On a reaction – diffusion – advection system : fixed boundary or free boundary

This paper is devoted to the asymptotic behaviors of the solution to a reaction–diffusion–advection system in a homogeneous environment with fixed boundary or free boundary. For the fixed boundary problem, the global asymptotic stability of nonconstant semi-trivial states is obtained. It is also shown that there exists a stable nonconstant co-existence state under some appropriate conditions. Numerical simulations are given not only to illustrate the theoretical results, but also to exhibit the advection-induced difference between the left and right boundaries as time proceeds. For the free boundary problem, the spreading–vanishing dichotomy is proved, i.e., the solution either spreads or vanishes finally. Besides, the criteria for spreading and vanishing are further established.

For a population growth model, u(x, t), v(x, t) in (1.1) respectively represent the densities of two species at location x and time t.d i > 0 denotes the random dispersal rate of the species; advection rate β i ∈ R is the moving speed of individuals towards their more favorable habitats.As noted in [12], β i > 0 means advection points towards larger x, while β i < 0 implies advection points towards smaller x.r i > 0 accounts for intrinsic growth rate; a i > 0 is intra-specific interaction rate; b i ∈ R is interspecific interaction rate.What described above means that system (1.1) is a competition model with b i > 0, see [16,17,29], and a predator-prey problem for b 1 > 0, b 2 < 0, see [24,26,27,30].For a information diffusion model, the meaning of u(x, t), v(x, t), d i , β i , r i , a i , b i can refer to [22,25].
The way to formulate boundary conditions for reaction-diffusion models is based on how the flux of individuals crosses a boundary.As mentioned in [5], the flux − → J = −d∇u + − → β u across the boundary at any given point is proportional to the density with constant of proportionality, that is where d > 0 is the diffusion rate, − → β is the advection velocity, − → n is the outward pointing normal vector, α is the proportionality coefficient.If α → ∞, then the boundary condition (1.2) becomes u = 0, which is a Dirichlet condition; if − → β = 0 and α = 0, we have (−d∇u) • − → n = 0, i.e., ∂u ∂ − → n = 0, which is known as a Neumann condition; if − → β = 0 and α < 0, then (1.2) is in the form of d ∂u ∂ − → n − αu = 0, which is referred to a Robin condition; if α = 0, then (1.2) becomes which is called a no-flux or reflecting boundary condition, since it means that individuals encountering the boundary are always reflected back so they do not leave the domain, that is, no individual crosses the boundary.
Here we focus on the no-flux boundary conditions d 1 u x (0, t) − β 1 u(0, t) = d 1 u x (L, t) − β 1 u(L, t) = 0, t ∈ (0, ∞), In fact, Lou et al. in [16] first qualitatively analyzed (1.1) with no-flux boundary (1.3) under It is shown that the movement with either smaller advection or no advection is eventually stable.Afterwards, based on the assumptions of (1.4), Zhou in [29] further investigated (1.1) with the addition of (1.3) to understand the joint effects of diffusion and advection on the outcome of competition.Overcoming the mathematical difficulties arising out of Zhou in [29] obtained much richer observations: the movement with smaller diffusion, smaller advection and smaller ratio of advection to diffusion, or with larger diffusion and smaller advection, wins the competition.However, for a more general model with d i , r i , a i ∈ R + , β i , b i ∈ R and without assumption (1.4), there have been no results so far.Due to this reason, in this paper, we study (1.1) with fixed boundary (1.3) and present a thorough understanding: for b i > 0, problem (1.1) and (1.3) may finally stabilize to a nonconstant semi-trivial steady state if β i > 0, but admit a stable coexistence state if β 1 • β 2 < 0; for b 1 > 0, b 2 < 0, the two semi-trivial steady semi-trivial steady states of (1.1) with the addition of (1.3) are both unstable.Furthermore, we give numerical simulations to find out that advection can induce great difference between the left and right boundaries as time goes on; and the problem with b 1 > 0, b 2 < 0 may have a co-existence state.
On the other hand, influenced by human activity, the habitat of some species often changes with time, which can be described by a free boundary.In this case, we go one step further and discuss the corresponding free boundary problem (1.5) Here µ, ρ and h 0 are given positive constants.x = h(t) is the free boundary, and the initial function u 0 (x), v 0 (x) ∈ Σ(h 0 ) for some h 0 > 0, where Wang et al. in [26] studied system (1.5) with β i = 0 and b i > 0 and obtained the long time behavior of two competing species spreading via a free boundary.Based on the assumptions of β i = 0, b 1 > 0, and b 2 < 0, Wang in [24] investigated system (1.5) to get a spreadingvanishing dichotomy and set the criteria for spreading and vanishing, moreover, Wang in [24] gave the estimation of asymptotic spreading speed when spreading successfully.
Motivated by the works in [24,26], we study system (1.5) with We prove that the spreading-vanishing dichotomy still holds, i.e. the solution to problem (1.5) is vanishing if h ∞ < +∞, on the other hand, it is spreading if h ∞ = +∞ under some proper conditions.Furthermore, we determine the criteria for spreading and vanishing.
The rest of this paper is organized as follows.In Section 2, the fixed boundary problem is analyzed, including the global asymptotic stability of nonconstant semi-trivial steady states, and the existence of a stable nonconstant co-existence state.Numerical simulations are presented to illustrate the results.Section 3 is devoted to the free boundary problem.The spreading-vanishing dichotomy is obtained and the criteria for spreading and vanishing are determined.

Existence of semi-trivial steady states
First, we consider the problem and the following statement is valid.Lemma 2.1.For any β 1 ∈ R, and d 1 , r 1 , a 1 ∈ R + , problem (2.1) admits a unique positive solution ũ.
Proof.For β 1 ≥ 0, we rewrite problem (2.1) as where L .After some simple computations, we have and (2.4) According to the definition of upper and lower solutions in [19], one can see that ū and u are upper and lower solutions to problem (2.2).Set The proof of uniqueness is similar to [16,Lemma 2.1].For the readers' convenience, we outline the main ideas.Suppose ũ1 and ũ2 are two different positive solutions to (2.2).We may assume ũ1 > ũ2 > 0. ũ1 and ũ2 satisfy respectively.Multiplying the first equation of (2.5) by e x ũ1 , subtract the resulting equations and integrate over [0, L], and then we get which contradicts ũ1 > ũ2 > 0. Therefore, we obtain the positive solution to (2.1) is unique.For β 1 < 0, we can make some minor modifications to get the existence and uniqueness of the positive solution to problem (2.1), so we omit the details.
Similarly, the problem also has a unique positive solution ṽ.
Thus, the following result follows from Lemma 2.1 directly.
Furthermore, as for ũ and ṽ, we have the following result, which is vital to our later analysis.

Local stability of semi-trivial steady states
In this subsection, assume d i , r i , a i , ∈ R + and β i , b i ∈ R, we focus on the local stability of the semi-trivial steady states of problem (1.1) with the addition of (1.3).Beginning with ( ũ, 0), and its stability is governed by the equations (2.12) The corresponding eigenvalue problem is (2.13) One can find that the second equation in (2.13) is decoupled from the first.As a result, we only consider the eigenvalue problem Similarly, in order to investigate the stability of (0, ṽ), we consider the eigenvalue problem (2.15) For convenience, the general formula of eigenvalue problems (2.14) and (2.15) is given as follows (2.16) Then we have the following statements.
), then the eigenvalue problem (2.16) has a simple principle eigenvalue σ 0 , and the corresponding eigenfunction δ 0 (x) can be chosen as δ 0 (x) 0. where has a simple principle eigenvalue σ 0 , and the corresponding eigenfunction δ 0 (x) can be chosen as δ 0 (x) 0. Thanks to [5, Corollary 2.13], we get that σ 0 is also the principle eigenvalue of Accordingly, Lemma 2.4 is established.
Taking derivative of h, we derive Note that (σ 0 , δ 0 ) satisfies (2.16), thanks to (2.23), then some direct computations yield Taking derivative of (2.24) in view of x, we obtain (2.25) As for the stability of semi-trivial steady states of system (1.1) with the addition of (1.3), the following result in [23] is of great concern in our subsequent analysis.Lemma 2.6.Suppose d i , r i , a i ∈ R + , and β i , b i ∈ R, then the semi-trivial steady state ( ũ, 0) is linearly stable (resp.unstable) if λ 0 is positive (resp.negative); the semi-trivial steady state (0, ṽ) is linearly stable (resp.unstable) if µ 0 is positive (resp.negative).
For clarity, we first state two propositions, which are vital to judge the stability of ( ũ, 0) and (0, ṽ). where (2.26) Note that (λ 0 , w 0 ) satisfies (2.28) Multiplying the first equation of (2.27) by e x ω 0 , and integrating the resulting equation over [0, L], we get By the boundary conditions of (2.27), (2.29) can be simplified to Multiplying the first equation of (2.28) by e x ũ, and integrating the resulting equation over [0, L], according to boundary conditions, we have x ũdx. (2.31) Combining (2.30) and ( 2.31), one can find (2.32) ) can be rewritten as Clearly, the sign of λ 0 is the same as that of λ * , which completes the proof of Proposition 2.7.
Similarly, we have the following proposition concerning µ 0 .
Proposition 2.8. where (2.34) Proof.By a similar method noted in the proof of Proposition 2.7, we can find ) can be simplified as which indicates that µ 0 and µ * have the same sign.Now, we can establish the local stability of ( ũ, 0) and (0, ṽ) respectively.
For part (iii), the case of d 1 = d 2 > 0 is easy to check, so we only verify the case of 0 By the similar argument to part (i), it is easy to find (2.40) Combining (2.40) with the conditions of 0 < r 1 < r 2 and 0 < b 2 < a 1 , we deduce which shows that ( ũ, 0) is unstable by Proposition 2.7 and Lemma 2.6.For β 2 > 0 > β 1 , combining this condition with part (i) of Lemma 2.3 and part (ii) of Lemma 2.5, we have For part (iv), in the case of b 2 < 0, again by part (i) of Lemma 2.3 and part (ii) of Lemma 2.5, we get Then the proof of part (iii) is completed by Proposition 2.7 and Lemma 2.6.
Proof.By using the similar argument to the one applied in the proof of Lemma 2.9, one can prove that µ Thus, part (i) and part (ii) directly follows from Proposition 2.7 and Lemma 2.6.
Part (iii), we only consider d 1 < d 2 , because the case of d 1 = d 2 > 0 can be verified in a similar argument. For Moreover, due to part (i) of Lemma 2.5, we deduce For β 2 > 0 > β 1 , we can use a similar argument to show (0, ṽ) is unstable.

The non-existence of coexistence steady state
In this subsection, we show the nonexistence of coexistence steady states under some proper conditions.
Lemma 2.11.Suppose that 0 ) with the addition of (1.3) has no coexistence steady state; ) with the addition of (1.3) has no coexistence steady state.
Proof.For part (i), arguing indirectly, we assume that system (1.1) with the addition of (1.3) has a coexistence steady state (U, V), then By the boundary conditions of (2.45), we have (2.48) In the following, we show 8 claims to finish the proof.Claim 1.
Claim 2 implies that f and g can not be identically zero in [0, L].Besides, f and g are real analytic.Therefore, all zero points of f and g are isolated.If g cannot change sign in [0, L], then g ≤, ≡ 0 in [0, L] by using Claim 2. Thus (2.56) Combining (2.45) with (2.56), we have x ∈ (0, y 1 ), (2.57) Multiplying the first equation of (2.57) by e x V and the second one by e x U, subtracting the resulting equations and then integrating over [0, y 1 ], we get (2.58) Thanks to our assumptions, we have From (2.56), it follows that x ∈ (0, y 1 ). (2.61) By part (ii) of Lemma 2.3, we get x ∈ (0, y 1 ). (2.62) Accordingly, it follows from (2.59), (2.60) and (2.62) that (2.63) On the other hand, due to the last equation of (2.57), one can find (2.64) (2.63) and (2.64) cause a contradiction to (2.58), which indicates that g must change sign in [0, y 1 ].In addition, g changes sign before f .Now, it's turn to show f also changes sign in [0, L].If not, we have f ≤, ≡ 0 in [0, L] by Claim 2. Denote the last zero point of g in (0, L) by x m−1 .Here x m−1 must exist because g changes sign in (0, L).Then (2.65) By some direct calculation, it follows from (2.45) and (2.65) that (2.66) According to (2.65) and our assumptions, one can see that the left side of (2.66) is nonpositive but the right is positive, which gives rise to a contradiction.Therefore, f also changes sign in [0, L].
Based on Claim 2 and Claim 3, g has at least one zero point in (0, L) such that the sign of g must change at each side of the point.Let x 2 be the first one.Obviously, either g ≤ 0 or g ≥ 0 in (0, x 2 ), so we consider these two cases: Case i: g ≤, ≡ 0 in (0, x 2 ); Case ii: g ≥, ≡ 0 in (0, x 2 ).In the following analysis, we show that f ≤ 0 in [0, L] in both cases.First, we consider Case i: g ≤, ≡ 0 in (0, x 2 ).By Claim 2 and Claim 3, there exists (2.67) If not, f has at least one zero point in (0, L) such that the sign of f must change at each side of the point.Let y 2 be the first one.Thus, (2.68) Combining (2.45) with (2.68), some direct calculations yield (2.69) By our assumptions and the last equation of (2.68), we have x UV[(r (2.71) The contradiction completes the proof of Claim 4. Claim 5. f ≤ 0 in (x 2 , x 3 ] with f (x 3 ) < 0 for Case i: g ≤, ≡ 0 in (0, x 2 ).First, we show f ≤ 0 in [x 2 , x 3 ].Otherwise, there is y 3 and z 2 with (2.72) (2.72) implies that there exists small 1 > 0 such that f is increasing in (y 3 , y 3 + 1 ).On the other hand, since g(x 3 ) = 0 and g(x) ≥ 0 in (x 2 , x 3 ), there exists small 2 > 0 such that g is diminishing in (x 3 − 2 , x 3 ).By Claim 1, f is also diminishing in (x 3 − 2 , x 3 ).Consequently, f must has at least one positive local maximum value point in (y 3 , x 3 ).Let z 3 be the closest to y 3 .Then we have and (2.75) By (2.73) and (2.74), we find (2.76) Due to (2.75) and (2.76), we deduce (2.78) From (2.78), we can obtain that T has a positive local maximum value point saying z 4 in (y 3 , z 3 ).By the first equation of (2.51), we get S(z 4 ) < 0. (2.79) On the other hand, since g(x) > 0 in (x 2 , x 3 ), (2.80) The contradiction caused by (2.79) and (2.80) shows that f (x) ≤ 0 in [x 2 , x 3 ].Considering f is diminishing in a small neighborhood of x 3 , we derive f (x 3 ) < 0. After x 3 , we can find the next zero point of g.Denote it by Actually, when g ≥ 0 in (x 3 , x 4 ] happens, we can deduce f ≤ 0 in (x 3 , x 4 ] with f (x 4 ) < 0 in the similar way to Claim 5. Next, we show f ≤ 0 in (x 3 , x 4 ] if g ≤ 0 in (x 3 , x 4 ] happens.If not, there exists y 4 ∈ (x 3 , y 4 ] such that By the similar method to Claim 4, we arrive at x UV[(r (2.84) The contradiction ends the proof of Claim 6. Claim 7. f ≤ 0 in [0, L] for g ≤, ≡ 0 in (0, x 2 ).Due to Claim 2, Claim 3 and the isolated properties of the zero points for g, g has finitely many zero points.Consequently, by repeating the above analysis, we can obtain that f ≤ 0 in [0, L] for g ≤, ≡ 0 in (0, x 2 ).
Actually, for g ≥, ≡ 0 in (0, x 2 ), we can show that f ≤ 0 in [0, L] by the similar method to the case of g ≤, ≡ 0 in (0, x 2 ), so Claim 8 can be verified directly.
f ≤ 0 in [0, L] contradicts Claim 3, which shows that the coexistence steady state (U, V) of system (1.1) with the addition of (1.3) For part (ii), we can apply the similar arguments to part (i) to show the non-existence of co-existence steady states if 0 so we omit the details.(1.1) with the addition of (1.3)In this subsection, we show the asymptotic behaviors of solution to problem (1.1) with the addition of (1.3).For convenience, we list the following conditions:
Proof.By using the theory of monotone dynamical system [23], part (i) of Theorem 2.12 directly follows from part (i) of lemma 2.9, part (i) of Lemma 2.10 and part (i) of Lemma 2.11; part (ii) of Theorem 2.12 directly follows from part (ii) of lemma 2.9, part (ii) of Lemma 2.10 and part (ii) of Lemma 2.11.
Actually, interchanging the labels of d 1 and d 2 , β 1 and β 2 , r 1 and r 2 , a 1 and a 2 , b 1 and b 2 , we can get the parallel results.
Again interchanging the labels of d 1 and d 2 , β 1 and β 2 , a 1 and a 2 , b 1 and b 2 , the parallel results for Theorem 2.15 is as follows.
Theorem 2.16.Suppose 0 < d 2 ≤ d 1 and C3 holds, then we have the same results as in Theorem 2.15.
Combining Theorem 2.15 with Theorem 2.16, we get the following more general conclusion.
Theorem 2.17.Assume d 1 , d 2 ∈ R + and C3 holds, then we have the same results as in Theorem 2.15.

Numerical simulations
Next we present some numerical simulations to indicate our results obtained above.Furthermore, for b 1 > 0, b 2 < 0, we observe that problem (1.1) with the addition of (1.3) may have a co-existence state.
Throughout this subsection, we fix and Figure 2.1: The asymptotic behaviors of u(x, t) in (a) and v(x, t) in (b) with Figure 2.2: The asymptotic behaviors of u(x, t) in (a) and v(x, t) in (b) with Figure 2.3: The asymptotic behaviors of u(x, t) in (a) and v(x, t) in (b) with Furthermore, from Figure 2.3, one can see that advection can induce great difference between the left and right boundaries as time proceeds.To illustrate the variation on u(x, t) and v(x, t) with x from the left boundary to the right, the curves of u(x) and v(x) with t = 3, 5, 10, 20, 30, 40 are presented in Figure 2.4.One can observe that u(x) in (a) and v(x) in (b) goes up and down as x increases when t is fixed on 3, 5 or 10; whereas u(x) in (c) grows and v(x) in (d) decreases with the increase of x when t is fixed on 20, 30 or 40.This is because the positive advection towards the right boundary but the negative advection towards the left boundary.
Example 2.20.Based on the the conditions in part (iv) of Lemma 2.9 and part (i) of Lemma 2.10, the numerical results are illustrated in Figure 2.5.We get that u(x, t) in (a) and v(x, t) in (b) both stabilize to positive nonconstant states as time goes on.This implies that ( ũ, 0) and (0, ṽ) are both unstable, which is consistent with part (iv) of Lemma 2.9 and part (i) of Lemma 2.10.
3 The free boundary problem

Existence and uniqueness
By a similar argument in [7,8], we have the following result.
Proof.Let ū be the unique solution of By [30, Theorem 2.4], lim x→∞ ū(x) = r 1 /a 1 .Similarly, the problem has a unique solution u, then lim x→∞ u(x) = a 2 r 1 −b 1 r 2 a 1 a 2 .Denote v is the unique solution of Since a 1 r 2 > b 2 r 1 , still by [30,Theorem 2.4], lim x→∞ v(x) = a 1 r 2 −b 2 r 1 a 1 a 2 .The above proof implies that u(x), v(x), ū(x) and v(x) are the coupled ordered lower and upper solutions of (3.2).Clearly, for any l > 0, u(x), v(x), ū(x) and v(x) are also the coupled ordered lower and upper solutions of By the standard upper and lower solutions method, we see that the problem has at least one positive solution (u l , v l ), satisfying According to the local estimation and compactness argument, we can conclude that (u l , v l ) → (u, v) in [C 2 loc ([0, ∞))] 2 , and (u, v) satisfies (3.2).
Next, we can obtain the similar result in the case: b 2 < 0.
where ū(x), v(x), u(x) and v(x) are the positive solutions of the following problems, respectively. (3.10)

Conditions for spreading and vanishing
Proof.The proof of this lemma is similar to [24, Theorem 3.1].We give the details for the readers' convenience.Define a transformation Let u(x, t) := U(y, t), v(x, t) := V(y, t) and set then the problem (1.5) becomes where Using the L p estimate and embedding theorem, there exists a positive constant K such that ≤ K for all n ≥ 0. Because these rectangles E n overlap and K Due to the Stefan condition and 0 < h (t) ≤ C 2 , we derive that h C In view of [27, Theorem 2.2] and Lemma 3.5, we get the following theorem.Theorem 3.6.Let (u, v, h) be the solution of problem (3.11)

Spreading case
Theorem 3.7.Suppose that d i > 0, where ū, v, u and v are given in the proof of Lemma 3.2.
The proof of this theorem is similar to the proof of [24,Theorem 3.5].
where ū, v, u and v are given in the proof of Lemma 3.3.

The criteria for spreading and vanishing
Here we first give the comparison principle.The proof is similar to the proof of [7,Lemma 3.5].

− β 1 d 1 x
ũ2 and the first equation of (2

Claim 3 .
f and g must change sign in [0, L].

Figure 2 . 4 :
Figure 2.4: The curves of u(x) and v(x) with respect to x when t is fixed.Here β 1 , β 2 , r 1 , r 2 , a 1 , a 2 , b 1 , b 2 take the same values as them in Figure 2.3.