3.1 Proof of global existence
In this section, we focus on proof of global existence in Theorem 1.1. Inspired by [30], we will first establish estimate on \(\int_{\Omega}u^{m+1}\, dx+\int_{\Omega}|\nabla v|^{4}\, dx\) to obtain estimate on \(\int_{\Omega}u^{p} \, dx\) for any \(p>1\). To achieve this purpose, we need a series of a priori estimates.
Lemma 3.1
Let the same assumptions as in Theorem
1.1
hold. Then the solution of (1.1) satisfies
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}u^{p} \,dx+\frac{p(p-1)c_{D}}{2}\int_{\Omega}u^{p+m-3}|\nabla u|^{2} \,dx \\& \quad \le\frac{p(p-1)\|\chi\|_{L^{\infty}(\Omega )}^{2}}{2c_{D}}\int_{\Omega}u^{p-m+1}|\nabla v|^{2} \,dx \\& \qquad {}+\kappa p\int_{\Omega}u^{p} \,dx-\mu p\int_{\Omega}u^{p+\tau}\,dx \end{aligned}$$
(3.1)
for any
\(p>m\)
and each
\(t\in(0,T^{*})\). Particularly, we have
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}u^{m+1} \,dx+\frac{m(m+1)c_{D}}{2}\int_{\Omega}u^{2m-2}|\nabla u|^{2} \,dx \\& \quad \le\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega )}^{2}}{2c_{D}}\int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \qquad {}+\kappa(m+1)\int_{\Omega}u^{m+1} \,dx-\mu(m+1)\int_{\Omega}u^{m+1+\tau}\,dx. \end{aligned}$$
(3.2)
Proof
Multiplying (1.1)1 by \(u^{p-1}\) and integrating the resulted equation over Ω, we obtain
$$\begin{aligned}& \frac{1}{p}\frac{d}{dt}\int_{\Omega}u^{p} \,dx+(p-1)\int_{\Omega}u^{p-2}D(u)| \nabla u|^{2} \,dx \\& \quad =(p-1)\int_{\Omega}u^{p-1} \chi(v)\nabla u\cdot \nabla v \,dx+\int_{\Omega}u^{p}f(u)\,dx. \end{aligned}$$
(3.3)
We now estimate the last three terms of the above equality. Indeed, by (1.2) we have
$$(p-1)\int_{\Omega}u^{p-2}D(u)|\nabla u|^{2} \,dx\ge(p-1)c_{D}\int_{\Omega}u^{p+m-3}|\nabla u|^{2} \,dx, $$
and by the Young inequality, we get
$$\begin{aligned}& (p-1)\int_{\Omega}u^{p-1}\chi(v)\nabla u\cdot\nabla v \,dx \\& \quad \le(p-1)\|\chi\| _{L^{\infty}(\Omega)}\int_{\Omega}u^{p-1}|\nabla u||\nabla v| \,dx \\& \quad \le\frac{(p-1)c_{D}}{2}\int_{\Omega}u^{p+m-3}|\nabla u|^{2} \,dx+\frac{(p-1)\| \chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}\int_{\Omega}u^{p-m+1}|\nabla v|^{2} \,dx. \end{aligned}$$
For the last term, in view of (1.5), we have
$$\int_{\Omega}u^{p}f(u)\,dx\le\kappa\int _{\Omega}u^{p}\,dx-\mu\int_{\Omega}u^{p+\tau}\,dx. $$
Summarily, we obtain
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}u^{p} \,dx+ \frac{p(p-1)c_{D}}{2}\int_{\Omega}u^{p+m-3}|\nabla u|^{2} \,dx \\& \quad \le\frac{p(p-1)\|\chi\|_{L^{\infty}(\Omega )}^{2}}{2c_{D}}\int_{\Omega}u^{p-m+1}|\nabla v|^{2} \,dx \\& \qquad {}+\kappa p\int_{\Omega}u^{p} \,dx-\mu p\int_{\Omega}u^{p+\tau}\,dx. \end{aligned}$$
Taking \(p=m+1\), one can easily deduce (3.2). This completes the proof of Lemma 3.1. □
We then establish a coupled estimate on \(\int_{\Omega}[(1+u)\ln (1+u)-u ]\,dx+\int_{\Omega}|\nabla v|^{2} \,dx\).
Lemma 3.2
Let the same assumptions as in Theorem
1.1
hold. Then there exists a positive constant
C
such that
$$ \int_{\Omega}\bigl[(1+u)\ln(1+u)-u \bigr]\,dx\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr) $$
(3.4)
and
$$ \int_{\Omega}|\nabla v|^{2} \,dx\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.5)
Proof
We shall divide the proof into three steps.
Step 1. First testing (1.1)1 against \(\ln(1+u)\), we have
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}\bigl[(1+u) \ln(1+u)-u \bigr] \,dx+\int_{\Omega}\frac {D(u)}{1+u}|\nabla u|^{2}\,dx \\& \quad =\int_{\Omega}\frac{u}{1+u}\chi(v)\nabla u\cdot \nabla v \,dx +\int_{\Omega}uf(u)\ln(1+u)\,dx \end{aligned}$$
(3.6)
for all \(t\in(0,T^{*})\). To estimate each term on the right side of (3.6), we first note that
$$ \ln(1+u)\le u \quad \text{for all } u\ge0. $$
Then utilizing the Young inequality and integration by parts, we get
$$\begin{aligned}& \int_{\Omega}\frac{u}{1+u}\chi(v)\nabla u \cdot\nabla v\,dx \\& \quad =\int_{\Omega}\bigl[\ln(1+u)-u \bigr]\chi(v) \Delta v\,dx+\int_{\Omega}\bigl[\ln(1+u)-u \bigr]\chi '(v)|\nabla v|^{2}\,dx \\& \quad \le\frac{\|\chi\|_{L^{\infty}}^{2}}{2}\int_{\Omega}\bigl[\ln(1+u)-u \bigr]^{2}\,dx+\frac{1}{2}\int_{\Omega}|\Delta v|^{2}\,dx \\& \quad \le\frac{\|\chi\|_{L^{\infty}}^{2}}{2}\int_{\Omega}u^{2}\,dx+ \frac{1}{2}\int_{\Omega}|\Delta v|^{2}\,dx\quad \text{for all } t\in\bigl(0,T^{*}\bigr), \end{aligned}$$
(3.7)
where we have used \(\chi'(v)\ge0\). For the last term on the right of (3.6), according to (1.4), we obtain
$$\begin{aligned}& \int_{\Omega}uf(u)\ln(1+u)\,dx \\& \quad \le\int _{\Omega}u\bigl(\kappa-\mu u^{\tau}\bigr)\ln (1+u)\,dx \\& \quad = \int_{\Omega}\bigl[\kappa u\ln(1+u)-\mu u^{1+\tau} \ln(1+u)\bigr]\,dx\quad \text{for all } t\in\bigl(0,T^{*}\bigr). \end{aligned}$$
(3.8)
Then combining (3.7) and (3.8) along with (3.6), we have
$$\begin{aligned}& \frac{d}{dt} \int_{\Omega}\bigl[(1+u) \ln(1+u)-u \bigr]\,dx+\int_{\Omega}\frac {D(u)}{1+u}|\nabla u|^{2} \,dx \\& \quad \le\frac{\|\chi\|_{L^{\infty}}^{2}}{2}\int_{\Omega}u^{2} \,dx+ \frac{1}{2}\int_{\Omega}|\Delta v|^{2}\,dx \\& \qquad {}+ \int _{\Omega}\bigl[ \kappa u\ln(1+u)-\mu u^{1+\tau}\ln (1+u) \bigr]\,dx \quad \text{for all } t\in\bigl(0,T^{*}\bigr). \end{aligned}$$
(3.9)
Step 2. In order to cancel \(\frac{1}{2}\int_{\Omega}|\Delta v|^{2}\,dx\) on the right of (3.9), we first test (1.1)2 against \(-\Delta v\) to find that
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int_{\Omega}|\nabla v|^{2}\,dx =&-\int_{\Omega}|\Delta v|^{2}\,dx+ \int_{\Omega}ug(v)\Delta v\,dx+\int_{\Omega}v\Delta v\,dx \\ &{}-\int_{\Omega}v\Delta v\,dx \quad \text{for all } t\in \bigl(0,T^{*}\bigr). \end{aligned}$$
Then by integration by parts and the Young inequality, we have
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int_{\Omega}| \nabla v|^{2}\,dx \le&-\int_{\Omega}|\Delta v|^{2}\,dx+\|g\|_{L^{\infty}}\int_{\Omega}u|\Delta v|\,dx \\ &{}-\int_{\Omega}|\nabla v|^{2}\,dx+ \frac{\int_{\Omega}|\Delta v|^{2}\,dx}{4}+\int_{\Omega}| v|^{2}\,dx \\ \le&-\frac{\int_{\Omega}|\Delta v|^{2}\,dx}{2}+\|g\|_{L^{\infty}}^{2}\int _{\Omega}u^{2}\,dx-\int_{\Omega}|\nabla v|^{2}\,dx \\ &{}+\|v_{0}\|_{L^{\infty}(\Omega)}^{2}|\Omega | \quad \text{for all } t\in\bigl(0,T^{*}\bigr), \end{aligned}$$
(3.10)
where (2.3) has been used.
Step 3. Adding (3.10) to (3.9) yields
$$\begin{aligned}& \frac{d}{dt} \biggl\{ \int_{\Omega}\bigl[(1+u)\ln(1+u)-u \bigr]\,dx+\frac{1}{2}\int_{\Omega}|\nabla v|^{2} \,dx \biggr\} \\& \qquad {}+\int_{\Omega}\frac {D(u)}{1+u}|\nabla u|^{2} \,dx+\int_{\Omega}|\nabla v|^{2}\,dx \\& \quad \le C_{2}\int_{\Omega}u^{2} \,dx+\int _{\Omega}\bigl[\kappa u\ln(1+u)-\mu u^{1+\tau }\ln(1+u)\bigr] \,dx \\& \qquad {}+\|v_{0}\|_{L^{\infty}(\Omega)}^{2}|\Omega| \quad \text{for all } t\in\bigl(0,T^{*}\bigr) \end{aligned}$$
with \(C_{2}=\frac{\|\chi\|_{L^{\infty}(\Omega)}^{2}+2\|g\|_{L^{\infty}(\Omega )}^{2}}{2}\), which is a positive constant according to (1.3) and (2.3). Adding \(\int_{\Omega}[(1+u)\ln(1+u)-u ]\,dx\) to both sides of this and dropping the nonnegative term \(\int_{\Omega}\frac {D(u)}{1+u}|\nabla u|^{2} \,dx\) on the left, we obtain
$$\begin{aligned}& \frac{d}{dt} \biggl\{ \int_{\Omega}\bigl[(1+u)\ln(1+u)-u \bigr]\,dx+\frac{1}{2}\int_{\Omega}| \nabla v|^{2}\,dx \biggr\} \\& \qquad {}+\int_{\Omega}|\nabla v|^{2}\,dx+\int_{\Omega}\bigl[(1+u)\ln(1+u)-u \bigr] \,dx \\& \quad \le C_{2}\int_{\Omega}u^{2} \,dx+\int _{\Omega}\bigl[(1+u)\ln(1+u)-u \bigr]\,dx \\& \qquad {}+ \int _{\Omega}\bigl[\kappa u\ln(1+u)-\mu u^{1+\tau}\ln(1+u)\bigr] \,dx+\|v_{0}\|_{L^{\infty}(\Omega)}^{2}|\Omega| \end{aligned}$$
(3.11)
for all \(t\in(0,T^{*})\). In view of Lemma 2.3, there exists a positive constant \(C^{*}\) such that
$$\int_{\Omega}\bigl\{ C_{2} u^{2}+\kappa u \ln(1+u)-\mu u^{1+\tau}\ln (1+u)+(1+u)\ln(1+u)-u \bigr\} \,dx\le C^{*}|\Omega|. $$
Thus \(y(t):=\int_{\Omega}[(1+u)\ln(1+u)-u ]\,dx+\frac{1}{2}\int_{\Omega}|\nabla v|^{2}\,dx\) satisfies
$$\frac{dy}{dt}+y\le C^{*}|\Omega|+\|v_{0}\|_{L^{\infty}(\Omega)}^{2}| \Omega | \quad \text{for all } t\in\bigl(0,T^{*}\bigr). $$
By standard ODE argument, we can derive
$$y(t)\le\max \bigl\{ y(0), C^{*}|\Omega|+\|v_{0}\|_{L^{\infty}(\Omega)}^{2}| \Omega | \bigr\} \quad \text{for all } t\in\bigl(0,T^{*}\bigr), $$
which implies (3.4) and (3.5). □
In order to obtain the coupled estimate on \(\int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx\), we then derive the following energy estimate on \(\int_{\Omega}|\nabla v|^{4}\,dx\).
Lemma 3.3
Let the same assumptions as in Theorem
1.1
hold. Then there exists a positive constant
\(C_{1}\)
such that the solution of (1.1) satisfies
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}|\nabla v|^{4}\,dx+\frac{1}{2}\int_{\Omega}\bigl\vert \nabla|\nabla v|^{2}\bigr\vert ^{2}\,dx \\& \quad \le(4+N)\|g \|_{L^{\infty}(\Omega)}^{2}\int_{\Omega}u^{2}| \nabla v|^{2} \,dx+C_{1}\quad \textit{for all } t\in \bigl(0,T^{*}\bigr). \end{aligned}$$
(3.12)
Proof
Differentiating equation (1.1)2, we obtain
$$\bigl(|\nabla v|^{2} \bigr)_{t}=2\nabla v\cdot\nabla\Delta v-2\nabla v\cdot \nabla\bigl(ug(v)\bigr), $$
which, together with the point-wise identity \(2\nabla v\cdot\nabla \Delta v=\Delta|\nabla v|^{2}-2|D^{2} v|^{2}\), yields
$$ \bigl(|\nabla v|^{2} \bigr)_{t}=\Delta|\nabla v|^{2}-2\bigl\vert D^{2} v\bigr\vert ^{2}-2 \nabla v\cdot \nabla\bigl(ug(v)\bigr). $$
(3.13)
Multiplying both sides of (3.13) by \(2|\nabla v|^{2}\) and integrating over Ω, we have
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}|\nabla v|^{4}\,dx+2\int_{\Omega}\bigl\vert \nabla |\nabla v|^{2}\bigr\vert ^{2}\,dx+4\int_{\Omega}| \nabla v|^{2}\bigl|D^{2} v\bigr|^{2}\,dx \\& \quad =-4\int_{\Omega}|\nabla v|^{2}\nabla v\cdot \nabla\bigl(ug(v)\bigr)\,dx+2\int_{\partial\Omega}|\nabla v|^{2}\frac{\partial|\nabla v|^{2}}{\partial\nu }\,dx\quad \text{for all } t\in\bigl(0,T^{*} \bigr). \end{aligned}$$
(3.14)
For the first term on the right-hand side of (3.14), we can use integration by parts and the Young inequality to obtain
$$\begin{aligned}& -4\int_{\Omega}|\nabla v|^{2}\nabla v \cdot\nabla\bigl(ug(v)\bigr)\,dx \\& \quad =4\int_{\Omega}ug(v)|\nabla v|^{2}\Delta v \,dx+4\int_{\Omega}ug(v)\nabla v\cdot \nabla |\nabla v|^{2}\,dx \\& \quad \le\frac{4}{N}\int_{\Omega}|\nabla v|^{2}| \Delta v|^{2}\,dx+\int_{\Omega}\bigl\vert \nabla| \nabla v|^{2}\bigr\vert ^{2}\,dx+(4+N)\int _{\Omega}u^{2}g^{2}(v)|\nabla v|^{2} \,dx \\& \quad \le4\int_{\Omega}|\nabla v|^{2}\bigl\vert D^{2} v\bigr\vert ^{2}\,dx+\int_{\Omega}\bigl\vert \nabla |\nabla v|^{2}\bigr\vert ^{2}\,dx+(4+N) \|g\|_{L^{\infty}(\Omega)}^{2}\int_{\Omega}u^{2}| \nabla v|^{2} \,dx, \end{aligned}$$
(3.15)
where we have used the fact \(|\Delta v|^{2}\le N|D^{2} v|^{2}\). For the second term on the right-hand side of (3.14), thanks to the boundedness of \(\int_{\Omega}|\nabla v|^{2} \,dx\), by the same procedure as (3.7)-(3.10) in [31], we deduce that there exists a positive constant \(C_{1}\) such that
$$ 2\int_{\partial\Omega}|\nabla v|^{2} \frac{\partial|\nabla v|^{2}}{\partial \nu}\,dx\le\frac{1}{2}\int_{\Omega}\bigl\vert \nabla|\nabla v|^{2} \bigr\vert ^{2} \,dx+C_{1}\quad \text{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.16)
Combining (3.15) and (3.16) with (3.14), we obtain (3.12) immediately. □
Corollary 3.1
Let the same assumptions as in Theorem
1.1
hold. Then for the same constant
\(C_{1}\)
as in Lemma
3.3, the solution of (1.1) carries the property
$$\begin{aligned}& \frac{d}{dt} \biggl\{ \int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx \biggr\} + \biggl\{ \int_{\Omega}u^{m+1} \,dx+\int _{\Omega}|\nabla v|^{4}\,dx \biggr\} \\& \qquad {}+\frac{m(m+1)c_{D}}{2}\int_{\Omega}u^{2m-2}| \nabla u|^{2} \,dx+\frac{1}{2}\int_{\Omega}\bigl\vert \nabla|\nabla v|^{2}\bigr\vert ^{2}\,dx \\& \quad \le \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)\int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \qquad {}+ \bigl(1+\kappa(m+1) \bigr)\int_{\Omega}u^{m+1} \,dx-\mu(m+1)\int_{\Omega}u^{m+1+\tau}\,dx+ \int_{\Omega}|\nabla v|^{4}\,dx+C_{1}. \end{aligned}$$
(3.17)
Proof
Adding (3.12) to (3.2) yields
$$\begin{aligned}& \frac{d}{dt} \biggl\{ \int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx \biggr\} \\& \qquad {}+\frac{m(m+1)c_{D}}{2}\int_{\Omega}u^{2m-2}|\nabla u|^{2} \,dx+\frac{1}{2}\int_{\Omega}\bigl\vert \nabla|\nabla v|^{2}\bigr\vert ^{2}\,dx \\& \quad \le \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)\int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \qquad {}+\kappa (m+1)\int_{\Omega}u^{m+1} \,dx -\mu(m+1)\int_{\Omega}u^{m+1+\tau} \,dx+C_{1}. \end{aligned}$$
(3.18)
Adding \(\int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx\) to both sides of this, then yields (3.17). □
Now we are ready to establish the estimate on \(\int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx\).
Lemma 3.4
Let
\(m>2-\frac{4}{N+2}\), \(N\ge1\). Then there exists a positive constant
C
such that
$$ \int_{\Omega}u^{m+1} \,dx\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr) $$
(3.19)
and
$$ \int_{\Omega}|\nabla v|^{4}\,dx\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.20)
Proof
In order to obtain the estimate on the couple of \(\int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx\), we need to estimate the integrals on the right of inequality (3.17). We first utilize the Gagliardo-Nirenberg inequality and (3.5) to estimate \(\int_{\Omega}|\nabla v|^{4}\,dx\):
$$\begin{aligned} \int_{\Omega}|\nabla v|^{4} \,dx &=\bigl\Vert |\nabla v|^{2}\bigr\Vert ^{2}_{L^{2}(\Omega)} \\ &\le c_{1}\bigl\Vert \nabla|\nabla v|^{2}\bigr\Vert ^{2\lambda}_{L^{2}(\Omega)}\bigl\Vert |\nabla v|^{2}\bigr\Vert ^{2(1-\lambda)}_{L^{1}(\Omega)}+c_{1}\bigl\Vert |\nabla v|^{2}\bigr\Vert ^{2}_{L^{1}(\Omega)} \\ &\le c_{2}\bigl\Vert \nabla|\nabla v|^{2}\bigr\Vert ^{2\lambda}_{L^{2}(\Omega)}+c_{2} \\ &\le\frac{1}{4}\bigl\Vert \nabla|\nabla v|^{2}\bigr\Vert ^{2}_{L^{2}(\Omega)}+c_{3}\quad \text{for all } t\in \bigl(0,T^{*}\bigr). \end{aligned}$$
(3.21)
Here in the last inequality we have used the Young inequality, because \(\lambda=\frac{N}{2+N}\in(0,1)\). Similarly, invoking the Young inequality with \(\epsilon_{2}>0\) to estimate \(\int_{\Omega}u^{2}|\nabla v|^{2} \,dx\), we have
$$\begin{aligned}& \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)\int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \quad \le\epsilon_{2}\int_{\Omega}\bigl(|\nabla v|^{2}\bigr)^{\frac{2N+2}{N}} \,dx+ \biggl( \frac {m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\|_{L^{\infty}(\Omega )}^{2} \biggr)^{\frac{2N+2}{N+2}} \\& \qquad {}\times\epsilon_{2}^{-\frac{N}{N+2}}\int_{\Omega}u^{\frac{4N+4}{N+2}}\,dx \quad \text{for all } t\in\bigl(0,T^{*}\bigr). \end{aligned}$$
(3.22)
Further utilize the Gagliardo-Nirenberg inequality and (3.5) to estimate
$$\begin{aligned} \epsilon_{2}\int_{\Omega}\bigl(|\nabla v|^{2}\bigr)^{\frac{2N+2}{N}} \,dx &=\epsilon_{2} \bigl\Vert |\nabla v|^{2}\bigr\Vert ^{\frac{2N+2}{N}}_{L^{\frac{2N+2}{N}}(\Omega)} \\ &\le\epsilon_{2} c_{1}\bigl\Vert \nabla|\nabla v|^{2}\bigr\Vert ^{2}_{L^{2}(\Omega)}\bigl\Vert |\nabla v|^{2}\bigr\Vert ^{\frac{2}{N}}_{L^{1}(\Omega)}+c_{1}\bigl\Vert |\nabla v|^{2} \bigr\Vert ^{\frac{2N+2}{N}}_{L^{1}(\Omega)} \\ &\le\epsilon_{2} c_{2}\bigl\Vert \nabla|\nabla v|^{2}\bigr\Vert ^{2}_{L^{2}(\Omega )}+c_{2} \quad \text{for all } t\in\bigl(0,T^{*}\bigr). \end{aligned}$$
(3.23)
Since \(m+\tau+1>\frac{4N+4}{N+2}\), utilizing the Young inequality we have
$$\begin{aligned}& \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)^{\frac{2N+2}{N+2}}\epsilon_{2}^{-\frac {N}{N+2}}\int _{\Omega}u^{\frac{4N+4}{N+2}}\,dx \\& \quad \le\frac{\mu(m+1)}{2}\int _{\Omega}u^{m+1+\tau}\,dx+c(\epsilon_{2}) \end{aligned}$$
(3.24)
for all \(t\in(0,T^{*})\). Taking \(\epsilon_{2}=\frac{1}{4c_{2}}\), then from (3.22)-(3.24), we have
$$\begin{aligned}& \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)\int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \quad \le\frac{1}{4}\bigl\Vert \nabla|\nabla v|^{2}\bigr\Vert ^{2}_{L^{2}(\Omega)}+\frac{\mu(m+1)}{2}\int_{\Omega}u^{m+1+\tau}\,dx+C_{2} \end{aligned}$$
(3.25)
for all \(t\in(0,T^{*})\). For the third term \([\kappa(m+1)+1]\int_{\Omega}u^{m+1} \,dx\), utilizing the Young inequality we have
$$ \bigl[\kappa(m+1)+1\bigr]\int_{\Omega}u^{m+1} \,dx\le\frac{\mu(m+1)}{2}\int_{\Omega}u^{m+1+\tau}\,dx+C_{3}\quad \text{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.26)
Since \(\frac{m(m+1)c_{D}}{2}\int_{\Omega}u^{2m-2}|\nabla u|^{2} \,dx\) is nonnegative, combining (3.24), (3.25), and (3.26) with (3.17) yields
$$ \frac{d}{dt} \biggl\{ \int_{\Omega}u^{m+1} \,dx+ \int_{\Omega}|\nabla v|^{4}\,dx \biggr\} + \biggl\{ \int _{\Omega}u^{m+1} \,dx+\int_{\Omega}| \nabla v|^{4}\,dx \biggr\} \le C_{4} $$
(3.27)
for all \(t\in(0,T^{*})\), where \(C_{4}=C_{1}+C_{2}+C_{3}\). Thus an ODE comparison shows that \(y(t):=\int_{\Omega}u^{m+1} \,dx+\int_{\Omega}|\nabla v|^{4}\,dx\) satisfies
$$ y(t)\le\max\bigl\{ y(0),C_{4}\bigr\} \quad \text{for all } t\in\bigl(0,T^{*}\bigr), $$
(3.28)
which implies (3.19) and (3.20). □
We note \(m>1\) if \(N=2\) in Lemma 3.4. However, if \(m=1\), \(N=2\), and \(\tau=1\), i.e., \(m+\tau+1=\frac{4N+4}{N+2}=3\), the Young inequality will fail to lead to (3.24). In this case, we need to utilize the following generalization of the Gagliardo-Nirenberg inequality for the general case when \(r > 0\) (cf. [32], Lemma A.5, for a detailed proof), which extends the standard case when \(r \ge1\) in [33], to build a bound for \(\int_{\Omega}u^{3} \,dx\).
Lemma 3.5
Let
\(\Omega\subset\mathbb{R}^{2}\)
be a bounded domain with smooth boundary, and let
\(p\in(1,\infty)\)
and
\(r\in(0, p)\). Then there exists
\(C > 0\)
such that for each
\(\eta> 0\)
one can pick
\(C_{\eta}> 0\)
with the property that
$$ \|u\|^{p}_{L^{p}(\Omega)}\le\eta\|\nabla u \|^{p-r}_{L^{2}(\Omega)}\bigl\Vert u\ln \vert u\vert \bigr\Vert ^{r}_{L^{r}(\Omega)}+C\|u\|^{p}_{L^{r}(\Omega)}+C_{\eta}$$
(3.29)
holds for all
\(u\in W^{1,2}(\Omega)\).
Lemma 3.6
Let
\(m=1\), \(N=2\). Then there exists a positive constant
C
such that
$$ \int_{\Omega}u^{2} \,dx\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr) $$
(3.30)
and
$$ \int_{\Omega}|\nabla v|^{4}\,dx\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.31)
Proof
By the same procedure as Lemma 3.4, we only need to handle
$$\biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)^{\frac{2N+2}{N+2}}\epsilon_{2}^{-\frac {N}{N+2}}\int _{\Omega}u^{\frac{4N+4}{N+2}}\,dx $$
in (3.22). Invoking Lemma 3.5 along with (3.4) and Lemma 2.2, we deduce
$$\begin{aligned}& \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)^{\frac{2N+2}{N+2}}\epsilon_{2}^{-\frac {N}{N+2}}\int _{\Omega}u^{\frac{4N+4}{N+2}}\,dx \\& \quad = \biggl(\frac{\|\chi\|_{L^{\infty}(\Omega)}^{2}}{c_{D}}+6\|g\|_{L^{\infty}(\Omega)}^{2} \biggr)^{\frac{3}{2}}\epsilon_{2}^{-\frac{1}{2}}\int _{\Omega}u^{3}\,dx \\& \quad \le \biggl(\frac{\|\chi\|_{L^{\infty}(\Omega)}^{2}}{c_{D}}+6\|g\|_{L^{\infty}(\Omega)}^{2} \biggr)^{\frac{3}{2}}\epsilon_{2}^{-\frac{1}{2}}\int _{\Omega}(u+1)^{3}\,dx \\& \quad \le \biggl(\frac{\|\chi\|_{L^{\infty}(\Omega)}^{2}}{c_{D}}+6\|g\|_{L^{\infty}(\Omega)}^{2} \biggr)^{\frac{3}{2}} \\& \qquad {}\times\epsilon_{2}^{-\frac{1}{2}} \bigl[\epsilon _{2}\|\nabla u\|^{2}_{L^{2}(\Omega)}\bigl\Vert (u+1) \ln(u+1)\bigr\Vert _{L^{1}(\Omega)}+C\|u+1\| ^{3}_{L^{1}(\Omega)}+C_{\epsilon_{2}} \bigr] \\& \quad \le c_{3}\sqrt{\epsilon_{2}}\|\nabla u \|^{2}_{L^{2}(\Omega)}+c_{4}(\epsilon _{2})\quad \text{for all } t\in\bigl(0,T^{*}\bigr). \end{aligned}$$
(3.32)
Taking \(\epsilon_{2}\le\min \{\frac{1}{4c_{2}},\frac{c_{D}^{2}}{4c_{3}^{2}} \}\), from (3.22), (3.23), and (3.32) we then deduce
$$\begin{aligned}& \biggl(\frac{m(m+1)\|\chi\|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}+(4+N)\|g\| _{L^{\infty}(\Omega)}^{2} \biggr)\int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \quad = \biggl(\frac{\|\chi\|_{L^{\infty}(\Omega)}^{2}}{c_{D}}+6\|g\|_{L^{\infty}(\Omega )}^{2} \biggr) \int_{\Omega}u^{2}|\nabla v|^{2} \,dx \\& \quad \le\frac{1}{4} \bigl\| \nabla|\nabla v|^{2} \bigr\| ^{2}_{L^{2}(\Omega)}+\frac {c_{D}}{2}\|\nabla u\|^{2}_{L^{2}(\Omega)}+c_{5} \quad \text{for all } t\in\bigl(0,T^{*}\bigr). \end{aligned}$$
(3.33)
Combing (3.21), (3.26), (3.33), with (3.17) yields
$$ \frac{d}{dt} \biggl\{ \int_{\Omega}u^{2} \,dx+ \int_{\Omega}|\nabla v|^{4}\,dx \biggr\} + \biggl\{ \int _{\Omega}u^{2} \,dx+\int_{\Omega}| \nabla v|^{4}\,dx \biggr\} \le C_{5} $$
(3.34)
for all \(t\in(0,T^{*})\), which implies (3.30) and (3.31). □
Lemma 3.4 and Lemma 3.6 result in the following useful corollary that will be used in the proof of Lemma 3.7 below.
Corollary 3.2
Let
\(N=2\), \(m\ge1\), and assume that the initial data
\((u_{0},v_{0})\)
satisfies (1.5). Then there exists a positive constant
C
such that for any constant
\(k>1\)
there exists
\(C(k)>0\)
such that
$$ \int_{\Omega}|\nabla v|^{k} \le C(k) \quad \textit{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.35)
Proof
Since \(m+1\ge N=2\), then by (3.19), (3.20), (3.30), (3.31), and the standard parabolic regularity theory (cf. [12], Lemma 4.1 or [34], Lemma 1), we can immediately obtain (3.35). □
Lemma 3.7
Let the same assumptions as in Theorem
1.1
hold. Then for any
\(p>1\)
there exists a positive constant
\(C(p)\)
such that
$$ \int_{\Omega}u^{p} \,dx\le C(p)\quad \textit{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.36)
Proof
Let us recall to (3.1) once again. Adding \(\int_{\Omega}u^{p} \,dx\) to the both sides of (3.1) and neglecting the nonnegative term \(\frac{p(p-1)c_{D}}{2}\int_{\Omega}u^{p+m-3}|\nabla u|^{2} \,dx\) on the left, we arrive at
$$\begin{aligned}& \frac{d}{dt}\int_{\Omega}u^{p} \,dx+\int_{\Omega}u^{p} \,dx \\& \quad \le\frac{p(p-1)\|\chi \|_{L^{\infty}(\Omega)}^{2}}{2c_{D}}\int _{\Omega}u^{p-m+1}|\nabla v|^{2} \,dx \\& \qquad {} +(\kappa p+1)\int_{\Omega}u^{p} \,dx-\mu p\int _{\Omega}u^{p+\tau}\,dx \end{aligned}$$
(3.37)
for any \(p>m\) and each \(t\in(0,T^{*})\). Utilizing (3.35) and the Young inequality we have
$$\begin{aligned}& \frac{p(p-1)\chi^{2}}{2c_{D}}\int_{\Omega}u^{p-m+1}| \nabla v|^{2} \,dx \\& \quad \le\frac {\mu p}{2}\int_{\Omega}u^{p+\tau} \,dx+c_{4}\int_{\Omega}|\nabla v|^{\frac {2(p+\tau)}{\tau+m-1}} \\& \quad \le\frac{\mu p}{2}\int_{\Omega}u^{p+\tau} \,dx+c_{2}C\biggl(\frac{2(p+\tau)}{\tau +m-1}\biggr)\quad \text{for all } t\in \bigl(0,T^{*}\bigr), \end{aligned}$$
(3.38)
with some \(c_{2}>0\) and \(C(\frac{2(p+\tau)}{\tau+m-1})\) defined by Corollary 3.2, as well as
$$ [\kappa p+1]\int_{\Omega}u^{p} \,dx \le\frac{\mu p}{2}\int_{\Omega}u^{p+\tau } \,dx+ \tilde{C}_{1} \quad \text{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.39)
Collecting (3.37)-(3.39), we thus deduce that \(y(t):=\int_{\Omega}u^{p} \,dx\) satisfies the differential inequality
$$ \frac{d}{dt}\int_{\Omega}u^{p} \,dx+\int_{\Omega}u^{p} \,dx\le C_{6}\quad \text{for all } t\in\bigl(0,T^{*}\bigr), $$
(3.40)
with \(C_{6}=c_{2}C(\frac{2(p+\tau)}{\tau+m-1})+\tilde{C}_{1}\). Upon an ODE comparison, this implies
$$ y(t)\le\max\bigl\{ y(0),C_{6}\bigr\} \quad \text{for all } t\in \bigl(0,T^{*}\bigr). $$
Thus (3.36) holds for any \(p>m\). Since \(\int_{\Omega}u \,dx\le M_{1}\), utilizing the interpolation inequality, (3.36) holds also for any \(1< p\le m\). This completes the proof of Lemma 3.7. □
Once the uniform estimate of \(\|u(\cdot, t)\|_{L^{k}(\Omega)}\) has been established, we can use the classical Alikakos iteration method to obtain the uniform bound of \(\|u(\cdot, t)\|_{L^{\infty}(\Omega)}\).
Lemma 3.8
Let the same assumptions as in Theorem
1.1
hold. Then there exists a positive constant
C
such that the solution component
u
of (1.1) satisfies
$$ \bigl\Vert u(\cdot, t)\bigr\Vert _{L^{\infty}(\Omega)}\le C \quad \textit{for all } t\in\bigl(0,T^{*}\bigr). $$
(3.41)
We are now in the position to prove the global existence in Theorem 1.1.
Proof of global existence
The existence of global classical solution to equations (1.1) is an immediate consequence of Lemma 3.8 and the extensibility criterion (2.1). □
3.2 Proof of decay property in Theorem 1.1
In this short subsection, we discuss the decay property in the limit case \(\kappa=0\). Our approach is inspired by that in [35].
Lemma 3.9
Suppose that
\(f(s)\)
satisfies (1.4) with
\(\kappa=0\). Then
$$ \int^{\infty}_{0}\int _{\Omega}u^{\tau+1} \, dx\, dt< \frac{1}{\mu} \|u_{0}\|_{L^{1}(\Omega)} $$
(3.42)
and
$$ \int_{\Omega}u(x,t)\,dx\le\|u_{0} \|_{L^{1}(\Omega)}|\Omega| \bigl( |\Omega|^{\tau}+\mu\tau t \|u_{0}\|^{\tau}_{L^{1}(\Omega)} \bigr)^{-\frac{1}{\tau}}\quad \textit{for all } t>0. $$
(3.43)
Proof
Integrating (1.1) in space we obtain under the assumption \(\kappa=0\),
$$\frac{d}{dt}\int_{\Omega}u\, dx=\int_{\Omega}uf(u)\,dx\le-\mu\int_{\Omega}u^{\tau+1}\,dx\le- \frac{\mu}{|\Omega|^{\tau}} \biggl(\int_{\Omega}u \,dx \biggr)^{1+\tau} \quad \text{for all } t>0, $$
which implies both (3.42) and (3.43). □
Lemma 3.10
Let
\(\kappa=0\), \(\mu>\frac{\|u_{0}\|_{L^{1}(\Omega)}}{(\tau+1)\|v_{0}\| _{L^{1}(\Omega)}}\), and assume that
$$ \int^{\infty}_{0}\int _{\Omega}g(x,t)^{\frac{\tau+1}{\tau}}\,dx< \frac{\tau +1}{\tau} \biggl( \|v_{0}\|_{L^{1}(\Omega)}-\frac{1}{(\tau+1)\mu} \|u_{0}\| _{L^{1}(\Omega)} \biggr). $$
(3.44)
Then there exists a positive constant
C
appropriately small such that
$$ \int_{\Omega}v(x,t)\,dx\ge C \quad \textit{for all } t>0. $$
(3.45)
Proof
Integrating the second equation in (1.1) in space and applying the Young inequality, yield
$$ \frac{d}{dt}\int_{\Omega}v\,dx=-\int _{\Omega}ug(v)\,dx\ge-\frac{1}{\tau+1}\int_{\Omega}u^{\tau+1}\,dx-\frac{\tau}{\tau+1}\int_{\Omega}g(v)^{\frac{\tau +1}{\tau}}\,dx. $$
(3.46)
In view of (3.42), (3.46) implies
$$ \int_{\Omega}v\,dx\ge\|v_{0} \|_{L^{1}(\Omega)}-\frac{1}{(\tau+1)\mu} \|u_{0}\| _{L^{1}(\Omega)}- \frac{\tau}{\tau+1}\int^{\infty}_{0}\int _{\Omega}g(v)^{\frac {\tau+1}{\tau}}\,dx \quad \text{for all } t>0. $$
(3.47)
Therefore, (3.44) asserts the positivity of the right-hand side of (3.47). We thus obtain the desired result (3.45). □
Lemma 3.11
There exist
\(\alpha\in(0,1)\)
and
\(C>0\)
such that
$$ \|u\|_{C^{\alpha,\frac{\alpha}{2}}(\bar{\Omega}\times[t,t+1])}\le C\quad \textit{for all } t>0. $$
(3.48)
Proof
Rewriting the first equation of (1.1) in the form
$$ u_{t}=\nabla\cdot \bigl(D(u)\nabla u-u\chi(v)\nabla v \bigr)+uf(u),\quad x\in\Omega,t>0. $$
(3.49)
Utilizing the Young inequality we can estimate
$$ \bigl(D(u)\nabla u-u\chi(v)\nabla v \bigr)\cdot\nabla u\ge\frac {D(u)}{2}| \nabla u|^{2}-\frac{u^{2}\chi^{2}(v)|\nabla v|^{2}}{2D(u)} $$
and evidently
$$ \bigl\vert D(u)\nabla u-u\chi(v)\nabla v\bigr\vert \le D(u)|\nabla u|+u\chi (v)|\nabla v| $$
in \(\Omega\times(0,\infty)\). As Lemma 3.8 and Corollary 3.2 imply u and ∇v are bounded in \(L^{\infty}((0,\infty);L^{k}(\Omega ))\) for any \(k\in(1,\infty)\), and that u is a bounded solution of (3.49), the Hölder continuity of u, i.e., (3.48), immediately results from a known result on parabolic Hölder regularity ([36], Theorem 1.3). □
Proof of decay property in Theorem 1.1
In view of the Hölder continuity of u, i.e., (3.48), Arzelà-Ascoli theorem asserts \((u(\cdot,t))_{t>1}\) is relatively compact in \(C^{0}(\bar{\Omega})\). Thus, the decay property (3.43) implies that \(\|u(\cdot,t)\|_{L^{\infty}(\Omega)}\rightarrow0\) as \(t\rightarrow\infty\). □
Remark 3.1
The decay property in Theorem 1.1 reveals the fact that if the proliferation of cells is ignored, then the cells will not survive. However, Lemma 3.10 shows that if the death rate of the cells is large and the consumption rate of oxygen is small enough, then there is always oxygen remaining.
Since the global existence and decay property both have been proved, we then have completed the proof of Theorem 1.1.