# The Hankel determinants for the generalized derangement polynomials of order
$r$

GHANIA GUETTAI  University Yahia Farès Médéa
Faculty of Science
urban pole, 26000
Médéa, Algeria<EMAIL_ADDRESS>,  DIFFALAH LAISSAOUI  University Yahia
Farès Médéa
Faculty of Science
urban pole, 26000
Médéa, Algeria<EMAIL_ADDRESS>and  MOURAD RAHMANI  Faculty of
Mathematics
USTHB
P.O. Box 32 El Alia 16111
Algiers
Algeria<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

This paper sets out to introduce the generalized derangement polynomials of
order $r$. It then proceeds to establish various identities associated with
these polynomials, along with providing recurrence relations for derangement
polynomials of order $r$. Additionally, the paper offers a probabilistic
approach for the generalized derangement polynomials of order $r$.
Furthermore, it furnishes a clear expression for the Hankel determinants
pertaining to these generalized derangement polynomials of order $r$, and
subsequently infers the Hankel determinants for derangement polynomials of the
same order, as well as for the count of cyclic derangements.

###### Key words and phrases:

Derangement number, derangement polynomials of order $r$, $r$-derangement
polynomials, probabilistic approach, Hankel determinants.

###### 2010 Mathematics Subject Classification:

11B83, 05A15, 60C05, 05A19

##  1\. Introduction

A derangement refers to a permutation where none of the elements occupy their
original positions. The derangement number, denoted as $D_{n}$, represents the
count of such permutations for a set of $n$ elements (also known as the number
of fixed point-free permutations). The task of enumerating derangements was
first tackled by Pierre Rémond de Montmort (1678-1719) and was referred to as
”le problème de rencontres” in French. The first few terms of the derangement
number are $D_{0}=1$, $D_{1}=0$, $D_{2}=1$, $D_{3}=2$, $D_{4}=9,\ldots$.
A simple explicit expression for $D_{n}$ is (see [2, 3])

(1.1) $D_{n}=n!\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{k!}.$

From (1.1), we note that the exponential generating function of derangement
numbers is given by

(1.2) $\frac{e^{-z}}{1-z}=\sum_{n\geq 0}D_{n}\frac{z^{n}}{n!}.$

A fixed point-free permutations on $n+r$ letters will be called fixed point-
free $r$-permutation if in its cycle decomposition the first $r$ letters
appear to be in distinct cycles. The number of fixed point-free
$r$-permutations denote by $D_{n}^{(r)}$ and call $r$-derangement number
$(r\in\mathbb{N}^{\ast})$. The first $r$ elements, as well as the cycles they
are contained in will be called distinguished. For any
$r\in\mathbb{N}^{\ast}$, the exponential generating function of the sequence
of $r$-derangements numbers is given by (see [7])

(1.3) $\frac{z{{}^{r}}e^{-z}}{\left(1-z\right)^{r+1}}=\sum_{n\geq
0}D_{n}^{(r)}\frac{z^{n}}{n!}.$

The derangement numbers $d_{n}^{(r)}$ of order $r$ $(n\geq 0)$, are defined by
the exponential generating function (see [4])

(1.4) $\frac{e^{-z}}{\left(1-z\right)^{r}}=\sum_{n\geq
0}d_{n}^{(r)}\frac{z^{n}}{n!}.$

Many generalizations and extensions have been published in recent years.
Firstly, the $r$-derangement polynomials $D_{n}^{(r)}(x)$, are defined by the
exponential generating function (see [7])

(1.5) $\frac{z{{}^{r}}e^{xz}}{\left(1-z\right)^{r+1}}=\sum_{n\geq
0}D_{n}^{(r)}(x)\frac{z^{n}}{n!}.$

Secondly, the derangement polynomials $d_{n}^{(r)}(x)$ of order $r$, are
defined by the exponential generating function (see [4])

(1.6) $\frac{e^{xz}}{\left(1-z\right)^{r}}=\sum_{n\geq
0}d_{n}^{(r)}(x)\frac{z^{n}}{n!}.$

Another generalization of the derangement number is the number of cyclic
derangements $d_{n,r}$, are defined by the exponential generating function
(see [1])

(1.7) $\frac{e^{-z}}{1-rz}=\sum_{n\geq 0}d_{n,r}\frac{z^{n}}{n!}.$

##  2\. The generalized derangement polynomials of order $r$

This section introduces a novel set of polynomials termed as the ”generalized
derangement polynomials of order $r$”. We will outline their characteristics
and elucidate their connection with derangement polynomials of the same order.
Following this, we will provide the recurrence relations governing these
derangement polynomials of degree $r$.

######  Definition 2.1.

We define the generalized derangement polynomials
$\mathbb{D}_{n}^{(r)}\left(x\right)$ of order $r$ by

(2.1) $\frac{e^{z}}{\left(1-xz\right)^{r}}=\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}.$

######  Theorem 2.2.

For $n\geq 0$, we have the explicit formula of the generalized derangement
polynomials of order $r$

(2.2)
$\mathbb{D}_{n}^{(r)}\left(x\right)=\sum_{k=0}^{n}\binom{n}{k}r^{\overline{k}}x^{k}.$

where $r^{\overline{k}}$ is the rising factorial, defined by
$r^{\overline{k}}=r(r+1)...(r+k-1)$, with $r^{\overline{0}}=1$.

###### Proof.

By using the exponential generating function (2.1), we obtain

$\displaystyle\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}$
$\displaystyle=\frac{e^{z}}{\left(1-xz\right)^{r}}$ $\displaystyle=\sum_{k\geq
0}\frac{r^{\overline{k}}(xz)^{k}}{k!}\sum_{n\geq 0}\frac{z^{n}}{n!}$
$\displaystyle=\sum_{k\geq 0}r^{\overline{k}}x^{k}\sum_{n\geq
0}\frac{z^{k}z^{n}}{k!n!}$ $\displaystyle=\sum_{k\geq
0}r^{\overline{k}}x^{k}\sum_{n\geq 0}\binom{n+k}{k}\frac{z^{n+k}}{(n+k)!}$
$\displaystyle=\sum_{k\geq 0}r^{\overline{k}}x^{k}\sum_{n\geq
k}\binom{n}{k}\frac{z^{n}}{n!}$ $\displaystyle=\sum_{n\geq
0}\left(\sum_{k=0}^{n}\binom{n}{k}r^{\overline{k}}x^{k}\right)\frac{z^{n}}{n!}.$

Equating the coefficients of $\frac{z^{n}}{n!}$ in both sides of the above
expression, we get the desired result. ∎

The recurrence relation of the generalized derangement polynomials
$\mathbb{D}_{n}^{(r)}\left(x\right)$ of order $r$ is given by

######  Theorem 2.3.

For $n\geq 0$, we have

(2.3)
$\mathbb{D}_{n+1}^{(r)}\left(x\right)=\mathbb{D}_{n}^{(r)}\left(x\right)+rx\mathbb{D}_{n}^{(r+1)}\left(x\right).$

###### Proof.

By derivation the exponential generating function (2.1) with respect to $z$,
we obtain

$\displaystyle\frac{d}{dz}\left(\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}\right)$
$\displaystyle=\frac{d}{dz}\left(\frac{e^{z}}{\left(1-xz\right)^{r}}\right)$
$\displaystyle=\frac{e^{z}}{\left(1-xz\right)^{r}}+\frac{rxe^{z}}{\left(1-xz\right)^{r+1}}$
$\displaystyle=\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}+rx\sum_{n\geq
0}\mathbb{D}_{n}^{(r+1)}\left(x\right)\frac{z^{n}}{n!}$
$\displaystyle=\sum_{n\geq
0}\left(\mathbb{D}_{n}^{(r)}\left(x\right)+rx\mathbb{D}_{n}^{(r+1)}\left(x\right)\right)\frac{z^{n}}{n!}.$

On the other hand, we have

$\frac{d}{dz}\left(\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}\right)=\sum_{n\geq
0}\mathbb{D}_{n+1}^{(r)}\left(x\right)\frac{z^{n}}{n!}.$

Equating the coefficients of $\frac{z^{n}}{n!}$ in both sides of the above
expression, we get the desired result. ∎

######  Theorem 2.4.

For $n\geq 0$, we have

(2.4)
$\mathbb{D}_{n+1}^{(r)}\left(x\right)=\mathbb{D}_{n}^{(r)}\left(x\right)+rx\sum_{k=0}^{n}\binom{n}{k}\mathbb{D}_{k}^{(r)}\left(x\right)x^{n-k}\left(n-k\right)!.$

###### Proof.

Using the derivative of the generating function (2.1) with respect to $z$, we
obtain

$\displaystyle\frac{d}{dz}\left(\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}\right)$
$\displaystyle=\frac{d}{dz}\left(\frac{e^{z}}{\left(1-xz\right)^{r}}\right)$
$\displaystyle=\frac{e^{z}}{\left(1-xz\right)^{r}}+\frac{rxe^{z}}{\left(1-xz\right)^{r+1}}$
$\displaystyle=\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}+rx\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}\sum_{n\geq
0}n!x^{n}\frac{z^{n}}{n!}.$

Thus,

$\sum_{n\geq
0}\mathbb{D}_{n+1}^{(r)}\left(x\right)\frac{z^{n}}{n!}=\sum_{n\geq
0}\left(\mathbb{D}_{n}^{(r)}\left(x\right)+rx\sum_{k=0}^{n}\binom{n}{k}\mathbb{D}_{k}^{(r)}\left(x\right)x^{n-k}\left(n-k\right)!\right)\frac{z^{n}}{n!}.$

Equating the coefficients of $\frac{z^{n}}{n!}$ in both sides of the above
expression, we get the desired result. ∎

Here we present some properties of the derangement polynomials
$d_{n}^{(r)}(x)$ of order $r$ given by the formula (1.6). We start with the
explicit form and then provide two recurrence relations.

######  Theorem 2.5.

For $n\geq 0$, we have the explicit formula of the derangement polynomials
$d_{n}^{(r)}(x)$ of order $r$

(2.5)
${d}_{n}^{(r)}\left(x\right)=\sum_{k=0}^{n}\binom{n}{k}r^{\overline{k}}x^{n-k}.$

where $r^{\overline{k}}$ is the rising factorial.

###### Proof.

By using the exponential generating function (1.6), we obtain

$\displaystyle\sum_{n\geq 0}{d}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!}$
$\displaystyle=\frac{e^{xz}}{\left(1-z\right)^{r}}$ $\displaystyle=\sum_{k\geq
0}\frac{r^{\overline{k}}z^{k}}{k!}\sum_{n\geq 0}\frac{(xz)^{n}}{n!}$
$\displaystyle=\sum_{k\geq 0}r^{\overline{k}}\sum_{n\geq
0}\frac{x^{n}z^{n+k}}{k!n!}$ $\displaystyle=\sum_{k\geq
0}r^{\overline{k}}\sum_{n\geq 0}\binom{n+k}{k}x^{n}\frac{z^{n+k}}{(n+k)!}$
$\displaystyle=\sum_{k\geq 0}r^{\overline{k}}\sum_{n\geq
k}\binom{n}{k}x^{n-k}\frac{z^{n}}{n!}$ $\displaystyle=\sum_{n\geq
0}\left(\sum_{k=0}^{n}\binom{n}{k}r^{\overline{k}}x^{n-k}\right)\frac{z^{n}}{n!}.$

Equating the coefficients of $\frac{z^{n}}{n!}$ in both sides of the above
expression, we get the desired result. ∎

Now, it is easy to show the relationship between the polynomials
$\mathbb{D}_{n}^{(r)}\left(x\right)$ and the derangement polynomials
$d_{n}^{\left(r\right)}\left(x\right)$ of order $r$. For $n\geq 0$ and $x\neq
0$, we have

(2.6)
$x^{n}\mathbb{D}_{n}^{(r)}\left(\frac{1}{x}\right)=d_{n}^{\left(r\right)}\left(x\right).$

Then, we obtain the following corollaries

######  Corollary 2.6.

For $n\geq 0$ and $x\neq 0$, we have

(2.7)
$d_{n+1}^{(r)}\left(x\right)=xd_{n}^{(r)}\left(x\right)+rd_{n}^{(r+1)}\left(x\right)\text{.}$

###### Proof.

Using the formula (2.3) and replacing $x$ by $\frac{1}{x}$, after that
multiplying the both sides of the new expression by $x^{n+1}$. Finally, by
using the formula (2.6), we obtain the desired formula (2.7). ∎

######  Corollary 2.7.

For $n\geq 0$ and $x\neq 0$, we have

(2.8)
$d_{n+1}^{(r)}\left(x\right)=xd_{n}^{(r)}\left(x\right)+r\sum_{k=0}^{n}\binom{n}{k}d_{k}^{(r)}\left(x\right)\left(n-k\right)!.$

##  3\. Probabilistic approach

In this section, we will present a probabilistic representation of the
generalized derangement polynomials of order $r$. Let $f_{X}(x)$ be the
probability density function of the continuous random variable $X$, and let
$h(x)$ be a real valued function. Then the expectation of $h(X)$, is defined
by (see [6])

$E(h(X))={\displaystyle\int\limits_{-\infty}^{+\infty}}h(x)f_{X}(x)dx.$

A continuous random variable $X$, is said to be the gamma random variable with
parameters $\alpha,\beta$ ; for some $\alpha>0$ and $\beta>0$, if its density
function is given by

$f_{X}(x)=\genfrac{\\{}{.}{0.0pt}{}{\dfrac{\beta^{\alpha}x^{\alpha-1}e^{-\beta
x}}{\Gamma(\alpha)}\text{ \ \ if }x\geq 0,}{0\text{ \ }else}$

where $\Gamma(\alpha)$ is the gamma function, is denoted
$X\sim\Gamma(\alpha,\beta)$.

######  Lemma 3.1.

Let $Y\sim\Gamma(1,1).$ Then, for all $x<1$, we have

$e^{z}\phi_{Y}(z)=\sum_{n\geq 0}D_{n}\frac{z^{n}}{n!},$

where $\phi_{Y}$ is the moment generating function of random variable $Y$.

######  Corollary 3.2.

Let $X_{1},X_{2},\ldots,X_{r}$ be independent and identically distributed
gamma random variables with shape $\alpha=1$ and rate $\beta=1$, we assume
$Y_{r}={\displaystyle\sum\limits_{l=1}^{r}}X_{l}$, then

$e^{z}\phi_{Y_{r}}(xz)=\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\left(x\right)\frac{z^{n}}{n!},$

where $\phi_{Y_{r}}$ is the moment generating function of random variable
$Y_{r}.$

######  Corollary 3.3.

Let $X_{1},X_{2},\ldots,X_{r}$ be independent and identically distributed
random variables with $X\sim\Gamma(1,1)$, we assume
$Y_{r}={\displaystyle\sum\limits_{l=1}^{r}}X_{l}$, then

$\displaystyle\mathbb{D}_{n}^{(r)}\left(x\right)$
$\displaystyle=\sum_{k=0}^{n}\binom{n}{k}x^{k}E(Y_{r}^{k})$

and

$d_{n}^{\left(r\right)}\left(x\right)=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}E(Y_{r}^{k}).$

######  Corollary 3.4.

Let $X_{1},X_{2},\ldots,X_{r}$ be independent and identically distributed
random variables with $X\sim\Gamma(1,1)$, then

$\displaystyle E\left(X_{1}+X_{2}+\cdots+X_{r}\right)^{k}$
$\displaystyle=r^{\overline{k}}.$

##  4\. On the Hankel determinant of $r-$derangement polynomials

The Hankel matrix of order $n+1$ of a sequence $a_{0},a_{1},\ldots,$ is the
$n+1$ by $n+1$ matrix whose $(i,j)$ entry is $a_{i+j},$ where the indices
range between $0$ and $n$. The Hankel determinant of order $n+1$ is the
determinant of the corresponding Hankel matrix, that is,

$\det(a_{i+j})_{0\leq i,j\leq
n}=\det\left(\begin{array}[c]{cccc}a_{0}&a_{1}&\cdots&a_{n}\\\
a_{1}&a_{2}&\cdots&a_{n+1}\\\ \vdots&\vdots&\ddots&\vdots\\\
a_{n}&a_{n+1}&\cdots&a_{2n}\end{array}\right).$

In this section, employing a methodology akin to that in [5], we construct a
comprehensive formulation for the Hankel determinant of the generalized
derangement polynomials in terms of order $r$. Subsequently, we derive the
Hankel determinant for the derangement polynomials of order $r$, as well as
for the count of cyclic derangements.

######  Lemma 4.1.

The following formula holds true

$\mathbb{D}_{n}^{(r)}\left(\frac{1}{1-z}\right)=\left(1-z\right)^{r}e^{-z}\frac{d^{n}}{dz^{n}}\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right).$

###### Proof.

According to Leibniz’s formula, we have the $n$-th derivative of the
generating function of $\mathbb{D}_{n}^{(r)}\left(1\right)$

$\displaystyle\frac{d^{n}}{dz^{n}}\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)$
$\displaystyle=\sum_{k=0}^{n}\binom{n}{k}\frac{d^{n-k}}{dz^{n-k}}\left(e^{z}\right)\frac{d^{k}}{dz^{k}}\left(\frac{1}{\left(1-z\right)^{r}}\right)$
$\displaystyle=e^{z}\sum_{k=0}^{n}\binom{n}{k}r^{\overline{k}}\frac{1}{\left(1-z\right)^{r+k}}$
$\displaystyle=\frac{e^{z}}{\left(1-z\right)^{r}}\sum_{k=0}^{n}\binom{n}{k}r^{\overline{k}}\frac{1}{\left(1-z\right)^{k}}$
$\displaystyle=\frac{e^{z}}{\left(1-z\right)^{r}}\mathbb{D}_{n}^{(r)}\left(\frac{1}{1-z}\right).$

∎

######  Theorem 4.2.

For $n\geq 1$ and $z\neq 1$, we have

(4.1) $\displaystyle\det\left(\frac{\partial^{i+j-2}}{\partial
z^{i+j-2}}\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)\right)_{1\leq
i,j\leq
n}=\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{n}\frac{r^{\overline{n-1}}{\displaystyle\prod\limits_{k=1}^{n-1}}r^{\overline{k-1}}k!}{\left(z-1\right)^{\left(n-1\right)n}}.$

###### Proof.

By induction, using Sylvestre’s formula, for $x=1$, we get the numbers
$\mathbb{D}_{n}^{(r)}\left(1\right)=\mathbb{D}_{n}^{(r)}$, according to lemma
4.1, the generating function of $\mathbb{D}_{n}^{(r)}$ is

$G\left(z\right)=\sum_{n\geq
0}\mathbb{D}_{n}^{(r)}\frac{z^{n}}{n!}=\frac{e^{z}}{\left(1-z\right)^{r}}.$

We denote that the Hankel determinant

$\left(\det\left(\frac{\partial^{i+j-2}}{\partial
z^{i+j-2}}G\left(z\right)\right)_{i,j=1,2,\ldots,n}\right)$

of the $i+j-2$-th derivative of the generating function $G\left(z\right)$ by
$\Delta_{n}\left(G\left(z\right)\right)$.
For $n=1$, we have

$\Delta_{1}\left(G\left(z\right)\right)=\frac{e^{z}}{\left(1-z\right)^{r}}.$

For $n=2$, we have

$\displaystyle\Delta_{2}\left(G\left(z\right)\right)$
$\displaystyle=\left|\begin{array}[c]{cc}G\left(z\right)&\frac{\partial}{\partial
z}G\left(z\right)\\\ \frac{\partial}{\partial
z}G\left(z\right)&\frac{\partial^{2}}{\partial
z^{2}}G\left(z\right)\end{array}\right|$
$\displaystyle=\left|\begin{array}[c]{cc}\frac{e^{z}}{\left(1-z\right)^{r}}&\frac{e^{z}}{\left(1-z\right)^{r}}\left[1+\frac{r}{\left(1-z\right)}\right]\\\
\frac{e^{z}}{\left(1-z\right)^{r}}\left[1+\frac{r}{\left(1-z\right)}\right]&\frac{e^{z}}{\left(1-z\right)^{r}}\left[1+\frac{2r}{\left(1-z\right)}+\frac{r\left(r+1\right)}{\left(1-z\right)^{2}}\right]\end{array}\right|$
$\displaystyle=\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{2}\frac{r}{\left(z-1\right)^{2}}.$

Suppose that, the formula (4.1) is true for $n=1,2,\ldots,m$ and we calculate
$\Delta_{m+1}\left(G\left(z\right)\right)$ by Sylvester’s formula

$\Delta_{m+1}\left(G\left(z\right)\right)=\frac{\Delta_{2}\left(\Delta_{m}\left(G\left(z\right)\right)\right)}{\Delta_{m-1}\left(G\left(z\right)\right)}.$

By induction hypothesis, we have

$\displaystyle\Delta_{m}\left(G\left(z\right)\right)$
$\displaystyle=\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{m}\frac{r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!}{\left(z-1\right)^{\left(m-1\right)m}}$
$\displaystyle=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)\frac{e^{mz}}{\left(1-z\right)^{mr+\left(m-1\right)m}}.$

To calculate $\Delta_{2}\left(\Delta_{m}\left(G\left(z\right)\right)\right)$,
we need to calculate the $1$st and the $2$nd derivative of
$\Delta_{m}\left(G\left(z\right)\right)$, for that we have the $1$st
derivative

$\frac{\partial}{\partial
z}\left(\Delta_{m}\left(G\left(z\right)\right)\right)=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)e^{mz}\\\
\left(\frac{m\left(1-z\right)^{mr+\left(m-1\right)m}+\left(mr+m\left(m-1\right)\right)\left(1-z\right)^{mr+m\left(m-1\right)-1}}{\left(1-z\right)^{2(mr+m\left(m-1\right))}}\right)\\\
=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)e^{mz}\times\alpha_{m}\left(z\right),$

with

$\displaystyle\alpha_{m}\left(z\right)$
$\displaystyle=\frac{m}{\left(1-z\right)^{m\left(r+m-1\right)}}+\frac{m\left(r+m-1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+1}}.$

The 2nd derivative of $\Delta_{m}\left(G\left(z\right)\right)$ is

$\displaystyle\frac{\partial^{2}}{\partial
z^{2}}\left(\Delta_{m}\left(G\left(z\right)\right)\right)$
$\displaystyle=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)e^{mz}\left[m\times\alpha_{m}\left(z\right)+\frac{\partial}{\partial
z}\alpha_{m}\left(z\right)\right]$
$\displaystyle=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)e^{mz}\times\beta_{m}\left(z\right),$

with

$\displaystyle\beta_{m}\left(z\right)$
$\displaystyle=\frac{m^{2}}{\left(1-z\right)^{m\left(r+m-1\right)}}+\frac{2m^{2}\left(r+m-1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+1}}$
$\displaystyle+\frac{m\left(r+m-1\right)\left(m\left(r+m-1\right)+1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+2}}.$

Therefore

$\Delta_{2}\left(\Delta_{m}\left(G\left(z\right)\right)\right)=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)^{2}e^{2mz}\\\
\times\left|\begin{array}[c]{c}\frac{1}{\left(1-z\right)^{m\left(r+m-1\right)}}\\\
\frac{m}{\left(1-z\right)^{m\left(r+m-1\right)}}+\frac{m\left(r+m-1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+1}}\end{array}\right.\\\
\left.\begin{array}[c]{c}\frac{m}{\left(1-z\right)^{m\left(r+m-1\right)}}+\frac{m\left(r+m-1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+1}}\\\
\frac{m^{2}}{\left(1-z\right)^{m\left(r+m-1\right)}}+\frac{2m^{2}\left(r+m-1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+1}}+\frac{m\left(r+m-1\right)\left(m\left(r+m-1\right)+1\right)}{\left(1-z\right)^{m\left(r+m-1\right)+2}}\end{array}\right|.$

Thus,

$\displaystyle\Delta_{2}\left(\Delta_{m}\left(G\left(z\right)\right)\right)$
$\displaystyle=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)^{2}e^{2mz}\frac{1}{\left(1-z\right)^{2m\left(r+m-1\right)}}$
$\displaystyle{\small\times\left|\begin{array}[c]{cc}1&m+\frac{m\left(r+m-1\right)}{1-z}\\\
m+\frac{m\left(r+m-1\right)}{1-z}&m^{2}+\frac{2m^{2}\left(r+m-1\right)}{1-z}+\frac{m\left(r+m-1\right)\left(m\left(r+m-1\right)+1\right)}{\left(1-z\right)^{2}}\end{array}\right|}$
$\displaystyle=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)^{2}e^{2mz}\frac{m\left(r+m-1\right)}{\left(1-z\right)^{2m\left(r+m-1\right)+2}}.$

According to Sylvester’s formula, we find

$\displaystyle\Delta_{m+1}\left(G\left(z\right)\right)$
$\displaystyle=\frac{\Delta_{2}\left(\Delta_{m}\left(G\left(z\right)\right)\right)}{\Delta_{m-1}\left(G\left(z\right)\right)}$
$\displaystyle=\frac{\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)^{2}e^{2mz}\frac{m\left(r+m-1\right)}{\left(1-z\right)^{2m\left(r+m-1\right)+2}}}{\left(\left(r\right)^{\overline{m-2}}{\displaystyle\prod\limits_{k=1}^{m-2}}r^{\overline{k-1}}k!\right)\frac{e^{(m-1)z}}{\left(1-z\right)^{\left(r+m-2\right)\left(m-1\right)}}}.$

Thus,

$\displaystyle\Delta_{m+1}\left(G\left(z\right)\right)$
$\displaystyle=\left(r^{\overline{m-1}}{\displaystyle\prod\limits_{k=1}^{m-1}}r^{\overline{k-1}}k!\right)r^{\overline{m-1}}m!e^{mz+z}\frac{\left(r+m-1\right)}{\left(1-z\right)^{mr+m^{2}+r+2m}}$
$\displaystyle=\left(r^{\overline{m}}{\displaystyle\prod\limits_{k=1}^{m}}r^{\overline{k-1}}k!\right)e^{\left(m+1\right)z}\frac{1}{\left(1-z\right)^{\left(m+1\right)r+\left(m+1\right)m}}$
$\displaystyle=\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{m+1}\frac{r^{\overline{m}}{\displaystyle\prod\limits_{k=1}^{m}}r^{\overline{k-1}}k!}{\left(z-1\right)^{\left(m+1\right)m}}.$

∎

######  Lemma 4.3.

For $n\geq 1$ and $z\neq 1$, we have

$\det\left(\frac{e^{z}}{\left(1-z\right)^{r}}\mathbb{D}_{i+j-2}^{(r)}\left(\frac{1}{1-z}\right)\right)_{1\leq
i,j\leq n+1}=\\\
\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{n+1}\det\left(\mathbb{D}_{i+j-2}^{(r)}\left(\frac{1}{1-z}\right)\right)_{1\leq
i,j\leq n+1}.$

######  Theorem 4.4.

The Hankel determinant of order $n+1$ of the generalized derangement
polynomials $\mathbb{D}_{n}^{(r)}\left(z\right)$ of order $r$, is given by

$\det\left(\mathbb{D}_{i+j-2}^{(r)}\left(z\right)\right)_{1\leq i,j\leq
n+1}=z^{n\left(n+1\right)}r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!.$

###### Proof.

By Lemma 4.3, we have

$\displaystyle\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{n+1}\det\left(\mathbb{D}_{i+j-2}^{(r)}\left(\frac{1}{1-z}\right)\right)$
$\displaystyle=\det\left(\frac{\partial^{i+j-2}}{\partial
z^{i+j-2}}\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)\right)$
$\displaystyle=\left(\frac{e^{z}}{\left(1-z\right)^{r}}\right)^{n+1}\frac{r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!}{\left(z-1\right)^{\left(n+1\right)n}},$

from where

$\displaystyle\det\left(\mathbb{D}_{i+j-2}^{(r)}\left(\frac{1}{1-z}\right)\right)$
$\displaystyle=\frac{r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!}{\left(z-1\right)^{\left(n+1\right)n}}$
$\displaystyle=\left(-1\right)^{n(n+1)}\left(\frac{1}{1-z}\right)^{{}^{\left(n+1\right)n}}r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!$
$\displaystyle=\left(\frac{1}{1-z}\right)^{{}^{\left(n+1\right)n}}r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!.$

Now let’s pose $T=\frac{1}{1-z}$, we obtain

$\det\left(\mathbb{D}_{i+j-2}^{(r)}\left(T\right)\right)=T^{n\left(n+1\right)}r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!.$

∎

######  Remark 4.5.

When $r=1$ and $x=-1$, the well-known Hankel determinants of order $n+1$ of
the matrices $[(i+j)!]$ and $[D_{i+j}]$, is given by

$det((i+j)!)_{0\leq i,j\leq n}=det(D_{i+j})_{0\leq i,j\leq
n}=\left({\displaystyle\prod\limits_{k=1}^{n}}k!\right)^{2}.$

######  Theorem 4.6.

The Hankel determinant of order $n+1$ of the derangement polynomials
$d_{n}^{(r)}\left(z\right)$ of order $r$, is given by

$\det\left(d_{i+j-2}^{(r)}\left(z\right)\right)_{1\leq i,j\leq
n+1}=r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!.$

###### Proof.

We have,

$\displaystyle\det\left(d_{i+j-2}^{(r)}\left(z\right)\right)$
$\displaystyle=\left|\begin{array}[c]{cccc}z^{0}\mathbb{D}_{0}^{(r)}\left(\frac{1}{z}\right)&z^{1}\mathbb{D}_{1}^{(r)}\left(\frac{1}{z}\right)&\ldots&z^{n}\mathbb{D}_{n}^{(r)}\left(\frac{1}{z}\right)\\\
z^{1}\mathbb{D}_{1}^{(r)}\left(\frac{1}{z}\right)&z^{2}\mathbb{D}_{2}^{(r)}\left(\frac{1}{z}\right)&\ldots&z^{n+1}\mathbb{D}_{n+1}^{(r)}\left(\frac{1}{z}\right)\\\
\ldotp&\ldotp&\ldotp&\ldotp\\\
z^{n}\mathbb{D}_{n}^{(r)}\left(\frac{1}{z}\right)&z^{n+1}\mathbb{D}_{n+1}^{(r)}\left(\frac{1}{z}\right)&\ldots&z^{2n}\mathbb{D}_{2n}^{(r)}\left(\frac{1}{z}\right)\end{array}\right|$
$\displaystyle=\left(z^{0}z^{1}z^{2}\cdots
z^{n}\right)^{2}\left|\begin{array}[c]{cccc}\mathbb{D}_{0}^{(r)}\left(\frac{1}{z}\right)&\mathbb{D}_{1}^{(r)}\left(\frac{1}{z}\right)&\ldots&\mathbb{D}_{n}^{(r)}\left(\frac{1}{z}\right)\\\
\mathbb{D}_{1}^{(r)}\left(\frac{1}{z}\right)&\mathbb{D}_{2}^{(r)}\left(\frac{1}{z}\right)&\ldots&\mathbb{D}_{n+1}^{(r)}\left(\frac{1}{z}\right)\\\
\ldotp&\ldotp&\ldotp&\ldotp\\\
\mathbb{D}_{n}^{(r)}\left(\frac{1}{z}\right)&\mathbb{D}_{n+1}^{(r)}\left(\frac{1}{z}\right)&\ldots&\mathbb{D}_{2n}^{(r)}\left(\frac{1}{z}\right)\end{array}\right|$
$\displaystyle=\left(z^{0}z^{1}z^{2}\cdots
z^{n}\right)^{2}\left(\frac{1}{z}\right)^{n\left(n+1\right)}r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!$
$\displaystyle=z^{n\left(n+1\right)-n\left(n+1\right)}r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!$
$\displaystyle=r^{\overline{n}}{\displaystyle\prod\limits_{k=1}^{n}}r^{\overline{k-1}}k!.$

∎

######  Corollary 4.7.

The Hankel determinant of order $n+1$ of the number of cyclic derangements
$d_{n,r}$, is given by

$\det\left(d_{i+j-2,r}\right)_{1\leq i,j\leq
n+1}=r^{n\left(n+1\right)}\left({\displaystyle\prod\limits_{k=1}^{n}}k!\right)^{2}.$

###### Proof.

The relationship between the generalized derangement polynomials
$\left(\mathbb{D}_{n}^{(r)}\left(z\right)\right)$ of order $r$ and the number
of cyclic derangements $d_{n,r}$ is given by

(4.2) $(-1)^{n}\mathbb{D}_{n}^{(1)}\left(-r\right)=d_{n,r}.$

Now, using the Theorem 4.4 and the formula (4.2), we get the desired result. ∎

## References

  * [1] S. H. Assaf, Cyclic derangements, Electron. J. Combin. 17 (2010), no. 1, Research Paper 163, 14 pp.
  * [2] M. Bóna, Combinatorics of permutations, Discrete Mathematics and its Applications (Boca Raton), CRC Press, Boca Raton, FL, second edition, (2012).
  * [3] L. Carlitz, The number of derangements of a sequence with given specification, Fibonacci Quart, 16(3) (1978), 255–258.
  * [4] T. Kim and D. S. Kim, Some identities on derangement and degenerate derangement polynomials, In Advances in mathematical inequalities and applications, Trends Math, Birkhäuser/Springer, Singapore, (2018), 265–277.
  * [5] C. Radoux, Déterminant de Hankel construit sur des polynômes liés aux nombres de dérangements, European J. Combin, 12(4) (1991), 327–329.
  * [6] S. M. Ross, Introduction to probability models, Harcourt/Academic Press, Burlington, MA, seventh edition, (2000).
  * [7] C. Wang, P. Miska, and I. Mezo, The $r$-derangement numbers, Discrete Math, 340(7) (2017), 1681–1692.


# Global existence and blow-up for the focusing inhomogeneous nonlinear
Schrödinger equation with inverse-square potential

JinMyong An, JinMyong Kim∗, RoeSong Jang
Faculty of Mathematics, Kim Il Sung University, Pyongyang, Democratic People’s
Republic of Korea
∗ Corresponding Author<EMAIL_ADDRESS>
###### Abstract

In this paper, we study the Cauchy problem for the focusing inhomogeneous
nonlinear Schrödinger equation with inverse-square potential

$iu_{t}+\Delta u-c|x|^{-2}u+|x|^{-b}|u|^{\sigma}u=0,\;u(0)=u_{0}\in
H_{c}^{1},\;(t,x)\in\mathbb{R}\times\mathbb{R}^{d},$

where $d\geq 3$, $0<b<2$, $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$ and
$c>-c(d):=-\left(\frac{d-2}{2}\right)^{2}$. We first establish the criteria
for global existence and blow-up of general (not necessarily radial or finite
variance) solutions to the equation. Using these criteria, we study the global
existence and blow-up of solutions to the equation with general data lying
below, at, and above the ground state threshold. Our results extend the global
existence and blow-up results of Campos-Guzmán (Z. Angew. Math. Phys., 2021)
and Dinh-Keraani (SIAM J. Math. Anal., 2021).

2020 Mathematics Subject Classification. 35Q55, 35A01, 35B44.

Key words and phrases. Inhomogeneous nonlinear Schrödinger equation, Inverse-
square potential, Global existence, Blow-up, Ground state.

##  1 Introduction

In this paper, we consider the Cauchy problem for the focusing inhomogeneous
nonlinear Schrödinger equation with inverse-square potential, denoted by INLSc
equation,

$\left\\{\begin{array}[]{l}{iu_{t}-P_{c}u+|x|^{-b}|u|^{\sigma}u=0,\;(t,x)\in\mathbb{R}\times\mathbb{R}^{d},}\\\
{u\left(0,\;x\right)=u_{0}(x)\in
H_{c}^{1}(\mathbb{R}^{d}),}\end{array}\right.$ (1.1)

where $d\geq 3$, $u:\mathbb{R}\times\mathbb{R}^{d}\to\mathbb{C}$,
$u_{0}:\mathbb{R}^{d}\to\mathbb{C}$, $0<b<2$, $\sigma>0$ and $P_{c}=-\Delta
u+c|x|^{-2}$ with $c>-c(d):=-\left(\frac{d-2}{2}\right)^{2}$.

The INLSc equation appears in a variety of physical settings, for example, in
nonlinear optical systems with spatially dependent interactions (see e.g. [4]
and the references therein). In particular, when $b=0$, the equation (1.1)
appears in various areas of physics, for instance in quantum field equations,
or in the study of certain black hole solutions of the Einstein equations (see
e.g. [5, 13]). When $c=0$, it appears in nonlinear optics, modeling
inhomogeneities in the medium in which the wave propagates (see e.g. [14]).

The case $b=c=0$ is the classic nonlinear Schrödinger (NLS) equation which has
been been widely studied over the last three decades (see e.g. [8, 17, 20] and
the references therein). The case $b=0$ and $c\neq 0$ is known as the NLS
equation with inverse-square potential, denoted by NLSc equation, which has
also been extensively studied in recent years (see e.g. [9, 15, 16, 18, 22]
and the references therein). Moreover, when $c=0$ and $b\neq 0$, we have the
inhomogeneous nonlinear Schrödinger equation, denoted by INLS equation, which
has also attracted a lot of interest in recent years (see e.g. [1, 2, 3, 6,
10] and the references therein).

The restriction on $c$ comes from the sharp Hardy inequality:

$c(d)\int_{\mathbb{R}^{d}}{|x|^{-2}\left|u(x)\right|^{2}dx}\leq\int_{\mathbb{R}^{d}}{\left|\nabla
u(x)\right|dx},\leavevmode\nobreak\ \forall u\in H^{1}(\mathbb{R}^{d}),$ (1.2)

which ensures that $P_{c}$ is a positive operator. Throughout the paper, for
$s\in\mathbb{R}$ and $1<p<\infty$, we denote by $H^{s,p}(\mathbb{R}^{d})$ and
$\dot{H}^{s,p}(\mathbb{R}^{d})$ the usual nonhomogeneous and homogeneous
Sobolev spaces associated to the Laplacian $-\Delta$. As usual, we abbreviate
$H^{s,2}(\mathbb{R}^{d})$ and $\dot{H}^{s,2}(\mathbb{R}^{d})$ as
$H^{s}(\mathbb{R}^{d})$ and $\dot{H}^{s}(\mathbb{R}^{d})$, respectively.
Similarly, we define Sobolev spaces in terms of $P_{c}$ via

$\left\|f\right\|_{\dot{H}_{c}^{s,p}(\mathbb{R}^{d})}=\left\|(P_{c})^{\frac{s}{2}}f\right\|_{L^{p}(\mathbb{R}^{d})},\leavevmode\nobreak\
\left\|f\right\|_{H_{c}^{s,p}(\mathbb{R}^{d})}=\left\|(1+P_{c})^{\frac{s}{2}}f\right\|_{L^{p}(\mathbb{R}^{d})}.$
(1.3)

We also abbreviate
$\dot{H}_{c}^{s}(\mathbb{R}^{d})=\dot{H}_{c}^{s,2}(\mathbb{R}^{d})$ and
$H_{c}^{s}(\mathbb{R}^{d})=H_{c}^{s,2}(\mathbb{R}^{d})$. If there is no
confusion, $\mathbb{R}^{d}$ will be omitted in various function spaces. Note
that by definition, we have

$\left\|f\right\|_{\dot{H}_{c}^{1}}=\int{|\nabla
u(x)|^{2}+c|x|^{-2}|u(x)|^{2}dx.}$ (1.4)

By sharp Hardy inequality (1.2), we see that

$\left\|f\right\|_{\dot{H}_{c}^{1}}\sim\left\|f\right\|_{\dot{H}^{1}}\leavevmode\nobreak\
\textrm{for}\leavevmode\nobreak\ c>-c(d).$

The INLSc equation (1.1) is invariant under the scaling,

$u_{\lambda}(t,x):=\lambda^{\frac{2-b}{\sigma}}u\left(\lambda^{2}t,\lambda
x\right),\leavevmode\nobreak\ \lambda>0.$

An easy computation shows that

$\left\|u_{\lambda}(0)\right\|_{\dot{H}^{s}}=\lambda^{s-\frac{d}{2}+\frac{2-b}{\sigma}}\left\|u_{0}\right\|_{\dot{H}^{s}},$

which implies that the critical Sobolev index is given by

$s_{c}:=\frac{d}{2}-\frac{2-b}{\sigma}.$

Note that, if $s_{c}=0$ (alternatively $\sigma=\frac{4-2b}{d}$) the problem is
known as the mass-critical or $L^{2}$-critical; if $s_{c}=1$ (alternatively
$\sigma=\frac{4-2b}{d-2}$) it is called energy-critical or
$\dot{H}^{1}$-critical. The problem is known as intercritical (mass-
supercritical and energy-subcritical) if $0<s_{c}<1$ (alternatively
$\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$). On the other hand, solutions to the
INLSc equation (1.1) conserve the mass and energy, defined respectively by

$M\left(u(t)\right):=\int_{\mathbb{R}^{d}}{\left|u(t,x)\right|^{2}dx},$
$E\left(u(t)\right):=\int_{\mathbb{R}^{d}}{\frac{1}{2}\left|\nabla
u(t,x)\right|^{2}+\frac{c}{2}|x|^{-2}\left|u(t,x)\right|^{2}-\frac{1}{\sigma+2}|x|^{-b}\left|u(t,x)\right|^{\sigma+2}dx}.$

For later uses, it is convenient to introduce the following exponents:

$\sigma_{c}:=\frac{1-s_{c}}{s_{c}}=\frac{4-2b-(d-2)\sigma}{d\sigma-4+2b},$
(1.5) $H_{G}:=E(Q)[M(Q)]^{\sigma_{c}},\leavevmode\nobreak\
K_{G}:=\left\|Q\right\|_{\dot{H}_{c}^{1}}\left\|Q\right\|_{L^{2}}^{\sigma_{c}},$
(1.6)

where $Q$ is a positive solution to the ground state equation:

$P_{c}Q+Q-|x|^{-b}|Q|^{\sigma}Q=0.$ (1.7)

Note that the existence of solution to the elliptic equation (1.7) was proved
by [7]. Moreover, it was proved that all possible solutions to (1.7) have the
same mass, the same $\dot{H}^{1}_{c}$ norm and the same energy. See Lemma 2.1.
For convenience, we also define the following quantities:

$M_{G}:=M(Q),\leavevmode\nobreak\ E_{G}:=E(Q),\leavevmode\nobreak\
S_{G}:=\left\|Q\right\|_{\dot{H}_{c}^{1}},\leavevmode\nobreak\
P_{G}:=\int{|x|^{-b}|Q|^{\sigma+2}dx},$ (1.8)

which are the same for all possible solutions to (1.7). The quantities
$H_{G}$, $K_{G}$, $M_{G}$, $E_{G}$, $S_{G}$ and $P_{G}$ defined in (1.6) and
(1.8) appear throughout the paper and play a important role in our work.

Let us recall the known results for the INLSc equation (1.1). Using the energy
method, Suzuki [19] showed that if 111 Note that the author in [19] considered
(1.1) with $c=-c(d)$. The authors in [7] pointed out that the proof for the
case $c>-c(d)$ is an immediate consequence of the previous one. $d\geq 3$,
$0<\sigma<\frac{4-2b}{d-2}$, $c>-c(d)$ and $0<b<2$, then the INLSc equation
(1.1) is locally well-posed in $H^{1}_{c}$. Recently, Campos-Guzmán [7]
established the sufficient conditions for global existence and blow-up of
$H^{1}_{c}$-solution to (1.1) with $d\geq 3$ and
$\frac{4-2b}{d}\leq\sigma<\frac{4-2b}{d-2}$. They also studied the local well-
posedness and small data global well-posedness under some assumption on $b$
and $c$ in the energy-subcritical case $\sigma<\frac{4-2b}{d-2}$ with $d\geq
3$, by using the standard Strichartz estimates combined with the fixed point
argument. Furthermore, they showed a scattering criterion and construct a wave
operator in $H^{1}_{c}$, for the intercritical case. Later, the authors in
[12] establish the local well-posedness as well as small data global well-
posedness and scattering in $H_{c}^{1}$ for the energy-critical INLSc equation
(1.1) with $d\geq 3$, $c>-\frac{(d+2-2b)^{2}-4}{(d+2-2b)^{2}}c(d)$,
$0<b<\frac{4}{d}$ and $\sigma=\frac{4-2b}{d-2}$. Furthermore, they establish
the blow-up criteria for radial or finite variance data.

In this paper, we study the global existence and blow-up of
$H_{c}^{1}$-solutions to the focusing INLSc equation (1.1) in the
intercritical case $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$. As mentioned
above, the global existence and blow-up for the focusing intercritical INLSc
equation (1.1) were already studied by [7]. More precisely, they proved that
the focusing intercritical INLSc equation (1.1) with $d\geq 3$, $0<b<2$ and
$c>-c(d)$ is globally well-posed in $H_{c}^{1}$ if $u_{0}\in H_{c}^{1}$
satisfies

$E(u_{0})[M(u_{0})]^{\sigma_{c}}<H_{G},\leavevmode\nobreak\
\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}<K_{G}.$

They also prove that if

$E(u_{0})[M(u_{0})]^{\sigma_{c}}<H_{G},\leavevmode\nobreak\
\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}>K_{G},$

and, in addition, if $|x|u_{0}\in L^{2}$ or $u_{0}$ is radial, then the
solution blows up in finite time.

As we can see, the global existence and blow up for the focusing intercritical
INLSc equation (1.1) were studied only for data below the ground state
threshold. In particular, the blow-up result was shown only for radial or
finite variance data. Recently, Dinh-Keraani [10] systematically studied the
long time dynamics of solutions to the focusing intercritical INLS equation
(i.e. (1.1) with $c=0$) with general (not necessarily radial or finite
variance) data lying below, at, and above the ground state threshold.

Motivated by [7, 10], this paper aims to study the global existence and blow-
up for the focusing intercritical INLSc equation with general (not necessarily
radial or finite variance) data lying below, at, and above the ground state
threshold. Our results can be seen as the extension of the global existence
and blow-up results of [10] for (1.1) with $c=0$ to the focusing intercritical
INLSc equation (1.1) with $c>-c(d)$. Our results also extend the ones of [7]
by treating the general data lying not only below the ground state threshold
but also at, and above the ground state threshold.

To this end, we have the following criteria for global existence and blow-up
of general solutions to (1.1).

######  Theorem 1.1.

Let $d\geq 3$, $0<b<2$, $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$ and
$c>-c(d)$. Let $u$ be the solution to (1.1) defined on the maximal forward
time interval of existence $[0,T^{*})$.

  1. 1.

(Global existence) If

$\sup_{t\in[0,T^{*})}{P(u(t))[M(u)]^{\sigma_{c}}}<P_{G}[M_{G}]^{\sigma_{c}},$
(1.9)

then $T^{*}=\infty$, where

$P(f):=\int{|x|^{-b}|f(x)|^{\sigma+2}dx}.$ (1.10)

  2. 2.

(Blow-up) Assume that

$\sup_{t\in[0,T^{*})}{G(u(t))}\leq-\delta$ (1.11)

for some $\delta>0$, where

$G(f):=\left\|f\right\|_{\dot{H}_{c}^{1}}^{2}-\frac{d\sigma+2b}{2(\sigma+2)}P(f).$
(1.12)

Then either $T^{*}<\infty$ or $T^{*}=\infty$ and there exists a time sequence
$t_{n}\to\infty$ such that
$\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to\infty$ as $n\to\infty$.
Moreover, if we assume in addition that $u$ has finite variance, i.e.
$|x|u(t)\in L^{2}$ for all $t\in[0,T^{*})$, then $T^{*}<\infty$.

A similar statement holds for negative times and we omit the details.

Using Theorem 1.1, we have the following criteria for global existence and
blow-up below the ground state threshold.

######  Theorem 1.2 (Below the ground state threshold).

Let $d\geq 3$, $0<b<2$, $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$ and
$c>-c(d)$. Let $u_{0}\in H_{c}^{1}$ satisfy

$E(u_{0})[M(u_{0})]^{\sigma_{c}}<H_{G}.$ (1.13)

  1. 1.

If $u_{0}$ satisfies

$\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}<K_{G},$
(1.14)

then the corresponding solution to (1.1) satisfies

$\sup_{t\in(-T_{*},\;T^{*})}{P(u(t))[M(u)]^{\sigma_{c}}}<P_{G}[M_{G}]^{\sigma_{c}}.$
(1.15)

In particular, the solution exists globally in time.

  2. 2.

If $u_{0}$ satisfies

$\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}>K_{G},$
(1.16)

then the corresponding solution to (1.1) satisfies

$\sup_{t\in(-T_{*},\;T^{*})}{G(u(t))}\leq-\delta,$ (1.17)

for some $\delta>0$. In particular, the solution either blows up in finite
time, or there exists a time sequence $(t_{n})_{n\geq 1}$ satisfying
$|t_{n}|\to\infty$ such that
$\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to\infty$ as $n\to\infty$.
Moreover, if we assume in addition that

     * •

$u_{0}$ has finite variance,

     * •

or $u_{0}$ is radially symmetric,

     * •

or $\sigma\leq 2$, and $u_{0}\in\Sigma_{d}$, where

$\Sigma_{d}:=\left\\{f\in H_{c}^{1}:\;f(y,x_{d})=f(|y|,x_{d}),\;x_{d}f\in
L^{2}\right\\}$ (1.18)

with $x=(y,x_{d})$, $y=(x_{1},\cdots,x_{d-1})\in\mathbb{R}^{d-1}$, and
$x_{d}\in\mathbb{R}$,

then the corresponding solution blows up in finite time, i.e.,
$T_{*},\;T^{*}<\infty$.

######  Remark 1.3.

The authors in [7] proved the finite time blow-up for radial or finite
variance data below the ground state threshold. Theorem 1.2 establishes the
blow-up criteria for general (not necessarily radial or finite variance) data.
In particular, the finite time blow-up for cylindrically symmetric data is
also proved.

We also have the following criteria for global existence and blow-up at the
ground state threshold.

######  Theorem 1.4 (At the ground state threshold).

Let $d\geq 3$, $0<b<2$, $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$ and
$c>-c(d)$. Let $u_{0}\in H_{c}^{1}$ satisfy

$E(u_{0})[M(u_{0})]^{\sigma_{c}}=H_{G}.$ (1.19)

  1. 1.

If $u_{0}$ satisfies

$\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}<K_{G},$
(1.20)

then the corresponding solution to (1.1) exists globally in time. Moreover,
the solution either satisfies

$\sup_{t\in\mathbb{R}}{P(u(t))[M(u)]^{\sigma_{c}}}<P_{G}[M_{G}]^{\sigma_{c}}.$
(1.21)

or there exists a time sequence $(t_{n})_{n\geq 1}$ satisfying
$|t_{n}|\to\infty$ such that

$u(t_{n})\to e^{i\theta}Q\leavevmode\nobreak\ \textrm{strongly
in}\leavevmode\nobreak\ H_{c}^{1}\leavevmode\nobreak\
\textrm{as}\leavevmode\nobreak\ n\to\infty,$ (1.22)

for some $\theta\in\mathbb{R}$ and some positive solution $Q$ to the ground
state equation (1.7).

  2. 2.

If $u_{0}$ satisfies

$\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}=K_{G},$
(1.23)

then $u(t,x)=e^{it}e^{i\theta}Q(x)$ for some $\theta\in\mathbb{R}$ and some
positive solution $Q$ to (1.7).

  3. 3.

If $u_{0}$ satisfies

$\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}\left\|u_{0}\right\|_{L^{2}}^{\sigma_{c}}>K_{G},$
(1.24)

then the corresponding solution to (1.1)

    1. (a)

either blows up forward in time, i.e., $T^{*}<\infty$

    2. (b)

or there exists a time sequence $t_{n}\to\infty$ such that
$\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to\infty$ as $n\to\infty$,

    3. (c)

or there exists a time sequence $t_{n}\to\infty$ such that (1.22) holds.

Moreover, if we assume in addition that

     * •

$u_{0}$ has finite variance,

     * •

or $u_{0}$ is radially symmetric,

     * •

or $\sigma\leq 2$, and $u_{0}\in\Sigma_{d}$,

then the possibility in Item (b) can be excluded.

Finally, we study the global existence and blow-up above the ground state
threshold.

######  Theorem 1.5 (Above the ground state threshold).

Let $d\geq 3$, $0<b<2$, $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$, $c>-c(d)$
and

$V(t):=\int{|x|^{2}\left|u(t,x)\right|^{2}dx}.$ (1.25)

Let $u_{0}\in H_{c}^{1}$ be such that $|x|u_{0}\in L^{2}$. Assume that

$E(u_{0})[M(u_{0})]^{\sigma_{c}}\geq H_{G},$ (1.26)
$\frac{E(u_{0})[M(u_{0})]^{\sigma_{c}}}{E_{G}[M_{G}]^{\sigma_{c}}}\left[1-\frac{(V^{\prime}(0))^{2}}{32E(u_{0})V(0)}\right]\leq
1.$ (1.27)

  1. 1.

If

$P(u_{0})[M(u_{0})]^{\sigma_{c}}<P_{G}[M_{G}]^{\sigma_{c}},$ (1.28)
$V^{\prime}(0)\geq 0,$ (1.29)

then the corresponding solution to (1.1) satisfies (1.9). In particular, the
solution exists globally in time.

  2. 2.

If

$P(u_{0})[M(u_{0})]^{\sigma_{c}}>P_{G}[M_{G}]^{\sigma_{c}},$ (1.30)
$V^{\prime}(0)\leq 0,$ (1.31)

then the corresponding solution to (1.1) blows up forward in finite time,
i.e., $T^{*}<\infty$.

######  Remark 1.6.

To our knowledge, Theorem 1.4 and Theorem 1.5 are the first global existence
and blow-up results for the focusing intercritical INLSc equation (1.1) with
data lying at, and above the ground state threshold.

######  Remark 1.7.

Theorem 1.3, Theorem 1.4 and Theorem 1.5 extend the global existence and blow-
up results of [7] for focusing INLS equation (i.e. (1.1) with $c=0$) to the
focusing INLSc equation (1.1) with $c>-c(d)$.

This paper is organized as follows. In Section 2, we prove Theorem 1.1. In
Section 3, we study the global existence and blow up of $H_{c}^{1}$-solutions
lying below, at, and above the ground state threshold.

##  2 Criteria for global existence and blow-up

In this section, we prove Theorem 1.1. Throughout the paper, $C>0$ will denote
positive universal constant, which can be different at different places.
$a\lesssim b$ means $a\leq Cb$ for some constant $C>0$. We also write $a\sim
b$ if $a\lesssim b\lesssim a$.

###  2.1 Variational Analysis

In this subsection, we recall the sharp Gagliardo–Nirenberg inequality related
the focusing INLSc equation (1.1). First of all, we recall the existence of a
ground state.

######  Lemma 2.1 (Existence of a ground state, [7]).

Let $d\geq 3$, $0<\sigma<\frac{4-2b}{d-2}$, $c>-c(d)$ and $0<b<2$. There
exists a positive solution to the elliptic equation

$P_{c}Q+Q-|x|^{-b}|Q|^{\sigma}Q=0$ (2.1)

in $H_{c}^{1}$. Moreover, all possible solutions have the same mass $M_{G}$,
the same $\dot{H}^{1}_{c}$ norm $S_{G}$ and the same energy $E_{G}$.

######  Lemma 2.2 (Sharp Gagliardo–Nirenberg inequality, [7]).

Let $d\geq 3$, $0<\sigma<\frac{4-2b}{d-2}$, $c>-c(d)$ and $0<b<2$. Then for
$f\in H_{c}^{1}$, we have

$P(f)\leq
C_{GN}\left\|f\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|f\right\|_{L^{2}}^{\frac{4-2b-\sigma(d-2)}{2}},$
(2.2)

where $P(f)$ is given in (1.10). The equality in (2.2) is attained by a
function $Q\in H_{c}^{1}$, which is a positive solution to the elliptic
equation (2.1).

######  Remark 2.3.

We also have the following Pohozaev identities:

$M_{G}=\frac{4-2b-(d-2)\sigma}{d\sigma+2b}[S_{G}]^{2}=\frac{4-2b-(d-2)\sigma}{2(\sigma+2)}P_{G}.$
(2.3)

In particular, we have

$C_{GN}=\frac{2(\sigma+2)}{d\sigma+2b}[K_{G}]^{-\frac{d\sigma-4+2b}{2}}.$
(2.4)

We also have

$E_{G}=\frac{d\sigma-4+2b}{2(d\sigma+2b)}[S_{G}]^{2}=\frac{d\sigma-4+2b}{4(\sigma+2)}P_{G},$
(2.5)

which implies

$H_{G}=\frac{d\sigma-4+2b}{2(d\sigma+2b)}[K_{G}]^{2}.$ (2.6)

###  2.2 Virial Identities

Given a real valued function $a$, we define the virial potential by

$V_{a}(t):=\int{a(x)\left|u(t,x)\right|^{2}dx.}$

A simple computation shows that the following result holds.

######  Lemma 2.4 ([9]).

Let $d\geq 3$ and $c>-c(d)$. If $u:I\times\mathbb{R}^{d}\to\mathbb{C}$ is a
smooth-in-time and Schwartz-in-space solution to $iu_{t}-P_{c}u=N(u)$, with
$N(u)$ satisfying $\textnormal{Im}(N(u)\bar{u})=0$, then we have for any $t\in
I$,

$\frac{d}{dt}V_{a}(t)=2\int{\nabla a(x)\cdot\textnormal{Im}(\bar{u}(t,x)\nabla
u(t,x))dx},$

and

$\displaystyle\begin{split}\frac{d^{2}}{dt^{2}}V_{a}(t)&=-\int{\Delta^{2}a(x)|u(t,x)|^{2}dx+4\sum_{j,k=1}^{d}{\int{\partial_{jk}^{2}a(x)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}}\\\
&+4c\int{\nabla a(x)\cdot\frac{x}{|x|^{4}}|u(t,x)|^{2}dx}+2\int{\nabla
a(x)\cdot\left\\{N(u),u\right\\}_{p}(t,x)dx},\end{split}$

where
$\left\\{f,g\right\\}_{p}:=\textnormal{Re}(f\nabla\bar{g}-g\nabla\bar{f})$ is
the momentum bracket.

Note that if $N(u)=-|x|^{-b}|u|^{\sigma}u$, then

$\left\\{N(u),u\right\\}_{p}=\frac{\sigma}{\sigma+2}\nabla(|x|^{-b}|u|^{\sigma+2})+\frac{2}{\sigma+2}\nabla(|x|^{-b})|u|^{\sigma+2}.$

Hence, we immediately have the following result.

######  Corollary 2.5.

If $u$ is a smooth-in-time and Schwartz-in-space solution to the focusing
INLSc equation, then we have for any $t\in I$,

$\displaystyle\begin{split}\frac{d^{2}}{dt^{2}}V_{a}(t)&=-\int{\Delta^{2}a(x)|u(t,x)|^{2}dx+4\sum_{j,k=1}^{d}{\int{\partial_{jk}^{2}a(x)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}}\\\
&+4c\int{\nabla
a(x)\cdot\frac{x}{|x|^{4}}|u(t,x)|^{2}dx}-\frac{2\sigma}{\sigma+2}\int{\Delta
a(x)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\ &+\frac{4}{\sigma+2}\int{\nabla
a(x)\cdot\nabla(|x|^{-b})|u(t,x)|^{\sigma+2}dx}.\end{split}$

We have the following standard virial identity.

######  Lemma 2.6 (Standard Virial Identity, [12]).

Let $d\geq 3$, $c>-c(d)$. Let $u_{0}\in H_{c}^{1}$ be such that $|x|u_{0}\in
L^{2}$ and $u:I\times\mathbb{R}^{d}\to\mathbb{C}$ be the corresponding
solution to the focusing INLSc equation. Then, $|x|u\in
C\left(I,\;L^{2}\right)$. Moreover, for any $t\in I$,

$\frac{d^{2}}{dt^{2}}\left\|xu(t)\right\|_{L^{2}}^{2}=8G(u(t)),$

where $G(\cdot)$ is given in (1.12).

Let us introduce a function $\theta:[0,\infty)\to[0,\infty)$ satisfying

$\theta(r)=\left\\{\begin{array}[]{l}{r^{2},\leavevmode\nobreak\
\textrm{if}\;0\leq r\leq 1,}\\\ {\textrm{0},\leavevmode\nobreak\
\textrm{if}\;r\geq 2,}\end{array}\right.\textrm{and}\leavevmode\nobreak\
\leavevmode\nobreak\ \theta^{\prime\prime}(r)\leq 2\leavevmode\nobreak\
\leavevmode\nobreak\ \textrm{for}\leavevmode\nobreak\ \leavevmode\nobreak\
r\geq 0.$ (2.7)

For $R>1$, we define the radial function
$\varphi_{R}:\mathbb{R}^{d}\to[0,\;\infty)$:

$\varphi_{R}(x)=\varphi_{R}(r):=R^{2}\theta(r/R),\;r=|x|.$ (2.8)

One can easily see that

$2-\varphi^{\prime\prime}_{R}(r)\geq
0,\;2-\frac{\varphi^{\prime}_{R}(r)}{r}\geq 0,\;2d-\Delta\varphi_{R}(x)\geq
0.$ (2.9)

We have the following localized virial estimate.

######  Lemma 2.7 (Localized Virial Estimate).

Let $d\geq 3$, $0<b<2$, $c>-c(d)$, $R>1$ and $\varphi_{R}$ be as in (2.8). Let
$u:I\times\mathbb{R}^{d}\to\mathbb{C}$ be a solution to the focusing INLSc
equation (1.1). Then for any $t\in I$,

$\frac{d^{2}}{dt^{2}}V_{\varphi_{R}}(t)\leq
8G(u(t))+CR^{-2}+CR^{-b}\left\|u(t)\right\|_{H_{c}^{1}}^{\sigma+2},$ (2.10)

where $G$ is given in (1.12).

###### Proof.

Applying Corollary 2.5 with $a(x)=\varphi_{R}(x)$, we have

$\displaystyle\begin{split}\frac{d^{2}}{dt^{2}}V_{\varphi_{R}}(t)&=-\int{\Delta^{2}\varphi_{R}(x)|u(t,x)|^{2}dx}+4\sum_{j,k=1}^{d}{\int{\partial_{jk}^{2}\varphi_{R}(x)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}\\\
&+4c\int{\nabla\varphi_{R}(x)\cdot\frac{x}{|x|^{4}}|u(t,x)|^{2}dx}-\frac{2\sigma}{\sigma+2}\int{\Delta\varphi_{R}(x)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&+\frac{4}{\sigma+2}\int{\nabla\varphi_{R}(x)\cdot\nabla(|x|^{-b})|u(t,x)|^{\sigma+2}dx}.\end{split}$
(2.11)

Noticing that

$\partial_{j}=\frac{x_{j}}{r}\partial_{r},\leavevmode\nobreak\
\partial^{2}_{jk}=\left(\frac{\delta_{jk}}{r}-\frac{x_{j}x_{k}}{r^{3}}\right)\partial_{r}+\frac{x_{j}x_{k}}{r^{2}}\partial^{2}_{r},$

we have

$\displaystyle\begin{split}&\sum_{j,k=1}^{d}{\int{\partial_{jk}^{2}\varphi_{R}(x)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
=\int{\frac{\varphi^{\prime}_{R}(r)}{r}\left|\nabla
u(t,x)\right|^{2}dx}+\int{\left(\frac{\varphi^{\prime\prime}_{R}(r)}{r^{2}}-\frac{\varphi^{\prime}_{R}(r)}{r^{3}}\right)\left|x\cdot\nabla
u(t,x)\right|^{2}dx.}\end{split}$ (2.12)

In view of (2.11) and (LABEL:GrindEQ__2_12_), we have

$\displaystyle\begin{split}\frac{d^{2}}{dt^{2}}V_{\varphi_{R}}(t)&=8G(u(t))-8\left\|u(t)\right\|_{\dot{H}_{c}^{1}}+\frac{4(d\sigma+2b)}{\sigma+2}\int{|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&-\int{\Delta^{2}\varphi_{R}(x)|u(t,x)|^{2}dx}+4\int{\frac{\varphi^{\prime}_{R}(r)}{r}\left|\nabla
u(t,x)\right|^{2}dx}\\\
&+4\int{\left(\frac{\varphi^{\prime\prime}_{R}(r)}{r^{2}}-\frac{\varphi^{\prime}_{R}(r)}{r^{3}}\right)\left|x\cdot\nabla
u(t,x)\right|^{2}dx}+4c\int{\frac{\varphi^{\prime}_{R}(r)}{r^{3}}|u(t,x)|^{2}dx}\\\
&-\frac{2\sigma}{\sigma+2}\int{\Delta\varphi_{R}(x)|x|^{-b}|u(t,x)|^{\sigma+2}dx}-\frac{4b}{\sigma+2}\int{|x|^{-b}\frac{\varphi^{\prime}(r)}{r}|u(t,x)|^{\sigma+2}dx}.\end{split}$
(2.13)

Since $\left\|\Delta^{2}\varphi_{R}\right\|_{L^{\infty}}\lesssim R^{-2}$, the
conservation of mass implies that

$\left|\int{\Delta^{2}\varphi_{R}(x)|u(t,x)|^{2}dx}\right|\lesssim
R^{-2}\left\|u(t)\right\|_{L^{2}}^{2}\lesssim R^{-2}.$ (2.14)

Using the conservation of mass and the following facts

$\left|x\cdot\nabla u\right|\leq r|\nabla u|,\leavevmode\nobreak\
\varphi^{\prime\prime}_{R}(r)\leq 2,\leavevmode\nobreak\
\frac{\varphi^{\prime}_{R}(r)}{r}\leq 2,$

we see that

$\displaystyle\begin{split}&4\int{\frac{\varphi^{\prime}_{R}(r)}{r}\left|\nabla
u(t)\right|^{2}dx}+4\int{\left(\frac{\varphi^{\prime\prime}_{R}(r)}{r^{2}}-\frac{\varphi^{\prime}_{R}(r)}{r^{3}}\right)\left|x\cdot\nabla
u(t)\right|^{2}dx}+4c\int{\frac{\varphi^{\prime}_{R}(r)}{r^{3}}|u(t)|^{2}dx}-8\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leq
4\int{\left(2-\frac{\varphi^{\prime}_{R}(r)}{r}\right)\left(-\left|\nabla
u(t)\right|^{2}+\frac{\left|x\cdot\nabla
u(t)\right|^{2}}{r^{2}}-c\frac{|u(t)|}{r^{2}}\right)dx}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
\leq-4c\int{\left(2-\frac{\varphi^{\prime}_{R}(r)}{r}\right)\frac{|u(t)|}{r^{2}}dx}=-4c\int_{|x|>R}{\left(2-\frac{\varphi^{\prime}_{R}(r)}{r}\right)\frac{|u(t)|}{|x|^{2}}dx}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
\leq\max\left\\{-4cS,0\right\\}\int_{|x|>R}{R^{-2}|u(t)|^{2}dx}\leq\max\left\\{-4cSM(u(t)),0\right\\}R^{-2}\lesssim
R^{-2},\end{split}$ (2.15)

where $S=\max_{r\geq 1}{2-\frac{\theta^{\prime}(r)}{r}}$. Furthermore, we have

$\displaystyle\begin{split}&\frac{4(d\sigma+2b)}{\sigma+2}\int{|x|^{-b}|u(t,x)|^{\sigma+2}dx}-\frac{2\sigma}{\sigma+2}\int{\Delta\varphi_{R}(x)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
-\frac{4b}{\sigma+2}\int{|x|^{-b}\frac{\varphi^{\prime}(r)}{r}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\
=\frac{2\sigma}{\sigma+2}\int{|x|^{-b}\left(2d-\Delta\varphi_{R}(x)\right)|u(t)|^{\sigma+2}dx}+\frac{4b}{\sigma+2}\int{|x|^{-b}\left(2-\frac{\varphi^{\prime}_{R}(r)}{r}\right)|u(t)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leq C\int_{|x|\geq
R}{|x|^{-b}|u(t)|^{\sigma+2}}\leq
CR^{-b}\left\|u(t)\right\|_{L^{\sigma+2}}^{\sigma+2}\leq
CR^{-b}\left\|u(t)\right\|_{H_{c}^{1}}^{\sigma+2},\end{split}$ (2.16)

where the last inequality follows from the Sobolev embedding and the fact
$H_{c}^{1}\sim H^{1}$ as $\sigma<\frac{4-2b}{d-2}$. In view of
(2.13)–(LABEL:GrindEQ__2_16_), we get the desired result. ∎

###  2.3 Proof of Theorem 1.1

First we prove Item 1. Let $u:[0,T^{*})\times\mathbb{R}^{d}\to\mathbb{C}$ be a
$H_{c}^{1}$-solution to (1.1) satisfying (1.9). It follows from (1.9) and the
conservation of mass and energy that

$\sup_{t\in[0,T^{*})}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}\leq M,$ (2.17)

for some $M<\infty$. This shows that $T^{*}=\infty$.

Next, we prove Item 2. Let $u:[0,T^{*})\times\mathbb{R}^{d}\to\mathbb{C}$ be a
solution to (1.1) satisfying (1.11). If $T^{*}<\infty$, then we are done. If
$T^{*}=\infty$, then we have to show that there exists $t_{n}\to\infty$ such
that $\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to\infty$ as $n\to\infty$.
Assume by contradiction that it doesn’t hold, i.e.
$\sup_{t\in[0,\infty)}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}\leq M_{0}$ for
some $M_{0}>0$. By the conservation of mass, we have

$\sup_{t\in[0,\infty)}{\left\|u(t)\right\|_{H_{c}^{1}}}\leq M_{1},$

for some $M_{1}>0$. Hence, it follows from Lemma 2.7 (localized virial
estimate) and (1.11) that

$V^{\prime\prime}_{\varphi_{R}}(t)\leq
8G(u(t))+CR^{-2}+CR^{-b}\left\|u(t)\right\|_{H_{c}^{1}}^{\sigma+2}\leq-8\delta+CR^{-2}+CR^{-b}M_{1}^{\sigma+2},$

for all $t\in[0,\infty)$. By taking $R>1$ large enough, we have for all
$t\in[0,\infty)$,

$V^{\prime\prime}_{\varphi_{R}}(t)\leq-4\delta.$

Integrating this estimate, there exists $t_{0}>0$ sufficiently large such that
$V_{\varphi_{R}}(t_{0})<0$ which is impossible. This completes the proof of
the first part of Item 2. Let us assume that $u$ has finite variance. Then it
follows from (1.11) and Lemma 2.6 that

$\frac{d^{2}}{dt^{2}}\left\|xu(t)\right\|_{L^{2}}^{2}=8G(u(t))\leq-8\delta.$

By the classical argument of Glassey [11] the solution must blow up in finite
time. This completes the proof of Theorem 1.1.

##  3 Global existence and blow up below, at, and above the ground state
threshold

In this section, we prove Theorem 1.2, Theorem 1.4 and Theorem 1.5.

###  3.1 Below the ground state threshold

In this subsection, we study the global existence and blow-up below the ground
state threshold.

###### Proof of Theorem 1.2.

(1.13) shows that there exists $\delta_{0}>0$ such that

$E(u_{0})[M(u_{0})]^{\sigma_{c}}\leq(1-\delta_{0})H_{G}.$ (3.1)

Proof of Item 1. Let $u_{0}\in H_{c}^{1}$ satisfy (1.13) and (1.14). It
follows from Lemma 3.5 of [7], (3.1) and the conservation of mass and energy
that there exists $\delta>0$ such that

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\left\|u(t)\right\|_{L^{2}}^{\sigma_{c}}\leq(1-\delta)K_{G},$
(3.2)

for all $t\in(-T_{*},T^{*})$. Lemma 2.2 and (3.2) yield

$\displaystyle\begin{split}P(u(t))[M(u(t))]^{\sigma_{c}}&\leq
C_{GN}\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|u(t)\right\|_{L^{2}}^{\frac{4-2b-(d-2)\sigma}{2}+2\sigma_{c}}\\\
&=C_{GN}\left(\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\left\|u(t)\right\|_{L^{2}}^{\sigma_{c}}\right)^{\frac{d\sigma+2b}{2}}\\\
&\leq
C_{GN}(1-\delta)^{\frac{d\sigma+2b}{2}}[K_{G}]^{\frac{d\sigma+2b}{2}},\end{split}$

for all $t\in(-T_{*},T^{*})$. By (2.4) and (2.3), we have

$P(u(t))[M(u(t))]^{\sigma_{c}}\leq\frac{2(\sigma+2)}{d\sigma+2b}(1-\delta)^{\frac{d\sigma+2b}{2}}[K_{G}]^{2}=(1-\delta)^{\frac{d\sigma+2b}{2}}P_{G}[M_{G}]^{\sigma_{c}},$

for all $t\in(-T_{*},T^{*})$ which shows (1.15). By Theorem 1.1, the solution
exists globally in time.

Proof of Item 2. Let $u_{0}\in H_{c}^{1}$ satisfy (1.13) and (1.16). Lemma 3.5
of [7] also shows that

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\left\|u(t)\right\|_{L^{2}}^{\sigma_{c}}>K_{G},$
(3.3)

for all $t\in(-T_{*},T^{*})$. By (2.6), (3.1), (3.3) and the conservation of
mass and energy, we have

$\displaystyle\begin{split}G(u(t))[M(u(t))]^{\sigma_{c}}&=\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{2}\left\|u(t)\right\|_{L^{2}}^{2\sigma_{c}}-\frac{d\sigma+2b}{2(\sigma+2)}P(u(t))[M(u(t))]^{\sigma_{c}}\\\
&=\frac{d\sigma+2b}{2}E(u(t))[M(u(t))]^{\sigma_{c}}-\frac{d\sigma-4+2b}{4}\left(\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\left\|u(t)\right\|_{L^{2}}^{\sigma_{c}}\right)^{2}\\\
&\leq\frac{d\sigma+2b}{2}(1-\delta_{0})H_{G}-\frac{d\sigma-4+2b}{4}[K_{G}]^{2}\\\
&=-\frac{d\sigma-4+2b}{4}\delta_{0}[K_{G}]^{2},\end{split}$

for all $t\in(-T_{*},T^{*})$. This shows (1.11) with

$\delta:=\frac{(d\sigma-4+2b)\delta_{0}[K_{G}]^{2}}{4[M(u_{0})]^{\sigma_{c}}}>0.$

By Theorem 1.1, the corresponding solution either blows up in finite time, or
there exists a time sequence $(t_{n})_{n\geq 1}$ satisfying $|t_{n}|\to\infty$
such that $\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to\infty$ as
$n\to\infty$. This completes the proof of the first part of Item 2.

Let us prove the second part of Item 2. If $u_{0}$ either has finite variance
or is radially symmetric, then the solution blows up in finite time. This
result was proved in [7]. It remains to consider the case of cylindrically
symmetric data. For $R>1$, we define the radial function
$\psi_{R}:\mathbb{R}^{d-1}\to[0,\infty)$:

$\psi_{R}(y)=\psi_{R}(r):=R^{2}\theta(r/R),\;r=|y|,$ (3.4)

where $\theta:[0,\infty)\to[0,\infty)$ is given in (2.7). We define the
function on $\mathbb{R}^{d}$:

$\phi_{R}(x)=\phi_{R}(y,x_{d}):=\psi_{R}(y)+x_{d}^{2}.$ (3.5)

Using the same argument as in the proof of Lemma 2.7, we have

$\displaystyle\begin{split}V^{\prime\prime}_{\phi_{R}}(t)&=-\int{\Delta_{y}^{2}\psi_{R}(y)|u(t,x)|^{2}dx}+4\sum_{j,k=1}^{d-1}{\int{\partial_{jk}^{2}\psi_{R}(y)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}\\\
&\leavevmode\nobreak\
+4c\int{\nabla_{y}\psi_{R}(y)\cdot\frac{y}{|x|^{4}}|u(t,x)|^{2}dx}-\frac{2\sigma}{\sigma+2}\int{\Delta_{y}\psi_{R}(y)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\
-\frac{4b}{\sigma+2}\int{|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}|x|^{-b-2}|u(t,x)|^{\sigma+2}dx}+8\left\|\partial_{d}u(t)\right\|_{L^{2}}^{2}-\frac{4\sigma}{\sigma+2}\int{|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\
-\frac{8b}{\sigma+2}\int{x_{d}^{2}|x|^{-b-2}|u(t,x)|^{\sigma+2}dx}+8c\int{x_{d}^{2}\frac{|u(t,x)|^{2}}{|x|^{4}}dx.}\\\
\end{split}$

We can rewrite it as

$\displaystyle\begin{split}V^{\prime\prime}_{\phi_{R}}(t)&=8G(u(t))-8\left\|\nabla_{y}u(t)\right\|_{L^{2}}^{2}+\frac{4((d-1)\sigma+2b)}{\sigma+2}P(u(t))\\\
&\leavevmode\nobreak\
-\int{\Delta_{y}^{2}\psi_{R}(y)|u(t,x)|^{2}dx}+4\sum_{j,k=1}^{d-1}{\int{\partial_{jk}^{2}\psi_{R}(y)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}\\\
&\leavevmode\nobreak\
+4c\int{|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}\frac{|u(t,x)|^{2}}{|x|^{4}}dx}-\frac{2\sigma}{\sigma+2}\int{\Delta_{y}\psi_{R}(y)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\
-\frac{4b}{\sigma+2}\int{|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}|x|^{-b-2}|u(t,x)|^{\sigma+2}dx}-\frac{8b}{\sigma+2}\int{x_{d}^{2}|x|^{-b-2}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\
+8c\int{x_{d}^{2}\frac{|u(t,x)|^{2}}{|x|^{4}}dx}-8c\int{|x|^{-2}|u(t,x)|^{2}dx}\\\
&=8G(u(t))-8\left\|\nabla_{y}u(t)\right\|_{L^{2}}^{2}+4\sum_{j,k=1}^{d-1}{\int{\partial_{jk}^{2}\psi_{R}(y)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}\\\
&\leavevmode\nobreak\
-\int{\Delta_{y}^{2}\psi_{R}(y)|u(t,x)|^{2}dx}+\frac{2\sigma}{\sigma+2}\int{\left(2(d-1)-\Delta_{y}\psi_{R}(y)\right)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\
+\frac{4b}{\sigma+2}\int{\left(2|x|^{2}-|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}-2x_{d}^{2}\right)|x|^{-b-2}|u(t,x)|^{\sigma+2}dx}\\\
&\leavevmode\nobreak\
-4c\int{\left(2|x|^{2}-|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}-2x_{d}^{2}\right)\frac{|u(t,x)|^{2}}{|x|^{4}}dx}.\end{split}$
(3.6)

Using the similar argument as in the proof of (LABEL:GrindEQ__2_15_), we can
get

$-8\left\|\nabla_{y}u(t)\right\|_{L^{2}}^{2}+4\sum_{j,k=1}^{d-1}{\int{\partial_{jk}^{2}\psi_{R}(y)\textnormal{Re}(\partial_{k}u(t,x)\partial_{j}\bar{u}(t,x))dx}}\leq
0$ (3.7) $\left|\int{\Delta_{y}^{2}\psi_{R}(y)|u(t,x)|^{2}dx}\right|\lesssim
R^{-2}.$ (3.8)

Similarly, we also have

$\left|\int{\left(2(d-1)-\Delta_{y}\psi_{R}(y)\right)|x|^{-b}|u(t,x)|^{\sigma+2}dx}\right|\lesssim\int_{|y|\geq
R}{|x|^{-b}|u(t,x)|^{\sigma+2}dx,}$ (3.9)
$\left|\int{\left(2|x|^{2}-|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}-2x_{d}^{2}\right)|x|^{-b-2}|u(t,x)|^{\sigma+2}dx}\right|\lesssim\int_{|y|\geq
R}{|x|^{-b}|u(t,x)|^{\sigma+2}dx},$ (3.10)
$\left|\int{\left(2|x|^{2}-|y|^{2}\frac{\psi^{\prime}_{R}(r)}{r}-2x_{d}^{2}\right)\frac{|u(t,x)|^{2}}{|x|^{4}}dx}\right|\lesssim\int_{|y|\geq
R}{|x|^{-2}|u(t,x)|^{2}dx}\lesssim R^{-2}.$ (3.11)

In view of (3.6)–(3.11), we have

$V^{\prime\prime}_{\phi}(t)\leq 8G(u(t))+CR^{-2}+CR^{-b}\int_{|y|\geq
R}{|u(t,x)|^{\sigma+2}dx.}$ (3.12)

Using the same argument as in the proof of (4.12) in [10], it follows from
(3.12) and the fact $\dot{H}_{c}^{1}\sim\dot{H}^{1}$ that

$V^{\prime\prime}_{\phi}(t)\leq
8G(u(t))+CR^{-2}+\left\\{\begin{array}[]{l}{CR^{-\frac{d-2}{2}-b}\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{2},\leavevmode\nobreak\
\textrm{if}\;\sigma=2,}\\\
{CR^{-\frac{(d-2)\sigma}{4}-b}\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{2}+CR^{-\frac{(d-2)\sigma}{4}-b},\leavevmode\nobreak\
\textrm{if}\;\sigma<2,}\end{array}\right.$ (3.13)

for all $t\in(-T_{*},T^{*})$. Lemma 3.6 of [7] and (3.2) also show that for
some $\eta>0$ and $\epsilon>0$,

$8G(u(t))+\epsilon\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{2}\leq-\eta.$ (3.14)

In view of (3.13) and (3.14), we take $R>1$ sufficiently large to get

$V^{\prime\prime}_{\phi}(t)\leq-\frac{\eta}{2}<0,$

for all $t\in(-T_{*},T^{*})$. This shows that $T_{*},T^{*}<\infty$. This
completes the proof. ∎

###  3.2 At the ground state threshold

In this subsection, we study the global existence and blow-up at the ground
state threshold.

######  Lemma 3.1 ([7]).

If $d\geq 3$, $c>-c(d)$, $0<b<2$ and $0<\sigma<\frac{4-2b}{d-2}$, then
$H_{c}^{1}$ is compactly embedded in $L^{\sigma+2}\left(|x|^{-b}dx\right)$.

######  Lemma 3.2.

Let $d\geq 3$, $c>-c(d)$, $0<b<2$ and $0<\sigma<\frac{4-2b}{d-2}$. Let
$(f_{n})_{n\geq 1}$ be a sequence of $H_{c}^{1}$ function satisfying

$M(f_{n})=M_{G},\leavevmode\nobreak\ E(f_{n})=E_{G},\leavevmode\nobreak\
\forall n\geq 1$

and

$\lim_{n\to\infty}\left\|f_{n}\right\|_{\dot{H}_{c}^{1}}=S_{G}.$

Then there exists a subsequence still denoted by $(f_{n})_{n\geq 1}$ such that

$f_{n}\to e^{i\theta}Q\leavevmode\nobreak\ \textrm{strongly
in}\leavevmode\nobreak\ H_{c}^{1}\leavevmode\nobreak\
\textrm{as}\leavevmode\nobreak\ n\to\infty$

for some $\theta\in\mathbb{R}$ and some positive solution $Q$ to the ground
state equation (1.7).

###### Proof.

Since $(f_{n})_{n\geq 1}$ is a bounded sequence in $H_{c}^{1}$, it follows
from Lemma 3.1 that there exists a subsequence still denoted by
$(f_{n})_{n\geq 1}$ and a function $f\in H_{c}^{1}$ such that $f_{n}\to f$
weakly in $H_{c}^{1}$ and $P(f_{n})\to P(f)$ as $n\to\infty$. Hence, it
follows from (2.5) that

$\displaystyle\begin{split}P(f)=\lim_{n\to\infty}{P(f_{n})}&=\lim_{n\to\infty}{(\sigma+2)\left(\frac{1}{2}\left\|f_{n}\right\|_{\dot{H}_{c}^{1}}^{2}-E(f_{n})\right)}\\\
&=(\sigma+2)\left(\frac{1}{2}[S_{G}]^{2}-E_{G}\right)\\\
&=\frac{2(\sigma+2)}{d\sigma+2b}[S_{G}]^{2}=P_{G}.\end{split}$

This shows that $f\neq 0$. Moreover, using Lemma 2.2, we have

$P(f)-C_{GN}\left\|f\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|f\right\|_{L^{2}}^{\frac{4-2b-\sigma(d-2)}{2}}\leq
0.$ (3.15)

By the lower continuity of weak convergence, we have

$\left\|f\right\|_{\dot{H}_{c}^{1}}\leq\liminf_{n\to\infty}{\left\|f_{n}\right\|_{\dot{H}_{c}^{1}}}.$

Hence, we have

$\displaystyle\begin{split}P(f)-C_{GN}\left\|f\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|f\right\|_{L^{2}}^{\frac{4-2b-\sigma(d-2)}{2}}&\geq\liminf_{n\to\infty}{P(f_{n})-C_{GN}\left\|f_{n}\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|f_{n}\right\|_{L^{2}}^{\frac{4-2b-\sigma(d-2)}{2}}}\\\
&=P_{G}-C_{GN}[S_{G}]^{\frac{d\sigma+2b}{2}}[M_{G}]^{\frac{4-2b-\sigma(d-2)}{4}}=0.\end{split}$
(3.16)

In view of (3.15) and (3.16), we see that $f$ is an optimizer for (2.2). We
also have

$\left\|f\right\|_{\dot{H}_{c}^{1}}=\lim_{n\to\infty}{\left\|f_{n}\right\|_{\dot{H}_{c}^{1}}},$

which implies that $f_{n}\to f$ strongly in $H_{c}^{1}$. We claim that there
exists $\theta\in\mathbb{R}$ such that $f(x)=e^{i\theta}g(x)$, where $g(x)$ is
a non-negative optimizer for (2.2). Indeed, since
$\left|\nabla|f|\right|\leq\left|\nabla f\right|$ for all $f\in H_{c}^{1}$, it
is clear that $|f|$ is also an optimizer for (2.2) and

$\left\||f|\right\|_{\dot{H}_{c}^{1}}=\left\|f\right\|_{\dot{H}_{c}^{1}}.$
(3.17)

Using (3.17) and the same argument as in the proof of Lemma 4.2 of [10], the
claim follows with $g(x)=|f(x)|$. Since $g$ is an optimizer for (2.2),
Euler–Lagrange equation for $g$ gives

$mP_{c}g+ng-\frac{\sigma+2}{C_{GN}}|x|^{-b}|g|^{\sigma}g=0,$

where

$m=\frac{d\sigma+2b}{2}\left\|f\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b-4}{2}}\left\|f\right\|_{L^{2}}^{\frac{4-2b-(d-2)\sigma}{2}},$
$n=\frac{4-2b-(d-2)\sigma}{2}\left\|f\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|f\right\|_{L^{2}}^{-\frac{2b+(d-2)\sigma}{2}}.$

By a change of variable $g(x)=\lambda Q(\mu x)$ with $\lambda,\mu>0$
satisfying

$\mu^{2}=\frac{n}{m},\lambda^{\sigma}=\frac{nC_{GN}}{\sigma+2}\mu^{-b},$

we can see that $Q$ solves (1.7). As
$\left\|g\right\|_{L^{2}}=\left\|Q\right\|_{L^{2}}$ and
$\left\|g\right\|_{\dot{H}_{c}^{1}}=\left\|Q\right\|_{\dot{H}_{c}^{1}}$, we
have $\lambda=\mu=1$. This completes the proof. ∎

###### Proof of Theorem 1.3.

First, we prove Item 1.

Proof of Item 1. Let $u_{0}\in H_{c}^{1}$ satisfy (1.19) and (1.20). Note that
(1.19) and (1.20) are invariant under the scaling

$u_{0}^{\lambda}(x):=\lambda^{\frac{2-b}{\sigma}}u_{0}(\lambda
x),\leavevmode\nobreak\ \lambda>0.$ (3.18)

Hence, by choosing a suitable scaling, we can assume that

$M(u_{0})=M_{G},\leavevmode\nobreak\ E(u_{0})=E_{G},\leavevmode\nobreak\
\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}<S_{G}.$ (3.19)

We claim that

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}<S_{G},$ (3.20)

for all $t\in(-T_{*},T^{*})$. Assume by contradiction that there exists
$t_{0}\in(-T_{*},T^{*})$ such that
$\left\|u(t_{0})\right\|_{\dot{H}_{c}^{1}}\geq S_{G}$. By continuity argument,
there exists $t_{1}\in(-T_{*},T^{*})$ such that
$\left\|u(t_{1})\right\|_{\dot{H}_{c}^{1}}=S_{G}$. It follows from (2.5) and
the conservation of energy that

$\displaystyle\begin{split}P(u(t_{1}))&=(\sigma+2)\left(\frac{1}{2}\left\|u(t_{1})\right\|_{\dot{H}_{c}^{1}}^{2}-E(u(t_{1}))\right)\\\
&=(\sigma+2)\left(\frac{1}{2}[S_{G}]^{2}-E_{G}\right)\\\
&=\frac{2(\sigma+2)}{d\sigma+2b}[S_{G}]^{2}=P_{G}.\end{split}$

Hence, we have

$P(u(t_{1}))-C_{GN}\left\|u(t_{1})\right\|_{\dot{H}_{c}^{1}}^{\frac{d\sigma+2b}{2}}\left\|u(t_{1})\right\|_{L^{2}}^{\frac{4-2b-\sigma(d-2)}{2}}=P_{G}-C_{GN}[S_{G}]^{\frac{d\sigma+2b}{2}}[M_{G}]^{\frac{4-2b-\sigma(d-2)}{4}}=0,$

which shows that $u(t_{1})$ is an optimizer for (2.2). Using the same argument
as in the proof of Lemma 3.2, we have $u(t_{1})=e^{i\theta}Q$ for some
$\theta\in\mathbb{R}$ and some positive solution $Q$ to (1.7). Moreover, by
the uniqueness of solution to (1.1), we have $u(t)=e^{i(t-t_{1})}e^{i\theta}Q$
which contradicts (3.19). This shows (3.20) and the solution exists globally
in time. We divide the study in two cases.
Case 1.1. Let us consider the case

$\sup_{t\in\mathbb{R}}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}<S_{G}.$

There exists $\rho>0$ such that

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\leq(1-\rho)S_{G},$ (3.21)

for all $t\in\mathbb{R}$. In view of (3.19) and (3.21), we have (3.2).
Repeating the same argument as in the proof of Item 1 of Theorem 1.2, we can
prove (1.21).
Case 1.2. Let us consider the case

$\sup_{t\in\mathbb{R}}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}=S_{G}.$

Then there exists a time sequence $(t_{n})_{n\geq 1}$ such that

$M(u(t_{n}))=M_{G},\leavevmode\nobreak\ E(u(t_{n}))=E_{G},\leavevmode\nobreak\
\lim_{n\to\infty}\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}=S_{G}.$

We claim that $|t_{n}|\to\infty$. Assume by contradiction that there exists
subsequence still denoted by $(t_{n})_{n\geq 1}$ and $t_{0}\in\mathbb{R}$ such
that $t_{n}\to t_{0}$ as $n\to\infty$. By continuity of solution, $u(t_{n})\to
u(t_{0})$ strongly in $H^{1}_{c}$. This shows that
$\left\|u(t_{0})\right\|_{\dot{H}_{c}^{1}}=S_{G}$. Using the same argument
above, we see that $u(t_{0})$ is an optimizer for (2.2) which is a
contradiction. Hence, it follows from Lemma 3.2 that there exists a
subsequence still denoted by $(t_{n})_{n\geq 1}$ such that

$u(t_{n})\to e^{i\theta}Q\leavevmode\nobreak\ \textrm{strongly
in}\leavevmode\nobreak\ H_{c}^{1}\leavevmode\nobreak\
\textrm{as}\leavevmode\nobreak\ n\to\infty,$

for some $\theta\in\mathbb{R}$ and some positive solution $Q$ to (1.7).

Proof of Item 2. Let $u_{0}$ satisfy (1.19) and (1.23). By scaling (3.18), we
can assume that

$M(u_{0})=M_{G},\leavevmode\nobreak\ E(u_{0})=E_{G},\leavevmode\nobreak\
\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}=S_{G}.$

This shows that $u_{0}$ is optimizer for (2.2) which implies that
$u_{0}=e^{i\theta}Q(x)$ for some $\theta\in\mathbb{R}$ and some positive
solution $Q$ to (1.7). By uniqueness of solution to (1.1), we have
$u(t)=e^{it}e^{i\theta}Q(x)$.

Proof of Item 3. Let $u_{0}$ satisfy (1.19) and (1.24). By scaling (3.18), we
can assume that

$M(u_{0})=M_{G},\leavevmode\nobreak\ E(u_{0})=E_{G},\leavevmode\nobreak\
\left\|u_{0}\right\|_{\dot{H}_{c}^{1}}>S_{G}.$ (3.22)

Using the same argument as in the proof of Item 1, we can prove that

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}>S_{G},$

for all $t\in(-T_{*},T^{*})$. We only consider the positive times. The
negative time is treated similarly. If $T^{*}<\infty$, then we are done. If
$T^{*}=\infty$, we have two possible cases.
Case 3.1. Let us consider the case

$\sup_{t\in[0,\infty)}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}>S_{G}.$

Then there exists $\delta>0$ such that

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\geq(1+\delta)S_{G}$ (3.23)

for all $t\in[0,\infty)$. It follows from (2.6), (3.22), (3.23) and the
conservation of mass and energy, we have

$\displaystyle\begin{split}G(u(t))[M(u(t))]^{\sigma_{c}}&=\frac{d\sigma+2b}{2}E(u(t))[M(u(t))]^{\sigma_{c}}-\frac{d\sigma-4+2b}{4}\left(\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\left\|u(t)\right\|_{L^{2}}^{\sigma_{c}}\right)^{2}\\\
&\leq\frac{d\sigma+2b}{2}H_{G}-\frac{d\sigma-4+2b}{4}\left((1+\delta)K_{G}\right)^{2}\\\
&=-\frac{d\sigma-4+2b}{4}\left((1+\delta)^{2}-1\right)[K_{G}]^{2},\end{split}$

for all $t\in[0,\infty)$. By Theorem 1.1, there exists a time sequence
$t_{n}\to\infty$ such that
$\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to\infty$ as $n\to\infty$.
Case 3.2. Let us consider the case

$\sup_{t\in[0,\infty)}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}=S_{G}.$

Then there exists a time sequence $(t_{n})_{n\geq 1}$ such that
$\left\|u(t_{n})\right\|_{\dot{H}_{c}^{1}}\to S_{G}$ as $n\to\infty$. Using
the same argument as in Case 1.2, we can see that there exists a subsequence
still denoted by $(t_{n})_{n\geq 1}$ such that $t_{n}\to\infty$ and

$u(t_{n})\to e^{i\theta}Q\leavevmode\nobreak\ \textrm{strongly
in}\leavevmode\nobreak\ H_{c}^{1}\leavevmode\nobreak\
\textrm{as}\leavevmode\nobreak\ n\to\infty$

for some $\theta\in\mathbb{R}$ and some positive solution $Q$ to (1.7). This
completes the proof of the first part of Item 3. Let us prove the second part.

  * •

Finite variance data. If $|x|u_{0}\in L^{2}$, then Case 3.1 is excluded.
Indeed, if Case 3.1 occurs, we have

$G(u(t))<-\eta,\leavevmode\nobreak\ \forall t\in[0,\infty)$

for some $\eta>0$. Then Lemma 2.6 shows that

$\frac{d^{2}}{dt^{2}}\left\|xu(t)\right\|_{L^{2}}^{2}=8G(u(t))<-8\eta<0,$

which contradicts $T^{*}=\infty$.

  * •

Radially symmetric data. If Case 3.1 occurs, we have (3.23). It follows from
(2.6), (3.22), (3.23) and the conservation of mass and energy that

$\displaystyle\begin{split}&[M(u(t))]^{\sigma_{c}}\left(8G(u(t))+\varepsilon\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\right)\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
=4(d\sigma+2b)E(u(t))[M(u(t))]^{\sigma_{c}}-(2d\sigma-8+4b-\varepsilon)\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{2}[M(u(t))]^{\sigma_{c}}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leq
4(d\sigma+2b)H_{G}-(2d\sigma-8+4b-\varepsilon)\left((1+\delta)K_{G}\right)^{2}\\\
&\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
=-2(d\sigma-4+2b)\left((1+\delta)K_{G}\right)^{2}\left[\frac{(1+\delta)^{2}-1}{(1+\delta)^{2}}-\frac{\varepsilon}{2(d\sigma-4+2b)}\right],\end{split}$

for all $t\in[0,\infty)$. Taking $\varepsilon>0$ sufficiently small, there
exists $\eta>0$ such that

$8G(u(t))+\varepsilon\left\|u(t)\right\|_{\dot{H}_{c}^{1}}\leq-\eta,$ (3.24)

for all $t\in[0,\infty)$. On the other hand, Lemma 4.6 of [12] shows that

$V^{\prime\prime}_{\varphi_{R}}(t)\leq
8G(u(t))+O\left(R^{-2}+\varepsilon^{-\frac{\sigma}{4-\sigma}}R^{-{\frac{2\left[(d-1)\sigma+2b\right]}{4-\sigma}}}+\varepsilon\left\|u(t)\right\|_{\dot{H}^{1}_{c}}^{2}\right),$
(3.25)

for any $\varepsilon>0$ and any $R>1$. Taking $\varepsilon>0$ small enough and
$R>1$ large enough, it follows from (3.23) and (3.24) that

$V^{\prime\prime}_{\varphi_{R}}(t)\leq-\frac{\eta}{2},$

for all $t\in[0,\infty)$. This is also impossible.

  * •

Cylindrically symmetric data. Using (3.13), (3.24) and repeating the same
argument above, we can easily prove that Case 3.1 is excluded. This completes
the proof.

∎

###  3.3 Above the ground state threshold

In this subsection, we study the global existence and blow-up above the ground
state threshold.

######  Lemma 3.3 ([21]).

Let $u\in H^{1}$ and $|x|u\in L^{2}$. Then we have

$\left\|u\right\|_{2}^{2}\leq\frac{2}{d}\left\||x|u\right\|_{2}\left\|\nabla
u\right\|_{2}.$

######  Lemma 3.4.

Let $d\geq 3$, $0<b<2$, $\frac{4-2b}{d}<\sigma<\frac{4-2b}{d-2}$ and
$c>-c(d)$. Assume $u\in H_{c}^{1}$ and $|x|u\in L^{2}$. Then we have

$\left[\textnormal{Im}\int{x\cdot\bar{u}\nabla
udx}\right]^{2}\leq\int{|x|^{2}|u|^{2}dx}\left[\left\|u\right\|_{\dot{H}_{c}^{1}}^{2}-\frac{[P(u)]^{\frac{4}{d\sigma+2b}}}{C_{GN}^{\frac{4}{d\sigma+2b}}[M(u)]^{\frac{4-2b-(d-2)\sigma}{d\sigma+2b}}}\right].$
(3.26)

###### Proof.

Noticing

$\int{|\nabla(e^{i\lambda|x|^{2}}u)|^{2}dx}=4\lambda^{2}\int{|x|^{2}|u|^{2}dx}+4\lambda\textnormal{Im}\int{x\cdot\bar{u}\nabla
udx}+\int{|\nabla u|^{2}dx},$

we have

$\left\|e^{i\lambda|x|^{2}}u\right\|_{\dot{H}_{c}^{1}}^{2}=4\lambda^{2}\int{|x|^{2}|u|^{2}dx}+4\lambda\textnormal{Im}\int{x\cdot\bar{u}\nabla
udx}+\left\|u\right\|_{\dot{H}_{c}^{1}}^{2}.$ (3.27)

On the other hand, Lemma 2.2 (sharp Gagliardo-Nirenberg inequality) implies
that

$[P(u)]^{\frac{4}{d\sigma+2b}}=[P(e^{i\lambda|x|^{2}}u)]^{\frac{4}{d\sigma+2b}}\leq
C_{GN}^{\frac{4}{d\sigma+2b}}\left\|e^{i\lambda|x|^{2}}u\right\|_{\dot{H}_{c}^{1}}^{2}\left\|u\right\|_{L^{2}}^{\frac{2[4-2b-(d-2)\sigma]}{d\sigma+2b}}$
(3.28)

It follows from (3.27) and (3.28) that

$4\lambda^{2}\int{|x|^{2}|u|^{2}dx}+4\lambda\textnormal{Im}\int{x\cdot\bar{u}\nabla{u}dx}+\left\|u\right\|_{\dot{H}_{c}^{1}}^{2}\geq\frac{[P(u)]^{\frac{4}{d\sigma+2b}}}{C_{GN}^{\frac{4}{d\sigma+2b}}[M(u)]^{\frac{4-2b-(d-2)\sigma}{d\sigma+2b}}},$

for $\lambda\in\mathbb{R}$, where the left-hand side is a polynomial in
$\lambda$. The discriminant of this polynomial in $\lambda$ must be negative
or null, which yields the conclusion of this lemma. ∎

###### Proof of Theorem 1.4..

Let $u_{0}\in H_{c}^{1}$ be such that $|x|u_{0}\in L^{2}$. Assume further that
$u_{0}$ satisfies (1.26) and (1.27). Using Lemma 2.6, we have

$\left\|u(t)\right\|_{\dot{H}_{c}^{1}}^{2}=\frac{4(d\sigma+2b)E(u(t))-V^{\prime\prime}(t)}{2(d\sigma+2b-4)},$
(3.29)
$P(u(t))=\frac{(16E(u(t))-V^{\prime\prime}(t))(\sigma+2)}{4(d\sigma+2b-4)}.$
(3.30)

(3.30) yields that $V^{\prime\prime}(t)\leq 16E(u(t))=16E(u_{0})$ for all $t$.
Using Lemma 2.4, Lemma 3.4, (3.29), (3.30) and the conservations of mass and
energy, we immediately have

$(V^{\prime}(t))^{2}\leq
16V(t)\left[\frac{4(d\sigma+2b)E(u_{0})-V^{\prime\prime}(t)}{2d\sigma+4b-8}-\frac{\left(\frac{(\sigma+2)(16E(u_{0})-V^{\prime\prime}(t))}{4d\sigma+8b-16}\right)^{\frac{4}{d\sigma+2b}}}{C_{GN}^{\frac{4}{d\sigma+2b}}[M(u_{0})]^{\frac{4-2b-(d-2)\sigma}{d\sigma+2b}}}\right].$
(3.31)

Putting $z(t):=\sqrt{V(t)}$, it follows from (3.31) that

$(z^{\prime}(t))^{2}\leq 4f(V^{\prime\prime}(t)),$ (3.32)

where

$f(x)=-\frac{x}{2d\sigma+4b-8}+\frac{(2d\sigma+4b)E(u_{0})}{d\sigma+2b-4}-\frac{\left(\frac{(\sigma+2)(16E(u_{0})-x)}{4d\sigma+8b-16}\right)^{\frac{4}{d\sigma+2b}}}{C_{GN}^{\frac{4}{d\sigma+2b}}[M(u_{0})]^{\frac{4-2b-(d-2)\sigma}{d\sigma+2b}}}$
(3.33)

with $x\leq 16E(u_{0})$. An easy computation shows that

$f^{\prime}(x)=-\frac{1}{2d\sigma+4b-8}+\frac{\frac{4}{d\sigma+2b}\left(\frac{\sigma+2}{4d\sigma+8b-16}\right)^{\frac{4}{d\sigma+2b}}\left(16E(u_{0})-x\right)^{\frac{4-d\sigma-2b}{d\sigma+2b}}}{C_{GN}^{\frac{4}{d\sigma+2b}}[M(u_{0})]^{\frac{4-2b-(d-2)\sigma}{d\sigma+2b}}}.$

Since $\sigma>\frac{4-2b}{d}$, $f(x)$ is decreasing on $(-\infty,x_{0})$,
increasing on $(x_{0},16E(u_{0})]$, where $x_{0}$ satisfies

$\frac{1}{2d\sigma+4b-8}=\frac{\frac{4}{d\sigma+2b}\left(\frac{\sigma+2}{4d\sigma+8b-16}\right)^{\frac{4}{d\sigma+2b}}\left(16E(u_{0})-x_{0}\right)^{\frac{4-d\sigma-2b}{d\sigma+2b}}}{C_{GN}^{\frac{4}{d\sigma+2b}}[M(u_{0})]^{\frac{4-2b-(d-2)\sigma}{d\sigma+2b}}}.$
(3.34)

In view of (3.33) and (3.34), we also have $f(x_{0})=\frac{x_{0}}{8}$. Using
(2.4), (2.6) and (3.34), we also have

$\left[\frac{M_{G}}{M(u_{0})}\right]^{\sigma_{c}}\frac{16E_{G}}{16E(u_{0})-x_{0}}=1.$
(3.35)

Hence, (1.26) is equivalent to

$x_{0}\geq 0.$ (3.36)

Noticing

$(z^{\prime}(t))^{2}=\frac{(V^{\prime}(t))^{2}}{4V(t)},$

and using (3.35), we can see that (1.27) is equivalent to

$(z^{\prime}(0))^{2}\geq 4f(x_{0})=\frac{x_{0}}{2}.$ (3.37)

Proof of Item 1. Let $u_{0}$ satisfy (1.28) and (1.29). The assumption (1.29)
means that

$z^{\prime}(0)\geq 0.$ (3.38)

Using (2.5) and (3.35), we can see that the assumption (1.28) is equivalent to

$\left[\frac{M_{G}}{M(u_{0})}\right]^{\sigma_{c}}\frac{4(\sigma+2)E_{G}}{(d\sigma+2b-4)P(u_{0})}>1=\left[\frac{M_{G}}{M(u_{0})}\right]^{\sigma_{c}}\frac{16E_{G}}{16E(u_{0})-x_{0}}.$
(3.39)

Using (3.30), we can see that (3.39) is equivalent to

$V^{\prime\prime}(0)>x_{0}.$ (3.40)

By (3.40), we can take $\delta_{1}>0$ such that

$V^{\prime\prime}(0)\geq x_{0}+2\delta_{1}.$

By continuity argument, we have

$V^{\prime\prime}(t)>x_{0}+\delta_{1},\leavevmode\nobreak\ \forall
t\in[0,t_{1}),$ (3.41)

for all $t_{1}>0$ sufficiently small. By reducing $t_{1}$ if necessary, we can
assume that

$z^{\prime}(t_{1})>2\sqrt{f(x_{0})}.$ (3.42)

In fact, using (3.37) and (3.38), we have $z^{\prime}(0)\geq
2\sqrt{f(x_{0})}$. If the inequality is strict, then (3.42) follows from the
continuity argument. If $z^{\prime}(0)=2\sqrt{f(x_{0})}$, it follows from
(3.40) that

$z^{\prime\prime}(0)=\frac{1}{z(0)}\left[\frac{V^{\prime\prime}(0)}{2}-(z^{\prime}(0))^{2}\right]>0.$

Hence, we can take $t_{1}>0$ small enough such that (3.42) holds. We then
choose $\varepsilon\in(0,1)$ small enough such that

$z^{\prime}(t_{1})\geq 2\sqrt{f(x_{0})}+2\varepsilon.$

We will prove by contradiction that

$z^{\prime}(t)>2\sqrt{f(x_{0})}+\varepsilon,\leavevmode\nobreak\ \forall t\geq
t_{1}.$ (3.43)

If (3.43) does not hold, there exists $t_{2}$ satisfying

$t_{2}=\inf\left\\{t\geq t_{1}:\leavevmode\nobreak\ z^{\prime}(t)\leq
2\sqrt{f(x_{0})}+\varepsilon\right\\}.$

By continuity of $z^{\prime}(t)$, we have

$z^{\prime}(t_{2})=2\sqrt{f(x_{0})}+\varepsilon,$ (3.44)

and

$z^{\prime}(t)\geq 2\sqrt{f(x_{0})}+\varepsilon,\leavevmode\nobreak\
\leavevmode\nobreak\ \forall t\in[t_{1},t_{2}].$ (3.45)

Using (3.45) and (3.32), we immediately have

$(2\sqrt{f(x_{0})}+\varepsilon)^{2}\leq(z^{\prime}(t))^{2}\leq
4f(V^{\prime\prime}(t)),\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
t\in[t_{1},t_{2}],$

which implies that $f(V^{\prime\prime}(t))>f(x_{0})$ for all
$t\in[t_{1},t_{2}]$. This means that $V^{\prime\prime}(t)\neq x_{0}$ for all
$t\in[t_{1},t_{2}]$. Using (3.40) and the continuity of $V^{\prime\prime}(t)$,
we have $V^{\prime\prime}(t)>x_{0}$ for all $t\in[t_{1},t_{2}]$. We claim that
there exists a constant $C>0$ (which is independent of $\varepsilon$ and $t$)
such that

$V^{\prime\prime}(t)\geq
x_{0}+\frac{\sqrt{\varepsilon}}{C},\leavevmode\nobreak\ \leavevmode\nobreak\
\forall t\in[t_{1},t_{2}].$ (3.46)

In fact, by the Taylor expansion of $f$ around $x=x_{0}$, we get

$x_{0}<x<x_{0}+\frac{16E(u_{0})-x_{0}}{2}\Rightarrow f(x)\leq
f(x_{0})+\frac{M}{2}(x-x_{0})^{2},$ (3.47)

where
$M:=\max\left\\{f^{\prime\prime}(x):x\in\left[x_{0},\frac{16E(u_{0})+x_{0}}{2}\right]\right\\}$.
It is obvious that $M>0$. If $V^{\prime\prime}(t)\geq
x_{0}+\frac{16E(u_{0})-x_{0}}{2}$, then we take
$C=C_{1}:=\frac{2}{16E(u_{0})-x_{0}}$. If
$x_{0}<V^{\prime\prime}(t)<x_{0}+\frac{16E(u_{0})-x_{0}}{2}$, then it follows
from (3.32), (3.45) and (3.47) that

$(2\sqrt{f(x_{0})}+\varepsilon)^{2}\leq(z^{\prime}(t))^{2}\leq
4f(V^{\prime\prime}(t))\leq 4f(x_{0})+2M(V^{\prime\prime}(t)-x_{0}),$

which implies that

$4\sqrt{f(x_{0})}\varepsilon+\varepsilon^{2}<2M(V^{\prime\prime}(t)-x_{0})^{2}.$

Hence, we can take $C=C_{2}:=\sqrt{M}(4f(x_{0}))^{-\frac{1}{4}}$. Therefore,
taking $C:=\max\left\\{C_{1},C_{2}\right\\}$, we have (3.46) for all
$t\in[t_{1},t_{2}]$. In view of (3.44) and (3.46), we have

$\displaystyle\begin{split}z^{\prime\prime}(t_{2})&=\frac{1}{z(t_{2})}\left[\frac{V^{\prime\prime}(t_{2})}{2}-(z^{\prime}(t_{2}))^{2}\right]\\\
&\geq\frac{1}{z(t_{2})}\left[\frac{x_{0}}{2}+\frac{\sqrt{\varepsilon}}{2C}-(2\sqrt{f(x_{0})}+\varepsilon)^{2}\right]\\\
&\geq\frac{1}{z(t_{2})}\left[\frac{\sqrt{\varepsilon}}{2C}-4\varepsilon\sqrt{f(x_{0})}-\varepsilon^{2}\right].\end{split}$
(3.48)

Taking $\varepsilon>0$ small enough in (3.48), we have
$z^{\prime\prime}(t_{2})>0$, which is contrary to (3.44) and (3.45). This
completes the proof of (3.43). Using (3.43) and the same argument as in the
proof of (3.46), we can prove

$V^{\prime\prime}(t)\geq
x_{0}+\frac{\sqrt{\varepsilon}}{C},\leavevmode\nobreak\ \leavevmode\nobreak\
\forall t\geq t_{1}.$ (3.49)

In view of (3.41) and (3.49), we have

$V^{\prime\prime}(t)\geq x_{0}+\delta,\leavevmode\nobreak\
\leavevmode\nobreak\ \forall t\geq 0,$ (3.50)

where $\delta=\min\left\\{\delta_{1},\frac{\sqrt{\varepsilon}}{C}\right\\}>0$.
Using (3.30), (3.50), (3.35), (2.5) and the conservations of mass and energy,
we have for any $t>0$

$\displaystyle\begin{split}P(u(t))[M(u(t))]^{\sigma_{c}}&=[M(u_{0})]^{\sigma_{c}}\frac{(16E(u_{0})-V^{\prime\prime}(t))(\sigma+2)}{4(d\sigma+2b)-16}\\\
&\leq[M(u_{0})]^{\sigma_{c}}\frac{(16E(u_{0})-x_{0}-\delta)(\sigma+2)}{4(d\sigma+2b)-16}\\\
&=\frac{4(\sigma+2)}{d\sigma-4+2b}[M_{G}]^{\sigma_{c}}E_{G}-\frac{\sigma+2}{4(d\sigma-4+2b)}\delta[M(u_{0})]^{\sigma_{c}}\\\
&=(1-\rho)P_{G}[M_{G}]^{\sigma_{c}},\end{split}$ (3.51)

where

$\rho:=\frac{\sigma+2}{4(d\sigma-4+2b)}\frac{[M(u_{0})]^{\sigma_{c}}}{P_{G}[M_{G}]^{\sigma_{c}}}\delta>0.$

This shows (1.9), and thus the solution exists globally in time.

Proof of Item 2. Let $u_{0}$ satisfy (1.30) and (1.31). Using the same
argument as in the proof of (3.38) and (3.40), we can prove that the
assumptions (1.31) and (1.30) are respectively equivalent to

$z^{\prime}(0)\leq 0,$ (3.52) $V^{\prime\prime}(0)<x_{0}.$ (3.53)

We will prove by contradiction that

$z^{\prime\prime}(t)<0,$ (3.54)

for any $t$ in the existence time. In fact, it follows from (3.37) and (3.53)
that

$z^{\prime\prime}(0)=\frac{1}{z(0)}\left[\frac{V^{\prime\prime}(0)}{2}-(z^{\prime}(0))^{2}\right]<\frac{1}{z(0)}\left(\frac{x_{0}}{2}-\frac{x_{0}}{2}\right)=0.$
(3.55)

Assume that (3.54) does not hold. Then there exists $t_{0}>0$ in the existence
time such that

$z^{\prime\prime}(t_{0})=0\leavevmode\nobreak\ \leavevmode\nobreak\
\textnormal{and}\leavevmode\nobreak\ \leavevmode\nobreak\
z^{\prime\prime}(t)<0,\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
t\in[0,t_{0}).$

In view of (3.37) and (3.52), we have

$z^{\prime}(t)<z^{\prime}(0)\leq-\sqrt{4f(x_{0})},\leavevmode\nobreak\
\leavevmode\nobreak\ \forall t\in(0,t_{0}],$ (3.56)

which implies that $(z^{\prime}(t))^{2}>4f(x_{0})$. Hence, using (3.32), we
immediately have

$f(V^{\prime\prime}(t))>f(x_{0}),\leavevmode\nobreak\ \leavevmode\nobreak\
\forall t\in(0,t_{0}].$ (3.57)

Using (3.57), (3.53) and the continuity of $V^{\prime\prime}(t)$, we can see
that

$V^{\prime\prime}(t)<x_{0},\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
t\in[0,t_{0}].$ (3.58)

In view of (3.56) and (3.58), we have

$z^{\prime\prime}(t_{0})=\frac{1}{z(t_{0})}\left[\frac{V^{\prime\prime}(t_{0})}{2}-(z^{\prime}(t_{0}))^{2}\right]<\frac{1}{z(t_{0})}\left(\frac{x_{0}}{2}-\frac{x_{0}}{2}\right)=0,$

which contradicts $z^{\prime\prime}(t_{0})=0.$ Hence, we have (3.54) for any
$t$ in the existence time. Using (3.52) and (3.54) and the fact
$z(t)=\sqrt{V(t)}$, we can see that there exists $T<\infty$ such that
$\lim_{t\rightarrow T}{V(t)}=0$. Noticing $\dot{H}_{c}^{1}\sim\dot{H}^{1}$, it
follows from Lemma 3.3 that $\lim_{t\rightarrow
T}{\left\|u(t)\right\|_{\dot{H}_{c}^{1}}}=\infty.$ This completes the proof. ∎

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78 0.71 78 0.71 63.6 1.14 64 0.71 70.8 0.76 71 1 18.4 1.14 18 0.71 ELMo ft 60
76.2 0.45 76.2 0.45 69 1.58 69.2 1.48 72.6 0.89 72.7 1 9.6 1.95 9.2 1.79 ELMo
+ BiLSTM 60 79 0.71 79 0.71 67 2.65 67 2.65 73 1.5 73 2 15.2 3.27 15.2 3.27
ELMo ft + BiLSTM 60 78.2 0.84 67.4 0.84 67.4 1.34 67.6 1.67 72.8 1.04 72.9 1
13.8 0.84 13.4 1.52 DistilBERT 60 79 0.71 79 0.71 66.8 3.96 67 3.54 72.9 2.27
73 2 15.4 4.51 15.2 4.09 DistlBERT ft 60 79 0.71 79 0.71 68 1.22 68 1.22 73.5
0.5 73.5 1 14.2 2.17 14.2 2.17 DistilBERT + BiLSTM 60 80 1 80 1 70.2 2.28 70.2
2.28 75.1 1.64 75.1 2 12.4 1.82 12.4 1.82 DistilBERT ft + BiLSTM 60 80 1 80 1
70.2 2.28 70.2 2.28 75.1 1.64 75.1 2 12.4 1.82 12.4 1.82 BERT 60 80.6 0.55
80.6 0.55 75 1.58 75 1.58 77.8 1.04 77.8 1.04 7 1.58 7 1.58 BERT ft 60 80.6
0.55 80.6 0.55 75 1.58 75 1.58 77.8 1.04 77.8 1.04 7 1.58 7 1.58 BERT + BiLSTM
60 82 1 82 1 71.8 3.56 71.8 3.56 76.9 2.13 76.9 2.13 12.6 3.91 12.6 3.91 BERT
ft + BiLSTM 60 82.6 0.55 82.6 0.55 75.6 1.34 75.8 1.64 79.1 0.89 79.2 1.04 8.6
1.52 8.4 1.82

##  Chapter B Hyper-parameter tuning results

Table B.1: Hyper-parameter tuning results.

task model opt best train F-score opt best dev F-score opt best epoch no opt
time (hrs:mins) drop l2 lrate lr decay relu PC ELMo 256 90 87 9 6:26 0.25 0
1e-3 0.5 TRUE PC ELMo 150 81 87 5 4:11 0 0.01 1e-3 0.5 TRUE PC ELMo ft 150 91
87 9 9:21 0 0 1e-3 0 TRUE PC ElMo + BiLSTM 256 84 84 7 8:37 0 0 5e-5 0.5 TRUE
PC ElMo + BiLSTM 150 89 86 9 4:11 0.5 0.01 1e-3 0 FALSE PC ElMo ft + BiLSTM
150 86 87 5 9:03 0 0.01 1e-3 0.5 FALSE PC DistilBERT 256 87 87 8 1:49 0 0 1e-3
0 TRUE PC DistilBERT 150 83 85 2 1:05 0 0 1e-3 0 TRUE PC DistilBERT ft 256 87
87 8 1:52 0.25 0 1e-3 0.5 TRUE PC DistilBERT ft 150 78 85 5 1:06 0.25 0 1e-1 0
FALSE PC DistilBERT + BiLSTM 256 88 84 7 2:24 0 0.01 1e-3 0 TRUE PC DistilBERT
+ BiLSTM 150 87 85 7 1:27 0.5 0 5e-5 0 FALSE PC DistilBERT ft + BiLSTM 256 88
86 6 2:27 0 0.01 1e-3 0 FALSE PC DistilBERT ft + BiLSTM 150 87 85 7 1:27 0.5 0
5e-5 0.5 FALSE PC BERT 150 85 86 9 4:31 0.5 0 1e-3 0 FALSE PC BERT ft 150 87
87 9 4:28 0.25 0 1e-3 0 TRUE PC BERT + BiLSTM 150 88 85 6 5:07 0.25 0.01 1e-3
0.5 TRUE PC BERT ft + BiLSTM 150 88 85 6 5:15 0.25 0.01 1e-3 0.5 TRUE SA ELMo
60 84 80 6 3:56 0 0 1e-3 0.5 TRUE SA ELMo ft 60 78 78 6 6:44 0.5 0 1e-3 0
FALSE SA ELMo + BiLSTM 60 83 83 2 3:06 0 0 1e-3 0 FALSE SA ELMo ft + BiLSTM 60
84 82 2 9:49 0 0 1e-3 0 TRUE SA DistilBERT 60 82 79 5 1:10 0 0 1e-3 0.5 TRUE
SA DistilBERT ft 60 82 79 5 1:32 0.25 0 1e-3 0.5 TRUE SA DistilBERT + BiLSTM
60 99 84 8 1:34 0 0 1e-3 0 FALSE SA DistilBERT ft + BiLSTM 60 99 84 8 1:08 0 0
1e-3 0 FALSE SA BERT 60 82 82 8 2:16 0.5 0 5e-5 0 TRUE SA BERT ft 60 82 82 8
4:36 0.5 0 5e-5 0 TRUE SA BERT + BiLSTM 60 88 84 2 2:33 0 0 1e-3 0 TRUE SA
BERT ft + BiLSTM 60 99 85 8 5:10 0.5 0 1e-3 0.5 FALSE