Long Time Behavior of Random and Nonautonomous Fisher–KPP Equations: Part I—Stability of Equilibria and Spreading Speeds

Abstract

In the current series of two papers, we study the long time behavior of nonnegative solutions to the following random Fisher–KPP equation,

$$\begin{aligned} u_t =u_{xx}+a(\theta _t\omega )u(1-u),\quad x\in {{\mathbb {R}}}, \end{aligned}$$

(1)

where $\omega \in \Omega$, $(\Omega , {\mathcal {F}},{\mathbb {P}})$ is a given probability space, $\theta _t$ is an ergodic metric dynamical system on $\Omega$, and $a(\omega )>0$ for every $\omega \in \Omega$. We also study the long time behavior of nonnegative solutions to the following nonautonomous Fisher–KPP equation,

$$\begin{aligned} u_t=u_{xx}+a_0(t)u(1-u),\quad x\in {{\mathbb {R}}}, \end{aligned}$$

(2)

where $a_0(t)$ is a positive locally Hölder continuous function. In this first part of the series, we investigate the stability of positive equilibria and the spreading speeds. Under some proper assumption on $a(\omega )$, we show that the constant solution $u=1$ of (1) is asymptotically stable with respect to strictly positive perturbations and show that (1) has a deterministic spreading speed interval $[2\sqrt{{\underline{a}}}, 2\sqrt{{\bar{a}}}]$, where ${\underline{a}}$ and ${\bar{a}}$ are the least and the greatest means of $a(\cdot )$, respectively, and hence the spreading speed interval is linearly determinate. It is shown that the solution of (1) with a nonnegative initial function which is bounded away from 0 for $x\ll -1$ and is 0 for $x\gg 1$ propagates at the speed $2\sqrt{{\hat{a}}}$, where ${\hat{a}}$ is the mean of $a(\cdot )$. Under some assumption on $a_0(\cdot )$, we also show that the constant solution $u=1$ of (2) is asymptotically stably and (2) admits a bounded spreading speed interval. It is not assumed that $a(\omega )$ and $a_0(t)$ are bounded above and below by some positive constants. The results obtained in this part are new and extend the existing results in literature on spreading speeds of Fisher–KPP equations. In the second part of the series, we will study the existence and stability of transition fronts of (1) and (2).

1 Introduction and Statements of the Main Results

The current series of two papers is concerned with the long time behavior of nonnegative solutions to the following random Fisher–KPP equation,

$$\begin{aligned} u_t =u_{xx}+a(\theta _t\omega )u(1-u),\quad x\in {{\mathbb {R}}}, \end{aligned}$$

(1.1)

where $\omega \in \Omega$, $(\Omega , {\mathcal {F}},{\mathbb {P}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$ is an ergodic metric dynamical system on $\Omega$, $a:\Omega \rightarrow (0,\infty )$ is measurable, and $a^\omega (t):=a(\theta _t\omega )$ is locally Hölder continuous for every $\omega \in \Omega$. It also considers the long time behavior of nonnegative solutions to the following nonautonomous Fisher–KPP equation,

$$\begin{aligned} u_t=u_{xx}+a_0(t)u(1-u), x\in {{\mathbb {R}}}, \end{aligned}$$

(1.2)

where $a_0:{{\mathbb {R}}}\rightarrow (0,\infty )$ is locally Hölder continuous. Among others, (1.1) and (1.2) are used to model the population growth of a species in biology. In such case, u(t, x) denotes the population density of the species. Thanks to the biological reason, we are only interested in nonnegative solutions of (1.1) and (1.2).

Observe that (1.1) [resp. (1.2)] with $a(\omega )\equiv 1$ (resp. with $a_0(t)\equiv 1$) becomes

$$\begin{aligned} u_t=u_{xx}+u(1-u),\quad x\in {{\mathbb {R}}}. \end{aligned}$$

(1.3)

Equation (1.3) is called in literature Fisher–KPP equation due to the pioneering works of Fisher [13] and Kolmogorov et al. [25] on traveling wave solutions and take-over properties of (1.3). It is clear that the constant solution $u=1$ of (1.3) is asymptotically stable with respect to strictly positive perturbations. Fisher in [13] found traveling wave solutions $u(t,x)=\phi (x-ct)$ of (1.3) $(\phi (-\infty )=1,\phi (\infty )=0, \phi (s)>0)$ of all speeds $c\ge 2$ and showed that there are no such traveling wave solutions of slower speed. He conjectured that the take-over occurs at the asymptotic speed 2. This conjecture was proved in [25] for some special initial distribution and was proved in [3] for general initial distributions. More precisely, it is proved in [25] that for the nonnegative solution u(t, x) of (1.3) with $u(0,x)=1$ for $x<0$ and $u(0,x)=0$ for $x>0$, $\lim _{t\rightarrow \infty }u(t,ct)$ is 0 if $c>2$ and 1 if $c<2$. It is proved in [3] that for any nonnegative solution u(t, x) of (1.3), if at time $t=0$, u is 1 near $-\infty$ and 0 near $\infty$, then $\lim _{t\rightarrow \infty }u(t,ct)$ is 0 if $c>2$ and 1 if $c<2$. In literature, $c^*=2$ is called the spreading speed for (1.3).

A huge amount of research has been carried out toward various extensions of traveling wave solutions and take-over properties of (1.3) to general time and space independent as well as time and/or space dependent Fisher–KPP type equations. See, for example, [2, 3, 11, 15, 24, 41, 48], etc., for the extension to general time and space independent Fisher–KPP type equations; see [4, 5, 7, 14, 22, 26,27,28,29, 31, 37, 38, 49, 50], and references therein for the extension to time and/or space periodic Fisher–KPP type equations; and see [5, 8,9,10, 16, 21, 30, 32,33,34,35,36, 43,44,47, 51, 52], and references therein for the extension to quite general time and/or space dependent Fisher–KPP type equations. The reader is referred to [12, 17, 53], etc. for the study of Fisher–KPP reaction diffusion equations with time delay.

All the existing works on (1.1) [resp. (1.2)] assumed $\inf _{t\in {{\mathbb {R}}}} a^\omega (t)>0$ and $a^\omega (\cdot )\in L^\infty ({{\mathbb {R}}})$ (resp. $\inf _{t\in {{\mathbb {R}}}} a_0(t)>0$ and $\sup _{t\in {{\mathbb {R}}}} a_0(t)<\infty$). The objective of the current series of two papers is to study the long time behavior, in particular, the stability of positive constant solutions, the spreading speeds, and the transition fronts of (1.1) [resp. (1.2)] without the assumption $\inf _{t\in {{\mathbb {R}}}} a^\omega (t)>0$ and $a^\omega (\cdot )\in L^\infty ({{\mathbb {R}}})$ (resp. without the assumption $\inf _{t\in {{\mathbb {R}}}} a_0(t)>0$ and $\sup _{t\in {{\mathbb {R}}}} a_0(t)<\infty$). It will also discuss the applications of the results established for (1.1) to Fisher–KPP equations whose growth rate and/or carrying capacity are perturbed by real noises.

In this first part of the series, we investigate the stability of positive constant solutions and the spreading speeds of (1.1) and (1.2). We first consider the stability of positive constant solutions and spreading speeds of (1.1) and then consider the stability of positive constant solutions and spreading speeds of (1.2). In the second part of the series, we will study the existence and stability of transition fronts of (1.1) and (1.2).

In the following, we state the main results of the current paper. Let

$$\begin{aligned} C_{\mathrm{unif}}^b({{\mathbb {R}}})=\{u\in C({{\mathbb {R}}})\,|\, u\,\, \text {is bounded and uniformly continuous}\} \end{aligned}$$

with norm $\Vert u\Vert _\infty =\sup _{x\in {{\mathbb {R}}}}|u(x)|$ for $u\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$. For given $u_0\in X:=C_{\mathrm{unif}}^b({{\mathbb {R}}})$ and $\omega \in \Omega$, let $u(t,x;u_0,\omega )$ be the solution of (1.1) with $u(0,x;u_0,\omega )=u_0(x)$. Note that, for $u_0\in X$ with $u_0\ge 0$, $u(t,x;u_0,\omega )$ exists for $t\in [0,\infty )$ and $u(t,x;u_0,\omega )\ge 0$ for all $t\ge 0$. Note also that $u\equiv 0$ and $u\equiv 1$ are two constant solutions of (1.1). Let

$$\begin{aligned} { {\hat{a}}_{\inf }(\omega )}=\liminf _{t-s\rightarrow \infty } \frac{1}{t-s}\int _s^ta(\theta _{\tau }\omega )d\tau :=\lim _{r\rightarrow \infty }\inf _{t-s\ge r}\frac{1}{t-s}\int _s^ta(\theta _{\tau }\omega )d\tau \end{aligned}$$

(1.4)

and

$$\begin{aligned} {{\hat{a}}_{\sup }(\omega )}=\limsup _{t-s\rightarrow \infty }\frac{1}{t-s} \int _s^ta(\theta _\tau \omega )d\tau :=\lim _{r\rightarrow \infty }\sup _{t-s\ge r}\frac{1}{t-s}\int _s^ta(\theta _{\tau }\omega )d\tau . \end{aligned}$$

(1.5)

Observe that

$$\begin{aligned} {\hat{a}}_{\inf }(\theta _t\omega )={\hat{a}}_{\inf }(\omega )\quad \mathrm{and}\quad {\hat{a}}_{\sup }(\theta _t\omega )={\hat{a}}_{\sup }(\omega ),\,\, \forall \,\, t\in {{\mathbb {R}}}, \end{aligned}$$

(1.6)

and that

$$\begin{aligned} {\hat{a}}_{\inf }(\omega )=\liminf _{t,s\in {{\mathbb {Q}}},t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t a(\theta _\tau )d\tau \,\, \,\, \mathrm{and}\,\, \,\, {\hat{a}}_{\sup }(\omega )=\liminf _{t,s\in {{\mathbb {Q}}},t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t a(\theta _\tau )d\tau . \end{aligned}$$

Then by the countability of the set ${{\mathbb {Q}}}$ of rational numbers, both ${\hat{a}}_{\inf }(\omega )$ and ${\hat{a}}_{\sup }(\omega )$ are measurable in $\omega$.

Throughout this paper, we assume that the following standing assumption holds.

(H1) $0< {\hat{a}}_{\inf }(\omega )\le {\hat{a}}_{\sup }(\omega )<\infty$ for a.e. $\omega \in \Omega$.

Note that (H1) implies that ${\hat{a}}_{\inf }(\cdot ),a(\cdot ),{\hat{a}}_{\sup }(\cdot )\in L^1 (\Omega , {\mathcal {F}},{\mathbb {P}})$, and that there are ${{\hat{a}}}, {\underline{a}}, {\bar{a}}\in {{\mathbb {R}}}^+$ and a measurable subset $\Omega _0\subset \Omega$ with ${{\mathbb {P}}}(\Omega _0)=1$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} \theta _t\Omega _0=\Omega _0\quad \forall \,\, t\in {{\mathbb {R}}},\\ \lim _{t\rightarrow \pm \infty }\frac{1}{t}\int _0^t a(\theta _\tau \omega )d\tau ={{\hat{a}}}\quad \forall \,\, \omega \in \Omega _0,\\ { {{\hat{a}}}_{\inf }(\omega )} ={\underline{a}}\quad \forall \,\,\omega \in \Omega _0,\\ { {{\hat{a}}}_{\sup }(\omega )}={\bar{a}}\quad \forall \,\, \omega \in \Omega _0 \end{array}\right. } \end{aligned}$$

(1.7)

(see Lemma 2.1). Throughout this paper, ${{\hat{a}}}$ is referred to as the mean or average of $a(\cdot )$, and ${\underline{a}}$ and ${\overline{a}}$ are referred to as the least mean and the greatest mean of $a(\cdot )$, respectively.

Our main result on the stability of the constant solution $u\equiv 1$ of (1.1) reads as follows.

Theorem 1.1

For every $u_0\in C^{b}_{\mathrm{uinf}}({{\mathbb {R}}})$ with $\inf _{x\in {{\mathbb {R}}}}u_0(x)>0$ and for every $\omega \in \Omega$, we have that

$$\begin{aligned} \Vert u(t,\cdot ;u_0,\omega )-1\Vert _{\infty }\le M(u_0) e^{-\int _0^ta(\theta _s\omega )ds}, \end{aligned}$$

(1.8)

where $M(u_0):=\max \{1,\Vert u_0\Vert _{\infty }\}\cdot \max \Big \{\Big |1-\frac{1}{\min \{1,\inf _{x\in {{\mathbb {R}}}} u_0(x)\}} \Big |, \Big |1-\frac{1}{\max \{1,\sup _{x\in {{\mathbb {R}}}} u_0(x)\}} \Big |\Big \}$. Hence if $\int _0^{\infty }a(\theta _s\omega )ds=\infty$, then

$$\begin{aligned} \lim _{t\rightarrow \infty } \Vert u(t,\cdot ;u_0,\omega )-1\Vert _{\infty }=0. \end{aligned}$$

In particular, if (H1) holds, then for every $0<{\tilde{a}}<{\underline{a}}$, every $u_0\in C^{b}_{\mathrm{uinf}}({{\mathbb {R}}})$ with $\inf _{x}u_0(x)>0$, and almost all $\omega \in \Omega$, there is positive constant $M>0$ such that

$$\begin{aligned} \Vert u(t,\cdot ;u_0,\theta _{t_0}\omega )-1\Vert _{\infty }\le Me^{-{\tilde{a}}t},\quad \forall \ t\ge 0, \ t_0\in {{\mathbb {R}}}. \end{aligned}$$

If $a(\theta _{\cdot }\omega )\in L^{1}(0,\infty )$, then the constant equilibrium solution, $u\equiv 1$, of (1.1) is not asymptotically stable.

To state our main results on the spreading speeds of (1.1), let

$$\begin{aligned} {\underline{c}}^*=2\sqrt{{\underline{a}}},\quad {\hat{c}}^*=2\sqrt{{\hat{a}}},\quad \mathrm{and}\quad {\overline{c}}^*=2\sqrt{{\overline{a}}}. \end{aligned}$$

(1.9)

Let

$$\begin{aligned} X_c^+=\{u\in C_{\mathrm{unif}}^b({{\mathbb {R}}})\,|\, u\ge 0,\,\, \mathrm{supp}(u)\,\,\, \text {is bounded and not empty}\}. \end{aligned}$$

Definition 1.1

For given $\omega \in \Omega$, let

$$\begin{aligned} C_{\sup }(\omega )=\{c\in {{\mathbb {R}}}^+\,|\, \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct}u(t,x;u_0,\theta _s\omega )=0\quad \forall \,\, u_0\in X_c^+\} \end{aligned}$$

and

$$\begin{aligned} C_{\inf }(\omega )=\{c\in {{\mathbb {R}}}^+\,|\, \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\le ct} |u(t,x;u_0,\theta _s\omega )-1|=0\quad \forall \,\, u_0\in X_c^+\}. \end{aligned}$$

Let

$$\begin{aligned} c_{\sup }^*(\omega )=\inf \{c\,|\, c\in C_{\sup }(\omega )\},\quad c_{\inf }^*(\omega )=\sup \{c\,|\, c\in C_{\inf }(\omega )\}. \end{aligned}$$

$[c_{\inf }^*(\omega ),c_{\sup }^*(\omega )]$ is called the spreading speed interval of (1.1) with respect to compactly supported initial functions.

The following theorem shows that the spreading speed interval of (1.1) with respect to compactly supported initial functions is deterministic and is linearly determinate, that is, $[c_{\inf }^*(\omega ),c_{\sup }^*(\omega )]=[{\underline{c}}^*,\bar{c}^*]$ for all $\omega \in \Omega _0$.

Theorem 1.2

Assume that (H1) holds. Then the following hold.

• (i) For any $\omega \in \Omega _0$, $c_{\sup }^*(\omega )=\bar{c}^*$.

• (ii) For any $\omega \in \Omega _0$, $c_{\inf }^{*}(\omega )={\underline{c}}^*$.

The above theorem concerns the spreading speeds of solutions of (1.1) with compactly supported nonnegative initial functions. To consider the spreading speeds of solutions of (1.1) with front-like initial functions, let

$$\begin{aligned} {\tilde{X}}_c^+=\{u\in C_{\mathrm{unif}}^b({{\mathbb {R}}})\,|\, u\ge 0,\,\, \liminf _{x\rightarrow -\infty }u_0(x)>0,\,\, u_0(x)=0\,\, \mathrm{for}\,\, x\gg 1\}. \end{aligned}$$

Definition 1.2

For given $\omega \in \Omega$, let

$$\begin{aligned} {{\tilde{C}}}_{\sup }(\omega )=\{c\in {{\mathbb {R}}}^+\,|\, \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},x\ge ct}u(t,x;u_0,\theta _s\omega )=0\quad \forall \,\, u_0\in {{\tilde{X}}}_c^+\} \end{aligned}$$

and

$$\begin{aligned} {{\tilde{C}}}_{\inf }(\omega )=\{c\in {{\mathbb {R}}}^+\,|\, \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},x\le ct} |u(t,x;u_0,\theta _s\omega )-1|=0\quad \forall \,\, u_0\in {{\tilde{X}}}_c^+\}. \end{aligned}$$

Let

$$\begin{aligned} {{\tilde{c}}}_{\sup }^*(\omega )=\inf \{c\,|\, c\in {{\tilde{C}}}_{\sup }(\omega )\},\quad {{\tilde{c}}}_{\inf }^*(\omega )=\sup \{c\,|\, c\in {{\tilde{C}}}_{\inf }(\omega )\}. \end{aligned}$$

$[{{\tilde{c}}}_{\inf }^*(\omega ),{{\tilde{c}}}_{\sup }^*(\omega )]$ is called the spreading speed interval of (1.1) with respect to front-like initial functions.

We have the following theorem on the spreading speeds of the solutions with front-like initial functions.

Theorem 1.3

Assume that (H1) holds. Then the following hold.

• (i) For any $\omega \in \Omega _0$, ${{\tilde{c}}}_{\sup }^*(\omega )=\bar{c}^*$.

• (ii) For any $\omega \in \Omega _0$, ${{\tilde{c}}}_{\inf }^{*}(\omega )={\underline{c}}^*$.

We also have the following theorem on the take-over property of the solutions of (1.1) with front-like initial functions and with the initial function $u_0^*(x)=1$ for $x< 0$ and $u_0^*(x)=0$ for $x>0$. Note that $u(t,x;u_0^*,\omega )$ exists for all $t>0$ (see [25, Theorem 1]).

Theorem 1.4

• (i) For a.e. $\omega \in \Omega$,

  $$\begin{aligned} \lim _{t\rightarrow \infty }\frac{x(t,\omega )}{t}= {\hat{c}}^*, \end{aligned}$$

  (1.10)

  where $x(t,\omega )$ is such that $u(t,x(t,\omega );u_0^*,\omega )=\frac{1}{2}$. Moreover,

  $$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{x\ge ({\hat{c}}^*+h)t}u(t,x;u_0^*,\omega )=0, \forall \ h>0, \ \text {a.e }\ \omega \end{aligned}$$

  (1.11)

  and

  $$\begin{aligned} \lim _{t\rightarrow \infty }\inf _{x\le ({\hat{c}}^*-h)t}u(t,x;u_0^*,\omega )=1, \forall \ h>0, \ \text {a.e }\ \omega . \end{aligned}$$

  (1.12)

• (ii) For any $u_0\in {{\tilde{X}}}_c^+$, it holds that

  $$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{x\ge ({\hat{c}}^*+h)t}u(t,x;u_0,\omega )=0, \forall \ h>0, \ \text {a.e }\ \omega \end{aligned}$$

  (1.13)

  and

  $$\begin{aligned} \lim _{t\rightarrow \infty }\inf _{x\le ({\hat{c}}^*-h)t}u(t,x;u_0,\omega )=1, \forall \ h>0, \ \text {a.e }\ \omega . \end{aligned}$$

  (1.14)

Consider now (1.2). Define ${\underline{a}}_0$ and ${\overline{a}}_0$ by

$$\begin{aligned} {\underline{a}}_0= \liminf _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t a_0(\tau )d\tau ,\quad {\overline{a}}_0=\limsup _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t a_0(\tau )d\tau . \end{aligned}$$

(1.15)

Let (H2) be the following standing assumption.

(H2) $0< {\underline{a}}_0\le {\overline{a}}_0<\infty$.

The assumption (H2) is the analogue of (H1). We will give some example for $a_0(\cdot )$ satisfying (H2) in Sect. 5. Assume (H2). Let

$$\begin{aligned} {\bar{c}}_0^*=2\sqrt{\bar{a}_0}\quad \mathrm{and}\quad {\underline{c}}_0^*=2\sqrt{{\underline{a}}_0}. \end{aligned}$$

(1.16)

For given $u_0\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$ with $u_0\ge 0$ and $s\in {{\mathbb {R}}}$, let $u(t,x;u_0,\sigma _s a_0)$ be the solution of

$$\begin{aligned} u_t=u_{xx}+\sigma _s a_0(t) u(1-u),\quad x\in {{\mathbb {R}}},\, t>0, \end{aligned}$$

with $u(0,x;u_0,\sigma _s a_0)=u_0(x)$, where $\sigma _s a_0(t)=a_0(s+t)$.

We have the following theorem on the spreading speeds of (1.2).

Theorem 1.5

Assume (H2). Then for every $u_0\in X_c^+$,

$$\begin{aligned} \liminf _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\le ct}|u(t,x;u_0,\sigma _s a_0)-1|=0, \quad \forall \ 0<c<{\underline{c}}^*_0:=2\sqrt{{\underline{a}}_0} \end{aligned}$$

(1.17)

and

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} u(t,x;u_0,\sigma _s a_0)=0, \quad \forall \ c>{\bar{c}}^*_0:=2\sqrt{{\bar{a}}_0}. \end{aligned}$$

(1.18)

We conclude the introduction with the following four remarks.

First, the results in Theorems 1.2–1.5 are new. If $a_0(t)$ is periodic with period T, then ${\underline{a}}_0={\bar{a}}_0={{\hat{a}}}_0:=\frac{1}{T}\int _0^T a_0(\tau )d\tau$ and hence ${\underline{c}}_0^*={\bar{c}}_0^*=2\sqrt{{{\hat{a}}}_0}$. More generally, if $a_0(t)$ in globally Hölder continuous and is uniquely ergodic in the sense that the space $H(a_0)$ is compact and the flow $(H(a_0),\sigma _t)$ is uniquely ergodic, where $H(a_0)=\mathrm{cl}\{\sigma _s a_0\,|\, s\in {{\mathbb {R}}}\}$ with open compact topology and $\sigma _s a_0(\cdot )=a_0(\cdot +s)$, then ${\underline{a}}_0={\bar{a}}_0={{\hat{a}}}_0:=\lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T a_0(\tau )d\tau$ and hence ${\underline{c}}_0^*={\bar{c}}_0^*=2\sqrt{{{\hat{a}}}_0}$. Therefore the existing results on spreading speeds of (1.2) in the time periodic and time almost periodic cases are recovered. The current paper provides a new and simpler proof in these special cases.

Second, by Theorems 1.2 and 1.3 ,

$$\begin{aligned}{}[c_{\mathrm{inf}}^*(\omega ),c_{\mathrm{sup}}^*(\omega )]=[{{\tilde{c}}}_{\mathrm{inf}}^*(\omega ),{{\tilde{c}}}_{\mathrm{sup}}^*(\omega )]=[{\underline{c}}^*,{\bar{c}}^*] \end{aligned}$$

for any $\omega \in \Omega _0$. Hence $[{\underline{c}}^*,{\bar{c}}^*]$ is called the spreading speed interval of (1.1), which is deterministic and is determined by the linearized equation of (1.1) at $u\equiv 0$. Theorem 1.4 is an extension of the take-over property proved in [3] and [25] for (1.3). In order to prove Theorem 1.4 we are first led to prove that $x(t,\omega )$ is a subadditive process (see Lemma 5.4 for more detail). The fact that $x(t,\omega )$ is a subadditive process is interesting. Its proof relies on comparison between various translation of the solution and on a zero-number argument enabling to bound the width of the interface. It is our belief that this result will open the way to other applications in the future.

Third, the results established for (1.1) and (1.2) can be applied to the following general random Fisher–KPP equation,

$$\begin{aligned} u_t=u_{xx}+u(r(\theta _t\omega )-\beta (\theta _t\omega ) u), \end{aligned}$$

(1.19)

where $r:\omega \rightarrow (-\infty ,\infty )$ and $\beta :\Omega \rightarrow (0,\infty )$ are measurable with locally Hölder continuous sample paths $r^\omega (t):=r(\theta _t\omega )$ and $\beta ^\omega (t):=\beta (\theta _t\omega )$, and to the following nonautonomous Fisher–KPP equation,

$$\begin{aligned} u_t=u_{xx}+u(r_0(t)-\beta _0(t) u), \end{aligned}$$

(1.20)

where $r_0:{{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ and $\beta _0:{{\mathbb {R}}}\rightarrow (0,\infty )$ are locally Hölder continuous. Note that (1.19) models the population growth of a species with random perturbations on its growth rate and carrying capacity, and (1.20) models the population growth of a species with deterministic time dependent perturbations on its growth rate and carrying capacity.

In fact, under some assumptions on $r(\omega )$ and $\beta (\omega )$, it can be proved that

$$\begin{aligned} u(t;\omega ):=Y(\theta _t\omega )=\frac{1}{\int _{-\infty }^0 e^{-\int _s ^0 r(\theta _{\tau +t}\omega )d\tau }\beta (\theta _{s+t}\omega )ds} \end{aligned}$$

is an random equilibrium of (1.19). Let ${{\tilde{u}}}=\frac{u}{Y(\theta _t\omega )}$ and drop the tilde, (1.19) becomes (1.1) with $a(\theta _t\omega )=\beta (\theta _t\omega )\cdot Y(\theta _t\omega )$, and then the results established for (1.1) can be applied to (1.19). For example, consider the following random Fisher–KPP equation,

$$\begin{aligned} u_t=u_{xx}+u(1+ \xi (\theta _t\omega ) -u),\quad x\in {{\mathbb {R}}}, \end{aligned}$$

(1.21)

where $\omega \in \Omega$, $(\Omega , {\mathcal {F}},{\mathbb {P}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$ is an ergodic metric dynamical system, $\xi :\Omega \rightarrow {{\mathbb {R}}}$ is measurable, and $\xi _t(\omega ):=\xi (\theta _t\omega )$ is locally Hölder continuous ($\xi _t$ denotes a real noise or a colored noise). Let ${\hat{\xi }}_{\inf }(\omega )$ and ${\hat{\xi }}_{\sup }(\omega )$ be defined as in (1.4) and (1.5) with $a(\cdot )$ being replaced by $\xi (\cdot )$, respectively. Assume that $\xi _t(\cdot )$ satisfies the following (H3).

(H3) $\xi :\Omega \rightarrow {{\mathbb {R}}}$ is measurable; $\int _\Omega |\xi (\omega )|d{\mathbb {P}}(\omega )<\infty$ and $\int _\Omega \xi (\omega )d{\mathbb {P}}(\omega )=0$; $-1<{ {\hat{\xi }}_{\inf }(\omega )}\le {{\hat{\xi }}_{\sup }(\omega )}<\infty$ and${ \inf _{t\in {{\mathbb {R}}}}\xi (\theta _{t}\omega )}>-\infty$ for a.e. $\omega \in \Omega$; and $\xi ^\omega (t):=\xi (\theta _t\omega )$ is locally Hölder continuous.

Assume (H3). By the arguments of Lemma 2.1, there are ${\underline{\xi }},{\overline{\xi }}\in {{\mathbb {R}}}$ such that ${\hat{\xi }}_{\inf }(\omega )={{\underline{\xi }}}$ and ${\hat{\xi }}_{\sup }(\omega )={{\overline{\xi }}}$ for a.e. $\omega \in \Omega$. It can be proved that

$$\begin{aligned} Y(\omega )=\frac{1}{\int _{-\infty }^0 e^{ s+\int _0^s \xi (\theta _\tau \omega )d\tau }ds} \end{aligned}$$

(1.22)

is a spatially homogeneous asymptotically stable random equilibrium of (1.21) (see Theorem 3.2 and Corollary 3.1). It can also be proved that for any $u_0\in X_c^+$,

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\le ct}|\frac{u(t,x;u_0,\theta _s\omega )}{Y(\theta _{t+s}\omega )}-1|=0, \quad \forall \ 0<c<2\sqrt{1+{\underline{\xi }}} \end{aligned}$$

and

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} \frac{u(t,x;u_0,\theta _s\omega )}{Y(\theta _{t+s}\omega )}=0, \quad \forall \ c>2\sqrt{1+{{\bar{\xi }}}}, \end{aligned}$$

for a.e. $\omega \in \Omega$. where $u(t,x;u_0,\theta _s\omega )$ is the solution of (1.21) with $\omega$ being replaced by $\theta _s\omega$ and $u(0,x;u_0,\theta _s\omega )=u_0(x)$ (see Corollary 4.1).

Fourth, it is interesting to study the spreading properties of (1.1) with (H1) being replaced by the following weaker assumption,

(H1)$'$ $0<{{\hat{a}}}:=\int _\Omega a(\omega )d{\mathbb {P}}(\omega )<\infty$.

We plan to study this general case somewhere else, which would have applications to the study of the spreading properties of the following stochastic Fisher–KPP equation,

$$\begin{aligned} d u=(u_{xx}+u(1-u))dt+\sigma u dW_t,\quad x\in {{\mathbb {R}}}, \end{aligned}$$

(1.23)

where $W_t$ denotes the standard two-sided Brownian motion ($dW_t$ is then the white noise). In fact, let $\Omega :=\{\omega \in C({{\mathbb {R}}},{{\mathbb {R}}})\ |\ \omega (0)=0\ \}$ equipped with the open compact topology, ${\mathcal {F}}$ be the Borel $\sigma -$field and ${\mathbb {P}}$ be the Wiener measure on $(\Omega , {\mathcal {F}})$. Let $W_t$ be the one dimensional Brownian motion on the Wiener space $(\Omega ,{\mathcal {F}},{\mathbb {P}})$ defined by $W_t(\omega )=\omega (t)$. Let $\theta _t\omega$ be the canonical Brownian shift: $(\theta _t\omega )(\cdot )=\omega (t+\cdot )-\omega (t)$ on $\Omega$. It is easy to see that $W_t(\theta _s\omega )=W_{t+s}(\omega )-W_s(\omega )$. If $\frac{\sigma ^2}{2}<1$, then it can be proved that

$$\begin{aligned} Y(\omega )=\frac{1}{\int _{-\infty }^0 e^{(1-\frac{\sigma ^2}{2})s+\sigma W_s (\omega )ds}} \end{aligned}$$

(1.24)

is a spatially homogeneous stationary solution process of (1.23). Let ${{\tilde{u}}}=\frac{u}{Y(\theta _t\omega )}$ and drop the tilde, (1.23) becomes (1.1) with $a(\theta _t\omega )= Y(\theta _t\omega )$. The reader is referred to [18,19,20, 23, 39, 40] for some study on the front propagation dynamics of (1.24). Note that Theorem 1.4 (i) is an analogue of [20, Theorem 1].

It is important to note that the authors of the work [10] studied the asymptotic spreading speeds for space-time heterogeneous equations of the form

$$\begin{aligned} u_t=\sum _{i,j=1}^Na_{i,j}(t,x)u_{x_ix_j} +\sum _{i=1}^Nq_i(t,x)u_{x_i} +f(t,x,u),\quad x\in {{\mathbb {R}}}^N, \end{aligned}$$

(1.25)

where $f(t,x,0)=f(t,x,1)=0$. We note that Theorem 1.5 improves [10, Proposition 3.9], since $\inf _{t\in {{\mathbb {R}}}}a_0(t)=0$ and $\sup _{t\in \mathbb {R}}a_0(t)=+\infty$ are allowed here. Moreover, the techniques developed in the current work are different from the ones in [10]. Certainly, it should be mentioned that (1.25) is more general than (1.2).

The rest of the paper is organized as follows. In Sect. 2, we present some preliminary lemmas, which will be used in the proofs of main results of the current paper in later sections. In Sect. 3, we establish some results about the stability of the positive constant equilibrium solution $u\equiv 1$ of (1.1) (resp. (1.2)) and prove Theorem 1.1. In Sect. 4, we study the spreading properties of solutions of (1.1) with nonnegative and compactly supported initial functions or front like initial functions and prove Theorems 1.2 and 1.3 . We investigate in Sect. 4 the take-over property of (1.1) and prove Theorem 1.4. We consider spreading properties of (1.2) in Sect. 5.

2 Preliminary Lemmas

In this section, we present some preliminary lemmas to be used in later sections of this paper as well as in the second part of the series.

Lemma 2.1

(H1) implies that ${\hat{a}}_{\inf }(\cdot ),a(\cdot ), { {{\hat{a}}}_{\sup }(\cdot )}\in L^1 (\Omega , {\mathcal {F}},{\mathbb {P}})$ and that there are ${\underline{a}}, {\bar{a}}, {{\hat{a}}}\in {{\mathbb {R}}}^+$ and a measurable subset $\Omega _0\subset \Omega$ with ${{\mathbb {P}}}(\Omega _0)=1$ such that $\theta _t\Omega _0=\Omega _0$ for all $t\in {{\mathbb {R}}}$, ${{\hat{a}}}_{\inf }(\omega ) ={\underline{a}}$ and ${{\hat{a}}}_{\sup }(\omega )={\bar{a}}$ for all $\omega \in \Omega _0$, and $\lim _{t\rightarrow \pm \infty }\frac{1}{t}\int _0^t a(\theta _\tau \omega )d\tau ={{\hat{a}}}$ for all $\omega \in \Omega _0$.

Proof

First, let

$$\begin{aligned} \Omega _n=\{\omega \in \Omega \, |\, {{\hat{a}}}_{\sup }(\omega )\le n\}\quad \forall \, n\in {{\mathbb {N}}}, \end{aligned}$$

and

$$\begin{aligned} \Omega _\infty =\{\omega \in \Omega \,|\, {{\hat{a}}}_{\sup }(\omega )=\infty \}. \end{aligned}$$

Then ${\Omega _\infty \cup } \cup _{n=1}^\infty \Omega _n=\Omega$. By (H1), there is ${\bar{n}}\in {{\mathbb {N}}}$ such that ${{\mathbb {P}}}(\Omega _{{\bar{n}}})>0$. By (1.6), $\theta _t\Omega _n=\Omega _n$ for all $t\in {{\mathbb {R}}}$ and $n\in {{\mathbb {N}}}$. Then by the ergodicity of the metric dynamical system $(\Omega , {\mathcal {F}},{\mathbb {P}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$, we have ${\mathbb {P}}(\Omega _{{\bar{n}}})=1$. This implies that ${\hat{a}}_{\sup }(\cdot )\in L^1 (\Omega , {\mathcal {F}},{\mathbb {P}})$, and then ${\hat{a}}_{\inf }(\cdot )\in L^1 (\Omega , {\mathcal {F}},{\mathbb {P}})$. Moreover, by (1.6),

$$\begin{aligned} {\hat{a}}_{\inf }(\omega )=\lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t{\hat{a}}_{\inf }(\theta _\tau \omega )d\tau =\int _\Omega {\hat{a}}_{\inf }(\omega ) d{\mathbb {P}}(\omega )\quad \mathrm{for}\quad a.e. \,\, \omega \in \Omega , \end{aligned}$$

and

$$\begin{aligned} {\hat{a}}_{\sup }(\omega )=\lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t{\hat{a}}_{\sup }(\theta _\tau \omega )d\tau =\int _\Omega {\hat{a}}_{\sup }(\omega ) d{\mathbb {P}}(\omega )\quad \mathrm{for}\quad a.e. \,\, \omega \in \Omega . \end{aligned}$$

It then follows that there are ${\underline{a}},{\bar{a}}\in {{\mathbb {R}}}$ and a measurable subset $\Omega _1\subset \Omega$ with ${{\mathbb {P}}}(\Omega _1)=1$ such that $\theta _t\Omega _1=\Omega _1$ for all $t\in {{\mathbb {R}}}$, and ${{\hat{a}}}_{\inf }(\omega )={\underline{a}}$ and ${{\hat{a}}}_{\sup }(\omega )={\bar{a}}$ for all $\omega \in \Omega _1$.

Next, for given $n\in {{\mathbb {N}}}$, let

$$\begin{aligned} a_n(\omega )=\min \{a(\omega ), n\}. \end{aligned}$$

Then $a_n(\cdot )\in L^1 (\Omega , {\mathcal {F}},{\mathbb {P}})$, $0<a_1(\omega )\le a_2(\omega )\le \cdots$, and $\lim _{n\rightarrow \infty } a_n(\omega )=a(\omega )$. By the ergodicity of the metric dynamical system $(\Omega , {\mathcal {F}},{\mathbb {P}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$, we have that for a.e. $\omega \in \Omega$,

$$\begin{aligned} \int _\Omega a_n(\omega )d{\mathbb {P}}(\Omega )=\lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t a_n(\theta _\tau \omega )d\tau \le {\hat{a}}_{\sup }(\omega )=\int _\Omega {\hat{a}}_{\sup }(\omega )d{\mathbb {P}}(\omega ). \end{aligned}$$

This together with the Monotone Convergence Theorem implies that

$$\begin{aligned} \int _\Omega a(\omega )d{\mathbb {P}}(\omega )=\lim _{n\rightarrow \infty }\int _\Omega a_n(\omega )d{\mathbb {P}}(\omega )\le \int _{\Omega }{{\hat{a}}}_{\sup }(\omega )d{\mathbb {P}}(\omega ). \end{aligned}$$

Therefore, $a(\cdot )\in L^1 (\Omega , {\mathcal {F}},{\mathbb {P}})$, and moreover, by the ergodicity of the metric dynamical system $(\Omega , {\mathcal {F}},{\mathbb {P}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$, there are ${{\hat{a}}}\in {{\mathbb {R}}}$ and a measurable subset $\Omega _2\subset \Omega$ with ${{\mathbb {P}}}(\Omega _2)=1$ such that $\theta _t \Omega _2=\Omega _2$ for all $t\in {{\mathbb {R}}}$, and

$$\begin{aligned} {{\hat{a}}}=\lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t a(\theta _\tau \omega )d\tau {=\lim _{t\rightarrow \infty }\frac{1}{t}\int _{-t}^0 a(\theta _\tau \omega )d\tau }= \int _\Omega a(\omega ) d{{\mathbb {P}}}(\omega )\quad \mathrm{for}\quad a.e. \,\,\omega \in \Omega . \end{aligned}$$

The lemma thus follows with $\Omega _0=\Omega _1\cap \Omega _2$. $\square$

Lemma 2.2

Suppose that $b\in C({{\mathbb {R}}}, (0,\infty ))$ and that $0<\underline{ b}\le \overline{ b}<\infty$, where

$$\begin{aligned} {\underline{b}}= \liminf _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t b(\tau )d\tau ,\quad {\overline{b}}=\limsup _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t b(\tau )d\tau . \end{aligned}$$

Then

$$\begin{aligned} {{{\underline{b}}}=}\sup _{B\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}})}\mathrm{essinf}_{\tau \in {{\mathbb {R}}}}(b(\tau )-{ B'}(\tau )). \end{aligned}$$

(2.1)

Proof

The proof of this lemma follows from a proper modification of the proof of [33, Lemma 3.2]. For the sake of completeness we give a proof here. Let $0<\gamma <{\underline{b}}$. By ${\overline{b}}<\infty$, there is $T>0$ such that

$$\begin{aligned} \gamma<\frac{1}{T}\int _s^{s+T}b(\tau )d\tau <2{\overline{b}}, \qquad \forall s\in {{\mathbb {R}}}. \end{aligned}$$

(2.2)

Define

$$\begin{aligned} B(t)= & {} \int _{kT}^{t}\Big ( b(\tau )-\varepsilon _k \Big )d\tau , \quad \forall t\in [kT, (k+1)T] \quad \text {where} \quad \varepsilon _k\\:= & {} \frac{1}{T}\int _{kT}^{(k+1)T}b(\tau )ds, \quad \forall \ k\in {\mathbb {Z}}. \end{aligned}$$

It is clear that $B\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}}) \cap L^\infty ({{\mathbb {R}}})$ with

$$\begin{aligned} \varepsilon _k=b(t)-B'(t) \quad \text {for}\ \ t\in (kT,(k+1)T). \end{aligned}$$

(2.3)

Furthermore, it follows from (2.2) that $\Vert B\Vert _{\infty }\le 2T{\overline{b}}$ and that $\gamma <\varepsilon _k$ for every $k\in {\mathbb {Z}}$. Hence (2.3) implies that

$$\begin{aligned} \gamma \le \sup _{B\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}})}\mathrm{essinf }_{t\in {{\mathbb {R}}}}(b(t)-B'(t)). \end{aligned}$$

Since $\gamma$ is arbitrarily chosen less than ${\underline{b}}$ we deduce that

$$\begin{aligned} {\underline{b}}\le \sup _{B\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}})}\mathrm{essinf }_{t\in {{\mathbb {R}}}}(b(t)-B'(t)). \end{aligned}$$

On the other hand for each given $B\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}})$ and $t>s$ we have

$$\begin{aligned} \frac{1}{t-s}\int _s^{t}b(\tau )d\tau\ge & {} \mathrm{essinf }_{\tau \in {{\mathbb {R}}}}(b(\tau )-B'(\tau ))+ \frac{(B(t)-B(s))}{t-s}\\\ge & {} \mathrm{essinf }_{\tau \in {{\mathbb {R}}}}(b(\tau )-B'(\tau ))- \frac{2\Vert B\Vert _{\infty }}{t-s}. \end{aligned}$$

Hence

$$\begin{aligned} {\underline{b}}=\liminf _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^{t}b(\tau )d\tau \ge \mathrm{essinf }_{\tau \in {{\mathbb {R}}}}(b(\tau )-B'(\tau )) \quad \forall B\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}}). \end{aligned}$$

This completes the proof of the lemma. $\square$

In the following, let $b\in C({{\mathbb {R}}}, (0,\infty ))$ be given and satisfy that $0<{\underline{b}}\le {\overline{b}}<\infty$. Consider

$$\begin{aligned} u_t=u_{xx}+b(t)u(1-u),\quad x\in {{\mathbb {R}}}. \end{aligned}$$

(2.4)

For given $u_0\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$ with $u_0\ge 0$, let $u(t,x;u_0,b)$ be the solution of (2.4) with $u(0,x;u_0,b)=u_0(x)$.

For every $0<\mu <{\underline{\mu }}^*:=\sqrt{{\underline{b}}}$, $x\in {{\mathbb {R}}}$, $t\in {{\mathbb {R}}}$ and $\omega \in \Omega$, let

$$\begin{aligned} c(t;b,\mu )=\frac{\mu ^2+b(t)}{\mu },\quad C(t;b,\mu )=\int _0^t c(\tau ;b,\mu )d\tau , \end{aligned}$$

(2.5)

and

$$\begin{aligned} \phi ^{\mu }(t,x;b)=e^{-\mu (x-C(t;b,\mu ))}. \end{aligned}$$

(2.6)

Then the function $\phi ^{\mu }$ satisfies

$$\begin{aligned} \phi ^{\mu }_t=\phi ^{\mu }_{xx} +b(t)\phi ^{\mu },\quad x\in {{\mathbb {R}}}. \end{aligned}$$

(2.7)

Lemma 2.3

Let

$$\begin{aligned} \phi _+^\mu (t,x;b)=\min \{1,\phi ^\mu (t,x;b)\}. \end{aligned}$$

Then

$$\begin{aligned} u(t,x;\phi _+^\mu (0,\cdot ;b),b)\le \phi _+^\mu (t,x;b)\quad \forall \,\, t>0,\,\, x\in {{\mathbb {R}}}. \end{aligned}$$

Proof

It follows directly from the comparison principle for parabolic equations. $\square$

Lemma 2.4

For every $\mu$ with $0<\mu<{\tilde{\mu }}<\min \{2\mu , {\underline{\mu }}^*\}$, there exist $\{t_k\}_{k\in {{\mathbb {Z}}}}$ with $t_k<t_{k+1}$ and $\lim _{k\rightarrow \pm \infty }t_k=\pm \infty$, $B_b\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}})$ with $B_b(\cdot )\in C^1((t_k,t_{k+1}))$ for $k\in {{\mathbb {Z}}}$, and a positive real number $d_b$ such that for every $d\ge d_{b}$ the function

$$\begin{aligned} \phi ^{\mu ,d,B_b}(t,x):=e^{-\mu (x-C(t;b,\mu ))}-de^{\big (\frac{{\tilde{\mu }}}{\mu }-1\big )B_b(t) -{\tilde{\mu }}(x-C(t;b,\mu ))} \end{aligned}$$

satisfies

$$\begin{aligned} \phi ^{\mu ,d,B_b}_t\le \phi ^{\mu ,d,B_b}_{xx}+b(t)\phi ^{\mu ,d,B_b}(1-\phi ^{\mu ,d,B_b}) \end{aligned}$$

for $t\in (t_k,t_{k+1})$, $x\ge C(t,b,\mu )+ \frac{\ln d}{{{\tilde{\mu }}}-\mu }+\frac{ B_b(t)}{\mu },\,\, k\in {{\mathbb {Z}}}$.

Proof

First of all, for given $0<\mu<{\tilde{\mu }}<\min \{2\mu , {\underline{\mu }}^*\}$, let $0<\delta \ll 1$ such that $(1-\delta ){\underline{b}}>{\tilde{\mu }}\mu$. It then follows from the arguments of Lemma 2.2 that there exist $T>0$ and $B_b\in W^{1,\infty }_{\mathrm{loc}}({{\mathbb {R}}})\cap L^{\infty }({{\mathbb {R}}})$ such that $B_b\in C^1((t_k,t_{k+1}))$, where $t_k=kT$ for $k\in {{\mathbb {Z}}}$, and

$$\begin{aligned} {\tilde{\mu }}\mu \le (1-\delta )b(t)+B_b'(t)\quad \text {for all}\ t\in (t_k,t_{k+1}),\,\, k\in {{\mathbb {Z}}}. \end{aligned}$$

Next, fix the above $\delta >0$ and $B_b(t)$. Let $d>1$ to be determined later. Let $\xi (t,x)=x-C(t;b,\mu )$. We have

$$\begin{aligned}&\phi ^{\mu ,d,B_b}_t-\Big ( \phi ^{\mu ,d,B_b}_{xx}+b(t)\phi ^{\mu ,d,B_b}(1-\phi ^{\mu ,d,B_b})\Big )\nonumber \\&\quad = d\Big [-(\frac{{\tilde{\mu }}}{\mu }-1)B_b'(t)+{\tilde{\mu }}^2 -{\tilde{\mu }}c(t;b,\mu )+b(t) \Big ]e^{(\frac{{\tilde{\mu }}}{\mu }-1)B_b(t)-{\tilde{\mu }}\xi (t,x)}\nonumber \\&\qquad + b(t)\Big [ e^{-2\mu \xi (t,x)}-2de^{(\frac{{\tilde{\mu }}}{\mu }-1)B_b(t)-(\mu +{\tilde{\mu }})\xi (t,x)} +d^2e^{2(\frac{{\tilde{\mu }}}{\mu }-1)B_b(t)-2{\tilde{\mu }}\xi (t,x)}\Big ]\nonumber \\&\quad = d\Big (\frac{{\tilde{\mu }}}{\mu }-1 \Big )\Big [ {\tilde{\mu }}\mu -b(t)-B_b'(t) \Big ]e^{(\frac{{\tilde{\mu }}}{\mu }-1)B_b(t)-{\tilde{\mu }}\xi (t,x)} +b(t)e^{-2\mu \xi (t,x)} \nonumber \\&\qquad -d\Big [2e^{-\mu \xi (t,x)}-de^{\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )B_b(t) -{\tilde{\mu }}\xi (t,x)} \Big ]e^{\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )B_b(t) -{\tilde{\mu }}\xi (t,x)}\nonumber \\&\quad = d\Big (\frac{{\tilde{\mu }}}{\mu }-1 \Big )\Big [ {\tilde{\mu }}\mu -(1-\delta )b(t) -B_b'(t) \Big ]e^{(\frac{{\tilde{\mu }}}{\mu }-1)B_b(t)-{\tilde{\mu }}\xi (t,x)} \nonumber \\&\qquad +\Big [e^{-(2\mu -{\tilde{\mu }})\xi (t,x)}-d\delta \Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )e^{\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )B_b(t)} \Big ]a(\theta _t\omega )e^{-{\tilde{\mu }} \xi (t,x)} \nonumber \\&\qquad +d\Big [-2e^{-\mu \xi (t,x)}+de^{\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )B_b(t)-{\tilde{\mu }}\xi (t,x)} \Big ]e^{\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )B_b(t)-{\tilde{\mu }}\xi (t,x)} \end{aligned}$$

(2.8)

for $t\in (t_k,t_{k+1})$.

Observe now that

$$\begin{aligned} d\delta \Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )e^{\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )B_b(t)}\ge 1, \qquad \forall \ d\ge \max \Big \{ \frac{e^{-\Big (\frac{{\tilde{\mu }}}{\mu }-1\Big )\Vert B_b\Vert _{\infty }}}{\delta \Big ( \frac{{\tilde{\mu }}}{\mu }-1\Big )}, e^{\Big ( \frac{{\tilde{\mu }}}{\mu }-1\Big )\Vert B_b\Vert _{\infty }} \Big \}. \end{aligned}$$

For this choice of d, if $\phi ^{\mu ,d,B_b}(t,x)\ge 0$, which is equivalent to $\xi (t,x)=x-C(t;b,\mu )\ge \frac{\ln d}{{{\tilde{\mu }}}-\mu }+\frac{ B_b(t)}{\mu }$, then $\xi (t,x)\ge 0$ and each term in the expression at the right hand side of (2.8) is less or equal to zero. The lemma thus follows. $\square$

Recall that $u_0^*(x)=1$ for $x< 0$ and $u_0^*(x)=0$ for $x>0$. By [25, Theorem 1], the solution of (2.4) with initial function $u_0^*$, denoted by $u(t,x;u_0^*,b)$, exists for $t>0$.

Lemma 2.5

Suppose that $u_\epsilon \in C_{\mathrm{unif}}^b({{\mathbb {R}}})$ with $u_\epsilon \ge 0$ and $\lim _{\epsilon \rightarrow 0}\int _{-\infty }^\infty |u_\epsilon (x)-u_0^*(x)|dx =0.$ Then for each $t>0$,

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} \Vert u(t,\cdot ;u_\epsilon ,b)-u(t,\cdot ;u_0^*,b)\Vert _\infty =0. \end{aligned}$$

Proof

See [25, Theorem 8]. $\square$

Lemma 2.6

For given $u_i\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$ with $u_i\ge 0$ $(i=1,2)$, if $u_1(x)-u_2(x)$ has exactly one simple zero $x_0$ and $u_1(x)>u_2(x)$ for $x<x_0$ and $u_1(x)<u_2(x)$ for $x>x_0$, then for any $t>0$, there is $\xi (t)\in [-\infty ,\infty ]$ such that

$$\begin{aligned} u(t,x;u_1,b)= {\left\{ \begin{array}{ll}>u(t,x;u_2,b) &{} x< \xi (t)\\ <u(t,x;u_2,b) &{} x> \xi (t). \end{array}\right. } \end{aligned}$$

Proof

Let $v(t,x)=u(t,x;u_1,b)-u(t,x;u_2,b)$. Then v(t, x) satisfies

$$\begin{aligned} v_t=v_{xx}+q(t,x) v,\quad x\in {{\mathbb {R}}}, \end{aligned}$$

where $q(t,x)=b(t)-b(t)(u(t,x;u_1,b)+u(t,x;u_2,b))$. Note that v(0, x) has exactly one simple zero $x_0$ and $v(0,x)>0$ for $x<x_0$, $v(x)<0$ for $x>x_0$. The lemma then follows from [1, Theorems A,B]. $\square$

Let x(t, b) and $x_+(t,b)$ be such that

$$\begin{aligned} u(t,x(t,b);u_0^*,b)=\frac{1}{2} \quad \mathrm{and}\quad u(t,x_+(t,b);\phi _+^\mu (0,\cdot ;b),b)=\frac{1}{2}. \end{aligned}$$

Lemma 2.7

For any $t>0$, there holds

$$\begin{aligned} u(t,x+x(t,b);u_0^*,b)){\left\{ \begin{array}{ll} \ge u(t,x+x_+(t,b);\phi _+^\mu (0,\cdot ;b),b)\quad x<0\\ \le u(t,x+x_+(t,b);\phi _+^\mu (0,\cdot ;b),b)\quad x>0. \end{array}\right. } \end{aligned}$$

(2.9)

Proof

First, let $\phi _n(x)=\min \{1-\frac{1}{n},\phi ^\mu (0,x;b)\}$. Then $\lim _{n\rightarrow \infty }\phi _n(x)=\phi _+^\mu (0,x;b)$ uniformly in $x\in {{\mathbb {R}}}$. Then for any given $t>0$,

$$\begin{aligned} u(t,x;\phi _+^\mu (0,\cdot ;b),b)=\lim _{n\rightarrow \infty } u(t,x;\phi _n,b) \end{aligned}$$

uniformly in $x\in {{\mathbb {R}}}$. Let $x_+^n(t,b)$ be such that $u(t,x_+^n(t,b);\phi _n,b)=\frac{1}{2}.$ We have

$$\begin{aligned} \lim _{n\rightarrow \infty } x_+^n(t,b)=x_+(t,b). \end{aligned}$$

Next, for given $n\ge 1$, let $u_\epsilon ^*(x)$ be a nonincreasing function such that $u_\epsilon ^*\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$; $u_\epsilon ^*(x)=1$ for $x\ll -1$ and $u_\epsilon ^*(x)=0$ for $x\gg 0$; $u_\epsilon ^*(x)-\phi _n(x+h)$ has exactly one simple zero for any $h\in {{\mathbb {R}}}$; and

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\int _{-\infty }^\infty |u_\epsilon ^*(x)-u_0^*(x)|dx=0. \end{aligned}$$

Let $x_\epsilon (t,b)$ be such that

$$\begin{aligned} u(t,x;u_\epsilon ^*,b)=\frac{1}{2}. \end{aligned}$$

By Lemma 2.6, for any $t>0$,

$$\begin{aligned} u(t,x+x_\epsilon (t,b),b){\left\{ \begin{array}{ll}>u(t,x+x_+^n(t,b);\phi _n,b)\quad x<0\\ <u(t,x+x_+^n(t,b);\phi _n,b)\quad x>0. \end{array}\right. } \end{aligned}$$

By Lemma 2.5, for any $t>0$,

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\Vert u(t,\cdot ;u_\epsilon ^*,b)-u(t,\cdot ;u_0^*,b)\Vert _\infty =0 \quad \mathrm{and}\quad \lim _{\epsilon \rightarrow 0} x_\epsilon (t,b)=x(t,b). \end{aligned}$$

Letting $\epsilon \rightarrow 0$, we get

$$\begin{aligned} u(t,x+x(t,b);u_0^*,b){\left\{ \begin{array}{ll} \ge u(t,x+x_+^n(t,b);\phi _n,b)\quad x<0\\ \le u(t,x+x_+^n(t,b);\phi _n,b)\quad x>0. \end{array}\right. } \end{aligned}$$

Letting $n\rightarrow \infty$, the lemma follows. $\square$

Lemma 2.8

Let $F: {{\mathbb {R}}}\times \Omega \rightarrow {{\mathbb {R}}}$ be measurable in $\omega \in \Omega$ and continuous hemicompact in $x\in {{\mathbb {R}}}$ (i.e for every $\omega \in \Omega$, $F(\cdot ,\omega )$ is continuous in x and any sequence $\{x_n\}_{n\ge 1}\subset {{\mathbb {R}}}$ with $|x_n-F(x_n,\omega )|\rightarrow 0$ as $n\rightarrow \infty$ has a convergent subsequence). Then F has a deterministic fixed point (i.e there is $X: \Omega \rightarrow {{\mathbb {R}}}$ such that $F(X(\omega ),\omega )=X(\omega )$) if and only if F has random fixed point (i.e there is a measurable function $X: \Omega \rightarrow {{\mathbb {R}}}$ such that $F(X(\omega ),\omega )=X(\omega )$).

Proof

See [42, Lemma 4.7] $\square$

Lemma 2.9

Let $f : {{\mathbb {R}}}\times \Omega \rightarrow (0,1)$ be a measurable function such that for every $\omega \in \Omega$ the function $f^{\omega }:=f(\cdot ,\omega ) : {{\mathbb {R}}}\rightarrow (0,1)$ is continuously differentiable and strictly decreasing. Assume that $\lim _{x\rightarrow -\infty }f^{\omega }(x)=1$ and $\lim _{x\rightarrow \infty }f^{\omega }(x)=0$ for every $\omega \in \Omega$. Then for every $a\in (0,1)$ the function $\Omega \ni \omega \mapsto f^{\omega ,-1}(a)\in {{\mathbb {R}}}$ is measurable, where $f^{\omega ,-1}$ denotes the inverse function of $f^{\omega }$.

Proof

Let $a\in (0,1)$ be given. Note that for every $\omega \in \Omega$, we have that $f^{\omega ,-1}(a)$ is the unique fixed point of the function

$$\begin{aligned} {{\mathbb {R}}}\ni x\mapsto F(x,\omega ):=f(x,w)+x-a. \end{aligned}$$

Note that

$$\begin{aligned} |x_n-F(x_n,\omega )|=|f(x_n,w)-a|\rightarrow 0 \ \text {as}\ n\rightarrow \infty \Rightarrow |x_n-f^{\omega ,-1}(a)|\rightarrow 0 \ \text {as}\ n\rightarrow \infty . \end{aligned}$$

Hence the function $F(x,\omega )$ is hemicompact in x. By Lemma 2.8, the function $\Omega \ni \omega \mapsto f^{\omega ,-1}(a)$ is measurable. The lemma is thus proved. $\square$

3 Stability of Positive Random Equilibrium Solutions

In this section, we establish some results about the stability of the positive constant equilibrium solution $u\equiv 1$ of (1.1) (resp. (1.2)). We also study the existence and stability of positive random equilibria of (1.21). The results obtained in this section will play a role in later sections for the investigation of spreading speeds and take-over property of solutions of (1.1) [resp. (1.2)].

3.1 Stability of the Positive Constant Equilibrium Solution $u\equiv 1$ of (1.1)

In this subsection, we establish some results about the stability of the positive constant equilibrium solution $u\equiv 1$ of (1.1) [resp. (1.2)]. Observe that $u(t,x)=v(t,x-C(t;\omega ))$ with $C(t;\omega )$ being differential in t solves (1.1) if and only if v(t, x) satisfies

$$\begin{aligned} v_t=v_{xx}+c(t;\omega )v_x+a(\theta _t\omega )v(1-v), \end{aligned}$$

(3.1)

where $c(t;\omega )=C'(t;\omega )$. In this subsection, we also study the stability of the positive constant equilibrium solution $u\equiv 1$ of (3.1).

We first prove Theorem 1.1.

Proof of Theorem 1.1

First, for given $u_0\in C^{b}_{\mathrm{uinf}}({{\mathbb {R}}})$ with $\inf _{x\in {\mathbb {R}}}u_0(x)>0$ and $\omega \in {\Omega }$, let ${\underline{u}}_0:=\min \{1, \inf _{x\in {\mathbb {R}}}u_0(x)\}$ and ${\overline{u}}_0:=\max \{1,\sup _{x\in {\mathbb {R}}}u_0(x)\}$. By the comparison principle for parabolic equations, we have that

$$\begin{aligned} {\underline{u}}_0\le u(t,x;{\underline{u}}_0,\omega )\le \min \{1, u(t,x;u_0,\omega )\},\quad \forall \ x\in {{\mathbb {R}}},\ \forall \ t\ge 0 \end{aligned}$$

(3.2)

and

$$\begin{aligned} \max \{1, u(t,x;u_0,\omega )\} \le u(t,x;{\overline{u}}_0,\omega )\le {\overline{u}}_0, \quad \forall \ x\in {{\mathbb {R}}},\ \forall \ t\ge 0. \end{aligned}$$

(3.3)

Since ${\underline{u}}_0$ and ${\overline{u}}_0$ are positive numbers, by the uniqueness of solutions of (1.1) and its corresponding ODE with a given initial function, we have that

$$\begin{aligned} u(t,x;{\underline{u}}_0,\omega )=u(t,0;{\underline{u}}_0,\omega )\quad \text {and}\quad u(t,x;{\overline{u}}_0,\omega )=u(t,0;{\overline{u}}_0,\omega )\quad \forall \ x\in {{\mathbb {R}}}, \ \forall t\ge 0. \end{aligned}$$

Next, let ${\underline{u}}(t)=\Big (\frac{1}{u(t,0;{\underline{u}}_0,\omega )} -1\Big )e^{\int _0^ta(\theta _s\omega )ds}$ and ${\overline{u}}(t)=\Big (1-\frac{1}{u(t,0;{\overline{u}}_0,\omega )}\Big ) e^{\int _0^ta(\theta _s\omega )ds}$. It can be verified directly that

$$\begin{aligned} \frac{d }{dt}{\underline{u}}=\frac{d}{dt}{\overline{u}}=0, \quad t>0. \end{aligned}$$

Hence,

$$\begin{aligned} {\underline{u}}(t)={\underline{u}}(0) \quad \text {and} \quad {\overline{u}}(t)={\overline{u}}(0),\quad \forall \ t\ge 0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} 1-u(t,x;{\underline{u}}_0,\omega )={\underline{u}}(0) u(t,x;{\underline{u}}_0,\omega )e^{-\int _0^ta(\theta _s\omega )ds} \end{aligned}$$

(3.4)

and

$$\begin{aligned} u(t,x;{\overline{u}}_0,\omega )-1={\overline{u}}(0)u(t,x;{\overline{u}}_0,\omega ) e^{-\int _0^ta(\theta _s\omega )ds}. \end{aligned}$$

(3.5)

Now, by (3.2)–(3.5), we have that

$$\begin{aligned} |u(t,x;u_0,\omega )-1|\le {\overline{u}}_0\max \{{\overline{u}}(0),{\underline{u}}(0)\} e^{-\int _0^ta(\theta _s\omega )ds}, \quad \forall x\in {{\mathbb {R}}},\ t\ge 0, \end{aligned}$$

which implies that inequality (1.8) holds. Taking $u_0$ to be a positive constant with $0<u_0<1$, it follows from (3.4) that

$$\begin{aligned} u(t,x;{\underline{u}}_0,\omega )=\frac{1}{1+(\frac{1}{{\underline{u}}_0}-1) e^{-\int _0^ta(\theta _s\omega )ds}}. \end{aligned}$$

If $\Vert a(\theta _{\cdot }\omega )\Vert _{L^{1}(0,\infty )}<\infty$, then $\lim _{t\rightarrow \infty }u(t,x;{\underline{u}}_0,\omega )=\frac{1}{1+(\frac{1}{{\underline{u}}_0}-1)e^{-\Vert a(\theta _{\cdot }\omega )\Vert _{L^{1}(0,\infty )}}}<1$, which completes the proof of the theorem. $\square$

Remark 3.1

• (1) Theorem 1.1 guarantees the exponential stability of the trivial constant equilibrium solution $u\equiv 1$ of (1.1) with respect to the solutions $u(t,x;u_0,\omega )$ with $\inf _{x\in {\mathbb {R}}^n}u_0(x)>0$ provided that (H1) holds. This result will be useful in the later sections.

• (2) Let $v(t,x;u_0,\omega )$ be the solution of (3.1) with $v(0,x;u_0,\omega )=u_0(x)$. The result in Theorem 1.1 also holds for $v(t,x;u_0,\omega )$.

Let

$$\begin{aligned} {\underline{c}}(\omega )= \liminf _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t c(\tau ;\omega )d\tau ,\quad {\overline{c}}(\omega )=\limsup _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t c(\tau ;\omega )d\tau . \end{aligned}$$

Next, we prove the following theorem about the stability of $u\equiv 1$.

Theorem 3.1

Assume (H1). Suppose that $v(t,x;\omega )$ with $0<v(t,x;\omega )<1$, is an entire solution of (3.1) which is nonincreaing in x. For given $\omega \in \Omega$ with $0<{{\underline{c}}(\omega )\le {\overline{c}}(\omega )}<\infty$, if there is $x^*\in {{\mathbb {R}}}$ such that $\inf _{t\in {{\mathbb {R}}}} v(t,x^*;\omega )>0$, then $\lim _{x\rightarrow -\infty } v(t,x;\omega )=1$ uniformly in $t\in {{\mathbb {R}}}$.

To prove the above theorem, we first prove a lemma.

Lemma 3.1

Let $u_0,u_n\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$ be such that $0\le u_n(x)\le u_0(x)\le 1$. Let $v(t,x;u_0,\theta _{t_0}\omega )$ (respectively $v(t,x;u_n,\theta _{t_0}\omega )$) denote the solution of (3.1) with $\omega$ being replaced by $\theta _{t_0}\omega$ and with initial function $u_0$ (respectively $u_n$). If $\lim _{n\rightarrow \infty } u_n(x)=u_0(x)$ locally uniformly in $x\in {{\mathbb {R}}}$, then for any fixed $t>0$ with $-\infty<\inf _{t_0\in {{\mathbb {R}}}}\int _{0}^{t} c(\tau +t_0;\omega )\le \sup _{t_0\in {{\mathbb {R}}}} \int _0^t c(\tau +t_0;\omega )<\infty$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty } v(t,x;u_n,\theta _{t_0}\omega )=v(t ,x;u_0,\theta _{t_0}\omega ) \end{aligned}$$

uniformly in $t_0\in {{\mathbb {R}}}$ and locally uniformly in $x\in {{\mathbb {R}}}$.

Proof

Fix $\omega \in \Omega$. For every $n\ge 1$, the function $v^n(t,x;t_0):=v(t,x;u_0,\theta _{t_0}\omega )-v(t,x;u_n,\theta _{t_0}\omega )$ is non-negative and satisfies

$$\begin{aligned} v^n_t&=v^n_{xx}{ + c(t+t_0;\omega )v^n_x} + a(\theta _{t_0+t}\omega ) (1-(v(t,x;u_0,\theta _{t_0}\omega )+v(t,x;u_n,\theta _{t_0}\omega ) ))v^n\\&\le v^n_{xx} + c(t+t_0;\omega )v^n_x + a(\theta _{t_0+t}\omega )v^n. \end{aligned}$$

It follows that, for every $n\ge 1$, ${{\tilde{v}}}^n(t,x;t_0):=v^n(t,x-\int _{t_0}^{t_0+t}c(\tau ;\omega )d\tau );t_0)$ satisfies

$$\begin{aligned} {{\tilde{v}}}^n(t,x;t_0)\le {{\tilde{v}}}^n_{xx}+a(\theta _{t+t_0}\omega ){{\tilde{v}}}^n, \end{aligned}$$

By the comparison principle for parabolic equations,

$$\begin{aligned} 0\le v^n(t,\cdot ;t_0)\le e^{\int _{t_0}^{t_0+t}a(\theta _{\tau }\omega )d\tau }e^{t\Delta } v^n(0,\cdot +\int _0^tc(\tau +t_0)d\tau ). \end{aligned}$$

Note that $\lim _{n\rightarrow \infty } \big ( e^{\int _{t_0}^{t_0+t}a(\theta _{\tau }\omega )d\tau }e^{t\Delta } v^n(0,\cdot +\int _0^tc(\tau +t_0)d\tau )\big )(x)=0$ locally uniformly in $x\in {{\mathbb {R}}}$ and uniformly in $t_0\in {{\mathbb {R}}}$. Hence $\lim _{n\rightarrow \infty } v^n(t,x;t_0)=0$ uniformly in $t_0\in {{\mathbb {R}}}$ and locally uniformly in $x\in {{\mathbb {R}}}$. $\square$

We now prove Theorem 3.1.

Proof of Theorem 3.1

Fix $\omega \in \Omega$ with $-\infty<{\underline{c}}(\omega )\le {\overline{c}}(\omega )<\infty$ and assume that there is $x^*\in {{\mathbb {R}}}$ such that $\inf _{t\in {{\mathbb {R}}}} v(t,x^*;\omega )>0$.

Consider the constant function $u_0\equiv \inf _{t\in {{\mathbb {R}}}} v(t,x^*;\omega )$. We first note from the hypotheses of Theorem 3.1 that $u_0>0$. Next, let ${{\tilde{u}}}_0(\cdot )$ be uniformly continuous, $0\le {{\tilde{u}}}_0(x)\le u_0$, ${{\tilde{u}}}_0(x)=u_0$ for $x\le x^*-1$, and ${{\tilde{u}}}_0(x)=0$ for $x\ge x^*$. For any $R>0$, it holds that ${\tilde{u}}_0(x-n)=u_0$ for every $|x|\le R$ and $n\ge R+1+|x^*|$. This shows that $\lim _{n\rightarrow \infty } {{\tilde{u}}}_0(x-n)=u_0$ locally uniformly in $x\in {{\mathbb {R}}}$. By (H1) and the arguments of Theorem 1.1,

$$\begin{aligned} \lim _{t\rightarrow \infty } v(t,x;u_0,\theta _{t_0}\omega )=1 \end{aligned}$$

uniformly in $t_0\in {{\mathbb {R}}}$ and $x\in {{\mathbb {R}}}$. Hence, for any $\epsilon >0$, there is $T>0$ such that

$$\begin{aligned} -\infty<\inf _{t_0\in {{\mathbb {R}}}}\int _{0}^{T} c(\tau +t_0;\omega )\le \sup _{t_0\in {{\mathbb {R}}}} \int _0^T c(\tau +t_0;\omega )<\infty \end{aligned}$$

and

$$\begin{aligned} 1>v(T,x; u_0, \theta _{t_0}\omega )>1-\epsilon \quad \forall \,\, t_0\in {{\mathbb {R}}},\,\, x\in {{\mathbb {R}}}. \end{aligned}$$

By Lemma 3.1, there is $N>1$ such that

$$\begin{aligned} 1>v(T,0;{{\tilde{u}}}_0(\cdot -N),\theta _{t_0}\omega )>1-2\epsilon \quad \forall \,\, t_0\in {{\mathbb {R}}}. \end{aligned}$$

This implies that

$$\begin{aligned} 1>v({ T},-N;{{\tilde{u}}}_0,\theta _{t_0}\omega )>1-2\epsilon \quad \forall \,\, t_0\in {{\mathbb {R}}}. \end{aligned}$$

Note that

$$\begin{aligned} v(t-T,x;\omega )\ge {{\tilde{u}}}_0(x)\quad \forall \, \, t\in {{\mathbb {R}}},\,\, x\in {{\mathbb {R}}}\end{aligned}$$

and

$$\begin{aligned} v(t,x;\omega )=v(T,x;v(t-T,\cdot ),\theta _{t-T}\omega ). \end{aligned}$$

Hence

$$\begin{aligned} 1>v(t,x;\omega )=v(T,x;v(t-T,\cdot ),\theta _{t-T}\omega )>1-2\epsilon \quad \forall \,\, t\in {{\mathbb {R}}},\,\, x\le -N. \end{aligned}$$

The theorem thus follows. $\square$

3.2 Existence and Stability of Positive Random Equilibria of (1.21)

In this subsection, we first study the existence and stability of positive random equilibria of (1.21), and then show that (1.21) can be transferred to (1.1).

To this end, we consider the following corresponding ODE,

$$\begin{aligned} \dot{u}=u(1+ \xi (\theta _t\omega )-u). \end{aligned}$$

(3.6)

Throughout this subsection, we assume that (H3) holds. For given $u_0\in {{\mathbb {R}}}$, let $u(t;u_0,\omega )$ be the solution of (3.6) with $u(0;u_0,\omega )=u_0$. It is known that

$$\begin{aligned} u(t;u_0,\omega )=\frac{u_0 e^{t+\int _0^ t\xi (\theta _\tau \omega )d\tau }}{1+u_0\int _0^t e^{ s+\int _0^s \xi (\theta _\tau \omega )d\tau }ds}. \end{aligned}$$

Theorem 3.2

$Y(\omega )=\frac{1}{\int _{-\infty }^0 e^{ s+\int _0^s \xi (\theta _\tau \omega )d\tau }ds}$ is a random equilibrium of (3.6), that is, $u(t;Y(\omega ),\omega )=Y(\theta _t\omega )$ for $t\in {{\mathbb {R}}}$ and $\omega \in \Omega$.

Proof

First, we note that

$$\begin{aligned} u(t;Y(\omega ),\omega )&=\frac{Y(\omega ) e^{t+\int _0^ t\xi (\theta _\tau \omega )d\tau }}{1+Y(\omega )\int _0^t e^{ s+\int _0^s \xi (\theta _\tau \omega )d\tau }ds}\\&=\frac{ e^{t+\int _0^ t\xi (\theta _\tau \omega )d\tau }}{\int _{-\infty }^0 e^{s +\int _0^ s\xi (\theta _\tau \omega )d\tau }ds +\int _0^t e^{ s+\int _0^s \xi (\theta _\tau \omega )d\tau }ds}\\&=\frac{ e^{t+\int _0^ t\xi (\theta _\tau \omega )d\tau }}{\int _{-\infty }^t e^{s +\int _0^ s\xi (\theta _\tau \omega )d\tau }ds}. \end{aligned}$$

Second, note that

$$\begin{aligned} Y(\theta _t\omega )&=\frac{1}{\int _{-\infty }^0 e^{ s+\int _0^s \xi (\theta _{t +\tau }\omega )d\tau }ds}=\frac{1}{\int _{-\infty }^t e^{(s-t)+\int _0^{s-t} \xi (\theta _{t+\tau }\omega )d\tau }ds}\\&=\frac{ e^{t+\int _0^ t\xi (\theta _\tau \omega )d\tau }}{\int _{-\infty }^t e^{s +\int _0^ s\xi (\theta _\tau \omega )d\tau }ds}. \end{aligned}$$

Hence $u(t;Y(\omega ),\omega )=Y(\theta _t\omega )$ and then $Y(\omega )$ is a random equilibrium of (3.6). $\square$

Observe that $0<Y(\omega )<\infty$. Let ${{\tilde{u}}}=\frac{u}{Y(\theta _t\omega )}$ and drop the tilde. We have

$$\begin{aligned} u_t=u_{xx}+Y(\theta _t\omega ) u(1-u). \end{aligned}$$

(3.7)

Clearly, (3.7) is of the form (1.1) with $a(\omega )=Y(\omega )$. Let $\hat{Y}_{\inf }(\omega )$ and $\hat{Y}_{\sup }(\omega )$ be defined as in (1.4) and (1.5) with $a(\cdot )$ being replaced by $Y(\cdot )$, respectively.

Lemma 3.2

$Y(\omega )$ satisfies the following properties.

• (1) For a.e. $\omega \in \Omega$, $0<{ \inf _{ t\in {{\mathbb {R}}}}Y(\theta _t\omega )\le \sup _{t\in {{\mathbb {R}}}}Y(\theta _t\omega )}<\infty$.

• (2) For a.e. $\omega \in \Omega$, $\lim _{t\rightarrow \infty } \frac{\ln Y(\theta _t\omega )}{t}=0$.

• (3) For a.e. $\omega \in \Omega$, $\lim _{t\rightarrow \infty } \frac{\int _0^t Y(\theta _s\omega )ds}{t}=1$.

• (4) ${\hat{Y}_{\inf }}(\omega )=1+{\underline{\xi }}>0$, and ${\hat{Y}_{\sup }}(\omega )=1+{\overline{\xi }}<\infty$ for a.e. $\omega \in \Omega$.

Proof

(1) First, note that

$$\begin{aligned} \frac{1}{Y(\theta _t\omega )}=&\int _{-\infty }^{-T}e^{s-\int _s^0\xi (\theta _{\tau +t}\omega )\tau }ds+ \int _{-T}^{0}e^{s-\int _s^0\xi (\theta _{\tau +t}\omega )\tau }ds, \quad \forall T>0, \forall \ t\in {{\mathbb {R}}}. \end{aligned}$$

(3.8)

By (H3), for every $\lambda \in (0,1)$ and a.e. $\omega \in \Omega$ there is $T_{\lambda }\gg 1$,

$$\begin{aligned} \lambda {\underline{\xi }}\le \frac{1}{T}\int _{0}^T\xi (\theta _{x+\tau }\omega )d\tau \le \frac{{\overline{\xi }}}{\lambda }, \quad \forall \ x\in {{\mathbb {R}}}, \forall \ T\ge T_{\lambda }. \end{aligned}$$

It then follows that

$$\begin{aligned} \int _{-\infty }^{-T_{\lambda }}e^{(1+\frac{{\overline{\xi }}}{\lambda })s}ds\le \int _{-\infty }^{-T_\lambda }e^{s-\int _s^0\xi (\theta _{\tau +t}\omega )\tau }ds\le \int _{-\infty }^{-T_{\lambda }}e^{(1+\lambda {\underline{\xi }})s}ds. \quad \end{aligned}$$

That is,

$$\begin{aligned} \frac{e^{-(1+\frac{{\overline{\xi }}}{\lambda })T_{\lambda }}}{(1+\frac{{\overline{\xi }}}{\lambda })}\le \int _{-\infty }^{-T_\lambda }e^{s-\int _s^0\xi (\theta _{\tau +t}\omega )\tau }ds\le \frac{e^{-(1+\lambda {\underline{\xi }})T_{\lambda }}}{(1+\lambda {\overline{\xi }})}. \end{aligned}$$

(3.9)

The first inequality of (3.9) combined with (3.8) yields that

$$\begin{aligned} \frac{1}{Y(\theta _t\omega )}\ge \frac{e^{-(1+\frac{{\overline{\xi }}}{\lambda })T_{\lambda }}}{(1+\frac{{\overline{\xi }}}{\lambda })}. \end{aligned}$$

Hence

$$\begin{aligned} Y(\theta _t\omega )\le (1+\frac{{\overline{\xi }}}{\lambda })e^{(1+\frac{{\overline{\xi }}}{\lambda }) T_{\lambda }}, \quad \ \forall \ t\in {{\mathbb {R}}}. \end{aligned}$$

(3.10)

Next, let $\xi _{\inf }(\omega )=\inf _{t\in {{\mathbb {R}}}}\xi (\theta _t\omega )$. Observe that

$$\begin{aligned} \int _{-T_{\lambda }}^{0}e^{s-\int _s^0\xi (\theta _{\tau +t}\omega )\tau }ds\le \int _{-T_{\lambda }}^{0}e^{s-\int _s^0\xi _{\inf }(\omega )d\tau }ds =\int _{-T_{\lambda }}^{0}e^{s(1+\xi _{\inf }(\omega ))}ds. \end{aligned}$$

This combined with the second inequality in (3.9) yield that

$$\begin{aligned} \frac{1}{Y(\theta _t\omega )}\le \frac{e^{-(1+\lambda {\underline{\xi }})T_{\lambda }}}{(1+\lambda {\overline{\xi }})}+\int _{-T_{\lambda }}^{0}e^{s(1+\xi _{\inf }(\omega ))}ds, \quad \forall \ t\in {{\mathbb {R}}},\,\, a.e.\, \omega \in \Omega . \end{aligned}$$

(3.11)

It easily follows from (3.10) and (3.11) that

$$\begin{aligned} { 0<\inf _{t\in {{\mathbb {R}}}} Y(\theta _t\omega )\le \sup _{t\in {{\mathbb {R}}}} Y(\theta _t\omega )<\infty } ,\quad a.e. \ \omega \in \Omega . \end{aligned}$$

The result (1) then follows.

(2) It follows from (1).

(3) Note that

$$\begin{aligned} \frac{{\dot{Y}}(\theta _t\omega )}{Y(\theta _t\omega )}=1+\xi (\theta _t\omega ) -Y(\theta _t\omega ). \end{aligned}$$

Integrating both sides with respect to t, we obtain that

$$\begin{aligned} \frac{1}{t-s}\int _s^tY(\theta _{\sigma }\omega )d\sigma +\frac{\ln (Y(\theta _{t}\omega )) -\ln (Y(\theta _{s}\omega ))}{t-s}=1+\frac{1}{t-s}\int _s^t \xi (\theta _{\sigma }\omega )d\sigma . \end{aligned}$$

(3.12)

The result (3) follows from (2) and the fact that $\lim _{t\rightarrow \infty }\frac{1}{t}\int _0^t\xi (\theta _s\omega )ds=0$ for a.e. $\omega \in \Omega$.

(4) Observe that (3.12) implies that

$$\begin{aligned} 1+{\underline{\xi }}\le&{ \hat{Y}_{\inf }(\omega )}+\limsup _{t-s\rightarrow \infty }\frac{\ln (Y(\theta _{t}\omega )) -\ln (Y(\theta _{s}\omega ))}{t-s} \end{aligned}$$

and

$$\begin{aligned} 1+{\underline{\xi }}\ge&{ \hat{Y}_{\inf }(\omega )}+\liminf _{t-s\rightarrow \infty }\frac{\ln (Y(\theta _{t}\omega )) -\ln (Y(\theta _{s}\omega ))}{t-s} \end{aligned}$$

for a.e. $\omega \in \Omega$. It follows from (1) that

$$\begin{aligned} \liminf _{t-s\rightarrow \infty }\frac{\ln (Y(\theta _{t}\omega ))-\ln (Y(\theta _{s}\omega ))}{t-s} =\limsup _{t-s\rightarrow \infty }\frac{\ln (Y(\theta _{t}\omega )) -\ln (Y(\theta _{s}\omega ))}{t-s}=0\quad \mathrm{for}\,\, a.e.\, \omega \in \Omega . \end{aligned}$$

Hence we have that ${\hat{Y}_{\inf }(\omega )}=1+{\underline{\xi }}>0$ for a.e. $\omega \in \Omega$. Similar arguments yield that ${\hat{Y}_{\sup }(\omega )}=1+{\overline{\xi }}$ for a.e. $\omega \in \Omega$. $\square$

Corollary 3.1

For given $u_0\in C^{b}_{\mathrm{uinf}}({{\mathbb {R}}})$ with $\inf _{x}u_0(x)>0$, for a.e. $\omega \in \Omega$,

$$\begin{aligned} \lim _{t\rightarrow \infty }\Vert \frac{u(t,\cdot ;u_0,\theta _{t_0}\omega )}{Y(\theta _t\theta _{t_0}\omega )}-1\Vert _{\infty }=0 \end{aligned}$$

uniformly in $t_0\in {{\mathbb {R}}}$, where $u(t,x;u_0,\theta _{t_0}\omega )$ is the solution of (1.21) with $u(0,x;u_0,\theta _{t_0}\omega )=u_0(x)$.

Proof

It follows from Theorems 1.1, 3.2, and Lemma 3.2. $\square$

4 Deterministic and Linearly Determinate Spreading Speed Interval

In this section, we discuss the spreading properties of solutions of (1.1) with nonempty compactly supported initials or front like initials and prove Theorems  1.2 and 1.3 .

We first prove some lemmas.

Lemma 4.1

Let $\omega \in \Omega _0$. If there is a positive constant $c(\omega )>0$ such that

$$\begin{aligned} \liminf _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}},|x|\le c(\omega )t}u(t,x;u_0,\theta _s\omega )>0,\quad \forall \ u_0\in X_c^+ \end{aligned}$$

(4.1)

then $c^*_{\inf }(\omega )\ge c(\omega )$. Therefore it holds that

$$\begin{aligned} c^*_{\inf }(\omega )=\sup \{c\in {{\mathbb {R}}}^+\ |\ \liminf _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}},|x|\le ct}u(t,x;u_0,\theta _s\omega )>0,\quad \forall \ u_0\in X_c^+ \}. \end{aligned}$$

(4.2)

Proof

Let $\omega \in \Omega _0$ and $c(\omega )$ satisfy (4.1). Let $0<c<c(\omega )$ and $u_0\in X_c^+$ be given. Choose ${\tilde{c}}\in (c, c(\omega ))$. It follows from (4.1) that

$$\begin{aligned} m_{{\tilde{c}}}:=\liminf _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}},|x|\le {\tilde{c}}t}u(t,x;u_0,\theta _s\omega )>0. \end{aligned}$$

There is $T\gg 1$ such that

$$\begin{aligned} \frac{m_{{\tilde{c}}}}{2}\le \min _{|x|\le {\tilde{c}}t}u(t,x;u_0,\theta _s\omega ),\quad \forall \ s\in {{\mathbb {R}}},\ t\ge T. \end{aligned}$$

(4.3)

Suppose by contradiction that there is $(s_n,t_n,x_n)\in {{\mathbb {R}}}\times {{\mathbb {R}}}^+\times {{\mathbb {R}}}$ with $|x_n|\le ct_n$ for every $n\ge 1$ and $t_n\rightarrow \infty$ such that

$$\begin{aligned} 0<\delta :=\inf _{n\ge 1}|u(t_n,x_n;u_0,\theta _{s_n}\omega )-1|. \end{aligned}$$

(4.4)

Let $0<\varepsilon <1$ be fixed. By (H1), Theorem 1.1 implies that there is ${\tilde{T}}_\varepsilon >T$ such that

$$\begin{aligned} \Vert u(t,\cdot ;\frac{m_{{\tilde{c}}}}{2},\theta _s\omega )-1\Vert _\infty +\Vert u(t,\cdot ;\Vert u_0\Vert _{\infty },\theta _s\omega )-1\Vert _\infty \le \varepsilon , \quad \forall t\ge {\tilde{T}}_\varepsilon , \ \forall \ s\in {{\mathbb {R}}}.\nonumber \\ \end{aligned}$$

(4.5)

Observe that $({\tilde{c}}-c)(t_n-{\tilde{T}}_\varepsilon )-2c{\tilde{T}}_\varepsilon \rightarrow \infty$ as $n\rightarrow \infty$. Then there is $n_\varepsilon$ such that

$$\begin{aligned} ({\tilde{c}}-c)(t_n-{\tilde{T}}_\varepsilon )-2c{\tilde{T}}_\varepsilon \ge T,\quad \forall \ \ n\ge n_\varepsilon . \end{aligned}$$

For every $n\ge n_\varepsilon$, let $u_{0n}\in C^{b}_{\mathrm{unif}}({{\mathbb {R}}})$ with $\Vert u_{0n}\Vert _{\infty }\le \frac{m_{{\tilde{c}}}}{2}$ and

$$\begin{aligned} u_{0n}(x)={\left\{ \begin{array}{ll} \frac{m_{{\tilde{c}}}}{2},\quad \ |x|\le ({\tilde{c}}-c)(t_n-{\tilde{T}}_\varepsilon )-2c{\tilde{T}}_\varepsilon ,\\ 0,\qquad |x|\ge ({\tilde{c}}-c)(t_n-{\tilde{T}}_\varepsilon )-c{\tilde{T}}_\varepsilon . \end{array}\right. } \end{aligned}$$

(4.6)

Since $|x|\le ({\tilde{c}}-c)(t_n-{\tilde{T}}_\varepsilon )-c{\tilde{T}}_\varepsilon$ implies that $|x+x_n|\le {\tilde{c}}(t_n-{\tilde{T}}_\varepsilon )$ for every $n\ge n_\varepsilon$, it follows from (4.3) to (4.6) that

$$\begin{aligned} u_{0n}(x)\le u(t_n-{\tilde{T}}_\varepsilon ,x+x_n;u_0,\theta _{s_n}\omega ),\quad \forall \ x\in {{\mathbb {R}}}, \ \forall \ n\ge n_\varepsilon . \end{aligned}$$

By the comparison principle for parabolic equations, we have

$$\begin{aligned} u(t,x;u_{0n},\theta _{{{\tilde{s}}}_n}\omega )\le u(t+t_n-{\tilde{T}}_\varepsilon ,x+x_n; u_0,\theta _{s_n}\omega ),\quad \forall \ x\in {{\mathbb {R}}}, \ t>0, \ n\ge n_\varepsilon ,\nonumber \\ \end{aligned}$$

(4.7)

where ${\tilde{s}}_n=s_n+t_n-{\tilde{T}}_\varepsilon$.

Observe from the definition of $u_{0n}$ that $u_{0n}(x)\rightarrow \frac{m_{{\tilde{c}}}}{2}$ as $n\rightarrow \infty$ locally uniformly in $x\in {{\mathbb {R}}}$. It then follows from Lemma 3.1 that for every $t>0$,

$$\begin{aligned} |u(t,x;u_{0n},\theta _{{{\tilde{s}}}_n}\omega )- u(t,x;\frac{m_{{\tilde{c}}}}{2},\theta _{{{\tilde{s}}}_n}\omega ) |\rightarrow 0\ \text {as}\ n\rightarrow \infty \ \text {locally uniformly in}\ x\in {{\mathbb {R}}}.\nonumber \\ \end{aligned}$$

(4.8)

By (4.5), we have that

$$\begin{aligned} 1-\varepsilon \le u({\tilde{T}}_\epsilon ,x;\frac{m_{{\tilde{c}}}}{2},\theta _{{{\tilde{s}}}_n}\omega ) ,\quad \forall \ x\in {{\mathbb {R}}}, \forall \ n\ge 1. \end{aligned}$$

This combined with (4.7) and (4.8) yields that

$$\begin{aligned} 1-\varepsilon \le \liminf _{n\rightarrow \infty }u({\tilde{T}}_\varepsilon ,0;u_{0n}, \theta _{{{\tilde{s}}}_n}\omega )\le \liminf _{n\rightarrow \infty }u(t_n,x_n;u_{0},\theta _{s_n}\omega ). \end{aligned}$$

(4.9)

On the other hand, since $\Vert u_0(\cdot +x_n)\Vert _{\infty }=\Vert u_0\Vert _{\infty }$ for every $n\ge 1$, it follows from the comparison principle for parabolic equations that

$$\begin{aligned} u(t_n,x;\Vert u_0\Vert _{\infty },\theta _{s_n}\omega )&\ge u(t_n,x;u_0(\cdot +x_n),\theta _{s_n}\omega ),\\&= u(t_n,x+x_n;u_0,\theta _{s_n}\omega ),\quad \forall \ x\in {{\mathbb {R}}}, \ t>0, \ n\ge 1. \end{aligned}$$

This together with (4.5) implies that

$$\begin{aligned} \limsup _{n\rightarrow \infty }u(t_n,x_n;u_0,\theta _{s_n}\omega )\le \limsup _{n\rightarrow \infty }\Vert u(t_n,\cdot ;\Vert u_0\Vert _{\infty },\theta _{s_n}\omega )\Vert _{\infty } \le 1+\varepsilon , \end{aligned}$$

which combined with (4.9) yields that

$$\begin{aligned} 1-\varepsilon \le \limsup _{n\rightarrow \infty }u(t_n,x_n;u_0,\theta _{s_n}\omega )\le \limsup _{n\rightarrow \infty }u(t_n,x_n;u_0,\theta _{s_n}\omega )\le 1+\varepsilon , \forall \ \varepsilon >0. \end{aligned}$$

Letting $\varepsilon \rightarrow 0$, we obtain that

$$\begin{aligned} \lim _{n\rightarrow \infty }|u(t_n,x_n;u_0,\theta _{s_n}\omega )-1|=0, \end{aligned}$$

which contradicts to (4.4). Thus we have that

$$\begin{aligned}\lim _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\le ct}|u(t,x;u_0,\theta _{s}\omega )-1|=0, \quad \forall u_0\in X_c^+,\ \forall 0<c<c(\omega ). \end{aligned}$$

This implies that $c^*_{\inf }(\omega )\ge c(\omega )$.

Therefore, we have that

$$\begin{aligned} c^{*}_{\inf }(\omega )\ge \sup \{c\in {{\mathbb {R}}}^+\ |\ \liminf _{t\rightarrow \infty }\inf _{|x|\le ct, s\in {{\mathbb {R}}}}u(t,x;u_0,\theta _s\omega )>0,\quad \forall \ u_0\in X_c^+ \}. \end{aligned}$$

On the other hand, it is clear from the definition of $C^{*}_{\sup }(\omega )$ that

$$\begin{aligned} c^{*}_{\inf }(\omega )\le \sup \{c\in {{\mathbb {R}}}^+\ |\ \liminf _{t\rightarrow \infty }\inf _{|x|\le ct, s\in {{\mathbb {R}}}}u(t,x;u_0,\theta _s\omega )>0,\quad \forall \ u_0\in X_c^+ \}. \end{aligned}$$

The lemma is thus proved. $\square$

Lemma 4.2

Let $b>0$ be a positive number and $v_0\in X_c^+$. Let $v(t,x;v_0,b)$ be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} v_t=v_{xx}+bv(1-v), \quad x\in {{\mathbb {R}}}\\ v(0,x)=v_0(x),\quad x\in {{\mathbb {R}}}. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} \lim _{t\rightarrow \infty }\min _{|x|\le ct}v(t,x;v_0,b)=1,\quad \forall \ 0<c<2\sqrt{b}. \end{aligned}$$

Proof

It follows from [3, Page 66, Corollary 1]. $\square$

Lemma 4.3

Assume (H1). Then for every $\omega \in \Omega _0$,

$$\begin{aligned} \liminf _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}},|x|\le ct}u(t,x;u_0,\theta _s\omega )>0, \quad \forall \ 0<c<2\sqrt{{\underline{a}}}, \ \forall \ u_0\in X_c^+. \end{aligned}$$

(4.10)

Therefore, $c^{*}_{\inf }(\omega )\ge 2\sqrt{{\underline{a}}}, \quad \forall \ \omega \in \Omega _0.$

Proof

First, fix $\omega \in \Omega _0$ and $u_0\in X_c^+$. Let $0<c<2\sqrt{{\underline{a}}}$ be given. Choose $b>c$ and $0<\delta <1$ such that $c<2\sqrt{b}<2\sqrt{\delta {\underline{a}}}.$ By the proof of Lemma 2.2, there are $\{t_{k}\}_{k\in {{\mathbb {Z}}}}$ with $t_k<t_{k+1}$, $t_{k}\rightarrow \pm \infty$ as $k\rightarrow \pm \infty$ and $A\in W^{1,\infty }_{loc}({{\mathbb {R}}})\cap L^\infty ({{\mathbb {R}}})$ such that $A\in C^1(t_k,t_{k+1})$ for every k and

$$\begin{aligned} b\le \delta a(\theta _t\omega )-A'(t),\quad \text {for}\,\, t\in (t_k,t_{k+1}),\,\, k\in {{\mathbb {Z}}}. \end{aligned}$$

Let $\sigma =\frac{(1-\delta )e^{-\Vert A\Vert _{\infty }}}{\Vert u_0\Vert _{\infty }+1}$ and $v(t,x;b)=v(t,x;u_0,b)$. By Lemma 4.2, we have that

$$\begin{aligned} \liminf _{t\rightarrow \infty }\min _{|x|\le ct}v(t,x;b)=1. \end{aligned}$$

(4.11)

Next, for given $s\in {{\mathbb {R}}}$, let ${\tilde{v}}(t,x;s)=\sigma e^{A(t+s)}v(t,x;b)$. By the comparison principle for parabolic equations, we have that

$$\begin{aligned} 0<v(t,x;b)\le \max \{\Vert u_0\Vert _{\infty },1\}<\Vert u_0\Vert _{\infty }+1, \quad \forall \ x\in {{\mathbb {R}}}, \ t\ge 0. \end{aligned}$$

Hence, it follows from the definition of $\sigma$ that

$$\begin{aligned} 0<{\tilde{v}}(t,x;s)\le \sigma e^{\Vert A\Vert _\infty }(\Vert u_0\Vert _\infty +1)=1-\delta , \quad \forall \ x\in {{\mathbb {R}}},\ \ t\ge 0,\\ s\in {{\mathbb {R}}}. \end{aligned}$$

Thus for any $s\in {{\mathbb {R}}}$,

$$\begin{aligned} {\tilde{v}}_t-{\tilde{v}}_{xx}-a(\theta _{s+t}\omega ){\tilde{v}}(1-{\tilde{v}})=&\left( A'(s+t)+b(1-v)-a(\theta _{s+t}\omega )(1-{\tilde{v}})\right) {\tilde{v}}(t,x)\\\le&\left( A'(s+t)+b(1-v)-\delta a(\theta _{s+t}\omega )\right) {\tilde{v}}(t,x)\\\le&\left( A'(s+t)+b-\delta a(\theta _{s+t}\omega )\right) {\tilde{v}}(t,x)\\\le&\; 0, \quad \ t\in (t_k,t_{k+1})\cap [0,\infty ), \ \ x\in {{\mathbb {R}}}. \end{aligned}$$

Note that

$$\begin{aligned} {\tilde{v}}(0,x;s)=\sigma e^{A(s)}u_0(x)\le u_0(x),\quad \forall \ x\in {{\mathbb {R}}}. \end{aligned}$$

By the comparison principle for parabolic equations again, we have that

$$\begin{aligned} \sigma e^{-\Vert A\Vert _{\infty }}v(t,x,b)\le {\tilde{v}}(t,x;s)\le u(t,x;u_0,\theta _s\omega ),\quad \forall \ x\in {{\mathbb {R}}},\ s\in {{\mathbb {R}}}, \ t\ge 0. \end{aligned}$$

This combined with (4.11) yields that

$$\begin{aligned} 0<\sigma e^{-\Vert A\Vert _{\infty }}\le \liminf _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}}|x|\le ct}u(t,x;u_0,\theta _s\omega ), \quad \forall \ 0<c<2\sqrt{{\underline{a}}}. \end{aligned}$$

Hence (4.10) holds. By (4.10) and Lemma 4.1, we have $c^{*}_{\inf }(\omega )\ge 2\sqrt{{\underline{a}}}, \quad \forall \ \omega \in \Omega _0.$ $\square$

Now, we prove Theorem 1.2.

Proof of Theorem 1.2

(i) We first prove $c^*_{\sup }(\omega )\le 2\sqrt{{\overline{a}}}$ for all $\omega \in \Omega _0.$

Suppose that $\mathrm{supp}(u_0)\subset (-R,R)$. For every $\mu >0$, let $C_\mu (t,s)=\int _s^{s+t}\frac{\mu ^2+a(\theta _\tau \omega )}{\mu }d\tau$ and $\phi ^\mu (x)=\Vert u_0\Vert _{\infty }e^{-\mu ( x-R)}$ and ${\tilde{\phi }}^{\mu }_{\pm }(t,x;s)=\phi ^\mu (\pm x-C_\mu (t,s))$ for every $x\in {{\mathbb {R}}}$ and $t\ge 0$. Then

$$\begin{aligned} \partial _t{\tilde{\phi }}_{\pm }^\mu -\partial _{xx}{\tilde{\phi }}_{\pm }^\mu -a(\theta _{s+t}\omega ){\tilde{\phi }}_{\pm }^\mu (1-{\tilde{\phi }}_{\pm }^\mu ) =a(\theta _{s+t}\omega )\left( {\tilde{\phi }}_{\pm }^\mu \right) ^2\ge 0, \ x\in {{\mathbb {R}}},\ t>0. \end{aligned}$$

and

$$\begin{aligned} u_0(x)\le {\tilde{\phi }}_{\pm }^\mu (0,x;s), \quad \forall x\in {{\mathbb {R}}},\ \forall \ s\in {{\mathbb {R}}}. \end{aligned}$$

By the comparison principle for parabolic equations, we have

$$\begin{aligned} u(t,x;u_0,\theta _s\omega )\le {\tilde{\phi }}_{\pm }^\mu (t,x;s)=\Vert u_0\Vert _{\infty }e^{-\mu ({\pm }x-R - C_\mu (t,s))},\quad \forall \ x,s\in {{\mathbb {R}}}, \forall t>0, \forall \mu >0. \end{aligned}$$

This implies that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} u(t,x;u_0,\theta _s\omega )=0\quad \forall \,\, \mu>0,\,\, c>\frac{\mu ^2+{\bar{a}}}{\mu }. \end{aligned}$$

For any $c>{\bar{c}}^*=2\sqrt{{\bar{a}}}=\inf _{\mu >0}\frac{\mu ^2+\sqrt{{\bar{a}}}}{\mu }$, choose $\mu >0$ such that $c>\frac{\mu ^2+\sqrt{{\bar{a}}}}{\mu }>{\bar{c}}^*$. By the above arguments, we have

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} u(t,x;u_0,\theta _s\omega )=0. \end{aligned}$$

Hence for any $\omega \in \Omega _0$,$c^*_{\sup }(\omega )\le 2\sqrt{{\overline{a}}}$.

Next, we prove that $c_{\sup }^*(\omega )\ge 2\sqrt{{\overline{a}}}$ for all $\omega \in \Omega _0.$ We prove this by contradiction.

Assume that there is $\omega \in \Omega _0$ such that $c_{\sup }^*(\omega )< 2\sqrt{{\overline{a}}}$. Then there is $0<\delta <1$ such that

$$\begin{aligned} c_{\sup }^*(\omega )< 2\sqrt{\delta {\overline{a}}}. \end{aligned}$$

Note that

$$\begin{aligned} \limsup _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s^t a(\theta _\tau \omega )d\tau ={\bar{a}}>\delta {\bar{a}}. \end{aligned}$$

Then there is $0<\delta ^{'}<1$ and $\{t_n\}$, $\{s_n\}$ such that $\lim _{n\rightarrow \infty } t_n-s_n=\infty$ and

$$\begin{aligned} \delta ^{'}\frac{1}{t_n-s_n}\int _{s_n}^{t_n}a(\theta _\tau \omega )d\tau >\delta {\bar{a}}. \end{aligned}$$

(4.12)

Choose $c\in (c^*_{\sup }(\omega ), 2\sqrt{\delta {\overline{a}}})$. Set $L=\frac{2\pi }{\sqrt{4{\bar{a}}\delta -c^2}}$ and

$$\begin{aligned} w^+(x)=e^{-\frac{c}{2}x}\sin \Big (\frac{\sqrt{4{\bar{a}}\delta -c^2}}{2}x\Big ). \end{aligned}$$

Then $w^+(x)$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} w^+_{xx}+cw^+_x+{\bar{a}}\delta w^+=0,\quad 0<x<L,\\ w^+(0)=w^+(L)=0, \end{array}\right. } \end{aligned}$$

(4.13)

and $0<w^+(x)<1$ for $0<x<L$.

For any given $u_0\in X_c^+$, by the assumption that $c>c_{\mathrm{sup}}^*(\omega )$,

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct}u(t,x;u_0,\theta _s\omega )=0. \end{aligned}$$

(4.14)

Hence there is $T>0$ such hat

$$\begin{aligned} u(t,x;u_0,\theta _s\omega )<1-\delta ^{'}\quad \forall \,\, t\ge T,\,\, |x|\ge ct,\,\, s\in {{\mathbb {R}}}, \end{aligned}$$

and then

$$\begin{aligned} u(t,x;u_0,\theta _s\omega ) (1-u(t,x;u_0,\theta _s\omega ))>\delta ^{'} u(t,x;u_0,\theta _s\omega )\quad \forall \,\, t\ge T,\,\, |x|\ge ct,\,\, s\in {{\mathbb {R}}}.\nonumber \\ \end{aligned}$$

(4.15)

Observe that $u(t,x;u_0,\theta _s\omega )\ge u(t,x;\frac{u_0}{1+\Vert u_0\Vert _{\infty }},\theta _s\omega )$ and

$$\begin{aligned} u_t(t,x;\frac{u_0}{1+\Vert u_0\Vert _{\infty }},\theta _{s}\omega )\ge u_{xx}(t,x;\frac{u_0}{1+\Vert u_0\Vert _{\infty }},\theta _{s}\omega ),\quad x\in {{\mathbb {R}}}. \end{aligned}$$

This implies that

$$\begin{aligned} \alpha :=\inf _{s\in {{\mathbb {R}}},0\le x\le L} u(T,x+cT;u_0,\theta _s\omega )\ge \inf _{s\in {{\mathbb {R}}},0\le x\le L} u(T,x+cT;\frac{u_0}{1+\Vert u_0\Vert _{\infty }},\theta _s\omega )>0.\nonumber \\ \end{aligned}$$

(4.16)

Let $v(t,x;s)=u(t,x+ct;u_0,\theta _{s-T}\omega )$. By (4.15),

$$\begin{aligned} v_t\ge v_{xx}+cv_x+\delta ^{'} a(\theta _{s-T+t}\omega ) v,\quad t\ge T,\,\, x\ge 0. \end{aligned}$$

Let $w(t,x;s)=e^{-\int _s ^{s-T+t} \big (\delta ^{'} a(\theta _\tau \omega )-\delta {\bar{a}}\big )d\tau } v(t,x;s)$. Then

$$\begin{aligned} w_t\ge w_{xx}+cw_x+\delta {\bar{a}} w,\quad t\ge T,\,\, x\ge 0. \end{aligned}$$

By (4.16) and the comparison principle for parabolic equations, we have

$$\begin{aligned} v(t,x;s)\ge \alpha e^{\int _s ^{s-T+t} \big (\delta ^{'} a(\theta _\tau \omega )-\delta {\bar{a}}\big )d\tau }w^+(x),\quad t\ge T,\, \, 0\le x\le L. \end{aligned}$$

This implies that for $0\le x\le L$,

$$\begin{aligned} u(t_n-s_n+T,x+c(t_n-s_n+T);u_0,\theta _{s_n-T}\omega )&\ge \alpha e^{\int _{s_n} ^{t_n} \big (\delta ^{'} a(\theta _\tau \omega ) -\delta {\bar{a}}\big )d\tau }w^+(x)\nonumber \\&\ge \alpha w^+(x)\quad \mathrm{(by \,\,\, (4.12))}. \end{aligned}$$

(4.17)

By (4.14),

$$\begin{aligned} \limsup _{n\rightarrow \infty }\sup _{0\le x\le L} u(t_n-s_n+T,x+c(t_n-s_n+T);u_0,\theta _{s_n-T}\omega )=0, \end{aligned}$$

which contradicts (4.17). Therefore, $c_{\mathrm{sup}}^*(\omega )\ge {\bar{c}}^*$ and then $c_{\mathrm{sup}}^*(\omega )={\bar{c}}^*$ for any $\omega \in \Omega _0$. (i) thus follows.

(ii) By Lemma 4.3, $c_{\inf }^*(\omega )\ge {\underline{c}}^*$ for every $\omega \in \Omega _0$. It then suffices to prove that $c_{\mathrm{inf}}^*(\omega )\le {\underline{c}}^*$ for every $\omega \in \Omega _0$. We prove this by contradiction.

Assume that there is $\omega \in \Omega _0$ such that $c_{\mathrm{inf}}^*(\omega )>{\underline{c}}^*$. Choose $c\in ({\underline{c}}^*,c_{\mathrm{inf}}^*(\omega ))$ and $\delta >1$ such that $c>2\sqrt{\delta {\underline{a}}}$. Then

$$\begin{aligned} \liminf _{t-s\rightarrow \infty }\frac{1}{t-s}\int _s ^t a(\theta _\tau \omega )d\tau <\delta {\underline{a}}. \end{aligned}$$

Hence there are $\{t_n\}$ and $\{s_n\}$ such that $\lim _{n\rightarrow \infty }t_n-s_n=\infty$ and

$$\begin{aligned} \frac{1}{t_n-s_n}\int _{s_n}^{t_n}a(\theta _\tau )d\tau <\delta {\underline{a}}\quad \forall \,\, n\ge 1. \end{aligned}$$

Let ${\underline{\mu }}=\sqrt{\delta {\underline{a}}}$. Then

$$\begin{aligned} 2\sqrt{\delta {\underline{a}}}=\frac{\delta {\underline{a}}+{\underline{\mu }}^2}{{\underline{\mu }}}<c. \end{aligned}$$

(4.18)

Choose $u_0\in X_c^+$ such that

$$\begin{aligned} 0\le u_0(x)<1,\quad u_0(x)\le e^{-{\underline{\mu }}x}\Vert u_0\Vert _\infty \quad \forall \,\, x\in {{\mathbb {R}}}. \end{aligned}$$

By the assumption that $c<c_{\mathrm{inf}}^*(\omega )$, there is $T>0$ such that for any $t\ge T$ and $s\in {{\mathbb {R}}}$,

$$\begin{aligned} \inf _{|x|\le ct}u(t,x;u_0,\theta _s\omega )\ge \Vert u_0\Vert _\infty . \end{aligned}$$

This implies that for any $n\ge 1$ with $t_n-s_n\ge T$,

$$\begin{aligned} \inf _{|x|\le c(t_n-s_n)}u(t_n-s_n,x;u_0,\theta _{s_n}\omega )\ge \Vert u_0\Vert _\infty . \end{aligned}$$

(4.19)

Observe that $u(t,x;u_0,\theta _{s_n}\omega )$ satisfies

$$\begin{aligned} u_t=u_{xx}+a(\theta _{s_n+t}\omega ) u(1-u)\le u_{xx}+a(\theta _{s_n+t}\omega ) u. \end{aligned}$$

It then follows from the comparison principle for parabolic equations that

$$\begin{aligned} u(t,x;u_0,\theta _{s_n}\omega )\le e^{-{\underline{\mu }} \Big ( x-\frac{1}{{\underline{\mu }}}\int _{s_n}^{s_n+t} (a(\theta _\tau \omega )+{\underline{\mu }}^2)d\tau \Big )}\Vert u_0\Vert _\infty \end{aligned}$$

and then for $x=c(t_n-s_n)$, we have

$$\begin{aligned} u(t_n-s_n,x;u_0,\theta _{s_n}\omega )&\le e^{-{\underline{\mu }} \Big ( x-\frac{1}{{\underline{\mu }}}\int _{s_n}^{t_n} (a(\theta _\tau \omega ) +{\underline{\mu }}^2)d\tau \Big )}\Vert u_0\Vert _\infty \\&\le e^{-{\underline{\mu }} \Big (x-\frac{1}{{\underline{\mu }}} (\delta {\underline{a}}+{\underline{\mu }}^2)(t_n-s_n)\Big )}\Vert u_0\Vert _\infty \\&=e^{-{\underline{\mu }} \Big ( c-\frac{1}{{\underline{\mu }}}(\delta {\underline{a}} +{\underline{\mu }}^2)\Big )(t_n-s_n)}\Vert u_0\Vert _\infty \\&<\Vert u_0\Vert _\infty \qquad \qquad \mathrm{(by (4.18))}, \end{aligned}$$

which contradicts to (4.19). Therefore $c_{\inf }^*(\omega )\le {\underline{c}}^*$ for any $\omega \in \Omega _0$ and (ii) follows. $\square$

The following corollary follows directly from Lemma 3.2 and Theorem 1.2.

Corollary 4.1

Assume (H3). Let $Y(\omega )$ be the random equilibrium solution of (1.21) given in (1.22). Then for any $u_0\in X_c^+$,

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\le ct}|\frac{u(t,x;u_0,\theta _s\omega )}{Y(\theta _{t+s}\omega )}-1|=0, \quad \forall \ 0<c<2\sqrt{1+{\underline{\xi }}} \end{aligned}$$

and

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} \frac{u(t,x;u_0,\theta _s\omega )}{Y(\theta _{t+s}\omega )}=0, \quad \forall \ c>2\sqrt{1+{{\bar{\xi }}}} \end{aligned}$$

for a.e. $\omega \in \Omega$. where $u(t,x;u_0,\theta _s\omega )$ is the solution of (1.21) with $\omega$ being replaced by $\theta _s\omega$ and $u(0,x;u_0,\theta _s\omega )=u_0(x)$.

Finally, we prove Theorem 1.3.

Proof

(i) It is clear that ${{\tilde{c}}}_{\mathrm{sup}}^*(\omega )\ge c_{\mathrm{sup}}^*(\omega )={\bar{c}}^*$ for any $\omega \in \Omega _0$. It then suffices to prove that ${{\tilde{c}}}_{\mathrm{sup}}^*(\omega )\le {\bar{c}}^*$ for any $\omega \in \Omega _0$.

To this end, fix $\omega \in \Omega _0$. For every $\mu >0$, let $C_\mu (t,s)=\int _s^{s+t}\frac{\mu ^2+a(\theta _\tau \omega )}{\mu }d\tau$ and ${\tilde{\phi }}^{\mu }_{+}(t,x;s)=e^{-\mu (x-C_\mu (t,s))}$ for every $x\in {{\mathbb {R}}}$ and $t\ge 0$. Note that for any $u_0\in {{\tilde{X}}}_c^+$, there is $M_0>0$ such that

$$\begin{aligned} u_0(x)\le M_0{\tilde{\phi }}_{+}^\mu (0,x;s), \quad \forall x\in {{\mathbb {R}}},\ \forall \ s\in {{\mathbb {R}}}. \end{aligned}$$

Note also that

$$\begin{aligned} \partial _tM_0{\tilde{\phi }}_{+}^\mu -\partial _{xx}M_0{\tilde{\phi }}_{+}^\mu -a(\theta _{s+t}\omega )M_0{\tilde{\phi }}_{+}^\mu (1-M_0{\tilde{\phi }}_{+}^\mu )= & {} a(\theta _{s+t}\omega )M_0^2 \big ( {\tilde{\phi }}_{+}^\mu \big )^2\\\ge & {} 0, \ x\in {{\mathbb {R}}},\ t>0. \end{aligned}$$

Hence, by the comparison principle for parabolic equations, we have that

$$\begin{aligned} u(t,x;u_0,\theta _s\omega )\le M_0{\tilde{\phi }}_{+}^\mu (t,x;s)=M_0e^{-\mu (x- C_\mu (t,s))},\quad \forall \ x,s\in {{\mathbb {R}}}, \forall t>0, \forall \mu >0. \end{aligned}$$

This implies that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},x\ge ct} u(t,x;u_0,\theta _s\omega )=0\quad \forall \,\, \mu>0,\,\, c>\frac{\mu ^2+{\bar{a}}}{\mu }. \end{aligned}$$

For any $c>{\bar{c}}^*=2\sqrt{{\bar{a}}}=\inf _{\mu >0}\frac{\mu ^2+\sqrt{{\bar{a}}}}{\mu }$, choose $\mu >0$ such that $c>\frac{\mu ^2+\sqrt{{\bar{a}}}}{\mu }>{\bar{c}}^*$. By the above arguments, we have

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},x\ge ct} u(t,x;u_0,\theta _s\omega )=0. \end{aligned}$$

Hence for any $\omega \in \Omega _0$, we have ${{\tilde{c}}}^*_{\sup }(\omega )\le 2\sqrt{{\overline{a}}}$. (i) thus follows.

(ii) First, it is clear that ${{\tilde{c}}}_{\inf }^*(\omega )\ge c_{\mathrm{inf}}^*(\omega )={\underline{c}}^*$. It then suffices to prove that ${{\tilde{c}}}_{\mathrm{inf}}^*(\omega )\le {\underline{c}}^*$ for any $\omega \in \Omega _0$. This can be proved by the similar arguments as those in Theorem 1.2 (ii). $\square$

The following corollary follows directly from Lemma 3.2 and Theorem 1.3.

Corollary 4.2

Assume (H3). Let $Y(\omega )$ be the random equilibrium solution of (1.21) given in (1.22). Then for any $u_0\in {\tilde{X}}_c^+$,

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},x\le ct}|\frac{u(t,x;u_0,\theta _s\omega )}{Y(\theta _{t+s}\omega )}-1|=0, \quad \forall \ 0<c<2\sqrt{1+{\underline{\xi }}} \end{aligned}$$

and

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},x\ge ct} \frac{u(t,x;u_0,\theta _s\omega )}{Y(\theta _{t+s}\omega )}=0, \quad \forall \ c>2\sqrt{1+{{\bar{\xi }}}} \end{aligned}$$

for a.e. $\omega \in \Omega$. where $u(t,x;u_0,\theta _s\omega )$ is the solution of (1.21) with $\omega$ being replaced by $\theta _s\omega$ and $u(0,x;u_0,\theta _s\omega )=u_0(x)$.

5 Take-Over Property

In this section, we investigate the take-over property of (1.1) and prove Theorem 1.4. We first prove some lemmas.

Recall that

$$\begin{aligned} u_0^*(x)={\left\{ \begin{array}{ll} 1,\quad x\le 0\\ 0,\quad x>0 \end{array}\right. } \end{aligned}$$

and that, for $t>0$, $x(t,\omega )\in {{\mathbb {R}}}$ is such that

$$\begin{aligned} u(t,x(t,\omega );u_0^*,\omega )=\frac{1}{2}. \end{aligned}$$

Note that, by Lemma  2.9, for each $t>0$, $x(t,\omega )$ is measurable in $\omega$. Note also that for $\omega \in \Omega$, the mapping $(t,t_0)\ni (0,\infty )\times {{\mathbb {R}}}\rightarrow u(t,\cdot ;u_0^*,\theta _{t_0}\omega )\in C_{\mathrm{unif}}^b({{\mathbb {R}}})$ is continuous and hence $x(t,\theta _{t_0}\omega )$ is continuous in $(t,t_0)\in (0,\infty )\times {{\mathbb {R}}}$.

Suppose that (H1) holds. Let $\omega \in \Omega _0$, and $0<\mu<{\tilde{\mu }}<\min \{2\mu ,{\underline{\mu }}^*\}$ be given, where ${\underline{\mu }}^*=\sqrt{{\underline{a}}}$. Let $b(t)=a(\theta _t\omega )$. Put

$$\begin{aligned} c(t;\omega ,\mu )=c(t;b,\mu ),\quad C(t;\omega ,\mu )=C(t;b,\mu ), \end{aligned}$$

and

$$\begin{aligned} A_\omega (t)=B_b(t),\quad d_\omega =d_b, \end{aligned}$$

where $c(t;b,\mu )$ and $C(t;b,\mu )$ are as in (2.5), and $B_b$ and $d_b$ are as in Lemma 2.4. Note that we can choose $d_{\theta _{t_0}\omega }=d_\omega$ and $A_{\theta _{t_0}\omega }(t)=A_\omega (t+t_0)$ for any $t_0\in {{\mathbb {R}}}$. Let

$$\begin{aligned} x_\omega (t)=C(t;\omega ,\mu )+\frac{\ln d_{\omega }+\ln {{\tilde{\mu }}}-\ln \mu }{{{\tilde{\mu }}}-\mu }+ \frac{A_\omega (t)}{\mu }. \end{aligned}$$

(5.1)

Note that for any given $t\in {{\mathbb {R}}}$,

$$\begin{aligned} \phi ^{\mu ,d_\omega ,A_\omega }(t,x_\omega (t))=\sup _{x\in {{\mathbb {R}}}} \phi ^{\mu ,d_\omega ,A_\omega }(t,x)=e^{-\mu \big (\frac{\ln d_\omega }{{{\tilde{\mu }}}-\mu }+\frac{A_\omega (t)}{\mu }\big )}e^{-\mu \frac{\ln {{\tilde{\mu }}}-\ln \mu }{{{\tilde{\mu }}}-\mu }}\big (1-\frac{\mu }{{{\tilde{\mu }}}}\big ). \end{aligned}$$

We introduce the following function

$$\begin{aligned} \phi _{-}^{\mu }(t,x;\theta _{t_0}\omega )={\left\{ \begin{array}{ll} { \phi ^{\mu ,d_{\omega },A_{\theta _{t_0}\omega }}(t,x)},\quad \text {if}\ x\ge x_{\theta _{t_0}\omega }(t), \\ \phi ^{\mu ,d_{\omega },A_{\theta _{t_0}\omega }}(t,x_{\theta _{t_0}\omega }(t)),\quad \ \text {if}\ x\le x_{\theta _{t_0}\omega }(t). \end{array}\right. } \end{aligned}$$

(5.2)

It is clear from Lemma 2.4, and the comparison principle for parabolic equations, that

$$\begin{aligned} 0<\phi _{-}^{\mu }(t,x;\theta _{t_0}\omega )< u(t,x;\phi _{+}^{\mu }(\cdot ,x;\theta _{t_0}\omega ),\theta _{t_0}\omega )\le 1, \forall \ t\in {{\mathbb {R}}},\,\, x\in {{\mathbb {R}}},\,\, t_0\in {{\mathbb {R}}}. \end{aligned}$$

(5.3)

Lemma 5.1

For every $\omega \in \Omega _0$, $\lim _{x\rightarrow -\infty } u(t,x+C(t,\theta _{t_0}\omega ,\mu ); \phi _+^\mu (0,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega )=1$ uniformly in $t>0$ and $t_0\in {{\mathbb {R}}}$, and $\lim _{x\rightarrow \infty } u(t,x+C(t,\theta _{t_0}\omega ,\mu );\phi _+^\mu (0,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega )=0$ uniformly in $t>0$ and $t_0\in \Omega$.

Proof

First, it follows from Lemma 2.3 that

$$\begin{aligned} \sup _{t>0,t_0\in {{\mathbb {R}}}}u(t,x+C(t,\theta _{t_0}\omega ,\mu ); \phi _+^\mu (0,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega )\le e^{-\mu x}\rightarrow 0 \ \text {as}\ x\rightarrow \infty . \end{aligned}$$

Second, define $v(t,x;\theta _{t_0}\omega )=u(t,x+C(t,\theta _{t_0}\omega ,\mu ); \phi _+^\mu (0,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega )$ and

$$\begin{aligned} x^*=\frac{\ln d_\omega +\ln {\tilde{\mu }}-\ln {\mu }}{{\tilde{\mu }}-\mu }-\frac{\Vert A_{\omega }\Vert _\infty }{\mu }. \end{aligned}$$

It follows from (5.1) and (5.3) that

$$\begin{aligned} 0<(1-\frac{\mu }{{\tilde{\mu }}})e^{-\mu \left( \frac{\ln d_\omega +\ln {\tilde{\mu }}-\ln \mu }{{\tilde{\mu }}-\mu }+\frac{\Vert A_\omega \Vert _\infty }{\mu }\right) }\le \inf _{t>0,t_0\in {{\mathbb {R}}}}v(t,x^*; \phi _+^\mu (0,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega ). \end{aligned}$$

Moreover, $x\mapsto v(t,x;\theta _{t_0}\omega )$ is decreasing and

$$\begin{aligned} v_t=v_{xx} +c(t;\theta _{t_0}\omega ,\mu )v_x +a(\theta _t\theta _{t_0}\omega )v(1-v), \end{aligned}$$

where $c(t;\omega ,\mu )=C'(t;\omega ,\mu )$. By the arguments of Theorem 3.1, we have that

$$\begin{aligned} v(t,x;\theta _{t_0}\omega )\rightarrow 1 \ \text {as}\ x\rightarrow -\infty \end{aligned}$$

uniformly in $t>0,t_0\in {{\mathbb {R}}}$. $\square$

Lemma 5.2

For each $t>0$, there is $m(t)\le n(t)\in {{\mathbb {R}}}$ such that

$$\begin{aligned} m(t)\le x(t,\omega ) \le n(t)\quad \,\, \text {for a.e }\ \omega \in \Omega , \end{aligned}$$

and hence $x(t,\omega )$ is integrable in $\omega$.

Proof

First, let

$$\begin{aligned} u_{0n}^*(x)=u_0^*(x-n), \quad x\in {{\mathbb {R}}},\,\, n\in {\mathbb {N}}. \end{aligned}$$

We have that $0\le u_{0n}^*(x)\le 1$ and $u_{0n}^*(x)\rightarrow 1$ as $n\rightarrow \infty$. By Lemma 3.1, for every $\omega \in \Omega _0$ and $t>0$

$$\begin{aligned} u(t,x;u_{0n}^*,\theta _{t_0}\omega )\rightarrow 1\quad \text {as}\ n\rightarrow \infty \end{aligned}$$

uniformly in $t_0\in {{\mathbb {R}}}$ and locally uniformly in $x\in {{\mathbb {R}}}$. Observe that

$$\begin{aligned} u(t,x;u_{0n}^*,\theta _{t_0}\omega )=u(t,x-n;u_{0}^*,\theta _{t_0}\omega ) \end{aligned}$$

and the mapping ${{\mathbb {R}}}\ni x\mapsto u(t,x;u_{0}^*,\theta _{t_0}\omega )$ is decreasing. Thus, there is $N(t,\omega )\in {\mathbb {N}}$ such that

$$\begin{aligned} u(t,x;u_{0}^*,\theta _{t_0}\omega )\ge \frac{3}{4}, \quad \forall x\le -N(t,\omega ), \quad \forall \ t_0\in {{\mathbb {R}}}. \end{aligned}$$

This implies that

$$\begin{aligned} -N(t,\omega )\le \inf _{t_0\in {{\mathbb {R}}}}x(t,\theta _{t_0}\omega ). \end{aligned}$$

Let

$$\begin{aligned} m(t,\omega ):=\inf _{t_0\in {{\mathbb {R}}}}x(t,\theta _{t_0}\omega )=\inf _{t_0\in {{\mathbb {Q}}}} x(t,\theta _{t_0}\omega ). \end{aligned}$$

We have that $\Omega _0\ni \omega \mapsto m(t,\omega )\in {{\mathbb {R}}}^+$ is measurable and $m(t,\theta _\tau \omega )=m(t,\omega )$ for any $\tau \in {{\mathbb {R}}}$. By the ergodicity of the metric dynamical system $(\Omega _0, {\mathcal {F}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$, we have that $m(t,\omega )=m(t) \quad \mathrm{for}\,\, a.e \ \text {in} \ \omega .$

Next, let ${\tilde{u}}_{0n}^*(x)=u_0^*(x+n)$. We have that $0\le u_{0n}^*(x)\le 1$ and ${\tilde{u}}_{0n}^*(x)\rightarrow 0$ as $n\rightarrow \infty$. By Lemma 3.1 again, for every $\omega \in \Omega _0$ and $t>0$,

$$\begin{aligned} u(t,x;{\tilde{u}}_{0n}^*,\theta _{t_0}\omega )\rightarrow 0\quad \text {as}\ n\rightarrow \infty \end{aligned}$$

uniformly in $t_0\in {{\mathbb {R}}}$ and locally uniformly in $x\in {{\mathbb {R}}}$. Observe that

$$\begin{aligned} u(t,x;{\tilde{u}}_{0n}^*,\theta _{t_0}\omega )=u(t,x+n;u_{0}^*,\theta _{t_0}\omega ) \end{aligned}$$

and the mapping ${{\mathbb {R}}}\ni x\mapsto u(t,x;u_{0}^*,\theta _{t_0}\omega )$ is decreasing. Thus, there is ${\tilde{N}}(t,\omega )\in {\mathbb {N}}$ such that

$$\begin{aligned} u(t,x;u_{0}^*,\theta _{t_0}\omega )\le \frac{1}{4}, \quad \forall x\ge {\tilde{N}}(t,\omega ), \quad \forall \ t_0\in {{\mathbb {R}}}. \end{aligned}$$

This implies that

$$\begin{aligned} N(t,\omega )\ge \sup _{t_0\in {{\mathbb {R}}}}x(t,\theta _{t_0}\omega ) \end{aligned}$$

(5.4)

Let

$$\begin{aligned} n(t,\omega ):=\sup _{t_0\in {{\mathbb {R}}}}x(t,\theta _{t_0}\omega ) =\sup _{t_0\in {{\mathbb {Q}}}} x(t,\theta _{t_0}\omega ). \end{aligned}$$

By (5.4), we have that $-\infty< x(t,\omega )\le n(t,\omega )\le N(t,\omega )<\infty$. Hence $\Omega _0\ni \omega \mapsto m(t,\omega )\in {{\mathbb {R}}}^+$ is measurable and $m(t,\theta _\tau \omega )=m(t,\omega )$ for any $\tau \in {{\mathbb {R}}}$. By the ergodicity of the metric dynamical system $(\Omega _0, {\mathcal {F}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$, we have that $m(t,\omega )=m(t) \quad \mathrm{for}\,\, a.e \ \text {in} \ \omega .$ $\square$

Let $x_+(t,\omega ,\mu )$ be such that

$$\begin{aligned} u(t,x+x_+(t,\omega ,\mu ) + C(t,\omega ,\mu );\phi _+^\mu (0,\cdot ;\omega ),\omega )=\frac{1}{2}. \end{aligned}$$

Lemma 5.3

For any $t>0$, there holds

$$\begin{aligned} u(t,x+x(t;\omega );u_0^*,\omega )){\left\{ \begin{array}{ll} \ge u(t,x+x_+(t,\omega ,\mu ) + C(t,\omega ,\mu );\phi _+^\mu (0,\cdot ;\omega ),\omega )\quad x<0\\ \le u(t,x+x_+(t,\omega ,\mu ) + C(t,\omega ,\mu );\phi _+^\mu (0,\cdot ;\omega ),\omega )\quad x>0. \end{array}\right. }\nonumber \\ \end{aligned}$$

(5.5)

Proof

It follows from Lemma 2.7. $\square$

Lemma 5.4

There is ${\hat{M}}>0$ such that

$$\begin{aligned} x(t,\omega )+x(s,\theta _t\omega )\le x(t+s,\omega )+{ {\hat{M}}} \end{aligned}$$

for all $t,s\ge 0$ and a.e. $\omega \in \Omega$.

Proof

First, let ${{\tilde{x}}}(t,\omega )$ and ${{\tilde{x}}}_+(t,\omega )$ be such that

$$\begin{aligned} u(t,{{\tilde{x}}}(t,\omega );u_0^*,\omega )=\frac{1}{4} \quad \mathrm{and}\quad u(t,{{\tilde{x}}}_+(t,\omega ,\mu ) +C(t,\omega ,\mu ); \phi _+^\mu (0,\cdot ;\omega ),\omega )=\frac{1}{4}, \end{aligned}$$

respectively. Since the function $x\mapsto u(t, x;u_0,\omega )$ is decreasing, we have

$$\begin{aligned} {{\tilde{x}}}(t,\omega )>x(t,\omega ). \end{aligned}$$

(5.6)

Moreover, for each $t>0$, ${{\tilde{x}}}(t,\omega )$ is measurable in $\omega$, and for each $\omega \in \Omega$, ${{\tilde{x}}}(t,\theta _{t_0}\omega )$ is continuous in $(t,t_0)\in (0,\infty )\times {{\mathbb {R}}}$. By Lemma 5.3,

$$\begin{aligned} {{\tilde{x}}}(t,\omega )-x(t,\omega )&\le ({{\tilde{x}}}_+(t,\omega ,\mu ) -C(t,\omega ,\mu ))-(x_+(t,\omega ,\mu ) -C(t,\omega ,\mu ))\nonumber \\&={\tilde{x}}_+(t,\omega ,\mu )-x_+(t,\omega ,\mu ),\,\, \forall \, t>0. \end{aligned}$$

(5.7)

Let

$$\begin{aligned} M(\omega )=\sup _{t>0,t_0\in {{\mathbb {R}}}}\big ({\tilde{x}}(t,\theta _{t_0}\omega )-x(t, \theta _{t_0}\omega )\big ) =\sup _{t\in (0,\infty )\cap {{\mathbb {Q}}}, t_0\in {{\mathbb {Q}}}} \big ({\tilde{x}}(t,\theta _{t_0}\omega )-x(t,\theta _{t_0}\omega )\big ). \end{aligned}$$

Note that

$$\begin{aligned} \frac{1}{2}&=u(t,x_+(t,\theta _{t_0}\omega ,\mu )+C(t,\theta _{t_0}\omega ,\mu );\phi ^\mu _ +(\cdot ,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega ),\quad \forall \ t>0, \forall \ t_0\in {{\mathbb {R}}}, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{4}=u(t,{\tilde{x}}_+(t,\theta _{t_0}\omega ,\mu )+C(t,\theta _{t_0} \omega ,\mu );\phi ^\mu _+(\cdot ,\cdot ;\theta _{t_0}\omega ),\theta _{t_0}\omega ),\quad \forall \ t>0, \forall \ t_0\in {{\mathbb {R}}}. \end{aligned}$$

By Lemma 5.1, there is a positive constant $K(\omega )$ such that

$$\begin{aligned} | {x}_+(t,\theta _{t_0}\omega ,\mu )|\le K(\omega ) \quad \text {and} \quad | {\tilde{x}}_+(t,\theta _{t_0}\omega ,\mu )|\le K(\omega ),\quad \forall \ t>0, \forall \ t_0\in {{\mathbb {R}}}. \end{aligned}$$

(5.8)

This combined with (5.7) implies that $M(\omega )<\infty$.

Note that the function $\Omega _0\ni \omega \mapsto M(\omega )\in {{\mathbb {R}}}^+$ is measurable and invariant. By the ergodicity of the metric dynamical system $(\Omega _0, {\mathcal {F}},\{\theta _t\}_{t\in {{\mathbb {R}}}})$, we have that there are an invariant measurable set ${\tilde{\Omega }}$ with ${{\mathbb {P}}}({{\tilde{\Omega }}})=1$ and a positive constant ${\hat{M}}$ such that

$$\begin{aligned} M(\omega )={{\hat{M}}}, \quad \forall \ \omega \in {\tilde{\Omega }}. \end{aligned}$$

(5.9)

Second, note that

$$\begin{aligned} u_0^*(x)\le 2 u(t,x+x(t,\omega );u_0^*,\omega ) \end{aligned}$$

Hence,

$$\begin{aligned} u(s,x;u_0^*,\theta _t\omega )&\le u(s,x;2u(t,\cdot +x(t,\omega );u_0^*,\omega ),\theta _t\omega )\\&\le 2u(s,x;u(t,\cdot +x(t,\omega );u_0^*,\omega ),\theta _t\omega )\\&=2 u(s,x+x(t,\omega );u_0^*,\omega ). \end{aligned}$$

This implies that

$$\begin{aligned} u(s,x(s,\theta _t\omega )+x(t,\omega );u_0^*,\omega )\ge \frac{1}{4}. \end{aligned}$$

It then follows from (5.9) that

$$\begin{aligned} x(s,\theta _t\omega )+x(t,\omega )\le {{\tilde{x}}}(t+s,\omega )\le x(t+s,\omega )+{\hat{M}}. \end{aligned}$$

The lemma follows. $\square$

We now prove Theorem 1.4.

Proof of Theorem 1.4

(i) We first prove that there is $c^*$ such that (1.10) holds with ${\hat{c}}^*$ being replaced by $c^*$. To this end, let $y(t,\omega )=-x(t,\omega )+{ {\hat{M}}}$ where ${\hat{M}}$ is given by Lemma 5.4. Then, by Lemma 5.4

$$\begin{aligned} y(t+s,\omega )=-x(t+s,\omega )+{\hat{M}}\le -x(t,\omega )-x(s,\theta _t\omega ) +2{\hat{M}} = y(t,\omega )+y(s,\theta _t\omega ) \end{aligned}$$

a.e in $\omega$. By Lemma 5.2, $y(t,\cdot )\in L^{1}(\Omega )$. It then follows from the subadditive ergodic theorem that there is $c^*\in {{\mathbb {R}}}$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{y(t,\omega )}{t}=c^*\quad \mathrm{for}\,\, a.e. \,\, \omega \in \Omega . \end{aligned}$$

Next, we claim that (1.11) and (1.12) hold with ${\hat{c}}^*$ being replaced by $c^*$. In fact, by (5.5), (5.8), and Lemma 5.1,

$$\begin{aligned} 0\le \sup _{x\ge (c^*+h)t}u(t,x;u_0^*,\omega )\le&u(t,(c^*+h)t;u_0*,\omega )\\&\le u(t,(c^*+h)t-x(t;\omega )+x_+(t,\omega ,\mu )\\&\quad +C(t,\omega ,\mu ); \phi _+^\mu (0,\cdot ;\omega ),\omega ) \\&\rightarrow 0\ \ \text {as}\ \ t\rightarrow \infty ,\ \forall h>0, \end{aligned}$$

and

$$\begin{aligned} 1\ge \inf _{x\le (c^*-h)t}u(t,x;u_0,\omega )\ge&u(t,(c^*-h)t;u_0,\omega )\\\ge&u(t,(c^*-h)t-x(t;\omega )\\&+x_+(t,\omega ,\mu )+C(t,\omega ,\mu ); \phi _+^\mu (0,\cdot ;\omega ),\omega ) \\\rightarrow&1\ \ \text {as}\ \ t\rightarrow \infty ,\ \forall h>0. \end{aligned}$$

Therefore, (1.11) and (1.12) hold with ${\hat{c}}^*$ being replaced by $c^*$.

Now, we prove that $c^*={\hat{c}}^*$. By the comparison principle for parabolic equations,

$$\begin{aligned} u(t,x;u_0^*,\omega )\le e^{-\mu (x-\frac{1}{\mu }\int _0^{t}(\mu ^2+a(\theta _{\tau }\omega )d\tau ))}, \forall \ t, \mu >0, \forall \ x\in {{\mathbb {R}}}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{1}{2}\le e^{-\mu (x(t,\omega )-\frac{1}{\mu }\int _0^{t}(\mu ^2+a(\theta _{\tau }\omega )d\tau ))}, \ \forall \ t, \mu >0. \end{aligned}$$

This implies that

$$\begin{aligned} \frac{x(t,\omega )}{t}-\frac{\ln (2)}{t\mu }\le \frac{1}{t\mu }\int _0^{t}(\mu ^2+a(\theta _{\tau }\omega )d\tau ). \end{aligned}$$

Letting $t\rightarrow \infty$, we obtain that

$$\begin{aligned} c^*\le \frac{\mu ^2+{\hat{a}}}{\mu }, \quad \forall \ \mu >0. \end{aligned}$$

Taking $\mu =\sqrt{{\hat{a}}}$, we obtain that

$$\begin{aligned} c^*\le {\hat{c}}^*=2\sqrt{{\hat{a}}}. \end{aligned}$$

It then remains to prove that

$$\begin{aligned} c^*\ge {\hat{c}}^*=2\sqrt{{\hat{a}}}. \end{aligned}$$

We prove this by contradiction.

Assume that $c^*< {\hat{c}}^*=2\sqrt{{\hat{a}}}$. Then there are $h>0$ and $0<\delta <1$ such that

$$\begin{aligned} c^*<c:=c^*+h<2\sqrt{\delta {{\hat{a}}}}. \end{aligned}$$

By (1.11), for a.e. $\omega \in \Omega$,

$$\begin{aligned} \lim _{t\rightarrow \infty } \sup _{x\ge ct} u(t,x;u_0^*,\omega )=0. \end{aligned}$$

Fix such $\omega$. There are $0<\delta ^{'}< 1$ and $T>0$ such that

$$\begin{aligned} \delta ^{'}\frac{1}{t}\int _0^t a(\theta _\tau \omega )d\tau >\delta { {\hat{a}}} \end{aligned}$$

and

$$\begin{aligned} u(t,x;u_0^*,\omega )\le 1-\delta ^{'}\quad \forall \,\, t\ge T,\,\, x\ge ct. \end{aligned}$$

As in the proof of Theorem 1.2(i), let $L=\frac{2\pi }{\sqrt{4{ {\hat{a}}}\delta -c^2}}$ and

$$\begin{aligned} w^+(x)=e^{-\frac{c}{2}x}\sin \Big (\frac{\sqrt{4{ {\hat{a}}}\delta -c^2}}{2}x\Big ). \end{aligned}$$

By the similar arguments as those in Theorem 1.2(i), we have

$$\begin{aligned} u(t,x+ct;u_0^*,\omega )&\ge \alpha e^{\int _T^t (\delta ^{'}a(\theta _\tau \omega )-\delta {{\hat{a}}})d\tau } w^+(x)\\&=\alpha e^{-\int _0^T (\delta ^{'}a(\theta _\tau \omega )-\delta {{\hat{a}}})d\tau }e^{\int _0^ t (\delta ^{'}a(\theta _\tau \omega )-\delta {{\hat{a}}})d\tau } w^+(x)\\&\ge \alpha e^{-\int _0^T (\delta ^{'}a(\theta _\tau \omega )-\delta {{\hat{a}}})d\tau } w^+(x) \end{aligned}$$

for $0\le x\le L$ and $t\ge T$, where $\alpha =\sup _{0\le x\le L}u(T,x+cT;u_0^*,\omega )$. This implies that

$$\begin{aligned} \lim _{t\rightarrow \infty } \sup _{x\ge ct} u(t,x;u_0^*,\omega )>0, \end{aligned}$$

which is a contradiction. Hence $c^*= {\hat{c}}^*=2\sqrt{{\hat{a}}}$.

(ii) For any given $u_0\in {{\tilde{X}}}_c^+$, there are $0<\alpha \le 1\le \beta$ and $x_-<x_+$ such that

$$\begin{aligned} \alpha u_0^*(x+x_+)\le u_0(x)\le \beta u_0^*(x+x_-)\quad \forall \,\, x\in {{\mathbb {R}}}. \end{aligned}$$

By the comparison principle for parabolic equations, we have

$$\begin{aligned} \alpha u(t,x;u_0^*(\cdot +x_+),\omega )\le u(t,x;u_0,\omega )\le \beta u(t,x;u_0^*(\cdot +x_-),\omega )\quad \forall \, t\ge 0,\,\, x\in {{\mathbb {R}}}. \end{aligned}$$

This together with (1.11) implies that there is a measurable set $\Omega _1\subset \Omega$ with ${{\mathbb {P}}}(\Omega _1)=1$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty } \sup _{x\ge ({\hat{c}}^*+h)t} u(t,x;u_0,\omega )=0,\quad \omega \in \Omega _1,\ \forall \ h>0, \end{aligned}$$

and

$$\begin{aligned} \liminf _{t\rightarrow \infty } \inf _{x\le ({\hat{c}}^*-h)t} u(t,x;u_0,\omega )\ge \alpha ,\quad \quad \omega \in \Omega _1,\ \forall \ h>0. \end{aligned}$$

(5.10)

We claim that

$$\begin{aligned} \liminf _{t\rightarrow \infty } \inf _{x\le (c^*-h)t}u(t,x;u_0,\omega )=1\quad \mathrm{for}\,\,\,\omega \in \Omega _1, \ \forall \ h>0. \end{aligned}$$

(5.11)

Indeed, let $\omega \in \Omega _1$ and $h>0$ be fixed. Let $\{x_n\}$ and $\{t_n\}$ with $t_n\rightarrow \infty$ and $x_n\le ({\hat{c}}^*-h)t_n$ be such that

$$\begin{aligned} \liminf _{t\rightarrow \infty } \inf _{x\le ({\hat{c}}^*-h)t}u(t,x;u_0,\omega )=\lim _{n\rightarrow \infty }u(t_n,x_n;u_0,\omega ). \end{aligned}$$

(5.12)

For every $0<\varepsilon \ll \frac{1}{2}$, Theorem 1.1 implies that there is $T_\varepsilon >0$ such that

$$\begin{aligned} 1-\varepsilon \le u(t,x;\frac{\alpha }{2},\theta _s\omega ),\quad \forall \ x\in {{\mathbb {R}}},\ s\in {{\mathbb {R}}},\ t\ge T_\varepsilon . \end{aligned}$$

(5.13)

Consider a sequence of $u_{0n}\in C^b_{\mathrm{unif}}({{\mathbb {R}}})$ satisfying that

$$\begin{aligned} u_{0n}(x)= {\left\{ \begin{array}{ll}\frac{\alpha }{2}, \quad x\le \frac{1}{2}ht_n-2({\hat{c}}^* -\frac{1}{2}h)T_{\varepsilon }\\ 0, \qquad x\ge \frac{1}{2}ht_n-({\hat{c}}^*-\frac{1}{2}h)T_{\varepsilon }. \end{array}\right. } \end{aligned}$$

Note that

$$\begin{aligned} x\le \frac{1}{2}ht_n-\left( {\hat{c}}^*-\frac{1}{2}h\right) T_{\varepsilon }\Rightarrow x+ x_n\le \left( {\hat{c}}^*-\frac{1}{2}h)(t_n-T_\varepsilon \right) . \end{aligned}$$

By (5.10), there is $N_1\gg 1$ such that

$$\begin{aligned} u(t_n-T_\varepsilon , x+x_n;u_0,\omega )\ge u_{0n}(x),\quad \forall \ x\in {{\mathbb {R}}}, n\ge N_1. \end{aligned}$$

By the comparison principle for parabolic equations, we then have that

$$\begin{aligned} u(t+t_n-T_\varepsilon ,x+x_n;u_0,\omega )\ge u(t,x;u_{0n},\theta _{t_n-T_\varepsilon }\omega ), \quad \forall x\in {{\mathbb {R}}}, \forall \ t\ge 0. \end{aligned}$$

In particular, taking $t=T_\varepsilon$ and $x=0$, we obtain

$$\begin{aligned} u(t_n,x_n;u_0,\omega )\ge u(T_\varepsilon ,x;u_{0n},\theta _{t_n-T_\varepsilon }\omega ). \end{aligned}$$

(5.14)

Note that $u_{0n}(x)\rightarrow \frac{\alpha }{2}$ as $n\rightarrow \infty$. Letting $t\rightarrow \infty$ in (5.14), it follows from (5.13) and Lemma 3.1 that

$$\begin{aligned} \lim _{n\rightarrow \infty }u(t_n,x_n;u_0,\omega )\ge 1-\varepsilon . \end{aligned}$$

Letting $\varepsilon \rightarrow 0$ in the last inequality, it follows from (5.12) that

$$\begin{aligned} \liminf _{t\rightarrow \infty } \inf _{x\le ({\hat{c}}^*-h)t}u(t,x;u_0,\omega )\ge 1, \quad \mathrm{for}\,\,\,\omega \in \Omega _1, \ \forall \ h>0. \end{aligned}$$

It is clear that

$$\begin{aligned} \liminf _{t\rightarrow \infty } \inf _{x\le ({\hat{c}}^*-h)t}u(t,x;u_0,\omega )\le 1, \quad \mathrm{for}\,\,\,\omega \in \Omega _1, \ \forall \ h>0. \end{aligned}$$

The Claim thus follows and (ii) is proved. $\square$

The following corollary follows directly from Lemma 3.2 and Theorem 1.4.

Corollary 5.1

Assume (H3). Let $Y(\omega )$ be the random equilibrium solution of (1.21) given in (1.22) and let $U^*_0(x;\omega )=Y(\omega )$ for $x<0$ and $U^*_0(x;\omega )=0$ for $x>0$. Then,

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{X(t,\omega )}{t}=2 \quad \text {for a.e}\ \omega \in \Omega , \end{aligned}$$

where $X(t,\omega )$ is such that $u(t,X(t,\omega );U^*_0(\cdot ;\omega ),\omega )=\frac{1}{2}Y(\omega )$, and

$$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{x\ge (2+h)t}\frac{u(t,x;U^*_0(\cdot ;\omega ),\omega )}{Y(\theta _{t}\omega )}=0, \quad \forall \ h>0, \ \text {a.e}\ \omega \in \Omega , \end{aligned}$$

and

$$\begin{aligned} \lim _{t\rightarrow \infty }\inf _{x\le (2- h)t}\frac{u(t,x;U^*_0(\cdot ;\omega ),\omega )}{Y(\theta _{t}\omega )}=1, \quad \forall \ h>0, \ \text {a.e}\ \omega \in \Omega , \end{aligned}$$

where $u(t,x;U^*_0(\cdot ;\omega ),\omega )$ is the solution of (1.21) with $u(0,x;U^*_0(\cdot ;\omega ),\omega )=U^*_0(x;\omega )$.

6 Spreading Speeds of Nonautonomous Fisher–KPP Equations

In this section we consider the nonautonomous Fisher–KPP equation (1.2) and prove Theorem 1.5.

Proof of Theorem 1.5

First, we prove (1.17). To this end, for given $0<c<2\sqrt{{\underline{a}}_0}$, choose $b>c$ and $0<\delta <1$ such that $c<2\sqrt{b}<2\sqrt{\delta {\underline{a}}_0}.$ By the proof of Lemma 2.2, there are $\{t_{k}\}_{k\in {{\mathbb {Z}}}}$ with $t_k<t_{k+1}$, $t_{k}\rightarrow \pm \infty$ as $k\rightarrow \pm \infty$ and $A\in W^{1,\infty }_{loc}({{\mathbb {R}}})\cap L^\infty ({{\mathbb {R}}})$ such that $A\in C^1(t_k,t_{k+1})$ for every k and

$$\begin{aligned} b\le \delta a_0(t)-A'(t),\quad \text {for}\,\, t\in (t_k,t_{k+1}),\,\, k\in {{\mathbb {Z}}}. \end{aligned}$$

Let $\sigma =\frac{(1-\delta )e^{-\Vert A\Vert _{\infty }}}{\Vert u_0\Vert _{\infty }+1}$ and v(t, x; b) be the solution of the PDE

$$\begin{aligned} {\left\{ \begin{array}{ll} v_t=v_{xx}+bv(1-v), \quad x\in {{\mathbb {R}}}, t>0,\\ v(0,x)=u_0(x), \quad x\in {{\mathbb {R}}}. \end{array}\right. } \end{aligned}$$

By Lemma 4.2, we have that

$$\begin{aligned} \liminf _{t\rightarrow \infty }\min _{|x|\le ct}v(t,x;b)=1. \end{aligned}$$

(6.1)

For given $s\in {{\mathbb {R}}}$, let ${\tilde{v}}(t,x;s)=\sigma e^{A(t+s)}v(t,x;b)$. By the similar arguments to those in Lemma 4.3, it can be proved that

$$\begin{aligned} \sigma e^{-\Vert A\Vert _{\infty }}v(t,x,b)\le {\tilde{v}}(t,x;s)\le u(t,x;u_0,\sigma _s a_0),\quad \forall \ x\in {{\mathbb {R}}},\ s\in {{\mathbb {R}}}, \ t\ge 0. \end{aligned}$$

This combined with (6.1) yields that

$$\begin{aligned} 0<\sigma e^{-\Vert A\Vert _{\infty }}\le \liminf _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}}, |x|\le ct}u(t,x;u_0,\sigma _s a_0), \quad \forall \ 0<c<2\sqrt{{\underline{a}}_0}. \end{aligned}$$

By the arguments in Lemma 4.1, it can be proved that

$$\begin{aligned} \lim _{t\rightarrow \infty }\inf _{s\in {{\mathbb {R}}},|x|\le ct}|u(t,x;u_0,\sigma _{s}a_0)-1|=0, \quad \forall u_0\in X_c^+,\ \forall 0<c<2\sqrt{{\underline{a}}_0}. \end{aligned}$$

(1.17) then follows.

Next, we prove (1.18). To this end, for any given $u_0\in X_c^+$, suppose that $\mathrm{supp}(u_0)\subset (-R,R)$. For every $\mu >0$, let $C_\mu (t,s)=\int _s^{s+t}\frac{\mu ^2+a_0(\tau \omega )}{\mu }d\tau$ and $\phi ^\mu (x)=\Vert u_0\Vert _{\infty }e^{-\mu ( x-R)}$ and ${\tilde{\phi }}^{\mu }_{\pm }(t,x;s)=\phi _{\pm }^\mu (\pm x-C_\mu (t,s))$ for every $x\in {{\mathbb {R}}}$ and $t\ge 0$. It is not difficult to see that

$$\begin{aligned} \partial _t{\tilde{\phi }}_{\pm }^\mu -\partial _{xx}{\tilde{\phi }}_{\pm }^\mu -a_0({s+t}){\tilde{\phi }}_{\pm }^\mu (1-{\tilde{\phi }}_{\pm }^\mu ) =a_0({s+t})\left( {\tilde{\phi }}_{\pm }^\mu \right) ^2\ge 0, \ x\in {{\mathbb {R}}},\ t>0, \end{aligned}$$

and

$$\begin{aligned} u_0(x)\le {\tilde{\phi }}_{\pm }^\mu (0,x;s), \quad \forall x\in {{\mathbb {R}}},\ \forall \ s\in {{\mathbb {R}}}. \end{aligned}$$

By the comparison principle for parabolic equations, we then have that

$$\begin{aligned} u(t,x;u_0,\sigma _s a_0)\le {\tilde{\phi }}_{\pm }^\mu (t,x;s)=\Vert u_0\Vert _{\infty }e^{-\mu ({\pm }x-R\mp C_\mu (t,s))},\quad \forall \ x,s\in {{\mathbb {R}}}, \forall t>0, \forall \mu >0. \end{aligned}$$

This implies that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} u(t,x;u_0,\sigma _s a_0)=0\quad \forall \,\, \mu>0,\,\, c>\frac{\mu ^2+{\bar{a}}_0}{\mu }. \end{aligned}$$

For any $c>2\sqrt{{\bar{a}}_0}=\inf _{\mu >0}\frac{\mu ^2+\sqrt{{\bar{a}}_0}}{\mu }$, choose $\mu >0$ such that $c>\frac{\mu ^2+\sqrt{{\bar{a}}_0}}{\mu }>{\bar{c}}^*$, we have

$$\begin{aligned} \limsup _{t\rightarrow \infty }\sup _{s\in {{\mathbb {R}}},|x|\ge ct} u(t,x;u_0,\theta _s\omega )=0. \end{aligned}$$

(1.18) then follows. $\square$

We conclude this section with some example of explicit function $a_0(t)$ satisfying (H2).

Define the sequences $\{l_{n}\}_{n\ge 0}$ and $\{L_n\}_{n\ge 0}$ inductively by

$$\begin{aligned} l_0=0, \quad L_n=l_n+\frac{1}{2^{2(n+1)}}, \quad l_{n+1}=L_n+n+1, \ \ n\ge 0. \end{aligned}$$

(6.2)

Define $a_0(t)$ such that $a_0(-t)=a_0(t)$ for $t\in {{\mathbb {R}}}$ and

$$\begin{aligned} a_0(t)={\left\{ \begin{array}{ll} f_n(t)\qquad \text {if}\ t\in [l_n,L_n]\\ g_n(t) \qquad \ \text {if}\ t\in [L_n,l_{n+1}]\\ \end{array}\right. } \end{aligned}$$

(6.3)

for $n\ge 0$, where $g_{2n}(t)=1$ and $g_{2n+1}(t)=2$ for $n\ge 0$, and $f_0(t)=1$, for $n\ge 1$, $f_{n}$ is Hölder’s continuous on $[l_{n},L_{n}]$, $f_n(l_n)=g_n(l_n)$, $f_{n}(L_{n})=g_n(L_{n})$, and satisfies

$$\begin{aligned} 1\le f_{2n}(t)\le 2^n, \quad { \max _{t\in [l_{2n},L_{2n}]}}f_{2n}(t)=2^n, f_{2n}(t)dt=\frac{1}{2^{n+3}} \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2^{n+1}}\le f_{2n+1}(t)\le 2, \quad { \min _{t\in [l_{2n+1},L_{2n+1}]}}f_{2n+1}(t)=2^{-(n+1)}. \end{aligned}$$

It is clear that $a_0(t)$ is locally Hölder’s continuous, $\inf _{t\in {{\mathbb {R}}}} a_{0}(t)=0$, and $\sup _{t\in {{\mathbb {R}}}} a_{0}(t)=\infty$. Moreover, it can be verified that

$$\begin{aligned} \underline{a_0}=1\quad \text {and} \quad {\overline{a}}_0=2. \end{aligned}$$

Hence $a_0(t)$ satisfies (H2).