Scholarly article on topic 'Existence of positive solutions of higher-order nonlinear neutral equations'

Existence of positive solutions of higher-order nonlinear neutral equations Academic research paper on "Mathematics"

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Academic research paper on topic "Existence of positive solutions of higher-order nonlinear neutral equations"

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Existence of positive solutions of higher-order nonlinear neutral equations

Tuncay Candan*

"Correspondence: tcandan@nigde.edu.tr Department of Mathematics, Faculty of Arts and Sciences, Nigde University, Nigde, 51200, Turkey

Abstract

In this work, we consider the existence of positive solutions of higher-order nonlinear neutral differential equations. In the special case, our results include some well-known results. In order to obtain new sufficient conditions for the existence of a positive solution, we use Schauder's fixed point theorem.

Keywords: neutral equations; fixed point; higher-order; positive solution

1 Introduction

The purpose of this article is to study higher-order neutral nonlinear differential equations of the form

[r(t) [x(t) - Pi(t)x(t - r )f-7 + (-l)"Qi(t)f(x(t - a )) = 0, [r(t)[x(t) - Pi(t)x(t - r )f-l)]' + (-l) J" Q2(t, £)f (x(t - £ )) d£ = 0

(l) (2)

x(t)- P2(t, £)x(t - £) d£

(«-l)n

Q2(t, £)f(x(t - £ )) d£ = 0,

where n > 2 is an integer, t > 0, a > 0, d > c > 0, b > a > 0, r, Pi e C([io, (0, to)), Pi e C([t0, to) x [a, b], (0,to)), Qi e C([t0, to),(0, to)), Q2 e C([t0, to) x [c,d],(0, to)),f e C(R, R), f is a nondecreasing function with xf (x) > 0, x =0.

The motivation for the present work was the recent work of Culakova et al. [i] in which the second-order neutral nonlinear differential equation of the form

[r(t)[x(t)-P(t)x(t - r)]']' + Q(t)f{x(t - a)) = 0

ft Spri

ringer

was considered. Note that when n = 2 in (i), we obtain (4). Thus, our results contain the results established in [i] for (i). The results for (2) and (3) are completely new.

Existence of nonoscillatory or positive solutions of higher-order neutral differential equations was investigated in [2-5], but in this work our results contain not only existence of solutions but also behavior of solutions. For books, we refer the reader to [6-ii].

Let pi = max{T, a}. By a solution of (i) we understand a function x e C([ti - pi, to), r), for some ti > t0, such that x(t) - Pi(t)x(t - t) is n - i times continuously differentiable,

©2013 Candan; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the originalworkis properly cited.

r(t)(x(t) - Pi(t)x(t - t))(n-1) is continuously differentiable on [ti, to) and (1) is satisfied for t > t1. Similarly, let p2 = max{T, d}. By a solution of (2) we understand a function x e C([t1 - p2, to), R), for some t1 > to, such that x(t) - P1(t)x(t - t)is n -1 times continuously differentiable, r(t)(x(t) - P1(t)x(t - t))(n-1) is continuously differentiable on [t1, to) and (2) is satisfied for t > t1. Finally, let p3 = max{b, d}. By a solution of (3) we understand a function x e C([t1 - p3, to), R), for some t1 > to, such that x(t) - P2(t, %)x(t - %) d% is n -1 times continuously differentiable, r(t)[x(t)- f^ P2(t, %)x(t - %) d%](n-1) is continuously differentiable on [t1, to) and (3) is satisfied for t > t1.

The following fixed point theorem will be used in proofs.

Theorem 1 (Schauder's fixed point theorem [9]) Let A be a closed, convex and nonempty subset of a Banach space ft. Let S : A ^ A be a continuous mapping such that SA is a relatively compact subset of ft. Then S has at least one fixed point in A. That is, there exists x e A such that Sx = x.

2 Main results

Theorem 2 Let

/ Qi(t) dt = to. (5)

Assume that 0 < k1 < k2 and there exists y > 0 such that k

f exp((k2-ki)jft0 Qi(t)d^j > 1, (6)

exp -kW Qi(s) ds) + exp kW Qi(s) ds

V Jt-T ' V Jto-Y

1 f to (s - t)n-2 fTO ( ( fu-a \\ X—(t -¡¡srJ. Qi(u)f{exp{-kiLQi(z)dv)duds

< P1 (t) < exp ^-k^ Q1 (s) ds^ + exp ^k1 Q1 (s) ds^ (7)

1 rTO (s - t)n-2 fTO ( ( fu-a \\

Xm-wL -¡asrJ. Qi(u)flexTk2LQi(z)dz^duds

t > t1 > t0 + max{T, a}. Then (1) has a positive solution which tends to zero.

Proof Let ft be the set of all continuous and bounded functions on [t0, to) with the sup norm. Then ft is a Banach space. Define a subset A of ft by

A = {x e ft : v1(t) < x(t) < v2(t), t > to},

where v1(t) and v2(t) are nonnegative functions such that

vi (t) = exp f -k2 f Qi (s) ds), V2 (t) = exp f-ki i Qi (s) ds), t > to. (8) \ Jto-Y / V Jto-Y '

It is clear that A is a bounded, closed and convex subset of ft. We define the operator S: A ft as

(Sx)(t)=lPl(t)x(t - -T ^ .T Qi(u)f(x(u - a)) du^ t > ti,

}(Sx)(ii) + v2(i)-v2(ii), t0 < t < ti.

We show that S satisfies the assumptions of Schauder's fixed point theorem. First, S maps A into A. For t > t1 and x e A, using (7) and (8), we have

1 rm (s - t)n-2 f (Sx)(t) < Pi(t)v2(t - t )- n-2yjt r(S) Js Qi(u)f{Mu - a )) duds

= Pi (t) exp f -ki i Qi (s) ds

\ Jto-Y

r-~ (s - t)n-2 '

(n -< V2(i)

1 f™ (s - t)n-2 t™ ( ( fu-a \\ -vil -¡trJ. Qi(u)/H-k2LQi(z)dz))duds

1 Z-™ (s - t)n-2 f™

(Sx)(t) > Pi(í)vi(í - r)- ¡(s) l Qi(u/(v*(u - ^)) duds

n-2 />™

(n -2)

= Pi(t) exp( -k2 Qi(s) ds

(n -> vi(t).

i r™ (s - t)n-2 (™ ( ( r-a w

-wL ~-tsrJ. Qi(u)/H-kiLQi(z)dz))duds

For t e [to, ti] and x e A,we obtain

(Sx)(t) = (SX)(ti) + v2(t) - v2(ti) < v2(t)

and in order to show (Sx)(t) > vi(t), consider

H(t) = V2(t) - V2(ti) - vi(t) + vi(ti).

By making use of (6), it follows that

H'(t) = v2(t) - vi(t) = -kiQi(t)v2(t) + k2Qi(t)vi(t)

= Qi(t)V2(t) = Qi(t)V2(t) < Qi(t)v2(t)

-ki + k2vi(t) expl ki I Qi(s) ds

\ Jto-Y

-ki + k2 exp (ki- k2) Qi(s) ds

\ Jto-Y

-ki + k2 exp (ki- k2) Qi(s) ds

\ Jto-Y

< o, to < t < ti.

Since H(ti) = 0 and H'(t) < 0 for t e [to, ti], we conclude that

H(t) = v2(t) - v2(ti) - vi(t) + vi(ti) > 0, to < t < ti.

Then t e [t0, ti] and for any x e A,

(Sx)(t) = (Sx)(ti) + v2(t) - v2(ti) > vi(ti) + v2(t) - v2(ti) > vi(t), to < t < ti.

Hence, S maps A into A.

Second, we show that S is continuous. Let {xi} be a convergent sequence of functions in A such that xi(t) — x(t) as i — to. Since A is closed, we have x e A.It is obvious that for t e [t0, ti] and x e A, S is continuous. For t > ti,

|(Sxi)(t)-(Sx)(t)|

< Pi(t)|xi(t - r)-x(t - r)|

Since f (xi(t - a))-f (x(t - a))| — 0 as i — to, by making use of the Lebesgue dominated convergence theorem, we see that

lim ||(Sxi)(t)-(Sx)(t)|| =0

t—>to 11

and therefore S is continuous.

Third, we show that SA is relatively compact. In order to prove that SA is relatively compact, it suffices to show that the family of functions {Sx: x e A} is uniformly bounded and equicontinuous on [t0, to). Since uniform boundedness of {Sx: x e A} is obvious, we need only to show equicontinuity. For x e A and any e > 0, we take T > ti large enough such that (Sx)(T) < |. For x e A and T2 > Ti > T, we have

< Pi(t)k(t - T) — x(t - T)

(Sx)(T2) — (Sx)(Ti)| < |(Sx)(T2)| + |(Sx)(Ti)| < t + J

Note that

Xn — Yn = (X — Y )(Xn—1 + Xn—2Y + ••• + XYn—2 + Yn—1) < n(X — Y)Xn—1, X > Y >0.

For x e A and ti < Ti < T2 < T, by using (9) we obtain

(Sx)(Tt) — (Sx)(Ti)|

< |Pi(Tt)x(T2 — t) — Pi(Ti)x(Ti — t)

1 fT2 (s - Tj)"-2

in-iy-ln^sT J. Ql{U)f{x{U - a^ dUdS

1 r~ (s - Ti)n-2-(s - T2)n-2 r (n '-2)\jT2

+ ,'-,./' -—Tl-^(S—T- Í Qi(u)f (x(u - a)) duds

)! JT2 r(s) Js

< |Pl(T2)x(T2 - r) -Pi(Ti)x(Ti - r)|

f i s"-2 f

t;<7<T4 (n - 2)! r(s)

+ ^max^ I --—,~7T Qi(u)f (x(u - a)) du^ (T2 - Ti)

i (s - Ti)n-3

! J --J Qi(u)f(x(u - a)) duds(T2 - Ti).

(n -3): ./T2 Thus there exits 5 >0 such that

|(Sx)T) - (Sx)(Ti)| < e if 0 < T2 - Ti < 5.

Finally, for x e A and t0 < Ti < T2 < ti, there exits 5 >0 such that

|(Sx)(T2) - (Sx)(Ti)| = |v2(Ti) - v2(T2)| < e if 0 < T2 - Ti < 5.

Therefore SA is relatively compact. In view of Schauder's fixed point theorem, we can conclude that there exists x e A such that Sx = x. That is, x is a positive solution of (i) which tends to zero. The proof is complete. □

Theorem 3 Let

/ Q 2(t) dt = to, (0)

where Q2(t) = fd Q2(t, £) d£. Assume that 0 < ki < k2 and there exists y > 0 such that

ki ex^(k2- ki) 0 Q2 (t) d^ > i, (ii)

exp ^-k2 £ Q2 (s) d^ + exp ^ k^ Q2 (s) ds^

i rm (s - t)n-2 rmrd / / r\\

x _/t J, IQ2(u,f<H-kiL Q2(z)dz))d£duds

< Pi (t) < exp ^-ki £ Q2 (s) d^ + exp ^ki £ Q2 (s) ds^

rm (s - t)n-2 rmrd / / ru-? _ \\

! J —r(s)—J J Q2(u,£)f(^ex^-k2 J Q2(z) dzjjd% duds,

(n -2)

t > ti > t0 + max{r, d}.

Then (2) has a positive solution which tends to zero.

Proof Let ft be the set of all continuous and bounded functions on [t0, to) with the sup norm. Then ft is a Banach space. Define a subset A of ft by

A = {x e ft : vi(t) < x(t) < v2(t), t > to},

where vi(t) and v2(t) are nonnegative functions such that

vi(t) = expl -k2 QQ 2 (s) ds J, V2(t) = expi -ki I QQ 2(s) ds\, t > to.

\ Jto-Y ' \ Jto-Y '

It is clear that A is a bounded, closed and convex subset of ft. We define the operator S: A —► ft as follows:

(Sx)(t)= (Pi(t)x(t - T) - (n-2)! ITO ^ ITO LdQ2(u, fx(u - f)) df duds, t > ti,

|(Sx)(ti) + V2(t)-V2(ti), to < t < ti.

Since the remaining part of the proof is similar to those in the proof of Theorem 2, it is omitted. Thus the theorem is proved. □

Theorem 4 Suppose that (i0) and (ii) hold. In addition, assume that

exp -/<2/ Q 2 (s) ds) + exp <<2/ Q 2 (s) ds \ Jt-a' \ Jto-Y

i r to (s - t)n-2 f to fd ( ( fu-f \\

X — Jt I Jc Q2(U, fexp(-ki Jtoy Q2(Z) dz))df duds

< P2 (t) < exp (-ki f Q2 (s) d^ + exp (ki i Q2 (s) d^ (i2)

\ Jt-b / \ Jto-Y '

i r to (s - t)n-2 f to fd ( ( fu-f \\

X (n^/t Js I Q2(u f)f{exA-«2 Jt0-Y Q2(Z) dZ))df dUdS,

t > ti > to + max{b, d},

where P2 (t) = fb P2 (t, f) df. Then (3) has a positive solution which tends to zero.

Proof Let ft be the set of all continuous and bounded functions on [to, to) with the sup norm. Then ft is a Banach space. Define a subset A of ft by

A = {x e ft : vi(t) < x(t) < v2(t), t > to},

where vi(t) and v2(t) are nonnegative functions such that

vi (t) = exp (-k2 i Q2 (s) ds), v2 (t) = exp f-«i i Q2 (s) ds), t > to. (i3) \ Jt§-Y/ \ Jto-Y '

It is clear that A is a bounded, closed and convex subset of Q. We define the operator S: A Q as

(Sx)(t) =

ft P2 (t, £)x(t - £) d£ -(n-mSr SL¡^f,00Jd Q2 (u, £)/(x(u - £)) d£ du ds,

t > tl,

(Sx)(ti) + v2(t) - v2(ti), to < t < tl.

We show that S satisfies the assumptions of Schauder's fixed point theorem.

First of all, S maps A into A. For t > t1 and x e A, using (12), (13), the decreasing nature of v2 and v1, we have

(Sx)(t) < /" P2(t, £)v2(t - £) d£ -

(s - t)

(« -2)!

t to ak-2 /• to /• d

/to (s - t)n-2 /-to /.d

( r(S) J J Q2(u, £)/(n(u - £)) d£ duds

t-b \ i

< P2 (t) expl -ki / Q2 (s) d^ -

Ao-y / (n -2)! /"to (s - t)n-2 /• to /•d ( / fu-£ „ \\

x y ——— J J Q2(u, £exp(^-k2 J Q2(2) dzjjd£ duds < V2(t)

(Sx)(t) > / P2(t, £)vi(t - £) d£ -■

(n -2)!

/TO (- — t)n-2 /• to /■ rf

-^-jf J Q2(u, f)f(v2(u - f )) (if rfurfs

> p2« exP(-fc/J^ 2(-)

X/ (- "r(-))" 21 I Q2 (u, f )f(ex^(-kijt ^Q2 (z) rfurf-

> Vi(t).

For t e [to, t1] and x e A,we obtain

(Sx)(t) = (Sx)(ti) + v2(t) - v2(ti) < v2(t)

and to show (Sx)(t) > v1(t), consider

H(t) = V2(t) - V2(ti) - vi(t) + vi(ti).

By making use of (11), it follows that

H'(t) = v2(t)-v1(t)

= -k1 (22(t)v2(t)+k2(22(t)v1(t)

= Q2(t)v2(t)

< Q2(t)v2(t)

-ki + «2 vi (t) expi ki I Q2 (s) ds

\ Jto-Y

-ki + «2 exp (ki- «2) Q 2 (s) ds

\ Jto-Y

< o, to < t < ti.

Since H(ti) = o and H'(t) < o for t e [to, ti], we conclude that

H(t) = v2(t) - v2(ti) - vi(t) + vi(ti) > o, to < t < ti.

Then t e [to, ti] and for any x e A,

(Sx)(t) = (Sx)(ti) + v2(t) - v2(ti) > vi(ti) + v2(t) - v2(ti) > vi(t), to < t < ti.

Hence, S maps A into A.

Next, we show that S is continuous. Let {xi} be a convergent sequence of functions in A such that xi(t) ^ x(t) as i ^to. Since A is closed, we have x e A.It is obvious that for t e [to, ti] and x e A, S is continuous. For t > ti,

|(Sxi)(t)-(Sx)(t)|

< i P2(t, f)|xi(t - f) -x(t - f)|df

i fto (s - t)n-2 rto 'd

(n -2)

/TO (s - t)n-2 /-TO pa

( r(s) J J Q2(u,f)f(xi(u-f))-f(x(u-f))| dfduds.

Since f (xi(t - f ))-f (x(t - f ))| ^ o as i ^to and f e [c, d], by making use of the Lebesgue dominated convergence theorem, we see that

tHm||(S*i)(t)-(S*)(t)|| =o. Thus S is continuous.

Finally, we show that SA is relatively compact. In order to prove that SA is relatively compact, it suffices to show that the family of functions {Sx: x e A} is uniformly bounded and equicontinuous on [to, to). Since uniform boundedness of {Sx: x e A} is obvious, we need only to show equicontinuity. For x e A and any e > o, we take T > ti large enough such that (Sx)(T) < |. For x e A and T2 > Ti > T, we have

|(Sx)(T2) - (Sx)(Ti)| < |(Sx)(T2)| + |(Sx)(Ti)| < 2 + 2 = e. For x e A and ti < Ti < T2 < T, by using (9) we obtain

|(Sx)(T2)-(Sx)(Ti)|

< f |P2(T2, f )x(T2 - f )- P2(Ti, f )x(Ti - f ) | df

i fT2 (s - Ti)n-2 fTO fd

(n -2)!

fT2 (s - T )n-2 fTO fd ~jT ( r(s)) Js J Q2(u, f)f(x(u - f)) df duds

1 f<x (, _ T )n-2 _(, _ T )n-2 fM rd

ñW» _ " _ 2' J. I Q2<", f № _ f»"f ^

í |P2(T2, f )x(T2 _ f )_P2(T1, f )x(Ti - f )\ df

Í 1 .n_2 PM Pd 1

+r^J, IQ2(uf)f{x(u_f))dfdu\(T2_Ti)

i fTO (s - T )n-3 pTO pd

+ (n-) JT ( r(s1)) Js J Q2(u,f )f (x(u - f )) df duds(T2 - Ti).

Thus there exits 5 > o such that

| (Sx)(T2) - (Sx)(Ti)| < e if o < T2 - Ti < 5.

For x e A and to < Ti < T2 < ti, there exits 5 > o such that

|(Sx)(T2) - (Sx)(Ti)| = |v2(Ti) - v2(T2)| < e if o < T2 - Ti < 5.

Therefore SA is relatively compact. In view of Schauder's fixed point theorem, we can conclude that there exists x e A such that Sx = x. That is, x is a positive solution of (i) which tends to zero. The proof is complete. □

Example 1 Consider the neutral differential equation

x(í)-Pi(í)x( t -2

- qx(t - 1) = 0, t > t0, (14)

where q e (0, m) and

exp(q[A^2(t + Y _ t - to)-ki(y - a - to)]) exp((-qki - 2)t) exp(-k2qr) +---

< P1(t) < exp(-k1 qT) + exp((-qk2 - 22) t)

k1 (k1q +¿)2

exp(q[k1(t + y - t - to) - /<2(y _ a - to)])

(«2q +2)2 '

Note that for ki = |, k2 = i,q = i and to = y = y ,we have

k- exp^(«2 - ki) 0 Qi(t) d^ = 2 exp^i jj 1 d^j = 5.8194 > i

(-3 \ 54 (-t-5\ , N 4 (-5t \

ex^ T j + 49 ex^(—g—) < Pi(t) < exp(-l) + 9 expl — J, t > 8.

If P1(t) fulfils the last inequality above, a straightforward verification yields that the conditions of Theorem 2 are satisfied and therefore (i4) has a positive solution which tends to zero.

Competing interests

The author declares that they have no competing Interests.

Received: 16 August 2013 Accepted: 11 November 2013 Published: 05 Dec 2013

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10.1186/1029-242X-2013-573

Cite this article as: Candan: Existence of positive solutions of higher-order nonlinear neutral equations. Journal of Inequalities and Applications 2013, 2013:573

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