输运方程在单能，纯吸收介质的固定源问题中有解析解。下面对解析解进行推导
单能纯吸收介质的解析解
$$\phi(r,\Omega)= \int s(r,Ω)exp(-\intσ_a(l’)dl’)dl$$
源区内
$$\phi(r,\Omega)= \int s(r,Ω)exp(-\int_0^lσ_a(l’)dl’)dl$$
假设只有一个源区，并且源区内的吸收截面为常数 $σ_{a,1}$，所以
$$\phi(r,\Omega)=\int s(r,Ω)exp(-σ_{a,1}l)dl = \frac{s(r,Ω)}{σ_{a,1}}(1-exp(-σ_{a,1}l_0(r,Ω))$$
其中 $l_0(r,Ω)$为$r$处沿着 $\Omega$方向到源区边界的距离。
源区外
$$\phi(r,\Omega)= \int s(r,Ω)exp(-\int_0^lσ_a(l’)dl’)dl$$
假设一共有$n$个区域，每个区域的吸收截面为 $σ_{a,i}$，则
$$\phi(r,\Omega)= \int_{l_0(r,Ω)}^{l_1(r,Ω)} s(r,Ω)exp(-\sigma_{a,1}(l-l_1)) \prod_{i=2}^{n}exp(-σ_{a,i}\delta l_i(r,Ω))dl$$
所以
$$\phi(r,\Omega) = \frac{s(r,Ω)}{σ_{a,1}}(1-exp(-σ_{a,1}δl_1(r,Ω))\prod_{i=2}^{n}exp(-σ_{a,i}\delta l_i(r,Ω))dl$$