January 01, 1970
This paper investigates spreading speeds for nonautonomous Fisher–KPP equations with nonlocal diffusion. The dispersal kernel depends on time and may be asymmetric in space. We assume exponential boundedness of the kernel and the existence of uniform mean values. Under these assumptions, the rightward and leftward spreading speeds are proved to be given by the linearly determined formula. The proof is based on comparison arguments, a refined maximum principle and proper sub- and super-solutions. The result includes, in particular, periodic, almost periodic, and uniquely ergodic time dependence. And it allows kernels whose support does not contain the neighborhood of the origin.
Key words. spreading speed; time-dependent kernel; asymmetric kernel; uniform mean value.
2020 Mathematical Subject Classification. 35K57; 35B40; 35B65; 92D25.
In this work, we study the Cauchy problem \[\label{Pb} \begin{cases} \displaystyle \partial_t u(t,x) = \int_{\mathbb{R}} K(t, x-y) [u(t,y) -u(t,x) ] \mathrm{d}y + u(t,x) f(t,u(t,x)), & t>0, x\in\mathbb{R}, \\ \displaystyle u(0,x) =u_0(x), & x\in \mathbb{R}. \end{cases}\tag{1}\] Here \(u(t,x)\) is the density of a population. The initial function \(0\leq u_0 \leq 1\) is continuous, compactly supported, and not identically zero. The reaction term is of Fisher–KPP type. The kernel \(K(t,x-y)\) describes the probability of the species jump from location \(y\) to location \(x\) at time \(t\). Here, we do not require \(K\) to be symmetric in spatial variable. This equation can arise in various contexts, such as phase transitions and combustion. In particular, it plays an important role in ecological and biological modeling. For instance, (1 ) can describe the spatio-temporal evolution of an invading population into some empty environments.
Propagation phenomena in local diffusion equations have been studied extensively in the literature. For the classical local Fisher–KPP equation \[\label{kpp} \partial_t u(t,x)=\partial_{xx}u(t,x)+u(t,x)(1-u(t,x)), \qquad t>0,\;x\in\mathbb{R},\tag{2}\] the propagation dynamics are well understood. The works of Fisher [1], Kolmogorov–Petrovskii–Piskunov [2], and Aronson–Weinberger [3] show that the solution with compactly supported data invades the unstable state \(u=0\) with a definite speed. Moreover, this speed coincides with the minimal speed of traveling fronts. Later works developed this relation in a more abstract form, see, for example, [4]–[6].
Nonlocal diffusion is used when dispersal is not well described by Brownian motion. One replaces \(\partial_{xx}\) by \[\label{nonlocal} \phi \mapsto \int_{\mathbb{R}}K(x-y)[\phi(y)-\phi(x)]\mathrm{d}y .\tag{3}\] This describes the long distance disperse. It is suitable for situations where seeds, pollen, individuals, or pathogens may move a long distance in a short time. From the analytical perspective, this replacement is not harmless. The nonlocal diffusion equation does not usually have the smoothing effect as parabolic equations, and the semiflow is not compact in the usual spaces. For this reason, proofs for local diffusion equations cannot be copied directly. Traveling waves and spreading speeds for nonlocal Fisher–KPP equations have been studied in [7]–[13]. In particular, asymmetric kernels and the difference between the two directions were discussed in [12]. Very recently, Du et al. [13] provided a new approach to study this problem, which based on the asymptotic limit of the principal eigenvalue. This methodology may applicable to our problem if one has the property of principal eigenvalue of the nonautonomous operator.
In applications, dispersal and growth may change with season, resource level, temperature, or other environmental factors. To make the model more realistic, the heterogeneous media are taken into consideration. For periodic media, there are many results for both local and nonlocal equations; see [5], [14]–[18]. For more general time dependence, such as almost periodic or uniquely ergodic coefficients, the usual periodic arguments are not enough. Generalized transition waves were introduced in [19]–[21]. We also refer to [22]–[26] for related propagation results in time/space heterogeneous media. For nonlocal equations, results on transition waves and spreading speed can be found in [27]–[33].
Now we give a rough introduction of our main results. Due to the KPP reaction term, the spreading speed may be linearly determined. Since the kernel itself depends on time, the linear selected speed should not be determined by the fixed exponential moment. Ansatz \(e^{-\lambda (x- \int_{0}^{t} c(s) \mathrm{d}s)}\) into the linearized equation. The following quantity appears, \[\int_{\mathbb{R}}K(t,y)(e^{\lambda y}-1)\mathrm{d}y+r(t), \qquad r(t):=f(t,0).\] We assume that this quantity has a uniform mean value in time. This assumption is weaker than periodicity and still gives enough information for studying the spreading behaviors of solutions. It includes the usual periodic case and also covers almost periodic and uniquely ergodic examples. Because \(K(t,y)\) may be nonsymmetric, the invasion speed to the right and to the left may be different. We therefore use two speeds, denoted by \(c^*_+\) and \(c^*_-\). Roughly speaking, our result says that these speeds are given by \[c^*_{+}=\inf_{\lambda>0} \frac{ \left\langle \int_{\mathbb{R}}K(\cdot,y)(e^{\lambda y}-1)\mathrm{d}y+r(\cdot)\right\rangle}{\lambda}, \qquad c^*_{-}=\inf_{\lambda<0} \frac{ \left\langle \int_{\mathbb{R}}K(\cdot,y)(e^{\lambda y}-1)\mathrm{d}y+r(\cdot)\right\rangle}{-\lambda},\] where the precise admissible range of \(\lambda\) is given below. This is the same linear determinacy principle as in the standard KPP case, but the exponential moment is replaced by its time average.
We state more explicitly what is allowed here. The function \(K(t,y)\) is only measurable in \(t\). It is not assumed to be symmetric in \(y\). We also do not assume that the support of \(K(t,\cdot)\) contains a neighborhood of the origin. This last point is important in the proof, since the usual strong maximum principle is no longer available. To overcome this difficulty, we prove a refined maximum principle adapted to such kernels. The upper estimate follows from exponential super-solutions and the uniform mean value formula. The lower estimate is based on localized exponential sub-solutions whose centers move with suitable average speeds. Some ideas are close to those in [12], [31], [32], but the time-dependent and nonsymmetric kernel requires several modifications.
We first define the rightward and leftward spreading speeds.
Definition 1. Two constants \(c^*_+\) and \(c^*_-\) are called the spreading speeds of (1 ) in the positive and negative directions respectively
provided the following two statements hold:
(i)For any nonnegative initial function \(u_0 \in L^\infty(\mathbb{R})\cap C(\mathbb{R})\) with a compact support, \[\displaystyle{\lim_{t\rightarrow+\infty}}\sup \limits_{x\geq (c^*_++\epsilon)
t}u(t,x)=\displaystyle{\lim_{t\rightarrow+\infty}}\sup \limits_{x\leq (-c^*_--\epsilon )t}u(t,x)=0, ~~\text{ for all small } \epsilon>0;\] (ii) There is some \(L>0\) such that for any nonnegative initial
function \(u_0 \in L^\infty(\mathbb{R})\cap C(\mathbb{R})\), if \(u_0(x)>0\) on an interval longer than \(L\), then \[\label{eq:limitinf} \displaystyle{\lim_{t\rightarrow+\infty}}\sup \limits_{(-c^*_-+\epsilon)t \leq x\leq(c^*_+-\epsilon )t}|u(t,x)-1|=0, ~~\text{ for all small } \epsilon>0.\qquad{(1)}\]
Remark 1. The above definition implies that \(c^*_{+}+c^*_{-}\geq0\). When equality holds, the interval in (?? ) is empty for every sufficiently small \(\epsilon>0\); in this degenerate case we say that the spreading speed is zero.
Before stating some assumptions, we first introduce the concept of uniform mean value which plays an important role in overcoming the difficulties induced by time heterogeneities, see [24]–[26]. We emphasize that most time heterogeneities admit a mean value, for instance periodic, almost periodic and uniquely ergodic media. Now we give the definition of uniform mean value.
Definition 2. A function \(h\in L^\infty(0, \infty)\) is said to have a uniform mean value if the following limit exists, \[\langle h \rangle := \lim\limits_{T\to +\infty} \frac{1}{T} \int_{0}^{T} h(t+s) \mathrm{d}t, \text{ uniformly for } s\geq 0.\] Herein, the quantity \(\langle h \rangle\) is called the uniform mean value of \(h\).
There are some equivalent and useful characterizations for the function \(h\) possessing a uniform mean value. As proved in [24], the uniform mean value \(\langle h \rangle\) has the following alternative representations: \[\label{reform} \langle h\rangle = \sup_{a\in W^{1, \infty}(0,\infty) }\mathop{\mathrm{ess\,inf}}_{t\in(0, \infty)} \left( a' + h\right) (t)= \inf_{a\in W^{1, \infty}(0, \infty) }\mathop{\mathrm{ess\,sup}}_{t \in (0, \infty)} \left( a' + h\right) (t).\tag{4}\]
Assumption 1 (KPP conditions). The function \(f:[0,+\infty)\times\mathbb{R}\to\mathbb{R}\) satisfies:
The function \(f(t,\cdot)\) is nonincreasing and Lipschitz continuous uniformly with respect to \(t\geq 0\), and \(f(\cdot,u) \in L^\infty (0,\infty)\) for each given \(u\in \mathbb{R}\);
\(f(t,1)\equiv 0\) for all \(t\geq 0\) and \(h(u):= \inf_{t\geq 0} f(t,u) >0, \forall u \in[0, 1).\) Moreover, the function \(r(t):= f(t,0)\) has a uniform mean value, denoted by \(\langle r\rangle\).
It is easy to see that \(\inf_{t\geq 0} r(t)=h(0)>0\). A typical example of Assumption 1 is given by \(f(t,u)=r(t) (1-u)\).
Next we state assumptions of kernel function \(K\).
Assumption 2. The kernel function \(K(t,y): [0, +\infty) \times \mathbb{R}\to [0, +\infty)\) satisfies:
The function \(K\) is Lebesgue measurable and \(K(\cdot, y) \in L^\infty (0, \infty)\) for a.e. \(y \in \mathbb{R}\);
The mean value \(\langle K(\cdot, y) \rangle\) exists for a.e. \(y \in \mathbb{R}\);
Denote \[\label{def:I} \mathcal{I} := \left\{ \lambda \in \mathbb{R}: \;\; \int_{\mathbb{R}} \| K(\cdot, y) \|_{L^\infty (0, \infty)} e^{\lambda y}\mathrm{d}y < +\infty \right\}.\qquad{(2)}\] There exists \(\eta>0\) such that \((-\eta,\eta)\subset\mathcal{I}\). Moreover, for each \(\lambda \notin \mathcal{I}\), we impose \[\label{blue} \liminf_{T\to +\infty} \frac{1}{T}\int_{0}^{T} \int_{\mathbb{R}}K(t,y)e^{\lambda y} \mathrm{d}y \mathrm{d}t =+\infty.\qquad{(3)}\]
Remark 2. Note that the kernel is integrable uniformly in time since \(0 \in \mathcal{I}\). The assumption (K3)* implies that the kernel is exponentially bounded uniformly in time. It should be noted that, under certain situations, the endpoints of the interval \(\mathcal{I}\), \(\sup \mathcal{I}\) and \(\inf \mathcal{I}\), may be elements of \(\mathcal{I}\). In the sequel, we denote \[\label{i-pm} i_+:= \sup \mathcal{I} \text{ and } i_{-}:= \inf\mathcal{I}.\tag{5}\] *
Condition (?? ) is imposed only at points outside the set \(\mathcal{I}\). It excludes the possibility that the averaged exponential moment remains finite at a boundary point not belonging to \(\mathcal{I}\), and it is used to handle the endpoint case in the variational formula of the spreading speed.
Define a map \(H(\lambda) : (i_-, i_+) \to \mathbb{R}\) as \[\label{H} H(\lambda):= \left \langle \int_{\mathbb{R}} K(\cdot,y) [e^{\lambda y} -1] \mathrm{d}y + r(\cdot) \right \rangle,\tag{6}\] and two constants \[\label{c42} c^*_{+} := \inf_{\lambda\in(0, i_+) } \frac{H(\lambda) }{\lambda} \;\; \text{ and } \;\; c^*_{-}:= \inf_{\lambda\in(i_{-}, 0) } \frac{ H(\lambda)}{-\lambda}.\tag{7}\]
Remark 3. Recalling Definition 2 and Assumption 2, the Fubini-Tonelli theorem and the Lebesgue dominated convergence theorem imply that for each \(\lambda \in (i_-, i_+)\), \[\label{Hchange} H(\lambda) = \left\langle \int_{\mathbb{R}} K(\cdot,y) [e^{\lambda y} -1] \mathrm{d}y \right \rangle+\left\langle r(\cdot) \right \rangle = \int_{\mathbb{R}} \left\langle K(\cdot,y) \right \rangle [e^{\lambda y} -1] \mathrm{d}y+\left\langle r(\cdot) \right \rangle.\qquad{(4)}\] Note that \[\lim_{\lambda \to 0^{\pm} } \frac{H(\lambda)}{\pm \lambda}= +\infty.\] Thus, \(c^*_{\pm}\) are well defined. The properties of \(H(\lambda)\) and \(c^*_\pm\) will be shown in Section 2.2.
Now we state the spreading speed result of (1 ).
Theorem 4. Let Assumptions 1 and 2 be satisfied. Then the spreading speeds of (1 ) are given by \(c^*_+\) and \(c^*_-\), which are defined in (7 ). Moreover, \(\text{sgn}(c^*_-+c^*_+)=\text{sgn} \left \langle\int_{\mathbb{R}} K(\cdot, y)\mathrm{d}y \right \rangle\).
The rest of this paper is organized as follows. In Section 2 we establish comparison principles, well-posedness, and elementary properties of the speed functions. Section 3 proves the upper and lower spreading estimates and hence Theorem 4. Section 4 records several examples and special cases illustrating the role of asymmetric kernels.
This section collects several facts used later: well-posedness of (1 ), comparison principles, some properties of \(H(\lambda)\), and the relation between the signs of \(c^*_\pm\) and positivity of solutions.
Throughout this subsection we only assume \(\overline{K}_0(\cdot):=\int_{\mathbb{R}}K(\cdot,y)\mathrm{d}y\in L^\infty_{loc}([0,+\infty))\). We prove well-posedness by an iteration argument. This avoids any continuity assumption on \(K(t,y)\) in the time variable.
Proposition 5. There exists a unique solution to (1 ) satisfying \(u(t,x)\in[0,1]\) on \((0,+\infty)\times\mathbb{R}\). Moreover, \(u(t,\cdot)\) is continuous for all \(t>0\).
Proof. We use the integral formulation. Let \(L_f\) be the Lipschitz constant of \(f\) and \(\tau\leq1\) be small to be chosen later. For any \(t\in[0,\tau]\), set \[u_1(t,x)=u_0(x)+\int_0^t\int_{\mathbb{R}}K(s,y)(u_0(x-y)-u_0(x))dyds+\int_0^tu_0(x) f(s,u_0(x))ds\] and for any \(n\in \mathbb{Z}_+,\) \[\label{eq:induction} u_{n+1}(t,x)=u_0(x)+\int_0^t\int_{\mathbb{R}}K(s,y)\big(u_n(s,x-y)-u_n(s,x)\big)dyds+\int_0^tu_n(s,x) f(s,u_n(s,x))ds.\tag{8}\] Recall that \(u_0\) is continuous, compactly supported, and satisfies \(0\leq u_0\leq1\). For any \((t,x)\in Q:=[0,\tau]\times\mathbb{R},\) \[|u_1(t,x)-u_0(x)|\leq2\int_0^{\tau}\int_{\mathbb{R}}K(s,y)dyds+\int_0^{\tau}u_0(x) f(s,u_0(x))ds\leq C\tau,\] where \(C=\big(2\|\overline{K}_0(\cdot)\|_{L^\infty([0,1])}+L_f\big)\) and \[\begin{align} &\quad |u_{n+1}(t,x)-u_n(t,x)|\\ &=\Big|\int_0^t\int_{\mathbb{R}}K(s,y)\big((u_n-u_{n-1})(s,x-y)-(u_n-u_{n-1})(s,x)\big)dyds+\int_0^tu_n f(s,u_n)-u_{n-1} f(s,u_{n-1})ds\Big|\\ &\leq\Big(2\int_0^{\tau}\int_{\mathbb{R}}K(s,y)dyds+L_f\tau\Big)\sup\limits_{(t,x)\in Q}|u_n(t,x)-u_{n-1}(t,x)|, \end{align}\] This gives \[\sup\limits_{(t,x)\in Q}|u_{n+1}(t,x)-u_n(t,x)|\leq C\tau\sup\limits_{(t,x)\in Q}|u_n(t,x)-u_{n-1}(t,x)|\leq(C\tau)^{n+1}\] Taking \(\tau<\min\{1/C,1\}\), we deduce that the sequence \(u_n(t,x)=u_0(x)+\sum_{i=1}^n\big(u_i(t,x)-u_{i-1}(t,x)\big)\) converges uniformly on \(Q\) to some limit function \(u(t,x)\). Letting \(n\to+\infty\) in (8 ), we have \[u(t,x)=u_0(x)+\int_0^t\int_{\mathbb{R}}K(s,y)\big(u(s,x-y)-u(s,x)\big)dyds+\int_0^tu(s,x) f(s,u(s,x))ds.\] Hence \(u(t,x)\) satisfies (1 ) on \(Q\). Similarly, we can find the solution \(\tilde{u}\) of (1 ) on \([\tau/2,3\tau/2]\times\mathbb{R}\) with initial value \(\tilde{u}(\tau/2,x)=u(\tau/2,x)\). We claim that \(\tilde{u}(t,x)=u(t,x)\) on \(Q_1=[\tau/2,\tau]\times\mathbb{R}\). From the integral equation we obtain, for any \((t,x)\in Q_1\), \[\begin{align} |u(t,x)-v(t,x)| &=\Big|\int_{\tau/2}^t\int_{\mathbb{R}}K(s,y)\big((u-v)(s,x-y)-(u-v)(s,x)\big)dyds+\int_{\tau/2}^tuf(s,u)-vf(s,v)ds\Big|\\ &\leq C\tau/2\sup\limits_{(t,x)\in Q_1}|u(t,x)-v(t,x)|. \end{align}\] Hence \(\tilde{u}(t,x)=u(t,x)\) on \(Q_1=[\tau/2,\tau]\times\mathbb{R}\). Now we obtain a unique bounded solution to (1 ) on \([0,3\tau/2]\times\mathbb{R}\). An induction argument yields a unique bounded solution to (1 ) on \([0,1]\times\mathbb{R}\) and thus on \([0,+\infty)\times\mathbb{R}\).
It remains to prove that the solution \(u(t,x)\) is continuous. Let \(\omega_0(\cdot)\) be the modulus of continuity of \(u_0\). For any \(x, \xi\in\mathbb{R}\) and \(t\in[0,\tau]\), we have \[\begin{align} &\quad|u_1(t,x)-u_1(t,\xi)|\\ &\leq|u_0(x)-u_0(\xi)|+\Big|\int_0^tu_0(x)f(s,u_0(x))-u_0(\xi)f(s,u_0(\xi))ds\Big|\\ &\quad+\Big|\int_0^t\int_{\mathbb{R}}K(s,y)\big(u_0(x-y)-u_0(\xi-y)+u_0(\xi)-u_0(x)\big)dyds\Big|\\ &\leq(1+C\tau)\omega_0(|x-\xi|). \end{align}\] This shows that \(u_1(t,\cdot)\) is uniformly continuous on \(\mathbb{R}\) uniformly with respect to \(t\in[0,\tau]\) and we denote its modulus of continuity by \(\omega_1(\cdot)\). The same argument gives \(\omega_{n+1}(|x-\xi|)\leq\omega_0(|x-\xi|)+C\tau\omega_n(|x-\xi|)\). Since \(u_n\to u\) uniformly on \(Q\), \(u(t,\cdot)\) is continuous.
Next we prove maximum principles.
Lemma 1 (Maximum Principle). Assume that \(g(t,x,\eta)/\eta\) is bounded if \(\eta\) is bounded and that \(v(t,x)\) is bounded and satisfies \[\begin{cases} \displaystyle \partial_t v(t,x) \geq \int_{\mathbb{R}} K(t, x-y) [v(t,y) -v(t,x) ] \mathrm{d}y + g(t,x,v(t,x)), & x\in\mathbb{R}, 0<t<T, \\ \displaystyle v(0,x) \geq0, & x\in \mathbb{R}. \end{cases}\] Then \(v(t,x)\geq 0\) for \((t,x)\in(0,T)\times\mathbb{R}\).
Proof. Setting \(V(t,x)=v(t,x)p(t,x)\) with \(p(t,x):=\text{exp}{(\int^t_0\overline{K}_0(s)-g(s,x,v(s,x))/v(s,x)ds)}\), we have \[\label{eq:V} \left\{\begin{array}{ll} \smallskip \partial_t V\ge \int_\mathbb{R}K(t,y)\frac{p(t,x)}{p(t,x-y)} V(t, x-y)dy, &(t, x)\in(0, T)\times\mathbb{R},\\ V(0, x)\ge0,&x\in\mathbb{R}. \end{array}\right.\tag{9}\] For any \(\phi(t,x)\) and \(k\in\mathbb{Z}_+\), we introduce the following notation: \[[R_k\phi]:=\int_0^{t_0}\cdots\int_0^{t_{k-1}}\int_{\mathbb{R}^k}\prod\limits_{j=1}^k\left[K(t_j,y_j)\frac{p(t_j,x-\displaystyle\sum_{i=0}^{j-1}y_i)}{p(t_j,x-\displaystyle\sum_{i=1}^{j}y_i)}\right]\phi(t_k, x-\displaystyle\sum_{i=1}^ky_i)dy_k\cdots dy_1dt_k\cdots dt_1,\] with convention that \(t_0=t\) and \(\displaystyle\sum_{i=0}^{j-1}y_i=0\) if \(j=1\). We have \(|[R_kV](t,x)|\le\frac{(tP_0)^k}{k!}\sup\limits_{s\in[0,T],x\in\mathbb{R}}|V(s,x)|\), where \(P_0:=\sup_{s\in[0,T],x,y\in\mathbb{R}}\frac{p(s,x)}{p(s,x-y)}\,\|\overline{K}_0(\cdot)\|_{L^\infty([0,T])}\). Integrating (9 ) from 0 to \(t\), we have \[\label{eq:int32V} V(t,x)\ge V(0,x)+ [R_1V](t,x),\;\forall (t, x)\in(0, T)\times\mathbb{R},\tag{10}\] This gives, for any \((t_1,x)\in(0,T)\times\mathbb{R}\) and \(y\in\mathbb{R}\), \[\begin{align} \label{eq:induction321} V(t_1,x-y_1)&\ge V(0,x-y_1)\nonumber\\ &\quad+\int_0^{t_1}\int_\mathbb{R}K(t_2,y_2)\frac{p(t_2,x-y_1)}{p(t_2,x-\displaystyle\sum_{i=1}^2y_i)} V(t_2, x-\displaystyle\sum_{i=1}^2y_i)dy_2dt_2. \end{align}\tag{11}\] Substituting this into (10 ), we have \[\begin{align} V(t,x)&\ge V(0,x)+ [R_1V(0,\cdot)](t,x) + [R_2V](t,x), \end{align}\] Replacing \(t_1, y_1\) and \(y_2\) by \(t_2\), \(\displaystyle\sum_{i=1}^2y_i\) and \(y_3\), respectively, in (11 ), and then plugging it into \([R_2V](t,x)\), we have \[\begin{align} V(t,x)&\ge V(0,x)+ [R_1V(0,\cdot)](t,x)+[R_2V(0,\cdot)](t,x)+[R_3V](t,x). \end{align}\] By induction, we have, for any \(k\in\mathbb{Z}_+\), \[V(t,x)\ge V(0,x)+\displaystyle\sum_{i=1}^k[R_iV(0,\cdot)](t,x)+[R_{k+1}V](t,x).\] Note that \([R_iV(0,\cdot)](t,x)\ge0\) and \(|[R_kV](t,x)|\le\frac{(tP_0)^k}{k!}\sup\limits_{s\in[0,T],x\in\mathbb{R}}|V(s,x)|\to0\) as \(k\to\infty\). Hence \(V(t,x)\ge0\) for all \((t, x)\in(0, T)\times\mathbb{R}\), and thus \(v\geq0\).
The maximum principle also gives the bound \(0\leq u(t,x)\leq1\) for bounded solutions of (1 ). Indeed, applying the lemma to \(v=1-u\) gives \(v\geq0\). We shall also need a comparison result on a moving spatial interval.
Lemma 2 (Maximum principle on a domain). Let \(T>0\) be given. Assume that \(a= a(t,x) \in L^\infty \big( (0, T) \times\mathbb{R}\big)\). Assume that \(X_1(t)\) and \(X_2(t)\) are two continuous functions defined in \([ 0, T]\) and \(X_1(t) < X_2(t)\) for all \(t\in (0, T]\). If \(u\) satisfies \[\begin{cases} \displaystyle \partial_t u(t,x) \geq \int_{\mathbb{R}} K(t, x-y) \left [ u(t, y) -u(t, x) \right] \mathrm{d}y+ a(t,x) u(t,x), \quad \forall t\in (0, T], \forall x \in \left(X_1(t), X_2(t) \right), \\ u(0, x) \geq 0,\forall x\in [X_1(0), X_2(0)],\\ u(t, x) \geq 0,~ \forall t\in (0, T], \forall x\in \mathbb{R}\setminus \left(X_1(t), X_2(t) \right), \end{cases}\] then \(u(t, x) \geq 0\) for all \(t\in [0, T]\) and \(x\in \left[X_1(t), X_2(t) \right]\).
Proof. Denote \[\Omega_T:=\{(t,x):0<t\leq T,\;X_1(t)<x<X_2(t)\}.\] Choose \(\mu>0\) so large that \[b(t,x):=\mu-\overline{K}_0(t)+a(t,x)\geq b_0>0,\qquad (t,x)\in\Omega_T,\] where \(\overline{K}_0 (t)=\int_\mathbb{R}K(t,y)\mathrm{d}y\). Set \(v(t,x)=e^{\mu t}u(t,x)\). Then \[v_t(t,x)\geq \int_\mathbb{R}K(t,x-y)v(t, y)\mathrm{d}y+b(t,x)v(t,x),\qquad (t,x)\in\Omega_T.\] For \(\delta>0\) and \(\alpha>0\), let \[v_\delta(t,x):=v(t,x)+\delta e^{\alpha t}.\] Choose \(\alpha>\sup_{t\in[0,T]}\overline{K}_0 (t)+\|b\|_{L^\infty(\Omega_T)}\). Then, one has \[(v_\delta)_t-\int_\mathbb{R}K(t,y)v_\delta(t,x-y)\mathrm{d}y-b(t,x)v_\delta(t,x) \geq \delta e^{\alpha t}\big(\alpha-\overline{K}(t)-b(t,x)\big)>0,~~ (t,x )\in \Omega_T.\] Moreover, \(v_\delta>0\) at \(t=0\) and on the parabolic boundary \(x\notin (X_1(t),X_2(t))\). If \(v_\delta\) became nonpositive somewhere in \(\Omega_T\), let \(t_0\) be the first time at which this happens and let \(x_0\in [X_1(t_0),X_2(t_0)]\) be a contact point. Since the boundary values are positive, \(x_0\) is an interior point. At \((t_0,x_0)\), we have \(v_\delta=0\), \(v_\delta(t_0,\cdot)\geq0\), and \((v_\delta)_t(t_0,x_0)\leq0\). The strict differential inequality above gives instead \[(v_\delta)_t(t_0,x_0)>\int_\mathbb{R}K(t_0,y)v_\delta(t_0,x_0-y)\mathrm{d}y\geq0,\] which is impossible. Hence \(v_\delta\geq0\) in \(\Omega_T\). Letting \(\delta\to0^+\) yields \(v\geq0\), and therefore \(u\geq0\), in \(\Omega_T\).
Lemma 3 (Refined maximum principle). Assume that the hypotheses of Lemma 1 hold. Suppose that there exist an interval \([a,b]\subset(0,\infty)\) and a number \(T_0>0\) such that \[\int_0^{T_0}\int_a^b K(t,y)\mathrm{d}y\mathrm{d}t>0.\] Then there exist \(R\in\mathbb{R}\), \(L>0\), and \(T>0\) such that, if \(v_0>0\) on an interval of length larger than \(L\), then \(v(t,x)>0\) for all \(t\geq T\) and \(x\geq R\). If instead \([a,b]\subset(-\infty,0)\) and the same integral positivity holds, then the analogous conclusion holds on a left half-line.
Proof. We give the proof for \([a,b]\subset(0,\infty)\). The proof for the negative direction is identical after mapping \(x\mapsto -x\). By Lemma 1, \(v\geq0\). Choose \(M>0\) so large that \(U(t,x):=e^{Mt}v(t,x)\) satisfies \[U(t,x)\geq U(0,x)+\int_0^t\int_\mathbb{R}K(s,y)U(s,x-y)\mathrm{d}y\mathrm{d}s,\qquad t\geq0.\] Assume, after a translation, that \(U(0,x)>0\) on \([R,R+l]\), where \(l\) is larger than \(a\) and \(b-a\). Let \[A(t):=\int_a^b K(t,y)\mathrm{d}y.\] By assumption, \(\int_0^{T_0}A(t)\mathrm{d}t>0\). Iterating the integral inequality gives the usual Duhamel series lower bound. In particular, for each \(n\geq1\), the \(n\)-fold term is strictly positive at time \(T_0\) for every \[x\in [R+na,\;R+ l+nb],\] because one may restrict all jumps to \([a,b]\) and the time integrals to the set on which \(A(t)>0\). Since \(l\) is chosen so that the intervals \([R+na,R+ l+nb]\) overlap and cover a right half-line, there is \(R_1\) such that \(U(T_0,x)>0\) for all \(x\geq R_1\). The first part of the proof of Lemma 1 then implies \(v(t,x)>0\) for all \(t\geq T_0\) and \(x\geq R_1\).
Remark 6. The previous lemma only uses positivity of the kernel on a set with positive measure in time and space. If this condition holds in both spatial directions, then the same argument gives positivity on the whole line after some time. In particular, if the support of the kernel contains a neighborhood of the origin, the usual hair-trigger property follows.
We show some elementary properties of the variational formula of linearly determined speed. We start with the speed in the positive direction. Define \[\label{cl} c(\lambda):= \frac{H(\lambda)}{\lambda}= \frac{ \left \langle \int_{\mathbb{R}} K(\cdot,y) [e^{\lambda y} -1] \mathrm{d}y + r(\cdot) \right \rangle}{\lambda}, \;\; \forall \lambda\in (0, i_+).\tag{12}\] Indeed, we have \[c(\lambda)= \frac{ \int_{\mathbb{R}}\left \langle K(\cdot,y) \right \rangle [e^{\lambda y} -1] \mathrm{d}y +\left \langle r(\cdot) \right \rangle}{\lambda}, \;\; \forall \lambda\in (0, i_+).\] Define \[\label{c4243} c^*_+ := \inf_{\lambda \in (0, i_+)} c(\lambda),\tag{13}\] and \[\label{Def-lam42} \lambda^*_+ : = \sup \left\{ k\in (0, i_+]: \;\; \lambda\mapsto c(\lambda) \text{ is decreasing on } (0, k) \right\}.\tag{14}\]
Proposition 7. Let Assumptions 1 and 2 hold. Let \(c(\lambda)\), \(c^*_+\) and \(\lambda^*_+\) be defined in (12 )-(14 ). Assume that \(\langle K(\cdot, y) \rangle\) has a support with positive measure. Then \(\lambda^*_+\) is well defined, \(c'(\lambda ) <0\) for all \(\lambda \in (0, \lambda^*_+)\) and \(\lambda^*_+\) is the unique root such that \(c(\lambda^*_+) = c^*_+\) if \(\lambda^*_+< +\infty\). In particular,
If \(\lambda^*_+ <i_+\), then \(c'(\lambda^*_+) = 0\).
If \(\lambda^*_+ = i_+ < +\infty\), then \(c^*_+ = c(i_+) :=\lim_{\lambda \to (i_+)^-}c(\lambda)\).
If \(\lambda^*_+= +\infty\), then \(c^*_+ = c(+\infty)= 0\).
Proof. Note that \(c(\lambda)\) is continuous on \((0, i_+)\) and \[c(\lambda) \to +\infty \text{ as } \lambda \to 0^+.\] Thus \(\lambda_+^*\) is well defined. Using (?? ) and (12 ), a direct computation gives \[c'(\lambda) =\frac{\lambda H'(\lambda) - H(\lambda)}{\lambda^2}= \frac{\lambda \int_{\mathbb{R}} \langle K(\cdot, y) \rangle e^{\lambda y} y \mathrm{d}y- \int_{\mathbb{R}} \langle K(\cdot,y) \rangle [e^{\lambda y} -1] \mathrm{d}y - \langle r \rangle }{\lambda^2}.\] By the definition of \(\lambda_+^*\), we have \(c'(\lambda_0)\le 0\) for all \(\lambda_0 \in (0, \lambda_+^*)\). We claim that the inequality is strict. Set \[g(\lambda):= \lambda H'(\lambda) - H(\lambda).\] Note that \(g(0)= -H(0)=- \langle r \rangle <0\). Since \(\langle K(\cdot, y) \rangle\) has a support with positive measure, we have \[g'(\lambda) = \lambda H''(\lambda)= \lambda \int_{\mathbb{R}} \langle K(\cdot, y) \rangle e^{\lambda y} y^2 \mathrm{d}y >0, \;\; \forall \lambda \in (0, i_+).\] Note that \(g(\lambda_0) = \lambda_0^2 c'(\lambda_0) \leq 0\). If \(g(\lambda_0) =0\) for some \(\lambda_0 \in (0, \lambda_+^*)\), then \(g(\lambda)>0\) for all \(\lambda \in ( \lambda_0, \lambda_+^*)\). Consequently \(c'(\lambda) >0\) for all \(\lambda \in ( \lambda_0, \lambda_+^*)\), which contradicts the fact that \(c(\lambda)\) is decreasing on \((0, \lambda_+^*)\). Therefore \(g(\lambda_0) <0\) for all \(\lambda \in (0, \lambda_+^*)\). Thus, \(c' (\lambda) <0\) for all \(\lambda \in (0, \lambda^*_+)\) and \(c'(\lambda^*_+)\leq 0\).
(i) If \(\lambda^*_+ <i_+\), then for some \(\delta>0\), \(c(\lambda)\) is nondecreasing on \((\lambda_{+}^*, \lambda^*_+ +\delta)\). From the property of \(g(\lambda)\), we have \(c'(\lambda^*_+)=0\) and \(\lambda^*_{+}\) is the unique point such that \(c(\lambda^*_+) = \inf_{\lambda \in (0, i_+)} c(\lambda)\).
(ii) If \(\lambda^*_{+} = i_+< +\infty\), then \(c(\lambda)\) is decreasing on \((0, i_+)\) and \(\inf_{\lambda \in (0, i_+) } c(\lambda) = \lim_{\lambda \to (i_+)^-}c(\lambda)\).
(iii) For \(\lambda_{+}^* = +\infty\), we claim that \(K(\cdot, y) =0\) for a.e. \(y>0\). Otherwise, we have \[\lim_{\lambda \to +\infty} \int_{\mathbb{R}}\left \langle K(\cdot,y) \right \rangle e^{\lambda y} \mathrm{d}y = +\infty.\] Thus, \(c(\lambda) \to +\infty\) as \(\lambda \to +\infty\). This gives \(\lambda^*_+ <+\infty\). This is a contradiction. This proves the claim. Then we have \(c^*_+= \lim_{\lambda\to +\infty} c(\lambda) =0\). The proof is complete.
For the negative direction we use the change of variables \(v(t,x)=u(t,-x)\). Then \[\partial_t v(t,x) = \int_{\mathbb{R}} K(t,y) [v(t,x+y) - v(t,x)] \mathrm{d}y + v(t,x) f(t, v(t,x)).\] The corresponding speed function is \[c(\lambda) = \frac{H(\lambda)}{- \lambda}, \;\; \forall \lambda \in (i_-, 0), \text{ and } c^*_-:= \inf_{\lambda \in (i_-, 0) } c(\lambda).\] and \[\label{Def-lam42-l} \lambda^*_-: = \inf \left\{ k\in [i_-, 0 ): \;\; \lambda\mapsto c(\lambda) \text{ is increasing on } (k, 0) \right\}.\tag{15}\]
The following analogue of Proposition 7 follows in the same way.
Proposition 8. Let Assumptions 1 and 2 hold. Assume that \(\langle K(\cdot, y) \rangle\) has a support with positive measure. Then \(\lambda^*_-\) is well defined, \(c'(\lambda ) >0\) for all \(\lambda \in ( \lambda^*_-, 0)\) and \(\lambda^*_-\) is the unique root such that \(c(\lambda^*_-) = c^*_-\) if \(\lambda^*_->- \infty\). In particular,
If \(\lambda^*_- >i_-\), then \(c'(\lambda^*_-) = 0\).
If \(\lambda^*_- = i_- >-\infty\), then \(c^*_- = c(i_-)\).
If \(\lambda^*_-= -\infty\), then \(c^*_- = c(-\infty)= 0\).
Proposition 9. We have \(\text{sgn}(c^*_-+c^*_+)=\text{sgn}\langle \int_{\mathbb{R}}K(\cdot, y) \mathrm{d}y\rangle=\text{sgn}\int_\mathbb{R}\langle K(\cdot, y) \rangle \mathrm{d}y.\)
Proof. First, \(\langle \int_{\mathbb{R}}K(\cdot, y) \mathrm{d}y\rangle= \int_\mathbb{R}\langle K(\cdot, y) \rangle \mathrm{d}y\) and \(\int_\mathbb{R}\langle K(\cdot, y) \rangle dy\geq0\). If \(\int_\mathbb{R}\langle K(\cdot, y) \rangle dy=0\), then \(\langle K(\cdot, y) \rangle =0\) almost everywhere. Hence \(H(\lambda)\equiv\langle r\rangle\) yield \(i_{\pm}= \pm \infty\). Hence \(c^*_-=c^*_+=0\).
We only prove the case \(\mathcal{I}=(-\infty, +\infty)\) Suppose now \(\int_\mathbb{R}\langle K(\cdot, y) \rangle dy>0\). Then either \(\int_0^{+\infty}\langle K(\cdot, y) \rangle dy>0\) or \(\int_{-\infty}^0\langle K(\cdot, y) \rangle dy>0.\) We prove \(c^*_-+c^*_+>0\) under the condition \(\int_0^{+\infty}\langle K(\cdot, y) \rangle dy>0\).
Choose \(y_0>0\) such that \(\int_{y_0}^\infty\langle K(\cdot, y) \rangle dy>0\) if \(\int_0^{+\infty}\langle K(\cdot, y) \rangle dy>0\). Hence \[\begin{align} \frac{H(\lambda)}{\lambda} &=\frac{\int_{\mathbb{R}} \langle K(\cdot, y) \rangle (e^{\lambda y}-1 )\mathrm{d}y +\langle r \rangle}{\lambda}\\ &\geq\frac{\int_{y_0}^{+\infty} \langle K(\cdot, y) \rangle e^{\lambda y}\mathrm{d}y-\int_{\mathbb{R}} \langle K(\cdot, y) \rangle \mathrm{d}y +\langle r \rangle}{\lambda}\\ &\geq\frac{\int_{y_0}^{+\infty} \langle K(\cdot, y) \rangle e^{\lambda {y_0}}\mathrm{d}y-\int_{\mathbb{R}} \langle K(\cdot, y) \rangle \mathrm{d}y +\langle r \rangle}{\lambda}\to+\infty\\ \end{align}\] as \(\lambda\to +\infty\). It follows from this and \(\lim\limits_{\lambda\to0^+}\frac{H(\lambda)}{\lambda}=+\infty\) that there exists \(\lambda_+\in(0,+\infty)\) such that \(c^*_+=\frac{H(\lambda_+)}{\lambda_+}\). Moreover, there exists a sequence \(\{\lambda_n\}_{n}\subset(-\infty,0)\) with \(\lim\limits_{n\to\infty}\lambda_n\in[-\infty,0)\) such that \(c^*_-=\lim\limits_{n\to\infty}\frac{H(\lambda_{n})}{-\lambda_n}\). Hence \[\begin{align} c^*_-+c^*_+ &=\frac{H(\lambda_+)}{\lambda_+}+\lim\limits_{n\to\infty}\frac{H(\lambda_{n})}{-\lambda_n}\\ &=\lim\limits_{n\to\infty}\frac{\lambda_+-\lambda_n}{-\lambda_+\lambda_n}\big(\frac{-\lambda_nH(\lambda_+)}{\lambda_+-\lambda_n}+\frac{\lambda_+ H(\lambda_n)}{\lambda_+-\lambda_n}\big)\\ &\geq\lim\limits_{n\to\infty}\frac{\lambda_+-\lambda_n}{-\lambda_+\lambda_n}H(0)\\ &\geq\lim\limits_{n\to\infty}\frac{\lambda_+-\lambda_n}{-\lambda_+\lambda_n}\langle r\rangle>0.\\ \end{align}\] The first inequality follows from the convexity of \(H(\lambda)\).
The proof of \(c^*_-+c^*_+>0\) under the condition \(\int_{-\infty}^0\langle K(\cdot, y) \rangle dy>0\) is similar. We omit the details.
Proposition 9 shows that, if \(\int_\mathbb{R}\langle K(\cdot,y)\rangle\mathrm{d}y>0\), then at least one of \(c^*_+\) and \(c^*_-\) is positive. Hence one of the following alternatives occurs: \[c^*_->0,\;c^*_+\leq0;\qquad c^*_-\leq0,\;c^*_+>0;\qquad c^*_->0,\;c^*_+>0.\] The next proposition relates these alternatives to strict positivity of solutions.
Proposition 10. If \(c^*_+>0\) (resp. \(c^*_->0\)), then there exist \(R\in\mathbb{R}\), \(T>0\) and \(L>0\) such that if \(u_0(x)>0\) on an interval longer than \(L\), then \(u(t,x)>0\) on \((T,+\infty)\times[R,+\infty)\) (resp. \((T,+\infty)\times(-\infty,R]\)). Moreover, if both \(c^*_- \text{ and }c^*_+\) are positive, then there is \(L>0\) such that if \(u_0(x)>0\) on an interval longer than \(L\), then \(u(t,x)>0\) on \((T,+\infty)\times\mathbb{R}\).
Proof. It remains only to verify the assumptions of Lemma 3.
Claim : If \(c^*_+>0\) (resp. \(c^*_->0\)), then there exist \(T_0>0\) and an interval \([a,b]\in(0,+\infty)\) (resp. \((-\infty,0)\)) such that \[\mu \left\{ t\in [0, T_0]: \; \int_{a}^{b} K(t,y) \mathrm{d}y >0 \right \} >0,\] where \(\mu\) is the Lebesgue measure. Set \(\mathcal{T} :=\left\{ t\in [0, T_0]: \; \int_{a}^{b} K(t,y) \mathrm{d}y >0 \right \}\). Suppose not. Then for any \([a,b] \subset (0, \infty)\) and \(T>0\), we have \[\mu \left\{ t\in [0, T]: \; \int_{a}^{b} K(t,y) \mathrm{d}y >0 \right \} =0.\] This implies that for each given \([a, b] \subset (0, \infty)\), \[\int_{a}^{b} K(t, y) \mathrm{d}y = 0, \text{ for a.e. } t\in [0, T].\] Since \(K\geq0\), we see that \(K(t,y) = 0\) for a.e. \(t\in[0,T]\) and \(y\in [a,b]\). Since \(a, b>0\), we have \(i_+=+\infty\) and \(\int_{a}^{b} K(t,y) e^{\lambda y} \mathrm{d}y =0\) for all \(\lambda\in (0, +\infty)\) and for a.e. \(t\in [0, T]\). Fix \(\lambda>0\) and \(\epsilon>0\). Choose \(a>0\) sufficiently small and \(b>0\) sufficiently large such that \[\begin{align} \int_{0}^{a} K(t,y) \mathrm{d}y \leq \int_{0}^{a} \| K(\cdot, y)\|_{L^\infty(\mathbb{R})} e^{\lambda y} \mathrm{d}y \leq \frac{\varepsilon}{2}, \\ \int_{b}^{\infty} K(t,y) \mathrm{d}y \leq \int_{b}^{\infty} \| K(\cdot, y)\|_{L^\infty(\mathbb{R})} e^{\lambda y} \mathrm{d}y \leq \frac{\varepsilon}{2}. \end{align}\] Note that \(\int_{-\infty}^{0} K(t,y) e^{\lambda y} \mathrm{d}y \leq \int_{-\infty}^{0} K(t,y) \mathrm{d}y\) for all \(\lambda >0\). Hence, for \([a,b] \in (0, \infty)\), we have \[\begin{align} \int_{\mathbb{R}} K(t, y)e^{\lambda y} \mathrm{d}y & \leq \int_{-\infty}^{0} K(t,y) \mathrm{d}y + \int_{ [0, a] \cup [b, \infty)} K(t,y) e^{\lambda y} \mathrm{d}y \\ & \leq \int_{\mathbb{R}} K(t,y) \mathrm{d}y + \epsilon. \end{align}\] We observe that \[\lim_{T\to +\infty} \frac{1}{T} \int_{0}^{T} \int_{\mathbb{R}} K(t, y) e^{\lambda y} \mathrm{d}y \mathrm{d}t \leq \left \langle \int_{\mathbb{R}} K(t,y) \mathrm{d}y \right \rangle + \epsilon.\] It follows that \[\begin{align} c^*_+ = \inf_{\lambda >0} \frac{H(\lambda)}{\lambda} \leq \inf_{\lambda >0 } \frac{\langle r \rangle +\epsilon}{\lambda} =0. \end{align}\] This contradicts \(c^*_+>0\). This proves the claim. We note that the proof of this claim does not require the existence of the uniform mean value of \(t\mapsto K(t,y)\). It suffices that the uniform mean value \(\int_{\mathbb{R}} K(t, y)\mathrm{d}y\) exists.
It follows from Lebesgue density theorem that there exists an increasing sequence \(\{t_n\}_{n\geq 1}\subset\mathcal{T}\). Combining this with Lemma 3, we finish the proof.
We now prove Theorem 4.
We begin with the upper estimate.
Theorem 11. Let Assumptions 1 and 2 be satisfied. Let \(c^*_{\pm}\) be defined in (7 ). Then the solution \(u(t,x)\) of (1 ) satisfies \[\displaystyle{\lim_{t\rightarrow+\infty}}\sup \limits_{x\geq (c^*_++\epsilon) t}u(t,x)=\displaystyle{\lim_{t\rightarrow+\infty}}\sup \limits_{x\leq (-c^*_--\epsilon )t}u(t,x)=0,\forall \epsilon>0.\] In particular, \(c^*_-=c^*_+=0\) is the spreading speed of (1 ) if \(\int_{\mathbb{R}} \langle K(\cdot, y) \rangle \mathrm{d}y=0.\)
Proof. We only prove the right-hand estimate, while the left-hand estimate is analogous. Fix \(\epsilon>0\). By the definition of \(c^*_+\), choose \(\lambda\in(0,i_+)\) such that \[\frac{H(\lambda)}{\lambda}<c^*_++\frac{\epsilon}{2}.\] Set \[m_\lambda(t):=\int_\mathbb{R}K(t,y)(e^{\lambda y}-1)\mathrm{d}y+r(t).\] Since \(m_\lambda\) has the uniform mean value \(H(\lambda)\), we have \[\int_0^t m_\lambda(s)\mathrm{d}s=H(\lambda)t+o(t)\qquad \text{as }t\to\infty.\] For \(A>0\), define \[\overline{u}(t,x):=A\exp\left\{-\lambda x+\int_0^t m_\lambda(s)\mathrm{d}s\right\}.\] Then \[\partial_t\overline{u}-\int_\mathbb{R}K(t,y)[\overline{u}(t,x-y)-\overline{u}(t,x)]\mathrm{d}y-r(t)\overline{u}=0.\] Since \(f(t,u)\leq f(t,0)=r(t)\) for \(u\geq0\), \(\overline{u}\) is a supersolution. Taking \(A\) large enough so that \(\overline{u}(0,\cdot)\geq u_0\), the comparison principle yields \(u\leq\overline{u}\). Therefore, for \(x\geq(c^*_++\epsilon)t\), \[u(t,x)\leq A\exp\left\{-\lambda(c^*_++\epsilon)t+H(\lambda)t+o(t)\right\}\to 0.\] This proves \[\lim_{t\to\infty}\sup_{x\geq(c^*_++\epsilon)t}u(t,x)=0.\] For the left-hand estimate, choose \(\lambda\in(i_-,0)\) such that \(H(\lambda)/(-\lambda)<c^*_-+\epsilon/2\) and use the same exponential supersolution \(A\exp\{-\lambda x+\int_0^t m_\lambda(s)\mathrm{d}s\}\). If \(x\leq-(c^*_-+\epsilon)t\), the exponent is bounded above by \(-(-\lambda)\epsilon t/2+o(t)\), and hence the solution converges uniformly to zero in that region.
If \(\int_\mathbb{R}\langle K(\cdot,y)\rangle\mathrm{d}y=0\), then \(\langle K(\cdot,y)\rangle=0\) for a.e. \(y\), so \(H(\lambda)=\langle r\rangle\). Consequently \(c^*_+=c^*_-=0\).
We next construct the subsolutions used for the lower estimate. Recall the definition of \(c^*_+\) in (7 ) and Proposition 7. For each \(l, B >0\) and \(\lambda\in (i_-, i_+)\), we define \[\label{clB} c_{l,B} (\lambda) (t) = \frac{2l}{\pi} \int_{-B}^{B} K(t,y) e^{\lambda y} \sin \frac{\pi y}{2l} \mathrm{d}y.\tag{16}\] Set \[K_B(t,y)= \begin{cases} K(t,y), & y\in [-B, B], t\geq 0, \\ 0, & \text{ otherwise. } \end{cases}\] For each \(\lambda\in (i_-, i_+)\), we define \[\label{Xt} X(\lambda) (t) :=\int_{0}^{t} c_{l, B} (\lambda) (s) \mathrm{d}s= \frac{2l}{\pi} \int_{0}^{t} \int_{-B}^{B} K (s, y) e^{\lambda y} \sin \frac{\pi y}{2l} \mathrm{d}y \mathrm{d}s.\tag{17}\]
Lemma 4. Recall \(\lambda^*_+\) given in Proposition 7. Let \(0< \lambda_1 < \lambda_2 < \lambda^*_+ \leq +\infty\) and \(B_0>0\) large enough be given. Let \(\theta_0>0\) small enough be given. Then there exists \(a(t) \in W^{1, \infty} (0, \infty)\), independent of \(\lambda\), such that for all \(\lambda\in [\lambda_1, \lambda_2]\) and \(l>B>B_0\), the function \[\phi (t, x) := \phi_\lambda(t,x)= \begin{cases} e^{a(t)} e^{- \lambda x} \cos \frac{\pi x }{2l}, & x\in [-l, l], t>0, \\ 0, & \text{ otherwise}, \end{cases}\] satisfies the following differential inequality for \(\theta\in(0, \theta_0)\), \(x\in [-l, l]\) and \(t>0\), \[\partial_t \phi (t,x) - c_{l,B}(\lambda) (t) \partial_x \phi (t,x)\leq \int_{\mathbb{R}} K(t,x-y) [\phi (t,y) - \phi (t,x)] \mathrm{d}y + ( r(t)- \theta) \phi(t,x).\] Moreover, there exists \(\eta >0\) (independent of \(\lambda \in [\lambda_1, \lambda_2]\)) small enough such that \(\underline{u} (t,x) := \eta \phi_\lambda (t,x-X(\lambda)(t))\) is a subsolution of (1 ) for all \(\lambda \in [\lambda_1, \lambda_2]\).
Proof. Set \[G_\lambda(t):= \lambda\int_{\mathbb{R}}K(t,y)e^{\lambda y}y\mathrm{d}y -\int_{\mathbb{R}}K(t,y)(e^{\lambda y}-1)\mathrm{d}y-r(t).\] Then \[\langle G_\lambda\rangle = \lambda H'(\lambda)-H(\lambda) = \lambda^2c'(\lambda).\] Since \(0<\lambda_1<\lambda_2<\lambda_+^*\), Proposition 7 implies that \(c'(\lambda)<0\) on \([\lambda_1,\lambda_2]\). In particular, \[\langle G_{\lambda_2}\rangle<0.\] Choose \(\theta_0>0\) so small that \[\langle G_{\lambda_2}\rangle<-3\theta_0.\] By the reform formula of the uniform mean value (4 ), there exists \(a\in W^{1,\infty}(0,\infty)\) such that \[\operatorname*{ess\,sup}_{t\ge0} \big(a'(t)+G_{\lambda_2}(t)\big)<-2\theta_0.\] Moreover, for \(\lambda>0\), \[\partial_\lambda G_\lambda(t) = \lambda\int_{\mathbb{R}}K(t,y)e^{\lambda y}y^2\mathrm{d}y\ge0.\] Hence \(G_\lambda(t)\le G_{\lambda_2}(t)\) for all \(\lambda\in[\lambda_1,\lambda_2]\) and a.e. \(t\ge0\). Consequently, \[\operatorname*{ess\,sup}_{t\ge0} \big(a'(t)+G_\lambda(t)\big)<-2\theta_0, \qquad \lambda\in[\lambda_1,\lambda_2].\] Thus the same function \(a(t)\) works for all \(\lambda\in[\lambda_1,\lambda_2]\).
From (?? ), the definition of \(\mathcal{I}\), we have \[\begin{align} \lim_{l\to +\infty }\lim_{B\to +\infty}\frac{2l }{\pi } \int_{\mathbb{R}} K_B(t,y) e^{\lambda y} \sin \frac{\pi y}{2l} \mathrm{d}y = \int_{\mathbb{R}} K(t, y) e^{\lambda y} y \, \mathrm{d}y,~~ &\forall \lambda \in [\lambda_1, \lambda_2] \\ \lim_{l\to +\infty } \lim_{B\to +\infty} \int_{\mathbb{R}} K_B(t,y) e^{\lambda y} \cos \frac{\pi y}{2l} \mathrm{d}y = \int_{\mathbb{R}} K(t, y) e^{\lambda y} \mathrm{d}y, ~~&\forall \lambda \in [\lambda_1, \lambda_2] . \end{align}\] By Assumption 2, the above convergence is uniform in \(t\ge0\) and \(\lambda\in[\lambda_1,\lambda_2]\) after first taking \(B\) large and then taking \(l\) sufficiently large. Hence there exist \(B_0>0\) and, for each \(B>B_0\), a number \(l_0(B)>B\) such that, for all \(l>l_0(B)\) and \(\lambda\in[\lambda_1,\lambda_2]\), \[\label{Less0} a'(t)+\lambda c_{l,B}(\lambda)(t) -\int_{\mathbb{R}}K_B(t,y)e^{\lambda y}\cos\frac{\pi y}{2l}\mathrm{d}y +\int_{\mathbb{R}}K(t,y)\mathrm{d}y-r(t) <-\theta_0, \quad \forall t>0.\tag{18}\]
By the similar idea in [7], [33], we observe that \[\begin{align} \int_{-l}^{l} K(t,x-y) e^{a(t)} e^{-\lambda y} \cos \frac{\pi y}{2l} \mathrm{d}y &\geq \int_{-l}^{l} K_B(t, x-y) e^{a(t)} e^{-\lambda y} \cos \frac{\pi y}{2l} \mathrm{d}y \\ & \geq \int_{\mathbb{R}} K_B(t, x-y) e^{a(t)} e^{-\lambda y} \cos \frac{\pi y}{2l} \mathrm{d}y \\ &= \int_{\mathbb{R}} K_B(t, y) e^{a(t)} e^{-\lambda (x-y)} \cos \frac{\pi (x-y)}{2l} \mathrm{d}y. \end{align}\] The last inequality can be derived from \(\cos \frac{\pi y}{2l} <0\) for \(l<|y|<2l\) and \(K_B(t,x-y)=0\) for \(|y| >2l\) and \(x\in [-l, l]\). By direct computation, we have for \(x\in [-l, l]\) and \(t>0\), \[\begin{align} &\partial_t \phi (t,x)- c_{l,B} (\lambda) (t)\partial_x \phi (t,x)- \int_{\mathbb{R}} K(t, x- y) [\phi (t,y) - \phi(t,x)] \mathrm{d}y- r(t) \phi (t,x) \\ & \leq e^{a(t) }e^{-\lambda x} \cos \frac{\pi x}{2l} \bigg\{ a'(t) + \lambda c_{l,B} (\lambda)(t) -\! \int_{\mathbb{R}} K_B(t, y) e^{\lambda y} \cos \frac{\pi y}{2l} \mathrm{d}y + \int_{\mathbb{R}}K(t,y)\mathrm{d}y - r(t) \bigg\} \\ & \quad + e^{a(t)} e^{-\lambda x} \sin\frac{\pi x}{2l} \bigg\{ \frac{\pi }{2l} c_{l, B} (\lambda) (t) - \int_{\mathbb{R}} K_B(t, y) e^{\lambda y} \sin\frac{\pi y}{2l} \mathrm{d}y \bigg\} \end{align}\] Recall the definition of \(c_{l, B}(\lambda)(t)\) in (16 ) and the inequality (18 ). We see that \(\phi(t,x )\) satisfies for \(\theta \in (0, \theta_0)\), \(x\in [-l, l]\) and \(t>0\), \[\partial_t \phi (t,x)- c_{R,B} (\lambda) (t)\partial_x \phi (t,x)- \int_{\mathbb{R}} K(t, x- y) [\phi (t,y) - \phi(t,x)]\mathrm{d}y - (r(t) - \theta) \phi (t,x) <0.\]
Next since \(u\mapsto f(\cdot, u)\) is Lipschitz continuous, there exists a constant \(M>0\) such that \(f(t,u) \geq r(t) - Mu\). It suffices to choose \(\eta> 0\) small enough such that \(M\underline{u}(t,x) \leq \theta_0\) for all \(t\geq 0\) and \(x\in \mathbb{R}\), that is \[\eta M \sup_{ \lambda\in [\lambda_1, \lambda_2]} e^{\|a\|_\infty} e^{\lambda l} \leq \theta_0.\] We obtain that \(\underline{u}\) is a subsolution of (1 ). This completes the proof.
If \(c^*_+=c^*_-=0\), Theorem 11 gives zero spreading speed. We now assume \(c^*_++c^*_->0\). The following lemma gives a positive lower bound in intervals with speeds close to \(c^*_+\).
Lemma 5. Assume that \(c^*_+ +c^*_->0\). For all small \(\delta>0\), for all \(c^*_+-\delta <c_1^+<c_2^+ <c^*_+\), we have \[\label{lim-interval-right} \lim_{t\to + \infty} \inf_{x\in [c_1^+t , c_2^+t]} u(t,x) >0.\qquad{(5)}\]
Proof. Step 1: If \(\lambda^*_+= +\infty\), the speed \(c^*_+=0\). This is the trivial case.
Now we consider the case \(\lambda_+^*\in (0, i_+)\). We first show there exist \(\lambda_1, \lambda_2 \in (0, \lambda^*_+)\) such that \([c_1^+t , c_2^+t ] \subset [X(\lambda_1)(t), X(\lambda_2)(t)]\). Recalling (17 ), we have \(\lambda\mapsto X(\lambda)(\cdot)\) is continuous. For all \(l>B\), we have \[\frac{\mathrm{d}X(\lambda) (t)}{\mathrm{d}\lambda} = \frac{2l}{\pi} \int_{0}^{t} \int_{-B}^{B} K(s,y) e^{\lambda y} y \sin\frac{\pi y}{2l} \mathrm{d}y \mathrm{d}s > 0, \;\; \forall t>0, \lambda \in (0, i_+).\] Hence, the function \(\lambda \mapsto X(\lambda)\) is increasing. From (16 ), we see that \(\lambda \mapsto \langle c_{l, B}(\lambda) \rangle\) is continuous and increasing with \(\lambda\in (0, \lambda^*_+)\). Since \(c'(\lambda^*_+)=0\) when \(\lambda^*_+ < i_+\), we have \[c^*_+ = c(\lambda^*_+) = \int_{\mathbb{R}} \langle K(\cdot, y ) \rangle e^{\lambda^*_+ y} y \mathrm{d}y.\] We also have \(\langle c_{l, B} (\lambda^*_+)\rangle < c^*_+\) and \[\lim_{l \to + \infty } \lim_{ B\to + \infty } \langle c_{l, B} (\lambda^*_+) \rangle = c^*_+.\] So there exists \(0<\lambda_1< \lambda_2<\lambda^*_+\) such that \(\langle c_{l, B} (\lambda_1) \rangle < c_1^+ < c_2^+ < \langle c_{l, B} (\lambda_2 ) \rangle\).
By the definition of the uniform mean value, for \(T_0>0\) large enough, we have \[X(\lambda_1) (t) < c_1^+ t < c_2^+ t <X(\lambda_2) (t), \quad \forall t\geq T_0.\] Assume that \(u_0(x)>0\) on an interval longer than \(L\). If \(c^*_+>0\), then Proposition 10 implies that \(u(t,x)>0\) on \([T_1, +\infty) \times [R, +\infty)\) for suitable \(T_1>0\) and \(R>0\). Note that the shifted function \(\phi (t,x-x_0)\) with \(x_0 \in \mathbb{R}\) is still a subsolution of (1 ). Choose \(\eta>0\) small enough and \(x_0= 2R\) with \(R>l\) such that \[\underline{u}(0, x-2R)= \eta \phi (0, x- 2R ) \leq u(T_1, x), \;\; \forall x\in \mathbb{R}.\] For the case of \(c^*_->0\), it suffices to choose proper \(x_0\) such that \(\underline{u}(0, x-x_0) \leq u(T_1, x)\) for all \(x\in \mathbb{R}\).
We next compare the solution \(u(t+T_1,x)\) with a subsolution constructed for the time-shifted equation. Set \[K^{T_1}(t,y):=K(t+T_1,y),\qquad f^{T_1}(t,u):=f(t+T_1,u),\qquad r^{T_1}(t):=f^{T_1}(t,0).\] Since the mean value in Definition 1.3 is uniform with respect to the time shift, we have \[\langle K^{T_1}(\cdot,y)\rangle=\langle K(\cdot,y)\rangle, \qquad \langle r^{T_1}\rangle=\langle r\rangle .\] Therefore the quantities \(H(\lambda)\), \(c^*_+\) and \(\lambda^*_+\) are unchanged for the shifted equation. Applying Lemma 3.2 to the coefficients \(K^{T_1}\) and \(f^{T_1}\), we obtain, for \(\lambda\in[\lambda_1,\lambda_2]\), a function \[\phi^{T_1}_\lambda(t,x) = \begin{cases} e^{a_{T_1}(t)}e^{-\lambda x}\cos\frac{\pi x}{2l}, & |x|\le l,\\ 0,& |x|>l, \end{cases}\] and \[X^{T_1}(\lambda)(t) := \int_0^t c^{T_1}_{l,B}(\lambda)(s)\,ds,\] where \(a_{T_1} \in W^{1, \infty} (0, \infty)\) and \[c^{T_1}_{l,B}(\lambda)(s) = \frac{2l}{\pi} \int_{-B}^{B} K(s+T_1,y)e^{\lambda y}\sin\frac{\pi y}{2l}\mathrm{d}y .\] Then \[\underline u_\lambda(t,x) := \eta\phi^{T_1}_\lambda \bigl(t,x-x_0-X^{T_1}(\lambda)(t)\bigr)\] is a subsolution of \[v_t = \int_{\mathbb{R}}K(t+T_1,y)\big[v(t,x-y)-v(t,x)\big]\mathrm{d}y + v f(t+T_1,v).\] On the other hand, \(v(t,x):=u(t+T_1,x)\) is a solution of this shifted equation. Choosing \(\eta>0\) sufficiently small and \(x_0\) properly, we may assume \[\underline u_\lambda(0,x)\le v(0,x)=u(T_1,x), \qquad x\in\mathbb{R},\quad \lambda\in[\lambda_1,\lambda_2].\] By the maximum principle, \[\underline u_\lambda(t,x)\le u(t+T_1,x), \qquad t\ge0,\;x\in\mathbb{R},\quad \lambda\in[\lambda_1,\lambda_2].\]
Due to the continuity of \(\lambda \mapsto X^{T_1}(\lambda) (t)\), for each \(\hat{t}>T_0\) and \(x-x_0\in [c_1^+\hat{t}, c_2^+\hat{t}]\), there exists \(\hat{\lambda}\in[\lambda_1,\lambda_2]\) such that \(x-x_0=X^{T_1}(\hat{\lambda})(\hat{t})\) and \[u(\hat{t}+T_1,x) \ge \underline u_{\hat{\lambda}}(\hat{t},x) = \eta\phi^{T_1}_{\hat{\lambda}} \bigl(\hat{t}, x-x_0-X^{T_1}(\hat{\lambda})(\hat{t}) \bigr) = \eta e^{a_{T_1}(\hat{t})} \ge \eta e^{-\|a_{T_1}\|_\infty}>0.\]
Hence, we have \[\lim_{t\to +\infty} \inf_{x\in [c_1^+t +x_0, \; c_2^+t +x_0]} u(t+T_1, x) >\eta e^{- \|a\|_\infty } >0.\] Since \(T_1\) and \(x_0\) are fixed, they do not affect the limit as \(t\to+\infty\). We obtain (?? ).
Step 2: We use the approximation method to prove the situation \(\lambda^*_+= i_+<+ \infty\). The proof is inspired by [13], [31].
In this case, we see that \[c^*_+ =\lim_{\lambda\to i_+-}\frac{H(\lambda)}{\lambda} = \inf_{\lambda \in (0, i_+)}c(\lambda) = c(i_+) \text{ and } \lim_{ \lambda \to i_+} c'(\lambda) \leq 0.\] The assumption (?? ) implies that \(\lambda^*_+= i_+\) is the unique minimum point of \(c(\lambda)\). For \(n>0\), set \[K_n(t,y):=K(t,y)\chi_{(-\infty,n]}(y)\] and define \[H_n(\lambda) := \left\langle \int_{\mathbb{R}}K_n(\cdot,y)e^{\lambda y}\mathrm{d}y - \int_{\mathbb{R}}K(\cdot,y)\mathrm{d}y + r(\cdot) \right\rangle , \qquad \lambda>0,\] and \[c^*_{n,+}:=\inf_{\lambda>0}\frac{H_n(\lambda)}{\lambda}.\] Equivalently, \[H_n(\lambda) = \int_{-\infty}^{n}\langle K(\cdot,y)\rangle e^{\lambda y}\mathrm{d}y - \int_{\mathbb{R}}\langle K(\cdot,y)\rangle\mathrm{d}y + \langle r\rangle .\]
For each fixed \(\lambda\in(0,i_+)\), the dominated convergence theorem gives \[H_n(\lambda)\to H(\lambda) \quad\text{as }n\to\infty.\] Moreover, for \(n\) large enough, \(H_n(0)>0\). Since \(K_n\) has bounded support on the right for large \(n\), we have \[\frac{H_n(\lambda)}{\lambda}\to+\infty \quad\text{as }\lambda\to0^+ \quad \text{and as }\lambda\to+\infty.\] Hence \(c^*_{n,+}\) is attained at some \(\lambda^*_{n,+}\in(0,+\infty)\).
We first prove \[\lim_{n\to\infty}c^*_{n,+}=c^*_+.\] For every fixed \(\lambda\in(0,i_+)\), \[\limsup_{n\to\infty}c^*_{n,+} \le \lim_{n\to\infty}\frac{H_n(\lambda)}{\lambda} = \frac{H(\lambda)}{\lambda}.\] Letting \(\lambda\to (i_+)^-\), we obtain \[\limsup_{n\to\infty}c^*_{n,+}\le c^*_+.\]
Then we prove the opposite inequality. Let \(\lambda^*_{n,+}\) be a minimizer of \(H_n(\lambda)/\lambda\). The previously upper bound implies that \(\{c^*_{n,+}\}\) is bounded above. We claim that \(\{\lambda^*_{n,+}\}\) is bounded and bounded away from zero. Indeed, since \(H_n(0)\to\langle r\rangle>0\), we see that \(H_n(\lambda)/\lambda\) is uniformly large as \(\lambda\to0^+\). On the other hand, since \(i_+<+\infty\), there exists an interval \([a,b]\subset(0,\infty)\) such that \[\int_a^b\langle K(\cdot,y)\rangle\mathrm{d}y>0.\] For \(n>b\), \[H_n(\lambda) \ge e^{\lambda a}\int_a^b\langle K(\cdot,y)\rangle\mathrm{d}y - \int_{\mathbb{R}}\langle K(\cdot,y)\rangle\mathrm{d}y + \langle r\rangle,\] and therefore \(H_n(\lambda)/\lambda\to+\infty\) as \(\lambda\to+\infty\), uniformly for large \(n\). Thus, up to a subsequence, \[\lambda^*_{n,+}\to\lambda_0\] for some \(\lambda_0\in(0,+\infty)\).
We next show that \(\lambda_0= i_+\). If \(\lambda_0>i_+\), choose \(\mu\) such that \(i_+<\mu<\lambda_0\). Then \(\lambda^*_{n,+}>\mu\) for large \(n\). Note that \[H'_n(\lambda^*_{n,+})=c^*_{n,+}.\] Thus we have \[c^*_{n,+} = H'_n(\lambda^*_{n,+}) \ge H'_n(\mu).\] But \(\mu\notin \mathcal{I}\), hence \[H'_n(\mu) = \int_{-\infty}^{n}\langle K(\cdot,y)\rangle e^{\mu y}y\mathrm{d}y \to+\infty, \text{ as } n \to +\infty,\] which contradicts the boundedness of \(c^*_{n,+}\). Therefore \(\lambda_0\le i_+\).
If \(\lambda_0<i_+\), then by dominated convergence, \[\lim_{n\to\infty}c^*_{n,+} = \lim_{n\to\infty}\frac{H_n(\lambda^*_{n,+})}{\lambda^*_{n,+}} = \frac{H(\lambda_0)}{\lambda_0}.\] Since in the present endpoint case the minimum of \(H(\lambda)/\lambda\) is attained only at \(\lambda=i_+\), we have \[\frac{H(\lambda_0)}{\lambda_0}>c^*_+,\] which contradicts the already proved inequality \[\limsup_{n\to\infty}c^*_{n,+}\le c^*_+.\] Hence \(\lambda_0=i_+\).
Finally, using \(K_n\le K\) and Fatou’s lemma, we get \[\liminf_{n\to\infty}H_n(\lambda^*_{n,+}) \ge H(i_+).\] Consequently, \[\liminf_{n\to\infty}c^*_{n,+} = \liminf_{n\to\infty} \frac{H_n(\lambda^*_{n,+})}{\lambda^*_{n,+}} \ge \frac{H(i_+)}{i_+} = c^*_+.\] Together with the upper bound, this proves \[\lim_{n\to\infty}c^*_{n,+}=c^*_+.\]
Now let \(u^n\) be the solution of \[\partial_t u^n = \int_{\mathbb{R}}K_n(t,y)u^n(t,x-y)\mathrm{d}y - \int_{\mathbb{R}}K(t,y)\mathrm{d}y\,u^n(t,x) + u^n f(t,u^n), \qquad u^n(0,x)=u_0(x).\] Since \(K_n\le K\), the comparison principle gives \[u^n(t,x)\le u(t,x),\qquad t\ge0,\;x\in\mathbb{R}.\]
For the approximate equation, the right endpoint of the interval \(\mathcal{I}\) is \(+\infty\). Meanwhile, \(c^*_{n,+}\) is attained at the interior point \(\lambda^*_{n,+}\). Hence Step 1 applies to \(u^n\). Let \(c^*_+-\delta<c^+_1<c^+_2<c^*_+\). Since \(c^*_{n,+}\to c^*_+\), for all large \(n\) we have \[c^+_2<c^*_{n,+}\] and \(c^+_1,c^+_2\) are still in the admissible interval in Step 1 for the approximate equation. Therefore, \[\liminf_{t\to\infty} \inf_{x\in[c^+_1t,c^+_2t]}u^n(t,x)>0.\] Using \(u^n\le u\), we conclude that \[\liminf_{t\to\infty} \inf_{x\in[c^+_1t,c^+_2t]}u(t,x)>0.\] This proves the endpoint case. The proof is complete.
Remark 12. In the endpoint case \(\lambda_+^*=i_+<+\infty\), the direct construction used in Step 1 may not work. The reason is that the subsolution moves with speed, in the average sense, close to \(H'(\lambda)\), while the linearly determined speed is \(H(\lambda)/\lambda\). When the minimum is attained at an interior point, these two quantities coincide at \(\lambda=\lambda_+^*\) because \(c'(\lambda_+^*)=0\). At the endpoint case, this identity is no longer available. Thus \(H'(\lambda)\) may fail to approach \(c_+^*\) as \(\lambda\to i_+-\). So we truncate the kernel function, which turns the endpoint minimum into an interior minimum for the approximating problems.
By the symmetric argument, we can obtain the similar lemma.
Lemma 6. Assume that \(c^*_+ +c^*_->0\). For all small \(\delta>0\), for all \(-c^*_- < -c_2^- <-c_1^- <-c^*_- + \delta\), we have \[\label{lim-interval-left} \lim_{t\to + \infty} \inf_{x\in [-c_2^- t , \; -c_1^- t]} u(t,x) >0.\qquad{(6)}\]
Next, we will adopt the idea in [31] to prove the inner spreading speed of our problem. We first use (?? ) and (?? ) to show the following lemma.
Lemma 7. Assume \(c^*_+ + c^*_->0\). Then we have \[\liminf_{t\to +\infty} \inf_{x\in [(-c_-^*+ \epsilon) t , (c_+^*- \epsilon) t]} u(t,x) >0, \quad \forall \epsilon>0.\]
Proof. Fix \(\varepsilon>0\). Choose \(c_l\) and \(c_r\) such that \[-c^*_-<c_l<-c^*_-+\varepsilon, \qquad c^*_+-\varepsilon<c_r<c^*_+.\] Let \(\theta>0\) be small such that \(c_l+\theta<c_r-\theta\). By Lemmas 5 and 6, there exist \(\delta >0\) and \(T_0>0\) such that \[u(t,x)\ge \delta\] for all \(t\ge T_0\) and \[x\in[c_lt,(c_l+\theta)t]\cup [(c_r-\theta)t,c_rt].\]
We claim that \[\liminf_{t\to+\infty}\inf_{x\in[c_lt,c_rt]}u(t,x)>0.\] Suppose not. Define \[m(t):=\min_{x\in[c_lt,c_rt]}u(t,x).\] Then \(\liminf_{t\to\infty}m(t)=0\). We choose a sequence \(t_n\to\infty\) such that \[m(t_n)=\min_{s\in[T_0,t_n]}m(s),\qquad m(t_n)\to0.\] Let \(x_n\in[c_lt_n,c_rt_n]\) satisfy \[u(t_n,x_n)=m(t_n).\] Since \(m(t_n)<\delta\) for large \(n\), we have \[x_n\in[(c_l+\theta)t_n,(c_r-\theta)t_n].\] Hence \[[x_n-\theta t_n,x_n+\theta t_n]\subset[c_lt_n,c_rt_n].\] Therefore \[u(t_n,y)\ge u(t_n,x_n), \qquad y\in[x_n-\theta t_n,x_n+\theta t_n].\]
Moreover, since \(x_n\) stays away from the moving boundaries, for all sufficiently small \(h>0\) and large \(n\), we have \(x_n\in[c_l(t_n-h),c_r(t_n-h)].\) Thus \[u(t_n-h,x_n)\ge m(t_n-h)\ge m(t_n)=u(t_n,x_n).\] Consequently the left time derivative at \((t_n,x_n)\) is nonpositive, in the sense of the lower left Dini derivative. At the differentiable point \((t_n,x_n)\), we obtain \[0\ge \partial_t u(t_n,x_n).\]
On the other hand, by the KPP assumption, \[f(t,u)\ge r(t)-M u,\qquad 0\le u\le1.\] Using the spatial minimality of \(u(t_n,x_n)\) on \([c_lt_n,c_rt_n]\), we get \[\begin{align} \int_{\mathbb{R}}K(t_n,x_n-y)u(t_n,y)\mathrm{d}y &\ge \int_{x_n-\theta t_n}^{x_n+\theta t_n} K(t_n,x_n-y)u(t_n,y)\mathrm{d}y \\ &\ge u(t_n,x_n) \int_{-\theta t_n}^{\theta t_n}K(t_n,y)\mathrm{d}y . \end{align}\] Therefore \[\label{contradicts} 0 \geq \partial_tu(t_n,x_n) \geq u(t_n,x_n) \left\{ \int_{-\theta t_n}^{\theta t_n}K(t_n,y)\mathrm{d}y -\overline{K}_0(t_n) + r(t_n) - Mu(t_n,x_n) \right\}.\tag{19}\] Here \(\overline{K}_0(t_n)=\int_{\mathbb{R}}K(t_n,y)\mathrm{d}y\). Since \(0\in \mathcal{I}\), namely \(\int_{\mathbb{R}}\|K(\cdot,y)\|_{L^\infty(0,\infty)}\mathrm{d}y<\infty,\) we see that \[0\le \overline{K}_0(t_n)-\int_{-\theta t_n}^{\theta t_n}K(t_n,y)\mathrm{d}y \le \int_{|y|>\theta t_n} \|K(\cdot,y)\|_{L^\infty(0,\infty)}\mathrm{d}y \to0.\] Since \(u(t_n,x_n)=m(t_n)\to0\) and \(\inf_{t\ge0}r(t)>0\), we obtain \[\liminf_{n\to\infty} \left\{ \int_{-\theta t_n}^{\theta t_n}K(t_n,y)\mathrm{d}y - K_0(t_n) + r(t_n) - Mu(t_n,x_n) \right\} \ge \inf_{t\ge0}r(t)>0.\] Since \(u(t_n,x_n)>0\), this contradicts the inequality (19 ). Hence \[\liminf_{t\to\infty}\inf_{x\in[c_lt,c_rt]}u(t,x)>0.\] Due to the choice of \(c_l\) and \(c_r\), the proof is complete.
We next prove convergence to \(1\) in the inner spreading region. We first need two elementary estimates for the following nonautonomous equation.
Lemma 8. Let \(\tau >0\) and \(a, b\in \mathbb{R}\) with \(a<b\). Let \(v_i(t,x)\), \(i=1,2\), be solutions of the following equation \[\label{Pbss} \begin{cases} \partial_t v_i (t,x) = \int_{\mathbb{R}} K(t, x- y) v_i(t,y) \mathrm{d}y - \overline{K} (t) v_i(t,x)+ v_i (t,x) f(t,v_i (t,x)), & t\geq \tau, x\in\mathbb{R}, \\ v_i (\tau, x) = v_0^i (x), & x\in\mathbb{R}, \end{cases}\qquad{(7)}\] where \(v_0^i\), \(i=1,2\), are continuous functions with \(0 \leq v_0^i \leq 1\). If \(v_0^1(x) = v_0^2(x)\) for \(x\in [a\tau , b\tau]\), then \[\big|v_1(\tau+t,z)-v_2(\tau+t,z)\big|\leq ce^{ct}e^{-\eta\theta\tau/2},\;\forall t\geq0,\;z\in [(a+\theta) \tau, (b- \theta) \tau],\] where \(\theta\in(0,({b-a})/{2})\), \(\eta\) is given in Assumption 2 and \(c\) is a constant depending on \(K \text{ and } f\). In particular, fixing \(T>0\), for any \(\sigma>0\), there is \(M_0>0\) such that if \(v_0^1(x) = v_0^2(x)\) for \(x\in [a\tau , b\tau]\), then \[\big|v_1(\tau+T,z)-v_2(\tau+T,z)\big|\leq\sigma,\;\forall\tau>M_0,\;z\in [(a+\theta) \tau, (b- \theta) \tau].\]
Proof. For a fixed \(z\in\mathbb{R}\), introduce the space \[X_z := \left\{ \phi \in C(\mathbb{R}) ; \sup_{ x \in \mathbb{R}} e^{-\eta|x-z|/2 } |\phi (x) | <\infty \right\}\] equipped with norm \(\| \phi \|_z = \sup_{x\in \mathbb{R}} e^{-\eta|x-z|/2} |\phi (x)|\).
Now we denote \(w(t,x)= v_1(t,x) - v_2(t,x)\). Note that \(w\) satisfies \[\label{eq:v1-v2} \begin{cases} \partial_t w(t,x) = \int_{\mathbb{R}} K(t, x-y) w(t,y) \mathrm{d}y + B(t,x)w(t,x), & t\geq \tau, x\in\mathbb{R},\\ w(\tau, x) = v_0^1(x) - v_0^2(x), & x\in \mathbb{R}. \end{cases}\tag{20}\] where \(B(t,x):= \frac{v_1f(t,v_1)-v_2f(t,v_2)}{v_1-v_2} - \overline{K} (t)\) if \(v_1\neq v_2\) and \(- \overline{K} (t)\) if \(v_1=v_2\). From (20 ), we have \[w(t,x)=w(\tau,x)+\int_\tau^t\int_{\mathbb{R}} K(s, y) w(s,x-y) \mathrm{d}yds +\int_\tau^t B(s,x)w(s,x)ds.\] Therefore, \[\label{eq:estimate32on32Xz} \begin{align} &\quad e^{-\eta|x-z|/2}w(t,x)\\ & \leq \|w(\tau,\cdot)\|_z+\int_\tau^t\int_{\mathbb{R}} K(s, x-y) e^{-\eta|x-z|/2}\big|w(s,y)\big| \mathrm{d}yds +\int_\tau^t \big|B(s,x)e^{-\eta|x-z|/2}w(s,x)\big|ds \\ &\leq \|w(\tau,\cdot)\|_z+\int_\tau^t\int_{\mathbb{R}} K(s, x-y) e^{\eta|y-z|/2-\eta|x-z|/2}\|w(s,\cdot)\|_z \mathrm{d}yds +\int_\tau^t \big|B(s,x)|\|w(s,\cdot)\|_zds \\ &\leq \|w(\tau,\cdot)\|_z+\int_\tau^t\big(\int_\mathbb{R}K(s,y)e^{\eta|y|/2}dy+|B(s,x)|\big)\|w(s,\cdot)\|_z ds. \end{align}\tag{21}\] By Assumption 2, \(\int_\mathbb{R}\|K(\cdot,y)\|_{L^\infty(0,\infty)}e^{\eta |y|/2}\mathrm{d}y<\infty\). Hence (21 ) gives \[\|w(t,\cdot)\|_z\leq\|w(\tau,\cdot)\|_z+M\int_\tau^t\|w(s,\cdot)\|_z ds, \quad \forall t>\tau,\;z\in\mathbb{R},\] where \(M\) is a constant depending only on \(K\) and \(f\). The Gronwall’s inequality implies that \[\|w(t,\cdot)\|_z \leq \| w(\tau, \cdot)\|_z e^{M(t-\tau)} \leq \| w(\tau, \cdot)\|_z e^{MT} , \forall t\in[\tau,\tau+T], z\in\mathbb{R}.\] That means \[\sup_{x\in \mathbb{R}} \big( e^{-\eta|x-z|/2} |v_1(t, x) -v_2(t,x)| \big) \leq \sup_{x\in \mathbb{R}} \big( e^{-\eta|x-z|/2} | v_0^1(x) - v_0^2(x) | \big) e^{MT},\;\forall t>\tau,\;z\in\mathbb{R}.\] Particularly, since \(v_0^1(x) = v_0^2(x)\) for all \(x\in [ a \tau, b\tau]\), we have \[\begin{align} |v_1(\tau +T, z) -v_2(\tau +T , z)| &\leq e^{MT} \sup_{x\in \mathbb{R}} \big( e^{-\eta|x-z|/2} | v_0^1(x) - v_0^2(x) | \big) \\ &= e^{MT} \sup_{x\in \mathbb{R}\setminus [a\tau, b\tau]} \big( e^{-\eta|x-z|/2} | v_0^1(x) - v_0^2(x) | \big). \end{align}\] Due to \(0\leq v_0^1, v_0^2\leq 1\), we see that for all \(z\in [ (a+\theta) \tau, (b-\theta)\tau]\) \[|v_1(\tau +T, z) -v_2(\tau +T, z)| \leq 2 e^{M T} \sup_{x\in \mathbb{R}\setminus [a\tau, b\tau] } e^{-\eta|x-z|/2} \leq 2e^{MT} e^{-\eta\theta \tau/2}.\] This completes the proof.
Next we state that a solution starting from a positive constant function will converge to \(1\).
Lemma 9. For all constant \(\alpha \in (0, 1)\), let \(\tau\geq 0\) and \(v\) be a solution of \[\label{Pbalpha} \begin{cases} \partial_t v (t,x) = \int_{\mathbb{R}} K(t, x- y) v(t,y) \mathrm{d}y - \overline{K} (t) v(t,x)+ v (t,x) f(t,v (t,x)), & t\geq \tau, x\in\mathbb{R}, \\ v (\tau,x) = \alpha, & x\in\mathbb{R}. \end{cases}\qquad{(8)}\] Then we have \[\lim_{t\to \infty} v(t,x) =1 \quad \text{ uniformly in } x\in \mathbb{R}.\]
Proof. From Assumption 1, we have \(h(v) := \inf_{t\geq 0} f(t,v) >0\) for all \(v \in [0, 1)\) and \(h(1)=0\). Let \(\underline{v}\) be the solution of \[\underline{v}' (t) = \underline{v}(t)h(\underline{v}(t)), \;\; \forall t\geq s, \;\; \text{ and } \; \underline{v}(\tau)=\alpha.\] By comparison principle, we have \[\underline{v} (t) \leq v(t,x) \leq 1, \quad \forall x\in \mathbb{R}, t\geq \tau.\] Note that \(\lim_{t\to +\infty } \underline{v} (t)=1\). This completes the proof.
The next lemma completes the proof of Theorem 4. Its proof follows the argument in [31].
Lemma 10. For all \(-c^*_- < c_l < c_r < c^*_+\), we have \[\lim_{t\to +\infty} \inf_{x\in [c_l t , c_r t]} u(t,x)=1.\]
Proof. Since \(0\leq u\leq1\), it is enough to prove that for any small \(\theta>0\), for any \(\varepsilon>0\), there exists \(T_0>0\) large enough such that \[u(t,x) \geq 1-\varepsilon, \forall t>T_0, \; x\in [ (c_l +2\theta )t , (c_r -2\theta)t].\] Set \(\alpha := \frac{1}{2} \liminf_{t\to +\infty} \inf_{x\in [c_l t , c_r t]} u(t,x) \in (0,1)\). Let \(w:= w(t,x; \tau, \alpha)\) be a solution of (?? ). The notation means that the initial time is \(\tau\). Lemma 9 gives \[\lim_{ s \to +\infty} w(s+ \tau,x; \tau, \alpha)=1, \text{ uniformly in } x\in \mathbb{R}.\] Hence, for each \(\sigma>0\), there is \(T>0\) such that \[w(s+\tau, x; \tau, \alpha) \geq 1- \sigma, \forall s\geq T.\] We define \[u_0^\tau (\cdot) := \min \{ \alpha, u(\tau, \cdot; 0, u_0) \}.\] Here \(u(\tau,\cdot;0,u_0)\) denotes the solution of (1 ) at time \(\tau\). Let \(\underline{u}^\tau ( t, \cdot; \tau, u_0^\tau(\cdot))\) be the solution of (?? ) equipped with initial data \(\underline{u}^\tau (\tau,\cdot) =u_0^\tau(\cdot)\). From the definition of \(\alpha\), note that \(u^\tau_0(x) =\alpha\) for all \(x\in [(c_l+\theta) \tau, (c_r- \theta)\tau ]\) and \(\tau>M\) large enough. Applying Lemma 8 to \(\underline{u}^\tau\) and \(w\), and increasing \(M\) if necessary, we have for all \(\tau>M\) \[|w(\tau + T, x; \tau , \alpha) - \underline{u}^\tau (\tau + T, x; \tau, u_0^\tau (x) ) | \leq \sigma, \quad \forall x\in [ (c_l+ \theta) \tau, (c_r- \theta) \tau].\] Hence, \(\underline{u}^\tau (\tau + T, x; \tau, u_0^\tau (\cdot) ) \geq w(\tau + T, x; \tau, \alpha) - \sigma \geq 1-2\sigma\) for all \(x\in [ (c_l+ \theta) \tau, (c_r- \theta) \tau]\). Increasing \(M\) once more, we may assume that for all \(t\geq M+T\), \[[(c_l +2\theta) t , (c_r-2\theta) t] \subset [(c_l +\theta) (t- T) , (c_r-\theta)(t-T)].\] Set \(\tau = t-T\). For all \(t\geq T+M\) large enough, we have \[\underline{u}^{t- T} (t, x; t- T, u_0^{t-T} (\cdot)) \geq 1-2\sigma, \quad \forall x\in [(c_l +2\theta) t , (c_r-2\theta)t].\] Note that \(u(t-T, x; 0,u_0) \geq u_0^{t-T}(x)\) for all \(x\in \mathbb{R}\). The comparison principle ensures that \[u(t,x; 0, u_0)= u(t, x; t-T, u(t-T, x; 0,u_0)) \geq \underline{u}^{t-T} (t,x; t-T, u_0^{t-T} ).\] Therefore, we obtain \[u(t,x; 0, u_0)\geq 1-2\sigma, \quad \forall t\geq T+M, x\in [(c_l +2\theta) t , (c_r-2\theta)t].\] This completes the proof.
We finish with a few simple cases covered by the assumptions above.
Example 1 (Separated time dependence). Let \(K(t,y)=d(t)J(y)\), where \(d\in L^\infty(0,\infty)\) is nonnegative and has a uniform mean value, and where \(J\geq0\) satisfies \(\int_\mathbb{R}J(y)e^{\lambda y}\mathrm{d}y<\infty\) for \(\lambda\) in a neighborhood of the origin. Then \[H(\lambda)=\langle d\rangle\int_\mathbb{R}J(y)(e^{\lambda y}-1)\mathrm{d}y+\langle r\rangle,\] and the rightward and leftward speeds are obtained from (7 ). If \(J\) is not symmetric, the two speeds are generally different.
Example 2 (One-sided kernels). If \(\langle K(\cdot,y)\rangle=0\) for a.e. \(y>0\), then the averaged kernel allows no positive jumps. In this case \(c^*_+=0\), whereas the leftward speed may be positive. The opposite conclusion holds when the averaged kernel is supported in \([0,\infty)\).
Corresponding author
E-mail addresses: jinzc@ustc.edu.cn (Z. Jin); tzhou910@ustc.edu.cn (T. Zhou)↩︎