Existence of positive solutions for first order discrete periodic boundary value problems with delay

Nonlinear Analysis 74 (2011) 4186–4191

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Existence of positive solutions for first order discrete periodic boundary value problems with delay Ruyun Ma ∗ , Chenghua Gao, Jia Xu Department of Mathematics, Northwest Normal University, Lanzhou 730070, PR China

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Article history: Received 17 November 2010 Accepted 29 March 2011 Communicated by Enzo Mitidieri

abstract Let T > 3 be a positive integer and T = {0, 1, 2, . . . , T − 1}. We prove the existence of positive periodic solutions of the nonlinear discrete boundary value problem

∆u(t ) = a(t )g (u(t ))u(t ) − λb(t )f (u(t − τ (t ))),

t ∈ T,

u(0) = u(T ).

MSC: 34B18 39A11

Our approach is based upon the global bifurcation techniques. © 2011 Elsevier Ltd. All rights reserved.

Keywords: Difference equations Eigenvalues Bifurcation Existence Positive periodic solutions

1. Introduction Let T > 3 be a positive integer and T = {0, 1, 2, . . . , T − 1}. Let R denote the real number set, Z the integer set. In this paper, we investigate the existence of positive periodic solutions for the following discrete periodic boundary value problems with delay

1u(t ) = a(t )g (u(t ))u(t ) − λb(t )f (u(t − τ (t ))), u(0) = u(T ),

t ∈ T,

(1.1)

where a, b : Z → [0, ∞) are T -periodic functions, τ : Z → Z is a T -periodic function. f , g ∈ C ([0, ∞), [0, ∞)), λ > 0 is a parameter. In recent years, there has been considerable interest in the existence of positive solutions of the following equation: x′ (t ) = a˜ (t )˜g (x(t ))x(t ) − λb˜ (t )f˜ (x(t − τ (t ))),

(1.2)

ω

ω

where a˜ , b˜ ∈ C (R, [0, ∞)) are ω-periodic functions, 0 a˜ (t )dt > 0, 0 b˜ (t )dt > 0 and τ is a continuous ω-periodic function. (1.2) has been proposed as a model for a variety of physiological processes and conditions including production of blood cells, respiration, and cardiac arrhythmias. See, for example, [1–12] and the references therein. So far, relatively little is known about the existence of positive periodic solutions of (1.1). To the best of our knowledge, only Raffoul [13] dealt with the special equations of (1.1) of the form

1x(t ) = α(t )x(t ) − λb(t )f (x(t − τ (t ))), ∗

Corresponding author. Tel.: +86 931 7971297. E-mail addresses: [email protected], [email protected] (R. Ma).

0362-546X/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2011.03.053

(1.3)

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with α(t ) = a(t ) − 1 > 0, and determined values of λ, for which there exist T -periodic positive solutions of (1.3). Zhou [14] proved the existence results of positive periodic solutions of first order difference equations (with no delay) by using Krasnosel’skii’s fixed point theorem. However, the conditions used in [13] are not sharp, and the main results in [13] give not any information about the global structure of the set of positive periodic solutions. It is the purpose of this paper to study the global structure of the set of positive periodic solutions of (1.1), and accordingly, to establish some existence results of the positive periodic solutions of (1.1). Our results are sharp. The main tool we used is the Dancer global bifurcation theorem (see [15, Corollary 15.2]). To obtain it, we make the assumptions: (H1) a, b : Z → [0, +∞) are T -periodic functions, and a(t ) ̸≡ 0, b(t ) ̸≡ 0 on t ∈ T, τ : Z → Z is a T -periodic function; (H2) f , g : [0, +∞) → [0, +∞) are continuous, 0 < l ≤ g (u) ≤ L < ∞ with l, L are positive constants, and f (u) > 0, for u > 0. Also, T −1

σl =

∏

T −1

(1 + a(s)l),

σL =

∏

s =0

(1 + a(s)L),

(1.4)

s=0

it is clear that 1 < σl ≤ σL . Let M (r ) = max {f (x)}, 0≤x≤r

m(r ) =

min

σl −1 (σL −1)σL ·r ≤x≤r

{f (x)}.

(1.5)

The rest of this paper is organized as follows. In Section 2, we give some notations and the main results. Section 3 is devoted to prove the main results, and finally, we give an example to illustrate our main result. 2. Notations and the main results

ˆ = {0, 1, 2, . . . , T }. Recall T = {0, 1, 2, . . . , T − 1} and let T Let ˆ → R|u(0) = u(T )} E = {u : T with the norm ‖u‖E = maxt ∈Tˆ |u(t )|. Then (E , ‖ · ‖E ) is a Banach space. Let Y = {u|u : T → R} with the norm ‖u‖Y = maxt ∈T |u(t )|. Then (Y , ‖ · ‖Y ) is a Banach space. It is easy to see that the operator χ : E → Y

χ (u(0), u(1), . . . , u(T − 1), u(T )) = (u(0), u(1), . . . , u(T − 1)) is a homomorphism. ˆ. By a positive solution of (1.1) we mean a pair (λ, u), where λ > 0 and u is a solution of (1.1) with u > 0 in T Let Σ ⊂ R+ × E be the closure of the set of positive solutions of (1.1). We extend the function f to a continuous function f˜ defined on R in such a way that f˜ > 0 for all s < 0. For λ > 0, we then look at arbitrary solutions u of the eigenvalue problem

1u(t ) = a(t )g (u(t ))u(t ) − λb(t )f˜ (u(t − τ (t ))), u(0) = u(T ).

t ∈ T,

It is well known that in [16, Theorem 1.1] that (2.1) is equivalent to u( t ) = λ

T −1 −

Gu (t , s)b(s)f (u(s − τ (s)))ds,

s=0

where T −1

∏ Gu (t , s) =

(1 + a(s)g (u(s)))

l=s+1 T −1

∏

,

s ∈ T.

(1 + a(s)g (u(s))) − 1

s=0

Notice that a(t ) ̸≡ 0 for all t ∈ T, 0 < l ≤ g (u) ≤ L < ∞, we have 1

σL − 1

≤ Gu (t , s) ≤

σL , σl − 1

s ∈ T,

(2.1)

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R. Ma et al. / Nonlinear Analysis 74 (2011) 4186–4191

and 0<

σl − 1 < 1. (σL − 1)σL

Define K be a cone in E by

 σl − 1 ‖u‖E , t ∈ Tˆ . (σL − 1)σL



u ∈ E |u(t ) ≥ 0, u(t ) ≥

K =

By the positivity of Green’s function Gu (t , s) and b(t ), f (u), such solutions are positive. Therefore, the closure of the set of nontrivial solutions (λ, u) of (2.1) in R+ × E is exactly Σ . Next, we consider the spectrum of the linear eigenvalue problem

−1u(t ) + ca(t )u(t ) = λb(t )u(t − τ (t )), u(0) = u(T ).

t ∈ T,

(2.2) (2.3)

Lemma 2.1. Let (H1) hold and c be a positive constant. Then the linear problem (2.2), (2.3) has a unique eigenvalue λ (c ), which is positive and simple, and the corresponding eigenfunction ψ(t ) is of one sign. Proof. If is a direct consequence of the Krein–Rutman Theorem [17, Theorem 19.3].



In the rest of the paper, we always assume that

‖ψ‖E = 1,

ψ(t ) > 0,

ˆ. t ∈T

(2.4)

Theorem 2.1. Let (H1)–(H2) hold. Assume that (H3) f0 , f∞ , f ∞ ∈ (0, ∞), where f0 = lim

f ( u)

u→0+

u

,

f∞ = lim inf u→∞

f (u) u

,

f ∞ = lim sup u→∞

f (u) u

.

Then (1.1) has a positive T -periodic solution if either

λ  ( L) f∞

<λ<

λ (g (0)) f0

,

(2.5)

.

(2.6)

or

λ (g (0)) f0

<λ<

λ (l) f∞

3. Proof of the main results Let g (s) = g (0) + ζ (s),

f (s) = f0 s + ξ (s).

(3.1)

Then lim ζ (s) = 0,

s→0+

lim

s→0+

ξ ( s) s

= 0.

(3.2)

Define a linear operator A˜ : E → Y , J : Y → E by

˜ )(t ) = −1u(t ) + a(t )g (0)u(t ), (Au

u ∈ E,

(3.3)

and J (u(0), u(1), . . . , u(T − 1)) = (u(0), u(1), . . . , u(T − 1), u(T )).

(3.4)

˜ Then A−1 : E → E is compact and continuous since E is finite dimensional. Let A := J ◦ A. Now (1.1) can be rewritten to the form u(t ) = λA−1 b(·)f0 u(· − τ (·)) (t ) + A−1 N (λ, u(·))(t ),





(3.5)

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where N (λ, u(·))(t ) := a(·)ζ (u(·))u(·) − λb(·)ξ (u(· − τ (·))) (t ).





It is easy to check that N (λ, u)

lim

‖u‖→0

‖ u‖ E

= 0,

uniformly on bounded λ intervals.

ˆ . Finally, let Let E = R × E under the product topology. Let S + denote the set of functions in E which are positive in T Φ+ = R × S+. The results of the Dancer global bifurcation theorem (see [15, Corollary 15.2]) for (3.5) can be stated as follows: there  λ (g (0)) exists a continuum C ⊆ Φ + of solutions of (3.5) joining ( f , 0) to infinity in Φ + . Moreover, C \ {( λ (fg (0)) , 0)} ⊂ Φ + . 0

0

Proof of Theorem 2.1. From (2.5) and (2.6), there exists a constant ϵ ∈ (0, f∞ ), such that

λ (L) λ (g (0)) <λ< , f∞ − ϵ f0

(3.6)

or

λ (g (0)) f0

<λ<

λ (l) . f∞ +ϵ

(3.7)

It is clear that any solution of (3.5) of the form (λ, u) yields a solution u of (1.1). We will show that (i) ProjR C ⊇ (ii) ProjR C ⊇

 

λ (l) λ (g (0)) f∞ −ϵ f0 λ (g (0)) λ (L) f f ∞ +ϵ

,

,

0



, if (3.6) holds;



, if (3.7) holds.

To do this, it is enough to show that C joins ( Let (µn , un ) ∈ C satisfy

λ (g (0)) f0

(l) λ (L) , 0) to [ fλ∞ +ϵ , f∞ −ϵ ] × {∞}. 



µn + ‖un ‖E → ∞.

(3.8)

We note that µn > 0 for all n ∈ N since (0, 0) is the only solution of (1.1) for λ = 0 and C ∩ ({0} × E ) = ∅. We claim that the sequence {µn } is bounded. Suppose on the contrary that µn → ∞ (after taking a subsequence and relabelling if necessary). Let

vn (t ) =

un ( t ) . ‖ un ‖ E

Then

1vn (t ) = a(t )g (un (t ))vn (t ) − µn b(t )

f (un (t − τ (t ))) un (t − τ (t ))

vn (t − τ (t )),

(3.9)

vn (0) = vn (T ). Since (H2) and (H3) imply that min{

1vn (t ) → −∞,

f (s) s

| s > 0} ≥ δ for some δ > 0, it follows from (3.9) that

ˆ, uniformly for t ∈ T

as n → ∞. However, this contradicts vn (0) = vn (T ). Therefore, {µn } is bounded. Now, (3.8) together with the fact that {µn } is bounded implies that

‖un ‖E → ∞.

(3.10)

Recall that



 σ L (1 − σ l ) K = y ∈ E |y(t ) ≥ ‖y‖E . 1 − σL

(3.11)

Then it is easy to verify that un ∈ K for all n ∈ N. Combining this with (3.10) and (3.11), we conclude that un (t ) → ∞,

ˆ uniformly for t ∈ T

(3.12)

as n → ∞. Recall that

1vn (t ) = a(t )g (un (t ))vn (t ) − µn b(t )

f (un (t − τ (t ))) un (t − τ (t ))

vn (t − τ (t )),

t ∈ T.

(3.13)

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R. Ma et al. / Nonlinear Analysis 74 (2011) 4186–4191 f (u (t −τ (t )))

Since {g (un (t ))}, {vn (t )}, { u n(t −τ (t )) }are bounded sets in E and {µn } is bounded, so is {1vn (t )}. By (3.12), ϵ ∈ (0, f∞ ) n satisfying (3.6) and (3.7), there exists Nϵ ∈ N, such that f (un (t − τ (t )))

f∞ − ϵ ≤

un (t − τ (t ))

≤ f ∞ + ϵ,

t ∈ T, n ≥ Nϵ .

Now, let ϵ and Nϵ be the above fixed constant. For each n ≥ Nϵ , let us compare µn of (3.13) with the principal eigenvalue λ(L, f∞ − ϵ) of the linear problem

1x(t ) = a(t )Lx(t ) − λb(t )[f∞ − ϵ]x(t − τ (t )), x(0) = x(T )

t ∈ T,

(3.14)

and the principal eigenvalue λ(l, f ∞ + ϵ) of the linear problem

1z (t ) = a(t )lz (t ) − λb(t )[f ∞ + ϵ]z (t − τ (t )), z (0) = z (T ),

t ∈ T,

(3.15)

respectively. Using the equivalent integral equations of (3.14) and (3.15) and applying [17, Theorem 19.3(d)], it follows

λ(l, f ∞ + ϵ) ≤ µn ≤ λ(L, f∞ − ϵ).

(3.16)

Notice that

λ (l) = λ(l, f ∞ + ϵ)(f ∞ + ϵ),

λ (L) = λ(L, f∞ − ϵ)(f∞ − ϵ).

(3.17)

This together with (3.16) implies that

λ (l) λ (L) ≤ µ ≤ . n f∞ +ϵ f∞ − ϵ

(3.18)

Thus, if (3.6) is true, then we deduce



 λ (l) λ (g (0)) , ; f∞ +ϵ f0

ProjR C ⊇

if (3.7) is true, then we deduce



λ (g (0))

ProjR C ⊇

f0

 λ  ( L) , .  f∞ − ϵ

Remark 3.1. The methods used in the proof of Theorem 2.1 have been used in the study of other kinds of boundary value problems; see [18–22] and the references therein. Remark 3.2. The conditions in Theorem 2.1 are sharp. Let us take g (s) ≡ 1,

b(t ) ≡ 1,

a(t ) ≡ a > 0,

λ = a,

f (s) = s + h(s),

τ (t ) ≡ 0.

Let 1

  

1 + u2 h(u) = 2  u 2

,

,

u ∈ [1, ∞), u ∈ [0, 1],

and consider the problem

 1u(t ) = au(t ) − a u(t ) + h(u)],

t ∈ T,

(3.19)

u(0) = u(T ). it is easy to see that

λ (g (0)) = λ (1) = a,

λ (L) = λ (l) = a,

f0 = 1 ,

f∞ = f ∞ = 1.

Since

λ (l) f∞

=a=

λ (g (0)) f0

,

(3.20)

R. Ma et al. / Nonlinear Analysis 74 (2011) 4186–4191

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and

λ (g (0)) f0

=a=

λ (L) f∞

.

(3.21)

(2.5) and (2.6) are not valid. In this case, (3.19) has no nontrivial solution. If fact, if u is a nontrivial solution of (3.19), then 0=

T −1 −

1u(t )dt = a

s=0

which is a contradiction.

T −1 −

h(u(t ))dt > 0,

s=0



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