Corrigendum: Dynamics of a susceptible—infected—susceptible epidemic reaction—diffusion model

Abstract

This corrigendum corrects a result in [1].

1. Uniqueness of EE

In Theorem 3.10 of [1], we claimed that the model has a unique EE if $\mathcal {R}_0>1$. However, there is an error in the proof of the case $d_S< d_I$. The correct statement and proof of Theorem 3.10 should be as follows. This correction has no impact on the results of the subsequent paper [2], which considered the same model.

Theorem If $\mathcal {R}_0>1$ and $d_S\ge d_I$, the model has a unique EE; if $\mathcal {R}_0>1$ and $d_S< d_I$, the model has an EE.

In the proof of Theorem 3.10, the following changes should be made:

• • Page 938, Line 2 from the bottom. ‘it suffices to show that problem (3.15), (3.16) has a unique positive solution’ should be ‘it suffices to study the positive solution of problem (3.15), (3.16)’.

• • Page 939, Line 9. ‘there exists a unique $\tau _0>0$’ should be ‘there exists $\tau _0>0$’.

We are able to prove the uniqueness of EE if $d_S< d_I$ with an additional assumption $N/|\Omega |\ge \gamma /\beta$. In the proof of the following result, for any $u, v\in C(\overline \Omega )$, we write $u\le v$ if $u(x)\le v(x)$ for all $x\in \overline \Omega$, $u<< v$ if $u(x)< v(x)$ for all $x\in \overline \Omega$ and $u< v$ if $u\le v$ but $u\neq v$.

Theorem Suppose $d_S< d_I$ and $N/|\Omega |\ge \gamma /\beta$. Then the EE of the model is unique.

Proof. By Proposition 3.2 and Remark 3.3, the assumption $N/|\Omega |\ge \gamma /\beta$ implies $\mathcal {R}_0>1$ and hence the model has at least one EE. To prove the uniqueness of the EE, by lemma 3.5, we only need to show that the positive solution of the following problem is unique:

(1.1)\begin{align} & d_I\Delta I + I\left( \frac{N}{|\Omega|}\beta-\gamma +d\frac{ \beta}{|\Omega|}\int_\Omega Idx-\frac{d_I \beta }{d_S} I \right)=0,\quad x\in\Omega,\end{align}

(1.2)\begin{align} & \frac{\partial I}{\partial n}=0, \quad x\in\partial\Omega, \end{align}

where $d=d_I/d_S-1$.

We first claim that for any positive solution $I$ of (1.1)–(1.2), it satisfies that

(1.3)\begin{equation} \frac{d_I}{d_S}I > \frac{d}{|\Omega|}\int_\Omega I dx. \end{equation}

To see this, we rewrite (1.1) as

(1.4)\begin{equation} d_I\Delta I + \beta I\left( \frac{N}{|\Omega|}-\frac{\gamma}{\beta} +\frac{ d}{|\Omega|}\int_\Omega Idx-\frac{d_I}{d_S} I \right)=0. \end{equation}

Let $I(x_0)=\min \{I(x): x\in \overline \Omega \}$ for some $x_0\in \overline \Omega$. Then similar to the proof of lemma 3.6, by using the maximum principle we can show

\[ \frac{d_I}{d_S}I(x_0) \ge \frac{N}{|\Omega|}-\frac{\gamma(x_0)}{\beta(x_0)}+\frac{d}{|\Omega|}\int_\Omega I dx. \]

Since $N/|\Omega |\ge \gamma /\beta$, we have

(1.5)\begin{equation} \frac{d_I}{d_S}I \ge \frac{d}{|\Omega|}\int_\Omega I dx. \end{equation}

If equality holds in (1.5), then $I$ is constant. However, since $d_I/d_S>d$, this is impossible. Thus the claim is valid.

Now suppose $I_1$ and $I_2$ are two distinct positive solutions of (1.1)–(1.2). Define

\[ k= \max\{\tilde k\ge 0: \tilde k I_1\le I_2 \}. \]

Interchanging $I_1$ and $I_2$ if necessary, we have $k\in (0, 1)$. Moreover, $kI_1 \le I_2$ and $kI_1(x_1)=I_2(x_1)$ for some $x_1\in \overline \Omega$.

Define a function $h:E\subset C_+(\overline \Omega ) \rightarrow C_+(\overline \Omega )$ by

\[ h(I)=(a-d_I\Delta)^{{-}1} I\left( a+ \frac{N}{|\Omega|}\beta-\gamma +d\frac{ \beta}{|\Omega|}\int_\Omega Idx-\frac{d_I \beta }{d_S} I\right) \]

for $I\in E$, where

\[ E=\left\{I\in C_+(\overline\Omega): \frac{d_I}{d_S}I \ge \frac{d}{|\Omega|}\int_\Omega I dx \text{ and } \|I\|\le C\right\}, \]

$a$ is a positive constant, and $C=\max \{\|I_i\|, i=1, 2\}$. We may choose $a$ large enough so that both $I_1$ and $I_2$ are fixed points of $h$. Moreover, since $d>0$, if $a$ is large, then $h$ is strictly increasing on $E$, i.e., $h(u) > > h(v)$ for $u> v$ and $u, v\in E$. For any $I\in E$, using (1.5), we can show $\tilde kh(I)\le h(\tilde kI)$ for any $\tilde k\in (0, 1)$. In addition, if (1.3) holds, then $\tilde kh(I)<< h(\tilde kI)$ for any $\tilde k\in (0, 1)$.

Hence, we have

\[ kI_1=kh(I_1)<{<} h(kI_1) \le h(I_2)=I_2, \]

which contradicts $kI_1(x_1)=I_2(x_1)$. This proves the uniqueness of the positive solution of (1.1)–(1.2).