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A new application of the Legendre reproducing kernel method

  • In this work, we apply the reproducing kernel method to coupled system of second and fourth order boundary value problems. We construct a novel algorithm to acquire the numerical results of the nonlinear boundary-value problems. We also use the Legendre polynomials. Additionally, we discuss the convergence analysis and error estimates. We demonstrate the numerical simulations to prove the efficiency of the presented method.

    Citation: Mohammad Reza Foroutan, Mir Sajjad Hashemi, Leila Gholizadeh, Ali Akgül, Fahd Jarad. A new application of the Legendre reproducing kernel method[J]. AIMS Mathematics, 2022, 7(6): 10651-10670. doi: 10.3934/math.2022594

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  • In this work, we apply the reproducing kernel method to coupled system of second and fourth order boundary value problems. We construct a novel algorithm to acquire the numerical results of the nonlinear boundary-value problems. We also use the Legendre polynomials. Additionally, we discuss the convergence analysis and error estimates. We demonstrate the numerical simulations to prove the efficiency of the presented method.



    In this work, we take into consideration the following nonlinear system of ordinary differential equation [21]:

    fS(xf+3f2ff)M2f=0,x(0,1), (1.1)
    θ+P1(2fθxθ)+P2(f2+12δ2f2)=0, (1.2)

    subject to the boundary conditions

    f(0)=0,f(0)βf(0)=0,θ(0)γθ(0)=0, (1.3)
    f(1)=12,f(1)+βf(1)=0,θ(1)+γθ(1)=1, (1.4)

    where S,P1,P2,δ,β and γ are real finite constants.

    We can see these problems in paper production, polymer extraction, aerodynamics, reaction-diffusion processes, fluid dynamics, biology and rheometry domains. These problems show up mainly due to the suction and injection effects on the unsteady magneto-hydrodynamic flow [24].

    Many methods have been improved for the analytical and approximate solution of nonlinear ordinary differential systems. These techniques contain finite-difference methods [5,31,32,33], Adams-Bashforth method [20,23], B-spline approximation method [8], Chebyshev finite difference method [28], finite element method [6], He's homotopy perturbation method [27], G/G- method [22], multi-step methods [14].

    In recent years, much attempt has been done to the newly developed methods to introduce an analytic and approximate solution of nonlinear boundary value problems [10,11,12,13,15,16,17,18,19,25]. For more details see [1,2,3,26,34,35,36,37]. In this work, we present an approximate-analytical technique for solving a coupled system of second and fourth order boundary value problems.

    The rest of this paper is organized as follows. In Section 2, an overview of shifted Legendre polynomials and their relevant properties required henceforward are presented. Also in this section, we will recall a brief review of the reproducing kernel spaces. In Section 3, we construct an orthogonal basis in the Legendre reproducing kernel space and construct a reproducing kernel space which includes boundary conditions. In Section 4, our method to approximate the solution of nonlinear system via shifted Legendre reproducing kernel basis function is considered. We present the convergence analysis and error estimation in Section 5. We demonstrate the numerical results in Section 6. We give the conclusion in the last section.

    In this section, we will recall some basic polynomial functionals and define some new reproducing kernel functions. The well-known shifted Legendre polynomials are described on [0, 1] and can be obtained by the following iterative formula

    P0(x)=1,P1(x)=2x1,(n+1)Pn+1(x)=(2n+1)(2x1)Pn(x)nPn1(x),n1. (2.1)

    The polynomials Pn(x) are orthogonal on [0,1] with ρ(x)=1, in the sense that

    10Pn(x)Pm(x)ρ(x)dx=γm,nδm,n, (2.2)

    where

    γm,n={0,ifmn,1,ifm=n=0,12n+1,ifm=n0.

    We use shifted Jacobi basis functions which provide the homogeneous boundary conditions as:

    f(0)=0andf(1)=0.

    Lemma 2.1. Let α,β1 and α,βZ. We have {aj} such that

    Jα,βn(x)=nj=nαβajPj(x),nα+β, (2.3)

    where Pj(x) are the shifted Legender polynomial of degree j and Jα,βn(x) is the shifted Jacobi polynomial on [0,1]. Then, we have

    J1,1n(x)=2(n1)2n1(Pn2(x)Pn(x)),n2. (2.4)

    Proof. For the proof of Lemma 2.1 (see [30], Lemma 1.4.3).

    Now, by utilizing the shifted Jacobi basis function and shifted Legendre functions, we will introduce a reproducing kernel Hilbert space method.

    f(0)=f(1)=0.

    Since Pn(1)=1 and Pn(0)=(1)n, we have

    J1,1n(0)=J1,1n(1)=0.

    Therefore, we describe

    un(x)=(n+2)(2n+3)(n+1)J1,1n+2(x),n=0,1,2,..., (2.5)

    and

    vn(x)=2n+12Pn(x),n=0,1,2,.... (2.6)

    Definition 2.2. [10] For a nonempty set E, let H be a Hilbert space of real value functions on some set E. A function K:E×ER is said to be the reproducing kernel function of H if and only if:

    (i) For every yE, K(,y)H.

    (ii) For every yE and fH,f(),K(,y)=f(y).

    Also, a Hilbert space of function H that possesses a reproducing kernel K is a reproducing kernel Hilbert space; we represent the reproducing kernel Hilbert space and it's kernel by HK(E) and Ky respectively.

    Theorem 2.3. [7] Let H be n-dimensional Hilbert space, {wi}ni=1 is an orthonormal basis of H, then the reproducing kernel of H as:

    Kn(x,y)=nj=0wj(x)wj(y),x,y[0,1]. (2.7)

    Theorem 2.4. ([29] Theorem 1.24) For the orthonormal system {wn}n=1, formula (2.7) yields the Christoffel-Darboux formula:

    Kn(x,y)=kn(wn+1(x)wn(y)wn(x)wn+1(y))kn+1(xy). (2.8)

    Where, kn>0 is the coefficient of xn in wn(x). We get

    Kn(x,x)=knkn+1(wn+1(x)wn(x)wn(x)wn+1(x)). (2.9)

    Definition 2.5. Let Hω,K1,n[0,1] be the weighted inner product space of Jacobi functions described as (2.5) on [0,1] with degree less than or equal to n. The inner product and norm are given respectively by

    u1,u2Hω,K1,n=10u1(x)u2(x)ω(x)dx,u1,u2Hω,K1,n[0,1],
    uHω,K1,n=u,u12Hω,K1,n,uHω,K1,n[0,1],

    where ω(x)=(1x)1(1+x)1 and K1,n(x,y) the reproducing kernel of Hω,K1,n[0,1] is constructed using (2.8) with wn(x):=un(x). From definition

    L2ω[0,1]={u:10|u(x)|2ω(x)dx<},

    for any fixed n, Hω,K1,n[0,1] is a subspace of L2ω[0,1] and

    u1,u2Hω,K1,n=u1,u2L2ω,u1,u2Hω,K1,n[0,1].

    From Definition 2.5, Hω,K1,n[0,1] is a finite dimensional inner product space. Every finite dimensional inner product space is a Hilbert space. Therefore, from this result and Theorem 2.3, Hω,K1,n[0,1] is a reproducing kernel Hilbert space.

    Definition 2.6. Let HK2,n[0,1] be the inner product space of Legendre functions described as (2.6) on [0,1] with degree less than or equal to n. The inner product and norm are given respectively by

    v1,v2HK2,n=10v1(x)v2(x)dx,v1,v2HK2,n[0,1],
    vHK2,n=v,v12HK2,n,vHK2,n[0,1],

    where the reproducing kernel K2,n(x,y) of HK2,n[0,1] is constructed using (2.8) with wn(x):=vn(x). From definition

    L2[0,1]={v:10|v(x)|2dx<},

    for any fixed n, HK2,n[0,1] is a subspace of L2[0,1] and

    v1,v2HK2,n=v1,v2L2,v1,v2HK2,n[0,1].

    Hω,R1[0,1] and HR2[0,1] are described by:

    Hω,R1[0,1]={f(x):f(x)Hω,K1,n[0,1],f(0)βf(0)=0,f(1)+βf(1)=0},HR2[0,1]={θ(x):θ(x)HK2,n[0,1],θ(0)γθ(0)=0,θ(1)+γθ(1)=0}.

    Clearly, Hω,R1[0,1] and HR2[0,1] are closed subspaces of Hω,K1,n[0,1] and HK2,n[0,1], respectively.

    Let T1f=f(0)βf(0) and T2f=f(1)+βf(1) be the boundary condition of function f(x). Put

    R1,1(x,y)=K1,n(x,y)T1,xK1,n(x,y)T1,yK1,n(x,y)T1,xT1,yK1,n(x,y), (3.1)

    and

    R1(x,y)=R1,1(x,y)T2,xR1,1(x,y)T2,yR1,1(x,y)T2,xT2,yR1,1(x,y), (3.2)

    where, the symbol T1,x shows that the operator T1 implements to the function of x.

    Theorem 3.1. If T1,xT1,yK1,n(x,y)0 and T2,xT2,yR1,1(x,y)0, then R1(x,y) given by (3.2) satisfies the boundary conditions T1f=0 and T2f=0 exactly.

    Proof. By applying the operator T1,x to R1,1(x,y) in Eq (3.1), we get

    T1,xR1,1(x,y)=T1,xK1,n(x,y)T1,xK1,n(x,y)T1,xT1,yK1,n(x,y)T1,xT1,yK1,n(x,y)=0. (3.3)

    Furthermore, by applying the operator T1,xT2,y to R1,1(x,y), we have

    T1,xT2,yR1,1(x,y)=T1,xT2,yK1,n(x,y)T2,yT1,xK1,n(x,y)T1,xT1,yK1,n(x,y)T1,xT1,yK1,n(x,y)=T1,xT2,yK1,n(x,y)T2,yT1,xK1,n(x,y)=0. (3.4)

    Then, by applying the operator T1,x to R1(x,y) in Eq (3.2) and using Eqs (3.3) and (3.4), we get

    T1,xR1(x,y)=T1,xR1,1(x,y)T2,xR1,1(x,y)T1,xT2,yR1,1(x,y)T2,xT2,yR1,1(x,y)=0.

    Also, by applying the operator T2,x to R1(x,y) in Eq (3.2), we have

    T2,xR1(x,y)=T2,xR1,1(x,y)T2,xR1,1(x,y)T2,xT2,yR1,1(x,y)T2,xT2,yR1,1(x,y)=0.

    Theorem 3.2. [9] If T1,xT1,yK1,n(x,y)0 and T2,xT2,yR1,1(x,y)0, then, we obtain

    R1(x,y)=R1,1(x,y)T2,xR1,1(x,y)T2,yR1,1(x,y)T2,xT2,yR1,1(x,y).

    Let T3θ=θ(0)γθ and T4θ=θ(1)+γθ(1)=0 be the boundary condition of function θ(x). Put

    R2,2(x,y)=K2,n(x,y)T3,xK2,n(x,y)T3,yK2,n(x,y)T3,xT3,yK2,n(x,y), (3.5)

    and

    R2(x,y)=R2,2(x,y)T4,xR2,2(x,y)T4,yR2,2(x,y)T4,xT4,yR2,2(x,y). (3.6)

    Theorem 3.3. If T3,xT3,yK2,n(x,y)0 and T4,xT4,yR2,2(x,y)0, then R2(x,y) given by (3.6) satisfies the boundary conditions T3θ=0 and T4θ=0 exactly.

    Proof. The proof of this theorem is similar to the proof of Theorem 3.1.

    Theorem 3.4. If T3,xT3,yK2,n(x,y)0 and T4,xT4,yR2,2(x,y)0, then HR2[0,1] is a reproducing kernel space and its reproducing kernel is

    R2(x,y)=R2,2(x,y)T4,xR2,2(x,y)T4,yR2,2(x,y)T4,xT4,yR2,2(x,y).

    Note that Rx(y)=R(x,y), R1,x(y)=R1(x,y) and R2,x(y)=R2(x,y). Henceforth and not to conflict unless stated otherwise, we denote H[0,1]=Hω,R1HR2, L[0,1]=L2ω[0,1]L2[0,1] and Rx(y)=(R1,x(y),R2,x(y))T.

    Definition 3.5. (a) The Hilbert space H[0,1] is described by:

    H[0,1]={z=(z1,z2)T:z1Hω,R1[0,1]andz2HR2[0,1]}.

    The inner product in H[0,1] is building as

    z,wH=z1,w1Hω,R1+z2,w2HR2

    and the norm is zH=z1Hω,Kn+z2HKn where z,wH[0,1].

    (b) The Hilbert space L[0,1] is described by:

    L[0,1]={z=(z1,z2)T:z1L2ω[0,1]andz2L2[0,1]}.

    The inner product in L[0,1] is building as

    z,wL=z1,w1L2ω+z2,w2L2

    and the norm is zL=z1L2ω+z2L2 where z,wL[0,1].

    We assume

    f(x)=F(x)+β1dβ(x) (4.1)

    and

    θ(x)=Θ(x)+γ1(x+γ), (4.2)

    where

    β1=110+24β+24β2,dβ(x)=2x3+(3+6β)x2+(6β+12β2)x

    and γ1=12γ+1, then Eqs (1.1) and (1.2) changes to the following problem:

    {FSxF+(2Sβ1dβ(x)3SM2)F+2Sβ1cβ(x)F=2SFF+g1(x),2P2β1cβ(x)F+24P2δ2β1dβ(x)F+2P1γ1F+Θ+P1(2β1dβ(x)x)Θ=2P1FΘ+P2F2+12P2δ2F2+g2(x), (4.3)

    and the boundary conditions changes to the following conditions:

    F(0)=0,F(0)βF(0)=0,Θ(0)γΘ(0)=0, (4.4)
    F(1)=0,F(1)+βF(1)=0,Θ(1)+γΘ(1)=0, (4.5)

    where

    cβ(x)=12x+12β+6,g1(x)=2Sβ21dβ(x)cβ(x)+12Sβ1x+M2β1cβ(x),g2(x)=P1γ1x2P1γ1β1dβ(x)P2β21c2β(x)12P2δ2β21d2β(x).

    Put

    L11F=FSxF+(2Sβ1dβ(x)3SM2)F+2Sβ1cβ(x)F,L12Θ=0,L21F=2P2β1cβ(x)F+24P2δ2β1dβ(x)F+2P1γ1F,L22Θ=Θ+P1(2β1dβ(x)x)Θ,L=(L11L12L21L22),N1(F,Θ)=2SFF,N2(F,Θ)=2P1FΘP2F212P2δ2F2,Φ=(F,Θ)T,Φ=(F,Θ)TandΦ=(F,0)T,

    then, the coupled differential systems of Eqs (1.1) and (1.2) can be written as follows:

    LΦ(x)=g(x)N(Φ(x),Φ(x),Φ(x)), (4.6)

    with boundary conditions:

    {(eT1Φ(0))e1=0,(eT1Φ(0))e1β(eT1Φ(0))e1=0,(eT2Φ(0))e2γ(eT2Φ(0))e2=0,(eT1Φ(1))e1=0,(eT1Φ(1))e1+β(eT1Φ(1))e1=0,(eT2Φ(1))e2+γ(eT2Φ(1))e2=0, (4.7)

    where g=(g1,g2)T, N=(N1,N2)T, ΦH[0,1], gNL[0,1], e1=(1,0)T, e2=(0,1)T and L:H[0,1]L[0,1].

    Here, (eT1Φ(i))e1=(F(i),0)T and (eT2Φ(i))e2=(0,Θ(i))T,i=0,1.

    Lemma 4.1. ([10], Lemma 4.1) The operators L22:HR2[0,1]L2[0,1] and Li1[0,1]:Hω,R1L2ω[0,1],i=1,2, are linear bounded operators.

    Theorem 4.2. The operator L:H[0,1]L[0,1] is bounded linear operator.

    Proof. For each ΦH[0,1], using Definition 3.5, we have

    LΦL=2i=1Li1F2Lω+L22Θ2L(2i=1Li1FHω,R1)2+(L22ΘHR2)2(L112+L212+L222)(F2Hω,R1+Θ2HR2)L112+L212+L222ΦH.

    The boundedness of L22 and Li1 for i=1,2, shows that L is bounded. The proof is complete.

    Let D={xi}i=1 is countable dense subset in the domain [0,1], then for any fixed xi[0,1], we have

    Ψij(x):=LR(x,xi)ej=LRxi(x)ej=LRxi(x),Rx(y)Hej=Rxi(x),LyRx(y)Lej=LyRx(y)ej|y=xi=LyRy(x)ej|y=xi,j=1,2,i=1,2,3,...,

    where L=(L11L210L22) is the adjoint operator of L and the subscript y in the operator Ly indicates that the operator L applied to the function y. For any fixed xi(0,1), Ψij(x)H[0,1].

    Theorem 4.3. If {xi}i=1 is distinct points dense on [0,1] and L1 is existent, then

    ImL=Hω,R1([0,1])HR2([0,1]),(KerL)=ImL=L2ω([0,1])L2([0,1]).

    Proof. Clearly ψij(x)Hω,R1([0,1])HR2([0,1]). For any Φ(ImL), since ψij(x)=LRxi(x)ej, we have

    Φ(x),ψij(x)H=0. (4.8)

    On the other hand,

    Φ(x)=F(x)e1+Θ(x)e2=2j=1Φ(.),Rx(.)ejHej.

    Thus, by Eq (4.8), we get

    LΦ(xi)=2j=1LΦ(y),Rx(y)ejHej=0,i=1,2,.

    Note that {xi}i=1 is dense on [0,1]. Hence (LΦ)(x)=0. So from the existence L1, we have Φ(x)=0. That is (ImL)=0. Therefore ImL=Hω,R1([0,1])HR2([0,1]). Similarly, we can show (KerL)=L2ω([0,1])L2([0,1]).

    Corollary 4.4. For Eqs (1.1)(1.4), if {xi}i=1 is distinct points dense on [0,1] and L1 is existent, then {ψij(x)}(,2)(i,j)=(1,1) is the complete function system of the space H([0,1]).

    By Gram-Schmidt process, we acquire an orthogonal basis {¯ψij(x)}(,2)(i,j)=(1,1) of H([0,1]), such that

    ¯ψij(x)=il=1jk=1αijlkψij(x),i=1,2,...,j=1,2,

    where αijlk represents orthogonal coefficients, which are given by the following relations [4]:

    α1j1k=1ψ1kH,αijlk=1aijlk,l=i1,αijlk=1aijlkl1s=icsjlkαijsk,l<i,

    such that aijlk=ψlk2Hl1s=i(csjlk)2 and csjlk=ψlk,¯ψsk2H.

    Lemma 4.5. Let {¯ψij(x)}(,2)(i,j)=(1,1) be an orthonormal basis of H then we have

    R(x,y)=i=12j=1¯ψij(x)¯ψij(y).

    Proof. Let gH, then

    g(y),R(x,y)H=g(y),i=12j=1¯ψij(x)¯ψij(y)H=i=12j=1g(y),¯ψij(y)H¯ψij(x)=g(x).

    Theorem 4.6. If {xi}i=1 is dense on [0,1] and L1 is existent, then the solution of Eq (4.6) satisfies the form

    Φ(x)=i=12j=1il=1jk=1αijlkGk(xl,Φ(xl),Φ(xl),Φ(xl))¯ψij(x) (4.9)

    where

    G(x,Φ(x),Φ(x),Φ(x))=g(x)N(Φ(x),Φ(x),Φ(x))=(G1,G2)T.

    Proof. Since {¯ψij(x)}(,2)(i,j)=(1,1) is orthonormal system, Φ(x) is expressed as

    Φ(x)=Φ(y),R(x,y)H=Φ(y),i=12j=1¯ψij(x)¯ψij(y)H=i=12j=1Φ(y),¯ψij(y)H¯ψij(x)=i=12j=1Φ(y),il=1jk=1αijlkψlk(y)H¯ψij(x)=i=12j=1il=1jk=1αijlkΦ(y),ψlk(y)H¯ψij(x)=i=12j=1il=1jk=1αijlkΦ(y),Lrlk(y)H¯ψij(x)=i=12j=1il=1jk=1αijlkLΦ(y),rlk(y)L¯ψij(x)=i=12j=1il=1jk=1αijlkGk(xl,Φ(xl),Φ(xl),Φ(xl))¯ψij(x),

    where rlk(y)=Rxl(y)ek.This completes the proof.

    Now, let

    HN[0,1]=Span{¯ψ11,¯ψ12,¯ψ21,¯ψ22,,¯ψN1,¯ψN2}.

    Define H[0,1]orthogonal projection TN:H[0,1]HN[0,1] such that for ΦH[0,1],

    TNΦΦ,ζH=0,ζHN[0,1],

    or equivalently,

    TNΦ=Ni=12j=1Φ,¯ψij¯ψij.

    Then, we get the approximate solution as:

    ΦN(x)=Ni=12j=1il=1jk=1αijlkGk(xl,Tl1Φ(xl),(Tl1Φ)(xl),(Tl1Φ)(xl))¯ψij(x). (4.10)

    Here, T0Φ(x) is any fixed function in H([0,1]).

    Theorem 5.1. Assume that the problem (4.6) has a unique solution. If {xi}i=1 is dense on [0,1], then ΦN(x) in (4.10) is convergence to the Φ(x) and for any fixed Φ0(x)H([0,1]),ΦN(x) is also represented by

    ΦN(x)=Ni=12j=1il=1jk=1αijlkGk(xl,Φ(xl),Φ(xl),Φ(xl))¯ψij(x). (5.1)

    Proof. We have

    LΦN(x)=Ni=12j=1βijL¯ψij(x),

    where

    βij=il=1jk=1αijlkGk(xl,Tl1Φ(xl),(Tl1Φ)(xl),(Tl1Φ)(xl)).

    Then for sN and p2, we have

    (LΦN)p(xs)=Ni=12j=1βijL¯ψij(x),rsp(x)H=Ni=12j=1βij¯ψij(x),Lrsp(x)H=Ni=12j=1βij¯ψij(x),ψsp(x)H.

    Therefore

    ss=1pp=1βijsp(LΦN)p(xs)=Ni=12j=1βij¯ψij(x),ss=1pp=1βijspψsp(x)H=Ni=12j=1βij¯ψij(x),¯ψsp(x)H=βsp.

    If s=1, we get

    (LΦN)j(x1)=Gj(x1,T0Φ(x1),(T0Φ)(x1),(T0Φ)(x1)),j=1,2,

    that is,

    LΦN(x1)=G(x1,T0Φ(x1),(T0Φ)(x1),(T0Φ)(x1)).

    For s=2, we have

    (LΦN)j(x2)=Gj(x2,T1Φ(x2),(T1Φ)(x2),(T1Φ)(x2)),j=1,2,

    that is,

    LΦN(x2)=G(x2,T1Φ(x2),(T1Φ)(x2),(T1Φ)(x2)).

    Hence it can be obtained by induction,

    LΦN(xn)=G(xn,Tn1Φ(xn),(Tn1Φ)(xn),(Tn1Φ)(xn)).

    Since {xn}n=1 is dense, for any x[0,1] there exists a subsequence {xni}i=1 such that xnix, as i. Then, we reach:

    limi+LΦN(xni)=limi+G(xni,Tni1Φ(xni),(Tni1Φ)(xni),(Tni1Φ)(xni))=G(x,Φ(x),Φ(x),Φ(x))=LΦ(x). (5.2)

    Moreover, according to (4.10) we have

    lims+LΦN(xns) (5.3)
    =lims+Ni=12j=1il=1jk=1αijlkGk(xl,Tl1Φ(xl),(Tl1Φ)(xl),(Tl1Φ)(xl))L¯ψij(xns)=+i=12j=1il=1jk=1αijlkGk(xl,Tl1Φ(xl),(Tl1Φ)(xl),(Tl1Φ)(xl))L¯ψij(x)=limN+LNi=12j=1il=1jk=1αijlkGk(xl,Tl1Φ(xl),(Tl1Φ)(xl),(Tl1Φ)(xl))L¯ψij(xns)=limN+LΦN(x). (5.4)

    So, from Eqs (5.2) and (5.3), we conclude that

    limN+LΦN(x)=LΦ(x). (5.5)

    Thus, we obtain

    limN+ΦN(x)=L1limN+(LΦN(x))=L1(Φ(x))=Φ(x).

    Theorem 5.2. Let Φn(x)=(Fn(x),Θn(x))T be approximate solution that has obtained from the present method in the space H[0,1] and Φ(x)=(F(x),Θ(x))T be exact solution for the differential equation (4.6) with boundary conditions (4.7). Also, assume that xnx(n), ΦnH is bounded and G(t,Φ(t),Φ(t),Φ(t)) is continuous for t[0,1], then

    G(xn,Φn1(xn),Φn1(xn),Φn1(xn))G(x,Φ(x),Φ(x),Φ(x)),

    as n.

    Proof. For any x[0,1], using the boundedness of ixR1(x,y)Hw,R1(i=0,1,2) and reproducing property of R1(x,y), we have

    F(i)n(x)F(i)(x)=(Fn(x)F(x))(i)=|ixFn(y)F(y),R1(x,y)Hw,R1|=|Fn(y)F(y),ixR1(x,y)Hw,R1|FnFHw,R1ixR1(x,y)Hw,R1αiFnFHw,R10, (5.6)

    where for i=0,1,2, αi are positive constants.

    Similarly, for each x[0,1] and i=0,1, we get

    Θ(i)n(x)Θ(i)(x)=(Θn(x)Θ(x))(i)=|ixΘn(y)Θ(y),R2(x,y)HR2|=|Θn(y)Θ(y),ixR2(x,y)HR2|ΘnΘHR2ixR2(x,y)HR2βiΘnΘHR20, (5.7)

    where for i=0,1,2, βi are positive constants.

    Furthermore, if ΦH[0,1], then Φ(x)=(F(x),Θ(x))T where F(x)Hw,R1[0,1] and Θ(x)HR2[0,1]. Thus for i=0,1,2, we have

    Φ(i)n(x)Φ(i)(x)=|F(i)n(x)F(i)(x)|2+|Θ(i)n(x)Θ(i)(x)|2=α2iF(i)n(x)F(i)(x)2Hw,R1+β2iΘ(i)n(x)Θ(i)(x)2HR20. (5.8)

    Note that, since ΦnH[0,1], exist a constant c1 such that

    |Φn1(x)|c1,x[0,1].

    Therefore

    |Φn1(xn)Φ(x)|=|Φn1(xn)Φn1(x)+Φn1(x)Φ(x)|=|Φn1(xn)Φn1(x)|+|Φn1(x)Φ(x)|=|Φn1(y1)||xnx|+|Φn1(x)Φ(x)|0,asn, (5.9)

    where y1 lies between xn and x. Now will show that Φn1(xn)Φ(x). Since Φn(x)H[0,1], exist a constant c2 such that |Φn1(x)|c2, so we get

    |Φn1(xn)Φ(x)|=|Φn1(xn)Φn1(x)+Φn1(x)Φ(x)|=|Φn1(xn)Φn1(x)|+|Φn1(x)Φ(x)|=|Φn1(y2)||xnx|+|Φn1(x)Φ(x)|0,asn, (5.10)

    where y2 lies between xn and x. Similarly, we can write

    |Φn1(xn)Φ(x)|0,asn.

    Now, from the continuation of G(t,Φ(t),Φ(t),Φ(t)), it is implies that

    G(xn,Φn1(xn),Φn1(xn),Φn1(xn))G(x,Φ(x),Φ(x),Φ(x)),asn.

    In this section, some illustrative examples demonstrate the applicability, efficiency and utility of the proposed technique. The computations associated with the examples were performed using Maple16 on a personal computer.

    Let us consider Eqs (1.1)–(1.4), using the shifted Legendre reproducing kernel Hilbert space method. We apply the technique on this problem with N=12 and

    xi=12cos(iπN)+12,i=0,1,2,...,N1.

    Table 1 demonstrate the obtained solutions of f(x) and at x=1 for various values of S, M, β and compares the results with homotopy analysis method (HAM) presented in [21]. Table 2 demonstrates the approximate solutions of velocity θ(x) at x=1 with N=12 and P1=M=P2=1.0, δ=0.1 for different values of S, β, γ and compares the result with the HAM presented in [21]. In [21], there is no analysis about the convergence or error estimate of results, whereas in the current work we discussed about the convergence of method and residual errors. Hence, we can claim from the error analysis that out obtained results are more accurate than [21]. For example, the result of θ(1) with respect to S=4.00 in [21] is 0.281319, but in the present techniqe we get 0.2880499297. It is evident that there is little difference between the obtained results, which the present method gives more accurate results.

    Table 1.  Values of f(1) for different values of S, M and β.
    S M β Hussain et al. [21] Present
    1.00 1.00 0.00 3.180310 3.1803102750
    0.05 2.414897 2.4148967196
    0.10 1.945943 1.9459433694
    0.15 1.629328 1.6293275195
    0.20 1.4012474830
    0.50 0.7614007594
    1.00 0.4322969902
    0.00 0.10 1.928044 1.9280441735
    2.00 1.997081 1.9970814583
    3.00 2.074769 2.0747689044
    2.00 1.00 1.994008 1.9940077672
    3.00 2.038898 2.0388977767
    0.10 1.8996168697
    0.50 1.9205892930

     | Show Table
    DownLoad: CSV
    Table 2.  Results of θ(1) for various values of S, β and γ with P1=M=P2=1.0 and δ=0.1.
    S β γ Hussain et al. [21] Present
    1.00 0.10 0.00 0.415935 0.4159351994
    0.01 0.396709 0.3967094545
    0.05 0.326735 0.3267347576
    0.10 0.252278 0.2522783590
    0.50 -0.0766615072
    0.00 0.10 -0.687930 -0.6879304850
    0.01 -0.514271 -0.5142710439
    0.02 -0.368998 -0.3689980708
    0.03 -0.246254 -0.2462544541
    0.10 0.2522783590
    0.50 0.7301443477
    1.00 0.7892670209
    2.00 0.10 0.263761 0.2637611894
    3.00 0.273318 0.2733180923
    4.00 0.281319 0.2880499297
    0.10 0.2399056923
    0.20 0.2413894057
    0.50 0.2456708266

     | Show Table
    DownLoad: CSV

    The effect of Hartmann number M on the radial velocity f(x) is exhibited in Figure 1. The radial velocity f(x), decreased for higher values of the Hartmann number on 0.24x0.76. The influence of parameter β on f(x) is plotted in Figure 2 with M=S=1.0. This Figure suggests that the f(x) show decreasing behavior with an increase in β. Figures 3 and 4 display the temperature profiles θ(x) for the various embedded parameters viz thermal slip parameter γ and Eckert number P2 on interval [0,1]. It is seen that when γ=0, which corresponds to no thermal slip, the temperature of the fluid and that of the disks surfaces is the same, which in this case is 0 and 1 for lower and upper disks, respectively.

    Figure 1.  Influence of magnetic parameter M on the radial velocity f(x) for values of S=1.0 and β=0.1.
    Figure 2.  Influence of parameter β on the radial velocity f(x) for values of S=1.0 and M=1.0.
    Figure 3.  Influence of thermal slip parameter γ on the temperature profile θ(x) for values of S=M=P1=1.0, β=δ=0.1 and P2=0.0.
    Figure 4.  Influence of parameter P2 on the temperature profile θ(x) for values of P1=S=M=1.0 and β=γ=δ=0.1.

    Since the exact solution of problems (1.1)–(1.4) is not known, we discuss the absolute residual error function which is a measure of how well the approximation satisfies the Eq (4.6) with S=P1=M=P2=1.0 and β=γ=δ=0.1 as

    Res(x)=|LΦ(x)+N(Φ(x),Φ(x),Φ(x))g(x)|.

    Note that the norm 2 of the residual function on the domain is

    Res2=(10|Res(x)|2dx)12,

    and it is employed in this paper to check the accuracy and the convergence of the proposed method. The absolute residual errors are plotted in Figure 5.

    Figure 5.  Residual errors f(x) and θ(x), with S=P1=M=P2=1.0 and β=γ=δ=0.1, respectively (from left to right).

    In this paper, the shifted Legendre reproducing kernel method is employed to compute approximate solutions of a nonlinear system of ordinary differential equation. In this approach, a truncated series based on shifted Legendre reproducing kernel functions with easily computable components. The convergence analysis and error estimation of the approximate solution using the proposed method are investigated. The validity and applicability of the method is demonstrated by solving several numerical examples. The main advantage of the present method lies in the lower computational cost and high accuracy. System of differential equations appear in various branches of science and technology. Results of current study show that the shifted Legendre reproducing kernel method is a reliable technique for the physical models in the system of differential equations form. Moreover, this method could be developed for systems of differential equations with fractional order derivatives or system of integro-differential equations.

    The authors wish to express their thanks to the referees for comments which improved the paper.

    The authors declare that they have no conflicts of interest.



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