left( \frac{df^2 }{d^2 x} \right)^2 dx = \int \frac{df^4}{d^4 x} dx $$

I would like help proving this statement.

Although I can show that

$$ \int \left( \frac{df^2 }{d^2 x} \right)^2 dx = \int \frac{df^4}{d^4 x} f dx $$

My attempts at "constructing" the functional derivative of this expression isn't dropping the term $f$. I'm not even sure this is the right way to go about solving the problem.

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1 Answer 1

Well, probably this is a case of confusing notation. Especially I am not sure where you want to have your exponents.

In my understanding, both equations you stated are just wrong. However, the first one may carry some truth: Consider the functional $J(f) = \int (\frac{d^2 f}{dx^2})^2 dx$. Then the functional derivative of $J$ (or first variation) of $J$ is (by integration by parts) $$ \delta J(f)(h) = \int 2\frac{d^2f}{dx^2}\frac{d^2h}{dx^2}dx = \int 2\frac{d^4 f}{dx^4} h dx. $$ Hence, one may say (if put in a proper framework of function spaces) that the derivative of $J$ is $$ J'(f) = 2\frac{d^4 f}{dx^4}. $$

To conclude: The pdf document you linked is very sloppy with the notation and probably you may consult a book one the calculus of variations to get more information here.

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Thanks. I get the first part, its not too different from my second equation, which BTW does not take the variation to account -- just an initial integration by parts. My problem is with the argument that leads to the second statement. Specifically, what do you mean by "if put in a proper framework of function spaces"? –  Olumide Apr 15 '11 at 14:43
@Dirk, I don't understand the solution. As far as I can see the E-L for this functional is different. Setting $ L \left( x, f, {f}_{x} \right) $ the E-L states that the derivative is $ {L}_{f} - \frac{d}{dx} {L}_{{f}_{x}} $. Since $ {L}_{f} $ vanishes we're left with $ {L}_{{f}_{x}} = 2 {f}_{xx} $ deriving by x again yields $ 2 {f}_{xxx} $ while you have $ 2 {f}_{xxxx} $. –  Drazick Dec 14 at 7:44

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