An approximation method for fractional integro

Open Phys. 2015; 13:370–376

Research Article

Open Access

Ibrahim Emiroglu*

An approximation method for fractional integro-differential equations DOI 10.1515/phys-2015-0049 Received November 9, 2015; accepted November 25, 2015

Abstract: In this work, an approximation method is proposed for fractional order linear Fredholm type integrodifferential equations with boundary conditions. The Sinc collocation method is applied to the examples and its efficiency and strength is also discussed by some special examples. The results of the proposed method are compared to the available analytic solutions. Keywords: Fractional Fredholm integro-differential equation, sinc-collocation method, Caputo derivative PACS: 02.60.Nm; 02.60.Cb; 02.70.Jn

1 Introduction The fractional order integral and differential equations has had an important role in science and engineering in the recent years. Recently many mathematicians and scientist have undertaken work on fractional calculus. One of these works, the Caputo-Fabrizo fractional derivative and its applications and simulations are given in the references [12– 15]. The aim of this work here is to overcome the problems in approximating solutions of the following fractional order integro-differential equations with the boundary conditions,

y′′ + p(x)y′ + q(x)Ca D αx y + r(x)y = f (x) + ∫︁b +λ K(x, u)y(u)du, 0 0 and k = 0, ±1, ±2, ... ⎧ (︀ )︀ (︁ x − kh )︁ ⎨ sin π x−kh h x ≠ kh π x−kh = S(k, h)(x) = sinc h h ⎩ 1 x = kh.

Definition 6. [19] Let F function be analytic in D E and F satisfies, ∫︁ |F(z)|dz →, as u = ∓∞,

Definition 4. If the function of f (x) is defined on the interval (−∞, ∞), and for h > 0 ∞ (︁ x − kh )︁ ∑︁ C(f , h)(x) = f (kh)sinc h

ψ(L+u)

where

{︁ π }︁ L = iy : |y| < d ≤ , 2 and the boundary conditions of D E satisfy ∫︁ |F(z)dz| < ∞. T(F) = ∂D E

k=−∞

is called the Whittaker cardinal expansion of f whenever this series converges [18]. Approximations are generally constructed for infinite, semi-infinite and finite intervals. To construct an approximation on the interval (a, b) the conformal map (︁ z − a )︁ (3) ϕ(z) = ln b−z is applied. This map has D E the eye-shaped domain in the z-plane ⃒ (︁ z − a )︁⃒ {︁ π }︁ ⃒ ⃒ D E = z = x + iy : ⃒arg . ⃒ 0 , ∫︁ ∫︁ ∞ ∑︁ F(z i ) F(z)k(ϕ, h)(z) i F(z)dz − h = dz ≡ I F (4) sin(πϕ(z)/h) ϕ′ (z i ) 2 i=−∞

Γ

∂D

where ]︀ ⃒ ⃒ [︀ iπϕ(z) ⃒ ⃒ h sgn(Imϕ(z)) ⃒

|k(ϕ, h)|z∈∂D = ⃒e

z∈∂D

=e

−πd h

.

The infinite quadrature algorithm must be truncated to a finite sum in the Sinc collocation method. The next theorem shows the exponential convergence results. Theorem 2. [19] If there exists constants α > 0, β > 0 and C > 0 such that {︃ ⃒ ⃒ e−α|ϕ(x)| x ∈ ψ((−∞, ∞)) ⃒ F(x) ⃒ (5) ⃒ ′ ⃒≤C ϕ (x) e−β|ϕ(x)| x ∈ ψ((0, ∞)). the equation (4) can be written as, N ⃒ ∫︁ ⃒ (︁ e−αMh e−βNh )︁ ∑︁ F(x i ) ⃒ ⃒ ≤ C + + |I F | (6) ⃒ F(x)dx − h ⃒ α β ϕ′ (x i ) Γ

i=−M

If we choose,

k = 0, ±1, ±2, ...

Unauthenticated Download Date | 1/13/16 8:19 PM

√︂ h= N≡

πd αM

[︁⌊︁ αM β

+1

⌋︁]︁

372 | Ibrahim Emiroglu √

Theorem 3. For 0 < α < 1 and h L = π/ L, the following contact exists

we have, ∫︁ F(x)dx = h

N (︁ )︁ ∑︁ F(x i ) −(παdM)1/2 + O e . ϕ′ (x i )

(7) C α a D x (y n (x))

i=−M

Γ

≈

N ∑︁

c k R(x)

(11)

k=−M

These theorems are used to approximate the integral in the fractional given in (1).

where R(x) =

Lemma 1. [20] Let ϕ be the conformal one-to-one mapping of the simply connected domain D E onto D S , given by (3). Then {︃ 1 i=k (0) δ ik = [S(i, h)oϕ(x)]|x=x k 0 i ≠ k.

⃒ d ⃒ =h [S(i, h)oϕ(x)]⃒ dϕ x=x k

{︃

⃒ d2 ⃒ [S(i, h)oϕ(x)] = h2 ⃒ dϕ2 x=x k

{︃

δ(1) ik

δ(2) ik

0

i=k

(−1)k−i k−i

and ξ is a conformal map for the interval [a, x]. Proof. By using Caputo fractional derivative given in (2), it can be written that

i ≠ k.

2

− π3

−2(−1) (k−i)2

N ∑︁

c k Ca D αx (S k (x))

where C α a D x (S k (x))

i=k k−i

=

k=−M

1 = Γ(1 − α)

∫︁x

The quadrature algorithm given in (7) can be used for computing the above integral which is divergent on the interval [a, x]. Then we have, (︁ u − a )︁ ξ (u) = ln x−u and x r = ξ −1 (rh L ) = √

c k S k (x),

n = M+N+1

(x − u)−α S′k (u)du

a

i ≠ k.

Consider the problem given in (1). Let y n (x) be the approximate solution of (1), which can be written by the finite expansion of Sinc basis functions, N ∑︁

r=−L

C α a D x (y n (x))

3 The sinc-collocation method

y n (x) =

L ∑︁ (x − x r )S′k (x r ) hL Γ(1 − α) ξ ′ (x r )

(8)

k=−M

where S k (x) is the function S(k, h)oϕ(x). The problem is thus to determine the unknown coefficients c k in (8). We use the Sinc-collocation method to determine the unknown coefficients. For this reason, we need the derivatives of y n (x). The first and second derivatives are given by :

a + xe rh L , 1 + e rh L

where h L = π/ L. Then, using the equality (7), we can write C α a D x (S k (x))

≈

L ∑︁ (x − x r )S′k (x r ) hL . Γ(1 − α) ξ ′ (x r )

(12)

r=−L

If we apply (7) to the kernel integral given in (1), we have the following lemma. Lemma 2. The following contact exists

N ∑︁ d d c k ϕ′ (x) y n (x) = S (x), dx dϕ k

(9)

∫︁b K(x, u)y(u)dt ≈ h

k=−M

a N (︁ ∑︁ d2 d y (x) = c ϕ′′ (x) S (x)+ n k dϕ k dx2 k=−M

)︁ d2 (ϕ ) S k (x) . 2 dϕ ′ 2

In a similar way, α order derivative of y n (x) for 0 < α < 1 is calculated by the following theorem.

N ∑︁ K(x, u k ) y ϕ′ (u k ) k

(13)

k=−M

where an approximate value of y(u k ) is denoted as y k . (10)

If the approximation given in (8),(10), (12), (13) is replaced in each term of (1) and then multiplying the resulting equation by {(1/ϕ′ )2 } and setting x = x i , the following nonlinear equation system is obtained,


An approximation method for fractional integro-differential equations

N ∑︁

]︃ [︃ (︁ 1 )︁ (︁ 1 )︁′ d d2 S + ck S + p ′ − dϕ k dϕ2 k ϕ ϕ′ k=−M }︃ (︁ 1 )︁2 (︁ 1 )︁2 K(x, t k ) (︁ 1 )︁2 q ′ R + r ′ S k − λh ′ (x i ) ϕ ϕ ϕ (t k ) ϕ′ (︃ )︃ (︁ 1 )︁2 = f (x i ), i = −M, ..., N. ϕ′ {︃

(︃ (︁ (︁ )︁2 )︁ (︁ (︁ )︁2 )︁ (x−M ), f ϕ1′ (x−M+1 ), ..., B= f ϕ1′ (︁ (︁ )︁2 )︁ f ϕ1′ (x N )

=

δ(0) ki ,

δ(1) ik

=

−δ(1) ki ,

δ(2) ik

=

)︃T .

C = (c−M , c−M+1 , ..., c N )T .

As it is known by Lemma 1: δ(0) ik

| 373

δ(2) ki ,

the following theorem follows. Theorem 4. If the solution of problem (1) is (8), the determination of the unknown coefficients {c k }Nk=−M are given by ]︃ [︃ {︃ N (︁ 1 )︁ ∑︁ 1 (2) 1 (︁ 1 )︁′ − p ′ (x i )δ(1) δ + ck ik + h ϕ′ h2 ik ϕ k=−M (︃ )︃ (︁ 1 )︁2 (︁ (︁ 1 )︁2 )︁ q ′ R (x i ) + r ′ (x i )δ(0) ik − ϕ ϕ (14) }︃ (︃ )︃ (︁ )︁ (︁ 1 )︁2 K(x , t ) 1 2 = f (x i ), λh ′ (x i ) ′ i k ϕ ϕ (t k ) ϕ′

Thus a nonlinear equation system is converted into linear equation system. This linear system has n equations and n unknown coefficients given by (15). Hence solving by (15), the unknown coefficients c k that are necessary for approximate solution in (8), can be found.

4 Computational examples Here, problems with boundary conditions are presented to test the proposed method. In the all examples, d = π/2, α = β = 1/2, N = M are assumed. Example 1. The solution of following linear fractional Fredholm integro-differential equation:

i = −M, ..., N. The matrix-vector form for the system (14) can be written as: Let D(y) denotes a diagonal matrix and its diagonal elements are y(x−M ), y(x−M+1 ), . . . , y(x N ) and non-diagonal elements are zero, let G = R(x i ) and E=

y′′ (x) +

1 ′ 1 y(x) = f (x) + y (x) + 0C D0.7 x x x2

I (i) = [δ(i) ik ],

with the boundary conditions y(0) = 0,

2

4

2

(1+6x )(π −36π +240) π6

where D, G, E, I (0) , I (1) and I (2) are square matrices of order n × n. Particularly, I (0) , I (1) and I (2) are the identity matrix, the skew-symmetric matrix and the symmetric matrix, respectively. For the calculation of unknown coefficients c k in linear system (14), this system should be rewritten by using the notations given in matrix-vector form as AC = B

y(1) = 0,

where f (x) = 20x3 − 7x2 − 4x +

i = 0, 1, 2

(15)

K(x, t)y(t)dt 0

K(x i , t k ) (ϕ′ (x i ))2 ϕ′ (t k )

denote a matrix and also let I (i) denotes the matrices

∫︁1

120 4.3 x Γ(5.3) 2

−

24 x3.3 Γ(4.3)

+

and K(x, t) = (6x + 1) cos πt. The analytic solution of this problem is y(x) = x4 (x − 1). Table 1 and Table 2 show analytic and numerical solutions for different values of L and M respectively. Figure 1 and Figure 2 show the results graphically. Example 2. The solution of following linear fractional Fredholm integro-differential equation: y

′′

(x) + 0C D0.5 x y(x) +

∫︁1 y(x) = f (x) + 2

K(x, t)y(t)dt 0

where

with the boundary conditions A= (︃ D q

1 (2) I h2

+ 1h D

(︁ )︁2 1 ϕ′

)︃

(︃ (︁ )︁′ 1 ϕ′

(︃ G+D r

−p

)︃ (︁ )︁

(︁ )︁2 1 ϕ′

1 ϕ′

I (1) +

y(0) = 0,

y(1) = 0, 0.5

)︃ I

(0)

− hλE.

2.5

√

x) −3.2x where f (x) = −6x + x(1 − x2 ) + 2x Γ(0.5) − 2(14−5e)(1+ e √ −t and K(x, t) = (1 + x)e . The analytic solution of this


374 | Ibrahim Emiroglu Table 1: Computational results for L = 5, M = 5

x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Analytic sol. 0 -0.00009 -0.00128 -0.00567 -0.01536 -0.03125 -0.05184 -0.07203 -0.08192 -0.06561 0

Approx Sol. 0 0.0136050 0.0264546 0.0198237 -0.0001862 -0.0265477 -0.0536647 -0.0758895 -0.0853275 -0.0676832 0

Error 0 1.36 × 10−2 2.77 × 10−2 2.54 × 10−2 1.51 × 10−2 4.70 × 10−3 1.82 × 10−3 3.85 × 10−3 3.40 × 10−3 2.07 × 10−3 0

Figure 2: The graphics of the analytic and approximate solutions for L = 30, M = 50 Table 3: Computational results for L = 5, M = 5

x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Table 2: Computational results for L = 30, M = 50

x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Analytic sol. 0 -0.00009 -0.00128 -0.00567 -0.01536 -0.03125 -0.05184 -0.07203 -0.08192 -0.06561 0

Approx Sol. 0 -0.0000024 -0.0011675 -0.0055478 -0.0152609 -0.0311556 -0.0517651 -0.0719892 -0.0819117 -0.0656166 0

Error 0 8.75 × 10−5 1.12 × 10−4 1.22 × 10−4 9.91 × 10−5 9.43 × 10−5 7.49 × 10−5 4.07 × 10−5 8.33 × 10−6 6.55 × 10−6 0

Approx Sol. 0 0.092975 0.189353 0.275658 0.338912 0.374984 0.381028 0.353044 0.284832 0.169001 0

Error 0 6.02 × 10−3 2.64 × 10−3 2.65 × 10−3 2.91 × 10−3 1.60 × 10−5 2.97 × 10−3 3.95 × 10−3 3.16 × 10−3 1.99 × 10−3 0


x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 1: The graphics of the analytic and approximate solutions for L = 5, M = 5

problem is y(x) = x(1 − x2 ). Table 3 and Table 4 show analytic and numerical solutions for different values of L and M respectively. Figure 3 and Figure 4 show the results graphically.

Analytic sol. 0 0.099 0.192 0.273 0.336 0.375 0.384 0.357 0.288 0.171 0

Analytic sol. 0 0.099 0.192 0.273 0.336 0.375 0.384 0.357 0.288 0.171 0

Approx Sol. 0 0.098998 0.191998 0.272998 0.335999 0.375001 0.384002 0.357004 0.288004 0.171003 0

Error 0 1.21 × 10−6 1.85 × 10−6 1.73 × 10−6 8.77 × 10−7 5.51 × 10−7 2.23 × 10−6 3.70 × 10−6 4.34 × 10−6 3.40 × 10−6 0

Example 3. The solution of following linear fractional Fedholm integro-differential equation: ′′

y (x) −

x2C0 D0.3 x y(x) +

∫︁1 xy(x) = f (x) −

K(x, t)y(t)dt 0


An approximation method for fractional integro-differential equations

| 375



x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Analytic sol. 0 0.0099 0.0384 0.0819 0.1344 0.1875 0.2304 0.2499 0.2304 0.1539 0

Approx Sol. 0 0.009899 0.038399 0.081899 0.134399 0.187499 0.230399 0.249899 0.230399 0.153899 0

Error 0 1.15 × 10−9 4.51 × 10−9 1.05 × 10−8 4.40 × 10−9 1.72 × 10−8 2.76 × 10−8 3.40 × 10−8 4.20 × 10−8 5.12 × 10−8 0


with the boundary conditions y(0) = 0,

y(1) = 0,

4x 68 2 24 where f (x) = −x5 + x3 −12x2 + 15 x5.7 − Γ(2.7) x3.7 + 35 + Γ(4.7)


and K(x, t) = 2x − t2 . The analytic solution of this problem is y(x) = x2 (1 − x2 ). Table 5 and Table 6 show analytic and numerical solutions for different values of L and M respectively. Figure 5 and Figure 6 show the results graphically. Table 5: Computational results for L = 5, M = 5

x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Analytic sol. 0 0.0099 0.0384 0.0819 0.1344 0.1875 0.2304 0.2499 0.2304 0.1539 0

Approx Sol. 0 0.007115 0.039353 0.087488 0.138538 0.185354 0.222019 0.240875 0.228574 0.157728 0

Error 0 2.78 × 10−3 9.53 × 10−4 5.58 × 10−3 4.13 × 10−3 2.14 × 10−3 8.38 × 10−3 9.02 × 10−3 1.82 × 10−3 2.82 × 10−3 0


5 Conclusion In this work, an approximation method for the fractional integro-differential equation is presented by using the Sinc collocation method. The effectiveness and accuracy of the method is illustrated by some examples. The results obtained are also compared to analytic results. The comparisons given in the related tables and graphical forms show that the approximate solutions converge to the analytic ones when increased the number of Sinc grid points N. The method presented here is a powerful tool for solving frac-


376 | Ibrahim Emiroglu tional integro-differential equations with boundary conditions.

[9] [10]

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