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I have encountered the following problem regarding the reconstruction of a particle density function $f$. It is possible to acquire "measurements" of the particle density function $f : [0,1]^2 \rightarrow \mathbb{R}^{>0} $. The discrete measurements $m_{i}=\int_{[0,1]^2}f(x) g_{i}(x)dx$ are given by a weighted integral of $f$ with the weighting functions $g_{i}$. The functions $g_{i}$ are known in advance and the task is to reconstruct $f$ numerically given the measurements $m_{i}$.

Unfortunately I cannot give much information about the weighting functions $g_{i}$. They are the output of a computer simulation and I have them sampled on a grid spanning the domain of $f$ (e.g. $[0,1]^2$). Generally speaking, the functions $g_{i}$ look a bit like the picture below where the red color corresponds to $g_{i}(x,y)=0$ and the blue/greenish color corresponds to $g_{i}(x,y)=1$ (please ignore the dark outlines):

Other sampling function are similar to rotated versions of $g_{i}$.

I am wondering if there is a general numerical approach to tackling this kind of problem?

Sampling function $g_{i}$. Red zones correspond to $g_{i}=0$.

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  • $\begingroup$ basically this is an underdetermined linear set of equations and can be solved using the pseudo inverse. $\endgroup$
    – user619894
    Aug 11, 2021 at 21:18

1 Answer 1

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The function $f(x, y)$ can be expressed as a linear combinations of the weighting functions:

$f(x, y) = \displaystyle \sum_j f_j g_j(x, y) $

Hence,

$m_i = \displaystyle \int f(x,y) g_i(x,y) = \sum_j f_j \int g_j(x,y) g_i(x,y) $

Define the square matrix $G = [ G_{ij} ]$, and $G_{ij} = \displaystyle \int g_i(x,y) g_j(x,y) $

Then we have the matrix equation

$ G \mathbf{f} = \mathbf{m} $

Hence, the vector $ \mathbf{f} = G^{-1} \mathbf{m} $

Now we can reconstruct $f(x, y)$ from the vector $\mathbf{f}$ and $\{ g_i(x,y) \}$.

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  • $\begingroup$ Notice that the problem is underdetermined: $f$ can be expanded into an infinite sum of basis functions. A judgment based on some domain specific assumptions must be done to decide what the "best" truncation is. $\endgroup$
    – user619894
    Aug 12, 2021 at 6:34

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