Boundedness and persistence of delay differential equations with mixed nonlinearity
aa r X i v : . [ m a t h . D S ] J un Boundedness and Persistence of Delay Differential Equations with MixedNonlinearity
Leonid Berezansky a , Elena Braverman b a Dept. of Math., Ben-Gurion University of the Negev, Beer-Sheva 84105, Israel b Dept. of Math. and Stats., University of Calgary,2500 University Drive N.W., Calgary, AB, Canada T2N 1N4;e-mail [email protected], phone 1-(403)-220-3956, fax 1-(403)–282-5150 (corresponding author)
Abstract
For a nonlinear equation with several variable delays˙ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) , where the functions f k increase in some variables and decrease in the others, we obtain conditionswhen a positive solution exists on [0 , ∞ ), as well as explore boundedness and persistence of solutions.Finally, we present sufficient conditions when a solution is unbounded. Examples include theMackey-Glass equation with non-monotone feedback and two variable delays; its solutions can beneither persistent nor bounded, unlike the well studied case when these two delays coincide. Keywords: nonlinear delay differential equations, a global positive solution, persistent,permanent and unbounded solutions, population dynamics models, Mackey-Glass equation
AMS Subject Classification:
1. Introduction
Many mathematical models of population dynamics can be written in the form of a scalarequation ˙ x ( t ) = f ( x ( t − τ )) − x ( t ) , (1.1)where f is a nonnegative continuous function describing reproduction or recruitment, τ is a positivenumber describing delay. Usually these models have a unique positive equilibrium K , and thereis a well-developed theory on the global stability of the positive equilibrium of (1.1). This theorywas applied to many well-known models described by Eq. (1.1) such as Nicholson’s blowflies delayequation and Mackey-Glass equations.Eq. (1.1) can be extended to the case when both the delay and the intrinsic growth rate arevariable ˙ x ( t ) = r ( t ) [ f ( x ( h ( t ))) − x ( t )] , (1.2)where h ( t ) ≤ t and r ( t ) > Preprint submitted to Applied Mathematics and Computation August 11, 2018 nother generalization of (1.1) is the model with several production terms and nonlinear mor-tality ˙ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) , (1.3)where f k , g are nonnegative continuous functions. This equation with some applications was studied,for example, in [3, 5, 8, 17, 25].For all mentioned above equations usual assumptions are the following: the function f k is eithermonotone or unimodal, g ( t, u ) is monotone increasing in u , there is only one delay involved in f k ,and a positive equilibrium is unique. However, it is possible to consider more general models, forexample, the modified Nicholson equation˙ x ( t ) = m X k =1 a k ( t ) x ( h k ( t )) e − λ k x ( g k ( t )) − b ( t ) x ( t ) , t ≥ , (1.4)and the modified Mackey-Glass type equation˙ x ( t ) = m X k =1 a k ( t ) x ( h k ( t ))1 + x n k ( p k ( t )) − (cid:18) b ( t ) − c ( t )1 + x n ( t ) (cid:19) x ( t ) , t ≥ . (1.5)There are also many generalizations of Eqs. (1.1)-(1.5) to the case of distributed delays andintegro-differential equations [6, 11, 12, 22, 26].Let us illustrate the idea that the presence of several delays instead of one delay can createa new type of dynamics. As Example 1.1 illustrates, an equation which was stable for coincidingdelays can become unstable, once the two delays are different. Example 1.1.
Consider the modified Mackey-Glass equation with two delays ˙ x ( t ) = 2 x ( h ( t ))1 + x ( g ( t )) − x ( t ) , t ≥ . (1.6) The unique positive equilibrium is x = 1 , the function f ( x ) = 2 x/ (1 + x ) is increasing on [0 , , soany positive solution of the equation ˙ x ( t ) = 2 x ( h ( t ))1 + x ( h ( t )) − x ( t ) , t ≥ satisfies lim t →∞ x ( t ) = 1 , see, for example, [10, 12]. Consider (1.6) with piecewise constant arguments h ( · ) , g ( · ) . Denote a = ln(59 / ≈ . and b = ln(134 / ≈ . and let ϕ ( t ) = 6 . − . e − ( t + a + b ) , t ∈ [ − a − b, − b ] , ϕ ( t ) = 117 + 6717 e − ( t + b ) , t ∈ [ − b, , then ϕ ( − a − b ) = 0 . , ϕ ( − b ) = 4 , ϕ (0) = + = 0 . . Assume for n = 0 , , , . . .h ( t ) = h ta + b i − b, t ∈ [ n ( a + b ) , n ( a + b ) + a ) , h ta + b i − a − b, t ∈ [ n ( a + b ) + a, ( n + 1)( a + b )) , here [ t ] is the integer part of t , g ( t ) = h ta + b i − a − b, t ∈ [ n ( a + b ) , n ( a + b ) + a ) , h ta + b i − b, t ∈ [ n ( a + b ) + a, ( n + 1)( a + b )) . Then the solution is ( a + b ) -periodic, the equation is ˙ x ( t ) = − x ( t ) on [ n ( a + b ) , n ( a + b ) + a ) , x ( n ( a + b )) = and ˙ x ( t ) = − x ( t ) on [ n ( a + b ) + a, ( n + 1)( a + b )) , x ( n ( a + b ) + a ) = 4 . Thus, withtwo delays, the equilibrium K = 1 of Eq. (1.6) is not globally asymptotically stable, unlike (1.7). As Example 1.1 illustrates, an equation which was stable for the coinciding delays can haveoscillating solutions with a constant amplitude which do not tend to the positive equilibrium.According to Example 5.8, two different delays can lead not only to sustainable oscillations butalso to unbounded solutions.The purpose of the present paper is to consider a general nonlinear delay equation which includes(1.4), (1.5) as particular cases and study the following properties of these equations: existence anduniqueness of a positive global solution, persistence, permanence, as well as existence of unboundedsolutions. To the best of our knowledge, equations with such mixed types of nonlinearities havenot been studied before.Compared to most of the previous publications, we consider two modifications: the productionfunction is a sum of several functions, and each f k involves several delays. The situation whenseveral (sometimes incomparable) delays are included, is quite common, for example, transmissionand translation delays in gene regulatory systems. Motivated by this, we apply the general resultsto some well-known population dynamics equations.The paper is organized as follows. After introducing some relevant assumptions and definitionsin Section 2, we justify existence of a global positive solution in Section 3. Section 4 deals withsufficient conditions when all positive solutions are bounded. In Section 5, we investigate persistenceof solutions and also consider their permanence. Section 6 explores positive unbounded solutions,and Section 7 involves brief discussion.
2. PreliminariesDefinition 2.1.
We will say that f ( t, u , . . . , u l ) is a Caratheodory function if in its domain itis continuous in u , . . . , u l for almost all t and is locally essentially bounded in t for any u , . . . , u l .The function f ( t, u , . . . , u l ) is a locally Lipschitz function if for any interval [ a, b ] thereexist positive constants α k ([ a, b ]) such that | f ( t, u , . . . , u l ) − f ( t, v , . . . , v l ) | ≤ l X k =1 α k ([ a, b ]) | u k − v k | , u k , v k ∈ [ a, b ] , k = 1 , . . . , l, t ≥ . In this paper we consider the scalar nonlinear equation with several delays˙ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) , t ≥ f k : [0 , ∞ ) × R l → [0 , ∞ ) , g : [0 , ∞ ) × R → [0 , ∞ ) are Caratheodory and locally Lipschitzfunctions, f k ( t, , . . . ,
0) = 0 , g ( t,
0) = 0;(a2) h j , j = 1 . . . , l, are Lebesgue measurable functions, h j ( t ) ≤ t, lim t →∞ h j ( t ) = ∞ .Together with Eq. (2.1) consider an initial condition x ( t ) = ϕ ( t ) , t ≤ , (2.2)where(a3) ϕ : ( −∞ , → R is a nonnegative Borel measurable bounded function, ϕ (0) > Definition 2.2.
The solution of problem (2.1),(2.2) is an absolutely continuous on [0 , ∞ ) functionsatisfying (2.1) almost everywhere for t ≥ and condition (2.2) for t ≤ . Instead of the initial point t = 0 we can consider any initial point t = t >
0. In Definition 2.2the interval [0 , ∞ ) can be substituted by the maximum interval (0 , c ), with c >
0, or ( t , c ), c > t where the solution exists. However, in the present paper we only consider the case when a globalpositive solution exists on [0 , ∞ ). Sufficient conditions for existence of a positive solution on [0 , ∞ )are discussed in the next section.
3. Existence of a Positive Solution on [0 , ∞ ) Let us first justify that if (a1)-(a3) are satisfied then the positive local solution of (2.1), (2.2)exists and is unique.Denote by L ([ t , t ]) the space of Lebesgue measurable real-valued functions x ( t ) such that Q = Z t t ( x ( t )) dt < ∞ , with the usual norm k x k L ([ t ,t ]) = √ Q , by C ([ t , t ]) the space ofcontinuous on [ t , t ] functions with the sup-norm.The following result from the book of Corduneanu [13, Theorem 4.5, p. 95] will be applied. Werecall that an operator N is causal (or Volterra ) if for any two functions x and y and each t thefact that x ( s ) = y ( s ), s ≤ t , implies ( N x )( s ) = ( N y )( s ), s ≤ t . Lemma 3.1. [13] Consider the equation y ′ ( t ) = ( L y )( t ) + ( N y )( t ) , t ∈ [ t , t ] , y ( t ) = y , (3.1) where L : C ([ t , t ]) → L ([ t , t ]) is a linear bounded causal operator, N : C ([ t , t ]) → L ([ t , t ]) is a nonlinear causal operator which satisfies kN x − N y k L ([ t ,t ]) ≤ λ k x − y k C ([ t ,t ]) for λ sufficiently small. Then there exists a unique absolutely continuous on [ t , t ] solution of(3.1). Let us note that in Lemma 3.1, L and N are causal operators and thus can include delays.They are defined on C ([ t , t ]) which corresponds to delay equations with the zero initial functionfor t < t . For an arbitrary initial function and t = 0, in the proof of Theorem 3.2 we reduce theproblem to the zero initial function and t ≥ Theorem 3.2.
Suppose (a1)-(a3) hold. Then there exists a unique local positive solution of (2.1),(2.2). roof. In order to reduce (2.1), (2.2) to the equation which will be considered for t ≥
0, we rewritethis problem as ˙ x ( t ) = m X k =1 f k ( t, x h ( t ) + ϕ h ( t ) , . . . , x h l ( t ) + ϕ h l ( t )) − g ( t, x ( t )) , (3.2)where x h j ( t ) = (cid:26) x ( h j ( t )) , h j ( t ) > , , h j ( t ) ≤ , , ϕ h j ( t ) = (cid:26) ϕ ( h j ( t )) , h j ( t ) ≤ , , h j ( t ) > . We can consider (3.2) for t ≥ L x )( t ) ≡
0, ( N x )( t ) = m X k =1 f k ( t, x h ( t ) + ϕ h ( t ) , . . . , x h l ( t ) + ϕ h l ( t )) − g ( t, x ( t )) . We havefor any t > kN x − N y k L ([0 ,t ]) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) m X k =1 f k ( · , x h ( · ) + ϕ h ( · ) , . . . , x h l ( · ) + ϕ h l ( · )) − m X k =1 f k ( · , y h ( · ) + ϕ h ( · ) , . . . , y h l ( · ) + ϕ h l ( · )) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ([0 ,t ]) + k g ( · , x ( · )) − g ( · , y ( · )) k L ([0 ,t ]) ≤ m X k =1 l X j =1 α jk k x h j ( · ) − y h j ( · ) k L ([0 ,t ]) + β k x − y k L ([0 ,t ]) , where α jk = α jk ([0 , t ]), β = β ([0 , t ]) are local Lipschitz constants for f k and g , respectively.Then kN x − N y k L ([0 ,t ]) ≤ m X k =1 l X j =1 α jk + β √ t k x − y k C ([0 ,t ]) . Hence if t is sufficiently small, the constant λ = ( P mk =1 P lj =1 α jk + β ) √ t is also sufficiently small.Thus by Lemma 3.1 there exists a unique solution of problem (2.1), (2.2) on [0 , t ]. Since x (0) > t this solution is positive, which concludes the proof. Theorem 3.3.
Suppose conditions (a1)-(a3) are satisfied and at least one of the following assump-tions holds: ( a ) for any [ a, b ] there exists ε > such that h j ( t ) ≤ b − ε for t ∈ [ a, b ] , j = 1 , . . . , l ; ( a ) 0 ≤ f k ( t, u , . . . , u l ) ≤ l X j =1 a kj ( t ) u j + b k ( t ) , where a kj , b k are locally integrable functions; ( a ) for x sufficiently large g ( t, x ) − m X k =1 f k ( t, u , . . . , u l ) ≥ a x > , x ≥ u j , j = 1 , . . . , l .Then problem (2.1), (2.2) has a unique positive global solution on [0 , ∞ ) .Proof. By Theorem 3.2 there exists a unique local positive solution of this problem. Suppose [0 , c ) isa maximum interval of existence for this solution. Since ˙ x ( t ) ≥ − g ( t, x ( t )), x (0) > g ( t,
0) = 0,5e have x ( t ) > z ( t ) ≡ t ∈ [0 , c ), where z is a solution of ˙ z ( t ) = − g ( t, z ( t )), z (0) = 0, and thesolution z is unique due to the local Lipschitz condition for g as a part of (a1).If c = + ∞ the theorem is proved. Suppose c < ∞ .Let us first verify that lim inf t → c − x ( t ) >
0. By (a1) and continuity of the solution on [0 , c ], thereexists
M > g ( t, x ( t )) ≤ M , t ∈ [0 , c ]. Following the above argument, we obtain ˙ x ( t ) ≥− g ( t, x ( t )) > − M and x ( t ) > x (0) e − Mc , which implies lim inf t → c − x ( t ) >
0. Thus lim sup t → c − x ( t ) =+ ∞ .In fact, assuming the contrary that lim sup t → c − x ( t ) < + ∞ then, by (a3), there exists M > ≤ x ( t ) ≤ M on [0 , c ) and ϕ ( t ) ≤ M . Since f k ( · , u , . . . , u l ) and g ( · , u ) are locallyLipschitz, they are locally essentially bounded for t ∈ [0 , c ], u, u j ∈ [0 , M ], thus ˙ x is also essentiallybounded on [0 , c ).The solution satisfies x ( t ) = x (0) + Z t ˙ x ( s ) ds, x ( c ) = x (0) + Z c ˙ x ( s ) ds, thus the solution can be defined for t ≥ c and [0 , c ) is not the maximum interval of existence. Thus,when justifying existence, we only need to prove boundedness of a solution on any finite interval.Consider now the three cases.1) Suppose ( a ) holds. Then there exists t < c such that h j ( t ) ≤ t , t ∈ [0 , c ), thus | x ( h j ( t )) | ≤ max t ∈ [0 ,t ] | x ( t ) | < ∞ , hence0 < x ( t ) ≤ | x (0) | + m X k =1 Z c sup ≤ s ≤ t f k ( s, x ( h ( s )) , . . . , x ( h l ( s ))) ds = A < ∞ , t ∈ [0 , c ) , and therefore lim sup t → c − x ( t ) = + ∞ is impossible.2) Suppose ( a ) holds. Then˙ x ( t ) ≤ m X k =1 l X j =1 a kj ( t ) x ( h j ( t )) + b k ( t ) . If lim t → c − x ( t ) = + ∞ then there is t ∈ (0 , c ) such that x ( t ) = max s ≤ t x ( s ) . (3.3)Since x ( t ) is positive on (0 , c ), on [ t , c ) it does not exceed the solution of the equation˙ z ( t ) = m X k =1 l X j =1 a kj ( t ) z ( h j ( t )) + b k ( t ) , z ( t ) = x ( t ) , t ≤ t . The function z ( t ) is monotone nondecreasing on [ t , c ), and from (3.3), z ( t ) = max ≤ s ≤ t z ( s ) , z ( t ) ≥ z ( h j ( t )) , t ∈ [ t , c ) , j = 1 , . . . , l. x ( t ) ≤ y ( t ), where y is a solution of the equation˙ y ( t ) = a ( t ) y ( t ) + b ( t ) , y ( t ) = x ( t ) , a ( t ) := m X k =1 l X j =1 a kj ( t ) , b ( t ) := m X k =1 b k ( t ) , t ∈ [ t , c ) . Hence x ( t ) ≤ y ( t ) ≤ Z ct b ( s ) exp (cid:26)Z ct a ( τ ) dτ (cid:27) ds + x ( t ) exp (cid:26)Z ct a ( τ ) dτ (cid:27) = A < ∞ , t ≤ c, since a kj and b k are integrable on [ t , c ], and therefore there is a positive solution on [0 , ∞ ).3) Suppose ( a ) holds. Since ( a ) is satisfied for x large enough, we can find A > a ) holds for x ≥ u j ≥ A , j = 1 , . . . , m . Let us choose M ≥ A , M ≥ A +sup t ≤ ϕ ( t ). The function g is locally Lipschitz, thus there is α > | g ( t, x ) − g ( t, y ) | ≤ α | x − y | , x, y ∈ [0 , M ], for any t . We recall that there is a A such that g ( t, x ) − P mk =1 f k ( t, u , . . . , u l ) ≥ a A > x ≥ u j ≥ A .Denote ε = min { a A / (2 α ) , A } then for x, y ∈ [0 , M ], | x − y | < ε , g ( t, y ) ≥ g ( t, x ) − | g ( t, x ) − g ( t, y ) | ≥ g ( t, x ) − α | x − y | ≥ g ( t, x ) − αε ≥ g ( t, x ) − a A . Therefore for y ≥ u j − ε , j = 1 , . . . , l , x, y ∈ [0 , M ], | x − y | ≤ εg ( t, y ) − m X k =1 f k ( t, u , . . . , u l ) ≥ a M / > . (3.4)If x ( t ) ≥ M for some t ∈ (0 , c ), denote t = inf { t ∈ [0 , c ] | x ( t ) = M } , t = sup { t ∈ [0 , t ] | x ( t ) = M − ε } . By definition t > t >
0, and from continuity of x , x ( t ) = M − ε , x ( t ) = M , x ( h j ( t )) < M , t < t , j = 1 , . . . , l and x ( t ) ∈ ( M − ε, M ), t ∈ ( t , t ). However, (3.4) implies˙ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≤ − a M / < , t ∈ ( t , t ) , thus x ( t ) > x ( t ), which contradicts to the assumption x ( t ) = M − ε < x ( t ) = M . Thus x ( t ) ≤ M , t ∈ [0 , c ]; in fact, the inequality is satisfied for any t . Hence a positive solution exists on[0 , ∞ ). Remark 3.4.
The conditions of Theorem 3.3 and [6, Theorem 2.2] are independent.
Example 3.5.
For the equation ˙ x ( t ) = x ( t − τ ) , τ > condition ( a ) holds and ( a ) , ( a ) fail. It is interesting to note that the equation ˙ x ( t ) = x with the initial condition x (0) = x > has the solution x ( t ) = 1 / ( x − − t ) which only exists on [0 , /x ) . or the equation ˙ x ( t ) = x ( t − | sin t | ) , (3.6) condition ( a ) holds and ( a ) , ( a ) fail.For the equation ˙ x ( t ) = x ( t − | sin t | )1 + x ( t ) − x ( t ) (3.7) condition ( a ) holds and ( a ) , ( a ) fail.By Theorem 3.3, problems for Eqs. (3.5)-(3.7) with an initial function satisfying (a3), have aunique positive global solution. For the rest of the paper, we everywhere assume that problem (2.1), (2.2) has a unique positiveglobal solution on [0 , ∞ ).
4. Boundedness of Solutions
Let us consider conditions under which all global solutions of (2.1), (2.2) are bounded.
Theorem 4.1.
Suppose conditions (a1)-(a3) hold. Let also one of the following conditions besatisfied:(a) f k ( t, u , . . . , u l ) are strictly monotone increasing in u , . . . , u l , lim sup u →∞ P mk =1 f k ( t, u, . . . , u ) g ( t, u ) < uniformly in t ;(b) f k ( t, u , . . . , u l ) are strictly monotone increasing in u j for some j ∈ { , , . . . , l } and lim sup u j →∞ P mk =1 f k ( t, u , . . . , u l ) g ( t, u j ) < uniformly in t , u , . . . , u j − , u j +1 . . . , u l .Then any solution of problem (2.1), (2.2) is bounded.If the following condition holds:(c) f k ( t, u , . . . , u l ) are strictly monotone increasing in u , . . . , u n for some n ∈ { , , . . . , l } andthere exists M > such that for any M ≥ M , . . . , M l − n ≥ M lim sup u →∞ P mk =1 f k ( t, u, . . . , u, M , . . . , M l − n ) g ( t, u ) < uniformly in t , then there is no solution x of problem (2.1),(2.2) such that lim t →∞ x ( t ) = ∞ . Proof.
Suppose that condition (a) holds and x is an unbounded solution of problem (2.1), (2.2).Let A > sup t ≤ ϕ ( t ) > σ > P mk =1 f k ( t, u, . . . , u ) ≤ (1 − σ ) g ( t, u ) for u > A . As x is unbounded, for any fixed M > A there exist points t such that x ( t ) ≥ M . Denote t = inf { t ≥ | x ( t ) ≥ M } , t > M > A > sup t ≤ ϕ ( t ) > x ( t ) < M for t ≤ t . Let t = sup { t ≤ t | x ( t ) ≤ A } . Since x ( t ) < A for t ≤
0, we have t >
0; by definition, t < t . Also, x ( t ) ≥ A on [ t , t ] with A = x ( t ) < x ( t ) = M .Since f k are increasing in u j and x ( h j ( t )) < M on ( −∞ , t ), we have m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) < m X k =1 f k ( t, M, . . . , M ) − g ( t, x ( t )) ≤ (1 − σ ) g ( t, M ) − g ( t, x ( t )) ≤ g ( t, M ) − g ( t, x ( t )) , t ∈ [ t , t ] . Thus the solution x ( t ) of (2.1),(2.2) on [ t , t ] does not exceed the solution of the initial valueproblem for the ordinary differential equation˙ y ( t ) = g ( t, M ) − g ( t, y ( t )) , y ( t ) = A < M, (4.1)i.e. x ( t ) ≤ y ( t ), t ∈ [ t , t ]. However, the solution of (4.1) satisfies y ( t ) < M , t ≥ t . In fact,assuming the contrary, we obtain that y ( t ∗ ) = M for some t ∗ > t , and there are two solutionsthrough ( t ∗ , M ): y and the one identically equal to M . This contradicts to the assumption of thelocal Lipschitz condition which implies uniqueness. Thus x ( t ) ≤ y ( t ) < M , and the contradictionwith x ( t ) = M proves boundedness of the solution x of (2.1),(2.2).If condition (b) holds, the proof is similar to the previous case. Let for some σ > P mk =1 f k ( t, u , . . . , u l ) ≤ (1 − σ ) g ( t, u j ) for u j > A . Defining A, M, t , t as previously, we obtain m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) < m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h j − ( t )) , M, x ( h j +1 ( t )) , x ( h l ( t ))) − g ( t, x ( t )) ≤ (1 − σ ) g ( t, M ) − g ( t, x ( t )) ≤ g ( t, M ) − g ( t, x ( t )) , t ∈ [ t , t ] . Again, comparing the solution x ( t ) of (2.1),(2.2) on [ t , t ] with the solution of (4.1) satisfying y ( t ) < M , we obtain the contradiction x ( t ) ≤ y ( t ) < M with the assumption x ( t ) = M .Finally, assume that condition (c) holds. Let x be a solution of problem (2.1),(2.2) satisfyinglim t →∞ x ( t ) = ∞ . Since lim t →∞ h k ( t ) = ∞ , k = 1 , . . . , l , there exists t ≥ x ( h n +1 ( t )) ≥ M , . . . , x ( h l ( t )) ≥ M for t ≥ t .In addition, there is a number A , A > sup t ≤ ϕ ( t ) > A > sup t ∈ [0 ,t ] x ( t ) > m X k =1 f k ( t, u, . . . , u, M , . . . , M l − n ) ≤ (1 − σ ) g ( t, u ) , u ≥ A, M ≥ M , . . . , M l − n ≥ M . M > A and choosing t > t > t as previously such that x ( t ) < M for x < t , x ( t ) = A and x ( t ) ∈ ( A, M ) for t ∈ [ t , t ], we notice that m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) < m X k =1 f k ( t, M, . . . , M, x ( h n +1 ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≤ (1 − σ ) g ( t, M ) − g ( t, x ( t )) ≤ g ( t, M ) − g ( t, x ( t )) , t ∈ [ t , t ] . Comparing the solution x ( t ) of (2.1),(2.2) on [ t , t ] with the solution of (4.1) satisfying y ( t ) < M ,we obtain a contradiction x ( t ) ≤ y ( t ) < M to the assumption x ( t ) = M . Thus there are nosolutions which tend to + ∞ as t → ∞ . Remark 4.2.
The proof of Theorem 4.1 implies that its conditions can be relaxed to m X k =1 f k ( t, u, . . . , u ) < g ( t, u ) for any t and u large enough in (a), m X k =1 f k ( t, u , . . . , u l ) < g ( t, u j ) for any t , u , . . . , u j − , u j +1 . . . , u l as mentioned in (b) and m X k =1 f k ( t, u, . . . , u, M , . . . , M l − n ) < g ( t, u ) for any t and for any u large enough in (c). Example 4.3.
Consider Eq. (1.5), where a k ( t ) ≥ , b ( t ) ≥ c ( t ) ≥ , b ( t ) − c ( t ) ≥ β > areLebesgue measurable bounded functions, for functions h k , g k condition (a2) holds, n k ≥ , n ≥ .Here condition ( a ) of Theorem 3.3 holds, thus there exists a global positive solution of problem(1.5), (2.2).Denote f k ( t, u, v ) = a k ( t ) u/ (1+ v n k ) , g ( t, u ) = b ( t ) u − c ( t ) u u n . The functions f k ( t, u, v ) are strictlymonotone increasing in u . We have m X k =1 f k ( t, u, v ) g ( t, u ) ≤ m X k =1 a k ( t ) b ( t ) − c ( t ) . Let lim sup t →∞ P mk =1 a k ( t ) b ( t ) − c ( t ) < , hen there exists t > such that sup t ≥ t P mk =1 a k ( t ) b ( t ) − c ( t ) < . Shifting in Theorem 4.1 (b) the initial pointto t and noticing that a continuous solution is bounded on [0 , t ] , we conclude that all solutions ofEq. (1.5) are bounded.Evidently, condition (c) of Theorem 4.1 holds without any additional conditions. Hence thereis no solution satisfying lim t →∞ x ( t ) = ∞ . Example 4.4.
Consider the equation ˙ x ( t ) = a ( t ) x ( t − h ) x ( t − g ) − (cid:18) b ( t ) − c ( t )1 + x n ( t ) (cid:19) x ( t ) , (4.2) where a ( t ) ≥ , b ( t ) ≥ c ( t ) ≥ , b ( t ) − c ( t ) ≥ β > , a, b, c are Lebesgue measurable boundedfunctions, h > , g > , n ≥ . Here condition ( a ) of Theorem 3.3 holds, thus there exists aglobal positive solution of problem (4.2), (2.2).Denote f ( t, u, v ) = a ( t ) uv , g ( t, u ) = (cid:16) b ( t ) − c ( t )1+ u n (cid:17) u . The function f is monotone increasingin both u and v . We have f ( t, u, u ) g ( t, u ) = a ( t ) b ( t ) − c ( t )1+ u n ≤ a ( t ) b ( t ) − c ( t ) . Hence if lim sup t →∞ a ( t ) b ( t ) − c ( t ) < then there exists t > such that sup t ≥ t P mk =1 a k ( t ) b ( t ) − c ( t ) < . Shifting in Theorem 4.1 (a) the initial pointto t and noticing that the solution is bounded on [0 , t ] , we conclude that all solutions of Eq. (4.2)are bounded. We will give another statement on boundedness where monotonicity is not required.
Theorem 4.5.
Suppose conditions (a1)-(a3) hold, g ( t, u ) ≥ a ( t ) u for all u ≥ and ≤ f k ( t, u , . . . , u l ) ≤ l X j =1 A kj ( t ) u j + B k for all u j ≥ , where a ( t ) ≥ , B k ≥ , A kj : [0 , ∞ ) → [0 , ∞ ) are locally essentially bounded functions.If the linear equation ˙ x ( t ) = − a ( t ) x ( t ) + m X k =1 l X j =1 A kj ( t ) x ( h j ( t )) (4.3) is exponentially stable, then any positive solution of problem (2.1), (2.2) is bounded.Proof. If x is a solution of (2.1), (2.2) then˙ x ( t ) ≤ − a ( t ) x ( t ) + m X k =1 l X j =1 A kj ( t ) x ( h j ( t )) + m X k =1 B k . x ( t ) ≤ y ( t ) by [1, Corollary 2.2], where y is a solution of the linear equation˙ y ( t ) = − a ( t ) y ( t ) + m X k =1 l X j =1 A kj ( t ) y ( h j ( t )) + m X k =1 B k , y ( t ) = x ( t ) , t ≤ . Since Eq. (4.3) is exponentially stable, y is a bounded function. Hence x is also a bounded function. Example 4.6.
Consider again Eq. (1.5) with the same conditions and notations as in Example 4.3.We have f k ( t, u, v k ) ≤ a k ( t ) u, g ( t, u ) ≥ ( b ( t ) − c ( t )) u , u ≥ . Hence if the linear equation ˙ x ( t ) = − ( b ( t ) − c ( t )) x ( t ) + l X k =1 a k ( t ) x ( h k ( t )) is exponentially stable, all solutions of Eq. (1.5) are bounded. In particular, the condition lim inf t →∞ ( b ( t ) − c ( t )) > , lim sup t →∞ P lk =1 a k ( t ) b ( t ) − c ( t ) < implies boundedness (see, for example, [4, Corollary 1.4]). Corollary 4.7.
Suppose conditions (a1)-(a3) hold, g ( t, u ) ≥ a u > for u > , and lim sup u j →∞ ,j =1 ,...,l f k ( t, u , . . . , u l ) ≤ B k . Then any solution of problem (2.1), (2.2) is bounded.
Example 4.8.
Consider the Mackey-Glass type equation ˙ x ( t ) = m X k =1 a k ( t ) | sin( x ( h k ( t ))) | x n ( h ( t )) + · · · + x n l ( h l ( t )) − (cid:18) b ( t ) + c ( t )1 + x n ( t ) (cid:19) x ( t ) , t ≥ , (4.4) where a k , b, c are nonnegative essentially bounded on [0 , ∞ ) functions, b ( t ) + c ( t ) ≥ β > , n j ≥ , n > . Here condition ( a ) of Theorem 3.3 holds, thus there exists a global positive solution ofproblem (4.4), (2.2). Denote f k ( t, u, u , . . . , u l ) = a k ( t ) | sin u | u n + · · · + u n l l , g ( t, u ) = b ( t ) u + c ( t ) u u n . Hence g ( t, u ) ≥ βu , and the functions f k ( t, u, u , . . . , u l ) are bounded. By Corollary 4.7, all solutionsof Eq. (4.4) are bounded.Let us note that Theorem 4.1 cannot be applied to (4.4), as the functions f k ( t, u, u , . . . , u l ) arenot monotone increasing in u .
5. Persistence of Solutions
We proceed now to persistence and permanence of solutions. As previously, we everywhereassume that problem (2.1), (2.2) has a unique positive global solution on [0 , ∞ ). Definition 5.1.
A positive solution x ( t ) is persistent if lim inf t →∞ x ( t ) > and is permanent if itis also bounded. heorem 5.2. Suppose that conditions (a1)-(a3) are satisfied.(a) If f k ( t, u , . . . , u l ) are strictly monotone increasing in u , . . . , u l and lim inf u → + P mk =1 f k ( t, u, . . . , u ) g ( t, u ) > uniformly on t ∈ [0 , ∞ ) then any solution x of (2.1), (2.2) satisfies lim inf t →∞ x ( s ) > .(b) If f k ( t, u , . . . , u l ) are strictly monotone increasing in u , . . . , u n for some n ∈ { , . . . , l } ,monotone decreasing in u n +1 , . . . , u l , and there exists M > such that for any < M , . . . , M l − n ≤ M we have lim inf u → + P mk =1 f k ( t, u, . . . , u, M , . . . , M l − n ) g ( t, u ) > uniformly on t ∈ [0 , ∞ ) , then there is no solution x of (2.1), (2.2) satisfying lim t →∞ x ( s ) = 0 .Proof. First assume that the assumption in (a) holds. Suppose that x is a solution of (2.1), (2.2)such that lim inf t →∞ x ( t ) = 0.The solution is positive, we can consider t such that h k ( t ) > t > t , and reduce ourselves to t > t . Then there exist σ > b > P mk =1 f k ( t, u, . . . , u ) ≥ (1+ σ ) g ( t, u )for u ∈ (0 , b ).Let us note that, as the solution is positive, min t ∈ [0 ,t ] x ( t ) > m < a < b such that also 0 < m < a < min t ∈ [0 ,t ] x ( t ).As lim inf t →∞ x ( t ) = 0, there exist points t such that x ( t ) ≤ m . Denote t = inf { t ≥ t | x ( t ) ≤ m } , then t > t and x ( t ) > m for t ∈ [0 , t ). Let t = sup { t ≤ t | x ( t ) ≥ a } . The inequality x ( t ) > a for t ∈ [0 , t ] implies t > t ; by definition, t < t . We have m ≤ x ( t ) ≤ a on [ t , t ] with a = x ( t ) > x ( t ) = m .Since f k are increasing in u j and x ( h j ( t )) > m on ( t , t ), m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) > m X k =1 f k ( t, m, . . . , m ) − g ( t, x ( t )) ≥ (1 + σ ) g ( t, m ) − g ( t, x ( t )) ≥ g ( t, m ) − g ( t, x ( t )) , t ∈ [ t , t ] . Thus the solution x ( t ) of (2.1),(2.2) on [ t , t ] is not less than the solution of the initial valueproblem for the ordinary differential equation˙ y ( t ) = g ( t, m ) − g ( t, y ( t )) , y ( t ) = a > m, (5.1)13.e. x ( t ) ≥ y ( t ), t ∈ [ t , t ]. However, the solution of (5.1) satisfies y ( t ) > m , t ≥ t . In fact,assuming the contrary, we obtain that y ( t ∗ ) = m for some t ∗ > t , and there are two solutionsthrough ( t ∗ , m ): y and the one identically equal to m . This is impossible as g is locally Lipschitzwhich implies uniqueness. Thus x ( t ) ≥ y ( t ) > m which contradicts to the assumption x ( t ) = m .Hence all solutions are persistent.Next, let us assume that the conditions in (b) hold and x ( t ) → t → ∞ . Let ¯ t be such that x ( t ) ≤ M for t ≥ ¯ t and t ≥ ¯ t such that h j ( t ) ≥ ¯ t for t ≥ t , j = 1 , . . . , l .Thus x ( h j ( t )) ≤ M for t ≥ t . Next, there are σ > a > P mk =1 f k ( t, u, . . . , u, M , . . . , M l − n ) ≥ (1 + σ ) g ( t, u ) for u ∈ (0 , a ) and any 0 < M , . . . , M l − n ≤ M .Let 0 < m < a ; as x ( t ) →
0, there is a t such that x ( t ) ≤ m . Introducing t = inf { t ≥ t | x ( t ) ≤ m } , t = sup { t ≤ t | x ( t ) ≥ a } , we notice that x ( h j ( t )) < M , m < x ( t ) < a for t ∈ [ t , t ], x ( t ) = a > x ( t ) = m . Therefore m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) > m X k =1 f k ( t, m, . . . , m, x n +1 ( h n +1 ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≥ (1 + σ ) g ( t, m ) − g ( t, x ( t )) ≥ g ( t, m ) − g ( t, x ( t )) , t ∈ [ t , t ] . Thus the solution x ( t ) of (2.1),(2.2) on [ t , t ] is not less than the solution of the initial valueproblem (5.1) which, as in case (a), satisfies y ( t ) > m . Hence x ( t ) ≥ y ( t ) > m , the contradictionwith x ( t ) = m yields that the solution does not tend to zero. Remark 5.3.
The proof of Theorem 5.2 implies that its conditions in fact can be relaxed to m X k =1 f k ( t, u, . . . , u ) > g ( t, u ) for any t ∈ [0 , ∞ ) and u > small enough in (a) and m X k =1 f k ( t, u, . . . , u, M , . . . , M l − n ) > g ( t, u ) for any t ∈ [0 , ∞ ) , u > small enough and < M , . . . , M l − n ≤ M in (b). Example 5.4.
Consider Eq. (1.5) with the same conditions as in Example 4.3. We also use thesame notations as in Example 4.3. For Eq. (1.5), we have lim inf u → + P mk =1 f k ( t, u, v ) g ( t, u ) ≥ m X k =1 a k ( t ) b ( t )(1 + v n k ) There exist M > and t ≥ such that the condition lim inf t →∞ m X k =1 a k ( t ) b ( t ) > implies m X k =1 a k ( t ) b ( t )(1 + M n k ) > or M ≤ M and t ≥ t . In Theorem 5.2 (b) we shift the initial point to t and notice that the boundsof the positive solution on [0 , t ] do not influence the asymptotics. Hence for lim inf t →∞ m X k =1 a k ( t ) b ( t ) > ,there is no solution x of Eq. (1.5) satisfying lim t →∞ x ( s ) = 0 . Further we illustrate in Example 5.8 that the conditions in Theorem 5.2 (b) are not sufficientto establish permanence of solutions.
Example 5.5.
Consider Eq. (4.2) with the same conditions as in Example 4.4. We also use thesame notations as in Example 4.4. For Eq. (4.2), we have lim inf u → f ( t, u, u ) g ( t, u ) ≥ a ( t ) b ( t ) . Let lim inf t →∞ a ( t ) b ( t ) > , then inf t ≥ t a ( t ) b ( t ) > for some t ≥ . Shifting the initial point to t inTheorem 5.2 (a) and noticing that the solution is positive on [0 , t ] , we conclude that lim inf t →∞ a ( t ) b ( t ) > implies that all solutions of Eq. (4.2) are persistent.Therefore if lim inf t →∞ a ( t ) b ( t ) > , lim sup t →∞ a ( t ) b ( t ) − c ( t ) < then Eq. (4.2) is permanent. Everywhere above, we only assumed that (a2) is satisfied, i.e. the arguments of x tend to ∞ as t → ∞ . In the following theorem we assume a stronger condition that the delays are bounded. Theorem 5.6.
Suppose conditions (a1)-(a3) are satisfied, f k ( t, u , . . . , u l ) are monotone increas-ing in u , . . . , u n for some n ∈ { , . . . , l } , monotone decreasing in u n +1 , . . . , u l , and there existconstants τ > , A > , µ > , M > , < β < B such that t − τ < h j ( t ) ≤ t , j = 1 , . . . , l , m X i =1 f i ( t, u , . . . , u l ) ≤ Au j , u j > , for some j ∈ { , . . . , n } , < βu ≤ g ( t, u ) ≤ Bu for u > . Ifthere exists M > such that lim sup t →∞ P mi =1 f i ( t, u, . . . , u, M, . . . , M ) g ( t, u ) < uniformly on u ∈ [ M, ∞ ) then any solution x of (2.1),(2.2) is bounded, with the upper bound lim sup t →∞ x ( t ) ≤ M e A + B ) τ . (5.3) If there exists µ > such that lim inf t →∞ P mi =1 f i ( t, u, . . . , u, µ, . . . , µ ) g ( t, u ) > uniformly on u ∈ [0 , µ ] then any solution x of (2.1), (2.2) is persistent, and lim inf t →∞ x ( t ) ≥ µe − Bτ . (5.5)15 roof. Let x be a solution of (2.1),(2.2). First we will prove that x is bounded and obtain aneventual upper estimate for x , then we justify permanence and present an eventual lower estimate.As a preliminary work, possible growth and decrease of x is estimated.Let us consider t ∗ large enough such that for some ˜ t , P mi =1 f i ( t, u, . . . , u, M, . . . , M ) g ( t, u ) ≤ α < , t ≥ ˜ t, u ≥ M, and h k ( t ) ≥ ˜ t for t ≥ t ∗ (we can take t ∗ = ˜ t + τ ). Denote t = t ∗ + τ , t j = t ∗ + jτ . The solutionis positive and continuous, so it is possible to introduce a series of maximum and minimum valueson [ t j − , t j ]: m j = min t ∈ [ t j − ,t j ] x ( t ) , M j = max t ∈ [ t j − ,t j ] x ( t ) , j ∈ N . Let x ( t ∗ j − ) = M j − , where t ∗ j − ∈ [ t j − , t j − ]. Also, ˙ x ( t ) ≥ − g ( t, x ( t )) ≥ − Bx ( t ), thus x ( t ) ≥ x ( t ∗ j − ) e − B ( t − t ∗ j − ) = M j − e − B ( t − t ∗ j − ) ≥ M j − e − Bτ , t ∈ [ t j − , t j ] , since t − t ∗ j − ≤ t j − t j − = 2 τ . Thus, m j ≥ M j − e − Bτ .Next, let us develop an upper estimate. By the assumptions of the theorem, the solution satisfies x ( t j − ) ≤ M j − and˙ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) < m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) ≤ Ax ( h j ( t )) ≤ A max (cid:26) M j − , max s ∈ [ t j − ,t ] x ( s ) (cid:27) , t ∈ [ t j − , t j ] . Hence x ( t ) is less than the solution of the initial value problem ˙ x ( t ) = Ax ( t ), x ( t j − ) = M j − ,which is M j − exp( A ( t − t j − )), therefore M j ≤ M j − e Aτ , x ( t ) ≤ M j − e Aτ , t ∈ [ t j , t j +1 ] . (5.6)For the sake of contradiction, let us assume that the solution x is unbounded, i.e. for any M > M e A + B ) τ , where M is described in the conditions of the theorem, there is an interval[ t j , t j +1 ] where the inequality x ( t ) ≥ M is attained for the first time. Hence there exists t ∗ where x ( t ∗ ) = M and ε > t ∈ [ t ∗ − ε, t ∗ ] ⊂ [ t j , t j +1 ] and x ( t ) = sup s ∈ [0 ,t ] x ( s ), t ∈ [ t ∗ − ε, t ∗ ].According to estimate (5.6), M > M e A + B ) τ implies M j − ≥ M e Bτ , while m j ≥ M j − e − Bτ yields that m j ≥ M and also x ( t ) ≥ M on [ t j − , t ∗ ]. Thus, all x ( h i ( t )) ≥ M for t ∈ [ t j , t j +1 ], i = 1 , . . . , l , and for t ∈ [ t ∗ − ε, t ∗ ],˙ x ( t ) = m X i =1 f i ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≤ m X i =1 f k ( t, x ( t ) , . . . , x ( t ) , M, . . . , M ) − g ( t, x ( t )) ≤ αg ( t, x ( t )) − g ( t, x ( t )) = − (1 − α ) g ( t, x ( t )) < , which contradicts to the assumption x ( t ∗ − ε ) ≤ x ( t ∗ ) = M . Thus, the solution is bounded withthe eventual upper bound of M e B + A ) τ . 16ext, let us proceed to persistence and assume that for t ≥ t ∗ − τ , P mi =1 f i ( t, u, . . . , u, µ, . . . , µ ) g ( t, u ) ≥ C > , t ≥ ˜ t, ≤ u ≤ µ, and introduce t j , m j and M j as previously. If lim inf t →∞ x ( t ) = 0 then there exist t ∗ large enoughand ε small enough such that x ( t ) = min s ∈ [0 ,t ] x ( s ) and x ( t ) < µe − Bτ on [ t ∗ − ε, t ∗ ] ⊂ [ t j , t j +1 ].As previously, we obtain x ( t ) < µ on [ t j − , t ∗ ], so x ( h i ( t )) < µ , i = 1 , . . . , l , t ∈ [ t ∗ − ε, t ∗ ]. On[ t ∗ − ε, t ∗ ], we have x ( h j ( t )) < µ , x ( t ) < µ and˙ x ( t ) = m X i =1 f i ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≥ m X i =1 f k ( t, x ( t ) , . . . , x ( t ) , µ, . . . , µ ) − g ( t, x ( t )) ≥ Cg ( t, x ( t )) − g ( t, x ( t )) = ( C − g ( t, x ( t )) > , which contradicts to the assumption x ( t ∗ − ε ) ≥ x ( t ∗ ). Thus, the solution is also persistent andsatisfies (5.5). Example 5.7.
Consider the Mackey-Glass equation ˙ x ( t ) = a ( t ) x ( h ( t ))1 + x n ( p ( t )) − b ( t ) x ( t ) , (5.7) where a and b are Lebesgue measurable bounded functions satisfying ≤ α ≤ a ( t ) ≤ A , < β ≤ b ( t ) ≤ B , t − h ( t ) ≤ τ , t − p ( t ) ≤ τ , n > . The bounds for a ( t ) and b ( t ) guarantee that lim sup t →∞ a ( t ) b ( t ) is finite. Thus inequality (5.2) is satisfied for any M > M , where M = , lim sup t →∞ a ( t ) b ( t ) ≤ , lim sup t →∞ (cid:16) a ( t ) b ( t ) − (cid:17) n , lim sup t →∞ a ( t ) b ( t ) > . Thus, all solutions of (5.7) are bounded, with the eventual upper bound of
M e A + B ) τ . Assumenow that in addition lim inf t →∞ a ( t ) b ( t ) > . (5.8) Inequality (5.4) is valid for any < µ < µ , where µ = lim inf t →∞ (cid:18) a ( t ) b ( t ) − (cid:19) n . Hence, if condition (5.8) holds, then any positive solution x is persistent with lim inf t →∞ x ( t ) ≥ µ e − Bτ . Moreover, condition (5.8) implies permanence of all positive solutions of Eq. (5.7).
Example 5.8.
Consider the equation ˙ x ( t ) = a ( t ) x ( h ( t ))1 + x ( g ( t )) − x ( t ) (5.9) with piecewise constant h ( t ) and g ( t ) . Let us note that the equation ˙ x + x ( t ) = A , x ( t ) = x hasthe solution x ( t ) = ( x − A ) exp {− ( t − t ) } + A , so for any B between A and x there is a finite t > t such that x ( t ) = B , t = t + ln(( x − A ) / ( B − A )) .Let t < t < . . . be a sequence of positive numbers such that a ( t ) = (cid:26) , t ∈ [ t k , t k +1 ) , , t ∈ [ t k +1 , t k +2 ) ,h ( t ) = (cid:26) t k − , t ∈ [ t k , t k +1 ) ,t k , t ∈ [ t k +1 , t k +2 ) , g ( t ) = (cid:26) t k , t ∈ [ t k , t k +1 ) ,t k − , t ∈ [ t k +1 , t k +2 ) , where t = 0 , x ( t ) = 1 , t − = − , x ( t − ) = ϕ ( −
1) = . We justify that we can find t i such that x ( t k ) = 2 k , x ( t k +1 ) = 2 − k − , k ∈ N . In fact, on [0 , t ] we have x ( h ( t )) = , x ( g ( t )) = 1 , a ( t ) = 2 , the initial value problem is ˙ x ( t )+ x ( t ) = , x (0) = 1 , so we can find t such that x ( t ) = .On [ t , t ] , x ( h ( t )) = 1 , x ( g ( t )) = , a ( t ) = 6 , the initial value problem is ˙ x ( t ) + x ( t ) =6 / (1 + 1 / > , x ( t ) = 1 / , so there is t such that x ( t ) = 2 .Let us proceed to the induction step. If x ( t k ) = 2 k , x ( t k +1 ) = 2 − k − then on [ t k , t k +1 ] wehave the initial value problem ˙ x ( t ) + x ( t ) = 2 · − k − k < − k − , k ∈ N , x ( t k ) = 2 k , thus there exists t k +1 such that x ( t k +1 ) = 2 − k − . On [ t k +1 , t k +2 ] , we have the initial valueproblem ˙ x ( t ) + x ( t ) = 6 · k − k − > k +1 , k ∈ N , x ( t k +1 ) = 2 − k − , hence there is t k +2 such that x ( t k +2 ) = 2 k +1 , which concludes the induction step. Here both b and a are bounded, separated from zero, a/b ≥ > , g and h satisfy (a2) but the solution is neitherbounded nor persistent. In this example, the delays h and g are unbounded.
6. Unbounded Solutions
Let us consider the case when positive solutions are unbounded.
Theorem 6.1.
Suppose f k ( t, u , . . . , u l ) , k = 1 , . . . , m are increasing functions in u , . . . , u l forany t , there is K > such that for any K ≥ K there exists a K > such that inf t ≥ " m X k =1 f k ( t, K, . . . , K ) − g ( t, K ) ≥ a K . or any K ≥ K , if ϕ ( t ) > K for t ≤ then the solution of (2.1), (2.2) satisfies lim sup t →∞ x ( t ) = + ∞ . Proof.
Suppose that x is a solution of Eq. (2.1) such that x ( t ) > K ≥ K , t ≤ x ( t ) > K for any t ≥
0. Assume that it is not so and there exists t > x ( t ) > K ≥ K , t ∈ [0 , t ), x ( t ) = K . In some left neighbourhood [ t − ε, t ) of t wehave K < x ( t ) and | g ( t, x ( t )) − g ( t, K ) | < a K /
2. Hence˙ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≥ m X k =1 f k ( t, K , . . . , K ) − g ( t, K ) + g ( t, K ) − x ( t, x ( t )) ≥ a K − a K / a K / > , t ∈ ( t − ε, t ) , which implies K = x ( t ) = x ( t − ε ) + Z t t − ε ˙ x ( s ) ds ≥ x ( t − ε ) + ( a K / ε > K . The contradiction proves x ( t ) > K for any t ≥ K = sup ( u > K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) inf t ≥ ,x ∈ [ K ,u ] " m X k =1 f k ( t, K , . . . , K ) − g ( t, x ) ≥ a K ) ,K ∗ = sup ( u > K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) inf t ≥ ,x ∈ [ K ,u ] " m X k =1 f k ( t, K , . . . , K ) − g ( t, x ) ≥ ) . Either K = + ∞ or K ∗ = + ∞ would imply that˙ x ( t ) ≥ m X k =1 f k ( t, K , . . . , K ) − g ( t, x ) > x ( t ) is increasing for any t ; if K ∗ = + ∞ then it is increasing with the guaranteed rate˙ x ( t ) ≥ a K , and the solution is obviously unbounded.By (a1), g is locally Lipschitz, hence there exists α = α ([ K , K ]) such thatinf t ≥ | g ( t, u ) − g ( t, y ) | ≤ α | u − y | , u, y ∈ [ K , K ] . Denote σ := min n a K α , K o . Thus, for x ∈ [ K , K + σ ] we haveinf t ≥ " m X k =1 f k ( t, K , . . . , K ) − g ( t, x ) ≥ inf t ≥ " m X k =1 f k ( t, K , . . . , K ) − g ( t, K ) − | g ( t, K ) − g ( t, x ) |≥ a K − α | x − K | ≥ a K − α a K α = a K − a K = a K . K ≥ K + σ . Similarly, K ∗ ≥ K + σ .As long as x ( t ) ∈ ( K , K ], we have ˙ x ( t ) ≥ a K , thus x ( t ) > K for some t ; moreover, x ( t ) > K for t large enough. In fact, assuming that there is an interval [ t − ε, t ) where x ( t ) ∈ ( K , K ∗ )while x ( t ) = K , we notice that, due to the fact that x ( t ) > K for t ≥ x ( t ) = m X k =1 f k ( t, x ( h ( t )) , . . . , x ( h l ( t ))) − g ( t, x ( t )) ≥ f ( t, K , . . . , K ) − g ( t, x ( t )) ≥ t ∈ [ t − ε, t ], which excludes the possibility x ( t − ε ) > x ( t ). Thus, x ( t ) > K for x largeenough, say, for t > t ∗ .Further, we consider t large enough such that h i ( t ) > t ∗ for any i = 1 , . . . , l . Denote K = sup ( u > K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) inf t ≥ ,x ∈ [ K ,u ] " m X k =1 f k ( t, K , . . . , K ) − g ( t, x ( t )) ≥ a K ) . Similarly to the previous argument we verify that x ( t ) > K for t large enough, denote K n , n ∈ N and repeat this procedure. Thus there is an increasing sequence of positive numbers K < K < · · · < K n < . . . , if finite, and points t ≤ t ≤ · · · ≤ t n ≤ . . . such that x ( t ) ≥ K n for x ≥ t n andinf t ≥ " m X k =1 f k ( t, K n , . . . , K n ) − g ( t, K n − ) = 12 a K n − . If at least one of K n is infinite, the solution tends to infinity, as explained earlier. In addition, forlim n →∞ K n = + ∞ , the solution is unbounded. Assuming that lim n →∞ K n = d < + ∞ and proceedingto the limit in n in the above inequality, we obtaininf t ≥ " m X k =1 f k ( t, d, . . . , d ) − g ( t, d ) = 12 a d , which contradicts to the assumption of the theorem that this infimum is not less than a d . Example 6.2.
Consider Eq. (4.2) with the same conditions as in Example 4.4 and in addition inf t ≥ [ a ( t ) − b ( t )] ≥ α > . (6.1) We also use the same notations f ( t, u, v ) = a ( t ) uv , g ( t, u ) = (cid:16) b ( t ) − c ( t )1+ u n (cid:17) u as in Example 4.4.This leads to f ( t, K, K ) − g ( t, K ) ≥ ( a ( t ) − b ( t )) K ≥ αK ≥ αK > for K ≥ K . Thus asolution of (4.2) with any positive initial function ϕ ( t ) ≥ K is unbounded by Theorem 6.1, for any K > .Consider a modification of (4.2) ˙ x ( t ) = a ( t ) x β ( t − h ) x γ ( t − g ) − (cid:18) b ( t ) − c ( t )1 + x n ( t ) (cid:19) x ( t ) , where a ( t ) ≥ , b ( t ) ≥ c ( t ) ≥ , b ( t ) − c ( t ) ≥ b > , a, b, c are Lebesgue measurable boundedfunctions, h > , g > , β, γ, n ≥ , β + γ ≥ , and (6.1) is satisfied. By the same calculations asbefore any solution with the initial function ϕ ( t ) ≥ K > is unbounded. xample 6.3. Consider the linear equation with several delays ˙ x ( t ) = m X k =1 a k ( t ) x ( h k ( t )) − b ( t ) x ( t ) , where a k ( t ) ≥ , b ( t ) ≥ β > , a k , b : [0 , ∞ ) → [0 , ∞ ) and h k ( t ) ≤ t are Lebesgue measurablebounded functions. Assume that lim inf t →∞ P mk =1 a k ( t ) b ( t ) > . Denote f k ( t, u ) = a k ( t ) u , g ( t, u ) = b ( t ) u . Then there exist t ≥ and α > such that P mk =1 a k ( t ) b ( t ) ≥ α for t ≥ t , and P mk =1 a k ( t ) − b ( t ) ≥ (1 − α ) b ( t ) ≥ (1 − α ) β for t ≥ t . Hence m X k =1 f k ( t, K ) − g ( t, K ) = m X k =1 a k ( t ) − b ( t ) ! K ≥ (1 − α ) βK for any K ≥ K > . Then a solution with any initial function, with a positive lower bound, isunbounded by Theorem 6.1.
7. Discussion
In the present paper, we have studied existence of global positive solutions for nonlinear equa-tion (1.3) with several delays, as well as boundedness and persistence of these solutions. Theresults were applied, for example, to the Mackey-Glass equation of population dynamics with non-monotone feedback [10]. However, they can also be applied to some other models, including theNicholson’s blowflies equation with two delays˙ x ( t ) = P ( t ) x ( h ( t )) e − x ( g ( t )) − δ ( t ) x ( t )which in the case when variable delays are equal h ( t ) = g ( t ) was studied, for example, in [7, 11, 12].Permanence of solutions of equations of type (1.3) was recently explored in [14] and [17]. Com-pared to [14, 17], we consider a more general model: in particular, it is not always assumed that f is increasing in all u -arguments, as well as continuity in t . Also, (H3) in [14, p. 86] is a special caseof conditions of the present paper. On the other hand, in [14, 17], solution bounds are obtainedand more advanced asymptotic properties, such as stability, are discussed.Equation (1.3) is a special case of the equation with a distributed delay˙ x ( t ) = m X k =1 f k t, Z th ( t ) x ( s ) d s R ( t, s ) , . . . , Z th l ( t ) x ( s ) d s R l ( t, s ) ! − g ( t, x ( t )) , (7.1)while the integro-differential equation˙ x ( t ) = m X k =1 f k t, Z th ( t ) K ( t, s ) x ( s ) ds, . . . , Z th l ( t ) K l ( t, s ) x ( s ) ds ! − g ( t, x ( t )) , (7.2)is another particular case of Eq. (7.1). All conditions for boundedness, persistence, permanenceand existence of unbounded solutions, obtained here for (1.3) can be extended to (7.1) and (7.2),using the ideas of the proofs of the present paper.21quations with several delays involved in a nonlinear function is a challenging object withproperties quite different from the case when these delays coincide, and we have presented severalexamples to outline this difference. However, so far only existence of a positive global solution,persistence and boundedness have been explored. It is interesting to investigate other qualitativeproperties for Eqs. (1.3), (7.1) and (7.2), such as oscillation, stability and existence of periodic oralmost periodic solutions.One of the main results in this paper is Theorem 5.6, where we obtain a priori estimations ofsolutions for equation (1.3). Such estimations were used in [9] to obtain global asymptotic stabilityresults for various types of nonlinear delay differential equations. We expect that this techniquecan be applied to obtain explicit global stability results for Eqs. (1.4) and (1.5).We conclude this discussion by noticing that there are many equations which have a differentform than (1.3), for example, the equation ˙ x ( t ) = f ( t, x ( h ( t ))) − g ( t, x ( r ( t ))) with the delay in thenegative term, and the logistic-type equation˙ x ( t ) = r ( t ) x ( h ( t ))[1 − x ( g ( t ))] . (7.3)Compared to (7.3), the Hutchinson equation, which is a standard delay-type logistic equation, has h ( t ) ≡ t . Another delay versions of the logistic equation were considered in [2, 3].However, it is known that Eq. (7.3) does not even necessarily have a global positive solution. Itwould be interesting to develop a technique to study such new classes of delay differential equationsincluding (7.3).
8. Acknowledgments
L. Berezansky was partially supported by Israeli Ministry of Absorption, E. Braverman waspartially supported by the NSERC research grant RGPIN-2015-05976. The authors are gratefulto the reviewers whose thoughtful comments significantly contributed to the presentation of theresults of the paper.
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