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Spreading properties for non-autonomous Fisher-KPP equations with nonlocal diffusion

Arnaud Ducrota,111Corresponding author.
E-mail: [email protected] (Arnaud Ducrot), [email protected] (Zhucheng Jin).
and Zhucheng Jina,b

aNormandie Univ., UNIHAVRE, LMAH, FR-CNRS-3335,
ISCN, 76600 Le Havre, France
bSchool of Mathematical Sciences, University of Science and Technology of China,
Hefei, Anhui 230026, China
Abstract

We investigate the large time behaviour of solutions to a non-autonomous Fisher-KPP equation with nonlocal diffusion, involving a thin-tailed kernel. In this paper, we are concerned with both compactly supported and exponentially decaying initial data. As far as general time heterogeneities are concerned, we provide upper and lower estimates for the location of the propagating front. As a special case, we derive a definite spreading speed when the time varying coefficients satisfy some averaging properties. This setting covers the cases of periodic, almost periodic and uniquely ergodic variations in time, in particular. Our analysis is based on the derivation of suitable regularity estimates (of uniform continuity type) for some particular solutions of a logistic equation with nonlocal diffusion. Such regularity estimates are coupled with the construction of appropriated propagating paths to derive spreading speed estimates, using ideas from the uniform persistence theory in dynamical systems.

Key words. Spreading speeds; Nonlocal diffusion; Time heterogeneity; Fisher-KPP equation; Uniform persistence.

2010 Mathematical Subject Classification. 35B40; 45K05; 35C07.

There is no conflict of interests.
There is no data associated to this work.

1 Introduction and main results

In this paper we study spreading properties for the solutions of the following non-autonomous and nonlocal one-dimensional equation

tu(t,x)=K(y)[u(t,xy)u(t,x)]dy+u(t,x)f(t,u(t,x)),\partial_{t}u(t,x)=\int_{\mathbb{R}}K(y)\left[u(t,x-y)-u(t,x)\right]\mathrm{d}y+u(t,x)f\left(t,u(t,x)\right), (1.1)

posed for time t0t\geq 0 and xx\in\mathbb{R}. This evolution problem is supplemented with an appropriated initial data, that will be discussed below. Here K=K(y)K=K(y) is a nonnegative dispersal kernel with thin-tailed (see Assumption 1.3 below). Let us set F(t,u):=uf(t,u)F(t,u):=uf(t,u). At the same time, F=F(t,u)F=F(t,u) stands for the nonlinear growth term, which depends on time tt and that will be assumed in this note to be of the Fisher-KPP type (see Assumption 1.5). The above problem typically describes the spatial invasion of a population (see for instance [6, 34] and the references therein) with the following features:
1) individuals exhibit long distance dispersal according to the kernel KK, in other words the quantity K(xy)K(x-y) corresponds to the probability for individuals to jump from yy to xx;
2) time varying birth and death processes modeled by the nonlinear Fisher-KPP type function F(t,u)F(t,u). The time variations may stand for seasonality and/or external events (see [24]).

When local diffusion is considered, the Fisher-KPP equation posed in a time homogeneous medium reads as

tu(t,x)=xxu(t,x)+F(u(t,x)).\partial_{t}u(t,x)=\partial_{xx}u(t,x)+F(u(t,x)). (1.2)

As mentioned above, this problem arises as a basic model in many different fields, in biology and ecology in particular. It can be used for instance to describe the spatio-temporal evolution of an invading species into an empty environment. The above equation (1.2) was introduced separately by Fisher [20] and Kolmogorov, Petrovsky and Piskunov [27], when the nonlinear function FF satisfies the Fisher-KPP conditions. Recall that a typical example of such Fisher-KPP nonlinearity is given by the logistic function F(u)=u(1u)F(u)=u(1-u).

There is a large amount of literature related to (1.2) and its generalizations. To study propagation phenomena generated by reaction diffusion equations, in addition to the existence of travelling wave solution, the asymptotic speed of spread (or spreading speed) was introduced and studied by Aronson and Weinberger in [4]. Roughly speaking if u0u_{0} is a nontrival and nonnegative initial data with compact support, then the solution of (1.2) associated with this initial data u0u_{0} spreads with the speed c>0c^{*}>0 (the minimal wave speed of the travelling waves) in the sense that

limtsup|x|ct|u(t,x)1|=0,c[0,c) and limtsup|x|ctu(t,x)=0,c>c.\lim_{t\to\infty}\sup_{|x|\leq ct}|u(t,x)-1|=0,\;\forall\;c\in[0,c^{*})\;\text{ and }\;\lim_{t\to\infty}\sup_{|x|\geq ct}u(t,x)=0,\;\forall\;c>c^{*}.

This concept of spreading speed has been further developed by several researchers in the last decades from different view points including PDE’s argument, dynamical systems theory, probability theory, mathematical biology, etc. Spreading speeds of KPP-type reaction diffusion equations in homogeneous and periodic media have been extensively studied (see [8, 17, 29, 30, 44, 45] and the references cited therein). There is also an extensive literature on spreading phenomena for reaction diffusion systems. We refer for instance [3, 15, 22] and the references cited therein.

Recently spreading properties for KPP-type reaction-diffusion equations in more general environments have attracted a lot of attention, see [7, 9, 37, 40] and the references cited therein. In particular, Nadin and Rossi [37] studied spreading properties for KPP equation with local diffusion and general time heterogeneities. Furthermore, they obtained a definite spreading speed when the coefficients share some averaging properties.

The spreading properties of nonlocal diffusion equation as (1.1) has attracted a lot of interest in the last decades. Since the semiflow generated by nonlocal diffusion equations does not enjoy any regularization effects, this brings additional difficulties. Fisher-KPP equations or monostable problems in homogeneous environments have been studied from various point of views: wave front propagation (see [12, 39] and the references cited therein), hair trigger effect and spreading speed (see [2, 10, 13, 18, 34, 47] and the references cited therein). For the thin-tailed kernel, we refer for instance to [34] and the recent work [47] where a new sub-solution has been constructed to provide a lower bound of the spreading speed. Note also that the aforementioned work deals with possibly non-symmetric kernel where the propagation speed on the left and the right-hand side of the domain can be different. For the fat-tailed dispersion kernels the propagating behaviour of the solutions can be very different from the one observed with thin-tailed kernel. Acceleration may occur. We refer to [19, 21] for fat-tailed kernel and to [10] for fractional Laplace type dispersion.

Recently, wave propagation and spreading speeds for nonlocal diffusion problem with time and/or space heterogeneities have been considered. Existence and nonexistence of generalized travelling wave solutions have been discussed in [16, 25, 32, 41] and the references cited therein. For spreading speed results, we refer the reader to [24, 25, 31, 42] and the references cited therein. We also refer to [5, 46, 48] for the analysis of the spreading speed for systems with nonlocal diffusion.

As far as monotone problem is concerned, one may apply the well developed monotone semiflow method to study the spreading speed for nonlocal diffusion problems. We refer the reader to [30, 44] and to [24, 25] for time periodic systems.
In this work, we provide a new approach which is based on the construction of suitable propagating paths (namely, functions tX(t)t\mapsto X(t) with lim inftu(t,X(t))>0\liminf_{t\to\infty}u(t,X(t))>0) coupled with what we call a persistence lemma (see Lemma 2.6 below) for uniformly continuous solutions, to obtain lower estimate for the propagating set. This lemma roughly states that controlling from below the solution at x=0x=0 and X(t)X(t) for t1t\gg 1 implies a control of the solution u=u(t,x)u=u(t,x) from below on the whole interval x[0,kX(t)]x\in[0,kX(t)] for some k(0,1)k\in(0,1) and t1t\gg 1. The proof of this lemma does not make use of the properties of the tail of the kernel, so that we expect our key persistence lemma to be applied for the study of acceleration phenomena for fat tailed dispersal kernel. However, the uniform continuity property for the solutions is important for our proof and this remains complicated to check. For the regularity results of some specific time global solutions to nonlocal diffusion equations, we refer the reader to [11, 32] for spatial heterogeneous case and to [16, 41] for time heterogeneous media. Here we are able to prove such a property for some specific initial data and logistic type nonlinearities.

Note that in [28] the authors consider the regularity problem. They show that when the nonlinear term satisfies Fu(u)<K¯F_{u}(u)<\overline{K} for any u0u\geq 0, where K¯=K(y)dy\overline{K}=\int_{\mathbb{R}}K(y)\mathrm{d}y, then solutions of the homogeneous problem inherit the Lipschitz continuity property from those of their initial data, with a control of the Lipschitz constant for all time t1t\gg 1. In this note, we prove the uniform continuity of some solutions when the above condition fails (see Assumption 1.5 (f4)(f4)). This point is studied in Section 3.1, where we provide a class of initial data for which the solutions (of the nonlocal logistic equation) are uniformly continuous on [0,)×[0,\infty)\times\mathbb{R}.

Now to state our results, we first introduce some notations and present our main assumptions. Let us define the important notion of the least mean for a bounded function.

Definition 1.1

Along this work, for any given function hL(0,;)h\in L^{\infty}(0,\infty;\mathbb{R}), we define

h:=limT+infs>01T0Th(t+s)dt.\lfloor h\rfloor:=\lim\limits_{T\to+\infty}\inf_{s>0}\frac{1}{T}\int_{0}^{T}h(t+s)\mathrm{d}t. (1.3)

In that case the quantity h\lfloor h\rfloor is called the least mean of the function hh (over (0,)(0,\infty)).

If hh admits a mean value h\langle h\rangle, that is, there exists

h:=limT+1T0Th(t+s)dt, uniformly with respect to s0.\langle h\rangle:=\lim\limits_{T\to+\infty}\frac{1}{T}\int_{0}^{T}h(t+s)\mathrm{d}t,\text{ uniformly with respect to }s\geq 0. (1.4)

Then h=h\lfloor h\rfloor=\langle h\rangle. Particularly, the time periodic, almost periodic and uniquely ergodic coefficients have the mean value. Here recall that a bounded and uniformly continuous function f:f:\mathbb{R}\to\mathbb{R} is called uniquely ergodic if, for any continuous map G:fG:\mathcal{H}_{f}\to\mathbb{R}, the following limit exists uniformly in ss\in\mathbb{R}:

limT+1Tss+TG(f(+τ))dτ,\lim\limits_{T\to+\infty}\frac{1}{T}\int_{s}^{s+T}G(f(\cdot+\tau))\mathrm{d}\tau,

where f:=cl{f(+τ),τ}\mathcal{H}_{f}:={\rm cl}\{f(\cdot+\tau),\;\tau\in\mathbb{R}\} is the closure of the translation set of ff under the local uniform topology.

Periodic, almost periodic and compactly supported functions are specific subclass of uniquely ergodic functions. A celebrated example of uniquely ergodic function is constructed from the Penrose tiling. For more examples and properties of almost periodic and uniquely ergodic functions, we refer the reader to [9, 33, 36].

An equivalent and useful characterization for the least mean of the function, as above, is given in the next lemma.

Lemma 1.2

[37, 38] Let hL(0,;)h\in L^{\infty}(0,\infty;\mathbb{R}) be given. Then one has

h=supaW1,(0,)inft>0(a+h)(t).\lfloor h\rfloor=\sup_{a\in W^{1,\infty}(0,\infty)}\inf_{t>0}\left(a^{\prime}+h\right)(t).

We are now able to present the main assumptions that will be needed in this note. First we assume that the kernel K=K(y)K=K(y) enjoys the following set of properties:

Assumption 1.3 (Kernel K=K(y)K=K(y))

We assume that the kernel K:[0,)K:\mathbb{R}\to[0,\infty) satisfies the following set of assumptions:

  • (i)

    The function yK(y)y\mapsto K(y) is non-negative, continuous and integrable;

  • (ii)

    There exists α>0\alpha>0 such that

    K(y)eαy𝑑y<.\int_{\mathbb{R}}K(y)e^{\alpha y}dy<\infty.
  • (iii)

    We also assume that K(0)>0K(0)>0.

Remark 1.4

Here we do not impose that the kernel function is symmetric. We focus on the propagation to the right-hand side of the spatial domain. Thus in (ii)(ii), we only assume the kernel is thin-tailed on the right-hand side.

Since K(y)K(y) is continuous and K(0)>0K(0)>0, then there exist δ>0\delta>0 and k:[0,)k:\mathbb{R}\to[0,\infty), continuous, even and compactly supported such that

suppk=[δ,δ],k(y)>0,y(δ,δ),k(y)K(y) and k(y)=k(y),y.\begin{split}&{\rm supp}\;k=[-\delta,\delta],\;k(y)>0,\;\forall y\in(-\delta,\delta),\\ &k(y)\leq K(y)\text{ and }k(y)=k(-y),\;\forall y\in\mathbb{R}.\end{split} (1.5)

This property will allow us to control the solution on bounded sets, around x=0x=0.

Now we discuss our Fisher-KPP assumptions for the nonlinear term F(t,u)=uf(t,u)F(t,u)=uf(t,u).

Assumption 1.5 (KPP nonlinearity)

Assume that the function f:[0,)×[0,1]f:[0,\infty)\times[0,1]\to\mathbb{R} satisfies the following set of hypotheses:

  • (f1)

    f(,u)L(0,;)f(\cdot,u)\in L^{\infty}(0,\infty;\mathbb{R}), for all u[0,1]u\in[0,1], and ff is Lipschitz continuous with respect to u[0,1]u\in[0,1], uniformly with respect to t0t\geq 0;

  • (f2)

    Let f(t,1)=0f(t,1)=0 for a.e. t0t\geq 0. Setting μ(t):=f(t,0)\mu(t):=f(t,0), we assume that μ()\mu(\cdot) is bounded and uniformly continuous. Also, we require that

    h(u):=inft0f(t,u)>0 for all u[0,1);h(u):=\inf_{t\geq 0}f(t,u)>0\text{ for all $u\in[0,1)$};
  • (f3)

    For almost every t0t\geq 0, the function uf(t,u)u\mapsto f(t,u) is nonincreasing on [0,1][0,1];

  • (f4)

    Set K¯:=K(y)dy\overline{K}:=\int_{\mathbb{R}}K(y)\mathrm{d}y. The least mean of the function μ\mu satisfies

    μ>K¯.\lfloor\mu\rfloor>\overline{K}.
Remark 1.6

Here we assume that the steady states are p=0p^{-}=0 and p+=1p^{+}=1. These assumptions can be relaxed by the change of variables to take into account p=p(t)p^{-}=p^{-}(t) and p+=p+(t)p^{+}=p^{+}(t). Indeed, under the conditions inft0p+(t)p(t)>0\inf_{t\geq 0}p^{+}(t)-p^{-}(t)>0 and p+(t)p(t)p^{+}(t)-p^{-}(t) is bounded, one can set

u~(t,x):=u(t,x)p(t)p+(t)p(t).\tilde{u}(t,x):=\frac{u(t,x)-p^{-}(t)}{p^{+}(t)-p^{-}(t)}.

This can reduce the equation heterogeneous steady states into the equation with steady states 0 and 11 as long as inft0p+(t)p(t)>0\inf_{t\geq 0}p^{+}(t)-p^{-}(t)>0 and p+(t)p(t)p^{+}(t)-p^{-}(t) is bounded.

Remark 1.7

From the above assumption, one can note that

inft0μ(t)=h(0)>0.\inf_{t\geq 0}\mu(t)=h(0)>0.

Next this assumption also implies that there exists some constant C>0C>0 such that for all u[0,1]u\in[0,1] and t0t\geq 0 one has

μ(t)f(t,u)μ(t)Cuμ(t)(1Hu),\mu(t)\geq f(t,u)\geq\mu(t)-Cu\geq\mu(t)(1-Hu), (1.6)

where we have set H:=supt0Cμ(t)=Ch(0)H:=\sup\limits_{t\geq 0}\frac{C}{\mu(t)}=\frac{C}{h(0)}.

Let us now define some notations related to the speed function that will be used in the following. We define σ(K)\sigma(K), the abscissa of convergence of KK, by

σ(K):=sup{γ>0:K(y)eγy𝑑y<}.\sigma\left(K\right):=\sup\left\{\gamma>0:\;\int_{\mathbb{R}}K(y)e^{\gamma y}dy<\infty\right\}.

Assumption 1.3 (ii)(ii) yields that σ(K)(0,]\sigma(K)\in(0,\infty]. We set

L(λ):=K(y)[eλy1]dy,λ[0,σ(K)),L(\lambda):=\int_{\mathbb{R}}K(y)[e^{\lambda y}-1]\mathrm{d}y,\;\lambda\in\left[0,\sigma(K)\right), (1.7)

as well for λ(0,σ(K))\lambda\in(0,\sigma(K)) and t0t\geq 0,

c(λ)(t):=λ1L(λ)+λ1μ(t).c(\lambda)(t):=\lambda^{-1}L(\lambda)+\lambda^{-1}\mu(t). (1.8)

For a given function aW1,(0,)a\in W^{1,\infty}(0,\infty), denote cλ,ac_{\lambda,a} the function given by

cλ,a(t):=c(λ)(t)+a(t),λ(0,σ(K)),t0.c_{\lambda,a}(t):=c(\lambda)(t)+a^{\prime}(t),\;\lambda\in(0,\sigma(K)),\;t\geq 0. (1.9)

Obviously, it follows from Definition 1.1 that cλ,a()=c(λ)()\lfloor c_{\lambda,a}(\cdot)\rfloor=\lfloor c(\lambda)(\cdot)\rfloor for each λ(0,σ(K))\lambda\in(0,\sigma(K)). Next note that

c(λ)()=λ1L(λ)+λ1μ.\lfloor c(\lambda)(\cdot)\rfloor=\lambda^{-1}L(\lambda)+\lambda^{-1}\lfloor\mu\rfloor.

Now we state some properties of c(λ)()\lfloor c(\lambda)(\cdot)\rfloor in the following proposition.

Proposition 1.8

Let Assumption 1.3 and 1.5 be satisfied. Then the following properties hold:

  • (i)

    The map λc(λ)()\lambda\mapsto\left\lfloor c(\lambda)(\cdot)\right\rfloor from (0,σ(K))(0,\sigma(K)) to \mathbb{R} is of class C1C^{1} from (0,σ(K))(0,\sigma(K)) into \mathbb{R}.

  • (ii)

    Set cr:=infλ(0,σ(K))c(λ)().\displaystyle c^{*}_{r}:=\inf_{\lambda\in(0,\sigma(K))}\lfloor c(\lambda)(\cdot)\rfloor. There exists λr(0,σ(K)]\lambda_{r}^{*}\in(0,\sigma(K)] such that

    limλ(λr)c(λ)()=cr.\lim_{\lambda\to(\lambda_{r}^{*})^{-}}\lfloor c(\lambda)(\cdot)\rfloor=c_{r}^{*}.

    Moreover, one has cr>0c_{r}^{*}>0 and the map λc(λ)()\lambda\mapsto\lfloor c(\lambda)(\cdot)\rfloor is decreasing on (0,λr)(0,\lambda_{r}^{*}).

  • (iii)

    Assume that λr<σ(K)\lambda_{r}^{*}<\sigma(K). One has

    cr=K(y)eλryydy.c^{*}_{r}=\int_{\mathbb{R}}K(y)e^{\lambda_{r}^{*}y}y\mathrm{d}y. (1.10)

The above Proposition 1.8 has been mostly proved in [16] (see Proposition 2.8 in [16]) with a more general kernel which depends on tt.

Here we only explain that cr>0c_{r}^{*}>0. To see this, note that for λ(0,σ(K))\lambda\in(0,\sigma(K)) one has

λc(λ)(t)=K(y)eλydy+μ(t)K¯,t0.\lambda c(\lambda)(t)=\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y+\mu(t)-\overline{K},\;\forall t\geq 0.

Next due to Assumption 1.5 (f4)(f4) and Lemma 1.2, there exists some function aW1,(0,)a\in W^{1,\infty}(0,\infty) such that μ(t)K¯+a(t)0\mu(t)-\overline{K}+a^{\prime}(t)\geq 0 for all t0t\geq 0. This yields for all λ(0,σ(K))\lambda\in(0,\sigma(K)) and t0t\geq 0,

λc(λ)(t)+a(t)=K(y)eλydy+μ(t)K¯+a(t)K(y)eλydy>0,\lambda c(\lambda)(t)+a^{\prime}(t)=\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y+\mu(t)-\overline{K}+a^{\prime}(t)\geq\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y>0,

that rewrites cr>0c_{r}^{*}>0 since a=0\lfloor a^{\prime}\rfloor=0. The result follows.

Remark 1.9

Let us point out that the assumption λr<σ(K)\lambda_{r}^{*}<\sigma(K) needed for (iii)(iii) to hold is satisfied for instance if we have

lim supλσ(K)L(λ)λ=+.\limsup_{\lambda\to\sigma(K)^{-}}\frac{L(\lambda)}{\lambda}=+\infty. (1.11)

Indeed, one can observe that

c(λ)()μλ+ as λ0+.\lfloor c(\lambda)(\cdot)\rfloor\sim\frac{\lfloor\mu\rfloor}{\lambda}\to+\infty\text{ as $\lambda\to 0^{+}$}.

In addition, if (1.11) holds, then the decreasing property of the map λc(λ)()\lambda\mapsto\lfloor c(\lambda)(\cdot)\rfloor on (0,λr)(0,\lambda_{r}^{*}) as stated in Proposition 1.8 (ii)(ii) ensures that λr<σ(K)\lambda_{r}^{*}<\sigma(K).

To state our spreading result, we impose in the following that the condition discussed in the previous remark is satisfied, that means λr\lambda_{r}^{*} is different from the convergence abscissa.

Assumption 1.10

In addition to Assumption 1.3, we assume that λr<σ(K)\lambda_{r}^{*}<\sigma(K).

Using the above properties for the speed function c(λ)()c(\lambda)(\cdot) and its least mean value, we are now able to state our main results.

Theorem 1.11 (Upper bounds)

Let Assumption 1.3, 1.5 and 1.10 be satisfied. Let u=u(t,x)u=u(t,x) denote the solution of (1.1) equipped with a continuous initial data u0u_{0}, with 0u0()10\leq u_{0}(\cdot)\leq 1 and u0()0u_{0}(\cdot)\not\equiv 0.
Then the following upper estimates for the propagation set hold: if u0(x)=O(eλx)u_{0}(x)=O(e^{-\lambda x}) as xx\to\infty for some λ>0\lambda>0, then one has

limtsupx0tc+(λ)(s)ds+ηtu(t,x)=0,η>0,\lim_{t\to\infty}\sup_{x\geq\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s+\eta t}u(t,x)=0,\;\forall\eta>0,

where the function c+(λ)()c^{+}(\lambda)(\cdot) is defined by

c+(λ)():={c(λr)() if λλr,c(λ)() if λ(0,λr).c^{+}(\lambda)(\cdot):=\begin{cases}c(\lambda_{r}^{*})(\cdot)&\text{ if $\lambda\geq\lambda_{r}^{*}$},\\ c(\lambda)(\cdot)&\text{ if $\lambda\in(0,\lambda_{r}^{*})$}.\end{cases}

For the lower estimates of the propagation set, we first state our result for a specific function f=f(t,u)f=f(t,u) of the form f(t,u)=μ(t)(1u).f(t,u)=\mu(t)(1-u). In other words, we are considering the following non-autonomous logistic equation

tu(t,x)=K(y)[u(t,xy)u(t,x)]dy+μ(t)u(t,x)(1u(t,x)).\partial_{t}u(t,x)=\int_{\mathbb{R}}K(y)\left[u(t,x-y)-u(t,x)\right]\mathrm{d}y+\mu(t)u(t,x)\left(1-u(t,x)\right). (1.12)

To enter the framework of Assumption 1.5, we assume that the function μ\mu satisfies following conditions:

tμ(t) is uniformly continuous and bounded with inft0μ(t)>0, and the least mean of μ() satisfies μ>K¯.\begin{split}&t\mapsto\mu(t)\text{ is uniformly continuous and bounded with }\inf_{t\geq 0}\mu(t)>0,\\ &\text{ and the least mean of }\mu(\cdot)\text{ satisfies }\lfloor\mu\rfloor>\overline{K}.\end{split} (1.13)

For this problem, our lower estimate of propagation set reads as follows.

Theorem 1.12 (Lower bounds)

Let Assumption 1.3, 1.10 be satisfied and assume furthermore that μ\mu satisfies (1.13). Let u=u(t,x)u=u(t,x) denote the solution of (1.12) equipped with a continuous initial data u0u_{0}, with 0u0()10\leq u_{0}(\cdot)\leq 1 and u0()0u_{0}(\cdot)\not\equiv 0. Then the following propagation occurs:

  1. (i)

    (Fast exponential decay case) If u0(x)=O(eλx)u_{0}(x)=O(e^{-\lambda x}) as xx\to\infty for some λλr\lambda\geq\lambda_{r}^{*}, then one has

    limtsupx[0,ct]|1u(t,x)|=0,c(0,cr);\lim_{t\to\infty}\sup_{x\in[0,ct]}\left|1-u(t,x)\right|=0,\;\forall c\in(0,c_{r}^{*});
  2. (ii)

    (Slow exponential decay case) If lim infxeλxu0(x)>0\displaystyle\liminf_{x\to\infty}e^{\lambda x}u_{0}(x)>0 for some λ(0,λr)\lambda\in(0,\lambda_{r}^{*}), then it holds that

    limtsupx[0,ct]|1u(t,x)|=0,c(0,c(λ)).\lim_{t\to\infty}\sup_{x\in[0,ct]}\left|1-u(t,x)\right|=0,\;\forall c\in\left(0,\lfloor c(\lambda)\rfloor\right).

Next as a consequence of the comparison principle, one obtains the following lower estimates of the propagation set to the right-hand side for more general nonlinearity satisfying Assumption 1.5.

Corollary 1.13 (Inner propagation)

Let Assumption 1.3, 1.5 and 1.10 be satisfied. Let u=u(t,x)u=u(t,x) denote the solution of (1.1) supplemented with a continuous initial data u0u_{0}, with 0u0()10\leq u_{0}(\cdot)\leq 1 and u0()0u_{0}(\cdot)\not\equiv 0. Then the following propagation result holds true:

  1. (i)

    (Fast exponential decay case) If u0(x)=O(eλx)u_{0}(x)=O(e^{-\lambda x}) as xx\to\infty for some λλr\lambda\geq\lambda_{r}^{*}, then one has

    lim inftinfx[0,ct]u(t,x)>0,c(0,cr);{\color[rgb]{1,0,0}\liminf_{t\to\infty}}\inf_{x\in[0,ct]}u(t,x)>0,\;\forall c\in(0,c_{r}^{*});
  2. (ii)

    (Slow exponential decay case) If lim infxeλxu0(x)>0\displaystyle\liminf_{x\to\infty}e^{\lambda x}u_{0}(x)>0 for some λ(0,λr)\lambda\in(0,\lambda_{r}^{*}), then one has

    lim inftinfx[0,ct]u(t,x)>0,c(0,c(λ)).{\color[rgb]{1,0,0}\liminf_{t\to\infty}}\inf_{x\in[0,ct]}u(t,x)>0,\;\forall c\in\left(0,\lfloor c(\lambda)\rfloor\right).
Remark 1.14

When the coefficients are periodic functions with period TT, from [25] one can note that 1T0Tc+(λ)(s)ds\frac{1}{T}\int_{0}^{T}c^{+}(\lambda)(s)\mathrm{d}s is the exact spreading speed for (1.1). In the periodic situation, our results are also sharp, in the sense that

limt1t0tc+(λ)(s)ds=c+(λ)=1T0Tc+(λ)(s)ds.\lim_{t\to\infty}\frac{1}{t}\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s=\lfloor c^{+}(\lambda)\rfloor=\frac{1}{T}\int_{0}^{T}c^{+}(\lambda)(s)\mathrm{d}s.

The two quantities limt1t0tc+(λ)(s)ds\lim_{t\to\infty}\frac{1}{t}\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s and c+(λ)\lfloor c^{+}(\lambda)\rfloor also coincide when c+(λ)(t)c^{+}(\lambda)(t) is a time almost periodic function. Therefore our results provide the exact spreading speed for nonlocal KPP equations in a time almost periodic environment.

In more general heterogeneous environment, for instance non-recurrent environment, one may have c+(λ)<lim inft1t0tc+(λ)(s)ds\displaystyle\lfloor c^{+}(\lambda)\rfloor<\liminf\limits_{t\to\infty}\frac{1}{t}\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s, see Example 1 in [37]. Our results provide the upper and lower estimates of the propagation set. For β(c+(λ),lim inft1t0tc+(λ)(s)ds)\beta\in\left(\lfloor c^{+}(\lambda)\rfloor,\liminf\limits_{t\to\infty}\frac{1}{t}\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s\right), the behaviour of u(t,βt)u(t,\beta t) for t1t\gg 1 is unknown. This open problem is similar to the non-autonomous Fisher-KPP equation with local diffusion [37].

In the above result we only consider the propagation to the right-hand side of the real line and obtain a propagation result on some interval of the form [0,ct][0,ct] for suitable speed cc and for t1t\gg 1. Note that the kernel is not assumed to be even, so that the propagation behaviours on the right and the left-hand sides can be different. For instance, different spreading speeds may arise at right and left-hand sides when the kernel is thin-tailed on both sides. To study the propagation behaviour of the left-hand side, it is sufficient to change xx to x-x in the above results.

The results stated in this section and more precisely the lower bounds for the propagation follows from the derivation of suitable regularity estimates for the solution. Here we show that the solutions of (1.12) with suitable initial data are uniformly continuous. Next Theorem 1.12 follows from the application of a general persistence lemma (see Lemma 2.6) for uniformly continuous solutions. This key lemma roughly ensures that if there is a uniformly continuous solution u=u(t,x)u=u(t,x) admitting a propagating path tX(t)t\mapsto X(t), then [0,kX(t)][0,kX(t)] with any k(0,1)k\in(0,1) is a propagating interval, that is uu stays uniformly far from 0 on this interval, in the large time. The idea of the proof of this lemma comes from the uniform persistence theory for dynamical systems for which we refer the reader to [23, 35, 43, 49] and references cited therein.

This paper is organized as follows. In Section 2, we recall comparison principles and derive our general key persistence Lemma. Section 3 is devoted to the derivation of some regularity estimates for the solutions of (1.12) with suitable initial data. With all these materials, we conclude the proofs of theorems and the corollary.

2 Preliminary and Key Lemma

This section is devoted to the statement of the comparison principle and a key lemma that will be used to prove the inner propagation theorem, namely Theorem 1.12.

2.1 Comparison principle and strong maximum principle

We start this section by recalling the following more general comparison principle.

Proposition 2.1

(See [16, Proposition 3.1])[Comparison principle] Let t0t_{0}\in\mathbb{R} and T>0T>0 be given. Let K:[0,)K:\mathbb{R}\to[0,\infty) be an integrable kernel and let F=F(t,u)F=F(t,u) be a function defined in [t0,t0+T]×[0,1][t_{0},t_{0}+T]\times[0,1] which is Lipschitz continuous with respect to u[0,1]u\in[0,1], uniformly with respect to tt. Let u¯\underline{u} and u¯\overline{u} be two uniformly continuous functions defined from [t0,t0+T]×[t_{0},t_{0}+T]\times\mathbb{R} into the interval [0,1][0,1] such that for each xx\in\mathbb{R}, the maps u¯(,x)\underline{u}(\cdot,x) and u¯(,x)\overline{u}(\cdot,x) both belong to W1,1(t0,t0+T)W^{1,1}(t_{0},t_{0}+T), satisfying u¯(t0,)u¯(t0,)\underline{u}(t_{0},\cdot)\leq\overline{u}(t_{0},\cdot), and for all xx\in\mathbb{R} and for almost every t(t0,t0+T)t\in(t_{0},t_{0}+T),

tu¯(t,x)K(y)[u¯(t,xy)u¯(t,x)]dy+F(t,u¯(t,x)),tu¯(t,x)K(y)[u¯(t,xy)u¯(t,x)]dy+F(t,u¯(t,x)).\begin{split}&\partial_{t}\overline{u}(t,x)\geq\int_{\mathbb{R}}K(y)\left[\overline{u}(t,x-y)-\overline{u}(t,x)\right]\mathrm{d}y+F(t,\overline{u}(t,x)),\\ &\partial_{t}\underline{u}(t,x)\leq\int_{\mathbb{R}}K(y)\left[\underline{u}(t,x-y)-\underline{u}(t,x)\right]\mathrm{d}y+F(t,\underline{u}(t,x)).\end{split}

Then u¯u¯\underline{u}\leq\overline{u} on [t0,t0+T]×[t_{0},t_{0}+T]\times\mathbb{R}.

We also need some comparison principle on moving domain as follows (this can be proved similarly as Lemma 5.4 in [1] and Lemma 4.7 in [48]).

Proposition 2.2

Assume that K:R[0,)K:R\to[0,\infty) is integrable. Let t0>0t_{0}>0 and T>0T>0 be given. Let b(t,x)b(t,x) be a uniformly bounded function from [t0,t0+T]×[t_{0},t_{0}+T]\times\mathbb{R}\to\mathbb{R}. Assume that u(t,x)u(t,x) is uniformly continuous defined from [t0,t0+T]×[t_{0},t_{0}+T]\times\mathbb{R} into the interval [0,1][0,1] such that for each xx\in\mathbb{R}, u(,x)W1,1(t0,t0+T)u(\cdot,x)\in W^{1,1}(t_{0},t_{0}+T). Assume that XX and YY are continuous functions on [t0,t0+T][t_{0},t_{0}+T] with X<YX<Y. If uu satisfies

{tuK(y)[u(t,xy)u(t,x)]dy+b(t,x)u,t[t0,t0+T],x(X(t),Y(t)),u(t,x)0,t(t0,t0+T],x(X(t),Y(t)),u(t0,x)0,x(X(t0),Y(t0)).\begin{cases}\partial_{t}u\geq\int_{\mathbb{R}}K(y)\left[u(t,x-y)-u(t,x)\right]\mathrm{d}y+b(t,x)u,&\!\forall t\in[t_{0},t_{0}+T],x\in(X(t),Y(t)),\\ u(t,x)\geq 0,&\!\forall t\in(t_{0},t_{0}+T],x\in\mathbb{R}\setminus(X(t),Y(t)),\\ u(t_{0},x)\geq 0,&\!\forall x\in(X(t_{0}),Y(t_{0})).\end{cases}

Then

u(t,x)0 for all t[t0,t0+T],x[X(t),Y(t)].u(t,x)\geq 0\text{ for all }t\in[t_{0},t_{0}+T],x\in[X(t),Y(t)].

We continue this section by the following strong maximum principle. We refer the reader to [26] for the proof of following proposition.

Proposition 2.3 (Strong maximum principle)

Let Assumption 1.3, 1.5 be satisfied. Let u=u(t,x)u=u(t,x) be the solution of (1.1) supplemented with some continuous initial data u0u_{0}, such that 0u010\leq u_{0}\leq 1 and u00u_{0}\not\equiv 0. Then u(t,x)>0u(t,x)>0 for all t>0,xt>0,x\in\mathbb{R}.

2.2 Key lemma

In this section, we derive an important lemma that will be used in the next section to prove our main inner propagation result, namely Theorem 1.12. In this section we only let Assumption 1.3 (i)(i), (iii)(iii) and Assumption 1.5 be satisfied.

Definition 2.4 (Limit orbits set)

Let u=u(t,x)u=u(t,x) be a uniformly continuous function on [0,)×[0,\infty)\times\mathbb{R} into [0,1][0,1], solution of (1.1). We define ω(u)\omega(u), the set of the limit orbits, as the set of the bounded and uniformly continuous functions u~:2\tilde{u}:\mathbb{R}^{2}\to\mathbb{R} where exist sequences (xn)n(x_{n})_{n}\subset\mathbb{R} and (tn)n[0,)(t_{n})_{n}\subset[0,\infty) such that tnt_{n}\to\infty as nn\to\infty and

u~(t,x)=limnu(t+tn,x+xn),\tilde{u}(t,x)=\lim_{n\to\infty}u(t+t_{n},x+x_{n}),

uniformly for (t,x)(t,x) in bounded sets of 2\mathbb{R}^{2}.

Let us observe that since uu is assumed to be bounded and uniformly continuous on [0,)×[0,\infty)\times\mathbb{R}, Arzelà-Ascoli theorem ensures that ω(u)\omega(u) is not empty. Indeed, for each sequence (tn)n[0,)(t_{n})_{n}\subset[0,\infty) with tnt_{n}\to\infty and (xn)(x_{n})\subset\mathbb{R} the sequence of functions (t,x)u(t+tn,x+xn)(t,x)\mapsto u(t+t_{n},x+x_{n}) is equi-continuous and thus has a converging subsequence with respect to the local uniform topology. In addition, it is a compact set with respect to the compact open topology, that is with respect to the local uniform topology.

Before going to our key lemma, we claim that the set ω(u)\omega(u) enjoys the following property:

Claim 2.5

Let u=u(t,x)u=u(t,x) be a uniformly continuous solution of (1.1). Let u~ω(u)\tilde{u}\in\omega(u) be given. Then one has:

Either u~(t,x)>0 for all (t,x)2 or u~(t,x)0 on 2.\text{Either $\tilde{u}(t,x)>0$ for all $(t,x)\in\mathbb{R}^{2}$ or $\tilde{u}(t,x)\equiv 0$ on $\mathbb{R}^{2}$}.

Proof. Note that due to Assumption 1.5 (see Remark 1.7), the function uu satisfies the following differential inequality for all t0t\geq 0 and xx\in\mathbb{R}

tu(t,x)Ku(t,)(x)K¯u(t,x)+u(t,x)(μ(t)Cu(t,x)).\partial_{t}u(t,x)\geq K\ast u(t,\cdot)(x)-\overline{K}u(t,x)+u(t,x)(\mu(t)-Cu(t,x)).

Since the function μ()\mu(\cdot) is bounded, for each u~ω(u)\tilde{u}\in\omega(u), there exists μ~=μ~(t)L()\tilde{\mu}=\tilde{\mu}(t)\in L^{\infty}(\mathbb{R}), a weak star limit of some shifted function μ(tn+)\mu(t_{n}+\cdot), for some suitable time sequence (tn)(t_{n}), such that u~\tilde{u} satisfies

tu~(t,x)Ku~(t,)(x)K¯u~(t,x)+u~(t,x)(μ~(t)Cu~(t,x))Ku~(t,)(x)+(K¯+inftμ~(t)C)u~(t,x),(t,x)2.\begin{split}\partial_{t}\tilde{u}(t,x)&\geq K\ast\tilde{u}(t,\cdot)(x)-\overline{K}\tilde{u}(t,x)+\tilde{u}(t,x)(\tilde{\mu}(t)-C\tilde{u}(t,x))\\ &\geq K\ast\tilde{u}(t,\cdot)(x)+\left(-\overline{K}+\inf_{t\in\mathbb{R}}\tilde{\mu}(t)-C\right)\tilde{u}(t,x),\;\forall(t,x)\in\mathbb{R}^{2}.\end{split}

Herein tu~\partial_{t}\tilde{u} is a weak star limit of tu(+tn,+xn)\partial_{t}u(\cdot+t_{n},\cdot+x_{n}) for some suitable sub-sequence of (xn)n(x_{n})_{n} and (tn)n(t_{n})_{n}. This is due to tuL([0,)×)\partial_{t}u\in L^{\infty}([0,\infty)\times\mathbb{R}).

Next the claim follows from the same arguments as for the proof of the strong maximum principle, see [26].

 

Using the above definition and its properties we are now able to state and prove the following key lemma.

Lemma 2.6

Let u=u(t,x):[0,)×[0,1]u=u(t,x):[0,\infty)\times\mathbb{R}\to[0,1] be a uniformly continuous solution of (1.1). Let tX(t)t\mapsto X(t) from [0,)[0,\infty) to [0,)[0,\infty) be a given continuous function. Let the following set of hypothesis be satisfied:

  • (H1)

    Assume that lim inftu(t,0)>0;\liminf\limits_{t\to\infty}u(t,0)>0;

  • (H2)

    There exists some constant ε~0>0\tilde{\varepsilon}_{0}>0 such that for all u~ω(u){0}\tilde{u}\in\omega(u)\setminus\{0\}, one has

    lim inftu~(t,0)>ε~0;\liminf\limits_{t\to\infty}\tilde{u}(t,0)>\tilde{\varepsilon}_{0};
  • (H3)

    The map tX(t)t\mapsto X(t) is a propagating path for uu, in the sense that

    lim inftu(t,X(t))>0.\liminf\limits_{t\to\infty}u(t,X(t))>0.

Then for any k(0,1)k\in(0,1), one has

lim inftinf0xkX(t)u(t,x)>0.\liminf\limits_{t\to\infty}\inf_{0\leq x\leq kX(t)}u(t,x)>0.
Remark 2.7

The above result holds without assuming that the convolution kernel is exponential bounded. We expect this key lemma may also be useful to study the spatial propagation for Fisher-KPP equation with fat-tailed dispersion kernel, where the solution may accelerate, see [10, 19, 21].

To prove the above lemma, we make use of ideas coming from uniform persistence theory, see[23, 35, 43]. This is somehow close to those developed in [14, 15].

Proof. To prove the lemma we argue by contradiction by assuming that there exists k(0,1)k\in(0,1), a sequence (tn)n[0,)(t_{n})_{n}\subset[0,\infty) with tnt_{n}\to\infty and a sequence (kn)(k_{n}) with 0knk0\leq k_{n}\leq k such that

u(tn,knX(tn))0 as n.u(t_{n},k_{n}X(t_{n}))\to 0\;\text{ as }\;n\to\infty. (2.1)

First we claim that one has

limnknX(tn)=.\lim_{n\to\infty}k_{n}X(t_{n})=\infty. (2.2)

To prove this claim we argue by contradiction by assuming that {knX(tn)}\{k_{n}X(t_{n})\} has a bounded subsequence. Hence there exists xx_{\infty}\in\mathbb{R} such that possibly along a subsequence still denoted with the index nn, one has knX(tn)xk_{n}X(t_{n})\to x_{\infty} as nn\to\infty.

Now let us consider the sequence of functions un(t,x):=u(t+tn,x)u_{n}(t,x):=u(t+t_{n},x). Since u=u(t,x)u=u(t,x) is uniformly continuous, possibly up to a sub-sequence still denoted with the same index nn, there exists uω(u)u_{\infty}\in\omega(u) such that

un(t,x)u(t,x) locally uniformly for (t,x)2.u_{n}(t,x)\to u_{\infty}(t,x)\text{ locally uniformly for $(t,x)\in\mathbb{R}^{2}$.}

Next since knX(tn)xk_{n}X(t_{n})\to x_{\infty}, then (2.1) ensures that

u(0,x)=limnu(tn,knX(tn))=0.u_{\infty}(0,x_{\infty})=\lim\limits_{n\to\infty}u(t_{n},k_{n}X(t_{n}))=0.

Since uω(u)u_{\infty}\in\omega(u), Claim 2.5 ensures that u(t,x)0u_{\infty}(t,x)\equiv 0. On the other hand, (H1)(H1) ensures that for all tt\in\mathbb{R}, one has

u(t,0)lim inftu(t,0)>0,u_{\infty}(t,0)\geq\liminf_{t\to\infty}u(t,0)>0,

a contradiction, so that (2.2) holds.

Now due to (2.2), there exists NN such that

X(0)<knX(tn),nN.X(0)<k_{n}X(t_{n}),\;\forall n\geq N.

Hence due to kn<1k_{n}<1 we have

X(0)<knX(tn)<X(tn),nN.X(0)<k_{n}X(t_{n})<X(t_{n}),\;\forall n\geq N.

And since tX(t)t\mapsto X(t) is continuous, then for each nNn\geq N there exists tn(0,tn)t_{n}^{\prime}\in(0,t_{n}) such that

X(tn)=knX(tn),nN.X(t_{n}^{\prime})=k_{n}X(t_{n}),\;\forall n\geq N.

Since knX(tn)k_{n}X(t_{n})\to\infty as nn\to\infty and tX(t)t\mapsto X(t) is continuous, then tnt_{n}^{\prime}\to\infty as nn\to\infty.

From the above definition of tnt_{n}^{\prime}, one has

u(tn,knX(tn))=u(tn,X(tn)),nN.u(t_{n}^{\prime},k_{n}X(t_{n}))=u(t_{n}^{\prime},X(t_{n}^{\prime})),\;\forall n\geq N.

So that (H3)(H3) ensures that for all nn large enough, there exists ε>0\varepsilon>0 such that

u(tn,knX(tn))=u(tn,X(tn))ε.u(t_{n}^{\prime},k_{n}X(t_{n}))=u(t_{n}^{\prime},X(t_{n}^{\prime}))\geq\varepsilon.

Recall that Assumption (H2)(H2). Now for all nn large enough, we define

tn′′:=inf{ttn;s(t,tn),u(s,knX(tn))min{ε~0,ε}2}(tn,tn).t_{n}^{\prime\prime}:=\inf\left\{t\leq t_{n};\;\forall s\in(t,t_{n}),\;u(s,k_{n}X(t_{n}))\leq\frac{\min\{\tilde{\varepsilon}_{0},\varepsilon\}}{2}\right\}\in(t_{n}^{\prime},t_{n}).

Since u(tn,knX(tn))0u(t_{n},k_{n}X(t_{n}))\to 0 as nn\to\infty, then one may assume that, for all nn large enough one has

{u(tn′′,knX(tn))=min{ε~0,ε}2,u(t,knX(tn))min{ε~0,ε}2,t[tn′′,tn],u(tn,knX(tn))1n.\begin{cases}u(t_{n}^{\prime\prime},k_{n}X(t_{n}))=\frac{\min\{\tilde{\varepsilon}_{0},\varepsilon\}}{2},\\ u(t,k_{n}X(t_{n}))\leq\frac{\min\{\tilde{\varepsilon}_{0},\varepsilon\}}{2},\;\forall t\in[t_{n}^{\prime\prime},t_{n}],\\ u(t_{n},k_{n}X(t_{n}))\leq\frac{1}{n}.\end{cases}

Next we claim that tntn′′t_{n}-t_{n}^{\prime\prime}\to\infty as nn\to\infty. Indeed, if (a subsequence of) tntn′′t_{n}-t_{n}^{\prime\prime} converges to σ\sigma\in\mathbb{R}, define the sequence of functions u~n(t,x):=u(t+tn′′,x+knX(tn))\tilde{u}_{n}(t,x):=u(t+t_{n}^{\prime\prime},x+k_{n}X(t_{n})), that converges, possibly along a subsequence, locally uniformly to some function u~=u~(t,x)ω(u)\tilde{u}_{\infty}=\tilde{u}_{\infty}(t,x)\in\omega(u) with

u~(0,0)=min{ε~0,ε}2>0,\tilde{u}_{\infty}(0,0)=\frac{\min\{\tilde{\varepsilon}_{0},\varepsilon\}}{2}>0,

and

u~(σ,0)=limnu~n(tntn′′,0)=limnu(tn,knX(tn))=0.\tilde{u}_{\infty}(\sigma,0)=\lim\limits_{n\to\infty}\tilde{u}_{n}(t_{n}-t_{n}^{\prime\prime},0)=\lim\limits_{n\to\infty}u(t_{n},k_{n}X(t_{n}))=0.

Since u~ω(u)\tilde{u}_{\infty}\in\omega(u), then the above two values of u~\tilde{u}_{\infty} contradict the dichotomy stated in Claim 2.5 and this proves that tntn′′t_{n}-t_{n}^{\prime\prime}\to\infty as nn\to\infty.

As a consequence one obtains that the function u~ω(u)\tilde{u}_{\infty}\in\omega(u) satisfies

u~(0,0)=min{ε~0,ε}2>0,\tilde{u}_{\infty}(0,0)=\frac{\min\{\tilde{\varepsilon}_{0},\varepsilon\}}{2}>0,

together with

u~(t,0)min{ε~0,ε}2,t0.\tilde{u}_{\infty}(t,0)\leq\frac{\min\{\tilde{\varepsilon}_{0},\varepsilon\}}{2},\;\forall t\geq 0. (2.3)

Due to Claim 2.5, the above equality yields u~ω(u){0}\tilde{u}_{\infty}\in\omega(u)\setminus\{0\} and (2.3) contradicts (H2)(H2). The proof is completed.  

3 Proof of spreading properties

In this section, we shall make use of the key lemma (see Lemma 2.6) to prove Theorem 1.12. To do this, we first derive some important regularity properties of the solutions of the Logistic equation (1.12) associated with suitable initial data. Next we prove Theorem 1.11 by constructing suitable exponentially decaying super-solutions for (1.1). Finally we turn to the proof of Theorem 1.12. As already mentioned we crucially make use of Lemma 2.6 and construct a suitable propagating path tX(t)t\mapsto X(t), that depends on the decay rate of the initial data u0=u0(x)u_{0}=u_{0}(x) for x1x\gg 1. As a corollary, we conclude the propagation results for (1.1).

3.1 Uniform continuity estimate

This subsection is devoted to giving some regularity estimates for the solutions of the following Logistic equation (recalling (1.12)) when endowed with suitable initial data,

tu(t,x)=K(y)u(t,xy)dyK¯u(t,x)+μ(t)u(t,x)(1u(t,x)).\partial_{t}u(t,x)=\int_{\mathbb{R}}K(y)u(t,x-y)\mathrm{d}y-\overline{K}u(t,x)+\mu(t)u(t,x)\left(1-u(t,x)\right).

Here we focus on two types of initial data, that will be used to prove Theorem 1.12: initial data with a compact support and initial data with support on a right semi-infinite interval and with some prescribed exponential decay on this right-hand side (that is for x1x\gg 1).

Our first lemma is concerned with the compactly supported case.

Lemma 3.1

Let Assumption 1.3 and (1.13) be satisfied. Let u=u(t,x)u=u(t,x) be the solution of (1.12) equipped with the initial data v0=v0(x)v_{0}=v_{0}(x), where v0v_{0} is Lipschitz continuous in \mathbb{R}, and 0<v0(x)<10<v_{0}(x)<1 for all x(0,A)x\in(0,A), for some constant A>0A>0 while v0=0v_{0}=0 outside of (0,A)(0,A). Then, the function (t,x)u(t,x)(t,x)\mapsto u(t,x) is uniformly continuous on [0,)×[0,\infty)\times\mathbb{R}.

Proof. Firstly, since 0u10\leq u\leq 1, then one has

tuL(+×)M:=2K¯+μ.\|\partial_{t}u\|_{L^{\infty}(\mathbb{R}^{+}\times\mathbb{R})}\leq M:=2\overline{K}+\|\mu\|_{\infty}. (3.1)

As a consequence, the map (t,x)u(t,x)(t,x)\mapsto u(t,x) is Lipchitz continuous for the variable t[0,)t\in[0,\infty), uniformly with respect to xx\in\mathbb{R}, that is

|u(t,x)u(s,x)|M|ts|,(t,s)[0,)2,x.|u(t,x)-u(s,x)|\leq M|t-s|,\;\forall(t,s)\in[0,\infty)^{2},\;\forall x\in\mathbb{R}. (3.2)

Next we investigate the regularity with respect to the spatial variable xx\in\mathbb{R}. To do so we claim that the following holds true:

Claim 3.2

For all h>0h>0 sufficiently small, there exists 0<σ(h)<10<\sigma(h)<1 such that σ(h)1\sigma(h)\to 1 as h0h\to 0 and

u(h,x)σ(h)v0(xh),x.u(\sqrt{h},x)\geq\sigma(h)v_{0}(x-h),\;\forall x\in\mathbb{R}.

Proof of Claim 3.2. Let us first observe that since u(t,.)>0u(t,.)>0 for all t>0t>0, it is sufficient to look at xh[0,A]x-h\in[0,A], that is hxA+hh\leq x\leq A+h.

Next to prove this claim, note that one has for all h>0h>0 and xx\in\mathbb{R}:

u(h,x)=v0(x)+0htu(l,x)dl=v0(x)+0h{K(y)[u(l,xy)u(l,x)]dy+μ(l)u(l,x)(1u(l,x))}dl\begin{split}u(\sqrt{h},x)&=v_{0}(x)+\int_{0}^{\sqrt{h}}\partial_{t}u(l,x)\mathrm{d}l\\ &=v_{0}(x)+\int_{0}^{\sqrt{h}}\left\{\int_{\mathbb{R}}K(y)\left[u(l,x-y)-u(l,x)\right]\mathrm{d}y+\mu(l)u(l,x)\left(1-u(l,x)\right)\right\}\mathrm{d}l\end{split}

Now coupling (3.2) and 0u10\leq u\leq 1, one gets, for all h>0h>0 small enough and uniformly for xx\in\mathbb{R}

u(h,x)v0(x)+0h{K(y)v0(xy)dyK¯v0(x)}dl+o(h),u(\sqrt{h},x)\geq v_{0}(x)+\int_{0}^{\sqrt{h}}\left\{\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y-\overline{K}v_{0}(x)\right\}\mathrm{d}l+o(\sqrt{h}),

that is

u(h,x)v0(x)(1K¯h)+h(K(y)v0(xy)dy+o(1)).u(\sqrt{h},x)\geq v_{0}(x)\bigg{(}1-\overline{K}\sqrt{h}\bigg{)}+\sqrt{h}\bigg{(}\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y+o(1)\bigg{)}.

Now observing Assumption 1.3 (see (i)(i) and (iii)(iii)), there exists ε>0\varepsilon>0 such that

minx[0,A]K(y)v0(xy)dy2ε,\min_{x\in[0,A]}\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y\geq 2\varepsilon,

so that for h>0h>0 small enough one has

minx[h,A+h]K(y)v0(xy)dyε,\min_{x\in[h,A+h]}\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y\geq\varepsilon,

Now to prove the claim, it is sufficiently to reach, for all h>0h>0 small enough and x[h,A+h]x\in[h,A+h],

v0(x)(1K¯h)+h(o(1)+ε)σ(h)v0(xh).v_{0}(x)\bigg{(}1-\overline{K}\sqrt{h}\bigg{)}+\sqrt{h}\left(o(1)+\varepsilon\right)\geq\sigma(h)v_{0}(x-h). (3.3)

Now set σ(h)=12K¯h\sigma(h)=1-2\overline{K}\sqrt{h} and let us show that Claim 3.2 follows.

Since v0v_{0} is Lipschitz continuous, then there exists some constant L>0L>0 such that

|v0(x)v0(xh)|Lh,x.|v_{0}(x)-v_{0}(x-h)|\leq Lh,\;\forall x\in\mathbb{R}.

Hence to obtain (3.3), it is sufficient to reach for all x[h,A+h]x\in[h,A+h] and all h>0h>0 small enough

K¯hv0(xh)+h(o(1)+ε)Lh(1K¯h).\overline{K}\sqrt{h}v_{0}(x-h)+\sqrt{h}\left(o(1)+\varepsilon\right)\geq Lh\bigg{(}1-\overline{K}\sqrt{h}\bigg{)}. (3.4)

Dividing by h\sqrt{h}, the above inequality holds whenever

K¯v0(xh)+(o(1)+ε)Lh(1K¯h),\overline{K}v_{0}(x-h)+\left(o(1)+\varepsilon\right)\geq L\sqrt{h}\bigg{(}1-\overline{K}\sqrt{h}\bigg{)}, (3.5)

which holds true for all h>0h>0 small enough. So the claim is proved.

 

Now we come back to the proof of Lemma 3.1. For each h>0h>0 small enough, let us introduce the following function

bh(t)=bh(0)exp{0t[μ(s+h)μ(s)]ds}, for all t0,b_{h}(t)=b_{h}(0)\exp\left\{\int_{0}^{t}\left[\mu(s+\sqrt{h})-\mu(s)\right]\mathrm{d}s\right\},\;\text{ for all }t\geq 0, (3.6)

where bh(0)b_{h}(0) is some constant depending on hh and that satisfies the following three conditions:

0<bh(0)σ(h)<1,0<b_{h}(0)\leq\sigma(h)<1,

bh(0)1b_{h}(0)\to 1 as h0h\to 0 and for all h>0h>0 small enough

bh(0)inft0μ(t)μ(t+h)exp{0t[μ(s)μ(s+h)]ds}.b_{h}(0)\leq\inf_{t\geq 0}\frac{\mu(t)}{\mu(t+\sqrt{h})}\exp\left\{\int_{0}^{t}\left[\mu(s)-\mu(s+\sqrt{h})\right]\mathrm{d}s\right\}.

For the later condition, one can observe that it is feasible since one has

|0t[μ(s+h)μ(s)]ds|=|ht+hμ(s)ds0tμ(s)ds|=|tt+hμ(s)ds0hμ(s)ds|2μh.\begin{split}\left|\int_{0}^{t}\left[\mu(s+\sqrt{h})-\mu(s)\right]\mathrm{d}s\right|&=\left|\int_{\sqrt{h}}^{t+\sqrt{h}}\mu(s)\mathrm{d}s-\int_{0}^{t}\mu(s)\mathrm{d}s\right|\\ &=\left|\int_{t}^{t+\sqrt{h}}\mu(s)\mathrm{d}s-\int_{0}^{\sqrt{h}}\mu(s)\mathrm{d}s\right|\\ &\leq 2\|\mu\|_{\infty}\sqrt{h}.\end{split}

As a consequence, recalling (1.13), μ()\mu(\cdot) is uniformly continuous and we end-up with

μ(t)μ(t+h)exp{0t[μ(s)μ(s+h)]ds}1, as h0, uniformly for t0.\frac{\mu(t)}{\mu(t+\sqrt{h})}\exp\left\{\int_{0}^{t}\left[\mu(s)-\mu(s+\sqrt{h})\right]\mathrm{d}s\right\}\to 1,\text{ as }h\to 0,\text{ uniformly for }t\geq 0.

Hence bh(0)b_{h}(0) is well defined and bh(t)1b_{h}(t)\to 1 as h0h\to 0 uniformly for t0t\geq 0.

Now, setting wh=wh(t,x)w_{h}=w_{h}(t,x) the function given by

wh(t,x):=u(t+h,x)bh(t)u(t,xh),w_{h}(t,x):=u(t+\sqrt{h},x)-b_{h}(t)u(t,x-h),

one obtains that it becomes a solution of the following equation

twh(t,x)=Kwh(t,x)K¯wh(t,x)+μ(t+h)[wh(t,x)+bh(t)u(t,xh)][1(wh(t,x)+bh(t)u(t,xh))]μ(t)bh(t)u(t,xh)[1u(t,xh)]bh(t)u(t,xh)=Kwh(t,x)K¯wh(t,x)+μ(t+h)wh(t,x)(1wh(t,x)2bh(t)u(t,xh))+bh(t)u(t,xh)(μ(t+h)μ(t)bh(t)bh(t))+bh(t)u2(t,xh)(μ(t)bh(t)μ(t+h)).\begin{split}\partial_{t}w_{h}(t,x)&=K\ast w_{h}(t,x)-\overline{K}w_{h}(t,x)\\ &\quad+\mu(t+\sqrt{h})\left[w_{h}(t,x)+b_{h}(t)u(t,x-h)\right]\left[1-\left(w_{h}(t,x)+b_{h}(t)u(t,x-h)\right)\right]\\ &\quad-\mu(t)b_{h}(t)u(t,x-h)\left[1-u(t,x-h)\right]-b^{\prime}_{h}(t)u(t,x-h)\\ &=K\ast w_{h}(t,x)-\overline{K}w_{h}(t,x)+\mu(t+\sqrt{h})w_{h}(t,x)\bigg{(}1-w_{h}(t,x)-2b_{h}(t)u(t,x-h)\bigg{)}\\ &\quad+b_{h}(t)u(t,x-h)\left(\mu(t+\sqrt{h})-\mu(t)-\frac{b_{h}^{\prime}(t)}{b_{h}(t)}\right)+b_{h}(t)u^{2}(t,x-h)\left(\mu(t)-b_{h}(t)\mu(t+\sqrt{h})\right).\end{split}

It follows from the definition of bh(t)b_{h}(t) (see (3.6) above) that wh(t,x)w_{h}(t,x) satisfies

twh(t,x)Kwh(t,x)K¯wh(t,x)+wh(t,x)μ(t+h)(1wh(t,x)2bh(t)u(t,xh)).\partial_{t}w_{h}(t,x)\geq K\ast w_{h}(t,x)-\overline{K}w_{h}(t,x)+w_{h}(t,x)\mu(t+\sqrt{h})\bigg{(}1-w_{h}(t,x)-2b_{h}(t)u(t,x-h)\bigg{)}.

The Claim 3.2 together with bh(0)<σ(h)b_{h}(0)<\sigma(h) ensure that wh(0,)0w_{h}(0,\cdot)\geq 0. Then the comparison principle applies and implies that wh(t,x)0w_{h}(t,x)\geq 0 for all t0,xt\geq 0,x\in\mathbb{R}, that rewrites as u(t+h,x)bh(t)u(t,xh)u(t+\sqrt{h},x)\geq b_{h}(t)u(t,x-h) for all t0,xt\geq 0,x\in\mathbb{R}, for h>0h>0 small enough. Recalling (3.2), for h>0h>0 sufficiently small, one has for all t0t\geq 0 and xx\in\mathbb{R},

u(t,xh)u(t,x)(1bh(t)1)u(t+h,x)+Mh(1bh(t)1)+Mh.u(t,x-h)-u(t,x)\leq\left(\frac{1}{b_{h}(t)}-1\right)u(t+\sqrt{h},x)+M\sqrt{h}\leq\left(\frac{1}{b_{h}(t)}-1\right)+M\sqrt{h}. (3.7)

Since for h>0h>0 small enough one has

minx[h,Ah]K(y)v0(xy)dyε,\min_{x\in[-h,A-h]}\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y\geq\varepsilon,

then one can similarly prove that for sufficiently small h>0h>0, there exists σ(h)=12K¯h\sigma(h)=1-2\overline{K}\sqrt{h} such that

u(h,x)σ(h)v0(x+h),x.u(\sqrt{h},x)\geq\sigma(h)v_{0}(x+h),\;\forall x\in\mathbb{R}.

This rewrites as

u(h,xh)σ(h)v0(x),x.u(\sqrt{h},x-h)\geq\sigma(h)v_{0}(x),\;\forall x\in\mathbb{R}.

Then as above one can choose a suitable function bh(t)b_{h}(t) and obtain that

u(t+h,xh)bh(t)u(t,x),t0,x.u(t+\sqrt{h},x-h)\geq b_{h}(t)u(t,x),\;\forall t\geq 0,x\in\mathbb{R}.

Recalling (3.2), for h>0h>0 sufficiently small, one obtains for all t0t\geq 0 and xx\in\mathbb{R},

u(t,x)u(t,xh)(1bh(t)1)u(t+h,xh)+Mh(1bh(t)1)+Mh.\begin{split}u(t,x)-u(t,x-h)&\leq\left(\frac{1}{b_{h}(t)}-1\right)u(t+\sqrt{h},x-h)+M\sqrt{h}\\ &\leq\left(\frac{1}{b_{h}(t)}-1\right)+M\sqrt{h}.\end{split} (3.8)

Since estimates (3.7) and (3.8) are uniform with respect to the spatial variable xx\in\mathbb{R}, one also obtains a similar estimates for u(t,x)u(t,x+h)u(t,x)-u(t,x+h) and u(t,x+h)u(t,x)u(t,x+h)-u(t,x). From these estimates one has reached that u=u(t,x)u=u(t,x) is uniformly continuous for all t0,xt\geq 0,x\in\mathbb{R}, which completes the proof of the lemma.  

In the following we derive regularity estimates for the solutions to (1.12) coming from an initial data with a prescribed exponential decay rate of the right, that for x1x\gg 1. To do this, we show that such solutions to (1.12) decay with the same rate as the initial data, at least in a short time.

Let us introduce some function spaces. Recalling that λr\lambda_{r}^{*} is defined in Proposition 1.8, for λ(0,λr)\lambda\in(0,\lambda_{r}^{*}) let us define the space BCλ()BC_{\lambda}(\mathbb{R}) by

BCλ():={ϕC():supxeλx|ϕ(x)|<},BC_{\lambda}(\mathbb{R}):=\left\{\phi\in C(\mathbb{R}):\;\sup_{x\in\mathbb{R}}e^{\lambda x}|\phi(x)|<\infty\;\right\},

equipped with the weighted norm

ϕBCλ:=supxeλx|ϕ(x)|.\|\phi\|_{BC_{\lambda}}:=\sup_{x\in\mathbb{R}}e^{\lambda x}|\phi(x)|.

Recall that BCλ()BC_{\lambda}(\mathbb{R}) is a Banach space when endowed with the above norm.

Define also the subset EE by

E:={ϕBCλ():0ϕ1},E:=\left\{\phi\in BC_{\lambda}(\mathbb{R}):0\leq\phi\leq 1\right\}, (3.9)

and let us observe that it is a closed subset of BCλ()BC_{\lambda}(\mathbb{R}).

Using these notations, we turn to the proof of the following lemma.

Lemma 3.3

Let Assumption 1.3 and 1.10 and (1.13) be satisfied. Let λ(0,λr)\lambda\in(0,\lambda_{r}^{*}) and u0Eu_{0}\in E be given. Then the solution of (1.12) with initial data u0u_{0}, denoted by u=u(t,x)u=u(t,x), satisfies

limt0+supxeλx|u(t,x)u0(x)|=0.\lim\limits_{t\to 0^{+}}\sup_{x\in\mathbb{R}}e^{\lambda x}|u(t,x)-u_{0}(x)|=0.

Proof. Fix α>K¯+2μ\alpha>\overline{K}+2\|\mu\|_{\infty}. Let us introduce for each ϕE\phi\in E and t0t\geq 0, the operator given by

Qt[ϕ]():=αϕ()+K(y)ϕ(y)dyK¯ϕ()+μ(t)ϕ()(1ϕ()).Q_{t}[\phi](\cdot):=\alpha\phi(\cdot)+\int_{\mathbb{R}}K(y)\phi(\cdot-y)\mathrm{d}y-\overline{K}\phi(\cdot)+\mu(t)\phi(\cdot)\left(1-\phi(\cdot)\right).

Note that one has

K(y)ϕ(y)dyBCλ=supx|K(y)eλyeλ(xy)ϕ(xy)dy||K(y)eλydy|ϕBCλ.\left\|\int_{\mathbb{R}}K(y)\phi(\cdot-y)\mathrm{d}y\right\|_{BC_{\lambda}}=\sup_{x\in\mathbb{R}}\left|\int_{\mathbb{R}}K(y)e^{\lambda y}e^{\lambda(x-y)}\phi(x-y)\mathrm{d}y\right|\leq\left|\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y\right|\left\|\phi\right\|_{BC_{\lambda}}.

Let us observe that |K(y)eλydy|<\left|\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y\right|<\infty due to 0<λ<λr<σ(K)0<\lambda<\lambda_{r}^{*}<\sigma(K). Since 0ϕ10\leq\phi\leq 1 then one has

Qt[ϕ]()BCλ(α+|K(y)eλydy|+K¯+μ)ϕBCλ<.\|Q_{t}[\phi](\cdot)\|_{BC_{\lambda}}\leq\left(\alpha+\left|\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y\right|+\overline{K}+\|\mu\|_{\infty}\right)\|\phi\|_{BC_{\lambda}}<\infty.

Thus for each ϕ()E\phi(\cdot)\in E, for all t0t\geq 0, Qt[ϕ]()BCλ()Q_{t}[\phi](\cdot)\in BC_{\lambda}(\mathbb{R}).

Next let us observe that Qt[ϕ]Q_{t}[\phi] is nondecreasing with respect to ϕE\phi\in E. Indeed, if for any ϕ,ψE\phi,\psi\in E and ϕ(x)ψ(x)\phi(x)\geq\psi(x) for all xx\in\mathbb{R}, then for each given t0,xt\geq 0,x\in\mathbb{R}

Qt[ϕ](x)Qt[ψ](x)=α(ϕ(x)ψ(x))+K(y)[ϕ(xy)ψ(xy)]dyK¯(ϕψ)(x)+μ(t)ϕ(x)(1ϕ(x))μ(t)ψ(x)(1ψ(x))(αK¯2μ)(ϕ(x)ψ(x))0.\begin{split}Q_{t}[\phi](x)-Q_{t}[\psi](x)&=\alpha(\phi(x)-\psi(x))+\int_{\mathbb{R}}K(y)[\phi(x-y)-\psi(x-y)]\mathrm{d}y-\overline{K}(\phi-\psi)(x)\\ &\quad+\mu(t)\phi(x)(1-\phi(x))-\mu(t)\psi(x)(1-\psi(x))\\ &\geq\left(\alpha-\overline{K}-2\|\mu\|_{\infty}\right)\left(\phi(x)-\psi(x)\right)\\ &\geq 0.\end{split}

The last inequality comes from α>K¯+2μ\alpha>\overline{K}+2\|\mu\|_{\infty}. So that for any t0t\geq 0, the map ϕQt[ϕ]\phi\mapsto Q_{t}[\phi] is nondecreasing on EE.

For each given u0Eu_{0}\in E and any fixed h>0h>0, we define the following space

W:={tu(t,)C([0,h],BCλ()): 0u1,u(0,x)=u0(x)}.W:=\left\{t\mapsto u(t,\cdot)\in C([0,h],BC_{\lambda}(\mathbb{R})):\;0\leq u\leq 1,u(0,x)=u_{0}(x)\right\}.

Let us rewrite (1.12) to

tu(t,x)+αu(t,x)=Qt[u(t,)](x),\partial_{t}u(t,x)+\alpha u(t,x)=Q_{t}[u(t,\cdot)](x),

then one has

u(t,)=eαtu0()+0teα(st)Qs[u(s,)]()ds=:T[u](t,).u(t,\cdot)=e^{-\alpha t}u_{0}(\cdot)+\int_{0}^{t}e^{\alpha(s-t)}Q_{s}[u(s,\cdot)](\cdot)\mathrm{d}s=:T[u](t,\cdot).

Next we show that for each uWu\in W, one has T[u]WT[u]\in W. Let uWu\in W be given. Firstly we show that Qt[u]()BCλ()Q_{t}[u](\cdot)\in BC_{\lambda}(\mathbb{R}) uniformly for t[0,h]t\in[0,h]. Since tu(t,)C([0,h],BCλ())t\mapsto u(t,\cdot)\in C([0,h],BC_{\lambda}(\mathbb{R})), then one has

supt[0,h]u(t,)BCλ<.\sup_{t\in[0,h]}\|u(t,\cdot)\|_{BC_{\lambda}}<\infty.

Thus one has

supt[0,h]Qt[u(t,)]()BCλ(α+|K(y)eλydy|+K¯+μ)supt[0,h]u(t,)BCλ<.\sup_{t\in[0,h]}\|Q_{t}[u(t,\cdot)](\cdot)\|_{BC_{\lambda}}\leq\left(\alpha+\left|\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y\right|+\overline{K}+\|\mu\|_{\infty}\right)\sup_{t\in[0,h]}\|u(t,\cdot)\|_{BC_{\lambda}}<\infty.

Moreover, one can observe that for each t[0,h]t\in[0,h],

T[u](t,)BCλu0BCλ+1αsupt[0,h]Qt[u(t,)]BCλ<.\|T[u](t,\cdot)\|_{BC_{\lambda}}\leq\|u_{0}\|_{BC_{\lambda}}+\frac{1}{\alpha}\sup_{t\in[0,h]}\|Q_{t}[u(t,\cdot)]\|_{BC_{\lambda}}<\infty.

That is T[u](t,)BCλ()T[u](t,\cdot)\in BC_{\lambda}(\mathbb{R}), for each t[0,h]t\in[0,h].

Then we show that tT[u](t,)t\mapsto T[u](t,\cdot) is continuous. To see this, fix t0[0,h]t_{0}\in[0,h] and observe that one has

T[u](t,)T[u](t0,)BCλ|eαteαt0|u0BCλ+supxeλx|0t0[eα(st)eα(st0)]Qs[u(s,)](x)ds|+supxeλx|t0teα(st)Qs[u(s,)](x)ds||eαteαt0|u0BCλ+|eαteαt0|sups[0,h]Qs[u(s,)]BCλ0t0eαsds+sups[0,h]Qs[u(s,)]BCλ|1eα(t0t)α|.\begin{split}\left\|T[u](t,\cdot)-T[u](t_{0},\cdot)\right\|_{BC_{\lambda}}&\leq\left|e^{-\alpha t}-e^{-\alpha t_{0}}\right|\|u_{0}\|_{BC_{\lambda}}\\ &\quad+\sup_{x\in\mathbb{R}}e^{\lambda x}\left|\int_{0}^{t_{0}}\left[e^{\alpha(s-t)}-e^{\alpha(s-t_{0})}\right]Q_{s}[u(s,\cdot)](x)\mathrm{d}s\right|\\ &\quad+\sup_{x\in\mathbb{R}}e^{\lambda x}\left|\int_{t_{0}}^{t}e^{\alpha(s-t)}Q_{s}[u(s,\cdot)](x)\mathrm{d}s\right|\\ &\leq\left|e^{-\alpha t}-e^{-\alpha t_{0}}\right|\|u_{0}\|_{BC_{\lambda}}\\ &\quad+\left|e^{-\alpha t}-e^{-\alpha t_{0}}\right|\sup_{s\in[0,h]}\|Q_{s}[u(s,\cdot)]\|_{BC_{\lambda}}\int_{0}^{t_{0}}e^{\alpha s}\mathrm{d}s\\ &\quad+\sup_{s\in[0,h]}\|Q_{s}[u(s,\cdot)]\|_{BC_{\lambda}}\left|\frac{1-e^{\alpha(t_{0}-t)}}{\alpha}\right|.\end{split}

So that tT[u](t,)C([0,h],BCλ())t\mapsto T[u](t,\cdot)\in C([0,h],BC_{\lambda}(\mathbb{R})) and T[u](0,)=u0()T[u](0,\cdot)=u_{0}(\cdot).

Also, note that due to for each t[0,h]t\in[0,h], Qt[u(t,)]Q_{t}[u(t,\cdot)] is nondecreasing with u(t,)Eu(t,\cdot)\in E, then we get

0T[u](t,)eαt+1α(1eαt)α1,t[0,h].0\leq T[u](t,\cdot)\leq e^{-\alpha t}+\frac{1}{\alpha}(1-e^{-\alpha t})\alpha\leq 1,\;\;\forall t\in[0,h].

Hence, for each uWu\in W, then T[u]WT[u]\in W.

For each u,vWu,v\in W and a given γ>0\gamma>0 large enough, we introduce a metric on WW defined by

d(u,v):=supt[0,h]supxeλx|u(t,x)v(t,x)|eγt.d(u,v):=\sup_{t\in[0,h]}\sup_{x\in\mathbb{R}}e^{\lambda x}|u(t,x)-v(t,x)|e^{-\gamma t}.

Note that

d(T[u],T[v])=supt[0,h]supxeλx|0teα(st)(Q[u](s,x)Q[v](s,x))ds|eγtsupt[0,h]supx|0te(α+γ)(st)[α+K(y)eλydy+K¯+3μ]eγseλx|u(s,x)v(s,x)|ds|[α+K(y)eλydy+K¯+3μ]supt[0,h]0te(α+γ)(st)dsd(u,v)α+K(y)eλydy+K¯+3μα+γd(u,v).\begin{split}&d(T[u],T[v])=\sup_{t\in[0,h]}\sup_{x\in\mathbb{R}}e^{\lambda x}\left|\int_{0}^{t}e^{\alpha(s-t)}\left(Q[u](s,x)-Q[v](s,x)\right)\mathrm{d}s\right|e^{-\gamma t}\\ &\leq\sup_{t\in[0,h]}\sup_{x\in\mathbb{R}}\left|\int_{0}^{t}e^{(\alpha+\gamma)(s-t)}\left[\alpha+\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y+\overline{K}+3\|\mu\|_{\infty}\right]e^{-\gamma s}e^{\lambda x}|u(s,x)-v(s,x)|\mathrm{d}s\right|\\ &\leq\left[\alpha+\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y+\overline{K}+3\|\mu\|_{\infty}\right]\sup_{t\in[0,h]}\int_{0}^{t}e^{(\alpha+\gamma)(s-t)}\mathrm{d}s\cdot d(u,v)\\ &\leq\frac{\alpha+\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y+\overline{K}+3\|\mu\|_{\infty}}{\alpha+\gamma}\cdot d(u,v).\end{split}

So that T[u]T[u] is a contraction map on WW endowed with the metric d=d(u,v)d=d(u,v), as long as γ>0\gamma>0 sufficiently large such that

α+K(y)eλydy+K¯+3μα+γ<1.\frac{\alpha+\int_{\mathbb{R}}K(y)e^{\lambda y}\mathrm{d}y+\overline{K}+3\|\mu\|_{\infty}}{\alpha+\gamma}<1.

Finally since (W,d)(W,d) is a complete metric space, by Banach fixed point theorem ensures that T[u]T[u] has a unique fixed point in WW which is the solution of (1.12) with u(0,)=u0()u(0,\cdot)=u_{0}(\cdot). Since tu(t,)C([0,h],BCλ())t\mapsto u(t,\cdot)\in C([0,h],BC_{\lambda}(\mathbb{R})), then one has obtained

limt0+supxeλx|u(t,x)u0(x)|=0,\lim\limits_{t\to 0^{+}}\sup_{x\in\mathbb{R}}e^{\lambda x}|u(t,x)-u_{0}(x)|=0,

that completes the proof of the lemma.  

Lemma 3.4

Let Assumption 1.3 and 1.10 and (1.13) be satisfied. Let u=u(t,x)u=u(t,x) be the solution of (1.12) supplemented with the initial data v0v_{0} satisfying the following properties:
assume v0v_{0} is Lipschitz continuous in \mathbb{R}, there is A>0A>0 large enough, α>0\alpha>0, p(0,1)p\in(0,1) and λ(0,λr)\lambda\in(0,\lambda_{r}^{*}) such that

v0(x)={increasing function,x[0,α],β:=peλA,x[α,A],peλx,x[A,),0,x(,0].v_{0}(x)=\begin{cases}\text{increasing function},\;&x\in[0,\alpha],\\ \beta:=pe^{-\lambda A},\;&x\in[\alpha,A],\\ pe^{-\lambda x},\;&x\in[A,\infty),\\ 0,\;&x\in(-\infty,0].\end{cases} (3.10)

Then the function u=u(t,x)u=u(t,x) is uniformly continuous on [0,)×[0,\infty)\times\mathbb{R}.

Proof. As in the proof of Lemma 3.1, u=u(t,x)u=u(t,x) also satisfies (3.2).
Now from the definition of v0v_{0}, for h>0h>0 small enough, for the given λ(0,λr)\lambda\in(0,\lambda_{r}^{*}), one can observe

v0(x)eλhv0(xh),x.v_{0}(x)\geq e^{-\lambda h}v_{0}(x-h),\;\forall x\in\mathbb{R}.

Let us show that the function vh(t,x):=eλhu(t,xh)v^{h}(t,x):=e^{-\lambda h}u(t,x-h) (with vh(0,x)=eλhv0(xh)v^{h}(0,x)=e^{-\lambda h}v_{0}(x-h)) is a sub-solution of (1.12). To see this, note that vh(t,x)v^{h}(t,x) satisfies

tvh(t,x)=K(y)vh(t,xy)dyK¯vh(t,x)+μ(t)vh(t,x)(1eλhvh(t,x))K(y)vh(t,xy)dyK¯vh(t,x)+μ(t)vh(t,x)(1vh(t,x)).\begin{split}\partial_{t}v^{h}(t,x)&=\int_{\mathbb{R}}K(y)v^{h}(t,x-y)\mathrm{d}y-\overline{K}v^{h}(t,x)+\mu(t)v^{h}(t,x)\left(1-e^{\lambda h}v^{h}(t,x)\right)\\ &\leq\int_{\mathbb{R}}K(y)v^{h}(t,x-y)\mathrm{d}y-\overline{K}v^{h}(t,x)+\mu(t)v^{h}(t,x)\left(1-v^{h}(t,x)\right).\end{split}

Hence vh(t,x)v^{h}(t,x) becomes a sub-solution of (1.12).

Since vh(0,)v0()v^{h}(0,\cdot)\leq v_{0}(\cdot), the comparison principle implies that

u(t,x)eλhu(t,xh),t0,x.u(t,x)\geq e^{-\lambda h}u(t,x-h),\;\forall t\geq 0,x\in\mathbb{R}.

Similarly as in (3.7), one also has, for all h>0h>0 sufficiently small,

u(t,xh)u(t,x)(1eλh)u(t,xh)1eλh,t0,x,u(t,x-h)-u(t,x)\leq\left(1-e^{-\lambda h}\right)u(t,x-h)\leq 1-e^{-\lambda h},\forall t\geq 0,x\in\mathbb{R}, (3.11)

and changing xx to x+hx+h yields for all h>0h>0 sufficiently small,

u(t,x)u(t,x+h)(1eλh)u(t,x)1eλh,t0,x.u(t,x)-u(t,x+h)\leq\left(1-e^{-\lambda h}\right)u(t,x)\leq 1-e^{-\lambda h},\;\forall t\geq 0,x\in\mathbb{R}. (3.12)

Next we show that there exists 0<α(h)<10<\alpha(h)<1, α(h)1\alpha(h)\to 1 as h0h\to 0 such that for all h>0h>0 small enough

u(h,x)α(h)v0(x+h),x.u(\sqrt{h},x)\geq\alpha(h)v_{0}(x+h),\;\forall x\in\mathbb{R}.

Since v0(x+h)=0v_{0}(x+h)=0 for xhx\leq-h, it is sufficiently to consider the above inequality for xhx\geq-h. As in the proof of Lemma 3.1, note that for all h>0h>0 sufficiently small and uniformly for xx\in\mathbb{R}, one has

u(h,x)v0(x)(1K¯h)+h(K(y)v0(xy)dy+o(1)).u(\sqrt{h},x)\geq v_{0}(x)\bigg{(}1-\overline{K}\sqrt{h}\bigg{)}+\sqrt{h}\bigg{(}\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y+o(1)\bigg{)}.

One may now observe that for all 2Axh2A\geq x\geq-h, there exists ε>0\varepsilon>0 such that

K(y)v0(xy)dyε>0.\int_{\mathbb{R}}K(y)v_{0}(x-y)\mathrm{d}y\geq\varepsilon>0.

As in the proof of Claim 3.2, set α1(h)=12K¯h\alpha_{1}(h)=1-2\overline{K}\sqrt{h}. Then one has

u(h,x)α1(h)v0(x+h),x2A.u(\sqrt{h},x)\geq\alpha_{1}(h)v_{0}(x+h),\;\forall x\leq 2A.

Let us now prove that there exists 0<α2(h)<10<\alpha_{2}(h)<1 and α2(h)1,\alpha_{2}(h)\to 1, as h0h\to 0 such that u(h,x)α2(h)v0(x+h)u(\sqrt{h},x)\geq\alpha_{2}(h)v_{0}(x+h) for x2Ax\geq 2A. From Lemma 3.3, one has

limh0+supx2Aeλx|u(h,x)peλx|=0.\lim\limits_{h\to 0^{+}}\sup_{x\geq 2A}e^{\lambda x}|u(\sqrt{h},x)-pe^{-\lambda x}|=0.

Set

γ(h):=supx2Aeλx|u(h,x)peλx|,\gamma(h):=\sup_{x\geq 2A}e^{\lambda x}|u(\sqrt{h},x)-pe^{-\lambda x}|,

and observe that, for hh sufficiently small, for all x2Ax\geq 2A, one has

(1γ(h)p)v0(x)=γ(h)eλx+peλxu(h,x)γ(h)eλx+peλx=(γ(h)p+1)v0(x).\begin{split}\left(1-\frac{\gamma(h)}{p}\right)v_{0}(x)=-\gamma(h)e^{-\lambda{x}}+pe^{-\lambda x}&\leq u(\sqrt{h},x)\\ &\leq\gamma(h)e^{-\lambda{x}}+pe^{-\lambda x}\\ &=\left(\frac{\gamma(h)}{p}+1\right)v_{0}(x).\end{split}

So that one can set α2(h):=1γ(h)p\alpha_{2}(h):=1-\frac{\gamma(h)}{p} to obtain 0<α2(h)<10<\alpha_{2}(h)<1, α2(h)1\alpha_{2}(h)\to 1 as h0h\to 0 and

u(h,x)α2(h)v0(x),x2A.u(\sqrt{h},x)\geq\alpha_{2}(h)v_{0}(x),\;\forall x\geq 2A.

Then since v0v_{0} is non-increasing for xAx\geq A, one has

u(h,x)α2(h)v0(x)α2(h)v0(x+h),x2A.u(\sqrt{h},x)\geq\alpha_{2}(h)v_{0}(x)\geq\alpha_{2}(h)v_{0}(x+h),\;\forall x\geq 2A.

Now, set α(h):=min{α1(h),α2(h)}\alpha(h):=\min\{\alpha_{1}(h),\alpha_{2}(h)\}. We get

u(h,x)α(h)v0(x+h),x.u(\sqrt{h},x)\geq\alpha(h)v_{0}(x+h),\;\forall x\in\mathbb{R}.

As in the proof of Lemma 3.1, one can also construct a function b~h(t)1\tilde{b}_{h}(t)\to 1 as h0h\to 0 uniformly for t0t\geq 0 with 0<b~h(0)<α(h)0<\tilde{b}_{h}(0)<\alpha(h) and such that for all h>0h>0 small enough one has

u(t+h,x)b~h(t)u(t,x+h),t0,x.u(t+\sqrt{h},x)\geq\tilde{b}_{h}(t)u(t,x+h),\;\forall t\geq 0,x\in\mathbb{R}.

With such a choice, for all h>0h>0 small enough, for all t0t\geq 0 and xx\in\mathbb{R}, one obtains that

u(t,x+h)u(t,x)(1b~h(t)1)u(t+h,x)+Mh(1b~h(t)1)+Mh.u(t,x+h)-u(t,x)\leq\left(\frac{1}{\tilde{b}_{h}(t)}-1\right)u(t+\sqrt{h},x)+M\sqrt{h}\leq\left(\frac{1}{\tilde{b}_{h}(t)}-1\right)+M\sqrt{h}. (3.13)

As well as, for all t0t\geq 0 and xx\in\mathbb{R}, one has

u(t,x)u(t,xh)(1b~h(t)1)u(t+h,xh)+Mh(1b~h(t)1)+Mh.u(t,x)-u(t,x-h)\leq\left(\frac{1}{\tilde{b}_{h}(t)}-1\right)u(t+\sqrt{h},x-h)+M\sqrt{h}\leq\left(\frac{1}{\tilde{b}_{h}(t)}-1\right)+M\sqrt{h}. (3.14)

Combined with (3.11) and (3.12), this ensures that uu is uniformly continuous on [0,)×[0,\infty)\times\mathbb{R} and completes the proof of the lemma.

 

Remark 3.5

Here we point out that problem (1.1) is invariant with respect to spatial translation, so that spatial shift on the initial data v0()v_{0}(\cdot), induces the same spatial shift on the solution and does not change the uniform continuity on [0,)×[0,\infty)\times\mathbb{R}.

3.2 Proof of Theorem 1.11

In this subsection, we construct a suitable exponentially decaying super-solution and prove Theorem 1.11.

Proof of Theorem 1.11. For each given λ>0\lambda>0 and sufficiently large A>0A>0, let us firstly construct the following function

u¯(t,x):={Aeλr(x0tc(λr)(s)ds), if λλr,Aeλ(x0tc(λ)(s)ds), if 0<λ<λr.\overline{u}(t,x):=\begin{cases}Ae^{-\lambda_{r}^{*}\left(x-\int_{0}^{t}c(\lambda_{r}^{*})(s)\mathrm{d}s\right)},&\text{ if }\lambda\geq\lambda_{r}^{*},\\ Ae^{-\lambda\left(x-\int_{0}^{t}c(\lambda)(s)\mathrm{d}s\right)},&\text{ if }0<\lambda<\lambda_{r}^{*}.\end{cases}

Here we let A>0A>0 large enough such that u¯(0,)u0()\overline{u}(0,\cdot)\geq u_{0}(\cdot) and recall that the speed function tc(λ)(t)t\mapsto c(\lambda)(t) is defined in (1.8).

Since f(t,u)μ(t)f(t,u)\leq\mu(t) for all t0t\geq 0 and u[0,1]u\in[0,1], then one readily obtains that u¯\overline{u} is super-solution of (1.1). So that the comparison principle implies that

limtsupx0tc+(λ)(s)ds+ηtu(t,x)limtsupx0tc+(λ)(s)ds+ηtu¯(t,x)=0,η>0.\lim\limits_{t\to\infty}\sup_{x\geq\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s+\eta t}u(t,x)\leq\lim\limits_{t\to\infty}\sup_{x\geq\int_{0}^{t}c^{+}(\lambda)(s)\mathrm{d}s+\eta t}\overline{u}(t,x)=0,\;\;\forall\eta>0.

This completes the proof of the upper estimate as stated in Theorem 1.11.

 

3.3 Proof of Theorem 1.12

In this section we first discuss some properties of the solution of the following autonomous Fisher-KPP equation:

tu(t,x)=k(y)u(t,xy)dyk¯u(t,x)+u(t,x)(mbu(t,x)),t0,x.\partial_{t}u(t,x)=\int_{\mathbb{R}}k(y)u(t,x-y)\mathrm{d}y-\bar{k}u(t,x)+u(t,x)(m-bu(t,x)),\;t\geq 0,x\in\mathbb{R}. (3.15)

Here k()k(\cdot) is a given symmetric kernel as defined in Remark 1.4, k¯=k(y)dy>0\bar{k}=\int_{\mathbb{R}}k(y)\mathrm{d}y>0 while mm and bb are given positive constants.

Define

c0:=infλ>0k(y)eλydyk¯+mλ.c_{0}:=\inf_{\lambda>0}\frac{\int_{\mathbb{R}}k(y)e^{\lambda y}\mathrm{d}y-\bar{k}+m}{\lambda}.

Note that c0>0c_{0}>0 since k()k(\cdot) is a symmetric function (see also [47] where the sign of the (right and left) wave speed is investigated). Next our first important result reads as follows.

Lemma 3.6

Let u=u(t,x)u=u(t,x) be the solution of (3.15) supplemented with a continuous initial data 0u0()mb0\leq u_{0}(\cdot)\leq\frac{m}{b} and u00u_{0}\not\equiv 0 with compact support. Let us furthermore assume that uu is uniformly continuous for all t0t\geq 0, xx\in\mathbb{R}. Then one has

limtsup|x|ct|mbu(t,x)|=0,c[0,c0).\lim\limits_{t\to\infty}\sup_{|x|\leq ct}\left|\frac{m}{b}-u(t,x)\right|=0,\;\forall c\in[0,c_{0}).
Remark 3.7

For the kernel function with supp(k)={\rm supp}(k)=\mathbb{R} and without the uniform continuity assumption, the above propagating behaviour is already known. We refer to [34, Theorem 3.2]. For the reader convenience, we give a short proof of Lemma 3.6, with the help of Theorem 3.3 in [47] and the additional regularity assumption of solution.

Proof. Let c[0,c0)c\in[0,c_{0}) be given and fixed. To prove the lemma let us argue by contradiction by assuming that there exists a sequence (tn,xn)(t_{n},x_{n}) and |xn|ctn|x_{n}|\leq ct_{n} such that

lim supnu(tn,xn)<mb.\limsup\limits_{n\to\infty}u(t_{n},x_{n})<\frac{m}{b}.

Denote for n0n\geq 0 the sequence of functions unu_{n} by un(t,x):=u(t+tn,x+xn)u_{n}(t,x):=u(t+t_{n},x+x_{n}). Since u=u(t,x)u=u(t,x) is uniformly continuous on [0,)×[0,\infty)\times\mathbb{R} and 0umb0\leq u\leq\frac{m}{b}, then Arzelà-Ascoli theorem applies and ensures that as nn\to\infty one has un(t,x)u(t,x)u_{n}(t,x)\to u_{\infty}(t,x) locally uniformly for (t,x)2(t,x)\in\mathbb{R}^{2}, for some function u=u(t,x)u_{\infty}=u_{\infty}(t,x) defined in 2\mathbb{R}^{2} and such that u(0,0)<mbu_{\infty}(0,0)<\frac{m}{b}.

Now fix c(c,c0)c^{\prime}\in(c,c_{0}). Recall that Theorem 3.3 in [47] ensures that there exists some constant qc(0,mb]q_{c^{\prime}}\in\left(0,\frac{m}{b}\right] such that

lim inftinf|x|ctu(t,x)qc.\liminf_{t\to\infty}\inf_{|x|\leq c^{\prime}t}u(t,x)\geq q_{c^{\prime}}.

Hence there exists T>0T>0 such that

inf|x|ctu(t,x)qc/2,tT.\inf_{|x|\leq c^{\prime}t}u(t,x)\geq q_{c^{\prime}}/2,\;\forall t\geq T.

This implies that for all n0n\geq 0 and tt\in\mathbb{R} such that t+tnTt+t_{n}\geq T one has

inf|x+xn|c(t+tn)u(t+tn,x+xn)qc/2.\inf_{|x+x_{n}|\leq c^{\prime}(t+t_{n})}u(t+t_{n},x+x_{n})\geq q_{c^{\prime}}/2.

Since one has |xn|ctn|x_{n}|\leq ct_{n} for all n0n\geq 0, this implies that for all n0n\geq 0 and tt\in\mathbb{R} with t+tnTt+t_{n}\geq T:

inf|x|(cc)tn+ctu(t+tn,x+xn)qc/2.\inf_{|x|\leq(c^{\prime}-c)t_{n}+c^{\prime}t}u(t+t_{n},x+x_{n})\geq q_{c^{\prime}}/2.

Finally since c>cc^{\prime}>c and tnt_{n}\to\infty as nn\to\infty, then one has u(t,x)qc/2>0u_{\infty}(t,x)\geq q_{c^{\prime}}/2>0 for all (t,x)2(t,x)\in\mathbb{R}^{2}.

Next, we consider U=U(t)U=U(t) with U(0)=qc/2>0U(0)=q_{c^{\prime}}/2>0 the solution of the ODE

U(t)=U(t)(mbU(t)),t0.U^{\prime}(t)=U(t)\left(m-bU(t)\right),\forall t\geq 0.

Since u(s,x)qc/2u_{\infty}(s,x)\geq q_{c^{\prime}}/2 for all (s,x)2(s,x)\in\mathbb{R}^{2}, then comparison principle implies that

u(t+s,x)U(t),t0,s,x.u_{\infty}(t+s,x)\geq U(t),\forall t\geq 0,s\in\mathbb{R},x\in\mathbb{R}.

So that

u(0,0)U(t),t0.u_{\infty}(0,0)\geq U(t),\forall t\geq 0.

On the other hand, since U(0)>0U(0)>0, one gets U(t)mbU(t)\to\frac{m}{b} as tt\to\infty. Hence this yields u(0,0)mbu_{\infty}(0,0)\geq\frac{m}{b}, a contradiction with u(0,0)<mbu_{\infty}(0,0)<\frac{m}{b}, which completes the proof.  

Now we apply the key lemma to prove our inner propagation result Theorem 1.12.

Proof of Theorem 1.12 (i). Here we assume that the initial data u0u_{0} has a fast decay rate and we aim at proving that

limtsupx[0,ct]|1u(t,x)|=0,c(0,cr).\lim\limits_{t\to\infty}\sup_{x\in[0,ct]}|1-u(t,x)|=0,\;\forall c\in(0,c_{r}^{*}).

One can construct a initial data v0v_{0} alike in Lemma 3.1, through choosing proper parameter and spatial shifting (see Remark 3.5) such that v0(x)u0(x)v_{0}(x)\leq u_{0}(x) for all xx\in\mathbb{R}. Let v(t,x)v(t,x) be the solution of (1.12) with initial data v0v_{0}. Lemma 3.1 ensures that v(t,x)v(t,x) is uniformly continuous for all t0,xt\geq 0,x\in\mathbb{R}. Since v0()u0()v_{0}(\cdot)\leq u_{0}(\cdot), then the comparison principle implies that v(t,x)u(t,x)v(t,x)\leq u(t,x) for all t0,xt\geq 0,x\in\mathbb{R}. Note that u(t,x)1u(t,x)\leq 1, it is sufficiently to prove that

limtinfx[0,ct]v(t,x)=1,c(0,cr).\lim_{t\to\infty}\inf_{x\in[0,ct]}v(t,x)=1,\;\forall c\in(0,c_{r}^{*}).

Firstly, let us prove that

lim inftinfx[0,ct]v(t,x)>0,c(0,cr).\liminf_{t\to\infty}\inf_{x\in[0,ct]}v(t,x)>0,\;\forall c\in(0,c_{r}^{*}).

To do this, for all B,R>0B,R>0, γ\gamma\in\mathbb{R}, we define cR,B(γ)c_{R,B}(\gamma) by

cR,B(γ):=2RπBBK(z)eγzsin(πz2R)dz.c_{R,B}(\gamma):=\frac{2R}{\pi}\int_{-B}^{B}K(z)\mathrm{e}^{\gamma z}\sin(\frac{\pi z}{2R})\mathrm{d}z. (3.16)

Note that γcR,B(γ)\gamma\mapsto c_{R,B}(\gamma) is continuous and recalling (1.10) one has

limγλrlimRBcR,B(γ)=cr.\lim\limits_{\gamma\to\lambda_{r}^{*}}\lim\limits_{\begin{subarray}{c}R\to\infty\\ B\to\infty\end{subarray}}c_{R,B}(\gamma)=c_{r}^{*}.

So for each c(c,cr)c^{\prime}\in(c,c_{r}^{*}), one can choose proper γ\gamma close to λr\lambda_{r}^{*} such that for R,B>0R,B>0 large enough,

ccR,B(γ).c^{\prime}\leq c_{R,B}(\gamma).

Then for all cc<k<1\frac{c}{c^{\prime}}<k<1,

ctkX(t):=cR,B(γ)t.\frac{ct}{k}\leq X(t):=c_{R,B}(\gamma)t.

Now, we apply Lemma 2.6 to show that

lim inft+inf0xkX(t)v(t,x)>0.\liminf_{t\to+\infty}\inf_{0\leq x\leq kX(t)}v(t,x)>0.

Note that tX(t)t\mapsto X(t) is continuous for t0t\geq 0, and Lemma 3.1 ensures that v=v(t,x)v=v(t,x) is uniformly continuous for all t0,xt\geq 0,x\in\mathbb{R}. We only need to check that v=v(t,x)v=v(t,x) satisfies the conditions (H1)(H3)(H1)-(H3) in Lemma 2.6.

To show (H1)(H1), recalling (1.5) and (1.6), one may observe that v=v(t,x)v=v(t,x) satisfies

tv(t,x)k(y)v(t,xy)dyK¯v(t,x)+v(t,x)(μ(t)Cv(t,x)).\partial_{t}v(t,x)\geq\int_{\mathbb{R}}k(y)v(t,x-y)\mathrm{d}y-\overline{K}v(t,x)+v(t,x)\left(\mu(t)-Cv(t,x)\right).

Recalling Assumption 1.5 (f4)(f4) and Lemma 1.2, there exists aW1,(0,)a\in W^{1,\infty}(0,\infty) such that μ(t)K¯+a(t)0\mu(t)-\overline{K}+a^{\prime}(t)\geq 0 for all t0t\geq 0. Setting w(t,x):=ea(t)v(t,x)w(t,x):=e^{a(t)}v(t,x) so that ww satisfies

tw(t,x)k(y)w(t,xy)dyk¯w(t,x)+w(t,x)(k¯+μ(t)K¯+a(t)Cea(t)w(t,x))k(y)w(t,xy)dyk¯w(t,x)+w(t,x)(mCeaw(t,x)),\begin{split}\partial_{t}w(t,x)&\geq\int_{\mathbb{R}}k(y)w(t,x-y)\mathrm{d}y-\bar{k}w(t,x)\\ &\quad+w(t,x)\left(\bar{k}+\mu(t)-\overline{K}+a^{\prime}(t)-Ce^{-a(t)}w(t,x)\right)\\ &\geq\int_{\mathbb{R}}k(y)w(t,x-y)\mathrm{d}y-\bar{k}w(t,x)+w(t,x)\left(m-Ce^{\|a\|_{\infty}}w(t,x)\right),\end{split}

where m:=inft0(k¯+μ(t)K¯+a(t))k¯>0m:=\inf\limits_{t\geq 0}\left(\bar{k}+\mu(t)-\overline{K}+a^{\prime}(t)\right)\geq\bar{k}>0. Now we consider w¯=w¯(t,x)\underline{w}=\underline{w}(t,x) the solution of following equation

tw¯(t,x)=kw¯(t,x)k¯w¯(t,x)+w¯(t,x)(mCeaw¯(t,x)).\partial_{t}\underline{w}(t,x)=k*\underline{w}(t,x)-\bar{k}\underline{w}(t,x)+\underline{w}(t,x)\left(m-Ce^{\|a\|_{\infty}}\underline{w}(t,x)\right). (3.17)

supplemented with the initial data w¯(0,x)=eav0(x)\underline{w}(0,x)=e^{-\|a\|_{\infty}}v_{0}(x). Thus note that one has w¯(0,x)w(0,x)\underline{w}(0,x)\leq w(0,x) for all xx\in\mathbb{R} and the comparison principle implies that

w(t,x)=ea(t)v(t,x)w¯(t,x),t0,x.w(t,x)=e^{a(t)}v(t,x)\geq\underline{w}(t,x),\;\;\forall t\geq 0,x\in\mathbb{R}.

Lemma 3.6 implies that there exists c~>0\tilde{c}>0 such that

limtsup|x|ct|w¯(t,x)mCea|=0,c(0,c~).\lim_{t\to\infty}\sup_{|x|\leq ct}\left|\underline{w}(t,x)-\frac{m}{Ce^{\|a\|_{\infty}}}\right|=0,\;\forall c\in(0,\tilde{c}). (3.18)

Since aW1,(0,)a\in W^{1,\infty}(0,\infty), we end-up with

lim inftv(t,0)limteaw¯(t,0)=mCe2a>0,\liminf\limits_{t\to\infty}v(t,0)\geq\lim\limits_{t\to\infty}e^{-\|a\|_{\infty}}\underline{w}(t,0)=\frac{m}{Ce^{2\|a\|_{\infty}}}>0,

and (H1)(H1) is fulfilled.

Next we verify assumption (H2)(H2). Recall that for all v~ω(v){0}\tilde{v}\in\omega(v)\setminus\{0\}, there exist sequences (tn)n(t_{n})_{n} with tnt_{n}\to\infty and (xn)n(x_{n})_{n} such that v~(t,x)=limnv(t+tn,x+xn)\tilde{v}(t,x)=\lim\limits_{n\to\infty}v(t+t_{n},x+x_{n}) where this limit holds locally uniformly for (t,x)2(t,x)\in\mathbb{R}^{2}. As in the proof of Claim 2.5, such a function v~\tilde{v} satisfies

tv~(t,x)k(y)v~(t,xy)dy+v~(t,x)(μ~(t)K¯Cv~(t,x)),(t,x)2,\partial_{t}\tilde{v}(t,x)\geq\int_{\mathbb{R}}k(y)\tilde{v}(t,x-y)\mathrm{d}y+\tilde{v}(t,x)(\tilde{\mu}(t)-\overline{K}-C\tilde{v}(t,x)),\;\forall(t,x)\in\mathbb{R}^{2},

where k(y)k(y) is defined in (1.5) and μ~=μ~(t)L()\tilde{\mu}=\tilde{\mu}(t)\in L^{\infty}(\mathbb{R}) is a weak star limit of some shifted function μ(tn+)\mu(t_{n}+\cdot). Similar to Definition 1.3 and Lemma 1.2, one can define the least mean of μ~\tilde{\mu} over \mathbb{R} as

μ~=limTinfs1T0Tμ~(t+s)dt.\lfloor\tilde{\mu}\rfloor=\lim_{T\to\infty}\inf_{s\in\mathbb{R}}\frac{1}{T}\int_{0}^{T}\tilde{\mu}(t+s)\mathrm{d}t.

Also, the least mean of μ~\tilde{\mu} satisfies

μ~=supaW1,()inft(a+μ~)(t).\lfloor\tilde{\mu}\rfloor=\sup_{a\in W^{1,\infty}(\mathbb{R})}\inf_{t\in\mathbb{R}}(a^{\prime}+\tilde{\mu})(t).

Assumption 1.5 (f4)(f4) implies that μ~K¯\lfloor\tilde{\mu}\rfloor\geq\overline{K} and the same argument as above yields

lim inftv~(t,0)mCe2b>0,\liminf\limits_{t\to\infty}\tilde{v}(t,0)\geq\frac{m}{Ce^{2\|b\|_{\infty}}}>0,

where bW1,()b\in W^{1,\infty}(\mathbb{R}) such that μ~(t)K¯+b(t)0\tilde{\mu}(t)-\overline{K}+b^{\prime}(t)\geq 0 for all tt\in\mathbb{R}. Hence the condition (H2)(H2) is satisfied.

Before proving (H3)(H3), we state a lemma related to a compactly supported sub-solution of (1.1). Since (1.12) is a special case of (1.1), one can construct the similar sub-solution of (1.12). The following lemma can be proved similarly to Lemma 6.1 in [16]. So that the proof is omitted.

Lemma 3.8

Let Assumption 1.3, 1.5 and 1.10 be satisfied. Let γ(0,λr)\gamma\in(0,\lambda_{r}^{*}) be given. Then there exist B0>0B_{0}>0 large enough and θ0>0\theta_{0}>0 such that for all B>B0B>B_{0} there exists R0=R0(B)>0R_{0}=R_{0}(B)>0 large enough enjoying the following properties: for all B>B0B>B_{0} and R>max(R0(B),B)R>\max(R_{0}(B),B), there exists some function aW1,(0,)a\in W^{1,\infty}(0,\infty) such that the function

uR,B(t,x)={ea(t)eγxcos(πx2R) if t0 and x[R,R],0 else,u_{R,B}(t,x)=\begin{cases}e^{a(t)}e^{-\gamma x}\cos(\frac{\pi x}{2R})&\text{ if $t\geq 0$ and $x\in[-R,R]$},\\ 0&\text{ else},\end{cases}

satisfies, for all θθ0\theta\leq\theta_{0}, for all x[R,R]x\in[-R,R] and for any t0t\geq 0,

tu(t,x)cR,B(γ)xu(t,x)K(xy)u(t,y)dy+(μ(t)θK¯)u(t,x).\partial_{t}u(t,x)-c_{R,B}(\gamma)\partial_{x}u(t,x)\leq\int_{\mathbb{R}}K(x-y)u(t,y)\mathrm{d}y+\left(\mu(t)-\theta-\overline{K}\right)u(t,x).

Herein the speed cR,B(γ)c_{R,B}(\gamma) is defined in (3.16). Furthermore, let

u¯(t,x):=ηuR,B(t,xX(t)),\underline{u}(t,x):=\eta u_{R,B}(t,x-X(t)),

where X(t)=cR,B(γ)tX(t)=c_{R,B}(\gamma)t and η>0\eta>0 small enough, then u¯(t,x)\underline{u}(t,x) is the sub-solution of (1.1).

Now with the help of Lemma 3.8 and the comparison principle, one can choose η>0\eta>0 small enough such that u¯(0,x)v0(x)\underline{u}(0,x)\leq v_{0}(x) and therefore one has

lim inftv(t,X(t))lim inftu¯(t,X(t))=lim inftηuR,B(t,0)>0,\liminf\limits_{t\to\infty}v(t,X(t))\geq\liminf\limits_{t\to\infty}\underline{u}(t,X(t))=\liminf\limits_{t\to\infty}\eta u_{R,B}(t,0)>0,

which ensures that (H3)(H3) is satisfied.

As a conclusion all the conditions of Lemma 2.6 are satisfied and this yields

lim inftinf0xkX(t)v(t,x)>0.\liminf\limits_{t\to\infty}\inf_{0\leq x\leq kX(t)}v(t,x)>0.

So that

lim inftinf0xctv(t,x)>0,c(0,cr).\liminf\limits_{t\to\infty}\inf_{0\leq x\leq ct}v(t,x)>0,\;\forall c\in(0,c_{r}^{*}). (3.19)

Finally, let us prove that

lim inftinf0xctv(t,x)=1,c(0,cr).\liminf\limits_{t\to\infty}\inf_{0\leq x\leq ct}v(t,x)=1,\;\forall c\in(0,c_{r}^{*}).

To do this, note that combining (3.18) and (3.19) yields

lim inftinfc1txctv(t,x)>0,0<c1<c~,c(0,cr).\liminf\limits_{t\to\infty}\inf_{-c_{1}t\leq x\leq ct}v(t,x)>0,\forall 0<c_{1}<\tilde{c},\forall c\in(0,c^{*}_{r}).

By the similar analysis to the proof of Lemma 3.6, one could show that the above limit is equal to 11. Hence the proof is completed.  

Next we prove Theorem 1.12 (ii)(ii). Firstly, we state a lemma about a sub-solution of (1.1). One can also construct the similar sub-solution for (1.12).

Lemma 3.9

Let Assumption 1.3, 1.5 and 1.10 be satisfied. For each given λ(0,λr)\lambda\in(0,\lambda_{r}^{*}), define that

φ(t,x)=eλ(x+a(t))eλa(t)+B0(t)+B1e(λ+h)x,t0,x,\varphi(t,x)=e^{-\lambda(x+a(t))}-e^{-\lambda a(t)+B_{0}(t)+B_{1}}e^{-(\lambda+h)x},\;\;t\geq 0,x\in\mathbb{R}, (3.20)

where a,B0W1,(0,)a,B_{0}\in W^{1,\infty}(0,\infty), B1>0B_{1}>0 and 0<h<min{λ,σ(K)λ}0<h<\min\left\{\lambda,\sigma(K)-\lambda\right\}. Then

ϕ¯(t,x):=max{0,φ(t,x0tcλ,a(s)ds)}\underline{\phi}(t,x):=\max\left\{0,\;\varphi\left(t,x-\int_{0}^{t}c_{\lambda,a}(s)\mathrm{d}s\right)\right\}

is the subsolution of (1.1).

Remark 3.10

Note that φ(t,x)\varphi(t,x) is positive when

x>B0(t)+B1h.x>\frac{\|B_{0}(t)\|_{\infty}+B_{1}}{h}.

We point out that this lemma can be proved similarly to [16, Theorem 2.9]. So we omit the proof.

Proof of Theorem 1.12(ii). As proof of Theorem 1.12 (i)(i), we can construct v0(x)v_{0}(x) alike in Lemma 3.4, through choosing proper parameter and spatial shifting (see Remark 3.5) such that v0(x)u0(x)v_{0}(x)\leq u_{0}(x) for all xx\in\mathbb{R}. Let v(t,x)v(t,x) be the solution of (1.12) equipped with initial data v0v_{0}. Lemma 3.4 ensures that v(t,x)v(t,x) is uniformly continuous for all t0,xt\geq 0,x\in\mathbb{R}.

Recalling (1.8) and (1.9), for each given λ(0,λr)\lambda\in(0,\lambda_{r}^{*}) and for all c<c<c(λ)c<c^{\prime}<\lfloor c(\lambda)\rfloor, one can choose a proper function aW1,(0,+)a\in W^{1,\infty}(0,+\infty) such that

c<cλ,a(t),t0.c^{\prime}<c_{\lambda,a}(t),\;\forall t\geq 0.

Then we define

X(t):=0tcλ,a(s)ds+P,X(t):=\int_{0}^{t}c_{\lambda,a}(s)\mathrm{d}s+P,

where P>B0(t)+B1h>0P>\frac{\|B_{0}(t)\|_{\infty}+B_{1}}{h}>0 and B0()B_{0}(\cdot), B1B_{1} and hh are given in Lemma 3.9. Note that for all cc<k<1\frac{c}{c^{\prime}}<k<1,

ctkX(t).ct\leq kX(t).

Next it is sufficiently to apply key Lemma 2.6 to show that

lim inftinf0xkX(t)v(t,x)>0.\liminf_{t\to\infty}\inf_{0\leq x\leq kX(t)}v(t,x)>0.

Note that for exponential decay initial data v0v_{0} on the right-hand side, that is x1x\gg 1, one can construct an initial data v¯0\underline{v}_{0} alike in Lemma 3.1 with compact support such that v¯0v0\underline{v}_{0}\leq v_{0}. Then comparison principle implies that (H1)(H1) and (H2)(H2) hold. To verify the condition (H3)(H3), by Lemma 3.9 and comparison principle, one has

lim inftv(t,X(t))lim inftϕ¯(t,X(t))=lim inftφ(t,P)>0.\liminf_{t\to\infty}v(t,X(t))\geq\liminf\limits_{t\to\infty}\underline{\phi}(t,X(t))=\liminf\limits_{t\to\infty}\varphi(t,P)>0.

So (H3)(H3) is satisfied. Hence the key Lemma 2.6 ensures that

lim inftinf0xkX(t)v(t,x)>0.\liminf_{t\to\infty}\inf_{0\leq x\leq kX(t)}v(t,x)>0.

Then one has

lim inftinf0xctv(t,x)>0,0<c<c(λ).\liminf_{t\to\infty}\inf_{0\leq x\leq ct}v(t,x)>0,\;\forall 0<c<\lfloor c(\lambda)\rfloor.

Similarly to the proof of Theorem 1.12 (i), one can show that

limtsupx[0,ct]|u(t,x)1|=0,0<c<c(λ).\lim_{t\to\infty}\sup_{x\in[0,ct]}|u(t,x)-1|=0,\;\forall 0<c<\lfloor c(\lambda)\rfloor.

The proof is completed.  

Finally, we prove Corollary 1.13.

Proof of Corollary 1.13. Recalling H>0H>0 given in Remark 1.7, let us consider

tv(t,x)=K(y)v(t,xy)dyK¯v(t,x)+μ(t)v(t,x)(1Hv(t,x)),t0,x.\partial_{t}v(t,x)=\int_{\mathbb{R}}K(y)v(t,x-y)\mathrm{d}y-\overline{K}v(t,x)+\mu(t)v(t,x)\left(1-Hv(t,x)\right),\;t\geq 0,x\in\mathbb{R}. (3.21)

By the same analysis, one can obtain that the similar result for (3.21) as in Theorem 1.12. For the reader convenience, we state it in the following.

Let v=v(t,x)v=v(t,x) be the solution of (3.21) equipped with a continuous initial data u0u_{0}, with 0u010\leq u_{0}\leq 1 and u00u_{0}\not\equiv 0. Then the following inner spreading occurs:

  1. (i)

    (fast exponential decay) If u0(x)=O(eλx)u_{0}(x)=O(e^{-\lambda x}) as xx\to\infty for some λλr\lambda\geq\lambda_{r}^{*} then one has

    limtsupx[0,ct]|v(t,x)1H|=0,c(0,cr);\color[rgb]{1,0,0}\lim_{t\to\infty}\sup_{x\in[0,ct]}\left|v(t,x)-\frac{1}{H}\right|=0,\;\forall c\in(0,c_{r}^{*});
  2. (ii)

    (slow exponential decay) If lim infxeλxu0(x)>0\displaystyle\liminf_{x\to\infty}e^{\lambda x}u_{0}(x)>0 for some λ(0,λr)\lambda\in(0,\lambda_{r}^{*}) then one has

    limtsupx[0,ct]|v(t,x)1H|=0,c(0,c(λ)).\color[rgb]{1,0,0}\lim_{t\to\infty}\sup_{x\in[0,ct]}\left|v(t,x)-\frac{1}{H}\right|=0,\;\forall c\in\left(0,\lfloor c(\lambda)\rfloor\right).

Denote that u(t,x)u(t,x) is a solution of (1.1) equipped with initial data u0u_{0}. Recall (1.6) so that v(t,x)v(t,x) is the sub-solution of (1.1). Then comparison principle implies that u(t,x)v(t,x)u(t,x)\geq v(t,x) for all t0,xt\geq 0,x\in\mathbb{R}. Hence the conclusion is proved.  

Acknowledgment: We are very grateful to two anonymous referees for their careful reading and helpful comments which led to improvements of our original manuscript. The second author Z. Jin would like to acknowledge the région Normandie for the financial support of his PhD thesis.

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