# On the blowup of solutions for a nonlocal multi-dimensional transport
equation

Wanwan Zhang1 1 School of Mathematics and Statistics, Jiangxi Normal
University, Nanchang, 330022, Jiangxi, P. R. China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we revisit the problem of finite-time blowup for a multi-
dimensional nonlocal transport equation studied in [Dong, Adv. Math. 264
(2014) 747-761]. Inspired by a one-dimensional analogous model considered in
[Li-Rodrigo, Adv. Math. 374 (2020) 1-26], we establish a new weighted
nonlinear inequality implying the blow-up by a completely real variable based
technique.

###### Key words and phrases:

Transport equation; Nonlocal velocity; Blow-up

###### 2000 Mathematics Subject Classification:

35Q35; 35B44; 76D05

∗Corresponding author

##  1\. Introduction and main results

This paper is concerned with a multi-dimensional transport equation with a
nonlocal velocity

$\left\\{\begin{aligned}
&\partial_{t}\theta+u\cdot\nabla\theta=0,~{}(x,t)\in\mathbb{R}^{n}\times\mathbb{R}_{+},\\\
&u=\nabla\Lambda^{-2+2\alpha}\theta,\\\
&\theta(x,0)=\theta_{0}(x),~{}x\in\mathbb{R}^{n},\end{aligned}\ \right.$ (1.1)

where $n\geq 2$ and $0<\alpha<1$. Here the unknown $\theta$ defined in
$\mathbb{R}^{n}\times\mathbb{R}_{+}$ is a scalar function, and the fractional
Laplacian $\Lambda^{s}=(-\Delta)^{\frac{s}{2}}$ with $s\in\mathbb{R}$ is
defined through the Fourier transform [34]:

$\displaystyle\widehat{\Lambda^{s}f}(\xi)=(2\pi|\xi|)^{s}\widehat{f}(\xi).$

The second equation in (1.1) makes this model nonlocal. The fractionally
dissipative version of (1.1) reads as

$\left\\{\begin{aligned}
&\partial_{t}\theta+u\cdot\nabla\theta+\Lambda^{\gamma}\theta=0,~{}(x,t)\in\mathbb{R}^{n}\times\mathbb{R}^{+}\\\
&u=\nabla\Lambda^{-2+2\alpha}\theta\\\
&\theta(x,0)=\theta_{0}(x),~{}x\in\mathbb{R}^{n}\end{aligned}\ \right.$ (1.2)

where $n\geq 1$, $0<\alpha<1$ and $0<\gamma<2$.

The nonlocal active scalar equations (1.1) and (1.2) were extensively studied.
When $n=1$ and $\alpha=\frac{1}{2}$, (1.2) reduces to the well-known Córdoba-
Córdoba-Fontelos model

$\displaystyle\partial_{t}\theta-H(\theta)\theta_{x}+\Lambda^{\gamma}\theta=0,$
(1.3)

where $H\theta$ is the Hilbert transform of $\theta$. It was first proposed by
Córdoba, Córdoba and Fontelos in [11] as a one-dimensional analogue of the
two-dimensional surface quasigeostrophic equation (SQG) [7]. There have been a
number of mathematical studies on the well-posedness for SQG. We will not
review here in detail the known results for the SQG and related equation. One
can refer to [4, 6, 8, 9, 10, 13, 14, 21, 22, 25, 26, 27, 36] and the
references therein for more details and the recent progresses. Here we briefly
summarize the progress related to the CCF model and related equations.
Concerning the inviscid CCF model, i.e., (1.1) with $n=1$ and
$\alpha=\frac{1}{2}$, Córdoba et al.[11] first obtained an ingenious nonlinear
weighed inequality for the Hilbert transform by the use of Meillin transform
and complex analysis, and proved smooth solutions must blow up in finite time
for a generic family of even initial data (see also [12] for another proof of
the blowup for the initial data not necessarily even). This blow-up phenomena
was later proved by Silvestre and Vicol in [33] via four essentially different
methods. Based on completely real-variable arguments, Li and Rodrigo [31]
recently provided a short proof of the nonlinear inequality first proved by
Córdoba et al. in [11], and obtained several new weighted inequalities for the
Hilbert transform and various nonlinear versions, which can be applied to show
the finite-time blow-up of smooth solutions to (1.1) with $n=1$ and
$\alpha\in(0,1)$. For the dissipative case, the authors in [11] also obtained
the global well-posedness of (1.3) for the positive $H^{2}$ initial data in
the subcritical case $1<\gamma<2$. Later in [15], by adapting the method of
continuity first used in [25], Dong established the global regularity of
solutions to (1.3) in the critical case $\gamma=1$ for arbitrary initial data
in appropriate critical Sobolev space. In [29], Li and Rodrigo proved the
finite-time blowup of smooth solutions to (1.3) in the supercritical case
$0<\gamma<\frac{1}{2}$ (see also [24, 31, 33] for different proofs for this
blow-up result). Recently, for each smooth nonnegative initial data, Ferreira
and Moitinho [18] obtained the existence of global classical solutions to
(1.3) for $\gamma\in(\gamma_{1},1)$ with $\gamma_{1}$ depending on the
$H^{\frac{3}{2}}$-norm of the initial data. For the remaining case
$\frac{1}{2}\leq\gamma<1$, whether smooth solutions to (1.3) may blow up in
finite time is currently still open. Some previously mentioned well-poseness
results on the dissipative CCF model (1.3) have been extended to the
interpolation $\alpha$-CCF model, i.e., (1.2) with $n=1$ and
$0<\alpha<\frac{1}{2}$, by Ferreira and Moitinho in a recent work [19].

A natural multi-dimensional generalization of the inviscid CCF equation, i.e.,
(1.1) with $n\geq 2$ and $\alpha=\frac{1}{2}$, was considered in [3], where
Balodis and Córdoba presented a weighted nonlinear inequality for the Riesz
transform by using Meillin transform and spherical harmonic expansion, and
obtained the blow-up of smooth solution for any nonnegative, not-identically
zero initial data. When $n=2$, such result was also proved for a similar
equation in [17] independently. For the two-dimensional case with fractional
dissipation, Li and Rodrigo [30] proved the finite-time blow-up of radial
smooth solutions to (1.2) for $\frac{1}{4}\leq\alpha<1$ and $0<\gamma<\alpha$.
Later, in [16], Dong was able to obtain the blow-up of smooth solutions to
(1.1) with full range $\alpha\in(0,1)$ and $n\geq 2$ for any smooth, radially
symmetric and nonnegative initial data with compact support and its positive
maximum attained at the origin. Recently, motivated by [33], Jiu and Zhang
[23] proved the finite time singularity of solutions to (1.1) with
$0<\alpha<1$ and $n\geq 2$ for smooth initial data $\theta_{0}$ with
$\displaystyle\sup_{x\in\mathbb{R}^{n}}\theta_{0}(x)>0$ via the De Giorgi
iteration technique. This iteration strategy in [33] has also been adapted by
Alonso-Orán and Martínez [1] to the proof of finite time blow-up for non-local
active scalar equations (1.1) on compact Riemannian manifolds. Very recently,
Li, Liu and Zhang [28] studied a related model, i.e., (1.1) with an additional
power type of damping term:

$\left\\{\begin{aligned}
&\partial_{t}\theta+u\cdot\nabla\theta+\kappa|\theta|^{\nu-1}\theta=0,~{}(x,t)\in\mathbb{R}^{n}\times\mathbb{R}_{+},\\\
&u=\nabla\Lambda^{-2+2\alpha}\theta,\\\
&\theta(x,0)=\theta_{0}(x),~{}x\in\mathbb{R}^{n},\end{aligned}\ \right.$ (1.4)

where $n\geq 1$, $0<\alpha<1$, $\kappa\in\mathbb{R}$ and $\nu>0$. In [28], by
some change of time variable to implement the iteration technique in [33], the
authors showed that the damping term can not avoid the singularity formation
in finite time and that for particular initial data depending usually on the
size of their $L^{1}$ and $L^{\infty}$-norm, the solutions to (1.4) must blow
up, independent of the value of $\kappa$.

In this paper, we revisit the problem of finite-time blowup for (1.1) with
$n\geq 2$ and $0<\alpha<1$. Our main result on (1.1) can be stated as

######  Theorem 1.1.

Let the initial data $\theta_{0}$ be a radial Schwartz function. There exists
a constant $A(n,\alpha)>0$ depending only on $n$ and $\alpha$ such that if

$\displaystyle\int_{\mathbb{R}^{n}}\frac{\theta_{0}(0)-\theta_{0}(x)}{|x|^{n}}e^{-|x|}dx\geq
A(n,\alpha)\|\theta_{0}\|_{L^{\infty}},$ (1.5)

then the smooth solution $\theta$ to (1.1) blows up in finite time.

######  Remark 1.1.

The blow-up result can be extended to the model with a fractional dissipation.
More precisely, by establishing a weighted inequality involving the
dissipative term along with the new weighted inequality for the nonlinear
term, i.e., Proposition 3.2, we can show the finite-time blowup of smooth
solutions to (1.2) for appropriately small $\gamma>0$. We will leave these
discussions in a forthcoming paper [35].

For the proof of Theorem 1.1, we consider the evolution of a weighted integral
of the solution and show that such quantity satisfies some ordinary
differential inequality and will blows up in finite time (see (4.1) and
(4.2)). We then obtain the blow-up of the solution. This type of strategy can
be, for instance, found in [11, 16, 29, 30, 31]. The key ingredient to adapt
to this approach is to derive a weighted nonlinear inequality (see (3.11)).
Inspired by [31], we prove the weighted nonlinear inequality (3.11) by some
completely real variable based techniques. It will be seen later that (3.11)
is reduced to the estimate of the integral $\mathcal{I}$ (see (3.12)). For the
case of $0<\alpha<\frac{1}{2}$, by discarding all the positive term in the
integral, a lower bound of $\mathcal{I}$ can be readily obtained with help of
the fact that the series $\displaystyle\sum^{\infty}_{k=0}a_{2k}(\alpha)$ is
convergent in this case (see Corollary 2.5). The remaining case
$\alpha\in[\frac{1}{2},1)$ is more involved and needs more refined arguments
(see (3.14)). The difficulty in both cases is that the series
$\displaystyle\sum^{\infty}_{k=0}a_{2k}(\alpha)$ diverges for
$\alpha\in[\frac{1}{2},1)$. By some detailed singular integral estimates and
the fact that the series
$\displaystyle\sum^{\infty}_{k=0}\frac{a_{2k}(\alpha)}{k}$ is convergent for
the full range $\alpha\in(0,1)$, we get the desired weighted inequality. In
comparison with [16], a different approach of the proof of the finite-time
blow-up of smooth solution to (1.1) is given in this paper.

Throughout this paper, we will use $C$ to denote a positive constant, whose
value may change from line to line, and write $C_{n,\alpha}$ or $C(n,\alpha)$
to emphasize the dependence of a constant on $n$ and $\alpha$. For
$p\in[1,\infty]$, we denote $L^{p}(\mathbb{R}^{n})$ the standard $L^{p}$-space
and its norm by $\|\cdot\|_{L^{p}(\mathbb{R}^{n})}$. For $s\geq 0$, we use the
notation $H^{s}(\mathbb{R}^{n})$ to denote the nonhomogeneous Sobolev space of
$s$ order, whose endowed norm is denoted by
$\|\cdot\|_{H^{s}(\mathbb{R}^{n})}=\|\cdot\|_{L^{2}(\mathbb{R}^{n})}+\|\Lambda^{s}(\cdot)\|_{L^{2}(\mathbb{R}^{n})}$
(see [2] for more details). For a sake of the convenience, the
$L^{p}(\mathbb{R}^{n})$-norm of a function $f$ is always abbreviated as
$\left\|f\right\|_{L^{p}}$ and the $\dot{H}^{s}(\mathbb{R}^{n})$-norm as
$\left\|f\right\|_{\dot{H}^{s}}$. The real Gamma function $\Gamma$ is defined
by

$\displaystyle\Gamma(s)=\int^{\infty}_{0}t^{s-1}e^{-t}dt,~{}{\rm for}~{}s>0.$

The Gamma function can be expressed as a limit of some sequence [32], that is,
for $s>0$,

$\displaystyle\Gamma(s)=\lim_{k\rightarrow+\infty}\frac{k^{s}k!}{s(s+1)\cdot\cdot\cdot(s+k)}.$
(1.6)

The related real Beta function $B$ is defined by

$\displaystyle B(p,q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}dt,~{}{\rm
for}~{}p>0,~{}q>0.$

It is well-known that

$\displaystyle B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)},~{}p>0,~{}q>0.$

Finally, let $\mathbb{S}^{n-1}$ be the unit sphere in $\mathbb{R}^{n}$, i.e.,
$\mathbb{S}^{n-1}=\\{x\in\mathbb{R}^{n}:|x|=1\\}$ and $\omega_{n-1}$ be its
surface area. We recall that

$\displaystyle\omega_{n-1}=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}.$

The remaining part of this paper is organized as follows. In Section 2, we
first recall the local well-posedness for the model (1.1), and then present
some useful facts utilized later in this paper. Section 3 is devoted to two
weighted nonlinear inequalities related to the model (1.1). The proof of
Theorems 1.1 is given in Section 4.

##  2\. Preliminaries

In this section, we will present some basic useful facts needed later. We
begin with the local well-posedness of (1.1) in the Sobolev space $H^{s}$ for
some appropriate $s>0$ , whose proof was given in [5].

######  Lemma 2.1.

(1) Let $n\geq 2$ and $\frac{1}{2}<\alpha<1$. Then for each $\theta_{0}\in
H^{s}$ with $s>\frac{n}{2}+2$, there exists a positive
$T=T(\|\theta_{0}\|_{H^{s}})$ such that the equation (1.1) admits a unique
solution $\theta$ in $C([0,T);H^{s})\cap{{\rm Lip}}((0,T);H^{s-1}).$

(2) Let $n\geq 2$ and $0<\alpha\leq\frac{1}{2}$. Then for each $\theta_{0}\in
H^{s}$ with $s>\frac{n}{2}+1$, there exists a positive
$T=T(\|\theta_{0}\|_{H^{s}})$ such that the equation (1.1) admits a unique
solution $\theta$ in $C([0,T);H^{s})\cap{\rm Lip}((0,T);H^{s-1})$.
Furthermore, if $T^{\ast}$ is the first time the solution cannot be continued
in $C([0,T^{\ast});H^{s})$, then there necessarily holds

$\int_{0}^{T^{\ast}}\|(\mathcal{R}\otimes\mathcal{R})\Lambda^{2\alpha}\theta(\cdot,t)\|_{L^{\infty}}dt=\infty,$

where
$\Big{(}(\mathcal{R}\otimes\mathcal{R})(f)\Big{)}_{jk}=\mathcal{R}_{j}\mathcal{R}_{k}(f)$
denotes the tensor product of the Riesz operators.

The next lemma shows that the radial symmetry of the initial data is preserved
by the solution to (1.1).

######  Lemma 2.2.

If $\theta$ is a smooth solution to (1.1) with the radially symmetric initial
data $\theta_{0}$, then $\theta(x,t)$ is radially symmetric for $t>0$.

Proof. Let $\mathbf{O}\in\mathbb{R}^{n\times n}$ be any orthogonal matrix. By
the uniqueness of solutions to (1.1) and the radial property of $\theta_{0}$,
it suffices to show that the function

$\displaystyle\theta_{\mathbf{O}}(x,t):=\theta(\mathbf{O}x,t)$

is also a solution to (1.1) with the initial data $\theta_{0}(\mathbf{O}x)$.
Indeed, standard computations give that

$\displaystyle(\partial_{t}\theta_{\mathbf{O}})(x,t)=\partial_{t}(\theta(\mathbf{O}x,t))=(\partial_{t}\theta)(\mathbf{O}x,t)$

and

$\displaystyle(\nabla_{x}\theta_{\mathbf{O}})(x,t)=\nabla_{x}(\theta(\mathbf{O}x,t))={\mathbf{O}}^{T}(\nabla_{x}\theta)(\mathbf{O}x,t).$

By the second equation in (1.1), the integral representation of the Riesz
potential and integration by parts, we can derive that

$\displaystyle u_{\mathbf{O}}(x,t)$ $\displaystyle=$ $\displaystyle
C_{n,\alpha}P.V.\int_{\mathbb{R}^{n}}\frac{x-y}{|x-y|^{n+2\alpha}}\theta(\mathbf{O}y,t)dy$
$\displaystyle=$ $\displaystyle
C_{n,\alpha}P.V.\int_{\mathbb{R}^{n}}\frac{x-{\mathbf{O}}^{-1}z}{|x-{\mathbf{O}}^{-1}z|^{n+2\alpha}}\theta(z,t)dz$
$\displaystyle=$
$\displaystyle{\mathbf{O}}^{-1}C_{n,\alpha}P.V.\int_{\mathbb{R}^{n}}\frac{\mathbf{O}x-z}{|\mathbf{O}x-z|^{n+2\alpha}}\theta(z,t)dz$
$\displaystyle=$ $\displaystyle{\mathbf{O}}^{-1}u(\mathbf{O}x,t),$

where

$\displaystyle
C_{n,\alpha}=-\frac{2\Gamma(\frac{n}{2}+\alpha)}{\pi^{\frac{n}{2}}2^{2-2\alpha}\Gamma(1-\alpha)}.$

Thus, we obtain

$\displaystyle(\partial_{t}\theta_{\mathbf{O}}+u_{\mathbf{O}}\cdot\nabla\theta_{\mathbf{O}})(x,t)$
$\displaystyle=$
$\displaystyle(\partial_{t}\theta)(\mathbf{O}x,t)+{\mathbf{O}}^{-1}u(\mathbf{O}x,t)\cdot{\mathbf{O}}^{T}(\nabla_{x}\theta)(\mathbf{O}x,t)$
$\displaystyle=$
$\displaystyle(\partial_{t}\theta)(\mathbf{O}x,t)+u(\mathbf{O}x,t)\cdot\mathbf{O}{\mathbf{O}}^{T}(\nabla_{x}\theta)(\mathbf{O}x,t)$
$\displaystyle=$
$\displaystyle(\partial_{t}\theta+u\cdot\nabla_{x}\theta)(\mathbf{O}x,t)=0.$

The proof of Lemma 2.2 is then finished.

We proceed to present a string of simple inequalities needed later.

######  Lemma 2.3.

Let $\frac{1}{2}<\alpha<1$. Given $f:[0,\infty)\mapsto\mathbb{R}$ a smooth and
bounded function. We have

$\displaystyle\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho\leq\epsilon\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha}}d\rho+\Big{(}\frac{1}{(2-2\alpha)\epsilon}+\frac{4}{2\alpha-1}\Big{)}\|f\|^{2}_{L^{\infty}},$
$\displaystyle\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho\leq\epsilon\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}d\rho+\Big{(}8+\frac{1}{\epsilon}\Big{)}\|f\|^{2}_{L^{\infty}}$

and

$\displaystyle\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}(1-e^{-\frac{\rho}{2}})d\rho\leq\epsilon\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}d\rho+\Big{(}4+\frac{1}{4\epsilon}\Big{)}\|f\|^{2}_{L^{\infty}},$

for any positive $\epsilon$.

Proof. Since $\frac{1}{2}<\alpha<1$, by Hölder’s and Young’s inequality, we
first derive that

$\displaystyle\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho$
$\displaystyle\leq$ $\displaystyle
2\|f\|_{L^{\infty}}\Big{(}\int^{1}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha}}d\rho\Big{)}^{\frac{1}{2}}\Big{(}\int^{1}_{0}\frac{d\rho}{\rho^{2\alpha-1}}\Big{)}^{\frac{1}{2}}+4\|f\|^{2}_{L^{\infty}}\int^{\infty}_{1}\frac{d\rho}{\rho^{2\alpha}}$
$\displaystyle\leq$
$\displaystyle\epsilon\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha}}d\rho+\Big{(}\frac{1}{(2-2\alpha)\epsilon}+\frac{4}{2\alpha-1}\Big{)}\|f\|^{2}_{L^{\infty}},$

where $\epsilon>0$ is any positive constant.

Similarly, also by the elementary inequalities

$\displaystyle e^{\frac{\rho}{2}}>\frac{\rho}{2}~{}{\rm
and}~{}1-e^{-\frac{\rho}{2}}<\frac{\rho}{2},~{}{\rm for}~{}\rho>0,$

we can obtain that

$\displaystyle\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho$ $\displaystyle\leq$ $\displaystyle
2\|f\|_{L^{\infty}}\int_{0}^{1}\frac{|f(\rho)-f(0)|}{\rho}d\rho+8\|f\|^{2}_{L^{\infty}}\int_{1}^{\infty}\frac{d\rho}{\rho^{2}}$
$\displaystyle\leq$ $\displaystyle
2\|f\|_{L^{\infty}}\Big{(}\int_{0}^{1}\frac{|f(\rho)-f(0)|^{2}}{\rho^{2}}d\rho\Big{)}^{\frac{1}{2}}+8\|f\|^{2}_{L^{\infty}}$
$\displaystyle\leq$
$\displaystyle\epsilon\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}d\rho+\Big{(}8+\frac{1}{\epsilon}\Big{)}\|f\|^{2}_{L^{\infty}}.$

and

$\displaystyle\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}(1-e^{-\frac{\rho}{2}})d\rho$
$\displaystyle\leq$
$\displaystyle\|f\|_{L^{\infty}}\int_{0}^{1}\frac{|f(\rho)-f(0)|}{\rho}d\rho+4\|f\|^{2}_{L^{\infty}}\int_{1}^{\infty}\frac{d\rho}{\rho^{2}}$
$\displaystyle\leq$
$\displaystyle\|f\|_{L^{\infty}}\Big{(}\int_{0}^{1}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}d\rho\Big{)}^{\frac{1}{2}}+4\|f\|^{2}_{L^{\infty}}$
$\displaystyle\leq$
$\displaystyle\epsilon\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}d\rho+\Big{(}4+\frac{1}{4\epsilon}\Big{)}\|f\|^{2}_{L^{\infty}}.$

We then finish the proof of Lemma 2.3.

Finally, we end with this section by introducing an auxiliary function defined
by

$\displaystyle g(\lambda,m,\beta)\triangleq\int_{0}^{\pi}\frac{\sin^{m}\mu
d\mu}{(1-2\lambda\cos\mu+\lambda^{2})^{\frac{m}{2}+\beta}},$

where $\lambda\in(-1,1)$, $m\in\mathbb{N}$ and $\beta>0$. The motivation for
this function will be clear later in the proof of Proposition 3.1. We can
readily see that $g$ is even, infinitely differentiable (and real analytic) in
$\lambda\in(-1,1)$ and its Taylor’s series at $0$ has the form:

$\displaystyle
g(\lambda,m,\beta)=\sum^{\infty}_{k=0}a_{2k}(m,\beta)\lambda^{2k}.$

The following lemma is mainly from [16]. Since it is vital for our arguments
and is also of independent interest, we include the proof here for the sake of
completeness.

######  Lemma 2.4.

For $k\geq 0$, it holds that $a_{2k}(m,\beta)>0$ and

$\displaystyle\displaystyle\lim_{k\rightarrow\infty}\frac{a_{2k}(m,\beta)}{k^{2\beta-2}}=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{m}{2}+\frac{1}{2})}{\Gamma(\beta)\Gamma(\frac{m}{2}+\beta)}.$

Proof. Direct differentiation yields that

$\displaystyle\frac{\partial}{\partial\lambda}g(\lambda,m,\beta)=-(m+2\beta)\int_{0}^{\pi}\frac{\sin^{m}\mu(\lambda-\cos\mu)}{(1-2\lambda\cos\mu+\lambda^{2})^{\frac{m}{2}+\beta+1}}d\mu$

and

$\displaystyle\frac{\partial^{2}}{\partial\lambda^{2}}g(\lambda,m,\beta)$
$\displaystyle=$ $\displaystyle-(m+2\beta)\int_{0}^{\pi}\frac{\sin^{m}\mu
d\mu}{(1-2\lambda\cos\mu+\lambda^{2})^{\frac{m}{2}+\beta+1}}$ $\displaystyle\
\ \
+(m+2\beta)(m+2\beta+2)\int_{0}^{\pi}\frac{\sin^{m}\mu(\lambda-\cos\mu)^{2}}{(1-2\lambda\cos\mu+\lambda^{2})^{\frac{m}{2}+\beta+2}}d\mu.$

Noting that

$\displaystyle(\lambda-\cos\mu)^{2}=(1-2\lambda\cos\mu+\lambda^{2})-\sin^{2}\mu,$

we further derive that

$\displaystyle\frac{\partial^{2}g(\lambda,m,\beta)}{\partial\lambda^{2}}$
$\displaystyle=$
$\displaystyle(m+2\beta)\Big{[}(m+2\beta+1)g(\lambda,m,\beta+1)$
$\displaystyle\ \ \ -(m+2\beta+2)g(\lambda,m+2,\beta+1)\Big{]},$

which implies that

$\displaystyle\frac{\partial^{2k}}{\partial\lambda^{2k}}g(0,m,\beta)$
$\displaystyle=$
$\displaystyle(m+2\beta)\Big{[}(m+2\beta+1)\frac{\partial^{2k-2}}{\partial\lambda^{2k-2}}g(0,m,\beta+1)$
(2.1) $\displaystyle\ \ \ \ \ \ \
-(m+2\beta+2)\frac{\partial^{2k-2}}{\partial\lambda^{2k-2}}g(0,m+2,\beta+1)\Big{]},$

for any integer $k\geq 1$. Since

$\displaystyle
a_{0}(m,\beta)=g(0,m,\beta)=2\int_{0}^{\frac{\pi}{2}}\sin^{m}\mu
d\mu=B\Big{(}\frac{1}{2},\frac{m+1}{2}\Big{)}>0,$

then, from the recurrence relation (2.1), by induction, we can obtain that,
for $k\geq 1$,

$\displaystyle\frac{\partial^{2k}}{\partial\lambda^{2k}}g(0,m,\beta)$
$\displaystyle=$ $\displaystyle
B\Big{(}\frac{1}{2},\frac{m+1}{2}\Big{)}(2k-1)!!\cdot
2\beta(2\beta+2)\cdot\cdot\cdot(2\beta+2k-2)$ $\displaystyle\ \ \ \
\cdot\frac{(m+2\beta)(m+2\beta+2)\cdot\cdot\cdot(m+2\beta+2k-2)}{(m+2)(m+4)\cdot\cdot\cdot(m+2k)}.$

It follows from

$\displaystyle
a_{2k}(m,\beta)=\frac{1}{(2k)!}\frac{\partial^{2k}}{\partial\lambda^{2k}}g(0,m,\beta)$

that, for $k\geq 1$,

$\displaystyle
a_{2k}(m,\beta)=B\Big{(}\frac{1}{2},\frac{m+1}{2}\Big{)}\frac{(\beta)_{k}(\frac{m}{2}+\beta)_{k}}{k!(\frac{m}{2}+1)_{k}}>0,$

where the Pochhammer’s symbol’s $(x)_{k}$ is defined by

$\displaystyle(x)_{k}=\begin{cases}1,&\mbox{if $k=0,$ }\\\
x(x+1)\cdot\cdot\cdot(x+k-1),&\mbox{if $k\geq 1.$ }\\\ \end{cases}$

It follows from (1.6) that

$\displaystyle\displaystyle\lim_{k\rightarrow\infty}\frac{a_{2k}(m,\beta)}{k^{2\beta-2}}$
$\displaystyle=$ $\displaystyle
B\Big{(}\frac{1}{2},\frac{m+1}{2}\Big{)}\displaystyle\lim_{k\rightarrow\infty}\Big{[}\frac{\frac{k^{\frac{m}{2}+1}k!}{(\frac{m}{2}+1)_{k+1}}}{\frac{k^{\beta}k!}{(\beta)_{k+1}}\cdot\frac{k^{\frac{m}{2}+\beta}k!}{(\frac{m}{2}+\beta)_{k+1}}}\cdot\frac{k(\frac{m}{2}+1+k)}{(\beta+k)(\frac{m}{2}+\beta+k)}\Big{]}$
$\displaystyle=$ $\displaystyle
B\Big{(}\frac{1}{2},\frac{m+1}{2}\Big{)}\frac{\Gamma(\frac{m}{2}+1)}{\Gamma(\beta)\Gamma(\frac{m}{2}+\beta)}$
$\displaystyle=$
$\displaystyle\frac{\Gamma(\frac{1}{2})\Gamma(\frac{m}{2}+\frac{1}{2})}{\Gamma(\beta)\Gamma(\frac{m}{2}+\beta)}.$

The proof of Lemma 2.4 is then complete.

For the convenience of our later application, we denote

$\displaystyle G_{\alpha}(\lambda)\triangleq\int_{0}^{\pi}\frac{\sin^{n}\mu
d\mu}{(1-2\lambda\cos\mu+\lambda^{2})^{\frac{n}{2}+\alpha}}=\sum^{\infty}_{k=0}a_{2k}(\alpha)\lambda^{2k},$
(2.2)

where $\lambda\in(-1,1)$, $n\geq 2$ and $\alpha\in(0,1)$. Then, for any
$\alpha\in(0,1)$,

$\displaystyle G_{\alpha}(0)=B\Big{(}\frac{1}{2},\frac{n+1}{2}\Big{)}$ (2.3)

and it is easy to check that, for any $\lambda>0$ and $\lambda\neq 1$ (notice
that $G_{\alpha}(\lambda)$ is not defined at 1 for
$\alpha\in[\frac{1}{2},1)$),

$\displaystyle
G_{\alpha}\Big{(}\frac{1}{\lambda}\Big{)}=\lambda^{n+2\alpha}G_{\alpha}(\lambda).$
(2.4)

As a corollary of Lemma 2.4, we immediately have

######  Corollary 2.5.

For $n\geq 2$ and $\alpha\in(0,1)$, $G_{\alpha}(\lambda)$,
$G^{\prime}_{\alpha}(\lambda)$ and $G^{\prime\prime}_{\alpha}(\lambda)$ are
all positive in $\lambda\in(0,1)$. Furthermore, the series
$\displaystyle\sum^{\infty}_{k=0}a_{2k}(\alpha)$ is convergent if and only if
$\alpha\in(0,\frac{1}{2})$, and
$\displaystyle\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}$ is convergent for
$\alpha\in(0,1)$.

Proof. By Lemma 2.4, we have, for $0<\alpha<1$,

$\displaystyle\displaystyle\lim_{k\rightarrow\infty}\frac{a_{2k}(\alpha)}{k^{2\alpha-2}}=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(\alpha)\Gamma(\frac{n}{2}+\alpha)},$

which implies that $\displaystyle\sum^{\infty}_{k=0}a_{2k}(\alpha)$ is
convergent if and only if $0<\alpha<\frac{1}{2}$, and that
$\displaystyle\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}$ is convergent for
$\frac{1}{2}\leq\alpha<1$. In addition, it follows that the radius of the
convergence of (2.2) is 1 for $0<\alpha<1$. Thus, by differentiation term by
term, we ge that, for $\lambda\in(-1,1)$,

$\displaystyle
G^{\prime}_{\alpha}(\lambda)=\sum^{\infty}_{k=1}2ka_{2k}(\alpha)\lambda^{2k-1}$
(2.5)

and

$\displaystyle
G^{\prime\prime}_{\alpha}(\lambda)=\sum^{\infty}_{k=1}2k(2k-1)a_{2k}(\alpha)\lambda^{2k-2},$

which along with $a_{2k}(\alpha)>0$ and (2.2) that $G_{\alpha}(\lambda)$,
$G^{\prime}_{\alpha}(\lambda)$ and $G^{\prime\prime}_{\alpha}(\lambda)$ are
all positive in $\lambda\in(0,1)$. The proof of Corollary 2.5 is then
finished.

##  3\. Two weighted inequalities

In this section, we will prove two nonlinear weighted inequalities for the
model (1.1). The one-dimensional analogous inequalities were established in
[31]. By abuse of notation, we will not distinguish $f(x)$ and $f(|x|)$ for a
radially symmetric function $f$.

######  Proposition 3.1.

Let $n\geq 2$ and $0<\alpha<1$. Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be
a radial Schwartz function. Then

$\displaystyle\int_{\mathbb{R}^{n}}\frac{\Lambda^{-2+2\alpha}\nabla
f(x)\cdot\nabla f(x)}{|x|^{n}}dx\geq\frac{\alpha
2^{2\alpha-1}\Gamma(\frac{n}{2}+\alpha)}{\Gamma(1-\alpha)\Gamma(\frac{n}{2}+1)}\int_{\mathbb{R}^{n}}\frac{(f(0)-f(x))^{2}}{|x|^{n+2\alpha}}dx.$

Proof. By the integral representation of the Riesz potential [34] and the
radial assumption on $f$, we have

$\displaystyle\Lambda^{-2+2\alpha}(\nabla f)(x)$ $\displaystyle=$
$\displaystyle c_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{\nabla
f(y)}{|x-y|^{n-2+2\alpha}}dy$ $\displaystyle=$ $\displaystyle
c_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f^{\prime}(|y|)\frac{y}{|y|}}{|x-y|^{n-2+2\alpha}}dy$
$\displaystyle=$ $\displaystyle
c_{n,\alpha}\int_{0}^{\infty}f^{\prime}(\rho)\rho^{n-1}\Big{(}\int_{\mathbb{S}^{n-1}}\frac{zd\sigma(z)}{|x-\rho
z|^{n-2+2\alpha}}\Big{)}d\rho,$

where

$\displaystyle
c_{n,\alpha}=\frac{\Gamma(\frac{n}{2}-1+\alpha)}{\pi^{\frac{n}{2}}2^{2-2\alpha}\Gamma(1-\alpha)}.$

For any $x\in\mathbb{R}^{n}\backslash\\{0\\}$, pick an orthogonal matrix
$\mathbf{O}\in\mathbb{R}^{n\times n}$ such that
$\mathbf{O}e_{1}=\frac{x}{|x|}$, where $e_{1}=(1,0,...,0)$. It follows that,
for any $x\in\mathbb{R}^{n}\backslash\\{0\\}$,

$\displaystyle\Lambda^{-2+2\alpha}\nabla f(x)\cdot\nabla f(x)$
$\displaystyle=$ $\displaystyle
c_{n,\alpha}f^{\prime}(|x|)\int_{0}^{\infty}f^{\prime}(\rho)\rho^{n-1}\Big{(}\int_{\mathbb{S}^{n-1}}\frac{z\cdot\frac{x}{|x|}d\sigma(z)}{|x-\rho
z|^{n-2+2\alpha}}\Big{)}d\rho$ $\displaystyle=$ $\displaystyle
c_{n,\alpha}f^{\prime}(|x|)\int_{0}^{\infty}f^{\prime}(\rho)\rho^{n-1}\Big{(}\int_{\mathbb{S}^{n-1}}\frac{z\cdot\mathbf{O}e_{1}d\sigma(z)}{||x|\mathbf{O}e_{1}-\rho
z|^{n-2+2\alpha}}\Big{)}d\rho$ $\displaystyle=$ $\displaystyle
c_{n,\alpha}f^{\prime}(|x|)\int_{0}^{\infty}f^{\prime}(\rho)\rho^{n-1}\Big{(}\int_{\mathbb{S}^{n-1}}\frac{e_{1}\cdot{\mathbf{O}}^{-1}zd\sigma(z)}{||x|e_{1}-\rho{\mathbf{O}}^{-1}z|^{n-2+2\alpha}}\Big{)}d\rho$
$\displaystyle=$ $\displaystyle
c_{n,\alpha}f^{\prime}(|x|)\int_{0}^{\infty}f^{\prime}(\rho)\rho^{n-1}\Big{(}\int_{\mathbb{S}^{n-1}}\frac{z_{1}d\sigma(z)}{||x|e_{1}-\rho
z|^{n-2+2\alpha}}\Big{)}d\rho.$

By a change of variables formula (see e.g., pp. 592 of [20]) and integration
by parts, we obtain that, for $\rho\neq|x|$,

$\displaystyle\int_{\mathbb{S}^{n-1}}\frac{z_{1}d\sigma(z)}{||x|e_{1}-\rho
z|^{n-2+2\alpha}}$ $\displaystyle=$
$\displaystyle\int_{-1}^{1}\int_{\sqrt{1-s^{2}}\mathbb{S}^{n-2}}\frac{sd\sigma(z)}{((|x|-\rho
s)^{2}+\rho^{2}|z|^{2})^{\frac{n}{2}-1+\alpha}}\frac{ds}{\sqrt{1-s^{2}}}$
$\displaystyle=$
$\displaystyle\omega_{n-2}\int_{-1}^{1}\frac{s(1-s^{2})^{\frac{n-3}{2}}ds}{(|x|^{2}-2|x|\rho
s+\rho^{2})^{\frac{n}{2}-1+\alpha}}$ $\displaystyle=$
$\displaystyle\omega_{n-2}\int_{0}^{\pi}\frac{\cos\mu\sin^{n-2}\mu
d\mu}{(|x|^{2}-2|x|\rho\cos\mu+\rho^{2})^{\frac{n}{2}-1+\alpha}}$
$\displaystyle=$
$\displaystyle\frac{(n-2+2\alpha)\omega_{n-2}}{n-1}\rho|x|\int_{0}^{\pi}\frac{\sin^{n}\mu
d\mu}{(|x|^{2}-2|x|\rho\cos\mu+\rho^{2})^{\frac{n}{2}+\alpha}}$
$\displaystyle=$
$\displaystyle\frac{(n-2+2\alpha)\omega_{n-2}\rho}{(n-1)|x|^{n-1+2\alpha}}\int_{0}^{\pi}\frac{\sin^{n}\mu
d\mu}{(1-2\frac{\rho}{|x|}\cos\mu+\frac{\rho^{2}}{|x|^{2}})^{\frac{n}{2}+\alpha}}$
$\displaystyle=$
$\displaystyle\frac{(n-2+2\alpha)\omega_{n-2}}{n-1}\frac{\rho}{|x|^{n-1+2\alpha}}G_{\alpha}\Big{(}\frac{\rho}{|x|}\Big{)},$

where $\omega_{n-2}=\frac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})}$ is the
surface area of $\mathbb{S}^{n-2}$ and the function $G_{\alpha}$ is defined as
(2.2). Therefore, integrating by parts along with the boundary conditions

$\displaystyle\lim_{\rho\rightarrow
0^{+}}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{|x|}\Big{)}(f(\rho)-f(|x|))=0,$
$\displaystyle\lim_{\rho\rightarrow+\infty}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{|x|}\Big{)}(f(\rho)-f(|x|))=|x|^{n+2\alpha}\lim_{\rho\rightarrow+\infty}G_{\alpha}\Big{(}\frac{|x|}{\rho}\Big{)}\frac{f(\rho)-f(|x|)}{\rho^{2\alpha}}=0,$

we have that, for any $x\in\mathbb{R}^{n}\backslash\\{0\\}$,

$\displaystyle\Lambda^{-2+2\alpha}\nabla f(x)\cdot\nabla f(x)$ (3.1)
$\displaystyle=$ $\displaystyle
c^{\prime}_{n,\alpha}\frac{f^{\prime}(|x|)}{|x|^{n-1+2\alpha}}\int_{0}^{\infty}f^{\prime}(\rho)\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{|x|}\Big{)}d\rho$
$\displaystyle=$ $\displaystyle
c^{\prime}_{n,\alpha}\frac{f^{\prime}(|x|)}{|x|^{n-1+2\alpha}}\int_{0}^{\infty}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{|x|}\Big{)}\frac{d}{d\rho}(f(\rho)-f(|x|))d\rho$
$\displaystyle=$
$\displaystyle-c^{\prime}_{n,\alpha}\frac{f^{\prime}(|x|)}{|x|^{n-1+2\alpha}}\int_{0}^{\infty}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{|x|}\Big{)}\Big{)}(f(\rho)-f(|x|))d\rho,$

with

$\displaystyle
c^{\prime}_{n,\alpha}=c_{n,\alpha}\frac{(n-2+2\alpha)\omega_{n-2}}{n-1}=\frac{2^{2\alpha-1}\Gamma(\frac{n}{2}+\alpha)}{\pi^{\frac{1}{2}}\Gamma(1-\alpha)\Gamma(\frac{n+1}{2})}.$

It then follows from Fubini’s theorem and integration by parts that

$\displaystyle\int_{\mathbb{R}^{n}}\frac{\Lambda^{-2+2\alpha}\nabla
f(x)\cdot\nabla f(x)}{|x|^{n}}dx$ (3.2) $\displaystyle=$ $\displaystyle
c^{\prime\prime}_{n,\alpha}\int_{0}^{\infty}\frac{f^{\prime}(r)}{r^{n+2\alpha}}\Big{(}-\int_{0}^{\infty}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}(f(\rho)-f(r))d\rho\Big{)}dr$
$\displaystyle=$
$\displaystyle\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\frac{\partial}{\partial
r}\Big{(}(f(\rho)-f(r))^{2}\Big{)}drd\rho$ $\displaystyle=$
$\displaystyle\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}(f(\rho)-f(r))^{2}\Big{)}\Big{|}^{\infty}_{r\rightarrow
0^{+}}d\rho$ $\displaystyle\ \ \
-\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\int_{0}^{\infty}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}(f(\rho)-f(r))^{2}d\rho
dr$

with

$\displaystyle c^{\prime\prime}_{n,\alpha}=c^{\prime}_{n,\alpha}\omega_{n-1}.$
(3.3)

By (2.3), we note that

$\displaystyle\lim_{r\rightarrow\infty}\frac{(f(\rho)-f(r))^{2}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}$
(3.4) $\displaystyle=$
$\displaystyle\lim_{r\rightarrow\infty}\frac{(f(\rho)-f(r))^{2}}{r^{n+2\alpha}}\Big{[}n\rho^{n-1}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}+\frac{\rho^{n}}{r}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{]}=0$

and

$\displaystyle\lim_{r\rightarrow
0^{+}}\frac{(f(\rho)-f(r))^{2}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}$
(3.5) $\displaystyle=$ $\displaystyle(f(\rho)-f(0))^{2}\lim_{r\rightarrow
0^{+}}\frac{\partial}{\partial\rho}\Big{(}\rho^{-2\alpha}G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle(f(\rho)-f(0))^{2}\lim_{r\rightarrow
0^{+}}\Big{(}-\frac{2\alpha}{\rho^{1+2\alpha}}G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}-\frac{r}{\rho^{2+2\alpha}}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle-\frac{2\alpha
G_{\alpha}(0)}{\rho^{1+2\alpha}}(f(\rho)-f(0))^{2}$ $\displaystyle=$
$\displaystyle-2\alpha
B\Big{(}\frac{1}{2},\frac{n+1}{2}\Big{)}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha}}.$

Thus,

$\displaystyle\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}(f(\rho)-f(r))^{2}\Big{)}\Big{|}^{\infty}_{r=0}d\rho$
(3.6) $\displaystyle=$ $\displaystyle\alpha
c^{\prime\prime}_{n,\alpha}B\Big{(}\frac{1}{2},\frac{n+1}{2}\Big{)}\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha}}d\rho$
$\displaystyle=$ $\displaystyle\frac{\alpha
c^{\prime\prime}_{n,\alpha}B(\frac{1}{2},\frac{n+1}{2})}{\omega_{n-1}}\int_{\mathbb{R}^{n}}\frac{(f(0)-f(x))^{2}}{|x|^{n+2\alpha}}dx$
$\displaystyle=$ $\displaystyle\frac{\alpha
2^{2\alpha-1}\Gamma(\frac{n}{2}+\alpha)}{\Gamma(1-\alpha)\Gamma(\frac{n}{2}+1)}\int_{\mathbb{R}^{n}}\frac{(f(0)-f(x))^{2}}{|x|^{n+2\alpha}}dx.$

We proceed to check that, for all $0<\rho,r<\infty$ and $\rho\neq r$,

$\displaystyle\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}<0.$
(3.7)

Indeed, direct calculations give that, for $0<\rho<r$,

$\displaystyle\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}=n\rho^{n-1}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}+\frac{\rho^{n}}{r}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}$

and

$\displaystyle\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}$
(3.8) $\displaystyle=$ $\displaystyle n\rho^{n-1}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}+\rho^{n}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+1+2\alpha}}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}$
$\displaystyle=$
$\displaystyle-\frac{n(n+2\alpha)\rho^{n-1}}{r^{n+1+2\alpha}}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}-\frac{(2n+1+2\alpha)\rho^{n}}{r^{n+2+2\alpha}}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}$
$\displaystyle\ \ \ \ \ \
-\frac{\rho^{n+1}}{r^{n+3+2\alpha}}G^{\prime\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}<0,$

where in the last inequality we have used Corollary 2.5 showing that
$G_{\alpha}$, $G^{\prime}_{\alpha}$ and $G^{\prime\prime}_{\alpha}$ are all
positive in $(0,1)$.

Furthermore, for $0<r<\rho$, by (2.4) and Corollary 2.5, we derive that

$\displaystyle\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial
r}\Big{(}\frac{\partial}{\partial\rho}\Big{(}\rho^{-2\alpha}G\Big{(}\frac{r}{\rho}\Big{)}\Big{)}\Big{)}$
(3.9) $\displaystyle=$ $\displaystyle\frac{\partial}{\partial
r}\Big{(}-\frac{2\alpha}{\rho^{1+2\alpha}}G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}-\frac{r}{\rho^{2+2\alpha}}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{)}$
$\displaystyle=$
$\displaystyle-\frac{1+2\alpha}{\rho^{2+2\alpha}}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}-\frac{r}{\rho^{3+2\alpha}}G^{\prime\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}<0.$

Combining (3.8) and (3.9) gives (3.7), which implies that

$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}(f(\rho)-f(r))^{2}d\rho
dr\leq 0.$

This negative integral along with (3.2) and (3.6) completes the proof of
Proposition 3.1.

######  Remark 3.1.

It should be remarked that Li and Rodrigo [31] established the following
weighted nonlinear inequality: for any $\delta\in(-2\alpha,2-2\alpha)$ and
radial decreasing Schwartz function $f$,

$\displaystyle\int_{\mathbb{R}^{n}}\frac{\Lambda^{-2+2\alpha}\nabla
f(x)\cdot\nabla f(x)}{|x|^{n+\delta}}dx\geq
C_{n,\alpha,\delta}\int_{\mathbb{R}^{n}}\frac{(f(0)-f(x))^{2}}{|x|^{n+2\alpha+\delta}}dx,$
(3.10)

which can be applied to show the finite-time blow-up of smooth solutions to
(1.1) for some class of radial decreasing initial data. The key observation
there is a pointwise lower bound for the nonlinear term, that is, for the
radial monotone function $f$ and for any
$x\in\mathbb{R}^{n}\backslash\\{0\\}$,

$\displaystyle\Lambda^{-2+2\alpha}\nabla f(x)\cdot\nabla f(x)\geq
C_{n,\alpha}\frac{f^{\prime}(|x|)}{|x|^{n-1+2\alpha}}\int_{0}^{|x|}f^{\prime}(\rho)\rho^{n}d\rho,$

which can be seen from the first equality in (3.1) and the fact that
$G_{\alpha}(\lambda)>G_{\alpha}(0)$ for $\lambda\in(0,1)$. Proposition 3.1 is
the special case $\delta=0$ in (3.10), but whose proof provided here does not
depend on the monotone decaying property of $f$. Unfortunately, from the
second equality in (3.5), we can see that the current method may not be
applied to prove more general inequality (3.10) for $\delta\in(0,2-2\alpha)$,
which is needed for the implication of blow-up. Finally, one can refer to [16]
for another proof of (3.10) for a general radial function $f$, where the main
tool is Meillin transform.

Since the nonlinear inequality in Proposition 3.1 is not directly useful for
implying blow-ups of (1.1) for general radial initial data, we establish the
next weighed inequality to show the finite-time blow-up. It should be remarked
that the one-dimensional analogous inequality was proved by Li and Rodrigo in
[31].

######  Proposition 3.2.

Let $n\geq 2$ and $0<\alpha<1$. Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be
a radial Schwartz function. Then

$\displaystyle\int_{\mathbb{R}^{n}}\frac{\Lambda^{-2+2\alpha}\nabla
f(x)\cdot\nabla f(x)}{|x|^{n}}e^{-|x|}dx\geq
C^{\prime}_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{(f(0)-f(x))^{2}}{|x|^{n+2\alpha}}dx-C^{\prime\prime}_{n,\alpha}\|f\|^{2}_{L^{\infty}},$
(3.11)

where

$\displaystyle C^{\prime}_{n,\alpha}=\frac{\alpha
2^{2\alpha-2}\Gamma(\frac{n}{2}+\alpha)}{\Gamma(1-\alpha)\Gamma(\frac{n}{2}+1)}$

and the constant $C^{\prime\prime}_{n,\alpha}$ depends only on $n$ and
$\alpha$.

Proof. We only need to modify the proof of Proposition 3.1. Similar to (3.2),
by (3.1), (3.4), (3.5) and (3.7), we can derive that

$\displaystyle\int_{\mathbb{R}^{n}}\frac{\Lambda^{-2+2\alpha}\nabla
f(x)\cdot\nabla f(x)}{|x|^{n}}e^{-|x|}dx$ (3.12) $\displaystyle=$
$\displaystyle
c^{\prime\prime}_{n,\alpha}\int_{0}^{\infty}\frac{f^{\prime}(r)e^{-r}}{r^{n+2\alpha}}\Big{(}-\int_{0}^{\infty}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}(f(\rho)-f(r))d\rho\Big{)}dr$
$\displaystyle=$
$\displaystyle\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-r}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\frac{\partial}{\partial
r}\Big{(}(f(\rho)-f(r))^{2}\Big{)}drd\rho$ $\displaystyle=$
$\displaystyle\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\Big{(}\frac{e^{-r}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}(f(\rho)-f(r))^{2}\Big{)}\Big{|}^{\infty}_{r\rightarrow
0^{+}}d\rho$ $\displaystyle\ \ \
-\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\int_{0}^{\infty}\frac{\partial}{\partial
r}\Big{(}\frac{e^{-r}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}(f(\rho)-f(r))^{2}d\rho
dr$ $\displaystyle=$ $\displaystyle\alpha
c^{\prime\prime}_{n,\alpha}B\Big{(}\frac{1}{2},\frac{n+1}{2}\Big{)}\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha}}d\rho$
$\displaystyle\ \ \ \
-\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\int_{0}^{\infty}e^{-r}\frac{\partial}{\partial
r}\Big{(}\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}(f(\rho)-f(r))^{2}d\rho
dr$ $\displaystyle\ \ \ \
+\frac{c^{\prime\prime}_{n,\alpha}}{2}\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-r}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}(f(\rho)-f(r))^{2}d\rho
dr$ $\displaystyle\geq$ $\displaystyle\frac{\alpha
2^{2\alpha-1}\Gamma(\frac{n}{2}+\alpha)}{\Gamma(1-\alpha)\Gamma(\frac{n}{2}+1)}\int_{\mathbb{R}^{n}}\frac{(f(0)-f(x))^{2}}{|x|^{n+2\alpha}}dx$
$\displaystyle\ \ \
+\frac{c^{\prime\prime}_{n,\alpha}}{2}\underbrace{\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-r}}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}(f(\rho)-f(r))^{2}d\rho
dr}_{\mathcal{I}},$

where the constant $c^{\prime\prime}_{n,\alpha}$ is defined as (3.3). We
proceed to estimate the integral $\mathcal{I}$. For $0<\rho<r$,
straightforward computation yields that

$\displaystyle\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}=\frac{\rho^{n-1}}{r^{n+2\alpha}}\Big{[}nG_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}+\frac{\rho}{r}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{]}.$

For $0<r<\rho$, by (2.4), we have that

$\displaystyle\frac{1}{r^{n+2\alpha}}\frac{\partial}{\partial\rho}\Big{(}\rho^{n}G_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{)}\Big{)}$
$\displaystyle=$
$\displaystyle\frac{\partial}{\partial\rho}\Big{(}\rho^{-2\alpha}G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{)}\Big{)}$
$\displaystyle=$
$\displaystyle-\frac{2\alpha}{\rho^{1+2\alpha}}G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}-\frac{r}{\rho^{2+2\alpha}}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}.$

Thus, we rewrite the integral $\mathcal{I}$ as

$\displaystyle\mathcal{I}$ $\displaystyle=$
$\displaystyle\iint\limits_{0<\rho<r}\frac{e^{-r}\rho^{n-1}(f(\rho)-f(r))^{2}}{r^{n+2\alpha}}\Big{[}nG_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}+\frac{\rho}{r}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}\Big{]}d\rho
dr$ (3.13) $\displaystyle\ \ \ \ \
-\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}\Big{[}2\alpha
G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}+\frac{r}{\rho}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{]}d\rho
dr.$

Case 1. $0<\alpha<\frac{1}{2}$. In this case, by Corollary 2.5, (2.2) and
(2.5) leads to

$\displaystyle\mathcal{I}$ $\displaystyle\geq$
$\displaystyle-\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}\Big{[}2\alpha
G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}+\frac{r}{\rho}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{]}d\rho
dr$ $\displaystyle=$
$\displaystyle-\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}\Big{[}\sum^{\infty}_{k=0}2(k+\alpha)a_{2k}(\alpha)\Big{(}\frac{r}{\rho}\Big{)}^{2k}\Big{]}d\rho
dr$ $\displaystyle=$
$\displaystyle-\sum^{\infty}_{k=0}2(k+\alpha)a_{2k}(\alpha)\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr$ $\displaystyle\geq$
$\displaystyle-4\|f\|^{2}_{L^{\infty}}\sum^{\infty}_{k=0}2(k+\alpha)a_{2k}(\alpha)\int^{\infty}_{0}e^{-r}r^{2k}\Big{(}\int^{\infty}_{r}\frac{d\rho}{\rho^{1+2\alpha+2k}}\Big{)}dr$
$\displaystyle=$
$\displaystyle-4\Gamma(1-2\alpha)\cdot\sum^{\infty}_{k=0}a_{2k}(\alpha)\cdot\|f\|^{2}_{L^{\infty}},$

which along with (3.12) implies the desired inequality (3.11).

Case 2. $\frac{1}{2}<\alpha<1$. In this case, by (3.13), Corollary 2.5 and a
change of variable, we derive that

$\displaystyle\mathcal{I}$ $\displaystyle\geq$
$\displaystyle\iint\limits_{0<\rho<r}\frac{e^{-r}\rho^{n-1}}{r^{n+2\alpha}}(f(\rho)-f(r))^{2}\frac{\rho}{r}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}d\rho
dr$ (3.14) $\displaystyle\ \ \ \ \
-\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}\Big{[}2\alpha
G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}+\frac{r}{\rho}G^{\prime}_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}\Big{]}d\rho
dr$ $\displaystyle=$
$\displaystyle-\underbrace{\iint\limits_{0<\rho<r}\Big{[}e^{-\rho}-\frac{e^{-r}\rho^{n-1}}{r^{n-1}}\Big{]}\frac{(f(\rho)-f(r))^{2}}{r^{1+2\alpha}}\frac{\rho}{r}G^{\prime}_{\alpha}\Big{(}\frac{\rho}{r}\Big{)}d\rho
dr}_{\mathcal{J}}$ $\displaystyle\ \ \ \ \
-2\alpha\underbrace{\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}G_{\alpha}\Big{(}\frac{r}{\rho}\Big{)}d\rho
dr}_{\mathcal{K}}.$

By (2.2), $\frac{1}{2}<\alpha<1$ and Corollary 2.5, the positive term
$\mathcal{K}$ can be estimated above as

$\displaystyle\mathcal{K}$ $\displaystyle=$
$\displaystyle\sum^{\infty}_{k=0}a_{2k}(\alpha)\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{1+2\alpha}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr$ (3.15) $\displaystyle\leq$
$\displaystyle\sum^{\infty}_{k=0}2a_{2k}(\alpha)\Big{[}\iint\limits_{0<r<\rho}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha+2k}}r^{2k}d\rho
dr+\iint\limits_{0<r<\rho}\frac{(f(r)-f(0))^{2}}{\rho^{1+2\alpha+2k}}r^{2k}d\rho
dr\Big{]}$ $\displaystyle=$
$\displaystyle\sum^{\infty}_{k=0}2a_{2k}(\alpha)\Big{[}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{1+2\alpha+2k}}\Big{(}\int_{0}^{\rho}r^{2k}dr\Big{)}d\rho$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+\int^{\infty}_{0}r^{2k}(f(r)-f(0))^{2}\Big{(}\int^{\infty}_{r}\frac{d\rho}{\rho^{1+2\alpha+2k}}\Big{)}dr\Big{]}$
$\displaystyle=$
$\displaystyle\sum^{\infty}_{k=0}a_{2k}(\alpha)\Big{(}\frac{2}{2k+1}+\frac{1}{k+\alpha}\Big{)}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho$
$\displaystyle\leq$
$\displaystyle\sum^{\infty}_{k=0}\frac{4a_{2k}(\alpha)}{2k+1}\cdot\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho.$

By (2.5), we further express the term $\mathcal{J}$ as

$\displaystyle\mathcal{J}$ $\displaystyle=$
$\displaystyle\iint\limits_{0<\rho<r}\Big{[}e^{-\rho}-e^{-r}\Big{(}\frac{\rho}{r}\Big{)}^{n-1}\Big{]}\frac{(f(\rho)-f(r))^{2}}{r^{1+2\alpha}}\sum^{\infty}_{k=1}2ka_{2k}(\alpha)\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ (3.16) $\displaystyle=$
$\displaystyle\sum^{\infty}_{k=1}2ka_{2k}(\alpha)\underbrace{\iint\limits_{0<\rho<r}\Big{[}e^{-\rho}-e^{-r}\Big{(}\frac{\rho}{r}\Big{)}^{n-1}\Big{]}\frac{(f(\rho)-f(r))^{2}}{r^{1+2\alpha}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{J}_{1}}.$

By the mean-value theorem, we derive that, for $0<\rho<r$,

$\displaystyle e^{-\rho}-e^{-r}\Big{(}\frac{\rho}{r}\Big{)}^{n-1}$
$\displaystyle=$
$\displaystyle\rho^{n-1}\Big{(}\frac{e^{-\rho}}{\rho^{n-1}}-\frac{e^{-r}}{r^{n-1}}\Big{)}$
(3.17) $\displaystyle\leq$ $\displaystyle
e^{-\rho}\rho^{n-1}\Big{[}\rho^{1-n}+\frac{n-1}{\rho^{n}}\Big{]}(r-\rho)$
$\displaystyle=$ $\displaystyle
e^{-\rho}(r-\rho)+(n-1)e^{-\rho}\Big{(}\frac{r}{\rho}-1\Big{)}.$

Then $\mathcal{J}_{1}$ can be bounded as

$\displaystyle\mathcal{J}_{1}$ $\displaystyle\leq$
$\displaystyle\underbrace{\iint\limits_{0<\rho<r}e^{-\rho}(r-\rho)\frac{(f(\rho)-f(r))^{2}}{r^{1+2\alpha}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{J}_{11}}$ (3.18) $\displaystyle\ \ \
+(n-1)\underbrace{\iint\limits_{0<\rho<r}e^{-\rho}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(r))^{2}}{r^{1+2\alpha}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{J}_{12}}.$

Since $\frac{1}{2}<\alpha<1$, we derive that, for any $k\geq 1$,

$\displaystyle\mathcal{J}_{11}$ $\displaystyle\leq$ $\displaystyle
4\|f\|^{2}_{L^{\infty}}\iint\limits_{0<\rho<r}e^{-\rho}\frac{r-\rho}{r^{1+2\alpha}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ (3.19) $\displaystyle=$ $\displaystyle
4\|f\|^{2}_{L^{\infty}}\int^{\infty}_{0}e^{-\rho}\rho^{2k}\Big{(}\int^{\infty}_{\rho}\frac{1-\frac{\rho}{r}}{r^{2k+2\alpha}}dr\Big{)}d\rho$
$\displaystyle=$ $\displaystyle
4\|f\|^{2}_{L^{\infty}}\Big{(}\frac{1}{2k+2\alpha-1}-\frac{1}{2k+2\alpha}\Big{)}\int^{\infty}_{0}e^{-\rho}\rho^{1-2\alpha}d\rho$
$\displaystyle=$
$\displaystyle\frac{4\Gamma(2-2\alpha)\|f\|^{2}_{L^{\infty}}}{(2k+2\alpha-1)(2k+2\alpha)}$
$\displaystyle<$
$\displaystyle\frac{\Gamma(2-2\alpha)}{k^{2}}\|f\|^{2}_{L^{\infty}}.$

Note that $\mathcal{J}_{12}$ can not be estimated in the way same as
$\mathcal{J}_{11}$. Utilizing the elementary inequality

$\displaystyle(a-b)^{2}\leq 2(a-c)^{2}+2(b-c)^{2}~{}{\rm
for}~{}a,b,c\in\mathbb{R},$

we estimate $\mathcal{J}_{12}$ in the following way, for any $k\geq 1$,

$\displaystyle\mathcal{J}_{12}$ $\displaystyle\leq$ $\displaystyle
2\iint\limits_{0<\rho<r}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(0))^{2}}{r^{1+2\alpha}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ $\displaystyle\ \ \ \ \ \
+2\iint\limits_{0<\rho<r}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(0)-f(r))^{2}}{r^{1+2\alpha}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ $\displaystyle=$ $\displaystyle
2\int_{0}^{\infty}\rho^{2k}(f(\rho)-f(0))^{2}\Big{(}\int^{\infty}_{\rho}\frac{\frac{r}{\rho}-1}{r^{2k+2\alpha+1}}dr\Big{)}d\rho$
$\displaystyle\
+2\int_{0}^{\infty}\frac{(f(r)-f(0))^{2}}{r^{2k+2\alpha+1}}\Big{[}\int^{r}_{0}\rho^{2k}\Big{(}\frac{r}{\rho}-1\Big{)}d\rho\Big{]}dr$
$\displaystyle=$ $\displaystyle
2\Big{(}\frac{1}{2k+2\alpha-1}-\frac{1}{2k+2\alpha}\Big{)}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho$
$\displaystyle\ \ \ \ \ \
+2\Big{(}\frac{1}{2k}-\frac{1}{2k+1}\Big{)}\int_{0}^{\infty}\frac{(f(r)-f(0))^{2}}{r^{2\alpha}}dr$
$\displaystyle=$
$\displaystyle\Big{[}\frac{2}{(2k+2\alpha-1)(2k+2\alpha)}+\frac{1}{k(2k+1)}\Big{]}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho$
$\displaystyle\leq$
$\displaystyle\frac{1}{k^{2}}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho,$

which follows from (3.19), (3.18) and (3.16) that

$\displaystyle\mathcal{J}$ $\displaystyle\leq$ $\displaystyle
2(n-1)\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}\cdot\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho$
$\displaystyle\
+2\Gamma(2-2\alpha)\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}\cdot\|f\|^{2}_{L^{\infty}}.$

This estimate along with (3.14) and (3.15) yields that

$\displaystyle\mathcal{I}$ $\displaystyle\geq$
$\displaystyle-\Big{[}2(n-1)\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}+8\alpha\sum^{\infty}_{k=0}\frac{a_{2k}(\alpha)}{2k+1}\Big{]}\cdot\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho^{2\alpha}}d\rho$
$\displaystyle\ \ \ \ \ \ \
-2\Gamma(2-2\alpha)\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}\|f\|^{2}_{L^{\infty}},$

Then, by utilizing (3.12) and the first inequality in Lemma 2.3 with

$\displaystyle\epsilon=\frac{\alpha
B(\frac{1}{2},\frac{n+1}{2})}{2(n-1)\displaystyle\sum^{\infty}_{k=1}\frac{a_{2k}(\alpha)}{k}+8\alpha\displaystyle\sum^{\infty}_{k=0}\frac{a_{2k}(\alpha)}{2k+1}},$

we can obtain the desired inequality (3.11).

Case 3. $\alpha=\frac{1}{2}$. In this case, for the conciseness, we use some
abbreviations of the notation as follows,

$\displaystyle
G_{\frac{1}{2}}(\lambda):=G(\lambda),~{}a_{2k}\Big{(}\frac{1}{2}\Big{)}:=a_{2k}$

for $\lambda\in(0,1)$ and $k=0,1,2\cdot\cdot\cdot$. Similar to (3.14), from
(3.13), we have that

$\displaystyle\mathcal{I}$ $\displaystyle\geq$
$\displaystyle-\underbrace{\iint\limits_{0<\rho<r}\Big{[}e^{-\rho}-\frac{e^{-r}\rho^{n-1}}{r^{n-1}}\Big{]}\frac{(f(\rho)-f(r))^{2}}{r^{2}}\frac{\rho}{r}G^{\prime}\Big{(}\frac{\rho}{r}\Big{)}d\rho
dr}_{\mathcal{\widetilde{J}}}$ (3.20) $\displaystyle\ \ \ \ \
-\underbrace{\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{2}}G\Big{(}\frac{r}{\rho}\Big{)}d\rho
dr}_{\mathcal{\widetilde{K}}}.$

By (2.2), we split the term $\mathcal{\widetilde{K}}$ into two parts as
follows

$\displaystyle\mathcal{\widetilde{K}}$ $\displaystyle=$
$\displaystyle\sum^{\infty}_{k=0}a_{2k}\iint\limits_{0<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{2}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr$ $\displaystyle=$
$\displaystyle\sum^{\infty}_{k=0}a_{2k}\underbrace{\iint\limits_{0<r\leq\frac{\rho}{2}}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{2}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr}_{\mathcal{\widetilde{K}}_{1}}$ $\displaystyle\ \ \ \
+\sum^{\infty}_{k=0}a_{2k}\underbrace{\iint\limits_{\frac{\rho}{2}<r<\rho}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{2}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr}_{\mathcal{\widetilde{K}}_{2}}.$

The term $\mathcal{\widetilde{K}}_{1}$ can be estimated as

$\displaystyle\mathcal{\widetilde{K}}_{1}$ $\displaystyle\leq$
$\displaystyle\Big{(}\frac{1}{2}\Big{)}^{2k}\iint\limits_{0<r<\frac{\rho}{2}}\frac{e^{-r}(f(\rho)-f(r))^{2}}{\rho^{2}}d\rho
dr$ $\displaystyle\leq$
$\displaystyle\frac{1}{2^{2k-1}}\iint\limits_{0<r<\frac{\rho}{2}}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}e^{r}}d\rho
dr+\frac{1}{2^{2k-1}}\iint\limits_{0<r<\frac{\rho}{2}}\frac{(f(r)-f(0))^{2}}{\rho^{2}e^{r}}d\rho
dr$ $\displaystyle=$
$\displaystyle\frac{1}{2^{2k-1}}\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}(1-e^{-\frac{\rho}{2}})d\rho+\frac{1}{2^{2k}}\int_{0}^{\infty}\frac{(f(r)-f(0))^{2}}{re^{r}}dr.$

For the term $\mathcal{\widetilde{K}}_{2}$, we derive that

$\displaystyle\mathcal{\widetilde{K}}_{2}$ $\displaystyle\leq$ $\displaystyle
2\iint\limits_{\frac{\rho}{2}<r<\rho}\frac{e^{-r}(f(\rho)-f(0))^{2}}{\rho^{2}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr+2\iint\limits_{\frac{\rho}{2}<r<\rho}\frac{e^{-r}(f(r)-f(0))^{2}}{\rho^{2}}\Big{(}\frac{r}{\rho}\Big{)}^{2k}d\rho
dr$ $\displaystyle\leq$ $\displaystyle
2\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2+2k}}\cdot\frac{e^{-\frac{\rho}{2}}\rho^{2k+1}}{2k+1}d\rho+2\int_{0}^{\infty}\frac{r^{2k}(f(r)-f(0))^{2}}{e^{r}}\cdot\frac{r^{-2k-1}}{2k+1}dr$
$\displaystyle=$
$\displaystyle\frac{2}{2k+1}\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho+\frac{2}{2k+1}\int_{0}^{\infty}\frac{(f(r)-f(0))^{2}}{re^{r}}dr$
$\displaystyle\leq$
$\displaystyle\frac{4}{2k+1}\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho.$

Thus, we obtain that

$\displaystyle\mathcal{\widetilde{K}}$ $\displaystyle\leq$
$\displaystyle\sum^{\infty}_{k=0}\frac{a_{2k}}{2^{2k-1}}\cdot\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}(1-e^{-\frac{\rho}{2}})d\rho$
(3.21) $\displaystyle\ \ \ \
+\sum^{\infty}_{k=0}\Big{(}\frac{a_{2k}}{2^{2k}}+\frac{4a_{2k}}{2k+1}\Big{)}\cdot\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho$ $\displaystyle\leq$
$\displaystyle\sum^{\infty}_{k=0}\frac{2a_{2k}}{2k+1}\cdot\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}(1-e^{-\frac{\rho}{2}})d\rho$
$\displaystyle\ \ \ \
+\sum^{\infty}_{k=0}\frac{5a_{2k}}{2k+1}\cdot\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho$

From (2.5) and (3.17), we can bound $\mathcal{\widetilde{J}}$ as

$\displaystyle\mathcal{\widetilde{J}}$ $\displaystyle=$
$\displaystyle\sum^{\infty}_{k=1}2ka_{2k}\iint\limits_{0<\rho<r}\Big{[}e^{-\rho}-e^{-r}\Big{(}\frac{\rho}{r}\Big{)}^{n-1}\Big{]}\frac{(f(\rho)-f(r))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ $\displaystyle\leq$
$\displaystyle\sum^{\infty}_{k=1}2ka_{2k}\underbrace{\iint\limits_{0<\rho<r}e^{-\rho}(r-\rho)\frac{(f(\rho)-f(r))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{\widetilde{J}}_{1}}$ $\displaystyle\
+(n-1)\sum^{\infty}_{k=1}2ka_{2k}\underbrace{\iint\limits_{0<\rho<r}e^{-\rho}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(r))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{\widetilde{J}}_{2}}.$

Similar to (3.19), $\mathcal{\widetilde{J}}_{1}$ can be estimated as, for any
$k\geq 1$,

$\displaystyle\mathcal{\widetilde{J}}_{1}$ $\displaystyle\leq$ $\displaystyle
4\|f\|^{2}_{L^{\infty}}\iint\limits_{0<\rho<r}e^{-\rho}\frac{r-\rho}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ (3.22) $\displaystyle=$ $\displaystyle
4\|f\|^{2}_{L^{\infty}}\int^{\infty}_{0}e^{-\rho}\rho^{2k}\Big{(}\int^{\infty}_{\rho}\frac{1-\frac{\rho}{r}}{r^{2k+1}}dr\Big{)}d\rho$
$\displaystyle=$ $\displaystyle
4\|f\|^{2}_{L^{\infty}}\Big{(}\frac{1}{2k}-\frac{1}{2k+1}\Big{)}\int^{\infty}_{0}e^{-\rho}d\rho$
$\displaystyle=$
$\displaystyle\frac{2\|f\|^{2}_{L^{\infty}}}{k(2k+1)}<\frac{1}{k^{2}}\|f\|^{2}_{L^{\infty}}.$

We split the term $\mathcal{\widetilde{J}}_{2}$ into two parts in the
following way

$\displaystyle\mathcal{\widetilde{J}}_{2}$ $\displaystyle=$
$\displaystyle\underbrace{\iint\limits_{0<\rho\leq\frac{r}{2}}e^{-\rho}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(r))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{\widetilde{J}}_{21}}$ $\displaystyle\ \ \
+\underbrace{\iint\limits_{\frac{r}{2}<\rho<r}e^{-\rho}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(r))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr}_{\mathcal{\widetilde{J}}_{22}}.$

We estimate the positive term $\mathcal{\widetilde{J}}_{21}$ as, for any
$k\geq 1$,

$\displaystyle\mathcal{\widetilde{J}}_{21}$ $\displaystyle\leq$
$\displaystyle\Big{(}\frac{1}{2}\Big{)}^{2k-1}\iint\limits_{0<\rho\leq\frac{r}{2}}e^{-\rho}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(r))^{2}}{r^{2}}\cdot\frac{\rho}{r}d\rho
dr$ $\displaystyle=$
$\displaystyle\Big{(}\frac{1}{2}\Big{)}^{2k-1}\iint\limits_{0<\rho\leq\frac{r}{2}}e^{-\rho}\Big{(}1-\frac{\rho}{r}\Big{)}\frac{(f(\rho)-f(r))^{2}}{r^{2}}d\rho
dr$ $\displaystyle\leq$
$\displaystyle\frac{1}{k^{2}}\iint\limits_{0<\rho\leq\frac{r}{2}}e^{-\rho}\frac{(f(\rho)-f(r))^{2}}{r^{2}}d\rho
dr$ $\displaystyle\leq$
$\displaystyle\frac{2}{k^{2}}\iint\limits_{0<\rho\leq\frac{r}{2}}e^{-\rho}\frac{(f(\rho)-f(0))^{2}}{r^{2}}d\rho
dr+\frac{2}{k^{2}}\iint\limits_{0<\rho\leq\frac{r}{2}}e^{-\rho}\frac{(f(0)-f(r))^{2}}{r^{2}}d\rho
dr$ $\displaystyle=$
$\displaystyle\frac{1}{k^{2}}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\rho}}d\rho+\frac{2}{k^{2}}\int^{\infty}_{0}\frac{(f(0)-f(r))^{2}}{r^{2}}(1-e^{-\frac{r}{2}})dr.$

We further estimate the term $\mathcal{\widetilde{J}}_{22}$ in the following
way,

$\displaystyle\mathcal{\widetilde{J}}_{22}$ $\displaystyle\leq$ $\displaystyle
2\iint\limits_{\frac{r}{2}<\rho<r}e^{-\frac{\rho}{2}}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(\rho)-f(0))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ $\displaystyle\ \ \ \ \ \
+2\iint\limits_{\frac{r}{2}<\rho<r}e^{-\frac{r}{2}}\Big{(}\frac{r}{\rho}-1\Big{)}\frac{(f(0)-f(r))^{2}}{r^{2}}\Big{(}\frac{\rho}{r}\Big{)}^{2k}d\rho
dr$ $\displaystyle=$ $\displaystyle
2\int_{0}^{\infty}e^{-\frac{\rho}{2}}\rho^{2k}(f(\rho)-f(0))^{2}\Big{(}\int^{2\rho}_{\rho}\frac{\frac{r}{\rho}-1}{r^{2k+2}}dr\Big{)}d\rho$
$\displaystyle\
+2\int_{0}^{\infty}e^{-\frac{r}{2}}\frac{(f(r)-f(0))^{2}}{r^{2k+2}}\Big{[}\int^{r}_{\frac{r}{2}}\rho^{2k}\Big{(}\frac{r}{\rho}-1\Big{)}d\rho\Big{]}dr$
$\displaystyle=$ $\displaystyle
4\Big{[}\frac{1-2^{-2k}}{2k}-\frac{1-2^{-2k-1}}{2k+1}\Big{]}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho$ $\displaystyle\leq$ $\displaystyle
4\Big{[}\frac{1}{2k}-\frac{1}{2k+1}\Big{]}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho$ $\displaystyle\leq$
$\displaystyle\frac{1}{k^{2}}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho.$

Thus, we obtain that, for any $k\geq 1$,

$\displaystyle\mathcal{\widetilde{J}}_{2}\leq\frac{2}{k^{2}}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho+\frac{2}{k^{2}}\int^{\infty}_{0}\frac{(f(0)-f(r))^{2}}{r^{2}}(1-e^{-\frac{r}{2}})dr.$

which along with (3.22) implies that

$\displaystyle\mathcal{\widetilde{J}}$ $\displaystyle\leq$ $\displaystyle
4(n-1)\sum^{\infty}_{k=1}\frac{a_{2k}}{k}\cdot\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho+2\sum^{\infty}_{k=1}\frac{a_{2k}}{k}\cdot\|f\|^{2}_{L^{\infty}}$
$\displaystyle\ \ \
+4(n-1)\sum^{\infty}_{k=1}\frac{a_{2k}}{k}\cdot\int^{\infty}_{0}\frac{(f(0)-f(r))^{2}}{r^{2}}(1-e^{-\frac{r}{2}})dr.$

It follows from (3.21) and (3.20) that

$\displaystyle\mathcal{I}$ $\displaystyle\geq$
$\displaystyle-\Big{[}4(n-1)\sum^{\infty}_{k=1}\frac{a_{2k}}{k}+\sum^{\infty}_{k=0}\frac{5a_{2k}}{2k+1}\Big{]}\int^{\infty}_{0}\frac{(f(\rho)-f(0))^{2}}{\rho
e^{\frac{\rho}{2}}}d\rho-2\sum^{\infty}_{k=1}\frac{a_{2k}}{k}\cdot\|f\|^{2}_{L^{\infty}}$
$\displaystyle\
-\Big{[}4(n-1)\sum^{\infty}_{k=1}\frac{a_{2k}}{k}+\sum^{\infty}_{k=0}\frac{2a_{2k}}{2k+1}\Big{]}\cdot\int_{0}^{\infty}\frac{(f(\rho)-f(0))^{2}}{\rho^{2}}(1-e^{-\frac{\rho}{2}})d\rho.$

Taking

$\displaystyle\epsilon=\frac{\alpha
B(\frac{1}{2},\frac{n+1}{2})}{2\Big{[}4(n-1)\displaystyle\sum^{\infty}_{k=1}\frac{a_{2k}}{k}+\displaystyle\sum^{\infty}_{k=0}\frac{5a_{2k}}{2k+1}\Big{]}}$

in the second inequality of Lemma 2.3 and another

$\displaystyle\epsilon=\frac{\alpha
B(\frac{1}{2},\frac{n+1}{2})}{2\Big{[}4(n-1)\displaystyle\sum^{\infty}_{k=1}\frac{a_{2k}}{k}+\displaystyle\sum^{\infty}_{k=0}\frac{2a_{2k}}{2k+1}\Big{]}}$

in the third inequality of Lemma 2.3, it follows from (3.12) that the desired
inequality (3.11). We then complete the proof of Proposition 3.2.

##  4\. Proof of Theorem 1.1

With Proposition 3.2 in hand, we are now ready to prove Theorem 1.1.

Proof of Theorem 1.1. We will argue by contradiction. Assume that the solution
$\theta$ to (1.1) starting from the initial data
$\theta_{0}\in\mathcal{S}(\mathbb{R}^{n})$, the Schwartz class, satisfying
(1.5) exists for all time. For our purpose, we introduce a quantity $J(t)$
given by

$\displaystyle
J(t)\triangleq\int_{\mathbb{R}^{n}}\frac{\theta(0,t)-\theta(x,t)}{|x|^{n}}e^{-|x|}dx.$
(4.1)

By Hölder’s inequality and Lemma 2.1, we have that

$\displaystyle|J(t)|$ $\displaystyle\leq$
$\displaystyle\|\nabla\theta(t)\|_{L^{\infty}}\int_{|x|\leq
1}\frac{dx}{|x|^{n-1}}+2\|\theta_{0}\|_{L^{\infty}}\int_{|x|>1}\frac{e^{-|x|}}{|x|^{n}}dx$
$\displaystyle=$
$\displaystyle\omega_{n-1}\|\nabla\theta\|_{L^{\infty}}+2\omega_{n-1}\Big{(}\int^{\infty}_{1}\frac{r}{e^{r}}dr\Big{)}\|\theta_{0}\|_{L^{\infty}}<+\infty,$

which shows that $J(t)$ is finite for any $t>0$. Next we prove that $J(t)$
will blow up at some finite time $T_{0}>0$ and then obtain a contradiction.

By Lemma 2.2, we know that the velocity at the origin is $0$, that is,

$\displaystyle u(0,t)$ $\displaystyle=$ $\displaystyle
c_{n,\alpha}P.V.\int_{\mathbb{R}^{n}}\frac{\nabla\theta(x,t)}{|x|^{n-2+2\alpha}}dx$
$\displaystyle=$ $\displaystyle
c_{n,\alpha}P.V.\int_{\mathbb{R}^{n}}\frac{x\theta^{\prime}(|x|,t)}{|x|^{n-1+2\alpha}}dx=0,$

which along with (1.1), Proposition 3.2 and the maximum principle
$\|\theta(t)\|_{L^{\infty}}\leq\|\theta_{0}\|_{L^{\infty}}$ implies that

$\displaystyle J^{\prime}(t)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}\frac{\Lambda^{-2+2\alpha}\nabla\theta\cdot\nabla\theta}{|x|^{n}}e^{-|x|}dx$
$\displaystyle\geq$ $\displaystyle
C^{\prime}_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{(\theta(0,t)-\theta(x,t))^{2}}{|x|^{n+2\alpha}}dx-C^{\prime\prime}_{n,\alpha}\|\theta_{0}\|^{2}_{L^{\infty}}.$

By (4.1) and Hölder’s inequality, we note that

$\displaystyle|J(t)|$ $\displaystyle\leq$
$\displaystyle\Big{(}\int_{\mathbb{R}^{n}}\frac{|\theta(0,t)-\theta(x,t)|^{2}}{|x|^{n+2\alpha}}dx\Big{)}^{\frac{1}{2}}\Big{(}\int_{\mathbb{R}^{n}}\frac{e^{-2|x|}}{|x|^{n-2\alpha}}dx\Big{)}^{\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\frac{\omega_{n-1}\Gamma(2\alpha)}{2^{2\alpha}}\Big{(}\int_{\mathbb{R}^{n}}\frac{|\theta(0,t)-\theta(x,t)|^{2}}{|x|^{n+2\alpha}}dx\Big{)}^{\frac{1}{2}},$

which gives that

$\displaystyle\int_{\mathbb{R}^{n}}\frac{|\theta(0,t)-\theta(x,t)|^{2}}{|x|^{n+2\alpha}}dx\geq\frac{2^{4\alpha}[J(t)]^{2}}{\omega^{2}_{n-1}\Gamma^{2}(2\alpha)}.$

Therefore, we derive that

$\displaystyle
J^{\prime}(t)\geq\frac{2^{4\alpha}C^{\prime}_{n,\alpha}}{\omega^{2}_{n-1}\Gamma^{2}(2\alpha)}[J(t)]^{2}-C^{\prime\prime}_{n,\alpha}\|\theta_{0}\|^{2}_{L^{\infty}}.$
(4.2)

Now, choosing sufficiently large $A(n,\alpha)>0$, by (1.5) and (4.2), we know
$J^{\prime}(0)>0$ and can derive that the differential inequality of the form

$\displaystyle J^{\prime}(t)\geq C[J(t)]^{2},$

for some constant $C>0$. Then $J$ goes to infinity in finite time. We have
obtained the desired contradiction and the proof of Theorem 1.1 is then
completed.

Acknowledgements. W. Zhang was partially supported by the Science and
Technology Research Project of Department of Education of Jiangxi Province
(No. GJJ2200363).

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