-
Some new implicit two-step multiderivativemethods for solving
the special second-order
I.V.P. and the wave equation.
Higinio Ramos, M. F. Patricio
July 3, 2013
Abstract
The aim of this paper is to select from the huge family of
implicittwo-step multiderivative methods those which have better
character-istic concerning the error (order and error constant),
stability andeasy of use (with most of the coefficients equal to
zero). A few ofthese multiderivative methods which are A0-stable
and A()-stablehave been obtained. This feature makes them useful
for solving stiffproblems. The stability intervals or stability
regions for the methodschosen are presented. Some numerical tests
are included. The numer-ical test concerning the wave equation
shows the good performanceof the methods of less order but
A0-stable, compared with the poorperformance of the methods of
higher orders.
1 Introduction
The numerical integration of the special second-order
initial-value problemof the form
y(t) = f(t, y(t)) , y(t0) = y0 , y(t0) = y0 (1)
has received a lot of attention in the past decades as several
physical problems(Schrodinger equations, orbital problems,
electronics, molecular dynamics,semi-spatial discretization of wave
equations, etc.) can be modelled by thisequation.
1
-
Two decades ago, several authors began to explore and develop
multi-step Obrechkoff methods, originally introduced in [12] to
integrate the prob-lem in (1) by using higher-order derivatives.
There is some literature aboutObrechkoff methods for solving
first-order initial-value problems ([8], [11],[1]), an also for
second-order problems ([2], [13], [14]), although not
veryextensive.
In [17] an analysis of P-stable symmetric two-step Obrechkoff
methodsof orders 2m, m = 1, 2, . . . has been conducted, but for
all the methodsconsidered there none of the coefficients is zero.
In this paper we providesome methods with good stability
characteristics, reasonable error and withas many coefficients as
possible equal to zero. In the following sections wemade an
analysis of the two-step multiderivative methods and present
themethods with the best characteristics according to the stated
criteria. Weapplied the obtained methods for solving a difficult
problem obtained afterdiscretizing by the method of lines a wave
partial differential equation. Thenumerical results show the good
performance of the proposed methods.
2 Multiderivative multistep methods for the
special second-order IVP
Consider the general linear multiderivative multistep methods of
the form
kj=0
j yn+j =
qi=1
h2ik
j=0
i j y(2i)n+j (2)
for solving the IVP
y = f(t, y) , y(t0) = y0 , y(t0) = y0
on a given interval [t0, tN ], where yn is an approximation to
y(tn), tn =
t0 + nh, being h > 0 the constant stepsize, and y(2i)n+j an
approximation for
the derivative y(2i)(tn+j) such that
y(2)(t) = f(t, y)
y(2i)(t) =d2y(2i2)(t)
dt2, i 2 .
2
-
Most of the definitions that follow may be found either in [4],
[8] or [5].The order of the method is defined to be p if for an
adequately smootharbitrary function z(t) it is
L[h, , ]z(t) = Cp+2 hp+2 z(p+2)(t) +O(hp+3) ,
with the error constant Cp+2 6= 0, where L[h, , ] is the linear
operatordefined by
L[h, , ]z(t) =k
j=0
j z(t+ jh)qi=1
h2ik
j=0
i j z(2i)(t+ jh) .
Considering the usual Taylor series expansion it can be written
as
L[h, , ]z(t) =p+1m=0
Cmhmz(m)(t) + Cp+2 h
p+2 z(p+2)(t) +O(hp+3) ,
where
C0 =k
j=0
j (3)
C1 =k
j=0
j j (4)
Cm =1
m!
kj=0
jm j ql=1
1
(m 2l)!k
j=0
l jjm2l ,m 2 (5)
and so, the conditions for the method to be of order p read
Cm = 0 , m = 0, 1, . . . , p+ 1 .
The coefficients {j}kj=0 and {i j} , i = 1, . . . , q ; j = 0, .
. . , k can bedetermined imposing that L[h, , ]l(t) = 0 for a set
of basis functions{l(t)}rl=0 , r k+(k+1)q. Traditionally, these
basis functions are chosen tobe monomials, l(t) = t
l, but other choices are possible, obtaining in this case
3
-
the so-called adapted Obrechkoff methods (in [18], [16]
exponentially-fittedsymmetric Obrechkoff methods are considered
while in [20] a trigonometric-fitted method of twelfth order was
presented).
Define (x) and i(x) to be the first and second characteristics
polyno-mials given by
(x) =k
j=0
j xj , i(x) =
kj=0
i j xj , i = 1, . . . , q
which we assume that have no common factors. Using these
polynomials themethod in (2) may be written as
(E) yn =
qi=1
h2ii(E) y(2i)n .
A method of the family in (2) is said to be zero-stable if no
root of the firstcharacteristic polynomial (x) has modulus greater
than one, and if everyroot of modulus one has multiplicity not
greater than two.
The linear stability theory for these methods is related to the
test equation
y(t) = y(t) .
The stability polynomial is defined by
(x, h) = (x)qi=1
hi i(x)
where h = h2 . A method of the family in (2) is said to be
absolutely stablefor a given h if all the roots of (x, h) satisfy
|ri| 1 , i = 1 . . . , k. The regionof the complex plane
D = {h C : the method is absolutely stable}is called stability
domain or region of absolute stability of the method. Whenthe
negative complex plane is included in D the method is said to be
A-stable.
A method of the family in (2) is said to be A()-stable with 0
< pi/2,if the angular sector
S = {z C : | arg(z)| < , z 6= 0}
4
-
is contained in the stability domain D.An interval [a, b] of the
real line is said to be an interval of absolute
stability if the method is absolutely stable for all h [a, b].
When theinterval of absolute stability consists in the negative
real axis the method issaid to be A0-stable.
The method is convergent if and only if p 1 and the method is
zero-stable (see Refs. [6, 7]).
The method is assumed to satisfy the following requirements:
k = 1 , |0|+qi=1
|i 0| 6= 0 , (normalization)
qi=0
i k 6= 0 (implicitness),
(x) and i(x) have no common factor (irreducibility), (1) = (1) =
0, [x(x)] (1) = 21(1) (consistency), the method is zero-stable (to
assure convergence).
The method is called symmetric if
j = kj , i j = i kj , for i = 1, . . . , q ; j = 0, . . . ,
bk/2c.Particular cases of the methods in (2) are the Stormer-Cowell
methods,
which are obtained for q = 1 and 0 = 1, 1 = 2, j = 0, j = 2, . .
. , k 1.There are no methods of type (2) of one step, and thus two
steps must beconsidered at least. Some examples of these methods
are the method
yn 2yn+1 + yn+2 = h2y(2)nwhich is an explicit method with order
p = 1, and the explicit method
yn 2yn+1 + yn+2 = h2y(2)n+1which has order p = 2 (this is an
Stormer-Cowell method). The formula
yn 2yn+1 + yn+2 =qi=1
h2iy(2i)n+1
i (2i 1)!is a two step method of the form in (2) of order p = 2q
with error constantC2q+2 = 1/((q + 1)(2q + 1)!) for solving (1)
that may be considered as theequivalent to the Taylor method for
first-order I.V.P.s. (the Taylor methodis a one-step explicit
Obrechkoff method for first-order I.V.P.s [8]).
5
-
3 Analysis of two-step implicit Obrechkoff meth-
ods
In this section we are going to study the implicit Obrechkoff
methods of theform in (2) for k = 2 and q = 1, 2, 3, and so the
methods read
2j=0
j yn+j =
qi=1
h2i(i 0y
(2i)n + i 1y
(2i)n+1 + i 2y
(2i)n+2
). (6)
We will denote the methods byM(q, p) where q refers to the
higher order 2q ofthe derivatives appearing in the method, and p
indicates the algebraic orderof the method. We have only considered
implicit methods due to their betterstability characteristics as we
are interested in applying them for solving stiffproblems.
Our goal consists in looking for optimal methods of the form in
(6)having the following properties:
1. high order with minimum absolute error constant
2. great stability regions or intervals of stability
3. reduced number of function evaluations, which means to have
as muchas possible of the coefficients i j equal to zero
Taking 2 = 1 in order to normalize the parameters, and imposing
C0 =C1 = 0 (necessary to guarantee the minimum order p = 1) results
that thefirst characteristic polynomial is the same for all the
methods, being
(x) = 1 2x+ x2 .
In what follows we will analyze different possibilities
according to thevalue of q.
3.1 Methods with q=1:
Let us consider q = 1 in (6). The most known method of this type
is theNumerov method [4] (the method named as M(1, 4) in Table
3.1), which hasthe highest order in this class.
6
-
There exist no values of the coefficients in (6) which cause the
order tobe p = 3.
A one-parameter family of symmetric zero-stable methods of
orderp = 2 whenever C4 6= 0 is given by
yn 2yn+1 + yn+2 = h2(1 0y
(2)n + (1 21 0)y(2)n+1 + 1 0y(2)n+2
)(7)
with error constant C4 =1121 0 (for 1 0 = 112 results the
Numerov method).
For 1 0 14 the absolute stability interval is (, 0]
(A0-stability), and thusour choice for this parameter will be on
this range. For 1 0 =
14we obtain
the method
yn 2yn+1 + yn+2 = h2
4
(y(2)n + 2y
(2)n+1 + y
(2)n+2
)(8)
known as Dahlquist method (it is the method obtained using the
(1, 1)-Padeapproximant in [22]), with error constant C4 =
16. Taking 1 0 =
12the
method in (7) reduces to
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
), (9)
with error constant C4 =512.
A two-parameter family of zero-stable methods of order p = 1
when-ever C3 6= 0 is given by
yn 2yn+1 + yn+2 = h2((1 1 1 1 2)y(2)n + 1 1y(2)n+1 + 1
2y(2)n+2
)with error constant C3 = 1 1 1 21 2. For 1 1 = 1/2, 1 2 = 1/4
is C3 = 0and we obtain the method in (8) and for 1 1 = 0, 1 2 = 1/2
we obtain themethod in (9), which are second order methods. In
fact, if C3 = 0,that is,1 1 = 1 21 2, we obtain the class of
second-order methods in (7).
Taking 1 1 = 0, 1 2 = 1 arises the method
yn 2yn+1 + yn+2 = h2 y(2)n+2 , (10)
which has order p = 1. The absolute stability region for this
method, thatincludes the negative real axis, is shown in Figure
1.
7
-
-2 0 2 4 6-4
-2
0
2
4
Figure 1: Absolute stability region for the method (10).
In Table 3.1 we include some of the characteristics of the
optimal methodsin the family M(1, p).
Method p 1,0 1,1 1,2 Stability Cp+2
M(1, 4) 41
12
5
6
1
12[-6,0] 1
240
M(1, 2) 21
20
1
2A0 5
12M(1, 1) 1 0 0 1 A() -1
Table 1: Characteristics of the optimal methods in the family
M(1, p).
3.2 Methods with q=2:
Now, let us consider q = 2 in (6). The method of higher order in
this classhas appeared in [17] (the method named as M(2, 8) in
Table 3.2) which has
8
-
p = 8, C10 =59
76204800, and stability interval [25.2, 0].
There are no values of the coefficients in (6) for which is
obtained amethod of order p = 7.
A class of zero-stable symmetric methods of sixth order with a
free pa-rameter a = 20 reads
yn 2yn+1 + yn+2 = h2((
1
30 12a
)(y(2)n + y
(2)n+2
)+
(24a+
14
15
)y(2)n+1
)
+h4(a(y(4)n + y
(4)n+2
)+
(10a+
1
20
)y(4)n+1
)with C8 = (13 + 15120a)/302400. We have found four methods of
thisclass with non empty stability interval and with some of the
coefficientsi j , i = 1, 2; j = 0, 1, 2 null:
1. for a = 7/180 the method has error constant C8 = 23/12096
andinterval of stability [6
47
(15 +
1165
), 0] ' [6.2721, 0]
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
) h
4
180
(7y(4)n + 61y
(4)n+1 + 7y
(4)n+2
)2. for a = 1/200 the method has error constant C8 = 313/1512000
and
interval of stability [23
(1309 47) , 0] ' [7.2132, 0]
yn 2yn+1 + yn+2 = h2
75
(7y(2)n + 61y
(2)n+1 + 7y
(2)n+2
) h
4
200
(y(4)n + y
(4)n+2
)3. for a = 1/360 the method has error constant C8 = 11/60480
and
interval of stability [12, 0]
yn 2yn+1 + yn+2 = h2 y(2)n+1 +h4
360
(y(4)n + 28y
(4)n+1 + y
(4)n+2
)9
-
4. for a = 0 the method has error constant C8 =13
302400and interval of
stability [20, 0]
yn 2yn+1 + yn+2 = h2
30
(y(2)n + 28y
(2)n+1 + y
(2)n+2
)+h4
20y(4)n+1
There are no values of the coefficients in (6) for which the
order is p = 5.A family of zero-stable symmetric fourth order
methods with two free
parameters, a = 20 , b = 21, is given by
yn 2yn+1 + yn+2 = h2(1 24a b
12(y(2)n + y
(2)n+2) +
24a+ 12b+ 5
6y(2)n+1
)
+h4(ay(4)n + by
(4)n+1 + ay
(4)n+2
)with error constant C6 = (1 200a + 20b)/240. We have found
threemethods of this class with non empty stability interval and
with some of thecoefficients i j , i = 1, 2; j = 0, 1, 2 zero:
1. for a = 1/24 , b = 0 the method has error constant C6
=7180
and
interval of stability [2(321) , 0] ' [3.1651, 0]
yn 2yn+1 + yn+2 = h2 y(2)n+1 +h4
24
(y(4)n + y
(4)n+2
)2. for a = 0 , b = 5/12 the method has error constant C6 =
7
180and
interval of stability [25
(3 +
69), 0] ' [4.5226, 0]
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
) 5h
4
12y(4)n+1
3. for a = 5/24 , b = 0 the method has error constant C6 =
61360
and
interval of stability (, 0]
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
) h
4
24
(5y(4)n + 5y
(4)n+2
)10
-
There is a class of zero-stable symmetric methods of third order
withthree free parameters, a = 20 , b = 21, c = 22 given by
yn 2yn+1 + yn+2 = h2
12
((1 d)
(y(2)n + y
(2)n+2
)+ 2(5 + d)y
(2)n+1
)+h4
(ay(4)n + by
(4)n+1 + cy
(4)n+2
)where d = 12(a+ b+ c), and with error constant C5 = a c. From
this classwe only have considered the method for a = 0, b = 0, c =
5/12
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
) h
4
125y
(4)n+2 (11)
which has the stability region shown in Figure 2.
-3 -2 -1 0 1 2 3-4
-2
0
2
4
Figure 2: Absolute stability region for the method (11).
In Table 3.2 the characteristics of the optimal methods of the
familyM(2, p) have been included.
11
-
Method p 1,0 1,1 1,2 2,0 2,1 2,2 Stability Cp+2
M(2, 8) 811
252
230
252
11
252
1315120
626
15120
1315120
[25.2, 0] 5976204800
M(2, 6) 61
30
28
30
1
300
1
200 [20, 0] 13
302400
M(2, 4) 41
20
1
2
524
0524
A061
360
M(2, 3) 31
20
1
20 0
512
A()5
12
Table 2: Characteristics of the optimal methods in the family
M(2, p).
3.3 Methods with q=3:
Finally, let us consider q = 3 in (6). The method of higher
order in this class
has also appeared in [17] which has p = 12, C14 =45469
1697361329664000and
stability interval [9.7954, 0].Although the authors in [17] say
that have no knowledge of any paper
in which this formula had been written down, this method had
previouslyappeared in [23], [24], and at the same time in [15].
In [1, 17, 22] there is also a sixth order symmetric method of
this class
with error constant C8 =1
50400and stability interval (, 0], but none of
the coefficients zero.There are no values of the coefficients in
(6) for which the order is p = 11.We present now different classes
of methods depending on some free pa-
rameters.There is a class of zero-stable symmetric methods of
tenth order with a
free parameter a = 30 given by
12
-
yn 2yn+1 + yn+2 = h2(1 + 45360a
39
(y(2)n + y
(2)n+2
)+37 90720a
39y(2)n+1
)
+ h4(17 3326400a
65520
(y(4)n + y
(4)n+2
)+1907 34776000a
32760y(4)n+1
)
+ h6(a(y(6)n + y
(6)n+2
)+59 3155040a
65520y(6)n+1
)(12)
with C12 = (127 39251520a)/43589145600. We have considered
differentvalues of the parameter a to get some of the coefficients
zero. The results forthese values appear in Table 3, where we have
included the error constantC12 and the stability intervals.
13
-
ja 0 1 2 C12 [, 0]
0 1 j139
3739
139
2.9135 109 [9.6275, 0]2 j
1765520
190732760
1765520
3 j 059
655200
593155040
1 j891878
850939
891878
1.3925 108 [28.6780, 0]2 j
19071577520
30257788760
19071577520
3 j59
31550400 59
3155040
173326400
1 j13660
317330
13660
7.5156 109 [9.4941, 0]2 j 0
7110
03 j
173326400
19071663200
173326400
190734776000
1 j6176900
28333450
6176900
4.6466 108 [26.7024, 0]2 j
72300
0 72300
3 j1907
347760003025717388000
190734776000
145360
1 j 0 1 0 2.2765 108 [9.2015, 0]2 j
1315120
6177560
1315120
3 j1
4536089
453601
45360
3790720
1 j12
0 12
3.6434 107 [25.4403, 0]2 j
31715120
28337560
31715120
3 j37
90720854536
3790720
Table 3: Some of the methods in (12) with order p = 10.
There are no values of the coefficients in (6) for which the
order is p = 9.A class of zero-stable symmetric methods of eighth
order depending on
two free parameters, a = 30, b = 31, is given in Table 4.
14
-
j 0 and 2 1
1 j50400a5040b+11
2525(10080a1008b23)
126
2 j282240a+10080b13
151201229760a+141120b+313
7560
3 j a b
Table 4: Coefficients for the class of eight-order methods with
k = 2 , q = 3taking two free parameters.
We can choose different values for the parameters in order to
have atleast two of the independent coefficients zero. There are 14
possibilities,from which some of them appear in Table 5. In
particular, for a = b = 0 itresults the higher order method with q
= 2 named as M(2, 8) in Table 3.2.
15
-
a jb 0 1 2 C10 [, 0]
120160
1 j 0 1 0 3.5824 106 [8.0792, 0]3
11202 j 0
112
03 j
120160
31120
120160
0 1 j 0 1 0 1.1022 106 [8.5893, 0]115040
2 j1
168023280
11680
3 j 0115040
01150400
1 j 0 1 0 9.8104 106 [11.3278, 0]0 2 j
92800
3234200
92800
3 j1150400
0 1150400
0 1 j12
0 12
2.0392 105 [20.4083, 0]231008
2 j9560
323840
9560
3 j 0231008
023
100801 j
12
0 12
9.3694 105 [7.3730, 0]0 2 j
731680
277840
731680
3 j23
100800 23
10080
0 1 j156
2728
156
3.3462 107 [9.1037, 0]13
100802 j 0
11168
03 j 0
1310080
013
2822401 j
27784
365392
27784
2.6812 106 [19.7195, 0]0 2 j 0
1152352
03 j
13282240
0 13282240
3131229760
1 j3233416
13851708
3233416
9.7634 106 [8.0133, 0]0 2 j
11520496
0 11520496
3 j313
12297600 313
1229760
Table 5: Some of the eight-order methods in Table (4) for
different values ofthe parameters.
There are no values of the coefficients in (6) for which the
order is p = 7.A class of zero-stable symmetric methods of sixth
order with three free
parameters, a = 30, b = 31, c = 20, is given in Table 6.
16
-
j 0 and 2 1
1 j720a360b360c+1
302(360a+180b+180c+7)
15
2 j c480a+240b+200c+1
20
3 j a b
Table 6: Coefficients for the class of sixth-order methods with
q = 3 takingthree free parameters.
From this family we have only considered the two methods that
follow,with some of the coefficients zero and absolute stability
interval (, 0]:
taking a = 1/720 , b = 0 , c = 0 we obtain the method
yn 2yn+1 + yn+2 = h2 y(2)n+1 +h4
12y(4)n+1 +
h6
720
(y(6)n + y
(6)n+2
)with C8 = 3/2240.
taking a = 61/720 , b = 0 , c = 5/24 we obtain the method
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
) 5h
4
24
(y(4)n + y
(4)n+2
)+61h6
720
(y(6)n + y
(6)n+2
)with C8 = 277/4032.
Finally, a family of fifth order methods with four free
parameters maybe considered. From this family we only take into
account the zero-stablemethods with some of the coefficients zero
and good regions of absolutestability (A()-stability). These
methods are:
the method with error constant C7 = 1/360
yn 2yn+1 + yn+2 = h2 y(2)n+1 +h4
12y(4)n+1 +
h6
360y(6)n+2 (13)
17
-
the method with error constant C7 = 61/360
yn 2yn+1 + yn+2 = h2
2
(y(2)n + y
(2)n+2
) 5h
4
24
(y(4)n + y
(4)n+2
)+61h6
360y(6)n+2 (14)
The absolute stability regions of the methods (13) and (14) are
plotted inFigure 3.
-20 -10 0 10 20 30 40 50-40
-20
0
20
40
-4 -2 0 2 4-4
-2
0
2
4
Figure 3: Absolute stability regions for the methods (13)[left]
and (14)[right].
In Table 7 we have included the characteristics of the optimal
methodsof the family M(3, p).
18
-
Method
p1,0
1,1
1,2
2,0
2,1
2,2
3,0
3,1
3,2
Stability
Cp+2
M(3,12)
12229
7788
7330
7788
229
7788
11
25960
1422
25960
11
25960
127
39251520
29230
39251520
127
39251520
[9.79,0]
45469
1697361329664000
M(3,10)
1089 1878
1700
1878
89 1878
1907
1577520
30257
1577520
1907
1577520
59
3155040
05
9
3155040
[28.67,0]
2923
209898501120
M(3,8)
811 252
230
252
11 252
13
15120
626
15120
13
15120
00
0[25.2,0]
59
76204800
M(3,6)
60
10
01 12
01 720
01 720
A0
5 12
M(3,5)
50
10
01 12
00
1 360
0A()
1 360
Table7:
Characteristicsoftheoptimalmethodsin
M(3,p).
-
4 Numerical time integration approach for
PDEs
For the integration of partial differential equations are used
different tech-niques, among which are the finite element method,
finite difference method,or the method of lines (MOL).
The latter procedure allows to obtain optimum results when the
spatialdiscretization is fine enough and the temporal integrator
has good propertiesof order, convergence and stability.
Let us consider the one-dimensional wave equation with initial
and bound-ary conditions given by
2u(x, t)
t2= c22u(x, t) , x [0, L] , t [0, tf ]
u(x, t)
t(x, 0) = f1(x) , u(x, 0) = f2(x)
u(0, t) = g1(t) , u(L, t) = g2(t)
(15)
The MOL is a very powerful technique which has been widely used
forthe solution of (15). The idea behind this approach consists in
taking on thespace domain a discrete mesh
:= {xi [0, L] : 0 = x0 x1 . . . xN+1 = L} ,
not necessarily evenly spaced, in such a way that for every xi ,
the spatialderivatives appearing in (15) are approximated by means
of finite differences,spectral methods or finite elements
techniques.
Even though having a uniform step size between points makes it
con-venient to write out the formulas, it is certainly not a
requirement. On auniform step size mesh with step size4x =
(xN+1x0)/(N+1) = L/(N+1),finite second order centered differences
for the first and second derivatives ofa function (x) are commonly
given by
(xi) =(xi+1) (xi1)
24x ,
(xi) =(xi+1) 2(xi) + (xi1)
(4x)2 . (16)
20
-
Thus, replacing the space derivative in (15) by the central
difference approx-imation
2u(x, t)
x2' 1
(4x)2 [u(x+4x, t) 2u(x, t) + u(x4x, t)] ,
and setting ui(t) = u(xi, t) for i = 1, . . . , N , with values
u0(t) = u(0, t) =g1(t) , uN+1(t) = u(L, t) = g2(t), from the
problem in (15) we obtain theinitial-value problem of the form
d2 u
d t2= c2 (Au(t) +B(t)) ,
u(0) = (u1(0), . . . , uN(0))T ,
u(0) = (f1(x1), . . . , f1(xN))T
(17)
where u(t) = (u1(t), . . . , uN(t))T , the N N matrix A is
A =
2(4x)2
1(4x)2 0 . . . 0 0 0
1(4x)2
2(4x)2
1(4x)2 . . . 0 0 0
0 1(4x)2
2(4x)2 . . . 0 0 0
......
......
......
0 0 0 . . . 2(4x)2
1(4x)2 0
0 0 0 . . . 1(4x)2
2(4x)2
1(4x)2
0 0 0 . . . 0 1(4x)2
2(4x)2
,
and the N -vector B(t) is
B(t) =
(u0(t)
(4x)2 , 0, . . . , 0,uN+1(t)
(4x)2)T
=
(g1(t)
(4x)2 , 0, . . . , 0,g2(t)
(4x)2)T
.
Note that the eigenvalues of A are (see [21])
j =2
(4x)2(cos
(j pi
N + 1
) 1), j = 1, . . . , N ,
which belong to the range (4 (N +1)2, 0), and so, for large
values of N thesystem becomes very stiff. We observe that for
solving the problem (17) wemay use some appropriate method of the
above section.
21
-
5 Numerical examples
We are going to apply the above methods for solving different
problems inorder to see their performance. Firstly we include a
scalar problem which willbe solved by two of the methods not
A0-stable, and later a problem relatedwith the wave equation.
Note that the application of any of the methods in (6) for
solving thesystem in (17) leads to a N N algebraic system, that
must be solved. Forthat purpose any of the Newton-type methods may
be used. In particular,for the problem (17) the resulting system is
linear, which makes easier itsresolution.
5.1 Problem 1
Let us consider the initial-value problem in [15] given by
y(t) + 25y(t) = 8(cos(t) +
2
3cos(3t)
), y(0) = 1 , y(0) = 0 (18)
whose exact solution is
y(t) =1
3(cosx+ cos(3x) + cos(5x)) .
We have solved over five periods the above problem with the
first of themethods in Table 3, which is a methodM(3, 10), and the
sixth of the methodsin Table 5, which is a methodM(3, 8). These
methods are given respectivelyby
yn 2yn+1 + yn+2 = h2
39
(y(2)n + 37y
(2)n+1 + y
(2)n+2
) h
4
65520
(17y(4)n 3814y(4)n+1 + 17y(4)n+2
)+
h6
65520
(59y
(6)n+1
)
22
-
M(3, 10) M(3, 8)h (1st in Table 3) (6th in Table 5)pi
108.22305 106 4.59128 104
pi
204.00879 109 1.35816 106
pi
403.71127 1012 2.76865 109
pi
801.75386 1015 5.22551 1012
Table 8: Maximum absolute errors for Problem 1.
and
yn 2yn+1 + yn+2 = h2
56
(y(2)n + 54y
(2)n+1 + y
(2)n+2
)+h4
168
(11y
(4)n+1
)+
h6
10080
(13y
(6)n+1
).
We observe that these methods have stability intervals very
similar.In Table 8 appear the maximum absolute errors taking
different values of
the stepsize h. The numerical solutions provided by the methods
are quiteaccurate, even for relatively large values of h. In Figure
4 the plot of theexact solution and the discrete one obtained with
the method M(3, 10) isshown (the plot with the method M(3, 8) is
similar).
23
-
5 10 15 20 25 30
-1.0
-0.5
0.5
1.0
Figure 4: Exact solution (continuous line) and discrete solution
in [0, 10pi]obtained with the method M(3, 10) taking h = pi/10 for
the problem (18).
5.2 Problem 2
Let us consider the problem of the form in (15) given by
2u(x, t)
t2=2u(x, t)
x2, x [0, L] , t [0, tf ]
u(x, t)
t(x, 0) = sin(x) , u(x, 0) = 0
u(0, t) = 0 , u(L, t) = sin(L) sin(t)
(19)
whose exact solution is given by
u(x, t) = sin(x) sin(t) .
After discretizing according to the guidelines in the previous
section we obtainthe differential system of the form in (17) with
initial values
u1(0) = 0, . . . , uN(0) = 0; u1(0) = sin(x1), . . . , u
N(0) = sin(xN) .
24
-
M(1,2) M(2,4) M(3,6)tf L x = 10
1 x = 102 x = 101 x = 102 x = 101 x = 102
2 2 2.9 103 2.4 103 5.0 104 1.5 105 4.9 104 4.9 10610 4.3 103
3.6 103 7.3 104 2.1 105 7.2 104 7.2 10630 4.3 103 3.6 103 7.3 104
2.1 105 7.2 104 7.2 106
10 2 1.9 103 1.6 103 3.0 104 9.0 106 3.0 104 2.9 10610 2.3 102
1.9 102 4.0 103 1.2 104 3.9 103 3.9 10530 1.9 102 1.6 102 3.3 103
9.9 105 3.2 103 3.2 105
Table 9: Maximum absolute errors for the problem in (19) using
differentoptimal methods.
We have solved this problem using some of the optimal two-step
mul-tiderivative methods in Section 3 (see Tables 3.1, 3.2 and 7).
As it wouldexpected, the methods that are not A0-stable are not
appropriate for solvingthe problem, except if a very small stepsize
4t = h is taken. The methodsof higher order M(1, 4) (Numerov), M(2,
8) and M(3, 12) produce very bigerrors for the stepsizes
considered. In Table 9 we have included the maximumabsolute errors
with the optimal methods M(1, 2),M(2, 4) and M(3, 6) fordifferent
values of L and tf taking x = 10
1, 102, 4t = 101.All the methods considered work well when there
is a decrease in the
space stepsize 4x, and also when the time length increases, tf =
2, 10. As itwas expected, the best accuracy is obtained with the
method M(3, 6), whichalso entails greater computational cost.
We have included in Fig. 5 the plots of the exact and discrete
solutions[up], together with the two plots overlapped [down],
showing the close agree-ment between them.
25
-
02
46
x
0
2
4
6
t-1
0
1
u
02
46
x
0
2
4
6
t-1
0
1
u
02
4
6
x
0
2
4
6
t
-1
0
1
u
Figure 5: Exact [up-left] and discrete [up-right] solutions of
the problem (19)in [0, 2pi] [0, 2pi] using the method M(3, 6)
taking 4x = pi/10,4t = 1/10.The two solutions overlapped
[down].
6 Conclusions
We have made an analysis of the two-step multiderivative methods
for solv-ing second order initial-value problems, looking for
methods of the form in(6) with better characteristics concerning
the error, the stability and lower
26
-
computational work. We have found that for q = 1, 2, 3, being 2q
the highestorder of the derivative appearing in the method, there
are methods A0-stablesof orders 2q, and methods A()-stables of
orders 2q 1. These are the mostappropriate methods to be used in
problems where stability is a requirement,such as stiff problems.
On the contrary, the methods of high order, which arethe Numerov
method namedM(1, 4) and the methodsM(2, 8) andM(3, 12),which had
appeared in [18], in general are not suitable for solving such
kindof problems.
The application of the methods M(1, 2),M(2, 4) and M(3, 6)
presentedin the article to a highly oscillatory problem and to
another concerning thewave equation shows the good performance of
the proposed methods.
Acknowledgements
The first author is grateful for the hospitality during his stay
at the Depart-ment of Mathematics, University of Coimbra, during
which this work wascarried out.
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29