Asymptotic Behavior of Solutions for Competitive Models with Free Boundaries

In this paper, we study a competitive model involving two species. When the competition is strong enough, the two species are separated by a free boundary. If the initial data has a positive bound at infinity. We prove that the solution will converge, as $t\rightarrow \infty$, to the traveling wave solution and the free boundary will move to infinity with a constant speed.

In population ecology, the appearance of regional partition of multi-species through strong competition is one interesting phenomena. In [15,16,17], Mimura, Yamada and Yotsutani used the following reaction-diffusion equations 0 < x < s(t), t > 0, Q t = d 2 Q xx + g(Q), s(t) < x < 1, t > 0, P (x, t) = Q(x, t) = 0, x = s(t), t > 0, s ′ (t) = −µ 1 P x (x, t) − µ 2 Q x (x, t), x = s(t), t > 0, P (0, t) = m 1 , Q(1, t) = m 2 , t > 0, s(0) = s 0 (0 < s 0 < 1), P (x, 0) = P 0 (x)(0 < x < s 0 ), Q(x, 0) = Q 0 (x)(s 0 < x < 1) to describe regional partition of two species, which are struggling on a boundary to obtain their own habitats. When m 1 , m 2 > 0 and f, g are monostable nonlinearities. They prove the global existence, uniqueness, regularity and asymptotic behavior of solutions for problem (1.2) in [15]. When m 1 , m 2 > 0 and f, g are bistable nonlinearities. The author establish the stability for stationary solutions for the free boundary problem and their argument is based on the notion of ω−limit set and the comparison principle in [16]. When m 1 , m 2 = 0, i.e. homogeneous Dirichlet boundary conditions are imposed, there is possibility that the free boundary may hit the fixed ends x = 0, 1 in a finite time. This is a very interesting phenomenon. Therefore, in [17], the author prove that if the free boundary hits the fixed boundary at a finite time t = T * , the free boundary stays there after T * .
Problem (1.2) is defined on finite interval [0,1]. A natural question is what will happen if the two species competitive model is defined on (−∞, ∞)? If the two species competitive model is defined on the entire space, the free boundary s(t) could move to infinity in different ways and the problem may become more complicated. This is why we are interested in studying problem (1.1). Possibly, the solution will develop into the traveling wave eventually if problem (1.2) is defined on unbounded domain. So, it is necessary to consider the traveling wave solution before we investigate the asymptotic behavior of solutions for problem (1.1). To study the traveling wave solution of problem (1.1) is equivalent to study the solution of the following ordinary differential equations Recently, in [3], Chang and Chen prove the existence of a traveling wave solution of (1.3) for logistic type nonlinearities. In [19], we extend the results in [3] to more general nonlinearities. We prove that if α > 0 is a given constant, then for any c ∈ (c * g ,ĉ f ), where c * g < 0 is the maximal speed when g is of (f M ) type, or the unique speed when g is of (f B ) type andĉ f > 0 depends only on α and f , there exists a unique β(c) > 0 such that (1.3) has a unique solution (φ, ψ, c). Moreover, β(c) is continuous and strictly decreasing in c ∈ (c * g ,ĉ f ) and If β > 0 is a given constant, we can obtain similar results in the same method.
The main purpose of this paper is to prove that for any initial data P 0 and Q 0 in problem (1.1), as long as they have positive lower bound at infinity, the solution of problem (1.1) will converge to a traveling wave as t → ∞. Our main result is as follows: Theorem 1.1. Assume f, g are of (f M ) type and the initial data P 0 , Q 0 satisfy then for some constant x * , the solution of problem (1.1) satisfies Remark 1.2. Similar results in Theorem 1.1 also holds when f and/or g are of (f B ) type; namely, (1.5) and (1.6) also hold if the initial data P, Q satisfy one of the following conditions: when f is of (f B ) type and g is of (f B ) type . Remark 1.3. If the diffusion constants d 1 = d 2 = 1, we can construct Lyapunov functional as in [11] to prove the asymptotic behavior of solutions for problem (1.1). However, through out our paper, we assume that the diffusion constants d 1 = d 2 . Thus, the method used in [11] does not work in this case. It is essentially different from the case d 1 = d 2 = 1. Therefore, in our paper, we use a different way to prove the asymptotic behavior of solutions.
The contents of this paper will be organized as follows: In section 2, we give some basic results on global existence of smooth solution for problem (1.1) and comparison principle. In section 3, we prove Theorem 1.1.

Preliminary results
2.1. The global existence of solutions. Theorem 2.1. If the initial data P 0 , Q 0 satisfy then problem (1.1) has a unique solution Moreover, The proof of Theorem 2.1 will be given in the Appendix. Proof. For any fix a ∈ (−∞, ∞), with no loss of generality, we may assume a ∈ [−l, l](l > 0), then we consider the following problem is the solution of (2.2). i.e., (U, V ) is a stationary solution of (1.1). Choose (U, V ) satisfy- Case 1: s(t) first moves across the point (a, 0) at x-axis from the right to the left, Q(x, t) must cross with V (x) at (a, 0). Therefore, Case 1-1: s(t) stop at the point (a, 0), we have |s ′ (t)| ≤ H; Case 1-2: s(t) moves backwards and then moves towards the left and cross the point (a, 0) again, this case can be discussed similarly as above; Case 1-3: s(t) moves across the point (a, 0), then P (x, t) must separate with U (x) temporarily. Even if s(t) moves towards the right and cross the point (a, 0) again, we can discuss this case as above, so |s ′ (t)| ≤ H.
Since a is arbitrary, so the conclusion of Lemma follows.
The proof of Lemma 2.3 is identical to that of Lemma 5.7 in [7], so we omit the details here.  In this section, f, g are always assumed to be of (f M ) or (f B ) type.
3.1. Converge to 1 uniformly at infinity. We consider the solution of , for each m ∈ (0, 1) (resp. m ∈ (θ, 1)), consider the trajectory Γ given by (1) If f, g are of (f M ) type and the initial data P 0 , Q 0 satisfy (1.4). P is the solution of the problem and Q is the solution of the problem (2) If f, g are of type (f B ) and the initial data P 0 , Q 0 satisfy (iii) in Remark 1.2. P is the solution of (3.2) and Q is the solution of (3.3), then for any p 0 , q 0 ∈ (0, 4 3 (1 −θ)), there exist T, M > 0 such that Since f (·) > 0 in (0, 1), for any p 0 ∈ (0, 1) and Let L ≫ R be a constant to be determined later(we may assume L < 9M 2 with no loss of generality) and w 0 (x) be a function satisfying By comparison principle, we have . By using the same method, we can repeat the above argument to prove P (x, On the other hand, since lim sup Then ξ(t) is an upper solution of (3.3). So Q(x, t) ≤ ξ(t) for all t ≥ 0. Since g(Q) < 0 for Q > 1, ξ(t) is a decreasing function converging to 1 as t → ∞. Thus, for any q 0 ∈ (0, 1), there exists Consequently, if we choose T = max{T 1 , T 2 , T 3 } and M > max{2M 2 , M 3 }, we obtain Since f (·) > 0 in (θ,θ), for any p 0 ∈ (0, 4 3 (1−θ)), and T 2 = Here, we fix R = Lθ +ǫ . The following proof is the same as (1), so we omit the details. This completes the proof.

3.2.
Precise estimates of the solutions in moving coordinates. In the following, we assume that (P (x, t), Q(x, t)) is the solution of (1.1). Denote u(z, t) = P (z+ct, t) = P (x, t), v(z, t) = Q(z + ct, t) = Q(x, t), which satisfies Under the assumption of Theorem 1.1, there exists T > 0 with some constants ρ 1 , ρ 2 , p 0 , q 0 and ϑ such that for all z and t ≥ T . Moreover, |s(t) − ct| is bounded for all t > 0.
Proof. Firstly, we prove the left part of (3.5) and the right part of (3.6).
By Lemma 3.1, we can select η(0) positive and large enough while ξ(0) negative such that Thus, we know that s(t) ≥ z(t) + ct, and lim inf Using the same method, we can construct an upper solution in the similar way. The proof is now completed.
for all z and all t ≥ T .

3.3.
Convergence of the solutions. By Lemma 3.2, we know that there exist C, T > 0 such that −C ≤ s(t) − ct ≤ C for t ≥ T. Here we assume that C > max{ρ 1 , ρ 2 }. Denote Obviously, C ≤ s(t) ≤ 3C for t ≥ T. By simple calculation, we know that ( u, v, s) satisfies Let t n → ∞ be an arbitrary sequence satisfying t n ≥ T for every n ≥ 1.
Lemma 3.4. Subject to a subsequence, Proof. By Lemma 2.2, we have |s ′ (t)| ≤ H, for all t > 0. So, there exists H > 0 such that | s ′ n (t)| ≤ H for t + t n large and every n ≥ 1.
then (û n (y, t),v n (y, t), s n (t)) satisfies For any given R 0 > 0 and T 0 ∈ R 1 , by using the interior-boundary L p estimates (see for all large n, where C R 0 is a constant depending on R 0 and p but independent of n and T 0 . Therefore, for any α ′ ∈ (0, 1), we can choose p > 1 large enough and use the Sobolev embedding theorem (see [13]) to obtain where C R 0 is a constant depending on R 0 and α ′ but independent of n and T 0 . From (3.9) and (3.10), we deduce that where C 1 is a constant independent of n and T 0 . Hence by passing to a subsequence, we obtain that, as n → ∞, where α ∈ (0, α ′ ). Moreover, by using (3.9), we know that (Û ,V , G) satisfies in the W 1,2 p sense (and hence classical sense by standard regularity theory), By the proof of Lemma 3.2, we have Proof. We prove it by contradiction, if not, we have R * > sup t∈R 1 G(t). Choose δ > 0 such that Step 1: satisfies the equation (3.8) 1 , by strong maximum principle, we conclude that U (x, t) ≡ φ(x− R * ) in D 0u := {(x, t) : x < G(t), t ≤ t 0 }, and this contradicts with Case 1 x 0 = R * . We have V (x 0 , t 0 ) = ψ(0) = 0, so x 0 = R * = G(t 0 ), and this contradiction implies the conclusion easily.
Step 2: By step 1, we know M u (x 10 ) and M v (x 20 ) can not be achieved at any finite t. Therefore, there exists s n ∈ R 1 with |s n | → ∞ such that ). Then we can use the same argument as in the proof of Lemma 3.4 to show that, by passing to a subsequence, ( U n , V n , G n ) → (U * , V * , G * ) with (U * , V * , G * ) satisfying Moreover, Since (φ(x − R * ), ψ(x − R * )) satisfies (3.11) with G * (t) replaced by R * , by strong maximum principle, we have , which is clearly impossible. On the other hand, If there exists τ n ∈ R with |τ n | → ∞ such that ψ(x 20 − R * ) = lim n→∞ V (x 20 , τ n ). In the same way, we can derive a contradiction. Therefore, the conclusion follows easily.
Since 1 is the an upper solution of (3.13) and (3.14), the unique solution of (3.13) and (3.14) are decreasing in t.
are the lower solution of (3.13) and (3.14). By comparison principle, we have Therefore, . We notice that (Φ(x), Ψ(x)) also satisfies (3.15). Moveover, , W 2 (R 20 ) = 0, and by the maximum principle we deduce, for any R 1 < R 10 , R 2 > R 20 , We now look at ( U (x, t), V (x, t)), since U (x, t) satisfies the first equation in (3.13), and for any In the same way, we know Consequently, we use the comparison principle to deduce that for all x > R 20 , t ∈ R 1 . By Step 2 and the continuity of Therefore we can combine with (3.16) to obtain for all small ǫ 1 ∈ (0, ǫ), which contradicts the definition of R u , R v . The proof is complete.
Lemma 3.6. There exists a sequence {s n } ⊂ R 1 such that Proof. There are two possibilities: (i) R * = sup t∈R 1 G(t) is achieved at some finite time t = s 0 , (ii) R * > G(t) for all t ∈ R 1 and G(s n ) → R * along some unbounded sequence s n .
On the other hand, we know Using the uniqueness of (3.8) with a given initial value, we conclude that for all x ≥ G(t), t ∈ R 1 , so we choose s n ≡ s 0 and the conclusion of the lemma holds. In case (ii), we consider the following sequence Using the same method as in the proof of Lemma 3.4, we can choose a subsequence such that loc (R 1 ), and (U, V, G) satisfies (3.8). Moreover, Hence we are back to case (i) and thus U( and G ≡ R * . The conclusion of the lemma follows easily. 3.4. The Proof of Theorem 1.1. We now apply Lemma 3.3, which indicate that once u is close to φ(x − R * ) and v is close to ψ(x − R * ) for some T k , it remains close for all later time. Thus Next, we prove s ′ (t) → c. If s ′ (t) does not converge to c, there exist a ǫ > 0 and a sequence {t k } ∞ k=1 such that |s ′ (t k )−c| > ǫ for k and t k → ∞ as k → ∞. With no loss of generality, we assume s ′ (t k ) > c + ǫ. Since |s(t) − ct − x * | → 0 as t → ∞, we can make estimate about u and v at However, this is a contradiction with the asymptotic behavior of u and v. Thus s ′ (t) → c. This completes the proof of Theorem 1.1.