Positive periodic solutions of systems of functional differential equations

POSITIVE PERIODIC SOLUTIONS OF SYSTEMS OF FUNCTIONAL DIFFERENTIAL EQUATIONS

Youssef N. Raffoul

Abstract. We apply a cone theoretic xed point theorem and obtain conditions for the existence of positive periodic solutions of the system of functional dierential equations

x

(t) = A(t)x(t) + λf (t, x(t − τ(t)).

1. Introduction

In this paper, we are concerned with determining values for λ so that the system of functional dierential equations

(1.1) x ^′ (t) = A(t)x(t) + λf (t, x(t − τ(t))

has a positive periodic solution. The matrix A(t) = diag[a ₁ (t), a ₂ (t), . . . , a _n (t)], a _j ∈ C(R, R), τ : R → R, are continuous and ω-periodic, j = 1, 2 . . . , n with ω > 0. The function f : R × R ⁿ + → R ⁿ + is continuous, where R ⁿ = (x 1 , x 2 , . . . , x n ) ^T and R ⁿ + = {(x 1 , x 2 , . . . , x n ) ^T ∈ R ⁿ : x j >

0, j = 1, 2, . . . , n }. We denote BC the normed vector space of bounded functions ϕ : R → R ⁿ with the norm ∥ϕ∥ = ∑ _n

j=1 sup _{t ∈R} |ϕ j (t)| where ϕ = (ϕ ₁ , ϕ ₂ , . . . , ϕ _n ) ^T . For each x = (x ₁ , x ₂ , . . . , x _n ) ^T ∈ R ⁿ , the norm of x is dened as |x| 0 = ∑ _n

j=1 |x j |, where we say that x is “positive”

whenever x ∈ R ⁿ + .

In this paper we not only carry the work of [17] to the continuous case, but we generalize it to systems. In this research, the set up of the map- ping is the same as in [10] in which λ = 1. In arriving at our results, we make use of Krasnosel’skii xed point theorem ([11]). The existence of positive periodic solutions of nonlinear functional dierential equations have been studied extensively, in recent years. For some appropriate

Received March 15, 2004.

2000 Mathematics Subject Classication: 34K13, 34B20.

Key words and phrases: cone theory, functional dierential equations, positive

periodic solution.

references we refer the reader to [1], [2], [3], [4], [5], [6], [7], [8], [9], [12], [13], [14], [15], [16], [19] and the references therein. This work is mainly motivated by the work of [9], [17], and [18].

In section 2, we state Krasnosel’skii xed point theorem ([11]), prove two Lemmas that are essential to this research and construct the cone of interest . In section 3, we present four theorems and a corollary. In each of the theorems and the corollary an open interval of eigenvalues is determined, which in returns, implies the existence of a positive periodic solution of (1.1), by appealing to Krasnosel’skii xed point theorem.

2. Preliminaries

Theorem 2.1. (Krasnosel’skii) Let B be a Banach space, and let P be a cone in B. Suppose 1 and 2 are bounded open subsets of B such that 0 ∈ 1 ⊂ 1 ⊂ 2 and suppose that

T : P ∩ ( 2 \ 1 ) → P is a completely continuous operator such that

(i) ∥T u∥ ≤ ∥u∥, u ∈ P ∩ ∂ 1 , and ∥T u∥ ≥ ∥u∥, u ∈ P ∩ ∂ 2 ; or (ii) ∥T u∥ ≥ ∥u∥, u ∈ P ∩ ∂ 1 , and ∥T u∥ ≤ ∥u∥, u ∈ P ∩ ∂ 2 . Then T has a xed point in P ∩ ( 2 \ 1 ).

We denote f = (f 1 , f 2 , . . . , f n ) ^T and assume (H1)

∫ _ω

0

a j (s)ds < 0 for j = 1, 2, . . . , n.

Definition 2.2. Let X be a Banach space and K be a closed, nonempty subset of X. K is a cone if

(i) αu + βv ∈ K for all u, v ∈ K and all α, β ≥ 0 (ii) u, −u ∈ K imply u = 0.

Dene the set C _ω by

C _ω = {x ∈ C(R, R ⁿ ) : x(t + ω) = x(t), t ∈ R}.

Then it is clear that C _ω ⊂ BC when it is endowed with supremum norm

||x|| = ∑ _n

j=1 ||x j || 0 , where ||x j || 0 = sup _{t ∈[0,ω]} |x j (t) |.

Next, we consider the scalar dierential equation (2.1) x ^′ (t) = a(t)x(t) + λf (t, x(t − τ(t)),

where λ is constant, a ∈ C(R, R), τ : R → R, are continuous and ω-

periodic with ω > 0. The function f : R × R → R is continuous and

ω-periodic in t. The proof of the next Lemma is trivial, and hence we omit it.

Lemma 2.3. x(t) ∈ C ω is a solution of (2.1) if and only if (2.2) x(t) = λ

∫ _t+ω

t

exp( ∫ _t

s a(u)du) exp( − ∫ _ω

0 a(u)du) − 1 f (s, x(s − τ(s))ds.

Now, we dene the cone K and the Green’s function G(t, s) for equation (1.1). For (t, s) ∈ R ² , j = 1, 2, . . . , n, we dene

σ := min {

exp( −2

∫ _ω

0

|a j (s) |ds), j = 1, 2, . . . , n } , (2.3)

G _j (t, s) = exp( ∫ _t

s a _j (ν)dν) exp( − ∫ _ω

0 a j (ν)dν) − 1 . (2.4)

We also dene

G(t, s) = diag[G 1 (t, s), G 2 (t, s), . . . , G n (t, s)].

It is clear that G(t, s) = G(t + ω, s + ω) for all (t, s) ∈ R ² and by (H1) and the assumption on f we have,

G _j (t, s) > 0, f _j (u, ϕ(u − τ(u)) > 0

for (t, s) ∈ R ² and (u, ϕ) ∈ R × BC(R, R ⁿ + ). Let K be the set dened by K = {x ∈ C ω : x j (t) ≥ σ∥x j ∥, t ∈ [0, ω], x = (x 1 , x 2 , . . . , x n ) ^T }.

It is straight forward to verify that K is a cone.

Now, we are in a position to dene an operator ψ : K → K as

(2.5) (ψx)(t) =

∫ _t+ω

t

G(t, s)f (s, x(s − τ(s)))ds

for x ∈ K, t ∈ R, where G(t, s) is dened following (2.4). We denote (ψx) =

(

ψ 1 x, ψ 2 x, . . . , ψ n x ) T

. Before we proceed any further we state the followings:

(2.6) A _j = e ^{− R}

^|a

^{(u) |du} e ^{− R}

^a

^(u)du − 1 and

(2.7) B j = e ^R

^|a

^{(u) |du}

e ^{− R}

^a

^(u)du − 1 ,

for j = 1, 2, . . . , n. It is easy to see that for for j = 1, 2, . . . , n, A _j ≤ G j (t, s) ≤ B j

for all s ∈ [t, t + ω].

If we set A = min

1 ≤j≤n A j and B = max

1 ≤j≤n B j , then A ≤ G j (t, s) ≤ B for j = 1, 2, . . . , n.

Lemma 2.4. If (ψx)(t) is given by (2.5), then ψ : K → K is com- pletely continuous.

Proof. For each x ∈ K, since f(t, x(t −τ(t))) is a continuous function of t, we have (ψx)(t) is continuous in t and

(ψx)(t + ω) =

∫ _t+2ω

t+ω

G(t + ω, s)f (s, x(s − τ(s)))ds

=

∫ _t+ω

t

G(t + ω, s + ω)f (s + ω, x(s + ω − τ(s + ω)))ds

=

∫ _t+ω

t

G(t, s)f (s, x(s − τ(s)))ds = (ψx)(t).

Thus, (ψx) ∈ C ω . Next we show that (ψx) is continuous. For θ, ϑ ∈ C ω ,

∥θ − ϑ∥ < δ imply sup

0 ≤s≤ω |f j (s, θ(s − τ(s))) − f j (s, ϑ(s − τ(s)))| < ε λnB j ω . If x, y ∈ K with ∥x − y∥ < δ, then

|(ψ j x)(t) − (ψ j y)(t) |

≤ λ

∫ _t+ω

t

|G j (t, s) ||f j (s, x(s − τ(s))) − f j (s, y(s − τ(s)))|ds

≤ λB j ω sup

0 ≤t≤ω |f j (t, θ(t − τ(t))) − f j (t, ϑ(t − τ(t)))|

< ε n

for all t ∈ [0, ω]. This yields to

||(ψ j x)(t) − (ψ j y)(t) || 0 < ε n . Thus,

||(ψx) − (ψy)|| < ε.

Hence, ψ is continuous. For x ∈ K, let (ψ _j x)(t) = λ

∫ _ω

0

G _j (t, s)f _j (s, x(s − τ(s)))ds.

Then,

(ψ _j x)(t) ≤ λB j

∫ _ω

0

|f j (s, x(s − τ(s)))|ds and

(ψ j x)(t) ≥ λA j

∫ _ω

0

|f j (s, x(s − τ(s)))|ds

≥ A _j

B j ∥ψ j x ∥ 0 = σ ∥ψ j x ∥ 0 , j = 1, 2, . . . , n.

Therefore, (ψx) ∈ K. The proof of ψ being completely continuous is similar to the proof of [10], and hence we omit it. This completes the prove.

3. Main results

Now we are ready to state and proof our results. But before we proceed we state the following.

(L1) lim

x

→0

f

(s,x(s −τ(s)))

x

= ∞,

(L2) lim

x

→∞

f (s,x(s −τ(s)))

x

= ∞,

(L3) lim

x

→0

f

(s,x(s −τ(s))) x

= 0, (L4) lim

x

→∞

f

(s,x(s −τ(s))) x

= 0, (L5) lim

x

→0

f

(s,x(s −τ(s)))

x

= l j uniformly in s with 0 < l j < ∞, and

(L6) lim

x

→∞

f

(s,x(s −τ(s)))

x

= L _j uniformly in s with 0 < L _j < ∞, for x ∈ R ⁿ . For notational convenience, we let

L M = max L j

1 ≤j≤n , L m = min L j

1 ≤j≤n , l M = max l j 1 ≤j≤n , l _m = min l _j

1≤j≤n and |G(t, s)| = max

1 ≤j≤n |G j (t, s) |.

Theorem 3.2. Assume that (H1), (L5), and (L6) hold. Then, for each λ satisfying

(3.1) 1

ωσAL m

< λ < 1 ωBl M

(1.1) has at least one positive periodic solution.

Proof. We construct the sets 1 and 2 in order to apply Theorem 2.1. Let λ be dened by (3.1), and choose ϵ > 0 such that

1

ωσA(L m − ϵ) ≤ λ ≤ 1 ωB(l M + ϵ) .

By condition (L5), there exists H ₁ > 0 such that f _j (t, y) ≤ (l j + ϵ)y _j ≤ (l M + ϵ)y j , for 0 < y j ≤ H 1 . Dene 1 = {x ∈ K : ∥x j ∥ 0 < H 1 , j = 1, · · · , n} and assume x ∈ K ∩ ∂ 1 . Then

(ψ _j x)(t) ≤ λB

∫ _ω

0

f _j (s, x(s − τ(s)))ds

≤ λBω(l j + ϵ)

∫ _ω

0

x _j (s, x(s − τ(s)))ds

≤ λBω(l j + ϵ) ∥x j ∥ 0

≤ λBω(l M + ϵ)∥x j ∥ 0

≤ ∥x j ∥ 0 . In particular,

||ψ j x|| 0 ≤ ||x j || 0

and

(3.2) ||ψx|| =

∑ n j=1

||ψ j x || 0 ≤

∑ n j=1

||x j || 0 = ||x|| for all x ∈ K ∩ ∂ 1 .

Next we construct the set ₂ . Considering (L6) there exists H ₂ such that f _j (t, y) ≥ (L j − ϵ)y j ≥ (L m − ϵ)y j , for all y _j ≥ H 2 . Let H ₂ = max {2H 1 , ^H _σ

} and set

2 = {x ∈ K : ∥x j ∥ 0 < H ₂ , j = 1, · · · , n}.

If x ∈ K with ||x|| ≥ H 2 , then

x _j ≥ σ||x j || ≥ H 2 . Thus

(ψ j x)(t) ≥ λA

∫ _ω

0

f j (s, x(s − τ(s)))ds ≥ λAωσ(L m − ϵ)||x j || 0 .

Hence

||ψ j x ∥ 0 ≥ ||x j || 0

and

(3.3) ||ψx|| =

∑ n j=1

||ψ j x|| 0 ≥

∑ n j=1

||x j || 0 = ||x|| for all x ∈ K ∩ ∂ 2 .

Applying (i) of Theorem 2.1 to (3.2) and (3.3) yields that ψ has a xed point x ∈ K ∩ ( 2 \ 1 ). The proof is complete.

Theorem 3.3. Assume that (H1), (L5), and (L6) hold. Then, for each λ satisfying

(3.4) 1

ωσAl _m < λ < 1 ωBL _M (1.1) has at least one positive periodic solution.

Proof. We construct the sets ₁ and ₂ in order to apply Theorem 2.1. Let λ be given as in (3.4), and choose ϵ > 0 such that

1

σA(l _m − ϵ) ≤ λ ≤ 1 B(L _M + ϵ) .

By condition (L5), there exists H 1 > 0 such that f j (t, y) ≥ (l j − ϵ)y j ≥ (l _m − ϵ)y j , for 0 < y _j ≤ H 1 . Dene ₁ = {x ∈ K : ∥x j ∥ 0 < H ₁ , j = 1, 2, · · · , ·, ·, n}, and assume x ∈ K ∩ ∂ 1 . Then

(ψ _j x)(t) ≥ λA

∫ _ω

0

f _j (s, x(s − τ(s)))ds

≥ λAω(l m − ϵ)x j (t − τ(t))

≥ λAσω(l m − ϵ)∥x j ∥ 0

≥ ∥x j ∥ 0 . In particular,

||ψ j x || 0 ≥ ||x j || 0 for all x ∈ K ∩ ∂ 1

and

(3.5) ||ψx|| =

∑ n j=1

||ψ j x || 0 ≥

∑ n j=1

||x j || 0 = ||x||, for all x ∈ K ∩ ∂ 1 .

Next we construct the set ₂ . Considering (L6) there exists H ₂ such that f _j (t, y) ≤ (L j + ϵ)y _j ≤ (L M + ϵ)y _j , for y _j ≥ H 2 .

We consider two cases; f _j (t, y) is bounded and f _j (t, y) is unbounded.

The case where f j (t, y) is bounded is straight forward. If f j (t, y) is bounded by Q > 0, set

H 2 = max{2H 1 , ωλQB}.

Then if x ∈ K and ||x|| 0 = H 2 , we have (ψ _j x)(t) ≤ λB

∫ _ω

0

f _j (s, x(s − τ(s)))ds

≤ ωλBQ ≤ ∥x j ∥ 0 .

Consequently, ||ψ j x || 0 ≤ ||x j || 0 , and hence ||ψx|| ≤ ||x||.

So, if we set

2 = {y ∈ K : ∥y j ∥ < H 2 , j = 1, 2, ·, ·, ·, n}, then

(3.6) ||ψx|| ≤ ||x||, for x ∈ K ∩ ∂ 2 .

When f is unbounded, we let H 2 > max{2H 1 , H 2 } be such that f j (t, y) ≤ f j (t, H 2 ), for 0 < y j ≤ H 2 . For x ∈ K with ||x j || 0 = H 2 ,

(ψ j x)(t) ≤ λB

∫ _ω

0

f j (s, x(s − τ(s)))ds

≤ λB

∫ _ω

0

f j (s, H 2 )ds

≤ λB

∫ _ω

0

(L j + ϵ)H 2 ds

≤ λBω(L M + ϵ) ∥x j ∥ 0

≤ ∥x j ∥ 0 .

Consequently, ||ψ j x|| ≤ ||x j || 0 , which implies that

||ψx|| =

∑ n j=1

||ψ j x || 0 ≤

∑ n j=1

||x j || 0 = ||x||.

So, if we set

2 = {x ∈ K : ∥x j ∥ 0 < H 2 , j = 1, 2, ·, ·, ·, n}, then

(3.7) ||ψx|| ≤ ||x||, for x ∈ K ∩ ∂ 2 .

Applying (ii) of Theorem 2.1 to (3.5) and (3.6) yields that T has a xed

point x ∈ K ∩ ( 2 \ 1 ). Also, applying (ii) of Theorem 2.1 to (3.5) and

(3.7) yields that ψ has a xed point x ∈ K ∩ ( 2 \ 1 ). The proof is

complete.

Theorem 3.4. Assume that (H1), (L1), and (L6) hold. Then, for each λ satisfying

(3.8) 0 < λ < 1

ωAL _M , (1.1)–(1.2) has at least one positive solution.

Proof. Apply (L1) and choose H 1 > 0 such that if 0 < x j < H 1 , then f j (t, x) ≥ x _j

λγA . Dene

1 = {x ∈ K : ∥x j ∥ 0 < H 1 }.

If x ∈ K ∩ ∂ 1 , then

(ψ j x)(t) ≥ λA

∫ _ω

0

f j (s, x(s − τ(s)))ds

≥ λA

∫ _ω

0

x _j (s, s − τ(s))

λσA ds

≥ λA

∫ _ω

0

σ ||x j || 0

λσA ds

= ∥x j ∥ 0 .

In particular, ||ψx∥ ≥ ||x||, for all x ∈ K ∩ ∂ 1 . In order to construct

2 , we let λ be given as in (3.8), and choose ϵ > 0 such that

0 ≤ λ ≤ 1

Bω(L _M + ϵ) .

The construction of ₂ follows along the lines of the construction of ₂ in Theorem 3.3, and hence we omit it. Thus, by (ii) of Theorem 2.1, equation (1.1) has at least one positive solution.

Theorem 3.5. Assume that (H1), (L2), and (L5) hold. Then, for each λ satisfying

(3.9) 0 < λ < 1

Bωl _M , (1.1)–(1.2) has at least one positive solution.

Proof. Assume (L5) holds. Then, we may take the set ₁ to be the one obtained for Theorem 3.2. That is,

1 = {x ∈ K : ∥x j ∥ 0 < H ₁ , j = 1, 2, ...n }.

Hence, we have

||ψx|| ≤ ||x||, for x ∈ K ∩ ∂ 1 .

Next, we assume (L2). Choose H 2 > 0 such that f j (t, x) ≥ _λσA ^x

, for x _j ≥ H 2 . Let H ₂ = max {2H 1 , ^H _σ

} and set

2 = {x ∈ K : ∥x j ∥ 0 < H 2 }.

If x ∈ K with ||x|| 0 = H 2 , (ψ j x)(t) ≥ λA

∫ _ω

0

f j (s, x(s − τ(s)))ds

≥ λA

∫ _ω

0

x _j (s, s − τ(s))

λσA ds

≥ λA

∫ _ω

0

σ ||x j || 0

λσA ds

= ∥x j ∥ 0 .

In particular, ||ψx∥ ≥ ||x||, for all x ∈ K ∩ ∂ 2 . Consequently,

||ψx|| ≥ ||x||, for x ∈ K ∩ ∂ 2 .

Applying (i) of Theorem 2.1 yields that ψ has a xed point x ∈ K ∩ ( 2 \ 1 ).

We state the next results as corollary, because by now, its proof can be easily obtained from the proofs of the previous results.

Corollary 3.6. Assume that (H1) hold. Also, if either (L3) and (L6) hold, or, (L4), and (L5) hold, then (1.1)–(1.2) has at least one positive solution if λ satises either 1/(σAL m ) < λ, or, 1/(σAl m ) < λ.

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Department of Mathematics University of Dayton

Dayton, OH 45469-2316, U.S.A

E-mail : [email protected]