Applied Mathematics and Computation 182 (2006) 498–508
www.elsevier.com/locate/amc
A Taylor collocation method for the numerical solutionof complex differential equations with mixed conditions
in elliptic domains
Mehmet Sezer *, Mustafa Gulsu, Bekir Tanay
Department of Mathematics, Faculty of Science, Mugla University, 48000 Mugla, Turkey
Abstract
An approximate method for solving higher-order linear complex differential equations in the elliptic domains is pro-posed. The approach is based on a Taylor collocation method, which consists of the matrix representation of expressionsin the differential equation and the collocation points defined in the elliptic domain. Illustrative examples are included todemonstrate the validity and applicability of the technique, and performed on the computer using a program written inMaple9.� 2006 Elsevier Inc. All rights reserved.
Keywords: Complex differential equations; Taylor polynomials and series; Taylor collocation methods
1. Introduction
When a mathematical model is formulated for a physical problem it is often represented by complex dif-ferential equations that are not solvable exactly by analytic techniques. Therefore one must resort to approx-imation and numerical methods. For example, the vibrations of a one-mass system with two degrees offreedom are mostly described using differential equation with a complex dependent variable. The differentialequation is usually linear as is shown in papers [1,2]. The solution of the differential equation clarifies the linearphenomena which occur in the system. The various methods for solving differential equations with complexdependent variable are introduced [2].
In recent years, the studies on complex differential equations, i.e., a geometric approach based on meromor-phic function in arbitrary domains [3], a topological description of solution of some complex differential equa-tions with multivalued coefficients [4], the zero distribution [5] and growth estimates [6] of linear complexdifferential equations, the rational and polynomial approximations of analytic functions in the complex plane[7,8], are developed very rapidly and intensively.
0096-3003/$ - see front matter � 2006 Elsevier Inc. All rights reserved.
doi:10.1016/j.amc.2006.04.011
* Corresponding author.E-mail addresses: [email protected] (M. Sezer), [email protected] (M. Gulsu).
M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508 499
On the other hand, some Taylor and Chebyshev (matrix and collocations) methods to solve linear differ-ential, integral, integro-differential, difference and integro-difference equations, have been presented in manypaper by Sezer et al. [9–16]. Our purpose in this study is to develop and to apply the mentioned methods aboveto the linear complex differential equation:
Xmk¼0
P kðzÞf ðkÞðzÞ ¼ gðzÞ; ð1Þ
which is a generalized case of the complex differential equations given in [5,6,17–19], with the mixedconditions:
Xm�1
k¼0
XJ
j¼0
arkf ðkÞðnjÞ� �
¼ kr; r ¼ 0; 1; . . . ;m� 1 ð2Þ
and to find the solution in terms of the Taylor polynomial form, at the point z = 0,
f ðzÞ ¼XN
n¼0
fnzn; f n ¼f ðnÞð0Þ
n!; z 2 D; N P m: ð3Þ
Here Pk(z) and g(z) are analytical functions in the elliptic domain
D ¼ z ¼ xþ iy; z 2 C; x; y 2 R; x ¼ a cos h; y ¼ b sin h;f0 < h 6 2p;�a 6 x 6 a;�b 6 y 6 b; a; b 2 Rþg;
ark and kr are appropriate complex or real constants; ni 2 D; the coefficients fn, n = 0,1, . . .,N are the Taylorcoefficients to be determined.
2. Determination of collocation points in elliptic domains
We first define the collocation points to be used in the solution method by
zpq ¼ xpq þ iypq ¼ ðxpq; ypqÞ 2 D;
xpq ¼aN
p coshN
q; ypq ¼bN
p sinhN
qð4Þ
so that
p; q ¼ 0; 1; . . . ;N ; 0 < h 6 2p; z0q ¼ z00 ¼ ð0; 0Þ:
Note that for h = 2p, p = 0, q = 0,1, . . .,N, z0q = z00 = 0 + 0i = (0,0); and forh ¼ 2p; p 6¼ 0; q ¼ 0 and q ¼ N ;
zp0 = zpN; that is z10 = z1N, z20 = z2N, . . .,zN0 = zNN.For example, the collocation points for a = 6, b = 3, N = 3, h = 2p are given in Fig. 1.Here for q = 0, q = 3 and p = 0,1,2,3,
z00 ¼ z01 ¼ z02 ¼ z03 ¼ ð0; 0Þ; z10 ¼ x10 þ iy10 ¼ ð2; 0Þ ¼ z13;
z20 ¼ x20 þ iy20 ¼ ð4; 0Þ ¼ z23; z30 ¼ x30 þ iy30 ¼ ð6; 0Þ ¼ z33;
for q = 1 and p = 1, 2, 3,
z11 ¼ x11 þ iy11 ¼ �1;
ffiffiffi3p
2
!; z21 ¼ x21 þ iy21 ¼ ð�2;
ffiffiffi3pÞ; z31 ¼ x31 þ iy31 ¼ �3;
3ffiffiffi3p
2
!;
for q = 2 and p = 1, 2, 3,
z12 ¼ x12 þ iy12 ¼ �1;�ffiffiffi3p
2
!; z22 ¼ x22 þ iy22 ¼ ð�2;�
ffiffiffi3pÞ; z32 ¼ x32 þ iy32 ¼ �3;
�3ffiffiffi3p
2
!:
y
z00 z10 z20
z11
z21
z31
z12
z22
z32
z30
x
Fig. 1. Collocation points for a = 6, b = 3, N = 3, h = 2p.
500 M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508
Another example, the collocation points for a = 6, b = 3, N = 4, h = 2p are
z0q ¼ x0q þ iy0q ¼ ð0; 0Þ; q ¼ 0; 1; 2; 3; 4;
z10 ¼ x10 þ iy10 ¼6
4; 0
� �¼ z14; z20 ¼ x20 þ iy20 ¼
12
4; 0
� �¼ z24;
z30 ¼ x30 þ iy30 ¼18
4; 0
� �¼ z34; z40 ¼ x40 þ iy40 ¼
24
4; 0
� �¼ z44;
z11 ¼ x11 þ iy11 ¼ 0;3
4
� �; z12 ¼
�6
4; 0
� �; z13 ¼ 0;
�3
4
� �;
z21 ¼ x21 þ iy21 ¼ 0;6
4
� �; z22 ¼
�12
4; 0
� �; z13 ¼ 0;
�6
4
� �;
z31 ¼ x31 þ iy31 ¼ 0;9
4
� �; z32 ¼ ð
�18
4; 0Þ; z13 ¼ 0;
�9
4
� �;
z41 ¼ x41 þ iy41 ¼ 0;12
4
� �; z42 ¼
�24
4; 0
� �; z13 ¼ 0;
�12
4
� �:
In the case h = p, N = 3 a = 6 and b = 3, the collocation points become,
zpq ¼ xpq þ iypq; p; q ¼ 0; 1; 2; 3;
xpq ¼ 2p cosp3
q; ypq ¼ p sinp3
q;
so that
z0q ¼ ð0; 0Þ ¼ z00:
In the case h = p, N = 4, a = 6 and b = 3 the collocation points become,
zpq ¼ xpq þ iypq; p; q ¼ 0; 1; 2; 3; 4;
xpq ¼3
2p cos
p4
q; ypq ¼3
4p sin
p4
q;
so that
z0q ¼ ð0; 0Þ ¼ z00:
M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508 501
3. Fundamental matrix relations
We first consider the solution f(z) and its derivative f(k)(z) in the forms
f ðzÞ ¼ f ð0ÞðzÞ ¼XN
n¼0
fnzn; f n ¼f ðnÞð0Þ
n!
and
f ðkÞðzÞ ¼XN
n¼0
f ðkÞn zn: ð5Þ
It is well known from [16] that the relation between the coefficients f ðkþ1Þn and f ðkÞnþ1 is
f ðkþ1Þn ¼ ðnþ 1Þf ðkÞnþ1; n; k ¼ 0; 1; 2; . . . ð6Þ
Then we convert the expressions in (5) to matrix forms, respectively,
½f ðzÞ� ¼ ZðzÞF ð7Þ
andf ðkÞðzÞ� �
¼ ZðzÞFðkÞ ð8Þ
so that
ZðzÞ ¼ 1 z z2 � � � zN� �
;
F ¼ ½ f0 f1 � � � fN �T;
FðkÞ ¼ ½ f ðkÞ0 f ðkÞ1 � � � f ðkÞN �T:
Note that
F ¼ Fð0Þ ¼ ½ f ð0Þ0 f ð0Þ1 � � � f ð0ÞN �T:
In addition, from the recurrence relation (6), it is obtained the matrix relation [16],
Fðkþ1Þ ¼MFðkÞ; k ¼ 0; 1; . . .
or
FðkÞ ¼MkF; ð9Þ
whereM ¼
0 1 0 � � � 0 0
0 0 2 � � � 0 0
� � � � � � � � � �� � � � � � � � � �� � � � � � � � � �0 0 0 � � � N
0 0 0 � � � 0 0
2666666666664
3777777777775; M0 ¼
1 0 0 � � � 0 0
0 1 0 � � � 0 0
� � � � � � � � � �� � � � � � � � � �� � � � � � � � � �0 0 0 � � � 0
0 0 0 � � � 0 1
2666666666664
3777777777775:
Substituting (9) into (8) we have the relation
f ðkÞðzÞ� �
¼ ZðzÞMkF: ð10Þ
For the collocation points z = zpq defined by (4), the matrix relation (10) becomes:
f ðkÞðzpqÞ� �
¼ ZðzpqÞMkF; ð11Þ
502 M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508
where
ZðzpqÞ ¼ ½ 1 zpq z2pq � � � zN
pq �; p; q ¼ 0; 1; . . . ;N :
For p = 0,1, . . .,N we can write the relation (11) in the form
f ðkÞðz0qÞ� �
¼ Zðz0qÞMkF;
f ðkÞðz1qÞ� �
¼ Zðz1qÞMkF;
� � �f ðkÞðzNqÞ� �
¼ ZðzNqÞMkF;
or briefly
FðkÞq ¼
f ðkÞðz0qÞf ðkÞðz1qÞ
..
.
f ðkÞðzNqÞ
2666664
3777775 ¼ ZqMkF; q ¼ 0; 1; . . . ;N ; ð12Þ
where
Zq ¼
Zðz0qÞZðz1qÞ
..
.
ZðzNqÞ
266664
377775 ¼
1 z0q z20q � � � zN
0q
1 z1q z21q � � � zN
1q
� � � �� � � �� � � �1 zNq z2
Nq � � � zNNq
26666666664
37777777775:
On the other hand, substituting the collocation points z = zpq into Eq. (1) we have,
Xm
k¼0
P kðzpqÞf ðkÞðzpqÞ ¼ gðzpqÞ; p; q ¼ 0; 1; . . . ;N : ð13Þ
By using the expressions (11)–(13), we obtain the fundamental matrix equation,
Xm
k¼0
XN
q¼0
PkqZqMkF ¼XN
q¼0
Gq; ð14Þ
so that
Gq ¼ ½ gðz0qÞ gðz1qÞ � � � gðzNqÞ �T
and
Pkq ¼
P kðz0qÞ 0 � � � 0
0 P kðz1qÞ � � � 0
� � �� � �� � �0 0 � � � P kðzNqÞ
2666666664
3777777775:
M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508 503
We can obtain the corresponding matrix form to the conditions (2) as follows. By means of the relation (10)we have the matrix equation,
Xm
k¼0
XJ
j¼0
arkZðnjÞMk
( )F ¼ ½kr�; r ¼ 0; 1; . . . ;m� 1; ð15Þ
where
ZðnjÞ ¼ ½ 1 nj n2j � � � nN
j �:
Briefly, the system of the matrix Eq. (15) can be written in the matrix form,
UrF ¼ ½kr�; ð16Þ
whereUr ¼Xm�1
k¼0
XJ
j¼0
arkZðnjÞMk � ½ ur0 ur1 � � � urN �; r ¼ 0; 1; . . . ;m� 1:
4. Method of solution
We now consider the fundamental matrix Eq. (14) corresponding to Eq. (1). We can write Eq. (14) in theform,
WF ¼ G; ð17Þ
whereW ¼ ½wst� ¼Xm
k¼0
XN
q¼0
PkqZqMk; s; t ¼ 0; 1; . . . ;N
and
G ¼XN
q¼0
Gq � ½ g0 g1 � � � gN �T
Gq is defined in (14). The augmented matrix of Eq. (17) becomes
½W; G� ¼ ½wst; gs�; s; t ¼ 0; 1; . . . ;N : ð18Þ
The augmented matrix of the Eq. (16) corresponding to condition (2) can be written in the form,½Ur; kr� ¼ ½ ur0 ur1 � � � urN; kr � ð19Þ
where Ur is defined in (16).Consequently, to find the unknown Taylor coefficients fn, n = 0,1, . . .,N, related with the approximate solu-
tion of the problem consisting of Eq. (1) and conditions (2), by replacing the m row matrices (19) by the last m
rows of the augmented matrix (18) [9], we have new augmented matrix,
½W�; G�� ¼
w00 w01 � � � w0N ; g0
w10 w11 � � � w1N ; g1
� � � � � � � � � ; � � �wN�m;0 wN�m;1 � � � wN�m;N ; gN�m
u00 u01 � � � u0N ; k0
u10 u11 � � � u1N ; k1
� � � � � � � � � ; � � �um�1;0 um�1;1 � � � um�1;N ; km�1
266666666666664
377777777777775
ð20Þ
504 M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508
or the corresponding matrix equation
W�F ¼ G�:
If detW* 5 0, we can write Eq. (20) as,
F ¼ W�ð Þ�1G�
and the matrix F is uniquely determined. Thus the mth-order linear complex differential equation with variablecoefficients (1) under the conditions (2) has a unique solution in the form (3).
Also we can easily check the accuracy of the obtained solutions as follows [13,15].Since the Taylor polynomial (3) is an approximate solution of Eq. (1), when the solutions f(z) and its deriv-
atives are substituted in Eq. (1), the resulting equation must be satisfied approximately; that is, for z = zj 2 D,j = 0,1,2, . . .,
EðzjÞ ¼Xm
k¼0
P kðzjÞf ðkÞðzjÞ � gðzjÞ�����
����� ffi 0
or
EðzjÞ 6 10�kj ðkj is any positive integerÞ:
If max ð10�kiÞ ¼ 10�k (k is any positive integer) is prescribed, then the truncation limit N is increased until thevalues E(zj) at each of the points zj becomes smaller than the prescribed 10�k.5. Illustrative examples
In this section, several numerical examples are given to illustrate the properties of the method and all ofthem were performed on the computer using a program written in Maple9. The absolute errors in tablesare the values of jf(z) � fN(z)j at selected points.
Example 1 [16, p. 299]. Let us first consider the second order complex differential equation (see Fig. 2):
f 00ðzÞ þ zf ðzÞ ¼ ez þ zez;
with f(0) = 1, f 0(0) = 1 and approximate the solution f(z) by the truncated Taylor series in the form:
f ðzÞ ¼X4
n¼0
fnzn; f n ¼f ðnÞð0Þ
n!;
so that a = 1, b = 1/2, P0(z) = z, P1(z) = 0, P2(z) = 1 and g(z) = ez + zez. For N = 4, and h = p we have thecollocation points
0.5
1
1.5
2
2.5
3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
N=6
exact
Fig. 2. Numerical and exact solution of the Example 1 for N = 6.
TableError
z
0.0 + 00.1 + 00.2 + 00.3 + 00.4 + 00.5 + 00.6 + 00.7 + 00.8 + 00.9 + 01.0 + 1
TableError
z
0.0 + 00.1 + 00.2 + 00.3 + 00.4 + 00.5 + 00.6 + 00.7 + 00.8 + 00.9 + 01.0 + 1
M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508 505
0 0 0 0 014
ffiffi2p
8þffiffi2p
16i 1
8i �
ffiffi2p
8þffiffi2p
16i �1
4
12
ffiffi2p
4þffiffi2p
8i 1
4i �
ffiffi2p
4þffiffi2p
8i � 1
2
34
3ffiffi2p
8þ 3
ffiffi2p
16i 3
8i � 3
ffiffi2p
8þ 3
ffiffi2p
16i �3
4
1ffiffi2p
2þffiffi2p
4i 1
2i �
ffiffi2p
2þffiffi2p
4i �1
8>>>>>>><>>>>>>>:
9>>>>>>>=>>>>>>>;
and the fundamental matrix equation
X2
k¼0
X4
q¼0
PkqZqMkF ¼X4
q¼0
Gq:
This equation has the solution
F ¼ ½ 1 1 0:5 0:1642306718þ 0:001125922559i 0:04292618462þ 0:003526627844i �T:
Therefore, we find the solution
f ðzÞ ¼ 1þ zþ 0:5z2 þ ð0:1642306718þ 0:001125922559iÞz3 þ ð0:04292618462þ 0:003526627844iÞz4:
The values of this solution are compared with the exact solution f(z) = ez for N = 5, 6 in Tables 1–3.
1analysis of Example 1 for the Re(z) value
Exact solution (Real) Present met (z0 = 0)
N = 5 (Real) Ne = 5 N = 6 (Real) Ne = 6
.0i 1.00000000 1.00000000 0.0000000 1.00000000 0.0000000
.1i 1.09964966 1.09965061 0.9500E�6 1.09964966 0.1600E�7
.2i 1.19705602 1.19706225 0.6238E�5 1.19705602 0.5450E�6
.3i 1.28956937 1.28957462 0.5247E�5 1.28956937 0.4566E�5
.4i 1.37406153 1.37400924 0.5229E�4 1.37406153 0.2039E�4
.5i 1.44688903 1.44660517 0.2838E�3 1.44688903 0.6384E�4
.6i 1.50385954 1.50295453 0.9050E�3 1.50385954 0.1583E�3
.7i 1.54020302 1.53793823 0.2264E�2 1.54020302 0.3317E�3
.8i 1.55054929 1.54566154 0.4887E�2 1.55054929 0.6078E�3
.9i 1.52891381 1.51938975 0.9524E�2 1.52891381 0.9909E�3
.0i 1.46869394 1.45148379 0.1721E�1 1.46869394 0.1441E�2
2analysis of Example 1 for the Im(z) value
Exact solution (Im) Present met (z0 = 0)
N = 5 (Im) Ne = 5 N = 6 (Im) Ne = 6
.0i 0.00000000 0.00000000 0.0000000 0.00000000 0.0000000
.1i 0.11033298 0.11033432 0.1336E�5 0.11033292 0.6510E�7
.2i 0.24265526 0.24267093 0.1566E�4 0.24265467 0.5889E�6
.3i 0.39891055 0.39899230 0.8175E�4 0.39890858 0.1968E�5
.4i 0.58094390 0.58123963 0.2957E�3 0.58094018 0.3712E�5
.5i 0.79043908 0.79129120 0.8521E�3 0.79043653 0.2544E�5
.6i 1.02884566 1.03094076 0.2095E�2 1.02885694 0.1127E�4
.7i 1.29729511 1.30187588 0.4580E�2 1.29735515 0.6003E�4
.8i 1.59650534 1.60565633 0.9150E�2 1.59669305 0.1877E�3
.9i 1.92667330 1.94369244 0.1701E�1 1.92714583 0.4725E�3
.0i 2.28735528 2.31722346 0.2986E�1 2.28839852 0.1043E�2
Table 3The absolute error of the considered method at various points for Example 1
z Exact solution ði ¼ffiffiffiffiffiffiffi�1p
Þ Ne = 4 Ne = 5 Ne = 6
0.0 + 0.0i 1.000000000 + 0.0000000000i 0.000000000 0.000000000 0.0000000000.1 + 0.1i 1.099649667 + 0.1103329887i 0.854850E�5 0.163957E�5 0.670373E�70.2 + 0.2i 1.197056021 + 0.2426552686i 0.721149E�4 0.168632E�4 0.802389E�60.3 + 0.3i 1.289569374 + 0.3989105538i 0.243685E�3 0.819207E�4 0.497217E�50.4 + 0.4i 1.374061539 + 0.5809439008i 0.570193E�3 0.300320E�3 0.207311E�40.5 + 0.5i 1.446889037 + 0.7904390832i 0.117417E�2 0.898160E�3 0.638936E�40.6 + 0.6i 1.503859541 + 1.0288456667i 0.246267E�2 0.228221E�2 0.158761E�30.7 + 0.7i 1.540203025 + 1.2972951132i 0.527614E�2 0.511006E�2 0.337166E�30.8 + 0.8i 1.550549297 + 1.5965053441i 0.108800E�2 0.103745E�1 0.636201E�30.9 + 0.9i 1.528913812 + 1.9266733004i 0.210188E�1 0.195027E�1 0.109785E�21.0 + 1.0i 1.468693940 + 2.2873552867i 0.380390E�1 0.344716E�1 0.177916E�2
506 M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508
Example 2. Consider the linear second order complex differential equation:
TableThe ab
z
0.0 + 00.1 + 00.2 + 00.3 + 00.4 + 00.5 + 00.6 + 00.7 + 00.8 + 00.9 + 01.0 + 1
ð1� z2Þf 00ðzÞ � 2zf 0ðzÞ þ 6f ðzÞ ¼ z2 sin z� 2z cos zþ 5 sin z� 3;
with f(0) = �1, f 0(0) = 1 and approximate the solution f(z) by the Taylor polynomial,
f ðzÞ ¼X6
n¼0
fnzn; f n ¼f ðnÞð0Þ
n!;
so that a = 1, b = 1/2, P0(z) = 6, P1(z) = �2z, P2(z) = 1 � z2, h = p and
gðzÞ ¼ z2 sin z� 2z cos zþ 5 sin z� 3:
It can be seen that the exact solution of this problem is f ðzÞ ¼ 32z2 þ sin z� 1.
When the presented method is applying to the problem, the fundamental matrix equation becomes:
X2
k¼0
X6
q¼0
PkqZqMkF ¼X6
q¼0
Gq:
Hence, The comparison of the absolute errors with different N is given in Table 4.
Example 3. Our last example is the linear complex differential equation:
f 000ðzÞ þ 2f 00ðzÞ þ f 0ðzÞ þ f ðzÞ ¼ 2z2 þ 6zþ 11;
4solute error of the considered method at various points for Example 2
Exact solution ði ¼ffiffiffiffiffiffiffi�1p
Þ Ne = 5 Ne = 6 Ne = 7
.0i �1.0000000000 + 0.0000000000i 0.000000000 0.000000000 0.00000000
.1i �0.8996670002 + 0.1296663335i 0.166725E�6 0.135844E�7 0.145845E�8
.2i �0.7973440203 + 0.3173226870i 0.160118E�5 0.107731E�5 0.237736E�7
.3i �0.6910813463 + 0.5609193480i 0.689432E�5 0.666524E�5 0.186524E�6
.4i �0.5790105890 + 0.8583279455i 0.228819E�4 0.182569E�4 0.492563E�5
.5i �0.4593873143 + 1.2073041532i 0.655500E�4 0.604268E�4 0.624268E�4
.6i �0.3306359870 + 1.6054528761i 0.166363E�3 0.337509E�3 0.527501E�4
.7i �0.1913979261 + 2.0501968182i 0.382185E�3 0.242106E�3 0.832109E�4
.8i �0.0405828954 + 2.5387493965i 0.811074E�3 0.111662E�3 0.661666E�3
.9i 0.12257513052 + 3.068093029i 0.161540E�2 0.112954E�2 0.712957E�3
.0i 0.29845758112 + 3.634963915i 0.305327E�2 0.282865E�2 0.492861E�3
M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508 507
with the condition f(0) = 1, f 0(0) = 2, f00(0) = 4. For N = 4, a = 1, b = 1/2 and h = p the matrix form of theproblem is defined by:
X3
k¼0
X4
q¼0
PkqZqMkF ¼X4
q¼0
Gq:
After the augmented matrices of the systems and conditions are computed, we obtain the solution:
F ¼ ½ 1 2 2 0 0 �T:
Therefore, we find the exact solution:
f ðzÞ ¼ 2z2 þ 2zþ 1:
6. Conclusions
High order linear complex differential equations are usually difficult to solve analytically. Then it is requiredto obtain the approximate solutions. For this reason, the present method has been proposed for approximatesolution and also analytical solution.
The method presented in this study is a method for computing the coefficients in the Taylor expansion ofthe solution of a linear complex differential equations, and is valid when the functions Pk(z) and g(z) aredefined in the elliptic domain D = {z 2 C, z = x + iy, x = acosh, y = b sinh, 0 6 h 6 2p, �a 6 x 6 a,� b 6 y 6 b, a, b 2 R}. The Taylor method is an effective method for the cases that the known functions havethe Taylor series expansion at z = z0. In this case, the Taylor polynomial solution f(z) and the values f(zj),zj 2 D can be easily evaluated at low-computation effort. In addition, an interesting feature of this methodis to find the analytical solutions if the equation has an exact solution that is a polynomial of degree N or lessthan N.
The method can also be extended to the system of linear complex differential equations with variable coef-ficients, but some modifications are required.
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