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 ) d n y d x n + ⋯ + a 1 ( x ) d y d x + a 0 ( x ) y = f ( x ) a_n(x)\frac{d^ny}{dx^n}+\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y = f(x) an(x)dxndny+⋯+a1(x)dxdy+a0(x)y=f(x)
The LHS is denoted as L y ( x ) Ly(x) Ly(x), then we got
L y ( x ) = f ( x ) Ly(x)=f(x) Ly(x)=f(x)
Suppose the Green function G ( x , z ) G(x,z) G(x,z) is the solution of such ODE obey a boundary conditions in the range a ≤ x ≤ b a \le x \le b a≤x≤b, which the solution would be given by
y ( x ) = ∫ a b G ( x , z ) f ( z ) d z y(x)=\int_a^{b}G(x,z)f(z)dz y(x)=∫abG(x,z)f(z)dz
So G ( x , z ) G(x,z) G(x,z) could be viewed as the response of a system to a unit impulse at x = z x =z x=z. Then we apply the linear operator on it:
L y ( x ) = ∫ a b L G ( x , z ) f ( z ) d z = f ( x ) Ly(x)=\int_a^bLG(x,z)f(z)dz=f(x) Ly(x)=∫abLG(x,z)f(z)dz=f(x)
Compared to the property of the Dirac Delta function:
∫ f ( t ) δ ( t − a ) d t = f ( a ) \int f(t)\delta(t-a)dt = f(a) ∫f(t)δ(t−a)dt=f(a)
Then, it could be find that:
L G ( x , z ) = δ ( z − x ) LG(x,z)=\delta(z-x) LG(x,z)=δ(z−x)
This means that the Green function satifies the ODE with RHS become the delta function.
An important fact is that
lim ϵ → 0 ∑ m = 0 n ∫ z − ϵ z + ϵ a m ( x ) d m G ( x , z ) d x m d x = lim ϵ → 0 ∫ z − ϵ z + ϵ δ ( x − z ) d x = 1 \lim_{\epsilon\rightarrow0}\sum_{m=0}^n\int_{z-\epsilon}^{z+\epsilon}a_m(x)\frac{d^mG(x,z)}{dx^m}dx=\lim_{\epsilon\rightarrow0}\int_{z-\epsilon}^{z+\epsilon}\delta(x-z)dx=1 ϵ→0limm=0∑n∫z−ϵz+ϵam(x)dxmdmG(x,z)dx=ϵ→0lim∫z−ϵz+ϵδ(x−z)dx=1
So d m G ( x , z ) d x m \frac{d^mG(x,z)}{dx^m} dxmdmG(x,z) must has derivative at x = z x=z x=z with infinite value (since RHS is a delta function), which means that d m − 1 G ( x , z ) d x ( m − 1 ) \frac{d^{m-1}G(x,z)}{dx^(m-1)} dx(m−1)dm−1G(x,z) 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:
lim ϵ → 0 ∑ m = 0 n ∫ z − ϵ z + ϵ a m ( x ) d m G ( x , z ) d x m d x = lim ϵ → 0 ∫ z − ϵ z + ϵ a n ( x ) d n G ( x , z ) d x n d x = lim ϵ → 0 [ a n ( x ) d n − 1 G ( x , z ) d x n − 1 ] z − ϵ z + ϵ − ∫ z − ϵ z + ϵ a n ( x ) ′ d n G ( x , z ) d x n d x = lim ϵ → 0 [ a n ( x ) d n − 1 G ( x , z ) d x n − 1 ] z − ϵ z + ϵ = 1 → a n ( z ) [ d n − 1 G ( x , z ) d x n − 1 ] ∣ x = z = 1 → [ d n − 1 G ( x , z ) d x n − 1 ] ∣ x = z = 1 a n ( z ) \begin{aligned} &\lim_{\epsilon\rightarrow0}\sum_{m=0}^n\int_{z-\epsilon}^{z+\epsilon}a_m(x)\frac{d^mG(x,z)}{dx^m}dx \\&=\lim_{\epsilon\rightarrow0}\int_{z-\epsilon}^{z+\epsilon}a_n(x)\frac{d^nG(x,z)}{dx^n}dx\\ &=\lim_{\epsilon\rightarrow0}[a_n(x)\frac{d^{n-1}G(x,z)}{dx^{n-1}}]_{z-\epsilon}^{z+\epsilon}-\int_{z-\epsilon}^{z+\epsilon}a_n(x)'\frac{d^{n}G(x,z)}{dx^{n}}dx\\ &=\lim_{\epsilon\rightarrow0}[a_n(x)\frac{d^{n-1}G(x,z)}{dx^{n-1}}]_{z-\epsilon}^{z+\epsilon}=1 \\ & \rightarrow a_n(z)[\frac{d^{n-1}G(x,z)}{dx^{n-1}}]|_{x=z}=1\\ & \rightarrow [\frac{d^{n-1}G(x,z)}{dx^{n-1}}]|_{x=z}=\frac{1}{a_n(z)} \end{aligned} ϵ→0limm=0∑n∫z−ϵz+ϵam(x)dxmdmG(x,z)dx=ϵ→0lim∫z−ϵz+ϵan(x)dxndnG(x,z)dx=ϵ→0lim[an(x)dxn−1dn−1G(x,z)]z−ϵz+ϵ−∫z−ϵz+ϵan(x)′dxndnG(x,z)dx=ϵ→0lim[an(x)dxn−1dn−1G(x,z)]z−ϵz+ϵ=1→an(z)[dxn−1dn−1G(x,z)]∣x=z=1→[dxn−1dn−1G(x,z)]∣x=z=an(z)1
The properties of Green’s function G ( x , z ) G(x,z) G(x,z) is summarised as following:
- G ( x , z ) G(x,z) G(x,z) obey the original ODE with f(x) set to δ ( x − z ) \delta(x-z) δ(x−z)
- Consider G ( x , z ) G(x,z) G(x,z) a function of x alone obeys the homogeneous boundarty conditions on y ( x ) y(x) y(x)
- The diravatives of G ( x , z ) G(x,z) G(x,z) w.r.t x up to order n-2 are continous at x = z x=z x=z, and n-1 order derivative has dicontinous of 1 a n ( z ) \frac{1}{a_n(z)} an(z)1 at this point.
E.g.
d 2 y d x 2 + y = c o s e c x s . t . y ( 0 ) = y ( π / 2 ) = 0 \frac{d^2y}{dx^2}+y=cosecx\ \ s.t.\ y(0)=y(\pi/2)=0 dx2d2y+y=cosecx s.t. y(0)=y(π/2)=0
Solution: Using the formula (6), we get
d 2 G ( x , z ) d x 2 + G ( x , z ) = δ ( z − x ) \frac{d^2G(x,z)}{dx^2}+G(x,z)=\delta(z-x) dx2d2G(x,z)+G(x,z)=δ(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
L u ( r ) = ρ ( r ) \mathcal{L}u(r)=\rho(r) Lu(r)=ρ(r)
Here, L \mathcal{L} L is linear paritial defferential operator. The solution to (1) satisfies some homogeneous boundary conditions on u ( r ) u(r) u(r) then the Green’s function G ( r , r 0 ) G(r,r_0) G(r,r0) for this problem is a solution of
L G ( r , r 0 ) = δ ( r − r 0 ) \mathcal{L}G(r,r_0)=\delta(r-r_0) LG(r,r0)=δ(r−r0)
where r 0 r_0 r0 lies in V. Then the solution to (1) is that
u ( r ) = ∫ G ( r , r 0 ) ρ ( r 0 ) d V ( r 0 ) u(r)=\int G(r,r_0)\rho(r_0)dV(r_0) u(r)=∫G(r,r0)ρ(r0)dV(r0)
3.Green function for in RTM
The stationary RTM is given by
Ω ⋅ ∇ I ( r , Ω ) + σ λ ( r , Ω ) I ( r , Ω ) = ∫ 4 π σ s ( r , Ω ′ , Ω ) I ( r , Ω ′ ) d Ω ′ + q ( r , Ω ) \begin{aligned} &\Omega\cdot\nabla I(r,\Omega)+\sigma_{\lambda}(r,\Omega)I(r,\Omega) \\&= \int_{4\pi}\sigma_{s}(r,\Omega',\Omega)I(r,\Omega')d\Omega'+q(r,\Omega) \end{aligned} Ω⋅∇I(r,Ω)+σλ(r,Ω)I(r,Ω)=∫4πσs(r,Ω′,Ω)I(r,Ω′)dΩ′+q(r,Ω)
with boundary condition
I ( r B , Ω ) = B ( r B , Ω ) , r B ∈ δ V , n ( r B ) ⋅ Ω < 0 I(r_B,\Omega)=B(r_B,\Omega), \ r_B\in\delta V,\ n(r_B)\cdot\Omega <0 I(rB,Ω)=B(rB,Ω), rB∈δV, n(rB)⋅Ω<0
We could find that the left side and the first term of (4) is linear operator on I ( r , Ω ) I(r,\Omega) I(r,Ω), that means if I 1 I_1 I1 and I 2 I_2 I2 is the solutions then I 1 + I 2 I_1 + I_2 I1+I2 is also the solution.
Then the Volumn Green function satisfies the equation:
Ω ⋅ ∇ G V ( r , Ω ; r ′ , Ω ′ ) + σ λ ( r , Ω ) G V ( r , Ω ; r ′ , Ω ′ ) = ∫ 4 π σ s ( r , Ω ′ ′ , Ω ) G V ( r , Ω ′ ′ ; r ′ , Ω ′ ) d Ω ′ ′ + δ ( Ω − Ω ′ ) δ V ( r − r ′ ) \begin{aligned} &\Omega\cdot\nabla G_V(r,\Omega;r',\Omega')+\sigma_{\lambda}(r,\Omega)G_V(r,\Omega;r',\Omega') \\&= \int_{4\pi}\sigma_{s}(r,\Omega'',\Omega)G_V(r,\Omega'';r',\Omega')d\Omega''+\delta(\Omega-\Omega')\delta_V(r-r') \end{aligned} Ω⋅∇GV(r,Ω;r′,Ω′)+σλ(r,Ω)GV(r,Ω;r′,Ω′)=∫4πσs(r,Ω′′,Ω)GV(r,Ω′′;r′,Ω′)dΩ′′+δ(Ω−Ω′)δV(r−r′)
with boundary condition
G V ( r B , Ω ; r ′ , Ω ′ = 0 ) r B ∈ δ V , Ω ⋅ n ( r B ) < 0 G_V(r_B,\Omega;r',\Omega'=0) \ \ r_B\in \delta V, \Omega \cdot n(r_B)<0 GV(rB,Ω;r′,Ω′=0) rB∈δV,Ω⋅n(rB)<0
And the surface Green’s function is the solution to RT with source q ( r , Ω ) = 0 q(r,\Omega)=0 q(r,Ω)=0 with boundary condition
G s ( r , Ω ; r B ′ , Ω ) = δ ( Ω − Ω ′ ) δ S ( r B , r B ′ ) r B ′ ∈ δ V , n ( r B ′ ) ⋅ Ω ′ < 0 G_s(r,\Omega;r_B',\Omega)=\delta(\Omega-\Omega')\delta_S(r_B,r'_B)\ \ r_B'\in\delta V, n(r'_B)\cdot\Omega'<0 Gs(r,Ω;rB′,Ω)=δ(Ω−Ω′)δS(rB,rB′) rB′∈δV,n(rB′)⋅Ω′<0
Using these two Green’s function, we could rewrite the solution to the RTE with arbitrary source q ( r , Ω ) q(r,\Omega) q(r,Ω) and boundart counditions with sources q B q_B qB on the non-reflecting boundary δ V \delta V δV as
I ( r , Ω ) = ∫ V d r ′ ∫ 4 π G V ( r , Ω ; r ′ , Ω ′ ) q ( r ′ , Ω ′ ) d Ω ′ + ∫ δ V d S ∫ n ( r B ′ ) ⋅ Ω ′ < 0 d Ω ′ G S ( r , Ω ; r B ′ , Ω ′ ) q B ( r B ′ , Ω ′ ) . \begin{aligned} I(r,\Omega)&=\int_Vdr'\int_{4\pi}G_V(r,\Omega;r',\Omega')q(r',\Omega')d\Omega'\\& + \int_{\delta V}dS\int_{n(r'_B)\cdot \Omega' <0}d\Omega'G_S(r,\Omega;r'_B,\Omega')q_B(r'_B,\Omega'). \end{aligned} I(r,Ω)=∫Vdr′∫4πGV(r,Ω;r′,Ω′)q(r′,Ω′)dΩ′+∫δVdS∫n(rB′)⋅Ω′<0dΩ′GS(r,Ω;rB′,Ω′)qB(rB′,Ω′).