In this section, we concentrate on the study of nonexistence of weak solutions to Problem (1.1). For convenience, we first state that the function \(f(s)\) and the coefficient \(b(x,t)\) satisfy the following conditions:
$$\begin{aligned}& b(x,t)\geqslant0,\qquad b_{t}(x,t)\leqslant0,\quad \forall(x,t)\in Q_{T}; \end{aligned}$$
(3.1)
$$\begin{aligned}& f(u)\in C(R),\qquad f(u)u- p^{+}G(u)\geqslant0,\quad \forall u\in \mathbb{R}, \end{aligned}$$
(3.2)
with \(G(u)=\int^{u}_{0}f(s)\,ds\). Before stating the main results, we give the definition of global solutions.
Definition 3.1
A function \(u(x,t)\) is called a global solution to Problem (1.1) if \(\forall T>0\) the following property holds:
$$\sup_{t\in(0,T)}\bigl\Vert u(x,t)\bigr\Vert _{L^{\infty}(\Omega)}< + \infty. $$
Otherwise, we say that Problem (1.1) does not admit global weak solutions.
First, we consider the case \(p(x,t)\equiv p(x)\). Our main result is as follows.
Theorem 3.1
Assume that
\(u(x,t)\in H(Q_{T})\cap L^{\infty}(0,T;L^{2}(\Omega))\), \(b(x,t)|\nabla u|\in L^{2}(0,T;L^{2}(\Omega))\)
is the local solution to Problem (1.1). If (3.1) and (3.2) are fulfilled and
\(u_{0}\in W^{1,p^{(}x)}_{0}(\Omega)\), \(p^{+}>2\), such that
$$\begin{aligned} \int_{\Omega}G(u_{0})\,dx >& \int_{\Omega}\frac{b(x,0)}{2}|\nabla u_{0}|^{2} \,dx+ \int_{\Omega}\frac{1}{p(x)}|\nabla u_{0}|^{p(x)} \,dx \\ &{}+\frac{4(p^{+}-1)}{Tp^{+}(p^{+}-2)^{2}} \int_{\Omega }|u_{0}|^{2}\,dx. \end{aligned}$$
(3.3)
Then there exists a
\(T^{*}\in(0,T]\)
such that
$$\lim_{t\rightarrow T^{*-}}\bigl\Vert u(\cdot,t)\bigr\Vert _{\infty,\Omega}=+\infty. $$
To prove Theorem 3.1, we need the following lemma.
Lemma 3.1
Assume that
\(u\in H(Q_{T})\)
is the solution to Problem (1.1), then
\(u(x,t)\)
satisfies the following relation:
$$\begin{aligned}& \int_{\Omega}\frac{1}{p(x)}\bigl\vert \nabla u(x,t)\bigr\vert ^{p(x)}\,dx+ \int_{\Omega}\frac{b(x,t)}{2}\bigl\vert \nabla u(x,t)\bigr\vert ^{2}\,dx- \int_{\Omega}G\bigl(u(x,t)\bigr)\,dx \\& \qquad {}+ \int^{t}_{0} \int_{\Omega }(u_{\tau})^{2}\,dx\,d\tau - \int^{t}_{0} \int_{\Omega}\frac{b_{\tau} \vert \nabla u\vert ^{2}}{2}\,dx\,d\tau \\& \quad = \int_{\Omega}\frac{1}{p(x)}\bigl\vert \nabla u_{0}(x)\bigr\vert ^{p(x)}\,dx+ \int_{\Omega}\frac{b(x,0)}{2}\bigl\vert \nabla u_{0}(x)\bigr\vert ^{2}\,dx- \int_{\Omega}G\bigl(u_{0}(x)\bigr)\,dx. \end{aligned}$$
(3.4)
Proof
Following the lines of the proof of Lemma 3.1 and Theorem 6.1 in [7], we know that \(u_{t}\in L^{2}(Q_{T})\). Noting that
$$\begin{aligned}& \frac{\partial}{\partial t}\biggl(\frac{|\nabla u|^{p(x)}}{p(x)}\biggr)=|\nabla u|^{p(x)-2} \nabla u\nabla u_{t},\qquad \frac{\partial}{\partial t}\bigl(G\bigl(u(x,t)\bigr) \bigr)=f\bigl(u(x,t)\bigr)u_{t}, \\& \frac{\partial}{\partial t}\biggl(b(x,t)\frac{|\nabla u|^{2}}{2}\biggr)=b(x,t)\nabla u\nabla u_{t}+b_{t}\frac{|\nabla u|^{2}}{2} \end{aligned}$$
and using the idea of the proof of Lemma 1 in [8], we arrive at the relation
$$ \frac{d}{dt} \int_{\Omega} \biggl(\frac{|\nabla u|^{p(x)}}{p(x)}+\frac{b(x,t)}{2}\bigl|\nabla u(x)\bigr|^{2}-G\bigl(u(x,t)\bigr) \biggr)\,dx- \int_{\Omega}\frac{b_{t}|\nabla u|^{2}}{2}\,dx=- \int_{\Omega}u_{t}^{2}\,dx. $$
After integrating over \((0,t)\), it is obvious that Lemma 3.1 holds. □
Proof of Theorem 3.1
Let
$$\begin{aligned}& \beta=\frac{2}{T(p^{+}-2)^{2}} \int_{\Omega}u_{0}^{2}\,dx,\qquad t_{0}=\frac {T(p^{+}-2)}{2}, \\& K(t)=\frac{1}{2} \int^{t}_{0} \int_{\Omega}u^{2}\,dx\,d\tau+(T-t) \int _{\Omega}\frac{1}{2}u_{0}^{2}\,dx+ \beta(t+t_{0})^{2}. \end{aligned}$$
Clearly
$$\begin{aligned}& K'(t) =\frac{1}{2} \int_{\Omega}u^{2}\,dx\,dt-\frac{1}{2} \int_{\Omega }u_{0}^{2}\,dx+2 \beta(t+t_{0}) \\& \hphantom{K'(t)}= \iint_{Q_{t}} \bigl(-b(x,\tau)|\nabla u|^{2}-|\nabla u|^{p(x)}+f(u)u \bigr)\,dx\,dt+2\beta(t+t_{0}); \\& K''(t) = \int_{\Omega} \bigl(-b(x,\tau)|\nabla u|^{2}-|\nabla u|^{p(x)}+f(u)u \bigr)\,dx+2\beta. \end{aligned}$$
By Hölder’s inequality, we have
$$\begin{aligned} \frac{1}{2} \int_{\Omega}u^{2}\,dx\,dt-\frac{1}{2} \int_{\Omega }u_{0}^{2}\,dx &= \frac{1}{2}\biggl\vert \int^{t}_{0} \int_{\Omega}\bigl(u^{2}\bigr)_{\tau}\,dx\,d\tau \biggr\vert \\ &\leqslant\biggl( \int^{t}_{0} \int_{\Omega}u^{2}\,dx\,d\tau\biggr)^{1/2} \biggl( \int ^{t}_{0} \int_{\Omega}|u_{\tau}|^{2}\,dx\,d\tau \biggr)^{1/2}. \end{aligned}$$
Thus by Schwarz’s inequality and the definition of \(K(t)\), we have
$$\begin{aligned} \bigl(K'(t)\bigr)^{2} \leqslant& \biggl( \int^{t}_{0} \int_{\Omega}u^{2}\,dx\,d\tau \biggr) \biggl( \int ^{t}_{0} \int_{\Omega}u_{\tau}^{2}\,dx\,d\tau \biggr) +4 \beta^{2}(t+t_{0})^{2} \\ &{} +4\beta(t+t_{0}) \biggl( \int^{t}_{0} \int_{\Omega}u^{2}\,dx\,d\tau \biggr)^{1/2} \biggl( \int^{t}_{0} \int_{\Omega}u_{\tau}^{2}\,dx\,d\tau \biggr)^{1/2} \\ \leqslant& \biggl( \int^{t}_{0} \int_{\Omega}u^{2}\,dx\,d\tau \biggr) \biggl( \int ^{t}_{0} \int_{\Omega}u_{\tau}^{2}\,dx\,d\tau \biggr) +4 \beta^{2}(t+t_{0})^{2} \\ &{} +2 \int^{t}_{0} \int_{\Omega}u_{\tau}^{2}\,dx\,d\tau \biggl[(T-t) \int _{\Omega}\frac{1}{2}u_{0}^{2}\,dx+ \beta(t+t_{0})^{2} \biggr] \\ &{} +4\beta^{2}(t+t_{0})^{2} \biggl( \int^{t}_{0} \int_{\Omega}\frac {1}{2}u^{2}\,dx\,d\tau \biggr) \biggl((T-t) \int_{\Omega}\frac{1}{2}u_{0}^{2}\,dx+ \beta(t+t_{0})^{2} \biggr)^{-1} \\ \leqslant& K(t) \biggl(2 \int^{t}_{0} \int_{\Omega}u_{\tau}^{2}\,dx\,d\tau +4\beta \biggr). \end{aligned}$$
Therefore by Lemma 3.1, we obtain the following inequality:
$$\begin{aligned}& K(t)K''(t)-\frac{p^{+}}{2} \bigl(K'(t)\bigr)^{2} \\& \quad \geqslant K(t) \biggl( \int_{\Omega }b(x,t) \biggl(\frac{p^{+}}{2}-1\biggr)|\nabla u|^{2}\,dx+ \int_{\Omega}\biggl(\frac{p^{+}}{p(x)}-1\biggr)|\nabla u|^{p(x)}\,dx \\& \qquad {}+ \int_{\Omega} \bigl(uf(u)-p^{+}G(u) \bigr)\,dx \biggr)-2 \beta\bigl(p^{+}-1\bigr) \\& \qquad {}+\frac{p^{+}}{2} \int_{\Omega} \biggl[2G(u_{0})-b(x,0)|\nabla u_{0}|^{2}-\frac{2}{p(x)}|\nabla u_{0}|^{p(x)} \biggr]\,dx \\& \qquad {}-\frac{p^{+}}{2} \int^{t}_{0} \int_{\Omega}\frac{b_{t}|\nabla u|^{2}}{2}\,dx\,dt. \end{aligned}$$
(3.5)
Noticing \(p^{+}>2\), \(K(t)>0\), we conclude from (3.1), (3.2), (3.3) that
$$ K(t)K''(t)-\frac{p^{+}}{2}\bigl(K'(t) \bigr)^{2}\geqslant0,\quad \mbox{for }t\in(0,T), $$
which implies
$$ \bigl(K^{1-\frac{p^{+}}{2}}(t)\bigr)''\leqslant0,\quad \mbox{for }t\in(0,T). $$
Noting that \(K^{1-\frac{p^{+}}{2}}(0)>0\), \((K^{1-\frac{p^{+}}{2}})'(0)\leqslant0\), then
$$ K^{1-\frac{p^{+}}{2}}\bigl(T^{*}\bigr)=0, \quad \mbox{for some }T^{*}\in\biggl(0,\frac{-K^{1-\frac {p^{+}}{2}}(0)}{(K^{1-\frac{p^{+}}{2}})'(0)}\biggr). $$
Here
$$ \frac{-K^{1-\frac{p^{+}}{2}}(0)}{(K^{1-\frac{p^{+}}{2}})'(0)}= \frac{T\int_{\Omega}u^{2}_{0}\,dx+2\beta t_{0}^{2}}{2(p^{+}-2)\beta t_{0}^{2}}\leqslant T. $$
Thus, \(0< T^{*}\leqslant T\) and
$$\lim_{t\rightarrow T^{*-}}\bigl\vert u(\cdot,t)\bigr\vert _{\infty,\Omega}=+\infty. $$
This completes the proof of Theorem 3.1. □
Next, we consider the case when p is dependent of t. Before stating the conclusion, we first give a useful lemma.
Lemma 3.2
Assume
\(u_{0}\in W^{1,p(x,0)}_{0}(\Omega)\), \(p^{+}>2\), \(p_{t}\leqslant0\). Then the solution of Problem (1.1) satisfies
$$\begin{aligned}& \int_{\Omega}\frac{1}{p(x,t)}\bigl\vert \nabla u(x,t)\bigr\vert ^{p(x,t)}\,dx+ \int_{\Omega}\frac{b(x,t)}{2}\bigl\vert \nabla u(x,t)\bigr\vert ^{2}\,dx- \int_{\Omega}G\bigl(u(x,t)\bigr)\,dx \\& \qquad {}+ \int^{t}_{0} \int_{\Omega }\vert u_{\tau} \vert ^{2}\,dx\,d \tau - \int^{t}_{0} \int_{\Omega}\frac{b_{\tau} \vert \nabla u\vert ^{2}}{2}\,dx\,d\tau \\& \quad \leqslant \int_{\Omega}\frac{1}{p(x,0)}\bigl\vert \nabla u_{0}(x)\bigr\vert ^{p(x,0)}\,dx+ \int_{\Omega}\frac{b(x,0)}{2}\bigl\vert \nabla u_{0}(x)\bigr\vert ^{2}\,dx- \int_{\Omega}G\bigl(u_{0}(x)\bigr)\,dx \\& \qquad {}+ \int_{\Omega} \biggl(\frac{1}{p(x,t)}-\frac{1}{p(x,0)} \biggr) \,dx. \end{aligned}$$
(3.6)
Proof
Following the lines of the proof of Lemmas 3.1 and Lemma 6.1 of [7], we know \(u_{t}\in L^{2}(Q_{T})\) and
$$\begin{aligned}& \frac{\partial}{\partial t}\biggl(\frac{|\nabla u|^{p(x,t)}}{p(x,t)}\biggr)=|\nabla u|^{p(x,t)-2} \nabla u\nabla u_{t}+\frac{p_{t}}{p^{2}}|\nabla u|^{p(x,t)} \bigl( \ln|\nabla u|^{p(x,t)}-1 \bigr), \\& \frac{\partial}{\partial t}\bigl(G\bigl(u(x,t)\bigr)\bigr)=f\bigl(u(x,t) \bigr)u_{t},\qquad \frac{\partial}{\partial t}\biggl(b(x,t)\frac{|\nabla u|^{2}}{2} \biggr)=b(x,t)\nabla u\nabla u_{t}+b_{t}\frac{|\nabla u|^{2}}{2}. \end{aligned}$$
On one hand, a simple analysis shows that
$$\begin{aligned}& \frac{d}{dt} \int_{\Omega}\biggl(\frac{\vert \nabla u\vert ^{p(x,t)}}{p(x,t)} +\frac{b(x,t)}{2}\bigl\vert \nabla u(x)\bigr\vert ^{2}-G\bigl(u(x,t)\bigr)\biggr)\,dx- \int_{\Omega}\frac{b_{t}\vert \nabla u\vert ^{2}}{2}\,dx \\& \quad = \int_{\Omega} \biggl[-u_{t}^{2}+ \frac {p_{t}}{p^{2}}\vert \nabla u\vert ^{p(x,t)} \bigl(\ln \vert \nabla u\vert ^{p(x,t)}-1 \bigr) \biggr]\,dx. \end{aligned}$$
(3.7)
On the other hand, we apply the condition \(p_{t}\leqslant0\) to obtain
$$\begin{aligned}& \int_{\{|\nabla u|^{p}\leqslant e\}} \frac{|\nabla u|^{p(x,t)}}{p^{2}(x,t)} \bigl(\ln|\nabla u|^{p(x,t)}-1 \bigr)p_{t}(x,t)\,dx \\& \quad \leqslant \int_{\{|\nabla u|^{p}\leqslant e\}} \frac{-p_{t}(x,t)}{p^{2}(x,t)}\,dx\leqslant \int_{\Omega}\frac{-p_{t}(x,t)}{p^{2}(x,t)}\,dx. \end{aligned}$$
(3.8)
The second inequality above follows from
$$ -\frac{1}{e}\leqslant s\ln s\leqslant0,\qquad 0\leqslant s\leqslant1. $$
Lemma 3.2 follows from (3.7) and (3.8). □
Our main result is as follows.
Theorem 3.2
Suppose that (3.1) and (3.2) hold and
\(p^{+}>2\), \(p_{t}\leqslant0\). If
\(u_{0}\in W^{1,p(x,0)}_{0}(\Omega)\)
satisfies
$$\begin{aligned} \int_{\Omega}G(u_{0})\,dx >& \int_{\Omega}\frac{b(x,0)}{2}|\nabla u_{0}|^{2} \,dx+ \int_{\Omega} \frac{1}{p(x,0)}|\nabla u_{0}|^{p(x,0)} \, dx \\ &{} +\frac{4(p^{+}-1)}{Tp^{+}(p^{+}-2)^{2}} \int_{\Omega}|u_{0}|^{2}\, dx+ \int_{\Omega} \biggl(\frac{2}{p^{-}}-\frac{1}{p(x,0)} \biggr) \,dx, \end{aligned}$$
(3.9)
then there exists
\(T^{*}\in(0,T]\)
such that
$$\lim_{t\rightarrow T^{*-}}\bigl\vert u(\cdot,t)\bigr\vert _{\infty,\Omega}=+\infty. $$
Proof
We argue by contradiction. Define
$$\begin{aligned}& \beta=\frac{2}{T(p^{+}-2)^{2}} \int_{\Omega}u_{0}^{2}\,dx,\qquad t_{0}=\frac {T(p^{+}-2)}{2}, \\& K(t)=\frac{1}{2} \int^{t}_{0} \int_{\Omega}u^{2}\,dx\,d\tau+(T-t) \int _{\Omega}\frac{1}{2}u_{0}^{2}\,dx+ \beta(t+t_{0})^{2}. \end{aligned}$$
It is easy to verify that
$$ \begin{aligned} &K'(t)=\frac{1}{2} \int_{\Omega}u^{2}\,dx\,dt-\frac{1}{2} \int_{\Omega }u_{0}^{2}\,dx+2 \beta(t+t_{0}) \\ &\hphantom{K'(t)}= \iint_{Q_{t}} \bigl(-b(x,\tau)|\nabla u|^{2}-|\nabla u|^{p(x,t)}+f(u)u \bigr)\,dx\,dt+2\beta(t+t_{0}); \\ &K''(t)= \int_{\Omega} \bigl(-b(x,\tau)|\nabla u|^{2}-|\nabla u|^{p(x,t)}+f(u)u \bigr)\,dx+2\beta; \\ &\bigl(K'(t)\bigr)^{2} \leqslant K(t) \biggl(2 \int^{t}_{0} \int_{\Omega}u_{\tau}^{2}\,dx\,d\tau +4\beta \biggr). \end{aligned} $$
(3.10)
Therefore by Lemma 3.2 and (3.10), we obtain the following inequality:
$$\begin{aligned}& K(t)K''(t)-\frac{p^{+}}{2} \bigl(K'(t)\bigr)^{2} \\& \quad \geqslant K(t) \biggl[ \int_{\Omega }b(x,t) \biggl(\frac{p^{+}}{2}-1\biggr)|\nabla u|^{2}\,dx+ \int_{\Omega}\biggl(\frac{p^{+}}{p(x,t)}-1\biggr)|\nabla u|^{p(x,t)}\, dx \\& \qquad {} + \int_{\Omega} \bigl(uf(u)-p^{+}G(u) \bigr)\,dx-2\beta \bigl(p^{+}-1\bigr) \\& \qquad {} +\frac{p^{+}}{2} \int_{\Omega} \biggl[2G(u_{0})-b(x,0)|\nabla u_{0}|^{2}-\frac{2}{p(x,0)}|\nabla u_{0}|^{p(x,0)} \biggr]\,dx \\& \qquad {} -\frac{p^{+}}{2} \int^{t}_{0} \int_{\Omega}\frac{b_{\tau}|\nabla u|^{2}}{2}\,dx\,d\tau+\frac{p^{+}}{2} \int_{\Omega} \biggl(\frac {1}{p(x,t)}-\frac{1}{p(x,0)} \biggr) \,dx \biggr]. \end{aligned}$$
(3.11)
In the rest of the proof, we follow the lines of the proof of Theorem 3.1 to finish the proof of this theorem. □
At the end of this paper, we give an example to illustrate that the condition on \(p_{t}(x,t)\) is weaker than that of [8, 9].
Example 3.1
Set \(p(x,y,z,t)=\frac{\sqrt{t}\cos x}{100}+\frac{5}{2}\), \(x\in(\frac{\pi}{2},\pi)\), \(y,z\in(0,\frac{\pi}{2})\), \(0< t<1\). A simple computation shows that
$$\begin{aligned}& p_{t}(x,y,z,t)=\frac{\cos x}{200\sqrt{t}}< 0, \\& \int_{0}^{1} \int_{\frac{\pi }{2}}^{\pi} \iint_{0}^{\frac{\pi}{2}}\bigl\vert p_{t}(x,y,z,t) \bigr\vert \, dx\, dy\, dz\, dt=\frac{\pi ^{2}}{400},\quad p_{t}(x,y,z,t) \notin L^{\infty}. \end{aligned}$$