Componentwise localization of positive periodic solutions for a general Bazykin-type system

Andra Malina Email: andra.malina@ubbcluj.ro Andrei Stan Affiliation: Email: andrei.stan@ubbcluj.ro Affiliation:

Abstract

In the present paper, we consider a general Bazykin-type model that describes the interaction of two populations. We provide sufficient conditions for the existence of a positive periodic solution where each population is localized independently within a distinct conical annular region. We illustrate how our conditions can be verified for certain particular cases and support these results with numerical simulations.

keywords

Bazykin model, localized solution, periodic solution

1 Introduction and Preliminaries

The interaction between two species (e.g., predator-prey interaction) is commonly described by Lotka–Volterra models (see [1] for a survey). Autonomous variants of such models are often formulated as Kolmogorov-type systems of the form (see [2, Section 5])

\begin{cases}u^{\prime}=uF(u,v),\\ v^{\prime}=vG(u,v),\end{cases}

which describe, for instance, the case of species in competition with respect to other populations (an increase in the size of either population tends to decrease the growth rate of the other),

F_{y}(x,y)<0,\quad G_{x}(x,y)<0,

or with respect to itself (an increase in the size of either population inhibits its own growth rate), that is,

F_{x}(x,y)<0,\quad G_{y}(x,y)<0.

Other models describing predator–prey interactions were proposed in the work of A. Bazykin (see, e.g., [3, 4, 5]). These models capture various behavioral scenarios of prey and predators, such as predator saturation, predator competition for resources (other than prey), nonlinear predation, and competition among prey. We recall below two of these models:

		$\displaystyle(a)\begin{cases}u^{\prime}=au-\dfrac{buv}{(1+Au)(1+Bv)},\\[6.0pt] v^{\prime}=-cv+\dfrac{duv}{(1+Au)(1+Bv)},\end{cases}$
		$\displaystyle(b)\begin{cases}u^{\prime}=au-\dfrac{buv}{(1+Au)(1+Bv)},\\[6.0pt] v^{\prime}=-cv+\dfrac{duv}{(1+Au)(1+Bv)}-\delta v^{2}.\end{cases}$

In models describing population dynamics, it is appropriate to replace the coefficients with periodic time-dependent functions, since such a modification captures repetitive external influences on the population, such as weather, food supply, or hunting (see [6]), and naturally leads to the search for periodic solutions. The existence of positive periodic solutions has been extensively studied in the literature for such models (see, e.g., [7, 8, 9, 10, 11, 12, 13, 14]). In particular, we mention the works [12, 13, 14], where localized positive periodic solutions are obtained.

For any domain $\mathcal{D}$ , let

C_{\omega}(\mathbb{R}\times\mathcal{D},\mathbb{R}):=\left\{u\in C(\mathbb{R}\times\mathcal{D},\mathbb{R}):u(\cdot,x)\text{ is $\omega$-periodic for each }x\in\mathcal{D}\right\}

denote the space of continuous $\omega$ -periodic real-valued functions defined on $\mathcal{D}$ .

We consider the following general non-autonomous Bazykin-type system:

\begin{cases}u^{\prime}=\mu(t)u\varrho(t,u,v)-f(t,u)-\varphi(t,u,v)uv,\\[6.0pt] v^{\prime}=-\eta(t)v-g(t,v)+\phi(t)\varphi(t,u,v)uv,\end{cases}

(1)

where $\omega>0$ is fixed, $\varrho,\varphi\in C_{\omega}(\mathbb{R}\times\mathbb{R}_{+}\times\mathbb{R}_{+},\mathbb{R}_{+})$ , $f,g\in C_{\omega}(\mathbb{R}\times\mathbb{R}_{+},\mathbb{R}_{+})$ , and $\mu,\eta,\phi\in C_{\omega}(\mathbb{R},\mathbb{R}_{+})$ satisfy

\int_{0}^{\omega}\mu(\tau)\,d\tau>0,\quad\int_{0}^{\omega}\eta(\tau)\,d\tau>0.

Our main purpose is to provide sufficient conditions for the existence of a positive $\omega$ -periodic solution $(u,v)$ to problem (1) such that

r_{1}\leq\|u\|\leq R_{1}\quad\text{and}\quad r_{2}\leq\|v\|\leq R_{2},

where $0<r_{i}<R_{i}<\infty$ , $i=1,2$ , are given constants, and

\|w\|:=\max_{t\in[0,\omega]}|w(t)|\quad\left(w\in C_{\omega}(\mathbb{R},\mathbb{R})\right).

The novelty of this paper is twofold. On the one hand, the localization of the solution is componentwise, that is, each component lies within a distinct conical annular region. We emphasize that in [14] (or [12]), the localization of a solution $(u,v)$ is performed with respect to the pair $(u,v)$ , i.e., the solution satisfies $r\leq\|u\|+\|v\|\leq R$ . On the other hand, the proposed model (1) encompasses both the system considered in [14] and the models introduced by Bazykin (e.g., models (a), (b)), while the system considered in [14] does not cover certain Bazykin-type models, for instance those involving quadratic terms.

We conclude this section with a fixed point result for operators defined on the Cartesian product of two Banach spaces, which will be crucial for establishing our localization result. For $i=1,2,$ let $(X_{i},|\cdot|_{i})$ be Banach spaces and let $K_{i}\subset X_{i}$ be cones. Additionally, let $r=(r_{1},r_{2}),R=(R_{1},R_{2})\in\mathbb{R}^{2}$ with $0<r_{i}<R_{i}$ , $i=1,2$ , and define the conical annular set

K_{rR}:=\left\{(u,v)\in K_{1}\times K_{2}\,:\,r_{1}<|u|_{1}<R_{1},\,r_{2}<|v|_{2}<R_{2}\right\}.

The following variant of Krasnosel’skiĭ’s fixed point theorem in cones was proved in [15, Theorem 3.1] (see also [16]). Since only the compressive–compressive case is needed in this paper, we state only this version and refer the reader to the above references for the remaining cases.

Theorem 1.

Let $\mathcal{T}=(T_{1},T_{2}):\overline{K}_{r,R}\to K_{1}\times K_{2}$ be a compact mapping. If the operator $T_{1}$ is compressive in the first variable and $T_{2}$ is compressive in the second one on $\overline{K}_{r,R}$ , that is,

(A)
1. (i)
  
  $|T_{1}(u,v)|_{1}\geq|u|_{1}$ if $|u|_{1}=r_{1}$ ;
2. (ii)
  
  $|T_{1}(u,v)|_{1}\leq|u|_{1}$ if $|u|_{1}=R_{1}$ ;
(B)
1. (i)
  
  $|T_{2}(u,v)|_{2}\geq|v|_{2}$ if $|v|_{2}=r_{2}$ ;
2. (ii)
  
  $|T_{2}(u,v)|_{2}\leq|v|_{2}$ if $|v|_{2}=R_{2}$ ,

then $\mathcal{T}$ has at least one fixed point $(u,v)\in\overline{K}_{r,R}$ .

Instead of applying Theorem 1 to obtain our localization result, one may employ other fixed point techniques, such as those from [17, 18, 19].

2 Main results

Let $0<r_{i}<R_{i}<\infty$ , $i=1,2$ , be given positive real numbers, and set $r=(r_{1},r_{2})$ and $R=(R_{1},R_{2})$ . In order to obtain a componentwise localized solution to problem (1), we apply Theorem 1. To this end, it is necessary to rewrite (1) as a fixed point problem, and for this purpose, we need the following auxiliary result.

Lemma 2.

Let $a,h\in C(\mathbb{R},\mathbb{R})$ be $\omega$ -periodic functions such that

\quad\int_{0}^{\omega}a(\tau)\,d\tau>0.

Then the equation

u^{\prime}(t)-a(t)u(t)=h(t),\quad t\in\mathbb{R},

(2)

has a unique $\omega$ -periodic solution given by

u(t)=\int_{t}^{t+\omega}H(t,s)(-h(s))\,ds,

where

H(t,s)=\frac{e^{-\int_{t}^{s}a(\tau)\,d\tau}}{1-e^{-\int_{0}^{\omega}a(\tau)\,d\tau}}>0.

Proof.

Equation (2) is clearly equivalent to

\left(u(t)\rho(t)\right)^{\prime}=\rho(t)h(t),

which implies

u(t)\rho(t)=u(0)+\int_{0}^{t}\rho(s)h(s)\,ds,

(3)

where we denoted

\rho(t)=e^{-\int_{0}^{t}a(\tau)\,d\tau}.

Since we are looking for $\omega$ periodic solutions, one needs $u(t+\omega)=u(t)$ for all $t\in\mathbb{R}$ . Thus, by (3) and using that $\rho(t+\omega)=\rho(t)\rho(\omega)$ , we obtain

\int_{t}^{t+\omega}\rho(s)h(s)\,ds=u(t+\omega)\rho(t+\omega)-u(t)\rho(t)=u(t)\rho(t)\left(\rho(\omega)-1\right),

that is,

u(t)=\int_{t}^{t+\omega}\frac{\rho(t)^{-1}\rho(s)}{\rho(\omega)-1}h(s)\,ds=\int_{t}^{t+\omega}H(t,s)(-h(s))\,ds.

∎

Next, let us consider the functions

	$\displaystyle\mathcal{F}_{1}(t,x,y)$	$\displaystyle:=\mu(t)x\bigl(1-\varrho(t,x,y)\bigr)+f(t,x)+\varphi(t,x,y)xy,$
	$\displaystyle\mathcal{F}_{2}(t,x,y)$	$\displaystyle:=-g(t,y)+\phi(t)\varphi(t,x,y)xy,$

and observe that by Lemma 2 (see also [20]), our problem (1) can be written equivalently as a fixed point equation

(u,v)=(N_{1}(u,v),N_{2}(u,v)),

(4)

where

N_{i}(u,v)(t)=\int_{t}^{t+\omega}H_{i}(t,s)\mathcal{F}_{i}\bigl(s,u(s),v(s)\bigr)\,ds,\quad t\in\mathbb{R},\,\,(i=1,2),

(5)

and

H_{1}(t,s)=\frac{e^{-\int_{t}^{s}\mu(\tau)d\tau}}{1-e^{-\int_{0}^{\omega}\mu(\tau)d\tau}}>0,\quad H_{2}(t,s)=\frac{e^{\int_{t}^{s}\eta(\tau)d\tau}}{e^{\int_{0}^{\omega}\eta(\tau)d\tau}-1}>0,\quad t,s\in\mathbb{R}.

Additionally, we introduce the constants

$\displaystyle m_{i}$	$\displaystyle:=\min\{H_{i}(t,s):t\in[0,\omega],\;s\in[t,t+\omega]\}>0,$	(6)
$\displaystyle M_{i}$	$\displaystyle:=\max\{H_{i}(t,s):t\in[0,\omega],\;s\in[t,t+\omega]\}>0,$	(7)
$\displaystyle C_{i}$	$\displaystyle:=\frac{m_{i}}{M_{i}}\in(0,1),$	(8)

and observe that

		$\displaystyle m_{1}=\frac{1}{e^{\int_{0}^{\omega}\mu(\tau)\,d\tau}-1},\quad M_{1}=\frac{1}{1-e^{-\int_{0}^{\omega}\mu(\tau)\,d\tau}},$		(9)
		$\displaystyle m_{2}=\frac{1}{e^{\int_{0}^{\omega}\eta(\tau)\,d\tau}-1},\quad M_{2}=\frac{1}{1-e^{-\int_{0}^{\omega}\eta(\tau)\,d\tau}},$
		$\displaystyle C_{1}=e^{-\int_{0}^{\omega}\mu(\tau)\,d\tau},\quad\quad\,\,\,\,C_{2}=e^{-\int_{0}^{\omega}\eta(\tau)\,d\tau}.$

Let $K_{0}$ be the cone of nonnegative functions in $C_{\omega}(\mathbb{R},\mathbb{R})$ , define the subcones $K_{i}$ of $K_{0}$ by

K_{i}:=\left\{w\in K_{0}:w(t)\geq C_{i}\|w\|\text{ for all }t\in\mathbb{R}\right\},\quad i=1,2,

(10)

and set

(K_{i})_{r_{i}R_{i}}:=\left\{w\in K_{i}:r_{i}<\|w\|<R_{i}\right\},\quad i=1,2,

and

K_{rR}:=(K_{1})_{r_{1}R_{1}}\times(K_{2})_{r_{2}R_{2}}.

Our first condition concerns the functions $g$ , $\phi$ , $\varrho,\varphi$ .

(H1)

One has,

\varrho(t,x,y)\leq 1,

(11)

and

\phi(t)\varphi(t,x,y)xy-g(t,y)\geq 0,

(12)

for all $(t,x,y)\in\mathbb{R}\times[C_{1}r_{1},R_{1}]\times[C_{2}r_{2},R_{2}]$ .

Under this condition, the following invariance result holds.

Lemma 3.

Let condition (H1) holds. Then, the image of the set $\overline{K}_{rR}$ under operators $N_{i}$ is included in $K_{i}$ , that is,

N_{i}\left(\overline{K}_{rR}\right)\subset K_{i},\quad i=1,2.

Proof.

Let $(u,v)\in\overline{K}_{rR}$ . By the definition of the sets $(K_{i})_{r_{i}R_{i}}$ , one has

\displaystyle C_{1}r_{1}\leq u(t)\leq R_{1},\quad C_{2}r_{2}\leq v(t)\leq R_{2}\quad\text{for all }t\in\mathbb{R}.

Using now conditions (11) and (12), for any $t\in\mathbb{R}$ , we have

1-\varrho(t,u(t),v(t))\geq 0,

and

-g(t,v(t))+\phi(t)\varphi(t,u(t),v(t))u(t)v(t)\geq 0,

so $\mathcal{F}_{i}(t,u(t),v(t))\geq 0$ , $i=1,2$ . Since the kernels $H_{i}$ are positive, it follows that

N_{i}(u,v)\in K_{0},\quad i=1,2.

Now, let us show that $N_{i}(u,v)\in K_{i}.$ To this end, set $w_{i}:=N_{i}(u,v)$ , and let $t_{i}^{\ast}\in[0,\omega]$ be such that $w_{i}(t_{i}^{\ast})=\|w_{i}\|$ , for $i=1,2$ . Then, we estimate,

	$\displaystyle w_{i}(t)$	$\displaystyle=\int_{t}^{t+\omega}H_{i}(t,s)\mathcal{F}_{i}\bigl(s,u(s),v(s)\bigr)\,ds$
		$\displaystyle\geq m_{i}\int_{t}^{t+\omega}\mathcal{F}_{i}\bigl(s,u(s),v(s)\bigr)\,ds$
		$\displaystyle=C_{i}M_{i}\int_{t}^{t+\omega}\mathcal{F}_{i}\bigl(s,u(s),v(s)\bigr)\,ds$
		$\displaystyle\geq C_{i}\int_{t}^{t+\omega}H_{i}(t_{i}^{\ast},s)\mathcal{F}_{i}\bigl(s,u(s),v(s)\bigr)\,ds$
		$\displaystyle=C_{i}\\|w_{i}\\|,$

so $w_{i}\in K_{i}$ , as wished. ∎

Before we state and prove our main result, let us denote

		$\displaystyle\underline{h}_{1}(t):=\min_{(x,y)\in[C_{1}r_{1},r_{1}]\times[C_{2}r_{2},R_{2}]}\mathcal{F}_{1}(t,x,y),$
		$\displaystyle\overline{h}_{1}(t):=\max_{(x,y)\in[C_{1}R_{1},R_{1}]\times[C_{2}r_{2},R_{2}]}\mathcal{F}_{1}(t,x,y),$
		$\displaystyle\underline{h}_{2}(t):=\min_{(x,y)\in[C_{1}r_{1},R_{1}]\times[C_{2}r_{2},r_{2}]}\mathcal{F}_{2}(t,x,y),$
		$\displaystyle\overline{h}_{2}(t):=\max_{(x,y)\in[C_{1}r_{1},R_{1}]\times[C_{2}R_{2},R_{2}]}\mathcal{F}_{2}(t,x,y),$

and assume that the following conditions are satisfied:

(H2)

For $i=1,2,$ one has

		$\displaystyle\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{i}(t,s)\underline{h}_{i}(s)ds\geq r_{i},$		(13)
		$\displaystyle\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{i}(t,s)\overline{h}_{i}(s)ds\leq R_{i}.$		(14)

Now we are ready to prove the main result of this paper.

Theorem 4.

Under conditions (H1)-(H2), problem (1) has at least one solution $(u,v)\in K_{1}\times K_{2}$ such that

r_{1}\leq\|u\|\leq R_{1}\quad\text{ and }\quad r_{2}\leq\|v\|\leq R_{2}.

Proof.

We apply Theorem 1 for $X_{i}:=C_{\omega}(\mathbb{R},\mathbb{R})$ $i=1,2$ and the cones $K_{i}$ defined in (10).

By standard application of the Arzelà–Ascoli theorem, the operators $N_{i}$ are completely continuous from $X\times X$ to $X$ (see, e.g., [21]). Also, by Lemma 3, we see that

N\left(\overline{K}_{rR}\right)\subset K_{1}\times K_{2}.

In what follows, we show that all the conditions in Theorem 1 hold.

Check of condition (A)(i). Let $(u,v)\in\overline{K}_{rR}$ with $\|u\|=r_{1}$ . By the definition of $K_{i}$ , $i=1,2$ , for any $t\in\mathbb{R}$ , we have

C_{1}r_{1}\leq u(t)\leq r_{1}\quad\text{ and }\quad C_{2}r_{2}\leq v(t)\leq R_{2},

which implies

\mathcal{F}_{1}(t,u(t),v(t))\geq\underline{h}_{1}(t).

Now, let $t^{\ast}\in[0,\omega]$ be such that

\int_{t^{\ast}}^{t^{\ast}+\omega}H_{1}(t^{\ast},s)\underline{h}_{1}(s)\,ds=\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{1}(t,s)\underline{h}_{1}(s)\,ds.

Then, by (5) and condition (H2), one has

\displaystyle\|N_{1}(u,v)\|\geq N_{1}(u,v)(t^{\ast})=\int_{t^{\ast}}^{t^{\ast}+\omega}H_{1}(t^{\ast},s)\mathcal{F}_{1}(s,u(s),v(s))\,ds\geq\int_{t^{\ast}}^{t^{\ast}+\omega}H_{1}(t^{\ast},s)\underline{h}_{1}(s)\,ds\geq r_{1},

(15)

as wished.

Check of condition (A)(ii). Let $(u,v)\in\overline{K}_{rR}$ satisfy $\|u\|=R_{1}.$ Since

C_{1}R_{1}\leq u(t)\leq R_{1}\quad\text{ for all }t\in\mathbb{R},

one clearly has

\mathcal{F}_{1}(t,u(t),v(t))\leq\overline{h}_{1}(t)\quad\text{ for all }t\in\mathbb{R}.

Thus, for any $t\in\mathbb{R}$ , we estimate

$\displaystyle N_{1}(u,v)(t)$	$\displaystyle=\int_{t}^{t+\omega}H_{1}(t,s)\mathcal{F}_{1}(s,u(s),v(s))ds$	(16)
	$\displaystyle\leq\int_{t}^{t+\omega}H_{1}(t,s)\overline{h}_{1}(s)ds$
	$\displaystyle\leq\max_{\tau\in[0,\omega]}\int_{\tau}^{\tau+\omega}H_{1}(\tau,s)\overline{h}_{1}(s)ds$
	$\displaystyle\leq R_{1},$

where the latter inequality follows by the second condition in (H2) for $i=1,$ so $\|N_{1}(u,v)\|\leq R_{1}$ .

Similar reasoning shows that conditions (B)(i) and (B)(ii) in Theorem 1 are also satisfied, so Theorem 1 applies and yields the conclusion. ∎

Remark 5.

The operators $N_{i}$ may fail to map the entire cone $K_{1}\times K_{2}$ into $K_{i}$ , $i=1,2$ . This can be seen already in model (b). Indeed, for elements from $K_{2}$ with sufficiently large values, the function $\mathcal{F}_{2}$ becomes negative. Therefore, in order to apply Theorem 1, one must restrict the domain so that its image under the operator $N_{2}$ remains in the cone $K_{2}$ , as shown in Lemma 3 under condition (H1).

Instead of condition (H2), we may assume the following:

(H2)^′

There exist continuous functions $\underline{k}_{i},\overline{k}_{i}$ and points $\overline{t}_{i}\in[0,\omega]$ , $i=1,2$ , such that

\underline{h}_{i}(t)\geq\underline{k}_{i}(t),\quad\overline{h}_{i}(t)\leq\overline{k}_{i}(t)\quad\text{for all }t\in\mathbb{R},

and

\displaystyle\int_{\overline{t}_{i}}^{\overline{t}_{i}+\omega}H_{i}(\overline{t}_{i},s)\underline{k}_{i}(s)\,ds\geq r_{i},\quad\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{i}(t,s)\overline{k}_{i}(s)\,ds\leq R_{i}.

Theorem 6.

Suppose that conditions (H1) and (H2)’ hold. Then the problem (1) has at least one solution $(u,v)\in K_{1}\times K_{2}$ such that

r_{1}\leq\|u\|\leq R_{1}\quad\text{ and }\quad r_{2}\leq\|v\|\leq R_{2}.

Proof.

The proof follows the arguments of Theorem 4, with minor modifications at inequalities (15) and (16). Instead of (15), we estimate

	$\displaystyle\\|N_{1}(u,v)\\|$	$\displaystyle\geq\int_{\overline{t}_{1}}^{\overline{t}_{1}+\omega}H_{1}(\overline{t}_{1},s)\mathcal{F}_{1}(s,u(s),v(s))\,ds$
		$\displaystyle\geq\int_{\overline{t}_{1}}^{\overline{t}_{1}+\omega}H_{1}(\overline{t}_{1},s)\underline{h}_{1}(s)\,ds$
		$\displaystyle\geq\int_{\overline{t}_{1}}^{\overline{t}_{1}+\omega}H_{1}(\overline{t}_{1},s)\underline{k}_{1}(s)\,ds$
		$\displaystyle\geq r_{1},$

and instead of (16), for all $t\in\mathbb{R}$ , we have

	$\displaystyle N_{1}(u,v)(t)$	$\displaystyle=\int_{t}^{t+\omega}H_{1}(t,s)\mathcal{F}_{1}(s,u(s),v(s))\,ds$
		$\displaystyle\leq\int_{t}^{t+\omega}H_{1}(t,s)\overline{h}_{1}(s)\,ds$
		$\displaystyle\leq\int_{t}^{t+\omega}H_{1}(t,s)\overline{k}_{1}(s)\,ds$
		$\displaystyle\leq\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{1}(t,s)\overline{k}_{1}(s)\,ds$
		$\displaystyle\leq R_{1}.$

Since the same estimates hold for $N_{2}$ , the conclusion follows by Theorem 1. ∎

Based on the previous Theorem 6, we immediately derive the following corollary.

Corollary 7.

Assume that condition (H1) holds. Suppose, in addition, that there exist $\overline{t}_{1},\overline{t}_{2}\in[0,\omega]$ such that

	$\displaystyle\frac{r_{1}}{\int_{\overline{t}_{1}}^{\overline{t}_{1}+\omega}H_{1}(\overline{t}_{1},s)\,ds}$	$\displaystyle\leq\mathcal{F}_{1}(t,x,y)\quad\text{for all }(t,x,y)\in[0,\omega]\times[C_{1}r_{1},r_{1}]\times[C_{2}r_{2},R_{2}],$
	$\displaystyle\frac{r_{2}}{\int_{\overline{t}_{2}}^{\overline{t}_{2}+\omega}H_{2}(\overline{t}_{2},s)\,ds}$	$\displaystyle\leq\mathcal{F}_{2}(t,x,y)\quad\text{for all }(t,x,y)\in[0,\omega]\times[C_{1}r_{1},R_{1}]\times[C_{2}r_{2},r_{2}],$

and for each $i=1,2,$

\displaystyle\mathcal{F}_{i}(t,x,y)\leq\frac{R_{i}}{\max\limits_{\tau\in[0,\omega]}\int_{\tau}^{\tau+\omega}H_{i}(\tau,s)\,ds},\quad\text{for all }(t,x,y)\in[0,\omega]\times[C_{1}r_{1},R_{1}]\times[C_{2}r_{2},R_{2}].

Then problem (1) has at least one solution $(u,v)\in K_{1}\times K_{2}$ satisfying

r_{1}\leq\|u\|\leq R_{1}\quad\text{and}\quad r_{2}\leq\|v\|\leq R_{2}.

Proof.

This follows from Theorem 6 with

\underline{k}_{i}:=\min_{t\in[0,\omega]}\underline{h}_{i}(t)\quad\text{ and }\quad\overline{k}_{i}:=\max_{t\in[0,\omega]}\overline{h}_{i}(t).

∎

If the functions $\varrho$ , $\phi$ , $f$ , and $g$ satisfy suitable monotonicity properties, then the result below holds.

Corollary 8.

(Monotonicity conditions) Suppose that, for each $t\in\mathbb{R}$ , the following monotonicity conditions hold:

(a): The mappings $(x,y)\mapsto\varrho(t,x,y)$ and $(x,y)\mapsto\varphi(t,x,y)$ are nonincreasing and nondecreasing, respectively, with respect to both $x$ and $y$ on $[C_{1}r_{1},R_{1}]\times[C_{2}r_{2},R_{2}]$ .
(b): The mappings $x\mapsto f(t,x)$ and $y\mapsto g(t,y)$ are nondecreasing on $[C_{1}r_{1},R_{1}]$ and $[C_{2}r_{2},R_{2}]$ , respectively.

Denote

		$\displaystyle\underline{k}_{1}(t)=\mu(t)C_{1}r_{1}(1-\varrho(t,C_{1}r_{1},C_{2}r_{2})+f(t,C_{1}r_{1})+C_{1}C_{2}r_{1}r_{2}\varphi(t,C_{1}r_{1},C_{2}r_{2}),$
		$\displaystyle\overline{k}_{1}(t)=\mu(t)R_{1}(1-\varrho(t,R_{1},R_{2})+f(t,R_{1})+R_{1}R_{2}\varphi(t,R_{1},R_{2}),$
		$\displaystyle\underline{k}_{2}(t)=-g(t,r_{2})+\phi(t)C_{1}C_{2}r_{1}r_{2}\varphi(t,C_{1}r_{1},C_{2}r_{2}),$
		$\displaystyle\overline{k}_{2}(t)=-g(t,C_{2}r_{2})+\phi(t)R_{1}R_{2}\varphi(t,R_{1},R_{2}),$

and suppose that conditions (13) and (14) hold with $\underline{k}_{i}$ in place of $\underline{h}_{i}$ and $\overline{k}_{i}$ in place of $\overline{h}_{i}$ ( $i=1,2$ ). Then problem (1) has at least one positive solution $(u,v)\in K_{1}\times K_{2}$ such that $r_{1}\leq\|u\|\leq R_{1}$ and $r_{2}\leq\|v\|\leq R_{2}$ .

Since conditions (H1)–(H2) are imposed on intervals closely related to the considered cone, if they are satisfied on disjoint conical sets, then multiple solutions are obtained.

Theorem 9.

Let $r_{i}^{j},R_{i}^{j}$ , $j=1,\ldots,m$ , $i=1,2$ , be sequences of positive real numbers such that

0<r_{i}^{1}<R_{i}^{1}<r_{i}^{2}<R_{i}^{2}<\cdots<r_{i}^{m}<R_{i}^{m}<\infty,

and assume that (H1)–(H2) hold for each pair $r^{j}=(r_{1}^{j},r_{2}^{j})$ and $R^{j}=(R_{1}^{j},R_{2}^{j})$ , $j=1,\ldots,m$ . Then problem (1) admits at least $m$ distinct solutions $(u^{j},v^{j})\in K_{1}\times K_{2}$ such that

r_{1}^{j}\leq\|u^{j}\|\leq R_{1}^{j},\quad r_{2}^{j}\leq\|v^{j}\|\leq R_{2}^{j},\quad j=1,\ldots,m.

The conclusion of Theorem 4 can be strengthened by replacing condition (H2) with a stronger condition.

(H2)^∗

For each $i=1,2$ , one has

\displaystyle m_{i}\int_{0}^{\omega}\underline{h}_{i}(\tau)\,d\tau\geq r_{i},\quad M_{i}\int_{0}^{\omega}\overline{h}_{i}(\tau)\,d\tau\leq R_{i},

where $m_{i},M_{i}$ are given in (6).

Theorem 10.

Under conditions (H1) and (H2)^∗, problem (1) has at least one solution $(u,v)\in K_{1}\times K_{2}$ such that

r_{1}\leq\min_{t\in\mathbb{R}}u(t),\quad\|u\|\leq R_{1},\quad\text{ and }\quad r_{2}\leq\min_{t\in\mathbb{R}}v(t),\quad\|v\|\leq R_{2}.

Proof.

Clearly, condition (H2)^∗ implies (H2), so by Theorem 4 there exists a fixed point $(u,v)\in\overline{K}_{rR}$ of the operator $(N_{1},N_{2})$ . On the other hand, for any $t\in\mathbb{R}$ , we have

		$\displaystyle u(t)=N_{1}(u,v)(t)\geq m_{1}\int_{t}^{t+\omega}\underline{h}_{1}(s)\,ds\geq r_{1},$
		$\displaystyle v(t)=N_{2}(u,v)(t)\geq m_{2}\int_{t}^{t+\omega}\underline{h}_{2}(s)\,ds\geq r_{2},$

so our (stronger) conclusion follows. ∎

3 Examples and numerical simulations

3.1 Non-autonomous Bazykin models of type (a)

Let us consider the following generalized Bazykin model of type (a):

\begin{cases}u^{\prime}=\mu(t)u-\varphi(t)\dfrac{b_{1}}{(1+Au)(1+Bv)}uv,\\ v^{\prime}=-\eta(t)v+\varphi(t)\dfrac{b_{2}}{(1+Au)(1+Bv)}uv,\end{cases}

(17)

where $A,B,b_{1},b_{2}>0$ and $\mu,\eta,\varphi\in C_{\omega}(\mathbb{R},\mathbb{R}_{+})$ for some $\omega>0$ . In system (1), with a slight abuse of notation, this corresponds to

\varrho\equiv 1,\quad\varphi(t,x,y)=\varphi(t)\frac{xy}{(1+Ax)(1+By)},\quad f\equiv 0,\quad g\equiv 0,\quad\phi(t)=\frac{b_{2}}{b_{1}}.

One clearly has

\mathcal{F}_{i}(t,x,y)=b_{i}\varphi(t)\frac{xy}{(1+Ax)(1+By)},\quad i=1,2,

whence

\displaystyle\underline{h}_{i}(t)=b_{i}\varphi(t)\frac{C_{1}C_{2}r_{1}r_{2}}{(1+AC_{1}r_{1})(1+BC_{2}r_{2})},\quad\overline{h}_{i}(t)=b_{i}\varphi(t)\frac{R_{1}R_{2}}{(1+AR_{1})(1+BR_{2})},\quad i=1,2.

Theorem 6 yields the subsequent existence and localization result for problem (17).

Theorem 11.

Assume that, for $i=1,2$ , one has

	$\displaystyle C_{1}C_{2}r_{1}r_{2}\,\frac{b_{i}}{(1+AC_{1}r_{1})(1+BC_{2}r_{2})}\min_{\tau\in[0,\omega]}\varphi(\tau)\max_{t\in[0,\omega]}\int_{0}^{\omega}H_{i}(t,s)\,ds$	$\displaystyle\geq r_{i},$		(18)
	$\displaystyle R_{1}R_{2}\,\frac{b_{i}}{(1+AR_{1})(1+BR_{2})}\max_{\tau\in[0,\omega]}\varphi(\tau)\max_{t\in[0,\omega]}\int_{0}^{\omega}H_{i}(t,s)\,ds$	$\displaystyle\leq R_{i}.$		(19)

Then system (17) has at least one positive and $\omega$ periodic solution $(u,v)$ such that

r_{1}\leq\|u\|\leq R_{1}\quad\text{and}\quad r_{2}\leq\|v\|\leq R_{2}.

Proof.

The conclusion is an immediate consequence of Corollary 7, since condition (H1) is automatically satisfied and (18)–(19) imply its hypotheses. ∎

Example 12.

In system (17), assume that $\mu\equiv\eta\equiv 1$ . Then

m_{1}=m_{2}=\frac{1}{e^{\omega}-1},\quad M_{1}=M_{2}=\frac{e^{\omega}}{e^{\omega}-1},\quad C_{1}=C_{2}=\frac{1}{e^{\omega}},

and

\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{1}(t,s)ds=\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{2}(t,s)ds=1.

We choose $A=B$ (to be specified later), set $b_{1}=b_{2}=1$ , and, for simplicity, look for radii $r_{1}=r_{2}$ and $R_{1}=R_{2}$ such that inequalities (18)–(19) are satisfied, that is,

\displaystyle\underline{\varphi}\left(\frac{\frac{r_{1}}{e^{\omega}}}{1+A\frac{r_{1}}{e^{\omega}}}\right)^{2}

\displaystyle\geq r_{1},\quad\text{ and }\quad\overline{\varphi}\left(\frac{R_{1}}{1+AR_{1}}\right)^{2}\leq R_{1},

(20)

where we denoted

\underline{\varphi}:=\min_{\tau\in[0,\omega]}\varphi(\tau),\quad\overline{\varphi}:=\max_{\tau\in[0,\omega]}\varphi(\tau).

Note that the second condition in (20) holds for $R_{1}$ sufficiently large, since the left-hand side has a finite limit as $R_{1}\to\infty$ . The first inequality is equivalent to

\frac{r_{1}}{e^{\omega}}\left(\underline{\varphi}\frac{1}{e^{\omega}}\,\frac{r_{1}}{e^{\omega}}\frac{1}{\left(1+A\frac{r_{1}}{e^{\omega}}\right)^{2}}-1\right)>0,

so it suffices to ensure positivity at some point of the function

\gamma(x)=\underline{\varphi}\frac{1}{e^{\omega}}\frac{x}{(1+Ax)^{2}}-1.

Clearly $\gamma$ has a unique critical point at $x_{0}=\frac{1}{A}$ , which is a maximum since

\gamma^{\prime\prime}(x_{0})=-\underline{\varphi}\frac{A}{8}\frac{1}{e^{\omega}}<0.

Therefore, the inequality

0<\gamma\!\left(\frac{1}{A}\right)=\underline{\varphi}\,\frac{1}{4Ae^{\omega}}-1

holds provided that

\underline{\varphi}\,\frac{1}{4Ae^{\omega}}-1>0,\quad\text{i.e.,}\quad A\in\left(0,\underline{\varphi}\frac{1}{4e^{\omega}}\right).

In particular, since

A<\underline{\varphi}\frac{1}{4e^{\omega}}<\overline{\varphi}\frac{1}{4},

both inequalities in (20) are satisfied if we take

r_{1}=\frac{e^{\omega}}{A}\quad\text{and}\quad R_{1}=(\geq)\frac{\overline{\varphi}-2A+\sqrt{\overline{\varphi}^{2}-4A\overline{\varphi}}}{2A^{2}}.

(21)

In the subsequent we choose values for $\omega,A$ and $\varphi$ , compute numerically a periodic solution of (17), and see on the graphic that the solutions are indeed localized by $r_{1}$ and $R_{1}$ given in (21).

Let

\varphi(t)=1+0.5\sin\!\left(\frac{2\pi t}{\omega}\right)\quad\text{ and }\quad\omega=1.2.

Then $\underline{\varphi}=0.5$ and $\overline{\varphi}=1.5$ , so by choosing $A=\frac{1}{8.01\,e^{\omega}}<\underline{\varphi}\frac{1}{4e}$ , from (21), we may take

r_{1}\approx 88.29,\quad R_{1}\approx 1006.98.

(22)

In Figure 1, we visualize inequalities (20), namely $\rho_{1}\!\left(\frac{r_{1}}{e^{\omega}}\right)>\frac{r_{1}}{e^{\omega}}$ and $\rho_{2}(R_{1})<R_{1}$ , where

\rho_{1}(x)=0.5e^{-\omega}\frac{x^{2}}{1+Ax^{2}},\quad\rho_{2}(x)=1.5\left(\frac{x}{1+Ax}\right)^{2}.

Subfigure (a) displays the graphical representation of the lines $y=x$ , $y=\rho_{1}(x)$ and $y=\rho_{2}(x)$ , whereas subfigure (b) provides a zoomed-in view for a better visualization of the two inequalities.

(a) Lines $y=x$ , $y=\rho_{1}(x)$ and $y=\rho_{2}(x)$

(b) Zoom-in of the inequalities

Figure 1: Inequalities

\rho_{1}\!\left(\frac{r_{1}}{e^{\omega}}\right)\geq\frac{r_{1}}{e^{\omega}}

and

\rho_{2}(R_{1})\leq R_{1}

In Figure 2, panel (a) shows that the solution satisfies $r_{1}\leq\|u\|,\ \|v\|\leq R_{1}$ , with $r_{1}$ and $R_{1}$ given in (22), while panel (b) provides a closer view of the solution.

(a) $r_{1}\leq\|u\|,\ \|v\|\leq R_{1}$

(b) Periodicity of the solution

Figure 2: Graph of a localized solution.

3.2 Non-autonomous Bazykin models of type (b)

Let us consider the following generalized Bazykin model of type (b):

\begin{cases}u^{\prime}=\mu(t)u-\varphi(t)\dfrac{b_{1}}{(1+Au)(1+Bv)}uv,\\ v^{\prime}=-\eta(t)v+\varphi(t)\dfrac{b_{2}}{(1+Au)(1+Bv)}uv-\delta v^{2},\end{cases}

(23)

where $A,B,b_{1},b_{2},\delta>0$ and $\mu,\eta,\varphi\in C_{\omega}(\mathbb{R},\mathbb{R}_{+})$ for some $\omega>0$ . In this case, we have

\mathcal{F}_{1}(t,x,y)=\varphi(t)\frac{b_{1}}{(1+Ax)(1+By)}xy,\quad\mathcal{F}_{2}(t,x,y)=\varphi(t)\frac{b_{2}}{(1+Ax)(1+By)}xy-\delta y^{2},

which immediately implies

		$\displaystyle\underline{h}_{1}(t)=b_{1}\varphi(t)\frac{C_{1}C_{2}r_{1}r_{2}}{(1+AC_{1}r_{1})(1+BC_{2}r_{2})},$
		$\displaystyle\overline{h}_{1}(t)=b_{1}\varphi(t)\frac{R_{1}R_{2}}{(1+AR_{1})(1+BR_{2})},$
		$\displaystyle\underline{h}_{2}(t)>\underline{k}_{2}(t):=b_{2}\varphi(t)\frac{C_{1}C_{2}r_{1}r_{2}}{(1+AC_{1}r_{1})(1+BC_{2}r_{2})}-\delta r_{2}^{2},$
		$\displaystyle\overline{h}_{2}(t)<\overline{k}_{2}(t):=b_{2}\varphi(t)\frac{R_{1}R_{2}}{(1+AR_{1})(1+BR_{2})}-\delta C_{2}^{2}r_{2}^{2}.$

Based on Theorem 6, the following existence and localization result holds for problem (23).

Theorem 13.

Assume that

\min_{t\in[0,\omega]}\varphi(t)\frac{b_{2}}{(1+AC_{1}r_{1})(1+AC_{2}r_{2})}C_{1}C_{2}r_{1}r_{2}-\delta R_{2}^{2}\geq 0,

(24)

and

		$\displaystyle b_{1}\frac{C_{1}C_{2}r_{1}r_{2}}{(1+AC_{1}r_{1})(1+BC_{2}r_{2})}\min_{\tau\in[0,\omega]}\varphi(\tau)\max_{t\in[0,\omega]}\int_{0}^{\omega}H_{1}(t,s)\,ds\geq r_{1},$		(25)
		$\displaystyle b_{1}\frac{R_{1}R_{2}}{(1+AR_{1})(1+BR_{2})}\max_{\tau\in[0,\omega]}\varphi(\tau)\max_{t\in[0,\omega]}\int_{0}^{\omega}H_{1}(t,s)\,ds\leq R_{1},$
		$\displaystyle\left(b_{2}\frac{C_{1}C_{2}r_{1}r_{2}}{(1+AC_{1}r_{1})(1+BC_{2}r_{2})}\min_{\tau\in[0,\omega]}\varphi(\tau)-\delta r_{2}^{2}\right)\max_{t\in[0,\omega]}\int_{0}^{\omega}H_{2}(t,s)\,ds\geq r_{2},$
		$\displaystyle\left(b_{2}\frac{R_{1}R_{2}}{(1+AR_{1})(1+BR_{2})}\max_{\tau\in[0,\omega]}\varphi(\tau)-\delta C_{2}^{2}r_{2}^{2}\right)\max_{t\in[0,\omega]}\int_{0}^{\omega}H_{2}(t,s)\,ds\leq R_{2}.$

Then the system (23) has at least one positive periodic solution $(u,v)\in K_{1}\times K_{2}$ such that $r_{1}\leq\|u\|\leq R_{1}$ and $r_{2}\leq\|v\|\leq R_{2}$ .

Proof.

From condition (24) we see that (H1) holds (note that $\varrho\equiv 1$ ), while conditions (25) ensures that (H2)’ is satisfied. Thus, Theorem 6 applies and yields the conclusion. ∎

Example 14.

In problem (23), let us take

\mu\equiv\eta=b_{1}=b_{2}=1,\quad\text{ and }\quad A=B,

which implies

\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{1}(t,s)ds=\max_{t\in[0,\omega]}\int_{t}^{t+\omega}H_{2}(t,s)ds=1,\quad C_{1}=C_{2}=\frac{1}{e^{\omega}}.

For simplicity, we look for radii $r_{i},R_{i}$ such that $r_{1}=r_{2}$ and $R_{1}=R_{2}$ . Then, denoting

\underline{\varphi}:=\min_{\tau\in[0,\omega]}\varphi(\tau),\quad\overline{\varphi}:=\max_{\tau\in[0,\omega]}\varphi(\tau),

conditions (24) and (25) become

\delta R_{1}^{2}\leq\underline{\varphi}\left(\frac{r_{1}}{e^{\omega}}\right)^{2}\frac{1}{(1+A\frac{r_{1}}{e^{\omega}})^{2}},

(26)

and

		$\displaystyle\underline{\varphi}\left(\frac{r_{1}}{e^{\omega}}\right)^{2}\frac{1}{(1+A\frac{r_{1}}{e^{\omega}})^{2}}\geq r_{1},$		(27)
		$\displaystyle\overline{\varphi}R_{1}^{2}\frac{1}{(1+AR_{1})^{2}}\leq R_{1},$		(28)
		$\displaystyle\underline{\varphi}\left(\frac{r_{1}}{e^{\omega}}\right)^{2}\frac{1}{(1+A\frac{r_{1}}{e^{\omega}})^{2}}-\delta r_{1}^{2}\geq r_{1},$		(29)
		$\displaystyle\overline{\varphi}R_{1}^{2}\frac{1}{(1+AR_{1})^{2}}-\delta\left(\frac{r_{1}}{e^{\omega}}\right)^{2}\leq R_{1}.$		(30)

Let us take

\varphi(t)=1+0.1\sin\!\left(\frac{2\pi t}{\omega}\right),\quad\omega=1.2,\quad\delta=0.0001,\quad\text{and}\quad A=0.005.

Then, we easily see that for $r_{1}\approx 70.4$ and $R_{1}\approx 400$ , all conditions (26)–(30) are satisfied. In Figure 3 below, we have a visual representation of the inequalities (27), (28) and (29), that is,

\rho_{1}\left(\frac{r_{1}}{e^{\omega}}\right)>\frac{r_{1}}{e^{\omega}},\quad\rho_{2}\left(R_{1}\right)<R_{1}\quad\text{ and }\quad\rho_{3}\left(\frac{r_{1}}{e^{\omega}}\right)>\frac{r_{1}}{e^{\omega}},

where

\displaystyle\rho_{1}(x)=e^{-\omega}\underline{\varphi}\frac{x^{2}}{(1+Ax)^{2}},\quad\rho_{2}(x)=\overline{\varphi}\frac{x^{2}}{(1+Ax)^{2}}\quad\text{ and }\quad\rho_{3}(x)=e^{-\omega}\underline{\varphi}\frac{x^{2}}{(1+Ax)^{2}}-\delta e^{\omega}x^{2}.

(a) Lines $y=x$ and $y=\rho_{i}(x)$ , $i=1,2,3$

(b) Zoom-in of the inequalities

Figure 3: Inequalities

\rho_{1}\left(\frac{r_{1}}{e^{\omega}}\right)>\frac{r_{1}}{e^{\omega}},\,\rho_{2}\left(R_{1}\right)<R_{1}

, and

\rho_{3}\left(\frac{r_{1}}{e^{\omega}}\right)>\frac{r_{1}}{e^{\omega}}

Also, Figure 4 shows that the periodic solution of problem (23) satisfies the localization stated in Theorem 23.

(a) $r_{1}\leq\|u\|,\ \|v\|\leq R_{1}$

(b) Periodicity of the solution

Figure 4: Graph of a localized solution.

Acknowledgments

The authors would like to thank Prof. Radu Precup for his insightful discussions on the subject and for his careful verification of the manuscript. The authors also thank the anonymous referees for their careful reading of the paper, for pointing out several mistakes in the notation of the indices, and for suggesting weaker assumptions that improved the results.

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Componentwise localization of positive periodic solutions for a general Bazykin-type system

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Componentwise localization of positive periodic solutions for a general Bazykin-type system

Abstract

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1 Introduction and Preliminaries

Theorem 1.

2 Main results

Lemma 2.

Proof.

Lemma 3.

Proof.

Theorem 4.

Proof.

Remark 5.

Theorem 6.

Proof.

Corollary 7.

Proof.

Corollary 8.

Theorem 9.

Theorem 10.

Proof.

3 Examples and numerical simulations

3.1 Non-autonomous Bazykin models of type (a)

Theorem 11.

Proof.

Example 12.

3.2 Non-autonomous Bazykin models of type (b)

Theorem 13.

Proof.

Example 14.

Acknowledgments

References

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