Applied Mathematics, 2011, 2, 1292-1296
doi:10.4236/am.2011.210179 Published Online October 2011 (http://www.SciRP.org/journal/am)
Existence and Uniqueness of Solution for a
Fractional Order Integro-Differential Equation
with Non-Local and Global Boundary
Conditions
Mehran Fatemi1, Nihan Aliev1, Sedaghat Shahmorad2
1
Department of Mathematics, Baku State University, Baku, Azerbaijan 2
Department of Applied Mathematics, University of Tabriz, Tabriz, Iarn E-mail: [email protected], [email protected]
Received February 24, 2011; revised September 5, 2011; accepted September 13, 2011
Abstract
In this paper, we prove an important existence and uniqueness theorem for a fractional order Fredholm – Volterra integro-differential equation with non-local and global boundary conditions by converting it to the corresponding well known Fredholm integral equation of second kind. The considered problem in this paper has been solved already numerically in [1].
Keywords:Fractional Order Integro-Differential Equation, Non-Local Boundary Conditions, Fundamental Solution
1. Introduction
Let’s consider a problem under boundary condition con-taining non-local and global terms for a fractional order integro-differential equation
1 , d 2 , d
1, , , ,
x b
q
a a
D y x f x K x t y t t K x t y t t
q m m x a b
,
(1)
1 1
=1
d ,
1, ,
b m
j j
ij ij i i
j a
y a y b H t y t t d
i m
where m q
0,1
, f x
, K1
x t, , K2
x t, and
i
H t , i= 1,m are continuous, real-valued functions,
ij
, ij, i and di, i= 1,m, j= 1,m are real
con-stants, and boundary conditions (2) are linearly independent.
2. Existence and Uniqueness of Solution
Theorem. Let the functions f x
, Kj
x t,
, j = 1, 2 and
i
H t , i= 1,m are continuous, ij, ij, i and di, . = 1
i j
(2)
,m are real constant, the boundary conditions (2) are linearly independent and condition (15) is satisfied. Then the boundary value problem (1)-(2) has unique solution.
Proof: Acting in Equation (1) by fractional order de-rivative operator Dm q [2], we get
1
, d 2
, dx b
m q q m q m q m q
a a
D D y x D f x D K x t y t tD K x t y t t
,,
d , since Dm q D y xq
=D y xm
then we get the equation
1
, d 2
,x b
m
a a
1293
where
1 1
2 2
d
= d
d !
d
, ,
d !
d
, ,
d !
q m x
m q
a q m x
t
q m x
a
x
F x D f x f
x q m
x
M x t K t
x q m
x
M x t K t
x q m
,
d ,
d .
(4)
Now, we write Equation (3) in the general from
,m
D y x G x y ,
(3.1)
and accept that is known, then the fundamen-tal solution (see [3]) is in the form
,G x y
1
.1 !
m
x
Y x x
m
(5)
where
1 2,1, > ,, 0, < ,x x
x x
d ,x
d (6)
is Heaviside’s unit function.
Now, we try to get some basic relations. The first of these relations is Lagrange’s formula. We multiply both ides of Equation (3) by fundamental solution (5) and integrate the obtained expression on (see [4,5]) to get
a b,
d
,
b b
m
a a
D y x Y x x G x y Y x
(7)where,
,
1
, d 2
,x b
a a
G x y F x
M x t y t t
M x t y t t (7.1) integrating by parts on the left hand side of expression (7) and taking into account that (5) is a fundamental solution of (3.1), give the first basic relation in the form
11
= =0
, ,
1 , d 1 1
, ,
2
b
m s b n
s
m s
x
x a
s a
y a
D y x Y x G x y Y x x
y a
b ,b. (8)Hence, the first expressions for the necessary conditions are obtained in the form
1
1 =0
1
1 =0
1
1 ,
2
1
1 ,
2
m b
m s b
s
m s
x
x a
s a
m b
m b
s m s s
x
x a
s a
y a D y x Y x a G x y Y x a x
y a D y x Y x b G x y Y x b x
d ,
d ,
(9)
It is easy to see that the second expression in (9) turns into an identity. Indeed, as it is seen from (5)-(6), the integral at the right side of the second condition contains the value of the function
x
, which is zero for=b
. For x=a the the summation in the second ex-pression contains the Heaviside function which is zero for s= 0,m1
.
Finally, the first summand contains positive degrees of
x for = 0,s m2 these terms become zero at= =
x b. Here, for =s m1, the the expression of fundamental solution for s=m1 yields the Heaviside function. For x=b, =b this becomes 1
2, therefore,
the second one of necessary conditions (9) turns into identity.
Now, we construct the second basic expression to get the second group of necessary conditions. For that, we multiply both sides of (3) by the derivative of (5) and integrate on
a b, [6,7]:
d
,
b b
m
x x
a a
D y x Y x x G x y Y x x
d .Integrating by parts on the left side of the obtained ex-pression and taking into account (5) and (6), we get the second basic relation as follows:
11
1 1
= =0
, ,
1 , d ( 1) 1
, ,
2
b
m s b
s
m s m
x x
x a
s a
,
,
y a b
D y x Y x G x y Y x x
y a b
(10)1294 M. FATEMI ET AL.
1 2 1 1 =0 1 2 1 1 =0 1 1 , 2 1 1 , 2 m bm s b
s m s x x x a s a m b m b
s m s s
x x
x a
s a
y a D y x Y x a G x y Y x a x
y b D y x Y x b G x y Y x b x
d , d . (11)Similar to the second expression of (9), we can show that the second expression of (11) turns into identity. If we continue this process, in order to get the m-th basic relation, we multiply (3) the
m1
b b
-th order derivative of (5) and integrate on
a b, to get:
1
1
d ,
m m
m
x x
a a
D y x Y x x G x y Y x x
d .Here, once integrating by parts on the left side of the
obtained expression gives
1 1 = 1 d, d .
b b m m m x x x a a b m x a
D y x Y x D y x Y x x
G x y Y x x
mThus, if we take into account that (5) is the fundamen-tal solution, the last relation (m-th) will be as follows:
1 1 1 1 ( 1) = , ,, d 1
, , 2 m b b m m m
x x m
x a a
y a
D y x Y x G x y Y x x
y a , . b b
(12)Therefore, the last group of necessary conditions will be in the form:
1 1 1 1
=
1 1 1 1
= 1 , d 2 1 , d 2 b b
m m m m
x x
x a a
b b
m m m m
x x
x a a
y a D y x Y x a G x y Y x a x
y b D y x Y x b G x y Y x b x
, , 1 m d , (13)here, as above, the second necessary condition turns into identity.
Now, we join to the given m linearly independent boundary condition (7), the necessary conditions in (9),
(11) and etc. (13) that are not identities, and write the system of 2m linear algebraic equations obtained with respect to the boundary values of the unknown function in the following way.
1 111 12 1 11 12 1 1 1 1
1 1
1 2 1 2
1
1 2
d ,
d ,
1 1 1 1
, d ,
b m m m m a b m m
m m mm m m mm m m m
a
m m m m m
b
a
y a y a y a y b y b y b d H t y t t
y a y a y a y b y b y b d H t y t t
y a y a y a y b Y b a y b Y b a y b
G x y Y x a x y a
1 1 1 2 2
( 1)
1 2 1 1 1
1 ( ) 1 1
1 , d ,
1 1 = ,
m m m m m
b m
a
b
m m m m m
a
y a y a y a y b y b Y b a y b
y b a y b G x y Y x a x
y a y b y b Y b a y b G x y Y x a x
1295
For solving the system (14) by the Cramer’s rule, it is necessary that its basic determinant differ from zero.
Accept that the following condition is satisfied
11 12 1 11 12 1
1 2 1 2
1 2
1
0,
1 0 0 1 1 1
...
0 0 ( 1) 0 0 1
m m
m m mm m m mm
m m m m
m
Y b a b a
Y b a
(15)
Then, from system (14), we get
1 , 1
=1 =1
1 , 1
=1 =1
1 1
d , d
1 1
d , d
b b
m m
k s k s
s s s
s a s a
b b
m m
k s m k s
s s s
s a s a
y a d H t y t t G x y Y x a x
y b d H t y t t G x y Y x a x
,
, ,
,
m s k
m s m k
(16)
where ( , )p q denotes the cofactor of the elements at the intersection of p-th row and q-th column of the
determi-nant . Calculate the following expression:
( )
1 2
1 2
1 2
, d , d , d ( )
d d , d d ,
d d , d d ,
b b x b
s s
a a a a
b b x b b
s s
s
a a a a a
b b b b b
s s
s
a a a a a
G x y Y x a x F x M x t y t t M x t y t t Y x a dx
d
d . F x Y x a x Y x a x M x t y t t Y x a x M x t y t t
F x Y x a x y t t Y x a M x t x y t t Y x a M x t x
Then, we get:
, d
b b
s
s s
a a
G x y Y xa xF M t y t t
d , (17)and so,
2 1
d ,
, d , d ,
b
s s
a
b b
s s
s
a a
F F x Y x a x
M t Y x a M x t x Y x a M x t
x
d ,t
(18)
Finally, coming back to (8), we take into account (16) and (17) and write the second kind Fredholm type
inte-gral equation [8] for which the boundary value problem (1)-(2) is reduced to:
,b
a
y A
B t y t (19) where
,2
1 1
1 ,2 1
1
0 1 0 1
,
1 1
1 ,
1
0 1 0 1
1 1
1 1
m k m s
m m m m
s m s k k m s s m s
k
s k s k
m k m s b
m m m m
s m s k k m s s m s
k
s k s k a
d
A Y b Y b F
d
Y a Y a F F x Y x x
d ,
1296 M. FATEMI ET AL.
,2
1 1
1 ,2
1
0 1 0 1
,
1 1
1 ,
1
0 1 0 1
1 2
, = 1 1
1 ( ) 1
, d , d .
m k m s
m m m m
s m s k k m s s m s
k k
s k s k
m k m s
m m m m
s m s k k m s s m s
k k
s k s k
b b
a a
d
B t Y b H t Y b M t
d
Y a H t Y a M t
Y x M x t x Y x M x t x
(21)
By the hypothesis of theorem on the functions f x
,
,j
K x t , j = 1, 2 and H ti
, i= 1,m the integralEquation (19) has unique solution and so in all con-ducted operations we can come back and we conclude that the solution of (19) is the unique solution of boundary value problem (1)-(2).
3. References
[1] D. Nazari and S. Shahmorad, “Application of Fractional Differential Transform Method to the Fractional Order Integro-Differential Equations with Nonlocal Boundary Conditions,” Journal of Computational and Applied Mathematics, Vol. 234, No. 3, 2010, pp. 883-891.
doi:10.1016/j.cam.2010.01.053
[2] S. G. Samko, A. A. Kilbas and O. I. Marichev, “Frac-tional Integrals and Derivatives,” Theory and Applica-tions, Cordon and Breach, Yverdon, 1993.
[3] C. J. Tranter, “Integral Transforms in Mathematical Phys-ics,” Methuen, London and New York, 1949.
[4] V. S. Vladimirov, “Equation of Mathematical Physics,” Mir Publication, Moscow, 1984.
[5] G. E. Shilov, “Mathematical Analysis. The Second Spe-cial Course,” Nauka, Moscow, 1965.
[6] S. M. Hosseini and N. A. Aliev, “Sufficient Conditions for the Reduction of a BVP for PDE with Non-Local and Global Boundary Conditions to Fredholm Integral Equa-tions (on a Rectangular Domain),” Applied Mathematics and Computation, Vol. 147, No. 3, 2004, pp. 669-685. [7] F. Bahrami, N. Aliev and S. M. Hosseini, “A Method for
the Reduction of Four Imensional Mixed Problems with General Boundary Conditions to a System of Second Kind Fredholm Integral Equations,” Italian Journal of Pure and Applied Mathematics, No. 17, 2005, pp. 91- 104.
