METHODS OF SOLVING FRACTIONAL ...

Bulletin of the Marathwada Mathematical Society Vol. 15, No. 2, December 2014, Pages 40–53

METHODS OF SOLVING FRACTIONAL DIFFERENTIAL EQUATIONS OF ORDER α (0 < α < 1) J.A.Nanware and Gunwant A.Birajdar Department of Mathematics, Dr. Babasaheb Ambedkar Marathwada University, Aurangabad - 431 004, INDIA. jag− skmg91@rediffmail.com/[email protected]

Abstract Our aim is to obtain solution of fractional differential equation of order α (0 < α < 1) with initial conditions by power series method and Laplace transform method. The methods are well illustrated by many examples. Graphical representation of solutions is given. 1

1

INTRODUCTION

The origin of fractional calculus goes back to 1695 when Leibniz considered the derivatives of order 12 . Since then, many famous mathematicians which includes Laplace, Fourier, Abel, Liouville, Riemann, Grunwald, Letnikov, Levy, Marchaud, Erdelyi, and Riesz have worked on this and related questions. However, for three centuries, the theory of fractional calculus was developed mainly as a purely theoretical field of mathematics. Many applications have been found for fractional calculus, some of which are discussed in Debnath [6], Kilbas [9] and Podlubny [23]. Fractional calculus plays a vital role in the analysis of scientific problems in a broad array of fields such as physics, engineering, biology and economics. There is no doubt that fractional calculus has become an existing new mathematical method of solution of diverse problems in Mathematics, Science and Engineering. Many researchers attracted towards fractional differential equations and obtained successfully existence and uniqueness results by using monotone method coupled with lower and upper solutions [7, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21]. Miller and Ross [14] and Oldham and Spanier [22] provided historical details on the fractional calculus. 1

Keywords: Euler Gamma function, Mittag-Leffler function, Riemann-Liouville fractional derivative, Fractional differential equations, power series method, Laplace transform method. @ Marathwada Mathematical Society, Aurangabad, INDIA, ISSN 0976-6049

40

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 41 Dhaigude and Birajdar [1, 2, 3, 4, 5] have developed new methods for obtaining solutions of linear and nonlinear fractional partial differential equations. We organize the paper as follows: In section 2, definitions of Euler Gamma function, Mittag-Leffler function and Riemann-Liouville fractional derivative and some properties are considered. Also definition of the Laplace transform of Riemann-Liouville derivative is given and listed Laplace transform of some functions. In section 3, solution of initial value problems (IVP) by power series method is obtained. In last section solution of fractional initial value problem (IVP) is obtained in terms of Mittag-Leffler function by Laplace transform method. Graphical representation of solutions is given in the Appendix.

2

PREREQUISITES

In this section we state some basic definitions, properties and formulae which are required in the later sections.

2.1

Euler Gamma Function [14]

The Euler Gamma function is denoted by Γ(z) and is defined as ∫

∞

Γ(z) =

tz−1 e−t dt,

(ℜ(z) > 0),

0

where tz−1 = e(z−1)log(t) and z ∈ C, set of complex numbers. For n ∈ R, Γ(n + 1) = n!. 2.1.1

Some Properties of Euler Gamma Function

• Γ(z + 1) = zΓ(z), • Γ(n + 1) = n!, • Γ(z)Γ(1 − z) = • Γ(n + 12 ) = • Γ( n2 ) =

z∈C

n∈N π sinπz ,

√

π(2n!) , 22n n!

non-integerz ∈ C n ∈ N.

√ (n−2)!! π (n−1)/2 2

where n!! is the double factorial:   n.(n − 2)...5.3.1 n!! = n.(n − 2)...6.4.2   1

n ≥ 0 odd n ≥ 0 even n = 0, −1

42

2.2

Nanware J.A., Birajdar G.A.

Mittag-Leffler Function [8]

We consider the definitons of one parameter Mittag-Leffler function, two parameter Mittag-Leffler function and state the convolution formula for two parameter MittagLeffler functions. (a) One Parameter Mittag-Leffler Function: One parameter Mittag-Leffler function is denoted by Eα (z) and is defined as Eα (z) =

∞ ∑ k=0

zk Γ(k + 1)

(2.1)

Note that when α = 1 and 2, one parameter Mittag-Leffler function is the exponential function ez and hyperbolic cosine function coshz respectively. (b) Two-Parameter Mittag-Leffler Function: Two parameter Mittag-Leffler function is denoted by Eα,β (z) and is defined as Eα,β (z) =

∞ ∑ k=0

zk , Γ(αk + β)

(α > 0, β > 0)

(2.2)

Note that exponential function ez and hyperbolic sine and cosine functions are particular cases of two parameter Mittag-Leffler function if α = 1, β = 1; α = 2, β = 1 respectively. (c) Convolution of Two Mittag-Leffler Fucntions: ∫ t τ β−1 Eα,β (aτ α )(t−τ )γ−1 Eα,γ (−a(t−τ )α )dτ = tβ+γ−1 E2α,β+γ (a2 τ 2α ), (β > 0, γ > 0) 0

2.3

Laplace transform of the function y(t)

∫∞ If y(t) is of exponential order α then 0 y(t)e−st dt exists for all Res > α. The Laplace transform [22] of y(t) is denoted by L{y(t)} and is defined as ∫ ∞ L{y(t)} = y¯(s) = y(t)e−st dt 0

We say that y(t) = L−1 {¯ y (s)} is the inverse Laplace transform of y¯(s). Laplace transform of tk eat is given by ∫ ∞ k! k at L{t e } = tk eat e−st dt = (Re(s) >| a |) (s − a)k+1 0 k (ztα ) is [23] More useful pair of Laplace transform of the function tαk+β−1 Eα,β

{ } 1 (k!)sα−β αk+β−1 k α L t Eα,β (at ) = α , (Re(s) >| a | a ) k+1 (s − a) } { α−β (k!)s k = tαk+β−1 Eα,β (atα ) L−1 (sα − a)k+1

(2.3) (2.4)

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 43

2.4

Riemann-Liouville Fractional Derivative and Riemann-Liouville Fractional Integral [9]

Let α ∈ R+ and n = [α], where [.] is the greatest integer function. The Riemann-Liouville Fractional Derivative of order α is denoted by Daα and is defined as ( )n ∫ t 1 d α Da y(t) = (t − τ )n−α−1 y(τ )dτ, for a ≤ t ≤ b. Γ(n − α) dt a Also, the Riemann-Liouville Fractional Integral of order α is denoted by Iaα and is defined as ∫ t 1 α Ia y(t) = (t − τ )α−1 y(τ )dτ. Γ(α) a 2.4.1

Illustrations

(i) 1 d D0 C = Γ( 21 ) dt 1 2

∫

t

(t − τ )

0

− 12

( √ ) t d C Cdτ = 2C =√ , dt π πt

(ii) 1 I0 C = Γ( 12 ) 1 2

2.5

∫

t

√ − 21

(t − τ )

0

Cdτ = 2C

t , π

C is any constant.

C is any constant.

Riemann-Liouville Fractional Derivative and Riemann-Liouville Fractional Integral of Power Function [9]

The Riemann-Liouville Fractional Derivative of order α of a Power function is given by Γ(1 + ν) ν−α α ν t , ν > −1, α > 0. (2.5) a Dt t = Γ(1 + ν − α) and the Riemann-Liouville Fractional Integral of order α of a Power function is given by Γ(1 + ν) ν+α α ν t , ν > −1, α > 0. a It t = Γ(1 + ν + α) 2.5.1

Illustrations

(i) 1 2

0 Dt

t2 =

Γ(1 + 2) 2− 1 Γ(3) 3 8√ 3 2 = 2 2 1 t 5 t = 3 πt Γ(1 + 2 − 2 ) Γ( 2 )

(ii) 1 2

0 It

t2 =

Γ(1 + 2) 2+ 1 Γ(3) 5 16 √ 5 2 = 2 2 1 t 7 t = 15 πt Γ(1 + 2 + 2 ) Γ( 2 )

44


2.6

Properties of Riemann-Liouville Fractional Derivatives

If y(t) is continuous for t ≥ a,then integration of arbitrary real order has the following properties. 1. a Dt−α [a Dt−β y(t)] =a Dt−α−β y(t) 2. a Dtα [a Dt−α y(t)] = y(t),

for α > a and t > a

3. a Dtα [λy1 (t) + µy2 (t)] = λ[a Dtα y1 (t)] + µ[a Dtα y2 (t)]

2.7

Laplace transform of the Riemann-Liouville fractional derivative

The Laplace transform of Riemann-Liouville fractional derivative of order α > 0 is L[a Dtα y(t); s] = sα Y (s) −

n−1 ∑

ska Dtα−k−1 y(0)

(2.6)

k=0

If we put α = 1, β = 1, k = 1, a = 3 in property 1, we get { } 1 2 α L t E1,2 (3t ) = s(s − 3) { −1 } s = tE1,2 (3t) L−1 s−3 The Laplace transforms of some functions are listed as under: Laplace transform pairs

Y (s)

y(t) = L−1 {Y (s)}

1 sα 1 (s+a)α 1 sα −a sα s(sα +a) a s(sα +a) 1 sα (s+a) sα−β sα −a 1 (s−a)(s−b)

tα−1 Γ(α) tα−1 −at Γ(α) e tα−1 Eα,α (atα ) Eα (−atα ) 1 − Eα (−atα ) tα E1,α+1 (at) tβ−1 Eα,β (atα ) 1 at bt a−b (e − e )

Definition 2.1 One-term equation [23] If the left hand side of fractional differential equation contains only one term then fractional differential equation is called as one-term equation. Definition 2.2 Radius of convergence of series [10] The radius of convergence R is a nonnegative real number or ∞ such that the series converges if |x − a| < R and diverges if |x − a| > R.

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 45

3

POWER SERIES METHOD

In this section power series method is used to obtain solution of fractional differential equations with different types of initial conditions. Power series method is the most transparent method of solution of fractional differential equations. The idea is to obtain the solution in the power series form whose coefficients are to be determined. The solution can be computed approximately as partial sum of series, that is why this method is frequently used for solving applied problems. One-term equation with zero, nonzero and [0 Dtα−1 y(t)]t=0 = B type initial conditions are considered.

3.1

One-term equation with zero initial condition

Consider the one-term fractional initial value problem α 0 Dt y(t)

= f (t),

y(0) = 0, t > 0,

Let y(t) =

∞ ∑

0 < α < 1.

an tn+α

(3.1)

(3.2)

n=0

be the power series solution of equation (3.1). Assume that the function f (t) can be expanded in the Taylor series converging for 0 ≤ t ≤ R, where R is the radius of convergence: f (t) =

∞ ∑ f (n) (0) n=0

n!

tn

(3.3)

Using equation (3.2) and (3.3) in equation (3.1), we have [∑ ] ∑ ∞ ∞ f (n) (0) n α n+α t an t = 0 Dt n! n=0

(3.4)

n=0

Applying formula (2.5), we obtain ∞ ∑ n=0

∞

Γ(1 + n + α) n ∑ f (n) (0) n t = t an Γ(1 + n) n! n=0

Equating coefficients of like powers of t, one can determine the coefficients by formula an =

Γ(1 + n) f (n) (0) f (n) (0) = , Γ(1 + n + α) n! Γ(1 + n + α)

n = 0, 1, 2, 3...

(3.5)

Substituting these coefficients in equation (3.2), the series solution of fractional initial value problem (3.1) is ∞ ∑ f (n) (0) y(t) = tn+α (3.6) Γ(1 + n + α) n=0

Clearly the series converges absolutely with region of convergence (0, ∞).

46


Example 3.1 Consider the one-term fractional initial value problem 1 2

0 Dt

y(t) = sint,

y(0) = 0

t>0

(3.7)

Now, we determine the coefficients an , n = 0, 1, 2... by (3.5) a0 =

f (0) , Γ( 32 )

a1 =

f ′ (0) , Γ( 52 )

a2 =

f ′′ (0) f ′′′ (0) a = , 3 Γ( 72 ) Γ( 92 )

..., an =

f (n) (0) , ... Γ(n + 32 )

Since f (t) = sint,

f ′ (t) = cost,

f ′′ (t) = −sint,

f (3) (t) = −cost,

f (4) (t) = sint, ...

so that f (0) = 0,

f ′ (0) = 1,

f ′′ (0) = 0,

f (3) (0) = −1,

f (4) (0) = 0, ...

Hence a0 = 0,

a1 =

1 , Γ( 52 )

a2 = 0,

a3 =

−1 , Γ( 92 )

a4 = 0,

a5 =

1 , ... Γ( 13 2 )

Thus the series solution of fractional initial value problem (3.7) is 7 11 15 4 3 16 64 256 √ t2 + √ t2 − √ t 2 + ... y(t) = √ t 2 − 3 π 105 π 10395 π 2027025 π

(3.8)

Remark 3.1 If the function f (t) on RHS of (3.1) has the form f (t) = tβ g(t), β > −1 then also we can discuss the power series method. Assume that g(t) can be expanded in the Taylor series converging for 0 ≤ t ≤ R, where R is the radius of convergence, as ∞ ∑ g (n) (0) n t (3.9) g(t) = n! n=0

The coefficients an ,

n = 0, 1, 2... are given by an =

Γ(1 + n + β) g (n) (0) Γ(1 + n + α + β) Γ(1 + n)

(3.10)

Then series solution of fractional initial value problem (3.1) is y(t) =

∞ ∑ n=0

Γ(1 + n + β) g (n) (0) n+α+β t . Γ(1 + n + α + β) Γ(1 + n)

(3.11)


0 Dt

y(t) = t2 sint,

y(0) = 0,

t > 0.

(3.12)

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 47 Determine the coefficients an , n = 0, 1, 2... from (3.10) as a0 = 0,

a1 =

Γ(4) , Γ( 92 )Γ(2)

a2 = 0,

a3 =

Γ(6) , Γ( 13 2 )Γ(4)

a4 = 0,

a5 =

Γ(8) , ... Γ( 17 2 )Γ(6)

Thus the series solution of fractional initial value problem (3.12) is y(t) =

3.2

7 11 15 96 1280 10572 √ t2 − √ t2 + √ t 2 − ... 105 π 10395 π 2027025 π

(3.13)

One-term equation with nonzero initial condition


= f (t),

y(0) = A,

Assume that

A ̸= 0,

t > 0,

0 < α < 1.

(3.14)

∞

f (t) =

∑ At−α + fn tn−α Γ(1 − α)

(3.15)

n=1

where the coefficients fn are known. Let y(t) =

∞ ∑

an tn

(3.16)

n=0

be the series solution of fractional initial value problem (3.14). Thus, the fractional initial value problem (3.14) reduces to [∑ ] ∞ ∞ ∑ At−α α n an t = + fn tn−α 0 Dt Γ(1 − α) n=0

(3.17)

n=1

Applying equation (2.5) to yield ∞ ∑

∞

an

n=0

∑ Γ(1 + n) n−α At−α t = + fn tn−α . Γ(1 + n − α) Γ(1 − α) n=1

Equating coefficients of like powers of t, we obtain a0 = A,

an =

Γ(1 + n − α) fn , Γ(1 + n)

n = 1, 2, 3, ...

(3.18)

Hence, the series solution of fractional initial value problem (3.14) is y(t) = A +

∞ ∑ Γ(1 + n − α) n=1

Γ(1 + n)

fn tn .

(3.19)


0 Dt

y(t) = f (t),

3 y(0) = , t > 0 2

and f (n) = n.

(3.20)

48


We determine the coefficients an , n = 0, 1, 2... as 3 a0 = , 2

a1 =

Γ( 32 ) 1, Γ(2)

a2 =

Γ( 25 ) 2, Γ(3)

a3 =

Γ( 72 ) 3, Γ(4)

a4 =

Γ( 92 ) 4, ... Γ(5)

Therefore, the series solution of fractional initial value problem (3.20) is 3 1√ 3 15 πt[1 + t + t2 + ...]. + 2 2 2 4

y(t) =

3.3

(3.21)

One term equation with [0 Dtα−1 y(t)]t=0 = B initial condition


= f (t),

[0 Dtα−1 y(t)]t=0 = B,

where t > 0, 0 < α < 1 andB is any constant. Let ∞ ∑ y(t) = an tn+α−1

(3.22)

(3.23)

n=0

be the solution of fractional initial value problem (3.22). Assume that f (t) can be expanded in the Taylor series as ∞ ∑ f (n) (0) n t , Γ(n + 1)

f (t) =

and

n=0

1 = 0. Γ(0)

(3.24)

Using equation (3.23) and (3.24) in equation (3.22) and then applying formula (2.5), one obtains ∞

∞

n=1

n=0

Γ(n + α + 1) n ∑ f (n) (0) n Γ(α) −1 ∑ t + an+1 t = t . a0 Γ(0) Γ(n + 1) Γ(n + 1) Equating the coefficients of like powers of t, to yield an+1 =

f (n) (0) , Γ(n + 1 + α)

n = 0, 1, 2...

(3.25)

To determine the coefficient a0 ,we use the initial condition [0 Dtα−1 y(t)]t=0 = B α−1 y(t) = 0 Dt

∞ ∑

an0 Dtα−1 tn+α−1 =

n=0

As n → 0, we obtain a0 = value problem (3.22) is

B Γ(α) .

∞ ∑ n=0

an

Γ(n + α) n t = B. Γ(n + 1)

Therefore, the series solution of fractional initial ∞

B α−1 ∑ f (n) (0) n+α−1 y(t) = t + t . Γ(α) Γ(n + α) n=1

(3.26)

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 49 Example 3.4 Consider one-term fractional initial value problem 1 2

0 Dt

y(t) = sint,

−1/2

[0 Dt

y(t)]t=0 = 1,

t > 0.

(3.27)

The coefficients can be determined as 1 a0 = √ , π

a1 =

1 , Γ( 52 )

−1 , Γ( 92 )

a2 = 0, a3 =

a4 = 0,

a5 =

1 , ... Γ( 13 2 )

Since f (t) = sint,

f ′ (t) = cost,

f ′′ (t) = −sint,

f (3) (t) = −cost,

f (4) (t) = sint...,

so that f (0) = 0,

f ′ (0) = 1,

f ′′ (0) = 0,

f (3) (0) = −1,

f (4) (0) = 0, ...

Hence the series solution of fractional initial value problem (3.27) is 1 16 5 64 9 4 1 y(t) = √ [1 + t 2 − t2 + t 2 − ...] 3 105 10395 π

4

(3.28)

LAPLACE TRANSFORM METHOD

In this section, we apply the Laplace transform to solve some fractional order initial value problems. The procedure is simple: Find the Laplace transform of the equation, solve for the transform of the unknown function and finally find the inverse Laplace to obtain desired solution. Consider the fractional differential equation α a Dt y(t)

+ a1 y(t) = 0

(4.1)

with initial condition α−1 y(0) a Dt

= ci ,

(4.2)

where c′i s and a1 are real constants and α ∈ (n − 1 < α ≤ n). This is called fractional initial value problem (IVP.) Theorem 4.1 The general solution of IVP (4.1) − (4.2) is given by y(t) =

n ∑

ci tα−j Eα,α−j+1 (−atα ),

1

for α ∈ (n − 1 < α ≤ n), where ci ’s are real constants.

50


Proof : Applying Laplace transform to equation (4.1), to get L{a Dtα y(t)} + L{a1 y(t)} = 0. Using initial conditions (4.2), we have s y¯(s) − α

n ∑

dj sj−1 + a1 y¯(s) = 0

j=1

where dj = 0 Dtα−j y(0),

(j = 1, 2, ..., n) i.e. y¯(s) =

n ∑

sj−1 . sα − a1

dj

j=1

(4.3)

Applying inverse Laplace transform to equation (4.3) and using equation (2.4), the solution is n ∑ y(t) = cj tα−j Eα,α−i+1 (−atα ) 1

Corollary 4.1 If α ∈ (0 < α ≤ 1) then solution of (4.1) − (4.2) is given by y(t) = c1 tα−1 Eα,α (−atα ) Example 4.1 Solve the following fractional IVP 1 2

0 Dt

y(t) − 2y(t) = 0 −1 2

0 Dt

y(0) = 1.

(4.4) (4.5)

Applying Laplace transform to (4.4) we get L{0 Dtα y(t)} − L{2y(t)} = 0 Using initial condition (4.5), we obtain 1

s 2 y¯(s) − 2¯ y (s) = 1. y¯(s) =

1

(4.6)

1 2

s −2

Applying inverse Laplace transform to (4.6), the solution of IVP (4.4) − (4.5) is y(t) = t

−1 2

1

E 1 , 1 (2t 2 ) 2 2

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 51 Example 4.2 Solve the following fractional IVP 1 2

0 Dt

−1 2

0 Dt

y(t) = sint

(4.7)

y(0) = 1

(4.8)

Applying Laplace transform to (4.7) to obtain 1

L{0 Dt2 y(t)} = L{sint}. Using the initial condition (4.8), we get ( y¯(s) = 1 +

) 1 1 . 2 s + 1 s 12

(4.9)

Applying inverse Laplace transform to (4.9), the solution of IVP (4.7) − (4.8) is 1 t− 2 2 1 + E 12 ,1 (−t ) Γ( 2 ) 1

y(t) = ACKNOWLEDGEMENTS

Authors are grateful to University Grants Commission-India for award of Teacher Fellowship under Faculty Development Programme (XI th Plan) and Research Fellowship under Meritorious Fellowship Scheme (XI th Plan) respectively. Authors are thankful to Prof.D.B.Dhaigude for providing valuable guidance while preparing the article. First author expresses sincere thanks to the authorities of Shrikrishna Education Society, Gunjoti, Dist.Osmanabad (M.S) and Principal of Shrikrishna Mahavidyalaya for granting study leave under UGC-FDP Scheme.

References [1] G.A. Birajdar, Numerical solution of time fractional Navier-Stokes equation by discrete Adomian decomposition method,Nonlinear Engineering, 3 (1), (2014), 21-26 [2] D.B. Dhaigude, G.A. Birajdar and V.R. Nikam,Adomain decomposition method for fractional Benjamin-Bona-Mahony-Burger’s equations, Int. J. Appl. Math. Mech., 8(12), (2012), 42-51. [3] D.B.Dhaigude and G.A. Birajdar, Numerical solution of system of fractional partial differential equations by discrete Adomian decomposition method, J.Frac.Cal.Appl.,3(12), (2012), 1-11 . [4] D.B.Dhaigude and G.A. Birajdar, Numerical solutions of fractional partial differential equations by discrete Adomian decomposition method, Adv. Appl. Math. and Mech., 6 (1), (2014), 107-119,.

52


[5] G.A. Birajdar and D.B.Dhaigude, An implicit numerical method for semilinear fractional diffusion equation, Int. Conf.Proce. Elsevier, (2014), 674-678. [6] Lokenath Debnath, Recent Applications of Fractional Calculus to Science and Engineering, IJMMS , 54, (2003), 3413-3442. [7] D.B.Dhaigude, J.A.Nanware and V.R.Nikam , Monotone Technique for System of Caputo Fractional Differential Equations with Periodic Boundary Conditions, Dyn. Conti., Discrete Impul. Sys., 19(5a), (2012), 575-584. [8] A.Erdelyi, W. Magnus, F. Oberhettinger and F.G. Tricomi, Higher Transendental Functions, Vol III , Mc.Graw-Hill, New York, (1955.) [9] A.A.Kilbas, H.M.Srivastava and J.J.Trujillo, Theory and Applications of Fractional Differential Equations, North Holland Mathematical Studies Vol.204 , Elsevier(North-Holland) Sciences Publishers, Amsterdam,(2006). [10] Erwin Kreyszig, Advanced Engineering Mathematics, Wiley India, (2005). [11] V.Lakshmikantham, A.S.Vatsala, Basic Theory of Fractional Differential Equations and Applications, Nonl. Anal. 69(8), (2008), 2677-2682. [12] V.Lakshmikantham, A.S.Vatsala , General Uniqueness and Monotone Iterative Technique for Fractional Differential Equations, Appl. Math. Lett., 21(8), (2008), 828-834 . [13] V.Lakshmikantham, S.Leela, Differential and Integral Inequalities, Vol.I, Academic Press, New York, (1969). [14] K.S.Miller and B.Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, Wiley, New York, (1993). [15] J.A.Nanware, Monotone Method In Fractional Differential Equations and Applications, Ph.D Thesis, Dr.Babasaheb Ambedkar Marathwada University, 2013),. [16] J.A.Nanware, Existence and Uniqueness Results for Fractional Differential Equations via Monotone Method, Bull.Marathwada Math. Soc.,14(1), (2013), 39-56. [17] J.A.Nanware, D.B.Dhaigude, Existence and Uniqueness of Solution of Differential Equations of Fractional Order with Integral Boundary Conditions, J. Nonlinear. Sci. Appl., 7, (2014), 246-254. [18] J.A.Nanware, D.B.Dhaigude , Existence and Uniqueness of solution of Riemann-Liouville Fractional Differential Equations with Integral Boundary Conditions, Int. J. Nonl. Sci., 14(4), (2012), 410-415. [19] J.A.Nanware, D.B.Dhaigude , Boundary Value Problems for Differential Equations of Non-integer Order Involving Caputo Fractional Derivative, Adv. Stu. Contem. Math., 24(3), (2014), 369-376.

METHODS OF SOLVING FRACTIONAL ... OF ORDER α (0 < α < 1) 53 [20] J.A.Nanware, D.B.Dhaigude , Monotone Iterative Scheme for System of Riemann-Liouville Fractional Differential Equations with Integral Boundary Conditions, Math. Modelling Scien.Comput., Springer-Verlag, 283, (2012), 395402. [21] J.A.Nanware, N.B.Jadhav, D.B.Dhaigude, Monotone Iterative Technique for Finite System of Riemann-Liouville Fractional Differential Equations with Integral Boundary Conditions, International Conference of Mathematical Sciences 2014, Elsevier, (2014), 235-238. [22] K.B.Oldham and J.Spanier , The Fractional Calculus, Academic Press, New York, (1974). [23] I.Podlubny , Fractional Differential Equations, Academic Press, San Diego, (1999).

APPENDIX Graphical representation of solutions:

(Ex.3.1)

(Ex.3.2)

(Ex.3.3)

(Ex.3.4)

(Ex.4.1)

(Ex.4.2)