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How do I perform the derivative of

$$-f\nabla^2f$$ with respect to $f$, i.e.

$$\frac{\delta(-f\nabla^2f)}{\delta f}~?$$

The answer is supposedly $-2\nabla^2f$.

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    Please elaborate on the physics context of this question and how it differs from pure [mathematics.se]. Thanks! – jng224 Mar 29 '21 at 18:47

1 Answers1

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That's not an ordinary derivative; it's a variational derivative. (In simpler notation): $$ \left(\partial_{kk}\frac{\partial}{\partial f_{,kk}}-\partial_{k}\frac{\partial}{\partial f_{,k}}+\frac{\partial}{\partial f}\right)\left(-ff_{,kk}\right)=\partial_{kk}\left(-f\right)+\left(-f_{,kk}\right)=-f_{,kk}-f_{,kk}=-2f_{,kk} $$

joigus
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  • it's a variational derivative With the functional implied! – G. Smith Mar 29 '21 at 20:00
  • I think it's also called "functional derivative"... Correct me if I'm wrong, @G.Smith. Thing is... in physics rarely ever people consider higher than first-order Lagrangians, and are not aware of general definition... For general expression $f_{,ijk...}$ you have to include factorial denominators to correct for over-counting. – joigus Mar 29 '21 at 20:05
  • Yes. Wikipedia has the general formula. But I don’t see the factorial denominators you mention. I think it’s hard for the OP to understand the negative sign in the second term unless they understand that it comes from integrating by parts in the (implied) functional integral. – G. Smith Mar 29 '21 at 20:20
  • One would if/when one had non-isotropic terms like $f_{,ijk}$. For indices that run from $1$ to $n$: $$\frac{\partial}{\partial f_{,i_{1}i_{2}\cdots i_{n}}}f_{,i_{1}i_{2}\cdots i_{n}}=\sum_{\pi\in S_{n}}\frac{\partial}{\partial f_{\pi\left(12\cdots n\right)}}f_{,\pi\left(12\cdots n\right)}=n!\frac{\partial}{\partial f_{,12\cdots n}}f_{,12\cdots n}$$. But it's not on Wikipedia, probably because we rarely ever consider those. Otherwise we would be overcounting $f_{123}$, $f_{231}$, etc. – joigus Mar 29 '21 at 21:08