Turing instabilities in prey–predator systems with dormancy of predators

Abstract

In this paper, we study the stationary and oscillatory Turing instabilities of a homogeneous equilibrium in prey–predator reaction–diffusion systems with dormant phase of predators. We propose a simple criterion which is useful in classifying these Turing instabilities. Moreover, numerical simulations reveal transient spatio-temporal complex patterns which are a mixture of spatially periodic steady states and traveling/standing waves. In this mixture, the steady part is the stable Turing pattern bifurcated primarily from the homogeneous equilibrium, while wave parts are unstable oscillatory solutions bifurcated secondarily from the same homogeneous equilibrium. Although our criterion does not exclude the occurrence of oscillatory Turing instability, we have not yet found stable traveling/standing waves due to oscillatory Turing instability in our simulations. These results suggest that dormancy of predators is not a generator but an enhancer of spatio-temporal Turing patterns in prey–predator reaction–diffusion systems.

Appendices

Appendix A: Proof of Theorem 3.1

Since $A$ is stable, we have

$$\begin{aligned} \mathrm{tr} A = s'(\bar{u}) - f'(\bar{u})\bar{v} - m'(\bar{v})\bar{v} < 0 \end{aligned}$$

(6.1)

and

$$\begin{aligned} \det A = - (s'(\bar{u}) - f'(\bar{u})\bar{v} ) m'(\bar{v}) \bar{v} + k f(\bar{u}) f'(\bar{u}) \bar{v} > 0. \end{aligned}$$

(6.2)

Let

$$\begin{aligned} \begin{array}{rcl} p_1 &{} := &{} - \mathrm{tr} B,\\ p_2 &{} := &{} \det B_{12} + \det B_{23} + \det B_{13},\\ p_3 &{} := &{} -\det B, \end{array} \end{aligned}$$

(6.3)

and

$$\begin{aligned} B_{12} = \left( \begin{array}{c@{\quad }c} b_{11} &{} b_{12}\\ b_{21} &{} b_{22} \end{array} \right) , \ \ B_{23} = \left( \begin{array}{c@{\quad }c} b_{22} &{} b_{23}\\ b_{32} &{} b_{33} \end{array} \right) , \ \ B_{13} = \left( \begin{array}{c@{\quad }c} b_{11} &{} b_{13}\\ b_{31} &{} b_{33} \end{array} \right) . \end{aligned}$$

Direct calculation shows that

$$\begin{aligned} \begin{array}{rcl} p_1 &{} = &{} - \left( s'(\bar{u}) - f'(\bar{u})\bar{v} - m'(\bar{v})\bar{v} - \alpha (\bar{u}) - k \nu (\bar{w})f(\bar{u}) \right) \\ &{} = &{} - \mathrm{tr} A + \alpha (\bar{u}) + k \nu (\bar{w})f(\bar{u}) > 0 \end{array} \end{aligned}$$

(6.4)

and

$$\begin{aligned} p_3 = \alpha (\bar{u}) \det A > 0 , \end{aligned}$$

(6.5)

by virtue of (6.1) and (6.2). Moreover, we have $\det B_{13} = - \alpha (\bar{u}) ( s'(\bar{u}) - f'(\bar{u})\bar{v} ), \det B_{23} = \alpha (\bar{u}) m'(\bar{v}) \bar{v}$ and

$$\begin{aligned} \begin{array}{rcl} \det B_{12}&= Y + \alpha '(\bar{u}) \bar{w} f(\bar{u}) , \end{array} \end{aligned}$$

where

$$\begin{aligned} Y = - ( s'(\bar{u}) \!-\! f'(\bar{u})\bar{v} )( k \nu (\bar{u}) f(\bar{u}) \!+\! m'(\bar{v})\bar{v} ) \!+\! k f(\bar{u}) \bar{v} \left( \mu '(\bar{u}) f(\bar{u}) \!+\! \mu (\bar{u})f'(\bar{u}) \right) \nonumber \\ \end{aligned}$$

(6.6)

is independent of $\alpha (\bar{u})$. Therefore, we have

$$\begin{aligned} \begin{array}{rcl} p_1 p_2 &{} = &{} \left( - { \mathrm tr} A + k \nu (\bar{u}) f(\bar{u}) + \alpha (\bar{u}) \right) \left( - \alpha (\bar{u}) \mathrm{tr} A + Y + \alpha '(\bar{u}) \bar{w} f(\bar{u})\right) \\ &{} > &{} \left( - { \mathrm tr} A + k \nu (\bar{u}) f(\bar{u}) \right) Y + \alpha (\bar{u}) Y \end{array} \end{aligned}$$

by virtue of (6.1) and $\alpha '(\bar{u}) \ge 0$. Hence, $p_1 p_2 - p_3 > 0$ holds for sufficiently small $\alpha (\bar{u})$ if $Y > 0$. Noting $\mu (\bar{u}) + \nu (\bar{u}) = 1$, we have

$$\begin{aligned} Y \!=\!k \mu '(\bar{u})f^2(\bar{u})\bar{v} \!-\! s'(\bar{u}) \left( m'(\bar{v})\bar{v} \!+\! k \nu (\bar{w})f(\bar{u}) \right) \!+\! \left( k f(\bar{u})f'(\bar{u})\bar{v} + m'(\bar{v})\bar{v}^2 f'(\bar{u}) \right) \end{aligned}$$

When $s'(\bar{u}) \le 0$, we have $Y >0$. On the other hand, when $s'(\bar{u}) > 0$, we have $Y > 0$ under the condition (3.5) because $\mu (\bar{u}) + \nu (\bar{u}) = 1$. Thus, $p_1 p_2 - p_3 > 0$ holds for sufficiently small $\alpha (\bar{u})$ under the condition (3.5). Therefore, it follows from the Routh–Hurwitz condition that $B$ is stable if $A$ is stable and (3.5) holds.

Appendix B: Proof of Theorem 3.3

By (Cross 1978, Theorem 4), Turing instability does not occur if and only if $B$ is strongly stable, i.e., all the signed principal minors of $B$ are nonnegative. Direct calculations show

$$\begin{aligned} \begin{array}{l} - b_{11} = - ( s'(\bar{u}) - f'(\bar{u})\bar{v} ) , \ \ - b_{22} = k \nu (\bar{u}) f(\bar{u}) + m'(\bar{v}) \bar{v}, \ \ - b_{33} = \alpha (\bar{u}),\\ \det B_{13} = - \alpha (\bar{u}) ( s'(\bar{u}) - f'(\bar{u})\bar{v} ), \ \ \det B_{23} = \alpha (\bar{u}) m'(\bar{v}) \bar{v},\\ \det B_{12} = Y + \alpha '(\bar{u}) \bar{w} f(\bar{u}) , \ \ - \det B = \alpha (\bar{u}) \det A \end{array} \end{aligned}$$

where $Y$ is given by (6.6). Noting (6.2) and that $Y \ge 0$ if $s'(\bar{u}) - f'(\bar{u})\bar{v} \le 0$, we see that $B$ is strongly stable if and only if $s'(\bar{u}) - f'(\bar{u})\bar{v} \le 0$.

Appendix C: Proof of Theorem 3.4

Since $B$ is stable, it follows from the Routh–Hurwitz condition that

$$\begin{aligned}&p_1 > 0,\end{aligned}$$

(8.1)

$$\begin{aligned}&p_3 > 0, \end{aligned}$$

(8.2)

and

$$\begin{aligned} p_1 p_2 - p_3 > 0, \end{aligned}$$

(8.3)

where $p_1, p_2$ and $p_3$ are given by (6.3). Notice that (8.1)–(8.3) imply

$$\begin{aligned} p_2 > 0, \end{aligned}$$

(8.4)

which plays a key role in the subsequent analysis.

We analyze the eigenvalue of $-\sigma D_B + B$, where $\sigma > 0$ and

$$\begin{aligned} D_B = \left( \begin{array}{c@{\quad }c@{\quad }c} d_u &{} 0 &{} 0\\ 0 &{} d_v &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right) . \end{aligned}$$

The characteristic equation of $-\sigma D_B + B$ is given by

$$\begin{aligned} \lambda ^3 + p_1(\sigma ) \lambda ^2 + p_2(\sigma ) \lambda + p_3(\sigma ) = 0, \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} p_1(\sigma ) := p_1 + ( d_u + d_v ) \sigma , \ \ p_2(\sigma ) := p_2 - \{ (\mathrm{tr} B_{23} ) d_u + ( \mathrm{tr} B_{13}) d_v \} \sigma + d_u d_v \sigma ^2,\\ p_3(\sigma ) := p_3 + \{ (\det B_{23}) d_u + (\det B_{13} ) d_v \} \sigma + \alpha (\bar{u}) d_u d_v \sigma ^2. \end{array} \end{aligned}$$

Since $p_1(\sigma ) > 0$ for all $\sigma >0$ by (8.1), it is impossible that both $p_3(\sigma ) = 0$ and $p_1(\sigma )p_2(\sigma ) - p_3(\sigma ) = 0$ hold due to the assumption of the simplicity of the eigenvalue $\lambda = \lambda (\sigma , D_B; B)$ (see, Assumption 2.2). Therefore, it follows from the Routh–Hurwitz condition that a pair of complex eigenvalues crosses the imaginary axis and enters into the right half plane if the sign of $p_1(\sigma )p_2(\sigma ) - p_3(\sigma )$ changes from positive to negative. Let

$$\begin{aligned} \begin{array}{l} F(\sigma ) := p_1(\sigma )p_2(\sigma ) - p_3(\sigma )\\ \ \ \ \ \ \ \ \ = p_1p_2 - p_3 + \{ \left( \det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) \right) d_u\\ \ \ \ \ \ \ \ \ \ \ \ \ \ +\, \left( \det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) \right) d_v \} \sigma \\ \ \ \ \ \ \ \ \ \ \ \ \ \ + \,\{ - (\mathrm{tr} B_{23}) d_u^2 - (\mathrm{tr} B_{13}) d_v^2 - 2(\mathrm{tr} B) d_u d_v \} \sigma ^2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ +\, d_u d_v ( d_u + d_v) \sigma ^3 . \end{array} \end{aligned}$$

(8.5)

Lemma 8.1

When $m'(\bar{v}) > 0$ and $s'(\bar{u}) - f'(\bar{u})\bar{v} < \alpha (\bar{u})$, there exists $\delta _2 >0$ independent of $\sigma$ such that $F(\sigma ) > \delta _2$ holds for all $\sigma >0$ and for all $d_u > 0$.

Proof

Let $I(d_u)$ and $J(d_u)$ denote the coefficients of $\sigma$ and $\sigma ^2$ in $F(\sigma )$, respectively, which are given by

$$\begin{aligned} \begin{array}{rcl} I(d_u) &{} = &{} \left( \det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) \right) d_u\\ &{} &{} \ +\, \left( \det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) \right) d_v \end{array} \end{aligned}$$

(8.6)

and

$$\begin{aligned} J(d_u) = - (\mathrm{tr} B_{23}) d_u^2 - (\mathrm{tr} B_{13}) d_v^2 - 2(\mathrm{tr} B) d_u d_v. \end{aligned}$$

(8.7)

We will prove that $I(d_u) >0$ and $J(d_u) >0$ hold for all $d_u >0$. In order to show $I(d_u) >0$, we have to prove $\det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) > 0$ and $\det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) > 0$. Assume that $\det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) \le 0$ holds. Then, we have

$$\begin{aligned} XY + Z - \bar{\alpha } X - \bar{\alpha } \bar{\nu } k \bar{f} + ( X+Y - \bar{\alpha } )( Y - \bar{\alpha }) \le 0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} XY + Z - \bar{\alpha } (X + Y) - \bar{\alpha } \bar{\nu } k \bar{f} + ( X+Y )Y - \bar{\alpha } (X + Y ) + \bar{\alpha }^2 \le 0, \end{aligned}$$

(8.8)

where $X:= - \nu (\bar{u}) k f(\bar{u}) - m'(\bar{v})\bar{v},\,Y:=s'(\bar{u}) - f'(\bar{u})\bar{v},\,Z:= f(\bar{u}) \{ (\mu '(\bar{u})f(\bar{u}) + \mu (\bar{u})f'(\bar{u}) )k \bar{v} + \alpha '(\bar{u})\bar{w} \},\,\bar{\alpha } = \alpha (\bar{u}),\,\bar{\nu } = \nu (\bar{u})$ and $\bar{f} = f(\bar{u})$. On the other hand, by (6.3) and (8.4), we have

$$\begin{aligned} XY + Z - \bar{\alpha } ( X + Y ) > \bar{\alpha } \bar{\nu } k \bar{f}. \end{aligned}$$

(8.9)

Therefore, it follows from (8.8) and (8.9) that

$$\begin{aligned} \begin{array}{rcl} 0 &{} \ge &{} \bar{\alpha } \bar{\nu } k \bar{f} - \bar{\alpha } \bar{\nu } k \bar{f} + ( X+Y )Y - \bar{\alpha } (X + Y ) + \bar{\alpha }^2\\ &{} = &{} (Y - \bar{\alpha }) ( X + Y ) + \bar{\alpha }^2 , \end{array} \end{aligned}$$

which leads to a contradiction because $Y < \bar{\alpha }$ and

$$\begin{aligned} X+Y < - m'(\bar{v})\bar{v} + s'(\bar{u}) - f'(\bar{u})\bar{v} < 0 \end{aligned}$$

(8.10)

by virtue of $0 < \mu (\bar{u}) < 1,\,f(\bar{u}) >0$ and (6.1). Therefore, $\det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) > 0$. Similarly, we can prove $\det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) > 0$. Hence, we see that $I(d_u) > 0$ holds.

Furthermore, we see that $J(d_u) > 0$ holds because $\mathrm{tr} B_{23} = X - \bar{\alpha } < 0,\,\mathrm{tr} B_{13} = Y - \bar{\alpha } < 0$, and $\mathrm{tr} B = -p_1 < 0$ by (8.1).

Thus, noting $\delta _2 := p_1p_2 -p_3 > 0$ by (8.3), we see that $F(\sigma ) > \delta _2$ holds for all $\sigma >0$ and for all $d_u >0$. $\square$

Lemma 8.2

When $m'(\bar{v}) > 0$ and $s'(\bar{u}) - f'(\bar{u})\bar{v} > \alpha (\bar{u})$, there exists $\delta _3 >0$ independent of $\sigma$ such that $F(\sigma ) > \delta _3$ holds for all $\sigma >0$ and for $d_u \ge d^*$, where $d^*$ is given by (3.7). Moreover, there exists $\delta _4 >0$ independent of $\sigma$ such that $F(\sigma ) < 0$ holds for $\sigma > \delta _4$ and for $d_u =0$.

Proof

In order to show the first assertion of this lemma, it suffices to prove that $I(d_u) >0$ and $J(d_u) >0$ hold for $d_u \ge d^*$, where $I(d_u)$ and $J(d_u)$ are given by (8.6) and (8.7), respectively. First, we will show that $\det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) > 0$ holds. Assume that $\det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) \le 0$, i.e.,

$$\begin{aligned} XY + Z - \bar{\alpha } Y + ( X+Y - \bar{\alpha } )( X - \bar{\alpha }) \le 0, \end{aligned}$$

(8.11)

where $X, Y, Z$ and $\bar{\alpha }$ are the same notations in the proof of Lemma 8.1. By using the same line of arguments as applied to (8.8), it follows from (8.9) and (8.11) that

$$\begin{aligned} \begin{array}{rcl} 0 &{} \ge &{} \bar{\alpha } \bar{\nu } k \bar{f} - \bar{\alpha }( X + Y ) + X(X+Y) + \bar{\alpha }^2\\ &{} > &{} \bar{\alpha } \bar{\nu } k \bar{f} - \bar{\alpha }( X + Y ) + \bar{\alpha }^2 \end{array} \end{aligned}$$

because $X < kf(\bar{u}) - m(\bar{v}) < 0$ and $X + Y < 0$ by (8.10), where $\bar{\nu }$ and $\bar{f}$ are the same notations in the proof of Lemma 8.1. Hence, we have $\bar{\alpha } < X + Y - \bar{\nu } k \bar{f} < 0$, which contradicts $\bar{\alpha } > 0$. Therefore, $\det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) > 0$.

Since $I(d_u) >0$ holds for all $d_u > 0$ if $\det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) \ge 0$, we consider the case that $\det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) <0$. In this case, $I(d_u) = 0$ has the positive root $d_1^*$ given by

$$\begin{aligned} d_1^* = -\displaystyle \frac{ \det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13})}{ \det B_{12} + \det B_{13} + (\mathrm{tr} B)(\mathrm{tr} B_{23}) } d_v > 0 . \end{aligned}$$

We will prove that $d_1^* \le d^*$ holds. Assume that $d_1^* > d^*$ holds. Then, noting (3.7) and $\mu (\bar{u}) + \nu (\bar{u}) = 1$, we have

$$\begin{aligned} \displaystyle \frac{ XY + Z - \bar{\alpha } X - \bar{\alpha } \bar{\nu } k \bar{f} + ( X+Y - \bar{\alpha } )( Y - \bar{\alpha })}{XY + Z - \bar{\alpha } Y + ( X+Y - \bar{\alpha } )( X - \bar{\alpha }) } < \displaystyle \frac{Y}{X + \bar{\nu } k \bar{f} } \ \end{aligned}$$

which implies that

$$\begin{aligned} W ( X - Y ) - \bar{\alpha } \bar{\nu } k \bar{f} X + \{ W + Y(X+Y) - \bar{\alpha } \bar{\nu } k \bar{f} \} \bar{\nu } k \bar{f} > 0, \end{aligned}$$

where $W := \bar{\alpha }^2 - 2\bar{\alpha } (X+Y) + XY + Z$. On the other hand, by $\det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) < 0$, we have $W + Y(X+Y) - \bar{\alpha } \bar{\nu } k \bar{f} < 0$. Therefore, we have $W(X-Y) - \bar{\alpha } \bar{\nu } k \bar{f} X > 0$. Since $X- Y < X < 0$ by $Y > \bar{\alpha } > 0$, we have

$$\begin{aligned} W < \displaystyle \frac{ \bar{\alpha } \bar{\nu } k \bar{f} X }{ X- Y } < \bar{\alpha } \bar{\nu } k \bar{f} , \end{aligned}$$

which implies

$$\begin{aligned} \begin{array}{rcl} \bar{\alpha } \bar{\nu } k \bar{f} &{} > &{} XY + Z - 2 \bar{\alpha } ( X + Y ) + \bar{\alpha }^2\\ &{} = &{} XY + Z - \bar{\alpha } ( X + Y ) - \bar{\alpha }( X + Y ) + \bar{\alpha }^2\\ &{} > &{} \bar{\alpha } \bar{\nu } k \bar{f} - \bar{\alpha } ( X + Y ) + \bar{\alpha }^2 \end{array} \end{aligned}$$

by virtue of (8.9). Hence, we have $\bar{\alpha } < X + Y < 0$ by (8.10), which contradicts $\bar{\alpha } >0$. Thus, we see that $d_1^* \le d^*$ holds, and hence $I(d_u) > 0$ holds for $d_u \ge d^*$.

Next, we will show that $J(d_u) > 0$ holds for $d_u \ge d^*$. Since $\mathrm{tr} B_{23} = X - \bar{\alpha } < 0$ and $\mathrm{tr} B_{13} = Y - \bar{\alpha } > 0$, we see that $J(d_u) > 0$ holds for $d_u > d_2^*$, where $d_2^*$ is the positive root of $J(d_u) = 0$. A direct calculation shows

$$\begin{aligned} d_2^* = \displaystyle \frac{ -( X + Y - \bar{\alpha } ) + \sqrt{ ( X - \bar{\alpha } + Y )^2 - ( X-\bar{\alpha } )(Y - \bar{\alpha }) } }{ X -\bar{\alpha } } d_v. \end{aligned}$$

Noting $X - \bar{\alpha } < 0,\, Y >0$ and $X + Y - \bar{\alpha } = \mathrm{tr} B = -p_1 < 0$ by (8.1), we have

$$\begin{aligned} d_2^* = \displaystyle \frac{ Y + \{ X - \bar{\alpha } - \sqrt{ ( X - \bar{\alpha } + Y )^2 - ( X-\bar{\alpha } )(Y - \bar{\alpha }) } \} }{ - ( X -\bar{\alpha } ) } d_v < \displaystyle \frac{ Y }{ -( X - \bar{\alpha } ) } d_v \end{aligned}$$

because

$$\begin{aligned} \begin{array}{l} ( X - \bar{\alpha } + Y )^2 - ( X-\bar{\alpha } )(Y - \bar{\alpha })\\ \ \ \ = ( X-\bar{\alpha } )^2 + ( X + Y - \bar{\alpha } )Y + \bar{\alpha } ( X-\bar{\alpha } ) < ( X-\bar{\alpha } )^2. \end{array} \end{aligned}$$

Since $X -\bar{\alpha } < X < - m'(\bar{v})\bar{v} < 0$ by $0 < \mu (\bar{u}) < 1$, we have $d_2^* < d^*$, where $d^*$ is given by (3.7). Thus, we see that $J(d_u) > 0$ holds for $d_u \ge d^*$.

The second assertion of this lemma follows from $\mathrm{tr} B_{13}= Y - \bar{\alpha } > 0$ and

$$\begin{aligned} F(\sigma ) = p_1p_2 - p_3 + \left( \det B_{12} + \det B_{23} + (\mathrm{tr} B)(\mathrm{tr} B_{13}) \right) d_v \sigma - (\mathrm{tr} B_{13}) d_v^2 \sigma ^2 \end{aligned}$$

for $d_u = 0$. $\square$

Remark 6

It should be noted that one cannot skip the proof of the first assertion of Lemma 8.2. For 3-component reaction–diffusion systems, some necessary and sufficient conditions for Turing instability were given by (Cross 1978; Satnoianu et al. 2000; Wang and Li 2001). However, they cannot distinguish the type of Turing instability. On the other hand, (White and Gilligan 1998) gave some necessary conditions for oscillatory Turing instability. To the best of our knowledge, convenient necessary and sufficient conditions for oscillatory Turing instability seem to be unknown.