Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2012, Article ID 489353,12 pages doi:10.1155/2012/489353

Research Article

Monotonic Positive Solutions of Nonlocal Boundary Value Problems for a Second-Order Functional Differential Equation

A. M. A. El-Sayed,1 E. M. Hamdallah,1 and Kh. W. El-kadeky2

1 Faculty of Science, Alexandria University, Alexandria, Egypt

2 Faculty of Science, Garyounis University, Benghazi, Libya

Correspondence should be addressed to Kh. W. El-kadeky, k_welkadeky@yahoo.com Received 13 October 2011; Accepted 5 December 2011 Academic Editor: Istvan Gyori

Copyright © 2012 A. M. A. El-Sayed et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study the existence of at least one monotonic positive solution for the nonlocal boundary value problem of the second-order functional differential equation x"(t) = f (t,x($(t))), t € (0,1), with the nonlocal condition ^^=1 akx(Tk) = x0, x'(0) + ^n=i bjx'(n.j) = x1, where Tk € (a,d) c (0,1), nj € (c,e) c (0,1), and x0,x; > 0. As an application the integral and the nonlocal conditions jd x(t)dt = x0, x'(0) + x(e) - x(c) = x; will be considered.

1. Introduction

The nonlocal boundary value problems of ordinary differential equations arise in a variety of different areas of applied mathematics and physics.

The study of nonlocal boundary value problems was initiated by Il'in and Moiseev [1, 2]. Since then, the non-local boundary value problems have been studied by several authors. The reader is referred to [3-22] and references therein.

In most of all these papers, the authors assume that the function f : [0,1] x R+ ^ R+ is continuous. They all assume that

!• f X 0

lim-= 0 or œ,

X ^œ X

1- f (X) 0

lim-= 0 or œ.

x ^ 0 X

These assumptions are restrictive, and there are many functions that do not satisfy these assumptions.

Here we assume that the function f : [0,1] x R+ ^ R+ is measurable in t e [0,1] for all x e R+ and continuous in x e R+ for almost all t e [0,1] is and there exists an integrable function a e L1 [0,1] and a constant b> 0 such that

|f (t,x)| < |a(t)| + b|x|, ^(t,x) e [0,1] x D. (1.2)

Our aim here is to study the existence of at least one monotonic positive solution for the nonlocal problem of the second-order functional differential equation

x"(t)= f (t,x($(t))), t e (0,1), (1.3)

with the nonlocal condition

^akx(rk) = x0, x'(0) + ^ bjx'( j = xi, (1.4)

k=1 j=1

where Tk e (a,d) c (0,e (c,e) c (0,1), and x0,xi > 0.

As an application, the problem with the integral and nonlocal conditions

x(t)dt = x0, x'(0) + x(e) - x(c) = x1, (1.5)

is studied.

It must be noticed that the nonlocal conditions

x(t)= x0, t e (a,d) , x'(0) + x'i^ = x1, n e (c,e),

^akx(Tk) = 0, Tk e (a, d), x'(0)^ bj

x\HÙ = ^ n e (c,e)'

k=l j=l ( d

x(t)dt = 0, x'(0) + x(e) = x(c)

are special cases of our the nonlocal and integral conditions.

2. Integral Equation Representation

Consider the functional differential equation (1.3) with the nonlocal condition (1.4) with the following assumptions.

(i) f : [0,1] x R+ ^ R+ is measurable in t e [0,1] for all x e R+ and continuous in x e R+ for almost all t e [0,1] and there exists an integrable function a e L1 [0,1], and a constant b> 0 such that

|f (t,x) | < |a(t)| + b|x|, V(t,x) e [0,1] x D. (2.1)

(ii) < : (0,1) ^ (0,1) is continuous.

(iii) b< 1/(3 - B), B = (j bj +1)-1.

^ak> 0, Vk = 1,2,...,m, > 0, Vj = 1,2,...,n. (2.2)

k=1 j=1

Now, we have the following Lemma. Lemma 2.1. The solution of the nonlocal problem (1.3)-(1.4) can be expressed by the integral equation

x(t) = ^xo - ^akj (Tk - s)f (s,x«(s)))ds

■B[ t - Ag akT^j |x1 - Yb^l f (s,x($(s))) ds ^ (2.3)

f (t - s)f (s,x(<(s)))ds, Jo

where A = (£m=1 ak)-1, B = (£n=1 bj + 1)-1. Proof. Integrating (1.3), we get

x'(t) = x'(0) + f f (s,x(<(s)))ds. (2.4) Jo

Integrating (2.4), we obtain

x(t) = x(0) + x'(0)t + Í (t - s)f (s,x(<(s)))ds. (2.5) Jo

Let t = Tk, in (2.5), we get

m n n m f Tk

^akx(Tk ) = X akx(0) + ^ akTkx' (0) + ^ ak I (Tk - s)f (s,x«(s))) ds, (2.6) k=1 k=1 k=1 k=1

and we deduce that

i m m fTk | fm\

x(0) = aJxo flfcTfcx'(0) ak\ (Tk - s)f(s,x($(s))) ds\, A =( ^k) . (2.7)

I k=1 k=1 J0 J \k=1 /

Substitute from (2.7) into (2.5), we obtain

{m T 1 / m

x0 - Vak \ (Tk - s)f(s,x($(s)))ds> + x'(0)( t - AS\akTk

k=i Jo J \ k=i

+ f (t - s)f (s,x($(s)))ds.

Let t = n, in (2.4), we obtain

n n n fnj

Ä^fa;) = X bjx'(0) + £ bj\ f(s,x($(s))) ds, j=1 j=1 j=1 Jo

n n f n

x1 - x'(0) = x'(0)^bj + ^bj\ f(s,x($(s)))ds, j=1 j=1 0

and we deduce that

(2.10)

x'(0) = Byx1 - ^kbjj (s,x($(s)))dsj, B = ^ ^bj + 1 Substitute from (2.10) into (2.8), we obtain

im fTk

xoak \ (Tk - s)f(s,x($(s))) ds

k=1 Jo

t - AgakTk^ |x1 - Ybij(s,x($(s)))ds(2.11)

f (t - s)f (s,x($(s)))ds,

which proves that the solution of the nonlocal problem (1.3)-(1.4) can be expressed by the integral equation (2.3). □

3. Existence of Solution

We study here the existence of at least one monotonic nondecreasing solution x e C[0,1] for the integral equation (2.3).

Theorem 3.1. Assume that (i)-(iv) are satisfied. Then the nonlocal problem (1.3)-(1.4) has at least one solution x e C [0,1 ].

Proof. Define the subset Qr c C(0,1) by Qr = {x e C : |x(t)| < r,r = (Axo + Bx1 + (3 -B)||a||)/(1 - (3 - B)b),r > 0}. Clear the set Qr which is nonempty, closed, and convex. Let H be an operator defined by

im rrk

x0ak\ (rk - s)f(s,x($(s))) ds

k=1 J0

m \ r n ftyj ^

■B(t - A g UkTkJ jxx - X byJo f(s,x(0(s))) dsj (3.1)

+ i (t - s)f (s,x($(s)))ds.

Let x e Qr, then

{m r-Tk 1

x0 + EakjQ (Tk - s)\f(s,x(0(s))) |ds

t - A^akTk)\x1 + jjb; f \f (s,x(^(s)))\ds L V k=1 / { j=1 J° j

+ f (t - s)\f (s,x(<£(s)))|ds J0

a|x0 + jjakj [|a(s)| + b\x($(s))\]dsj

+ bJ xx + jrb; I" [|a(s)| + b\x(#(s))\]ds

[ ;=1 j0

+ I" [|a(s)| + b\x(^(s))\]ds J0

< Ax0 + ||a|| + bsup\x($(t)) \ + Bxi + B^bj||a||

teI j=1

+ bB^bj sup\x(0(t)) \ + ||a|| + b sup\x(0(t)) \

;=1 teI teI

< Axo + Bxi + 2\\a\\ + 2b\\x\\ + (1 - B)\\a\\ + b(1 - B)\\x\\

< Ax0 + Bx1 + (3 - B)\\a\\ + (3 - B)br < r,

then H : Qr ^ Qr and (Hx(t)} is uniformly bounded in Qr. Also for t1, t2 e [0,1] such that t1 < t2, we have

(Hx)(t2) - (Hx)(h) = B(t2 - Amakrk)\ x1 - jtbj P f (s,x($(t)))ds I

\ k=1 / [ j=1 J° j

+ (t2 - s)f(s,x(0(t)))ds

t1 - AmakTk)\x1 - ^bj f f (s,x(tp(t)))ds\ K k=1 J y j=1 Jo j

- f1 (t1 - s)f (s,x($(t)))ds (3.3)

B(t2 - t1)\ x1 - fbj Cf (s,x($(t)))ds I [ j-1 Jo j

+ (t2 - t1 )f(s,x(4>(t)))ds

+ (t2 - s)f(s,x($(t)))ds.

|(Hx)(t2) - (Hx)(t1)| < B|t2 - t1\\ x1 + fbj f1 [|a(s)| + b|x(0(s)) |]ds I

I ^ Jo j

+ |t2 - t1M [|a(s)| + b|x(#(s)) |]ds

+ (t2 - s) [|a(s)| + b|x(0(s)) |]ds

< B\t2 - tilxi + £bj[\\a\\ + br]

+ \t2 - i1\[yay + br] + i \\a\\ds + br[t2 - ti]

The above inequality shows that

|(Hx)(t2) - (Hx)(t1)| —> 0 as t2 t1. (3.5)

Therefore {Hx(t)} is equicontinuous. By the Arzela-Ascoli theorem, {Hx(t)} is relatively compact.

Since all conditions of the Schauder theorem hold, then H has a fixed point in Qr which proves the existence of at least one solution x e C[0,1] of the integral equation (2.3), where

^limx(f) = A j xo - g ak J (rk - s)/(s,x($(s))) ds

m I n BAy\akrk< xi - Vbj \ /(s,x($(s)))ds > = x(0),

k=i I j=i Jo

lim x(t) = A x0 - V ak (rk - s)/(s,x($(s))) ds\ (3.6)

t^i- I k=i Jo I

■B( i - AYakTk) \ xi - Yb; f'/(s,x($(s)))ds I

k=i / | j=i jo j

I" (i - s)/(s,x($(s)))ds = x(i).

To complete the proof, we prove that the integral equation (2.3) satisfies nonlocal problem (1.3)-(1.4). Differentiating (2.3), we get

x'(t) = bJ xi - Yb; f'/(s,x(^(s)))ds I + I" /(s,x(^(s)))ds, (3.7)

,=i Jo Jo

]Tb' f/(s,x($(s)))ds I + f.

^ j=i Jo j Jo

x"(t)= / (t,x($(t))). (3.8)

Let t = Tk in (2.3), we obtain

im fTk 1 fTk

xo -Y. ak\ (Tk - s)f (s,x($(s))) ds\ + (Tk - s)f (s,x($(s))) ds, (3.9)

k=1 Jo J Jo

which proves

£akx(Tk) = xo. (3.10)

Also let t = ni in (3.7), we obtain

x!(Vj]) = bJ x1 - ^Lbjjn,f (s,x($(s)))dsi +jH'f (s,x($(s)))ds, (3.11)

" " [ " fn, ] " fn,

Ybxin,) = B^bA x1 - £b, f (s,x(0(s)))ds\ + £b, f (s,x(0(s)))ds. (3.12)

1=1 1=1 y 1=1 J0 j 1=1 J0

Let t = 0 in (3.7), we obtain

x'(0) = bJ x1 - Yb1 C' f (s,x($(s)))ds\. (3.13)

y 1=1 h !

Adding (3.12) and (3.13), we obtain

x'(0) + £ bx'i 1 = x1. (3.14)

This implies that there exists at least one solution x e C[0,1] of the nonlocal problem (1.3) and (1.4). This completes the proof. □

Corollary 3.2. The solution of the problem (1.3)-(1.4) is monotonic nondecreasing. Proof. Let t1 < t2, we deduce from (2.3) that

Îm çrk

xoUk\ (Tfc - s)f(s,x($(s))) ds

k=i Jo

B\ ti - AZakrk

n Cn> 1

jxi- Zbi)o f(s,x(0(s)))ds j fh

+ (t1 - s)f (s,x(0(s)))ds

{m fTk

xo - ^akj (Tk - s)f(s,x($(s)))ds

m \ | n f n j 1

t2 - A^akTk)! xi bj f(s,x($(s)))ds ^

k=i / [ j=i J0 j

+ (t2 - s)f (s,x(0(s)))ds = x(t2),

(3.i5)

which proves that the solution x of the problem (i.3)-(i.4) is monotonic nondecreasing. □ 3.1. Positive Solution

Let bj = 0, j = i, 2,...n and xi = 0, then the nonlocal problem condition (i.4) will be

J^akx(Tk ) = xo, x'(0) = 0. (3.i6)

Theorem 3.3. Let the assumptions (i)-(iv) of Theorem 3.1 be satisfied. Then the solution of the nonlocal problem (1.3)—(3.16) is positive t e [d, 1].

Proof. Let bj = 0,j = 1,2,...n and x1 = 0 in the integral equation (2.3) and the nonlocal condition (1.4), then the solution of the nonlocal problem (1.3)—(3.16) will be given by the integral equation

m çTk çt

x(t) = A\x0-Y.ak\ (Tk - s)f(s,x($(s)))ds\ + (t - s) f (s,x($(s)))ds, (3.i7)

I k=i J0 I J0

where A = (^i ak)-1.

Let t e [d, 1], then

{Tk ft

(Tk - s)/(s,x($(s)))ds < (t - s)/(s,x($(s)))ds, Tk < t, Jo Jo

m ^Tk m ^ t

^a^ (Tk - s)/(s,x($(s))) ds ak \ (t - s)/(s,x($(s))) ds.

k=1 Jo k=i Jo

(3.18)

Multiplying by A = ™=1 ak) we obtain

m /^Tk m z^t

^^a^ (Tk - s)/(s,x(#(s)))ds < ^^a^ (t - s)/(s,x($(s)))ds k=1 Jo k=1 Jo

= f (t - s)/(s,x($(s)))ds, Jo

(3.19)

and the solution x of the nonlocal problem (1.3) and (3.16), given by the integral equation (3.17), is positive for f e [d, 1]. This complete the proof. □

Example 3.4. Consider the nonlocal problem of the second-order functional differential equation (1.3) with two-point boundary condition

x'(0) = 0, x(n) = xo, n e (a,d) c (0,1). (3.20)

Applying our results here, we deduce that the two-point boundary value problem (1.3)-(3.20) has at least one monotonic nondecreasing solution x e C[0,1] represented by the integral equation

x(f) = x0 - C(n - s)f (s,x($(s)))ds + i(f - s)f (s,x($(s)))ds. (3.21)

This the solution is positive with f > n 4. Nonlocal Integral Condition

Let x e C[0,1] be the solution of the nonlocal problem (1.3) and (1.4).

Let ak = fk - fk-1,Tk e (ffc_1,ffc) c (a,d) c (0,1) and let bj = lj - ij-1,nj e (lj-\,lj) c (c, e) c (0,1), then

^(tk - tk-1)x(Tk) = xo, x'(o) + Yj {Ij - ¿j-1)x'( j = x1. (4.1)

k=1 j=1

From the continuity of the solution x of the nonlocal problem (1.3) and (1.4), we obtain

lim V(tk - ifc-i)x(Tfc) = x(s)ds,

k=1 J a

'(0) + lim £ & - ¿/-1)x'inù = x'(0) + x'(s)ds,

and the nonlocal condition (1.4) transformed to the integral condition

x(s)ds = x0, x'(0) + x(e) - x(c) = x1, (4.3)

and the solution of the integral equation (2.3) will be

(t - s)/(s,x($(s)))dsdt

+ ((b - c) + 1)-1(t - 1)j xi - /(s,x(#(s))) dsdt\ (4.4)

/(s,x($(s))) ds.

Now, we have the following theorem.

Theorem 4.1. Let the assumptions (i)-(iv) of Theorem 3.1 be satisfied. Then the nonlocal problem

x"(t) = /(t,x(#(t))), t e (0,1),

d (4.5)

x(s)ds = x0, x'(0) + x(e) - x(c) = x1

has at ¡east one monotonie nondecreasing solution x e C[0,1] represented by (4.4).

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