Skip to main content
7 of 12
edited tags
Mirar
  • 350
  • 1
  • 7

Equivalent action of convolution of directional derivative

I have asked this question a while back on StackExchange but have not received any answer/comment. I received a suggestion to post the same question in here which is more research oriented.

Let $k*f(x)=\int_{}^{}k(x,t)f(x-t)dt$, where $x,t\in \mathbb{R}^{3}$, $f:\mathbb{R}^{3}\to\mathbb{R^{+}}$ and $k:\mathbb{R}^{3}\to\mathbb{R^{+}}$ for a given $t$. Additionally, let $v(x):\mathbb{R}^{3}\to\mathbb{R}^{3}$ and $D_{v(x)}(f(x))$ as directional derivative of $f(x)$ along $v(x)$. Is it possible to write $k*D_{v}(f)$ as a function of $k*f$:

$k*D_{v}(f)=F(k*f)$,

where $k*D_{v}(f)=\int{}^{}k(x,t)v(x-t)\nabla^{T}f(x-t)dt.$ Or is there any map/operator $F$ that generates $k*D_{v}(f)$ by acting on $k*f$ ?

In other words, the objective is finding $F$ in a way that path (1) and (2), in the following diagram, lead to the same function. enter image description here

Note1: In my problem, $f$ is unknown but $k$, $v$ and $k*f$ are known numerically. Therefore, in addition to $k*f$, the map $F$ can be dependent to $k$ and/or $v$ and/or $k*v$ and/or their derivatives but there should not be any dependency to $f$ or its derivatives.

Note2: I must say that I am not a mathematician and would like to apologize beforehand if my mathemtical language here and in the comment section is impercise. I welcome any comments for improvement. Thank you.

Mirar
  • 350
  • 1
  • 7