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2 Multidimensional ARKN Methods and the Corresponding Order Conditions

# 2 Multidimensional ARKN Methods and the Corresponding Order Conditions

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10.2 Multidimensional ARKN Methods and the Corresponding Order Conditions

213

Definition 10.1 An s-stage multidimensional ARKN scheme for the numerical integration of the initial value problem (10.1) is defined by

s

a¯ i j f (Y j , Y j ) − MY j ,

Yi = yn + hci yn + h 2

i = 1, . . . , s,

j=1

s

ai j f (Y j , Y j ) − MY j ,

Yi = yn + h

i = 1, . . . , s,

j=1

(10.3)

s

2

y

=

φ

(V

)y

+

(V

)y

+

h

b¯i (V ) f (Yi , Yi ),

n+1

0

n

1

n

i=1

s

y

=

φ

(V

)y

h

(V

)y

+

h

bi (V ) f (Yi , Yi ),

0

1

n

⎩ n+1

n

i=1

where a¯ i j and ai j are real constants, the weight functions bi (V ) : Rd×d → Rd×d and

b¯i (V ) : Rd×d → Rd×d , i = 1, . . . , s, in the updates are matrix-valued functions of

V = h 2 M. The scheme (10.3) can also be denoted by the following Butcher tableau:

c

A

b (V ) b¯ (V )

c1 a11 . . . a1s

..

..

..

..

.

.

.

.

cs as1 · · · ass

=

a¯ 11

..

.

a¯ s1

. . . a¯ 1s

..

..

.

.

· · · a¯ ss

b1 (V ) · · · bs (V ) b¯1 (V ) · · · b¯s (V )

In the special case where the right-hand-side function of (10.1) is independent of

y , the scheme (10.3) reduces to the corresponding ARKN method for the special

oscillatory system [17],

y + M y = f (y),

y(t0 ) = y0 ,

t ∈ [t0 , T ],

(10.4)

y (t0 ) = y0 ,

namely,

s

a¯ i j f (Y j ) − MY j ,

Yi = yn + hci yn + h 2

i = 1, . . . , s,

j=1

s

b¯i (V ) f (Yi ),

i=1

s

bi (V ) f (Yi ).

⎩ yn+1 = φ0 (V )yn − h Mφ1 (V )yn + h

yn+1 = φ0 (V )yn + hφ1 (V )yn + h 2

i=1

(10.5)

214

10 Multidimensional ARKN Methods for General Multi-frequency …

The scheme (10.5) can also be denoted by the following Butcher tableau:

c

c1 a¯ 11 . . . a¯ 1s

..

..

..

..

.

.

.

.

cs a¯ s1 · · · a¯ ss

b¯ (V )

b (V )

=

b¯1 (V ) · · · b¯s (V )

b1 (V ) · · · bs (V )

A multidimensional ARKN method (10.3) is said to be of order p if, for a sufficiently smooth problem (10.1), the conditions

en+1 := yn+1 − y(tn + h) = O(h p+1 )

en+1 := yn+1 − y (tn + h) = O(h p+1 )

(10.6)

are satisfied simultaneously, where y(tn + h) and y (tn + h) are the respective exact

solution and its derivative of (10.1) at tn + h, whereas yn+1 and yn+1 are the onestep numerical approximations obtained by the method from the local assumptions:

yn = y(tn ) and yn = y (tn ). The order conditions, established in [19] for the method

(10.3), are given below.

Theorem 10.1 (Wu et al. [19]) A necessary and sufficient condition for a multidimensional ARKN method (10.3) to be of order p is given by

¯ ) − ρ(τ )! φρ(τ )+1 (V ) = O(h p−ρ(τ ) ), ρ(τ ) = 1, . . . , p − 1,

⎨ Φ(τ ) ⊗ Im b(V

γ (τ )

ρ(τ )!

⎩ Φ(τ ) ⊗ Im b(V ) −

φρ(τ ) (V ) = O(h p+1−ρ(τ ) ), ρ(τ ) = 1, . . . , p,

γ (τ )

where τ is the Nyström tree associated with the elementary differential F (τ )(yn , yn )

of the function f˜(y, y ) = f (y, y ) − M y at (yn , yn ), the entries Φ j (τ ), j =

1, . . . , s, of Φ(τ ) are weight functions defined in [8], and φρ(τ ) (V ), ρ(τ ) =

1, . . . , p, are given by (4.7).

10.3 ARKN Methods for General Multi-frequency

and Multidimensional Oscillatory Systems

This section constructs three explicit multidimensional ARKN schemes for the general multi-frequency and multidimensional oscillatory system (10.1) and analyses

the stability and phase properties of the schemes.

10.3 ARKN Methods for General Multi-frequency …

215

10.3.1 Construction of Multidimensional ARKN Methods

First, consider an explicit three-stage ARKN method of order three. Since f depends

on both y and y in (10.1), at least three stages are required to satisfy all the order

conditions. The explicit three-stage ARKN methods for (10.1) can be expressed by

the following Butcher tableau:

c

A

b (V ) b¯ (V )

=

c1 0

c2 a21

c3 a31

0

0

a32

0

0

0

0

a¯ 21

a¯ 31

0

0

a¯ 32

0

0

0

b1 (V ) b2 (V ) b3 (V ) b¯1 (V ) b¯2 (V ) b¯3 (V )

From Theorem 10.1, a three-stage ARKN method is of order three if its coefficients

satisfy

(e ⊗ I )b(V ) = φ1 (V ) + O (h 3 ),

(c ⊗ I )b(V ) = φ2 (V ) + O (h 2 ),

2

¯ ) = φ2 (V ) + O (h 2 ),

(c ) ⊗ I b(V ) = 2φ3 (V ) + O (h), (e ⊗ I )b(V

¯

¯

( Ae) ⊗ I b(V ) = φ3 (V ) + O (h),

(c ⊗ I )b(V ) = φ3 (V ) + O (h),

¯ ) = φ3 (V ) + O (h), (Ae) ⊗ I b(V ) = φ2 (V ) + O (h 2 ),

(Ae) ⊗ I b(V

(Ac) ⊗ I b(V ) = φ3 (V ) + O (h), (c · Ae) ⊗ I b(V ) = 2φ3 (V ) + O (h),

(A2 e) ⊗ I b(V ) = φ3 (V ) + O (h), (Ae · Ae) ⊗ I b(V ) = 2φ3 (V ) + O (h),

(10.7)

where e = (1, 1, 1) .

1

Choosing c1 = 0, c2 = , c3 = 1, and solving all the equations in (10.7), we

2

obtain

1

⎨ a21 = , a31 = −1, a32 = 2,

2

(10.8)

⎩ a¯ 21 = 1 , a¯ 31 = 1 , a¯ 32 = 0,

8

2

and

⎨ b1 (V ) = φ1 (V ) − 3φ2 (V ) + 4φ3 (V ),

b2 (V ) = 4φ2 (V ) − 8φ3 (V ),

⎩ b (V ) = −φ (V ) + 4φ (V ),

3

2

3

3

b¯1 (V ) = φ2 (V ) − φ3 (V ),

2

b¯2 (V ) = φ3 (V ),

1

b¯3 (V ) = φ3 (V ).

2

(10.9)

Equations (10.8)–(10.9) give a three-stage ARKN method of order three. The Taylor

expansions of the coefficients bi (V ) and b¯i (V ) of this method are given by

216

10 Multidimensional ARKN Methods for General Multi-frequency …

⎪ b1 (V ) =

b2 (V ) =

⎨ b3 (V ) =

3

5

1

7

1

I− V+

V2 −

V3 +

V4 + ··· ,

6

40

1008

51840

492800

1

1 2

2

1

1

I− V+

V −

V3 +

V4 + ··· ,

3

10

252

12960

1108800

1

1

1

1

1

I+

V−

V2 +

V3 −

V4 + ··· ,

6

120

1680

72576

5702400

1

1

19

7

11

b¯1 (V ) = I −

V+

V2 −

V3 +

V4 + ··· ,

4

240

10080

48348

79833600

1

1

1

1

1

V+

V2 −

V3 +

V4 + ··· ,

b¯2 (V ) = I −

6

120

5040

362880

39916800

1

1

⎩ b¯3 (V ) = 1 I − 1 V + 1 V 2 −

V3 +

V4 + ··· .

12

240

10080

725760

79833600

(10.10)

This method is denoted by ARKN3s3.

We next consider the explicit multidimensional ARKN methods of order four for

(10.1). In this case, four stages are needed. An explicit four-stage ARKN method

can be denoted by the following Butcher tableau:

c

A

b (V ) b¯ (V )

c1 0

c2 a21

c3 a31

c

= 4 a41

0

0

a32

a42

0

0

0

a43

0

0

0

0

0

a¯ 21

a¯ 31

a¯ 41

0

0

a¯ 32

a¯ 42

0

0

0

a¯ 43

0

0

0

0

b1 (V ) b2 (V ) b3 (V ) b4 (V ) b¯1 (V ) b¯2 (V ) b¯3 (V ) b¯4 (V )

The simplifying conditions Ae = c, A¯ = A2 are used to reduce the order to those

listed below,

(e ⊗ I )b(V ) = φ1 (V ) + O (h 4 ),

(c ⊗ I )b(V ) = φ2 (V ) + O (h 3 ),

2

2

2

⎪ (c 2) ⊗ I b(V ) = 2φ3 (V ) + O (h ), (Ac)2 ⊗ I b(V ) = φ3 (V ) + O (h ),

(A c) ⊗ I b(V ) = φ4 (V ) + O (h), (Ac ) ⊗ I b(V ) = 3φ4 (V ) + O (h),

⎪ (c3 ) ⊗ I b(V ) = 6φ4 (V ) + O (h), (c · Ac) ⊗ I b(V ) = 2φ4 (V ) + O (h),

¯ ) = φ2 (V ) + O (h 3 ),

¯ ) = φ3 (V ) + O (h 2 ),

(c ⊗ I )b(V

⎪ (e ⊗ I )b(V

⎩ 2

¯ ) = 2φ4 (V ) + O (h), (Ac) ⊗ I b(V

¯ ) = φ4 (V ) + O (h),

(c ) ⊗ I b(V

(10.11)

where e = (1, 1, 1, 1) .

10.3 ARKN Methods for General Multi-frequency …

217

1

1

Choosing c1 = 0, c2 = , c3 = , c4 = 1, and solving the above equations, we

2

2

obtain

1

1

⎨ a21 = , a31 = 0, a32 = , a41 = 0, a42 = 0, a43 = 1,

2

2

(10.12)

1

1

⎩ a¯ 21 = 0, a¯ 31 = , a¯ 32 = 0, a¯ 41 = 0, a¯ 42 = , a¯ 43 = 0,

4

2

and

b1 (V ) = φ1 (V ) − 3φ2 (V ) + 4φ3 (V ),

b2 (V ) = 2φ2 (V ) − 4φ3 (V ),

b (V ) = 2φ2 (V ) − 4φ3 (V ),

⎩ 3

b4 (V ) = −φ2 (V ) + 4φ3 (V ),

b¯1 (V ) = φ2 (V ) − 3φ3 (V ) + 4φ4 (V ),

b¯2 (V ) = 2φ3 (V ) − 4φ4 (V ),

b¯3 (V ) = 2φ3 (V ) − 4φ4 (V ),

b¯4 (V ) = −φ3 (V ) + 4φ4 (V ).

(10.13)

Equations (10.12)–(10.13) give a four-stage ARKN method of order four. The Taylor

expansions of the coefficients bi (V ) and b¯i (V ) of this method are

b1 (V ) =

⎪ b2 (V ) =

b3 (V ) =

⎨ b4 (V ) =

b¯1 (V ) =

b¯2 (V ) =

⎪ b¯3 (V ) =

⎩ b¯4 (V ) =

3

5

1

7

I− V+

V2 −

V3 + ··· ,

6

40

1008

51840

1

1 2

1

1

I− V+

V −

V3 + ··· ,

3

20

504

25920

1

1 2

1

1

I− V+

V −

V3 + ··· ,

3

20

504

25920

1

1

1

1

I+

V−

V2 +

V3 + ··· ,

6

120

1680

72576

1

1

1

1

I− V+

V2 −

V3 + ··· ,

6

45

1120

56700

1

1

1

1

I− V+

V2 −

V3 + ··· ,

6

90

3360

226800

1

1

1

1

I− V+

V2 −

V3 + ··· ,

6

90

3360

226800

1

1

1

V−

V2 +

V3 + ··· .

360

10080

604800

(10.14)

This method is denoted by ARKN4s4.

In what follows, the construction of the ARKN methods of order five for (10.1) is

considered. In this case, at least six stages are required. Thus, consider the following

Butcher tableau:

218

c1

c2

c3

c4

c5

c6

10 Multidimensional ARKN Methods for General Multi-frequency …

0

a21

a31

a41

a51

a61

0

0

a32

a42

a52

a62

0

0

0

a43

a53

a63

0

0

0

0

a54

a64

0

0

0

0

0

a65

0

0

0

0

0

0

0

a¯ 21

a¯ 31

a¯ 41

a¯ 51

a¯ 61

0

0

a¯ 32

a¯ 42

a¯ 52

a¯ 62

0

0

0

a¯ 43

a¯ 53

a¯ 63

0

0

0

0

a¯ 54

a¯ 64

0

0

0

0

0

a¯ 65

0

0

0

0

0

0

b1 (V ) b2 (V ) b3 (V ) b4 (V ) b5 (V ) b6 (V ) b¯1 (V ) b¯2 (V ) b¯3 (V ) b¯4 (V ) b¯5 (V ) b¯6 (V )

The order conditions up to order five are imposed using the simplifying assumptions Ae = c, A¯ = A2 and the coefficients must then satisfy

(e ⊗ I )b(V ) = φ1 (V ) + O (h 5 ),

(c ⊗ I )b(V ) = φ2 (V ) + O (h 4 ),

2

3

(Ac) ⊗ I b(V ) = φ3 (V ) + O (h 3 ),

(c ) ⊗ I b(V ) = 2φ3 (V ) + O (h ),

((A2 c) ⊗ I )b(V ) = φ4 (V ) + O (h 2 ),

((Ac2 ) ⊗ I )b(V ) = 2φ4 (V ) + O (h 2 ),

3

2

(c · Ac) ⊗ I b(V ) = 3φ4 (V ) + O (h 2 ),

⎨ (c ) ⊗ I b(V ) = 6φ4 (V ) + O (h ),

3

(c · A2 c) ⊗ I b(V ) = 4φ5 (V ) + O (h),

(A c) ⊗ I b(V ) = 6φ5 (V ) + O (h),

4

⎪ (c ) ⊗ I b(V ) = 24φ5 (V ) + O (h),

(c2 · Ac) ⊗ I b(V ) = 12φ5 (V ) + O (h),

(Ac · Ac) ⊗ I b(V ) = 6φ5 (V ) + O (h), (c · Ac2 ) ⊗ I b(V ) = 8φ5 (V ) + O (h),

(A(c · Ac)) ⊗ I b(V ) = 3φ5 (V ) + O (h),

(A2 c2 ) ⊗ I b(V ) = 2φ5 (V ) + O (h),

3

(Ac ) ⊗ I b(V ) = 6φ5 (V ) + O (h),

(10.15)

and

¯ ) = φ2 (V ) + O (h 4 ),

(e ⊗ I )b(V

⎨ (c2 ) ⊗ I b(V

¯ ) = 2φ4 (V ) + O (h 2 ),

¯ ) = φ5 (V ) + O (h),

(A2 c) ⊗ I b(V

⎩ 3

¯ ) = 6φ5 (V ) + O (h),

(c ) ⊗ I b(V

¯ ) = φ3 (V ) + O (h 3 ),

(c ⊗ I )b(V

¯ ) = φ4 (V ) + O (h 2 ),

(Ac) ⊗ I b(V

¯ ) = 2φ5 (V ) + O (h),

(Ac2 ) ⊗ I b(V

¯ ) = 3φ5 (V ) + O (h),

(c · Ac) ⊗ I b(V

(10.16)

where e = (1, 1, 1, 1, 1, 1) .

1

1

1

2

Choosing c1 = 0, c2 = , c3 = , c4 = , c5 = and c6 = 1, and solving the

6

3

2

3

above equations, we obtain

a21

a43

⎨ a61

a¯ 21

a¯ 43

⎩ a¯ 61

1

1

, a41 = − ,

3

4

1

2

1

= 0,

a51 = − , a52 = , a53 = ,

27

9

3

2

3

27

, a63 =

, a64 = −4,

= − , a62 =

11

11

11

1

1

, a¯ 32 = 0, a¯ 41 = ,

= 0,

a¯ 31 =

18

8

2

= 0,

a¯ 51 = 0,

a¯ 52 = , a¯ 53 = 0,

9

21

18

9

4

=

, a¯ 62 = − , a¯ 63 =

, a¯ 64 =

,

22

11

11

11

=

1

,

6

a31 = 0,

a32 =

3

,

4

4

a54 =

,

27

27

,

a65 =

11

a42 =

a¯ 42 = 0,

a¯ 54 = 0,

a¯ 65 = 0,

(10.17)

10.3 ARKN Methods for General Multi-frequency …

and

15

φ2 (V ) + 40φ3 (V ) − 135φ4 (V ) + 216φ5 (V ),

b1 (V ) = φ1 (V ) −

2

⎪ b (V ) = 0,

2

b

3 (V ) = 27 φ2 (V ) − 9φ3 (V ) + 39φ4 (V ) − 72φ5 (V ) ,

b

(V ) = −32 φ2 (V ) − 11φ3 (V ) + 54φ4 (V ) − 108φ5 (V ) ,

⎪ 4

27

b5 (V ) =

φ2 (V ) − 12φ3 (V ) + 66φ4 (V ) − 144φ5 (V ) ,

2

b6 (V ) = −φ2 (V ) + 13φ3 (V ) − 81φ4 (V ) + 216φ5 (V ),

64

b¯1 (V ) = φ2 (V ) − 5φ3 (V ) +

φ4 (V ) − 13φ5 (V ),

5

b¯2 (V ) = 0,

⎪ b¯3 (V ) = 9φ3 (V ) − 171 φ4 (V ) + 45φ5 (V ),

5

64

¯

b

(V

)

=

−4φ

(V

)

+

φ4 (V ) − 16φ5 (V ),

4

3

5

54

b¯5 (V ) =

φ4 (V ) − 27φ5 (V ),

5

⎩ b¯ (V ) = − 11 φ (V ) + 11φ (V ).

6

4

5

5

219

(10.18)

Equations (10.17)–(10.18) give a six-stage ARKN method of order five. The Taylor

expansions of the coefficients (10.18) are given by

11

259

3

25 2

3

⎪ b1 (V ) = 120 I − 70 V + 8064 V − 2851200 V + · · · ,

b2 (V ) = 0,

27

17

99

9 2

⎪ b3 (V ) =

I−

V+

V −

V3 + ··· ,

40

560

896

70400

8

19

4

1 2

b (V ) = − I +

V−

V +

V3 + ··· ,

⎪ 4

15

35

126

89100

27

3

9

3 2

b (V ) =

I−

V+

V −

V3 + ··· ,

⎪ 5

40

140

896

35200

11

47

1

1

I+

V−

V2 +

V3 + ··· ,

⎨ b6 (V ) =

120

336

4480

7983360

2839

⎪ b¯ (V ) = 11 I − 383 V + 1231 V 2 −

V3 + ··· ,

1

120

25200

1814400

199584000

⎪ b¯2 (V ) = 0,

61

⎪ b¯ (V ) = 9 I − 51 V + 107 V 2 −

V3 + ··· ,

3

20

1400

100800

3696000

4

197

59

59

⎪¯

b4 (V ) = − I +

V−

V2 +

V3 + ··· ,

15

3150

113400

24948000

9

17

27

13

b¯5 (V ) =

I−

V+

V2 −

V3 + ··· ,

40

2280

67200

7392000

1

⎩ b¯6 (V ) = 11 V − 11 V 2 +

V3 + ··· .

12600

453600

3024000

This method is denoted by ARKN6s5.

(10.19)

220

10 Multidimensional ARKN Methods for General Multi-frequency …

10.3.2 Stability and Phase Properties of Multidimensional

ARKN Methods

This section is concerned with the stability and phase properties of multidimensional ARKN methods (10.3). In the case of the classical RKN methods, the stability

properties are analysed using the second-order homogeneous linear test equation

y (t) = −λ2 y(t), with λ > 0.

(10.20)

Since the ARKN methods integrate y + M y = 0 exactly, it is pointless to consider

the stability and phase properties of ARKN methods on the basis of the conventional

linear test equation (10.20).

For the stability analysis of multidimensional ARKN methods, we use the following revised test equation [13, 16]:

y (t) + ω2 y(t) = −εy(t), with ε + ω2 > 0,

(10.21)

where ω represents an estimate of the dominant frequency λ and ε = λ2 − ω2 is the

error of that estimation. Applying an ARKN integrator to (10.21), we obtain

z = εh 2 ,

V = h 2 ω2 ,

⎨ Y = eyn + chyn − (V + z)AY,

y

= φ0 (V )yn + φ1 (V )hyn − z b¯ (V )Y,

⎩ n+1

hyn+1 = −V φ1 (V )yn + φ0 (V )hyn − zb (V )Y.

(10.22)

It follows from (10.22) that

yn+1

hyn+1

= R(V, z)

yn

hyn

,

where the stability matrix R(V, z) is given by

R(V, z) =

φ0 (V ) − z b¯ (V )N −1 e φ1 (V ) − z b¯ (V )N −1 c

−V φ1 (V ) − zb (V )N −1 e φ0 (V ) − zb (V )N −1 c

,

with N = I + (V + z)A and e = (1, 1, . . . , 1) .

The spectral radius ρ(R(V, z)) represents the stability of an ARKN method. Since

the stability matrix R(V, z) depends on the variables V and z, the characterization of

stability is determined by two-dimensional regions in the (V, z)-plane. Accordingly,

we have the following definitions of stability for an ARKN method:

(i) Rs = {(V, z)| V > 0, z > 0 and ρ(R) < 1} is called the stability region of an

ARKN method.

(ii) R p = {(V, z)| V > 0, z > 0, ρ(R) = 1 and tr(R)2 < 4det(R)} is called the

periodicity region of an ARKN method.

10.3 ARKN Methods for General Multi-frequency …

Stability Region of Method ARKN3s3

10

8

6

4

Stability Region of Method ARKN6s5

Stability Region of Method ARKN4s4

10

5

0

5

10

V

15

20

z

z

z

2

0

−2

−4

−6

−8

−10

10

8

6

4

2

0

−2

−4

−6

−8

−10

221

0

−5

−10

0

5

10

V

15

20

0

5

10

V

15

20

Fig. 10.1 Stability regions of the methods ARKN3s3 (left), ARKN4s4 (middle), and ARKN6s5

(right)

(iii) If Rs = (0, ∞) × (0, ∞), the ARKN method is called A-stable.

(iv) If R p = (0, ∞) × (0, ∞), the ARKN method is called P-stable.

The stability regions based on the test equation (10.21) for the methods derived

in this section are depicted in Fig. 10.1.

For the integration of oscillatory problems, it is common practice to consider the

phase properties (dispersion order and dissipation order) of the numerical methods.

Definition 10.2 The quantities

tr(R)

, d(H ) = 1 −

2 det(R)

φ(H ) = H − arccos

det(R)

are, respectively, called

√ the dispersion error and the dissipation error of ARKN methods, where H = V + z. Then, a method is said to be dispersive of order q and

dissipative of order r , if φ(H ) = O(H q+1 ) and d(H ) = O(H r +1 ). If φ(H ) = 0

and d(H ) = 0, then the method is said to be zero dispersive and zero dissipative.

The dissipation errors and the dispersion errors for the methods derived in

Sect. 10.3.1 are

• ARKN3s3:

d(H ) =

ε

H 4 + O(H 6 ),

96(ε + ω2 )

φ(H ) = −

• ARKN4s4:

ε

H 5 + O(H 7 );

480(ε + ω2 )

d(H ) =

ε(4ε + 3ω2 ) 6

H + O(H 8 ),

576(ε + ω2 )2

φ(H ) =

ε

H 5 + O(H 7 );

120(ε + ω2 )

222

10 Multidimensional ARKN Methods for General Multi-frequency …

• ARKN6s5:

d(H ) = −

φ(H ) =

ε(154ε + 129ω2 ) 6

H + O(H 8 ),

110880(ε + ω2 )2

ε(44ε + 27ω2 ) 7

H + O(H 9 ).

36960(ε + ω2 )2

10.4 Numerical Experiments

In order to show the remarkable efficiency of the ARKN methods derived in Sect. 10.3

in comparison with some existing methods in the scientific literature, four problems

are considered. The methods we select for comparison are

ARKN3s3: the three-stage ARKN method of order three;

ARKN4s4: the four-stage ARKN method of order four;

ARKN6s5: the six-stage ARKN method of order five;

RKN4s4: the classical four-stage RKN method of order four (see, e.g. II.14 of

[8]);

• RKN6s5: the six-stage RKN method of order five obtained from ARKN6s5 with

V = 0.

For each experiment, we display the efficiency curves of accuracy versus computational cost as measured by the number of function evaluations required by each

method.

Problem 10.1 The linear problem:

2

⎨ y (t) + ω y(t) = −δy (t),

⎩ y(0) = 1, y (0) = − δ .

2

The analytic solution of this initial value problem is given by

2

δ

δ

t⎠ .

y(t) = exp − t cos ⎝ ω2 −

2

4

In this test, the parameter values ω = 1, δ = 10−3 are chosen. The problem is

integrated on the interval [0, 100] with the stepsizes h = 1/2 j , j = 1, 2, 3, 4. The

numerical results are displayed in Fig. 10.2 (left).

10.4 Numerical Experiments

Problem 1: The efficiency curves

−1

ARKN3s3

ARKN4s4

ARKN6s5

RKN4s4

RKN6s5

−2

−1

−2

−3

−4

−5

−6

−7

−8

−9

−10

−11

2.6 2.8

Problem 2: The efficiency curves

ARKN3s3

ARKN4s4

ARKN6s5

RKN4s4

RKN6s5

log10(GE)

−3

log10(GE)

223

−4

−5

−6

−7

−8

−9

−10

2.6

2.8

3.0 3.2 3.4 3.6 3.8 4.0

log10(Function evaluations)

3.0 3.2 3.4 3.6 3.8 4.0

log10(Function evaluations)

Fig. 10.2 Results for Problems 10.1 and 10.2: The log–log plot of maximum global error against

number of function evaluations

Problem 10.2 The van de Pol equation:

2

⎨ y (t) + y(t) = δ(1 − y(t) )y (t),

⎩ y(0) = 2 + 1 δ 2 + 1033 δ 4 + 1019689 δ 6 , y (0) = 0,

96

552960

55738368000

with δ = 0.8 × 10−4 . Integrate the problem on the interval [0, 100] with the stepsizes

h = 1/2 j , j = 1, 2, 3, 4. In order to evaluate the error for each method, a reference

numerical solution is obtained by the method RKN4s4 in II.14 of [8] with a very

small stepsize. The numerical results are displayed in Fig. 10.2 (right).

Problem 10.3 The oscillatory initial value problem

⎨ y (t) +

⎩ y(0) =

12ε

13 −12

y(t) =

−12 13

5

−4

ε

,

, y (0) =

6

ε

with

f 1 (t) =

3 2

−2 −3

y (t) + ε2

f 1 (t)

, 0 < t ≤ tend ,

f 2 (t)

36

24

sin(t) + 24 sin(5t), f 2 (t) = − sin(t) − 36 sin(5t).

5

5

The analytic solution to this problem is given by

y(t) =

sin(t) − sin(5t) + ε cos(t)

.

sin(t) + sin(5t) + ε cos(5t)

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