Abstract
Smith [66] reasoned that a food-limited population in its growing stage requires food for both maintenance and growth, whereas, when the population has reached saturation level, food is needed for maintenance only. On the basis of these assumptions, Smith derived a model of the form
which is called the “food limited” population . Here N, r, and K are the mass of the population, the rate of increase with unlimited food, and the value of N at saturation, respectively. The constant 1∕c is the rate of replacement of mass in the population at saturation. Since a realistic model must include some of the past history of the population, Gopalsamy, Kulenovic and Ladas introduced the delay in (5.1) and considered the equation
as the delay “food-limited” population model, where r, K, c, and τ are positive constants.
If a nonnegative quantity was so small that is smaller than any given one, then it certainly could not be anything but zero. To those who ask what the infinity small quantity in mathematics is, we answer that it is actually zero. Hence there are not so many mysteries hidden in this concept as they are usually believed to be.
Leonhard Euler (1707–1783)
The real end of science is the honor of the human mind.
Gustav J. Jacobi (1804–1851)
Access provided by Autonomous University of Puebla. Download chapter PDF
Similar content being viewed by others
Keywords
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
Smith [66] reasoned that a food-limited population in its growing stage requires food for both maintenance and growth, whereas, when the population has reached saturation level, food is needed for maintenance only. On the basis of these assumptions, Smith derived a model of the form
which is called the “food limited” population . Here N, r, and K are the mass of the population, the rate of increase with unlimited food, and the value of N at saturation, respectively. The constant 1∕c is the rate of replacement of mass in the population at saturation. Since a realistic model must include some of the past history of the population, Gopalsamy, Kulenovic and Ladas introduced the delay in (5.1) and considered the equation
as the delay “food-limited” population model, where r, K, c, and τ are positive constants.
In this chapter we discuss autonomous and nonautonomous “food-limited” population models with delay times.
5.1 Oscillation of Delay Models
Motivated by the model
in this section we consider
with the following assumptions:
- (A1) :
-
r(t) and s(t) are Lebesgue measurable locally essentially bounded functions such that r(t) ≥ 0 and s(t) ≥ 0.
- (A2) :
-
h, g: [0, ∞) → R are Lebesgue measurable functions such that h(t) ≤ t, \(g(t) \leq t,\mathop{\lim }\limits_{t \rightarrow \infty }h(t) = \infty \), and \(\mathop{\lim }\limits_{t \rightarrow \infty }g(t) = \infty \).
Note the oscillation (or nonoscillation) of N about K is equivalent to oscillation (nonoscillation) of (5.3) about zero (let x = N∕K − 1).
One could also consider for each t 0 ≥ 0 the problem
$$\displaystyle{ x^{{\prime}}(t) = -r(t)x(h(t)) \frac{1 + x(t)} {1 + s(t)[1 + x(g(t))]},\text{ }t \geq t_{0}, }$$(5.4)with the initial condition
$$\displaystyle{ x(t) =\varphi (t),\text{ }t < t_{0},\ x(t_{0}) = x_{0}. }$$(5.5)We also assume that the following hypothesis holds:
- (A3) :
-
\(\varphi: (-\infty,t_{0}) \rightarrow \mathbf{R}\) is a Borel measurable bounded function.
An absolutely continuous function x(: R → R) on each interval [t 0, b] is called a solution of problems (5.4) and (5.5), if it satisfies (5.4) for almost all t ∈ [t 0, ∞) and the equality (5.5) for t ≤ t 0. Equation (5.3) has a nonoscillatory solution if it has an eventually positive or an eventually negative solution. Otherwise, all solutions of (5.3) are oscillatory. The results in this section can be found in [10]. In the following, we assume that (A1)–(A3) hold and we consider only such solutions of (5.3) for which the following condition holds:
The proof of the following lemma follows a standard argument (see the proof in Theorem 2.4.1 and see Lemma 2.6.1).
Lemma 5.1.1.
Let (A1) and (A2) hold for the equation
Then the following hypotheses are equivalent:
-
(1)
The differential inequality
$$\displaystyle{ x^{{\prime}}(t) + r(t)x(h(t)) \leq 0,\text{ }t \geq 0 }$$(5.8)has an eventually positive solution.
-
(2)
There exists t 0 ≥ 0 such that the inequality
$$\displaystyle{ u(t) \geq r(t)\exp \left \{\int _{h(t)}^{t}u(s)ds\right \},\text{ }t \geq t_{ 0},\,\,u(t) = 0,t < t_{0} }$$(5.9)has a nonnegative locally integrable solution.
-
(3)
Equation (5.7) has a nonoscillatory solution.
If
$$\displaystyle{ \mathop{\lim }\limits_{t \rightarrow \infty }\sup \int _{h(t)}^{t}r(s)ds < \frac{1} {e}, }$$(5.10)then (5.7) has a nonoscillatory solution. If
$$\displaystyle{ \mathop{\lim }\limits_{t \rightarrow \infty }\inf \int _{h(t)}^{t}r(s)ds > \frac{1} {e}, }$$(5.11)then all the solutions of (5.7) are oscillatory.
Lemma 5.1.2.
Let x(t) be a nonoscillatory solution of (5.3) and suppose that
Then limt→∞ x(t) = 0.
Proof.
Suppose first x(t) > 0, t ≥ t 1. Then there exists t 2 ≥ t 1 such that
Let
Then u(t) ≥ 0, t ≥ t 2 and
Substituting this into (5.3) we obtain
where h(t) ≤ t, g(t) ≤ t, for t ≥ t 2, and c = x(t 2) > 0. Hence
From (5.12) we have \(\int _{t_{2}}^{\infty }u(t)dt = \infty \).
Now suppose − 1 < x(t) < 0, t ≥ t 1. Then there exists t 2 ≥ t 1 such that (5.13) holds for t ≥ t 2. With u(t) denoted in (5.14) and c = x(t 2) we have u(t) ≥ 0 and − 1 < c < 0. Substituting (5.15) into (5.3) and using (5.16), we have
Thus \(\int _{t_{2}}^{\infty }u(t)dt = \infty \). Equation (5.15) implies that lim t → ∞ x(t) = 0. The proof is complete. ■
Theorem 5.1.1.
Suppose (5.12) holds and for some \(\varepsilon > 0\) , all solutions of the linear equation
are oscillatory. Then all solutions of (5.3) are oscillatory.
Proof.
First suppose x(t) is an eventually positive solution of (5.3). Lemma 5.1.2 implies that there exists t 1 ≥ 0 such that \(0 < x(t) <\varepsilon\) for t ≥ t 1. We suppose (5.13) holds for t ≥ t 2 ≥ t 1. For t ≥ t 2, we have
Equation (5.3) implies
Lemma 5.1.1 yields that (5.19) has a nonoscillatory solution. We have a contradiction.
Now suppose \(-\varepsilon < x(t) < 0\) for t ≥ t 1 and (5.13) holds for t ≥ t 2 ≥ t 1. Then for t ≥ t 2
Hence, (5.19) has a nonoscillatory solution and we again obtain a contradiction which completes the proof. ■
Corollary 5.1.1.
If
then all solutions of (5.3) are oscillatory.
Theorem 5.1.2.
Suppose for some \(\varepsilon > 0\) there exists a nonoscillatory solution of the linear delay differential equation
Then there exists a nonoscillatory solution of (5.3) .
Proof.
Lemma 5.1.1 implies that there exists t 0 ≥ 0 such that
and
Suppose \(0 < c <\varepsilon\) and consider two sequences:
and
where w 0 is as defined above and \(\upsilon _{0}(t) \equiv 0\). We have
from (5.25). Clearly \(\upsilon _{1}(t) \geq \upsilon _{0}(t)\) and \(w_{0}(t) \geq \upsilon _{0}(t)\). Hence by induction
There exist pointwise limits of the nonincreasing nonnegative sequence w n (t) and of the nondecreasing sequence ν n (t). Let
Then by the Lebesgue Convergence Theorem , we conclude that
and
We fix b ≥ t 0 and define the operator \(T: L_{\infty }[t_{0},b] \rightarrow L_{\infty }[t_{0},b]\) by
For every function u from the interval \(\upsilon \leq u \leq w\), we have \(\upsilon \leq Tu \leq w\). One can also check that T is a completely continuous operator on the space L ∞ [t 0, b]. Then by Schauder’s Fixed Point Theorem there exists a nonnegative solution of equation u = Tu. Let
and then x(t) is a nonoscillatory solution of (5.3) which completes the proof. ■
The results in this section apply to (5.2). For example by applying Theorem 5.1.1 we have the following result.
Theorem 5.1.3.
Suppose (5.12) holds and for some \(\varepsilon > 0\) , all solutions of the linear equation
are oscillatory. Then all solutions of (5.2) are oscillatory about K.
5.2 Oscillation of Impulsive Delay Models
In this section we consider the impulsive “food-limited” population model
here \(N(t_{k}) = N(t_{k}^{-})\). In this section, we will assume that the following assumptions hold:
- (A1) :
-
\(0 \leq t_{0} < t_{1} < t_{2} <\ldots < t_{k} <\ldots\) are fixed points with
\(\lim _{k\rightarrow \infty }t_{k} = \infty \),
- (A2) :
-
b k > −1, k = 1, 2, …, K is a positive constant,
- (A3) :
-
r(t) and p i , i = 1, 2, …, m, are Lebesgue measurable locally
essentially bounded functions, in each finite interval [0, b], r(t) ≥ 0 and p i (t) ≥ 0, for i = 1, 2, …, m,
- (A4) :
-
h, g i : [0, ∞) → R are Lebesgue measurable functions, h(t) ≤ t, g i (t) ≤ t, lim t → ∞ h(t) = ∞, lim t → ∞ g i (t) = ∞, i = 1, 2, …, m.
In this section (motivated by (5.31) with \(y(t) = \frac{N(t)} {K} - 1\)) we consider the delay model with impulses
where b k > −1 and r, h, p i for m = 1, 2, … are nonnegative real-valued functions. We consider (5.32) with the initial condition
Here for any T 0 ≥ 0, \(T^{-} =\min _{1\leq i\leq m}\inf _{t\geq T_{0}}(g_{i}(t),h(t))\), and \(\varphi: [T^{-},T_{0}] \rightarrow \mathbf{R}_{+}\) is a Lebesgue measurable function .
For any T 0 ≥ 0 and \(\varphi (t)\), a function y: [T −, ∞] → R is said to be a solution of (5.32) on [T, ∞] satisfying the initial value condition (5.33), if the following conditions are satisfied:
-
1.
y(t) satisfies (5.33);
-
2.
y(t) is absolutely continuous in each interval \((T_{0},t_{k}),(t_{k},t_{k+1}),t_{k} > T_{0},k \geq k_{0},y(t_{k}^{+}),y(t_{k}^{-})\) exist and \(y(t_{k}^{-}) = y(t_{k}),k > k_{0};\)
-
3.
y(t) satisfies the former equation of (5.32) in [T, ∞)∖{t k } and satisfies the latter equation for every t = t k , k = 1, 2, ….
For any t ≥ 0, consider the nonlinear delay differential equation
where
The results in this section are adapted from [77] (in fact as we see below it is easy to extend the theory in the nonimpulsive case in Sect. 5.1 to the impulsive case).
Lemma 5.2.1.
Assume that (A1)–(A4) hold. Then the solution N(t) of (5.31) oscillates about K if and only if the solution y(t) of (5.32) oscillates about zero.
The proof (which is elementary and straightforward) of the next lemma can be found in [81].
Lemma 5.2.2.
Assume that (A1)–(A4) hold. For any T 0 ≥ 0,y(t) is a solution of (5.32) on [T 0 ,∞) if and only if
is a solution of the nonimpulsive delay differential equation (5.34) .
From Lemmas 5.2.1 and 5.2.2 we see that the solution N(t) of (5.31) is oscillatory about K if and only if the solution y(t) of (5.32) is oscillatory.
We consider only such solutions of (5.32) for which the following condition holds:
and hence, in view of (5.35),
With \(y(t) = \frac{N(t)} {K} - 1\) then from (5.36) and (5.37), we see that
Thus for the initial condition \(N(t) =\varphi (t): [T^{-},T_{0}] \rightarrow \mathbf{R}_{+},\varphi (T_{0}) > 0\), the solution of (5.31) is positive on [T 0, ∞).
Lemma 5.2.3.
Assume that (A1)–(A4) hold,
and
If y(t) is a nonoscillatory solution of (5.32) , then limt→∞ y(t) = 0.
Proof.
Suppose first y(t) > 0 for t ≥ T 1 ≥ 0. From (5.35) and (A1), x(t) > 0 for t ≥ T 1. Then there exists T 2 ≥ T 1 such that
Let
Then u(t) ≥ 0 for t ≥ T 2 and
Setting c = x(T 2), we have
Then from (5.38) and (5.39), \(\mathop{\int }\limits _{T_{2}}^{\infty }u(t)dt = \infty \).
Now suppose − 1 < y(t) < 0. Hence in view of (5.36),
Then there exists T 2 > T 1 such that (5.40) holds for t > T 2. With u(t) denoted in (5.41) and c = x(T 2), then from (5.37) u(t) ≥ 0, −1 < c < 0, and we obtain
Then by (5.37)–(5.39), we have \(\mathop{\int }\limits _{T_{2}}^{\infty }u(t)dt = \infty \). Equation (5.42) implies lim t → ∞ x(t) = 0. Use (5.35), and then we have lim t → ∞ y(t) = 0. The proof is complete. ■
Theorem 5.2.1.
Assume that (A1) and (A2), (5.38) hold and for some ε > 0, all solutions of the linear equation
are oscillatory. Then all solutions of (5.32) are oscillatory.
Proof.
Suppose y(t) is an eventually positive solution of (5.32). Then x(t) is an eventually positive solution of (5.34). Lemma 5.2.3 implies that there exists T 1 ≥ 0, such that
We suppose (5.40) holds for t ≥ T 2, and we have
Equation (5.34) implies
This implies that the (5.43) has a positive solution, which is a contradiction.
Now, we suppose
and (5.38) holds for t ≥ T 2 ≥ T 1. Then for t ≥ T 2, we also get
Thus (5.43) has a nonoscillatory solution and we again obtain a contradiction. The proof is complete. ■
Theorem 5.2.2.
Assume that (A1) and (A2) hold and
Moreover, for some ε > 0 if there exists a nonoscillatory solution of the linear delay differential equation
then there exists a nonoscillatory solution of (5.32) .
Proof.
Suppose that x(t) > 0 for t > T 0 is a solution of (5.48). Then by (5.34) there exist T 0 ≥ 0 and ω 0(t) ≥ 0, t ≥ T 0, \(\omega _{0}(t) = 0,T_{0}^{-}\leq t \leq T_{0}\) such that
Since \(\mathop{\prod }\limits _{T_{0}\leq t_{k}<g_{i}(t)}(1 + b_{k})\) is convergent, there exists a positive constant c such that
Consider the two sequences:
where ω 0 is defined above and \(\upsilon _{0} \equiv 0\). Thus we have
Clearly \(\upsilon _{1}(t) \geq \upsilon _{0}(t),\omega _{0}(t) \geq \upsilon _{0}(t)\). Hence by induction
There exist pointwise limits of the nonincreasing nonnegative sequence ω n (t) and of the nondecreasing sequence \(\upsilon _{n}(t)\). Let ω(t) = lim n → ∞ ω n (t) and \(\upsilon (t) =\lim _{n\rightarrow \infty }\upsilon _{n}(t)\). Then by the Lebesgue Convergence Theorem , we deduce that
We fix b ≥ T 0 and define the operator \(T: L_{\infty }[T_{0},b] \rightarrow L_{\infty }[T_{0},b]\) by the following
For every function u from the interval \(\upsilon \leq u \leq \omega\), we have \(\upsilon \leq Tu \leq \omega\). Also T is a completely continuous operator on the space L ∞ [T 0, b], and then by the Schauder Fixed Point Theorem there exists a nonnegative solution of the equation u = Tu. Let
Then x(t) is a nonoscillatory solution of (5.34). Thus by Lemma 5.2.1
is a nonoscillatory solution of (5.32) which completes the proof of Theorem 5.2.2. ■
The results in this section apply to (5.31).
5.3 \(\frac{3} {2}\)-Global Stability
In this section we examine the global attractivity of the “food-limited” population model
where
We consider solutions of (5.55) which correspond to the initial condition
Motivated by (5.55) in this section, we will study the global stability of the general equation
where \(F(t,\varphi )\) is a continuous functional on [0, ∞) × C t , such that F(t, 0) = 0 for t ≥ 0 and satisfies a York-type condition
where g: [0, ∞) → (−∞, ∞) is a nondecreasing continuous function with g(t) < t for t ≥ 0 and lim t → ∞ g(t) = ∞, \(M_{t}(\varphi ) =\max \{ 0,\sup _{s\in [g(t),t]}\varphi (s)\}\), c ∈ (0, ∞) and r ∈ C([0, ∞), (0, ∞)). The class C t is the set of all continuous functions \(\varphi: [g(t),t] \rightarrow [-1,\infty )\) with the sup-norm \(\left \Vert \varphi \right \Vert _{t} =\sup _{s\in [g(t),t]}\left \vert \varphi (s)\right \vert\).
Let τ = −g(0). We consider solutions of (5.57) which correspond to the initial condition
In the following, we will establish a 3∕2-global attractivity condition for (5.57), and then apply this condition on equation (5.55) to establish a 3∕2-global attractivity condition. The results in this section are adapted from [73]. To prove the results, we need the following results (whose proofs are standard; for Lemma 5.3.7 see Lemma 5.7.3 with c = 1).
Lemma 5.3.1.
Assume that c ∈ (0,1]. Then for any v ∈ [0,1)
Lemma 5.3.2.
Assume that c ∈ (0,1]. Then for any u ∈ [0,∞)
Lemma 5.3.3.
Assume that c ∈ (0,1] and v ∈ (0,1). Then for any x ∈ [0,∞)
Lemma 5.3.4.
Assume that c ∈ (0,1]. Then for \(0 < v < \left [1-\frac{c} {2}+\sqrt{\frac{2(1- c)} {3} +\frac{c^{2}} {4}} \right ]^{-1}\)
Lemma 5.3.5.
Assume that c ∈ (0,1]. Then for any x ∈ [0,∞)
Lemma 5.3.6.
Assume that c ∈ (0,1] and
Then
Lemma 5.3.7.
The system of inequalities
has only a unique solution x=y=0 in the region {(x,y): 0 ≤ x ≤ 1, 0 ≤ y < 1∕c}.
Theorem 5.3.1.
Assume that (5.58) holds. Then the solution \(x(t,0,\varphi )\) of (5.57), (5.59) exists on [0,∞) and satisfies \(-1 < x(t,0,\varphi ) < 1/c\) .
Theorem 5.3.2.
Assume that (5.58) holds and there exists a function r ∗ ∈ C([0,∞),(0,∞)) such that for each \(\varepsilon > 0\) there is a \(\eta =\eta (\varepsilon ) > 0\) satisfying
and
Then every nonoscillatory solution of IVP (5.57) and (5.59) tends to zero.
Theorem 5.3.3.
Assume that (5.58), (5.60) , and (5.61) hold. If there exists a constant M such that
then the solutions of (5.57), (5.59) satisfy
We now prove our main result in this section.
Theorem 5.3.4.
Assume that (5.58)–(5.61) hold, and
Then every solution of (5.57), (5.59) tends to zero.
Proof.
Let x(t) be a solution of (5.57) and (5.59) (note also Theorem 5.3.1 so − 1 < x(t) < < 1∕c, t ≥ 0). By Theorem 5.3.2, we only consider the case when x(t) is oscillatory. First assume that 0 < c ≤ 1. Set
By Theorem 5.3.3, 0 ≤ u < ∞ and 0 ≤ v < 1. It suffices to prove that u = v = 0. For any \(0 <\varepsilon < 1 - v\), by (5.64) and (5.65) there exists a \(t_{0} = t_{0}(\varepsilon ) > g^{-2}(0)\) such that
From (5.57), (5.58), and (5.67), we have
and
Let {l n } be an increasing infinite sequence of real numbers such that g(l n ) > t 0, \(x(l_{n}) > 0,x^{{\prime}}(l_{n}) = 0\), and lim n → ∞ x(l n ) = u. We may assume that l n is a left local maximum point of x(t). It is easy to show that there exists \(\zeta _{n} \in [g(l_{n}),l_{n})\) such that x(ζ n ) = 0 and x(t) > 0 for t ∈ (ζ n , l n ]. By (5.68), we have
and [see also (5.57) and (5.58)] for ζ n ≤ t ≤ l n we have
which together with (5.68) yields for ζ n ≤ t ≤ l n
There are two cases to consider.
Case 1.
\(\int _{\zeta _{ n}}^{l_{n}}r(s)ds \leq -\frac{1} {v_{1}} \ln \frac{(1+c)e^{-cv_{1}(1-cv_{1}/2)}-1} {c} \equiv A\)
Then by (5.66) and (5.70), we have
If \(\int _{\zeta _{n}}^{l_{n}}r(s)ds \leq A \leq \delta _{0} = \frac{3} {2}(1 + c)\), then by Lemmas 5.3.1 and 5.3.3
If \(\int _{\zeta _{n}}^{l_{n}}r(s)ds \leq \delta _{0} = \frac{3} {2}(1 + c) < A\), then
From Lemma 5.3.4 we have that
Hence from (5.71), Lemmas 5.3.5 and 5.3.6, we have
Case 2.
\(A <\int _{ \zeta _{n}}^{l_{n}}r(s)ds \leq \delta _{0}\)
Choose \(\eta _{n} \in (\zeta _{n},l_{n})\) such that \(\int _{\eta _{n}}^{l_{n}}r(s)ds = A\). Then by (5.66), (5.70), and Lemma 5.3.1 we have
Combining the above cases we see that
Letting n → ∞ and \(\varepsilon \rightarrow 0\), we have
Now, we show that
Let {s n } be an increasing infinite sequence of real numbers such that g(s) > t 0, \(x(s_{n}) < 0,x^{{\prime}}(s_{n}) = 0\) and lim n → ∞ x(s n ) = −v. We may assume that s n is a left local minimum point of x(t). It is easy to show that there exists \(\eta _{n} \in [g(s_{n}),s_{n})\) such that x(η n ) = 0 and x(t) < 0 for t ∈ (η n , s n ]. By (5.69), we get
which together with (5.58) yields
Note that u 1 is bounded and note
We consider two cases.
Case I.
\(\int _{\eta _{n}}^{s_{n}}r(s)ds < \frac{3(1+c)} {2} - \frac{1} {u_{1}} \ln \frac{(1+c)e^{cu_{1}(1+cu_{1}/2)}-1} {c} \equiv B\).
From (5.69) and Lemma 5.3.2, we have
Case II.
\(B <\int _{ \eta _{n}}^{s_{n}}r(s)ds < \frac{3(1+c)} {2}\)
Choose \(h_{n} \in (\eta _{n},s_{n})\) such that \(\int _{\eta _{n}}^{h_{n}}r(s)ds = B\). Then by (5.69) and (5.74) we have
Combining these two cases we have
Letting n → ∞ and \(\varepsilon \rightarrow 0\) we see that (5.73) holds. In view of Lemma 5.3.7, we see from (5.72) to (5.73) that u = v = 0.
Next assume that c > 1. Set y(t) = −cx(t). Then (5.57) reduces to
where c ∗ = 1∕c ∈ (0, 1) and \(F^{{\ast}}(t,\varphi ) = -cF(t,-\frac{1} {c}\varphi )\) satisfies the York-type condition
Note for large t that
so we have lim t → ∞ y(t) = 0, and this implies that lim t → ∞ x(t) = 0. The proof is complete. ■
Applying Theorem 5.3.4 on (5.55) we have the following result.
Theorem 5.3.5.
Assume that
and
where c 0 = inf {c(t): t ≥ 0}. Then every solution of (5.55), (5.56) tends to 1.
5.4 \(\frac{3} {2}\)-Uniform Stability
In this section we discuss the uniform stability of the “food-limited” population model
where r(t) and s(t) are positive functions, l, τ > 0 are positive constants, and k 1∕l is the unique positive equilibrium point of (5.79). The results in this section are adapted from [67].
Motivated by (5.79) (let x(t) = (N(t)∕k 1∕l) − 1) in this section we examine
We consider solutions of (5.80), which correspond to the initial condition for any t 0 ≥ 0
The zero solution of (5.80) is said to be uniformly stable if, for \(\varepsilon > 0\), there exists a \(\delta (\varepsilon )\) such that t 0 > 0 and \(\left \Vert \phi \right \Vert =\sup _{s\in [t_{0}-\tau,t_{0}]}\left \vert \varphi (s)\right \vert <\delta\) imply \(\left \vert y(t;t_{0},\varphi )\right \vert <\varepsilon\) for all t ≥ t 0 where \(y(t;t_{0},\varphi )\) is a solution of (5.80) with the initial value \(\varphi\) at t 0.
Theorem 5.4.1.
If
then the zero solution of (5.80) is uniformly stable .
Proof.
Since \(\alpha < \frac{3} {2}\), there exist α 1 > 1 and 0 < p < 1, such that
and
For \(0 <\varepsilon < p\), we choose a \(\delta =\delta (\varepsilon ) > 0\) sufficiently small so that δ < p,
where
Clearly, \(\delta < p_{1} < p_{2} <\varepsilon\). Consider a solution \(x(t) = x(t;t_{0},\varphi )\) of (5.80) with initial condition \(\varphi\) at t 0, where t 0 ≥ 0 and \(\left \Vert \varphi \right \Vert =\sup _{s\in [t_{0}-\tau,t_{0}]}\left \vert \varphi (s)\right \vert <\delta\). We need to prove that
For t ∈ [t 0, t 0 +τ], we have
since
and
Hence
It follows that
and so
Repeating the previous argument, we have \(\left \vert x(t)\right \vert < p_{2} <\varepsilon\) for all \(t \in [t_{0}+\tau,t_{0} + 2\tau ]\) and thus
There are two cases to consider.
Case 1.
x(t) has no zeros on [t 0 +τ, t 0 + 2τ].
Without loss of generality, we assume that x(t) > 0 for t ∈ [t 0 +τ, t 0 + 2τ] (the case when x(t) < 0 is similar). Then by (5.80)
If x(t) > 0 for all t ≥ t 0 +τ, then x ′(t) < 0 for all t ≥ t 0 + 2τ and
Now let t 1 be the smallest zero of x(t) on (t 0 + 2τ, ∞). Clearly, 0 < x(t) < p 2 for t ∈ [t 0 + 2τ, t 1) since x(t) is decreasing on [t 0 + 2τ, t 1). Thus \(\left \vert x(t)\right \vert < p_{2}\) for t ∈ [t 0, t 1]. Assume that (5.84) does not hold. Then there must exist t 2 > t 1 such that \(\left \vert x(t_{2})\right \vert = p_{2}\) and \(x(t_{2})x^{{\prime}}(t_{2}) \geq 0\) and \(\left \vert x(t)\right \vert < p_{2},\) for t 0 ≤ t < t 2. By (5.80), we have that x(t) has a zero in [t 2 −τ, t 2], which we call ξ. Since
we have for t ∈ [ξ, t 2] that
and so
Thus, we get for t ∈ [ξ, t 2] that
and therefore
where
There are two possibilities.
Case I.
Then
since
and
Using the fact that \(\frac{3} {2}az -\frac{1} {2}z^{2}\) (here a > 0) is an increasing function for \(0 < z < \frac{3} {2}a\), we have
which is a contradiction.
Case II.
Choose η ∈ (ξ, t 2) such that
Then
which is a contradiction.
This shows that if x(t) has no zero in [t 0 +τ, t 0 + 2τ], then \(\left \vert x(t)\right \vert < p_{2} <\varepsilon\) for all t ≥ t 0.
Case 2.
x(t) has a zero \(\overline{t} \in [t_{0}+\tau,t_{0} + 2\tau ]\).
We prove that
In fact, if (5.85) does not hold, then there must be a point \(t^{{\ast}} > \overline{t}\) such that \(\left \vert x(t^{{\ast}})\right \vert = p_{2},\;x(t^{{\ast}})\;x^{{\prime}}(t^{{\ast}}) \geq 0\) and \(\left \vert x(t)\right \vert < p_{2}\) for \(t \in [t_{0},t^{{\ast}})\). Following the reasoning in Case 1 we derive a similar contradiction. The proof of Theorem 5.4.1 is now complete. ■
Theorem 5.4.2.
Assume that
If (5.82) holds, then the zero solution of (5.80) is uniformly and asymptotically stable.
Proof.
In view of Theorem 5.4.1, it suffices to prove that there exists a δ 0 > 0 such that the solution of (5.80) with the initial condition \(\left \Vert \varphi \right \Vert =\sup _{t\in [t_{0}-\tau,t_{0}]}\vert \varphi (t)\vert <\delta _{0}\) satisfies
Let α 1 > 1 and 0 < p < 1 be such that
and
Since the zero solution of (5.80) is uniformly stable, it follows that for \(0 <\varepsilon < p\), there exists δ 0 > 0 such that
provided \(\left \Vert \varphi \right \Vert =\sup _{t\in [t_{0}-\tau,t_{0}]}\vert \varphi (t)\vert <\delta _{0}\). Set
Clearly \(0 \leq \Delta <\varepsilon\). We prove that \(\Delta = 0\).
If x(t) is eventually nonnegative, then by (5.80), x(t) is eventually decreasing and hence \(\lim _{t\rightarrow \infty }x(t) = \Delta _{1}\) exists. Suppose \(\Delta _{1} > 0\). Then there exists t 1 > t 0 such that
By (5.80), we have for t ≥ t 1 +τ that
Using (5.86), we have
which contradicts \(\Delta _{1} > 0\). Hence \(\lim _{t\rightarrow \infty }x(t) = \Delta _{1} = 0\). Similarly, one can show that if x(t) is eventually nonpositive then lim t → ∞ x(t) = 0.
Now assume that x(t) is oscillatory. For any \(0 <\eta <\varepsilon -\Delta \), by (5.87) there exists t 2 > t 0 such that \(\left \vert x(t)\right \vert < \Delta +\eta\) for t ≥ t 2. Let {t n ∗} be an increasing sequence such that \(t_{n}^{{\ast}}\geq t_{2} + 2\tau,\;x^{{\prime}}(t_{n}^{{\ast}}) = 0,\;\lim _{n\rightarrow \infty }\left \vert x(t_{n}^{{\ast}})\right \vert = \Delta \) and t n ∗ → ∞ as n → ∞. By (5.80), x(t n ∗−τ) = 0. Thus, we have
This yields
Consequently,
since \(\left \vert \ln (1 + z)\right \vert \leq a\) implies \(\left \vert z\right \vert \leq e^{a} - 1\). Thus for \(t \in [t_{n}^{{\ast}}-\tau,t_{n}^{{\ast}}]\)
which implies for \(t \in [t_{n}^{{\ast}}-\tau,t_{n}^{{\ast}}]\) that
where
There are three cases to consider:
Case I.
Then
since the function
is increasing for 0 ≤ z ≤ α ∗ and
Thus,
Case II.
Then
or
Case III.
Choose h ∈ (0, τ) such that
Then by (5.89)
and so
Combining all the three cases, we have
where
Since
and
it follows that there exists α 0 < 1 such that, for sufficiently small \(\varepsilon > 0\), we have
and
Thus by (5.90), we get
Letting n → ∞ and η → 0, we have
which, together with α 0 < 1, implies \(\Delta = 0\). The proof is now complete. ■
5.5 Models with Periodic Coefficients
The variation of the environment plays an important role in many biological and ecological dynamical systems . The assumption of periodicity of the parameters in the system (in a way) incorporates the periodicity of the environment. It is realistic to assume that the parameters in the models are periodic functions of period ω. We consider the nonautonomous “food-limited” population model
In this section we discuss (5.91) when K is a periodic function. The results in this section are adapted from [28]. We first consider the nondelay case.
Theorem 5.5.1.
Suppose r,c, and K are continuous and positive periodic function of period ω. Then there exists a unique ω-periodic solution N ∗ (t) of the periodic differential equation
such that all other positive solutions of (5.92) satisfy
Proof.
Let N(t, 0, N 0) denote the unique solution of (5.92) through the initial point (0, N 0). Let
Then it follows from (5.92) that
and in particular
Define the function
by
As N(t; 0, N 0) depends continuously on N 0, it follows that f is a continuous function mapping [K ∗, K ∗] into itself. Therefore f has a fixed point N 0 ∗. In view of the ω-periodic of r, c, and K, it follows that the unique solution \(N^{{\ast}}(t) \equiv N(t,0,N_{0}^{{\ast}})\) of (5.92) through the initial point (0, N 0 ∗) is positive and ω-periodic. This completes the proof of the existence of a positive and ω-periodic solution N ∗(t) of (5.92).
Let N(t) be an arbitrary positive solution of (5.92). We let
and note
where
By the mean-value theorem of differential calculus, we can rewrite (5.95) in the form
where
and ξ(t) lies between N ∗(t) and N ∗(t)e x(t). Define a Lyapunov function V for (5.96) in the form
Calculating the rate of change of V along the solutions of (5.96) we obtain for x(t) ≠ 0 that
One can easily see that every positive solution of this equation is bounded. Therefore x(t) is also bounded. As r, K, and N ∗ are positive functions and ξ(t) lies between N ∗(t) and N ∗(t)e x(t), it follows from (5.97) that there exists a positive number μ such that
Thus from (5.98) we have
so
Hence
Since x(t) and \(\mathop{x}\limits^{.}(t)\) are bounded in [0, ∞), it follows from Barbalats’ Theorem (see Sect. 1.4) that
Thus x(t) → 0 as t → ∞ and the result follows from (5.94). This completes the proof. ■
Now we consider the periodic delay differential equation (5.91), namely
together with the initial condition
Note the unique positive periodic solution N ∗(t) of (5.92) is also a periodic solution of (5.99).
For convenience, we introduce the notations
Theorem 5.5.2.
If N(t) is a solution of the initial value problems (5.99) and (5.100) then there exists a number \(T = T(\varphi )\) such that
Proof.
We note that any solution of (5.99) satisfies the differential inequality
Solutions of (5.104) can be either oscillatory or nonoscillatory about K ∗.
First, suppose that N(t) is oscillatory about K ∗. Then there exists a sequence {t n }, t n → ∞ as n → ∞ of zeros of N(t) − K ∗ such that N(t) − K ∗ takes both positive and negative values on (t n , t n+1) for n = 1, 2, …. Let N(t n ∗) denote a local maximum of N(t) on (t n , t n+1). Then from (5.104), we obtain
which implies that
This shows that there exists a point \(\xi \in [t_{n}^{{\ast}}- m\omega,t_{n}^{{\ast}}]\) such that N(ξ) = K ∗. Integrating (5.104) over [ξ, t n ∗] we obtain
and
Since the right side of (5.105) is independent of t n , we conclude that
Next assume that N(t) is non oscillatory about K ∗. Then it is easily seen that for every \(\varepsilon > 0\) there exists a \(T_{1} = T_{1}(\varepsilon )\) such that
This and (5.106) imply that there exists a \(T = T(\varphi )\) such that
In a similar way we can derive a lower bound for positive solutions of (5.99). In fact from (5.99) we find
Let N(t) be an oscillatory solution about K ∗ and let {s n } → ∞ as n → ∞ be such that
and N(t) − K ∗ takes both positive and negative values on (t n , t n+1). Let s n ∗ be such that N(s n ∗) is a local minimum of N(t). Then from (5.107), we obtain
which implies that
This show that there exists a point \(\eta \in [s_{n}^{{\ast}}- m\omega,s_{n}^{{\ast}}]\) such that N(η) = K ∗. Integrating (5.107) over [η, s n ∗] we find
and
Hence
Next, assume that N(t) is nonoscillatory about K ∗. One can easily show in this case that for every positive \(\varepsilon\) there exists a \(T_{2} = T_{2}(\varepsilon )\) such that
This and (5.108) imply that there exists a \(T_{2} = T_{2}(\varphi )\) such that
The proof is complete. ■
We will derive sufficient conditions for the global attractivity of N ∗(t) with respect to all other positive solutions of (5.99) and (5.100). As before we set
in (5.99) and note that
where
We can rewrite (5.110) in the form
where
and ζ(t) lies between N ∗(t) and N(t − mω). Clearly
Theorem 5.5.3.
Assume that the positive periodic functions r(t),K(t), and c(t) satisfy the condition
Then every solution of (5.99) and (5.100) satisfies
Proof.
It suffices to prove that every solution x of (5.112) and (5.113) satisfies
Consider V (t) = V (x(t)) given by
which in view of (5.112) yields
Using the inequality
and simplifying (5.119) we obtain
It follows from (5.115) that V is eventually nonincreasing say for t ≥ T. Clearly all solutions of (5.99) are bounded and so by (5.109) and (5.110), x is uniformly continuous on [0, ∞). Integrating (5.120) over [T, t] and taking into account the inequality (5.115), we get
Hence \(x^{2} \in L_{1}(T,\infty )\) and by Barbalat’s Theorem (see Sect. 1.4)
The proof is complete. ■
5.6 Global Stability of Models with Impulses
In this section, we are concerned with the global stability of “food-limited” population models with impulsive effects . We consider the model
where p ∈ C[0, ∞) with p > 0, λ ∈ (0, ∞), τ > 0, b k > −1 for all k ∈ N. The aim in this section is to establish some sufficient conditions which ensure that every solution of (5.121) tends to 1 as t → ∞. The results in this section are adapted from [41]. Let the sequence t k (k ∈ N) be fixed and satisfy the condition,
We only consider solutions of (5.121) with initial conditions of the form
Lemma 5.6.1.
Suppose that any ε > 0 there exists an integer N such that
If in addition
then every non-oscillatory solution of
tends to zero as t tends to infinity.
Proof.
Without loss of generality, suppose that x(t) is an eventually positive solution of (5.125). Then there is a T 1 ≥ 0 such that x(t −τ) > 0 for t ≥ T 1, t ≠ t k . Thus (5.125) implies that x(t) is decreasing in (t k , t k+1] with t k ≥ T 1. Let
Then α ≥ 0. First we prove α = 0. Since x(t k ) is a left locally minimum value of x(t), there is a subsequence \(\{x(t_{k_{j}})\}\) such that
If α ≠ 0, then α > 0. Choose ε > 0 such that α −ε > 0. Again there is a T > T 1, T ≠ t k such that x(t −τ) > α −ε, for t ≥ T. Hence (5.125) implies
Integrating the above inequality from T to \(t_{k_{j}},\) we get
Let either
and it follows that ∞ ≤ −∞ or − x(T) ≤ −∞, a contradiction. Then α = 0.
Now for any t ≥ T, there is a \(t_{k_{j}}\) such that \(t_{k_{j}} \leq t < t_{k_{j+1}}\). Suppose that \(t_{k_{j}} < t_{k_{j}+1} <\ldots < t_{k_{j}+l} \leq t\). Then
From (5.123), there is a constant A > 0 such that \(\mathop{\prod }\limits _{s=0}^{l}(1 + b_{k_{j}+s}) \leq A\) for any l and any k j . Thus \(0 < x(t) \leq Ax(t_{k_{j}})\). Then lim t → +∞ x(t) = 0. The proof is complete. ■
Lemma 5.6.2.
Suppose that (5.123), (5.124) hold and there is a constant M > 0 such that
Then every oscillatory solution of (5.125) is bounded.
Proof.
Let x(t) be oscillatory solution of (5.125). Equation (5.125) implies
Choose a sequence {c n } such that
Let
It suffices to prove that \(\{\hat{x}_{i}\}\) and \(\{\tilde{x}_{i}\}\) are bounded. First, we prove that \(\{\hat{x}_{i}\}\) is bounded above. In this step, there are two cases to consider.
Case 1.
\(\hat{x}_{i}\) is the maximum value of x(t) in [c 2i−1, c 2i ].
In this case, there is a c ∈ (c 2i−1, c 2i ) such that \(\hat{x}_{i} = x(c) > 0,\;x^{{\prime}}(c) \geq 0\). Equation (5.125) implies x(t −τ) ≤ 0. Then there is a ξ ∈ (c −τ, c) such that x(ξ) = 0. Integrating (5.127) from ξ to c, we get
Case 2.
\(\hat{x}_{i}\) is not the maximum value of x(t) in [c 2i−1, c 2i ].
In this case, there is a \(t_{k+l} \in (c_{2i-1},c_{2i})\) such that \(\hat{x}_{i} = x(t_{k+l}^{+})\). We suppose that
There are two cases to consider.
- Subcase 2.1: :
-
\(x(t_{k+j-1}^{+}) \geq x(t_{k+j})\), j = 2, …, l
Then x(t) has maximum x(c) in [c 2i−1, t k+1]. By Case 1 we have x(c) ≤ M. Hence
- Subcase 2.2: :
-
There is an integer j ⋆ ∈ { 2, …, l} with \(x(t_{k+j^{\star }-1}^{+}) < x(t_{t+j^{\star }})\) and \(x(t_{k+j-1}^{+}) \geq x(t_{k+j})\), j = j ⋆ + 1, …, l.
Then x(t) has maximum x(c) in \([t_{k+j^{\star }-1},t_{k+j^{\star }}]\). By Case 1 we have x(c) ≤ M. Hence
From condition (5.123), from Cases 1 and 2, one gets that there is a constant A > 0 such that
Next, we prove that \(\{\tilde{x}_{i}\}\) is bounded below. From (5.128), there is a constant B > 0 such that x(t) ≤ B, for all t ≥ 0. Equation (5.125) implies
Using a method similar to that in Cases 1 and 2, we get
or
This shows that \(\{\tilde{x}_{i}\}\) is bounded below. The proof is complete. ■
The following result is well known.
Lemma 5.6.3.
The system of inequalities
has only a unique solution u = v = 0 in the region −∞ < u ≤ 0 ≤ v < +∞.
Lemma 5.6.4.
Suppose that λ ∈ (0,1] and (5.123), (5.124) hold. If
then every oscillatory solution of (5.125) tends to zero as t tends to infinity.
Proof.
Let x(t) be an oscillatory solution of (5.125). By Lemma 5.6.2, x(t) is bounded. Let
Then
For any ε > 0, (5.123) implies that there is a N > 0 such that
In addition, for this ε there is a T > t N such that
and
Then (5.125) implies
and
Choose a sequence {c n } such that \(x(c_{n}) = 0,\;T < c_{1} < c_{2} <\ldots,\;c_{n} \rightarrow +\infty,\;x(t) \geq 0,\) for t ∈ (c 2i−1, c 2i ) and x(t) ≤ 0 for t ∈ (c 2i , c 2i+1). Let
Then
We divide the proof into two steps.
Case 1.
\(\hat{x}_{i}\) is the maximum value of x(t) in (c 2i−1, c 2i ).
In this case, there is a c ∈ (c 2i−1, c 2i ) such that \(\hat{x}_{i} = x(c) > 0,\;x^{{\prime}}(c) \geq 0\), and x(t −τ) ≤ 0. Then there is a ξ ∈ (c −τ, c) such that x(ξ) = 0. Integrating (5.131) from ξ to c, we get
Case 2.
\(\hat{x}_{i}\) is not the maximum value of x(t) in (c 2i−1, c 2i ).
In this case, there is a \(t_{k+l} \in (c_{2i-1},c_{2i})\) such that \(\hat{x}_{i} = x(t_{k+l}^{+})\). Suppose \(c_{2i-1} < t_{k+1} <\ldots < t_{k+l}\). As in Case 2 in Lemma 5.6.2, there is a c ∈ (c 2i−1, t k+l ) such that x(c) is a left locally maximum value of x(t), and we have that there is a j ∈ { 1, 2, …, l} such that
Then by (5.123), we get
Let i → +∞, ε → 0, and we get
Similarly, we have
From Lemma 5.6.3, we get from (5.132) and (5.133) that u = v = 0. Then lim t → +∞ x(t) = 0. This completes the proof. ■
Lemma 5.6.5.
Suppose that λ > 1 and (5.123), (5.124) , and (5.130) hold. Then every oscillatory solution of (5.125) tends to zero as t tends to infinity.
Proof.
Since λ ∈ (1, +∞), let \(M(t) = \dfrac{1} {N(t)},\) and (5.121) becomes
We note \(\dfrac{1} {\lambda } \in (0,1)\). Then by Lemma 5.6.4, we get Lemma 5.6.5. The proof is complete. ■
Lemma 5.6.6.
Suppose that λ ∈ (0,1], and (5.123), (5.124) holds. If
then every oscillatory solution of (5.125) tends to zero as t → +∞.
Proof.
Let x(t) be an oscillatory solution of (5.125). By Lemma 5.6.2, x(t) is bounded. Let
Then
From (5.123), for any ε > 0, there is a N such that
Again for this ε > 0, there is a T ≥ t N such that
Then (5.125) implies
Choose a sequence {c n } such that \(x(c_{n}) = 0,\;T < c_{1} < c_{2} <\ldots,\;c_{n} \rightarrow +\infty,n \rightarrow +\infty,x(t) \geq 0\) for t ∈ (c 2i−1, c 2i ) and x(t) ≤ 0 for t ∈ (c 2i , c 2i+1). Let
Then
We first prove
or
There are two cases to be considered.
Case 1.
\(\hat{x}_{i}\) is the maximum value of x(t) in (c 2i−1, c 2i ).
In this case, there is a c ∈ (c 2i−1, c 2i ) such that \(\hat{x}_{i} = x(c) > 0,\;x^{{\prime}}(c) \geq 0\). By (5.125) we have x(t −τ) ≤ 0. Then there is a ξ ∈ (c −τ, c) such that x(ξ) = 0. If t ∈ [ξ, c], then t −τ ≤ ξ. Integrating (5.137) from t −τ to ξ, one gets
Equation (5.125) implies for t ≥ 0 that
Integrating (5.141) from ξ to c and noting that \(\dfrac{1 - e^{x}} {1 +\lambda e^{x}}\) is decreasing, we get
- Subcase 1.1: :
-
$$\displaystyle\begin{array}{rcl} \mathop{\int }\limits _{\xi }^{c}p(t)\mathop{\prod }\limits _{ t\leq t_{k}<c}(1 + b_{k})dt& \leq &-\dfrac{1} {A}\ln \dfrac{(1+\lambda )e^{-\lambda A(1- \frac{\lambda }{ A})}} {\lambda } (1+\epsilon ) {}\\ & \equiv &\alpha (1+\epsilon ) \leq \delta (1+\epsilon ). {}\\ \end{array}$$
By the monotone property of the function
and using \(\lambda e^{-A\alpha } = (1+\lambda )e^{-\lambda A(1-\frac{\lambda A} {2} )} - 1\), we get that
Then Lemma 5.3.3 gives us that
Then from Lemma 5.3.1
i.e.,
- Subcase 1.2: :
-
$$\displaystyle{ \mathop{\int }\limits _{\xi }^{c}p(t)\mathop{\prod }\limits _{ t\leq t_{k}<c}(1 + b_{k})dt \leq \delta (1+\epsilon ) <\alpha (1+\epsilon ). }$$
In this case \(\alpha > \frac{3} {2}(1+\lambda )\), i.e.,
From Lemma 5.3.4 we have that
Integrating (5.141) from ξ to c, we get
By a method similar to that in Lemmas 5.3.5 and 5.3.6, we get
i.e.,
- Subcase 1.3: :
-
$$\displaystyle{ \delta (1+\epsilon ) \geq \mathop{\int }\limits _{\xi }^{c}p(t)\mathop{\prod }\limits _{ t\leq t_{k}<c}(1 + b_{k})dt >\alpha (1+\epsilon ). }$$
Choose η ∈ (ξ, c) such that
Integrating (5.137) from ξ to η, one gets
Integrating (5.137) from η to c, we get
By deleting x(η) and noting
we have
Using the monotone property of the function
and by Lemma 5.3.1, it follows that
where
i.e.,
Case 2.
\(\hat{x}_{i}\) is not the maximum value of x(t) in (c 2i−1, c 2i ).
In this case, there is a \(t_{k+l} \in (c_{2i-1},c_{2i})\) such that \(\hat{x}_{i} = x(t_{k+l}^{+})\). Suppose \(c_{2i-1} < t_{k+1} <\ldots < t_{k+l}\). As in Case 2 in Lemma 5.6.2, there is a c ∈ (c 2i−1, t k+l ) such that x(c) is a locally maximum value of x(t), and there is a j ∈ { 1, 2, …, l} such that
where x(c) satisfies (5.138). Then by (5.123), we get
Let i → +∞, ε → 0 in (5.138) and (5.139) to obtain
Next we prove
Let \(B = \dfrac{1 - e^{v}} {1 +\lambda e^{v}}\). Then by (5.125), we have
There are two cases to consider.
Case 1.
\(\tilde{x}_{i}\) is the minimum value of x(t) in (c 2i , c 2i+1).
In this case, there is a c ∈ (c 2i , c 2i+1) such that \(x(c) =\tilde{ x}_{i} < 0,\;x^{{\prime}}(c) \leq 0,\) and then there is a ξ ∈ (c −τ, c) such that x(ξ) = 0. If t ∈ [ξ, c], then t −τ ≤ ξ. Integrating (5.137) from t −τ to c, we get
Then, we get for t ∈ [ξ, c], t ≠ t k , that
We consider two subcases.
- Subcase 1.1: :
-
$$\displaystyle{ \mathop{\int }\limits _{\xi }^{c}p(t)\mathop{\prod }\limits _{ t\leq t_{k}<c}(1 + b_{k})dt \leq (1+\epsilon )\left (\delta + \dfrac{1} {B}\ln \dfrac{(1+\lambda )e^{-\lambda B(1-\frac{\lambda B} {2} )} - 1} {\lambda } \right ). }$$
In this case, it is easy to see that
Then by Lemma 5.3.2, we get
Integrating (5.147) from ξ to c, one gets
Then
- Subcase 1.2: :
-
$$\displaystyle\begin{array}{rcl} \delta (1+\epsilon )& \geq &\mathop{\int }\limits _{\xi }^{c}p(t)\mathop{\prod }\limits _{ t\leq t_{k}<c}(1 + b_{k})dt {}\\ & >& (\delta + \dfrac{1} {B}\ln \dfrac{(1+\lambda )e^{-\lambda B(1-\frac{\lambda B} {2} )} - 1} {\lambda } )(1+\epsilon ). {}\\ \end{array}$$
Choose η ∈ (ξ, c) such that
Integrating (5.147) from ξ to η, integrating (5.148) from η to c, and deleting x(η), we get
Using the monotone property of the function
we get
By Lemma 5.3.2, we get
Case 2.
\(\tilde{x}_{i}\) is not the minimum value of x(t) in (c 2i , c 2i+1).
In this case, there is a \(t_{k+l} \in (c_{2i-1},c_{2i})\) such that \(\tilde{x}_{i} = x(t_{k+l}^{+})\). Suppose \(c_{2i} < t_{k+1} <\ldots < t_{k+l}\). As in Case 2 in Lemma 5.6.2, there is a c ∈ (c 2i−1, t k+l ) such that x(c) is a locally minimum value of x(t), and x(c) satisfies (5.149) [(5.150)]. Then there is a j ∈ { 1, 2, …, l} such that
By (5.123), we have
Let i → +∞, ε → 0 in (5.149) and (5.151) and we get (5.146). Let
Then (5.145) and (5.146) become
By Lemma 5.3.7, then x = y = 0. Thus u = v = 0. Then x(t) tends to zero as t tends to infinity. The proof is complete. ■
Lemma 5.6.7.
Suppose that λ ∈ (1,∞) and (5.123), (5.130) holds. Then every oscillatory solution of (5.125) tends to zero as t tends to infinity.
Theorem 5.6.1.
Assume − 1 < b k ≤ 0 for every k ∈ N and \(\sum _{k=1}^{\infty }b_{k} = -\infty \) . In addition if
is bounded, then every positive solution of (5.121) tends to 1 as t tends to infinity.
Proof.
It follows from − 1 < b k ≤ 0 and\(\mathop{\int }\limits _{t-\tau }^{t}p(s)\mathop{\prod }\limits _{s\leq t_{k}<t}(1 + b_{k})ds\) is bounded that (5.123) holds. Let
An argument similar to that in the proof of Lemma 5.6.2 yields that y(t) is bounded. If − 1 < b k ≤ 0, then \(\mathop{\prod }\limits _{k=1}^{\infty }(1 + b_{k}) = 0\), if and only if \(\sum _{k=1}^{\infty }b_{k} = -\infty \). Hence
and the conditions of this theorem imply that x(t) tends to zero as t tends to infinity. This completes the proof. ■
Theorem 5.6.2.
Suppose (5.123), (5.124) , and (5.135) hold. Then every positive solution of (5.121) tends to 1 as t tends to infinity.
5.7 Global Stability of Generalized Models
In this section we establish some global attractivity conditions of the generalized “food-limited” population model
where
and α is a ratio of two odd positive integers so that α ≥ 1. The results in this section are adapted from [39]. We consider solutions of (5.153) under the initial condition
Lemma 5.7.1.
For any v ∈ [0,1),
and for any u ∈ [0,∞),
Proof.
Let
It is easy to see that
and
It follows that f(v) ≥ f(0) = 1 for v ∈ [0, 1). The other assertion can be similarly proved. The proof is complete. ■
Lemma 5.7.2.
Assume that v ∈ (0,1). Then for any x ∈ [0,∞),
Proof.
Set
and
Note
and
Since α ≤ 1, it follows that \(f^{^{{\prime\prime}} }(x) \leq 0\) for x ≥ 0. By the mean-value theorem and the fact that
we have
The proof is complete. ■
The following result follows the usual argument in the literature (for completeness we include it here; see also Lemma 5.3.7).
Lemma 5.7.3.
The system of inequalities
has a unique solution (u,v) = (0,0) in the region {(u,v): −1 < v ≤ 0 ≤ u < 1}.
Proof.
Set
and
Observe that h(1) = 0,
and for x > 1
It follows that h(x) > h(1) = 0 for x > 1, and so f ′(x) > 0 for x > 1. This shows that f(x) > f(1) = 0 for x > 1. From (5.156), we have
where
If u > 0, then μ > 1, and so
This contradiction implies that u = v = 0. The proof is complete. ■
The following result follows the usual argument.
Lemma 5.7.4.
Suppose that
Then every solution of (5.153) and (5.154) that does not oscillate about 1 tends to 1 as t →∞.
Lemma 5.7.5.
Suppose 0 < λ(t) ≤ 1 for t ≥ 0 and
Let N(t) = N(t;0,ϕ) be a solution of (5.153) and (5.154) which is oscillatory about 1. Then N(t) is bounded above and is strictly bounded below by 0.
Proof.
Let t 0 be large enough so that
Let t ∗ be a local maximum point of N(t) for t ≥ t 0 +τ. Then
Integrating (5.153) from t ∗−τ to t ∗ yields
Consequently,
Next, let t ∗ be a local minimum point of N(t) for t ≥ t 0 + 3τ. Then \(N^{{\prime}}(t_{{\ast}}) = 0\) and N(t ∗−τ) = 1. Proceeding as before and using the fact that
for t ≥ t 0 +τ, we have
Hence
The proof is complete. ■
The proof of next result is similar to the proof of Lemma 5.7.5 and is thus omitted.
Lemma 5.7.6.
Assume that λ(t) ≥ 1 for t ≥ 1 and
Let N(t) = N(t,0,ϕ) be a solution of (5.153) and (5.154) which is oscillatory about 1. Then N(t) is bounded above and strictly bounded below by 0.
Theorem 5.7.1.
Suppose 0 < λ(t) ≤ 1, for t ≥ 0, and (5.157) holds. If (5.158) holds, then every solution of (5.153) and (5.154) tends to 1 as t tends to + ∞.
Proof.
Let
Then by Lemma 5.7.5, 0 < v ≤ 1 and u ≥ 1. It suffices to show that u = v = 1. For any \(\varepsilon \in (0,v),\) choose \(t_{0} = t_{0}(\varepsilon )\) such that
and
Note that
and
Thus
and
Consequently,
and
Let R(t) = r(t)∕λ α(t). Let {p n } be an increasing sequence such that p n ≥ t 0 +τ
By (5.153), N(p n −τ) = 1. For p n −τ ≤ t ≤ p n , by integrating (5.164) from t −τ to p n −τ, we get
Substituting this into (5.153), if N(t −τ) ≤ 1, we have
If N(t −τ) > 1, by (5.153), N ′(t) < 0, and thus
If t ∈ (p n −τ, p n ), we have
where
Since
it follows from Lemma 5.7.1 that
and so
There are two possibilities.
Case 1.
where \(v_{0} = (1 - v_{1})/(1 + v_{1})\).
Then
Note that the function
is increasing in \([0,3+\varepsilon ]\) and we have by Lemmas 5.7.1 and 5.7.2, that
Case 2.
Choose \(\xi _{n} \in (p_{n}-\tau,p_{n})\) such that
Then by (5.166) and Lemma 5.7.1,
where
and we have used the fact that the function
is increasing on \([0,3+\varepsilon ]\).
In either cases, we have proved that
Letting n → ∞ and \(\varepsilon \rightarrow 0\), we have
Next, let {q n } be an increasing sequence such that \(q_{n} \geq t_{0}+\tau,\;\mathop{\lim }\limits_{n \rightarrow \infty }q_{n} = +\infty,\;N^{{\prime}}(q_{n}) = 0\), and \(\lim _{n\rightarrow \infty }N(q_{n}) = -v\). By (5.153), N(q n −τ) = 1. For q n −τ ≤ t ≤ p n , integrating (5.165) from t −τ to q n −τ, we have
Substituting this into (5.153), if N(t −τ) ≥ 1, we have
for q n −τ ≤ t ≤ q n . If N(t −τ) < 1, then by (5.153), N ′(t) > 0, and thus
where \(u_{0} = (1 - u_{1})/(1 + u_{1})\). Thus
for q n −τ ≤ t ≤ q n . Note that 0 < −u 0 < 1, and one can easily see that
There are two cases to consider.
Case 1.
By (5.168) and Lemma 5.7.1,
Case 2.
We choose \(\eta _{n} \in (q_{n}-\tau,q_{n})\) such that
Then by (5.155) and Lemma 5.7.1, we have
where we have used the fact that
is increasing on \([0,3+\varepsilon ]\).
In either cases, we have proved that \(-\ln N(p_{n}) \leq -(2+\varepsilon )u_{0}\) for n = 1, 2, …. Letting n → ∞ and \(\varepsilon \rightarrow 0\), we have
Let
and
then in view of (5.167), (5.169), and Lemma 5.7.3, we get x = y = 0. This shows that u = v = 1. The proof is complete. ■
By methods similar to those in the proof of Theorem 5.7.1, and by noting that if λ ≥ 1, then
and
one can prove the next result. The details are omitted.
Theorem 5.7.2.
Suppose λ(t) ≥ 1 for t ≥ 0, (5.157) , and (5.159) hold. Then every solution of (5.153) and (5.154) tends to 1 as t tends to + ∞.
5.8 Existence of Periodic Solutions
In this section, we consider the equation
and establish some sufficient condition which ensures the existence of periodic solutions . Here a, b, c, d, k, r are continuous ω-periodic functions with r > 0, k > 0, a > 0, b ≥ 0, c ≥ 0, and d ≥ 0. The results in this section are adapted from [22]. Considering the biological significance of system (5.170), we always assume that N(0) > 0. The main results will be proved by applying Theorem 1.4.11. To prove the main results we present some useful lemmas.
Let f be a ω-periodic function and define
Lemma 5.8.1.
There exists a unique u ∗ > 0 such that
Proof.
Let
It is clear that
and then from the zero point theorem, it follows that there exists a \(u^{{\ast}}\in \left (0, \dfrac{r^{u} + 1} {a^{l} + b^{l}}\right )\) such that f(u ∗) = 0. Moreover,
that is, f(u) is monotonically decreasing with respect to u, and hence u ∗ is unique. The proof is complete. ■
Theorem 5.8.1.
Equation (5.170) has at least one positive periodic solution of period ω
Proof.
Let N(t) = exp{x(t)}. Then (5.170) may be reformulated as
In order to apply Theorem 1.4.11 to (5.171), we first let
and
Then \(\mathbb{X}\) and \(\mathbb{Y}\) are Banach spaces with the norm \(\left \Vert.\right \Vert\). Let
Then it follows that
and P, Q are continuous projectors such that
Therefore, L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (of L)
is
Also
and
By the Arzela–Ascoli Theorem , it is easy to see that \(\overline{K_{P}(I - Q)N(\overline{\Omega )}}\) is compact for any open bounded subset \(\Omega \) of \(\mathbb{X}\) and \(QN(\overline{\Omega })\) is bounded. Thus, N is L-compact on \(\overline{\Omega }\) for any open bounded set \(\Omega \in \mathbb{X}\).
Consider the operator equation L x = λ N x, λ ∈ (0, 1), that is,
Let \(x = x(t) \in \mathbb{X}\) be a solution of (5.172) for a certain λ ∈ (0, 1). Integrating (5.172) with respect to t over the interval [0, ω] yields
and therefore
which together with (5.172) implies
From (5.173) and the mean-value theorem for integral, we see that there exists ξ ∈ [0, ω] such that
and therefore
Since \(x(t) \in \mathbb{X}\), there exist \(t_{1},t_{2} \in [0,\omega ]\) such that \(x(t_{1}) = x^{l},\;x(t_{2}) = x^{u},\) and then from (5.175) it follows that
from which we derive
and hence
Clearly, B 1 is independent of the choice of λ. Take B = B 1 + B 2, where B 2 > 0 is taken sufficiently large such that \(\left \vert \ln (u^{{\ast}})\right \vert < B_{2}\) and define
When \(x \in \partial \Omega \cap Ker\;L = \partial \Omega \cap \mathbb{R},\;x = B\) or x = −B, and then
Furthermore, a direct calculation reveals that
here J is the identity mapping since ℑ P = KerL. Thus all the requirements in Theorem 1.4.11 are satisfied. Hence (5.171) has at least one solution \(x^{{\ast}}(t) \in Dom\;L \cap \overline{\Omega }\). Set \(N^{{\ast}}(t) =\exp \{ x^{{\ast}}(t)\}\). Then N ∗(t) is a positive ω-periodic solution of (5.170). The proof is complete. ■
References
L. Berezansky, E. Braverman, Oscillation of a food-limited population model with delay. Abstr. Appl. Anal. 1, 55–66 (2003)
M. Fan, K. Wang, Periodicity in a food limited population model with toxicants and time delays. Acta Math. Appl. Sin. 18, 309–314 (2002)
A. Friedman, What is mathematical biology and how useful is it? Notices Am. Math. Soc. 57, 851–857 (2010)
W. Krawcewicz, J. Wu, Theory of Degrees with Applications to Bifurcations and Differential Equations (Wiley, New York, 1997)
B.S. Lalli, B.G. Zhang, On a periodic delay population model. Q. Appl. Math. 52, 35–42 (1994)
G. Seifret, Periodic solutions of a certain logistic equations with piecewise constant delays and almost periodic time dependence. J. Differ. Equat. 164, 451–458 (2000)
G. Seifret, Almost Periodic solutions of differential equation with discontinuous delays. Comm. Appl. Anal. 6, 235–240 (2002)
J.W.-H. So, J.S. Yu, Global stability in a logistic equation with piecewise constant arguments. Hokkaido Math. J. 24, 269–286 (1995)
X. Tang, J. Yu, 3/2 global attractivity of the zero solution of the “food-limited” type functional differential equations. Sci. China Ser. A. 44, 610–618 (2001)
J. Wang, J. Yan, On Oscillation of a food-limited population model with impulse and delay. J. Math. Anal. Appl. 334, 349–357 (2007)
Author information
Authors and Affiliations
Rights and permissions
Copyright information
© 2014 Springer International Publishing Switzerland
About this chapter
Cite this chapter
Agarwal, R.P., O’Regan, D., Saker, S.H. (2014). Food-Limited Population Models. In: Oscillation and Stability of Delay Models in Biology. Springer, Cham. https://doi.org/10.1007/978-3-319-06557-1_5
Download citation
DOI: https://doi.org/10.1007/978-3-319-06557-1_5
Published:
Publisher Name: Springer, Cham
Print ISBN: 978-3-319-06556-4
Online ISBN: 978-3-319-06557-1
eBook Packages: Mathematics and StatisticsMathematics and Statistics (R0)