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Solving a class of variable order nonlinear fractional integral differential equations by using reproducing kernel function

  • In this paper, reproducing kernel interpolation collocation method is explored for nonlinear fractional integral differential equations with Caputo variable order. In order to testify the feasibility of this method, several examples are studied from the different values of parameters. In addition, the influence of the parameters of the Jacobi polynomial on the numerical results is studied. Our results reveal that the present method is effective and provide highly precise numerical solutions for solving such fractional integral differential equations.

    Citation: Zhi-Yuan Li, Mei-Chun Wang, Yu-Lan Wang. Solving a class of variable order nonlinear fractional integral differential equations by using reproducing kernel function[J]. AIMS Mathematics, 2022, 7(7): 12935-12951. doi: 10.3934/math.2022716

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  • In this paper, reproducing kernel interpolation collocation method is explored for nonlinear fractional integral differential equations with Caputo variable order. In order to testify the feasibility of this method, several examples are studied from the different values of parameters. In addition, the influence of the parameters of the Jacobi polynomial on the numerical results is studied. Our results reveal that the present method is effective and provide highly precise numerical solutions for solving such fractional integral differential equations.



    Fractional order models [1,2,3,4,5] have important applications in materials sciences, information science, and so on [6,7,8,9,10,11,12,13,14]. Reproducing kernel functions [15,16] in reproducing kernel Hilbert spaces [17,18,19,20,21] and related theory have important application in stochastic processes, signal analysis, machine learning and pattern recognition [22,23,24,25,26,27,28,29,30,31,32]. the reproducing kernel method [6,7,8,9,10,11,12,13,14] can not only obtain the exact solution in the form of series but also obtain the approximate solution with higher accuracy, the method has been widely used in linear and nonlinear problems, integral and differential equations, fractional partial differential equation and so on [22,23,24,25,26,27,28,29,30,31,32]. In [33], we use reproducing kernel interpolation collocation method to solve the linear integro differential equations of fractional order.

    In this paper, reproducing kernel method with reproducing kernel function in the form of Jacobi polynomials is applied to solving the following variable fractional order nonlinear integral differential equations:

    {Dα(x)u1(x)+t0k11(x,t)u1(t)+k12(x,t)u2(t)dt=f1(x,u1(x),u2(x)),Dβ(x)u2(x)+t0k21(x,t)u1(t)+k22(x,t)u2(t)dt=f2(x,u1(x),u2(x)),u1(0)=0,u2(0)=0,0<x,t1, (1.1)

    where 0<α(x)1, 0<β(x)1, fn(x,u1(x),u2(x)),n=1,2 and kij(x,t), i,j=1,2 are given functions. Dα(x)u1(x) indicates the α(x) is the Caputo fractional derivative defined of u1(x), Dβ(x)u2(x) indicates the β(x) is the Caputo fractional derivative defined of u2(x).

    Definition 1.1. The Caputo fractional derivative operator of variable order 0<α(x)1 is defined as

    Dα(x)u(t)={1Γ(1α(x))t0(tτ)α(x)u(τ)τdτ,0<α(x)<1,u(t)t,α(x)1. (1.2)

    The well-known Jacobi polynomials are defined on the interval [1,1] and can be generated with the aid of the following recurrence formula:

    Jμ,νi(t)=(μ+ν+2i1){μ2ν2+t(μ+ν+2i)(μ+ν+2i2)}2i(μ+ν+i)(μ+ν+2i2)Jμ,νi1(t)(μ+i1)(ν+i1)(μ+ν+2i)i(μ+ν+i)(μ+ν+2i2)Jμ,νi2(t),i=2,3,,

    and

    Jμ,ν0(t)=1,Jμ,ν1(t)=(μ+ν+2)2t+(μν)2.

    The weight function of Jacobi polynomials is ω(t)=(1t)μ(1+t)ν,t[1,1]. If μ=ν=0, Jacobi polynomials are Legendre polynomials, if μ=ν=12, Jacobi polynomials are the first chebyshev polynomials, if μ=ν=12, Jacobi polynomials are the second kind of Chebyshev polynomials. Some polynomials are shown in Figures 13.

    Figure 1.  Legendre polynomial.
    Figure 2.  Chebyshev polynomial of first kind.
    Figure 3.  Chebyshev polynomial of second kind.

    In order to use these polynomials on the interval x[0,1]. Let t=2x1,t[1,1], the shifted Jacobi polynomials are denoted by Jμ,νi(x). Then Jμ,νi(x) can be generated from:

    Jμ,νi(x)=(μ+ν+2i1){μ2ν2+(2x1)(μ+ν+2i)(μ+ν+2i2)}2i(μ+ν+i)(μ+ν+2i2)Jμ,νi1(x)(μ+i1)(ν+i1)(μ+ν+2i)i(μ+ν+i)(μ+ν+2i2)Jμ,ν1,i2(x),i=2,3,, (2.1)

    and

    Jμ,ν0(x)=1,Jμ,ν1(x)=(μ+ν+2)2(2x1)+(μν)2. (2.2)

    The analytic form of the shifted Jacobi polynomials Jμ,νi(x) of degree i is given by

    Jμ,νi(x)=ik=0(1)(ik)Γ(i+ν+1)Γ(i+k+1+μ+ν)Γ(k+1+ν)Γ(i+μ+ν+1)(ik)!k!xk, (2.3)

    and

    Jμ,νi(0)=(1)iΓ(i+ν+1)Γ(1+ν)i!,Jμ,νi(1)=Γ(i+μ+1)Γ(1+μ)i!. (2.4)

    Definition 2.1. Let

    Hn[0,1]=Span{Jμ,ν0(x),Jμ,ν1(x),...,Jμ,νn(x)}. (2.5)

    The inner product and norm are defined as:

    u(x),v(x)=10ω(x)u(x)v(x)dx,u(x)Hn=u(x),u(x)Hn, (2.6)

    where ω(x)=xν(1x)μ is a weight function. So, the shifted Jacobi polynomials have the following properties:

    Pμ,νi(x),Pμ,νj(x)=10ω(x)Pμ,νi(x)Pμ,νj(x)dx=hi, (2.7)

    where

    hi={Γ(i+α+1)Γ(i+β+1)(2i+α+β+1)k!Γ(i+α+β+1),i=j,0,ij. (2.8)
    Pμ,νi(x)2Hn=Pμ,νi(x),Pμ,νi(x)Hn=Γ(i+μ+1)Γ(i+ν+1)(2i+μ+ν+1)i!Γ(i+μ+ν+1). (2.9)

    Definition 2.2. Let

    ˉHn[0,1]={u|uHn[0,1],u(0)=0}.

    Its the norm as same as the norm of Hn[0,1]. From [33], we can prove that Hn[0,1] and ˉHn[0,1] are two reproducing kernel Hilbert spaces. The reproducing kernel of Hn[0,1] is

    Rn(x,y)=nk=0(2k+μ+ν+1)k!Γ(k+μ+ν+1)Γ(k+μ+1)Γ(k+ν+1)Jμ,νk(x)Jμ,νk(y), (2.10)

    The reproducing kernel of ˉHn[0,1] is

    K(x,y)=Kx(y)=R(x,y)R(0,x)R(y,0)R(0,0)2. (2.11)

    Definition 2.3. The inner product space is defined as:

    ˉHn[0,1]ˉHn[0,1]={U(x)=[u1(x),u2(x)]T|u1(x),u2(x)ˉHn[0,1]},

    its inner product and norm are defined by

    U(x),V(x)=2i=1ui(x),vi(x)ˉHn[0,1]ˉHn[0,1],U(x)2=2i=1ui(x)ˉHn[0,1]ˉHn[0,1]. (2.12)

    It is easy to verify that ˉHn[0,1]ˉHn[0,1] is a Hilbert space with the definition of inner product (2.12).

    Some reproducing kernels are shown in Table 1 and Figures 47.

    Table 1.  Some reproducing kernel functions.
    μ,ν n K(x,y)
    μ=ν=0 3 15xy(2060y+42y2+7x2(620y+15y2)4x(1548y+35y2))
    μ=ν=12 3 128xy15π(297924y+672y2+32x2(2172y+56y2)12x(77254y+192y2))
    μ=ν=1 3 84xy(2580y+60y210x(827y+21y2)+6x2(1035y+28y2))
    μ=0,ν=1 3 6xy(28x2(1030y+21y2)35x(1235y+24y2)+10(1542y+28y2))
    μ=0,ν=12 3 33128xy(65x2(63198y+143y2)78x(77234y+165y2)+21(99286y+195y2))
    μ=ν=2 3 630xy(2170y+55y2+11x2(518y+15y2)2x(35122y+99y2))

     | Show Table
    DownLoad: CSV
    Figure 4.  Reproducing kernel R13(x,y) at different μ=ν.
    Figure 5.  Contour plot of R13(x,y) at different μ=ν.
    Figure 6.  Reproducing kernel R20(x,y) at different μ=ν.
    Figure 7.  Contour plot of R20(x,y) at different μ=ν.

    To solve Eq (1), let

    {l11u1=Dα(x)u1(x)+t0k11(x,t)u1(t)dt,l12u2=t0k12(x,t)u2(x,t)dt,l21u1=t0k21(x,t)u1(t)dt,l22u2=Dβ(x)u2(x)+t0k22(x,t)u2(t)dt. (3.1)

    So, Eq (1) can be turn into Eq (3.2).

    LU(x)=F(x,u1(x),u2(x)), (3.2)

    where

    L=[l11l12l21l22],U(X)=[u1(x)u2(x)],F(x,u1(x),u2(x))=[f1(x,u1(x),u2(x))f2(x,u1(x),u2(x))]. (3.3)

    The operator L: ˉHn[0,1]ˉHn[0,1]H1[0,1]H1[0,1] is a bounded linear operator.

    Assuming that {xi}i=1 is dense on the interval [0,1], put ϕijk=lijkxk(x), where lij is the adjoint operator of lij. From [33], we have

    ϕijk(x)=lijKx(xk).i,j=1,2,k=1,2,.

    Putting

    Ψi1(x)=(ϕ11i(x),ϕ12i(x))T,Ψi2(x)=(ϕ21i(x),ϕ22i(x))T,i,=1,2,.

    Theorem 3.1. For each fixed n, {Ψij}(n,2)(1,1) is linearly independent in ˉHn[0,1]ˉHn[0,1].

    Theorem 3.2. {Ψij}(,2)(1,1)is complete in space in ˉHn[0,1]ˉHn[0,1].

    Proof. See of Theorems 3.1 and 3.2 in [33].

    Using Theorems 3.1 and 3.2, the exact solution of Eq (1) can be expressed as

    U(X)=i=12j=1cijΨij(x), (3.4)

    and truncate the infinite series of the analytic solution, we obtain the approximate solution of Eq (1).

    Um(X)=mi=12j=1cijΨij(x). (3.5)

    If we can obtain the coefficients of each Ψij(x), the approximate solution Um(x) can be obtained also. Use Ψij(x) to do the inner products with both sides of Eq (3.5) and let u1,0(x)=u2,0(x)=0, we have

    {mi=1ci1Ψi1,Ψn1+mj=1cj2Ψj2,Ψn1=f1(xk,u1,n1(xk),u2,n1(xk)),mi=1ci1Ψi1,Ψn2+mj=1cj2Ψj2,Ψn2=f2(xk,u1,n1(xk),u2,n1(xk)),k=1,2,,m. (3.6)

    Letting

    L2m=[Ψi1,Ψm1Ψj2,Ψm1Ψi1,Ψm2Ψj2,Ψm2]i,j,n=1,2,,m,
    F=(f1(x1),,f1(xm),f2(x1),,f1(xm))T.

    It is obvious that the inverse of A2m exists by Theorem 3.1. So, we have

    (c11,c12,,c1m,c21,c22,,c2m)T=L12mF.

    Theorem 3.3. Let UˉHn[0,1]ˉHn[0,1] be the exact solution of Eq (1), Um,n be the solution of (3.6). If

    fi(x,y,z)fi(x,ˉy,z)<c|yˉy|

    and

    fi(x,y,z)fi(x,yˉz)<c|zˉz|,

    0<c<1, then Um,n converges uniformly to U [17].

    Example 1. We consider the following nonlinear integro-differential equations of fractional order [34].

    {Dα(x)u1(x)x0(x+t)u2(t)dt=f1(x,u1(x),u2(x)),Dβ(x)u2(x)x0(xt)u1(t)dt=f2(x,u1(x),u2(x)),u1(0)=0,u2(0)=0.0<x,t1, (4.1)

    (1) Where

    α(x)=13,β(x)=23,
    f1(x,u1(x),u2(x))=x2/3Γ(5/3)+5x36,
    f2(x,u1(x),u2(x))=x1/3Γ(4/3)x2+13.

    The exact solution u1(x)=x, u2(x)=x, The numerical results which of Example 1 for m=10,μ=ν=0,n=2 are given in Table 2, and the absolute errors of this example for m=10,n=2 are given in Figures 8 and 9.

    Table 2.  Comparison of the absolute errors for Example 1.
    x Absolute errors of u1Absolute errors of u2Absolute errors of u1Absolute errors of u2
    Ref. [34]Ref.[34]Present methodPresent method
    05.8392E56.9382E500
    196.9402E56.7372E51.6653E169.7145E16
    296.3829E57.9300E51.1102E162.7756E16
    397.3428E58.0320E54.9960E162.7756E16
    497.8230E58.9324E54.4409E166.1062E16
    598.8492E59.2803E54.4409E164.4409E16
    698.3723E59.7832E57.7716E163.3307E16
    799.8402E52.8943E43.3307E162.2205E16
    893.4829E43.7231E41.1102E163.3307E16

     | Show Table
    DownLoad: CSV
    Figure 8.  Absolute errors of u1 for Example 1.
    Figure 9.  Absolute errors of u2 for Example 1.

    (2) Where

    α(x)=2x3,β(x)=x3,
    f1(x,u1(x),u2(x))=3x12x3(32x)Γ(12x3)+5x36,
    f2(x,u1(x),u2(x))=3x1x3(x3)Γ(1x3)x36.

    The exact solution u1(x)=x, u2(x)=x. The absolute errors of Example 1 for m=10,μ=ν=0,n=2 are given in Figures 10 and 11.

    Figure 10.  Absolute errors of u1 with μ=ν=0, α(x)=2x3,β(x)=x3 for Example 1.
    Figure 11.  Absolute errors of u2 with μ=ν=0, α(x)=2x3,β(x)=x3 for Example 1.

    (3) Where

    α(x)=cosx,β(x)=sinx,
    f1(x,u1(x),u2(x))=x1cosx(1cosx)Γ(1cosx)+5x36,
    f2(x,u1(x),u2(x))=x1sin(x)(1sinx)Γ(1sinx)x36.

    The exact solution u1(x)=x, u2(x)=x. The absolute errors of Example 1 for m=10,μ=ν=0,n=2 are given in Figures 12 and 13.

    Figure 12.  Absolute errors of u1 with μ=ν=0,α(x)=cosx,β(x)=sinx for Example 1.
    Figure 13.  Absolute errors of u2 with μ=ν=0,α(x)=cosx,β(x)=sinx for Example 1.

    Example 2. We consider the following linear integro-differential equations of fractional order [34].

    {Dαu1(x)10xt2u1(t)dtx0(x2+t)u2(t)dt=f1(x),Dβu2(x)10(x+t2)u1(t)dtx0x2tu2(t)dt=f2(x),u1(0)=0,u2(0)=0.0<x,t1, (4.2)

    Where α=β=12,

    f1(x)=8x1.53πx5+x4(4x+3)12,
    f2(x)=8x1.53π+x64x315.

    The exact solution u1(x)=x2, u2(x)=x2. The absolute errors of Example 1 with m=10,n=5 are given in Tables 3 and 4 and Figures 1419.

    Table 3.  Comparison of the numerical result of u1(x) in Example 2 with μ=ν=0.
    Ref.[34]Present methodPresent method
    Exact solutionNumerical solutionNumerical solutionAbsolute errors
    0.10.01000.00990.01002.1511E16
    0.30.09000.08970.09008.3267E16
    0.50.25000.24960.25001.1102E15
    0.70.49000.48940.49005.9397E15
    0.90.81000.80890.81001.1102E14

     | Show Table
    DownLoad: CSV
    Table 4.  Comparison of the numerical result of u2(x) in Example 2 with μ=ν=0.
    Ref.[34]Present methodPresent method
    Exact solutionNumerical solutionNumerical solutionAbsolute errors
    0.10.01000.01030.01001.6168E15
    0.30.09000.09050.09007.7716E16
    0.50.25000.25080.25005.5511E17
    0.70.49000.49130.49005.7732E15
    0.90.81000.81170.81001.9984E14

     | Show Table
    DownLoad: CSV
    Figure 14.  Absolute errors of u1 for Example 2 with μ=ν=0.
    Figure 15.  Absolute errors of u2 for Example 2 with μ=ν=0.
    Figure 16.  Absolute errors of u1 for Example 2.
    Figure 17.  Absolute errors of u2 for Example 2.
    Figure 18.  Absolute errors of u1 for Example 2 with μ=0.5.
    Figure 19.  Absolute errors of u2 for Example 2 with μ=0.5.

    In this paper, fractional nonlinear integro-differential equations of variable order have been solved by reproducing kernel interpolation collocation method with reproducing kernel function in the form of Jacobi polynomials. By comparing the obtained numerical solutions to the exact solutions of the fractional nonlinear integro-differential equations of variable order, it is indicated that our approach is powerful for variable order time fractional nonlinear integro-differential equations. All computations are performed by the Mathematica 7.0.

    The work is supported by the Natural Science Foundation of Inner Mongolia [2021MS01009].

    The authors declare that there are no conflicts of interest regarding the publication of this article.



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