It is well known that the equation in (1.1) is degenerate if m>1 and singular if 0<m<1, and therefore there is no classical solution in general. To state the definition of the weak solution, we first define the class of nonnegative testing functions
\mathcal{F}=\{\xi :\xi \in C({\overline{Q}}_{T})\cap {C}^{2,1}({Q}_{T}),{\xi}_{t},\mathrm{\Delta}\xi \in {L}^{2}({Q}_{T});\xi \ge 0,\xi {}_{\partial \mathrm{\Omega}\times (0,T)}=0\},
where {Q}_{T}=\mathrm{\Omega}\times (0,T).
Definition 2.1 A function u\in {L}^{\mathrm{\infty}}({Q}_{T}) is called a subsolution (supersolution) of Problem (1.1) in {Q}_{T} if the following conditions hold:

(i)
u(x,0)\le (\ge ){u}_{0}(x) in Ω,

(ii)
u(x,t)\le (\ge )0 on \partial \mathrm{\Omega}\times (0,T),

(iii)
for almost every t\in (0,T) and every \xi \in \mathcal{F},
\begin{array}{r}{\int}_{\mathrm{\Omega}}u(x,t)\xi (x,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x{\int}_{\mathrm{\Omega}}{u}_{0}(x)\xi (x,0)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+b{\int}_{0}^{t}{\int}_{\mathrm{\Omega}}{u}^{r}(x,s)\xi (x,s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \phantom{\rule{1em}{0ex}}\le (\ge ){\int}_{0}^{t}{\int}_{\mathrm{\Omega}}\{u{\xi}_{s}+{u}^{m}\mathrm{\Delta}\xi +a{\int}_{\mathrm{\Omega}}{u}^{q}(y,s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\xi (x,s)\}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.\end{array}
(2.1)
A function u(x,t) is called a local solution of (1.1) if it is both a subsolution and a supersolution for some T>0 and u(x,t) is called a solution of (1.1) if it is a local solution of (1.1) in {Q}_{T} for any T>0.
Local existence of weak solutions of (1.1) can be obtained by utilizing the methods of standard regularization (see [9]) and the continuity of the solutions can be derived by the arguments similar to that in [18]. Moreover, Problem (1.1) admits global solutions when the initial data are small (see [1]). Since the regularization procedure is crucial in what follows, we shall sketch the outline. Consider the regularized problem
\{\begin{array}{cc}{u}_{t}=\mathrm{\Delta}{u}^{m}+a{\int}_{\mathrm{\Omega}}{u}^{q}(y,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}yb{u}^{r},\hfill & x\in \mathrm{\Omega},0<t<T,\hfill \\ u(x,t)=1/k,\hfill & x\in \partial \mathrm{\Omega},0<t<T,\hfill \\ u(x,0)={u}_{0}(x)+1/k,\hfill & x\in \mathrm{\Omega},\hfill \end{array}
(2.2)
where T>0 may be chosen sufficiently small in such a way that there exists a solution {u}_{k} of (2.2) on {Q}_{T} for every k\in \mathbb{N}, and {\parallel {u}_{k}\parallel}_{\mathrm{\infty}} is bounded independently of k. Furthermore, 1/l\le {u}_{l}\le {u}_{k} for k<l, and a supersolution (subsolution) comparison theory holds for (2.2) (see [1, 19]).
Since {u}_{k} is monotone in k, we may define U(x,t)\equiv {lim}_{k\to \mathrm{\infty}}{u}_{k}(x,t), and it is easy to see that U(x,t) is a solution of (1.1). Furthermore, if u is a solution of (1.1), then we have
\begin{array}{r}{\int}_{\mathrm{\Omega}}(u{u}_{k})\xi (x,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\\ \phantom{\rule{1em}{0ex}}={\int}_{0}^{t}{\int}_{\mathrm{\Omega}}\{(u{u}_{k}){\xi}_{s}+({u}^{m}{u}_{k}^{m})\mathrm{\Delta}\xi +a{\int}_{\mathrm{\Omega}}({u}^{q}(y,s){u}_{k}^{q}(y,s))\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\xi (x,s)\}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \phantom{\rule{2em}{0ex}}b{\int}_{0}^{t}{\int}_{\mathrm{\Omega}}({u}^{r}(x,s){u}_{k}^{r}(x,s))\xi (x,s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+\frac{1}{{k}^{m}}{\int}_{0}^{t}{\int}_{\mathrm{\Omega}}\frac{\partial \xi}{\partial n}\phantom{\rule{0.2em}{0ex}}\mathrm{d}{S}_{x}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\frac{1}{k}{\int}_{\mathrm{\Omega}}\xi (x,0)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\\ \phantom{\rule{1em}{0ex}}\le {\int}_{0}^{t}{\int}_{\mathrm{\Omega}}\{(u{u}_{k}){\xi}_{s}+({u}^{m}{u}_{k}^{m})\mathrm{\Delta}\xi +a{\int}_{\mathrm{\Omega}}({u}^{q}(y,s){u}_{k}^{q}(y,s))\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\xi (x,s)\}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\end{array}
where we use the fact \partial \xi /\partial n\le 0 on ∂ Ω to derive this inequality. With {\mathrm{\Phi}}_{k} and {F}_{k} defined so that
(u{u}_{k}){\mathrm{\Phi}}_{k}={u}^{m}{u}_{k}^{m},\phantom{\rule{2em}{0ex}}(u{u}_{k}){R}_{k}={u}^{r}{u}_{k}^{r}
and
(u{u}_{k}){F}_{k}={u}^{q}{u}_{k}^{q},
we have
\begin{array}{rcl}{\int}_{\mathrm{\Omega}}(u{u}_{k})\xi (x,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x& \le & {\int}_{0}^{t}{\int}_{\mathrm{\Omega}}(u{u}_{k})\{{\xi}_{s}+{\mathrm{\Phi}}_{k}\mathrm{\Delta}\xi b{R}_{k}\xi \}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ +a{\int}_{0}^{t}{\int}_{\mathrm{\Omega}}\xi (x,s)\{{\int}_{\mathrm{\Omega}}(u{u}_{k}){F}_{k}(y,s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y\}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.\end{array}
Thus, we can choose the appropriate test function ξ as in [1, 19] to obtain u\le {u}_{k}. If u is a subsolution of (1.1), the above argument shows that u\le {u}_{k}. Thus U(x,t) is the maximal solution of (1.1), and this solution satisfies a subsolution comparison principle.
Before proving our main results, we give a comparison principle for the solution of Problem (1.1), which is similar to Proposition 2.3 in [9] and can be proved by modifying the above arguments (see also [1, 10, 19]).
Proposition 2.1 Let u and v be a nonnegative bounded subsolution and a nonnegative supersolution of (1.1), respectively. If either q\ge 1 and u is bounded from the above or 0<q<1 and v has a positive lower bound, then u(x,t)\le v(x,t) in {Q}_{T} if {u}_{0}(x)\le {v}_{0}(x) in Ω.
Proof of Theorem 1.2 Case (i): q=m with a\mu <1. For any bounded smooth domain {\mathrm{\Omega}}^{\prime} such that {\mathrm{\Omega}}^{\prime}\supset \supset \mathrm{\Omega}, let \varphi (x) be the unique solution of the following elliptic problem:
\{\begin{array}{cc}\mathrm{\Delta}\varphi (x)=1,\hfill & x\in {\mathrm{\Omega}}^{\prime},\hfill \\ \varphi (x)=0,\hfill & x\in \partial {\mathrm{\Omega}}^{\prime}.\hfill \end{array}
(2.3)
By the comparison principle for linear elliptic problem we know \phi (x)\le \varphi (x) in Ω. Set {\mu}_{1}={\int}_{\mathrm{\Omega}}\varphi (x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x, {M}_{1}={max}_{x\in {\overline{\mathrm{\Omega}}}^{\prime}}\varphi (x) and \delta ={min}_{x\in \overline{\mathrm{\Omega}}}\varphi (x). It is well known from the strong maximum principle that \delta >0.
By continuity, we can choose a suitable domain {\mathrm{\Omega}}^{\prime} with {\mathrm{\Omega}}^{\prime}\supset \supset \mathrm{\Omega} such that a{\mu}_{1}<1. Define v(x,t)=g(t){\varphi}^{\frac{1}{m}}(x), where g(t) satisfies
\{\begin{array}{cc}{g}^{\prime}(t){M}_{1}^{\frac{1}{m}}+(1a{\mu}_{1}){g}^{m}(t)=0,\hfill & t>0,\hfill \\ g(0)=A\ge {\delta}^{\frac{1}{m}}{\parallel {u}_{0}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})}.\hfill \end{array}
(2.4)
Since 0<m<1, it follows from the theory in ODEs that g(t) is nonincreasing and g(t)=0 for all
t\ge {T}^{\ast}=\frac{{M}_{1}^{\frac{1}{m}}}{(1a{\mu}_{1})(1m)}{A}^{1m}.
Then it can be verified that v(x,t) is a supersolution of (1.1). In fact, because q=m and {g}^{\prime}(t)\le 0, we know that v(x,t) satisfies the following inequalities (in the weak sense):
\begin{array}{r}\frac{\partial v}{\partial t}\mathrm{\Delta}{v}^{m}a{\int}_{\mathrm{\Omega}}{v}^{m}(y,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y+b{v}^{r}\\ \phantom{\rule{1em}{0ex}}={g}^{\prime}(t){\varphi}^{\frac{1}{m}}(x)+{g}^{m}(t)a{\mu}_{1}{g}^{m}(t)+b{g}^{r}(t){\varphi}^{\frac{r}{m}}(x)\\ \phantom{\rule{1em}{0ex}}\ge {g}^{\prime}(t){M}_{1}^{\frac{1}{m}}+(1a{\mu}_{1}){g}^{m}(t)\\ \phantom{\rule{1em}{0ex}}=0.\end{array}
(2.5)
In addition, v(x,t)\ge 0 on \partial \mathrm{\Omega}\times (0,T), for any 0<T<{T}^{\ast}, and v(x,0)\ge {u}_{0}(x) by the choice of A. Moreover, there exists a positive constant {C}_{1} such that v(x,t)\ge {C}_{1} in {Q}_{T}. Therefore, by applying Proposition 2.1 to (1.1) we see that u(x,t)\le v(x,t) for (x,t)\in {Q}_{T}, which implies u(x,T)\le v(x,T). The arbitrariness of T<{T}^{\ast} and v(x,{T}^{\ast})=0 ensure that u(x,{T}^{\ast})=0. Furthermore, let \tilde{u}(x,t)=u(x,t+{T}^{\ast}), then \tilde{u}(x,t) satisfies (1.1) with the initial condition \tilde{u}(x,0)=0. By the aforementioned proof, we see that \tilde{u}(x,t)\le v(x,t) with any A>0. From the relation of the extinction time {T}^{\ast} of v(x,t) to A, it follows that \tilde{u}(x,t)=0 for any t>0, i.e. u(x,t)=0 for any t\ge {T}^{\ast}.
Case (ii): q>m. Let ϕ, {M}_{1} be the same as Case (i) and denote {k}_{0}=max\{1,\frac{2{M}_{1}^{\frac{qm}{m}}\mu}{{\int}_{\mathrm{\Omega}}{\varphi}^{\frac{q}{m}}(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x}\}. Set v(x,t)=k{\varphi}^{\frac{1}{m}}(x) with k={(\frac{1}{a{k}_{0}{\int}_{\mathrm{\Omega}}{\varphi}^{\frac{q}{m}}(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x})}^{\frac{1}{qm}}, then it is easy to verify that v(x,t) is a supersolution of (1.1) when {u}_{0}(x) is sufficiently small such that {u}_{0}(x)\le k{\varphi}^{\frac{1}{m}}(x) in Ω. Applying Proposition 2.1 to Problem (1.1) in {Q}_{T} for any T>0 we obtain u(x,t)\le v(x,t) in {Q}_{T}, which implies that u(x,t)\le k{M}_{1}^{\frac{1}{m}}. Therefore, u(x,t) satisfies
{u}_{t}\mathrm{\Delta}{u}^{m}+b{u}^{r}\le a{\left(k{M}_{1}^{\frac{1}{m}}\right)}^{qm}{\int}_{\mathrm{\Omega}}{u}^{m}(y,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}y,\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega},t>0.
By the choice of k and {k}_{0} it is easily verified that a{(k{M}_{1}^{\frac{1}{m}})}^{qm}\mu \le \frac{1}{2}<1. Thus, by the results of Case (i), we can conclude that the solution u(x,t) vanishes in finite time when the initial data are suitably small. The proof of this theorem is complete. □
Proof of Theorem 1.3 We first prove the case r=q<1 with b>a\mathrm{\Omega}. Set v(x,t)=g(t) where g(t) satisfies the following ordinary differential equation:
\{\begin{array}{cc}{g}^{\prime}(t)+(ba\mathrm{\Omega}){g}^{q}(t)=0,\hfill & t>0,\hfill \\ g(0)={\parallel {u}_{0}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})}.\hfill \end{array}
(2.6)
Since 0<q<1 and b>a\mathrm{\Omega}, we know by integrating the ODE that g(t) vanishes at some finite time {T}^{\ast}. Moreover, as in the proof of Theorem 1.2, it can be verified that g(t) is a supersolution of (1.1). Thus, by applying Proposition 2.1 to u(x,t) and g(t) for any 0<T<{T}^{\ast} we know that u(x,t) also vanishes at {T}^{\ast}.
In the case r<min\{q,1\}, let g(t) satisfy the following ODE:
\{\begin{array}{cc}{g}^{\prime}(t)+(ba\mathrm{\Omega}{g}^{qr}(t)){g}^{r}(t)=0,\hfill & t>0,\hfill \\ g(0)={g}_{0},\hfill \end{array}
(2.7)
where 0<{g}_{0}<{(\frac{b}{a\mathrm{\Omega}})}^{\frac{1}{qm}}. Similar to the first case, it is well known that g(t) vanishes in finite time since r<1 and g(t) is a supersolution of (1.1) provided that {u}_{0}(x) is small enough such that {\parallel {u}_{0}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})}\le {g}_{0}. Applying Proposition 2.1 to u(x,t) and g(t) guarantees the finite time extinction of u(x,t). This completes the proof of Theorem 1.3. □
Proof of Theorem 1.4 (i) Consider first the case q<r\le m. Let {\lambda}_{1} be the first eigenvalue of the following eigenvalue problem:
\{\begin{array}{cc}\mathrm{\Delta}\psi (x)=\lambda \psi (x),\hfill & x\in \mathrm{\Omega},\hfill \\ \psi (x)=0,\hfill & x\in \partial \mathrm{\Omega},\hfill \end{array}
(2.8)
and {\psi}_{1}(x)>0 (x\in \mathrm{\Omega}) be the corresponding eigenfunction. We may normalize {\psi}_{1}(x) such that {\parallel {\psi}_{1}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})}=1. Denote \gamma ={\int}_{\mathrm{\Omega}}{\psi}_{1}^{\frac{q}{m}}(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x and let g(t) satisfy the ODE problem
\{\begin{array}{cc}{g}^{\prime}(t)={\lambda}_{1}{g}^{m}(t)+a\gamma {g}^{q}(t)b{g}^{r}(t),\hfill & t>0,\hfill \\ g(0)=0,\hfill \\ g(t)>0,\phantom{\rule{1em}{0ex}}t>0.\hfill \end{array}
(2.9)
It is easy to check that g(t) is nondecreasing and bounded from above by min\{{(\frac{a\gamma}{{\lambda}_{1}})}^{\frac{1}{mq}},{(\frac{a\gamma}{b})}^{\frac{1}{rq}}\}. Set v(x,t)=\rho g(t){\psi}_{1}^{\frac{1}{m}}(x). We shall show that v(x,t) is a subsolution of (1.1) when \rho >0 is sufficiently small. In fact, simple computations show that
\frac{\partial v}{\partial t}=\rho ({\lambda}_{1}{g}^{m}(t)+a\gamma {g}^{q}(t)b{g}^{r}(t)){\psi}_{1}^{\frac{1}{m}}(x)
and
\begin{array}{r}\mathrm{\Delta}{v}^{m}+a{\int}_{\mathrm{\Omega}}{v}^{q}(y,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}yb{v}^{r}\\ \phantom{\rule{1em}{0ex}}={\lambda}_{1}{\rho}^{m}{g}^{m}{\psi}_{1}+a\gamma {\rho}^{q}{g}^{q}b{\rho}^{r}{g}^{r}{\psi}_{1}^{\frac{r}{m}}.\end{array}
For v(x,t) to be a subsolution of (1.1), it suffices to show that
{\lambda}_{1}{g}^{m}(t){\rho}^{m}+b{g}^{r}(t){\rho}^{r}\le a\gamma {g}^{q}(t)({\rho}^{q}\rho ),
which follows from
C({\rho}^{mq}+{\rho}^{rq})\le a\gamma (1{\rho}^{1q}),
(2.10)
where C={max}_{t>0}\{{\lambda}_{1}{g}^{mq}(t),b{g}^{rq}(t)\}<+\mathrm{\infty}. It is easy to see that (2.10) is valid for sufficiently small \rho >0 since q<r\le m<1.
Next, we turn our attention to construct a supersolution of (1.1). Set w(x,t)=L, where L=max\{{\parallel {u}_{0}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})},{(\frac{a\mathrm{\Omega}}{b})}^{\frac{1}{rq}},\rho {max}_{t\ge 0}g(t)\}. Then it is not hard to see that w(x,t) is a supersolution and v(x,t)\le w(x,t). Therefore, by an iteration process, one can obtain a solution of Problem (1.1), which satisfies v(x,t)\le u(x,t)\le w(x,t). Indeed, define {u}_{1}(x,t)=w(x,t) and {\{{u}_{k}(x,t)\}}_{k=2}^{\mathrm{\infty}} iteratively to be a solution of the problem
{u}_{kt}\mathrm{\Delta}{u}_{k}^{m}+b{u}_{k}^{r}=a{\int}_{\mathrm{\Omega}}{u}_{k1}^{q}(x,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x,\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega},t>0,
subject to the boundary and initial conditions as that in (1.1). By applying the comparison technique used in the proof of Lemma 2.1 in [1, 12] we know that the function u(x,t)={lim}_{k\to \mathrm{\infty}}{u}_{k}(x,t), for every x\in \overline{\mathrm{\Omega}} and t>0, is a solution of (1.1). Because v(x,t) does not vanish, neither does u(x,t).

(ii)
The case q<m<r can be treated similarly to Case (i).

(iii)
Finally we consider the case r=q<m with a\gamma >b. Let g(t) satisfy the following ODE:
\{\begin{array}{cc}{g}^{\prime}(t)={\lambda}_{1}{g}^{m}(t)+(a\gamma b){g}^{q}(t),\hfill & t>0,\hfill \\ g(0)=0,\hfill \\ g(t)>0,\phantom{\rule{1em}{0ex}}t>0.\hfill \end{array}
(2.11)
Then g(t) is nondecreasing and satisfies g(t)\le {(\frac{a\gamma b}{{\lambda}_{1}})}^{\frac{1}{mq}}. (The upper bound of g(t) can be obtained by contradiction arguments and the monotonicity of g(t) follows immediately as the upper bound is derived.) As in the proof of Case (i), we can construct a nonextinction subsolution v(x,t)=\rho g(t){\psi}^{\frac{1}{m}}(x) with \rho >0 sufficiently small.
To construct a supersolution, consider the following eigenvalue problem:
\{\begin{array}{cc}\mathrm{\Delta}\psi (x)=\lambda \psi (x),\hfill & x\in \tilde{\mathrm{\Omega}},\hfill \\ \psi (x)=0,\hfill & x\in \partial \tilde{\mathrm{\Omega}},\hfill \end{array}
where \tilde{\mathrm{\Omega}}\supset \supset \mathrm{\Omega} is a bounded domain with smooth boundary \partial \tilde{\mathrm{\Omega}}. Let {\tilde{\lambda}}_{1} and {\tilde{\psi}}_{1}(x)>0 (x\in \tilde{\mathrm{\Omega}}) be its first eigenvalue and the corresponding eigenfunction, respectively. We may normalize {\tilde{\psi}}_{1}(x) such that {\parallel {\tilde{\psi}}_{1}\parallel}_{{L}^{\mathrm{\infty}}(\tilde{\mathrm{\Omega}})}=1. Denote \tilde{\gamma}={\int}_{\mathrm{\Omega}}{\tilde{\psi}}_{1}^{\frac{q}{m}}(x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x and \tilde{\delta}={min}_{x\in \overline{\mathrm{\Omega}}}{\tilde{\psi}}_{1}(x)>0. Set w(x,t)=k{\tilde{\psi}}_{1}^{\frac{1}{m}}(x), then we shall show that w(x,t) is a supersolution of (1.1) provided that k>0 is suitably large. Indeed, if k=max\{{(\frac{a\tilde{\gamma}}{{\tilde{\lambda}}_{1}\tilde{\delta}})}^{\frac{1}{mq}},{\tilde{\delta}}^{\frac{1}{m}}{\parallel {u}_{0}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})},\rho {(\frac{a\gamma b}{{\lambda}_{1}})}^{\frac{1}{mq}}\}, we know that w(x,t)\ge 0 on \partial \mathrm{\Omega}\times (0,\mathrm{\infty}), w(x,0)\ge {u}_{0}(x) in Ω and w(x,t) satisfies the following inequalities (in the weak sense):
\begin{array}{r}\frac{\partial w}{\partial t}\mathrm{\Delta}{w}^{m}a{\int}_{\mathrm{\Omega}}{w}^{q}(x,t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+b{w}^{q}\\ \phantom{\rule{1em}{0ex}}={k}^{m}{\tilde{\lambda}}_{1}{\tilde{\psi}}_{1}(x)a{k}^{q}\tilde{\gamma}+b{k}^{q}{\tilde{\psi}}_{1}^{\frac{q}{m}}(x)\\ \phantom{\rule{1em}{0ex}}\ge {k}^{q}({k}^{mq}{\tilde{\lambda}}_{1}\tilde{\delta}a\tilde{\gamma})\\ \phantom{\rule{1em}{0ex}}\ge 0.\end{array}
Moreover, v(x,t)\le w(x,t) by the choice of k. Therefore, by applying the monotonicity iteration process we can obtain a nonextinction solution u(x,t) of (1.1) satisfying v(x,t)\le u(x,t)\le w(x,t). The proof of Theorem 1.4 is complete. □
Proof of Theorem 1.5 The proof of this theorem is similar to that of Theorem 1.4, so we only sketch the outline here. Set v(x,t)=\rho g(t){\phi}^{\frac{1}{m}}(x) where \phi (x) is defined in (1.2) and g(t) satisfies the following ODE problem:
\{\begin{array}{cc}{g}^{\prime}(t)={M}^{\frac{1}{m}}\{(a\mu 1){g}^{m}(t)b{M}^{\frac{r}{m}}{g}^{r}(t)\},\hfill & t>0,\hfill \\ g(0)=0,\hfill \\ g(t)>0,\phantom{\rule{1em}{0ex}}t>0.\hfill \end{array}
(2.12)
Since a\mu >1 and m<r, it is well known that g(t) is nondecreasing and bounded above by {(\frac{a\mu 1}{b{M}^{\frac{r}{m}}})}^{\frac{1}{rm}}. Then v(x,t) is a subsolution of (1.1) if \rho >0 is sufficiently small. On the other hand, the supersolution w(x,t) can be chosen to be a large positive constant L satisfying L\ge max\{{\parallel {u}_{0}\parallel}_{{L}^{\mathrm{\infty}}(\mathrm{\Omega})},{(\frac{a\mathrm{\Omega}}{b})}^{\frac{1}{rq}},\rho {M}^{\frac{1}{m}}{max}_{t\ge 0}g(t)\}. It can be observed that (v,w) is a pair of subsolution and supersolution of (1.1) satisfying v(x,t)\le w(x,t). Therefore, by monotonicity iteration, we know that (1.1) admits at least one solution u(x,t) such that v(x,t)\le u(x,t)\le w(x,t). Since v(x,t)>0 in \mathrm{\Omega}\times (0,+\mathrm{\infty}), u(x,t) cannot vanish at any finite time. The proof of Theorem 1.5 is complete. □
Proof of Theorem 1.6 (i) Let u(x,t) be any solution of (1.1). It can be verified that, for the case m=q<r<1, a sufficiently large constant L is a supersolution of (1.1). Therefore, we know that u(x,t)\le L in \mathrm{\Omega}\times (0,+\mathrm{\infty}). For convenience, in the following proof, we assume that the weak solution is appropriately smooth; otherwise, we can consider the corresponding regularized problem, and the same result can also be obtained through an approximate process (see [15]). Multiplying equation (1.1) by {u}^{m} and integrating by parts over Ω yield the identity
\frac{1}{m+1}\frac{\mathrm{d}}{\mathrm{d}t}{\int}_{\mathrm{\Omega}}{u}^{m+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+{\int}_{\mathrm{\Omega}}{\left\mathrm{\nabla}{u}^{m}\right}^{2}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+b{\int}_{\mathrm{\Omega}}{u}^{m+r}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x=a{\left({\int}_{\mathrm{\Omega}}{u}^{m}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)}^{2}.
(2.13)
Recall the embedding theorem
{\int}_{\mathrm{\Omega}}{\left\mathrm{\nabla}{u}^{m}\right}^{2}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\ge {\lambda}_{1}{\int}_{\mathrm{\Omega}}{u}^{2m}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x.
Combining this result with (2.13) and using Hölder’s inequality on the right hand side of (2.13) one obtains
\frac{1}{m+1}\frac{\mathrm{d}}{\mathrm{d}t}{\int}_{\mathrm{\Omega}}{u}^{m+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+{\lambda}_{1}{\int}_{\mathrm{\Omega}}{u}^{2m}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+b{\int}_{\mathrm{\Omega}}{u}^{m+r}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\le a\mathrm{\Omega}{\int}_{\mathrm{\Omega}}{u}^{2m}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x.
(2.14)
Noticing that a\mathrm{\Omega}\le {\lambda}_{1} and u(x,t)\le L, we see from (2.14) that
\frac{\mathrm{d}}{\mathrm{d}t}{\int}_{\mathrm{\Omega}}{u}^{m+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x+b(m+1){L}^{r1}{\int}_{\mathrm{\Omega}}{u}^{m+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\le 0,
which implies
{\int}_{\mathrm{\Omega}}{u}^{m+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\le {e}^{b(m+1){L}^{r1}t}{\int}_{\mathrm{\Omega}}{u}_{0}^{m+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x.
This shows that {\parallel u(\cdot ,t)\parallel}_{m+1} tends to 0 exponentially as t\to +\mathrm{\infty}.

(ii)
Let w(x,t)=g(t){\phi}^{\frac{1}{m}}(x), where g(t) satisfies
\{\begin{array}{cc}{g}^{\prime}(t)+b{M}^{\frac{1r}{m}}{g}^{r}(t)=0,\hfill & t>0,\hfill \\ g(0)=A>0.\hfill \end{array}
(2.15)
Since 0<r<1, g(t) is nonincreasing and g(t)=0 for t\ge {T}^{\ast}=\frac{{A}^{1r}}{b{M}^{\frac{1r}{m}}(1r)}. Noticing m=q and a\mu =1, one can see that w(x,t) is a supersolution of (1.1) provided that {u}_{0}(x)\le A{\phi}^{\frac{1}{m}}(x) in Ω. By using the arguments similar to that of the proof of Case (i) of Theorem 1.2 we can show that any solution u(x,t) of Problem (1.1) vanishes in finite time.

(iii)
Finally we consider the case q=m<1\le r. First we construct a nonextinction subsolution of (1.1). Set v(x,t)={h}_{0}{e}^{\alpha t}{\phi}^{\frac{1}{m}}(x), where {h}_{0}, α are two positive constants to be determined. Noticing that a\mu =1, it is easily verified that when r=1, v(x,t) is a subsolution of (1.1) if \alpha \ge b and if {h}_{0} is so small such that {h}_{0}{\phi}^{\frac{1}{m}}(x)\le {u}_{0}(x). When r>1, for v(x,t) to be subsolution of (1.1) it is reasonable to choose first {h}_{0} so small such that {h}_{0}{\phi}^{\frac{1}{m}}(x)\le {u}_{0}(x) and then \alpha \ge b{h}_{0}^{r1}{M}^{\frac{r1}{m}}. Next, since r>q and v(x,t) is bounded, we can choose a sufficiently large constant L\ge v(x,t) to be a supsolution of (1.1). Therefore, by monotonicity iteration, we can obtain a solution of (1.1) satisfying v(x,t)\le u(x,t)\le L. Since v(x,t) does not vanish at any finite time, neither does u(x,t). The proof of Theorem 1.6 is complete. □