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Combination of Laplace transform and residual power series techniques to solve autonomous n-dimensional fractional nonlinear systems

  • Marwan Alquran EMAIL logo , Maysa Alsukhour , Mohammed Ali and Imad Jaradat
Published/Copyright: October 8, 2021
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Abstract

In this work, a new iterative algorithm is presented to solve autonomous n-dimensional fractional nonlinear systems analytically. The suggested scheme is combination of two methods; the Laplace transform and the residual power series. The methodology of this algorithm is presented in details. For the accuracy and effectiveness purposes, two numerical examples are discussed. Finally, the impact of the fractional order acting on these autonomous systems is investigated using graphs and tables.

1 Introduction

Extending the ordinary-partial differential equations into fractional differential equations has been attracted by many researchers since the fractional derivatives are more general, applicable and more efficient for real world phenomena, especially when the dynamics of a given mathematical model is affected by constraints inherent to the system [1]. Since it is difficult to find explicit solutions to these fractional problems, it is necessary to use, alternative methods, numerical and approximate techniques.

The most popular numerical techniques for solving fractional problems are the Collocation methods whose basic functions are of types: Haar functions, Legendre wavelets, Bernoulli polynomials, B-spline functions, Chebyshev polynomials, and others [2, 3, 4, 5, 6]. On the other side, many effective analytical schemes were developed to treat nonlinear problems involving fractional derivatives, such schemes are: generalized power series [13, 16, 22, 23], residual power series method [17, 26, 27, 28, 29], differential transform method [19, 21, 24, 25], homotopy perturbation method [14, 15, 18, 20] and others.

In this work, we are interested in introducing a new analytical scheme to solve nonlinear fractional autonomous dynamical systems. Dynamical systems describe the prediction of future states follow from the current state. Different numerical and analytical algorithms were used to solve fractional dynamical systems, such as, Homotopy analysis method, the variational iteration method, Ritz method and the explicit one-step method [7, 8, 9, 10]. In this context, we present a new algorithm constructed by combining the Laplace transform method and the residual power series method (LRPS). This new technique has been recently proposed for the first time in [11] and used in [12].

The organization of the paper is the following: In Section 2 we present the steps of applying the LRPS method to solve n × n autonomous fractional dynamical systems. In Section 3 we study two examples of order 2 × 2 and 3 × 3, and also provide graphical analysis. Finally, the conclusion is given in Section 4.

2 Description of LRPS

In this section, we present in details the steps of applying the LRPS scheme in solving the following autonomous fractional system:

(1) {Dtαν1(t)=h1(t,ν1(t),ν2(t),...,νn(t)),Dtαν2(t)=h2(t,ν1(t),ν2(t),...,νn(t)),...Dtανn(t)=hn(t,ν1(t),ν2(t),...,νn(t)),

subject to the initial conditions

(2) νm(0)=am,m=1,2,3,,n,

where 0 < α ≤ 1, 0 ≤ t ≤ 1, hm are suitable functions, and Dtα is the Caputo-derivative.

Applying the Laplace transform to (1) we get

(3) Vm(s)=ams+Hm(s)sα,

where

Vm(s)=[νm(t)],Hm(s)=[hm(t,ν1(t),ν2(t),...,νn(t))].

We assume that Vm(s), m = 1, 2, 3, . . ., n have fractional power series representation, i.e.,

(4) Vm(s)=r=0cmrsrα+1.

Next, we let Vmk(s) denote the k-th truncated series of Vm(s), i.e.,

(5) Vmk(s)=r=0kcmrsrα+1.

By condition (2), the 0-th LRPS approximate solution of Vm(s) is νm(0)=limssVm(s)=c0r=am . Thus,

(6) Vmk(s)=ams+r=1kcmrsrα+1.

Now, we define the Laplace residual function and the k-th Laplace residual function to (3), respectively, as:

(7) LResm(s)=Vm(s)amsHm(s)sα,

(8) LResmk(s)=Vmk(s)amsHm(s)sα.

To determine the coefficients cmr, m = 1, 2, 3, . . ., n and r = 1, 2, 3, . . ., k, we substitute (6) into (8), then multiply the resulting equation by s+1, and next we solve the following iterative equation

(9) lims(skα+1LResmk(s))=0,

for the unknowns cmr : k = 1, 2, 3, .... Finally, we apply the Laplace inverse to Vmk(s) to obtain the k-th approximate solution νmk(t) .

3 Numerical Problems and concluding remarks

In this section we present two examples of fractional nonlinear autonomous systems of order 2 × 2 and 3 × 3. The steps of implementing the LRPS will be clarified, and the effectiveness of the proposed method will be tested by providing a graphical analysis.

3.1 Example 1

Consider the following 2×2 nonlinear autonomous system [7, 8]

(10) {Dαν1(t)=ν1(t)2,Dαν2(t)=ν12(t)+ν2(t),

subject to

(11) ν1(0)=1,ν2(0)=0.

Applying the Laplace transform to (10)(11), we get

(12) V1(s)=1s+V1(s)2sα,V2(s)=1sα([1[V1(s)]2]+V2(s)).

We assume that both V1(s) and V2(s) have fractional power series representation as

(13) V1(s)=n=0ansnα+1,V2(s)=n=0bnsnα+1,

and we assume that the k-th truncated series of V1(s) and V2(s) are

(14) V1k(s)=n=0kansnα+1=1s+n=1kansnα+1,V2k(s)=n=0kbnsnα+1=n=1kbnsnα+1.

It is clear that the Laplace residual function for both V1k(s) and V2k(s) are

(15) LRes1(s)=V1(s)1sV1(s)2sα,LRes2(s)=V2(s)1sα([1[V1(s)]2]+V2(s)).

Accordingly, the k-th Laplace residual functions, LResk, are

(16) LRes1k(s)=V1k(s)1sV1k(s)2sα,LRes2k(s)=V2k(s)1sα([1[V1k(s)]2]+V2k(s)).

To determine a1 and b1, we consider

(17) LRes11(s)=V11(s)1sV11(s)2sα,LRes21(s)=V21(s)1sα([1[V11(s)]2]+V21(s)).

As V11(s)=1s+a1sα+1 and V21(s)=b1sα+1 , we get

(18) LRes11(s)=a1sα+112sα(1s+a1sα+1),LRes21(s)=b1sα+11sα([1[1s+a1sα+1]2]+b1sα+1)=b1sα+11sα([(1+a1tαΓ(1+α))2]+b1sα+1)=b1sα+11sα([(1+2a1tαΓ(1+α)+a12t2αΓ2(1+α))]+b1sα+1)=b1sα+11sα(1s+2a1s1+α+a12Γ(1+2α)Γ2(1+α)s1+2α+b1sα+1).

Multiply (18) by sα+1, we obtain that

(19) sα+1LRes11(s)=a112(1+a1sα),sα+1LRes21(s)=2a1sαa12Γ(2α+1)Γ2(α+1)s2α+b1(11sα)1.

Finally, we solve the following system

(20) lims(sα+1LRes11(s))=0,lims(sα+1LRes21(s))=0,

which gives that

(21) a1=12,b1=1.

In a similar manner, to find a2 and b2, we consider

(22) LRes12(s)=V12(s)1sV12(s)2sα,LRes22(s)=V22(s)1sα([1[V12(s)]2]+V22(s)).

As V12(s)=1s+12sα+1+a2s2α+1 and V22(s)=1sα+1+b2s2α+1 , we obtain

(23) LRes12(s)=a2s2α+114s2α+1a22s3α+1,LRes22(s)=1sα+1+b2s2α+11sα([1[1s+12sα+1+a2s2α+1]2]+1sα+1+b2s2α+1).

Multiply (23) by s2α+1, and then solve

(24) lims(s2α+1LRes12(s))=0,lims(s2α+1LRes22(s))=0,

we deduce that

(25) a2=14,b2=2.

Hence, the 2nd-approximate LRPS solution of V12(s) and V22(s) are

(26) V12(s)=1s+12sα+1+14s2α+1,V22(s)=1sα+1+12s2α+1.

Proceeding as the above illustrated steps in determining the unknown functions ak and bk, one can easily reach the following results:

(27) a3=18,a4=116,a5=132,

and

(28) b3=52+Γ(1+2α)4Γ2(1+α),b4=14(11+Γ(1+2α)Γ2(1+α)+Γ(1+3α)Γ(1+α)Γ(1+2α)),b5=116(46+4Γ(1+2α)Γ2(1+α)+Γ(1+4α)Γ2(1+2α)+1Γ(1+α)(4Γ(1+3α)Γ(1+2α)+2Γ(1+4α)Γ(1+3α))).

Therefore,

(29) V1(s)=1s+12sα+1+14s2α+1+18s3α+1+116s4α+1+132s5α+1+...,V2(s)=1sα+1+12s2α+1+1s3α+1(52+Γ(1+2α)4Γ2(1+α))+1s4α+1(14(11+Γ(1+2α)Γ2(1+α)+Γ(1+3α)Γ(1+α)Γ(1+2α)))+1s5α+1(116(46+4Γ(1+2α)Γ2(1+α)+Γ(1+4α)Γ2(1+2α)+1Γ(1+α)(4Γ(1+3α)Γ(1+2α)+2Γ(1+4α)Γ(1+3α))))+....

Consequently, the solution of (10)(11) is

(30) ν1(t)=1+tα2Γ(1+α)+t2α4Γ(1+2α)+t3α8Γ(1+3α)+t4α16Γ(1+4α)+t5α32Γ(1+5α)+...,ν2(t)=2t2αΓ(1+2α)+10t3α4Γ(1+3α)++tα4Γ2(1+α)(4Γ(1+α)+t2αΓ(1+2α)Γ(1+3α))+t4α(11Γ2(1+α)Γ(1+2α)+Γ2(1+2α)+Γ(1+α)Γ(1+3α))4Γ2(1+α)Γ(1+2α)Γ(1+4α)+....

We point out that the exact solutions to the system in Example 1 for the case of α = 1 are ν1(t)=et2 and ν2(t) = tet [7, 8]. We consider ϕ1(t)=i=06aitiα to be the LRPS approximation of ν1(t), and ϕ2(t)=i=06bitiα to be the LRPS approximation of ν2(t). In Figure 1, the first sub-figure represents the values of φ1(t) for different values of the fractional order α and values of ν1(t), while the second sub-figure represents the absolute error 1(t) − φ1(t)| when α = 1. In Figure 2, the first sub-figure represents the values of φ2(t) for different values of the fractional order α and values of ν2(t), while the second sub-figure represents the absolute error 2(t) − φ2(t)| when α = 1. From these plots, we observe that the curves of φi(t) : i = 1, 2, are mapping continuously and gradually as α varies from 0 to 1 and converges to νi(t) : i = 1, 2, when α = 1. Also, we observe that the approximations φi(t) : i = 1, 2, are in excellent agreement with νi(t) : i = 1, 2, when α = 1.

Figure 1 Exact, profile approximate solutions and absolute error regarding ν1 of Example 1.
Figure 1

Exact, profile approximate solutions and absolute error regarding ν1 of Example 1.

Figure 2 Exact, profile approximate solutions and absolute error regarding ν2 of Example 1.
Figure 2

Exact, profile approximate solutions and absolute error regarding ν2 of Example 1.

On the other side, as shown in Table 1, we provide numerical investigations on the accuracy of LRPS applied to Example 1. While as, in Table 2, we present the impact of the fractional order α acting on the values of the unknowns field functions νi(t) : i = 1, 2.

Table 1

Numerical values of φ1(t) and φ2(t) for α = 0.5, 0.7, 0.9 to Example 1.

t α = 0.5 α = 0.7 α = 0.9
ν1(t) ν2(t) ν1(t) ν2(t) ν1(t) ν2(t)
0.2 1.312141902 1.184140706 1.201599157 0.580240824 1.130764596 0.320222598
0.4 1.486564677 2.444376604 1.354985531 1.290827616 1.259270690 0.756805142
0.6 1.6445934256 4.006862318 1.506375497 2.249749087 1.396080779 1.365217628
0.8 1.7972927210 5.900827955 1.661915587 3.516856974 1.543797927 2.198213783
Table 2

Absolute errors: 1(t) − ν1(t)| and 2(t) − ν2(t)| to Example 1.

t 1(t) − ν1(t)| 2(t) − ν2(t)|
0.2 1.408981153×10−9 5.516320339×10−7
0.4 9.149350321×10−8 3.65457231×10−5
0.6 1.057576003×10−6 4.31280234×10−4
0.8 6.030974603×10−6 2.51274279×10−3

3.2 Example 2

Let us consider the following 3 × 3 nonlinear autonomous system [9, 10]

(31) {Dαν1(t)=ν1(t),Dαν2(t)=2(ν1(t))2,Dαν3(t)=3ν1(t)ν2(t);

subject to

(32) ν1(0)=1,ν2(0)=1,ν3(0)=0.

Apply the Laplace transform to (31)(32), we get

(33) V1(s)=1s+V1(s)sα,V2(s)=1s+2sα([1[V1(s)]2]),V3(s)=3[1[V1(s)]1[V2(s)]]sα,

Assume that V1(s), V2(s) and V3(s) have fractional power series as

(34) V1(s)=n=0ansnα+1,V2(s)=n=0bnsnα+1,V3(s)=n=0cnsnα+1,

with the k-th truncated series

(35) V1k(s)=n=0kansnα+1=1s+n=1kansnα+1,V2k(s)=n=0kbnsnα+1=1s+n=1kbnsnα+1,V3k(s)=n=0kcnsnα+1=n=1kcnsnα+1,respectively,

whose the Laplace residual functions for V1k(s) , V2k(s) and V3k(s) are

(36) LRes1(s)=V1(s)1sV1(s)sα,LRes2(s)=V2(s)1s1sα([1[V1(s)]2]),LRes3(s)=V3(s)3[1[V1(s)]1[V2(s)]]sα.

Accordingly, the k-th Laplace residual functions, LResk, are

(37) LRes1k(s)=V1k(s)1sV1k(s)sα,LRes2k(s)=V2k(s)1s1sα([1[V1k(s)]2]),LRes3k(s)=V3k(s)3[1[V1k(s)]1[V2k(s)]]sα,

To determine a1, b1 and c1, we substitute (35) in (37) with k = 1 to get

(38) LRes11(s)=a1sα+11sα(1s+a1sα+1),LRes21(s)=b1sα+11sα([1[1s+a1sα+1]2]),LRes31(s)=c1sα+13sα([1[1s+a1sα+1]1[1s+b1sα+1]]),

Multiply (38) by sα+1,

(39) sα+1LRes11(s)=a1(1+a1sα),sα+1LRes21(s)=b1s([1[1s+a1sα+1]2]),sα+1LRes31(s)=c13s([1[1s+a1sα+1]1[1s+b1sα+1]]).

It is clear that when we solve lims(sα+1LResm1(s))=0 , m = 1, 2, 3, we directly obtain

(40) a1=1,b1=2,c1=3.

Similarly, when we substitute (35) in (37) with k = 2 and then use the fact that lims(sα+1LResm2(s))=0 , m = 1, 2, 3, we reach

(41) a2=1,b2=4,c2=9.

Hence, the 2nd-approximate LRPS solution of V12(s) , V22(s) and V32(s) are

(42) V12(s)=1s+1sα+1+1s2α+1,V22(s)=1s+2sα+1+4s2α+1,V32(s)=3sα+1+9s2α+1.

Proceeding as the above illustrated steps in determining the unknown functions ak, bk and ck, one can easily verify the following results:

(43) a3=1,a4=1,a5=1,

(44) b3=4+2Γ(1+2α)Γ2(1+α),b4=4+4Γ(1+3α)Γ(1+α)Γ(1+2α),b5=4+2(1Γ2(1+2α)+2Γ(1+α)Γ(1+3α))Γ(1+4α),

(45) c3=15+6Γ(1+2α)Γ2(1+α),c4=15+6(Γ2(1+2α)+3Γ(1+α)Γ(1+3α))Γ2(1+α)Γ(1+2α),c5=3(5+4Γ(1+4α)Γ2(1+2α)+2Γ(1+2α)Γ(1+4α)Γ3(1+α)Γ(1+3α)+1Γ(1+α)(4Γ(1+3α)Γ(1+2α)+6Γ(1+4α)Γ(1+3α))).

Therefore,

(46) V1(s)=1s+1sα+1+1s2α+1+1s3α+1+1s4α+1+1s5α+1+...,V2(s)=1s+2sα+1+4s2α+1+1s3α+1(4+2Γ(1+2α)Γ(1+α)2)+1s4α+1(4+4Γ(1+3α)Γ(1+α)Γ(1+2α))+1s5α+1(4+2(1Γ(1+2α)2+2Γ(1+α)Γ(1+3α))Γ(1+4α))+...,V3(s)=3sα+1+9s2α+1+1s3α+1(15+6Γ(1+2α)Γ(1+α)2)+1s4α+1(15+6(Γ(1+2α)2+3Γ(1+α)Γ(1+3α))Γ(1+α)2Γ(1+2α))+3(5+4Γ(1+4α)Γ(1+2α)2+2Γ(1+2α)Γ(1+4α)Γ(1+α)3Γ(1+3α)+1Γ(1+α)(4Γ(1+3α)Γ(1+2α)+6Γ(1+4α)Γ(1+3α)))s5α+1+....

Consequently, the solution of (31)(32) is

(47) ν1(t)=1+tαΓ(1+α)+t2αΓ(1+2α)+t3αΓ(1+3α)+t4αΓ(1+4α)+t5αΓ(1+5α)+...,ν2(t)=1+4t2αΓ(1+2α)+4t3αΓ(1+3α)+2t3αΓ(1+2α)Γ(1+α)2Γ(1+3α)+2tα+4t4αΓ(1+3α)Γ(1+2α)Γ(1+4α)Γ(1+α)+4t4αΓ(1+4α)+1Γ(1+5α)(t5α(4+2(1Γ(1+2α)2+2Γ(1+α)Γ(1+3α))Γ(1+4α)))+...,ν3(t)=tα(3(tα(3Γ(1+2α)+5tα(1Γ(1+3α)+tαΓ(1+4α)))+1Γ(1+α)2(2t2αΓ(1+2α)(1Γ(1+3α)+1Γ(1+4α)))+1+6t3αΓ(1+3α)Γ(1+2α)Γ(1+4α)Γ(1+α)))+....

It is worth mentioning that the exact solutions to the system in Example 2 for the case α = 1 are ν1(t) = et, ν2(t) = e2t and ν3(t) = e3t − 1 [9, 10]. We consider ψ1(t)=i=06aitiα to be the LRPS approximation of ν1(t), ψ2(t)=i=06bitiα to be the LRPS approximation of ν2(t) and ψ3(t)=i=06citiα to be the LRPS approximation of ν3(t). In Figure 3, the first sub-figure represents the values of ψ1(t) for different values of the fractional order α and values of ν1(t), while the second sub-figure represents the absolute error 1(t)− ψ1(t)| when α = 1. In Figure 4, the first sub-figure represents the values of ψ2(t) for different values of the fractional order α and values of ν2(t), while the second sub-figure represents the absolute error 2(t) − ψ2(t)| when α = 1. In Figure 5, the first sub-figure represents the values of ψ3(t) for different values of the fractional order α and values of ν3(t), while the second sub-figure represents the absolute error 3(t) − ψ3(t)| when α = 1. For this 3 × 3 system, one can observe the same findings depicted for Example 1. Finally, as shown in Table 3, we provide numerical investigations on the accuracy of LRPS applied to Example 2.

Figure 3 Exact, profile approximate solutions and absolute error regarding ν1 of Example 2.
Figure 3

Exact, profile approximate solutions and absolute error regarding ν1 of Example 2.

Figure 4 Exact, profile approximate solutions and absolute error regarding ν2 of Example 2.
Figure 4

Exact, profile approximate solutions and absolute error regarding ν2 of Example 2.

Figure 5 Exact, profile approximate solutions and absolute error regarding ν3 of Example 2.
Figure 5

Exact, profile approximate solutions and absolute error regarding ν3 of Example 2.

Table 3

Absolute errors: 1(t) − ν1(t)|, 2(t) − ν2(t)| and 3(t) − ν3(t)| to Example 2.

t 1(t) − ν1(t)| 2(t) − ν2(t)| 3(t) − ν3(t)|
0.2 9.149350315×10−8 6.030974603×10−6 7.080039050 ×10−5
0.4 6.030974603×10−6 4.102618258×10−4 4.980922736 ×10−3
0.6 7.080039050×10−5 4.980922736×10−3 6.278346441 ×10−2
0.8 4.102618258×10−4 2.991775772×10−2 3.932243806 ×10−1

4 Conclusion

A combination of two schemes; the Laplace transform and the residual power series, is adapted to solve nonlinear Caputo-fractional autonomous dynamical systems. The methodology, reliability and the accuracy of the new technique are introduced by solving 2 × 2 and 3 × 3 systems. The role of the fractional derivative is investigated by using graphical analysis. Finally, the advantage of the current method was depicted as converting the whole fractional problem into pure algebraic computational scheme which can be executed using any available computational softwares.

As a future work, the authors plan to extend the use of LRPS to solve multi-dimensional various fractional problems arising in Engineering and Science.

  1. Funding information: The authors state no funding involved.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors state no conflict of interest.

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Received: 2021-04-30
Accepted: 2021-08-03
Published Online: 2021-10-08

© 2021 Marwan Alquran et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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