Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 126, pp. 1–6.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
UNIQUENESS OF POSITIVE SOLUTIONS FOR AN ELLIPTIC
SYSTEM
WENSHU ZHOU, XIAODAN WEI
Abstract. We prove the uniqueness of positive solutions for an elliptic system
that appears in the study of solutions for a degenerate predator-prey model in
the strong-predator case.
1. Introduction
This article is devoted to showing the uniqueness of positive solutions for the
elliptic system
−∆u = λu − buv in Ω,
v
in Ω,
−∆v = µv 1 − ξ
(1.1)
u
∂ν u = ∂ν v = 0 on ∂Ω,
where Ω ⊂ RN is a smooth bounded domain, ν is the outward unit normal vector
∂
on ∂Ω, ∂ν = ∂ν
, λ, b, µ and ξ are positive constants.
Problem (1.1) appears in the study of positive solutions of the degenerate predator-prey model in the strong-predator case
−∆u = λu − a(x)u2 − βuv in Ω,
v
in Ω,
−∆v = µv 1 −
u
∂ν u = ∂ν v = 0 on ∂Ω,
(1.2)
where β is a positive constant, and a(x) is a continuous function satisfying a(x) = 0
on Ω0 and a(x) > 0 in Ω \ Ω0 , where Ω0 is a smooth domain with Ω0 ⊂ Ω. Recently, problem (1.2) has been studied in [2, 3]. Under the condition µ > λ ≥ λ1 ,
where λ1 denotes the first eigenvalue of the Laplace equation on Ω0 with homogenous Dirichlet boundary condition, Du and Wang [3] described spatial patterns
of positive solutions of problem (1.2) by studying asymptotic behavior of positive
solutions as β → 0+ (weak-predator), β → +∞ (strong-predator) and µ → +∞
(small-predator diffusion), respectively. For related work on problem (1.2), please
refer to [8].
2000 Mathematics Subject Classification. 35J57, 92D25.
Key words and phrases. Predator-prey model; strong-predator; positive solution; uniqueness.
c
2011
Texas State University - San Marcos.
Submitted April 19, 2011. Published September 29, 2011.
Supported by grants 10901030 and 11071100 from the National Natural Science Foundation
of China, 2009A152 from the Department of Education of Liaoning Province.
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W. ZHOU, X. WEI
EJDE-2011/126
λ
Clearly, problem (1.1) has a positive solution (u, v) = ( ξλ
b , b ). In [3, Remark
3.2], the authors pointed out that when the spatial dimension N = 1, the positive
solution of problem (1.1) is unique for any µ > 0 by a simple variation of the
arguments in [6]. In the present paper, we prove the uniqueness for all sufficiently
large µ in the high dimensional case, which can be stated as follows
Theorem 1.1. Let N ≥ 2. Then there exists a positive constant µ0 depending
only on λ and Ω such that problem (1.1) admits a unique positive solution for any
µ ≥ µ0 .
Remark 1.2. The proof to Theorem 1.1 is based on the fact that (û, v̂) is a positive
solution of problem (1.1) if and only if ( ξb û, bv̂) is a positive solution of
−∆u = u(λ − v) in Ω,
v
−∆v = µv 1 −
in Ω,
u
∂ν u = ∂ν v = 0 on ∂Ω.
(1.3)
Remark 1.3. As a result of Theorem 1.1 and [3, Remarks 3.1-3.2], one can prove
that if (uβ , vβ ) is a solution of problem (1.2), then for any µ ≥ µ0 , we have, as
β → +∞,
u
vβ β
,
* (1, 1) in [H 1 (Ω)]2 ,
kuβ k∞ kvβ k∞
u
vβ β
,
→ (1, 1) in [Lp (Ω)]2 , ∀p > 1.
kuβ k∞ kvβ k∞
2. Proof of Theorem 1.1
First recall several preliminary results.
Lemma 2.1 (Harnack Inequality [5]). Let w ∈ C 2 (Ω)∩C 1 (Ω) be a positive solution
to ∆w(x) + c(x)w(x) = 0, where c ∈ C(Ω), satisfying the homogeneous Neumann
boundary condition. Then there exists a positive constant C which depends only on
B where kck∞ ≤ B such that maxΩ w ≤ C minΩ w.
Lemma 2.2 (Maximum Principle [7]). Suppose that g ∈ C 1 (Ω × R1 ). Then
(i) if w ∈ C 2 (Ω) ∩ C 1 (Ω) satisfies ∆w(x) + g(x, w) ≥ 0 in Ω, ∂ν w ≤ 0 on ∂Ω,
and w(x0 ) = maxΩ w, then g(x0 , w(x0 )) ≥ 0.
(ii) if w ∈ C 2 (Ω) ∩ C 1 (Ω) satisfies ∆w(x) + g(x, w) ≤ 0 in Ω, ∂ν w ≥ 0 on ∂Ω,
and w(x0 ) = minΩ w, then g(x0 , w(x0 )) ≤ 0.
The following lemma can be inferred from [2, Lemma 3.7] (see also [8]).
Lemma 2.3. Let {un } ⊂ H 1 (Ω) satisfy, in the weak sense,
−∆un ≤ Aun ,
un ≥ 0,
∂ν un |∂Ω = 0,
kun k∞ ≤ B, ∀n ≥ 1,
where A and B are positive constants. Then there exists a subsequence of {un },
still denoted by {un }, and a nonnegative function u ∈ H 1 (Ω) ∩ Lp (Ω) for all p > 1,
such that
un * u in H 1 (Ω), un → u in Lp Ω).
If we further assume that kun k∞ ≥ δ > 0 for all n ≥ 1, then u 6= 0.
The following lemma gives the uniform bounds of the positive solutions for problem (1.3).
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UNIQUENESS OF POSITIVE SOLUTIONS
3
Lemma 2.4. Let (uµ , vµ ) be a positive solution of problem (1.3). Then there exist
a positive constant µ0 = µ0 (λ, Ω) and two positive constants C2 , C1 independent of
µ such that for all µ ≥ µ0 ,
C1 ≤ uµ ,
vµ ≤ C2
on Ω.
(2.1)
Moreover, as µ → +∞,
uµ → λ
in C 1 (Ω).
(2.2)
Proof. By Lemma 2.2 and the definition of vµ , it follows that
max uµ ≥ max vµ ,
Ω
min vµ ≥ min uµ .
Ω
Ω
(2.3)
Ω
Hence, to prove (2.1), it suffices to show that there exist a positive constant µ0 =
µ0 (λ, Ω) and two positive constants C2 , C1 independent of µ such that
C1 ≤ min uµ ,
Ω
max uµ ≤ C2 ,
∀µ ≥ µ0 .
(2.4)
Ω
We first prove the second inequality of (2.4). Assume on the contrary that there
exist a sequence {µn } converging to +∞ and the corresponding solution (uµn , vµn ),
such that
kuµn k∞ → +∞ as n → +∞.
Denote
ûµn =
uµn
,
kuµn k∞ + kvµn k∞
v̂µn =
vµn
.
kuµn k∞ + kvµn k∞
1
2
Then ûµn and v̂µn satisfy kûµn k∞ + kv̂µn k∞ = 1, kûµn k∞ ≥
−∆ûµn = ûµn (λ − vµn )
by (2.3), and
in Ω,
v̂µn in Ω,
ûµn
= 0 on ∂Ω.
−∆v̂µn = µn v̂µn 1 −
∂ν ûµn = ∂ν v̂µn
(2.5)
In particular, we have
− ∆ûµn ≤ λûµn
in Ω,
∂ν ûµn = 0
on ∂Ω.
(2.6)
By Lemma 2.3 and kv̂µn k∞ ≤ 1, there exist a subsequence of {(ûµn , v̂µn )}, still
denoted by itself, and a pair of non-negative functions (û, v̂) ∈ H 1 (Ω) ∩ Lp (Ω) ×
L∞ (Ω) for all p > 1, û 6= 0, such that
ûµn * û
in H 1 (Ω),
ûµn → û
in Lp (Ω),
in L2 (Ω).
v̂µn * v̂
Integrating the first equation of (2.5) over Ω yields
Z
Z
Z
λ
ûµn dx =
vµn ûµn dx = (kuµn k∞ + kvµn k∞ )
ûµn v̂µn dx.
Ω
Ω
Ω
From kuµn k∞ → +∞ (n → +∞), we have
Z
Z
ûµn v̂µn dx = lim
ûv̂dx = lim
Ω
n→+∞
Ω
λ
n→+∞ kuµn k∞ + kvµn k∞
Z
ûµn dx = 0. (2.7)
Ω
By the second equation in (2.5), v̂µn is a positive solution of
− ∆w + µn
v̂µn
w = µn w
ûµn
in Ω,
∂ν w = 0
on ∂Ω.
(2.8)
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W. ZHOU, X. WEI
EJDE-2011/126
From the variational characterization of the first eigenvalue it follows that
Z
Z
Z
v̂µn 2
|∇φ|2 dx + µn
φ dx ≥ µn
φ2 dx
û
µ
Ω
Ω
Ω
n
for any φ ∈ {w ∈ H 2 (Ω); ∂ν w = 0 on ∂Ω} (cf. [1]). Taking φ = ûµn yields
Z
Z
Z
1
2
|∇ûµn | dx +
v̂µn ûµn dx ≥
û2µn dx.
µn Ω
Ω
Ω
R
Passing to the limit and using (2.7), we obtain Ω û2 dx = 0, so û = 0, which is a
contradiction. Thus there exist a positive constant µ0 = µ0 (λ, Ω) and a positive
constant C2 independent of µ such that
max uµ ≤ C2 , ∀µ ≥ µ0 .
(2.9)
Ω
Next we prove the first inequality in (2.4). Suppose that this is not so. Then
there exist {µn } converging to +∞ and the corresponding solution (uµn , vµn ) such
that
lim min uµn = 0.
(2.10)
n→+∞
Ω
Now rewrite the equation of uµn as
∆uµn + f (x)uµn = 0
in Ω,
∂ν uµn = 0
on ∂Ω,
where f (x) = λ − vµn . By the first estimate of (2.3) and (2.9), we have, for all
sufficiently large n,
kf k∞ ≤ λ + kvµn k∞ ≤ λ + C2 ,
by Lemma 2.1, there exists a positive constant C3 independent of n such that for
all sufficiently large n,
max uµn ≤ C3 min uµn .
Ω
Ω
Therefore, it follows from (2.10) and the first estimate of (2.3) that
lim max uµn = 0,
n→+∞
Ω
lim max vµn = 0.
n→+∞
(2.11)
Ω
Denote ũµn = uµn /kuµn k∞ . Then ũµn satisfies kũµn k∞ = 1, and
−∆ũµn = ũµn (λ − vµn )
in Ω,
∂ν ũµn = 0
on ∂Ω.
By (2.3), (2.9) and the definition of ũµn , both {−∆ũµn } and {ũµn } are bounded
sets in L∞ (Ω). By the standard elliptic theory (cf. [4, Theorem 9.9]), {ũµn } is
bounded in W 2,p (Ω) for any p > 1. Therefore, there exist a subsequence of {ũµn },
still denoted by itself, and a nonnegative function ũ ∈ C 1 (Ω) with kũk∞ = 1, such
that
ũµn → ũ in C 1 (Ω),
by (2.11) and the definition of ũµn , we derive that
−∆ũ = λũ
in Ω,
∂ν ũ = 0
on ∂Ω.
This implies ũ = 0, which is a contradiction. This proves (2.1).
Next we show (2.2). By (2.1) and the equation of uµ , {−∆uµ }µ≥µ0 , {uµ }µ≥µ0
and {vµ }µ≥µ0 are bounded sets in L∞ (Ω). By the standard elliptic theory, there
exist a sequence {µn } converging to +∞, the corresponding solution (uµn , vµn ) of
problem (1.1) and a pair of functions (u, v) ∈ C 1 (Ω) × L∞ (Ω) with C1 ≤ u, v ≤ C2 ,
such that
uµn → u in C 1 (Ω), vµn * v in L2 (Ω).
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UNIQUENESS OF POSITIVE SOLUTIONS
5
Clearly, (u, v) satisfies, in the weak sense,
−∆u = u(λ − v)
in Ω,
∂ν u = 0
on ∂Ω.
Multiplying the equation of vµn by φ ∈ C0∞ (Ω) and integrating over Ω, we get
Z
Z
1
vµ −
vµn ∆φdx =
vµn 1 − n φdx.
µn Ω
uµn
Ω
Passing to the limit yields
Z
v 1−
Ω
v
u
v
φdx = 0,
u
which implies that v 1 −
= 0. Since v 6= 0, we must have v = u. By the
regularity theory of elliptic equation, u ∈ C 2 (Ω) and satisfies
−∆u = u(λ − u)
in Ω,
∂ν u = 0
on ∂Ω.
Then u = λ. The proof is complete.
Proof of Theorem 1.1. Let (uµ , vµ ) be a positive solution of problem (1.3). By
(2.2), there exists a constant µ0 = µ0 (λ, Ω) such that for all µ ≥ µ0 ,
uµ ≤ 2λ
Multiplying the equations of uµ and vµ by
Z
−2λ
Ω
|∇uµ |2
dx +
u3µ
Z
Ω
on Ω.
λ−uµ
u2µ
|∇uµ |2
dx =
u2µ
(2.12)
and
Z
Ω
1 λ−vµ
µ vµ ,
respectively, we obtain
(λ − uµ )(λ − vµ )
dx,
uµ
and
λ
−
µ
Z
Ω
|∇vµ |2
dx =
vµ2
Z
Ω
Z
=
Ω
(uµ − vµ )(λ − vµ )
dx
uµ
(uµ − λ)(λ − vµ )
dx +
uµ
Z
Ω
(λ − vµ )2
dx.
uµ
Adding these two equalities yields
Z
Z
Z
Z
|∇uµ |2
|∇uµ |2
λ
|∇vµ |2
(λ − vµ )2
− 2λ
dx
+
dx
−
dx
=
dx.
3
2
2
uµ
uµ
µ Ω vµ
uµ
Ω
Ω
Ω
(2.13)
Noting (2.12), for all µ ≥ µ0 , we obtain
Z
Z
Z
|∇uµ |2
|∇uµ |2
|∇uµ |2
−2λ
dx
+
dx
=
(u
−
2λ)
dx ≤ 0,
µ
u3µ
u2µ
u3µ
Ω
Ω
Ω
R (λ−v )2
which and (2.13) implies that Ω uµµ dx ≤ 0, hence vµ = λ for all µ ≥ µ0 , so
uµ = λ for all µ ≥ µ0 . Combining this and Remark 1.2 completes the proof.
Acknowledgments. The authors thank the anonymous referee for his/her insightful comments.
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W. ZHOU, X. WEI
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Wenshu Zhou
Department of Mathematics, Dalian Nationalities University, Dalian 116600, China
E-mail address: pdezhou@126.com, wolfzws@163.com
Xiaodan Wei
School of Computer Science, Dalian Nationalities University, Dalian 116600, China
E-mail address: weixiaodancat@126.com