VLIDORT User's Guide

of the single scatter albedo ω and the matrix of expansion coefficients Β l , and its (real-variable)~linearization L (Γ)is easy to establish from chain-rule differentiation:~ ~ + ~ − ~ + ~ −L ( Γ)= L(S ) S + S L(S ) ; (2.48)L(⎡L(ω)LM~ ± ⎧ ~~ ⎫ ± ~ T ~ ~ −1S ) = ⎢∑⎨Π(l,m)Bl+ Π(l,m)L(Bl) ⎬AΠ ( l,m)Ω= 22⎥Μl m⎣⎩ω⎭⎤⎦. (2.49)In Eq. (2.49), L (ω)= U and L (Bl) = Zlare the linearized optical property inputs (Eq. (2.33)).Next, we differentiate both the left and right eigensystems (2.37) to find:~ ⊥ ~ ~ ⊥ ~~ ⊥ 2 ~ ⊥L(Xα) Γ + XαL(Γ)= 2k αL(kα) Xα+ kαL(Xα) ; (2.50)~ ~ ~ ~~ 2 ~ΓL(Xα) + L(Γ)Xα= 2k αL(kα) Xα+ kαL(Xα) . (2.51)⊥We form a dot product by pre-multiplying (2.51) with the transpose vector X ~ α , rearranging toget:~ ⊥ ~ ~ ⊥ ~ ~ 2 ~ ⊥ ~ ~ ⊥ ~ ~2k αL(kα)〈Xα, Xα〉 − 〈 Xα, L(Γ)Xα〉 = kα〈 Xα, L(Xα)〉− 〈 Xα, ΓL(Xα)〉. (2.52)From the definitions in Eq. (2.37), we have:~ ⊥ ~ ~ ~ ⊥ ~ ~2 ~ ⊥ ~〈 Xα, ΓL(Xα)〉= 〈 XαΓ,L(Xα)〉= kα〈 Xα, L(Xα)〉, (2.53)and hence the right hand side of (2.52) is identically zero. We thus have:~ ⊥ ~〈 Xα, L(Γ~ ) Xα〉L ( kα) = ~ ⊥ ~ . (2.54)2k〈 X , X 〉αααNext, we substitute Eq. (2.54) in (2.52) to obtain the following 4N x 4N linear algebra problemfor each eigensolution linearization:~ ~ ~ΗαL ( Xα) = Cα; (2.55)~ ~ 2~Hα = Γ − k αE ; (2.56)~~ ~Cα= 2k αL(kα) Xα− L(Γ~ ) Xα. (2.57)Implementation of Eq. (2.55) “as is” is not possible due to the degeneracy of the eigenproblem,and we need additional constraints to find the unique solution for L ( X~α) . The treatment for realand complex solutions is different.~ ~Real solutions. The unit-modulus eigenvector normalization can be expressed as 〈 X , X 〉 α α= 1 indot-product notation. Linearizing, this yields one equation:~ ~ ~ ~L ( Xα) Xα+ XαL(Xα) = 0. (2.58)The solution procedure uses 4N −1 equations from (2.55), along with Eq. (2.58) to form aslightly modified linear system of rank 4N. This system is then solved by standard means usingthe DGETRF and DGETRS LU-decomposition routines from the LAPACK suite.22