Abstract

Little seems to be known about the ergodicity of random dynamical systems with multiplicative nonlinear noise. This paper is devoted to discern asymptotic behavior dynamics through the stochastic coral reefs model with multiplicative nonlinear noise. By support theorem and Hörmander theorem, the Markov semigroup corresponding to the solutions is to prove the Foguel alternative. Based on boundary distributions theory, the required conservative operators related to the solutions are further established to ensure the existence a stationary distribution. Meanwhile, the density of the distribution of the solutions either converges to a stationary density or weakly converges to some probability measure.

1. Introduction

The coral reefs equation is one of the most famous ecosystem models [1]:where represents the cover of macroalgae; represents the cover of corals.(i)is the rate that corals recruit to and overgrow algal turfs;(ii) is the natural mortality rate of corals;(iii) is the rate that corals are overgrown by macroalgae;(iv) is the rate that macroalgae spread vegetatively over algal turfs;(v) is the grazing rate that parrotfish graze macroalgae without distinction from algal turfs.

By the results in Li et al. [2], they discuss all kinds of dynamical behaviors. Recently, system (1) was studied extensively that it exhibits complex dynamical phenomena, including chaos, bifurcation, stability, and attractiveness [110].

However, ecosystem in the real world is very often subject to environmental noise due to uncertainty and unknown factors [1116]. From a biological point of view and the generality of the models considered [1721], these systems can appear very formal. This paper studies a stochastic coral reefs model where the intrinsic growth rate of the cover of macroalgae, , and the one of the cover of corals, , are perturbed stochastically and . In the paper, we only consider that stochastic coral reefs model can be described bywhere is a two-sided canonical Brownian motion. and represent the intensity of random noise and the differential and are to be understood in the sense of multiplicative nonlinear noise. Since the drift and diffusion coefficients of (2) satisfy locally Lipschitz continuous condition, we can apply standard theorems that provide both existence and uniqueness of the positive solution of (2) (see [20]), for any given initial value. However, since the diffusion term of (2) is not linear but nonlinear, the existing powerful classical results [1121] fail to work here. Nevertheless, this paper discusses the asymptotic behavior of the stochastic coral reefs model with multiplicative nonlinear noise using Fokker-Planck equations. To overcome the difficulty from the diffusion term, based on boundary distributions theory and conservative operators, we show that the density of the distribution of the solutions either converges to a stationary density or weakly converges to some probability measure.

This paper is organized as follows. In Section 2, we study the global attractiveness of the solution for stochastic coral reefs model with multiplicative nonlinear noise. In Section 3, we discuss the ergodicity of the solution for stochastic coral reefs model with multiplicative nonlinear noise.

2. Global Attractiveness

In the section, we study the global attractiveness of the solution for stochastic coral reefs model with multiplicative nonlinear noise.

Proposition 1. Suppose that , hold.(I)If the cover of macroalgae is absent, then the cover of corals dies with probability one.(II)If the cover of corals is absent, the quantity of the preys oscillates between and , and there exists a unique stationary distribution with the density where is a constant.

Proof. Denoting , , we replace system (2) byLetThen, system (4) becomesor Stratonovich stochastic differential equationLet denote the generator of diffusion (4); that is,Then Fokker-Planck equation (FPE) of (4) can be described by(I) If the cover of macroalgae is absent, the quantity of the cover of corals satisfiesFix It is easy to see that and . Then, we can obtainIt implies that without the cover of macroalgae, the cover of corals dies with probability one.
(II) If the cover of corals is absent, the quantity of the cover of macroalgae satisfiesLet It easily shows that and . We haveIt implies that, without the predators, the quantity of the preys oscillates between and . Furthermore, there exists a stationary distribution of system (13) with the density satisfying the FPESolving (16), then we get where and are real numbers. With the conditions andit means thatIt easily shows that where is standard normal distribution function. Thus, we getBy ergodic theorem, if is a solution of system (13), then we haveFurthermore, converges in probability to when . DefineIt is easy to see that

Theorem 2. Suppose that and hold. Then there exists a constant such thatholds with probability 1, where and are the solution of (2) with the initial condition .

Proof. Define the Lyapunov function on Applying Itô’s formula to function (26), we getThat is,whereis a local martingale with quadratic form:Fix . For any , by martingale inequality, we haveBy using Borel-Cantelli theorem, we can choose a set with and for any there is a such that It implies thatfor and . Substituting (33) into (28), we get for and for almost and . Moreover, there exists a constant satisfyingBy inequality (35), we get it means thatMoreover, there exists a positive constant satisfyingTherefore, we getThen, we havefor any , , and
If with , then we getIt means thatorThe proof is completed.

Theorem 3. Suppose that and hold. Then, with probability 1, we have

Proof. Define -function:Applying Itô’s formula to function (44), we haveThat is,whereis a local martingale with quadratic form:By using Borel-Cantelli theorem and martingale inequality, for , , and , for almost , there is a such that , and we haveBy (47) and (50), we getMoreover, there exists a constant satisfyingBy inequality (52), it easily shows thatMoreover, there exists a positive constant satisfyingCombining (51), (52), and (53), we getfor any .
It is easy to see that there exists a real number independent of satisfyingFrom (55) and (56), we getfor any Then, we haveIf and , we obtainLetting , we getBy (60), for every , , and , then by letting , , and , we obtainThe proof is completed.

Theorem 4. Suppose that , hold. Let denote the solution of the following the equation:with the initial value . Let denote the solution of the following the equation:with the initial value .
Then, with probability 1, there exist and .

Proof. Let , . By Itô’s formula, we have By using comparison theorem, we have for all a.s. It implies that for all . It is easy to see that we show the second assertion by a similar way for all . The proof is completed.

Theorem 5. Suppose that , hold. Then the following assertions are true:

Proof. From Theorem 3, we getTherefore,For any , by martingale inequality, we have By using Borel-Cantelli theorem, for almost all , there is a real constant satisfying, for all and ,It means that for Then, we getCombining (67) and (71), it yieldsMoreover, by Theorem 4, we getTherefore,Letting , we have The first assertion is proven. Using similar way, the second assertion is also proven. The proof is completed.

3. Ergodicity

In the section, we discuss the ergodicity of the solution for stochastic coral reefs model with multiplicative nonlinear noise.

Theorem 6. Suppose that , hold.(I)Then, the transition probability function of system (4), that is, for an , results in a density for all (II)Then, system (4) results in integral Markov semigroup.

Proof. Let and denote vector fields on ; then the Lie bracket denotes a vector field: Denote Then it is easy to see that where Assume that there is a point and so the vectors , , and can not span the space . Then, the vectors and are parallel; the vectors and are also parallel. Therefore, we getIt is easy to check that equality (80) is impossible.
It implies that the vectors , , span at any point . Therefore, we obtain the Hörmander condition.()For every , the vectorsspan the space .By Hypothesis () and Hörmander theorem [2229], then the transition probability function results in a density and . Thus, the first assertion has been proved.
From the first assertion, it easily shows that for any , of system (4) results in the density satisfying the FPE (9). Furthermore, we getDefining the operator isfor any , By using continuation theorem of operator and assertion (I), it easily shows that the operator is an integral Markov semigroup. The proof is completed.

Theorem 7. Suppose that , hold. Then there is no more than three solution curves such that satisfying rank if .

Proof. Let and . We consider the following system:Denote System (84) becomes We show that denote the solution of system (84) with the initial value , and defined as . By using the perturbation method, we get the Frechet derivative of . Let and . denote the fundamental matrix of the following differential equation:That is, and for . Then, we havewhere . Let and if and if By using Taylor formula, we get as . Then, we have Let , , and . By using mean value theorem of integration, we get It is easy to calculate that we show Therefore, we obtain If the two vectors and are not linearly dependent, then we get the rank . Since the between and is linear dependence, thus, it is easy to see that and denote the solution of the following differential equation: That is,It easily knows that there is no more than three solution curves satisfying (94), and the graph of a function represents each solution curve. The proof is completed.

Theorem 8. Suppose that , , hold. Hypothesis condition : there is a point satisfying for all and . The following assertions are true.(I)If Hypothesis condition does not hold, then system (84) is controllable in .(II)If Hypothesis condition holds, then system (84) is controllable in (), where

Proof. LetThen, system (84) can be replaced by the following differential equations:where DenoteBy (96), it is easy to see that for any . There is a point satisfying for all and . It is easily know that there exists ; letStep  1. Fix . Due to uniformly in , it shows that there is satisfying for any . We takewhere denotes the solution ofIt implies that system (97) results in the solution and . Due to whenever , we can choose a satisfying .
Step  2. Fix . Due to uniformly in , we can choose satisfying for any . By using similar way to Step 1, it easily shows that there is a function and such that (97) results in a solution , , and .
Step  3. Let andSince , it easily knows that there is real number and satisfying for any and . In the case, we can choose a control function satisfying , , and for some .
Step  4. If Hypothesis condition holds. It is easy to see that . Based on the definition of , for any , there are and satisfying the property: if such that , one can choose a such that for and . Hence, we can choose a control function satisfying , , and for some .
Step  5. If Hypothesis condition does not hold. Then there are and satisfying the property: if such that and , one can choose a such that for and . By the same way as before, we can choose a control function satisfying , , and for some .
Step  6. Let , , , and and Denoteand . For every , it easily knows that system (97) with , results in the solution such that, for all ,From (97), we getIt is easy to calculate that the solution of the equation isBy using comparison theorem, we have, for any ,Thus,Therefore, (I) term of (106) can be proven if we find a constant satisfyingIt is obvious that (111) is equivalent toand (112) is equivalent toObviously, for is small enough, we getIn other words, we can find such that (I) term of (106) is true. For the second assertion, we get, for all ,From (106), it easily shows that for there is a and a satisfying and . By using similar proofs, . Therefore, for any , we obtain that there is a and a satisfying and . The proof is completed.

Theorem 9. Suppose that , , hold. The following assertions are true.(I)If Hypothesis condition does not hold, then there is a constant satisfying for each and for almost every .(II)If Hypothesis condition holds, then there is a constant such that for any point and almost every such that and in   (), where , is a measurable function.

Proof. By using continuity theory, we can find a continuity function . If there exists such that the derivative is the rank 2, then we have . Therefore, the proof is completed.

Theorem 10. Suppose that , , hold. If Hypothesis condition holds. Then, , where is defined in (100).

Proof. System (7) becomesTo prove s.a, we assume that for all with . Due to for all , we get for and . Generally, with , the comparison theorem fails to work here since the diffusion term is nonlinear. By , we havewhere denote the solution ofwith . Since on for all , it is easy to see that . Hence, we get for , where is the solution ofDenoteIt is obvious that , . Thus, we get a.s. Hence, we have on . Moreover, by the definition of , there exists a satisfying for all , . Therefore, for any , there is a such that for , it means that . By using the second equation of (117), we get , which contradicts our assumption that on .
Next, we prove that a.s. It easily knows thatwhere is a polynomial of order 3 of the variable . Based on the definition of , there is no more than one point satisfying . Then, we have for all that for every , for all or . By continuity theorem, we can choose an and a “rectangle” such thatwith satisfying for all . By using the Markov property, we get a.s. The proof is complete.

Theorem 11. Suppose that , , hold. If Hypothesis condition holds, then is an invariant set; namely, if , then for all .

Proof. System (117) results in a solution satisfying and with some , . Based on the continuity of the path , it is easy to show that there is satisfyingThe Property (P). For any there are two constants and satisfying for all , whereThe property (P) has been proven in Theorem 10. Since the equation , (125) and the property (P) that for any , we get . From (125), it easily shows that there is a decreasing sequence satisfying and for any It is obvious that the result contradicts the property (P). Therefore, the proof is complete.

Theorem 12. Suppose that , , , hold, where is defined by (23). Then ; the distribution of the Markov process weakly converges to the probability measure with the density when , where the Markov process denotes a solution of (4) with .

Proof. By Theorem 4, it is easy to see that ; then we getTherefore,Based on the proof of Theorem 5, we haveMoreover, by ergodic theorem, we getFrom (23) and (128)–(130), we haveThenThus, for sufficiently small , it is easy to see that there exist and a set satisfying Prob and for and . From the inequalitiesit easily shows that the distribution of the Markov process weakly converges to the probability measure which possesses the density . The proof is complete.

Theorem 13. Suppose that , , , hold; then there is a stationary distribution in system (4).

Proof. By using the former random variables and . It is easy to see that is a Markov process on and , whereMoreover, Therefore, by using the second assertion of Theorem 5, we getFurthermore, by Theorem 5, we haveFrom (136) and (137), we getBy Theorem 5 again, we haveMoreover, we get that the inequalityThen there are two constants and such thatFrom the inequalityit is easy to see thatBy Holder’s inequality, we getFor , there is a constant satisfyingThus, we getThat is,Furthermore, there is a constant such thatIt means thatwhere is the semigroup related to the random variable andFrom (149) and Theorem  1 in [29, pp 36], there is a stationary distribution in for the random variable satisfying . Because the boundary is invariant under and if , then we get that . Thus, there is a stationary distribution on ; that is, there is a stationary distribution related to the random variable . The proof is complete.

Theorem 14. Suppose that , , hold. Then the distribution of the random variable exists a density satisfying (8). Furthermore, the following assertions are true(I)Assume ; if Hypothesis condition does not hold, then the Markov semigroup is asymptotically stable on ; that is, there is a stationary density of (8) satisfying (II)Assume ; if Hypothesis condition holds, then is a transient set and is an invariant set. Furthermore, the integral Markov semigroup is asymptotically stable on . It implies that support and (III)Assume ; then a.s. and the distribution of random variable weakly converges to the probability measure with the density when .

Proof. From Theorem 6, then the distribution of the random variable results in a density satisfying (8). By Theorem 9, for every , we getBy Corollary  1 in [23, pp 248], it is easy to see that the integral Markov semigroup is asymptotically stable or sweeping. Based on Theorems 10, 11, and 13, it is easy to see that (I) assertion and (II) assertion are directly proven. By Theorem 12, (III) assertion is directly proven. The proof is complete.

Competing Interests

The author declares that he has no competing interests.

Acknowledgments

This research was supported by the National Natural Science Foundation of China (no. 11301090) and Guangxi Science Research and Technology Development Project (1599005-2-13).