An N-barrier maximum principle for autonomous systems of n species and its application to problems arising from population dynamics Chiun-Chuan Chen, Li-Chang Hung and Chen-Chih Lai ∗ 6 1 0 Department of Mathematics, National Taiwan University, Taiwan 2 r L.-C. Hung dedicates this work to Mach Nguyet Minh p A 9 ] P A Abstract . h t We show that the N-barrier maximum principle (NBMP) remains true for n a (n > 2)species. Inaddition,astrongerlowerboundinNBMPisgivenbyemploying m an improved tangent line method. As an application of NBMP, we establish a [ nonexistence result for traveling wave solutions to the four species Lotka-Volterra 2 system. v 4 0 3 1 Introduction 0 0 . The main purpose of the present paper is to establish the N-barrier maximum principle 2 0 (NBMP) for n(n > 2) species, the case n = 2 having been considered previously ([3],[8]). 6 To be more specific, we study the autonomous system of n species 1 : v d (u ) +θ(u ) +umif (u ,u ,...,u ) = 0, x R, i = 1,2,...,n, (1.1) i i i xx i x i i 1 2 n ∈ X where u = u (x), d ,m > 0, and f (u ,u ,...,u ) C0(R+ R+ ... R+) for i = r i i i i i 1 2 n a 1,2,...,n; θ R. Throughout, we assume, unless o∈therwise s×tated,×that×the following ∈ hypothesis on f (u ,u ,...,u ) is satisfied: i 1 2 n [H] For i = 1,2,...,n, there exist u¯ > u > 0 such that i i ¯ f (u ,u ,...,u ) 0 whenever (u ,u ,...,u ) ; i 1 2 n 1 2 n ≥ ∈ R¯ f (u ,u ,...,u ) 0 whenever (u ,u ,...,u ) ¯, i 1 2 n 1 2 n ≤ ∈ R ∗Corresponding author’s email address: [email protected] 2010 Mathematics Subject Classification. Primary 35B50; Secondary 35C07, 35K57. Key words and phrases. Maximum principles, travelingwave solutions, reaction-diffusionequations. 1 where n u i = (u ,u ,...,u ) 1, u ,u ,...,u 0 ; R¯ 1 2 n u ≤ 1 2 n ≥ i n (cid:12) Xi=1 ¯ o (cid:12) n ¯ = (u ,u ,...,u ) (cid:12) ui 1, u ,u ,...,u 0 . 1 2 n 1 2 n R u¯ ≥ ≥ i n (cid:12) Xi=1 o (cid:12) We couple (1.1) with the prescribed Dirich(cid:12)let conditions at x = : ±∞ (u ,u ,...,u )( ) = e , (u ,u ,...,u )( ) = e , (1.2) 1 2 n − 1 2 n + −∞ ∞ where e ,e (u ,u ,...,u ) umif (u ,u ,...,u ) = 0 (i = 1,2,...,n),u ,u ,...,u 0 − + ∈ 1 2 n i i 1 2 n 1 2 n ≥ n (cid:12) (o1.3) (cid:12) are the equilibria of (1.1) w(cid:12)hich connect the solution (u ,u ,...,u )(x) at x = and 1 2 n −∞ x = . This leads to the boundary value problem of (1.1) and (1.2): ∞ d (u ) +θ(u ) +umif (u ,u ,...,u ) = 0, x R, i = 1,2,...,n, i i xx i x i i 1 2 n ∈ (BVP) (u ,u ,...,u )( ) = e , (u ,u ,...,u )( ) = e . 1 2 n − 1 2 n + −∞ ∞ (BVP)arises from the study of traveling waves in the following reaction-diffusion system of n species ([16],[18]): (ω ) = d (ω ) +ωmif (ω ,ω ,...,ω ), y R, t > 0, i = 1,2,...,n, (1.4) i t i i yy i i 1 2 n ∈ where ω (y,t) (i = 1,2,...n) is the density of the i-th species and d (i = 1,2,...n) repre- i i sents the diffusion rate of the i-th species. A special solution (u (x),u (x),...,u (x)) = 1 2 n (ω (y,t),ω (y,t),...,ω (y,t)), x = y θt, where θ is the propagation speed of the travel- 1 2 n − ing wave, is a traveling wave solution of (1.4). It is easy to see that the traveling wave solution of such form satisfies (1.1). Our main result is that (BVP) enjoys the following N-barrier maximum principle. Theorem 1.1 (NBMP for n Species). Assume that [H] holds. Given any set of α > 0 (i = 1,2,...,n), suppose that (u (x),u (x),...,u (x)) is a nonnegative C2 solution i 1 2 n to (BVP). Then n λ α u (x) λ¯, x R, (1.5) i i ¯ ≤ ≤ ∈ i=1 X where −1 λ¯ = max α u¯ max d min d , (1.6) i i i i i=1,2,...,n i=1,2,...,n i=1,2,...,n (cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17)−1 λ = min α u min d max d χ, (1.7) i i i i ¯ i=1,2,...,n ¯ i=1,2,...,n i=1,2,...,n (cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17) with χ defined by 0, if e = (0,...,0) or e = (0,...,0), + − χ = (1.8) 1, otherwise. 2 We note that both the lower bound λ and the upper bound λ¯ in NBMP do not depend ¯ explicitly on the propagation speed θ. To illustrate Theorem 1.1, we present an example. For n = 3, suppose that m = 1 and f (u ,u ,u ) = u (σ c u c u c u ) for i i 1 2 3 i i i1 1 i2 2 i3 3 − − − i = 1,2,3. Then (BVP) becomes d (u ) +θ(u ) +u (σ c u c u c u ) = 0, x R, 1 1 xx 1 x 1 1 11 1 12 2 13 3 − − − ∈ d (u ) +θ(u ) +u (σ c u c u c u ) = 0, x R, (LV3) 2 2 xx 2 x 2 2 − 21 1 − 22 2 − 23 3 ∈ d3(u3)xx +θ(u3)x +u3(σ3 c31u1 c32u2 c33u3) = 0, x R, − − − ∈ (u1,u2,u3)( ) = e−, (u1,u2,u3)( ) = e+, −∞ ∞ where e ,e (u ,u ,u ) u (σ c u c u c u ) = 0 (i = 1,2,3),u ,u ,u 0 . − + 1 2 3 i i i1 1 i2 2 i3 3 1 2 3 ∈ − − − ≥ n (cid:12) (1o.9) (cid:12) The parameters d , σ(cid:12), c (i = 1,2,3), and c (i,j = 1,2,3 with i = j), which i i ii ij 6 are all positive constants, stand for the diffusion rates, intrinsic growth rates, intra- specific competition rates, and inter-specific competition rates, respectively. A solution (u (x),u (x),u (x)) to (LV3) is a traveling wave solution which solves the competitive 1 2 3 Lotka-Volterra systems of three competing species: (ω ) = d (ω ) +ω (σ c ω c ω c ω ), y R, t > 0, i = 1,2,3, (1.10) i t i i yy i i i1 1 i2 2 i3 3 − − − ∈ where (ω (y,t),ω (y,t),ω (y,t)) = (u (x),u (x),u (x)), x = y θt. When the diffusion 1 2 3 1 2 3 − termsareabsent, (1.10)isthecelebrated May-Leonardmodel ([13])under theassumption that σ = c = 1 (i = 1,2,3), c = c = c = µ > 0 and c = c = c = µ > 0 i ii 12 23 31 1 13 21 32 2 (ω ) = ω (1 ω µ ω µ ω ), t > 0, 1 t 1 1 1 2 2 3 − − − (ω ) = ω (1 µ ω ω µ ω ), t > 0, (1.11) 2 t 2 − 2 1 − 2 − 1 3 (ω ) = ω (1 µ ω µ ω ω ), t > 0, 3 t 3 1 1 2 2 3 − − − where (ω ,ω ,ω ) =(ω (t),ω (t),ω (t)). 1 2 3 1 2 3 Fromtheviewpoint of thestudy of competitive exclusion ([1], [6], [7], [10], [14], [17]) or competitor-mediated coexistence ([2], [12], [15]), (LV3) or (1.10) arises from investigating problems where one exotic competing species (say, u ) invades the ecological system of 3 two native species (say, u and u ) that are competing in the absence of u . As indicated 1 2 3 in [9, 11], when u (x) is absent in (LV3) with e = (σ1,0) and e = (0, σ2 ), (LV3) 3 − c11 + c22 under the condition of strong competition (i.e. σ1 > σ2 and σ2 > σ1) admits solutions c11 c21 c22 c12 (u (x),u (x)) having profiles with u (x) being monotonically decreasing and u (x) being 1 2 1 2 monotonically increasing. Since u (x) and u (x) dominate the neighborhoods of x = 1 2 −∞ and x = , respectively, we are led to expect that the profile of u (x) must be pulse-like 3 (we call u∞(x) a pulse if u ( ) = u ( ) = 0 and u (x) > 0 for x R) if it exists since 3 3 3 3 −∞ ∞ ∈ u will prevail only when u and u are not dominant. It turns out that this conjecture 3 1 2 is true under certain assumptions on the parameters. In [4, 5], we established existence 3 of this type of solution by finding exact traveling wave solutions in addition to numerical experiments. To the best of our knowledge, however, a priori estimates for the parameter depen- denceofsolutionsto(LV3)havenotyetbeenfound. Corollary1.2providesanaffirmative answer to the following question: Q: Can upper and lower bounds of u +u +u can be given in terms of the parameters 1 2 3 in (LV3)? The above question arises in attempts to understand the ecological capacity of the inhabitant of the three competing species u , u , and u . Due to limited resources, 1 2 3 the investigation of the total density of the three species is of interest. More generally, estimates of α u +α u +α u , where α > 0 (i = 1,2,3), are given in Corollary 1.2. 1 1 2 2 3 3 i Corollary 1.2 (NBMP for Lotka-Volterra systems of three competing species). Assume that (u(x),v(x),w(x)) is a nonnegative C2 solution to (LV3). For any set of α > 0 (i = 1,2,3), we have i λ α u (x)+α u (x)+α u (x) λ¯, x R, (1.12) 1 1 2 2 3 3 ¯ ≤ ≤ ∈ where σ −1 λ¯ = max α max j max d min d , (1.13) i i i i=1,2,3 j=1,2,3 cji i=1,2,...,n i=1,2,...,n (cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17) σ −1 j λ = min α min min d max d χ, (1.14) i i i ¯ i=1,2,3 j=1,2,3 cji i=1,2,...,n i=1,2,...,n (cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17) with χ defined by 0, if e = (0,0,0) or e = (0,0,0), + − χ = (1.15) 1, otherwise. Proof. We apply Theorem 1.1 to prove Corollary 1.2. Taking σ j u¯ = max ; (1.16) i j=1,2,3cji σ j u = min . (1.17) i ¯ j=1,2,3cji It can be verified that [H] is satisfied. Indeed, we have n u i = (u ,u ,...,u ) 1, u ,u ,...,u 0 ; 1 2 n σ 1 2 n R¯ j ≤ ≥ ( (cid:12) min ) (cid:12)(cid:12) Xi=1 j=1,2,3 cji (cid:12) n (cid:12) u ¯ = (u ,u ,...,u ) i 1, u ,u ,...,u 0 . 1 2 n σ 1 2 n R j ≥ ≥ ( (cid:12) max ) (cid:12)(cid:12) Xi=1 j=1,2,3 cji σ σ (cid:12) Since min j ( max j, respectivel(cid:12)y) is the smallest (largest, respectively) u -intercept i j=1,2,3 cji j=1,2,3 cji of the three hyperplanes σ c u c u c u = 0 (i = 1,2,3), we see that i i1 1 i2 2 i3 3 − − − σ c u c u c u 0 whenever (u ,u ,...,u ) ; i i1 1 i2 2 i3 3 1 2 n − − − ≥ ∈ R¯ σ c u c u c u 0 whenever (u ,u ,...,u ) ¯, i i1 1 i2 2 i3 3 1 2 n − − − ≤ ∈ R 4 for each i = 1,2,3. The desired result follows from Theorem 1.1. NBMP for the diffusive Lotka-Volterra system of two competing species was estab- lished in[3], where itwas also shown that under additional restrictions onthe parameters, a lower bound stronger than the one given in Proposition 1.3 can be found by employing the tangent line method. Proposition 1.3 ([3]). Let a > 1 and a > 1. Suppose that (u(x),v(x)) is C2, nonneg- 1 2 ative, and satisfies the following differential inequalities and asymptotic behavior: u +θu +u(1 u a v) 0, x R, xx x 1 − − ≤ ∈ dv +θv +kv(1 a u v) 0, x R, (1.18) xx x − 2 − ≤ ∈ (u,v)( ) = (1,0), (u,v)(+ ) = (0,1), −∞ ∞ where d, k, a1, a2 are positive constants. For any α,β > 0, we have α β αu(x)+βv(x) min , min[1,d2], x R. (1.19) ≥ a d a ∈ (cid:20) 2 1(cid:21) We show in Section 3 that the tangent line method can be improved so that the additional parameter restrictions for giving a stronger lower bound than the one given in Proposition 1.3 are no longer needed and, additionally, this lower bound holds for a > 1 1 and a > 1. 2 Theremainderofthepaperisorganizedasfollows. NBMPfornspecies(Theorem1.1) is proved in Section 2. As an application of Corollary 1.2, we establish in Section 4 a nonexistence result for traveling wave solutions of the Lotka-Volterra system for four competing species d (u ) +θ(u ) +u (σ c u c u c u c u ) = 0, x R, i = 1,...,4, i i xx i x i i i1 1 i2 2 i3 3 i4 4 − − − − ∈ (LV4) (u ,u ,u ,u )( ) = (σ1,0,0,0), (u ,u ,u ,u )( ) = (0, σ2,0,0), 1 2 3 4 −∞ c11 1 2 3 4 ∞ c22 where d , σ , and c (i,j = 1,2,3,4) are positive constants; θ R is the propagation i i ij ∈ speed of the traveling wave. 2 Proof of Theorem 1.1 In this section, we prove Theorem 1.1. To this end, we first show in Proposition 2.1 that the lower bound given in Theorem 1.1 holds when (u ,u ,...,u )(x) is an upper solution 1 2 n of (BVP) by constructing an appropriate N-barrier. Proposition 2.1 (Lower bound in NBMP). Suppose that u (x) C2(R) with u (x) i i ∈ ≥ 0 (i = 1,2,...,n)and satisfythe followingdifferentialinequalities and asymptotic behavior: d (u ) +θ(u ) +umif (u ,u ,...,u ) 0, i = 1,2,...,n, x R, i i xx i x i i 1 2 n ≤ ∈ (Upper) (u ,u ,...,u )( ) = e , (u ,u ,...,u )( ) = e , 1 2 n − 1 2 n + −∞ ∞ 5 where e and e are given by (1.3). If the hypothesis − + [H] For i = 1,2,...,n, there exist u > 0 such that i ¯ f (u ,u ,...,u ) 0 whenever (u ,u ,...,u ) , i 1 2 n 1 2 n ≥ ∈ R¯ where is as defined in [H] R¯ holds, then we have for any α > 0 (i = 1,2,...,n) i n min(d ,d ,...,d ) α u (x) min α u ,α u ,...,α u 1 2 n χ, x R, (2.1) i i 1 1 2 2 n n ≥ ¯ ¯ ¯ max(d ,d ,...,d ) ∈ 1 2 n i=1 X (cid:0) (cid:1) where χ is defined as in (1.8). Proof. For the case where e = (0,...,0) or e = (0,...,0), a trivial lower bound 0 + − of n α u (x) is obvious. It suffices to show (2.1) for the case e = (0,...,0) and i=1 i i + 6 e = (0,...,0). To this end, we let − P6 n p(x) = α u (x); (2.2) i i i=1 X n q(x) = α d u (x). (2.3) i i i i=1 X Adding the n equations in (Upper), we obtain a single equation involving p(x) and q(x) d2q(x) dp(x) +θ +F(u (x),u (x),...,u (x)) 0, x R, (2.4) dx2 dx 1 2 n ≤ ∈ where F(u ,u ,...,u ) = n α umif (u ,u ,...,u ). First of all, we treat the case of 1 2 n i=1 i i i 1 2 n d = d at least for some i,j 1,2,...,n . i j 6 P∈ { } Determining an appropriate N-barrier is crucial in establishing (2.1). The construc- tion of the N-barrier consists of determining λ , η, and λ such that the three hyperplanes 2 1 n α d u = λ , n α u = η and n α d u = λ satisfy the relationship i=1 i i i 2 i=1 i i i=1 i i i 1 P P P , (2.5) Qλ1 ⊂ Pη ⊂ Qλ2 ⊂ R¯ where n = (u ,u ,...,u ) α u η, u ,u ,...,u 0 ; (2.6) η 1 2 n i i 1 2 n P ≤ ≥ n (cid:12) Xi=1 o (cid:12) n (cid:12) = (u ,u ,...,u ) α d u λ, u ,u ,...,u 0 . (2.7) λ 1 2 n i i i 1 2 n Q ≤ ≥ n (cid:12) Xi=1 o (cid:12) We follow the three steps below to con(cid:12)struct the N-barrier: 1. Taking λ = min α d u ,α d u ,...α d u , the hyperplane n α d u = λ 2 { 1 1¯1 2 2¯2 n n¯n} i=1 i i i 2 λ λ λ has the n intercepts ( 2 ,0,...,0), (0, 2 ,0,...,0),..., and (0,0,...,0, 2 ). It α d α d P α d 1 1 2 2 n n λ is readily seen that 2 u for i = 1,2,...,n, which gives ; αidi ≤ ¯i Qλ2 ⊂ R¯ 6 2. Taking η = λ min 1 , 1 ,..., 1 , the hyperplane n α u = η has the n in- 2 d d d i=1 i i η 1 η2 n η tercepts ( ,0,...,0)n, (0, ,0,...,0o),..., and (0,0,...,0, ). It is readily seen that α α Pα 1 2 n η λ 2 for i = 1,2,...,n, which gives ; αi ≤ αidi Pη ⊂ Qλ2 3. Taking λ = η min d ,d ,...,d , the hyperplane n α d u = λ has the n 1 { 1 2 n} i=1 i i i 1 λ λ λ intercepts ( 1 ,0,...,0), (0, 1 ,0,...,0),..., and (0,0,...,0, 1 ). It is readily α d α d P α d 1 1 2 2 n n λ η seen that 1 for i = 1,2,...,n, which gives . αidi ≤ αi Qλ1 ⊂ Pη The three hyperplanes n α d u = λ , n α u = η and n α d u = λ con- i=1 i i i 2 i=1 i i i=1 i i i 1 structed above form the N-barrier. From the above three steps, it follows immediately P P P that λ is given by 1 min(d ,d ,...,d ) λ = min α d u ,α d u ,...,α d u 1 2 n . (2.8) 1 1 1¯1 2 2¯2 n n¯n max(d ,d ,...,d ) 1 2 n (cid:0) (cid:1) We claim that q(x) λ , x R. This proves (2.1) since the α > 0 (i = 1,2,...,n) are ar- 1 i bitrary. Supposetha≥t, contr∈arytoourclaim, thereexistsz Rsuchthatq(z) < λ . Since 1 ∈ u,v C2(R) and (u1,u2,...,un)( ) = e±, we may assume minx∈Rq(x) = q(z). We de- ∈ ±∞ noterespectively byz andz thefirstpointsatwhichthesolution(u (x),u (x),...,u (x)) 2 1 1 2 n intersects the hyperplane n α d u = λ when x moves from z towards and . i=1 i i i 2 ∞ −∞ For the case where θ 0, we integrate (2.4) with respect to x from z to z and obtain 1 ≤ P z ′ ′ q (z) q (z )+θ(p(z) p(z ))+ F(u (x),u (x),...,u (x))dx 0. (2.9) 1 1 1 2 n − − ≤ Zz1 On the other hand we have: due to minx∈Rq(x) = q(z), q′(z) = 0; • q(z ) = λ follows from the fact that z is on the hyperplane n α d u = λ . • 1 2 1 i=1 i i i 2 Since z is the first point for q(x) taking the value λ when x moves from z to , 1 2 we conclude that q(z +δ) λ for z z > δ > 0 and q′(z ) P0; −∞ 1 2 1 1 ≤ − ≤ p(z) < η since z is below the hyperplane n α u = η; p(z ) > η since z is above • the hyperplane n α u = η; i=1 i i 1 1 i=1 i i P let = (u ,Pu ,...,u ) F(u ,u ,...,u ) > 0,u ,u ,...,u 0 . Due to the + 1 2 n 1 2 n 1 2 n • factFthat (u{ (z ),u (z ),...|,u (z )) is on the hyperplane n≥α }d u = λ and 1 1 2 1 n 1 i=1 i i i 2 (u (z),u (z),...,u (z)) , (u (z ),u (z ),...,u (z )), (u (z),u (z),...,u (z)) 1by (2.25). Becanuse of∈[HQ]λa1nd F1 (u1 ,u2,..1.,u ) =n 1n αP1umif (2u ,u ,...,nu ), ∈it R¯ 1 2 n i=1 i i i 1 2 n is easy to see that (u (x),u (x),...,u (x)) z x z . Therefore we 1 2 n 1 + z { | ≤ ≤P} ⊂ R¯ ⊂ F have F(u (x),u (x),...,u (x))dx > 0. z1 1 2 n CombinRing the above arguments, we obtain z q′(z) q′(z )+θ(p(z) p(z ))+ F(u (x),u (x),...,u (x))dx > 0, (2.10) 1 1 1 2 n − − Zz1 7 which contradicts (2.9). Therefore when θ 0, q(x) λ for x R. For the case where 1 ≤ ≥ ∈ θ 0, integrating (2.4) with respect to x from z to z yields 2 ≥ z2 q′(z ) q′(z)+θ(p(z ) p(z))+ F(u (x),u (x),...,u (x))dx 0. (2.11) 2 2 1 2 n − − ≤ Zz In a similar manner, it can be shown that q′(z ) 0, q′(z) = 0, p(z ) > η, p(z) < η, and 2 2 z2 F(u (x),u (x),...,u (x))dx > 0. These tog≥ether contradict (2.11). Consequently, z 1 2 n (2.1) is proved for the case of d = d at least for some i,j 1,2,...,n . Now we turn i j R 6 ∈ { } to the case of d = d for all i = 1,2,...,n. In this case, q(x) = dp(x) and (2.4) becomes i d2p(x) dp(x) d +θ +F(u (x),u (x),...,u (x)) 0, x R. (2.12) dx2 dx 1 2 n ≤ ∈ We take λ = λ = d min α u ,α u ,...,α u and η = min α u ,α u ,...,α u . It 1 2 1 1 2 2 n n 1 1 2 2 n n follows that the three hype{rpla¯nes ¯n α d ¯u }= λ , n α u{ =¯η and¯ n α ¯d }u = i=1 i i i 2 i=1 i i i=1 i i i λ coincide. Analogously to the previous case, we assume that there exists zˆ R such 1 that p(zˆ) < λ1 and minx∈Rp(x) =Pp(zˆ). Due to minxP∈Rp(x) = p(zˆ), wePhave ∈p′(zˆ) = 0 and p′′(zˆ) 0. Since (u (zˆ),u (zˆ),...,u (zˆ)) is in the interior of , which is contained in 1 2 n ≥ R¯ the interior of , we have F(u (zˆ),u (zˆ),...,u (zˆ)) > 0. These together give dp′′(zˆ) + + 1 2 n F θp′(zˆ) + F(u (zˆ),u (zˆ),...,u (zˆ)) > 0, which contradicts (2.12). As a result, p(x) λ 1 2 n 1 ≥ for x R when d = d for all i = 1,2,...,n. The proof is completed. i ∈ When(u ,u ,...,u )(x)isalower solutionof (BVP),theupperboundinTheorem1.1 1 2 n can be proved in a similar manner. Proposition 2.2 (Upper bound in NBMP). Suppose that u (x) C2(R) with u (x) i i ∈ ≥ 0 (i = 1,2,...,n)and satisfythe followingdifferentialinequalities and asymptotic behavior: d (u ) +θ(u ) +umif (u ,u ,...,u ) 0, i = 1,2,...,n, x R, i i xx i x i i 1 2 n ≥ ∈ (Lower) (u ,u ,...,u )( ) = e , (u ,u ,...,u )( ) = e , 1 2 n − 1 2 n + −∞ ∞ where e and e are given by (1.3). If the hypothesis − + [H¯] For i = 1,2,...,n, there exist u¯ > 0 such that i f (u ,u ,...,u ) < 0 whenever (u ,u ,...,u ) ¯, i 1 2 n 1 2 n ∈ R where ¯ is as defined in [H] R holds, then we have for any α > 0 (i = 1,2,...,n) i n max(d ,d ,...,d ) α u (x) max α u¯ ,α u¯ ,...,α u¯ 1 2 n . (2.13) i i 1 1 2 2 n n ≤ min(d ,d ,...,d ) 1 2 n i=1 X (cid:0) (cid:1) Proof. The proof lies in the fact that an appropriate N-barrier for the upper bound (2.13) can be constructed. Let n = (u ,u ,...,u ) α u η, u ,u ,...,u 0 ; (2.14) η 1 2 n i i 1 2 n P ≥ ≥ n (cid:12) Xi=1 o (cid:12) n (cid:12) = (u ,u ,...,u ) α d u λ, u ,u ,...,u 0 . (2.15) λ 1 2 n i i i 1 2 n Q ≥ ≥ n (cid:12) Xi=1 o (cid:12) We determine λ2, η, and λ1 in the foll(cid:12)owing steps: 8 1. Taking λ = max α d u¯ ,α d u¯ ,...α d u¯ , the hyperplane n α d u = λ 2 { 1 1 1 2 2 2 n n n} i=1 i i i 2 λ λ λ has the n intercepts ( 2 ,0,...,0), (0, 2 ,0,...,0),..., and (0,0,...,0, 2 ). It α d α d P α d 1 1 2 2 n n λ follows that 2 u¯ for i = 1,2,...,n, which gives ; αidi ≥ i Qλ2 ⊃ R¯ 2. Taking η = λ max 1 , 1 ,..., 1 , the hyperplane n α u = η has the n 2 d d d i=1 i i η 1 η2 n η η intercepts ( ,0,...,0n), (0, ,0,...,o0),..., and (0,0,...,0, ). It follows that α α Pα α 1 2 n i ≥ λ 2 for i = 1,2,...,n, which gives ; αidi Pη ⊃ Qλ2 3. Taking λ = η max d ,d ,...,d , the hyperplane n α d u = λ has the n in- 1 { 1 2 n} i=1 i i i 1 λ λ λ tercepts ( 1 ,0,...,0), (0, 1 ,0,...,0),..., and (0,0,...,0, 1 ). It follows that α d α d P α d 1 1 2 2 n n λ η 1 for i = 1,2,...,n, which gives . αidi ≥ αi Qλ1 ⊃ Pη The three hyperplanes n α d u = λ , n α u = η and n α d u = λ con- i=1 i i i 2 i=1 i i i=1 i i i 1 structed above form the N-barrier which satisfies the property P P P ¯. (2.16) Qλ1 ⊃ Pη ⊃ Qλ2 ⊃ R It follows immediately that λ is given by 1 max(d ,d ,...,d ) λ = max α u¯ ,α u¯ ,...,α u¯ 1 2 n . (2.17) 1 1 1 2 2 n n min(d ,d ,...,d ) 1 2 n (cid:0) (cid:1) We claim that q(x) = n α d u (x) λ , x R by contradiction as we have done i=1 i i i ≤ 1 ∈ in Propositions 2.1. The detailed proof of the claim is omitted here for brevity. This P completes the proof. We are now in the position to prove Theorem 1.1. Proof of Theorem 1.1. In Propositions 2.1 and 2.2, we obtain a lower and upper bound for n α u (x), respectively. Combining the results in Propositions 2.1 and 2.2, we i=1 i i immediately establish Theorem 1.1. P 3 Improved tangent line method In [3], it is shown that under certain restrictions on the parameters, the lower bound in Proposition 1.3 can be improved by means of the tangent line method. In this section, we show that an improved lower bound can be given without additional conditions on the parameters. To achieve this, let us denote by the quadratic curve αu(1 u a v)+ 1 L − − βkv(1 a u v) = 0 in the first quadrant of the uv-plane, i.e. 2 − − = (u,v) H(u,v) = 0,u 0,v 0 , (3.1) L ≥ ≥ n (cid:12) o (cid:12) (cid:12) 9 where H(u,v) := αu(1 u a v)+ βkv(1 a u v). Under the bistable condition 1 2 − − − − a > 1 and a > 1, we observe that H(u,v) = 0 is a hyperbola with one branch through 1 2 (1,0) and (0,1) and the other branch through (0,0). Let Q = (u,v) αu+dβv λ, u,v 0 ; (3.2) λ ≤ ≥ n (cid:12) o R = (u,v) (cid:12) H(u,v) 0, u,v 0 . (3.3) (cid:12) ≥ ≥ n (cid:12) o It is readily seen from the proof of Pr(cid:12)oposition 2.1 that a stronger lower bound can be (cid:12) found if we determine λ in the first step for the construction of the N-barrier by 2 λ = sup λ. (3.4) 2 Qλ⊂R To determine λ given by (3.4), we find the tangent line with the slope α to at a 2 −dβ L given point on . To this end, we first solve v = v(u) from H(u,v) = 0 to get L (αa u+βk(a u 1))+ (αa u+βk(a u 1))2 4αβku(u 1) 1 2 1 2 v(u) = − − − − − . (3.5) 2βk p A straightforward calculation yields (αa +βka )+ (αa1u+βk(a2u−1))(αa1+βka2)−2αβk(2u−1) dv(u) − 1 2 √(αa1u+βk(a2u−1))2−4αβku(u−1) = . (3.6) du 2βk It immediately follows that the slope of the tangent line to H(u,v) = 0 at the point (0,1) (respectively, (1,0)) is dv(0) = −α(a1−1)−βka2 (respectively, dv(1) = −α ). Under du βk du αa1+βk(a2−1) the bistable condition a > 1 and a > 1, we easily verify −α(a1−1)−βka2 < −α . 1 2 βk αa1+βk(a2−1) Noting that the slope of the line αu+dβv = λ is α , we are led to the following three 2 −dβ cases: (i) When −α −α(a1−1)−βka2, we determine the line αu+dβv = λ so that it passes dβ ≤ βk 2 through (0,1) and hence λ = dβ. Note that in this case z may be + in the 2 2 ∞ proof of Proposition 1.3 (Figure 3.1 (i)). (ii) When −α −α , we determine the line αu+ dβv = λ so that it passes dβ ≥ αa1+βk(a2−1) 2 through (1,0) and hence λ = α. Note that in this case z may be in the proof 2 1 −∞ of Proposition 1.3 (Figure 3.1 (ii)). (iii) When −α(a1−1)−βka2 < −α < −α , we determine the line αu+dβv = λ so βk dβ αa1+βk(a2−1) 2 that it is tangent to the curve v = v(u) at some point in the first quadrant of the uv-plane. By (3.6), we have (αa +βka )+ (αa1u+βk(a2u−1))(αa1+βka2)−2αβk(2u−1) − 1 2 √(αa1u+βk(a2u−1))2−4αβku(u−1) α = − (3.7) 2βk dβ or Au2 +Bu+C = G, (3.8) Du2 +Eu+J 10