Mathematical Analysis of Spontaneous Emergence of Cell Polarity

Abstract

Cell polarization, in which intracellular substances are asymmetrically distributed, enables cells to carry out specialized functions. While cell polarity is often induced by intracellular or extracellular spatial cues, spontaneous polarization (the so-called symmetry breaking) may also occur in the absence of spatial cues. Many computational models have been used to investigate the mechanisms of symmetry breaking, and it was proved that spontaneous polarization occurs when the lateral diffusion of inactive signaling molecules is much faster than that of active signaling molecules. This conclusion leaves an important question of how, as observed in many biological systems, cell polarity emerges when active and inactive membrane-bound molecules diffuse at similar rates while cycling between cytoplasm and membrane takes place. The recent studies of Rätz and Röger showed that, when the cytosolic and membrane diffusion are very different, spontaneous polarization is possible even if the membrane-bound species diffuse at the same rate. In this paper, we formulate a two-equation non-local reaction-diffusion model with general forms of positive feedback. We apply Turing stability analysis to identify parameter conditions for achieving cell polarization. Our results show that spontaneous polarization can be achieved within some parameter ranges even when active and inactive signaling molecules diffuse at similar rates. In addition, different forms of positive feedback are explored to show that a non-local molecule-mediated feedback is important for sharping the localization as well as giving rise to fast dynamics to achieve robust polarization.

Appendix

1.1 Proofs of Lemmas

In this section, we will state three lemmas, which will be used in the next section for the proofs of Theorems 1–4. First, we define a function $f_{y}(a)$ used in the lemmas:

$$\begin{aligned} f_{y}(a) = (\gamma _1 + \gamma _2 a^n)(1-\gamma _3 y) -\gamma _4 a - D(a-y). \end{aligned}$$

(45)

where $n>1$, $\gamma _1,\gamma _2, \gamma _4> 0$, $\gamma _3\ge 1$ and $0\le D<\gamma _1\gamma _3$.

Lemma 1

The function $f_{y}$ in (45) has the following properties:

• 1. $\min \limits _{a\ge 0}f_{y}(a)$ equals to

  $$\begin{aligned} \gamma _1-(\gamma _1\gamma _3-D)y- \frac{n-1}{n^{\frac{n}{n-1}}}\frac{(\gamma _4+D)^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}(1-\gamma _3 y)^{-\frac{1}{n-1}}, \end{aligned}$$

  which is strictly decreasing with respect to $y$ for $y\in (0,1/\gamma _3)$.

• 2. For each $y$, there exist at most two solutions in $\{a|a\ge 0\}$ satisfying $f_{y}(a)=0$.

• 3. There exists a number $y_m$ in $[0, 1/\gamma _3)$ such that two smooth functions $a_1(y)$, $a_2(y)$ can be well defined in the domain $[y_m, 1/\gamma _3)$ and the following properties hold:

  • (a) $\min \limits _{a\ge 0}f_{y}(a)\le 0$ for any $y\in [y_m, 1/\gamma _3)$;

  • (b) $f_y(a_1(y)) = f_y(a_2(y))= 0$ for any $y\in [y_m, 1/\gamma _3)$;

  • (c) $a_1(y)> a_2(y)\ge 0$ for any $y\in (y_m, 1/\gamma _3)$;

  • (d) $a_1'(y)>0$ and $a_2'(y)<0$ for any $y\in (y_m, 1/\gamma _3)$;

  • (e) $\lim \limits _{y\rightarrow 1/\gamma _3}a_1(y)=\infty$ and $\lim \limits _{y\rightarrow 1/\gamma _3}a_2(y)=0$;

  • (f) $\frac{\mathrm{d}f_y}{\mathrm{d}a}\left| _{a=a_1(y)}\right. >0$ and $\frac{df_y}{da}\left| _{a=a_2(y)}\right. <0$ for any $y\in (y_m, 1/\gamma _3)$;

  • (g) if there is at least one solution in $a\ge 0$ for $f_0(a)=0$, then $y_m=0$;

  • (h) if there is no solution in $a\ge 0$ for $f_0(a)=0$, then $a_1(y_m)=a_2(y_m)$, $\frac{df_{y_m}}{da}\left| _{a=a_1(y_m)}\right. =\frac{df_{y_m}}{da}\left| _{a=a_2(y_m)}\right. =0$ and $\min \limits _{a\ge 0}f_{y_m}(a)= 0$.

Proof

1.:

First we consider the first and second derivatives of $f_y$,

$$\begin{aligned} \frac{df_{y}(a)}{da}&= n\gamma _2 a^{n-1}(1-\gamma _3 y) -\gamma _4 - D,\end{aligned}$$

(46)

$$\begin{aligned} \frac{d^2f_{y}(a)}{da^2}&= n(n-1)\gamma _2 a^{n-2}(1-\gamma _3 y). \end{aligned}$$

(47)

By Eq. (47), we show that the minimum point in $\{a|a\ge 0\}$, with $y\in (0,1/\gamma _3)$, is at

$$\begin{aligned} a =\left( \frac{\gamma _4+D}{n\gamma _2(1-\gamma _3y)}\right) ^{\frac{1}{n-1}} \end{aligned}$$

with

$$\begin{aligned} \min \limits _{a\ge 0}f_{y}(a)=\gamma _1-(\gamma _1\gamma _3-D)y- \frac{n-1}{n^{\frac{n}{n-1}}}\frac{(\gamma _4+D)^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}(1-\gamma _3 y)^{-\frac{1}{n-1}}. \end{aligned}$$

By the given condition $D<\gamma _1\gamma _3$, it is easy to show that $\min \limits _{a\ge 0}f_{y}(a)$ is strictly decreasing with respect to $y$.

2.:

Suppose that $y$ is a fixed number. If $\min \limits _{a\ge 0}f_{y}(a)>0$, there is no solution $a\ge 0$ satisfying $f_y(a)=0$.

If $\min \limits _{a\ge 0}f_{y}(a)=0$, the minimum point

$$\begin{aligned} \bar{a} = \left( \frac{\gamma _4+D}{n\gamma _2(1-\gamma _3y)}\right) ^{\frac{1}{n-1}} \end{aligned}$$

is one of the roots for $f_y(a)$. As $\frac{df_{y}(a)}{da}>0$ for $a > \bar{a}$ and $\frac{df_{y}(a)}{da}<0$ for $0\le a < \bar{a}$, $f_y(a)>f_y(\bar{a})$ for any $a\ne \bar{a}$, and therefore $\bar{a}$ is the only solution of $f_y(a)=0$.

If $\min \limits _{a\ge 0}f_{y}(a)<0$, by the fact that $f_y(0)>0$, $\lim \limits _{a\rightarrow \infty }f_y(a)>0$ and the intermediate value theorem, we know that there are at least two solutions satisfying $f_y(a)=0$. As $\frac{df_{y}(a)}{da}>0$ for $a > \bar{a}$ and $\frac{df_{y}(a)}{da}<0$ for $0\le a < \bar{a}$, $f_y(a)>f_y(\bar{a})$ for any $a\ne \bar{a}$. So there are only two roots of $f_y(a)$: one is in $\left[ 0, \bar{a}\right)$, and the other is in $\left( \bar{a}, \infty \right)$.

3.:

By the result of part 1, $\min \limits _{a\ge 0}f_{y}(a)$ tends to $-\infty$ as $y$ is close to $1/\gamma _3$. If $\min \limits _{a\ge 0}f_{y}(a)> 0$ for $y=0$, according to the intermediate value theorem, we can find $y_m$ such that $\min \limits _{a\ge 0}f_{y_m}(a)$ equals zero; if $\min \limits _{a\ge 0}f_{y}(a)\le 0$ for $y=0$, we define $y_m=0$.

Since $\min \limits _{a\ge 0}f_{y}(a)$ is strictly decreasing with respect to $y$, and according to the results of part 2, $f_y(a)=0$ has two solutions $a$ for any $y\in (y_m, 1/\gamma _3)$, so we can define two functions $a_1(y)$ and $a_2(y)$ that satisfy $f_y(a_1(y))=f_y(a_2(y))=0$ and $a_1(y)>a_2(y)$ for any $y\in (y_m,1/\gamma _3)$, that is,

$$\begin{aligned} a_1(y) = \max \{a\ge 0|f_y(a)=0\},\qquad a_2(y) = \min \{a\ge 0|f_y(a)=0\}. \end{aligned}$$

The derivative of $f_y(a)$ with respect to $y$ is $-\gamma _1\gamma _3+D - \gamma _2\gamma _3 a^n$, which is always negative, and $f_y(a)$ is a smooth function with respect to $y$ and $a$, and therefore we can apply the inverse function theorem to show that $a_1(y)$ and $a_2(y)$ are smooth functions. By the definitions and the proof of part 2, it is easy to verify the properties (a, b, c, f, g, h).

By property (b), we have $f_y(a_1(y))=0$ and $f_y(a_2(y))=0$. When differentiating these two equations with respect to $y$ on both sides, we have $-\gamma _1\gamma _3+D - \gamma _2\gamma _3 a_1(y)^n + \frac{df_y}{da}(a_1(y))a_1'(y)=0$ and $-\gamma _1\gamma _3+D - \gamma _2\gamma _3 a_2(y)^n + \frac{df_y}{da}(a_1(y))a_2'(y)=0$. Hence we obtain

$$\begin{aligned} a_1'(y)&= -\frac{-\gamma _1\gamma _3+D - \gamma _2\gamma _3 a_1(y)^n}{ \frac{df_y}{da}(a_1(y))},\\ a_2'(y)&= -\frac{-\gamma _1\gamma _3+D - \gamma _2\gamma _3 a_2(y)^n}{ \frac{df_y}{da}(a_2(y))}. \end{aligned}$$

By property (f) and $\gamma _1\gamma _3>D$, we show that $a_1'(y)>0$ and $a_2'(y)<0$, which completes the proof of property (d).

From the proof of part 2, we have $a_2\in \left[ 0, \left( \frac{\gamma _4+D}{n\gamma _2(1-\gamma _3y)}\right) ^\frac{1}{n-1}\text { }\right)$ and $a_1\in \left( \text {}\left( \frac{\gamma _4+D}{n\gamma _2(1-\gamma _3y)}\right) ^\frac{1}{n-1}, \infty \right)$. So we know that $a_1(y)$ tends to infinity as $y$ goes to $1/\gamma _3$. Since $a=0$ is the solution for $f_{1/\gamma _3}(a)=0$, we have $\lim \limits _{y\rightarrow 1/\gamma _3}a_2(y)=0$, which completes the proof of property (e).$\square$

Lemma 2

If

$$\begin{aligned} \left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4+D}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right) >\frac{\gamma _1\gamma _3}{\gamma _1\gamma _3-\frac{1}{n}D+ \frac{n-1}{n}\gamma _4} \end{aligned}$$

(48)

is satisfied, then $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. >0$ holds for any solution $a_0$ satisfying $f_{a_0}(a_0)=0$.

For the proofs of Lemmas 2 and 3, we define two functions $S_1$, $S_2$ in the domain $[y_m, 1/\gamma _3)$:

$$\begin{aligned} S_1(y)&= a_1(y)-y,\\ S_2(y)&= a_2(y)-y, \end{aligned}$$

where $a_1$, $a_2$ and $y_m$ are defined in Lemma 1.

Proof

There are two parts in the proof:

• 1. Prove that if $S_1(y_m) <0$, $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. >0$ holds for any solution $a_0\ge 0$ satisfying $f_{a_0}(a_0)=0$.

• 2. Prove that condition (48) implies $S_1(y_m) <0$.

By combining these two results, we can prove that if the condition (48) is satisfied, then $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. >0$ holds for any solution $a_0\ge 0$ satisfying $f_{a_0}(a_0)=0$. $\square$

Proof of part 1

Suppose that $S_1(y_m) <0$. Since $a_1(y)\ge a_2(y)$, we get $S_2(y_m)\le S_1(y_m)<0$. By $a_2'(y)<0$ (Lemma 1(3c)), we have $S_2'<0$, which means that $S_2$ is a decreasing function. Since $S_2(y_m) <0$ and $S_2$ is a decreasing function, $S_2(y)<0$ for all $y\in [y_m, 1/\gamma _3)$, and there is no solution to $S_2(y)= 0$.

According to Lemma 1 and the definitions of $S_1$ and $S_2$, all solutions $a_0\ge 0$ for $f_{a_0}(a_0)=0$ have to satisfy $S_1(a_0) = 0$ or $S_2(a_0) = 0$. Since $S_1(y_m) <0$ implies that there is no solution satisfying $S_2(y)= 0$, all solutions $a_0\ge 0$ for $f_{a_0}(a_0)$ have to satisfy $S_1(a_0) = 0$ and therefore $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. >0$ according to Lemma 1(3f).$\square$

Proof of part 2

Suppose that condition (48) is satisfied, by Lemma 1(1), we have

$$\begin{aligned} \min \limits _{a\ge 0}f_{y}(a)=\gamma _1-(\gamma _1\gamma _3-D)y- \frac{n-1}{n^{\frac{n}{n-1}}}\frac{(\gamma _4+D)^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}(1-\gamma _3 y)^{-\frac{1}{n-1}}. \end{aligned}$$

If $0<\gamma _3y<1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4+D}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}$, we have

$$\begin{aligned} \gamma _1(1-\gamma _3y)^{\frac{n}{n-1}}&> \frac{n-1}{n^{\frac{n}{n-1}}}\frac{(\gamma _4+D)^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}, \end{aligned}$$

and therefore

$$\begin{aligned} \gamma _1-(\gamma _1\gamma _3-D)y&> \frac{n-1}{n^{\frac{n}{n-1}}}\frac{(\gamma _4+D)^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}(1-\gamma _3 y)^{-\frac{1}{n-1}},\\ \min \limits _{a\ge 0}f_{y}(a)&> 0. \end{aligned}$$

$\square$

Lemma 1(3a) implies that $y_m$ is larger than $\frac{1}{\gamma _3}\left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4+D}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right)$, that is,

$$\begin{aligned} y_m>\frac{1}{\gamma _3}\left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4+D}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right) >0. \end{aligned}$$

(49)

Then we apply Lemma 1(3h) to show that there is no solution with $a\ge 0$ such that $f_0(a)=0$.

By Lemma 1(3b, h), we know that ($y_m, a_1(y_m))$ satisfies the following two equations:

$$\begin{aligned} f_{y_m}(a_m)&= (\gamma _1 + \gamma _2 a_m^n)(1-\gamma _3 y_m) -\gamma _4 a_m - D(a_m-y_m)=0,\end{aligned}$$

(50)

$$\begin{aligned} \frac{df_{y_m}}{da}\left| _{a=a_m}\right.&= n\gamma _2 a_m^{n-1}(1-\gamma _3 y_m) -\gamma _4 - D=0, \end{aligned}$$

(51)

where $a_m=a_1(y_m)$.

After multiplying (50) and (51) by $n$ and $a_m$, respectively, we have

$$\begin{aligned} n\gamma _1(1-\gamma _3 y_m) + n \gamma _2 a_m^n(1-\gamma _3 y_m) -n\gamma _4 a_m - nDa_m+ nDy_m&= 0,\end{aligned}$$

(52)

$$\begin{aligned} n\gamma _2 a_m^{n}(1-\gamma _3 y_m) -\gamma _4a_m - Da_m&= 0. \end{aligned}$$

(53)

Substracting (52) by (53), we obtain

$$\begin{aligned} n\gamma _1 (1-\gamma _3 y_m) - (n-1)(\gamma _4+D)a_m+ nDy_m=0, \end{aligned}$$

which leads to

$$\begin{aligned} a_m= \frac{n}{n-1}\frac{1}{\gamma _4+D}(\gamma _1-(\gamma _3\gamma _1-D)y_m). \end{aligned}$$

(54)

By substituting (54) into $S_1(y_m)$, we obtain

$$\begin{aligned} S_1(y_m) = a_m -y_m = \frac{n}{n-1}\frac{1}{\gamma _4+D}\left( \gamma _1-\left( \gamma _1\gamma _3-\frac{1}{n}D+ \frac{n-1}{n}\gamma _4\right) y_m\right) . \end{aligned}$$

(55)

By applying (49) and condition (48),

$$\begin{aligned} y_m&> \frac{1}{\gamma _3}\left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4+D}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right) \\&> \frac{\gamma _1}{\gamma _1\gamma _3-\frac{1}{n}D+ \frac{n-1}{n}\gamma _4}, \end{aligned}$$

which, coupled with (55), implies that $S_1(y_m)<0$.

Lemma 3

Suppose $D=0$, and if

$$\begin{aligned} \left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right) <\frac{\gamma _1\gamma _3}{\gamma _1\gamma _3+ \frac{n-1}{n}\gamma _4}. \end{aligned}$$

(56)

holds, then there exists a solution $a_0$ satisfying $f_{a_0}(a_0)=0$ and $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. <0$.

Proof

There are two parts in the proof:

• 1. Prove that if $S_2(y_m) \ge 0$, there exists $a_0\ge 0$ such that $f_{a_0}(a_0)=0$ and $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. <0$.

• 2. Prove that condition (56) implies $S_2(y_m) \ge 0$.

By combining these two results, we can prove that if the condition (56) is satisfied, there exists $a_0\ge 0$ satisfying $f_{a_0}(a_0)=0$ and $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. <0$.

Proof of part 1

Suppose $S_2(y_m) >0$, as we know that $S_2(1/\gamma _3)= -1/\gamma _3<0$, then by the intermediate value theorem, there exists a solution $a_0$ satisfying $S_2(a_0) = 0$ ($f_{a_0}(a_0)=0$), and therefore $\frac{df_{a_0}}{da}\left| _{a=a_0}\right. <0$, according to Lemma 1(3f).

Proof of part 2

Suppose that condition (56) is satisfied.

If $y_m=0$, we have $S_2(y_m)= a_2(y_m) \ge 0$, which completes the proof of part 2. Otherwise if $y_m>0$, by Lemma 1(1), we have

$$\begin{aligned} \min \limits _{a\ge 0}f_{y}(a)=\gamma _1-\gamma _1\gamma _3y- \frac{n-1}{n^{\frac{n}{n-1}}}\frac{\gamma _4^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}(1-\gamma _3 y)^{-\frac{1}{n-1}}. \end{aligned}$$

If $\gamma _3y>1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}$, we have

$$\begin{aligned} \gamma _1(1-\gamma _3y)^{\frac{n}{n-1}}&< \frac{n-1}{n^{\frac{n}{n-1}}}\frac{\gamma _4^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}, \end{aligned}$$

and therefore

$$\begin{aligned} \gamma _1-\gamma _1\gamma _3y&< \frac{n-1}{n^{\frac{n}{n-1}}}\frac{\gamma _4^{\frac{n}{n-1}}}{\gamma _2^{\frac{1}{n-1}}}(1-\gamma _3 y)^{-\frac{1}{n-1}},\\ \min \limits _{a\ge 0}f_{y}(a)&< 0. \end{aligned}$$

Lemma 1(3h) implies that

$$\begin{aligned} y_m\le \frac{1}{\gamma _3}\left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right) . \end{aligned}$$

(57)

By Lemma 1(3b, h), ($y_m, a_1(y_m))$ satisfies the following two equations:

$$\begin{aligned} f_{y_m}(a_m)&= (\gamma _1 + \gamma _2 a_m^n)(1-\gamma _3 y_m) -\gamma _4 a_m=0,\end{aligned}$$

(58)

$$\begin{aligned} \frac{df_{y_m}}{da}\left| _{a=a_m}\right.&= n\gamma _2 a_m^{n-1}(1-\gamma _3 y_m) -\gamma _4 =0, \end{aligned}$$

(59)

where $a_m=a_2(y_m)$.

After multiplying (58) and (59) by $n$ and $a_m$, respectively, we have

$$\begin{aligned} n\gamma _1(1-\gamma _3 y_m) + n \gamma _2 a_m^n(1-\gamma _3 y_m) -n\gamma _4 a_m&= 0,\end{aligned}$$

(60)

$$\begin{aligned} n\gamma _2 a_m^{n}(1-\gamma _3 y_m) -\gamma _4a_m&= 0. \end{aligned}$$

(61)

Then subtracting (60) by (61), one obtains

$$\begin{aligned} n\gamma _1 (1-\gamma _3 y_m) - (n-1)\gamma _4a_m=0, \end{aligned}$$

which leads to

$$\begin{aligned} a_m= \frac{n}{n-1}\frac{1}{\gamma _4}(\gamma _1-\gamma _3\gamma _1y_m). \end{aligned}$$

(62)

After substituting (62) into $S_2(y_m)$, we get

$$\begin{aligned} S_2(y_m) = a_m -y_m = \frac{n}{n-1}\frac{1}{\gamma _4+D}\left( \gamma _1-\left( \gamma _1\gamma _3+ \frac{n-1}{n}\gamma _4\right) y_m\right) . \end{aligned}$$

(63)

By applying (57) and condition (56),

$$\begin{aligned} y_m&\le \frac{1}{\gamma _3}\left( 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{\gamma _4}{\gamma _2^{\frac{1}{n}}\gamma _1^{\frac{n-1}{n}}}\right) \\&< \frac{\gamma _1}{\gamma _1\gamma _3+ \frac{n-1}{n}\gamma _4}, \end{aligned}$$

which, coupled with (63), implies that $S_2(y_m)\ge 0$.

1.2 Proofs of Theorems 1–4

1.2.1 Theorem 1

Proof

First, we set $\gamma _1=k_3$, $\gamma _2=k_4$, $\gamma _3=1$, $\gamma _4=k_\mathrm{{off}}$, and $D=D^*$ in the lemmas. By applying Lemma 2, we obtain that if

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}+D^*}{k_4^{\frac{1}{n}}k_3^{\frac{n-1}{n}}}>\frac{k_3}{k_3-\frac{1}{n}D^*+\frac{n-1}{n}k_\mathrm{{off}}} \end{aligned}$$

(64)

then

$$\begin{aligned} nk_4a_0^{n-1} (1-a_0)- k_\mathrm{{off}} - D^*>0 \end{aligned}$$

(65)

holds for any $a_0$ satisfying

$$\begin{aligned} (k_3+k_4a_0^{n})(1-a_0)- k_\mathrm{{off}}a_0 =0. \end{aligned}$$

(66)

By (30), we know that $a_0$ is a homogeneous steady state solution for $a$ in system (1)–(2) with the cooperative feedback (28) if and only if $a_0$ satisfies (66). Also, by $D^*>{ \sigma _{1}D_\mathrm{m}}+\frac{{ \sigma _{1}D_\mathrm{m}}}{{ \sigma _{1}D_\mathrm{m}}+g_\mathrm{{off}}}({ k_{11}}+{ k_{12}}a_0)$, inequality (65) implies inequality (26).

By the result obtained from Lemma 2, we have proved that if (64) is satisfied, then all possible homogeneous steady state solutions satisfy inequality (26). Since at least one homogeneous steady state solution satisfies inequality (25), we have proved that if

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}+D^*}{k_4^{\frac{1}{n}}k_3^{\frac{n-1}{n}}}>\frac{k_3}{k_3-\frac{1}{n}D^*+\frac{n-1}{n}k_\mathrm{{off}}} \end{aligned}$$

then there exists a homogeneous steady state solution satisfying (25) and (26). In addition, since all possible homogeneous steady state solutions satisfy inequality (26) in this case, locally stable homogeneous steady state solution does not exist.$\square$

1.2.2 Theorem 2

Proof

Let $\gamma _1=k_3$, $\gamma _2=k_4$, $\gamma _3=1$, $\gamma _4=k_\mathrm{{off}}$, and $D=0$ in the lemmas. By applying Lemma 3, we obtain that if

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}}{k_4^{\frac{1}{n}}k_3^{\frac{n-1}{n}}}<\frac{k_3}{k_3+\frac{n-1}{n}k_\mathrm{{off}}}, \end{aligned}$$

there exists a solution $a_0$ satisfying

$$\begin{aligned} (k_3+k_4a_0^{n})(1-a_0)- k_\mathrm{{off}}a_0 =0 \end{aligned}$$

and

$$\begin{aligned} nk_4a_0^{n-1} (1-a_0)- k_\mathrm{{off}} <0. \end{aligned}$$

(67)

By (30), we know that $a_0$ is a homogeneous steady state solution for $a$ in the system (1)–(2) with the cooperative feedback (28) if and only if $a_0$ satisfies (66). Also, inequality (67) implies that a homogeneous steady state solution is locally stable for perturbations with any nonnegative wavenumber [satisfying the condition (25) but not satisfying (26)]. So we have proved that if the condition

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}}{k_4^{\frac{1}{n}}k_3^{\frac{n-1}{n}}}<\frac{k_3}{k_3+\frac{n-1}{n}k_\mathrm{{off}}} \end{aligned}$$

is satisfied, there exists a locally stable homogeneous steady state solution.$\square$

1.2.3 Theorem 3

Proof

First, we set $\gamma _1=k_5$, $\gamma _2=k_6$, $\gamma _3=\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}$, $\gamma _4=k_\mathrm{{off}}$ and $D=D^+$ in the lemmas. By applying Lemma 2, we obtain that if

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}+D^*}{k_6^{\frac{1}{n}}k_5^{\frac{n-1}{n}}}>\frac{\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}k_5}{\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}k_5-\frac{1}{n}D^++\frac{n-1}{n}k_\mathrm{{off}}}, \end{aligned}$$

(68)

then

$$\begin{aligned} nk_6a_0^{n-1} \left( 1-\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}a_0\right) - k_\mathrm{{off}} - D^+>0 \end{aligned}$$

(69)

holds for any $a_0$ satisfying

$$\begin{aligned} (k_5+k_6a_0^{n})\left( 1-\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}a_0\right) - k_\mathrm{{off}}a_0 =0. \end{aligned}$$

(70)

By (41), we know that $a_0$ is a homogeneous steady state solution for $a$ in the system (1)–(2) with the feedback form (39) if and only if $a_0$ satisfies (69).

By equation (70) with $k_5={ k_{21}}k_\mathrm{{on}}^*$ and $k_6={ k_{22}}k_\mathrm{{on}}^*$, we have

$$\begin{aligned} 1+{ k_{21}}+{ k_{22}}a_0^n = \frac{1-a_0}{1-\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}a_0}. \end{aligned}$$

(71)

Now we substitute the feedback form (39) into inequality (26), we obtain

$$\begin{aligned} -{ \sigma _{1}D_\mathrm{m}}+ \frac{nk_6a_0^{n-1}}{1+{ k_{21}}+{ k_{22}}a_0^n} (1-a_0)- k_\mathrm{{off}}- \frac{{ \sigma _{1}D_\mathrm{m}}}{{ \sigma _{1}D_\mathrm{m}}+g_\mathrm{{off}}} k_\mathrm{{on}}\frac{{ k_{21}}+{ k_{22}}a_0^n}{1+{ k_{21}}+{ k_{22}}a_0^n}>0, \end{aligned}$$

and by using (71), this inequality can be rewritten as

$$\begin{aligned}&-{ \sigma _{1}D_\mathrm{m}}+ nk_6a_0^{n-1} \left( 1-\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}a_0\right) \nonumber \\&\quad - k_\mathrm{{off}}- \frac{{ \sigma _{1}D_\mathrm{m}}}{{ \sigma _{1}D_\mathrm{m}}+g_\mathrm{{off}}} k_\mathrm{{on}}\frac{{ k_{21}}+{ k_{22}}a_0^n}{1+{ k_{21}}+{ k_{22}}a_0^n}>0. \end{aligned}$$

Since $D^+>{ \sigma _{1}D_\mathrm{m}}+\frac{{ \sigma _{1}D_\mathrm{m}}}{{ \sigma _{1}D_\mathrm{m}}+g_\mathrm{{off}}}\frac{{ k_{21}}+{ k_{22}}a_0^n}{1+{ k_{21}}+{ k_{22}}a_0^n}$, inequality (69) implies inequality (26).

By the result obtained from Lemma 2, we have proved that if (68) is satisfied, then all possible homogeneous steady state solutions satisfy inequality (26). Since at least one homogeneous steady state solution satisfies (25), we have proved that if

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}+D^*}{k_6^{\frac{1}{n}}k_5^{\frac{n-1}{n}}}>\frac{k_5}{k_5-\frac{1}{n}\frac{k_\mathrm{{on}}^*}{k_\mathrm{{on}}^*+k_\mathrm{{off}}}D^++\frac{n-1}{n}\frac{k_\mathrm{{on}}^*}{k_\mathrm{{on}}^*+k_\mathrm{{off}}}k_\mathrm{{off}}}, \end{aligned}$$

there exists a homogeneous steady state solution satisfying (25) and (26). In addition, since all possible homogeneous steady state solutions satisfy (26) in this case, locally stable homogeneous steady state solution does not exist.$\square$

1.2.4 Theorem 4

Proof

Let $\gamma _1=k_5$, $\gamma _2=k_6$, $\gamma _3=\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}$, $\gamma _4=k_\mathrm{{off}}$ and $D=0$ in the lemmas. By applying Lemma 3, we obtain that if

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}}{k_6^{\frac{1}{n}}k_5^{\frac{n-1}{n}}}<\frac{k_5}{k_5+\frac{n-1}{n}\frac{k_\mathrm{{on}}^*}{k_\mathrm{{on}}^*+k_\mathrm{{off}}}k_\mathrm{{off}}}, \end{aligned}$$

then there exists a solution $a_0$ satisfying

$$\begin{aligned} (k_5+k_6a_0^{n})\left( 1-\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}a_0\right) - k_\mathrm{{off}}a_0 =0 \end{aligned}$$

and

$$\begin{aligned} nk_6a_0^{n-1} \left( 1-\frac{k_\mathrm{{on}}^*+k_\mathrm{{off}}}{k_\mathrm{{on}}^*}a_0\right) - k_\mathrm{{off}} <0. \end{aligned}$$

(72)

By (41), we know that $a_0$ is a homogeneous steady state solution for $a$ in the system (1)–(2) with the feedback form (39) if and only if $a_0$ satisfies (69).

By (71), (72) can be written as

$$\begin{aligned} \frac{nk_6a_0^{n-1}}{1+{ k_{21}}+{ k_{22}}a_0^n} (1-a_0)- k_\mathrm{{off}} <0, \end{aligned}$$

which implies that a homogeneous steady state solution is locally stable for perturbations with any nonnegative wavenumber (satisfying the condition (25) but not (26)). Thus, we have proved that if the condition

$$\begin{aligned} 1-\frac{(n-1)^{\frac{n-1}{n}}}{n}\frac{k_\mathrm{{off}}}{k_6^{\frac{1}{n}}k_5^{\frac{n-1}{n}}}<\frac{k_5}{k_5+\frac{n-1}{n}\frac{k_\mathrm{{on}}^*}{k_\mathrm{{on}}^*+k_\mathrm{{off}}}k_\mathrm{{off}}} \end{aligned}$$

is satisfied, there exists a locally stable homogeneous steady state solution.$\square$