遥感辐射传输方程中的格林函数

Green Function

1.Green function for ODE

Consider a linear ODE:
$$a_n(x)\frac{d^{n}y}{d{x}^n}+\cdots +a_1(x)\frac{dy}{dx}+a_0(x)y=f(x)$$
The LHS is denoted as $Ly(x)$, then we got
$$Ly(x)=f(x)$$
Suppose the Green function $G(x,z)$ is the solution of such ODE obey a boundary conditions in the range $a\le x\le b$, which the solution would be given by
$$y(x)={\int }_a^{b}G(x,z)f(z)dz$$
So $G(x,z)$ could be viewed as the response of a system to a unit impulse at $x=z$. Then we apply the linear operator on it:
$$Ly(x)={\int }_a^{b}LG(x,z)f(z)dz=f(x)$$
Compared to the property of the Green function:
$$\int f(t)\delta (t-a)dt=f(a)$$
Then, it could be find that:
$$LG(x,z)=\delta (z-x)$$
This means that the Green function satifies the ODE with RHS become the delta function.

An important fact is that
$$\underset{ϵ\to 0}{\lim }\sum _{m=0}^n{\int }_{z-ϵ}^{z+ϵ}{a}_m(x)\frac{d^{m}G(x,z)}{d{x}^m}dx=\underset{ϵ\to 0}{\lim }{\int }_{z-ϵ}^{z+ϵ}\delta (x-z)dx=1$$
So $\frac{d^{m}G(x,z)}{d{x}^m}$ must has derivative at $x=z$ with infinite value (since RHS is a delta function), which means that $\frac{d^{m-1}G(x,z)}{d{x}^{(}m-1)}$ must have a finite discontinuity, and then lower order derivative must be continous. Then, they are 0 in the intergral. By intetration by parts, we get that:
$$\begin{array}{rl}& \underset{ϵ\to 0}{\lim }\sum _{m=0}^n{\int }_{z-ϵ}^{z+ϵ}{a}_m(x)\frac{d^{m}G(x,z)}{d{x}^m}dx\\ & =\underset{ϵ\to 0}{\lim }{\int }_{z-ϵ}^{z+ϵ}{a}_n(x)\frac{d^{n}G(x,z)}{d{x}^n}dx\\ & =\underset{ϵ\to 0}{\lim }[a_n(x)\frac{d^{n-1}G(x,z)}{d{x}^{n-1}}{]}_{z-ϵ}^{z+ϵ}-{\int }_{z-ϵ}^{z+ϵ}{a}_n(x{)}^{\prime }\frac{d^{n}G(x,z)}{d{x}^n}dx\\ & =\underset{ϵ\to 0}{\lim }[a_n(x)\frac{d^{n-1}G(x,z)}{d{x}^{n-1}}{]}_{z-ϵ}^{z+ϵ}=1\\ & \to a_n(z)[\frac{d^{n-1}G(x,z)}{d{x}^{n-1}}]{|}_{x=z}=1\\ & \to [\frac{d^{n-1}G(x,z)}{d{x}^{n-1}}]{|}_{x=z}=\frac{1}{a_n(z)}\end{array}$$
The properties of Green’s function $G(x,z)$ is summarised as following:

• $G(x,z)$ obey the original ODE with f(x) set to $\delta (x-z)$
• Consider $G(x,z)$ a function of x alone obeys the homogeneous boundarty conditions on $y(x)$
• The diravatives of $G(x,z)$ w.r.t x up to order n-2 are continous at $x=z$, and n-1 order derivative has dicontinous of $\frac{1}{a_n(z)}$ at this point.

E.g.
$$\frac{d^{2}y}{d{x}^2}+y=cosecxs.t.y(0)=y(\pi /2)=0$$

Solution: Using the formula (6), we get
$$\frac{d^{2}G(x,z)}{d{x}^2}+G(x,z)=\delta (z-x)$$
Then solve the (10) and using (3) could find th solution.

2.Green function for in homogeneous PDE

A linear PDE is given by
$$\mathcal{L}u(r)=\rho (r)$$
Here, $\mathcal{L}$​ is linear paritial defferential operator. The solution to (1) satisfies some homogeneous boundary conditions on $u(r)$ then the Green’s function $G(r,r_0)$ for this problem is a solution of
$$\mathcal{L}G(r,r_0)=\delta (r-r_0)$$
where $r_0$ lies in V. Then the solution to (1) is that
$$u(r)=\int G(r,r_0)\rho (r_0)dV(r_0)$$

3.Green function for in RTM

The stationary RTM is given by
$$\begin{array}{rl}& \Omega \cdot \nabla I(r,\Omega )+{\sigma }_{\lambda }(r,\Omega )I(r,\Omega )\\ & ={\int }_{4\pi }{\sigma }_s(r,{\Omega }^{\prime },\Omega )I(r,{\Omega }^{\prime })d{\Omega }^{\prime }+q(r,\Omega )\end{array}$$
with boundary condition
$$I(r_B,\Omega )=B(r_B,\Omega ),r_B\in \delta V,n(r_B)\cdot \Omega <0$$
We could find that the left side and the first term of (4) is linear operator on $I(r,\Omega )$, that means if $I_1$ and $I_2$ is the solutions then $I_1+I_2$ is also the solution.

Then the Volumn Green function satisfies the equation:
$$\begin{array}{rl}& \Omega \cdot \nabla G_V(r,\Omega ;r^{\prime },{\Omega }^{\prime })+{\sigma }_{\lambda }(r,\Omega )G_V(r,\Omega ;r^{\prime },{\Omega }^{\prime })\\ & ={\int }_{4\pi }{\sigma }_s(r,{\Omega }^{\prime \prime },\Omega )G_V(r,{\Omega }^{\prime \prime };r^{\prime },{\Omega }^{\prime })d{\Omega }^{\prime \prime }+\delta (\Omega -{\Omega }^{\prime }){\delta }_V(r-r^{\prime })\end{array}$$
with boundary condition
$$G_V(r_B,\Omega ;r^{\prime },{\Omega }^{\prime }=0)r_B\in \delta V,\Omega \cdot n(r_B)<0$$
And the surface Green’s function is the solution to RT with source $q(r,\Omega )=0$​ with boundary condition
$$G_s(r,\Omega ;r_B^{\prime },\Omega )=\delta (\Omega -{\Omega }^{\prime }){\delta }_S(r_B,r_B^{\prime })r_B^{\prime }\in \delta V,n(r_B^{\prime })\cdot {\Omega }^{\prime }<0$$
Using these two Green’s function, we could rewrite the solution to the RTE with arbitrary source $q(r,\Omega )$ and boundart counditions with sources $q_B$ on the non-reflecting boundary $\delta V$ as
$$\begin{array}{rl}I(r,\Omega )& ={\int }_{V}d{r}^{\prime }{\int }_{4\pi }{G}_V(r,\Omega ;r^{\prime },{\Omega }^{\prime })q(r^{\prime },{\Omega }^{\prime })d{\Omega }^{\prime }\\ & +{\int }_{\delta V}dS{\int }_{n(r_B^{\prime })\cdot {\Omega }^{\prime }<0}d{\Omega }^{\prime }{G}_S(r,\Omega ;r_B^{\prime },{\Omega }^{\prime })q_B(r_B^{\prime },{\Omega }^{\prime }).\end{array}$$