Blowup and scattering criteria above the threshold for the focusing inhomogeneous nonlinear Schrödinger equation

Abstract

We consider the inhomogeneous nonlinear Schrödinger equation (INLS) in $\mathbb {R}^N$

$$\begin{aligned} i \partial _t u + \Delta u + |x|^{-b} |u|^{p-1}u = 0, \end{aligned}$$

with initial data $u_0\in H^1({\mathbb {R}}^N)$ having finite variance. We extend the dichotomy between scattering and blow-up for solutions above the mass-energy threshold (and with arbitrarily large energy). We also show two other blow-up criteria, which are valid in any mass-supercritical setting, given there is local well-posedness.

1 Introduction

We consider the initial value problem associated to the inhomogeneous nonlinear Schrödinger equation (INLS)

$$\begin{aligned} \left\{ \begin{array}{l} i\partial _t u +\triangle u +|x|^{-b}|u|^{p-1}u=0,\,\,\,\,t>0, \,\, x\in \mathbb {R}^N,\\ u(\cdot ,0)=u_0\in H^{1}(\mathbb {R}^N). \end{array} \right. \end{aligned}$$

(1.1)

This model arises naturally as a limiting problem in nonlinear optics for the propagation of laser beams. The case $b = 0$ is the classical nonlinear Schrödinger equation (NLS), extensively studied in recent years (see Sulem-Sulem [27], Bourgain [2], Cazenave [6], Linares-Ponce [25], Fibich [15] and the references therein).

The lower Sobolev index where one can expect well-posedness for this model is given by scaling. If u(x, t) is a solution to (1.1), so is $u_{\lambda }(x, t)=\lambda ^{\frac{2-b}{p-1}}u(\lambda x,\lambda ^2 t)$, with initial data $u_{0,\lambda }(x),$ for all $\lambda > 0$. Computing the homogeneous Sobolev norm, we get

$$\begin{aligned} \Vert u_{0,\lambda }\Vert _{\dot{H}^{s}}=\lambda ^{s-\frac{N}{2}+\frac{2-b}{p-1}}\Vert u_0\Vert _{\dot{H}^{s}}. \end{aligned}$$

Thus, the scale-invariant Sobolev norm is $\dot{H}^{s_c}({\mathbb {R}}^N)$, where

$$\begin{aligned} s_c=\frac{N}{2}-\frac{2-b}{p-1} \end{aligned}$$

is called the critical Sobolev index.

In this paper, we are interested in the case $s_c > 0$, known as mass-supercritical. Rewriting this condition in terms of p, we obtain

$$\begin{aligned} p >1+ \frac{2(2-b)}{N}. \end{aligned}$$

The local well-posedness for the INLS equation was first studied by Genoud-Stuart in [20] (see also Genoud [17]) by the abstract theory of Cazenave [6], without relying on Strichartz type inequalities. They analyzed the IVP (1.1) in the sense of distributions, that is, $i\partial _tu+\Delta u+|x|^{-b}|u|^{p-1}u=0$ in $H^{-1}(\mathbb {R}^N)$, $N \ge 1$, with $0< b < \min \{2,N\}$, and showed it is well-posed

• locally if $1< p < p^*_b$ ($s_c < 1$);

• globally for any initial data in $H^1({\mathbb {R}}^{N})$ if $p <1+ \frac{2(2-b)}{N}$ ($s_c < 0$);

• globally for sufficiently small initial data if $1+\frac{2(2-b)}{N} \le p < p_b^*$ ($0 \le s_c < 1$),

where

$$\begin{aligned} p^*_b = {\left\{ \begin{array}{ll} \infty , &{} N \le 2, \\ 1+\frac{2(2-b)}{N-2}, &{} N \ge 3. \\ \end{array}\right. } \end{aligned}$$

More recently, Guzmán [23], Cho and Lee [7] and Dinh [8] established local well-posedness of the INLS in $H^s({\mathbb {R}}^{N})$ based on Strichartz estimates. In particular, they proved that, for $N \ge 2$, $1< p < p_b^*$ and $0< b < \min \{N/2,2\}$, the initial value problem (1.1) is locally well-posed in $H^1({\mathbb {R}}^N)$.

Note that the results of Guzmán [23] and Dinh [8] do not treat the case $N = 1$, and the ranges of b are more restrictive than those in the results of Genoud-Stuart [20]. However, Guzmán and Dinh give more detail information on the solutions, showing that there exists $T(\Vert u_0\Vert _{H^1})>0$ such that $u \in L^q\left( [-T,T];W^{1,r}({\mathbb {R}}^N)\right)$ for any $L^2$-admissible pair (q, r) satisfying

$$\begin{aligned} \frac{2}{q}=\frac{N}{2}-\frac{N}{r}, \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{array}{cl} 2\le &{} r \le \frac{2N}{N-2}\text {if}\;\;\; N\ge 3,\\ 2 \le &{} r < +\infty \; \text {if}\;\; \;N=2,\\ 2 \le &{} r \le + \infty \; \text {if}\;\;\;N=1. \end{array}\right. \end{aligned}$$

The solutions to (1.1) have the following conserved quantities

$$\begin{aligned}&M\left[ u(t) \right] = \int |u(t)|^2 dx = M[u_0], \\&E\left[ u(t) \right] = \frac{1}{2}\int |\nabla u(t)|^2 dx - \frac{1}{p+1} \int |x|^{-b}|u(t)|^{p+1} dx = E[u_0]. \end{aligned}$$

The blow-up theory in the INLS equation is related to the concept of ground state, which is the unique positive radial solution of the elliptic problem

$$\begin{aligned} \Delta Q - Q + |x|^{-b}|Q|^{p-1} Q = 0. \end{aligned}$$

The existence of the ground state is proved by Genoud-Stuart [16, 20] for dimension $N \ge 2$, and by Genoud [17] for $N = 1$. Uniqueness was proved in dimension $N \ge 3$ by Yanagida [30] (see also Genoud [16]), in dimension $N = 2$ by Genoud [18] and in dimension $N = 1$ by Toland [28]. The existence and uniqueness hold for $0< b < \min \{2,N\}$ and $1< p < p^*_b$.

The ground state satisfies the following Pohozaev’s identities (see relations (1.9)-(1.10) in Farah [12])

$$\begin{aligned} \Vert \nabla Q\Vert ^{2}_{L^2}=\frac{N(p-1)+2b}{2(p+1)}\int |x|^{-b}|Q|^{p+1}\,dx, \end{aligned}$$

and

$$\begin{aligned} E[Q]=\frac{(p-1)s_c}{2(p+1)}\int |x|^{-b}|Q|^{p+1}\,dx. \end{aligned}$$

(1.2)

Genoud [19] and Farah [12] proved the following sharp Gagliardo-Nirenberg inequality which is valid for $N \ge 1$, $0< b < \min \{2, N\},$ and $1< p < p^*_b$

$$\begin{aligned} \int _{\mathbb {R}^N}|x|^{-b}|f(x)|^{p+1}\,dx\le C_{p,N}\Vert \nabla f\Vert _{L^2(\mathbb {R}^N)}^{\frac{N(p-1)+2b}{2}}\Vert f\Vert _{L^2(\mathbb {R}^N)}^{p+1-\frac{(N(p-1)+2b)}{2}}, \end{aligned}$$

(1.3)

where $C_{p,N}>0$ is the sharp constant. More precisely,

$$\begin{aligned} C_{p,N}=\left( \frac{2(p+1)}{N(p-1)+2b}\right) ^{\frac{N(p-1)+2b}{4}}\frac{\left( \displaystyle \int |x|^{-b}|Q|^{p+1}\right) ^{1-\frac{N(p-1)+2b}{4}}}{\Vert Q\Vert ^{p+1-\frac{N(p-1)+2b}{2}}_{L^2(\mathbb {R^N})}}. \end{aligned}$$

(1.4)

This inequality can be seen as an extension to the case $b > 0$ of the classical Gagliardo-Nirenberg inequality.

If u is a solution to (1.1) and $u_0 \in \Sigma = \left\{ f \in H^1(\mathbb {R}^N) ; |x|f \in L^2(\mathbb {R}^N)\right\}$, we define its variance at time t as

$$\begin{aligned} V(t)=\int |x|^2|u(x,t)|^2\,dx. \end{aligned}$$

The variance satisfies the virial identities (see Farah [12, Proposition 4.1])

$$\begin{aligned} V_t(t)&= 4 \text { Im}\int x\cdot \nabla u(x,t) \overline{u}(x,t)\,dx \end{aligned}$$

(1.5)

and

$$\begin{aligned} V_{tt}(t)&= 4(N(p-1)+2b)E[u]-2(N(p-1)+2b-4)\Vert \nabla u\Vert ^{2}_{L^2({\mathbb {R}}^N)}. \end{aligned}$$

(1.6)

Together with the variance, a scale-invariant quantity which plays an important role in the global behavior is $M[u_0]^{\frac{1-s_c}{s_c}}E[u_0]$, which we normalize (for $0< s_c < 1$) as

$$\begin{aligned} {\mathcal {M}}{\mathcal {E}}[u] = \mathcal {ME}[u_0]\ = \frac{ M[u_0]^\frac{1-s_c}{s_c}E[u_0]}{M[Q]^\frac{1-s_c}{s_c}E[Q]} \end{aligned}$$

and call it the mass-energy.

Remark 1.1

From the identity (1.6), if $u_0 \in \Sigma$, $p >1+ \frac{2(2-b)}{N}$ and $E\left[ u_0 \right] < 0$, then the graph of $t \mapsto \int |x|^2 |u|^2$ lies below an inverted parabola, which becomes negative in finite time. Therefore, the solution cannot exist globally and blows up in finite time. Recently, [9] extended this blow-up result to the radial case, and to the case $N = 1$ without symmetry or decaying assumptions. Since we only consider initial data in $\Sigma$, in this work we focus on data with non-negative energy, but avoid rising the energy to a fractional power in the definition of mass-energy so that this quantity makes sense in any setting.

Other useful scale-invariant quantities are the mass-potential-energy

$$\begin{aligned} \mathcal {MP}[u(t)]\ = \frac{ M[u_0]^\frac{1-s_c}{s_c}\displaystyle \int |x|^{-b}|u(t)|^{p+1}}{M[Q]^\frac{1-s_c}{s_c}\displaystyle \int |x|^{-b}|Q|^{p+1}} \end{aligned}$$

and the mass-kinetical-energy

$$\begin{aligned} \mathcal {MK}[u(t)]= \frac{ M[u_0]^\frac{1-s_c}{s_c}\displaystyle \int |\nabla u(t)|^2}{M[Q]^\frac{1-s_c}{s_c}\displaystyle \int |\nabla Q|^2}. \end{aligned}$$

1.1 Dichotomy above the mass-energy threshold

In previous works, [9, 14] and [4] studied the global behavior of solutions to (1.1) below the mass-energy threshold, i.e, in the case $\mathcal {ME}[u_0] < 1$. They proved a dichotomy between blow-up and scattering, depending on the quantity $\mathcal {MK}[u_0]$.

We summarize the global behavior of solutions to (1.1) with $\mathcal {ME}[u_0] < 1$ in the following theorem

Theorem 1.2

([3, 4, 9, 14]). Let u(x, t) be a solution of (1.1) and $0< s_c < 1$. Assume $\mathcal {ME}[u_0] < 1$. Then

• (i) If $\mathcal {MP}[u_0] > 1$, and either $u_0\in \Sigma ,$ or $u_0$ is radial, or $N = 1$, then the solution blows up in finite time, in both time directions.

• (ii) If $\mathcal {MP}[u_0] < 1$, $N \ge 2$, $b < \min \left\{ \frac{N}{2},2\right\}$, then u scatters in both time directions in $H^1({\mathbb {R}}^N)$.

Remark 1.3

In [14] and [9], Theorem 1.2 was proven using $\mathcal {MK}[u_0]$ instead of $\mathcal {MP}[u_0]$. We show the equivalence, in the case $\mathcal {ME}[u_0] \le 1$, in Proposition 2.1. Therefore, by Theorem 1.2 and Theorem 1.5 below, as in the case $\mathcal {ME}[u_0] > 1$ the equivalence does not hold, the quantity that governs the dichotomy between blow-up and scattering is, in any case, $\mathcal {MP}[u_0]$.

Remark 1.4

The case $\mathcal {MP}[u_0] = 1$ cannot occur if $\mathcal {ME}[u_0] < 1$ (see [14, Lemma 4.2, item (ii)]).

We are interested here in criteria that include initial data above the threshold $\mathcal {ME}[u_0] = 1$. The first theorem we prove is a dichotomy

Theorem 1.5

Let u be a solution of (1.1), where $1+\frac{2(2-b)}{N}< p < p^*_b$. Assume $N \ge 2$, $u_0\in \Sigma$ and

$$\begin{aligned} \mathcal {ME}[u_{0}]\left( 1-\displaystyle \frac{(V_t(0))^2}{32E[u_0]V(0)}\right) \le 1. \end{aligned}$$

(1.7)

• (i) (Blow-up) If

  $$\begin{aligned} \mathcal {MP}[u_0] > 1 \end{aligned}$$

  (1.8)

  and

  $$\begin{aligned} V_t(0)\le 0, \end{aligned}$$

  (1.9)

  then u(t) blows-up in finite positive, $T_{+}<\infty .$

• (ii) (Boundedness and scattering) If

  $$\begin{aligned} \mathcal {MP}[u_0] < 1 \end{aligned}$$

  (1.10)

  and

  $$\begin{aligned} V_t(0)\ge 0, \end{aligned}$$

  (1.11)

  then

  $$\begin{aligned}&\limsup _{t\rightarrow T_+(u)}M[u_0]^{1-s_c}\left( \int |x|^{-b}|u(t)|^{p+1}\right) ^{s_c}\nonumber \\&\quad <M[Q]^{1-s_c}\left( \int |x|^{-b}|Q|^{p+1}\right) ^{s_c}. \end{aligned}$$

  (1.12)

  In particular, $T_+=+\infty$. Moreover, u scatters forward in time in $H^1({\mathbb {R}}^N)$.

Remark 1.6

If $\mathcal {ME}[u_0] < 1$, the conclusion of Theorem 1.5 follows from Theorem 1.2. Theorem is new only in the case $\mathcal {ME} [u_0]\ge 1$.

Remark 1.7

The proof of Theorem 1.5 shows that there are two disjoint subsets (defined by (1.7), (1.8) and (1.9); and by (1.7), (1.10) and (1.11)) that are stable under the INLS flow and contain solutions with arbitrary mass and energy (see, for example, Remark 1.11 below).

Remark 1.8

We prove in Sect. 3 that any solution of (1.1) that satisfies (1.12) scatters for positive time. Replacing $\mathcal {MP}[u_0]$ by $\mathcal {MK}[u_0]$, this result is already known (see [14]). Due to the one-sided implication (2.1), our assumption is weaker. Therefore, Theorem 1.5 improves known results.

Remark 1.9

The scattering statement of Theorem 1.5 is optimal in the following sense: If $u_0 \in H^1({\mathbb {R}}^N)$ has finite variance and scatters forward in time, then there exists $t_0 \ge 0$ such that (1.7), (1.10) and (1.11) are satisfied by u(t) for all $t \ge t_0$. In fact, if u(t) scatters forward in time, then $\displaystyle \int |x|^{-b}|u(t)|^{p+1}\rightarrow 0$. This implies $E[u_0] > 0$ and, by (1.6),

$$\begin{aligned} V_t(t) \approx 16 E[u_0]t\quad \quad \text {and } V(t) \approx 8 E[u_0] t^2 \end{aligned}$$

which implies

$$\begin{aligned} \mathcal {ME}[u_{0}]\left( 1-\displaystyle \frac{(V_t(t))^2}{32E[u_0]V(t)}\right) \rightarrow 0, \text { as }t \rightarrow +\infty . \end{aligned}$$

As a consequence of Theorem 1.5, we obtain

Corollary 1.10

Let $\gamma \in {\mathbb {R}}\backslash \{0\}$, $v_0 \in \Sigma$ such that $\mathcal {ME}[v_0] < 1$, and $u^\gamma$ be the solution of (1.1) with initial data

$$\begin{aligned} u_0^\gamma = e^{i\gamma |x|^2}v_0. \end{aligned}$$

• (i) If $\mathcal {MP}[v_0] > 1$, then for any $\gamma < 0$, $u^\gamma$ blows up in finite positive time;

• (ii) If $\mathcal {MP}[v_0] < 1$, then for any $\gamma > 0$, $u^\gamma$ satisfies (1.12). Moreover, $u^\gamma$ scatters forward in time in $H^1({\mathbb {R}}^N)$.

Remark 1.11

With the above corollary, we can predict the behavior of a class of solutions with arbitrarily large energy. If $\mathcal {ME}[v_0] < 1$, then

$$\begin{aligned} E[u_0^{\gamma }] = 4\gamma ^2 \Vert xv_0\Vert _{L^2}^2+4\gamma \text { Im } \int x \cdot \nabla v_0 \bar{v}_0 + E[v_0] \end{aligned}$$

and $E[u_0^\gamma ] \rightarrow +\infty$ as $\gamma \rightarrow \pm \infty$.

Remark 1.12

Note that the statement of Theorem 1.5 is not symmetric in time as the statement of Theorem 1.2. Indeed, Corollary 1.13 below shows solutions with different behaviors in positive and negative times.

Corollary 1.13

Let $\gamma \in {\mathbb {R}}$ and $Q^\gamma$ be the solution to (1.1) with initial data

$$\begin{aligned} Q_0^\gamma = e^{i\gamma |x|^2}Q. \end{aligned}$$

• (i) If $\gamma > 0$, then $Q^\gamma$ is globally defined on $[0,+\infty )$, scatters forward in time and blows up backwards in time.

• (ii) If $\gamma < 0$, then $Q^\gamma$ is globally defined on $(-\infty ,0]$, scatters backward in time and blows up forward in time.

1.2 Blow-up criteria

The blow up criterion of [29], Zakharov [31] and Glassey [21] for the NLS use the second derivative of the variance V(t) to show that finite variance, negative energy solutions blow up in finite time. The second derivative of the variance is also used in [26], but with an approach based on classical mechanics, resulting in a finer blow-up criterion. This and another criteria were proven in [24] for the 3D cubic NLS. The argument was extended in [10] to the focusing mass-supercritical NLS in any dimension, and examples were given to show that these new criteria are not equivalent to the previous ones. We extend these criteria for the focusing, mass-supercritical INLS equation in any dimension:

Theorem 1.14

Suppose that $u_0\in \Sigma$ and $N \ge 1$. The following inequality is a sufficient condition for blow-up in positive finite time for solutions to (1.1) with $0< s_c < 1$ and $E[u_0] > 0$

$$\begin{aligned} \frac{V_t(0)}{M[u_0]} < \sqrt{8Ns_c} g \left( \frac{4}{Ns_c}\frac{E[u_0]V(0)}{M[u_0]^2}\right) , \end{aligned}$$

where

$$\begin{aligned} g(x) = \left\{ \begin{array}{l} \sqrt{\frac{1}{kx^k}+x-(1+\frac{1}{k})}\, \text{ if } 0 < x\le 1\\ -\sqrt{\frac{1}{kx^k}+x-(1+\frac{1}{k})}\, \text{ if } x\ge 1 \end{array} \right. \text{ with } k = \frac{(p -1)s_c}{2}. \end{aligned}$$

(1.13)

Theorem 1.15

Suppose that $u_0\in \Sigma$ and $N\ge 1$. The following inequality is a sufficient condition for blow-up in positive finite time for solutions to (1.1) with $0< s_c < 1$ and $E[u_0]>0$

$$\begin{aligned} \frac{V_t(0)}{M[u_0]}<\frac{4\sqrt{2} M[u_0]^{\frac{1}{2}-\frac{p+1}{N(p-1)+2b}}E[u_0]^{\frac{s_c}{N}}}{C}g\left( C^2\frac{E[u_0]^{\frac{4}{N(p-1)+2b}}V(0)}{M[u_0]^{1+\frac{2(p+1)}{N(p-1)+2b}}}\right) , \end{aligned}$$

where g is defined in (1.13),

$$\begin{aligned} C=\left( \frac{2(p+1)}{s_c(p-1)}(C_{p,N})^{\frac{N(p-1)+2b}{2}+(p+1)}\right) ^{\frac{2}{N(p-1)+2b}} \end{aligned}$$

and $C_{p,N}$ the a sharp constant in the interpolation inequality (1.3).

Remark 1.16

For real-valued initial data, Theorem 1.15 is an improvement of Theorem 1.14 if

$$\begin{aligned} \mathcal {ME}[u_0] > \left( \frac{Ns_c C^2}{4}\right) ^{\frac{N(p-1)+2b}{N(p-1)+2b-4}}, \end{aligned}$$

since in this case the right-hand side of Theorem 1.15 is bigger.

Remark 1.17

In both theorems, the restriction $s_c < 1$ is only needed to ensure the local well-posedness.

This paper is structured as follows: In Sect. 2, we prove the boundedness and blow-up part of Theorem 1.5. The scattering part is proven in Sect. 3. In Sect. 4, we show two non-equivalent blow-up criteria for the INLS (Theorems 1.14 and 1.15).

2 Boundedness and blow-up

We start this section with the proof of the equivalence between using $\mathcal {MK}[u_0]$ and $\mathcal {MP}[u_0]$ in the dichotomy when $\mathcal {ME}[u_0] \le 1$.

Proposition 2.1

If $f \in H^1({\mathbb {R}}^N)$, then

$$\begin{aligned} \mathcal {MK}[f]< 1 \Longrightarrow \mathcal {MP}[f]<1. \end{aligned}$$

(2.1)

Furthermore, assume $\mathcal {ME}[f] \le 1$. Then

$$\begin{aligned} \mathcal {MK}[f]< 1 \Longleftrightarrow \mathcal {MP}[f]<1. \end{aligned}$$

(2.2)

Proof

We write the sharp Gagliardo-Nirenberg inequality (1.3) as

$$\begin{aligned} \left( \mathcal {MP}[f]\right) ^\frac{4}{N(p-1)+2b}\le \mathcal {MK}[f], \end{aligned}$$

and (2.1) follows. Now, if $\mathcal {MP}[f]< 1$ and $\mathcal {ME}[f] \le 1$, then

$$\begin{aligned} M[Q]^\frac{1-s_c}{s_c}E[Q] \ge M[f]^\frac{1-s_c}{s_c}E[f] >&\frac{1}{2} M[f]^\frac{1-s_c}{s_c}\int |\nabla f|^2\,dx\\&-\frac{1}{p+1}M[Q]^\frac{1-s_c}{s_c}\int |x|^{-b}|Q|^{p+1}\,dx \end{aligned}$$

taking the first and last member, we conclude $\mathcal {MK}[f]< 1$. $\square$

We also point out that the inequalities in (2.2) can be replaced by equalities: we can scale f so that $M[f] = M[Q]$. By similar arguments as the ones used in proving (2.1) and (2.2), $\mathcal {MP}[f] = 1$ or $\mathcal {MK}[f] = 1$ in the case $\mathcal {ME}[f]\le 1$, implies $\mathcal {MP}[f] = \mathcal {MK}[f] = \mathcal {ME}[f] = 1$. In this case, f is equal to Q up to scaling and phase.

We now turn to the proof of Theorem 1.5. Start rewriting the Gagliardo-Nirenberg inequality (1.3) as

$$\begin{aligned} \left( \int |x|^{-b}|f|^{p+1}\,dx\right) ^{\frac{4}{N(p-1)+2b}}\le C_QM[f]^{\kappa }\int \, |\nabla u|^2\,dx, \end{aligned}$$

(2.3)

where

$$\begin{aligned} \kappa =\frac{2(p+1)}{N(p-1)+2b}-1 \end{aligned}$$

and

$$\begin{aligned} C_Q&:=(C_{p,N})^{\frac{4}{N(p-1)+2b}}=\frac{2(p+1)}{N(p-1)+2b}\frac{\left( \displaystyle \int |x|^{-b}|Q|^{p+1}\,dx\right) }{M[Q]^{\kappa }}^{\frac{4}{N(p-1)+2b}-1}\\&\\&=\left( \frac{8(p+1)}{A}\right) ^{\frac{4}{N(p-1)+2b}}\frac{s_c(p-1)}{N(p-1)+2b}\cdot \frac{E[Q]}{M[Q]^{\kappa }}^{\frac{4}{N(p-1)+2b}-1} \end{aligned}$$

and

$$\begin{aligned} A:=2(N(p-1)+2b-4)=4(p-1)s_c. \end{aligned}$$

We use the following Cauchy-Schwarz inequality, proved by Banica [1]. We include the proof here for the sake of completeness.

Lemma 2.2

Let $f\in H^1({\mathbb {R}}^N)$ such that $|x|f\in L^2({\mathbb {R}}^N)$. Then,

$$\begin{aligned} \left( \text {Im}\int x\cdot \nabla f\, \overline{f}\,dx\right) ^2\le&\int |x|^2|f|^2\,dx \Bigg [\int |\nabla f|^2\,dx\\&-\frac{1}{C_QM^\kappa }\left( \int |x|^{-b}|f|^{p+1}\,dx\right) ^{\frac{4}{N(p-1)+2b}}\Bigg ]. \end{aligned}$$

Proof

Given $f \in H^1({\mathbb {R}}^N)$ and $\lambda >0$, we have

$$\begin{aligned} \nabla \left( e^{i\lambda |x|^2}f\right) =2i\lambda e^{i\lambda |x|^2}xf+e^{i\lambda |x|^2}\nabla f=e^{i\lambda |x|^2}(2i\lambda \,x f+\nabla f). \end{aligned}$$

Thus,

$$\begin{aligned} \int \left| \nabla \left( e^{i\lambda |x|^2}f\right) \right| ^2\,dx&=\int e^{i\lambda |x|^2}(2i\lambda \,x f+\nabla f)e^{-i\lambda |x|^2}(-2i\lambda \,x \overline{f}+\nabla \overline{f})\,dx\\&=4\lambda ^2\int |x|^2|f|^{2}\,dx+4\lambda \text { Im}\int x\cdot \nabla f\,\overline{f}\,dx\\&\quad \quad +\int |\nabla f|^2\,dx \end{aligned}$$

and from the Gagliardo-Niremberg inequality (2.3), for all $\lambda \in {\mathbb {R}}$ we get

$$\begin{aligned} C_QM[f]^\kappa \left[ 4\lambda ^2\int |x|^2|f|^2\,dx+4\lambda \text { Im}\int x\cdot \nabla f\,\overline{f}\,dx+\int |\nabla f|^2\,dx\right]&\\ -\left( \int |x|^{-b}|f|^{p+1}\,dx\right) ^{\frac{4}{N(p-1)+2b}}&\ge 0. \end{aligned}$$

Note that the left-hand side of inequality above is a quadratic polynomial in $\lambda$ . The discriminant of this polynomial is non-positive, wich yields the conclusion of the lemma. $\square$

Proof of Theorem 1.5

We will assume

$$\begin{aligned} \mathcal {ME}[u_0]\ge 1, \end{aligned}$$

(2.4)

as the case $\mathcal {ME}[u_0] < 1$ has been proven by [14]. By (1.6), we have

$$\begin{aligned} \int |\nabla u|^2\,dx=\frac{4(N(p-1)+2b)E[u_0]-V_{tt}}{A}. \end{aligned}$$

(2.5)

Furthermore,

$$\begin{aligned} \int |x|^{-b}|u|^{p+1}\,dx&=(p+1)\frac{8\Vert \nabla u\Vert ^2_2-V_{tt}}{4(N(p-1)+2b)}\\&=(p+1)\frac{16E[u_0]-V_{tt}}{4(N(p-1)+2b)}\\&\quad \quad +\frac{16}{4(N(p-1)+2b)}\int |x|^{-b}|u|^{p+1}\,dx. \end{aligned}$$

Solving the equality above for $\displaystyle \int |x|^{-b}|u|^{p+1}\,dx$, we have

$$\begin{aligned} \int |x|^{-b}|u|^{p+1}\,dx&=(p+1)\frac{16E[u_0]-V_{tt}}{2A}. \end{aligned}$$

(2.6)

Note that the expression (2.6) implies that $V_{tt}\le 16E[u_0]$ for all t. In view of the equation (1.5), the derivative of variance V(t), and Lemma 2.2 we get,

$$\begin{aligned} (V_t(t))^2&=16\left( \text {Im}\int x\cdot \nabla u(t)\,\overline{u}(t)\,dx\right) ^2\nonumber \\&\le 16\int V(t)\Bigg [\int |\nabla u(t)|^2\,dx\nonumber \\&\quad \quad \quad \quad \quad -\frac{1}{C_QM[u_0]^\kappa }\left( \int |x|^{-b}|u(t)|^{p+1}\,dx\right) ^{\frac{4}{N(p-1)+2b}}\Bigg ]. \end{aligned}$$

(2.7)

If $z(t)=\sqrt{V(t)}$, then

$$\begin{aligned} z_t(t)=\frac{1}{2}\frac{V_t(t)}{\sqrt{V(t)}}. \end{aligned}$$

Dividing (2.7) by V(t), using (2.5), (2.6) and (2.7), we have

$$\begin{aligned} z_t^2(t)&=\frac{1}{4}\frac{(V_t(t))^{2}}{V(t)}\\&\le 4\Bigg [\frac{4(N(p-1)+2b)E[u_0]-V_{tt}}{A}\\&\quad \quad -\frac{1}{C_QM[u_0]^\kappa }\left( \frac{(p+1)(16E[u_0]-V_{tt})}{2A}\right) ^{\frac{4}{N(p-1)+2b}}\Bigg ], \end{aligned}$$

that is,

$$\begin{aligned} z_t^2(t)\le 4\varphi (V_{tt}), \end{aligned}$$

(2.8)

where

$$\begin{aligned} \varphi (\alpha )&=\frac{4(N(p-1)+2b)E[u_0]-\alpha }{A}\\&\quad -\frac{1}{C_QM[u_0]^\kappa }\left( \frac{(p+1)(16E[u_0]-\alpha )}{2A}\right) ^{\frac{4}{N(p-1)+2b}} \end{aligned}$$

is defined for $\alpha \in (-\infty ,16E[u_0]]$. We have

$$\begin{aligned} \varphi '(\alpha )=-\frac{1}{A}+\frac{4(16E[u_0]-\alpha )^{\frac{4}{N(p-1)+2b}-1}}{C_QM[u_0]^{\kappa }(N(p-1)+2b)}\left( \frac{p+1}{2A}\right) ^{\frac{4}{N(p-1)+2b}}. \end{aligned}$$

Consider $\alpha _m\in (-\infty , 16E[u_0])$ such that $\varphi '(\alpha _m)=0$, that is,

$$\begin{aligned} \frac{1}{A}=\frac{4(16E[u_0]-\alpha _m)^{\frac{4}{N(p-1)+2b}-1}}{C_QM[u_0]^{\kappa }(N(p-1)+2b)}\left( \frac{p+1}{2A}\right) ^{\frac{4}{N(p-1)+2b}}. \end{aligned}$$

(2.9)

Since $s_c>0$,

$$\begin{aligned} \frac{4}{N(p-1)+2b}-1=\frac{4-N(p-1)-2b}{N(p-1)+2b}=-\frac{2s_c}{(p-1)(N(p-1)+2b)}<0, \end{aligned}$$

therefore $\varphi$ is decreasing on $(-\infty , \alpha _m)$ and increasing on $(\alpha _m,16E[u_0]]$. Note that (2.9) implies

$$\begin{aligned} \frac{\alpha _m}{8}=\frac{(\alpha _m-16E)(N(p-1)+2b)}{4A}+\frac{4(N(p-1)+2b)E}{A}-\frac{\alpha _m}{A}= \varphi (\alpha _m). \end{aligned}$$

Using (2.9) and (1.4), we have

$$\begin{aligned} \frac{E[Q]}{M[Q]^{\kappa }}^{\frac{4}{N(p-1)+2b}-1}=\frac{\left( E[u_0]-\frac{\alpha _m}{16}\right) }{M[u_0]^\kappa }^{\frac{4}{N(p-1)+2b}-1}, \end{aligned}$$

hence raising both sides to $\frac{2(p-1)}{N(p-1)+2b}$, we get

$$\begin{aligned} \left( \frac{M[u_0]}{M[Q]}\right) ^{\frac{1-s_c}{s_c}}\frac{E[u_0]-\frac{\alpha _m}{16}}{E[Q]}=1. \end{aligned}$$

(2.10)

As a consequence of (2.4)

$$\begin{aligned} \left( \frac{M[u_0]}{M[Q]}\right) ^{\frac{1-s_c}{s_c}}\frac{E[u_0]-\frac{\alpha _m}{16}}{E[Q]}=1 \le \mathcal {ME}[u_0]=\left( \frac{M[u_0]}{M[Q]}\right) ^{\frac{1-s_c}{s_c}}\frac{E[u_0]}{E[Q]}, \end{aligned}$$

i.e,

$$\begin{aligned} \alpha _m\ge 0, \end{aligned}$$

and by (1.7) and (2.10),

$$\begin{aligned} (z_t(0))^2&=-\left( 1-\frac{(V_t(0))^2}{32E[u_0]V(0)}\right) \frac{8E[u_0]\mathcal {ME}[u_0]}{\mathcal {ME}[u_0]}+8E[u_0]\nonumber \\&\ge -\frac{8E[u_0]}{\mathcal {ME}[u_0]}\left( \frac{M[u_0]}{M[Q]}\right) ^{\frac{1-s_c}{s_c}}\frac{E[u_0]-\frac{\alpha _m}{16}}{E[Q]}+8E[u_0]\nonumber \\&= \frac{\alpha _m}{2}=4\varphi (\alpha _m). \end{aligned}$$

(2.11)

We first prove case (i) of Theorem 1.5. Suppose that $u\in \Sigma$ satisfies (1.8) and (1.9). Note that (1.9) is equivalent to

$$\begin{aligned} z_t(0)=\frac{V_t(0)}{2\sqrt{V(0)}}\le 0. \end{aligned}$$

(2.12)

In view of (1.2), the assumption (1.8) means

$$\begin{aligned} \left( \frac{M[u_0]}{M[Q]}\right) ^{\frac{1-s_c}{s_c}}\frac{A\displaystyle \int |x|^{-b}|u_0|^{p+1}\,dx}{(p+1)E[Q]}=\left( \frac{M[u_0]}{M[Q]}\right) ^{\frac{1-s_c}{s_c}}\frac{\displaystyle \int |x|^{-b}|u_0|^{p+1}\,dx}{\displaystyle \int |x|^{-b}|Q|^{p+1}\,dx}>1 \end{aligned}$$

and consequently, from (2.6)

$$\begin{aligned} V_{tt}(0)=-\frac{2A}{p+1}\int |x|^{-b}|u_0|^{p+1}+16E[u_0]<\alpha _m. \end{aligned}$$

(2.13)

Note that, for all $t>0$

$$\begin{aligned} z_{tt}(t)&=\frac{d}{dt}\left[ \frac{V_t(t)}{2\sqrt{V(t)}}\right] =\frac{V_{tt}(t)}{2\sqrt{V(t)}}-\frac{(V_t(t))^2}{4\sqrt{V(t)^3}}\nonumber \\&=\frac{1}{z(t)}\left( \frac{V_{tt}(t)}{2}-z_t^2(t)\right) . \end{aligned}$$

(2.14)

Hence from (2.11) and (2.13), we have

$$\begin{aligned} z_{tt}(0)=\frac{1}{z(0)}\left( \frac{V_{tt}(0)}{2}-z_t^2(0)\right) <\frac{1}{z(0)}\left( \frac{\alpha _m}{2}-\frac{\alpha _{m}}{2}\right) =0. \end{aligned}$$

Suppose that $z_{tt}(\tilde{t})\ge 0$ for some $\tilde{t}$ belonging to $[0,T_+(u))$. Then, as $z_{tt}$ is continuous on $[0,T_+(u))$, by the intermediate value theorem there exists $t_0\in (0,T_+(u))$ such that

$$\begin{aligned} \forall t \in [0,t_0),\,\,z_{tt}(t)<0\, \text{ and } z_{tt}(t_0)=0. \end{aligned}$$

Thus for (2.11) and (2.12)

$$\begin{aligned} \forall t\in (0,t_0],\,\,z_{t}(t)<z_t(0)\le -\sqrt{4\varphi (\alpha _m)}. \end{aligned}$$

We have, thus,

$$\begin{aligned} \forall t\in (0,t_0],\,\,z_{t}^{2}(t)> 4\varphi (\alpha _m). \end{aligned}$$

Using the inequality above and (2.8),

$$\begin{aligned} \forall t\in (0,t_0],\,\,4\varphi (V_{tt}(t))\ge z_{t}^{2}(t)>4\varphi (\alpha _m). \end{aligned}$$

Therefore, $V_{tt}(t)\ne \alpha _m$ for $t\in (0,t_0]$. Since $V_{tt}(0)<\alpha _m$ and by the continuity of $V_{tt}$,

$$\begin{aligned} \forall t\in [0,t_0],\,\,V_{tt}(t)<\alpha _m. \end{aligned}$$

(2.15)

Since $V_{tt}(t)\ne \alpha _m$ and by (2.15), we get

$$\begin{aligned} z_{tt}(t_0)=\frac{1}{z(t_0)}\left( \frac{V_{tt}(t_0)}{2}-z_t^{2}(t_0)\right) <\frac{1}{z(t_0)}\left( \frac{\alpha _m}{2}-\frac{\alpha _m}{2}\right) , \end{aligned}$$

contradicting the definition of $t_0$. Therefore,

$$\begin{aligned} z_{tt}<0 \text{ for } \text{ all } t\in [0,T_+(u)). \end{aligned}$$

(2.16)

By contradiction, suppose that $T_+(u)=+\infty$. From (2.12) and (2.16),

$$\begin{aligned} \forall t>0,\,\, z_t(t)<z_t(0)\le 0, \end{aligned}$$

a contradiction with nonnegativity of z(t).

We now prove case (ii) of Theorem 1.5. We assume, besides the conditions (1.7) and (2.4), that (1.10) and (1.11) hold. That implies, in the same way as we did in case (i),

$$\begin{aligned} z_t(0)\ge 0 \end{aligned}$$

(2.17)

and

$$\begin{aligned} V_{tt}(0)>\alpha _m. \end{aligned}$$

(2.18)

We affirm that there is $t_0\ge 0$ such that

$$\begin{aligned} z_t(t_0)>2\sqrt{\varphi (\alpha _m)}. \end{aligned}$$

(2.19)

Indeed, by (2.11) and (2.17),

$$\begin{aligned} z_t(0)\ge 2\sqrt{\varphi (\alpha _m)}. \end{aligned}$$

(2.20)

If $z_t(0)> 2\sqrt{\varphi (\alpha _m)}$, then choose $t_0=0$ and we have the result. If not,

$$\begin{aligned} z_{tt}(0)=\frac{1}{z(0)}\left( \frac{V_{tt}(0)}{2}-z_t^{2}(0)\right) >\frac{1}{z(0)}\left( \frac{\alpha _m}{2}-\frac{\alpha _m}{2}\right) =0, \end{aligned}$$

by (2.18) and (2.20). Hence, there is a small $t_0>0$ satisfying (2.19).

Let $\varepsilon _0$ be a positive small number and assume

$$\begin{aligned} z_t(t_0)\ge 2\sqrt{\varphi (\alpha _m)}+2\varepsilon _0. \end{aligned}$$

(2.21)

We will show that, for all $t\le t_0$

$$\begin{aligned} z_t(t)>2\sqrt{\varphi (\alpha _m)}+\varepsilon _0. \end{aligned}$$

(2.22)

Suppose (2.22) is false, and define

$$\begin{aligned} t_1=\inf \{t\ge t_0;\, z_t(t)\le 2\sqrt{\varphi (\alpha _m)}+\varepsilon _0\}. \end{aligned}$$

By (2.21) $t_1>t_0$. By continuity of $z_t$,

$$\begin{aligned} z_t(t_1)=2\sqrt{\varphi (\alpha _m)}+\varepsilon _0 \end{aligned}$$

(2.23)

and

$$\begin{aligned} \forall \in [t_0,t_1],\,\,\,\,z_t(t)\ge 2\sqrt{\varphi (\alpha _m)}+\varepsilon _0. \end{aligned}$$

(2.24)

In view of (2.8),

$$\begin{aligned} \forall t\in [t_0,t_1],\,\, (2\sqrt{\varphi (\alpha _m)}+\varepsilon _0)^2\le z_t^{2}(t)\le 4\varphi (V_{tt}(t)). \end{aligned}$$

(2.25)

Hence, $\varphi (V_{tt}(t))>\varphi (\alpha _m)$ for all $t\in [t_0,t_1]$, so, $V_{tt}(t)\ne \alpha _m$ and by continuity $V_{tt}(t)>\alpha _m$ for $t\in [t_0,t_1]$. Using the Taylor expansion of $\varphi$ around $\alpha =\alpha _m$, there exists $a>0$ such that, if $|\alpha -\alpha _m|\le 1$, then

$$\begin{aligned} \varphi (\alpha )\le \varphi (\alpha _m)+a(\alpha -\alpha _m)^2. \end{aligned}$$

(2.26)

We show that there exists a universal constant $D>0$ such that

$$\begin{aligned} \forall \,t\in [t_0,t_1]\,\, V_{tt}(t)\ge \alpha _m +\frac{\sqrt{\varepsilon _0}}{D}. \end{aligned}$$

(2.27)

Consider two cases:

• a) If $V_{tt}(t)\ge \alpha _m+1$, then for $D>0$ large, we get (2.27)

• b) If $\alpha _m<V_{tt}(t)\le \alpha _m+1$, then by (2.25) and (2.26), we obtain

  $$\begin{aligned} (2\sqrt{\varphi (\alpha _m)}+\varepsilon _0)^2\le (z_t(t))^2\le 4\varphi (V_{tt}(t))\le 4\varphi (\alpha _m)+4a(V_{tt}(t)-\alpha _m)^2. \end{aligned}$$

  Thus,

  $$\begin{aligned} 4\sqrt{\varphi (\alpha _m)}\varepsilon _0<4\sqrt{\varphi (\alpha _m)}\varepsilon _0+\varepsilon _0^2\le 4a(V_{tt}-\alpha _m)^2, \end{aligned}$$

  and choosing $D=\sqrt{a}(\varphi (\alpha _m))^{-\frac{1}{4}}$, (2.27) holds.

Furthermore, by (2.14) and (2.24)

$$\begin{aligned} z_{tt}(t_1)&=\frac{1}{z(t_1)}\left( \frac{V_{tt}(t_1)}{2}-z_{t}^{2}(t_1)\right) \\&\ge \frac{1}{z(t_1)}\left( \frac{\alpha _m}{2}+\frac{\sqrt{\varepsilon }_0}{2D}-(2\sqrt{\varphi (\alpha _m)}+\varepsilon _0)^2\right) \\&\ge \frac{1}{z(t_1)}\left( \frac{\sqrt{\varepsilon _0}}{2D}-4\varepsilon \sqrt{\varphi (\alpha _m)}-\varepsilon _0^2\right) >0, \end{aligned}$$

if $\varepsilon _0$ is small enough. That is, $z_t$ is increasing close to $t_1$, contradicting (2.23) and(2.24). This shows (2.22). Note that we have also shown that the inequality (2.27) holds for all $t\in [t_0,T_+(u))$. Hence, by (2.6), (1.2) and (2.10)

$$\begin{aligned} M[u_0]^{1-s_c}\left( \int |x|^{-b}|u(t)|^{p+1}\, dx\right) ^{s_c}&=M[u_0]^{1-s_c}\left[ \frac{p+1}{2A}(16E[u_0]-V_{tt}(t))\right] ^{s_c}\\&<M[u_0]^{1-s_c}\left[ \frac{p+1}{2A}\left( 16E[u_0]-\alpha _m\right) \right] ^{s_c}\\&=M[u_0]^{1-s_c}\left[ \frac{8(p+1)}{A}E[Q]\right] ^{s_c}\\&=M[Q]^{1-s_c}\left[ \int |x|^{-b}|Q|^{p+1}\, dx\right] ^{s_c}. \end{aligned}$$

$\square$

2.1 Dichotomy for quadratic phase initial data

We now prove Corollary 1.10, except for the scattering statement, which will follow from the results in Sect. 3.

Proof of Corollary 1.10

Let $v_0$ satisfy $\mathcal {ME}[v_0] < 1$, $\gamma \in {\mathbb {R}}\backslash \{0\}$ and u be the solution with initial data $u_0 = e^{i\gamma |x|^2}v_0$. We assume

$$\begin{aligned} \mathcal {ME}[u_0] \ge 1 \end{aligned}$$

(otherwise the result follows from Theorem 1.2).

We will now show that $u_0$ satisfies the assumption of Theorem 1.5. We need to calculate

$$\begin{aligned} E[u_0] = E[v_0] + 2 \gamma \text{ Im }\int x \cdot \nabla v_0 \bar{v}_0\, dx + 2 \gamma ^2 \int |x|^2|v_0|^2\,dx \end{aligned}$$

(2.28)

and

$$\begin{aligned} \text{ Im }\int \bar{u}_0\,x\cdot \nabla u_0\,dx = \text{ Im }\int \bar{v}_0\, x \cdot \nabla v_0\,dx + 2 \gamma \int |x|^2|v_0|^2\,dx. \end{aligned}$$

Rewriting the above equations,

$$\begin{aligned} E[u_0] - \frac{\left( \text{ Im }\displaystyle \int \bar{u}_0\,x\cdot \nabla u_0\,dx\right) ^2}{2\displaystyle \int |x|^2|u_0|^2\,dx}= & {} E[v_0] - \frac{\left( \text{ Im }\displaystyle \int \bar{v}_0\,x\cdot \nabla v_0\,dx\right) ^2}{2\displaystyle \int |x|^2|v_0|^2\,dx}\nonumber \\\le & {} E[v_0], \end{aligned}$$

(2.29)

or,

$$\begin{aligned} \mathcal {ME}[u_0]\left[ 1-\frac{\left( \text{ Im }\displaystyle \int \bar{u}_0x\cdot \nabla u_0\right) ^2}{2E[u_0]\int |x|^2|u_0|^2}\right] = \mathcal {ME}[v_0] \le 1. \end{aligned}$$

(2.30)

Therefore, the assumption (1.7) follows from (1.5) and (2.30).

We will assume here $\gamma > 0$ and $\mathcal {MP}[v_0] < 1$, as the proof of the other case is very similar. First note that, since $\mathcal {ME}[v_0] < 1$ and $\int |x|^2|v_0|^2 > 0$, there is only one positive solution of

$$\begin{aligned} M[v_0]^\frac{1-s_c}{s_c}&\Bigg (E[v_0] + 2 \gamma \text{ Im }\int x \cdot \nabla v_0 \bar{v}_0\,dx\nonumber \\&\quad \quad \quad \quad \quad + 2 \gamma ^2 \int |x|^2|v_0|^2\,dx\Bigg )=M[Q]^\frac{1-s_c}{s_c} E[Q]. \end{aligned}$$

(2.31)

Now, since $\mathcal {ME}[u_0] \ge 1$ and $\gamma >0$, (2.28), we have $\gamma \ge \gamma _c^+$, where $\gamma _c^+$ is the positive solution of (2.31). Rewriting (2.31), we have

$$\begin{aligned} \frac{\gamma _c^+ \text{ Im }\int x \cdot \nabla v_0 \bar{v}_0 \, dx + (\gamma _c^+)^2 \int |x|^2|v_0|^2\, dx}{M[Q]^\frac{1-s_c}{s_c}E[Q]}=\frac{1-\mathcal {ME}[v_0]}{2M[v_0]^\frac{1-s_c}{s_c}} > 0, \end{aligned}$$

which implies

$$\begin{aligned} \text{ Im }\int x \cdot \nabla v_0 \bar{v}_0 \, dx + \gamma _c^+\int |x|^2|v_0|^2 \, dx> 0. \end{aligned}$$

Using that $\gamma \ge \gamma _c^+$, we see that

$$\begin{aligned} \text{ Im }\int x \cdot \nabla u_0 \bar{u}_0 \, dx = \text{ Im }\int x \cdot \nabla v_0 \bar{v}_0 \, dx + \gamma \int |x|^2|v_0|^2 \, dx> 0, \end{aligned}$$

which yields (1.11). Since Theorem 1.5 applies, we conclude the proof. $\square$

We next prove prove Corollary 1.13, except for the scattering statement.

Proof of Corollary 1.13

Given that $\bar{u}(x,-t)$ is a solution of (1.1) if u(x, t) is a solution, we can assume $\gamma > 0$. We only need to prove that

$$\begin{aligned}&\text{ Im }\displaystyle \int x\cdot \nabla Q^\gamma (t_0)\overline{Q^\gamma }(t_0)\,dx \ge 0, \\&\mathcal {MP}[Q^\gamma (t_0)] < 1 \end{aligned}$$

and

$$\begin{aligned} ME[Q^{\gamma }(t_0)]\left( 1-\frac{(V_t(t_{0}))^2}{32E[Q^\gamma (t_0)]V(t_{0})}\right) \le 1, \end{aligned}$$

for some $t_0>0$, where $V(t) = \displaystyle \int |x|^2|Q^\gamma (x,t)|^2\,dx$. First note that, for $Q^\gamma _0=e^{i\gamma |x|^2}Q$, we have

$$\begin{aligned} \nabla Q^\gamma _0=(2i\gamma x Q+\nabla Q)e^{i\gamma |x|^2} \end{aligned}$$

and

$$\begin{aligned} \triangle Q^{\gamma }_0=e^{i\gamma |x|^2}(2iN\gamma Q+4i\gamma x\cdot \nabla Q-4\gamma ^2|x|^2Q+\triangle Q). \end{aligned}$$

(2.32)

Thus,

$$\begin{aligned} \text{ Im }\int x\cdot \nabla Q_0^{\gamma }Q_0^\gamma \,dx&= \text{ Im }\int x\cdot (2i\gamma x Q+\nabla Q)e^{i\gamma |x|^2}e^{-i\gamma |x|^2}Q\, dx\nonumber \\&= \text{ Im }\int x\cdot (2i\gamma x Q+\nabla Q)Q\,dx\nonumber \\&=2\gamma \int |x|^2Q^2\,dx>0. \end{aligned}$$

(2.33)

which shows $\text{ Im }\displaystyle \int x\cdot \nabla Q^\gamma (t_0)\overline{Q^\gamma }(t_0)\,dx >0$ for sufficiently small $t_0$. Moreover, using the fact that $Q^{\gamma }$ is a solution to (1.1), we have

$$\begin{aligned} \frac{d}{dt}\int |x|^{-b}|Q^{\gamma }|^{p+1}\,dx&=(p+1) \text{ Re }\int |x|^{-b}(\partial _tQ^{\gamma }\overline{Q^{\gamma }})|Q^{\gamma }|^{p-1}\,dx\\&=(p+1) \text{ Re }\int |x|^{-b}(i\triangle Q^{\gamma }\overline{Q^{\gamma }})|Q^{\gamma }|^{p-1}\,dx\\&=-(p+1) \text{ Im }\int |x|^{-b}|Q^{\gamma }|^{p-1}\triangle Q^\gamma \overline{Q^\gamma }\,dx. \end{aligned}$$

Consequently, from (2.32),

$$\begin{aligned} \left[ \frac{d}{dt}\int |x|^{-b}|Q^{\gamma }|^{p+1}\,dx\right] \Bigg |_{t=0}&=\left[ -(p+1) \text{ Im }\int |x|^{-b}|Q^{\gamma }|^{p-1}\triangle Q^\gamma \overline{Q^\gamma }\,dx\right] \Bigg |_{t=0}\\&=-2N\gamma (p-1)\int |x|^{-b}Q^{p+1}\,dx<0. \end{aligned}$$

Since

$$\begin{aligned} M[Q^{\gamma }_0]^{\frac{1-s_c}{s_c}}\int |x|^{-b}|Q^{\gamma }_0|^{p+1}\,dx=M[Q]^{\frac{1-s_c}{s_c}}\int |x|^{-b}|Q|^{p+1}\,dx, \end{aligned}$$

we get, for sufficiently small $t_0$

$$\begin{aligned} \mathcal {MP}[Q^{\gamma }(t_0)]<1. \end{aligned}$$

Now, define the function F as

$$\begin{aligned} F(t)&=M[Q^{\gamma }]^{\frac{1-s_c}{s_c}}\left[ E[Q^\gamma ]-\frac{\left( \text { Im}\displaystyle \int x\cdot \nabla Q^\gamma (t)\overline{Q^{\gamma }}(t)\,dx\right) ^2}{2\displaystyle \int |x|^2|Q^{\gamma }(t)|^2\,dx}\right] \nonumber \\&\quad -M[Q]^{\frac{1-s_c}{s_c}}E[Q]. \end{aligned}$$

(2.34)

In view of (2.29), with $v_0=Q$, we conclude $F(0)=0$. We just need to check that $F(t)\le 0$ for small positive t. Let

$$\begin{aligned} V(t)=\int |x|^2|Q^{\gamma }(x,t)|^2\,dx,\,\,\,\,\,\,\quad z(t)=\sqrt{V(t)}. \end{aligned}$$

We can rewrite (2.34) as

$$\begin{aligned} F(t)=M[Q^{\gamma }]^{\frac{1-s_c}{s_c}}\left( E[Q^\gamma ]-\frac{1}{8}(z_t(t)^2)\right) -M[Q]^{\frac{1-s_c}{s_c}}E[Q], \end{aligned}$$

and thus,

$$\begin{aligned} F_t(t)=-\frac{1}{4}M[Q^{\gamma }]^{\frac{1-s_c}{s_c}}z_t(t)z_{tt}(t). \end{aligned}$$

Using (1.5), (1.6) and the fact that Gagliardo-Nirenberg inequality (1.3) is an equality for $f = Q = e^{-i\gamma |x|^2}Q^\gamma _0$, we conclude that $z_{tt}(0) = 0$. Therefore,

$$\begin{aligned} F_{tt}(0)&=-\frac{1}{4}M[Q^{\gamma }]^{\frac{1-s_c}{s_c}}\left( z_t(0)z_{ttt}(0)+(z_{tt}(0))^2\right) \\&=-\frac{1}{4}M[Q^{\gamma }]^{\frac{1-s_c}{s_c}}z_t(0)z_{ttt}(0). \end{aligned}$$

On the other hand,

$$\begin{aligned} V_{tt}=2(z_t)^2+2zz_{tt}\quad \text{ and } \quad V_{ttt}=6z_tz_{tt}+2zz_{ttt}. \end{aligned}$$

Thus, $V_{ttt}(0)=2z(0)z_{ttt}(0)$. Hence, $F_{tt}(0)$ and $-V_{ttt}(0)$ have the same sign, but from (2.33) $z_t(0)>0$. By (2.6), we get that this sign is the same as the one of

$$\begin{aligned} \left[ \frac{d}{dt}\int |x|^{-b}|Q^\gamma |^{p+1}\,dx\right] \Bigg |_{t=0}= -\frac{(p+1)}{2A} V_{ttt}(0). \end{aligned}$$

Therefore, $F_{tt}(0)<0$, which shows that F(t) is negative for small $t> 0$. This completes the proof. $\square$

3 Scattering

We now prove the scattering part of theorem 1.5. We start with a lemma:

Lemma 3.1

Let $0< a< A < \left( \displaystyle \int |x|^{-b}|Q|^{p+1}\right) ^{s_c}M[Q]^{1-s_c}$. Then, there exists $\epsilon _0 = \epsilon _0(a,A)$ such that for all $f \in H^1(\mathbb {R}^N)$ with

$$\begin{aligned} a \le \left( \int |x|^{-b}|f|^{p+1}\,dx\right) ^{s_c}M[f]^{1-s_c} \le A, \end{aligned}$$

one has

$$\begin{aligned} \int \left| \nabla f\right| ^2\,dx-\frac{N(p-1)+2b}{2(p+1)}\int |x|^{-b}|f|^{p+1}\,dx \ge \epsilon _0 M[f]^{1-\frac{1}{s_c}} \end{aligned}$$

(3.1)

and

$$\begin{aligned} E[f] \ge \frac{\epsilon _0}{2}M[f]^{1-\frac{1}{s_c}}. \end{aligned}$$

(3.2)

Proof

Recalling the sharp Gagliardo-Nirenberg inequality, we have:

$$\begin{aligned}&M[f]^{\frac{1}{s_c}-1}\left[ \int |\nabla f|^2\,dx-\frac{N(p-1)+2b}{2(p+1)}\int |x|^{-b}|f|^{p+1}\,dx\right] \nonumber \\&\quad \quad \quad \ge \frac{1}{c_Q}M[f]^{\frac{1}{s_c}-1-\kappa }\left( \int |x|^{-b}|f|^{p+1}\,dx\right) ^{\frac{4}{N(p-1)+2b}} \nonumber \\&\quad \quad \quad \quad \quad \quad - M[f]^{\frac{1}{s_c}-1}\frac{N(p-1)+2b}{2(p+1)}\int |x|^{-b}|f|^{p+1}\,dx\nonumber \\&\quad \quad \quad = \frac{y^{\frac{4}{N(p-1)+2b}}}{c_Q}-\frac{N(p-1)+2b}{2(p+1)}y. \end{aligned}$$

(3.3)

where $y=M[f]^{\frac{1}{s_c}-1}\displaystyle \int |x|^{-b}|f|^{p+1}\,dx$. One can check, by direct calculations, that the function $y \mapsto \frac{y^{\frac{4}{N(p-1)+2b}}}{c_Q}-\frac{N(p-1)+2b}{2(p+1)}y$ has only one zero $y^*$ on $(0,+\infty )$ and is positive in $(0,y^*)$. Since the inequality (3.3) is an equality when $f = Q$, $y^*$ is exactly $M[Q]^{\frac{1}{s_c}-1}\displaystyle \int |x|^{-b}|Q|^{p+1}\,dx$, and (3.1) follows. Noting that

$$\begin{aligned} E[f] \ge \frac{1}{2}\left( \int |\nabla f|^2\,dx-\frac{N(p-1)+2b}{2(p+1)}\int |x|^{-b}|f|^{p+1}\,dx\right) , \end{aligned}$$

we get (3.2), because $\frac{N(p-1)+2b}{4} \ge 1$. $\square$

Definition 3.2

If $N \ge 1$ and $s \in (0,1)$, the pair (q, r) is called $\dot{H}^s$-admissible if it satisfies the condition

$$\begin{aligned} \frac{2}{q} = \frac{N}{2}-\frac{N}{r}-s, \end{aligned}$$

where

$$\begin{aligned} 2 \le q,r \le \infty , \text { and } (q,r,N) \ne (2,\infty ,2). \end{aligned}$$

Also, considering the following closed subset of $H^s$-admissible pairs

$$\begin{aligned} \mathcal {A}_s = \left\{ (q,r)\text { is } \dot{H}^{s}\text {-admissible} \left| \, {\left\{ \begin{array}{ll} \left( \frac{2N}{N-2s}\right) ^+ \le r \le \left( \frac{2N}{N-2}\right) ^- , &{} N \ge 3\\ \left( \frac{2}{1-s}\right) ^+ \le r \le \left( \left( \frac{2}{1-s}\right) ^+\right) ', &{} N = 2\\ \frac{2}{1-2s} \le r \le \infty , &{}N = 1 \end{array}\right. } \right. \right\} \end{aligned}$$

where $a^+ = a + \epsilon$, for a fixed, small $\epsilon >0$ and $(a^+)'$ is defined as the number such that

$$\begin{aligned} \frac{1}{a} = \frac{1}{a^+} + \frac{1}{(a^+)'}, \end{aligned}$$

we define the scattering norm

$$\begin{aligned} \Vert u\Vert _{S(\dot{H}^{s_c})} = \sup _{(q,r) \in \mathcal {A}_{s_c}}\Vert u\Vert _{L^q_t L^r_x}. \end{aligned}$$

It is already known that scattering follows from the uniform boundedness of the $H^1$ norm and the finiteness of the $S(\dot{H}^{s_c})$ norm (see [14, Proposition 1.4]). The following proposition was proved in [5] and covers for the broken translation symmetry and lack of momentum conservation.

Proposition 3.3

Suppose $N\ge 2$, $0<b<\min \{\tfrac{N}{2},2\}$, and $\tfrac{4-2b}{N}<\alpha <\tfrac{4-2b}{N-2}$. Let $\psi \in H^1(\mathbb {R}^N)$. Suppose that $t_n\equiv 0$ or $t_n\rightarrow \pm \infty$ and that $|x_n|\rightarrow \infty$. Then for all n sufficiently large, there exists a global solution $v_n$ to (1.1) with

$$\begin{aligned} v_n(0)=\psi _n:=e^{it_n\Delta } \psi (x-x_n) \end{aligned}$$

that scatters in $H^1$ and obeys

$$\begin{aligned} \Vert v_n\Vert _{S(\dot{H}^{s_c})}+\Vert v_n\Vert _{S(L^2)}+\Vert \nabla v_n\Vert _{S(L^2)}\le C \end{aligned}$$

for some $C=C(\Vert \psi \Vert _{H^1})$. Moreover, for any $\varepsilon > 0$, there exists $K>0$ and $\phi \in C_c^\infty (\mathbb {R}\times \mathbb {R}^N)$ such that

$$\begin{aligned} \Vert v_n - \phi (\cdot - x_n, \cdot + t_n)\Vert _{S(\dot{H}^{s_c})} < \varepsilon \;\;\text {for}\;\; n \ge K. \end{aligned}$$

We now have all tools to upgrade the global existence to scattering.

Proposition 3.4

Define S(L, A) as the supremum of $\Vert u\Vert _{S(\dot{H}^{s_c})}$ such that u is a solution to (1.1) on $[0, +\infty )$ with

$$\begin{aligned} \mathcal {ME}[u_0] \le L \end{aligned}$$

(3.4)

and

$$\begin{aligned} \sup _{t \in [0,+\infty )} \left( \int |x|^{-b}|u(t)|^{p+1}\,dx\right) ^{s_c}M[u]^{1-s_c} \le A. \end{aligned}$$

(3.5)

If $A < \left( \displaystyle \int |x|^{-b}Q^{p+1}\,dx\right) ^{s_c}M[Q]^{1-s_c}$, then $S(L,A) < +\infty$.

Proof

The proof goes along the spirit of [10, 14] and (see also [22]). We give an outline of the proof, highlighting the main differences.

First we note that, if $0<L<1$, by Theorem 1.2, then $S(L,A) < +\infty$. Assume, by contradiction, that $S(L,A) = +\infty$ for some $L \in \mathbb {R}$. Note that, if a nonzero u satisfies the equation (3.5), with $A < \left( \!\displaystyle \int |x|^{-b}Q^{p+1}\,dx\!\right) ^{s_c}M[Q]^{1-s_c}$, then by Lemma 3.1, $E[u] >0$. Thus, the quantity $L_c$ given by

$$\begin{aligned} L_c = L_c(A) := \inf \left\{ L \in \mathbb {R} \text { s.t. } S(L,A) = +\infty \right\} \end{aligned}$$

is well-defined and positive.

Moreover, there exists a sequence $\{u_n\}$ of (global) solutions such that

$$\begin{aligned}&M[u_n] = 1, \\&\left\| u_n\right\| _{S\left( \dot{H}^{s_c}\right) } \rightarrow +\infty , \\&\mathcal {ME}[u_n] \searrow L_c, \end{aligned}$$

and

$$\begin{aligned} \sup _{t \in [0, +\infty )}\int |x|^{-b}|u_n|^{p+1} \, dx \le A. \end{aligned}$$

Therefore, using the linear profile decomposition ([11, Theorem 5.1]) for the initial conditions $u_{n,0}$ (note that $\{u_{n,0}\}$ is bounded in $H^1({\mathbb {R}}^N)$), the existence of wave operators for large times (see [14, 22]) and Proposition 3.3, to deal with the unbounded translation parameters, we obtain for each $M \in \mathbb {N}$ (passing, if necessary, to a subsequence) a nonlinear profile decomposition of the form:

$$\begin{aligned} u_{n,0} = \sum _{j=1}^M \tilde{u}^j_n(0)+\tilde{W}_n^M \end{aligned}$$

(3.6)

and an approximate solution

$$\begin{aligned} {\tilde{u}}_{n} (t) = \sum _{j=1}^M \tilde{u}_n^j(t), \end{aligned}$$

where, for each pair (j, n), $\tilde{u}^j_n$ is a solution to (1.1) and

• 1. for each (j, n), there exists $T^j_n > 0$ such that $[0,T^j_n)$ is the maximal (positive) interval of existence of $\tilde{u}^j_n$;

• 2. $\displaystyle \lim _{M \rightarrow +\infty } \left[ \lim _{n\rightarrow +\infty }\left\| e^{it\Delta }\tilde{W}_n^M\right\| _{S\left( \dot{H}^{s_c}\right) }\right] = 0$;

• 3. The profiles can be ordered in such a way that

  • (a) The first nonlinear profile $\tilde{u}^1_n$ (corresponding to bounded space and time translation parameters in the linear profile decomposition), may or may not be global in time, but we can assume that it is independent of n and often write it as ${\tilde{u}}^1$ and its maximal time of existence as $[0,T^1)$;

  • (b) $\tilde{u}_n^j$, for $2 \le j \le M_1$ corresponding to bounded space translation, but unbounded time translation, are obtained from the wave operators, therefore scattering forward in time in $H^1$ to their correspondent linear profile and satisfying, $\Vert \tilde{u}^j_n\Vert _{S(\dot{H}^{s_c},[0,+\infty ))} \rightarrow 0$ as $n \rightarrow \infty$;

  • (c) $\tilde{u}_n^j$, for $M_1+1 \le j \le M$, corresponding to unbounded space translation, are obtained from Lemma 3.3, scattering in both time directions in $H^1$ and satisfying the global space-time bounds $\Vert \nabla \tilde{u}^j_n\Vert _{L^\infty _t L^2_x} \le \Vert \nabla \tilde{u}^j_n\Vert _{S(L^2)} \le C \Vert \nabla \tilde{u}^j_n(0)\Vert _{L^2}+o_n(1)$;

• 4. for fixed $M \in \mathbb {N}$ and any $0\le s \le 1$, the asymptotic Pythagorean expansion holds for the $\dot{H}^s$ norm

  $$\begin{aligned} \left\| u_{n,0}\right\| ^2_{\dot{H}^s} = \sum _{j = 1}^M \left\| \tilde{u}^j_n\left( 0\right) \right\| ^2_{\dot{H}^s}+\left\| \tilde{W}_n^M\right\| ^2_{\dot{H}^s}+o_n(1) \end{aligned}$$

  and for the energy

  $$\begin{aligned} E[u_{n,0}] = \sum E[{\tilde{u}}_n^j] + E[{\tilde{W}}^M_n] + o_n(1). \end{aligned}$$

These items follow from the usual approach, as in [5, 13] and [14]. Items 1-4 follow from the construction of the nonlinear profiles together with the so-called asymptotic orthogonality of the space and time translation parameters from the linear profile decomposition. The major difference is that we do not have information for $t > T^1$ because it is not clear whether all the nonlinear profiles evolve into global solutions, since the quantity $\mathcal {ME}[\tilde{u}^1_n(0)]$ may not be small. $\square$

Thus, in order to prove that all $\tilde{u}^j_n(t)$ exist on $[0, +\infty )$, we need to track $\Vert \nabla \tilde{u}^j_n (t)\Vert _{L^2}$. Denoting by $\text {INLS}(t)v_0$ the evolution of the datum $v_0$ under the flow of (1.1), we prove the following:

Lemma 3.5

(Pythagorean expansion along the bounded INLS flow). Suppose $u_{n,0}$ is a bounded sequence in $H^1({\mathbb {R}}^N)$. Let $T \in (0, +\infty )$ be a fixed time. Assume that $u_n(t) = \text{ INLS }(t)u_{n,0}$ exists up to time T for all n and $\displaystyle \lim _n \left\| \nabla u_n(t)\right\| _{L ^\infty _{[0,T]}L^2_x} < +\infty$. Consider the nonlinear profile decomposition (3.6) and write ${\tilde{W}}^M_n(t) = \text{ INLS }(t) {\tilde{W}}^M_n$. Then given any $T>0$, for all j, the nonlinear profiles $\tilde{u}^j_n(t)$ exist up to time T and for all $t \in [0,T]$,

$$\begin{aligned} \Vert \nabla u_n(t)\Vert _{ L^2_x}^2 = \sum _{j=1}^M \Vert \nabla {\tilde{u}}^j_n(t)\Vert _{ L^2_x}^2 + \Vert \nabla {\tilde{W}}^M_n(t)\Vert _{ L^2_x}^2+o_{M,n}(1), \end{aligned}$$

where $o_{M,n}(1) \rightarrow 0$ uniformly on $0 \le t \le T$.

Proof

For fixed $T>0$, define $B = \max \{1, \displaystyle \lim _n \Vert \nabla u_n(t)\Vert _{L^\infty _{[0,T]}L^2_x}\}$ and let $\tilde{T}^1$ be the maximal time of existence of $\tilde{u}_n^1$ such that $\tilde{T^1} \le T$ and $\Vert \nabla \tilde{u}^1_n\Vert _{L^\infty _{[0,\tilde{T}^1] }L^2_x} \le 2B$. This is the only possibly “ill-behaved” profile, and we aim to show the converse inequality $\tilde{T}^1 \ge T$. From the items 1-4 above, we estimate

$$\begin{aligned} \Vert \nabla \tilde{u}_n\Vert _{L^\infty _{[0,\tilde{T}^1]}L^2_x}&\le \Vert \nabla \tilde{u}^1_n\Vert _{L^\infty _{[0,\tilde{T}^1]}L^2_x} + \sum _{j=2}^{M_1}\Vert \nabla \tilde{u}^j_n\Vert _{L^\infty _{[0,\tilde{T}^1]}L^2_x}\\&\quad + \sum _{j=M_1+1}^{M}\Vert \nabla \tilde{u}^j_n\Vert _{L^\infty _{[0,\tilde{T}^1]}L^2_x}\\&\le 2B + 2\sum _{j=2}^{M_1} \Vert \nabla \tilde{u}^j_n(0)\Vert _{L^2_x} +C\sum _{M_1+1}^{M} \Vert \nabla \tilde{u}^j_n(0)\Vert _{L^2_x}+o_n(1)\\&\le 2B + C\Vert \nabla u_{n,0}\Vert _{L^2} +o_n(1), \end{aligned}$$

by interpolation and Sobolev embedding,

$$\begin{aligned} \Vert \tilde{u}_n\Vert _{S(\dot{H}^{s_c},[0,\tilde{T}^1])}&\le c \Vert \tilde{u}_n\Vert _{L^\infty _{[0,\tilde{T}^1]}L^\frac{2N}{N-2s_c}} + c\Vert \tilde{u}_n\Vert _{L^\frac{2}{1-s_c}{[0,\tilde{T}^1]}L^\frac{2N}{N-2}} \\&\le c(1+({\tilde{T}}^1)^\frac{1-s_c}{2})\Vert \tilde{u}_n\Vert _{L^\infty _{[0,\tilde{T}^1]}H^1_x}, \end{aligned}$$

and, by construction,

$$\begin{aligned} \Vert e^{it\Delta }[u_n(0)-\tilde{u}_{n}(0)]\Vert _{S(\dot{H}^{s_c})} = \Vert e^{it\Delta }\tilde{W}_n^M \Vert _{S(\dot{H}^{s_c})} = o_{M,n}(1). \end{aligned}$$

Defining the error of the approximation

$$\begin{aligned} e_n^M&= (i\partial _t+\Delta ){\tilde{u}}_n + |x|^{-b}f({\tilde{u}}_n) = |x|^{-b}\left[ f\left( \sum _{j=1}^M \tilde{u}_n^j\right) - \sum _{j=1}^M f(\tilde{u}_n^j) \right] , \end{aligned}$$

where $f(z)=|z|^\alpha z$, we have

$$\begin{aligned} \Vert e_n^M\Vert _{S'(\dot{H}^{-s_c},[0,T^1])}+\Vert e_n^M\Vert _{S'(L^2,[0,T^1])}+\Vert \nabla e_n^M\Vert _{S'(L^2,[0,T^1])}= o_{M,n}(1). \end{aligned}$$

These estimates are obtained from the pointwise linear estimates of the difference in the right-hand side, also making use of the asymptotic orthogonality and the individual space-times bounds of each $\tilde{u}^j_n$ on $[0,T^1)$.

Note all the profiles are defined at least for $t \in [0,T^1)$, since the only profile with possibly finite time of existence is $\tilde{u}^1_n$. By using long-time perturbation and interpolation,

$$\begin{aligned} \sup _{t \in [0,\tilde{T}^1]}\int |x|^{-b}|u_n(t)-{\tilde{u}}_n(t)|^{p+1}\, dx&\le c\Vert u_n-{\tilde{u}}_n\Vert _{L^\infty _{[0,{\tilde{T}}^1]}L^\frac{2N}{N-2s_c}}^{p-1} \Vert \nabla u_n\Vert _{L^\infty _{[0,{\tilde{T}}^1]}L^2}^2\\&= o_{M,n}. \end{aligned}$$

(Recall that $T>0$ is fixed and ${\tilde{T}}^1\le T$). Thus, by the asymptotic orthogonality, we get

$$\begin{aligned} \int |x|^{-b} |u_{n}(t)|^{p+1}\, dx = \displaystyle \sum _{j = 1}^M \int |x|^{-b} |{\tilde{u}}_{n}^j(t)|^{p+1}\, dx +o_{M,n}(1). \end{aligned}$$

Now, energy conservation and the Pythagorean expansion for the energy at $t=0$ gives

$$\begin{aligned} 0 = E[u_{n,0}]-E[u_{n}(t)] = \sum _{j=1}^M E[{\tilde{u}}_n^j(t)] + E[{\tilde{W}}^M_n(t)] -E[u_{n}(t)] + o_n(1), \end{aligned}$$

which in turn proves

$$\begin{aligned} \Vert \nabla u_n\Vert _{L^\infty _{[0,{\tilde{T}}^1]} L^2_x}^2 = \sum _{j=1}^M \Vert \nabla {\tilde{u}}^j_n\Vert _{L^\infty _{[0,{\tilde{T}}^1]} L^2_x}^2 + \Vert \nabla {\tilde{W}}^M_n\Vert _{L^\infty _{[0,{\tilde{T}}^1]} L^2_x}^2+o_{M,n}(1). \end{aligned}$$

(3.7)

The last bound shows that $\Vert \nabla {\tilde{u}}^1_n(t)\Vert _{ L^2_x} \le B+o_{M,n}(1)$ for all $t \in [0,{\tilde{T}}^1]$ which in turn, by maximality of ${\tilde{T}}^1$, shows that $\tilde{T}^1\ge T$. In particular, $u^1_n(t)$ exists up to time T, and we can replace ${\tilde{T}}^1$ by T in (3.7), finishing the proof of the lemma. $\square$

Invoking (3.4) and (3.5) and using the orthogonality along the INLS flow, one is able to prove that ${\tilde{u}}^j_n(t)$ is defined on $[0, +\infty )$ as well, and satisfies, for every j and every large n,

$$\begin{aligned}&M[{\tilde{u}}^j_n] \le 1, \\&\mathcal {ME}[{\tilde{u}}^j_n] \le L_c \end{aligned}$$

and

$$\begin{aligned} \sup _{t \in [0,+\infty )} \left( \int |x|^{-b}|{\tilde{u}}^j_n(t)|^{p+1}\,dx\right) ^{s_c}M[\tilde{u}^j_n]^{1-s_c} \le A. \end{aligned}$$

The rest of the proof follows the same lines as [10] and [14], using the criticality of $L_c$ to show that only the first profile, ${\tilde{u}}^1$, can be non-zero, and letting $u_c(t) ={\tilde{u}}^1(t)$. This criticality also shows that $M[u_c] = 1$ and $\mathcal {ME}[u_c] = L_c$. Long-time perturbation theory yields $\left\| u_c\right\| _{S\left( \dot{H}^{s_c}\right) }= +\infty$. At this point, the classical compactness lemma follows.

Lemma 3.6

(Compactness). Assume that there exists $L_0 \in \mathbb {R}$ and a positive number

$$\begin{aligned} A < \left( \displaystyle \int |x|^{-b}|Q|^{p+1}\,dx\right) ^{s_c}M[Q]^{1-s_c} \end{aligned}$$

such that $S(L_0,A) = +\infty$. Then there exists a global solution $u_c$ of (1.1) such that the set

$$\begin{aligned} K = \displaystyle \left\{ u_c(t), t \in [0, +\infty )\right\} \end{aligned}$$

has a compact closure in $H^1(\mathbb {R}^N)$.

Using this compactness lemma and the virial identity (1.6), we also have the classic rigidity lemma.

Lemma 3.7

(Rigidity). There exists no solution $u_c$ of (1.1) satisfying the conclusion of Lemma 3.6.

The proof goes on the same lines as in [10] and [14]. $\square$

4 Proof of the blowup criteria

In this section we prove two criteria for blow up in finite time. The first one is a generalization of Lushnikov’s criterion in [26] and of Holmer-Platte-Roudenko criteria in [24] for the INLS, and the second one is the modification of the first approach, where the generalized uncertainty principle is replaced by the interpolation inequality (4.10). The two criteria are the INLS versions of the ones proved by Duyckaerts and Roudenko in [10].

Proof of Theorem 1.14

Integrating by parts,

$$\begin{aligned} \Vert u\Vert ^{2}_{L^2}&=\int |u|^2\,dx=\frac{1}{N}\sum ^{N}_{j=1}\int \partial _jx_j|u|^2\,dx=-\frac{1}{N}\sum _{j=1}^{N}\int x_j\partial _j(|u|^2)\,dx\\&=-\frac{1}{N}\sum _{j=1}^{N}\int x_j(\partial _ju\overline{u}+u\partial _j\overline{u})\,dx=-\frac{2}{N}\sum _{j=1}^{N}\text {Re}\,\int x_j\partial _ju\overline{u}\,dx\\&=-\frac{2}{N}\text {Re}\,\int (x\cdot \nabla u)\overline{u}\,dx. \end{aligned}$$

Since $|z|^2=|\text {Re}\,z|^2+|\text {Im}\,z|^2$, using Hölder’s inequality

$$\begin{aligned} \Vert xu\Vert ^{2}_{L_2}\Vert \nabla u\Vert ^{2}_{L^2}&\ge \left| \int (x\cdot \nabla u)\overline{u}\,dx\right| ^2\\&=\left| \text {Re}\,\int (x\cdot \nabla u)\overline{u}\,dx\right| ^2+\left| \text {Im}\,\int (x\cdot \nabla u)\overline{u}\,dx\right| ^2\\&=\frac{N^2}{4}\Vert u\Vert ^4_{L^2}+\left| \text {Im}\,\int (x\cdot \nabla u)\overline{u}\,dx\right| ^2. \end{aligned}$$

From the definition of variance and the identity for the first derivative of the variance (1.5), we get the uncertainty principle

$$\begin{aligned} \frac{N^2}{4}\Vert u_0\Vert ^2_{L^2}+\left| \frac{V_t}{4}\right| ^2\le V(t)\Vert \nabla u(t)\Vert ^{2}_{L^2}. \end{aligned}$$

(4.1)

Using the equation (1.6) for the second derivative of the variance, we obtain

$$\begin{aligned} V_{tt}(t)=4(N(p-1)+2b)E[u_0]-4(p-1)s_c\Vert \nabla u(t)\Vert ^{2}_{L^2}. \end{aligned}$$

(4.2)

Substituting (4.2) in the uncertainty principle (4.1), we have

$$\begin{aligned} V_{tt}(t)\le 4(N(p-1)+2b)E[u_0]-N^2(p-1)s_c\frac{(M[u_0])^2}{V(t)}-\frac{(p-1)s_c}{4}\frac{|V_t(t)|^2}{V(t)}. \end{aligned}$$

(4.3)

Now, we rewrite equation (4.3) in order to cancel the term $V_t^2$. For this, define

$$\begin{aligned} V=B^{\frac{1}{\alpha +1}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha =\frac{(p-1)s_c}{4}=\frac{N(p-1)-4+2b}{8}. \end{aligned}$$

(4.4)

Then,

$$\begin{aligned} V_t=\frac{1}{\alpha +1}B^{-\frac{\alpha }{\alpha +1}}\,\,\,\,\,\,\,\, \text{ and } \,\,\,\,\,\,\,\,V_{tt}=-\frac{\alpha }{(\alpha +1)^2}B^{-\frac{2\alpha +1}{\alpha +1}}B_t^2+\frac{1}{\alpha +1}B^{-\frac{\alpha }{\alpha +1}}B_{tt}, \end{aligned}$$

which gives

$$\begin{aligned} B_{tt}\le 4(\alpha +1)N(p-1)E[u_0]B^{\frac{\alpha }{\alpha +1}}-(\alpha +1)N^2(p-1)s_c(M[u_0])^2B^{\frac{\alpha -1}{\alpha +1}}, \end{aligned}$$

that is, for all $t\in [0,T_+(u))$

$$\begin{aligned} B_{tt}\le \frac{N(p-1)(N(p-1)+4+2b)}{2}&\Bigg (E[u_0]B^{\frac{N(p-1)-4+2b}{N(p-1)+4+2b}}\\&-\frac{Ns_c}{4}(M[u_0])^{2}B^{\frac{N(p-1)-12+2b}{N(p-1)+4+2b}}\Bigg ). \end{aligned}$$

In order to further simplify the inequality, let us make a rescaling. Define $B(t)=\mu \Phi (\lambda t)$, with

$$\begin{aligned} \mu =\left( \frac{Ns_c(M[u_0])^2}{4E[u_0]}\right) ^{\frac{N(p-1)+4+2b}{8}},\,\,\,\,\,\,\,\,\,\lambda =\frac{8\sqrt{2}}{\sqrt{Ns_c}}\frac{E[u_0]}{M[u_0]}. \end{aligned}$$

(4.5)

Then letting $s=\lambda t$, we obtain

$$\begin{aligned} \omega \Phi _{ss}\le \Phi ^{\gamma }-\Phi ^{\delta }, \,\,\,s\in [0,T_+/a), \end{aligned}$$

(4.6)

where

$$\begin{aligned}&\gamma =\frac{N(p-1)-4+2b}{N(p-1)+4+2b},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\delta =\frac{N(p-1)-12+2b}{N(p-1)+4+2b}=2\gamma -1, \\&\omega =\frac{64}{N(p-1)(N(p-1)+4+2b)} \end{aligned}$$

and since $p>1+\frac{4}{N}$,

$$\begin{aligned} 0<\gamma<1,\,\,\,\,\,-1<\delta <\gamma . \end{aligned}$$

We rewrite (4.6) as

$$\begin{aligned} \omega \Phi _{ss}+\frac{\partial U}{\partial \Phi }\le 0, \end{aligned}$$

(4.7)

for $t\in [0,T_+/a)$, where $U(\Phi )=\frac{\Phi ^{\delta +1}}{\delta +1}-\frac{\Phi ^{\gamma +1}}{\gamma +1}$. Define the energy of the particle

$$\begin{aligned} \mathcal {E}(s)=\frac{\omega }{2}\Phi _s^2(s)+U(\Phi (s)) \end{aligned}$$

which is conserved for solutions of

$$\begin{aligned} \omega \Phi _{ss}+\frac{\partial U}{\partial \Phi }=0. \end{aligned}$$

Based on the ideas of Lushnikov [26], Duyckaerts and Roudenko [10] studied this model and showed the following proposition $\square$

Proposition 4.1

Let $\Phi$ be a nonnegative solution of (4.7) such that one of the following holds:

(A):

$\mathcal {E}(0)<U_{max} \text{ and } \Phi (0)<1,$

(B):

$\mathcal {E}(0)>U_{max} \text{ and } \Phi _s(0)<0,$

(C):

$\mathcal {E}(0)=U_{max}, \Phi _s(0)<0 \text{ and } \Phi (0)<1,$

where $U_{max}$ is the absolute maximum of U on the interval $[0,+\infty )$. Then we have $T_+<\infty$.

Proof

For the sake of completeness of this work, we will give the proof of the proposition. Multiplying equation (4.7) by $\Phi _s$, we get

$$\begin{aligned} \Phi _s(s)>0\Rightarrow \mathcal {E}_s(s)<0,\,\,\,\,\,\,\Phi _s(s)<0\Rightarrow \mathcal {E}_s(s)>0. \end{aligned}$$

(4.8)

We argue by contradiction, assuming $T_+=T_+(u)=+\infty$.

We first assume (A). Let us prove by contradiction that

$$\begin{aligned} \exists \,s>0,\,\,\,\Phi _s(s)<0. \end{aligned}$$

If not, $\Phi _s(s)\ge 0$ for all s,  and (4.8) implies that the energy decays. By (A), $\mathcal {E}(s)\le \mathcal {E}(0)<U_{max}$ for all s. Thus, $|\Phi (s)-1|\ge \varepsilon _0$ (where $\varepsilon _0>0$ depends on $\mathcal {E}(0))$ for all s. Since by (A) $\Phi (0)<1$, we obtain by continuity of $\Phi$ that $\Phi (s)\le 1-\varepsilon _0$ for all s. By equation (4.6), we deduce $\Phi _{ss}\le -\varepsilon _1$ for all s, where $\varepsilon _1>0$ depends on $\varepsilon _0.$ Thus, $\Phi$ is strictly concave, a contradiction with the fact that $\Phi$ is positive and $T_+=+\infty .$

We have proved that there exists $s> 0$ such that $\Phi _s(s)< 0$. Letting

$$\begin{aligned} t_1=\inf \{s> 0; \Phi _s(s)< 0\}, \end{aligned}$$

we get by (4.8) that the energy is nonincreasing on $[0,t_1]$. Thus, $\mathcal {E}(s)<\mathcal {E}(0)\le U_{max}$ on $[0,t_1]$, which proves that $\Phi (s)\ne 1$ on $[0,t_1]$. Since $\Phi (0)<1,$ we deduce by the intermediate value theorem that $\Phi (t_1)<1$ and by (4.6) that $\Phi _{ss}(t_1)<0$. Since $\Phi _s(t_1)\le 0$, an elementary bootstrap argument, together with equation (4.6) shows that $\Phi (s)\le 1-\varepsilon _0,\,\Phi _s(s)<0$ and $\Phi _{ss}(s)\le -\varepsilon _1$ for $s>t_1$, for some positive constants $\varepsilon _0, \varepsilon _1$. This is again a contradiction with the positivity of $\Phi$.

We next assume (B). Let $t_1$ be such that $\Phi _s(s)<0$ on $[0,t_1]$. By (4.8), $\mathcal {E}$ is nondecreasing on $[0,t_1]$, and thus, $\mathcal {E}(s)\ge \mathcal {E}(0)>U_{max}$ for all s on $[0,t_1]$. As a consequence, $\frac{1}{2}\Phi _s(s)^2\ge \mathcal {E}(0)-U_{max}>0$ for all s in $[0,t_1]$, which shows that the inequality $\Phi _s(s)\le -\sqrt{\mathcal {E}(0)-U_{max}}$ holds on $[0,t_1]$. Finally, an elementary bootstrap argument shows that the inequality $\Phi _s(s)\le -\sqrt{\mathcal {E}(0)-U_{max}}$ is valid for all $s\ge 0$, a contradiction with the positivity of $\Phi$.

Finally, we assume (C). By bootstrap again, $\Phi _s(s)<0$, $\Phi (s)<1$ and $\Phi _{ss}(s)<0$ for all positive s, proving again that $\Phi$ is a strictly concave function, a contradiction. $\square$

Since

$$\begin{aligned} \alpha =\frac{(p-1)s_c}{4}=\frac{N(p-1)-4+2b}{8}, \end{aligned}$$

we have

$$\begin{aligned}&2\alpha +1=\frac{N(p-1)+2b}{4}, \quad \alpha +1=\frac{N(p-1)+4+2b}{8}, \\&(\alpha +1)(\delta +1)=2\alpha ,\quad (\alpha +1)(\gamma +1)=2\alpha +1 \text { and } \omega =\frac{2}{(2\alpha +1)(\alpha +1)}. \end{aligned}$$

By making $\Phi = v^{\alpha +1}$, then

$$\begin{aligned} \mathcal {E}=\frac{\omega }{2}\Phi _s^2(s)+U(\Phi (s))=\frac{\alpha +1}{2\alpha +1}(v')^2v^{2\alpha }+\frac{\alpha +1}{2\alpha }v^{2\alpha }-\frac{\alpha +1}{2\alpha +1}v^{2\alpha +1} \end{aligned}$$

and

$$\begin{aligned} U_{max}= \frac{1}{2\alpha }\frac{\alpha +1}{2\alpha +1}. \end{aligned}$$

Consider the function f given for

$$\begin{aligned} f(x)=\sqrt{\frac{1}{kx^k}+x-\left( 1+\frac{1}{k}\right) }, \end{aligned}$$

(4.9)

where $k=\frac{(p-1)s_c}{2}=2\alpha$. Hence, if $v_s(0)$ satisfies the condition

$$\begin{aligned} v_s(0)< \left\{ \begin{array}{ll} +f(v(0)),\,\,\,&{}\text{ if } v(0)<1,\\ -f(v(0)),\,\,\,&{}\text{ if } v(0)\ge 1, \end{array} \right. \end{aligned}$$

then $\Phi =v^{\alpha +1}$ satisfies the conditions of Proposition 4.1. Indeed, the condition $\mathcal {E}<U_{max}$ is equivalent to

$$\begin{aligned} 2\alpha (v')^2v^{2\alpha }+(2\alpha +1)v^{2\alpha }-2\alpha v^{2\alpha +1}<1 \end{aligned}$$

that is,

$$\begin{aligned} |v_s|<f(v). \end{aligned}$$

Hence, the condition (A) means

$$\begin{aligned} v(0)<1\,\,\,\,\, \text{ and } \,\,\,\,\,-f(v(0))<v_s(0)<f(v(0)) \end{aligned}$$

and the condition (B) holds if and only if

$$\begin{aligned} |v_s(0)|>f(v(0))\,\,\,\,\, \text{ and } \,\,\,\,\,v_s(0)<0. \end{aligned}$$

More precisely,

$$\begin{aligned} v_s(0)<-f(v(0)) \end{aligned}$$

and the condition (C) is equivalent to

$$\begin{aligned} v(0)<1\,\,\,\,\, \text{ and } \,\,\,\,\,v_s(0)=-f(v(0)). \end{aligned}$$

Therefore, from (4.4), (4.5) and from the definition of v, we have

$$\begin{aligned} V(0)&=(\mu \Phi (\lambda t))^{\frac{1}{\alpha +1}}\Bigg |_{t=0}=\mu ^{\frac{8}{N(p-1)+4+2b}}v\left( \frac{8\sqrt{2}}{\sqrt{Ns_c}}\frac{E[u_0]}{M[u_0]}t\right) \Bigg |_{t=0}\\&=\mu ^{\frac{8}{N(p-1)+4+2b}}v(0)=\frac{Ns_cM^2}{4E[u_0]}v(0) \end{aligned}$$

and

$$\begin{aligned} V_t(0)&=\mu ^{\frac{8}{N(p-1)+4+2b}}\frac{8\sqrt{2}}{\sqrt{Ns_c}}\frac{E[u_0]}{M[u_0]}v_s(0)\\&=\frac{Ns_cM^2}{4E[u_0]}\frac{8\sqrt{2}}{\sqrt{Ns_c}}\frac{E[u_0]}{M[u_0]}v_s(0)\\&=M[u_0]\sqrt{8Ns_c}v_s(0). \end{aligned}$$

Furthermore,

$$\begin{aligned} \frac{V_t(0)}{M[u_0]}=\sqrt{8Ns_c}v_s(0)<\sqrt{8Ns_c}g(v(0))=\sqrt{8Ns_c}g\left( \frac{4}{Ns_c}\frac{V(0)E[u_0]}{M[u_0]^2}\right) , \end{aligned}$$

which completes the proof of Theorem 1.14. $\square$

We now proceed to the proof of Theorem 1.15. For that, we consider the following proposition.

Proposition 4.2

Let $p>1$ and $N\ge 1$. Then, the following inequality

$$\begin{aligned} \Vert u\Vert ^{2}_{L^2}\le C_{p,N}\left( \Vert xu\Vert _{L^2}^{\frac{N(p-1)+2b}{2}}\Vert |\cdot |^{\frac{-b}{p+1}}u\Vert ^{p+1}_{L^{p+1}}\right) ^{\frac{2}{N(p-1)+2(p+1)+2b}} \end{aligned}$$

(4.10)

holds with the sharp constant $C_{p,N}$ (depending on the nonlinearity p and dimension N) given by (4.14). Moreover, the equality occurs if and only if there exists $\beta \ge 0$, $\alpha \le 0$ such that $|u(x)|=\beta \phi (\alpha x)$, where

$$\begin{aligned} \phi (x)= \left\{ \begin{array}{ll} |x|^{\frac{b}{p-1}}(1-|x|^2)^{\frac{1}{p-1}}&{} \text{ if } 0\le |x|<1,\\ 0&{} \text{ if } |x|>1. \end{array} \right. \end{aligned}$$

The proof of Proposition 4.2 follows the ideas of [10].

Proof

Let $R>0$ to be specified later. Split the mass of u as follows

$$\begin{aligned} \int |u(x)|^2\,dx=\frac{1}{R^2}\int _{|x|\le R}(R^2-|x|^2)|u(x)|^2\,dx&+\frac{1}{R^2}\int _{|x|\le R}|x|^2|u(x)|^2\,dx\\&+\int _{|x|\ge R}|u(x)|^2\,dx. \end{aligned}$$

By Hölder inequality we have

$$\begin{aligned}&\frac{1}{R^2}\int (R^2-|x|^2)|u(x)|^2\,dx\nonumber \\&\quad \le \frac{1}{R^2}\left( \int _{|x|\le R}|x|^{\frac{2b}{p-1}}(R^2-|x|^2)^{\frac{p+1}{p-1}}\,dx\right) ^{\frac{p-1}{p+1}}\nonumber \\&\qquad \times \left( \int |x|^{-b}|u(x)|^{p+1}\,dx\right) ^{\frac{2}{p+1}}\nonumber \\&\quad \le \frac{1}{R^2}\left( \int _{|x|\le 1}R^{\frac{2b}{p-1}}|y|^{\frac{2b}{p-1}}(R^2-R^2|y|^2)^{\frac{p+1}{p-1}}R^N\,dy\right) ^{\frac{p-1}{p+1}}\nonumber \\&\qquad \times \left( \int |x|^{-b}|u(x)|^{p+1}\,dx\right) ^{\frac{2}{p+1}}\nonumber \\&\quad =R^{\frac{N(p-1)+2b}{p+1}}D_{p,N} \left\| |\cdot |^{-\frac{b}{p+1}}u\right\| ^{2}_{p+1}, \end{aligned}$$

(4.11)

where

$$\begin{aligned} D_{p,N}=\left( \int _{|y|\le 1}|y|^{\frac{2b}{p-1}}(1-|y|^2)^{\frac{p+1}{p-1}}\,dy\right) ^{\frac{p-1}{p+1}}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \frac{1}{R^2}\int _{|x|\le R}|x|^2|u(x)|^{2}\,dx+\int _{|x|\ge R}|u(x)|^2\,dx\le \frac{1}{R^2}\int |x|^2|u(x)|^2\,dx.\nonumber \\ \end{aligned}$$

(4.12)

Combining (4.11) and (4.12), we get

$$\begin{aligned} \forall R>0,\,\,\,\, \Vert u\Vert ^{2}_{L^2}\le D_{p,N}\left\| |\cdot |^{-\frac{b}{p+1}}u\right\| _{L^{p+1}}^2R^{\frac{N(p-1)+2b}{p+1}}+\frac{1}{R^2}\Vert xu\Vert ^{2}_{L^2}. \end{aligned}$$

(4.13)

Let $F:(0,+\infty )\rightarrow {\mathbb {R}}$ given by $F(R)=AR^{\alpha }+BR^{-2}$, where $A,B>0 \text{ and } \alpha >0$. The minimum value of F is reached at $R=\left( \frac{2B}{\alpha A}\right) ^{\frac{1}{\alpha +2}}$ and

$$\begin{aligned} F\left( \left( \frac{2B}{\alpha A}\right) ^{\frac{1}{\alpha +2}}\right) =A\left( \frac{2B}{\alpha A}\right) ^{\frac{\alpha }{\alpha +2}}+B\left( \frac{\alpha A}{2B}\right) ^{\frac{2}{\alpha +2}}=\frac{2+\alpha }{\alpha }(\alpha A)^{\frac{2}{\alpha +2}}(2B)^{\frac{\alpha }{\alpha +2}}. \end{aligned}$$

Thus, by taking

$$\begin{aligned} R=\left( \frac{p+1}{N(p-1)+2b}\frac{2\Vert xu\Vert ^2_{L^2}}{D_{p,N}\left\| |\cdot |^{-\frac{b}{p+1}}u\right\| _{L^{p+1}}^2}\right) ^{\frac{p+1}{N(p-1)+2(p+1)+2b}} \end{aligned}$$

in (4.13), we have

$$\begin{aligned} \Vert u\Vert ^{2}_{L^2}\le C_{p,N}^2\left\| |\cdot |^{-\frac{b}{p+1}}u\right\| _{L^{p+1}}^{\frac{4(p+1)}{N(p-1)+2(p+1)+2b}}\Vert xu\Vert _{L^2}^{\frac{2N(p-1)+4b}{N(p-1)+2(p+1)+2b}}, \end{aligned}$$

where

$$\begin{aligned} C_{p,N}&=\left( \frac{N(p-1)+2(p+1)+2b}{2N(p-1)+4b}\right) ^{\frac{1}{2}}\nonumber \\&\quad \times \left( \frac{N(p-1)+2b}{p+1}D_{p,N}\right) ^{\frac{(p+1)}{N(p-1)+2(p+1)+2b}} \nonumber \\&\quad \times 2^{\frac{N(p-1)+2b}{2N(p-1)+4(p+1)+4b}}. \end{aligned}$$

(4.14)

Note that equality in (4.10) holds if and only if there exists $R > 0$ such that (4.13) is an equality. This is equivalent to the fact that for some $R > 0$, both (4.11) and (4.12) are equalities. The inequality (4.11) is an equality if and only if, for $|x| < R$, $|x|^{-b}|u(x)|^{p+1} = c|x|^{\frac{2b}{p-1}}(R^2-|x|^2)^{\frac{p+1}{p-1}}$ for some constant $c \ge 0$, and inequality (4.12) is an equality if and only if $u(x) = 0$ for $|x| \ge R$. This completes the proof of Proposition 4.2. $\square$

Proof of Theorem 1.15

Since the energy is

$$\begin{aligned} E[u_0]=\frac{1}{2}\Vert \nabla u(t)\Vert ^{2}_{L^2}-\frac{1}{p+1}\left\| |\cdot |^{-\frac{b}{p+1}}u(t)\right\| ^{p+1}_{L^{p+1}}, \end{aligned}$$

from (1.6), we obtain

$$\begin{aligned} V_{tt}(t)&=4(N(p-1)+2b)E[u_0]-2(N(p-1)+2b-4)\Vert \nabla u(t)\Vert ^{2}_{L^2({\mathbb {R}}^N)}\\&=16E[u_0]-\frac{8(p-1)s_c}{p+1}\left\| |\cdot |^{-\frac{b}{p+1}}u(t)\right\| ^{p+1}_{L^{p+1}}. \end{aligned}$$

Using the sharp interpolation inequality (4.10)

$$\begin{aligned} V_{tt}(t)\le 16E[u_0]-\frac{8(p-1)s_c}{(p+1)(C_{p,N})^{\frac{N(p-1)}{2}+(p+1)+b}}\frac{M[u_0]^{\frac{N(p-1)}{4}+\frac{(p+1)}{2}+\frac{b}{2}}}{V(t)^\frac{N(p-1)+2b}{4}},\nonumber \\ \end{aligned}$$

(4.15)

with $C_{p,N}$ from (4.10). As done in the proof of Proposition 1.14, take v(s) with $s=at$ such that

$$\begin{aligned} V(t)=\mu v(\lambda t),\,\,\,\,\lambda =\sqrt{\frac{32E[u_0]}{\mu }}, \end{aligned}$$

where

$$\begin{aligned} \mu =\left( \frac{s_c(p-1)}{2(p+1)}\right) ^{\frac{4}{N(p-1)+2b}}\frac{M[u_0]^{1+(p+1)\left( \frac{2}{N(p-1)+2b}\right) }}{(C_{p,N})^{2+(p+1)\left( \frac{4}{N(p-1)+2b}\right) }E[u_0]^{\frac{4}{N(p-1)+2b}}}. \end{aligned}$$

Hence, applying in the inequality (4.15), we have

$$\begin{aligned} v_{ss}(s)\le \frac{1}{2}\left( 1-v^{-\frac{N(p-1)+2b}{4}}(s)\right) . \end{aligned}$$

If the inequality in the above expression is replaced by an equality, then we have that the following energy is conserved

$$\begin{aligned} \mathcal {E}(s)=\frac{k}{1+k}\left( (v(s))^2-v(s)-\frac{1}{kv(s)^k}\right) , \end{aligned}$$

where as before $k=\frac{(p-1)s_c}{2}=\frac{N(p-1)+2b}{4}-1$. The maximum of the function

$$\begin{aligned} f(x)=\frac{k}{1+k}\left( x+\frac{1}{kx^k}\right) , \end{aligned}$$

attained at $x=1$, is $-1$. As we did to (A), (B) and (C), we identify the three sufficient conditions for blow-up in finite time.

($A^*$):

$\mathcal {E}(0)<-1$ and $v(0)<1,$

($B^*$):

$\mathcal {E}(0)>-1$ and $v_s(0)<0,$

($C^*$):

$\mathcal {E}(0)=-1$, $v_s(0)<0$ and $v(0)<1.$

If $v_s(0)$ satisfies the condition

$$\begin{aligned} v_s(0)< \left\{ \begin{array}{ll} +f(v(0)),\,\,\,&{}\text{ if } v(0)<1\\ -f(v(0)),\,\,\,&{}\text{ if } v(0)\ge 1, \end{array} \right. \end{aligned}$$

then v satisfies one of the conditions (A*), (B*) and (C*). Indeed, recalling the function f from (4.9) and using the definition of $\mathcal {E}$, we obtain

• a) $\mathcal {E}<-1$ if and only if $|v_s|<f(v).$

• b) $\mathcal {E}\ge -1$ if and only if $|v_s|\ge f(v).$

Then the previous conditions can be written in the following form:

• $(A^*) \Leftrightarrow v(0)<1 \text{ and } -f(v(0))<v_s(0)<f(v(0)),$

• $(B^*) \Leftrightarrow v_s(0)<-f(v(0)),$

• $(C^*) \Leftrightarrow v_s(0)=-f(v(0)),\,\,\,v(0)<1.$

Substituting back V(t), we obtain

$$\begin{aligned} \frac{V_t(0)}{\lambda \mu }<g\left( \frac{V(0)}{\mu }\right) , \end{aligned}$$

where g is defined in (1.13). Hence,

$$\begin{aligned}&\frac{V_t(0)}{4\sqrt{2}} \left( \frac{2(p+1)}{s_c(p-1)}(C_{p,N})^{\frac{N(p-1)+2b}{2}+(p+1)}\right) ^{\frac{2}{N(p-1)+2b}}\\&\qquad \times \frac{(C_{p,N})^{1+(p+1)\left( \frac{2}{N(p-1)+2b}\right) }}{E[u_0]^{\frac{s_c}{N}}M[u_0]^{\frac{1}{2}+(p+1)\left( \frac{1}{N(p-1)+2b}\right) }}<g(\theta ), \end{aligned}$$

with

$$\begin{aligned} \theta&=\left( \frac{2(p+1)}{s_c(p-1)}(C_{p,N})^{\frac{N(p-1)+2b}{2}+(p+1)}\right) ^{\frac{4}{N(p-1)+2b}}\\&\quad \times \frac{E[u_0]^{\frac{4}{N(p-1)+2b}}}{M[u_0]^{1+(p+1)\left( \frac{2}{N(p-1)+2b}\right) }}V(0). \end{aligned}$$

This completes the proof of Theorem 1.15. $\square$