Spatiotemporal dynamics of a reaction-diffusion model of pollen tube tip growth

Abstract

A reaction-diffusion model is proposed to describe the mechanisms underlying the spatial distributions of ROP1 and calcium on the pollen tube tip. The model assumes that the plasma membrane ROP1 activates itself through positive feedback loop, while the cytosolic calcium ions inhibit ROP1 via a negative feedback loop. Furthermore it is proposed that lateral movement of molecules on the plasma membrane are depicted by diffusion. It is shown that bistable or oscillatory dynamics could exist even in the non-spatial model, and stationary and oscillatory spatiotemporal patterns are found in the full spatial model which resemble the experimental data of pollen tube tip growth.

Appendix

Proof of Proposition 3.1

In this proposition, we study the number of roots of equation (3.4) with $1<\alpha <2$. The function f(R) has the properties that

$$\begin{aligned} \lim _{R \rightarrow 0^+} f(R) = -\infty \quad \text {and} \quad \lim _{R \rightarrow \infty } f(R) = \infty . \end{aligned}$$

(6.1)

Also, we have the first derivative of f(R) as

$$\begin{aligned} f'(R)=\alpha R^{\alpha -1}-(\alpha -1)R^{\alpha -2}+k_3^2(\alpha -2)R^{\alpha -3}-k_3^2(\alpha -3)R^{\alpha -4}. \end{aligned}$$

(6.2)

Let

$$\begin{aligned} h(R)=\alpha R^3-(\alpha -1)R^2+k_3^2(\alpha -2)R-k_3^2(\alpha -3). \end{aligned}$$

(6.3)

So we have $f'(R)=R^{\alpha -4}h(R)$.

Step 1. There exists a unique $x_2>0$ such that $h'(R)<0$ for $0<R<x_2$, $h'(R)>0$ for $R>x_2$, and h(R) reaches the global minimum in $(0,\infty )$ at $R=x_2$.

Note that the function h(R) has the properties that

$$\begin{aligned} h(0)=-k_3^2(\alpha -3)>0 \quad \text {and} \quad \lim _{R \rightarrow \infty } h(R) = \infty , \end{aligned}$$

(6.4)

and we have the first derivative of h(R) as

$$\begin{aligned} h'(R)=3\alpha R^2-2(\alpha -1)R+k_3^2(\alpha -2). \end{aligned}$$

(6.5)

Since $\alpha \in (1,2)$, for (6.5), we have the discriminant $\bigtriangleup _1=4(\alpha -1)^2-12k_3^2\alpha (\alpha -2)>0$. In this case, $h'(R)=0$ must have two roots in $(-\infty ,\infty )$. Notice that $\dfrac{2(\alpha -1)}{3\alpha }>0$ and $\dfrac{k_3^2(\alpha -2)}{3\alpha }<0$ so $h'(R)$ must have one negative root and one positive root. Let $x_2$ be the positive root of $h'(R)=0$. Since $h'(0)=k_3^2(\alpha -2)<0$, we have $h'(R)<0$ for $0<R<x_2$, $h'(R)>0$ for $R>x_2$. Therefore, h(R) decreases for $0<R<x_2$, and increases for $R>x_2$. That is to say, if $h(x_2)\ge 0$, then $h(R)\ge 0$ for any $R>0$, while $h(R)=0$ has two positive solutions if $h(x_2)<0$.

Step 2. There exist $k_{31},k_{32}>0$ such that

$$\begin{aligned} h(x_2) \left\{ \begin{array}{cc} \ge 0, \quad &{} \text {if } k_{31}<k_3<k_{32}. \quad \quad \quad \quad \quad \\<0, \quad &{} \text {if } 0<k_3<k_{31} \text { or } k_3>k_{32}. \end{array} \right. \end{aligned}$$

(6.6)

Since $h'(R)$ is an quadratic function and the relationship between h(R) and $h'(R)$, we can have following two facts:

$$\begin{aligned} x_2&=\dfrac{(\alpha -1)+\sqrt{(\alpha -1)^2-3k_3^2\alpha (\alpha -2)}}{3\alpha }, \end{aligned}$$

(6.7)

$$\begin{aligned} h(x_2)=&\left( \dfrac{x_2}{3}-\dfrac{\alpha -1}{9\alpha }\right) h'(x_2)+\dfrac{6k_3^2\alpha (\alpha -2)-2(\alpha -1)^2}{9\alpha }x_2 \nonumber \\&+\dfrac{k_3^2(\alpha -1)(\alpha -2)-9k_3^2\alpha (\alpha -3)}{9\alpha }. \end{aligned}$$

(6.8)

Therefore, we have

$$\begin{aligned}&h(x_2)\ge 0 \end{aligned}$$

(6.9)

$$\begin{aligned} \Leftrightarrow \quad&x_2\le \dfrac{(\alpha -1)(\alpha -2)-9\alpha (\alpha -3)}{2(\alpha -1)^2-6k_3^2\alpha (\alpha -2)}k_3^2 \end{aligned}$$

(6.10)

$$\begin{aligned} \Leftrightarrow \quad&\dfrac{(\alpha -1)+\sqrt{(\alpha -1)^2-3k_3^2\alpha (\alpha -2)}}{3\alpha }\le \dfrac{(\alpha -1)(\alpha -2)-9\alpha (\alpha -3)}{2(\alpha -1)^2-6k_3^2\alpha (\alpha -2)}k_3^2 \end{aligned}$$

(6.11)

$$\begin{aligned} \Leftrightarrow \quad&Ak_3^4+Bk_3^2+C\ge 0, \end{aligned}$$

(6.12)

where

$$\begin{aligned} A=&\alpha (\alpha -2)^3<0, \end{aligned}$$

(6.13)

$$\begin{aligned} B=&2(\alpha -1)^2(\alpha -2)^2-18(\alpha -1)(\alpha -2)+27>0, \end{aligned}$$

(6.14)

$$\begin{aligned} C=&(\alpha -1)^3(\alpha -3)<0 . \end{aligned}$$

(6.15)

Notice that the discriminant of the quadratic function $Ay^2+By+C$ is

$$\begin{aligned} \bigtriangleup _2&=[2(\alpha -1)^2(\alpha -2)^2-18(\alpha -1)(\alpha -2){+}27]^2-4\alpha (\alpha -1)^3(\alpha -2)^3(\alpha -3) \end{aligned}$$

(6.16)

$$\begin{aligned}&=[2(\alpha -1)^2(\alpha -2)^2-18(\alpha -1)(\alpha -2)+27]^2 \nonumber \\&\qquad -4(\alpha -1)^4(\alpha -2)^4 +8(\alpha -1)^3(\alpha -2)^3 \end{aligned}$$

(6.17)

$$\begin{aligned}&>-18(\alpha -1)(\alpha -2)[4(\alpha -1)^2(\alpha -2)^2 \nonumber \\&\qquad -18(\alpha -1)(\alpha -2)+27]+8(\alpha -1)^3(\alpha -2)^3 \end{aligned}$$

(6.18)

$$\begin{aligned}&=-2(\alpha -1)(\alpha -2)[32(\alpha -1)^2(\alpha -2)^2-162(\alpha -1)(\alpha -2)+243]. \end{aligned}$$

(6.19)

Since the discriminant of the quadratic function $32y^2-162y+243$ is $\bigtriangleup _3=162^2-4\times 32\times 243=-4860<0$, then $32(\alpha -1)^2(\alpha -2)^2-162(\alpha -1)(\alpha -2)+243>0$ for any $\alpha$. Therefore

$$\begin{aligned} \bigtriangleup _2>-2(\alpha -1)(\alpha -2)[32(\alpha -1)^2(\alpha -2)^2-162( \alpha -1)(\alpha -2)+243]>0. \end{aligned}$$

(6.20)

So the quadratic equation $Ay^2+By+C=0$ has two real-valued solutions $k_{31}^*<k_{32}^*$. Because $k_{31}^*+k_{32}^*=-\dfrac{B}{A}>0$ and $k_{31}^*k_{32}^*=\dfrac{C}{A}>0$, then $k_{31}^*$ and $k_{32}^*$ are both positive. Moreover we can have that $Ak_3^4+Bk_3^2+C\ge 0$ if and only if $k_{31}^*\le k_3^2\le k_{32}^*$. Now let $k_{31}=\sqrt{k_{31}^*}$ and $k_{32}=\sqrt{k_{32}^*}$, we reach the conclusion in (6.6).

Step 3. We consider the number of roots of equation $f(R)=0$ in (3.4) for each case in (6.6). In the case where $h(x_2)\ge 0$, we would have $h(R)\ge 0$ for any $R>0$ because h(R) decreases for $0<R<x_2$, and increases for $R>x_2$. That is to say, $f'(R)=R^{\alpha -4}h(R)>0$ for any $R>0$. So f(R) increases for all $R>0$. According to property (6.1), $f(R)=0$ has one unique positive root.

On the other hand, when $h(x_2)<0$, $h(R)=0$ has two positive solutions. Let $0<r_1<x_2<r_2$ be the solutions of $h(R)=0$. Then $h(R)>0$ if $R\in [0,r_1)\cup (r_2,+\infty )$ and $h(R)<0$ if $R\in (r_1,r_2)$. That is to say,

$$\begin{aligned} f'(R)=R^{\alpha -4}h(R) \left\{ \begin{array}{c@{\qquad }c}>0, &{} \text {if } R \in (0,r_1), \\ <0, &{} \text {if } R \in (r_1,r_2), \\ >0, &{} \quad \text {if } R \in (r_2,+\infty ) . \end{array} \right. \end{aligned}$$

(6.21)

Therefore, f(R) increases for $0<R<r_1$, decreases for $r_1<R<r_2$, and increases when $R>r_2$. Then we know that

• 1. If $f(r_1)f(r_2)>0$, then $f(R)=0$ has one unique positive solution.

• 2. If $f(r_1)f(r_2)=0$, then $f(R)=0$ has two positive solutions.

• 3. If $f(r_1)f(r_2)<0$, then $f(R)=0$ has three positive solutions.

Define

$$\begin{aligned} l(R)=R^{\alpha -3}(1-R)(R^2+k_3^2). \end{aligned}$$

(6.22)

Then $f(r_1)f(r_2)=[k_2-l(r_1)][k_2-l(r_2)]$. Since $r_1$ and $r_2$ are solutions of $h(R)=0$, $r_1$ and $r_2$ only depends on $\alpha$ and $k_3$. So there exists $k_{21}$, $k_{22}$ which only depends on $\alpha$, $k_3$ and are defined as

$$\begin{aligned} k_{21}=r_1^{\alpha -3}(1-r_1)(r_1^2+k_3^2), \;\; k_{22}=r_2^{\alpha -3}(1-r_2)(r_2^2+k_3^2), \end{aligned}$$

(6.23)

such that

• 1. If $k_2<k_{21}$ or $k_2>k_{22}$, then $f(R)=0$ has one unique positive solution.

• 2. If $k_2=k_{21}$ or $k_2=k_{22}$, then $f(R)=0$ has two positive solutions.

• 3. If $k_{21}<k_2<k_{22}$, then $f(R)=0$ has three positive solutions.

We claim that $0<k_{21}<k_{22}$ for $0<k_3<k_{31}$, while $k_{21}<k_{22}<0$ for $k_3>k_{32}$. This is equivalent to $r_1<r_2<1$ for $0<k_3<k_{31}$, while $1<r_1<r_2$ for $k_3<k_{32}$. Notice that $h(1)=1+k_3^2>0$, and that h(R) decreases for $0<R<x_2$ and increases for $R>x_2$. So we only need to prove that $h'(1)=\alpha +2+k_3^2(\alpha -2)<0$ for $0<k_3<k_{31}$, while $h'(1)>0$ for $k_3>k_{32}$. In fact, $k_{31}^2$ and $k_{32}^2$ are two positive roots of equation $Ay^2+By+C=0$, where A, B, C are defined as (6.13), (6.14) and (6.15) respectively. Notice that $A<0$ and

$$\begin{aligned} A\left( \dfrac{2+\alpha }{2-\alpha }\right) ^2+B\left( \dfrac{2+\alpha }{2-\alpha }\right) +C =32\left( \alpha -\dfrac{1}{4}\right) ^2+54\dfrac{\alpha }{2-\alpha }>0, \quad \text {for } 1<\alpha <2. \end{aligned}$$

(6.24)

So $k_{31}^2<\dfrac{2+\alpha }{2-\alpha }<k_{32}^2$. Therefore, we have

$$\begin{aligned} h'(1)&=\alpha +2+k_3^2(\alpha -2)>\alpha +2+k_{31}^2(\alpha -2)>0, \quad \text {if }0<k_3<k_{31}, \end{aligned}$$

(6.25)

$$\begin{aligned} h'(1)&=\alpha +2+k_3^2(\alpha -2)<\alpha +2+k_{32}^2(\alpha -2)<0, \quad \text {if }k_3>k_{32}. \end{aligned}$$

(6.26)

So we have proved that $0<k_{21}<k_{22}$ for $0<k_3<k_{31}$, while $k_{21}<k_{22}<0$ for $k_3>k_{32}$. Therefore we reach the conclusion about number of solution of $f(R)=0$. $\square$

Proof of Proposition 3.2

By the Center Manifold Theorem (Page 116 in Perko (2001)), we can compute the center manifold near the equilibrium (0, 0):

$$\begin{aligned} C=\vartheta (R)=\frac{1}{k_1(2-\alpha )}R^{2-\alpha }+o(R^{2-\alpha }). \end{aligned}$$

(6.27)

Then, by substituting (6.27) into the first equation of the kinetic system (3.1), we obtain the following scalar system which gives the flow of Eq. (3.1) on the center manifold:

$$\begin{aligned} R_t=k_{1}R^{\alpha }(1-R)-k_1k_{2}\dfrac{R\vartheta ^2(R)}{\vartheta ^2(R)+k_3^2}>0, \;\; \text {for}~0<R<\delta . \end{aligned}$$

(6.28)

Thus, we know that the flow on the center manifold is moving away from the origin and it is an unstable orbit.

Next we show that there is an invariant region near $R=0$, $C>0$ for Eq. (3.1). Define

$$\begin{aligned} O=\left\{ (R,C): 0\le R\le \left( \frac{k_2}{2(k_3^2+\delta ^2)}\right) ^{\frac{1}{\alpha -1}}C^{\frac{2}{\alpha -1}},~0\le C\le \delta \right\} . \end{aligned}$$

It is obvious that $R=0$ is invariant for (3.1). Then, if $C=\delta ,~0\le R\le \left( \dfrac{k_2}{2(k_3^2+\delta ^2)}\right) ^{\frac{1}{\alpha -1}}\delta ^{\frac{2}{\alpha -1}}$, since $1<\alpha <2$ then $\frac{2}{\alpha -1}>1$, so one can choose $\delta >0$ small enough so that $\left( \dfrac{k_2}{2(k_3^2+\delta ^2)}\right) ^{\frac{1}{\alpha -1}}\delta ^{\frac{2}{\alpha -1}}\le \delta$. By using $C=\delta$, we have $C^{\prime }<0$.

On the boundary $R=\left( \dfrac{k_2}{2(k_3^2+\delta ^2)}\right) ^{\frac{1}{\alpha -1}}C^{\frac{2}{\alpha -1}}$, we have

$$\begin{aligned} \begin{aligned} \frac{d}{d t}\left( \dfrac{R^{\frac{\alpha -1}{2}}}{C}\right)&=\dfrac{\frac{\alpha -1}{2}C R^{\frac{\alpha -3}{2}}R^{\prime }-R^{\frac{\alpha -1}{2}}C^{\prime }}{C^2}=\dfrac{R^{\frac{\alpha -3}{2}}}{C^2}\left( \frac{\alpha -1}{2}C R^{\prime }-R C^{\prime }\right) \\&=\dfrac{R^{\frac{\alpha -3}{2}}}{C^2}\left[ \frac{\alpha -1}{2} Ck_1\left( R^{\alpha }-R^{\alpha +1}-k_2\frac{R C^2}{C^2+k_3^2}\right) -R^2+C R\right] \\&\le \dfrac{R^{\frac{\alpha -3}{2}}}{C}\left[ \frac{\alpha -1}{2}k_1\left( R^{\alpha }-k_2\frac{R C^2}{C^2+k_3^2}\right) +R\right] \\&\le \dfrac{R^{\frac{\alpha -3}{2}}}{C}\left[ (\alpha -1)k_1R^{\alpha }-\frac{(\alpha -1)k_1k_2R C^2}{2(C^2+k_3^2)}\right] \\&\le \dfrac{R^{\frac{\alpha -3}{2}}}{C}\left[ (\alpha -1)k_1\left( R^{\alpha }-\frac{(\alpha -1)k_1k_2R C^2}{2(\delta ^2+k_3^2)}\right) \right] \\&=(\alpha -1)R^{\frac{\alpha -1}{2}}k_1 C^2\left[ \frac{R^{\alpha -1}}{C^2}-\frac{k_2}{2(\delta ^2+k_3^2)}\right] =0, \end{aligned} \end{aligned}$$

and the first inequality holds for R small enough: $R<\frac{\alpha -1}{2}k_1R^{\alpha }$. The above calculation implies that the dynamics of (3.1) is inward on $R=\left( \dfrac{k_2}{2(k_3^2+\delta ^2)}\right) ^{\frac{1}{\alpha -1}}C^{\frac{2}{\alpha -1}}$. This shows that O is an invariant region for Eq. (3.1), and any orbit in O converges to the origin. It is also clear that when $R=0,~C>0$ (the positive C-axis), we have $(R_t,C_t)=(0,-C)$. So we know that all the solutions starting from $R=0,~C>0$ will always stay on this curve and eventually converge to the origin. One can choose a maximum orbit $R=h_s(C)$ so that all orbits such that $0\le R\le h_s(C)$ converge to the origin. Then other trajectories exhibits saddle behavior near the origin. $\square$

Proof of Theorem 3.3

First, we look at the determinant of the Jacobian matrix $J(R_j,R_j)$: $\text {Det}(J(R_j,R_j))=k_1R_j(f_2'(R_j)-f_1'(R_j))$, where $f_1,f_2$ are defined in (3.6). From (3.6), we have

$$\begin{aligned}&f_2(R)-f_1(R)=\dfrac{k_2R^2}{R^2+k_3^2}-R^{\alpha -1}(1-R) \end{aligned}$$

(6.29)

$$\begin{aligned} \Leftrightarrow&f(R)=k_2-R^{\alpha -3}(1-R)(R^2+k_3^2)=\dfrac{R^2+k_3^2}{R^2}(f_2(R)-f_1(R)) \end{aligned}$$

(6.30)

$$\begin{aligned} \Leftrightarrow&f'(R)=-\dfrac{2k_3^2}{R^3}(f_2(R)-f_1(R))+\dfrac{R^2+k_3^2}{R^2}(f_2'(R)-f_1'(R)). \end{aligned}$$

(6.31)

From (6.30), we know that $f_2(R)-f_1(R)=0$ when $f(R)=0$. Since positive steady states $(R_j,R_j)$ ($j=1,2,3$) satisfy $f(R_j)=0$, we have $f_2(R_j)-f_1(R_j)=0$. Therefore, from (6.31), we know

$$\begin{aligned}&f'(R_j)>0 \Leftrightarrow f_2'(R_j)-f_1'(R_j)>0 \Leftrightarrow \text {Det}(J(R_j,R_j))>0; \\&f'(R_j)=0 \Leftrightarrow f_2'(R_j)-f_1'(R_j)=0 \Leftrightarrow \text {Det}(J(R_j,R_j))=0; \\&f'(R_j)<0 \Leftrightarrow f_2'(R_j)-f_1'(R_j)<0 \Leftrightarrow \text {Det}(J(R_j,R_j))<0. \end{aligned}$$

According to the proof of Proposition 3.1, we have the following result of $\text {Det}(J(R_j,R_j))$: there exists a constant $k_{31}>0$ such that

• 1. If $0<k_3<k_{31}$, then there exists $r_1$ and $r_2$, which are two positive solutions of $h(R)=0$, such that

  • (a) $f'(R)>0\Rightarrow \text {Det}(J(R,R))>0$ for $R \in (0,r_1) \cup (r_2,1)$;

  • (b) $f'(R)<0\Rightarrow \text {Det}(J(R,R))<0$ for $R \in (r_1,r_2)$.

• 2. If $k_{31}<k_3$, then for any $0<R<1$, we always have $f'(R)>0\Rightarrow \text {Det}(J(R,R))>0$.

Here we want to point out that from (6.23), we have $k_2(r_1)=k_{21}$ and $k_2(r_2)=k_{22}$.

Next we look at the trace of Jacobian matrix (3.5): $\text {Tr}(J(R_j,C_j))=k_1R_jf_1'(R_j)-1$. Define a new function

$$\begin{aligned} g(R)=Rf_1'(R)=R^{\alpha -1}[(\alpha -1)-\alpha R]. \end{aligned}$$

(6.32)

We observe that g(R) has the following properties:

$$\begin{aligned} g(0)=0, \quad g\left( \dfrac{\alpha -1}{\alpha }\right) =0, \quad \text {and} \quad \lim _{R \rightarrow \infty } g(R) = \infty . \end{aligned}$$

(6.33)

Also we have the first derivative of g(R) as

$$\begin{aligned} g'(R)=R^{\alpha -2}[(\alpha -1)^2-\alpha ^2R]. \end{aligned}$$

(6.34)

Hence the function g(R) increases for $0<R<\displaystyle \left( \dfrac{\alpha -1}{\alpha }\right) ^2$ and decreases for $R>\displaystyle \left( \dfrac{\alpha -1}{\alpha }\right) ^2$. So g(R) achieves its maximum at $R=\displaystyle \left( \dfrac{\alpha -1}{\alpha }\right) ^2$ with $g\left( \left( \dfrac{\alpha -1}{\alpha }\right) ^2\right) =\left( \dfrac{\alpha -1}{\alpha }\right) ^{2\alpha -1}$.

Therefore we conclude that

• 1. If $k_1<\left( \dfrac{\alpha }{\alpha -1}\right) ^{2\alpha -1}$, then

  $$\begin{aligned} \text {Tr}(J(R,R))=k_1g(R)-1<\left( \dfrac{\alpha }{\alpha -1}\right) ^{2\alpha -1}\left( \dfrac{\alpha -1}{\alpha }\right) ^{2\alpha -1}-1=0. \end{aligned}$$

  (6.35)

• 2. If $k_1>\left( \dfrac{\alpha }{\alpha -1}\right) ^{2\alpha -1}$, then there exists $0<{\tilde{R}}_1<{\tilde{R}}_2$, such that $g({\tilde{R}}_1)=g({\tilde{R}}_2)=\dfrac{1}{k_1}$. Therefore,

  • (a) If ${\tilde{R}}_1<R<{\tilde{R}}_2$, then $\text {Tr}(J(R,R))=k_1g(R)-1>k_1g({\tilde{R}}_1)=0.$

  • (b) If $0<R<{\tilde{R}}_1$ or ${\tilde{R}}_2<R<1$, then $\text {Tr}(J(R,R))=k_1g(R)-1<k_1g({\tilde{R}}_1)=0.$

$\square$

Proof of Proposition 3.4

From the proof of Theorem 3.3, we can easily get Part 1 in Proposition 3.4. So here we only discuss Part 2: the case that $0<k_3<k_{31}$. Since $\text {Det}(J(R_2,C_2))<0$, the steady state $(R_2,C_2)$ is always a saddle point. So we focus on the positive steady states $(R_1,C_1)$ and $(R_3,C_3)$. To prove the results in Proposition 3.4, we need to determine the order of the possible bifurcation points: $r_1$, $r_2$, ${\tilde{R}}_1$ and ${\tilde{R}}_2$, where $r_1$ and $r_2$ are the steady state bifurcation points satisfying $h(r_1)=h(r_2)=0$ with h(R) defined in (6.3), and ${\tilde{R}}_1$, ${\tilde{R}}_2$ are possible Hopf bifurcation points satisfying $g({\tilde{R}}_1)=g({\tilde{R}}_2)=1/k_{1}$. Then, by the results of Theorem 3.3, we can obtain the stability of each steady state.

First, we prove that $g(r_1)>g(r_2)$ always holds. From the definition of h(R), we know that

$$\begin{aligned} h(r_1)=\alpha r_1^3-(\alpha -1)r_1^2+k_3^2(\alpha -2)r_1-k_3^2(a\alpha -3)=0. \end{aligned}$$

(6.36)

Multiplying (6.36) by $r_1^{\alpha -3}$, we have

$$\begin{aligned} \alpha r_1^{\alpha }-(\alpha -1)r_1^{\alpha -1}+k_3^2(\alpha -2)r_1^{\alpha -2}-k_3^2(a\alpha -3)r_1^{\alpha -3}=0, \end{aligned}$$

(6.37)

which together with $g(r_1)=-\alpha r_1^{\alpha }+(\alpha -1)r_1^{\alpha -1}$ from (6.32) implies that

$$\begin{aligned} g(r_1)=k_3^2(\alpha -2)r_1^{\alpha -2}-k_3^2(\alpha -3)r_1^{\alpha -3}. \end{aligned}$$

(6.38)

Define

$$\begin{aligned} G(R)=k_3^2(\alpha -2)R^{\alpha -2}-k_3^2(\alpha -3)R^{\alpha -3},~R\in (0,1),~\alpha \in (1,2). \end{aligned}$$

(6.39)

By direct calculation, we have $G^{\prime }(R)=k_3^2(\alpha -2)^{2}R^{\alpha -3}-k_3^2(\alpha -3)^{2}R^{\alpha -4}$ and $G^{\prime }(R)<0$ for $R\in \left( 0, \left( \frac{\alpha -3}{\alpha -2}\right) ^2\right) \supset (0,1)$. Therefore, G(R) is strictly decreasing for $R\in (0,1)$. By the fact that $0<r_1<r_2<1$, immediately we reach the conclusion that $g(r_1)>g(r_2)$.

Now we consider the case that $0<k_3<k_{31}$ which implies the existence of multiple steady states. For the convenience of discussion, we define

$$\begin{aligned} {\tilde{g}}(R)=g(R)-1/k_1, \end{aligned}$$

(6.40)

then we know that ${\tilde{g}}$ has two zeros ${\tilde{R}}_1$ and ${\tilde{R}}_2$. For the order of $r_1$, $r_2$, ${\tilde{R}}_1$ and ${\tilde{R}}_2$, we have the following six possible situations:

• (i) $r_1<r_2<{\tilde{R}}_1<{\tilde{R}}_2$. We show that this case will not happen. By the property of h(R), it is not difficult to verify that $h\left( \left( \frac{\alpha -1}{\alpha }\right) ^{2}\right) >0=h(r_2)$, so we know that $\left( \frac{\alpha -1}{\alpha }\right) ^{2}<r_2$. Because $\left( \frac{\alpha -1}{\alpha }\right) ^{2}$ is the maximum point of ${\tilde{g}}(R)$ and ${\tilde{R}}_1$ is the smallest root of ${\tilde{g}}(R)$, so we have ${\tilde{R}}_1<r_2$ which is a contradiction to the assumption.

• (ii) $r_1<{\tilde{R}}_1<r_2<{\tilde{R}}_2$. By the fact that ${\tilde{g}}(R)>0$ for $R\in ({\tilde{R}}_1,{\tilde{R}}_1)$ and $g(R)<0$ for $R\in (0,{\tilde{R}}_1)\cup ({\tilde{R}}_2,1)$, it is easy to obtain that ${\tilde{g}}(r_1)<0$ since $r_1<{\tilde{R}}_1$, which is equivalent to $k_1<1/g(r_1)$. Also, by ${\tilde{g}}(r_2)>0$, we have $k_1>1/g(r_2)$. However, it has been proved that $g(r_1)>g(r_2)$, so the set $(1/g(r_2),1/g(r_1))$ is empty, which means that this case cannot happen.

• (iii) ${\tilde{R}}_1<r_1<r_2<{\tilde{R}}_2$ (see Fig. 7c). Because that $r_1,~r_2\in ({\tilde{R}}_1,{\tilde{R}}_2)$, so we have ${\tilde{g}}(r_1)>0$ and ${\tilde{g}}(r_2)>0$, which is equivalent to $k_1>1/g(r_2)$. In this case, by Theorem 3.3, we know that Hopf bifurcations occur at both of $(R_1,R_1)$ and $(R_3,R_3)$.

• (iv) ${\tilde{R}}_1<r_1<{\tilde{R}}_2<r_2$ (see Fig. 7d). By similar argument, since $r_1\in ({\tilde{R}}_1,{\tilde{R}}_2)$ and $r_2>{\tilde{R}}_2$, we can obtain that ${\tilde{g}}(r_1)>0$ and ${\tilde{g}}(r_2)<0$ which imply that $k_1\in (1/g(r_1),1/g(r_2))$. In this case, from ${\tilde{R}}_1<r_1<{\tilde{R}}_2<r_2$ and Theorem 3.3, a Hopf bifurcation only occurs at $(R_3,R_3)$ and does not occur at $(R_1,R_1)$.

• (v) $r_1<{\tilde{R}}_1<{\tilde{R}}_2<r_2$ (see Fig. 7e). Similarly, we have ${\tilde{g}}(r_1)<0$ and ${\tilde{g}}(r_2)<0$, then it can be inferred that $k_{11}<k_1<1/g(r_1)$. In this case, no Hopf bifurcation can occur. Also, we have $\left( \frac{\alpha -1}{\alpha }\right) ^{2}>r_1$ in this case, which will be used later.

• (vi) ${\tilde{R}}_1<{\tilde{R}}_2<r_1<r_2$ (see Fig. 7f). In this case, we still have $k_{11}<k_1<1/g(r_1)$, but the difference with case (v) is that $\left( \frac{\alpha -1}{\alpha }\right) ^{2}<r_1$. In this case, two Hopf bifurcations occur at $(R_1,R_1)$.

So in order to distinguish the last two cases, we define ${\tilde{k}}_3$ to be the value of $k_3$ such that $r_1=\left( \frac{\alpha -1}{\alpha }\right) ^{2}$, and it is easy to calculate that ${\tilde{k}}_3$ is given by (3.16). So case (v) is for $0<k_3<{\tilde{k}}_3$ which is equivalent to $r_1<\left( \frac{\alpha -1}{\alpha }\right) ^{2}$ and case (vi) is for ${\tilde{k}}_3<k_3<k_{31}$ which implies $r_1>\left( \frac{\alpha -1}{\alpha }\right) ^{2}$. Also we must have that ${\tilde{k}}_3<k_{31}$. Suppose not, then first we assume that $k_3>k_{32}$, then $h(R)=0$ has two positive solutions, but both of them should be bigger than 1 and here we have $0<R<1$ which is a contradiction. If $k_{31}<k_3<k_{32}$, then $h(R)=0$ has no roots, so it contradicts with the fact that $h(R)=0$ has one of positive roots at $(\dfrac{\alpha -1}{\alpha })^2<1$ when $k_3={\tilde{k}}_3$. Therefore, we can conclude that ${\tilde{k}}_3<k_{31}$. Finally if $0<k_1<k_{11}$, then ${\tilde{g}}(R)$ has no zeros, $(R_1,R_1)$ and $(R_3,R_3)$ are both always linearly stable and Hopf bifurcation will not occur, which is similar to (v) above.

In summary the case (c) is implied by (iii) above, case (d) is implied by (iv) above, case (e) is implied by (v) and the case of $0<k_1<k_{11}$, and case (f) is implied by (vi) above. The proof is completed. $\square$