In the next two sections, we prove Theorem 1.2 with five lemmas. This section deals with the conditions of u blowing up while v remains bounded.
First of all, we consider the single equation problem
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} u_{t}=u_{xx}-\lambda u^{m}, & (x,t)\in(0,1)\times(0,T), \\ u_{x}(1,t)=u^{\alpha}(1,t)e^{ph(t)},\qquad u_{x}(0,t)=0, & t\in(0,T), \\ u(x,0)=u_{0}(x), & x\in[0,1], \end{array}\displaystyle \right . $$
(3.1)
where the parameters m, p, α, λ agree with system (1.1), and \(h(t)\) is a continuous, bounded, and non-decreasing function, with \(h(t)\geq\delta>0\).
Now we prove a lemma, using similar steps to Theorem 1.2 in [16]. C is a positive constant independent of t, and it may change during the proof and at different places.
Lemma 3.1
Assume
u
is a solution of (3.1), and one of the following conditions holds:
$$\begin{aligned} \begin{aligned} (\mathrm{i}) &\quad \alpha>\mu, \\ (\mathrm{ii}) &\quad \alpha=\mu>1\quad \textit{with} \quad \delta>\frac{1}{2p}\log \biggl(\frac{\lambda}{\alpha}\biggr). \end{aligned} \end{aligned}$$
Then
u
may blow up for sufficient large initial value, and furthermore
$$ u(1,t)=\max_{[0,1]}u(\cdot,t)\leq C(T-t)^{\frac{-1}{2(\alpha-1)}}, \quad 0< t< T, $$
(3.2)
where
T
is the blow-up time.
Proof
Assume w is a solution of the following problem:
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {w_{t}=w_{xx}-\lambda w^{m}}, & (x,t)\in(0,1)\times(0,T), \\ {w_{x}(1,t)=w^{\alpha}(1,t)e^{p\delta}}, \qquad w_{x}(0,t)=0, & t\in(0,T), \\ {w(x,0)=u_{0}(x)}, & x\in[0,1]. \end{array}\displaystyle \right . $$
(3.3)
In the light of the comparison principle, \(w\leq u\), in \((0,1)\times [0,T)\). Obviously the two parameter conditions in the lemma are, respectively, the same as
$$\begin{aligned} (\mathrm{i})&\quad \left \{ \textstyle\begin{array}{l} m\leq1, \\ \alpha>1 \end{array}\displaystyle \right . \quad \text{or}\quad \left \{ \textstyle\begin{array}{l} m>1, \\ \alpha>\frac{m+1}{2} \end{array}\displaystyle \right . \end{aligned}$$
and
$$\begin{aligned} (\mathrm{ii})&\quad \left \{ \textstyle\begin{array}{l} m>1, \\ \alpha=\frac{m+1}{2}, \\ \delta>(\frac{\lambda}{\alpha})^{\frac{1}{2p}}. \end{array}\displaystyle \right . \end{aligned}$$
From Theorem 4.2 in [22], there exist initial data \(u_{0}\) such that w blows up. Thus, u blows up at a finite time T.
Similarly to [20, 23], we utilize the scaling method to obtain the blow-up rate estimate (3.2). Let \(M(t)=\max_{[0,1]}u(\cdot ,t)=u(1,t)\), for \(t\in(0,T)\), and assume
$$\varphi_{k}(y,s)=\frac{1}{M(t)}u\bigl(ky+1,k^{2}s+t \bigr), \quad [-1/k,0]\times\bigl(-t/k^{2},0 \bigr], $$
it is easy to check
$$0\leq\varphi_{k}\leq1, \qquad (\varphi_{k})_{s} \geq0, \qquad \varphi_{k}(0,0)=1. $$
Let \(k=M^{-\alpha+1}(t)\), \(\varphi_{k}\) solves the following problem:
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {(\varphi_{k})_{s}=(\varphi_{k})_{yy}-\lambda M^{m-2\alpha +1}(t)(\varphi_{k})^{m}}, & (y,s)\in(-1/k,0)\times(-t/k^{2},0], \\ {(\varphi_{k})_{y}(0,s)=(\varphi_{k})^{\alpha}(0,s)\cdot e^{ph(k^{2} s+t)}}, & s\in(-t/k^{2},0], \\ {(\varphi_{k})_{y}(-1/k,s)=0}, & s\in(-t/k^{2},0]. \end{array}\displaystyle \right . $$
(3.4)
Now we will prove that there exist constants \(c>0\) and \(C>0\), such that
$$ c\leq(\varphi_{k})_{s}(0,0)\leq C, $$
(3.5)
for every k small.
First of all, we prove case (i), when \(\alpha>\mu\), it easy to check \(M^{m-2\alpha+1}\rightarrow0 \) as \(t\rightarrow T\). For given \(\{\varphi _{k_{j}}\}\), there exist a continuous function φ and, maybe, a subsequence of it, which is also denoted by \(\{\varphi_{k_{j}}\}\), such that \(\varphi_{k_{j}}\rightarrow\varphi\), \(k_{j}\rightarrow0\), as \(M\rightarrow\infty\).
As \(h(t)\) is bounded, we conclude that \(\varphi_{k_{j}}\) is uniformly bounded. Using the Schauder estimates
$$\|\varphi_{k_{j}}\|_{C^{2+\alpha,1+\alpha/2}}\leq C. $$
The upper estimate in (3.5) follows immediately.
To get the lower bound of (3.5), and prove it by contradiction, we assume that the lower bound does not hold and that there is a sequence \(\varphi_{k_{j}}\) such that \((\varphi_{k_{j}})_{s}(0,0)\rightarrow0\). As \(\varphi_{k_{j}}\rightarrow\varphi\), \(0\leq\varphi\leq1\), \(\varphi (0,0)=1\), \(\varphi_{s}\geq0\), \(\varphi_{s}(0,0)=0\), and φ satisfies
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {\varphi_{s}=\varphi_{yy}}, & (y,s)\in(-\infty,0)\times (-\infty,0], \\ {\varphi_{y}(0,s)=\varphi^{\alpha}(0,s)\cdot e^{ph(T)}}, & s\in (-\infty,0]. \end{array}\displaystyle \right . $$
(3.6)
Set \(\psi=\varphi_{s}\), thus ψ satisfies
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {\psi_{s}=\psi_{yy}}, & (y,s)\in(-\infty,0)\times(-\infty ,0], \\ {\psi_{y}(0,s)=\alpha\cdot\varphi^{\alpha-1}(0,s)\cdot e^{ph(T)}\cdot\varphi_{s}(0,s)\geq0}, & s\in(-\infty,0]. \end{array}\displaystyle \right . $$
(3.7)
Meanwhile \((\varphi_{k_{j}})_{s}(0,0)\rightarrow\varphi_{s}(0,0)\), thus \(\psi (0,0)=\varphi_{s}(0,0)=0\), as \(\psi=\varphi_{s}\geq0\), hence ψ has a minimum at \((0,0)\). By Hopf’s lemma \(\psi\equiv0\), that is to say, ψ is independent of s. Thus φ satisfies the ODE problem
$$ \left \{ \textstyle\begin{array}{l} {\varphi_{yy}=0}, \\ {\varphi_{y}(0)=e^{ph(T)}\geq e^{\delta p}}, \\ {\varphi(0)=1}, \end{array}\displaystyle \right . $$
(3.8)
then \(\varphi(y)\geq e^{\delta p}\cdot y+1\), which make a contradiction of the fact that \(0\leq\varphi\leq1\), hence (3.5) holds, and (3.2) follows immediately.
Now, we prove case (ii). The proof of upper bound in (3.5) is similar, we omit it. The main difference of case (ii) is the critical case \(\alpha=\frac {m+1}{2}>1\). Thus (3.6) becomes
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {\varphi_{s}=\varphi_{yy}-\lambda\varphi^{m}}, & (y,s)\in (-\infty,0)\times(-\infty,0], \\ {\varphi_{y}(0,s)=\varphi^{\alpha}(0,s)\cdot e^{ph(T)}}, & s\in (-\infty,0]. \end{array}\displaystyle \right . $$
(3.9)
To obtain a contradiction, we set \(\psi=\varphi_{s}\), thus
$$\left \{ \textstyle\begin{array}{l@{\quad}l} {\psi_{s}=\psi_{yy}-\lambda\cdot m \cdot\varphi^{m-1}\cdot \psi}, & (y,s)\in(-\infty,0)\times(-\infty,0], \\ {\psi_{y}(0,s)=\alpha\cdot\varphi^{\alpha-1}(0,s)\cdot e^{ph(T)}\cdot\varphi_{s}(0,s)\geq0}, & s\in(-\infty,0]. \end{array}\displaystyle \right . $$
By Hopf’s lemma, \(\psi\equiv0\), φ satisfies
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {\varphi_{yy}-\lambda\varphi^{m}=0}, \\ {\varphi_{y}(0)=e^{ph(T)}\geq e^{\delta p}}, \\ {\varphi(0)=1}. \end{array}\displaystyle \right . $$
(3.10)
The first equation in (3.10) we multiply by \(\varphi_{y}\), and we integrate from 0 to y. Since \(\alpha=\frac{m+1}{2}\), we derive
$$\varphi_{y}^{2}=\varphi_{y}^{2}(0)+ \frac{\lambda}{\alpha}\varphi^{2\alpha}-\frac {\lambda}{\alpha}, $$
so
$$\varphi_{y}=\biggl[\varphi_{y}^{2}(0)+ \frac{\lambda}{\alpha}\varphi^{2\alpha}-\frac {\lambda}{\alpha}\biggr]^{\frac{1}{2}}, $$
where \(\varphi_{y}\geq0\). Furthermore
$$y= \int_{0}^{y}\varphi_{y}\biggl[ \varphi_{y}^{2}(0)+\frac{\lambda}{\alpha}\varphi ^{2\alpha}- \frac{\lambda}{\alpha}\biggr]^{-\frac{1}{2}}\, dy= \int_{\varphi (0)}^{\varphi(y)}\biggl[e^{2ph(T)}+ \frac{\lambda}{\alpha}\varphi^{2\alpha }-\frac{\lambda}{\alpha}\biggr]^{-\frac{1}{2}}\, d\varphi. $$
Noticing \(\delta>\frac{1}{2p}\log(\frac{\lambda}{\alpha})\), we deduce
$$y\geq-\biggl(e^{2\delta p}-\frac{\lambda}{\alpha}\biggr)^{-\frac{1}{2}}, $$
which contradicts \(y\in(-\infty,0)\). We conclude (3.5) is true, and (3.2) follows. □
Lemma 3.2
Let
u
be a solution of following problem:
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {u_{t}=u_{xx}-\lambda u^{m}}, & (x,t)\in(0,1)\times(0,T), \\ {u_{x}(1,t)\leq Lu^{\alpha}(1,t)}, \qquad u_{x}(0,t)=0, & t\in(0,T), \\ {u(x,0)=u_{0}(x)}, & x\in[0,1], \end{array}\displaystyle \right . $$
(3.11)
with
\(\lambda>0\), α, \(m\geq0\). If
u
blows up at a finite time
T, then either
\(\alpha>\mu\)
or
\(\alpha=\mu>1\), and there exists a positive constant
C, such that
$$u(1,t)=\max_{[0,1]}u(\cdot,t)\geq C(T-t)^{\frac{-1}{2(\alpha-1)}}, \quad \textit{as } t\rightarrow T. $$
Proof
The proof of Lemma 3.2 is similar to Lemma 3.1 in [23], and we omit it here. □
Lemma 3.3
There exist initial data such that
u
blows up while
v
remains bounded if and only if
$$\left \{ \textstyle\begin{array}{l} {\alpha>q+1}, \\ {\alpha>\mu \quad \textit{or} \quad \alpha=\mu>1}. \end{array}\displaystyle \right . $$
Proof
We first prove the sufficient condition.
To find suitable initial data \((u_{0}, v_{0})\), such that u blows up while v remains bounded, fix \(v_{0}\) so that \(v_{0}(x)\geq\delta>\frac {1}{2p}\log(\frac{\lambda}{\alpha})\), and denote \(K=\max_{[0,1]}v_{0}=v_{0}(1)\), \(N=\frac{e^{2\beta K}}{K}+3\). Assume \(w(x,t)\) is a solution of (3.3), then \(w(x,t)\) is a sub-solution of u. There must exist sufficiently large initial data \(u_{0}\) such that w blows up at a finite time \(T\leq\varepsilon\), under the assumption \(\delta>\frac {1}{2p}\log(\frac{\lambda}{\alpha})\).
We claim that there exists \(u_{0}\) large enough such that \(v<2K\). If it is not true, there must exist \(t_{0}< T\), which is the first time \(\max_{[0,1]}v(\cdot,t_{0})=v(1,t_{0})=2K\). We denote the cutoff function
$$ \tilde{v}(x,t)= \left \{ \textstyle\begin{array}{l@{\quad}l} {v(x,t)}, & (x,t)\in[0,1]\times[0,t_{0}], \\ {2K}, & (x,t)\in[0,1]\times[t_{0} ,T]. \end{array}\displaystyle \right . $$
(3.12)
Meanwhile, assume \(\tilde{u}(x,t)\) solves
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {\tilde{u}_{t}=\tilde{u}_{xx}-\lambda\tilde {u}^{m}}, & (x,t)\in(0,1)\times(0,\widetilde{T}), \\ {\tilde{u}_{x}(1,t)=\tilde{u}^{\alpha }(1,t)e^{p\tilde{v}(1,t)}}, \qquad \tilde{u}_{x}(0,t)=0, & t\in (0,\widetilde{T}), \\ {\tilde{u}(x,0)=u_{0}(x) }, & x\in[0,1], \end{array}\displaystyle \right . $$
(3.13)
then the blow-up time of ũ, which is denoted by T̃, is later than T. We conclude by Lemma 3.1
$$\tilde{u}(1,t)=\max_{[0,1]}\tilde{u}(\cdot,t)\leq C(\widetilde {T}-t)^{\frac{-1}{2(\alpha-1)}}, \quad 0< t< \widetilde{T}. $$
Thus,
$$ u(1,t)=\tilde{u}(1,t)\leq C(\widetilde{T}-t)^{\frac{-1}{2(\alpha -1)}}\leq C(T-t)^{\frac{-1}{2(\alpha-1)}}, \quad 0< t< t_{0}. $$
(3.14)
Let \(\Gamma(x,t)\) be the fundamental solution of the heat equation in \([0,1]\), so
$$\Gamma(x,t)=\frac{1}{2\sqrt{\pi t}}\exp\biggl\{ \frac{-x^{2}}{4t}\biggr\} . $$
From the classical theory of the heat equation, we know that Γ satisfies (see [24])
$$\begin{aligned}& \int^{1}_{0} \Gamma(x-y,t-z)\, dy\leq1, \\& \int^{t}_{z} \Gamma(0,t-\tau)\, d\tau= \frac{1}{\sqrt{\pi}}\sqrt{t-z},\qquad \int ^{t}_{z} \Gamma(1,t-\tau)\frac{1}{2(t-\tau)}\, d \tau\leq C^{*}\sqrt{t-z}, \\& \frac{\partial\Gamma}{\partial\eta_{y}}(x-y,t-\tau)=\frac{x-y}{2(t-\tau )}\Gamma(x-y,t-\tau), \quad x,y \in[0,1], 0\leq z< t. \end{aligned}$$
We deduce from the Green’s identity of (1.1) for v that
$$\begin{aligned} v(x,t) =& { \int^{1}_{0}\Gamma(x-y,t-z)v(y,z)\, dy+ \int^{t}_{z}\frac{\partial v}{\partial x}(1,\tau)\Gamma(x-1,t- \tau)\, d\tau} \\ &{}- { \int^{t}_{z}\frac{\partial\Gamma}{\partial\eta_{y}}(x-1,t-\tau )v(1,\tau) \, d\tau+ \int^{t}_{z}\frac{\partial\Gamma}{\partial\eta_{y}}(x,t-\tau )v(0,\tau)\, d\tau}, \end{aligned}$$
(3.15)
with \(0\leq z< t< T\), \(0< x<1\). Set \(z=0\), and \(x\rightarrow1\), one derives that
$$\begin{aligned} \begin{aligned} v(1,t)={}& { \int^{1}_{0}\Gamma(1-y,t)v(y,0)\, dy+ \int^{t}_{0} u^{q}(1,\tau )e^{\beta v(1,\tau)}\Gamma(0,t-\tau)\,d\tau} \\ &{}+ { \int^{t}_{0} v(0,\tau)\Gamma(1,t-\tau) \frac{1}{2(t-\tau)}\,d\tau}. \end{aligned} \end{aligned}$$
Combining with (3.14), we have
$$ v(1,t_{0})\leq v_{0}(1)+C_{0} e^{\beta v(1,t_{0})} \int_{0}^{t_{0}}(t_{0}-\tau)^{-\frac {q}{2(\alpha-1)}-\frac{1}{2}}\, d\tau+C^{*} \sqrt{t_{0}}\cdot v(1,t_{0}). $$
(3.16)
As \(q<\alpha-1\),
$$\int_{0}^{t_{0}}(t_{0}-\tau)^{-\frac{q}{2(\alpha-1)}-\frac{1}{2}}\, d\tau< \frac{1}{NC_{0}}, $$
and we can choose enough large \(u_{0}\) such that T sufficiently small, it implies \(\sqrt{t_{0}}\leq\sqrt{T}\leq\frac{1}{NC^{*}}\). Therefore
$$\frac{N-1}{N}v(1,t_{0})\leq v_{0}(1)+ \frac{1}{N}e^{\beta v(1,t_{0})}. $$
Thus
$$\frac{2N-1}{N}K\leq K+\frac{1}{N}e^{2K\beta}, $$
so
$$N\leq\frac{e^{2K\beta}}{K}+2, $$
which is a contradiction.
Next, we prove the necessary condition.
If u blows up, v remains bounded. Then \(v\leq K\) for \((x,t)\in [0,1]\times[0,T)\), so
$$\left \{ \textstyle\begin{array}{l@{\quad}l} {u_{t}=u_{xx}-\lambda u^{m}}, & (x,t)\in(0,1)\times(0,T), \\ {u_{x}(1,t)\leq u^{\alpha}(1,t)e^{pK}}, \qquad u_{x}(0,t)=0, & t\in(0,T), \\ {u(x,0)=u_{0}(x)}, & x\in[0,1]. \end{array}\displaystyle \right . $$
Thus we can conclude from Lemma 3.2 that \(\alpha>\mu\) or \(\alpha=\mu >1\). Furthermore,
$$u(1,t)=\max_{[0,1]}u(\cdot,t)\geq C(T-t)^{\frac{-1}{2(\alpha-1)}}, \quad \text{as } t\rightarrow T. $$
Now, we will show the requirement of \(q<\alpha-1\). Let \(x\rightarrow1\) in (3.15), we deduce
$$v(1,t)\geq \int_{z}^{t} u^{q}(1,\tau)\cdot e^{\beta v(1,\tau)}\Gamma(0,t-\tau )\,d\tau\geq C_{1} \int_{z}^{t}(T-\tau)^{-\frac{q}{2(\alpha-1)}-\frac{1}{2}}\,d\tau. $$
Then we get \(q<\alpha-1\), in order to ensure the boundedness of \(v(1,t)\) as \(t\rightarrow T\). □