# Strongly damped wave equations with mass-like
terms of the logarithmic-Laplacian

Alessandra Piske and Ruy Coimbra Charão
Graduate Program in Pure and Applied Mathematics
Department of Mathematics
Federal University of Santa Catarina
88040-900, Florianopolis, Brazil,
and
Ryo Ikehata
Department of Mathematics, Division of Educational Sciences
Graduate School of Humanities and Social Sciences
Hiroshima University
Higashi-Hiroshima 739-8524, Japan<EMAIL_ADDRESS><EMAIL_ADDRESS>author<EMAIL_ADDRESS>
###### Abstract

We consider strongly damped wave equations with logarithmic mass-like terms
with a parameter $\theta\in(0,1]$. This research is a part of a series of wave
equations that was initiated by Charão-Ikehata [6], Charão-D’Abbicco-Ikehata
considered in [5] depending on a parameter $\theta\in(1/2,1)$ and Piske-
Charão-Ikehata [26] for small parameter $\theta\in(0,1/2)$. We derive a
leading term as $t\to\infty$ of the solution, and by using it, a growth and a
decay property of the solution itself can be precisely studied in terms of
$L^{2}$-norm. An interesting aspect appears in the case of $n=1$, roughly
speaking, a small $\theta$ produces a diffusive property, and a large $\theta$
gives a kind of singularity, expressed by growth rates.

000Keywords and Phrases: Strongly damped wave equation; logarithmic-mass;
$L^{2}$-decay; asymptotic profile; optimal estimates, growth rates.0002020
Mathematics Subject Classification. Primary 35B40; 35L05; Secondary 35B20,
35R12, 35S05.

##  1 Introduction

We consider in this work strongly damped wave equations under effects of a
logarithmic mass-like term dependng on a parameter $\theta$:

$\displaystyle u_{tt}-\Delta u+m^{2}L_{\theta}u-\Delta u_{t}=0,\quad t>0,\quad
x\in{\bf R}^{n},$ (1.1) $\displaystyle u(0,x)=u_{0}(x),\quad
u_{t}(0,x)=u_{1}(x),\quad x\in{\bf R}^{n},$ (1.2)

where $\theta\in(0,1]$ and $m>0$. In this connection, $\theta=0$ case is
already considered by D’Abbicco-Ikehata [12], so one should restrict the
parameter $\theta$ to the case $\theta>0$. In this sense, this study is a kind
of generalization of [12] to the general $\theta\in(0,1)$ through the
logarithmic Laplacian type of mass-like term. Asymptotic behavior of the
solution and its optimal decay/growth property has been already discussed in
[7, 8, 26] in the case when the logarithmic Laplacian with various physical
sense can be included in the equation itself, and it motivates us to study
(1.1) as one more topic.

We first define several function spaces related to the logarithmic Laplace
operator $L_{\theta}$ introduced by Charão-Ikehata [6] (for $\theta=1$) and
Charão-D’Abbicco-Ikehata [5] such that for $\delta\geq 0$

$\displaystyle Y_{\theta}^{\delta}:=\\{f\in L^{2}({\bf R}^{n}):\int_{{\bf
R}_{\xi}^{n}}(1+\log(1+|\xi|^{2\theta}))^{\delta}|\hat{f}(\xi)|^{2}d\xi<\infty\\}$
(1.3)

with its natural norm

$\|f\|_{Y_{\theta}^{\delta}}:=\left(\int_{{\bf
R}_{\xi}^{n}}(1+\log(1+|\xi|^{2\theta}))^{\delta}|\hat{f}(\xi)|^{2}d\xi\right)^{1/2},\quad
f\in Y_{\theta}^{\delta},$ (1.4)

and its corresponding inner product.

###### Remark 1.1

Due to the fact that $\log(1+|\xi|^{2\theta})\leq|\xi|$ for all $\xi\in{\bf
R}^{n}$ satisfying $|\xi|\gg 1$, one notices $H^{s}({\bf R}^{n})\subset
Y_{\theta}^{s}\subset L^{2}$ for $s\geq 0$.

The linear operator

$L_{\theta}:D(L_{\theta})\subset L^{2}({\bf R}^{n})\to L^{2}({\bf
R}^{n}),\quad\theta>0,$

is defined as follows:

$D(L_{\theta}):=\left\\{f\in L^{2}({\bf R}^{n})\,\bigm{|}\,\int_{{\bf
R}^{n}}(\log(1+|\xi|^{2\theta}))^{2}|\hat{f}(\xi)|^{2}d\xi<+\infty\right\\}=Y_{\theta}^{2},$

and for $f\in D(L_{\theta})$,

$(L_{\theta}f)(x):={\cal F}_{\xi\to
x}^{-1}\left(\log(1+|\xi|^{2\theta})\hat{f}(\xi)\right)(x).$

Here, one has just denoted the Fourier transform ${\cal F}_{x\to\xi}(f)(\xi)$
of $f(x)$ by

${\cal F}_{x\to\xi}(f)(\xi)=\hat{f}(\xi):=\displaystyle{\int_{{\bf
R}^{n}}}e^{-ix\cdot\xi}f(x)dx,\quad\xi\in{\bf R}^{n},$

as usual with $i:=\sqrt{-1}$, and ${\cal F}_{\xi\to x}^{-1}$ expresses its
inverse Fourier transform.

Since $L_{\theta}$ is constructed by a nonnegative-valued multiplication
operator it is nonnegative and self-adjoint in $L^{2}({\bf R}^{n})$ (see [6]),
the square root

$L_{\theta}^{1/2}:D(L_{\theta}^{1/2})\subset L^{2}({\bf R}^{n})\to L^{2}({\bf
R}^{n})$

can be defined, and is also nonnegative and self-adjoint with its domain

$D(L_{\theta}^{1/2})=\left\\{f\in L^{2}({\bf R}^{n})\,\bigm{|}\,\int_{{\bf
R}^{n}}\log(1+|\xi|^{2\theta})|\hat{f}(\xi)|^{2}d\xi<+\infty\right\\}=Y_{\theta}^{1}.$

Note that $Y_{\theta}^{1}$ becomes Hilbert space with its graph norm

$\|v\|_{D(L_{\theta}^{1/2})}:=\left(\|v\|^{2}+\|L_{\theta}^{1/2}v\|^{2}\right)^{1/2},$

where to simplify the notation we define the $L^{2}({\bf R}^{n})$-norm by

$\|\cdot\|:=\|\cdot\|_{L^{2}({\bf R}^{n})}.$

We also note that

$H^{s}({\bf R}^{n})\hookrightarrow Y_{\theta}^{1}\hookrightarrow L^{2}({\bf
R}^{n})$

for any $s>0$. Symbolically writing, one can see

$L_{\theta}=\log(I+(-\Delta)^{\theta}),$

where $\Delta$ is the usual Laplace operator defined on $H^{2}({\bf R}^{n})$.
For more detailed and recent studies concerning the logarithmic Laplacian, one
can cite [3, 4], [10], [23] and [28].

The associated energy identity is given by

$E_{u}(t)+\int_{0}^{t}\|\nabla u_{s}(s,\cdot)\|^{2}ds=E_{u}(0),\quad t>0,$
(1.5)

where

$E_{u}(t):=\frac{1}{2}\left(\|u_{t}(t,\cdot)\|_{L^{2}}^{2}+\|\nabla
u(t,\cdot)\|_{L^{2}}^{2}+m^{2}\|L_{\theta}^{1/2}u(t,\cdot)\|_{L^{2}}^{2}\right).$

The identity (1.5) implies that the total energy is a non-increasing function
in time because of the existence of the strong damping term $-\Delta u_{t}$ in
(1.1).

Let us recall several known historical facts related to our dissipative model
(1.1), which has double dispersion.
First of all, one should cite two celebrated works due to [25] and [27], which
studies the diffusive aspect and $L^{p}$-$L^{q}$ estimates of the solution to
the strongly damped wave equation:

$u_{tt}-\Delta u-\Delta u_{t}=0.$ (1.6)

After these works, so many precise analysis can be applied to (1.6) through
the papers due to [17, 18, 19, 20], where the authors investigate the
asymptotic profile and optimal decay and/or non-decay property of the solution
itself. Precise decay estimates and its application to nonlinear problem can
be discussed in [13]. $L^{1}$-estimates of the solution can be derived deeply
in [24]. Higer order asymptotic expansions of the solution itself and the
square of the $L^{2}$-norm of the solution as $t\to\infty$ can be well-
investigated in [1, 2] and [22]. Abstract theories concrning the regularity
and decay property of (1.6) can be constructed in [14], [20] and [21].

Next, a generalization of (1.6) to the so-called Klein-Gordon equation with
strong damping such that

$u_{tt}-\Delta u+m^{2}u-\Delta u_{t}=0$ (1.7)

can be considered in [12], and the effect of oscillation in the low frequency
part of the solution captured to (1.6) has been disappeared in the equation
(1.7). This implies that the additional mass-like term $m^{2}u$ makes the
structure of the solution itself change. Thus, it is quite natural to consider
the following generalization of (1.7):

$u_{tt}+(-\Delta)^{\sigma}u+m^{2}(-\Delta)^{\theta}u+(-\Delta)^{\delta}u_{t}=0,$
(1.8)

where $\sigma>0$, $\theta>0$ and $\delta>0$. To the best of authors’
knowledge, this type of equation is studied at least by two papers [11] nd
[9]. In fact, in the case of $m>0$, $\sigma>\theta>0$, and $\delta>0$ the
author in [11] studies several decay rates of the solution and the anomalous
diffusion phenomenon, while the authors in [9] treat the case of $m=1$,
$\sigma=2$, $\theta=1$, and $\delta=1$, and investigates $L^{p}$-$L^{q}$
estimates fo $1\leq p\leq q\leq\infty$ of the solution, and inviscid limits
for the first and the second order profiles together with some nonlinear
effect, however, it seems that these authors are not conscious to observe the
growth property of the solution itself at this stage.

In this connection, in the case when $m>0$, $\sigma=1$, $\theta=0$, and
$\delta=0$ in [29] the author studied the exponential decay of the
corresponding total energy $E_{m}(t)$ such that

$E_{m}(t):=\frac{1}{2}\left(\|u_{t}(t,\cdot)\|^{2}+\|\nabla
u(t,\cdot)\|^{2}+m^{2}\|u(t,\cdot)\|^{2}\right).$

In fact, the author of [29] considered the nonlinear problem in a more general
framework.

Under these previous developments it seems to be little investigations on the
topic to observe the growth property (as $t\to\infty$) of the solution itself.
Our main concern is to consider the equation (1.1) by replacing the mass-like
term $m^{2}(-\Delta)^{\theta}u$ of (1.8) with $\sigma=\delta=1$ by logarithmic
mass-like one $m^{2}\log(I+(-\Delta)^{\theta})u$ to soften the regularity of
the solution, and to capture the growth property of the solution itself coming
from an oscillation effect in the low frequency region of the solution. In
this situation, one wants to observe the effect of the mass-like term brought
by a new one $\log(I+(-\Delta)^{\theta})$ on the decay structure of the
equation. To catch information on the time behavior of the $L^{2}$-nom of the
solution itself is very important to observe some singularity (near low
frequency part) included in the solution itself to problem (1.1)-(1.2). Such a
singularity reflects on the quantity of the function $t\mapsto\|u(t,\cdot)\|$,
not on the time and/or spatial derivatives of the solutions such as total
energy.

Now, we introduce the asymptotic profile as $t\to\infty$ of the solution to
problem (1.1)-(1.2):

$\varphi(t,\xi)=P_{1}e^{-\frac{|\xi|^{2}}{2}t}\frac{\sin\big{(}t\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)}}{\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}},$
(1.9)

where the moment of the initial velocity $P_{1}\in{\bf R}$ is defined by

$P_{1}:=\int_{{\bf R}^{n}}u_{1}(x)dx.$

Then, our main result reads as follows.

###### Theorem 1.1

Let $n\geq 1$, and $0<\theta\leq 1$. Let $\varphi$ be the function defined in
(1.9). Choose $u_{0}=0$, and $u_{1}\in L^{1,\theta}({\bf R}^{n})\cap
L^{2}({\bf R}^{n})$. Then there exists a positive constant
$C=C_{n,\delta_{0},\theta}$ such that the mild solution $u\in
C([0,\infty);H^{1}({\bf R}^{n}))\cap C^{1}([0,\infty);L^{2}({\bf R}^{n}))$ to
problem (1.1)-(1.2) satisfies

$\displaystyle\int_{{\bf R}^{n}}|\hat{u}(t,\xi)-\varphi(t,\xi)|^{2}d\xi$
$\displaystyle\leq
C_{n,\delta_{0},\theta}\Big{(}m^{-2}P^{2}_{1}t^{-\frac{n+8-6\theta}{2}}+m^{-2}P^{2}_{1}t^{-\frac{n+4-4\theta}{2}}+m^{-2}||u_{1}||^{2}_{1,\theta}t^{-\frac{n}{2}}+P_{1}^{2}e^{-\gamma
t}$ $\displaystyle+||u_{1}||^{2}_{1}t^{2}e^{-\gamma
t}+||u_{1}||^{2}_{1}t^{2}e^{-\alpha t}+||u_{1}||^{2}_{1}t^{2}e^{-\beta
t}\Big{)},\quad t\geq 0,\quad$

where $\alpha>0$, $\beta>0$ and $\gamma>0$ are generous constants.

As a reult of Theorem 1.1, one can state the following optimal $L^{2}$-decay
and/or growth estimates of the solution itself to problem (1.1)-(1.2)

###### Theorem 1.2

Under the same assumptions as in Theorem 1.1 the following statements are
true:

  * (i)

for $n=1$ with $0<\theta<\frac{1}{2}$, $n=2$ with $0<\theta<1$ and $n\geq 3$
with $0<\theta\leq 1$, there exist positive constants $C_{1},C_{2}$ depending
on $n$, $\theta$ and $m>0$ such that

$C_{1}|P_{1}|t^{-\frac{n-2\theta}{4}}\leq\|u(t,\cdot)\|\leq
C_{2}\Big{(}\|u_{1}\|_{1}+\|u_{1}\|_{1,\theta}\Big{)}t^{-\frac{n-2\theta}{4}},\quad
t\gg 1,$

  * (ii)

for $n=1$ and $\frac{1}{2}<\theta\leq 1$, there exist positive constants
$C_{1},C_{2}$ such that

$\frac{C_{1}}{(1+m^{2})^{\frac{1}{4\theta}}}|P_{1}|t^{\frac{2\theta-1}{2\theta}}\leq\|u(t,\cdot)\|\leq\frac{C_{2}}{m}\frac{1}{\sqrt{2\theta-1}}\Big{(}\|u_{1}\|_{1}+\|u_{1}\|_{1,\theta}\Big{)}t^{\frac{2\theta-1}{2\theta}},\quad
t\gg 1,$

  * (iii)

for $n=1$ and $\theta=\frac{1}{2}$ or $n=2$ and $\theta=1$, there exist
positive constants $C_{1},C_{2}$ such that

$\frac{C_{1}}{\sqrt{2+m^{2}}}|P_{1}|\sqrt{\log
t}\leq\|u(t,\cdot)\|\leq\frac{C_{2}}{m}\Big{(}\|u_{1}\|_{1}+\|u_{1}\|_{1,\theta}\Big{)}\sqrt{\log
t},\quad t\gg 1.$

Let us reconsider the role of mass-like term from Theorem 1.2. The result of
(i) in Theorem 1.2 implies the decay estimate, and this comes from the
stronger effect with decay order $t^{-\frac{n}{4}}$ of the Gauss kernel
$e^{-t|\xi|^{2}/2}$ in the low frequency zone, and the oscillation effect
comimg from the mass-like term $m^{2}\log(1+(-\Delta)^{\theta})u\sim
m^{2}(-\Delta)^{\theta}u$ gives a small role with growth order
$t^{\frac{\theta}{2}}$. Totally, one can get the optimal decay rate
$t^{-\frac{n}{4}+\frac{\theta}{2}}=t^{-\frac{n-2\theta}{4}}$. Recall that in
case of $n\geq 3$ the optimal decay order of the $L^{2}$-norm of the solution
to (1.6) is $t^{-\frac{n-2}{2}}$ (see [17]). In this case, one can see that
$\frac{n-2}{2}<\frac{n-2\theta}{2}$ for $\theta\in[0,1]$. In particular, in
the case of $\theta=0$, the decay order coincides with that derived in [12],
and the result corresponds to the strongly damped Klein-Gordon equation (1.7).
Thus, one can understand that the additional term
$m^{2}\log(1+(-\Delta)^{\theta})u$ plays a role to make the decay rate more
fast. In this sense, the name of ”mass-like” term seems suitable. While, (ii)
of Theorem 1.2 reflects a stronger effect with growth order $t$ of the
oscillation factor in the low frequency part:

$\frac{\sin\big{(}t\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)}}{\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}}\sim\frac{\sin(mtr^{\theta})}{mr^{\theta}}\quad(r\to+0),$

and the diffusion part is no-effective. Roughly observing, for $n=1$ one can
see that

$\left(\int_{0}^{1}\frac{\sin^{2}(mtr^{\theta})}{(mr^{\theta})^{2}}dr\right)^{1/2}\sim
t\left(\int_{0}^{m^{-\frac{1}{\theta}}t^{-\frac{1}{\theta}}}dr\right)^{1/2}\sim
t\cdot t^{-\frac{1}{2\theta}}\quad(t\to\infty),$

since $|\displaystyle{\frac{\sin\eta}{\eta}}|\sim 1$ when $\eta\to+0$. As a
result one can get infinite time blow up rate $t\cdot
t^{-\frac{1}{2\theta}}=t^{\frac{2\theta-1}{2\theta}}$ when $t\to\infty$ as
shown in (ii). In some sense, in (ii) of Theorem 1.2 the dominant part in the
low frequency zone as $t\to\infty$ of the equation (1.1) will be

$u_{tt}+m^{2}(-\Delta)^{\theta}u=0.$

(iii) of Thorem 1.2 is just an intermediate case between (i) and (ii). These
reflect a removable singularity in the low frequency zone included in the
solution itself. This kind of growth property as shown in (ii) and (iii) has
been newly discovered to the equation (1.1) itself. In this connection, a
regularity aspect which corresponds to high frequency asymptotics of the
solution to (1.1) can be dominated by (1.6) even if $\theta>0$ is large enough
(cf., [9]).

Notation. Throughout this paper, $\|\cdot\|_{q}$ stands for the usual
$L^{q}({\bf R}^{n})$-norm. For simplicity of notation, in particular, we use
$\|\cdot\|$ instead of $\|\cdot\|_{2}$. By $(\cdot,\cdot)$ we mean the usual
$L^{2}$-inner product. Furthermore, we denote $\|\cdot\|_{H^{l}}$ as the usual
$H^{l}({\bf R}^{n})$-norm. We also define a relation $f(t)\sim g(t)$ as
$t\to\infty$ by: there exist constant $C_{j}>0$ ($j=1,2$) such that

$C_{1}g(t)\leq f(t)\leq C_{2}g(t)\quad(t\gg 1).$

For $\Omega\subset{\bf R}^{n}$ we denote $f\approx g$ on $\Omega$, if and only
if there are constants $K_{1},K_{2}$ such that

$K_{1}f(y)\leq g(y)\leq K_{2}f(y),\;\;\mbox{for all}\;y\in\Omega.$

We also introduce the following weighted functional spaces for $\gamma>0$:

$L^{1,\gamma}({\bf R}^{n}):=\left\\{f\in L^{1}({\bf
R}^{n})\;\bigm{|}\;\|f\|_{L^{1,\gamma}}:=\int_{{\bf
R}^{n}}(1+|x|^{\gamma})|f(x)|dx<+\infty\right\\}.$

${\rm Re}z$ and ${\rm Im}z$ denote the real and the imaginary parts of $z$,
respectively. Finally, we denote the surface area of the $n$-dimensional unit
ball by $\omega_{n}:=\displaystyle{\int_{|\omega|=1}}d\omega$.

##  2 Basic preliminary results

In this section we shall collect important lemmas to derive precise estimates
of the several quantities related to the solution to problem (1.1)-(1.2).
These are already studied and developed in our previous works (see [6, 5]).

The following two lemmas are well known.

###### Lemma 2.1

Let $k>-n$, $\alpha>0$ and $C>0$. Then there exists a constant $K>0$ depending
only on $n,\alpha,k$ such that

$\displaystyle\int_{|\xi|\leq
1}e^{-C|\xi|^{\alpha}t}|\xi|^{k}d\xi=\omega_{n}\int_{0}^{1}e^{-Cr^{\alpha}t}r^{n-1+k}dr\leq
Kt^{-\frac{n+k}{\alpha}},$

for all $t\geq 1$.

###### Lemma 2.2

$\frac{\sinh x}{x}\leq e^{x}$

for all $x>0$.

One also needs the following decomposition of the Fourier transform of a
function $f\in L^{1}({\bf R}^{n})$ as follows:

$\hat{f}(\xi)=A_{f}(\xi)-iB_{f}(\xi)+P_{f},$ (2.1)

for all $\xi\in{\bf R}^{n}$ , where

  * $\bullet$

$A_{f}(\xi)=\displaystyle{\int_{{\bf R}^{n}}}{(\cos(x\cdot\xi)-1)f(x)}dx,$

  * $\bullet$

$B_{f}(\xi)=\displaystyle{\int_{{\bf R}^{n}}}{\sin(x\cdot\xi)f(x)}dx,$

  * $\bullet$

$P_{f}=\displaystyle{\int_{{\bf R}^{n}}}{f(x)}dx.$

Then, the next lemma has been prepared in [16] (see Notation for the
definition of $L^{1,\kappa}({\bf R}^{n})$).

###### Lemma 2.3

  * i)

If $f\in L^{1}({\bf R}^{n})$ then for all $\xi\in{\bf R}^{n}$ it is true that

$|A_{f}(\xi)|\leq L\|f\|_{L^{1}}\quad\text{ and }\quad|B_{f}(\xi)|\leq
N\|f\|_{L^{1}}.$

  * ii)

If $0<\kappa\leq 1$ and $f\in L^{1,\kappa}({\bf R}^{n})$ then for all
$\xi\in{\bf R}^{n}$ it is true that

$|A_{f}(\xi)|\leq K|\xi|^{\kappa}\|f\|_{L^{1,\kappa}}\quad\text{ and
}\quad|B_{f}(\xi)|\leq M|\xi|^{\kappa}\|f\|_{L^{1,\kappa}}$

with $L$, $N$, $K$ and $M$ positive constants depending only on the dimension
$n$ or $n$ and $\kappa$.

In order to justify the existence of a unique solution to problem (1.1)-(1.2)
we prepare the following theorem (cf. [15, Theorem 6.4]).

###### Theorem 2.1

Let $X$ be a Hilbert space and $A\subset X:D(A)\rightarrow X$ be a generator
of a $C^{0}$-semigroup in $X$. If a linear operator $B:X\rightarrow X$ is
bounded, then the operator $A+B$ generates a $C^{0}$-semigroup on $X$.

##  3 Existence and Uniqueness

The total energy associated to the equation (1.1) is defined by

$\displaystyle E(t):=\frac{\|u_{t}\|^{2}+\|\nabla
u\|^{2}+\|L_{\theta}^{\frac{1}{2}}u\|^{2}}{2},\quad t\geq 0,$

and it satisfies the following energy identity

$\displaystyle\frac{d}{dt}E(t)+\|\nabla u_{t}\|^{2}=0,\quad t>0.$

So it is natural to consider the following energy space

$\displaystyle X=(H^{1}({\bf R}^{n})\cap D(L_{\theta}^{1/2}))\times L^{2}.$
(3.1)

Here, one can observe that $H^{1}({\bf R}^{n})\subset D(L_{\theta}^{1/2}))$
since $0\leq\theta\leq 1$, and due to the facts that
$0\leq\log(1+r^{2\theta})\leq\log 2<1$ for $0\leq r\leq 1$, and because of
$\log(1+r^{2\theta})\leq r^{2}$ for $r\geq 1$. Thus, in the case
$0\leq\theta\leq 1$ the energy space can be sufficiently chosen as

$\displaystyle X=H^{1}({\bf R}^{n})\times L^{2}({\bf R}^{n}).$ (3.2)

We remember that the space $H^{1}({\bf R}^{n})$ is a Hilbert space with the
natural norm given by

$\|f\|_{H^{1}}:=\left(\int_{{\bf
R}^{n}}(1+|\xi|)^{2}|\hat{f}(\xi)|^{2}d\xi\right)^{\frac{1}{2}},$

but we may observe that this norm is equivalent to the following norm

$\||f\||_{{H^{1}}}:=\left(\int_{{\bf
R}^{n}}(1+m^{2}\log(1+|\xi|^{2\theta})+|\xi|^{2})|\hat{f}(\xi)|^{2}d\xi\right)^{\frac{1}{2}},\quad
f\in H^{1}({\bf R}^{n}),$

since $0\leq\theta\leq 1$. This second norm can be defined by the following
inner product

$((f,g))_{H^{1}}:=\int_{{\bf
R}^{n}}(1+m^{2}\log(1+|\xi|^{2\theta})+|\xi|^{2})\hat{f}(\xi)\overline{\hat{g}(\xi)}d\xi.$
(3.3)

Therefore, in this section we consider the Hilbert space $H^{1}({\bf R}^{n})$
with the inner product $((\cdot,\cdot))_{H^{1}}$ just defined in (3.3). In
order to prove the existence and uniqueness, we reduce the original equation
(1.1) to the first order system as usual.
We set $v=u_{t}$, and $U=\begin{pmatrix}u\\\ v\end{pmatrix}$ and, we have (at
least formally)

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}U$
$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}u\\\
v\end{pmatrix}=\frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}u\\\
u_{t}\end{pmatrix}=\begin{pmatrix}u_{t}\\\ \Delta u_{t}+\Delta
u-m^{2}L_{\theta}u\end{pmatrix}$ $\displaystyle=\begin{pmatrix}v\\\
\Delta(v+u)-m^{2}L_{\theta}u\end{pmatrix}$
$\displaystyle=\begin{pmatrix}0&I\\\ \Delta-m^{2}L_{\theta}-I&\Delta\\\
\end{pmatrix}\begin{pmatrix}u\\\ v\end{pmatrix}+\begin{pmatrix}0\\\
u\end{pmatrix}$ $\displaystyle=:BU+FU,$ (3.4)

where

$\displaystyle BU:=\begin{pmatrix}v\\\
\Delta(u+v)-m^{2}L_{\theta}u-u\end{pmatrix}.$ (3.5)

for $U:=(u,v)\in D(B)$, and

$\displaystyle FU=\begin{pmatrix}0\\\ u\end{pmatrix}$ (3.6)

for $U:=(u,v)\in X$. In this case, we consider the domain $D(B)$ of the linear
operator $B$ as

$\displaystyle D(B)=\\{(u,v)\in H^{1}({\bf R}^{n})\times H^{1}({\bf
R}^{n})\;\;\mbox{such that }u+v\in H^{2}({\bf R}^{n})\\}.$ (3.7)

This idea is coming from [20], and it is crucial to prove Lemma 3.2. The
domain $D(B)$ is dense in the energy space $X=H^{1}({\bf R}^{n})\times
L^{2}({\bf R}^{n})$.

Let us see that the linear operator $B$ is dissipative in $X$, and satisfies
$(I-B)(D(B))=X$.

###### Lemma 3.1

The linear operator $B$ is dissipative in $X$.

Proof. In the space $X=H^{1}({\bf R}^{n})\times L^{2}({\bf R}^{n})$, we
consider the natural inner product

$\left((u_{1},v_{1}),(u_{2},v_{2})\right)_{X}=((u_{1},u_{2}))_{H^{1}}+(v_{1},v_{2})$

for $(u_{j},v_{j})\in H^{1}({\bf R}^{n})\times L^{2}({\bf R}^{n})$, ($j=1,2$),
where $((\cdot,\cdot))_{H^{1}}$ is given by (3.3) and $(\cdot,\cdot)$ denotes
the usual inner product in $L^{2}$.

Let $(u,v)\in D(B)$. Then, the definition of the equivalent inner product
(3.3) in $H^{1}({\bf R}^{n})$ implies that

$\displaystyle\left(B(u,v),(u,v)\right)_{X}$
$\displaystyle=\left((v,\Delta(u+v)-m^{2}L_{\theta}u-u),(u,v)\right)_{X}$
$\displaystyle=((v,u))_{H^{1}}+(\Delta(u+v)-m^{2}L_{\theta}u-u,v)$
$\displaystyle=\int_{{\bf
R}^{n}}(1+m^{2}\log(1+|\xi|^{2\theta})+|\xi|^{2})\hat{v}\overline{\hat{u}}d\xi$
$\displaystyle-\int_{{\bf
R}^{n}}(1+m^{2}\log(1+|\xi|^{2\theta})+|\xi|^{2})\hat{u}\overline{\hat{v}}d\xi-\int_{{\bf
R}^{n}}|\xi|^{2}|\hat{v}|^{2}d\xi$ $\displaystyle=2i\int_{{\bf
R}^{n}}(1+m^{2}\log(1+|\xi|^{2\theta})+|\xi|^{2})\text{Im}(\hat{v}\overline{\hat{u}})d\xi-\int_{{\bf
R}^{n}}|\xi|^{2}|\hat{v}|^{2}d\xi.$

Thus, one can get

$\text{Re}\left(B(u,v),(u,v)\right)_{X}=-\int_{{\bf
R}^{n}}|\xi|^{2}|\hat{v}|^{2}d\xi\leq 0,$

and this concludes the desired statement. $\Box$

###### Lemma 3.2

$(I-B)(D(B))=X$.

Proof. At first, the inclusion $(I-B)D(B)\subset X$ is easily checked. So, it
suffices to check $X\subset(I-B)(D(B))$. For this, let $(f,g)\in X=H^{1}({\bf
R}^{n})\times L^{2}({\bf R}^{n})$. We need to show that there exists a pair
$(u,v)\in D(B)$ that satisfies the following two identities

$\displaystyle\;u-v=f,$ (3.8)
$\displaystyle-\Delta(u+v)+m^{2}L_{\theta}u+u+v=g.$ (3.9)

Since we need to find $u,v$ such that $v=u-f$ and $u$ satisfying (3.9) it is
sufficient to obtain $u\in H^{1}$ such that

$\displaystyle-2\Delta u+2u+m^{2}L_{\theta}u=g-\Delta f+f.$ (3.10)

Note that $g-\Delta f+f\in H^{-1}({\bf R}^{n})$ at this stage. Then we may
apply the Lax-Milgram Theorem to obtain the existence of a unique function
$u\in H^{1}({\bf R}^{n})$ satisfying (3.10) in weak sense. Since $v=u-f$ and
$f\in H^{1}({\bf R}^{n})$ it follows that $(u,v)\in H^{1}({\bf R}^{n})\times
H^{1}({\bf R}^{n})$.

Now from (3.9) we have

$-\Delta(u+v)+u+v=g-m^{2}L_{\theta}u\in L^{2}({\bf R}^{n})$

due to the assumption on $g$ and $u\in H^{1}({\bf R}^{n})$. Thus, the elliptic
regularity theorem says that $u+v\in H^{2}({\bf R}^{n})$. Therefore $(u,v)\in
D(B)$ and the operator $I-B$ is surjective. $\Box$

###### Theorem 3.1

$B:D(B)\rightarrow X$ is an infinitesimal generator of a $C^{0}$-semigroup of
contractions.

Proof. It follows from the facts that $D(B)$ is dense in $X$, $B$ is
dissipative according to Lemma 3.1, and $R(I-B)=X$ by Lemma 3.2. The result
follows from the Lumer-Phillips theorem.

$\Box$

In order to conclude the existence and uniqueness of solutions to problem
(1.1)-(1.2), we observe that the operator $F$ given in (3.6) with domain

$D(F)=H^{1}({\bf R}^{n})\times L^{2}({\bf R}^{n})$

satisfies the following fact.

###### Lemma 3.3

$F:X\rightarrow X$ is a bounded linear operator.

Proof. Clearly the operator $F$ is linear. Let $(u,v)\in X=H^{1}({\bf
R}^{n})\times L^{2}({\bf R}^{n})$. Then,

$\displaystyle\|F(u,v)\|_{X}$
$\displaystyle=\|(0,u)\|_{X}=\|u\|\leq\|(u,v)\|_{X},\quad(u,v)\in X.$

Therefore, the operator $F$ is bounded in $X$. $\Box$

Since $B:D(B)\rightarrow X$ is an infinitesimal generator of a
$C^{0}$-semigroup of contractions and $F:D(F)\rightarrow X$ is a bounded
linear operator with $D(B)\subset D(F)=X$, we may conclude the following
result from Theorem 2.1.

###### Theorem 3.2

$B+F:D(B)(\subset X)\rightarrow X$ is an infinitesimal generator of a
$C^{0}$-semigroup in $X$.

As a result, one can arrive at the following theorem concerning the existence
and uniqueness to problem (1.1)-(1.2).

###### Theorem 3.3

Let $n\geq 1$ and $(u_{0},u_{1})\in D(B)$, with $D(B)$ given by (3.7) Then the
problem (1.1)-(1.2) admits a unique solution $u$ in the class

$C^{1}([0,\infty),H^{1}({\bf R}^{n}))\cap C^{2}([0,\infty),L^{2}({\bf
R}^{n})).$

Moreover, for initial data $(u_{0},u_{1})\in H^{1}({\bf R}^{n})\times
L^{2}({\bf R}^{n})$ the problem admits a unique weak solution in the class

$C([0,\infty),H^{1}({\bf R}^{n}))\cap C^{1}([0,\infty),L^{2}({\bf R}^{n})).$

##  4 Cauchy problem in the Fourier space

The associated Cauchy problem in the Fourier space to problem (1.1)-(1.2), is
given by

$\displaystyle\hat{u}_{tt}+|\xi|^{2}\hat{u}+m^{2}\log(1+|\xi|^{2\theta})\hat{u}+|\xi|^{2}\hat{u}_{t}=0,\quad
t>0,\quad\xi\in{\bf R}^{n}$ (4.1)
$\displaystyle\hat{u}(0,\xi)=\hat{u}_{0},\quad\hat{u}_{t}(0,\xi)=\hat{u}_{1},\quad\xi\in{\bf
R}^{n}.$ (4.2)

The characteristic equation associated to problem (4.1)–(4.2) is given by

$\lambda^{2}+|\xi|^{2}\lambda+(|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta}))=0,$

and its characteristics roots are

$\lambda_{\pm}=\frac{-|\xi|^{2}\pm\sqrt{|\xi|^{4}-4(|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta}))}}{2}.$

We need to analyse the sign of the expression
$|\xi|^{4}-4(|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta}))$, for $\xi\in{\bf R}^{n}$
to decide whether the characteristics roots are real or complex.

To do that we consider the function

$f(r)=r^{4}-4r^{2}-4m^{2}\log(1+r^{2\theta}),\quad r\geq 0,$ (4.3)

where $\theta\geq 0$. We set

$g(r):=r^{2}(r^{2}-4),\quad h(r):=k\log(1+r^{2\theta}),$

where $k:=4m^{2}>0$. Then, one has

$f(r)=g(r)-h(r),$

and $g(0)=g(2)=0$, $\displaystyle{\lim_{r\to\infty}}g(r)=\infty$, $h(0)=0$,
and if $r\in[0,2]$, then $g(r)\leq 0$, and $g(r)>0$ in case of $r>2$.
Furthermore, $g(r)$ is monotone increasing on $r\in(\sqrt{2},\infty)$, and is
monotone decreasing on $[0,\sqrt{2}]$. Furthermore, $h(r)$ is monotone
(slowly) increasing on $r\in[0,\infty)$. By these observations on the drown
graphs of such two functions $g(r)$ and $h(r)$, the graphs on $r>0$ of $g(r)$
and $h(r)$ have only one common point $\delta>2$, so that such only one point
$\delta>2$ plays a role to satisfy the following property:

$\displaystyle
f(r)=g(r)-h(r)=r^{4}-4r^{2}-4m^{2}\log(1+r^{2\theta}))<0,\quad\mbox{for}\;\;0<r<\delta,$
$\displaystyle f(r)=g(r)-h(r)=r^{4}-4r^{2}-4m^{2}\log(1+r^{2\theta}))\geq
0,\quad\mbox{for}\;\;r\geq\delta.$ (4.4)

Note that thse arguments above hold good for each fixed $\theta\geq 0$, and
the uniquely determined constant $\delta>2$ depends on each fixed $\theta\geq
0$. Therefore, we conclude that the characteristics roots are complex on the
zone $\\{|\xi|<\delta\\}$ and are given by

$\lambda_{\pm}=a(\xi)\pm ib(\xi),\quad i=\sqrt{-1},$

where

$\displaystyle a(\xi)=-\frac{|\xi|^{2}}{2},\quad
b(\xi)=\frac{\sqrt{4|\xi|^{2}+4m^{2}\log(1+|\xi|^{2\theta})-|\xi|^{4}}}{2},$
(4.5)

and they are real on the region $\\{|\xi|\geq\delta\\}$, and can be defined by

$\lambda_{\pm}=a(\xi)\pm d(\xi),$ $\displaystyle
d(\xi)=\frac{\sqrt{|\xi|^{4}-4|\xi|^{2}-4m^{2}\log(1+|\xi|^{2\theta})}}{2}.$
(4.6)

###  4.1 Solution formula for low frequency and asymptotic profile

At this point, without loss of generality, we may assume that the initial data
$u_{0}=0$.

Then, the solution formula to the Cauchy problem (4.1)–(4.2) for low
frequencies

$\displaystyle\hat{u}(t,\xi)=e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}\hat{u}_{1}+\frac{a(\xi)}{b(\xi)}e^{-a(\xi)t}\sin(b(\xi)t)\hat{u}_{0}+e^{-a(\xi)t}\cos(b(\xi)t)\hat{u}_{0}$

becomes

$\displaystyle\hat{u}(t,\xi)=e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}\hat{u}_{1},\;t\geq
0,$ (4.7)

where $a(\xi)$ and $b(\xi)$ are given in (4.5) for $0\leq\theta$.

Our interest in this section is to find a candidate to be an asymptotic
profile of the solution (4.7).

In that direction we consider the elementary decomposition of the initial data
$\hat{u}_{1}$ such that

$\hat{u}_{1}(\xi)=A(\xi)-iB(\xi)+P_{1},$

where

$A(\xi)=\int_{{\bf R}^{n}}u_{1}(x)(\cos(\xi x)-1)dx,\quad B(\xi)=\int_{{\bf
R}^{n}}u_{1}(x)\sin(\xi x)dx,\quad P_{1}=\int_{{\bf R}^{n}}u_{1}(x)dx.$ (4.8)

Thus,

$\displaystyle\hat{u}(t,\xi)=P_{1}e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}+\big{(}A(\xi)-iB(\xi)\big{)}e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}.$

Now, we apply the mean value theorem to write that

$\displaystyle\sin(b(\xi)t)$
$\displaystyle=\sin\big{(}t\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)}$
$\displaystyle+t\big{(}b(\xi)-\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)}\cos(\mu(\xi)t)$

with
$\mu(\xi)=(1-\eta)b(\xi)+\eta\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}$
for some $0<\eta<1$, $t>0$. The above expression for $\sin(b(\xi)t)$ implies
that

$\displaystyle\hat{u}(t,\xi)$
$\displaystyle=P_{1}e^{-a(\xi)t}\frac{\sin\big{(}t\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)}}{b(\xi)}$
$\displaystyle+tP_{1}e^{-a(\xi)t}\frac{b(\xi)-\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}}{b(\xi)}\cos(\mu(\xi)t)$
$\displaystyle+\big{(}A(\xi)-iB(\xi)\big{)}e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}.$

Now, in order to obtain a suitable expression of an asymptotic profile we
observe that

$\frac{1}{b(\xi)}=\frac{1}{\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}}+R(|\xi|),$

where

$R(|\xi|)=\frac{|\xi|^{4}}{{2}b(\xi)\big{(}|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})\big{)}\left(2+\sqrt{4-\frac{|\xi|^{4}}{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}}\right)}$
(4.9)

for $0<|\xi|<\delta$.

Then one can rewrite $\hat{u}(t,\xi)$ as

$\displaystyle\hat{u}(t,\xi)=\varphi(t,\xi)+F_{1}(t,\xi)+F_{2}(t,\xi)+F_{3}(t,\xi),$
(4.10)

where

$\displaystyle F_{1}(t,\xi)=$ $\displaystyle
P_{1}e^{-a(\xi)t}R(|\xi|)\sin\big{(}t\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)},$
$\displaystyle F_{2}(t,\xi)=$ $\displaystyle
tP_{1}e^{-a(\xi)t}\frac{b(\xi)-\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}}{b(\xi)}\cos(\mu(\xi)t),$
(4.11) $\displaystyle F_{3}(t,\xi)=$
$\displaystyle\big{(}A(\xi)-iB(\xi)\big{)}e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)},$

and the function $\varphi(t,\xi)$ is the asymptotic profile, which can be
given by

$\varphi(t,\xi)=P_{1}e^{-a(\xi)t}\frac{\sin\big{(}t\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}\big{)}}{\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}}.$
(4.12)

Next we need to estimate the $L^{2}$-norms of the remainder terms
$F_{i}(t,\xi),i=1,2,3,$ on the zone of low frequency $\\{|\xi|\leq\delta\\}$.

Before beginning to estimate them, in order to avoid a singularity of
$|\xi|^{4}/b(\xi)=|\xi|^{4}/b(|\xi|)$ in $|\xi|=\delta$ we need to separate
the zone of low frequency into two parts. To do that we note the limit

$\lim_{r\rightarrow+0}\frac{r^{\theta}}{b(r)}=\lim_{r\rightarrow+0}\frac{r^{\theta}}{\sqrt{4r^{2}+4m^{2}\log(1+r^{2\theta})-r^{4}}}<+\infty.$

Thus one can take a small number $\delta_{0},\;0<\delta_{0}<\delta$ such that

$0\leq\frac{r^{\theta}}{b(r)}=\frac{r^{\theta}}{\sqrt{4r^{2}+4m^{2}\log(1+r^{2\theta})-r^{4}}}\leq
C$ (4.13)

for all $|\xi|\leq\delta_{0}$, $0\leq\theta$, where $C$ is a positive constant
depending only on $\theta$ and $\delta_{0}$ because $b(r)>0$ on the interval
$(0,\delta_{0}]$. Therefore we need to consider seeral estimates on the low
frequency zone $\\{|\xi|\leq\delta_{0}\\}$, and on the intermediate frequency
zone, that is, the middle zone $\\{\delta_{0}\leq|\xi|\leq\delta\\}$.

Now, we first consider $F_{1}$-estimates in the low frequency zone
$|\xi|\leq\delta_{0}$.

$\displaystyle\int_{|\xi|\leq\delta_{0}}|F_{1}(t,\xi)|^{2}d\xi$
$\displaystyle\leq\int_{|\xi|\leq\delta_{0}}|P_{1}|^{2}e^{-2a(\xi)t}|R(|\xi|)|^{2}d\xi$
$\displaystyle=\omega_{n}P_{1}^{2}\int_{0}^{\delta_{0}}r^{n-1}e^{-tr^{2}}|R(r)|^{2}dr.$
(4.14)

Here, we need to get an estimate to $|R(r)|$. According to its definition in
(4.9) and the estimate (4.13) it follows that for $|\xi|\leq\delta_{0}$,

$\displaystyle|R(r)|$
$\displaystyle=\dfrac{r^{4}}{2\big{|}b(r)\big{|}\big{(}r^{2}+m^{2}\log(1+r^{2\theta})\big{)}\left(2+\sqrt{4-\dfrac{r^{4}}{r^{2}+m^{2}\log(1+r^{2\theta})}}\right)}$
$\displaystyle\leq
C\dfrac{r^{4-\theta}}{\big{(}r^{2}+m^{2}\log(1+r^{2\theta})\big{)}\left(2+\sqrt{4-\dfrac{r^{4}}{r^{2}+m^{2}\log(1+r^{2\theta})}}\right)}$
$\displaystyle\leq
C\dfrac{r^{4-\theta}}{2\big{(}r^{2}+m^{2}\log(1+r^{2\theta})\big{)}},$ (4.15)

because $\sqrt{4-\dfrac{r^{4}}{r^{2}+m^{2}\log(1+r^{2\theta})}}>0$ for
$|\xi|\leq\delta_{0}$, where $C>0$ is a generous constant (if necessarily).

Combining the estimates (4.1) and (4.1) we arrive at

$\displaystyle\int_{|\xi|\leq\delta_{0}}|F_{1}(t,\xi)|^{2}d\xi$
$\displaystyle=C^{2}\omega_{n}P_{1}^{2}\int_{0}^{\delta_{0}}r^{n-1}e^{-tr^{2}}\dfrac{r^{8-2\theta}}{4\big{(}r^{2}+m^{2}\log(1+r^{2\theta})\big{)}^{2}}dr$
$\displaystyle\leq
C^{2}\omega_{n}P_{1}^{2}\int_{0}^{\delta_{0}}e^{-tr^{2}}\dfrac{r^{n+7-2\theta}}{4m^{4}\log^{2}(1+r^{2\theta})}dr$
$\displaystyle\leq
C_{n,\delta_{0},\theta}\frac{P_{1}^{2}}{m^{2}}\int_{0}^{\delta_{0}}e^{-tr^{2}}r^{n+7-6\theta}dr$
$\displaystyle\leq
C_{n,\delta_{0},\theta}\frac{P_{1}^{2}}{m^{2}}t^{-\frac{n+8-6\theta}{2}},\quad
t>0,$ (4.16)

due to Lemma 2.1, and the fact that the function
$\dfrac{r^{4\theta}}{\log^{2}(1+r^{2\theta})}$ is bounded in the interval
$(0,\delta_{0}]$ because
$\displaystyle{\lim_{r\rightarrow+0}}\dfrac{r^{2\theta}}{\log(1+r^{2\theta})}$
is finite. The constant $C_{n,\delta_{0},\theta}$ depends only on $n$ ,
$\delta_{0}$ and $\theta$.

A next step is to estimate $F_{2}(t,\xi)$.

$\displaystyle\int_{|\xi|\leq\delta_{0}}|F_{2}(t,\xi)|^{2}d\xi$
$\displaystyle=\int_{|\xi|<\delta_{0}}t^{2}P_{1}^{2}e^{-2a(\xi)t}\frac{|b(\xi)-\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}|^{2}}{|b(\xi)|^{2}}\cos^{2}(\mu(\xi)t)d\xi$
$\displaystyle\leq
t^{2}P_{1}^{2}\int_{|\xi|<\delta_{0}}e^{-t|\xi|^{2}}\frac{|b(\xi)-\sqrt{|\xi|^{2}+m^{2}\log(1+|\xi|^{2\theta})}|^{2}}{|b(\xi)|^{2}}d\xi$
$\displaystyle\leq\omega_{n}t^{2}P_{1}^{2}\int_{0}^{\delta_{0}}r^{n-1}e^{-tr^{2}}\frac{\big{|}b(r)-\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}\big{|}^{2}}{|b(\xi)|^{2}}dr.$
(4.17)

To estimate the last integral above we note that

$|b(\xi)-\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}|=\frac{r^{4}}{4b(\xi)+4\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}}\leq\frac{r^{4}}{\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}}$

for $0<r\leq\delta_{0}$ because $b(\xi)$ is positive on this interval. Then,
as in (4.1) based on the estimate (4.13) we may conclude that

$\frac{|b(r)-\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}|}{|b(r)|}\leq\frac{Cr^{4-\theta}}{\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}}\leq\frac{C}{m}\frac{r^{4-\theta}}{\sqrt{\log(1+r^{2\theta})}}\leq\frac{C}{m}r^{4-2\theta}$

for $0<|\xi|\leq\delta_{0}$, with $C>0$ depending only on $\delta_{0}$ and
$\theta$. Combining the above estimate with (4.1) one can obtain

$\displaystyle\int_{|\xi|\leq\delta_{0}}|F_{2}(t,\xi)|^{2}d\xi$
$\displaystyle\leq\frac{C^{2}}{m^{2}}\omega_{n}t^{2}P_{1}^{2}\int_{0}^{\delta_{0}}r^{n-1}e^{-tr^{2}}r^{8-4\theta}dr$
$\displaystyle\leq
C_{n,\delta_{0},\theta}\;\omega_{n}m^{-2}t^{2}P_{1}^{2}t^{-\frac{n+8-4\theta}{2}}$
$\displaystyle\leq
C_{n,\delta_{0},\theta}\;\omega_{n}m^{-2}P_{1}^{2}\;t^{-\frac{n+4-4\theta}{2}}.$
(4.18)

###### Remark 4.1

Due to the estimate (4.1), to get a suitable decay rate for the function
$F_{2}(t,\xi)$ it suffices to constrain $\theta$, at least, to the interval
$0<\theta\leq 1$ (see [12] for the case $\theta=0$).

A step now is to estimate the function $F_{3}(t,\xi)$ on the low frequency
zone $|\xi|\leq\delta_{0}$. Applying Lemma 2.3 with
$\kappa=\theta,\;0\leq\theta\leq 1$, we have

$\displaystyle\int_{|\xi|\leq\delta_{0}}|F_{3}(t,\xi)|^{2}d\xi$
$\displaystyle=\int_{|\xi|<\delta_{0}}\Big{|}\Big{(}A(\xi)-iB(\xi)\Big{)}e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}\Big{|}^{2}d\xi$
$\displaystyle\leq\int_{|\xi|\leq\delta_{0}}\big{|}A(\xi)-iB(\xi)\big{|}^{2}e^{-t|\xi|^{2}}\frac{1}{b(\xi)^{2}}d\xi$
$\displaystyle\leq\int_{|\xi|\leq\delta}\big{(}|A(\xi)|+|B(\xi)|\big{)}^{2}e^{-t|\xi|^{2}}\frac{1}{b(\xi)^{2}}d\xi$
(4.19)
$\displaystyle\leq(K+M)^{2}\|u_{1}\|_{1,\theta}^{2}\int_{|\xi|\leq\delta_{0}}e^{-t|\xi|^{2}}\frac{|\xi|^{2\theta}}{b(\xi)^{2}}d\xi$
$\displaystyle=(K+M)^{2}\omega_{n}\|u_{1}\|_{1,\theta}^{2}\int_{0}^{\delta_{0}}e^{-tr^{2}}\frac{4r^{2\theta}}{4r^{2}+4m^{2}\log(1+r^{2\theta})-r^{4}}r^{n-1}dr.$

Since $0<\theta\leq 1$, we have

$\displaystyle\lim_{r\rightarrow
0}\frac{4r^{2\theta}}{4r^{2}+4m^{2}\log(1+r^{2\theta})-r^{4}}=\lim_{r\rightarrow
0}\frac{4r^{2\theta}}{r^{2\theta}\left(4r^{2-2\theta}+4m^{2}\frac{\log(1+r^{2\theta})}{r^{2\theta}}-r^{4-2\theta}\right)}=\frac{1}{m^{2}}.$

Then, due to $\delta_{0}<\delta$, there exists a constant $C>0$ such that

$\displaystyle\Big{|}\frac{4r^{2\theta}}{4r^{2}+4m^{2}\log(1+r^{2\theta})-r^{4}}\Big{|}\leq\frac{C}{m^{2}},\quad
0<r\leq\delta.$ (4.20)

Using estimate (4.20) in (4.1), we obtain with the help of Lemma 2.1 the
following estimate

$\displaystyle\int_{|\xi|\leq\delta_{0}}|F_{3}(t,\xi)|^{2}d\xi$
$\displaystyle\leq
C(K+M)^{2}m^{-2}\omega_{n}\|u_{1}\|_{1,\theta}^{2}\int_{0}^{1}e^{-tr^{2}}r^{n-1}dr$
$\displaystyle\leq C_{n}m^{-2}\|u_{1}\|_{1,\theta}^{2}\;t^{-\frac{n}{2}}$
(4.21)

with $C_{n}>0$ a constant depending on the space dimension $n$.

##  5 Estimate for middle frequency $\delta_{0}\leq|\xi|<\delta$

In this zone, the solution is given by

$\displaystyle\hat{u}(t,\xi)=e^{-a(\xi)t}\frac{\sin(b(\xi)t)}{b(\xi)}\hat{u}_{1},\;t\geq
0.$

We remember that $b(\xi)=b(|\xi|)=0$ for $|\xi|=\delta$. But the singularity
of $\hat{u}(t,\xi)$ in $|\xi|=\delta$ is removable. In fact, we know that
$|\sin a|\leq a$ for all $a\geq 0$, thus one can get

$\displaystyle\int_{\delta_{0}\leq|\xi|<\delta}|\hat{u}(t,\xi)|^{2}d\xi$
$\displaystyle=\int_{\delta_{0}\leq|\xi|<\delta}e^{-2a(\xi)t}\frac{\sin^{2}(b(\xi)t)}{b(\xi)^{2}}|\hat{u}_{1}(\xi)|^{2}d\xi$
$\displaystyle\leq
t^{2}\int_{\delta_{0}\leq|\xi|\leq\delta}e^{-2|\xi|^{2}t}|\hat{u}_{1}(\xi)|^{2}d\xi$
$\displaystyle\leq
t^{2}e^{-2\delta_{0}^{2}t}\int_{\delta_{0}\leq|\xi|\leq\delta}|\hat{u}_{1}(\xi)|^{2}d\xi$
$\displaystyle\leq t^{2}e^{-2\delta_{0}^{2}\;t}\|u_{1}\|^{2},\quad t>0,$ (5.1)

that is, the solution $\hat{u}(t,\xi)$ decays exponentially on the middle
zone. This estimate holds for $0\leq\theta\leq 2$.

###  5.1 Estimate on high frequency $|\xi|\geq\delta$

In this zone the characteristics roots are real and given for $0\leq\theta$ by

$\lambda_{\pm}=\frac{-|\xi|^{2}\pm\sqrt{|\xi|^{4}-4|\xi|^{2}-4m^{2}\log(1+|\xi|^{2\theta})}}{2},$

and the solution formula in the high frequency region is

$\displaystyle\hat{u}(t,\xi)=e^{-a(\xi)t}\frac{\sinh(b(\xi)t)}{b(\xi)}\hat{u}_{1},$
(5.2)

where

$\displaystyle a(\xi)=\frac{|\xi|^{2}}{2},\quad
b(\xi)=\frac{\sqrt{|\xi|^{4}-4|\xi|^{2}-4m^{2}\log(1+|\xi|^{2\theta})}}{2}.$
(5.3)

We recall that $\delta>0$ is such that $f(\delta)=0$ ($\delta>2$ is the unique
positive zero of $f(r)$) and $f(r)$ is a increasing positive function for
$r>\delta$ defined by

$f(r)=r^{4}-4r^{2}-4m^{2}\log(1+r^{2\theta}).$

Then there exists $\delta_{1}>\delta$ such that

$\displaystyle
0<\sqrt{1-\frac{4}{\delta_{1}^{2}}-4\frac{m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}<1,$
(5.4)

and, in particular,

$\displaystyle\sqrt{1-\frac{4}{\delta_{1}^{2}}-4\frac{m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}\geq\sqrt{1-\frac{4}{r^{2}}-4\frac{m^{2}}{r^{4}}\log(1+r^{2\theta})}\geq
0$ (5.5)

for $\delta\leq r\leq\delta_{1}.$

Now we are able to estimate the $L^{2}$-norm of $\hat{u}$ on the region of
high frequency. We do that into two steps. The first step is to estimate on
the middle frequency zone $\\{\delta\leq|\xi|\leq\delta_{1}\\}$:

$\displaystyle\int_{\delta\leq|\xi|\leq\delta_{1}}|\hat{u}(t,\xi)|^{2}d\xi$
$\displaystyle\leq\int_{\delta\leq|\xi|\leq\delta_{1}}|\hat{u}_{1}(t,\xi)|^{2}e^{-2ta(\xi)}\frac{\sinh^{2}(b(\xi)t)}{b^{2}(\xi)}d\xi$
$\displaystyle\leq\|u_{1}\|_{1}^{2}\int_{\delta\leq|\xi|\leq\delta_{1}}e^{-t|\xi|^{2}}\frac{\sinh^{2}(b(\xi)t)}{b^{2}(\xi)}d\xi$
$\displaystyle\leq\|u_{1}\|_{1}^{2}\int_{\delta\leq|\xi|\leq\delta_{1}}e^{-t|\xi|^{2}}t^{2}e^{2b(\xi)t}d\xi$
$\displaystyle\leq\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta}^{\delta_{1}}\omega_{n}r^{n-1}e^{-tr^{2}}e^{2b(r)t}dr$
$\displaystyle\leq\omega_{n}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta}^{\delta_{1}}r^{n-1}e^{-tr^{2}}e^{\sqrt{r^{4}-4r^{2}-4m^{2}\log(1+r^{2\theta})}t}dr$
$\displaystyle\leq\omega_{n}\delta_{1}^{n-1}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta}^{\delta_{1}}e^{-tr^{2}\big{(}1-{\sqrt{1-\frac{4}{r^{2}}-4\frac{m^{2}}{r^{4}}\log(1+r^{2\theta})}\big{)}}}dr,$

which holds for any space dimensions $n\geq 1$ and $t>0$. In the above
estimates we have just used the elementary inequality Lemma 2.2.

By using (5.5) in the last estimate above, one can get

$\displaystyle\int_{\delta\leq|\xi\leq\delta_{1}}|\hat{u}(t,\xi)|^{2}d\xi$
$\displaystyle\leq\omega_{n}\delta_{1}^{n-1}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta}^{\delta_{1}}e^{-tr^{2}\big{(}1-{\sqrt{1-\frac{4}{\delta_{1}^{2}}-4\frac{m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}\big{)}}}dr$
$\displaystyle\leq\omega_{n}\delta_{1}^{n-1}\|u_{1}\|_{1}^{2}\;t^{2}(\delta_{1}-\delta)e^{-t\delta^{2}\big{(}1-{\sqrt{1-\frac{4}{\delta_{1}^{2}}-4\frac{m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}\big{)}}}$
$\displaystyle\leq C_{n}\|u_{1}\|_{1}^{2}\;t^{2}e^{-\alpha t},\quad t>0,$
(5.6)

where
$\alpha=\delta^{2}\big{(}1-{\sqrt{1-\frac{4}{\delta_{1}^{2}}-4\frac{m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}}\big{)}>0$
(due to (5.4)), and $C_{n}$ a positive constant depending only on $n$ and
$\theta$ because $\delta$ depends on $\theta$.

Now, to complete the estimate on the zone of high frequency we proceed with
the second step, that is, we estimate $\hat{u}(t,\cdot)$ on
$|\xi|\geq\delta_{1}$. Then, because of Lemma 2.2 one can obtain

$\displaystyle\int_{|\xi\geq\delta_{1}}$
$\displaystyle|\hat{u}(t,\xi)|^{2}d\xi\leq\int_{|\xi|\geq\delta_{1}}|\hat{u}_{1}(t,\xi)|^{2}e^{-2ta(\xi)}\frac{\sinh^{2}(b(\xi)t)}{b^{2}(\xi)}d\xi$
$\displaystyle\leq\|u_{1}\|_{1}^{2}\int_{|\xi|\geq\delta_{1}}e^{-2t|\xi|^{2}}\;t^{2}e^{2b(\xi)t}d\xi$
$\displaystyle\leq\omega_{n}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta_{1}}^{\infty}r^{n-1}e^{-tr^{2}}e^{t\sqrt{r^{4}-4r^{2}-4m^{2}\log(1+r^{2\theta})}}dr$
$\displaystyle\leq\omega_{n}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta_{1}}^{\infty}r^{n-1}e^{-tr^{2}\big{(}1-\sqrt{1-\frac{4}{r^{2}}-\frac{4m^{2}}{r^{4}}\log(1+r^{2\theta})}\big{)}}dr$
$\displaystyle\leq\omega_{n}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta_{1}}^{\infty}r^{n-1}e^{-\frac{tr^{2}}{2}\big{(}1-\sqrt{1-\frac{4}{r^{2}}-\frac{4m^{2}}{r^{4}}\log(1+r^{2\theta})}\big{)}}e^{-\frac{tr^{2}}{2}\big{(}1-\sqrt{1-\frac{4}{r^{2}}-\frac{4m^{2}}{r^{4}}\log(1+r^{2\theta})}\big{)}}dr$
$\displaystyle\leq\omega_{n}\|u_{1}\|_{1}^{2}\;t^{2}\int_{\delta_{1}}^{\infty}r^{n-1}e^{-\frac{r^{2}}{2}\big{(}1-\sqrt{1-\frac{4}{r^{2}}-\frac{4m^{2}}{r^{4}}\log(1+r^{2\theta})}\big{)}}e^{-\frac{{\bf
t}\delta_{1}^{2}}{2}\big{(}1-\sqrt{1-\frac{4}{\delta_{1}^{2}}-\frac{4m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}\big{)}}dr$
$\displaystyle\leq\omega_{n}\|u_{1}\|_{1}^{2}\;t^{2}\Big{(}\int_{\delta_{1}}^{\infty}r^{n-1}e^{-\frac{r^{2}}{2}\big{(}1-\sqrt{1-\frac{4}{{}^{2}}-\frac{4m^{2}}{r^{4}}\log(1+r^{2\theta})}}\big{)}dr\Big{)}e^{-\frac{{\bf
t}\delta_{1}^{2}}{2}\big{(}1-\sqrt{1-\frac{4}{\delta_{1}^{2}}-\frac{4m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}\big{)}}$
$\displaystyle\leq\omega_{n}C_{n,\theta}\|u_{1}\|_{1}^{2}\;t^{2}e^{-\beta t},$
(5.7)

which holds for $t\geq 1$. The constant
$\beta=\frac{\delta_{1}^{2}}{2}\big{(}1-{\sqrt{1-\frac{4}{\delta_{1}^{2}}-4\frac{m^{2}}{\delta_{1}^{4}}\log(1+\delta_{1}^{2\theta})}}\big{)}$
is a positive constant (due to (5.4)) depending only on $n$ and $\theta$
because $\delta_{1}$ depends on $\theta$.

The estimates (5.1) and (5.1) prove the exponential decay of the $L^{2}$-norm
of $\hat{u}$ on the zone of high frequency.

##  6 Estimates to the asymptotic profile $\varphi(t,\xi)$

In this section one shall study the $L^{2}$-estimates of the profile
$\varphi(t,x)$ itself in order to get sharp $L^{2}$-estimates of the solution
itself. This is important to observe a singularity included in the solution
$u(t,x)$. Such a singularity reflects on a “growth rate” as $t\to\infty$ of
the quantity $\|u(t,\cdot)\|$ as stated in Theorem 1.2.

###  6.1 Decay estimates

In this subsection we present several estimates on the asymptotic profile of
the solution to problem (1.1)–(1.2).

To begin with, for low frequency parameter $|\xi|\leq 1$ one has

$\displaystyle\int_{|\xi|\leq\delta_{0}}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=P_{1}^{2}\int_{0}^{\delta_{0}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}r^{n-1}dr$
$\displaystyle\leq
P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{r^{2\theta}}{r^{2}+m^{2}\log(1+r^{2\theta})}r^{n-1-2\theta}dr,$

where $\delta_{0}>0$ is a number already defined in previous section 4 (see
(4.13)). Since $0<\theta\leq 1$, we have

$\displaystyle\lim_{r\rightarrow+0}\frac{r^{2\theta}}{r^{2}+m^{2}\log(1+r^{2\theta})}=\lim_{r\rightarrow+0}\frac{r^{2\theta}}{r^{2\theta}\left(r^{2-2\theta}+m^{2}\frac{\log(1+r^{2\theta})}{r^{2\theta}}\right)}=\frac{1}{m^{2}}$

so that there is a constant $C>0$ such that

$\frac{r^{2\theta}}{r^{2}+m^{2}\log(1+r^{2\theta})}\leq\frac{C}{m^{2}},\quad
0<r\leq 1.$

Thus, we conclude that

$\displaystyle\int_{|\xi|\leq\delta_{0}}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle\leq\frac{C}{m^{2}}P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}r^{n-1-2\theta}dr.$

Applying Lemma 2.1 the next statement holds.

###### Lemma 6.1

Let $0<\theta\leq 1$ such that $n-2\theta>0$. Then, it holds that

$\displaystyle\int_{|\xi|\leq\delta_{0}}|\varphi(t,\xi)|^{2}d\xi\leq
C_{n,\theta}\frac{P_{1}^{2}}{m^{2}}t^{-\frac{n-2\theta}{2}},\quad t\gg 1.$
(6.1)

###### Remark 6.1

The decay estimate (6.1) holds in the following cases:

  * (i)

$n=1$ with $0<\theta<\frac{1}{2}$;

  * (ii)

$n=2$ with $0<\theta<1$;

  * (iii)

$n\geq 3$ and $0<\theta\leq 1$.

Note that the cases $n=1$ with $\theta\in[\frac{1}{2},1]$ and and $n=2$ with
$\theta=1$ are not considered in the classification above. These missing cases
are studied in the next Subsection 6.2.

The effect on the decay estimates of the function $\varphi(t,\xi)$ in the zone
of high frequency is exponentially small according to the next lemma.

###### Lemma 6.2

There exists a positive constant $C_{\delta_{0},n,\theta}$, $0\leq\theta\leq
1$, such that

$\displaystyle\int_{|\xi|\geq\delta_{0}}|\varphi(t,\xi)|^{2}d\xi\leq
C_{\delta_{0},n,\theta}P_{1}^{2}\;e^{-\frac{\delta_{0}^{2}}{2}\;t},\quad t\geq
1.$ (6.2)

Proof.

$\displaystyle\int_{|\xi|\geq\delta_{0}}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=\omega_{n}P^{2}_{1}\int_{\delta_{0}}^{\infty}e^{-r^{2}t}r^{n-1}\frac{\sin^{2}\big{(}{t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}}\big{)}}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle\leq\omega_{n}P^{2}_{1}\int_{\delta_{0}}^{\infty}e^{-r^{2}t}r^{n-1}\frac{1}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle\leq\omega_{n}P^{2}_{1}\int_{\delta_{0}}^{\infty}e^{-r^{2}t}r^{n-1}\frac{1}{\delta_{0}^{2}}dr$
$\displaystyle\leq\omega_{n}\frac{P^{2}_{1}}{\delta_{0}^{2}}\;e^{-\frac{\delta_{0}^{2}}{2}t}\int_{\delta_{0}}^{\infty}e^{-\frac{r^{2}}{2}}r^{n-1}dr,\quad
t\geq 1.$

This estimate proves the lemma because the last integral is finite. $\Box$

###### Remark 6.2

The Lemmas 6.1 and 6.2 imply that

$\displaystyle\int_{{\bf R}^{n}}|\varphi(t,\xi)|^{2}d\xi\leq
C_{n,\theta,\delta_{0}}\frac{P_{1}^{2}}{m^{2}}t^{-\frac{n-2\theta}{2}},\quad
t>0,$ (6.3)

with $C_{n,\theta,\delta_{0}}$ a positive constant independent of $t$.

The estimate obtained in (6.3) may be optimal, but to check it, one needs more
precise estimates from below. The lemma below is one of main difficulties to
be proved in the paper.

###### Lemma 6.3

For $0<\theta\leq 1$ with $n-2\theta>0$, there exist positive constants
$C_{n,m,\theta}$ and $t_{0}$ such that

$\displaystyle\int_{{\bf R}^{n}}|\varphi(t,\xi)|^{2}d\xi\geq
C_{n,m,\theta}P_{1}^{2}t^{-\frac{n-2\theta}{{2}}},\quad t\geq t_{0}.$ (6.4)

Proof. In order to capture the precise estimate from below we need to estimate
the $L^{2}$-norm of the profile only on the low frequency zone due to that on
the high frequency the $L^{2}$-norm of the profile decays exponential,
according to Lemma (6.2).

We first consider the change of variable
$w=\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}$. Then

$dw=\frac{r+\theta
m^{2}\frac{r^{2\theta-1}}{1+r^{2\theta}}}{\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}}dr,$

and

$\displaystyle\int_{|\xi|\leq 1}$
$\displaystyle|\varphi(t,\xi)|^{2}d\xi=\omega_{n}P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}r^{n-1}dr$
$\displaystyle=\omega_{n}P_{1}^{2}\int_{0}^{1}\frac{e^{-r^{2}t}r^{n-1}\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}\Big{(}r+\theta
m^{2}\frac{r^{2\theta-1}}{1+r^{2\theta}}\Big{)}}\frac{r+\theta
m^{2}\frac{r^{2\theta-1}}{1+r^{2\theta}}}{\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}}dr.$

For $0\leq r\leq 1$ and $0\leq\theta\leq 1$, we have

$\displaystyle\frac{m^{2}}{2}r^{2\theta}\leq
w^{2}=r^{2}+m^{2}\log(1+r^{2\theta})\leq(1+m^{2})r^{2\theta},$ (6.5)

that implies the important equivalence $r\approx w^{\frac{1}{\theta}}$, for
$0\leq r\leq 1$. We also notice that

$r+\theta m^{2}\frac{r^{2\theta-1}}{1+r^{2\theta}}\approx r^{2\theta-1},\quad
0<r\leq 1,\quad 0\leq\theta\leq 1.$

Thus, the above change of variable give us the estimate

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle\approx\omega_{n}P_{1}^{2}\int_{0}^{\eta}e^{-w^{\frac{2}{\theta}}t}\frac{\sin^{2}(tw)}{w}w^{\frac{n-2\theta}{\theta}}dw$
$\displaystyle=\omega_{n}P_{1}^{2}\int_{0}^{\eta}e^{-w^{\frac{2}{\theta}}t}\sin^{2}(tw)w^{\frac{n-2\theta}{\theta}-1}dw$
$\displaystyle=\omega_{n}P_{1}^{2}\int_{0}^{\eta}e^{-w^{\frac{2}{\theta}}t}\sin^{2}\Big{(}t^{1-\frac{\theta}{2}}t^{\frac{\theta}{2}}w\Big{)}w^{\frac{n-3\theta}{\theta}}dw,$

where $\eta=\sqrt{1+m^{2}\log 2}$.

Now we set $y=t^{\frac{\theta}{2}}w$. Then for $t\gg 1$

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle\approx\omega_{n}P_{1}^{2}t^{-\frac{\theta}{2}}\int_{0}^{\eta}e^{-w^{\frac{2}{\theta}}t}\sin^{2}\Big{(}t^{1-\frac{\theta}{2}}t^{\frac{\theta}{2}}w\Big{)}w^{\frac{n-3\theta}{\theta}}t^{\frac{\theta}{2}}dw$
$\displaystyle=\omega_{n}P_{1}^{2}t^{-\frac{n-3\theta}{2}}t^{-\frac{\theta}{2}}\int_{0}^{\eta
t^{\frac{\theta}{2}}}e^{-y^{\frac{2}{\theta}}}\sin^{2}\Big{(}t^{1-\frac{\theta}{2}}y\Big{)}y^{\frac{n-3\theta}{\theta}}dy$
$\displaystyle\geq\omega_{n}P_{1}^{2}t^{-\frac{n-2\theta}{2}}\int_{0}^{1}e^{-y^{\frac{2}{\theta}}}\sin^{2}\Big{(}t^{1-\frac{\theta}{2}}y\Big{)}y^{\frac{n-3\theta}{\theta}}dy$
$\displaystyle=\omega_{n}\frac{P_{1}^{2}}{2}t^{-\frac{n-2\theta}{2}}\left(\int_{0}^{1}e^{-y^{\frac{2}{\theta}}}y^{\frac{n-3\theta}{\theta}}dy-\int_{0}^{1}e^{-y^{\frac{2}{\theta}}}y^{\frac{n-3\theta}{\theta}}\cos\Big{(}2t^{1-\frac{\theta}{2}}y\Big{)}dy\right)$
$\displaystyle=:\omega_{n}\frac{P_{1}^{2}}{2}t^{-\frac{n-2\theta}{2}}(A_{n}-F_{n}(t)),$

where

$\displaystyle
A_{n}:=\int_{0}^{1}e^{-y^{\frac{2}{\theta}}}y^{\frac{n-3\theta}{\theta}}dy,$
$\displaystyle
F_{n}(t):=\int_{0}^{1}e^{-y^{\frac{2}{\theta}}}y^{\frac{n-3\theta}{\theta}}\cos\Big{(}2t^{1-\frac{\theta}{2}}y\Big{)}dy.$

For $0<\theta\leq 1$ and $n-2\theta>0$ the function $y\mapsto
e^{-y^{\frac{2}{\theta}}}y^{\frac{n-3\theta}{\theta}}$ is integrable on the
interval $[0,1]$. Thus, it follows from the Riemann-Lebesgue theorem that

$\lim_{t\rightarrow\infty}F_{n}(t)=0,$

due to $t^{1-\frac{\theta}{2}}\rightarrow\infty$ as $t\rightarrow\infty$.
Therefore, there exists $T>0$ such that

$-\frac{A_{n}}{2}\leq F_{n}(t)\leq\frac{A_{n}}{2},\quad t\geq T,$

so that

$A_{n}-F_{n}(t)\geq\frac{A_{n}}{2},\quad t\geq T.$

Therefore

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle\geq\omega_{n}\frac{A_{n}P_{1}^{2}}{4}t^{-\frac{n-2\theta}{2}},\quad
t\geq T,$

which implies the desired statement. $\Box$

###  6.2 Infinite time blow-up estimates

The omitted cases $n=1$ with $\frac{1}{2}\leq\theta\leq 1$ and $n=2$ with
$\theta=1$ are more delicate and we discuss each of them in this subsection
through some lemmas. We prove that the asymptotic profile of the solution
blows up in infinite time. This is a quite novel property which reflects a
singularity hidden in the solution itself.

#### (I) Estimates for $n=1$ and $\theta\in(1/2,1]$.

To obtain sharp estimates to this cases, we first prepare the following fact.

###### Lemma 6.4

Let $c>0$ and $\alpha>0$ be constants. Then it holds that

$\rm{erf}(ct^{-\alpha})\sim t^{-\alpha},\quad t\geq c^{\frac{1}{\alpha}},$

where $\rm{erf}(\cdot)$ is the Gauss error function defined by

${\rm erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-w^{2}}dw.$

Proof. Using the change of variable $v=\frac{t^{\alpha}w}{c}$, we have

$\displaystyle\text{erf}(ct^{-\alpha})$
$\displaystyle=\frac{2}{\sqrt{\pi}}\int_{0}^{ct^{-\alpha}}e^{-w^{2}}dw=\frac{2c}{\sqrt{\pi}t^{\alpha}}\int_{0}^{1}e^{-\frac{c^{2}v^{2}}{t^{2\alpha}}}dv.$
(6.6)

Then, we can get the following upper estimate

$\displaystyle\text{erf}(ct^{-\alpha})$
$\displaystyle\leq\frac{2c}{\sqrt{\pi}t^{\alpha}}\int_{0}^{1}dv=\frac{2c}{\sqrt{\pi}}t^{-\alpha}$

for $t>0$. On the other hand, from (6.6) and assuming that $t\geq
c^{\frac{1}{\alpha}}$, we have the following lower estimate

$\text{erf}(ct^{-\alpha})\geq\frac{2c}{\sqrt{\pi}t^{\alpha}}\int_{0}^{1}e^{-v^{2}}dv=c\;{\rm
erf}(1)t^{-\alpha}.$ (6.7)

This completes the desired estimates. $\Box$

In order to prove the following lemma we observe that for $n=1$

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr.$
(6.8)

In the proof of the following lemma, we use the fact that

$\displaystyle\frac{m^{2}}{2}r^{2\theta}\leq
r^{2}+m^{2}\log(1+r^{2\theta})\leq(1+m^{2})r^{2\theta},$ (6.9)

due to

$\displaystyle\frac{1}{2}r^{2\theta}\leq\log(1+r^{2\theta})\leq r^{2\theta},$
$\displaystyle r^{2}\leq r^{2\theta}$

for $0\leq r\leq 1$ and $0\leq\theta\leq 1$.

###### Lemma 6.5

Let $n=1$ and $\frac{1}{2}<\theta\leq 1$. There exist positive constants
$C_{1},C_{2}$ such that

$\frac{C_{1}}{(1+m^{2})^{\frac{1}{2\theta}}}P_{1}^{2}t^{\frac{2\theta-1}{\theta}}\leq\int_{|\xi|\leq
1}|\varphi(t,\xi)|^{2}d\xi\leq\frac{C_{2}}{m^{2}}\frac{1}{2\theta-1}t^{\frac{2\theta-1}{\theta}},\quad
t\gg 1.$

Proof. First we get the upper estimate.

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=P_{1}^{2}\int_{0}^{t^{-\frac{1}{\theta}}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle+P_{1}^{2}\int_{t^{-\frac{1}{\theta}}}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle=:P_{1}^{2}\Big{(}K_{1}(t)+K_{2}(t)\Big{)}.$

Set $L:=\displaystyle{\sup_{\nu\neq 0}|\frac{\sin\nu}{\nu}|}<+\infty$. Then,
one can estimate as follows:

$\displaystyle K_{1}(t)$ $\displaystyle\leq
L^{2}t^{2}\int_{0}^{t^{-\frac{1}{\theta}}}e^{-r^{2}t}dr\leq
L^{2}t^{2}\int_{0}^{t^{-\frac{1}{\theta}}}dr$ $\displaystyle\leq
L^{2}t^{2}t^{-\frac{1}{\theta}}$
$\displaystyle=L^{2}t^{\frac{2\theta-1}{\theta}}.$ (6.10)

Due to $|\sin a|\leq 1$, using (6.9) and integration by parts, assuming that
$\theta\neq\frac{1}{2}$ one can get an upper bound estimate to $K_{2}(t)$ as
follows:

$\displaystyle K_{2}(t)$
$\displaystyle=\int_{t^{-\frac{1}{\theta}}}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle\leq\int_{t^{-\frac{1}{\theta}}}^{1}e^{-r^{2}t}\frac{1}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle\leq\frac{2}{m^{2}}\int_{t^{-\frac{1}{\theta}}}^{1}\frac{1}{r^{2\theta}}dr$
$\displaystyle=\frac{2}{m^{2}}\left(\frac{1}{2\theta-1}t^{\frac{2\theta-1}{\theta}}-\frac{1}{2\theta-1}\right)$
$\displaystyle\leq\frac{2}{m^{2}}\frac{1}{2\theta-1}t^{\frac{2\theta-1}{\theta}}.$
(6.11)

From the estimates above one can observe that the Gauss kernel $e^{-tr^{2}}$
is no effective on the growth rates.

From (6.10) and (6.2), we prove that if $\theta>\frac{1}{2}$ there exists a
positive constant $C$ such that

$\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi\leq
P_{1}^{2}\frac{C}{m^{2}}\frac{1}{2\theta-1}t^{\frac{2\theta-1}{\theta}},\quad
t\gg 1.$

In order to obtain the lower estimate we define

$r_{0}(t):=\frac{2^{\frac{1}{\theta}}}{(1+m^{2})^{\frac{1}{2\theta}}}t^{-\frac{1}{\theta}}.$

Then for $0\leq r\leq r_{0}(t)<1$ we have

$\displaystyle t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})}\leq
t\sqrt{1+m^{2}}r^{\theta}\leq 2,\quad t\gg 1,$ (6.12)

due to (6.9). We observe that there exist a constant $C>0$ such that
$C\leq\displaystyle{\frac{\sin^{2}a}{a^{2}}}\leq L^{2}$ for $0<|a|\leq 2$.
Thus for $0\leq r\leq r_{0}(t)$

$\displaystyle
C\leq\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{t^{2}(r^{2}+m^{2}\log(1+r^{2\theta}))}\leq
L^{2}.$ (6.13)

Therefore,

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle\geq
P_{1}^{2}\int_{0}^{r_{0}(t)}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle\geq Ct^{2}P_{1}^{2}\int_{0}^{r_{0}(t)}e^{-r^{2}t}dr$
$\displaystyle=Ct^{\frac{3}{2}}P_{1}^{2}\int_{0}^{\sqrt{t}r_{0}(t)}e^{-w^{2}}dw$
$\displaystyle=CP_{1}^{2}t^{\frac{3}{2}}\text{erf}\left(\frac{2^{\frac{1}{\theta}}}{(1+m^{2})^{\frac{1}{2\theta}}}t^{\frac{1}{2}-\frac{1}{\theta}}\right)$
$\displaystyle\geq
K\frac{P_{1}^{2}}{(1+m^{2})^{\frac{1}{2\theta}}}t^{\frac{2\theta-1}{\theta}},\quad
t\gg 1,$ (6.14)

with some generous constant $K>0$, where we just used the estimate (6.7). Note
that (6.2) makes sense even if one chooses $m=0$. $\Box$

#### (II) Estimate for $n=1$ and $\theta=1/2$.

In this part we shall study the critical case of $\theta=1/2$ just omitted in
the statement of Lemma 6.5. For this, we prepare the following lemma.

###### Lemma 6.6

There exists a positive constant $C$ such that

$\displaystyle\int_{0}^{\infty}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw\geq
C\log t,\quad t\gg 1.$ (6.15)

Proof. The whole idea comes from [18].
Let

$\nu_{j}=\left(\frac{1}{4}+j\right)\frac{\pi}{\sqrt{t}},\qquad\nu^{\prime}_{j}=\left(\frac{3}{4}+j\right)\frac{\pi}{\sqrt{t}},\quad
j=0,1,2,3,\cdots$

and observe that

$\sin^{2}(\sqrt{t}w)\geq\frac{1}{2},\forall w\in[\nu_{j},\nu^{\prime}_{j}].$

Then

$\displaystyle\int_{0}^{\infty}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw$
$\displaystyle\geq\sum_{j=0}^{\infty}\int_{\nu_{j}}^{\nu^{\prime}_{j}}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw$
$\displaystyle\geq\frac{1}{2}\sum_{j=0}^{\infty}\int_{\nu_{j}}^{\nu^{\prime}_{j}}\frac{e^{-w^{2}}}{w}dw.$
(6.16)

The function $(0,\infty)\ni w\mapsto\displaystyle{\frac{e^{-w^{2}}}{w}}$ is
decreasing and the intervals $[\nu_{j},\nu^{\prime}_{j}]$ and
$[\nu^{\prime}_{j},\nu_{j+1}]$ have the same length, then

$\displaystyle
2\sum_{j=0}^{\infty}\int_{\nu_{j}}^{\nu^{\prime}_{j}}\frac{e^{-w^{2}}}{w}dw\geq\sum_{j=0}^{\infty}\int_{\nu_{j}}^{\nu^{\prime}_{j}}\frac{e^{-w^{2}}}{w}dw+\sum_{j=0}^{\infty}\int_{\nu^{\prime}_{j}}^{\nu_{j+1}}\frac{e^{-w^{2}}}{w}dw=\int_{\frac{\pi}{4\sqrt{t}}}^{\infty}\frac{e^{-w^{2}}}{w}dw.$
(6.17)

From (6.2) and (6.17), for large $t>0$ it follows that

$\displaystyle\int_{0}^{\infty}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw$
$\displaystyle\geq\frac{1}{4}\int_{\frac{\pi}{4\sqrt{t}}}^{\infty}\frac{e^{-w^{2}}}{w}dw$
$\displaystyle\geq\frac{1}{4}\int_{\frac{\pi}{4\sqrt{t}}}^{1}\frac{e^{-w^{2}}}{w}dw$
$\displaystyle\geq\frac{1}{4}e^{-1}\int_{\frac{\pi}{4\sqrt{t}}}^{1}\frac{1}{w}dw$
$\displaystyle=\frac{1}{4}e^{-1}\left(\frac{1}{2}\log
t-\log\frac{\pi}{4}\right).$ (6.18)

This implies the desired estimate. $\Box$

The next lemma state the $\log$-growth rate of the critical case for
$\theta=1/2$ and $n=1$.

###### Lemma 6.7

Let $n=1$ and $\theta=\frac{1}{2}$. There exist positive constants
$C_{1},C_{2}$ such that

$\frac{C_{1}}{m^{2}+2}P_{1}^{2}\log t\leq\int_{|\xi|\leq
1}|\varphi(t,\xi)|^{2}d\xi\leq\frac{C_{2}}{m^{2}}P_{1}^{2}\log t,\quad t\gg
1.$

Proof. For $n=1$ and $\theta=1/2$ and $t>1$, we have

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2\theta})})}{r^{2}+m^{2}\log(1+r^{2\theta})}dr$
$\displaystyle=P_{1}^{2}\int_{0}^{\frac{1}{t^{2}}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr$
$\displaystyle=P_{1}^{2}\int_{0}^{\frac{1}{t^{2}}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr$
$\displaystyle+P_{1}^{2}\int_{\frac{1}{t^{2}}}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr$
$\displaystyle=:K_{1}(t)+K_{2}(t).$ (6.19)

One first considers the upper bound estimates of (6.19).
Due to $\sin^{2}a\leq a^{2}$ for $a\geq 0$, one has

$\displaystyle
K_{1}(t)=P_{1}^{2}\int_{0}^{\frac{1}{t^{2}}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr\leq
t^{2}P_{1}^{2}\int_{0}^{\frac{1}{t^{2}}}dr=P_{1}^{2}.$ (6.20)

In order to get an upper estimate to $K_{2}(t)$, we again remember that for
$0\leq r\leq 1$,

$\displaystyle\frac{m^{2}}{2}r\leq r^{2}+m^{2}\log(1+r)\leq(1+m^{2})r,$ (6.21)

as just observed in (6.9) with $2\theta=1$. Therefore

$\displaystyle K_{2}(t)$
$\displaystyle=P_{1}^{2}\int_{\frac{1}{t^{2}}}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr\leq
P_{1}^{2}\int_{\frac{1}{t^{2}}}^{1}\frac{1}{r^{2}+m^{2}\log(1+r)}dr$
$\displaystyle\leq\frac{2}{m^{2}}P_{1}^{2}\int_{\frac{1}{t^{2}}}^{1}\frac{1}{r}dr=\frac{2}{m^{2}}P_{1}^{2}(1-\log
t^{-2})=\frac{2}{m^{2}}P_{1}^{2}(1+2\log t)$
$\displaystyle\leq\frac{C}{m^{2}}P_{1}^{2}\log t,\quad t\gg 1.$ (6.22)

(6.19), (6.20) and (6.22) imply the desired statement

$\displaystyle\int_{|\xi|\leq
1}|\varphi(t,\xi)|^{2}d\xi\leq\frac{C}{m^{2}}P_{1}^{2}\log t,\quad t\gg 1$
(6.23)

with some constant $C>0$.

One next derives the lower bound estimates of (6.19) in the case when
$2\theta=1$ and $n=1$.
One starts with the following inequality:

$\displaystyle\int_{|\xi|\leq 1}$
$\displaystyle|\varphi(t,\xi)|^{2}d\xi=P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr$
$\displaystyle\geq
P_{1}^{2}\int_{0}^{1}e^{-t(r^{2}+m^{2}\log(1+r))}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{r^{2}+m^{2}\log(1+r)}dr$
$\displaystyle=2P_{1}^{2}\int_{0}^{1}\frac{e^{-t(r^{2}+m^{2}\log(1+r))}\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r)})}{\sqrt{t}\sqrt{r^{2}+m^{2}\log(1+r)}\left(\frac{m^{2}}{r+1}+2r\right)}\frac{\sqrt{t}\left(\frac{m^{2}}{r+1}+2r\right)}{2\sqrt{r^{2}+m^{2}\log(1+r)}}dr.$
(6.24)

Set $\alpha=\sqrt{1+m^{2}\log 2}$. We apply the change of variable
$w=\sqrt{t}\sqrt{r^{2}+m^{2}\log(1+r)}$ in (6.2). Then, one has

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$ $\displaystyle\geq
C^{*}P_{1}^{2}\int_{0}^{\alpha\sqrt{t}}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw,$
(6.25)

where one has used the estimate:

$2\left(\frac{m^{2}}{r+1}+2r\right)^{-1}\geq 2(m^{2}+2)^{-1}=:C^{*}\quad(0\leq
r\leq 1).$

Now, for $t\geq 1$, we observe that

$\displaystyle\int_{\alpha\sqrt{t}}^{\infty}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw$
$\displaystyle\leq\frac{1}{\alpha}t^{-\frac{1}{2}}\int_{\alpha\sqrt{t}}^{\infty}e^{-w^{2}}dw$
$\displaystyle\leq t^{-\frac{1}{2}}\int_{0}^{\infty}e^{-w^{2}}dw$
$\displaystyle=\frac{\sqrt{\pi}}{2}t^{-\frac{1}{2}}.$ (6.26)

Thus, applying Lemma 6.6 and the estimate just above, one obtains

$\displaystyle\int_{0}^{\alpha\sqrt{t}}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw$
$\displaystyle=\int_{0}^{\infty}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw-\int_{\alpha\sqrt{t}}^{\infty}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw$
$\displaystyle\geq C\log t-\frac{\sqrt{\pi}}{2}t^{-\frac{1}{2}},$ (6.27)

which implies

$\displaystyle\int_{0}^{\alpha\sqrt{t}}\frac{e^{-w^{2}}\sin^{2}(\sqrt{t}w)}{w}dw\geq\frac{C}{2}\log
t,\quad t\gg 1.$ (6.28)

(6.25) and (6.28) imply the desired estimate from below with a constant $C>0$:

$\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi\geq\frac{CC^{*}}{2}P_{1}^{2}\log
t,\quad t\gg 1.$

$\Box$

#### (III) Estimates for $n=2$ and $\theta=1$.

Finally one treats the critical case for $n=2$ and $\theta=1$ to observe the
growth property of the solution itself. In this case, one can get the
following statement.

###### Lemma 6.8

Let $n=2$ and $\theta=1$. There exist positive constants $C_{1},C_{2}$ such
that

$\frac{C_{1}}{2+m^{2}}P_{1}^{2}\log t\leq\int_{|\xi|\leq
1}|\varphi(t,\xi)|^{2}d\xi\leq\frac{C_{2}}{1+m^{2}}P_{1}^{2}\log t,\quad t\gg
1.$

Proof.  For $t>0$, for $n=2$ and $\theta=1$ we observe that

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle=P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2})})}{r^{2}+m^{2}\log(1+r^{2})}rdr$
(6.29)
$\displaystyle=P_{1}^{2}\int_{0}^{\frac{1}{t}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2})})}{r^{2}+m^{2}\log(1+r^{2})}rdr$
$\displaystyle+P_{1}^{2}\int_{\frac{1}{t}}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2})})}{r^{2}+m^{2}\log(1+r^{2})}rdr$
$\displaystyle=:P_{1}^{2}(M_{1}(t)+M_{2}(t)).$ (6.30)

First of all, one knows that $|\sin a|\leq a$ for all $a\geq 0$. Thus

$\displaystyle M_{1}(t)$
$\displaystyle=\int_{0}^{\frac{1}{t}}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2})})}{r^{2}+m^{2}\log(1+r^{2})}rdr$
$\displaystyle\leq t^{2}\int_{0}^{\frac{1}{t}}e^{-r^{2}t}rdr\leq
t^{2}\int_{0}^{\frac{1}{t}}rdr$ $\displaystyle=\frac{1}{2}.$ (6.31)

In order to estimate $M_{2}(t)$, we observe that

$\frac{r}{r^{2}+m^{2}\log(1+r^{2})}\approx\frac{1}{1+m^{2}}\frac{1}{r},\quad
0<r\leq 1.$

Then, the same steps to obtain estimates for $K_{2}(t)$ in (6.22) can be
applied to get

$\displaystyle M_{2}(t)\leq\frac{C}{1+m^{2}}\log t,\quad t\gg 1.$

Applying the estimates to $M_{1}(t)$ and $M_{2}(t)$ in (6.30), we can prove
the upper estimate

$\displaystyle\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi$
$\displaystyle\leq\frac{C}{1+m^{2}}P_{1}^{2}\log(t),\quad t\gg 1.$ (6.32)

To obtain the lower bound estimate we set $\alpha=\sqrt{1+m^{2}\log 2}$. Then,
similarly to (6.25) one can estimate as follows:

$\displaystyle\int_{|\xi|\leq 1}$
$\displaystyle|\varphi(t,\xi)|^{2}d\xi=P_{1}^{2}\int_{0}^{1}e^{-r^{2}t}\frac{\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2})})}{r^{2}+m^{2}\log(1+r^{2})}rdr$
$\displaystyle\geq
P_{1}^{2}\int_{0}^{1}\frac{e^{-t\Big{(}r^{2}+m^{2}\log(1+r^{2})\Big{)}}\sin^{2}(t\sqrt{r^{2}+m^{2}\log(1+r^{2})})}{\sqrt{t}\sqrt{r^{2}+m^{2}\log(1+r^{2})}\Big{(}2+\frac{m^{2}}{1+r^{2}}\Big{)}}\frac{\sqrt{t}r\Big{(}2+\frac{m^{2}}{1+r^{2}}\Big{)}}{\sqrt{r^{2}+m^{2}\log(1+r^{2})}}dr$
$\displaystyle\geq
C_{0}P_{1}^{2}\int_{0}^{\alpha\sqrt{t}}\frac{e^{-w^{2}}\sin(\sqrt{t}w)}{w}dw,$
(6.33)

where we have just used the change of variable
$w=\sqrt{t}\sqrt{r^{2}+m^{2}\log(1+r^{2})}$ and the fact

$\left(2+\frac{m^{2}}{1+r^{2}}\right)^{-1}\geq(2+m^{2})^{-1}=:C_{0}\quad(0\leq
r\leq 1).$

As we have already proved in (6.28), we have the desired estimate:

$\int_{|\xi|\leq 1}|\varphi(t,\xi)|^{2}d\xi\geq C_{0}P_{1}^{2}\log t,\quad
t\gg 1.$

$\Box$

Now, let us check the validity of the statement of Theorem 1.1.
Proof of Theorem 1.1. The proof of Theorem 1.1 is a direct consequence of
(4.10), (4.1), (4.1), (4.1), (5), (5.1), (5.1) and (6.2). $\Box$

Therefore, based on Theorem 1.1, the Plancherel Theorem and the standard
inequality

$\|\varphi(t,\cdot)\|-\|\hat{u}(t,\cdot)-\varphi(t,\cdot)\|\leq\|\hat{u}(t,\cdot)\|\leq\|\hat{u}(t,\cdot)-\varphi(t,\cdot)\|+\|\varphi(t,\cdot)\|,\;\;t\geq
1,$ (6.34)

one can prove Theorem 1.2.
Proof of theorem 1.2. The proof of item (i) follows by using (6.34), Theorem
1.1, (6.3), (6.4). Similarly, the proof of items (ii) and (iii) follows from
Theorem 1.1 combined with Lemmas 6.5, 6.7 and 6.8. $\Box$

Acknowledgements. The work of the first author (A. Piske) was financed in part
by the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior - Brasil
(CAPES) - Finance Code 001. The work of the second author (R. C. Charão) was
partially supported by PRINT/CAPES-Process 88881.310536/2018-00. The work of
the third author (R. Ikehata) was supported in part by Grant-in-Aid for
Scientific Research (C) 20K03682 of JSPS.

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