A nonlinear age-dependent model with spatial diffusion

A nonlinear age-dependent model with spatial diffusion ? Manuel Delgado, M´onica Molina-Becerra and Antonio Su´arez 1 Dpto Ecuaciones Diferenciales y An´alisis Num´erico, Universidad de Sevilla, Apdo de Correos 1160, 41080 Sevilla, Spain Abstract The main goal of this paper is to study the existence and uniqueness of positive solution for a nonlinear age-dependent equation with spatial diffusion. For that, we mainly use properties of an eigenvalue problem related to the equation and the subsupersolution method. We justify that this method works for this kind of equation, in which appears a potential blowing-up and a non-local initial condition. Key words: Sub-supersolution method, non-local initial condition, logistic age-dependent. ?Supported by the Spanish Ministry of Science and Technology under Grants BFM2000-0797 and BFM2003-06446. Email addresses: [email protected] (Manuel Delgado), [email protected] (M´onica Molina-Becerra), [email protected] (Antonio Su´arez). 1Corresponding author. Fax:(34) 95 455 28 98 Preprint submitted to Journal of Mathematical Analysis and Applications11 March 2005 1 Introduction In this paper we study the existence of solution of the nonlinear problem                        ua−∆u+q(x, a)u=g(x, a, u) in Q:= Ω ×(0, A), u= 0 on Σ := ∂Ω×(0, A), u(x, 0) = ZA 0β(x, a)u(x, a)da in Ω, (1) where Ω is a bounded and regular domain of IRN,A > 0, βa bounded, nonnegative and nontrivial function, qis a measurable function blowing up at a=A and gis a measurable function with assumptions that will be detailed below. For that, we are going to employ the sub-supersolution method. Observe that, mathematically, apart from the nonlinearity, (1) has two main difficulties: the coefficient qis not bounded and the initial datum is non-local. So, we can not apply the classical sub-supersolution method for parabolic problem (see for instance [12] and [6]). We assure that assuming the existence of an ordered pair of sub-supersolution of (1), there exists a solution between the sub and the supersolution provided of, basically, gis a lipschitz function in the variable u. Hence, our result generalizes the classical ones for parabolic problem in the two ways mentioned above. We would like to point out that although comparison results have been used in this framework (see for instance Lemma 2 in [7] and Lemma 4.5 in [9]), we have not found a sub-supersolution method developed for this kind of problem. We apply the above result to study the existence and uniqueness of positive 2 solution of the logistic equation                        ua−∆u+q(x, a)u=λu −u2in Q, u= 0 on Σ, u(x, 0) = ZA 0β(x, a)u(x, a)da in Ω, (2) with λ∈IR. Equ. (2) provides us with the steady-state solutions of the corresponding time dependent problem, and so, u(x, a) represents the density population of a species of age alocalized in x, being thus Ω the habitat of the species. Observe that we are assuming that Ω is surrounded by inhospitable areas because the homogenous Dirichlet boundary condition. Ais the maximal age for the species, λ−q(x, a) is the birth (when positive) or death (when negative) rate function. In this context, the fact that q(x, a) blows up at a=A it will assure that the species dies out at the age of A. Finally, βis the rate fertility function. In general, the study of the structure of positive solutions set of a problem similar to (2) is far from to be easy. In fact, to our knowledge only linear problem in uhas been analyzed in [9], although in this case the equation also depends on the total population P(x) = RA 0u(x, a)da. Specifically in [9], the reaction term is the trivial function and qand βverify q(x, a) = q1(a) + q2(P), β(x, a) = β1(a), and moreover Pis the positive solution of the classical logistic elliptic equation, and so it is known (see Theorem 3.5 in [9]). Under these assumptions, the author proved that only separable solutions exist and he looked at them giving the explicit solution. Finally, we want to mention that nonlinear problems in u without diffusion have been studied previously, see for example [13] and [14]. For the nonexistence and uniqueness results of (2) we will study an eigenvalue problem related to (2) with the coefficient qdepending of aand x. Specifically, 3 the following eigenvalue problem is analyzed                        ua−∆u+q(x, a)u=λu in Q, u= 0 on Σ, u(x, 0) = ZA 0β(x, a)u(x, a)da in Ω, (3) For that, we follow the main idea of [8] but we have given another point of view to their results and some of them will be shaped. We prove that there exists a unique principal eigenvalue (in the sense that it is the unique with positive eigenfunction associated) denoted by λ0(q). Observe that great difference that there exists between problem (3) and the classical parabolic one (3) with u(x, 0) = u0(x)>0 instead of the non-local initial condition, where a unique positive solution exists for all λ∈IR. We apply all the above results to show that (2) possesses a positive solution if, and only if, λ > λ0(q). Moreover, if λ≤λ0(q) the only nonnegative solution of (2) is the trivial one and if λ > λ0(q) there exists a unique positive solution. Again, a drastic change occurs with respect to the problem (2) with u(x, 0) = u0(x), which possesses a unique positive solution for all λ∈IR. An outline of the work is as follows: in Section 2 we analyze the eigenvalue problem related to (2), study the linear case and establish a strong maximum principle; in Section 3 we prove that the sub-supersolution method works and in the last section we apply these results to the logistic equation (2). 2 The eigenvalue problem In this section we study the eigenvalue problem (3) assuming that 4 (Hq)qis a function such that q∈L∞(Ω ×(0, r)) for r < A and Zr 0qM(a)da < ∞,ZA 0qL(a)da = +∞,(4) being qL(a) := infx∈Ωq(x, a) and qM(a) := supx∈Ωq(x, a). (Hβ)β∈L∞(Q), β≥0, nontrivial and mes{a∈[0, A] : βL(a) := inf x∈Ω β(x, a)>0}>0. Remark 1 Condition (4) is necessary to have that lima↑Au(x, a)≡0, for u solution of (3), see Remark 4 below. Definition 2 λis an eigenvalue of (3) if there exists u∈L2(0, A;H1 0(Ω)), ua+qu ∈L2(0, A;H−1(Ω)) with u6= 0 solution of (3) in the sense that ∀v∈L2(0, A;H1 0(Ω)): ZA 0< ua+qu, v > da +ZQ∇u·∇v da dx =λZQuv da dx, u(x, a) = 0 on Σ, u(x, 0) = ZA 0β(x, a)u(x, a)da, in Ω, where <, > denotes the duality pairing between H1 0(Ω) and H−1(Ω). We say that λis a principal eigenvalue if u > 0in Q. Before studying (3) we need to analyze the autonomous case, i.e.,                        ua−∆u+m(a)u=λu in Q, u= 0 on Σ, u(x, 0) = ZA 0γ(a)u(x, a)da in Ω, (5) where 5 (Hm)m∈L∞(0, r) for r < A and ZA 0m(a)da = +∞.(6) (Hγ)γ∈L∞(0, A), γ≥0 and nontrivial. Theorem 3 Assume (Hm)and (Hγ). Then, (5) possesses a positive solution if, and only if, λ=λ1+rm, where rmis defined by 1 = ZA 0γ(a)erma−Ra 0m(s)dsda, (7) and λ1is the principal eigenvalue of the −∆under homogeneous Dirichlet boundary condition. Moreover, in this case the solution is ϕ0(x, a) = erma−Ra 0m(s)dsϕ1(x), being ϕ1a positive eigenfunction associated to λ1. PROOF. First, thanks to Theorem 3.5 of [9], any solution of (5) is separable. Observe that in the cited result, A=∞, but we can adapt the proof to the case A < ∞. Take u(x, a) = p(a)ϕ(x). Then, pa+m(a)p=rp, r ∈IR; and so, p(a) = p0era−Ra 0m(s)ds. It is not hard to show that psatisfies the initial condition p(0) = ZA 0γ(a)p(a)da, if r=rm. On the other hand, −∆ϕ= (λ−rm)ϕin Ω, ϕ= 0 on ∂Ω, 6 and thus, λ−rm=λ1and ϕ=ϕ1. This completes the proof. Remark 4 (1) Observe that thanks to (6) we get that lima↑Aϕ0(x, a) = 0 for all x∈Ω. (2) A related result was proved in Theorem 1 of [4], using properties of the infinitesimal generator associated to (5). To end the autonomous case, we establish a strong maximum principle. Definition 5 Denote by L2 +(Ω) := {f∈L2(Ω) : f(x)≥0a. e. x∈Ω}. We say that u∈L2 +(Ω) is quasi-interior point of L2 +(Ω), and we write uÀ0, if ZΩu(x)f(x)dx > 0,for all f∈L2 +(Ω) and nontrivial. Lemma 6 Assume (Hm)and that uis solution of                        ua−∆u+m(a)u=f(x, a)in Q, u= 0 on Σ, u(x, 0) = φ(x)in Ω, (8) and f≥0,φ≥0and some of the inequalities strict. Then, uÀ0. If f≡φ≡0, then u≡0. PROOF. Observe that if uis solution of (8), then v:= ueRa 0m(s)ds, is the solution of the equation                        va−∆v=g(x, a) := f(x, a)eRa 0m(s)ds in Q, v= 0 on Σ, v(x, 0) = φ(x) in Ω. (9) 7 By [10] (see also [2]), vÀ0 and so uÀ0. Similarly, if f≡φ≡0, then v≡0, and so that u. We state now a result for the non-autonomous linear case. Consider the problem                        ua−∆u+q(x, a)u=f(x, a) in Q, u= 0 on Σ, u(x, 0) = φ(x) in Ω, (10) where qsatisfies (Hq) and φ∈L2(Ω). Lemma 7 Suppose that f∈L2(Q). Then, there exists a unique solution u of (10) such that u∈L2(0, A;H1 0(Ω)) and ua+q(x, a)u∈L2(0, A;H−1(Ω)). Moreover, for each 0< A0< A we have that u∈C([0, A0]; L2(Ω)). Furthermore, we have the following comparison principles: (1) If f≥0and φ≥0, then u≥0. If some of the inequalities is strict, we deduce that uÀ0. (2) If f1≥f2≥0,φ1≥φ2≥0and q1≤q2in their respective domains, then u1≥u2, where ui,i= 1,2, is the solution of (10) with f=fi,φ=φi and q=qi. PROOF. Under the change of variable w=e−kau, k > 0, 8 wsatisfies                        wa−∆w+ (q+k)w=g:= fe−ka in Q, w= 0 on Σ, w(x, 0) = φ(x) in Ω, (11) and so by (Hq), we can take klarge such that q+k/3≥0. We study now (11) instead of (10). Define qn:= min{q, n}, n ∈IN, and consider the problem                        wa−∆w+ (qn(x, a) + k)w=g(x, a) in Q, w= 0 on Σ, w(x, 0) = φ(x) in Ω. (12) Now, for each n∈IN, since qn+kis bounded, there exists a unique wnsolution of (12) with wn∈L2(0, A;H1 0(Ω)) and (wn)a∈L2(0, A;H−1(Ω)). Multiplying (12) by wnand integrating we obtain 1 2 d da ZQ|wn|2+ZQ|∇wn|2+ZQ(qn+k)w2 n=ZQgwn, and so, applying that 2ab ≤(ε2a2+ (1/ε2)b2) we get 1 2 d da ZQ|wn|2+ZQ|∇wn|2+ZQ(qn+k/3)w2 n+ (k/3)w2 n≤C. Now, we can extract a sequence (wn) such that wn* w in L2(0, A;H1 0(Ω)), qqn+ (k/3)wn* h in L2(Q), (wn)a+ (qn+k/3)wn* z in L2(0, A;H−1(Ω)). 9 (a) For all v∈L2(0, A;H1 0(Ω)), v ≥0 ZA 0< ua+qu, v > da +ZQ∇u·∇v da dx ≥ZQg(x, a, u)v da dx, (b) u≥0on Σ, (c) u(x, 0) ≥ZA 0β(x, a)u(x, a)da in Ω. Similar definition for a subsolution, interchanging the inequalities. Theorem 13 Assume (Hβ),(Hq)and that |g(x, a, s1)−g(x, a, s2)| ≤ L|s1−s2|,for a.e. x∈Ω,a∈(0, A),s1, s2∈IR. (18) Then, if there exists a pair of sub-supersolution of (1) such that u≤uthere exists a minimal u∗and maximal u∗solutions of (1), in the sense that for any other solution u∈[u, u] := {u∈L2(Q) : u≤u≤u}, it holds that u≤u∗≤u≤u∗≤u. PROOF. Take M > 0 a positive constant to be chosen later, and define the sequence unas u0=uand for n≥1                        (un)a−∆(un) + q(x, a)un+Mun=g(x, a, un−1) + Mun−1in Q, un= 0 on Σ, un(x, 0) = ZA 0β(x, a)un−1(x, a)da in Ω, (19) u0=uand undefined by                        (un)a−∆(un) + q(x, a)un+Mun=g(x, a, un−1) + Mun−1in Q, un= 0 on Σ, un(x, 0) = ZA 0β(x, a)un−1(x, a)da in Ω. (20) 16 First, we show that unis well-defined. Since g(x, a, u0) = g(x, a, u)∈L2(Q), we can apply Lemma 7 and conclude the existence of u1. Moreover, since −L|u1−u0|+g(x, a, u0)≤g(x, a, u1)≤L|u1−u0|+g(x, a, u0), it follows that g(x, a, u1)∈L2(Q), and so the existence of u2, and analogously un. Similarly, it can proved the existence of un. We will show that un(resp. un) is increasing (resp. decreasing) and that u≤. . . ≤un≤un+1 ≤un+1 ≤un≤. . . ≤u. (21) Indeed, taking w:= u1−u0, it satisfies                        wa−∆w+q(x, a)w+Mw ≥0 in Q, w≥0 on Σ, w(x, 0) ≥0 in Ω. (22) Using Lemma 7 we conclude that w≥0, i.e., u=u0≤u1. Now assume that un−1≤un. Observe that g(x, a, un)−g(x, a, un−1) + M(un−un−1)≥(M−L)(un−un−1)≥0, for M > L. Then w:= un+1 −unsatisfies                        wa−∆w+q(x, a)w+Mw ≥0 in Q, w= 0 on Σ, w(x, 0) = ZA 0β(x, a)(un−un−1)(x, a)da ≥0 in Ω, (23) again Lemma 7 shows that un≤un+1. Similarly, it can be proved the rest of inequalities of (21). 17 Now, we multiply (19) by unand obtain 1 2 d da ZQ|un|2+ZQ|∇un|2+ZQ(q+M)u2 n=ZQ(g(x, a, un−1) + Mun−1)un. On the other hand, observe that g(x, a, un−1)≤L(un−1−u) + g(x, a, u)≤L(u−u) + g(x, a, u), and g(x, a, un−1)≥g(x, a, u)−L(un−1−u)≥g(x, a, u)−L(u−u), and so, 1 2 d da ZQ|un|2+ZQ|∇un|2+ZQ(qn+M)u2 n≤C, with Cindependent of n. With a similar reasoning to the used in Lemma 7, we can extract a subsequence (un) such that un* u∗in L2(0, A;H1 0(Ω)), √qnun* w in L2(Q), (un)a+qnun* z in L2(0, A;H−1(Ω)). By the monotony of unand the Monotone Convergence Theorem, we can conclude un→u∗in L2(Q). (24) Now, using that −L(un−u∗) + g(x, a, u∗)≤g(x, a, un)≤L(un−u∗) + g(x, a, u∗), and so, g(x, a, un)* g(x, a, u∗) weakly in L2(Q), it follows that u∗a solution of (1). Finally, the continuity of the trace application on a= 0 and (24) imply that u∗(x, a) = ZA 0β(x, a)u∗(x, a)da. 18 That u∗is the minimal solution of (1) is not difficult to show. Indeed, if uis a solution of (1) such that u∈[u, u], it can be shown that the sequence un built in (19) satisfies that u≤un≤u. So, un↑u∗≤u. Similarly, we can reason with the sequence unand conclude the existence of a maximal solution u∗of (1). This ends the proof. 4 Application to the logistic equation The main result of this section is: Theorem 14 Problem (2) possesses a positive solution if, and only if, λ > λ0(q). Moreover, in the case that the solution exists, then it is unique. PROOF. Suppose that u > 0 is solution of (2). Then, we can write the equation (2) as ua−∆u+ (q(x, a) + u(x, a)−λ)u= 0, u(x, 0) >0, with q+u−λsatisfying (Hq). Hence, by Lemma 7, uÀ0, and so by Theorem 8, and taking into account that ua−∆u+ (q(x, a) + u(x, a))u=λu, it follows that λ=λ0(q+u) (25) and by the monotony of the map q7→ λ0(q), λ=λ0(q+u)> λ0(q). Assume now that λ > λ0(q). Take u:= εϕ(x, a) 19 with ε > 0 sufficiently small and ϕa positive eigenfunction associated to λ0(q). It is not difficult to show that uis subsolution of (2). Indeed, u= 0 on Σ and u(x, 0) = εϕ(x, 0) = εZA 0β(x, a)ϕ(x, a)da =ZA 0β(x, a)u(x, a)da. Finally, ua−∆u+q(x, a)u≤λu −u2, provided that εϕ(x, a)≤λ−λ0(q), which is true taking εsufficiently small (observe that ϕis bounded, cf. Theorem 8). We will build a supersolution. Define Fµ(a) := µa −Za 0qL(s)ds, µ ∈IR, and take µ∈IR sufficiently large so that ZA 0eFµ(a)da ≥1 β.(26) where β:= supQβ(x, a). Consider the function G(x) := ZA 0 eFµ(a) 1 + xRa 0eFµ(s)dsda. Observe that Gis a continuous function and by (26) we have that lim x↓0G(x)≥1 β,lim x→+∞G(x) = 0, and so, there exists y0>0 such that G(y0) = 1/β, i.e., ZA 0 eFµ(a) 1 + y0Ra 0eFµ(s)dsda =1 β.(27) Define Y(a) the unique solution of the differential equation ya+qL(a)y=µy −y2, y(0) = y0; 20 where y0is defined by (27). Solving the above equation, we get that Y(a) = eFµ(a) 1 y0+Ra 0eFµ(s)ds.(28) Take u(a) := KY (a), with Ka positive constant large. It can be proved that uis a supersolution of (2) for µlarge. Indeed, u > 0 on Σ and ua−∆u+q(x, a)u≥λu −u2, provided that µ−λ+q(x, a)−qL(a) + (K−1)Y≥0, which is satisfied if µ≥λ, and K≥1. On the other hand, using (27) we have that ZA 0β(x, a)u(x, a)da ≤Ky0βZA 0 eFµ(a) 1 + y0Ra 0eFµ(s)dsda =Ky0=u(x, 0). Now, it is clear that we can choose ε > 0 and K > 0 such that u≤u. This completes the proof of the existence of positive solution. For the uniqueness we assume that there exist two different positive solutions u1and u2. Define w:= u2−u16= 0. It is clear that wsatisfies                        wa−∆w+ (q(x, a) + u1+u2)w=λw in Q, w= 0 on Σ, w(x, 0) = ZA 0β(x, a)w(x, a)da in Ω, (29) and so, since w6= 0, by (13), we have that λ≥λ0(q+u1+u2)> λ0(q+u1), 21 which is an absurdum. Indeed, since u1is positive solution of (1) we have that λ=λ0(q+u1), see (25). Remark 15 (1) Observe that the unique solution uof (2) satisfies that lim a↑Au(x, a) = 0,for x∈Ω. 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