4
$\begingroup$

Let $p$ be a scalar field $p: \mathbb R^n \to \mathbb R$. I encountered the problem of reconstructing an unknown density $p$ from its integral values $$I(t,z) = \int_{V_t} p(x) dS$$ along a one-parameter family $V_t$ of manifolds.

I am interested in conditions $V_t$, so that the reconstruction problem admits a unique solution, but I would be happy with any kind of hints, remarks, references on this problem.

$\endgroup$
2
  • $\begingroup$ I'm confused about your notation. Do you mean $(H \Phi_t)^{-1}({\{z\}})=\{x\in\mathbb R^n;H(x)\Phi_t(x)=z\}$ where $z\in\mathbb R^n$ and $t\geq0$ are fixed? This set is typically zero dimensional, so I don't quite understand the integral. $\endgroup$ Nov 29, 2014 at 8:35
  • $\begingroup$ I mean $(H \circ \Phi)^{-1}(\{z\})$. Sorry for the confusion. $\endgroup$
    – user45183
    Nov 29, 2014 at 8:37

1 Answer 1

1
$\begingroup$

There are counterexamples to your problem. If the flow $\Phi_t$ preserves the level sets of $H$, then all you can recover are the integrals over level sets of $H$.

This information does not determine $p$: Let for example $H(x)=|x|^2$ and let $p$ be any compactly supported, smooth, odd ($p(-x)=-p(x)$) function. Then $I(z,t)=0$ for all $z,t$. If you want to require $p\geq0$, choose your favorite positive function and this example to it; you will get two different positive functions with the same $I$.

It seems that even if the flow $\Phi_t$ and the function $H$ are nice, reconstruction should only be possible in dimension two. (I assume you are not interested in the case $n=1$.) The reason is that you want to recover a function $p$ that depends on $n$ variables and you know a function $I$ that depends on two variables. Dimension counting arguments are not foolproof, but they are very reliable in finding situations where something is impossible.

Here is an example in two dimensions where recovery is possible if $t$ is allowed to range in $\mathbb R$ and $H$ has a mild singularity. (I assume one could find an example in the exact original setting, but want to keep the example simple.) Suppose $p$ is continuous and compactly supported (actually $L^1$ seems to be enough, but regularity is not the main point here). Let $F$ be any constant nonzero vector field. Let $H(x,y)$ be the angle the vector $(x,y)$ makes with the positive $x$ axis; for example when $x>0$, we have $H(x,y)=\arctan(x/y)$. At zero assign any value to $H$.

Now $I(0,\alpha)+I(0,\alpha+\pi)$ is the integral of $p$ over the line that goes through the origin at angle $\alpha$. Therefore $I(t,\alpha)+I(t,\alpha+\pi)$ is the integral over the same line translated by the vector $tF$. This means that you can recover the integrals of $p$ over all lines except those parallel to $F$ that do not meet the origin. But you can approximate these lines with other ones, so using the regularity of $f$ you can indeed recover the integrals over all lines. This means that you know the Radon transform of $f$, and since the Radon transform is injective, you can recover $f$. In this case you can even write down a reconstruction formula if you want.

Your full problem is a generalized Radon transform, but I don't know a reference for just this kind of situation.

$\endgroup$
3
  • $\begingroup$ It holds that $(H \circ \Phi_t)^{-1}(\{z\}) = \Phi_{-t} (H^{-1}(\{z\}))$! $\endgroup$
    – user45183
    Nov 29, 2014 at 8:56
  • $\begingroup$ Thanks! I edited my answer. I plan to write more when I have more time later on. $\endgroup$ Nov 29, 2014 at 9:10
  • $\begingroup$ Thanks. I was aware of the case when the flow leaves the level sets of $H$ invariant. Loosely speaking, the flow needs to break symmetries of $H$ and vice versa. $\endgroup$
    – user45183
    Nov 29, 2014 at 9:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.