# Complex valued semi-linear heat equations in super-critical spaces
$E^{s}_{\sigma}$

Jie Chen, Baoxiang Wang111Corresponding author., Zimeng Wang

###### Abstract

We consider the Cauchy problem for the complex valued semi-linear heat
equation

$\partial_{t}u-\Delta u-u^{m}=0,\ \ u(0,x)=u_{0}(x),$

where $m\geq 2$ is an integer and the initial data belong to super-critical
spaces $E^{s}_{\sigma}$ for which the norms are defined by

$\|f\|_{E^{s}_{\sigma}}=\|\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}(\xi)\|_{L^{2}},\
\ \sigma\in\mathbb{R},\ s<0.$

If $s<0$, then any Sobolev space $H^{r}$ is a subspace of $E^{s}_{\sigma}$,
i.e., $\cup_{r\in\mathbb{R}}H^{r}\subset E^{s}_{\sigma}$. We obtain the global
existence and uniqueness of the solutions if the initial data belong to
$E^{s}_{\sigma}$ ($s<0,\ \sigma\geq d/2-2/(m-1)$) and their Fourier transforms
are supported in the first octant, the smallness conditions on the initial
data in $E^{s}_{\sigma}$ are not required for the global solutions. Moreover,
we show that the error between the solution $u$ and the iteration solution
$u^{(j)}$ is $C^{j}/(j\,!)^{2}$. Similar results also hold if the nonlinearity
$u^{m}$ is replaced by an exponential function $e^{u}-1$.

2020 MSC: 35K58.

##  1 Introduction

Let $\mathscr{S}$ be the Schwartz space and $\mathscr{S}^{\prime}$ be its dual
space. We write

$p_{\lambda}(f)=\sup_{x\in\mathbb{R}^{d}}e^{\lambda|x|}|f(x)|,\quad
q_{\lambda}(f)=\sup_{\xi\in\mathbb{R}^{d}}e^{\lambda|\xi|}|\widehat{f}(\xi)|,$
$\mathscr{S}_{1}:=\\{f\in\mathscr{S}:p_{\lambda}(f)+q_{\lambda}(f)<\infty,~{}\forall~{}\lambda>0\\}.$

$\mathscr{S}_{1}$ equipped with the system of semi-norms
$\\{p_{\lambda}+q_{\lambda}\\}_{\lambda>0}$ is a complete locally convex
linear topological space, which is said to be the Gelfand-Shilov space, cf.
[18]. We denote by $\mathscr{S}_{1}^{\prime}$ the dual space of
$\mathscr{S}_{1}$. One easily sees that

$\mathscr{S}_{1}\subset\mathscr{S},\ \
\mathscr{S}^{\prime}\subset\mathscr{S}^{\prime}_{1}.$

$\mathscr{S}_{1}$ contains the translations and modulations of Gaussian
$e^{-\mathrm{i}mx}e^{-|x-n|^{2}/2}$ and their linear combinations, which are
dense in any Sobolev spaces $H^{\sigma}$, cf. [18, 21]. The Fourier transforms
on $\mathscr{S}_{1}^{\prime}$ can be defined by duality (cf. [14]), namely,
for any $f\in\mathscr{S}^{\prime}_{1}$, its Fourier transform
$\mathscr{F}f=\widehat{f}$ satisfies

$\langle\mathscr{F}f,\,\varphi\rangle=\langle f,\,\mathscr{F}\varphi\rangle,\
\ \forall\ \varphi\in\mathscr{S}_{1}.$

######  Definition 1.1.

Let $s,\sigma\in\mathbb{R}$. Denote

$E^{s}_{\sigma}=E^{s}_{\sigma}(\mathbb{R}^{d}):=\\{f\in\mathscr{S}_{1}^{\prime}:\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}(\xi)\in
L^{2}(\mathbb{R}^{d})\\}$

for which the norm is given by

$\displaystyle\|f\|_{E^{s}_{\sigma}}=\left\|\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}(\xi)\right\|_{L^{2}(\mathbb{R}^{d})}.$
(1)

$E^{s}_{\sigma}$ is a Banach space. Let us denote $E^{s}:=E^{s}_{0}$. In the
case $s>0$, $E^{s}$ as an infinitely smooth function space, was introduced in
Björck [2]. $E^{s}_{\sigma}$ can be regarded as a generalization of Sobolev
spaces $H^{\sigma}:=(I-\Delta)^{-\sigma/2}L^{2}(\mathbb{R}^{d})$ for which the
norm is defined by

$\displaystyle\|f\|_{H^{\sigma}}:=\|\langle\xi\rangle^{\sigma}\widehat{f}\|_{L^{2}(\mathbb{R}^{d})}.$
(2)

At first glance, we find that the norm on $E^{s}_{\sigma}$ can be obtained by
appending an exponential regularity weight $2^{s|\xi|}$ in the Sobolev norm
(2) and $H^{\sigma}=E^{0}_{\sigma}$. However, $E^{s}_{\sigma}$ is a rather
rough function space. In fact, one has that

######  Proposition 1.2.

Let $s<0$ and $\sigma,r\in\mathbb{R}$. Then we have $H^{r}\subset
E^{s}_{\sigma}$.

###### Proof.

Since $2^{s|\xi|}\lesssim\langle\xi\rangle^{r-\sigma}$ for $s<0$ and
$\sigma,r\in\mathbb{R}$, one easily sees that $H^{r}\subset E^{s}_{\sigma}$. ∎

By Proposition 1.2, we see that $\cup_{r\in\mathbb{R}}H^{r}$ is a subset of
$E^{s}_{\sigma}$ if $s<0$. It is easy to see that $E^{s}\subset
E^{s}_{\sigma}$ for $\sigma\leq 0$, and $E^{s}_{\sigma}\subset E^{s}$ for
$\sigma\geq 0$.

The aim of this paper is to study the Cauchy problem for a class of semi-
linear heat equations with initial data in $E^{s}_{\sigma}$ with $s\leq 0$.
Generally speaking, if a nonlinear evolution equation has a critical Sobolev
space $H^{\sigma_{c}}$, which means that it is well-posed in $H^{\sigma}$ for
$\sigma>\sigma_{c}$ and ill posed in $H^{\sigma}$ for $\sigma<\sigma_{c}$,
then $H^{\sigma}$ with $\sigma>\sigma_{c}$ ($\sigma<\sigma_{c}$) is said to be
the subcritical (supercritical) Sobolev space. Any Banach function space $X$
satisfying $H^{\sigma}\subset X$ for some $\sigma<\sigma_{c}$ is said to be a
supercritical space. For a nonlinear evolution equation, if it has a critical
Sobolev space $H^{s_{c}}$, then Proposition 1.2 implies that any supercritical
Sobolev space $H^{r}$ ($r<\sigma_{c}$) is a subspace of $E^{s}_{\sigma}$ if
$\sigma<0$. So, $E^{s}_{\sigma}$ with $s<0$ is also a supercritical space.

Recently, Navier-Stokes equation in supercritical modulation spaces
$E^{s}_{p,1}$ was studied in [14], where a general idea to handle a class of
supercritical initial data was presented. In this paper, we will consider the
semi-linear heat equation (SLH)

$\partial_{t}u-\Delta u-f(u)=0,\ \ u(0,x)=u_{0}(x)$ (3)

in the supercritical spaces $E^{s}_{\sigma}$ with $s<0$, where $f(u)=u^{m}$
(or $f(u)=e^{u}-1$), $u$ is a complex-valued distribution function of
$(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d}$, $m\geq 2$ is a natural number.

For the real valued SLH, Fujita [15] showed that there does not exist a global
solution for any nontrivial nonnegative and suitable smooth initial data in
the case $m<1+2/d$. Hayakawa [22], Kobayashi, Sino and Tanaka [32], Weissler
[55] considered the case $m=1+2/d$ and showed that (3) has no nontrivial
global nonnegative solutions. In the case $m>1+2/d$ and the initial data are
suitable small in $L^{p}$ with $p=d(m-1)/2>1$, then (3) has a unique global
solution in $L^{p}$, cf. Weissler [55] and Giga [19], and their arguments are
also adapted to the complex-valued cases. So, $m=1+2/d$ is a critical power
for the global solutions of (3) with non-negative initial data. The blowup
behavior of the solutions of (3) were studied in [20, 33, 35, 47, 48] (see
also [43]). It seems that the sign-change solutions are more complicated.
Mizoguchi and Yanagida [34] found that $m=1+2/(k+1)$ is the critical index in
1D if $u_{0}\in H^{1}_{\rho}(\mathbb{R})$ changes signs $k$ times, where
$H^{1}_{\rho}$ denote the weighted $H^{1}$ spaces with the weight
$\rho=e^{x^{2}/4}$. In the cases $1<m<1+2/d$, Cazenave, Dickstein and Weissler
[5] constructed some initial value $\psi\in C_{0}(\mathbb{R}^{d})\cap
L^{1}(\mathbb{R}^{d})$ such that (3) has a unique global solution; but for
initial data $u_{0}=c\psi$, the unique solution of (3) blows up in finite time
if $c>0$ is either sufficiently small or sufficiently large.

For the complex-valued SLH, it has a strong relation with the viscous
Constantin–Lax–Majda equation [9, 46],

$\displaystyle v_{t}=v_{xx}+vHv,$ (4)

which is a one dimensional model for the vorticity equation. Here $v$ is a
real valued function and $H$ is the Hilbert transform defined by

$[Hf](x)=\frac{1}{\pi}\,p.v.\,\int_{\mathbb{R}}\frac{f(y)}{x-y}dy.$

It is easy to see that $Hv$ satisfies

$(Hv)_{t}=(Hv)_{xx}+\frac{1}{2}((Hv)^{2}-v^{2}).$

Denote $u=\frac{1}{2{\rm i}}(v+iHv)$, we see that $u$ satisfies

$\displaystyle u_{t}=\Delta u+u^{2}.$ (5)

Noticing that $\widehat{Hv}(\xi)=-{\rm i}\,{\rm sgn}(\xi)\widehat{v}(\xi)$, we
have222$\chi_{E}$ denote the characteristic function on $E$.

$\displaystyle\widehat{u}=-{\rm i}\chi_{[0,\infty)}\widehat{v},$ (6)

which means that $\widehat{u}$ is supported in the half line
$\mathbb{R}_{+}=[0,\infty)$. So, it is natural to consider the solution of (5)
whose Fourier transform is supported in $\mathbb{R}_{+}$. Guo, Ninomiya,
Shimojo and Yanagida [13] gave a systematical study to (5) for the continuous
and bounded initial data, they show that if the convex hull of the image of
the initial data does not intersect the positive real axis, then the solution
exists globally in time and converges to the trivial steady state. In one-
dimensional space, they provide some solutions with nontrivial imaginary parts
that blow up simultaneously. If the initial data are asymptotically constants,
they also obtained some global existence and blowup results. Chouichi, Otsmane
and Tayachi [7] considered a class of continuous, bounded and decaying initial
data like $\mathfrak{Re}u_{0}\sim|x|^{-2\alpha}$,
$\mathfrak{Im}u_{0}\sim|x|^{-2\beta}$ (as $|x|\to\infty$) and they obtained
the existence and the asymptotic behavior of global solutions. Harada [23, 24]
constructed the non-simultaneous blow-up solutions in one spatial dimension.
For the general complex-valued SLH (3), Chouichi, Majdoub and Tayachi [8]
generalized the results in [7] to any power nonlinearity $u^{m}$, $2\leq
m\in\mathbb{N}$. The asymptotic behavior of the blow-up profiles at blow-up
time were constructed in Nouaili and Zaag [38] and Duong [11] for $m=2$ and
for $m\geq 2$, respectively. For the irrational power $m>1$, an interesting
blowup complex solution with one blowup point was constructed in Duong [12]
and its asymptotic profiles were also obtained. The local or global solutions
obtained in [7, 8, 11, 12, 13, 23, 24, 38] are in the space
$L^{\infty}(\mathbb{R}^{d})\cap C(\mathbb{R}^{d})$ and the blowing up means
that

$\lim\sup_{t\to T}\|u(t)\|_{L^{\infty}}=\infty$

for some $0<T<\infty.$

Now we state our main results and indicate the crucial ideas. One of our main
results is the following: Let $f(u)=u^{m}$, $m\geq 2$, $s<0,\,\sigma\geq
d/2-2/(m-1)$. Assume that $u_{0}\in E^{s}_{\sigma}$ with $\widehat{u}_{0}$
supported in the first octant away from the origin.333Such a kind of initial
data have a direct relation with (6) and they are complex-valued. Then there
exists $s_{0}\leq s$ such that SLH (3) for has a unique solution $u\in
C([0,\infty);E^{s_{0}}_{\sigma})\cap\widetilde{L}^{m}(0,\infty;E^{s_{0},\sigma+2/m}_{2,2})$,
where $\widetilde{L}^{m}(0,\infty;E^{s_{0},\sigma+2/m}_{2,2})$ is a resolution
space defined in (11). It seems that such a kind of results are a bit
surprising, since we have no smallness condition on initial data and moreover,
the initial data can be rougher than those in any Sobolev space $H^{r}$ with
negative index $r<0$.

One of the main difficulties to solve SLH in supercritical Sobolev spaces
$H^{\sigma}$ lies in the fact that one cannot make self-contained nonlinear
mapping estimates and the ill-posedness occurs in supercritical Sobolev spaces
$H^{\sigma}$. However, $E^{s}_{\sigma}$ type spaces have some good algebraic
structures when the frequency is localized in the first octant, so that the
nonlinear estimates become available in $E^{s}_{\sigma}$ type spaces. Using
the exponential decay for the semigroup $e^{t\Delta}$, one can get a global
well-posedness result of SLH for sufficiently small initial data in
$E^{s}_{\sigma}$ for which their Fourier transforms are supported in the first
octant away from the origin.

Recall that $\dot{H}^{d/2-2/(m-1)}$
($\dot{H}^{\sigma}=(-\Delta)^{-\sigma/2}L^{2}$) is the scaling critical
Sobolev space of SLH, which means that the scaling solution
$u_{\lambda}(t,x)=\lambda^{2/(m-1)}u(\lambda^{2}t,\lambda x)$ is invariant in
$\dot{H}^{d/2-2/(m-1)}$ for all $\lambda>0$. Let us observe that
$u_{\lambda}|_{t=0}$ in the supercritical space $\dot{H}^{\sigma}$, for any
$\sigma<d/2-2/(m-1)$,

$\displaystyle\|u_{\lambda}|_{t=0}\|_{\dot{H}^{\sigma}}=\lambda^{2/(m-1)+\sigma-d/2}\|u_{0}\|_{\dot{H}^{\sigma}}\to
0,\ \ \lambda\to\infty.$

It follows that the scaling solution can have very small initial data in
supercritical Sobolev spaces. The above observation is also adapted to the
supercritical space $E^{s}_{\sigma}$ ($s<0$), $u_{\lambda}$ will vanish in
$E^{s}_{\sigma}$ when $\lambda\to\infty$, which means that any large data in
$E^{s}_{\sigma}$ can become small data by the scaling argument. This is why we
can handle the large data in $E^{s}_{\sigma}$ ($s<0$).

###  1.1 Some notations and prelimilaries

We denote by $L^{p}_{x}$ the Lebesgue space on $x\in\mathbb{R}^{d}$, i.e.,

$\|f\|_{p}:=\|f\|_{L^{p}_{x}}=\left(\int_{\mathbb{R}^{d}}|f(x)|^{p}dx\right)^{1/p}.$

For any function $g$ of $(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d}$, we
denote

$\|g\|_{L^{\gamma}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}=\|\|g\|_{L^{p}_{x}(\mathbb{R}^{d})}\|_{L^{\gamma}_{t}(\mathbb{R}_{+})},$

where $L^{\gamma}_{t}$ can be defined in a similar way as $L^{p}_{x}$ by
replacing $\mathbb{R}^{d}$ with $\mathbb{R}_{+}$. If there is no confusion, we
will write
$L^{\gamma}_{t}L^{p}_{x}=L^{\gamma}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})$.
Let us denote
$\langle\nabla\rangle^{s}=\mathscr{F}^{-1}\langle\xi\rangle^{s}\mathscr{F}$,
$2^{s|\nabla|}=\mathscr{F}^{-1}2^{s|\xi|}\mathscr{F}$. We will use the
frequency-uniform decomposition techniques, which were first applied to
nonlinear PDE in [54], see also some recent works [3, 6, 25, 27, 28, 41, 42,
49, 51, 53] and their references in the study for a variety of nonlinear
evolution equations. For any $k\in\mathbb{Z}^{d}$, we denote

$\displaystyle\Box_{k}=\mathscr{F}^{-1}\chi_{k+[0,1)^{d}}\mathscr{F},$ (7)

where $\chi_{A}$ denote the characteristic function on
$A\subset\mathbb{R}^{d}$. $\Box_{k}$ ($k\in\mathbb{Z}^{d}$) are said to be the
frequency-uniform decomposition operators. In view of Plancherel’s identity
and the orthogonality of $\Box_{k}$, we see that

$\displaystyle\|u\|_{E^{s}_{\sigma}}\sim\left(\sum_{k\in\mathbb{Z}^{d}}2^{2s|k|}\langle
k\rangle^{2\sigma}\|\Box_{k}u\|^{2}_{2}\right)^{1/2}.$ (8)

If $1<p\leq q\leq\infty$, we have (cf. [14, 54])

$\displaystyle\|\Box_{k}u\|_{L^{q}_{x}}\lesssim\|\Box_{k}u\|_{L^{p}_{x}}$ (9)

holds for all $k\in\mathbb{Z}^{d}$. Since $\widehat{\Box_{k}u}$ is supported
on $k+[0,1)^{d}$, we see that

$\displaystyle\Box_{k}(\Box_{k_{1}}u...\Box_{k_{m}}u)=0,\ \
|k-k_{1}-...-k_{m}|_{\infty}>m+1.$ (10)

This is an important fact for the frequency-uniform decomposition, which will
be repeatedly applied in the paper. For convenience, we denote for any
$\mathbb{A}\subset\mathbb{Z}^{d}$,

$\displaystyle\|u\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma}_{p,q}(\mathbb{A}))}=\left(\sum_{k\in\mathbb{A}}2^{s|k|q}\langle
k\rangle^{\sigma
q}\|\Box_{k}u\|^{q}_{L^{\gamma}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}\right)^{1/q}$
(11)

and
$\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma}_{p,q}):=\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma}_{p,q}(\mathbb{Z}^{d}))$,
$\widetilde{L}^{\gamma}(0,\infty;E^{s}_{p,q}):=\widetilde{L}^{\gamma}(0,\infty;E^{s,0}_{p,q})$.

Throughout this paper, we write $|x|=|x(1)|+...+|x(d)|$,
$|x|_{\infty}=\max_{1\leq i\leq d}|x(i)|$ and $\langle
x\rangle=(1+x(1)^{2}+...+x(d)^{2})^{1/2}$ for
$x=(x(1),...,x(d))\in\mathbb{R}^{d}$. We will use the following notations.
$C\geq 1,\ c\leq 1$ will denote constants which can be different at different
places, we will use $A\lesssim B$ to denote $A\leqslant CB$; $A\sim B$ means
that $A\lesssim B$ and $B\lesssim A$, $A\vee B=\max(A,B)$. We denote by
$\mathscr{F}^{-1}f$ the inverse Fourier transform of $f$. For any $1\leq
p\leq\infty$, we denote by $p^{\prime}$ the dual number of $p$, i.e.,
$1/p+1/p^{\prime}=1$, $\ell^{p}$ stands for the sequence Lebesgue space. The
following inequality will be used in this paper (cf. [1, 52]).

######  Proposition 1.3.

(Multiplier estimate) Let $1\leq r\leq\infty,\ L\geq[d/2]+1$ and $\rho\in
H^{L}$. Then we have

$\displaystyle\|\mathscr{F}^{-1}\rho\|_{1}\lesssim\|\rho\|^{1-d/2L}_{2}\left(\sum^{d}_{i=1}\|\partial^{L}_{x_{i}}\rho\|_{2}\right)^{d/2L}.$
(12)

The paper is organized as follows. In Sections 2 and 3 we consider the global
existence and uniqueness of solutions in $E^{s}_{\sigma}$ for the semi-linear
heat equations with the nonlinearity $u^{m}$, $m\geq 2$ by assuming that the
Fourier transforms of initial data supported in the first octant and away from
$0$, where Fujita’s critical or subcritical powers are contained in 1D and 2D.
If $m\geq 2\vee(1+4/d)$, we can remove the condition that the Fourier
transforms of initial data are away from the origin. In Section 4, we show the
global existence and uniqueness of solutions in $E^{s}_{\sigma}$ for the semi-
linear heat equations with an exponential nonlinearity $e^{u}-1$. In Section 5
we can generalize the above results to the spaces $E^{s}_{2,1}$. Finally, we
give an example to describe the solutions of the semi-linear heat equations.

##  2 SLH in $E^{s}_{\sigma}$

In this section, we will consider (3) with $f(u)=u^{m}$ in the super-critical
space $E^{s}_{\sigma}$. Denote

$\mathbb{R}^{d}_{I}:=\\{\xi\in\mathbb{R}^{d}:\ \xi(j)\geq 0,\ j=1,...,d\\},\ \
\mathbb{Z}^{d}_{I}=\mathbb{R}^{d}_{I}\cap\mathbb{Z}^{d}.$

We have the following

######  Theorem 2.1.

Let $d\geq 1$, $f(u)=u^{m}$, $m\in\mathbb{N}\setminus\\{1\\}$. Let $s\leq 0,\
\sigma\geq d/2-2/(m-1)$, $u_{0}\in E^{s}_{\sigma}$ with
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}$. We
have the following results.

  * (i)

If $s<0$, then there exists $s_{0}\leq s$ such that SLH (3) has a unique
solution $u\in\widetilde{L}^{m}(0,\infty;\,E^{s_{0},\sigma+2/m}_{2,2})$
satisfying the equivalent integral equation

$\displaystyle
u(t)=e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau$ (13)

and

$\displaystyle u\in
C([0,\infty);E^{s_{0}}_{\sigma})\cap\widetilde{L}^{\infty}(0,\infty;\,E^{s_{0},\sigma}_{2,2})\cap\widetilde{L}^{1}(0,\infty;\,E^{s_{0},2+\sigma}_{2,2}).$
(14)

Moreover, condition ${\rm supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}$ is
sharp in the sense that, for any $s_{0}\leq s$, the solution map $u_{0}\to u$
from $E^{s}_{\sigma}$ into $E^{s_{0}}_{\sigma}$ is not $C^{m}$ for certain
initial data whose Fourier transforms are supported in
$\mathbb{R}^{d}_{I}\cup(-\mathbb{R}^{d}_{I})$.

  * (ii)

Let $s=0$. Assume that $\|u_{0}\|_{E^{0}_{\sigma}}$ is sufficiently small,
then (13) has a unique solution $u$ satisfying (14) for $s_{0}=0$. Moreover,
$\sigma\geq\sigma_{c}=d/2-2/(m-1)$ is optimal in the sense that the solution
map $u_{0}\to u$ is not $C^{m}$ in $E^{0}_{\sigma,+}:=\\{f\in
E^{0}_{\sigma}:\,{\rm
supp}\widehat{f}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}\\}$ if
$\sigma<\sigma_{c}$.

  * (iii)

Let $s<0$ and $s_{0}$ be as in (i). Let $\\{u^{(j)}\\}$ be the sequence of the
iteration solutions

$\displaystyle
u^{(j+1)}(t)=e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u^{(j)}(\tau)^{m}d\tau,\
\ u^{(0)}=0.$ (15)

Then for any $\tilde{s}_{0}<s_{0}$, there exists $C>1$ such that

$\|u^{(j)}(t)-u(t)\|_{E^{\tilde{s}_{0}}}\leq\frac{C^{j}}{(j\,!)^{2}},\ \
\forall\ t\geq 0,\ j\in\mathbb{N}.$

Theorem 2.1 needs several remarks.

  * (i)

From condition
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}$, we
see that there exists $\varepsilon_{0}>0$ such that
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{\xi:\,|\xi|<\varepsilon_{0}\\}$.
If $s<0$,
$s_{0}:=s_{0}(\varepsilon_{0},m,\sigma,s,\|u_{0}\|_{E^{s}_{\sigma}})\leq s$
comes from the scaling argument of the solutions. For example, in the scaling
critical case $\sigma=d/2-2/(m-1)$, we can take

$s_{0}=s-(C+\log_{2}^{1+\|u_{0}\|_{E^{s}_{\sigma}}})\varepsilon^{-1}_{0},$

see Remark 2.9 for details. However, if $\|u_{0}\|_{E^{s}_{\sigma}}$ is
sufficiently small and ${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus[0,1)^{d}$, then we
can take $s_{0}=s$ in (14).

  * (ii)

For the global existence and uniqueness results in Theorem 2.1, we have no
condition on the size of $u_{0}\in E^{s}_{\sigma}$ and as indicated in
Proposition 1.2, $E^{s}_{\sigma}$ is a rather rough space. Recall that
condition
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}$
implies that $u_{0}$ is a complex valued function, such a kind of initial data
are different from the nonnegative data in [15, 22, 55], also different from
the initial data $\psi$ in [5]. For instance, let $d=1$, for any
$k\in\mathbb{Z}_{+}$,

$\displaystyle u_{0}$
$\displaystyle=A\,e^{\mathrm{i}x}\frac{d^{k}}{dx^{k}}\left(\delta(x)+\frac{2\,\mathrm{i}}{x}\right),\
A\in\mathbb{C},$ (16) $\displaystyle u_{0}$
$\displaystyle=e^{\mathrm{i}x}\sum^{\infty}_{m=0}\frac{\lambda^{m}}{m!}(-{\rm
i})^{m}\frac{d^{k}}{dx^{k}}\left(\delta(x)+\frac{2\,\mathrm{i}}{x}\right),\ \
|\lambda|<|s|.$ (17)

satisfy the condition of Theorem 2.1, where $\delta$ is the Dirac measure.

In [13], Guo, Ninomiya, Shimojo and Yanagida considered the SLH with
nonlinearity $u^{2}$ and they showed that if $u_{0}\in
L^{\infty}(\mathbb{R}^{d})\cap C(\mathbb{R}^{d})$,
$\mathfrak{Re}\,u_{0}(x)<A\,\mathfrak{Im}\,u_{0}(x)$ for some $A>0$ and for
all $x\in\mathbb{R}^{d}$, then SLH has a unique global solution in $\in
L^{\infty}(\mathbb{R})\cap C(\mathbb{R})$. In 1D case, they obtained the
blowup solution if either $\mathfrak{Re}\widehat{u}_{0}(\xi)>0$ and
$\mathfrak{Im}\widehat{u}_{0}(\xi)=0$ for all $\xi\in\mathbb{R}$; or
$\mathfrak{Re}u$ even, $\mathfrak{Im}u$ odd and $\mathfrak{Im}\,u_{0}(x)>0$
for all $x>0$. In Theorem 2.1 we need $\widehat{u}_{0}(\xi)=0$ for any
$\xi\in\mathbb{R}^{d}\setminus\mathbb{R}^{d}_{I}$, which has no implicit
relations with pointwise condition
$\mathfrak{Re}\,u_{0}<A\,\mathfrak{Im}\,u_{0}$ in [13] and the decaying
condition $\mathfrak{Re}u_{0}\sim c|x|^{-2\alpha}$, $\mathfrak{Im}u_{0}\sim
c|x|^{-2\beta}$ (as $|x|\to\infty$) in [7, 8]. Moreover, one easily sees that
$\widehat{u}_{0}$ supported in $\mathbb{R}^{d}_{I}$ does not contradict the
blowup conditions in [13].

  * (iii)

Recall that $E^{0}_{\sigma_{c}}=H^{\sigma_{c}}$ with $\sigma_{c}=d/2-2/(m-1)$
is the scaling critical space of the SLH. If ${\rm supp}\,\widehat{u}_{0}$ can
be any subset of the whole line $\mathbb{R}$ and $m\geq 3$, Molinet, Ribaud
and Youssfi [36] showed that SLH in 1D is locally well-posed in $H^{\sigma}$
for $\sigma\geq\sigma_{c}$ and ill-posed in $H^{\sigma}$ for
$\sigma<\sigma_{c}$. In the case $m=2$, they obtained that SLH in 1D is local
well-posed in $H^{\sigma}$ for $\sigma>-1$ and ill-posed in $H^{\sigma}$ for
$\sigma<-1$, and the local well-posedness in the critical case $\sigma=-1$ was
shown by Molinet and Tayachi [37], where the fractional NLH was also studied.
However, in the case $m=2$ in 1D, our results indicate that the critical space
is $H^{-3/2}$ if the Fourier transforms of solutions are supported in
$\mathbb{R}^{d}_{I}$, which is different from $H^{-1}$ in [36, 37] (see also
[25]). If $m\geq 1+4/d$, the existence of small data global solutions in (ii)
of Theorem 2.1 seems known, which can be derived by following the global well-
posedness in $L^{r}$ in Giga [19] together with the standard regularity
arguments.

###  2.1 Linear and multi-linear estimates

In Theorem 2.1, one needs that the Fourier transforms of initial data and
solutions supported in $\mathbb{R}^{d}_{I}\setminus\\{0\\}$. By scaling
argument, we can first assume that

$\displaystyle{\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I,+}:=\left\\{\xi\in\mathbb{R}^{d}_{I}:\max_{1\leq
j\leq d}\xi(j)\geq 1\right\\}.$ (18)

The semi-group $e^{t\Delta}$ has very good regularity estimates in Besov and
Triebel spaces, see for instance Iwabuchi and Nakamura [26], Ogawa and Shimizu
[39, 40], Kozono, Okada and Shimizu [29, 30] and in modulation spaces [25,
54]. Concerning the very rough data, we have

######  Lemma 2.2.

Let $d\geq 1$, $s\leq 0$, $\sigma\in\mathbb{R}$,
$1\leq\gamma_{1}\leq\gamma\leq\infty$. Suppose that
$\mathrm{supp}~{}\mathscr{F}_{x}f(t),\,{\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I,+}$. Then we have

$\displaystyle\|e^{t\Delta}u_{0}\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})}$
$\displaystyle\lesssim\|u_{0}\|_{E^{s}_{\sigma}},$ (19)
$\displaystyle\left\|\int_{0}^{t}e^{(t-\tau)\Delta}f(\tau)~{}d\tau\right\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})}$
$\displaystyle\lesssim\|f\|_{\widetilde{L}^{\gamma_{1}}(0,\infty;E^{s,\sigma-2/\gamma^{\prime}_{1}}_{2,2})}.$
(20)

###### Proof.

By Plancherel’s identity and
supp$\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I,+}$, we have

$\displaystyle\|\Box_{k}e^{t\Delta}u_{0}\|_{2}\leq
e^{-t|k|^{2}}\|\Box_{k}u_{0}\|_{2}$ (21)

for all $k\in\mathbb{Z}^{d}_{I}$. Hence, in view of (21) and Hölder’s
inequality,

$\displaystyle\|\Box_{k}e^{t\Delta}u_{0}\|_{L^{\gamma}_{t}L_{x}^{2}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim|k|^{-2/\gamma}\|\Box_{k}u_{0}\|_{2},\ \
k\in\mathbb{R}^{d}_{I,+}\cap\mathbb{Z}^{d}_{I}.$ (22)

(22) is multiplied by $2^{s|k|}\langle k\rangle^{2/\gamma+\sigma}$ and then
taken $\ell^{2}$-norm on $k\in\mathbb{Z}^{d}$, we have (19). For the
inhomogenous part, by (21) and Young’s inequality, we have

$\displaystyle\left\|\Box_{k}\int_{0}^{t}e^{(t-\tau)\Delta}f(\tau)~{}d\tau\right\|_{L^{\gamma}_{t}L^{2}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim\left\|\int_{0}^{t}e^{-(t-\tau)|k|^{2}}\|\Box_{k}f(\tau)\|_{L^{2}}d\tau\right\|_{L^{\gamma}_{t}(\mathbb{R}_{+})}$
$\displaystyle\lesssim|k|^{-2/\gamma-2/\gamma^{\prime}_{1}}\|\Box_{k}f\|_{L^{\gamma_{1}}_{t}L_{x}^{2}(\mathbb{R}_{+}\times\mathbb{R}^{d})}.$
(23)

(23) is multiplied by $2^{s|k|}\langle k\rangle^{2/\gamma+\sigma}$ and then
taken $\ell^{2}$-norm on $k\in\mathbb{Z}^{d}$, we have (20). ∎

######  Corollary 2.3.

Let $d\geq 1$, $s\leq 0$, $\sigma\in\mathbb{R}$, $1\leq\gamma<\infty$. Suppose
that $\mathrm{supp}\mathscr{F}_{x}f(t),\,{\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I,+}$. Then we have

$\displaystyle\|e^{t\Delta}u_{0}\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}$
$\displaystyle\lesssim\|u_{0}\|_{E^{s}_{\sigma}},$ (24)
$\displaystyle\left\|\int_{0}^{t}e^{(t-\tau)\Delta}f(\tau)~{}d\tau\right\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}$
$\displaystyle\lesssim\|f\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2})}.$
(25)

######  Lemma 2.4.

Let $d\geq 1$, $s\leq 0$, $\sigma\in\mathbb{R}$, $1\leq\gamma<\infty$. Then we
have
$\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})\subset\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})$
and

$\displaystyle\|u\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})}\leq\|u\|^{1/\gamma}_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2})}\|u\|^{1-1/\gamma}_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}.$
(26)

###### Proof.

It is a consequence of Hölder’s inequality. ∎

######  Lemma 2.5.

Let $m\geq 2$, $s\leq 0$, $\sigma\geq d/2-2/(m-1)$. Suppose that
$\mathrm{supp}\mathscr{F}_{x}u_{j}(t)\subset\mathbb{R}^{d}_{I},~{}1\leq j\leq
m,\ t\geq 0$. Then we have, for some $C_{m}>0$,

$\displaystyle\|u_{1}u_{2}\cdots
u_{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2})}\leq
C_{m}\prod^{m}_{j=1}\|u_{j}\|_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})},$
(27)

where $C_{m}\leq C^{m}m^{m/2}$ if $\sigma\geq d/2$, $C$ depends only on
$s,d,\sigma$.

###### Proof.

By definition and the support set property of $\mathscr{F}u_{j}(t)$, we have

$\displaystyle\|u_{1}u_{2}\cdots
u_{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2})}=\left(\sum_{k\in\mathbb{Z}^{d}_{I}}\langle
k\rangle^{2\sigma}2^{2s|k|}\|\Box_{k}(u_{1}u_{2}...u_{m})\|^{2}_{L^{1}_{t}L^{2}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}\right)^{1/2}.$
(28)

Using the definition of $\Box_{k}$, we see that

$\displaystyle\|\Box_{k}(u_{1}u_{2}\cdots u_{m})\|_{L_{t}^{1}L_{x}^{2}}$
$\displaystyle\leq\sum_{k_{1},...,k_{m}\in\mathbb{Z}^{d}_{I}}\|\Box_{k}(\Box_{k_{1}}u_{1}\Box_{k_{2}}u_{2}\cdots\Box_{k_{m}}u_{m})\|_{L_{t}^{1}L_{x}^{2}}.$

By the orthogonality of $\Box_{k}$,

$\Box_{k}(\Box_{k_{1}}u_{1}\Box_{k_{2}}u_{2}\cdots\Box_{k_{m}}u_{m})\neq 0,$

implies that $(k_{1},...,k_{m})$ belongs to

$\Lambda_{k}(k_{1},...,k_{m}):=\left\\{(k_{1},...,k_{m}):\,-1\leq
k(j)-k_{1}(j)-...-k_{m}(j)\leq m,\ \ j=1,...,d\right\\}.$

It follows that

$\displaystyle\|\Box_{k}(u_{1}u_{2}\cdots u_{m})\|_{L_{t}^{1}L_{x}^{2}}$
$\displaystyle\leq\sum_{k_{1},...,k_{m}\in\mathbb{Z}^{d}_{I}}\|\Box_{k_{1}}u_{1}\Box_{k_{2}}u_{2}\cdots\Box_{k_{m}}u_{m}\|_{L_{t}^{1}L_{x}^{2}}\chi_{\Lambda_{k}(k_{1},...,k_{m})}.$
(29)

Let us further write

$(\mathbb{Z}^{d}_{I})^{m,i}:=\left\\{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m}:\,|k_{i}|=\max_{1\leq
j\leq m}|k_{j}|\right\\}.$

One has that

$\displaystyle\|\Box_{k}(u_{1}u_{2}\cdots u_{m})\|_{L_{t}^{1}L_{x}^{2}}$
$\displaystyle\leq\sum^{m}_{i=1}\sum_{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m,i}}\|\Box_{k_{1}}u_{1}\Box_{k_{2}}u_{2}\cdots\Box_{k_{m}}u_{m}\|_{L_{t}^{1}L_{x}^{2}}\chi_{\Lambda_{k}(k_{1},...,k_{m})}$
$\displaystyle:=\sum^{m}_{i=1}L_{i}.$ (30)

By symmetry, it suffices to estimate $L_{m}$. By Hölder’s inequality and
$\|\Box_{k}u\|_{\infty}\leq\|\Box_{k}u\|_{2}$,

$\displaystyle L_{m}$
$\displaystyle\leq\sum_{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m,m}}\prod^{m}_{j=1}\|\Box_{k_{j}}u_{j}\|_{L^{m}_{t}L^{\infty}_{x}}\|\Box_{k_{m}}u_{m}\|_{L_{t}^{m}L_{x}^{2}}\chi_{\Lambda_{k}(k_{1},...,k_{m})}$
$\displaystyle\leq\sum_{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m,m}}\prod^{m}_{j=1}\|\Box_{k_{j}}u_{j}\|_{L^{m}_{t}L^{2}_{x}}\chi_{\Lambda_{k}(k_{1},...,k_{m})}.$
(31)

Let us write

$(\mathbb{Z}^{d}_{I})^{m-1}_{r}:=\left\\{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}:\,|k_{1}|,...,|k_{m-1}|\leq
r\right\\}.$

Then for $l_{m}=k-k_{1}-...-k_{m-1}$, $\mathbb{A}=\\{l\in\mathbb{Z}^{d}:\
-1\leq l(j)\leq m,\,j=1,...,d\\}$,

$\displaystyle L_{m}$
$\displaystyle\leq\sum_{l\in\mathbb{A}}\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\|\Box_{k_{j}}u_{j}\|_{L^{m}_{t}L^{2}_{x}}\|\Box_{l_{m}-l}u_{m}\|_{L^{m}_{t}L^{2}_{x}}\chi_{\\{l_{m}-l\in\mathbb{Z}^{d}_{I}\\}}.$
(32)

Using the Cauchy-Schwarz inequality, one has that

$\displaystyle L_{m}\leq$
$\displaystyle\sum_{l\in\mathbb{A}}\left(\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{2(\sigma+2/m)}2^{2s|k_{j}|}\|\Box_{k_{j}}u_{j}\|^{2}_{L^{m}_{t}L^{2}_{x}}\right)^{1/2}$
$\displaystyle\times\left(\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}2^{-2s|k_{j}|}\|\Box_{l_{m}-l}u_{m}\|^{2}_{L_{t}^{m}L_{x}^{2}}\chi_{\\{l_{m}-l\in\mathbb{Z}^{d}_{I}\\}}\right)^{1/2}$
$\displaystyle\lesssim$ $\displaystyle\
2^{|s|dm}2^{-s|k|}\prod^{m-1}_{j=1}\|u_{j}\|_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})}$
$\displaystyle\times\sum_{l\in\mathbb{A}}\left(\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\left(2^{s|l_{m}-l|}\|\Box_{l_{m}-l}u_{m}\|_{L_{t}^{m}L_{x}^{2}}\right)^{2}\right)^{1/2}.$
(33)

By (33), we have for $l_{m}=k-k_{1}-...-k_{m-1}$,

$\displaystyle\left(\sum_{k\in\mathbb{Z}^{d}_{I}}\langle
k\rangle^{2\sigma}2^{2s|k|}L_{m}^{2}\right)^{1/2}$ $\displaystyle\lesssim\
2^{|s|dm}\sum_{l\in\mathbb{A}}\left(\sum_{k\in\mathbb{Z}^{d}_{I}}\langle
k\rangle^{2\sigma}\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}2^{2s|l_{m}-l|}\|\Box_{l_{m}-l}u_{m}\|_{L_{t}^{m}L_{x}^{2}}^{2}\right)^{1/2}$
$\displaystyle\quad\quad\times\prod^{m-1}_{j=1}\|u_{j}\|_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})}.$
(34)

Noticing that $|l|\leq dm$ for $l\in\mathbb{A}$, we have

$\displaystyle\sum_{k\in\mathbb{Z}^{d}_{I}}$ $\displaystyle\langle
k\rangle^{2\sigma}\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}2^{2s|l_{m}-l|}\|\Box_{l_{m}-l}u_{m}\|_{L_{t}^{m}L_{x}^{2}}^{2}$
$\displaystyle\leq\sum_{k_{m}\in\mathbb{Z}^{d}_{I}}\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\left\langle\sum^{m}_{j=1}k_{j}+l\right\rangle^{2\sigma}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\left(2^{s|k_{m}|}\|\Box_{k_{m}}u_{m}\|_{L_{t}^{m}L_{x}^{2}}\right)^{2}$
$\displaystyle\lesssim
m^{2|\sigma|}\sum_{k_{m}\in\mathbb{Z}^{d}_{I}}\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\langle k_{m}\rangle^{-4/m}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\left(\langle
k_{m}\rangle^{\sigma+2/m}2^{s|k_{m}|}\|\Box_{k_{m}}u_{m}\|_{L_{t}^{m}L_{x}^{2}}\right)^{2}.$
(35)

If $d/2-2/(m-1)\leq\sigma<d/2-2/m$, then we easily see that, for
$\sigma_{c}=d/2-2/(m-1)$,

$\displaystyle\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\langle k_{m}\rangle^{-4/m}$
$\displaystyle\leq\frac{C^{m-1}}{(d-2(\sigma+2/m))^{m-1}}\langle
k_{m}\rangle^{(m-1)(d-2\sigma_{c})-4}$
$\displaystyle\leq\frac{C^{m-1}}{|d-2(\sigma+2/m)|^{m-1}}.$ (36)

If $\sigma=d/2-2/m$, we have

$\displaystyle\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\langle k_{m}\rangle^{-4/m}$
$\displaystyle\leq\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-d-4/m(m-1)}$ $\displaystyle\leq C^{m-1}(m(m-1))^{m-1}.$ (37)

If $d/2-2/m<\sigma<d/2$, we have

$\displaystyle\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\langle k_{m}\rangle^{-4/m}$
$\displaystyle\leq\frac{C^{m-1}}{|d-2(\sigma+2/m)|^{m-1}}.$ (38)

If $\sigma=d/2$, then we have

$\displaystyle\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\langle k_{m}\rangle^{-4/m}$
$\displaystyle\leq(Cm)^{m-1}.$ (39)

If $\sigma>d/2$, then we have

$\displaystyle\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|k_{m}|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}\langle k_{m}\rangle^{-4/m}$ $\displaystyle\leq
C^{m}.$ (40)

So, (35) has a bound

$\displaystyle\sum_{k\in\mathbb{Z}^{d}_{I}}$ $\displaystyle\langle
k\rangle^{2\sigma}\sum_{(k_{1},...,k_{m-1})\in(\mathbb{Z}^{d}_{I})^{m-1}_{|l_{m}-l|}}\prod^{m-1}_{j=1}\langle
k_{j}\rangle^{-2(\sigma+2/m)}2^{2s|l_{m}-l|}\|\Box_{l_{m}-l}u_{m}\|_{L_{t}^{m}L_{x}^{2}}^{2}$
$\displaystyle\lesssim
m^{2|\sigma|}C^{m}A_{m}\|u_{m}\|^{2}_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})},$
(41)

where $A_{m}$ is the number in the right hand sides of (36)–(40) corresponding
to different $\sigma\geq d/2-2/(m-1)$. Inserting (41) into (34) and noticing
that $m^{2|\sigma|}\leq C^{m}$, $\mathbb{A}$ has at most $d^{m}$ many indices,
one has that for $C_{m}=C^{m}A_{m}^{1/2}$,

$\displaystyle\left(\sum_{k\in\mathbb{Z}^{d}_{I}}\langle
k\rangle^{2\sigma}2^{2s|k|}L_{m}^{2}\right)^{1/2}\leq
C_{m}\prod^{m}_{j=1}\|u_{j}\|_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})}.$
(42)

In an analogous way to (42), one can estimate

$\displaystyle\left(\sum_{k\in\mathbb{Z}^{d}_{I}}\langle
k\rangle^{2\sigma}2^{2s|k|}L_{i}^{2}\right)^{1/2}\leq
C_{m}\prod^{m}_{j=1}\|u_{j}\|_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})},\
\ i=1,...,m-1.$ (43)

By (39), (40), we see that $C_{m}\leq C^{m}m^{m/2}$ if $\sigma\geq d/2$.
Combining (28), (30), (42) and (43), we obtain the result, as desired. ∎

######  Remark 2.6.

From the proof of Lemma 2.5, we see that if we remove the lower frequency part
of $u_{1}...u_{m}$, we can obtain that

$\displaystyle\|u_{1}u_{2}\cdots u_{m}$
$\displaystyle-\Box_{0}u_{1}\,\Box_{0}u_{2}...\Box_{0}u_{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}\setminus\\{0\\}))}$
$\displaystyle\leq
C_{m}\sum^{m}_{i=1}\|u_{i}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}\setminus\\{0\\}))}\prod^{m}_{j=1,\,j\neq
i}\|u_{j}\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})},$ (44)

where the constant $C_{m}=C^{m}m^{m/2}$ for $\sigma=d/2$, and $C_{m}=C^{m}$
for $\sigma\neq d/2$. Indeed, we write

$(\mathbb{Z}^{d}_{I})^{m}_{+}=\\{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m}:\,\max_{1\leq
j\leq m}|k_{j}|_{\infty}\geq 1\\}.$

By the definition, we have

$\displaystyle\|u_{1}$ $\displaystyle u_{2}\cdots
u_{m}-\Box_{0}u_{1}\,\Box_{0}u_{2}...\Box_{0}u_{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}$
$\displaystyle=\left(\sum_{k\in\mathbb{Z}^{d}_{I}\setminus\\{0\\}}\langle
k\rangle^{2\sigma}2^{2s|k|}\|\Box_{k}(u_{1}u_{2}...u_{m}-\Box_{0}u_{1}\,\Box_{0}u_{2}...\Box_{0}u_{m})\|^{2}_{L^{1}_{t}L^{2}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}\right)^{1/2}.$
(45)

Observing that

$\displaystyle u_{1}u_{2}\cdots u_{m}$
$\displaystyle-\Box_{0}u_{1}\,\Box_{0}u_{2}...\Box_{0}u_{m}=\sum_{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m}_{+}}\Box_{k_{1}}u_{1}\,\Box_{k_{2}}u_{2}...\Box_{k_{m}}u_{m},$
(46)

we have

$\displaystyle\|\Box_{k}$ $\displaystyle(u_{1}u_{2}\cdots
u_{m}-\Box_{0}u_{1}\Box_{0}u_{2}...\Box_{0}u_{m})\|_{L_{t}^{1}L_{x}^{2}}$
$\displaystyle\leq\sum^{m}_{i=1}\sum_{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m,i}_{+}}\|\Box_{k_{1}}u_{1}\Box_{k_{2}}u_{2}\cdots\Box_{k_{m}}u_{m}\|_{L_{t}^{1}L_{x}^{2}}\chi_{\Lambda_{k}(k_{1},...,k_{m})}$
$\displaystyle:=\sum^{m}_{i=1}L_{i}.$ (47)

where
$(\mathbb{Z}^{d}_{I})^{m,i}_{+}=\\{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m}:\,|k_{i}|_{\infty}=\max_{1\leq
j\leq m}|k_{j}|_{\infty}\geq 1\\}$. One can estimate $L_{m}$ by

$\displaystyle L_{m}$
$\displaystyle\leq\sum_{(k_{1},...,k_{m})\in(\mathbb{Z}^{d}_{I})^{m,m}_{+}}\prod^{m-1}_{j=1}\|\Box_{k_{j}}u_{j}\|_{L^{\infty}_{t}L^{2}_{x}}\|\Box_{k_{m}}u_{m}\|_{L^{1}_{t}L^{2}_{x}}\chi_{\Lambda_{k}(k_{1},...,k_{m})},$
(48)

In this way we have removed the lower frequency part of $u_{m}$ in the
estimate of $L_{m}$. Using a similar way as in the proof of Lemma 2.5, we can
get (44).

######  Lemma 2.7.

Assume $s\leq 0$, $\varphi\in E^{s}_{\sigma}(\mathbb{R}^{d})$. Write
$D_{\lambda}:\,\varphi\to\varphi_{\lambda}(\cdot)=\varphi(\lambda\cdot)$.
Assume that ${\rm supp}\,\varphi\subset\\{\xi:|\xi|\geq\varepsilon_{0}\\}$ for
some $\varepsilon_{0}>0$. Then for any $\lambda>1$, we have

$\displaystyle\|D_{\lambda}\varphi\|_{E^{s}_{\sigma}(\mathbb{R}^{d})}\lesssim\lambda^{-d/2}2^{s(\lambda-1)\varepsilon_{0}}\|\varphi\|_{E^{s}_{\sigma}(\mathbb{R}^{d})},\
\ \sigma\leq 0;$
$\displaystyle\|D_{\lambda}\varphi\|_{E^{s}_{\sigma}(\mathbb{R}^{d})}\lesssim\lambda^{-d/2+\sigma}2^{s(\lambda-1)\varepsilon_{0}}\|\varphi\|_{E^{s}_{\sigma}(\mathbb{R}^{d})},\
\ \sigma>0.$

For any $\mu<1$, we have

$\displaystyle\|D_{\mu}\varphi\|_{E^{s/\mu}_{\sigma}(\mathbb{R}^{d})}\lesssim\mu^{-d/2+\sigma}\|\varphi\|_{E^{s}_{\sigma}(\mathbb{R}^{d})},\
\ \sigma\leq 0;$
$\displaystyle\|D_{\mu}\varphi\|_{E^{s/\mu}_{\sigma}(\mathbb{R}^{d})}\lesssim\mu^{-d/2}\|\varphi\|_{E^{s}_{\sigma}(\mathbb{R}^{d})},\
\ \sigma>0.$

###### Proof.

By the definition of $E^{s}_{\sigma}(\mathbb{R}^{d})$, we have

$\displaystyle\|\varphi_{\lambda}\|_{E^{s}_{\sigma}}=\lambda^{-d/2}\|\langle\lambda\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{\varphi}(\xi)\|_{L^{2}_{\xi}}.$

If ${\rm supp}\,\widehat{\varphi}\subset\\{\xi:|\xi|\geq\varepsilon_{0}\\}$
and $\sigma>0$, then

$\displaystyle\|\varphi_{\lambda}\|_{E^{s}_{\sigma}}$
$\displaystyle\leq\lambda^{-d/2+\sigma}\|\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{\varphi}(\xi)\|_{L^{2}_{\xi}}\leq\lambda^{-d/2+\sigma}2^{s(\lambda-1)\varepsilon_{0}}\|\varphi\|_{E^{s}_{\sigma}}.$

If ${\rm supp}\,\widehat{\varphi}\subset\\{\xi:|\xi|\geq\varepsilon_{0}\\}$
and $\sigma\leq 0$, then

$\displaystyle\|\varphi_{\lambda}\|_{E^{s}_{\sigma}}$
$\displaystyle\leq\lambda^{-d/2}\|\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{\varphi}(\xi)\|_{L^{2}_{\xi}}\leq\lambda^{-d/2}2^{s(\lambda-1)\varepsilon_{0}}\|\varphi\|_{E^{s}_{\sigma}}.$

Hence, we have the results for $\lambda>1$. For $\mu<1$, we have

$\displaystyle\|\varphi_{\mu}\|_{E^{s/\mu}_{\sigma}}=\mu^{-d/2}\|\langle\mu\xi\rangle^{\sigma}2^{s|\xi|}\widehat{\varphi}(\xi)\|_{L^{2}_{\xi}}$

Using $\langle\mu\xi\rangle\geq\mu\langle\xi\rangle$ for $\sigma\leq 0$, and
$\langle\mu\xi\rangle\leq\langle\xi\rangle$ for $\sigma>0$, we have the
results for $\mu<1$. ∎

By Lemma 2.7, we see that a great advantage for the scaling in
$E^{s}_{\sigma}$ is that $\varphi_{\lambda}\in E^{s}_{\sigma}$ vanishes as
$\lambda\to\infty$ if the support set of $\widehat{\varphi}$ is away from the
origin.

###  2.2 Proof of Theorem 2.1

Step 1. First, we consider the case $s\leq 0$ and ${\rm
supp}\,\widehat{u}_{0}\subset\\{\xi:|\xi|_{\infty}\geq 1\\}$ and
$\|u_{0}\|_{E^{s}_{\sigma}}$ is sufficiently small. Let us consider the
mapping

$\displaystyle\mathcal{T}:u(t)\to
e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau$

in the space

$\displaystyle\mathcal{D}=\\{u:\,{\rm
supp}\,\widehat{u(t)}\subset\\{\xi:|\xi|_{\infty}\geq 1\\},\
\|u\|_{X_{s,\sigma}}\leq\delta\\},$

where

$\displaystyle\|u\|_{X_{s,\sigma}}=\|u\|_{\widetilde{L}^{m}(0,\infty;E^{s,\sigma+2/m}_{2,2})}$

By Corollary 2.3 and Lemma 2.5, we have

$\displaystyle\|\mathcal{T}u\|_{X_{s,\sigma}}\leq
C\|u_{0}\|_{E^{s}_{\sigma}}+C\|u\|^{m}_{X_{s,\sigma}},$

and similarly,

$\displaystyle\|\mathcal{T}u-\mathcal{T}v\|_{X_{s,\sigma}}\leq
C(\|u\|^{m-1}_{X_{s,\sigma}}+\|v\|^{m-1}_{X_{s,\sigma}})\|u-v\|_{X_{s,\sigma}}.$

Taking $\delta=2C\|u_{0}\|_{E^{s}_{\sigma}}$, we see that for sufficiently
small $\|u_{0}\|_{E^{s}_{\sigma}}$ satisfying

$\displaystyle C\delta^{m-1}\leq 1/100,$ (49)

we have $\mathcal{T}:\mathcal{D}\to\mathcal{D}$ is a contraction mapping and
it has a unique fixed point $u\in X_{s,\sigma}$, i.e.,

$\displaystyle
u(t)=e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau$

in the space $X_{s,\sigma}$. Moreover, by Corollary 2.3 and Lemma 2.5,

$\displaystyle\|u\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}\leq
2C\|u_{0}\|_{E^{s}_{\sigma}}.$ (50)

Furthermore, we can show that $u\in C([0,\infty);E^{s}_{\sigma})$. In fact, By
Corollary 2.3

$\displaystyle\left\|\int_{t_{0}}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau\right\|_{L^{\infty}(t_{0},t_{1};E^{s}_{\sigma})}$
$\displaystyle\leq\left\|\int_{t_{0}}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau\right\|_{\widetilde{L}^{\infty}(t_{0},t_{1};E^{s,\sigma}_{2,2})}$
$\displaystyle\lesssim\|u^{m}\|_{\widetilde{L}^{1}(t_{0},t_{1};E^{s,\sigma}_{2,2})}.$

By Lemma 2.5,

$\displaystyle\|u^{m}\|_{\widetilde{L}^{1}(t_{0},t_{1};E^{s,\sigma}_{2,2})}\lesssim\|u\|^{m}_{\widetilde{L}^{m}(t_{0},t_{1};E^{s,\sigma+2/m}_{2,2})}\to
0,\ t_{0}\to t_{1}.$

It follows that $u\in C([0,\infty);E^{s}_{\sigma})$.

Now, we consider the global solution in the case $s<0$ and ${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}$. Then there
exists $\varepsilon_{0}>0$ such that
$\\{\xi:|\xi|_{\infty}\geq\varepsilon_{0}\\}$. It suffices to consider the
case that $\|u_{0}\|_{E^{s}_{\sigma}}$ is large. Denote

$\displaystyle u_{0,\lambda}(x)=\lambda^{2/(m-1)}u_{0}(\lambda x).$ (51)

In view of Lemma 2.7, we have

$\|u_{0,\lambda}\|_{E^{s}_{\sigma}}\leq\lambda^{2/(m-1)-d/2+0\vee\sigma}2^{s(\lambda-1)\varepsilon_{0}}\|u_{0}\|_{E^{s}_{\sigma}}.$

Noticing that $s<0$, we see that

$\lim_{\lambda\to\infty}\lambda^{2/(m-1)-d/2+0\vee\sigma}2^{s(\lambda-1)\varepsilon_{0}}=0.$

Hence, $\|u_{0,\lambda}\|_{E^{s}_{\sigma}}$ is sufficiently small if
$\lambda\gg 1$. Moreover,
$\widehat{u}_{0,\lambda}\subset\\{\xi:|\xi|_{\infty}\geq\varepsilon_{0}\lambda\\}\subset\\{\xi:|\xi|_{\infty}\geq
1\\}$ if $\lambda>1/\varepsilon_{0}$. Using the above result, we get a unique
solution $u_{\lambda}$ of the integral equation

$\displaystyle
v(t)=e^{t\Delta}u_{0,\lambda}+\int_{0}^{t}e^{(t-\tau)\Delta}v(\tau)^{m}~{}d\tau$

in the space $X_{s,\sigma}$. Put

$u(t,x)=\lambda^{-2/(m-1)}u_{\lambda}(\lambda^{-2}t,\lambda^{-1}x).$

We need to consider the scaling property in $X_{s,\sigma}$.

######  Lemma 2.8.

Assume $s\leq 0$,
$f\in\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})$.
Write $f_{1/\lambda}(t,x)=f(t/\lambda^{2},x/\lambda)$. Then we have

$\displaystyle\|f_{1/\lambda}\|_{\widetilde{L}^{1}(0,\infty;E^{s\lambda,\sigma+2}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s\lambda,\sigma}_{2,2})}\leq
2^{C\lambda}\|f\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})},\
\ \lambda>1.$

###### Proof.

By the definition of $X_{s,\sigma}$ and Lemma 2.4, we have

$\displaystyle\|f_{1/\lambda}\|_{\widetilde{L}^{1}(0,\infty;E^{s\lambda,\sigma+2}_{2,2})\cap\widetilde{L}^{\infty}(0,\infty;E^{s\lambda,\sigma}_{2,2})}\leq$
$\displaystyle\left(\sum_{k\in\mathbb{Z}^{d}}\langle
k\rangle^{2\sigma}2^{2s\lambda|k|}\|\chi_{k+[0,1)^{d}}\widehat{f_{1/\lambda}}\|^{2}_{L^{\infty}_{t}L^{2}_{x}}\right)^{1/2}$
$\displaystyle+\left(\sum_{k\in\mathbb{Z}^{d}}\langle
k\rangle^{2(2+\sigma)}2^{2s\lambda|k|}\|\chi_{k+[0,1)^{d}}\widehat{f_{1/\lambda}}\|^{2}_{L^{1}_{t}L^{2}_{x}}\right)^{1/2}.$

Since

$\displaystyle\langle k\rangle^{\sigma}$ $\displaystyle
2^{s\lambda|k|}\|\chi_{k+[0,1)^{d}}\widehat{f_{1/\lambda}}\|_{L^{\infty}_{t}L^{2}_{x}}$
$\displaystyle\leq\lambda^{d/2+0\vee(-\sigma)}2^{|s|d\lambda}\|\langle\xi\rangle^{\sigma}2^{s|\xi|}\chi_{\lambda
k+[0,\lambda)^{d}}\widehat{f}\|_{L^{\infty}_{t}L^{2}_{x}}$
$\displaystyle\leq\lambda^{d/2+0\vee(-\sigma)}2^{|s|d\lambda}\sum_{\ell\in\mathbb{Z}^{d}\cap\lambda
k+[0,\lambda)^{d}}\|\langle\xi\rangle^{\sigma}2^{s|\xi|}\chi_{\ell+[0,1)^{d}}\widehat{f}\|_{L^{\infty}_{t}L^{2}_{x}}$
$\displaystyle\leq\lambda^{d+0\vee(-\sigma)}2^{|s|d\lambda}\left(\sum_{\ell\in\mathbb{Z}^{d}\cap\lambda
k+[0,\lambda)^{d}}\|\langle\xi\rangle^{\sigma}2^{s|\xi|}\chi_{\ell+[0,1)^{d}}\widehat{f}\|^{2}_{L^{\infty}_{t}L^{2}_{x}}\right)^{1/2},$

one has that

$\displaystyle\left(\sum_{k\in\mathbb{Z}^{d}}\langle
k\rangle^{2\sigma}2^{2s\lambda|k|}\|\chi_{k+[0,1)^{d}}\widehat{f_{1/\lambda}}\|^{2}_{L^{\infty}_{t}L^{2}_{x}}\right)^{1/2}\leq
2^{C\lambda}\|f\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}.$

Similarly,

$\langle
k\rangle^{2+\sigma}2^{s\lambda|k|}\|\chi_{k+[0,1)^{d}}\widehat{f_{1/\lambda}}\|_{L^{1}_{t}L^{2}_{x}}\leq\lambda^{2+d/2+0\vee(-2-\sigma)}2^{|s|d\lambda}\|\langle\xi\rangle^{2+\sigma}2^{s|\xi|}\chi_{\lambda
k+[0,\lambda)^{d}}\widehat{f}\|_{L^{1}_{t}L^{2}_{x}}.$

It follows that

$\displaystyle\left(\sum_{k\in\mathbb{Z}^{d}}\langle
k\rangle^{2(2+\sigma)}2^{2s\lambda|k|}\|\chi_{k+[0,1)^{d}}\widehat{f_{1/\lambda}}\|^{2}_{L^{1}_{t}L^{2}_{x}}\right)^{1/2}\leq
2^{C\lambda}\|f\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2})}.$

We get the result, as desired. ∎

By Lemmas 2.7 and 2.8, $u\in X_{s\lambda,\sigma}$ and it satisfies

$\displaystyle
u(t)=e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau.$

Taking

$\displaystyle s_{0}=\lambda s,$ (52)

we have shown the global existence and uniqueness of solutions in (i) of
Theorem 2.1.

######  Remark 2.9.

We can give an explanation to $s_{0}=\lambda s$. Recall that $\lambda$ can be
chosen from (49). In order to (49) holds, by Lemma 2.7, it suffices to take a
$\lambda>1$ satisfying

$C(\lambda^{2/(m-1)-d/2+0\vee\sigma}2^{s(\lambda-1)\varepsilon_{0}}\|u_{0}\|_{E^{s}_{\sigma}})^{m-1}\leq
1/100,$

where $C$ is a constant that only depends on the upper bound of the
inequalities in Corollary 2.3 and Lemma 2.5. So,
$\lambda=\lambda(\varepsilon_{0},m,\sigma,s,\|u_{0}\|_{E^{s}_{\sigma}})$. For
example, in the case $\sigma=d/2-2/(m-1)$, we can take

$s_{0}=s-(C+\log_{2}^{1+\|u_{0}\|_{E^{s}_{\sigma}}})\varepsilon^{-1}_{0}.$

Step 2. Now we consider the small data in $E^{0}_{\sigma}=H^{\sigma}$, we have
shown the global existence and uniqueness of solutions in
$C([0,\infty);E^{0}_{\sigma})\cap\widetilde{L}^{\infty}(0,\infty;\,E^{0,\sigma}_{2,2})\cap\widetilde{L}^{1}(0,\infty;\,E^{0,2+\sigma}_{2,2})$
if initial data $u_{0}\in H^{\sigma}$ small enough and

${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\cap\\{\xi:\,|\xi|_{\infty}\geq
1\\}.$

Recall that for the initial data ${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}$, there exists
$\varepsilon_{0}>0$ such that

${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\cap\\{\xi:\,|\xi|_{\infty}\geq\varepsilon_{0}\\}.$

Again, by the scaling argument, we see that $u_{0,\lambda}$ defined in (51)
satisfying

${\rm
supp}\,\widehat{u}_{0,\lambda}\subset\mathbb{R}^{d}_{I}\cap\\{\xi:\,|\xi|_{\infty}\geq\varepsilon_{0}\lambda\\}$

and

$\|{u}_{0,\lambda}\|_{H^{\sigma}}\leq
C\lambda^{\sigma-d/2+2/(m-1)}\|u_{0}\|_{H^{\sigma}}.$

Taking $\lambda=1/\varepsilon_{0}$ and

$C\|u_{0}\|_{H^{\sigma}}\leq\delta\varepsilon_{0}^{\sigma-d/2+2/(m-1)}.$

for some small $\delta>0$ as in (49), we see that (13) with initial data
$u_{0,\lambda}$ has a unique global solution $u_{\lambda}\in
C([0,\infty);H^{\sigma})\cap\widetilde{L}^{\infty}(0,\infty;\,E^{0,\sigma}_{2,2})\cap\widetilde{L}^{1}(0,\infty;\,E^{0,2+\sigma}_{2,2})$.
Then $u(t,x)=\lambda^{-2/(m-1)}u_{\lambda}(\lambda^{-2}t,\lambda^{-1}x).$ is
the unique global solution of (13) with initial data $u_{0}$ in the spaces
$C([0,\infty);H^{\sigma})\cap\widetilde{L}^{\infty}(0,\infty;\,E^{0,\sigma}_{2,2})\cap\widetilde{L}^{1}(0,\infty;\,E^{0,2+\sigma}_{2,2})$.
So, we finish the proof of the part of existence and uniqueness of solutions
in (ii) of Theorem 2.1.

Step 3.  We consider the ill-posedness issues of SLH

$\partial_{t}u-\Delta u-u^{m}=0,\ \ u(0,x)=u_{0}(x),$ (53)

Let $s<0$. Put

$\varphi=\sum_{k\in
2^{4\mathbb{N}},k>2d}\mathscr{F}^{-1}(2^{-sd|k|/2}(\chi_{[(m-1)k,(m-1)k+1/2)^{d}}+\chi_{[-k-1/2(m-1),-k)^{d}})),\
\ u_{0}=\delta\varphi,$ (54)

we have $\|u_{0}\|_{E^{s}_{\sigma}}\leq C$. Let $u_{\delta}$ be the solution
of (53) with initial datum $u_{0}=\delta\varphi$. Then

$\displaystyle\left.\frac{\partial
u_{\delta}(t)}{\partial\delta}\right|_{\delta=0}=e^{t\Delta}\varphi,...,\
\left.\frac{\partial^{m}u_{\delta}(t)}{\partial\delta^{m}}\right|_{\delta=0}=m!\int_{0}^{t}e^{(t-\tau)\Delta}(e^{\tau\Delta}\varphi)^{m}d\tau.$

######  Lemma 2.10.

Let $s<0$. Suppose that $u_{0}$ is defined by (54). Then, for any $s_{0}\leq
s$,

$\displaystyle\left\|\int_{0}^{t}e^{(t-\tau)\Delta}(e^{\tau\Delta}\varphi)^{m}d\tau\right\|_{E^{s_{0}}_{\sigma}}=\infty,\quad\forall~{}t>0.$

###### Proof.

Let us observe that for $\xi=\xi_{1}+...+\xi_{m}$, $e=(1,1,...,1)$,

$\displaystyle\widehat{e^{\tau\Delta}\varphi}*...*\widehat{e^{\tau\Delta}\varphi}(\xi)$
$\displaystyle\geq\int_{\mathbb{R}^{d}}e^{-\tau(|\xi_{1}|^{2}+...+|\xi_{m}|^{2})}\widehat{\varphi}(\xi_{1})...\widehat{\varphi}(\xi_{m})d\xi_{1}...d\xi_{m-1}$
$\displaystyle\geq
2^{-smdk/2}\int_{\mathbb{R}^{d}}e^{-\tau(|\xi_{1}|^{2}+...+|\xi_{m}|^{2})}\prod^{m-1}_{j=1}\chi_{[-k-1/2(m-1),-k)^{d}}(\xi_{j})$
$\displaystyle\quad\quad\quad\quad\quad\quad\times\chi_{[(m-1)k,(m-1)k+1/2)^{d}}(\xi_{m})d\xi_{1}...d\xi_{m-1}$
$\displaystyle=2^{-smdk/2}\int_{\mathbb{R}^{d}}e^{-\tau(|\xi_{1}-ke|^{2}+...+|\xi_{m-1}-ke|^{2}+|\xi_{m}+(m-1)ke|^{2})}$
$\displaystyle\quad\quad\quad\quad\quad\quad\times\prod^{m-1}_{j=1}\chi_{[-1/2(m-1),0)^{d}}(\xi_{j})\chi_{[0,1/2)^{d}}(\xi_{m})d\xi_{1}...d\xi_{m-1}$
$\displaystyle\geq 2^{-smdk/2}e^{-\tau
Cd^{2}m^{2}k^{2}}\int_{\mathbb{R}^{d}}\prod^{m-1}_{j=1}\chi_{[-1/2(m-1),0)^{d}}(\xi_{j})\chi_{[0,1/2)^{d}}(\xi_{m})d\xi_{1}...d\xi_{m-1}$
$\displaystyle:=2^{-smdk/2}e^{-\tau Cd^{2}m^{2}k^{2}}\psi(\xi)$

for all $k\in 2^{4\mathbb{N}}$. Note that ${\rm
supp}\,\psi\subset[-1/2,1/2)^{d}$. So, we have

$\displaystyle\mathscr{F}\int_{0}^{t}e^{(t-\tau)\Delta}(e^{\tau\Delta}\varphi)^{m}dt$
$\displaystyle=\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}(\widehat{e^{\tau\Delta}\varphi}*...*\widehat{e^{\tau\Delta}\varphi})(\xi)d\tau$
$\displaystyle\gtrsim 2^{-smdk/2}e^{-t|\xi|^{2}}\psi\int_{0}^{t}e^{-\tau
Cd^{2}m^{2}k^{2}}d\tau$ $\displaystyle\gtrsim
2^{-smdk/2}\frac{e^{-td}}{k^{2}}(1-e^{-tCd^{2}m^{2}k^{2}})\psi.$

Hence, we immediately obtain that

$\left\|\int_{0}^{t}e^{(t-\tau)\Delta}(e^{\tau\Delta}u_{0})^{m}dt\right\|_{E^{s_{0}}_{\sigma}}=\infty$

by taking $k\to\infty$. So, the mapping $u_{0}\to u$ is not $C^{m}$ from
$E^{s}_{\sigma}$ into $E^{s_{0}}_{\sigma}$ for any $s_{0}\leq s$. ∎

We have shown a stronger result, i.e., the corresponding integral equation of
(53) has no iteration solution in any $E^{s_{0}}_{\sigma}$ for the initial
data as in (54). In fact, let us observe the iteration sequence

$u^{(j+1)}(t)=\delta
e^{t\Delta}\varphi+\int_{0}^{t}e^{(t-\tau)\Delta}u^{(j)}(\tau)^{m}d\tau,\ \
u^{(0)}=0.$

It follows that

$\widehat{u^{(j+1)}(t)}=\delta
e^{t|\xi|^{2}}\widehat{\varphi}+\int_{0}^{t}e^{(t-\tau)|\xi|^{2}}\widehat{u^{(j)}(\tau)}*...*\widehat{u^{(j)}(\tau)}d\tau,\
\ \widehat{u^{(0)}}=0.$

Since $\widehat{\varphi}(\xi)\geq 0$ for all $\xi\in\mathbb{R}^{d}$, we see
that

$\widehat{u^{(2)}(t)}\geq\delta
e^{t|\xi|^{2}}\widehat{\varphi}+\delta^{m}\int_{0}^{t}e^{(t-\tau)|\xi|^{2}}\widehat{e^{\tau\Delta}\varphi}*...*\widehat{e^{\tau\Delta}\varphi}d\tau.$

Hence we have

$\|u^{(2)}(t)\|_{E^{s_{0}}_{\sigma}}=\infty,\ \ \forall\ t>0.$

Since $\widehat{u^{(j)}}$ is nonnegative for any $j\geq 1$, we immediately
have $\|u^{(j)}(t)\|_{E^{s_{0}}_{\sigma}}=\infty$ for any $t>0$. It follows
that SLH (3) has no iteration solution in $E^{s_{0}}_{\sigma}$ for any
$s_{0}\leq s$.

Now we consider the ill-posedness for the solutions in $H^{\sigma}$,
$\sigma<d/2-2/(m-1)$. Put

$\psi(\xi)=\prod^{d}_{j=1}\chi_{[1/2d,1/d)}(\xi(j)),\ \
\varphi=N^{-\sigma-d/2}\psi(\xi/N),\ \ u_{0}=\delta\varphi,$ (55)

we have $\|\varphi\|_{H^{\sigma}}\sim 1$. Let $u_{\delta}$ be the solution of
(53) with initial datum $u_{0}=\delta\varphi$. Then

$\displaystyle\left.\frac{\partial
u_{\delta}(t)}{\partial\delta}\right|_{\delta=0}=e^{t\Delta}\varphi,\
\left.\frac{\partial^{m}u_{\delta}(t)}{\partial\delta^{m}}\right|_{\delta=0}=m!\int_{0}^{t}e^{(t-\tau)\Delta}(e^{\tau\Delta}\varphi)^{m}d\tau.$

We have for $\xi_{m}=\xi-\xi_{1}-...-\xi_{m-1}$,

$\displaystyle\left.\widehat{\frac{\partial^{m}u_{\delta}(t)}{\partial\delta^{m}}}\right|_{\delta=0}$
$\displaystyle=m!\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}(\widehat{e^{\tau\Delta}\varphi}*...*\widehat{e^{\tau\Delta}\varphi})d\tau$
$\displaystyle=m!N^{-(\sigma+d/2)m}e^{-t|\xi|^{2}}\int_{0}^{t}\int
e^{\tau(|\xi|^{2}-\sum^{m}_{j=1}|\xi_{j}|^{2})}\varphi\left(\frac{\xi_{1}}{N}\right)...\varphi\left(\frac{\xi_{m}}{N}\right)d\tau
d\xi_{1}...d\xi_{m-1}$
$\displaystyle=m!N^{-(\sigma+d/2)m}e^{-t|\xi|^{2}}\int\frac{e^{t(|\xi|^{2}-\sum^{m}_{j=1}|\xi_{j}|^{2})}-1}{|\xi|^{2}-\sum^{m}_{j=1}|\xi_{j}|^{2}}\varphi\left(\frac{\xi_{1}}{N}\right)...\varphi\left(\frac{\xi_{m}}{N}\right)d\xi_{1}...d\xi_{m-1}.$

Taking $t_{N}\sim 1/N^{2}$, one sees that

$\frac{e^{t_{N}(|\xi|^{2}-\sum^{m}_{j=1}|\xi_{j}|^{2})}-1}{|\xi|^{2}-\sum^{m}_{j=1}|\xi_{j}|^{2}}=\frac{e^{2t_{N}\sum_{i<j}\xi_{i}\xi_{j}}-1}{2\sum_{i<j}\xi_{i}\xi_{j}}\gtrsim\frac{1}{N^{2}}$

in the support set of $\varphi(\xi_{1}/N)...\varphi(\xi_{m}/N)$. So, we have

$\displaystyle\left\|\left.\frac{\partial^{m}u_{\delta}(t_{N})}{\partial\delta^{m}}\right|_{\delta=0}\right\|_{H^{\sigma}}$
$\displaystyle=\left\|\langle\xi\rangle^{\sigma}\left.\widehat{\frac{\partial^{m}u_{\delta}(t_{N})}{\partial\delta^{m}}}\right|_{\delta=0}\chi_{[mN/2,mN]}(|\xi|)\right\|_{2}$
$\displaystyle\gtrsim
N^{-m(\sigma+d/2)-2+\sigma}\left\|\int\varphi\left(\frac{\xi_{1}}{N}\right)...\varphi\left(\frac{\xi_{m-1}}{N}\right)\varphi\left(\frac{\xi-\xi_{1}-...-\xi_{m-1}}{N}\right)d\xi_{1}...d\xi_{m-1}\right\|_{2}$
$\displaystyle\gtrsim N^{(m-1)(d/2-\sigma)-2}.$

Noticing that $\sigma<d/2-2/(m-1)$, we see that $(m-1)(d/2-\sigma)-2>0$.
Taking $N\to\infty$, we immediately obtain that

$\displaystyle\left\|\left.\frac{\partial^{m}u_{\delta}(t_{N})}{\partial\delta^{m}}\right|_{\delta=0}\right\|_{H^{\sigma}}\to\infty.$

So, we have shown the ill-posedness part of (ii) in Theorem 2.1.

Step 4. We show the error estimates for the iteration solutions. The iteration
equations are

$u^{(j+1)}=e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}(u^{(j)}(\tau))^{m}d\tau,\
\ u^{(0)}=0.$

Denote $\widehat{u^{(j)}}=v^{j}$, $\widehat{u}_{0}=v_{0}$. For convenience, we
denote $*^{m}a=\underset{m}{\underbrace{a*...*a}}$. It follows that $v^{j}$
satisfies

$\displaystyle
v^{j+1}=e^{-t|\xi|^{2}}v_{0}+\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}(*^{m}v^{j})(\tau)d\tau,\
\ v^{0}=0.$ (56)

By induction we have

######  Lemma 2.11.

Let $v^{j}$ be the iteration solution of (56). Then we have for all $j\geq 1$,

${\rm supp}\,(v^{j+1}-v^{j})\subset\\{\xi\in\mathbb{R}^{d}_{I}:\,|\xi|\geq
j(m-1)\varepsilon_{0}\\}.$

###### Proof.

It is easy to see that

$v^{1}=e^{-t|\xi|^{2}}v_{0},\ \
v^{2}=e^{-t|\xi|^{2}}v_{0}+\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}(*^{m}v^{1})(\tau)d\tau.$

So, we have ${\rm
supp}\,(v^{2}-v^{1})\subset\\{\xi\in\mathbb{R}^{d}_{I}:\,|\xi|\geq(m-1)\varepsilon_{0}\\}$
and ${\rm
supp}\,v^{2}\subset\\{\xi\in\mathbb{R}^{d}_{I}:\,|\xi|\geq\varepsilon_{0}\\}$.
Let us observe that

$\displaystyle
v^{j+1}-v^{j}=\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}\left((*^{m}v^{j})-(*^{m}v^{j-1})\right)(\tau)d\tau.$
(57)

Using the identity

$(*^{m}v^{j})-(*^{m}v^{j-1})=(v^{j}-v^{j-1})*\left(\sum^{m-1}_{\ell=0}(*^{m-1-\ell}v^{j})*(*^{\ell}v^{j-1})\right),$

by induction we have the result, as desired. ∎

Using the fact that

$v^{j+r}=v^{j}+\sum^{r}_{\ell=1}(v^{j+\ell}-v^{j+\ell-1})$

and Lemma 2.11, we have for any $r\in\mathbb{N}$,

$v^{j+r}(t,\xi)\chi_{\\{\xi\in\mathbb{R}^{d}_{I}:|\xi|<(m-1)j\varepsilon_{0}\\}}=v^{j}(t,\xi)\chi_{\\{\xi\in\mathbb{R}^{d}_{I}:|\xi|<(m-1)j\varepsilon_{0}\\}}.$

By (i) of Theorem 2.1, we see that for any $t>0$,

$\|u(t)-u^{(j)}(t)\|_{E^{s_{0}}_{\sigma}}\to 0,\ \ j\to\infty.$

It follows that

$v(t,\xi)\chi_{\\{\xi\in\mathbb{R}^{d}_{I}:\,|\xi|<(m-1)j\varepsilon_{0}\\}}=v^{j}(t,\xi)\chi_{\\{\xi\in\mathbb{R}^{d}_{I}:\,|\xi|<(m-1)j\varepsilon_{0}\\}},$

which means that

$\displaystyle{\rm
supp}\,(v-v^{j})\subset\left\\{\xi\in\mathbb{R}^{d}_{I}:|\xi|\geq(m-1)j\varepsilon_{0}\right\\}.$
(58)

######  Lemma 2.12.

Let $1\leq p,q\leq\infty$, $\sigma,\tilde{s}\in\mathbb{R}$, $A\gg 1$. Suppose
that ${\rm supp}\,\widehat{f}\subset\\{\xi\in\mathbb{R}^{d}:|\xi|_{\infty}\geq
A\\}$. Then we have

$\left\|\int^{t}_{0}e^{(t-\tau)\Delta}f(\tau)d\tau\right\|_{\widetilde{L}^{\infty}(\mathbb{R}_{+},E^{\tilde{s},\sigma}_{p,q})}\lesssim\frac{1}{A^{2}}\|f\|_{\widetilde{L}^{\infty}(\mathbb{R}_{+},E^{\tilde{s},\sigma}_{p,q})}.$

###### Proof.

Let us recall the decay estimate (cf. [54])

$\|\Box_{k}e^{t\Delta}u_{0}\|_{p}\lesssim
e^{-ct|k|^{2}}\|\Box_{k}u_{0}\|_{p},\ k\in\mathbb{Z}^{d}_{I},\
|k|_{\infty}\geq 1.$

It follows that

$\left\|\Box_{k}\int^{t}_{0}e^{(t-\tau)\Delta}f(\tau)d\tau\right\|_{p}\lesssim\int^{t}_{0}e^{-c(t-\tau)|k|^{2}}\|\Box_{k}f(\tau)\|_{p}d\tau.$

Using Young’s inequality, we have for $|k|_{\infty}\geq A$

$\displaystyle\left\|\Box_{k}\int^{t}_{0}e^{(t-\tau)\Delta}f(\tau)d\tau\right\|_{L^{\infty}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim\|e^{-ct|k|^{2}}\|_{L^{1}_{t}}\|\Box_{k}f(\tau)\|_{L^{\infty}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim\frac{1}{|k|^{2}}\|\Box_{k}f(\tau)\|_{L^{\infty}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim\frac{1}{A^{2}}\|\Box_{k}f(\tau)\|_{L^{\infty}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}.$
(59)

(59) is multiplied by $2^{\tilde{s}|k|}\langle k\rangle^{\sigma}$ and then
equipped with the $\ell^{q}$-norm, we have the result, as desired. ∎

Using the integral equation,

$\displaystyle
v-v^{j}=\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}(*^{m}v-*^{m}v^{j-1})(\tau)d\tau,$
(60)

and taking $p=2,\,q=1,\sigma=0,\tilde{s}=\tilde{s}_{0}<s_{0}$ in Lemma 2.12,
we have

$\displaystyle\sum_{l\in\mathbb{Z}^{d}}$ $\displaystyle
2^{\tilde{s}_{0}|l|}\|v-v^{j}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$
$\displaystyle\lesssim\frac{1}{|j|^{2}}\sum_{l\in\mathbb{Z}^{d}}2^{\tilde{s}_{0}|l|}\left\|(v-v^{j-1})*\sum^{m-1}_{\ell=0}(*^{m-1-\ell}v)*(*^{\ell}v^{j-1})\right\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}.$
(61)

To estimate the right hand side of (2.2), we need the following

######  Lemma 2.13.

Let $\tilde{s}_{0}\leq 0$. Suppose that ${\rm
supp}\,f_{j}\subset\mathbb{R}^{d}_{I}$. Then we have

$\sum_{l\in\mathbb{Z}^{d}}2^{\tilde{s}_{0}|l|}\|f_{1}*...*f_{m}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}\lesssim\prod^{m}_{j=1}\sum_{l\in\mathbb{Z}^{d}}2^{\tilde{s}_{0}|l|}\|f_{j}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}.$

###### Proof.

By Young’s and Hölder’s inequality, we have

$\displaystyle\sum_{l\in\mathbb{Z}^{d}}$ $\displaystyle
2^{\tilde{s}_{0}|l|}\|f_{1}*...*f_{m}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$
$\displaystyle\leq\sum_{l\in\mathbb{Z}^{d}_{I}}2^{\tilde{s}_{0}|l|}\sum_{l_{1},...,l_{m}\in\mathbb{Z}^{d}_{I}}\left\|\chi_{l+[0,1)^{d}}\left((\chi_{l_{1}+[0,1)^{d}}f_{1})*\cdots*(\chi_{l_{m}+[0,1)^{d}}f_{m})\right)\right\|_{L_{t}^{\infty}L_{\xi}^{2}}$
$\displaystyle\leq\sum_{l\in\mathbb{Z}^{d}_{I}}2^{\tilde{s}_{0}|l|}\sum_{l_{1},...,l_{m}\in\mathbb{Z}^{d}_{I}}\|(\chi_{l_{1}+[0,1)^{d}}f_{1})*\cdots*(\chi_{l_{m}+[0,1)^{d}}f_{m})\|_{L_{t}^{\infty}L_{\xi}^{2}}\chi_{\\{|l-l_{1}-...-l_{m}|\leq
m\\}}$
$\displaystyle\lesssim\sum_{l_{1},...,l_{m}\in\mathbb{Z}^{d}_{I}}2^{\tilde{s}_{0}|l_{1}+...+l_{m}|}\|(\chi_{l_{1}+[0,1)^{d}}f_{1})*\cdots*(\chi_{l_{m}+[0,1)^{d}}f_{m})\|_{L_{t}^{\infty}L_{\xi}^{2}}$
$\displaystyle\leq\sum_{l_{1},...,l_{m}\in\mathbb{Z}^{d}_{I}}\prod^{m-1}_{j=1}2^{\tilde{s}_{0}|l_{j}|}\|\chi_{l_{j}+[0,1)^{d}}f_{j}\|_{L_{t}^{\infty}L_{\xi}^{1}}2^{\tilde{s}_{0}|l_{m}|}\|\chi_{l_{m}+[0,1)^{d}}f_{m}\|_{L_{t}^{\infty}L_{\xi}^{2}}$
$\displaystyle\leq\sum_{l_{1},...,l_{m}\in\mathbb{Z}^{d}_{I}}\prod^{m}_{j=1}2^{\tilde{s}_{0}|l_{j}|}\|\chi_{l_{j}+[0,1)^{d}}f_{j}\|_{L_{t}^{\infty}L_{\xi}^{2}}.$

We have the result, as desired. ∎

Using the result of (i) of Theorem 2.1, we see that
$u\in\widetilde{L}^{\infty}(0,\infty;\,E^{s_{0},\sigma}_{2,2})$ and

$\left\|u^{(j)}-u\right\|_{\widetilde{L}^{\infty}(0,\infty;\,E^{s_{0},\sigma}_{2,2})}\to
0,\ \ k\to\infty.$

For any $\tilde{s}_{0}<s_{0}$, by Hölder’s inequality, we have

$\|u\|_{\widetilde{L}^{\infty}(0,\infty;\,E^{\tilde{s}_{0},0}_{2,1})}\leq\|\\{2^{(\tilde{s}_{0}-s_{0})|k|}\langle
k\rangle^{-\sigma}\\}\|_{\ell^{2}}\|u\|_{\widetilde{L}^{\infty}(0,\infty;\,E^{{s}_{0},\sigma}_{2,2})}\lesssim\|u\|_{\widetilde{L}^{\infty}(0,\infty;\,E^{{s}_{0},\sigma}_{2,2})}$

So, by Lemma 2.13 we obtain that

$\displaystyle\sum_{l\in\mathbb{Z}^{d}}2^{\tilde{s}_{0}|l|}$
$\displaystyle\left\|(v-v^{j-1})\sum^{m-1}_{\ell=0}(*^{m-1-\ell}v)*(*^{\ell}v^{j-1})\right\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$
$\displaystyle\lesssim$
$\displaystyle\left(\sum_{l\in\mathbb{Z}^{d}}2^{\tilde{s}_{0}|l|}\left(\|v^{j-1}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}+\|v\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}\right)\right)^{m-1}$
$\displaystyle\times\sum_{l\in\mathbb{Z}^{d}_{I}}2^{\tilde{s}_{0}|l|}\|v-v^{j-1}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$
$\displaystyle\lesssim$
$\displaystyle\left(\|u^{(j)}\|_{\widetilde{L}^{\infty}(0,\infty;\,E^{s_{0},\sigma}_{2,2})}+\|u\|_{\widetilde{L}^{\infty}(0,\infty;\,E^{s_{0},\sigma}_{2,2})}\right)^{m-1}$
$\displaystyle\times\sum_{l\in\mathbb{Z}^{d}_{I}}2^{\tilde{s}_{0}|l|}\|v-v^{j-1}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$

Combining the above estimate with (2.2), we have

$\displaystyle\sum_{l\in\mathbb{Z}^{d}}$ $\displaystyle
2^{\tilde{s}_{0}|l|}\|v-v^{j}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$
$\displaystyle\leq\frac{C}{|j|^{2}}\sum_{l\in\mathbb{Z}^{d}}2^{\tilde{s}_{0}|l|}\|(v-v^{j-1})\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}$
(62)

By induction, we have

$\displaystyle\|u^{(j)}(t)-u(t)\|_{E^{\tilde{s}_{0}}}\leq\sum_{l\in\mathbb{Z}^{d}}$
$\displaystyle
2^{\tilde{s}_{0}|l|}\|v-v^{j}\|_{L^{\infty}_{t}L^{2}_{\xi}(\mathbb{R}_{+}\times(l+[0,1)^{d}))}\leq\frac{C^{j}}{(j!)^{2}}.$
(63)

This proves (iii) of Theorem 2.1.

##  3 SLH for $m\geq 1+4/d$

In the case $m\geq 1+4/d$, we can improve the condition ${\rm
supp}\,\widehat{u}_{0}\subset\\{\xi\in\mathbb{R}^{d}_{I}:\,|\xi(j)|_{\infty}\geq\varepsilon_{0}\\}$
for some $\varepsilon_{0}>0$ by a weak condition ${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}$. For example, let $d=1$ and

$\displaystyle u_{0}$
$\displaystyle=A\,\frac{d}{dx}\left(\delta(x)+\frac{2\,\mathrm{i}}{x}\right),\
A\in\mathbb{C},$ (64)

we have $\widehat{u}_{0}=i\xi\chi_{\\{\xi:\,\xi\geq 0\\}}$ and ${\rm
supp}\widehat{u}_{0}\subset[0,\infty)$. Comparing (64) with (16), we see that
the imaginary part of $u_{0}$ in (16) changes signs for infinite times and in
(64) never changes signs.

###  3.1 Scaling in the critical cases

Recalling that for the initial data $u_{0}\in E^{s}_{\sigma}$ with ${\rm
supp}\,\widehat{u}_{0}\subset\\{\xi:\,|\xi|\geq\varepsilon_{0}\\}$, we have
shown that $\|u_{0,\lambda}\|_{E^{s}_{\sigma}}$ vanishes as $\lambda\to\infty$
by using the exponential decay of $2^{s\lambda\varepsilon_{0}}$. However, in
the critical case $\sigma=d/2-2/(m-1)$, it still holds
$\lim_{\lambda\to\infty}\|u_{0,\lambda}\|_{E^{s}_{\sigma}}=0$ if ${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}$.

######  Lemma 3.1.

Let $s<0$, $\sigma\geq 0$. Denote $f_{\lambda}=\lambda^{a}f(\lambda\,\cdot)$.
Assume that $\sigma+a=d/2$. Then for any $f\in E^{s}_{\sigma}$, we have

$\displaystyle\lim_{\lambda\to+\infty}\|f_{\lambda}\|_{E^{s}_{\sigma}}=0.$
(65)

###### Proof.

Noticing that $\sigma+a=d/2$, We have for any $\lambda>1$

$\displaystyle\|f_{\lambda}\|_{E^{s}_{\sigma}}=\lambda^{-d+a}\|\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}(\xi/\lambda)\|_{2}\leq\|\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{f}\|_{2}.$
(66)

Since $f\in E^{s}_{\sigma}$, we have

$\lim_{\delta\to
0}\|\chi_{\\{|\xi|\leq\delta\\}}\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}\|_{2}=0.$

Hence, for any $\varepsilon>0$, there exists $\delta_{0}>0$ such that

$\displaystyle\|\chi_{\\{|\xi|\leq\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{f}\|_{2}\leq\|\chi_{\\{|\xi|\leq\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}\|_{2}<\varepsilon/2.$
(67)

On the other hand,

$\displaystyle\|\chi_{\\{|\xi|>\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{f}\|_{2}\leq
2^{s(\lambda-1)\delta_{0}}\|\chi_{\\{|\xi|\leq\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s|\xi|}\widehat{f}\|_{2}\leq
2^{s(\lambda-1)\delta_{0}}\|f\|_{E^{s}_{\sigma}}.$

Choosing $\lambda_{0}$ sufficiently large such that

$2^{s(\lambda-1)\delta_{0}}\|f\|_{E^{s}_{\sigma}}<\varepsilon/2,$

we have for any $\lambda>\lambda_{0}$,

$\displaystyle\|\chi_{\\{|\xi|>\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{f}\|_{2}<\varepsilon/2.$
(68)

In view of (66)–(68), we have for $\lambda>\lambda_{0}$,

$\displaystyle\|f_{\lambda}\|_{E^{s}_{\sigma}}\leq\|\chi_{\\{|\xi|\leq\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{f}\|_{2}+\|\chi_{\\{|\xi|>\delta_{0}\\}}\langle\xi\rangle^{\sigma}2^{s\lambda|\xi|}\widehat{f}\|_{2}<\varepsilon,$
(69)

which implies the result. ∎

###  3.2 Result in the case $m\geq 1+4/d$

For convenience, we denote

$\displaystyle\|u\|_{X^{s,\sigma}}=$
$\displaystyle\|u\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))\cap\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}$
$\displaystyle+\|\Box_{0}u\|_{L^{\infty}_{t}L^{2}_{x}\cap
L^{m}_{t}L^{2m}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})},$ (70)

######  Theorem 3.2.

Let $d\geq 1$, $f(u)=u^{m}$, $m\in\mathbb{N}$, $m\geq 1+4/d$,
$\sigma=d/2-2/(m-1)$. Let $s<0$, $u_{0}\in E^{s}_{\sigma}$ with
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}$. Then there exists
$s_{0}:=s_{0}(m,u_{0})\leq s$ such that SLH (3) has a unique solution $u\in
X^{s_{0},\sigma}$ satisfying the following equivalent integral equation:

$\displaystyle
u(t)=e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u(\tau)^{m}~{}d\tau.$ (71)

Moreover, $u\in C([0,\infty);E^{s_{0}}_{\sigma})$.

Theorem 3.2 needs several remarks.

  * (i)

Let $u_{0,\lambda}$ be as in (51). From the proof of Lemma 3.1, we see that
$\lim_{\lambda\to\infty}\|u_{0,\lambda}\|_{E^{s}_{\sigma}}=0$ and the
vanishing process of $\|u_{0,\lambda}\|_{E^{s}_{\sigma}}$ depends on both
$u_{0}$ and its norm. It follows that $s_{0}\leq s$ also depends on $u_{0}$
and its norm in $E^{s}_{\sigma}$. However, we can take $s=s_{0}$ if $u_{0}\in
E^{s}_{\sigma}$ with
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}$ is sufficiently
small.

  * (iii)

By the same reason as in Theorem 2.1, condition ${\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}$ is necessary for the
existence of the global solutions.

###  3.3 Proof of Theorem 3.2

Step 1. We prove the result for sufficiently small initial data $u_{0}\in
E^{s}_{\sigma}$. We need a lemma to control the lower frequency part of the
solution, see [52].

######  Lemma 3.3.

Let $2\leq p<\infty$, $2/\gamma\leq d(1/2-1/p)$. Then we have

$\displaystyle\|\Box_{0}e^{t\Delta}u_{0}\|_{L^{\gamma}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim\|\Box_{0}u_{0}\|_{2},$ (72)
$\displaystyle\left\|\Box_{0}\int^{t}_{0}e^{(t-\tau)\Delta}f(\tau)d\tau\right\|_{L^{\gamma}_{t}L^{p}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
$\displaystyle\lesssim\left\|\Box_{0}f\right\|_{L^{1}_{t}L^{2}_{x}(\mathbb{R}_{+}\times\mathbb{R}^{d})}$
(73)

Let us consider the mapping

$\displaystyle\mathcal{T}:u(t)\to
e^{t\Delta}u_{0}+\int_{0}^{t}e^{(t-\tau)\Delta}u^{m}(\tau)~{}d\tau$

in the space

$\displaystyle\mathcal{D}=\\{u:\,{\rm
supp}\,\widehat{u(t)}\subset\mathbb{R}^{d}_{I},\
\|u\|_{X^{s,\sigma}}\leq\delta\\},$

where $\|u\|_{X^{s,\sigma}}$ is defined in (70). By Lemma 3.3,

$\displaystyle\|\Box_{0}\mathcal{T}u\|_{L^{\infty}_{t}L^{2}_{x}\cap
L^{m}_{t}L^{2m}_{x}}\leq\,$
$\displaystyle\,\|\Box_{0}e^{t\Delta}u_{0}\|_{L^{\infty}_{t}L^{2}_{x}\cap
L^{m}_{t}L^{2m}_{x}}$
$\displaystyle\,+\left\|\Box_{0}\int^{t}_{0}e^{(t-\tau)\Delta}u(\tau)^{m}d\tau\right\|_{L^{\infty}_{t}L^{2}_{x}\cap
L^{m}_{t}L^{2m}_{x}}$ $\displaystyle\lesssim\,$
$\displaystyle\|u_{0}\|_{2}+\left\|\Box_{0}(u^{m})\right\|_{L^{1}_{t}L^{2}_{x}}.$
(74)

Using $\sum_{k}\Box_{k}=I$, we have

$\displaystyle\left\|\Box_{0}(u^{m})\right\|_{L^{1}_{t}L^{2}_{x}}\leq\sum_{k_{1},...,k_{m}\in\mathbb{Z}^{d}_{I}}\|\Box_{0}(\Box_{k_{1}}u\Box_{k_{2}}u...\Box_{k_{m}}u)\|_{L^{1}_{t}L^{2}_{x}}.$
(75)

Since ${\rm supp}\,\widehat{u(t)}\subset\mathbb{R}^{d}_{I}$, we see that

$\Box_{0}(\Box_{k_{1}}u\Box_{k_{2}}u...\Box_{k_{m}}u)=0$

if there exists $k_{j}$ such that $|k_{j}|_{\infty}\geq 1$. Hence

$\displaystyle\left\|\Box_{0}(u^{m})\right\|_{L^{1}_{t}L^{2}_{x}}=\|\Box_{0}(\Box_{0}u)^{m}\|_{L^{1}_{t}L^{2}_{x}}\leq\|(\Box_{0}u)^{m}\|_{L^{1}_{t}L^{2}_{x}}\leq\|\Box_{0}u\|^{m}_{L^{m}_{t}L^{2m}_{x}}.$
(76)

It follows that

$\displaystyle\|\Box_{0}\mathcal{T}u\|_{L^{\infty}_{t}L^{2}_{x}\cap
L^{m}_{t}L^{2m}_{x}}$
$\displaystyle\lesssim\|\Box_{0}u_{0}\|_{2}+\|\Box_{0}u\|^{m}_{L^{m}_{t}L^{2m}_{x}}$
$\displaystyle\lesssim\|u_{0}\|_{E^{s}_{\sigma}}+\|u\|^{m}_{X^{s,\sigma}}.$
(77)

Next, by Corollary 2.3, we can estimate

$\displaystyle\|\mathcal{T}u$
$\displaystyle\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))\cap\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}$
$\displaystyle\leq\ $
$\displaystyle\|e^{t\Delta}u_{0}\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))\cap\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}$
$\displaystyle\,+\left\|\int^{t}_{0}e^{(t-\tau)\Delta}u(\tau)^{m}d\tau\right\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))\cap\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}$
$\displaystyle\lesssim$ $\displaystyle\
\|u_{0}\|_{E^{s}_{\sigma}}+\|u^{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}.$
(78)

One has that

$\displaystyle\|u^{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}$
$\displaystyle\leq\left(\sum_{k\in\mathbb{Z}^{d}\setminus\\{0\\}}2^{2s|k|}\langle
k\rangle^{2\sigma}\|\Box_{k}(\Box_{0}u)^{m}\|_{L^{1}_{t}L^{2}_{x}}\right)^{1/2}$
$\displaystyle\quad\quad+\left(\sum_{k\in\mathbb{Z}^{d}\setminus\\{0\\}}2^{2s|k|}\langle
k\rangle^{2\sigma}\|\Box_{k}(u^{m}-(\Box_{0}u)^{m})\|_{L^{1}_{t}L^{2}_{x}}\right)^{1/2}$
$\displaystyle:=I+II.$ (79)

Noticing that $\Box_{k}(\Box_{0}u)^{m}\neq 0$ implies that $|k|_{\infty}\leq
m$, we have

$\displaystyle I$ $\displaystyle\lesssim
m^{d/2+\sigma}\|\Box_{0}u\|^{m}_{L^{m}_{t}L^{2m}_{x}}.$ (80)

Applying Remark 2.6,

$\displaystyle II\leq
C_{m}\|u\|^{m-1}_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}\|u\|_{\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}.$
(81)

It is easy to see that

$\displaystyle\|u\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2})}\leq\|\Box_{0}u\|_{L^{\infty}_{t}L^{2}_{x}}+\|u\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}\leq\|u\|_{X^{s,\sigma}}.$
(82)

By (81) and (82),

$\displaystyle II\leq C_{m}\|u\|^{m}_{X^{s,\sigma}}.$ (83)

Then, by (78), (79), (80) and (83),

$\displaystyle\|\mathcal{T}u\|_{\widetilde{L}^{\infty}(0,\infty;E^{s,\sigma}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))\cap\widetilde{L}^{1}(0,\infty;E^{s,\sigma+2}_{2,2}(\mathbb{Z}^{d}_{I}\setminus\\{0\\}))}\lesssim$
$\displaystyle\ \|u_{0}\|_{E^{s}_{\sigma}}+C_{m}\|u\|^{m}_{X^{s,\sigma}}.$
(84)

By (3.3) and (84)

$\displaystyle\|\mathcal{T}u\|_{X^{s,\sigma}}\lesssim$ $\displaystyle\
\|u_{0}\|_{E^{s}_{\sigma}}+C_{m}\|u\|^{m}_{X^{s,\sigma}}.$ (85)

So, it follows from (85) that SLH (3) has a unique solution in $X^{s,\sigma}$
if $\|u_{0}\|_{E^{s}_{\sigma}}$ is sufficiently small.

Step 2. We consider the large initial data in $E^{s}_{\sigma}$. The idea is
the same as in the proof of Theorem 2.1. By Lemma 3.1,
$u_{0,\lambda}=\lambda^{2/(m-1)}u(\lambda\,\cdot)$ satisfies

$\lim_{\lambda\to\infty}\|u_{0,\lambda}\|_{E^{s}_{\sigma}}=0.$

So, for any $u_{0}\in E^{s}_{\sigma}$, we can choose a $\lambda\gg 1$ such
that

$\|u_{0,\lambda}\|_{E^{s}_{\sigma}}\ll 1.$

It follows that

$\displaystyle
v(t)=e^{t\Delta}u_{0,\lambda}+\int_{0}^{t}e^{(t-\tau)\Delta}v(\tau)^{m}~{}d\tau.$
(86)

has a unique solution $u_{\lambda}\in X^{s,\sigma}$. Taking
$u=\lambda^{-2/(m-1)}u(t/\lambda^{2},x/\lambda)$, we can repeat the procedures
as in the proof of Theorem 2.1 to get the result of Theorem 3.2.

##  4 SLH with an exponential nonlinearity

The heat equation with exponential nonlinearities has been studied in [16, 17,
31, 50], where the global existence and blowup behavior for the radial
solutions were considered in the real-valued cases. We will study the Cauchy
problem for the heat equation with an exponential nonlinearity

$u_{t}-\Delta u=e^{u}-1,\ \ u(0,x)=u_{0}(x),$ (87)

where $u(t,x)$ is a complex-valued function of
$(t,x)\in[0,\infty)\times\mathbb{R}^{d}$, $u_{0}\in{E}^{s}_{\sigma}$ with
$s<0$ and $\widehat{u}_{0}$ is supported in the first octant and away from the
unit ball, say

$\displaystyle{\rm supp}\,\widehat{u}_{0}\subset\\{\xi\in\mathbb{R}^{d}:\
\xi_{i}\geq 0,\ i=1,...,d,\ |\xi|_{\infty}\geq 2\\}.$ (88)

The main result of this section is the following

######  Theorem 4.1.

Let $s<0$, $\sigma\geq d/2$, and $u_{0}\in E^{s}_{\sigma}$ satisfy (88) Then
there exists $s_{0}\leq s$ such that (87) has a unique solution

$u\in
C(\mathbb{R}_{+},E^{s_{0}}_{\sigma})\cap\widetilde{L}^{\infty}(0,\infty;E^{s_{0},\sigma}_{2,2})\cap\widetilde{L}^{1}(0,\infty;E^{s_{0},\sigma+2}_{2,2})$

in the sense that $u$ satisfies the equivalent integral equation

$\displaystyle
u(t)=e^{t(I+\Delta)}u_{0}+\int^{t}_{0}e^{(t-\tau)(I+\Delta)}(e^{u(\tau)}-u(\tau)-1)d\tau.$
(89)

Recalling that

$\displaystyle e^{u}=\sum^{\infty}_{m=0}\frac{u^{m}}{m!},$ (90)

we see that Fujita’s critical and subcritical powers are contained in the
Taylor expansion of $e^{u}$ in the cases $m=2,3$ for $d=1$, and $m=2$ for
$d=2$. Moreover, the linear part $u$ contained in the Taylor’s expansion of
$e^{u}$ plays a bad role which prevents the global existence of solutions.
Indeed, let us observe the semi-group $e^{t(I+\Delta)}$, its lower frequency
part has no time-decay and one easily sees that

$e^{t/2}\|P_{\leq 1/2}u_{0}\|_{2}\leq\|e^{t(I+\Delta)}P_{\leq
1/2}u_{0}\|_{2}\leq e^{t}\|u_{0}\|_{2},\ \ P_{\leq
1/2}=\mathscr{F}^{-1}\chi_{|\xi|\leq 1/2}\mathscr{F}.$

So, the lower frequency part of the free solutions tends to infinity as
$t\to\infty$. This is why we assume that the support set of the Fourier
transform of initial data is away from the unit ball in (88).

###  4.1 Linear Estimates

Now, let us make a scaling to (87). Denote

$u_{\lambda}(t,x)=u(\lambda^{2}t,\lambda x).$

If $u$ is a solution of (87), we easily see that $u_{\lambda}$ solves

$u_{t}-\Delta u=\lambda^{2}(e^{u}-1),\ \ u(0,x)=u_{0}(\lambda x).$ (91)

(91) can be rewritten as

$u_{t}-(\lambda^{2}+\Delta)u-\lambda^{2}(e^{u}-u-1)=0,\ \ u(0,x)=u_{0}(\lambda
x),$

which is essentially equivalent to the following integral equation

$\displaystyle
u(t)=e^{t(\lambda^{2}I+\Delta)}u_{0}(\lambda\,\cdot)+\lambda^{2}\int^{t}_{0}e^{(t-\tau)(\lambda^{2}I+\Delta)}(e^{u(\tau)}-u(\tau)-1)d\tau.$
(92)

We can first solve (92). For the sake of convenience, we write

$H_{\lambda}(t):=e^{t(\lambda^{2}I+\Delta)}=\mathscr{F}^{-1}e^{-t(|\xi|^{2}-\lambda^{2})}\mathscr{F}.$

Let us consider the estimate of $H_{\lambda}(t)$. As indicated in the above,
$H_{\lambda}(t)$ has no time-decay for $|\xi|\leq\lambda$.

######  Lemma 4.2.

Let $1\leq p\leq\infty$, $\lambda\gg 1$. Assume that $k\in\mathbb{Z}^{d}$ and
$|k|_{\infty}\geq 2\lambda$. Then we have

$\displaystyle\|\Box_{k}H_{\lambda}(t)u_{0}\|_{p}\leq
Ce^{-ct|k|^{2}}\|\Box_{k}u_{0}\|_{p},$ (93)

where $C$ is independent of $\lambda$ and $k$.

Proof. Let $\sigma$ be a smooth cut-off function verifying $\sigma(\xi)=1$ for
$\xi\in[0,1]^{d}$ and $\sigma(\xi)=0$ if $\xi\not\in[-1/4,5/4]^{d}$. Put
$\sigma_{k}=\sigma(\cdot-\xi)$. By Young’s inequality,

$\|\Box_{k}H_{\lambda}(t)u_{0}\|_{p}\leq\left\|\mathscr{F}^{-1}\left(\sigma_{k}e^{t(\lambda^{2}-|\xi|^{2})}\right)\right\|_{1}\|\Box_{k}u_{0}\|_{p}.$

Using the translation invariance,

$\left\|\mathscr{F}^{-1}\left(\sigma_{k}e^{t(\lambda^{2}-|\xi|^{2})}\right)\right\|_{1}=e^{t(\lambda^{2}-|k|^{2})}\left\|\mathscr{F}^{-1}\left(\sigma
e^{-t(|\xi|^{2}+2k\xi)}\right)\right\|_{1}.$

Noticing that ${\rm supp}\,\sigma$ is compact, in view of Proposition 1.3 we
have

$\left\|\mathscr{F}^{-1}\left(\sigma
e^{-t(|\xi|^{2}+2k\xi)}\right)\right\|_{1}\lesssim e^{Ct|k|}.$

Noticing that $|k|_{\infty}\geq 2\lambda$, one has the result as desired.
$\hfill\Box$

######  Lemma 4.3.

Let $1\leq\gamma,\gamma_{1},p\leq\infty$, $\gamma_{1}\leq\gamma$, $\lambda\gg
1$. Assume that

$\displaystyle{\rm supp}\,\widehat{u}_{0},{\rm
supp}\,\widehat{f(t)}\subset\mathbb{R}^{d}_{I,\lambda}:=\\{\xi\in\mathbb{R}^{d}_{I}:|\xi|_{\infty}\geq
2\lambda\\}.$ (94)

Then we have

$\displaystyle\|\Box_{k}H_{\lambda}(t)u_{0}\|_{L^{\gamma}_{t}L^{p}_{x}}$
$\displaystyle\leq C|k|^{-2/\gamma}\|\Box_{k}u_{0}\|_{p},$ (95)
$\displaystyle\left\|\int^{t}_{0}\Box_{k}H_{\lambda}(t-\tau)f(\tau)d\tau\right\|_{L^{\gamma}_{t}L^{p}_{x}}$
$\displaystyle\leq
C|k|^{-2/\gamma-2/\gamma^{\prime}_{1}}\left\|\Box_{k}f\right\|_{L^{\gamma^{\prime}_{1}}_{t}L^{p}_{x}},$
(96)

where $C$ is independent of $\lambda$ and $k$.

Proof. Taking the $L^{\gamma}$ norm in both sides of (93), we immediately have
(95). By (93),

$\displaystyle\left\|\int^{t}_{0}\Box_{k}H_{\lambda}(t-\tau)f(\tau)d\tau\right\|_{p}\leq\int^{t}_{0}e^{-c(t-\tau)|k|^{2}}\|\Box_{k}f(\tau)\|_{p}d\tau.$
(97)

Taking $L^{\gamma}$ norm in both sides of (97) and using Young’s inequality,
we obtain (96). $\hfill\Box$

######  Corollary 4.4.

Let $s\leq 0$, $\sigma\in\mathbb{R}$, $1\leq\gamma_{1}\leq\gamma\leq\infty$.
Suppose that $\mathrm{supp}~{}\mathscr{F}_{x}f,\,{\rm
supp}\,\widehat{u}_{0}\subset\mathbb{R}^{d}_{I,\lambda}$. Then there exists
$C>0$ which is independent of $\lambda\gg 1$ such that

$\displaystyle\|H_{\lambda}(t)u_{0}\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})}$
$\displaystyle\leq C\|u_{0}\|_{E^{s}_{\sigma}},$ (98)
$\displaystyle\left\|\int_{0}^{t}H_{\lambda}(t-\tau)f(\tau)~{}d\tau\right\|_{\widetilde{L}^{\gamma}(0,\infty;E^{s,\sigma+2/\gamma}_{2,2})}$
$\displaystyle\leq
C\|f\|_{\widetilde{L}^{\gamma_{1}}(0,\infty;E^{s,\sigma-2/\gamma^{\prime}_{1}}_{2,2})}.$
(99)

###### Proof.

Taking $p=2$, multiplying $2^{s|k|}\langle k\rangle^{\sigma+2/\gamma}$ and
then equipping $\ell^{2}$ norms in both sides of (95) and (96), we have the
results, as desired. ∎

###  4.2 Proof of Theorem 4.1

Let us consider the scaling of initial data
$u_{0,\lambda}=u_{0}(\lambda\,\cdot)$. By Lemma 2.7 we see that

$\|u_{0,\lambda}\|_{E^{s}_{\sigma}}\leq
C2^{s(\lambda-1)}\lambda^{\sigma-d/2}\|u_{0}\|_{E^{s}_{\sigma}}\to 0,\ \
\lambda\to+\infty.$

Let $\lambda\in\mathbb{N}$ be a sufficiently large number. Similar to the
above argument, Taking
$\delta=2C2^{s(\lambda-1)}\lambda^{\sigma-d/2}\|u_{0}\|_{E^{s}_{\sigma}}$ and

$X_{s,\sigma,\lambda}=\left\\{u:\,\|u\|_{X_{s,\sigma,\lambda}}=\|u\|_{\widetilde{L}^{\infty}(\mathbb{R}_{+};E^{s,\sigma}_{2,2})\cap\widetilde{L}^{1}(\mathbb{R}_{+};E^{s,\sigma+2}_{2,2})}\leq\delta,\
{\rm supp}\,\widehat{u(t)}\subset\mathbb{R}^{d}_{I,\lambda}\right\\},$

where $\mathbb{R}^{d}_{I,\lambda}$ is as in (94). Let us consider the mapping

$\displaystyle\mathscr{T}_{\lambda}:u(t)\to
e^{t(\lambda^{2}I+\Delta)}u_{0,\lambda}+\lambda^{2}\int^{t}_{0}e^{(t-\tau)(\lambda^{2}I+\Delta)}(e^{u(\tau)}-u(\tau)-1)d\tau.$

Applying Corollary 2.3, Lemmas 2.4 and 2.5, we can get for any $u\in
X_{s,\lambda}$, $\sigma\geq d/2$,

$\displaystyle\|\mathscr{T}_{\lambda}u\|_{X_{s,\sigma,\lambda}}$
$\displaystyle\leq
C\|u_{0}\|_{E^{s}_{\sigma}}+\lambda^{2}C\|e^{u}-u-1\|_{\widetilde{L}^{1}(\mathbb{R}_{+};E^{s,\sigma}_{2,2})}$
$\displaystyle\leq
C\|u_{0,\lambda}\|_{E^{s}_{\sigma}}+C\lambda^{2}\sum^{\infty}_{m=2}\frac{1}{m!}\|u^{m}\|_{\widetilde{L}^{1}(\mathbb{R}_{+};E^{s,\sigma}_{2,2})}$
$\displaystyle\leq
C\|u_{0,\lambda}\|_{E^{s}_{\sigma}}+C\lambda^{2}\sum^{\infty}_{m=2}\frac{C^{m}m^{m/2}}{m!}\|u\|^{m}_{\widetilde{L}^{m}(\mathbb{R}_{+};E^{s,\sigma+2/m}_{2,2})}$
$\displaystyle\leq
C\|u_{0,\lambda}\|_{E^{s}_{\sigma}}+C\lambda^{2}\sum^{\infty}_{m=2}\frac{C^{m}m^{m/2}}{m!}\|u\|^{m}_{X_{s,\sigma,\lambda}}$
$\displaystyle\leq\delta/2+C\lambda^{2}\delta^{2}\sum^{\infty}_{m=2}\frac{C^{m}m^{m/2}}{m!}\delta^{m-2}.$

If $\lambda$ is large enough, then we have $\delta\leq 1$ and

$\displaystyle\sum^{\infty}_{m=2}\frac{C^{m}m^{m/2}}{m!}\delta^{m-2}\leq C.$

Moreover, we can assume that

$C^{2}\lambda^{2}\delta\leq\frac{1}{2}.$

Hence, we have

$\displaystyle\|\mathscr{T}_{\lambda}u\|_{X_{s,\sigma,\lambda}}\leq\delta.$

Similarly

$\displaystyle\|\mathscr{T}_{\lambda}u-\mathscr{T}_{\lambda}v\|_{X_{s,\sigma,\lambda}}\leq\frac{1}{2}\|u-v\|_{X_{s,\sigma,\lambda}}.$

Hence, $\mathscr{T}_{\lambda}:X_{s,\sigma,\lambda}\to X_{s,\sigma,\lambda}$
has a fixed point $u_{\lambda}\in X_{s,\sigma,\lambda}$, which solves

$\displaystyle
u(t)=e^{t(\lambda^{2}I+\Delta)}u_{0,\lambda}+\lambda^{2}\int^{t}_{0}e^{(t-\tau)(\lambda^{2}I+\Delta)}(e^{u(\tau)}-u(\tau)-1)d\tau.$

Taking

$u(t,x)=u_{\lambda}\left(\frac{t}{\lambda^{2}},\frac{x}{\lambda}\right),$

by the scaling argument we see that $u$ satisfies

$\displaystyle
u(t)=e^{t(I+\Delta)}u_{0}+\int^{t}_{0}e^{(t-\tau)(I+\Delta)}(e^{u(\tau)}-u(\tau)-1)d\tau.$

Following Lemma 2.8 we have

$u\in\widetilde{L}^{\infty}(0,\infty;E^{s_{0},\sigma}_{2,2})\cap\widetilde{L}^{1}(0,\infty;E^{s_{0},\sigma+2}_{2,2}),\
\ s_{0}=\lambda s.$

$u\in C(\mathbb{R}_{+},E^{s_{0}}_{\sigma})$ can be obtained by following the
same way as in Theorem 2.1 and the details are omitted.

##  5 Initial data in $E^{s}_{2,1}$

Denote $Q_{k}:=k+[0,1)^{d}$ for $k\in\mathbb{Z}^{d}$, Let $s<0$. Define (cf.
[14])

$E^{s}_{2,1}:=\left\\{f\in\mathscr{S}_{1}^{\prime}:\|f\|_{E^{s}_{2,1}}=\sum_{k\in\mathbb{Z}^{d}}2^{s|k|}\|\widehat{f}\|_{L^{2}(Q_{k})}<\infty\right\\}.$

By the inclusion $l^{1}\subset l^{2}$, we see that $E^{s}_{2,1}\subset
E^{s}_{\sigma}$ for any $\sigma\leq 0$. By (8) and Cauchy-Schwarz inequality,
$E^{s}_{\sigma}\subset E^{s}_{2,1}$ for any $\sigma>d/2$. It follows that

$\displaystyle E^{s}_{\sigma_{1}}\subset E^{s}_{2,1}\subset
E^{s}_{\sigma_{2}},\ \ \forall\ \sigma_{1}>d/2,\ \sigma_{2}\leq 0$ (100)

and the above embeddings are sharp in the sense that

$\displaystyle E^{s}_{d/2}\not\subset E^{s}_{2,1}\not\subset
E^{s}_{\varepsilon},\ \ \forall\ \varepsilon>0.$ (101)

$E^{s}_{2,1}$ has algebraic structures if it is localized in the first octant
of frequency spaces. So, we can use a similar way as in the cases of initial
data in $E^{s}_{\sigma}$ to obtain the following

######  Theorem 5.1.

Let $d\geq 1$, $f(u)=u^{m}$, $m\in\mathbb{N}\setminus\\{1\\}$. Let $s<0$,
$u_{0}\in E^{s}_{2,1}$ with
$\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I}\setminus\\{0\\}$.
Then there exists $s_{0}\leq s$ such that SLH (3) has a unique solution

$u\in
C([0,\infty);E^{s_{0}}_{2,1})\cap\widetilde{L}^{\infty}(0,\infty;E^{s_{0}}_{2,1})\cap\widetilde{L}^{1}(0,\infty;E^{s_{0}}_{2,1})$

in the sense that $u$ satisfies the integral equation (13).

For the Cauchy problem of the semi-linear heat equation with an exponential
nonlinearity, we have

######  Theorem 5.2.

Let $s<0$ and $u_{0}\in E^{s}_{2,1}$ satisfy (88) Then there exists $s_{0}\leq
s$ such that (87) has a unique solution

$u\in
C(\mathbb{R}_{+},E^{s_{0}}_{2,1})\cap\widetilde{L}^{\infty}(0,\infty;E^{s_{0}}_{2,1})\cap\widetilde{L}^{1}(0,\infty;E^{s_{0}}_{2,1})$

in the sense that $u$ satisfies the integral equation (89).

By (100) and (101), we see that Theorems 5.1 and 5.2 contain a class of
initial data not covered by the spaces $E^{s}_{\sigma}$ as in Theorems 2.1,
3.2 and 4.1. Now we sketch their proofs.

######  Lemma 5.3.

Assume $s<0$, $f\in E^{s}_{2,1}$ and
$\mathrm{supp}~{}\widehat{f}\subset\mathbb{R}^{d}_{I,+}$, $\lambda\gg 1$. Then

$\displaystyle\|f_{\lambda}\|_{E^{s}_{2,1}}\leq
C2^{s(\lambda-1)}\|f\|_{E^{s}_{2,1}}.$

###### Proof.

It suffices to consider the case
$\mathrm{supp}~{}\widehat{f}\subset\\{\xi:\xi_{1}\geq
1\\}\cap\mathbb{R}^{d}_{I,+}$. By the definition of $E^{s}_{2,1}$ and the
support property of $\widehat{f}$, we have from Cauchy-Schwarz inequality that

$\displaystyle\|f_{\lambda}\|_{E^{s}_{2,1}}$
$\displaystyle=\sum_{k}2^{s|k|}\lambda^{-d}\|\widehat{f}(\lambda^{-1}\xi)\|_{L^{2}_{\xi\in
k+[0,1]^{d}}}$
$\displaystyle=\sum_{k}\lambda^{-d/2}2^{s|k|}\|\widehat{f}(\xi)\|_{L^{2}_{\xi\in\lambda^{-1}(k+[0,1]^{d})}}$
$\displaystyle\leq\sum_{l:l_{1}\geq
1}\lambda^{-d/2}\sum_{k:\lambda^{-1}(k+[0,1]^{d})\cap
l+[0,1]^{d}\not=\emptyset}2^{s|k|}\|\widehat{f}(\xi)\|_{L^{2}_{\xi\in\lambda^{-1}(k+[0,1]^{d})}}$
$\displaystyle\lesssim\sum_{l:l_{1}\geq
1}2^{s\lambda|l|}\|\widehat{f}(\xi)\|_{L^{2}_{\xi\in l+[0,1]^{d}}}$
$\displaystyle\lesssim 2^{s(\lambda-1)}\|f\|_{E^{s}_{2,1}}.$

We immediately have the result, as desired. ∎

######  Lemma 5.4.

Assume
$\mathrm{supp}\,\mathscr{F}_{x}f(t,\cdot),\mathrm{supp}~{}\widehat{u}_{0}\subset\mathbb{R}^{d}_{I,+}$
for all $t>0$. Then,

$\displaystyle\|e^{t\Delta}u_{0}\|_{\widetilde{L}^{1}(0,\infty;E^{s}_{2,1})\cap\widetilde{L}^{\infty}(0,\infty;E^{s}_{2,1})}$
$\displaystyle\lesssim\|u_{0}\|_{E^{s}_{2,1}},$
$\displaystyle\left\|\int_{0}^{t}e^{(t-\tau)\Delta}f(\tau)~{}d\tau\right\|_{\widetilde{L}^{1}(0,\infty;E^{s}_{2,1})\cap\widetilde{L}^{\infty}(0,\infty;E^{s}_{2,1})}$
$\displaystyle\lesssim\|f\|_{\widetilde{L}^{1}(0,\infty;E^{s}_{2,1})}.$

######  Lemma 5.5.

Let $s<0$. Assume that ${\rm supp}\,\widehat{u}\subset\mathbb{R}^{d}_{I}$.
Then we have

$\displaystyle\|u^{m}\|_{\widetilde{L}^{1}(0,\infty;E^{s}_{2,1})}\leq
C^{m}m^{d}\|u\|^{m-1}_{\widetilde{L}^{\infty}(0,\infty;E^{s}_{2,1})}\|u\|_{\widetilde{L}^{1}(0,\infty;E^{s}_{2,1})}$
(102)

Lemmas 5.4 and 5.5 can be obtained by following similar ways as in Section 2
and we omit the details of their proofs. Then we can imitate the procedures as
in Sections 2 and 4 to get the proofs of Theorems 5.1 and 5.2, respectively.

##  6 An example of solutions for SLH

Let $u$ be the solution of

$\partial_{t}u-\Delta u-u^{2}=0,\ \ u(0,x)=\delta u_{0}(x).$ (103)

Let $d=1$, $v=\widehat{u}$, and $\widehat{u}_{0}(\xi)=v_{0}(\xi)$. Then $v$
satisfies

$v(\xi,t)=\delta
e^{-t|\xi|^{2}}v_{0}+\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}v(\tau)*v(\tau)d\tau$

Let $\mathbb{R}^{d}_{I,+}$ be as in (18). Suppose that ${\rm
supp}\,v_{0}\subset\mathbb{R}^{d}_{I,+}$. Using the support set of $v_{0}$ and
similar to (58), one sees that for any $K\in\mathbb{N}$,

$v(t,\xi)=\sum^{K}_{k=1}\frac{\delta^{k}}{k!}\,\frac{\partial^{k}v}{\partial\delta^{k}}\Big{|}_{\delta=0}(t,\xi),\
\ |\xi|<K,$ (104)

where

$\displaystyle\frac{\partial v}{\partial\delta}\Big{|}_{\delta=0}(t,\xi)$
$\displaystyle=e^{-t\xi^{2}}v_{0}(\xi),$
$\displaystyle\frac{\partial^{k}v}{\partial\delta^{k}}\Big{|}_{\delta=0}(t,\xi)$
$\displaystyle=\int_{0}^{t}e^{-(t-\tau)\xi^{2}}\sum^{k-1}_{i=1}\binom{k}{i}\left(\frac{\partial^{k-i}v}{\partial\delta^{k-i}}\Big{|}_{\delta=0}\right)*\left(\frac{\partial^{i}u}{\partial\delta^{i}}\Big{|}_{\delta=0}\right)(\tau,\xi)d\tau.$
(105)

######  Example 6.1.

Let $H(\xi-1)$ be the translation of Heaviside function,

$H(\xi)=\left\\{\begin{array}[]{ll}1,&\xi\geq 0,\\\
0,&\xi<0.\end{array}\right.$

Let $v_{0}(\xi)=e^{\xi}H(\xi-1)$. We have

$\displaystyle\frac{\partial
v}{\partial\delta}\Big{|}_{\delta=0}(t_{1},\xi_{1})$
$\displaystyle=e^{-t_{1}\xi^{2}_{1}+\xi_{1}}H(\xi_{1}-1),$
$\displaystyle\frac{\partial^{2}v}{\partial\delta^{2}}\Big{|}_{\delta=0}(t_{2},\xi_{2})$
$\displaystyle=2e^{-t_{2}\xi^{2}_{2}+\xi_{2}}H(\xi_{2}-2)\int^{t_{2}}_{0}\int^{\xi_{2}-1}_{1}e^{2t_{2}\xi_{1}(\xi_{2}-\xi_{1})}d\xi_{1}dt_{1},$
$\displaystyle\frac{\partial^{3}v}{\partial\delta^{3}}\Big{|}_{\delta=0}(t_{3},\xi_{3})$
$\displaystyle=2\cdot 3!\
e^{-t_{3}\xi^{2}_{3}+\xi_{3}}H(\xi_{3}-3)\int^{t_{3}}_{0}\int^{\xi_{3}-1}_{2}e^{2t_{2}\xi_{2}(\xi_{3}-\xi_{2})}$
$\displaystyle\ \ \ \ \ \ \ \ \ \
\times\int^{t_{2}}_{0}\int^{\xi_{2}-1}_{1}e^{2t_{1}\xi_{1}(\xi_{2}-\xi_{1})}d\xi_{1}dt_{1}d\xi_{2}dt_{2}.$

We see that $(\frac{\partial
v}{\partial\delta}+\frac{1}{2}\frac{\partial^{2}v}{\partial\delta^{2}}+\frac{1}{6}\frac{\partial^{3}v}{\partial\delta^{3}})\Big{|}_{\delta=0}(\xi,t)$
is the exact solution of

$v(\xi,t)=e^{-t|\xi|^{2}}v_{0}+\int_{0}^{t}e^{-(t-\tau)|\xi|^{2}}v(\tau)*v(\tau)d\tau$

in the region $\\{\xi:\xi\leq 3\\}.$

Claims. On behalf of all authors, the corresponding author states that there
is no conflict of interest.

Acknowledgments. The second named author is very grateful to Professor Y. Giga
for his enlightening discussions and for his valuable suggestions on the
paper. Also, he would like to thank the reviewers for their valuable
suggestions, which enable us to give an essential improvement to Theorem 2.1.
The authors are supported in part by the NSFC, grants 11771024, 12171007.

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Jie Chen and Baoxiang Wang: School of Sciences, Jimei University, Xiamen,
361021 and School of Mathematical Sciences, Peking University, Beijing 100871,
P. R. of China.

E-mail address<EMAIL_ADDRESS>and<EMAIL_ADDRESS>
Zimeng Wang: School of Mechanical and Aerospace Engineering, Queen University,
Belfast, Northern Ireland, BT9 5AH.

E-mail address<EMAIL_ADDRESS>