Asymptotic Behavior of Eigen’s Quasispecies Model

Abstract

We study Eigen’s quasispecies model in the asymptotic regime where the length of the genotypes goes to $\infty$ and the mutation probability goes to 0. A limiting infinite system of differential equations is obtained. We prove convergence of trajectories, as well as convergence of the equilibrium solutions. We give analogous results for a discrete-time version of Eigen’s model, which coincides with a model proposed by Moran.

Appendix: Lemmas on Linear ODEs

We give here some lemmas concerning linear ODEs, and specially their longtime behavior.

Lemma A.1

Let $\alpha :[0,+\infty [\,\longrightarrow [0,+\infty [\,$ and $\beta :[0,+\infty [\,\longrightarrow {\mathbb {R}}$ be Lipschitz functions, and let $(z(t),t\ge 0)$ be the solution of the differential equation

$$\begin{aligned} z'(t) = \alpha (t)+\beta (t)z(t). \end{aligned}$$

If $z(0)\ge 0$, then $z(t)\ge 0$ for all $t\ge 0$.

Proof

The trajectory $(z(t),t\ge 0)$ is continuous. If there exists $t^*\ge 0$ such that $z(t^*)=0$, then

$$\begin{aligned} z'(t^*) = \alpha (t^*)\,\ge \,0, \end{aligned}$$

and thus $z(t)\ge 0$ for all $t\ge 0$. $\square$

Lemma A.2

Let $\alpha ,{\widetilde{\alpha }}:[0,+\infty [\,\longrightarrow [0,+\infty [\,$ and $\beta ,{\widetilde{\beta }}:[0,+\infty [\,\longrightarrow {\mathbb {R}}$ be Lipschitz functions satisfying

$$\begin{aligned} \forall \,t\ge 0,\quad \alpha (t) \le {\widetilde{\alpha }}(t),\quad \beta (t) \le {\widetilde{\beta }}(t). \end{aligned}$$

Let $(y(t),t\ge 0)$ and $(z(t),t\ge 0)$ be the solutions of the ODEs

$$\begin{aligned} y'(t) = \alpha (t)+\beta (t)y(t),\quad z'(t) = {\widetilde{\alpha }}(t)+{\widetilde{\beta }}(t)z(t). \end{aligned}$$

If $z(0)\ge y(0)\ge 0$, then $z(t)\ge y(t)$ for all $t\ge 0$.

Proof

We have

$$\begin{aligned} z'(t)-y'(t) = {\widetilde{\alpha }}(t)-\alpha (t)+\left( {\widetilde{\beta }}(t)-\beta (t)\right) z(t) +{\widetilde{\beta }}(t)\left( z(t)-y(t)\right) . \end{aligned}$$

From the previous lemma, $z(t)\ge 0$ for all $t\ge 0$. Thus, applying the previous lemma once again, $z(t)-y(t)\ge 0$ for all $t\ge 0$. $\square$

Lemma A.3

Let $\alpha ,\beta :[0,+\infty [\,\longrightarrow [0,+\infty [\,$ be Lipschitz functions, and suppose that there exist $\alpha ^*,\beta ^*\in \,]0,+\infty [\,$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty }\alpha (t) = \alpha ^*,\quad \lim _{t\rightarrow \infty }\beta (t) = \beta ^*. \end{aligned}$$

Let $(y(t),t\ge 0)$ be the solution of the differential equation

$$\begin{aligned} y'(t) = \alpha (t)-\beta (t)y(t). \end{aligned}$$

Then, for every initial condition $y(0)\in {\mathbb {R}}$,

$$\begin{aligned} \lim _{t\rightarrow \infty }y(t) = \frac{\alpha ^*}{\beta ^*}. \end{aligned}$$

Proof

Let $\varepsilon >0$ be small enough so that $\alpha ^*-\varepsilon ,\beta ^*-\varepsilon >0$. Let $T\ge 0$ be large enough so that

$$\begin{aligned} \forall \,t\ge T,\quad |\alpha (t)-\alpha ^*|<\varepsilon ,\quad |\beta (t)-\beta ^*|<\varepsilon . \end{aligned}$$

Let $({\underline{y}}(t),t\ge 0)$ and $({\overline{y}}(t),t\ge 0)$ be the solutions of the differential equations

$$\begin{aligned} {\underline{y}}'(t) = (\alpha ^*-\varepsilon )-(\beta ^*+\varepsilon ){\underline{y}}'(t),\quad {\overline{y}}'(t) = (\alpha ^*+\varepsilon )-(\beta ^*-\varepsilon ){\overline{y}}'(t), \end{aligned}$$

with ${\underline{y}}(0)= {\overline{y}}(0)=y(T)$. From the previous lemma, for all $t\ge 0$,

$$\begin{aligned} {\underline{y}}(t) \le y(T+t) \le {\overline{y}}(t). \end{aligned}$$

Yet, ${\underline{y}}(t)$ and ${\overline{y}}(t)$ converge:

$$\begin{aligned} \lim _{t\rightarrow \infty }{\underline{y}}(t) = \frac{\alpha ^*-\varepsilon }{\beta ^*+\varepsilon },\quad \lim _{t\rightarrow \infty }{\overline{y}}(t) = \frac{\alpha ^*+\varepsilon }{\beta ^*-\varepsilon }. \end{aligned}$$

We conclude that

$$\begin{aligned} \frac{\alpha ^*-\varepsilon }{\beta ^*+\varepsilon } \le \liminf _{t\rightarrow \infty } y(t) \le \limsup _{t\rightarrow \infty }y(t) \le \frac{\alpha ^*+\varepsilon }{\beta ^*-\varepsilon }. \end{aligned}$$

We send $\varepsilon$ to 0, and we obtain the desired result. $\square$