A Mathematical Model of the Inflammatory Response to Pathogen Challenge

Abstract

The human body’s immune response to bacterial challenge, even when successful in controlling the infection, can result in negative consequences for the host, including reduced functionality of associated tissues. We present and analyze a low-dimensional mathematical model of this immune response to pathogen invasion, incorporating the coordinated actions of active immune cells, and both pro- and anti-inflammatory cytokines. The model simulates both the positive (pathogen reduction) and negative (local tissue dysfunction) effects of the immune response and includes the important role of immunologic memory in the process of a return to stasis. This differential equation-based model is sufficiently general to be applicable to a wide range of human tissues and organs.

Appendix

Proof of Theorem 4

Letting $\epsilon >0$, we will use the shorthand notation $P_\epsilon \equiv P_{\mathrm{max}}-\epsilon$, $\phi _\epsilon \equiv \phi (P_\epsilon )$, $\delta _\epsilon \equiv \delta (P_\epsilon )$, $\psi _\epsilon \equiv \psi (P_\epsilon ,0)$, and $\beta _\epsilon \equiv \beta (P_\epsilon ,0)$.

Note that $P_{\mathrm{null}}(P)$ attains an absolute minimum $M_\epsilon >0$ on $0\le P\le P_\epsilon$. By Property 3(b) of Lemma 3, it suffices to demonstrate the existence of $q>0$ and $c>0$ for which

$$\begin{aligned} I_+(P_\epsilon ,0) < M_\epsilon . \end{aligned}$$

To this end, we can choose q to satisfy $\displaystyle {0< q < \frac{r}{\lambda ^{\frac{3}{2}}}\left( M_\epsilon +B\right) }$, from which it follows that

$$\begin{aligned} \frac{\sqrt{\lambda }q}{M_\epsilon +B} < \frac{r}{\lambda }, \end{aligned}$$

and, consequently,

$$\begin{aligned} -\frac{\delta _\epsilon }{M_\epsilon +B} = \frac{\sqrt{\lambda }q}{M_\epsilon +B}\frac{P_\epsilon }{K+P_\epsilon } < \frac{r}{\lambda } + ue^{-\chi P_\epsilon } = \phi _\epsilon , \end{aligned}$$

so that

$$\begin{aligned} \phi _\epsilon + \frac{\delta _\epsilon }{M_\epsilon +B} > 0, \end{aligned}$$

thus making it possible to choose c to satisfy $\displaystyle {0< c < \frac{1}{\sqrt{\lambda }}\left( \frac{K+P_\epsilon }{P_\epsilon }\right) M_\epsilon \left( \phi _\epsilon + \frac{\delta _\epsilon }{M_\epsilon +B}\right) }$. Then, we have

$$\begin{aligned} -\frac{\psi _\epsilon }{B}&= \sqrt{\lambda }c\frac{P_\epsilon }{K+P_\epsilon }< M_\epsilon \left( \phi _\epsilon + \frac{\delta _\epsilon }{M_\epsilon +B}\right) ,\\&\implies -\psi _\epsilon \left( \frac{M_\epsilon }{B}+1\right)< M_\epsilon \phi _\epsilon (M_\epsilon +B) + M_\epsilon \delta _\epsilon ,\\&\implies -\psi _\epsilon< M_\epsilon \left( \phi _\epsilon M_\epsilon + \phi _\epsilon B + \frac{\psi _\epsilon }{B} + \delta _\epsilon \right) = \phi _\epsilon M_\epsilon ^2 + \beta _\epsilon M_\epsilon ,\\&\implies \beta _\epsilon ^2-4\phi _\epsilon \psi _\epsilon < 4\phi _\epsilon ^2 M_\epsilon ^2 + 4\phi _\epsilon \beta _\epsilon M_\epsilon + \beta _\epsilon ^2 = \left( 2\phi _\epsilon M_\epsilon + \beta _\epsilon \right) ^2. \end{aligned}$$

Taking positive square roots yields

$$\begin{aligned} \sqrt{\beta _\epsilon ^2-4\phi _\epsilon \psi _\epsilon } < 2\phi _\epsilon M_\epsilon + \beta _\epsilon , \end{aligned}$$

so that

$$\begin{aligned} I_+(P_\epsilon ,0) = \frac{-\beta _\epsilon + \sqrt{\beta _\epsilon ^2-4\phi _\epsilon \psi _\epsilon }}{2\phi _\epsilon } < M_\epsilon , \end{aligned}$$

as desired. $\square$