From the sharp Gagliardo-Nirenberg inequality (
2.4) and the Young inequality, we deduce that
$$\begin{aligned} E\bigl((u,u_{t})\bigr)&=\frac{1}{2} \int|u_{t}|^{2}\,dx+\frac{1}{2} \int \bigl(|\nabla u|^{2}+|u|^{2} \bigr)\,dx- \frac{1}{4} \int|u|^{4}\,dx \\ &\geq\frac{1}{2}\|\nabla u\|_{2}^{2}+\frac{1}{2} \|u\|_{2}^{2}-\frac{1}{\sqrt{3}\| \nabla Q\|_{2}^{2}}\|u\|_{2}\| \nabla u \|_{2}^{3} \\ &\geq\frac{1}{2}\|\nabla u\|_{2}^{2}+\frac{1}{2} \|u\|_{2}^{2}-\frac{1}{2}\|u\|_{2}^{2}- \frac{1}{6\|\nabla Q\| _{2}^{4}}\|\nabla u\|_{2}^{6} \\ &\geq\frac{1}{2}\|\nabla u\|_{2}^{2}- \frac{1}{6\|\nabla Q\|_{2}^{4}}\|\nabla u\|_{2}^{6}. \end{aligned}$$
(3.17)
Define a function
\(f(y)\) on
\([0,+\infty)\) by
\(f(y)=\frac{1}{2}y^{2}-\frac{1}{6\| \nabla Q\|_{2}^{4}}y^{6}\). We see that
$$ f^{\prime}(y)=y-\frac{1}{ \|\nabla Q\| _{2}^{4}}y^{5}=y \biggl(1- \frac{1}{ \|\nabla Q\|_{2}^{4}}y^{4} \biggr). $$
(3.18)
It is obvious that there are two roots for the equation
\(f^{\prime}(y)=0\):
\(y_{1}=0\),
\(y_{2}= \|\nabla Q\|_{2}\). Hence,
\(y_{1}\) and
\(y_{2}\) are two minimizers of
\(f(y)\), and
\(f(y)\) is increasing on the interval
\([y_{1},y_{2})\) and decreasing on the interval
\([y_{2},+\infty)\). Note that
\(f(y_{1})=0\) and
\(f(y_{2})=\frac{1}{3} \|\nabla Q\|_{2}^{2}\). By the conservation of energy and (
1.3), we get
$$ f \bigl(\|\nabla u\|_{2} \bigr)\leq E(u)=E \bigl((u_{0},u_{1}) \bigr) < \frac{1}{3} \|\nabla Q\|_{2}^{2}=f(y_{2}). $$
(3.19)
Therefore, using the convexity and monotony of
\(f(y)\) and the conservation of energy, one obtains two invariant evolution flows generated by the Cauchy problem (
1.1)-(
1.2), as follows. Let
u be the solution of equation (
1.1).
$$\begin{aligned}& K_{1}:= \biggl\{ u\in H^{1}\setminus\{0\} \Bigm| \|\nabla u \|_{2}< \|\nabla Q\|_{2}, 0< E \bigl((u,u_{t}) \bigr)< \frac{ \|\nabla Q\|_{2}^{2}}{4} \biggr\} , \\& K_{2}:= \biggl\{ u\in H^{1}\setminus\{0\} \Bigm| \|\nabla u \|_{2}> \|\nabla Q\|_{2}, 0< E \bigl((u,u_{t}) \bigr)< \frac{ \|\nabla Q\|_{2}^{2}}{4} \biggr\} . \end{aligned}$$
Indeed, if the initial data is such that
\(E((u_{0},u_{1}))<\frac{ \|\nabla Q\|_{2}^{2}}{4}\), then from (
2.1), the corresponding solution satisfies
\(E((u,u_{t}))<\frac{ \|\nabla Q\|_{2}^{2}}{4}\). If
\(u_{0}\in K_{1}\),
i.e.
\(\|\nabla u_{0}\|_{2}^{2} <\|\nabla Q\|_{2}^{2}\), then
\(\|\nabla u_{0}\|_{2} < y_{2}\). Then, by the bootstrap and continuity argument, we claim that the solution
\(u(t,x)\) is such that, for all
\(t\in I\),
$$ \bigl\| \nabla u(t) \bigr\| _{2}^{2} < \|\nabla Q \|_{2}^{2}. $$
(3.20)
This implies that
\(K_{1}\) is invariant. Indeed, if (
3.20) is not true for all
\(t\in I\), then there exists
\(t_{1}\in I\) such that
\(\|\nabla u(t_{1})\|_{2}^{2} \geq\|\nabla Q\|_{2}^{2}=y_{2}^{2}\). But from the fact that the corresponding solution
\(u(t,x)\) of the Cauchy problem (
1.1)-(
1.2) is continuous with respect to
t, there exists
\(0< t_{0}\leq t_{1}\) such that
\(\|\nabla u(t_{0})\|_{2}^{2}= \|\nabla Q\|_{2}^{2}=y_{2}^{2}\). Inject this fact into (
3.19) and take
\(t=t_{0}\). We see that
$$f(y_{2})=f \bigl( \bigl\| \nabla u(t_{0}) \bigr\| _{2} \bigr) \leq E \bigl((u_{0},u_{1}) \bigr)< \frac{\|\nabla Q\|_{2}^{2}}{4}< f(y_{2}). $$
This is a contradiction because
\(f(y)\) is increasing on the interval
\([0,y_{2})\). Moreover, by the same argument, we can give the proof of the invariant of
\(K_{2}\) (see also the proof of Theorem
1.1. Here, we omit the detailed proof).
Now, we return to the proof the Theorem
3.1. From (
3.13) and (
3.14), we get
\(u_{0}\in K_{1}\). Applying the invariant of
\(K_{1}\) and the local well-posedness, we deduce that (
3.15) is true and the solution
\(u(t,x)\) of the Cauchy problem (
1.1)-(
1.2) exists globally by the local well-posedness theory (see Remark
2.2). This completes part (i) of the proof. Next, we give the proof of (ii). By (
3.13) and (
3.16), we see that
\(u_{0}\in K_{2}\). Applying the invariant of
\(K_{2}\), we have
$$ \bigl\| \nabla u(t) \bigr\| _{2}^{2} >\|\nabla Q \|_{2}^{2} \quad\mbox{for all } t\in I. $$
(3.21)
From (
2.1) and (
3.13), we get
$$ 2\|u\|_{4}^{4}>-2\| \nabla Q \|_{2}^{2}+4\|u_{t}\|_{2}^{2}+4 \|\nabla u\|_{2}^{2}+4\|u\|_{2}^{2}. $$
(3.22)
Let
\(J(t):=\int|u(t,x)|^{2}\,dx\). By some basic computations, we deduce that
\(J^{\prime}(t)=2 \int uu_{t}\,dx\) and
$$ J^{\prime\prime}(t)= 2 \int \bigl(|u_{t}|^{2}+|u|^{4}-|\nabla u|^{2}-|u|^{2} \bigr)\,dx. $$
(3.23)
It follows from (
3.21)-(
3.23) that
$$ J^{\prime\prime}(t)\geq 6\|u_{t}\|_{2}^{2}-2 \|\nabla Q\|_{2}^{2}+2\|\nabla u\|_{2}^{2}+2 \| u\|_{2}^{2}>6\|u_{t}\|_{2}^{2}. $$
(3.24)
Multiplying (
3.24) with
\(J(t)\) and injecting
\(J^{\prime}(t)^{2}\leq4\|u\|_{2}^{2}\|u_{t}\|_{2}^{2}\), we get
$$ J(t)J^{\prime\prime}(t)>6\|u_{t}\|_{2}^{2} \|u\|_{2}^{2}>\frac {3}{2}J^{\prime}(t)^{2}. $$
(3.25)
From (
3.24) and (
3.25), there exists a
\(t_{0}>0\) such that for
\(t>t_{0}\)
$$ J^{\prime}(t)>K J(t)^{\frac{3}{2}}, $$
(3.26)
where
\(K>0\). Since
\(\frac{3}{2}>1\), by the same argument as the proof of Theorem
1.1, we can deduce that the solution
\(u(t,x)\) of the Cauchy problem (
1.1)-(
1.2) blows up in the finite time
\(0< T<+\infty\). □