From the introduction: Let \(X\subset\mathbb{R}^n\), \(n\geq 2\), be an open bounded set with \(C^1\) boundary \(\partial X\) and let \(V\subset\mathbb{R}^n\) be open. Denote \(\Gamma_\pm= \{(x,v)\in \partial X\times V;\pm n(x) \cdot v>0\}\), denote by \(f\) the solution (if exists) to the stationary linear transport (Boltzmann) equation: \[ \begin{cases} -v\cdot \nabla_xf(x,v)-\sigma_a(x,v) f(x,v)+ \int_V k(x,v',v) f(x,v')dv'= 0\text{ in }X\times V,\\ f|_{\Gamma_-}=f_-. \end{cases} \tag{1} \] Here \(f_-\) is a given function on \(\Gamma_-\). We assume that the pair \((\sigma_a,k)\) is admissible, i.e.
(i) \(0\leq \sigma_a\in L^\infty (X\times V)\),
(ii) \(0\leq k(x,v',\cdot)\in L^1(V)\) for a.e. \((x,v')\in X \times V\) and \(\sigma_p (x,v'):= \int_Vk(x,v',v)dv\) belongs to \(L^\infty (X \times V)\).
If the direct problem (1) is solvable, one can define the albedo operator \({\mathcal A}:f_-\mapsto f|_{\Gamma_+}\), that maps the incoming flux on the boundary into the outgoing one. We are interested in the following inverse problem:
(IP) Does the albedo operator \({\mathcal A}\) determine uniquely the coefficients \(\sigma_a(x,v)\), \(k(x,v',v)\)?
In general, the direct problem (1) may not be uniquely solvable, so we consider two physically important situations where (1) is well posed. First we assume that \[ \|\tau \sigma_a\|_{L^\infty} <\infty,\;\|\tau\sigma_p |_{L^\infty } <\infty\text{ and }\|\tau \sigma_p\|_{L^\infty} <1.\tag{2} \] Note that (2) holds if in particular \(\||v|^{-1}\sigma_a \|_{L^\infty} <\infty, \text{diam} (X)\||v|^{-1}\sigma_p \|_{L^\infty}<1\). The second situation is when \[ \sigma_a(x,v)-\sigma_p (x,v)\geq v>0 \text{ for a.e. }(x,v) \in X\times V \tag{3} \] with some \(\nu>0\). In other words, (3) says that the absorption rate is greater than the production rate. The main result of this paper is the following:
Let \((\sigma_a,k)\), \((\widehat\sigma_a, \widehat k)\) be two admissible pairs with \(\sigma_a=\sigma_a(x,|v|)\), \(\widehat \sigma_a =\widehat \sigma_a(x,|v|)\) and assume that they satisfy either (2) or (3). Assume that the corresponding albedo operators \({\mathcal A}\) and \(\widehat {\mathcal A}\) coincide. Then
(a) if \(n\geq 3\), then \(\sigma_a= \widehat \sigma_a\), \(k=\widehat k\);
(b) if \(n=2\), then \(\sigma_a= \widehat \sigma_a\).
Our proof is constructive, and we obtain explicit formulas for \(\sigma_a\) and \(k\).