Research article Special Issues

Haar wavelet method for solution of variable order linear fractional integro-differential equations

  • Received: 22 October 2021 Revised: 14 December 2021 Accepted: 20 December 2021 Published: 06 January 2022
  • MSC : 34K05, 34K30

  • In this paper, we developed a computational Haar collocation scheme for the solution of fractional linear integro-differential equations of variable order. Fractional derivatives of variable order is described in the Caputo sense. The given problem is transformed into a system of algebraic equations using the proposed Haar technique. The results are obtained by solving this system with the Gauss elimination algorithm. Some examples are given to demonstrate the convergence of Haar collocation technique. For different collocation points, maximum absolute and mean square root errors are computed. The results demonstrate that the Haar approach is efficient for solving these equations.

    Citation: Rohul Amin, Kamal Shah, Hijaz Ahmad, Abdul Hamid Ganie, Abdel-Haleem Abdel-Aty, Thongchai Botmart. Haar wavelet method for solution of variable order linear fractional integro-differential equations[J]. AIMS Mathematics, 2022, 7(4): 5431-5443. doi: 10.3934/math.2022301

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  • In this paper, we developed a computational Haar collocation scheme for the solution of fractional linear integro-differential equations of variable order. Fractional derivatives of variable order is described in the Caputo sense. The given problem is transformed into a system of algebraic equations using the proposed Haar technique. The results are obtained by solving this system with the Gauss elimination algorithm. Some examples are given to demonstrate the convergence of Haar collocation technique. For different collocation points, maximum absolute and mean square root errors are computed. The results demonstrate that the Haar approach is efficient for solving these equations.



    Due to its well-established applications in various scientific and technical fields, fractional calculus has gained prominence during the last three decades. Many pioneers have shown that when adjusted by integer-order models, fractional-order models may accurately represent complex events [1,2]. The Caputo fractional derivatives are nonlocal in contrast to the integer-order derivatives, which are local in nature [1]. In other words, the integer-order derivative may be used to analyze changes in the area around a point, but the Caputo fractional derivative can be used to analyze changes in the whole interval. Senior mathematicians including Riemann [4], Caputo [5], Podlubny [6], Ross [7], Liouville [8], Miller and others, collaborated to create the fundamental foundation for fractional order integrals and derivatives. The theory of fractional-order calculus has been related to real-world projects, and it has been applied to chaos theory [9], signal processing [10], electrodynamics [11], human diseases [12,13], and other areas [14,15,16].

    Due to the numerous applications of fractional differential equations in engineering and science such as electrodynamics [17], chaos ideas [18], accounting [19], continuum and fluid mechanics [20], digital signal [21] and biological population designs [22] fractional differential equations are now more widely known. For such issues to be resolved, efficient tools are needed [23,24,25]. Because of this, we will attempt to apply an efficient analytical technique to solve nonlinear arbitrary order differential equations in this article. Many strategies in collaboration fields may be delightfully and even more accurately analyzed using fractional differential equations. Various strategies have been developed in this regard, some of them are as follows, such as the fractional Reduced differential transformation technique [26], Adomian decomposition technique [27], the fractional Variational iteration technique [28], Elzaki decomposition technique [29,30], iterative transformation technique [31], the fractional natural decomposition method (FNDM) [32], and the fractional homotopy perturbation method [33].

    The power series solution is used to solve some classes of the differential and integral equations of fractional or non-fractional order, and it is based on assuming that the solution of the equation can be expanded as a power series. RPS is an easy and fast technique for determining the coefficients of the power series solution. The Jordanian mathematician Omar Abu Arqub created the residual power series method in 2013, as a technique for quickly calculating the coefficients of the power series solutions for 1st and 2nd-order fuzzy differential equations [34]. Without perturbation, linearization, or discretization, the residual power series method provides a powerful and straightforward power series solution for highly linear and nonlinear equations [35,36,37,38]. The residual power series method has been used to solve an increasing variety of nonlinear ordinary and partial differential equations of various sorts, orders, and classes during the past several years. It has been used to make non-linear fractional dispersive partial differential equation have solitary pattern results and to predict them [39], to solve the highly nonlinear singular differential equation known as the generalized Lane-Emden equation [40], to solve higher-order ordinary differential equations numerically [41], to approximate solve the fractional nonlinear KdV-Burger equations, to predict and represent the RPSM differs from several other analytical and numerical approaches in some crucial ways [42]. First, there is no requirement for a recursion connection or for the RPSM to compare the coefficients of the related terms. Second, by reducing the associated residual error, the RPSM offers a straightforward method to guarantee the convergence of the series solution. Thirdly, the RPSM doesn't suffer from computational rounding mistakes and doesn't use a lot of time or memory. Fourth, the approach may be used immediately to the provided issue by selecting an acceptable starting guess approximation since the residual power series method does not need any converting when transitionary from low-order to higher-order and from simple linearity to complicated nonlinearity [43,44,45]. The process of solving linear differential equations using the LT method consists of three steps. The first step depends on transforming the original differential equation into a new space, called the Laplace space. In the second step, the new equation is solved algebraically in the Laplace space. In the last step, the solution in the second step is transformed back into the original space, resulting in the solution of the given problem.

    In this article, we apply the Laplace residual power series method to achieve the definitive solution of the fractional-order nonlinear partial differential equations. The Laplace transformation efficiently integrates the residual power series method for the renewability algorithmic technique. This proposed technique produces interpretive findings in the sense of a convergent series. The Caputo fractional derivative operator explains quantitative categorizations of the partial differential equations. The offered methodology is well demonstrated in modelling and enumeration investigations. The exact-analytical findings are a valuable way to analyze the problematic dynamics of systems, notably for computational fractional partial differential equations.

    Definition 2.1. The fractional Caputo derivative of a function u(ζ,t) of order α is given as [46]

    CDαtu(ζ,t)=Jmαtum(ζ,t),m1<αm,t>0, (2.1)

    where mN and Jαt is the fractional integral Riemann-Liouville (RL) of u(ζ,t) of order α is given as

    Jσtu(ζ,t)=1Γ(α)t0(tτ)α1u(φ,τ)dτ (2.2)

    Definition 2.2. The Laplace transformation (LT) of u(ζ,t) is given as [46]

    u(ζ,s)=Lt[u(ζ,t)]=0estu(ζ,t)dt,s>α, (2.3)

    where the Laplace transform inverse is defined as

    u(ζ,t)=L1t[u(ζ,s)]=l+iliestu(ζ,s)ds,l=Re(s)>l0. (2.4)

    Lemma 2.1. Suppose that u(ζ,t) is piecewise continue term and U(ζ,s)=Lt[u(ζ,t)], we get

    (1) Lt[Jαtu(ζ,t)]=U(ζ,s)sα,α>0.

    (2) Lt[Dαtu(ζ,t)]=sσU(ζ,s)m1k=0sαk1uk(ζ,0),m1<αm.

    (3) Lt[Dnαtu(ζ,t)]=snαU(ζ,s)n1k=0s(nk)α1Dkαtu(ζ,0),0<α1.

    Proof. For proof see Refs. [46].

    Theorem 2.1. Let u(ζ,t) be a piecewise continuous function on I×[0,) with exponential order ζ. Assume that the fractional expansion of the function U(ζ,s)=Lt[u(ζ,t)] is as follows:

    U(ζ,s)=n=0fn(ζ)s1+nα,0<α1,ζI,s>ζ. (2.5)

    Then, fn(ζ)=Dnσtu(ζ,0).

    Proof. For proof see Refs. [46].

    Remark 2.1. The inverse Laplace transform of the Eq (2.5) is represented as [46]

    u(ζ,t)=i=0Dαtu(ζ,0)Γ(1+iα)ti(ζ),0<ζ1,t0. (2.6)

    Consider the fractional order partial differential equation,

    DαtU(ζ,t)+3U(ζ,t)tζ24U(ζ,t)t2ζ2+4U(ζ,t)ζ4+a(2U(ζ,t)ζ2)2b(2U(ζ,t)t2)3+cU(ζ,t)=0. (3.1)

    Applying LT of Eq (3.1), we get

    U(ζ,s)+f0(ζ,s)s+1sα[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+aL(L1t(2U(ζ,s)ζ2))2bL(L1t(2U(ζ,s)t2))3+cU(ζ,s)]=0. (3.2)

    Suppose that the result of Eq (3.2), we get

    U(ζ,s)=n=0fn(ζ,s)snα+1. (3.3)

    The kth-truncated term series are

    U(ζ,s)=f0(ζ,s)s+kn=1fn(ζ,s)snα+1,k=1,2,3,4. (3.4)

    Residual Laplace function (RLF) is given as

    LtResu(ζ,s)=U(ζ,s)+f0(ζ,s)s+1sα[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+aL(L1t(2U(ζ,s)ζ2))2bL(L1t(2U(ζ,s)t2))3+cU(ζ,s)]. (3.5)

    And the kth-LRFs as

    LtResk(ζ,s)=Uk(ζ,s)+f0(ζ,s)s+1sσ[3Uk(ζ,s)tζ24Uk(ζ,s)t2ζ2+4U(ζ,s)ζ4+aL(L1t(2Uk(ζ,s)ζ2))2bL(L1t(2Uk(ζ,s)t2))3+cUk(ζ,s)]. (3.6)

    To illustrate a few facts, the following LRPSM features are provided:

    (1) LtRes(ζ,s)=0 and limjLtResk(ζ,s)=LtResu(ζ,s) for each s>0.

    (2) limssLtResu(ζ,s)=0limssLtResu,k(ζ,s)=0.

    (3) limsskα+1LtResu,k(ζ,s)=limsskα+1LtResu,k(ζ,s)=0,0<α1,k=1,2,3,.

    To calculate the coefficients using fn(ζ,s), gn(ζ,s), hn(ζ,s) and ln(ζ,s), the following system is recursively solved:

    limsskα+1LtResu,k(α,s)=0,k=1,2,. (3.7)

    In finally inverse Laplace transform to Eq (3.4), to get the kth analytical result of uk(ζ,t).

    Example 4.1. Consider the fractional partial differential equations [47],

    Dαtu(ζ,t)3u(ζ,t)tζ24u(ζ,t)t2ζ2+4u(ζ,t)ζ4+19(2u(ζ,t)ζ2)21216(2u(ζ,t)t2)3+16u(ζ,t)=0, where 2<α3, (4.1)

    with the following IC's:

    u(x,0)=ζ4,tu(ζ,0)=0,2t2u(ζ,0)=0. (4.2)

    Using Laplace transform to Eq (4.1), we get

    U(ζ,s)+ζ4s+1sα[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+19L(L1t(2U(ζ,s)ζ2))21216L(L1t(2U(ζ,s)t2))3+16U(ζ,s)]=0, (4.3)

    and so the kth-truncated term series are

    ζu(ζ,s)=ζ4s+kn=1fn(ζ,s)snα+1,k=1,2,3,4. (4.4)

    Residual Laplace function is given as

    LtResu(ζ,s)=U(ζ,s)+ζ4s+1sα[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+19L(L1t(2U(ζ,s)ζ2))21216L(L1t(2U(ζ,s)t2))3+16U(ζ,s)], (4.5)

    and the kth-LRFs as:

    LtResk(ζ,s)=Uk(ζ,s)+ζ4s+1sσ[3Uk(ζ,s)tζ24Uk(ζ,s)t2ζ2+4U(ζ,s)ζ4+19L(L1t(2Uk(ζ,s)ζ2))21216L(L1t(2Uk(ζ,s)t2))3+16Uk(ζ,s)]. (4.6)
    Table 1.  Comparison of the exact and proposed technique solution and various fractional-orders α and t=0.25 for Example 4.1.
    ζ α=2.5 α=2.7 α=2.9 α=3 HPM [47] Exact
    0 0.0222397 0.0111683 0.00553658 0.00388069 0.00388069 0.0038812
    0.2 0.0206397 0.00956834 0.00393658 0.00228069 0.00228069 0.0022812
    0.4 -0.00336028 -0.0144317 -0.0200634 -0.0217193 -0.0217193 -0.0217188
    0.6 -0.10736 -0.118432 -0.124063 -0.125719 -0.125719 -0.125719
    0.8 -0.38736 -0.398432 -0.404063 -0.405719 -0.405719 -0.405719
    1.0 -0.97776 -0.988832 -0.994463 -0.996119 -0.996119 -0.996119

     | Show Table
    DownLoad: CSV

    Now, we calculate fk(ζ,s), k=1,2,3,, substituting the kth-truncate series of Eq (4.4) into the kth residual Laplace term Eq (4.6), multiply the solution equation by skα+1, and then solve recursively the link lims(skα+1LtResu,k(ζ,s))=0, k=1,2,3,. Following are the first some term:

    f1(ζ,s)=24,f2(ζ,s)=384,f3(ζ,s)=6144. (4.7)

    Putting the value of fk(ζ,s), k=1,2,3,, in Eq (4.4), we get

    U(ζ,s)=ζ4s+24sα+1384s2α+1+6144s3α+1+. (4.8)

    Using inverse LT, we get

    u(ζ,t)=ζ4+24tαΓ(α+1)384t2αΓ(2α+1)+6144t3αΓ(3α+1)+, (4.9)

    and the exact solution are

    u=ζ4+4t3. (4.10)

    In Figure 1, the exact and LRPSM solutions for u(ζ,t) at α=3 at ζ and t=0.3 of Example 4.1. In Figure 2, analytical solution for u(ζ,t) at different value of α=2.8 and 2.6 at ζ and t=0.3. In Figure 3, analytical solution for u(ζ,t) at various value of α at t=0.3 of Example 4.1.

    Figure 1.  The actual and LRPSM results for u(ζ,t) at α=3 at ζ and t=0.3.
    Figure 2.  Analytical solution for u(ζ,t) at different value of α=2.8 and 2.6 at ζ and t=0.3.
    Figure 3.  Analytical solution for u(ζ,t) at various value of α at t=0.3.

    Example 4.2. Consider the fractional partial differential equations [47]:

    DαtU(ζ,t)3U(ζ,t)tζ24U(ζ,t)t2ζ2+4U(ζ,t)ζ4+(2U(ζ,t)ζ2)2(2U(ζ,t)t2)2+2U2(ζ,t)=0, where 2<α3, (4.11)

    with the following IC's:

    U(ζ,0)=eζ,tU(ζ,0)=eζ,2t2U(ζ,0)=eζ. (4.12)

    Using Laplace transform to Eq (4.11), we get

    U(ζ,s)eζseζs2eζs3+1sα[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+Lt(L1t(2U(ζ,s)ζ2))2Lt(L1t(2U(ζ,s)t2))2+2Lt(L1t(U(ζ,s)))2]=0. (4.13)
    Table 2.  Comparison of the exact and proposed technique solution and various fractional-orders α and t=0.099 for Example 4.2.
    ζ α=2.5 α=2.7 α=2.9 α=3 Exact
    0 1.08911 1.09052 1.09122 1.09142 1.09199
    0.2 1.32968 1.33169 1.33268 1.33297 1.33376
    0.4 1.62325 1.62612 1.62754 1.62795 1.62905
    0.6 1.98139 1.98553 1.98759 1.98818 1.98973
    0.8 2.41823 2.42422 2.42721 2.42807 2.43026
    1 2.95086 2.95959 2.96394 2.96519 2.96833

     | Show Table
    DownLoad: CSV

    Residual Laplace function is given as

    LtResu(ζ,s)=U(ζ,s)eζseζs2eζs3+1sα[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+Lt(L1t(2U(ζ,s)ζ2))2Lt(L1t(2U(ζ,s)t2))2+2Lt(L1t(U(ζ,s)))2], (4.14)

    and so the kth-truncated term series are

    u(ζ,s)=eζs+eζs2+eζs3+kn=1fn(ζ,s)snα+1,k=1,2,3,4, (4.15)

    and the kth-LRFs as:

    LtResk(ζ,s)=Uk(ζ,s)eζseζs2eζs3+1sσ[3Uk(ζ,s)tζ24Uk(ζ,s)t2ζ2+4Uk(ζ,s)ζ4+Lt(L1t(2Uk(ζ,s)ζ2))2Lt(L1t(2Uk(ζ,s)t2))2+2Lt(L1t(Uk(ζ,s)))2]. (4.16)

    Now, we calculate fk(ζ,s), k=1,2,3,, substituting the kth-truncate series of Eq (4.15) into the kth residual Laplace term Eq (4.16), multiply the solution equation by skα+1, and then solve recursively the link lims(skα+1LtResu,k(ζ,s))=0, k=1,2,3,. Following are the first some term:

    f1(ζ,s)=(eζ+3e2ζ),f2(ζ,s)=eζ+54e2ζ+36e3ζ,f3(ζ,s)=(eζ+870e2ζ+3564e3ζ+792e4ζ).   (4.17)

    Putting the value of fk(ζ,s), k=1,2,3,, in Eq (4.15), we get

    U(ζ,s)=eζs+eζs2+eζs3eζ+3e2ζsα+1eζ+54e2ζ+36e3ζs2α+1eζ+870e2ζ+3564e3ζ+792e4ζs3α+1+. (4.18)

    Using inverse LT, we get

    u(ζ,t)=eζ+eζt+eζt2(eζ+3e2ζ)tαΓ(α+1)+(eζ+54e2ζ+36e3ζ)t2αΓ(2α+1)+(eζ+870e2ζ+3564e3ζ+792e4ζ)t3αΓ(3α+1)+, (4.19)

    and the exact solution are

    u=eζ+t. (4.20)

    In Figure 4, the exact and LRPSM solutions for u(ζ,t) at α=3 at ζ and t=0.3 of Example 4.2. In Figure 5, LRPSM solutions for u(ζ,t) at α=2.5 and α=2.8 and t=0.3 of Example 4.2.

    Figure 4.  Exact and LRPSM solutions for u(ζ,t) at α=3 at ζ and t=0.3.
    Figure 5.  LRPSM solutions for u(ζ,t) at α=2.5 and α=2.8 and t=0.3.

    Example 4.3. Consider the fractional partial differential equations [47]:

    Dαtu(ζ,t)3u(ζ,t)tζ24u(ζ,t)t2x2+4u(ζ,t)ζ4(2u(ζ,t)t2)(u(ζ,t)ζ)u(ζ,t)(u(ζ,t)t)=0, where 2<α3, (4.21)

    with the following IC's:

    U(ζ,0)=cosζ,tU(ζ,0)=sinζ,2t2U(ζ,0)=cosζ. (4.22)
    Table 3.  Comparison of the exact and proposed technique solution and various fractional-orders α and t=0.22 for Example 4.3.
    ζ α=2.5 α=2.7 α=2.9 α=3 Exact
    0 0. 0.97178 0.973463 0.974025 0.975897
    0.2 -0.04370738 0.908702 0.910351 0.910903 0.913089
    0.4 -0.085672 0.809397 0.810946 0.811465 0.813878
    0.6 -0.124221 0.677823 0.679212 0.679677 0.682221
    0.8 -0.157818 0.519227 0.5204 0.520792 0.523366
    1 -0.185124 0.339931 0.34084 0.341145 0.343646

     | Show Table
    DownLoad: CSV

    Using Laplace transform to Eq (4.21), we get

    U(ζ,s)cosζs+sinζs2+cosζs3+1sσ[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+L(L1t(2u(ζ,s)t2)L1t(u(ζ,s)ζ))L(L1t(U(ζ,s))L1t(U(ζ,s)t))]=0. (4.23)

    Residual Laplace function is given as

    LtResu(ζ,s)=U(ζ,s)cosζs+sinζs2+cosζs3+1sσ[3U(ζ,s)tζ24U(ζ,s)t2ζ2+4U(ζ,s)ζ4+L(L1t(2u(ζ,s)t2)L1t(u(ζ,s)ζ))L(L1t(U(ζ,s))L1t(U(ζ,s)t))], (4.24)

    and so the kth-truncated term series are

    u(ζ,s)=cosζs+sinζs2+cosζs3+kn=1fn(ζ,s)snα+1,k=1,2,3,4, (4.25)

    and the kth-LRFs as:

    LtResk(ζ,s)=Uk(ζ,s)cosζs+sinζs2+cosζs3+1sα[3Uk(ζ,s)tζ24Uk(ζ,s)t2ζ2+4Uk(ζ,s)ζ4+Lt(L1t(2Uk(ζ,s)t2)L1t(Uk(ζ,s)ζ))Lt(L1t(Uk(ζ,s))L1t(Uk(ζ,s)t))]. (4.26)

    Now, we calculate fk(ζ,s), k=1,2,3,, substituting the kth-truncate series of Eq (4.25) into the kth residual Laplace term Eq (4.26), multiply the solution equation by skα+1, and then solve recursively the link lims(skα+1LtResu,k(ζ,s))=0, k=1,2,3,. Following are the first some term:

    f1(ζ,s)=cosζ,f2(ζ,s)=cosζ,f3(ζ,s)=cosζ. (4.27)

    Putting the value of fk(x,s), k=1,2,3,, in Eq (4.25), we get

    U(ζ,s)=cosζssinζs2cosζs3cosζsα+1+cosζs2α+1cosζs3α+1+. (4.28)

    Using inverse LT, we get

    u(ζ,t)=cosζtsinζt2cosζ2tαcosζΓ(α+1)+t2αcosζΓ(2α+1)t3αcosζΓ(3α+1)+, (4.29)

    and the exact solution are

    u=cos(ζ+t). (4.30)

    In Figure 6, exact and LRPSM solutions for u(ζ,t) at α=3 and t=0.3 of Example 4.3. Figure 7, LRPSM solutions for u(ζ,t) at α=2.5, α=2.8, and t=0.3.

    Figure 6.  Exact and LRPSM solutions for u(ζ,t) at α=3 at and t=0.3.
    Figure 7.  LRPSM solutions for u(ζ,t) at α=2.5 α=2.8, and t=0.3.

    In this article, the fractional partial differential equation has been solved analytically by employing the Laplace residual power series method in conjunction with the Caputo operator. To demonstrate the validity of the recommended method, we analyzed three distinct partial differential equation problems. The simulation results demonstrate that the outcomes of our method are in close accordance with the exact answer. The new method is highly straightforward, efficient, and suitable for getting numerical solutions to partial differential equations. The primary advantage of the proposed approach is the series form solution, which rapidly converges to the exact answer. We can therefore conclude that the suggested approach is quite methodical and efficient for a more thorough investigation of fractional-order mathematical models.

    The authors declare no conflicts of interest.



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