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A free boundary problem of a predator–prey model with higher dimension and heterogeneous environment
Nonlinear Analysis: Real World Applications 16 (2014) 250–263
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A free boundary problem of a predator–prey model with higher dimension and heterogeneous environment✩ Jingfu Zhao a,b , Mingxin Wang a,∗ a
Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, PR China
b
Department of Mathematics, Shaanxi University of Technology, Hanzhong 723000, PR China
article
info
Article history: Received 15 February 2013 Accepted 7 October 2013
abstract This paper is concerned with a free boundary problem for a prey–predator model in higher space dimensions and heterogeneous environment. Such a model may be used to describe the spreading of an invasive or new predator species in which the free boundary represents the spreading front of the predator species and is described by Stefan-like condition. For simplicity, we assume that the environment and solutions are radially symmetric. We prove a spreading–vanishing dichotomy for this model, namely the predator species either successfully spreads to infinity as t → ∞ and survives in the new environment, or it fails to establish and dies out in the long run while the prey species stabilizes at a positive equilibrium state. The criteria for spreading and vanishing are given. © 2013 Elsevier Ltd. All rights reserved.
1. Introduction The spreading of a new or invasive species is one of the most important topics in mathematical ecology. A lot of mathematicians have made efforts to develop various invasion models and investigated them from a viewpoint of mathematical ecology. In this paper we consider a free boundary problem for a prey–predator model in higher space dimensions and heterogeneous environment. In the real world, the following two kind of phenomenon may happen: (i) At the initial state, one kind of prey species (for example, some pest species) occupied the entire space or a large area. In order to control such prey species we put one kind of predator species (natural enemies) in some bounded area or a small area (initial habitat). (ii) There is some kind of species (prey) in the entire space or a large area, and at some time (initial time) another type species (the new or invasive species, predator) enters some bounded area or a small area (initial habitat). In general, the predator has a tendency to emigrate from the boundary to obtain their new habitat, i.e., it will move outward along the unknown curves (free boundaries) as time increases. It is assumed that speeds of free boundaries are proportional to the gradient of predator. We want to realize the dynamics/variations of prey, predator and free boundary. For simplicity, we assume that the environment and solution are radially symmetric. Thus, the model we are concerned here becomes
✩ This work was supported by NSFC Grants 11071049 and 11371113.
∗
Corresponding author. Tel.: +86 15145101503. E-mail address: [email protected] (M. Wang).
1468-1218/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.nonrwa.2013.10.003
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
ut − ∆u = u [a(r ) − b(r )u + c (r )v ] , u(t , r ) ≡ 0, vt − d∆v = v [α(r ) − β(r )u − γ (r )v ] , ur (t , 0) = 0, vr (t , 0) = 0, u(t , h(t )) = 0, h′ (t ) = −µur (t , h(t )), h(0) = h0 , u(0, r ) = u0 (r ), v(0, r ) = v0 (r ),
t t t t t r r
> 0, 0 < r < h(t ), > 0, r ≥ h(t ), > 0, r > 0, > 0, > 0, ∈ [0, h0 ], ∈ [0, ∞),
251
(1.1)
where ∆u = urr + n−r 1 ur , ∆v = vrr + n−r 1 vr (r = |x|, x ∈ Rn , n ≥ 1), d, µ and h0 are given positive constants, functions a, b, c , α, β, γ ∈ C ν0 ([0, ∞)) for some ν0 ∈ (0, 1). The sphere r = h(t ) is the moving boundary to be determined. Throughout this paper we always assume that there are positive constants κ1 ≤ κ2 such that
κ1 ≤ a(r ), b(r ), c (r ), α(r ), β(r ), γ (r ) ≤ κ2 ,
∀r ∈ [0, ∞),
(1.2)
and the initial functions u0 (r ) and v0 (r ) satisfy
u0 ∈ C 2 ([0, h0 ]), u0 (h0 ) = 0, u0 (r ) > 0 v0 ∈ Cb ([0, ∞)), v0 (r ) > 0
in (0, h0 ), in [0, ∞).
(1.3)
The free boundary problem (1.1) describes the spreading of a new or invasive predator with population density u(t , |x|) over an n-dimensional habitat, which is radially symmetric but heterogeneous. The initial function u0 (|x|), which occupies a ball Bh0 , stands for the populations of the new or invasive predator in the very early stage of its introduction, and v0 (|x|), which occupies the whole space, stands for the populations of the prey in the very early stage of the predator introduction. Here and in what follows, we use BR to stand for the ball with center at 0 and radius R. The spreading front is represented by the free boundary |x| = h(t ), which is the sphere ∂ Bh(t ) whose radius h(t ) grows at a speed that is proportional to the population gradient of predator at the front: h′ (t ) = −µur (t , h(t )). The ecological background and derivation of the free boundary conditions can also refer to [1,2]. Recently, Wang and Zhao [3] studied a similar free boundary problem to (1.1) with double free boundaries in which n = 1 and the coefficients a(r ), b(r ), c (r ), α(r ), β(r ) and γ (r ) are all positive constants. Wang [4] discussed some similarly models over the half spatial line with zero Dirichlet boundary condition or zero Neumann boundary condition. They all established the spreading–vanishing dichotomy, long time behavior of solution and sharp criteria for spreading and vanishing. The main purpose of this paper is to show that the most of those results of [3,4] continue to hold in the more realistic situation of higher space dimensions and heterogeneous environment. In the absence of v , the system is reduced to a Stefan problem which was studied by Du and Guo in [5,2]. Free boundary problems similar to the one in [5] have been studied by many authors, please refer to [2,6–10] and references cited therein. These aforementioned works present a new approach to describe the front propagation for population species, which is different from the classical approach of traveling wave solutions of the diffusive logistic equation on the entire space Rn with n ≥ 1 (Refs. [11–13]): ut − d∆u = u(a − bu),
t > 0, x ∈ Rn .
Similar free boundary problems to (1.1) in the one dimension and homogeneous environment case (the coefficients are all positive constants) have been used in two-species models in several earlier papers; see, for example, [14–19] over a bounded spatial interval, and [20] over the half spatial line for the competition model. The well-known Stefan condition has been used in the modeling of a number of applied problems. Except the above cited papers, for example, it was used to describe the melting of ice in contact with water [21], the modeling of oxygen in the muscle [22], the wound healing [23], the tumor growth [24], and so on. There is a vast literature on the Stefan problem, and some important theoretical advances can be found in [25,22] and the references therein. The organization of this paper is as follows. In Section 2, we first give the local existence and uniqueness of the solution to (1.1) and then show that the solution exists for all time t ∈ (0, ∞). Section 3 is devoted to the proof of the spreading–vanishing dichotomy. Our arguments are based on the comparison principle and the construction of suitable upper and lower solutions of (1.1). Section 4 is a brief discussion. 2. Existence and uniqueness In this section, we prove the existence and uniqueness of the solution to (1.1) for all t > 0. The proof of the local existence and uniqueness can be done by modifying the arguments of [5,3]. So we omit the details. Theorem 2.1. For any given (u0 , v0 ) satisfying (1.3) and any α ∈ (0, 1), there is a T > 0 such that problem (1.1) admits a unique solution
(u, v, h) ∈ C
1+α ,1+α 2
α
(D¯ T ) × CT × C 1+ 2 ([0, T ]).
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Moreover,
∥ u∥
1+α ,1+α C 2 (D¯ T )
+ ∥ h∥
C
1+ α 2
([0,T ])
≤ C,
where DT = (t , r ) ∈ R2 : t ∈ (0, T ], r ∈ (0, h(t )) ,
CT = Cb ([0, T ] × [0, ∞))
1+ α ,2+α 2
Cloc
((0, T ] × [0, ∞)),
positive constants C and T only depend on h0 , α , ∥u0 ∥C 2 ([−h0 ,h0 ]) and ∥v0 ∥Cb (R) . To show that the local solution obtained in Theorem 2.1 can be extended to all t > 0, we need the following estimate. Lemma 2.1. Let (u, v, h) be a solution to problem (1.1) defined for t ∈ (0, T ) for some T ∈ (0, ∞]. Then there exist constants M1 , M2 and M3 independent of T such that 0 < u(t , r ) ≤ M1 ,
for 0 < t ≤ T , 0 < r < h(t ),
0 < v(t , r ) ≤ M2 , for 0 < t ≤ T , 0 < r < ∞, 0 < h′ (t ) ≤ M3 , for 0 < t ≤ T . Proof. Using the strong maximum principle, we are easy to see that u > 0 in (0, T ] × (0, h(t )) and v > 0 in (0, T ] × [0, ∞) as long as the solution exists. Since v(t , r ) satisfies
vt − d∆v = v [α(r ) − β(r )u − γ (r )v ] , v(0, r ) = v0 (r ) > 0,
t > 0, 0 < r < ∞, r ≥ 0,
and κ1 ≤ α(r ), γ (r ) ≤ κ2 , we have v(t , r ) ≤ max {∥v0 ∥∞ , κ2 /κ1 } := M2 . Similarly, as u(t , r ) satisfies ut − ∆u = u [a(r ) − b(r )u + c (r )v ] , ur (t , 0) = u(t , h(t )) = 0, u(0, r ) = u0 (r ) > 0,
t > 0, 0 < r < h(t ), t > 0, 0 < r < h0
and κ1 ≤ a(r ), b(r ), c (r ) ≤ κ2 , we also have u(t , r ) ≤ max {∥u0 ∥∞ , (κ2 + κ2 M2 )/κ1 } := M1 . Moreover, the strong maximum principle yields the inequalities ur (t , h(t )) < 0, and therefore h′ (t ) > 0 in (0, T ]. It remains to show that h′ (t ) ≤ M3 for all t ∈ (0, T ) with some M3 independent to T . To this end, let M be a positive constant, define
ΩM := {(t , r ) : 0 < t < T , h(t ) − 1/M < r < h(t )} and construct an auxiliary function
w(t , r ) = M1 [2M (h(t ) − r ) − M 2 (h(t ) − r )2 ]. We will choose M so that w(t , r ) ≥ u(t , r ) holds over ΩM . Direct calculations show that, for (t , r ) ∈ ΩM ,
wt = 2M1 Mh′ (t ) [1 − M (h(t ) − r )] ≥ 0, −wr = 2M1 M [1 − M (h(t ) − r )] ≥ 0, −wrr = 2M1 M 2 , u [a(r ) − b(r )u + c (r )v ] ≤ M1 κ2 (1 + M2 ). It follows that
wt − wrr −
n−1 r
wr ≥ 2M1 M 2 ≥ u [a(r ) − b(r )u + c (r )v ]
in ΩM ,
if M 2 ≥ κ2 (1 + M2 )/2. On the other hand,
w(t , h(t ) − M −1 ) = M1 ≥ u(t , h(t ) − M −1 ),
w(t , h(t )) = 0 = u(t , h(t )).
Note that u0 ( r ) = −
h0
u′0 (s)ds ≤ (h0 − r )∥u′0 ∥C [0,h0 ]
in [h0 − M −1 , h0 ],
r
w(0, r ) = M1 [2M (h0 − r ) − M 2 (h0 − r )2 ] ≥ M1 M (h0 − r ) in [h0 − M −1 , h0 ], we know that if MM1 ≥ ∥u′0 ∥C [0,h0 ] , then u0 (r ) ≤ (h0 − r )∥u′0 ∥C [0,h0 ] ≤ w(0, r ) in [h0 − M −1 , h0 ].
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
253
Let
M = max
κ2 (1 + M2 ) ∥u′0 ∥C [0,h0 ] , 2
M1
.
Applying the maximum principle to w − u over ΩM we can deduce that u(t , r ) ≤ w(t , r ) for (t , r ) ∈ ΩM . It would then follow that ur (t , h(t )) ≥ wr (t , h(t )) = −2M1 M , The proof is complete.
h′ (t ) = −µux (t , h(t )) ≤ 2µM1 M := M3 .
Theorem 2.2. The solution of problem (1.1) exists and is unique for all t ∈ (0, ∞). Proof. Let [0, Tmax ) be the maximal time interval in which the solution exists. By Theorem 2.1, Tmax > 0. It remains to show that Tmax = ∞. Arguing indirectly, we assume that Tmax < ∞. By Lemma 2.1, there exist positive constants M1 , M2 and M3 independent of Tmax such that 0 ≤ u(t , r ) ≤ M1
in [0, Tmax ) × [0, h(t )],
0 ≤ h′ (t ) ≤ M3 ,
0 ≤ h(t ) − h0 ≤ M3 t
0 ≤ v(t , r ) ≤ M2
in [0, Tmax ) × [0, ∞),
in [0, Tmax ).
By the standard Lp estimates and the Sobolev embedding theorem, we can find a constant C > 0 depending only on Mi (i = 1, 2, 3) such that ∥u(t , ·)∥C 1+α/2 ([0,h(t )]) ≤ C and v is continuous for (t , r ) ∈ [0, Tmax ) × [0, ∞). It then follows from the proof of Theorem 2.1 that there exists a τ > 0 depending only on C and Mi (i = 1, 2, 3) such that the solution of problem (1.1) with initial time Tmax − τ /2 can be extended uniquely to the time Tmax − τ /2 + τ . But this contradicts the assumption. The proof is now complete. 3. The spreading and vanishing In this section, we prove the spreading–vanishing dichotomy. Though our approach here mainly follows the lines of [3], considerable changes in the proofs are needed. Since the situation here is more general and difficult, it follows from Lemma 2.1 that r = h(t ) is monotonic increasing, therefore, there exists h∞ ∈ (0, ∞] such that limt →∞ h(t ) = h∞ . Let λ1 (d, p, R) be the principal eigenvalue of the problem
−d∆φ = λp(|x|)φ, φ = 0,
x ∈ BR , x ∈ ∂ BR ,
where d > 0 is a constant, p(r ) > κ1 /2 for r ∈ [0, ∞) and p ∈ C 1 ([0, ∞)). It is well known that λ1 (d, p, R) is a strictly decreasing and continuous function in R, and limR→0+ λ1 (d, p, R) = +∞, limR→+∞ λ1 (d, p, R) = 0 (cf. Section 2.2 of [26]). Therefore, for fixed d > 0 and p ∈ C 1 ([0, ∞)), there is a unique R∗ (d, p) such that
λ1 d, p, R∗ (d, p) = 1 and
λ1 (d, p, R) < 1 for R > R∗ (d, p);
λ1 (d, p, R) > 1 for R < R∗ (d, p).
Since λ1 (d, p, R) is a strictly decreasing continuous function in p and R, we have that R∗ (d, p) is a strictly decreasing continuous function in p. In order to investigate asymptotic properties of solutions for (1.1), we will derive a priori estimate for any global solution. Lemma 3.1. Let (u, v, h) be any solution of (1.1). If h∞ < ∞, then there exists a constant K > 0, such that
∥u(t , ·)∥C 1 ([0,h(t )]) = ∥u(t , ·)∥C 1 (Bh(t ) ) < K ,
∀t ≥ 1.
(3.1)
Moreover, lim h′ (t ) = 0.
(3.2)
t →∞
Proof. We introduce new transformation
(t , x) → (t , y),
with x = h(t )y, y ∈ Rn ,
and functions w(t , y) and z (t , y) with the following definitions
w(t , y) = u(t , h(t )|y|),
z (t , y) = v(t , h(t )|y|).
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Clearly, w and z are radially symmetric and satisfy the following initial boundary value problem in the domain B1 with fixed boundary ∂ B1 :
n wt = ϕ(t )∆w + ψi (t , y)wyi + w [a(h(t )|y|) − b(h(t )|y|)w + c (h(t )|y|)z] , i=1
w(t , y) = 0, w(0, y) = u0 (h0 |y|),
t > 0, |y| < 1, (3.3)
t > 0, |y| = 1, |y| ≤ 1,
where
ϕ(t ) =
1 h(t )
2
,
ψi (t , y) =
h′ (t )yi h( t )
.
By Proposition A, we can obtain that there exists a positive constant K0 such that
∥w∥
1+α ,1+α C 2 ([1,∞)×B1 )
< K0 .
(3.4)
It follows from uxi (t , |x|) = h(1t ) wyi (t , y) and (3.4) that there exists a constant K such that (3.1) holds. n ′ ′ Now we prove (3.2). Note ur (t , h(t )) = h(1t ) i=1 wyi (t , y0 )yi and h (t ) = −µur (t , h(t )), by virtue of 0 < h (t ) < M3 α and ∥wyi (·, y)∥ 2 < K0 for |y| = 1 and i = 1, . . . , n, we can obtain C
′
∥h ∥
([1,∞))
α C 2 ([1,∞))
< L,
(3.5)
where L depends on K0 and M3 . (3.2) can be deduced from h′ (t ) > 0, h∞ < ∞ and (3.5). The proof is complete.
Theorem 3.1. Let (u, v, h) be any solution of (1.1). If h∞ < ∞, then lim ∥u(t , ·)∥C ([0,h(t )]) = 0,
(3.6)
lim v(t , r ) = V (r ) uniformly in any bounded subset of [0, ∞),
(3.7)
t →∞
t →∞
here V (|x|) is the unique positive (radial) solution of the equation
− d∆v = v[α(|x|) − γ (|x|)v],
x ∈ Rn .
(3.8)
This result shows that if the predator cannot spread into the whole space, then it will die out eventually.
∞
Proof. Step 1: Proof of (3.6). On the contrary we assume that there exist σ > 0 and (tj , xj ) j=1 , with xj ∈ Bh(tj ) and tj → ∞ as j → ∞, such that
u(tj , |xj |) ≥ σ ,
j = 1, 2, . . . .
(3.9)
Since xj ∈ Bh∞ , there is a subsequence, noted by itself, and x0 ∈ Bh∞ , such that xj → x0 as j → ∞. If |x0 | = h∞ , then we j
have ||xj | − h(tj )| → 0 as j → ∞. There exists x¯ j in the line connecting xj and h(tj ) |xxj | for every j such that ur (tj , |¯xj |) =
u(tj , |xj |) − u(tj , h(tj ))
|xj | − h(tj )
→ ∞ as j → ∞.
This is a contradiction to (3.1). So |x0 | ̸= h∞ holds, and hence x0 ∈ Bh∞ . By the use of (3.1) and (3.9), there exists δ > 0 such that Bδ (x0 ) ⊂ Bh∞ and u(tj , |x|) ≥ σ /2,
∀x ∈ Bδ (x0 )
for all large enough j. Since h(tj ) → h∞ as j → ∞, without loss of generality we may think that h(tj ) > |x0 | + δ for all large j. Let gj (t ) = δ + t − tj and tj∗ = inf t > tj : h(t ) = gj (t ) + |x0 | ,
j
x0 . |x0 |
x0 = |x0 | + gj (tj∗ )
It is easy to obtain that tj∗ < h∞ + tj − δ − |x0 |. This implies gj (t ) + |x0 | ≤ h(t ),
t ∈ [tj , tj∗ ].
(3.10) j
Moreover, ∂ Bh(t ∗ ) and ∂ Bgj (t ∗ ) (x0 ) tangent at the point x0 . j
j
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Define yj (t , x) = (π − ε) uj (t , x) =
σ
|x−x0 | gj (t )
and
e−k(t −tj ) cos yj (t , x) + cos ε ,
4
255
( t , x ) ∈ Ω tj ,
where ε (ε < π /8) and k are positive constants to be chosen later, and
Ωtj = {(t , x) : tj < t < tj∗ , |x − x0 | < gj (t )}. It is obvious that uj (t , x) = 0 for |x − x0 | = gj (t ), and |yj (t , x)| ≤ π − ε for (t , x) ∈ Ω tj , the latter implies uj (t , x) ≥ 0 in Ωtj . We want to compare u(t , |x|) and uj (t , x) in Ω tj . Thanks to (3.10), it follows that u(t , |x|) ≥ 0 = uj (t , x),
t ∈ [tj , tj∗ ], x ∈ ∂ Bgj (t ) (x0 ).
On the other hand, it is obvious that u(tj , |x|) ≥
σ 2
≥ uj (tj , x),
x ∈ Bδ (x0 ).
Thus, if the positive constants ε and k can be chosen independent of j such that ujt − ∆uj − uj a(|x|) − b(|x|)uj ≤ 0,
( t , x ) ∈ Ω tj ,
(3.11)
it can be deduced that uj (t , |x|) ≤ u(t , x) for (t , x) ∈ Ωtj by applying the maximum principle to u − uj over Ωtj . Since j
j
u(tj∗ , h(tj∗ )) = 0 = uj (tj∗ , x0 ) and h(tj∗ ) = |x0 |, it follows that
∂ uj ∂ν
(tj∗ , xj0 ) ≥
∂u ∗ j (t , |x0 |) = ur (tj∗ , h(tj∗ )). ∂ν j
The direct computation gives
∂ uj ∂ν
(tj∗ , xj0 ) = −
7σ π −kh∞ σ (π − ε) −k(tj∗ −tj ) e sin(π − ε) ≤ − e sin ε. ∗ 4(δ + tj − tj ) 32h∞
7σ π −kh∞ e Note the boundary condition −µur (tj∗ , h(tj∗ )) = h′ (tj∗ ), we have h′ (tj∗ ) ≥ µ 32h sin ε . This implies lim supt →∞ ∞
h′ (t ) > 0 since limj→∞ tj∗ → ∞. This is a contradiction to (3.2). So limt →∞ ∥u(t , ·)∥C [0,h(t )] = 0 is obtained. We shall prove that if ε and k satisfy
ε< k≥
π 8
,
sin ε ≤
9δ 2 π 8π h2∞ (3h∞ + 4(n − 1))
,
(3.12)
π (h∞ + (n − 1)π ) σ κ2 π 2 + + , δ 2 (cos ε − cos 2ε) 2 δ
(3.13)
then (3.11) holds. Direct computations yield ujt − ∆uj − uj a − buj = −kuj −
≤
2
≤
σ κ2 σ κ2
σ 4
+
e
−k(t −tj )
n
yjt sin yj +
y2jxi − k uj −
i =1
σ 4
σ
4
e
−k(t −tj )
sin yj ∆yj + cos yj
e−k(t −tj ) cos ε
δ
y2jxi −
i =1
−k(t −tj )
π −ε
2
cos ε
By (3.13), one has σ b1 /2 + (π /δ)2 − k < 0. 2 Since 0 ≤ yj ≤ π − ε when (t , x) ∈ Ω tj , we can decompose Ωtj = Ωt1j Ωtj , where
Ωt1j = (t , x) ∈ Ωtj : tj < t < tj∗ , π − 2ε < yj (t , x) < π − ε , Ωt2j = (t , x) ∈ Ωtj : tj < t < tj∗ , 0 ≤ yj (t , x) ≤ π − 2ε .
n
y2jxi
− uj a − buj
i =1 n
− k uj − e 4 h∞ σ π −k(t −tj ) |x − x0 | n−1 e sin yj + + 4δ δ |x − x0 | , I. 2
+
π 2
σ
σ 4
e−k(t −tj ) sin yj (yjt − ∆yj )
256
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
It is obvious that sin yj ≤ sin 2ε when (t , x) ∈ Ωt1j , and cos yj ≥ − cos 2ε when (t , x) ∈ Ωt2j . In virtue of ε < π /8, we can obtain
|x − x0 | n−1 3h∞ + 4(n − 1) + ≤ δ |x − x0 | 3δ when (t , x) ∈ Ωt1j , and
|x − x0 | n−1 + δ | x − x0 |
sin yj when (t , x) ∈ I ≤
σ 4
Ωt2j .
≤
h∞ + (n − 1)π
δ
Note that uj (t , x) ≥ 0 in Ωtj , in view of (3.12) and (3.13), we conclude
−k(t −tj )
3π 2
−
e
4h2∞
π (3h∞ + 4(n − 1)) cos ε + sin 2ε < 0 3δ 2
when (t , x) ∈ Ωt1j , and I ≤
σ 4
−k(t −tj )
e
σ κ2 π 2 π (h∞ + (n − 1)π ) −k + + (cos ε − cos 2ε) + <0 2 δ δ2
when (t , x) ∈ Ωt2j . Therefore, (3.11) holds. Step 2: Proof of (3.7). The existence and uniqueness of a positive solution of (3.8) follows from Theorem 2.3 of [27]. It must be radially symmetric since (3.8) is invariant under rotations around the origin of Rn . To show (3.7), we use a squeezing argument introduced in [28]. We first consider the Dirichlet problem
−d∆z = z [α(r )(1 − ε) − γ (r )z ],
|x| < R;
z (R) = 0,
and the boundary blow-up problem
−d∆w = [α(r ) − γ (r )w],
|x| < R;
w(R) = +∞.
When ε is small and R is large, it is well known that these problems have positive radial solutions zRε and wR , respectively. By the comparison principle given in [28], as ε → 0+ and R → +∞, zRε increases to the unique positive solution V of (3.8) and wR decreases to V . Choose a decreasing sequence {εi } and an increasing sequence {Ri } such that εi → 0+ , Ri → +∞ as i → ∞, and κ α(r ) 1 > λ1 (d, α(1 − εi ), Ri ) for all i. In view of (3.6), for each εi , there exists Ti > 0 such that u(t , r ) < κ1 εi ≤ β(r ) εi for all 2 t ≥ Ti and r ∈ [0, ∞). For each i, by comparison principle we have
v(t , r ) ≥ zi (t , r ),
(t , r ) ∈ [Ti , ∞) × [0, Ri ],
where zi (t , r ) is the solution of the following logistic problem with fixed boundary zt − d∆z = z (α(r )(1 − εi ) − γ (r )v), zr (t , 0) = 0, z (t , Ri ) = 0, z (Ti , r ) = v(Ti , r ),
t > Ti , r ∈ [0, Ri ], t > Ti , r ∈ [0, Ri ].
(3.14)
Since 1 > λ1 (d, α(1 − εi ), Ri ), it is well known (see, for example, Proposition 3.3 in [26]) that (3.14) admits a unique positive ε solution zi (t , r ). Moreover, limt →∞ zi (t , r ) = zRii (r ) uniformly for r ∈ [0, Ri ]. It follows that ε
lim inf v(t , x) ≥ zRii (r ) uniformly for r ∈ [0, Ri ]. t →∞
Sending i → ∞, we obtain lim inf v(t , x) ≥ V (r ) uniformly in any compact subset of [0, ∞). t →∞
(3.15)
Analogously, by arguments similar to those in the proof of Theorem 4.1 of [28], we see that lim sup v(t , x) ≤ wRi (r ) uniformly for r ∈ [0, Ri ], t →∞
which implies (by sending i → ∞) lim sup v(t , x) ≤ V (r ) uniformly in any compact subset of [0, ∞). t →∞
From (3.15) and (3.16) we see that (3.7) holds.
(3.16)
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
257
The symbol V defined by (3.8) will be always used in the sequel. In the following theorem, we will give a sufficient condition of spreading. Theorem 3.2. Let (u, v, h) be any solution of (1.1). If h∞ < ∞, then h∞ ≤ R∗ (1, a + cV ).
(3.17)
Proof. Theorem 3.1 implies that, if h∞ < ∞, then (3.6) and (3.7) hold. We assume h∞ > R∗ (1, a + cV ) to get a contradiction. Note that (3.7) and R∗ (d, p) is a strictly decreasing continuous function in p, it is easy to see that for any given 0 < ε ≪ 1 there exists τ ≫ 1 such that h(τ ) > max h0 , R∗ (1, a + cV − ε) ,
v(t , r ) ≥ V (r ) −
ε , c (r )
∀t ≥ τ , r ∈ [0, h∞ ].
Set R = h(τ ), then R > R∗ (1, a + cV − ε). Let w = w(t , r ) be the positive solution of the following initial boundary value problem with fixed boundary:
wt = ∆w + w [a(r ) − b(r )w + c (r )V (r ) − ε ] , wr (t , 0) = w(t , R) = 0, w(τ , r ) = u(τ , r ),
t > τ , 0 < r < R, t > τ, 0 ≤ r ≤ R.
By the comparison principle,
w(t , r ) ≤ u(t , r ),
∀t ≥ τ , 0 ≤ r ≤ R.
Since
λ1 (1, a + cV − ε, R) < λ1 1, a + cV − ε, R∗ (1, a + cV − ε) = 1, it is well known that w(t , r ) → w ∗ (r ) as t → ∞ uniformly for r ∈ [0, R], where w ∗ is the unique positive solution of
∆w ∗ + w ∗ a(r ) + c (r )V − ε − b(r )w ∗ = 0, wr∗ (0) = w∗ (R) = 0.
r ∈ (0, R),
Hence, lim inft →∞ u(t , r ) ≥ limt →∞ w(t , r ) = w ∗ (x) > 0 in [0, R]. This is a contradiction to the fact that (3.6). Therefore, (3.17) holds. We now present some comparison principles which can be used in the following to estimate the solution (u(t , x), v(t , x)) and the free boundary r = h(t ). The following Lemmas 3.2 and 3.3 can be proved as in Lemma 3.1 of [3]. We omit the details. ∗
Lemma 3.2 (Comparison Principle). Suppose that T ∈ (0, ∞), h¯ ∈ C 1 ([0, T ]), u¯ ∈ C (DT ) C 1,2 (D∗T ) with D∗T = {(t , x) ∈ R2 : 0 < t ≤ T , 0 < r < h¯ (t )}, v¯ ∈ C 1,2 (GT ) with GT = (0, T ] × [0, ∞). Assume further that (¯u, v¯ , h¯ ) satisfies
u¯ − ∆u¯ ≥ u¯ [a(r ) − b(r )¯u + c (r )¯v ] , t v¯ t − d∆v¯ ≥ v¯ (α(r ) − γ (r )¯v ), u¯ r (t , 0) = 0, v¯ r (t′ , 0) = 0, u¯ (t , h¯ (t )) = 0, h¯ (t ) ≥ −µ¯ur (t , h¯ (t )),
t t t t
> 0, 0 < r < h¯ (t ), > 0, 0 < r < ∞, > 0, > 0,
(3.18)
and h¯ (0) ≥ h0 ,
and u¯ (0, r ) ≥ 0 in [0, h¯ (0)],
u0 (r ) ≤ u¯ (0, r ) in [0, h0 ],
v0 (r ) ≤ v¯ (0, r ) in [0, ∞).
Then the solution (u, v, h) of the free boundary problem (1.1) satisfies u(t , r ) ≤ u¯ (t , r ) on DT ,
v(t , r ) ≤ v¯ (t , r ) on GT ,
h(t ) ≤ h¯ (t ) on [0, T ],
where DT as in Theorem 2.1. Lemma 3.3 (Comparison Principle). Let T ∈ (0, ∞), h ∈ C 1 ([0, T ]) with h(t ) > 0 for all t ∈ [0, T ], and u ∈ C (OT ) with OT = {(t , r ) ∈ R2 : 0 < t ≤ T , 0 < r < h(t )}. Suppose that (u, h) satisfies
ut − ∆u ≤ u a(r ) − b(r )u , u (t , 0) = u(t , h(t )) = 0, h′r(t ) ≤ −µu (t , h(t )), r
t > 0, 0 < r < h(t ), t > 0, t > 0,
C 1,2 (OT )
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
and h(0) ≤ h0 ,
and
u0 (r ) ≥ u(0, r ) in [0, h(0)].
Then the solution (u, v, h) of the free boundary problem (1.1) satisfies u(t , r ) ≥ u(t , r ) on OT ,
h(t ) ≥ h(t )
on [0, T ].
The pair (¯u, v¯ , h¯ ) in Lemma 3.2 is called an upper solution of the problem (1.1). Similarly, the pair (u, 0, h) in Lemma 3.3 is called a lower solution of the problem (1.1). Lemma 3.4. If h∞ = ∞, then U1 (r ) ≤ lim inf u(t , r ) ≤ lim sup u(t , r ) ≤ U2 (r ) t →∞
t →∞
uniformly in any compact subset of [0, ∞), where U1 (r ) and U2 (r ) are the unique positive (radial) solutions of the equations
− ∆u = u [a(|x|) − b(|x|)u] ,
x ∈ Rn
(3.19)
and
−∆u = u [a(|x|) − b(|x|)u + c (|x|)V ] ,
x ∈ Rn
respectively. Proof. Since h∞ = ∞, there exists T > 0 such that h(T ) > R∗ (1, a). Choose a function u0 (r ) satisfying u0 ∈ C 2 [0, h(T )], u0 (r ) ≤ u0 (T , r ) in [0, h(T )], u0 (r ) > 0 in (0, h(T )) and u0 (h(T )) = 0. Consider the following problem
ut − ∆u = u a(r ) − b(r )u , ur (t , 0) = u(t , h(t )) = 0, h′ (t ) = −µur (t , h(t )), u(T , r ) = u0 (r ), h(T ) = h(T ).
t > T , 0 < r < h(t ), t > T, t > T, 0 ≤ r ≤ h(T ),
By Theorem 2.1 of [5], this problem has a unique solution (u, h) and exists for all t ≥ T . In view of Lemma 3.4, we have u(t , r ) ≥ u(t , r ) for t ≥ T , 0 ≤ r ≤ h(t );
h(t ) ≥ h(t ) for [T , ∞).
(3.20)
Using Theorem 2.5 of [5] firstly and Lemma 2.3 of [5] secondly, we can see that lim inf u(t , r ) = U1 (r ) uniformly in any compact subset of [0, ∞).
(3.21)
t →∞
It follows (3.20) and (3.21) that lim inf u(t , r ) ≥ U1 (r ) uniformly in any compact subset of [0, ∞). t →∞
On the other hand, we define H =
κ2 ∥v0 ∥∞ , κ1
θ=
κ12 , κ2
v¯ (t , r ) = (1 + He−θ t )V (r ).
(3.22) κ
κ
Because of V (r ) satisfying (3.8) and κ1 ≤ α(r ), γ (r ) ≤ κ2 , by the comparison principle we have κ1 ≤ V (r ) ≤ κ2 for 2 1 r ∈ [0, ∞). Direct calculations yield
v¯ t − d∆v¯ − v¯ [α(r ) − γ (r )¯v ] = He−θ t V (r )[−θ + (1 + He−θ t )γ (r )V (r )] κ2 ≥ He−θ t V (r ) −θ + 1 κ2 = 0, and v¯ (0, r ) = (1 + H )V (r ) > ∥v0 ∥∞ ≥ v0 (r ). Since limt →∞ v¯ (t , r ) = V (r ) uniformly in [0, ∞), for any given ε > 0 small, there exists Tε > T , such that
v¯ (t , r ) ≤ V (r )(1 + ε),
∀t ≥ Tε , r ∈ [0, ∞).
Consider the auxiliary problem
u¯ − ∆u¯ = u¯ [a(r ) − b(r )¯u + c (r )V (r )(1 + ε)] , t u¯ r (t , 0) = 0, u¯ (t , h¯ (t )) = 0, h¯ ′ (t ) = −µ¯ur (t , h¯ (t )), u¯ (Tε , r ) = u(Tε , r ), h¯ (Tε ) = h(Tε ),
t > Tε , 0 < r < h¯ (t ), t > Tε , t > Tε , 0 ≤ r ≤ h(Tε ).
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
259
It is easy to deduce that (¯u, V (r )(1 + ε), h¯ ) satisfies (3.18) for t > Tε . By Lemma 3.2, h¯ (t ) ≥ h(t ),
u¯ (t , r ) ≥ u(t , r ),
∀t > Tε , 0 < r < h(t ).
Hence, h¯ (∞) = ∞. Using Lemma 2.3 of [5], we easily see that lim sup u(t , r ) ≤ lim u¯ (t , r ) = U ε (r ) t →∞
t →∞
uniformly for r in any compact subset of [0, ∞), where U ε (r ) is the unique positive solution of the equation
−∆u = u [a(|x|) − b(|x|)u + c (|x|)V (|x|)(1 + ε)] ,
x ∈ Rn .
Note that U ε (r ) is continuously dependent on ε and the arbitrariness of ε , we have lim sup u(t , r ) ≤ U2 (r ) t →∞
uniformly in the compact subset of [0, ∞). The proof is completed.
Combining Theorems 3.1 and 3.2 and Lemma 3.4, we have the following dichotomy theorem. Theorem 3.3. Let (u, v, h) be any solution of (1.1). Then, the following alternative holds: Either (i) spreading: h∞ = ∞ and U1 (r ) ≤ lim inf u(t , r ) ≤ lim sup u(t , r ) ≤ U2 (r ) t →∞
t →∞
uniformly in any compact subset of [0, ∞); or (ii) vanishing: h∞ < R∗ (1, a + cV ) and limt →∞ ∥u(t , ·)∥C [0,h(t )] = 0. Now we decide exactly when each of the two alternative occurs. The discussion will be divided into two cases: (a) h0 ≥ R∗ (1, a + cV ),
(b) h0 < R∗ (1, a + cV ).
For the case (a), due to h′ (t ) > 0 for t > 0, we must have h∞ > R∗ (1, a + cV ). Hence Theorem 3.2 and Lemma 3.4 imply the following result. Theorem 3.4. If h0 ≥ R∗ (1, a + cV ), then h∞ = ∞. Next we discuss the case (b). Lemma 3.5. Suppose h0 < R∗ (1, a + cV ). Then there exists µ0 > 0 depending on u0 such that h∞ = ∞ if µ ≥ µ0 . Proof. Consider the following auxiliary problem
u − ∆u = u a(r ) − b(r )u , t ur (t , 0) = u(t , h(t )) = 0, ′ h (t ) = −µur (t , h(t )), u(0, r ) = u0 (r ),
t > 0, 0 < r < h(t ), t > 0, t > 0, 0 ≤ r ≤ h0
with h(0) = h0 . By Lemma 3.3, h(t ) ≤ h(t ),
u(t , r ) ≤ u(t , r ),
∀t > 0, 0 < r < h(t ).
Note that h0 < R (1, a + cV ) < R (1, a) and µ ≥ µ0 , by Lemma 2.8 of [5], we have h(∞) = ∞. Therefore, h∞ = ∞. The proof is finished. ∗
∗
Lemma 3.6. Suppose h0 < R∗ (1, a + cV ), then there exists µ0 > 0 depending on (u0 , v0 ) such that h∞ < ∞ if µ ≤ µ0 . Proof. We are going to construct a suitable upper solution to (1.1) and then apply Lemma 3.2. We define v¯ (t , r ) as in (3.22), then v¯ (t , r ) satisfies
v¯ t − d∆v¯ − v¯ [α(r ) − γ (r )¯v ] ≥ 0, v¯ (0, r ) ≥ v0 (r ),
t > 0, 0 < r < ∞, 0 < r < ∞.
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Inspired by Du and Guo [5] and Wang and Zhao [3], for t > 0 and r ∈ [0, η(t )], we define
t H κ22 −θ s −ρ + f (t ) = M exp e ds , t > 0; κ1 0 1/2 t δ2 2 f (s)ds , t > 0; η(t ) = h0 (1 + δ) + ∞ f (s)ds 0 0 h0 r u¯ (t , r ) = f (t )W , t > 0, 0 ≤ r ≤ η(t ), η(t ) where H , θ defined as in (3.22), ρ, δ and M are positive constants to be chosen later and W (|x|) is the first eigenfunction of the problem
−∆W = λ1 (1, a + cV , h0 )[a(r ) + c (r )V (r )]W , W = 0,
x ∈ Bh0 , x ∈ ∂ Bh0 ,
with W > 0 in Bh0 and ∥W ∥∞ = 1. Since h0 < R∗ (1, a + cV ), we have λ1 (1, a + cV , h0 ) > 1. We also observe that W ′ (0) = 0 and
−(r n−1 W ′ )′ = r n−1 λ1 (1, a + cV , h0 )[a(r ) + c (r )V (r )]W > 0,
∀ 0 < r < h0 .
It follows that W ′ (r ) < 0,
∀ 0 < r ≤ h0 .
On the other hand, we have η′ (t ) > 0 for t ≥ 0 and f ′ (t ) f (t )
= −ρ +
H κ22 −θ t e ,
κ1
∀t > 0.
Hκ2
t
Note that −ρ + κ 2 e−θ t < 0 for t large enough, it is deduced that 0 f (s)ds is uniformly bounded in [0, ∞). 1 Set σ (t ) = η(t )/h0 , then η(t ) = h0 σ (t ). Direct computations yield u¯ t − ∆u¯ − u¯ [a(r ) − b(r )¯u + c (r )¯v ] = f ′ W − fr σ −2 σ ′ W ′ + f
≥ fW
r
− fW [a(r ) − b(r )fW + c (r )¯v ] σ r r f′ λ1 (1, a + cV , h0 ) r + a + c V f σ2 σ σ σ −∆W
− [a(r ) + c (r )(1 + He−θ t )V (r )] −ρ+
≥ fW
H κ22 −θ t e − He−θ t c (r )V (r )
κ1
λ1 (1, a + cV , h0 ) a σr + − 1 a( r ) (1 + 2δ)2 a(r ) λ1 (1, a + cV , h0 ) c σr V σr + − 1 c (r )V (r ) . (1 + 2δ)2 c (r )V (r ) We easily see that h0 (1 + δ) ≤ η(t ) ≤ h0 (1 + 2δ),
1 + δ ≤ σ (t ) ≤ 1 + 2δ.
Hence, due to λ1 (1, a + cV , h0 ) > 1, we can choose δ > 0 sufficiently small such that
λ1 (1, a + cV , h0 ) a σr − 1 > 0, ρ1 := min t >0,r ∈[0,η(t )] (1 + 2δ)2 a( r ) r r λ1 (1, a + cV , h0 ) c σ V σ ρ2 := min − 1 > 0. t >0,r ∈[0,η(t )] (1 + 2δ)2 c (r )V (r ) Setting ρ = ρ1 κ1 + ρ2 κ12 /κ2 , we deduce u¯ t − ∆u¯ − u¯ [a(r ) − b(r )¯u + c (r )¯v ] ≥ 0,
t > 0, r ∈ [0, η(t )].
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
261
We now choose M > 0 sufficiently large such that u0 (r ) ≤ MW
r 1+δ
= u¯ (0, r ),
∀r ∈ [0, h0 ].
The direct calculation yields
η′ (t ) =
h20 δ 2 f (t ) 2η(t )
∞ 0
f (s)ds
,
−µ¯ur (t , η(t )) =
µh0 f (t )|Wr (h0 )| . η(t )
Therefore, if we take
µ0 =
h0 δ 2 2|Wr (h0 )|
∞ 0
f (s)ds
,
then for any 0 < µ ≤ µ0 , there holds η′ (t ) ≥ −µ¯ur (t , η(t )). Thus, (¯u, v¯ , η) satisfies
u¯ t − ∆u¯ ≥ u¯ [a(r ) − b(r )¯u + c (r )¯v ] , v¯ t − d∆v¯ ≥ v¯ [α(r ) − γ (r )¯v ] , u¯ r (t , 0) = 0, v¯ r (t , 0) = 0, ′ u¯ (t , η(t )) = 0, η (t ) ≥ −µ¯ur (t , η(t )), η(0) = (1 + δ)h0 .
t t t t
> 0, 0 < r < η(t ), > 0, 0 < r < ∞, > 0, > 0,
We can apply Lemma 3.2 to conclude that h(t ) ≤ η(t ), u(t , r ) ≤ u¯ (t , r ) for (t , r ) ∈ D∞ and v(t , r ) ≤ v¯ (t , r ) for (t , r ) ∈ G∞ . Therefore, h∞ ≤ limt →∞ η(t ) ≤ h0 (1 + 2δ) < ∞. Theorem 3.5. Suppose h0 < R∗ (1, a + cV ). Then there exist µ∗ ≥ µ∗ > 0 depending on u0 (r ), v0 (r ) and h0 such that h∞ ≤ R∗ (1, a + cV ) if µ ≤ µ∗ , and h∞ = ∞ if µ > µ∗ . Proof. The proof is similar to that of Theorem 5.2 of [3] and Theorem 4.11 of [9]. For the convenience to the readers we shall give the details. We will write (uµ , vµ , hµ ) in place of (u, v, h) to clarify the dependence of the solution of (1.1) on µ. Set
Σ ∗ = µ > 0 : hµ,∞ ≤ R∗ (1, a + cV ) . By Lemma 3.6 and Theorem 3.2, (0, µ0 ] ⊂ Σ ∗ . In view of Lemma 3.5, Σ ∗ [µ0 , ∞) = ∅. Therefore, µ∗ := sup Σ ∗ ∈ [µ0 , µ0 ]. By this definition and Theorem 3.2 we find that hµ,∞ = ∞ when µ > µ∗ . We will show that µ∗ ∈ Σ ∗ . Otherwise, hµ∗ ,∞ = ∞. We can find T > 0 such that hµ∗ (T ) > R∗ (1, a + cV ). By the continuous dependence of (uµ , vµ , hµ ) on µ, there is ε > 0 such that hµ (T ) > R∗ (1, a + cV ) for µ ∈ (µ∗ − ε, µ∗ + ε). It follows that for all such µ,
lim hµ (t ) > hµ (T ) > R∗ (1, a + cV ).
t →∞
This implies that [µ∗ − ϵ, µ∗ + ϵ] ∩ Σ ∗ = ∅, and sup Σ ∗ ≤ µ∗ − ϵ . This contradicts the definition of µ∗ . Define
Σ∗ = ν : ν ≥ µ0 such that hµ,∞ ≤ R∗ (1, a + cV ) for all µ ≤ ν , where µ0 is given by Lemma 3.6. Then µ∗ := sup Σ∗ ≤ µ∗ and (0, µ∗ ) ⊂ Σ∗ . Similar to the above, we can prove that µ∗ ∈ Σ∗ . The proof is completed. We should remark before ending this section that, if the coefficients a(r ), b(r ), c (r ), α(r ), β(r ) and γ (r ) are all positive constants, then the corresponding long time behaviors to Theorems 4.3 and 4.4 of [3] are all true for our present problem (1.1). 4. Discussion We have examined a free boundary problem of a predator–prey model with higher space dimensions and heterogeneous environment for the special case that the environment and solution are radially symmetric. If the environment or solution is not radially symmetric, then the boundary of the spreading domain would not still be a sphere and the Stefan condition h′ (t ) = −µur (t , h(t )) would become very complicated. Similar to the classical Stefan problem, smooth solutions to these free boundary problems need not exist even if the initial data are smooth. It is necessary to make use of other methods to discuss these problems. In this paper, the dynamic behaviors are discussed. It has been proved that: (i) If the size of the initial region Bh0 occupied by predator satisfies h0 ≥ R∗ (1, a + cV ), or h0 < R∗ (1, a + cV ) but the moving parameter/coefficient of free boundaries µ > µ∗ (it depends on the initial data (u0 , v0 ) and h0 ), then h(t ) → ∞
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
as t → ∞ and lim inft →∞ u(t , r ) ≥ U1 (r ) (the predator will fill the whole space eventually). Here V (r ) and U1 (r ) are respectively the unique positive solutions of (3.8) and (3.19). (ii) While if h0 < R∗ (1, a + cV ) and µ ≤ µ∗ (it also depends on the initial data (u0 , v0 ) and h0 ), then limt →∞ h(t ) < ∞ (the region occupied eventually by predator is finite), ∥u(t , x)∥C [g (t ),h(t )] → 0 and limt →∞ v(t , r ) = V (r ) uniformly in any bounded subset of [0, ∞). These results tell us that in order to control the prey species (pest species) we should put predator species (natural enemies) at the initial state at least in one of three ways: (i) increase the initial targets of predator species, (ii) increase the moving parameter/coefficient of free boundaries, (iii) increase the initial density of the predator species. At last we should remark that our methods can be used to deal with the general predator–prey models with Holling or other functional responses. Appendix. Global estimate of the solution w to (3.3) Proposition A. Let (u, v, h) be any solution of (1.1) and assume h∞ < ∞, if w(t , y) is the solution of (3.3), then there exists a constant K0 such that
∥w∥
< K0 .
1+α ,1+α C 2 ([1,∞)×B1 )
(A.1)
Proof. The proof is similar to that of Proposition A.1 in [3]. For the readers’ convenience, we give the details. Inspired by [29, Theorem A2]. we denote ϕn (t ) = ϕ(t + n), gn = g (t + n, y), where g may be one of the following functions ψi , a, b, c , z and w . Let w(t + n, y) = ηn (t , y) + θ n (t , y), where ηn and θ n are solutions of
n ηt = ϕn (t )∆ηn , ηn (t , y) = 0, ηn (0, y) = w(n, y),
t > 0, y ∈ B1 , t > 0, y ∈ ∂ B1 y ∈ B1
and
n θt = ϕn (t )∆θ n + ψin θyni + ψin ηyni + wn (an − bn wn + cn zn ), θ n (t , y) = 0, n θ (0, y) = 0,
t > 0, y ∈ B1 , t > 0, y ∈ ∂ B1 y ∈ B1 ,
respectively. Let 0 < λ1 ≤ λ2 ≤ · · · be the eigenvalues of the problem
−∆φ = λφ,
φ(y) = 0,
y ∈ B1 ;
y ∈ ∂ B1
and let φ1 , φ2 , . . . be the corresponding set of orthonormal eigenfunctions. We may express ηn as
η n (t , y) =
t
ϕn (s)ds wkn φk ,
exp −λk 0
k≥1
where wkn = (φk , w(n, ·))L2 . In view of 0 < u ≤ M1 (cf. Lemma 2.1), it follows that
j n 2 ∆ η (T ) 2
L (−1,1)
=
ϕn (s)ds (wkn )2
0
k≥1
≤ sup ℓ2j exp
ℓ≥0
≤
T
(λk )2j exp −2λk
2M12
h2∞ j 4T
2j
−8T ℓ h2∞
∥w(n, x)∥2L2 (−1,1)
e−2j .
By Sobolev’s theorem, we have that ∥ηn (t )∥C 2 (B1 ) ≤ K1 (1 + t −j ), where K1 is independent of n provided that j ≥ 1 + 4n . From 2 −j this last estimate and the differential equation satisfied by ηn , we obtain ∥ηtn (t )∥C (B1 ) ≤ K1 h− 0 (1 + t ). Hence, there exists
positive constant K2 such that ∥ηn ∥
1+α ,1+α C 2 (E1 )
< K2 , where E1 = [ 12 , 2] × B1 and K2 depends only on K1 and E1 .
1 Next we estimate θ n . It is obvious that the function ρ n = e− t θ n satisfies
n ρt = ϕn (t )∆ρ n + ψin ρyni + fn (t , y), ρ n (t , y) = 0, n ρ (0, y) = 0,
t > 0, y ∈ B1 , t > 0, y ∈ ∂ B1 y ∈ B1 ,
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
263
where fn (t , y) =
1 t
1
e− t wn + 2
1 1 1 ψin − 2 e− t ηyni + e− t wn (an − bn wn + cn zn ),
0,
t
t >0 t = 0.
1 Note that limt →0+ t −j e− t = 0 for any j ≥ 0, we have fn (t , y) is continuous in E = [0, 3] × B1 and ∥fn ∥C (E ) ≤ K3 where K3
is dependent on K1 and independent of n. By using [30, Theorem 4, p. 191], we can obtain that ∥ρ n ∥ K˜ 3 depends on K3 . It follows that ∥θ n ∥
∥w∥
1+α ,1+α (En ) C 2
≤ ∥ηn ∥
1+α ,1+α C 2 (E1 )
1+α ,1+α C 2 (E1 )
+ ∥θ n ∥
< K4 . We therefore have that 1+α ,1+α C 2 (E1 )
1+α ,1+α C 2 (E1 )
< K˜ 3 where
< K2 + K4 = K0 ,
where En = [n + 21 , n + 2] × B1 . It easily to get (A.1) since the intervals En overlap and K0 is independent of n. References [1] G. Bunting, Y. Du, K. Krakowski, Spreading speed revisited: analysis of a free boundary model, Netw. Heterog. Media 7 (2012) 583–603. Special issue dedicated to H. Matano. [2] Y.H. Du, Z. Guo, The Stefan problem for the Fisher–KPP equation, J. Differential Equations 253 (3) (2012) 996–1035. [3] M.X. Wang, J.F. Zhao, A free boundary problem for a predator–prey model with double free boundaries, 2012, Preprint. [4] M.X. Wang, On some free boundary problems of the Lotka–Volterra type prey–predator model, 2012, Preprint. [5] Y.H. Du, Z. Guo, Spreading-vanishing dichotomy in the diffusive logistic model with a free boundary II, J. Differential Equations 250 (2011) 4336–4366. [6] Y.H. Du, Z.M. Guo, R. Peng, A diffusive logistic model with a free boundary in time-periodic environment, J. Funct. Anal. 265 (2013) 2089–2142. [7] Y.H. Du, Z.G. Lin, Spreading-vanishing dichotomy in the diffusive logistic model with a free boundary, SIAM J. Math. Anal. 42 (2010) 377–405. [8] Y.H. Du, B, D. Lou, Spreading and vanishing in nonlinear diffusion problems with free boundaries, J. Eur. Math. Soc. (2013) in preparation. arXiv:1301.5373 [math.AP]. [9] Y. Kaneko, Y. Yamada, A free boundary problem for a reaction diffusion equationa appearing in ecology, Adv. Math. Sci. Appl. 21 (2) (2011) 467–492. [10] R. Peng, X.Q. Zhao, The diffusive logistic model with a free boundary and seasonal succession, Discrete Contin. Dyn. Syst. A 33 (5) (2013) 2007–2031. [11] D.G. Aronson, H.F. Weinberger, Nonlinear diffusion in population genetics, combustion, and nerve pulse propagation, in: Partial Differential Equations and Related Topics, in: Lecture Notes in Math., vol. 446, Springer, Berlin, 1975, pp. 5–49. [12] D.G. Aronson, H.F. Weinberger, Multidimensional nonlinear diffusions arising in population genetics, Adv. Math. 30 (1978) 33–76. [13] R.A. Fisher, The wave of advance of advantageous genes, Ann. Eugenics 7 (1937) 335–369. [14] D. Hilhorst, M. Iida, M. Mimura, H. Ninomiya, A competition-diffusion system approximation to the classical two-phase Stefan problem, Jpn. J. Ind. Appl. Math. 18 (2) (2001) 161–180. [15] D. Hilhorst, M. Mimura, R. Schätzle, Vanishing latent heat limit in a Stefan-like problem arising in biology, Nonlinear Anal. RWA 4 (2003) 261–285. [16] Z.G. Lin, A free boundary problem for a predator–prey model, Nonlinearity 20 (2007) 1883–1892. [17] M. Mimura, Y. Yamada, S. Yotsutani, A free boundary problem in ecology, Japan J. Appl. Math. 2 (1985) 151–186. [18] M. Mimura, Y. Yamada, S. Yotsutani, Stability analysis for free boundary problems in ecology, Hiroshima Math. J. 16 (1986) 477–498. [19] M. Mimura, Y. Yamada, S. Yotsutani, Free boundary problems for some reaction diffusion equations, Hiroshima Math. J. 17 (1987) 241–280. [20] J.S. Guo, V.H. Wu, On a free boundary problem for a two-species weak competition system, J. Dynam. Differential Equations 24 (2012) 873–895. [21] L.I. Rubinstein, The Stefan Problem, American Mathematical Society, Providence, RI, 1971. [22] J. Crank, Free and Moving Boundary Problem, Clarendon Press, Oxford, 1984. [23] X.F. Chen, A. Friedman, A free boundary problem arising in a model of wound healing, SIAM J. Math. Anal. 32 (2000) 778–800. [24] X.F. Chen, A. Friedman, A free boundary problem for an elliptic–hyperbolic system: an application to tumor growth, SIAM J. Math. Anal. 35 (2003) 974–986. [25] L. Caffarelli, S. Salsa, A Geometric Approach to Free Boundary Problems, in: Grad. Stud. Math., vol. 68, Amer. Math. Soc., Providence, RI, 2005. [26] R.S. Cantrell, C. Cosner, Spatial Ecology via Reaction–Diffusion Equations, John Wiley & Sons Ltd., 2003. [27] Y.H. Du, L.S. Liu, Remarks on the uniqueness problem for the logistic equation on the entire space, Bull. Aust. Math. Soc. 73 (2006) 129–137. [28] Y.H. Du, L. Ma, Logistic type equations on RN by a squeezing method involving boundary blow-up solutions, J. Lond. Math. Soc. 64 (2) (2001) 107–124. [29] K.J. Brown, D.C. Dunne, R.A. Candner, A semilinear parabolic system arising in the theory of super conductivity, J. Differential Equations 40 (1981) 232–252. [30] A. Friedman, Partial Differential Equations of Parabolic Type, Prentice-Hall, Englewood Cliffs, NJ, 1964.
Contents lists available at ScienceDirect
Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa
A free boundary problem of a predator–prey model with higher dimension and heterogeneous environment✩ Jingfu Zhao a,b , Mingxin Wang a,∗ a
Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, PR China
b
Department of Mathematics, Shaanxi University of Technology, Hanzhong 723000, PR China
article
info
Article history: Received 15 February 2013 Accepted 7 October 2013
abstract This paper is concerned with a free boundary problem for a prey–predator model in higher space dimensions and heterogeneous environment. Such a model may be used to describe the spreading of an invasive or new predator species in which the free boundary represents the spreading front of the predator species and is described by Stefan-like condition. For simplicity, we assume that the environment and solutions are radially symmetric. We prove a spreading–vanishing dichotomy for this model, namely the predator species either successfully spreads to infinity as t → ∞ and survives in the new environment, or it fails to establish and dies out in the long run while the prey species stabilizes at a positive equilibrium state. The criteria for spreading and vanishing are given. © 2013 Elsevier Ltd. All rights reserved.
1. Introduction The spreading of a new or invasive species is one of the most important topics in mathematical ecology. A lot of mathematicians have made efforts to develop various invasion models and investigated them from a viewpoint of mathematical ecology. In this paper we consider a free boundary problem for a prey–predator model in higher space dimensions and heterogeneous environment. In the real world, the following two kind of phenomenon may happen: (i) At the initial state, one kind of prey species (for example, some pest species) occupied the entire space or a large area. In order to control such prey species we put one kind of predator species (natural enemies) in some bounded area or a small area (initial habitat). (ii) There is some kind of species (prey) in the entire space or a large area, and at some time (initial time) another type species (the new or invasive species, predator) enters some bounded area or a small area (initial habitat). In general, the predator has a tendency to emigrate from the boundary to obtain their new habitat, i.e., it will move outward along the unknown curves (free boundaries) as time increases. It is assumed that speeds of free boundaries are proportional to the gradient of predator. We want to realize the dynamics/variations of prey, predator and free boundary. For simplicity, we assume that the environment and solution are radially symmetric. Thus, the model we are concerned here becomes
✩ This work was supported by NSFC Grants 11071049 and 11371113.
∗
Corresponding author. Tel.: +86 15145101503. E-mail address: [email protected] (M. Wang).
1468-1218/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.nonrwa.2013.10.003
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
ut − ∆u = u [a(r ) − b(r )u + c (r )v ] , u(t , r ) ≡ 0, vt − d∆v = v [α(r ) − β(r )u − γ (r )v ] , ur (t , 0) = 0, vr (t , 0) = 0, u(t , h(t )) = 0, h′ (t ) = −µur (t , h(t )), h(0) = h0 , u(0, r ) = u0 (r ), v(0, r ) = v0 (r ),
t t t t t r r
> 0, 0 < r < h(t ), > 0, r ≥ h(t ), > 0, r > 0, > 0, > 0, ∈ [0, h0 ], ∈ [0, ∞),
251
(1.1)
where ∆u = urr + n−r 1 ur , ∆v = vrr + n−r 1 vr (r = |x|, x ∈ Rn , n ≥ 1), d, µ and h0 are given positive constants, functions a, b, c , α, β, γ ∈ C ν0 ([0, ∞)) for some ν0 ∈ (0, 1). The sphere r = h(t ) is the moving boundary to be determined. Throughout this paper we always assume that there are positive constants κ1 ≤ κ2 such that
κ1 ≤ a(r ), b(r ), c (r ), α(r ), β(r ), γ (r ) ≤ κ2 ,
∀r ∈ [0, ∞),
(1.2)
and the initial functions u0 (r ) and v0 (r ) satisfy
u0 ∈ C 2 ([0, h0 ]), u0 (h0 ) = 0, u0 (r ) > 0 v0 ∈ Cb ([0, ∞)), v0 (r ) > 0
in (0, h0 ), in [0, ∞).
(1.3)
The free boundary problem (1.1) describes the spreading of a new or invasive predator with population density u(t , |x|) over an n-dimensional habitat, which is radially symmetric but heterogeneous. The initial function u0 (|x|), which occupies a ball Bh0 , stands for the populations of the new or invasive predator in the very early stage of its introduction, and v0 (|x|), which occupies the whole space, stands for the populations of the prey in the very early stage of the predator introduction. Here and in what follows, we use BR to stand for the ball with center at 0 and radius R. The spreading front is represented by the free boundary |x| = h(t ), which is the sphere ∂ Bh(t ) whose radius h(t ) grows at a speed that is proportional to the population gradient of predator at the front: h′ (t ) = −µur (t , h(t )). The ecological background and derivation of the free boundary conditions can also refer to [1,2]. Recently, Wang and Zhao [3] studied a similar free boundary problem to (1.1) with double free boundaries in which n = 1 and the coefficients a(r ), b(r ), c (r ), α(r ), β(r ) and γ (r ) are all positive constants. Wang [4] discussed some similarly models over the half spatial line with zero Dirichlet boundary condition or zero Neumann boundary condition. They all established the spreading–vanishing dichotomy, long time behavior of solution and sharp criteria for spreading and vanishing. The main purpose of this paper is to show that the most of those results of [3,4] continue to hold in the more realistic situation of higher space dimensions and heterogeneous environment. In the absence of v , the system is reduced to a Stefan problem which was studied by Du and Guo in [5,2]. Free boundary problems similar to the one in [5] have been studied by many authors, please refer to [2,6–10] and references cited therein. These aforementioned works present a new approach to describe the front propagation for population species, which is different from the classical approach of traveling wave solutions of the diffusive logistic equation on the entire space Rn with n ≥ 1 (Refs. [11–13]): ut − d∆u = u(a − bu),
t > 0, x ∈ Rn .
Similar free boundary problems to (1.1) in the one dimension and homogeneous environment case (the coefficients are all positive constants) have been used in two-species models in several earlier papers; see, for example, [14–19] over a bounded spatial interval, and [20] over the half spatial line for the competition model. The well-known Stefan condition has been used in the modeling of a number of applied problems. Except the above cited papers, for example, it was used to describe the melting of ice in contact with water [21], the modeling of oxygen in the muscle [22], the wound healing [23], the tumor growth [24], and so on. There is a vast literature on the Stefan problem, and some important theoretical advances can be found in [25,22] and the references therein. The organization of this paper is as follows. In Section 2, we first give the local existence and uniqueness of the solution to (1.1) and then show that the solution exists for all time t ∈ (0, ∞). Section 3 is devoted to the proof of the spreading–vanishing dichotomy. Our arguments are based on the comparison principle and the construction of suitable upper and lower solutions of (1.1). Section 4 is a brief discussion. 2. Existence and uniqueness In this section, we prove the existence and uniqueness of the solution to (1.1) for all t > 0. The proof of the local existence and uniqueness can be done by modifying the arguments of [5,3]. So we omit the details. Theorem 2.1. For any given (u0 , v0 ) satisfying (1.3) and any α ∈ (0, 1), there is a T > 0 such that problem (1.1) admits a unique solution
(u, v, h) ∈ C
1+α ,1+α 2
α
(D¯ T ) × CT × C 1+ 2 ([0, T ]).
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Moreover,
∥ u∥
1+α ,1+α C 2 (D¯ T )
+ ∥ h∥
C
1+ α 2
([0,T ])
≤ C,
where DT = (t , r ) ∈ R2 : t ∈ (0, T ], r ∈ (0, h(t )) ,
CT = Cb ([0, T ] × [0, ∞))
1+ α ,2+α 2
Cloc
((0, T ] × [0, ∞)),
positive constants C and T only depend on h0 , α , ∥u0 ∥C 2 ([−h0 ,h0 ]) and ∥v0 ∥Cb (R) . To show that the local solution obtained in Theorem 2.1 can be extended to all t > 0, we need the following estimate. Lemma 2.1. Let (u, v, h) be a solution to problem (1.1) defined for t ∈ (0, T ) for some T ∈ (0, ∞]. Then there exist constants M1 , M2 and M3 independent of T such that 0 < u(t , r ) ≤ M1 ,
for 0 < t ≤ T , 0 < r < h(t ),
0 < v(t , r ) ≤ M2 , for 0 < t ≤ T , 0 < r < ∞, 0 < h′ (t ) ≤ M3 , for 0 < t ≤ T . Proof. Using the strong maximum principle, we are easy to see that u > 0 in (0, T ] × (0, h(t )) and v > 0 in (0, T ] × [0, ∞) as long as the solution exists. Since v(t , r ) satisfies
vt − d∆v = v [α(r ) − β(r )u − γ (r )v ] , v(0, r ) = v0 (r ) > 0,
t > 0, 0 < r < ∞, r ≥ 0,
and κ1 ≤ α(r ), γ (r ) ≤ κ2 , we have v(t , r ) ≤ max {∥v0 ∥∞ , κ2 /κ1 } := M2 . Similarly, as u(t , r ) satisfies ut − ∆u = u [a(r ) − b(r )u + c (r )v ] , ur (t , 0) = u(t , h(t )) = 0, u(0, r ) = u0 (r ) > 0,
t > 0, 0 < r < h(t ), t > 0, 0 < r < h0
and κ1 ≤ a(r ), b(r ), c (r ) ≤ κ2 , we also have u(t , r ) ≤ max {∥u0 ∥∞ , (κ2 + κ2 M2 )/κ1 } := M1 . Moreover, the strong maximum principle yields the inequalities ur (t , h(t )) < 0, and therefore h′ (t ) > 0 in (0, T ]. It remains to show that h′ (t ) ≤ M3 for all t ∈ (0, T ) with some M3 independent to T . To this end, let M be a positive constant, define
ΩM := {(t , r ) : 0 < t < T , h(t ) − 1/M < r < h(t )} and construct an auxiliary function
w(t , r ) = M1 [2M (h(t ) − r ) − M 2 (h(t ) − r )2 ]. We will choose M so that w(t , r ) ≥ u(t , r ) holds over ΩM . Direct calculations show that, for (t , r ) ∈ ΩM ,
wt = 2M1 Mh′ (t ) [1 − M (h(t ) − r )] ≥ 0, −wr = 2M1 M [1 − M (h(t ) − r )] ≥ 0, −wrr = 2M1 M 2 , u [a(r ) − b(r )u + c (r )v ] ≤ M1 κ2 (1 + M2 ). It follows that
wt − wrr −
n−1 r
wr ≥ 2M1 M 2 ≥ u [a(r ) − b(r )u + c (r )v ]
in ΩM ,
if M 2 ≥ κ2 (1 + M2 )/2. On the other hand,
w(t , h(t ) − M −1 ) = M1 ≥ u(t , h(t ) − M −1 ),
w(t , h(t )) = 0 = u(t , h(t )).
Note that u0 ( r ) = −
h0
u′0 (s)ds ≤ (h0 − r )∥u′0 ∥C [0,h0 ]
in [h0 − M −1 , h0 ],
r
w(0, r ) = M1 [2M (h0 − r ) − M 2 (h0 − r )2 ] ≥ M1 M (h0 − r ) in [h0 − M −1 , h0 ], we know that if MM1 ≥ ∥u′0 ∥C [0,h0 ] , then u0 (r ) ≤ (h0 − r )∥u′0 ∥C [0,h0 ] ≤ w(0, r ) in [h0 − M −1 , h0 ].
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
253
Let
M = max
κ2 (1 + M2 ) ∥u′0 ∥C [0,h0 ] , 2
M1
.
Applying the maximum principle to w − u over ΩM we can deduce that u(t , r ) ≤ w(t , r ) for (t , r ) ∈ ΩM . It would then follow that ur (t , h(t )) ≥ wr (t , h(t )) = −2M1 M , The proof is complete.
h′ (t ) = −µux (t , h(t )) ≤ 2µM1 M := M3 .
Theorem 2.2. The solution of problem (1.1) exists and is unique for all t ∈ (0, ∞). Proof. Let [0, Tmax ) be the maximal time interval in which the solution exists. By Theorem 2.1, Tmax > 0. It remains to show that Tmax = ∞. Arguing indirectly, we assume that Tmax < ∞. By Lemma 2.1, there exist positive constants M1 , M2 and M3 independent of Tmax such that 0 ≤ u(t , r ) ≤ M1
in [0, Tmax ) × [0, h(t )],
0 ≤ h′ (t ) ≤ M3 ,
0 ≤ h(t ) − h0 ≤ M3 t
0 ≤ v(t , r ) ≤ M2
in [0, Tmax ) × [0, ∞),
in [0, Tmax ).
By the standard Lp estimates and the Sobolev embedding theorem, we can find a constant C > 0 depending only on Mi (i = 1, 2, 3) such that ∥u(t , ·)∥C 1+α/2 ([0,h(t )]) ≤ C and v is continuous for (t , r ) ∈ [0, Tmax ) × [0, ∞). It then follows from the proof of Theorem 2.1 that there exists a τ > 0 depending only on C and Mi (i = 1, 2, 3) such that the solution of problem (1.1) with initial time Tmax − τ /2 can be extended uniquely to the time Tmax − τ /2 + τ . But this contradicts the assumption. The proof is now complete. 3. The spreading and vanishing In this section, we prove the spreading–vanishing dichotomy. Though our approach here mainly follows the lines of [3], considerable changes in the proofs are needed. Since the situation here is more general and difficult, it follows from Lemma 2.1 that r = h(t ) is monotonic increasing, therefore, there exists h∞ ∈ (0, ∞] such that limt →∞ h(t ) = h∞ . Let λ1 (d, p, R) be the principal eigenvalue of the problem
−d∆φ = λp(|x|)φ, φ = 0,
x ∈ BR , x ∈ ∂ BR ,
where d > 0 is a constant, p(r ) > κ1 /2 for r ∈ [0, ∞) and p ∈ C 1 ([0, ∞)). It is well known that λ1 (d, p, R) is a strictly decreasing and continuous function in R, and limR→0+ λ1 (d, p, R) = +∞, limR→+∞ λ1 (d, p, R) = 0 (cf. Section 2.2 of [26]). Therefore, for fixed d > 0 and p ∈ C 1 ([0, ∞)), there is a unique R∗ (d, p) such that
λ1 d, p, R∗ (d, p) = 1 and
λ1 (d, p, R) < 1 for R > R∗ (d, p);
λ1 (d, p, R) > 1 for R < R∗ (d, p).
Since λ1 (d, p, R) is a strictly decreasing continuous function in p and R, we have that R∗ (d, p) is a strictly decreasing continuous function in p. In order to investigate asymptotic properties of solutions for (1.1), we will derive a priori estimate for any global solution. Lemma 3.1. Let (u, v, h) be any solution of (1.1). If h∞ < ∞, then there exists a constant K > 0, such that
∥u(t , ·)∥C 1 ([0,h(t )]) = ∥u(t , ·)∥C 1 (Bh(t ) ) < K ,
∀t ≥ 1.
(3.1)
Moreover, lim h′ (t ) = 0.
(3.2)
t →∞
Proof. We introduce new transformation
(t , x) → (t , y),
with x = h(t )y, y ∈ Rn ,
and functions w(t , y) and z (t , y) with the following definitions
w(t , y) = u(t , h(t )|y|),
z (t , y) = v(t , h(t )|y|).
254
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Clearly, w and z are radially symmetric and satisfy the following initial boundary value problem in the domain B1 with fixed boundary ∂ B1 :
n wt = ϕ(t )∆w + ψi (t , y)wyi + w [a(h(t )|y|) − b(h(t )|y|)w + c (h(t )|y|)z] , i=1
w(t , y) = 0, w(0, y) = u0 (h0 |y|),
t > 0, |y| < 1, (3.3)
t > 0, |y| = 1, |y| ≤ 1,
where
ϕ(t ) =
1 h(t )
2
,
ψi (t , y) =
h′ (t )yi h( t )
.
By Proposition A, we can obtain that there exists a positive constant K0 such that
∥w∥
1+α ,1+α C 2 ([1,∞)×B1 )
< K0 .
(3.4)
It follows from uxi (t , |x|) = h(1t ) wyi (t , y) and (3.4) that there exists a constant K such that (3.1) holds. n ′ ′ Now we prove (3.2). Note ur (t , h(t )) = h(1t ) i=1 wyi (t , y0 )yi and h (t ) = −µur (t , h(t )), by virtue of 0 < h (t ) < M3 α and ∥wyi (·, y)∥ 2 < K0 for |y| = 1 and i = 1, . . . , n, we can obtain C
′
∥h ∥
([1,∞))
α C 2 ([1,∞))
< L,
(3.5)
where L depends on K0 and M3 . (3.2) can be deduced from h′ (t ) > 0, h∞ < ∞ and (3.5). The proof is complete.
Theorem 3.1. Let (u, v, h) be any solution of (1.1). If h∞ < ∞, then lim ∥u(t , ·)∥C ([0,h(t )]) = 0,
(3.6)
lim v(t , r ) = V (r ) uniformly in any bounded subset of [0, ∞),
(3.7)
t →∞
t →∞
here V (|x|) is the unique positive (radial) solution of the equation
− d∆v = v[α(|x|) − γ (|x|)v],
x ∈ Rn .
(3.8)
This result shows that if the predator cannot spread into the whole space, then it will die out eventually.
∞
Proof. Step 1: Proof of (3.6). On the contrary we assume that there exist σ > 0 and (tj , xj ) j=1 , with xj ∈ Bh(tj ) and tj → ∞ as j → ∞, such that
u(tj , |xj |) ≥ σ ,
j = 1, 2, . . . .
(3.9)
Since xj ∈ Bh∞ , there is a subsequence, noted by itself, and x0 ∈ Bh∞ , such that xj → x0 as j → ∞. If |x0 | = h∞ , then we j
have ||xj | − h(tj )| → 0 as j → ∞. There exists x¯ j in the line connecting xj and h(tj ) |xxj | for every j such that ur (tj , |¯xj |) =
u(tj , |xj |) − u(tj , h(tj ))
|xj | − h(tj )
→ ∞ as j → ∞.
This is a contradiction to (3.1). So |x0 | ̸= h∞ holds, and hence x0 ∈ Bh∞ . By the use of (3.1) and (3.9), there exists δ > 0 such that Bδ (x0 ) ⊂ Bh∞ and u(tj , |x|) ≥ σ /2,
∀x ∈ Bδ (x0 )
for all large enough j. Since h(tj ) → h∞ as j → ∞, without loss of generality we may think that h(tj ) > |x0 | + δ for all large j. Let gj (t ) = δ + t − tj and tj∗ = inf t > tj : h(t ) = gj (t ) + |x0 | ,
j
x0 . |x0 |
x0 = |x0 | + gj (tj∗ )
It is easy to obtain that tj∗ < h∞ + tj − δ − |x0 |. This implies gj (t ) + |x0 | ≤ h(t ),
t ∈ [tj , tj∗ ].
(3.10) j
Moreover, ∂ Bh(t ∗ ) and ∂ Bgj (t ∗ ) (x0 ) tangent at the point x0 . j
j
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Define yj (t , x) = (π − ε) uj (t , x) =
σ
|x−x0 | gj (t )
and
e−k(t −tj ) cos yj (t , x) + cos ε ,
4
255
( t , x ) ∈ Ω tj ,
where ε (ε < π /8) and k are positive constants to be chosen later, and
Ωtj = {(t , x) : tj < t < tj∗ , |x − x0 | < gj (t )}. It is obvious that uj (t , x) = 0 for |x − x0 | = gj (t ), and |yj (t , x)| ≤ π − ε for (t , x) ∈ Ω tj , the latter implies uj (t , x) ≥ 0 in Ωtj . We want to compare u(t , |x|) and uj (t , x) in Ω tj . Thanks to (3.10), it follows that u(t , |x|) ≥ 0 = uj (t , x),
t ∈ [tj , tj∗ ], x ∈ ∂ Bgj (t ) (x0 ).
On the other hand, it is obvious that u(tj , |x|) ≥
σ 2
≥ uj (tj , x),
x ∈ Bδ (x0 ).
Thus, if the positive constants ε and k can be chosen independent of j such that ujt − ∆uj − uj a(|x|) − b(|x|)uj ≤ 0,
( t , x ) ∈ Ω tj ,
(3.11)
it can be deduced that uj (t , |x|) ≤ u(t , x) for (t , x) ∈ Ωtj by applying the maximum principle to u − uj over Ωtj . Since j
j
u(tj∗ , h(tj∗ )) = 0 = uj (tj∗ , x0 ) and h(tj∗ ) = |x0 |, it follows that
∂ uj ∂ν
(tj∗ , xj0 ) ≥
∂u ∗ j (t , |x0 |) = ur (tj∗ , h(tj∗ )). ∂ν j
The direct computation gives
∂ uj ∂ν
(tj∗ , xj0 ) = −
7σ π −kh∞ σ (π − ε) −k(tj∗ −tj ) e sin(π − ε) ≤ − e sin ε. ∗ 4(δ + tj − tj ) 32h∞
7σ π −kh∞ e Note the boundary condition −µur (tj∗ , h(tj∗ )) = h′ (tj∗ ), we have h′ (tj∗ ) ≥ µ 32h sin ε . This implies lim supt →∞ ∞
h′ (t ) > 0 since limj→∞ tj∗ → ∞. This is a contradiction to (3.2). So limt →∞ ∥u(t , ·)∥C [0,h(t )] = 0 is obtained. We shall prove that if ε and k satisfy
ε< k≥
π 8
,
sin ε ≤
9δ 2 π 8π h2∞ (3h∞ + 4(n − 1))
,
(3.12)
π (h∞ + (n − 1)π ) σ κ2 π 2 + + , δ 2 (cos ε − cos 2ε) 2 δ
(3.13)
then (3.11) holds. Direct computations yield ujt − ∆uj − uj a − buj = −kuj −
≤
2
≤
σ κ2 σ κ2
σ 4
+
e
−k(t −tj )
n
yjt sin yj +
y2jxi − k uj −
i =1
σ 4
σ
4
e
−k(t −tj )
sin yj ∆yj + cos yj
e−k(t −tj ) cos ε
δ
y2jxi −
i =1
−k(t −tj )
π −ε
2
cos ε
By (3.13), one has σ b1 /2 + (π /δ)2 − k < 0. 2 Since 0 ≤ yj ≤ π − ε when (t , x) ∈ Ω tj , we can decompose Ωtj = Ωt1j Ωtj , where
Ωt1j = (t , x) ∈ Ωtj : tj < t < tj∗ , π − 2ε < yj (t , x) < π − ε , Ωt2j = (t , x) ∈ Ωtj : tj < t < tj∗ , 0 ≤ yj (t , x) ≤ π − 2ε .
n
y2jxi
− uj a − buj
i =1 n
− k uj − e 4 h∞ σ π −k(t −tj ) |x − x0 | n−1 e sin yj + + 4δ δ |x − x0 | , I. 2
+
π 2
σ
σ 4
e−k(t −tj ) sin yj (yjt − ∆yj )
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
It is obvious that sin yj ≤ sin 2ε when (t , x) ∈ Ωt1j , and cos yj ≥ − cos 2ε when (t , x) ∈ Ωt2j . In virtue of ε < π /8, we can obtain
|x − x0 | n−1 3h∞ + 4(n − 1) + ≤ δ |x − x0 | 3δ when (t , x) ∈ Ωt1j , and
|x − x0 | n−1 + δ | x − x0 |
sin yj when (t , x) ∈ I ≤
σ 4
Ωt2j .
≤
h∞ + (n − 1)π
δ
Note that uj (t , x) ≥ 0 in Ωtj , in view of (3.12) and (3.13), we conclude
−k(t −tj )
3π 2
−
e
4h2∞
π (3h∞ + 4(n − 1)) cos ε + sin 2ε < 0 3δ 2
when (t , x) ∈ Ωt1j , and I ≤
σ 4
−k(t −tj )
e
σ κ2 π 2 π (h∞ + (n − 1)π ) −k + + (cos ε − cos 2ε) + <0 2 δ δ2
when (t , x) ∈ Ωt2j . Therefore, (3.11) holds. Step 2: Proof of (3.7). The existence and uniqueness of a positive solution of (3.8) follows from Theorem 2.3 of [27]. It must be radially symmetric since (3.8) is invariant under rotations around the origin of Rn . To show (3.7), we use a squeezing argument introduced in [28]. We first consider the Dirichlet problem
−d∆z = z [α(r )(1 − ε) − γ (r )z ],
|x| < R;
z (R) = 0,
and the boundary blow-up problem
−d∆w = [α(r ) − γ (r )w],
|x| < R;
w(R) = +∞.
When ε is small and R is large, it is well known that these problems have positive radial solutions zRε and wR , respectively. By the comparison principle given in [28], as ε → 0+ and R → +∞, zRε increases to the unique positive solution V of (3.8) and wR decreases to V . Choose a decreasing sequence {εi } and an increasing sequence {Ri } such that εi → 0+ , Ri → +∞ as i → ∞, and κ α(r ) 1 > λ1 (d, α(1 − εi ), Ri ) for all i. In view of (3.6), for each εi , there exists Ti > 0 such that u(t , r ) < κ1 εi ≤ β(r ) εi for all 2 t ≥ Ti and r ∈ [0, ∞). For each i, by comparison principle we have
v(t , r ) ≥ zi (t , r ),
(t , r ) ∈ [Ti , ∞) × [0, Ri ],
where zi (t , r ) is the solution of the following logistic problem with fixed boundary zt − d∆z = z (α(r )(1 − εi ) − γ (r )v), zr (t , 0) = 0, z (t , Ri ) = 0, z (Ti , r ) = v(Ti , r ),
t > Ti , r ∈ [0, Ri ], t > Ti , r ∈ [0, Ri ].
(3.14)
Since 1 > λ1 (d, α(1 − εi ), Ri ), it is well known (see, for example, Proposition 3.3 in [26]) that (3.14) admits a unique positive ε solution zi (t , r ). Moreover, limt →∞ zi (t , r ) = zRii (r ) uniformly for r ∈ [0, Ri ]. It follows that ε
lim inf v(t , x) ≥ zRii (r ) uniformly for r ∈ [0, Ri ]. t →∞
Sending i → ∞, we obtain lim inf v(t , x) ≥ V (r ) uniformly in any compact subset of [0, ∞). t →∞
(3.15)
Analogously, by arguments similar to those in the proof of Theorem 4.1 of [28], we see that lim sup v(t , x) ≤ wRi (r ) uniformly for r ∈ [0, Ri ], t →∞
which implies (by sending i → ∞) lim sup v(t , x) ≤ V (r ) uniformly in any compact subset of [0, ∞). t →∞
From (3.15) and (3.16) we see that (3.7) holds.
(3.16)
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
257
The symbol V defined by (3.8) will be always used in the sequel. In the following theorem, we will give a sufficient condition of spreading. Theorem 3.2. Let (u, v, h) be any solution of (1.1). If h∞ < ∞, then h∞ ≤ R∗ (1, a + cV ).
(3.17)
Proof. Theorem 3.1 implies that, if h∞ < ∞, then (3.6) and (3.7) hold. We assume h∞ > R∗ (1, a + cV ) to get a contradiction. Note that (3.7) and R∗ (d, p) is a strictly decreasing continuous function in p, it is easy to see that for any given 0 < ε ≪ 1 there exists τ ≫ 1 such that h(τ ) > max h0 , R∗ (1, a + cV − ε) ,
v(t , r ) ≥ V (r ) −
ε , c (r )
∀t ≥ τ , r ∈ [0, h∞ ].
Set R = h(τ ), then R > R∗ (1, a + cV − ε). Let w = w(t , r ) be the positive solution of the following initial boundary value problem with fixed boundary:
wt = ∆w + w [a(r ) − b(r )w + c (r )V (r ) − ε ] , wr (t , 0) = w(t , R) = 0, w(τ , r ) = u(τ , r ),
t > τ , 0 < r < R, t > τ, 0 ≤ r ≤ R.
By the comparison principle,
w(t , r ) ≤ u(t , r ),
∀t ≥ τ , 0 ≤ r ≤ R.
Since
λ1 (1, a + cV − ε, R) < λ1 1, a + cV − ε, R∗ (1, a + cV − ε) = 1, it is well known that w(t , r ) → w ∗ (r ) as t → ∞ uniformly for r ∈ [0, R], where w ∗ is the unique positive solution of
∆w ∗ + w ∗ a(r ) + c (r )V − ε − b(r )w ∗ = 0, wr∗ (0) = w∗ (R) = 0.
r ∈ (0, R),
Hence, lim inft →∞ u(t , r ) ≥ limt →∞ w(t , r ) = w ∗ (x) > 0 in [0, R]. This is a contradiction to the fact that (3.6). Therefore, (3.17) holds. We now present some comparison principles which can be used in the following to estimate the solution (u(t , x), v(t , x)) and the free boundary r = h(t ). The following Lemmas 3.2 and 3.3 can be proved as in Lemma 3.1 of [3]. We omit the details. ∗
Lemma 3.2 (Comparison Principle). Suppose that T ∈ (0, ∞), h¯ ∈ C 1 ([0, T ]), u¯ ∈ C (DT ) C 1,2 (D∗T ) with D∗T = {(t , x) ∈ R2 : 0 < t ≤ T , 0 < r < h¯ (t )}, v¯ ∈ C 1,2 (GT ) with GT = (0, T ] × [0, ∞). Assume further that (¯u, v¯ , h¯ ) satisfies
u¯ − ∆u¯ ≥ u¯ [a(r ) − b(r )¯u + c (r )¯v ] , t v¯ t − d∆v¯ ≥ v¯ (α(r ) − γ (r )¯v ), u¯ r (t , 0) = 0, v¯ r (t′ , 0) = 0, u¯ (t , h¯ (t )) = 0, h¯ (t ) ≥ −µ¯ur (t , h¯ (t )),
t t t t
> 0, 0 < r < h¯ (t ), > 0, 0 < r < ∞, > 0, > 0,
(3.18)
and h¯ (0) ≥ h0 ,
and u¯ (0, r ) ≥ 0 in [0, h¯ (0)],
u0 (r ) ≤ u¯ (0, r ) in [0, h0 ],
v0 (r ) ≤ v¯ (0, r ) in [0, ∞).
Then the solution (u, v, h) of the free boundary problem (1.1) satisfies u(t , r ) ≤ u¯ (t , r ) on DT ,
v(t , r ) ≤ v¯ (t , r ) on GT ,
h(t ) ≤ h¯ (t ) on [0, T ],
where DT as in Theorem 2.1. Lemma 3.3 (Comparison Principle). Let T ∈ (0, ∞), h ∈ C 1 ([0, T ]) with h(t ) > 0 for all t ∈ [0, T ], and u ∈ C (OT ) with OT = {(t , r ) ∈ R2 : 0 < t ≤ T , 0 < r < h(t )}. Suppose that (u, h) satisfies
ut − ∆u ≤ u a(r ) − b(r )u , u (t , 0) = u(t , h(t )) = 0, h′r(t ) ≤ −µu (t , h(t )), r
t > 0, 0 < r < h(t ), t > 0, t > 0,
C 1,2 (OT )
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
and h(0) ≤ h0 ,
and
u0 (r ) ≥ u(0, r ) in [0, h(0)].
Then the solution (u, v, h) of the free boundary problem (1.1) satisfies u(t , r ) ≥ u(t , r ) on OT ,
h(t ) ≥ h(t )
on [0, T ].
The pair (¯u, v¯ , h¯ ) in Lemma 3.2 is called an upper solution of the problem (1.1). Similarly, the pair (u, 0, h) in Lemma 3.3 is called a lower solution of the problem (1.1). Lemma 3.4. If h∞ = ∞, then U1 (r ) ≤ lim inf u(t , r ) ≤ lim sup u(t , r ) ≤ U2 (r ) t →∞
t →∞
uniformly in any compact subset of [0, ∞), where U1 (r ) and U2 (r ) are the unique positive (radial) solutions of the equations
− ∆u = u [a(|x|) − b(|x|)u] ,
x ∈ Rn
(3.19)
and
−∆u = u [a(|x|) − b(|x|)u + c (|x|)V ] ,
x ∈ Rn
respectively. Proof. Since h∞ = ∞, there exists T > 0 such that h(T ) > R∗ (1, a). Choose a function u0 (r ) satisfying u0 ∈ C 2 [0, h(T )], u0 (r ) ≤ u0 (T , r ) in [0, h(T )], u0 (r ) > 0 in (0, h(T )) and u0 (h(T )) = 0. Consider the following problem
ut − ∆u = u a(r ) − b(r )u , ur (t , 0) = u(t , h(t )) = 0, h′ (t ) = −µur (t , h(t )), u(T , r ) = u0 (r ), h(T ) = h(T ).
t > T , 0 < r < h(t ), t > T, t > T, 0 ≤ r ≤ h(T ),
By Theorem 2.1 of [5], this problem has a unique solution (u, h) and exists for all t ≥ T . In view of Lemma 3.4, we have u(t , r ) ≥ u(t , r ) for t ≥ T , 0 ≤ r ≤ h(t );
h(t ) ≥ h(t ) for [T , ∞).
(3.20)
Using Theorem 2.5 of [5] firstly and Lemma 2.3 of [5] secondly, we can see that lim inf u(t , r ) = U1 (r ) uniformly in any compact subset of [0, ∞).
(3.21)
t →∞
It follows (3.20) and (3.21) that lim inf u(t , r ) ≥ U1 (r ) uniformly in any compact subset of [0, ∞). t →∞
On the other hand, we define H =
κ2 ∥v0 ∥∞ , κ1
θ=
κ12 , κ2
v¯ (t , r ) = (1 + He−θ t )V (r ).
(3.22) κ
κ
Because of V (r ) satisfying (3.8) and κ1 ≤ α(r ), γ (r ) ≤ κ2 , by the comparison principle we have κ1 ≤ V (r ) ≤ κ2 for 2 1 r ∈ [0, ∞). Direct calculations yield
v¯ t − d∆v¯ − v¯ [α(r ) − γ (r )¯v ] = He−θ t V (r )[−θ + (1 + He−θ t )γ (r )V (r )] κ2 ≥ He−θ t V (r ) −θ + 1 κ2 = 0, and v¯ (0, r ) = (1 + H )V (r ) > ∥v0 ∥∞ ≥ v0 (r ). Since limt →∞ v¯ (t , r ) = V (r ) uniformly in [0, ∞), for any given ε > 0 small, there exists Tε > T , such that
v¯ (t , r ) ≤ V (r )(1 + ε),
∀t ≥ Tε , r ∈ [0, ∞).
Consider the auxiliary problem
u¯ − ∆u¯ = u¯ [a(r ) − b(r )¯u + c (r )V (r )(1 + ε)] , t u¯ r (t , 0) = 0, u¯ (t , h¯ (t )) = 0, h¯ ′ (t ) = −µ¯ur (t , h¯ (t )), u¯ (Tε , r ) = u(Tε , r ), h¯ (Tε ) = h(Tε ),
t > Tε , 0 < r < h¯ (t ), t > Tε , t > Tε , 0 ≤ r ≤ h(Tε ).
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
259
It is easy to deduce that (¯u, V (r )(1 + ε), h¯ ) satisfies (3.18) for t > Tε . By Lemma 3.2, h¯ (t ) ≥ h(t ),
u¯ (t , r ) ≥ u(t , r ),
∀t > Tε , 0 < r < h(t ).
Hence, h¯ (∞) = ∞. Using Lemma 2.3 of [5], we easily see that lim sup u(t , r ) ≤ lim u¯ (t , r ) = U ε (r ) t →∞
t →∞
uniformly for r in any compact subset of [0, ∞), where U ε (r ) is the unique positive solution of the equation
−∆u = u [a(|x|) − b(|x|)u + c (|x|)V (|x|)(1 + ε)] ,
x ∈ Rn .
Note that U ε (r ) is continuously dependent on ε and the arbitrariness of ε , we have lim sup u(t , r ) ≤ U2 (r ) t →∞
uniformly in the compact subset of [0, ∞). The proof is completed.
Combining Theorems 3.1 and 3.2 and Lemma 3.4, we have the following dichotomy theorem. Theorem 3.3. Let (u, v, h) be any solution of (1.1). Then, the following alternative holds: Either (i) spreading: h∞ = ∞ and U1 (r ) ≤ lim inf u(t , r ) ≤ lim sup u(t , r ) ≤ U2 (r ) t →∞
t →∞
uniformly in any compact subset of [0, ∞); or (ii) vanishing: h∞ < R∗ (1, a + cV ) and limt →∞ ∥u(t , ·)∥C [0,h(t )] = 0. Now we decide exactly when each of the two alternative occurs. The discussion will be divided into two cases: (a) h0 ≥ R∗ (1, a + cV ),
(b) h0 < R∗ (1, a + cV ).
For the case (a), due to h′ (t ) > 0 for t > 0, we must have h∞ > R∗ (1, a + cV ). Hence Theorem 3.2 and Lemma 3.4 imply the following result. Theorem 3.4. If h0 ≥ R∗ (1, a + cV ), then h∞ = ∞. Next we discuss the case (b). Lemma 3.5. Suppose h0 < R∗ (1, a + cV ). Then there exists µ0 > 0 depending on u0 such that h∞ = ∞ if µ ≥ µ0 . Proof. Consider the following auxiliary problem
u − ∆u = u a(r ) − b(r )u , t ur (t , 0) = u(t , h(t )) = 0, ′ h (t ) = −µur (t , h(t )), u(0, r ) = u0 (r ),
t > 0, 0 < r < h(t ), t > 0, t > 0, 0 ≤ r ≤ h0
with h(0) = h0 . By Lemma 3.3, h(t ) ≤ h(t ),
u(t , r ) ≤ u(t , r ),
∀t > 0, 0 < r < h(t ).
Note that h0 < R (1, a + cV ) < R (1, a) and µ ≥ µ0 , by Lemma 2.8 of [5], we have h(∞) = ∞. Therefore, h∞ = ∞. The proof is finished. ∗
∗
Lemma 3.6. Suppose h0 < R∗ (1, a + cV ), then there exists µ0 > 0 depending on (u0 , v0 ) such that h∞ < ∞ if µ ≤ µ0 . Proof. We are going to construct a suitable upper solution to (1.1) and then apply Lemma 3.2. We define v¯ (t , r ) as in (3.22), then v¯ (t , r ) satisfies
v¯ t − d∆v¯ − v¯ [α(r ) − γ (r )¯v ] ≥ 0, v¯ (0, r ) ≥ v0 (r ),
t > 0, 0 < r < ∞, 0 < r < ∞.
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
Inspired by Du and Guo [5] and Wang and Zhao [3], for t > 0 and r ∈ [0, η(t )], we define
t H κ22 −θ s −ρ + f (t ) = M exp e ds , t > 0; κ1 0 1/2 t δ2 2 f (s)ds , t > 0; η(t ) = h0 (1 + δ) + ∞ f (s)ds 0 0 h0 r u¯ (t , r ) = f (t )W , t > 0, 0 ≤ r ≤ η(t ), η(t ) where H , θ defined as in (3.22), ρ, δ and M are positive constants to be chosen later and W (|x|) is the first eigenfunction of the problem
−∆W = λ1 (1, a + cV , h0 )[a(r ) + c (r )V (r )]W , W = 0,
x ∈ Bh0 , x ∈ ∂ Bh0 ,
with W > 0 in Bh0 and ∥W ∥∞ = 1. Since h0 < R∗ (1, a + cV ), we have λ1 (1, a + cV , h0 ) > 1. We also observe that W ′ (0) = 0 and
−(r n−1 W ′ )′ = r n−1 λ1 (1, a + cV , h0 )[a(r ) + c (r )V (r )]W > 0,
∀ 0 < r < h0 .
It follows that W ′ (r ) < 0,
∀ 0 < r ≤ h0 .
On the other hand, we have η′ (t ) > 0 for t ≥ 0 and f ′ (t ) f (t )
= −ρ +
H κ22 −θ t e ,
κ1
∀t > 0.
Hκ2
t
Note that −ρ + κ 2 e−θ t < 0 for t large enough, it is deduced that 0 f (s)ds is uniformly bounded in [0, ∞). 1 Set σ (t ) = η(t )/h0 , then η(t ) = h0 σ (t ). Direct computations yield u¯ t − ∆u¯ − u¯ [a(r ) − b(r )¯u + c (r )¯v ] = f ′ W − fr σ −2 σ ′ W ′ + f
≥ fW
r
− fW [a(r ) − b(r )fW + c (r )¯v ] σ r r f′ λ1 (1, a + cV , h0 ) r + a + c V f σ2 σ σ σ −∆W
− [a(r ) + c (r )(1 + He−θ t )V (r )] −ρ+
≥ fW
H κ22 −θ t e − He−θ t c (r )V (r )
κ1
λ1 (1, a + cV , h0 ) a σr + − 1 a( r ) (1 + 2δ)2 a(r ) λ1 (1, a + cV , h0 ) c σr V σr + − 1 c (r )V (r ) . (1 + 2δ)2 c (r )V (r ) We easily see that h0 (1 + δ) ≤ η(t ) ≤ h0 (1 + 2δ),
1 + δ ≤ σ (t ) ≤ 1 + 2δ.
Hence, due to λ1 (1, a + cV , h0 ) > 1, we can choose δ > 0 sufficiently small such that
λ1 (1, a + cV , h0 ) a σr − 1 > 0, ρ1 := min t >0,r ∈[0,η(t )] (1 + 2δ)2 a( r ) r r λ1 (1, a + cV , h0 ) c σ V σ ρ2 := min − 1 > 0. t >0,r ∈[0,η(t )] (1 + 2δ)2 c (r )V (r ) Setting ρ = ρ1 κ1 + ρ2 κ12 /κ2 , we deduce u¯ t − ∆u¯ − u¯ [a(r ) − b(r )¯u + c (r )¯v ] ≥ 0,
t > 0, r ∈ [0, η(t )].
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
261
We now choose M > 0 sufficiently large such that u0 (r ) ≤ MW
r 1+δ
= u¯ (0, r ),
∀r ∈ [0, h0 ].
The direct calculation yields
η′ (t ) =
h20 δ 2 f (t ) 2η(t )
∞ 0
f (s)ds
,
−µ¯ur (t , η(t )) =
µh0 f (t )|Wr (h0 )| . η(t )
Therefore, if we take
µ0 =
h0 δ 2 2|Wr (h0 )|
∞ 0
f (s)ds
,
then for any 0 < µ ≤ µ0 , there holds η′ (t ) ≥ −µ¯ur (t , η(t )). Thus, (¯u, v¯ , η) satisfies
u¯ t − ∆u¯ ≥ u¯ [a(r ) − b(r )¯u + c (r )¯v ] , v¯ t − d∆v¯ ≥ v¯ [α(r ) − γ (r )¯v ] , u¯ r (t , 0) = 0, v¯ r (t , 0) = 0, ′ u¯ (t , η(t )) = 0, η (t ) ≥ −µ¯ur (t , η(t )), η(0) = (1 + δ)h0 .
t t t t
> 0, 0 < r < η(t ), > 0, 0 < r < ∞, > 0, > 0,
We can apply Lemma 3.2 to conclude that h(t ) ≤ η(t ), u(t , r ) ≤ u¯ (t , r ) for (t , r ) ∈ D∞ and v(t , r ) ≤ v¯ (t , r ) for (t , r ) ∈ G∞ . Therefore, h∞ ≤ limt →∞ η(t ) ≤ h0 (1 + 2δ) < ∞. Theorem 3.5. Suppose h0 < R∗ (1, a + cV ). Then there exist µ∗ ≥ µ∗ > 0 depending on u0 (r ), v0 (r ) and h0 such that h∞ ≤ R∗ (1, a + cV ) if µ ≤ µ∗ , and h∞ = ∞ if µ > µ∗ . Proof. The proof is similar to that of Theorem 5.2 of [3] and Theorem 4.11 of [9]. For the convenience to the readers we shall give the details. We will write (uµ , vµ , hµ ) in place of (u, v, h) to clarify the dependence of the solution of (1.1) on µ. Set
Σ ∗ = µ > 0 : hµ,∞ ≤ R∗ (1, a + cV ) . By Lemma 3.6 and Theorem 3.2, (0, µ0 ] ⊂ Σ ∗ . In view of Lemma 3.5, Σ ∗ [µ0 , ∞) = ∅. Therefore, µ∗ := sup Σ ∗ ∈ [µ0 , µ0 ]. By this definition and Theorem 3.2 we find that hµ,∞ = ∞ when µ > µ∗ . We will show that µ∗ ∈ Σ ∗ . Otherwise, hµ∗ ,∞ = ∞. We can find T > 0 such that hµ∗ (T ) > R∗ (1, a + cV ). By the continuous dependence of (uµ , vµ , hµ ) on µ, there is ε > 0 such that hµ (T ) > R∗ (1, a + cV ) for µ ∈ (µ∗ − ε, µ∗ + ε). It follows that for all such µ,
lim hµ (t ) > hµ (T ) > R∗ (1, a + cV ).
t →∞
This implies that [µ∗ − ϵ, µ∗ + ϵ] ∩ Σ ∗ = ∅, and sup Σ ∗ ≤ µ∗ − ϵ . This contradicts the definition of µ∗ . Define
Σ∗ = ν : ν ≥ µ0 such that hµ,∞ ≤ R∗ (1, a + cV ) for all µ ≤ ν , where µ0 is given by Lemma 3.6. Then µ∗ := sup Σ∗ ≤ µ∗ and (0, µ∗ ) ⊂ Σ∗ . Similar to the above, we can prove that µ∗ ∈ Σ∗ . The proof is completed. We should remark before ending this section that, if the coefficients a(r ), b(r ), c (r ), α(r ), β(r ) and γ (r ) are all positive constants, then the corresponding long time behaviors to Theorems 4.3 and 4.4 of [3] are all true for our present problem (1.1). 4. Discussion We have examined a free boundary problem of a predator–prey model with higher space dimensions and heterogeneous environment for the special case that the environment and solution are radially symmetric. If the environment or solution is not radially symmetric, then the boundary of the spreading domain would not still be a sphere and the Stefan condition h′ (t ) = −µur (t , h(t )) would become very complicated. Similar to the classical Stefan problem, smooth solutions to these free boundary problems need not exist even if the initial data are smooth. It is necessary to make use of other methods to discuss these problems. In this paper, the dynamic behaviors are discussed. It has been proved that: (i) If the size of the initial region Bh0 occupied by predator satisfies h0 ≥ R∗ (1, a + cV ), or h0 < R∗ (1, a + cV ) but the moving parameter/coefficient of free boundaries µ > µ∗ (it depends on the initial data (u0 , v0 ) and h0 ), then h(t ) → ∞
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J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
as t → ∞ and lim inft →∞ u(t , r ) ≥ U1 (r ) (the predator will fill the whole space eventually). Here V (r ) and U1 (r ) are respectively the unique positive solutions of (3.8) and (3.19). (ii) While if h0 < R∗ (1, a + cV ) and µ ≤ µ∗ (it also depends on the initial data (u0 , v0 ) and h0 ), then limt →∞ h(t ) < ∞ (the region occupied eventually by predator is finite), ∥u(t , x)∥C [g (t ),h(t )] → 0 and limt →∞ v(t , r ) = V (r ) uniformly in any bounded subset of [0, ∞). These results tell us that in order to control the prey species (pest species) we should put predator species (natural enemies) at the initial state at least in one of three ways: (i) increase the initial targets of predator species, (ii) increase the moving parameter/coefficient of free boundaries, (iii) increase the initial density of the predator species. At last we should remark that our methods can be used to deal with the general predator–prey models with Holling or other functional responses. Appendix. Global estimate of the solution w to (3.3) Proposition A. Let (u, v, h) be any solution of (1.1) and assume h∞ < ∞, if w(t , y) is the solution of (3.3), then there exists a constant K0 such that
∥w∥
< K0 .
1+α ,1+α C 2 ([1,∞)×B1 )
(A.1)
Proof. The proof is similar to that of Proposition A.1 in [3]. For the readers’ convenience, we give the details. Inspired by [29, Theorem A2]. we denote ϕn (t ) = ϕ(t + n), gn = g (t + n, y), where g may be one of the following functions ψi , a, b, c , z and w . Let w(t + n, y) = ηn (t , y) + θ n (t , y), where ηn and θ n are solutions of
n ηt = ϕn (t )∆ηn , ηn (t , y) = 0, ηn (0, y) = w(n, y),
t > 0, y ∈ B1 , t > 0, y ∈ ∂ B1 y ∈ B1
and
n θt = ϕn (t )∆θ n + ψin θyni + ψin ηyni + wn (an − bn wn + cn zn ), θ n (t , y) = 0, n θ (0, y) = 0,
t > 0, y ∈ B1 , t > 0, y ∈ ∂ B1 y ∈ B1 ,
respectively. Let 0 < λ1 ≤ λ2 ≤ · · · be the eigenvalues of the problem
−∆φ = λφ,
φ(y) = 0,
y ∈ B1 ;
y ∈ ∂ B1
and let φ1 , φ2 , . . . be the corresponding set of orthonormal eigenfunctions. We may express ηn as
η n (t , y) =
t
ϕn (s)ds wkn φk ,
exp −λk 0
k≥1
where wkn = (φk , w(n, ·))L2 . In view of 0 < u ≤ M1 (cf. Lemma 2.1), it follows that
j n 2 ∆ η (T ) 2
L (−1,1)
=
ϕn (s)ds (wkn )2
0
k≥1
≤ sup ℓ2j exp
ℓ≥0
≤
T
(λk )2j exp −2λk
2M12
h2∞ j 4T
2j
−8T ℓ h2∞
∥w(n, x)∥2L2 (−1,1)
e−2j .
By Sobolev’s theorem, we have that ∥ηn (t )∥C 2 (B1 ) ≤ K1 (1 + t −j ), where K1 is independent of n provided that j ≥ 1 + 4n . From 2 −j this last estimate and the differential equation satisfied by ηn , we obtain ∥ηtn (t )∥C (B1 ) ≤ K1 h− 0 (1 + t ). Hence, there exists
positive constant K2 such that ∥ηn ∥
1+α ,1+α C 2 (E1 )
< K2 , where E1 = [ 12 , 2] × B1 and K2 depends only on K1 and E1 .
1 Next we estimate θ n . It is obvious that the function ρ n = e− t θ n satisfies
n ρt = ϕn (t )∆ρ n + ψin ρyni + fn (t , y), ρ n (t , y) = 0, n ρ (0, y) = 0,
t > 0, y ∈ B1 , t > 0, y ∈ ∂ B1 y ∈ B1 ,
J. Zhao, M. Wang / Nonlinear Analysis: Real World Applications 16 (2014) 250–263
263
where fn (t , y) =
1 t
1
e− t wn + 2
1 1 1 ψin − 2 e− t ηyni + e− t wn (an − bn wn + cn zn ),
0,
t
t >0 t = 0.
1 Note that limt →0+ t −j e− t = 0 for any j ≥ 0, we have fn (t , y) is continuous in E = [0, 3] × B1 and ∥fn ∥C (E ) ≤ K3 where K3
is dependent on K1 and independent of n. By using [30, Theorem 4, p. 191], we can obtain that ∥ρ n ∥ K˜ 3 depends on K3 . It follows that ∥θ n ∥
∥w∥
1+α ,1+α (En ) C 2
≤ ∥ηn ∥
1+α ,1+α C 2 (E1 )
1+α ,1+α C 2 (E1 )
+ ∥θ n ∥
< K4 . We therefore have that 1+α ,1+α C 2 (E1 )
1+α ,1+α C 2 (E1 )
< K˜ 3 where
< K2 + K4 = K0 ,
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