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Publicly Available Published by De Gruyter March 9, 2016

High-Order Algorithms for Riesz Derivative and their Applications (III)

  • Hengfei Ding EMAIL logo and Changpin Li

Abstract

Numerical methods for fractional calculus attract increasing interest due to its wide applications in various fields such as physics, mechanics, etc. In this paper, we focus on constructing high-order algorithms for Riesz derivatives, where the convergence orders cover from the second order to the sixth order. Then we apply the established schemes to the Riesz type turbulent diffusion equation (or, Riesz space fractional turbulent diffusion equation). Numerical experiments are displayed which support the theoretical analysis.

MSC: 65M06; 65M12

1. Introduction

In recent years, fractional calculus has attracted increasing interests due to its applications in physics, mechanics, etc. For more details, see the recent publications [1, 2, 3, 5, 6, 7, 8, 9, 10, 14, 15, 17, 19, 20, 21, 22], and references cited therein. The Riemann-Liouville (R-L) derivative and Caputo derivative are commonly used, respectively defined below:

RLDa,xαf(x)=1Γ(nα)dndxnax(xξ)nα1f(ξ)dξ,x(a,b)

and

cDa,xαf(x)=1Γ(nα)ax(xξ)nα1f(n)(ξ)dξ,x(a,b)

where n − 1 < α < nZ+.

A linear combination of the left R-L derivative (defined above) and the right R-L derivative given as

RLDx,bαf(x)=(1)nΓ(nα)dndxnxb(ξx)nα1f(ξ)dξ,x(a,b)

defines a new derivative, i.e., the Riesz derivative,

dαf(x)d|x|α=12 cos (πα2)(RLDa,xα+RLDx,bα)f(x),n1<α<nZ+.

The left R-L derivative reflects the dependence on the history, while the right R-L derivative the dependence upon the future. So the Riesz derivative value off(x) at x relies on the whole space (a, b), but the values are different at differentx ∈(a, b). From an angle of application, the case with α ∈ (0, 2) is mostly attracted attention.

From the studies available, the high-order numerical algorithms for RL derivatives were firstly established in [16]. The high-order algorithms for Caputo derivatives were firstly constructed in [12]. Shortly after, some other high-order algorithms for Caputo derivatives were also appeared, for example see [1, 3, 14] and references cited therein. And the high-order algorithms for Riesz derivatives were derived in [5, 6]. In [5], Ding et al. constructed the fourth-order schemes for the Riesz derivative and applied them to the space Riesz fractional diffusion equation. In that paper they established the fourth-order schemes for R-L derivatives [5] from which some similar fourth-order schemes were derived by other people. In [6], Ding et al. continued to establish the sixth-order, eighth-order, tenth-order and twelfth-order schemes for the Riesz derivative by using the Fourier analysis, where the fourth-order compact scheme for R-L derivative was especially highlighted. Then they used the two schemes, among them to the Riesz space fractional reaction-dispersion equation, where the rigorous error analysis was given. The other odd-order (third-order, fifth-order, seventh-order, ninth-order, eleventh-order) schemes can be established from [6] by choosing the different parameters therein. In spite of these, it is absolutely necessary to construct intuitionist and straightforward schemes for the Riesz derivatives. Here we find an interesting and enlightening way to establish high-order schemes (from 2nd-order to 6th-order) by using the corresponding generating functions [16]. Then we use these schemes to solve the Riesz space fractional turbulent diffusion equation.

In the following, we briefly introduce fractional modelling in this respect. From the known first Fick’s law

J(x,t)=d1u(x,t)d2u(x,t)x,d1>0,d2>0,

one obtain the following advection-diffusion equation,

u(x,t)t=d1u(x,t)x+d22u(x,t)x2.

If the advection-diffusion process at any position x ∈(a, b) relies on the whole space (a, b) (i.e., long-range interactions), then the classical Fick’s law does not work well yet. However, the fractional derivative can well characterize such long-range interactions. Now we generalize the typical Fick’s law to a fractional version,

Jα(x,t)=d1u(x,t)d2u(x,t)xdαα1u(x,t)|x|α1,α(0,1)

where d1, d2, dα are positive constants. It follows that the Riesz space fractional turbulent diffusion equation (or, Riesz type turbulent diffusion equation) is obtained,

u(x,t)t=d1u(x,t)x+d22u(x,t)x2+dααu(x,t)|x|α.

If there is a source term, then one has

(1.1)u(x,t)t=d1u(x,t)x+d22u(x,t)x2+dααu(x,t)|x|α+s(x,t),

where the Riesz partial derivative with order α ∈ (0, 1) is given by

(1.2)αu(x,t)|x|α=12 cos (πα2)(RLDa,xα+RLDx,bα)u(x,t),0<α<1

Equation (1.1) is subject to the following initial value condition,

u(x,O)=φ(x),x[a,b]

and the boundary value conditions (here choosing the homogeneous condition for brevity),

u(a,t)=0,u(b,t)=0,t0

The remainder of this paper is outlined as follows. In Section 2, five kinds of high-order (2nd-order, 3rd-order, ···, 6th-order) algorithms for the Riesz derivatives are developed. In the next Section 3, we apply the derived schemes to the Riesz-type turbulent diffusion equation (1.1). Here we only use the 2nd-order, 4th-order, 6th-order schemes to (1.1). The convergence orders are 𝒪(τ2 + h2), 𝒪(τ2 + h4), and 𝒪(τ2 + h6), where τ and h are temporal and spatial stepsizes, respectively. In Section 4, numerical examples are presented which support the theoretical analysis. The last Section5 concludes this article. In Appendices A and B we provide the detailed proofs of the main results from Section 2.

2. High-order numerical schemes

If f(k)(a+) = 0 (k = 0, 1,...,p − 1), then it follows from [16] that the left R-L derivative has the following approximations

(2.1)RLDa,xαf(x)=1hα=0ϖp,(α)f(xh)+O(hp)

in which h is the steplength. Here we only show interests in p = 2, 3, 4, 5, 6.

The convolution (or weight) coefficients

in the above equations are those of the Taylor series expansions of the corresponding generating functions
,

Wp(α)(z)==0ϖp,(α)z,α(0,2)

where

W2(α)(z)=(322z+12z2)α,W3(α)(z)=(1163z+32z213z3)α,   W4(α)(z)=(25124z+3z243z3+14z4)α  W5(α)(z)=(137605z+5z2103z3+54z415z5)αW6(α)(z)=(147606z+152z2203z3+154z465z5+16z6)α

By tedious but direct calculations, one has

ϖ2,(α)=(32)α1=0(13)1ϖ1,1(α)ϖ1,1(α),ϖ3,(α)=(116)α1=02=0[121](711)12(27)2(1)2(12)!2!(122)!ϖ1,1(α)ϖ1,12(α)ϖ4,(α)=(2512)α1=02=0[231]3= max {0,221}[122](2325)12(1323)23(313)3×(1)2(12)!3!(223)!(1+322)!ϖ1,1(α)ϖ1,12(α)ϖ5,(α)=(13760)α1=02=0[341]3=0=max{0,221}[232]4=0=max{0,232}[123](163137)12×(137163)23(63137)34(421)4×(1)2(12)!ϖ1,1(α)ϖ1,12(α)4!(324)!(1+322)!(2+423)!

and

ϖ6,l(α)=(14760)αl1=0ll2=0[45l1]l3=max{0,2l2l1}[34l2]l4=max{0,2l3l2}[23l3]l5=max{0,2l4l3}[12l4](213147)l1l2(237213)l2l3(163237)l3l4(62163)l4l5(531)l5×(1)l2(l1l2)!ϖ1,ll1(α)ϖ1,l1l2(α)l5!(l42l5)!(l1+l32l2)!(l2+l42l3)!(l3+l52l4)!,l=0,1,

Here

is the first order coefficients defined by
,If j≥2 then
for a ∈ (0,1) whilst
for a ∈ (1,2). See 13 for more information.

On the other hand, if f(k)(b−) = 0(k = 0, 1,...,p−1), then one has the approximations below,

(2.2)RLDx,bαf(x)=1hα=0ϖp,(α)f(x+h)+O(hp),p=2,,6

where h is the stepsize.

Based on (2.1) and (2.2), if f(x), together with its derivatives, has homogeneous boundary value conditions, one easily gets

(2.3)αf(x)|x|α=12 cos (πα2)=0ϖp,(α)f(xh)+f(x+h)hα+O(hp)

Here, we limit our interests in a ∈ (0, 1). The case a ∈ (1, 2) can be similarly studied. When

is the trivial case so is omitted here.

The properties of convolution coefficients

are very important for constructing effective numerical algorithms for R-L time fractional differential equations and R-L (or Riesz) space fractional differential equations.

For the space fractional differential equations, we use the following properties of coefficients

, to show the stability and convergence of the derived algorithms.

Theorem 2.1.

For 0 < α < 1, then the following inequalities hold:

=0ϖp,(α) cos (θ)0,θ[π,π],p=2,3,5,6.
Proof.

We only prove p = 2, the rest cases can be almost similarly shown. Let

f1(α,θ)==0ϖ2,(α) cos (θ)

which can be expanded as

f1(α,θ)==0ϖ2,(α) cos (θ)=12=0ϖ2,(α)( exp (iθ)+ exp (iθ))=12[(1 exp (iθ))α(3212 exp (iθ))α+(1 exp (iθ))α(3212 exp (iθ))α]

Note that f1(α, θ) is a real-value and even function, so we need only consider θ ∈ [0, π].

Using the following equations

(1 exp (±iθ))α=(2 sin θ2)α exp (±iα(θπ2))

and

(xyi)α=(x2+y2)α2 exp (iαϕ),ϕ= arctan yx,

we can rewrite f1(α, θ) as

f1(α,θ)=(2 sin θ2)α(λ12(θ)+μ12(θ))α2 cos α(θπ2+ϕ1)

where

λ1(θ)=3 cos θ,μ1(θ)= sin θ,ϕ1= arctan μ1(θ)λ1(θ)

Let

z(θ)=θπ2+ϕ1,0θπ

Then

z(θ)=(θπ2+ϕ1)'=3sin2(θ2)1+3sin2(θ2)0

Hence z(θ) is an increasing function in [0 ] and

z min (θ)=z(0)=π2,z max (θ)=z(π)=0.

Evidently α ∈ (0, 1) and θ ∈ [0 ] imply

. Furthermore, one has

f1(α,θ)=(2 sin θ2)α(λ12(θ)+μ12(θ))α2 cos α(θπ2+ϕ1)0.

All this ends the proof. □

For p = 4, α can not attain to 1. But we have following theorem.

Theorem 2.2.

If

then the following inequality holds:

=0ϖ4,(α) cos (θ)0,θ[π,π].
Proof.

Let

. By almost the same reasoning as those in Theorem 2.1 , we can get

f2(α,θ)=(2 sin θ2)α(λ22(θ)+μ22(θ))α2 cos α(θπ2+ϕ2)

where

λ2(θ)=2523 cos θ+13 cos 2θ3 cos 3θμ2(θ)=23 sin θ13 sin 2θ+3 sin 3θ,ϕ2= arctan μ2(θ)λ2(θ)

Since

λ2(θ)=14( cos (θ)12)2+24cos2(θ) sin 2(θ2)+172>0

and

μ2(θ)=[12( cos (θ)1312)2+7124] sin (θ)0

it immediately follows that

. We need only consider 0 ≤ θπ, therefore
.

Obviously, if

, then f2(α, θ) ≥ 0. A sufficient condition for
is

π2 min α(θπ2+ϕ2)0

i.e.,

0<α min {ππθ2ϕ2}.

Let y(θ) = πθ − 2 φ2, then

y(θ)=1920(5 cos θ1)sin4(θ2)a22(θ)+b22(θ)

It is clear that

is a unique maximum point of y(θ) when θ ∈ [0, π], i.e.,

y max (θ)=y max ( arccos 15)=π arccos 15+2 arctan 1916317

it follows that

 min {ππθ2φ2}=ππ arccos 15+2 arctan 19163170.8439,

i.e.,

0<α0.8439

This finishes the proof. □

Theorems2.1 and2.2 are very suitable for numerically analyzing R-L space fractional partial differential equations. But for numerically analyzing R-L time fractional partial differential equations, the monotonicity of the coefficients

(p = 1, 2, 3, 4, 5, 6) is often necessary. As far as we know, only the monotonicity of the coefficients
, for α ∈ (0, 1) and α ∈ (1, 2),
for α ∈ (0, 1) are available [13]. In the following, we study the rest cases.

Theorem 2.3.

The second-order coefficients

satisfy:
  1. ϖ2,0(α)=(32)α>0,ϖ2,1(α)=4α3(32)α<0ϖ2,2(α)=α(8α3)9(32)αϖ2,3(α)=4α(α1)(8α7)81(32)αϖ2,4(α)=α(α1)(64α2176α+123)486(32)α
  2. When 0 < α < 1,

    and
    hold for ≥ 4,

  3. When 1 < α < 2,

    and
    hold for ≥ 5,

P r o o f. See Appendix A. □

From the results of the above theorem, one can see that: If α = 1, then

,for ℓ = 0, 1, 2, and
, for ℓ ≥ 3; If α ∈ (0, 1), then
for ℓ ≥ 0,
and
for ℓ ≥ 4. Similarly, if α = 2, then
for = 0, 1 ,..., 4, and
for ℓ ≥ 5; Ifα ∈ (1, 2), then
for ℓ ≥ 0,
and
for < ≥ 5. That is to say, the monotonicity of
appears from ℓ = 3 which is just the number of weigh coefficients for approximating
with order 2. The case of α ∈ (1, 2) has similar explanation.

Besides their monotonicity, studying bounds of these coefficients is also of importance, which can be used to analyze the stability and convergence for time fractional differential equations. In [4], Dimitrov gave the bounds for first-order

(0 < α < 1) below.

Theorem 2.4.

The first-order coefficients

(0 < α < 1) satisfy:
  1. , ℓ ≥ 3

    where

  2. where

In this paper, we can give tighter estimates for the lower bounds. See the following theorem.

Theorem 2.5.

The first-order coefficients

(0 < α < 1) satisfy:
  1. , ℓ ≥ 3

    where

  2. ℓ ≥ 3.

    where

Next, we list the comparison theorem for

and
and
, respectively

Theorem 2.6.

The following inequalities hold:

  1. for

    for

    for 0 < α < 1 and ℓ ≥ 5;

  2. for 0 < α < 1 and ℓ ≥ 3.

In the following, we give the bounds for the corresponding coefficients.

Theorem 2.7.

The first-order coefficients

, (0 < α < 1) satisfy:
  1. , ℓ ≥ 4, where

    B¯(1+α,)=(1α)α(1+α)6(3)2(2+α)B¯(1+α,)=α(1+α)2(3+1)2+α
  2. where

    S¯(1+α,)=(1α)α(1+α)2(3+2α)(3)3+2α,S¯(1+α,)=3α2(3)1+α

Now we start to show the bounds of the second-order coefficients.

Theorem 2.8.

The second-order coefficients

, and
for 0 < α < 1 satisfy:
  1. where

    B2L(α,)=(32)α[(1+(13))α(1α)2(2)2(1+α)(1(13)1)α222α+11+(α+1)],
    B2L(α,)=(32)α[(1+(13))α2α+1(+1)α+1α2(1α)242α+12(1(13)1)(2)4(α+1)]
  2. where

    B¯(1+α,)=(32)1+α[(1+(13))(1α)α(1+α)6(3)2(2+α)+(1α)2α2(1+α)2216(1(13)3)(6)4(2+α)α(1+α)22(13+(13)1)(3)2+α],B¯(1+α,)=(32)1+α[(1+(13))α(α+1)3α+22(+1)α+2+(1(13)3)α2(1+α)232(2+α)24(1+(2+α))(1α)α(1+α)26(13+(13)1)(31)2(2+α)]

All proofs for Theorems 2.4-2.8 are given in Appendix B.

3. Numerical methods for the Riesz-type turbulent diffusion equation

Define tk = , k = 0, 1 ,...,N, and

for a given T > 0,
is the equidistant grid size in space, xj = a + jh , j = 0, 1,...,M.

3.1. The 2nd-order scheme in space

Firstly, using the Crank-Nicolson method for the Riesz space fractional turbulent diffusion equation (1.1) in time direction, we obtain

(3.1)u(xj,tk+1)u(xj,tk)τ=12(d22u(xj,tk+1)x2d1u(xj,tk+1)x
(3.2)+dααu(xj,tk+1)|x|α+d22u(xj,tk)x2d1u(xj,tk)x+dααu(xj,tk)|x|α)+s(xj,tk+12)+O(τ2)

Secondly, for the first- and second-order derivatives, we use the following approximations, respectively

(3.3)u(xj,tk)x=μxδxu(xj,tk)+O(h2)

and

(3.4)2u(xj,tk)x2=δx2u(xj,tk)+O(h2)

where μx and

are defined by

μxδxu(xj,tk)=u(xj+1,tk)u(xj1,tk)2h

and

δx2u(xj,tk)=u(xj+1,tk)2u(xj,tk)+u(xj1,tk)h2.

Next, we choose the second-order formula to approximate Riesz derivative,

(3.5)αu(x,t)|x|α=12cos(πα2)hα(=0ϖ2,(α)u(xh,t)+=0ϖ2,(α)u(x+h,t))+O(h2)

Substituting (3.3), (3.4) and (3.5) into (3.2) and removing the high-order term yield

(3.6)(d2h2+d12h)uj1k+1(2τ+2d2h2)ujk+1+(d2h2d12h)uj+1k+1ν[=0jϖ2,(α)ujlk+1+=0Mjϖ2,(α)uj+lk+1]=(d2h2+d12h)uj1k2τ2d2h2)ujk(d2h2d12h)uj+1k+l[=0jϖ2,(α)ujlk+=0Mjϖ2,(α)uj+lk]2sjk+12,j=1,,M1,k=0,1,,N1,

where

Next we discuss the stability and convergence of scheme (3.6).

Theorem 3.1.

The numerical scheme (3.6) is unconditionally stable and convergent with order 𝒪(τ2 + h2).

Proof.

Assume that the solution of equation (1.1) can be zero extended to the whole real line R. Suppose

is the approximation solution of equation (3.6) and let
, then the error equation reads as

(3.7)(d2h2+d12h)j1k+1(2τ+2d2h2)jk+1+(d2h2d12h)j+1k+1v[=0ϖ2,(γ)jlk+1+=0ϖ2,(γ)j+lk+1]=(d2h2+d12h)j1k(2τ2d2h2)jk(d2h2d12h)j+1k+U[=0ϖ2,(γ)jlk+=0ϖ2,(γ)j+lk],j=1,,M1,k=0,1,,N1.

Let

be the solution of equation (3.7),
, where θ ∈ [−π, π] is called the phase angle. The stability condition of scheme (3.6) is (θ)| ≤ 1 for all θ ∈ [−π, π].

Substituting

into (3.7) gives

ξ(θ)=(2hτ4d2hsin2(θ2)2νh=0ϖ2,(γ)cos(θ))id1sin(θ)(2hτ+4d2hsin2(θ2)+2νh=0ϖ2,(γ)cos(θ))+id1sin(θ).

According to Theorem 2.1, we easily obtain |ξ (θ)| ≤ 1. So scheme (3.6) is unconditionally stable. It is easy to show the convergence order of scheme (3.6) is 𝒪 (τ2 +h2). In effect, the proof is almost the same as that of [5]. □

3.2. The 4th-order scheme in space

Let us first consider the following differential equation

(3.8)d22u(x,t)x2d1u(x,t)x=g(x,t)

Applying the technique presented in [18], we can obtain a fourth-order difference scheme for solving the above equation

(3.9)4u(xj,t)=4g(xj,t)+O(h4)

Where

4=(d2+d12h212d2)δx2d1μxδx,Δ˜4=I+h212(δx2d1d2μxδx)

in which I is a unit operator.

Combing (3.2) with (3.8) and (3.9) leads to

(3.10)2τΔ˜4.(u(xj,tk+1)u(xj,tk))=4(u(xj,tk+1)+u(xj,tk))+dαΔ˜4.(αu(xj,tk+1)|x|α+u(xj,tk)|x|α)+2Δ˜4s(xj,tk+12)+O(τ2+h4).

For the Riesz derivative in equation (3.10), we use the fourth-order numerical scheme

αu(x,t)|x|α=12cos(πα2)hα=0ϖ4,(α)((u(xh,t)+u(x+h,t))+O(h4).

Substituting it into (3.10) and ignoring the truncation error term give

(3.11)(2a1τb1)uj1k+1+(2a2τb2)ujk+1+(2a3τb3)uj+1k+1+v[=0jϖ4,(α)(a1ujl1k+1+a2ujlk+1+a3ujl+1k+1+=0Mjϖ4,(α)(a1uj+l1k+1+a2uj+lk+1+a3uj+l+1k+1)=(2a1τ+b1)uj1k+(2a2τ+b2)ujk+(2a3τ+b3)uj+1ku[=0jϖ4,(α)(a1ujl1k+a2ujlk+a3ujl+1k)+=0Mjϖ4,(α)(a1uj+l1k+a2uj+lk+a3uj+l+1k)]+2a1sjk+112+2a2sjk+12+2a3sj+k+112,j=1,,M1,k=0,1,,N1.

Here, the parameters in (3.11) are given below,

v=dα2cos(πα2)hα,a1=112+d1h24d2,a2=56,a3=112d1h24d2,b1=d2h2+d1212d2+d12h,b2=2(d2h2+d1212d2),b3=d2h2+d1212d2d12h.
Theorem 3.2.

When 0 < α ≤ 0.8439, the numerical scheme (3.11) is unconditionally stable and convergent with order 𝒪(τ2 +h4).

Proof.

Similar to the above subsection, we can get the error equation of equation (3.11) and let

, θ ∈ [−π, π]. Substituting it into the error equation gives

ξ(θ)=(s1s2s3v)is4vsin(θ)(s1+s2+s3v)+iS4vsin(θ),

where

s1=2τ(113sin2(θ2)),s2=2sin2(θ2)(2d2h2+d126d2),s3=2(113sin2(θ2))=0ϖ4,(α)cos(θ),s4=(d1h6τd2+d1h)d1h6d2=0ϖ4,(α)cos(θ).

Note that ν, s2, s3 ≥ 0. It follows from Theorem 2.2 that |ξ (θ)| ≤ 1 if 0 < α ≤ 0.8439. So scheme (3.11) is unconditionally stable if 0 < α ≤ 0.8439. For such α ∈ (0, 0.8439], the convergence order of scheme (3.11) is 𝒪(τ2 + h4). □

3.3. The 6th-order scheme in space

From Taylor expansion, we have

(3.12)u(xj,t)x=μxδx(Ih26δx2)u(xj,t)+h4305u(xj,t)x5+O(h6)
(3.13)2u(xj,t)x2=δx2(Ih212δx2)u(xj,t)+h4906u(xj,t)x6+O(h6),
(3.14)3u(xj,t)x3=μxδx3u(xj,t)+O(h2),
(3.15)4u(xj,t)x4=δx4u(xj,t)+O(h2),

According to (3.8), one gets

(3.16)5u(xj,t)x5=d1d24u(xj,t)x4+1d23g(xj,t)x3,

and

(3.17)6u(xj,t)x6=d12d224u(xj,t)x4+1d24g(xj,t)x4+d1d223g(xj,t)x3.

From equations (3.12)-(3.17), one has

(3.18)u(xj,t)x=μxδx(Ih26δx2)u(xj,t)+d1h430d2δx4u(xj,t)+h430d2μxδx3g(xj,t)+O(h6)

and

(3.19)2u(xj,t)x2=δx2(Ih212δx2)u(xj,t)+d12h490d22δx4u(xj,t)+h490d2δx4g(xj,t)+h4d190d22μxδx3g(xj,t)+O(h6)

Substituting (3.18) and (3.19) into (3.8) gives

(3.20)6u(xj,t)=Δ˜6g(xj,t)+O(h6)

where

6=d2δx2(Ih212δx2)d1μxδx(Ih26δx2)d12h445d2δx4,Δ˜6=Ih490δx2(δx22d1d2μxδx).

Combining (3.2) with (3.8) and (3.20), we can obtain

(3.21)2τΔ˜6.(u(xj,tk+1)u(xj,tk))=6(u(xj,tk+1)+u(xj,tk)),+dγΔ6˜(u(xj,tk+1)|x|γ+u(xj,tk)|x|γ)+2Δ˜6s(xj,tk+12),+O(τ2+h6).

For the Riesz derivative, we apply the sixth-order numerical scheme

αu(x,t)|x|α=12cos(πα2)hα=0ϖ6,(α)(u(xh,t)+u(x+h,t))+O(h6).

Substitution in (3.21) gives

(3.22)(2c1τ+e1)uj2k+1+(2c2τ+e2)uj1k+1+(2c3τ+e3)ujk+1+(2c4τ+e4)uj+1k+1+(2c5τ+e5)uj+2k+1+v[=0jϖ6,(α)(c1ujl2k+1+c2ujl1k+1+c3ujlk+1+c4ujl+1k+1+c5ujl+2k+1)+=0Mjϖ6,(α)(c1uj+l2k+1+c2uj+l1k+1+csuj+lk+1+c4uj+l+1k+1+c5uj+l+2k+1)]=(2c1τe1)uj2k+(2c2τe2)uj1k+(2c3τe3)ujk+(2c4τe4)uj+1k+(2c5τe5)uj+2kv[=0jϖ6,(α)(c1ujl2k+c2ujl1k+csujlk+1+c4ujl+1k+c5ujl+2k)+=0Mjϖ6,(α)(c1uj+l2k+c2uj+l1k+csuj+lk+c4uj+l+1k+c5uj+l+2k)]+2c1fj2k+12+2c2fj1k+12+2c3fjk+12+2c4fj+1k+12+2c5fj+2k+12,j=2,,M2,k=0,1,,N1.

Here

v=dα2cos(πα2)hα,c1=190(1+d1hd2),c2=190(4+2d1hd2),c3=1415,c4=190(42d1hd2),c5=190(1d1hd2),e1=d212h2+d112h+d1245d2,e2=(4d23h2+2d13h+4d1245d2),e3=5d22h2+2d1215d2,e4=(4d23h22d13h+4d1245d2),e5=d212h2d112h+d1245d2.
Theorem 3.3.

The numerical scheme (3.22) is unconditionally stable and convergent with order 𝒪 (τ2 +h6).

Proof.

Let

θ ∈ [−π, π]. Substituting it into the error equation of the equation (3.22)

(2c1τ+e1)j2k+1+(2c2τ+e2)j1k+1+(2c3τ+e3)jk+1+(2c4τ+e4)j+1k+1+(2c5τ+e5)j+2k+1+ν[=0jϖ6,(α)(c1jl2k+1+c2jl1k+1+c3jlk+1+c4jl+1k+1+c5jl+2k+1)+=0Mjϖ6,(α)(c1j+l2k+1+c2j+l1k+1+c3j+lk+1+c4j+l+1k+1+c5j+l+2k+1)]=(2c1τe1)j2k+(2c2τe2)j1k+(2c3τe3)jk+(2c4τe4)j+1k+(2c5τe5)j+2kv[=0jϖ6,(α)(c1jl2k+c2jl1k+c3jlk+1+c4jl+1k+c5jl+2k)+=0Mjϖ6,(α)(c1j+l2k+c2j+l1k+c3j+lk+c4j+l+1k+c5j+l+2k)],j=2,,M2,k=0,1,,N1,

leads to

ξ(θ)=(w1w2w3ν)iw4sin(θ)(w1+w2+w3ν)+iw4sin(θ)

Here

w1=245τ[458sin4(θ2)],w2=2d23h2(7cosθ)sin2(θ2)+16d1245d2sin4(θ2),w3=245[458sin4(θ2)]=0ϖ6,(α)cos(θ),w4=8d1h45τd2sin2(θ2)+d13h(4cosθ)+8d1hl45d2sin2(θ2)=0ϖ6,(α)cos(θ).

It is clear that ν, w2, w3 0. By using Theorem2.1 , one also has

|ξ(θ)|1.

So scheme (3.22) is unconditionally stable. It can be shown that the convergence order of scheme (3.22) for equation (1.1) is 𝒪 (τ2 + h6).□

In the next section, we present several numerical examples.

4. Numerical examples

We now test the higher-order schemes for Riesz derivatives.

Example 1.

Consider the function fp(x) = xp(1 − x) p, x ∈ [0, 1], p = 2,3,4,5,6.

The Riesz derivative of the above function is analytically expressed as

αfp(x)|x|α=12cos(πα2)=0p(1)p!(p+l)!!(pl)!Γ(p++1α)[xp+α+(1x)p+α].

We numerically solve fp(x) by using numerical scheme (2.3). The numerical results are presented in Tables 1-5. From these tables, the experimental orders are in line with the theoretical orders p (p = 2,3,4,5,6).

Table 1.

The absolute error, convergence order of Example 1 by numerical scheme (2.3) with p = 2.

αhthe absolute errorthe convergence order
0.2
2.381267e-004
5.900964e-0052.0127
1.460639e-0052.0144
3.628491e-0062.0092
9.039358e-0072.0051
0.4
7.097814e-004
1.696639e-0042.0647
4.123703e-0052.0407
1.014980e-0052.0225
2.516782e-0062.0118
0.6
1.638369e-003
3.728453e-0042.1356
8.824023e-0052.0791
2.141683e-0052.0427
5.272483e-0062.0222
0.8
3.782418e-003
7.826735e-0042.2728
1.747182e-0042.1634
4.102708e-0052.0904
9.923174e-0062.0477
Table 2.

The absolute error, convergence order of Example 1 by numerical scheme (2.3) with p = 3.

αhthe absolute errorthe convergence order
0.2
3.146678e-007
1.576085e-0071.7052
7.991483e-0082.3608
4.501080e-0082.5726
2.761993e-0082.6786
0.4
4.349687e-006
1.491902e-0062.6391
6.709309e-0072.7779
3.560502e-0072.8394
2.108270e-0072.8742
0.6
2.194879e-005
6.996810e-0062.8196
3.050074e-0062.8861
1.590859e-0062.9169
9.316704e-0072.9347
0.8
1.067572e-004
3.282279e-0052.9088
1.407258e-0052.9439
7.270149e-0062.9598
4.231299e-0062.9688

Next we test the numerical schemes for the equations which have the form of equation (1.1).

Example 2.

Consider the following equation

u(x,t)t=2u(x,t)x2u(x,t)x+αu(x,t)|x|α+s(x,t),0x1,0t1,

in which

Table 3.

The absolute error, convergence order of Example 1 by numerical scheme (2.3) with p = 4.

αhthe absolute errorthe convergence order
0.2
3.254967e-006
1.421712e-0063.7121
7.061638e-0073.8381
2.279637e-0073.9509
0.4
1.307893e-005
5.433994e-0063.9362
2.610192e-0064.0217
1.395202e-0064.0635
8.089264e-0074.0821
0.6
4.165885e-005
1.647646e-0054.1569
7.642179e-0064.2137
3.980841e-0064.2309
2.261840e-0064.2336
0.8
1.466022e-004
5.460563e-0054.4258
2.420431e-0054.4625
1.216174e-0054.4647
6.707129e-0064.4568

f(x,t)=exp(t)x4(1x)4(x4+10x3149x2+138x30)+exp(t)2cos(π2α){Γ(7)Γ(7α)[x6α+(1x)6α]6Γ(8)Γ(8α)[x7α+(1x)7α]+15Γ(9)Γ(9α)[x8α+(1x)8α]20Γ(10)Γ(10α)[x9α+(1x)9α]+15Γ(11)Γ(11α)[x10α+(1x)10α]6Γ(12)Γ(12α)[x11α+(1x)11α]+Γ(13)Γ(13α)[x12α+(1x)12α]}.
Table 4.

The absolute error, convergence order of Example 1 by numerical scheme (2.3) with p = 5.

αhthe absolute errorthe convergence order
0.2
3.254967e-006
1.421712e-0063.7121
7.061638e-0073.8381
3.863551e-0073.9123
2.279637e-0073.9509
0.4
8.731739e-010
3.348011e-0104.2959
1.473288e-0104.5023
7.232096e-0114.6160
3.867341e-0114.6877
0.6
5.385482e-009
1.900398e-0094.6680
7.985621e-0104.7554
3.806168e-0104.8071
1.994064e-0104.8412
0.8
3.005433e-008
1.023908e-0084.8256
4.211445e-0094.8727
1.978515e-0094.9008
1.025811e-0094.9192

Its analytical solution is u (x, t) = exp(t)x6(1 − x)6 and satisfy the corresponding initial and boundary values conditions.

We solve this problem with the numerical schemes (3.6) and (3.11) for different values of τ ,h and α. The absolute error, temporal and spatial convergence orders are listed in Tables 6 and 7, which display that the numerical results are in line with our theoretical analysis.

In the following, we give a slightly different example.

Example 3.

Consider the following equation

u(x,t)t=2u(x,t)x22u(x,t)x+α2αu(x,t)|x|α+s(x,t),0x1,0t1,

in which

Table 5.

The absolute error, convergence order of Example 1 by numerical scheme (2.3) with p = 6.

αhthe absolute errorthe convergence order
0.2
3.783855e-008
2.310553e-0094.0335
4.258651e-0115.7617
6.690552e-0135.9921
1.035841e-0146.0133
0.4
3.116503e-007
1.031745e-0084.9168
1.651582e-0105.9651
2.418476e-0126.0936
3.582209e-0146.0771
0.6
1.564617e-006
3.647148e-0085.4229
5.062291e-0106.1708
6.799919e-0126.2181
9.539537e-0146.1555
0.8
8.311643e-006
1.432632e-0075.8584
1.663072e-0096.4287
1.942938e-0116.4195
2.433601e-0136.3190

s(x,t)=x6(1x)6[cos(t)(x42x3+x2)+sin(t)(32x3288x2+256x56)]+α22sin(t)sec(π2α){Γ(9)Γ(9α)[x8α+(1x)8α]8Γ(10)Γ(10α)[x9α+(1x)9α]+28Γ(11)Γ(11α)[x10α+(1x)10α]56Γ(12)Γ(12α)[x11α+(1x)11α]+70Γ(13)Γ(13α)[x12α+(1x)12α]56Γ(14)Γ(14α)[x13α+(1x)13α]+28Γ(15)Γ(15α)[x14α+(1x)14α]8Γ(16)Γ(16α)[x15α+(1x)15α]+Γ(17)Γ(17α)[x16α+(1x)16α]}.
Table 6.

The absolute errors, temporal and spatial convergence orders of Example 2 by difference scheme (3.6).

temporalspatial
αthe maximum errorsconvergence ordersconvergence orders
0.2
2.581219e-005
6.217660e-0062.05362.0536
1.536085e-0062.01712.0171
3.844480e-0071.99841.9984
0.3
2.514418e-005
6.061411e-0062.05252.0525
1.498825e-0062.01582.0158
3.755193e-0071.99691.9969
0.4
2.416676e-005
5.842791e-0062.04832.0483
1.448271e-0062.01232.0123
3.636282e-0071.99381.9938
0.5
2.270557e-005
5.532646e-0062.03702.0370
1.379247e-0062.00412.0041
3.477777e-0071.98761.9876
0.6
2.043415e-005
5.079755e-0062.00822.0082
1.283408e-0061.98481.9848
3.264917e-0071.97491.9749
0.7
1.664449e-005
4.378341e-0061.92661.9266
1.145019e-0061.93501.9350
2.972789e-0071.94551.9455

Its exact solution is u (x, t) = sin(t)x8(1 − x)8 and satisfy the corresponding initial and boundary values conditions.

The absolute error, temporal and spatial convergence orders are listed in Table8 by numerical scheme (3.22). The numerical results agree with the theoretical results.

Table 7.

The maximum errors, temporal and spatial convergence orders of Example 2 by difference scheme (3.11).

αthe maximum errorstemporal convergence ordersspatial convergence orders
0.2
1.151043e-004
4.361384e-0062.36104.7220
2.346706e-0072.10814.2161
1/322.036847e-0081.76313.5262
0.3
1.128702e-004
4.362181e-0062.34684.6935
2.461975e-0072.07364.1472
2.396154e-0081.68053.3610
0.4
1.088961e-004
4.433621e-0062.30914.6183
2.870612e-0071.97463.9491
2.910816e-0081.65093.3019
0.5
1.018053e-004
4.654112e-0062.22564.4512
3.607253e-0071.84473.6895
3.674011e-0081.64783.2955
0.6
8.897936e-005
5.201584e-0062.04824.0964
4.980729e-0071.69233.3845
4.862623e-0081.67833.3566
0.7
6.521192e-005
6.540139e-0061.65883.3177
7.724068e-0071.54103.0819
6.867979e-0081.74573.4914

5. Conclusions

In paper, we construct high-order (from 2nd-order to 6th-order) numerical schemes to approximate the Riesz derivatives. Next, we develop three kinds of difference schemes for the Riesz space fractional turbulent diffusion equation. The stability of the derived numerical algorithms are shown by the Fourier method. The convergence orders are 𝓞(τ2 + h2), 𝓞(τ2 + h4) and 𝓞(τ2 + h6), respectively. Finally, numerical results confirm the theoretical analysis.

Table 8.

The maximum errors, temporal and spatial convergence orders of Example 3 by difference scheme (3.22).

αthe maximum errorstemporal convergence ordersspatial convergence orders
0.2
1.360207e-007
2.071201e-0092.01246.0372
3.348089e-0111.98375.9510
5.235085e-0131.99975.9990
0.3
1.356431e-007
2.092867e-0092.00616.0182
3.254863e-0112.00226.0067
4.855887e-0132.02226.0667
0.41/81.348600e-007
1/162.146379e-0091.99115.9734
2.972961e-0112.05806.1739
3.852828e-0132.08996.2698
0.51/81.335205e-007
1/162.246228e-0091.96455.8934
2.200601e-0112.22456.6735
2.816274e-0132.09606.2880
0.61/81.322258e-007
1/162.372915e-0091.93345.8002
1.827817e-0112.34017.0204
4.506292e-0131.78075.3420
0.7
1.357968e-007
1/163.805230e-0091.71915.1573
6.671019e-0111.94465.8339
1.819328e-0121.73215.1964

Appendix A

Proof of Theorem 2.3.

(1) Direct calculations can finish it so we omit the details.

(2) See [13] for details.

(3) Now we show the case α ∈ (1,2). We firstly show that

for l ≥ 5. For convenience, denote α = 1 + γ, where 0 < γ < 1. Lengthy calculations give

ϖ2,l(a)=(32)1+γm=0l(13)mϖ1,m(1+γ)ϖ1,lm(1+γ)=(32)1+γ[ϖ1,0(1+γ)ϖ1,(1+γ)+13ϖ1,1(1+γ)ϖ1,1(1+γ)+19ϖ1,2(1+γ)ϖ1,2(1+γ)+(13)lϖ1,0(1+γ)ϖ1,(1+γ)+(13)l1ϖ1,1(1+γ)ϖ1,1(1+γ)]+(32)1+γm=3l2(13)mϖ1,m(1+γ)ϖ1,m(1+γ)=(32)1+γ[1γ+13ll2γ+γ(γ+1)18l(l1)(l2γ)(l3γ)+(13)l(13l(γ+1)l2γ]ϖ1,m1+γϖ1,lm1+γ(32)1+γ[24424328(γ+1)81ll2γ+γ(γ+1)18x(x1)(x2γ)2,ϖ1,l1+γ+(32)1+γm=3l1(13)mϖ1,m1+γϖ1,lm1+γl5

Let

F(x,γ)=24424328(γ+1)81x(x2γ)+γ(γ+1)18x(x1)(x2γ)2,x5,

and

G(x,γ)=486(x2γ)2F(x,γ)=488(x2γ)2168(γ+1)x(x2γ)+27γ(γ+1)x(x1),x5.

Then

Gx(x,γ)=976(x2γ)168(γ+1)(2x2γ)+27γ(γ+1)(2x1)

and

Gxx(x,γ)=54γ2282γ+640.

Obviously,

Gxx(x,γ)0for0<γ<1,

it immediately follows that Gx(x, γ) is an increasing function and Gx(x, γ) ≥ Gx(5,γ) for x ≥ 5.

Note that

Gx(5,γ)=411γ21909γ+1585>0,0<γ<1.

Hence G(x, γ) is an increasing function too, and G (x, γ) ≥ G (5,γ) if x ≥ 5. Simple calculations yields

G(5,γ)=1868γ23068γ+1872Gmin(5,γ)=G(5,767934)=612131467,

so, G(x, γ) ≥ 0. Therefore, the following inequality holds

F(x,γ)=G(x,γ)486(x2γ)20,

which means

for l ≥ 5. Here we used the positivity of the coeffcient
.

Next, we show that

for l ≥ 5. Note that

ϖ2,lαϖ2,l+1(α)=(2+γ)(32)1+γ(13)ll1=0l3l1l1+1ϖ1,l1(1+γ)ϖ1,ll1(1+γ)(32)1+γ(13)l+1(12+γl+1)ϖ1,l(1+γ)=(32)1+γ(13)l[(2+γ)l1=0l3l1l1+1ϖ1,l1(1+γ)ϖ1,ll1(1+γ)13ϖ1,l(1+γ)]+2+γl+1(32)1+γ(13)1+lϖ1,l(1+γ)(2+γ)(32)1+γ(13)l1=0l3l1l1+1ϖ1,1(1+γ)ϖ1,1(1+γ)16ϖ1,(1+γ)]+2+γ+1(32)1+γ(13)+1ϖ1,(1+γ)=(2+γ)(32)1+γ(13)P(,γ)ϖ1,(1+γ)+2+γ+1(32)1+γ(13)+1ϖ1,(1+γ)+(2+γ)(32)1+γ(13)l1=0l3l1l1+1ϖ1,1(1+γ)ϖ1,1(1+γ)

Here

P(,γ)=[563(γ+1)2(2γ)+3γ(γ+1)(1)2(2γ)(3γ)]+3+1[1(+1)(γ+1)3(2γ)+γ(γ+1)(+1)18(2γ)(3γ)]

Obviously, the last two terms in the right-hand side of the last equality are both nonnegative, so we only need prove that the factor P (l, γ) in the first term is nonnegative.

Let

P1(,γ)=563(γ+1)2(2γ)+3γ(γ+1)(1)2(2γ)(3γ),P2(,γ)=1(+1)(γ+1)3(2γ)+γ(γ+1)(+1)18(2γ)(3γ),

then

P(,γ)=P1(,γ)+3+1P2(,γ)

If l = 5, then

P2(5,γ)=12(γ+1)(3γ)+5γ(γ+1)3(3γ)(2γ)>0.

Now we consider the case l ≥ 6. Let

Q(x,γ)=18(x2γ)(x2.5γ)P3(x,γ),x[6,)

where

P3(x,γ)=1(x+1)(γ+1)3(x2γ)+γ(γ+1)x(x+1)18(x2γ)(x2.5γ).

By simple calculations, one has

Qxx(x,γ)=2γ210γ+24>0,0<γ<1.

So Qx(x, γ) is an increasing function and

Qx(x,γ)Qx(6,γ)=19γ280γ+72>0,x6,0<γ<1.

It immediately follows that Q(x, γ) is an increasing function with respect to x and

Q(x,γ)Q(6,γ)=102γ2198γ+105>0,0<γ<1,

i.e., P3(x, γ) > 0. Noticing P2(l, γ) > P3(l, γ) yields

P2(,γ)>0,[5,)

Therefore,

P(,γ)=P1(,γ)+3+1P2(,γ)P1(,γ)+812P2(,γ)=12433(γ+1)(10+9)2(2γ)+3γ(γ+1)(5+1)4(2γ)(3γ)>12433(γ+1)(10+9)2(2γ)+3γ(γ+1)(5+1)4(2γ)2.

When l = 5, we easily know that P(l, γ) > 0 by direct calculations. Next, we discuss the case l ≥ 6. Let

P4(,γ)=12433(γ+1)(10+9)2(2γ)+3γ(γ+1)(5+1)4(2γ)2,

and

R(x,γ)=12(x2γ)2P4(x,γ),x6.

Differentiating twice with respect to x gives

Rxx(x,γ)=90γ2270γ+632,

which is positive when γ ∈ (0, 1).

So, Rx(x, γ) is an increasing function and

Rx(x,γ)>Rx(6,γ)=729γ22225γ+2006>0.

Furthermore, R(x, γ) is an increasing function as well and

R(x,γ)>R(6,γ)=3412γ26020γ+2968>0,x[6,)

So, P4(l, γ) > 0 implies P(l, γ) > 0. It follows that

for l ≥ 5.

All this completes the proof. □

Appendix B

First, we list a lemma which comes from [4,11].

Lemma A

  1. 1 − x < exp (−x) holds for 0 < x < 1.

  2. 1 − x > exp (−2x) holds for 0 < x ≤ 0.7968.

  3. (1 + x)α ≥ 1 + αx holds for α ≤ 0 or α ≥ 1,

  4. (1 + x1 + x2 + x3 + ··· + xn) ≤ (1 + x1)(1 + x2)(1 + x3) ··· (1 + xs), where xm ≥ −1 and sign(xm) = sign(xn) for ∀ 1 ≤ m, ns.

Proof of Theorem 2.5.

(1) In view of (ii) of Lemma A, one has

|ϖ1,(α)|=(1α+1)|ϖ1,1(α)|=(1α+1)(1α+11)(1α+13)|ϖ1,2(α)|exp(2(α+1))exp(2(α+1)1)exp(2(α+1)3)|ϖ1,2(α)|=exp(2(α+1)k=3l1k)|ϖ1,2(α)|.

Note that

k=3l1k<21xdx=ln2,

so we get

|ϖ1,(α)|>exp(2(α+1)ln2)|ϖ1,2(α)|=α(1α)2(2)2(α+1)

From [4], we know that

|ϖ1,(α)|<α2α+1(+1)α+1,

i.e.,

B1L(α,)<|ϖ1,(α)|<B1R(α,)whereB1L(α,l)=α(1α)2(2l)2(α+1),B1R(α,)=α(2+1)α+13.

(2) From (1), we have

k=l|ϖ1,k(α)|>α(1α)22α+1k=l1k2(α+1).

Because of

k=l1k2(α+1)>1x2(α+1)dx=1(2α+1)2α+1,

we get

k=l|ϖ1,k(α)|>α(1α)2α+1(2)2α+1.

In addition, we also have [4]

k=|ϖ1,k(α)|<2(2)α

it immediately follows that

S1L(α,l)<k=l|ϖ1,k(α)|<B1R(α,l),whereS1L(α,l)=α(1α)2α+1(2l)2α+1,B1R(α,l)=2(2l)α,l3.
Proof of Theorem 2.6.

(1) Let

α=12ln22π2151.

Then from α < αl, we can get

(2)1+αexp((α+1)2(π2654))>1,

i.e.,

B1L(α,)B¯(α,)<1.

Obviously, the conditions for the above inequality are

α30.0267,α40.7551,andα552ln521>941>1.

(2) Let

f(α,)=B¯(α,)B1L(α,)=2α+15α(2)1+α

and

g(α,)=lnf(α,)=ln(2α+1)ln(5α)+(1+α)ln2.

Then

gα(α,3)=22α+11α+ln32,

and

gαα(α,3)=4α+1α2(2α+1)2>0,

that is to say that the function g(α, 3) has a minimum value and from the equation gα(α, 3) = 0 we know that the minimum point is

α=1+1+8ln324.

Therefore,

f(α,)>f(α,3)f(α,3)=exp(g(α,3))=2α+15α(32)1+α1.3443>1,

i.e.,

. □

Proof Theorem 2.7

is almost the same as that of Theorem 2.5.

Proof Theorem 2.8

(1) Note that

for l ≥ 3 and
for l ≥ 3 and
for l ≥ 1 (0 < α < 1), then

|ϖ2,(α)|=(32)α|1=0(13)1ϖ1,1(α)ϖ1,1(α)=(32)α|(1+(13))ϖ1,(α)+1=11(13)1ϖ1,1(α)ϖ1,1(α)=(32)α[(1+(13))|ϖ1,(α)|1=11(13)1|ϖ1,1(α)||ϖ1,1(α)|]

From Lemma A, we know that

(1+1)1+α(1+(1))1+α(1+(α+1)1)(1+(α+1)(1))1+(α+1)

and

12(1+α)(1)2(1+α)((1+(1)2)2)2(1+α)=(2)4(α+1)

So,

|ϖ2,(α)|=(32)α[(1+(13)))|ϖ1,(α)|1=11(13)1|ϖ1,1(α)||ϖ1,1(α)|](32)α[(1+(13))α2α+1(+1)α+11=11(13)1α2(1α)242α+1(1)2(α+1)12(α+1)](32)α[(1+(13))α2α+1(+1)α+1α2(1α)242α+12×(1(13)1)(2)4(α+1)],×(1(13)1)(2)4(α+1)],

and

|ϖ2,(α)|=(32)α[(1+(13))|ϖ1,(α)|1=11(13)1|ϖ1,1(α)||ϖ1,1(α)|](32)α[(1+(13))α(1α)2(2)2(1+α)α222(α+1)1=11(13)11(1+1)α+1(1+1)α+1](32)α[(1+(13))α(1α)2(2)2(1+α)(1(13)1)α222α+11+(α+1)]

(2) Note that

for l ≥ 4 and
for l ≥ 2 (0 < α < 1), then

|ϖ2,(1+α)|=(32)1+α|1=0(13)1ϖ1,1(1+α)ϖ1,1(1+α)|=(32)1+α[(1+(13))|ϖ1,1+1=22(13)1(1+α)|ϖ1,1(1+α)||ϖ1,1(1+α)|(13+(13)1)(1+α)|ϖ1,1(1+α)|]

So,

|ϖ2,(1+α)|(32)1+α[(1+(13))α(α+1)3α+22(+1)α+2+1=22(13)1α2(1+α)232(2+α)4(1+1)α+2(1+1)2+α(1α)α(1+α)26(31)2(2+α)(13+(13)1)](32)1+α[(1+(13))α(α+1)3α+22(+1)α+2+(1(13)3)α2(1+α)232(2+α)24(1+(2+α))(1α)α(1+α)26(13+(13)1)(31)2(2+α)]

and

|ϖ2,(1+α)|(32)1+α[(1+(13))(1α)α(1+α)6(3)2(2+α)+(1α)2α2(1+α)2216(1(13)3)(6)4(2+α)α(1+α)22(13+(13)1)(3)2+α]

The proof is thus complete. □

Acknowledgements

The work was partially supported by the National Natural Science Foundation of China under Grant Nos. 11372170 and 11561060, Key Program of Shanghai Municipal Education Commission under Grant No. 12ZZ084, the grant of “The First-class Discipline of Universities in Shanghai”, the Scientific Research Program for Young Teachers of Tianshui Normal University under Grant No. TSA1405, and Tianshui Normal University Key Construction Subject Project (Big data processing in dynamic image).

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  1. Please cite to this paper as published in:

    Fract. Calc. Appl. Anal., Vol. 19, No 1 (2016), pp. 19–55, DOI: 10.1515/fca-2016-0003

Received: 2015-4-2
Published Online: 2016-3-9
Published in Print: 2016-2-1

© 2016 Diogenes Co., Sofia

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