On the existence of dead cores for degenerate Lotka-Volterra models

ON THE EXISTENCE OF DEAD CORES FOR DEGENERATE LOTKA-VOLTERRA MODELS Manuel DELGADO and An onio SU´ AREZ Dp o. de Ecuaciones Di e enciales y An´alisis Num´e ico, Facul ad de Ma em´a icas, C/ Ta ia s/n, Uni e sidad de Se illa, 41012-Se illa, Spain. e-mail: delgado@nume .us.es and sua ez@nume .us.es Abs ac In his wo k we s udy he exis ence, uniqueness and quali a i e p ope ies o nonnega i e solu ions o he Lo ka-Vol e a models wi h nonlinea di usion unde homogeneous Di ichle bounda y condi ions. We conside he h ee ypical in e ac ions: p ey-p eda o , compe i ion and symbiosis. Unlike he linea di usion models, non i ial nonnega i e solu ions can exis which a e no s ic ly posi i e. Su icien condi ions in e ms o he coe icien s in ol ed in he se ing o he models a e gi en assu ing ha one species (o bo h) does no su i e on a se o i s habi a (called “dead co e”) o posi i e measu e. AMS Classi ica ion Codes: 35B50, 35B99, 35J55, 35K57, 92D25. Keywo ds and ph ases: degene a e Lo ka-Vol e a models, coexis ence s a es, dead co e se . 1 1 In oduc ion In his pape we a e in e es ed in nonnega i e solu ions o he ollowing sys em      −d1∆Um=U(A−BU ±CV ) in Ω, −d2∆Vn=V(D−EV ±FU) in Ω, U=V= 0 on ∂Ω, (1) whe e Ω is a bounded domain o IRN,N≥1, wi h a smoo h bounda y ∂Ω and m, n, d1, d2, B, C,E,Fa e posi i e cons an s wi h m, n ≥1 and A, D ∈IR. The p oblem (1) models he in- e ac ions be ween wo species, wi h popula ion densi ies U(x) and V(x), inhabi ing he egion Ω. Mo eo e , we a e assuming ha Ω is ully su ounded by inhospi able a eas, because bo h popula ion densi ies a e subjec o homogeneous Di ichle bounda y condi ions. Nonlinea di - usion a ises mainly in he po ous media equa ion and i was in oduced in popula ion dynamics in [16]. We e e o [12] and he e e ences he ein o de ails abou he model. Fo he p esen model Uand Vcan in e ac in h ee di e en ways indica ed by he sign o he las e ms in he equa ions: i bo h o hem a e nega i e hen Uand Vcompe e; i hey a e posi i e hen U and Vco-ope a e; and i , o example, he sign is posi i e in he i s equa ion and nega i e in he second one hen Uand V ep esen he p eda o and p ey popula ions espec i ely. To s udy (1), we make an app opia e change o a iables (see [12]) and ob ain      −∆wm=w(λ−w±bz) in Ω, −∆zn=z(µ−z±cw) in Ω, w=z= 0 on ∂Ω, (2) whe e λ, µ ∈IR and b, c > 0. When m=n= 1, (2) is he classical Lo ka-Vol e a model which has been widely s udied in he las yea s: see [4], [6], [8], [9], [15], [23], [26], [28] in compe i ion, [4], [9], [10], [21], [25] in p eda o -p ey and [11], [19], [22], [24], [27] in symbiosis, o ins ance. When m, n > 1 he e exis s an impo an change in he beha iou o he solu ions o (2). Mo e p ecisely, i is possible ha one species (o bo h) is nonnega i e bu no s ic ly posi i e, i.e. he e exis some subse s o Ω wi h s ic ly posi i e measu es whe e he species does no su i e. We call hem “dead co es” (see [14]). Fo ha , we dis inguish wo ypes o non i ial nonnega i e solu ions o (2); hose wi h bo h componen s posi i e, he coexis ence s a es, and hose whe e a leas one componen could ha e a dead co e, he semi-coexis ence s a es. We now desc ibe he dis ibu ion and he con en s o his wo k. In Sec ion 2 we s udy he single bounda y alue p oblem (−∆wm=w(a(x)−dw) in Ω, w= 0 on ∂Ω, (3) which appea s when one o he species is ze o and whe e a∈Cα(Ω), α∈(0,1), m≥1 and d≥0. I m= 1 i is known ha (3) admi s a unique posi i e solu ion i and only i σΩ 1(−a)<0, whe e σΩ 1(q) s ands o he p incipal eigen alue o he p oblem (−∆w+q(x)w=σw in Ω, w= 0 on ∂Ω, 2 wi h q∈L∞(Ω). The pa icula case m > 1 and d= 0 has been s udied in [1], [2], [3] and [17]. Thus, we ocus on he case m > 1 and d > 0 which has been p e iously analyzed in [17] and [30]. To s a e ou main esul s, we need he ollowing no a ion. Fo any a∈Cα(Ω) we deno e aL:= min x∈Ω a(x)aM:= max x∈Ω a(x). In Theo em 1 we ob ain some esul s conce ning exis ence and uniqueness o nonnega i e solu- ions o (3) which can be summa ized as ollows: •I aL≥0 and 1 < m ≤2 hen he e exis s a unique posi i e solu ion o (3). •I aM>0 hen he e exis s a leas a nonnega i e solu ion o (3). •I aM≤0 hen (3) does no admi a nonnega i e solu ion. Pa icula a en ion is paid o he exis ence o a dead co e o solu ions o (3). We show: •I ei he aL>0 o aL= 0 and 1 < m ≤2, any solu ion o (3) is posi i e and so i has no a dead co e. •I aL= 0, m > 2 and A0:= in {x∈Ω : a(x)=0} 6=∅, he e exis s a posi i e cons an C=C(N, d, ε) such ha i aM< C hen any nonnega i e solu ion o (3) has a dead co e, whe e ε > 0 is he adius o a ball con ained in A0. When achanges sign, we w i e a (x) := a+(x)− a−(x) whe e a±(x) := max{0,±a(x)}and o any > 0 and R > 0 we conside he se s N(R, ) := {x∈Ω : a−(x)≥R }, M(R, ) := {x∈N(R, ) : dis (x, ∂N(R, ) ∂Ω) ≥CR}, wi h CR:= sN 2Z(a+ M d)m 0 ds [Zs 0 (Rν1/m +dν2/m)dν]1/2. Wi h his no a ion, we ob ain ha i M(R, )6=∅ o some > 0 and R > 0 hen any nonnega i e solu ion o (3) has a dead co e. Mo eo e , we conside as pa ame e and we ob ain in o ma ion abou he quali a i e beha iou o he nonnega i e solu ions o (3). Ou esul s in his di ec ion can be s a ed in he ollowing gene al e ms: •The e exis s 0>0 such ha i ≥ 0 hen any nonnega i e solu ion o (3) has a dead co e. •I 1 < m ≤2 he e exis s 1>0 such ha i 0 < ≤ 1 hen any nonnega i e solu ion o (3) is posi i e. •I m > 2 he e exis d0>0 and 0(d0)>0 such ha i 0 < ≤ 0(d0) hen any nonnega i e solu ion o (3) has a dead co e. 3 These esul s imp o e and ex end hose in [30] (See Rema k 5 and Sec ion 2). In Sec ion 3 we analyze he sys em (2) in he h ee classical in e ac ions: p eda o -p ey, com- pe i ion and symbiosis. The p ey-p eda o case has been in es iga ed in [7], [18], [20], [29], [31]; compe i ion one in [7], [29]; and las ly, he symbiosis one in [7]. We conside λand µas p incipal pa ame e s and we s udy he se o alues o (λ, µ) o which (2) admi s a semi-coexis ence o coexis ence s a e. On he o he hand, we ob ain a p io i bounds o nonnega i e solu ions o (2) which a e he key o ge he non-exis ence egions o nonnega i e solu ions in he (λ, µ)-plane. We should no ice ha ou esul s imp o e in some sense hose in he abo e men ioned wo ks (see Rema ks 8, 9 and 11). Now, we a e in e es ed in he exis ence o a dead co e o any nonnega i e solu ion o (2). This has been p e iously s udied in he cases o p ey-p eda o and compe i ion in [18] and [29]. To ge esul s conce ning his subjec , we apply he esul s o Sec ion 2 and ob ain o he p ey-p eda o case: •Fo ixed λ > 0, he e exis s µ0(λ)>0 such ha i 0 < µ < µ0(λ) any nonnega i e solu ion o (2) has a dead co e. •Fo ixed µ > 0, he e exis λ0(µ)<0 and ε > 0 such ha i λ≤λ0(µ) he e is no nonnega i e solu ion o (2) and i λ∈(λ0(µ), λ0(µ) + ε) any nonnega i e solu ion o (2) has a dead co e. •Fo ixed λ, µ, b > 0, he e exis s c0>0 such ha i c > c0any nonnega i e solu ion o (2) has a dead co e. In he case o compe i ion, we ha e: •Fo ixed λ > 0 ( esp. µ > 0), he e exis s µ0(λ)>0 ( esp. λ0(µ)>0) such ha i 0< µ < µ0(λ) ( esp. 0 < λ < λ0(µ)) any nonnega i e solu ion o (2) has a dead co e. •Fo ixed λ, µ > 0 and c > 0 ( esp. b > 0), he e exis s b0>0 ( esp. c0>0) such ha i b > b0( esp. c > c0) any nonnega i e solu ion o (2) has a dead co e. Mo eo e , in his Sec ion we will gi e a biological in e p e a ion o hese esul s. 2 The logis ic equa ion wi h nonlinea di usion In his pape we use he ollowing no a ion: Ω is a bounded domain in IRNwi h a smoo h enough bounda y ∂Ω. Fo ixed α > 0, we conside he spaces U:= {w∈C2,α(Ω) : w= 0 on ∂Ω} and V:= Cα(Ω) o de ed by hei cones o nonnega i e unc ions PU:= {w∈ U :w≥0}and PV:= {w∈ V :w≥0}. We will w i e ≥gi −g∈P, > g i −g∈P−{0}and Àg i −g∈˙ P, whe e ˙ Pdeno es he in e io o P. Mo eo e , o any ∈ V we deno e M:= max x∈Ω (x) L:= min x∈Ω (x). 4 Le a∈ V,m > 1 and d > 0, we conside he logis ic equa ion wi h nonlinea di usion, namely (−∆wm=a(x)w−dw2in Ω, w= 0 on ∂Ω. (4) Gi en q∈L∞(Ω), σΩ 1(q) s ands o he i s eigen alue o (−∆w+q(x)w=σw in Ω, w= 0 on ∂Ω, (5) whose co esponding associa ed eigen unc ion ϕΩ 1[q] can be chosen such ha ϕΩ 1[q]À0 and kϕΩ 1[q]k∞= 1. Mo eo e , due o he s ong maximun p inciple, we ha e ∂ϕΩ 1[q] ∂n <0 (6) whe e nis he ou wa d uni no mal a ∂Ω. Finally, we w i e σΩ 1:= σΩ 1(0) and ϕΩ 1:= ϕΩ 1[0]. To s udy (4), we pe o m he change o a iables wm=uand we ob ain (−∆u=a(x)u1/m −du2/m in Ω, u= 0 on ∂Ω. (7) Fo he exis ence and uniqueness o nonnega i e solu ions o (7), he main esul is he ollowing one: Theo em 1 Le a∈ V,d > 0and m > 1. The ollowing asse ions a e ue: 1. I aÀ0, hen he e exis s a solu ion u∈ U o (7) wi h uÀ0. Mo eo e , i m≤2 he solu ion is unique. 2. I a > 0and m≤2, hen he e exis s a unique solu ion u∈ U o (7) wi h uÀ0. 3. I he e exis s x0∈Ωsuch ha a(x0)>0, hen he e exis s a leas a solu ion u∈ U o (7) wi h u > 0. Rema k 1 I a≤0 hen u= 0 is he unique nonnega i e solu ion o (7) owing o he maximum p inciple. On he o he hand, obse e ha he unc ion acan change o sign in case 3. P oo . We will apply he sub-supe solu ion me hod. By he egula i y o a he e exis x0∈Ω and > 0 such ha a(x)≥a0>0, o x∈B(x0, ), whe e B(x0, ) is he ball o adius > 0 cen e ed a x0. Now, we de ine he unc ion φ(x) := (ϕΩ0 1(x) i x∈Ω0, 0 i x∈Ω Ω0, 5 wi h Ω0= Ω in he case 1 and Ω0=B(x0, ) in he o he cases. We will show ha u=ρφ is a subsolu ion o (7) wi h ρa posi i e cons an . Indeed, we can ake ρ > 0 such ha −∆(ρϕΩ0 1)≤a(x)(ρϕΩ0 1)1/m −d(ρϕΩ0 1)2/m in Ω0. By (6) we can use Lemma I.1 o [3] and conclude ha uis a subsolu ion o (7). As a supe solu ion we pick u=M, wi h M > 0 a su icien ly la ge cons an . Mo eo e , i is easy o see ha kuk∞≤M:= µaM d¶m (8) o any solu ion uo (7). So, he e exis s a leas a solu ion u∈ U such ha u≤u≤u. Clea ly uÀ0 in he i s case. Now, we show ha uÀ0 in he case 2 by applying he s ong maximum p inciple. Le u > 0 be a solu ion o (7); i su ices o ind a cons an K > 0 such ha (−∆ + K)u > 0, o equi alen ly K > du2/m−1−a(x)u1/m−1,wi h u∈(0, M],(9) which ollows om hypo heses and so uÀ0. Fo he uniqueness we use [5]. We de ine he map 7→ g(x, ) := a(x) 1/m −d 2/m =a(x) 1/m−1−d 2/m−1. I is no ha d o p o e ha his applica ion is dec easing in he cases aÀ0 and m≤2 as well as a > 0 and m < 2. I emains o p o e he uniqueness in he case m= 2 and a > 0. Le u1and u2be wo a bi a y solu ions o (7). Now, he unc ion gis noninc easing, so by he Rema k 1 o [5], i ollows ha he e exis s a cons an Csuch ha u1=Cu2. Hence, a(x)u1/2 1−du1=−∆u1=−∆(Cu2) = C(a(x)u1/2 2−du2) = Ca(x)u1/2 2−du1 om which a(x)C1/2u1/2 2(1 −C1/2) = 0, and he e o e C= 1. This comple es he p oo . • Rema k 2 1.- When aL>0 he e exis s a unique posi i e solu ion o (7) p o ided ha m > 2and aL aM >m−2 m−1. Indeed, i can be p o ed ha he unc ion 7→ g(x, )de ined in he p oo o Theo em 1 is dec easing. 2.- In he pa icula case a(x)≡λ=c e i is known, c . [13], ha he e exis s a unique posi i e solu ion o (7) i and only i λ > 0. 3.- When aÀ0and m > 1, any solu ion o (7) is posi i e in Ω. In his case, i is easy o p o e he exis ence o he cons an Ksa is ying (9). 6 ¿F om he p io i bound (8) ollows he exis ence o a non i ial maximal solu ion o (7) which we deno e by θ[a,d,m]. We will ake θ[a,d,m]= 0 i a≤0. The nex esul will be e y use ul o compa e posi i e solu ions o di e en logis ic bounda y alue p oblems. P oposi ion 1 Le a, b ∈ V. 1. I uis a subsolu ion o (7), hen u≤θ[a,d,m]. 2. I a≤b, hen θ[a,d,m]≤θ[b,d,m]. P oo . Le ube a subsolu ion o (7). As su icien ly la ge cons an s, say K > 0, a e supe solu- ions, hen he e exis s a solu ion uo (7) such ha u≤u≤K. F om he maximali y o θ[a,d,m] i ollows ha u≤θ[a,d,m]. Fo he second pa i is enough o p o e ha θ[a,d,m]is a subsolu ion o (−∆w=b(x)w1/m −dw2/m in Ω, w= 0 on ∂Ω, and apply he p e ious esul . • 2.1 Exis ence o a “dead co e” In his subsec ion we shall conside he exis ence o a “dead co e” o solu ions uo (7), i.e., we will show ha he se Ω0:= {x∈Ω : u(x) = 0}has a s ic ly posi i e measu e unde sui able easily checked hypo heses on a,d,m, Ω and N. Assume ha a > 0 and A0:= in {x∈Ω : a(x) = 0} 6=∅. Theo em 1 ensu es ha i m≤2 any nonnega i e solu ion o (7) is posi i e. Ou p incipal esul is: Theo em 2 Le m > 2,a > 0and A0=in {x∈Ω : a(x)=0} 6=∅. Le x0∈A0and ε > 0be such ha B(x0,2ε)⊂A0and assume ha aM<dm−1 m−2ε2 m−2 [q(q−1 + N−1 2)] 1 m−2 ,(10) wi h q= 2m/(m−2). Then he e exis s a dead co e o any u > 0solu ion o (7). Mo eo e , we ha e Ω0={x∈Ω : u(x) = 0} ⊃ B(x0, ε). 7 P oo . Le x0∈A0and ε > 0 be such ha B(x0,2ε)⊂A0. We will build a unc ion which is ze o in B(x0, ε)⊂Ω and bigge han he maximal solu ion o (7). Wi hou loss o gene ali y we can suppose ha x0= 0. We conside he unc ion Ψ(x) :=      0 i x∈B(0, ε), (|x|−ε)q εqi x∈B(0,2ε) B(0, ε), 1 i x∈Ω B(0,2ε). I is clea ha Ψ ∈H1(Ω). We will show ha he unc ion u0:= KΨ sa is ies θ[a,d,m]≤u0in Ω, wi h an app opia e cons an K. Taking K≥(aM d)m(11) we ha e u0≥θ[a,d,m]in Ω B(0,2ε) om (8). We will p o e now ha (θ[a,d,m]−u0)+= 0 in B(0,2ε). I is su icien o p o e ha ZB(0,2ε)|∇(θ[a,d,m]−u0)+|2≤0,(12) because (θ[a,d,m]−u0)+= 0 on ∂B(0,2ε). Le φ∈H1 0(B(0,2ε)), φ ≥0. Then ZB(0,2ε)∇(θ[a,d,m]−u0)·∇φ=ZB(0,2ε)∇θ[a,d,m]·∇φ−KZB(0,2ε) B(0,ε)∇Ψ·∇φ= ZB(0,2ε) θ1/m [a,d,m](a(x)−dθ1/m [a,d,m])φ+KZB(0,2ε) B(0,ε) ∆Ψ ·φ−KZ∂B(0,ε) ∂Ψ ∂n φ. Now using ha ∂Ψ/∂n = 0 on ∂B(0, ε) and a(x) = 0 in B(0,2ε), we ind ha ZB(0,2ε)∇(θ[a,d,m]−u0)·∇φ≤ −ZB(0,2ε) dθ2/m [a,d,m]φ+KZB(0,2ε) B(0,ε) ∆Ψ ·φ= =−ZB(0,ε) dθ2/m [a,d,m]φ+ZB(0,2ε) B(0,ε) (−dθ2/m [a,d,m]+K(q εq(|x|−ε)q−2(q−1 + N−1 |x|(|x|−ε))))φ. Taking φ= (θ[a,d,m]−u0)+, we ob ain ZB(0,2ε)|∇(θ[a,d,m]−u0)+|2≤ ≤ZB(0,2ε) B(0,ε) (−dθ2/m [a,d,m]+K(q εq(|x|−ε)q−2(q−1 + N−1 |x|(|x|−ε))))(θ[a,d,m]−u0)+. (13) Using (13) and no ing ha q−2 = 2q/m, o p o e (12) i is su icien o show ha K1−2/m q ε2(q−1 + N−1 |x|(|x|−ε)) −d≤0x∈B(0,2ε) B(0, ε). 8 We deno e K1:= K1−2/m ε2q(q−1) −dand K2:= K1−2/m ε2q(N−1). We mus p o e ha K1+K2≤εK2 |x| o ε < |x|<2ε. (14) Assume ha N≥2. I K1+K2≤0, (14) is i ial because K2>0. I K1+K2>0, (14) is equi alen o |x| ≤ εK2 K1+K2 . This inequali y is ue in B(0,2ε) B(0, ε) i 2K1+K2≤0.(15) I N= 1 hen K2= 0, and so (15) is equi alen o be K1≤0. I is no ha d o p o e om (10) he exis ence o a cons an Ksa is ying (11) and (15). This inishes he p oo . • Rema k 3 We ema k ha he cons an C:= C(d, ε, m, N) = dm−1 m−2ε2 m−2 [q(q−1 + N−1 2)] 1 m−2 sa is ies C↑+∞as d↑+∞o ε↑+∞. Then, i a > 0is gi en, Ω0exis s i dis la ge enough. On he o he hand, i dis gi en, Ω0exis s i aMis su icien ly small o i A0is la ge. Now we conside he case whe e achanges sign. Le ∈IR+and we de ine a (x) := a+(x)− a−(x) whe e a±(x) := max{0,±a(x)}and we suppose ha a+6≡ 0. The aim is now o s udy he exis ence o a dead co e o solu ions o (−∆u=a (x)u1/m −du2/m in Ω, u= 0 on ∂Ω, (16) wi h as a pa ame e . To p o e he main esul we need he ollowing Lemma mo i a ed by Lemma 7 in [30]. Lemma 1 Le R > 0, R0:= sN 2Z+∞ 0 ds [Zs 0 (Rν1/m +dν2/m)dν]1/2 9 Theo em 6 The ollowing asse ions a e ue: 1. Assume λ > 0. Then he e exis s ρ(λ)>0such ha i 0<µ<ρ(λ)any nonnega i e solu ion o (20) has a dead co e. 2. Assume µ > 0. Then he e exis s λ0(µ)<0and ε > 0such ha i λ∈(λ0(µ), λ0(µ) + ε) any nonnega i e solu ion o (20) has a dead co e. 3. Assume µ > 0and λ > 0. Then he e exis s c0>0such ha i c > c0any nonnega i e solu ion o (20) has a dead co e. P oo . Using P oposi ions 1 and 2 i is easy o p o e ha o any nonnega i e solu ion (u, ) o (20) we ha e u≤θ[λ+bθ1/n [µ,n],m] ≤θ[µ−cθ1/m [λ,m],n],(26) and so, Ωu,0:= {x∈Ω : u(x) = 0} ⊃ {x∈Ω : θ[λ+bθ1/n [µ,n],m](x) = 0}, Ω ,0:= {x∈Ω : (x) = 0} ⊃ {x∈Ω : θ[µ−cθ1/m [λ,m],n](x) = 0}. To p o e he i s pa we will use Theo em 3 and Rema k 5. In his case a(x) = µ−cθ1/m [λ,m](x). Hence aM=µand A−={x∈Ω : µ−cθ1/m [λ,m](x)<0}.Le x0∈Ω and R > 0 be such ha θ1/m [λ,m](x0) = (θ1/m [λ,m])Mand c(θ1/m [λ,m])M−R > 0. We de ine B(x) := cθ1/m [λ,m](x)−R, and so, N(R, 1) := N(µ) = {x∈Ω : µ≤B(x)}. I is easily seen ha i µ1≤µ2 hen N(µ2)⊂N(µ1).(27) Le δ > 0 be such ha 0 < δ < c(θ1/m [λ,m])M−Rand de ine µ0:= c(θ1/m [λ,m])M−R−δ. Wi h hese choices we can see ha x0∈N(µ0) and om he con inui y o B(x) he e exis s µ0>0 such ha B(x0, µ0)⊂N(µ0). Mo eo e , om (27) we ge B(x0, µ0)⊂N(µ0)⊂N(µ) o µ∈[0, µ0]. On he o he hand, i µ↓0 hen aM=µ↓0, and so CR↓0. Thus, he e exis s µ1>0 such ha i 0 < µ < µ1 hen CR< µ0. Finally, we ake ρ:= min{µ0, µ1}. Thus, o µ∈(0, ρ) CR< µ0and B(x0, µ0)⊂N(µ) 16 and so, dis (x0, ∂N(µ) ∂Ω) ≥dis (x0, ∂B(x0, µ0) ∂Ω) = µ0> CR. The e o e M(R, 1) 6=∅,and so he i s pa ollows om Theo em 3. In o de o p o e he second pa le a(x) = λ+bθ1/n [µ,n](x) and so, aM=λ+b(θ1/n [µ,n])M. We ake λ0=−b(θ1/n [µ,n])M. I is clea ha i λ > λ0 hen aM>0. Le Rand δbe such ha 0 < R < δ < −λ0. Thus, o λ∈(λ0,−δ] he se N(R, 1) = {x∈Ω : λ≤ −R−bθ1/n [µ,n](x)} 6=∅. Obse e ha i λ↓λ+ 0 hen aM↓0, and so CR↓0. Now, easoning as in he abo e pa ag aph, he esul ollows. To p o e he las pa we ix λ, µ > 0. We ake R=c1/2and deno e by N(R, 1) := N(c). Thus, N(c) = {x∈Ω : µ+c1/2 c≤θ1/m [λ,m](x)}. Obse e ha he unc ion (c) := µ+c1/2 c is dec easing and ends o 0 as c↑ ∞. Mo eo e , i c1≤c2 hen N(c1)⊂N(c2).Hence, o c su icien ly la ge we ge N(c)6=∅. Finally, obse e ha CR(c)↓0 as c↑ ∞. We can comple e he p oo by using an a gumen simila o ha used p e iously. • Rema k 7 We can gi e a biological in e p e a ion o he abo e esul . In he pa 1 we ix he p eda o ’s g ow h a e. Then i he p ey’s g ow h a e is small, he p ey do no li e in all he space Ω, i.e., he e exis some sub egions o Ωwhe e he p ey is d i en o ex inc ion by he p eda o . On he o he hand, in pa 3 i he a e a which he p ey is consumed by he p eda o ( he pa ame e cin he se ing o he sys em (20)) is su icien ly la ge, hen he e exis some a eas whe e he p ey do no exis . In pa 2, om (26) obse e ha i λ≤ −b(θ1/n [µ,n])M:= λ0 hen he e a e no semi-coexis ence s a es o (20). Theo em 6ensu es ha (20) does no admi a coexis ence s a e i λ∈(λ0, λ0+ε). Rema k 8 Ou esul s a e imp o emen on p e ious esul s pape s. Indeed, in [20] he au ho s ob ained unde condi ion (25) he exis ence o a coexis ence s a e o (20) bu only in he pa icula case 1< m =n < 2. In [7] he au ho s showed he exis ence o a semi-coexis ence s a e when λ and µsa is y (25). I is clea ha condi ion (22) is weake han (25). 17 3.2 Compe i ion Conside he ollowing model whose solu ions ep esen he s eady s a es o dynamical models o compe ing popula ions      −∆u=λu1/m −u2/m −bu1/m 1/n in Ω, −∆ =µ 1/n − 2/n −c 1/nu1/m in Ω, u= = 0 on ∂Ω. (28) Conce ning he exis ence o semi-coexis ence s a es, we ha e he ollowing esul : Theo em 7 (28) possesses a semi-coexis ence s a e i and only i λ > 0and µ > 0. P oo . Le (u, ) be a semi-coexis ence s a e o (28). Then he maximum p inciple ensu es ha λand µha e o be posi i e. Assume now λ, µ > 0. I is clea ha B1:= {x∈Ω : λ−bθ1/n [µ,n](x)>0} 6=∅, B2:= {x∈Ω : µ−cθ1/m [λ,m](x)>0} 6=∅.(29) We ha e o ind a couple (u, u)−( , ) o sub-supe solu ions o (28), see [12], i.e. u, u, , ∈ H2(Ω) ∩L∞(Ω), u≤uand ≤ in Ω and u≤0≤uand ≤0≤ on ∂Ω and such ha          −∆u≤λu1/m −u2/m −bu1/m 1/n in Ω, −∆u≥λu1/m −u2/m −bu1/m 1/n in Ω, −∆ ≤µ 1/n − 2/n −c 1/nu1/m in Ω, −∆ ≥µ 1/n − 2/n −c 1/nu1/m in Ω. (30) We pick (u, u) = (θ[λ−bθ1/n [µ,n],m], θ[λ,m]) ( , ) = (θ[µ−cθ1/m [λ,m],n], θ[µ,n]). Using (29) and Theo em 1, i can be shown ha u > 0, > 0. Mo eo e , om P oposi ion 1, u≤uand ≤ in Ω. I is no ha d o p o e ha his couple sa is ies he inequali ies in (30) and his comple es he p oo . • The p oo s o he ollowing esul s a e simila o he p eda o -p ey ones, so we omi hem. P oposi ion 4 Le (u, )be a semi-coexis ence s a e o (28). Then u≤θ[λ,m] ≤θ[µ,n].(31) Wi h espec o he exis ence o coexis ence s a es we ha e he ollowing esul whose p oo can be ound in [12]. Theo em 8 Assume bc < 1and λ > bµ and µ > cλ. (32) Then he e exis s a leas one coexis ence s a e o (28). 18 Finally, conce ning condi ions on λand µunde which any coexis ence s a e has no a dead co e, we ob ain: P oposi ion 5 Assume bc < 1and ha λand µsa is y (32). Then any nonnega i e solu ion o (28) has no a dead co e. The ollowing esul gi es us condi ions which ensu e ha any nonnega i e solu ion o (28) has a dead co e. Theo em 9 The ollowing asse ions a e ue: 1. Assume λ > 0. Then he e exis s µ(λ)>0such ha i 0< µ < µ(λ)any nonnega i e solu ion o (28) has a dead co e. 2. Assume µ > 0. Then he e exis s λ(µ)>0such ha i 0< λ < λ(µ)any nonnega i e solu ion o (28) has a dead co e. 3. Assume λ, µ > 0and b > 0. Then he e exis s c0>0such ha i c > c0any nonnega i e solu ion o (28) has a dead co e. 4. Assume λ, µ > 0and c > 0. Then he e exis s b0>0such ha i b > b0any nonnega i e solu ion o (28) has a dead co e. P oo . We will p o e he i s and hi d pa s. The o he ones ollow simila ly. We claim ha o any nonnega i e solu ion (u, ) o (28), we ha e u≤θ[λ−bθ1/n [µ−cθ1/m [λ,m],n] ,m], ≤θ[µ−cθ1/m [λ−bθ1/n [µ,n],m] ,n].(33) Indeed, using P oposi ion 1 i is su icien o show ha uis subsolu ion o   −∆w= (λ−bθ1/n [µ−cθ1/m [λ,m],n])w1/m −w2/m in Ω, w= 0 on ∂Ω, which is ue i θ[µ−cθ1/m [λ,m],n]≤ . ¿F om P oposi ions 4 and 1, i ollows ha =θ[µ−cu1/m,n]≥θ[µ−cθ1/m [λ,m],n]. This p o es he i s ela ion o (33). The second one ollows analogously. So, om (33) we ha e Ωu,0={x∈Ω : u(x) = 0} ⊃ {x∈Ω : θ[λ−bθ1/n [µ−cθ1/m [λ,m],n] ,m](x) = 0}, Ω ,0={x∈Ω : (x) = 0} ⊃ {x∈Ω : θ[µ−cθ1/m [λ−bθ1/n [µ,n],m] ,n](x) = 0}. We now p o e he i s pa . Fo ixed λ > 0, we conside a(x) = µ−cθ1/m [λ−bθ1/n [µ,n],m](x), 19 and so aM=µ. As λ > 0, he se {x∈Ω : λ−bθ1/n [µ,n](x)>0} 6=∅and hus θ[λ−bθ1/n [µ,n],m]>0 o µ≥0. Le a(x, µ) := cθ1/m [λ−bθ1/n [µ,n],m](x). ¿F om P oposi ion 1, i µ1≤µ2 hen a(x, µ2)≤a(x, µ1).(34) On he o he hand, a(x, 0) = cθ1/m [λ,m]À0 and so he e exis s µ0>0 such ha o 0 ≤µ<µ0 we ha e 0≤µ < (a(x, µ0))M≤(a(x, µ))M. By he con inui y o a(x, µ0) he e exis s R0>0 such ha o µ < µ0 he se {x∈Ω : µ≤a(x, µ0)−R0} 6=∅. Hence, N(µ) := N(R0,1) = {x∈Ω : µ≤a(x, µ)−R0} ⊃ {x∈Ω : µ≤a(x, µ0)−R0} 6=∅. Finally, om (34) i ollows easily ha i µ1≤µ2 hen N(µ2)⊂N(µ1).Now we can eason as in he i s pa o Theo em 6. We will p o e he hi d pa . Fo ixed λ, µ > 0 and b > 0, we conside he same unc ion a(x) as abo e. In his case, we ha e N(R, 1) := N(c) = {x∈Ω : µ+R c≤θ1/m [λ−bθ1/n [µ,n],m](x)}, and again a guing as o he second pa o Theo em 6 he p oo ollows. • Rema k 9 1. Theo em 9has a biological in e p e a ion simila o Theo em 6gi en in Re- ma k 7. 2. The exis ence o semi-coexis ence s a es o (28) has been s udied in [7]. The au ho s ob- ained a semi-coexis ence s a e unde he condi ion (32), which is s onge han λ, µ > 0. I seems ha he exis ence o coexis ence s a es o (32) has no been analyzed p e iously. 3.3 Symbiosis In his case we conside he coope a i e sys em      −∆u=λu1/m −u2/m +bu1/m 1/n in Ω, −∆ =µ 1/n − 2/n +c 1/nu1/m in Ω, u= = 0 on ∂Ω. (35) Ou main esul abou he exis ence o semi-coexis ence s a es is he ollowing one: 20 Theo em 10 Assume bc < 1and ha he se s P1:= {x∈Ω : µ+cθ1/m [λ,m](x)>0} 6=∅P2:= {x∈Ω : λ+bθ1/n [µ,n](x)>0} 6=∅. Then he e exis s a leas one semi-coexis ence s a e o (35). Rema k 10 An obse a ion abou he way in which he abo e condi ions should be in e p e ed is app op ia e. I o example λ≤0 hen θ[λ,m]= 0, and so P16=∅is equi alen o µ > 0. P oo . Again we use he sub-supe solu ion me hod o sys ems. In his case a couple (u, u)−( , ) is sub-supe solu ion o (35) i , see [12], u, u, , ∈H2(Ω) ∩L∞(Ω), u≤uand ≤ in Ω and u≤0≤uand ≤0≤ on ∂Ω and such ha          −∆u≤λu1/m −u2/m +bu1/m 1/n in Ω, −∆u≥λu1/m −u2/m +bu1/m 1/n in Ω, −∆ ≤µ 1/n − 2/n +c 1/nu1/m in Ω, −∆ ≥µ 1/n − 2/n +c 1/nu1/m in Ω. (36) We conside wo cases: 1. Assume λ≤0. As we ha e men ioned abo e, he condi ion P16=∅is equi alen o µ > 0. We ake (u, u) = (θ[λ+bθ1/n [µ,n],m], M) ( , ) = (θ[µ,n], N), wi h M, N > 0 cons an s s ill o be chosen. Since P26=∅ hen u > 0. I is easy o show ha Mand Nsa is y (36) p o ided ha λ−M1/m +bN1/n ≤0, µ −N1/n +cM1/m ≤0.(37) Since bc < 1 he e exis Mand Nsa is ying (37). Thus he p oo o case 1 is comple e. 2. Assume λ > 0. In his case, P2= Ω. Now we pick (u, u) = (θ[λ,m], M) ( , ) = (θ[µ+cθ1/m [λ,m],n], N), wi h Mand Nas abo e. This comple es he p oo . • Again he ollowing esul s do no need any p oo . P oposi ion 6 Assume bc < 1and le (u, )be a nonnega i e solu ion o (35). Then θ[λ,m]≤u≤µλ+bµ 1−bc ¶m θ[µ,n]≤ ≤µµ+cλ 1−bc ¶n .(38) As an immedia e consequence o P oposi ion 6, we ha e 21 Co olla y 5 Assume bc < 1and ei he λ+bµ ≤0o µ+cλ ≤0. Then (35) does no admi a semi-coexis ence s a e. We ob ain he ollowing esul on coexis ence s a es Theo em 11 Assume bc < 1and (λ, µ)∈IR+×IR+ {(0,0)}.(39) Then (35) possesses a leas one coexis ence s a e. P oo . Assume λ, µ > 0. We ake (u, u) = (θ[λ,m], M) ( , ) = (θ[µ,n], N) wi h Mand Nsa is ying (37). I is no di icul o show ha his couple is a sub-supe solu ion o (35). On he o he hand, i o example λ= 0 and µ > 0 we pick (u, u) = (θ[bθ1/n [µ,n],m], M) ( , ) = (θ[µ,n], N). In his case, Mand Nha e o sa is y he ollowing inequali ies −M1/m +bN1/n ≤0, µ −N1/n +cM1/m ≤0. We can eason analogously in he emaining case. This concludes he p oo . • The ollowing esul can be p o ed simila ly o P oposi ion 3. P oposi ion 7 Assume (39). Then any nonnega i e solu ion o (35) is posi i e, i.e., i has no a dead co e. The ollowing esul says ha he se o (λ, µ)∈IR2whe e (35) has a leas a semi-coexis ence s a e is connec ed. Theo em 12 Assume bc < 1. Le us deno e Γ := {(λ, µ)∈IR2: (35) has a leas a semi-coexis ence s a e in (λ, µ)}. Then Γis connec ed. P oo . Le (u0, 0) be a semi-coexis ence s a e in (λ0, µ0). We ix λ0. Le µ∈[µ0,∞) and de ine (u, u) = (u0, M) ( , ) = ( 0, N) wi h Mand Nsa is ying (37). I is no ha d o show ha his couple is a sub-supe solu ion o (35) in (λ0, µ). This comple es he p oo . • Rema k 11 To ou knowledge, (35) has been only s udied p e iously in [7]. In his wo k he au ho s ob ained he exis ence o a semi-coexis ence s a e unde he condi ion λ > 0and µ > 0. F om Theo em 10,(35) possesses a semi-coexis ence s a e e en when λo µis nega i e. 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