Abstract
In this paper, we investigate the existence of entire solutions for a delayed lattice competitive system. Here the entire solutions are the solutions that exist for all (n,t)in mathbb{Z}times mathbb{R}. In order to prove the existence, we firstly embed the delayed lattice system into the corresponding larger system, of which the traveling front solutions are identical to those of the delayed lattice system. Then based on the comparison theorem and the sup–sub solutions method, we construct entire solutions which behave as two opposite traveling front solutions moving towards each other from both sides of x-axis and then annihilating. Moreover, our conclusions extend the invading way, which the superior species invade the inferior ones from both sides of x-axis and then the inferior ones extinct, into the lattice and delay case.
Highlights
We study the following delayed lattice competitive system:
The aim of this paper is to investigate the existence of an entire solution for (1.1) which converges to two monotone fronts with opposite speeds
Based on the known results of traveling front solutions for (1.3) in [3, 7] and by using the similar methods in [2] and [4], they have proved the existence of two-front entire solutions to (1.3) for the cases (i) and (iii)
Summary
We study the following delayed lattice competitive system:. where a, b, d, k, h are all positive numbers and τi > 0 (i = 1, 2) are the maturation time for the species. We study the following delayed lattice competitive system:. Where a, b, d, k, h are all positive numbers and τi > 0 (i = 1, 2) are the maturation time for the species. Un = un(t) and vn = vn(t), t ∈ R, denote the population density of two competitive species at time t and niches n, respectively. We consider that both un(t) and vn(t) are nonnegative. It is obvious that there are four equilibria of (1.1),. When τ1 = τ2 = 0, as stated in [5], the solution of (1.1) has the following asymptotic behaviors depending on h and k as t → ∞:. (iii) If k, h > 1, (un(t), vn(t)) → (1, 0) or (0, 1) depending on the initial condition. (i) If 0 < h < 1 < k, (un(t), vn(t)) → (0, 1) (vn wins). (ii) If 0 < k < 1 < h, (un(t), vn(t)) → (1, 0) (un wins). (iii) If k, h > 1, (un(t), vn(t)) → (1, 0) or (0, 1) depending on the initial condition. (iv) If 0 < k, h < 1, (un(t), vn(t)) → (k1, k2) (un and vn coexist)
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