Lecture 3 Numerical Solutions to the Transport Equation

Lecture 3 Numerical Solutions to the Transport Equation

Introduction I There are numerous methods for solving the transport problem numerically. First we must recognize that we need to solve two problems: The formal solution Integral Methods Feautrier Method Characteristic Methods The scattering problem Λ iteration Variable Eddington Factor Method Accelerated Λ iteration Today we will study the Feautrier Method, Λ iterations, the Variable Eddington Factor Method, and Accelerated Λ Iteration

Λ Iterations I Let s first start out with what to avoid. First off let s be clear that the solution of the RTE is a solution for J ν. Once we have that we just have to do a formal solution. The existance of scattering terms is what makes the solution of the RTE so difficult. Physically these terms couple regions that are spatial separate and hence couple regions with vastly different temperatures causing large departures of J ν from B ν even for τ ν >> 1. This leads to the failure of Λ iteration. Consider S ν = (1 ɛ ν )J ν + ɛ ν B ν µ di ν dτ ν = I ν S ν J ν = Λ τν [S ν ] = Λ τν [ɛ ν B ν ] + Λ τν [(1 ɛ ν )J ν ] if ɛ ν = 1 the answer is exact. So the obvious thing to do is to set J ν = B ν (which is true at great depth) and then iterate J (n) ν = Λ τν [S ν (n) ] = Λ τν [ɛ ν B ν ] + Λ τν [(1 ɛ ν )J ν (n 1) ]

Λ Iterations II But but for large τ Λ τν [f (t)] = 1/2 0 E 1 e τ τ f (t)e 1 ( t τ ν ) dt so information about J can only propagate of ( τ = 1). If we start with J ν = B ν then we need 1 ɛν iterations to allow the outer boundary to be felt by the solution. For lines ɛ ν 10 8 10 4 iterations. In practice J ν (n) J ν (n 1) tends to stabilize leading to apparent convergence even though J ν is still far from the true solution.

Convergence Rate in Static Atmosphere

Geometry for Solution to Plane-Parallel Equation

Variable Eddington Factor Method I Feautrier Solution The Plane-parallel transfer can be written as or where di ω dτ ω ±µ di dτ = I S = I ω S ω di ω dτ ω = I ω S ω dτ ω dτ/µ If we assume S ω = S ω Then we can define j ω = 1/2(I ω + I ω )

Variable Eddington Factor Method II Feautrier Solution h ω = 1/2(I ω I ω ) Then adding the two equations we get and subtracting gives or dj ω dτ ω = h ω dh ω dτ ω = j ω S ω d 2 j ω dτ 2 ω = j ω S ω for each ray and each frequency. Okay we need two boundary conditions I ω (0) = 0 and I ω (τ ) = I BC so we can write these in terms of the Feautrier variables as

Variable Eddington Factor Method III Feautrier Solution at τ = 0 and at τ = τ h ω = 1/2(I ω I ω ) + 1/2I ω 1/2I ω = j ω I ω Moment Equations h ω = 1/2(I ω I ω ) + 1/2I ω 1/2I ω = I + ω j ω dτ ω = dτ/µ so we have µ dh dτ = j S µ dj dτ = h

Variable Eddington Factor Method IV Feautrier Solution then or J = H = K = 1 0 1 0 1 0 j dµ hµ dµ jµ 2 dµ dh dτ = J S d 2 K dτ 2 dk dτ = H = J S

Variable Eddington Factor Method V Feautrier Solution Let s define the Eddington factor: f K = K /J = 1 0 jµ2 dµ 1 0 j dµ and at the surface, we ll define f H = H/J = 1 0 jµ dµ 1 0 j dµ where we used the BC at the surface and assumed no incoming radiation. Okay now we can solve the whole problem numerically, if we assume that the Eddington factors are known functions of depth. Then we get d 2 (f K J) dτ 2 = J S

Variable Eddington Factor Method VI Feautrier Solution with BCs where d(f K J) dτ d(f K J) dτ = f H J + H at τ = 0 = H + f H J at τ = τ H = H + = 1 0 1 0 µi dµ µi + dµ These follow directly from the Feautrier BCs Okay now we have the tools to solve the scattering problem: We start with the Feautrier Equations µ 2 d 2 j dτ 2 = j S

Variable Eddington Factor Method VII Feautrier Solution and BCs µ dj dτ = j I at τ = 0 µ dj dτ = I+ j at τ = τ Now we introduce a grid and Finite Difference so our transfer equation becomes df dx = f j+1 f j x d 2 f dx 2 = f j+1 2f j + f j 1 x 2 µ 2 2 [j d+1 2j d + j d 1 ] = S d

Variable Eddington Factor Method VIII Feautrier Solution with BCs or µ [j 2 j 1 ] = j 1 I 1 µ [j D j D 1 ] = j D I + D µ j 2 + (1 + µ )j 1 = I 1 µ j D 1 + (1 + µ )j D = I + D

Variable Eddington Factor Method IX Feautrier Solution 1 + µ µ µ2 1 + 2 µ2 µ2 2 2 2 µ2 1 + 2 µ2 µ2 2 2 2................................................................................................................................................ µ 1 + µ I S 2 S 3 =.......... S D 1 I + D j 1 j 2 j 3... j D

Variable Eddington Factor Method X Feautrier Solution So given S d I can solve this by solving the tri-diagonal matrix.

Solution to a Tri-diagonal Matrix Equation I In general inverting an N N matrix requires N 3 operations, so even with a supercomputer you can t invert a very big matrix N < few thousand But consider a system of equations of the form A j u j+1 + B j u j + C j u j 1 = D j (1) Then we can solve this for the vector u with (N) operations as follows: We seek two quantities E j and F j such that We assume the boundary conditions require u j = E j u j+1 + F j (2) u 0 = 0 and u N = 0 which implies that E 0 = F 0 = 0

Solution to a Tri-diagonal Matrix Equation II then re-writing equation 2 as u j 1 = E j 1 u j + F j 1 and plugging into equation 1 we obtain u j = from which we can read off and A j B j + C j E j 1 u j+1 + D j C j F j 1 B j + C j E j 1 A j E j = B j + C j E j 1 F j = D j C j F j 1 B j + C j E j 1 and we sweep through the grid twice, first to get the E and F starting at j = 1 and then backwards to get the u j, starting with the BC value u N = 0.

Putting together VEF I So now we can consider the scattering problem S = (1 ɛ)j + ɛb J = The Eddington factor Equation is with BCs 1 0 j(µ) dµ d 2 (f K J) dτ 2 = J S = ɛ(j B) d(f K J) dτ d(f K J) dτ Again we finite difference and obtain = f H J + H at τ = 0 = H + f H J at τ = τ

Putting together VEF II or with BCs f K d+1 J d+1 2f K d J d + f K d 1 J d 1 2 f K d+1 J d+1 2 = ɛ d J d ɛ d B d (ɛ d + 2f d K 2 )J d f d 1 K J d 1 2 (f H 1 + f K 1 )J 1 f K 2 J 2 = H 1 = ɛ d B d f K D 1 J D 1 + (f H D + f K D )J D = H + D

Putting together VEF III And again we have a tridiagonal matrix f 1 H + f 1 K f 2 k f 1 K ɛ 2 2 + 2 f 2 K f 2 3 K 2 f 2 K ɛ 2 3 + 2 f 3 K f 2 4 K 2..................................................................................................................................................................... f D 1 K fd H + f D K H 1 ɛ 2 B 2 ɛ 3 B 3 =.................. ɛ D 1 B D 1 H + D J 1 J 2 J 3...... J D

Putting together VEF IV So given f K d, f H 1, and f H D we can solve for J d and then given J d we have S d = (1 ɛ d )J d + ɛ d B d. But then given S d we get get j d at each µ and from that we can calculate the Eddington factors: f K d, f H 1, and f H D. As a first approximation we can take f K d = 1/3, f H 1 = 1/ 3, and f H D = 0 and repeat the whole thing until it converges

Accelerated Λ Interation I We have an idea already how to constuct the Λ τ operator using the Exponential Integrals. Numerically we will not construct the operator that way and I leave the details to papers by Olson & Kunasz; Hauschildt; Hauschildt & Baron. Let s for the moment return to the Plane-Parallel static RTE µ di dz = χi + κb + σj J = 1/2 1 1 χ = κ + σ dτ = χdz I dµ µ di dτ = I κb + σj χ = I S S = ɛb + (1 ɛ)j

Accelerated Λ Interation II ɛ = κ χ And we ve seen in Lecture 2, that if S is known then I can be computed by numerical integration J = Λ[S] Formal solutions are numerically cheap and we don t need an explicit expression for Λ in order to obtain the formal solution. Problems: 1. Stability of numerical integration (relatively easy to beat down) 2. S depends on J

Accelerated Λ Interation III If we knew Λ numerically then solution is simple: but J = Λ[ɛB] + Λ[(1 ɛ)j] [1 Λ[(1 ɛ)]j = Λ[ɛB] J = [1 Λ[(1 ɛ)] 1 Λ[ɛB] 1. Numerical computation of Λ is expensive 2. Numerical inversion of Λ may also be expensive Straight forward Λ iteration J new = Λ[S old ] S new = (1 ɛ)j new + ɛb Will always converge

Accelerated Λ Interation IV Formal solutions are cheap Needs (1/ ɛ) iterations for convergence Mathematically Λ iterations are totally stable and will converge Eigenvalues < 1, but Eigenvalues of (1 ɛ). So convergence is extremely slow. Idea: Accelerate convergence Technically: Reduce Eigenvalues of Amplification matrix Practically: Introduce approximate Lambda operator Λ Now operator split iteration so Λ = Λ + (Λ Λ )

Accelerated Λ Interation V J new = Λ [S new ] + (Λ Λ )S old = Λ [(1 ɛ)j new ] + (Λ[S old ] Λ [(1 ɛ)j old ]) = Λ [(1 ɛ)j new ] + J FS Λ [(1 ɛ)j old ] J FS = Λ[S old ] J new = [1 Λ [(1 ɛ)] 1 [J FS Λ [(1 ɛ)j old ]] Now if Λ has a simple form inversion is not too expensive. We want that the eigenvalues of Λ Λ << Eigenvalues of Λ Solution: Choose Λ as bands of Λ including diagonal. Diagonal: Core saturation (Rybicki & Hummer, Scharmer) Tri-Diag: Olson & Kunasz Bands: Hauschildt et al. Why use bands? Easy to invert

Accelerated Λ Interation VI Eigenvalues significantly reduced Easy to evaluate Practically: Exists a tradeoff between band-width of Λ and number of iterations Tridiag is often a good choice Can be Ng accelerated

Convergence Rate in Static Atmosphere

Spherical Geometry This is standard method for spherical symmetry

Spherical Geometry Question But why not do it this way?