Modeling the Bistable Dynamics of the Innate Immune System

Abstract

The size of primary challenge with lipopolysaccharide induces changes in the innate immune cells phenotype between pro-inflammatory and pro-tolerant states when facing a secondary lipopolysaccharide challenge. To determine the molecular mechanisms governing this differential response, we propose a mathematical model for the interaction between three proteins involved in the immune cell decision making: IRAK-1, PI3K, and RelB. The mutual inhibition of IRAK-1 and PI3K in the model leads to bistable dynamics. By using the levels of RelB as indicative of strength of the immune responses, we connect the size of different primary lipopolysaccharide doses to the differential phenotypical outcomes following a secondary challenge. We further predict under what circumstances the primary LPS dose does not influence the response to a secondary challenge. Our results can be used to guide treatments for patients with either autoimmune disease or compromised immune system.

Appendix

Here, we investigate the positivity and boundness of the following system’s solutions [system (1)].

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t}&= \frac{c_x+a_x L}{b_x^m+y^m} - d_x x, \\ \frac{\mathrm{d}y}{\mathrm{d}t}&= \frac{c_y+a_y L}{b_y+ x} - d_y \frac{y}{b_v+y}, \\ \frac{\mathrm{d}z}{\mathrm{d}t}&= \frac{a_z y}{b_z+y} - d_z z \frac{x}{b_z+ y}, \\ \frac{\mathrm{d}L}{\mathrm{d}t}&= -d_L L. \end{aligned} \end{aligned}$$

(11)

Assume that $b_x^m=b_y=b_z=b_v=1$ and all other parameters $\text {parm}=\{a_x, d_x, c_y, a_y, d_y, a_z, d_z, d_L, m\}$ are positive. Let $w(t) := [x(t),y(t),z(t),L(t)]^\mathrm{T}$ be the solution vector and $u(t) := [x(t),y(t),z(t)]^\mathrm{T}$ be the vector that considers the first three variables. Consider the following functions

$$\begin{aligned}&g:\mathbb {R}_+^4 \rightarrow \mathbb {R}^4 \,\text {given by}\nonumber \\&\quad g(t,w) = \begin{pmatrix} g_1\\ g_2\\ g_3\\ g_4 \end{pmatrix} = \begin{pmatrix} \frac{c_x+a_x L}{b_x^m+ y^m} - d_x x, \\ \frac{c_y+a_y L}{b_y+ x} - d_y \frac{y}{b_v+y} \\ \frac{a_z y}{b_z+y} - d_z z \frac{x}{b_z+y} \\ -d_L L \end{pmatrix}, \end{aligned}$$

(12)

$$\begin{aligned}&f: \mathbb {R}_+^3\rightarrow \mathbb {R}^3 \text {given by}\nonumber \\&\quad f(t,u) = \begin{pmatrix} g_1\\ g_2\\ g_3 \end{pmatrix}, \end{aligned}$$

and the following initial value problem

$$\begin{aligned} \begin{aligned}&w'(t) = g(t,w), \\&\qquad \text {subject to}\\&x(0) = x_0>0, \ y(0) = y_0>0, \ z(0)= z_0>0, \ L(0) = L_0>0. \end{aligned} \end{aligned}$$

(13)

We want to show that the solutions of the initial value problem (13) are positive and bounded. The fourth variable L(t) yields

$$\begin{aligned} L(t) = L_0 e^{-d_L t}\in (0,L_0], \quad \text {for all } t \in [0,\infty ). \end{aligned}$$

(14)

The initial value problem (13) reduces to

$$\begin{aligned} \begin{aligned}&u'(t) = f(t,u ),\\&\qquad \text { subject to }\\&x(0) = x_0>0, \ y(0) = y_0>0, \ z(0)= z_0 >0, \end{aligned} \end{aligned}$$

(15)

where $L(t) = L_0 e^{-d_L t}$.

Proposition 1

There exists a positive number $\beta >0$ such that system (13) has a unique positive solution on $[0, \beta )$.

Proof

Since f is continuously differentiable on $\mathbb {R}_+^3$, it is locally Lipschitz on $\mathbb {R}_3^+$. By Thm. 2, there exists a maximal value $\beta >0$ such that (15) has a unique solution on the interval $[0, \beta )$ with values in $\mathbb {R}_+^3$. $\square$

Proposition 2

The solution of (15) exists and is positive in $\mathbb {R}_+^3$. Furthermore, if $c_y < d_y$ the solution is bounded.

Proof

Assume that $\beta$ found in Proposition 1 is finite. Since $x(t) ,y(t) ,z(t) ,L(t) >0$ and L is decreasing on $[0,\beta )$, the following inequalities hold for all $t \in [0, \beta )$

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{c_x + a_x L}{1+y^m} - d_xx< c_x + a_x L_0, \\&\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{c_y + a_yL}{1+x}-d_y\frac{y}{1+y}< c_y + a_y L_0,\\&\frac{\mathrm{d}z}{\mathrm{d}t} = a_z \frac{y}{1+y} - d_z\frac{xz}{1+y} < a_z. \end{aligned} \end{aligned}$$

(16)

This yields

$$\begin{aligned} \begin{aligned}&x(t)< x_0 + (c_x+a_x L_0) t< x_0 + (c_x+a_x L_0) \beta :=x_\mathrm{max}, \\&y(t)< y_0 + (c_y + a_y L_0) t< y_0 + (c_y + a_y L_0) \beta :=y_\mathrm{max},\\&z(t)< z_0 + a_z t < z_0 + a_z \beta :=z_\mathrm{max}, \end{aligned} \end{aligned}$$

(17)

for all $t \in [0,\beta )$. Thus, x, y and z are bounded from above on $[0,\beta )$. Using

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{c_x+a_x L}{1+y^m} - d_xx \ge \frac{c_x}{1+y_\mathrm{max}^m} - d_xx,\\&\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{c_y + a_yL}{1+x}-d_y\frac{y}{1+y} \ge \frac{c_y }{1+x_\mathrm{max}}-d_y y, \end{aligned} \end{aligned}$$

(18)

we find that if $x \le \frac{c_x}{d_x (1+y_\mathrm{max}^m)}$, then $\frac{\mathrm{d}x}{\mathrm{d}t} \ge 0$. Similarly, if $y \le \frac{c_y}{d_y(1+x_\mathrm{max})}$ then $\frac{\mathrm{d}y}{\mathrm{d}t} \ge 0$. Therefore

$$\begin{aligned} x(t) \ge \min \left\{ x_0, \frac{c_x}{d_x (1+y_\mathrm{max}^m)}\right\} := x_\mathrm{min} >0, \quad \text {for all } t \in [0,\beta ), \end{aligned}$$

(19)

and

$$\begin{aligned} y(t) \ge \min \left\{ y_0, \frac{c_y}{d_y(1+x_\mathrm{max})}\right\} := y_\mathrm{min} >0, \quad \text {for all } t \in [0,\beta ). \end{aligned}$$

(20)

Lastly, since

$$\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{a_z y}{1+y} - d_z \frac{xz}{1+y} \ge \frac{a_z y_\mathrm{min}}{1+y_\mathrm{max}} - d_z \frac{x_\mathrm{max}}{1+y_\mathrm{min}} z, \end{aligned}$$

(21)

for all $t \in [0,\beta )$, we have that if $z \le (\frac{a_z y_\mathrm{min}}{1+y_\mathrm{max}})/(d_z \frac{x_\mathrm{max}}{1+y_\mathrm{min}})$ then $\frac{\mathrm{d}z}{\mathrm{d}t} \ge 0$. Thus

$$\begin{aligned} z(t) \ge \min \left\{ z_0, \left( \frac{a_z y_\mathrm{min}}{1+y_\mathrm{max}}\right) /\left( d_z \frac{x_\mathrm{max}}{1+y_\mathrm{min}}\right) \right\} := z_\mathrm{min} >0, \quad \text { for all } t \in [0,\beta ). \end{aligned}$$

(22)

Therefore, x, y and z are bounded from below on $[0,\beta )$. If $\beta$ is finite, there are positive lower and upper bounds for x, y and z on $[0, \beta )$, i.e., $u = [x,y,z]^\mathrm{T}$ is bounded on $[0,\beta )$. Since f is continuous, f(u) is bounded on $[0,\beta )$. By $3^{\circ }$ of Thm. 2 (Bourbaki 2004) $\lim _{t \rightarrow \beta } x(t) = 0$, or $\lim _{t \rightarrow \beta } y(t) = 0$, or $\lim _{t \rightarrow \beta } z(t) = 0$. This contradicts the positive lower bounds of x, y and z. Thus $\beta = \infty$.

It remains to show that for $c_y < d_y$, the solutions are bounded on $[0,\infty )$. We know that $x,y,z >0$ for $t \in [0,\infty )$. We have

$$\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{c_x + a_xL}{1+y^m}-d_xx \le c_x + a_x L_0 - d_x x,\quad \text {for all } t \ge 0, \end{aligned}$$

(23)

which yields that if $x \ge \frac{c_x + a_x L_0}{d_x}$ then $\frac{\mathrm{d}x}{\mathrm{d}t} \le 0$. Hence

$$\begin{aligned} x(t) \le \max \left\{ x_0, \frac{c_x + a_x L_0}{d_x}\right\} =:x_\mathrm{max}^1, \quad \text {for all } t>0. \end{aligned}$$

(24)

Since $\lim _{t\rightarrow \infty } L(t) =0$ and $c_y<d_y$, there exists a constant $t_1>0$ such that

$$\begin{aligned} c_y + a_y L(t) < d_y, \end{aligned}$$

(25)

and

$$\begin{aligned} L(t) \le L(t_1) =:L_1, \end{aligned}$$

(26)

for $t > t_1$. Then

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{c_y + a_yL}{1+x}-d_y\frac{y}{1+y} \le c_y + a_y L_1 - d_y \frac{y}{1+y},\quad \text {for all } t \ge t_1. \end{aligned}$$

(27)

Therefore, using $c_y + a_y L_1< d_y$ we obtain that for $t \in [t_1,\infty )$, if $y \ge \frac{c_y + a_y L_1}{d_y - c_y - a_y L_1}$, then $\frac{\mathrm{d}y}{\mathrm{d}t} \le 0$. Furthermore, y is continuous, therefore bounded on the closed interval $[0,t_1]$. Thus, for $t \in [0,\infty )$

$$\begin{aligned} y(t) \le \max \left\{ \max _{t \in [0,t_1]} y(t), \frac{c_y + a_y L_1}{d_y - c_y - a_y L_1} \right\} =:y_\mathrm{max}^1. \end{aligned}$$

(28)

As in Eq. (19)

$$\begin{aligned} x(t) \ge \min \left\{ x_0, \frac{c_x}{d_x \left( 1+{y^1_\mathrm{max}}^m\right) }\right\} =:x_\mathrm{min}^1>0,\quad \text {for all } t>0. \end{aligned}$$

(29)

As in Eq. (20), we have

$$\begin{aligned} y(t) \ge \min \left\{ y_0, \frac{c_y}{d_y\left( 1+x_\mathrm{max}^1\right) }\right\} = :y_\mathrm{min}^1>0, \quad \text {for all } t>0. \end{aligned}$$

(30)

To find an upper bound of z we use

$$\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{a_z y}{1+y} - d_z \frac{zx}{1+y} \le a_z - d_z \frac{x_\mathrm{min}^1}{1+y_\mathrm{max}^1} z. \end{aligned}$$

(31)

This yields

$$\begin{aligned} z(t) \le \max \left\{ z_0, \frac{a_z}{d_z \frac{x_\mathrm{min}^1}{1+y_\mathrm{max}^1}} \right\} =:z_\mathrm{max}^1, \quad t \ge 0. \end{aligned}$$

(32)

Lastly, to find a lower bound for z(t) we use

$$\begin{aligned} z(t) \ge \min \left\{ z_0, \left( \frac{a_z y_\mathrm{min}^1}{1+y_\mathrm{max}^1}\right) /\left( d_z \frac{x_\mathrm{max}^1}{1+y_\mathrm{min}^1}\right) \right\} = :z_\mathrm{min}^1 >0, \quad \text {for all } t \ge 0. \end{aligned}$$

(33)

Thus, we have shown that there is a unique solution of (13) on $[0,\infty )$ that is positive and bounded. $\square$

Proposition 3

An equilibrium solution of system (1) with $b_x=b_y=b_z=1$ is locally asymptotically stable if and only if $\frac{AB}{m}>\bar{y}^{m+1} \bar{x}^2$, where $A= \frac{c_y}{d_y}$ and $B = \frac{c_x}{d_x}$.

Proof

Let $(\bar{x}, \bar{y}, \bar{z}, \bar{L})$ be an equilibrium solution of system (1). From (14) it follows that $\lim _{t \rightarrow \infty } L(t) = 0$, therefore we set $\bar{L}=0$. Further, it follows from the proof of Proposition 2 that $\bar{x}, \bar{y}, \bar{z} >0$. The Jacobian of system (1) evaluated at $(\bar{x}, \bar{y}, \bar{z}, \bar{L})$ is given by

$$\begin{aligned} J = \begin{pmatrix} -d_x &{}\quad -\frac{c_x m \bar{y}^{m+1}}{(1+\bar{y}^m)^2} &{}\quad 0 &{}\quad -\frac{a_x}{1+\bar{y}^m}\\ -\frac{c_y}{(1+\bar{x})^2} &{}\quad -\frac{d_y}{(1+\bar{y})^2} &{}\quad 0 &{}\quad \frac{a_y}{1+\bar{x}} \\ - \frac{d_z \bar{z}}{1+\bar{y}} &{}\quad \frac{a_z}{(1+\bar{y})^2}+\frac{d_z \bar{z} \bar{x}}{(1+\bar{y})^2} &{}\quad -\frac{d_z \bar{x}}{1+\bar{y}} &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{} \quad -d_L \end{pmatrix}. \end{aligned}$$

(34)

Two eigenvalues of J are given by $\lambda _1 = -d_L<0$ and $\lambda _2 = -\frac{d_z \bar{x}}{1+\bar{y}}<0$. The remaining two eigenvalues $\lambda _3$ and $\lambda _4$ satisfy the equation

$$\begin{aligned} (\lambda +a)(\lambda +d)-cb=0, \end{aligned}$$

(35)

where $a=d_x, \ b=\frac{c_x m \bar{y}^{m-1}}{(1+\bar{y}^m)^2}, \ c=\frac{c_y}{(1+\bar{x})^2}$, and $d=\frac{d_y}{(1+\bar{y})^2}$. Since $a, b, c, d>0$ this implies that

$$\begin{aligned} \lambda _{3,4} = -\frac{a+d}{2} \pm \sqrt{\frac{(a+d)^2}{4}-(ad-bc)} \end{aligned}$$

(36)

have negative real parts iff $ad>bc$, which is equivalent to

$$\begin{aligned} \begin{aligned}&\frac{d_x d_y}{(1+\bar{y})^2}>\frac{c_x c_y m \bar{y}^{m-1}}{(1+\bar{y}^m)^2(1+\bar{x})^2}\\&\quad \Longleftrightarrow \frac{d_x d_y}{(1+\bar{y})^2}> c_y c_x m \bar{y}^{m-1} \left( \frac{1}{1+ \bar{y}^m} \right) ^2 \left( \frac{1}{1+ \bar{x}} \right) ^2\\&\quad \Longleftrightarrow \frac{d_x d_y}{(1+\bar{y})^2}> c_y c_x m \bar{y}^{m-1} \left( \frac{d_x}{c_x} \bar{x} \right) ^2 \left( \frac{d_y}{c_y} \frac{\bar{y}}{1+ \bar{y}} \right) ^2\\&\quad \Longleftrightarrow \frac{c_x c_y}{d_x d_y m}> \bar{y}^{m+1} \bar{x}^2\\&\quad \Longleftrightarrow \frac{A B}{m} > \bar{y}^{m+1} \bar{x}^2. \end{aligned} \end{aligned}$$

(37)

We have shown that all eigenvalues of J have negative real part, hence an equilibrium is locally asymptotically stable, iff $\frac{A B}{m} > \bar{y}^{m+1} \bar{x}^2$. $\square$

Proposition 4

If $0<m \le 1$ then system (1) has at most two positive equilibria.

Proof

We find that if $E=(\bar{x}, \bar{y}, \bar{z}, \bar{L})$ is an equilibrium of system (1), then it satisfies

$$\begin{aligned} \begin{aligned}&\bar{L} =0, \\&\bar{x} = \frac{B}{1+\bar{y}^m} ,\\&\bar{z} = \frac{a_z}{d_z} \cdot \frac{\bar{y}}{\bar{x}}, \end{aligned} \end{aligned}$$

(38)

where $\bar{y}$ satisfies the equation

$$\begin{aligned} \frac{B}{1+\bar{y}^m} - \frac{A(1+\bar{y})-\bar{y}}{\bar{y}}= 0, \end{aligned}$$

and $A = \frac{c_y}{d_y}$ and $B= \frac{c_x}{d_x}$. Therefore system (1) has as many positive equilibria as there are roots of the function $g(y) = \frac{B}{1+\bar{y}^m} - \frac{A(1+\bar{y})-\bar{y}}{\bar{y}}$ in $(0, \infty )$.

We find $g(y) = \frac{By-[A(1+y)-y](1+y^m)}{(1+y^m)y}$. Hence, $g(y)=0$ if and only if $f(y)=0$, where $f(y)= By-[A(1+y)-y](1+y^m)$ is the numerator of g(y). Expanding f(y) and taking terms with the same powers of y together yields

$$\begin{aligned} f(y) = y^{m+1}(1-A)-Ay^m+(B-A+1)y-A. \end{aligned}$$

Note that $A = \frac{c_y}{d_y} <1$ because $c_y < d_y$. f is a smooth function on $(0, \infty )$. Hence, for f to have at least three roots in $(0,\infty )$ its second derivative needs to have a root in $(0,\infty )$. Using

$$\begin{aligned} \begin{aligned} f''(y)&= (1-A)(m+1)my^{m-1}-Am(m-1)y^{m-1}\\&=y^{m-2}[(1-A)(m+1)my-Am(m-1)]=0\\ \Longleftrightarrow y&= \frac{Am(m-1)}{(1-A)(m+1)m} \end{aligned} \end{aligned}$$

Since $A \in (0,1)$, the equation $f''(y)=0$ has one solution in $(0,\infty )$ if $m>1$ and no solution in that interval otherwise. Therefore, f and hence g can have at most three solutions in $(0,\infty )$ if $m>1$ and at most two solution in $(0,\infty )$ if $m \le 1$. $\square$

Proposition 5

If $m >1$ then system (1) has

• 1. either exactly one locally asymptotically stable equilibrium or

• 2. exactly two locally asymptotically stable and one unstable equilibrium or

• 3. exactly one locally asymptotically stable equilibrium and one equilibrium that is not locally asymptotically stable.

Proof

Let g be defined as in proposition 4. In the proof of proposition 4, we have shown that g has at most three roots in $(0, \infty )$. Since g is a smooth function on $(0, \infty )$, $\lim _{y \rightarrow 0^+} g(y)= -\infty$ and $\lim _{y \rightarrow \infty } g(y) = \infty$, we find that g has

• 1. either one root $y_1 \in (0,\infty )$ with $g'(y_1)>0$ and no other roots in $(0,\infty )$ or

• 2. three distinct roots $y_1, y_2, y_3 \in (0,\infty )$ with $y_1< y_2< y_3$ and $g'(y_1), g'(y_3)>0$ and $g'(y_2)<0$ and no other roots in $(0,\infty )$ or

• 3. one root $y_1 \in (0,\infty )$ with $g'(y_1)>0$ and one root $y_2 \in (0,\infty )$ with $g'(y_2)=0$ and no other roots in $(0,\infty )$.

We find

$$\begin{aligned} g'(y) = \frac{-Bmy^{m+1}+A(1+y^m)^2}{[(1+y^m)y]^2}, \end{aligned}$$

and therefore $g'(y)>0$ if and only if

$$\begin{aligned} A(1+y^m)^2 > Bmy^{m+1}. \end{aligned}$$

(39)

for y in $(0,\infty )$. Let $\bar{y}$ be a root of g. Then, $\bar{y}$ is the y-value of an equilibrium of system (1) and the corresponding x-value is given by $\bar{x} = \frac{B}{1+\bar{y}^m}$. Using this in (39) we obtain

$$\begin{aligned} \frac{AB}{m}>\bar{y}^{m+1} \bar{x}^2, \end{aligned}$$

(40)

which implies stability of the equilibrium corresponding to the root $\bar{y}$ of g. Similarly, we can show that if $\bar{y}$ is a root of g with $g'(\bar{y})<0$, then the equilibrium defined by $\bar{y}$ is unstable. This implies that statements (1)–(3) are equivalent to the three statements in the formulation of the proposition. $\square$