on a uniformly second order numerical method for the one ...

ON A UNIFORMLY SECOND ORDER NUMERICAL METHOD FOR THE ONE-DIMENSIONAL DISCRETE-ORDINATE TRANSPORT EQUATION AND ITS DIFFUSION LIMIT WITH INTERFACE∗ SHI JIN† , MIN TANG‡ , AND HOUDE HAN§ Abstract. In this paper, we study a uniformly second order numerical method for the discreteordinate transport equation in the slab geometry in the diffusive regimes with interfaces. At the interfaces, the scattering coefficients have discontinuities, so suitable interface conditions are needed to define the unique solution. We first approximate the scattering coefficients by piecewise constants determined by their cell averages, and then, following the work of De Barros and Larsen [12], obtain the analytic solution at each cell, using which to piece together the numerical solution with the neighboring cells using the interface conditions. We show that this method is asymptotic-preserving, which preserves the discrete diffusion limit with the correct interface condition. Moreover, we show that this method is quadratically convergent uniformly in the diffusive regime, even with the boundary layers. This is 1) the first sharp uniform convergence result for linear transport equations in the diffusive regime, a problem that involves both transport and diffusive scales; and 2) the first uniform convergence valid up to the boundary even if the boundary layers exist, so the boundary layer does not need to be resolved numerically. Numerical examples are presented to justify the uniform convergence. Key words. Linear transport equation, discrete-ordinate method, diffusion limit, interface, asymptotic preserving, uniform numerical convergence, boundary layer AMS subject classifications. 65D25, 70B05, 65L12

1. Introduction. The transport equation with discontinuous coefficients is important since it arises in many important applications, ranging from neutron transport, radiative transfer, high frequency waves in heterogeneous and random media, semiconductor device simulation, to quantum mechanics. We consider particles in a bounded domain that interact with a background through absorption and scattering processes. In highly diffusive media, the mean free path (the average distance a particle travels between interaction with the background media) is small compared to the typical length scales. This small ratio is embodied by the introduction of a dimensionless parameter ǫ into the transport equation. In one space dimension, with zL , zR being the left and right boundaries respectively, the phase space density Ψ(z, µ) over [zL , zR ] × [−1, 1] is governed by a scaled linear transport equation: σ 1 Z 1 σT T Ψ(z, µ′ )dµ′ + ǫq, (1.1a) µ∂z Ψ + Ψ= − ǫσa ǫ ǫ 2 −1 with boundary conditions Ψ(zL , µ) = ΨL (µ),

µ > 0;

Ψ(zR , µ) = ΨR (µ),

µ < 0.

(1.1b)

Here σT , σa , q are the total scattering and absorption coefficients and the source respectively. They are O(1) and have O(1) derivatives with respect to z, except at ∗ This research was supported by NSF grant DMS-0608720, NSFC project 10676017, and a Van Vleck Distinguished Research Prize at the University of Wisconsin-Madison. † Department of Mathematics, University of Wisconsin, Madison, WI 53706, USA; Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China([email protected]). ‡ Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China([email protected]). § Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China([email protected]).

1

2

S. Jin, M. Tang AND H. Han

finite number of points. Namely, they are piecewise smooth. Their discontinuities correspond to interfaces between different materials or media. Clearly (1.1) does not admit a unique solution when the coefficients σT , σa , q are discontinuous. Interface conditions are needed to select the unique solution. Assume the interface is located at z = 0. In most physical applications, Ψ is continuous crossing the interface: =0 (1.2) [Ψ(·, µ)] z=0

where we use [Ψ] to present the jump of Ψ at a point. The starting point of numerical methods for solving the transport equation is the discrete-ordinate method [8, 28, 32], which uses a quadrature formula to approximate the scattering term. Let V = {−M, · · · , −1, 1, · · · , M }.

The discrete-ordinates approximation to (1.1) is  X σT (z) σT (z) dψm (z)+ ψm (z) = −ǫσa (z) ψn (z)wn +ǫq(z) , µm dz ǫ ǫ

m ∈ V (1.3a)

n∈V

with boundary conditions ψm (zL ) = ψLm = ΨL (µm );

ψ−m (zR ) = ψRm = ΨR (−µm ) ,

µm > 0,

(1.3b)

and the interface condition [ψm ]

z=0

= 0.

(1.3c)

Here {µm , wm |m ∈ V } is an order-2M quadrature set with weight wm normalized as X X wn = 1 , wn µ2n = 1/3 . (1.4) n∈V

n∈V

µm are ordered as 1 > µM > µM −1 > · · · > µ1 > 0 > µ−1 > µ−2 > · · · > µ−M +1 > µ−M > −1

(1.5)

and are symmetric: µm = −µ−m ,

wm = w−m ,

m ∈ V.

(1.6)

The quadrature set is studied in [19, 20] and the commonly used {µm , wm } satisfies conditions (1.4)-(1.6). In this paper we will concentrate on the discrete ordinate equations (1.3) and always use ψ to denote (ψ−M , · · · , ψM )T . Even when the scattering coefficients are smooth, numerical solutions to the transport equation or the discrete-ordinate equations are challenging when ǫ 0; x(2) = σT (s) ds, ǫ 0 ǫ 0

z < 0,

to (1.3a) respectively gives (1) (1) ψm (z) = ψm (x ) + O(ǫ2 ),

z > 0;

(2) (2) ψm (z) = ψm (x ) + O(ǫ2 ),

z < 0,

where ψ i satisfies (i) (i) µm ∂x(i) ψm + ψm −

X

wn ψn(i) = 0,

in X (i) × V.

(2.11)

n∈V

Here i = 1, 2, X (1) = R+ , X (2) = R− . The interface condition for (2.11) corresponding to (2.10) is (1) − (2) + ψm (0 ) = ψm (0 ),

m ∈ V.

(2.12)

The differential equation (2.11) on X (i) , being a transport equation with constant coefficients, can be solved exactly [19]: (i)

(i)

(i) (i) ψm (x ) = c1 + c2 (x(i) − µm ) +

X

(k) A(i,k) lm exp(−ξk x(i) ),

(2.13)

1≤|k|≤M −1 (k)

where the eigenvalue ξk and corresponding eigenfunction lm are defined in (2.5) and (2.6).

6

S. Jin, M. Tang AND H. Han (i)

(i)

(i)

The coefficients c1 , c2 , A(i,k) can be determined by ψm (0) through the following boundary conditions: X (i) (i) (i) ± (k) ψm (0 ) = c1 − µm c2 + A(i,k) lm . 1≤|k|≤M −1

The interface condition (2.12) implies (1)

(2)

(1)

(2)

(c1 − c1 ) − µm (c2 − c2 ) +

X

1≤|k|≤M −1


(k) A(1,k) − A(2,k) lm = 0.

(k)

Since the eigenfunctions 1, µm , lm (1 ≤ |k| ≤ M − 1) are linearly independent (this will be proved in the Appendix), one gets (1)

(2)

c1 = c1 ,

(1)

(2)

A(1,k) = A(2,k) ,

c2 = c2 ,

for 1 ≤ |k| ≤ M − 1.

(2.14)

Furthermore, since the diffusion approximation contains no growing exponentials on both sides, one must have A(1,n) = 0,

A(2,n) = 0,

−M + 1 ≤ n ≤ −1;

1 ≤ n ≤ M − 1.

(2.15)

Thus (i)

(i)

(i) (i) ψm (x ) = c1 + c2 (x(i) − µm ).

(2.16)

We now match these solution with the interior (diffusion) solution. Away from the interface, the solution of (1.3a) can be expressed as µm (i) ∂z φ(i) (z) + O(ǫ2 ). ψm (z) = φ(i) (z) − ǫ σT Comparing this with (2.16) at, say, x(i) = O(ǫβ ) with β > 1 on both side of the interface, and note that ∂x = σǫT ∂z , gives ([19]) (1)

(2)

c1 = φ(1) (0+ ), c1 = φ(2) (0− ); ǫ ǫ (1) (2) c2 = ∂z φ(1) (0+ ), c1 = ∂z φ(2) (0− ). σT (0+ ) σT (0− ) Now (2.14) yields the interface conditions of the diffusion limit φ(1) (0+ ) = φ(2) (0− ),

1 1 ∂z φ(1) (0+ ) = ∂z φ(2) (0− ). σT (0+ ) σT (0− )

(2.17)

In summary the diffusion limit with interface at x = 0 is −

d 1 ( φz ) + σ a φ = q dz 3σT

(2.18a)

with the boundary condition PM φ − ǫ σλT ∂z φ = m=1 ψLm αm , z=z L P−M = m=−1 ψRm αm , φ + ǫ σλT ∂z φ

(2.18b)

z=zR

and the interface condition

[φ] x=0 =



 1 ∂z φ = 0. σT x=0

(2.18c)

A uniform second order Method for Transport Equation with Interfaces

7

3. The numerical method and its AP property. 3.1. Derivation of the method. Assume σa , σT , q are discontinuous at some discrete set of points. Generate a set of grid points zL = z0 < z1 < · · · < zN = zR , so all discontinuities of σa , σT , q are grid points. Let △z = maxi=0,··· ,N −1 |zi+1 − zi |. Define the cell averages Z zi+1 Z zi+1 1 1 σa (z)dz, σT i = σT (z)dz, σai = zi+1 − zi zi zi+1 − zi zi

qi =

1 zi+1 − zi

Z

zi+1

q(z)dz,

i = 0, · · · , N − 1,

zi

and σ ˜a , σ ˜T , q˜ to be piecewise constant functions: σ ˜a (z) = σai ,

σ ˜T (z) = σT i ,

i = 0, . . . , N − 1. (3.1) We get an approximation of the discrete ordinate equation (1.3a): σ ˜T (z) ˜ dψ˜m ψm (z) = (z) + µm dz ǫ



q˜(z) = qi ,

z ∈ (zi , zi+1 ],

X σ ˜T (z) ψ˜n (z)wn + ǫ˜ q (z), − ǫ˜ σa (z) ǫ

m ∈ V, (3.2)

n∈V

with the same boundary condition (1.3b). The rest of the method pretty much follows the framework of “spectral Green’s function” method by De Barros and Larsen [12] (see also later works [3, 13], except that we are using the explicit eigenvalues and eigenfunctions and the corresponding discrete W -functions derived in [19]. Note that on [zi , zi+1 ], (3.2) is the constant coefficient discrete-ordinate equation, thus can be solved exactly like in [19]. In order to find the solution of X  dψm σT i σT i µm (z) + ψm (z) = − ǫσai ψn (z)wn + ǫqi , (3.3) dz ǫ ǫ n∈V

firstly we seek the general solution of the homogeneous equation X  σT i σT i dψm ψn (z)wn , z ∈ [zi , zi+1 ]. (z) + ψm (z) = − ǫσai µm dz ǫ ǫ

(3.4)

n∈V

Inserting ψm (z) = lm e−

σT i ǫ

ξz

into (3.4) gives the eigenfunction equation  σai  X (1 − µm ξ)lm = 1 − ǫ2 ln wn . σT i n∈V

Hence lm must have the form lm =

constant , 1 − µm ξ

(3.5)

(3.6)

8

S. Jin, M. Tang AND H. Han

where the ’constant’ is independent of m but depends on i. Substituting this lm into (3.6), one gets the characteristic equation X wn 1 . (3.7) σai = 2 1 − ǫ σT i 1 − µn ξ n∈V

It has been shown by Chandrasekha [8] that when σai 6= 0, all the eigenvalues are different from each other and appear in positive/negative pairs. When σai = 0, 0 is a double root and the other eigenvalues occur in positive/negative pairs. Let ξn = −ξ−n ≥ 0, for n = 1, · · · , M . When σai = 0, ξ−M = ξM = 0. Now, because the general solutions have different forms, we consider the two different cases: (i) When σai 6= 0, there are 2M eigenfunctions of the form (3.5). Hence the general solution of (3.4) can be expressed as X
σT i (n) ξn z , z ∈ [zi , zi+1 ], (3.8) ψm (z) = A(n) lm exp − ǫ n∈V

(n)

where A(n) are undetermined and lm are the eigenfunctions corresponding to the nonzero eigenvalues ξn which is given by (n) lm =

1 . 1 − µm ξn

In the Appendix we will prove that, if σai 6= 0, for 1 ≤ |n|, |k| ≤ M ,  X 0 n 6= k (k) (n) wm µm lm lm = , c(k) n = k

(3.9)

m∈V

where c(k) is some nonzero constant. In addition, it is easy to check that ψm =

qi σai

(3.10)

is a special solution of the inhomogeneous equation (3.3). Then the general solution of (3.3) has the following form X
σT i qi (n) ψm (z) = A(n) lm exp − ξn z + , z ∈ [zi , zi+1 ]. (3.11) ǫ σai 1≤|n|≤M

˜ i ) and ψ(z ˜ i+1 ) By the aid of these expressions, we can find the relation between ψ(z as follows: A(n) can be determined by ψ˜m (zi ) from X
qi σT i (n) ξn zi + . ψ˜m (zi ) = A(n) lm exp − ǫ σai n∈V

(k)

Multiplying both sides of the above equation by wm µm lm and summing over V give X (k) ˜ wm µm lm ψm (zi ) m∈V

=

X

m∈V

=

X

n∈V


σT i ξn zi + ǫ n∈V X
σ Ti (k) (n) wm µm lm lm exp − ξn zi + ǫ

(k) wm µm lm

A(n)

m∈V

X

(n) A(n) lm exp −

qi X (k) wm µm lm σai m∈V qi X (k) wm µm lm . σai m∈V

A uniform second order Method for Transport Equation with Interfaces

˜ i ) as By using (3.9), A(k) can be given by ψ(z 
qi X σT i 1 X (k) ˜ (k) ψm (zi ) − exp wm µm lm wm µm lm ξk zi . A(k) = (k) σai ǫ c m∈V m∈V

9

(3.12a)

˜ i+1 ) as Similarly A(k) can also be expressed by ψ(z 
σT i 1 X qi X (k) (k) ˜ exp A(k) = (k) wm µm lm ξk zi+1 . (3.12b) wm µm lm ψm (zi+1 ) − σai ǫ c m∈V m∈V Equating the two equations of (3.12), the relations between ψ˜m (zi ) and ψ˜m (zi+1 ) are: X 
qi X σT i (k) ˜ (k) wm µm lm ψm (zi ) − exp wm µm lm ξk (zi − zi+1 ) σai ǫ m∈V m∈V X X qi (k) ˜ (k) = wm µm lm ψm (zi+1 ) − wm µm lm , k>0 (3.13a) σai m∈V m∈V  X
σT i qi X (k) (k) ˜ exp wm µm lm ξk (zi+1 − zi ) wm µm lm ψm (zi+1 ) − σai ǫ m∈V m∈V X qi X (k) ˜ (k) = wm µm lm ψm (zi ) − wm µm lm , k < 0. (3.13b) σai m∈V

m∈V

The reason why we express the relations by (3.13a) and (3.13b) is to guarantee that the coefficients would not become exponentially large when ǫ ≪ 1. Since the interface condition (2.10) requires ψ˜ to be continuous, i.e. two neighboring intervals [zi−1 , zi ], ˜ i ), (3.13a)(3.13b) are in fact a particular finite difference [zi , zi+1 ] use a common ψ(z scheme. (ii) When σai = 0, the general solution of the homogeneous equation (3.4) is  σ X
σT i Ti (k) A(k) lm exp − z − µm + ξk z . (3.14) ψm (z) = c1 + c2 ǫ ǫ 1≤|k|≤M −1

Here c1 , c2 , A(n) are undetermined, the function 1 and σTǫ i z − µm are solutions of the (k) eigenfunction equation (3.6) corresponding to the double root zero and lm are the eigenfunctions corresponding to the nonzero eigenvalues ξk which is given by (2.6). The special solution of the inhomogeneous equation (3.3), when σai = 0, is 3 ǫ2 qi ψm = − σT i qi z 2 + 3ǫqi µm z − (3µ2m − 1). 2 σT i Thus the general solution of (3.3) becomes  σ Ti z − µm + ψm (z) = c1 + c2 ǫ

X

(n) A(n) lm exp −

1≤|n|≤M −1

(3.15)


σT i ξn z ǫ

ǫ2 qi 3 (3µ2m − 1). − σT i qi z 2 + 3ǫqi µm z − 2 σT i

(3.16)

˜ i ) and ψ(z ˜ i+1 ) can be derived as follows: The relation between ψ(z At z = zi , (3.16) gives σ  X
σT i Ti (n) ψ˜m (zi ) = c1 + c2 zi − µm + ξn zi A(n) lm exp − ǫ ǫ 1≤|n|≤M −1

3 ǫ2 qi − σT i qi zi2 + 3ǫqi µm zi − (3µ2m − 1). 2 σT i

(3.17)

10

S. Jin, M. Tang AND H. Han (k)

For 1 ≤ |k| ≤ M − 1, multiplying both sides of (3.17) by wm µm lm , summing over V (n) and using the properties of lm for σai = 0 in (2.8)(2.7), one gets A(k) =

1

exp c(k)


σT i ξk zi ǫ

X

(k) ˜ wm µm lm ψm (zi ) + 3ǫ2

 qi X (m) . (3.18) wm µ3m lk σT i m∈V

m∈V

Then multiplying (3.17) by wm µm and wm µ2m respectively and summing over V , we obtain qi X qi 3 wm µ4m − ǫ2 c1 = − σT i qi zi2 + 9ǫ2 2 σT i σT i m∈V X
σT i +3 wm µm zi + µm ψ˜m (zi ), ǫ m∈V X c2 = 3ǫqi zi − 3 wm µm ψ˜m (zi ).

(3.19) (3.20)

m∈V

˜ i+1 ) and we consequently Similarly, A(k) , c1 , c2 can also be expressed in terms of ψ(z ˜ ˜ get the relations between ψ(zi ) and ψ(zi+1 ): exp


σT i ξk (zi − zi+1 ) ǫ

X

(k) ˜ wm µm lm ψm (zi ) + 3ǫ2

 qi X (k) wm µ3m lm σT i m∈V

m∈V

qi X (k) (k) ˜ wm µ3m lm , k>0 (3.21a) = wm µm lm ψm (zi+1 ) + 3ǫ2 σT i m∈V m∈V   X
X σT i 3 (k) (k) ˜ 2 qi exp ξk (zi+1 − zi ) wm µm lm wm µm lm ψm (zi+1 ) + 3ǫ ǫ σT i m∈V m∈V X qi X (k) (k) ˜ wm µ3m lm , k 0, σa (z) > 0, q(z) are piecewise C 2 functions. Let σ ˜T (z), σ ˜a (z), q˜(z) be the piecewise cell-average constants defined in (3.1) and ψ˜ = (ψ˜−M , · · · , ψ˜−1 , ψ˜1 · · · , ψ˜M )T be the solution of (3.2) with the same boundary condition (1.3b). Then there exists an ǫ0 > 0, s.t. if ǫ ≤ ǫ0 , we have n  o ˜m | ˜ L∞ ([z ,z ]×V ) = max < Ctra ∆z 2 . max |ψ − ψ kψ − ψk m L R z∈[zL ,zR ]

m∈V

Here Ctra is independent of ǫ and ∆z. Proof. Let the error em = ψm − ψ˜m . (1.3a) minus (3.2) gives X  σ ˜T σ ˜T dem + em − − ǫ˜ σa en wn µm dz ǫ ǫ n∈V  X
σT − σ ˜T σT − σ ˜T =− ψm + − ǫ σa − σ ˜a ) ψn wn + ǫ q − q˜ ǫ ǫ n∈V

≡ Qm ,

(4.1)

while the boundary conditions become emL = em (zL ) = 0,

µm > 0 ;

emR = em (zR ) = 0,

µm < 0.

Step 1: We try to derive some moment equations of (4.1). Introduce some moments: P (k) e˜k = Pm∈V µm wm lm em , for 1 ≤ P|k| ≤ M − 1, e˜−M = m∈V wm µm em , e˜M = m∈V wm µ2m em ,

(4.2)

(4.3)

(k)

where lm are defined as in (2.6). Let e = (e−M , · · · , eM )T ,

˜ = (˜ e e−M , · · · , e˜M )T .

(4.4)

˜ satisfies, we have to express e in terms of e ˜. With In order to find the equation that e c(k) (1 ≤ |k| ≤ M − 1) given in (2.8) and c(−M ) = c(M ) = 1/3, from (2.9), we have for n∈V, X

1≤|k|≤M −1

X

=

=

1 c(M )

X

m∈V

+

µn e˜ (−M ) −M c

+

1

e˜M c(M )

1

1≤|k|≤M −1

+

1 (k) ln e˜k (k) c

X µn X (k) (k) l µ w l e + wm µm em m m m n m c(k) c(−M ) m∈V m∈V

X

(4.5)

wm µ2m em

m∈V

wm µm em



X

1≤|k|≤M −1

µn µm  (k) (k) = en . l l + + n m c(k) c(−M ) c(M ) 1

(4.6)

A uniform second order Method for Transport Equation with Interfaces

13

For m ∈ V , let (−M ) lm = µm ,

(M ) lm = 1.

(4.7)

(4.6) then is en =

X 1 l(k) e˜k . (k) n c k∈V

(4.8)

Write (4.1) in the following form: µm

X σ ˜T σ ˜T X dem + em − en wn = −ǫ˜ σa en wn + Qm . dz ǫ ǫ n∈V

(4.9)

n∈V

(k)

For1 ≤ |k| ≤ M − 1, multiplying both sides of (4.9) by wm lm and summing over V give X X σ ˜T X σ ˜T X (k) (k) (k) dem + wm lm em − wm lm en wn wm µm lm dz ǫ ǫ m∈V m∈V n∈V m∈V X X X (k) (k) = −ǫ˜ σa wm lm en wn + wm lm Qm . (4.10) m∈V

n∈V

m∈V

From (2.5)(2.6), X σ ˜T X σ ˜T X (k) (k) wm lm em − wm lm en wn ǫ ǫ m∈V m∈V n∈V  X σ ˜T X σ ˜T  X (k) (k) wm lm em − em wm = ξk µm wm lm em . = ǫ ǫ m∈V

m∈V

m∈V

Substituting this into (4.10) gives σ ˜T X dem (k) + ξk µm wm lm em dz ǫ m∈V m∈V X X (k) = −ǫ˜ σa em wm + wm lm Qm . X

(k) wm µm lm

m∈V

(4.11)

m∈V

Similarly, multiplying both sides of (4.9) by wm and wm µm , summing over V , one gets X X X dem = −ǫ˜ σa em wm + wm Qm , (4.12) wm µm dz m∈V m∈V m∈V X X σ ˜T X dem + wm µm em = wm µm Qm , (4.13) wm µ2m dz ǫ m∈V

m∈V

m∈V

Substituting (4.3)(4.8) into (4.11)-(4.13) gives a system of equations that e˜k satisfy: X X 1 X σ ˜T d˜ ek (n) (k) + ξk e˜k = −ǫ˜ σa wm lm e˜n + wm lm Qm (n) dz ǫ c m∈V n∈V m∈V X X 1 (k) e ˜ + wm lm Qm , = −ǫ˜ σa m c(m) m∈V m∈V \{M }

(4.14a)

14

S. Jin, M. Tang AND H. Han

d˜ e−M = −ǫ˜ σa dz

1

X

m∈V \{M }

e˜m c(m)

+

X

wm Qm ,

(4.14b)

m∈V

X d˜ eM σ ˜T + e˜−M = wm µm Qm . dz ǫ

(4.14c)

m∈V

˜ defined by (4.14), including its values Step 2: Write the integral formulations of e at the boundaries. Let Z z σ ˜T (x)dx. (4.15) α ˜ (z) = zL

Using the method of integrating factor, for k = 1, · · · , M − 1, multiplying both sides
˜ ξ and integrating from zL to z gives of (4.14a) by exp α(z) k ǫ e˜k (z) (4.16) Z z  α    X ˜ (z) α ˜ (x) − α(z) ˜ 1 = exp − exp e˜ dx ˜a ξk e˜k (zL ) − ǫ ξk σ (m) m ǫ ǫ c zL m∈V \{−M } Z z α ˜ (x) − α ˜ (z)  X (k) ξk wm lm Qm dx. (4.17) exp + ǫ zL m∈V

Similarly, for k = −M + 1, · · · , −1, integrate from z to zR : α ˜ (zR ) − α ˜ (z)  e˜k (z) = exp ξk e˜k (zR ) ǫ Z zR α  X ˜ (x) − α(z) ˜ 1 exp +ǫ e˜m dx ˜a ξk σ (m) ǫ c z m∈V \{−M } Z zR
X α ˜ (x) − α(z) ˜ (k) exp − ξk wm lm Qm dx. ǫ z

(4.18)

m∈V

Integrating (4.14b)(4.14c) from zL to z gives Z z X σ ˜a e˜−M (z) = e˜−M (zL ) − ǫ e˜M (z) = −

Z

z zL

z X wm Qm dx,(4.19) e ˜ dx + m c(m) zL m∈V zL m∈V \{−M } Z z X σ ˜T (x) wm µm Qm dx. (4.20) e˜−M (x) dx + e˜M (zL ) + ǫ zL

1

Z

m∈V

Like in (4.17)–(4.19), for (4.20), we would like to express e˜M (z) by its boundary data. Substituting (4.19) into (4.20) gives Z x Z z X α ˜ (z) 1 e˜M (z) + σ ˜a e˜ dsdx σ ˜T e˜−M (zL ) − (m) m ǫ c zL zL m∈V \{−M } Z z X Z Z z σ ˜T x X wm µm Qm dx. (4.21) wm Qm dsdx + e˜M (zL ) + =− zL zL zL ǫ m∈V

m∈V

A uniform second order Method for Transport Equation with Interfaces

15

In particular, when z = zR , (4.21) is

α ˜ (zR ) e˜−M (zL ) e˜M (zR ) − e˜M (zL ) + ǫ Z zR Z x X 1 σ ˜a σ ˜T = e˜m dsdx (m) c zL zL m∈V \{−M } Z z X Z Z zR σ ˜T x X wm µm Qm dx. wm Qm dsdx + − ǫ zL zL zL

(4.22)

m∈V

m∈V

From (4.22) we can get an expression for e˜−M (zL )/ǫ. Substituting it into (4.21), one gets

α ˜ (z) e˜M (z) + α ˜ (zR ) Z x Z z σ ˜a σ ˜T − zL

=

zL

Z

zR

σ ˜T

Z

x

σ ˜a

zL

zL

X

m∈V \{−M }

1

X

m∈V \{−M }

1

e˜m c(m)

e˜m c(m)

dsdx

dsdx

 α(z) ˜ α ˜ (z)  e˜M (zL ) e˜M (zR ) + 1 − α ˜ (zR ) α ˜ (zR ) Z z Z zR Z x X Z σ ˜T x X α(z) ˜ σ ˜T + wm Qm dsdx − wm Qm dsdx α ˜ (zR ) zL ǫ zL zL ǫ zL m∈V m∈V Z z X Z zR X α(z) ˜ wm µm Qm dx. (4.23) wm µm Qm dx + − α ˜ (zR ) zL zL m∈V

m∈V

Formally, (4.17)(4.18)(4.19)(4.23) is the integral formulation of (4.14). Note that the left side of the combination of these equations can be viewed as an ǫ-independent ˜ and all the perturbations of σT , σa , q occur in Qm . In order to operator acting on e ˜(zL ), e ˜(zR ) by Qm prove the convergence we have to find the explicit expressions of e and some operator acting on e.

16

S. Jin, M. Tang AND H. Han

Particularly, let z = zR , (4.17)(4.18)(4.19) become  α ˜ (zR )  ξk e˜k (zL ) e˜k (zR ) − exp − ǫ Z zR α X ˜ (z) − α ˜ (zR )  1 e˜m dz σ ˜a exp = −ǫ ξk (m) ǫ c zL m∈V \{−M } Z zR α ˜ (z) − α ˜ (zR )
X (k) exp + ξk wm lm Qm dz, k = 1, · · · , M − 1 ǫ zL m∈V

(4.24a) α ˜ (zR )
ξk e˜k (zR ) − e˜k (zL ) exp Z zR ǫ X
α(z) ˜ 1 σ ˜a exp = −ǫ e˜m dz ξk (m) ǫ c zL m∈V \{−M } Z zR  α(z) X ˜ (k) exp + ξk wm lm Qm dz, k = −M + 1, · · · , −1. ǫ zL m∈V

e˜−M (zR ) − e˜−M (zL ) Z zR X σ ˜a = −ǫ zL

m∈V \{−M }

(4.24b) 1 c(m)

e˜m dz +

Z

zR

zL

X

wm Qm dz.

(4.24c)

m∈V

Furthermore, write (4.22) as follows ǫ˜ eM (zR ) − ǫ˜ eM (zL ) + α ˜ (zR )˜ e−M (zL ) Z zR Z x X 1 σ ˜a σ ˜T =ǫ e˜ dsdx (m) m c zL zL m∈V \{−M } Z z X Z zR Z x X wm µm Qm dx. wm Qm dsdx + ǫ σ ˜T −

(4.24d)

zL m∈V

zL m∈V

zL

and the boundary conditions of e using (4.8) imply 1 (k) l e˜k (zL ) c(k) n 1 (k) ˜k (zR ) k∈V c(k) ln e

P Pk∈V

= 0, = 0,

for k = 1, · · · , M . for k = −M, · · · , −1

(4.24e)

Now (4.24) is a system of 4M linear equations with 4M unkonwns. Denote the right hand side vector by Λ. It is obvious that e˜k (zL ), e˜k (zR ), k ∈ V can be determined by Λ through some linear operators LLk and LRk (which are part of the inverse of the coefficient matrix of (4.24)): ˜k (zL ) = LLk Λ, e

˜k (zR ) = LRk Λ. e

Λ can be decomposed into two parts, one only contains the fluxes of Qm and the ˜. That is other is some linear operator acting on e Λ = ǫΛ(1) (˜ e) + Λ(2) , where (1)

(1)

Λ(1) = (Λ−M , · · · , ΛM ),

(2)

(2)

Λ(2) = (Λ−M , · · · , ΛM ),

17

A uniform second order Method for Transport Equation with Interfaces (1)

(2)

and Λk , Λk are given as follows: When k = 1, · · · , M − 1, let Z

(1)

Λk (f ) = − (2)

Λk =

zR zL

zR

Z

α(z) ˜ −α ˜ (zR )
ξk ǫ

σ ˜a exp

exp

zL

α ˜ (z) − α(z ˜ R) ξk ǫ


X

X

m∈V \{−M }

1 fm dz, c(m)

(k) wm lm Qm dz,

m∈V

where f = (f−M , · · · , f−1 , f1 , · · · , fM ) ∈ R2M . When k = −M + 1, · · · , −1, (1)

Λk (f ) = − (2) Λk

=

Z

zR

Z

α ˜ (z)
ξk ǫ

σ ˜a exp

zL

zR

α(z) ˜ ǫ ξk

e

zL

X

X

m∈V \{−M }

1 fm dz c(m)

(k) wm lm Qm dz.

m∈V

Moreover (1) Λ−M (f )

=−

Z

zR

σ ˜a zL

m∈V \{−M }

(1)

ΛM (f ) =

(2)

ΛM = −

Z

1

X

zR

zL

σ ˜T

Z

zR

σ ˜T

z

Z

z

σ ˜a

zL

zL

Z

fm dz, c(m)

X

=

Z

zR

zL

X

wm Qm dz,

m∈V

1

X

m∈V \{−M }

wm Qm dxdz + ǫ

zL m∈V

(2) Λ−M

Z

fm dxdz, c(m)

zR zL

X

wm µm Qm dz.

m∈V

Then e˜k (zL ) = LLk Λ = ǫLLk Λ(1) (˜ e) + LLk Λ(2) ,

e˜k (zR ) = LRk Λ = ǫLRk Λ(1) (˜ e) + LRk Λ(2) . (4.25) Now we have found the expressions of e˜k (zL ), e˜k (zR ) through the linear operators LLk , LRk . The combination of (4.17)(4.18)(4.19)(4.23) can be written in the vector form Π˜ e(z) = Θ(z) + ǫΩ˜ e(z), where   k = −M, · · · , M − 1  fk , R z Rx P 1 fM − zL σ ˜T zL σ ˜a m∈V \{−M } c(m) fm dsdx (Πf )k = R R P  z x α ˜ (z) R 1 T  fm dsdx, + α(z ˜T zL σ ˜a m∈V \{−M } c(m) ˜ R ) zL σ

, k=M (4.26)

18

S. Jin, M. Tang AND H. Han



P Rz (k) ˜ ˜ α(z) ˜ (2)  exp − α(z) + zL exp α(x)− ξk  m∈V wm lm Qm dx, ǫ ξk LLk Λ ǫ    k = 1, · · · , M − 1  

P R zR  (k)  α(x)− ˜ α(z) ˜ α(z ˜ R )−α(z) ˜ (2)  exp ξ L Λ − ξ exp  k Rk k m∈V wm lm Qm dx, ǫ ǫ z    k = −M + 1, · · · , −1 Rz P (2) , Θk (z) = L Λ + w Q dx, k = −M m L(−M ) zL m∈V   m   α(z) ˜ α(z) ˜ (2) (2)   + 1 − α(z  α(z ˜ R ) LRM Λ ˜ ) LLM Λ  R zR σ˜T R x P R R zR P   α(z) ˜ α(z) ˜  − α(z + α(z  m∈V wm µm Qm dx m∈V wm Qm dsdx ˜ R ) zRL P ǫ ˜ R ) zL zL  R R P  z  − z σ˜T x w Q dsdx + w µ Q k=M m m m m m dx, m∈V m∈V zL zL ǫ zL (4.27) and
 ˜ (1) (f ) exp − α(z)  ǫ ξk LLk Λ 
P R  z α(x)− ˜ α(z) ˜ 1   − exp fm dx, ξ ˜a m∈V \{−M } c(m) k σ  ǫ z L    k = 1, ··· ,M − 1  
 ˜ R )−α(z) ˜ (1)  exp α(z ξk LRk Λ (f )
P R zR ǫ α(x)− (Ωf )k (z) = . ˜ α(z) ˜ 1 fm dx, ξk σ ˜a m∈V \{−M } c(m) + exp  ǫ z     k = −M + 1, · · · , −1  Rz P  1   LL(−M ) Λ(1) (f ) − zL σ ˜a m∈V \{−M } c(m) fm dx, k = −M       α(z) ˜ α(z) ˜ (1) (1) (f ) + 1 − α(z (f ), k=M α(z ˜ R ) LRM Λ ˜ R ) LLM Λ (4.28) Π, Ω are operators. It is obvious that Π is invertible (since it is diagonal except the last row) and independent of ǫ, thus ˜(z) = Π−1 Θ(z) + ǫΠ−1 Ω˜ e e(z).

(4.29)

Step 3: A Neumann series can be constructed corresponding to (4.29): ˜(z) = e

∞ X

˜(i) (z) e

(4.30)

i=0

with ˜(0) (z) = Π−1 Θ(z) e and
˜(i) (z) = (ǫΠ−1 Ω˜ e e(i−1) )(z) = ǫi Π−1 Ω)i Θ(z) .

(4.31)

If the sum in (4.30) is convergent, (4.30) is just the solution of (4.29) [6]. In order to prove the convergence of the Neumann series, we study the properties of the terms in the Neumann series. We first establish several lemmas. Lemma 4.2. For an interval [zi , zi+1 ], assume ∆z = |zi+1 −zi |, f ∈ C 2 ([zi , zi+1 ]), g ∈ C 1 ([zi , zi+1 ]), and η ∈ [zi , zi+1 ] satisfies Z zi+1 1 ˜ f dz. f (η) = f = ∆z zi

19

A uniform second order Method for Transport Equation with Interfaces

Then, for x ≥ zi+1 R zi+1


f − f˜ g dz ∼ O(∆z 3 ),
α(z)− ˜ α(x) ˜ ξ ǫ f − f˜ g dz ∼ ξǫ O(∆z 3 ), e zi
R zi+1 α(z) ˜ e ǫ ξ f − f˜ g dz ∼ ξǫ O(∆z 3 ), zi zi

R zi+1

ξ>0

(4.32)

ξ 0 we have Z zi+1
α(z) ˜ −α ˜ (x)
ξ f − f˜ g dz exp ǫ zi Z
α ˜ (zi+1 ) − α ˜ (x)
zi+1 σT i = exp exp − ξ ξ(zi+1 − z) ǫ ǫ zi

′ 2 · f (η)(z − η) + O(∆z ) g(η) + O(∆z) dz 
σT i ˜ (x)
α ˜ (zi+1 ) − α ξ exp − ξ(zi+1 − η) = exp ǫ ǫ 1 
· f ′ (η)g(η) zi + zi+1 − 2η + O(∆z 2 ) ∆z 2 Z zi+1 

 σT i σT i ξ(zi+1 − z) − exp − ξ(zi+1 − η) exp − +f ′ (η)g(η) ǫ ǫ zi 
· (z − η) + O(∆z 2 ) dz ∼ exp

α ˜ (η) − α ˜ (x)
ξ ξ ξ O(∆z 3 ) + O(∆z 3 ) ∼ O(∆z 3 ). ǫ ǫ ǫ

The result for ξ < 0 can be obtained similarly. Lemma 4.3. For the linear operator LLm , LRm defined in (4.25), we have kLLm k∞ ≤ Cl ,

kLRm k∞ ≤ Cl ,

where Cl is a constant independent of ǫ and ∆z.

∀m ∈ V

20

S. Jin, M. Tang AND H. Han

Proof. The linear operators LLk , LRk defined in (4.25) arise in inverting the coefficient matrix of the following system of linear equations: e˜k (zR ) − exp − exp

α ˜ (zR )
ξk e˜k (zL ) = bk , ǫ

α ˜ (zR )
ξk e˜k (zR ) − e˜k (zL ) = bk , ǫ

e˜−M (zR ) − e˜−M (zL ) = b−M , X 1 (n) l e˜n (zL ) = 0, c(n) k n∈V

1 ≤ k ≤ M − 1,

(4.33a)

−M + 1 ≤ k ≤ −1

(4.33b)

ǫ˜ eM (zR ) − ǫ˜ eM (zL ) + α ˜ (zR )e−M (zL ) = bM , (4.33c) X 1 (n) l e˜n (zR ) = 0, c(n) k n∈V

k > 0,

k < 0,

with b = (b−M , · · · , bM )T being any vector. Assume X 1 (n) X 1 (n) l e ˜ (z ) = a , k < 0, l e˜n (zR ) = ak , n L k k c(n) c(n) k n∈V n∈V

(4.33d)

k > 0.

Using (4.33d)(2.8)(2.7)(4.8), we have, for 1 ≤ |m| ≤ M − 1, X X X 1 (n) (m) (m) l e˜n (zR ) wk µk lk ak = wk µk lk c(n) k n∈V k>0 k∈V X 1 X (m) (n) = e˜n (zR ) wk µk lk lk = e˜m (zR ), (n) c n∈V k∈V X

(m)

X 1 (n) l e˜n (zL ) c(n) k n∈V k∈V X 1 X (m) (n) = e˜n (zL ) wk µk lk lk = e˜m (zL ), (n) c n∈V k∈V

wk µk lk ak =

k0 k∈V

X

wk µk ak = e˜−M (zL ),

k0

X

wk µ2k

k∈V

Inserting these into (4.33) gives X α ˜ (zR )
X (m) (m) exp ξm wk µk lk ak − wk µk lk ak = bm , ǫ k>0

(m)

wk µk lk ak − exp −

X

wk µk ak −

k>0

ǫ

α ˜ (zR )
X (m) ξm wk µk lk ak = bm , ǫ k0

X

wk µk ak = b−M ,

k0

and e˜−M (zR ) ≤ kB −1 k∞ kbk∞

X

|wk µk |,

e˜M (zR ) ≤ kB −1 k∞ kbk∞

k>0

X

|wk µ2k |.

k>0

For any vector b, we have kLRm k∞ ≤

X

(m)

|wk µk lk |kB −1 k∞ .

k>0

Now what remains is to prove kB −1 k∞ is uniformly bounded with respect to ǫ. Firstly, we recall a basic lemma in matrix theory (see for [16]). Assume D1 , D2 ∈ Rn×n . If D1 is invertible and kD1−1 k∞ ≤ β1 , kD1 − D2 k∞ ≤ β2 , β1 β2 < 1, then β1 . For the matrix B given in (4.35), we can D2 is invertible and kD2−1 k∞ ≤ 1−β 1 β2 decompose it into B = B1 + B2 , where   −(wk µk )k0  − w µ l(m)
 0 k k k   −M +1≤m≤−1,k0  −˜ α(zR )(wk µk )k 0

(4.37b)

γRm = ψRm − φ(zR ) − ǫ

µm ∂z φ(zL ), σT (zL )

for m > 0

(4.37c)

where x=

1 α(z) = ǫ ǫ

Z

z

σT (s)ds.

(4.38)

zL

The next lemma establishes the uniform boundedness of gm . Lemma 4.4. For gm defined in (4.36), we have kgm kL∞ ([zL ,zR ]×V ) < Cg , where Cg is independent of ǫ. Proof. It has been shown in Theorem 3.2 in [14] that, when there is no interface, kgm kL∞ ([zL ,zR ]×V ) is uniformly bounded with respect to ǫ. With an interface at z = 0, since there is no sharp interface layer on both sides of the interface, by the same analysis as in [14], if

 1 d k

φ ∞ ,

σT dz L ([zL ,zR ]\0)

k = 1, 2, 3, 4

(4.39)

are uniformly bounded, we have kgm kL∞ ([zL ,zR ]×V ) uniformly bounded. The maximum of φ in [zL , zR ] is at the boundary, or the interface or the interior of the domain. If it is at the boundary, we have ∂z φ|zL < 0 or ∂z φ|zR > 0. From (2.18b), kφkL∞ [zL ,zR ] < max{

M X

m=1

ψLm wm ,

−M X

ψRm wm } ≤ max(kψL k, kψR k) .

m=−1

If the maximum is at the interface z = 0, since ∂z φ(0− ), ∂z φ(0+ ) have the same sign by the interface condition (2.18c), φ will decrease or increase simultaneously on

23

A uniform second order Method for Transport Equation with Interfaces

both sides of the interface. Then ∂z φ(0+ ) = ∂z φ(0− ) = 0 and ∂z2 φ(0+ ), ∂z2 φ(0− ) < 0. Thus (2.18a) gives kφkL∞ [zL ,zR ] < k σqa kL∞ [zL ,zR ] . This also holds when the maximum occurs at neither boundary nor interface points. In summary, kφkL∞ [zL ,zR ] is bounded independent of ǫ. Use the same discussion as in Lemma 2.1 in [14], we have that

 1 d k

, k = 1, 2, 3, 4 φ ∞

σT dz L ([zL ,zR ]\0) are also uniformly bounded. The lemma is proved. Lemma 4.5. For Θ defined in (4.27), we have

kΘkL∞ ([zL ,zR ]×V ) < Cθ ∆z 2 , where Cθ is some constant independent of ǫ, ∆z. (2) Proof. RIt is obvious that the integrals in the definition of Λk can be divided into zi+1 the sum of zi , i = 0, 1, · · · , N −1. So do the integrals in the definition of Θk (4.27), Rz by assuming z ∈ (zj , zj+1 ], can be divided into the sum of zii+1 , i = 0, · · · , j − 1 and Rz (2) . When 1 ≤ k ≤ M − 1, all integrals in Λk and Θk defined in the interval [zi , zi+1 ] zj can be controlled by some constant multiplying Z zi+1 α(z)− X ˜ α(z ˜ i+1 ) ξk (k) ǫ wm lm Qm dz. (4.40) e zi

m∈V

Let ∆z = |zi+1 − zi |. Inserting (4.36) into the definition of Qm in (4.1) gives Qm  µm dφ σT − σ ˜T  φ−ǫ + ǫ2 gm + γLm + γRm + ǫ(q − q˜) =− ǫ σT dz   X X X
σT − σ ˜T + − ǫ(σa − σ ˜a ) φ + ǫ2 wn gn + wm γLm + wm γRm ǫ n∈V n∈V n∈V   X dφ σT − σ ˜T µm − ǫ(σa − σ ˜a ) φ + ǫ2 wn gn = σT dz n∈V X  wn gn − gm + ǫ(q − q˜) +ǫ(σT − σ ˜T ) n∈V

 X X σT − σ ˜T  γLm − wn γLn + γRm − wn γRn − ǫ n∈V

(4.41)

n∈V

Using the eigenfunction (2.5), the definition of (2.6) and (2.7), one gets  X X σT − σ ˜T  X (k) (k) (k) wm lm Qm = − ξk wm µm lm γLm + wm µm lm γRm ǫ m∈V m∈V m∈V   X +ǫ (σT − σ ˜T ) wm gm − (σa − σ ˜a )φ + (q − q˜) m∈V

3

−ǫ (σa − σ ˜a )

X

wm gm − ǫ(σT − σ ˜T )

m∈V

X

(k) wm lm gm .

m∈V

In order to estimate (4.40), firstly consider Z zi+1 α X ˜T ˜ (x) − α ˜ (zi+1 )  σT − σ (k) ξk ξk wm µm lm γLm . exp − ǫ ǫ zi m∈V

24

S. Jin, M. Tang AND H. Han

Similar to the derivation from (4.9) to (4.11), multiplying both sides of (4.37a) by (k) wm lm and summing over V gives X X
(k) (k) ∂x wm µm lm γLm + ξk wm µm lm γLm = 0. m∈V

m∈V

Then X

(k) wm µm lm γLm = C exp(−ξx) = C exp −

m∈V


ξk α(z) , ǫ

where C is a constant depending only on the boundary data and α(z) is given in (4.38). From the definition of σ ˜T in (3.1), α(z) ˜ (4.15), α(z) (4.38), we have σT (z) = ∂z α(z), σ ˜T (z) = ∂z α ˜ (z) and α(z ˜ j ) = α(zj ), ∀j. Thus, zi+1

σ − σ X ˜T T (k) ξk wm µm lm γLm dz ǫ ǫ zi m∈V Z  α(z  α(z) ˜ i+1 )  zi+1 σT − σ ˜ − α(z)  ˜T = −C exp − ξk ξk exp ξk dz ǫ ǫ ǫ zi α α  α(z ˜ (zi+1 ) − α(zi+1 )  ˜ (zi ) − α(zi )  ˜ i+1 )  ξk exp ξk − exp ξk = C exp − ǫ ǫ ǫ = 0. (4.42) −

Z

exp

α ˜ (z) − α ˜ (z

i+1 )

ξk

Note the cancelation in (4.42) is exactly due to the definition of σ˜T (x) as in (3.1). From Lemma 4.2, we have Z zi+1 α ˜ (x) − α(z ˜ i+1 )  X (k) exp ξk wm lm Qm dx ∼ O(δz 3 ) , (4.43) ǫ zi m∈V

by using (4.42). The same results hold for −M + 1 ≤ k ≤ −1. Moreover, we also need to estimate Z zi+1 X Z zi+1 X wm µm Qm dsdx, wm Qm dsdx zi

zi

m∈V

m∈V

(2)

that appear in the integrals in ΛM and ΘM when confined in [zi , zi+1 ]. Multiplying both sides of (4.37a) by wm and summing over V gives X  ∂x wm µm γm = 0. m∈V

Then, since γm decays exponentially to zero, from (4.41), Z

zi+1

Z

zi+1

zi

=

zi

X

P

m∈V

wm µm γm = 0 over [0, ∞), thus,

wm µm Qm dsdx

m∈V

 σ − σ X ˜T dφ T − ǫ(σT − σ ˜T ) wm µm gm dsdx ∼ O(∆z 3 ), 3σT dz m∈V

(4.44)

25

A uniform second order Method for Transport Equation with Interfaces

Z

zi+1

Z

zi+1

zi

=

zi

X

wm Qm dsdx

m∈V



   X (σa − σ ˜a ) φ + ǫ2 wn gn − (q − q˜) dsdx ∼ ǫO(∆z 3 ).

(4.45)

n∈V

(2)

In summary Λk is O(∆z 2 ) for 1 ≤ |k| ≤ M , namely, kΛ(2) k∞ ∼ O(∆z 2 ). And (2) Lemma 4.3 implies LLm ∼ ∆z 2 , LRm Λ(2) ∼ ∆z 2 . When z ∈ [zj , zj+1 ], the R zΛ analogous estimate of zj ∼ O(∆z 2 ) or ǫO(∆z 2 ) is easy to obtain. Then by using (4.43)-(4.45), a similar discussion shows that the integrals appearing in the definition of Θk in (4.27) are also of order ∆z 2 . Therefore we prove our result.

Now we are ready to prove Theorem 4.1. From the definition of Ω, Π, Λ(1) and Lemma 4.3, it is obvious that kΩk∞ < Cω ,

kΠ−1 k∞ < Cπ

(4.46)

where Cω , Cπ are constants independent of ∆z and ǫ. (4.31) implies

(i)

e˜ (z) ∞ = ǫi (Π−1 Ω)i Θ(z) L∞ ([zL ,zR ]×V ) L ([zL ,zR ]×V ) ≤ ǫi kΠ−1 Ωki∞ kΘkL∞ ([zL ,zR ]×V ) .

Assume when ǫ < ǫ0 , ǫkΠ−1 Ωk∞ ≤ ǫkΠ−1 k∞ kΩk∞ < 1. Using the Neumann series (4.30) and Lemma 4.5, one gets k˜ ek∞
0; 2

ψm (1) = 0,

µm < 0

σT = 1 + (10z) , σa = 1 + (10z) , q = 0, ε1 = 0.01, σT = 1, σa = z, q = 2, ε2 = 1. z ∈ [0.2, 1] .

z ∈ [0, 0.2] ;

In this example the mean free paths are different at the two sides of the interface. We can see clearly in Figure 5.3 that this method still gives a good result and a good description of the internal layer and of the boundary layer, even if the boundary layer is not resolved numerically. The error between the exact solution and the numerical solutions computed by this method with different ∆z for different ǫ are shown in the Table 3. Here the ’exact’ solution refers to the result computed by this numerical method when ∆z = 1/200. 6. Conclusion. In this paper, we study a numerical method for the discreteordinate linear transport equation with boundary and interface. The main idea of the

28

S. Jin, M. Tang AND H. Han ǫ

1

∆z 2−2 2−3 2−4 2−5

0.1

0.01

0.27615 0.00776 0.0945 0.0865 0.0197 0.0238 0.0230 0.0050 0.0059 0.0056 0.0012 0.0014 Table 5.2 Example 2. The L∞ norm of the errors between the “exact” solution and the numerical solutions computed with different ∆z for different ǫ, where the ’exact solution’ is calculated by the same numerical method with ∆z = 1/256.

∆z 1/5 1/10 1/20 1/40

kErrork∞ 2.854 ∗ 10−2 7.268 ∗ 10−3 1.813 ∗ 10−3 4.400 ∗ 10−4 Table 5.3 Example 3. The error between the “exact” solution and the numerical solutions computed by the numerical method with different ∆z , where the ’exact’ solution refers to the result computed with ∆z = 1/200

method is to first approximate the scattering coefficients using their cell averages, and then solve the local equation in each cell excatly [12]. Numerical solutions at different cells are pieced together using the interface condition which requires the density distribution to be continuous. This method is shown to be asymptotic-preserving, thus captures the correct diffusion limit with the correct interface condition. Moreover, we prove the this method converges quadratically, uniformly in the mean free path. This result is sharp, in constract to the previous analysis [14] in the same direction. Moreover, the method converges even if the boundary layer is not numerically resolved. In the future we will study this method in two space dimensions [3], and some other related transport equations. Acknowledgments. The second author thanks Prof. Zhongyi Huang for many helps during the preparation of this work. Appendix A. Theorem A.1. Consider the equation X

n∈V

1 wn = . 1 − µn ξ r

(A.1)

(i) When r < 1, (A.1) has 2M simple roots occur in positive/negative pairs, let (n) them be ξn (1 ≤ |n| ≤ M ). Assume lm = 1−µ1m ξn , we have  X 0 n 6= k (k) (n) wm µm lm lm = , (A.2) c(k) n = k m∈V

where c(k) satisfy  X 1 (k) (k) l l = c(k) m n k∈V

0 1 wn µn

m 6= n . m=n

(A.3)

A uniform second order Method for Transport Equation with Interfaces

29

4

3.5

ρ

3

2.5

2

1.5

1

0

0.1

0.2

0.3

0.4

0.5 z

0.6

0.7

0.8

0.9

1

Fig. 5.3. Example 3. Numerical solutions with ∆z = 1/10 are represented by circles, while the “exact” solution is represented by the solid lines.

(ii) When r = 1, (A.1) has 2M −2 simple roots appear in positive/negative pairs while 0 is a double root. Assume ξn (1 ≤ n ≤ M − 1) is the unique (positive, simple) (n) root in (1/µn+1 , 1/µn ), ξ−n = −ξn and lm = 1−µ1m ξn . (A.2) still holds for k, n ∈ P {−M + 1, · · · , M − 1}. Moreover, defining c(−M ) = c(M ) = m∈V wm µ2m = 1/3, we have  X µn µm 1 (k) (k) 0 m 6= n l l + + = , (A.4) 1 m=n c(k) n m c(−M ) c(M ) wn µn 1≤|k|≤M −1

and X

(k) µm wm lm = 0,

X

(k) µ2m wm lm = 0.

(A.5)

m∈V

m∈V

Proof. Let 

w−M  .. W = . w−M

··· .. . ···

 wM ..  , .  wM





µ−M

 U =

..

. µM

 ,

A = U −1 (I − rW ), where I is the identity matrix of order 2M . Assume l(n) is the eigenvector of A corresponding to ξn . Then U −1 (I − rW )l(n) = ξn l(n) holds and (n) lm

=

r

P

(n)

wk lk , 1 − ξn µm k∈V

m ∈ V.

(A.6)

30

S. Jin, M. Tang AND H. Han (n)

Multiplying lm by wm and summing over V give the eigenfunction as follows: X

m∈V

1 wm = , 1 − ξµm r

which is just (A.1). The properties of (A.1)’s roots when r < 1 and r = 1 as shown in the theorem have already been studied in [8]. From (A.6), let (n) lm =

1 , 1 − ξn µm

(n)

(n)

l(n) = (l−M , · · · , lM )

(A.7)

is the eigenvector corresponding to ξn . (i) When r < 1, A is diagonalizable. Let the eigenvalues satisfy ξn = ξ−n , for n = 1, · · · , M . Assume A = P −1 DP , where D = diag{ξ−M , · · · , ξ−1 , ξ1 , · · · , ξM }. Then the nth column of P −1 is the eigenvector corresponding to ξn and P −1 can be written as    h
i 1 −1 diag{c−M , · · · , c−1 , c1 , · · · , cM } . (A.8) P = mn 1 − ξn µm mn Here ci are some constants, [(aij )mn ] denotes the matrix whose element in row m and column n is aij . Now consider AT , the eigenvalues of AT are the same as A. Assuming ι(n) is the eigenvector of AT corresponding to ξn , we have (I − rW T )U −1 ι(n) = ξn ι(n) . Thus ι(n) m =

rµm wm X ιk , 1 − µm ξn µk k∈V

and ι(n) m =

µm wm 1 − µm ξn

is an eigenvector corresponding to ξn . Since AT = P T D(P T )−1 , the nth column of P T is the eigenvector of AT corresponding to ξn and P can be written as    i h 1 diag{w−M µ−M , · · · , wM µM }. (P )mn = diag{c′−M , · · · , c′M } 1 − ξm µn mn (A.9) From (A.8)(A.9), h h
i
i (n) ln(m) mn diag{w−M µ−M , · · · , wM µM } lm mn is equal to an diagonal matrix n diag

1 1 o 1 1 . , , · · · , , · · · , c−M c′−M c−1 c′−1 c1 c′1 cM c′M

31

A uniform second order Method for Transport Equation with Interfaces

Let c(k) = h

1 ck c′k ,

i h i n (n) (ln(m) )mn diag{w−M µ−M , · · · , wM µM } (lm )mn diag

h

i n (n) (lm )mn diag

1 c(−M )

,··· ,

1 o ,

c(M )

i 1 oh (m) , · · · , (l ) diag{w−M µ−M , · · · , wM µM } mn n c(−M ) c(M ) 1

are identity matrices. This gives (A.2)(A.3). (ii) When r = 1 in A, A is singular. We have to consider its Jordan Canonical Form. Suppose A = P −1 JP , where     ξ−M +1     ..     .   (A.10) J = , ξM −1       0 1 0 The eigenvector corresponding to ξn , 1 ≤ |n| ≤ M − 1, is still l(n) defined in (A.7). Let i = (1, · · · , 1)T2M ×1 and u = (µ−M , · · · , µ−1 , µ1 , · · · , µM )T . It is easy to check that Ai = 0 and Au = i, then P −1 can be written as P −1 = [l(−M +1) , · · · , l(M −1) , u, i]. As for AT , we can get P = diag{c(−M +1) , · · · , c(M −1) , c(−M ) , c(M ) }[l(−M +1) , · · · , l(M −1) , i, u]T ×diag{w−M µ−M , · · · , wM µM } by a similar discussion. Since P −1 P , P P −1 are identity matrices, (A.4)(A.5) are obtained. REFERENCES [1] G. Bal, L. Ryzhik, Diffusion approximation of radiative transfer problems with interfaces, SIAM J. App. Math. ,60(2000), pp. 1887–1912 [2] C. Bardos, R. Santos and R. Sentis,Diffusion approximation and computation of the critical size, Trans. Amer. Math. Soc., 284(1984), pp. 617–649 [3] R. C. De Barros, F. C. Da Silva and H. A. Filho, Recent advances in spectral nodal methods for x,y-geometry discrete ordinates deep penetration and eigenvalue problems, Progress in Nuclear Energy, 35(1999), pp. 293–331 [4] A. E. Berger,J. M. Solomon and M. Ciment, An analysis of a uniformly accurate difference Method for a singular perturbation problem, Math. of Comput., 37(155)(1981),pp. 79–94 [5] A. E. Berger, H. Han and R. B. Kellogg, A priori estimates and analysis of a numerical method for a turning point problem, Math. of Comput., 42(166)(1984), pp. 465–492 [6] K. M. Case and P. F. Zweifel, Existence and uniqueness theorems for the neutron transport equation, J. Math. Phys., 4(11)(1963) [7] R. Caflisch, S. Jin and G. Russo, Uniformly accurate schemes for hyperbolic systems with relaxations, SIAM J. Num. Anal., 34(1997), pp. 246–281 [8] S. Chandrasekhar, Radiative Transfer,Dover, New York, (1960) [9] P. Crispel, P. Degond and M.-H. Vignal, An asymptotic preserving scheme for the two-fluid Euler-Poisson model in the quasineutral limit, J. Comp. Phys., 223(2007), pp. 208–234, [10] P. Degond, S. Jin and J. G. Liu, Mach-number uniform asymptotic-preserving gauge schemes for compressible flows, Bulletin of the Institute of Mathematics, Academia Sinica, New Series, 2(4)(2007), pp. 851–892,

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