Appendix
Proof of Theorem 1
Let us set \( \max _{\bar{\Omega }_T}\,B=B({\mathbf x}_1, t_1)\). We have to distinguish two cases.
-
(1)
If \(({\mathbf x}_1,t_1)\) belongs to the interior of \(\Omega _T\), then (1)\(_3\) and \(\Vert I\Vert _\infty \le N_0\) imply that
$$\begin{aligned} \left[ \frac{\partial B}{\partial t}-eN_0+(\mu _B-\pi _B)B-\gamma _3\Delta B\right] _{({\mathbf x}_1,t_1)}<0. \end{aligned}$$
(53)
Since
$$\begin{aligned} \left[ \frac{\partial B}{\partial t}\right] _{({\mathbf x}_1,t_1)}=0,\qquad \left[ \Delta B\right] _{({\mathbf x}_1,t_1)}<0, \end{aligned}$$
then (53) can hold if and only if
$$\begin{aligned} -eN_0+(\mu _B-\pi _B)B({\mathbf x}_1,t_1)<0 \end{aligned}$$
and hence if and only if \(B({\mathbf x}_1,t_1)< \frac{eN_0}{\mu _B-\pi _B}\).
-
(2)
If \(({\mathbf x}_1, t_1)\in \Gamma _T\), in view of the regularity of the domain \(\Omega \), since \(\Omega \) verifies in any point \({\mathbf x}_0\in \partial \Omega \) the interior ball condition, there exists an open ball \(B^*\subset \Omega \) with \({\mathbf x}_0\in \partial B^*\). If \(B({\mathbf x}_1, t_1)> \frac{eN_0}{\mu _B-\pi _B}\), on choosing the radius of \(B^*\) sufficiently small, it follows that
$$\begin{aligned} \gamma _3\Delta B- \frac{\partial B}{\partial t}>0,\quad \hbox {in}\,\, B^* \end{aligned}$$
and by virtue of Hopf’s Lemma (Protter and Weinberger 1967), one obtains that
$$\begin{aligned} \left( \frac{dB}{d{\mathbf n}}\right) _{({\mathbf x}_1,t_1)}>0. \end{aligned}$$
Since \(\frac{dB}{d{\mathbf n}}=0\) on \(\partial \Omega \times {\mathbb R}^+\), (10) follows.
Proof of Theorem 2
Substituting \((\bar{S},\,\bar{I},\,\bar{B},\,\bar{R})=(N_0,0,0,0)\) in (17), one has that
$$\begin{aligned} {\left\{ \begin{array}{ll} b_{11}=-(\mu +\bar{\alpha }\gamma _1),\quad b_{13}=- \frac{\mu _3\beta N_0}{\mu _1K_B},\quad b_{21}=0,\\ b_{22}=-(\sigma +\mu +\bar{\alpha }\gamma _2),\quad b_{23}= \frac{\mu _3\beta N_0}{\mu _2K_B},\\ b_{32}= \frac{\mu _2}{\mu _3}e,\quad b_{33}=-(\mu _B-\pi _B+\bar{\alpha }\gamma _3). \end{array}\right. } \end{aligned}$$
(54)
Hence
$$\begin{aligned} A^{*}&= (\sigma +\mu +\bar{\alpha }\gamma _2)(\mu _B-\pi _B+\bar{\alpha }\gamma _3)- \frac{\beta N_0 e}{K_B}\nonumber \\&= (\sigma + \mu )(\mu _B - \pi _B)\left[ 1 - R_0 + \frac{\bar{\alpha }[\gamma _2(\mu _B - \pi _B) + \gamma _3(\sigma + \mu ) + \bar{\alpha }\gamma _2\gamma _3]}{(\sigma +\mu )(\mu _B-\pi _B)}\right] \nonumber \\&= (\sigma +\mu )(\mu _B-\pi _B)(R^*_0-R_0) \end{aligned}$$
(55)
and
$$\begin{aligned} A^{*}_1=0,\quad A^{*}_2=-b_{11}(b_{11}+\mathtt{I }^{*})<0. \end{aligned}$$
(56)
In view of (56), it follows that
$$\begin{aligned} \max \{A^{*}_1,A^*_2\}=0. \end{aligned}$$
Hence (24) is verified if and only if
$$\begin{aligned} A^{*}>0, \end{aligned}$$
(57)
i.e., by virtue of (55), if and only if (27) holds.
Proof of Theorem 3
Substituting \((\bar{S},\,\bar{I},\,\bar{B},\,\bar{R})=(S_2,\,I_2,\,B_2,\, R_2)\) in (17), one has that
$$\begin{aligned} {\left\{ \begin{array}{ll} b_{11}=- \dfrac{\mu (\beta +\mu )R_0+\bar{\alpha }\gamma _1(\beta +\mu R_0)}{\beta +\mu R_0},\\ b_{13}=- \dfrac{\mu _3(\sigma +\mu )(\mu _B-\pi _B)(\beta +\mu )}{\mu _1e(\beta +\mu R_0)},\\ b_{21}= \dfrac{\mu _1\beta \mu (R_0-1)}{\mu _2(\beta +\mu R_0)},\quad b_{22}=-(\sigma +\mu +\bar{\alpha }\gamma _2),\\ b_{23}=- \dfrac{\mu _1}{\mu _2}b_{13},\quad b_{32}= \dfrac{\mu _2}{\mu _3}e,\quad b_{33}=-(\mu _B-\pi _B+\bar{\alpha }\gamma _3).\end{array}\right. } \end{aligned}$$
(58)
Hence
$$\begin{aligned} A^{*}&= (\sigma + \mu + \bar{\alpha }\gamma _2)(\mu _B - \pi _B + \bar{\alpha }\gamma _3) - \frac{(\sigma + \mu )(\mu _B - \pi _B)(\beta + \mu )}{\beta +\mu R_0}\nonumber \\&= \frac{\mu (\sigma + \mu )(\mu _B - \pi _B)(R_0 - 1)}{\beta +\mu R_0} + \bar{\alpha }\left[ \gamma _2(\mu _B - \pi _B) + \gamma _3(\sigma + \mu ) + \bar{\alpha }\gamma _2\gamma _3\right] ,\nonumber \\ \end{aligned}$$
(59)
and
$$\begin{aligned} {\left\{ \begin{array}{ll} A^{*}_1=- \dfrac{\beta \mu (\beta +\mu )(\sigma +\mu )(\mu _B-\pi _B)(R_0-1)}{(\beta +\mu R_0)[\mu R_0(\beta +\mu )+\bar{\alpha }\gamma _1(\beta +\mu R_0)]},\\ A^*_2=- \dfrac{K_1 R^2_0+K_2R_0+K_3}{(\beta +\mu R_0)^2[\sigma +\mu +\mu _B-\pi _B+\bar{\alpha }(\gamma _2+\gamma _3)]}(<\!0), \end{array}\right. } \end{aligned}$$
(60)
where \(K_i,\,(i=1,2,3)\) are positive constants given by
$$\begin{aligned} K_1&= \mu ^2(\sigma + \mu + \mu _B - \pi _B + \bar{\alpha }(\gamma _2 + \gamma _3))\left\{ \bar{\alpha }\gamma _1[\sigma + \mu + \mu _B - \pi _B \right. \\&+\, \bar{\alpha }(\gamma _1 + \gamma _2 + \gamma _3)] + (\beta +\mu ) \left[ \beta +\mu +\sigma +\mu +\mu _B-\pi _B\right. \\&+\left. \left. \bar{\alpha }(\gamma _1+\gamma _2+\gamma _3)+\bar{\alpha }\gamma _1\right] \right\} , \end{aligned}$$
$$\begin{aligned} K_2&= \beta \mu \left\{ \left[ \sigma +\mu +\mu _B-\pi _B+\bar{\alpha }(\gamma _2+\gamma _3)\right] \left[ \bar{\alpha }(\beta +\mu )(\gamma _1+\gamma _2+\gamma _3)\right. \right. \\&+\left. \bar{\alpha }\gamma _1(\beta +\mu )+2\bar{\alpha }\gamma _1(\sigma +\mu +\mu _B-\pi _B+\bar{\alpha }(\gamma _1+\gamma _2+\gamma _3))\right] \\&+\!\left. (\beta +\mu )(\mu _B-\pi _B)^2+(\beta +\mu )(\sigma +\mu +\mu _B-\pi _B)[\sigma +\mu +\bar{\alpha }(\gamma _2+\gamma _3)]\right\} \end{aligned}$$
and
$$\begin{aligned} K_3&= \beta ^2\bar{\alpha }\gamma _1[\sigma \!+\! \mu \!+\! \mu _B \!-\! \pi _B \!+\! \bar{\alpha }(\gamma _2 + \gamma _3)] [\sigma + \mu + \mu _B - \pi _B + \bar{\alpha }(\gamma _1 + \gamma _2 + \gamma _3)]\\&+\beta \mu (\mu _B-\pi _B)(\sigma +\mu )(\beta +\mu ). \end{aligned}$$
Since \((S_2,\,I_2,\,B_2,\,R_2)\) exists if and only if \(R_0>1\), then, from (59) and (60)\(_1\), it turns out that
$$\begin{aligned} A^{*}>0,\quad A^{*}_1<0. \end{aligned}$$
(61)
In view of (60)\(_2\) and (61), it follows that (24) is always satisfied.
Proof of Lemma 2
By virtue of Remark 4, the linear stability of the biologically meaningful equilibria guarantees that \(A^*>0\). Hence, the proof of (i) can be obtained on following the same procedure by Rionero (2011a, b), Capone et al. (2013, 2014). As concerns (ii), for the disease-free equilibrium, since \(b_{21}=0\), one has that
$$\begin{aligned} \Phi ^{*}=b_{13}\langle U_1,U_3 \rangle \le \frac{\mu _3^3\beta ^2N^2_0}{2\mu ^2_1K^2_B|I^*A^*|}\Vert U_1\Vert ^2+ \frac{1}{2}|I^*A^*|(\Vert U_2\Vert ^2+\Vert U_3\Vert ^2). \end{aligned}$$
On choosing
$$\begin{aligned} \frac{\mu ^2_3}{\mu ^2_1}= \frac{|b_{11}I^*A^*|K^2_B}{\beta ^2N^2_0}, \end{aligned}$$
(62)
(ii) is obtained. For the endemic equilibrium, since \(b_{21}>0\) and \(b_{13}<0\), on choosing
$$\begin{aligned} \frac{\mu ^2_1}{\mu _2\mu _3}=- \frac{(\sigma +\mu )(\mu _B-\pi _B)(\beta +\mu )}{A_3e\beta \mu (R_0-1) }>0, \end{aligned}$$
(63)
it follows that \(b_{13}-A_3b_{21}=0\) and
$$\begin{aligned} \Phi ^*\!=\! \frac{\mu _1\beta \mu (R_0-1)A_1}{\mu _2(\beta +\mu R_0)} \langle U_1,U_2 \rangle \!\le \! \frac{\mu _1^2\beta ^2\mu ^2(R_0\!-\!1)^2A^2_1}{2|I^*A^*|\mu _2^2(\beta +\mu R_0)^2}\Vert U_1\Vert ^2+ \frac{1}{2}|I^*A^*|\Vert U_2\Vert ^2. \end{aligned}$$
Hence, on taking
$$\begin{aligned} \frac{\mu ^2_1}{\mu ^2_2}= \frac{|b_{11}I^*A^*|(\beta +\mu R_0)^2}{\beta ^2\mu ^2(R_0-1)^2A^2_1}, \end{aligned}$$
(64)
(34) is proved.
As concerns (iii), by virtue of (6), (11), (15)\(_1\), the following inequalities hold a.e. in \(\Omega \)
$$\begin{aligned} \theta _1\mu _1U_1+\bar{S}=\theta _1X_1+\bar{S}=\theta _1(S-\bar{S})+\bar{S}=\theta _1S+(1-\theta _1)\bar{S}\le N_0 \end{aligned}$$
and
$$\begin{aligned} K_B+\theta _1\mu _3U_3+\bar{B}=K_B+\theta _1B+(1-\theta _1)\bar{B}>K_B. \end{aligned}$$
Hence, from (32)\(_7\), it turns out that
$$\begin{aligned} \Phi _2&< \mu _3\left[ c_1 \frac{\mu _3}{\mu _1} \int _{\Omega }|U_1|U^2_3\,d \Omega +c_2 \int _{\Omega }U^2_1|U_3|\,d \Omega \right. \\&\quad +\left. c_3 \frac{\mu _3}{\mu _2} \int _{\Omega }|U_2|U_3^2\,d \Omega +c_4 \frac{\mu _1}{\mu _2} \int _{\Omega }|U_1U_2U_3|\,d \Omega \right. \\&\quad \left. +c_5 \frac{\mu _3}{\mu _2} \int _{\Omega }|U_3|^3\,d \Omega +c_6 \frac{\mu _1}{\mu _2} \int _{\Omega }|U_1|U_3^2\,d \Omega \right] . \end{aligned}$$
From (6), (10) and (15)\(_1\) one obtains
$$\begin{aligned} \Vert U_i\Vert _\infty = \frac{1}{\mu _i}\Vert X_i\Vert _\infty \le \frac{2N_0}{\mu _i},\quad i=1,2 \end{aligned}$$
and
$$\begin{aligned} \Vert U_3\Vert _\infty = \frac{1}{\mu _3}\Vert X_3\Vert _\infty = \frac{1}{\mu _3}\Vert B-\bar{B}\Vert _\infty \le \frac{2M}{\mu _3}. \end{aligned}$$
Hence
$$\begin{aligned} \Phi _2&< \mu _3\left[ \frac{2c_1M}{\mu _1} \int _\Omega |U_1||U_3|d\Omega + \frac{2c_2N_0}{\mu _1} \int _\Omega |U_1||U_3|d\Omega + \frac{2c_3M}{\mu _2} \int _\Omega |U_2||U_3|d\Omega \right. \\&+ \left. \frac{2c_4N_0}{\mu _2} \int _\Omega |U_2||U_3|d\Omega + \frac{2c_5M}{\mu _2} \int _\Omega U^2_3d\Omega + \frac{2c_6N_0}{\mu _2} \int _\Omega U^2_3d\Omega \right] . \end{aligned}$$
On applying the Cauchy–Schwarz, (35) is obtained.