# On the Solutions Fractional Riccati Differential Equation with Modified Riemann-Liouville Derivative

Authors: Mehmet Merdan
Journal: International Journal of Differential Equations
 (2012)

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## Abstract

Fractional variational iteration method (FVIM) is performed to give an approximate analytical solution of nonlinear fractional Riccati differential equation. Fractional derivatives are described in the Riemann-Liouville derivative. A new application of fractional variational iteration method (FVIM) was extended to derive analytical solutions in the form of a series for these equations. The behavior of the solutions and the effects of different values of fractional orderα are indicated graphically. The results obtained by the FVIM reveal that the method is very reliable, convenient, and effective method for nonlinear differential equations with modified Riemann-Liouville derivative


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## Body

## 1. Introduction

In recent years, fractional calculus used in many areas such as electrical networks, control theory of dynamical systems, probability and statistics, electrochemistry of corrosion, chemical physics, optics, engineering, acoustics, viscoelasticity, material science and signal processing can be successfully modelled by linear or nonlinear fractional order differential equations [1–10]. As it is well known, Riccati differential equations concerned with applications in pattern formation in dynamic games, linear systems with Markovian jumps, river flows, econometric models, stochastic control, theory, diffusion problems, and invariant embedding [11–17]. Many studies have been conducted on solutions of the Riccati differential equations. Some of them, the approximate solution of ordinary Riccati differential equation obtained from homotopy perturbation method (HPM) [18–20], homotopy analysis method (HAM) [21], and variational iteration method proposed by He [22]. The He’s homotopy perturbation method proposed by He [23–25] the variational iteration method [26] and Adomian decomposition method (ADM) [27] to solve quadratic Riccati differential equation of fractional order.The variational iteration method (VIM), which proposed by He [28, 29], was successfully applied to autonomous ordinary and partial differential equations and other fields. He [30] was the first to apply the variational iteration method to fractional differential equations. In recent years, a new modified Riemann-Liouville left derivative is suggested by Jumarie [31–35]. Recently, the fractional Riccati differential equation is solved with help of new homotopy perturbation method (HPM) [23].In this paper, we extend the application of the VIM in order to derive analytical approximate solutions to nonlinear fractional Riccati differential equation:(1.1)D*αy(x)=A(x)+B(x)y(x)+C(x)y2(x),x∈R,0<α≤1,t>0,
subject to the initial conditions(1.2)y(k)(0)=dk,k=0,1,2,…,n-1,
where α is fractional derivative order, n is an integer, A(x), B(x), and C(x) are known real functions, and dk is a constant.The goal of this paper is to extend the application of the variational iteration method to solve fractional nonlinear Riccati differential equations with modified Riemann-Liouville derivative.The paper is organized as follows: In Section2, we give definitions related to the fractional calculus theory briefly. In Section 3, we define the solution procedure of the fractional variational iteration method to show inefficiency of this method, we present the application of the FVIM for the fractional nonlinear Riccati differential equations with modified Riemann-Liouville derivative and numerical results in Section 4. The conclusions are then given in the final Section 5.


## 2. Basic Definitions

Here, some basic definitions and properties of the fractional calculus theory which can be found in [31–35].Definition 2.1.
Assumef:R→R,x→f(x) denote a continuous (but not necessarily differentiable) function, and let the partition h>0 in the interval [0,1]. Jumarie’s derivative is defined through the fractional difference [34]:
(2.1)Δ(α)=(FW-1)αf(x)=∑k=0∞(-1)k(αk)f[x+(α-k)h],
where FWf(x)=f(x+h). Fractional derivative is defined as the following limit form [1, 7]:
(2.2)f(α)=lim⁡h→0Δα[f(x)-f(0)]hα.
This definition is close to the standard definition of derivatives (calculus for beginners), and as a direct result, the αth derivative of a constant, 0<α<1, is zero.Definition 2.2.
The left-sided Riemann-Liouville fractional integral operator of orderα≥0, of a function f∈Cμ,μ≥-1 is defined as
(2.3)Jaαf(x)=1Γ(α)∫ax(x-τ)α-1f(τ)dτ,forα>0,x>0,Ja0f(x)=f(x).
The properties of the operator Jα can be found in [1, 7, 36].Definition 2.3.
The modified Riemann-Liouville derivative [33, 34] is defined as
(2.4)Dαx0f(x)=1Γ(n-α)dndxn∫x0(x-τ)n-α(f(τ)-f(0))dτ,
where x∈[0,1],n-1≤α<n,andn≥1.
In addition, we want to give as in the following some properties of the fractional modified Riemann-Liouville derivative.
Fractional Leibniz product law:(2.5)Dαx0(uv)=u(α)v+uv(α).
Fractional Leibniz Formulation:(2.6)Iαx0Dαx(uv)=f(x)-f(0),0<α≤1.
Fractional the integration of part:(2.7)Iαba(u(α)v)=(uv)|ab-Iαba(uv(α)).Definition 2.4.
Fractional derivative of compounded functions [33, 34] is defined as
(2.8)dαf≅Γ(1+α)df,0<α<1.Definition 2.5.
The integral with respect to(dx)α [33, 34] is defined as the solution of the fractional differential equation:
(2.9)dy≅f(x)(dx)α,x≥0,y(0)=0,0<α<1.Lemma 2.6.
Letf(x) denote a continuous function [33, 34] then the solution of (2.5) is defined as
(2.10)y=∫0xf(τ)(dτ)α=α∫0x(x-τ)αf(τ)dτ,0<α<1.
For example, f(x)=xβ in (2.10) one obtains
(2.11)∫0xτβ(dτ)α=Γ(β+1)Γ(α+1)Γ(α+β+1)xβ+α,0<α<1.Definition 2.7.
Assume that the continuous functionf:R→R,x→f(x) has a fractional derivative of order kα, for any positive integer k and any α, 0<α≤1; then the following equality holds, which is
(2.12)f(x+h)=∑k=0∞hαkαk!fαk(x),0<α≤1.
On making the substitution h→x and x→0, we obtain the fractional Mc-Laurin series:
(2.13)f(x)=∑k=0∞xαkαk!fαk(0),0<α≤1.


## 3. Fractional Variational Iteration Method

To describe the solution procedure of the fractional variational iteration method [31–35], we consider the following fractional Riccati differential equation:(3.1)D*αy(x)=A(x)+B(x)y(x)+C(x)y2(x),x∈R,0<α≤1,t>0.
According to the VIM, we can build a correct functional for (3.1) as follows:(3.2)yn+1(x)=yn(x)+Iα[λ(τ)(dαyn(τ)dτα-(A(τ)+B(τ)y(τ)+C(τ)y2(τ)))]yn+1(x)=yn(x)+1Γ(α)∫0x(x-τ)α-1λ(τ)(dαyn(τ)dτα-(A(τ)+B(τ)y(τ)+C(τ)y2(τ)))dτ.
Using (2.3), we obtain a new correction functional:(3.3)yn+1(x)=yn(x)+1Γ(α+1)∫0xλ(τ)(dαyn(τ)dτα-(A(τ)+B(τ)y(τ)+C(τ)y2(τ)))(dτ)α.
It is obvious that the sequential approximations yk, k≥0 can be established by determining λ, a general Lagrange’s multiplier, which can be identified optimally with the variational theory. The function ỹn is a restricted variation which means δỹn=0. Therefore, we first designate the Lagrange multiplier λ that will be identified optimally via integration by parts. The successive approximations yn+1(x),n≥0 of the solution y(x) will be readily obtained upon using the obtained Lagrange multiplier and by using any selective function y0. The initial values are usually used for choosing the zeroth approximation y0. With λ determined, then several approximations yk, k≥0 follow immediately [37]. Consequently, the exact solution may be procured by using(3.4)y(x)=lim⁡n→∞yn(x).


## 4. Applications

In this section, we present the solution of two examples of the Riccati differential equations as the applicability of FVIM.Example 4.1.
Let usconsider the fractional Riccati differential equation, we get(4.1)dαydxα=-y2(x)+1,0<α≤1,
with initial conditions:
(4.2)y(0)=0.
Construction the following functional:
(4.3)yn+1(x)=yn(x)+1Γ(α+1)∫0xλ(τ){dαyndτα+yn2(τ)-1}(dτ)α,
we have
(4.4)δyn+1(x)=δyn(x)+1Γ(α+1)δ∫0xλ(τ){dαyndτα+yn2(τ)-1}(dτ)α=δyn+λδyn|τ=x-1Γ(α+1)∫0xdαλ(τ)dταδyn(τ)(dτ)α.
Similarly, we can get the coefficients of δyn to zero:
(4.5)1+λ(τ)|τ=x=0,dαλ(τ)dτα=0.
The generalized Lagrange multiplier can be identified by the above equations:
(4.6)λ(x)=-1.
substituting (4.6) into (4.3) produces the iteration formulation as follows:
(4.7)yn+1(x)=yn(x)-1Γ(α+1)∫0x{dαydτα+y2(τ)-1}(dτ)α.
Taking the initial value y0(x)=0, we can derive
(4.8)y1(x)=y0(x)-1Γ(α+1)∫0x{dαy0dτα+y02(τ)-1}(dτ)α=xαΓ(α+1),y2(x)=y1(x)-1Γ(α+1)∫0x{dαy1dτα+y12(τ)-1}(dτ)α=xαΓ(α+1)-1Γ(α+1)∫0x{1+x2αΓ2(α+1)-1}(dτ)α=xαΓ(α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1),y3(x)=xαΓ(α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+2Γ(2α+1)Γ(4α+1)x5αΓ3(α+1)Γ(3α+1)Γ(5α+1)-Γ2(2α+1)Γ(6α+1)x7αΓ4(α+1)Γ2(3α+1)Γ(7α+1),y4(x)=xαΓ(α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+2Γ(2α+1)Γ(4α+1)x5αΓ3(α+1)Γ(3α+1)Γ(5α+1)-Γ2(2α+1)Γ(6α+1)x7αΓ4(α+1)Γ2(3α+1)Γ(7α+1)-4Γ(2α+1)Γ(4α+1)Γ(6α+1)x7αΓ4(α+1)Γ(3α+1)Γ(5α+1)Γ(7α+1)+2Γ(2α+1)Γ(6α+1)Γ(8α+1)x9αΓ5(α+1)Γ2(3α+1)Γ(7α+1)Γ(9α+1)+4Γ2(2α+1)Γ(4α+1)Γ(8α+1)x9αΓ5(α+1)Γ2(3α+1)Γ(5α+1)Γ(9α+1)-2Γ3(2α+1)Γ(6α+1)Γ(10α+1)x11αΓ6(α+1)Γ3(3α+1)Γ(7α+1)Γ(11α+1)-4Γ2(2α+1)Γ2(4α+1)Γ(10α+1)x11αΓ6(α+1)Γ2(3α+1)Γ2(5α+1)Γ(11α+1)+4Γ3(2α+1)Γ(4α+1)Γ(6α+1)Γ(12α+1)x13αΓ7(α+1)Γ3(3α+1)Γ(5α+1)Γ(7α+1)Γ(13α+1)-Γ4(2α+1)Γ2(6α+1)Γ(14α+1)x15αΓ8(α+1)Γ4(3α+1)Γ2(7α+1)Γ(15α+1)⋮
Then, the approximate solutions in a series form are
(4.9)y(x)=Limn→∞yn(x)=xαΓ(α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+2Γ(2α+1)Γ(4α+1)x5αΓ3(α+1)Γ(3α+1)Γ(5α+1)-Γ2(2α+1)Γ(6α+1)x7αΓ4(α+1)Γ2(3α+1)Γ(7α+1)-4Γ(2α+1)Γ(4α+1)Γ(6α+1)x7αΓ4(α+1)Γ(3α+1)Γ(5α+1)Γ(7α+1)+⋯.
As α=1 is
(4.10)y(x)=Limn→∞yn(x)=x-x33+2x515-17x763+⋯.
The exact solution of (4.1) is y(x)=(e2x-1)/(e2x+1), when α=1.
Figure1 indicates the solution obtained using FVIM versus the exact solution when α=1. Figure 2 is plotted for approximate solution of time-fractional Riccati differential equation for α=0.7, 0.8, 0.9, and 1. Equation (4.1) is solved by using the homotopy perturbation method (HPM) [24]. FVIM solutions indicate that the present algorithm performs extreme efficiency, simplicity, and reliability. The results obtained from FVIM are fully compatible with those of the HPM.
Table1 shows the approximate solutions for (4.1) obtained for different values of α using the variational iteration method and HPM [24]. From the numerical results in Table 1, it is clear that the approximate solutions are in high agreement with the exact solutions, when α=1, and the solution continuously depends on the time-fractional derivative. Example 4.1 has been solved using HAM [21], ADM [27], VIM [26], and HPM [23–25].Table 1
Approximate solutions for (4.1).
tRef. [24]α=0.5y3(x)α=0.5[24]α=0.75y3(x)α=0.75[24]α=1y3(x)α=1y(x)exact000000000.10.2738750.0865130.1847950.1901020.0996680.0996670.0996670.20.4541250.1615840.3137950.3100330.1973750.1973750.1973750.30.5739320.2382560.4145620.4050620.2913120.2913200.2913120.40.6444220.3215230.4928890.4834790.3799440.3800050.3799480.50.6741370.4136820.4621170.5504700.4620780.4623750.4621170.60.6719870.5154450.5973930.6103440.5368570.5379230.5370490.70.6480030.6264030.6317720.6669610.6036310.6067680.6043670.80.6133060.7452780.6604120.7237600.6617060.6696950.6640360.90.5796410.8700740.6879600.7836380.7099190.7281390.7162971.00.5585570.9981760.7182600.8487830.7460320.7841260.761594Figure 1
The graph indicates the solutiony(x) for (4.1), when α=1.Figure 2
Plots of approx. solutiony3(x) for different values of α.Example 4.2.
Let usconsider the fractional Riccati differential equation, we get(4.11)dαydxα=2y(x)-y2(x)+1,0<α≤1,
with initial conditions
(4.12)y(0)=0.
Construction the following functional:
(4.13)yn+1(x)=yn(x)+1Γ(α+1)∫0xλ(τ){dαyndτα-2yn(τ)+yn2(τ)-1}(dτ)α.
we have
(4.14)δyn+1(x)=δyn(x)+1Γ(α+1)δ∫0xλ(τ){dαyndτα-2yn(τ)+yn2(τ)-1}(dτ)α=δyn+λδyn|τ=x-1Γ(α+1)∫0xdαλ(τ)dταδyn(τ)(dτ)α.
Similarly, we can get the coefficients of δyn to zero:
(4.15)1+λ(τ)|τ=x=0,dαλ(τ)dτα=0.
The generalized Lagrange multiplier can be identified by the above equations:
(4.16)λ(x)=-1,
substituting (4.16) into (4.13) produces the iteration formulation as follows:
(4.17)yn+1(x)=yn(x)-1Γ(α+1)∫0x{dαyndτα-2yn(τ)+yn2(τ)-1}(dτ)α.
Taking the initial value y0(x)=0, we can derive
(4.18)y1(x)=y0(x)-1Γ(α+1)∫0x{dαy0dτα-2y0(τ)+y02(τ)-1}(dτ)α=xαΓ(α+1),y2(x)=y1(x)-1Γ(α+1)∫0x{dαy1dτα+y12(τ)-1}(dτ)α=xαΓ(α+1)-1Γ(α+1)∫0x{1-2ταΓ(α+1)+τ2αΓ2(α+1)-1}(dτ)α=xαΓ(α+1)+2x2αΓ(2α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1),y3(x)=xαΓ(α+1)+2x2αΓ(2α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+4x3αΓ(3α+1)-4Γ(3α+1)x4αΓ(α+1)Γ(2α+1)Γ(4α+1)-2Γ(2α+1)x4αΓ2(α+1)Γ(4α+1)-4Γ(4α+1)x5αΓ2(2α+1)Γ(5α+1)+2Γ(2α+1)Γ(4α+1)x5αΓ3(α+1)Γ(3α+1)Γ(5α+1)-Γ2(2α+1)Γ(6α+1)x7αΓ4(α+1)Γ2(3α+1)Γ(7α+1)⋮
Then, the approximate solutions in a series form are
(4.19)y(x)=Limn→∞yn(x)=xαΓ(α+1)+2x2αΓ(2α+1)-Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+4x3αΓ(3α+1)-4Γ(3α+1)x4αΓ(α+1)Γ(2α+1)Γ(4α+1)-2Γ(2α+1)x4αΓ2(α+1)Γ(4α+1)-4Γ(4α+1)x5αΓ2(2α+1)Γ(5α+1)+2Γ(2α+1)Γ(4α+1)x5αΓ3(α+1)Γ(3α+1)Γ(5α+1)-Γ2(2α+1)Γ(6α+1)x7αΓ4(α+1)Γ2(3α+1)Γ(7α+1)+⋯.
As α=1 is
(4.20)y(x)=Limn→∞yn(x)=x-x33+2x515-x763+⋯.
The exact solution of (4.11) is y(x)=1+2tanh(2t+(1/2)log((2-1)/(2+1))), when α=1.
Figure3 is plotted for approximate solution of time-fractional Riccati differential equation found in Example 4.2. In Figure 4, we have shown the graphic of approximate solution of (4.11) for α=0.7,0.8,0.9,and1. Figures 2 and 4 show that a decrease in the fractional order α causes an increase in the function.
Table2 indicates the approximate solutions for (4.11) obtained for different values of α using the variational iteration method and HPM [24]. From the numerical results in Table 2, it is clear that the approximate solutions are in high agreement with the exact solutions, when α=1, and the solution continuously depends on the time-fractional derivative.Table 2
Approximate solutions for (4.11).
tRef. [24]α=0.5y3(x)α=0.5[24]α=0.75y3(x)α=0.75[24]α=1y3(x)α=1y(x)exact000000000.10.3217300.5774310.2168660.2444600.1102940.1102660.1102950.20.6296660.9126540.4288920.4697090.2419650.2415850.2419760.30.9409411.1662530.6546140.6987180.3951060.3935150.3951040.41.2507371.3535490.8914040.9243190.5681150.5640130.5678120.51.5494391.4826331.1327631.1379520.7575640.7495280.7560140.61.8254561.5596561.3702401.3314620.9582590.9451550.9535660.72.0665231.5899841.5942781.4976001.1634591.1448261.1529460.82.2606331.5785591.7948791.6302341.3652401.3415521.3463630.92.3968391.5300281.9622391.7244391.5549601.5276901.5269111.02.4660041.4488052.0873841.7765421.7238101.6952381.689498Figure 3
The graph indicates the solutiony(x) for (4.11), when α=1.Figure 4
Plots of approx. solutiony3(x) for different values of α.Example 4.3.
Let usconsider the fractional Riccati differential equation, we get(4.21)dαydxα=x2+y2(x),0<α≤1,
with initial conditions:
(4.22)y(0)=1.
Construction the following functional:
(4.23)yn+1(x)=yn(x)+1Γ(α+1)∫0xλ(τ){dαyndτα-τ2-yn2(τ)}(dτ)α.
we have
(4.24)δyn+1(x)=δyn(x)+1Γ(α+1)δ∫0xλ(τ){dαyndτα-τ2-yn2(τ)}(dτ)α=δyn+λδyn|τ=x-1Γ(α+1)∫0xdαλ(τ)dταδyn(τ)(dτ)α.
Similarly, we can get the coefficients of δyn to zero:
(4.25)1+λ(τ)|τ=x=0,dαλ(τ)dτα=0.
The generalized Lagrange multiplier can be identified by the above equations:
(4.26)λ(x)=-1.
substituting (4.26) into (4.23) produces the iteration formulation as follows:
(4.27)yn+1(x)=yn(x)-1Γ(α+1)∫0x{dαyndτα-τ2-yn2(τ)}(dτ)α.
Taking the initial value y0(x)=1, we can derive
(4.28)y1(x)=y0(x)-1Γ(α+1)∫0x{dαy0dτα-τ2-y02(τ)}(dτ)α=1+xαΓ(α+1)+2x2+αΓ(α+3),y2(x)=y1(x)-1Γ(α+1)∫0x{dαy1dτα-y12(τ)-τ2}(dτ)α=1+xαΓ(α+1)+2x2+αΓ(α+3)+2x2αΓ(2α+1)+4x2+2αΓ(2α+3)+Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+4Γ(2α+3)x2+3αΓ(α+1)Γ(α+3)Γ(3α+3)+4Γ(2α+5)x4+3αΓ2(α+3)Γ(3α+5),y3(x)=1+xαΓ(α+1)+2x2+αΓ(α+3)+2x2αΓ(2α+1)+4x2+2αΓ(2α+3)+Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+4Γ(2α+3)x2+3αΓ(α+1)Γ(α+3)Γ(3α+3)+4Γ(2α+5)x4+3αΓ2(α+3)Γ(3α+5)+8x2+3αΓ(3α+3)+16Γ(5α+3)Γ(2α+3)x2+6αΓ(2α+1)Γ(α+1)Γ(α+3)Γ(3α+3)Γ(6α+3)+16Γ(3α+5)x4+4αΓ(α+3)Γ(2α+3)Γ(4α+5)+4Γ(4α+1)x5αΓ2(2α+1)Γ(5α+1)+2Γ(2α+1)x4αΓ2(α+1)Γ(4α+1)+2Γ(2α+1)Γ(4α+1)x5αΓ3(α+1)Γ(3α+1)Γ(5α+1)+8Γ(3α+3)x2+4αΓ(α+1)Γ(α+3)Γ(3α+4)+8Γ(3α+3)x2+4αΓ(α+3)Γ(2α+1)Γ(4α+3)+4x3αΓ(3α+1)+Γ2(2α+1)Γ(4α+1)x5αΓ4(α+1)Γ2(3α+1)Γ(7α+1)+4Γ(5α+1)x6αΓ(3α+1)Γ2(α+1)Γ(6α+1)+8Γ(2α+5)x4α+4Γ2(α+3)Γ(4α+5)+4Γ(3α+1)x4αΓ(α+1)Γ(2α+1)Γ(4α+1)+16Γ(4α+5)x4+5αΓ2(2α+3)Γ(5α+5)+8Γ(2α+3)x2+4αΓ(α+1)Γ(α+3)Γ(4α+3)+8Γ(2α+3)Γ(4α+3)x5α+2Γ2(α+1)Γ(α+3)Γ(3α+3)Γ(5α+3)+8Γ(2α+5)Γ(4α+5)x6α+2Γ(α+1)Γ2(α+3)Γ(3α+5)Γ(5α+5)+4Γ(2α+1)Γ(4α+3)x5α+2Γ(3α+1)Γ(α+3)Γ(3α+1)Γ(5α+3)+16Γ(2α+3)Γ(4α+5)x5α+4Γ(α+1)Γ2(α+3)Γ(3α+3)Γ(5α+5)+8Γ(2α+1)Γ(5α+3)x6α+2Γ(3α+1)Γ2(α+1)Γ(2α+3)Γ(6α+3)+16Γ(2α+5)Γ(5α+5)x6α+4Γ(2α+1)Γ2(α+3)Γ(3α+5)Γ(6α+5)+32Γ(5α+5)x6α+4Γ(α+1)Γ(α+3)Γ(3α+3)Γ(6α+5)+32Γ(2α+5)Γ(5α+7)x6α+6Γ(2α+3)Γ2(α+3)Γ(3α+5)Γ(6α+7)+16Γ2(2α+3)Γ(6α+5)x7α+4Γ2(α+1)Γ2(α+3)Γ2(3α+3)Γ(7α+5)+16Γ(4α+3)x2+5αΓ(2α+1)Γ(2α+3)Γ(5α+3)+16Γ(2α+5)Γ(4α+7)x5α+6Γ3(α+3)Γ(3α+5)Γ(5α+7)+16Γ2(2α+5)Γ(6α+9)x8+7αΓ4(α+3)Γ(7α+9)Γ2(3α+5)+8Γ(2α+1)Γ(2α+3)Γ(6α+3)x7α+2Γ3(α+1)Γ(3α+1)Γ(3α+3)Γ(α+3)Γ(7α+3)+8Γ(2α+1)Γ(2α+5)Γ(6α+5)x7α+4Γ2(α+1)Γ(3α+1)Γ(3α+5)Γ2(α+3)Γ(7α+5)+32Γ(2α+3)Γ(2α+5)Γ(6α+7)x7α+6Γ(α+1)Γ(3α+3)Γ(3α+5)Γ3(α+3)Γ(7α+7)⋮
Then, the approximate solutions in a series form are
(4.29)y(x)=Limn→∞yn(x)=1+xαΓ(α+1)+2x2+αΓ(α+3)+2x2αΓ(α+1)Γ(2α+1)+4x2+2αΓ(2α+3)+Γ(2α+1)x3αΓ2(α+1)Γ(3α+1)+4Γ(2α+3)x2+3αΓ(α+1)Γ(α+3)Γ(3α+3)+4Γ(2α+5)x4+3αΓ2(α+3)Γ(3α+5)+⋯.
As α=1 is
(4.30)y(x)=Limn→∞yn(x)=1+x+x2+5x36+5x46+8x515+29x690+47x7315+41t8360+299t911340+4t10525+184t1151975+t122268+4t1312285+t1559535+⋯.
The exact solution of (4.21) is
(4.31)y(x)=t(J-3/4(t2/2)Γ(1/4)+2J3/4(t2/2)Γ(3/4))J1/4(t2/2)Γ(1/4)-2J-1/4(t2/2)Γ(3/4),
where Jv(t) is the Bessel function of first kind, when α=1.
Figure5 is plotted for approximate solution of time-fractional Riccati differential equation found in Example 4.3. In Figure 6, we have shown the graphic of approximate solution of (4.21) for α=0.5,0.65,0.75,and1. Figures 2, 4, and 6 show that a decrease in the fractional order α causes an increase in the function.
Table3 indicates the approximate solutions for (4.21) obtained for different values of α using the HPM [23]. From the numerical results in Table 3, it is clear that the approximate solutions are in substantial agreement with the exact solutions, when α=1, and the solution continuously depends on the time-fractional derivative.Table 3
Approximate solutions for (4.21).
tRef. [23]α=0.50y3(x)α=0.5[23]α=0.75y3(x)α=0.75[23]α=1y3(x)α=1y(x)exact011111110.11.4128711.6710551.2191541.2498631.1114631.1109221.1114500.21.6932872.3421311.4265281.5134631.2529971.2481931.2525330.32.0359653.2415011.6847351.8522181.4392931.4208181.4354500.42.5007054.4685082.0314142.3066991.6927491.6417421.6752000.53.1616716.1443892.5173162.9289962.0473551.9294321.9895830.64.1217328.4296663.2150233.7896432.5536422.3101012.3992000.75.52324611.540664.2271994.9853543.2847902.8208612.9274500.87.55899915.770475.6957006.6493244.3439403.5141503.6005330.910.4839721.517007.8118048.9653765.8727874.4639924.4474501.014.6281829.3209810.8277012.187538.0615075.7748175.500000Figure 5
The graph indicates the solutiony(x) for (4.21), when α=1.Figure 6
Plots of approx. solutiony3(x) for different values of α.


## 5. Conclusions

In this paper, variational iteration method having integral w.r.t.(dτ)α has been successfully implemented to finding approximate analytical solution of fractional Riccati differential equations. Variational iteration method known as very powerful and an effective method for solving fractional Riccati differential equation. It is also a promising method to solve other nonlinear equations. In this paper, we have discussed modified variational iteration method having integral w.r.t. (dτ)α used for the first time by Jumarie. The obtained results indicate that this method is powerful and meaningful for solving the nonlinear fractional differential equations. Three examples indicate that the results of variational iteration method having integral w.r.t. (dτ)α are in excellent agreement with those obtained by HPM, ADM, and HAM, which is available in the literature.



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*Source: 346089-2012-05-15.xml*