In this section, we assume that all assumptions in Theorem 1 are satisfied. As in [4, 5, 15], we derive a priori estimates of the smooth approximate solutions \(u(t)\), and our argument will be justified through such an approximate procedure.
Proof of Theorem 1
First, we take \(f_{n}(s)\) (\(n=1,2,\ldots \)) such that \(f_{n}(s)\to f(s)= \vert s \vert ^{q-2}s\) uniformly in \(\mathbb{R}^{1} \) as \(n\to \infty \).
For \(1< q<2\), we choose \(f_{n}^{+}(s)=a_{n}s^{2}+b_{n}s\) if \(0\le ns\le 1\) and \(f_{n}^{+}(s)=s^{q-1}\) if \(ns\ge 1\), where \(a_{n}=(q-2)n^{3-q}\), \(b_{n}=(3-q)n^{2-q}\). Further, let \(f_{n}(s)\) be the odd extension of \(f_{n}^{+}(s)\) in \(\mathbb{R}^{1}\).
If \(q\ge 2\), then we take \(f_{n}(s)= \vert s \vert ^{q-2}s\). For \(q=1\), we let
$$\begin{aligned} f_{n}(s)= \textstyle\begin{cases} 1, & s\ge 1/n, \\ ns(2-ns), & 0\le s \le 1/n, \\ -ns(2+ns), &-1/n\le s \le 0, \\ -1,& s< -1/n. \end{cases}\displaystyle \end{aligned}$$
(3.1)
Then we easily verify that \(f_{n}(s)\in C^{1}(\mathbb{R}^{1})\), \(f_{n}(s)\to f(s)= \vert s \vert ^{q-2}s\) uniformly in \(\mathbb{R}^{1}\) as \(n\to \infty \).
Let \(\varphi _{n}^{+}(s)=s^{\beta -1}\) if \(ns\ge 1\), \(\varphi _{n}^{+}(s)=A_{n}s +B_{n}\) if \(0\le ns\le 1\), where \(A_{n}=(\beta -1)n^{2-\beta }\), \(B_{n}=(2-\beta )n^{1-\beta }\). Further, let \(\varphi _{n}(s)\) be the even extension of \(\varphi _{n}^{+}(s)\) in \(\mathbb{R}^{1}\). Obviously, \(\varphi _{n}(s)\in C^{1}( \mathbb{R}^{1} )\), and \(\varphi _{n}(s)\to \varphi (s)= \vert s \vert ^{\beta -1}\) uniformly in \(\mathbb{R}^{1} \) as \(n\to \infty \).
Let \(u_{0,n}\in C_{0}^{2}(\Omega )\) and \(u_{0,n}\to u_{0}\) in \(L^{q}(\Omega )\) as \(n\to \infty \). We take the approximate problem of (2.3) of the form
$$\begin{aligned} \textstyle\begin{cases} \varphi _{i}(u)u_{t}= \operatorname{div}(a(x,t, \vert u \vert ^{\beta -1}u,\nabla ( \vert u \vert ^{ \beta -1}u))) & \text{in } \Omega \times (0, \infty ), \\ u(x,t)=0 & \text{on }\partial \Omega \times (0, \infty ), \\ u(x,0)=u_{0,i}(x) & \text{in }\Omega , \end{cases}\displaystyle \end{aligned}$$
(3.2)
for \(i=1,2,\ldots \) .
Then problem (3.2) has a unique smooth solution \(u_{i}(x,t)\); see [14]. We further always write u instead of \(u_{i}\) and \(u^{p}\) for \(\vert u \vert ^{p-1}u\) when \(p>0\).
Multiplying the equation in (3.2) by \(f_{k}(u)\varphi _{i}^{-1}(u)\), we obtain
$$\begin{aligned} \begin{aligned} & \int _{\Omega }f_{k}(u)u_{t}\,dx \\ &\quad =- \int _{\Omega }a \bigl(x,t, \vert u \vert ^{\beta -1}u, \nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr) \nabla u \bigl(f'_{k}(u)\varphi _{i}(u)-\varphi ' _{i}(u)f_{k}(u) \bigr)\varphi _{i}^{-2}(u)\,dx, \end{aligned} \end{aligned}$$
(3.3)
where
$$ f'_{k}(u)\varphi _{i}(u)-\varphi '_{i}(u)f_{k}(u)\geq 0. $$
By \((H_{1})\) we have
$$\begin{aligned} \begin{aligned} &a \bigl(x,t, \vert u \vert ^{\beta -1}u,\nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr) \nabla u \\ &\quad =\beta ^{-1}a \bigl(x,t, \vert u \vert ^{\beta -1}u,\nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr)\nabla \bigl( \vert u \vert ^{ \beta -1}u \bigr) \vert u \vert ^{1-\beta } \\ &\quad \geq \alpha _{0}\beta ^{-1} \vert u \vert ^{\beta r} \bigl\vert \nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr\vert ^{m} \vert u \vert ^{1- \beta }\geq 0. \end{aligned} \end{aligned}$$
(3.4)
Hence from (3.3) and (3.4) it follows that
$$\begin{aligned} \begin{aligned} \int _{\Omega }f_{k}(u)u_{t}\,dx\leq 0. \end{aligned} \end{aligned}$$
(3.5)
Letting \(k\to \infty \) in (3.5) gives
$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert _{q}\le \Vert u_{0} \Vert _{q},\quad t\ge 0. \end{aligned}$$
(3.6)
We now derive an \(L^{\infty }\) decay estimate for the solution \(u_{i}(t)\) of (3.2). Multiplying the equation in (3.2) by \(\varphi _{i}^{-1}(u) \vert u \vert ^{p-2}u\), \(p\geq 2\), we have
$$\begin{aligned} \begin{aligned} \frac{1}{p} \frac{d}{dt} \Vert u \Vert _{p}^{p}+ \int _{\Omega }a \bigl(x,t, \vert u \vert ^{ \beta -1}u, \nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr)\nabla u E_{i}[u]\,dx=0, \end{aligned} \end{aligned}$$
(3.7)
where
$$\begin{aligned} \begin{aligned} E_{i}[u]= \bigl((p-1) \vert u \vert ^{p-2}\varphi _{i}(u)-\varphi _{i}^{-1}(u) \vert u \vert ^{p-2}u \bigr) \varphi _{i}^{-2}(u)\geq \frac{p-\beta }{4} \vert u \vert ^{p-\beta -1}. \end{aligned} \end{aligned}$$
(3.8)
Noting that \(\beta =(m-1)/(r+m-1)\), from (3.4) we get that
$$\begin{aligned} \begin{aligned} a \bigl(x,t, \vert u \vert ^{\beta -1}u,\nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr) \nabla u &\geq \beta ^{-1}\alpha _{0} \vert u \vert ^{\beta r} \bigl\vert \nabla \bigl( \vert u \vert ^{\beta -1}u \bigr) \bigr\vert ^{m} \vert u \vert ^{1- \beta } \\ &=\alpha _{0}\beta ^{m-1} \vert \nabla u \vert ^{m}. \end{aligned} \end{aligned}$$
(3.9)
Hence from (3.7)–(3.9) it follows that
$$\begin{aligned} \frac{1}{p}\frac{d}{dt} \Vert u \Vert _{p}^{p}+C_{1} p \biggl(\frac{m}{p+M} \biggr)^{m} \int _{\Omega } \bigl\vert \nabla u^{\frac{p+M}{m}} \bigr\vert ^{m}\,dx\leq 0, \end{aligned}$$
(3.10)
where \(M=m-1-\beta >0\). Then (3.10) implies that
$$\begin{aligned} \frac{d}{dt} \bigl\Vert u(t) \bigr\Vert _{p}^{p}+C_{1} p^{2-m} \bigl\Vert \nabla u^{\frac{p+M}{m}} \bigr\Vert _{m}^{m}\le 0,\quad \forall t>0. \end{aligned}$$
(3.11)
Let C, \(C_{j}\) be general constants independent of p, i, n changeable from line to line. We now employ Moser’s technique as in [4, 5, 15]. Set \(R>1+M/q\), \(p_{1}=q\), \(p_{n}=Rp_{n-1}-M\), \(\theta _{n}=RN(1-p_{n-1}p_{n}^{-1})(m+N(R-1))^{-1}\), \(\beta _{n}=(p_{n}+M)\theta _{n}^{-1}\), \(n=2,3,\ldots \) .
From Lemma 2 we see that
$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert _{p_{n}}\leq C^{\frac{m}{p_{n}+M}} \Vert u \Vert _{p_{n-1}}^{1- \theta _{n}} \bigl\Vert \nabla u^{\frac{p_{n}+M}{m}} \bigr\Vert _{m}^{m\theta _{n}/(p_{n}+M)}. \end{aligned}$$
(3.12)
Inserting this into (3.11) \((p=p_{n})\) yields
$$\begin{aligned} \frac{d}{dt} \bigl\Vert u(t) \bigr\Vert _{p_{n}}+C_{1}C^{\frac{-m}{\theta _{n}}} p_{n}^{2-m} \Vert u \Vert _{p_{n-1}}^{M-\beta _{n}} \Vert u \Vert _{p_{n}}^{1+\beta _{n}}\leq 0, \quad \forall t>0. \end{aligned}$$
(3.13)
We claim that there exist bounded sequences \(\{\xi _{n}\}\) and \(\{\lambda _{n}\}\) such that
$$\begin{aligned} \begin{aligned} \bigl\Vert u(t) \bigr\Vert _{p_{n}} \leq \xi _{n}t^{-\lambda _{n}}, \quad \forall t>0, \end{aligned} \end{aligned}$$
(3.14)
where \(\lambda _{n}=(1+\lambda _{n-1}(\beta _{n}-M))/\beta _{n}\). It is not difficult to show that \(\lambda _{n}\to \lambda =\frac{N}{MN+mq}\) as \(n\to \infty \).
In fact, let \(\xi _{1}= \Vert u_{0} \Vert _{q}\) and \(\lambda _{1}=0\). If (3.14) is true for \(n-1\), the from (3.13) it follows that
$$\begin{aligned} \frac{d}{dt} \bigl\Vert u(t) \bigr\Vert _{p_{n}}+C_{1}C^{\frac{-m}{\theta _{n}}} p_{n}^{1-m} \xi _{n}^{M-\beta _{n}}t^{\lambda _{n-1}(\beta _{n}-M)} \Vert u \Vert _{p_{n}}^{1+ \beta _{n}}\leq 0, \quad \forall t>0. \end{aligned}$$
(3.15)
An application of Lemma 1 to (3.15) yields
$$\begin{aligned} \begin{aligned} \bigl\Vert u(t) \bigr\Vert _{p_{n}} &\leq \bigl(C_{1} C^{\frac{-m}{\theta _{n}}} p_{n}^{1-m} \xi _{n-1}^{M-\beta _{n}}\beta _{n}/ \bigl(1+\lambda _{n-1}(\beta _{n}-M) \bigr) \bigr)^{-1/ \beta _{n}}t^{-(1+\lambda _{n-1}(\beta _{n}-\mu )) / \beta _{n}} \\ &= \bigl(C_{1} C^{\frac{-m}{\theta _{n}}} \bigr)^{-1/ \beta _{n}}\lambda _{n}^{1/ \beta _{n}} p_{n}^{(m-1)/\beta _{n}} \xi _{n-1}^{(\beta _{n}-M)/ \beta _{n}} t^{-\lambda _{n}}. \end{aligned} \end{aligned}$$
(3.16)
Since
$$\begin{aligned} \lim_{n\to \infty }\frac{p_{n}}{\beta _{n}}=\frac{M+2}{N(M+1)}, \end{aligned}$$
we see that there exists a constant \(\lambda _{0}>0\), independent of n, such that
$$\begin{aligned} \begin{aligned} \bigl\Vert u(t) \bigr\Vert _{p_{n}}\leq (\lambda _{0}p_{n})^{\lambda _{0}/p_{n}} \xi _{n-1}^{1-M/\beta _{n}} t^{-\lambda _{n}}, \quad t>0. \end{aligned} \end{aligned}$$
(3.17)
Hence we define \(\xi _{n}\) inductively by
$$\begin{aligned} \xi _{n}=(\lambda _{0}p_{n})^{\lambda _{0}/p_{n}} \xi _{n-1}^{1-M/ \beta _{n}} \end{aligned}$$
(3.18)
for \(n=2,3,\ldots \) with \(\xi _{1}= \Vert u_{0} \Vert _{q}\). Here, setting \(\omega _{n}=mp_{n}+MN\), \(p_{1}=q\), and \(p_{n}=Rp_{n-1}-M\), by direct calculation we get
$$\begin{aligned} \begin{aligned} \frac{\beta _{n}-M}{\beta _{n}}= \frac{\omega _{n}}{p_{n}}\cdot \frac{p_{n-1}}{\omega _{n-1}} \end{aligned} \end{aligned}$$
(3.19)
and
$$\begin{aligned} \begin{aligned} \prod _{k=2}^{n}\frac{\beta _{k}-M}{\beta _{k}}= \frac{\omega _{n}}{p_{n}} \cdot \frac{p_{1}}{\omega _{1}}= \frac{MN+p_{n}m}{p_{n}}\cdot \frac{q}{mq+MN}. \end{aligned} \end{aligned}$$
(3.20)
It is easy to show that
$$\begin{aligned} \begin{aligned} \lim_{n\to \infty } \prod_{k=2}^{n} \frac{\beta _{k}-M}{\beta _{k}}= \mu = \frac{mq}{mq+MN}. \end{aligned} \end{aligned}$$
(3.21)
On the other hand, the definition of \(\xi _{n}\) gives
$$\begin{aligned} \begin{aligned} \log \xi _{n}&= \frac{\lambda _{0}}{p_{n}}(\log \lambda _{0}+ \log p_{n})+ \biggl(1- \frac{M}{\beta _{n}} \biggr)\log \xi _{n-1} \\ &=\frac{\lambda _{0}}{p_{n}}(\log \lambda _{0}+\log p_{n})+ \biggl(1- \frac{M}{\beta _{n}} \biggr) \biggl(\frac{\lambda _{0}}{p_{n}}(\log \lambda _{0}+ \log p_{n-1}) \\ &\quad {}+ \biggl(1-\frac{M}{\beta _{n-1}} \biggr)\log \xi _{n-2} \biggr) \\ &\leq \lambda _{0}\sum_{k=2}^{n} \frac{\log \lambda _{0}+\log p_{k}}{p_{k}}+\prod_{k=2}^{n} \biggl(1- \frac{M}{\beta _{k}} \biggr)\log \xi _{1}. \end{aligned} \end{aligned}$$
(3.22)
Hence
$$\begin{aligned} \begin{aligned} \log \xi _{n}\leq C_{0}+\frac{MN+p_{n}m}{p_{n}}\cdot \frac{q}{m q+MN}\log \xi _{1} \end{aligned} \end{aligned}$$
(3.23)
with some \(C_{0}>0\) independent of n. Then
$$\begin{aligned} \begin{aligned} \log \xi _{n}\leq C_{0}+\mu \log \xi _{1} \end{aligned} \end{aligned}$$
(3.24)
and
$$\begin{aligned} \begin{aligned} \xi _{n}\leq e^{C_{0}}\xi _{1} ^{\mu }=C_{1} \Vert u_{0} \Vert _{q}^{ \mu } \quad t>0. \end{aligned} \end{aligned}$$
(3.25)
Then, letting \(n\to \infty \) in (3.14), we obtain (2.7) and finish the proof of Theorem 1. □