Timeline for Equivalent action of convolution of directional derivative
Current License: CC BY-SA 4.0
8 events
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| Mar 4, 2022 at 10:47 | comment | added | Carlo Beenakker | well, since a non-invertible $k$ is of measure zero that should not be a problem; think of it like one big matrix; a zero eigenvalue will require fine tuning of the matrix elements, it will not happen generically. | |
| Mar 4, 2022 at 10:19 | comment | added | Mirar | Thank you for the answer. We know that $k*f$ exists and equals to $I$. Intuitively speaking, $q$ may not exist but it still should be possible to find a non-vanishing $k*f$ for some $k$ because it is just a convolution process, especially since $k$ is spatially variant. I am probably missing something here. Additionally, even if $k*f$ vanishes, $k*D_{v}(f)$ may also vanish (?). Would not that be OK as long as finding $F$ is the objective? | |
| Mar 4, 2022 at 9:57 | comment | added | Carlo Beenakker | if $q$ does not exist it means that $k\ast f$ may vanish and then the reconstruction of $k\ast D_v f$ from $k\ast f$ will fail. | |
| Mar 4, 2022 at 9:22 | comment | added | Mirar | Please allow me to ask another related question. Is it given that $q$ exists? | |
| Mar 4, 2022 at 9:15 | comment | added | Mirar | We initially tried to calculate $f$ by constructing a matrix equation $K.vec(f)=I$, where $vec(f)$ is the vectorized form of $f$. However, we faced some difficulties. Since $f$ is a 3D data, the matrix $K$ is very large ($N_{x} \times N_{y} \times N_{z})^{2}$) and since, in our case, $k$ has a large spatial support, $K$ is not sparse and it takes a large memory space which makes it impractical to use. We even tried to downsample the data but did not help. | |
| Mar 3, 2022 at 22:03 | comment | added | Carlo Beenakker | you are right, Fourier transformation does not simplify the calculation of the inverse kernel; what I would try is to calculate it numerically: discretize $x$ and $t$, so that $k(x,t)$ is a finite-dimensional matrix, then $q$ is the inverse matrix. | |
| Mar 3, 2022 at 21:42 | comment | added | Mirar | Thank you Carlo, I really appreciate it. I have two questions: 1. Are we garanteed that the inverse of $k$ exists even if it is numerically known? 2. I persume that you are referring to the convolution theorem when saying to calculate $q$ by Fourier transformation. I understand that if the convolution kernel is space/time invariant, the Fourier transform of $q$ is one over the Fourier transform of $k$. However, the kernel here is space/time variant. Does the theorem apply? | |
| Mar 3, 2022 at 20:44 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |