# Lucas Polynomial Approach for System of High-Order Linear Differential Equations and Residual Error Estimation

Authors: Muhammed Çetin; Mehmet Sezer; Coşkun Güler
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

An approximation method based on Lucas polynomials is presented for the solution of the system of high-order linear differential equations with variable coefficients under the mixed conditions. This method transforms the system of ordinary differential equations (ODEs) to the linear algebraic equations system by expanding the approximate solutions in terms of the Lucas polynomials with unknown coefficients and by using the matrix operations and collocation points. In addition, the error analysis based on residual function is developed for present method. To demonstrate the efficiency and accuracy of the method, numerical examples are given with the help of computer programmes written inMaple and Matlab.


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## Body

## 1. Introduction

The systems of differential equations with variable coefficients have been encountered in many scientific and technological problems. Some of these differential equation systems do not have analytic solutions, so numerical methods are required. The systems of linear differential equations have been solved by many mathematicians and engineers by using the various methods such as variational iteration method [1], the differential transform method [2–5], the Adomian decomposition method [6, 7] and the linearizability criteria [8, 9], finite difference method [10], and Adomian-Pade technique [11].Taylor, Chebyshev, Legendre, Berstein, Hermite, Laguerre, and Bessel matrix methods are used for solving differential and integral equations, integrodifferential-difference equations, and their systems in [12–20]. In this paper, by means of the above-mentioned methods and the Lucas polynomials, we have developed a new method called Lucas collocation method to solve the system of high-order linear differential equations with variable coefficients in the form
(1)Lyi(x)=∑n=0m∑j=1kpi,jn(x)yj(n)(x)=gix,i=1,2,…,k,0≤a≤x≤b
under the mixed conditions
(2)∑n=0m-1ai,njyjna+bi,njyjnb=cj,i,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1,
where yj(0)(x)=yj(x) is an unknown function, pi,jn(x) and gi(x) are the known continuous functions defined on interval [a,b], and coefficients ai,nj, bi,nj and cj,i are the real constants.In addition, by improving the present method with the help of the residual error function used in [21–25], we obtain the corrected approximate solutions of the system (1) expressed in the truncated Lucas series form
(3)yj,N,Mx=yj,Nx+ej,N,Mx,j=1,2,…,k,
where
(4)yj≅yj,N(x)=∑n=0Naj,nLn(x)
is the Lucas polynomial solution and
(5)ej,N,M(x)=∑n=0Maj,n*Ln(x),(M>N)
is the solution of the error problem obtained with the aid of the residual error function. Here aj,n,  (n=0,1,2,…,N) and aj,n*,  (n=0,1,2,…,N) are the unknown Lucas coefficients, and Ln(x),  n=0,1,…,N are the Lucas polynomials defined by
(6)L0x=2;Lnx=∑k=0n/2nn-kn-kkxn-2k,n≥1,n2=n2,neven(n-1)2,nodd
[26–28]. The purpose of this study is to improve the approximate solutions for high-order systems of ODEs by means of the residual error function and to give an efficient and useful error estimation via the error problem.In order to find solutions of the system (1), with the mixed conditions (2), we can use the collocation points defined by
(7)xi=a+b-aNi,i=0,1,…,N,0≤a≤x≤b.


## 2. Fundamental Matrix Relations

The Lucas polynomialsLn(x) can be written in the matrix form as
(8)Lx=XxDT,
where
(9)Lx=L0xL1xL2x⋯LNx,Xx=1xx2⋯xN,
and if N is odd,(10)D=20000⋯001110000⋯021110222000⋯003221033300⋯042220433104440⋯0⋮⋮⋮⋮⋮⋮⋮n-1(n-1)/2(n-1)2(n-1)20n-1(n+1)/2(n+1)2(n-3)20n-1(n+3)/2n+32n-5200n(n+1)/2(n+1)2(n-1)20n(n+3)/2(n+3)2(n-3)20⋯nnn0.If N is even,(11)D=20000⋯001110000⋯021110222000⋯003221033300⋯042220433104440⋯0⋮⋮⋮⋮⋮⋮⋮0n-1n/2n2(n-2)20n-1(n+2)/2(n+2)2(n-4)20⋯0nn/2n2n20n(n+2)/2(n+2)2(n-2)20n(n+4)/2(n+4)2(n-4)2⋯nnn0.We can write the approximate solutionsyj,N(x) given by (4) in the matrix form
(12)yj,Nx=LxAj,j=1,2,…,k,
where
(13)Aj=aj,0aj,1aj,2⋯aj,NT.From (8) and (12), we obtain the matrix relation
(14)yj,Nx=XxDTAj.Also, the relation between the matrixX(x) and its derivatives X(k)(x) is
(15)Xkx=XxBk,
where
(16)B=010⋯0002⋯0⋮⋮⋮⋱⋮000⋯N000⋯0
and B0=I(N+1)×(N+1) is the unit matrix.By using the relations (14) and (15), we obtain the following relations:
(17)yj,Nix=XxBiDTAj,i=0,1,…,m,j=1,2,…,k.Hence, we can write the matrix relations as(18)Yix=X¯xB¯iD¯A,i=0,1,…,m,
where
(19)Y(i)(x)=y1,N(i)(x)y2,N(i)(x)⋮yk,N(i)(x),X¯(x)=X(x)0⋯00X(x)⋯0⋮⋮⋱⋮00⋯X(x),B¯=B0⋯00B⋯0⋮⋮⋱⋮00⋯B,D¯=DT0⋯00DT⋯0⋮⋮⋱⋮00⋯DT,A=A1A2⋮Ak.


## 3. Method for Solution

Firstly, we can write the system (1) in the matrix form
(20)∑i=0mPi(x)Y(i)(x)=Gx,
where
(21)Pix=p1,1ixp1,2ix⋯p1,kixp2,1ixp2,2ix⋯p2,kix⋮⋮⋱⋮pk,1ixpk,2ix⋯pk,kix,Yix=y1,Nixy2,Nix⋮yk,Nix,Gx=g1xg2x⋮gkx.By substituting the collocation points (7) into (20) we obtain the system of matrix equations
(22)∑i=0mPi(xs)Y(i)(xs)=Gxs,s=0,1,…,N,
or the compact form
(23)∑i=0mPiY(i)=G,
where
(24)Pi=Pi(x0)0⋯00Pi(x1)⋯0⋮⋮⋱⋮00⋯Pi(xN),Y(i)=Y(i)(x0)Y(i)(x1)⋮Y(i)(xN),G=G(x0)G(x1)⋮G(xN).From the relation (18) and the collocation points (7), we have
(25)Yixs=X¯xsB¯iD¯A,s=0,1,…,N,
or, briefly,
(26)Y(i)=XB¯iD¯A,
where
(27)X=X¯(x0)X¯(x1)⋮X¯(xN),X¯xs=Xxs0⋯00Xxs⋯0⋮⋮⋱⋮00⋯Xxs.By substituting (26) into (23), we obtain the fundamental matrix equation as
(28)∑i=0mPiXB¯iD¯A=G.In (28) the full dimensions of the matrices Pi, X, B¯, D¯, A, and G are k(N+1)×k(N+1), k(N+1)×k(N+1), k(N+1)×k(N+1), k(N+1)×k(N+1), k(N+1)×1, and k(N+1)×1, respectively.The fundamental matrix equation (28) corresponding to (1) can be written in the form
(29)WA=GorW;G.This is a linear system ofk(N+1) algebraic equations in the k(N+1) unknown Lucas coefficients such that
(30)W=∑i=0mPiXB¯iD¯=wp,q,p,q=1,2,…,kN+1.By using the conditions (2) and the relations (18), the matrix form for the conditions is obtained as
(31)∑j=0m-1ajX¯(a)+bjX¯(b)B¯jD¯A=C,
where
(32)aj=aj10⋯00aj2⋯0⋮⋮⋱⋮00⋯ajk,bj=bj10⋯00bj2⋯0⋮⋮⋱⋮00⋯bjk,C=c1c2⋮ck,aji=a0,jia1,ji⋮am-1,ji,bji=b0,jib1,ji⋮bm-1,ji,ci=ci,0ci,1⋮ci,m-1,i=0,1,…,k.Hence, the fundamental matrix form for conditions is(33)UA=CorU;C
such that
(34)U=∑j=0m-1ajX¯(a)+bjX¯(b)B¯jD¯.Consequently, by replacing the row matrices (33) by last rows of the matrix (29), we obtain the new augmented matrix
(35)W~;G~.We do not have to change the last rows of the matrix equation given by (29). If the matrix W~ is singular, then rows of the matrix (33) can be replaced with any rows of the matrix (29). If rank⁡⁡W~=rank⁡⁡W~;G~=N+1, then we can write
(36)A=W~-1G~.Hence, the unknown Lucas coefficients matrixA is determined. We can find the Lucas polynomial solutions
(37)yj,Nx=∑n=0Naj,nLnx,j=1,2,…,k.


## 4. Residual Correction and Error Estimation

In this section, we will give an error estimation for the Lucas polynomial solutions (4) with the residual error function [21–25]. Moreover, we will improve the solution (4) by means of the residual error function. Firstly, we can define the residual function of the method as
(38)Ri,Nx=Lyj,Nx-gix,i=1,2,…,k,j=1,2,…,k.
Here, yj,N(x) represent the Lucas polynomial solutions given by (4) of the problem (1) and (2). Hence, yj,N(x) satisfies the problem
(39)Lyj,N(x)=∑n=0m∑j=1kpi,jn(x)yj,N(n)(x)=gix+Ri,Nx,i=1,2,…,k,j=1,2,…,k,n=0,1,…,m,∑n=0m-1ai,njyj,N(n)(a)+bi,njyj,N(n)(b)=cj,i,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1.Also, the error functionej,N(x) can be defined as
(40)ej,Nx=yjx-yj,Nx,
where yj(x) are the exact solutions of the problem (1) and (2). From (1), (2), (38), and (40), we obtain the error differential equation system
(41)Lej,N(x)=Lyj(x)-Lyj,N(x)=-Ri,N(x)
with the homogeneous mixed conditions
(42)∑n=0m-1ai,njej,N(n)(a)+bi,njej,N(n)(b)=0,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1
or, openly, the error problem
(43)∑n=0m∑j=1kpi,jn(x)ej,N(n)(x)=-Ri,Nx,i=1,2,…,k,j=1,2,…,k,n=0,1,…,m,∑n=0m-1ai,njej,N(n)(a)+bi,njej,N(n)(b)=0,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1.
Here, note that the nonhomogeneous mixed conditions
(44)∑n=0m-1ai,njyj(n)(a)+bi,njyj(n)(b)=cj,i,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1,∑n=0m-1ai,njyj,N(n)(a)+bi,njyj,N(n)(b)=cj,i,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1
are reduced to homogeneous mixed conditions
(45)∑n=0m-1ai,njej,N(n)(a)+bi,njej,N(n)(b)=0,i=0,1,…,m-1,j=1,2,…,k,n=0,1,…,m-1.The error problem (43) can be solved by using the procedure in Section 3. Thus, we obtain the approximation ej,N,M(x) to ej,N(x) as follows:
(46)ej,N,Mx=∑n=0Maj,n*Lnx,M>N,j=1,2,…,k.Consequently, the corrected Lucas polynomial solutionyj,N,M(x)=yj,N(x)+ej,N,M(x) is obtained by means of the polynomials yj,N(x) and ej,N,M(x),M>N. Also, we construct the error function ej,N(x)=yj(x)-yj,N(x), the estimated error function ej,N,M(x), and the corrected error function Ej,N,M(x)=ej,N(x)-ej,N,M(x)=yj(x)-yj,N,M(x).


## 5. Numerical Examples

In this section, the several numerical examples are given to demonstrate the efficiency and applicability of our method. The computations related to the examples are calculated by using a computer programme which is calledMaple and the figures are drawn inMatlab. In tables and figures, we calculate the values of the Lucas polynomial solution yj,N(x), the corrected Lucas polynomial solution yj,N,M(x)=yj,N(x)+ej,N,M(x), the actual absolute error function ej,N(x)=yj(x)-yj,N(x), and the estimated absolute error function ej,N,M(x).Example 1.
Let us consider the system of second-order linear differential equations given by(47)y12x+xy1x+xy2x=2,y22(x)+2xy2(x)+2xy1(x)=-2,hhhhhhhhhhhhhhhhhhh0≤x≤1
with the boundary conditions
(48)y1(0)=0,y1(1)=0,y2(0)=0,y2(1)=0
which has the exact solutions y1(x)=x2-x and y2(x)=-x2+x [30]. In this problem k=2, m=2, p1,12=1, p1,11=0, p1,10=x, p1,22=0, p1,21=0, p1,20=x, g1(x)=2, p2,12=0, p2,11=0, p2,10=2x, p2,22=1, p2,21=0, p2,20=2x, and g2(x)=-2.
The approximate solutionsy1,2(x) and y2,2(x) for N=2 are given by
(49)yi,2x=∑n=02ai,nLnx,i=1,2.
The set of the collocation points given by (7) for a=0, b=1, and N=2 is calculated as
(50)x0=0,x1=12,x2=1.
From (28), the fundamental matrix equation of the problem (47) is written as
(51)P0XD¯+P1XB¯D¯+P2XB¯2D¯A=G.
By applying the procedure in Section3, we obtain the Lucas polynomial solutions for N=2 as y1,2(x)=x2-x and y2,2(x)=-x2+x, which are the exact solutions.Example 2.
Let us consider the linear differential equations system given by(52)y11(x)+y21(x)+y2(x)=x-e-x,y11x+4y21x+y1x=1+2e-x,hhhhhhhhhhhhhhhhhhhhh0≤x≤1
with the initial conditions
(53)y1(0)=1,y2(0)=0
which has the exact solutions y1(x)=e-x+3e-x/3-3 and y2(x)=-e-x/2+3e-x/3/2-1+x [12, 29]. In this problem k=2, m=1, p1,11=1, p1,10=0, p1,21=1, p1,20=1, g1(x)=x-e-x, p2,11=1, p2,10=1, p2,21=4, p2,20=0, and g2(x)=1+2e-x.
We can write the fundamental matrix equation of the problem (52) from (28) as
(54)P0XD¯+P1XB¯D¯A=G.
By using our method, the approximate solutions of the problem (52) for N=6 are obtained as
(55)y1,6x=-1.999999999999999904x+0.6666586621368288521x2-0.185113238503384888x3+(0.429421011617410024e-1)x4-(0.7937368973290355e-2)x5+(0.9237777698612969e-3)x6+1.0000000000000000808,y2,6x=1.000000000000000105x-0.1666626772075632049x2+(0.74038219115823775e-1)x3-(0.199282827658208166e-1)x4+(0.3866617014883018e-2)x5-(0.4569099838690821e-3)x6+(0.108e-16).
In order to calculate the corrected Lucas polynomial solutions, let us consider the error problem(56)e1,61x+e2,61x+e2,6x=-R1,6x,e1,6(1)(x)+4e2,6(1)(x)+e1,6(x)=-R2,6(x)
such that e1,6(0)=0, e2,6(0)=0, and the residual functions are
(57)R1,6x=y1,61x+y2,61x+y2,6x-x+e-x,R2,6(x)=y1,6(1)(x)+4y2,6(1)(x)+y1,6(x)-1-2e-x.
By solving the error problem (56) for M=8, the estimated Lucas error functions e1,6,8(x) and e2,6,8(x) to e1,6(x) and e2,6(x) are obtained as
(58)e1,6,8x=-0.1210e-17x-(0.178885917374467780e-3)x7+(0.142216618722969108e-4)x8+(0.79326892394531528e-5)x2-(0.71083092941970440e-4)x3+(0.2633108714690733660e-3)x4-(0.486294856240169128e-3)x5+(0.4502363215270527764e-3)x6-0.6384e-18,e2,6,8x=(0.6913e-18)x+(0.892342831286434209e-4)x7-(0.710875043057030262e-5)x8-(0.39530091035351939e-5)x2+(0.354285130160153836e-4)x3-0.1312587470335659604e-3x4+(0.2424513686179994193e-3)x5-(0.22451354772782758896e-3)x6-(0.18124e-18).
Thus, we can calculate the corrected Lucas polynomial solutionsy1,6,8(x) and y2,6,8(x) as
(59)y1,6,8x=-2x+0.6666665948x2-0.1851843216x3+(0.4320541203e-1)x4-(0.8423663829e-2)x5+(0.1374014091e-2)x6+1-(0.178885917374467780e-3)x7+0.142216618722969108e-4x8,y2,6,8x=x-0.1666666302x2+(0.7407364763e-1)x3-(0.2005954152e-1)x4+(0.4109068384e-2)x5-(0.6814235316e-3)x6+(0.1061876e-16)+(0.892342831286434209e-4)x7-(0.710875043057030262e-5)x8.It is seen from Table1 and Figures 1(a) and 1(b) that the accuracy of solution increases when the values of N and M increase.Table 1
Numerical results of the exact solutionsyi(x) and the approximate solutions yi,N(x),yi,N,M(x) for i=1,2, N=6 and M=8,11 of problem (52).
x
i
Exact solution
Lucas polynomial solutions
y
1
(
x
i
)
y
1,6
(
x
i
)
y
1,6
,
8
(
x
i
)
y
1,6
,
11
(
x
i
)
0.1
0.806485719482
0.806485689143
0.806485719287
0.806485719292
0.2
0.625251708173
0.625251667103
0.625251707990
0.625251708032
0.5
0.145975834384
0.145975783291
0.145975834235
0.145975834133
0.8
−0.252886020789
−0.252886103976
−0.252886020930
−0.252886021210
1.0
−0.482526627107
−0.482526066408
−0.482526628764
−0.482526626284
x
i
y
2
(
x
i
)
y
2,6
(
x
i
)
y
2,6
,
8
(
x
i
)
y
2,6
,
11
(
x
i
)
0.1
0.098405441705
0.098405456828
0.098405441810
0.098405441788
0.2
0.193895101008
0.193895121487
0.193895101121
0.193895101070
0.5
0.466457257480
0.466457282978
0.466457257577
0.466457257606
0.8
0.724228025488
0.724228067006
0.724228025625
0.724228025747
1.0
0.890857245275
0.890856966173
0.890857246295
0.890857244962(a) Comparison of the exact solutiony1(x) and the approximate solutions y1,N(x) and y1,N,M(x). (b) Comparison of the exact solution y2(x) and the approximate solutions y2,N(x) and y2,N,M(x).
(a)
(b)Table2 and Figures 2(a) and 2(b) display that the actual and estimated errors are very close to zero and almost identical.Table 2
Comparison of the actual and estimated absolute errors forN=6 and M=8,11 of the problem (52).
x
i
Actual absolute errors
Estimated absolute errors
e
1,6
(
x
i
)
=
y
1
(
x
i
)
-
y
1,6
(
x
i
)
e
1,6
,
8
(
x
i
)
e
1,6
,
11
(
x
i
)
0.1
3.0339
e
-
8
3.0144
e
-
8
3.0149
e
-
8
0.2
4.1070
e
-
8
4.0888
e
-
8
4.0930
e
-
8
0.5
5.1093
e
-
8
5.0951
e
-
8
5.0851
e
-
8
0.8
8.3187
e
-
8
8.3065
e
-
8
8.2796
e
-
8
1.0
5.6070
e
-
7
5.6232
e
-
7
5.5982
e
-
7
x
i
e
2,6
(
x
i
)
=
y
2
(
x
i
)
-
y
2,6
(
x
i
)
e
2,6
,
8
(
x
i
)
e
2,6
,
11
(
x
i
)
0.1
1.5123
e
-
8
1.5017
e
-
8
1.5040
e
-
8
0.2
2.0479
e
-
8
2.0367
e
-
8
2.0418
e
-
8
0.5
2.5498
e
-
8
2.5404
e
-
8
2.5377
e
-
8
0.8
4.1518
e
-
8
4.1390
e
-
8
4.1270
e
-
8
1.0
2.7910
e
-
7
2.8011
e
-
7
2.7877
e
-
7(a) Comparison of the actual and estimated absolute errors fory1(x). (b) Comparison of the actual and estimated absolute errors for y2(x).
(a)
(b)Table3 and Figures 3(a) and 3(b) show that when the value of N increases, the accuracy of solution increases.Table 3
Numerical results of the exact solutionsy1(x),y2(x) and the Lucas polynomial solutions y1,N(x),y2,N(x) for N=6,8,11 of problem (52).
x
i
Exact solution
Lucas polynomial solutions
y
1
(
x
i
)
y
1,6
(
x
i
)
y
1,8
(
x
i
)
y
1,11
(
x
i
)
0.1
0.806485719482
0.806485689143
0.806485719432
0.806485719479
0.2
0.625251708173
0.625251667103
0.625251708124
0.625251708169
0.5
0.145975834384
0.145975783291
0.145975834317
0.145975834379
0.8
−0.252886020789
−0.252886103976
−0.252886020880
−0.252886020798
1.0
−0.482526627107
−0.482526066408
−0.482526625954
−0.482526627122
x
i
y
2
(
x
i
)
y
2,6
(
x
i
)
y
2,8
(
x
i
)
y
2,11
(
x
i
)
0.1
0.098405441705
0.098405456828
0.098405441730
0.098405441706
0.2
0.193895101008
0.193895121487
0.193895101033
0.193895101010
0.5
0.466457257480
0.466457282978
0.466457257512
0.466457257479
0.8
0.724228025488
0.724228067006
0.724228025534
0.724228025484
1.0
0.890857245275
0.890856966173
0.890857244698
0.890857245264(a) Comparison of the exact solutiony1(x) and the approximate solutions y1,N(x). (b) Comparison of the exact solution y2(x) and the approximate solutions y2,N(x).
(a)
(b)Table4 and Figures 4(a) and 4(b) show that the value of N is increased; the actual absolute errors decrease rapidly.Table 4
Comparison of the absolute error functionse1,N(x),e2,N(x) for N=5,6,8,11 of problem (52).
Lucas collocation method
x
i
e
1,5
(
x
i
)
e
1,6
(
x
i
)
e
1,8
(
x
i
)
e
1,11
(
x
i
)
0.1
5.9187
e
-
7
3.0339
e
-
8
4.9550
e
-
11
2.3912
e
-
12
0.2
1.0110
e
-
6
4.1070
e
-
8
4.9303
e
-
11
2.8583
e
-
12
0.5
1.0978
e
-
6
5.1093
e
-
8
6.4840
e
-
11
4.8214
e
-
12
0.8
9.0094
e
-
7
8.3187
e
-
8
9.1255
e
-
11
8.8867
e
-
12
1.0
1.3659
e
-
5
5.6070
e
-
7
1.1536
e
-
9
1.4366
e
-
11
x
i
e
2,5
(
x
i
)
e
2,6
(
x
i
)
e
2,8
(
x
i
)
e
2,11
(
x
i
)
0.1
2.9327
e
-
7
1.5123
e
-
8
2.4764
e
-
11
1.5621
e
-
12
0.2
5.0134
e
-
7
2.0479
e
-
8
2.4641
e
-
11
1.5043
e
-
12
0.5
5.4596
e
-
7
2.5498
e
-
8
3.2407
e
-
11
2.5725
e
-
12
0.8
4.5116
e
-
7
4.1518
e
-
8
4.5610
e
-
11
6.7968
e
-
12
1.0
6.7555
e
-
6
2.7910
e
-
7
5.7653
e
-
10
1.3288
e
-
11(a) Comparison of the absolute error functionse1,Nx. (b) Comparison of the absolute error functions e2,Nx.
(a)
(b)In addition, this problem was solved by Akyüz-Daşoğlu and Sezer [12] and Davies and Crann [29]. Now, let us compare our method (LCM) with the other methods (Chebyshev method and Stehfest method) given by [12, 29]. Table 5 indicates this comparison.Table 5
Comparison of the absolute error functionse1,N(x) and e2,N(x) for different methods of problem (52).
x
i
Chebyshev method in [12]
Stehfest Method in [29]
Present method
Present method
e
1,5
(
x
i
)
e
1,5
(
x
i
)
e
1,11
(
x
i
)
0.1
4.5105
e
-
5
6.7614
e
-
5
5.9187
e
-
7
2.3912
e
-
12
0.2
7.9850
e
-
5
8.4949
e
-
5
1.0110
e
-
6
2.8583
e
-
12
0.5
9.7191
e
-
5
3.1897
e
-
3
1.0978
e
-
6
4.8214
e
-
12
0.8
8.0060
e
-
5
5.2028
e
-
3
9.0094
e
-
7
8.8867
e
-
12
1.0
1.0677
e
-
4
1.1938
e
-
2
1.3659
e
-
5
1.4366
e
-
11
x
i
Chebyshev method in [12]
Stehfest Method in [29]
Present method
Present method
e
2,5
(
x
i
)
e
2,5
(
x
i
)
e
2,11
(
x
i
)
0.1
2.2477
e
-
5
8.4086
e
-
6
2.9327
e
-
7
1.5621
e
-
12
0.2
3.9847
e
-
5
1.9575
e
-
5
5.0134
e
-
7
1.5043
e
-
12
0.5
4.8907
e
-
5
2.2420
e
-
4
5.4596
e
-
7
2.5725
e
-
12
0.8
4.0642
e
-
5
4.6470
e
-
4
4.5116
e
-
7
6.7968
e
-
12
1.0
5.3904
e
-
5
4.7100
e
-
4
6.7555
e
-
6
1.3288
e
-
11It is seen from Table5 that the present method gives better approximations than the other methods given by [12, 29].Example 3.
Let us consider the linear differential equations system given by(60)y11x-y3x=-cos⁡x,y21(x)-y3(x)=-ex,y31(x)-y1(x)+y2(x)=0,hhhhhhhhhhhhhh0≤x≤1
with the initial conditions
(61)y1(0)=1,y2(0)=0,y3(0)=2
which has the exact solutions y1(x)=ex, y2(x)=sin(x), and y3(x)=ex+cos⁡(x) [2]. By using the method, the approximate solutions of the problem (60) for N=2,5 are obtained as
(62)y1,2x=1+x+0.67061360741023699830x2,y2,2x=-0.1e-18+x-0.1005251013995184324x2,y3,2x=2+x+0.19278467720243885767x2,y1,5x=0.99999999999999999824+1.0000000000000000125x+0.49976858483072423438x2+0.16822545527123026922x3+(0.3772109990725761363e-1)x4+0.12531436624829093170e-1x5,y2,5x=+1.000000000000000009x+(0.45327625709752e-4)x2-0.166988949339599508x3+(0.782824607296818e-3)x4+0.7640927371818304e-2x5+0.6e-17,y3,5x=1.9999999999999999978+1.0000000000000000004x-(0.938977776578073e-4)x2+0.1672264869687396031x3+(0.820919229141536686e-1)x4+(0.93438587611964760e-2)x5.It is seen from Table6 and Figures 5(a), 5(b), and 5(c) that the accuracy increases as the N increase.Table 6
Numerical results of the exact solutions and the approximate solutions forN=2,5,9 of problem (60).
x
i
Exact solution
Lucas polynomial solutions
y
1
(
x
i
)
y
1,2
(
x
i
)
y
1,5
(
x
i
)
y
1,9
(
x
i
)
0.1
1.105170918076
1.106706136074
1.105169808728
1.105170918072
0.2
1.221402758160
1.226824544296
1.221400910855
1.221402758157
0.5
1.648721270700
1.667653401853
1.648719504255
1.648721270696
0.8
2.225540928492
2.229192708743
2.225540191066
2.225540928487
1.0
2.718281828459
2.670613607410
2.718246576634
2.718281828322
y
2
(
x
i
)
y
2,2
(
x
i
)
y
2,5
(
x
i
)
y
2,9
(
x
i
)
0.1
0.099833416646
0.098994748986
0.099833619019
0.099833416647
0.2
0.198669330795
0.195978995944
0.198669599126
0.198669330795
0.5
0.479425538604
0.474868724650
0.479425418757
0.479425538603
0.8
0.717356090899
0.735663935104
0.717355091659
0.717356090898
1.0
0.841470984808
0.899474898600
0.841480130265
0.841470984845
y
3
(
x
i
)
y
3,2
(
x
i
)
y
3,5
(
x
i
)
y
3,9
(
x
i
)
0.1
2.100175083354
2.101927846772
2.100174590140
2.100175083352
0.2
2.201469336001
2.207711387088
2.201468393096
2.201469335999
0.5
2.526303832591
2.548196169301
2.526302577195
2.526303832587
0.8
2.922247637840
2.923382193410
2.922246514015
2.922247637835
1.0
3.258584134327
3.192784677202
3.258568370866
3.258584134265(a) Comparison of the exact solutiony1(x) and the approximate solutions y1,N(x). (b) Comparison of the exact solution y2(x) and the approximate solutions y2,N(x). (c) Comparison of the exact solution y3(x) and the approximate solutions y3,N(x).
(a)
(b)
(c)Table7 shows that while the value of N is increased, the errors decrease rapidly. Now, we compare the present method with the differential transform method given by [2].Table 7
Comparison of the actual absolute errorse1,N(x),e2,N(x) for N=2,5,9 of problem (60).
x
i
e
1,2
(
x
i
)
e
1,5
(
x
i
)
e
1,9
(
x
i
)
0.1
1.5352
e
-
3
1.1093
e
-
6
3.5633
e
-
12
0.2
5.4218
e
-
3
1.8473
e
-
6
3.1738
e
-
12
0.5
1.8932
e
-
2
1.7664
e
-
6
4.0266
e
-
12
0.8
3.6518
e
-
3
7.3743
e
-
7
5.3975
e
-
12
1.0
4.7668
e
-
2
3.5252
e
-
5
1.3694
e
-
10
x
i
e
2,2
(
x
i
)
e
2,5
(
x
i
)
e
2,9
(
x
i
)
0.1
8.3867
e
-
4
2.0237
e
-
7
6.1484
e
-
13
0.2
2.6903
e
-
3
2.6833
e
-
7
2.8246
e
-
13
0.5
4.5568
e
-
3
1.1985
e
-
7
4.6422
e
-
13
0.8
1.8308
e
-
2
9.9924
e
-
7
1.5204
e
-
12
1.0
5.8004
e
-
2
9.1455
e
-
6
3.7266
e
-
11
x
i
e
3,2
(
x
i
)
e
3,5
(
x
i
)
e
3,9
(
x
i
)
0.1
1.7528
e
-
3
4.9321
e
-
7
1.7336
e
-
12
0.2
6.2421
e
-
3
9.4291
e
-
7
1.8815
e
-
12
0.5
2.1892
e
-
2
1.2554
e
-
6
3.0301
e
-
12
0.8
1.1346
e
-
3
1.1238
e
-
6
4.3129
e
-
12
1.0
6.5799
e
-
2
1.5763
e
-
5
6.2478
e
-
11It is seen from Table8 that the present method (LCM) is very effective compared to the differential transform method (DTM) for problem (60).Table 8
Comparison of the errors of the methods for problem (60).
x
i
Differential transformmethod in [2]
Present methode1,9(xi)
0.1
2.0092
e
-
11
3.5633
e
-
12
0.2
2.6046
e
-
9
3.1738
e
-
12
0.5
1.6526
e
-
6
4.0266
e
-
12
0.8
4.6173
e
-
5
5.3975
e
-
12
1.0
2.2627
e
-
4
1.3694
e
-
10
x
i
Differential transformmethod in [2]
Present methode2,9(xi)
0.1
1.9839
e
-
11
6.1484
e
-
13
0.2
2.5383
e
-
9
2.8246
e
-
13
0.5
1.5447
e
-
6
4.6422
e
-
13
0.8
4.1242
e
-
5
1.5204
e
-
12
1.0
1.9568
e
-
4
3.7266
e
-
11
x
i
Differential transformmethod in [2]
Present methode3,9(xi)
0.1
2.0340
e
-
11
1.7336
e
-
12
0.2
2.6681
e
-
9
1.8815
e
-
12
0.5
1.7493
e
-
6
3.0301
e
-
12
0.8
5.0305
e
-
5
4.3129
e
-
12
1.0
2.5080
e
-
4
6.2478
e
-
11Figures6(a), 6(b), and 6(c) display the actual absolute error functions obtained by present method for N=9 and the differential transform method. These figures display that the results gained by the present method are better than those obtained by the differential transform method.(a) Comparison of the absolute error functions fory1(x). (b) Comparison of the absolute error functions for y2(x). (c) Comparison of the absolute error functions for y3(x).
(a)
(b)
(c)


## 6. Conclusions

It is known that solving the high-order linear differential equations system is usually very difficult analytically. In this case, it is required to approximate solutions. In this paper, a new method based on the Lucas polynomials with the help of the residual error function for solving system of high-order linear differential equations numerically is presented. When the obtained results are investigated in examples, it can be seen that the developed method is very effective compared to the others. Also, it can be seen from the tables and the figures that the accuracy increased when the value ofN is increased. The approximate solutions are obtained in a short time with computer programmes such asMaple,Mathematica, andMatlab. We have used theMaple andMatlab for computations and graphics, respectively. Additionally, the presented method can be applied to the other system of linear integral and integrodifferential equations.



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