辐射传输方程的分解

Decomposition of the Boundary Value Problem for Radiative Transfer Equation of MODIS and MISR instruments

我才发现公式的变化并没有能够带入进来，读者可自己编号。

0.Notions

Let $L$ be the straming-collision operator, and $S$ is scattering operator:
$$\begin{gathered} LI=\Omega \cdot \nabla I(r,\Omega )+\sigma (r,\Omega )I(r,\Omega ) \\ S={\int }_{4\pi }{\sigma }_s(r,{\Omega }^{\prime },\Omega )I(r,{\Omega }^{\prime })d{\Omega }^{\prime } \end{gathered}$$
To descirbe reflective properties of boundary $\delta V$, the scattering operator defined on the boundary $\delta V$ for radiation exiting the regine $V$ is indroduced as
$$\mathcal{R}{I}^{+}=\frac{1}{\pi }{\int }_{\delta V}d{r}_B^{\prime }{\int }_{n(r_B^{\prime })\cdot {\Omega }^{\prime }>0}{\rho }_B(r_B^{\prime },{\Omega }^{\prime };r_B,\Omega )|n(r_B^{\prime })\cdot {\Omega }^{\prime }|I^{+}(r_B^{\prime },{\Omega }^{\prime })d{\Omega }^{\prime }$$
Here, we use $I^{+}$ to represent the radiance exiting and $I^{-}$ entering domina V. And then, the stationary radiative transfer equation could be expressed as
$$LI=SI+q{I}^{-}=\mathcal{R}I+q_B$$
And if $\mathcal{R}$ is 0 and $q_B$ is 0, the porblem is called standard porblem and the straming-collision oprator would be write as $L_0$.

1.Black Soil Problem

The RTE is
$$LI=SI+q{I}^{-}=\mathcal{R}I+q_B.$$
As in Green function blog before, we have
$$I(r,\Omega )=I_{bs}(r,\Omega )+I_{rest}(r,\Omega )$$
The first question is black soil problem, means that the bottom of canopy is non-reflecting surface (soil absorption is 1, black soil), so its the solution of
$$\begin{array}{rl}& L{I}_{bs}=S{I}_{bs}\\ & I_{bs}^{(}{r}_t,\Omega )=c_T\delta (\Omega ,{\Omega }_0)+d(r_t,\Omega )\\ & I_{bs}^{-}(r_l,\Omega )=0\\ & I_{bs}^{-}(r_b,\Omega )=0\end{array}$$
Here, we assumte the lateral effects is zero.

2.Canopy-Surface Interation

For notion, we use $I_r$ for the $I_{rest}$. We have that
$$L{I}_r=S{I}_r$$
and the boundary conditions is
$$\begin{array}{rl}& I_r^{-}(r_t,\Omega )=0\\ & I_r^{-}(r_l,\Omega )=0\\ & I_r^{-}(r_b,\Omega )=\mathcal{R}{I}^{+}\end{array}$$
The solution $I_r$ of this boundary value problem depend on the solution of “complete tranport problem” $I=I_{bs}+I_r$ (since $\mathcal{R}{I}^{+}$) is also not known. It need further decompose.

The lower boundary conditions can be rewritten as
$$I_r^{-}(r_b,\Omega )=\frac{\mathcal{R}{I}^{+}}{T}T$$
Here, $T$​ is downward radiation flux density at canopy bottom, i.e.,
$$T(r_b)={\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}I(r_b,{\Omega }^{\prime })|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime },r_b\in \delta V$$
The ratio $\frac{\mathcal{R}{I}^{+}}{T}$ is a factor of $\pi$ smaller than HDRF (since there is no $1/\pi$ in the dominator compared to HDRF).

And then ground BHR could be writen as
$${\rho }_{eff}(r_b)=\frac{{\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}{\int }_{n(r_b)\cdot \Omega <0}|n(r_b)\cdot {\Omega }^{\prime }|{\pi }^{-1}{\rho }_b(r_b,{\Omega }^{\prime },\Omega )|\Omega \cdot n(r_b)|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }d\Omega }{{\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}|{\Omega }^{\prime }\cdot n(r_b)|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }}$$
Here, ${\rho }_b$ is the BRF of canopy ground. So we could find that the denomator is actually the $T$, which means that ${\rho }_{eff}=\frac{something}{T}$.

For horizontally inhomogeneous vegetation canopy, $T$ vary a lot, but BHR would not so.

Now, we introduce the effective ground anisotropy:
$$\begin{array}{rl}{d}_b(r_b,\Omega )& =\frac{\mathcal{R}{I}^{+}}{T{\rho }_{eff}}\\ & =\frac{1}{{\rho }_{eff}}\cdot \frac{{\int }_{2\pi +}{\rho }_b(r_b,{\Omega }^{\prime },\Omega )|n(r_b)\cdot {\Omega }^{\prime }|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }}{\pi {\int }_{2\pi -}|{\Omega }^{\prime }\cdot n(r_b)|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }}\end{array}$$
This value seems strange. But if we use ${\rho }_{eff}=\frac{something}{T}$​, we could find that this is actually tha ratio of reflected flux density and radiance. So its cosine-weighted intergral over upward directions is unity:
$${\int }_{2\pi +}{d}_b(r_b,\Omega )|n(r_b)\cdot \Omega |d\Omega =1.$$
So this term descibed the inhomogenous of the surface., its something with denonimator a flux density and numenator is reflected radiance with some direction.

By using the Equation (12), we obtain
$$I_r^{-}(r_b,\Omega )=\mathcal{R}{I}^{+}={\rho }_{eff}T(r_b)d_b(r_b,\Omega )$$
Using Green function method with $q=I_r^{-}$, we obtain
$$I_r(r,\Omega )={\int }_{\delta V}d{r}_b^{\prime }{\int }_{n(r_b^{\prime })\cdot {\Omega }^{\prime }<0}d{\Omega }^{\prime }{G}_S(r,\Omega ;r_b^{\prime },{\Omega }^{\prime }){\rho }_{eff}(r_b^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })T(r_b^{\prime })$$

Substituing this into Equation (5) we get that
$$I(r,\Omega )=I_{bs}(r,\Omega )+{\int }_{\delta V_b}d{r}_b^{\prime }{\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })J(r,\Omega ;r_b^{\prime })$$
where
$$J(r,\Omega ;r_b^{\prime })={\int }_{n(r_b^{\prime })\cdot {\Omega }^{\prime }<0}d{\Omega }^{\prime }{G}_S(r,\Omega ;r^{\prime },{\Omega }^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })$$
By the theory of Green function, $J$ is the intensity of radiation field at r with derection $\Omega$ generated by a point anisotropic source $d_b(r_b^{\prime },{\Omega }^{\prime })\delta (r_b,r_b^{\prime })$ located at $r_b^{\prime }$.

With $T(r_b)={\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}I(r_b,{\Omega }^{\prime })|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }$ and (16), (17), we obtain
$$\begin{array}{rl}T(r_b)=& {\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}(I_{bs}(r_b,{\Omega }^{\prime })\\ & +{\int }_{\delta V}d{r}_b^{\prime }{\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })J(r_b,{\Omega }^{\prime };r_b^{\prime }))|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }\end{array}$$
By letting
$$T_{bs}={\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}{I}_{bs}(r_b,\Omega )|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }$$
and
$$\begin{array}{rl}{G}_d(r_b,r_b^{\prime })& ={\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}d{\Omega }^{\prime }J(r_b,{\Omega }^{\prime };r_b^{\prime })|{\Omega }^{\prime }\cdot n(r_b^{\prime })|\\ & ={\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}|{\Omega }^{\prime }\cdot n(r_b^{\prime })|d{\Omega }^{\prime }{\int }_{n(r_b^{\prime })\cdot {\Omega }^{\prime }<0}d{\Omega }^{\prime }{G}_S(r,\Omega ;r^{\prime },{\Omega }^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })\end{array}$$
Noting that $G_d$ is downward radiation flux density at $r_b$ due to point anisotropic source $d_b(r_b,{\Omega }^{\prime })\delta (r_b,r_b^{\prime })$.

The (18) becomes
$$T(r_b)=T_{bs}+{\int }_{\delta V}{G}_d(r,r^{\prime }){\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })d{r}_b^{\prime }$$
Since the $J$ is the radiance generated by $d_b(r_b^{\prime },{\Omega }^{\prime })\delta (r_b,r_b^{\prime })$, so $G_d$​ is actually the flux density generated also by it.

Now, we note that
$$\begin{array}{rl}{I}_S(r,\Omega )& ={\int }_{\delta V}d{r}_b^{\prime }J(r,\Omega ;r_b^{\prime })\\ & ={\int }_{\delta V}d{r}_b^{\prime }{\int }_{n(r_b^{\prime })\cdot {\Omega }^{\prime }<0}d{\Omega }^{\prime }{G}_S(r,\Omega ;r^{\prime },{\Omega }^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })\end{array}$$
is the intensity of radiation field in canopy generated by anisotropic and heterogeneoussource source $d_b$ located at canopy bottom. It satiesfy the equation
$$L{I}_S=S{I}_S$$
and boundary condtions $I_S^{-}(r_t,\Omega )=0$, $I_S^{-}(r_l,\Omega )=0$ and $I^{-}(r_b,\Omega )=d_b(r_b,\Omega )$.

This is the second “basic problem”, “S problem”. It follows that
$${\int }_{{\delta }_V}{G}_d(r,r_b^{\prime })d{r}_b^{\prime }={\int }_{n(r_b)\cdot \Omega >0}{I}_S(r_b,\Omega )|\Omega \cdot n(r_b)|d\Omega$$
The integral of $G_d$ over boundary is downward flux at $r_b$ which acount for contribution from all anisotropic sources at canopy bottom.

Rethinking the equation (16), the J is actually a weight to weight the all reflected flux density.

The ground BHR (Equation (11)) and the effective ground anisotropy is independent on vegetation canopy and are known.

Once $G_d$ is given, one can solve Equation (18) and substitude it into (15).

Solution of horizontally homogeneous vegetation canopy with Lambertian surface

If we assume that the gournd BRF is homegeneous Lambertian surface, then, $T_{bs}$​ and T is independent to (x,y) . And gournd BHR become
$$\begin{array}{rl}{\rho }_{eff}& =\frac{{\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}{\int }_{n(r_b)\cdot \Omega <0}|n(r_b)\cdot {\Omega }^{\prime }|{\pi }^{-1}{\rho }_b(r_b,{\Omega }^{\prime },\Omega )|\Omega \cdot n(r_b)|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }d\Omega }{{\int }_{n(r_b)\cdot {\Omega }^{\prime }>0}|{\Omega }^{\prime }\cdot n(r_b)|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }}\\ & =\frac{{\rho }_b{\int }_{n(r_b)\cdot \Omega <0}d\Omega |\Omega \cdot n(r_b)|}{\pi }\\ & ={\rho }_b\end{array}$$
and effective ground anisotropy is
$$\begin{array}{rl}{d}_b(r_b,\Omega )& =\frac{\mathcal{R}{I}^{+}}{T{\rho }_{eff}}\\ & =\frac{1}{{\rho }_{eff}}\cdot \frac{{\int }_{2\pi +}{\rho }_b(r_b,{\Omega }^{\prime },\Omega )|n(r_b)\cdot {\Omega }^{\prime }|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }}{\pi {\int }_{2\pi -}|{\Omega }^{\prime }\cdot n(r_b)|I(r_b,{\Omega }^{\prime })d{\Omega }^{\prime }}\\ & =\frac{{\rho }_b}{{\rho }_{eff}\pi }=\frac{1}{\pi }\end{array}$$
Then the Equation (21) become
$$\begin{array}{rl}T& =T_{bs}+{\int }_{\delta V}{G}_d(r_b,r_b^{\prime }){\rho }_{eff}Td{r}_b^{\prime }\\ & =T_{bs}+{\rho }_{b}T{\int }_{\delta V}{G}_d(r_b,r_b^{\prime })d{r}_b^{\prime }\\ & \to T(1-{\rho }_{b}T{R}_S)=T_{bs}\\ & \to T=\frac{T_{bs}}{1-{\rho }_b{R}_s}\end{array}$$
where
$$R_S={\int }_{\delta V}{G}_d(r_b,r_b^{\prime })d{r}_b^{\prime }$$
is the downward flux density at $r_b$ generated by source $\frac{1}{\pi }$ (since $d_b$ has become $1/\pi$ and combine the Equation (24) and (23)).

Substituting Eq. (22) into Eq. (16), we obtain:
$$I(r,\Omega )=I_{bs}(r,\Omega )+I_S(r,\Omega ){\rho }_{eff}T$$
Using ${\rho }_{eff}={\rho }_b$, Then, by Eq.(27), we obtain:
$$I(r,\Omega )=I_{bs}(r,\Omega )+\frac{T_{bs}{\rho }_b}{1-{\rho }_b{R}_S}{I}_S(r,\Omega )$$
So, the solution of original boundary problem is decomposed into the solution of black-soil problem and S problem.

Solution of horizontally inhomogeneous vegetation canopy

In this case, the solution of Eq. (21) is not directly Eq. (27), and approximation is performed. Consider the ratio
$$R_S(r_b)=\frac{{\int }_{\delta V}{G}_d(r_b,r_b^{\prime }){\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })d{r}_b^{\prime }}{{\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })}$$
which is BHR calculated for canopy illuminated by anisotropic sources ${\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })$ from below. To understand this, we could think that the numerator is the responsed flux density at $r_b$ due to radiance ${\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })$ by Eq. (24), and the denomator is
$${\int }_{n(r_b^{\prime })\cdot {\Omega }^{\prime }<0}{\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })d(r_b^{\prime }{\Omega }^{\prime })|n(r_b^{\prime })\cdot {\Omega }^{\prime }|d{\Omega }^{\prime }$$
which is equal to ${\rho }_{eff}(r_b^{\prime })T(r_b^{\prime })$.

Since the T varies, so the $R_s$ varies. But linear operator theory has the conclusion that a continous positive linear operator $B$ minimum $m_n$ and maximum $M_n$ values of the function ${\eta }_n=(\sqrt{B^{n}u}{)}^{1/n}$ converge to mazimum eigenvalue $\rho (B)$ for any chosen positive function $u$ as n tend to infinity.

Here, the numerator in Eq. (35) can be treated as a positive intergral operator $B$ with a kernel $G_d$. We have $n=1$, and we use the maximum eigenvalue $\overline{R_S}$ of $B$ with kernel $G_d$.

Then , we rewrite the Eq. (25) as
$$T(r_b)=T_{bs}(r_b)+\overline{R_S}{\rho }_{eff}(r_b)T(r_b).$$
Then we have
$$T(r_b)=\frac{T_{bs}(r_b)}{1-\overline{R_S}{\rho }_{eff}(r_b)}$$
Substitude this into Eq. (16), we obtain
$$I(r,\Omega )=I_{bs}(r,\Omega )+{\int }_{{\delta }_V}d{r}_b^{\prime }J(r_b^{\prime };r_b^{\prime },{\Omega }^{\prime }){\rho }_{eff}(r_b^{\prime })\frac{T_{bs}(r_b^{\prime })}{1-\overline{R_S}{\rho }_{eff}(r_b^{\prime })}$$
where J has already mentioned below as
$$J(r,\Omega ;r_b^{\prime })={\int }_{n(r_b^{\prime })\cdot {\Omega }^{\prime }<0}d{\Omega }^{\prime }{G}_S(r,\Omega ;r^{\prime },{\Omega }^{\prime })d_b(r_b^{\prime },{\Omega }^{\prime })$$
For futher simlification, we replace the $J$​ with its mean value over the canopy bottom, then this becomes
$$I(r,\Omega )=I_{bs}(r,\Omega )+{\int }_{{\delta }_V}d{r}_b^{\prime }J(r_b^{\prime };r_b^{\prime },{\Omega }^{\prime })\frac{\overline{{\rho }_{eff}}\overline{T_{bs}}}{1-\overline{R_S}\overline{{\rho }_{eff}}}$$
Then the integral of $J$ is $I_S$​ by Eq. (22), then we have
$$I(r,\Omega )=I_{bs}(r,\Omega )+\frac{\overline{{\rho }_{eff}}\overline{T_{bs}}}{1-\overline{R_S}\overline{{\rho }_{eff}}}{I}_S(r,\Omega )$$
Then, the results of original boundary problem become the solution of black soil problem and S problem.

Now, we could rewrite the HDRF of the surface :
$$\begin{array}{rl}HDRF& =\frac{<I(\Omega ){>}_0}{\frac{1}{\pi }{\int }_{2\pi -}<I({\Omega }^{\prime }){>}_0|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }}\\ & =\frac{<I_{bs}(r,\Omega )+\frac{\overline{{\rho }_{eff}}\overline{T_{bs}}}{1-\overline{R_S}\overline{{\rho }_{eff}}}{I}_S(r,\Omega )>}{\frac{1}{\pi }{\int }_{2\pi -}<I({\Omega }^{\prime }){>}_0|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }}\\ & =R_{bs}(\Omega )+\frac{\overline{{\rho }_{eff}}}{1-\overline{R_S}\overline{{\rho }_{eff}}}{t}_{bs}{T}_S(\Omega )\end{array}$$
Here, transmitance is
$$t_{bs}=\frac{\overline{T_{bs}}}{{\int }_{2\pi -}<I({\Omega }^{\prime }){>}_0|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }}$$
and
$$T_s=\frac{<I_S(\Omega ){>}_0}{\frac{1}{\pi }{\int }_{2\pi -}<d_b({\Omega }^{\prime }){>}_0|{\Omega }^{\prime }\cdot n(r_b)|d{\Omega }^{\prime }}$$
which is the radiance generated by the isotropic source located at canopy bottom, noting that
$${\int }_{2\pi +}{d}_b(r_b,\Omega )|n(r_b)\cdot \Omega |d\Omega =1.$$
So its actually
$$T_s=<I_S(\Omega ){>}_0\pi$$