In this section, we will prove Theorem 1.2. First of all, we will introduce some useful lemmas.
Lemma 2.1
Assume the function
\(u\not\equiv0\)
satisfying
\(u^{m}\in H_{0}^{1}(\Omega )\). Then there exists a unique positive value
\(\mu_{*}\)
defined as
$$ \mu_{*}=\sqrt[p-m-1]{\frac{\int_{\Omega} \vert \nabla u^{m}\vert ^{2}\,dx}{\int_{\Omega }\vert u\vert ^{m+p-1}\,dx}} $$
(2.1)
such that
\(E(\mu u)\)
is strictly increasing for
\(0<\mu<\mu_{*}\), strictly decreasing for
\(\mu_{*}<\mu<\infty\).
Proof
From
$$ E(\mu u)=\mu^{2m} \biggl[\frac{1}{2m} \bigl\Vert \nabla u^{m} \bigr\Vert _{2}^{2}-\frac{\mu^{p-m-1}}{m+p-1} \Vert u\Vert _{m+p-1}^{m+p-1} \biggr] $$
and \(p>m+1\) we get \(\lim_{\mu\rightarrow0}E(\mu u)=0\), \(\lim_{\mu \rightarrow+\infty}E(\mu u)=-\infty\). Furthermore, since \(\mu=\mu _{*}\) is the unique positive root of the equation \(\frac{dE(\mu u)}{d\mu }=0\), the conclusion follows. □
Lemma 2.2
Let
\(\mathcal {S}\), \(\mathcal {B}\), \(\partial \mathcal {S}\), and
\(\partial \mathcal {B}\)
be the sets defined as (1.12) and (1.13).
-
(i)
If
\(u \in\mathcal{S}\)
and
\(\Vert \nabla u^{m}\Vert _{2}\neq0\), then
\(\Vert \nabla u^{m}\Vert _{2}^{2}>\Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\).
-
(ii)
If
\(u\in \partial \mathcal{S}\), then
\(\Vert \nabla u^{m}\Vert _{2}^{2} \geq \Vert u^{m}\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}} \).
-
(iii)
If
\(\Vert \nabla u^{m}\Vert _{2}^{2} < \Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}} \), then
\(u\in\mathcal{B}\).
-
(iv)
If
\(\Vert \nabla u^{m}\Vert _{2}^{2} \leq \Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\)
and
\(\Vert \nabla u^{m}\Vert _{2}\neq0 \), then
\(u\in\mathcal {B}\cup \partial \mathcal{B}\).
Proof
(i) Since \(u\in\mathcal{S}\), we get from (1.9) and (1.10)
$$\bigl\Vert \nabla u^{m} \bigr\Vert _{2}< \biggl( \frac {2m(m+p-1)}{p-1-m}d \biggr)^{\frac{1}{2}}= C^{\frac{-(m+p-1)}{p-1-m}} \leq \biggl( \frac{\Vert u^{m}\Vert _{\frac{m+p-1}{m}}}{\Vert \nabla u^{m}\Vert _{2}} \biggr)^{\frac{-(m+p-1)}{p-1-m}}, $$
which implies \(\Vert \nabla u^{m}\Vert _{2}> \Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\).
(ii) From \(u\in \partial \mathcal{S}\) we get
$$\bigl\Vert \nabla u^{m} \bigr\Vert _{2}= \biggl( \frac {2m(m+p-1)}{p-1-m}d \biggr)^{\frac{1}{2}}\neq0. $$
Then in the same way as the proof of (i), \(\Vert \nabla u^{m}\Vert _{2}^{2}\geq \Vert u^{m}\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}}\) holds.
(iii) By (1.10) and \(\Vert \nabla u^{m}\Vert _{2}^{2}<\Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\), we have
$$\bigl\Vert \nabla u^{m} \bigr\Vert _{2}^{2}< \bigl\Vert u^{m} \bigr\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}}\leq C^{\frac{m+p-1}{m}} \bigl\Vert \nabla u^{m} \bigr\Vert _{2}^{\frac{m+p-1}{m}}, $$
which is equivalent to \(\Vert \nabla u^{m}\Vert _{2}>C^{\frac {-(m+p-1)}{p-1-m}}\). So \(u\in \mathcal {B}\).
(iv) In the same way as the proof of (iii), we have
$$\bigl\Vert \nabla u^{m} \bigr\Vert _{2}\geq C^{\frac{-(m+p-1)}{p-1-m}}, $$
which implies \(u\in\mathcal{B}\cup \partial \mathcal{B}\). □
Lemma 2.3
Let
u
be a solution of (1.5). Then the functional
\(E(u(t))\)
defined as (1.6) is non-increasing in
t. Moreover,
$$ \frac{4}{(m+1)^{2}} \int_{0}^{t} \bigl\Vert \vert x\vert ^{-\frac{s}{2}} \bigl(u^{\frac{m+1}{2}}(x,\tau) \bigr)_{\tau}\bigr\Vert _{2}^{2}\,d\tau+E \bigl(u(t) \bigr)=E(u_{0}). $$
(2.2)
Proof
Multiplying the first equation of (1.5) with \(\frac {1}{m}(u^{m})_{t}\) and integrating over \(\Omega \times(0,t)\), we get (2.2) and then that \(E(u(t))\) is non-increasing in t follows. □
Lemma 2.4
Let
u
be the solution of (1.5) with initial value
\(u_{0}\)
such that
\(u_{0}^{m}\in H_{0}^{1}(\Omega )\)
and
\(E(u_{0})\leq d\). Then
-
(i)
\(\Vert \nabla u^{m}\Vert _{2}^{2}> \Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\)
if and only if
\(0<\Vert \nabla u^{m}\Vert _{2}< (\frac {2m(m+p-1)}{p-1-m}d )^{\frac{1}{2}} \);
-
(ii)
\(\Vert \nabla u^{m}\Vert _{2}^{2}<\Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\)
if and only if
\(\Vert \nabla u^{m}\Vert _{2}> (\frac {2m(m+p-1)}{p-1-m}d )^{\frac{1}{2}}\).
Proof
By (1.6), (2.2) and \(E(u_{0})\leq d\) we have
$$ \begin{aligned}[b] E \bigl(u(t) \bigr)&= \frac{p-1-m}{2m(m+p-1)} \bigl\Vert \nabla u^{m} \bigr\Vert _{2}^{2}+ \frac{1}{m+p-1} \bigl( \bigl\Vert \nabla u^{m} \bigr\Vert _{2}^{2}- \bigl\Vert u^{m} \bigr\Vert _{\frac{m+p-1}{m}}^{\frac {m+p-1}{m}} \bigr) \\ &\leq E(u_{0})\leq d. \end{aligned} $$
(2.3)
Then we can easily get (i) and (ii) from Lemma 2.2 and (2.3). □
Lemma 2.5
Let
u
be the solution of (1.5) with initial value
\(u_{0}\)
such that
\(u_{0}^{m}\in H_{0}^{1}(\Omega )\)
and
\(E(u_{0})\leq d\). Then:
-
(i)
\(u(t)\in \mathcal {S}\)
for
\(t\in[0,T)\)
if
\(u_{0}\in \mathcal {S}\);
-
(ii)
\(u(t)\in \mathcal {B}\)
for
\(t\in[0,T)\)
if
\(u_{0}\in \mathcal {B}\);
where
\(\mathcal {S}\)
and
\(\mathcal {B}\)
are the sets defined in (1.12).
Proof
(i) If the conclusion (i) is false, there must exist a time \(t_{0}\in(0,T)\) such that \(u(t_{0})\in \partial \mathcal {S}\) and \(u(t)\in \mathcal {S}\) for \(0\leq t< t_{0}\). Hence
$$ \bigl\Vert \nabla u^{m}(t_{0}) \bigr\Vert _{2}= \biggl(\frac {2m(m+p-1)}{p-1-m}d \biggr)^{\frac{1}{2}} $$
(2.4)
and
$$ \bigl\Vert \nabla u^{m}(t) \bigr\Vert _{2}< \biggl(\frac {2m(m+p-1)}{p-1-m}d \biggr)^{\frac{1}{2}}, \quad t \in[0,t_{0}). $$
(2.5)
From (1.6), the second conclusion of Lemma 2.2 and (2.4), we obtain
$$\begin{aligned} E \bigl(u(t_{0}) \bigr) =& \frac{p-1-m}{2m(m+p-1)}\bigl\Vert \nabla u^{m}(t_{0}) \bigr\Vert _{2}^{2} +\frac{1}{m+p-1} \bigl(\bigl\Vert \nabla u^{m}(t_{0})\bigr\Vert _{2}^{2}- \bigl\Vert u^{m}(t_{0})\bigr\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}} \bigr) \\ \geq&\frac{p-1-m}{2m(m+p-1)} \bigl\Vert \nabla u^{m}(t_{0}) \bigr\Vert _{2}^{2}=d. \end{aligned}$$
(2.6)
By (2.4) and (2.5) we know that \(\int_{0}^{t_{0}}\Vert \vert x\vert ^{-\frac{s}{2}} (u^{\frac{m+1}{2}} )_{t}\Vert _{2}^{2}\,dt>0\). Then it follows from (2.2) and (2.6) that \(E(u_{0})>E(u(t_{0}))\geq d\), which contradicts \(E(u_{0})\leq d\).
(ii) The conclusion can be proved in the same way as (i). □
Based on above preparations, we are ready to prove Theorem 1.2.
Proof of Theorem 1.2 (global existence part)
We see from \(E(u_{0})=d\) and (1.6) that \(\Vert \nabla u_{0}^{m}\Vert _{2}>0\), which combines with \(u_{0}\in \mathcal {S}\) and the first conclusion of Lemma 2.2 implies
$$ \bigl\Vert \nabla u^{m}_{0} \bigr\Vert _{2}^{2}> \bigl\Vert u^{m}_{0} \bigr\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}. $$
(2.7)
Let \(\lambda_{n}=1-\frac{1}{n}\) and \(u_{0n}=\lambda_{n}u_{0}\) for \(n=2,3,\ldots \) . Then it follows from (2.7), \(\lambda _{n}<1\), and \(m-p+1<0\) that
$$\begin{aligned}& \begin{aligned}[b] \bigl\Vert \nabla u^{m}_{0n} \bigr\Vert _{2}^{2}&= \lambda _{n}^{2m} \bigl\Vert \nabla u^{m}_{0} \bigr\Vert _{2}^{2}> \lambda _{n}^{2m} \bigl\Vert u^{m}_{0} \bigr\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}}= \lambda _{n}^{m-p+1} \bigl\Vert u^{m}_{0n} \bigr\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}} \\ &> \bigl\Vert u^{m}_{0n} \bigr\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}},\quad n=2,3,\ldots, \end{aligned} \end{aligned}$$
(2.8)
$$\begin{aligned}& \begin{aligned}[b] E(u_{0n})&= \frac{p-1-m}{2m(m+p-1)} \bigl\Vert \nabla u_{0n}^{m} \bigr\Vert _{2}^{2}+\frac{1}{m+p-1} \bigl( \bigl\Vert \nabla u_{0n}^{m} \bigr\Vert _{2}^{2}- \bigl\Vert u_{0n}^{m} \bigr\Vert _{\frac{m+p-1}{m}}^{\frac {m+p-1}{m}} \bigr) \\ &> 0,\quad n=2,3,\ldots. \end{aligned} \end{aligned}$$
(2.9)
Furthermore, by Lemma 2.1, there exists an integer \(n_{*}\) such that \(E(\lambda _{n}u_{0})\) is strictly increasing for \(n\leq n_{*}\), which means
$$ E(u_{0n})=E(\lambda _{n} u_{0})< \lim _{n\rightarrow+\infty}E(\lambda _{n} u_{0})=E(u_{0})=d, \quad n=n_{*},n_{*}+1,\ldots. $$
(2.10)
Equations (2.8)-(2.10) imply \(u_{0n}\in\Sigma_{1}\), where \(\Sigma_{1}\) is defined as (1.11). Let \(u_{n}\) be the solution of (1.5) with initial value \(u_{0n}\), then Theorem 1.1 implies \(u_{n}\) exists globally such that
$$ \begin{aligned} u_{n}^{m}(t)\in L^{\infty}\bigl(0,+\infty;H_{0}^{1}(\Omega ) \bigr),\quad n=n_{*},n_{*}+1,\ldots. \end{aligned} $$
(2.11)
Similar to (2.3), for \(0\leq t<+\infty\), \(n=n_{*},n_{*}+1,\ldots\) , we get
$$ \begin{aligned}[b] d&>E(u_{0n})= \frac{4}{(m+1)^{2}} \int_{0}^{t} \bigl\Vert \vert x\vert ^{-\frac {s}{2}} \bigl(u_{n}^{\frac{m+1}{2}}(x,\tau) \bigr)_{\tau}\bigr\Vert _{2}^{2}\,d\tau+E \bigl(u_{n}(t) \bigr) \\ &=\frac{4}{(m+1)^{2}} \int_{0}^{t} \bigl\Vert \vert x\vert ^{-\frac{s}{2}} \bigl(u_{n}^{\frac{m+1}{2}}(x,\tau) \bigr)_{\tau}\bigr\Vert _{2}^{2}\,d\tau \\ &\quad {}+\frac{p-1-m}{2m(m+p-1)} \bigl\Vert \nabla u_{n}^{m} \bigr\Vert _{2}^{2}+\frac {1}{m+p-1} \bigl( \bigl\Vert \nabla u_{n}^{m} \bigr\Vert _{2}^{2}- \bigl\Vert u_{n}^{m} \bigr\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}} \bigr). \end{aligned} $$
(2.12)
Next, we will prove \(\Vert \nabla u_{n}^{m}(t)\Vert _{2}^{2}>\Vert u_{n}^{m}(t)\Vert _{\frac{m+p-1}{m}}^{\frac {m+p-1}{m}}\) for \(0\leq t<+\infty\). If not, it follows from (2.8) that there exists \(t_{*}>0\) such that \(\Vert \nabla u_{n}^{m}(t_{*})\Vert _{2}^{2}=\Vert u_{n}^{m}(t_{*})\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}}\). Then it follows from (1.9) that \(E(u_{n}(t_{*}))\geq d\), which contradicts \(E(u_{n}(t_{*}))< d\) by (2.12). Then from (2.12), we obtain
$$\begin{aligned}& \int_{0}^{t} \bigl\Vert \vert x\vert ^{-\frac{s}{2}} \bigl(u_{n}^{\frac {m+1}{2}}(x,\tau) \bigr)_{\tau}\bigr\Vert _{2}^{2}\,d\tau< \frac {d(m+1)^{2}}{4}, \\& \quad 0\leq t< +\infty, n=n_{*},n_{*}+1,\ldots, \end{aligned}$$
(2.13)
$$\begin{aligned}& \bigl\Vert u_{n}^{m}(t) \bigr\Vert _{\frac{m+p-1}{m}}^{\frac {m+p-1}{m}}\leq \bigl\Vert \nabla u_{n}^{m}(t) \bigr\Vert _{2}^{2}\leq\frac {2m(m+p-1)}{p-1-m}d, \\& \quad 0\leq t< + \infty, n=n_{*},n_{*}+1,\ldots. \end{aligned}$$
(2.14)
From (2.13), (2.14), and the compactness method in [9], it follows that there exist u and a subsequence \(\{u_{k}\}\) of \(\{u_{n}\}\) such that for all \(T>0\)
-
1.
\(u\in L^{\infty}(0,T;H_{0}^{1}(\Omega ) )\) and \(\int _{0}^{T}\Vert \vert x\vert ^{-\frac{s}{2}} (u^{\frac{m+1}{2}}(x,t) )_{t}\Vert _{2}^{2}\,dt\leq\frac{d(m+1)^{2}}{4}\),
-
2.
\(u_{k}\rightarrow u\) a.e. on \(\Omega \times(0,T)\),
-
3.
\(u_{k}^{m}\rightarrow u^{m}\) weakly star in \(L^{\infty}(0,T;H_{0}^{1}(\Omega ) )\),
-
4.
\(u_{k}\rightarrow u\) weakly star in \(L^{\infty}(0,T; L^{m+p-1}(\Omega ) )\),
-
5.
\(\vert x\vert ^{-\frac{s}{2}} (u_{k}^{\frac{1+m}{2}} )_{t}\rightarrow \vert x\vert ^{-\frac{s}{2}} (u^{\frac{1+m}{2}} )_{t}\) weakly in \(L^{2}(0,T;L^{2}(\Omega))\).
Then it follows from the construction of \(u_{n}\) that u is a global solution of (1.5) and \(u(t)\in\bar{\mathcal{S}}\) for \(0\leq t<\infty\). □
Proof of Theorem 1.2 (blow-up part)
Let \(u(t)\) be the solution of problem (1.5) with initial value \(u_{0}\) satisfying \(E(u_{0})=d\) and \(u_{0}\in\mathcal{B}\). We need to show that the maximal existence time T of u is finite. We assume \(T=+\infty\) and prove the conclusion by contradiction. Let
$$f(t)=\frac{1}{m+1} \int_{0}^{t} \int_{\Omega} \vert x\vert ^{-s} \bigl\vert u(x,\tau ) \bigr\vert ^{m+1}\,dx\,d\tau. $$
Then
$$ f''(t)= \int_{\Omega} \vert x\vert ^{-s}u^{m} u_{t}\,dx=- \bigl\Vert \nabla u^{m} \bigl\Vert _{2}^{2}+ \bigr\Vert u^{m} \bigr\Vert _{\frac {m+p-1}{m}}^{\frac {m+p-1}{m}}. $$
(2.15)
From (2.2), (2.15), and
$$ E \bigl(u(t) \bigr)=\frac{p-1-m}{2m(m+p-1)} \bigl\Vert \nabla u^{m}(t) \bigr\Vert _{2}^{2}+ \frac{1}{m+p-1} \bigl( \bigl\Vert \nabla u^{m}(t) \bigr\Vert _{2}^{2}- \bigl\Vert u^{m}(t) \bigr\Vert _{\frac{m+p-1}{m}}^{\frac {m+p-1}{m}} \bigr) $$
(2.16)
we get
$$ \begin{aligned}[b] f''(t)&= \frac{p-1-m}{2m} \bigl\Vert \nabla u^{m} \bigr\Vert _{2}^{2}-(m+p-1)E(u_{0}) \\ &\quad {}+ \frac{4(m+p-1)}{(m+1)^{2}} \int_{0}^{t} \bigl\Vert \vert x\vert ^{-\frac{s}{2}} \bigl(u^{\frac{m+1}{2}}(x,\tau) \bigr)_{\tau}\bigr\Vert _{2}^{2}\,d\tau. \end{aligned} $$
(2.17)
By \(u_{0}\in\mathcal{B}\) and Lemma (2.5), we obtain \(u(t)\in \mathcal {B}\) for \(0\leq t<+\infty\), i.e.,
$$ \bigl\Vert \nabla u^{m}(t) \bigr\Vert _{2}> \biggl(\frac {2m(m+p-1)}{p-1-m}d \biggr)^{\frac{1}{2}}, \quad 0\leq t< + \infty. $$
(2.18)
From (2.17), (2.18) and \(E(u_{0})=d\) we obtain \(f''(t)> \frac{4(m+p-1)}{(m+1)^{2}}\int_{0}^{t}\Vert \vert x\vert ^{-\frac {s}{2}} (u^{\frac{m+1}{2}}(x,\tau) )_{\tau} \Vert _{2}^{2}\,d\tau\). The remaining part of the proof is the same as that in [5]. □