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On some dynamic inequalities of Hilbert's-type on time scales

  • † These two authors contributed equally to this work and are co-first authors.
  • In this article, we will prove some new conformable fractional Hilbert-type dynamic inequalities on time scales. These inequalities generalize some known dynamic inequalities on time scales, unify and extend some continuous inequalities and their corresponding discrete analogues. Our results will be proved by using some algebraic inequalities, conformable fractional Hölder inequalities, and conformable fractional Jensen's inequalities on time scales.

    Citation: Ahmed A. El-Deeb, Dumitru Baleanu, Nehad Ali Shah, Ahmed Abdeldaim. On some dynamic inequalities of Hilbert's-type on time scales[J]. AIMS Mathematics, 2023, 8(2): 3378-3402. doi: 10.3934/math.2023174

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  • In this article, we will prove some new conformable fractional Hilbert-type dynamic inequalities on time scales. These inequalities generalize some known dynamic inequalities on time scales, unify and extend some continuous inequalities and their corresponding discrete analogues. Our results will be proved by using some algebraic inequalities, conformable fractional Hölder inequalities, and conformable fractional Jensen's inequalities on time scales.



    Fractional calculus (FC) is a subject that dates back to 1695 and is regarded to be as old as ordinary calculus. Ordinary calculus made it impossible to model nonlinear real-world phenomenon in nature, hence fractional calculus became popular among researchers. The fractional derivative is the derivative of arbitrary order in applied mathematics and mathematical analysis. In the nineteenth century, Riemann Liouville [1] implemented the fractional derivative when simulating real-world problems. Some of the Nobel contributions of mathematicians are listed here, Caputo [2], Kemple and Beyer [3], Abbasbandy [4], Jafari and Seifi [5, 6], Miller and Ross [7], Podlubny [8], Kilbas and Trujillo [9], Diethelm et al. [10], Hayat et al. [11], Debanth [12], Momani and Shawagfeh [13], etc. Fractional calculus has become one of the fascinating science and engineering research areas in recent years. Viscoelasticity and damping, diffusion and wave propagation, electromagnetism and heat transfer, biology, signal processing, robotics system classification, physics, mechanics, chemistry, and control theory are the most important scientific fields that use fractional calculus at the moment.

    Researchers are interested in FC because of its wide applications in physics, engineering, and real-life sciences. Fractional differential equations accurately represent these physical facts. Fractional differential equations are significantly greater generalizations of integer-order differential equations. Fractional differential equations (FDEs) have generated much interest in recent years. As a result of their frequent appearance in diverse applications, such as quantum mechanics [14], chaotic dynamics [15], plasma physics [16, 17], theory of long-range interaction [18], mechanics of non-Hamiltonian systems [19], physical kinetics [20], anomalous diffusion and transport theory [21], mechanics of fractional media [22], astrophysics [23], and so on. FDEs have been the subject of numerous investigations. Many works have been dedicated to developing efficient methods for solving FDEs, but it is important to remember that finding an analytical or approximate solution is difficult, therefore, accurate methods for obtaining FDE solutions are still being researched. In the literature, there are several analytical and numerical approaches for solving FDEs. For example, the generalized differential transform method (GDTM) [24], adomian decomposition method (ADM) [25], homotopy analysis method (HAM) [26], variational iteration method (VIM) [27], homotopy perturbation method (HPM) [28], Elzaki transform decomposition method (ETDM) [29], iterative Laplace transform method (ILTM) [30], fractional wavelet method (FWM) [31, 32], residual power series method (RPSM) [33, 34].

    In this paper, we used two powerful techniques with the aid of the Antagana-Baleanu fractional derivative operator and Laplace transform for solving time fractional NWSEs. The two well-known methods that we implement are LTDM and VITM. The suggested techniques give series form solutions having quick convergence towards the exact solutions. Four non-linear NWSEs case study issues are resolved using the given methodology. Newell and Whitehead [35] developed the non-linear NWSE. The diffusion term's influence interacts with the reaction term's nonlinear effect in the Newell-Whitehead-Segel equation model. The fractional NWSE is written in the following way:

    Dτμ(ξ,τ)=kD2ξμ(ξ,τ)+gμhμr, (1.1)

    where r is a positive integer and g,h are real numbers with k>0. The first term Dτμ(ξ,τ) on the left hand side in (1.1) shows the deviations of μ(ξ,τ) with time at a fixed location, while the right hand side first term D2ξμ(ξ,τ) shows the deviations with spatial variable ξ of μ(ξ,τ) at a specific time and the right hand side remaining terms gμhμr, is the source terms. μ(ξ,τ) is a function of the spatial variable ξ and the temporal variable τ in (1.1), with ξR and τ0. The function μ(ξ,τ) might be considered the (nonlinear) temperature distribution in an infinitely thin and long rod or as the fluid flow velocity in an infinitely long pipe with a small diameter. Many researchers find the analytical solution of NWSEs [36, 37] due to their wide range of applications in mechanical and chemical engineering, ecology, biology, and bioengineering.

    Some basic definitions related to fractional calculus are expressed here in this section.

    Definition 2.1. The Caputo fractional-order derivative is given as

    LCDτ{g(τ)}=1(n)τ0(τk)n1gn(k)dk, (2.1)

    where n<n+1.

    Definition 2.2. The Caputo fractional-order derivative via Laplace transformation LCDτ{g(τ)} is defined as

    L{LCDτ{g(τ)}}(ω)=1ωn[ωnL{g(ξ,τ)}(ω)ωn1g(ξ,0)gn1(ξ,0)]. (2.2)

    Definition 2.3. The Atangana-Baleanu derivative in Caputo manner is given as

    ABCDτ{g(τ)}=A()1τag(k)E[1(1k)]dk, (2.3)

    where A(γ) is a normalization function such that A(0)=A(1)=1,gH1(a,b),b>a, [0,1] and Eγ represent the Mittag-Leffler function.

    Definition 2.4. The Atangana-Baleanu derivative in Riemann-Liouville manner is given as

    ABCDτ{g(τ)}=A()1ddττag(k)E[γ1(1k)]dk. (2.4)

    Definition 2.5. The Laplace transform connected with the Atangana-Baleanu operator is define as

    ABDτ{g(τ)}(ω)=A(γ)ωL{g(τ)}(ω)ω1g(0)(1)(ω+1γ). (2.5)

    Definition 2.6. Consider 0<<1, and g is a function of , then the fractional-order integral operator of is given as

    ABCIτ{g(τ)}=1A()g(τ)+A()Γ()τag(k)(τk)1dk. (2.6)

    The solution by LTDM for partial differential equations having fractional-order is described in this section.

    Dτμ(ξ,τ)+ˉG1(ξ,τ)+N1(ξ,τ)=F(ξ,τ),0<1, (3.1)

    with some initial sources

    μ(ξ,0)=ξ(ξ)andτμ(ξ,0)=ζ(ξ),

    where Dτ=τ is the fractional-order AB operator having order ,ˉG1 is linear operator and N1 is non-linear and F(ξ,τ) indicates the source term.

    Employing the Laplace transform to (3.1), and we acquire

    L[Dτμ(ξ,τ)+ˉG1(ξ,τ)+N1(ξ,τ)]=L[F(ξ,τ)]. (3.2)

    By the virtue of Laplace differentiation property, we have

    L[μ(ξ,τ)]=Θ(ξ,ω)ω+(1)ωL[ˉG1(ξ,τ)+N1(ξ,τ)], (3.3)

    where

    Θ(ξ,ω)=1ω+1[ωg0(ξ)+ω1g1(ξ)++g1(ξ)]+ω+(1)ωF(ξ,τ).

    Now, applying inverse Laplace transform yields (3.3) into

    μ(ξ,τ)=Θ(ξ,ω)L1{ω+(1)ωL[ˉG1(ξ,τ)+N1(ξ,τ)]}, (3.4)

    where Θ(ξ,ω) demonstrates the terms occurring from source factor. LTDM determines the solution of the infinite sequence of μ(ξ,τ)

    μ(ξ,τ)=m=0μm(ξ,τ). (3.5)

    and decomposing the nonlinear operator N1 as

    N1(ξ,τ)=m=0Am, (3.6)

    where Am are Adomian polynomials given as

    Am=1m![mm{N1(k=0kξk,k=0kτk)}]=0. (3.7)

    Putting (3.5) and (3.7) into (3.4), gives

    m=0μm(ξ,τ)=Θ(ξ,ω)L1{ω+(1)ωL[ˉG1(m=0ξm,m=0τm)+m=0Am]}. (3.8)

    The following terms are described:

    μ0(ξ,τ)=Θ(ξ,ω) (3.9)
    μ1(ξ,τ)=L1{ω+(1)ωL[ˉG1(ξ0,τ0)+A0]}. (3.10)

    Thus all components for m1 are calculated as

    μm+1(ξ,τ)=L1{ω+(1)ωL[ˉG1(ξm,τm)+Am]}. (3.11)

    The VITM solution for FPDEs is defined in this section.

    Dτμ(ξ,τ)+Mμ(ξ,τ)+Nμ(ξ,τ)P(ξ,τ)=0,m1<m, (4.1)

    with initial source

    μ(ξ,0)=g1(ξ), (4.2)

    where Dτ=τ is stand for fractional-order AB operator, M is a linear operator and N is nonlinear term and P indicates the source term.

    The Laplace transform is applied to Eq (4.1), we have

    L[Dτμ(ξ,τ)]+L[Mμ(ξ,τ)+Nμ(ξ,τ)P(ξ,τ)]=0. (4.3)

    By the property of LT differentiation, we get

    L[μ(ξ,τ)]=ωω+(1)L[Mμ(ξ,τ)+Nμ(ξ,τ)P(ξ,τ)]. (4.4)

    The iteration technique for (4.4) as

    μm+1(ξ,τ)=μm(ξ,τ)+(s)[ωω+(1)L[Mμ(ξ,τ)+Nμ(ξ,τ)P(ξ,τ)]], (4.5)

    where (s) is Lagrange multiplier and

    (s)=ω+(1)ω, (4.6)

    with the application of inverse Laplace transform, (4.5) series form solution is given by

    μ0(ξ,τ)=μ(0)+L1[(s)L[P(ξ,τ)]]μ1(ξ,τ)=L1[(s)L[Mμ(ξ,τ)+Nμ(ξ,τ)]]μn+1(ξ,τ)=L1[(s)L[M[μ0(ξ,τ)+μ1(ξ,τ)++μn(ξ,τ)]]+N[μ0(ξ,τ)+μ1(ξ,τ)++μn(ξ,τ)]].

    Here we discuss uniqueness and convergence analysis.

    Theorem 5.1. The result of (3.1) is unique for LTDMABC, when

    0<(θ1+θ2)(1+τμΓ(μ+1))<1.

    Proof. Let H=(C[J],||.||) with the norm ||ϕ(τ)||=maxτJ|ϕ(τ)| be the Banach space, for all continuous function on J. Let I:HH is a non-linear mapping, where

    μCl+1=μC0+L1[p(,υ,ω)L[L(μl(ξ,τ))+N(μl(ξ,τ))]],l0.

    Suppose that |L(μ)L(μ)|<θ1|μμ| and |N(μ)N(μ)|<θ2|μμ|, where μ:=μ(ξ,τ) and μ:=μ(ξ,τ) are two different function values and θ1,θ2 are Lipschitz constants.

    ||IμIμ||maxtJ|L1[q(,υ,ω)L[L(μ)L(μ)]+q(,υ,ω)L[N(μ)N(μ)]|]maxtJ[θ1L1[q(,υ,ω)L[|μμ|]]+θ2L1[q(,υ,ω)L[|μμ|]]]maxtJ(θ1+θ2)[L1[q(,υ,ω)L|μμ|]](1+θ2)[L1[q(,υ,ω)L||μμ||]]=(θ1+θ2)(1+τΓ+1)||μμ||, (5.1)

    where I is contraction as

    0<(θ1+θ2)(1+τΓ+1)<1.

    From Banach fixed point theorem, the result of (3.1) is unique.

    Theorem 5.2. The LTDMABC result of (3.1) is convergent.

    Proof. Let μm=mr=0μr(ξ,τ). To show that μm is a Cauchy sequence in H. For nN, let

    ||μmμn||=maxτJ|mr=n+1μr|maxτJ|L1[q(,υ,ω)L[mr=n+1(L(μr1)+N(μr1))]]|=maxτJ|L1[q(,υ,ω)L[m1r=n+1(L(μr)+N(ur))]]|maxτJ|L1[q(,υ,ω)L[(L(μm1)L(μn1)+N(μm1)N(μn1))]]|θ1maxτJ|L1[q(,υ,ω)L[(L(μm1)L(μn1))]]|+θ2maxτJ|L1[p(,υ,ω)L[(N(μm1)N(μn1))]]|=(θ1+θ2)(1+τΓ(+1))||μm1μn1||. (5.2)

    Let m=n+1, then

    ||μn+1μn||θ||μnμn1||θ2||μn1μn2||θn||μ1μ0||, (5.3)

    where

    θ=(θ1+θ2)(1+τΓ(+1)).

    Similarly, we have

    ||μmμn||||μn+1μn||+||μn+2μn+1||++||μmμm1||(θn+θn+1++θm1)||μ1μ0||θn(1θmn1θ)||μ1||. (5.4)

    As 0<θ<1, we get 1θmn<1. Therefore, we have

    ||μmμn||θn1θmaxtJ||μ1||. (5.5)

    Since ||μ1||< and ||μmμn||0, when n. As a result, μm is a Cauchy sequence in H, implying that the series μm is convergent.

    Four cases of nonlinear NWSEs are presented to demonstrate the suggested technique's capability and reliability.

    Example 1. The NWSE has given in (1.1) for g=2,h=3,k=1 and r=2 becomes

    Dτμ(ξ,τ)=D2ξμ(ξ,τ)+2μ(ξ,τ)3μ2(ξ,τ),0<θ1, (6.1)

    with initial source μ(ξ,0)=Υ.

    Applying Laplace transform to (6.1), we get

    ωL[μ(ξ,τ)]ω1μ(ξ,0)ω+(1ω)=L[D2ξμ(ξ,τ)+2μ(ξ,τ)3μ2(ξ,τ)]. (6.2)

    On taking Laplace inverse transform, we get

    μ(ξ,τ)=Υ+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+2μ(ξ,τ)3μ2(ξ,τ)]]. (6.3)

    Assume that the solution, μ(ξ,τ) in the form of infinite series given by

    μ(ξ,τ)=m=0μm(ξ,τ), (6.4)

    where μ2=m=0Am are the so-called Adomian polynomials that represent the nonlinear terms, thus (6.3) having certain terms are rewritten as

    m=0μm(ξ,τ)=Υ+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+2μ(ξ,τ)3m=0Am]]. (6.5)

    According to (3.7), the decomposition of nonlinear terms by Adomian polynomials is defined as

    A0=μ20,A1=2μ0μ1,A2=2μ0μ2+(μ1)2. (6.6)

    Thus, on comparing both sides of (6.5)

    μ0(ξ,τ)=Υ.

    For m=0, we have

    μ1(ξ,τ)=Υ(23Υ)[τΓ(+1)+(1)].

    For m=1, we have

    μ2(ξ,τ)=2Υ(23Υ)(13Υ)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2].

    The approximate series solution is expressed as

    μ(ξ,τ)=m=0μm(ξ,τ)=μ0(ξ,τ)+μ1(ξ,τ)+μ2(ξ,τ)+.

    Therefore, we have

    μ(ξ,τ)=Υ+Υ(23Υ)[τΓ(+1)+(1)]+2Υ(23Υ)(13Υ)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+.

    Particularly, putting =1, we get the exact solution

    μ(μ,τ)=23Υexp(2τ)23+ΥΥexp(2τ). (6.7)

    The analytical results by VITM:

    The iteration formulas for (6.1), we have

    μm+1(ξ,τ)=μm(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμm(ξ,τ)+2μm(ξ,τ)3μ2m(ξ,τ)}], (6.8)

    where μ0(ξ,τ)=Υ.

    For m=0,1,2,, we have

    μ1(ξ,τ)=μ0(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ0(ξ,τ)+2μ0(ξ,τ)3μ20(ξ,τ)}]μ1(ξ,τ)=Υ(23Υ)[τΓ(+1)+(1)], (6.9)
    μ2(ξ,τ)=μ1(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ0(ξ,τ)+2μ0(ξ,τ)3μ20(ξ,τ)}]μ2(ξ,τ)=2Υ(23Υ)(13Υ)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2], (6.10)

    Therefore, we obtain

    μ(ξ,τ)=m=0μm(ξ,τ)=Υ+Υ(23Υ)[τΓ(+1)+(1)]+2Υ(23Υ)(13Υ)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+. (6.11)

    Particularly, putting =1, we get the exact solution, see Figure 1 and Table 1.

     μ(ξ,τ)=23Υexp(2τ)23+ΥΥexp(2τ). (6.12)
    Figure 1.  The solution graph of Example 1, the exact solution and the analytical solution at =1.
    Table 1.  μ(ξ,τ) Comparison of the exact solution, our methods solution and absolute error (AE) of Example 1.
    τ=0.0001 Exact solution Our methods solution AE of our methods AE of our methods AE of our methods
    Υ =1 =1 =1 =0.9 =0.8
    0 0.000000000000000 0.000000000000000 0.0000000000E+00 0.0000000000E+00 0.0000000000E+00
    0.1 0.100000170000000 0.100000100000000 7.0000000000E-08 2.4390000000E-07 1.5317000000E-06
    0.2 0.200000280000000 0.200000200000000 8.0000000000E-08 5.4790000000E-07 3.1233000000E-06
    0.3 0.300000330000000 0.300000300000000 3.0000000000E-08 9.1180000000E-07 4.7750000000E-06
    0.4 0.400000320000000 0.400000400000000 8.0000000000E-08 1.3357000000E-06 6.4866000000E-06
    0.5 0.500000250000000 0.500000500000000 2.5000000000E-07 1.8197000000E-06 8.2582000000E-06
    0.6 0.600000120000000 0.600000600000000 4.8000000000E-07 2.3636000000E-06 1.0089700000E-05
    0.7 0.699999930000000 0.700000700000000 7.7000000000E-07 2.9675000000E-06 1.1981300000E-05
    0.8 0.799999680000000 0.800000800000000 1.1200000000E-06 3.6315000000E-06 1.3932800000E-05
    0.9 0.899999370000000 0.900000900000000 1.5300000000E-06 4.3554000000E-06 1.5944400000E-05
    1.0 0.999999000000000 1.000001000000000 2.0000000000E-06 5.1390000000E-06 1.8016000000E-05

     | Show Table
    DownLoad: CSV

    Example 2. The NWSE has given in (1.1) for g=1,h=1,k=1 and r=2 becomes

    Dτμ(ξ,τ)=D2ξμ(ξ,τ)+μ(ξ,τ)(1μ(ξ,τ)),0<1, (6.13)

    with initial source

    μ(ξ,0)=1(1+exp(ξ6))2.

    Applying Laplace transform to (6.13), we get

    ωL[μ(ξ,τ)]ω1μ(ξ,0)ω+(1ω)=L[D2ξμ(ξ,τ)+μ(ξ,τ)(1μ(ξ,τ))]. (6.14)

    On taking Laplace inverse transform, we get

    μ(ξ,τ)=1(1+exp(ξ6))2+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+μ(ξ,τ)(1μ(ξ,τ))]]. (6.15)

    Assume that the solution, μ(ξ,τ) in the form of infinite series given by

    μ(ξ,τ)=m=0μm(ξ,τ), (6.16)

    where μ2=m=0Am are the so-called Adomian polynomials that represent the nonlinear terms, thus (6.15) having certain terms are rewritten as

    m=0μm(ξ,τ)=1(1+exp(ξ6))2+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+μ(ξ,τ)m=0Am]]. (6.17)

    Applying the proposed analytical approach and the nonlinear terms can be obtained with the aid of Adomian's polynomials stated in (3.7), we acquire

    μ0(ξ,τ)=1(1+exp(ξ6))2.

    For m=0, we have

    μ1(ξ,τ)=53exp(ξ6)(1+exp(ξ6))3[τΓ(+1)+(1)].

    For m=1, we have

    μ2(ξ,τ)=2518(exp(ξ6)(1+2exp(ξ6))(1+exp(ξ6))4)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2].

    The approximate series solution is expressed as

    μ(ξ,τ)=m=0μm(ξ,τ)=μ0(ξ,τ)+μ1(ξ,τ)+μ2(ξ,τ)+.

    Therefore, we obtain

    μ(ξ,τ)=1(1+exp(ξ6))2+53exp(ξ6)(1+exp(ξ6))3[τΓ(+1)+(1)]+2518(exp(ξ6)(1+2exp(ξ6))(1+exp(ξ6))4)×[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+.

    Particularly, putting =1, we get the exact solution

    μ(μ,τ)=(11+exp(ξ656τ))2. (6.18)

    The analytical results by VITM:

    The iteration formulas for (6.13), we have

    μm+1(ξ,τ)=μm(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμm(ξ,τ)+μm(ξ,τ)(1μm(ξ,τ))}], (6.19)

    where

    μ0(ξ,τ)=1(1+exp(ξ6).

    For m=0,1,2,, we have

    μ1(ξ,τ)=μ0(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ0(ξ,τ)+μ0(ξ,τ)(1μ0(ξ,τ))}]μ1(ξ,τ)=53exp(ξ6)(1+exp(ξ6))3[τΓ(+1)+(1)] (6.20)
    μ2(ξ,τ)=μ1(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ1(ξ,τ)+μ1(ξ,τ)(1μ1(ξ,τ))}]μ2(ξ,τ)=2518(exp(ξ6)(1+2exp(ξ6))(1+exp(ξ6))4)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2] (6.21)

    Therefore, we obtain

    μ(ξ,τ)=m=0μm(ξ,τ)=1(1+exp(ξ6))2+53exp(ξ6)(1+exp(ξ6))3[τΓ(+1)+(1)]+2518(exp(ξ6)(1+2exp(ξ6))(1+exp(ξ6))4)[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+. (6.22)

    Particularly, putting =1, we get the exact solution, see Figures 2, 3 and Table 2.

    μ(ξ,τ)=(11+exp(ξ656τ))2. (6.23)
    Figure 2.  The exact solution and analytical solution of Example 2 at =1.
    Figure 3.  The two and three dimensional different fractional order graphs of example 2.
    Table 2.  μ(ξ,τ) Comparison of exact solution, our methods solution and Absolute Error (AE) of Example 2.
    τ=0.0001 Exact solution Our methods solution AE of our methods AE of our methods AE of our methods
    ξ =1 =1 =1 =0.9 =0.8
    0 0.250020833800000 0.250020833700000 1.0000000000E-10 3.3580800000E-05 1.2032390000E-04
    0.1 0.239919747900000 0.239919747800000 5.0941566800E-11 3.2881658260E-05 1.1781851670E-04
    0.2 0.230035072700000 0.230035072600000 8.1583004800E-11 3.2156851240E-05 1.1522127190E-04
    0.3 0.220374241800000 0.220374241700000 1.3295068340E-10 3.1408781520E-05 1.1254074080E-04
    0.4 0.210943948200000 0.210943948100000 1.7637132220E-10 3.0639941880E-05 1.0978578190E-04
    0.5 0.201750129200000 0.201750129000000 1.5669454130E-10 2.9852894210E-05 1.0696543270E-04
    0.6 0.192797956100000 0.192797956000000 1.0301314750E-10 2.9050141560E-05 1.0408873760E-04
    0.7 0.184091829000000 0.184091828700000 2.3924337590E-10 2.8234000470E-05 1.0116457590E-04
    0.8 0.175635376500000 0.175635376200000 2.9445508530E-10 2.7407274000E-05 9.8202291510E-05
    0.9 0.167431461100000 0.167431460800000 3.1297190840E-10 2.6572425380E-05 9.5210825130E-05
    1.0 0.159482188800000 0.159482188600000 2.6399135400E-10 2.5731952810E-05 9.2199048510E-05

     | Show Table
    DownLoad: CSV

    Example 3. The NWSE has given in (1.1) for g=1,h=1,k=1 and r=4 becomes

    Dτμ(ξ,τ)=D2ξμ(ξ,τ)+μ(ξ,τ)μ4(ξ,τ),0<1 (6.24)

    with initial source

    μ(ξ,0)=1(1+exp(3ξ10))23.

    Applying Laplace transform to (6.24), we get

    ωL[μ(ξ,τ)]ω1μ(ξ,0)ω+(1ω)=L[D2ξμ(ξ,τ)+μ(ξ,τ)μ4(ξ,τ)]. (6.25)

    On taking Laplace inverse transform, we get

    μ(ξ,τ)=1(1+exp(3ξ10))23+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+μ(ξ,τ)μ4(ξ,τ)]]. (6.26)

    Assume that the solution, μ(ξ,τ) in the form of infinite series given by

    μ(ξ,τ)=m=0μm(ξ,τ), (6.27)

    where μ4=m=0Am are the so-called Adomian polynomials that represent the nonlinear terms, thus (6.26) having certain terms are rewritten as

    m=0μm(ξ,τ)=1(1+exp(3ξ10))23+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+μ(ξ,τ)m=0Am]]. (6.28)

    According to (3.7), the decomposition of nonlinear terms by Adomian polynomials is defined as

    A0=μ40,A1=4μ30μ1,A2=4μ30μ2+6μ20μ21. (6.29)

    Thus, on comparing both sides of (6.30)

    μ0(ξ,τ)=1(1+exp(3ξ10))23.

    For m=0, we have

    μ1(ξ,τ)=75exp(3ξ10)(1+exp(3ξ10))53[τΓ(+1)+(1)].

    For m=1, we have

    μ2(ξ,τ)=4950(2exp(3ξ10)3)exp(3ξ10)(1+exp(3ξ10))83[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2].

    The approximate series solution is expressed as

    μ(ξ,τ)=m=0μm(ξ,τ)=μ0(ξ,τ)+μ1(ξ,τ)+μ2(ξ,τ)+.

    Therefore, we obtain

    μ(ξ,τ)=1(1+exp(3ξ10))23+75exp(3ξ10)(1+exp(3ξ10))53[τΓ(+1)+(1)]+4950(2exp(3ξ10)3)exp(3ξ10)(1+exp(3ξ10))83[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+.

    Particularly, putting =1, we get the exact solution

     μ(ξ,τ)=12tanh(3210(ξ710τ)). (6.30)

    The analytical results by VITM:

    The iteration formulas for (6.24), we have

    μm+1(ξ,τ)=μm(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμm(ξ,τ)+μm(ξ,τ)μ4m(ξ,τ)}], (6.31)

    where

    μ0(ξ,τ)=1(1+exp(3ξ10))23.

    For m=0,1,2,, we have

    μ1(ξ,τ)=μ0(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ0(ξ,τ)+μ0(ξ,τ)μ40(ξ,τ)}],μ1(ξ,τ)=75exp(3ξ10)(1+exp(3ξ10))53[τΓ(+1)+(1)], (6.32)
    μ2(ξ,τ)=μ1(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ1(ξ,τ)+μ1(ξ,τ)μ21(ξ,τ)}],μ2(ξ,τ)=4950(2exp(3ξ10)3)exp310ξ(1+exp(3ξ10))83[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]. (6.33)

    Therefore, we obtain

    μ(ξ,τ)=m=0μm(ξ,τ)=1(1+exp(3ξ10))23+75exp(3ξ10)(1+exp(3ξ10))53[τΓ(+1)+(1)]+4950(2exp(3ξ10)3)exp(3ξ10)(1+exp(3ξ10))83[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+. (6.34)

    Particularly, putting =1, we get the exact solution, see Figures 4, 5 and Table 3.

    μ(ξ,τ)=12tanh(3210(ξ710τ)). (6.35)
    Figure 4.  The exact solution and analytical solution graph of Example 3 at =1.
    Figure 5.  The error graph of Example 3 at =1.
    Table 3.  μ(ξ,τ) Comparison of exact solution, our methods solution and Absolute Error (AE) of Example 3.
    τ=0.0001 Exact solution Our methods solution AE of our methods AE of our methods AE of our methods
    ξ =1 =1 =1 =0.9 =0.8
    0 0.630004621400000 0.630004623000000 1.6000000000E-09 7.1080200000E-05 2.5467930000E-04
    0.1 0.609938405400000 0.609938406700000 1.3721312690E-09 7.2077200130E-05 2.5825129220E-04
    0.2 0.589628520300000 0.589628521400000 1.1413782460E-09 7.2815436450E-05 2.6089584380E-04
    0.3 0.569148496800000 0.569148497700000 8.7201231790E-10 7.3288425700E-05 2.6259010230E-04
    0.4 0.548573070400000 0.548573071000000 6.9266503090E-10 7.3494089020E-05 2.6332630440E-04
    0.5 0.527977019100000 0.527977019700000 5.8687018460E-10 7.3434320340E-05 2.6311129590E-04
    0.6 0.507434030100000 0.507434030300000 1.9085441870E-10 7.3114619290E-05 2.6196573810E-04
    0.7 0.487015628200000 0.487015628200000 1.4619007300E-11 7.2545024110E-05 2.5992429560E-04
    0.8 0.466790202200000 0.466790202100000 1.0115621760E-10 7.1738176370E-05 2.5703268760E-04
    0.9 0.446822151300000 0.446822151000000 3.7765613140E-10 7.0709457140E-05 2.5334660600E-04
    1.0 0.427171171700000 0.427171171200000 5.2356035770E-10 6.9477036140E-05 2.4893040880E-04

     | Show Table
    DownLoad: CSV

    Example 4. The NWSE has given in (1.1) for g=3,h=4,k=1 and r=3 becomes

    Dτμ(ξ,τ)=D2ξμ(ξ,τ)+3μ(ξ,τ)4μ3(ξ,τ),  0<1, (6.36)

    with initial source

    μ(ξ,0)=34exp(6ξ)exp(6ξ)+exp(62ξ).

    Applying Laplace transform to (6.36), we get

    ωL[μ(ξ,τ)]ω1μ(ξ,0)ω+(1ω)=L[D2ξμ(ξ,τ)+3μ(ξ,τ)4μ3(ξ,τ)]. (6.37)

    On taking Laplace inverse transform, we get

    μ(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ)+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+3μ(ξ,τ)4μ3(ξ,τ)]]. (6.38)

    Assume that the solution, μ(ξ,τ) in the form of infinite series given by

    μ(ξ,τ)=m=0μm(ξ,τ), (6.39)

    where μ3=m=0Am are the so-called Adomian polynomials that represent the nonlinear terms, thus (6.38) having certain terms are rewritten as

    m=0μm(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ)+L1[ω+(1ω)ωL[D2ξμ(ξ,τ)+3μ(ξ,τ)4m=0Am]]. (6.40)

    According to (3.7), the decomposition of nonlinear terms by Adomian polynomials is defined as,

    A0=μ30,A1=3μ20μ1,A2=3μ20μ2+3μ0μ21. (6.41)

    Thus, on comparing both sides of (6.40)

    μ0(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ).

    For m=0, we have

    μ1(ξ,τ)=9234exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))2[τΓ(+1)+(1)].

    For m=1, we have

    μ2(ξ,τ)=81434exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))(exp(6ξ)+exp(62ξ))3×[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2].

    The approximate series solution is expressed as

    μ(ξ,τ)=m=0μm(ξ,τ)=μ0(ξ,τ)+μ1(ξ,τ)+μ2(ξ,τ)+.

    Therefore, we obtain

    μ(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ)+9234exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))2[τΓ(+1)+(1)]+81434exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))(exp(6ξ)+exp(62ξ))3×[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+.

    Particularly, putting =1, we get the exact solution

     μ(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ92τ). (6.42)

    The analytical results by VITM:

    The iteration formulas for (6.36), we have

    μm+1(ξ,τ)=μm(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμm(ξ,τ)+3μm(ξ,τ)4μ3m(ξ,τ)}], (6.43)

    where

    μ0(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ).

    For m=0,1,2,, we have

    μ1(ξ,τ)=μ0(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ0(ξ,τ)+3μ0(ξ,τ)4μ30(ξ,τ)}],μ1(ξ,τ)=9234exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))2[τΓ(+1)+(1)], (6.44)
    μ2(ξ,τ)=μ1(ξ,τ)L1[ω+(1ω)ωL{ωω+(1ω)D2ξμ1(ξ,τ)+3μ1(ξ,τ)4μ31(ξ,τ)}],μ2(ξ,τ)=81434exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))(exp(6ξ)+exp(62ξ))3×[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]. (6.45)

    Therefore, we obtain

    μ(ξ,τ)=m=0μm(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ)+9234exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))2[τΓ(+1)+(1)]+81434exp(6ξ)exp(62ξ)(exp(6ξ)+exp(62ξ))(exp(6ξ)+exp(62ξ))3×[2τ2Γ(2+1)+2(1)τΓ(+1)+(1)2]+. (6.46)

    Particularly, putting =1, we get the exact solution, see Figures 6, 7 and Table 4.

    μ(ξ,τ)=34exp(6ξ)exp(6ξ)+exp(62ξ92τ). (6.47)
    Figure 6.  The exact solution and analytical solution graph of Example 4 at =1.
    Figure 7.  The error graph of Example 4 at =1.
    Table 4.  μ(ξ,τ) Comparison of Exact solution, Our methods solution and Absolute Error (AE) of example 4.
    τ=0.0001 Exact solution Our methods solution AE of our methods AE of our methods AE of our methods
    ξ =1 =1 =1 =0.9 =0.8
    0 0.433022444800000 0.433022445000000 1.7320508080E-10 7.9386816730E-06 2.2292879540E-05
    0.1 0.459505816600000 0.459505816600000 3.2194421390E-11 7.9089032580E-06 2.2209048130E-05
    0.2 0.485791725000000 0.485791725200000 1.6732056440E-10 7.8207979170E-06 2.1961331750E-05
    0.3 0.511688623000000 0.511688622800000 2.1377960370E-10 7.6762722350E-06 2.1556125910E-05
    0.4 0.537016284000000 0.537016284300000 3.7853826310E-10 7.4809626080E-06 2.1006564010E-05
    0.5 0.561610593500000 0.561610593600000 1.3316432970E-10 7.2385076780E-06 2.0326137420E-05
    0.6 0.585327390000000 0.585327389900000 4.4316889200E-11 6.9562540670E-06 1.9533837800E-05
    0.7 0.608045207500000 0.608045207500000 4.4341099960E-11 6.6414869210E-06 1.8649762960E-05
    0.8 0.629666845500000 0.629666845900000 4.6131349780E-10 6.3016518320E-06 1.7694707350E-05
    0.9 0.650119804000000 0.650119803700000 7.6816056310E-10 5.9421274000E-06 1.6687344050E-05
    1.0 0.669355693000000 0.669355693500000 4.9676049840E-10 5.5740946070E-06 1.5651582380E-05

     | Show Table
    DownLoad: CSV

    Figure 1, show the behavior of the exact and proposed methods solution at =1 in (AB fractional derivative) manner of Example 1. The comparison of the exact and analytical solution of Example 2 is shown Figure 2, whereas the graphical view for various fractional orders is demonstrated with the help of figures. In Figure 3, the two and three dimensional different fractional order graphs of Example 2. The figures show that our solution approaches the exact solution as the fractional order goes towards the integer-order. Figure 4, demonstrate the layout of the exact and analytical solution while Figure 5 shows the error comparison of the exact and analytical results of Example 3. The error confirms the efficiency of the suggested techniques. The graphical view of Example 4 for exact and our solution can be seen in Figure 6, however, Figure 7 shows the error comparison of both results. Furthermore, the behavior of the exact and proposed method solution with the aid of absolute error at different orders of is shown in Tables 14. Finally, it is clear from the figures and tables that the proposed methods have a sufficient degree of accuracy and quick convergence towards the exact solution.

    The LTDM and VITM were used for solving time fractional Newell-Whitehead-Segel equation. The solution we obtained is a series that quickly converges to exact solutions. Four cases are studied, which shows that the proposed methods solutions strongly agree with the exact solution. It is found that the suggested techniques are easy to implement and need a small number of calculations. This shows that LTDM and VITM are very efficient, effective, and powerful mathematical tools easily applied in finding approximate analytic solutions for a wide range of real-world problems arising in science and engineering.

    This research received funding support from the National Science, Research and Innovation Fund (NSRF), Thailand.

    The authors declare that they have no competing interests.



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