Green's Function fo the Optical Path Difference In this paper (1), the authors claim that the solution of the following Poisson equation
$$
k\nabla^2 T(\mathbf r) = P\delta(\mathbf r)
$$
is
$$
T(r)=T_0+\frac{P_0}{4\pi k r}=T_0+\frac{P_0}{4\pi k }G_T(r)
$$
where they say that $G_T$ is the green function for the temperature $T$.
Now, they are interested in the optical path difference due to the increase of temperature. The two quantities are related by the formula
$$
\Delta \ell=\int_0^h\beta T(r) dz
$$
where $h$ is the thickness of a material. Performing the calculation they obtain
$$
\Delta \ell=\frac{P_0\beta}{4\pi k }\sinh^{-1}(h/\rho)=\frac{P_0}{4\pi k }G_{\ell}(\rho)
$$
where $\rho$ is the radius on the x-y plane. So, they claim that $\beta \sinh^{-1}(h/\rho)$ is the Green Function for the quantity $\Delta \ell$, so it is possible to calculate any $\Delta \ell$, given any source term just by convolution. Is this claim fair? If so, why?
(1): Baffou, Guillaume & Bon, Pierre & Savatier, Julien & Polleux, Julien & Zhu, Min & Merlin, Marine & Rigneault, Hervé & Monneret, Serge. (2012). Thermal Imaging of Nanostructures by Quantitative Optical Phase Analysis. ACS nano. 6. 2452-8. 10.1021/nn2047586.
 A: The heat diffusion equation is linear WRT the sources, meaning that the solution for any distribution of sources can be obtained from the convolution of the source distribution with the Green function corresponding to a point source. If you assume that the optical index varies linearly with $T$, then you can write the path difference as a certain integral over $T(r)$. The point is, because this operation is also linear, the final results remains linear WRT the sources, so you can write a Green function for the optical path difference and convolve it with your souce distribution, just as you did when solving for $T$.
I am also assuming that the charges are all located at $z=0$ for the Green function to depend only on the distance in the $xy$ plane.

To convince yourself, you can also write explicitely the dependency of the optical path on the distribution of sources. From the Green's function for the temperature $T$:
$$T(r) = \int G_T (|r-r'|) S(r') dr'$$
And from $\Delta l = \beta \int T(r) dz$, you find:
$$\Delta l = \beta \int \limits_{z} \int \limits_{r'} G_T (|r-r'|) S(r') dr' dz = \int \limits_{r'} \left( \beta \int \limits_{z}  G_T (|r-r'|)  dz \right) S(r') dr' = \int \limits_{\rho} G_{\Delta l}(\rho) S(\rho) d \rho,$$
where $G_{\Delta l}(\rho) = \beta \int G_T (\sqrt{\rho^2 + z^2}) dz$ is the Green function for optical path difference (assuming all sources are located at $z=0$ so that it only depends on $\rho$).
