In this paper the optimal homotopy asymptotic method (OHAM) and multistage optimal homotopy asymptotic method (MOHAM) are applied to obtain an analytic approximate solution to a time-fractional Klein-Fock-Gordon (FKFG) equation. The FKFG equation plays an important role in characterizing the relativistic electrons. The MOHAM relies on OHAM to obtain analytic approximate solutions, it actually applies OHAM in each subinterval and we show that it achieves better results than OHAM over the large intervals; this is one of the advantages of this method which can be used for large intervals and to obtain good results. The convergence of the method is also addressed.

Optimal homotopy asymptotic method, Multistage Optimal homotopy asymptotic method, Convergence, Time-fractional Klein-Fock-Gordon (FKFG) equation

Fractional calculus (FC) is a part of mathematical analysis that studies the derivation of integrals and derivatives of rational orders [1]. The concept of fractional calculus (integrals and derivatives of any rational order) is established over 300 years ago, and nowadays is a very important subject. Gradually, researchers in different fields of sciences have discovered that fractional differential models have much better descriptors for different phenomena. Fractional calculus has widespread applications of in physics, chemistry, economics, dynamic systems, medical engineering, biological sciences, imaging, etc. On the other hand, physicists Klein, Fock, and Gordon have developed an equation called Klein-Fock-Gordon, which describes the relativity of electrons. This equation is a kind of the wave equation, also relates to the Schrödinger equation, and is a quantitative version of the relation of relative energy and motion. This equation is theoretically similar to the Dirac equation.

Consider the time-fractional Klein-Fock-Gordon equation as follows

Dαu(x,t) = uxx(x,t)+au(x,t)+bun(x,t), 1 < α≤2, (1)

Subject to the following initial conditions

u(x,0) = f(x) and ux(x,0) = g(x), (2)

Where a and b are real constants and n is a positive integer.

The initial guess to the solution is as follows

u0(x,t) = f(x)+g(x)t.

Various methods have been developed to obtain approximate solutions of the fractional time differential equations such as, fractional Jacobi collocation method [2], homotopy perturbation transform method [3], homotopy analysis transform method [4], Adomian decomposition method [5], variational iteration [6], and homotopy analysis method [7]. Recently, Optimal homotopy analysis method (OHAM) was proposed by Marinca, et al. [8-12], which was used to obtain analytic approximate solutions for some nonlinear problems [13-17].

OHAM results in to satisfactory solutions on short domains, but when the interval becomes longer, the accuracy of the method decreases, so a new approach was proposed by Anakira, et al. which is called multistage optimal homotopy asymptotic method (MOHAM) that suitable for analytic approximate solutions for large intervals [18].

Finally, the approximate solutions obtained from both methods are compared with the exact solution.

In this section, some basic definitions of fractional calculus are explained briefly [1,19].

Definition 4.1: A real-valued function f(t) with t>0 can be defined on the space Cμ, μ∈R, if there is a real number such ρ>μ f(t)= t p f 1 (t) , where f1(t)∈(0,+∞) and it is defined on the space Cnμ, if and only if fn(t)∈Cμ, for n∈N.

Definition 4.2: The Riemann-Liouville's integral of fractional order α>0 for a continuous function f∈Cμ with μ>−1 is defined as follows

Jαf(t) = 1Γ(α)∫0t(t−s)α−1f(s)ds, α>0, (3)

J0f(t) = f(t). (4)

By considering f∈Cμ, μ>−1, α,β>0 and γ>−1, the main properties of the operator Jα are listed as follows

(1) JαJβf(t) = Jα+βf(t), (5)

(2) JαJβf(t) = JβJαf(t), (6)

(3) Jαtγ = Γ(γ+1)Γ(α+γ+1)tα+γ. (7)

Definition 4.3: The fractional-order derivative of f(t), in Caputo sense, is defined as follows

Dαf(t) = 1Γ(m−α)∫0t(t−s)m−α−1fm(s)ds, (8)

for m−1 < α≤m, m∈N, t>0, f∈Cm−1.

Lemma 4.1: By assuming m−1 < α≤m, m∈N, f∈Cm−1, and μ≥−1, the following properties will be valid.

(1) DαJαf(t) = f(t), (9)

(2) (JαDα)f(t) = f(t)−∑k=0m−1f(k)(0+)tkk! (10)

A short introduction to the methods that will be used in this research.

OHAM

Let's consider the following fractional equation with the boundary conditions

L(u(ξ))+N(u(ξ))+f(ξ) = 0, B(u(ξ),du(ξ)dξ) = 0, (11)

Where ξ is an independent variable, L, f, N , and u(ξ) represent, respectively a linear operator, a nonlinear operator, a known function, and an unknown function. Also, B(u(ξ)) shows a boundary operator.

An optimal convex homotopy H(φ(ξ;p,ci)):Ω×[0,1]→R is constructed as follows,

H(φ(ξ;p,ci)) = (1−p)[L(φ(ξ;p,ci))+f(ξ)] = (12) H(ξ;p,ci)[L(φ(ξ;p,ci))+f(ξ)+N(φ(ξ;p,ci))], B(φ(ξ;p,ci)) = 0.

A deformation equation of zero-order as the following

(1−p)[L(φ(ξ,p))+f(ξ)] = H(p)[L(φ(ξ,p))+f(ξ)+N(φ(ξ,p))], B(φ(ξ,p),dφ(ξ,p)dξ) = 0, (13)

Where p is an embedding parameter in the interval [0,1], H(p,ξ), is an auxiliary function with non-zero and zero outputs for p≠0 and p = 0, respectively and u0(ξ) represents the initial condition of u(ξ), and φ(ξ,p) is an unknown function. By inserting p = 0 and 1 into Eq. (13), the following functions are obtained

φ(ξ,0) = u0(ξ), (14)

φ(ξ,1) = u(ξ). (15)

Therefore, φ(ξ,p) will change continuously from the initial guess to the solution, u0(ξ) , to u(ξ) when p increase from 0 up to 1.

By putting p = 0 into Eq. (13), the initial solution u0(ξ) is determined as a solution for the problem

L(u0(ξ))+f(ξ) = 0, B(u0,du0dξ) = 0. (16)

Next, choose an auxiliary function H(p) in the following form

H(p) = p c1+p2c2+p3c3+... , (17)

Where c1,c2,c3,... are called the convergence control parameters, that will be determined later. The auxiliary function H(p) can be expressed in other forms see the paper by Herisanu and Marinca [8].

By expanding φ(ξ;p,ci) in Taylor's series about p, the following expansion will be obtained

φ(ξ;p,ci) = u0(ξ)+∑m=1∞um(ξ,c1,c2,...,cm)pm (18)

By putting Eqs. (14-18) into (13), and equating the coefficients of the terms with identical powers of p, one will obtain the governing equation of the initial approximation u0(ξ), given by Eq.(16), and then the governing equation of the first order problem is defined as

L(u1(ξ))+f(ξ) = c1N0(u0(ξ)), B(u1,du1dξ) = 0. (19)

And the governing equation of the mth order is defined as

L(um(ξ))−L(um−1(ξ)) = cmN0(u0(ξ))+∑j=1m−1cj[L(um−j(ξ))+

Nm−j(u0(ξ),u1(ξ),...,um−1(ξ))], B(um,dumdξ) = 0, m = 2,3,... (20)

Where Nm(u0(ξ),u1(ξ),...,um(ξ)) is the coefficient of pm in the expansion of N(φ(ξ,p)) about the embedding parameter p

N(φ(ξ,p)) = N0(u0(ξ))+∑m=1∞Nm(u0(ξ),u1(ξ),...,um(ξ))pm. (21)

Solving Eq. (20) gives various approximates solutions um(ξ,c1,c2,...,cm), but there exist still m unknowns, auxiliary parameters, c1,c2,c3,..,cm in the obtained solutions. It is assumed that the auxiliary parameters (c1,c2,c3,..,cm), the linear operator L, and the deformation equation of the zero order (16) are appropriately determined in order to ensure the convergence of the series (18) at p = 1. Hence, putting Eqs. (14) and (15) into Eq. (18) for p = 1 gives the solution u(ξ) as

u(ξ,c1,c2,...) = u0(ξ)+∑m=1∞um(ξ,c1,c2,...,cm) (22)

The approximate solution of Eq. (11) can be calculated as

u(ξ,ci) = u0(ξ)+∑k=1mui(ξ,c1,c2,...,ck). (23)

Substitution of Eq. (23) in Eq. (11), results in the following residual

R(ξ,ci) = L(u(ξ,ci))+f(ξ), i = 1,2,3,... (24)

By supposing that R(ξ,ci) = 0, the exact solution will be u(ξ,ci). However, such a case could not be true for a nonlinear equation. By least squares technique the functional J(ci) should be minimized

J(ci) = ∫abR2(ξ,ci)dξ, (25)

Where a and b are two values relating to the problem in hand. The optimal values of the unknown coefficients ci (i = 1,2,...,m) can be determined based on the following conditions

∂J∂c1 = ∂J∂c2 = ∂J∂c3 = ... = ∂J∂cm (26)

In order to get an analytic approximate solution at the level m, the obtained optimal coefficients will be substituted in Equation (23).

Although the OHAM is used to obtain approximate solutions of nonlinear problems. It has some disadvantage in nonlinear problems with large domain. To control this drawback, we introduce in this section a multistage OHAM to obtain the nonlinear problem with long of the domain. A simple way to confirm the validity of the approximate solutions of large T is by dividing the interval [0, T] in to subinterval as [t0, t1),...,[tj−1, tj], where tj = T and utilizing the MOHAM solution on each subinterval. The initial approximation in each interval is given from the solution in prior interval. First, by considering the following initial condition

yi(tj) = αi. (27)

then we can obtain the initial approximation yi,0(t) = α and the following zero-order equation

(1−p)[L(φi(t,p))−yi,0(t)] = Hi(p)[L(φi(t,p))+fi(t)+Ni(φi(t,p))]. (28)

Next, we pick out the auxiliary function Hi(p) in the form

Hi(p) = c1,jp+c2,jp2+c3,jp3+..., (29)

Or

Hi(p) = (c1,j+c2,j+c3,j+...)p. (30)

Then, the first, second and mth order-approximate solutions can be generated subject to initial condition yi,1(tj) = yi,2(tj) = ... = yi,m(tj) = 0, and the approximate solution as follows,

yi(t,c1,j,c2,j,...,cm,j) = yi,0(t)+∑k=1myi,k(t,c1,j,c2,j,...,ck,j). (31)

Substituting Eq. (31) into Eq. (11) yields the following residual

Ri(t,c1,j,c2,j,...,cm,j) = L(yi(t,c1,j,c2,j,...,cm,j))+fi(t)+Ni(yi(t,c1,j,c2,j,...,cm,j)). (32)

If Ri=0, then yi will be the exact solution. Generally, such a case will not arise for nonlinear problems, but we can minimize the function

Ji(c1,j,c2,j,...,cm,j) = ∫tjtj+hR2i(t,c1,j,c2,j,...,cm,j)dt, (33)

Where h is the length of the subinterval [tj,tj+1] and N = [T/h] the number of subinterval. Now, we can solve Eq. (33) for j = 0,1,...,N with changing the initial approximation αi in each subinterval from the solution in the last point of the prior interval. The unknown convergence control parameters ci,j(i = 1,2,3,...,m, j = 1,2,3,...,N) can be obtained by solving the following system of equations

∂Ji∂c1,j = ∂Ji∂c2,j = ... = ∂Ji∂cm,j = 0. (34)

Thus, the analytic approximate solution will be as follows

y(t) = ⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪y1(t), t0 ≤ t < t1,y2(t), t1≤t < t2,⋮yN(t), tN−1≤t≤T. (35)

By this way, we successfully gain the solution of the initial value problem for a large interval T. It should be noted that if J = 0 the MOHAM expresses the OHAM. One of the benefits of MOHAM is that it provides a simple way to control convergence and regulate convergence region and adjust the convergence region though the auxiliary function Hi(p) involving several convergent control parameters ci,j,s. In general, this method eliminates the difficulty of finding approximate solutions in large ranges.

If the series (23) converges to u(ξ), where un(ξ)∈L(R+) is produced by Eqs. (16)-(19), and the satisfies the nth order deformation equation (20), then u(ξ) is the exact solution of (11) [18].

Proof:

Since the series ∑n=1∞un(ξ,c1,c2,...,cn) is convergent, it can be written as follows

S(ξ) = ∑n=1∞un(ξ,c1,c2,...,cn) (36)

so, we have

limn→∞un(ξ,c1,c2,...,cn) = 0. (37)

The left hand-side of (20) satisfies

u1(ξ,c1)+∑j=2nuj(ξ,c→j)−∑j=2nuj−1(ξ,c→j−1) = un(ξ,c→n). (38)

According to (38) and the limit, we have

u1(ξ,c1)+∑j=2∞uj(ξ,c→j)−∑j=2∞uj−1(ξ,c→j−1) = limn→∞ un(ξ,c→n) = 0. (39)

Applying the linear operator

L(u1(ξ,c1))+L(∑j=2∞uj(ξ,c→j))−L(∑j=2∞uj−1(ξ,c→j−1)) = 0. (40)

Eq.(40) can be written as following

L(u1(ξ,c1))+∑j=2∞(L(uj(ξ,c→j))−L(uj−1(ξ,c→j−1))) =

c1N0(u0)+f(ξ)+∑j=2∞(cjN0(u0)+∑k=1j−1ck[L(uj−k)+Nj−k(u→j−1)]) =

c1N0(u0)+f(ξ)+∑j=2∞cjN0(u0)+∑j=2∞∑k=1j−1(ck[L(uj−k)+Nj−k(u¯j−1)]) =

∑j=2∞∑k=1j−1ck[L(uj−k)]+∑j=2∞∑k=1j−1(cjN0(u0)+ck[Nj−k(u¯j−1)])+f(ξ) = 0. (41)

So by choosing the optimal ck, k = 1,2,3,..., Eq.(41) is converted to the following

L(u(ξ))+N(u(ξ))+f(ξ) = 0. (42)

Which is the exact solution of the problem.

We consider one of the nonlinear cases of FKFG equation, for a = −34,b = 32, and n = 2, therefore Eq. (1) becomes

Dαtu(x,t) = uxx(x,t)−34u(x,t)+32u3(x,t), 1 < α≤2. (43)

With the initial condition

u(x,0) = sech(x) and ut(x,0) = 12sech(x)tanh(x). (44)

and the exact solution

u(x,t) = −sech(x+0.5t).

Now, we solve Eq. (43) by OHAM and MOHAM.

Solution of FKFG equation by OHAM Having the linear operator L = Dαt and nonlinear operator

N = −uxx(x,t)+34u(x,t)−32u3(x,t) and using Eq. (16), we have

u0(x,t) = 12sech(x)tanh(x)t+sech(x), (45)

The following problems are resulted from Eqs. (19) - (20)

The First-order problem:

Dαt(u1(x,t)) = c1[−Dxx(u0(x,t))+34u0(x,t)−32u30(x,t)], (46)

The Second-order problem:

Dαt(u2(x,t))−Dαt(u1(x,t)) = c1[Dαtu1(x,t)−Dxxu1(x,t)+34u1(x,t)−

32(3u20(x,t)u1(x,t))]+c2[−Dxxu0(x,t)+34u0(x,t)−32u30(x,t)], (47)

The Third-order problem:

Dαtu3(x,t)−Dαtu2(x,t) = c1[Dαtu2(x,t)−Dxxu2(x,t)+34u2(x,t)−32(3u20(x,t)u2(x,t)+

3u21(x,t)u0(x,t))]+c2[Dαtu1(x,t)−Dxxu1(x,t)+34u1(x,t)−32(3u20(x,t)u1(x,t))]+

c3[−Dxxu0(x,t)+34u0(x,t)−32u30(x,t)], (48)

By solving the Eqs. (46) - (48), analytic approximate solutions will be obtained.

In this research, a four terms approximation for u(x,t) is considered. By substituting the solutions of the zero-order, first-order, second-order and third-order into Eq. (43) and using the least squares technique, the parameters c1,c2 and c3 are determined for different values of α, as in Table 1.

In this section, we utilize MOHAM to Eq. (43). We will consider the auxiliary function H(p,t) as the following

H(p,t) = (c1,j+c2,jt+c3,jt2)p. (49)

This approach leads to the following sequence of equations

p0: Dαt(u0j(x,t)) = αj, (50)

p1: Dαt(u1j(x,t)) = c1j[−Dxxu0j(x,t)+34u0j(x,t)−32u30j(x,t)], (51)

p2: Dαt(u2j(x,t)) = −Dαt(u1j(x,t)) = c1j[Dαtu1j(x,t)−Dxxu1j(x,t)+34u1j(x,t)−,

32(3u20j(x,t)u1j(x,t))]+c2j[−Dxxu0j(x,t)+34u0j(x,t)−32u30j(x,t)]. (52)

p3: Dαtu3j(x,t)−Dαtu2j(x,t) = c1j[Dαtu2j(x,t)−Dxxu2j(x,t)+34u2j(x,t)−

32(3u20j(x,t)u2j(x,t)+3u21j(x,t)u0j(x,t))]+c2j[Dαtu1j(x,t)−Dxxu1j(x,t)+34u1j(x,t)−

32(3u20j(x,t)u1j(x,t))]+c3j[−Dxxu0j(x,t)+34u0j(x,t)−32u30j(x,t)], (53)

As in OHAM, we obtain two approximate solutions to Equation (43)

uj(x,t) = u0j(x,t)+u1j(x,t). (54)

By applying the least squares technique, the parameters c1j,c2j and c3j are determined for different values of α where h = 0.2, and t0 = 0 up to t2 = T = 1 as in Table 2.

Approximate solution for α = 2 is as the following form

u(x,t) = ⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪0.5000000000sech(x)tanh(x)t−(sech(x)−0.007922060576)sech3(x)tanh3(x)t7+(0.003558142171sech3(x)tanh3(x)+0.06654530884sech3(x)tanh2(x))t6+..., 0≤x < 0.21cosh6(x)⎛⎝ (−0.008882058897sinh(x)cosh(x)2+0.008882058897sinh(x))t7+(0.07461cosh3(x)+0.0048sinh(x)cosh2(x)−0.0746cosh(x)−0.004833sinh(x))t6+...⎞⎠ 0.2≤x < 0.4 1cosh6(x)( (0.00546237854sinh(x)cosh2(x)−0.00546237854sinh(x))t7+(−0.045884cosh3(x)−0.12506588sinh(x)cosh2(x)+0.045884cosh(x)+0.0125sinh(x))t6+...) 0.4≤x < 0.6 1cosh6(x)( (0.002476348954sinh(x)cosh2(x)−0.002476348954sinh(x))t7+(−0.02080cosh3(x)−0.006727sinh(x)cosh2(x)+0.0208cosh(x)+0.006727sinh(x))t6+...) 0.6≤x < 0.8 1cosh6(x)( (0.001751944671sinh(x)cosh2(x)−0.001751944671sinh(x))t7+(−0.014716cosh3(x)−0.006178sinh(x)cosh2(x)+0.0147cosh(x)+0.006178sinh(x))t6+...) 0.8≤x ≤ 1

From Figure 1, Figure 2, Figure 3 and Figure 4 one can see that the solutions obtained by OHAM and MOHAM are nearly identical with the exact solution Table 3.

In this study OHAM and MOHAM are used to derive an analytic approximate solution for the time-fractional Klein-Fock-Gordon (FKFG) equation. The results obtained from these methods show that MOHAM converges better than OHAM. One observes that the results agree very well with the exact solution. The MOHAM by dividing the interval [o, T] can obtain better solution than OHAM. As far as the authors are aware, MOHAM has not been used to solve fractional partial differential equations, so far, the method has been tested on fractional-PDE and yields to satisfactory results. The Figures and Tables expose that good results are obtained by MOHAM and more accurate solution as compared to OHAM. The convergence of the method is proved.