# On blow up for a class of radial Hartree type equations

Shumao Wang  Academy of Mathematics and Systems Science, the Chinese Academy
of Sciences, Beijing 100190, China<EMAIL_ADDRESS>
###### Abstract.

We study a class of Hartree type equations and prove a quantitative blow up
rate for their blow up solutions. This is an analogue of the result by Merle
and Raphaël on 3d NLS.

###### Key words and phrases:

Hartree type equations; blow up rate.

##  1\. INTRODUCTION

We consider the Cauchy problem for the focusing Hartree equation for $d=3$:

(1.1) $\begin{cases}i\partial_{t}u+\triangle
u=-(V*|u|^{2})u,&(t,x)\in\mathbb{R}\times\mathbb{R}^{3},\\\
u\mid_{t=0}=u_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}.\end{cases}$

Here $V(x)$ is a real valued function, problem (1.1) has three conservation
laws:

  * •

Mass:

(1.2) $M(u(t))=M(u_{0}):=\int_{R^{3}}|u(t)|^{2}dx,$

  * •

Energy:

(1.3) $E(u(t))=E(u_{0}):=\frac{1}{2}\int|\nabla
u(t,x)|^{2}dx-\frac{1}{4}\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx,$

  * •

Momentum:

(1.4) $P(u(t))=P(u_{0}):=Im\int\bar{u}\nabla u(t)dx.$

We will focus on radial $V$ and radial initial data $u_{0}$ throughout the
article.

###  1.1. Setting of the problem and statement of the main result

For the equation (1.1), it describes the dynamics of the mean-field limits of
many-body quantum systems, such as coherent states, condensates. In
particular, it provides effective model for quantum systems with long- range
interactions. Readers can refer to [24, 15, 12] for more information about the
physical background of the equation. Besides its physical importance, a lot of
mathematical interest for equation (1.1) lies in its connection and similarity
to cubic NLS,

(1.5) $\begin{cases}i\partial_{t}u+\triangle
u=-|u|^{2}u.&(t,x)\in\mathbb{R}\times\mathbb{R}^{3},\\\
u\mid_{t=0}=u_{0}.\end{cases}$

In particular, when $V(x)=\delta(x)$, then (1.1) formally becomes (1.5). From
this perspective, if one obtains some result for (1.5), it is reasonable to
expect that a similar result holds for (1.1). It is also well-expected that
one can generalize a result to a cubic NLS from a Hartree type model. And this
is the main purpose of the current article. However, the Hartree equation
differs from cubic NLS mainly in two ways,

  * •

Equation (1.1) does not necessarily enjoy the scaling symmetry111If $u(t,x)$
solves (1.5), so does
$u_{\lambda}(t,x):=\frac{1}{\lambda}u(\frac{t}{\lambda^{2}},\frac{x}{\lambda})$.,
which is one of the most important property for (1.5).

  * •

Due to the convolution structure in (1.1), it is non-local.

The study of blow up problem for focusing NLS (and other nonlinear dispersive
PDEs) has been an active research field. Classical virial argument by Glassey
in [14] implies the existence of many blow up solutions. Constructive blow up
solutions and universality of blow up solutions under certain regime have
attracted a lot of researchers, see for examples [31, 29, 32, 27, 39]. But
very few can be said about general blow up solutions, i.e. solutions without a
size constraint, and it is very hard. In [33], Merle and Raphaël consider the
nonlinear Schrödinger equation (1.5), and proved that,

######  Theorem 1.1.

Let $u_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}$ be radial. Assume that the
corresponding solution to (1.5) blows up in finite time $T$, then there exists
a constant $\gamma>0$ such that 222They obtain (1.6) for
$\gamma=\frac{1}{12}$.

(1.6) $||u(t)||_{L^{3}}\geq C(u_{0})|\log(T-t)|^{\gamma}$

for $t$ close enough to $T$.

######  Remark 1.2.

Merle and Raphaël have proved a more general result for a larger class of
Schrödinger equations. Indeed they cover all the mass-supercritical and
energy-subcritical cases for $d\geq 3$.

######  Remark 1.3.

$\dot{H}^{s_{c}}$ with $s_{c}=\frac{1}{2}$ is the critical norm of (1.5), i.e.
this norm is invariant under the natural scaling of (1.5). Note that one has,
via Sobolev embedding, $\dot{H}^{\frac{1}{2}}(\mathbb{R}^{3})\hookrightarrow
L^{3}(\mathbb{R}^{3})$.

When the scaling index $s_{c}=0$, we say it is mass-critical; $s_{c}=1$
corresponds to the energy-critical case; in this article we focus on the case
with $0<s_{c}<1$, i.e. mass supercritical and energy subcritical case.

######  Remark 1.4.

The estimate (1.6) may be rephrased as: there is no type II333a type II blow
up solution means its critical norm remains bounded when $t$ tends to the blow
up time $T$, i.e. $\limsup_{t\nearrow T}||u(t)||_{\dot{H}^{s_{c}}}<+\infty$,
and a blow up solution is type I if it is not type II. blow up solution to
(1.5), for radial initial data in $\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}$.

Thus, all such finite time blow up solutions do not fall into the regime of
Soliton Resolution Conjecture, which formally predicts all type II blow up
solutions to nonlinear dispersive equation will decouple into solitary wave
living at different scales and a regular term.

The main result of the current article is to prove an analogous result for a
class of Hartree type equations.

######  Theorem 1.5.

Consider the equation

(1.7) $\begin{cases}i\partial_{t}u+\triangle
u=-(V*|u|^{2})u.&(t,x)\in\mathbb{R}\times\mathbb{R}^{3},\\\
u\mid_{t=0}=u_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}.\end{cases}$

where $u_{0}$ is a radial function. Assume $V(x)$ is a radial function
satisfying

(1.8) $\sum x_{j}\partial_{x_{j}}V(x)\leq-\alpha V(x)$

for some $\alpha\in(2,+\infty)$, and assume the integrability condition

(1.9) $V(x)\in L^{1}(\mathbb{R}^{3}),\;\sum x_{j}\partial_{x_{j}}V(x)\in
L^{1}(\mathbb{R}^{3}),$

and the following pointwise bound

(1.10) $|\sum x_{j}\partial_{x_{j}}V(x)|\leq\frac{C}{|x|^{3}}.$

If $u(t)$ blows up in finite time $T$ , then there exists a constant
$\gamma>0$ such that 444Note that (1.11) implies
$||u(t)||_{\dot{H}^{\frac{1}{2}}}\geq C(u_{0})|\log(T-t)|^{\gamma}$ in
$\dot{H}^{\frac{1}{2}}$ setting and our method could give
$\gamma=\frac{2}{31+\varepsilon}$, $\forall\varepsilon>0$.

(1.11) $||u(t)||_{L^{3}}\geq C(u_{0})|\log(T-t)|^{\gamma}$

for $t$ close enough to $T$.

######  Remark 1.6.

Assumption (1.9) seems natural. Assumptions (1.8) and (1.10) are due to
technical reasons.

######  Remark 1.7.

In order to connect virial identity with energy, we have assumed the condition
(1.8) holds. From (1.8), we note that either

(1.12) $V(x)\geq\frac{c_{V}}{|x|^{\alpha}}$

at a neighborhood of the origin or $V(x)$ is non-positive. Note that $V(x)$
can not be non-positive because in this case there exist no blow up solutions
due to energy conservation law.

######  Remark 1.8.

Under the assumption (1.8), by the following virial identity ,

(1.13) $\frac{d^{2}}{dt^{2}}\int|x|^{2}|u|^{2}=8\int|\nabla u|^{2}+4\int
x_{j}|u|^{2}\partial_{j}(V*|u|^{2})dx,$

one could easily derive blow up solutions with negative energy. For the
completeness of the article, we have supplemented the proof of this part in
Appendix A.

######  Remark 1.9.

The assumptions on $V(x)$ can be relaxed. Indeed, (1.8) and $V(x)\in
L^{1}(\mathbb{R}^{3})$ imply $\sum x_{j}\partial_{x_{j}}V(x)\in
L^{1}(\mathbb{R}^{3})$.

######  Remark 1.10.

We give an example which satisfies the conditions (1.8), (1.9), and (1.10)
i.e. the object of the above analysis is not Ø,

(1.14) $V(Z)=\frac{1}{|Z|^{3}|\log Z|^{\alpha}}\chi(Z),\;for\;\alpha>1,$

with

$\chi(Z)=\begin{cases}1&Z\leq\delta\\\ 0&Z\geq
2\delta\end{cases}for\;some\;\delta\;small\;enough.$

We review a series of related work regarding Hartree type equations and NLS,

  * •

mass-critical case

In the 1990s, Merle’s work [27] had given the characteristic in $H^{1}$ for
the blow-up solutions with minimal mass in nonlinear Schrödinger
equation555The ground state associated to (1.15) is the unique positive
solution to  $\triangle Q+Q^{1+\frac{4}{d}}=Q,$ which supply a stationary
solution to (1.15) with $u(t,x)=e^{it}Q(x)$.,

(1.15) $\begin{cases}i\partial_{t}u+\triangle
u=-|u|^{\frac{4}{d}}u.&(t,x)\in\mathbb{R}\times\mathbb{R}^{d},\\\
u\mid_{t=0}=u_{0}.\end{cases}$

And Dodson improved this result in $L^{2}$ setting in [6, 7]. The parallel
result in Hartree-equation setting can be found in [23] and [35].

When the mass is below the ground state $Q(x)$, Weinstein proved the solution
is global and scattering in $H^{1}$ in [41] and Dodson improved this result to
$L^{2}$ setting in [4].

When the mass is beyond the ground state, Bourgain and Wang in [1] constructed
a type of blow up solutions with the blow up rate $||\nabla
u(t)||_{L^{2}}\sim\frac{1}{T-t}$. Besides, another type of blow up solutions,
the log-log blow up solutions, are suggested numerically by Landman,
Papanocolaou, Sulem, Sulem in [21]. This kind of solutions blow up in finite
time with the rate $||\nabla
u(t)||_{L^{2}}\sim(\frac{\log|\log(T-t)|}{T-t})^{\frac{1}{2}}$ . Perelman
firstly constructed this kind of solutions in her work [37]. After that, the
log-log blow-up solutions have been studied in depth and comprehensively by
Merle and Raphaël in a series of work [31, 29, 32, 28, 30, 39]. They give a
more complete portrayal of the blow up rate in the vicinity of the ground
state solution and a classification of the blow up solutions.

We should remark, for the mass-critical focusing NLS, when the mass is much
larger than threshold, Merle constructed a $k$-points blow up solution in
[26], and readers can also refer to [38, 11] for the study of weakly
interacting multi bubbles blow up dynamics for NLS. Besides, Martel and
Raphaël in [25] constructed a multi-bubbles blow up solution, with the rate
$||\nabla u(t)||_{L^{2}}\sim\frac{|\log(T-t)|}{T-t}$ due to strong
interactions.

  * •

mass-supercritical and energy-subcritical case

Between this range, for the focusing 3d cubic NLS, Merle and Raphaël in [33]
gave a universal blow-up lower bound in the radial case. Besides, towards this
model, there is a series of work concerning the following
quantity,666$M(u):=\int|u|^{2}$ and $E(u):=\frac{1}{2}\int|\nabla
u|^{2}-\frac{1}{4}\int|u|^{4}$. And $Q$ is the unique $H^{1}$ radial positive
solution of  $\triangle Q-\frac{1}{2}Q+|Q|^{2}Q=0.$

$M\varXi:=\frac{M[u]E[u]}{M[Q]E[Q]},$

readers can refer to for [16, 8, 17] when $M\varXi\in(0,1)$, [9] for
$M\varXi=1$ and [36, 10] in the case $M\varXi>1$.

For the focusing Hartree equation, readers may also refer to [13] in this
range. Our result is also a step toward understanding some universal blow up
behaviour and the connection between NLS and Hartree type equations.

  * •

energy-critical case

For the focusing energy-critical Schrödinger equation, Kenig and Merle in [9]
used a concentration compactness argument and a rigidity theorem to prove any
radial solutions $u(t)$ in $d=3,4,5$ which satisfy $E(u_{0})<E(W)$ and
$||u_{0}||_{\dot{H}^{1}}<||W||_{\dot{H}^{1}}$777The $W(x)$ is the unique
radial positive solution in $\dot{H}^{1}$ to  $\triangle
W+|W|^{\frac{4}{d-2}}W=0,$ with $d\geq 3$. must be global and scatter. For the
nonradial case, readers can refer to [5] in $d=4$ and [19] $d\geq 5$.

Returning to the Hartree-equation setting, [22] treated the focusing case, and
proved a parallel result with [19].

###  1.2. A review of Merle and Raphael’s work in [33] and the connection with
our result

Since our work rely on the method developed by Merle and Raphaël in [33], let
us review some points and highlight the main quantities in their analysis.
There is also some interesting work related to this topic, one can refer to
[40] for Navier Stokes equation, [20] for focusing nonlinear Klein-Gordon
equation and [25] for inhomogeneous nonlinear Schrödinger equation.

In this subsection, in order to make the idea of the article more concise and
clear, we only review a weaker version of Merle and Raphaël. More precisely,
we impose the following non-positive energy assumption for Theorem 1.1,

$E(u_{0})\leq 0,$

and the critical norm under consideration becomes $\dot{H}^{\frac{1}{2}}$.
Towards the nonlinear Schrödinger equation (1.5), for $t$ close enough to the
blow up time $T$, we renormalize $u(t)$ with its $\dot{H}^{1}$ norm, i.e. let

(1.16) $\lambda_{u}(t)=(\frac{1}{||\nabla u(t)||_{L^{2}}})^{2},$

and define the renormalization of $u(t)$:

(1.17)
$v(\tau,x)=\lambda_{u}(t)\bar{u}(t-\lambda_{u}(t)^{2}\tau,\lambda_{u}(t)x).$

From local theory of NLS, we note that

(1.18) $\lambda_{u}(t)\lesssim\sqrt{T-t}.$

Then a long time behaviour of blow up dynamics has been transformed into a
Cauchy problem that satisfies the following special conditions:

(1.19) $\begin{cases}i\partial_{\tau}v+\triangle
v(\tau,x)=|v|^{2}v(\tau,x),&(\tau,x)\in[0,\frac{1}{\lambda_{u}(t)}]\times\mathbb{R}^{3},\\\
v(\tau,x)\mid_{\tau=0}=\lambda_{u}(t)\bar{u}(t,\lambda_{u}(t)x),\end{cases}$

with

(1.20) $||v(0,x)||_{\dot{H}^{\frac{1}{2}}}=||u(t)||_{\dot{H}^{\frac{1}{2}}},$
(1.21) $||v(0,x)||_{\dot{H}^{1}}=1,$

and the non-positive energy

(1.22) $E(v_{0})\leq 0.$

Let

(1.23) $N(t)=-\log\lambda_{u}(t),$

we could conclude, Merle and Raphaël converted Theorem 1.1 into proving the
following conclusion, there exists a universal constant $\gamma>0$ such that

(1.24) $||v(0,x)||_{\dot{H}^{\frac{1}{2}}}\geq N^{\gamma}.$

The above analysis reduces the difficulty of the problem to some extent. We
know that the blow up behaviour of the solution is a long-time dynamic
behaviour, and there are few tools that can be directly applied to
characterize the explosion. However, if we are now looking at the local
behaviour of the solution, then a rich local theory of the solutions can be
applied, making the problem possible.

From the above analysis, it is also a very natural thing to give the
connection between the lower bounds of the blow up rate of the Schrödinger
equation and Hartree equation, which also gives the possibility to apply the
local theory.

To further explain the work by Merle and Raphaël, we now introduce the
following notations, the scaling invariant Morrey-Campanato norm

(1.25) $\rho(u,R)=\sup_{R^{\prime}\geq
R}\frac{1}{R^{\prime}}\int_{R^{\prime}\leq|x|\leq 2R^{\prime}}|u(x)|dx,$

a quantity related to the initial data,

$M_{0}=\frac{4||v(0)||_{\dot{H}^{\frac{1}{2}}}}{C_{GN}},$

where $C_{GN}$ is a universal constant related to Gagliardo-Nirenberg
inequality and a similar definition to (1.16),

(1.26) $\lambda_{v}(\tau)=\frac{1}{||v(\tau)||_{\dot{H}^{1}}^{2}}.$

.

In order to achieve the goal (1.24), there are three steps need to be
implemented,

Step 1. Uniform control of the $\rho$ norm and the dispersive estimate.

To achieve the first goal, we give the following proposition,

######  Proposition 1.11.

Let $v(\tau)\in C([0,e^{N}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ be a
radially symmetric solution to (1.5) where $N$ is a sufficiently large number
and (1.22), (1.21) hold. Then there exist universal constants $C_{1}$,
$\alpha_{1}$ and $\alpha_{2}$ such that the following hold,
$\forall\tau_{0}\in[0,e^{N}]$,

(i) the uniform control of the $\rho$ norm:

(1.27) $\rho(v(\tau_{0}),C_{1}M_{0}^{\alpha_{1}}\sqrt{\tau_{0}})\leq
C_{1}M_{0}^{2}.$

(ii) the dispersive estimate:

(1.28)
$\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||v(\tau)||_{\dot{H}^{1}}^{2}d\tau\leq
C_{1}M_{0}^{\alpha_{2}}\tau_{0}^{\frac{3}{2}}.$

Moreover, if we assume $M_{0}^{\alpha_{2}}<e^{\frac{\sqrt{N}}{2}}$, there
exist a sequence of $\\{\tau_{i}\\}$ with $\tau_{i}\in[0,e^{i}]$ and
$i\in\\{\sqrt{N},\sqrt{N}+1,...,N\\}$ with the following bounds holds,

(1.29) $\frac{\sqrt{\tau_{i}}}{\lambda_{v}(\tau_{i})}\leq
C_{1}M_{0}^{\alpha_{2}}\;\;\;\;\lambda_{v}(\tau_{i})\in[\frac{1}{10C_{1}M_{0}^{\alpha_{2}}}e^{\frac{i-1}{2}},\frac{10}{C_{1}M_{0}^{\alpha_{2}}}e^{\frac{i}{2}}].$

There are two points we should note,

######  Remark 1.12.

$\begin{cases}\alpha_{1}=1,\\\ \alpha_{2}=5\end{cases}$ is an allowable value
of the above proposition, and it is related to the blow up rate in Theorem
1.1, we could see it more clearly to achieve step 2.

######  Remark 1.13.

The conclusion (1.29) in proposition 1.11 is a direct inference of (1.28) with
the help of (1.21). Since (1.28) only gives an estimate of the decay of
$||v(\tau)||_{\dot{H}^{1}}$ in the average sense, we can only get (1.29) at
some special time rather than a pointwise estimate.

Step 2. Lower bound on a weighted local $L^{2}$ norm of $v(0)$.

Now we are ready to give a uniform lower bound on a weighted local $L^{2}$
norm of $v(0)$,

######  Proposition 1.14.

Let $v(\tau)$ satisfies the conditions in proposition (1.11) and
$\\{\tau_{i}\\}$ are chosen in proposition (1.11), then the following weighted
$L^{2}$ norm on a sufficiently large ball holds,

(1.30) $\frac{1}{\lambda_{v}(\tau_{i})}\int_{|x|\leq
M_{0}^{2+2\alpha_{2}}\lambda_{v}(\tau_{i})}|v(0)|^{2}\geq c_{3},$

where $c_{3}$ is a universal constant.

######  Remark 1.15.

The idea of proof is the following: first we derive the lower bound at time
$\tau=\tau_{i}$ with the aid of non-positive energy constraint, i.e.

(1.31) $\frac{1}{\lambda_{v}(\tau_{i})}\int_{|x|\leq
M_{0}^{2+2\alpha_{2}}\lambda_{v}(\tau_{i})}|v(\tau_{i})|^{2}\geq 2c_{3}.$

Second, thanks to (1.27) and (1.28), we conclude the difference between
$\tau=0$ and $\tau=\tau_{i}$ is sufficiently small, i.e.
$\exists\;\varepsilon>0$ small enough, such that

(1.32) $|\frac{1}{\lambda(\tau_{i})}\int_{|x|\leq
M_{0}^{2+2\alpha_{2}}\lambda_{v}(\tau_{i})}|v(\tau_{i})|^{2}-\frac{1}{\lambda(\tau_{i})}\int_{|x|\leq
M_{0}^{2+2\alpha_{2}}\lambda_{v}(\tau_{i})}|v(0)|^{2}|<\varepsilon.$

Combining (1.31) and (1.32), we derive the final conclusion (1.30).

######  Remark 1.16.

The lower bound (1.30) has implied the choice of $\gamma$ in Theorem 1.1. A
straightforward algebraic computation of our analysis implies

$\gamma=\frac{1}{4+2\alpha_{2}+\varepsilon},$

for $\forall\varepsilon>0$, i.e we could derive
$\gamma=\frac{1}{14+\varepsilon}$ in this setting while Merle and Raphaël in
[33] give

(1.33) $\gamma=\frac{1}{12}.$

Step 3. The construction of $N^{\gamma}$ disjoint annuli.

Now we are going to construct sufficiently many disjoint annuli. We need to
pay attention to the following two points,

  * •

These $N^{\gamma}$ annuli are disjoint.

  * •

At each annulus, the similar lower bound still holds in (1.30).

A direct computation by Hölder inequality implies the following choice of the
size of annuli satisfies the second point,

$\mathscr{C}_{i}:=\\{x\in\mathbb{R}^{3}\mid\frac{\lambda_{v}(\tau_{i})}{M_{0}^{2+\alpha_{2}}}\leq|x|\leq\lambda_{v}(\tau_{i})M_{0}^{2+\alpha_{2}}\\}.$

In order to satisfy the first point above, we need to choose the number $p$
such that

$\frac{\lambda_{v}(\tau_{i+p})}{M_{0}^{2+\alpha_{2}}}\geq\lambda_{v}(\tau_{i})M_{0}^{2+\alpha_{2}}.$

Thanks to (1.29), we can choose $p$ such that $e^{\frac{p}{2}}\thickapprox
M_{0}^{4+2\alpha_{2}}$ which gives the derived choice of $N^{\gamma}$ disjoint
annuli.

Based on the above three steps, we have the following technical remark,

######  Remark 1.17.

The condition (1.22) can be weaken by another three assumptions, and the
stronger version will be stated in section 3. Here we are only using condition
(1.22) to state this proposition for convenience.

Now we want to talk about the relation in this paper between NLS and Hartree
equation,

  * •

Using the above techniques for proving Theorem 1.1, we can obtain result that
is parallel in the Hartree equation, which is also the purpose of our article.

  * •

If we get the conclusion in Hartree equation directly, then we can approximate
$\delta$ function by doing a suitable scaling transformation of the potential
function $V(x)$888We should note that the conditions (1.8), (1.9), (1.10) is
invariant under the transformation (1.34).,

(1.34)
$V_{\varepsilon}(x)=\frac{1}{\varepsilon^{3}}V(\frac{x}{\varepsilon}),\;\forall\varepsilon>0.$

After using the mature local theory999The above three steps are local versions
for the dynamics of the solutions, so we could apply local theory., we prove
that the Schrödinger equation is also valid in this case.

The above two points give a connection between NLS and Hartree equation, and
we supply complete proof in section 3.

###  1.3. Strategy and structure of the paper

We use the robust strategy by Merle and Raphaël in [33]. We should note that
there are two key points:

(1) The additional error terms.

The condition (1.9) is natural assumption in our problem setting, and the role
of the condition (1.8) is to connect the energy (1.3) with the virial identity
(1.13). Another assumption (1.10) is technical, because we need to control
additional error terms which do not exist in the Schrödinger-equation setting.
As in the Schrödinger equation, the “potential function” is $\delta(x)$.
However, in our setting, $V(x)$ belongs to $L^{1}(\mathbb{R}^{3})$. So, when
the blow-up phenomenon happens, there are some additional error terms we
should treat. We give a priori control on two channels which helps us to
overcome this difficulty. For more details, one can see step 2 in Proposition
2.2.

(2) The difficulty caused by the no-scaling property of $V(x)$.

Since we treat the no-scaling case for the potential function $V(x)$, we
should check carefully the constants chosen in Proposition 2.2 and 2.4. After
the renormalization of $u(t)$,
$v^{\lambda(t)}(\tau,x):=\lambda_{u}(t)\bar{u}(t-\lambda_{u}^{2}(t)\tau,\lambda_{u}(t)x)$
at different time $t$ satisfy different equations

(1.35) $\begin{cases}i\partial_{\tau}v^{\lambda(t)}+\triangle
v^{\lambda(t)}=-(V_{\lambda(t)}*|v^{\lambda(t)}|^{2})v^{\lambda(t)}.&(\tau,x)\in[0,\frac{t}{\lambda(t)^{2}})\times\mathbb{R}^{3},\;V_{\lambda(t)}(x)=\lambda(t)^{3}V(\lambda(t)x),\\\
v^{\lambda(t)}\mid_{\tau=0}=\lambda(t)\bar{u}(t,\lambda(t)x)\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1},\end{cases}$

where $\lambda_{u}(t)=(\frac{1}{||\nabla u(t)||_{L^{2}}})^{2}$. For these
$v^{\lambda(t)}(\tau,x)$, we should give a uniform estimate which is
independent of the time $t$. We will deal with this very carefully both in
Proposition 2.2 and 2.4, which is also the core of our analysis. Here, we
explain from two aspects why the potential function does not have scaling
invariant property will not have an essential impact for our analysis.

(i) The renormalization of $u(t)$

$v^{\lambda(t)}(\tau,x)$ defined above satisfies equation (1.35), and it also
follows the law of conservation of energy:

(1.36) $\displaystyle E^{\lambda}(v^{\lambda(t)}(\tau))$
$\displaystyle=E^{\lambda}(v^{\lambda(t)}(0))=\frac{1}{2}\int|\nabla
v^{\lambda(t)}(0,x)|^{2}dx-\frac{1}{4}\int(V_{\lambda(t)}*|v^{\lambda(t)}|^{2})(0,x)|v^{\lambda(t)}|^{2}(0,x)dx$
$\displaystyle=\lambda(t)[\frac{1}{2}\int|\nabla
u(t,x)|^{2}dx-\frac{1}{4}\int(V*|u|^{2})|u|^{2}(t,x)dx]$
$\displaystyle=\lambda(t)E(u(t)).$

(1.36) implies we still obtain the scaling invariant property for the energy
conservation law although $V(x)$ does not own it.

(ii) The scaling for the potential function $V(x)$.

A direct computation implies

(1.37) $|\sum x_{j}\partial_{x_{j}}V_{\lambda}(x)|=\lambda^{3}|\sum\lambda
x_{j}\partial_{\lambda x_{j}}V(\lambda
x)|\leq\lambda^{3}\cdotp\frac{C}{|\lambda x|^{3}}=\frac{C}{|x|^{3}},$

under the assumption (1.10) which is a universal upper bound independent of
$\lambda$. Besides, we only use $L^{1}$ norm for the potential function $V(x)$
which is also a scaling invariant norm under the renormalization, i.e.
$||V_{\lambda}(x)||_{L^{1}}=||V(x)||_{L^{1}}$.

The paper is organized as follows. In section 2, we will prove Theorem 1.5.
More specifically, we will give a uniform control of the $\rho$ norm, which is
a suitable scaling invariant Morrey-Campanato norm, and a lower bound on a
weight local $L^{2}$ norm, then we use a blackbox by applying Proposition 2.2
and 2.4 to finish the proof for Theorem 1.5. In section 3, we supply the
connection between Schrödinger equation and Hartree equation. In Appendix A,
for readers’ convenience, we give some examples of blow up solutions and some
direct observations for the blow up rate for $V(x)$ with higher integrability
conditions. . However, we can not give any special examples $V(x)$ which lead
to this kind of blow up solutions. Appendix B is devoted to give some standard
result of stability theory needed in Section 3.

##  2\. BLOW-UP RATE FOR THE BLOW-UP SOLUTION

In this section, we focus on the proof of Theorem 1.5, i.e. we consider the
equation

(2.1) $\begin{cases}i\partial_{t}u+\triangle
u=-(V*|u|^{2})u.&(t,x)\in\mathbb{R}\times\mathbb{R}^{3},\\\
u\mid_{t=0}=u_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}.\end{cases}$

with $V(x)$ satisfies the conditions in Theorem 1.5. We will give a lower
bound for the blow-up rate.

First, we prove the main propositions at the heart of the proof of Theorem
1.5. Let $u_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}$ with radial symmetry
and assume that the corresponding solution $u(t)$ to (2.1) blows up in finite
time $0<T<+\infty$. We can pick $t$ close enough to $T$. Let

(2.2) $\lambda_{u}(t)=(\frac{1}{||\nabla u(t)||_{L^{2}}})^{2},$

then from the local theory,

(2.3) $\lambda_{u}(t)\lesssim\sqrt{T-t}.$

We define the renormalization of $u(t)$ by

(2.4)
$v^{(t)}(\tau,x)=\lambda_{u}(t)\bar{u}(t-\lambda_{u}^{2}(t)\tau,\lambda_{u}(t)x).$

In this subsection, to clarify the notations, we omit the dependence of
$\lambda$ on $u(t)$ and define $v^{\lambda}:=v^{(t)}(\tau,x)$.

Then it is not hard to check that $v^{\lambda}$ satisfies the following
equation

(2.5) $\begin{cases}i\partial_{\tau}v^{\lambda}+\triangle
v^{\lambda}=-(V_{\lambda}*|v^{\lambda}|^{2})v^{\lambda}.&(\tau,x)\in[0,\frac{t}{\lambda^{2}})\times\mathbb{R}^{3},\;V_{\lambda}(x)=\lambda^{3}V(\lambda
x),\\\ v^{\lambda}\mid_{\tau=0}=\lambda\bar{u}(t,\lambda
x)\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}.\end{cases}$

Because of the no-scaling property of the potential function $V(x)$, the
renormalization $v^{\lambda}$ satisfies different equation at different time
$t$. So we need some universal properties for these $v^{\lambda}$ which are
independent of $\lambda$.

We first give needed definitions in our following propositions and recall some
elementary inequalities.

Define a smooth radially symmetric cut-off function,

(2.6) $\psi(x)=\begin{cases}\frac{|x|^{2}}{2},&|x|\leq 2,\\\ 0,&|x|\geq
3,\end{cases}$

with $\psi_{R}(x):=R^{2}\psi(\frac{x}{R})$ and we state the following lemma in
[33]:

######  Lemma 2.1.

(Radial Gagliardo-Nirenberg inequality)

(i) There exists a universal constant $C>0$ such that for all $u\in L^{3}$,

(2.7) $\forall R>0,\;\frac{1}{R}\int_{|y|\leq R}|u|^{2}dy\leq
C||u||_{L^{3}}^{2}.$

(ii) For all $\eta>0,$ there exists a constant $C_{\eta}>0$ such that for all
$u\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}$ with radial symmetry, for all
$R>0,$

(2.8) $\int_{|x|\geq R}|u|^{4}\leq\eta||\nabla u||_{L^{2}(|x|\geq
R)}^{2}+\frac{C_{\eta}}{R}[\rho(u,R)^{2}+\rho(u,R)^{3}].$

And from the definition of energy, we know

(2.9) $\displaystyle E^{\lambda}(v^{\lambda})$
$\displaystyle:=\frac{1}{2}\int|\nabla
v^{\lambda}|^{2}dx-\frac{1}{4}\int(V_{\lambda}*|v^{\lambda}|^{2})|v^{\lambda}|^{2}dx$
$\displaystyle\geq\frac{1}{2}(1-(\frac{||v^{\lambda}||_{L^{3}}}{c_{V}})^{2})\int|\nabla
v^{\lambda}|^{2}dx,$

for some $c_{V}>0$ which is independent of $\lambda$.

With these prepared knowledge, we prove the following,

######  Proposition 2.2.

Let $\tau_{*}>0$ and $v^{\lambda}(\tau)\in
C([0,\tau_{*}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ be a radially symmetric
solution to (2.5) and assume

(2.10) $\tau_{*}^{\frac{1}{2}}max(E^{\lambda}(v_{0}^{\lambda}),0)<1,$

and

(2.11) $M_{0}^{\lambda}:=\frac{4||v_{0}^{\lambda}||_{L^{3}}}{c_{V}}\geq 2,$

then there exist $C_{1},\alpha_{1},\alpha_{2}>0$ which are independent of
$\lambda$ such that

(2.12)
$\rho(v^{\lambda}(\tau_{*}),(M_{0}^{\lambda})^{\alpha_{1}}\sqrt{\tau_{*}})\leq
C_{1}(M_{0}^{\lambda})^{2},$

and

(2.13) $\int_{0}^{\tau_{*}}(\tau_{*}-\tau)||\nabla
v^{\lambda}(\tau)||_{L_{x}^{2}}^{2}d\tau\leq(M_{0}^{\lambda})^{\alpha_{2}}\tau_{*}^{\frac{3}{2}}.$

###### Proof.

Step 1. Local radial virial estimate.

We define

(2.14) $V_{a}(\tau):=\int_{R^{3}}a(x)|v^{\lambda}(\tau,x)|^{2}dx,$

and

(2.15) $P_{a}(\tau):=\frac{d}{d\tau}V_{a}(\tau)=2Im\int_{R^{3}}\nabla
a(x)\nabla v^{\lambda}(\tau,x)\bar{v^{\lambda}}(\tau,x)dx.$

The a direct computation gives

$\displaystyle\frac{1}{4}\frac{d}{d\tau}P_{a}(\tau)$ $\displaystyle\leq\alpha
E^{\lambda}(v^{\lambda}(\tau))-(\frac{\alpha}{2}-1)\int|\nabla
v^{\lambda}|^{2}dx+\frac{C}{R^{2}}\int_{2R\leq|x|\leq 3R}|v^{\lambda}|^{2}dx$
$\displaystyle+\frac{1}{4}\int\int[(\partial_{x_{j}}\psi_{R}(x)-\partial_{y_{j}}\psi_{R}(y))-(x_{j}-y_{j})]\partial_{x_{j}}V_{\lambda}(x-y)|v^{\lambda}(\tau,y)|^{2}|v^{\lambda}(\tau,x)|^{2}dydx$
$\displaystyle=\alpha
E^{\lambda}(v^{\lambda}(\tau))-(\frac{\alpha}{2}-1)\int|\nabla
v^{\lambda}|^{2}dx+\frac{C}{R^{2}}\int_{2R\leq|x|\leq
3R}|v^{\lambda}|^{2}dx+\Phi(\tau),$

where we substitute $a(x)$ with $\psi_{R}(x)$ and use the condition (1.8) with

$\displaystyle|\Phi(\tau)|$
$\displaystyle=\frac{1}{4}|\int\int[(\partial_{x_{j}}\psi_{R}(x)-\partial_{y_{j}}\psi_{R}(y))-(x_{j}-y_{j})]\partial_{x_{j}}V_{\lambda}(x-y)|v^{\lambda}(\tau,y)|^{2}|v^{\lambda}(\tau,x)|^{2}dydx|$
$\displaystyle\lesssim\iint_{|x|\geq R,|y|\geq
R}|[(\partial_{x_{j}}\psi_{R}(x)-\partial_{y_{j}}\psi_{R}(y))-(x_{j}-y_{j})]\partial_{x_{j}}V_{\lambda}(x-y)||v^{\lambda}(\tau,y)|^{2}|v^{\lambda}(\tau,x)|^{2}dxdy$
$\displaystyle+\iint_{|x|\leq R,|y|\geq
2R}|[(\partial_{x_{j}}\psi_{R}(x)-\partial_{y_{j}}\psi_{R}(y))-(x_{j}-y_{j})]\partial_{x_{j}}V_{\lambda}(x-y)||v^{\lambda}(\tau,y)|^{2}|v^{\lambda}(\tau,x)|^{2}dxdy$
$\displaystyle+\iint_{|y|\leq R,|x|\geq
2R}|[(\partial_{x_{j}}\psi_{R}(x)-\partial_{y_{j}}\psi_{R}(y))-(x_{j}-y_{j})]\partial_{x_{j}}V_{\lambda}(x-y)||v^{\lambda}(\tau,y)|^{2}|v^{\lambda}(\tau,x)|^{2}dxdy$
$\displaystyle:=(I)+(II)+(III).$

We should point out, the term $\Phi(\tau)$ is different from the error term in
Schrödinger-equation setting and this is the reason to assume the additional
condition (1.10). We summarize the estimate of $\Phi(\tau)$ as follows,

######  Lemma 2.3.

Assume (1.9), (1.10) in Theorem 1.5 hold, then $\forall\eta>0,\;0<R_{1}<R,$
there exists $C(\eta)>0$ such that

(2.16) $\displaystyle|\Phi(\tau)|$ $\displaystyle\leq\eta||\nabla
v^{\lambda}(\tau)||_{L_{x}^{2}}^{2}+C(\eta)R^{-1}[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]$
$\displaystyle+||\nabla
v^{\lambda}||_{L^{2}}^{2}\cdotp\frac{R_{1}^{2}}{R^{2}}\cdotp\rho(v^{\lambda}(\tau),R).$

If we assume Lemma 2.3 holds, then we conclude

(2.17) $\displaystyle\frac{1}{4}\frac{d}{dt}P_{a}(\tau)$
$\displaystyle\leq\alpha
E^{\lambda}(v^{\lambda}(\tau))-\frac{1}{2}(\frac{\alpha}{2}-1)\int|\nabla
v^{\lambda}|^{2}dx$
$\displaystyle+C(\eta)R^{-1}[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]+||\nabla
v^{\lambda}||_{L^{2}}^{2}\cdotp\frac{R_{1}^{2}}{R^{2}}\cdotp\rho(v^{\lambda}(\tau),R).$

Now we return to the proof of Lemma 2.3,

###### Proof.

For $(I)$,

(2.18) $\displaystyle(I)$ $\displaystyle\leq||x\cdotp\nabla_{x}V(\lambda
x)||_{L^{1}}||v^{\lambda}(\tau,x)\cdotp\chi_{\\{|x|\geq
R\\}}||_{L^{4}}^{2}||v^{\lambda}(\tau,y)\cdotp\chi_{\\{|y|\geq
R\\}}||_{L^{4}}^{2}$
$\displaystyle\leq||v^{\lambda}(\tau,x)\cdotp\chi_{\\{|x|\geq
R\\}}||_{L^{4}}^{4}$ $\displaystyle\lesssim\eta||\nabla
v^{\lambda}||_{L^{2}}^{2}+\frac{C_{\eta}}{R}[\rho(v^{\lambda},R)^{2}+\rho(v^{\lambda},R)^{3}]$

by Lemma 2.1.

Since $(II)$ and $(III)$ are symmetric, we only estimate $(II)$. In fact, this
term does not appear in the Schrödinger equation. We should divide the area
$\\{|x|\leq R,|y|\geq 2R\\}$ into $\\{|x|\leq R_{1},|y|\geq 2R\\}$ and
$\\{R_{1}\leq|x|\leq R,|y|\geq 2R\\}$ and choose $R_{1}$ carefully in next
step. The term in the area $\\{|x|\leq R_{1},|y|\geq 2R\\}$ can be controlled
by $\frac{R_{1}^{2}}{R^{2}}||\nabla
v^{\lambda}||_{L^{2}}^{2}\rho(v^{\lambda}(\tau),R)$ and the other is
controlled by
$\frac{1}{R}\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)$. The
details are the followings.

In the area $\\{|x|\leq R,|y|\geq 2R\\},$ using the property (1.10), we
conclude

(2.19)
$|[(\partial_{x_{j}}\psi_{R}(x)-\partial_{y_{j}}\psi_{R}(y))-(x_{j}-y_{j})]\partial_{x_{j}}V_{\lambda}(x-y)|\lesssim\lambda^{3}\cdotp\frac{1}{(\lambda|x-y|)^{3}}=\frac{1}{|x-y|^{3}}\lesssim\frac{1}{|y|^{3}},$

which leads to

(2.20) $\displaystyle(II)$ $\displaystyle\lesssim|\iint_{|x|\leq R,|y|\geq
2R}\frac{1}{|y|^{3}}|v^{\lambda}(\tau,y)|^{2}|v^{\lambda}(\tau,x)|^{2}dxdy|$
$\displaystyle\lesssim\int_{|x|\leq
R}|v^{\lambda}(\tau,x)|^{2}dx[\sum_{j=0}^{+\infty}\frac{1}{(2^{j}R)^{3}}\int_{2^{j}R\leq|y|\leq
2^{j+1}R}|v^{\lambda}(\tau,y)|^{2}dy]$ $\displaystyle\lesssim[\int_{|x|\leq
R_{1}}|v^{\lambda}(\tau,x)|^{2}dx+\int_{R_{1}\leq|x|\leq
R}|v^{\lambda}(\tau,x)|^{2}dx]\cdotp\frac{1}{R^{2}}\rho(v^{\lambda}(\tau),R)$
$\displaystyle\lesssim[R_{1}^{2}||\nabla
v^{\lambda}||_{L^{2}}^{2}+R\rho(v^{\lambda}(\tau),R_{1})]\cdotp\frac{1}{R^{2}}\cdotp\rho(v^{\lambda}(\tau),R).$

From (2.18) and (2.20), we finish the proof of Lemma 2.3. ∎

Step 2. A priori control of the $\rho$ norm on parabolic space time interval.

In this step, we need to choose $R_{1}$ appropriately. On the one hand, we
should choose $R_{1}$ small enough so that
$\frac{R_{1}^{2}}{R^{2}}\cdotp\rho(v^{\lambda}(\tau),R)\ll\frac{1}{2}$. On the
other hand, we can not choose $R_{1}$ too small otherwise we can not control
$R^{-1}\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)$. However, the
above analysis is not self-contradictory. A natural idea is, we choose $R_{1}$
with $\frac{R_{1}^{2}}{R^{2}}\lesssim\varepsilon^{c_{1}}$ such that
$\frac{R_{1}^{2}}{R^{2}}\cdotp\rho(v^{\lambda}(\tau),R)\ll\frac{1}{2}$. At the
same time, let $R\geq R_{1}\gg(M_{0}^{\lambda})^{\frac{1}{2\varepsilon}}$ so
that $R^{-1}\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)$ still can
be treated as an error term. Once this difficulty is overcome, the rest of the
analysis is similar to the one in [33].

Let $\varepsilon,\delta>0$ be a small enough constant to be chosen later. Let

(2.21)
$G_{\varepsilon}=(M_{0}^{\lambda})^{\frac{1}{\varepsilon}},\;A_{\varepsilon_{1}}=(\frac{\varepsilon
G_{\varepsilon}}{(M_{0}^{\lambda})^{2}})^{\frac{1}{3}},\;A_{\varepsilon_{2}}=(\frac{\varepsilon
G_{\varepsilon}}{(M_{0}^{\lambda})^{2}})^{\frac{1}{3}}\cdotp(\frac{\delta}{M_{0}^{\lambda}}).$

Recall from Lemma 2.1, there exists a universal constant $C>0$ such that

$\forall R>0,\;\forall u\in L^{3},\;\rho(u,R)\leq C||u||_{L^{3}}^{2}.$

From the regularity of the flow $v^{\lambda}\in
C([0,\tau^{*}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ and the definition of
$M_{0}^{\lambda}$, we may consider the largest time $\tau_{1}\in[0,\tau_{*}]$
such that

(2.22)
$\forall\tau_{0}\in[0,\tau_{1}],\;[M_{1}^{\lambda}(A_{\varepsilon_{1}},\tau_{1})]^{2}=\max_{\tau\in[0,\tau_{1}]}\rho(v^{\lambda}(\tau),A_{\varepsilon_{1}}\sqrt{\tau})\leq
2\frac{(M_{0}^{\lambda})^{2}}{\varepsilon},$

(2.23)
$[M_{2}^{\lambda}(A_{\varepsilon_{2}},\tau_{1})]^{2}=\max_{\tau\in[0,\tau_{1}]}\rho(v^{\lambda}(\tau),A_{\varepsilon_{2}}\sqrt{\tau})\leq
2\frac{(M_{0}^{\lambda})^{5}}{\varepsilon^{10}}$

and

(2.24) $\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}(\tau)||_{L^{2}}^{2}d\tau\leq
G_{\varepsilon}\tau_{0}^{\frac{3}{2}}.$

We claim that :

(2.25)
$\forall\tau_{0}\in[0,\tau_{1}],\;[M_{1}^{\lambda}(A_{\varepsilon_{1}},\tau_{1})]^{2}=\max_{\tau\in[0,\tau_{1}]}\rho(v^{\lambda}(\tau),A_{\varepsilon_{1}}\sqrt{\tau})\leq\frac{(M_{0}^{\lambda})^{2}}{\varepsilon},$

(2.26)
$[M_{2}^{\lambda}(A_{\varepsilon_{2}},\tau_{1})]^{2}=\max_{\tau\in[0,\tau_{1}]}\rho(v^{\lambda}(\tau),A_{\varepsilon_{2}}\sqrt{\tau})\leq\frac{(M_{0}^{\lambda})^{5}}{\varepsilon^{10}},$

and

(2.27) $\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}(\tau)||_{L^{2}}^{2}d\tau\leq\frac{G_{\varepsilon}}{2}\tau_{0}^{\frac{3}{2}}$

provided $\varepsilon,\delta>0$ has been chosen small enough, and (2.12),
(2.13) follow.

Proof of (2.25) and (2.26). $\forall\tau_{0}\in[0,\tau_{1}]$, we integrate
twice in time between $0$ and $\tau_{0}$ from (2.17) and get :

(2.28)
$\displaystyle\int\psi_{R}|v^{\lambda}(\tau_{0})|^{2}+C^{\prime}_{1}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}(\tau)||_{L^{2}}^{2}d\tau$
$\displaystyle\lesssim\int\psi_{R}|v^{\lambda}(0)|^{2}+\tau_{0}[Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(0)\bar{v^{\lambda}}(0)dx)+E(v_{0}^{\lambda})\tau_{0}]$
$\displaystyle+\frac{1}{R}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]d\tau$
$\displaystyle+\frac{R_{1}^{2}}{R^{2}}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}||_{L^{2}}^{2}\rho(v^{\lambda}(\tau),R)d\tau.$

We let $R\geq A_{\varepsilon_{1}}\sqrt{\tau_{0}}$ for
$\forall\tau_{0}\in[0,\tau_{1}]$ and choose $R_{1}$ as
$R_{1}:=\frac{\delta}{M_{0}^{\lambda}}R$, then a direct computation yields

(2.29)
$\displaystyle\frac{R_{1}^{2}}{R^{2}}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}||_{L^{2}}^{2}\rho(v^{\lambda}(\tau),R)d\tau$ $\displaystyle\leq
2(\frac{\delta}{M_{0}^{\lambda}})^{2}\frac{(M_{0}^{\lambda})^{2}}{\varepsilon}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}||_{L^{2}}^{2}d\tau$ $\displaystyle\leq
2\frac{\delta^{2}}{\varepsilon}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}||_{L^{2}}^{2}d\tau.$

Now we can choose $\delta$ small enough such that
$2\frac{\delta^{2}}{\varepsilon}\ll\frac{C_{1}^{\prime}}{2}$. For the sake of
simplicity, if we define $\delta(\varepsilon)=\varepsilon$ and $\varepsilon$
small enough, substitute (2.29) into (2.28) and we get

(2.30)
$\displaystyle\int\psi_{R}|v^{\lambda}(\tau_{0})|^{2}+\frac{C^{\prime}_{1}}{2}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}(\tau)||_{L^{2}}^{2}d\tau$
$\displaystyle\lesssim\int\psi_{R}|v^{\lambda}(0)|^{2}+\tau_{0}[Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(0)\bar{v^{\lambda}}(0)dx)+E^{\lambda}(v_{0}^{\lambda})\tau_{0}]$
$\displaystyle+\frac{1}{R}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]d\tau.$

We directly estimate the right hand side of (2.30) except the second term:

(2.31)
$\int\psi_{R}|v^{\lambda}(\tau_{0})|^{2}\leq(\int(\psi_{R})^{3}dx)^{\frac{1}{3}}(\int|v_{0}^{\lambda}|^{3}dx)^{\frac{2}{3}}\lesssim
R^{3}(M_{0}^{\lambda})^{2},$

(2.32)
$\displaystyle\frac{1}{R}\int_{0}^{\tau_{0}}(\tau_{0}-\tau)[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]d\tau$
$\displaystyle\leq\frac{\tau_{0}^{2}}{R}(\frac{(M_{0}^{\lambda})^{6}}{\varepsilon^{3}}+\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}})$
$\displaystyle\leq
2\frac{\tau_{0}^{2}}{R}\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}}.$

For the second term, we claim the key estimate:
$\forall\tau_{0}\in[0,\tau_{1}],\;\forall A\geq A_{\varepsilon_{1}},$ let
$R=A\sqrt{\tau_{0}}$ , then

(2.33) $Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(0)\overline{v^{\lambda}(0)}+E^{\lambda}(v_{0}^{\lambda})\tau_{0}\leq
C\frac{(M_{0}^{\lambda})^{2}A^{3}}{\varepsilon^{\frac{2}{3}}}\tau_{0}^{\frac{1}{2}}.$

In step 2 and step 3, we assume (2.33) holds and derive the desired upper
bounds (2.25), (2.26) and (2.27). In step 4, we give the proof of (2.33) and
thus finish the whole proof.

Since $R\geq A_{\varepsilon_{1}}\sqrt{\tau_{0}}$ and the definition,
$R_{1}=\frac{\varepsilon}{M_{0}^{\lambda}}R$, we deduce $R_{1}\geq
A_{\varepsilon_{2}}\sqrt{\tau_{0}}$.

We divide (2.30) by $R^{3}$, and get

(2.34) $\frac{1}{R}\int_{R\leq|x|\leq
2R}|v^{\lambda}(\tau_{0})|^{2}dx\lesssim(M_{0}^{\lambda})^{2}+\frac{(M_{0}^{\lambda})^{2}}{\varepsilon^{\frac{2}{3}}}+\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}}\frac{1}{A_{\varepsilon_{1}}^{4}}\leq\frac{(M_{0}^{\lambda})^{2}}{\varepsilon},$

by the definition of $G_{\varepsilon}$ and $A_{\varepsilon_{1}}$.

Similarly, we divide (2.30) by $R_{1}^{3}$, and get

(2.35) $\displaystyle\frac{1}{R_{1}}\int_{R_{1}\leq|x|\leq
2R_{1}}|v^{\lambda}(\tau_{0})|^{2}dx$
$\displaystyle\leq(\frac{R}{R_{1}})^{3}(M_{0}^{\lambda})^{2}+\frac{(M_{0}^{\lambda})^{5}}{\varepsilon^{3+\frac{2}{3}}}+\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}}\frac{1}{A_{\varepsilon_{1}}A_{\varepsilon_{2}}^{3}}$
$\displaystyle\leq\frac{(M_{0}^{\lambda})^{5}}{\varepsilon^{3}}+\frac{(M_{0}^{\lambda})^{5}}{\varepsilon^{3+\frac{2}{3}}}+\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}}\frac{1}{A_{\varepsilon_{1}}A_{\varepsilon_{2}}^{3}}$
$\displaystyle\leq\frac{(M_{0}^{\lambda})^{5}}{\varepsilon^{10}}.$

By standard continuity argument, we conclude (2.12) holds. We should note that
the choice of $C_{1}$ and $\alpha_{1}$ is independent of $\lambda$ from (2.22)
and the definition of $A_{\varepsilon_{1}}$.

Step 3. Self similar decay of the gradient.

We substitute (2.25), (2.26) and (2.33) at $A=A_{\varepsilon_{1}}$ into
(2.30), then we derive

(2.36) $\displaystyle\int_{0}^{\tau_{0}}(\tau_{0}-\tau)||\nabla
v^{\lambda}(\tau)||_{L^{2}}^{2}d\tau$ $\displaystyle\lesssim
R^{3}(M_{0}^{\lambda})^{2}+C\frac{(M_{0}^{\lambda})^{2}A_{\varepsilon_{1}}^{3}}{\varepsilon^{\frac{2}{3}}}\tau_{0}^{\frac{3}{2}}+\frac{\tau_{0}^{2}}{R}\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}}$
$\displaystyle\leq\frac{\tau_{0}^{\frac{3}{2}}(M_{0}^{\lambda})^{2}A_{\varepsilon_{1}}^{3}}{\varepsilon^{\frac{2}{3}}}\leq\frac{G_{\varepsilon}}{2}\tau_{0}^{\frac{3}{2}},$

which finishes the proof of (2.13). From (2.21), we know the choice of
$\alpha_{2}$ in (2.13) is also independent of $\lambda$.

Step 4. Proof of the momentum estimate (2.33).

We can not directly use the interpolation estimate,

(2.37) $|Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(0)\bar{v^{\lambda}}(0)dx|\lesssim||v_{0}^{\lambda}||_{\dot{H}^{\frac{1}{2}}}^{2}R,$

since we only know the information about $||v_{0}^{\lambda}||_{L^{3}}$ rather
than $||v_{0}^{\lambda}||_{\dot{H}^{\frac{1}{2}}}$. On the other hand, if we
use Hölder inequality to estimate

$\displaystyle|Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(0)\bar{v^{\lambda}}(0)dx)|$ $\displaystyle\leq
R^{\frac{3}{2}}||\nabla
v_{0}^{\lambda}||_{L^{2}}(\frac{1}{R^{3}}\int\psi_{R}|v_{0}^{\lambda}|^{2})^{\frac{1}{2}}$
$\displaystyle\leq R^{\frac{3}{2}}||\nabla
v_{0}^{\lambda}||_{L^{2}}M_{0}^{\lambda}.$

Although we have used the information of $||v_{0}^{\lambda}||_{L^{3}}$, we can
not take full advantage of the information from (2.13) at the initial time
$\tau=0$. It is because that we hope $||\nabla
v^{\lambda}(\tilde{\tau})||_{L^{2}}$ has the asymptotic behaviour with

(2.38) $||\nabla
v^{\lambda}(\tilde{\tau})||_{L^{2}}\leq\frac{C}{\tilde{\tau}^{\frac{1}{4}}},$

for some special large enough time $\tilde{\tau}$ from (2.13). With the above
analysis, a natural idea is to choose a time $\tilde{\tau}$ carefully to
obtain the decay estimate (2.38) at $\tilde{\tau}$. Then we use (2.15) to
conclude the difference between $Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(0)\bar{v^{\lambda}}(0)dx$ and $Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(\tilde{\tau})\bar{v^{\lambda}}(\tilde{\tau})dx$ is suitably small,
which helps us to finish the proof (2.33). In order to explain the above
content more clearly, we have divided the proof into three parts.

Part I: Choose a suitable time to obtain the decay estimate (2.38).

Let $\tau_{0}\in[0,\tau_{1}]\;A\geq A_{\varepsilon_{1}}$ and
$R=A\sqrt{\tau_{0}}$. First, we should choose the proper $\tilde{\tau}$ as
above. To be more specific, we claim the following fact: there exists a
universal constant $K>0$ which is independent of $\lambda$ and
$\tilde{\tau}_{0}$ such that

(2.39)
$\tilde{\tau}_{0}\in[\frac{\varepsilon^{\frac{2}{3}}}{4}\tau_{0},\frac{\varepsilon^{\frac{2}{3}}}{2}\tau_{0}]\;with\;||\nabla
v^{\lambda}(\tilde{\tau}_{0})||_{L^{2}}^{2}\leq\frac{KG_{\varepsilon}}{\tilde{\tau}_{0}^{\frac{1}{2}}}.$

Proof of (2.39). By contradiction, let
$\tilde{\tau}=\varepsilon^{\frac{2}{3}}\tau_{0}$, then

(2.40) $\int_{\frac{\tilde{\tau}}{4}}^{\frac{\tilde{\tau}}{2}}||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma\geq
KG_{\varepsilon}\int_{\frac{\tilde{\tau}}{4}}^{\frac{\tilde{\tau}}{2}}\frac{d\sigma}{\sigma^{\frac{1}{2}}}\geq
CKG_{\varepsilon}\tilde{\tau}^{\frac{1}{2}}.$

Moreover,
$\tilde{\tau}=\varepsilon^{\frac{2}{3}}\tau_{0}\leq\tau_{0}\leq\tau_{1}$ and
(2.24) implies:

(2.41)
$G_{\varepsilon}\tilde{\tau}^{\frac{3}{2}}\geq\int_{0}^{\tilde{\tau}}(\tilde{\tau}-\sigma)||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma\geq\frac{\tilde{\tau}}{2}\int_{\frac{\tilde{\tau}}{4}}^{\frac{\tilde{\tau}}{2}}||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma\geq
CKG_{\varepsilon}\tilde{\tau}^{\frac{3}{2}}.$

From (2.41), we conclude a contradiction for $K>0$ large enough.

Part II: Derive the desired upper bound for the quantity
$Im(\int\nabla\psi_{R}(x)\nabla
v^{\lambda}(\tilde{\tau}_{0})\bar{v^{\lambda}}(\tilde{\tau}_{0})dx$.

Define $R=A\sqrt{\tau_{0}}=A_{1}\sqrt{\tilde{\tau}_{0}}$ and thus

(2.42)
$\frac{\varepsilon^{\frac{1}{3}}}{16}\leq\frac{A}{A_{1}}\leq\varepsilon^{\frac{1}{3}}.$

We claim:

(2.43) $|Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(\tilde{\tau}_{0})\overline{v^{\lambda}(\tilde{\tau}_{0})}|\leq
C\frac{(M_{0}^{\lambda})^{2}A^{3}}{\varepsilon^{\frac{2}{3}}}\tau_{0}^{\frac{1}{2}}.$

Proof of (2.43). Before we derive (2.43), we show the following inequality

(2.44) $\frac{1}{R^{3}}\int\psi_{R}|v^{\lambda}(\tilde{\tau}_{0})|^{2}\leq
C[(M_{0}^{\lambda})^{2}+\frac{G_{\varepsilon}}{A_{1}^{3}}].$

In fact, from (2.15) and Cauchy-Schwarz inequality, we obtain

$|\frac{d}{d\tau}\int\psi_{R}|v^{\lambda}|^{2}|=2|Im(\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}\bar{v^{\lambda}})|\leq C||\nabla
v^{\lambda}||_{L^{2}}(\int\psi_{R}|v^{\lambda}|^{2})^{\frac{1}{2}}.$

Therefore, we integrate this differential inequality from $0$ to
$\tilde{\tau}_{0}$ and get:

(2.45) $\displaystyle\int\psi_{R}|v^{\lambda}(\tilde{\tau}_{0})|^{2}$
$\displaystyle\leq
C[\int\psi_{R}|v^{\lambda}(0)|^{2}+(\int_{0}^{\tilde{\tau}_{0}}||\nabla
v^{\lambda}(\sigma)||_{L^{2}}d\sigma)^{2}]$ $\displaystyle\leq
C[R^{3}(M_{0}^{\lambda})^{2}+\tilde{\tau}_{0}\int_{0}^{\tilde{\tau}_{0}}||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma].$

Since
$2\tilde{\tau}_{0}\leq\varepsilon^{\frac{2}{3}}\tau_{0}\leq\tau_{0}\leq\tau_{1}$
and using (2.24), we obtain

(2.46) $\int_{0}^{\tilde{\tau}_{0}}||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma\leq\frac{1}{\tilde{\tau}_{0}}\int_{0}^{2\tilde{\tau}_{0}}(2\tilde{\tau}_{0}-\sigma)||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma\leq
CG_{\varepsilon}\tilde{\tau}_{0}^{\frac{1}{2}}.$

Thus, recalling that $R=A_{1}\sqrt{\tilde{\tau}_{0}}$, we combine (2.45) and
(2.46) to get

$\int\psi_{R}|v^{\lambda}(\tilde{\tau}_{0})|^{2}\leq
C[R^{3}(M_{0}^{\lambda})^{2}+G_{\varepsilon}\tilde{\tau}_{0}^{\frac{3}{2}}]=CR^{3}[(M_{0}^{\lambda})^{2}+\frac{G_{\varepsilon}}{A_{1}^{3}}]$

and concludes the proof of (2.44).

Now we control the virial quantity (2.43) at time $\tilde{\tau}_{0}$:

$\displaystyle|Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(\tilde{\tau}_{0})\overline{v^{\lambda}(\tilde{\tau}_{0})}|$
$\displaystyle\leq R^{\frac{3}{2}}||\nabla
v^{\lambda}(\tilde{\tau}_{0})||_{L^{2}}(\frac{1}{R^{3}}\int|v^{\lambda}(\tilde{\tau}_{0})|^{2})^{\frac{1}{2}}$
$\displaystyle\leq
CR^{\frac{3}{2}}\frac{G_{\varepsilon}^{\frac{1}{2}}}{\tilde{\tau}_{0}^{\frac{1}{4}}}[(M_{0}^{\lambda})^{2}+\frac{G_{\varepsilon}}{A_{1}^{3}}]^{\frac{1}{2}}$
$\displaystyle\leq
CAA_{1}^{\frac{1}{2}}\tau_{0}^{\frac{1}{2}}G_{\varepsilon}^{\frac{1}{2}}[M_{0}^{\lambda}+\frac{G_{\varepsilon}^{\frac{1}{2}}}{A_{1}^{\frac{3}{2}}}]$

from $A\geq A_{\varepsilon_{1}}$ and the choice of $\tilde{\tau}_{0}$. From
the definition of $G_{\varepsilon},A_{\varepsilon_{1}}$ and $A_{1}$ in (2.21)
and (2.42), we derive

$\displaystyle|Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(\tilde{\tau}_{0})\overline{v^{\lambda}(\tilde{\tau}_{0})}|$
$\displaystyle\leq
C(M_{0}^{\lambda})^{2}A^{3}\tau_{0}^{\frac{1}{2}}\frac{1}{\varepsilon^{\frac{1}{2}}\cdotp\varepsilon^{\frac{1}{6}}}=\frac{C(M_{0}^{\lambda})^{2}A^{3}}{\varepsilon^{\frac{2}{3}}}\tau_{0}^{\frac{1}{2}},$

which finishes the proof of (2.43).

Part III: Prove the initial control on the virial quantity (2.33).

Using (2.17), we derive the crude estimate which connects $\tau=0$ with
$\tau=\tilde{\tau}_{0}$:

(2.47) $\displaystyle|\frac{d}{d\tau}Im(\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}\overline{v^{\lambda}})|$ $\displaystyle\leq
C\\{|E^{\lambda}(v_{0}^{\lambda})|+\int|\nabla v^{\lambda}|^{2}dx$
$\displaystyle+R^{-1}[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]\\}.$

Since

$\displaystyle
R^{-1}[\rho(v^{\lambda}(\tau),R)+\rho(v^{\lambda}(\tau),R)^{3}+\rho(v^{\lambda}(\tau),R_{1})\rho(v^{\lambda}(\tau),R)]$
$\displaystyle\leq\frac{C}{R}\frac{(M_{0}^{\lambda})^{7}}{\varepsilon^{11}}\leq\frac{1}{\tilde{\tau}_{0}^{\frac{1}{2}}}$

for $\varepsilon$ small enough, we conclude from (2.47)

(2.48) $|\frac{d}{d\tau}Im(\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}\overline{v^{\lambda}})|\leq
C\\{|E^{\lambda}(v_{0}^{\lambda})|+\int|\nabla
v^{\lambda}|^{2}dx+\frac{1}{\tilde{\tau}_{0}^{\frac{1}{2}}}\\}.$

We integrate (2.48) in time from 0 to $\tilde{\tau}_{0}$ and get

(2.49) $\displaystyle|Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(0)\overline{v^{\lambda}(0)}|$
$\displaystyle\leq|Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(\tilde{\tau}_{0})\overline{v^{\lambda}(\tilde{\tau}_{0})}|$
$\displaystyle+C[\int_{0}^{\tilde{\tau}_{0}}||\nabla
v^{\lambda}(\sigma)||_{L^{2}}^{2}d\sigma+|E^{\lambda}(v_{0}^{\lambda})|\tilde{\tau}_{0}+\tilde{\tau}_{0}^{\frac{1}{2}}].$

Thanks to (2.43) and (2.46), we conclude

(2.50) $\displaystyle Im\int\nabla\psi_{R}\cdotp\nabla
v^{\lambda}(0)\overline{v^{\lambda}(0)}+E^{\lambda}(v_{0}^{\lambda})\tau_{0}$
$\displaystyle\leq\frac{C(M_{0}^{\lambda})^{2}A^{3}}{\varepsilon^{\frac{2}{3}}}\tau_{0}^{\frac{1}{2}}+CG_{\varepsilon}\tilde{\tau}_{0}^{\frac{1}{2}}$
$\displaystyle+E^{\lambda}(v_{0})\tau_{0}+C|E^{\lambda}(v_{0})|\tilde{\tau}_{0}.$

Our main task is to estimate the right-hand side in (2.50) term by term. Since
the first term has been controlled, we estimate the second term

(2.51) $G_{\varepsilon}\tilde{\tau}_{0}^{\frac{1}{2}}\leq
CA_{\varepsilon_{1}}^{3}(M_{0}^{\lambda})^{2}\frac{\varepsilon^{\frac{1}{3}}}{\varepsilon}\tau_{0}^{\frac{1}{2}}\leq\frac{C(M_{0}^{\lambda})^{2}A^{3}}{\varepsilon^{\frac{2}{3}}}\tau_{0}^{\frac{1}{2}},$

where we use the definition of $G_{\varepsilon},A_{\varepsilon_{1}}$ in (2.21)
and the choice of $\tilde{\tau}_{0}$ in (2.39). Lastly, to control the
remaining two terms in (2.50), we observe that

$\displaystyle
E^{\lambda}(v_{0}^{\lambda})\tau_{0}+C|E^{\lambda}(v_{0}^{\lambda})|\tilde{\tau}_{0}$
$\displaystyle\leq[E^{\lambda}(v_{0}^{\lambda})+C\varepsilon^{\frac{2}{3}}|E^{\lambda}(v_{0}^{\lambda})|]\tau_{0}$
$\displaystyle\leq C\max[E^{\lambda}(v_{0}^{\lambda}),0]\tau_{0}\leq
C\tau_{0}^{\frac{1}{2}}$

for $\varepsilon>0$ small enough and we have used the condition (2.10).
Summing up the above estimates, we finish the desired proof (2.33). ∎

The following proposition implies a nontrivial repartition of the $L^{2}$ mass
of the initial data. This repartition helps us to get uniform lower bound on
sufficient disjoint annuli.

######  Proposition 2.4.

(Lower bound on a weighted local $L^{2}$ norm of $v^{\lambda}(0))$

Let $\tau_{*}>0$ and $v^{\lambda}(\tau)\in
C([0,\tau_{*}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ be a radially symmetric
solution to (2.5) and assume (2.10) and (2.11) in Prop 2.2 hold. Then there
exist universal constants $\alpha_{3},c_{3}>0$ which are independent of
$\lambda$ such that the following holds true. Let

(2.52) $\tilde{\lambda}_{v}(\tau)=(\frac{1}{||\nabla
v^{\lambda}(\tau)||_{L^{2}}})^{2},$

and let

(2.53) $\tau_{0}\in[0,\frac{\tau_{*}}{2}],$

(2.54)
$\tilde{\lambda}_{v}(\tau_{0})E^{\lambda}(v_{0}^{\lambda})\leq\frac{1}{4}.$

Let

(2.55) $F_{*}=\frac{\sqrt{\tau_{0}}}{\tilde{\lambda}_{v}(\tau_{0})},$

and

(2.56) $D_{*}=(M_{0}^{\lambda})^{\alpha_{3}}max[1,F_{*}^{3}],$

then

(2.57) $\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int_{|x|\leq
D_{*}\tilde{\lambda}_{v}(\tau_{0})}|v^{\lambda}(0)|^{2}\geq c_{3}.$

###### Proof.

Step 1. Energy constraint and lower bound on $v^{\lambda}(\tau_{0}).$

We claim that there exist universal constants $C_{3},c_{3}>0$ which are
independent of $\lambda$ such that:

(2.58) $\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int_{|x|\leq
A_{*}\tilde{\lambda}_{v}(\tau_{0})}|v^{\lambda}(\tau_{0})|^{2}\geq c_{3}$

with

(2.59)
$A_{*}=C_{3}\max[(M_{0}^{\lambda})^{\alpha_{1}}F_{*},(M_{0}^{\lambda})^{6}].$

Now we prove the claim (2.58). Consider a renormalization of
$v^{\lambda}(\tau_{0})$ :

(2.60)
$w(x)=\tilde{\lambda}_{v}(\tau_{0})v^{\lambda}(\tau_{0},\tilde{\lambda}_{v}(\tau_{0})x),$

then from (2.52), (2.54) and the conservation of the energy:

(2.61) $||\nabla w||_{L^{2}}=1$

and

(2.62) $\displaystyle E^{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}(w)$
$\displaystyle=\frac{1}{2}\int|\nabla
w|^{2}dx-\frac{1}{4}\int(V_{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}*|w|^{2})(x)|w(x)|^{2}dx$
$\displaystyle=\tilde{\lambda}_{v}(\tau_{0})[\frac{1}{2}\int|\nabla
v^{\lambda}|^{2}dx-\frac{1}{4}\int(V_{\lambda}*|v^{\lambda}|^{2})(x)|v^{\lambda}(x)|^{2}dx]$
$\displaystyle=\tilde{\lambda}_{v}(\tau_{0})E^{\lambda}(v^{\lambda}(\tau_{0}))=\tilde{\lambda}_{v}(\tau_{0})E^{\lambda}(v{}_{0}^{\lambda})\leq\frac{1}{4},$

and thus

$\iint_{R^{3}\times
R^{3}}V_{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}(x-y)|w(x)|^{2}|w(y)|^{2}dxdy=4(\frac{1}{2}||\nabla
w||_{L_{x}^{2}}^{2}-E^{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}(w))\geq 1.$

Pick now $\varepsilon>0$ small enough and let

(2.63)
$A_{\varepsilon}=C_{\varepsilon}\max[(M_{0}^{\lambda})^{\alpha_{1}}F_{*},(M_{0}^{\lambda})^{6}]$

for $C_{\varepsilon}$ large enough to be chosen. First we observe that

$\displaystyle\rho(w,A_{\varepsilon})$
$\displaystyle\leq\rho(w,(M_{0}^{\lambda})^{\alpha_{1}}F_{*})=\rho(v^{\lambda}(\tau_{0}),(M_{0}^{\lambda})^{\alpha_{1}}\tilde{\lambda}_{v}(\tau_{0})F_{*})$
$\displaystyle=\rho(v^{\lambda}(\tau_{0}),(M_{0}^{\lambda})^{\alpha_{1}}\sqrt{\tau_{0}})\leq
C_{1}(M_{0}^{\lambda})^{2}.$

Thus, from (1.9), (2.8)

(2.64) $\displaystyle\iint_{|x|\geq A_{\varepsilon},|y|\geq
A_{\varepsilon}}V_{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}(x-y)|w(x)|^{2}|w(y)|^{2}dxdy\leq||w(x)\cdotp\chi_{\\{|x|\geq
A_{\varepsilon}\\}}||_{L^{4}}^{2}||w(y)\cdotp\chi_{\\{|y|\geq
A_{\varepsilon}\\}}||_{L^{4}}^{2}$
$\displaystyle\leq||w(x)\cdotp\chi_{\\{|x|\geq
A_{\varepsilon}\\}}||_{L^{4}}^{4}$ $\displaystyle\leq\varepsilon||\nabla
w||_{L_{x}^{2}}^{2}+\frac{C(\varepsilon)}{A_{\varepsilon}}[\rho(w,A_{\varepsilon})+\rho(w,A_{\varepsilon})^{3}]$
$\displaystyle\leq 2\varepsilon$

for $C_{\varepsilon}$ large enough and

(2.65) $\iint_{|x|\leq A_{\varepsilon},|y|\leq A_{\varepsilon}}+\iint_{|x|\geq
A_{\varepsilon},|y|\leq A_{\varepsilon}}+\iint_{|x|\leq
A_{\varepsilon},|y|\geq
A_{\varepsilon}}V_{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}(x-y)|w(x)|^{2}|w(y)|^{2}dxdy\geq\frac{1}{2}.$

By the Pigeon house principle, there must exist at least one term that is
bigger than $\frac{1}{6}$. Without loss of generality, we assume

$\iint_{|x|\leq A_{\varepsilon},|y|\geq
A_{\varepsilon}}V_{\lambda\cdotp\tilde{\lambda}_{v}(\tau_{0})}(x-y)|w(x)|^{2}|w(y)|^{2}dxdy\geq\frac{1}{6}.$

Then by the Hölder inequality, Young inequality and Sobolev embedding theory,

$\frac{1}{6}\leq C||\chi_{\\{|x|\leq
A_{\varepsilon}\\}}w||_{L_{x}^{2}}||\nabla w||_{L_{y}^{2}}^{3},$

which implies that

$\int_{|x|\leq A_{\varepsilon}}|w(x)|^{2}dx\geq c_{3}>0$

for some constant $c_{3}>0$ which is independent of $\lambda$. By (2.60), this
is

$\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int_{|x|\leq
A_{*}\tilde{\lambda}_{v}(\tau_{0})}|v^{\lambda}(\tau_{0})|^{2}\geq c_{3}$

which finishes the claim (2.58).

Step 2. Backwards integration of the $L^{2}$ fluxes.

We claim: $\forall\varepsilon>0$, there exists $\tilde{C}_{\varepsilon}>0$
such that $\forall D\geq D_{\varepsilon}$ with

(2.66)
$D_{\varepsilon}=\tilde{C}_{\varepsilon}\max[F_{*},F_{*}^{3}]\cdotp\max[(M_{0}^{\lambda})^{\alpha_{1}},(M_{0}^{\lambda})^{2+\alpha_{2}}],$

let

(2.67) $\tilde{R}=\tilde{R}(D,\tau_{0})=D\tilde{\lambda}_{v}(\tau_{0}),$

and $\chi_{\tilde{R}}(r)=\chi(\frac{r}{R})$ for some smooth radially symmetric
cut-off function $\chi(r)=1$ for $r\leq 1,\chi(r)=0$ for $r\geq 2$, then :

(2.68)
$|\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int\chi_{\tilde{R}}|v^{\lambda}(\tau_{0})|^{2}-\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int\chi_{\tilde{R}}|v^{\lambda}(0)|^{2}|<\varepsilon.$

(2.58) and (2.68) now imply (2.57).

Proof of (2.68). Pick $\varepsilon>0$. We compute the $L^{2}$ fluxes from
(2.15) with $\chi_{\tilde{R}}$ :

$\displaystyle|\frac{d}{d\tau}\int\chi_{\tilde{R}}|v^{\lambda}|^{2}|$
$\displaystyle=2|Im(\int\nabla\chi_{\tilde{R}}\cdotp\nabla
v^{\lambda}\bar{v^{\lambda}})|$ $\displaystyle\leq\frac{C}{\tilde{R}}||\nabla
v^{\lambda}(\tau)||_{L^{2}}(\int_{\tilde{R}\leq|x|\leq
2\tilde{R}}|v^{\lambda}(\tau)|^{2})^{\frac{1}{2}}$
$\displaystyle\leq\frac{C}{\tilde{R}^{\frac{1}{2}}}||\nabla
v^{\lambda}(\tau)||_{L^{2}}(\rho(v^{\lambda}(\tau),\tilde{R}))^{\frac{1}{2}}.$

Now observe from (2.55), (2.66) and (2.67) that :

$\forall\tau\in[0,\tau_{0}],\;\tilde{R}=D\tilde{\lambda}_{v}(\tau_{0})\geq
D_{\varepsilon}\frac{\tilde{\lambda}_{v}(\tau_{0})}{\sqrt{\tau_{0}}}\sqrt{\tau_{0}}\geq\frac{D_{\varepsilon}}{F_{*}}\sqrt{\tau}\geq(M_{0}^{\lambda})^{\alpha_{1}}\sqrt{\tau},$

and thus (2.12) and the monotonicity of $\rho$ ensure :

(2.69)
$\forall\tau\in[0,\tau_{0}],\;\rho(v^{\lambda}(\tau),\tilde{R})\leq\rho(v^{\lambda}(\tau),(M_{0}^{\lambda})^{\alpha_{1}}\sqrt{\tau})<C_{1}(M_{0}^{\lambda})^{2}.$

Now we derive that

$\forall\tau\in[0,\tau_{0}],\;|\frac{d}{d\tau}\int\chi_{\tilde{R}}|v^{\lambda}(\tau)|^{2}|\leq\frac{CM_{0}^{\lambda}}{\tilde{R}^{\frac{1}{2}}}||\nabla
v^{\lambda}(\tau)||_{L^{2}}.$

We integrate this between 0 and $\tau_{0}$, divide by $\tilde{R}$ and use
(2.13) to get :

$\displaystyle|\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int\chi_{\tilde{R}}|v^{\lambda}(\tau_{0})|^{2}-\frac{1}{\tilde{\lambda}_{v}(\tau_{0})}\int\chi_{\tilde{R}}|v^{\lambda}(0)|^{2}|$
$\displaystyle\leq\frac{CM_{0}^{\lambda}\cdotp
D}{\tilde{R}^{\frac{3}{2}}}\int_{0}^{\tau_{0}}||\nabla
v^{\lambda}(\tau)||_{L^{2}}d\tau\leq\frac{CM_{0}^{\lambda}\cdotp
D}{D^{\frac{3}{2}}\cdotp\tilde{\lambda}_{v}(\tau_{0})^{\frac{3}{2}}}(\tau_{0}\int_{0}^{\tau_{0}}||\nabla
v^{\lambda}(\tau)||_{L^{2}}^{2}d\tau)^{\frac{1}{2}}$
$\displaystyle\leq\frac{CM_{0}^{\lambda}}{D^{\frac{1}{2}}\cdotp\tilde{\lambda}_{v}(\tau_{0})^{\frac{3}{2}}}(\int_{0}^{2\tau_{0}}(2\tau_{0}-\tau)||\nabla
v(\tau)||_{L^{2}}^{2}d\tau)^{\frac{1}{2}}\leq\frac{CM_{0}^{\lambda}}{D^{\frac{1}{2}}\cdotp\tilde{\lambda}_{v}(\tau_{0})^{\frac{3}{2}}}(M_{0}^{\lambda})^{\frac{\alpha_{2}}{2}}\cdotp\tau_{0}{}^{\frac{3}{4}}$
$\displaystyle\leq\frac{C(M_{0}^{\lambda})^{1+\frac{\alpha_{2}}{2}}F_{*}^{\frac{3}{2}}}{D_{\varepsilon}^{\frac{1}{2}}}\leq\varepsilon$

for $\tilde{C}_{\varepsilon}$ large enough which ends the proof. ∎

In the end of this section, we state a proposition as a blackbox to finish the
proof of Theorem 1.5,

######  Proposition 2.5.

Let $v_{\varepsilon}(\tau,x)\in
C([0,e^{N}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$101010$N$ is a large enough
fixed number. be a radial solution to (2.1) with the potential function

$V_{\varepsilon}(x):=\frac{1}{\varepsilon^{3}}V(\frac{x}{\varepsilon}),\;\forall\varepsilon>0,$

where $V(x)$ is a fixed potential function satisfying the conditions in
Theorem 1.5 and initial data satisfies

(2.70) $||v_{\varepsilon}(0,x)||_{\dot{H}^{1}}=1,$

Besides, we assume the following
conditions111111$E^{V_{\varepsilon}}(v_{0}):=\frac{1}{2}\int|\nabla
v_{\varepsilon}(0)|^{2}-\frac{1}{4}\int(V_{\varepsilon}*|v_{\varepsilon}(0)|^{2})|v_{\varepsilon}(0)|^{2}.$

(2.71) $e^{\frac{N}{2}}\cdotp\max(E^{V}(v_{\varepsilon}(0)),0)<1,$

(2.72) $M_{0}:=\frac{4||v_{\varepsilon}(0)||_{L^{3}}}{c_{V}}\geq 2,$

and

(2.73)
$\forall\tau_{0}\in[0,e^{N}],\;\frac{E^{V_{\varepsilon}}(v_{\varepsilon}(0))}{||v_{\varepsilon}(\tau_{0})||_{\dot{H}^{1}}^{2}}\leq\frac{1}{4},$

then there exists a universal constant $\gamma_{1}>0$ which is independent of
$\varepsilon$ such that

$||v_{\varepsilon}(0)||_{L^{3}}\geq N^{\gamma_{1}}.$

Using this proposition to equation (2.5) with $\tau=0$, it is direct to see

$||u(t,x)||_{L^{3}}=||v^{\lambda}(0,x)||_{L^{3}}\geq|\log(T-t)|^{\gamma},$

and we finish the proof of Theorem 1.5.

##  3\. The connection between Schr0dinger equation and Hartree equation

In section 2, we have proved Theorem 1.5. In this section, we aim to show the
our result also implies Theorem 1.1 as said in section 1.2. Before our
analysis, we state a stronger version proved by Merle and Raphaël in [33],
which is also claimed in Remark 1.17,

######  Proposition 3.1.

Let $v(\tau,x)\in C([0,e^{N}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ be a
radial solution to equation (1.5) with the initial data satisfying

(3.1) $||v(0,x)||_{\dot{H}^{1}}=1,$

and satisfies the following conditions121212$E(v_{0}):=\frac{1}{2}\int|\nabla
v_{0}|^{2}-\frac{1}{4}\int|v_{0}|^{4}.$

(3.2) $e^{\frac{N}{2}}\cdotp\max(E(v_{0}),0)<1,$ (3.3)
$M_{0}:=\frac{4||v_{0}||_{L^{3}}}{C_{GN}}\geq 2,$

and

(3.4)
$\forall\tau_{0}\in[0,e^{N}],\;\frac{E(v_{0})}{||v(\tau_{0})||_{\dot{H}^{1}}^{2}}\leq\frac{1}{4},$

then there exists a universal constant $\gamma>0$ such that

(3.5) $||v_{0}||_{L^{3}}\geq N^{\gamma}.$

Now we want to derive proposition 3.1 by applying our conclusion, proposition
2.5. First we state a result of standard stability theory, and supply the
proof in Appendix B,

######  Lemma 3.2.

We consider the nonlinear Schrödinger equation (1.5) and Hartree equation
(2.1) with the same radial initial data
$v_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1}$, and assume the normalization
condition

(3.6) $||V(x)||_{L^{1}}=||V_{\varepsilon}(x)||_{L^{1}}=1.$

We define $T_{max}$ as the lifetime of $v(\tau)$ and claim the following holds
true, $\forall\delta>0,\;\forall
T\in[0,T_{max}),\;\exists\varepsilon^{*}=\varepsilon^{*}(\delta,T)>0,$ such
that $\forall 0<\varepsilon<\varepsilon^{*}$ ,

(3.7)
$||v_{\varepsilon}(\tau)-v(\tau)||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}\leq\delta.$

Combining Lemma 3.2 and Proposition 2.5, we will prove Proposition 3.1. The
main task is to verify the above conditions in Proposition 3.1 hold true item
by item.

###### Proof.

From the standard analysis knowledge and the definition of
$E(v_{0}),E^{V_{\varepsilon}}(v_{0})$, $\forall m_{1}>0,$ $\exists m_{2}>0$,
such that $\forall 0<\varepsilon<m_{2}$,

(3.8) $|E(v_{0})-E^{V_{\varepsilon}}(v_{0})|<m_{1}.$

Since $\tau_{*}$ is a fixed number, condition (2.71) is naturally established.
The assumption (2.73) is verified by condition (3.4). Lastly, combining Lemma
3.2 and (3.8), we conclude (2.73) holds. The conclusion of the Proposition 2.5
then leads directly to the conclusion (3.5) as we want. ∎

Acknowledgement:

The author is very grateful to his advisers, Ping Zhang and Chenjie Fan for
their constant encouragements, guidance and very helpful discussions during
preparing this work. The author thanks Ning Liu and Hequn Zhang, for some
helpful comments and discussions. The author is supported by National Key R&D
Program of China under grant 2021YFA1000800.

##  Appendix A A virial argument and The applications of local theory

In this section, we give complete analysis claimed in Remark 1.8 and some
applications of local theory. First, through a classical virial argument, we
prove the existence of the blow up solutions, and state it as follows,

######  Proposition A.1.

Let $v(t,x)\in C([0,T),\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ be a solution to
equation (1.1) where $V(x)$ satisfies the condition (1.8), if the initial data
$v_{0}$ satisfies

(A.1) $E^{V}(v_{0})=\frac{1}{2}\int|\nabla
v_{0}|^{2}-\frac{1}{4}\int(V*|v_{0}|^{2})|v_{0}|^{2}<0,$

then $v(t,x)$ blows up in finite time, i.e. $T<+\infty.$

###### Proof.

Since $V(x)$ is a radial function, then

$2\int x_{j}|v(t,x)|^{2}\partial_{x_{j}}\int
V(x-y)|v(t,y)|^{2}dydx=\int\int(x_{j}-y_{j})\partial_{x_{j}}V(x-y)|v(t,x)|^{2}|v(t,y)|^{2}dydx.$

So we can rewrite the virial identity (1.13) as the follows

$\displaystyle\frac{d^{2}}{dt^{2}}\int|x|^{2}|v|^{2}$
$\displaystyle=8\int|\nabla v|^{2}+4\int
x_{j}|v|^{2}\partial_{j}(V*|v|^{2})dx$ $\displaystyle=8\int|\nabla
v|^{2}+2\int\int(x_{j}-y_{j})\partial_{x_{j}}V(x-y)|v(t,x)|^{2}|v(t,y)|^{2}dydx$
$\displaystyle=16(\frac{1}{2}\int|\nabla
v|^{2}+\frac{1}{8}\int\int(x_{j}-y_{j})\partial_{x_{j}}V(x-y)|v(t,x)|^{2}|v(t,y)|^{2}dydx)$
$\displaystyle:=16K^{V}(v(t)).$

Using the condition (1.8), we conclude

$K^{V}(v(t))\leq E^{V}(v(t))=E^{V}(v_{0})<0,$

and we finish the proof. ∎

######  Proposition A.2.

Assume $u(t)$ is a solution in $C([0,T),\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$
to equation (1.1) with

$\begin{cases}u_{0}\in\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1},\\\
||u_{0}||_{\dot{H}^{1}}\leq M,\end{cases}$

and $V(x)\in L^{1}(\mathbb{R}^{3})\cap L^{\frac{3}{2}}(\mathbb{R}^{3}),$ then,
if $u(t)$ blows up in finite time $T<\infty$ , there holds

(A.2) $\lim_{t\nearrow T}||u(t)||_{\dot{H}^{\frac{1}{2}}}=+\infty.$

###### Proof.

step 1. The energy is well-defined.

From the energy conservation law,

$E^{V}(u(t))=E^{V}(u_{0})=\frac{1}{2}\int|\nabla
u(t,x)|^{2}dx-\frac{1}{4}\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx.$

Also, with the aid of Sobolev embedding theorem in $\mathbb{R}^{3}$,
$\dot{H}^{\frac{1}{2}}\hookrightarrow L^{3}$ and $\dot{H}^{1}\hookrightarrow
L^{6}$,

(A.3)
$\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx\lesssim||(V*|u|^{2})||_{L_{x}^{3}}||u||_{L_{x}^{3}}^{2}\lesssim||V||_{L_{x}^{\frac{3}{2}}}||u||_{L_{x}^{3}}^{4}\lesssim||u||_{\dot{H}^{\frac{1}{2}}}^{4}.$

So $E^{V}(u(t))$ is well defined.

step 2 . Proof by contradiction.

From step 1,

(A.4) $E^{V}(u(t))\geq\frac{1}{2}\int|\nabla
u(t,x)|^{2}dx-C||u||_{\dot{H}^{\frac{1}{2}}}^{4}.$

If $u(t)$ blows up at $T$, then

$\lim_{t\nearrow T}||u(t)||_{\dot{H}^{1}}=+\infty.$

(Otherwise we can continues the solution to $[t,t+C(M)]$ with $t+C(M)>T$,
which is a contradiction.)

The fact $E^{V}(u(t))=E^{V}(u_{0})<+\infty$ shows that

$\lim_{t\nearrow T}||u(t)||_{\dot{H}^{\frac{1}{2}}}=+\infty.$

∎

######  Proposition A.3.

Assume the assumptions in Proposition A.2 hold, and $u_{0}\in
H^{1}(\mathbb{R}^{3})$ , $V(x)\in L^{1}(\mathbb{R}^{3})\cap
L^{\infty}(\mathbb{R}^{3})$. Then $u(t)$ can not blow up in finite time
$T<\infty$.

###### Proof.

Recall

(A.5) $E^{V}(u(t))=E^{V}(u_{0})=\frac{1}{2}\int|\nabla
u(t,x)|^{2}dx-\frac{1}{4}\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx<+\infty$

and

$\displaystyle\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx$
$\displaystyle\lesssim||(V*|u|^{2})||_{L_{x}^{\infty}}|||u|^{2}||_{L_{x}^{1}}$
$\displaystyle\lesssim||V||_{L_{x}^{\infty}}||u||_{L_{x}^{2}}^{4}$
$\displaystyle\lesssim||u_{0}||_{L_{x}^{2}}^{4}$ $\displaystyle\lesssim
C(u_{0}),$

where we also use mass conservation.

The above gives

(A.6) $\int|\nabla
u(t,x)|^{2}dx\leq\frac{1}{2}\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx+2E^{V}(u_{0})\leq
C(u_{0}),$

which is a contradiction to $\lim_{t\nearrow
T}||u(t)||_{\dot{H}^{1}}=+\infty.$ ∎

In fact, we also have a stronger characterization about the blow-up rate of
critical norm, and state it as follows:

######  Proposition A.4.

Under the assumptions in proposition A.2 and assume $u(t)$ blows up in finite
time $T<+\infty$, then $\exists C=C(u_{0})\;s.t.$

(A.7) $||u(t)||_{\dot{H}^{1}}\geq\frac{C}{(T-t)^{\frac{1}{4}}},$

and

(A.8) $||u(t)||_{\dot{H}^{\frac{1}{2}}}\geq\frac{C}{(T-t)^{\frac{1}{8}}}$

for $t$ close enough to $T$.

###### Proof.

If not, $\exists\\{t_{n}\\}$ with $\lim_{n\nearrow+\infty}t_{n}=T$ such that
$||u(t_{n})||_{\dot{H}^{1}}\leq\frac{1}{n(T-t)^{\frac{1}{4}}}:=M_{n}.$

At the time $t_{n}$,

(A.9) $\begin{cases}i\partial_{t}u+\triangle
u=-(V*|u|^{2})u.&(t,x)\in\mathbb{R}\times\mathbb{R}^{3},\\\
u\mid_{t=t_{n}}=u(t_{n},x)\in\dot{H}^{1}\cap\dot{H}^{\frac{1}{2}}.\end{cases}$

From local theory, $u(t)\in
C([t_{n},t_{n}+T_{n}],\dot{H}^{\frac{1}{2}}\cap\dot{H}^{1})$ with
$T_{n}\geq\frac{C}{M_{n}^{4}}\gtrsim n^{4}(T-t_{n}).$

However, $t_{n}+n^{4}(T-t_{n})>T$ for $n$ large enough, which is a
contradiction.

From the energy conservation law,

$\displaystyle E^{V}(u(t))$ $\displaystyle=E^{V}(u_{0})=\frac{1}{2}\int|\nabla
u(t,x)|^{2}dx-\frac{1}{4}\int(V*|u|^{2})(t,x)|u(t,x)|^{2}dx$
$\displaystyle\geq\frac{1}{2}||\nabla
u(t)||_{L_{x}^{2}}^{2}-C||u(t)||_{\dot{H}^{\frac{1}{2}}}^{4}.$

From the (A.9) , $||u(t)||_{\dot{H}^{1}}\geq\frac{C}{(T-t)^{\frac{1}{4}}}$ ,
so $||u(t)||_{\dot{H}^{\frac{1}{2}}}\geq\frac{C}{(T-t)^{\frac{1}{8}}}$. ∎

######  Remark A.5.

In proposition A.4, we have used the fact $V(x)\in
L^{\frac{3}{2}}(\mathbb{R}^{3}).$

######  Remark A.6.

Although we can give some characterizations about the blow-up rate, we can not
give a concrete example of the blow-up solution. This is a weakness for
proposition A.2-A.4.

##  Appendix B The proof of Lemma 3.2

###### Proof.

We define $u_{\varepsilon}:=v_{\varepsilon}-v$ , i.e.
$v_{\varepsilon}(\tau,x)=u_{\varepsilon}(\tau,x)+v(\tau,x),$ then
$u_{\varepsilon}(\tau,x)$ satisfies

(B.1) $\begin{cases}i\partial_{t}u_{\varepsilon}(\tau,x)+\triangle
u_{\varepsilon}(\tau,x)=-(V_{\varepsilon}*|v_{\varepsilon}(\tau,x)|^{2})v_{\varepsilon}(\tau,x)+|v|^{2}v.&(\tau,x)\in\mathbb{R}\times\mathbb{R}^{3},\\\
u_{\varepsilon}(\tau,x)\mid_{\tau=0}=0.\end{cases}$

From the integral equation,

(B.2)
$u_{\varepsilon}(\tau,x)=i\int_{0}^{\tau}e^{i(\tau-s)\triangle}[(V_{\varepsilon}*|v_{\varepsilon}|^{2})v_{\varepsilon}-|v|^{2}v](s)ds,$

which implies

$\displaystyle||u_{\varepsilon}(\tau,x)||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}$
$\displaystyle=||\int_{0}^{\tau}e^{i(\tau-s)\triangle}[(V_{\varepsilon}*|v_{\varepsilon}|^{2})v_{\varepsilon}-|v|^{2}v](s)ds||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}$
$\displaystyle\leq||\int_{0}^{\tau}e^{i(\tau-s)\triangle}\nabla[(V_{\varepsilon}*|v_{\varepsilon}|^{2})v_{\varepsilon}-(V_{\varepsilon}*|v_{\varepsilon}|^{2})v](s)ds||_{L_{\tau}^{\infty}([0,T],L^{2})}$
$\displaystyle+||\int_{0}^{\tau}e^{i(\tau-s)\triangle}\nabla([(V_{\varepsilon}*|v_{\varepsilon}|^{2})-|v|^{2}]v(s))ds||_{L_{\tau}^{\infty}([0,T],L^{2})}$
$\displaystyle:=(I)+(II).$

Next we will estimate these two terms.

$\displaystyle(I)$
$\displaystyle\lesssim||\nabla[(V_{\varepsilon}*|v_{\varepsilon}|^{2})v_{\varepsilon}-(V_{\varepsilon}*|v_{\varepsilon}|^{2})v]||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}$
$\displaystyle\leq||V_{\varepsilon}*(\nabla|v_{\varepsilon}|^{2})(v_{\varepsilon}(s)-v(s))||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}+||V_{\varepsilon}*(|v_{\varepsilon}|^{2})\nabla(v_{\varepsilon}(s)-v(s))||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}$
$\displaystyle\leq||V_{\varepsilon}*(\nabla|v_{\varepsilon}|^{2})||_{L_{\tau}^{\infty}([0,T],L_{x}^{\frac{3}{2}})}||(v_{\varepsilon}(s)-v(s))||_{L_{\tau}^{\infty}([0,T],L_{x}^{6})}\cdotp
T^{\frac{1}{2}}$
$\displaystyle+||V_{\varepsilon}*(|v_{\varepsilon}|^{2})||_{L_{\tau}^{\infty}([0,T],L_{x}^{3})}||\nabla(v_{\varepsilon}(s)-v(s))||_{L_{\tau}^{\infty}([0,T],L_{x}^{2})}\cdotp
T^{\frac{1}{2}}$ $\displaystyle\leq||\nabla
v_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],L_{x}^{2})}^{2}||u_{\varepsilon}(\tau,x)||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}T^{\frac{1}{2}}$
$\displaystyle\leq(||u_{\varepsilon}(\tau,x)||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}+||v(\tau,x)||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2})||u_{\varepsilon}(\tau,x)||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}\cdotp
T^{\frac{1}{2}}.$

Before we estimate part $(II)$, let us recall some elementary knowledge,

(B.3) $\forall f\in
L^{p}(\mathbb{R}^{3}),\;p\in(1,+\infty),\;\lim_{\varepsilon\searrow
0^{+}}||V_{\varepsilon}*f-f||_{L^{p}(\mathbb{R}^{3})}=0.$

From the local theory, $v(\tau,x)\in C([0,T],\dot{H}^{1})$, so
$\forall\tilde{\delta}>0,\;\exists\delta_{1}>0,$ s.t.
$\forall\tau_{1},\tau_{2}\in[0,T]$ with $|\tau_{2}-\tau_{1}|<\delta_{1}$ ,
$||v(\tau_{1},x)-v(\tau_{2},x)||_{\dot{H}^{1}}<\tilde{\delta}$ . Similarly, we
have $||v(\tau_{1},x)-v(\tau_{2},x)||_{L^{6}}<\tilde{\delta}$ and
$||v(\tau_{1},x)\cdotp\nabla v(\tau_{1},x)-v(\tau_{2},x)\cdotp\nabla
v(\tau_{2},x)||_{L^{\frac{3}{2}}}<\tilde{\delta}$.

Now we estimate the part $(II)$:

$\displaystyle(II)$
$\displaystyle=||\int_{0}^{\tau}e^{i(\tau-s)\triangle}\nabla([(V_{\varepsilon}*|v_{\varepsilon}|^{2})-|v|^{2}]v(s))ds||_{L_{\tau}^{\infty}([0,T],L^{2})}$
$\displaystyle\leq||\nabla([(V_{\varepsilon}*|v|^{2})-|v|^{2}]v(s))||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}$
$\displaystyle+||\nabla[(V_{\varepsilon}*(\bar{v}u_{\varepsilon}+v\bar{u}_{\varepsilon}))v(s)]||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}$
$\displaystyle+||\nabla[(V_{\varepsilon}*|u_{\varepsilon}|^{2})v(s)]||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}$
$\displaystyle:=(i)+(ii)+(iii).$

For the term $(ii)$,

(B.4)
$(ii)\leq||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}||v||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}T^{\frac{1}{2}},$

and the computation is also direct for $(iii)$,

(B.5)
$(iii)\leq||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}||v||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}T^{\frac{1}{2}}.$

Finally, we estimate term $(i)$:

$\displaystyle(i)$
$\displaystyle\leq||[(V_{\varepsilon}*|v|^{2})-|v|^{2}]\nabla
v(s)||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}+||[(V_{\varepsilon}*\nabla|v|^{2})-\nabla|v|^{2}]v(s))||_{L_{\tau}^{2}([0,T],L_{x}^{\frac{6}{5}})}$
$\displaystyle\leq||[(V_{\varepsilon}*|v|^{2})-|v|^{2}]||_{L_{\tau}^{\infty}([0,T],L_{x}^{3})}||\nabla
v(s)||_{L_{\tau}^{\infty}([0,T],L_{x}^{2})}T^{\frac{1}{2}}$
$\displaystyle+||[(V_{\varepsilon}*\nabla|v|^{2})-\nabla|v|^{2}]||_{L_{\tau}^{\infty}([0,T],L_{x}^{\frac{3}{2}})}||v(s)||_{L_{\tau}^{\infty}([0,T],L_{x}^{6})}T^{\frac{1}{2}}$
$\displaystyle\leq\delta_{2}||\nabla
v(s)||_{L_{\tau}^{\infty}([0,T],L_{x}^{2})}T^{\frac{1}{2}},$

where we used the prepared knowledge in the last inequality and $\delta_{2}$
is a small enough constant to be defined later. Summing up the above three
estimations, we conclude,

$\forall$ fixed $T\in[0,T_{max}),$ $\forall\delta_{2}>0$,
$\exists\varepsilon_{1}>0$, such that
$\forall\varepsilon\in(0,\varepsilon_{1}),$

(B.6)
$\displaystyle||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}$
$\displaystyle\leq
C\\{||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{3}T^{\frac{1}{2}}+||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}||v||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}T^{\frac{1}{2}}$
$\displaystyle+||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}||v||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}T^{\frac{1}{2}}+\delta_{2}||v||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}T^{\frac{1}{2}}\\}.$

Dual to the fact $v(t)\in C([0,T],\dot{H}^{1})$, there exists $M=M(T)$ such
that $||\nabla v(\tau)||_{L_{\tau}^{\infty}([0,T],L_{x}^{2})}<M$. We can also
choose $\delta_{2}$ small enough such that $\delta_{3}$ small and
$\delta_{3}:=\delta_{2}M^{2}.$ Then (B.6) can be rewrote as

(B.7) $||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}\leq
C\\{||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{3}T^{\frac{1}{2}}+||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}^{2}MT^{\frac{1}{2}}+||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}M^{2}T^{\frac{1}{2}}+\delta_{3}T^{\frac{1}{2}}\\}.$

We can divide $T$ into $N(T)$ intervals and the length of each interval is
$T_{1}$ such that $CT_{1}^{\frac{1}{2}}\ll
1,\;CMT_{1}^{\frac{1}{2}}\ll\frac{1}{100}\;and\;CM^{2}T_{1}^{\frac{1}{2}}\ll\frac{1}{100}.$

From each interval, (B.7) leads to

(B.8)
$||u_{\varepsilon}||_{L_{\tau}^{\infty}([\tau_{i},\tau_{i+1}],\dot{H}^{1})}\leq||u_{\varepsilon}(\tau_{i})||_{\dot{H}^{1}}+\frac{1}{100}||u_{\varepsilon}||_{L_{\tau}^{\infty}([\tau_{i},\tau_{i+1}],\dot{H}^{1})}^{3}+\frac{1}{100}||u_{\varepsilon}||_{L_{\tau}^{\infty}([\tau_{i},\tau_{i+1}],\dot{H}^{1})}^{2}+\delta_{3}.$

On the one hand, we choose $\delta_{3}$ such that $2^{N(T)}\delta_{3}\ll 1$.
On the other hand, with the aid of standard continuity argument, we adjust
$\delta_{3}$ such that

(B.9) $||u_{\varepsilon}||_{L_{\tau}^{\infty}([0,T],\dot{H}^{1})}\leq
2^{N(T)}\delta_{3}\leq\delta,$

which ends the proof. ∎

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