Research

# Global behavior for a diffusive predator-prey system with Holling type II functional response

Yanzhong Zhao

Author Affiliations

Department of Basic Research, Qinghai University, Xining, 810016, China

Boundary Value Problems 2012, 2012:111  doi:10.1186/1687-2770-2012-111

 Received: 6 April 2012 Accepted: 25 September 2012 Published: 9 October 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

A strongly coupled self- and cross-diffusion predator-prey system with Holling type II functional response is considered. Using the energy estimate, Sobolev embedding theorem and bootstrap arguments, the global existence of non-negative classical solutions to this system in which the space dimension is not more than five is obtained.

MSC: 35K50, 35K55, 35K57, 92D40.

##### Keywords:
predator-prey system; cross-diffusion; global solution

### 1 Introduction

In this paper, we consider the global existence of non-negative classical solutions to the following diffusion predator-prey system with Holling type II functional response:

{ u t = Δ ( d 1 u + α 11 u 2 ) + α u ( 1 u K ) β m u v 1 + a m u , x Ω , t > 0 , v t = Δ ( d 2 v + α 21 u v + α 22 v 2 ) r v + c β m u v 1 + a m u , x Ω , t > 0 , η u ( x , t ) = η v ( x , t ) = 0 , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) 0 , v ( x , 0 ) = v 0 ( x ) 0 , x Ω , (1.1)

where Ω is a bounded region in R n ( n 1 ) with a smooth boundary Ω; η is the outward normal on Ω, η = / η ; u 0 ( x ) and v 0 ( x ) are non-negative smooth functions and are not identically zero; u and v denote the population densities of predator and prey, respectively; α, β, r, a, K, m and c are positive constants, and m ( 0 , 1 ] ; d 1 and d 2 are the diffusion rates of the two species; α i j ( i , j = 1 , 2 ) are given non-negative constants, α 11 and α 22 are self-diffusion rates; α 21 is the cross-diffusion rate. It means that the diffusion is from one species of high-density areas to the other species of low-density areas. See [1,2] for more details on the ecological backgrounds of this system.

Obviously, the non-negative equilibrium solutions of system (1.1) are ( K , 0 ) and ( u , v ) = ( r m ( c β a r ) , α c K m ( c β a r ) α c r K m 2 ( c β a r ) 2 ) . For the reaction-diffusion problem of system (1.1), i.e., α i i = 0 ( i = 1 , 2 ), the global attraction, persistence and stability of non-negative equilibrium solutions are studied in [3]. The main result can be summarized as follows:

(1) If m < r K ( c β a r ) , a semi-trivial solution ( K , 0 ) is globally asymptotically stable;

(2) If r K ( c β a r ) < m r K ( c β a r ) + 1 K a and c β > a r , a unique positive constant solution ( u , v ) is globally asymptotically stable;

(3) If r K ( c β a r ) < m < r K ( c β a r ) + c β K a ( c β a r ) and c β > a r , a positive constant solution ( u , v ) is locally asymptotically stable.

In view of the study of dynamic behavior of a predator-prey reaction-diffusion system with Holling type II functional response, a natural problem is what the global behavior for a predator-prey cross-diffusion system (1.1) is. To the best of our knowledge, the existing results are very few. In this paper, we consider the space dimension to be less than six, and initial function u 0 ( x ) and v 0 ( x ) under some smooth conditions, using the energy estimate, Sobolev embedding theorem and bootstrap arguments, we consider the global existence of non-negative classical solutions for system (1.1).

We denote Q T = Ω × ( 0 , T ) . u W q 2 , 1 ( Q T ) means that u, u x i , u x i x j ( i , j = 1 , , n ) and u t are in L q ( Q T ) . u L p , q ( Q T ) = [ 0 T ( Ω | u ( x , t ) | p d x ) q p d t ] 1 q . u V 2 ( Q T ) = sup 0 t T u ( , t ) L 2 ( Ω ) + u L 2 ( Q T ) .

### 2 Auxiliary results

Lemma 2.1Let ( u , v ) be the solution of (1.1). There exists a positive constant M 0 ( 1 ) such that

0 u M 0 , 0 v , t 0 . (2.1)

Proof Firstly, the existence of local solutions for system (1.1) is given in [4-6]. Roughly speaking, if u 0 , v 0 W p 1 ( Ω ) , p > 1 , there exists the maximum T + such that system (1.1) admits a unique non-negative solution

u , v C ( [ 0 , T ) , W p 1 ( Ω ) ) C ( ( 0 , T ) , C ( Ω ) ) . (2.2)

If

sup { u ( , t ) w p 1 ( Ω ) , v ( , t ) w p 1 ( Ω ) : 0 < t < T } < ,

then T = + .

Choose M 0 = max { K , u 0 L ( Ω ) } . By use of the maximum principle, the non-negative solution of system (1.1) can be derived from the maximum principle, i.e., u , v 0 for all t 0 . This completes the proof of Lemma 2.1. □

Lemma 2.2Let X = ( d 1 + α 11 u ) u , u L ( Q T ) for the solution to the following equation:

u t = Δ [ ( d 1 + α 11 u ) u ] + α u ( 1 u K ) β m u v 1 + a m u , ( x , t ) Ω × ( 0 , T ) , η u = 0 , ( x , t ) Ω × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) 0 , x Ω ,

where d 1 , α 11 are positive constants and 0 u L 2 ( Q T ) . Then there exists a positive constant C ( T ) , depending on u 0 W 2 1 ( Ω ) and u 0 L ( Ω ) , such that

X W 2 2 , 1 ( Q T ) C ( T ) . (2.3)

Furthermore,

X V 2 ( Q T ) , u L 2 ( n + 2 ) n ( Q T ) . (2.4)

Proof From X = ( d 1 + α 11 u ) u , it is easy to find that

X t = ( d 1 + 2 α 11 u ) Δ X + C 1 C 2 v , (2.5)

where C 1 = d 1 α u + ( 2 α 11 α d 1 α K ) u 2 2 α 11 α K u 3 and C 2 = β m u 1 + a m u ( d 1 + 2 α 11 u ) . C 1 and C 2 are bounded in Q T from (2.1). Multiplying (2.5) by Δ X and integrating by parts over Q t yields

(2.6)

Using the Hölder inequality and Young inequality to estimate the right-hand side of (2.6), we have

C 1 + C 2 v L 2 ( Q T ) Δ X L 2 ( Q T ) m 1 ( 1 + v L 2 ( Q T ) ) Δ X L 2 ( Q T ) m 1 2 ( 1 + M 3 ) 2 2 d 1 + d 1 2 Δ X L 2 ( Q T ) 2 (2.7)

with some m 1 > 0 . Substituting (2.7) into (2.6), we obtain

sup 0 t T Ω | X ( x , t ) | 2 d x + d 1 Q t | Δ X | 2 d x d t m 2 ,

where m 2 depends on u 0 W 2 1 ( Ω ) and u 0 L ( Ω ) . So, we know X V 2 ( Q T ) . Since X L 2 ( Q T ) , it follows from the elliptic regularity estimate [[7], Lemma 2.3] that

Q T | X x i x j | 2 d x d t m 3 , i , j = 1 , , n .

From (2.5), we have X t L 2 ( Q T ) . Hence, X W 2 2 , 1 ( Q T ) C ( T ) . Moreover, (2.4) can be obtained by use of the Sobolev embedding theorem. □

Lemma 2.3Assume that w W P 2 , 1 ( Q T ) C 2 , 1 ( Ω ¯ × [ 0 , T ) ) is a bounded function satisfying

w t a ( x , t , w ) Δ w + f ( x , t ) in  Q T

with the boundary condition w v 0 on Q T , where f L P ( Q T ) . ThenWis in L 2 p ( Q T ) .

The proof of the above lemma can be found in [[8], Proposition 2.1].

The following result can be derived from Lemma 2.3 and Lemma 2.4 of [9].

Lemma 2.4Let p > 1 , p ˜ = 2 + 4 p n ( p + 1 ) . If

sup 0 t T w L 2 p p + 1 ( Ω ) + w L 2 ( Q T ) < ,

and there exist positive constants β ( 0 , 1 ) and C T such that Ω | w ( , t ) | β d x C T ( t T ), there exists a positive constant M , independent ofwbut possibly depending onn, Ω, p, βand C T , such that

w L p ˜ ( Q T ) M { 1 + ( sup 0 t T w ( t ) L 2 p p + 1 ( Ω ) ) 4 p n ( p + 1 ) p ˜ w L 2 ( Q T ) 2 p ˜ } .

Finally, one proposes some standard embedding results which are important to obtain the L and C 2 + α , 1 + α 2 ( Q ¯ T ) normal estimates of the solution for (1.1).

Lemma 2.5There exists a constant C 3 ( T ) such that u L 4 ( Q T ) C 3 ( T ) .

Proof Let δ = α 11 / d 1 , X = ( 1 + δ u ) u . By Lemma 2.1, u is bounded. Therefore, X is also bounded. By Lemma 2.2, we have X W 2 2 , 1 ( Q T ) . Moreover, X satisfies

X t d 1 ( 1 + 2 δ u ) Δ X + α u ( 1 + 2 δ u ) = d 1 2 + 4 δ d 1 X Δ X + α u ( 1 + 2 δ u ) .

By Lemma 2.3 with p = 2 , a ( x , t , ξ ) = d 1 2 + 4 δ d 1 ξ , f ( x , t ) = α u ( x , t ) ( 1 + 2 δ u ( x , t ) ) , we obtain the desired result. □

Lemma 2.6Let Ω R n be a fixed bounded domain and Ω C 2 . Then for all u W q 2 , 1 ( Q T ) with q 1 , one has

(1) u L p ( Q T ) C u W q 2 , 1 ( Q T ) , 1 p ( n + 2 ) q n + 2 q , q < n + 2 ;

(2) u L p ( Q T ) C u W q 2 , 1 ( Q T ) , 1 p , q = n + 2 ;

(3) u C α , α 2 ( Q T ) C u W q 2 , 1 ( Q T ) , 1 n + 2 q α 1 , q > n + 2 ,

whereCis a positive constant dependent onq, n, Ω, T.

### 3 The existence of classical solutions

The main result about the global existence of non-negative classical solutions for the cross-diffusion system (1.1) is given as follows.

Theorem 3.1Assume that u 0 > 0 and v 0 > 0 satisfy homogeneous Neumann boundary conditions and belong to C 2 + α ( Ω ¯ ) for some α ( 0 , 1 ) . Then system (1.1) has a unique non-negative solution ( u , v ) C 2 + α , 1 + α 2 ( Ω ¯ × [ 0 , ) ) if the space dimension is n 5 .

Proof When n = 1 , the proof is similar to the methods of [10-12]. So, we just give the proof of Theorem 3.1 for n = 2 , 3 , 4 , 5 . The proof is divided into three parts.

(i) L 1 -, L 2 -estimate and L q -estimate for v.

Firstly, integrating the first equation of (1.1) over Ω, we have

d d t Ω u d x = Ω u ( α α u K β m v 1 + a m u ) d x Ω α u d x α K Ω u 2 d x α Ω u d x α K | Ω | ( Ω u d x ) 2 .

Thus, for all t 0 , we can obtain

Ω u d x M 1 ,

where M 1 = max { K | Ω | , Ω u 0 d x } .

Furthermore,

u L 1 ( Q T ) 0 T M 1 d t M 1 . (3.1)

Secondly, linear combination of the second and first equations of (1.1) and integrating over Ω yields

d d t Ω ( c u + v ) d x = c Ω α u d x c α K Ω u 2 d x r Ω v d x c ( r + α ) Ω u d x c α K | Ω | ( Ω u d x ) 2 r Ω ( c u + v ) d x c ( r + α ) M 1 r Ω ( c u + v ) d x .

So, we get

Ω ( c u + v ) d x max { c ( r + α ) M 1 r , Ω ( c u 0 ( x ) + v 0 ( x ) ) d x } M 2 , t 0 . (3.2)

Further,

v L 1 ( Q T ) 0 T M 1 d t M 2 . (3.3)

Then multiplying both sides of the second equation of system (1.1) by v and integrating over Ω, we obtain

1 2 d d t Ω v 2 d x = Ω v [ d 2 v + α 21 u v + α 22 v 2 ] d x r Ω v 2 d x + c m β Ω u v 2 1 + a m u d x d 2 Ω | v | 2 d x α 21 Ω v v u d x 2 α 22 Ω v | v | 2 d x + c β a Ω v 2 d x .

Integrating the above expression in [ 0 , 1 ] yields

(3.4)

Estimating the first term on the right-hand side of (3.4),

| Q t u v v d x d t | = | Q t u v 1 2 v v 1 2 d x d t | ε 1 Q t v | u | 2 d x d t + 1 4 ε 1 Q t v | v | 2 d x d t ε 2 Q t v 2 d x d t + 1 4 ε 2 Q t | u | 4 d x d t + 1 4 ε 1 Q t v | v | 2 d x d t . (3.5)

Substituting (3.5) into (3.4) yields

(3.6)

Select ε 1 > α 21 8 α 22 and denote C 4 = 4 α 22 α 21 2 ε 1 . Notice the positive equilibrium point of (1.1) exists under condition c β > a r , then

(3.7)

By Lemma 2.5, u L 4 ( Q T ) C 3 ( T ) . Integrating the above inequality and using the Gronwall inequality, we get

sup 0 < t < T Ω v 2 d x C ( T ) .

Hence, there exists a positive constant M 3 such that Ω v 2 d x M 3 . Furthermore, we have

v L 2 ( Q T ) 0 T M 2 d t M 3 . (3.8)

Secondly, multiplying both sides of the second equation of system (1.1) by q v q 1 ( q > 1 ) and integrating over Ω, we have

d d t Ω v q d x 4 ( q 1 ) d 2 q Ω | ( v q 2 ) | 2 d x 8 q ( q 1 ) α 22 ( q + 1 ) 2 Ω | ( v q + 1 2 ) | 2 d x q ( q 1 ) α 21 Ω v q 1 u v d x + q Ω v q ( r + c β m u 1 + a m u ) d x .

Integrating the above equation over [ 0 , t ] ( t T ), it is clear that

(3.9)

By Lemma 2.2, it can be found that u L 2 ( n + 2 ) n ( Q T ) . According to the Hölder inequality and Young inequality, we get

(3.10)

Choose an appropriate number ε satisfying C 3 ε 2 8 q ( q 1 ) α 22 ( q + 1 ) 2 . Substituting (3.10) into (3.9) and taking v ¯ = v q + 1 2 , we have

(3.11)

Let

E sup 0 < t < T Ω v ¯ 2 q q + 1 ( x , t ) d x + Q T | v ¯ | 2 d x d t .

From (3.11), we know

E C 4 ( 1 + v ¯ L ( q 1 ) ( n + 2 ) q + 1 ( Q T ) 2 ( q 1 ) q + 1 + v ¯ L 2 q q + 1 ( Q T ) 2 q q + 1 ) .

When q < n ( n + 4 ) n 2 4 , it is easy to find that 2 q q + 1 < 2 < q ˜ and ( q 1 ) ( n + 2 ) q + 1 < q ˜ = 2 + 4 q n ( q + 1 ) . So,

E C 5 ( 1 + v ¯ L q ˜ ( Q T ) 2 ( q 1 ) q + 1 + v ¯ L q ˜ ( Q T ) 2 q q + 1 ) . (3.12)

Set β = 2 q + 1 ( 0 , 1 ) . It follows from the L 1 ( Ω ) -estimate for v that

v ¯ L β ( Ω ) = ( Ω | v ¯ ( , t ) | β d x ) 1 β = v L 1 ( Ω ) 1 β M 2 1 β , t T .

By Lemma 2.4 and (3.12), we know

E C 5 [ 1 + ( M + M sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 4 q n ( q + 1 ) q ˜ v ¯ L 2 ( Q T ) 2 q ˜ ) 2 ( q 1 ) q + 1 + ( M + M sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 4 q n ( q + 1 ) q ˜ v ¯ L 2 ( Q T ) 2 q ˜ ) 2 q q + 1 ] C 6 [ 1 + ( sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 2 q q + 1 ) 4 ( q 1 ) n ( q + 1 ) q ˜ ( v ¯ L 2 ( Q T ) 2 ) 2 ( q 1 ) ( q + 1 ) q ˜ + ( sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 2 q q + 1 ) 4 ( q 1 ) n ( q + 1 ) q ˜ ( v ¯ L 2 ( Q T ) 2 ) 2 q ( q + 1 ) q ˜ ] C 6 ( 1 + E 4 ( q 1 ) n ( q + 1 ) q ˜ E 2 ( q 1 ) ( q + 1 ) q ˜ + E 4 q n ( q + 1 ) q ˜ E 2 q ( q + 1 ) q ˜ ) . (3.13)

Obviously, 4 ( q 1 ) n ( q + 1 ) q ˜ + 2 ( q 1 ) ( q + 1 ) q ˜ ( 0 , 1 ) and 4 q n ( q + 1 ) q ˜ + 2 q ( q + 1 ) q ˜ ( 0 , 1 ) . It is easy to know that E is bounded by use of reduction to absurdity. Since q < n ( n + 4 ) n 2 4 , ( q + 1 ) q ˜ 2 ( 1 , 2 ( n + 1 ) n 2 ) . So, v ¯ L q ˜ ( Q T ) is bounded, i.e., v L ( q + 1 ) q ˜ 2 ( Q T ) . Denote ( q + 1 ) q ˜ 2 still as q. So,

v L q ( Q T ) , q ( 1 , 2 ( n + 1 ) n 2 ) . (3.14)

Finally, when n = 2 , 3 , 4 , 5 , ( n 2 4 ) q < n 2 + 4 n with q = 2 . For n 5 , taking q = 2 in (3.9), it follows from (3.8) that there exists a positive constant M 4 such that

v V 2 ( Q T ) M 4 . (3.15)

By embedding theorem, we get

v L 2 ( n + 2 ) n ( Q T ) M 4 .

(ii) L -estimate for v.

The second equation of system (1.1) can be written as the following divergence form:

v t = i , j = 1 n x i ( a i j ( x , t ) v x j ) + i = 1 n x i ( a i ( x , t ) v ) + v ( r + c β m u 1 + a m u ) , (3.16)

where a i j ( x , t ) = ( d 2 + α 21 u + 2 α 22 v ) δ i j , a i ( x , t ) = α 21 u x j and δ i j is the Kronecker sign.

In order to apply the maximum principle [13] to (3.16), we need to prove the following conditions:

(1) v V 2 ( Q T ) is bounded;

(2) i , j = 1 n a i j ( x , t ) ξ i ξ j ν i = 1 n ξ i 2 ;

(3) i = 1 n a i 2 ( x , t ) , v ( r + c β m u 1 + a m u ) L p , r ( Q T ) μ 1 ,

where ν, μ 1 are positive constants and

1 r + n 2 p = 1 χ , 0 < χ < 1 , p [ n 2 ( 1 χ ) , + ) , r [ 1 1 χ , + ) , n 2 . (3.17)

Next, we will show that the above conditions (1) to (3) are satisfied for (3.16). When n 5 , it is easy to find that the condition (1) is satisfied by use of (3.15). Since i , j = 1 n a i j ( x , t ) ξ i ξ j d 3 | ξ | 2 for all ξ R n , the condition (2) is verified. In view of the condition (3), we take appropriate q and r. Rewrite the first equation of system (1.1) as

u t = [ ( d 1 + 2 α 11 u ) u ] + α u ( 1 u K ) β m u v 1 + a m u . (3.18)

When n = 2 , 3 , 4 , 5 , n + 2 2 < 2 ( n + 1 ) n 2 , it is clear that d 1 + 2 α 11 u has an upper bound over Q ¯ T by Lemma 2.1. Set

q ( n + 2 2 , 2 ( n + 1 ) n 2 ) .

From (3.14), we have α u ( 1 u K ) β m u v 1 + a m u L q ( Q T ) . Therefore, all conditions of the Hölder continuity theorem [[5], Theorem 10.1] hold for (3.18). Hence,

u C β , β 2 ( Q ¯ T ) , β ( 0 , 1 ) . (3.19)

We will discuss (2.5) which is the corresponding form of (3.18). It follows from (2.1) and (3.14) that C 1 C 2 v L q ( Q T ) , q ( n + 2 2 , 2 ( n + 1 ) n 2 ) . From (3.19), we obtain d 1 + α 11 u C β , β 2 ( Q ¯ T ) . Thus, according to the parabolic regularity result of [[5], pp.341-342, Theorem 9.1], we can conclude that

X W q 2 , 1 ( Q T ) , q ( n + 2 2 , 2 ( n + 1 ) n 2 ) , (3.20)

which implies that X L ( n + 2 ) q n + 2 q ( Q T ) by Lemma 2.6.

Since X = ( d 1 + 2 α 11 u ) u , we have u = ( d 1 + 2 α 11 u ) 1 X , i.e., u L ( n + 2 ) q n + 2 q ( Q T ) . It means that | u | 2 , | v | 2 L ( n + 2 ) q 2 ( n + 2 q ) ( Q T ) . So, i = 1 n a i 2 ( x , t ) L ( n + 2 ) q 2 ( n + 2 q ) ( Q T ) . From (2.1) and (3.14), v ( r + c β m u 1 + a m u ) L q ( Q T ) .

Then the condition (3) and (3.17) are satisfied by choosing p = r = ( n + 2 ) p 2 ( n + 2 p ) . According to the maximum principle [[13], p.181, Theorem 7.1], we can conclude that v L ( Q T ) . From (2.1), there exists a positive constant M 5 such that

u L ( Q T ) , v L ( Q T ) M 5 , T > 0 . (3.21)

Therefore, the global solution to the problem (1.1) exists.

(iii) The existence of classical solutions.

Under the conditions of Theorem 3.1, we consider above global solutions ( u , v ) to be classical. By (3.20) and Lemma 2.6, we know X C α , α 2 ( Q ¯ T ) , α ( 0 , 1 ) . It follows from Lemma 3.3 in [13] that X C 1 + α , 1 + α 2 ( Q ¯ T ) . Since X = ( d 1 + α 11 u ) u , we have u = d 1 d 1 2 + 4 α 11 X 2 α 11 .

So,

u C 1 + α , 1 + α 2 ( Q ¯ T ) , α ( 0 , 1 ) . (3.22)

Rewrite the second equation of system (1.1) as

v t = [ ( d 2 + α 21 u + 2 α 22 v ) v + α 21 u v ] + v ( r + c β m u 1 + a m u ) .

Therefore, we can conclude that v ( r + c β m u 1 + a m u ) L ( Q T ) , u, v, ∇u and ∇v are all bounded. By the Schauder estimate [13], there exists α ( 0 , 1 ) such that

v C α , α 2 ( Q ¯ T ) . (3.23)

Furthermore, by the Schauder estimate, we obtain

u C 2 + σ , 1 + σ 2 ( Q ¯ T ) , σ = min { α , α } . (3.24)

Next, the regularity of v will be discussed. Set v ¯ = ( d 2 + α 21 u + α 22 v ) v . So, v ¯ satisfies

v ¯ t = ( d 2 + α 21 u + 2 α 22 v ) Δ v ¯ + f ( x , t ) , (3.25)

where f ( x , t ) = ( d 2 + α 21 u + 2 α 22 v ) v ( r + c β m u 1 + a m u ) + α 21 u t v . According to (3.22) to (3.24), we have d 2 + α 21 u + 2 α 22 v , f ( x , t ) C σ , σ 2 ( Q ¯ T ) . Applying the Schauder estimate to (3.25), we know

v ¯ C 2 + σ , 1 + σ 2 ( Q ¯ T ) .

From v = ( d 3 + α 21 u ) + ( d 2 + α 21 u ) 2 + 4 α 22 v ¯ 2 α 22 , we can see

v C 2 + σ , 1 + σ 2 ( Q ¯ T ) , σ = min { α , α } . (3.26)

Combining (3.24) and (3.26), we get

u , v C 2 + σ , 1 + σ 2 ( Q ¯ T ) .

Therefore, the result of Theorem 3.1 can be obtained for α < α , namely σ = α . When α > α , namely σ < α , we have C 2 + σ , 1 + σ 2 ( Q ¯ T ) C α , α 2 ( Q ¯ T ) . (3.24) and (3.26) are obtained by repeating the above bootstrap argument and the Schauder estimate. This completes the proof of Theorem 3.1. □

### Competing interests

The author declares that he has no competing interests.

### Authors’ contributions

The work was realized by the author.

### Acknowledgements

The author thanks the anonymous referee for a careful review and constructive comments. This work was supported by the Qinghai Provincial Natural Science Foundation (2011-Z-915).

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