1 Introduction

Suppose a compressible fluid flows in a homogeneous isotropic rigid porous medium. Then the volumetric moisture content \(\theta(x)\), the macroscopic velocity V⃗ and the density of the fluid ρ are governed by the following equation [1, 2]:

$$ \theta(x)\frac{\partial\rho}{\partial t}+\operatorname {div}(\rho\vec {V} )-f(\rho)=0, $$
(1.1)

where \(f(u)\) is the source. From Darcy’s law, one has the following relation:

$$ \rho\vec{V}=-\lambda \nabla P, $$
(1.2)

where ρV⃗ and P denote the momentum velocity and pressure, respectively, \(\lambda >0\) is some physical constant.

If the fluid considered is the polytropic gas, then the pressure and density satisfy the following equation of the state:

$$ P=c\rho^{\gamma}, $$
(1.3)

where \(c>0\), \(\gamma>0\) are some constants. Thus, it follows from (1.1)-(1.3) that

$$ \theta(x)\frac{\partial\rho}{\partial t}=c\lambda \Delta \bigl(\rho ^{\gamma}\bigr)+f(\rho). $$
(1.4)

In this paper, we consider (1.4) with \(\theta(x)=\vert x\vert ^{-\delta}\) and \(f(\rho)=\rho^{\sigma}\). Furthermore, we incorporate zero boundary condition to this problem. Then we get the following initial-boundary problem after changing variables and notations:

$$ \textstyle\begin{cases} \vert x\vert ^{-s} \frac{\partial u}{\partial t}-\Delta u^{m}=u^{p-1},\quad (x,t)\in\Omega\times (0,T), \\ u(x,t)=0, \quad (x,t)\in \partial \Omega \times(0,T), \\ u(x,0)=u_{0}(x),\quad x\in\Omega, \end{cases} $$
(1.5)

where \(u_{0} \in H^{1}_{0}(\Omega)\) is a nonnegative and nontrivial function, \(T \in(0,\infty] \), Ω is a bounded domain in \(\mathbb {R}^{N}\) (\(N\geq3\)) with smooth boundary ∂Ω, \(m\geq1\), \(0\leq s\leq1+1/m\leq2\), \(m< p-1\leq\frac{(N+2)m}{N-2}\).

Problem (1.5) and the related models were studied in [2–8], in order to introduce the main results of [5], we need the following functionals and sets, which were given in [5].

  • A function u is called a solution of (1.5) if

    $$u^{m}\in L^{\infty}\bigl(0,T;H_{0}^{1}( \Omega ) \bigr),\quad \int_{0}^{T} \bigl\Vert \vert x\vert ^{-\frac{s}{2}} \bigl(u^{\frac{m+1}{2}} \bigr)_{t} \bigr\Vert _{2}^{2}\,dt< +\infty, $$

    and u satisfies (1.5) in the distribution sense.

  • The energy functional related to the stationary equation

    $$ E(u)=\frac{1}{2m} \int_{\Omega }\bigl\vert \nabla u^{m} \bigr\vert ^{2}\,dx-\frac {1}{m+p-1} \int_{\Omega }\vert u\vert ^{m+p-1}\,dx, \quad u^{m}\in H_{0}^{1}(\Omega ). $$
    (1.6)

  • The Nehari functional

    $$ H(u)= \int_{\Omega }\bigl\vert \nabla u^{m} \bigr\vert ^{2}\,dx- \int_{\Omega }\vert u\vert ^{m+p-1}\,dx, \quad u^{m}\in H_{0}^{1}(\Omega ). $$
    (1.7)

  • The Nehari manifold

    $$ K= \bigl\{ u:u^{m}\in H_{0}^{1}( \Omega ), H(u)=0, u\neq0 \bigr\} . $$
    (1.8)

  • The potential depth

    $$\begin{aligned} d =&\inf \Bigl\{ \sup_{\lambda \geq0}E(\lambda u): u^{m} \in H_{0}^{1}(\Omega ), u\neq0 \Bigr\} \\ =&\inf_{u\in K}E(u)= \frac{p-1-m}{2m(m+p-1)}C^{\frac {-2(m+p-1)}{p-1-m}}, \end{aligned}$$
    (1.9)

    where C is the optimal constant of the Sobolev embedding \(H_{0}^{1}(\Omega )\subset L^{\frac{m+p-1}{m}}(\Omega )\). Particularly we have

    $$ \bigl\Vert u^{m} \bigr\Vert _{\frac{m+p-1}{m}}\leq C \bigl\Vert \nabla u^{m} \bigr\Vert _{2} $$
    (1.10)

    for \(u^{m}\in H_{0}^{1}(\Omega )\) since \(m< p-1\leq\frac{(N+2)m}{N-2}\), where \(\Vert \cdot \Vert _{r}\) denotes the norm of \(L^{r}(\Omega )\).

  • The sets related to global existence and blow-up

    $$ \begin{aligned} &\Sigma_{1}= \bigl\{ u:u^{m}\in H_{0}^{1}(\Omega ), E(u)< d, H(u)>0 \bigr\} \cup \{0\}, \\ &\Sigma_{2}= \bigl\{ u:u^{m}\in H_{0}^{1}( \Omega ), E(u)< d, H(u)< 0 \bigr\} . \end{aligned} $$
    (1.11)

The solution \(u(x,t)\) of problem (1.5) is called blow-up at finite time T if \(\Vert u\Vert _{L^{\infty}(\Omega)}\rightarrow +\infty\) as \(t\rightarrow T_{-}\). Otherwise, we say \(u(x,t)\) exists globally. The following are the main results of [5].

Theorem 1.1

If \(u_{0}\in\Sigma_{1}\), then the solution u to the problem (1.5) exists globally; if \(u_{0}\in\Sigma_{2}\), then u blows up at finite time.

In view of the above results, we may ask if the solution of u of the problem (1.5) blows up or exists globally when \(E(u_{0})\geq d\). The main task of this paper is to answer the question for \(E(u_{0})=d\). In order to give the main results of the present paper, we introduce two sets as follows:

$$ \begin{aligned} &\mathcal{S}= \biggl\{ u: u^{m} \in H_{0}^{1}(\Omega ), \bigl\Vert \nabla u^{m} \bigr\Vert _{2}< \biggl(\frac{2m(m+p-1)}{p-1-m}d \biggr)^{\frac {1}{2}} \biggr\} , \\ &\mathcal{B}= \biggl\{ u: u^{m}\in H_{0}^{1}(\Omega ), \bigl\Vert \nabla u^{m} \bigr\Vert _{2}> \biggl( \frac{2m(m+p-1)}{p-1-m}d \biggr)^{\frac {1}{2}} \biggr\} . \end{aligned} $$
(1.12)

Then

$$ \partial \mathcal{S}=\partial \mathcal{B}= \biggl\{ u: u^{m}\in H_{0}^{1}(\Omega ), \bigl\Vert \nabla u^{m} \bigr\Vert _{2}= \biggl(\frac{2m(m+p-1)}{p-1-m}d \biggr)^{\frac {1}{2}} \biggr\} . $$
(1.13)

The main results of this paper are the following theorem.

Theorem 1.2

Assume \(E(u_{0})=d\), then we have

  1. 1.

    if \(u_{0}\in\mathcal{S}\), then the problem (1.5) admits a global solution u such that \(u^{m}(t)\in L^{\infty}(0,+\infty; H_{0}^{1}(\Omega))\) and \(u(t)\in\bar{\mathcal{S}}=\mathcal{S}\cup \partial \mathcal{S}\) for \(0\leq t<+\infty\);

  2. 2.

    if \(u_{0}\in\mathcal{B}\), then the solution of problem (1.5) will blow up at finite time.

2 Proof of Theorem 1.2

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).

  1. (i)

    If \(u \in\mathcal{S}\) and \(\Vert \nabla u^{m}\Vert _{2}\neq0\), then \(\Vert \nabla u^{m}\Vert _{2}^{2}>\Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}}\).

  2. (ii)

    If \(u\in \partial \mathcal{S}\), then \(\Vert \nabla u^{m}\Vert _{2}^{2} \geq \Vert u^{m}\Vert _{\frac {m+p-1}{m}}^{\frac{m+p-1}{m}} \).

  3. (iii)

    If \(\Vert \nabla u^{m}\Vert _{2}^{2} < \Vert u^{m}\Vert _{\frac{m+p-1}{m}}^{\frac{m+p-1}{m}} \), then \(u\in\mathcal{B}\).

  4. (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

  1. (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 \(0<\Vert \nabla u^{m}\Vert _{2}< (\frac {2m(m+p-1)}{p-1-m}d )^{\frac{1}{2}} \);

  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:

  1. (i)

    \(u(t)\in \mathcal {S}\) for \(t\in[0,T)\) if \(u_{0}\in \mathcal {S}\);

  2. (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. 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. 2.

    \(u_{k}\rightarrow u\) a.e. on \(\Omega \times(0,T)\),

  3. 3.

    \(u_{k}^{m}\rightarrow u^{m}\) weakly star in \(L^{\infty}(0,T;H_{0}^{1}(\Omega ) )\),

  4. 4.

    \(u_{k}\rightarrow u\) weakly star in \(L^{\infty}(0,T; L^{m+p-1}(\Omega ) )\),

  5. 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]. □

3 Conclusion

In this paper, we study a singular porous medium equation considered in [5], where the global existence and blow-up conditions were got for the case of subcritical initial energy \(E(u_{0})< d\). We complete the results by studying the global existence and blow-up conditions for the case of critical initial energy \(E(u_{0})=d\).