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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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2 Answers 2

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 '14 at 7:44
@Drazick, your formula for the EL equation is based on a function $L$ that depends on $x$, $f$, and $f_x$; but the functional in the original post depends on the second derivative $f_{xx}$. So the formula doesn't apply. Dirk's answer gives a direct way of showing what the correct expression is. – Mark Peletier Dec 22 '14 at 10:05
@MarkPeletier, In the case above $ L = { \left( {f}_{xx} \right) }^{2} $. Let's define $ g = {f}_{x} $. Now we have $ L = {\left( {g}_{x} \right)}^{2} $. Now apply the above formula I wrote (Since now $ L $ is a function of $ {g}_{x} $) and the result is different. I'm not saying I'm right, I just don't understand why the contradiction? – Drazick Dec 22 '14 at 14:29
@Dirk, Can you please explain your moves? Thank You. – Drazick Dec 25 '14 at 7:35

Consider the functional $ F \left( u \right) = \int { \left( \frac{{d}^{2} u}{d{x}^{2}} \right) }^{2} dx $.

defining $ v = {u}_{x} $ the following is given:

$$ F \left( u \right) = \int_{\Omega} {{u}_{xx}}^{2} dx = \int_{\Omega} {{v}_{x}}^{2} dx = G \left( v \right) $$

Now, using the Gateaux derivative definition and $ L \left( x, v, {v}_{x} \right) = {{v}_{x}}^{2} $: $$ {G}' \left( v \right) = \int_{\Omega} \left( \frac{\partial}{\partial v} L \left( x, v, {v}_{x} \right) - \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) \right) h dx $$

Hence the critical point happens at $ \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) = 0 $ since $ \frac{\partial}{\partial v} L \left( x, v, {v}_{x} \right) $ vanishes.
This implies the E-L is given by $ \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) = \frac{d}{dx} 2 {v}_{x} = 2 {v}_{xx} = 0 $ which means $ {u}_{xxx} = 0 $.

This is probably a wrong answer. Yet it rose from a discussion with @Mark Peletier, hence I mark it as Community Wiki so people will be able to see why this property of the E-L can not be used here.

Thank You.

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