Abstract
We establish the existence of traveling wave solutions and small amplitude traveling wave train solutions for a reaction-diffusion system based on a predator-prey model incorporating a prey refuge. By using the shooting argument, invariant manifold theory, and the Hopf bifurcation theorem, we analyze the dynamic behavior of this model in the three-dimensional phase space. Numerical results are also presented to illustrate the theoretical results.
1. Introduction
In mathematical biology, one interesting and dominant theme is the dynamic relationship between predators and their prey [1–3]. Predator-prey models have been studied mathematically since the pioneering work of Lotka and Volterra. In recent years, Leslie-Gower model [4, 5], an important predator-prey model, has been extensively modified and studied by many authors [6–11]. A modified Leslie-Gower predator-prey model is read as where function values and represent prey and predator population densities, respectively, at any time . The model parameters , , , , , and are positive constants. describes the growth rate of prey . measures the strength of competition among individuals of species . measures the extent to which environment provides protection to prey . is the growth rate of predators . is the maximum value of per capita reduction of due to . has a similar meaning to .
As the authors of [6] said, we live in a spatial world, and spatial component of ecological interaction has been identified as an important factor in how ecological communities are shaped. Mite predator-prey interactions often exhibit spatial refugia, which means the prey received some degree of protection from predation and reduces the chance of extinction due to predation [6, 9–15]. A great deal of researches on the effects of prey refuges on the population dynamic has been studied. Kar [12] indicated that the increasing refuge can increase prey densities and lead to population outbreaks. Chen et al. [9] showed that the prey refuge could greatly influence the densities of both prey and predator species, while it has no influence on the species’ persistence property. In [13–15] it was obtained that the refuges protecting a constant number of prey have a stronger stabilizing effect on population dynamic than the refuges protecting a constant proportion of prey.
On the other hand, the existence of traveling solutions has been wildly studied by many researchers [16–24]. A traveling wave solution is a spatial translation invariant solution of differential equations with spatial-diffusion. Dunbar [16] proved the existence of traveling wave solutions of diffusive Lotka-Volterra and used the methods of a shooting argument and a Lyapunov function. Zhang [19] showed the existence of traveling wave solutions in a modified vector-disease model by using the geometric singular perturbation theory. Hou and Leung [20] used the method of upper-lower solutions to prove the existence of traveling solutions of a competitive reaction-diffusive system. Ahmad et al. [21, 22] used only functional analysis, without constructing a Lyapunov function, to prove the existence of such solutions for a class of reaction-diffusion equations. Huang et al. [23] and Li and Wu [24] used Dunbar’ method to study the existence of traveling solutions of diffusive predator-prey models with Holling type-II and Holling type-III, respectively.
In this paper, based on the above discussion, we are interested in the existence of traveling wave solutions of a reaction-diffusion Leslie-Gower-type model incorporating a prey refuge, which is modified from model (1). Taking , and dropping the stars on , we will extend model (1) by incorporating a prey refuge into the following system: where is the usual Laplacian operator in two-dimensional space. and are the diffusion coefficients of prey and predator, respectively. is constant. is a refuge protecting of the prey, which means of prey available to the predator. To ensure system (2) has a positive equilibrium point, we require that . Obviously, system (2) has four equilibrium points: where The equilibrium point corresponds to absence of both species, corresponds to the prey at the environment carrying capacity in the absence of the predator, means the extinct of prey, and corresponds to coexistence of the two species. From [6], we know and are two saddle points and is globally asymptotical stable when , which indicates that system (2) may have traveling waves.
For mathematical simplicity, we assume that (considered as the is sufficient small which indicates the prey disperse very slowly relative to the mobile herbivore predator [16]). Then system (2) can be converted to the system: We will establish the existence of traveling wave solutions and small amplitude traveling wave train solutions of this system. The method used here is a shooting argument in together with a Lyapunov function, LaSalle’s invariant principle, and Hopf bifurcation theorem.
Remark that although the methods we use to prove the existence are similar to these in [16, 23, 24], there are several differences. For one thing, it is a different model, a modified Leslie-Gower model incorporating a prey refuge. For the other thing, we construct a different Wazewski set and a new Lyapunov function [25–27].
The rest of the paper is organized as follows. In Section 2, main results on the existence of traveling wave solutions and small amplitude wave train solutions are stated. In Section 3, we give the proofs of the main results. In Section 4, some numerical results are presented.
2. Main Results
A traveling wave solution is a spatial translation invariant solution. In order to establish the existence of traveling wave solutions of system (5), we assume the system has a solution of the special form , , where parameter is the wave speed. Substituting , , into (5), the corresponding wave equations become Here denotes the differentiation with respect to the traveling wave variable . Due to ecological motivation, we require that the traveling wave solutions and are nonnegative and satisfying the following boundary conditions: Rewrite the system (6) as a system of first order equation in :
Lemma 1. Let , and then and has two real roots when (i.e., ). Furthermore, the following results hold:(a)if , then ;(b)if , then .
Now we state the main results as follows.
Theorem 2. (i) If , then there are no nonnegative solutions of system (8) satisfying the boundary conditions (7).
(ii) If , , then there are nonnegative solutions of system (8) satisfying the boundary conditions (7), which correspond to traveling wave solutions of system (5).
Theorem 3. Let , where .(a)If , then , spreads to the positive equilibrium point , nonmonotonously for traveling wave variable .(b)If , then , spreads to the positive equilibrium point , monotonously for traveling wave variable .
Theorem 4. Let and . If then, as the parameter crosses the bifurcation curve at in the -parameter plane, system (8) undergoes a Hopf bifurcation to a small amplitude periodic solution at the equilibrium point , which corresponds to a small amplitude traveling wave train solution of system (5).
3. Proofs of the Main Results
3.1. Proof of Theorem 2
In this section, we subdivide the proof into several Sections 3.1.1–3.1.4 for convenience. In Section 3.1.1, we recall some notations used throughout this section and state the well-known Wazewski Theorem. Section 3.1.2 contains a Wazewski set and the exit set . In Section 3.1.3, the behavior of trajectories on the strongly unstable manifold at is presented by some technical lemmas. In Section 3.1.4, we finish the proof of existence of traveling wave solutions by constructing a Lyapunov function.
3.1.1. Recall the Wazewski Theorem [16, 17]
Consider the differential equation: where is a continuous function and satisfying a Lipschitz condition. Let be the unique solution of (10) satisfying . For convenience, set . Let be the set of points , where and .
Given , define is called the immediate exit set of . Given , let For , define is called an exit time. Note that and if and only if . The notation cl() denotes the closure of .
Lemma 5. Suppose that(i)if and , then ;(ii)if and , then there is an open set about disjoint from ;(iii), is a compact set and intersects a trajectory of (10) only once. Then the mapping is a homeomorphism from to its image on .
A set satisfying the conditions and is called a Wazewski set.
3.1.2. Construct and
Evaluating the Jacobin of system (8) at the equilibrium gives
The corresponding eigenvalues of (14) are If , then and are a pair of complex conjugate eigenvalues with positive real part. By Theorems 6.1 and 6.2 in [25], there exists a 2-dimensional unstable manifold based at , the point is a spiral point on this unstable manifold, and the trajectory approaching as must have for some . This violates the requirement that the traveling wave solution must be nonnegative. So the first part of Theorem 2 is proved.
We only need to account for the case in the following. It is obvious that , the eigenvectors , , associated with , , , respectively, are where . Applying Theorems 6.1 and 6.2 in [25], we get a one-dimension strongest unstable manifold tangent to at and a two-dimension strongly unstable manifold tangent to the span of , at point . In a small neighborhood of , points on are parametrically represented by a function (): and points on also could be represented by a function (): Obviously, .
The motivation and method of constructing the Wazewski set are similar to that in Dunbar [17]: it will be the complement of two blocks of and the two blocks are chosen so that has the same sign as . Thus, solutions would not have as when entering these blocks. In this paper, the Wazewski set is defined as follows: where Note that and is a closed set. We obtain where Obviously, is not a connected set. Actually, one component of is and the other is .
As the details of proving that is the set described above are tedious, we just prove the portion of to show why the set must be excluded from to .(1), , . Then we have Then the trajectory enters .(2), , . Then Thus, is decreasingly entering .(3), , . Then (i), and thus and the trajectory enters the .(ii), then . This implies , , . The trajectory does not enter and .(iii), and then , ; furthermore, . This implies the trajectory does not enter the inner of .(4), , . This is a singular point not in the immediate exit set.(5), , . Then , and which implies and both decrease. The trajectory enters .(6), , . Then Hence, the trajectory enters .(7), , . Then (i) , and then . This implies , , . The trajectory does not enter and .(ii), and then , . This implies the trajectory does not enter the inner of .(iii), and then Hence, it implies , , , which ensures the trajectory enters the .
Based on the above analysis, and must be excluded from to .
3.1.3. Construct the Set
We need to construct the set before using Lemma 5. By a series of lemmas (Lemmas 5–9), we obtain set will be an arc of a sufficient small circle surrounding on the unstable manifold . Furthermore, one endpoint of the arc is the intersection of the circle with the strongly unstable manifold , and the other endpoint is the intersection of the circle with the plane defined by . Lemmas also show that the first endpoint is carried by the strongly unstable manifold into while the other is carried into .
We take a notation .
Lemma 6. Let . Any solutions of (8) having a point such that , , and will have and for all . This is particularly true for trajectories on the branch of strongly unstable manifold in the octant .
Proof. Take without loss of generality. Suppose, on the contrary, that there exists an such that . Let
For , and , so . Also and for . Thus (i.e., . Then, from (8), we have
Then
Since , we have .
It must be the case that . The plane defined by is an invariant manifold, so is obvious. We just verify that . If this is not true, then there exists such that and . But then
so for . So , which is a contradiction. Thus for all . Then also for all .
A trajectory on the branch of the strongly unstable manifold in the octant approaches tangent to . From subset , the second and third components of this tangent vector satisfy . Thus there exists a point on the trajectory whose components satisfy , , and . This completes the proof.
Lemma 7. Assume that ; then a trajectory on the portion of the strongly unstable manifold in the octant must satisfy for all .
Proof. A trajectory on the portion of the strongly unstable manifold in the octant could be written as , where
Lemma 8. Let be a fixed number. A solutions of (8) having a point such that , , and will have for all such that . In particular, this is true for trajectories on branch of the strongly unstable manifold in the octant .
The proof is similar to that of Lemma 6, so it is omitted.
Lemma 9. If a solution of (8) has a point, taking to without loss of generality, such that , , and , then for all , as long as , the trajectory must have that . In particular, this is true for a trajectory on the branch of the strongly unstable manifold in the octant .
Proof. We first prove that for all such that . If this is not true, then there exists a first time such that , and . But then,
This is a contradiction. Thus for all such that .
Now we show that for all as long as and . Let . Suppose on the contrary that there exists a first time such that , , but . Then . By Lemma 8, for all such that . Then
For and , we have
which is a contradiction. This completes the proof.
Now combine all the results of Lemmas 6–9 to follow the trajectory of a solution of (8) on the strongly unstable manifold . Let Then the trajectory of a solution of (8) on the strongly unstable manifold is contained in . Since , we obtain This shows the region lies in the region defined by and . Then, on the strongly unstable manifold , . So, for a solution of (8) on , decreases until for some finite . And at the time , we have Thus the trajectory of this solution hits on the face , , and . Therefore, the vector field shows that the solution of (8) on enters the region at some finite time.
Lemma 10. In a sufficient small neighborhood of the two-dimensional unstable manifold intersects the plane defined by in a smooth curve , given by , , where
Proof. The proof is similar to Lemma 5 in [16] and is omitted.
Remark 11. The portion of the curve is in the region . Obviously, the component of points along the curve satisfies from Lemma 10. From the direction of the vector filed on the quarter plane, , , and , any trajectory passing through a point of near will immediately enter the region .
Now, we place a sufficiently small circle about on the two-dimensional unstable manifold . The circle is contained in the neighborhood of given in Lemma 10 and satisfies the conditions of Lemmas 6–9. Then the circle intersects the curve . Define to be arc of this circle contained in the octant whose endpoints are the intersections of the circle with and the curve .
3.1.4. Proof of (ii) of Theorem 2
In this section, we firstly use Lemma 5 to produce a trajectory which remains in the region W. Second, we construct a Lyapunov function to demonstrate that the trajectory approaches . For simplicity, we denote , .
Lemma 12. There exists a point such that the solution of (8) remains in the region for all .
Proof. It is obvious that the set is closed satisfying the of Lemma 5. Before using Lemma 5 to prove this conclusion, we also need to check the conditions and of it. Suppose . Then . As , we easily verify that
Moreover, as is an invariant manifold,
Thus and there exists an open set around disjoint from . So of Lemma 5 is satisfied.
From the previous 5 lemmas, we know that the image of one endpoint of lies in the portion of ; and the image of the other endpoint is in the component of . Thus is compact, intersects any trajectory of (8) only once, and is simple connected. If , then would be a homeomorphism of the connected set to its image in the disconnected set . This is impossible. So . Thus there exists some point such that for all .
Lemma 13. The solution remains in the region for all , where
Proof. Firstly, must have for all , as is an invariant manifold.
Secondly, we prove . If it is not true, then enters region . Let . Then and , so . As is an invariant manifold, . And for . From (8), , which means is increasing for . Then the solution enters
Obviously, in , , so is decreasing. Thus, we have
So increases to in the finite time ; that is, . Then also , . So enter . This is a contradiction. Therefore, for all time.
By Lemma 9, we know
As , so for all .
Suppose, on the contrary, there exists such that for , where . Take
Then , , and . Then either or immediately enter , which is impossible. So for .
At last, we prove . is obvious. Because a trajectory starting on approaches tangent to or has or . Since , from Lemma 8, we know for all s. We only need to prove . Suppose, on the contrary, that there exists a such that ; then for all . If this is not true, there exists a such that , and thus . Then from (8) we have
Then after some calculation, we obtain
this is a contradiction. So if , then for all . Thus,
for all . So for all . Thus and bounded away from zero by . Therefore for some finite , which is a contradiction. So .
This completes the proof.
Lemma 14. The trajectory as .
Proof. Define following Lyapunov function:
Then is continuous and bounded below on , and
where is defined. Obviously, in and . According to Lemma 1, when , the following result always holds:
Therefore, the is always nonpositive in . Moreover, if and only if , , and ; the largest invariant subset of this segment is the single point . By LaSalle’s Invariance Principle, as . This completes the proof.
3.2. Proof of Theorem 3
The Jacobin of system (8) at the equilibrium is
Let ; then the corresponding characteristic equation of (57) is given by In order to get the sign of the roots of characteristic equation (58), we will use Routh-Hurwitz analysis [25]. The Routh-Hurwitz range of (58) is where In the above range, we easily know that , . When , (i) if , then no matter the sigh of , the sigh of the first arrange of (59) will change once, and the no row of (59) is full zero. So character equation (59) always has a real root and two complex roots with negative real part; if , we obtain with , and then . Thus, the sigh of the first arrange of (59) will change once and the no row of (59) is full zero. So character equation (58) has a real root and two complex roots with negative real part.
Therefore, there is a 2-dimensional stable manifold and 1-dimensional unstable manifold based at when .
The differentiation of (58) is Let ; then we obtain Thus, get the maximum at , get the minimum at , and . So we just consider If , (58) has two negative roots and a positive root. If , (58) has a negative root and a positive root. If , (58) has a positive root and two complex roots with negative real part. So the solution of (8) satisfying (7) spreads to the positive equilibrium monotonously when , and it spreads to the positive equilibrium nonmonotonously when .
3.3. Proof of Theorem 4
In order to prove Theorem 4, we take , , , , and as fixed and and as parameters. It means we only allow the predator effectiveness to vary. We search for purely imaginary roots of the characteristic equation where , , , and .
It is easy to see that , and . Substituting into (64) and simplifying it, we have Thus, a pair of imaginary eigenvalues exists if the parameters and satisfy the condition
Regarding as a function of and differentiating the characteristic equation (64) with respect to , we obtain Here denotes the differentiation with respect to . Substituting into (66), we have After some calculation, we have that the sign of the real part of is determined by the sign of From (64), we know . Thus, it is obvious that So the sign of is opposite to that of . In fact, , while Define function , and then if , where denotes the differentiation with respect to . So is increasing with respect to . Thus, So if we have Then we have Therefore, By the Hopf bifurcation Theorem, we obtain that when the parameter crosses the bifurcation curve at in the parameter plane, system (8) undergoes a Hopf bifurcation to a small amplitude periodic solution at the equilibrium point . It corresponds to a small amplitude traveling wave train solution of system (5). This completes the proof.
4. Numerical Simulations
In this section, we will give numerical examples to illustrate the results of Theorems 2 and 3. All the numerical simulations are under the Neumann boundary conditions.
Figure 1 shows that there exists traveling wave solution and it from tends to monotonously. In Figure 1, we consider the following parameters , , , , , , and . Figure 2 shows that there exists traveling wave solution and it from tends to nonmonotonously with the parameters , , , , , , and .
(a) The traveling wave of prey
(b) The traveling wave of predator
(c) The traveling wave of predator at different time
(d) The traveling wave of predator at different time
(a) The traveling wave of prey
(b) The traveling wave of predator
(c) The traveling wave of predator at different time
(d) The traveling wave of predator with different time
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.