Nonlinear Dynamics of the Movement of the Venus Flytrap

Abstract

The Venus flytrap has long been regarded as one of the most amazing examples of movement in the plant kingdom. The trapping ability of the flytrap consists of three unique features. First, trap closure represents one of the fastest movements in the plant kingdom. Second, a decision-making stage allows the plant to “decide” whether to completely close or open the trap, based on stimuli provided from the trapped object. Finally, the Venus flytrap contains a “memory function” that requires two mechanical stimuli within about 30 seconds to initiate trap closure. The movement involved in trap closure consists of nonlinear dynamics that have not been well understood. By understanding the movement, through nonlinear dynamics analysis, it will be possible to better understand this biological process. A mathematical model describing the movement of the Venus flytrap was first proposed by the authors in Yang et al., Plant Signal. Behav. 5(8), 968–978 (2010). In the current work, the earlier research has been advanced and an in-depth nonlinear and control analysis of the dynamic process has been provided.

Appendices

Appendix A: Proof of Main Theorems

1.1 A.1 Proof of Theorem 3.2

Because of the existence of stable invariant manifold W _s by Theorem 3.1, system (4) can be restricted onto W _s and rewritten as

$$ \dot{X}_I=f_0(X_I)= \displaystyle \frac{\alpha X_I^p}{(c-X_I)^p+X_I^p}-\mu X_I. $$

(13)

Then $f_{0}^{\prime }(0)=f_{0}^{\prime }(c)=-\mu<0$ and $f_{0}^{\prime }(\frac{c}{2})=\mu(p-1)>0$ implies that $S_{1}^{0}$ and $S_{3}^{0}$ are stable and $S_{2}^{0}$ unstable. Set Z=X _I −X _o and then

$$ \dot{Z}=(A_0-\mu)Z,\quad A_0= \frac{\displaystyle \alpha (X_I^p-X_o^p)}{(X_I-X_o)(X_o^p+X_I^p)}. $$

(14)

Note that A ₀(X _o ,X _I ) is continuous in X _o and X _I as p≥1. Particularly when p>1 is an integer, we have ${A}_{0}=\alpha \sum_{i=0}^{p-1}X_{o}^{i}X_{I}^{p-i-1}/(X_{o}^{p}+X_{I}^{p})$. Because Z=0 is an equilibrium of Eq. (14) and as $X_{o}=X_{I}<\frac{pc}{2}$,

$$A_0(X_I, X_I)=\displaystyle \lim_{X_o\rightarrow {X}_I}A_0(X_o, X_I)= \frac{\alpha p}{2X_I}>\mu, $$

one can see that W _u is an invariant manifold and $W_{u}^{0} \subset W_{u}$ is an unstable submanifold of (4). And W _u plays the role of separatrix for bistability.

1.2 A.2 Proof of Theorem 3.4

It follows from (6) that

$$\dot{Z}= A_1Z,\quad A_1=A_0-2k_tH(X_I-X_o)- \mu $$

and

$$\begin{array}{rcl} A_1\bigl(X_I^-\bigr)&=&\displaystyle \lim_{X_o\rightarrow {X}_I^-} A_1=\frac{\alpha p}{2X_o}-2k_t-\mu ,\\ A_1\bigl(X_I^+\bigr)&=&\displaystyle \lim_{X_o\rightarrow {X}_I^+} A_1=\frac{\alpha p}{2X_o}-\mu . \end{array} $$

Obviously $A_{1}(X_{I}^{-})<A_{1}(X_{I}^{+})$, $A_{1}(X_{I}^{+})>0$ as $X_{I}<\frac{pc}{2}$ and $A_{1}(X_{I}^{-})<0$ as $X_{I}>\frac{\alpha p}{2(2k_{t}+\mu)}$. Thus, $W_{u}^{1}$ is locally stable in W _o (where X _I >X _o ) and unstable in W _c (where X _I <X _o ). Moreover, $S_{2}^{0}\in W_{u}^{1}$ if and only if

$$\frac{\alpha p}{2(2k_t+\mu)}<\frac{c}{2}\iff p<1+\frac{2k_t}{\mu}. $$

Therefore, $S_{2}^{0}=W_{s}\cap W_{u}^{1}$ is stable in W _o and unstable in W _c .

When system (6) is restricted onto W _s , we have

$$ \dot{X}_I=f_1(X_I)=f_0(X_I)-k_t(2X_I-c)H(2X_I-c), $$

(15)

for which obviously X _I =c is not a solution, and hence $S_{1}^{0}$ is not a steady state of (6). Additionally when $X_{I}\in(\frac{c}{2}, c)$, set $\beta =\frac{k_{t}}{\mu}$ and $x=\frac{X_{o}}{X_{I}}\in[0, 1]$, then

$$\begin{array}{l} \displaystyle \frac{\alpha X_I^p}{X_o^p+X_I^p}-k_t(X_I-X_o)-\mu X_I=0\\[8pt] \quad{}\iff\quad cX_I^p-X_I\bigl(X_o^p+X_I^p\bigr)=\beta (X_I-X_o)\bigl(X_o^p+X_I^p\bigr) \\[4pt] \quad\iff\quad (c-X_I)X_I^p-X_IX_o^p=\beta (X_I-X_o)\bigl(X_o^p+X_I^p\bigr) \\[4pt] \quad\iff\quad X_oX_I^p-X_IX_o^p=\beta (X_I-X_o)\bigl(X_o^p+X_I^p\bigr) \\[4pt] \quad\iff\quad x-x^p=\beta (1-x)\bigl(1+x^p\bigr) \\[4pt] \quad\iff\quad \displaystyle \beta =g(x)=\frac{x-x^p}{(1-x)(1+x^p)}. \end{array} $$

Because g(x) is an increasing function as x∈[0,1] (see Appendix B),

$$\displaystyle \beta _M=\max_{x\in[0,1]}g(x)=g(1)= \frac{1}{2}(p-1). $$

Therefore, if β>β _M or equivalently $p<1+\frac{2k_{t}}{\mu}$, then Eq. (15) has no steady state to satisfy X _I >X _o . When X _I <X _o , system (15) is the same as (13), thus system (6) has only two steady states $S_{1}^{1}=S_{2}^{0}$ and $S_{2}^{1}=S_{3}^{0}$.

1.3 A.3 Proof of Theorem 3.5

The proof is similar to that of Theorem 3.4. By the fact that $p>1+\frac{2k_{t}}{\mu}$ is equivalent to β<β _M , (15) has only one steady state at which X _I >X _o , denoted by $S_{3}^{1}$. And its stability follows directly from the positive invariance of W _s ∩W _o that contains a unique steady state $S_{3}^{1}$.

Note that for X _I >X _o ,

$$\begin{array}{c} f_1^\prime (X_I)=\displaystyle \frac{\alpha pcX_I^{p-1}(c-X_I)^{p-1}}{[X_I^p+(c-X_I)^p]^2}-(2k_t+\mu) =\displaystyle \frac{\alpha pcX_o^{p-1}X_I^{p-1}}{(X_I^p+X_o^p)^2}-(2k_t+\mu). \end{array} $$

It follows from $f_{1}^{\prime }(\frac{c}{2}-)>f_{1}^{\prime }(\frac{c}{2}+)>0$ that $S_{1}^{1}$ is unstable in both W _o and W _c . The stability of $S_{2}^{1}$ is trivial.

1.4 A.4 Proof of Theorem 3.7

By restricting on the stable manifold W _s , system (8) is reduced to

$$ \dot{X}_I=f_2(X_I)= \displaystyle {f}_1(X_I)-k_fX_I. $$

(16)

When (X _o ,X _I )∈W _c , that is X _I <X _o or $X_{I}<\frac{c}{2}$,

$$f_2(X_I)=\displaystyle \frac{\alpha X_I^p}{X_I^p+ (c-X_I )^p}-(k_f+ \mu)X_I. $$

Obviously X _I =0 is a root of f ₂(X _I )=0. Set $\gamma =\frac{k_{f}}{\mu}$ and for $X_{I}\in(0, \frac{c}{2})$,

$$f_2(X_I)=0\Rightarrow \displaystyle \frac{cX_I^p}{X_I^p+ (c-X_I )^p}= \frac{c}{1+ (\frac {c}{X_I}-1 )^p}=(1+\gamma )X_I. $$

Let $y=\frac{c}{X_{I}}>2$ and $a=\frac{1}{1+\gamma }<1$, we have that

$$h(y)=1+(y-1)^p-ay=0. $$

Note that for y>2,

$$\begin{array}{l} h^\prime (y)=p(y-1)^{p-1}-a>p-a>0 \\[4pt] \quad \Rightarrow \quad h(y)>h(2)=2-2a>0, \end{array} $$

which implies that h(y)=0 has no root for y>2 and consequently f ₂(X _I )=0 has no root other than X _I =0, so that X _I <X _o . Therefore, system (8) has a unique steady state (c,0) in W _c . It is easy to see that (c,0) is stable.

1.5 A.5 Proof of Theorem 3.7

Similar to the proof of Theorems 3.4 and 3.5, if X _I >X _o , i.e., $X_{I}\in(\frac{c}{2}, c)$, let $\beta =\frac{k_{t}}{\mu}$, $\gamma =\frac{k_{f}}{\mu}$ and $x=\frac{X_{o}}{X_{I}}<1$, then we have

$$\begin{array}{l} \displaystyle \frac{\alpha X_I^p}{X_o^p+X_I^p}-k_t(X_I-X_o)-(k_f+\mu) X_I=0\\[8pt] \quad\iff\quad cX_I^p-(1+\gamma )X_I \bigl(X_o^p+X_I^p\bigr)=\beta (X_I-X_o)\bigl(X_o^p+X_I^p\bigr) \\[4pt] \quad\iff\quad X_oX_I^p-X_IX_o^p-\gamma X_I\bigl(X_o^p+X_I^p\bigr)=\beta (X_I-X_o)\bigl(X_o^p+X_I^p\bigr) \\[4pt] \quad\iff\quad x-x^p-\gamma \bigl(1+x^p\bigr)=\beta (1-x)\bigl(1+x^p\bigr), \\[4pt] \quad\iff\quad \displaystyle \frac{x-x^p}{1+x^p}-\gamma =\beta (1-x), \\[4pt] \quad\iff\quad h_0(x)=\displaystyle \frac{1+x}{1+x^p}+\beta {x}=1+\gamma +\beta ,\quad x\in[0, 1]. \\ \end{array} $$

Note that

$$h_0^\prime (x)=\beta +\frac{h_1(x)}{(1+x^p)^2},\quad h_1(x)=1+x^p-px^{p-1}(1+x) $$

and

$$\begin{array}{rcl} h_1^\prime (x)&=&px^{p-1}-p\bigl[(p-1)x^{p-2}+px^{p-1}\bigr]\\[4pt] &=&p\bigl[-(p-1)x^{p-2}+(1-p)x^{p-1}\bigr]<0.\\ \end{array} $$

Obviously, the monotonicity of h ₁(x) implies the monotonicity of $\frac{h_{1}(x)}{(1+x^{p})^{2}}$ and

$$\frac{1-p}{2}\leq\frac{h_1(x)}{(1+x^p)^2}\leq1. $$

If p−1≤2β, i.e., $p\leq1+\frac{2k_{t}}{\mu}$, then $h_{0}^{\prime }(x)\geq0$ for all x∈[0,1] and hence

$$1=h_0(0)\leq h_0(x)\leq h_0(1)=1+\beta <1+\gamma +\beta , $$

which implies that h ₀(x)=1+γ+β has no root in [0,1], i.e., f ₂(X _o ) has no root such that $X_{o}>\frac{c}{2}$.

If p−1>2β, i.e., $p>1+\frac{2k_{t}}{\mu}$, then there exists a unique $\bar{x}\in(0, 1)$ such that $h_{0}^{\prime }(\bar{x})=0$. If $\gamma \geq\frac{p-1-2\beta }{p+1}$, then

$$1+x+\beta {x}>\frac{1+x}{1+x^p}+\beta {x}=1+\gamma +\beta \Rightarrow x>\frac{\gamma +\beta }{1+\beta }. $$

On the other hand, an equivalent form of h ₀(x)=0 yields

$$\begin{array}{l} \displaystyle \frac{x-x^p}{1+x^p}=\gamma +\beta (1-x) \\[4pt] \quad \Rightarrow \quad \displaystyle \beta +\frac{\gamma }{1-x}=\frac{x-x^p}{(1-x)(1+x^p)}=g(x)<\frac{p-1}{2} \end{array} $$

which implies that $x<1-\frac{2\gamma }{p-1-2\beta }$. This is contradictory to the fact that $x>\frac{\gamma +\beta }{1+\beta }$ under the assumption $\gamma \geq\frac{p-1-2\beta }{p+1}$. Therefore, as $\gamma \geq\frac{p-1-2\beta }{p+1}$ and p>1+2β, h ₀(x)=1+γ+β has no root in [0,1]. This completes the proof.

1.6 A.6 Proof of Theorem 3.8

By Theorems 3.6 and 3.7, the closed state (c,0) is the unique steady state of system (8) and it is stable, thus it is a global attractor of system (8). Then there exists $\bar{T}>0$ such that $X_{o}(\bar{T})=X_{I}(\bar{T})=\frac {c}{2}$, when X _I (0)=c.

Note that both u _h (t) and u _a (t) are present as $t\leq T_{0}^{2}$ in system (7). If $T_{0}^{2}>\bar{T}$, then $X(T_{0}^{2})\in W_{c}$. As $t>T_{0}^{2}$, there is no stimulus present in system (7) and there are three steady states for the system without stimulus. Because $X(T_{0}^{2})$ is lying in the stability region W _c of closed state $S_{3}^{0}=(c, 0)$, eventually X will approach the closed state $S_{3}^{0}$.

In contrast, if $T_{0}^{2}<\bar{T}$, then $X(T_{0}^{2})$ remains in W _o that is the stability region of open state $S_{1}^{0}=(0, c)$. Hence, eventually X moves back to open state again.

1.7 A.7 Proof of Theorem 3.9

Obviously there exists $X_{I}^{*}<\frac{c}{2}$ such that $f_{0}^{\prime }(X_{I}^{*})=0$ and

$$\min_{x\in[0, c]}\bigl\{f_0(x)\bigr\}=f_0 \bigl(X_I^*\bigr)<0. $$

Let $\bar{k}_{d}=-f_{0}(X_{I}^{*})>0$. If $k_{d}>\bar{k}_{d}$, then f ₃(X _I )>0 for all X _I ∈[0,c]. Thus, system (11) has no steady state and X _I will increase in [0,c]. If $k_{d}\leq\bar{k}_{d}$, then f ₃(X _I )=0 has at least one root in $[0, \frac {c}{2}]$ and thus (11) has at least one steady state in W _c . Hence, X _I can never go beyond W _c if it starts from the closed state X _I =0.

Moreover, when $k_{d}>\bar{k}_{d}$, there may exist $\tilde {T}_{2}>\tilde {T}_{1}>0$ (dependent on k _d and $\bar{k}_{d}$) such that with X _I (T _start)=0, $X_{I}(T_{\mathrm{start}}+\tilde {T}_{1})=\frac{c}{2}$, $X_{I}(T_{\mathrm{start}}+\tilde {T}_{2})=c$ and X _I (T _start+t) is in between as $t\in(\tilde {T}_{1}, \tilde {T}_{2})$ since X _I is increasing. Therefore, in system (10), if $T_{D}\in(\tilde {T}_{1}, \tilde {T}_{2})$ and X _I (T _start)=0, then $X_{I}(T_{\mathrm{start}}+T_{D})\in(\frac{c}{2}, c)$ as $T_{D}\in(\tilde {T}_{1}, \tilde {T}_{2})$. As shown in Fig. 6b, when t>T _start+T _D , X _I (t) will eventually approach the open state X _I =c because X _I (T _start+T _D ) is in the stability region of open state.

Appendix B: Monotonicity of g(x)

In the following, we will prove that function

$$g(x)=\frac{x-x^p}{(1-x)(1+x^p)} $$

is an increasing function as x∈[0,1]. First, one has

$$g^\prime (x)=\frac{g_1(x)}{(1-x)^2(1+x^p)^2},\quad g_1(x)=1-px^{p-1}+px^{p+1}-x^{2p}, $$

and

$$g_1^\prime (x)=-p(p-1)x^{p-2}+p(p+1)x^{p}-2px^{2p-1}=px^{p-2}g_2(x), $$

where

$$g_2(x)=-(p-1)+(p+1)x^2-2x^{p+1}. $$

Note that for x∈[0,1],

$$g_2^\prime (x)=2(p+1)x-2(p+1)x^p=2(p+1)x \bigl(1-x^{p-1}\bigr)>0, $$

thus

$$\begin{array}{l} g_2(x)\leq g_2(1)=0\Rightarrow g_1^\prime (x)\leq0 \\[4pt] \quad \Rightarrow \quad g_1(x)\geq g_1(1)=0\Rightarrow g^\prime (x)\geq0 \\ \end{array} $$

which implies that g(x) is increasing in x∈[0,1].

Appendix C: Charge Accumulation

Consider the equation for charge accumulation

$$ \begin{array}{l} \dot{C}_m=-k_cC_m+k_au_e^m(t),\quad C_m(0)=0, \\ u_e^m(t)=\displaystyle \sum_{i=0}^m u_t(t-\tau_i)H(t-\tau_i), \end{array} $$

(17)

where u _e (t) denotes a sequence of external stimuli with delay τ _i such that 0=τ ₀<τ ₁<⋯<τ _m , H(t) is the Heaviside function.

Lemma C.1

Let C _m (t) be the solution to Eq. (17), then {C _m } is an increasing sequence of bounded functions. Suppose that C _m attains its maximum $\bar{C}_{m}$ at t _m , that is, $\bar{C}_{m}=C_{m}(t_{m})$, then t _m ∈(τ _m ,t ₀+τ _m ] and $\{\bar{C}_{m}\}$ is a strictly increasing sequence.

Proof

Obviously C _m (t) is bounded for $u_{e}^{m}(t)$ being bounded and k _c >0. And it follows from comparison principle that C _m (t)≤C _m+1(t) for all t≥0 and hence $\bar{C}_{m}\leq\bar {C}_{m+1}$, because $u_{e}^{m}(t)\leq u_{e}^{m+1}(t)$ for all t≥0. More precisely, one can have the exact solution

$$\begin{array}{c} C_m(t)=\displaystyle \sum_{i=0}^m C_0(t-\tau_i)H(t-\tau_i), \\ \end{array} $$

where $C_{0}(t)=\frac{Ak_{a}}{B-k_{c}} (-e^{-Bt}+e^{-k_{c}t} )$ is the solution of Eq. (2), increases as t<t ₀ and decreases as t>t ₀ and attains its maximum at $t_{0}=\frac{\ln(\frac{B}{k_{c}})}{B-k_{c}}>0$, which can be solved from $\dot{C}_{0}=\frac{Ak_{a}}{B-k_{c}} (Be^{-Bt}-k_{c}e^{-k_{c}t} )=0$.

Since $C_{1}(t_{0}+\tau_{1})=C_{0}(t_{0}+\tau_{1})+C_{0}(t_{0})>\bar{C}_{0}$, one can readily see that $\bar{C}_{1}>\bar{C}_{0}$. Note that as t≤τ ₁,

$$C_{1}(t)=C_{0}(t)\leq\bar{C}_0< \bar{C}_1, $$

which implies that t _m >τ _m . By induction, one has that $\bar{C}_{m+1}>\bar{C}_{m}$ and t _m >τ _m for m≥0.

On the other hand, as t>t ₀+τ _m ,

$$\dot{C}_{m}(t)=\dot{C}_{m-1}(t)+\dot{C}_0(t- \tau_m)<0 $$

and hence t _m ≤t ₀+τ _m . Therefore, t _m ∈(τ _m ,t ₀+τ _m ].

Moreover, if t ₀>τ ₁, then as t∈(τ ₁,t ₀],

$$\dot{C}_1(t)=\dot{C}_0(t)+\dot{C}_0(t- \tau_1)>0, $$

which implies that t ₁>t ₀. Thus t ₁∈(max{t ₀,τ ₁},t ₀+τ ₁]. This completes the proof. □

Lemma C.2

For any C _T >0, there exist an integer m>0 and an increasing finite sequence $\{\tau_{i}\}_{i=1}^{m}$ such that the solution of (17) satisfies $\bar{C}_{m}>C_{T}$.

Proof

For fixed A,B,k _a ,k _c >0, C ₀(t)>0 for all t>0. For any fixed L>0, set c _L =C ₀(t ₀+L)>0 and m to be such that (m+1)c _L ≥C _T . Define $\tau_{i}=\frac{Ti}{m}$ for 0≤i≤m, then

$$\begin{array}{rcl} \bar{C}_m &\geq& \displaystyle \lim_{t\rightarrow {t}_0+\tau_m^+}C_m(t)=\displaystyle \sum_{i=0}^{m}C_0(t_0+\tau_m-\tau_i)\\ &\geq& \displaystyle \sum_{i=0}^{m}C_0(t_0+L)=(m+1)c_L\geq C_T. \\ \end{array} $$

 □

Lemma C.3

For fixed C _T >0 and integer m>1, suppose that $\bar{C}_{0}<C_{T}$. Then there exists d>0 such that for any increasing sequence $\{\tau_{i}\}_{i=0}^{m}$ satisfying τ ₀=0 and τ _i+1−τ _i ≥d, $\bar{C}_{m}<C_{T}$.

Proof

Set $D=\frac{1}{2}(C_{T}-\bar{C}_{0})$. Because of the exponential decay of C ₀(t), there exists d>t ₀ sufficiently large such that $C_{0}(t)<\frac{D}{m}$ as t≥d. By Lemma C.1, C _m (t) attains its maximum $\bar{C}_{m}$ at t _m ∈(τ _m ,t ₀+τ _m ].

Then for any t∈(τ _m ,t ₀+τ _m ] and 1≤i≤m−1, one has t−τ _i >τ _m −τ _m−1≥d>t ₀ and thus

$$\begin{array}{rcl} C_m(t) & = & \displaystyle \sum_{i=0}^{m-1}C_0(t-\tau_i)+C_0(t-\tau_m)\\ &\leq& \displaystyle \displaystyle \sum_{i=0}^{m-1}C_0(\tau_m-\tau_i)+\bar{C}_0 \\ &\leq& \displaystyle \displaystyle \sum_{i=0}^{m-1}C_0(d)+\bar{C}_0=mC_0(d)+\bar{C}_0 \\ &\leq& D+\bar{C}_0=\frac{1}{2}(\bar{C}_0+C_T), \end{array} $$

from which it follows immediately that

$$\bar{C}_m=\displaystyle \max_{t\in(\tau_m, t_0+\tau_m]}\bigl\{C_m(t) \bigr\}\leq\frac{1}{2}(\bar {C}_0+C_T)<C_T. $$

 □

Remark C.4

Lemmas C.1 and C.2 show that the charge is always possible to be accumulated to reach the threshold C _T regardless of how high its value is. However, if the hairs can be touched only finite times (m times) and the time delay between any two consecutive hair triggering is larger than certain value d, then the charge accumulation can never reach C _T , as shown in Lemma C.3. The exact value of d is not analytically available but can be estimated from $C_{0}(d)<\frac{D}{m}$, or an improved estimate from $\sum_{i=0}^{m-1}C_{0}((i+1)d)<D$.

Theorem C.5

The maximum $\bar{C}_{m}$ of C _m (t) is given by

$$ \bar{C}_m=\displaystyle {A}k_a \Biggl( \frac{1}{B}\displaystyle \sum_{i=0}^m e^{k_c\tau_i} \Biggr)^{\frac {B}{B-k_c}} \Biggl(\frac{1}{k_c}\displaystyle \sum_{i=0}^m e^{B\tau_i} \Biggr)^{\frac{-k_c}{B-k_c}}. $$

(18)

Proof

By Lemma C.1, t _m >τ _m and then

(19)

On the other hand, $\dot{C}_{m}(t_{m})=0$ implies that

(20)

By (19) and (20), one also has

$$ \bar{C}_m=\displaystyle \frac{Ak_a}{B}\sum _{i=0}^m e^{-k_c(t_m-\tau_i)}. $$

(21)

Then it follows from solving (20) and (21) that

$$\bar{C}_m\bigl(\{\tau_i\}\bigr)= {A}k_a \biggl[\frac{ (\frac{1}{B} \sum_{i=0}^m e^{k_c\tau_i} )^B}{ (\frac{1}{k_c} \sum_{i=0}^m e^{B\tau_i} )^{k_c}} \biggr]^{\frac{1}{B-k_c}}. $$

 □

To reach the threshold C _T , one only need the time delays {τ _i } be such that $\bar{C}_{m}(\{\tau_{i}\})\geq C_{T}$. In the following, we will consider a simple case of m=1, where we let τ=τ ₁ for simplicity.

Corollary C.6

Suppose that B>k _c , m=1 and $\bar{C}_{0}>\frac{C_{T}}{2}$. Then there exists a unique $\bar{\tau}>0$ such that $\bar{C}_{1}>C_{T}$ as $\tau\in(0, \bar{\tau})$ and $\bar{C}_{1}<C_{T}$ as $\tau>\bar{\tau}$. And $\bar{\tau}$ is increasing in  $\frac{Ak_{a}}{C_{T}}$.

Proof

As m=1, consider

$$\bar{C}_1=\frac{Ak_ak_c^{\frac{k_c}{B-k_c}}}{B^{\frac{B}{B-k_c}}} \bigl(1+e^{k_c\tau} \bigr)^{\frac{B}{B-k_c}} \bigl(1+e^{B\tau} \bigr)^{\frac {-k_c}{B-k_c}}\geq C_T. $$

Then

$$\frac{ (1+e^{k_c\tau} )^B}{ (1+e^{B\tau} )^{k_c}}\geq \biggl(\frac{B^B}{k_c^{k_c}} \biggr) \biggl( \frac{C_T}{Ak_a} \biggr)^{B-k_c} $$

or equivalently

$$f(\tau)=B\ln \bigl(1+e^{k_c\tau} \bigr)-k_c\ln \bigl(1+e^{B\tau} \bigr)-c_f\geq0, $$

where $c_{f}=B\ln B-k_{c}\ln k_{c}+(B-k_{c})\ln (\frac{C_{T}}{Ak_{a}} )$. Note that

$$f^\prime (\tau)=Bk_c \biggl(\frac{e^{k_c\tau}}{1+e^{k_c\tau}}- \frac{e^{B\tau }}{1+e^{B\tau}} \biggr)<0 $$

as B>k _c . Note that f(0)>0 if and only if $2\bar{C}_{0}>C_{T}$ and lim _τ→∞ f(τ)=−c _f <0. Thus, there exists unique $\bar {\tau}>0$ such that $f(\bar{\tau})=0$ and f(τ)>0 as $\tau\in(0, \bar{\tau})$. In other words, $\bar{C}_{1}= C_{T}$ as $\tau=\bar{\tau}$, $\bar{C}_{1}>C_{T}$ as $\tau\in(0, \bar{\tau})$ and $\bar{C}_{1}<C_{T}$ as $\tau>\bar{\tau}$.

Furthermore, $\bar{\tau}$ can be solved from f(τ)=0 and it depends on B,k _c and $\frac{C_{T}}{Ak_{a}}$. Because $\frac{\partial f}{\partial (\frac{C_{T}}{Ak_{a}})}<0$ as B>k _c , by implicit differentiation one knows that $\bar{\tau}$ depends decreasingly on $\frac{C_{T}}{Ak_{a}}$ or increasingly on $\frac{Ak_{a}}{C_{T}}$ as a whole. □