Proof. First, we suppose that the solution exists on some maximal interval $[0, t_{max})$ , $t_{\max }\in (0,\infty ]$ and prove the boundedness of it. Let $M_1=\max \{\|S_0\|_{L^\infty (\Omega )}, \ r_M\}$ . By the first equation of (3.1), $S(x, t)\le M_1$ for all $x\in \bar \Omega$ and $0\le t\lt t_{\max }$ . Since $\int _\Omega Idx\le \int _\Omega (S+I)dx\le N$ for all $t\in [0,t_{\max })$ , by [Reference Alikakos1, Theorem 3.1] (or Lemma 2.3), there exists $M_2\gt 0$ depending on $\|I_0\|_{L^\infty (\Omega )}$ and $M_1$ such that $\|I(\cdot, t)\|_{L^\infty (\Omega )}\le M_2$ for all $0\le t\lt t_{\max }$ . This proves (3.2).

The integral form of (3.1) is

(3.3) \begin{equation} \begin{cases} S(\cdot, t)=e^{-at}S_0+\int _0^t e^{-a(t-s)}(aS(\cdot,s)-\beta S(\cdot, s)I(\cdot, s)+\gamma I(\cdot, s))ds, &t\gt 0,\\[6pt] I(\cdot, t)=e^{t(d_I\Delta -a)}I_0+\int _0^t e^{(t-s)(d_I\Delta -a)}(a+\beta S(\cdot, s)-\gamma )I(\cdot, s)ds,&t\gt 0, \end{cases} \end{equation}

where $a\gt 0$ is chosen to be sufficiently large. Using Banach fixed point theory, one can show that (3.3) has a unique nonnegative solution $(S, I)\in [C([0, T], C(\overline{\Omega }))]^2$ for some $T\gt 0$ . By the first equation of (3.3), $S\in C^1([0, T], C(\overline{\Omega }))$ . By [Reference Pazy44, Theorem 4.3.1 and Corollary 4.3.3], $I\in C([0, T], C(\overline{\Omega }))\cap C^1((0, T], \textrm{Dom}_{\infty }(\Delta ))$ . The a priori bound (3.2) enables us to extend the solution globally.

Proof. Define

\begin{equation*} J(x,t)=\int _0^{t}I(x,\tau )d\tau,\quad \forall \ (x, t)\in \bar {\Omega }\times [0, \infty ). \end{equation*}

It is clear that $J(x,t)$ is strictly monotone increasing in $t$ for each $x\in \bar{\Omega }$ . Hence, we can define

\begin{equation*} \hat{\!J}(x)\,:\!=\,\int _{0}^{\infty }I(x,\tau )d\tau =\lim _{t\to \infty }J(x,t) \in (0,\infty ],\quad \forall \ x\in \bar {\Omega }. \end{equation*}

Now, we distinguish two cases.

Case 1. $\hat{\!J}(x_0)\lt \infty$ for some $x_0\in \overline{\Omega }$ . By (3.2), we have $\sup _{t\ge 0}\|\beta S(\cdot,t)-\gamma \|_{\infty }\lt \infty$ . So by Lemma 2.3, there is a positive constant $c_1$ such that

(3.5) \begin{equation} I_M(\cdot,t)\le c_1I_{m}(\cdot,t),\quad \forall \ t\ge 1. \end{equation}

It follows that

\begin{equation*} \hat{\!J}(x)=J(x,1)+\int _{1}^{\infty }I(x,\tau )d\tau \le J(x,1)+c_1\int _{1}^{\infty }I(x_0,\tau )d\tau \le J(x,1)+c_1J(x_0),\quad \forall \ x\in \bar {\Omega }. \end{equation*}

This shows that $\hat{\!J}(x)\lt \infty$ for all $x\in \bar{\Omega }$ and $\|\hat{\!J}\|_{L^{\infty }(\Omega )}\lt \infty$ . Moreover, by (3.5),

\begin{equation*} \|\hat{\!J}-J(\cdot,t)\|_{L^{\infty }(\Omega )}\le c_1\int _{t}^{\infty }I(x_0,\tau )d\tau \to 0 \quad \text{as}\quad t\to \infty. \end{equation*}

Hence, $\hat{\!J}\in C(\bar{\Omega })$ . Next, observing that

\begin{equation*} \partial _t(S-r)=\partial _tS=-\beta (S-r)I, \qquad \forall \ (x, t)\in \bar \Omega \times (0, \infty ), \end{equation*}

we have

(3.6) \begin{equation} S(x,t)-r(x)=(S_0(x)-r(x))e^{-\beta (x)J(x,t)},\quad \forall \ (x, t)\in \bar \Omega \times (0, \infty ). \end{equation}

This implies

\begin{equation*} S(\cdot, t)\to S^*\,:\!=\,r+e^{-\beta \hat{\!J}}(S_0-r) \ \text{uniformly on}\ \bar \Omega \ \text{as}\ t\to \infty. \end{equation*}

Let $\lambda ^*=e^{-\beta \hat{\!J}}$ . Then $0\lt \lambda ^*\lt 1$ , $\lambda ^*\in C(\bar{\Omega })$ and $S^*=\lambda ^*S_0+(1-\lambda ^*)r$ . Let $\varphi ^*\gt 0$ be the eigenfunction associated with $\sigma (d_I,\beta (S^*-r))$ satisfying $\varphi ^*_{M}=1$ . Then it holds that

\begin{align*} \frac{d}{dt}\int _{\Omega }\varphi ^*Idx&=d_I\int _{\Omega }\varphi ^*\Delta Idx +\int _{\Omega }\beta (S-r)\varphi ^*Idx\\[5pt] =&\sigma (d_I,\beta (S^*-r))\int _{\Omega }\varphi ^*Idx+\int _{\Omega }\beta (S-S^*)\varphi ^*Idx\\[5pt] \ge & \Big (\sigma (d_I,\beta (S^*-r))-\beta _M\|S-S^*\|_{L^{\infty }(\Omega )}\Big )\int _{\Omega }\varphi ^*Idx. \end{align*}

Hence, we have

\begin{equation*} N\ge \int _{\Omega }\varphi ^*I(x,t)dx\ge e^{\int _{0}^t(\sigma (d_I,\beta (S^*-r))-\beta _M\|S-S^*\|_{L^{\infty }(\Omega )})d\tau }\int _{\Omega }\varphi ^*I_0dx,\quad \forall \ t\gt 0. \end{equation*}

This implies that

\begin{equation*} \frac {\beta _M}{t}\int _{0}^{t}\|S(\cdot,\tau )-S^*\|_{L^{\infty }(\Omega )}d\tau +\frac {\ln\! (N)-\ln\! (\|\varphi ^*I_0\|_{L^1(\Omega )})}{t}\ge \sigma (d_I,\beta (S^*-r)),\quad t\gt 0. \end{equation*}

Observing that the left-hand side of this inequality tends zeo as $t\to \infty$ , we conclude that $\sigma (d_I,\beta (S^*-r))\le 0$ .

Finally, by (3.2) and the parabolic estimates and the Sobolev emdedding theorem, $I$ is Hölder continuous on $\bar \Omega \times [1, \infty )$ . Hence, by Lemma 2.2 and the fact that $\hat{\!J}(x_0)\lt \infty$ , we obtain from (3.5) that

\begin{equation*} \|I(\cdot,t)\|_{L^\infty (\Omega )}\le c_1I(x_0,t)\to 0\quad \text{as} \quad t\to \infty, \end{equation*}

which in turn implies that

\begin{equation*}\int _{\Omega }S^*dx=\lim _{t\to \infty }\int _{\Omega }Sdx=\lim _{t\to \infty }\int _{\Omega }(S+I)dx=N.\end{equation*}

This completes the proof of (i).

Case 2. $\hat{\!J}(x)=\infty$ for all $x\in \bar{\Omega }$ . Fix $x_1\in \bar \Omega$ . Then $\int _1^{t}I(x_1,\tau )d\tau \to \infty$ as $t\to \infty$ . This together with (3.5)–(3.6) implies that

\begin{equation*} \|S(\cdot,t)-r\|_{L^{\infty }(\Omega )}\le \|S_0-r\|_{L^{\infty }(\Omega )}e^{-\frac {\beta _m}{c_1}\int _1^{t}I(x_1,\tau )d\tau }\to 0\quad \text{as}\quad t\to \infty. \end{equation*}

As a result,

(3.7) \begin{equation} \lim _{t\to \infty }\int _{\Omega }Idx=N-\int _{\Omega }rdx. \end{equation}

This shows that, in the current case, we must have that $N\ge \int _{\Omega }rdx$ . By (3.2), the parabolic estimates and the Sobolev embedding theorem, the orbit $\{I(\cdot,t)\}_{t\ge 1}$ is precompact in $C(\bar \Omega )$ . Noticing $S(\cdot, t)\to r$ in $C(\bar \Omega )$ as $t\to \infty$ , the $\omega$ limit set $\omega (S_0, I_0)=\cap _{t\ge 1} \overline{\cup _{s\ge t}\{(S(\cdot, t), I(\cdot, s))\}}$ is well defined, where the completion is in $[C(\bar \Omega )]^2$ . Fix $(S^*,I^*)\in \omega (S_0,I_0)$ . Since $S(\cdot,t)\to r$ in $C(\bar{\Omega })$ as $t\to \infty$ , then $S^*=r^*$ . Next, we show that $I^*(x)=(N-\int _{\Omega }r)/|\Omega |$ for all $x\in \Omega$ . To this end, since $\omega (I_0,S_0)$ is invariant under the semiflow of the solution operator induced by (3.1) and the orbit $\{I(\cdot,t)\}_{t\ge 1}$ is precompact in $C(\bar{\Omega })$ , we can employ standard parabolic regularity arguments to the equation of $I(\cdot,t)$ , and coupled with the fact that $S(\cdot,t)\to r$ in $C(\bar{\Omega })$ as $t\to \infty$ , to conclude that there is a bounded entire solution $(\tilde{S}(x,t),\tilde{I}(x,t))$ of (3.1) fulfilling $\tilde{I}(\cdot,0)=I^*$ and $\tilde{S}(\cdot,t)=r$ for all $t\in \mathbb{R}$ . Hence, $\tilde{I}(x,t)$ satisfies

\begin{equation*} \begin {cases} \partial _t\tilde {I}=d_I\Delta \tilde {I}, & x\in \Omega, \ t\in \mathbb {R},\\ \partial _{\nu }\tilde {I}=0, & x\in \partial \Omega,\ t\in \mathbb {R},\\ \tilde {I}(x,0)=I^*(x), & x\in \bar {\Omega }. \end {cases} \end{equation*}

So, $\tilde{I}(\cdot,t)=\int _{\Omega }I^*(x)dx/|\Omega |$ for all $t\in \mathbb{R}$ . However by (3.7), $\int _{\Omega }I^*dx=\big(N-\int _{\Omega }rdx\big)/|\Omega |$ . Hence, $I^*$ is the constant function $\big(N-\int _\Omega r dx\big)/|\Omega |$ . This shows that $\omega (S_0,I_0)=\big\{r,\big(N-\int _{\Omega }rdx\big)/|\Omega |\big\}$ and completes the proof of (ii).

It is easy to see that if $N\le \int _\Omega rdx$ , then (i) holds. Note that if (i) holds then $N=\int _{\Omega }(\lambda ^*S_0+(1-\lambda ^*)r)dx$ for some $\lambda ^*\in C(\overline{\Omega })$ satisfying $0\lt \lambda ^*\lt 1$ and $\sigma (d_I,\beta \lambda ^*(S_0-r))\le 0$ . Hence, we must have $N\le N^*_{S_0,r}$ . So alternative (ii) must hold whenever $N\gt N^*_{S_0,r}$ .

Proof. (1) Taking $\lambda ^*\equiv 0$ in (3.4), we have the desired result.

(2) For any $\lambda ^* \in C(\bar \Omega ;\, [0, 1])$ , we have $\int _\Omega (\lambda ^*S_0+(1-\lambda ^*)r)dx\le \max \{S_0, r\}$ . By the definition of $N^*_{S_0,r}$ , we have $N^*_{S_0,r}\leq \int _{\Omega }\max \{S_0,r\}dx$ .

(3) Suppose that $\|(S_0-r)_+\|_{L^\infty (\Omega )}\gt 0$ and we prove $N^*_{S_0,r}\lt \int _{\Omega }\max \{S_0,r\}dx$ by contradiction. Suppose to the contrary that there is $\lambda ^*_k\in C(\overline{\Omega };\,[0,1])$ satisfying ${\sigma \big(d_I,\beta \lambda ^*_k(S_0-r)\big)}\le 0$ for each $k\ge 1$ such that

\begin{equation*} \int _{\Omega }\big(\lambda ^*_kS_0+\big(1-\lambda ^*_k\big)r\big)dx\to \int _{\Omega }\max \{S_0,r\}dx\quad \text{as}\quad k\to \infty. \end{equation*}

Since $\|\lambda ^*_k\|_{L^{\infty }(\Omega )}\le 1$ , possibly after passing to a subsequence and using the Banach–Alaoglu theorem, there is $\lambda ^*\in L^{\infty }(\Omega )$ satisfying $0\le \lambda ^*\le 1$ almost everywhere on $\Omega$ , such that $\lambda ^*_k\to \lambda ^*$ weakly star in $L^\infty (\Omega )$ as $k\to \infty$ . So we have

\begin{equation*} \int _{\Omega }\big(\lambda ^*_kS_0+\big(1-\lambda ^*_k\big)r\big)dx\to \int _{\Omega }(\lambda ^*S_0+(1-\lambda ^*)r)dx\quad \text{as} \quad k\to \infty. \end{equation*}

It follows that

\begin{equation*} \int _{\Omega }(\lambda ^*S_0+(1-\lambda ^*)r)dx=\int _{\Omega }\max \{S_0,r\}dx, \end{equation*}

which yields that $\max \{S_0,r\}=\lambda ^*S_0+(1-\lambda ^*)r$ almost everywhere on $\Omega$ . So, $\beta \lambda ^*(S_0-r)=\beta \big(\!\max \{S_0,r\}-r\big)$ almost everywhere on $\Omega$ . Therefore, by the assumption $\|(S_0-r)_+\|_{\infty }\gt 0$ ,

(3.8) \begin{equation} \int _{\Omega }\beta \lambda ^*(S_0-r)dx=\int _{\Omega }\beta \big(\!\max \{S_0,r\}-r\big)dx=\int _{\{S_0\gt r\}}\beta (S_0-r)dx\gt 0. \end{equation}

However, since ${\sigma \big(d_I,\beta \lambda ^*_k(S_0-r)\big)}\le 0$ , we have $\int _{\Omega }\beta \lambda ^*_k(S_0-r))dx\le 0$ for any $k\ge 1$ by Lemma 2.4. Letting $k\to \infty$ , we get $\int _{\Omega }\beta \lambda ^*(S_0-r)dx\le 0$ , which contradicts with (3.8).

Proof. Let $N^*_{S_0,r}$ be defined as in (3.4). (1) If $\beta$ is constant, then for any $\lambda ^*\in C(\overline{\Omega };\,[0,1])$ , $\sigma (d_I,\beta \lambda ^*(S_0-r))\le 0$ implies that $\int _{\Omega }\lambda ^*(S_0-r)dx\le 0$ . In this case,

\begin{equation*} \int _{\Omega }(\lambda ^*S_0+(1-\lambda ^*)r)dx=\int _{\Omega }\lambda ^*(S_0-r)dx+\int _{\Omega }rdx\le \int _{\Omega }rdx,\quad \forall \ \lambda ^*\in C(\overline {\Omega };\,[0,1]). \end{equation*}

So, we have $N^*_{S_0,r}=\int _{\Omega }rdx$ .

(2) If $S_0\le r$ , then $N^*_{S_0,r}=\int _{\Omega }rdx$ .

(3) If $S_0\ge r$ , then $N^*_{S_0,r}\le \int _{\Omega }S_0dx$ .

In cases (1)–(3), we have $N^*_{S_0,r}\le \max\! \big\{\!\int _{\Omega }rdx,\int _{\Omega }S_0dx\big\}$ . By hypothesis (A2), we always have $N\gt \int _{\Omega } S_0dx$ . Therefore, if either of these scenarios holds, $N\gt \int _{\Omega }rdx$ implies $N\gt N^*_{S_0,r}$ . The conclusion now follows from Theorem 3.3.

Remark 3.6. Let $N^*_{S_0,r}$ be defined as in (3.4). Alternative (ii) of Theorem 3.3 always holds when $N\gt \int _{\Omega }rdx$ if $N^*_{S_0,r}\le \max\! \big\{\!\int _{\Omega }S_0dx,\int _{\Omega }rdx\big\}$ . Sufficient conditions ensuring the validity of the latter inequality are given in Corollary 3.5. It remains an open problem to know whether alternative (i) of Theorem 3.3 may hold for some initial data satisfying $\max\! \big\{\!\int _{\Omega }S_0dx,\int _{\Omega }rdx\big\}\lt N\lt N^*_{S_0,r}$ . Note that it is possible to construct examples of positive and continuous functions $S_0$ satisfying $\max\! \big\{\!\int _{\Omega }S_0dx,\int _{\Omega }rdx\big\}\lt N^*_{S_0,r}$ . Whenever such $S_0$ is fixed, we can always select $I_0$ to be small enough such that $N=\int _{\Omega }(S_0+I_0)dx$ satisfies $\max\! \big\{\!\int _{\Omega }S_0dx,\int _{\Omega }rdx\big\}\lt N\lt N^*_{S_0,r}$ .

Proof. Suppose that a nonnegative solution exists. Let $M_1=\max \{\|S_0\|_{L^\infty (\Omega )}, \ r_M\}$ . By the first equation of (3.9) and the comparison principle, $0\le S(x, t)\le M_1$ for all $x\in \bar \Omega$ and $t\gt 0$ , which implies that

\begin{equation*} 0\le I(x, t)= I_0(x) e^{\int _0^t (\beta (x) S(x, s)-\gamma (x))ds}\le I_0(x)e^{\beta _MM_1t}, \ \ \forall x\in \bar \Omega, t\ge 0. \end{equation*}

This gives $\|I(\cdot, t)\|_{L^\infty (\Omega )}\le \|I_0\|_{L^\infty (\Omega )} e^{\beta _{M}M_1t}$ for all $t\ge 0$ . The local existence of the solution of (3.9) can be proved similar to Proposition 3.1. The a prior bound of the solution ensures the global existence of it.

Remark 3.8. We point out that it is very easy to draw a false conclusion using the above Lyapunov function: first by the term $\int _\Omega \beta (S-r)^2Idx$ in (3.10), one may conclude that either $S\to r$ or $I\to 0$ as $t\to \infty$ ; then by the term $\int _\Omega |\triangledown S|^2dx$ , $\triangledown S\to 0$ and so $I\to 0$ if $r$ is not constant. We will show that this intuition is indeed false in Theorem 3.11. Actually, it is possible that $S$ converges to some constant $\bar S$ with $r_m\le \bar S\le r_M$ and $I$ converges to some measure supported at $\{x\in \bar \Omega \,:\, r(x)=\bar S\}$ .

Proof. Setting $Z=S(\cdot,t)-\overline{S(\cdot,t)}$ and $F(\cdot,t)=\beta (r-S(\cdot,t))I(\cdot,t)-\overline{\beta (r-S(\cdot,t))I(\cdot,t)}$ , it holds that $Z(\cdot,t), F(\cdot,t)\in \mathcal{Z}$ for all $t\ge 0$ and

\begin{equation*} \begin {cases} \partial _tZ=d_S\Delta Z+F(\cdot,t), & t\gt 0,\ x\in \Omega,\\ \partial _{\nu }Z=0,\ & t\gt 0,\ x\in \partial \Omega. \end {cases} \end{equation*}

Hence by the variation of constant formula, we have

\begin{equation*} Z(\cdot,t)=e^{td_S\Delta _{|\mathcal {Z}}}Z(\cdot,0)+\int _{0}^{t}e^{(t-\tau )\Delta _{|\mathcal {Z}}}F(\cdot,\tau )d\tau, \quad \forall \ t\gt 0. \end{equation*}

By Proposition 3.7 with $M\,:\!=\,\max \{\|S_0\|_{L^{\infty }(\Omega )},r_M\}$ , it holds that

(3.11) \begin{equation} \|F(\cdot,t)\|_{L^1(\Omega )}\le 2\beta _{M}M\|I(\cdot,t)\|_{L^1(\Omega )}\le M_1\,:\!=\,2NM\beta _M,\quad \forall \ t\ge 0. \end{equation}

(i) Fix $0\le \tilde{\alpha }\lt \tilde{\alpha }+\alpha \lt 1$ . For any $t\ge 1$ and $h\gt 0$ , by (2.1)–(2.3) and (3.11),

(3.12)

In particular, if $\tilde{\alpha }=0$ , we get that

\begin{equation*} \|Z(t+h)-Z(t)\|_{L^1(\Omega )}\le {M}_{\alpha,\tilde {\alpha }}(h^{\alpha }+h),\quad \forall \ t\ge 1, \ h\gt 0, \end{equation*}

which yields the first assertion of (i). On the other, if $ n=1$ , choosing $\alpha \in (\frac{3}{4},1)$ , it follows from [Reference Henry20, Theorem 1.6.1] that $\mathcal{Z}^{\alpha }$ is continuously embedded in $W^{1,2}(\Omega )$ . Hence, the last assertion of (i) also follows from (3.12).

(ii) Suppose $ n=1$ . Let $G(t)\,:\!=\,\int _{\Omega }\beta (S-r)^2Idx$ for $t\ge 0$ . Since $ n=1$ , $W^{1,2}(\Omega )$ is compactly embbedded in $ C(\overline{\Omega })$ . Since the mapping $[1,\infty )\ni t\mapsto Z(\cdot,t)\in{W^{1,2}(\Omega )}$ is Hölder continuous by (i), the mapping $[1,\infty )\ni t\mapsto Z(\cdot,t)\in{C(\overline{\Omega })}$ is also Hölder continuous. This together with the fact that

\begin{equation*} \sup _{t\gt 0}\bigg |\frac {d}{dt}\int _{\Omega }S(x,t)dx \bigg |=\sup _{t\gt 0}\bigg |\int _{\Omega }\beta (S(x,t)-r(x))I(x,t)dx \bigg |\le MN\beta _M \end{equation*}

implies that the mapping $[1,\infty )\ni t\mapsto S(\cdot,t)\in C(\overline{\Omega })$ is also Hölder continuous. Thus, there exist $0\lt \tau \lt 1$ and $c\gt 0$ such that

\begin{equation*} \|S(\cdot,t+h)-S(\cdot,t)\|_{L^{\infty }(\Omega )}\le c|h|^{\tau },\quad t\ge 1. \end{equation*}

So for any $t\ge 1$ and $h\gt 0$ , we have

\begin{align*} |G(t+h)-G(t)|\le & \beta _M\|(S(\cdot,t+h)-r)^2-(S(\cdot,t)-r)^2\|_{\infty }\int _{\Omega }I(x,t+h)dx\\[5pt] &+\beta _M\|(S(\cdot,t)-r)^2\|_{L^{\infty }(\Omega )}\int _{\Omega }|I(x,t+h)-I(x,t)|dx\\[5pt] \le & 2cMNh^{\tau }\beta _M+M^2\beta _M\int _{\Omega }\int _{0}^h\beta |S(x,t+s)-r(x)|I(x,t+s)dsdx\\[5pt] \le & 2cMNh^{\tau }\beta _M+M^3\beta _M^2\int _{\Omega }\int _{0}^hI(x,t+s)dsdx\\[5pt] =&2cMNh^{\tau }\beta _M+M^3\beta _M^2\int _{0}^h\int _{\Omega }I(x,t+s)dxds\\[5pt] \le & 2cMNh^{\tau }\beta _M+M^3\beta _M^2Nh\le \left(c+M^2\beta _M\right)M_1(h^{\tau }+h), \end{align*}

which yields the desired result.

Proof. Integrating (3.10) over $(0, \infty )$ and by $\int _\Omega I dx\le N$ and Proposition 3.7, we have

(3.13) \begin{equation} \int _0^\infty \|\triangledown S(\cdot, t)\|_{L^2(\Omega )}^2 dt\lt \infty \ \text{and} \ \int _0^\infty \int _\Omega \beta (S(x, t)-r)^2I(x, t) dxdt\lt \infty. \end{equation}

Then the claim follows from Lemmas 2.2 and 3.9.

Proof. By the Poincaré inequality, there is a positive constant $\lambda _0\gt 0$ such that

(3.16) \begin{equation} \|S(\cdot,t)-\overline{S(\cdot,t)}\|_{L^2(\Omega )}\le \lambda _0\|\nabla S(\cdot,t)\|_{L^2(\Omega )},\quad \forall \ t\gt 0. \end{equation}

Hence using Hölder’s inequality and recalling (3.10), we get that

\begin{align*} \int _0^{\infty }\|S(\cdot,t)-\overline{S(\cdot,t)}\|_{L^1(\Omega )}^2dt\le & |\Omega |\int _{0}^{\infty }\|S(\cdot, t)-\overline{S(\cdot,t)}\|_{L^2(\Omega )}^2dt\\ \le & \lambda _0^2|\Omega |\int _{0}^{\infty }\|\nabla S(\cdot, t)\|_{L^2(\Omega )}^2dt\lt \infty. \end{align*}

Therefore, by Lemmas 2.2 and 3.9, we obtain $\|S(\cdot, t)-\overline{S(\cdot,t)}\|_{L^1(\Omega )}^2\to 0$ as $t\to \infty$ . By $\sup _{t\ge 1}$ $\|S(\cdot,t)\|_{\infty }\lt \infty$ and Hölder’s inequality, (3.14) holds.

From this point, we shall suppose that $ n=1$ and complete the proof of the theorem. In view of Lemma 3.10 and inequality (3.16), we have

\begin{equation*} \lim _{t\to \infty }\|S(\cdot,t)-\overline {S(\cdot,t)}\|_{W^{1,2}(\Omega )}=0. \end{equation*}

Since $ n=1$ , $W^{1,2}(\Omega )$ is compactly embedded into $C(\bar{\Omega })$ . Therefore,

(3.17) \begin{equation} \lim _{t\to \infty }\big \| S(\cdot,t)-\overline{S(\cdot,t)}\big \|_{L^\infty (\Omega )}=0. \end{equation}

Fix $\varepsilon \gt 0$ . By $N=\int _{\Omega }(S+I)dx$ for all $t\ge 0$ and (3.17), there is $t_{\varepsilon }\gt 0$ such that

(3.18) \begin{equation} S(x,t)\le \frac{1}{|\Omega |}\Big (N+\varepsilon -\int _{\Omega }I(x,t)dx\Big ),\quad \forall \ (x, t)\in \bar \Omega \times [t_{\varepsilon }, \infty ). \end{equation}

Let $K\subset H^{-}$ be a compact set if $H^-$ is not empty. By the definition of $H^-$ , we have $\min _{x\in K}r(x)\gt N/|\Omega |$ . If $0\lt \varepsilon \ll 1$ is chosen such that $\eta _{\varepsilon }\,:\!=\,(N+\varepsilon )/|\Omega |-\min _{x\in K}r(x)\lt 0$ , then

\begin{equation*} \partial _tI(x,t)\le \beta (x)\left(\frac {N+\varepsilon }{|\Omega |}-\min _{x\in K}r(x)\right)I(x,t)\le \beta _m\eta _{\varepsilon }I(x,t), \quad (x, t)\in K\times [t_{\varepsilon }, \infty ). \end{equation*}

It follows that

\begin{equation*} \|I(\cdot,t)\|_{L^{\infty }(K)}\le e^{(t-t_\varepsilon )\beta _m\eta _{\varepsilon }}\|I(\cdot,t_{\varepsilon })\|_{L^{\infty }(K)}\to 0,\quad \text{as}\ t\to \infty. \end{equation*}

This together with the fact that $I(x,t)=0$ for all $t\ge 0$ and $x\in \{I_0=0\}$ completes the proof of (3.15).

To prove (i)–(ii), we claim that

(3.19) \begin{equation} \limsup _{t\to \infty }\|I(\cdot,t)\|_{L^1(\Omega )}\le \left(N-|\Omega |\tilde{r}_{m}\right)_+. \end{equation}

Since $\int _{\{I_0=0\}}I(x,t)dx=0$ for all $t\ge 0$ , it suffices to show $\limsup _{t\to \infty }\|I(\cdot,t)\|_{L^1(\{I_0\gt 0\})}\le (N-|\Omega |\tilde{r}_{m})_+$ . To see this, observe from (3.18) that

(3.20) \begin{equation} \partial _tI\le \frac{\beta }{|\Omega |}\left( (N-|\Omega |\tilde{r}_{m})_++\varepsilon -\|I(\cdot,t)\|_{L^1(\{I_0\gt 0\})}\right)I,\quad \forall \ t\ge t_{\varepsilon },\ x\in \{I_0\gt 0\}, \end{equation}

where $\tilde{r}_m=\inf _{x\in \{I_0\gt 0\}}r(x)$ . Let

(3.21) \begin{equation} F(t)\,:\!=\,\frac{\int _{\{I_0\gt 0\}}\beta I(x,t)dx}{\int _{\{I_0\gt 0\}}I(x,t)dx}, \ \ \forall \ t\ge 1. \end{equation}

Then $F(t)\,:\, [1, \infty )\to \mathbb{R}_+$ is Locally Lipschitz continuous with $\beta _m\le F(t)\le \beta _M$ , $t\ge 1$ . Integrating (3.20) over $\{I_0\gt 0\}$ , for any $t\gt t_{\varepsilon }$ , we get

\begin{align*} \frac{d}{dt}\|I\|_{L^1(\{I_0\gt 0\})}\le & \frac{1}{|\Omega |} \left((N-|\Omega |\tilde{r}_m)_++\varepsilon -\|I\|_{L^1(\{I_0\gt 0\})}\right)\int _{\{I_0\gt 0\}}\beta Idx\\[6pt] =& \frac{F(t)}{|\Omega |}\left((N-|\Omega |\tilde{r}_{m})_++\varepsilon -\|I\|_{L^1(\{I_0\gt 0\})}\right)\|I\|_{L^1(\{I_0\gt 0\})}, \end{align*}

where $F$ is defined by (3.21). Let $v(t)$ be the solution of

\begin{equation*} \left \{ \begin {array}{lll} v^{\prime}(t)&=& \displaystyle \frac {F(t)}{|\Omega |}\left((N-|\Omega |\tilde {r}_{m})_++\varepsilon -v(t)\right)v(t),\ \ t\gt t_\varepsilon,\\[16pt] v(t_\varepsilon )&=& \|I(\cdot, t_\varepsilon )\|_{L^1(\{I_0\gt 0\})}. \end {array} \right. \end{equation*}

By the comparison principle, we know that

\begin{equation*} \|I(\cdot,t)\|_{L^1(\{I_0\gt 0\})}\le v(t),\quad \forall t\ge t_{\varepsilon }. \end{equation*}

Since $v(t)\to (N-|\Omega |\tilde{r}_{m})_++\varepsilon$ as $t\to \infty$ (because $F\ge \beta _m\gt 0$ and $v(t_\varepsilon )\gt 0$ ) and $\varepsilon$ is arbitrarily chosen, (3.19) holds.

(i) Suppose that $H^+\cap \{I_0\gt 0\}=\emptyset$ . Then we have $(N-|\Omega |\tilde{r}_{m})_+=0$ . By (3.19), $\|I(\cdot,t)\|_{L^1(\Omega )}\to 0$ as $t\to \infty$ . This together with the fact that $\int _{\Omega }Sdx=N-\int _{\Omega }Idx$ for all $t\gt 0$ , yields $\int _{\Omega }S(x,t)dx\to N$ as $t\to \infty$ . It then follows from (3.17) that $\|S(\cdot,t)-{N}/{|\Omega |}\|_{L^\infty (\Omega )}\to 0$ as $t\to \infty$ .

(ii) Suppose that $H^+\cap \{I_0\gt 0\}\ne \emptyset$ . Then we have $N/|\Omega |\gt \tilde{r}_m$ . Since $\int _{\Omega }Sdx=N-\int _{\Omega }Idx$ , we conclude from (3.19) that

\begin{equation*} \liminf _{t\to \infty }\int _{\Omega }S(x,t)dx\ge |\Omega |\tilde {r}_{\min }, \end{equation*}

which in view of (3.17) implies that

(3.22) \begin{equation} \liminf _{t\to \infty }\min _{x\in \overline{\Omega }}S(x,t)\ge \tilde{r}_m. \end{equation}

Now, we claim that

(3.23) \begin{equation} \liminf _{t\to \infty }\min _{x\in \overline{\Omega }}S(x,t)=\tilde{r}_m. \end{equation}

We proceed by contradiction. Suppose to the contrary that (3.23) is false. Thanks to (3.22), there exist $0\lt \tilde{\varepsilon }\ll 1$ and $\tilde{t}_0\gt 0$ such that

\begin{equation*} S(x,t)\ge \tilde {r}_{m}+\tilde {\varepsilon },\quad \forall \ t\ge \tilde {t}_0. \end{equation*}

Since $r$ is continuous on $\bar{\Omega }$ and $\tilde{r}_m=\inf _{x\in \{I_0\gt 0\}}r(x)$ , there is an open set $\mathcal{O}\subset \{I_0\gt 0\}$ such that $r(x)\lt \tilde{r}_{m}+\tilde{\varepsilon }/{2}$ for all $x\in \mathcal{O}$ . Hence, we have

\begin{equation*} \partial _{t}I(x,t)=\beta (S-r)I\ge \frac {\tilde {\varepsilon }}{2}\beta _mI(x,t),\quad t\gt \tilde {t}_0,\ x\in \mathcal {O}. \end{equation*}

An integration of this inequality yields

\begin{equation*} N\ge \int _{\mathcal {O}}I(x,t)dx\ge e^{\frac {\tilde {\varepsilon }}{2}\beta _m(t-\tilde t_0)}\int _{\mathcal {O}}I(x,\tilde {t}_0)dx,\quad t\gt \tilde {t}_0. \end{equation*}

This is clearly impossible since $ \int _{\mathcal{O}}I(x,\tilde{t}_0)dx\gt 0$ . Therefore, (3.23) must hold.

By (3.23), there is a sequence $\{t_k\}_{k\ge 1}$ converging to infinity such that $\min _{x\in \overline{\Omega }}S(x,t_k)\to \tilde{r}_m$ as $k\to \infty$ . Hence, since

\begin{align*} \|\tilde{r}_m-S(\cdot,t_k)\|_{L^\infty (\Omega )}\le & |\tilde{r}_m-\overline{S(\cdot,t_k)}|+ \|S(\cdot,t_k)-\overline{S(\cdot,t_k)}\|_{L^\infty (\Omega )} \\ \le & |\tilde{r}_m-\min _{x\in \overline{\Omega }}S(x,t_k)|+|\min _{x\in \overline{\Omega }}S(x,t_k)-\overline{S(\cdot,t_k)}|+\|S(\cdot,t_k)-\overline{S(\cdot,t_k)}\|_{L^\infty (\Omega )} \\ \le & |\tilde{r}_m-\min _{x\in \overline{\Omega }}S(x,t_k)|+2\|S(\cdot,t_k)-\overline{S(\cdot,t_k)}\|_{L^{\infty }(\Omega )},\quad \forall \ k\ge 1, \end{align*}

we conclude from (3.17) that $\|S(\cdot,t_k)-\tilde{r}_m\|_{L^\infty (\Omega )}\to 0$ as $k\to \infty$ . This in turn implies that $\int _{\Omega }I(x,t_k)dx\to N-|\Omega |\tilde{r}_m$ as $k\to \infty$ . However, by Lemma 3.10, we know that $\int _{\{I_0\gt 0\}}(S(x,t_k)-r)^2I(x,t_k)dx=\int _{\Omega }(S(x,t_k)-r)^2I(x,t_k)dx\to 0$ as $k\to \infty$ . Therefore, possibly after passing to a subsequence, $I(\cdot,t_k)\to 0$ as $k\to \infty$ almost everywhere on $\{ x\in \Omega\,:\, r(x)\ne \tilde{r}_m\}$ . Finally, since $\{I(\cdot, t_k)\}$ is bounded in $L^1(\Omega )$ and by Riesz representation theorem, passing to a subsequence if necessary, $I(\cdot, t_k)\to (N-|\Omega |\tilde{r}_m)\mu$ weakly as $k\to \infty$ for some probability Radon measure $\mu$ . Since

\begin{equation*} \int _{\{I_0\gt 0\}} \beta (S(x, t_k)-r)^2I(x, t_k) dx\to \big(N-|\Omega |\tilde {r}_m\big)\int _{\{I_0\gt 0\}} \beta (r_m-r)^2d\mu =0\ \ as \ {k\to \infty }, \end{equation*}

$\mu$ is supported in $\overline{\{I_0\gt 0\}}\cap \mathcal{M}$ . In particular if $\mathcal{M}=\{x_1,\cdots,x_L\}\subset \{I_0\gt 0\}$ , then $\mu =\sum _{i=1}^Lc_i\delta _{x_i}$ for some $0\le c_i\le 1$ with $\sum _{i=1}^Lc_i=1$ .

Remark 3.12. We conjecture that Theorem 3.11 holds for any $n\ge 1$ and $\|S(\cdot,t)-\tilde{r}_m\|_{L^\infty (\Omega )}\to 0$ as $t\to \infty$ in (ii).

Proof. If we define $SI/(S+I)=0$ when $(S, I)=(0, 0)$ , then $SI/(S+I)$ is Lipschitz in the first quadrant. So the existence and uniqueness of nonnegative local solution $(S, I)$ can be proved using the Banach fixed point theorem, where $S\in C^1([0, T), C(\overline{\Omega }))$ and $I\in C([0, T), C(\overline{\Omega }))\cap C^1((0, T), \textrm{Dom}_\infty (\Delta ))$ for some $T\gt 0$ .

Since the right-hand side of the second equation of (4.1) has linear growth rate (i.e. $\beta SI/(S+I)- \gamma I\le C I$ for some constant $C\gt 0$ ) and $\int _\Omega I dx\le N$ for all $t\gt 0$ , by [Reference Alikakos1, Theorem 3.1], there exists $M\gt 0$ such that $\|I(\cdot, t)\|_{L^\infty (\Omega )}\le M$ for all $t\gt 0$ . By the first equation of (4.1), we have $\partial _t S\le \gamma I$ , which implies

\begin{equation*} \|S(\cdot, t)\|_{L^\infty (\Omega )}\le \|S_0\|_{L^\infty (\Omega )}+M\gamma _Mt, \ \ t\ge 0. \end{equation*}

Hence, we can extend the solution globally.

Remark 4.3. The only case not covered by Theorem 4.2 is when $\beta \ge \gamma$ and $H^0$ is not empty, which we will deal with later in this section. In the case $H^-\ne \emptyset$ and $\mathcal{R}_0\gt 1$ , it has been shown in [Reference Lou and Salako34, Theorem 2.5] that $J^*\,:\!=\,\{x\in \bar \Omega\, :\, S^*(x)=0\}$ is a subset of $H^+$ such that both $J^*$ and $\Omega \backslash J^*$ have positive measure, where $\mathcal{R}_0$ , defined as

(4.4) \begin{equation} \mathcal{R}_0=\sup \left \{\frac{\int _\Omega \beta \phi ^2dx}{\int _\Omega (d_I|\triangledown \phi |^2+\gamma \phi ^2) dx}\,:\, \phi \in H^1(\Omega )\backslash \{0\}\right \}, \end{equation}

is the basic reproduction number of the diffusive epidemic model (1.1) with standard infection incidence mechanism and diffusion rates $d_S\gt 0$ and $d_I\gt 0$ .

Proof. (i) By Lemma 2.3, there exists $C\gt 1$ such that

(4.6) \begin{equation} \max _{x\in \bar \Omega } I(x, t)\le C \min _{x\in \bar \Omega } I(x, t), \ \ x\in \bar \Omega, t\ge 1. \end{equation}

Fix $x_0\in H^0$ . Let $x\in \bar \Omega$ . Then $\partial _t S=\frac{(-\beta +\gamma )S+\gamma I}{S+I}I\le \gamma I^2/(S+I)$ . We consider the following problem:

(4.7) \begin{equation} \begin{cases} \displaystyle \bar S^{\prime}=\gamma (x) \frac{I^2(x, t)}{\bar S+I(x, t)},&t\gt 1,\\[10pt] \bar S(1)=S(x, 1). & \end{cases} \end{equation}

Then we have $S(x, t)\le \bar S(t)$ for all $t\ge 1$ . We claim that $KS(x_0, t)$ is an upper solution of (4.7) if $K\gt 0$ is large enough. To see it, it suffices to check

\begin{equation*} K\partial _tS(x_0, t)=K\gamma (x_0) \frac {I^2(x_0, t)}{S(x_0, t)+I(x_0, t)}\ge \gamma (x) \frac {I^2(x, t)}{KS(x_0, t)+I(x, t)}. \end{equation*}

Noticing (4.6), we only need to check

\begin{equation*} K\gamma (x_0) \frac {I^2(x, t)/C^2}{S(x_0, t)+{\frac {I(x,t)}{C}}}\ge \gamma (x) \frac {I^2(x, t)}{KS(x_0, t)+I(x, t)}, \end{equation*}

which is equivalent to $(K^2\gamma (x_0)-C^2\gamma (x))S(x_0, t)+(K\gamma (x_0)-{C}\gamma (x))I(x, t)\ge 0$ . So we can choose $K$ large independent of $x\in \bar \Omega$ such that the inequality holds. Hence, we have $KS(x_0, t)\ge \bar S(t)\ge S(x, t)$ for all $t\ge 1$ . Moreover, interchanging the role of $x_0$ and $x$ , we have

(4.8) \begin{equation} S(x_0, t)/K\le S(x, t)\le KS(x_0, t), \ \ \ \forall x\in H^0, \ t\ge 1. \end{equation}

By (4.8), $N\ge \int _\Omega S(x, t)dx\ge \int _{H^0} S(x, t)dx\ge \int _{H^0} S(x_0, t)/Kdx=|H^0| S(x_0, t)/K$ for all $t\ge 1$ . Therefore, we have $S(x_0, t)\le KN/|H^0|$ and $S(x, t)\le K^2N/|H^0|$ for all $x\in \bar \Omega$ and $t\ge 1$ .

The convergence of $(S, I)$ can be proved similar to [Reference Lou and Salako34, Lemma 5.6], and we include it for completeness. By Proposition 4.1, we have $0\le S(x, t), I(x, t)\le M$ for all $x\in \bar \Omega$ and $t\ge 0$ . It follows from the equation of $S$ that

\begin{equation*} \partial _t S=\frac {\gamma I^2}{S+I}\ge \frac {\gamma _m}{2M} \min _{y\in H^0} I^2(y, t), \ \ \forall \ x\in H^0, { t\gt 0}. \end{equation*}

Integrating the above inequality over $H^0\times (0, \infty )$ and noticing that $H^0$ has positive measure, we see that $\int _0^\infty \min _{x\in H^0} I^2(x, t) dx\lt \infty$ . By (4.6), it holds that

(4.9) \begin{equation} \int _{0}^{\infty }\|I(\cdot,t)\|^2_{L^\infty (\Omega )}dt\lt \infty. \end{equation}

By (4.2) and the $L^p$ estimate, $I$ is Hölder continuous on $\bar \Omega \times [1, \infty )$ . Therefore, Lemma 2.2 and (4.9) imply that $I(x, t)\to 0$ uniformly on $\bar \Omega$ as $t\to \infty$ .

Since $\partial _t S=\gamma I^2/(S+I)$ , $S(x,t)$ is strictly increasing in $t\in (0,\infty )$ for every $x\in H^{0}$ and

(4.10) \begin{align} \int _{1}^{\infty }\|S_t(\cdot,t)\|_{L^{\infty }(H^0)}dt & \le \gamma _M\int _{1}^{\infty }\frac{\|I(\cdot,t)\|_{L^\infty (\Omega )}^2}{\min _{x\in H^0}S(x,t)}dt\nonumber\\[12pt]& \le \frac{\gamma _M}{\min _{x\in H^0}S(x,1)}\int _{1}^{\infty }\|I(\cdot,t)\|^2_{L^\infty (\Omega )}dt\lt \infty. \end{align}

Whence, $S(\cdot,t)\to S^*_{|H^0}\,:\!=\,S(\cdot,0)+\int _0^{\infty }S_t(\cdot,t)dt\in C(H^0)$ uniformly on $H^0$ as $t\to \infty$ .

Next, we discuss the convergence of $S(x,t)$ as $t\to \infty$ for $x\in H^+$ . To this end, let $\kappa \,:\!=\,(\beta -\gamma )/\beta$ and define

\begin{equation*} V(S, I)=\frac {1}{2}\int _\Omega \left (\kappa S^2+I^2 \right )dx. \end{equation*}

It is easy to check that

(4.11) \begin{equation} \frac{d}{dt}V(S, I)=-d_I\int _\Omega |\triangledown I|^2dx-\int _\Omega \gamma \frac{(\kappa S-I)^2}{S+I}Idx. \end{equation}

Integrating (4.11) over $(0,t)$ and taking $t\to \infty$ , we obtain

(4.12) \begin{equation} \int _0^{\infty }\int _{\Omega }\gamma \frac{(\kappa S-I)^2}{S+I}Idxdt\lt \infty. \end{equation}

On the other hand, we have

\begin{align*} \frac{1}{2}\big(\kappa S^2\big)_t=\gamma \frac{(I-\kappa S)\kappa S I}{S+I}=\gamma \frac{(I-\kappa S)(\kappa S-I+I)I}{S+I}=-\gamma \frac{(I-\kappa S)^2I}{S+I}+\gamma \frac{(I-\kappa S)}{S+I}I^2. \end{align*}

Hence, by (4.9) and (4.12), we have that

\begin{align*} \int _{1}^{\infty }\Big \|\frac{\big(\kappa S^2\big)_t}{2}\Big \|_{L^{1}(H^+)}dt\le & \int _1^{\infty }\int _{\Omega }\gamma \frac{(I-\kappa S)^2I}{S+I}dxdt+\int _{1}^{\infty }\int _{\Omega }\gamma \frac{|I-\kappa S|}{S+I}I^2dxdt\\[7pt] \le & \int _1^{\infty }\int _{\Omega }\gamma \frac{(I-\kappa S)^2I}{S+I}dxdt+\gamma _M(1+\|\kappa \|_{\infty })|\Omega |\int _{1}^{\infty }\|I\|_{L^{\infty }(\Omega )}^2dt\\[7pt] \lt &\infty. \end{align*}

Therefore, there is a measurable subset $\mathcal{N}\subset H^+$ with ${\text{meas}(\mathcal{N})}=0$ such that

\begin{equation*} \int _{1}^{\infty }\Big |\frac {\big(\kappa S^2\big)_t}{2}\Big |dt\lt \infty, \quad \forall \ x\in H^{+}\setminus \mathcal {N}. \end{equation*}

As a result, we have that $ \kappa (x) S^2(x,t)\to \kappa (x)(S_0^2(x)+\int _0^{\infty }(S^2)_tdt)$ as $t\to \infty$ for $x\in H^{+}\setminus \mathcal{N}$ . So, $S(x,t)\to S^*_{H^+}(x)\,:\!=\,\sqrt{S^2_0(x)+\int _0^{\infty }(S^2)_tdt}$ as $t\to \infty$ for $x\in H^+\setminus \mathcal{N}$ . Since $\|S(\cdot,t)\|_{L^{\infty }(\Omega )}\le M$ for all $t\ge 1$ , then $S^*_{|H^+}\in L^{\infty }(H^+)$ , and by the Lebesgue dominated theorem and the fact that $\text{meas}(\mathcal{N})=0$ , $\|S(\cdot,t)-S^*_{|H^+}\|_{L^1(H^+)}\to 0$ as $t\to \infty$ . Now, taking $S^*\,:\!=\,S^*_{|H^0}\chi _{H^0}+S^*_{|H^{+}}\chi _{H^{+}}$ , then $S^*\in L^{\infty }(\Omega )$ , $S^*\in C({H^0})$ , $S^*\gt 0$ on $H^0$ , and $\|S(\cdot,t)-S^*\|_{L^{\infty }(H^0)}+\|S(\cdot,t)-S^*\|_{L^1(H^+)}\to 0$ as $t\to \infty$ .

Finally, we suppose that $H^+\ne \emptyset$ and proceed by contradiction to show that

(4.13) \begin{equation} \text{meas}(\{x\in H^{+}\, |\, S^*(x)=0\})\gt 0. \end{equation}

So, suppose to the contrary that (4.12) does not hold. Consider the function

\begin{equation*} F(x,t)=\frac {1}{t}\int _0^t\frac {I(x,s)}{S(x,s)+I(x,s)}ds,\quad \forall \ x\in \Omega, \ t\gt 0. \end{equation*}

It is clear that

\begin{equation*} 0\le F(x,t)\le 1,\quad \forall \ x\in \Omega,\ t\gt 0. \end{equation*}

Moreover, for each $x\in \Omega$ satisfying $S^*(x)\gt 0$ , it holds that

\begin{equation*} \lim _{t\to \infty } F(x, t)=\lim _{t\to \infty }\frac {I(x,t)}{S(x,t)+I(x,t)}=0. \end{equation*}

Hence, since $S^*\gt 0$ on $H^0$ and $ \text{meas}(\{x\in H^{+}\ |\ S^*(x)=0\})=0$ , it follows from the Lebesgue dominated convergence theorem that

(4.14) \begin{equation} \lim _{t\to \infty }\int _{\Omega }F(x,t)dx=0. \end{equation}

Let $\varphi$ be the positive eigenfunction associated with $\sigma (d_I,\beta -\gamma )$ satisfying $\max _{x\in \bar{\Omega }}\varphi (x)=1$ . By the second equation of (4.1) and (4.6), we have

\begin{align*} \frac{d}{dt}\int _{\Omega }\varphi Idx=&\int _{\Omega }d_I\varphi \Delta I dx+\int _{\Omega }\left(\frac{\beta S}{S+I}-\gamma \right)\varphi Idx\\[6pt] =& d_I\int _{\Omega }I(\cdot,t)\Delta \varphi dx+\int _{\Omega }\left(\frac{\beta S}{S+I}-\gamma \right)\varphi Idx\\[6pt] =& \sigma (d_I,\beta -\gamma )\int _{\Omega }\varphi I dx+\int _{\Omega }\beta \left(\frac{S}{S+I}-1\right)\varphi Idx\\[6pt] =& \sigma (d_I,\beta -\gamma )\int _{\Omega }\varphi Idx-\int _{\Omega }\beta \frac{I}{S+I}\varphi Idx\\[6pt] \ge & \sigma (d_I,\beta -\gamma )\int _{\Omega }\varphi Idx-\beta _M\varphi _{M}\|I\|_{L^\infty (\Omega )}\int _{\Omega }\frac{I}{S+I}dx\\[6pt] \ge & \left(\sigma (d_I,\beta -\gamma )-\frac{\beta _MC\varphi _M}{\varphi _{m}}\int _{\Omega }\frac{I}{S+I}dx\right)\int _{\Omega }\varphi Idx. \end{align*}

By the comparison principle for the ODE, we obtain that

(4.15) \begin{equation} \int _{\Omega }\varphi (x) I(x,t)dx\ge e^{t\left(\sigma (d_I,\beta -\gamma )-M^*\int _{\Omega }F(x,t)dx\right)}\int _{\Omega }\varphi (x) I_0(x)dx,\quad t\gt 0, \end{equation}

where $M^*\,:\!=\,{\beta _MC\varphi _M}/{\varphi _{m}}$ . Note that $ \sigma (d_I,\beta -\gamma )\gt 0$ , since $\beta \ge \gamma$ and $H^+\ne \emptyset$ . Hence, it follows from (4.14)–(4.15) that $\int _{\Omega }\varphi I(\cdot,t)dx\to \infty$ as $t\to \infty$ . This contradicts the fact that $\sup _{t\ge 1}\|I(\cdot,t)\|_{L^{\infty }(\Omega )}\lt \infty$ . Therefore, (4.13) holds.

(ii) Integrating (4.11) over $(0, t)$ and taking $t\to \infty$ , we find that

(4.16) \begin{equation} \int _0^\infty \int _\Omega |\triangledown I|^2dxdt\lt \infty. \end{equation}

By (4.2), the parabolic estimates and the Sobolev embedding theorem, $I\in C^{1+\alpha, 1+\alpha/2}(\bar \Omega \times [1, \infty )])$ . So by (4.16) and Lemma 2.2, $\|\triangledown I(\cdot, t)\|_{L^2(\Omega )}\to 0$ as $t\to \infty$ . Moreover, let $\omega _I\,:\!=\,\cap _{t\ge 1} \overline{\cup _{s\ge t} I(\cdot, s))}$ , where the completion is in $C(\bar \Omega )$ . Then $\omega _I$ is well defined, compact and consists with constants.

Suppose to the contrary that $0\not \in \omega _I$ . By the compactness of $\omega _I$ , there exists $\varepsilon _0\gt 0$ such that

(4.17) \begin{equation} \liminf _{t\to \infty }\min _{x\in \bar \Omega } I(x, t)\gt \varepsilon _0. \end{equation}

By the first equation of (4.1), we have

(4.18) \begin{equation} \partial _t S=\frac{(\gamma I-(\beta -\gamma )S)I}{S+I}. \end{equation}

This combined with (4.17) implies that $\liminf _{t\to \infty } S(x, t)\ge \varepsilon _0\gamma/(\beta -\gamma )$ pointwise for $x\in \Omega \backslash H^0$ as $t\to \infty$ . By Fatou’s Lemma, we have

\begin{equation*} \int _{\Omega \backslash H^0} \left (\frac {\varepsilon _0\gamma }{\beta -\gamma }+\varepsilon _0\right )dx\le \int _\Omega \liminf _{t\to \infty } (S+I)dx\le \liminf _{t\to \infty } \int _\Omega (S+I)dx=N. \end{equation*}

Since $H^0$ has measure zero and $\int _\Omega 1/(\beta -\gamma )dx=\infty$ , the first term in the above inequality equals infinity. This is a contradiction, and therefore $0\in \omega _I$ . Hence, there exists $\{t_k\}$ converging to infinity such that $I(\cdot, t_k)\to 0$ in $C(\bar \Omega )$ as $k\to \infty$ . Thus, $\int _{\Omega }S(\cdot,t_k)dx=N-\int _{\Omega }I(\cdot,t_k)dx\to N$ as $k\to \infty$ .

Proof. Similar to Proposition 4.1, (4.19) has a unique local nonnegative solution $(S, I)$ , where $S\in C([0, T), C(\overline{\Omega }))\cap C^1((0, T), \textrm{Dom}_{\infty }(\Delta ))$ and $I\in C^1([0, T), C(\overline{\Omega }))$ for some $T\gt 0$ . It remains to show the boundedness of the solution.

We claim that for any nonnegative integer $k$ there exists $C\gt 0$ depending on initial data such that $\|S(\cdot, t)\|_{L^{2^{k}}(\Omega )}, \|I(\cdot, t)\|_{L^{2^{k}}(\Omega )}\le C$ for all $t\ge 0$ . We prove this claim by induction. It is easy to see that the claim holds for $k=0$ . Now we assume that the claim holds for $k$ and will show that it holds for $k+1$ . To see it, multiplying both sides of the second equation of (4.19) by $I^{2^{k+1}-1}$ and integrating over $\Omega$ , we obtain

(4.21) \begin{eqnarray} \frac{1}{2^{k+1}}\frac{d}{dt}\int _\Omega I^{2^{k+1}}dx&\le & \beta _M \int _\Omega \frac{S I^{{2^{k+1}}-1} I}{S+I}dx-\gamma _m\int _\Omega I^{2^{k+1}}dx \nonumber \\[5pt] &\le & C_0\int _\Omega S^{2^{k+1}}dx-\frac{\gamma _m}{2}\int _\Omega I^{2^{k+1}}dx, \end{eqnarray}

where we have used Young’s inequality in the last step.

Multiplying both sides of the first equation of (4.19) by $S^{2^{k+1}-1}$ and integrating over $\Omega$ , we obtain

(4.22) \begin{eqnarray} \frac{1}{2^{k+1}}\frac{d}{dt}\int _\Omega S^{2^{k+1}}dx&=&-\frac{2^{k+1}-1}{2^{k+1}}d_S\int _\Omega |\triangledown S^{2^k}|^2dx- \int _\Omega \frac{\beta S^{2^{k+1}}I}{S+I} pdx+\int _\Omega \gamma S^{2^{k+1}-1}I dx\nonumber \\[5pt] &\le & -C_1 \int _\Omega |\triangledown S^{2^k}|^2dx+C_2\left (\int _\Omega S^{2^{k+1}} dx+\int _\Omega I^{2^{k+1}} dx\right ), \end{eqnarray}

where we have used Young’s inequality in last step.

Multiplying (4.22) by $\displaystyle C_3\,:\!=\,\frac{\gamma _m}{4C_2}$ and summing up with (4.21), we have

\begin{equation*} \frac {d}{dt}\int _\Omega \left (C_3S^{2^{k+1}}+I^{2^{k+1}}\right )dx \le -C_4 \int _\Omega |\triangledown S^{2^k}|^2dx+ C_5\int _\Omega S^{2^{k+1}} dx-C_6\int _\Omega I^{2^{k+1}} dx. \end{equation*}

By the following interpolation inequality

\begin{eqnarray*} \|u\|_{L^2(\Omega )}^2\le \epsilon \|\triangledown u\|^2_{L^2(\Omega )}+C_\epsilon \|u\|_{L^1(\Omega )}^2, \ \ \forall u\in H^1(\Omega ), \end{eqnarray*}

we obtain

(4.23) \begin{equation} \frac{d}{dt}\int _\Omega \left (C_3S^{2^{k+1}}+I^{2^{k+1}}\right )dx \le C_7\left (\int _\Omega S^{2^{k}} dx\right )^2-C_8\int _\Omega \left (C_3S^{2^{k+1}}+I^{2^{k+1}}\right ) dx. \end{equation}

By the assumption that the claim holds for $k$ and (4.23) there exists $C\gt 0$ such that $\|S(\cdot, t)\|_{L^{2^{k+1}}(\Omega )}, \|I(\cdot,t)\|_{L^{2^{k+1}}(\Omega )}\le C$ . This proves the claim.

Fixing $p\gt N+2$ , by the claim and the parabolic $L^p$ estimate, there exists $C\gt 0$ such that $\|S\|_{W^{2, 1}_p(\Omega \times (\tau, \tau +1))}\lt C$ for any $\tau \gt 0$ . Since $W^{2, 1}_p(\Omega \times (\tau, \tau +1))$ can be embedded into $C(\bar \Omega \times [\tau, \tau +1])$ , we obtain the boundedness of $S$ . We rewrite the equation of $I$ as

(4.24) \begin{equation} \partial _t I=\frac{((\beta -\gamma )S-\gamma I)I}{S+I}. \end{equation}

Hence, $\|I(\cdot, t)\|_{L^\infty (\Omega )}\le \max _{0\le s\le t}\{\|I_0\|_{L^\infty (\Omega )}, \ \|(\beta +\gamma )S(\cdot, s)\|_{L^\infty (\Omega )}/\gamma _m\}$ for any $t\ge 0$ . This proves the boundedness of $I$ .

Proof. (i) Note that (4.26) follows from Proposition 4.5. Next, observe that

\begin{align*} \frac{d}{dt}\int _{\Omega }Sdx=&\int _{\Omega }\gamma Idx-\int _{\Omega }\beta \frac{I}{I+S}Sdx\\[5pt] \ge & \gamma _m\int _{\Omega }Idx-\beta _M\int _{\Omega }Sdx\\[5pt] = &\gamma _m \left (N-\int _{\Omega }Sdx\right )-\beta _M\int _{\Omega }Sdx\\[5pt] =&\gamma _mN-(\beta _M+\gamma _m)\int _{\Omega }Sdx. \end{align*}

Therefore, by $\int _{\Omega }S_0dx\ge 0$ and the comparison principle, we have that

(4.30) \begin{equation} \int _{\Omega }S(x, t)dx\ge \frac{\gamma _mN}{\beta _M+\gamma _m}\left(1-e^{-t(\beta _M+\gamma _m)}\right),\quad \forall \ t\ge 0. \end{equation}

Next by (4.19), we see that

\begin{equation*} \begin {cases} \partial _tS\ge d_S\Delta S-\beta _MS, & x\in \Omega, t\gt 0, \\[5pt] \partial _{\nu }S=0, & x\in \partial \Omega,\ t\gt 0. \end {cases} \end{equation*}

For any $t_0\gt 0$ , it follows from the comparison principle for parabolic equations that

(4.31) \begin{equation} S(\cdot, t+t_0)\ge e^{-t\beta _M}e^{td_S\Delta }S(\cdot, t_0),\quad \forall \ t\gt 0. \end{equation}

Thanks to the Harnack’s inequality (see Lemma 2.3), there is a positive constant $c_0$ such that

(4.32) \begin{equation} e^{td_S\Delta }S(\cdot, t_0)(x)\ge c_0e^{td_S\Delta }S(\cdot, t_0)(y),\quad \forall \ td_S\ge 1, \ x,y\in \bar \Omega,\ t_0\gt 0. \end{equation}

This in turn together with (4.30) implies that

\begin{align*} \min _{x\in \overline{\Omega }}e^{d_St\Delta }S(\cdot,t_0)(x)\ge & \frac{c_0}{|\Omega |}\int _{\Omega }e^{td_S\Delta }S(\cdot,t_0)(y)dy\\[5pt] = & \frac{c_0}{|\Omega |}\int _{\Omega }S(y,t_0)dy\\[5pt] \ge & \frac{\gamma _m Nc_0}{|\Omega |(\beta _M+\gamma _m)}\left(1-e^{-t_0(\beta _M+\gamma _m)}\right),\quad \quad t\ge \frac{1}{d_S},\ t_0\gt 0. \end{align*}

As a result, it follows from (4.31) that

\begin{equation*} S\left (x,\frac {1}{d_S}+t_0\right )\ge \frac {\gamma _{m}Nc_0 e^{-\frac {\beta _M}{d_S}}}{|\Omega |(\beta _M+\gamma _{m})}\left(1-e^{-t_0(\beta _M+\gamma _m)}\right),\quad \quad \ x\in \bar \Omega, t_0\gt 0. \end{equation*}

Therefore, taking $t_1=\frac{1}{d_S}+\frac{\ln (2)}{\beta _M+\gamma _{m}}$ , (4.27) holds with $\underline{S}\,:\!=\,\frac{\gamma _mNc_0 e^{-\frac{\beta _M}{d_S}}}{2|\Omega |(\beta _M+N\gamma _{m})}\gt 0$ , where $\underline{S}$ and $t_1$ are independent of the initial data.

Next, we show that (4.28) holds. To this end, we first note that

\begin{align*} I_t= \beta \Big ((1-r)-\frac{I}{I+S}\Big )I\le & \beta \Big ((1-r)-\frac{I}{I+\overline{S}}\Big )I\\[5pt] =& \gamma \Big ((R-1)\overline{S}-I\Big )\frac{I}{I+\overline{S}}, \ \quad t\gt 0, \end{align*}

which in view of the comparison principle for ordinary differential equations implies that $\limsup _{t\to \infty }I(x,t)\le (R-1)_{+}\overline{S}$ . Recalling that $I(x,t)=0$ for all $t\gt 0$ whenever $I_0(x)=0$ , we then conclude that $\limsup _{t\to \infty }I(x,t)\le (R-1)_{+}\overline{S}\chi _{\{I_0\gt 0\}}$ .

Similarly, observing that

\begin{align*} I_t= \beta \left(\left(1-\frac{\gamma }{\beta }\right)-\frac{I}{I+S}\right)I\ge & \beta \left((1-r)-\frac{I}{I+\underline{S}}\right)I\\ =& \gamma \left((R-1)\underline{S}-I\right)\frac{I}{I+\underline{S}},\quad t\gt t_1, \end{align*}

we can proceed as in the previous case to establish that $\liminf _{t\to \infty }I(x,t)\ge (R-1)_{+}\underline{S}\chi _{\{I_0\gt 0\}}$ , which completes the proof of (i).

(ii) By the equation of $I$ , we have

\begin{equation*} \int _0^t\int _{H^-\cup H^0\cup \{I_0=0\}} \left (\beta \frac {SI}{S+I}-\gamma I\right ) dxds=\int _{H^-\cup H^0\cup \{I_0= 0\}} I(x, t)dx-\int _{H^-\cup H^0\cup \{I_0= 0\}} I_0dx. \end{equation*}

Note that

\begin{equation*} \beta \frac {SI}{S+I}-\gamma I=\frac {((\beta -\gamma )S-{\gamma }I)I}{S+I}\le 0, \ \ \forall x\in H^-\cup H^0\cup \{I_0=0\}. \end{equation*}

We have

\begin{equation*} 0\le \int _0^\infty \int _{H^-\cup H^0\cup \{I_0=0\}} \left (-\beta \frac {SI}{S+I}+\gamma I\right ) dxds\lt \infty, \end{equation*}

which together with (4.26) yields that

\begin{equation*} 0\le \int _0^\infty \int _{H^-\cup H^0\cup \{I_0=0\}} S\left (-{\beta }\frac {SI}{S+I}+\gamma I\right )dxdt\lt \infty. \end{equation*}

Hence, integrating (4.25) over $(0, \infty )$ , we obtain that

(4.33) \begin{equation} \int _0^\infty \int _\Omega |\triangledown S|^2dxdt\lt \infty \end{equation}

and

(4.34) \begin{equation} \int _0^\infty \int _{H^+\cap \{I_0\gt 0\}} (\beta -\gamma )_+I\frac{(S-\kappa I)^2}{S+I}dxdt\lt \infty. \end{equation}

By (4.26), we have

\begin{equation*} \sup _{t\ge 0}\left\|\frac {\beta SI}{S+I}-\gamma I\right\|_{L^\infty (\Omega )}\lt \infty. \end{equation*}

So by the regularity theory for parabolic equations and Sobolev embedding theorem, the mappings $\nabla S(x,t)$ and $S$ are Hölder continuous on $\overline{\Omega }\times [1,\infty )$ , and $\{S(\cdot,t)\}_{t\ge 1}$ is precompact in $C^{1+\alpha }(\Omega )$ , $0\lt \alpha \lt 1$ . Then by (4.33) and Lemma 2.2, $\int _{\Omega }|\nabla S|^2dx\to 0$ as $t\to \infty$ . Hence, the set $w_S\,:\!=\,\cap _{t\ge 1}\overline{\cup _{s\ge t}\{S(\cdot,s)\}}$ consists of positive constant functions. Furthermore, since $ \sup _{t\ge 0}\|\partial _t I(\cdot,t)\|_{L^\infty (\Omega )}\lt \infty$ and $S$ is Hölder continuous on $\overline{\Omega }\times [1,\infty )$ , the mapping

\begin{equation*} t\mapsto \int _{H^+\cap \{I_0\gt 0\}} (\beta -\gamma )_+\frac {(S-\kappa I)^2}{S+I}Idx \end{equation*}

is Hölder continuous on $[1,\infty )$ . Therefore, by Lemma 2.2 and (4.34), we have

(4.35) \begin{equation} \lim _{t\to \infty }\int _{H^+\cap \{I_0\gt 0\}} (\beta -\gamma )_+\frac{(S-\kappa I)^2}{S+I}Idx=0. \end{equation}

Now, let $S^*\in w_S$ . Then there is a sequence $t_k\to \infty$ such that $S(\cdot,t_k)\to S^*$ uniformly on $\bar \Omega$ as $k\to \infty$ . Since (4.35) holds, after passing to a subsequence if necessary, we may suppose that

(4.36) \begin{equation} \lim _{k\to \infty }I(x,t_k)\frac{(S^*-\kappa (x)I(x,t_k))^2}{S^*+I(x,t_k)}=0 \quad \text{ a.e. on }\ H^+\cap \{I_0\gt 0\}. \end{equation}

However, when $x\in H^+\cap \{I_0\gt 0\}$ , we have from (4.28) that $\liminf _{t\to \infty }I(x,t)\ge \underline{S}(\beta (x)-\gamma (x))\gt 0$ . Therefore, we conclude from (4.36) that

\begin{equation*} \lim _{{k}\to \infty }I(x,t_k)=\frac {S^*}{\kappa (x)}=\frac {(\beta (x)-\gamma (x))S^*}{\gamma (x)} \quad \text{ a.e. on }\ H^+\cap \{I_0\gt 0\}. \end{equation*}

By (4.28) again, $\lim _{t\to \infty }I(x,t)=0$ almost everywhere for $x\in \Omega \setminus (H^+\cap \{I_0\gt 0\})$ . So by the dominated convergence theorem, we have

\begin{eqnarray*} N&=&\lim _{k\to \infty }\int _{\Omega }(S(\cdot,t_k)+I(\cdot,t_k))dx\\ &=&|\Omega |S^*+S^*\int _{H^+\cap \{I_0\gt 0\}}\frac{1}{\kappa }dx=\left(|\Omega |+\int _{H^+\cap \{I_0\gt 0\}}\frac{\beta -\gamma }{\gamma }dx\right)S^*. \end{eqnarray*}

This yields that

\begin{equation*} S^*=\frac {N}{|\Omega |+\int _{H^+\cap \{I_0\gt 0\}}\frac {\beta -\gamma }{\gamma }dx}. \end{equation*}

Since $S^*$ is independent of the chosen subsequence, we have

\begin{equation*} w_S=\left\{\frac {N}{|\Omega |+\int _{H^+\cap \{I_0\gt 0\}}\frac {\beta -\gamma }{\gamma }dx} \right\}, \end{equation*}

and therefore (4.29) holds. Since $S(\cdot,t)\to S^*$ uniformly on $\Omega$ as $t\to \infty$ , we obtain from (4.24) that $I(\cdot,t)\to \frac{S^*(\beta -\gamma )^+}{\gamma }\chi _{H^+\cap \{I_0\gt 0\}}$ uniformly on $\bar \Omega$ as $t\to \infty$ .