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Steady-state solution for a general Schnakenberg model
Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
Contents lists available at ScienceDirect
Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa
Steady-state solution for a general Schnakenberg model✩ Yan Li ∗ Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, PR China College of Mathematics and Computational Science, China University of Petroleum, Dongying 257061, PR China
article
abstract
info
Article history: Received 23 July 2010 Accepted 22 December 2010
This paper is concerned with a general Schnakenberg model subject to the homogeneous Neumann boundary condition. Various existence and non-existence results are obtained in terms of self-diffusion coefficients. © 2011 Elsevier Ltd. All rights reserved.
Keywords: Schnakenberg model Implicit function theorem Topological degree
1. Introduction The Schnakenberg model was introduced in 1979 by Schnakenberg [1] as a model for a chemical reaction with limit cycle behavior. It consists of the following three reaction steps: A X,
B → X,
2X + Y → 3X .
After some scalings and changes of variables, the mathematical model corresponding to the Schnakenberg system is
ut − d1 1u = a − u + u2 v, vt − d2 1v = b − u2 v,
x ∈ Ω , t > 0, x ∈ Ω , t > 0,
(1.1)
subject to the homogeneous Neumann boundary condition. Here Ω ⊂ RN is a smooth and bounded domain, the unknowns u, v represent the concentrations of the reactants X and Y having the diffusion rates d1 , d2 > 0, and a, b > 0 are the concentrations of A, B. For more details of the background for (1.1), please refer to [1] and the references therein. Our attention here is paid to the positive steady-state solutions to the general system of (1.1), and this leads us to deal with the elliptic system:
p −d1 1u = a − u p+ u v, −d2 1v = b − u v, ∂ u = ∂v = 0, ∂ν ∂ν
x ∈ Ω, x ∈ Ω,
(1.2)
x ∈ ∂Ω.
Here p > 0. It is easy to check that system (1.2) has only one positive constant steady-state solution (˜u, v˜ ) = (a + b, (a+bb)p ). For our purpose, we shall use some of the mathematical techniques used by Wang, Peng and Shi in papers [2–5]. First of all, we have to establish a priori estimates for any positive solution to (1.2). Then based on these estimates, we will use
✩ This work was supported by PRC Grants NSFC 11071049, and the Fundamental Research Funds for the Central Universities 09CX04063A.
∗
Corresponding address: Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, PR China. E-mail address: [email protected].
1468-1218/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2010.12.014
1986
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
the implicit function theorem and some topological degree arguments to obtain some conclusions concerning the existence and non-existence of positive non-constant steady-state solutions to (1.2). The organization of this paper is as follows. In Section 2, we give a priori estimates for positive solutions to (1.2). In Sections 3 and 4, we study the non-existence and existence of positive non-constant steady-state solution to system (1.2) respectively. Throughout this paper, we denote by 0 = µ0 < µ1 < µ2 < · · · < µn < · · · the eigenvalues of −△ in Ω with the homogeneous Neumann boundary condition. For any k ≥ 0, we also denote the multiplicity of µk by m(µk ). 2. A priori estimates We first state the following lemma which is due to Lou and Ni [6]. Lemma 2.1. Let g ∈ C 1 (Ω × R). (i) If w ∈ C 2 (Ω ) ∩ C 1 (Ω ) satisfies
1w + g (x, w) ≥ 0, x ∈ Ω , ∂w ≤ 0, x ∈ ∂Ω, ∂ν and w(x0 ) = maxΩ¯ w , then g (x0 , w(x0 )) ≥ 0. (ii) If w ∈ C 2 (Ω ) ∩ C 1 (Ω ) satisfies 1w + g (x, w) ≤ 0, x ∈ Ω , ∂w ≥ 0, x ∈ ∂Ω, ∂ν and w(x0 ) = minΩ¯ w , then g (x0 , w(x0 )) ≤ 0. Theorem 2.1. Let a, b, d, d∗ > 0 be fixed. Then there exist two positive constants C1 (a, b, d, d∗ ) and C2 (a, b, d, d∗ ) such that for all d1 ≥ d,
0 ≤ d2 ≤ d∗ ,
any positive solution (u, v) of (1.2) satisfies C1 < u, v < C2 . Proof. Let u(x0 ) = minΩ¯ u, by Lemma 2.1, it follows a − u(x0 ) + up (x0 )v(x0 ) ≤ 0, which implies u(x0 ) ≥ a. Hence u ≥ a. Let v(x1 ) = maxΩ¯ v , by Lemma 2.1, it follows b − up (x1 )v(x1 ) ≥ 0, which implies v(x1 ) ≤
b . ap
Hence v ≤
−1ϕ = a + b − u in Ω ,
b . ap
Let ϕ = d1 u + d2 v . Adding the first two relations in (1.2), we have
∂ϕ = 0 on ∂ Ω . ∂ν
Let ϕ(x2 ) = maxΩ¯ ϕ , according to Lemma 2.1, we have a + b − u(x2 ) ≥ 0, which implies u(x2 ) ≤ a + b. So we have d1 u ≤ ϕ(x) ≤ ϕ(x2 ) = d1 u(x2 ) + d2 v(x2 ) ≤ d1 (a + b) + d2 which yields u≤a+b+
d2 b d1
ap
≤a+b+
d∗ b dap
.
Let v(x3 ) = minΩ¯ v , again by Lemma 2.1, we have b − up (x3 )v(x3 ) ≤ 0, which implies v(x3 ) ≥
b up (x3 )
≥
b
∗ b p p)
(a+b+ dd
a
, that is,
b ap
,
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
b
v≥
a+b+
1987
.
d∗ b p d ap
The conclusion is complete.
Furthermore, by standard elliptic arguments and Theorem 2.1, we obtain. Theorem 2.2. Let a, b, d, d∗ > 0 be fixed. Then there exists a positive constant C (d, d∗ ) such that for all d1 ≥ d and 0 ≤ d2 ≤ d∗ , any positive solution (u, v) of (1.2) satisfies
‖u, v‖C 2+α (Ω¯ ) ≤ C (d, d∗ ). 3. Non-existence results In this section, we will obtain the non-existence of positive solution to system (1.2) as d1 is large enough. Lemma 3.1. Let (u, v) be a positive solution of (1.2), then we have lim (u, v) = (˜u, v˜ )
d1 −→∞
¯ )]2 . in [C 2 (Ω
Proof. Let (uj , vj ) be a positive solution of (1.2) with d1j −→ ∞, By Theorem 2.1, (uj , vj ) is bounded. Hence, by passing to a ¯ )]2 to some (u, v), we obtain (u, v) satisfies the following equation subsequence if necessary, (uj , vj ) converges in [C 2+α (Ω
−1u = 0, −d2 1v = b − up v, ∂ u = ∂v = 0, ∂ν ∂ν
x ∈ Ω, x ∈ Ω, x ∈ ∂Ω.
Obviously u is a non-negative constant. Integrating the first two equations in (1.2) over Ω and adding them up, we have udx = ( a + b)|Ω |. Since u is a constant, it follows u = a + b = u˜ . So v satisfies Ω
−d2 1v = b − (a + b)p v, ∂v = 0, ∂ν
x ∈ Ω, x ∈ ∂Ω.
Multiplying the above equation with b − (a + b)p v and then integrating over Ω , we obtain 0≤
∫
d2
(a + b)p
Ω
|∇(b − (a + b) v)| dx = − p
2
Hence v = (a+bb)p = v˜ and the proof follows.
∫ Ω
[b − (a + b)p v]2 dx ≤ 0.
Theorem 3.1. Let a, b, d2 > 0 be fixed. Then there exists a positive constant dˆ1 such that system (1.2) has no non-constant positive solutions for all d1 ≥ dˆ1 . Proof. Define
α
¯): X = u ∈ C (Ω Let ρ =
∫ Ω
udx = 0 ,
Y =
u∈C
2+α
¯): (Ω
∂ u =0 , ∂ν ∂ Ω
Z = X ∩ Y.
, u = h + z , h ∈ R, z ∈ Z . Define ∫ 1 f (ρ, h, z , v) = [a − (h + z ) + (h + z )p v]dx, 1 |Ω | Ω f2 (ρ, h, z , v) = 1z + ρ[a − (h + z ) + (h + z )p v] − ρ f1 , p f3 (ρ, h, z , v) = d2 1v + b − (h + z ) v, T F (ρ, h, z , v) = (f1 , f2 , f3 ) (ρ, h, z , v), 1 d1
¯ ), and for any given ρ > 0, (u, v) is a positive solution of (1.2) if and only if then F : R2 × Z × Y −→ R × X × C α (Ω F (ρ, h, z , v) = 0. Obviously F (ρ, u˜ , 0, v˜ ) = 0 for any given ρ. Let Ψ be the Frèchet derivation of F (ρ, h, z , v) with variable (h, z , v) at (0, u˜ , 0, v˜ ). Direct computations show that Ψ is given by ∫ 1 ∫ u˜ p p−1 (−1 + pu˜ v˜ )(h + z )dx + v dx |Ω | Ω Ψ (h, z , v) = |Ω | Ω . 1z d2 1v − u˜ p v − pu˜ p−1 v˜ (h + z )
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Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
Next we will prove that Ψ is a one-to-one and surjective map. It suffices to show the the equation Ψ (h, z , v) = (h0 , z0 , v0 ) ¯ ) or equivalently the following system has a unique solution (h, z , v) ∈ R × Z × Y for any given (h0 , z0 , v0 ) ∈ R × X × C α (Ω
∫ 1 ∫ u˜ p p−1 ˜ (− 1 + p u v ˜ )( h + z ) dx + v dx = h0 , |Ω | Ω |Ω | Ω ∫ ∂ z 1z = z , x ∈ Ω, = 0, zdx = 0, 0 (3.1) ∂ν ∂ Ω Ω p p−1 d2 1v − u˜ v − pu˜ v˜ (h + z ) = v0 , x ∈ Ω , ∂v = 0, x ∈ ∂ Ω , ∂ν has a unique solution. In fact, because of z0 ∈ X , the equation 1z = z0 has a unique solution z ∈ Z . Considering the third equation of (3.1) d2 1v − u˜ p v = pu˜ p−1 v˜ (h + z ) + v0 ,
∂v = 0, ∂ν
Because of
− u˜ p
Ω
x ∈ Ω, x ∈ ∂Ω.
(3.2)
zdx = 0, system (3.2) has the unique solution v ∈ Y satisfying
∫ Ω
v dx = (v0 + pu˜ p−1 v˜ h)|Ω |.
(3.3)
Putting (3.3) into the first equation of (3.1) and computing, we obtain h = −v0 − h0 . So h is determined uniquely. From the above, we can conclude that Ψ is a one-to-one and surjective map. By virtue of the Banach inverse operator theorem, Ψ −1 exists and is a bounded linear operator. According to the implicit function theorem, there exists a constant δ > 0 such that the equation F (ρ, h, z , v) = 0 has a unique solution in the neighborhood of (˜u, 0, v˜ ) for all 0 < ρ < δ , and the unique solution is (˜u, 0, v˜ ). So system (1.2) has only one positive constant solution (˜u, v˜ ) as d1 is large enough. The proof is complete. 4. Existence results Throughout this section, we will assume that p > 1 and we will show the existence of positive non-constant solution to system (1.2). Let w = (u, v)T , w ˜ = (˜u, v˜ )T , system (1.2) is translated into
−1w = G(w),
1 (a − u + up v) with G(w) = d1 1 . p (b − u v) d2
Obviously −1w = G(w) is equivalent to F (w) = 0, here
F (w) = w − (I − ∆)−1 [w + G(w)].
(4.1)
In order to compute the index(I − F (·), w) ˜ , we need the sign of H (a, b, d1 , d2 , µ). Define H (a, b, d1 , d2 , µ) = det(µI − Gw (w)). ˜ Directly computing, we have
1 bp 1 −1 + (a + b)p a+b d1 Gw (w) ˜ = d1 , bp 1 p − − (a + b) d 2 ( a + b) d2 ] [ p ( a + b ) 1 bp (a + b)p H (a, b, d1 , d2 , µ) = µ2 + + − µ+ = 0. d2 d1 d 1 ( a + b) d1 d2 If bp
(a + b)d1
>
1 d1
+
( a + b) p d2
2 ,
(4.2)
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
1989
then the equation H (a, b, d1 , d2 , µ) = 0 has two positive solutions µ± (a, b, d1 , d2 ) given by
µ (a, b, d1 , d2 ) = ±
1 2
θ (a, b, d1 , d2 ) − 4
θ (a, b, d1 , d2 ) ±
2
( a + b) p d1 d2
,
where
θ (a, b, d1 , d2 ) =
bp
(a + b)d1 Note: Assume that bp > a + b. If
−
1 d1
−
(a + b)p d2
.
(1) d1 is small enough; (2) d2 is large enough then condition (4.2) holds. Example. Since p > 1, let p − 1 = ε > 0, and a = 2ε b, then bp > a + b holds for any p > 1, which means condition (4.2) is not empty. Theorem 4.1. Assume that condition (4.2) holds and there exist i > j ≥ 0 such that (i) µi < µ+ (a, b, d1 , d2 ) < µi+1 and µj < µ− (a, b, d1 , d2 ) < µj+1 ; (ii)
∑i
k=j+1
m(µk ) is odd.
Then system (1.2) has at least one positive non-constant solution. Proof. We assume that system (1.2) has no positive non-constant solution. By Theorem 3.1, we can fix D > dˆ1 such that (a) system (1.2) with diffusion coefficients D and d2 has no non-constant positive solutions; (b) H (a, b, D, d2 , µ) > 0 for all µ ≥ 0. Define
u +
Φ (t , w) = (I − ∆)−1
(a − u + up v) D d1 , 1 p v + (b − u v)
1−t
+
t
d2
¯ ) × C (Ω ¯ ) : C1 < u, v < C2 , x ∈ Ω }, B = {(u, v) ∈ C (Ω ¯ ) × C (Ω ¯ ). Φ : [0, 1] × B −→ C (Ω By virtue of (4.1), we have I − Φ (1, ·) = F . It is easy to see that solving system (1.2) is equivalent to find the fixed points of Φ (1, ·) in B. Moreover, Φ (t , w) has no fixed points on ∂ B for all 0 ≤ t ≤ 1, so deg(I − Φ (t , ·), B, 0) is well defined. By the homotopy invariance of degree, we have deg(I − Φ (1, ·), B, 0) = deg(I − Φ (0, ·), B, 0).
(4.3)
By (a) and (b), we obtain deg(I − Φ (0, ·), B, 0) = 1.
(4.4)
On the contrary, we assume that system (1.2) has no positive non-constant solution, by virtue of conditions (i) and (ii), it follows i ∑
deg(I − Φ (1, ·), B, 0) = index(I − F , w) ˜ = (−1)
k=j+1
m(µk )
= −1.
(4.5)
Thus (4.3)–(4.5) arrives a contradiction, which implies that system (1.2) has at least one positive non-constant solution. So the proof is complete. The next results provide existence of non-constant positive solutions to system (1.2) with respect to diffusion coefficients d1 and d2 . Corollary 4.1. Let a, b, d2 > 0 be fixed. Assume that bp > a + b and
−
m(µk )
is odd.
k>0,µ− <µk <µ+
Then there exists A > 0 such that system (1.2) has at least one positive non-constant solution for any 0 < d1 < A.
(4.6)
1990
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
Proof. By our assumption we see that condition (4.2) holds for d1 > 0 small enough. So we have that, as d1 → 0,
µ− (a, b, d1 , d2 ) →
(a + b)p+1 , d2 [bp − (a + b)]
µ+ (a, b, d1 , d2 ) → +∞.
Therefore, for 0 < d1 < A, condition (4.6) implies that (i) and (ii) in Theorem 4.1 hold. This concludes the proof.
Corollary 4.2. Let a, b, d1 > 0 be fixed. Assume that bp > a + b and
− bp k>0, 0<µk < (a+b)d 1
m(µk ) is odd.
(4.7)
− d1 1
Then there exists C > 0 such that system (1.2) has at least one positive non-constant solution for d2 > C . Proof. It is easy to see condition (4.2) holds for d2 > 0 large enough. So we have that, as d2 → ∞
µ− (a, b, d1 , d2 ) → 0,
µ+ (a, b, d1 , d2 ) →
bp
(a + b)d1
−
bp
1
1 d1
,
and 0 < µ− (a, b, d1 , d2 ) < µ+ (a, b, d1 , d2 ) <
(a + b)d1
−
d1
.
Therefore for d2 > C , condition (4.7) implies that (i) and (ii) in Theorem 4.1 hold. This concludes the proof.
Acknowledgements The author is grateful to the referee for his helpful comments and suggestions. References [1] [2] [3] [4]
J. Schnakenberg, Simple chemical reaction system with limit cycle behavior, J. Theor. Biol. (81) (1979) 389–400. R. Peng, Qualitative analysis of steady states to the Sel’kov model, J. Differential Equations 154 (1999) 157–190. R. Peng, J.P. Shi, M.X. Wang, Stationary pattern of a ratio-dependent food chain model with diffusion, SIAM J. Appl. Math 67 (2007) 1479–1503. R. Peng, M.X. Wang, Positive steady-state solutions of the Noyes–Field model for Belousov–Zhabotinskii reaction, Nonlinear Anal.TMA 56 (3) (2004) 451–464. [5] M.X. Wang, Non-constant positive steady states of the Sel’kov model, J. Differential Equations 190 (2) (2003) 600–620. [6] Y. Lou, W.M. Ni, Diffusion, self-diffusion and cross-diffusion, J. Differential Equations 131 (1996) 79–131.
Contents lists available at ScienceDirect
Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa
Steady-state solution for a general Schnakenberg model✩ Yan Li ∗ Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, PR China College of Mathematics and Computational Science, China University of Petroleum, Dongying 257061, PR China
article
abstract
info
Article history: Received 23 July 2010 Accepted 22 December 2010
This paper is concerned with a general Schnakenberg model subject to the homogeneous Neumann boundary condition. Various existence and non-existence results are obtained in terms of self-diffusion coefficients. © 2011 Elsevier Ltd. All rights reserved.
Keywords: Schnakenberg model Implicit function theorem Topological degree
1. Introduction The Schnakenberg model was introduced in 1979 by Schnakenberg [1] as a model for a chemical reaction with limit cycle behavior. It consists of the following three reaction steps: A X,
B → X,
2X + Y → 3X .
After some scalings and changes of variables, the mathematical model corresponding to the Schnakenberg system is
ut − d1 1u = a − u + u2 v, vt − d2 1v = b − u2 v,
x ∈ Ω , t > 0, x ∈ Ω , t > 0,
(1.1)
subject to the homogeneous Neumann boundary condition. Here Ω ⊂ RN is a smooth and bounded domain, the unknowns u, v represent the concentrations of the reactants X and Y having the diffusion rates d1 , d2 > 0, and a, b > 0 are the concentrations of A, B. For more details of the background for (1.1), please refer to [1] and the references therein. Our attention here is paid to the positive steady-state solutions to the general system of (1.1), and this leads us to deal with the elliptic system:
p −d1 1u = a − u p+ u v, −d2 1v = b − u v, ∂ u = ∂v = 0, ∂ν ∂ν
x ∈ Ω, x ∈ Ω,
(1.2)
x ∈ ∂Ω.
Here p > 0. It is easy to check that system (1.2) has only one positive constant steady-state solution (˜u, v˜ ) = (a + b, (a+bb)p ). For our purpose, we shall use some of the mathematical techniques used by Wang, Peng and Shi in papers [2–5]. First of all, we have to establish a priori estimates for any positive solution to (1.2). Then based on these estimates, we will use
✩ This work was supported by PRC Grants NSFC 11071049, and the Fundamental Research Funds for the Central Universities 09CX04063A.
∗
Corresponding address: Natural Science Research Center, Harbin Institute of Technology, Harbin 150080, PR China. E-mail address: [email protected].
1468-1218/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2010.12.014
1986
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
the implicit function theorem and some topological degree arguments to obtain some conclusions concerning the existence and non-existence of positive non-constant steady-state solutions to (1.2). The organization of this paper is as follows. In Section 2, we give a priori estimates for positive solutions to (1.2). In Sections 3 and 4, we study the non-existence and existence of positive non-constant steady-state solution to system (1.2) respectively. Throughout this paper, we denote by 0 = µ0 < µ1 < µ2 < · · · < µn < · · · the eigenvalues of −△ in Ω with the homogeneous Neumann boundary condition. For any k ≥ 0, we also denote the multiplicity of µk by m(µk ). 2. A priori estimates We first state the following lemma which is due to Lou and Ni [6]. Lemma 2.1. Let g ∈ C 1 (Ω × R). (i) If w ∈ C 2 (Ω ) ∩ C 1 (Ω ) satisfies
1w + g (x, w) ≥ 0, x ∈ Ω , ∂w ≤ 0, x ∈ ∂Ω, ∂ν and w(x0 ) = maxΩ¯ w , then g (x0 , w(x0 )) ≥ 0. (ii) If w ∈ C 2 (Ω ) ∩ C 1 (Ω ) satisfies 1w + g (x, w) ≤ 0, x ∈ Ω , ∂w ≥ 0, x ∈ ∂Ω, ∂ν and w(x0 ) = minΩ¯ w , then g (x0 , w(x0 )) ≤ 0. Theorem 2.1. Let a, b, d, d∗ > 0 be fixed. Then there exist two positive constants C1 (a, b, d, d∗ ) and C2 (a, b, d, d∗ ) such that for all d1 ≥ d,
0 ≤ d2 ≤ d∗ ,
any positive solution (u, v) of (1.2) satisfies C1 < u, v < C2 . Proof. Let u(x0 ) = minΩ¯ u, by Lemma 2.1, it follows a − u(x0 ) + up (x0 )v(x0 ) ≤ 0, which implies u(x0 ) ≥ a. Hence u ≥ a. Let v(x1 ) = maxΩ¯ v , by Lemma 2.1, it follows b − up (x1 )v(x1 ) ≥ 0, which implies v(x1 ) ≤
b . ap
Hence v ≤
−1ϕ = a + b − u in Ω ,
b . ap
Let ϕ = d1 u + d2 v . Adding the first two relations in (1.2), we have
∂ϕ = 0 on ∂ Ω . ∂ν
Let ϕ(x2 ) = maxΩ¯ ϕ , according to Lemma 2.1, we have a + b − u(x2 ) ≥ 0, which implies u(x2 ) ≤ a + b. So we have d1 u ≤ ϕ(x) ≤ ϕ(x2 ) = d1 u(x2 ) + d2 v(x2 ) ≤ d1 (a + b) + d2 which yields u≤a+b+
d2 b d1
ap
≤a+b+
d∗ b dap
.
Let v(x3 ) = minΩ¯ v , again by Lemma 2.1, we have b − up (x3 )v(x3 ) ≤ 0, which implies v(x3 ) ≥
b up (x3 )
≥
b
∗ b p p)
(a+b+ dd
a
, that is,
b ap
,
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
b
v≥
a+b+
1987
.
d∗ b p d ap
The conclusion is complete.
Furthermore, by standard elliptic arguments and Theorem 2.1, we obtain. Theorem 2.2. Let a, b, d, d∗ > 0 be fixed. Then there exists a positive constant C (d, d∗ ) such that for all d1 ≥ d and 0 ≤ d2 ≤ d∗ , any positive solution (u, v) of (1.2) satisfies
‖u, v‖C 2+α (Ω¯ ) ≤ C (d, d∗ ). 3. Non-existence results In this section, we will obtain the non-existence of positive solution to system (1.2) as d1 is large enough. Lemma 3.1. Let (u, v) be a positive solution of (1.2), then we have lim (u, v) = (˜u, v˜ )
d1 −→∞
¯ )]2 . in [C 2 (Ω
Proof. Let (uj , vj ) be a positive solution of (1.2) with d1j −→ ∞, By Theorem 2.1, (uj , vj ) is bounded. Hence, by passing to a ¯ )]2 to some (u, v), we obtain (u, v) satisfies the following equation subsequence if necessary, (uj , vj ) converges in [C 2+α (Ω
−1u = 0, −d2 1v = b − up v, ∂ u = ∂v = 0, ∂ν ∂ν
x ∈ Ω, x ∈ Ω, x ∈ ∂Ω.
Obviously u is a non-negative constant. Integrating the first two equations in (1.2) over Ω and adding them up, we have udx = ( a + b)|Ω |. Since u is a constant, it follows u = a + b = u˜ . So v satisfies Ω
−d2 1v = b − (a + b)p v, ∂v = 0, ∂ν
x ∈ Ω, x ∈ ∂Ω.
Multiplying the above equation with b − (a + b)p v and then integrating over Ω , we obtain 0≤
∫
d2
(a + b)p
Ω
|∇(b − (a + b) v)| dx = − p
2
Hence v = (a+bb)p = v˜ and the proof follows.
∫ Ω
[b − (a + b)p v]2 dx ≤ 0.
Theorem 3.1. Let a, b, d2 > 0 be fixed. Then there exists a positive constant dˆ1 such that system (1.2) has no non-constant positive solutions for all d1 ≥ dˆ1 . Proof. Define
α
¯): X = u ∈ C (Ω Let ρ =
∫ Ω
udx = 0 ,
Y =
u∈C
2+α
¯): (Ω
∂ u =0 , ∂ν ∂ Ω
Z = X ∩ Y.
, u = h + z , h ∈ R, z ∈ Z . Define ∫ 1 f (ρ, h, z , v) = [a − (h + z ) + (h + z )p v]dx, 1 |Ω | Ω f2 (ρ, h, z , v) = 1z + ρ[a − (h + z ) + (h + z )p v] − ρ f1 , p f3 (ρ, h, z , v) = d2 1v + b − (h + z ) v, T F (ρ, h, z , v) = (f1 , f2 , f3 ) (ρ, h, z , v), 1 d1
¯ ), and for any given ρ > 0, (u, v) is a positive solution of (1.2) if and only if then F : R2 × Z × Y −→ R × X × C α (Ω F (ρ, h, z , v) = 0. Obviously F (ρ, u˜ , 0, v˜ ) = 0 for any given ρ. Let Ψ be the Frèchet derivation of F (ρ, h, z , v) with variable (h, z , v) at (0, u˜ , 0, v˜ ). Direct computations show that Ψ is given by ∫ 1 ∫ u˜ p p−1 (−1 + pu˜ v˜ )(h + z )dx + v dx |Ω | Ω Ψ (h, z , v) = |Ω | Ω . 1z d2 1v − u˜ p v − pu˜ p−1 v˜ (h + z )
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Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
Next we will prove that Ψ is a one-to-one and surjective map. It suffices to show the the equation Ψ (h, z , v) = (h0 , z0 , v0 ) ¯ ) or equivalently the following system has a unique solution (h, z , v) ∈ R × Z × Y for any given (h0 , z0 , v0 ) ∈ R × X × C α (Ω
∫ 1 ∫ u˜ p p−1 ˜ (− 1 + p u v ˜ )( h + z ) dx + v dx = h0 , |Ω | Ω |Ω | Ω ∫ ∂ z 1z = z , x ∈ Ω, = 0, zdx = 0, 0 (3.1) ∂ν ∂ Ω Ω p p−1 d2 1v − u˜ v − pu˜ v˜ (h + z ) = v0 , x ∈ Ω , ∂v = 0, x ∈ ∂ Ω , ∂ν has a unique solution. In fact, because of z0 ∈ X , the equation 1z = z0 has a unique solution z ∈ Z . Considering the third equation of (3.1) d2 1v − u˜ p v = pu˜ p−1 v˜ (h + z ) + v0 ,
∂v = 0, ∂ν
Because of
− u˜ p
Ω
x ∈ Ω, x ∈ ∂Ω.
(3.2)
zdx = 0, system (3.2) has the unique solution v ∈ Y satisfying
∫ Ω
v dx = (v0 + pu˜ p−1 v˜ h)|Ω |.
(3.3)
Putting (3.3) into the first equation of (3.1) and computing, we obtain h = −v0 − h0 . So h is determined uniquely. From the above, we can conclude that Ψ is a one-to-one and surjective map. By virtue of the Banach inverse operator theorem, Ψ −1 exists and is a bounded linear operator. According to the implicit function theorem, there exists a constant δ > 0 such that the equation F (ρ, h, z , v) = 0 has a unique solution in the neighborhood of (˜u, 0, v˜ ) for all 0 < ρ < δ , and the unique solution is (˜u, 0, v˜ ). So system (1.2) has only one positive constant solution (˜u, v˜ ) as d1 is large enough. The proof is complete. 4. Existence results Throughout this section, we will assume that p > 1 and we will show the existence of positive non-constant solution to system (1.2). Let w = (u, v)T , w ˜ = (˜u, v˜ )T , system (1.2) is translated into
−1w = G(w),
1 (a − u + up v) with G(w) = d1 1 . p (b − u v) d2
Obviously −1w = G(w) is equivalent to F (w) = 0, here
F (w) = w − (I − ∆)−1 [w + G(w)].
(4.1)
In order to compute the index(I − F (·), w) ˜ , we need the sign of H (a, b, d1 , d2 , µ). Define H (a, b, d1 , d2 , µ) = det(µI − Gw (w)). ˜ Directly computing, we have
1 bp 1 −1 + (a + b)p a+b d1 Gw (w) ˜ = d1 , bp 1 p − − (a + b) d 2 ( a + b) d2 ] [ p ( a + b ) 1 bp (a + b)p H (a, b, d1 , d2 , µ) = µ2 + + − µ+ = 0. d2 d1 d 1 ( a + b) d1 d2 If bp
(a + b)d1
>
1 d1
+
( a + b) p d2
2 ,
(4.2)
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
1989
then the equation H (a, b, d1 , d2 , µ) = 0 has two positive solutions µ± (a, b, d1 , d2 ) given by
µ (a, b, d1 , d2 ) = ±
1 2
θ (a, b, d1 , d2 ) − 4
θ (a, b, d1 , d2 ) ±
2
( a + b) p d1 d2
,
where
θ (a, b, d1 , d2 ) =
bp
(a + b)d1 Note: Assume that bp > a + b. If
−
1 d1
−
(a + b)p d2
.
(1) d1 is small enough; (2) d2 is large enough then condition (4.2) holds. Example. Since p > 1, let p − 1 = ε > 0, and a = 2ε b, then bp > a + b holds for any p > 1, which means condition (4.2) is not empty. Theorem 4.1. Assume that condition (4.2) holds and there exist i > j ≥ 0 such that (i) µi < µ+ (a, b, d1 , d2 ) < µi+1 and µj < µ− (a, b, d1 , d2 ) < µj+1 ; (ii)
∑i
k=j+1
m(µk ) is odd.
Then system (1.2) has at least one positive non-constant solution. Proof. We assume that system (1.2) has no positive non-constant solution. By Theorem 3.1, we can fix D > dˆ1 such that (a) system (1.2) with diffusion coefficients D and d2 has no non-constant positive solutions; (b) H (a, b, D, d2 , µ) > 0 for all µ ≥ 0. Define
u +
Φ (t , w) = (I − ∆)−1
(a − u + up v) D d1 , 1 p v + (b − u v)
1−t
+
t
d2
¯ ) × C (Ω ¯ ) : C1 < u, v < C2 , x ∈ Ω }, B = {(u, v) ∈ C (Ω ¯ ) × C (Ω ¯ ). Φ : [0, 1] × B −→ C (Ω By virtue of (4.1), we have I − Φ (1, ·) = F . It is easy to see that solving system (1.2) is equivalent to find the fixed points of Φ (1, ·) in B. Moreover, Φ (t , w) has no fixed points on ∂ B for all 0 ≤ t ≤ 1, so deg(I − Φ (t , ·), B, 0) is well defined. By the homotopy invariance of degree, we have deg(I − Φ (1, ·), B, 0) = deg(I − Φ (0, ·), B, 0).
(4.3)
By (a) and (b), we obtain deg(I − Φ (0, ·), B, 0) = 1.
(4.4)
On the contrary, we assume that system (1.2) has no positive non-constant solution, by virtue of conditions (i) and (ii), it follows i ∑
deg(I − Φ (1, ·), B, 0) = index(I − F , w) ˜ = (−1)
k=j+1
m(µk )
= −1.
(4.5)
Thus (4.3)–(4.5) arrives a contradiction, which implies that system (1.2) has at least one positive non-constant solution. So the proof is complete. The next results provide existence of non-constant positive solutions to system (1.2) with respect to diffusion coefficients d1 and d2 . Corollary 4.1. Let a, b, d2 > 0 be fixed. Assume that bp > a + b and
−
m(µk )
is odd.
k>0,µ− <µk <µ+
Then there exists A > 0 such that system (1.2) has at least one positive non-constant solution for any 0 < d1 < A.
(4.6)
1990
Y. Li / Nonlinear Analysis: Real World Applications 12 (2011) 1985–1990
Proof. By our assumption we see that condition (4.2) holds for d1 > 0 small enough. So we have that, as d1 → 0,
µ− (a, b, d1 , d2 ) →
(a + b)p+1 , d2 [bp − (a + b)]
µ+ (a, b, d1 , d2 ) → +∞.
Therefore, for 0 < d1 < A, condition (4.6) implies that (i) and (ii) in Theorem 4.1 hold. This concludes the proof.
Corollary 4.2. Let a, b, d1 > 0 be fixed. Assume that bp > a + b and
− bp k>0, 0<µk < (a+b)d 1
m(µk ) is odd.
(4.7)
− d1 1
Then there exists C > 0 such that system (1.2) has at least one positive non-constant solution for d2 > C . Proof. It is easy to see condition (4.2) holds for d2 > 0 large enough. So we have that, as d2 → ∞
µ− (a, b, d1 , d2 ) → 0,
µ+ (a, b, d1 , d2 ) →
bp
(a + b)d1
−
bp
1
1 d1
,
and 0 < µ− (a, b, d1 , d2 ) < µ+ (a, b, d1 , d2 ) <
(a + b)d1
−
d1
.
Therefore for d2 > C , condition (4.7) implies that (i) and (ii) in Theorem 4.1 hold. This concludes the proof.
Acknowledgements The author is grateful to the referee for his helpful comments and suggestions. References [1] [2] [3] [4]
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