# A Weak Comparison Principle for Reaction-Diffusion Systems

Authors: José Valero
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We prove a weak comparison principle for a reaction-diffusion system without uniqueness of solutions. We apply the abstract results to the Lotka-Volterra system with diffusion, a generalized logistic equation, and to a model of fractional-order chemical autocatalysis with decay. Moreover, in the case of the Lotka-Volterra system a weak maximum principle is given, and a suitable estimate in the space of essentially bounded functionsL∞
is proved for at least one solution of the problem.


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## Body

## 1. Introduction

Comparison results for parabolic equations and ordinary differential equations are well known in the literature (see, e.g., [1–4] among many others). One of the important applications of such kind of results is the theory of monotone dynamical systems, which leads to a more precise characterization of ω-limit sets and attractors. In the last years, several authors have been working in this direction (see, e.g., [4–8] for the deterministic case, and [9–12] for the stochastic case). In all these papers, it is considered the classical situation where the initial-value problem possesses a unique solution.However, the situation is more complicated when we consider a differential equation for which uniqueness of the Cauchy problem fails (or just it is not known to hold). Let us consider an abstract parabolic problem:(1.1)dudt=A(t,u(t)),τ≤t≤T,u(τ)=uτ,
for which we can prove that for every initial data in the phase space X (with a partial order ≤) there exists at least one solution.If we try to compare solutions of (1.1) for two ordered initial data uτ1≤uτ2, then we can consider a strong comparison principle and a weak one.The strong version would imply the existence of a solutionu1 with u1(τ)=uτ1 such that
(1.2)u1(t)≤u2(t)fort∈[τ,T],
for any solution u2 with u2(τ)=uτ2, and, viceversa, the existence of a solution u2 with u2(τ)=uτ2 such that (1.2) is satisfied for any solution u1 with u1(τ)=uτ1. This kind of result is established in [13] for a delayed ordinary differential equations, defining then a multivalued order-preserving dynamical system.The weak version of the comparison principle says that ifuτ1≤uτ2, then there exist two solutions u1,u2 of (1.1) such that u1(τ)=uτ1, u2(τ)=uτ2, and (1.2) hold.There is in fact an intermediate version of the comparison principle, which says that if we fix a solutionu1 of (1.1) with u1(τ)=uτ1, then there exists a solution u2 with u2(τ)=uτ2 such that (1.2) is satisfied (and vice versa). This is proved in [14] for a differential inclusion generated by a subdifferential map.In this paper, we establish a weak comparison principle for a reaction-diffusion system in which the nonlinear term satisfies suitable dissipative and growth conditions, ensuring existence of solutions but not uniqueness. This principle is applied to several well-known models in physics and biology. Namely, a weak comparison of solutions is proved for the Lotka-Volterra system, the generalized logistic equation and for a model of fractional-order chemical autocatalysis with decay. Moreover, in the case of the Lotka-Volterra system a weak maximum principle is given, and a suitable estimate in the space of essentially bounded functionsL∞ is proved for at least one solution of the problem.We note that in the papers [15, 16] the existence of a global attractor is proved for such kind of reaction-diffusion systems. In the near future, we will apply these results to obtain theorems concerning the structure of the global attractor.


## 2. Comparison Results for Reaction-Diffusion Systems

We shall denote by|·| and (·,·) the norm and scalar product in the space ℝm, m≥1. Let d>0 be an integer and Ω⊂ℝN be a bounded open subset with smooth boundary. Consider the problem:
(2.1)∂u∂t-aΔu+f(t,u)=h(t,x),(t,x)∈(τ,T)×Ω,u|x∈∂Ω=0,u|t=τ=uτ(x),
where τ,T∈ℝ, T>τ, x∈Ω, u=(u1(t,x),…,ud(t,x)),f=(f1,…,fd), a is a real d×d matrix with a positive symmetric part (a+at)/2≥βI,β>0, h∈L2(τ,T;(L2(Ω))d). Moreover, f=(f1(t,u),…,fd(t,u)) is jointly continuous on [τ,T]×ℝd and satisfies the following conditions:
(2.2)∑i=1d‍|fi(t,u)|pi/(pi-1)≤C1(1+∑i=1d‍|ui|pi),(f(t,u),u)≥α∑i=1d‍|ui|pi-C2,
where pi≥2, α,C1,C2>0.LetH=(L2(Ω))d, V=(H01(Ω))d, and let V′ be the dual space of V. By ∥·∥, ∥·∥V we denote the norm by H and V, respectively. For p=(p1,…,pd), we define the spaces
(2.3)Lp(Ω)=Lp1(Ω)×⋯×Lpd(Ω),Lp(τ,T;Lp(Ω))=Lp1(τ,T;Lp1(Ω))×⋯×Lpd(τ,T;Lpd(Ω)).
We take q=(q1,…,qd), where (1/pi)+(1/qi)=1.We say that the functionu(·) is a weak solution of (2.1) if u∈Lp(τ,T;Lp(Ω))∩L2(τ,T;V)∩C([τ,T];H), du/dt∈L2(τ,T;V′)+Lq(τ,T;Lq(Ω)), u(τ)=uτ, and
(2.4)∫τT‍〈dudt,ξ〉dt+∫τT‍∫Ω‍(∇(au),∇ξ)dxdt+∫τT‍∫Ω‍(f(t,u),ξ)dxdt=∫τT‍∫Ω‍(h,ξ)dxdt,
for all ξ∈Lp(τ,T;Lp(Ω))∩L2(τ,T;V), where 〈·,·〉 denotes pairing in the space V′+Lq(Ω), and (∇u,∇v)=∑i=1d‍(∇ui,∇vi).Under conditions (2.2), it is known [17, page 284] that for any uτ∈H there exists at least one weak solution u=u(t,x) of (2.1), and also that the function t↦∥u(t)∥2 is absolutely continuous on [τ,T] and (d/dt)∥u(t)∥2=2〈du/dt,u〉 for a.a. t∈(τ,T).Denoter=(r1,…,rd),ri=max{1;N(1/qi-1/2)}. Any weak solution satisfies (du/dt)∈Lq(τ,T;H-r(Ω)) and
(2.5)Lq(0,T;H-r(Ω))=Lq1(0,T;H-r1(Ω))×⋯×Lqd(0,T;H-rd(Ω)).If, additionally, we assume that thatf(t,u) is continuously differentiable with respect to u for any t∈[τ,T],u∈ℝd, and
(2.6)(fu(t,u)w,w)≥-C3(t)|w|2,∀w,u∈ℝd,
where C3(·)∈L1(τ,T), C3(t)≥0, the weak solution of (2.1) is unique. Here, fu denotes the Jacobian matrix of f.We consider also the following assumption: there existsR0>0 such that
(2.7)fi(t,u)≥fi(t,v),
for any t∈[τ,T] and any u,v∈ℝd such that ui=vi and uj≤vj if j≠i, and |u|,|v|≤R0, which means that the systems is cooperative in the ball with radius R0 centered at 0.Consider the two problems:(2.8)∂u∂t-aΔu+f1(t,u)=h1(t,x),(t,x)∈(τ,T)×Ω,u|x∈∂Ω=0,u|t=τ=uτ(x),(2.9)∂u∂t-aΔu+f2(t,u)=h2(t,x),(t,x)∈(τ,T)×Ω,u|x∈∂Ω=0,u|t=τ=uτ(x),
where fj are jointly continuous on [τ,T]×ℝd. Among conditions (2.2) and (2.6)-(2.7), we shall consider the following:
(2.10)h1(t,x)≤h2(t,x),fora.a.(t,x),f1i(t,u)≥f2i(t,u),∀t,u.Lemma 2.1.
Iffj satisfy (2.2) and (2.10), then the constants pi have to be the same for f1 and f2.Proof.
Denote bypij, αj,C1j, and C2j the constants corresponding to fj in (2.2). By contradiction let, for example, p12>p11. Take the sequence un=(un1,0,…,0), where un1→+∞ as n→∞. Then by (2.2), (2.10), and Young’s inequality, we have
(2.11)α2|un1|p12-C22≤(f2(t,un),un)≤(f1(t,un),un)=f11(t,un)un1≤(C11(1+|un1|p11))(p11-1)/p11u1n≤C(1+|un1|p11).
But p12>p11 implies the existence of n such that α2|un1|p12-C22>C(1+|un1|p11), which is a contradiction. Hence, p12≤p11.
Conversely, letp12<p11. Then we take un=(un1,0,…,0) with un1→-∞ as n→∞ so that
(2.12)α1|un1|p11-C21≤(f1(t,un),un)≤(f2(t,un),un)=f21(t,un)un1≤(C12(1+|un1|p12))(p12-1)/p12u1n≤C(1+|un1|p12).
As before, we obtain a contradiction, so p12=p11.
Repeating similar arguments for the otherpij, we obtain that pi1=pi2 for i=1,…,d.We recall [15] that under conditions (2.2) any solution u(·) of (2.8) satisfies the inequality:
(2.13)∥u(t)∥2+2β∫st∥∇u(τ)∥2dτ+α∑i=1d‍∫st∥ui(r)∥Lpi(Ω)pidr≤∥u(s)∥2+C∫st(∥h1(r)∥2+1)dr,
for some constant C>0. Of course, the same is valid for any solution of (2.9). From (2.13), for any T>τ we obtain
(2.14)∥u(t)∥2≤∥uτ∥2+C∫τT(∥h1(r)∥2+1)dr=K2(∥uτ∥,τ,T)∀τ≤t≤T.We shall denote byu1(t) the solution of (2.8) corresponding to the initial data uτ1 and by u2(t) the solution of (2.9) corresponding to the initial data uτ2. Also, we take u+=max{u,0} for u∈ℝ.We obtain the following comparison result.Theorem 2.2.
Assume thatfj,hj satisfy (2.2), (2.6), and (2.10). If uτ1≤uτ2 and we suppose that f2 satisfies (2.7) with R02≥2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)}, where K(∥uτj∥,τ,T) is taken from (2.14), we have u1(t)≤u2(t), for all t∈[τ,T].Remark 2.3.
The results remain valid if, instead,f1 satisfies (2.7) with R02≥K2(∥uτ1∥,T).Remark 2.4.
Iff2 satisfies (2.7) for an arbitrary R0>0 (i.e., in the whole space ℝd), then the result is true for any initial data uτ1≤uτ2.Proof.
Letg2(t,u)=f2(t,u)+C3(t)u. The function g2(t,·) satisfies (2.6) with C3≡0. For any u1,u2∈ℝd define v2(u1,u2) by
(2.15)v2i={u2i,ifu1i≥u2i,u1i,ifu1i<u2i.
Note that u1-v2=(u1-u2)+ and v2-u2=-(u2-u1)+, so v2i≤u2i for all i. For the function (u1-u2)+, we can obtain by (2.10) and the mean value theorem that
(2.16)12ddt∥(u1-u2)+∥2+a∥(u1-u2)+∥V2≤-∫Ω‍(f1(t,u1)-f2(t,u2),(u1-u2)+)dx≤-∫Ω‍(f2(t,u1)-f2(t,v2),(u1-u2)+)dx-∫Ω‍(f2(t,v2)-f2(t,u2),(u1-u2)+)dx=-∫Ω‍(g2u(t,v(t,x,u1,u2))(u1-v2),(u1-u2)+)dx+C3(t)∥(u1-u2)+∥2-∫Ω‍(f2(t,v2)-f2(t,u2),(u1-u2)+)dx=-∫Ω‍(g2u(t,v(t,x,u1,u2))(u1-u2)+,(u1-u2)+)dx+C3(t)∥(u1-u2)+∥2-∫Ω‍(f2(t,v2)-f2(t,u2),(u1-u2)+)dx≤C3(t)∥(u1-u2)+∥2-∫Ω‍(f2(t,v2)-f2(t,u2),(u1-u2)+)dx.
For allt, we have
(2.17)(f2(t,v2)-f2(t,u2),(u1-u2)+)=∑i∈J‍(f2i(t,v2)-f2i(t,u2))(u1i-u2i)+,
where u1i-u2i>0, for i∈J, and u1i-u2i≤0 if i∉J. For any i∈J, we have that v2i=u2i, and then by v2≤u2, |v2|2≤|u1|2+|u2|2, (2.14), and (2.7), we get
(2.18)f2i(t,v2)-f2i(t,u2)≥0.
By Gronwall’s lemma, we get(2.19)∥(u1(t)-u2(t))+∥2≤∥(uτ1-uτ2)+∥2e∫τt‍2C3(s)ds=0.
Thus ∥(u1-u2)+∥=0, which means that u1i(x,t)-u2i(x,t)≤0, for a.a. x∈Ω, and all i∈{1,…,d}, t∈[τ,T].Remark 2.5.
In the scalar case, that is,d=1, condition (2.7) is trivially satisfied.When condition (2.6) fails to be true, we will obtain a weak comparison principle.Define a sequence of smooth functionsψk:ℝ+→[0,1] satisfying
(2.20)ψk(s)={1,if0≤s≤k,0≤ψk(s)≤1,ifk≤s≤k+1,0,ifs≥k+1.For everyk≥1 we put fki(t,u)=ψk(|u|)fi(t,u)+(1-ψk(|u|))gi(u), where gi(u)=|ui|pi-2ui. Then fk∈ℂ([τ,T]×ℝd;ℝd) and for any A>0,
(2.21)supt∈[τ,T]sup|u|≤A|fk(t,u)-f(t,u)|→0,ask→∞.Letρε:ℝd→ℝ+ be a mollifier, that is, ρϵ∈ℂ0∞(ℝd;ℝ), suppρϵ⊂Bϵ={x∈ℝd:|x|<ϵ}, ∫ℝd‍ρϵ(s)ds=1 and ρϵ(s)≥0 for all s∈ℝd. We define the functions
(2.22)fkϵ(t,u)=∫ℝd‍ρϵ(s)fk(t,u-s)ds.
Since for any k≥1fk is uniformly continuous on [0,T]×[-k-1,k+1], there exist ϵk∈(0,1) such that for all u satisfying |u|≤k, and for all s for which |u-s|<ϵk we have
(2.23)supt∈[τ,T]|fk(t,u)-fk(t,s)|≤1k.
We put fk(t,u)=fkϵk(t,u). Then fk(t,·)∈ℂ∞(ℝd;ℝd), for all t∈[τ,T],k≥1.For further arguments we need the following technical result [16, Lemma 2].Lemma 2.6.
Letf satisfy (2.2). For all k≥1, the following statements hold:
(2.24)supt∈[τ,T]sup|u|≤A|fk(t,u)-f(t,u)|→0,ask→∞,∀A>0,(2.25)∑i=1d‍|fki(t,u)|pi/(pi-1)≤D1(1+∑i=1d‍|ui|pi),(fk(t,u),u)≥γ∑i=1d‍|ui|pi-D2,(2.26)(fuk(t,u)w,w)≥-D3(k)|w|2,∀u,w,
where D3(k) is a nonnegative number, and the positive constants D1, D2≥C2, γ do not depend on k.Consider first the scalar case.Theorem 2.7.
Letd=1. Assume that fj,hj satisfy (2.2) and (2.10). If uτ1≤uτ2, there exist two solutions u1,u2 (of (2.8) and (2.9), resp.) such that u1(t)≤u2(t) for all t∈[τ,T].Proof.
For the functionsfj we take the approximations fjk (defined in Lemma 2.6), which satisfy (2.24)–(2.26), and consider the problems
(2.27)∂u∂t-aΔu+fjk(t,u)=hj(t,x),(t,x)∈(τ,T)×Ω,u|x∈∂Ω=0,u|t=τ=uτ,
for j=1,2. Problem (2.27) has a unique solution for any initial data uτ∈H. In view of Lemma 2.1, the constant p is the same for f1 and f2. We note that
(2.28)f1k(t,u)=ψk(|u|)f1(t,u)+(1-ψk(|u|))|u|p-2u≥ψk(|u|)f2(t,u)+(1-ψk(|u|))|u|p-2u=f2k(t,u).
Then, it is clear that f1k(t,u)≥f2k(t,u) for every (t,u).
By Theorem2.2 we know that as uτ1≤uτ2, we have u1k(t)≤u2k(t), for all t∈[τ,T], for the corresponding solutions of (2.27).
In view of Lemma2.6, one can obtain in a standard way that (2.13) is satisfied for the solutions of (2.27) with a constant C not depending on k and replacing α by γ. Hence, the sequences ujk(·) are bounded in L∞(τ,T;H)∩L2(τ,T;V)∩Lp(τ,T;Lp(Ω)). It follows from (2.25) that fjk(·,ujk(·)) are bounded in Lq(τ,T;Lq(Ω)) and also that {(duk/dt)(·)} is bounded in Lq(τ,T;H-r(Ω)), where ri=max{1;(1/2-1/pi)N}. By the Compactness Lemma [18], we have that for some functions uj=uj(t,x), j=1,2:
(2.29)ujk→ujweaklystarinL∞(τ,T;H),(2.30)ujk→ujinL2(τ,T;H),ujk(t)→uj(t)inHfora.a.t∈(τ,T),(2.31)ujk(t,x)→uj(t,x)fora.a.(t,x)∈(τ,T)×Ω,(2.32)ujk→ujweaklyinL2(τ,T;V),(2.33)dujkdt→dujdtweaklyinLq(τ,T;H-r(Ω)),(2.34)ujk→ujweaklyinLp(τ,T;Lp(Ω)).
Also, arguing as in [19, page 3037] we obtain
(2.35)ujk(t)→uj(t)weaklyinH∀t∈[τ,T].
Moreover, by (2.24) and (2.31) we have fjk(t,uk(t,x))→fj(t,u(t,x)) for a.a. (t,x)∈(τ,T)×Ω and then the boundedness of fjk(·,ujk(·)) in Lq(τ,T;Lq(Ω)) implies that fjk(·,ujk(·)) converges to f(·,u(·)) weakly in Lq(τ,T;Lq(Ω)) [18]. It follows that u1(·),u2(·) are weak solutions of (2.8) and (2.9), respectively, with u1(τ)=uτ1, u2(τ)=uτ2.
Moreover, one can prove that(2.36)ujk(t)→uj(t)stronglyinH∀t∈[τ,T].
Indeed, we define the functions Jjk(t)=∥ujk(t)∥2-C∫τt(∥hj(s)∥2+1)ds, Jj(t)=∥uj(t)∥2-C∫τt(∥hj(s)∥2+1)ds, which are nonincreasing in view of (2.13). Also, from (2.30) we have Jjk(t)→Jj(t) for a.a. t∈(τ,T). Then one can prove that limsupk→∞Jjk(t)≤Jj(t) for all t∈[τ,T] (see [15, page 623] for the details). Hence, limsupk→∞∥ujk(t)∥≤∥uj(t)∥. Together with (2.35) this implies (2.36) (see again [15, page 623] for more details).
Hence, passing to the limit we obtain(2.37)u1(t)≤u2(t),∀t∈[τ,T].Further, let us prove the general case for an arbitraryd∈ℕ.Theorem 2.8.
Assume thatfj,hj satisfy (2.2) and (2.10). Also, suppose that either f1 or f2 satisfies (2.7) for an arbitrary R0>0. If uτ1≤uτ2, there exist two solutions u1,u2 (of (2.8) and (2.9), resp.) such that u1(t)≤u2(t), for all t∈[τ,T].Proof.
Letf1 be the function which satisfies (2.7). We take the approximations f1k,f2k (defined in Lemma 2.6), which satisfy (2.24)–(2.26). Then, we consider problems (2.27).
In view of Lemma2.1, the constants pi are the same for f1 and f2. We note that
(2.38)f1ki(t,u)=ψk(|u|)f1i(t,u)+(1-ψk(|u|))|ui|pi-2ui≥ψk(|u|)f2i(t,u)+(1-ψk(|u|))|ui|pi-2ui=f2ki(t,u).
Then, it is clear that f1k(t,u)≥f2k(t,u) for every (t,u).
Using Lemma2.6 it is standard to obtain estimate (2.14) with a constant C not depending on k. Hence, the solutions ujk(·) of (2.27) satisfy
(2.39)∥ujk(t)∥2≤∥uτj∥2+C∫τT(∥hj(r)∥2+1)dr=K2(∥uτj∥,τ,T).
We note that
(2.40)f1k(t,u)=f1(t,u),
if |u|≤k, since in such a case ψk(|u|)=1. Hence, if k2≥2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)}, the functions f1k satisfy condition (2.7) with R0=k. Therefore, for any t∈[τ,T] and any u,v∈ℝd such that ui=vi and uj≤vj if j≠i, and |u|,|v|≤k-1, we have
(2.41)f1k(t,u)=∫ℝd‍ρϵk(s)f1k(t,u-s)ds≥∫ℝd‍ρϵk(s)f1k(t,v-s)ds=f2k(t,u).
Thus, if (k-1)2≥2max{K2(∥uτ1∥,T),K2(∥uτ2∥,T)}, the functions f1k satisfy condition (2.7) with R0=k-1.
By Theorem2.2 we know that as uτ1≤uτ2, we have u1k(t)≤u2k(t), for all t∈[τ,T], k≥1+(2max{K2(∥uτ1∥,T),K2(∥uτ2∥,T)})1/2, for the corresponding solutions of (2.27).
Repeating the same proof of Theorem2.7, we obtain that the sequences u1k,u2k converge (up to a subsequence) in the sense of (2.29)–(2.36) to the solutions u1,u2 of problems (2.8) and (2.9), respectively. Also, it holds
(2.42)u1(t)≤u2(t),∀t∈[τ,T].In the applications we need to generalize this theorem to the case where the constantα can be negative. We shall do this when f1,f2 have sublinear growth (i.e., pi=2 for all i). Consider for (2.1) the following conditions:
(2.43)|f(t,u)|≤C1(1+|u|),(f(t,u),u)≥α|u|2-C2,
where α∈ℝ, and C1,C2>0.Letf1,f2 satisfy (2.43) with constants αj,C1j,C2j,j=1,2. Then if min{α1,α2}≤0, we make in (2.1) the change of variable v=e-βtu, where β>-min{α1,α2}. Hence, multiplying (2.8) and (2.9) by e-βt we have
(2.44)∂v∂t-aΔv+e-βtf1(t,eβtv)+βv=e-βth1(t,x),(t,x)∈(τ,T)×Ω,v|x∈∂Ω=0,v|t=τ=e-βτuτ(x),(2.45)∂v∂t-aΔv+e-βtf2(t,eβtv)+βv=e-βth2(t,x),(t,x)∈(τ,T)×Ω,v|x∈∂Ω=0,v|t=τ=e-βτuτ(x).It is easy to check that ifv(t) is a weak solution of (2.44), then u(t)=eβtv(t) is a weak solution of (2.8) (and the same is true, of course, for (2.45) and (2.9)). Conversely, if u(t) is a weak solution of (2.8), then v(t)=e-βtu(t) is a weak solution of (2.44).The functionsf~j(t,v)=e-βtfj(t,eβtv)+βv satisfy (2.2) with pi=2 for all i. Indeed,
(2.46)|f~j(t,v)|≤e-βtC1j(1+eβt|v|)+β|v|≤C~1j(1+|v|),(f~j(t,v),v)≥e-2βt(fj(t,eβtv),eβtv)+β|v|2≥(αj+β)|v|2-C2j,
where αj+β>0.Then, we obtain the following.Theorem 2.9.
Assume thatfj,hj satisfy (2.43) and (2.10). Also, suppose that either f1 or f2 satisfies (2.7) for an arbitrary R0>0. If uτ1≤uτ2, there exist two solutions u1,u2 (of (2.8) and (2.9), resp.) such that u1(t)≤u2(t), for all t∈[τ,T].Proof.
We consider problems (2.44) and (2.45). In view of (2.46) f~j(t,v)=e-βtfj(t,eβtv)+βv satisfy (2.2). Also, defining h~j(t,x)=e-βthj(t,x) it is clear that (2.10) holds. Finally, if, for example, f1 satisfies (2.7) for any R0>0, then it is obvious that for f~1 is true as well.
Hence, by Theorem2.8 there exist two solutions v1,v2 (of (2.44) and (2.45), resp.), with vj(τ)=e-βτuτj such that v1(t)≤v2(t) for all t∈[τ,T]. Thus
(2.47)u1(t)=eβtv1(t)≤eβtv2(t)=u2(t),fort∈[τ,T],
and u1,u2 are solutions of (2.8) and (2.9), respectively such that uj(τ)=uτj.Remark 2.10.
Iffj satisfy (2.6), then the solutions u1,u2 given in Theorem 2.9 are unique for the corresponding initial data.


## 3. Comparison for Positive Solutions

Denoteℝ+d={u∈ℝd:ui≥0}. Let us consider the previous results in the case where the solutions have to be positive. Consider now the following conditions. 						(3.1)Thematrixaisdiagonal,(3.2)hi(t,x)-fi(t,u1,…,ui-1,0,ui+1,…,ud)≥0,
for all i, a.e. (t,x)∈(τ,T)×Ω and uj≥0 if j≠i. Obviously, in the scalar case these conditions just mean that
(3.3)h(t,x)-f(t,0)≥0,
for a.e. (t,x)∈(τ,T)×Ω.It is well known (see [16, Lemma 5] for a detailed proof) that if we assume conditions (2.2) only for u∈ℝ+d, and also (2.6) and (3.1)-(3.2), then for any uτ≥0 there exists a unique weak solution u(·) of (2.1). Moreover, u(·) is such that u(t)≥0 for all t∈[τ,T].On the other hand, if we assume these conditions except (2.6), then there exists at least one weak solution u(·) of (2.1) such that u(t)≥0 for all t∈[τ,T] [16, Theorem 4]. Moreover, we can prove the following.Lemma 3.1.
Assume conditions (2.2), (2.6) only for u∈ℝ+d, and also (3.1)-(3.2). Then there exists a weak solution u(·) of (2.1), which is unique in the class of solutions satisfying u(t)≥0 for all t∈[τ,T].Proof.
Letu1,u2 be two solutions with ui(τ)=uτ, i=1,2 such that ui(t)≥0 for all t. Denote w(t)=u1(t)-u2(t). Then in a standard way by the mean value theorem, we obtain
(3.4)12ddt∥w(t)∥2≤-∫Ω‍(f(t,u1(t,x))-f(t,u2(t,x)),w(t,x))dx=-∫Ω‍(fu(t,v(t,x,u1,u2))w(t,x),w(t,x))dx≤C3(t)∥w(t)∥2,
where v(t,x,u1,u2)∈L(u1(t,x),u2(t,x))={αu1(t,x)+(1-α)u2(t,x):α∈0,1]} so that v(t,x,u1,u2)≥0. The uniqueness follows from Gronwall’s lemmaWe prove also a result, which is similar to Lemma2.1. Denote by pij, αj,C1j, and C2j the constants corresponding to fj in (2.2) for problems (2.8) and (2.9), respectively. Arguing as in the proof of Lemma 2.1 we obtain the following lemma.Lemma 3.2.
Iffj satisfy (2.2) and (2.10) for u∈ℝd+, then pi1≥pi2 for all i.Theorem 3.3.
Letfj,hj satisfy (2.6) and (3.1)-(3.2). Assume that fj,hj satisfy (2.2) and (2.10) for u∈ℝd+. If 0≤uτ1≤uτ2 and one supposes that f2 satisfies (2.7) for u∈ℝ+d with R02≥2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)}, where K(∥uτj∥,τ,T) is taken from (2.14), one has  0≤u1(t)≤u2(t) for all t∈[τ,T], where u1(·),u2(·) are the solutions corresponding to uτ1 and uτ2, respectively.Proof.
As the solutionsu1(·), u2(·) corresponding to uτ1 and uτ2 are nonnegative, repeating exactly the same steps of the proof of Theorem 2.2 we obtain the desired result.Remark 3.4.
The results remain valid if, instead,f1 satisfies (2.7) with the same R0.Remark 3.5.
Iff2 satisfies (2.7) for an arbitrary R0>0 (i.e., in the whole space ℝd), then the result is true for any initial data 0≤uτ1≤uτ2.We shall need also the following modification of Theorem3.3.Theorem 3.6.
Letfj,hj satisfy (2.6) and (3.1)-(3.2). Assume that fj,hj satisfy (2.2) and (2.10) for u∈ℝ+d. Let 0≤uτ1≤uτ2.  One supposes that f2 satisfies
(3.5)f2i(t,u)≥f2i(t,v)-ε,
for any t∈[τ,T] and any u,v∈ℝ+d such that ui=vi and uj≤vj if j≠i, and |u|,|v|≤R0 with R02≥2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)}, where K(∥uτj∥,τ,T) is taken from (2.14).
Then there exists a constantC(τ,T) such that
(3.6)∥(u1(t)-u2(t))+∥≤C(τ,T)ε,∀t∈[τ,T],
where u1(·),u2(·) are the solutions corresponding to uτ1 and uτ2, respectively.Proof.
Arguing as in the proof of Theorem2.2 we obtain the inequality
(3.7)12ddt∥(u1-u2)+∥2+a∥(u1-u2)+∥V2≤C3(t)∥(u1-u2)+∥2-∫Ω‍(f2(t,v2)-f2(t,u2),(u1-u2)+)dx,
where v2 is defined in (2.15).
Using (2.17), v2≤u2,|v2|2≤|u1|2+|u2|2, (2.14), and (3.5), we get
(3.8)f2i(t,v2)-f2i(t,u2)≥-ε.
Thus
(3.9)ddt∥(u1-u2)+∥2≤2C3(t)∥(u1-u2)+∥2+2ε∫Ω‍∑i∈J‍(u1i-u2i)+dx≤(2C3(t)+1)∥(u1-u2)+∥2+Kε2,
for some constant K>0. By Gronwall’s lemma, we get
(3.10)∥(u1(t)-u2(t))+∥2≤∥(uτ1-uτ2)+∥2e∫τt‍(2C3(s)+1)ds+Kε2∫τt‍e∫rt‍(2C3(s)+1)dsdr≤C2(τ,T)ε2.Let us consider now the multivalued case. We will obtain first some auxiliary statements.We shall define suitable approximations. For anyn≥1 we put fni(t,u)=ψn(|u|)fi(t,u)+(1-ψn(|u|))gi(t,u), where gi(t,u)=|ui|pi-2ui+fi(t,0,…,0), and ψn was defined in (2.20). Then fn∈ℂ([τ,T]×ℝd;ℝd) and for any A>0,
(3.11)supt∈[τ,T]sup|u|≤A|fn(t,u)-f(t,u)|→0,asn→∞.
We will check first that fn satisfy conditions (2.2) for u∈ℝ+d, where the constants do not depend on n.Lemma 3.7.
Letf satisfy (2.2) for u∈ℝ+d. For all n≥1 one has
(3.12)∑i=1d‍|fni(t,u)|pi/(pi-1)≤D1(1+∑i=1d‍|ui|pi),(fn(t,u),u)≥γ∑i=1d‍|ui|pi-D2,
for u∈ℝd+, where the positive constants D1, D2, and γ do not depend on n.
If|u|>n+1, then for any w∈ℝd one has
(3.13)(fnu(t,u)w,w)≥0.
Moreover, if f,h satisfy (3.2), then fn,h also satisfies this condition.Proof.
In view of (2.2) we get
(3.14)(fn(t,u),u)=ψn(|u|)(f(t,u),u)+(1-ψn(|u|))(g(u),u)≥ψn(|u|)(α∑i=1d‍|ui|pi-C2)+(1-ψn(|u|))∑i=1d‍|ui|pi+(1-ψn(|u|))∑i=1d‍fi(t,0,…,0)ui≥ψn(|u|)α∑i=1d‍|ui|pi-C2+(1-ψn(|u|))12∑i=1d‍|ui|pi-K1(1-ψn(|u|))∑i=1d‍|fi(t,0,…,0)|pi/(pi-1)≥α~∑i=1d‍|ui|pi-C2-K1C1,
where α~=min{1/2,α}, for some constant K1>0. Also,
(3.15)∑i=1d‍|fni(t,u)|pi/(pi-1)≤K2(∑i=1d‍|fi(t,u)|pi/(pi-1)+∑i=1d‍|gi(u)|pi/(pi-1))≤K3(C1(1+∑i=1d‍|ui|pi)+∑i=1d‍|ui|pi+∑i=1d‍|fi(t,0,…,0)|pi/(pi-1))≤K4(∑i=1d‍|ui|pi+1),
for some constant K4>0. Thus, for D1=K4, D2=C2+K1C1, α~=min{1/2,α} we have (3.12) for the functions fn.
Moreover, if|u|>n+1, then for any w∈ℝd,
(3.16)(fnu(t,u)w,w)=(gu(t,u)w,w)=∑i=1d‍(pi-1)|ui|pi-2wi2≥0.
Finally, if (3.2) is satisfied, then
(3.17)hi(t,x)-fni(t,u)=ψn(|u|)(hi(t,x)-fi(t,u))+(1-ψn(|u|))(hi(t,x)-fi(t,0,…,0))≥0,
for all i, a.e. (t,x)∈(τ,T)×Ω and u such that ui=0 and uj≥0 if j≠i.Let2≤qi≤pi, i=1,…,d. We define also the following approximations f~ni(t,u)=ψn(|u|)fi(t,u)+(1-ψn(|u|))g~i(t,u), where g~i(t,u)=|ui|pi-2ui+|ui|qi-2ui+fi(t,0,…,0). Then (3.11) holds. We check that f~n satisfy conditions (2.2) for u∈ℝ+d, where the constants do not depend on n.Lemma 3.8.
Letf satisfy (2.2) for u∈ℝ+d. For all n≥1 one has
(3.18)∑i=1d‍|f~ni(t,u)|pi/(pi-1)≤D1(1+∑i=1d‍|ui|pi),(f~n(t,u),u)≥γ∑i=1d‍|ui|pi-D2,
for u∈ℝd+, where the positive constants D1, D2, and γ do not depend on n.
If|u|>n+1, then for any w∈ℝd one has
(3.19)(f~nu(t,u)w,w)≥0.
Moreover, iff,h satisfy (3.2), then f~n,h also satisfy this condition.Proof.
In view of (3.12), we have
(3.20)(f~n(t,u),u)=(fn(t,u),u)+(1-ψn(|u|))∑i=1d‍|ui|qi≥γ∑i=1d‍|ui|pi-D2,∑i=1d‍|f~ni(t,u)|pi/(pi-1)≤K1(∑i=1d‍|fni(t,u)|pi/(pi-1)+∑i=1d‍|ui|pi(qi-1)/(pi-1))≤K1(D1(1+∑i=1d‍|ui|pi)+∑i=1d‍|ui|qi)≤K2(1+∑i=1d‍|ui|pi),
where we have used that pi≥qi implies pi/(pi-1)≤qi/(qi-1). Finally, (3.19) and condition (3.2) are proved in the same way as in Lemma 3.7.For everyn≥1 consider the sequence fnε(t,u) defined by fnϵ(t,u)=∫ℝd‍ρϵ(s)bn(t,u-s)ds, where either bn=fn or bn=f~n, defined before. Since any bn are uniformly continuous on [τ,T]×[-k-1,k+1], for any k≥1, there exist ϵk,n∈(0,1) such that for all u satisfying |u|≤k, and for all s for which |u-s|<ϵk,n we have
(3.21)supt∈[τ,T]|bn(t,u)-bn(t,s)|≤1k.
We put fnk(t,u)=fnϵk,n(t,u). Then, fnk(t,·)∈ℂ∞(ℝd;ℝd) for all t∈[τ,T],k,n≥1. Since for any compact subset A⊂ℝd and any n we have fnk→bn uniformly on [τ,T]×A, we obtain the existence of a sequence δnk∈(0,1) such that δnk→0, as k→∞, and |fnki(t,u)-bni(t,u)|≤δnk, for any i,n and any u satisfying |u|≤n+2. We define the function Fnk=(Fnk1,…,Fnkd) given by
(3.22)Fnki(t,u)=fnki(t,u)-pδnk,
where p∈ℕ.Lemma 3.9.
Letf satisfy (2.2) for u∈ℝ+d. For all n,k≥1 we have
(3.23)∑i=1d‍|Fnki(t,u)|pi/(pi-1)≤D3(1+∑i=1d‍|ui|pi),(Fnk(t,u),u)≥ν∑i=1d‍|ui|pi-D4,
for u∈ℝ+d, where the positive constants D3,D4, and   ν do not depend neither on n nor k.
Moreover, iff,h satisfy (3.2), then Fnk,h also satisfy this condition if |u|≤n+2.Proof.
Sincefn satisfy (3.12) and f~n satisfies (3.18), we have
(3.24)∑i=1d‍|Fnki(t,u)|pi/(pi-1)≤R1(∑i=1d‍|fnki(t,u)|pi/(pi-1)+1)≤R1(∑i=1d‍((∫ℝd‍ρϵk(s)ds)1/(pi-1)∫ℝd‍ρϵk(s)|bni(t,u-s)|pi/(pi-1)ds)+1)≤R2(∑i=1d‍∫ℝd‍ρϵk(s)(1+|ui-si|pi)ds+1)≤R3(∑i=1d‍∫ℝd‍ρϵk(s)(|ui|pi+ϵkpi)ds+1)≤R4(1+∑i=1d‍|ui|pi),
for some constant R4>0.
On the other hand,(3.25)(Fnk(t,u),u)=∫ℝd‍ρϵk(s)(bn(t,u-s),u-s)ds+∫ℝd‍ρϵk(s)(bn(t,u-s),s)ds-pδnk∑i=1n‍ui≥∫ℝd‍ρϵk(s)(γ∑i=1d‍|ui-si|pi-D2)ds-∫ℝd‍ρϵk(s)∑i=1d‍(γ2D1|bni(t,u-s)|pi/(pi-1)+R5|si|pi)ds-pδnk∑i=1n‍ui≥∫ℝd‍ρϵk(s)(γ∑i=1d‍|ui-si|pi-D2)ds-∫ℝd‍ρϵk(s)(γ2∑i=1d‍|ui-si|pi+R6)ds-pδnk∑i=1n‍ui≥γ2∫ℝd‍ρϵk(s)∑i=1d‍|ui-si|pids-R7-pδnk∑i=1n‍ui≥ν∑i=1d‍|ui|pi-R8,
for some constants ν,R8>0, where in the last inequality we have used that for some D>0,
(3.26)|ui|pi=|ui-si+si|pi≤D(|ui-si|pi+|si|pi)≤D(|ui-si|pi+ϵkpi).
Hence, (3.23) holds.
In view of Lemmas3.7 and 3.8 the functions bn,h satisfy (3.2). Hence, |fnki(t,u)-bni(t,u)|≤δnk, for any i,n and any u satisfying |u|≤n+2, implies that
(3.27)hi(t,x)-Fnki(t,u)=hi(t,x)-fnki(t,u)+pδnk≥hi(t,x)-bni(t,u)≥0,
for u such that ui=0, uj≥0, j≠i, and |u|≤n+2.Define a smooth functionϕn:ℝ+→[0,1] satisfying
(3.28)ϕn(s)={1,if0≤s≤n+1+γ¯,0≤ϕn(s)≤1,ifn+1+γ¯≤s≤n+2,0,ifs≥n+2,
where 0<γ¯<1 is fixed. Let lnk(t,u) be given by
(3.29)lnk(t,u)=ϕn(|u|)Fnk(t,u)+(1-ϕn(|u|))bn(t,u).
Since for any compact subset A⊂ℝd and any n we have fnk→bn uniformly on [τ,T]×A as k→∞, it is clear that
(3.30)supt∈[τ,T]sup|u|≤A|lnk(t,u)-bn(t,u)|→0,ask→∞.Lemma 3.10.
Letf satisfy (2.2) for u∈ℝ+d. For all n,k≥1 we have
(3.31)∑i=1d‍|lnki(t,u)|pi/(pi-1)≤D5(1+∑i=1d‍|ui|pi),(lnk(t,u),u)≥λ∑i=1d‍|ui|pi-D6,
for u∈ℝ+d, where the positive constants D5,D6, and λ do not depend neither on n nor k. Also,
(3.32)(lnuk(t,u)w,w)≥-D7(k,n)|w|2,∀u,w,
where D7(k,n) is a nonnegative number.
Moreover, iff,h satisfy (3.2), then lnk,h also satisfies this condition.Proof.
The inequalities given in (3.31) are an easy consequence of (3.12), (3.18), and (3.23).
On the other hand, ifu is such that ui=0, uj≥0, j≠i, then in view of Lemmas 3.7, 3.8, and 3.9 we have
(3.33)ϕn(|u|)(hi(t,x)-Fnki(t,u))≥0,(1-ϕn(|u|))(hi(t,x)-bni(t,u))≥0,
as ϕn(|u|)=0, for |u|≥n+2. Hence, hi(t,x)-lnki(t,u)≥0 so that condition (3.2) holds.
It is also clear thatlnk(t,u) is continuously differentiable with respect to u for any t and u. We obtain the existence of D7(k,n) such that (3.32) holds. Indeed, if |u|≤n+1+γ¯, then lnk(t,u)=Fnk(t,u) so that
(3.34)|(lnuk(t,u)w,w)|=|(Fnuk(t,u)w,w)|=|(fnuk(t,u)w,w)|≤|w|2∫ℝd‍|∇ρϵk(u-s)||bn(s,u)|ds≤R1(k,n)|w|2.
If|u|≥n+2, then lnk(t,u)=bn(t,u) so that by (3.13), (3.19) we have (lnuk(t,u)w,w)=(bnu(t,u)w,w)≥0. Finally, if n+1+γ¯<|u|<n+2, we have
(3.35)(lnuk(t,u)w,w)=ϕn(|u|)(Fnuk(t,u)w,w)+(1-ϕn(|u|))(bnu(t,u)w,w)+∑i,j=1d‍∂∂ujϕn(|u|)Fnki(t,u)wiwj-∑i,j=1d‍∂∂ujϕn(|u|)bni(t,u)wiwj≥-R2(k,n)|w|2,
where we have used similar arguments as in (3.34), (3.13),  and (3.19) and also that
(3.36)|∂∂ujϕn(|u|)|≤R3(n),|bni(t,u)|≤R4(n),|Fnki(t,u)|≤R5(k,n),
for any u satisfying n+1+γ¯<|u|<n+2,t∈[τ,T] and any i,j.Now we are ready to obtain the weak comparison principle for positive solutions.Theorem 3.11.
Letfj,hj satisfy (3.1)-(3.2). Assume that fj,hj satisfy (2.2) and (2.10) for u∈ℝ+d. One supposes that either f1 or f2 satisfies (2.7) for u∈ℝ+d and for an arbitrary R0>0. If 0≤uτ1≤uτ2, there exist two solutions u1,u2 of (2.8) and (2.9), respectively, with u1(τ)=uτ1, u2(τ)=uτ2 such that 0≤u1(t)≤u2(t) for all t∈[τ,T].Proof.
Forfj let us consider the approximations f~1n,f2n defined before with qi=pi2≤pi1, where the last inequality follows from Lemma 3.2. Then by Lemmas 3.7 and 3.8 we have that f~1n,f2n satisfy (2.2) for u∈ℝ+d with constants not depending on n. Also, (3.13), (3.19) hold and (3.2) is satisfied in both cases. Moreover, by (2.10) for u∈ℝ+d, we have
(3.37)f~1ni(t,u)=ψn(|u|)f1i(t,u)+(1-ψn(|u|))(|ui|pi1-1+|ui|pi2-1+f1i(t,0,…,0))≥ψn(|u|)f2i(t,u)+(1-ψn(|u|))(|ui|pi2-1+f2i(t,0,…,0))=f2ni(t,u),
if u∈ℝ+d.
As explained before, we can choose a sequenceδnk∈(0,1) such that δnk→0, as k→∞, and |f1nki(t,u)-f~1ni(t,u)|≤δnk,|f2nki(t,u)-f2ni(t,u)|≤δnk, for any i,n and any u satisfying |u|≤n+2. Further, we consider the functions
(3.38)F1nki(t,u)=f1nki(t,u)-δnk,F2nki(t,u)=f2nki(t,u)-3δnk.
By Lemma 3.9 we know that Fjnk satisfy (2.2) for u∈ℝd+ with constants not depending neither on k nor n, and condition (3.2) for |u|≤n+2, as well. Moreover, by (3.37) we have
(3.39)F2nki(t,u)=f2nki(t,u)-3δnk≤f2ni(t,u)-2δnk≤f1ni(t,u)-2δnk≤f1nki(t,u)-δnk=F1nki(t,u),
if u∈ℝ+d and |u|≤n+2.
Suppose, for example, thatf2 satisfies condition (2.7) for u∈ℝ+d. For any t∈[τ,T] and any u,v∈ℝ+d such that ui=vi and uj≤vj if j≠i, |u|,|v|≤n, we have
(3.40)F2nki(t,u)=f2nki(t,u)-3δnk≥f2n(t,u)-4δnk=f2i(t,u)-4δnk≥f2i(t,v)-4δnk=f2ni(t,v)-4δnk≥f2nki(t,v)-5δnk=F2nki(t,v)-2δnk.
Hence, (3.5) is satisfied with R0=n and ε=2δnk.
Now, we will define the following  functions:(3.41)l1nk(t,u)=ϕn(|u|)F1nk(t,u)+(1-ϕn(|u|))f~1n(t,u),l2nk(t,u)=ϕn(|u|)F2nk(t,u)+(1-ϕn(|u|))f2n(t,u).
By Lemma 3.10, these functions satisfy (2.2) for u∈ℝ+d with constants not depending neither on k nor n, inequality (3.32), and condition (3.2).
In view of (3.37), (3.39), and ϕn(|u|)=0, if |u|≥n+2, we obtain
(3.42)ϕn(|u|)F1nki(t,u)≥ϕn(|u|)F2nki(t,u),(1-ϕn(|u|))f~1ni(t,u)≥(1-ϕn(|u|))f2ni(t,u)
and then if u∈ℝ+d,
(3.43)l1nk(t,u)≥l2nk(t,u).
On the other hand, sincel2nk(t,u)=F2nk(t,u) if |u|≤n+1+γ¯, (3.40) implies
(3.44)l2nk(t,u)≥l2nk(t,v)-2δnk
for any t∈[τ,T] and any u,v∈ℝ+d such that ui=vi and uj≤vj if j≠i,|u|,|v|≤n. Thus, (3.5) is satisfied with R0=n and ε=2δnk.
We consider now the follwoing  problems:(3.45)∂u∂t-aΔu+ljnk(t,u)=hj(t,x),(t,x)∈(τ,T)×Ω,u|x∈∂Ω=0,u|t=τ=uτ,(3.46)∂u∂t-aΔu+bjn(t,u)=hj(t,x),(t,x)∈(τ,T)×Ω,u|x∈∂Ω=0,u|t=τ=uτ,
where b1n=f~1n and b2n=f2n.
In view of Lemma3.10 and [16, Lemma 5], problem (3.45) has a unique weak solution unk(·) such that unk(t)≥0 for all t∈[τ,T]. Let ujnk(·) be the solutions of (3.45) corresponding to the initial data uτj, where j=1,2. Using Lemma 3.10 it is standard to obtain estimate (2.14) with a constant D not depending neither on n nor k. Then, the solutions ujnk(·) of (3.45) satisfy
(3.47)∥ujnk(t)∥2≤∥uτj∥2+D∫τT(∥hj(r)∥2+1)dr=K2(∥uτj∥,τ,T).
Thus, since l2nk satisfy condition (3.5) with R02=n2≥2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)}, by Theorem 3.6 we know that as uτ1≤uτ2, we have
(3.48)∥(u1nk(t)-u2nk(t))+∥≤2C(τ,T)δnk
for all t∈[τ,T], all k and n≥(2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)})1/2.
Arguing as in the proof of Theorem2.7 we obtain that for j=1,2 the sequence ujnk converges (up to a subsequence) in the sense of (2.29)–(2.36) to a solution ujn of problem (3.46) with initial data uτj. In particular, as k→∞ we have
(3.49)ujnk(t)→ujn(t)inH∀t∈[τ,T].
Fix n≥(2max{K2(∥uτ1∥,τ,T),K2(∥uτ2∥,τ,T)})1/2,i∈{1,…,d} and take any t∈[τ,T]. Denote by Ωnkit the set
(3.50)Ωnkit={x∈Ω:u1nki(t,x)-u2nki(t,x)≥0}.
By (3.48) as k→∞ we have
(3.51)∫Ωnkt‍(u1nki(t,x)-u2nki(t,x))2dx=∫Ω‍((u1nki(t)-u2nki(t))+)2dx≤4C2(τ,T)δnk2→0.
Hence, (3.49) implies
(3.52)∫Ω∖Ωnkt‍(u1nki(t,x)-u2nki(t,x))2dx→∫Ω‍(u1ni(t,x)-u2ni(t,x))2dxask→∞.
Define the sequence
(3.53)wnki(t,x)={u1nki(t,x)-u2nki(t,x),ifx∈Ω∖Ωnkit,0,ifx∈Ωnkit.
Then, it is clear that ∥wnki(t)∥L2(Ω)→∥u1nki(t)-u2nki(t)∥L2(Ω) as k→∞. We note also that wnki(t)→u1ni(t)-u2ni(t) weakly in L2(Ω), as for any ξ∈L2(Ω), (3.51) gives
(3.54)∫Ω‍wnki(t,x)ξ(x)dx=∫Ω‍(u1nki(t,x)-u2nki(t,x))ξ(x)dx-∫Ωnkt‍(u1nki(t,x)-u2nki(t,x))ξ(x)dx→0.
Therefore, wnki(t)→u1ni(t)-u2ni(t) strongly in L2(Ω) so that wnki(t,x)→u1ni(t,x)-u2ni(t,x) for a.a. x∈Ω. Since wnki(t,x)≤0, for a.a. Ω, we obtain
(3.55)0≤u1n(t)≤u2n(t)∀t∈[τ,T].
Arguing again as in the proof of Theorem2.8 we obtain that the sequences u1n,u2n converge (up to a subsequence) in the sense of (2.29)–(2.36) to the solutions u1,u2 of problems (2.8) and (2.9), respectively. Also, it holds
(3.56)0≤u1(t)≤u2(t),∀t∈[τ,T].As in the previous section, we shall generalize this theorem to the case where the constantα can be negative. We note that if fj,hj satisfy (3.2), then f~j(t,v)=e-βtfj(t,eβtv)+βv, h~j=e-βthj also satisfy (3.2). Arguing as in Theorem 2.9, we obtain the following.Theorem 3.12.
Letfj,hj satisfy (3.1)-(3.2). Assume that fj,hj satisfy (2.43) and (2.10) for u∈ℝ+d. Also, suppose that either f1 or f2 satisfies (2.7) for u∈ℝ+d for an arbitrary R0>0. If uτ1≤uτ2, there exist two solutions u1,u2 (of (2.8) and (2.9), resp.), with u1(τ)=uτ1, u2(τ)=uτ2 such that 0≤u1(t)≤u2(t), for all t∈[τ,T].But in this case we can consider another interesting situation, as the valuespi1 and pi2 are not necessarily equal.Letf1 satisfy (2.2) for u∈ℝ+d, whereas f2 satisfy (2.43) for u∈ℝ+d, with constants αj,C1j,C2j,j=1,2. Then if α2<0, we make in (2.1) the change of variable v=e-βtu, where β>-α2. We obtain problems (2.44)-(2.45).Ifv(t) is a weak solution of (2.44), then u(t)=eβtv(t) is a weak solution of (2.8) (and the same is true, of course, for (2.45) and (2.9)).The functionf~1(t,v)=e-βtf1(t,eβtv)+βv satisfies (2.2) for v∈ℝ+d with the same pi1 as f1. Indeed, as eβt(pi1-2)≥1, we get
(3.57)∑i=1d‍|f~1i(t,v)|pi1/(pi1-1)≤K1(∑i=1d‍|f1i(t,eβtv)|pi1/(pi1-1)+∑i=1d‍|vi|pi1/(pi1-1))≤K2(1+∑i=1d‍|vi|pi1),(f~1(t,v),v)≥e-2βt(α1∑i=1d‍eβtpi|vi|pi1-C21)+β|v|2≥α1∑i=1d‍|vi|pi1-C~21.
Then, we obtain the following.Theorem 3.13.
Letfj,hj satisfy (3.1)-(3.2). Let f1 satisfy (2.2) for u∈ℝ+d, whereas f2 satisfy (2.43) for u∈ℝ+d, with constants αj,C1j,C2j,j=1,2. Assume that fj,hj satisfy (2.10). Also, suppose that either f1 or f2 satisfies (2.7) for u∈ℝ+d for an arbitrary R0>0. If uτ1≤uτ2, there exist two solutions u1,u2 (of (2.8) and (2.9), resp.), with u1(τ)=uτ1, u2(τ)=uτ2 such that 0≤u1(t)≤u2(t) for all t∈[τ,T].Proof.
We consider problems (2.44) and (2.45). In view of (2.46) and (3.57), f~2(t,v)=e-βtfj(t,eβtv)+βv satisfy (2.2) for v∈ℝ+d with pi2=2, and f~1 satisfies (2.2) for v∈ℝ+d with the same pi1 as f1. Also, defining h~j(t,x)=e-βthj(t,x) it is clear that (2.10) and (3.2) hold. Finally, if, for example, f1 satisfies (2.7) for u∈ℝ+d for any R0>0, then it is obvious that for f~1 is true as well.
Hence, by Theorem3.11 there exist two solutions v1,v2 (of (2.44) and (2.45), resp.), with vj(τ)=e-βτuτj such that 0≤v1(t)≤v2(t) for all t∈[τ,T]. Thus,
(3.58)0≤u1(t)=eβtv1(t)≤eβtv2(t)=u2(t),fort∈[τ,T],
and u1,u2 are solutions of (2.8) and (2.9), respectively such that uj(τ)=uτj.Remark 3.14.
Iffj satisfy (2.6), then the solutions u1,u2 given in Theorems 3.12 and 3.13 are unique for the corresponding initial data.Remark 3.15.
All the result proved so far are true if instead of Dirichlet boundary conditions we consider Neumann boundary conditions(3.59)∂u∂ν=0in∂Ω,
where ν is the unit outward normal. In such a case, the space V will be (H1(Ω))d. The proofs remain the same.


## 4. Applications

We shall apply now the previous results to some model of physical and biological interest.


### 4.1. The Lotka-Volterra System

We also study the Lotka-Volterra system with diffusion:(4.1)∂u1∂t=D1Δu1+u1(a1(t)-u1-a12(t)u2-a13(t)u3),∂u2∂t=D2Δu2+u2(a2(t)-u2-a21(t)u1-a23(t)u3),∂u3∂t=D3Δu3+u3(a3(t)-u3-a31(t)u1-a32(t)u2),
with either Dirichlet or Neumann boundary conditions, where ui=ui(x,t)≥0 and the functions ai(t),aij(t) are positive and continuous. Also, Di are positive constants and Ω⊂ℝ3. The initial data uτ belongs to (L2(Ω))3.In this case, the functionsf1,h1 are given by
(4.2)f1(t,u)=(-u1(a1(t)-u1-a12(t)u2-a13(t)u3)-u2(a2(t)-u2-a21(t)u1-a23(t)u3)-u3(a3(t)-u3-a31(t)u1-a32(t)u2)),h1(t)≡0.Uniqueness of the Cauchy problem for this system has been proved only if we consider solutions confined in an invariant region (e.g., in a parallelepiped𝒟={(u1,u2,u3):0≤ui≤ki} when the parameters do not depend on t) (see [20, 21]). However, in the general case for initial data just in (L2(Ω))3 it is an open problem so far.System (4.1) satisfies conditions (2.2) with p1=p2=p3=3 foru∈ℝ+3 [16, page 263]. Also, it is clear that (3.1)-(3.2) hold.We shall compare with the following system:(4.3)∂u1∂t=D1Δu1+u1(a1(t)-u1),∂u2∂t=D2Δu2+u2(a2(t)-u2),∂u3∂t=D3Δu3+u3(a3(t)-u3),
which is a system of three uncoupled logistic equations. The functions f2,h2 are given by
(4.4)f2(t,u)=(-u1(a1(t)-u1)-u2(a2(t)-u2)-u3(a3(t)-u3)),h2(t)≡0.It is easy to see that system (4.3) satisfies conditions (2.2) with p1=p2=p3=3 foru∈ℝ3. Also, it is clear that (3.1)-(3.2) hold, and that condition (2.7) is trivially satisfied. Moreover,
(4.5)f1i(t,u)≥f2i(t,u),
for all t, u∈ℝ+3 and i so that (2.10) holds for u∈ℝ+3.Also, we have(4.6)(f2u(t,u)w,w)≥-a1(t)w12-a2(t)w22-a3(t)w32+2(u1w12+u2w22+u3w32)≥-C|w|2,
for all u∈ℝ+3, t∈[τ,T] and w∈ℝ3, where C>0.Thus, by Theorem3.11, Lemma 3.1, and Remark 3.15, we obtain the following theorem.Theorem 4.1.
If0≤uτ1≤uτ2, there exists a solution u1 for (4.1) with u1(τ)=uτ1 such that 0≤u1(t)≤u2(t), for all t∈[τ,T], where u2 is the unique solution with u2(τ)=uτ2 of (4.3) in the class of solutions satisfying u2(t)≥0 for all t.This theorem says that there exists at least one solution for the Lotka-Volterra system which is dominated by the unique nonnegative solution of the uncoupled logistic system (4.3).Further, we shall compare with the uncoupled linear system(4.7)∂u1∂t=D1Δu1+u1a1(t),∂u2∂t=D2Δu2+u2a2(t),∂u3∂t=D3Δu3+u3a3(t).
Hence,
(4.8)f2(t,u)=(-u1a1(t)-u2a2(t)-u3a3(t)),h2(t)≡0.Obviously, system (4.7) satisfies conditions (2.43) foru∈ℝ3. Also, it is clear that (3.1)-(3.2) and condition (2.7) are trivially satisfied. Moreover,
(4.9)f1i(t,u)≥f2i(t,u),
for all t, u∈ℝ+3 and i so that (2.10) holds for u∈ℝ+3. Also, we have
(4.10)(f2u(t,u)w,w)≥-a1(t)w12-a2(t)w22-a3(t)w32≥-C|w|2,
for all u∈ℝ3, t∈[τ,T] and w∈ℝ3, where C>0.Thus, by Theorem3.13 and Remark 3.15, we have the following.Theorem 4.2.
If0≤uτ1≤uτ2, there exist a solution u1 for (4.1) with u1(τ)=uτ1 such that 0≤u1(t)≤u2(t) for all t∈[τ,T], where u2 is the unique solution of (4.7) with u2(τ)=uτ2.Let us consider now the autonomous case, that is,ai(t)≡ai>0 and τ=0, with Diriclet boundary conditions.By the changes of variablevi(t)=e-aitui(t), system (4.7) becomes
(4.11)∂vi∂t=DiΔvi,i=1,2,3,vi|x∈∂Ω=0,
with initial data v0=u0.The operatorA=-DiΔ in the space L2(Ω) with domain D(A)=H2(Ω)∩H01(Ω) is sectorial [22]. Moreover, since the eigenvalues of A are 0<Diλ1≤Diλ2≤⋯, we have that the minimum eigenvalue is strictly greater than 0. Denote by e-At the analytic semigroup generated by the operator A. Then, v(t)=e-Atv0 is the unique solution of (4.11) with v(0)=v0.It is well known [22] that the operator A generates a scale of interpolation spaces Xα=D(Aα) with the norm ∥v∥α=∥Aαv∥L2(Ω), α≥0, where Xα⊂H2α(Ω) with continuous embedding. Take 0<δ<D1λ1. Then, by Theorem  1.4.3 in [22] we obtain
(4.12)∥Aαe-Atv0i∥L2(Ω)≤Cαt-αe-δt∥v0i∥L2(Ω)fort>0.
Since this is true for every δ<Diλ1, we obtain that
(4.13)∥Aαe-Atv0i∥L2(Ω)≤Cαt-αe-Diλ1t∥v0i∥L2(Ω),∥e-Atv0i∥H2α(Ω)≤C~αt-αe-Diλ1t∥v0i∥L2(Ω)fort>0.
We note that H2α(Ω)⊂L∞(Ω) with continuous embedding if α>3/4. Since the constants Cα are bounded for α in compact sets, we obtain the existence of C such that
(4.14)∥e-Atv0i∥L∞(Ω)≤Ct-3/4e-Dλ1t∥v0i∥L2(Ω)fort>0,
where D=min{D1,D2,D3}. Then the unique solution of (4.7) with u(0)=u0 satisfies
(4.15)∥u(t)∥(L∞(Ω))3≤Ct-3/4e(a-Dλ1)t∥u0∥fort>0,
where a=max{a1,a2,a3}.Joining (4.15) and Theorem 4.2, we obtain the following result.Theorem 4.3.
There exist at least one solutionu(t) for the autonomous system (4.1) with u(0)=u0 and Dirichlet boundary conditions such that
(4.16)∥u(t)∥(L∞(Ω))3≤Ct-3/4e(a-Dλ1)t∥u0∥fort∈(0,T],
where C>0 and D=min{D1,D2,D3},a=max{a1,a2,a3}.We shall obtain also a weak maximum principle for the autonomous Lotka-Volterra system with Dirichlet boundary conditions.By the maximum principle for the heat equation it is well known (see [23]) that for any t≥0 the unique solution of (4.11) satisfies
(4.17)0≤vi(t,x)≤supΩv0ifora.a.x∈Ω.
Hence, the unique solution of (4.7) satisfies
(4.18)0≤ui(t,x)≤eaitsupΩv0ifora.a.x∈Ω.By (4.18) and Theorem 4.2, we obtain the following weak maximum principle.Theorem 4.4.
There exist at least one solutionu(t) for the autonomous system (4.1) with u(0)=u0 and Dirichlet boundary conditions such that for any t∈[0,T],
(4.19)0≤ui(t,x)≤eaitsupΩu0ifora.a.x∈Ω,i=1,2,3.
In particular, ifu0∈(L∞(Ω))3, then
(4.20)∥u(t)∥(L∞(Ω))3≤eat∥u0∥(L∞(Ω))3fort∈[0,T],
where a=max{a1,a2,a3}.


### 4.2. A Model of Fractional-Order Chemical Autocatalysis with Decay

Consider the following scalar problem:(4.21)∂u∂t=∂2u∂x2+(1-u)um-kur,∂u∂x(0,t)=∂u∂x(a,t)=0,u|t=0=u0(x),
where u≥0, N=1, d=1,Ω=(0,L), and k>0,0<m,r<1. The initial data u0 belongs to L2(0,a). This equation models an isothermal chemical autocatalysis (see [24]). In [24], the authors study the travelling waves of the equation in the case where Ω=(0,+∞) with Neumann boundary conditions at x=0. The variable u is nonnegative, since it represents a chemical concentration.The functionsf,h are given by f(u)=(u-1)um+kur, h≡0. Clearly, conditions (2.2) hold (withp=m+2) for u≥0. In this case (3.1)-(3.2) and (2.7) are trivially satisfied.We takef1=f2=f and apply Theorem 3.11 and Remark 3.15 to obtain the following.Theorem 4.5.
If0≤uτ1≤uτ2, there exist solutions u1,u2 for (4.21) with u1(0)=u01,u2(0)=u02 such that 0≤u1(t)≤u2(t) for all t∈[0,T].


### 4.3. A Generalized Logistic Equation

Consider the following scalar problem:(4.22)∂u∂t=∂2u∂x2+(1-uq)ur,∂u∂x(0,t)=∂u∂x(a,t)=0,u|t=0=u0(x),
where u≥0, N=1, Ω=(0,L), r,q>0, and r+q≥1. The initial data u0 belongs to L2(0,a).This kind of nonlinearities for the logistic equation (instead of the classical(1-u)u) has been considered in [25, Chapter 11].The functionsf,h are given by f(u)=(uq-1)ur, h≡0. Clearly, conditions (2.2) hold (with p=r+q+1) for u≥0. In this case, (3.1)-(3.2) and (2.7) are trivially satisfied.We takef1=f2=f and apply Theorem 3.11 and Remark 3.15 to obtain the following.Theorem 4.6.
If0≤uτ1≤uτ2, there exist two solutions u1,u2 for (4.22) with u1(0)=u01,u2(0)=u02 such that 0≤u1(t)≤u2(t) for all t∈[0,T].


## 4.1. The Lotka-Volterra System

We also study the Lotka-Volterra system with diffusion:(4.1)∂u1∂t=D1Δu1+u1(a1(t)-u1-a12(t)u2-a13(t)u3),∂u2∂t=D2Δu2+u2(a2(t)-u2-a21(t)u1-a23(t)u3),∂u3∂t=D3Δu3+u3(a3(t)-u3-a31(t)u1-a32(t)u2),
with either Dirichlet or Neumann boundary conditions, where ui=ui(x,t)≥0 and the functions ai(t),aij(t) are positive and continuous. Also, Di are positive constants and Ω⊂ℝ3. The initial data uτ belongs to (L2(Ω))3.In this case, the functionsf1,h1 are given by
(4.2)f1(t,u)=(-u1(a1(t)-u1-a12(t)u2-a13(t)u3)-u2(a2(t)-u2-a21(t)u1-a23(t)u3)-u3(a3(t)-u3-a31(t)u1-a32(t)u2)),h1(t)≡0.Uniqueness of the Cauchy problem for this system has been proved only if we consider solutions confined in an invariant region (e.g., in a parallelepiped𝒟={(u1,u2,u3):0≤ui≤ki} when the parameters do not depend on t) (see [20, 21]). However, in the general case for initial data just in (L2(Ω))3 it is an open problem so far.System (4.1) satisfies conditions (2.2) with p1=p2=p3=3 foru∈ℝ+3 [16, page 263]. Also, it is clear that (3.1)-(3.2) hold.We shall compare with the following system:(4.3)∂u1∂t=D1Δu1+u1(a1(t)-u1),∂u2∂t=D2Δu2+u2(a2(t)-u2),∂u3∂t=D3Δu3+u3(a3(t)-u3),
which is a system of three uncoupled logistic equations. The functions f2,h2 are given by
(4.4)f2(t,u)=(-u1(a1(t)-u1)-u2(a2(t)-u2)-u3(a3(t)-u3)),h2(t)≡0.It is easy to see that system (4.3) satisfies conditions (2.2) with p1=p2=p3=3 foru∈ℝ3. Also, it is clear that (3.1)-(3.2) hold, and that condition (2.7) is trivially satisfied. Moreover,
(4.5)f1i(t,u)≥f2i(t,u),
for all t, u∈ℝ+3 and i so that (2.10) holds for u∈ℝ+3.Also, we have(4.6)(f2u(t,u)w,w)≥-a1(t)w12-a2(t)w22-a3(t)w32+2(u1w12+u2w22+u3w32)≥-C|w|2,
for all u∈ℝ+3, t∈[τ,T] and w∈ℝ3, where C>0.Thus, by Theorem3.11, Lemma 3.1, and Remark 3.15, we obtain the following theorem.Theorem 4.1.
If0≤uτ1≤uτ2, there exists a solution u1 for (4.1) with u1(τ)=uτ1 such that 0≤u1(t)≤u2(t), for all t∈[τ,T], where u2 is the unique solution with u2(τ)=uτ2 of (4.3) in the class of solutions satisfying u2(t)≥0 for all t.This theorem says that there exists at least one solution for the Lotka-Volterra system which is dominated by the unique nonnegative solution of the uncoupled logistic system (4.3).Further, we shall compare with the uncoupled linear system(4.7)∂u1∂t=D1Δu1+u1a1(t),∂u2∂t=D2Δu2+u2a2(t),∂u3∂t=D3Δu3+u3a3(t).
Hence,
(4.8)f2(t,u)=(-u1a1(t)-u2a2(t)-u3a3(t)),h2(t)≡0.Obviously, system (4.7) satisfies conditions (2.43) foru∈ℝ3. Also, it is clear that (3.1)-(3.2) and condition (2.7) are trivially satisfied. Moreover,
(4.9)f1i(t,u)≥f2i(t,u),
for all t, u∈ℝ+3 and i so that (2.10) holds for u∈ℝ+3. Also, we have
(4.10)(f2u(t,u)w,w)≥-a1(t)w12-a2(t)w22-a3(t)w32≥-C|w|2,
for all u∈ℝ3, t∈[τ,T] and w∈ℝ3, where C>0.Thus, by Theorem3.13 and Remark 3.15, we have the following.Theorem 4.2.
If0≤uτ1≤uτ2, there exist a solution u1 for (4.1) with u1(τ)=uτ1 such that 0≤u1(t)≤u2(t) for all t∈[τ,T], where u2 is the unique solution of (4.7) with u2(τ)=uτ2.Let us consider now the autonomous case, that is,ai(t)≡ai>0 and τ=0, with Diriclet boundary conditions.By the changes of variablevi(t)=e-aitui(t), system (4.7) becomes
(4.11)∂vi∂t=DiΔvi,i=1,2,3,vi|x∈∂Ω=0,
with initial data v0=u0.The operatorA=-DiΔ in the space L2(Ω) with domain D(A)=H2(Ω)∩H01(Ω) is sectorial [22]. Moreover, since the eigenvalues of A are 0<Diλ1≤Diλ2≤⋯, we have that the minimum eigenvalue is strictly greater than 0. Denote by e-At the analytic semigroup generated by the operator A. Then, v(t)=e-Atv0 is the unique solution of (4.11) with v(0)=v0.It is well known [22] that the operator A generates a scale of interpolation spaces Xα=D(Aα) with the norm ∥v∥α=∥Aαv∥L2(Ω), α≥0, where Xα⊂H2α(Ω) with continuous embedding. Take 0<δ<D1λ1. Then, by Theorem  1.4.3 in [22] we obtain
(4.12)∥Aαe-Atv0i∥L2(Ω)≤Cαt-αe-δt∥v0i∥L2(Ω)fort>0.
Since this is true for every δ<Diλ1, we obtain that
(4.13)∥Aαe-Atv0i∥L2(Ω)≤Cαt-αe-Diλ1t∥v0i∥L2(Ω),∥e-Atv0i∥H2α(Ω)≤C~αt-αe-Diλ1t∥v0i∥L2(Ω)fort>0.
We note that H2α(Ω)⊂L∞(Ω) with continuous embedding if α>3/4. Since the constants Cα are bounded for α in compact sets, we obtain the existence of C such that
(4.14)∥e-Atv0i∥L∞(Ω)≤Ct-3/4e-Dλ1t∥v0i∥L2(Ω)fort>0,
where D=min{D1,D2,D3}. Then the unique solution of (4.7) with u(0)=u0 satisfies
(4.15)∥u(t)∥(L∞(Ω))3≤Ct-3/4e(a-Dλ1)t∥u0∥fort>0,
where a=max{a1,a2,a3}.Joining (4.15) and Theorem 4.2, we obtain the following result.Theorem 4.3.
There exist at least one solutionu(t) for the autonomous system (4.1) with u(0)=u0 and Dirichlet boundary conditions such that
(4.16)∥u(t)∥(L∞(Ω))3≤Ct-3/4e(a-Dλ1)t∥u0∥fort∈(0,T],
where C>0 and D=min{D1,D2,D3},a=max{a1,a2,a3}.We shall obtain also a weak maximum principle for the autonomous Lotka-Volterra system with Dirichlet boundary conditions.By the maximum principle for the heat equation it is well known (see [23]) that for any t≥0 the unique solution of (4.11) satisfies
(4.17)0≤vi(t,x)≤supΩv0ifora.a.x∈Ω.
Hence, the unique solution of (4.7) satisfies
(4.18)0≤ui(t,x)≤eaitsupΩv0ifora.a.x∈Ω.By (4.18) and Theorem 4.2, we obtain the following weak maximum principle.Theorem 4.4.
There exist at least one solutionu(t) for the autonomous system (4.1) with u(0)=u0 and Dirichlet boundary conditions such that for any t∈[0,T],
(4.19)0≤ui(t,x)≤eaitsupΩu0ifora.a.x∈Ω,i=1,2,3.
In particular, ifu0∈(L∞(Ω))3, then
(4.20)∥u(t)∥(L∞(Ω))3≤eat∥u0∥(L∞(Ω))3fort∈[0,T],
where a=max{a1,a2,a3}.


## 4.2. A Model of Fractional-Order Chemical Autocatalysis with Decay

Consider the following scalar problem:(4.21)∂u∂t=∂2u∂x2+(1-u)um-kur,∂u∂x(0,t)=∂u∂x(a,t)=0,u|t=0=u0(x),
where u≥0, N=1, d=1,Ω=(0,L), and k>0,0<m,r<1. The initial data u0 belongs to L2(0,a). This equation models an isothermal chemical autocatalysis (see [24]). In [24], the authors study the travelling waves of the equation in the case where Ω=(0,+∞) with Neumann boundary conditions at x=0. The variable u is nonnegative, since it represents a chemical concentration.The functionsf,h are given by f(u)=(u-1)um+kur, h≡0. Clearly, conditions (2.2) hold (withp=m+2) for u≥0. In this case (3.1)-(3.2) and (2.7) are trivially satisfied.We takef1=f2=f and apply Theorem 3.11 and Remark 3.15 to obtain the following.Theorem 4.5.
If0≤uτ1≤uτ2, there exist solutions u1,u2 for (4.21) with u1(0)=u01,u2(0)=u02 such that 0≤u1(t)≤u2(t) for all t∈[0,T].


## 4.3. A Generalized Logistic Equation

Consider the following scalar problem:(4.22)∂u∂t=∂2u∂x2+(1-uq)ur,∂u∂x(0,t)=∂u∂x(a,t)=0,u|t=0=u0(x),
where u≥0, N=1, Ω=(0,L), r,q>0, and r+q≥1. The initial data u0 belongs to L2(0,a).This kind of nonlinearities for the logistic equation (instead of the classical(1-u)u) has been considered in [25, Chapter 11].The functionsf,h are given by f(u)=(uq-1)ur, h≡0. Clearly, conditions (2.2) hold (with p=r+q+1) for u≥0. In this case, (3.1)-(3.2) and (2.7) are trivially satisfied.We takef1=f2=f and apply Theorem 3.11 and Remark 3.15 to obtain the following.Theorem 4.6.
If0≤uτ1≤uτ2, there exist two solutions u1,u2 for (4.22) with u1(0)=u01,u2(0)=u02 such that 0≤u1(t)≤u2(t) for all t∈[0,T].



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*Source: 679465-2012-09-11.xml*