J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012)
Principle of Contraction Mapping on Quadratic Integral Equation B. D. KARANDE1, F. I. MOMIN2 and S. HANDIBAG3 1
Department of Mathematics, M. U. Mahavidyalata, Udgir-413517, Maharashtra, INDIA. 2 Departments of Mathematics, Milliya Arts and Science College, Beed-413517, Maharashtra, INDIA. 3 Departments of Mathematics, S. M. B. Mahavidyalata, Latur-413517, Maharashtra, INDIA. (Received on: July 30, 2012) ABSTRACT In this paper, we the existence of unique solution of a quadratic integral equation is proved by principle of contraction mapping and the Classical method of successive approximations (Picard method) which consists the construction of a sequence of functions such that the limit of this sequence of functions in the sense of uniform convergence is the solution of a quadratic integral equation. Keywords: quadratic integral equation, Principle of contraction mapping, Picard method, continuous unique solution, convergence analysis. Mathematical Subject Classification: 38B20, 44B20.
1. INTRODUCTION Quadratic integral equations are often applicable in the theory of radioactive transfer, kinetic theory of gases, in the theory of neutron transport and in the traffic theory. The quadratic integral equations can be very often encountered in many applications1-15.
Given a Closed and bounded interval J= [0, 1] of the real line R , Consider the nonlinear quadratic integral equation t
x ( t ) = q ( t ) + f ( t , x ( t ) ) ∫ g ( t , s , x ( s ) ) ds 0
(1.1) for all t ∈ J
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B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012)
where q : J → R+ = [ 0, +∞ ) , x : J → R+ ,
f : J × R+ → R+ , g : J × J × R+ → R+ By a solution of the quadratic integral equation (1.1) we mean a function x ∈ C ( J , R+ ) that satisfies (1.1) on J, where
C ( J , R+ ) is the space of all
continuous function x : J → R+ . In recent years, the topic of nonlinear quadratic integral equations is received the attention of several authors and at present, there is a considerable literature available in this direction. See C. Cordunean9, A. M. A. El-Sayed, H. H. G. Hashen1-3,8 and there references therein. In this paper we will prove the existence and the uniqueness of continuous solution for nonlinear quadratic integral equation (1.1) by using principle of contraction mapping and Picard method for proving our existence results. 2. AUXILIARY RESULTS In this section, we will see general principle, called the Contraction Mapping Principle or Banach’s Fixed Point point theorem. But first, we need some preparation. Definition 2.1: Let (X, d) be a metric space. A mapping f : X → X is Called a
Contraction if there exists a number α < 1 such that d f ( x ) , f ( y ) ≤ α d ( x, y )
(
for all x, y ∈ X .
)
Remark 2.1: A Contraction is automatically Lipschitz Continuous, but the condition is stronger than Lipschitz continuity, because we require that α < 1 .
Definition 2.2: X is a set and f : X → X a mapping. A point x ∈ X is called fixed point of f if f ( x ) = x Theorem 2.1: (Contraction Mapping Theorem). Let (X, d) be a complete metric space and f : X → X a contraction. Then there exists a unique fixed point of f. Proof: Choose an arbitrary point x1 ∈ X and define the sequence
{ xn}n∈N
recursively as follows:
xn +1 = f ( xn ) , n=1, 2, 3… We claim that this is a Cauchy sequence. In order to prove the claim, note first that we have a number α < 1 such that for all n ∈ N ,
d ( xn +2 , xn +1 ) = d ( f ( xn +1 ) , f ( xn ) ) , Because f is a contraction. Defining
c = d ( x2 , x1 ) , we obtain d ( x3 , x2 ) ≤ cα d ( x4 , x3 ) ≤ cα 2 :
d ( xn +1 , xn ) ≤ cα n −1 : : If m, n ∈ N with n < m , then
d ( xm , xm ) ≤ d ( xn , xn+1 ) + d ( xn+1, xn+2 ) + ..... + d ( xm−1, xm )
Journal of Computer and Mathematical Sciences Vol. 3, Issue 4, 31 August, 2012 Pages (422-497)
B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012)
≤ cα
n −1
+ cα
n
≤c
+ ..... + c α ∞
∑
αk =
k = n −1
m−2
466
It is a unique fixed point, because for every other fixed point y0 ∈ X , we have
cα . 1−α n −1
In the last step we have used the fact that α < 1 (but α ≥ 0 , because the inequality in definition 2.1 cannot be true for a negative number). For the same reason, we have
d ( x0 , y0 ) = d ( f ( x0 ) , f ( y0 ) ) ≤αd ( x0 , y0 ) . Because α < 1 , it follows that d ( f ( x0 ) , x0 ) = 0 , and so x0 = y0 .
cα n −1 → 0 as n → ∞ . 1−α
3. MAIN THEOREM
Hence for any ε > 0 , there exists an N such
In this section, we shall state and prove our main results.
cα n−1 < ε for n ≥ N . 1−α For m, n ≥ N , we d ( xm , xn ) < ε .
that
then
have
Thus { xn }n∈N is a Cauchy sequence.
Our first result is based on the following assumptions.
( H0 )
q : J → R + = [0, + ∞ ]
is continuous
on J, where J = [ 0 , 1]
Because we work in a complete metric space, every Cauchy sequence converges. In particular, there exists a point x0 ∈ X such
( H1 )
that xn → x0 as n → ∞. Now for n ∈ N ,
there exist Positive constant M1 and M2 such that f ( t, x ) ≤ M1 and g ( t, s, x ( s ) ) ≤ M 2 on R .
d ( f ( x0 ) , x0 ) ≤ d ( f ( x0 ) , xn +1 ) + d ( xn +1 , x0 ) = d ( f ( x 0 ) , f ( x n ) ) + d ( x n +1 , x 0 )
≤ α d ( x 0 , x n ) + d ( x n +1 , x 0 ) . The right- hand side converges to 0 as n → ∞ , and it follows that
d
( f ( x ), x ) = 0 0
0
Therefore, we have f ( x0 ) = x0 .that is, the point x0 is a fixed point.
f : J × D ⊂ R+ → R+ and
g : J × J × D ⊂ R+ → R+ are continuous and
( H3 )
f , g satisfy Lipschitz condition with
Lipschitz constants L1 and L2 such that
f ( t , x ) − f ( t , y ) ≤ L1 x − y g ( t , s, x ( s ) ) − g ( t , s, y ( s ) ) ≤ L2 x − y Theorem 3.1: Assume that hypotheses
( H 0 ) - ( H 3 ) hold. Further if
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B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012)
{
}
h = ( L1M 2 + L2 M 1 ) < 1 , then the nonlinear
S = x ∈ C : x ( t ) − q ( t ) ≤ K , K = M 1M 2 .
quadratic integral equation (1.1) has a unique positive solution x ∈ C . Proof: Let C = C ( J , R ) be the space of all
Then the operator A map S in to S, since for x ∈ S
x (t ) − q (t ) ≤ f (t , x (t ))
real valued functions which are continuous on J.
t
∫ g ( t , s, x ( s ) ) ds 0
Define the operator A as
t
≤ M 1 M 2 ∫ ds = M 1 M 2 t = M 1 M 2
t
Ax ( t ) = q ( t ) + f ( t , x ( t ) ) ∫ g ( t , s , x ( s ) ) ds
0
0
∀ x ∈C
Moreover it is easy to see that S is a closed subset of C. In order to show that A is a contraction we compute
Now define a subset S of C as t
t
0
0
Ax ( t ) − Ay ( t ) = f ( t , x ( t ) ) ∫ g ( t , s , x ( s ) )ds − f ( t , y ( t ) ) ∫ g ( t , s , y ( t ) )ds t
t
0
0
= f ( t , x ( t ) ) ∫ g ( t , s , x ( s ) )d s − f ( t , y ( t ) ) ∫ g ( t , s , y ( s ) )d s t
t
+ f ( t , x ( t ) ) ∫ g ( t , s , y ( s ) )d s − f ( t , x ( t ) ) ∫ g ( t , s , y ( s ) )d s 0
0 t
= f ( t , x ( t ) ) − f ( t , y ( t ) ) ∫ g ( t , s , y ( s ) )ds 0
+ f ( t , x ( t ) ) ∫ g ( t , s , x ( s ) ) − g ( t , s , y ( s ) ) ds 0 t
A x (t ) − A y (t ) ≤ f (t , x (t )) − f (t , y (t ))
t
∫ g (t , s , g ( s )) d s 0
+ f (t , x (t ))
t
∫ g ( t , s , x ( s ) ) − g ( t , s , y ( s ) ) ds 0 t
≤ L1 M 2 x − y + L 2 M 1 ∫ x ( s ) − y ( s ) ds 0
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B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012)
Ax ( t ) − Ay ( t ) =
M t∈ J
Ax ( t ) − Ay ( t )
≤ L1 M 2 x − y + L 2 M 1 x − y
≤ ( L1 M 2 + L 2 M 1 ) x − y ≤h x− y Since h = ( L1 M 2 + L 2 M 1 ) < 1 Then A is a Contraction and A has a unique fixed point in S, thus there exists a unique solution for (1.1) 4. METHOD OF SUCCESSIVE APPROXIMATIONS (PICARD METHOD) Applying Picads method to the quadratic integral equation (1.1), the solution is constructed by the sequence t
x n ( t ) = q ( t ) + f ( t , x n −1 ( t ) ) ∫ g ( t , s , x n −1 ( s ) )ds
n=1, 2, 3, 4, ….
0
x0 (t ) = q (t )
(4.1)
All the functions xn ( t ) are continuous functions and xn can be written as a sum of successive differences
x n = x 0 + ∑ ( x j − x j −1 ) n
j =1
This means that convergence of the sequence xn is equivalent to convergence of the infinite series
∑(x
j
− x j −1 ) and the solution will be x ( t ) = limxn ( t )
That is if the infinite series
∑(x
n →∞
j
− x j −1 ) converges, then the sequence xn ( t ) will
converge to x (t). Step I: To prove the uniform convergence of
{ x ( t )} n
, we shall consider the associated
∞
series
∑ x ( t ) − x ( t ) n =1
n
n −1
From (4.1) for n=1 we get t
x1 ( t ) − x 0 ( t ) = f ( t , x 0 ( t ) ) ∫ g ( t , s , x 0 ( s ) )ds 0
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B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012) t
x1 ( t ) − x0 ( t ) ≤ M 1M 2 ∫ ds = M 1M 2t
And
(4.2)
0
Now, we shall obtain an estimate for xn ( t ) − xn −1 ( t ) , n ≥ 2 t
xn ( t ) − xn −1 ( t ) ≤ q ( t ) + f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds 0 t
− q ( t ) + f ( t , xn − 2 ( t ) ) ∫ g ( t , s, xn − 2 ( s ) )ds 0 t
t
≤ f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds ≤ f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds 0
0 1
1
0
0
− f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds + f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds t
≤ f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) ) − g ( t , s, xn − 2 ( s ) ) ds 0 t
+ f ( t , xn −1 ( t ) ) − f ( t , xn − 2 ( t ) ) ∫ g ( t , s, xn − 2 ( t ) ) ds Using assumption ( H1 ) and ( H 3 ) we get
0
t
t
0
0
xn ( t ) − xn −1 ( t ) ≤ M 1 L2 ∫ xn −1 ( s ) − xn − 2 ( s ) ds + M 2 L1 xn −1 ( t ) − xn − 2 ( t ) ∫ ds Put n=2, then using (4.2), we get t
x2 ( t ) − x1 ( t ) ≤ M 1 L2 ∫ x1 ( s ) − x0 ( s ) ds + M 2 L1 x1 ( t ) − x0 ( t ) 0
2
t + M 1M 2 2 L1t 2 2 1 ≤ M 1M 2 M 1 L2 + M 2 L1 t 2 2
x2 − x1 ≤ M 12 M 2 L2
t
t
0
0
x3 − x2 ≤ M 1 L2 ∫ x2 ( t ) − x1 ( t ) ds + M 2 L1 x2 ( t ) − x1 ( t ) ∫ ds 1 1 ≤ M 1 L2 ∫ M 1M 2 M 1 L2 + M 2 L1 t 2 ds + M 2 L1 M 1M 2 M 1M 2 + M 2 L1 t 2 ∫ ds 2 2 0 0 t
t
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B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012)
470
3 1 t 1 ≤ M M 2 L2 M 1 L2 + M 2 L1 + M 1M 22 L2 M 1 L2 + M 2 L1 t 3 2 3 2 1 1 ≤ M 1M 2 M 1 L2 + M 2 L1 M 1 L2 + M 2 L1 t 3 2 3 2 1
Repeating this technique, we obtain the general estimate for the terms of the series.
1 1 1 xn − xn −1 ≤ M 1M 2 M 1 L2 + M 2 L1 M 1 L2 + M 2 L1 × ....... × M 1 L2 + M 2 L1 t n 2 3 n 1 1 1 ≤ M 1M 2 M 1 L2 + M 2 L1 M 1 L2 + M 2 L1 × .... × M 1 L2 + M 2 L1 2 3 n ( Q 0 ≤ t ≤1 ) ≤ ( M 1 L2 + M 2 L1 )( M 1 L2 + M 2 L1 ) × .... × ( M 1 L2 + M 2 L1 ) n times. ( Q M 1M 2 ≤ ( M 1 L2 + M 2 L1 ) )
≤ ( M 1 L2 + M 2 L1 )
n
∞
Since ( M 1 L2 + M 2 L1 ) < 1 , then the uniform convergence of
∑ x ( t ) − x ( t ) is proved n =1
{ } f ( t , x ) and g ( t , s, x ( t ) ) are continuous in x, then
and so the sequence xn ( t ) is uniformly convergent. Since
n
n −1
t
x ( t ) = lim f ( t , x n ( t ) ) ∫ f ( t , s , x n ( t ) )ds n→∞
0
t
= f ( t , x ( t ) ) ∫ f ( t , s, x ( s ) ) ds 0
Thus, the existence of a solution is proved. Step II: To prove the uniqueness, let y ( t ) be a continuous solution of (1.1). Then t
y ( t ) = q ( t ) + f ( t , y ( t ) ) ∫ g ( t , s, y ( t ) ) ds
t ∈ [ 0,1]
0
And t
t
o
0
y ( t ) − xn ( t ) ≤ f ( t , y ( t ) ) ∫ g ( t , s, y ( s ) ) ds − f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) ) ds Journal of Computer and Mathematical Sciences Vol. 3, Issue 4, 31 August, 2012 Pages (422-497)
471
B. D. Karande, et al., J. Comp. & Math. Sci. Vol.3 (4), 464-472 (2012) t
t
0 t
0
− f ( t , y ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds + f ( t , y ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds = f ( t , y ( t ) ) ∫ g ( t , s, y ( s ) ) − g ( t , s, xn −1 ( s ) ) ds 0 t
+ f ( t , y ( t ) ) − f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) ) ds Using assumption ( H1 ) and ( H 3 ) , we get
y ( t ) − xn ( t ) ≤ f ( t , y ( t ) )
0
t
∫ g ( t, s, y ( s ) ) − g ( t, s, x ( s ) ) ds n −1
0 1
+ f ( t , y ( t ) ) − f ( t , xn −1 ( t ) ) ∫ g ( t , s, xn −1 ( s ) )ds 0 t
t
0
0
≤ M 1 L2 × ∫ y ( s ) − xn −1 ( s ) ds + M 2 L1 y ( t ) − xn −1 ( t ) × ∫ ds
(4.3)
y ( t ) − q ( t ) ≤ M 1 M 2t
But And using (4.3)
y ( t ) − xn ( t ) ≤ ( M 1 L2 + M 2 L1 ) .
Hence
lim x ( t ) = y ( t ) = x ( t ) ,
n
n→ ∞
n
Which completes the proof. When g ( t , x ) = 1 , then M 1 = 1 and L1 = 0 and we obtain the original Picard theorem. Corollary 4.1: Let the assumptions of theorem 1 (with g ( t , x ) = 1 ) be satisfied. If L2 < 1 , t
∫ (
)
then the integral equation x ( t ) = x0 ( t ) + g t , s, x ( s ) ds has a unique continuous solution. 0
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