Critical Fujita exponents to a class of non-Newtonian filtration equations with fast diffusion

Abstract

We consider the Cauchy problem to a class of fast-diffusion non-Newtonian filtration equations. Besides the usual degeneracy in the fast-diffusion non-Newtonian filtration, the equation is degenerate or singular at infinity, depending on the sign of the parameter related to the coefficient of diffusion. Fujita type theorems are established and the critical Fujita exponent is determined. Specially, we also prove that the nontrivial solution blows up in a finite time on the critical situation.

1 Introduction

The purpose of this paper is to investigate the critical Fujita exponent for the following initial value problem:

$$\begin{aligned} &\bigl(\vert x\vert +1\bigr)^{\mu_{1}}\frac{\partial u}{\partial t}= \operatorname{div} \bigl(\vert \nabla u\vert ^{q-1}\nabla u \bigr)+ \bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p}, \quad x\in\mathbb {R}^{n}, t>0, \end{aligned}$$

(1)

$$\begin{aligned} &u(x,0)=u_{0}(x),\quad x\in\mathbb {R}^{n}, \end{aligned}$$

(2)

where $p>1$, $0< q<1$, $\max\{-n,(n-1)/q-(n+1)\}<\mu_{1}\le\mu_{2}<p\mu_{1}+(p-1)n$, and $0\le u_{0}\in C_{0}(\mathbb {R}^{n})$.

The study of critical exponents began in 1966 by Fujita in [1], where it was proved for the initial value problem of

$$\begin{aligned} \frac{\partial u}{\partial t}=\Delta u+u^{p}, \quad x\in\mathbb {R}^{n}, t>0 \end{aligned}$$

that the problem admits no nontrivial nonnegative global solution if $1< p< p_{c}=1+2/n$, whereas if $p>p_{c}$, it admits both global (with small data) and non-global (with large initial data) solutions. Later, in 1981, Weissler [2] proved that the critical case $p=p_{c}$ is still a blow-up case.

In Fujita’s work, the new phenomenon of nonlinear parabolic equations was discovered. From then on, there has been a lot of work on the critical Fujita exponents for various nonlinear evolution equations and systems (see, e.g., the survey papers [3, 4] and the references therein, and also [5–15]). Among those, the Fujita type theorems for the slow-diffusion non-Newtonian filtration equation

$$\begin{aligned} \frac{\partial u}{\partial t}=\operatorname{div} \bigl(\vert \nabla u\vert ^{q-1}\nabla u \bigr)+u^{p}, \quad x\in\mathbb {R}^{n}, t>0 \end{aligned}$$

(3)

was investigated by Galaktionov in [16, 17], where $p, q>1$. He proved that $p_{c}=q+(q+1)/n$ by blow-up subsolutions and global supersolutions. Recently, the same problem for an interesting variant of (3) is studied by the authors [13]. The non-Newtonian filtration equations with fast diffusion were considered by Qi and Wang in [18], where the critical Fujita exponent was determined for the Cauchy problem of the equation

$$\begin{aligned} \frac{\partial u}{\partial t} =\operatorname{div} \bigl(\vert \nabla u\vert ^{q-1}\nabla u \bigr)+\vert x\vert ^{\sigma}u^{p}, \quad x\in\mathbb {R}^{n}, t>0 \end{aligned}$$

(4)

with $p>1$, $(n-1)/(n+1)< q<1$, and $\sigma>n(1-q)-q-1$. It is shown that $p_{c}=q+(q+1+\sigma)/n$ by energy functions. Obviously, they did not cover the portion $0< q\le(n-1)/(n+1)$ of the fast-diffusion range.

In the present paper, we study the problem (1), (2) and formulate the critical Fujita exponent as

$$\begin{aligned} p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1}) \end{aligned}$$

and the critical situation $p=p_{c}$ is still the blow-up case. The range of m considered in this paper is $0< m<1$, the whole fast-diffusion range of (1). Like the non-Newtonian filtration equation with fast diffusion, (1) is singular at points where $\vert \nabla u\vert =0$. In addition, (1) is degenerate at $\vert x\vert =+\infty$ for $\mu_{1}>0$ and singular for $\mu_{1}<0$, different from both (3) and (4). Inspired by [11, 18, 19], to prove the solutions’ blow-up, we analyze the interaction between the nonlinear source and nonlinear diffusion via precise estimates through constructing energy functions by use of the normalized principal eigenfunction of −Δ in the unit ball $B_{1}$ of $\mathbb {R}^{n}$ with homogeneous initial-boundary condition, rather than constructing subsolutions as the author did in [16, 17]. This method for equation (1) and its special case (4) basically depends upon the nonincreasing properties in the spatial variant of solutions, which is trivial with $\mu_{1}=\mu_{2}$, while it may be invalid if $\mu_{1}<\mu_{2}$. For all these reasons, we have to overcome some technical difficulties.

This paper is arranged as follows. Some preliminaries are introduced in Section 2, including the local existence theorem, the comparison principle, and a property of solutions from propagation of disturbances. The Fujita type theorems are established in Section 3. Finally in Section 4, the critical case will be concerned.

2 Preliminaries

Throughout this paper, we use $B_{r}$ to indicate the ball in $\mathbb {R}^{n}$ with radius r and center at the origin. The solution considered here is taken in the following sense.

Definition 2.1

We call

$$\begin{aligned} 0\le u\in C \bigl([0,T );L^{\infty} \bigl(\mathbb {R}^{n} \bigr)\bigr) \cap L_{\mathrm{loc}}^{q+1} \bigl(0,T;W_{\mathrm{loc}}^{1,q+1} \bigl(\mathbb {R}^{n} \bigr) \bigr) \end{aligned}$$

a solution to the problem (1), (2) in $(0,T)$ with $0< T\le+\infty$ if

$$\begin{aligned} \int_{0}^{T} \int_{\mathbb {R}^{n}} \biggl(\bigl(\vert x\vert +1 \bigr)^{\mu_{1}}u\frac{\partial\phi }{\partial t} -\vert \nabla u\vert ^{q-1} \nabla u\cdot\nabla\phi+\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \phi \biggr)\,dx\,dt=0 \end{aligned}$$

holds for any $\phi\in C^{\infty}_{0}(\mathbb {R}^{n}\times(0,T))$ and

$$\begin{aligned} \lim_{t\to0^{+}} \int_{\mathbb {R}^{n}}u(x,t)\zeta(x)\,dx = \int_{\mathbb {R}^{n}}u_{0}(x)\zeta(x)\,dx \end{aligned}$$

for any $\zeta\in C^{\infty}_{0}(\mathbb {R}^{n})$.

Like the non-Newtonian filtration equation, it is not hard to prove the well-posedness to the problem (1), (2), one can see, e.g., [20].

Next, we will prove the following proposition on a property of solutions from propagation of disturbances.

Proposition 2.1

Assume that u is a solution to the problem (1), (2) with $0\le u_{0}\in C_{0}(\mathbb {R}^{n})$ nontrivial, then $u(0,t_{0})>0$ for some $t_{0}>0$.

Proof

That $u_{0}$ is nontrivial shows that there exists $0\neq x_{0}\in\mathbb {R}^{n}$ and $\kappa, \rho>0$ such that

$$\begin{aligned} u_{0}(x)\ge\kappa \biggl(1- \biggl(\frac{\vert x-x_{0}\vert ^{q+1}}{\rho^{q+1}} \biggr)^{1/q} \biggr)_{+}^{2},\quad x\in\mathbb {R}^{n}, \end{aligned}$$

where $s_{+}=\max\{s,0\}$. Let

$$\begin{aligned} &\varPhi(x,t)=\frac{\kappa\rho^{(q+1)\xi}}{R^{\xi}(t)} \biggl(1- \biggl(\frac{\vert x-x_{0}\vert ^{q+1}}{R(t)} \biggr)^{1/q} \biggr)_{+}^{2},\quad x\in\mathbb {R}^{n}, t>0, \\ &D= \biggl\{ (x,t)\in\mathbb {R}^{n+1}\times\mathbb {R}_{+}: \vert x-x_{0}\vert < 2\vert x_{0}\vert , \vert x-x_{0}\vert ^{q+1}< R(t), 0< t< \frac{\kappa^{1-q}\rho^{q+1}}{\xi} \biggr\} , \end{aligned}$$

with $R(t)=\kappa^{q-1}t+\rho^{q+1}$, and $\xi>1$ independent of κ and ρ to be chosen later.

Denote

$$\begin{aligned} \Vert z\Vert =\frac{\vert x-x_{0}\vert ^{q+1}}{R(t)}, \qquad H=1-\Vert z\Vert ^{1/q}. \end{aligned}$$

A direct calculation within D shows

$$\begin{aligned} &\frac{\partial\varPhi}{\partial t}=-\frac{\xi\kappa^{q}\rho ^{(q+1)\xi}}{R^{\xi+1}(t)}H^{2} +\frac{2}{q} \frac{\kappa^{q}\rho^{(q+1)\xi}}{R^{\xi+1}(t)}H\Vert z\Vert ^{1/q}, \\ &\operatorname{div} \bigl(\vert \nabla\varPhi \vert ^{q-1}\nabla \varPhi \bigr) =- \biggl(\frac{2(q+1)}{q} \biggr)^{q} \frac{1}{R(t)} \biggl( \frac{\kappa\rho^{(q+1)\xi}}{R^{\xi}(t)}H \biggr)^{q} \biggl(n-(q+1)\frac{\Vert z\Vert ^{1/q}}{H} \biggr). \end{aligned}$$

Setting

$$\mathscr{L}[\varPhi]=\frac{R^{\xi+1}(t)}{\kappa^{q}\rho^{(q+1)\xi }H} \biggl(\bigl(\vert x\vert +1 \bigr)^{\mu_{1}} \frac{\partial\varPhi}{\partial t}-\operatorname{div} \bigl(\vert \nabla \varPhi \vert ^{q-1}\nabla \varPhi \bigr) \biggr),\quad(x,t)\in D, $$

then

$$\begin{aligned} \mathscr{L}[\varPhi] ={}&{-}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}\xi H+ \frac {2}{q}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} \Vert z\Vert ^{1/q} \\ &{}+ \biggl(\frac{2(q+1)}{q} \biggr)^{q} \biggl(\frac{\rho ^{(q+1)\xi}}{R^{\xi}(t)}H \biggr)^{q-1} \biggl(n-(q+1)\frac{\Vert z\Vert ^{1/q}}{H} \biggr). \end{aligned}$$

Divide D into two sets

$$\begin{aligned} D^{(1)}= \bigl\{ (x,t)\in D:H< \delta \bigr\} \quad\mbox{and}\quad D^{(2)}= \bigl\{ (x,t)\in D:H\ge\delta \bigr\} \end{aligned}$$

with $\delta>0$ satisfying

$$\begin{aligned} \delta^{q-1} \biggl((q+1) \biggl(\frac{1}{\delta}-1 \biggr)-n \biggr)\ge\frac{2}{q}\varLambda_{1}, \end{aligned}$$

(5)

where

$$\begin{aligned} \varLambda_{1}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} (3\vert x_{0}\vert +1)^{\mu_{1}}, &\mu_{1}\ge0,\\ 1, &\mu_{1}< 0. \end{array}\displaystyle \right . \end{aligned}$$

Then in $D^{(1)}$,

$$\begin{aligned} \mathscr{L}[\varPhi]&\le\frac{2}{q}{\varLambda_{1}}+ \biggl( \frac {2(q+1)}{q} \biggr)^{q} \delta^{q-1} \biggl(n-(q+1) \biggl(\frac{1}{\delta}-1 \biggr) \biggr) \\ &\le\frac{2}{q}{\varLambda_{1}}+\delta^{q-1} \biggl(n+q+1-\frac {q+1}{\delta}\biggr) \\ &\le0. \end{aligned}$$

For the chosen $\delta>0$, we have in $D^{(2)}$

$$\begin{aligned} \mathscr{L}[\varPhi]\le-\varLambda_{2}\xi\delta+\frac{2}{q}{ \varLambda_{1}}+n \biggl(\frac {2(q+1)}{q} \biggr)^{q} \biggl(\frac{\rho^{(q+1)\xi}}{R^{\xi}(t)}\delta \biggr)^{q-1}, \end{aligned}$$

where

$$\begin{aligned} \varLambda_{2}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1, &\mu_{1}\ge0,\\ (3\vert x_{0}\vert +1)^{\mu_{1}}, &\mu_{1}< 0. \end{array}\displaystyle \right . \end{aligned}$$

Due to

$$\begin{aligned} \biggl(\frac{\rho^{(q+1)\xi}}{R^{\xi}(t)}\delta \biggr)^{q-1}\le \biggl(1+ \frac{1}{\xi}\biggr)^{\xi(1-q)}\delta^{q-1} \le \biggl( \frac{\mathrm{e}}{\delta}\biggr)^{1-q}, \quad(x,t)\in D, \end{aligned}$$

we know

$$\begin{aligned} \mathscr{L}[\varPhi] \le-\varLambda_{2}\xi\delta+\frac{2}{q}{ \varLambda_{1}}+n \biggl(\frac {2(q+1)}{q} \biggr)^{m} \biggl(\frac{\mathrm{e}}{\delta}\biggr)^{1-q},\quad(x,t)\in D. \end{aligned}$$

So for fixed $\delta>0$ satisfying (5) and $\xi>1$ satisfying

$$\begin{aligned} \varLambda_{2}\xi\delta\ge\frac{2}{q}{\varLambda_{1}}+n \biggl(\frac {2(q+1)}{q} \biggr)^{q} \biggl(\frac{\mathrm{e}}{\delta}\biggr)^{1-q}, \end{aligned}$$

we obtain

$$\bigl(\vert x\vert +1\bigr)^{\mu_{1}}\frac{\partial\varPhi}{\partial t}-\operatorname {div} \bigl(\vert \nabla \varPhi \vert ^{q-1}\nabla\varPhi \bigr)\le0, \quad x\in\mathbb {R}^{n}, 0< t< \frac{\kappa^{1-q}\rho^{q+1}}{\xi}. $$

Clearly, the constant $\xi>1$ is independent of κ and ρ. The comparison principle implies

$$\begin{aligned} u(x,t)\ge\varPhi(x,t),\quad(x,t)\in D. \end{aligned}$$

In particular,

$$\begin{aligned} u(x,t_{1})>0,\quad x\in\varGamma_{1} \end{aligned}$$

with $t_{1}=\frac{\kappa^{1-q}\rho^{q+1}}{\xi}$ and

$$\begin{aligned} \varGamma_{1}= \biggl\{ x\in\mathbb {R}^{n}:\vert x-x_{0}\vert < 2\vert x_{0}\vert , \vert x-x_{0}\vert ^{q+1}< \frac{\xi+1}{\xi}\rho^{q+1} \biggr\} . \end{aligned}$$

If $0\in\varGamma_{1}$, the proof is complete. Otherwise,

$$\begin{aligned} u(x,t_{1})\ge\varPhi(x,t_{1})=\kappa_{1} \biggl(1- \biggl(\frac {\vert x-x_{0}\vert ^{q+1}}{\rho_{1}^{q+1}} \biggr)^{1/q} \biggr)_{+}^{2}, \quad x \in\mathbb {R}^{n}, \end{aligned}$$

where

$$\begin{aligned} \kappa_{1}=\kappa \biggl(\frac{\xi}{\xi+1} \biggr)^{\xi},\qquad \rho_{1}=\rho \biggl(\frac{\xi+1}{\xi}\biggr)^{1/(q+1)}. \end{aligned}$$

From the above argument, we have

$$\begin{aligned} u(x,t)\ge\varPhi_{1}(x,t),\quad(x,t)\in D_{1}, \end{aligned}$$

where

$$\begin{aligned} &\varPhi_{1}(x,t)=\frac{\kappa_{1}\rho_{1}^{(q+1)\xi}}{R_{1}^{\xi}(t)} \biggl(1- \biggl( \frac{\vert x-x_{0}\vert ^{q+1}}{R_{1}(t)} \biggr)^{1/q} \biggr)_{+}^{2},\quad x\in \mathbb {R}^{n}, t>t_{1}, \\ &D_{1}= \biggl\{ (x,t)\in\mathbb {R}^{n}\times\mathbb {R}_{+}: \vert x-x_{0}\vert < 2\vert x_{0}\vert , \vert x-x_{0}\vert ^{q+1}< R_{1}(t), t_{1}< t< t_{1}+\frac{\kappa_{1}^{1-q}\rho_{1}^{q+1}}{\xi} \biggr\} , \end{aligned}$$

with $R_{1}(t)=\kappa_{1}^{q-1}(t-t_{1})+\rho_{1}^{q+1}$. In particular,

$$\begin{aligned} u(x,t_{2})>0,\quad x\in\varGamma_{2}, \end{aligned}$$

with $t_{2}=t_{1}+\frac{\kappa_{1}^{1-q}\rho_{1}^{q+1}}{\xi}$ and

$$\begin{aligned} \varGamma_{2}= \biggl\{ x\in\mathbb {R}^{n}:\vert x-x_{0}\vert < 2\vert x_{0}\vert , \vert x-x_{0}\vert ^{q+1}< \frac{\xi+1}{\xi} \rho_{1}^{q+1} \biggr\} . \end{aligned}$$

If $0\in\varGamma_{2}$, the proof is complete. Otherwise, repeat the above procedure. We get the conclusion in finite steps. □

3 Fujita type theorems

Let us establish the Fujita type theorems.

Definition 3.1

We call u the blow-up solution to equation (1) if there exists some $0< T_{*}<+\infty$ such that

$$\begin{aligned} \lim_{t\to T_{*}^{-}} \bigl\Vert u(\cdot,t)\bigr\Vert _{L^{\infty}(\mathbb {R}^{n})} =\lim_{t\to T_{*}^{-}}\sup_{x\in\mathbb {R}^{n}}u(x,t)=+ \infty. \end{aligned}$$

Theorem 3.1

Assume that $1< p< p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$ and $0\le u_{0}\in C_{0}(\mathbb {R}^{n})$ is nontrivial. Then the problem (1), (2) admits a blow-up solution.

Proof

Due to Proposition 2.1, we may assume $u_{0}(0)>0$. By the comparison principle, we only need to prove the conclusion for radial and nonincreasing $u_{0}(x)$, i.e.,

$$\begin{aligned} u_{0}(x)=h_{0}\bigl(\vert x\vert \bigr),\quad x\in \mathbb {R}^{n}, \end{aligned}$$

where $h_{0}\in C^{1}_{0}([0,+\infty))$ satisfies $h'_{0}(0)=0$ and $h'_{0}(r)\le0$ for $r>0$. With such initial data, the solution u is also radial, namely

$$\begin{aligned} u(x,t)=h\bigl(\vert x\vert ,t\bigr), \quad x\in\mathbb {R}^{n}, t\ge0. \end{aligned}$$

(6)

If $\mu_{1}=\mu_{2}$, it is easy to know that u is also nonincreasing by a standard regularization argument and the maximum principle. However, this method is invalid if $\mu_{1}<\mu_{2}$. In the following discussion, we will first of all consider a nonincreasing u, namely $h(r,t)$ is nonincreasing with respect to $r\in[0,+\infty)$ for any $t\ge0$, and we treat the general case finally.

Let

$$\begin{aligned} \psi(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1,&0\le \vert x\vert \le1,\\ f(\vert x\vert -1), &1< \vert x\vert < 2,\\ 0,&\vert x\vert \ge2, \end{array}\displaystyle \right . \end{aligned}$$

where f is the principal eigenfunction of −Δ in the unit ball $B_{1}$ of $\mathbb {R}^{n}$ with homogeneous initial-boundary condition, normalized by $\Vert f\Vert _{L^{\infty}(B_{1})}=1$. For $l>1$, define

$$\begin{aligned} \psi_{l}(x)=\psi(x/l), \quad x\in\mathbb {R}^{n}. \end{aligned}$$

Then

$$\begin{aligned} \vert \nabla\psi_{l}\vert \le\frac{M_{0}}{l}, \qquad \vert \Delta \psi_{l}\vert \le\frac{M_{0}}{l^{2}}, \qquad\frac{\vert \Delta\psi_{l}\vert }{\psi_{l}}\le \frac {M_{0}}{l^{2}}, \quad x\in B_{2l}\setminus B_{l}, \end{aligned}$$

with $M_{0}>0$ independent of l. Set

$$\begin{aligned} \eta_{l}(t)= \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u \psi_{l} \,dx,\quad t\ge0. \end{aligned}$$

(7)

Definition 2.1 gives

$$\begin{aligned} \frac{d\eta_{l}}{dt} =- \int_{B_{2l}}\vert \nabla u\vert ^{q-1}\nabla u\cdot \nabla \psi_{l} \,dx+ \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx. \end{aligned}$$

For radial and nonincreasing $u(x,t)$, one has

$$\begin{aligned} \int_{B_{2l}}\vert \nabla u\vert ^{q-1}\nabla u\cdot \nabla \psi_{l} \,dx &= \int_{B_{2l}}\vert \nabla u\vert ^{q}\vert \nabla \psi_{l}\vert \,dx \\ &\le \biggl( \int_{B_{2l}}\vert \nabla u\vert \cdot \vert \nabla \psi_{l}\vert \,dx \biggr)^{q} \biggl( \int_{B_{2l}}\vert \nabla\psi_{l}\vert \,dx \biggr)^{1-q} \\ &= \biggl( \int_{B_{2l}}\nabla u\cdot\nabla\psi_{l}\,dx \biggr)^{q} \biggl( \int_{B_{2l}}\vert \nabla\psi_{l}\vert \,dx \biggr)^{1-q} \\ &\le M_{0}l^{(n-1)(1-q)} \biggl( \int_{B_{2l}}\nabla u\cdot\nabla\psi_{l}\,dx \biggr)^{q} \end{aligned}$$

and

$$\begin{aligned} 0\le \int_{B_{2l}}\nabla u\cdot\nabla\psi_{l}\,dx = \int_{\partial B_{2l}}u\nabla\psi_{l}\cdot\boldsymbol {\nu}\,d\sigma- \int_{B_{2l}} u\Delta\psi_{l} \,dx \le- \int_{B_{2l}} u\Delta\psi_{l} \,dx, \end{aligned}$$

where ν is the unit outer normal to $\partial B_{2l}$. Hence

$$\begin{aligned} \frac{d\eta_{l}}{dt}\ge-M_{0}l^{(1-q)(n-1)} \biggl\vert \int_{B_{2l}} u\Delta\psi_{l} \,dx \biggr\vert ^{q} + \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx. \end{aligned}$$

(8)

The Hölder inequality yields

$$\begin{aligned} \biggl\vert \int_{B_{2l}} u\Delta\psi_{l} \,dx \biggr\vert ^{q} \le{}& \biggl( \int_{B_{2l}\setminus B_{l}} u\vert \Delta\psi_{l}\vert \,dx \biggr)^{q} \\ \le{}& \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{-\mu_{2}/(p-1)}\vert \Delta\psi_{l}\vert ^{p/(p-1)} \psi_{l}^{-1/(p-1)}\,dx \biggr)^{q(p-1)/p} \\ &{} \cdot \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi _{l} \,dx \biggr)^{q/p} \\ \le{}& M_{1}l^{q(n-2)-q(n+\mu_{2})/p} \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx \biggr)^{q/p} \end{aligned}$$

with $M_{1}>0$ independent of l, which, together with (8), implies

$$\begin{aligned} \frac{d\eta_{l}}{dt} \ge{}& \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx \biggr)^{q/p} \\ &{} \cdot \biggl\{ -M_{0}M_{1}l^{n-q-1-q(n+\mu_{2})/p}+ \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx \biggr)^{(p-q)/p} \biggr\} . \end{aligned}$$

(9)

By the Hölder inequality,

$$\begin{aligned} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u \psi_{l}\,dx \le{}& \biggl( \int_{B_{2l}}\bigl(\vert x\vert +1\bigr)^{(p\mu_{1}-\mu_{2})/(p-1)} \psi_{l}\,dx \biggr)^{(p-1)/p} \\ &{} \cdot \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx \biggr)^{1/p} \\ \le{}& M_{2}l^{(n+\mu_{1})-(n+\mu_{2})/p} \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx \biggr)^{1/p}, \end{aligned}$$

and hence

$$\begin{aligned} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p} \psi_{l} \,dx \ge M_{2}^{-p}l^{-p(n+\mu_{1})+(n+\mu_{2})} \eta_{l}^{p} \end{aligned}$$

(10)

with $M_{2}>0$ independent of l. Equations (9) and (10) show that

$$\begin{aligned} \frac{d\eta_{l}}{dt} \ge{}& \bigl(M_{2}^{-p}l^{-p(n+\mu_{1})+(n+\mu_{2})} \bigr)^{q/p}\eta_{l}^{q} \\ &{} \cdot \bigl\{ -M_{0}M_{1}l^{n-q-1-q(n+\mu_{2})/p} + M_{2}^{q-p}l^{[-p(n+\mu_{1})+(n+\mu_{2})](p-q)/p}\eta_{l}^{p-q} \bigr\} . \end{aligned}$$

(11)

We mention that the above discussion holds provided that $p>1$.

If $p< p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$, then

$$\begin{aligned} n-q-1-(n+\mu_{2})q/p< \bigl[-p(n+\mu_{1})+(n+ \mu_{2}) \bigr](p-q)/p. \end{aligned}$$

Notice that $\eta_{l}$ is nondecreasing with respect to $l\in(1,+\infty)$ and $\sup\{\eta_{l}(0):l\in(1,+\infty)\}>0$, and from (11) one shows that, for $l>1$ large enough, there exists a constant $\delta>0$ such that

$$\begin{aligned} \frac{d\eta_{l}}{dt} \ge{}& \bigl(M_{2}^{-p}l^{-p(n+\mu_{1})+(n+\mu_{2})} \bigr)^{q/p}\eta_{l}^{q} \\ &{} \cdot \biggl(\frac{1}{2}M_{2}^{q-p}l^{[-p(n+\mu_{1})+(n+\mu _{2})](p-q)/p} \eta_{l}^{p-q} \biggr) \\ \ge{}&\delta\eta_{l}^{p}. \end{aligned}$$

So there exists some $0< T_{*}<+\infty$ such that

$$\begin{aligned} \lim_{t\to T_{*}^{-}}\eta_{l}(t)=+\infty. \end{aligned}$$

Due to $\operatorname{supp} \psi_{l}=B_{2l}$, we obtain

$$\begin{aligned} \lim_{t\to T_{*}^{-}}\bigl\Vert u(\cdot,t)\bigr\Vert _{L^{\infty}(\mathbb {R}^{n})}=+\infty. \end{aligned}$$

Next, for the general case without the assumption that $u(x,t)$ is nonincreasing, define

$$\begin{aligned} \underline{u}(x,t)=\min_{0\le r\le \vert x\vert } h(r,t),\quad x \in\mathbb {R}^{n}, t\ge0. \end{aligned}$$

(12)

Then $\underline{u}(x,t)$ is nonincreasing,

$$\begin{aligned} 0\le\underline{u}(x,t)\le u(x,t), \quad x\in\mathbb {R}^{n}, t\ge0, \end{aligned}$$

(13)

and

$$\begin{aligned} \frac{d}{dt} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} \underline{u} \psi_{l} \,dx \ge- \int_{B_{2l}}\vert \nabla\underline{u}\vert ^{q-1} \nabla\underline {u}\cdot\nabla\psi_{l} \,dx + \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}} \underline{u}^{p} \psi_{l} \,dx. \end{aligned}$$

(14)

From the above argument, we get

$$\begin{aligned} \lim_{t\to\tilde{T}_{*}^{-}}\bigl\Vert \underline{u}(\cdot,t)\bigr\Vert _{L^{\infty}(\mathbb {R}^{n})}=+\infty \end{aligned}$$

for some $0<\tilde{T}_{*}<+\infty$, and (13) ensures that u is a blow-up solution. □

Let us turn to the case $p>p_{c}$. Suppose that

$$\begin{aligned} U(x,t)={(t+1)^{-\alpha}}V \bigl((t+1)^{-\beta}\bigl( \vert x\vert +1\bigr) \bigr), \end{aligned}$$

(15)

where

$$\begin{aligned} \alpha=\frac{q+1+\mu_{2}}{\mu_{1}(p-q)+\mu _{2}(q-1)+(q+1)(p-1)},\qquad \beta=\frac{p-q}{q+1+\mu_{2}}\alpha, \end{aligned}$$

(16)

is a self-similar solution to (1). It is easy to show that $V(r)$ solves

$$\begin{aligned} \bigl(\bigl\vert V'\bigr\vert ^{q-1}V' \bigr)'+\frac{n-1}{r}\bigl\vert V'\bigr\vert ^{q-1}V'+ \beta r^{\mu_{1}+1}V'+\alpha r^{\mu_{1}}V+r^{\mu_{2}}V^{p}=0, \quad r>0. \end{aligned}$$

(17)

Lemma 3.1

Assume that $p>p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$. Then for $\varepsilon>0$ small enough, the function

$$\begin{aligned} V(r)=\varepsilon \bigl(1+\rho(\varepsilon)r^{\lambda}\bigr)^{-\gamma}, \quad r>0, \end{aligned}$$

where

$$\begin{aligned} \lambda=1+\frac{\mu_{1}+1}{q},\qquad \gamma=\frac{q}{1-q}, \qquad\rho (\varepsilon)= \frac{1}{\gamma\lambda}\beta^{1/q}\varepsilon^{(1-q)/q}, \end{aligned}$$

is a supersolution to equation (17), i.e.

$$\begin{aligned} \bigl(\bigl\vert V'\bigr\vert ^{q-1}V' \bigr)'+ \frac{n-1}{r}\bigl\vert V'\bigr\vert ^{q-1}V'+\beta r^{\mu_{1}+1}V'+\alpha r^{\mu_{1}}V+r^{\mu_{2}}V^{p} \le0, \quad r>0. \end{aligned}$$

Proof

It is not hard to show that it suffices to verify

$$\begin{aligned} &q(\gamma+1)\rho(\varepsilon)\lambda \bigl[\varepsilon\gamma \lambda\rho( \varepsilon) \bigr]^{q} \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-q(\gamma+1)-1}r^{(q+1)(\lambda-1)} \\ &\quad{} -q(\lambda-1) \bigl[\varepsilon\gamma\lambda\rho (\varepsilon) \bigr]^{q} \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-q(\gamma+1)}r^{q(\lambda-1)-1} \\ &\quad{} -(n-1) \bigl[\varepsilon\gamma\lambda\rho(\varepsilon) \bigr]^{q} \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-q(\gamma+1)}r^{q(\lambda-1)-1} \\ &\quad{} -\beta\varepsilon\gamma\lambda\rho(\varepsilon) \bigl[1+\rho( \varepsilon)r^{\lambda}\bigr]^{-(\gamma+1)}r^{\mu_{1}+\lambda} +\alpha \varepsilon \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-\gamma }r^{\mu_{1}} \\ &\quad{} +\varepsilon^{p} \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-p\gamma }r^{\mu_{2}} \le0,\quad r>0, \end{aligned}$$

namely

$$\begin{aligned} & \bigl\{ \bigl[\varepsilon\gamma\lambda\rho (\varepsilon) \bigr]^{q}-\beta\varepsilon \bigr\} \gamma\lambda\rho(\varepsilon) \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-(\gamma+1)}r^{\mu _{1}+\lambda} \\ &\quad{} - \bigl\{ \bigl[q(\lambda-1)+n-1 \bigr]\varepsilon ^{-1} \bigl[ \varepsilon\gamma\lambda\rho(\varepsilon) \bigr]^{q} -\alpha- \varepsilon^{p-1} \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-(p-1)\gamma}r^{\mu_{2}-\mu_{1}} \bigr\} \\ &\quad{} \cdot\varepsilon \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-\gamma }r^{\mu_{1}} \le0, \quad r>0. \end{aligned}$$

(18)

From the definition of $[\varepsilon\gamma\lambda\rho(\varepsilon )]^{q}=\beta\varepsilon$, we have

$$\begin{aligned} & \bigl[q(\lambda-1)+n-1 \bigr]\varepsilon^{-1} \bigl[\varepsilon \gamma \lambda\rho(\varepsilon) \bigr]^{q} -\alpha-\varepsilon^{p-1} \bigl[1+\rho(\varepsilon)r^{\lambda}\bigr]^{-(p-1)\gamma}r^{\mu _{2}-\mu_{1}} \\ &\quad= (n+\mu_{1})\beta-\alpha-\varepsilon^{p-1} \bigl[1+ \rho(\varepsilon)r^{\lambda}\bigr]^{-(p-1)\gamma}r^{\mu_{2}-\mu_{1}} \\ &\quad\ge(n+\mu_{1})\beta-\alpha-\varepsilon^{p-1} \bigl[ \rho(\varepsilon) \bigr]^{-(p-1)\gamma}r^{\mu_{2}-\mu_{1}-\lambda (p-1)\gamma}. \end{aligned}$$

Due to $p>p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$ and $(n-1)/q-(n+1)<\mu_{1}\le\mu_{2}$,

$$\begin{aligned} (n+\mu_{1})\beta>\alpha, \quad\lambda(p-1)\gamma\ge\mu_{2}- \mu_{1}\ge0. \end{aligned}$$

Hence,

$$\begin{aligned} (n+\mu_{1})\beta-\alpha-\varepsilon^{p-1} \bigl[\rho( \varepsilon) \bigr]^{-(p-1)\gamma}r^{\mu_{2}-\mu_{1}-\lambda (p-1)\gamma}>0,\quad r>0 \end{aligned}$$

holds for sufficiently small $\varepsilon>0$, and (18) is obtained. □

Theorem 3.2

Assume that $p>p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$. Then the solution to the problem (1), (2) exists globally with small $u_{0}$, or blows up with large $u_{0}$.

Proof

Let $V(r)$ be a supersolution to equation (17) in Lemma 3.1. From

$$\begin{aligned} V'(r)< 0, \quad r>0, \end{aligned}$$

one can show that $U(x,t)$ given in (15) is a supersolution to equation (1) with α and β given in (16). Therefore, the comparison principle implies the problem (1), (2) has a nontrivial global solution with small $u_{0}$.

Let us turn to the case of large $u_{0}$. Denote by the radial u a solution to the problem (1), (2). Temporarily suppose u is nonincreasing. Then (11) holds with $\eta_{l}$ defined by (7). If $u_{0}$ is so large that

$$\begin{aligned} M_{0}M_{1}l_{0}^{n-q-1-(n+\mu_{2})q/p} < \frac{1}{2}M_{2}^{q-p}l_{0}^{[-p(n+\mu_{1})+(n+\mu_{2})](p-q)/p} \eta_{l_{0}}^{p-q}(0) \end{aligned}$$

holds for some $l_{0}>1$, then from (11), we get

$$\begin{aligned} \eta_{l_{0}}(t)\ge\eta_{l_{0}}(0),\quad t>0, \end{aligned}$$

and

$$\begin{aligned} \frac{d\eta_{l_{0}}}{dt} \ge{}& \bigl(M_{2}^{-p}l_{0}^{-p(n+\mu _{1})+(n+\mu_{2})} \bigr)^{q/p}\eta_{l_{0}}^{q} \\ &{} \cdot \biggl(\frac{1}{2} M_{2}^{q-p}l_{0}^{[-p(n+\mu_{1})+(n+\mu_{2})](p-q)/p} \eta_{l_{0}}^{p-q} \biggr) \\ ={}&\delta_{0}\eta_{l_{0}}^{p} \end{aligned}$$

for some $\delta_{0}>0$. Therefore, u is a blow-up solution.

For the general case without the assumption that $u(x,t)$ is nonincreasing, considering a new function just as in the proof of Theorem 3.1, one can also see that u is a blow-up solution. □

4 Critical case

Now, let us deal with the critical case $p=p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$. Let $\psi_{l}$, $\eta_{l}$, $M_{1}$, $M_{2}$ be defined as in the previous section.

Lemma 4.1

Assume that $p=p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$ and u is a nontrivial, global, radial, and nonincreasing solution to the problem (1), (2). Then

$$\begin{aligned} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,t)\,dx\le M, \quad t>0, \end{aligned}$$

(19)

holds for some constant $M>0$ independent of t.

Proof

$p=p_{c}$ yields

$$\begin{aligned} n-q-1-(n+\mu_{2})q/{p_{c}}= \bigl[-p_{c}(n+ \mu_{1})+(n+\mu_{2}) \bigr](p_{c}-q)/{p_{c}}. \end{aligned}$$

Then, for the global, radial and nonincreasing solution u, from (11), we have

$$\begin{aligned} M_{2}^{q-p_{c}}\eta_{l}^{p_{c}-q}(t) \le2M_{0}M_{1},\quad l>1, t>0. \end{aligned}$$

Otherwise, u blows up in a finite time. Therefore (19) holds for some constant $M>0$ owing to

$$\begin{aligned} \lim_{l\to+\infty}\eta_{l}(t) = \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,t)\,dx. \end{aligned}$$

 □

Lemma 4.2

Assume that $p=p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$, u be a nontrivial, radial, and nonincreasing solution to the problem (1), (2) and $0<\theta<1$. Then

$$\begin{aligned} \frac{d\eta_{l}}{dt}\ge{}& M_{2}^{-q(1-\theta)} l^{(n+\mu_{2})-p_{c}(n+\mu_{1})}\eta_{l}^{q(1-\theta)} \\ &{} \cdot \biggl\{ -M_{3} \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u \psi_{l} \,dx \biggr)^{q\theta}+ M_{2}^{q(1-\theta)-p_{c}} \eta_{l}^{p_{c}-q(1-\theta)} \biggr\} \end{aligned}$$

(20)

holds for any $l>1$ with a constant $M_{3}>0$ independent of l.

Proof

For any $l>1$, the Hölder inequality yields

$$\begin{aligned} & \biggl\vert \int_{B_{2l}} u\Delta\psi_{l}\,dx \biggr\vert ^{q} \\ &\quad\le \biggl( \int_{B_{2l}\setminus B_{l}} u\vert \Delta\psi_{l}\vert \,dx \biggr)^{q} \\ &\quad\le \biggl( \int_{B_{2l}\setminus B_{l}} \bigl(\vert x\vert +1\bigr)^{-(\theta p_{c}\mu_{1}+(1-\theta)\mu_{2})/[(p_{c}-1)(1-\theta)]} \\ &\qquad{} \cdot \vert \Delta\psi_{l}\vert ^{p_{c}/[(p_{c}-1)(1-\theta)]} \psi _{l}^{-(1-\theta+p_{c}\theta)/[(p_{c}-1)(1-\theta)]}\,dx \biggr)^{q(p_{c}-1)(1-\theta)/{p_{c}}} \\ &\qquad{}\cdot \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}} u^{p_{c}} \psi_{l} \,dx \biggr)^{q(1-\theta)/{p_{c}}} \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u \psi_{l} \,dx \biggr)^{q\theta} \\ &\quad\le Ml^{q(n-2)-q(n+\mu_{2})/{p_{c}} +q\theta(\mu_{2}-p_{c}\mu_{1}-(p_{c}-1)n)/{p_{c}}} \\ &\qquad{} \cdot \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}} u^{p_{c}} \psi_{l} \,dx \biggr)^{q(1-\theta)/{p_{c}}} \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u \psi_{l} \,dx \biggr)^{q\theta}, \end{aligned}$$

with $M>0$ a constant independent of l, which, together with (10) and (8), implies

$$\begin{aligned} \frac{d\eta_{l}}{dt} \ge{}& \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p_{c}} \psi_{l} \,dx \biggr)^{q(1-\theta)/{p_{c}}} \\ &{} \cdot \biggl\{ -M_{0}Ml^{(n-1)(1-q)+q(n-2)-q(n+\mu_{2})/{p_{c}} +q\theta[\mu_{2}-p_{c}\mu_{1}-(p_{c}-1)n]/{p_{c}}} \\ &{} \cdot \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u \psi_{l} \,dx \biggr)^{q\theta} + \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p_{c}} \psi_{l} \,dx \biggr)^{[p_{c}-q(1-\theta)]/{p_{c}}} \biggr\} \\ \ge{}& \bigl(M_{2}^{-p_{c}}l^{-p_{c}(n+\mu_{1})+(n+\mu_{2})} \bigr)^{q(1-\theta)/{p_{c}}} \eta_{l}^{q(1-\theta)} \\ &{} \cdot \biggl\{ -M_{0}Ml^{(n-1)(1-q)+q(n-2)-q(n+\mu_{2})/{p_{c}} +q\theta[\mu_{2}-p_{c}\mu_{1}-(p_{c}-1)n]/{p_{c}}} \\ &{} \cdot \biggl( \int_{B_{2l}\setminus B_{l}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u \psi_{l} \,dx \biggr)^{q\theta} \\ &{} + M_{2}^{q(1-\theta)-p_{c}}\eta_{l}^{p_{c}-q(1-\theta)}l^{[-p_{c}(n+\mu _{1})+(n+\mu_{2})] [p_{c}-q(1-\theta)]/{p_{c}}} \biggr\} . \end{aligned}$$

Then (20) holds due to

$$\begin{aligned} &(n-1) (1-q)+q(n-2)-q(n+\mu_{2})/{p_{c}} +q\theta \bigl[ \mu_{2}-p_{c}\mu_{1}-(p_{c}-1)n \bigr]/{p_{c}} \\ &\quad= \bigl[-p_{c}(n+\mu_{1})+(n+\mu_{2}) \bigr] \bigl[p_{c}-q(1-\theta) \bigr]/{p_{c}}. \end{aligned}$$

 □

Lemma 4.3

Assume that $p=p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$, u be a nontrivial, radial and nonincreasing solution to the problem (1), (2). Then

$$\begin{aligned} \frac{d\eta_{l}}{dt}\ge-{M_{4}}l^{[p_{c}(n-q-1)-q(n+\mu _{2})]/(p_{c}-q)},\quad l>1 \end{aligned}$$

(21)

holds for some constant $M_{4}>0$ independent of l.

Proof

From (9), we have

$$\begin{aligned} \frac{d\eta_{l}}{dt} \ge{}& \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p_{c}} \psi_{l} \,dx \\ &{} -Ml^{(n-1)(1-q)+q[n-2-(n+\mu_{2})/p_{c}]} \biggl( \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p_{c}} \psi_{l}\,dx \biggr)^{q/{p_{c}}}. \end{aligned}$$

Then the Young inequality gives

$$\begin{aligned} \frac{d\eta_{l}}{dt} \ge{}& \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p_{c}} \psi_{l} \,dx-\frac {q}{p_{c}} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{2}}u^{p_{c}} \psi_{l}\,dx \\ &{} -\frac{p_{c}-q}{p_{c}}M^{{p_{c}}/(p_{c}-q)} l^{[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)} \\ \ge{}&{-}{M_{4}}l^{[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)}. \end{aligned}$$

 □

We are ready to prove the blow-up theorem of Fujita type for the critical case $p=p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$.

Theorem 4.1

Assume that $p=p_{c}=q+(q+1+\mu_{2})/(n+\mu_{1})$ and u is a solution to the problem (1), (2) with $0\le u_{0}\in C_{0}(\mathbb {R}^{n})$ nontrivial. Then the problem (1), (2) admits a blow-up solution.

Proof

Similarly to the proof of Theorem 3.1, at first assume $u_{0}$ is radial and nonincreasing. Then u is radial, given by (6). Denote

$$\begin{aligned} \varLambda=\sup_{l>1,t>0}\eta_{l}(t)=\sup _{t>0} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,t)\,dx. \end{aligned}$$

(22)

From (19) and the nontriviality of u, $0<\varLambda<+\infty$. For any $0<\sigma<\varLambda$, due to (22) and $\eta_{l}$ being nondecreasing with respect to $l\in(1,+\infty)$, there exist $\omega_{0}\ge0$ and $l_{0}>2$ such that

$$\begin{aligned} \eta_{{l_{0}}/2}(\omega_{0})\ge\varLambda-\sigma. \end{aligned}$$

Then it follows from (21) that

$$\begin{aligned} & \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,s) \psi_{{l_{0}}/2}(x)\,dx \\ &\quad\ge \int_{\mathbb {R}^{n}}\vert x\vert ^{\mu_{1}}u(x, \omega_{0}) \psi_{{l_{0}}/2}(x)\,dx -{M_{4}}({l_{0}}/2)^{[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)}(s- \omega_{0}) \\ &\quad\ge\varLambda-\sigma-{M_{4}}({l_{0}}/2)^{[p_{c}(n-q-1)-q(n+\mu _{2})]/(p_{c}-q)}(s- \omega_{0}),\quad s\ge\omega_{0}. \end{aligned}$$

Thus

$$\begin{aligned} & \int_{B_{2l_{0}}\setminus B_{l_{0}}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,s) \psi_{l_{0}}(x)\,dx \\ &\quad\le \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,s)\,dx- \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,s) \psi_{{l_{0}}/2}(x)\,dx \\ &\quad\le\sigma+{M_{4}}({l_{0}}/2)^{[p_{c}(n-q-1)-q(n+\mu _{2})]/(p_{c}-q)}(s- \omega_{0}),\quad s\ge\omega_{0}. \end{aligned}$$

Choosing $l=l_{0}$ in (20), we have

$$\begin{aligned} \frac{d\eta_{l_{0}}}{dt} \ge{}& M_{2}^{-q(1-\theta)}l_{0}^{(n+\mu _{2})-p_{c}(n+\mu_{1})} \eta_{l_{0}}^{q(1-\theta)} \\ &{} \cdot \biggl\{ -M_{3} \biggl( \int_{B_{2l_{0}}\setminus B_{l_{0}}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u \psi_{l_{0}}\,dx \biggr)^{q\theta}+ M_{2}^{q(1-\theta)-p_{c}} \eta_{l_{0}}^{p_{c}-q(1-\theta)} \biggr\} \\ \ge{}& M_{2}^{-q(1-\theta)}l_{0}^{(n+\mu_{2})-p_{c}(n+\mu_{1})}\eta _{l_{0}}^{q(1-\theta)} \\ &{} \cdot \bigl\{ -M_{3} \bigl(\sigma +{M_{4}}({l_{0}}/2)^{[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)}(s- \omega_{0}) \bigr\} \bigr)^{q\theta} \\ &{} +M_{2}^{q(1-\theta)-p_{c}}\eta_{l_{0}}^{p_{c}-q(1-\theta )} \},\quad t>\omega_{0}. \end{aligned}$$

Fix $\sigma_{0}\in(0,\varLambda)$ and $M_{5}>0$, independent of $l_{0}$, such that

$$\begin{aligned} M_{3} (\sigma_{0}+M_{5} )^{q\theta}\le \frac{1}{2} M_{2}^{q(1-\theta)-p_{c}} (\varLambda- \sigma_{0})^{p_{c}-q(1-\theta)}. \end{aligned}$$

Then

$$\begin{aligned} \frac{d\eta_{l_{0}}}{dt} \ge\frac{1}{2} M_{2}^{-p_{c}}l_{0}^{(n+\mu _{2})-p_{c}(n+\mu_{1})} \eta_{l_{0}}^{p_{c}},\quad \omega_{0}< t< \omega_{1}, \end{aligned}$$

where

$$\begin{aligned} \omega_{1}=\omega_{0}+\frac{M_{5}}{M_{4}}({l_{0}}/2)^{-[p_{c}(n-q-1)-q(n+\mu _{2})]/(p_{c}-q)}. \end{aligned}$$

Hence

$$\begin{aligned} \eta_{l_{0}}(\omega_{1}) \ge{}&\eta_{l_{0}}( \omega_{0})+\frac{1}{2} M_{2}^{-p_{c}}l_{0}^{(n+\mu_{2})-p_{c}(n+\mu_{1})} (\varLambda-\sigma_{0})^{p_{c}}(\omega_{1}- \omega_{0}) \\ \ge{}&\eta_{l_{0}}(\omega_{0})+\frac{1}{2} M_{2}^{-p_{c}}l_{0}^{(n+\mu _{2})-p_{c}(n+\mu_{1})} (\varLambda- \sigma_{0})^{p_{c}} \\ &{} \cdot\frac{M_{5}}{M_{4}}({l_{0}}/2)^{-[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)}. \end{aligned}$$

Notice

$$\begin{aligned} (n+\mu_{2})-p_{c}(n+\mu_{1})- \bigl[p_{c}(n-q-1)-q(n+\mu_{2}) \bigr]/(p_{c}-q)=0, \end{aligned}$$

one gets

$$\begin{aligned} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u(x, \omega_{1})\,dx \ge\eta_{l_{0}}(\omega_{1}) \ge \eta_{l_{0}}( \omega_{0})+\delta_{0} \ge \varLambda-\sigma_{0}+ \delta_{0}, \end{aligned}$$

with a positive constant

$$\begin{aligned} \delta_{0}=\frac{M_{2}^{-p_{c}}M_{5}}{2M_{4}}(\varLambda-\sigma _{0})^{p_{c}}2^{[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)} \end{aligned}$$

independent of $l_{0}$. Similarly, we reason

$$\begin{aligned} \eta_{(2l_{0})/2}(\omega_{1})=\eta_{l_{0}}( \omega_{1})\ge\varLambda-\sigma_{0}+\delta_{0} \ge\varLambda-\sigma_{0}. \end{aligned}$$

The same argument yields

$$\begin{aligned} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x,t_{2})\,dx \ge \eta_{2l_{0}}(\omega_{2})\ge\eta_{2l_{0}}( \omega_{1})+\delta_{0} \ge\varLambda-\sigma_{0}+2 \delta_{0} \end{aligned}$$

with

$$\begin{aligned} \omega_{2}=\omega_{1}+\frac{M_{5}}{M_{4}}{l_{0}}^{-[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)}. \end{aligned}$$

Repeating the procedure, one can show that

$$\int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}}u(x, \omega_{k})\,dx \ge\eta_{2^{k-1}l_{0}}(\omega_{k}) \ge \eta_{2^{k-1}l_{0}}(\omega _{k-1})+\delta_{0} \ge\varLambda- \sigma_{0}+k \delta_{0} $$

with

$$\begin{aligned} \omega_{k}=\omega_{k-1}+\frac {M_{5}}{M_{4}} \bigl(2^{k-2}l_{0} \bigr)^{-[p_{c}(n-q-1)-q(n+\mu_{2})]/(p_{c}-q)},\quad k=1, 2, \ldots. \end{aligned}$$

Therefore

$$\begin{aligned} \sup_{t>0} \int_{\mathbb {R}^{n}}\bigl(\vert x\vert +1\bigr)^{\mu_{1}} u(x,t)\,dx=+\infty, \end{aligned}$$

which contradicts (19).

Now, for the general case without the assumption that $u(x,t)$ is nonincreasing, consider $\underline{u}(x,t)$ defined by (12), which is nonincreasing and satisfies (13) and (14). Therefore, the conclusions of Lemmas 4.1-4.3 are all valid for $\underline{u}$. Similar to the above argument, one can show that $\underline{u}$ blows up in some $0< T_{*}<+\infty$, and thus u is a blow-up solution. □