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:
I The formal solution
I
I
I
I
Integral Methods
Feautrier Method
Characteristic Methods
The scattering problem
I
I
I
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 Jν . 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 Jν from Bν
even for τν >> 1. This leads to the failure of Λ–iteration.
Consider
Sν = (1 − ν )Jν + ν Bν
µ
dIν
= Iν − Sν
dτν
Jν = Λτν [Sν ] = Λτν [ν Bν ] + Λτν [(1 − ν )Jν ]
if ν = 1 the answer is exact. So the obvious thing to do is to set
Jν = Bν (which is true at great depth) and then iterate
Jν(n) = Λτν [Sν(n) ] = Λτν [ν Bν ] + Λτν [(1 − ν )Jν(n−1) ]
Λ–Iterations II
But
Z
∞
f (t)E1 (|t − τν |) dt
Λτν [f (t)] = 1/2
0
but for large ∆τ
E1 ∼
e−∆τ
∆τ
so information about J can only propagate of (∆τ = 1). If we start
with Jν = Bν then we need √1ν iterations to allow the outer boundary
to be felt by the solution.
(n)
(n−1)
For lines ν ∼ 10−8 → 104 iterations. In practice Jν − Jν
tends to
stabilize leading to apparent convergence even though Jν 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
±µ
or
dI
=I−S
dτ
dI−ω
= −I−ω − S−ω
dτω
dIω
= Iω − Sω
dτω
where
dτω ≡ dτ /µ
If we assume
Sω = S−ω
Then we can define
jω = 1/2(Iω + I−ω )
Variable Eddington Factor Method II
Feautrier Solution
hω = 1/2(Iω − I−ω )
Then adding the two equations we get
djω
= hω
dτω
and subtracting gives
dhω
= jω − Sω
dτω
or
d 2 jω
= jω − S ω
dτω2
for each ray and each frequency. Okay we need two boundary
conditions
I−ω (0) = 0 and Iω (τ ∗ ) = IBC
so we can write these in terms of the Feautrier variables as
Variable Eddington Factor Method III
Feautrier Solution
at τ = 0
hω = 1/2(Iω − I−ω ) + 1/2I−ω − 1/2I−ω = jω − I−ω
and at τ = τ ∗
hω = 1/2(Iω − I−ω ) + 1/2Iω − 1/2Iω = Iω+ − jω
Moment Equations
dτω = dτ /µ
so we have
µ
dh
=j −S
dτ
dj
µ
=h
dτ
Variable Eddington Factor Method IV
Feautrier Solution
1
Z
J=
j dµ
0
1
Z
H=
hµ dµ
0
Z
K =
1
jµ2 dµ
0
then
or
dH
=J −S
dτ
dK
=H
dτ
d 2K
=J −S
dτ 2
Variable Eddington Factor Method V
Feautrier Solution
Let’s define the Eddington factor:
R1
jµ2 dµ
fK = K /J = 0R 1
0 j dµ
and at the surface, we’ll define
R1
jµ dµ
fH = H/J = R0 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 (fK J)
=J −S
dτ 2
Variable Eddington Factor Method VI
Feautrier Solution
with BCs
d(fK J)
= f H J + H− at τ = 0
dτ
d(fK J)
= H+ − f H J at τ = τ ∗
dτ
where
Z
1
H− =
µI − dµ
0
Z
H+ =
1
µI + dµ
0
These follow directly from the Feautrier BCs
Okay now we have the tools to solve the scattering problem: We start
with the Feautrier Equations
µ2
d 2j
=j −S
dτ 2
Variable Eddington Factor Method VII
Feautrier Solution
and BCs
dj
= j − I − at τ = 0
dτ
dj
µ
= I + − j at τ = τ ∗
dτ
Now we introduce a grid and Finite Difference
µ
fj+1 − fj
df
=
dx
∆x
fj+1 − 2fj + fj−1
d 2f
=
dx 2
∆x 2
so our transfer equation becomes
µ2
[jd+1 − 2jd + jd−1 ] = Sd
∆2
Variable Eddington Factor Method VIII
Feautrier Solution
with BCs
µ
[j2 − j1 ] = j1 − I1−
∆
µ
[jD − jD−1 ] = −jD − ID+
∆
or
µ
µ
j2 + (1 + )j1 = I1−
∆
∆
µ
µ
− jD−1 + (1 + )jD = ID+
∆
∆
−
Variable Eddington Factor Method IX
Feautrier Solution
µ
µ
1+ ∆
−∆
2
µ2
µ2
 −µ
1 + 2∆
−∆
 ∆2
2
2

2
µ
µ2
µ2

− ∆2
1 + 2∆
−∆
2
2

 ................................................

 ................................................

 ................................................
µ
µ
1+ ∆
−∆
 − 
I
 S2 


 S3 



=
 ..... 
 ..... 


 SD−1 
ID+












j1
j2
j3
.
.
.
jD










Variable Eddington Factor Method X
Feautrier Solution
So given Sd 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
Aj uj+1 + Bj uj + Cj uj−1 = Dj
(1)
Then we can solve this for the vector ~u with (N) operations as
follows:
We seek two quantities Ej and Fj such that
uj = Ej uj+1 + Fj
We assume the boundary conditions require
u0 = 0 and uN = 0
which implies that
E0 = F0 = 0
(2)
Solution to a Tri-diagonal Matrix Equation II
then re-writing equation 2 as
uj−1 = Ej−1 uj + Fj−1
and plugging into equation 1 we obtain
uj = −
Dj − Cj Fj−1
Aj
uj+1 +
Bj + Cj Ej−1
Bj + Cj Ej−1
from which we can read off
Ej = −
and
Fj =
Aj
Bj + Cj Ej−1
Dj − Cj Fj−1
Bj + Cj Ej−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 uj , starting with the BC value
uN = 0.
Putting together VEF I
So now we can consider the scattering problem
S = (1 − )J + B
Z
J=
1
j(µ) dµ
0
The Eddington factor Equation is
d 2 (fK J)
= J − S = (J − B)
dτ 2
with BCs
d(fK J)
= f H J + H− at τ = 0
dτ
d(fK J)
= H+ − f H J at τ = τ ∗
dτ
Again we finite difference and obtain
Putting together VEF II
K J
K
K
fd+1
d+1 − 2fd Jd + fd−1 Jd−1
= d Jd − d Bd
∆2
or
K J
K J
fd−1
−fd+1
2fdK
d−1
d+1
−
(
+
)J
−
= d Bd
d
d
∆2
∆2
∆2
with BCs
(f1H +
−
f1K
fK
)J1 − 2 J2 = H1−
∆
∆
K
fD−1
fK
JD−1 + (fDH + D )JD = HD+
∆
∆
Putting together VEF III
And again we have a tridiagonal matrix

f1K
fk
f1H + ∆
− ∆2
 fK
fK
fK
 − 1
2 + 2 ∆2 2 − ∆3 2
 ∆2

fK
fK
fK

− ∆2 2
3 + 2 ∆3 2 − ∆4 2

 .......................................................

 .......................................................

 .......................................................

fK
fDK
− D−1
fDH + ∆
∆


H1−


2 B2




3 B3



=  ......... 

 ......... 


 D−1 BD−1 
HD+














J1
J2
J3
..
..
..
JD










Putting together VEF IV
So given fdK , f1H , and fDH we can solve for Jd and then given Jd we have
Sd = (1 − d )Jd + d Bd . But then given Sd we get get jd at each µ and
K
H
H
from that we can calculate the Eddington factors: f√
d , f1 , and fD . As a
first approximation we can take fdK = 1/3, f1H = 1/ 3, and fDH = 0 and
repeat the whole thing until it converges
Accelerated Λ–Interation I
We have an idea already how to constuct the Λτ 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
= −χI + κB + σJ
dz
Z 1
J = 1/2
I dµ
−1
χ=κ+σ
dτ = −χdz
µ
dI
κB + σJ
=I−
=I−S
dτ
χ
S = B + (1 − )J
Accelerated Λ–Interation II
=
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:
J = Λ[B] + Λ[(1 − )J]
[1 − Λ[(1 − )]J = Λ[B]
J = [1 − Λ[(1 − )]−1 Λ[B]
but
1. Numerical computation of Λ is expensive
2. Numerical inversion of Λ may also be expensive
Straight forward Λ–iteration
Jnew = Λ[Sold ]
Snew = (1 − )Jnew + B
I
Will always converge
Accelerated Λ–Interation IV
I
I
Formal solutions are cheap
√
Needs (1/ ) iterations for convergence
Mathematically
Λ–iterations are totally stable and will converge → Eigenvalues < 1,
√
but Eigenvalues of (1 − ). So convergence is extremely slow.
Idea: Accelerate convergence
Technically: Reduce Eigenvalues of Amplification matrix
Practically: Introduce approximate Lambda–operator Λ∗
Λ = Λ∗ + (Λ − Λ∗ )
Now operator split iteration so
Accelerated Λ–Interation V
Jnew = Λ∗ [Snew ] + (Λ − Λ∗ )Sold
= Λ∗ [(1 − )Jnew ] + (Λ[Sold ] − Λ∗ [(1 − )Jold ])
= Λ∗ [(1 − )Jnew ] + JFS − Λ∗ [(1 − )Jold ]
JFS = Λ[Sold ]
Jnew = [1 − Λ∗ [(1 − )]−1 [JFS − Λ∗ [(1 − )Jold ]]
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.
I
Diagonal: Core saturation (Rybicki & Hummer, Scharmer)
I
Tri-Diag: Olson & Kunasz
I
Bands: Hauschildt et al.
Why use bands?
I
Easy to invert
Accelerated Λ–Interation VI
I
Eigenvalues significantly reduced
I
Easy to evaluate
I
Practically: Exists a tradeoff between band-width of Λ∗ and
number of iterations
I
Tridiag is often a good choice
I
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?