Global existence and blow-up to the solutions of a singular porous medium equation with critical initial energy

Abstract

This paper is devoted to the study of a singular porous medium equation, which was studied extensively in recent years. We obtain the global existence and blow-up condition at the critical initial energy $E(u_{0})=d$, while the previous papers only considered the case $E(u_{0})< d$, where d is a positive constant which will be given in the main part of this paper.

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. 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. 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).

• (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]. □

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 $E(u_{0})=d$.