Solvability of a nonlinear Neumann problem for systems arising from a burglary model

1. The problem Let L > 0, η > 0, let A0 : [0, L] → R be a positive function of class C2 such that (A0)′(0) = 0 = (A0)′(L), (1) and let A1 : [0, L] → R be a positive continuous function. Consider the Neumann problem η[A − A0(x)]′′ − A + A0(x) + NA = 0, A(0) = 0 = A(L), (2)  N ′ − 2N A A ′ − NA + A1(x) − A0(x) = 0, N (0) = 0 = N (L). (3) A solution of (2)–(3) is a couple of real functions (A,N) ∈ C2([0, T ]) × C2([0, T ]) such that A(x) > 0 for all x ∈ [0, L] which satisfies the system and the boundary conditions. We are interested in positive solutions of this problem, i.e. in solutions (A,N) such that A(x) > 0 and N(x) > 0 for all x ∈ [0, L]. This problem is a one dimensional version of a problem that arises in the pioneering work of [1] where a very successful model for burglary of houses was obtained by Short et al. See also the related papers [2–7]. The model of [1] was derived by first considering an agent based statistical model to study the formation of hot spots taking two major sociological effects into account: the ‘‘broken window effect’’ and the ‘‘repeat near-repeat effect’’. In a second step, by taking a suitable limit of the equations for the discretemodel derived in the first step, a continuousmodel for the two unknowns (A,N)was obtained: ✩ The three authors were partially supported by FONDECYT GRANT 1110268. RM was also partially supported by Basal-CMM-Conicyt, Milenio grantP05-004F, and Anillo grant ACT 87-Conicyt. ∗ Corresponding author. Tel.: +56 223544511. E-mail addresses:[email protected], [email protected] (M. Garcia-Huidobro), [email protected] (R. Manásevich), [email protected] (J. Mawhin). 0893-9659/$ – see front matter© 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.aml.2013.11.003 M. Garcia-Huidobro et al. / Applied Mathematics Letters 35 (2014) 102–108 103 A representing attractiveness for a house to be burglarized, and N representing density of burglars. By the definitions of A and N , the restrictions A > 0 and N > 0 appear as natural. When A0, A1 are positive constants, system (2)–(3) admits the unique positive constant solution A = A1, N = A1 − A0 A1 under the condition that A1 > A0. In [1] and for the PDE case, a linear stability analysis of the (corresponding) constant solutionwas performed, while in [3], a study of global bifurcation of solutions from this constant solution is done. With a view towards extending some of these results, a natural question to start with is to know if a positive (non constant) solution still exists when A0 and A1 are no longer constant. For the one dimensional model that we consider here, this means A0 and A1 depend on x. This question is answered in this paper by a combination of estimates based upon maximum or minimum properties, on L1-estimates of the type introduced by Ward [8] in some periodic problems, and by the use of the Leray–Schauder degree results. A similar problem was recently considered in [4] for a variant of the burglary model in which the linear part of the differential system was invertible. This is not the case here, which makes the fixed point reduction more complicated and requires a more sophisticated version of the Leray–Schauder theory. 2. The homotopy and a priori estimates Let us associate to (2)–(3) the homotopy, with λ ∈ (0, 1], η[A − A0(x)]′′ = λ  A − A0(x) − NA  , A(0) = 0 = A(L), (4)  N ′ − 2λN A A ′ = λ[NA − A1(x) + A0(x)], N (0) = 0 = N (L). (5) For λ = 1, (4)–(5) reduces to (2)–(3). For any B ∈ L1(0, L) we denote by B its mean value L−1  L 0 B(x) dx, and for any B ∈ C([0, L]) we set max B := max[0,L] B and min B := min[0,T ] B. Lemma 1. If (A,N) is any possible solution of (4)–(5) for some λ ∈ (0, 1], then A = A1. (6) Proof. Add Eqs. (4) and (5), integrate both members over [0, L] and use the boundary conditions. Lemma 2. If (A,N) is any possible solution of (4)–(5) for some λ ∈ (0, 1], then NA = A1 − A0. (7) Proof. Integrate Eq. (4) over [0, L], use the boundary conditions and (6). Remark 1. Notice that (7) implies that a necessary condition for the existence of a positive solution (A,N) is that A1 > A0. (8) We shall assume, from now on, the stronger condition: A1(x) > A0(x) (x ∈ [0, L]), (9) which is necessary to show that N cannot have a minimum equal to zero. Lemma 3. If (A,N) is any possible positive solution of (4)–(5) for some λ ∈ (0, 1], then, for all x ∈ [0, L], |A(x)| ≤ Lmax |(A0)′′| + 2LA1 η . (10) A(x) ≤ L2 max |(A0)′′| +  1 + 2L2 η  A1 := A2. (11) 104 M. Garcia-Huidobro et al. / Applied Mathematics Letters 35 (2014) 102–108 Proof. Let (A,N) be a possible positive solution of (4)–(5) for some λ ∈ (0, 1]. From Eq. (4) we obtain, for all x ∈ [0, L], |A(x)| ≤ |(A0)′′(x)| + λ η A(x) − A0(x) − N(x)A(x) ≤ |(A0)′′(x)| + 1 η [A(x) + A0(x) + N(x)A(x)]. Hence, for any x ∈ [0, L], using (6) and (7) and the boundary conditions, |A(x)| =  x 0 A(y) dy  ≤  x