Iğdır Üniversitesi Fen Bilimleri Enstitüsü Dergisi Iğdır University Journal of the Institute of Science and Technology
Araştırma Makalesi / Research Article
Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech. 7(4): 189-201, 2017
Numerical Solution for High-Order Linear Complex Differential Equations By Hermite Polynomials Faruk DÜŞÜNCELİ1, Ercan ÇELİK2 ABSTRACT: In this paper, the numerical solutions of complex differential equations are provided by the Hermite Polynomials and carried on two problems. As a result, the exact solutions and numerical one’s have compared by tables and graphs that the method is practical, reliable and functional. Keywords: Hermite polynomials, linear complex differential equations, numerical solution
Yüksek Mertebeden Lineer Kompleks Diferansiyel Denklemlerin Hermite Polinomları ile Nümerik Çözümleri ÖZET: Bu makalede lineer kompleks diferansiyel denklemleri hermite polinomları vasıtasıyla nümerik çözümünü sağladık ve iki test problemine uyguladık. Tam çözümler ile nümerik çözümleri tablo ve grafikler ile karşılaştırdık. Sonuç olarak metodumuzun güvenilir, pratik ve kullanışlı olduğunu gördük.
ISSN: 2146-0574, e-ISSN: 2536-4618 DOI:
Cilt/Volume: 7, Sayı/Issue: 4, Sayfa/pp: 189-201, 2017
Anahtar Kelimeler:Hermite polinomları, lineer kompleks diferansiyel denklemler, nümerik çözüm
Faruk DÜŞÜNCELİ (0000-0002-2368-7963), Mardin Artuklu Üniversitesi, Mimarlık Fakültesi, Mimarlık Bölümü, Mardin, Türkiye Ercan ÇELİK (0000-0002-1402-1457), Atatürk Üniversitesi, Fen Fakültesi, Matematik Bölümü, Erzurum, Türkiye Sorumlu yazar/Corresponding Author: Faruk DÜŞÜNCELİ,
[email protected] 1 2
Geliş tarihi / Received: 23.05.2017 Kabul tarihi / Accepted: 19.07.2017
20
Different type
21
Yalçınbaş, 2009), Bessel (Yüzbaşı et al., 2011), laguerre (Gülsu et al., 2011), hermite
22 23
of differential equations have been solved with taylor (Sezer and
Faruk DÜŞÜNCELİ ve Ercan ÇELİK
(Yüzbaşı et al., 2011), legendre (Tohidi, 2012; Düşünceli and Çelik, 2015) and Fibonacci polynomials (Düşünceli and Çelik, 2017).Düşünceli and Çelik, 2015) and Fibonacci polynomials INTRODUCTION
(Düşünceli and Çelik, 2017). In this paper, the matrix Different type of differential equations have been between the Hermite and their 24 solved In this the matrix operates between Hermite polynomials andpolynomials their with paper, taylor (Sezer and Yalçınbaş, 2009), Bessel theoperates derivatives, we utilized the Hermite method to solve (Yüzbaşı et al., 2011), laguerre (Gülsu et al., 2011), 25 hermite derivatives, we utilized the Hermite method to solvecomplex linear differential complex equation. differential (Yüzbaşı et al., 2011), legendre (Tohidi, 2012; linear
26
equation. (𝑛𝑛) ∑𝑚𝑚 (𝑧𝑧) = 𝑔𝑔(𝑧𝑧) 𝑛𝑛=0 𝑃𝑃𝑛𝑛 (𝑧𝑧)𝑓𝑓
(1)
𝑓𝑓 (𝑡𝑡) (𝛼𝛼) = 𝜗𝜗𝑡𝑡 𝑡𝑡 = 0,1, … . , 𝑚𝑚 − 1
(2)
27 28
with the initial conditions
29
31
We accept 𝑓𝑓(𝑧𝑧) is unknown function,𝑃𝑃𝑛𝑛 (𝑧𝑧) and 𝑔𝑔(𝑧𝑧) are analytical functions in the
32
appropriate complex or real constant.
33
Suppose that the solution of (1) under the initial conditions (2) is approximated
30
circular
domain
𝑓𝑓(𝑧𝑧) = ∑𝑁𝑁 𝑛𝑛=0 𝑎𝑎𝑛𝑛 𝐻𝐻𝑛𝑛 (𝑧𝑧), 𝑧𝑧 ∈ 𝐷𝐷
34 35 36
which𝐷𝐷 = {𝑧𝑧 = 𝑥𝑥 + 𝑖𝑖𝑖𝑖, 𝑧𝑧 ∈ 𝐶𝐶, |𝑧𝑧| ≤ 𝑟𝑟, 𝑟𝑟 ∈ 𝑅𝑅 + };
(3)
Hermite coefficients to be determined.Hermite polynomials defined by 𝑛𝑛 2
𝐻𝐻𝑛𝑛 (𝑧𝑧) = 𝑛𝑛! ∑
𝑚𝑚=0
Where[ 2] = 𝑛𝑛
𝑛𝑛 2
if 𝑛𝑛 is even and [ 2] = 𝑛𝑛
𝑟𝑟
𝑖𝑖𝑖𝑖
(−1)𝑚𝑚 (2𝑧𝑧)𝑛𝑛−2𝑚𝑚 𝑚𝑚! (𝑛𝑛 − 2𝑚𝑚)!
𝑛𝑛−1 2
if 𝑛𝑛 is odd and we use the collocation points 3
𝑧𝑧𝑝𝑝𝑝𝑝 = 𝑁𝑁 𝑝𝑝𝑒𝑒 𝑁𝑁 𝑝𝑝 , 0 < 𝜃𝜃 ≤ 2𝜋𝜋, 𝑟𝑟 ∈ 𝑅𝑅 + , 𝑝𝑝 ∈ 0,1, … , 𝑁𝑁
38 39
MATERIAL AND METHOD
40
We can write the desired solution 𝑓𝑓(𝑧𝑧) of Equation (3)
190
41
𝜗𝜗𝑡𝑡 is
which is the Hermite series of the unknown function 𝑓𝑓(𝑧𝑧), where all of 𝑎𝑎𝑛𝑛 are the [ ]
37
𝛼𝛼 ∈ 𝐷𝐷,
(4)
Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.
𝑓𝑓(𝑧𝑧) = 𝐻𝐻(𝑧𝑧)𝐴𝐴
(5)
37
Where[ 2] = 𝑛𝑛
𝑛𝑛 2
if 𝑛𝑛 is even and [ 2] = 𝑛𝑛
𝑛𝑛−1 2
𝑖𝑖𝑖𝑖
if 𝑛𝑛 is odd and we use the collocation points
Solution Hermite Polynomials 𝑧𝑧𝑝𝑝𝑝𝑝 = 𝑁𝑁 𝑝𝑝𝑒𝑒 𝑁𝑁 𝑝𝑝 Numerical , 0 < 𝜃𝜃 ≤ 2𝜋𝜋, 𝑟𝑟for∈High-Order 𝑅𝑅 + , 𝑝𝑝 ∈ Linear 0,1, …Complex , 𝑁𝑁 Differential Equations By (4) 𝑟𝑟
38 39
MATERIAL AND METHOD
40
We can write the desired solution 𝑓𝑓(𝑧𝑧) of Equation (3)
41 42
43
44 45 46
47
where
𝐴𝐴 = [𝑎𝑎0
𝑎𝑎1
… 𝐻𝐻𝑁𝑁 (𝑧𝑧)]
… 𝑎𝑎𝑁𝑁 ]𝑇𝑇
The Hermite polynomials𝐻𝐻𝑛𝑛 (𝑧𝑧) can be formed in matrix form as
𝑍𝑍(𝑧𝑧) = [1
and whether N is odd,
[
(6)
𝐻𝐻(𝑧𝑧) = 𝑍𝑍(𝑧𝑧)𝐵𝐵𝑇𝑇
where
=
49
𝐻𝐻(𝑧𝑧) = [𝐻𝐻0 (𝑧𝑧) 𝐻𝐻1 (𝑧𝑧)
and
𝐵𝐵
48
(5)
𝑓𝑓(𝑧𝑧) = 𝐻𝐻(𝑧𝑧)𝐴𝐴
0!
(−1)0 0 2 0! 0!
2!
(−1)1 0 2 1! 0!
4!
(−1)2 0 2 2! 0! ⋮
0
0
0
1! 3!
𝑛𝑛!
0
(−1)0 0! 1! 0
(−1)1 1! 1! 0 ⋮
(−1)( (
𝑛𝑛−1 2
21 21
𝑛𝑛−1 ) 2
) ! 1!
21
𝑧𝑧 2
𝑧𝑧
0
0
2!
(−1)0 2 2 0! 2!
4!
(−1)1 2 2 1! 2! ⋮
0
0
⋯
𝑧𝑧 𝑁𝑁 ]
0
0
0
0
0
3!
(−1)0 0! 3! 0 ⋮
(−1)(
23
𝑛𝑛−3 ) 2
𝑛𝑛! 4 𝑛𝑛−3 23 ( ) ! 3! 2
4!
…
0
⋯
0
0
⋯
0
⋯
(−1)0 4 2 0! 4! ⋮ 0
⋯ ⋱
0
0
0 ⋮
(−1)0 𝑛𝑛 ⋯ 𝑛𝑛! 2 0! 𝑛𝑛!
whether N is even,
191
Cilt / Volume: 7, Sayı / Issue: 4, 2017
0!
(−1)0 0 2 0! 0!
0
]𝑁𝑁+1xN+1
0
0
…
0
0
48 49
0
[
𝑛𝑛!
(−1)( (
𝑛𝑛−1 2
𝑛𝑛−1 ) 2
) ! 1!
1
2
0
𝑛𝑛!
Faruk DÜŞÜNCELİ ve Ercan ÇELİK
[
0!
(−1)0 0 2 0! 0!
2!
(−1)1 0 2 1! 0!
4!
(−1)2 0 2 2! 0! ⋮
𝑛𝑛!
0
0
𝑛𝑛 2
( )
(−1) 𝑛𝑛
( ) ! 0! 2
20
the
1! 3!
0
(−1)0 1 2 0! 1! 0
(−1)1 1! 1! 0
21
⋮
0
relation
2!
(−1)0 2 2 0! 2!
4!
(−1)1 2 2 1! 2! ⋮
𝑛𝑛!
(−1) (
𝑛𝑛−2 2
(
0
0
…
0
0
⋯
0
0
⋯ 𝑛𝑛!
3!
22
between
the
⋯
0
(−1)0 0! 3! 0
23
⋮
0
0
]𝑁𝑁+1xN+1
where 0 1 0 0 𝐾𝐾 = 0 0 ⋮ 0 [0
0 0 2 0 0 0 ⋮ 0 0
0 0 0 3 0 0 ⋮ 0 0
0 0 05 0 4 0 ⋮ 0 0
0 0 0 0 0 5 ⋮ 0 0
0
⋯
0
0
⋱
⋮
⋮
⋯ ⋯ 𝑛𝑛!
matrix
0
(−1)0 0! 𝑛𝑛!
𝐻𝐻(𝑧𝑧)
𝐻𝐻 ′ (𝑧𝑧) = 𝐻𝐻(𝑧𝑧)𝐾𝐾 𝑇𝑇 𝐻𝐻 (2) (𝑧𝑧) = 𝐻𝐻(𝑧𝑧)(𝐾𝐾 𝑇𝑇 )2 ⋮ 𝐻𝐻 (𝑛𝑛) (𝑧𝑧) = 𝐻𝐻(𝑧𝑧)(𝐾𝐾 𝑇𝑇 )𝑛𝑛
52
0
2𝑛𝑛
0
]𝑁𝑁+1xN+1
and
its
(7)
… 0 ⋯ 0 ⋯ 0 ⋯ 0 ⋯ 0 ⋯ 0 ⋱ ⋮ ⋯ 𝑁𝑁 ⋯ 0 ]𝑁𝑁+1xN+1
By using the relations (6) and (7) we obtain the relation
𝑓𝑓 (𝑛𝑛) (𝑧𝑧) = 𝐻𝐻 (𝑛𝑛) (𝑧𝑧)𝐾𝐾 𝑇𝑇 𝐴𝐴 = 𝐻𝐻(𝑧𝑧)(𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴 = 𝑍𝑍(𝑧𝑧)𝐵𝐵 𝑇𝑇 (𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴
55
59
𝑛𝑛−2 ) 2
) ! 2!
derivatives𝐻𝐻 ′ (𝑧𝑧), 𝐻𝐻 (2) (𝑧𝑧), … , 𝐻𝐻 (𝑛𝑛) (𝑧𝑧)are
58
) ! 3!
(−1)0 𝑛𝑛 2 0! 𝑛𝑛!
23
0
0
51
57
2
0
Then,
56
𝑛𝑛−3
0
50
54
(
𝑛𝑛−3 ) 2
whether N is even,
𝐵𝐵 =
53
(−1)(
(8)
By changing the collocation points 𝑧𝑧 = 𝑧𝑧𝑝𝑝𝑝𝑝 into the relation (8), we get the following matrix equations 192
Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.
𝑓𝑓 (𝑛𝑛) (𝑧𝑧𝑝𝑝𝑝𝑝 ) = 𝑍𝑍(𝑧𝑧𝑝𝑝𝑝𝑝 )𝐵𝐵𝑇𝑇 (𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴, 𝑝𝑝 ∈ 0,1, … , 𝑁𝑁
For 𝑝𝑝 = 0,1, . . . , 𝑁𝑁, we can write the relation (9)
(9)
𝑓𝑓
55 56 57
60
61
𝑓𝑓 (𝑛𝑛) (𝑧𝑧𝑝𝑝𝑝𝑝 ) = 𝑍𝑍(𝑧𝑧𝑝𝑝𝑝𝑝 )𝐵𝐵𝑇𝑇 (𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴, 𝑝𝑝 ∈ 0,1, … , 𝑁𝑁
66
67 68
𝑓𝑓 (𝑛𝑛) (𝑧𝑧00 ) = 𝑍𝑍(𝑧𝑧00 )𝐵𝐵 𝑇𝑇 (𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴 𝑓𝑓 (𝑛𝑛) (𝑧𝑧11 ) = 𝑍𝑍(𝑧𝑧11 )𝐵𝐵 𝑇𝑇 (𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴 ⋮ (𝑛𝑛) 𝑓𝑓 (𝑧𝑧𝑁𝑁𝑁𝑁 ) = 𝑍𝑍(𝑧𝑧𝑁𝑁𝑁𝑁 )𝐵𝐵 𝑇𝑇 (𝐾𝐾 𝑇𝑇 )𝑛𝑛 𝐴𝐴
where
𝑍𝑍0 (𝑧𝑧00 ) 𝑍𝑍00 𝑍𝑍 𝑍𝑍 (𝑧𝑧 ) 𝑍𝑍 = [ 11 ] = [ 0 11 ⋮ ⋮ 𝑍𝑍𝑁𝑁𝑁𝑁 𝑍𝑍0 (𝑧𝑧𝑁𝑁𝑁𝑁 )
71 72
𝑍𝑍1 (𝑧𝑧00 ) 𝑍𝑍1 (𝑧𝑧11 ) ⋮ 𝑍𝑍1 (𝑧𝑧𝑁𝑁𝑁𝑁 )
𝑍𝑍2 (𝑧𝑧00 ) 𝑍𝑍2 (𝑧𝑧11 ) ⋮ 𝑍𝑍2 (𝑧𝑧𝑁𝑁𝑁𝑁 )
Let us modify the collocation points (4) into equation(1),
⋯ ⋯ ⋮ ⋯
𝑍𝑍𝑁𝑁 (𝑧𝑧00 ) 𝑍𝑍𝑁𝑁 (𝑧𝑧11 ) ] ⋮ 𝑍𝑍𝑁𝑁 (𝑧𝑧𝑁𝑁𝑁𝑁 ) (10)
(𝑛𝑛) ∑𝑚𝑚 (𝑧𝑧𝑝𝑝𝑝𝑝 )𝐴𝐴 = 𝑔𝑔(𝑧𝑧𝑝𝑝𝑝𝑝 ) 𝑛𝑛=0 𝑃𝑃𝑛𝑛 (𝑧𝑧𝑝𝑝𝑝𝑝 )𝑓𝑓
We attain the basic matrix equation of the relations (8)–(10),
where
(11)
𝑁𝑁 𝑁𝑁 𝑇𝑇 𝑇𝑇 𝑛𝑛 ∑𝑚𝑚 𝑛𝑛=0 ∑𝑝𝑝=0 𝑃𝑃𝑛𝑛 (𝑧𝑧𝑝𝑝𝑝𝑝 )𝑍𝑍(𝑧𝑧𝑝𝑝𝑝𝑝 )𝐵𝐵 (𝐾𝐾 ) A = ∑𝑝𝑝=0 𝐺𝐺𝑝𝑝
𝑃𝑃𝑛𝑛 (𝑧𝑧00 ) 0 𝑃𝑃𝑛𝑛 = [ ⋮ 0
6
0 𝑃𝑃𝑛𝑛 (𝑧𝑧11 ) ⋮ 0
⋯ ⋯ ⋮ ⋯
0 𝑔𝑔(𝑧𝑧00 ) 0 𝑔𝑔(𝑧𝑧11 ) ]and𝐺𝐺𝑝𝑝 = [ ] ⋮ ⋮ 𝑃𝑃𝑛𝑛 (𝑧𝑧𝑁𝑁𝑁𝑁 ) 𝑔𝑔(𝑧𝑧𝑁𝑁𝑁𝑁 )
Since the A is unknown and should be determined that the matrix equation (11) could be rewritten in the subsequent form:
69 70
where,
𝑊𝑊𝑊𝑊 = 𝐺𝐺or[𝑊𝑊; 𝐺𝐺] = [𝑤𝑤𝑝𝑝𝑝𝑝 ; 𝑔𝑔𝑝𝑝 ] 𝑝𝑝, 𝑞𝑞 = 0,1, … , 𝑁𝑁
𝑁𝑁 𝑇𝑇 𝑇𝑇 𝑛𝑛 𝑎𝑎 𝑊𝑊 = ∑𝑚𝑚 𝑛𝑛=0 ∑𝑝𝑝=0 𝑃𝑃𝑛𝑛 (𝑧𝑧𝑝𝑝𝑝𝑝 )𝑍𝑍(𝑧𝑧𝑝𝑝𝑝𝑝 )𝐵𝐵 (𝐾𝐾 ) and𝐴𝐴 = [ 0
𝑎𝑎1
…
𝑎𝑎𝑁𝑁 ]𝑇𝑇
We write the matrix shape of the initial conditions (2) by the aid of (8), Cilt / Volume: 7, Sayı / Issue: (𝑡𝑡)4, 2017
73
(9)
For 𝑝𝑝 = 0,1, . . . , 𝑁𝑁, we can write the relation (9)
64 65
(8)
Numerical Solution for High-Order Linear Complex Differential Equations By Hermite Polynomials
matrix equations
62 63
(𝑧𝑧)𝐾𝐾 𝐴𝐴 = 𝐻𝐻(𝑧𝑧)(𝐾𝐾 ) 𝐴𝐴 = 𝑍𝑍(𝑧𝑧)𝐵𝐵 (𝐾𝐾 ) 𝐴𝐴
By changing the collocation points 𝑧𝑧 = 𝑧𝑧𝑝𝑝𝑝𝑝 into the relation (8), we get the following
58 59
(𝑧𝑧) = 𝐻𝐻
𝑓𝑓
(𝛼𝛼) = 𝐻𝐻(𝛼𝛼)(𝐾𝐾 𝑇𝑇 )𝑡𝑡 𝐴𝐴 = 𝜗𝜗𝑡𝑡 𝑡𝑡 = 0,1, … . , 𝑚𝑚 − 1
In another hand the matrix shape of the initial conditions could be reformed as
(12)
or real constant. 193
69 70 71 72
73
74
75
where,
78 79 80
𝑁𝑁 𝑇𝑇 𝑇𝑇 𝑛𝑛 𝑎𝑎 𝑊𝑊 = ∑𝑚𝑚 𝑛𝑛=0 ∑𝑝𝑝=0 𝑃𝑃𝑛𝑛 (𝑧𝑧𝑝𝑝𝑝𝑝 )𝑍𝑍(𝑧𝑧𝑝𝑝𝑝𝑝 )𝐵𝐵 (𝐾𝐾 ) and𝐴𝐴 = [ 0
In another hand the matrix shape of the initial conditions could be reformed as
87 88 89
𝑈𝑈𝑡𝑡 𝐴𝐴 = 𝜗𝜗𝑡𝑡 𝑡𝑡 = 0,1, … . , 𝑚𝑚 − 1
where
𝑈𝑈𝑡𝑡 = 𝐻𝐻(𝛼𝛼)(𝐾𝐾 𝑇𝑇 )𝑡𝑡 𝑡𝑡 = 0,1, … . , 𝑚𝑚 − 1
the augmented form of these equations are
[𝑈𝑈𝑡𝑡 ; 𝜗𝜗𝑡𝑡 ] = [𝑢𝑢𝑡𝑡0 , 𝑢𝑢𝑡𝑡1 , … , 𝑢𝑢𝑡𝑡𝑡𝑡 ; 𝜗𝜗𝑡𝑡 ] 𝑡𝑡 = 0,1, … . , 𝑚𝑚 − 1
(13)
Finally, to find the unknown Hermite coefficients𝑎𝑎𝑛𝑛 , 𝑛𝑛 = 0, 1, . . . , 𝑁𝑁,connected to the
approximate solution of the problem (1) under the initial conditions (2), we require to
Finally, to find the unknown Hermite require to replace the𝑚 rows of (13) by the last 𝑚 rows coefficientsconnected to the approximate solution of the augmented matrix (12) hence we have new replace the𝑚𝑚 rows of (13) by the last 𝑚𝑚 rows of theofaugmented matrix (12) andand hence the problem (1) under the initial conditions (2), we augmented matrix
we have new augmented matrix
̃ ; 𝐺𝐺̃ ] = [𝑊𝑊
𝑤𝑤00 𝑤𝑤10 ⋮
𝑤𝑤𝑁𝑁−𝑚𝑚 0 𝑢𝑢00 𝑢𝑢10 ⋮ [ 𝑢𝑢𝑚𝑚−1 0
𝑤𝑤01 𝑤𝑤11 ⋮
𝑤𝑤𝑁𝑁−𝑚𝑚 1 𝑢𝑢01 𝑢𝑢11 ⋮ 𝑢𝑢𝑚𝑚−1 1
⋯ 𝑤𝑤0𝑁𝑁 ⋯ 𝑤𝑤1𝑁𝑁 ⋮7 ⋮ ⋯ 𝑤𝑤𝑁𝑁−𝑚𝑚 𝑁𝑁 ⋯ 𝑢𝑢0𝑁𝑁 ⋯ 𝑢𝑢1𝑁𝑁 ⋯ ⋮ ⋯ 𝑢𝑢𝑚𝑚−1 𝑁𝑁
; 𝑔𝑔0 ; 𝑔𝑔1 ⋮ ⋮ ; 𝑔𝑔𝑁𝑁 ; 𝜗𝜗0 ; 𝜗𝜗1 ⋮ ⋮ ; 𝜗𝜗𝑚𝑚−1 ]
(14)
or the matrix equation ̃ 𝐴𝐴 = 𝐺𝐺̃ 𝑊𝑊
84
86
… 𝑎𝑎𝑁𝑁 ]𝑇𝑇
𝑓𝑓 (𝑡𝑡) (𝛼𝛼) = 𝐻𝐻(𝛼𝛼)(𝐾𝐾 𝑇𝑇 )𝑡𝑡 𝐴𝐴 = 𝜗𝜗𝑡𝑡 𝑡𝑡 = 0,1, … . , 𝑚𝑚 − 1
82
85
𝑎𝑎1
We write the matrix shape of the initial conditions (2) by the aid of (8),
81
83
(12)
Faruk DÜŞÜNCELİ ve Ercan ÇELİK
76 77
𝑊𝑊𝑊𝑊 = 𝐺𝐺or[𝑊𝑊; 𝐺𝐺] = [𝑤𝑤𝑝𝑝𝑝𝑝 ; 𝑔𝑔𝑝𝑝 ] 𝑝𝑝, 𝑞𝑞 = 0,1, … , 𝑁𝑁
(15)
̃ ) ≠ 0 can rewrite (15) in the form 𝐴𝐴 = 𝑊𝑊 ̃ −1 𝐺𝐺̃ and the Ais uniquely set. The If 𝑑𝑑𝑑𝑑𝑑𝑑(𝑊𝑊 solution is given by the Hermite series are determined. And so on, the 𝑚𝑚th order linear
If can rewrite (15) in the form and the Ais uniquely an approximated. We can control the precision of set. The solution is given by theunder Hermite the acquired Since We the can Hermite series are complex differential equation theseries initialare conditions has an solutions. approximated. determined. And so on, the 𝑚th order linear complex numerical solution of (1). So the equation should be control theequation precision of the solutions. the Hermiteefficient series for are numerical differential under the acquired initial conditions hasSince approximately
solution of (1). So the equation should be approximately efficient for Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.
194
𝑧𝑧 = 𝑧𝑧𝑗𝑗 ∈ 𝐷𝐷, 𝑗𝑗 = 0, 1, 2, . ..
87
complex differential equation under the initial conditions has an approximated. We can
88
control the precision of the acquired solutions. Since the Hermite series are numerical
89
solution of (1). So the equation should be approximately efficient for
Numerical Solution for High-Order Linear Complex Differential Equations By Hermite Polynomials
𝑧𝑧 = 𝑧𝑧𝑗𝑗 ∈ 𝐷𝐷, 𝑗𝑗 = 0, 1, 2, . ..
(𝑛𝑛) 𝐸𝐸(𝑧𝑧𝑗𝑗 ) = |∑𝑚𝑚 (𝑧𝑧𝑗𝑗 ) − 𝑔𝑔(𝑧𝑧𝑗𝑗 )| ≅ 0 𝑛𝑛=0 𝑃𝑃𝑛𝑛 (𝑧𝑧𝑗𝑗 )𝑓𝑓
90 91
or
92
𝐸𝐸(𝑧𝑧𝑗𝑗 ) ≤ 10−𝑙𝑙𝑗𝑗 (𝑙𝑙𝑗𝑗 is any positive integer).
(16)
94
If 𝑚𝑚𝑚𝑚𝑚𝑚10−𝑙𝑙𝑗𝑗 = 10−𝑙𝑙 is described, then the truncation limit N is put on until the values
95
𝐸𝐸(𝑧𝑧𝑗𝑗 ) at eachof the points 𝑧𝑧𝑗𝑗 becomes smaller than the prescribed 10−𝑙𝑙 . RESULTS AND DISCUSSION
96
In this part, two examples are given to illustrate the accuracy and effectiveness of the
97
proposed way and all of them are complemented on a computer by using programs
98
typed in Matlab. Therefore, we have recorded in tables, the values of the exact solution
93
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥(573/9223372036854775808 + (21𝑖𝑖)/144115188075855872) + 𝑦𝑦(− 21/144115188075855872
+ (573𝑖𝑖)/9223372036854775808) 8
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)2 (− 18014398509481987/36028797018963968
+ (53𝑖𝑖)/576460752303423488)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)3 (− 159/2305843009213693952
− (7𝑖𝑖)/36028797018963968)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)4 (1485316528861191/36028797018963968 − (11906580003201𝑖𝑖)/36028797018963968)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)5 (1666248956492659/2305843009213693952
− (41586247975829𝑖𝑖)/36028797018963968)
+ 144115188075855857/144115188075855872 110
For N=10,
− (19𝑖𝑖)/2305843009213693952
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195
Faruk DÜŞÜNCELİ ve Ercan ÇELİK
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥(13882071/147573952589676412928
− (6532991𝑖𝑖)/18446744073709551616) + 𝑦𝑦(6532991/18446744073709551616
+ (13882071𝑖𝑖)/147573952589676412928)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)2 (− 2305843009202906759/4611686018427387904
− (3280233𝑖𝑖)/2305843009213693952)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)3 (− 1814333/18446744073709551616 + (735701𝑖𝑖)/2305843009213693952)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)4 (47986826942415845/1152921504606846976 + (38497718388899𝑖𝑖)/576460752303423488)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)5 (23264094237931/18446744073709551616 − (34829294771𝑖𝑖)/2305843009213693952)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)6 (− 792491809524217/576460752303423488 − (2823403318151𝑖𝑖)/288230376151711744)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)7 (248913431329/4611686018427387904
− (159064944217𝑖𝑖)/576460752303423488)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)8 (13170887529351/576460752303423488 + (378968311233𝑖𝑖)/288230376151711744)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)9 (− 658775106425/9223372036854775808
+ (4512177617𝑖𝑖)/1152921504606846976)
+ (𝑥𝑥 + 𝑦𝑦𝑦𝑦)10 (− 29038783659/288230376151711744 − (23378868389𝑖𝑖)/144115188075855872)
+ 9223372036851657523/9223372036854775808
+ (1994837𝑖𝑖)/4611686018427387904 196
11
Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.
Numerical Solution for High-Order Linear Complex Differential Equations By Hermite Polynomials
111
The solutions of the linear complex differential equation for N = 3,5 and 10 are
112
obtained. The absolute errors shown in Tables 1,2 and in Figures 1,2.
113
Table 1 Comparison of real parts of the exact solution and numerical one’s for some values 𝑧𝑧𝑗𝑗
114 115 116
-.9(1+i) -.8(1+i) -.7(1+i) -.6(1+i) -.5(1+i) -.4(1+i) -.3(1+i) -.2(1+i) -.1(1+i) 0(1+i) .1(1+i) .2(1+i) .3(1+i) .4(1+i) .5(1+i) .6(1+i) .7(1+i) .8(1+i) .9(1+i)
.8908207824 .9317999001 .9600062080 .9784066646 .9895848832 .9957335935 .9986500259 .9997333347 .9999833333 1 .9999833333 .9997333347 .9986500259 .9957335935 .9895848832 .978406646 .960006208 .931799900 .890820782
N=3
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
N=5
.6216215573 .6966493057 .7647768674 .8252866747 .8775540316 .9210479812 .9553321731 .9800657301 .9950041154 1 .9950041298 .9800661925 .9553356851 .9210627805 .8775991953 .8253990566 .7650197690 .6971228821 .6224749574
N=10
.8908892455 .9318501705 .9600393451 .9895949000 .9895949000 .9957378882 .9986514289 .9997336172 .9999833512 1 .9999833511 .9997336139 .9986514034 .9957377771 .9895945485 .9784252442 .9600372394 .9318457566 .8908805837
Table 2 Comparison of imaginer parts of the exact solution and numerical one’s for some
values 𝑧𝑧𝑗𝑗
-.9(1+i)
117
Exact solution(Real)
-.8(1+i) -.7(1+i) -.6(1+i) -.5(1+i) -.4(1+i) -.3(1+i) -.2(1+i) -.1(1+i) 0(1+i) .1(1+i) .2(1+i) .3(1+i) .4(1+i) .5(1+i) .6(1+i) .7(1+i) .8(1+i) .9(1+i)
Exact solution(Im.) -.8040981746
-.6370882357 - .4886930379 - .3594816533 - .2498263975 - .1599544898 - .0899919000 - .0399992888 - .0099999888 0 -.0099999888 -.0399992888 -.0899919000 -.1599544898 -.2498263975 -.3594816533 -.4886930379 -.6370882357 -.8040981746
Cilt / Volume: 7, Sayı / Issue: 4, 2017
N=3
N=5
N=10
-.81
-.8101521967
-.8043179038
-.64 -.49 - .36 - .25 - .16 - .09 - .04 - .01 0 -.01 -.04 -.09 -.16 -.25 -.36 -.49 -.64 -.81
12
- .6400242974 - .4899727890 - .3599629365 - .2499713353 - .1599838390 - .0899934881 - .0399984374 - .0099998850 0 -.009999850 -.0399973324 -.0899850972 -.1599484800 -.2498634277 -.3596944279 -.4893924366 -.6388928054 -.8081132113
-.6372208337 - .4887680105 - .3595206898 - .2498445938 - .1599617215 - .0899941321 - .0399997218 - .0100000001 0 -.0100000000 -.0399997251 -.0899941575 -.1599618328 -.2498449507 -.3595216345 -.4887702076 -.6372254971 -.8043271515
197
Faruk DÜŞÜNCELİ ve Ercan ÇELİK
118 119
Figure 1. The real parts of the absolute errors functions
120
121 122 123
Figure 2. The imaginer parts of the absolute errors functions Example 2:Finally, consider the linear complex differential equation 198
13
Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.
Numerical Solution for High-Order Linear Complex Differential Equations By Hermite Polynomials
𝑓𝑓 ′′ (𝑧𝑧) + 2𝑓𝑓 ′ (𝑧𝑧) + 𝑧𝑧𝑧𝑧(𝑧𝑧) = z 3 + 𝑒𝑒 𝑧𝑧 (𝑧𝑧 + 3) + 6𝑧𝑧 + 2
125
with initial conditions 𝑓𝑓(0) = 3
126
𝑒𝑒 𝑧𝑧 (𝑧𝑧 + 3) + 6𝑧𝑧 + 2. The next step of our method, we obtain the numerical solution for N
= 3,5,10. The values of the numerical solution in the issue of N = 3,5,10 for both parts
127
of real and imaginary together with the exact solution and absolute errors are supported
128
in figures 3,4,5,6 as follows.
124
129 130 131
, 𝑓𝑓 ′ (0) = 1. The exact solution is 𝑓𝑓(𝑧𝑧) = 𝑧𝑧 2 +
Figure 3. The real parts of the exact solution and numerical one’s
Cilt / Volume: 7, Sayı / Issue: 4, 2017
14
199
Faruk DÜŞÜNCELİ ve Ercan ÇELİK
132 133
Figure 4. The imaginer parts of the exact solution and numerical one’s
134 135
136 137
Figure 5. The real parts of the absolute errors functions 200
15
Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.
Numerical Solution for High-Order Linear Complex Differential Equations By Hermite Polynomials
138
139 140 141 142
Figure 6. The imaginer parts of the absolute errors functions CONCLUSIONS
CONCLUSIONS is righteous. It may be concluded that the method is an 143 This way is established on reckoning the coefficients in the Hermite numerical series extension efficient way to discover solutionsoffor linear This way is established on reckoning the coefficients complex differential equations. On the other hand, the in144 the Hermite series extension of the solution of a linear equations the solution of a linear complex differential providing the circular domain is results are quite reliable and good-agreement with the complex differential equations providing the circular 145 isidentified is righteous. It may be concluded domain identifiedby by the the functions functions 𝑃𝑃 and . . and This𝐺𝐺(𝑧𝑧). way . This exactway solutions. 𝑛𝑛 (𝑧𝑧)
146
that the method is an efficient way to discover numerical solutions for linear complex
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