Let $\mu a(r,\lambda )$ be the absorption, $\mu s\u2032(r,\lambda )$ the reduced scattering, and $\kappa (r,\lambda )={3[\mu a(r,\lambda )+\mu s\u2032(r,\lambda )]}\u22121$ the diffusion coefficient for a wavelength $\lambda $ at position $r$ in a domain $\Omega $. In a diffuse range, which is valid for highly scattering ($\mu s\u2032\u226b\mu a$) and weakly anisotropic media, photon density is approximated by the diffusion equation^{5}. Hence, in fDOT, the excitation photon density $\Phi ex(r,\lambda ex)$ at the excitation wavelength $\lambda ex$ and emission photon density $\Phi em(r,\lambda em)$ at the emission wavelength $\lambda em$ are given by the solution of a pair of coupled diffusion equations.^{6}^{,}^{21} The excitation photon density is emitted by an external source $q0(rs)$ at a location $rs\u2208\Omega $, and the emission photon intensity comes from a fluorescent region characterized by a fluorescent yield $F(rfl)$, which accounts for its quantum efficiency, absorption parameter and the concentration of fluorophore. Assuming a time-stationary problem (frequency $w=0$) and that the absorption parameter is not affected by the presence of the fluorophore, excitation and emission photon densities are given by Display Formula
$\u2212\u2207\xb7\kappa (r,\lambda ex)\u2207\Phi ex(r,\lambda ex)+\mu a(r,\lambda ex)\Phi ex(r,\lambda ex)=q0(rs)$(1)
Display Formula$\u2212\u2207\xb7\kappa (r,\lambda em)\u2207\Phi em(r,\lambda em)+\mu a(r,\lambda em)\Phi em(r,\lambda em)=F(rfl)\Phi ex(r,\lambda ex).$(2)