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If $u: \mathbb{R}^m \rightarrow \mathbb{R}$ then what is the following solution to: $$ \dfrac{\partial u}{\partial \nabla u} = \hspace{2mm} ? $$

Where $\dfrac{\partial u}{\partial \nabla u}$ is the derivative of a scalar with respect to a vector.

Edit

The first version of the question did not really make much sense. I guess what I meant to say was, what is $$ \dfrac{df}{df'} = \hspace{5mm} ? $$ in an attempt to solve the next problem. I have since updated the question to fit the definitions.

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  • $\begingroup$ How do you define these expressions? $\endgroup$ – gerw Oct 12 '17 at 8:05
  • $\begingroup$ @gerw I noticed that my first question does not really make sense. The second expression $\partial u/ \partial \nabla u$, however, is defined as the derivative of a scalar with respect to a vector. $\endgroup$ – Carlos Brito Oct 13 '17 at 16:45
  • $\begingroup$ I have updated the question with the proper definitions. $\endgroup$ – Carlos Brito Oct 13 '17 at 16:51
  • $\begingroup$ What possible purpose doa striking out text serves? If the text useful, keep it — if it is not, delete it. $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '17 at 16:53
  • $\begingroup$ @MarianoSuárez-Álvarez as you wish $\endgroup$ – Carlos Brito Oct 13 '17 at 17:18
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Fix $x$ and let $\gamma(x)$ be some path from $0$ to $x$. Then $f(x) = \int_{\gamma(x)} \nabla f \cdot dr.$ The functional derivative of this w.r.t. $\nabla f$ is defined as the linear functional (often a distribution) $\delta u$ given by $$\langle u, \phi \rangle = \left. \frac{d}{d\lambda} \int_{\gamma(x)} \nabla (f+\lambda\phi) \cdot dr \right|_{\lambda=0}$$ Now, the right hand side equals $$\left. \int_{\gamma(x)} \frac{\partial}{\partial\lambda} \nabla (f+\lambda\phi) \cdot dr \right|_{\lambda=0} = \left. \int_{\gamma(x)} \nabla \phi \cdot dr \right|_{\lambda=0} = \int_{\gamma(x)} \nabla \phi \cdot dr $$ Thus, $$\langle u, \phi \rangle = \int_{\gamma(x)} \nabla \phi \cdot dr = \phi(x) - \phi(0) = \langle \delta(t-x) - \delta(t), \phi(t) \rangle$$ so the functional derivative is $$\frac{\partial f(x)}{\partial \nabla f(t)} = \delta(t-x) - \delta(t)$$ Luckily this result doesn't depend on the choice of $\gamma(x).$

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  • $\begingroup$ Thank you! This is perfect. One small question though, if the domain of integration is fixed (ie. does not depend on $x$) then what would happen? Would the functional derivative equal 1? I'm sorry if this doesn't make sense, but I'm not far into the study of functionals. $\endgroup$ – Carlos Brito Oct 13 '17 at 18:57
  • $\begingroup$ @CarlosBrito. I'm not sure I get your question, but notice that I never vary $x$, it's kept fixed. $\endgroup$ – md2perpe Oct 13 '17 at 19:04

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