1 Introduction
In this paper, we study the blow-up and global solutions for the following nonlinear reaction-diffusion equations under Neumann boundary conditions: where \(D\subset\mathbb{R}^{N}\) (\(N\geq2\)) is a bounded domain with smooth boundary ∂D, \(\partial/\partial n\) represents the outward normal derivative on ∂D, \(u_{0}\) is the initial value, T is the maximal existence time of u, and D̅ is the closure of D. In order to study the blow-up problem of (1.1) by using maximum principles, we make the following assumptions about the functions a, b, f, g, and \(u_{0}\). Set \(\mathbb{R}^{+}:=(0,+\infty)\). Throughout the paper, we assume that \(a(s)\) is a positive \(C^{2}(\mathbb{R}^{+})\) function, \(b(x)\) is a positive \(C^{1}(\overline{D})\) function, \(f(x,s)\) is a nonnegative \(C^{1}(D\times\mathbb{R}^{+})\) function, \(g(s)\) is a \(C^{3}(\mathbb{R}^{+})\) function, \(g'(s)>0\) for any \(s\in\mathbb{R}^{+}\), and \(u_{0}(x)\) is a positive \(C^{2}(\overline{D})\) function. Under these assumptions, the classical parabolic equation theory ensures that there exists a unique classical solution \(u(x,t)\) for problem (1.1) with some \(T>0\) and the solution is positive over \(\overline{D}\times[0,T)\). Moreover, by regularity theorem [1], \(u\in C^{3}(D\times(0,T))\cap C^{2}(\overline{D}\times[0,T))\).
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (g(u) )_{t} =\nabla\cdot(a(u)b(x)\nabla u)+f(x,u) & \mbox{in } D\times(0,T), \\ \frac{\partial u}{\partial n}=0 &\mbox{on } \partial D\times(0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.1)
During the past decades, the problems of the blow-up and global solutions for nonlinear reaction-diffusion equations have received considerable attention. The contributions in the filed can be found in [2‐8] and the references therein. Many authors discussed the blow-up and global solutions for nonlinear reaction-diffusion equations under Neumann boundary conditions and obtained a lot of interesting results; we refer the reader to [9‐19]. Some particular cases of (1.1) have been investigated already. Lair and Oxley [20] studied the following problem: where D is a bounded domain of \(\mathbb{R}^{N}\) (\(N\geq2\)) with smooth boundary ∂D. Necessary and sufficient conditions characterized by functions a and f were given for the existence of blow-up and global solutions. Zhang [21] discussed the following problem: where D is a bounded domain of \(\mathbb{R}^{N}\) (\(N\geq2\)) with smooth boundary ∂D. Sufficient conditions were developed there for the existence of blow-up and global solutions. Ding and Guo [22] considered the following problem: where D is a bounded domain of \(\mathbb{R}^{N}\) (\(N\geq2\)) with smooth boundary ∂D. Sufficient conditions were given there for the existence of blow-up and global solutions. Meanwhile, an upper bound of the ‘blow-up time’, an upper estimate of ‘blow-up rate’, and an upper estimate of the global solution were also obtained.
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} u_{t}=\nabla\cdot(a(u)\nabla u)+f(u) & \mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 & \mbox{on } \partial D\times (0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.2)
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (g(u))_{t}=\Delta u+f(u) & \mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 & \mbox{on } \partial D\times (0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.3)
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (g(u))_{t}=\nabla\cdot(a(u)\nabla u)\Delta u+f(u) & \mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 & \mbox{on } \partial D\times (0,T), \\ u(x,0)=u_{0}(x)>0 & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
(1.4)
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The object of this paper is the blow-up and global solutions for problem (1.1). Since the reaction function \(f(x,u)\) contains not only the concentration variable u but also the space variable x, it seems that the methods of [20‐22] are not applicable to problem (1.1). In this paper, we investigate problem (1.1) by constructing auxiliary functions completely different from those in [20‐22] and technically using maximum principles and a first-order differential inequality technique. We obtain some existence theorems for a blow-up solution, an upper bound of ‘blow-up time’, an upper estimate of ‘blow-up rate’, existence theorems for a global solution, and an upper estimate of the global solution. Our results can be considered as extensions and supplements of those obtained in [20‐22].
2 Blow-up solution
Our main result for the blow-up solution is stated in the following theorem.
Theorem 2.1
Let
u
be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are satisfied:
Then the solution
u
to problem (1.1) must blow up in a finite
T, and
where
and
\(H^{-1}\)
is the inverse function of
H.
(i)
for any
\(s\in\mathbb{R}^{+}\),
$$ \biggl(\frac{a(s)}{g'(s)} \biggr)'\geq0,\quad \biggl[ \frac{1}{a(s)} \biggl(\frac{a(s)}{g'(s)} \biggr)' + \frac{1}{g'(s)} \biggr]'+ \biggl[\frac{1}{a(s)} \biggl( \frac {a(s)}{g'(s)} \biggr)' +\frac{1}{g'(s)} \biggr]\geq0; $$
(2.1)
(ii)
for any
\((x,s)\in D\times\mathbb{R}^{+}\),
$$ \biggl(\frac{f(x,s)g'(s)}{a(s)} \biggr)_{s} -\frac{f(x,s)g'(s)}{a(s)} \geq0; $$
(2.2)
(iii)
$$ \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\, {\mathrm{d}}s< + \infty,\qquad M_{0}:=\max_{\overline{D}}u_{0}(x); $$
(2.3)
(iv)
$$ \beta:=\min_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{u_{0}}}>0. $$
(2.4)
$$\begin{aligned}& T\leq\frac{1}{\beta} \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\, {\mathrm{d}}s, \end{aligned}$$
(2.5)
$$\begin{aligned}& u(x,t)\leq H^{-1} \bigl(\beta(T-t) \bigr), \quad \forall (x,t)\in {\overline{D}}\times[0,T), \end{aligned}$$
(2.6)
$$ H(z):= \int^{+\infty}_{z}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s, \quad z>0, $$
(2.7)
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Proof
Consider the auxiliary function For brevity of notation, we write g in place of \(g(u)\), suppressing the symbol u. We find that
and It follows from (2.10) and (2.11) that By (1.1) we have Substituting (2.13) into (2.12), we get In view of (2.9), we have Substitution of (2.15) into (2.14) results in With (2.8), we have Substituting (2.17) into (2.16), we obtain By assumptions (2.1) and (2.2) the right-hand side of (2.18) is nonpositive, that is, Now by (2.4) we have It follows from (1.1) that The assumptions concerning the functions a, b, f, g, and \(u_{0}\) in Section 1 imply that we can use maximum principles to (2.19)-(2.21). Combining (2.19)-(2.21) and applying maximum principles [23], it follows that the minimum of Ψ in \(\overline{D}\times[0,T)\) is zero. Thus, we have that is, the differential inequality Suppose that \(x_{0}\in\overline{D}\) and \(u_{0}(x_{0})=M_{0}\). At the \(x_{0}\), integrate (2.22) over \([0,t]\) to get which implies that u must blow up in finite time. Actually, if u is a global solution of (1.1), then for any \(t>0\), it follows from (2.23) that Letting \(t\rightarrow+\infty\) in (2.24) yields which contradicts with assumption (2.3). This shows that u must blow up in a finite time \(t=T\). Furthermore, letting \(t\rightarrow T\) in (2.23), we have Integrating inequality (2.22) over \([t,s]\) (\(0< t< s< T\)) yields, for each fixed x, that Passing to the limit as \(s\rightarrow T^{-}\) gives Since H is a decreasing function, we have The proof is complete. □
$$ \Psi(x,t):=g'(u)u_{t}-\beta{ \mathrm{e}}^{u}. $$
(2.8)
$$\begin{aligned}& \nabla\Psi=g''u_{t}\nabla u+g'\nabla u_{t}-\beta{\mathrm{e}}^{u}\nabla u, \end{aligned}$$
(2.9)
$$\begin{aligned}& \Delta\Psi=g'''u_{t}| \nabla u|^{2}+2g''\nabla u\cdot\nabla u_{t}+g''u_{t}\Delta u +g'\Delta u_{t}-\beta{\mathrm{e}}^{u}|\nabla u|^{2}-\beta{\mathrm{e}}^{u}\Delta u, \end{aligned}$$
(2.10)
$$\begin{aligned} \Psi_{t} =&g''(u_{t})^{2}+g'(u_{t})_{t}- \beta{\mathrm{e}}^{u}u_{t} \\ =&g''(u_{t})^{2}+g' \biggl(\frac{ab}{g'}\Delta u+\frac{a'b}{g'}|\nabla u|^{2} + \frac{a}{g'}\nabla b\cdot\nabla u+\frac{f}{g'} \biggr)_{t}- \beta{\mathrm{e}}^{u}u_{t} \\ =&g''(u_{t})^{2}+ \biggl(a'b-\frac{abg''}{g'} \biggr)u_{t}\Delta u +ab \Delta u_{t}+ \biggl(a''b-\frac{a'bg''}{g'} \biggr)u_{t}|\nabla u|^{2} \\ &{}+2a'b (\nabla u\cdot\nabla u_{t} ) + \biggl(a'- \frac{ag''}{g'} \biggr)u_{t} (\nabla b\cdot\nabla u )+a (\nabla b \cdot\nabla u_{t} ) \\ &{} + \biggl(f_{u}-\frac{fg''}{g'}-\beta{\mathrm{e}}^{u} \biggr)u_{t}. \end{aligned}$$
(2.11)
$$\begin{aligned} \frac{ab}{g'}\Delta\Psi-\Psi_{t} =& \biggl( \frac{abg'''}{g'}+\frac{a'bg''}{g'}-a''b \biggr)u_{t}|\nabla u|^{2} + \biggl(2\frac{abg''}{g'}-2a'b \biggr) (\nabla u\cdot\nabla u_{t} ) \\ &{}+ \biggl(2\frac{abg''}{g'}-a'b \biggr)u_{t}\Delta u -\beta\frac{ab{\mathrm{e}}^{u}}{g'}|\nabla u|^{2} -\beta\frac{ab{\mathrm{e}}^{u}}{g'}\Delta u-g''(u_{t})^{2} \\ &{}+ \biggl(\frac{ag''}{g'}-a' \biggr)u_{t} (\nabla b \cdot\nabla u ) -a (\nabla b\cdot\nabla u_{t} ) + \biggl( \frac{fg''}{g'}-f_{u}+\beta{\mathrm{e}}^{u} \biggr)u_{t}. \end{aligned}$$
(2.12)
$$ \Delta u=\frac{g'}{ab}u_{t}-\frac{a'}{a}| \nabla u|^{2}-\frac{1}{b} (\nabla b\cdot\nabla u )- \frac{f}{ab}. $$
(2.13)
$$\begin{aligned} \frac{ab}{g'}\Delta\Psi-\Psi_{t} =& \biggl( \frac{abg'''}{g'}-\frac{a'bg''}{g'}-a''b+ \frac {(a')^{2}b}{a} \biggr)u_{t}|\nabla u|^{2} + \biggl(2 \frac{abg''}{g'}-2a'b \biggr) (\nabla u\cdot\nabla u_{t} ) \\ &{}-\frac{(g')^{2}}{a} \biggl(\frac{a}{g'} \biggr)'(u_{t})^{2} -\frac{ag''}{g'}u_{t} (\nabla b\cdot\nabla u ) + \biggl( \frac{a'f}{a}-\frac{fg''}{g'}-f_{u} \biggr)u_{t} \\ &{}+ \biggl(\beta\frac{a'b{\mathrm{e}}^{u}}{g'}-\beta\frac{ab{\mathrm{e}}^{u}}{g'} \biggr)|\nabla u|^{2} +\beta\frac{a{\mathrm{e}}^{u}}{g'} (\nabla b\cdot\nabla u ) \\ &{}+\beta \frac{f{\mathrm{e}}^{u}}{g'}-a (\nabla b\cdot\nabla u_{t} ). \end{aligned}$$
(2.14)
$$ \nabla u_{t}=\frac{1}{g'}\nabla\Psi- \frac{g''}{g'}u_{t}\nabla u+\beta \frac{{\mathrm{e}}^{u}}{g'}\nabla u. $$
(2.15)
$$\begin{aligned}& \frac{ab}{g'}\Delta\Psi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Psi- \Psi_{t} \\& \quad = \biggl(\frac{abg'''}{g'}+\frac{a'bg''}{g'}-a''b+ \frac{(a')^{2}b}{a} -2\frac{ab(g'')^{2}}{(g')^{2}} \biggr)u_{t}|\nabla u|^{2} \\& \qquad {} + \biggl(2\beta\frac{abg''{\mathrm{e}}^{u}}{(g')^{2}}-\beta\frac{a'b{\mathrm{e}}^{u}}{g'} -\beta \frac{ab{\mathrm{e}}^{u}}{g'} \biggr)|\nabla u|^{2} -\frac{(g')^{2}}{a} \biggl( \frac{a}{g'} \biggr)'(u_{t})^{2} \\& \qquad {} + \biggl(\frac{a'f}{a}-\frac{fg''}{g'}-f_{u} \biggr)u_{t} +\beta\frac{f{\mathrm{e}}^{u}}{g'}. \end{aligned}$$
(2.16)
$$ u_{t}=\frac{1}{g'}\Psi+\beta\frac{{\mathrm{e}}^{u}}{g'}. $$
(2.17)
$$\begin{aligned}& \frac{ab}{g'}\Delta\Psi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Psi \\& \qquad {} + \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} + \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Psi- \Psi_{t} \\& \quad = -\beta ab{\mathrm{e}}^{u} \biggl\{ \biggl[\frac{1}{a} \biggl(\frac {a}{g'} \biggr)'+\frac{1}{g'} \biggr]' + \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)'+\frac{1}{g'} \biggr] \biggr\} |\nabla u|^{2} -\frac{(g')^{2}}{a} \biggl(\frac{a}{g'} \biggr)'(u_{t})^{2} \\& \qquad {}-\beta\frac{a{\mathrm{e}}^{u}}{(g')^{2}} \biggl[ \biggl(\frac{fg'}{a} \biggr)_{u}-\frac{fg'}{a} \biggr]. \end{aligned}$$
(2.18)
$$\begin{aligned}& \frac{ab}{g'}\Delta\Psi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Psi \\& \quad {}+ \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} + \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Psi- \Psi _{t}\leq0 \quad \mbox{in } D\times(0,T). \end{aligned}$$
(2.19)
$$\begin{aligned} \min_{\overline{D}}\Psi(x,0) =&\min _{\overline{D}} \bigl\{ g'(u_{0}) (u_{0})_{t}-\beta{\mathrm{e}}^{u_{0}} \bigr\} \\ =&\min_{\overline{D}} \bigl\{ \nabla\cdot \bigl(a(u_{0})b(x) \nabla u_{0} \bigr)+f(x,u_{0})-\beta{\mathrm{e}}^{u_{0}} \bigr\} \\ =&\min_{\overline{D}} \biggl\{ {\mathrm{e}}^{u_{0}} \biggl[ \frac{\nabla\cdot(a(u_{0})b(x)\nabla u_{0})+f(x,u_{0})}{{\mathrm{e}}^{u_{0}}}-\beta \biggr] \biggr\} =0. \end{aligned}$$
(2.20)
$$ \frac{\partial\Psi}{\partial n}=g''u_{t} \frac{\partial u}{\partial n} +g'\frac{\partial u_{t}}{\partial n} -\beta{\mathrm{e}}^{u} \frac{\partial u}{\partial n} =g' \biggl(\frac{\partial u}{\partial n} \biggr)_{t}=0 \quad \mbox{on } \partial D\times(0,T). $$
(2.21)
$$\Psi\geq0 \quad \mbox{in } \overline{D}\times[0,T), $$
$$ \frac{g'(u)}{{\mathrm{e}}^{u}}u_{t}\geq\beta. $$
(2.22)
$$ \int^{t}_{0}\frac{g'(u)}{{\mathrm{e}}^{u}}u_{t}\,{ \mathrm{d}}t = \int^{u(x_{0},t)}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s\geq \beta t, $$
(2.23)
$$ \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s \geq \int^{u(x_{0},t)}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s \geq \beta t. $$
(2.24)
$$\int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s=+ \infty, $$
$$T\leq\frac{1}{\beta} \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s. $$
$$\begin{aligned} H \bigl(u(x,t) \bigr) \geq& H \bigl(u(x,t) \bigr) -H \bigl(u(x,s) \bigr) = \int^{+\infty}_{u(x,t)}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s- \int ^{+\infty}_{u(x,s)}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s \\ =& \int^{u(x,s)}_{u(x,t)}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s= \int^{s}_{t}\frac {g'(u)}{{\mathrm{e}}^{u}}u_{t}\,{ \mathrm{d}}t\geq\beta(s-t). \end{aligned}$$
$$H\bigl(u(x,t)\bigr)\geq\beta(T-t). $$
$$u(x,t)\leq H^{-1} \bigl(\beta(T-t) \bigr). $$
3 Global solution
The following theorem is the main result for the global solution.
Theorem 3.1
Let
u
be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are satisfied:
Then the solution
u
of (1.1) must be a global solution, and
where
and
\(G^{-1}\)
is the inverse function of
G.
(i)
for any
\(s\in\mathbb{R}^{+}\),
$$ \biggl(\frac{a(s)}{g'(s)} \biggr)'\leq0,\quad \biggl[ \frac{1}{a(s)} \biggl(\frac{a(s)}{g'(s)} \biggr)' - \frac{1}{g'(s)} \biggr]'- \biggl[\frac{1}{a(s)} \biggl( \frac {a(s)}{g'(s)} \biggr)' -\frac{1}{g'(s)} \biggr]\leq0; $$
(3.1)
(ii)
for any
\((x,s)\in D\times\mathbb{R}^{+}\),
$$ \biggl(\frac{f(x,s)g'(s)}{a(s)} \biggr)_{s} +\frac{f(x,s)g'(s)}{a(s)} \leq0; $$
(3.2)
(iii)
$$ \int^{+\infty}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s=+ \infty,\qquad m_{0}:=\min_{\overline{D}}u_{0}(x); $$
(3.3)
(iv)
$$ \alpha:=\max_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{-u_{0}}}>0. $$
(3.4)
$$ u(x,t)\leq G^{-1} \bigl(\alpha t+G\bigl(u_{0}(x,t) \bigr) \bigr), \quad \forall (x,t)\in\overline{D}\times\overline{\mathbb{R}}^{+}, $$
(3.5)
$$ G(z):= \int^{z}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s, \quad z\geq m_{0}, $$
(3.6)
Proof
Construct the auxiliary function By using the same reasoning process with that of (2.9)-(2.18), we have From assumptions (3.1) and (3.2) we see that the right-hand side of (3.8) is nonnegative, that is, By (3.4) we have Repeating the arguments for (2.21), we have Combining (3.9)-(3.11) and applying the maximum principles again, we get that the maximum of Φ in \(\overline{D}\times[0,T)\) is zero. Hence, we have that is, the differential inequality For each fixed \(x\in\overline{D}\), integrate (3.12) over \([0,t]\) to produce which shows that u must be a global solution. In fact, suppose that u blows up at finite time T, that is, Passing to the limit as \(t\rightarrow T^{-}\) in (3.13) gives and which is a contradiction. This shows that u is global. Moreover, (3.13) implies that Since G is an increasing function, we have The proof is complete. □
$$ \Phi(x,t):=g'(u)u_{t}-\alpha{ \mathrm{e}}^{-u}. $$
(3.7)
$$\begin{aligned}& \frac{ab}{g'}\Delta\Phi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Phi \\& \qquad {} + \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} - \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Phi- \Phi_{t} \\& \quad = -\alpha ab{\mathrm{e}}^{-u} \biggl\{ \biggl[\frac{1}{a} \biggl(\frac {a}{g'} \biggr)'-\frac{1}{g'} \biggr]' - \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)'-\frac{1}{g'} \biggr] \biggr\} |\nabla u|^{2} -\frac{(g')^{2}}{a} \biggl(\frac{a}{g'} \biggr)'(u_{t})^{2} \\& \qquad {} -\alpha\frac{a{\mathrm{e}}^{-u}}{(g')^{2}} \biggl[ \biggl(\frac{fg'}{a} \biggr)_{u}+\frac{fg'}{a} \biggr]. \end{aligned}$$
(3.8)
$$\begin{aligned}& \frac{ab}{g'}\Delta\Phi+ \biggl[2b \biggl(\frac{a}{g'} \biggr)'\nabla u +\frac{a}{g'}\nabla b \biggr]\cdot\nabla\Phi \\& \quad {}+ \biggl\{ ab \biggl[\frac{1}{a} \biggl(\frac{a}{g'} \biggr)' \biggr]'|\nabla u|^{2} - \frac{a}{(g')^{2}} \biggl(\frac{fg'}{a} \biggr)_{u} \biggr\} \Phi- \Phi_{t} \geq0 \quad \mbox{in } D\times(0,T). \end{aligned}$$
(3.9)
$$\begin{aligned} \max_{\overline{D}}\Phi(x,0) =& \max _{\overline{D}} \bigl\{ g'(u_{0}) (u_{0})_{t}-\alpha{\mathrm{e}}^{-u_{0}} \bigr\} \\ =&\max_{\overline{D}} \bigl\{ \nabla\cdot \bigl(a(u_{0})b(x) \nabla u_{0} \bigr)+f(x,u_{0})-\alpha{\mathrm{e}}^{-u_{0}} \bigr\} \\ =&\max_{\overline{D}} \biggl\{ {\mathrm{e}}^{-u_{0}} \biggl[ \frac{\nabla\cdot(a(u_{0})b(x)\nabla u_{0})+f(x,u_{0})}{{\mathrm{e}}^{-u_{0}}}-\alpha \biggr] \biggr\} =0. \end{aligned}$$
(3.10)
$$ \frac{\partial\Phi}{\partial n}=0 \quad \mbox{on } \partial D\times(0,T). $$
(3.11)
$$\Phi\leq0 \quad \mbox{in } \overline{D}\times[0,T), $$
$$ \frac{g'(u)}{{\mathrm{e}}^{-u}}u_{t}\leq\alpha. $$
(3.12)
$$ \int^{t}_{0}\frac{g'(u)}{{\mathrm{e}}^{-u}}u_{t}\,{ \mathrm{d}}t = \int^{u(x,t)}_{u_{0}(x)} \frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s \leq \alpha t, $$
(3.13)
$$\lim_{t\rightarrow T^{-}}u(x,t)=+\infty. $$
$$\int^{+\infty}_{u_{0}(x)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s\leq \alpha T $$
$$\int^{+\infty}_{m_{0}} \frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s = \int^{u_{0}(x)}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s + \int^{+\infty}_{u_{0}(x)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s \leq \int^{u_{0}(x)}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s+ \alpha T< +\infty, $$
$$\int^{u(x,t)}_{u_{0}(x)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s = \int_{m_{0}}^{u(x,t)}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s - \int^{u_{0}(x)}_{m_{0}}\frac{g'(s)}{{\mathrm{e}}^{-s}}\,{\mathrm{d}}s =G \bigl(u(x,t) \bigr)-G\bigl(u_{0}(x)\bigr) \leq\alpha t. $$
$$u(x,t)\leq G^{-1} \bigl(\alpha t+G\bigl(u_{0}(x)\bigr) \bigr). $$
4 Applications
When \(g(u)\equiv u\), \(b(x)\equiv1\), and \(f(x,u)\equiv f(u)\), problem (1.1) is problem (1.2) studied by Lair and Oxley [20]. When \(a(u)\equiv1\), \(b(x)\equiv1\), and \(f(x,u)\equiv f(u)\), problem (1.1) is problem (1.3) discussed by Zhang [21]. When \(b(x)\equiv1\) and \(f(x,u)\equiv f(u)\), problem (1.1) is problem (1.4) considered by Ding and Guo [22]. In these three cases, the conclusions of Theorems 2.1 and 3.1 still hold. In this sense, our results extend and supplement the results of [20‐22].
Example 4.1
Let u be a solution of the following problem: where \(D= \{x=(x_{1},x_{2},x_{3}) \mid \|x\| ^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}<1 \}\) is the unit ball of \(\mathbb{R}^{3}\). Now we have In order to determine the constant β, we assume that Then \(0\leq s\leq1\) and It is easy to check that (2.1)-(2.3) hold. By Theorem 2.1, u must blow up in a finite time T, and
$$\left \{ \textstyle\begin{array}{l@{\quad}l} (2{\mathrm{e}}^{\frac{u}{2}}+u )_{t} =\nabla\cdot ( (1+{\mathrm{e}}^{\frac{u}{2}} ) (1+\| x\|^{2} )\nabla u ) +7{\mathrm{e}}^{u}-\|x\|^{2} &\mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 &\mbox{on } \partial D\times(0,T), \\ u(x,0)=1+ (1-\|x\|^{2} )^{2} & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
$$\begin{aligned}& g(u)=2{\mathrm{e}}^{\frac{u}{2}}+u, \qquad a(u)=1+{\mathrm{e}}^{\frac{u}{2}}, \qquad b(x)=1+\|x\|^{2}, \\& f(x,u)=7{\mathrm{e}}^{u}-\|x\|^{2}, \qquad u_{0}(x)=1+ \bigl(1-\|x\|^{2} \bigr)^{2}. \end{aligned}$$
$$s=\|x\|^{2}. $$
$$\begin{aligned} \beta =&\min_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{u_{0}}} \\ =&\min_{\overline{D}} \bigl\{ \bigl({\mathrm{e}}^{-1-(1-\|x\|^{2})^{2}}+{ \mathrm{e}}^{-\frac{1}{2}-\frac{1}{2}(1-\|x\|^{2})^{2}} \bigr) \bigl(-12+28\|x\|^{2} \bigr) \\ &{}+8{\mathrm{e}}^{-\frac{1}{2}-\frac{1}{2} (1-\|x\| ^{2} )^{2}}\|x\|^{2} \bigl(1+\|x\|^{2} \bigr) \bigl(1-\|x\|^{2} \bigr)^{2} +7-\|x\|^{2}{ \mathrm{e}}^{-1- (1-\|x\|^{2} )^{2}}\bigr\} \\ =&\min_{0\leq s\leq1} \bigl\{ \bigl({\mathrm{e}}^{-1-(1-s)^{2}}+{ \mathrm{e}}^{-\frac{1}{2}-\frac{1}{2}(1-s)^{2}} \bigr) (-12+28s) \\ &{}+8{\mathrm{e}}^{-\frac{1}{2}-\frac{1}{2}(1-s)^{2}}s(1+s) (1-s)^{2}+7-s{ \mathrm{e}}^{-1-(1-s)^{2}}\bigr\} \\ =&0.9614. \end{aligned}$$
$$\begin{aligned}& T\leq\frac{1}{\beta} \int^{+\infty}_{M_{0}}\frac{g'(s)}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s = \frac{1}{0.9614} \int^{+\infty}_{1}\frac{{\mathrm{e}}^{\frac {s}{2}}+1}{{\mathrm{e}}^{s}}\,{\mathrm{d}}s =1.4025, \\& u(x,t)\leq H^{-1} \bigl(\beta(T-t) \bigr) =\ln\frac{1}{ (\sqrt{1+0.9614(T-t)}-1 )^{2}}. \end{aligned}$$
Example 4.2
Let u be a solution of the following problem: where \(D= \{x=(x_{1},x_{2},x_{3}) \mid \|x\| ^{2}=x_{2}^{2}+x_{2}^{2}+x_{3}^{2}<1 \}\) is the unit ball of \(\mathbb{R}^{3}\). Now we have By setting we have \(0\leq s\leq1\) and Again, it is easy to check that (3.1)-(3.3) hold. By Theorem 3.1, u must be a global solution, and
$$\left \{ \textstyle\begin{array}{l@{\quad}l} (\ln ({\mathrm{e}}^{u}-1 )-u )_{t} =\nabla\cdot (\frac{1}{{\mathrm{e}}^{u}-1} (1+\|x\|^{2} )\nabla u ) +{\mathrm{e}}^{-u} (1+\|x\|^{2} ) &\mbox{in }D\times(0,T), \\ \frac{\partial u}{\partial n}=0 &\mbox{on } \partial D\times(0,T), \\ u(x,0)=1+ (1-\|x\|^{2} )^{2} & \mbox{in } \overline{D}, \end{array}\displaystyle \right . $$
$$\begin{aligned}& g(u)=\ln \bigl({\mathrm{e}}^{u}-1 \bigr)-u, \qquad a(u)= \frac{1}{{\mathrm{e}}^{u}-1}, \qquad b(x)=1+\|x\|^{2}, \\& f(x,u)={\mathrm{e}}^{-u} \bigl(1+\|x\|^{2} \bigr), \qquad u_{0}(x)=1+ \bigl(1-\| x\|^{2} \bigr)^{2}. \end{aligned}$$
$$s=\|x\|^{2}, $$
$$\begin{aligned} \alpha =&\min_{\overline{D}} \frac{\nabla\cdot (a(u_{0})b(x)\nabla u_{0} )+f(x,u_{0})}{{\mathrm{e}}^{-u_{0}}} \\ =&\min_{\overline{D}} \biggl\{ \frac{1}{ ({\mathrm{e}}^{1+(1-\|x\|^{2})^{2}}-1 )^{2}} \bigl[ \bigl(-12+28 \|x\|^{2} \bigr){\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}} \bigl({\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}}-1 \bigr) \\ &{}-16\|x\|^{2} \bigl(1+\|x\|^{2} \bigr) \bigl(1-\|x \|^{2} \bigr)^{2}{\mathrm{e}}^{2+2 (1-\|x\|^{2} )^{2}} + \bigl(1+\|x \|^{2} \bigr) \bigl({\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}}-1 \bigr)^{2} \bigr]\biggr\} \\ =&\min_{0\leq s\leq1} \biggl\{ \frac{1}{ ({\mathrm{e}}^{1+(1-s)^{2}}-1 )^{2}} \bigl[ (-12+28s ){ \mathrm{e}}^{1+ (1-s )^{2}} \bigl({\mathrm{e}}^{1+ (1-s )^{2}}-1 \bigr) \\ &{}-16s (1+s ) (1-s )^{2}{\mathrm{e}}^{2+2 (1-s )^{2}} + (1+s ) \bigl({ \mathrm{e}}^{1+ (1-s )^{2}}-1 \bigr)^{2}\bigr]\biggr\} \\ =&27.3116. \end{aligned}$$
$$u(x,t)\leq G^{-1} \bigl(\alpha t+G\bigl(u_{0}(x)\bigr) \bigr) =\ln \bigl[1+{\mathrm{e}}^{27.3116t} \bigl({\mathrm{e}}^{1+ (1-\|x\|^{2} )^{2}}-1 \bigr) \bigr]. $$
Acknowledgements
This work was supported by the National Natural Science Foundation of China (Nos. 61473180 and 61174082).
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Competing interests
The author declares that he has no competing interests.