Hyperbolic Conservation Laws
with Source Terms:
Errors of the Shock Location
P. Klingenstein
Research Report No. 94-07
August 1994
Seminar fur Angewandte Mathematik
Eidgenossische Technische Hochschule
CH-8092 Zurich
Switzerland
Hyperbolic Conservation Laws with Source
Terms:
1
Errors of the Shock Location
P. Klingenstein
Seminar fur Angewandte Mathematik
Eidgenossische Technische Hochschule
CH-8092 Zurich
Switzerland
Research Report No. 94-07
August 1994
Abstract
We investigate the error of the shock location which occurs in
numerical solutions of hyperbolic conservation laws with source terms.
For our theoretical analysis we consider a scalar Riemann problem. We
compute its solution using a splitting method. This means that in each
time step the homogeneous conservation law and an ODE (modeling
the source term) are solved separately. We show that the local error of
the shock location can be considered to consist of two parts: one part
introduced by the splitting and another occurring because of smearedout shock proles.
Numerical examples show that these error-estimates can be used
to adapt the step size so that the error of the shock location remains
suciently small. The numerical examples include one-dimensional
systems.
Talk at the Fifth International Conference on Hyperbolic Problems, June, 13-17, 1994,
University at Stony Brook, New York.
1
1 Introduction
When computing numerical solutions of hyperbolic conservation laws with source terms,
one may obtain spurious solutions | solutions which seem to be correct but which are
totally unphysical such as shock waves moving with wrong speeds 3]1]. Therefore it is
important to know how errors of the shock location can be estimated. This is the aim
of our investigation in the rst part of this paper. We are interested in the errors of the
shock location of numerical solutions which are computed using a splitting method. This
means that in each time step the homogeneous conservation law and an ODE (modeling
the source term) are solved separately. We consider a scalar Riemann problem in Section 2
to get error-estimates. First we analyze the local error considering solutions with sharp
shock proles and then we look at solutions with smeared-out shock proles. In Section 3
these error-estimates are used to construct an adaptation of the step size so that the
error of the shock location remains suciently small. The numerical examples which
include a scalar problem (Burgers' equation with a linear source term) and the simplied
combustion model introduced by A. Majda 1] show that the adaptation based on the
error-estimates for the scalar Riemann problem derived in Section 2 gives satisfactory
results.
2 Theoretical analysis
In this section we investigate the local error of the shock location considering a onedimensional scalar Riemann problem:
t + f (u)x
u
( 0) =
(
u x
l
ur u = ( )
q u
0
0
x < x
x > x
:
and are assumed to be smooth functions of where ( ) : + ! and
=: ~(n) with ~(n) = (1) = 0 1 . This restriction on is not necessary to get
the error-estimates | it just determines the higher order terms. The conditions for the
existence of a shock (at least in the time interval n n+1]) should be fullled.
f
q
(n)
q
u
q
q
O
n
u x t
:::
IR
IR
IR
q
t t
Solution operators:
The solution operator for the homogeneous conservation law t + ( )x = 0 is called
' f ' and the one for the ODE t = ( ) ' q '. The numerical solution is computed using
the Strang splitting which is dened as follows:
u
L
u
q u
u
n+1
f u
L
= ( 21
1
q Lf 2 Lq )un :
(1)
L
Here 21 q signies that q is applied for half a time step and n is the numerical solution
at time n. In a Riemann problem a step function ( ) is given. The values of ( )
on the left resp. right side of the discontinuity only depend on . We denote them by l( )
resp. r( ).
L
L
u
t
u x t
u x t
t
u
t
1
u
t
If we apply
q then q only changes the quantities l and r . Of course, in solutions
with smeared-out shock proles, q would inuence the intermediate values, too.
L
L
u
u
L
u
u
ul
Lq
uLq.eps
104 24 mm
ur
x
If we apply
x
f then just the discontinuity is propagated. If f produces smearedout shock proles, we can determine the shock location using the 'equal area rule'
due to conservation.
L
L
u
u
Lf
uLf.eps
104 23 mm
x
x
2.1 Solutions with sharp shock proles
Now f and q are considered to be the exact solution operators . Then the shock proles
are not smeared out and the quantities l( ) and r ( ) are exact. In this case the local
error of the shock location can be considered to be the local splitting error because the
solution operators are exact.
To investigate this local splitting error, we assume the shock location n of the numerical solution at time n to be correct. We use the Rankine Hugoniot jump condition
( r ( )) =: (
_ ( ) = ( l(())) ;
(2)
l r)
l ; r( )
where ( ) is the shock location of the exact solution. The function is the shock speed
and we denote n to be n := ( nl nr).
Dening 4 as the dierence 4 := n+1 ; n and noticing that
L
L
u
t
u
t
t
f u
t
t
u
f u
t
u
t
t
h u u
t
h
h
h
h u u
4
n+1
n
n+1
n
n+1
n+1
num ; 4exact = num ; num ] ; exact ; exact ] = num ; exact
we write the local error as 4 num ; 4
nd that local splitting error is
Espl := 4
exact.
By means of the Taylor series expansion we
num ; 4exact
= ( 12 ; )4 2 t
n q (un)
ul
l
h
where 2 (0 1)
+
n q (un ) ]
ur
r
h
:
2
+ ( 12 ; ) ( 24 3)
O t
(3)
Remarks:
If q is not the exact solution operator but one with a local truncation error of the
order : n+1 ; ( n+1) = (( 4 )p) then an additional error ( p 4 p+1) occurs
in Espl.
If f is any consistent solution operator in conservation form then no additional
L
p
u
u t
O
t
O t
L
error occurs.
2.2 Solutions with smeared-out shock proles
To complete the analysis we now consider numerical solutions with smeared-out shock
proles. Therefore we assume f to be a conservative solution operator which produces
smearing. q and n are exact.
L
L
Let us look at one time step of the Strang splitting procedure:
We start with the function
n
u
with a smeared-out shock prole:
n
u
u
un.eps
94 38 mm
x
n
The shock location is determined by the equal area rule.
Then a half time step q is applied. We get n+ 12 :
L
u
u
u
n
n+ 21
u
usd1.eps
141 51 mm
n
n+
4
x
1
2
1
works on the values on the left and on the right sides of the shock but it also
works on the intermediate values. This causes a change in the shock prole and
therefore, in general, a change in the shock location: 4 1.
L
q
3
Now one time step
f
L
is applied and we get n+ 12 :
u
u
n+ 21
u
usdf.eps
162 40 mm
n+ 12
u
x
n+ 12
4
n+ 12
f
f changes the shock location due to conservation | according to the (exact) values
n+ 21 n+ 12
, ur : The change in the shock location is denoted by 4f .
ul
Again a half time step Lq is applied. We get un+1:
L
u
n+ 12
u
usd2.eps
n+1
164 43 umm
x
n+
1
2
n+1
4
2
The change in the shock location is denoted by 4 2.
The numerical shock location changes over one time step by
4 num = 4 1 + 4 f + 4 2
and the formula for the local error of the shock location calculates as
E := 4 num ; 4 exact = 4 1 + 4 2 + 4
| f ;{z4 exact}
(4)
Espl
where the dierence of the last two terms is already known: It is the error that occurs in
the solution with sharp discontinuities. So only the quantities 4 1 and 4 2 are left to
be determined.
Inuence of smearing:
Considering the numerical solution at time
correct and determine it by the equal area rule:
Z n +
n ;
n
u dx
n
t
we assume the shock location to be
= ( nl + nr)
u
u
:
Here is assumed to be the width of the smearing of the shock prole.
4
u
n
u
x
n
1
2 Lq
eps2.eps
142 80 mm
u
n+ 21
u
41
s
l
r
n
+ 12
n
1 L we
2 q
n
+ n+ 1
After applying
Z
n ;
get
2 dx
u
=
n+ 1
l l 2
u
By denition of we have 4
+
n+ 12
and dening
n+ 1
r r 2
u
1 < .
u
Z n +
= ( + 4 1)
n+ 12
l
u
+ ( ; 4 1)
n+ 1
r 2:
u
Taking into account that
2
; n = 42 ( n) + 48 0( n) ( n) + ( 3 4 3)
t
u
t
q u
q
u
q u
O t
( ) := 42 ( ) + 48 0( ) ( )
t
u
we get
x
(un+ 12
; un)dx =
Z n +
2
q
u q u ( n) + ( 34 3) = 2 ( n ) + (
u
n ;
n ;
n
n
n
u 2 (maxx u (x) minx u (x)).
t
q u
dx
O t
u
3 4t3)
O (5)
where
Rearranging terms and summing up all the higher
order terms in the expression ( ), we get the following result:
n
n
n
4 1 = 2n (; )n;+ (( ln));; (( rn)) + ( 34 3)
(6)
l
r
l
r
O :::
If we replace
n
u
u
u
u
u
u
u
u
O t
:
by n+ 12 , we get 4 2. So the error analysis is complete.
u
3 Numerical applications
In this section we test by means of numerical examples how far the error-estimates derived
in the previous section can be used to adapt the step size so that the error of the shock
location remains suciently small. We show two examples | a scalar problem and a
simplied combustion model.
5
3.1 The scalar example
We consider Burgers' equation with a linear source term:
1 2) = ; ( ; )
t+(
2 x
where
(
0 25
:= 10
0 25
u
u
a
u
u
:
u <
:
a
:
The initial data dene a Riemann problem
(
( 0) = 10
u x
x <
x >
0
0
:
The source term shifts values of which are greater than 0.25 towards 1 and values
which are less than 0.25 towards 0. With the given initial data the source is equal to zero
on both sides of the shock. The exact solution is
u
0.50
x
ref58.eps
42 30 mm
u=1
( ) = ( ; 21 0) .
u x t
u=0
u x
t
t
1.00
3.1.1 Numerical wave speeds
The numerical solution is computed using the Strang splitting. The homogeneous conservation law is solved with an upwind scheme and the ODE is solved with the implicit
Euler scheme. Because the solution does not depend on 4 but on 4 , the step sizes
are xed and only the relative time rel := 4 is varied. We get the following results
(see Fig. 1):
t
T
t
t
For rel small the shock speed is correct.
If we increase rel, the shock speed becomes wrong.
If rel is large enough, two phenomena occur:
{ For ratios of the step sizes 44xt 0 5 the discontinuity does not move at all.
{ For 44xt 0 5 the discontinuity moves one grid cell per time step.
T
T
T
<
>
:
:
These phenomena occur because of the smearing of the shock prole. Let us look at
the single steps of the Strang splitting procedure (see Fig. 2):
In the rst step nothing happens:
( ni) = 0 8 .
q u
6
i
Figure 1: Shock curves of the numerical solutions for dierent
0 55, and 4 = 0 01.
:
t
:
4t
4x
T
4t
rel, 4x
= 0 45 or
:
4t
4x
=
= 0 45:
:
t
1.00
4
1
2
3
0.75
6ges.eps
81 57 mm
0.50
0.25
0.00
-0.0
4t
4x
0.5
1.0
1.5
2.0
x
= 0 55:
:
t
1.00
2
1
4
3
0.75
5ges.eps
81 57 mm
0.50
0.25
0.00
-0.0
1:
rel
T
= 0 00001
:
0.5
2:
rel
T
1.0
= 0 01
:
1.5
3:
2.0
x
rel = 1 4:
T
rel
T
= 100.
In4t the second step the upwind scheme produces one intermediate value. For ratios
0 5 this value is less than 0.25 and for 44xt 0 5 it is greater than 0.25.
We consider the rst case.
For 44xt 0 5 the intermediate value is less than 0.25 so q shifts this value towards
zero.
In the next time step 12 q is applied rst. The intermediate value is shifted again
towards zero.
4x
<
:
<
>
:
:
L
L
If rel is large enough, the intermediate value is shifted so close to zero that after one
time step the discontinuity has not moved at all. The case 44xt 0 5 behaves analogously.
T
>
7
:
Figure 2: The three steps of the Strang splitting (over one time step) for
4t
4x = 0 45.
rel
T
= 100 and
:
1.00
1.00
1.00
0.75
0.75
0.75
1n.eps
44 65 mm
0.50
n.eps///burgers
44 65 mm
0.50
0.25
0.25
-0.00
0.25
-0.00
-0.100
-0.050
0.000
0.050
0.100
21n.eps
44 65 mm
0.50
-0.00
-0.100
-0.050
0.000
0.050
0.100
-0.100
-0.050
0.000
0.050
0.100
3.1.2 Local errors
First we have to decide how the changes of the shock location over one time step are to
be computed in this example. For the exact solution we have
4 exact = 12 4
and for the numerical solution we can write
4
num
t
=4
x
X
i
(
n+1 ; un )
i
i
u
where ni is the numerical solution at time n and location i. The local error of the shock
location is the dierence of these two expressions
u
t
E := 4
x
num ; 4exact = 4x
X
i
(
u
n+1 ; un ) ;
i
i
14
2
t:
(7)
Now we want to test if the error-estimates derived in Section 2 are suciently exact
when the remaining terms of higher order are neglected. The local error can be written
as in (4). It reduces in the considered problem to
E =4 1+4
2
because with the special choice of the source term Espl = 0. Also 4 1 reduces to
4 1 = 2 ( n ) + ( 3 4 3)
u
O 8
t
as ( nl) = ( nr) = 0 and nl ; nr = 1. The same holds for 4 2 (replacing
To compute 4 1 and 4 2 we remember (5)
u
u
u
u
Because of
a
Z n +
Dening
n ;
n
i
u
u
n
(u )dx =
x;
u
( n)
u
to be the cell average
n
i
u
we get
a
dx
(4t)2 ;
for any
by n+ 12 ).
u
x
X
x
i
i; 12
1 4
2
;
x
Zx 1
= 41 x i+ 2 (
4 1+4 2=4
1
( )=( ; ) 8
Z x+
n
Z n +
n
2(u ) = n (un)dx:
;
u
( depending on ) we have
u
t
n;
and
n+
+
x
:
n)dx
u x t
n+ 12
n
)
(ui ) + (
ui
+ ( 3 4 3)
O t
(8)
:
Dropping the higher order terms, we calculate the estimated error as
Eest := 4
x
X
i
( ni) + (ni + 2 )
1
u
u
(9)
:
In our numerical computations the relative time rel is varied. The step sizes 4 and
4 are xed and we look at the solutions at time = N = 1. The relative (global) error
j N ; ( N )j
:=
rel
j ( N )j
the maximum of the local errors, and the maximum of the dierences between the estimated and the exact local error are shown in the table below:
T
x
T
E
rel
T
0.1 E-02
0.2 E-02
0.3 E-02
0.4 E-02
0.5 E-02
0.6 E-02
0.7 E-02
0.8 E-02
0.9 E-02
1.0 E-02
rel
E
0.159448E-02
0.317808E-02
0.476176E-02
0.633019E-02
0.806836E-02
0.929683E-02
0.110235E-01
0.125525E-01
0.138511E-01
0.157083E-01
t
t
t
t
max(E )
max(Eest )
Dierence
0.146037E-04
0.288762E-04
0.435261E-04
0.577346E-04
0.731136E-04
0.842295E-04
0.101773E-03
0.116287E-03
0.128711E-03
0.144855E-03
0.146037E-04
0.288762E-04
0.435261E-04
0.577346E-04
0.731135E-04
0.842294E-04
0.101773E-03
0.116287E-03
0.128710E-03
0.144854E-03
0.608730E-12
0.481385E-11
0.163281E-10
0.385089E-10
0.762074E-10
0.126439E-09
0.207968E-09
0.310409E-09
0.434888E-09
0.604318E-09
9
The errors depend linearly on rel | this is obvious as the spatial step size is not
rened. All the dierences between the estimated and the exact local error are so small
that we conclude that the estimation formula is suciently exact.
T
3.1.3 Adaptation
Now we are ready to test if an adaptation based on Eest gives satisfactory results. This
adaptation works as follows: The estimated local error of the shock location Eest is expected to be smaller than a certain upper bound . If this assumption is not satised,
the step sizes 4 and 4 are bisected so that the ratio 44xt stays constant.
We dene the upper bound for the local error to be
8
>
0
3
<1
:= 0 5 4 0 0 01
4
8
0 := > 0
: 0 + 1 9 00
B
t
x
B
:
t
c
:
c
b
b b
b
b
where 0 is the number of bisections. Notice that the factor 0.5 is the shock speed. We
choose = 0 1. As before we look at solutions at time = 1. In the table below you
see the various step sizes 4 0 at time = 0, the number of bisections and time steps, the
resulting smallest 4 , and the relative error of the shock location.
b
:
T
t
t
t
4 0 #(bisections) #(time steps) Resulting 4
t
0.01
0.10
0.20
1.00
2.00
4
8
8
11
12
1601
2554
1278
2020
1901
t
0.625000E-03
0.396250E-03
0.761250E-03
0.488281E-03
0.488281E-03
rel
E
0.100392E-01
0.676327E-02
0.129742E-01
0.945125E-02
0.128107E-01
The resulting step sizes are all about the same size and the relative error remains less
than 1.3%.
3.2 The combustion model
We consider a simplied model for the inviscid reacting compressible Euler equations in
one space dimension. This model is a (2 2) - system, given by Burgers' equation coupled
to a chemical kinetics equation. It has analogues in the Z-N-D theory and the structure
of the reacting shock proles (see 1] or 4] for more details). Therefore we use it as a rst
test example to investigate if an adaptation based on the estimation formula (4) would
also work for combustion problems.
The model equations are given by
1
t + ( 2 u2 ; q0 Z )x
Zx
u
10
=0
= !( )
u Z
with initial data
(
0
( 0) = ;10 07
0
We use ignition temperature kinetics so that the function !( ) is given as
u x
(
: x >
: x <
u
!( ) = 10
u
:
0
u
u <
0
:
is the mass fraction of unburned gas with limx!1 ( ) = 1. We set 0 = 0 935 where
0 is the heat release.
The initial data are chosen such that the speed of the combustion wave at time = 0
is _ (0) = 0 15 and a traveling wave solution evolves with a xed speed _ ( ) = 0 7. The
solution exhibits a combustion spike. In Figure 3 the solution proles of and at
time = 10 and also the shock curve are shown.
Z
Z x t
q
:
q
t
:
t
u
:
u
Z
T
Figure 3: Reference solution of the simplied combustion model showing ,
= 10 and the shock curve.
u
T
u
2.0
1.5
up001.eps
84 59 mm
1.0
0.5
0.0
-0.5
-5.0
-2.5
0.0
2.5
5.0
7.5
x
Z
1.0
zp001.eps
84 59 mm
0.5
-0.0
-10
-5
0
11
5
x
Z
at time
t
10.0
7.5
sp001neu.eps
84 59 mm
5.0
2.5
0.0
0.0
2.5
x
5.0
3.2.1 Numerical wave speeds
The numerical solution is computed using the Strang splitting where the equations
1 2) = 0
(10)
t+(
2 x
(11)
x = !( )
(12)
t = 0!( )
u
u
Z
u Z
u
q
u Z
are solved separately in each time step. Equation (10) is solved with an upwind scheme
as in the case of the scalar example, equation (11) is solved by trapezoidal approximation
of the integral in the exact solution formula, and equation (12) is solved exactly. For
dierent step sizes 4 (with 44xt =
) the solution shows the following behaviour for
the combustion wave speeds (see Fig. 4):
t
const:
For 4 small the shock speed is correct.
If we increase 4 , the shock speed becomes slower than the correct speed.
If 4 is large enough, the shock speed remains unchanged: _ ( ) _ (0) = 0 15.
t
t
t
t
:
This phenomenon occurs because the combustion spike of the solution decreases when the
step size is increased.
12
Figure 4: Numerical solution of the simplied combustion model showing after 1000
time steps and the shock curve for dierent 4 .
u
t
u
t
u
10.0
2.0
1.5
7.5
up01.eps
66 46 mm
1.0
0.5
sp01neu.eps
66 46 mm
5.0
0.0
2.5
-0.5
0.0
2.5
5.0
7.5
0.0
0.0
x
2.5
x
5.0
4 = 0 01
t
t
u
100
2.0
1.5
75
up1.eps
66 46 mm
1.0
0.5
sp1neu.eps
66 46 mm
50
0.0
25
-0.5
0
0
25
50
75
0
x
10
20
30
50
60
70 x
500
600
700 x
5000
6000
7000 x
40
4 =01
t
:
t
u
1000
2.0
1.5
750
u1p.eps
66 46 mm
1.0
0.5
s1pneu.eps
66 46 mm
500
0.0
250
-0.5
0
0
250
500
750
0
x
100
200
300
400
4 =1
t
t
u
10000
2.0
1.5
7500
u10p.eps
66 46 mm
1.0
0.5
s10pneu.eps
66 46 mm
5000
0.0
2500
-0.5
0
0
2500
5000
7500
0
x
4 = 10
t
13
1000
2000
3000
4000
:
3.2.2 Local errors
Again we have to decide how the local error of the shock location (given by (4))
E := 4
num ; 4exact = 41 + 42 + 4
| f ;{z4exact}
Espl
is to be approximated. Let us rst consider Espl. In (3) we drop the higher order terms.
Equation (10) denes
( l r) = 21 ( l + r ) with ul = ur = 12
Equation (12) gives
h u u
u
u
( ) = 0!( )
q u
q
h
with
u Z
h
( r) = 0
q u
:
( l) =
q u
0
q
where, for the sake of simplicity, is assumed to not depend locally on . Dropping higher
order terms in (3) we set the following inequality for E~spl:
jE~splj := jEspl + ( 12 ; ) ( 2 4 3)j
Z
t
O t
= j( 21 ; )4 2 12 0j
t
(13)
q
1 q 4t2 :
4 0
<
To approximate 4 1 and 4 2 we dene the region of the smearing (
L := xkL L R
k k
R)
as
R := xkR
x
where
L
x x
x
are dened such that
L
k
:
kL
u
= max
i
and
i
u
R
k
:= minf j i = r g
i
u
Using 0 = 0 we get
u
:
q
( ) = 21 4 !( ) 0
For ( ) we have 1 and we approximate by
smearing ( L R). Then (14) gives
t
u
x
t
u q Z:
Z
x x
Z
(
1
( ) = 02 04
u
q
t
u
0
u <
0
x
2(
Z
L
(14)
1 over the region of the
R ):
(15)
x x
We use the equal area rule to compute the shock location := k (where the values of
on the left and right sides of the shock are kL and kR and 2 ). We dene 0 to
be the largest index for which i 0.
u
u
u
14
u
x
k
IR
k
Now we are ready to compute 4 1 and 4 2 as
1 ( n ; n) 4 4
1 ( n+ 12 ; n+ 12 ) 4 4
0
0
4 1 = 2 n ; n + 1 4 4 2 = 2 n0+ 12 n+ 12 01
kL
kR 2 0
kL ; kR + 2 0 4
Summing up the above results we estimate the local error of the shock location by
Eest := j4 1 + 4 2j + 14 04 2
(16)
k
u
k
q
u
x
q
t
t
k
k
u
s
s
x
q
u
q
q
t
:
t
t :
3.2.3 Adaptation
Now we test if an adaptation based on Eest gives satisfactory results. If Eest is not smaller
than a certain upper bound , the step sizes 4 and 4 are bisected so that the ratio
4t
4x stays constant.
We set the upper bound for the local error to be
:= 0 01 4
We look at solutions at time = N = 10. Listed in the table below are the various step
sizes 4 0 at time = 0, the number of bisections and time steps, the resulting smallest
4 , and the relative (global) error of the shock location which is now dened by
PN ;1
n
j
:= n=0 (4 num ; 4 exact) j
B
t
B
T
t
:
x
t:
t
t
t
rel
E
.
jN j
:
4 0 #(bisections) #(time steps) Resulting 4
t
0.1
1.0
2.0
5.0
10.
20.
5
8
9
11
12
13
3199
2560
2560
4090
4090
4090
t
0.312500E-02
0.390625E-02
0.390625E-02
0.244141E-02
0.244141E-02
0.244141E-02
rel
E
0.550740E-02
0.694970E-02
0.694970E-02
0.432801E-02
0.432801E-02
0.432801E-02
The resulting step sizes are all about the same size and the relative errors remain less
than 0.7%.
4 Conclusions
We have analyzed the local error of the shock location for a scalar one-dimensional Riemann problem and computed two numerical examples. We have estimated the local errors
of the shock location in the numerical computations and shown that an adaptation of the
step size based on these estimates works well.
Using our experiences with the simplied combustion model we are next going to
test an adaptation for the inviscid reacting compressible Euler equations in one space
dimension.
15
References
1] P. Colella, A. Majda, and V. Roytburd. Theoretical and numerical structure for
reacting shock waves. SIAM J. Sci. Stat. Comput., 4:1059{1080, 1986.
2] R. J. LeVeque. Numerical Methods for Conservation Laws. Lectures in Mathematics,
ETH Zurich, Birkhauser, 1992.
3] R. J. LeVeque and H. C. Yee. A study of numerical methods for hyperbolic conservation laws with sti source terms. J. Comput. Phys., 86:187 { 210, 1990.
4] A. Majda. A qualitative model for dynamic combustion. SIAM J. Appl. Math., 41:50
{ 93, 1981.
5] G. Strang. On the construction and comparison of dierence schemes. SIAM J.
Numer. Anal., 5:506{517, 1968.
6] H. C. Yee. A class of high-resolution explicit and implicit shock-capturing methods.
NASA Technical Memorandum 101088, 1989.
16