Kt2418201822

K.L. Bondar, A.B. Jadhao, S.T. Patil / International Journal of Engineering Research and
Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 2, Issue4, July-August 2012, pp.1820-1822
Existence Of Solutions For Nonlinear Difference Equations With
Nonlocal Conditions
K.L. Bondar 1, A.B. Jadhao 2 and S.T. Patil3
1
P.G. Dept. of Mathematics, N.E.S. Science College, Nanded – 431 605 (MS) India
2
DSM College, Parbhani (MS) INDIA
3
D.Y. Patil College of Engineering, Akurdi, Pune INDIA

Abstract 2. Preliminary Notes and Hypothesis
Existence of mild solutions of initial Let E be a real Banach space and
value problem for nonlinear difference equation J = {0,1, 2,..., N}, C(J, E) denote the Banach space
with nonlocal conditions is obtained. The proof of functions y : J→E equipped with the norm
relay on the fixed point theorem for compact
mapping. ||y||∞ = sup{||y(t)|| : t  J}
AMS Subject Classification: 39A10; 47A10.
and B(E) denotes the Banach space of bounded
Keywords: Nonlocal condition; Mild solution; linear operators from E into E with norm
Fixed point..
|| T ||B(E) = sup{|| T (y) || : || y || = 1}.
1. Introduction:
Due to wide application in many fields Let us list the following hypothesis:
such as science, economics, neural network,
ecology, the theory of nonlinear difference (H1) A : J × E → B(E), (t, v) : → A(t, v) is a
equations has been widely studied since 1970; see, continuous function with respect to v such that
for example, [1, 3, 15, 16]. At the same time, for r > 0 there exists r1 > 0 such that
boundary value problems and initial value problems
of difference equations have received much ||v|| ≤ r → ||A(t, v)|| B(E) ≤ r1
attentions from many authors; see
[2,4,7,9,10,13,19]. for all t  J and v E.
In this paper, we consider an initial value
problem for a nonlinear difference equation with (H2) f : J × E ! E, (t; v) :! f(t;
nonlocal initial conditions. More precisely we
consider the IVP v) is a continuous function with
respect to v.
y(t )  A(t , y(t )) y(t )  f (t , y(t )) (1.1)
(H3) A function g : C (J, E) × E is continuous and
there exists a constant L > 0 such that
y (0)  g ( y )  y0 (1.2)
|| g(y) || ≤ L for each y  E.

where t  J  {0, 1, 2,...,N } , A, f : J × E → E are (H4) There exists constant K such that
two continuous functions, g : C(J, E)→E, y0  E || f (t, u) || ≤ K for all t  J, u E .
and E is a real Banach space with the norm || . ||. The
existence of solutions for IVP (1.1)-(1.2) is obtained For each u  C(J, E), define a function
by using the classical fixed point theorem for Uu : J × J → B(E) such that
compact map due to Smart [18].
Such problems with classical initial t 1

conditions have been studied by Anichini [5], U u (t , s )  I   Au ( w)U u ( w, s ) (2.1)
Anichini and Conti [6], Conti [11], Conti and w s
Iannacci [12], Kartsatos [14], Mario and Pietramala where I stands for an identity operator on E and
[17]. Au(t) = A(t, u(t)).
The nonlocal conditions, which are From (2.1), one has
generalization of the classical initial conditions was
motivated by physical problems. The pioneering U u (t , s)  I , s  t, u  E (2.2)
work on nonlocal conditions is due to Byszewski and
[8].

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∆ Uy(t, s) = A(t, y(t))Uy(t, s).
 tU u (t , s)  A(t , u (t ))U u (t , s) (2.3) Put
t
In the list of hypothesis, assume z (t ) =  U y (t , s ) f ( s  1, y ( s  1)) .
s 1
(H5) M = sup{||Uy(t, s)||B(E) : (t, s)  J × J , Therefore,
y  B(E)}. t
Remark 2.1 From (H1), it follows that for z (t ) = { tU y (t , s) f (s  1, y(s  1))}  U y (t  1, t  1) f (t , y(t ))
u  C(J, E), Au  C(J, B(E)). s 1

Remark 2.2 Suppose {un} is a sequence in C(J, E)
converging to u*  C(J, E). Then (H1) implies t
Aun → Au* i.e. A(t, un(t)) → A(t, u*(t)) for all t  J. = A(t , y (t ))  U y (t , s) f ( s  1, y ( s  1))  f (t , y (t )).
s 1
Now we prove the following lemma.
Lemma 2.1 If (H1) holds then for each u  C(J, E); Thus equation (2.5) reduces to
Uu : J × J → B(E) defined by (2.1) is continuous
with respect to u. y(t) = Uy(t, 0)y0 - Uy(t, 0)g(y) + z(t).
Proof: From (2.1) by putting t = s+1 and using
Uu(s, s) = I, we get Therefore,

Uu(s+1, s) = I + A(s, u(s))Uu(s, s) ∆ y(t) = ∆ Uy(t, 0)y0 - ∆ Uy(t, 0)g(y) + ∆ z(t)
= I +A(s, u(s)). = A(t, y(t))y(t) + f(t, y(t)).
For t = s+2, This completes the proof.
Uu(s+2, s) = I+A(s, u(s))Uu(s, s)+ A function y(t) given by (2.5) is called mild
A(s+1, u(s+1))Uu(s+1, s) solution of IVP (1.1)-(1.2).
= (I + A(s, u(s)))(I + A(s+1, u(s+1))). 3. Existence Theorems:
Continuing the process, we obtain In this section we establish the existence of solution
of IVP (1.1)-(1.2).
t 1
U u (t , s )   ( I  A( w, u ( w)))
The following Lemma is crucial in the proof of main
(2.4) theorem.
w s
Lemma 3.1: (18) Let X be a Banach space and let
Now, suppose {un} is a sequence in C(J, E) T : X → X be a continuous compact map. If the set
converging to u*  C(J, E). From (2.4), one has Ω = {y  X : λy = T(y), for some λ >1}
t 1
U un (t , s )   ( I  A( w, u n ( w)))
is bounded, then T has a fixed point.
Theorem 3.1 : Assume that hypothesises (H1)-(H5)
w s
are satisfied. Then the problem (1.1)-(1.2) has at
and least one mild solution on J.
t 1
U u* (t , s )   ( I  A( w, u* ( w)))
Proof: We transform the problem (1.1)-(1.2) into a
fixed point problem. Consider the mapping
w s
T : C(J, E) → C(J, E) defined by
The conclusion follows from (H1) and Remark 2.2.

Theorem 2.1: A function y  C(J, E) given by (Ty)(t ) = U y (t , 0) y0 - U y (t , 0) g ( y) 
t . (3. 1)
y(t) = U y (t , 0) y0 - U y (t , 0) g ( y)   U (t, s) f (s  1, y(s  1))
s 1
y

t (2.5)
U (t, s) f (s  1, y(s  1))
It is clear that the fixed points of T are mild
y solutions of (1.1)-(1.2).
s 1
We shall show that T is a continuous
is a solution of IVP (1.1)-(1.2). compact mapping. The continuity of T follows from
Proof: It follows from (2.2) and (2.5) that Lemma 2.1 and hypothesises (H2), (H3). Now we
prove that T maps bounded sets into relatively
y(0) = Uy(0, 0)y0-Uy(0, 0)g(y) compact sets. i.e. T is a compact mapping. Let
i.e. y(0) + g(y) = y0.
Br = {y  C(J, E) : ||y|| ≤ r}.
From (2.3), we have
Then for each t  J and y  Br, we have
∆ Uy(t, 0) = A(t, y(t))Uy(t, 0)

and

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(Ty)(t ) = U y (t , 0) y0 - U y (t , 0) g ( y) 
t .
 U y (t, s) f (s  1, y(s  1))
s 1
By (H3)-(H5) we have
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