How Spatial Heterogeneity Affects Transient Behavior in Reaction–Diffusion Systems for Ecological Interactions?

Abstract

Most studies of ecological interactions study asymptotic behavior, such as steady states and limit cycles. The transient behavior, i.e., qualitative aspects of solutions as and before they approach their asymptotic state, may differ significantly from asymptotic behavior. Understanding transient dynamics is crucial to predicting ecosystem responses to perturbations on short timescales. Several quantities have been proposed to measure transient dynamics in systems of ordinary differential equations. Here, we generalize these measures to reaction–diffusion systems in a rigorous way and prove various relations between the non-spatial and spatial effects, as well as an upper bound for transients. This extension of existing theory is crucial for studying how spatially heterogeneous perturbations and the movement of biological species involved affect transient behaviors. We illustrate several such effects with numerical simulations.

Appendices

Appendices

Proof of Theorem 2.2

1.1 Proof When (40) Holds

In the Appendix, we first show the proof of Theorem 2.2, where the estimated upper bound of $\ln \rho (t)$ is given.

Proof

Because the steady state $\varvec{\omega }^{*}$ remains asymptotically stable, from (37) and (40), we immediately have that $\lambda _{1,i}, \lambda _{2,i}$ are real eigenvalues and

$$\begin{aligned} \lambda _{2,i}<\lambda _{1,i}<0 \quad \text{ for } \quad i=1,2,\ldots . \end{aligned}$$

(A.1)

Then, (39) can be further simplified as

$$\begin{aligned}&\Vert \omega _1(x,t) \Vert ^2=\sum _{i=1}^{\infty } \big | k_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,e^{\lambda _{2,i}\,t} \big |^2, \nonumber \\&\Vert \omega _2(x,t) \Vert ^2=\sum _{i=1}^{\infty } \big | k_{1,i}\,c_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,c_{2,i}\,e^{\lambda _{2,i}\,t} \big |^2. \end{aligned}$$

(A.2)

To estimate the profile $\rho (t)$ from (38) is equivalent to maximize

$$\begin{aligned} \begin{aligned}&\Vert \omega _1(x,t) \Vert ^2+ \Vert \omega _2(x,t) \Vert ^2 \\&\quad =\sum _{i=1}^{\infty } \big | k_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,e^{\lambda _{2,i}\,t} \big |^2+\sum _{i=1}^{\infty } \big | k_{1,i}\,c_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,c_{2,i}\,e^{\lambda _{2,i}\,t} \big |^2 \end{aligned} \end{aligned}$$

(A.3)

subject to the constraint

$$\begin{aligned} \sum _{i=1}^{\infty }(k_{1,i}+k_{2,i})^2+\sum _{i=1}^{\infty }(k_{1,i}\,c_{1,i}+k_{2,i}\,c_{2,i})^2=1. \end{aligned}$$

(A.4)

Because of the infinite sum in (A.3) and (A.4), it is almost impossible to find the exact maximum value of (A.3). Instead, we can find an upper bound of (A.3) subject to (A.4). To simplify notations, we let

$$\begin{aligned} M_{1,i}=\int _{\Omega } \omega _{10}(x) \phi _i(x) dx,\quad M_{2,i}=\int _{\Omega } \omega _{20}(x) \phi _i(x) dx. \end{aligned}$$

Then, the constraint (A.4) is equivalent to

$$\begin{aligned} \sum _{i=1}^{\infty }\left( M_{1,i}^2+M_{2,i}^2\right) =1. \end{aligned}$$

(A.5)

Reorganizing (A.3) and using (41), we obtain

$$\begin{aligned}&\Vert \omega _1(x,t) \Vert ^2+ \Vert \omega _2(x,t) \Vert ^2 \nonumber \\&\quad =\sum _{i=1}^{\infty }\left[ (k_{1,i}+k_{2,i})^2\left( e^{2\lambda _{1,i}\,t}+e^{2\lambda _{2,i}\,t}\right) -\left( k_{1,i}\,e^{\lambda _{2,i}\,t}-k_{2,i}\,e^{\lambda _{1,i}\,t}\right) ^2 \right. \nonumber \\&\qquad \left. -2k_{1,i}\,k_{2,i} \left( e^{2\lambda _{1,i}\,t}{+}e^{2\lambda _{2,i}\,t}\right) \right] \nonumber \\&\qquad +\sum _{i=1}^{\infty }\Big [(k_{1,i}\,c_{1,i}{+}k_{2,i}\,c_{2,i})^2\left( e^{2\lambda _{1,i}\,t}{+}e^{2\lambda _{2,i}\,t}\right) {-}\left( k_{1,i}\,c_{1,i}\,e^{\lambda _{2,i}\,t}-k_{2,i}\,c_{2,i}\,e^{\lambda _{1,i}\,t}\right) ^2 \nonumber \\&\qquad -2\,k_{1,i}\,k_{2,i}\,c_{1,i}\,c_{2,i} \left( e^{2\lambda _{1,i}\,t}+e^{2\lambda _{2,i}\,t}\right) \Big ] \nonumber \\&\quad \le \sum _{i=1}^{\infty }\Big \{\Big [\left( k_{1,i}+k_{2,i}\right) ^2+\left( k_{1,i}\,c_{1,i}+k_{2,i}\,c_{2,i}\right) ^2\Big ] \left( e^{2\lambda _{1,i}\,t}+e^{2\lambda _{2,i}\,t}\right) \Big \} \nonumber \\&\qquad -2\sum _{i=1}^{\infty } \left( 1+c_{1,i}\,c_{2,i}\right) k_{1,i}\,k_{2,i}\left( e^{2\lambda _{1,i}\,t}+e^{2\lambda _{2,i}\,t}\right) \nonumber \\&\quad \le 2 \sum _{i=1}^{\infty }\Big [\left( k_{1,i}+k_{2,i}\right) ^2+\left( k_{1,i}\,c_{1,i}+k_{2,i}\,c_{2,i}\right) ^2\Big ] e^{2\lambda _{1,i}\,t}\nonumber \\&\qquad -2\sum _{i=1}^{\infty } \left( 1+c_{1,i}\,c_{2,i}\right) k_{1,i}\,k_{2,i}\left( e^{2\lambda _{1,i}\,t}+e^{2\lambda _{2,i}\,t}\right) . \nonumber \\&\quad \le 2\,e^{2\lambda _1^{*}t}\sum _{i=1}^{\infty }\left( M_{1,i}^2+M_{2,i}^2\right) +2\bigg | \sum _{i=1}^{\infty } \left( 1+c_{1,i}\,c_{2,i}\right) k_{1,i}\,k_{2,i}\left( e^{2\lambda _{1,i}\,t}+e^{2\lambda _{2,i}\,t}\right) \bigg | \nonumber \\&\quad \le 2\,e^{2\lambda _1^{*}t}+4\sum _{i=1}^{\infty }\big |\left( 1+c_{1,i}c_{2,i}\right) k_{1,i}k_{2,i}\big |e^{2\lambda _{1,i}t} \nonumber \\&\quad \le 2\,e^{2\lambda _1^{*}t}+4\,e^{2\lambda _1^{*}t}\sum _{i=1}^{\infty }\big |\left( 1+c_{1,i}c_{2,i}\right) k_{1,i}k_{2,i}\big |, \end{aligned}$$

(A.6)

where $\lambda _{1,i}\le \lambda _1^{*}$ for $i=1,2,\ldots$

Direct calculations show that

$$\begin{aligned} 1+c_{1,i}c_{2,i}=\frac{J_{12}-J_{21}}{J_{12}}, \end{aligned}$$

by using (34). It follows that

$$\begin{aligned}&\sum _{i=1}^{\infty }\left| \left( 1+c_{1,i}c_{2,i}\right) k_{1,i}k_{2,i}\right| =\frac{\vert J_{12}-J_{21}\vert }{\vert J_{12} \vert }\sum _{i=1}^{\infty }\vert k_{1,i}k_{2,i} \vert \nonumber \\&\quad =\left| J_{12}\left( J_{12}-J_{21}\right) \right| \sum _{i=1}^{\infty } \bigg | \frac{\left( -c_{2,i}M_{1,i}+M_{2,i}\right) \left( c_{1,i}M_{1,i}-M_{2,i}\right) }{A_1 \mu _i^2+A_2 \mu _i+A_3}\bigg | \nonumber \\&\quad =\left| J_{12}-J_{21} \right| \nonumber \\&\qquad \times \sum _{i=1}^{\infty } \frac{\left| M_{1,i}M_{2,i}\left[ \mu _i \left( d_1-d_2\right) +\left( J_{22}-J_{11}\right) \right] -J_{12}^2M_{2,i}^2+J_{12}J_{21}M_{1,i}^2 \right| }{A_1 \mu _i^2+A_2 \mu _i+A_3} \nonumber \\&\quad \le \left| J_{12}-J_{21} \right| \nonumber \\&\qquad \times \sum _{i=1}^{\infty } \frac{\frac{1}{2}\vert \mu _i \left( d_1-d_2\right) +\left( J_{22}-J_{11}\right) \vert \left( \vert M_{1,i} \vert ^2+\vert M_{2,i} \vert ^2\right) +\vert J_{12}^2 \vert M_{2,i}^2+\vert J_{12}J_{21} \vert M_{1,i}^2 }{A_1 \mu _i^2+A_2 \mu _i+A_3} \nonumber \\&\quad =\left| J_{12}-J_{21}\right| \left( \sum _{i=1}^{\infty }f_1(\mu _i) M_{1,i}^2 +\sum _{i=1}^{\infty } f_2(\mu _i) M_{2,i}^2 \right) , \end{aligned}$$

(A.7)

where $f_1(\mu _i), f_2(\mu _i)$ are given in (44) and (45), respectively. By Lemma 2.2, we have $f_1(\mu _i) \le A_1^{*}$ and $f_2(\mu _i) \le A_2^{*}$ and $A_1^{*}>0, A_2^{*}>0.$ Then, we continue the estimate of (A.7) by using the above inequalities as

$$\begin{aligned} \begin{aligned}&\vert J_{12}-J_{21}\vert \left( \sum _{i=1}^{\infty }f_1(\mu _i) M_{1,i}^2 +\sum _{i=1}^{\infty } f_2(\mu _i) M_{2,i}^2 \right) \\&\quad \le \vert J_{12}-J_{21} \vert \left( A_1^{*} \sum _{i=1}^{\infty } M_{1,i}^2+A_2^{*}\sum _{i=1}^{\infty }M_{2,i}^2\right) \\&\quad \le \vert J_{12}-J_{21} \vert \max \left\{ A_1^{*},A_2^{*}\right\} \sum _{i=1}^{\infty }\left( M_{1,i}^2+M_{2,i}^2\right) =\vert J_{12}-J_{21} \vert \max \left\{ A_1^{*},A_2^{*}\right\} , \end{aligned} \end{aligned}$$

(A.8)

where the last equality comes from (A.5). Substituting (A.8) back to (A.6) gives

$$\begin{aligned} \rho (t) \le e^{\lambda _1^{*}t}\sqrt{2+4 \vert J_{12}-J_{21} \vert \max \left\{ A_1^{*},A_2^{*}\right\} }. \end{aligned}$$

(A.9)

We take $\ln$ for (A.9) at both sides of the inequality and obtain

$$\begin{aligned} \ln \rho (t) \le \lambda _1^{*} t+ \ln \sqrt{2+4 \vert J_{12}-J_{21} \vert \max \{A_1^{*},A_2^{*}\}}. \end{aligned}$$

(A.10)

Notice that the above estimate lacks accuracy for $\ln \rho (t)$ when $t \rightarrow 0$ because $\lambda _1^{*} t$ decreases linearly while perturbations may grow initially. Hence, we give another estimate for $\ln \rho (t)$ when t is close to 0.

We show that $\ln \rho (t) \le \sigma t$ for $t \ge 0,$ where $\sigma$ is the reactivity defined in (13). Suppose not, then there exists at least a $t_1>0$ such that $\ln \rho (t_1)>\sigma t_1.$ Moreover, we have $\ln \rho (t_0)-\sigma t_0=0$ if $t_0=0.$ This leads to

$$\begin{aligned} \frac{\hbox {d} \ln \rho (t) }{\hbox {d}t}\Big |_{t=t_1}>\frac{\hbox {d} (\sigma t)}{\hbox {d}t}\Big |_{t=t_1}, \end{aligned}$$

which directly gives

$$\begin{aligned} \frac{1}{\rho (t_1)}\frac{\hbox {d} \rho (t)}{\hbox {d}t}\Big |_{t=t_1}>\sigma . \end{aligned}$$

(A.11)

By the definition of $\rho (t)$ in (27), rearranging (A.11) leads to

$$\begin{aligned} \max _{\Vert \varvec{\omega }_0(x) \Vert = 1}{\left( \frac{1}{\Vert \varvec{\omega }(x,t) \Vert }\frac{\hbox {d} \Vert \varvec{\omega }(x,t) \Vert }{\hbox {d}t}\right) \Big |_{t=t_1}}>\sigma . \end{aligned}$$

This contradicts the definition of $\sigma$ in (13) because $\sigma$ is chosen as the maximum and $t=0$ can be chosen arbitrarily. Therefore, when t is close to 0,  $\ln \rho (t)$ is bounded above by $\sigma t.$ This together with (A.10) gives (46) and thus completes the proof. $\square$

1.2 Proof When (40) is Relaxed

Now, we relax assumption (40) such that for some index j,  $h_1^2(\mu _j)-4h_2(\mu _j)<0$ holds. It follows that for these $\mu _j,$ we have

$$\begin{aligned} \lambda _{1,j}=a_1(\mu _j)+\mathrm {i} b_1(\mu _j),\quad \lambda _{2,j}=a_1(\mu _j)-\mathrm {i}b_1(\mu _j), \end{aligned}$$

(A.12)

where

$$\begin{aligned} a_1(\mu _j)=\frac{-h_1(\mu _j)}{2},\quad b_1(\mu _j)=\frac{\sqrt{4h_2(\mu _j)-h_1^2(\mu _j)}}{2}. \end{aligned}$$

Following (39), we have

$$\begin{aligned} \Vert \omega _1(x,t) \Vert ^2=&\sum _{\begin{array}{c} i=1\\ i \ne j \end{array}}^{\infty }\left( k_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,e^{\lambda _{2,i}\,t}\right) ^2 \nonumber \\&+\sum _{j \ne i}\overline{\left( k_{1,j}\,e^{\lambda _{1,j}\,t}+ k_{2,j}\,e^{\lambda _{2,j}\,t}\right) }\left( k_{1,j}\,e^{\lambda _{1,j}\,t}+ k_{2,j}\,e^{\lambda _{2,j}\,t}\right) , \nonumber \\ \Vert \omega _2(x,t) \Vert ^2=&\sum _{\begin{array}{c} i=1\\ i\ne j \end{array}}^{\infty } \left( k_{1,i}\,c_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,c_{2,i}\,e^{\lambda _{2,i}\,t} \right) ^2 \nonumber \\&\quad +\sum _{j \ne i}\overline{\left( k_{1,j}\,c_{1,j}\,e^{\lambda _{1,j}\,t}+ k_{2,j}\,c_{2,j}\,e^{\lambda _{2,j}\,t}\right) }\nonumber \\&\times \left( k_{1,j}\,c_{1,j}\,e^{\lambda _{1,j}\,t}+ k_{2,j}\,c_{2,j}\,e^{\lambda _{2,j}\,t}\right) . \end{aligned}$$

(A.13)

In the above (A.13), we have

$$\begin{aligned}&k_{1,j}e^{\lambda _{1,j}t}+k_{2,j}e^{\lambda _{2,j}t}\nonumber \\&\quad =\frac{J_{12}}{\lambda _{1,j}-\lambda _{2,j}} \left\langle -\frac{\lambda _{2,j}+d_1\mu _j-J_{11}}{J_{12}} \omega _{10}(x)+\omega _{20}(x),\phi _j(x)\right\rangle e^{\lambda _{1,j}t}\nonumber \\&\qquad +\frac{J_{12}}{\lambda _{1,j}-\lambda _{2,j}} \left\langle \frac{\lambda _{1,j}+d_1\mu _j-J_{11}}{J_{12}}\omega _{10}(x)-\omega _{20}(x),\phi _j(x)\right\rangle e^{\lambda _{2,j}t}. \end{aligned}$$

(A.14)

To simplify the inner product in (A.14), let

$$\begin{aligned}&f_1(x)=\frac{-a_1(\mu _j)-d_1\mu _j+J_{11}}{J_{12}} \omega _{10}(x)+\omega _{20}(x),\nonumber \\&f_2(x)=\frac{b_1(\mu _j)}{J_{12}} \omega _{10}(x). \end{aligned}$$

(A.15)

Then, by (A.12) and (A.15), we have

$$\begin{aligned}&\left\langle -\frac{\lambda _{2,j}+d_1\mu _j-J_{11}}{J_{12}}\omega _{10}(x)+\omega _{20}(x),\phi _j(x)\right\rangle e^{\lambda _{1,j}t} \nonumber \\&\quad =e^{a_1(\mu _j)t}\left\langle \left( f_1(x)+f_2(x)\mathrm {i}\right) e^{b_1(\mu _j)t\mathrm {i}},\phi _j(x)\right\rangle \nonumber \\&\quad =e^{a_1(\mu _j)t}\left\langle f_3(x)+f_4(x)\mathrm {i},\phi _j(x)\right\rangle , \end{aligned}$$

(A.16)

where

$$\begin{aligned}&f_3(x)=f_1(x)\cos (b_1(\mu _j)t)-f_2(x)\sin (b_1(\mu _j)t), \\&f_4(x)=f_1(x)\sin (b_1(\mu _j)t)+ f_2(x)\cos (b_1(\mu _j)t). \end{aligned}$$

Similar calculations show that

$$\begin{aligned}&\left\langle \frac{\lambda _{1,j}+d_1\mu _j-J_{11}}{J_{12}} \omega _{10}(x)-\omega _{20}(x),\phi _j(x)\right\rangle e^{\lambda _{2,j}t}\nonumber \\&\quad =e^{a_1(\mu _j)t}\left\langle -f_3(x)+f_4(x) \mathrm {i},\phi _j(x)\right\rangle . \end{aligned}$$

(A.17)

Furthermore, by (A.12), it is obvious that $\lambda _{1,j}-\lambda _{2,j}=2b_1(\mu _j)\mathrm {i}.$ Then, substituting (A.16) and (A.17) back to (A.14) gives

$$\begin{aligned}&k_{1,j}e^{\lambda _{1,j}t}+k_{2,j}e^{\lambda _{2,j}t}=\frac{J_{12}e^{a_1 (\mu _j) t}}{2b_1(\mu _j)}\left\langle 2f_4(x),\phi _j(x)\right\rangle , \end{aligned}$$

(A.18)

which is a real number. Hence, we obtain

$$\begin{aligned} k_{1,j}e^{\lambda _{1,j}t}+k_{2,j}e^{\lambda _{2,j}t}= \overline{k_{1,j}e^{\lambda _{1,j}t}+k_{2,j}e^{\lambda _{2,j}t}}, \end{aligned}$$

which further gives

$$\begin{aligned} \Vert \omega _1(x,t) \Vert ^2=&\sum _{\begin{array}{c} i=1\\ i \ne j \end{array}}^{\infty }\left( k_{1,i}\,e^{\lambda _{1,i}\,t}+ k_{2,i}\,e^{\lambda _{2,i}\,t}\right) ^2 \nonumber \\&+\sum _{j \ne i}\overline{\left( k_{1,j}\,e^{\lambda _{1,j}\,t}+ k_{2,j}\,e^{\lambda _{2,j}\,t}\right) }\left( k_{1,j}\,e^{\lambda _{1,j}\,t}+ k_{2,j}\,e^{\lambda _{2,j}\,t}\right) \nonumber \\ =&\sum _{i=1}^{\infty }\left( k_{1,i}e^{\lambda _{1,i}t}+k_{2,i}e^{\lambda _{2,i}t}\right) ^2. \end{aligned}$$

(A.19)

Similar calculations show that $\Vert \omega _2(x,t) \Vert ^2$ from (A.13) can be reduced to the equivalent form in (A.2). Therefore, as long as (37) is satisfied, even if some eigenvalues $\lambda _{1,j},\lambda _{2,j}$ have imaginary parts, expressions in (A.3) and (A.4) remain the same, which leads to similar results for the upper bound of the amplification envelope $\rho (t).$

Proof of Proposition 2.3

The following gives the proof of Proposition 2.3.

Proof

Assume that (47), (49), (51) and (52) are satisfied. Direct calculations give

$$\begin{aligned} \frac{d\lambda _{1,i}}{d \mu _i}&=\frac{\left[ (d_1-d_2)^2\mu _i+(d_2-d_1)(J_{11}-J_{22})-(d_1+d_2)\sqrt{h_1^2(\mu _i)-4h_2(\mu _i)}\right] }{2\sqrt{h_1^2(\mu _i)-4h_2(\mu _i)}} \nonumber \\&:=\frac{1}{2\sqrt{h_1^2(\mu _i)-4h_2(\mu _i)}}g(\mu _i). \end{aligned}$$

(B.1)

From (B.1), it is clear that

$$\begin{aligned} \frac{d\lambda _{1,i}}{d\mu _i}<0 \Leftrightarrow g(\mu _i)<0. \end{aligned}$$

(B.2)

Let $\bar{\mu }=-(J_{11}-J_{22})/(d_2-d_1)>0,$ by (49). If $\mu _i<\bar{\mu },$ then $g(\mu _i)<0$ is satisfied. If $\mu _i>\bar{\mu },$$g(\mu _i)<0$ leads to

$$\begin{aligned} a_1(\mu _i)^2+a_2(\mu _i)+a_3>0, \end{aligned}$$

(B.3)

where

$$\begin{aligned} \begin{aligned}&a_1=4d_1d_2(d_1-d_2)^2,\quad a_2=8d_1d_2(d_2-d_1)(J_{11}-J_{22}),\\&a_3=4 \left\{ J_{12}J_{21}\left( d_1^2+d_2^2\right) +d_1d_2\left[ 2J_{12}J_{21}+(J_{11}-J_{22})^2\right] \right\} ,\\&a_2^2-4a_1a_3=-64d_1d_2(d_1-d_2)^2J_{12}J_{21}(d_1+d_2)^2. \end{aligned} \end{aligned}$$

(B.4)

From (B.4), it is clear that $a_1>0, a_2<0, a_3>0$ and $a_2^2-4a_1a_3<0$ when (47), (49), (51) and (52) hold. This shows that (B.3) is satisfied for all $\mu _i \ge 0,$ which further leads to $(d \lambda _{1,i})/(d \mu _i)<0$.

Therefore, $\lambda _{1,i}$ is a decreasing function of $\mu _i$ and is maximized at $\lambda _{1,1}, \text{ i.e., } \,\mu _i=0,$ which in fact agrees with the largest eigenvalue corresponding to the spatial mode zero. We notice that the condition $J_{12}J_{21}>0$ in (49) implies that the system describes competitive or cooperative interactions. This suggests that for such type of systems, the upper bound of the amplification envelope is, in fact, regulated by the largest eigenvalue of the non-spatial model.

Now assume that (47), (48), (51) and (53) are satisfied. Following similar calculations as (B.1), we obtain the same (B.2), (B.3) and (B.4). Condition (53) directly gives $J_{12}J_{21}<0,$ which further leads to $a_1>0, a_2>0$ and $a_2^2-4a_1a_3>0$ but the sign of $a_3$ is undetermined.

If $a_3>0$, then (B.3) holds for all $\mu _i \ge 0.$ This again shows that $\lambda _{1,i}$ is a decreasing function of $\mu _i$ and is maximized at $\lambda _{1,1}$. If $a_3<0,$ a unique positive root of $a_1(\mu _i)^2+a_2(\mu _i)+a_3=0$ exists and we denote the root by $\bar{\mu }_1.$ Direct calculations show that

$$\begin{aligned} \bar{\mu }_1=\frac{-a_2+\sqrt{a_2^2-4a_1a_3}}{2a_1}. \end{aligned}$$

(B.5)

Following (B.5), if $\mu _i<\bar{\mu }_1,$ then (B.3) is violated, which further leads to $d\lambda _{1,i}/d\mu _i>0.$ If $\mu _i>\bar{\mu }_1,$ then (B.3) holds, which gives $d\lambda _{1,i}/d\mu _i<0.$ Hence, by the above discussion, $\mu _i=\bar{\mu }_1$ is a local maximum and maximizes $\lambda _{1,i}$ at some intermediate i. Still in the analysis, $\mu _i$ is regarded as a continuous variable, but $\bar{\mu }_1$ provides an approximation for the $\mu _i$ which maximizes $\lambda _{1,i}$. In fact, when $\mu _i$ is discrete, it is straightforward to choose $\mu _j$ and $\mu _{j+1}$ which satisfy

$$\begin{aligned} \mu _{j}<\bar{\mu }_1<\mu _{j+1}, \end{aligned}$$

and compare $\lambda _{1,j},\lambda _{1,j+1}$ to find $\lambda _1^{*}=\max \left\{ \lambda _{1,j},\lambda _{1,j+1}\right\} .$

We notice that $J_{12}J_{21}<0$ above indicates that the system is a predator–prey-type model. Different from the previous case where $\lambda _1^{*}=\lambda _{1,1}$ if the system is a competitive or cooperative type, the upper bound of $\rho (t)$ for a predator–prey-type system needs not to be the largest eigenvalue corresponding to spatial mode 0 but may be regulated by the largest eigenvalue corresponding to some intermediate spatial mode.

We finally discuss the case where (47), (48), (51) and (52) hold. Following the same steps in the previous analysis, we obtain the same (B.3) if $d\lambda _{1,i}/d\mu _i<0.$ By the assumptions (47), (48), (51) and (52), we have $a_1>0, a_2>0$ but the signs of $a_3$ and $a_2^2-4a_1a_3$ are undetermined. Still, it is straightforward that $a_2^2-4a_1a_3>0$ if and only if $J_{12}J_{21}<0$. If $a_3>0,$ then (B.3) is satisfied regardless of the sign of $a_2^2-4a_1a_3$. This implies that $\lambda _{1,i}$ is a decreasing function of $\mu _i$ and is maximized at $\lambda _{1,1}$. If $a_3<0,$ we have $a_2^2-4a_1a_3>0$ and therefore $J_{12}J_{21}<0$ as a necessary condition. This results in a similar case to the above one, where $\lambda _{1,i}$ is maximized at some intermediate $\lambda _{1,n}$ for $n \ge 2.$ The analysis is similar and is thus omitted. $\square$