Download A numerical solution of a two-dimensional transport equation.

CEJM 2(2) 2004 191–198
A numerical solution of a two-dimensional transport
equation.
Olga Martin
∗
Department of Mathematics,
University “Politehnica” of Bucharest,
Splaiul Independentei 313, Bucharest 16, Romania
Received 22 September 2003; revised 25 March 2004; accepted 5 April 2004
Abstract: In this paper we present a variational method for approximating solutions of the
Dirichlet problem for the neutron transport equation in the stationary case. Error estimates
from numerical examples are used to evaluate an approximation of the solution with respect to
the steps of two grids.
c Central European Science Journals. All rights reserved.
Keywords: transport equation, variational calculus, difference scheme, Euler-Lagrange equation
MSC (2000): 35J99, 65N99
1
Introduction
In a reactor, neutrons are produced by the fission of a nucleus and are called rapid
neutrons if their average speeds are 2·107 m/s. Rapid neutrons are subjected to a slowness
process, decreasing their energy until they are in a state of equilibrium with the other
atoms in the environment. When the reactor is in a stationary state, the atoms have a
tendency to move from a region with a high density to another with a low density, thus
obtaining a uniform density. This process is called diffusion.
The main problem in nuclear reactor theory is to find the distribution of neutrons in a
reactor, hence to find the reactor’s density. The result is a scalar function that depends on
the following variables: the position vector of the neutron in a datum coordinate system,
the neutron speed and the time.
The neutron density is the solution of an integral-differential equation called the neutron transport equation. Many authors treat this problem and its applications [1-3, 4-6].
∗
E-mail: omartin [email protected]
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O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
In this paper we provide a solution for the two-dimensional stationary problem, where
the solution u is the neutron density and where f from (1) is the source function. The
neutron moves in a direction that makes an angle α with the Ox axis and an angle β
with the Oy axis. We let µ=cosα and ν = cosβ. In order to find the solution u of the
Dirichlet problem for a transport equation, we use the variational calculus.
Let ∆1 be a square grid with step k for D1 = [-1, 1]×[-1, 1]. For every value (µi , νj ) ∈
∆1 , we define u by an interpolation polynomial that both satisfies the boundary conditions
and minimizes a given functional J(u). The approximate solutions of numerical examples
are compared with the accurate solution, thereby producing an estimation of the error.
2
Problem Formulation
Let us now consider the integral-differential equation of the transport theory:
µ ∂u
(x, y, µ, ν) + ν ∂u
(x, y, µ, ν) + u(x, y, µ, ν) =
∂x
∂y
u(x, y, µ , ν )dµ dν + f (x, y, µ, ν)
=
(1)
D1
with
(µ, ν) ∈ D1 = [ - 1, 1] × [ - 1, 1] ,
and u(x, y, µ, ν) |∂D2 = 0,
(x, y) ∈ D2 = [0, 1] × [0, 1]
∂D2 − the boundary of D2 (2)
(2)
where u(x, y, µ, ν) is the density of neutrons and where f is the source function. Let
u be continuous on D1 (u ∈ C(D1 )) and have continuous second derivatives on D2 (u ∈
C 2 (D2 )).
In order to solve the problem (1)–(2) using variational methods, we replace the problem with the equation obtained by addition of the partial derivatives of (1) with respect
to the variables x and y, multiplied by µ and ν, respectively
2
2
2
∂ u
µ2 ∂∂xu2 + 2νµ ∂x∂y
+ ν 2 ∂∂yu2 + µ ∂u
+ ν ∂u
=
∂x
∂y
∂u ∂u dµ dν + ν
dµ dν + µ ∂f
+ ν ∂f
=µ
∂x
∂y
∂x
∂y
D1
(3)
D1
or
µ2
2
∂2u
∂2u
2∂ u
+
2νµ
− u = F (x, y, µ, ν)
+
ν
∂x2
∂x∂y
∂ y2
(4)
where
F (x, y, µ, ν) = µ
−
D1
D1
∂u
(x, y, µ , ν )dµ dν + ν
∂x
u(x, y, µ , ν ) dµ dν + µ
D1
∂u
(x, y, µ , ν )dµ dν −
∂y
∂f
∂f
(x, y, µ, ν) + ν
(x, y, µ, ν) − f (x, y, µ, ν)
∂x
∂y
(5)
O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
193
in accordance with (1).
Next, we present a variational method using the Ritz-Galerkin method.
Let us define an approximate formulation of our problem (4)–(2). For this, we consider
a square grid ∆1 for D1 with step k = 1/4, ∆1 = { (µi , νj ) ∈ D1 |{ i ∈ {0, 1, ..., 8}, j ∈
{0, 1, ..., 8}}, and a square grid with step h, ∆2 = { (xi , yj ) ∈ D2 | i ∈ {0, 1, ..., N + 1}, j ∈
{0, 1, ..., N + 1}}.
Then, as shown in Fig. 1, each square of the h side from D2 is split into two isosceles
triangles (N = 3). Consider the approximate solution of equation (4) given by
u (x, y) =
N
αmn (µ, ν)ωmn (x, y)
(6)
m,n=1
where
ηm , ξn − constants.
αmn (µ, ν) = µν + ηm µ + ξn ν,
(7)
Dklh
x2=y
2
3
1
4
Pk l
6
5
w
1
x2
Pkl
0
x1
0
x1=x
Fig. 1
Fig. 2
From (5) and (7), we obtain:
∂ f (x, y, µ, ν)
∂ f (x, y, µ, ν)
+ν ·
− f (x, y, µ, ν)
(8)
∂x
∂y
In order to obtain a Ritz-Galerkin approximation for the solution of equation (4), we
minimize the functional:
F (x, y, µ, ν) = µ ·
µ
J(u) =
D2
2
∂u
∂x
2
∂u ∂u
+ 2µν
·
+ ν2
∂x ∂y
∂u
∂y
2
2
+ u (x, y, µ, ν) dxdy+2
F u dxdy
D2
(9)
over the space of functions of the following form:
uij (x, y) =
N
αmn (µi , νj )ωmn (x, y)
(10)
m,n=1
where the ωmn (x, y) are the pyramidal functions that correspond to the N hexagonal
regions, Dmn , each of which is centered at the point Pmn of the network ∆2 . Each hexagonal sub-domain is the union of six triangles, {Dmn,,j }, j ∈{1,2,. . . ,6}, the numbering of
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O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
which is shown in Fig. 1. The geometric interpretation of the function ωmn is a pyramid
with a height equal to unity and with projection P of its apex (Fig. 2). The function
ωmn can be written in the following form:


h

1 − h1 (xm − x) − h1 (yn − y), if (x, y) ∈ Dnm,1





h

1 - h1 (xm − x)
if (x, y) ∈ Dnm,2






h

1 + h1 (yn − y), if (x, y) ∈ Dnm,3



h
ωmn (x, y) = 1 + h1 (xm − x) + h1 (yn − y), if (x, y) ∈ Dnm,4




h


1 + h1 (xm − x),
if (x, y) ∈ Dnm,5





h

1 − h1 (yn − y), if (x, y) ∈ Dnm,6





 0, in rest.
(11)
In order to minimize the functional J(u), we find coefficients αmn as solutions of the
system of equations given by
∂J(uh )
= 0,
∂αij
i, j ∈ {1, 2, · · · , N }
(12)
or, in matrix form, given by
A·α=g
(13)
where the elements of matrix A form a band along the main diagonal. The elements of
A are
N
∂ωkl ∂ωmn
aN (k−1)+l,N (m−1)+n =
Cst
+ ωkl ωmn dxdy , k, l, m, n = 1, 2, · · · , N
∂xs ∂xt
D2
s,t=1
(14)
where x1 = x, x2 = y and the coefficients Cst are given by equation (11), expressed in
the form
2
∂2u
Cst
− u = F.
(15)
∂
x
∂
x
s
t
s,t=1
The vector g = (g1 , g2 , · · · , gN 2 ) with gN (k−1)+l = gkl , k, l, ∈ {1, 2, ..., N } has
gkl = −
F ωkl dxdy, k, l = 1, 2, · · · , N
(16)
h
Dkl
but α = (α1 , α2 , · · · , αN 2 )t ( t – transpose matrix) is the unknown vector with
αN (k−1)+l = αkl ,
k, l, ∈ {1, 2, ..., N } .
(17)
O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
3
195
Numerical results
Let us consider the transport equation
µ
∂u
∂u
(x, y, µ, ν) + ν (x, y, µ, ν) + u(x, y, µ, ν) =
∂x
∂y
u(x, y, µ , ν )dµ dν + f (x, y, µ, ν)
=
(18)
D1
with the conditions from (2) and
f (x, y, µ, ν) = µν(µπ cos πx sin πy + νπ sin πx cos πy + sin πx sin πy).
(19)
For the domain D1 = [-1,1] × [-1,1] we use the network ∆1 with the step k = 1/4 and
for D2 = [0,1] × [0,1] we have h = 1/3 (Fig. 3).
y(x2)
(0,1)
(0,2/3)
(0,1/3)
0
12
22
11
21
(1/3,0)
(2/3,0)
(1,0)
x (x1)
Fig. 3
We begin the construction of an approximate solution by considering the sum (N =
2)
u(x, y, µ, ν) = α11 (µ, ν)ω11 (x, y) + α12 (µ, ν)ω12 (x, y)+
(20)
+α21 (µ, ν)ω21 (x, y) + α22 (µ, ν)ω22 (x, y)
with
α11 (µ, ν) = η1 µ + ξ1 ν + µν,
α21 (µ, ν) = η2 µ + ξ1 ν + µν,
α12 (µ, ν) = η1 µ + ξ2 ν + µν,
α22 (µ, ν) = η2 µ + ξ2 ν + µν.
(21)
In accordance with (8), we obtain
F (x, y, µi , νj ) = −µi νj (π 2 (µ2i + νj2 ) + 1) sin πx sin πy − 2π 2 µ2i νj2 cos πx cos πy.
We find A from (14). For example, when k = 1, l= 1, m= 1 and n =2, we have
(22)
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O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
a12 =
1/3
dx
0
+
2/3
2
11 ∂ω12
Cst ∂ω
∂xs ∂xt
s,t=1
2/3−x
2
2/3
1−x
dx
1/3
1/3
s,t=1
− ω11 ω12
11 ∂ω12
Cst ∂ω
∂xs ∂xt
dxdy+
− ω11 ω12
dxdy
corresponding to D11,3 , D11,4 , D12,1 , andD12,6 . Thus, we determine the matrices A and g
of equation (13) in the form


2(µ2i + µi νi + νi2 ) + 0.054 −(µi νi + νi2 ) + 0.01

 −(µ ν + ν 2 ) + 0.01
 ii i
A=
 −(µ2i + µi νi ) + 0.01

2(µ2i + µi νi + νi2 ) + 0.054 µi νi + 0.01
0

−(µ2i + µi νi ) + 0.01
0
−(µ2i + µi νi ) + 0.01
µi νi + 0.01
2(µ2i + µi νi + νi2 ) + 0.54 −(µi νi + νi2 ) + 0.01
−(µ2i + µi νi ) + 0.01
−(µi νi + νi2 ) + 0.01






2(µ2i + µi νi + νi2 ) + 0.54

g
 11 


 g12 


g=
,


 g21 


g22
with
gij = −
F (x, y, µi , νj ) ωij dxdy
(23)
D2
h
and, because ωij =0 only for (x, y) ∈ Dij
, we obtain
F (x, y, µi , νj )ωij dxdy.
gij = −
(24)
Dij
In the following tables, we present the values of u(xk , yl , µi , νj ) where (xk , yl ) corresponds to the nodes of (11). From (10) and (11), it follows that
u(xk , yl , µi , νj ) = αkl .
(25)
The approximate solution u is then compared with the exact solution ū(x, y, µ, ν) =
µν sin πx sin πy.
O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
µ =-3/4
µ =-1/4
µ = 1 /2
ν = -3/4
ν = -1/2
ν = -1/4
ν = 1/4
ν = 1/2
ν = 3/3
ν = -3/4
ν = -1/2
ν = -1/4
ν = 1/4
ν = 1/2
ν = 3/3
ν = -3/4
ν= -1/2
ν = -1/4
ν = 1/4
ν = 1/2
ν= 3/3
U
ū
-0.458
-0.312
-0.164
0.194
0.41
0.623
-0.563
-0.375
-0.188
0.188
0.37
0.563
U
ū
-0.48
-0.329
-0.174
0.214
0.406
0.58
-0.563
-0.375
-0.188
0.188
0.375
0.563
U
ū
-0.614
-0.417
-0.203
0.164
0.313
0.47
-0.56
-0.375
-0.188
0.188
0.375
0.563
µ =-1/2
µ = 1/4
µ = 3/4
ν = -3/4
ν = -1/2
ν = -1/4
ν =1/4
ν = 1/2
ν = 3/3
ν = -3/4
ν = -1/2
ν = -1/4
ν = 1/4
ν = 1/2
ν = 3/3
ν = -3/4
ν= -1/2
ν = -1/4
ν = 1/4
1/2
3/3
197
U
ū
-0.47
-0.406
-0.164
0.203
0.417
0.614
-0.563
-0.375
-0.188
0.188
0.375
0.563
U
ū
-0.58
0.407
-0.214
0.171
0.329
0.48
-0.563
-0.375
-0.188
0.188
0.375
0.563
U
ū
-0.623
-0.41
-0.194
0.164
0.312
0.458
-0.563
-0.375
-0.188
0.188
0.375
0.563
Table 1
4
Conclusions
In solving equation (13) we find angle α for the values (µi ,ν j ) ∈ ∆1 , hence we find
u(xk , yl , µi , νj ), where (xk , yl ) ∈ ∆2 . Note that all functions of the form (10) are not
generalized solutions of (4), but are Ritz–Galerkin type approximations of the generalized
solutions.
As we have demonstrated in [4], an approximate solution u of the elliptic equation (4)
with u ∈ C 2 (D2 ) verifies the following inequality:
u − ū < C1 h
(26)
where ū is the exact solution and h is the step of a square grid defined over the domain
D2 . In analyzing both the exact and numerical solutions, we find that C1 = 5k where k is
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O. Martin / Central European Journal of Mathematics 2(2) 2004 191–198
the step of a square grid defined over the domain D1 and, hence, that the approximation
of the solution with respect to steps h and k has the form
u − ū ≤ 5hk.
(27)
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