Persistence and Spreading Speeds of Integro-Difference Equations with an Expanding or Contracting Habitat

Abstract

We study an integro-difference equation model that describes the spatial dynamics of a species in an expanding or contracting habitat. We give conditions under which the species disperses to a region of poor quality where the species eventually becomes extinct. We show that when the species persists in the habitat, the rightward and leftward spreading speeds are determined by c, the speed at which the habitat quality increases or decreases in time, as well as $c^*(\infty )$, $c^*_-(\infty )$, $c^*(-\infty )$, and $c^*_-(-\infty )$, which are formulated in terms of the dispersal kernel and species growth rates in both directions. We demonstrate that in the case that the species grows everywhere in space, the rightward spreading speed is $c^*(\infty )$ if c is relatively small and is $c^*(-\infty )$ if c is large, and the leftward spreading speed is one of $-c$, $c^*(-\infty )$, or $c^*_-(-\infty )$. We also show that it is possible for a solution to form a two-layer wave, with the propagation speeds of the two layers analytically determined.

Appendix

In this section, we provide the proofs for the theorems given in Sect. 3. We observe that if $u_n(x)$ is a solution of (1) then $\hat{u}_n(x)=u_n(x+nc)$ satisfies

$$\begin{aligned} \hat{u}_{n+1}(x)=\int _{-\infty }^{\infty }k(x+c-y)g(y, \hat{u}_n(y))\hbox {d}y. \end{aligned}$$

(6)

This equation can be written as

$$\begin{aligned} \hat{u}_{n+1}(x)=\hat{Q}[\hat{u}_n](x) \end{aligned}$$

with $\hat{Q}$ the operator determined by the right-hand side of (6). The model

$$\begin{aligned} {u}_{n+1}(x)=\int _{-\infty }^{\infty }k(x-y)g(y, {u}_n(y))\hbox {d}y \end{aligned}$$

(7)

was studied in Li et al. (2014a). We extend the results provided for (7) in Li et al. (2014a) to study (6) and (1). We particularly use Lemma 3.3 and Lemma 3.4 in Li et al. (2014a) to construct useful upper and lower solutions for (1).

1.1 Upper and Lower Solutions

1.1.1 An Upper Solution

We introduce the following lemma regarding an upper solution for (1).

Lemma 2

Assume that Hypotheses 1 are satisfied. Let $\bar{u}_n(x)$ be the solution of $\bar{u}_{n+1}(x)=\hat{Q}[\bar{u}_n](x)$ with $\bar{u}_0(x)\equiv \beta (\infty )$.

• (i) Each $\bar{u}_n(x)$ is continuous and nondecreasing in x, $0 \le \bar{u}_{n+1}(x) \le \bar{u}_{n}(x)\le \beta (\infty )$ for $n \ge 0$ and all $x\in {\mathbb {R}}$, and the sequence $\bar{u}_{n}(x)$ converges point-wise to a nondecreasing function $\bar{u}(x)$ with $\bar{u}(-\infty )=\beta (-\infty )$.

• (ii) If $u_n(x)$ is a solution of (1) with $u_0(x)\le \beta (\infty )$, then $u_n(x)\le \bar{u}_n(x-nc)$.

Proof

$\bar{u}_0(x)\equiv \beta (\infty )$ is continuous in x. If $\bar{u}_n(x)$ is continuous, then continuity of $\bar{u}_{n+1}(x)$ in x directly follows from that $\bar{u}_{n+1}(x)$ is given by an integral with the integrand discontinuous at most at a finite number of numbers. Induction shows that $\bar{u}_n(x)$ is continuous in x for all n.

Since $\bar{{u}}_0(x)\equiv {\beta }(\infty )$ and ${g}(x, {u})$ is nonnegative and nondecreasing in x and u, $\bar{{u}}_1(x)=\int _{-\infty }^{\infty }{k}(y+c){g}(x-y, \bar{{u}}_0(x-y))\hbox {d}y$ is nondecreasing in x. Induction shows that $u_n(x)$ is nondecreasing in x. On the other hand, since ${\beta }(-\infty ) \le \bar{{u}}_0(x)\equiv {\beta }(\infty )$, ${\beta }(-\infty ) \le \bar{{u}}_1(x)=\int _{-\infty }^{\infty }{k}(y+c){g}(x-y, \bar{{u}}_0(x-y))\hbox {d}y\le \bar{{u}}_0 (x)$. Induction and monotonicity of ${g}(x, {u})$ in u show that ${\beta }(-\infty ) \le \bar{{u}}_{n+1}(x) \le \bar{{u}}_n(x)\le {\beta }(\infty )$. It follows that $\bar{{u}}_n(x)$ converges point-wise to a nondecreasing function $\bar{{u}}(x)$ with $\bar{{u}}(x)$ nondecreasing in x and ${\beta }(-\infty ) \le \bar{{u}}(x)\le {\beta }(\infty )$ for all x.

By taking the limit $n\rightarrow \infty$ in $\bar{u}_{n+1}(x)=\hat{Q}[\bar{u}_n](x)$ and using the dominated convergence theorem, we have

$$\begin{aligned} \bar{{u}}(x)=\int _{-\infty }^{\infty }{k}(y+c ){g}(x-y, \bar{{u}}(x-y))\hbox {d}y. \end{aligned}$$

We then take the limit $x\rightarrow -\infty$ and use the dominated convergence theorem to obtain

$$\begin{aligned} {\beta }(-\infty )\le \bar{{u}}(-\infty )= & {} \int _{-\infty }^{\infty }{k}(y+c){g}(-\infty , \bar{{u}} (-\infty ))\hbox {d}y \\= & {} g(-\infty , \bar{{u}}(-\infty ))\\\le & {} \min \{\beta (\infty ), \frac{\partial {g(-\infty , 0)}}{\partial u}\bar{{u}}(-\infty )\}. \end{aligned}$$

If $\frac{\partial {g(-\infty , 0)}}{\partial u}<1$, we have $\bar{{u}}(-\infty )=\beta (-\infty )=0$. If $\frac{\partial {g(-\infty , 0)}}{\partial u}>1$, Hypothesis 2 implies $\bar{{u}}(-\infty )=\beta (-\infty )>0$. This proves the statement (i).

Since $\hat{u}_n(x)=u(x+nc)$ satisfies (6) and $\hat{u}_0(x)\le \bar{u}_0(x)$, Lemma 1 implies that $\hat{u}_n(x) \le \bar{u}_n(x)$, which indicates $u_n(x)\le \bar{u}_n(x-nc).$ This proves the statement (ii). $\square$

Lemma 2 shows that $\bar{u}_n(x-nc)$ is an upper solution of (1), and $\bar{u}_n(x)$ converges to a nondecreasing function with the limit $\beta (-\infty )$ at $-\infty$. It is useful in determining nonpersistence and the asymptotic behavior of solutions near $-\infty$ under appropriate conditions.

1.1.2 Lower Solutions

We now construct lower solutions for (1) by extending the work in Li et al. (2014a). We first need the following lemma.

Lemma 3

Assume that Hypotheses 1 are satisfied. The following statements hold:

• (a) Let $c^*(\infty )>c$. If $u_0(x) > 0$ on a closed interval $[x_1, x_2]$ where $g(x, u) > 0$ for $u > 0$, then there exist $a_2> a_1 > 0$ such that for all positive n, $u_n(x+nc) > 0$ for $x\in [x_1 + na_1, x_2 + na_2]$.

• (b) Let $\frac{\partial g(-\infty , 0)}{\partial u}>1$. If $u_0(x) > 0$ on a closed interval $[x_1, x_2]$, then there exist real numbers $a_2 > a_1$ such that for all positive n, $u_n(x+nc) > 0$ for $x \in [x_1 +na_1, x_2 + na_2]$.

Proof

Consider $\hat{u}_n(x)=u(x+nc)$ that satisfies (6). The condition $c^*(\infty )>0$ with $c^*(\infty )$ defined by (3.1) in Li et al. (2014a) is equivalent to $c^*(\infty )-c>0$ with $c^*(\infty )$ given by (3) in the current paper. Lemma 3.4 (a) in Li et al. (2014a) implies that there exists $a_2>a_1>0$ such that $\hat{u}_n(x)=u_n(x+nc)>0$ for $x\in [x_1+na_1, x_2+na_2]$. This proves the statement (a).

In the case of $\frac{\partial g(-\infty , 0)}{\partial u}>1$, $g(x, u)>0$ for any real x and $u>0$. Since k(x) is continuous and $\int _{-\infty }^{\infty }k(x)\hbox {d}x =1$, there exit real numbers ${a}_2>{a}_1$ such that $k(x+c)>0$ for $x\in [ {a}_1, {a}_2]$. The last part of the proof of Lemma 3. 4 (a) in Li et al. (2014a) shows that $\hat{u}_n(x)=u_n(x+nc)>0$ for $x\in [x_1+na_1, x_2+na_2]$. The proof is complete.

This lemma shows that the solution of (1) becomes positive in a moving interval whose length increases to $\infty$ as n approaches $\infty$. It is essential to build useful lower solutions for (1).

To construct lower solutions for (1), we recall the functions $v ({\mu };x)$ and $z({\mu }; \gamma )$ used in Li et al. (2014a). The function $v ({\mu };x)$ is given by

$$\begin{aligned} v ({\mu };x)= \left\{ \begin{array}{ll} \alpha e^{-\mu x}\sin \gamma x, &{}\quad {\mathrm{if }}\; 0\le x\le \pi /\gamma ,\\ 0, &{}\quad \mathrm{elsewhere}, \end{array} \right. \end{aligned}$$

where $\alpha$, $\mu$, and $\gamma$ are positive numbers. (This function is called v(s) in Weinberger (1982).) The maximum of $v (\mu ; x)$ occurs at

$$\begin{aligned} \sigma (\mu )=(1/\gamma )\tan ^{-1}(\gamma /\mu ). \end{aligned}$$

(8)

$\sigma (\mu )$ is a positive and strictly decreasing function of $\mu$. Clearly $\sigma (\mu ) <\pi /\gamma$.

The function $z({\mu }; \gamma )$ is given by

$$\begin{aligned} z({\mu }; \gamma )=\frac{1}{\gamma }\tan ^{-1}\frac{ \int _{-\infty }^{\infty }k(y)e^{\mu y}\sin \gamma y \hbox {d}y}{ \int _{-\infty }^{\infty }k(y)e^{\mu y}\cos \gamma y\hbox {d}y}. \end{aligned}$$

We also need

$$\begin{aligned} v_{-}(\mu ; x)=v(\mu ; -x), \end{aligned}$$

and

$$\begin{aligned} z_-({\mu }; \gamma )=-\frac{1}{\gamma }\tan ^{-1}\frac{ \int _{-\infty }^{\infty }k(y)e^{-\mu y}\sin \gamma y \hbox {d}y}{ \int _{-\infty }^{\infty }k(y)e^{-\mu y}\cos \gamma y\hbox {d}y}. \end{aligned}$$

The work in Weinberger (1982) shows that

$$\begin{aligned} \lim _{\gamma \rightarrow 0}z({\mu }; \gamma )=\psi (\mu ), \ \ \lim _{\gamma \rightarrow 0}z_-({\mu }; \gamma )=\psi _-(\mu ). \end{aligned}$$

(9)

Let $\epsilon$ be a small positive number and L be a number with $L>\frac{4\pi }{\gamma }$. Let $\mu _1$ and $\mu _2$ be positive numbers.

The following function was used in Li et al. (2014a).

$$\begin{aligned}&u^{(n)}_{r}(\epsilon , \mu _1, \mu _2; x)\\&\quad = \left\{ \begin{array}{l@{\quad }l} v({\mu _1}; x-nz(\mu _1; \gamma )), &{} {\mathrm{if }}\; nz(\mu _1; \gamma )\le x\le \sigma (\mu _1) +nz(\mu _1; \gamma ), \\ \epsilon , &{} {\mathrm{if }}\; \sigma (\mu _1)+nz (\mu _1; \gamma ) \le x \le \sigma (\mu _2)+L \\ &{}\quad -\frac{\pi }{\gamma } +nz (\mu _2; \gamma ), \\ dv({\mu _2}; x-L+ \frac{\pi }{\gamma }-nz(\mu _2; \gamma )), &{} {\mathrm{if}}\; \sigma (\mu _2)+L-\frac{\pi }{\gamma }+nz(\mu _2; \gamma ) \le x \le L \\ &{}\quad +nz(\mu _2; \gamma ), \\ 0, &{} \mathrm{elsewhere,} \end{array} \right. \end{aligned}$$

where $\sigma (\mu )$ is given by (8), $\alpha$ and d satisfy $v({\mu _1};\sigma (\mu _1))=dv({\mu _2};\sigma (\mu _2))=\epsilon$, and $z(\mu _2; \gamma )>z(\mu _1; \gamma )$. (A translation of this function is given by (6.16) in Li et al. (2014a).) The proof of Theorem 1 in Li et al. (2014a) shows that for k(x) with compact support, small positive $\alpha$, $\epsilon$ and $\gamma$, large L, and appropriate $\mu _1$ and $\mu _2$, a proper translation of $u^{(n)}_{r}(\epsilon , \mu _1, \mu _2; x)$ is a lower solution of (1) for $c=0$. $u^{(n)}_{r}(\epsilon , \mu _1, \mu _2; x)$ is $\epsilon$ for x in the interval $[\sigma (\mu _1) +nz(\mu _1; \gamma ), \sigma (\mu _2)+L-\pi /\gamma +nz(\mu _2; \gamma )]$ with the end points shifting rightward at speeds $z(\mu _1; \gamma )$ and $z(\mu _2; \gamma )$, respectively, as n increases.

We introduce two additional functions:

$$\begin{aligned}&u^{(n)}(\epsilon , \mu _1, \mu _2; x)\nonumber \\&\quad = \left\{ \begin{array}{l@{\quad }l} v_-({\mu _1}; x+nz_-(\mu _1; \gamma )), &{} {\mathrm{if}}\; -nz_-(\mu _1; \gamma )-\frac{\pi }{\gamma }\le x\le -nz_-(\mu _1; \gamma ) \\ &{}\quad -\sigma (\mu _1),\\ \\ \epsilon , &{} {\mathrm{if}}\; -nz_- (\mu _1; \gamma ) - \sigma (\mu _1) \le x \le \sigma (\mu _2)+L \\ &{}\quad -\frac{\pi }{\gamma } +nz (\mu _2; \gamma ), \\ dv({\mu _2}; x-L+ \frac{\pi }{\gamma }-nz(\mu _2; \gamma )), &{} {\mathrm{if}} \sigma (\mu _2)+L-\frac{\pi }{\gamma }+nz(\mu _2; \gamma ) \le x \le L \\ &{}\quad +\,nz(\mu _2; \gamma ), \\ 0, &{} { \mathrm elsewhere,} \end{array} \right. \nonumber \\ \end{aligned}$$

(10)

and

$$\begin{aligned}&u^{(n)}_{l}(\epsilon , \mu _1, \mu _2; x)\nonumber \\&\quad = \left\{ \begin{array}{l@{\quad }l} v_-({\mu _1}; x+nz_-(\mu _1; \gamma )), &{} {\mathrm{if}}\; -nz_-(\mu _1; \gamma )-\frac{\pi }{\gamma } \le x\le -nz_-(\mu _1; \gamma ) \\ &{}\quad -\sigma (\mu _1), \\ \epsilon , &{} {\mathrm{if}} -nz_- (\mu _1; \gamma )-\sigma (\mu _1) \le x \le L- nz_- (\mu _2; \gamma ) \\ &{}\quad - \sigma (\mu _2), \\ dv_-({\mu _2}; x-L+nz_-(\mu _2; \gamma )), &{} {\mathrm{if}}\; L-nz_-(\mu _2; \gamma )-\sigma (\mu _2) \le x \le L \\ &{}\quad -nz_-(\mu _2; \gamma ), \\ 0, &{} { \mathrm elsewhere.} \end{array} \right. \nonumber \\ \end{aligned}$$

(11)

In (10) $\sigma (\mu )$ is given by (8), $\alpha$ and d satisfy $v_-({\mu _1};-\sigma (\mu _1))=dv({\mu _2};\sigma (\mu _2))=\epsilon$, and $z_-(\mu _1; \gamma ))+z(\mu _2; \gamma )>0$. In (11) $\sigma (\mu )$ is given by (8), $\alpha$ and d satisfy $v_-({\mu _1};-\sigma (\mu _1))=dv_-({\mu _2};-\sigma (\mu _2))=\epsilon$, and $z_-(\mu _1; \gamma )>z_-(\mu _2; \gamma )$.

$u^{(n)}(\epsilon , \mu _1, \mu _2; x)$ and $u^{(n)}_{l}(\epsilon , \mu _1, \mu _2; x)$ may be viewed as extensions of $u^{(n)}_r(\epsilon , \mu _1, \mu _2; x)$. $u^{(n)}(\epsilon , \mu _1, \mu _2; x)$ is $\epsilon$ in the interval $[-nz_-(\mu _1; \gamma )-\sigma (\mu _1), \sigma (\mu _2)+L-\pi /\gamma + nz(\mu _2; \gamma )]$ with the left-hand end point shifting leftward at speed $z_-(\mu _1; \gamma )$ and the right-hand end point shifting rightward at speed $z(\mu _2; \gamma )$, as n increases. $u^{(n)}_{l}(\epsilon , \mu _1, \mu _2; x)$ is $\epsilon$ in the interval $[-nz_-(\mu _1; \gamma )-\sigma (\mu _1), L-nz_-(\mu _2; \gamma )-\sigma (\mu _2)]$ with the end points shifting leftward at speeds $z_-(\mu _1; \gamma )$ and $z_-(\mu _2; \gamma )$, respectively, as n increases.

The following lemma shows that under appropriate conditions, proper translations of $u^{(n)}(\epsilon , \mu _1, \mu _2;$ x), $u^{(n)}_r(\epsilon , \mu _1, \mu _2; x)$ and $u^{(n)}_{l}(\epsilon , \mu _1, \mu _2; x)$ are lower solutions of (1).

Lemma 4

  Assume that Hypotheses 1 are satisfied and that there exists $b>0$ such that $k(x)\equiv 0$ for $|x|\ge b$. Suppose that the continuous initial function $u_0 (x)$ is positive at a number x where $g(x, u)>0$ for $u>0$, and $0\le {u}_0(x)\le {\beta }(\infty )$ for all x. Then for any small positive number $\varepsilon$, there exist small positive numbers $\alpha$, $\epsilon$, and $\gamma$, a large number $L>4\pi /\gamma$, positive numbers $\mu _1$ and $\mu _2$, a real number $\tilde{x}$, and a positive integer $n_0$, such that the following statements hold:

• (i) If $c>c^*(\infty )$ and $\frac{\partial {g(-\infty , 0)}}{\partial u} >1$, for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ where $z_-(\mu _1; \gamma )=c^*_-(-\infty )-\varepsilon /2$, $z(\mu _2; \gamma )=c^*(-\infty )-\varepsilon /2$.

• (ii) If $c^*(\infty )> c \ge \psi (0)$, for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}_r(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ where $z(\mu _1; \gamma ) =c+\varepsilon /2$, $z(\mu _2; \gamma )=c^*(\infty )-\varepsilon /2$.

• (iii) If $\psi (0)>c$, for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ where $z_-(\mu _1; \gamma ) =\min \{-c, c^*_-(\infty )\} -\varepsilon /2$ and $z(\mu _2; \gamma )=c^*(\infty )-\varepsilon /2$.

• (iv) If $- c < \psi _-(0)$ and $\frac{\partial {g(-\infty , 0)}}{\partial u} >1$, for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ where $z_-(\mu _1; \gamma ) =c^*_-(-\infty )-\varepsilon /2$ and $z(\mu _2; \gamma )=\min \{c, c^*(-\infty )\} -\varepsilon /2$.

• (v) If $c^*_-(-\infty )>-c\ge \psi _-(0)$ and $\frac{\partial {g(-\infty , 0)}}{\partial u} >1$, for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}_l(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ where $z_-(\mu _1; \gamma ) =c^*_-(-\infty )-\varepsilon /2$ and $z_-(\mu _2; \gamma )=-c+\varepsilon /2$.

Proof

Define

$$\begin{aligned} {\hat{k}}(\mu )=\int _{-\infty }^{\infty }k(x+c)e^{\mu x}\hbox {d}x, \end{aligned}$$

and

$$\begin{aligned} {\hat{k}}(-\mu )=\int _{-\infty }^{\infty }k(x+c)e^{-\mu x}\hbox {d}x. \end{aligned}$$

It is easily seen that ${\hat{k}}(\mu )=\bar{k}(\mu )e^{-\mu c}$ and ${\hat{k}}(-\mu )=\bar{k}(-\mu )e^{\mu c}$ where $\bar{k}(\mu )$ is given in Hypothesis 2.1. iv. b.

For $0<\delta <1$, define

$$\begin{aligned} \hat{\phi }_{\delta }(\infty ; \mu )=(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(\infty , 0)}}{\partial u}{\hat{k}}(\mu )\right) =-c+(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(\infty , 0)}}{\partial u}\bar{k}(\mu )\right) , \end{aligned}$$

and

$$\begin{aligned} \hat{\phi }_{\delta -}(\infty ; \mu )=(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(\infty , 0)}}{\partial u}{\hat{k}}(-\mu )\right) =c+(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(\infty , 0)}}{\partial u}\bar{k}(-\mu )\right) . \end{aligned}$$

If $\frac{\partial {g(-\infty , 0)}}{\partial u}>1$, define

$$\begin{aligned} \hat{\phi }_{\delta }(-\infty ; \mu )=(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(-\infty , 0)}}{\partial u}{\hat{k}}(\mu )\right) =-c+(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(-\infty , 0)}}{\partial u}\bar{k}(\mu )\right) , \end{aligned}$$

and

$$\begin{aligned} \hat{\phi }_{\delta -}(-\infty ;\mu )=(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(-\infty , 0)}}{\partial u}{\hat{k}}(-\mu )\right) =c+(1/\mu )\ln \left( (1-\delta )\frac{\partial {g(-\infty , 0)}}{\partial u}\bar{k}(-\mu )\right) . \end{aligned}$$

In what follows, we use $\hat{u}_n(x)$ to denote $u_n(x+nc)$ and use $\hat{z}(\mu ; \gamma )$ and $\hat{z}_-(\mu ; \gamma )$ to denote ${z}(\mu ; \gamma )-c$ and ${z}_-(\mu ; \gamma )+c$, respectively. We also use $\hat{u}^{(n)}(\epsilon , \mu _1, \mu _2; x)$ to denote ${u}^{(n)}(\epsilon , \mu _1, \mu _2; x)$ with $z_-(\mu _1;\gamma )$ replaced by $\hat{z}_-(\mu _1;\gamma )$ and $z(\mu _2;\gamma )$ replaced by $\hat{z}(\mu _2;\gamma )$, and $\hat{u}^{(n)}_r(\epsilon , \mu _1, \mu _2; x)$ to denote ${u}^{(n)}_r(\epsilon , \mu _1, \mu _2; x)$ with $z(\mu _i;\gamma )$ replaced by $\hat{z}(\mu _i;\gamma )$, i=1,2, and $\hat{u}^{(n)}_l(\epsilon , \mu _1, \mu _2; x)$ to denote ${u}^{(n)}_l(\epsilon , \mu _1, \mu _2; x)$ with $z_-(\mu _i;\gamma )$ replaced by $\hat{z}_-(\mu _i;\gamma )$, i=1,2. Note that $\hat{u}^{(n)}(\epsilon , \mu _1, \mu _2; x)={u}^{(n)}(\epsilon , \mu _1, \mu _2; x+nc)$, $\hat{u}^{(n)}_r(\epsilon , \mu _1, \mu _2; x)={u}^{(n)}_r(\epsilon , \mu _1, \mu _2; x+nc)$, and $\hat{u}^{(n)}_l(\epsilon , \mu _1, \mu _2; x)={u}^{(n)}_l(\epsilon , \mu _1, \mu _2; x+nc)$.

We now prove the statement (i). Define

$$\begin{aligned} \hat{c}^*_{\delta }(-\infty )=\inf _{\mu >0} \hat{\phi }_{\delta }(-\infty ; \mu ), \end{aligned}$$

and

$$\begin{aligned} \hat{c}^*_{\delta -}(-\infty )=\inf _{\mu >0} \hat{\phi }_{\delta -}(-\infty ; \mu ). \end{aligned}$$

As $\delta \rightarrow 0$, $\hat{c}^*_{\delta }(-\infty )\rightarrow {c}^*(-\infty )-c$ and $\hat{c}^*_{\delta -}(-\infty ) \rightarrow c^*_-(-\infty )+c$. For any small positive $\varepsilon$, choose $\delta >0$ sufficiently small such that

$$\begin{aligned}&(1-\delta )\frac{\partial g(-\infty , 0)}{\partial u}>1,\\&\hat{c}^*_{\delta }(-\infty )> {c}^*(-\infty )-c-\varepsilon /2, \end{aligned}$$

and

$$\begin{aligned} \hat{c}^*_{\delta -}(-\infty )> c^*_-(-\infty )+c-\varepsilon /2. \end{aligned}$$

Observe that

$$\begin{aligned} \hat{\psi }(\mu )=\frac{\partial (\mu \hat{\phi }_{\delta }(-\infty ; \mu ))}{\partial \mu }=\psi (\mu )-c, \ \hat{\psi }_-(\mu )=\frac{\partial (\mu \hat{\phi }_{\delta -}(-\infty ; \mu ))}{\partial \mu }=\psi _-(\mu )+c \end{aligned}$$

where $\psi (\mu )$ and $\psi _-(\mu )$ are given by (4).

Since $g(x, u)\ge g(-\infty , u)$ for $u\ge 0$ and all $x \in {\mathbb {R}}$, there exists a small positive ${\omega }$ such that $g(x, u)\ge (1-\delta )\frac{\partial g(-\infty , 0)}{\partial u}u$ for $0\le u \le {\omega }$ and $x \in {\mathbb {R}}$.

Define

$$\begin{aligned} \hat{M}^-[{u}](x)=\min \left\{ {\omega }, (1-\delta )\int _{-\infty }^{\infty }{k}(x+c-y)\frac{\partial {g(-\infty , 0)}}{\partial u}{u}(y) \hbox {d}y\right\} . \end{aligned}$$

It is easily seen that for $u(x)\ge 0$,

$$\begin{aligned} \hat{Q}[u](x)\ge \hat{M}^-[u](x). \end{aligned}$$

(12)

Let ${{\mu }}^-_{\delta }$ denote the smallest positive number at which $\hat{\phi }_{\delta }(-\infty ; \mu )$ attains its infimum, and ${{\mu }}^-_{\delta -}$ denote the smallest positive number at which $\hat{\phi }_{\delta -}(-\infty ; \mu )$ attains its infimum. The results from Weinberger (1982) show that for $0< \mu < {{\mu }}^-_{\delta }$, $\hat{\phi }_{\delta }(-\infty ; \mu )$ is strictly decreasing, $\hat{\psi } (\mu )$ is strictly increasing, and $\hat{\phi }_{\delta }(-\infty ; {\mu }^-_{\delta })=\hat{\psi }({{\mu }}^-_{\delta })=\psi ({{\mu }}^-_{\delta })-c =c^*_{\delta }-c$. Similar properties hold for $\hat{\phi }_{\delta -}(-\infty ; \mu )$, $\hat{\psi }_-(\mu )$, ${\mu }^-_{\delta -}$, and particularly $\hat{\phi }_{\delta -}(-\infty ; {\mu }^-_{\delta })=\hat{\psi }_-({{\mu }}^-_{\delta })=\psi _-({{\mu }}^-_{\delta })+c =c^*_{\delta -}+c$. Due to (9), for small positive $\varepsilon$, one can choose $0<\mu _1<{{\mu }}^-_{\delta -}$, $0<\mu _2<{{\mu }}^-_{\delta }$, and $\gamma$ sufficiently small such that $\hat{z}_-(\mu _1; \gamma )=c^*_-(-\infty )+c-\varepsilon /2$, $\hat{z}(\mu _2; \gamma )=c^*(-\infty )-c-\varepsilon /2$, $\hat{z}_-(\mu _1; \gamma )+\hat{z}(\mu _2; \gamma ) >0,$ and $|\hat{z}_-(\mu _1; \gamma )|+|\hat{z}(\mu _2; \gamma )| < \pi /\gamma$. The work in Weinberger (1982, page 387) shows that for $x\in {\mathbb {R}}$,

$$\begin{aligned}&\hat{ M}^-[{v}_-(\mu _1; \cdot )](x)\ge {v}_-(\mu _1; x+\hat{z}_-(\mu _1; \gamma )), \nonumber \\&\hat{M}^-[{v}(\mu _2; \cdot )](x)\ge {v}(\mu _2; x-\hat{z}(\mu _2; \gamma )). \end{aligned}$$

(13)

Choose $L>4\pi /\gamma$. Since $u^{(0)}(\epsilon , \mu _1, \mu _2; x)\ge v_-({\mu _1}; x-\ell _1)$ for $0\le \ell _1\le L-\pi /\gamma$, (13) shows that $\hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge v_-({\mu _1}; x+\hat{z}_-(\mu _1; \gamma )-\ell _1)$ for $0\le \ell _1\le L-\pi /\gamma$, and consequently $\hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge v_-({\mu _1}; x+\hat{z}_-(\mu _1; \gamma ))$ and $\hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge \epsilon$ for $x \in [-\sigma (\mu _1)-\hat{z}_-(\mu _1; \gamma ), \ L-\pi /\gamma -\sigma (\mu _1)-\hat{z}_-(\mu _1; \gamma )]$. On the other hand since $u^{(0)}(x)\ge dv({\mu _2}; x-\ell _2)$ for $0\le \ell _2\le L-\pi /\gamma$, (13) shows that $\hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge dv({\mu _2}; x-\hat{z}(\mu _2; \gamma )-\ell _2)$ for $0\le \ell _2\le L-\pi /\gamma$, and thus $\hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge dv({\mu _2}; x-\hat{z}(\mu _2; \gamma )-(L-\pi /\gamma ))$ and $\hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge \epsilon$ for $x\in [\sigma (\mu _2)+\hat{z}(\mu _2; \gamma ), \ L-\pi /\gamma +\sigma (\mu _2)+ \hat{z}(\mu _2; \gamma )]$. Since $L>4\pi /\gamma$, $\hat{z}_-(\mu _1; \gamma )+\hat{z}(\mu _2; \gamma ) >0,$ and $|\hat{z}_-(\mu _1; \gamma )|+|\hat{z}(\mu _2; \gamma )| < \pi /\gamma$, the two intervals $[-\sigma (\mu _1)-\hat{z}_-(\mu _1; \gamma ), \ L-\pi /\gamma -\sigma (\mu _1)-\hat{z}_-(\mu _1; \gamma )]$ and $[\sigma (\mu _2)+\hat{z}(\mu _2; \gamma ), \ L-\pi /\gamma +\sigma (\mu _2)+ \hat{z}(\mu _2; \gamma )]$ overlap. It follows that for $x\in {\mathbb {R}}$,

$$\begin{aligned} \hat{ M}^-[u^{(0)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge \hat{u}^{(1)}(\epsilon , \mu _1, \mu _2; x). \end{aligned}$$

Induction shows that for $x\in {\mathbb {R}}$,

$$\begin{aligned} \hat{M}^-[\hat{u}^{(n)}(\epsilon , \mu _1, \mu _2; \cdot )](x)\ge \hat{u}^{(n+1)}(\epsilon , \mu _1, \mu _2; x). \end{aligned}$$

By Lemma 3 (b), for $L>4\pi /\gamma$ there exist $\tilde{x}$ and a positive integer $n_0$ such that $u_{n_0}(x+n_0c)>0$ for $x\in [\tilde{x}-\pi /\gamma , \tilde{x}+L ]$. One can choose $\epsilon$ sufficiently small so that $u_{n_0}(x+n_0c)\ge \epsilon$ on $[\tilde{x}-\pi /\gamma , \tilde{x}+L]$. Since $\hat{u}^{(0)}(\epsilon , \mu _1, \mu _2; x-\tilde{x})\le \epsilon$ for $x\in [\tilde{x}-\pi /\gamma , \tilde{x}+L ]$ and $\hat{u}^{(0)}(\epsilon , \mu _1, \mu _2; x-\tilde{x})\equiv 0$ outside this interval, $u_{n_0}(x+n_0c)\ge u^{(0)}(\epsilon , \mu _1, \mu _2; x-\tilde{x})$ or equivalently $\hat{u}_{n_0}(x)\ge \hat{u}^{(0)}(\epsilon , \mu _1, \mu _2; x-\tilde{x})$ for $x\in {\mathbb {R}}$.

The inequality (12) and induction show that for $n\ge n_0$, $\hat{u}_n(x)\ge \hat{u}^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-\tilde{x})$ for $x\in {\mathbb {R}}$, which is equivalent to $u_n(x+nc)\ge \hat{u}^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-\tilde{x}) ={u}^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x+ (n-n_0)c-\tilde{x})$ for $x\in {\mathbb {R}}$. We therefore have $u_n(x)\ge {u}^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ for $x\in {\mathbb {R}}$. This completes the proof of the statement (i).

We next prove the statement (ii). Since $\lim _{x\rightarrow \infty }g(x, u)=g(\infty , u)$, for any small positive numbers $\delta$ and $\omega$, there exists a sufficiently large number $x_{1}$ such that for $x\ge x_{1}$ and $0\le u \le {\omega }$,

$$\begin{aligned} g(x_1, u)\ge (1-\delta )\frac{\partial g(\infty , 0)}{\partial u}u. \end{aligned}$$

We choose $\delta$ sufficiently small such that for the given small positive $\varepsilon$,

$$\begin{aligned} \hat{c}^*_{\delta }=\inf _{\mu>0} \hat{\phi }_{\delta }(\infty ; \mu )> {c}^*(\infty )-c-\varepsilon /2. \end{aligned}$$

(14)

Observe that

$$\begin{aligned} \frac{\partial (\mu \hat{\phi }_{\delta }(\infty ; \mu ))}{\partial \mu }=\hat{\psi } (\mu )=\psi (\mu )-c. \end{aligned}$$

Let ${{\mu }}_{\delta }$ denote the smallest positive number at which $\hat{\phi }_{\delta }(\infty ; \mu )$ attains its infimum. For $0< \mu < {{\mu }}_{\delta }$, $\hat{\phi }_{\delta }(\infty ; \mu )$ is strictly decreasing, $\hat{\psi } (\mu )$ is strictly increasing, and $\hat{\phi }_{\delta }(\infty ; {{\mu }}_{\delta })=\hat{\psi }({{\mu }}_{\delta })=\psi ({{\mu }}_{\delta })-c =c^*_{\delta }-c$. Since $c^*(\infty )>c\ge \psi (0)$ and $\hat{z}(\mu _i; \gamma )={z}(\mu _i; \gamma )-c$ for $i=1,2$, by (9), for small positive $\varepsilon$, there exist $0<\mu _1<\mu _2<{{\mu }}_{\delta }$ and a small positive $\gamma$ such that $\hat{z}(\mu _1; \gamma )= \varepsilon /2$, $\hat{z}(\mu _2; \gamma )=c^*(\infty )-c-\varepsilon /2$, $\hat{z}(\mu _2; \gamma ) > \hat{z}(\mu _1; \gamma )$, and $| \hat{z}(\mu _1; \gamma )| + |\hat{z}(\mu _2; \gamma ) |<\pi /\gamma$.

Choose $L>4\pi /\gamma$. Since $c^*(\infty )>c$, by Lemma 3 (a), there exists a positive integer $n_0$ such that $u_{n_0}(x+n_0c)>0$ on $[{x}_1, {x}_1+b+L]$. We choose $\epsilon$ sufficiently small such that $u_{n_0}(x+n_0c)\ge \epsilon$ on $[{x}_1, {x}_1+b+L]$. Define

$$\begin{aligned} \hat{M}[{u}](x)=\min \{{\omega }, (1-\delta )\int _{-\infty }^{\infty }{k}(x+c-y)\frac{\partial {g(\infty , 0)}}{\partial u}{u}(y) \hbox {d}y\}. \end{aligned}$$

We have that for $u(x)\ge 0$ and $x\ge x_1+b$,

$$\begin{aligned} \hat{Q}[u](x)\ge \hat{M}[u](x). \end{aligned}$$

(15)

Let $\tilde{x}=x_1+b$. The work in Weinberger (1982, page 387) shows that for $x\in {\mathbb {R}}$,

$$\begin{aligned}&\hat{ M}[{v}(\mu _1; \cdot -\tilde{x})](x)\ge v(\mu _1; x-\tilde{x}-\hat{z}(\mu _1; \gamma )), \\&\hat{M}[v (\mu _2; \cdot -\tilde{x})](x)\ge v(\mu _2; x-\tilde{x}-\hat{z}(\mu _2;\gamma )). \end{aligned}$$

The definitions of $u^{(0)}_r(\epsilon , \mu _1, \mu _2; x)$ and $v({\mu _1}; x)$ show that $u^{(0)}_r(\epsilon , \mu _1, \mu _2; x-\tilde{x})\ge v({\mu _1}; x-\tilde{x}-\ell _1)$ for $0\le \ell _1\le L-2\pi /\gamma$, and thus $\hat{ M}[u^{(0)}_r(\epsilon , \mu _1, \mu _2; \cdot -\tilde{x})](x)\ge v({\mu _1}; x-\tilde{x}-\hat{z}(\mu _1; \gamma ))$ and $\hat{ M}[u^{(0)}_r(\epsilon , \mu _1, \mu _2; \cdot -\tilde{x})](x)\ge \epsilon$ for $x \in [\sigma (\mu _1)+\hat{z}(\mu _1; \gamma )+\tilde{x}, \ L-2\pi /\gamma +\sigma (\mu _1)+\hat{z}(\mu _1; \gamma )+\tilde{x}]$. We also have that $u^{(0)}(x)\ge dv({\mu _2}; x-\tilde{x}-\ell _2)$ for $\pi /\gamma \le \ell _2\le L-\pi /\gamma$, and thus $\hat{ M}[u^{(0)}_r(\epsilon , \mu _1, \mu _2; \cdot -\tilde{x})](x)\ge dv({\mu _2}; x-\tilde{x}-\pi /\gamma -\hat{z}(\mu _1; \gamma ))$ and $\hat{ M}[u^{(0)}_r(\epsilon , \mu _1, \mu _2; \cdot -\tilde{x})](x)\ge \epsilon$ for $x \in [\pi /\gamma +\sigma (\mu _2)+\hat{z}(\mu _2; \gamma )+\tilde{x}, \ L-\pi /\gamma +\sigma (\mu _2)+\hat{z}(\mu _2; \gamma )+\tilde{x}]$. On the other hand, since $L>4\pi /\gamma$, $\sigma (\mu _2)<\sigma (\mu _1)$, $\hat{z}(\mu _2; \gamma ) > \hat{z}(\mu _1; \gamma )$, and $| \hat{z}(\mu _1; \gamma )| + |\hat{z}(\mu _2; \gamma ) |<\pi /\gamma$, the two intervals $[\sigma (\mu _1)+\hat{z}(\mu _1; \gamma )+\tilde{x}, \ L-2\pi /\gamma +\sigma (\mu _1)+\hat{z}(\mu _1; \gamma )+\tilde{x}]$ and $[\pi /\gamma +\sigma (\mu _2)+\hat{z}(\mu _2; \gamma )+\tilde{x}, \ L-\pi /\gamma +\sigma (\mu _2)+\hat{z}(\mu _2; \gamma )+\tilde{x}]$ overlap. It follows that $\hat{ M}^-[u^{(0)}_r(\epsilon , \mu _1, \mu _2; \cdot -\tilde{x})](x)\ge \hat{u}^{(1)}_r(\epsilon , \mu _1, \mu _2; x-\tilde{x})$. Induction shows that for $x\in {\mathbb {R}}$,

$$\begin{aligned} \hat{M}[\hat{u}^{(n)}_r(\epsilon , \mu _1, \mu _2; \cdot -\tilde{x})](x)\ge \hat{u}^{(n+1)}_r(\epsilon , \mu _1, \mu _2; x-\tilde{x}). \end{aligned}$$

Since $\hat{u}_{n_0}(x)\ge \hat{u}^{(0)}_r(\epsilon , \mu _1, \mu _2; x-\tilde{x})$ for $x\in {\mathbb {R}}$, (15) and induction show that for $n\ge n_0$, $\hat{u}_n(x)\ge \hat{u}^{(n-n_0)}_r(\epsilon , \mu _1,$ $\mu _2; x-\tilde{x})$ for $x\in {\mathbb {R}}$, which is equivalent to $u_n(x+nc)\ge {u}^{(n-n_0)}_r(\epsilon , \mu _1, \mu _2; x+(n-n_0)c-\tilde{x})$ for $x\in {\mathbb {R}}$. We therefore have $u_n(x)\ge {u}^{(n-n_0)}_r(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ for $x\in {\mathbb {R}}$. This completes the proof of the statement (ii).

To prove the statement (iii), one need only modify the proof of the statement (ii). Choose $\delta$ sufficiently small such that $\hat{c}^*_{\delta -}(\infty )=\inf _{\mu>0}\hat{\phi }_{\delta -}(\infty ; \mu )> c^*_-(\infty )+c-\varepsilon /2$ and (14) hold . We use $\mu _{\delta -}$ to denote the smallest positive number where the infimum is attained. Since $-c>-\psi (0)=\psi _-(0)$ and since $c^*(\infty )>\psi (0)$, for any small positive number $\varepsilon$, by (9), there exist $0<\mu _1<\mu _{\delta -}$, $0<\mu _2<{{\mu }}_{\delta }$, and small positive number $\gamma$ such that $\hat{z}_-(\mu _1; \gamma )=\min \{0, c^*_-(\infty )+c\}-\varepsilon /2$ and $\hat{z}(\mu _2; \gamma )=c^*(\infty )-c-\varepsilon /2$. The proof of the statement (ii) with $\hat{u}^{(n)}_r(\epsilon , \mu _1, \mu _2; x)$ replaced by $\hat{u}^{(n)}(\epsilon , \mu _1, \mu _2; x)$ incorporating these $\hat{z}_-(\mu _1; \gamma )$ and $\hat{z}(\mu _2; \gamma )$ works to show the statement (iii).

If $-c<\psi _-(0)$ and $\frac{\partial {g(-\infty , 0)}}{\partial u} >1$, (9) shows that for any small positive number $\varepsilon$ there exist $0<\mu _1<{{\mu }}^-_{\delta -}$ and $0<\mu _2<\mu _{\delta -}$ such that $\hat{z}_-(\mu _1; \gamma )=c^*_-(-\infty )+c-\varepsilon /2$ and $\hat{z}(\mu _2; \gamma )=\min \{0, c^*(-\infty )-c\}-\varepsilon /2$. The proof of the statement (i) with these $\hat{z}_-(\mu _1; \gamma )$ and $\hat{z}(\mu _2; \gamma )$ proves the statement (iv).

If $-c\ge \psi _-(0)$ and $\frac{\partial {g(-\infty , 0)}}{\partial u} >1$, due to (9), for any small positive number $\varepsilon$, one can choose $0<\mu _1<\mu _2<{{\mu }}^-_{\delta -}$ and $\gamma$ sufficiently small such that $\hat{z}_-(\mu _1; \gamma )=c^*_-(-\infty )+c-\varepsilon /2$ and $\hat{z}_-(\mu _2; \gamma )=\varepsilon /2$. The proof of the statement (i) with $\hat{u}^{(n)}(\epsilon , \mu _1, \mu _2; x)$ replaced by $\hat{u}^{(n)}_l(\epsilon , \mu _1, \mu _2; x)$ incorporating these $\hat{z}_-(\mu _1; \gamma )$ and $\hat{z}_-(\mu _2; \gamma )$ proves the statement (v). The proof is complete. $\square$

1.2 Proofs of Theorems

We now provide the proofs for the theorems. Theorem 1 is proven by employing Lemma 2 and the assumption $g(x, u)\le \ell (x) u$ with $\ell (-\infty )=\frac{\partial {g(-\infty , 0)}}{\partial u}$ and $\ell (\infty )=\frac{\partial {g(\infty , 0)}}{\partial u}$ given in Hypotheses 1 (iii) (b). In the proofs of Theorem 1–Theorem 4, this assumption and Lemma 3.3 and its proof in Li et al. (2014a) are used to find upper bounds for the spreading speeds, while Lemma 4 is used to establish lower bounds for the spreading speeds. Lemma 2 is also used to determine the asymptotic behavior of solutions near $-\infty$ in the proofs of Theorem 2 (ii) (b) and Theorem 3 (i) (b).

1.2.1 Proof of Theorem 1

Proof

Hypotheses 1 (i) and (iii) (b) show that $g(x-nc, u_n)\le g(\infty , u_n)\le \frac{\partial {g(\infty , 0)}}{\partial u} u_n$. It follows from (1) that $u_n(x)$ satisfies

$$\begin{aligned} u_{n+1}(x)\le \int _{-\infty }^{\infty }k(x-y)\frac{\partial {g(\infty , 0)}}{\partial u}u_n(y)\hbox {d}y. \end{aligned}$$

(16)

Choose $\epsilon$ with $0<\epsilon < c-c^*(\infty )$. Let ${\mu }_1$ denote the smallest positive solution of $\phi (\infty ; \mu )=c^*(\infty )+\epsilon /2$. Choose $\rho$ sufficiently large such that $u_0(x)\le \rho e^{-{\mu }_1 x}$. Using (16) and induction,

$$\begin{aligned} u_n(x)\le \rho e^{-{\mu }_1(x-n(c^*(\infty )+\epsilon /2))}. \end{aligned}$$

It follows that for any $\varepsilon >0$, there exists a positive integer $N_1$ such that for $n\ge N_1$ and $x\ge n(c^*(\infty )+\epsilon )$,

$$\begin{aligned} u_n(x) < \varepsilon . \end{aligned}$$

(17)

Let $\bar{u}_n(x)$ be the sequence satisfying (6) with $\bar{u}_0(x)\equiv \beta (\infty )$. Lemma 2 (i) shows that $\bar{u}_n(x)$ decreases point-wise to $\bar{u}(x)$, and $\bar{u}(-\infty )=0$ due to $\frac{\partial {g(-\infty , 0)}}{\partial u}<1$. For the given $\varepsilon >0$, there exists $x_0$ such that $\bar{u}(x_0)<\varepsilon /2$. On the other hand, there exists $N_2$ such that for $n\ge N_2$, $0\le \bar{u}_n(x_0)-\bar{u} (x_0)\le \varepsilon /2$. It follows that for $n>N_2$, $0\le \bar{u}_n(x_0) \le \bar{u}_n(x_0)-\bar{u} (x_0)+\bar{u} (x_0)<\varepsilon$. Since $\bar{u}_n(x)$ is nondecreasing in x, we have that for $n>N_2$ and $x\le x_0$, $\bar{u}_n(x)<\varepsilon$. In view of Lemma 2 (ii), $u_n(x+nc)\le \bar{u}_n(x).$ We therefore have that for $n\ge N_2$ and $x<x_0+nc$,

$$\begin{aligned} u_n(x)<\varepsilon . \end{aligned}$$

(18)

Since $c>c^*(\infty )+\epsilon$, there exists a positive integer $N_3$ such that $x_0+nc >n(c^*(\infty )+\epsilon ).$ It follows from (17) and (18) that for $n\ge \max \{ N_1, N_2, N_3 \}$ and for all x, $u_n(x)<\varepsilon$. The proof is complete. $\square$

1.2.2 Proof of Theorem 2

Proof

Choose $\rho _1 >0$ such that $u_0(x)\le \rho _1 e^{-\hat{\mu }_1 x}$ where $\hat{\mu }_1$ is the smallest solution of $\phi (\infty ; \mu )=c^*(\infty )+\varepsilon /2$. The inequality (16) and induction show that

$$\begin{aligned} {u}_n(x)\le \rho _1 e^{-\hat{\mu }_1 (x- n(c^*(\infty )+\varepsilon /2))}. \end{aligned}$$

The statement (i)(a) immediately follows.

We now prove the statement (i) (b). Since $c^*_-(\infty )>\psi _-(0)=-\psi (0)$, in the case of $c\ge \psi (0)$, $\min \{-c, c^*_-(\infty )\}=-c$. We first consider this case, and assume that there exists $b>0$ such that $k(x)\equiv 0$ for $|x|\ge b$. Since $c^*(\infty )>c\ge \psi (0)$, Lemma 4 (ii) shows that for any small positive number $\varepsilon$, there exist small positive numbers $\alpha$, $\epsilon$, and $\gamma$, a large number $L>4\pi /\gamma$, positive numbers $\mu _1$ and $\mu _2$, a real number $\tilde{x}$, and a positive integer $n_0$, such that for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}_r(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ with $z(\mu _1; \gamma ) =c+\varepsilon /2$ and $z(\mu _2; \gamma )=c^*(\infty )-\varepsilon /2$.

Hypotheses 1 i–iii imply that for any small positive $\eta >0$, there exists a large number $\hat{x}$ such that for small positive u,

$$\begin{aligned} g(\hat{x}, u)>u, \ \text{ and } \beta (\hat{x})>\beta (\infty )-\eta . \end{aligned}$$

Since $z (\mu _1; \gamma )>c$, there is $N_0$ such that for $n\ge N_0$ and $x\ge \bar{x}+\sigma (\mu _1)+(n-n_0)z(\mu _1; \gamma )$,

$$\begin{aligned} x-nc \ge \hat{x}. \end{aligned}$$

We use $g^1(x, \epsilon )$ to denote $g(x, \epsilon )$ and $g^n(x, \epsilon )$ to denote $g(x, g^{(n-1)}(x))$ for $n>1$. Since $\epsilon$ is small, there exists $N_1$ such that

$$\begin{aligned} g^{N_1}(\hat{x}, \epsilon )>\beta (\infty )-2\eta . \end{aligned}$$

On the other hand, since $z(\mu _2; \gamma )>z(\mu _1; \gamma )$, there exists $N_2$ such that such that for $n\ge N_2$,

$$\begin{aligned}&\bar{x}+\sigma (\mu _1)+(n-n_0)z (\mu _1; \gamma )+N_1b <\bar{x}+\sigma (\mu _2)+L-\frac{\pi }{\gamma }\\&\quad +(n-n_0)z (\mu _2; \gamma )-N_1b. \end{aligned}$$

Observe that for $n\ge n_0$,

$$\begin{aligned} u_{n}(x)\ge u^{(n-n_0)}_r(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x}) =\epsilon \end{aligned}$$

for $x\in [\bar{x}+\sigma (\mu _1)+(n-n_0)z(\mu _1; \gamma ), \bar{x}+\sigma (\mu _2)+L-\pi /\gamma +(n-n_0)z(\mu _2; \gamma )]$ with $\bar{x}=n_0c+\tilde{x}$. It follows from (1) that for $n\ge N_0+N_1+N_2$ and $x\in [\bar{x}+\sigma (\mu _1)+(n-n_0)z (\mu _1; \gamma )+bN_1, \; \bar{x} +\sigma (\mu _2)+L-\frac{\pi }{\gamma } +(n-n_0)z (\mu _2; \gamma )-bN_1]$,

$$\begin{aligned} u_{n+1}(x)\ge g^{N_1}(\hat{x}, \epsilon )\ge {\beta }(\infty )-2\eta . \end{aligned}$$

Since $c^*(\infty )>c$ and $\varepsilon$ is small, $z(\mu _2; \gamma )=c^*(\infty )-\varepsilon /2>c+\varepsilon >z(\mu _1; \gamma ) =c+\varepsilon /2$. It follows that there exists an integer $N_3>N_0+N_1+N_2$ such that for $n\ge N_3$,

$$\begin{aligned}&\bar{x}+b+\sigma (\mu _1)+(n-n_0)z (\mu _1; \gamma )+N_1b<n (c+\varepsilon )<n(c^*(\infty )-\varepsilon ) \\&\quad <\bar{x}+b +\sigma (\mu _2)+L-\frac{\pi }{\gamma } +(n-n_0)z (\mu _2; \gamma )-N_1b. \end{aligned}$$

We therefore have that for $n\ge N_3$,

$$\begin{aligned} \inf _{n(c+\varepsilon ) \le x\le n(c^* (\infty )-\varepsilon )} {u}_n(x)\ge {\beta }(\infty )-2\eta . \end{aligned}$$

(19)

Since $\eta$ is arbitrary and $u_n(x)\le \beta (\infty )$ for all n and x, the statement (i) (b) holds.

If k does not have bounded support, we choose a smooth nonincreasing function $\zeta$ with the property

$$\begin{aligned} \zeta (s)= \left\{ \begin{array}{ll} 1, &{}\quad {\mathrm{if}}\; s\le 1/2,\\ 0, &{}\quad {\mathrm{if}}\; s \ge 1. \end{array} \right. \end{aligned}$$

We approximate the kernel k by

$$\begin{aligned} k_m(x)=\zeta \left( \frac{|x|}{m}\right) k(x) \end{aligned}$$

which has bounded support. We define the number $c^*_m(\infty )$ by replacing k by $k_m$ in (3). The work in Weinberger (1982 see (9.18)) shows that if $\bar{k}(\mu )$ is convergent for all $\mu >0$,

$$\begin{aligned} \lim _{m\rightarrow \infty } c^*_m(\infty )=c^*(\infty ). \end{aligned}$$

(20)

If $\bar{k}(\mu )$ is divergent at a positive number $\mu$, the proof of Theorem 2.1 in Weinberger and Zhao (2010) essentially works to show that (20) is still valid. We define the operator ${Q}_{m,n}$ and the function $\psi _m(\mu )$ by replacing k by $k_m$ in the definitions of ${Q}_n$ and $\psi (\mu )$, respectively. For any small positive $\varepsilon$, we choose m sufficiently large such that $c^*_m(\infty )>c$ and $(\int _{-\infty }^{\infty }k_m(y)\hbox {d}y) g(\infty , u)=u$ has a positive root. We use ${\beta }_m(\infty )$ to denote the smallest of such roots. Then for large m the statement (i) (b) holds for the solution $u_{m,n}(x)$ of the recursion defined by $Q_{m,n}$ with $\psi (0)$ replaced by $\psi _m(0)$ and $\beta (\infty )$ replaced by $\beta _m(\infty )$. That is, if the continuous initial function $u_{m,0}(x)>0$ at a number x where $g(x, u)>0$ for $u>0$ and $u_{m,0}(x)\le \beta (\infty )$, then for any small positive $\varepsilon$,

$$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \sup _{n(c+\varepsilon )\le x\le n(c^*_m (\infty )-\varepsilon )} |\beta _m(\infty )-{u}_{m,n}(x)|\right] =0. \end{aligned}$$

(21)

Let $u_n(x)$ be a solution of (1) with $u_0(x)\ge u_{m,0}(x)$. Since $Q_n[u](x)\ge Q_{m,n}[u](x)$ for $u(x)\ge 0$, induction shows $u_n(x)\ge u_{m,n}(x)$ for $n>1$. Since $\beta _m(\infty )$ increases to ${\beta }(\infty )$ and $\psi _m(0) \rightarrow \psi (\mu )$ as $m\rightarrow \infty$, $0\le u_{m,n} (x)\le {\beta }_m(\infty )$, and $0\le u_n(x) \le {\beta }(\infty )$, (21) implies that the statement (i) (b) holds for $u_n(x)$. We have proven the statement (i) (b) if $c\ge \psi (0)$.

If $c < \psi (0)$ or equivalently $-c>-\psi (0)=\psi _-(0)$, Lemma 4 (iii) shows that for any small positive number $\varepsilon$, there exist small positive numbers $\alpha$, $\epsilon$, and $\gamma$, a large number L, positive numbers $\mu _1$ and $\mu _2$, a real number $\tilde{x}$, and a positive integer $n_0$, such that for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ with $z_-(\mu _1; \gamma )=\min \{-c, c^*_-(\infty )\}-\varepsilon /2$ and $z(\mu _2; \gamma )=c^*(\infty )-\varepsilon /2$. The statement (i) (b) can then be shown by using an argument similar to what is used above for the case of $c\ge \psi (0)$. We omit the details here.

We next show the statement (ii) (a). By using an argument similar to the proof of (i) (a), we have that for any small $\epsilon >0$, there exist $\rho _1 >0$ such that

$$\begin{aligned} {u}_n(x)\le \rho _1 e^{-{\mu }_1 (x- n(c^*(\infty )+\epsilon ))} \end{aligned}$$

(22)

where ${\mu }_1$ is the smallest solution of $\phi (\infty ; \mu )=c^*(\infty )+\epsilon .$

Since

$$\begin{aligned} \lim _{\delta \rightarrow 0}\inf _{\mu >0} (1/\mu )\ln \left[ \left( \frac{\partial g(-\infty , 0)}{\partial u}+\delta \right) \bar{k}(\mu )\right] =c^*(-\infty ), \end{aligned}$$

for the small positive $\epsilon$, there exists a small $\delta >0$ such that $(1/\mu ) \ln [(\frac{\partial g(-\infty , 0)}{\partial u}+\delta ) \bar{k}(\mu )] = c^*(-\infty )+ \epsilon /2$ has a positive root. We use ${\mu }_2$ to denote the smallest of such roots. Hypotheses 1 (i) and (iii) imply that there exists a number $x_0$ such that for $x\le x_0$,

$$\begin{aligned} \max \{\ell (x), \ \frac{\partial {g(-\infty , 0)}}{\partial u}+\delta \}= \frac{\partial {g(-\infty , 0)}}{\partial u}+\delta . \end{aligned}$$

(23)

The proof of (6.2) in Li et al. (2014a) with $\ell (x-y)$ replaced by $\ell (x-y-nc)$ shows that for the given $\epsilon >0$ and ${\mu }_2>0$, there exists $T>0$ such that for any real number x,

$$\begin{aligned} \int _{-\infty }^{\infty }k(y) {\ell (x-y-nc)}e^{{\mu }_2 (x-y)}\hbox {d}y \le \left( 1+\frac{\mu _2\epsilon }{2}\right) \int _{-T}^{\infty }k(y) {\ell (x-y-nc)}e^{{\mu }_2 (x-y)}\hbox {d}y. \end{aligned}$$

(24)

We choose T sufficiently large so that $T>|c^*(-\infty ) |$. Choose $\rho _2 >0$ such that

$$\begin{aligned} {u}_0(x)\le \rho _2 e^{-{\mu }_2 x} \end{aligned}$$

for all x. We choose $\epsilon$ sufficiently small so that $\mu _1\approx \mu ^*$, $\mu _2\approx \mu ^*_-$, $\mu _1>\mu _2$, and

$$\begin{aligned} c> \frac{\mu _1(c^*(\infty )+\epsilon )-\mu _2(c^*(-\infty )+\epsilon )}{\mu _1-\mu _2}. \end{aligned}$$

(25)

We choose large $\rho _1$ and $\rho _2$ so that

$$\begin{aligned} \rho _1/\rho _2\le e^{(\mu _1-\mu _2)(x_0-T-c)}. \end{aligned}$$

(26)

It follows from (16), (23), and (24) that for $x\le x_0-T$,

$$\begin{aligned} \begin{aligned} u_1(x)&=\int _{-\infty }^{\infty }k(y) g(x-y, u_0(x-y))\hbox {d}y \\&\le \left( 1+\frac{\mu _2\epsilon }{2}\right) \rho _2 \int _{-T}^{\infty }k(y) {\ell (x-y)}e^{-{\mu }_2 (x-y)}\hbox {d}y \\&\le \left( 1+\frac{\mu _2\epsilon }{2}\right) \rho _2 \int _{-T}^{\infty }k(y) \left( \frac{\partial {g(-\infty , 0)}}{\partial u}+\delta \right) e^{-{\mu }_2 (x-y)}\hbox {d}y \\&\le \left( 1+\frac{\mu _2\epsilon }{2}\right) \rho _2 \int _{-\infty }^{\infty }k(y) \left( \frac{\partial {g(-\infty , 0)}}{\partial u}+\delta \right) e^{-{\mu }_2 (x-y)}\hbox {d}y \\&=\left( 1+\frac{\mu _2\epsilon }{2}\right) \rho _2 e^{-{\mu }_2 (x-(c^*(-\infty )+\epsilon /2))} \\&\le \rho _2 e^{-{\mu }_2 (x-(c^*(-\infty )+\epsilon ))}. \end{aligned} \end{aligned}$$

(27)

On the other hand, (25) and (26) show that

$$\begin{aligned} \rho _1 e^{-\mu _1 (x-(c^*(\infty )+\epsilon ))} \le \rho _2 e^{-{\mu }_2 (x-(c^*(-\infty )+\epsilon ))} \end{aligned}$$

for $x> x_0-T$. It follows from this, (22), and (27) that

$$\begin{aligned} u_1(x) \le \rho _2 e^{-{\mu }_2 (x-(c^*(-\infty )+\epsilon ))} \end{aligned}$$

for all x. Assume that for some n,

$$\begin{aligned} u_n(x) \le \rho _2 e^{-{\mu }_2 (x-n(c^*(-\infty )+\epsilon ))} \end{aligned}$$

(28)

for all x. Then from (16), (23), and (24) we have

$$\begin{aligned} \begin{aligned} u_{n+1}(x)&=\int _{-\infty }^{\infty }k(y) g(x-y-nc, u_n(x-y))\hbox {d}y \\&\le \left( 1+\frac{\mu _2\epsilon }{2}\right) \rho _2 \int _{-T}^{\infty }k(y) {\ell (x-y-nc)}{u} e^{-{\mu }_2 (x-y-n(c^*(-\infty )+\epsilon /2))} \hbox {d}y \\&\le \rho _2 e^{-{\mu }_2 (x-(n+1)(c^*(-\infty )+\epsilon ))}. \end{aligned} \end{aligned}$$

(29)

for $x\le x_0-T+nc$. It follows from (22), (25), and (26) that

$$\begin{aligned} u_{n+1}(x)\le \rho _2 e^{-{\mu }_2 (x-(n+1)(c^*(-\infty )+\epsilon ))} \end{aligned}$$

(30)

for $x> x_0-T+nc$. This and (29) show that (30) is valid for all x. By induction, (28) holds for all n and all x. For any small $\varepsilon >0$, choose $\epsilon =\varepsilon /2$. Then (28) yields the statement (ii) (a).

To prove (ii) (b), we first assume that there exists $b>0$ such that $k(x)\equiv 0$ for $|x|\ge b$. Since $c>c^*(-\infty )>\psi (0)=-\psi _-(0)$, Lemma 4 (iv) shows that for any small positive number $\varepsilon$, there exist small positive numbers $\alpha$, $\epsilon$, and $\gamma$, a large number L, positive numbers $\mu _1$ and $\mu _2$, a real number $\tilde{x}$, and a positive integer $n_0$, such that for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ with $z_-(\mu _1; \gamma ) =c^*_-(-\infty )-\varepsilon /2$ and $z(\mu _2; \gamma )=c^*(-\infty )-\varepsilon /2$. It follows that for $n\ge n_0$,

$$\begin{aligned} u_{n}(x)\ge \epsilon \end{aligned}$$

for $x\in [\bar{x}-\sigma (\mu _1)-(n-n_0)z_-(\mu _1; \gamma ), \bar{x}+\sigma (\mu _2)+L-\pi /\gamma +(n-n_0)z(\mu _2; \gamma )]$ with $\bar{x}=n_0c+\tilde{x}$. Note that $\beta (x)\ge \beta (-\infty )$ for all x. For any small $\eta >0$, an argument similar to what is used to prove (19) shows that there exists a positive integer N such that for $n\ge N$,

$$\begin{aligned} \inf _{ -n(c^*_-(-\infty ) - \varepsilon )\le x\le n(c^*(-\infty )-\varepsilon )} {u}_n(x)\ge {\beta }(-\infty )-\eta . \end{aligned}$$

(31)

If ${k}$ does not have bounded support, an argument similar to the last part of the proof of the statement (i) (b) works to show that (31) still holds.

On the other hand, $u_n(x+nc)\le \bar{u}_n (x)$ where $\bar{u}_n(x)$ is defined in Lemma 2. It follows that for $x\le n(c^*(-\infty )-\epsilon )$, $u_n(x)\le \bar{u}_n ( n(c^*(-\infty )-\epsilon -c))$. Since $c>c^*(\infty )>c^*(-\infty )$ and $\epsilon$ is positive, $c^*(-\infty )-\epsilon -c<0$. Lemma 2 (i) implies that $\bar{u}_n ( n(c^*(-\infty )-\epsilon -c))$ converges to $\beta (-\infty )$ as $n \rightarrow \infty$. We therefore have that for the given $\eta >0$ there exists $N_2>N_1$ such that for $n>N_2$ and $x\le n(c^*(-\infty )-\epsilon )$,

$$\begin{aligned} u_n(x)\le \beta (-\infty )+\eta . \end{aligned}$$

This and (31) show that the statement (ii) (b) holds. The proof is complete. $\square$

1.2.3 Proof of Theorem 3

Proof

We first prove the statement (i). Recall that $\hat{u}_n(x)=u_n(x+nc)$ satisfies (6). Since $c^*_-(-\infty )+c>0$, Lemma 3.3 (b) in Li et al. (2014a) with k(x) replaced by $k(x+c)$ and $c^*_-(-\infty )$ replaced by $c^*_-(-\infty )+c$ shows that for any positive $\varepsilon$ there exist positive numbers $\rho$ and $\hat{\mu }$ such that for all n,

$$\begin{aligned} u_n(x+nc)\le \min \{\rho e^{\hat{\mu }(x+n(c^*_-(-\infty )+c+\varepsilon /2))}, \beta (\infty )\}. \end{aligned}$$

It follows that $u_n(x)\le \min \{\rho e^{\hat{\mu }(x+n(c^*_-(-\infty )+\varepsilon /2))}, \beta (\infty )\}.$ This leads to the statement (i) (a).

To prove the statement (i) (b), we first consider the case of $c^*_-(-\infty )>-c\ge \psi _-(0)$ and assume that there exists $b>0$ such that $k(x)\equiv 0$ for $|x|\ge b$. Lemma 4 (v) shows that for any small positive number $\varepsilon$, there exist small positive numbers $\alpha$, $\epsilon$, and $\gamma$, a large number L, positive numbers $\mu _1$ and $\mu _2$, a real number $\tilde{x}$, and a positive integer $n_0$, such that for $n\ge n_0$, $u_n(x)\ge u^{(n-n_0)}_l(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ with $z_-(\mu _1; \gamma )=c^*_-(-\infty )-\varepsilon /2$ and $z_-(\mu _2, \gamma )=-c+\varepsilon /2$. The proof of Theorem 2 (ii) (b) with $u^{(n)}(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ replaced by $u^{(n)}_l(\epsilon , \mu _1, \mu _2; x-n_0c-\tilde{x})$ incorporating these $z_-(\mu _1; \gamma )$ and $z_-(\mu _2, \gamma )$ works to show the validity of the statement (i) (b). The proof for the case of $-c< \psi _-(0)$ is the proof of Theorem 2 (ii) (b) with $z(\mu _2; \gamma )$ now given by $z(\mu _2; \gamma )=\min \{c, c^*(-\infty )\}-\varepsilon /2$.

To prove the statement (ii) (a), recall that $\hat{u}_n(x)=u_n(x+nc)$ satisfies (6). Since $c^*_-(-\infty )+c<0$, Lemma 3.3 (c) in Li et al. (2014a) with k(x) replaced by $k(x+c)$ shows that for any positive $\varepsilon$ there exist positive numbers $\hat{\rho }$ and $\hat{\mu }$ such that for all n,

$$\begin{aligned} u_n(x+nc)\le \min \{\hat{\rho } e^{\hat{\mu } x}, \beta (\infty )\}. \end{aligned}$$

It follows that

$$\begin{aligned} u_n(x)\le \min \{\hat{\rho } e^{\hat{\mu }(x-nc)}, \beta (\infty )\}. \end{aligned}$$

This leads to the statement (ii) (a). Since $-c^*_-(\infty )\le c$, $\min \{-c, c^*_-(\infty )\}=-c$, the statement (ii)(b) follows from Theorem 2 (i) (b).

We finally prove the statement (iii). Choose $\rho _1>0$ such that $u_0(x)\le \rho _1e^{\mu _1 x}$ where $\mu _1$ is the smallest root of $\phi _-(\infty ; \mu )=c^*_-(\infty )+\varepsilon /2.$ The inequality (16) and induction show that

$$\begin{aligned} {u}_n(x)\le \rho _1 e^{\mu _1 (x+ n(c^*_-(\infty )+\varepsilon /2))}. \end{aligned}$$

(32)

It follows immediately the statement (iii) (a).

Since $c<-c^*_-(\infty )$ implies $c^*(\infty )>c$, the statement (iii) (b) follows from Theorem 2 (i) (b). The proof is complete. $\square$

1.2.4 Proof of Theorem 4

Proof

It follows from (32) that for any $\eta >0$ there exist $N_1>0$ such that

$$\begin{aligned} u_n(x)<\eta , \text{ for } n>N_1 \text{ and } x\le -n(c^*(-\infty )+\varepsilon ). \end{aligned}$$

(33)

On the other hand, Lemma 2 shows that $\beta (-\infty )=0$, and that for any $\eta >0$ there exist a sufficiently negative $\tilde{x}$ and $N>0$ such that

$$\begin{aligned} u_{n}(x)<\eta , \text{ for } n>N_2 \text{ and } x\le \tilde{x}+nc. \end{aligned}$$

(34)

Since $x\le -n(-c+\varepsilon )$ implies $x\le \tilde{x}+nc$ for large n, (33) and (34) show that there exists $N_3>\max \{N_1, N_2 \}$ such that for $n>N_3$, $u_{n}(x)<\eta$ for $x\le -n(\min \{-c, \ c^*_-(\infty )\}+\varepsilon )$. This proves the statement (a).

The statement (b) directly follows from Theorem 2 (i) (b). The proof is complete. $\square$

1.2.5 Proof of Corollary 1

Proof

Statements (a) and (b) follow from Theorem 2 (i). Statement (c) follows from Theorem 3 (i) (b). Statement (d) follows from Theorem 3 (i) (a). $\square$