A free boundary problem of a predator–prey model with advection in heterogeneous environment

Applied Mathematics and Computation 289 (2016) 22–36

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A free boundary problem of a predator–prey model with advection in heterogeneous environment Ling Zhou a,∗, Shan Zhang b, Zuhan Liu a a b

School of Mathematical Science, Yangzhou University, Yangzhou 225002, China School of Mathematical Science, Nanjing normal University, Nanjing 210023, China

a r t i c l e

i n f o

MSC: 35K51 35R35 92B05 35B40 Keywords: A free boundary problem Spreading–vanishing dichotomy Heterogeneous environment Advection Spreading speed

a b s t r a c t This paper is concerned with a system of reaction–diffusion–advection equations with a free boundary, which arises in a predator–prey ecological model in heterogeneous environment. The evolution of the free boundary problem is discussed. Precisely, we prove a spreading–vanishing dichotomy, namely both prey and predator either survive and establish themselves successfully in the new environment, or they fail to establish and vanishes eventually. Furthermore, when spreading occurs, we obtain an upper bound of the asymptotic spreading speed, which is smaller than the minimal speed of the corresponding traveling wave problem. © 2016 Elsevier Inc. All rights reserved.

1. Introduction The spatial behavior of populations in homogeneous or heterogeneous environments is a central topic in biology and ecology, and the spreading of the species (subject to laws of diffusion, reaction, advection, and interaction etc.) is a crucial quantity in the study of biological invasions and disease spread. In this paper we consider the following free boundary problem for the Lotka–Volterra type predator–prey modal with an advection term in heterogeneous environments:

⎧ ⎪ ⎪ut − uxx + β ux = u(m(x ) − u − av ), ⎪ ⎪ vt − dvxx = v(c − v + bu ), ⎪ ⎪ ⎨u(t, 0 ) = v(t, 0 ) = 0, u(t, h(t )) = v(t, h(t )) = 0, ⎪ ⎪ ⎪ ⎪ h (t ) = −μ[ux (t , h(t )) + ρvx (t , h(t ))], ⎪ ⎪ ⎩ u ( 0, x ) = u0 ( x ), v ( 0, x ) = v0 ( x )

t > 0, 0 < x < h(t ), t > 0, 0 < x < h(t ), t > 0, t > 0, t > 0, 0 ≤ x ≤ h0 = h ( 0 ),

(1.1)

where β , a, b, c, d, μ and ρ are given positive constants. Biologically, u and v represent, respectively, the spatial densities of prey and predator species that are interacting and migrating in a one dimensional habitat, m(x) accounts for the local growth rate of the prey, the free boundary x = h(t ) represents the expanding fronts of the species. Here β ux is the advection term, which means individuals are confronted with unidirectional drift that drives them out of the system and thus induces decline in population. Such advection term may appear due to some external environmental forces such as water flow ∗

Corresponding author. Tel.: +86 13773598226. E-mail address: [email protected] (L. Zhou).

http://dx.doi.org/10.1016/j.amc.2016.05.008 0 096-30 03/© 2016 Elsevier Inc. All rights reserved.

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

23

[22,24], wind [2,3] and gravity [11,23]. The ecological background of free boundary condition in (1.1) can refer to [1,20]. Such kind of free boundary condition has also been used in [12,26,29]. When β = 0 (i.e., there is no advection in the environment), the analysis of the evolution of the corresponding free boundary problems has been undertaken. Wang [26] studied system (1.1) in homogeneous environment, they proved a spreading–vanishing dichotomy, namely the two species either successfully spread to the entire space as time t goes to infinity and survive in the new environment, or they fail to establish and die out in the long run. Moreover, when spreading successfully, they obtained an estimate to show that the spreading speed (if exists) cannot be faster than the minimal speed of traveling wavefront solutions for the prey–predator model on the whole real line without a free boundary. various two species competition models with free boundaries have been studied in [12,13,29,32,33] and reference therein. Wang and Zhao [30] also studied the similar free boundary problems to (1.1) with double free boundaries in which the prey lives in the whole space but the predator lives in the region enclosed by the free boundary. In particular, in [31], the authors dealt with the higher dimension and heterogeneous environment case. In the absence of v, problem (1.1) with β = 0 is reduced to the one phase Stefan problem, which was studied by many authors. Starting from the work of Du and Lin [8], Kaneko and Yamada [17] in homogeneous environment, free boundary problems for the logistic type model, including the higher dimension case, heterogeneous environment case and with seasonal succession, have been extensively studied in [5–7,9,10,17,18,21,25,28,34]. When β > 0, which means that the species can move up along the gradient of the density (see [14,15] and the reference therein). Due to the appearance of the advection term, new difficulties are introduced mathematically. Currently, there have been few analytical results about single-species free boundary models in the literature. Gu et al. [14–16] studied the corresponding logistic modal with small advection, they discussed how advection term (β ux ) affects the asymptotic spreading speeds when spreading occurs. Up to our knowledge, the multispecies with advection case in the heterogeneous environment is not treated in the literature. In this paper, we discuss the free boundary problem (1.1) for a reaction–diffusion–advection model in heterogeneous  environment. Throughout our paper, the positive function m(x) satisfies m(x ) ∈ C 1 ([0, ∞ )) L∞ ([0, ∞ )) and

m0 := inf m(x ) > 0, x≥0

m1 := sup m(x ) x≥0

and m∞ := lim m(x ). x→+∞

(1.2)

The initial functions u0 , v0 satisfy

u0 ,

v0 ∈ C 2 ([0, h0 ] ), u0 (0 ) = v0 (0 ) = u0 (h0 ) = v0 (h0 ) = 0, u0 (x ), v0 (x ) > 0 in (0, h0 ).

The main purpose of the present paper is to analyze the effect of the advection term and the heterogeneous environment on the criteria for spreading or vanishing and the asymptotic spreading speeds as spreading occurs. In this sense, the present paper can be regarded as the generalization of [26]. The most challenging issue lies in the discussion of the following corresponding elliptic problem in half line



−du + β u = u( f (x ) − λu ), 0 < x < ∞, u ( 0 ) = 0.

(1.3)

To overcome the difficulty induced by the advection term, we rewrite (1.3) as





β

−d e− d x u u ( 0 ) = 0.



β

= e− d x u( f (x ) − λu ), 0 < x < +∞,

(1.4)

When β = 0 and f(x) is a monotonous function of x, Wang [26] proved the convergence result

lim u(x ) =

x→+∞

1

lim f (x ).

λ x→+∞

In contrast to problem (1.3) (β = 0) discussed in [26], the function f(x) here is influenced by the coefficient m(x) which describes the intrinsic rate of growth and is not monotonous in x. The conclusion in [26] cannot be used directly in this situation. However, by construction monotonous functions and using comparison principle, we also prove the convergence result that limx→+∞ u(x ) = λ1 limx→+∞ f (x ) (see Theorem 2.4). Moreover, the long time behavior of solution and criteria for spreading and vanishing are also obtained. When spreading happens, we also provide the estimate

lim sup t→∞



 √ h(t ) ≤ max 2 cd, 2 m1 + β t

to show that the spreading speed (if exists) cannot be faster than the minimal speed of traveling wavefront solutions for the prey–predator model on the whole real line without a free boundary. By a similar argument as in [8,26], we have the following basic results. Theorem 1.1. Problem (1.1) has a unique global solution, and for any α ∈ (0, 1) and T > 0,

(u, v, h ) ∈ [C (1+α )/2,1+α (DT )]2 × C 1+α/2 ([0, T ] ), where

DT = {(t, x ) ∈ R2 : t ∈ (0, T ], x ∈ (0, h(t ))}.

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L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

Furthermore, there exists a positive constant M such that

0 < u(t, x ), v(t, x ) ≤ M for 0 < t < ∞, 0 < x < h(t ), 0 < h (t ) ≤ M for 0 < t < ∞.

The organization of this paper is as follows. In Section 2 we discuss the stationary solutions. Section 3 is devoted to the long time behavior of (u, v ) and establishing a spreading–vanishing dichotomy. The criteria for spreading and vanishing will be given in Section 4. In Section 5, we study the estimation of asymptotic spreading speed. 2. Positive solutions of the corresponding elliptic problems in half line In this section, we discuss the stationary solutions of the problem (1.1), which is the following elliptic problem in the half line

⎧  ⎨−u + β u = u(m(x ) − u − av ), 0 < x < +∞, −dv = v(c − v + bu ), 0 < x < +∞, ⎩ u ( 0 ) = v ( 0 ) = 0.

(2.1)

To this purpose, we will study the existence and uniqueness of positive solution to the following problem:



−du + β u = u( f (x ) − λu ), 0 < x < +∞, u ( 0 ) = 0,

(2.2)

where d and λ are positive constants, f(x) > 0 is a positive function. We rewrite (2.2) as



β

β

−d (e− d x u ) = e− d x u( f (x ) − λu ), 0 < x < +∞, u ( 0 ) = 0.

(2.3)

First we give priori estimates for the positive solution of problem (2.2). Lemma 2.1. Assume that f satisfies α ([0, ∞ )) with 0 < α < 1, inf f (x ) := f > 0, f ∈ Cloc 0 x≥0

f ∞ < ∞.

(2.4)

If u is a bounded positive solution of (2.2), then

u ( 0 ) > 0,

sup u(x ) ≤ x≥0

1

λ

f ∞

(2.5)

and there exists some positive constant τ , depending on d, λ and f, such that

u (x ) ≥ τ ,

∀x ≥ 1 .

(2.6)

Proof. Since u(x) > 0 in (0, ∞), we have u (0) ≥ 0 and u (0 ) = 0. If u (0 ) = 0, it follows from (2.3) that u (x) < 0 as 0 < x 1. Consequently, u(x) < 0 as 0 < x 1. This is a contradiction. Hence u (0) > 0. Moreover we consider two impossible cases: (a) if u(x1 ) > λ1 f ∞ and u (x1 ) ≥ 0 for some x1 ∈ (0, ∞), then u(x) must approach infinity. This is impossible, since u(x) is bounded in [0, ∞). f (b) if u(x1 ) < λ0 and u (x1 ) ≤ 0 for some x1 ∈ (0, ∞), then u(x) must vanish at some finite x2 . This is also impossible as u(x) is positive in (0, ∞). Next we claim that supx≥0 u(x ) ≤ λ1 f ∞ . If not, there exists x0 ∈ (0, ∞) such that u(x0 ) > λ1 f ∞ . We also have u (x0 ) < 0, since Case (a) is impossible. We deduce that there exists x1 ∈ (0, x0 ) such that

u(x1 ) = max u(x ) ≥ u(x0 ) > [0,x0 ]

1

λ

f ∞ and u (x1 ) = 0,

which is a contradiction. Finally, we prove (2.6) also by contradiction. We assume that for every N > 0, there exists xN > 1 such that u(xN ) < 1 N . That is, there exists a sequence {xN } such that limN→+∞ u (xN ) = 0. Note that u(x) > 0 for 0 < x < +∞ and Case (b) is impossible, we deduce that there exists a subsequence, still denoted by {xN }, such that xN → +∞ as N → +∞, {u(xN )} is decreasing and converges to zero, and for any N,

u ( xN ) <

f0

λ

, u ( xN ) > 0.

Thus there exists x¯N ∈ (xN , xN+1 ) such that u (x¯N ) ≤ 0 and u(x¯N ) < u(xN ) < f0 /λ, which is also a contradiction.

(2.7) 

In order to study problem (2.1) and also for later applications, we need a comparison principle which is similar to Proposition 2.1 in [26].

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

25

Proposition 2.2 (Comparison principle). Assume f1 (x), f2 (x) satisfy (2.4) and f1 (x) ≤ f2 (x) for all x ≥ 0. Let u1 (x), u2 (x) be positive solutions of problem (2.2) with f (x ) = f1 (x ) and f (x ) = f2 (x ) respectively, then we have

u1 ( x ) ≤ u2 ( x ),

∀ x ≥ 0.

(2.8)

Proof. First by Lemma 2.1, we have that ui (0 ) > 0 and 0 < ui (x ) ≤ λ1 supx≥0 fi (x ) < ∞ for all x > 0, and ui (x) ≥

τ for all x ≥ 1 and some positive constant τ (i = 1, 2). Hence, there exists a constant k ≥ 1 such that u1 (x) ≤ ku2 (x) for all x ≥ 0. Let k0 = inf{k > 0 : u1 (x ) ≤ ku2 (x ), u (x ) supx≥0 u1 (x ) . 2

i.e. k0 :=

∀x ≥ 0},

Then u1 (x) ≤ k0 u2 (x) for all x ≥ 0.

If k0 ≤ 1, our conclusion is true. Next, we assume that k0 > 1 and derive a contradiction. Let ϕ (x ) = k0 u2 (x ) − u1 (x ). Then ϕ (x) ≥ 0 for all x ≥ 0, and ϕ (x) satisfies β

β

β

−d (e− d x ϕ  ) = k0 e− d x u2 ( f2 (x ) − λu2 ) − u1 e− d x ( f1 (x ) − λu1 ) β

β

β

x

≥ e− d x f1 (x )ϕ − λ(k0 u22 − u21 )





= e− d x f1 (x )ϕ − λ(k0 u2 + u1 )ϕ + λk0 u22 (k0 − 1 ) , > e− d ≥

f 1 ( x )ϕ − λ ( k0 u2 + u1 )ϕ

1 f 1 ( 0 )ϕ , 2



0
0 < x 1,

(2.9)

since f1 (0) > 0 and ui (x) → 0 as x → 0 for i = 1, 2. Thus we derive β

( e− d x ϕ  ) < 0 0 < x 1.

(2.10)

Note that ϕ (0 ) = 0, we have ϕ  (0) ≥ 0. If

ϕ  (0 ) = 0, it follows from (2.10) < 0 as 0 < x 1. This is a contradiction. Hence ϕ  (0) > 0, i.e., k0 u2 (0 ) >

that ϕ  (x) < 0 as 0 < x 1. Consequently, ϕ (x) u (0 ). Remember this fact and the definition of

k0 , it is easily seen that at least one of the following happens:

1

(i) There exists some x0 ∈ (0, ∞) such that k0 u2 (x0 ) = u1 (x0 ), (ii) k0 u2 (x ) − u1 (x ) > 0 for all x > 0 and lim infx→∞ [k0 u2 (x ) − u1 (x )] = 0. When Case (i) happens, then ϕ (x0 ) = ϕ  (x0 ) = 0 and ϕ   (x0 ) ≥ 0. It is derived from (2.9) that

0 ≥ −dϕ  (x0 ) ≥ λk0 u22 (x0 )(k0 − 1 ).

(2.11)

Thus k0 ≤ 1. This is a contradiction. When Case (ii) occurs, then lim infx→∞ ϕ (x ) = 0. We provide the following claim: Claim. For every M > 0, there exists xM ∈ [M, +∞ ) such that β

( e− d x ϕ  ) ( xM ) ≥ 0. Proof of Claim: If not, there exists M0 > 0 such that β

( e− d x ϕ  ) ( x ) < 0 ∀ x ∈ [M0 , +∞ ).

(2.12)

β

Thus e− d x ϕ  (x ) is decreasing for x ∈ [M0 , +∞ ). Note that lim infx→∞ ϕ (x ) = 0, there exists a increasing sequence {xn }∞ n=1 such that xn → +∞,

ϕ (xn ) → 0 as n → ∞ and ϕ (xn ) > 0, ϕ  (xn ) < 0 ∀ n. β

Suppose that xN > M0 and −δ = e− d xN ϕ  (xN ) < 0. Then β

e− d x ϕ  (x ) < −δ < 0,

∀ x > xN

⇒

ϕ  (x ) < −C δ, ∀ x > xN ,

where C is determined by β and d. It is a contradiction with lim infx→∞ ϕ (x ) = 0.



Due to the above claim and ϕ (x) > 0 for all x > 0, it is not hard to prove that there exists a sequence {xn } with xn → ∞ such that β

(e− d x ϕ  ) (xn ) ≥ 0 and

lim

n→∞

ϕ ( xn ) = 0.

By passing to a subsequence, we may assume that

u2 ( xn ) → σ

as n → ∞

for some positive constants σ . It follows from (2.9) that

0 ≥ f1 (xn )ϕ (xn ) − λ(k0 u2 (xn ) + u1 (xn ))ϕ (xn ) + λk0 u22 (xn )(k0 − 1 ).

(2.13)

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L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

Letting n → ∞, we get

0 ≥ λk0 σ 2 (k0 − 1 ) > 0, which is also a contradiction. The proof is completed.



Lemma 2.3. Assume that f satisfies (2.4). If u(x) is a unique positive solution of the problem (2.2), we have (i) if f(x) is increasing in x, so is u(x) and limx→∞ u(x ) = λ1 limx→∞ f (x ); (ii) if f(x) is decreasing in x, then either u(x) is increasing in x, or there exists x0 > 0 such that u(x) is increasing in (0, x0 ) and u(x) is decreasing in (x0 , ∞). Therefore, limx→∞ u(x ) = λ1 limx→∞ f (x ). Proof. The proof is similar to that of Theorem 2.1 in [26], and we omit it here.



Next we use Lemma 2.1, Proposition 2.2 and Lemma 2.3 to prove the existence, uniqueness and convergence result for the positive solution of problem (2.2). In contrast to Theorem 2.1 in [26], f(x) here is influenced by the function m(x) and is not monotonous in x. The following theorem can be viewed as a generalization of Theorem 2.1 in [26]. Theorem 2.4. Assume f satisfies (2.4), and



0 ≤ β < 2 d f0 . Then the problem (2.2) has a unique positive solution u(x). Moreover, if

lim f (x ) = ζ > 0,

x→+∞

we also have

lim u(x ) =

x→+∞

1

ζ.

λ

Proof. Let λ1 (l) := λ1 (l, d, β ) be the first eigenvalue of the problem



−dφ  + βφ  = λ1 (l )φ , 0 < x < l, φ ( 0 ) = φ ( l ) = 0.



With the assumption 0 ≤ β < 2 d f0 , it is easy to provide that

λ1 (l ) =

β2 4d

+

π 2d

and

l2

lim λ1 (l ) =

l→∞

β2 4d

< f0 .

Thus, if we choose l large enough such that λ1 (l) < f0 , it is well known that the problem



−du + β u = u( f (x ) − λu ), 0 < x < l, u (0 ) = u (l ) = 0

(2.14)

has a unique positive solution, denoted by ul , and ul satisfies

sup ul (x ) ≤

0≤x≤l

1

λ

f ∞ .

By the comparison principle ([4], Lemma 5.2), we have that ul (x) is increasing in l. In view of the regularity theory and 2 ([0, ∞ )) as l → ∞, u solves compactness argument, it follows that there exists a positive function u, such that ul → u in Cloc (2.2). The uniqueness is followed from Proposition 2.2. Next, we prove the convergence result. Note that limx→+∞ f (x ) = ζ . For any positive ε < ζ , there exists xε > 0 such that

ζ − ε < f ( x ) < ζ + ε , ∀ x > xε . Let

 ζ + ε , x > xε f1,ε (x ) =

f ∞ , x < xε − 1,

 f2,ε (x ) =

ζ − ε , x > xε f0 ,

x < xε − 1

and f1, ε , f2, ε are smooth and monotonous functions and f2, ε (x) ≤ f(x) ≤ f1, ε (x). Let u1, ε (x), u2, ε (x) be positive solutions of problem (2.2) with f (x ) = f1,ε (x ) and f (x ) = f2,ε (x ) respectively. Thus, using Lemma 2.3, we have

lim u1,ε (x ) =

x→+∞

lim u2,ε (x ) =

x→+∞

1

lim f1,ε (x ) =

λ x→+∞ 1

lim f2,ε (x ) =

λ x→+∞

1

λ 1

λ

( ζ + ε ), ( ζ − ε ).

(2.15)

On the other hand, using Proposition 2.2, we have

u2,ε (x ) ≤ u(x ) ≤ u1,ε (x ),

∀x ≥ 0.

(2.16)

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

27

Combining (2.15) and (2.16), we obtain

1

λ

(ζ − ε ) = lim u2,ε (x ) ≤ limx→∞ u(x ) ≤ limx→∞ u(x ) ≤ lim u1,ε (x ) = x→+∞

x→+∞

1

λ

( ζ + ε ).

(2.17)

Since ε > 0 can be arbitrarily small, this implies that

lim u(x ) =

x→∞

1

λ

ζ. 

Theorem 2.5 (Global stability). Assume that d and λ are positive constants and f(x) satisfies (2.4). Suppose that φ ≡ 0 is a bounded, continuous and nonnegative function. Let u(t, x) be the unique solution of the following parabolic problem

⎧ ⎨ut − duxx + β ux = u( f (x ) − λu ), t > 0, 0 < x < ∞, u(t, 0 ) = 0, t > 0, ⎩ u ( 0, x ) = φ ( x ), 0 ≤ x < ∞.

Then

lim u(t, x ) = uˆ (x ) uniformly in any compact subset of [0, ∞ ),

(2.18)

t→∞

where uˆ (x ) is the unique positive solution of (2.2). Proof. The proof is similar to that of Theorem 2.2 in [26], and we omitted it here.



Finally, we give the existence of positive solution to (2.1) with the help of Theorem 2.4. Theorem 2.6. Assume that

√ 0 ≤ β < 2 m0

and a(bm1 + c ) < m∞ ,

then problem (2.1) has a positive solution (u(x ), v(x )). Moreover, any positive solution (u, v ) of (2.1) satisfies

u ( x ) ≤ u ( x ) ≤ u ( x ),

v ( x ) ≤ v ( x ) ≤ v ( x ), ∀x ≥ 0,

(2.19)

where u, v, u and v will be given in the following proof. Proof. Let u be the unique positive solution of



−u + β u = u(m1 − u ), 0 < x < ∞, u (0 ) = 0

(2.20)

√ with 0 ≤ β < 2 m0 . By Theorem 2.4, we have u (x ) > 0 and limx→+∞ u(x ) = m1 . Moreover, the problem



−dv = v(c − v + bu ), 0 < x < ∞, v (0 ) = 0

(2.21)

√ has a unique solution, denoted by v(x ). Then limx→+∞ v(x ) = bm1 + c. Since 0 ≤ β < 2 m0 and a(bm1 + c ) < m∞ , still by Theorem 2.4, the problem



−u + β u = u(m(x ) − av − u ), 0 < x < ∞, u (0 ) = 0

(2.22)

has a unique positive solution u (x). Then u (x) < m1 and limx→∞ u(x ) = m∞ − a(bm1 + c ). By Theorem 2.4 once again, the problem



−dv = v(c − v + bu ), 0 < x < ∞, v (0 ) = 0

(2.23)

has a unique positive solution, denoted by v(x ). Then v(x ) < bm1 + c. Applying Proposition 2.2 we have that u(x ) ≤ u(x ) for all x ≥ 0. Consequently, v(x ) ≤ v(x ) for all x ≥ 0. The above proof shows that u, v, u and v are the coupled ordered lower and upper solutions of (2.1). For any given l > 0, it is obvious that u, v, u and v are also the coupled ordered lower and upper solutions of the following problem

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L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

⎧  −u + β u = u(m(x ) − u − av ), 0 < x < l, ⎪ ⎪ ⎨−dv = v(c − v + bu ), 0 < x < l, u ( 0 ) = u ( 0 ) , v ( 0 ) = v ( 0 ) , ⎪ ⎪ ⎩ u ( l ) = u ( l ), v ( l ) = v ( l ).

(2.24)

By the standard upper and lower solutions method we deduce that the above problem has at least one positive solution, denoted by (ul , vl ), and

u ( x ) ≤ ul ≤ u ( x ),

v(x ) ≤ vl ≤ v(x ), ∀ 0 ≤ x ≤ l.

Applying the local estimation and compactness argument, it can be concluded that there exists a pair (u, v ), such that 2 ([0, ∞ ))]2 , and (u, v ) solves (2.1). It is obvious that (u, v ) satisfies (2.19).  (ul , vl ) → (u, v ) in [Cloc In the end of this section we provide two propositions that will be used in the study of the long time behavior of solution to the problem (1.1). The proof is similar to those of Propositions 2.2 and 2.3 in [26] and we omitted the proof here. Proposition 2.7. Assume that d and λ are positive constants, f(x) satisfies (2.4). For any given constant K > λ1 f ∞ and any l  1, let u¯ l (x ) be the unique positive solution of



−du + β u = u( f (x ) − λu ), 0 < x < l, u(0 ) = 0, u(l ) = K.

Then we have

lim u¯ l (x ) = uˆ (x ) uniformly in any compact subset of [0, ∞ ),

l→∞

where uˆ (x ) is the unique positive solution of (2.2). Proposition 2.8. Assume that d and λ are positive constants, f(x) satisfies (2.4). Let 0 < ε 1 and u± ε (x ) be the unique positive solution of



−du + β u = u( f (x ) ± ε − λu ), 0 < x < ∞, u ( 0 ) = 0.

Then

lim u¯ ± ε (x ) = uˆ (x ) uniformly in any compact subset of [0, ∞ ),

ε →0

where uˆ (x ) is the unique positive solution of (2.2). 3. Long time behavior of (u, v ) and spreading–vanishing dichotomy In this section, we are going to study the long time behavior of (u, v ) and obtain a spreading–vanishing dichotomy. Due to Theorem 1.1, x = h(t ) is monotonic increasing. Thus limt→∞ h(t ) = h∞ ∈ (0, ∞]. First we derive an estimate. Its proof is similar to that of Theorem 4.1 of [30]. We omit the details. Theorem 3.1. Let (u, v, h ) be the solution of (1.1). If h∞ < ∞, then there exists a constant C > 0, such that

u(t, · ), v(t, · ) C1 [0,h(t )] ≤ C, ∀t > 1. Moreover,

lim h (t ) = 0.

t→+∞

3.1. Vanishing case To discuss the asymptotic behavior of u and v, we first introduce a general result which is proved by Wang [25]. Let d, μ and g0 be positive constants and C ∈ R. Assume that w0 ∈ C 2 ([0, g0 ] ) and satisfies w0 (0 ) = 0 (or w0 (0 ) = 0), w0 (g0 ) = 0 and w0 (x ) > 0 in (0, h0 ). Let

g ∈ C 1+α /2 ([0, ∞ )),

w ∈ C (1+α )/2,1+α ([0, ∞ ) × [0, g(t )] )

for some α > 0, and satisfy g(t) > 0, w(t, x ) > 0 for all 0 ≤ t < ∞ and 0 < x < g(t). Proposition 3.2 ([26] Proposition 3.1). Under the above conditions, we further suppose that

lim g(t ) = g∞ < ∞,

t→∞

lim g (t ) = 0,

t→∞

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

29

and

w(t, · ) C1 [0,g(t )]  M, ∀ t > 1 for some constant M > 0. If (w, g) satisfies

⎧ wt − dwxx + β wx  w(C − w ), ⎪ ⎪ ⎪ ⎪ ⎨wx = 0(or w = 0 ), w = 0, g (t )  −μwx , ⎪ ⎪ w ( 0, x ) = w0 ( x ), ⎪ ⎪ ⎩ g( 0 ) = g 0 ,

t > 0, 0 < x < g(t ), t > 0, x = 0, t > 0, x = g(t ), x ∈ [0, g0 ],

then

max w(t, x ) = 0.

lim

t→∞ 0xg(t )

In view of Theorem 3.1 and Proposition 3.2 we have the following theorem. Theorem 3.3. Let (u, v, h ) be any solution of (1.1). If h∞ < ∞, then

lim u(t, · ), v(t, · ) C ([0,h(t )]) = 0.

t→∞

(3.1)

Remark 3.1. Theorem 3.3 shows that if both prey and predator cannot spread into the infinity, then they will die out eventually. 3.2. Spreading case √ Theorem 3.4. Assume the h∞ = ∞. If 0 ≤ β < 2 m0 and a(bm1 + c ) < m∞ , then the solution (u(t, x ), v(t, x )) of problem (1.1) satisfies

lim inf u(t, x ) ≥ u(x ), lim sup u(t, x ) ≤ u(x ) uniformly in any compact subset of [0, +∞ ),

(3.2)

lim inf v(t, x ) ≥ v(x ), lim sup v(t, x ) ≤ v(x ) uniformly in any compact subset of [0, +∞ ),

(3.3)

t→∞

t→∞

t→∞

t→∞

where u, v, u and v are given in the proof of Theorem 2.6. Proof. The proof is similar to that of Theorem 3.5 in [25]. Step 1. Define



φ (x ) =

u0 ( x ), 0 ≤ x ≤ h0 , 0, x ≥ h0 ,

and let w(t, x ) be the unique positive solution of

⎧ ⎨wt − wxx + β wx = w(m1 − w ), t > 0, 0 < x < ∞, w(t, 0 ) = 0, t > 0, ⎩ w ( 0, x ) = φ ( x ), x ≥ 0.

By the comparison principle, u(t, x ) ≤ w(t, x ) for all t > 0 and 0 ≤ x ≤ h(t). In view of Theorem 2.5, limt→∞ w(t, x ) = u¯ (x ) uniformly in any compact subset of [0, ∞), where u¯ (x ) is the unique positive solution of (2.20). Note that h(∞ ) = ∞, we get the second limit of (3.2). Step 2. For any given 0 < ε 1 and l  1, there exists a large T, such that

h(t ) > l, u(t, x ) < u¯ + ε , Let

vl (t, x )

∀ 0 ≤ x ≤ l, t ≥ T .

be the unique positive solution of

⎧ ⎨vt − dvxx = v(c − v + b(u¯ + ε )), t > T , 0 < x < l, v(t, 0 ) = 0, v(t, l ) = K, t > T, ⎩ v(T , x ) = K, 0 ≤ x ≤ l,

where K > max{M, c + b(m1 + ε )} and M is given in Theorem 1.1. Since v(t, x ) ≤ M, by the comparison principle we have

v(t, x ) ≤ vl (t, x ), ∀ 0 ≤ x ≤ l, t ≥ T .

(3.4)

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L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

Let vε be the unique positive solution of (2.2) with β = 0, λ = 1 and f (x ) = c + b(u¯ + ε ). and let vl (x ) be the unique positive solution of



−dv = v(c − v + b(u¯ + ε )), v(0 ) = 0, v(l ) = K.

0 < x < l,

(3.5)

Note that vε (x ) < K, the comparison principle asserts

vε (x ) ≤ vl (x ), vε (x ) ≤ vl (t, x ), ∀ t ≥ T , 0 ≤ x ≤ l. Because K is an upper solution of (3.5), it follows that the limit limt→∞ vl (t, x ) exists and is a positive solution of (3.5). By the uniqueness of vl (x ) we have that limt→∞ vl (t, x ) = vl (x ) and this limit holds uniformly for x ∈ [0, l]. In view of (3.4), it yields

lim sup v(t, x ) ≤ vl (x ) uniformly on [0, l].

(3.6)

t→∞

By Proposition 2.2 in [25], limt→∞ vl (x ) = vε (x ) uniformly in any compact subset of [0, ∞). By Proposition 2.3 in [25], limε→0 vε (x ) = v¯ (x ) uniformly in any compact subset of [0, ∞), where v¯ is the unique positive solution of (2.21). These facts and (3.6) imply the second limit of (3.3). Step 3. Since a(bm1 + c ) < m∞ , choose ε 0 > 0 such that a(bm1 + c + ε0 ) < m∞ . For any given 0 < ε < ε 0 and l  1, there exists a large T such that

h(t ) > l,

v(t, x ) < v¯ (t, x ) + ε ,

∀ 0 ≤ x ≤ l, t ≥ T .

Moreover, when l  1, the problem



−u + β u = u[m1 − u − a(bm1 + c + ε )], 0 < x < l, u (0 ) = 0 = u (l )

has a unique positive solution, denoted by u∗l (x ). Note that ux (T, 0) > 0 and u(T, l) > 0, there is a positive constant σ < 1 such that u(T , x ) ≥ σ u∗l (x ) for all 0 ≤ x ≤ l. As v¯ (x ) ≤ bm1 + c, it is easy to see that σ u∗l (x ) is a lower solution of the problem



−u + β u = u[m1 − u − a(v¯ (x ) + ε )], 0 < x < l, u ( 0 ) = 0 = u ( l ).

(3.7)

Let ul (t, x) be the unique solution of

⎧ ⎨ut − uxx + β ux = u[m1 − u − a(v¯ (x ) + ε )], t > T , 0 < x < l, u(t, 0 ) = 0 = u(t, l ), t > T, ⎩ ∗ u ( T , x ) = σ ul ( x ), 0 ≤ x ≤ l.

Then u(t, x) ≥ ul (t, x) for 0 ≤ x ≤ l and t ≥ T, and ul (t, x) is increasing in t. Similar to the above, it can be deduced that the limit limt → ∞ ul (t, x) := u l (x) exists and is the unique positive solution of (3.7). Moreover, such limit holds uniformly on [0, l]. Hence,

lim inf u(t, x ) ≥ ul (x ) uniformly on [0, l]. t→∞

(3.8)

Similar to the argument of Step 2, let l → ∞ firstly and ε → 0 secondly, and apply Propositions 2.7 and 2.8 successively, we can prove that

lim ul (x ) = u(x ) uniformly in any compact subset of [0, ∞ ).

l→∞

(3.9)

This fact combined with (3.8) allows us to derive the first limit of (3.2). Similarly, we can prove the first limit of (3.3). The proof is finished.  4. The criteria governing spreading and vanishing First we introduce two comparison principle. The proof is same to the proof of Lemmas 4.1 and 4.2 in [26], and we omit the proof here.  Lemma 4.1 (Comparison principle). Let h¯ ∈ C 1 ([0, ∞ )) and h¯ (t ) > 0 in [0, ∞). Let u¯ , v¯ ∈ C (D¯ ) C 1,2 (D ), with D := {(t, x ) : t > 0, 0 < x < h¯ (t )}. Assume that (u¯ , v¯ , h¯ ) satisfies

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

⎧ u¯ t − u¯ xx + β u¯ x ≥ u¯ (m1 − u¯ ), ⎪ ⎪ ⎪ ⎪ ⎨v¯ t − dv¯ xx = v¯ (c − v¯ + bu¯ ), u¯ (t, 0 ) ≥ 0, v¯ (t, 0 ) ≥ 0, ⎪ ⎪ ⎪ u¯ (t, h¯ (t )) = v¯ (t, h¯ (t )) = 0, ⎪ ⎩  h¯ (t ) ≥ −μ[u¯ x (t , h¯ (t )) + ρ v¯ x (t , h¯ (t ))],

31

t > 0, 0 < x < h¯ (t ), t > 0, 0 < x < h¯ (t ), t > 0,

(4.1)

t > 0, t > 0.

If h¯ (0 ) ≥ h0 , u¯ (0, x ), v¯ (0, x ) ≥ 0 on [0, h¯ (0 )] and u¯ (0, x ) ≥ u0 (x ), v¯ (0, x ) ≥ v0 (x ) on [0, h0 ]. Then the solution (u, v, h ) of (1.1) satisfies h(t ) ≤ h¯ (t ) on [0, ∞), and u ≤ u¯ , v ≤ v¯ on D, where D = {(t, x ) : t > 0, 0 < x < h(t )}.  Lemma 4.2. (Comparison principle) Let h ∈C1 ([0, ∞)) and h (t) > 0 in [0, ∞). Let v ∈ C (D¯1 ) C 1,2 (D1 ), with D1 := {(t, x ) : t > 0, 0 < x < h(t )}. Assume that (v, h ) satisfies

⎧ t > 0, 0 < x < h(t ), ⎨vt − dvxx = v(c − v ), v(t, 0 ) = v(t , h(t )) = 0, t > 0, ⎩  h (t ) ≥ −μρvx (t , h(t )), t > 0.

(4.2)

If 0 < h (0) ≤ h0 , 0 ≤ v(0, x ) ≤ v0 (x ) on [0, h (0)]. Then the solution (u, v, h ) of (1.1) satisfies h(t) ≥ h (t) on [0, ∞), and v(t, x ) ≥ v(t, x ) on D1 . Let λ1 (l) be the principle eigenvalue of the problem



−φ  + βφ  = λ1 (l )φ , x ∈ (0, l ), φ ( 0 ) = φ ( l ) = 0.

(4.3)

We have the following result. √ Theorem 4.3. Assume 0 ≤ β < 2 m0 . Then λ1 (l) is a strictly decreasing and continuous function in l, and

lim

l→0+

λ1 (l ) = +∞,

lim

l→+∞

λ1 (l ) =

β2 4

< m0 .

Thus, for any x ≥ m0 , there exists a unique constant Lx > 0 such that

λ1 (Lx ) = x and

λ1 (l ) < x for l > Lx ; λ1 (l ) > x for l < Lx . Proof. From the proof of Theorem 2.4.10 in [27], we obtain that λ1 (l) is strictly decreasing and continuous in l. Moveover, the converging results

lim

l→0+

λ1 (l ) = +∞,

lim

l→+∞

λ1 (l ) =

β2 4

< m0

can be derived directly by comparison principle for eigenvalues, we refer to Section 2.4.3 of [27] for further details. √ Theorem 4.4. Assume 0 ≤ β < 2 m0 . If h∞ < ∞, then h∞ ≤ 1 := min{π Theorem 4.3. Thus h0 ≥ 1 implies h∞ = ∞.





d/c, Lm0 }, where Lm0 is determined in

Proof. If h∞ > Lm0 , then there exists ε > 0 such that h∞ > Lm0 −aε . For such ε , there exists T  1 such that h(T ) > Lm0 −aε and

v(t, x )  ε , ∀ t  T , x ∈ [0, h(T )]. Set l = h(T ) and let w = w(t, x ) be the unique positive solution of the following initial boundary value problem with fixed boundary:

⎧ ⎨wt − wxx + β wx = w(m(x ) − w − aε ), t > T , 0 < x < l, w(t, 0 ) = w(t, l ) = 0, t > T, ⎩ w ( T , x ) = u ( T , x ), 0  x  l.

By the comparison principle,

w(t, x )  u(t, x ),

∀t  T , 0  x  l.

Since l > Lm0 −aε , it is well known that w(t, x ) → W (x ) as t → ∞ uniformly in any compact subset of (0, l), where W is the unique positive solution of

32

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36



−W  + β W  = W (m(x ) − aε − W ), W ( 0 ) = W ( l ) = 0.

0 < x < l,

Hence, lim inf t→∞ w (t, x ) = W (x ) > 0 in (0, l). This is a contradiction  to (3.1). If h∞ > π d/c, then there exists T  1 such that h(T ) > π d/c. Set l = h(T ) and let z = z(t, x ) be the unique positive solution of the following initial boundary value problem with fixed boundary:

⎧ ⎨zt = dzxx + z(c − z ), t > T , 0 < x < l, z(t, 0 ) = z(t, l ) = 0, t > T , ⎩ z ( T , x ) = v ( T , x ), 0  x  l.

By the comparison principle,

∀ t  T , 0  x  l.

z(t, x )  v(t, x ), Since l > π



d/c, similarly to the above, we can get a contradiction to (3.1).

Next we discuss the case h0 < 1 . √ Lemma 4.5. Suppose that 0 ≤ β < 2 m0 and h0 < 1 . If

μ≥μ

0

  π 2d



1 d := max 1, v0 ∞ c ρ

c

 

−

h20

h0

2 0



−1 xv0 (x )dx

,

then h∞ = ∞. Proof. We consider the following auxiliary problem

⎧ vt − dvxx = v(c − v ), ⎪ ⎪ ⎨v(t, 0 ) = v(t , h(t )) = 0,  h (t ) = −μρvx (t , h(t )), ⎪ ⎪ ⎩ v ( 0, x ) = v0 ( x ),

t > 0, 0 < x < h(t ), t > 0, t > 0, 0 ≤ x ≤ h ( 0 ) = h0 .

It follows from Lemma 4.2 that

h(t ) ≤ h(t ),

∀ t > 0, 0 < x < h(t ).

v(t , x ) ≤ v(t, x ),

Recall that h0 <  ≤ π



d/c, and μ ≥ μ0 , in view of the Proposition 4.8 in [17], it yields h∞ = ∞. Therefore, h∞ = ∞.



 √ Lemma 4.6. Assume that 0 ≤ β < 2 m0 and h0 < 2 := min{π d/c, Lm1 } ≤ 1 , where Lm1 < Lm0 is also determined in Theorem 4.3. Then there exists μ0 > 0, depending also on u0 and v0 , such that h∞ < ∞ when μ ≤ μ0 . Proof. First let (u¯ , v¯ , h¯ ) be the solution of the problem

⎧ u¯ t − u¯ xx + β u¯ x = u¯ (m1 − u¯ ), ⎪ ⎪ ⎪ ⎪ ⎪ v¯ t − dv¯ xx = v¯ (c − v¯ + bu¯ ), ⎪ ⎪ ⎨u¯ (t, 0 ) ≥ 0, v¯ (t, 0 ) ≥ 0, ⎪u¯ (t, h¯ (t )) = v¯ (t, h¯ (t )) = 0, ⎪ ⎪ ⎪ ⎪h¯  (t ) = −μ[u¯ x (t , h¯ (t )) + ρ v¯ x (t , h¯ (t ))], ⎪ ⎪ ⎩ u¯ (0, x ) = u0 (x ), v¯ (0, x ) = v0 (x ),

t > 0, 0 < x < h¯ (t ), t > 0, 0 < x < h¯ (t ), t > 0,

(4.4)

t > 0, t>0 0 < x < h¯ 0 = h0 .

It follows from Lemma 4.1 that

h(t ) ≤ h¯ (t ),

v(t , x ) ≤ v¯ (t , x ), u(t , x ) ≤ u¯ (t, x )

∀ t > 0, 0 < x < h(t ).

(4.5)

Next, inspired by Du and Lin [8] and Wang [26], we are going to construct a suitable upper solution to (4.4) and then apply Lemma 4.1.  Recall that h0 < min{π d/c, Lm1 } and

λ1 (l ) =

β2 4

+

π2 l2

> m1

for

l < Lm1

by Theorem 4.3. We can verify that exist two positive constants δ , γ 1 such that

1 h0 ( 1 + δ )



π2

h0 ( 1 + δ )



− δγ h0

+

β2 4

− γ − m1 > 0,

(4.6)

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

1 h0 ( 1 + δ )



dπ 2 − δγ h0 h0 ( 1 + δ )

33

 − γ − c > 0.

(4.7)

For such fixed δ and γ , we define



δ

σ (t ) = h0 1 + δ − e



−γ t

2

and β

uˆ (t, x ) = Me 2 x−γ t V

V (y ) = sin(π y ), 0 ≤ y ≤ 1,

, t ≥ 0;

 x   x  , vˆ (t, x ) = bMe−γ t V , t ≥ 0, 0 ≤ x ≤ σ (t ), σ (t ) σ (t )

where M is a positive constants to be chosen later. It is obvious that

uˆ (t, 0 ) = vˆ (t, 0 ) = uˆ (t , σ (t )) = vˆ (t , σ (t )) = 0,

∀ t ≥ 0,

(4.8)

and

∀ 0 ≥ x ≥ σ ( 0 ) = h0 ( 1 + δ /2 ),

uˆ (0, x ) ≥ u0 (x ), vˆ (0, x ) ≥ vˆ 0 (x ),

(4.9)

provided that M  1. Moreover, for such fixed constants δ , γ and M, since σ (t ) ≥ h0 (1 + δ /2 ), it is easy to see that there exists 0 < μ0 1 such that

σ  (t ) + μ(uˆx + ρ vˆ x )|x=σ (t ) = e−γ t

 δγ h

− μM π e

0

2

provided that 0 < μ ≤ μ0 . Direct computation yield

βσ (t ) 2



β



uˆt − uˆxx + β uˆx − uˆ (m1 − uˆ ) = Me 2 x−γ t −γ V − xσ  σ −2V  −

 +β

1  V +

σ

β



≥ Me 2 x−γ t

 = uˆ −γ +

β 2



β

4

+



β2 4



V+

β

− V m1 − Me 2 x−γ t V

V

−γ + 2

1 σ (t )

(1 + bρ )

β2



4

V − xσ  σ −2V  −

> 0,

∀t≥0

β  1  V + 2V σ σ  1

σ

V  − m1V 2

π δγ h0 −γ t π y − m1 − e cos π y σ 2 sin π y σ2 2

(4.10)



 

Since cos π y ≤ 0 for 1/2 ≤ y ≤ 1, and σ (t) is increasing, we have



uˆt − uˆxx + β uˆx − uˆ (m1 − uˆ ) ≥ uˆ −γ +

 ≥ uˆ −γ +

β2 4

β2 4

+ +

π2 − m1 σ2



π2

h20 (1 + δ )2

 − m1

∀ σ (t )/2 ≤ x ≤ σ (t )

> 0,

by (4.6). Remember that 0 ≤ cos π y ≤ 1, y ≤ π2 sin π y for all 0 ≤ y ≤ 1/2, and e−γ t ≤ 1 for all t ≥ 0. We have that for all t > 0 and 0 ≤ x ≤ σ /2,

δγ h0 −γ t π y δγ h0 e cos π y ≤ σ (t ) 2 sin π y σ (t ) It follows that, for all t > 0 and 0 ≤ x ≤



1 2 σ (t ),

 π2 δγ h0 uˆt − uˆxx + β uˆx − uˆ (m1 − uˆ ) ≥ uˆ −γ + + 2 − m1 − 4 σ (t ) σ    π2 β2 1 ≥ uˆ − δγ h0 + − γ − m1 h0 ( 1 + δ ) h0 ( 1 + δ ) 4 > 0, ∀ σ (t )/2 ≤ x ≤ σ (t ) β2

by (4.6). In conclusion, we have

uˆt − uˆxx + β uˆx − uˆ (m1 − uˆ ) > 0,

∀ t > 0, 0 ≤ x ≤ σ (t ).

(4.11)

For vˆ (t, x ), in view of (4.7), similar to the above, we can verify that

vˆ t − dvˆ xx − vˆ (c − vˆ + buˆ ) > 0,

∀ t > 0, 0 ≤ x ≤ σ (t ).

(4.12)

34

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

Notice that (4.8)–(4.12), by virtue of Lemma 4.1, we have σ (t ) ≥ h¯ (t ) ≥ h(t ). Taking t → ∞ we have h∞ ≤ σ (∞ ) = h0 (1 + δ ) < ∞. The proof is complete. 



Theorem 4.7. Suppose that d/c ≤ Lm1 < Lm0 , thus 1 = 2 . Then there exist μ∗ ≥ μ∗ > 0 depending on u0 , v0 and h0 , such that h∞ = ∞ if μ > μ∗ , and h∞ ≤  if μ ≤ μ∗ or μ = μ∗ . Proof. The proof is similar to that of Theorem 3.9 in [8] and Theorem 5.2 in [26].



5. Asymptotic spreading speed In this section, under some suitable conditions, we provide an upper bound for lim supt→+∞ h(tt ) . It shows that the

√ √ asymptotic spreading speed (if exists) cannot be faster that max 2 cd, 2 m1 + β . To show this result, we should consider the traveling wave front of the correspondence system



ut − uxx + β ux = u(m1 − u − av ), t > 0, x ∈ R, vt − dvxx = v(c − v + bu ), t > 0, x ∈ R.

(5.1)

√ √ In view of Theorem 5.17 of [19], for any given s > max{2 cd, 2 m1 + β}, the corresponding traveling wave problem

⎧ ⎨(s − β )φ  + φ  + φ (m1 − φ ) = 0, sψ  + dψ  + ψ (c − ψ + bφ ) = 0, x ∈ R, (φ , ψ )(−∞ ) = (1, b + c ), (φ , ψ )(∞ ) = (0, 0 ), ⎩  φ < 0, ψ  < 0, x ∈ R.

(5.2)

has a solution (φ (ξ ), ψ (ξ )) with ξ = x − st. Further, (φ (ξ ), ψ (ξ )) satisfies

lim

ξ →+∞

φ (ξ )eλ1 ξ = lim ψ (ξ )eλ2 ξ = 1

where

λ1 =

(5.3)

ξ →+∞

(s − β ) +



( s − β )2 − 4m1 2

> 0,

λ2 =

s+

√ s2 − 4cd > 0. 2d

(5.4)

√ √ Theorem 5.1. Let (u, v, h ) be the solution of the problem (1.1) and h∞ = ∞. If for any given s > max 2 cd, 2 m1 + β , the problem (5.2) has a solution (φ (ξ ), ψ (ξ )) satisfying

ψ ( ξ ) ≥ γ φ ( ξ ),

∀ ξ ∈R

(5.5)

for some positive constant γ depending on s. Then we have

lim sup t→∞



 √ h(t ) ≤ max 2 cd, 2 m1 + β . t

√ √ Remark 5.1. Let λ1 and λ2 be given by (5.4). For any given s > max 2 cd, 2 m1 + β , if c s ≤d≤ , s−β m1 we have λ1 ≥ λ2 by carefully computation. In view of (5.3) and the limits:

lim

ξ →−∞

φ ( ξ ) = 1,

lim

ξ →−∞

ψ (ξ ) = b + c,

It can be seen that there exists a positive constant γ such that (5.5) holds. Proof. The idea of this proof comes from [12,26]. First we consider the problem

⎧ u¯ t − u¯ xx + β u¯ x = u¯ (m1 − u¯ ), ⎪ ⎪ ⎪ ⎪ ⎪ v¯ t − dv¯ xx = v¯ (c − v¯ + bu¯ ), ⎪ ⎪ ⎨u¯ (t, 0 ) ≥ 0, v¯ (t, 0 ) ≥ 0, u¯ (t, h¯ (t )) = v¯ (t, h¯ (t )) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ h¯  (t ) = −μ[u¯ x (t , h¯ (t )) + ρ v¯ x (t , h¯ (t ))], ⎪ ⎪ ⎩ u¯ (0, x ) = u0 (x ), v¯ (0, x ) = v0 (x ),

t > 0, 0 < x < h¯ (t ), t > 0, 0 < x < h¯ (t ), t > 0, t > 0,

(5.6)

t>0 0 < x < h¯ 0 = h0 .

It follows from Lemma 4.1 that

h(t ) ≤ h¯ (t ),

v(t , x ) ≤ v¯ (t , x ), u(t , x ) ≤ u¯ (t, x ) ∀ t > 0, 0 < x < h(t ).

(5.7) √ For any given s > max 2 cd, 2 m1 + β , let (φ (ξ ), ψ (ξ )) be the solution of (5.2) satisfying (5.5). Choose g > h  1 such that

√

L. Zhou et al. / Applied Mathematics and Computation 289 (2016) 22–36

35

gγ ≥ 2bh, which implies (g − 1 )γ ≥ 2b(h − 1 ), hφ (ξ ) > u0 ∞ , gψ (ξ ) > v0 ∞ ,

(5.8)

∀ ξ ∈ [0, h0 ].

For such fixed g and h, recall that (φ (ξ ), ψ (ξ )) → 0 and (φ  (ξ ), ψ  (ξ )) → 0 as ξ → ∞, there exists σ 0 > h0 such that



φ (σ0 ) < min

0≤x≤h0



u (x ) , φ (x ) − 0 h

  v0 ( x ) ψ (σ0 ) < min ψ (x ) − ,

(5.9)

g

0≤x≤h0

φ (σ0 ) < 1 − 1/h, ψ (σ0 ) < 2c(g − 1 )/(g + 3g2 ),

(5.10)



−μ hφ  (σ0 ) + gρψ  (σ0 ) < s.

(5.11)

Set σ (t ) = σ0 + st and

u¯ (t, x ) = hφ (x − st ) − hφ (σ0 ),

v¯ (t, x ) = gψ (x − st ) − gψ (σ0 ).

It is clear that

u¯ (t, σ (t )) = v¯ (t, σ (t )) = 0,

∀ t ≥ 0.

Since φ  < 0, ψ  < 0, we see that

u¯ (t, 0 ) > 0, v¯ (t, 0 ) > 0, u¯ x (t, 0 ) < 0, v¯ x (t, 0 ) < 0,

∀ t ≥ 0.

It is deduced from (5.9) that

u¯ (0, x ) > u0 (x ),

v¯ (0, x ) > v0 (x ), ∀0 ≤ x ≤ h0 .

By the first inequality of (5.10),



  2 hψ (σ0 ) h − 1 − hφ (σ0 ) + φ ( σ0 ) ≥ 0. u¯ t − u¯ xx + β u¯ x − u¯ (1 − u¯ ) = h (h − 1 ) φ − h−1

h−1

(5.12)

Applying the second inequality of (5.10), (5.5) and (5.8) we have

1 [v¯ t − dv¯ xx − v(c − v¯ + bu¯ )] g g−1 = 2



ψ

2g − ψ ( σ0 ) g−1

2



g( 1 + 3 g ) + ψ ( σ0 ) c − ψ ( σ0 ) 2 (g − 1 )

 +

1 ψ [ ( g − 1 ) ψ − 2 b( h − 1 ) φ ] 2

1 ψ (σ0 )(gψ (σ0 ) − 2bhφ (σ0 )) + bh(φψ (σ0 ) + φ (σ0 )ψ ) 2 1 1 > ψφ [(g − 1 )γ − 2b(h − 1 )] + ψ (σ0 )φ (σ0 )(gγ − 2bh ) 2 2 ≥ 0.

(5.13)

It follows from (5.11) that

σ  (t ) = s > −μ[hφ  (σ0 ) + ρ gψ (σ0 )] = −μ[u¯ x (t , σ (t )) + ρ v¯ x (t , σ (t ))]. In view of Lemma 4.1, σ (t) ≥ h(t). Therefore

lim sup t→∞

h(t ) ≤ lim t→∞ t

σ (t ) t

= s.

√ √ By the arbitrariness of s > max 2 cd, 2 m1 + β and (5.7), we obtain lim sup t→∞



 √ h(t ) h¯ (t ) ≤ lim sup ≤ max 2 cd, 2 m1 + β . t t t→∞ 

Acknowledgments The work is partially supported by PRC Grant NSFC 11371310 and 11401515, CPSF 2014M551621, and the postdoctoral Research Project of Jiangsu Province no. 1302043C.

36

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