# The higher order nonlinear Schrödinger equation with quadratic nonlinearity
on the real axis

Andrei V. Faminskii  Peoples’ Friendship University of Russia (RUDN
University), 6 Miklukho-Maklaya Street, Moscow, 117198, Russian Federation
<EMAIL_ADDRESS>
###### Abstract.

The initial value problem is considered for a higher order nonlinear
Schrödinger equation with quadratic nonlinearity. Results on existence and
uniqueness of weak solutions are obtained. In the case of an effective at
infinity additional damping large-time decay of solutions without any
smallness assumptions is also established. The main difficulty of the study is
the non-smooth character of the nonlinearity.

The work was supported by the Ministry of Science and Higher Education of the
Russian Federation: agreement no. 075-03-2020-223/3 (FSSF-2020-0018).

##  1\. Introduction. Notation. Description of main results

The nonlinear Schödinger equation (NLS) in one spatial dimension

$iu_{t}+au_{xx}+\lambda|u|^{p}u=0,$

where $u=u(t,x)$ is a complex-valued function, $p>0$, $a,\lambda$ are real
non-zero constants, and the Korteweg–de Vries equation (KdV)

$u_{t}+bu_{x}+u_{xxx}+\beta uu_{x}=0,$

where $u=u(t,x)$ is a real-valued function, $b,\beta$ are real constants,
$\beta\neq 0$, the modified KdV equation

$u_{t}+bu_{x}+u_{xxx}+\beta u^{2}u_{x}=0$

or more generally the $k$-generalized KdV equation (k-gKdV)

$u_{t}+bu_{x}+u_{xxx}+\beta u^{k}u_{x}=0$

are ones of the most famous examples of nonlinear dispersive equations (see,
for example, [22]).

Each of these equations has its own physical meaning, but, moreover, certain
common mathematical properties. Consider, for example, the initial value
problem with the initial function $u_{0}(x)$. Multiply the NLS equation by
$2\bar{u}(t,x)$, take the imaginary part and integrate over $\mathbb{R}$, then
since

$\int\operatorname{{Im}}2iu_{t}\bar{u}\,dx=\int(\bar{u}u_{t}+\bar{u}_{t}u)\,dx=\frac{d}{dt}\int|u|^{2}\,dx,$
$\int\operatorname{{Im}}2au_{xx}\bar{u}\,dx=\frac{a}{i}\int(u_{xx}\bar{u}-\bar{u}_{xx}u)\,dx=0,$
$\operatorname{{Im}}2\lambda|u|^{p}u\bar{u}=0,$

where here and further we drop the limits in integrals over the whole real
line $\mathbb{R}$, it follows the conservation law

$\|u(t,\cdot)\|_{L_{2}(\mathbb{R})}=\|u_{0}\|_{L_{2}(\mathbb{R})}.$

But the same conservation law holds for the KdV equation and its
generalizations and it can be obtained by the multiplication of the equation
by $2u(t,x)$ and consequent integration over $\mathbb{R}$.

Moreover, for the NLS and k-gKdV equations the following quantities, which are
usually referred to as the energy, are also preserved by the solution flow:

$\int\Bigl{(}|u_{x}|^{2}-\frac{2\lambda}{a(p+2)}|u|^{p+2}\Bigr{)}\,dx,\quad\int\Bigl{(}u_{x}^{2}-\frac{2}{(k+1)(k+2)}u^{k+2}\Bigr{)}\,dx.$

Next, consider the corresponding linear analogs of the NLS and the KdV
equations:

$iu_{t}+au_{xx}=0,\qquad u_{t}+bu_{x}+u_{xxx}=0.$

Then the images of the Fourier transform of the solutions to the initial value
problem are written as follows:

$\widehat{u}(t,\xi)=\widehat{u}_{0}(\xi)e^{ia\xi^{2}t},\quad\widehat{u}(t,\xi)=\widehat{u}_{0}(\xi)e^{i(\xi^{3}-b\xi)t},$

that is in the form

$\widehat{u}(t,\xi)=\widehat{u}_{0}(\xi)e^{iP(\xi)t},$

where $P(\xi)$ is a real polynomial with $P^{\prime\prime}(\xi)\neq 0$, which
shows the dispersive nature of these equations.

Therefore, one can combine these equations and consider the following one:

$iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}+\lambda|u|^{p}u+i\beta(|u|^{p}u)_{x}+i\gamma(|u|^{p})_{x}u=0,$

where $u=u(t,x)$ is a complex-valued function, $a$, $b$, $\lambda$, $\beta$,
$\gamma$ are real constants. This equation is called the higher order
nonlinear Schödinger equation (HNLS) and also has the physical meaning. It has
applications in the propagation of femtosecond optical pulses in a monomode
optical fiber, accounting for additional effects such as third order
dispersion, self-steeping of the pulse, and self-frequency shift (see [15, 17,
18, 20], and the references therein).

Similarly to the NLS and k-gKdV equations it can be shown that the norm in
$L_{2}(\mathbb{R})$ of a solution to the initial value problem is also
preserved. On the contrary, if one tries to derive the analogue of the
conservation law for the energy, the following identity is obtained:

$\frac{d}{dt}\int\Bigl{[}|u_{x}|^{2}+\frac{i}{\beta+\gamma}\Bigl{(}\lambda-\frac{a(3\beta+2\gamma)}{3}\Bigr{)}u\bar{u}_{x}-\frac{2(3\beta+2\gamma)}{3(p+2)}|u|^{p+2}\Bigr{]}\,dx\\\
-\frac{\gamma}{3}\int\bigl{(}|u|^{p}\bigr{)}_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx=0.$

Therefore, the integral of the expression in the square brackets is preserved
only if either $\gamma=0$ or $p=2$, since

$\int_{\mathbb{R}}\bigl{(}|u|^{2}\bigr{)}_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx=0.$

Another serious obstacle in the study of such an equation is the non-
smoothness of the function $|u|^{p}$ except the special cases of even natural
values of $p$. Just in the case $p=2$, that is for the equation

$iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}+\lambda|u|^{2}u+i\beta\bigl{(}|u|^{2}u\bigr{)}_{x}+i\gamma(|u|^{2})_{x}u=0,$

were previously obtained results on local and global well-posedness of the
initial value problem. In particular, in [21] local well-posedness was proved
for the initial data $u_{0}$ in $H^{s}(\mathbb{R})$, $s>3/4$, and global well-
posedness in $H^{s}(\mathbb{R})$, $s\geq 1$, if $\beta+\gamma\neq 0$. In [24]
the local result was improved up to $s\geq 1/4$. In [7] the same result was
obtained for the analogous equation with variable (depending on $t$)
coefficients. Under certain relation between the coefficients the local result
was extended to the global one in [6] if $s>1/4$. In [8] global well-posedness
was proved in certain weighted subspaces of $H^{2}(\mathbb{R})$, where power
weights were effective at $\pm\infty$ (looking ahead, in the present article
the weights are effective only at $+\infty$); the argument used global
estimates from [21]. A unique continuation property was obtained in [9].

No results for the initial value problem for the HNLS equation with other
values of $p$ have been previously established.

In [10] an initial-boundary value problem on a bounded interval $I$ for the
HNLS equation was considered. In the case $p\in[1,2]$ and the initial function
$u_{0}\in H^{s}(I)$, $0\leq s\leq 3$, results on global existence and
uniqueness of mild solutions were obtained. For $u_{0}\in L_{2}(I)$ the result
on global existence was extended either to $p\in(2,3)$ or $p\in(2,4)$,
$\gamma=0$. Moreover, after addition to the equation of the damping term
$id(x)u$, where the non-negative function $d$ was strictly positive on a
certain sub-interval, large-time decay of solutions was established. Certain
preceding results for the truncated version of the HNLS equation
($\beta=\gamma=0$) can be found in [1, 2, 3, 4, 12, 13].

Note that the value $p=2$ in the nonlinearity in the HNLS equation corresponds
in the KdV case to the modified KdV equation, while to the KdV equation itself
corresponds the value $p=1$, and this is object of the study of the present
article. We consider the initial value problem for an equation

$iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}+\lambda|u|u+i\beta(|u|u)_{x}+id(x)u=0$ (1.1)

with the initial data

$u(0,x)=u_{0}(x),\quad x\in\mathbb{R}.$ (1.2)

Here $u=u(t,x)$ and $u_{0}$ are complex-valued function, $a$, $b$, $\lambda$,
$\beta$ are real constants. In comparison with the general form of the HNLS
equation it is assumed that $\gamma=0$, that is the nonlinearity has the
divergent form and, in particular, the conservation law for the analog of the
energy holds. The main results consist of theorems on global existence and
uniqueness of weak solutions. Moreover, the presence of the damping term
$id(x)u$, where the non-negative function $d$ is strictly positive at
infinity, gives an opportunity to establish large-time decay of solutions.
Similar assumptions, which mean that the damping is effective only at
infinity, were previously used in [11] for the case of the KdV equation
itself. For existence and uniqueness results the presence of the damping term
is irrelevant.

The initial function $u_{0}$ is assumed to be from the following weighted at
$+\infty$ $L_{2}$ space: $(1+x_{+})^{3/4}u_{0}\in L_{2}(\mathbb{R})$ (here and
further $x_{+}=\max(x,0)$). Previously the corresponding theory was developed
for the KdV equation itself in [19]. The result on global well-posedness of
the initial value problem obtained there was not optimal with respect to the
properties of the initial function. Later, for example, in the paper [5]
global well-posedness was established for $u_{0}\in L_{2}(\mathbb{R})$ without
any additional weights and even for $u_{0}\in H^{s}(\mathbb{R})$ for negative
values of $s$ (see, for example, [16]). However, the methods of that papers
required the use of the Bourgain spaces, where a very sophisticated harmonic
analysis was applied, in particular, closely related to the concrete
nonlinearity $(u^{2})_{x}$. Possibility to apply the technique of the Bourgain
spaces to the nonlinearity $(|u|u)_{x}$ is not evident and is an open problem.
On the contrary, the class of well-posedness in [19] was very simple, did not
require any smoothness and consisted of functions $u(t,x)$ such that
$(1+x_{+})^{3/4}\in L_{\infty}(0,T;L_{2}(\mathbb{R}))$. It turned out that the
ideas of the paper [19] can be applied to equation (1.1). Although the methods
of the last paper look like rough and archaic in comparison with [5], they are
effective for the considered problem. However, a lot of preliminary work had
to be done.

Let $\Pi_{T}=(0,T)\times\mathbb{R}$. Let $L_{p}=L_{p}(\mathbb{R})$,
$H^{s}=H^{s}(\mathbb{R})$, $C_{b}=C_{b}(\mathbb{R})$ (the subscript $b$ means
bounded functions), $C_{b}^{k}=C_{b}^{k}(\mathbb{R})$,
$\mathcal{S}=\mathcal{S}(\mathbb{R})$,
$\mathcal{S}^{\prime}=\mathcal{S}^{\prime}(\mathbb{R})$. The notation
$C_{w}([0,T];\dots)$ means a weakly continuous map.

Let $\psi(x)\not\equiv\text{const}$ be a non-negative continuous on
$\mathbb{R}$ function. Define a special weighted space:

$L_{p}^{\psi(x)}=L_{p}^{\psi(x)}(\mathbb{R})=\\{\varphi(x):\varphi\psi^{1/2}\in
L_{p}\\},$

endowed with the natural norm. Let $L_{p}^{0}=L_{p}$ and for $\alpha\neq 0$

$L_{p}^{\alpha}=L_{p}^{\alpha}(\mathbb{R})=L_{p}^{(1+x_{+})^{2\alpha}}=\\{\varphi(x):(1+x_{+})^{\alpha}\varphi(x)\in
L_{p}\\}.$

The notion of a weak solution of the considered problem is understood in the
following sense.

######  Definition 1.1.

Let $T>0$, $u_{0}\in L_{2}$, $d\in L_{2}$. A function $u\in
L_{\infty}(0,T;L_{2})$ is called a weak solution to problem (1.1), (1.2) if
for any function $\phi\in C^{1}([0,T];L_{2})\cap C([0,T];H^{3})$,
$\phi\big{|}_{t=T}=0$, the following equality holds:

$\iint_{\Pi_{T}}\bigl{(}iu\phi_{t}-au\phi_{xx}+ibu\phi_{x}+iu\phi_{xxx}-\lambda|u|u\phi+i\beta|u|u\phi_{x}-id(x)u\phi\bigr{)}\,dxdt\\\
+\int u_{0}\phi\big{|}_{t=0}\,dx=0.$ (1.3)

Note that under the hypothesis of Definition 1.1 $|u|u\in
L_{\infty}(0,T;L_{1})$, $\phi\in C([0,T];C^{2}_{b})$ and so the integral in
the left-hand side of (1.3) exists.

######  Definition 1.2.

We say that the function $d(x)$ satisfies Condition A, if it is non-negative
on $\mathbb{R}$ and there exist positive constants $d_{0}$ and $R_{0}$ such
that

$d(x)\geq d_{0}\qquad\text{for}\quad|x|\geq R_{0}.$ (1.4)

The main result of the paper is the following theorem.

######  Theorem 1.3.

Let $u_{0}\in L_{2}^{3/4}$, $d\in L_{\infty}$. Then for any $T>0$ in the strip
$\Pi_{T}$ there exists a unique weak solution $u\in C_{w}([0,T];L_{2}^{3/4})$
to problem (1.1), (1.2). If in addition the function $d(x)$ satisfies
Condition A, then there exists a positive constant $\gamma$ and for any $M>0$
a positive constant $c(M)$ such that if $\|u_{0}\|_{L_{2}}\leq M$

$\|u(t,\cdot)\|_{L_{2}}^{2}\leq c(M)e^{-\gamma t}\quad\forall\ t\geq 0.$ (1.5)

Introduce certain auxiliary notation. For $\alpha\geq 0$ let

$H^{1,\alpha}=\\{\varphi(x):\varphi,\varphi^{\prime}\in L_{2}^{\alpha}\\}$

endowed with the natural norm.

Let $\eta(x)$ denotes a cut-off function, namely, $\eta$ is an infinitely
smooth non-decreasing function on $\mathbb{R}$ such that $\eta(x)=0$ for
$x\leq 0$, $\eta(x)=1$ for $x\geq 1$, $\eta(x)+\eta(1-x)\equiv 1$.

Define the following weight function $\rho_{\alpha,\epsilon}(x)$ for
$\alpha\geq 0$, $\epsilon>0$: let $\rho_{\alpha,\epsilon}\in
C^{\infty}(\mathbb{R})$ be an increasing function such that
$\rho_{\alpha,\epsilon}(x)=e^{2\epsilon x}$ for $x\leq-1$,
$\rho_{\alpha,\epsilon}(x)=(1+x)^{2\alpha}$ if $\alpha>0$ and
$\rho_{0,\epsilon}=2-\ln^{-1}(x+e)$ for $x\geq 0$,
$\rho^{\prime}_{\alpha,\epsilon}(x)>0$ for $x\in(-1,0)$. Note that if either
$\rho(x)\equiv\rho_{\alpha,\epsilon}(x)$ or
$\rho(x)\equiv\rho^{\prime}_{\alpha,\epsilon}(x)$ then for any natural $j$

$|\rho^{(j)}(x)|\leq c(j,\alpha,\epsilon)\rho(x)\quad\forall x\in\mathbb{R}.$
(1.6)

######  Definition 1.4.

For any $T>0$ define a class $X(\Pi_{T})$ of functions $u(t,x)$ such that

$\displaystyle u\in C([0,T];H^{1}),\quad$ $\displaystyle u_{x}\in
L_{6}(0,T;C_{b}),$ $\displaystyle u_{xx}\in
C_{b}(\mathbb{R};L_{2}(0,T)),\quad$ $\displaystyle u\in
L_{2}(\mathbb{R};C[0,T]).$

For $\delta\geq 0$ and $\theta\geq 0$ define a function

$g_{\delta}(\theta)\equiv(\theta+\delta)^{1/2}.$ (1.7)

The paper is organized as follows. In Section 2 an auxiliary linear problem is
considered. In Section 3 we present the results on existence of solutions to
the original problem, and in Section 4 — on uniqueness and continuous
dependence on initial data. Section 5 contains certain results on internal
regularity of weak solutions, which, in particular, are used in Section 6
devoted to large-time decay of solutions. Certain technical items related to
the properties of the fundamental solution of the corresponding linear
operator and differentiation of $|u|$ are presented in Appendixes A and B.

##  2\. Auxiliary linear problem

Consider a linear problem

$\displaystyle iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}=f(t,x),$ (2.1) $\displaystyle
u(0,x)=u_{0}(x),\quad x\in\mathbb{R}.$ (2.2)

######  Lemma 2.1.

Let $u_{0}\in\mathcal{S}$, $f\in C^{\infty}([0,T];\mathcal{S})$, then there
exists a unique solution to problem (2.1), (2.2) $u\in
C^{\infty}([0,T];\mathcal{S})$.

###### Proof.

The unique solution to the considered problem in the considered space is
constructed via the Fourier transform and is given by a formula

$\widehat{u}(t,\xi)=\widehat{u}_{0}(\xi)e^{i(\xi^{3}-a\xi^{2}-b\xi)t}-i\int_{0}^{t}\widehat{f}(\tau,\xi)e^{i(\xi^{3}-a\xi^{2}-b\xi)(t-\tau)}\,d\tau,$
(2.3)

where

$\widehat{u}(t,\xi)\equiv\int e^{-ix\xi}u(t,x)\,dx$

with similar notation for $\widehat{u}_{0}$ and $\widehat{f}$. ∎

######  Definition 2.2.

Let $u_{0}\in\mathcal{S}^{\prime}$, $f\equiv f_{1}+f_{2x}$, where
$f_{1},f_{2}\in\bigl{(}C^{\infty}([0,T];S)\bigr{)}^{\prime}$. A function
$u\in\bigl{(}C^{\infty}([0,T];S)\bigr{)}^{\prime}$ is called a generalized
solution to problem (2.1), (2.2) in a strip $\Pi_{T}$ if for any function
$\phi\in C^{\infty}([0,T];S)$, $\phi(T,x)\equiv 0$, it verifies an equality

$\langle u,i\phi_{t}-a\phi_{xx}+ib\phi_{x}+i\phi_{xxx}\rangle+\langle
f_{1},\phi\rangle-\langle f_{2},\phi_{x}\rangle+\langle
u_{0},\phi\big{|}_{t=0}\rangle=0.$ (2.4)

######  Remark 2.3.

Any weak solution to problem (2.1), (2.2) in the sense of Definition 1.1 (with
corresponding changes) is, of course, a generalized solution.

######  Lemma 2.4.

The generalized solution to problem (2.1), (2.2) is unique.

###### Proof.

This lemma succeeds from Lemma 2.1 by the standard Hölmgren argument. ∎

Now consider the differential operator

$\mathcal{L}(\partial_{t},\partial_{x})=\partial_{t}+\partial_{x}^{3}-ia\partial_{x}^{2}+b\partial_{x}.$
(2.5)

The general theory of linear differential operators provides that its
fundamental solution is written in a form

$G(t,x)=\theta(t)\mathcal{F}^{-1}\bigl{[}e^{it(\xi^{3}-a\xi^{2}-b\xi)}\bigr{]}(x),$
(2.6)

where $\theta$ is the Heaviside function.

######  Theorem 2.5.

Let $u_{0}\in L_{2}^{1/4+\varepsilon}$, $f_{11}\in
L_{1}(0,T;L_{2}^{1/4+\varepsilon})$, $f_{12}\in L_{1}(0,T;L_{1}^{-1/4})$,
$f_{2}\in L_{1}(0,T;L_{1}^{1/4})$ for certain $\varepsilon>0$ and $T>0$,
$f_{1}\equiv f_{11}+f_{12}$, $f\equiv f_{1}+f_{2x}$. Then there exists a
generalized solution $u(t,x)$ to problem (2.1), (2.2) in the strip $\Pi_{T}$
such that at any point $(t,x)\in\Pi_{T}$

$u(t,x)=\int G(t,x-y)u_{0}(y)\,dy-i\int_{0}^{t}\\!\\!\int
G(t-\tau,x-y)f_{1}(\tau,y)\,dyd\tau\\\ -i\int_{0}^{t}\\!\\!\int
G_{x}(t-\tau,x-y)f_{2}(\tau,y)\,dyd\tau.$ (2.7)

###### Proof.

First let $u_{0}\in\mathcal{S}$, $f_{1},f_{2}\in
C^{\infty}([0,T];\mathcal{S})$. Then Lemma 2.1 yields that a solution $u\in
C^{\infty}([0,T];\mathcal{S})$ to the considered problem exists. According to
(2.3)

$u(t,x)=\mathcal{F}^{-1}\Bigl{[}e^{i(\xi^{3}-a\xi^{2}-b\xi)t}\widehat{u}_{0}(\xi)\Bigr{]}(x)-i\int_{0}^{t}\mathcal{F}^{-1}\Bigl{[}e^{i(\xi^{3}-a\xi^{2}-b\xi)(t-\tau)}\widehat{f}(\tau,\xi)\Bigr{]}(x)\,d\tau$
(2.8)

and with the use of (2.6) we obtain (2.7) in the smooth case.

Inequalities (A.26) yield that

$\Bigl{|}\int
G(t,x-y)u_{0}(y)\,dy\Bigr{|}\leq\frac{c(T)}{t^{1/3}}\Bigl{[}\int_{x}^{+\infty}(1+y-x)^{-1/4}|u_{0}(y)|\,dy\\\
+\int_{-\infty}^{x}e^{-c_{0}(x-y)^{3/2}T^{-1/2}}|u_{0}(y)|\,dy\Bigr{]}$

whence, since
$(1+y-x)^{-1/4}=(1+y-x)^{-1/2-\varepsilon}(1+y-x)^{1/4+\varepsilon}$ and
$(1+y-x)\leq(1+y_{+})(1+x_{-})$, it follows that

$\Bigl{|}\int
G(t,x-y)u_{0}(y)\,dy\Bigr{|}\leq\frac{c(T,\varepsilon)}{t^{1/3}}\Bigl{[}\Bigl{(}\int_{x}^{+\infty}(1+y-x)^{1/2+2\varepsilon}|u_{0}(y)|^{2}\,dy\Bigr{)}^{1/2}\\\
+\Bigl{(}\int_{-\infty}^{x}|u_{0}(y)|^{2}\,dy\Bigr{)}^{1/2}\Bigr{]}\leq\frac{c(T,\varepsilon)}{t^{1/3}}(1+x_{-})^{1/4+\varepsilon}\|u_{0}\|_{L_{2}^{1/4+\varepsilon}},$
(2.9)

where $x_{-}=\max(-x,0)$. Similarly,

$\displaystyle\int_{0}^{T}\Bigl{|}\int_{0}^{t}\\!\\!\int
G(t-\tau,x-y)f_{11}(\tau,y)\,dyd\tau\Bigr{|}\,dt\\\ \leq
c(T,\varepsilon)(1+x_{-})^{1/4+\varepsilon}\int_{0}^{T}\\!\\!\int_{0}^{t}\frac{\|f_{11}(\tau,\cdot)\|_{L_{2}^{1/4+\varepsilon}}}{(t-\tau)^{1/3}}\,d\tau
dt\\\ \leq
c_{1}(T,\varepsilon)(1+x_{-})^{1/4+\varepsilon}\|f_{11}\|_{L_{1}(0,T;L_{2}^{1/4+\varepsilon})}.$
(2.10)

Next, since $1+y_{+}\leq(1+x_{+})(1+y-x)$ if $x\leq y$, it follows that

$\int_{0}^{T}\Bigl{|}\int_{0}^{t}\\!\\!\int
G(t-\tau,x-y)f_{12}(\tau,y)\,dyd\tau\Bigr{|}\,dt\\\ \leq
c(T)\int_{0}^{T}\\!\\!\int_{0}^{t}\frac{1}{(t-\tau)^{1/3}}\Bigl{[}\int_{x}^{+\infty}(1+y-x)^{-1/4}|f_{12}(\tau,y)|\,dy\\\
+\int_{-\infty}^{x}e^{-c_{0}(x-y)^{3/2}T^{-1/2}}|f_{12}(\tau,y)|\,dy\Bigr{]}\,d\tau
dt\leq c_{1}(T)(1+x_{+})^{1/4}\|f_{12}\|_{L_{1}(0,T;L_{1}^{-1/4})}.$ (2.11)

Finally,

$\int_{0}^{T}\Bigl{|}\int_{0}^{t}\\!\\!\int
G_{x}(t-\tau,x-y)f_{2}(\tau,y)\,dyd\tau\Bigr{|}\,dt\\\ \leq
c(T)\int_{0}^{T}\\!\\!\int_{0}^{t}\frac{1}{(t-\tau)^{3/4}}\Bigl{[}\int_{x}^{+\infty}(1+y-x)^{1/4}|f_{2}(\tau,y)|\,dy\\\
+\int_{-\infty}^{x}e^{-c_{0}(x-y)^{3/2}T^{-1/2}}|f_{2}(\tau,y)|\,dy\Bigr{]}\,d\tau
dt\leq c_{1}(T)(1+x_{-})^{1/4}\|f_{2}\|_{L_{1}(0,T;L_{1}^{1/4})}.$ (2.12)

Estimates (2.9)–(2.12) provide an inequality

$\|u\|_{L_{1}(0,T;L_{\infty}^{(1+|x|)^{-1/2-2\varepsilon}})}\leq
c(T,\varepsilon)\bigl{[}\|u_{0}\|_{L_{2}^{1/4+\varepsilon}}+\|f_{11}\|_{L_{1}(0,T;L_{2}^{1/4+\varepsilon})}\\\
+\|f_{12}\|_{L_{1}(0,T;L_{1}^{-1/4})}+\|f_{2}\|_{L_{1}(0,T;L_{1}^{1/4})}\bigr{]}.$
(2.13)

Now consider the general case. Approximate the functions $u_{0}$, $f_{11}$,
$f_{12}$ and $f_{2}$ in the corresponding spaces by the smooth ones. Then on
the basis of estimate (2.13) the function $u(t,x)$ defined by formula (2.7)
can be obtained by the passage to the limit in the space
$L_{1}(0,T;L_{\infty}^{(1+|x|)^{-1/2-2\varepsilon}})\subset\bigl{(}C^{\infty}([0,T];S)\bigr{)}^{\prime}$
of the sequence of smooth solutions. Moreover, writing down corresponding
equalities (2.4) for smooth solutions and passing to the limit we obtain that
the function $u$ also verifies equality (2.4). Note, that estimates
(2.9)–(2.12) also hold in the non-smooth case. ∎

######  Lemma 2.6.

Let $u_{0}\in L_{2}$, $f\in L_{1}(0,T;L_{2})$ for certain $T>0$. Then there
exists a weak solution to problem (2.1), (2.2) $u\in C([0,T];L_{2})$,

$\|u\|_{C([0,T];L_{2})}\leq\|u_{0}\|_{L_{2}}+\|f\|_{L_{1}(0,T;L_{2})}$ (2.14)

and for a.e. $t\in(0,T)$

$\frac{d}{dt}\int|u|^{2}\,dx=2\operatorname{{Im}}\int f\bar{u}\,dx.$ (2.15)

Moreover, for any non-negative non-decreasing function $\psi\in C^{3}_{b}$,
$\psi^{\prime}\not\equiv 0$, $u_{x}\in L_{2}(0,T;L_{2}^{\psi^{\prime}(x)})$,

$\|u_{x}\|_{L_{2}(0,T;L_{2}^{\psi^{\prime}(x)})}\leq
c(T,\psi)\bigl{(}\|u_{0}\|_{L_{2}}+\|f\|_{L_{1}(0,T;L_{2})}\bigr{)}$ (2.16)

and for a.e. $t\in(0,T)$

$\frac{d}{dt}\int|u(t,x)|^{2}\psi(x)\,dx+3\int|u_{x}|^{2}\psi^{\prime}\,dx=2a\operatorname{{Im}}\int
u_{x}\bar{u}\psi^{\prime}\,dx\\\
+\int|u|^{2}\psi^{\prime\prime\prime}\,dx+b\int|u|^{2}\psi^{\prime}\,dx+2\operatorname{{Im}}\int
f\bar{u}\psi\,dx.$ (2.17)

###### Proof.

First assume that the functions $u_{0}$ and $f$ satisfy the hypothesis of
Lemma 2.1 and consider the corresponding smooth solution $u(t,x)$. Let
$\psi\in C^{3}_{b}$. Multiply equation (2.1) by $2\bar{u}(t,x)\psi(x)$,
integrate over $\mathbb{R}$ and take the imaginary part, then

$\int\operatorname{{Im}}2iu_{t}\bar{u}\psi\,dx+a\int
2\operatorname{{Im}}u_{xx}\bar{u}\psi\,dx+b\int\operatorname{{Im}}2iu_{x}\bar{u}\psi\,dx+\int\operatorname{{Im}}2iu_{xxx}\bar{u}\psi\,dx\\\
=2\operatorname{{Im}}\int f\bar{u}\psi\,dx.$ (2.18)

Here

$\int\operatorname{{Im}}2iu_{t}\bar{u}\psi\,dx=\int\operatorname{{Re}}2u_{t}\bar{u}\psi\,dx=\int(u_{t}\bar{u}+u\bar{u}_{t})\psi\,dx=\frac{d}{dt}\int|u|^{2}\psi\,dx,$
$\int
2\operatorname{{Im}}u_{xx}\bar{u}\psi\,dx=-2\int\operatorname{{Im}}(|u_{x}|^{2}\psi+u_{x}\bar{u}\psi^{\prime})\,dx=-2\operatorname{{Im}}\int
u_{x}\bar{u}\psi^{\prime}\,dx,$
$\int\operatorname{{Im}}2iu_{x}\bar{u}\psi\,dx=\int
2\operatorname{{Re}}u_{x}\bar{u}\psi\,dx=\int(u_{x}\bar{u}+u\bar{u}_{x})\psi\,dx=-\int|u|^{2}\psi^{\prime}\,dx,$
$\int\operatorname{{Im}}2iu_{xxx}\bar{u}\psi\,dx=\int
2\operatorname{{Re}}u_{xxx}\bar{u}\psi\,dx=\int(u_{xxx}\bar{u}+u\bar{u}_{xxx})\psi\,dx\\\
=-\int(u_{xx}\bar{u}_{x}+u_{x}\bar{u}_{xx})\psi\,dx-\int(u_{xx}\bar{u}+u\bar{u}_{xx})\psi^{\prime}\,dx\\\
=-\int(|u_{x}|^{2})_{x}\psi\,dx+2\int|u_{x}|^{2}\psi^{\prime}\,dx+\int(|u|^{2})_{x}\psi^{\prime\prime}\,dx\\\
=3\int|u_{x}|^{2}\psi^{\prime}\,dx-\int|u|^{2}\psi^{\prime\prime\prime}\,dx.$

Choose in (2.18) $\psi(x)\equiv 1$, then equality (2.13) follows for smooth
solutions which, in turn, provides estimate (2.14) and for any $t\in[0,T]$ an
equality

$\int|u(t,x)|^{2}\,dx=\int|u_{0}|^{2}\,dx+2\operatorname{{Im}}\int_{0}^{t}\int
f(\tau,x)\bar{u}(\tau,x)\,dxd\tau.$ (2.19)

Applying closure on the basis of estimate (2.14) for smooth solutions we
obtain the result on existence in the space $C([0,T];L_{2})$, estimate (2.14)
for such solutions and equality (2.19). Since $\int f(t,x)\bar{u}(t,x)\,dx\in
L_{1}(0,T)$, the right-hand side of equality (2.19) is absolutely continuous
with respect to $t$ and equality (2.15) succeeds.

Now let $\psi$ be non-negative, non-decreasing and $\psi\not\equiv 0$. Then
equality (2.18) yields equality (2.17) for smooth solutions. With the use of
estimate (2.14) this equality insures estimate (2.16). Closure finishes the
proof. ∎

######  Lemma 2.7.

Let $u\in L_{\infty}(0,T;L_{2}^{\alpha})$, $\alpha\geq 3/4$, be a weak
solution to problem (2.1), (2.2) in a certain strip $\Pi_{T}$, where
$u_{0}\equiv 0$, $f\equiv f_{1}+f_{2x}$, $f_{1}\in
L_{p_{1}}(0,T;L_{1}^{\alpha+1/4})$, $f_{2}\in
L_{p_{2}}(0,T;L_{1}^{\alpha+3/4})$ for certain $p_{1}>3/2$, $p_{2}>4$. Then
for any $\epsilon>0$

$\|u\|_{L_{\infty}(0,T;L_{2}^{\rho_{\alpha,\epsilon}(x)})}\leq
c(T,\alpha,\epsilon,p_{1},p_{2})\bigl{[}\|f_{1}\|_{L_{p_{1}}(0,T;L_{1}^{\rho_{\alpha,\epsilon}(x)(1+x_{+})^{1/2}})}\\\
+\|f_{2}\|_{L_{p_{2}}(0,T;L_{1}^{\rho_{\alpha,\epsilon}(x)(1+x_{+})^{3/2}})}.$
(2.20)

###### Proof.

First of all note that $u\in
L_{\infty}(0,T;L_{2}^{\rho_{\alpha,\epsilon(x)}})$. Apply Theorem 2.5 and
Lemma A.3, then equality (2.7) and inequality (A.30) (for $f_{1}$ in the case
$n=0$ and for $f_{2}$ in the case $n=1$) imply that for any $t\in(0,T]$

$\|u(t,\cdot)\|_{L_{2}^{\rho_{\alpha,\epsilon}(x)}}=\|u(t,\cdot)\rho^{1/2}_{\alpha,\epsilon}\|_{L_{2}}\leq\int_{0}^{t}\\!\\!\int\|G(t-\tau,\cdot-y)\rho^{1/2}_{\alpha,\epsilon}\|_{L_{2}}|f_{1}(\tau,y)|\,dyd\tau\\\
+\int_{0}^{t}\\!\\!\int\|G_{x}(t-\tau,\cdot-y)\rho^{1/2}_{\alpha,\epsilon}\|_{L_{2}}|f_{2}(\tau,y)|\,dyd\tau\\\
\leq
c\int_{0}^{t}\frac{1}{(t-\tau)^{1/3}}\int|f_{1}(\tau,y)|\rho_{\alpha,\epsilon}^{1/2}(y)(1+y_{+})^{1/4}\,dyd\tau\\\
+c\int_{0}^{t}\frac{1}{(t-\tau)^{3/4}}\int|f_{2}(\tau,y)|\rho_{\alpha,\epsilon}^{1/2}(y)(1+y_{+})^{3/4}\,dyd\tau\\\
\leq
c_{1}\|f_{1}\|_{L_{p_{1}}(0,T;L_{1}^{\rho_{\alpha,\epsilon}(x)(1+x_{+})^{1/2}})}+c_{1}\|f_{2}\|_{L_{p_{2}}(0,T;L_{1}^{\rho_{\alpha,\epsilon}(x)(1+x_{+})^{3/2}})}.$

∎

######  Lemma 2.8.

Let $u_{0}\in L_{2}^{\alpha}$, $f\in L_{1}(0,T;L_{2}^{\alpha})$ for certain
$\alpha>0$, $T>0$. Then there exists a weak solution $u(t,x)$ to problem
(2.1), (2.2) such that $u\in C([0,T];L_{2}^{\alpha})$, $u_{x}\in
L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})$ for any $\epsilon>0$,

$\|u\|_{C([0,T];L_{2}^{\alpha})}+\|u_{x}\|_{L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})}\leq
c(T,\alpha,\epsilon)\bigl{[}\|u_{0}\|_{L_{2}^{\alpha}}+\|f\|_{L_{1}(0,T;L_{2}^{\alpha})}\bigr{]}$
(2.21)

and for a.e $t\in(0,T)$ and $\rho(x)\equiv\rho_{\alpha,\epsilon}(x)$

$\frac{d}{dt}\int|u(t,x)|^{2}\rho(x)\,dx+3\int|u_{x}|^{2}\rho^{\prime}\,dx=2a\operatorname{{Im}}\int
u_{x}\bar{u}\rho^{\prime}\,dx\\\
+\int|u|^{2}\rho^{\prime\prime\prime}\,dx+b\int|u|^{2}\rho^{\prime}\,dx+2\operatorname{{Im}}\int
f\bar{u}\rho\,dx.$ (2.22)

###### Proof.

First assume that the functions $u_{0}$ and $f$ satisfy the hypothesis of
Lemma 2.1 and consider the corresponding smooth solution $u(t,x)$. Multiply
equation (2.1) by $2\bar{u}(t,x)\rho(x)$, integrate over $\mathbb{R}$ and take
the imaginary part, then equality (2.18) follows, where $\psi$ is substituted
by $\rho$, whence equality (2.22) follows. In turn, with the use of properties
(1.6) this equality provides estimate (2.21). The end of the proof is similar
to Lemma 2.6. ∎

######  Lemma 2.9.

Let $u_{0}\in H^{1}$, $f\in L_{1}(0,T;H^{1})$ for certain $T>0$. Then there
exists a unique solution to problem (2.1), (2.2) $u\in X(\Pi_{T})$,

$\|u\|_{X(\Pi_{T})}\leq
c(T)\left(\|u_{0}\|_{H^{1}}+\|f\|_{L_{1}(0,T;H^{1})}\right)$ (2.23)

and for a.e $t\in(0,T)$

$\frac{d}{dt}\int|u_{x}|^{2}\,dx=2\operatorname{{Im}}\int
f_{x}\bar{u}_{x}\,dx,$ (2.24)
$\frac{2}{3}\frac{d}{dt}\int|u|^{3}\,dx+2a\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx+3\int|u|_{x}|u_{x}|^{2}\,dx-\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx\\\
=2\operatorname{{Im}}\int f|u|\bar{u}\,dx,$ (2.25) $i\frac{d}{dt}\int
u\bar{u}_{x}\,dx=2\operatorname{{Re}}\int f\bar{u}_{x}\,dx.$ (2.26)

Moreover, for any non-negative non-decreasing function $\psi\in C^{3}_{b}$,
$\psi^{\prime}\not\equiv 0$, $u_{xx}\in L_{2}(0,T;L_{2}^{\psi^{\prime}(x)})$,

$\|u_{xx}\|_{L_{2}(0,T;L_{2}^{\psi^{\prime}(x)})}\leq
c(T,\psi)\bigl{(}\|u_{0}\|_{H^{1}}+\|f\|_{L_{1}(0,T;H^{1})}\bigr{)}$ (2.27)

and for a.e. $t\in(0,T)$

$\frac{d}{dt}\int|u_{x}(t,x)|^{2}\psi(x)\,dx+3\int|u_{xx}|^{2}\psi^{\prime}\,dx=2a\operatorname{{Im}}\int
u_{xx}\bar{u}_{x}\psi^{\prime}\,dx\\\
+\int|u_{x}|^{2}\psi^{\prime\prime\prime}\,dx+b\int|u_{x}|^{2}\psi^{\prime}\,dx+2\operatorname{{Im}}\int
f_{x}\bar{u}_{x}\psi\,dx,$ (2.28)
$\frac{2}{3}\frac{d}{dt}\int|u|^{3}\psi\,dx+2a\operatorname{{Im}}\int\bigl{(}|u|\psi\bigr{)}_{x}u\bar{u}_{x}\,dx+3\int\bigl{(}|u|\psi\bigr{)}_{x}|u_{x}|^{2}\,dx\\\
-\int\bigl{(}|u|\psi\bigr{)}_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx-\frac{2}{3}b\int|u|^{3}\psi^{\prime}\,dx=2\operatorname{{Im}}\int
f|u|\bar{u}\psi\,dx.$ (2.29)

###### Proof.

Similarly to (2.14)

$\|u_{x}\|_{C([0,T];L_{2})}\leq\|u_{0}^{\prime}\|_{L_{2}}+\|f_{x}\|_{L_{1}(0,T;L_{2})}.$

Next, apply estimates from [14]: if $f\equiv 0$

$\displaystyle\|u\|_{L_{6}(0,T;L_{\infty})}$ $\displaystyle\leq
c(T)\|u_{0}\|_{L_{2}},$
$\displaystyle\|u_{x}\|_{L_{\infty}(\mathbb{R};L_{2}(0,T))}$
$\displaystyle\leq c(T)\|u_{0}\|_{L_{2}},$
$\displaystyle\|u\|_{L_{2}(\mathbb{R};L_{\infty}(0,T))}$ $\displaystyle\leq
c(T)\|u_{0}\|_{H^{1}}.$

Note that in the smooth case all $L_{\infty}$ can be changed to $C_{b}$ norms
and then by closure the same can be done in the general case. After that by
the standard argument we consider the non-homogeneous case and obtain the
result on existence in the space $X(\Pi_{T})$ with estimate (2.23).

In order to prove other properties as in the previous proof first consider
smooth solutions. Let $\psi\in C^{3}_{b}$. Multiply (2.1) by
$-2\bigl{(}\bar{u}_{x}\psi(x)\bigr{)}_{x}$, integrate over $\mathbb{R}$ and
take the imaginary part, then

$-\int\operatorname{{Im}}2iu_{t}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx-a\int
2\operatorname{{Im}}u_{xx}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx-b\int\operatorname{{Im}}2iu_{x}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx\\\
-\int\operatorname{{Im}}2iu_{xxx}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx=-2\operatorname{{Im}}\int
f\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx.$ (2.30)

Here

$-\int\operatorname{{Im}}2iu_{t}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx=\int\operatorname{{Re}}2u_{tx}\bar{u}_{x}\psi\,dx\\\
=\int(u_{tx}\bar{u}_{x}+u_{x}\bar{u}_{tx})\psi\,dx=\frac{d}{dt}\int|u_{x}|^{2}\psi\,dx,$
$-\int 2\operatorname{{Im}}u_{xx}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx=-\int
2\operatorname{{Im}}|u_{xx}|^{2}\psi\,dx-2\operatorname{{Im}}\int
u_{xx}\bar{u}_{x}\psi^{\prime}\,dx\\\ =-2\operatorname{{Im}}\int
u_{xx}\bar{u}_{x}\psi^{\prime}\,dx,$
$-\int\operatorname{{Im}}2iu_{x}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx=-\int
2\operatorname{{Re}}u_{x}\bar{u}_{xx}\psi\,dx-2\int|u_{x}|^{2}\psi^{\prime}\,dx\\\
=-\int\bigl{(}|u_{x}|^{2}\bigr{)}_{x}\psi\,dx-2\int|u_{x}|^{2}\psi^{\prime}\,dx=-\int|u_{x}|^{2}\psi^{\prime}\,dx,$
$-\int\operatorname{{Im}}2iu_{xxx}\bigl{(}\bar{u}_{x}\psi\bigr{)}_{x}\,dx=-\int
2\operatorname{{Re}}u_{xxx}(\bar{u}_{xx}\psi+\bar{u}_{x}\psi^{\prime})\,dx\\\
=-\int(u_{xxx}\bar{u}_{xx}+\bar{u}_{xxx}u_{xx})\psi\,dx+2\int|u_{xx}|^{2}\psi^{\prime}\,dx+2\int\operatorname{{Re}}u_{xx}\bar{u}_{x}\psi^{\prime\prime}\,dx\\\
=3\int|u_{xx}|^{2}\psi^{\prime}\,dx-\int|u_{x}|^{2}\psi^{\prime\prime\prime}\,dx.$

Choosing either $\psi(x)\equiv 1$ or $\psi\not\equiv 0$ non-negative and non-
decreasing, we derive from (2.30) equalities (2.24) and (2.28) for smooth
solutions. Then estimate (2.23) and equality (2.28) provide estimate (2.27).

Next, multiply (2.1) by $2|u(t,x)|\bar{u}(t,x)\psi(x)$, integrate over
$\mathbb{R}$ and take the imaginary part, then

$\int\operatorname{{Im}}2iu_{t}|u|\bar{u}\psi\,dx+a\int
2\operatorname{{Im}}u_{xx}|u|\bar{u}\psi\,dx+b\int\operatorname{{Im}}2iu_{x}|u|\bar{u}\psi\,dx\\\
+\int\operatorname{{Im}}2iu_{xxx}|u|\bar{u}\psi\,dx=2\operatorname{{Im}}\int
f|u|\bar{u}\psi\,dx.$ (2.31)

Here

$\int\operatorname{{Im}}2iu_{t}|u|\bar{u}\psi\,dx=\int\operatorname{{Re}}2|u|u_{t}\bar{u}\psi\,dx=\int|u|(u_{t}\bar{u}+\bar{u}_{t}u)\psi\,dx\\\
=\int\bigl{(}|u|^{2}\bigr{)}^{1/2}\bigl{(}|u|^{2}\bigr{)}_{t}\psi\,dx=\frac{2}{3}\int\Bigl{(}\bigl{(}|u|^{2}\bigr{)}^{3/2}\Bigr{)}_{t}\psi\,dx=\frac{2}{3}\frac{d}{dt}\int|u|^{3}\psi\,dx,$
$\int 2\operatorname{{Im}}u_{xx}|u|\bar{u}\psi\,dx=-\int
2\operatorname{{Im}}\bigl{(}|u|\psi\bigr{)}_{x}\bar{u}u_{x}\,dx-\int
2\operatorname{{Im}}|u|u_{x}\bar{u}_{x}\psi\,dx\\\
=2\operatorname{{Im}}\int\bigl{(}|u|\psi\bigr{)}_{x}u\bar{u}_{x}\,dx,$
$\int\operatorname{{Im}}2iu_{x}|u|\bar{u}\psi\,dx=\int\operatorname{{Re}}|u|(u_{x}\bar{u}+\bar{u}_{x}u)\psi\,dx=\int(|u|^{2})^{1/2}\bigl{(}|u|^{2}\bigr{)}_{x}\psi\,dx\\\
=-\frac{2}{3}\int|u|^{3}\psi^{\prime}\,dx,$
$\int\operatorname{{Im}}2iu_{xxx}|u|\bar{u}\psi\,dx=\int\operatorname{{Re}}2u_{xxx}|u|\bar{u}\psi\,dx\\\
=-\int\operatorname{{Re}}2\bigl{(}|u|\psi\bigr{)}_{x}\bar{u}u_{xx}\,dx-\int\operatorname{{Re}}2|u|\bar{u}_{x}u_{xx}\psi\,dx\\\
=-\int\bigl{(}|u|\psi\bigr{)}_{x}(\bar{u}u_{xx}+\bar{u}_{xx}u)\,dx-\int|u|\psi(\bar{u}_{x}u_{xx}+u_{x}\bar{u}_{xx})\,dx\\\
=-\int\bigl{(}|u|\psi\bigr{)}_{x}\bigl{[}\bigl{(}|u|^{2}\bigr{)}_{xx}-2|u_{x}|^{2}\bigr{]}\,dx-\int|u|\psi\bigl{(}|u_{x}|^{2}\bigr{)}_{x}\,dx\\\
=3\int\bigl{(}|u|\psi\bigr{)}_{x}|u_{x}|^{2}\,dx-\int\bigl{(}|u|\psi\bigr{)}_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx,$

and similarly to (2.30) equality (2.31) ensures equalities (2.25) and (2.29)
for smooth solutions.

Finally, multiply (2.1) by $2\bar{u}_{x}(t,x)$, integrate over $\mathbb{R}$
and take the real part, then

$\int\operatorname{{Re}}2iu_{t}\bar{u}_{x}\,dx+a\int
2\operatorname{{Re}}u_{xx}\bar{u}_{x}\,dx+b\int\operatorname{{Re}}2iu_{x}\bar{u}_{x}\,dx\\\
+\int\operatorname{{Re}}2iu_{xxx}\bar{u}_{x}\,dx=2\operatorname{{Re}}\int
f\bar{u}_{x}\,dx.$ (2.32)

Here

$\int\operatorname{{Re}}2iu_{t}\bar{u}_{x}\,dx=-\int\operatorname{{Im}}2u_{t}\bar{u}_{x}\,dx=i\int(u_{t}\bar{u}_{x}-u_{x}\bar{u}_{t})\,dx\\\
=i\int(u_{t}\bar{u}_{x}+u\bar{u}_{tx})\,dx=i\frac{d}{dt}\int
u\bar{u}_{x}\,dx,$ $\int
2\operatorname{{Re}}u_{xx}\bar{u}_{x}\,dx=\int(u_{xx}\bar{u}_{x}+\bar{u}_{xx}u_{x})\,dx=0,$
$\int\operatorname{{Re}}2iu_{x}\bar{u}_{x}\,dx=0,$
$\int\operatorname{{Re}}2iu_{xxx}\bar{u}_{x}\,dx=i\int(u_{xxx}\bar{u}_{x}-\bar{u}_{xxx}u_{x})\,dx=0,$

and, therefore, (2.32) ensures equality (2.26) for smooth solutions.

The general case is obtained via closure on the basis of estimates (2.23) and
(2.27). ∎

##  3\. Existence

Alongside with equation (1.1) consider an equation with the regularized
nonlinearity

$iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}+\lambda
g_{\delta}(|u|^{2})u+i\beta\bigl{(}g_{\delta}(|u|^{2})u\bigr{)}_{x}+id(x)u=0,$
(3.1)

where $\delta\in(0,1]$ and the function $g_{\delta}$ is defined in (1.7). The
notion of a weak solution is similar to Definition 1.1.

Note that $g_{\delta}\in C^{\infty}(\mathbb{R}_{+})$
($C^{\infty}(\overline{\mathbb{R}}_{+})$ if $\delta>0$) and for $\theta>0$
uniformly with respect to $\delta\in[0,1]$

$|g_{\delta}(\theta)|\leq\bigl{(}\theta^{1/2}+1\bigr{)},\quad|g_{\delta}^{\prime}(\theta)|\leq\frac{1}{2}\theta^{-1/2},\quad\quad|g^{\prime\prime}_{\delta}(\theta)|\leq\frac{1}{4}\theta^{-3/2}.$
(3.2)

######  Lemma 3.1.

Let $d\in H^{1}$, $T>0$, $u\in X(\Pi_{T})$, $\delta\in[0,1]$, define

$F(u(t,x))\equiv-\lambda
g_{\delta}(|u|^{2})u-i\beta\bigl{(}g_{\delta}(|u|^{2})u\bigr{)}_{x}-id(x)u.$
(3.3)

Then $F\in L_{2}(0,T;H^{1})$ and there exists a positive constant
$c=c(T,\|d\|_{H^{1}},|\lambda|,|\beta|)$ (non-decreasing with respect to its
arguments) such that uniformly with respect to $\delta$

$\|F\|_{L_{2}(0,T;H^{1})}\leq
c\|u\|_{X(\Pi_{T})}\bigl{(}1+\|u\|_{X(\Pi_{T})}\bigr{)}.$ (3.4)

###### Proof.

From (3.2) it follows that for $\delta\in[0,1]$

$|g_{\delta}(|u|^{2})u|,|g_{\delta}(|u|^{2})u_{x}|,|g_{\delta}(|u|^{2})u_{xx}|\leq(1+|u|)(|u|+|u_{x}|+|u_{xx}|).$
(3.5)

Next, from (B.7) it follows that for $\delta\in[0,1]$

$\displaystyle\bigl{|}\bigl{(}g_{\delta}(|u|^{2})\bigr{)}_{x}u\bigr{|}\leq
c|uu_{x}|,$ (3.6)
$\displaystyle\bigl{|}\bigl{(}g_{\delta}(|u|^{2})\bigr{)}_{x}u_{x}\bigl{|},\bigl{(}\bigl{(}g_{\delta}(|u|^{2})\bigr{)}_{x}u\bigr{)}_{x}\bigr{|}\leq
c\bigl{(}|u_{x}|^{2}+|uu_{xx}|\bigr{)}.$ (3.7)

Here

$\iint_{\Pi_{T}}|u|^{2}|u_{xx}|^{2}\,dxdt\leq\int\sup\limits_{t\in(0,T)}|u|^{2}\,dx\cdot\sup\limits_{x\in\mathbb{R}}\int_{0}^{T}|u_{xx}|^{2}\,dt\leq
c\|u\|^{4}_{X(\Pi_{T})},$ (3.8)
$\iint_{\Pi_{T}}|u_{x}|^{4}\,dxdt\leq\sup\limits_{t\in(0,T)}\int|u_{x}|^{2}\,dx\cdot\int_{0}^{T}\sup\limits_{x\in\mathbb{R}}|u_{x}|^{2}\,dt\leq
c(T)\|u\|^{4}_{X(\Pi_{T})}.$ (3.9)

Other terms are estimated in a similar and more obvious way. ∎

######  Lemma 3.2.

Let $u_{0}\in H^{1}$, $d\in H^{1}$, $\delta\in(0,1]$. Then there exists
$T_{0}>0$, depending on $\|u_{0}\|_{H^{1}}$ and uniform with respect to
$\delta$, such that problem (3.1), (1.2) has a unique solution $u_{\delta}\in
X(\Pi_{T_{0}})$, moreover, uniformly with respect to $\delta$ for certain
$R>0$, depending on $\|u_{0}\|_{H^{1}}$,

$\|u_{\delta}\|_{X(\Pi_{T_{0}})}\leq R.$ (3.10)

###### Proof.

Without loss of generality we assume that $T_{0}\leq 1$. Let
$\|u_{0}\|_{H^{1}}\leq M$.

For any $\delta\in(0,1]$ the functions $g_{\delta}$ verify inequalities (3.2)
uniformly with respect to $\delta$. Therefore, for any $T\in(0,1]$ and $u\in
X(\Pi_{T})$ estimate (3.4) is verified uniformly with respect to $T$ and
$\delta$.

Let $R>0$. Define on a set $\widetilde{X}_{R}(\Pi_{T})=\\{v\in
X(\Pi_{T}):\|v\|_{X(\Pi_{T})}\leq R,v\big{|}_{t=0}=u_{0}\\}$ a map
$u=\Lambda_{0}v$, where $u\in X(\Pi_{T})$ is a solution to a linear problem

$iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}=F(v(t,x)),\quad u\big{|}_{t=0}=u_{0},$ (3.11)

where the function $F$ is given by formula (3.3) and the function $u$ is
substituted by a function $v\in\widetilde{X}_{R}(\Pi_{T})$.

By virtue of Lemma 3.1 $F(v)\in L_{2}(0,T;H^{1})$ and uniformly with respect
to $\delta$

$\|F(v)\|_{L_{2}(0,T;H^{1})}\leq cR(1+R).$ (3.12)

Then by virtue of Lemma 2.9 the map $\Lambda_{0}$ exists and according to
(2.23)

$\|u\|_{X(\Pi_{T})}\leq c_{0}M+c_{0}T^{1/2}R(1+R),$ (3.13)

for certain constant $c_{0}$ uniform with respect to $\delta$. Choose

$R\geq 2c_{0}M,\quad
T_{0}=\min\left(1,\frac{1}{16c_{0}^{2}(1+2R)^{2}}\right).$ (3.14)

Then $\Lambda_{0}$ maps $\widetilde{X}_{R}(\Pi_{T})$ into itself for any
$T\leq T_{0}$.

Next we show that for any $\delta\in(0,1]$ there exists
$T(\delta)\in(0,T_{0}]$ such that for any $T\in(0,T(\delta)]$ on the set
$\widetilde{X}_{R}(\Pi_{T})$ the map $\Lambda_{0}$ is a contraction. To this
end one has to estimate $F(v)-F(\widetilde{v})$ and
$\bigl{(}F(v)-F(\widetilde{v})\bigr{)}_{x}$. In particular, consider
$\bigl{(}g_{\delta}(|v|^{2})v\bigr{)}_{xx}-\bigl{(}g_{\delta}(|\widetilde{v}|^{2})\widetilde{v}\bigr{)}_{xx}$.
Note that

$\bigl{(}g_{\delta}(|v|^{2})v\bigr{)}_{xx}=g_{\delta}(|v|^{2})v_{xx}+\bigl{(}g_{\delta}(|v|^{2})\bigr{)}_{x}v_{x}+\bigl{(}(g_{\delta}(|v|^{2}))_{x}v\bigr{)}_{x}.$

Since

$\bigl{|}g_{\delta}(|v|^{2})-g_{\delta}(|\widetilde{v}|^{2})\bigr{|}\leq|v-\widetilde{v}|$

it follows similarly to (3.5) that

$\bigl{|}g_{\delta}(|v|^{2})v_{xx}-g_{\delta}(|\widetilde{v}|^{2})\widetilde{v}_{xx}\bigr{|}\leq(1+|v|)|\cdot|v_{xx}-\widetilde{v}_{xx}|+|\widetilde{v}_{xx}|\cdot|v-\widetilde{v}|.$

In comparison with (3.6) and (3.7) other terms are not estimated uniformly
with respect to $\delta$. Since

$\Bigl{|}\frac{1}{(|v|^{2}+\delta)^{1/2}}-\frac{1}{(|\widetilde{v}|^{2}+\delta)^{1/2}}\Bigr{|}\leq\frac{|v-\widetilde{v}|}{(|v|^{2}+\delta)^{1/2}(|\widetilde{v}|^{2}+\delta)^{1/2}}$

we derive with the use of (B.5) that

$\bigl{|}\bigl{(}g_{\delta}(|v|^{2})\bigr{)}_{x}v_{x}-\bigl{(}g_{\delta}(|\widetilde{v}|^{2})\bigr{)}_{x}\widetilde{v}_{x}\bigr{|}\leq
c\Bigl{[}\bigl{(}|v_{x}|+|\widetilde{v}_{x}|\bigr{)}|v_{x}-\widetilde{v}_{x}|+\frac{1}{\delta^{1/2}}\bigl{(}|v_{x}|^{2}+|\widetilde{v}_{x}|^{2}\bigr{)}|v-\widetilde{v}|\Bigr{]}.$

Next, since

$\Bigl{|}\frac{1}{(|v|^{2}+\delta)^{3/2}}-\frac{1}{(|\widetilde{v}|^{2}+\delta)^{3/2}}\Bigr{|}\leq\frac{3|v-\widetilde{v}|\bigl{(}|v|^{2}+\delta+|\widetilde{v}|^{2}+\delta\bigr{)}}{2(|v|^{2}+\delta)^{3/2}(|\widetilde{v}|^{2}+\delta)^{3/2}}$

we derive with the use of (B.6) that

$\bigl{|}\bigl{(}(g_{\delta}(|v|^{2}))_{x}v\bigr{)}_{x}-\bigl{(}(g_{\delta}(|\widetilde{v}|^{2}))_{x}\widetilde{v}\bigr{)}_{x}\bigr{|}\leq
c\Bigl{[}\bigl{(}|v|+|\widetilde{v}|\bigr{)}|v_{xx}-\widetilde{v}_{xx}|+\bigl{(}|v_{x}|+|\widetilde{v}_{x}|\bigr{)}|v_{x}-\widetilde{v}_{x}|\\\
+\bigl{(}|v_{xx}|+|\widetilde{v}_{xx}|\bigr{)}|v-\widetilde{v}|+\frac{1}{\delta^{1/2}}\bigl{(}|v_{x}|+|\widetilde{v}_{x}|\bigr{)}\bigl{(}|v|+|\widetilde{v}|\bigr{)}|v_{x}-\widetilde{v}_{x}|+\frac{1}{\delta^{1/2}}\bigl{(}|v_{x}|^{2}+|\widetilde{v}_{x}|^{2}\bigr{)}|v-\widetilde{v}|\\\
+\frac{1}{\delta}\bigl{(}|v_{x}|^{2}+|\widetilde{v}_{x}|^{2}\bigr{)}\bigl{(}|v|+|\widetilde{v}|\bigr{)}|v-\widetilde{v}|\Bigr{]}.$

Other terms are estimated in a similar and more obvious way. Then similarly to
(3.8), (3.9) and with the use of the embedding $C([0,T];H^{1})\subset
C_{b}(\overline{\Pi}_{T})$ it follows that for
$v,\widetilde{v}\in\widetilde{X}_{R}(\Pi_{T})$

$\|F(v)-F(\widetilde{v})\|_{L_{2}(0,T;H^{1})}\leq
c\Bigl{[}(1+2R)+\frac{(2R)^{3}}{\delta}\Bigr{]}\|v-\widetilde{v}\|_{\widetilde{X}_{R}(\Pi_{T})},$
(3.15)

whence similarly to (3.13) follows that

$\|u-\widetilde{u}\|_{X(\Pi_{T})}\leq
c_{1}T^{1/2}\Bigl{[}(1+2R)+\frac{(2R)^{3}}{\delta}\Bigr{]}\|v-\widetilde{v}\|_{X(\Pi_{T})}.$
(3.16)

where the constant $c_{1}$ does not depend on $\delta$. Then inequality (3.16)
ensures that for
$T(\delta)=\min\bigl{[}T_{0},\bigl{(}2c_{1}(1+2R+(2R)^{3}/\delta)\bigr{)}^{-2}\bigr{]}$

$\|\Lambda
v-\Lambda\widetilde{v}\|_{X(\Pi_{T})}\leq\frac{1}{2}\|v-\widetilde{v}\|_{X_{(\Pi_{T}})}.$
(3.17)

.

Next, assume that for certain $T<T_{0}$ there constructed a unique solution
$u\in\widetilde{X}_{R}(\Pi_{T})$ to problem (3.1), (1.2). Choose

$t_{0}=\min(T_{0}-T,T(\delta)),\quad T_{1}=T+t_{0}.$

Let $\Pi_{T,T_{1}}=(T,T_{1})\times\mathbb{R}$, define the space
$X(\Pi_{T,T_{1}})$ similarly to $X(\Pi_{T})$ ($0$ is substituted by $T$, $T$ —
by $T_{1}$) and let $\widetilde{X}_{R}(\Pi_{T,T_{1}})=\\{v\in
X(\Pi_{T,T_{1}}):\|v\|_{X(\Pi_{T,T_{1}})}\leq
R,v\big{|}_{t=T}=u\big{|}_{t=T}\\}$. For any
$v\in\widetilde{X}_{R}(\Pi_{T,T_{1}})$ let

$V(t,x)\equiv\begin{cases}u(t,x),\quad&t\in[0,T],\\\
v(t,x),\quad&t\in(T,T_{1}].\end{cases}$

Note that $V\in\widetilde{X}_{2R}(\Pi_{T_{1}})$. On the strip $\Pi_{T_{1}}$
consider a map $\Lambda_{1}$, where $U=\Lambda_{1}V\in X(\Pi_{T_{1}})$ is a
solution to problem (3.11), where $v$ is substituted by $V$. By virtue of
Lemmas 2.9 and 3.1 such a map exists and

$\|U\|_{X(\Pi_{T_{1}})}\leq c_{0}M+c_{0}T_{0}^{1/2}2R(1+2R)\leq R.$

Note that $U(t,x)=u(t,x)$ for $t\leq T$ because of uniqueness. Define a map
$\Lambda$ on the set $\widetilde{X}_{R}(\Pi_{T,T_{1}})$ in the following way:
let $\Lambda v$ be the restriction of $\Lambda_{1}V$ on the strip
$\Pi_{T,T_{1}}$. Note that $\Lambda v\in\widetilde{X}_{R}(\Pi_{T,T_{1}})$.
Moreover, similarly to (3.15)–(3.17)

$\|\Lambda v-\Lambda\widetilde{v}\|_{X(\Pi_{T,T_{1}})}\leq
c_{1}t_{0}^{1/2}\Bigl{[}(1+2R)+\frac{(2R)^{3}}{\delta}\Bigr{]}\|v-\widetilde{v}\|_{X(\Pi_{T,T_{1}})}\\\
\leq\frac{1}{2}\|v-\widetilde{v}\|_{X(\Pi_{T,T_{1}})}.$

As a result, the map $\Lambda$ is a contraction. Let $v_{0}$ be the unique
fixed point of the map $\Lambda$. Extend the function $u$ to the strip
$\Pi_{T_{1}}$ by the function $v_{0}$. Then this extended function $u$ is the
fixed point of the map $\Lambda_{1}$, in particular,
$u\in\widetilde{X}_{R}(\Pi_{T_{1}})$.

Moving step by step we obtain the solution
$u=u_{\delta}\in\widetilde{X}_{R}(\Pi_{T_{0}})$ to problem (3.1), (1.2). ∎

######  Theorem 3.3.

Let $u_{0}\in H^{1}$, $d\in H^{1}$. Then there exist $T_{0}>0$, depending on
$\|u_{0}\|_{H^{1}}$, and a solution $u\in X(\Pi_{T_{0}})$ to problem (1.1),
(1.2).

###### Proof.

Let $T_{0}$ be the same value as obtained in Lemma 3.2.

For any $\delta\in(0,1]$ consider the solution $u_{\delta}\in X(\Pi_{T_{0}})$
to problem (3.1), (1.2). According to (3.10) this set is bounded in the space
$X(\Pi_{T_{0}})$. In particular, for any interval $I_{n}=(-n,n)$,
$n\in\mathbb{N}$, uniformly with respect to $\delta$

$\|u_{\delta}\|_{L_{2}(0,T;H^{2}(I_{n}))}\leq c(n).$ (3.18)

Then since

$\bigl{(}g_{\delta}(|u|^{2})\bigr{)}_{x}=g_{\delta}^{\prime}(|u|^{2})(u\bar{u}_{x}+u_{x}\bar{u})$
(3.19)

with the use of equation (3.1) itself and inequalities (3.2) we derive that
uniformly with respect to $\delta$

$\|u_{\delta t}\|_{L_{2}(0,T;H^{-1}(I_{n}))}\leq c(n).$ (3.20)

Then by the standard argument we select a sequence $\delta_{k}\to+0$ such that

$\displaystyle u_{\delta_{k}}\rightharpoonup u$
$\displaystyle\quad*-\text{weakly in}\quad L_{\infty}(0,T_{0};H^{1}),$
$\displaystyle u_{\delta_{k}x}\rightharpoonup u_{x}$
$\displaystyle\quad*-\text{weakly in}\quad L_{6}(0,T_{0};L_{\infty}),$
$\displaystyle u_{\delta_{k}xx}\rightharpoonup u_{xx}$
$\displaystyle\quad*-\text{weakly in}\quad
L_{\infty}(\mathbb{R};L_{2}(0,T_{0})),$ $\displaystyle
u_{\delta_{k}}\rightharpoonup u$ $\displaystyle\quad*-\text{weakly in}\quad
L_{2}(\mathbb{R};L_{\infty}(0,T_{0})),$ $\displaystyle
u_{\delta_{k}}\rightarrow u$ $\displaystyle\quad\text{strongly in}\quad
L_{2}(0,T_{0};H^{1}(I_{n}))\quad\forall n\in\mathbb{N}.$

Write the corresponding integral identities for the functions $u_{\delta}$:

$\int_{0}^{T}\\!\\!\int\bigl{[}iu_{\delta}\phi_{t}-au_{\delta}\phi_{xx}+ibu_{\delta}\phi_{x}+iu_{\delta}\phi_{xxx}-\lambda
g_{\delta}(|u_{\delta}|^{2}|)u_{\delta}\phi+i\beta
g_{\delta}(|u_{\delta}|^{2})u_{\delta}\phi_{x}\\\
-idu_{\delta}\phi\bigr{]}\,dxdt+\int u_{0}\phi\big{|}_{t=0}\,dx=0,$ (3.21)

where $\phi\in C^{1}([0,T];L_{2})\cap C([0,T];H^{3})$, $\phi\big{|}_{t=T}=0$.
Assume first that the function $\phi$ has a compact support with respect to
$x$. We have:

$g_{\delta}(|u_{\delta}|^{2})-|u|=(|u_{\delta}|^{2}+\delta)^{1/2}-|u_{\delta}|+|u_{\delta}|-|u|,$

where

$\bigl{|}(|u_{\delta}|^{2}+\delta)^{1/2}-|u_{\delta}|\bigr{|}\leq\delta^{1/2},\quad\bigl{|}|u_{\delta}|-|u|\bigr{|}\leq|u_{\delta}-u|,$

and the passage to the limit in (3.21) when $\delta=\delta_{k}$ is obvious. As
a result we obtain the corresponding integral identity of (1.3)-type for the
functions $\phi$ with a compact support. Approximating general test functions
from Definition 1.1 by such ones and passing to the limit we derive the
desired integral identity in its general form.

Finally, note that according to Lemma 3.1 $F(u(t,x))\in L_{1}(0,T_{0};H^{1})$
and by virtue of Lemma 2.9 and uniqueness of a solution to a linear problem
$u\in X(\Pi_{T_{0}})$. ∎

######  Lemma 3.4.

Let $u_{0}\in H^{1}$, $d\in H^{1}$ and for certain $T>0$ a function $u\in
X(\Pi_{T})$ be a solution in $\Pi_{T}$ to problem (1.1), (1.2). Then for any
$t\in(0,T]$

$\int|u(t,x)|^{2}\,dx+2\int_{0}^{t}\\!\\!\int
d(x)|u(\tau,x)|^{2}\,dxd\tau=\int|u_{0}|^{2}\,dx.$ (3.22)

###### Proof.

Consider $u$ as the solution to problem (2.1), (2.2), where $f(t,x)\equiv
F(u(t,x))$ is defined by formula (3.3) for $\delta=0$. Then Lemma 3.1 implies
that $f\in L_{2}(0,T;H^{1})$ and according to Lemma 2.6 equality (2.15) holds:

$\frac{d}{dt}\int|u|^{2}\,dx+2\operatorname{{Im}}\int\bigl{[}\lambda|u|u+i\beta\bigl{(}|u|u\bigr{)}_{x}+id(x)u\bigr{]}\bar{u}\,dx=0.$
(3.23)

Obviously,

$\operatorname{{Im}}\int|u|u\bar{u}\,dx=0,$ $\operatorname{{Im}}i\int
2|u|u_{x}\bar{u}\,dx=\int
2|u|\operatorname{{Re}}(\bar{u}u_{x})\,dx=\frac{2}{3}\int\bigl{(}|u|^{3}\bigr{)}_{x}\,dx=0,$
$\operatorname{{Im}}i\int 2|u|_{x}u\bar{u}\,dx=-\int
2|u|\bigl{(}|u|^{2}\bigr{)}_{x}\,dx=-\frac{4}{3}\int\bigl{(}|u|^{3}\bigr{)}_{x}\,dx=0$

and (3.23) yields that

$\frac{d}{dt}\int|u|^{2}\,dx+2\int d(x)|u|^{2}\,dx=0.$

∎

######  Theorem 3.5.

Let $u_{0}\in H^{1,\alpha}$ for certain $\alpha\geq 0$, $d\in H^{1}$,
$\beta\neq 0$. Then there exists a function $u(t,x)$ such that $u\in
X(\Pi_{T})\cap C_{w}([0,T];H^{1,\alpha})$ for any $T>0$ and is a solution to
problem (1.1), (1.2) in $\Pi_{T}$; moreover, $u_{xx}\in
L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})$ for any $\epsilon>0$.

###### Proof.

Estimate the following a priori estimate. Let for certain $T^{\prime}\in(0,T]$
a function $u\in X(\Pi_{T^{\prime}})$ be a solution to problem (1.1), (1.2).
Then there exists a constant, depending on $T$, $\|u_{0}\|_{H^{1}}$ and
$\|d\|_{H^{1}}$, such that

$\|u\|_{C([0,T^{\prime}];H^{1})}\leq c.$ (3.24)

First note that equality (3.22) yields that

$\|u\|_{C[0,T^{\prime}];L_{2})}\leq
c(T,\|u_{0}\|_{L_{2}},\|d\|_{L_{\infty}}).$ (3.25)

Next, let the function $F(u)$ be defined by formula (3.3) for $\delta=0$. Then
Lemma 3.1 provides that $F\in L_{2}(0,T^{\prime};H^{1})$ and, therefore, Lemma
2.9 for $f\equiv F$ can be applied. Write down equality (2.24):

$\frac{d}{dt}\int|u_{x}|^{2}\,dx+2\operatorname{{Im}}\int\bigl{[}\lambda|u|u+i\beta\bigl{(}|u|u\bigr{)}_{x}+id(x)u\bigr{]}_{x}\bar{u}_{x}\,dx=0.$
(3.26)

Here

$2\operatorname{{Im}}\int\bigl{(}|u|u\bigr{)}_{x}\bar{u}_{x}\,dx=2\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx+2\operatorname{{Im}}\int|u|u_{x}\bar{u}_{x}\,dx\\\
=2\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx,$
$2\operatorname{{Im}}i\int\bigl{(}|u|u_{x}\bigr{)}_{x}\bar{u}_{x}\,dx=2\int|u|_{x}|u_{x}|^{2}\,dx+2\operatorname{{Re}}\int|u|u_{xx}\bar{u}_{x}\,dx\\\
=2\int|u|_{x}|u_{x}|^{2}\,dx+\int|u|\bigl{(}|u_{x}|^{2}\bigr{)}_{x}\,dx=\int|u|_{x}|u_{x}|^{2}\,dx,$
$2\operatorname{{Im}}i\int(|u|_{x}u)_{x}\bar{u}_{x}\,dx=2\operatorname{{Re}}\int(|u|_{x}u)_{x}\bar{u}_{x}\,dx=-2\operatorname{{Re}}\int|u|_{x}u\bar{u}_{xx}\,dx\\\
=-\int|u|_{x}(u\bar{u}_{xx}+u_{xx}\bar{u})\,dx=-\int|u|_{x}\bigl{[}\bigl{(}|u|^{2}\bigr{)}_{xx}-2|u_{x}|^{2}\bigr{]}\,dx\\\
=-\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx+2\int|u|_{x}|u_{x}|^{2}\,dx.$

As a result it follows from (3.26) that

$\frac{d}{dt}\int|u_{x}|^{2}\,dx+2\lambda\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx+3\beta\int|u|_{x}|u_{x}|^{2}\,dx\\\
-\beta\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx+2\int
d(x)|u_{x}|^{2}\,dx+2\operatorname{{Re}}\int d^{\prime}(x)u\bar{u}_{x}\,dx=0.$
(3.27)

Write down equality (2.26):

$i\frac{d}{dt}\int
u\bar{u}_{x}\,dx+2\operatorname{{Re}}\int\bigl{[}\lambda|u|u+i\beta\bigl{(}|u|u\bigr{)}_{x}+id(x)u\bigr{]}\bar{u}_{x}\,dx=0.$
(3.28)

Here

$2\operatorname{{Re}}\int|u|u\bar{u}_{x}\,dx=\int|u|\bigl{(}|u|^{2}\bigr{)}_{x}\,dx=0,$
$2\operatorname{{Re}}i\int|u|u_{x}\bar{u}_{x}\,dx=-2\operatorname{{Im}}\int|u||u_{x}|^{2}\,dx=0,$
$2\operatorname{{Re}}i\int|u|_{x}u\bar{u}_{x}\,dx=-2\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx.$

As a result it follows from (3.28) that

$i\frac{d}{dt}\int
u\bar{u}_{x}\,dx-2\beta\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx-2\operatorname{{Im}}\int
d(x)u\bar{u}_{x}\,dx=0.$ (3.29)

Finally, write down equality (2.25):

$\frac{2}{3}\frac{d}{dt}\int|u|^{3}\,dx+2a\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx+3\int|u|_{x}|u_{x}|^{2}\,dx\\\
-\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx+2\operatorname{{Im}}\int\bigl{[}\lambda|u|u+i\beta\bigl{(}|u|u\bigr{)}_{x}+id(x)u\bigr{]}|u|\bar{u}\,dx=0.$
(3.30)

Here

$2\operatorname{{Im}}\int|u|^{2}u\bar{u}\,dx=0,$
$2\operatorname{{Im}}i\int|u|^{2}u_{x}\bar{u}\,dx=\int|u|^{2}\bigl{(}|u|^{2}\bigr{)}_{x}\,dx=0,$
$2\operatorname{{Im}}i\int|u|_{x}|u|u\bar{u}\,dx=\int\bigl{(}|u|^{2}\bigr{)}_{x}|u|^{2}\,dx=0.$

As a result, it follows from (3.30) that

$\frac{2}{3}\frac{d}{dt}\int|u|^{3}\,dx+2a\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\,dx+3\int|u|_{x}|u_{x}|^{2}\,dx\\\
-\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx+2\int d(x)|u|^{3}\,dx=0.$ (3.31)

Sum equality (3.27) with (3.29), multiplied by
$\displaystyle\left(\frac{\lambda}{\beta}-a\right)$, and with (3.31),
multiplied by $\displaystyle-\beta$, then

$\frac{d}{dt}\left[\int|u_{x}|^{2}\,dx+i\left(\frac{\lambda}{\beta}-a\right)\int
u\bar{u}_{x}\,dx-\frac{2\beta}{3}\int|u|^{3}\,dx\right]\\\ +2\int
d(x)|u_{x}|^{2}\,dx+2\operatorname{{Re}}\int
d^{\prime}(x)u\bar{u}_{x}\,dx-2\left(\frac{\lambda}{\beta}-a\right)\operatorname{{Im}}\int
d(x)u\bar{u}_{x}\,dx\\\ -2\beta\int d(x)|u|^{3}\,dx=0.$ (3.32)

Note that

$\Bigl{|}2\operatorname{{Re}}\int d^{\prime}(x)u\bar{u}_{x}\,dx\Bigr{|}\leq
2\sup\limits_{x\in\mathbb{R}}|u|\cdot\|d^{\prime}\|_{L_{2}}\Bigl{(}\int|u_{x}|^{2}\,dx\Bigr{)}^{1/2}\\\
\leq
c\Bigl{(}\int|u_{x}|^{2}\,dx\Bigr{)}^{3/4}\Bigl{(}\int|u|^{2}\,dx\Bigr{)}^{1/4},$
$\int|u|^{3}\,dx\leq\sup\limits_{x\in\mathbb{R}}|u|\int|u|^{2}\,dx\leq
c\Bigl{(}\int|u_{x}|^{2}\,dx\Bigr{)}^{1/4}\Bigl{(}\int|u|^{2}\,dx\Bigr{)}^{5/4}\\\
\leq\varepsilon\int|u_{x}|^{2}\,dx+c(\varepsilon)\Bigl{(}\int|u|^{2}\,dx\Bigr{)}^{5/3},$

where $\varepsilon>0$ can be chosen arbitrarily small. Then equality (3.32)
with the use of the already obtained estimate (3.25) provides estimate (3.24).

Global estimate (3.24) together with the results on local well-posedness from
Theorem 3.3 provides existence of the solution to problem (1.1), (1.2) $u\in
X(\Pi_{T})$ $\forall T>0$.

Next, establish the corresponding properties of the solution $u$ in weighted
spaces. Let $\psi(x)\in C_{b}^{3}$, $\psi,\psi^{\prime}\geq 0$. Write down
equality (2.17) for $f\equiv F$, then

$\frac{d}{dt}\int|u(t,x)|^{2}\psi\,dx+3\int|u_{x}|^{2}\psi^{\prime}\,dx=2a\operatorname{{Im}}\int
u_{x}\bar{u}\psi^{\prime}\,dx\\\
+\int|u|^{2}\psi^{\prime\prime\prime}\,dx+b\int|u|^{2}\psi^{\prime}\,dx-2\operatorname{{Im}}\int\bigl{[}\lambda|u|u+i\beta\bigl{(}|u|u\bigr{)}_{x}+id(x)u\bigr{]}\bar{u}\psi\,dx.$
(3.33)

Here

$\operatorname{{Im}}\int|u|u\bar{u}\psi\,dx=0,$ (3.34)
$-\operatorname{{Im}}i\int 2\bigl{(}|u|u\bigr{)}_{x}\bar{u}\psi\,dx=-\int
2\operatorname{{Re}}\bigl{(}|u|u\bigr{)}_{x}\bar{u}\psi\,dx=\int|u|2\operatorname{{Re}}(u\bar{u}_{x})\psi\,dx\\\
+2\int|u|^{3}\psi^{\prime}\,dx=\int|u|\bigl{(}|u|^{2}\bigr{)}_{x}\psi\,dx+2\int|u|^{3}\psi^{\prime}\,dx=\frac{4}{3}\int|u|^{3}\psi^{\prime}\,dx.$
(3.35)

Note that

$\int|u|^{3}\psi^{\prime}\,dx\leq\sup\limits_{x\in\mathbb{R}}\bigl{(}|u|\sqrt{\psi^{\prime}}\bigr{)}\Bigl{(}\int|u|^{2}\psi^{\prime}\,dx\int|u|^{2}\,dx\Bigr{)}^{1/2},$
(3.36)

where

$\sup\limits_{x\in\mathbb{R}}|u|\sqrt{\psi^{\prime}}\leq\Bigl{(}\int\bigl{|}(u\sqrt{\psi^{\prime}})_{x}u\sqrt{\psi^{\prime}}\bigr{|}\,dx\Bigr{)}^{1/2}\leq\Bigl{(}\int\bigl{(}|u_{x}u|\psi^{\prime}+\frac{1}{2}|u^{2}\psi^{\prime\prime}|\bigr{)}\,dx\Bigr{)}^{1/2}\\\
\leq\Bigl{(}\int|u_{x}|^{2}\psi^{\prime}\,dx\int|u|^{2}\psi^{\prime}\,dx\Bigr{)}^{1/4}+\Bigl{(}\int|u|^{2}|\psi^{\prime\prime}|\,dx\Bigr{)}^{1/2}.$
(3.37)

As a result,

$\int|u|^{3}\psi^{\prime}\,dx\leq\varepsilon\int|u_{x}|^{2}\psi^{\prime}\,dx+c(\varepsilon)\Bigl{(}\int|u|^{2}\,dx\Bigr{)}^{2/3}\int|u|^{2}\psi^{\prime}\,dx\\\
+c\Bigl{(}\int|u|^{2}\,dx\Bigr{)}^{1/2}\int|u|^{2}(\psi^{\prime}+|\psi^{\prime\prime}|)\,dx,$
(3.38)

where $\varepsilon>0$ can be chosen arbitrarily small. Then equality (3.33)
and estimate (3.25) imply that

$\frac{d}{dt}\int|u(t,x)|^{2}\psi\,dx+\int|u_{x}|^{2}\psi^{\prime}\,dx\leq
c\int|u|^{2}\bigl{(}\psi+\psi^{\prime}+|\psi^{\prime\prime}|+|\psi^{\prime\prime\prime}|\bigr{)}\,dx,$
(3.39)

where the constant $c$ depends on $T$, $\|u_{0}\|_{L_{2}}$,
$\|d\|_{L_{\infty}}$, $a$, $b$, $\beta$ and does not depend on $\psi$.

Approximate the function $\rho_{\alpha,\epsilon}(x)$ in such a way: for any
$r>0$ let

$\rho_{r,\alpha,\epsilon}(x)\equiv\rho_{\alpha,\epsilon}(x)\eta(x+r-1)+\rho_{\alpha,\epsilon}(r+1)\eta(x-r),$
(3.40)

where the cut-off function $\eta$ is defined in Section 1.

Obviously, $\rho_{r,\alpha,\epsilon}(x)=\rho_{\alpha,\epsilon}(x)$ for $x\leq
r$, $\rho_{r,\alpha,\epsilon}(x)=(2+r)^{2\alpha}$ for $x\geq r+1$. The
function $\rho_{r,\alpha,\epsilon}\in C_{b}^{\infty}$,
$\rho^{(j)}_{r,\alpha,\epsilon}(x)\to\rho^{(j)}_{\alpha,\epsilon}(x)$ when
$r\to+\infty$ and $|\rho^{(j)}_{r,\alpha,\epsilon}(x)|\leq
c(\alpha,\epsilon,j)\rho_{\alpha,\epsilon}(x)$ for any $x\in\mathbb{R}$ and
any non-negative integer number $j$,
$0\leq\rho^{\prime}_{r,\alpha,\epsilon}(x)\leq
c(\alpha)\rho^{\prime}_{\alpha,\epsilon}(x)$ for any $x\in\mathbb{R}$.

Set in (3.39) $\psi(x)\equiv\rho_{r,\alpha,\epsilon}(x)$. then uniformly with
respect to $r$

$\frac{d}{dt}\int|u(t,x)|^{2}\rho_{r,\alpha,\epsilon}\,dx+\int|u_{x}|^{2}\rho_{r,\alpha,\epsilon}^{\prime}\,dx\leq
c\int|u|^{2}\rho_{r,\alpha,\epsilon}\,dx,$ (3.41)

whence it follows that

$\|u\|_{L_{\infty}(0,T;L_{2}^{\alpha})}+\|u_{x}\|_{L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})}\leq
c(T,\alpha,\epsilon,\|u_{0}\|_{L_{2}^{\alpha}},\|d\|_{L_{\infty}},a,b,\beta).$
(3.42)

Again let $\psi(x)\equiv\rho_{r,\alpha,\epsilon}(x)$. Write down corresponding
equality (2.28). Note that

$F_{x}=-\lambda\bigl{(}|u|u\bigr{)}_{x}-i\beta\bigl{(}|u|u\bigr{)}_{xx}-i\bigl{(}d(x)u\bigr{)}_{x},$
(3.43)

then

$\frac{d}{dt}\int|u_{x}(t,x)|^{2}\psi\,dx+3\int|u_{xx}|^{2}\psi^{\prime}\,dx-2a\operatorname{{Im}}\int
u_{xx}\bar{u}_{x}\psi^{\prime}\,dx\\\
-\int|u_{x}|^{2}\psi^{\prime\prime\prime}\,dx-b\int|u_{x}|^{2}\psi^{\prime}\,dx+2\lambda\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\psi\,dx\\\
+2\beta\operatorname{{Re}}\int\bigl{(}|u|u\bigr{)}_{xx}\bar{u}_{x}\psi\,dx+2\operatorname{{Re}}\int\bigl{(}d(x)u\bigr{)}_{x}\bar{u}_{x}\psi\,dx=0.$

Here

$2\operatorname{{Re}}\int\bigl{(}|u|u\bigr{)}_{xx}\bar{u}_{x}\psi\,dx=2\operatorname{{Re}}\int\bigl{(}|u|u_{x}\bigr{)}_{x}\bar{u}_{x}\psi\,dx+2\operatorname{{Re}}\int\bigl{(}|u|_{x}u\bigr{)}_{x}\bar{u}_{x}\psi\,dx\\\
=2\int|u|_{x}|u_{x}|^{2}\psi\,dx+2\operatorname{{Re}}\int|u|u_{xx}\bar{u}_{x}\psi\,dx-2\operatorname{{Re}}\int|u|_{x}u\bar{u}_{xx}\psi\,dx\\\
-2\operatorname{{Re}}\int|u|_{x}u\bar{u}_{x}\psi^{\prime}\,dx=2\int|u|_{x}|u_{x}|^{2}\psi\,dx+\int|u|\bigl{(}|u_{x}|^{2}\bigr{)}_{x}\psi\,dx\\\
-\int|u|_{x}\bigl{(}(|u|^{2})_{xx}-2|u_{x}|^{2}\bigr{)}\psi\,dx-2\operatorname{{Re}}\int|u|_{x}u\bar{u}_{x}\psi^{\prime}\,dx\\\
=3\int|u|_{x}|u_{x}|^{2}\psi\,dx-\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\psi\,dx-\int|u||u_{x}|^{2}\psi^{\prime}\,dx-\int|u|_{x}(|u|^{2})_{x}\psi^{\prime}\,dx.$

As a result,

$\frac{d}{dt}\int|u_{x}(t,x)|^{2}\psi(x)\,dx+3\int|u_{xx}|^{2}\psi^{\prime}\,dx-2a\operatorname{{Im}}\int
u_{xx}\bar{u}_{x}\psi^{\prime}\,dx\\\
-\int|u_{x}|^{2}\psi^{\prime\prime\prime}\,dx-b\int|u_{x}|^{2}\psi^{\prime}\,dx+2\lambda\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\psi\,dx\\\
+3\beta\int|u|_{x}|u_{x}|^{2}\psi\,dx-\beta\int|u|_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\psi\,dx-\beta\int|u||u_{x}|^{2}\psi^{\prime}\,dx\\\
-\beta\int|u|_{x}(|u|^{2})_{x}\psi^{\prime}\,dx+2\int
d(x)|u_{x}|^{2}\psi\,dx+2\operatorname{{Re}}\int
d^{\prime}(x)u\bar{u}_{x}\psi\,dx=0.$ (3.44)

Next write down corresponding equality (2.29):

$\frac{2}{3}\frac{d}{dt}\int|u|^{3}\psi\,dx+2a\operatorname{{Im}}\int\bigl{(}|u|\psi\bigr{)}_{x}u\bar{u}_{x}\,dx+3\int\bigl{(}|u|\psi\bigr{)}_{x}|u_{x}|^{2}\,dx\\\
-\int\bigl{(}|u|\psi\bigr{)}_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx-\frac{2}{3}b\int|u|^{3}\psi^{\prime}\,dx+2\lambda\operatorname{{Im}}\int|u|^{4}\psi\,dx\\\
+2\beta\operatorname{{Re}}\int\bigl{(}|u|u\bigr{)}_{x}|u|\bar{u}\psi\,dx+2\int
d(x)|u|^{3}\psi\,dx=0.$

Here

$2\operatorname{{Re}}\int\bigl{(}|u|u\bigr{)}_{x}|u|\bar{u}\psi\,dx=2\operatorname{{Re}}\int|u|^{2}u_{x}\bar{u}\psi\,dx+2\int|u|_{x}|u|^{3}\psi\,dx\\\
=2\int|u|^{2}\bigl{(}|u|^{2}\bigr{)}_{x}\psi\,dx=-\int|u|^{4}\psi^{\prime}\,dx.$

As a result,

$\frac{2}{3}\frac{d}{dt}\int|u|^{3}\psi\,dx+2a\operatorname{{Im}}\int\bigl{(}|u|\psi\bigr{)}_{x}u\bar{u}_{x}\,dx+3\int\bigl{(}|u|\psi\bigr{)}_{x}|u_{x}|^{2}\,dx\\\
-\int\bigl{(}|u|\psi\bigr{)}_{x}\bigl{(}|u|^{2}\bigr{)}_{xx}\,dx-\frac{2}{3}b\int|u|^{3}\psi^{\prime}\,dx-\beta\int|u|^{4}\psi^{\prime}\,dx+2\int
d(x)|u|^{3}\psi\,dx=0.$ (3.45)

Sum equality (3.44) with equality (3.45) multiplied by $-\beta$, then

$\frac{d}{dt}\int\left[|u_{x}|^{2}-\frac{2\beta}{3}|u|^{3}\right]\psi\,dx+3\int|u_{xx}|^{2}\psi^{\prime}\,dx-2a\operatorname{{Im}}\int
u_{xx}\bar{u}_{x}\psi^{\prime}\,dx\\\
-\int|u_{x}|^{2}\psi^{\prime\prime\prime}\,dx-b\int|u_{x}|^{2}\psi^{\prime}\,dx+2\bigl{(}\lambda-a\beta\bigr{)}\operatorname{{Im}}\int|u|_{x}u\bar{u}_{x}\psi\,dx\\\
-2a\beta\operatorname{{Im}}\int|u|u\bar{u}_{x}\psi^{\prime}\,dx-4\beta\int|u||u_{x}|^{2}\psi^{\prime}\,dx+\beta\int|u|\bigl{(}|u|^{2}\bigr{)}_{xx}\psi^{\prime}\,dx\\\
-\beta\int|u|_{x}(|u|^{2})_{x}\psi^{\prime}\,dx+\frac{2b\beta}{3}\int|u|^{3}\psi^{\prime}\,dx+\beta^{2}\int|u|^{4}\psi^{\prime}\,dx\\\
+2\int d(x)|u_{x}|^{2}\psi\,dx+2\operatorname{{Re}}\int
d^{\prime}(x)u\bar{u}_{x}\psi\,dx-2\beta\int d(x)|u|^{3}\psi\,dx=0.$ (3.46)

The already obtained estimate (3.24) yields, in particular, that

$\|u\|_{L_{\infty}(\Pi_{T})}\leq c.$ (3.47)

Therefore, uniformly with respect to $r$ by virtue of (3.42)

$\sup\limits_{t\in(0,T)}\int|u|^{3}\psi\,dx\leq
c,\quad\int|u|^{2}|u_{xx}|\psi^{\prime}\,dx\leq\varepsilon\int|u_{xx}|^{2}\psi^{\prime}\,dx+c(\varepsilon),$

where $\varepsilon>0$ can be chosen arbitrarily small; similarly to (3.37)

$\Bigl{|}\operatorname{{Re}}\int
d^{\prime}(x)u\bar{u}_{x}\psi\,dx\Bigr{|}\leq\sup\limits_{x\in\mathbb{R}}\bigl{(}|u|\sqrt{\psi}\bigr{)}\Bigl{(}\int(d^{\prime}(x))^{2}\,dx\int|u_{x}|^{2}\psi\,dx\Bigr{)}^{1/2}\\\
\leq c\int|u_{x}|^{2}\psi\,dx.$

Therefore, it follows from (3.46) that uniformly with respect to $r$

$\sup\limits_{t\in(0,T)}\int|u_{x}|^{2}\psi\,dx+\int_{0}^{T}\\!\\!\int|u_{xx}|^{2}\psi^{\prime}\,dxdt\leq
c,$

whence passing to the limit when $r\to+\infty$ we obtain that

$\|u_{x}\|_{L_{\infty}(0,T;L_{2}^{\alpha})}+\|u_{xx}\|_{L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})}\leq
c.$ (3.48)

The end of the proof is standard. ∎

######  Theorem 3.6.

Let $u_{0}\in L_{2}^{\alpha}$ for certain $\alpha\geq 0$, $d\in L_{\infty}$,
then there exists a function $u(t,x)$ such that for any $T>0$ it is a weak
solution to problem (1.1), (1.2), $u\in C_{w}([0,T];L_{2}^{\alpha})$,
verifying properties

$\sigma(u;T)=\sup\limits_{x_{0}\in\mathbb{R}}\int_{0}^{T}\\!\int_{x_{0}}^{x_{0}+1}|u_{x}|^{2}\,dxdt\leq
c(T,\|u_{0}\|_{L_{2}},\|d\|_{L_{\infty}})$ (3.49)

and $u_{x}\in L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})$ for any
$\epsilon>0$.

###### Proof.

Apply the mollifying procedure to approximate the functions $u_{0}$ and $d$ by
smooth ones: $u_{0h},d_{h}\in H^{1}$, $u_{0h}\to u_{0}$ in $L_{2}^{\alpha}$,
$d_{h}\to d$ in $L_{2}$ when $h\to+0$,
$\|d_{h}\|_{L_{\infty}}\leq\|d\|_{L_{\infty}}$. Consider a set of regularized
problems

$\displaystyle
iu_{t}+au_{xx}+ibu_{x}+iu_{xxx}+\lambda|u|u+i\beta_{h}\bigl{(}|u|u\bigr{)}_{x}+id_{h}(x)u=0,$
(3.50) $\displaystyle u(0,x)=u_{0h}(x),\quad x\in\mathbb{R}.$ (3.51)

where $\beta_{h}=\beta$ if $\beta\neq 0$, $\beta_{h}=h$ if $\beta=0$.
According to Theorem 3.5 for any $h\in(0,1]$ there exists a function $u_{h}\in
X(\Pi_{T})\cap C_{w}([0,T];H^{1,\alpha})$ for any $T>0$ which satisfies
(3.50), (3.51) in $\Pi_{T}$. Equality (3.22) yields that for any $T>0$
uniformly with respect to $h$

$\|u_{h}\|_{L_{\infty}(0,T;L_{2})}\leq c(T).$ (3.52)

Write for the functions $u_{h}$ corresponding inequality (3.39) for
$\psi(x)\equiv\rho_{0,1}(x-x_{0})$ for any $x_{0}\in\mathbb{R}$, then the
properties of the function $\rho_{0,1}$ and estimate (3.52) provide that
uniformly with respect to $h$

$\sigma(u_{h};T)\leq c(T).$ (3.53)

Moreover, using in (3.39) $\psi(x)\equiv\rho_{r,\alpha,\epsilon}(x)$ similarly
to (3.41), (3.42) we derive that uniformly with respect to $h$

$\|u_{h}\|_{L_{\infty}(0,T;L_{2}^{\alpha})}+\|u_{hx}\|_{L_{2}(0,T;L_{2}^{\rho^{\prime}_{\alpha,\epsilon}(x)})}\leq
c(T,\alpha,\epsilon).$ (3.54)

It follows from the well–known embedding $L_{1}\subset H^{-1}$, estimate
(3.52) and equation (3.50) itself that uniformly with respect to $h$

$\|u_{ht}\|_{L_{\infty}(0,T;H^{-3})}\leq c(T).$ (3.55)

Then by the standard argument we select a sequence $h_{k}\to+0$ such that for
any $T>0$

$\displaystyle u_{h_{k}}\rightharpoonup u$ $\displaystyle\quad*-\text{weakly
in}\quad L_{\infty}(0,T;L_{2}),$ $\displaystyle u_{h_{k}x}\rightharpoonup
u_{x}$ $\displaystyle\quad\text{weakly in}\quad L_{2}(0,T;L_{2}(I_{n})),$
$\displaystyle u_{h_{k}}\rightarrow u$ $\displaystyle\quad\text{strongly
in}\quad L_{2}(0,T;L_{2}(I_{n}))\quad\forall n\in\mathbb{N},I_{n}=(-n,n).$

Write down the corresponding integral identities for the functions $u_{h}$ and
certain $T>0$:

$\iint_{\Pi_{T}}\bigl{[}iu_{h}\phi_{t}-au_{h}\phi_{xx}+ibu_{h}\phi_{x}+iu_{h}\phi_{xxx}-\lambda|u_{h}|u_{h}\phi+i\beta_{h}|u_{h}|u_{h}\phi_{x}\\\
-d_{h}(x)u_{h}\phi\bigr{]}\,dxdt+\int u_{0h}\phi\big{|}_{t=0}\,dx=0,$ (3.56)

where $\phi\in C^{1}([0,T];L_{2})\cap C([0,T];H^{3})$, $\phi\big{|}_{t=T}=0$.
Assume first that the function $\phi$ has a compact support with respect to
$x$. Then passage to the limit in (3.56) when $h=h_{k}\to+0$ is obvious and we
obtain (1.3) for such test functions. Approximating general test functions
from Definition 1.1 by such ones and passing to the limit we derive integral
identity (1.3) in its general form.

The end of the proof is standard. ∎

######  Remark 3.7.

Let $u_{0}\in H^{1}$, $d\in H^{1}$ and $u\in X(\Pi_{T})$ be a solution to
problem (1.1), (1.2) for certain $T>0$. Let $|\lambda|,|\beta|\leq A$ for
certain $A>0$. Then similarly to (3.52), (3.53) one can obtain that

$\|u\|_{L_{\infty}(0,T;L_{2})}+\sigma(u;T)\leq
c(T,\|u_{0}\|_{L_{2}},\|d\|_{L_{\infty}},A).$ (3.57)

##  4\. Uniqueness and continuous dependence

######  Theorem 4.1.

Let $u_{0},\widetilde{u}_{0}\in L_{2}^{3/4}$, $d\in L_{\infty}$,
$u,\widetilde{u}\in L_{\infty}(0,T;L_{2}^{3/4})$ – be solutions to problem
(1.1), (1.2) (for $\widetilde{u}$ the initial function in (1.2) is equal to
$\widetilde{u}_{0}$) in the strip $\Pi_{T}$ for certain $T>0$. Assume that
$\|u\|_{L_{\infty}(0,T;L_{2}^{3/4})},|\widetilde{u}\|_{L_{\infty}(0,T;L_{2}^{3/4})}\leq
M$ for certain $M>0$ Then for any $\epsilon>0$ there exists a constant
$c=c(T,M,\epsilon)$ such that

$\operatorname*{ess\,sup}\limits_{t\in(0,T)}\int\bigl{|}u(t,x)-\widetilde{u}(t,x)\bigr{|}^{2}\rho_{3/4,\epsilon}\,dx\leq
c\int\bigl{|}u_{0}-\widetilde{u}_{0}\bigr{|}^{2}\rho_{3/4,\epsilon}\,dx.$
(4.1)

In particular a weak solution to problem (1.1), (1.2) in unique in the space
$L_{\infty}(0,T;L_{2}^{3/4})$.

###### Proof.

Let $w\equiv u-\widetilde{u}$, $w_{0}\equiv u_{0}-\widetilde{u}_{0}$,
$f_{11}\equiv-idw$,
$f_{12}\equiv-\lambda\bigl{(}|u|u-|\widetilde{u}|\widetilde{u}\bigr{)}$,
$f_{2}\equiv-i\beta\bigl{(}|u|u-|\widetilde{u}|\widetilde{u}\bigr{)}$. Then
the function $w\in L_{\infty}(0,T;L_{2}^{3/4})$ is a weak solution to problem

$iw_{t}+aw_{xx}+ibw_{x}+iw_{xxx}=f_{11}+f_{12}+f_{2x},\quad
w\big{|}_{t=0}=w_{0}.$ (4.2)

Consider a linear problem

$iw_{1t}+aw_{1xx}+iw_{1x}+iw_{1xxx}=f_{11},\quad w_{1}\big{|}_{t=0}=w_{0}.$
(4.3)

Note that $w_{0}\in L_{2}^{3/4}$, $f_{11}\in L_{\infty}(0,T;L_{2}^{3/4})$.
Then according to Lemma 2.8 there exists a solution to this problem $w_{1}\in
C([0,T];L_{2}^{3/4})$, $w_{1x}\in
L_{2}(0,T;L_{2}^{\rho^{\prime}_{3/4,\epsilon}(x)})$ for any $\epsilon>0$.
Write down corresponding equality (2.22) for
$\rho(x)\equiv\rho_{3/4,\epsilon}(x)$:

$\frac{d}{dt}\int|w_{1}(t,x)|^{2}\rho(x)\,dx+3\int|w_{1x}|^{2}\rho^{\prime}\,dx=2a\operatorname{{Im}}\int
w_{1x}\bar{w}_{1}\rho^{\prime}\,dx\\\
+\int|w_{1}|^{2}\rho^{\prime\prime\prime}\,dx+b\int|w_{1}|^{2}\rho^{\prime}\,dx-2\operatorname{{Re}}\int
dw\bar{w}_{1}\rho\,dx.$ (4.4)

It follows from (4.4) that

$\int|w_{1}(t,x)|^{2}\rho\,dx\leq
c\int|w_{0}|^{2}\rho\,dx+c\int_{0}^{t}\\!\\!\int|w|^{2}\rho\,dxd\tau.$ (4.5)

Let $w_{2}\equiv w-w_{1}$, then $w_{2}\in L_{\infty}(0,T;L_{2}^{3/4})$ and is
a solution to a problem

$iw_{2t}+aw_{2xx}+ibw_{2x}+iw_{2xxx}=f_{12}+f_{2x},\quad
w_{2}\big{|}_{t=0}=0.$ (4.6)

Note that $|f_{12}|,|f_{2}|\leq c\bigl{(}|u|+|\widetilde{u}|\bigr{)}|w|$; in
particular, $f_{12},f_{2}\in L_{\infty}(0,T;L_{1}^{3/2})$. Write corresponding
inequality (2.20), then uniformly with respect to $t\in(0,T)$ for
$\rho(x)\equiv\rho_{3/4,\epsilon}(x)$ and $p>4$

$\operatorname*{ess\,sup}\limits_{\tau\in(0,t)}\Bigl{(}\int|w_{2}(\tau,x)|^{2}\rho\,dx\Bigr{)}^{1/2}\\\
\leq
c\Bigl{[}\int_{0}^{t}\Bigl{(}\int\bigl{(}|u|+|\widetilde{u}|\bigr{)}|w|\rho^{1/2}(1+x_{+})^{3/4}\,dx\Bigl{)}^{p}d\tau\Bigr{]}^{1/p}\\\
\leq
c_{1}\left(\|u\|_{L_{\infty}(0,T;L_{2}^{3/4})}+\|\widetilde{u}\|_{L_{\infty}(0,T;L_{2}^{3/4})}\right)\Bigl{[}\int_{0}^{t}\Bigl{(}\int|w|^{2}\rho\,dx\Bigl{)}^{p/2}d\tau\Bigr{]}^{1/p}.$
(4.7)

Inequalities (4.5) and (4.7) yield that

$\operatorname*{ess\,sup}\limits_{\tau\in(0,t)}\Bigl{(}\int|w(\tau,x)|^{2}\rho\,dx\Bigr{)}^{1/2}\leq
c\Bigl{(}\int|w_{0}|^{2}\rho\,dx\Bigr{)}^{1/2}\\\
+c\Bigl{[}\int_{0}^{t}\int|w|^{2}\rho\,dxd\tau\Bigr{]}^{1/2}+c\Bigl{[}\int_{0}^{t}\Bigl{(}\int|w|^{2}\rho\,dx\Bigl{)}^{p/2}d\tau\Bigr{]}^{1/p},$
(4.8)

whence (4.1) follows by the standard argument. ∎

##  5\. Internal regularity

######  Theorem 5.1.

Let $u_{0}\in L_{2}^{\alpha}$ for certain $\alpha>1/4$, $d\in L_{\infty}$,
$u\in L_{\infty}(0,T;L_{2}^{\alpha})$ be a solution to problem (1.1), (1.2)
for certain $T>0$. Then (after possible correction on a zero measure set)
$u\in C_{b}(\overline{\Pi}_{T}^{t_{0},x_{0}})$ for any $t_{0}\in(0,T)$,
$x_{0}\in\mathbb{R}$, where
$\Pi_{T}^{t_{0},x_{0}}=(t_{0},T)\times(x_{0},+\infty)$.

###### Proof.

Consider the function $u$ as a solution to problem (2.1), (2.2), where
$f\equiv F$ and the function $F$ is given by formula (3.3) for $\delta=0$. Let

$f_{11}\equiv-id(x)u,\quad f_{12}\equiv-\lambda u|u|,\quad f_{2}\equiv-i\beta
u|u|,$ (5.1)

$f_{1}\equiv f_{11}+f_{12}$. Then $f=f_{1}+f_{2x}$. Note that

$\|f_{11}\|_{L_{\infty}(0,T;L_{2}^{\alpha})}\leq\|d\|_{L_{\infty}}\|u\|_{L_{\infty}(0,T;L_{2}^{\alpha})},$
(5.2)
$\|f_{12}\|_{L_{\infty}(0,T;L_{1})},\|f_{2}\|_{L_{\infty}(0,T;L_{1}^{1/4})}\leq
c\|u\|^{2}_{L_{\infty}(0,T;L_{2}^{\alpha})}.$ (5.3)

Thus the hypothesis of Theorem 2.5 is verified. Write down equality (2.7). For
the first term in the right-hand side of (2.7) apply estimate (2.9). Next,
similarly to (2.10)

$\displaystyle\Bigl{|}\int_{0}^{t}\\!\\!\int
G(t-\tau,x-y)f_{11}(\tau,y)\,dyd\tau\Bigr{|}\\\ \leq
c(T,\alpha)(1+x_{-})^{\alpha}\int_{0}^{t}\frac{\|f_{11}(\tau,\cdot)\|_{L_{2}^{\alpha}}}{(t-\tau)^{1/3}}\,d\tau\leq
c_{1}(T,\alpha)(1+x_{-})^{\alpha}\|f_{11}\|_{L_{\infty}(0,T;L_{2}^{\alpha})},$
(5.4)

similarly to (2.11)

$\Bigl{|}\int_{0}^{t}\\!\\!\int
G(t-\tau,x-y)f_{12}(\tau,y)\,dyd\tau\Bigr{|}\\\ \leq
c(T)\int_{0}^{t}\frac{1}{(t-\tau)^{1/3}}\Bigl{[}\int_{x}^{+\infty}(1+y-x)^{-1/4}|f_{12}(\tau,y)|\,dy\\\
+\int_{-\infty}^{x}e^{-c_{0}(x-y)^{3/2}T^{-1/2}}|f_{12}(\tau,y)|\,dy\Bigr{]}\,d\tau\leq
c_{1}(T)\|f_{12}\|_{L_{\infty}(0,T;L_{1})},$ (5.5)

similarly to (2.12)

$\Bigl{|}\int_{0}^{t}\\!\\!\int
G_{x}(t-\tau,x-y)f_{2}(\tau,y)\,dyd\tau\Bigr{|}\\\ \leq
c(T)\int_{0}^{t}\frac{1}{(t-\tau)^{3/4}}\Bigl{[}\int_{x}^{+\infty}(1+y-x)^{1/4}|f_{2}(\tau,y)|\,dy\\\
+\int_{-\infty}^{x}e^{-c_{0}(x-y)^{3/2}T^{-1/2}}|f_{2}(\tau,y)|\,dy\Bigr{]}\,d\tau\leq
c_{1}(T)(1+x_{-})^{1/4}\|f_{2}\|_{L_{\infty}(0,T;L_{1}^{1/4})}.$ (5.6)

Moreover, the considered integrals are continuous with respect to
$(t,x)\in\Pi_{T}$. ∎

######  Theorem 5.2.

Let $u_{0}\in L_{2}^{\alpha}$ for certain $\alpha\geq 1/2$, $d\in H^{1}$.
Consider the function $u(t,x)$, constructed in Theorem 3.6, which is the weak
solution to problem (1.1), (1.2) in any strip $\Pi_{T}$. Then $u_{x}\in
C_{w}([t_{0},T];L_{2}^{\rho_{\alpha-1/2,\epsilon}(x)})$, $u_{xx}\in
L_{2}(t_{0},T;L_{2}^{\rho^{\prime}_{\alpha-1/2,\epsilon}(x)})$ for any
$t_{0}\in(0,T)$, $\epsilon>0$.

###### Proof.

Approximate the function $u_{0}$ by functions $u_{0h}\in H^{1,\alpha}$, let
$\beta_{h}$ be the same as in the proof of Theorem 3.6 and consider the
corresponding solutions $u_{h}(t,x)$ to problem (1.1), (1.2) constructed in
Theorem 3.5: $u_{h}\in X(\Pi_{T})\cap C_{w}([0,T];H^{1,\alpha})$, $u_{xx}\in
L_{2}(0,T;L_{2}^{\rho_{\alpha-1/2,\epsilon}(x)})$ $\forall T>0$,
$\forall\epsilon>0$. Estimate (3.54) is verified uniformly with respect to
$h$. Fix $t_{0}\in(0,T)$. In the consequent argument temporarily drop the
index $h$.

Let $\rho(x)\equiv\rho_{\alpha-1/2,\epsilon}(x)$, for $r>0$ let
$\psi(x)\equiv\rho_{r,\alpha-1/2,\epsilon}(x)$. Write down corresponding
equality (3.46) and multiply it by $\eta^{2}(2t/t_{0}-1)$. Let $v(t,x)\equiv
u(t,x)\eta(2t/t_{0}-1)$ (for simplicity in the consequent argument
$\varphi\equiv\eta(2t/t_{0}-1)$). Note that

$\varphi^{2}\frac{d}{dt}\int|u_{x}|^{2}\psi\,dx=\frac{d}{dt}\int|v_{x}|^{2}\psi\,dx-2\int|u_{x}|^{2}\psi\varphi^{\prime}\varphi\,dx,$
$\varphi^{2}\frac{d}{dt}\int|u|^{3}\psi\,dx=\frac{d}{dt}\int|uv^{2}|\psi\,dx-2\int|u|^{3}\psi\varphi^{\prime}\varphi\,dx.$

Then

$\frac{d}{dt}\int\left[|v_{x}|^{2}-\frac{2\beta}{3}|uv^{2}|\right]\psi\,dx+3\int|v_{xx}|^{2}\psi^{\prime}\,dx-2a\operatorname{{Im}}\int
v_{xx}\bar{v}_{x}\psi^{\prime}\,dx\\\
-\int|v_{x}|^{2}\psi^{\prime\prime\prime}\,dx-b\int|v_{x}|^{2}\psi^{\prime}\,dx+2\bigl{(}\lambda-a\beta\bigr{)}\operatorname{{Im}}\int
u|v|_{x}\bar{v}_{x}\psi\,dx\\\
-2a\beta\operatorname{{Im}}\int|u|v\bar{v}_{x}\psi^{\prime}\,dx-4\beta\int|u||v_{x}|^{2}\psi^{\prime}\,dx+\beta\int|u|\bigl{(}|v|^{2}\bigr{)}_{xx}\psi^{\prime}\,dx\\\
-\beta\int|v|_{x}|uv|_{x}\psi^{\prime}\,dx+\frac{2b\beta}{3}\int|uv^{2}|\psi^{\prime}\,dx+\beta^{2}\int|u^{2}v^{2}|\psi^{\prime}\,dx\\\
+2\int d(x)|v_{x}|^{2}\psi\,dx+2\operatorname{{Re}}\int
d^{\prime}(x)v\bar{v}_{x}\psi\,dx-2\beta\int d(x)|uv^{2}|\psi\,dx\\\
-2\int|u_{x}|^{2}\psi\varphi^{\prime}\varphi\,dx+\frac{4\beta}{3}\int|u|^{3}\psi\varphi^{\prime}\varphi\,dx=0.$
(5.7)

Integrate this equality with respect to $t$, pass to the limit when
$r\to+\infty$ and again differentiate with respect to $t$, then for a.e.
$t\in(0,T)$ equality (5.7) holds, where $\psi(x)$ is substituted by $\rho(x)$.
Uniformly with respect to $h$ similarly to (3.36)–(3.38)

$\int|uv^{2}|\rho\,dx\leq\sup\limits_{x\in\mathbb{R}}\bigl{(}|v|\sqrt{\rho}\bigr{)}\Bigl{(}\int|u|^{2}\,dx\int|v|^{2}\rho\,dx\Bigr{)}^{1/2}\leq\varepsilon\int|v_{x}|^{2}\rho\,dx+c(\varepsilon),$

where $\varepsilon>0$ can be chosen arbitrarily small. Next, note that
$\rho(x)/\rho^{\prime}(x)\leq c(1+x_{+})^{2}$ (in fact, it is obvious for
$x\leq 0$, while for $x\geq 0$ if $\alpha>0$ then
$\rho(x)/\rho^{\prime}(x)=c(1+x)$ and if $\alpha=0$ then
$\rho(x)/\rho^{\prime}(x)\leq c(\varepsilon)(1+x)^{1+\varepsilon}$ for any
$\varepsilon>0$). Therefore, since $\alpha\geq 1/2$ (here and further
uniformly with respect to $h$)

$\int|uv_{x}^{2}|\rho\,dx\leq\sup\limits_{x\in\mathbb{R}}\bigl{(}|v_{x}|(\rho^{\prime}\rho)^{1/4}\bigr{)}\Bigl{(}\int|v_{x}|^{2}\rho\,dx\int|u|^{2}(\rho/\rho^{\prime})^{1/2}\,dx\Bigr{)}^{1/2}\\\
\leq
c\Bigl{(}\int\bigl{(}|v_{xx}|^{2}\rho^{\prime}+|v_{x}|^{2}\rho\bigr{)}\,dx\Bigr{)}^{1/4}\Bigr{(}\int|v_{x}|^{2}\rho\,dx\Bigr{)}^{1/2}\Bigl{(}\int(1+x_{+})|u|^{2}\,dx\Bigr{)}^{1/2}\\\
\leq\varepsilon\int|v_{xx}|^{2}\rho^{\prime}\,dx+c(\varepsilon)\int|v_{x}|^{2}\rho\,dx,$

where $\varepsilon>0$ can be chosen arbitrarily small. Next,

$\int|uvv_{xx}|\rho^{\prime}\,dx\leq\sup\limits_{x\in\mathbb{R}}\bigl{(}|v|\sqrt{\rho^{\prime}}\bigr{)}\bigl{(}\int|u|^{2}\,dx\int|v_{xx}|^{2}\rho^{\prime}\,dx\Bigr{)}^{1/2}\\\
\leq\varepsilon\int|v_{xx}|^{2}\rho^{\prime}\,dx+c(\varepsilon)\int|v_{x}|^{2}\rho\,dx+c(\varepsilon),$

where $\varepsilon>0$ also can be chosen arbitrarily small. Finally,

$\iint_{\Pi_{T}}|u_{x}|^{2}\rho\,dxdt\leq
c\iint_{\Pi_{T}}|u_{x}|^{2}\rho^{\prime}_{\alpha,\epsilon}\,dxdt\leq c_{1},$
$\iint_{\Pi_{T}}|u|^{3}\rho\,dxdt\leq\int_{0}^{T}\sup\limits_{x\in\mathbb{R}}\bigl{(}|u|\sqrt{\rho}\bigr{)}\Bigl{(}\int|u|^{2}\,dx\int|u|^{2}\rho\,dx\Bigr{)}^{1/2}dt\\\
\leq c\iint_{\Pi_{T}}|u_{x}|^{2}\rho\,dxdt+c\leq c_{1}.$

As a result, since $\eta(2t/t_{0}-1)=1$ for $t\geq t_{0}$ we find that
uniformly with respect to $h$

$\|u_{hx}\|_{L_{\infty}(t_{0},T;L_{2}^{\rho_{\alpha-1/2,\epsilon}(x)})}+\|u_{hxx}\|_{L_{2}(t_{0},T;L_{2}^{\rho^{\prime}_{\alpha-1/2,\epsilon}(x)})}\leq
c(T,t_{0}).$ (5.8)

Application of the closure finishes the proof. ∎

######  Theorem 5.3.

Let $u_{0}\in L_{2}^{\alpha}$ for certain $\alpha>3/4$, $d\in H^{1}$, $u\in
L_{\infty}(0,T;L_{2}^{\alpha})$ be a solution to problem (1.1), (1.2) for
certain $T>0$. Then (after possible correction on a zero measure set)
$u,u_{x}\in C_{b}(\overline{\Pi}_{T}^{t_{0},x_{0}})$ for any $t_{0}\in(0,T)$,
$x_{0}\in\mathbb{R}$.

###### Proof.

The considered solution lies in the class of uniqueness, therefore, according
to Theorem 5.2 $u\in C_{w}([0,T];L_{2}^{\alpha})$, $u_{x}\in
C_{w}([t_{0},T];L_{2}^{\rho_{\alpha-1/2,\epsilon}(x)})$, $u_{xx}\in
L_{2}(t_{0},T;L_{2}^{\rho^{\prime}_{\alpha-1/2,\epsilon}(x)})$ for any
$t_{0}\in(0,T)$, $\epsilon>0$.

Fix $t_{0}\in(0,T)$, $x_{0}\in\mathbb{R}$. Let $v(t,x)\equiv
u_{x}(t,x)\eta(x-x_{0}+2+b_{+}T)\eta(2t/t_{0}-1)$ (in the consequent argument
for simplicity
$\eta\equiv\eta(x-x_{0}+2+b_{+}T),\varphi\equiv\eta(2t/t_{0}-1)$), then the
function $v\in C_{w}([0,T];L_{2}^{\alpha-1/2})$ is a weak solution to problem
(2.1), (2.2) for $u_{0}\equiv 0$ and $f\equiv f_{1}+f_{2x}$, where

$f_{1}=f_{11}\equiv
iu_{x}\eta\varphi^{\prime}+a(2u_{xx}\eta^{\prime}+u_{x}\eta^{\prime\prime})\varphi+ibu_{x}\eta^{\prime}\varphi+iu_{x}\eta^{\prime\prime\prime}\varphi\\\
+\lambda|u|u\eta^{\prime}\varphi+i\beta\bigl{(}|u|u\bigr{)}_{x}\eta^{\prime}\varphi-
id(x)u\eta^{\prime}\varphi,$ (5.9)
$f_{2}\equiv-\bigl{[}\lambda|u|u+i\beta\bigl{(}|u|u\bigr{)}_{x}+id(x)u\bigr{]}\eta\varphi+3iu_{xx}\eta^{\prime}\varphi.$
(5.10)

Note that $\operatorname*{supp}\eta^{\prime}=[x_{0}-2-b_{+}T,x_{0}-1-b_{+}T]$,
$\operatorname*{supp}\varphi=[t_{0}/2,+\infty)$. Then $f_{1}\in
L_{2}(0,T;L_{2}^{\alpha-1/2})$. Moreover, $f_{2}\in
L_{2}(0,T;L_{1}^{\alpha})$, since for $t\geq t_{0}$

$\int|u_{x}u|(1+x_{+})^{\alpha}\eta\,dx\leq\int|u_{x}u|(1+x_{+})^{2\alpha-1/2}\eta\,dx\\\
\leq\Bigl{(}\int_{x_{0}-2-b_{+}T}^{+\infty}|u_{x}|^{2}(1+x_{+})^{2\alpha-1}\,dx\int|u|^{2}(1+x_{+})^{2\alpha}\,dx\Bigr{)}^{1/2}\
\leq c(t_{0},x_{0}),$
$\int|d(x)u|(1+x_{+})^{\alpha}\eta\,dx\leq\Bigl{(}\int|d(x)|^{2}\,dx\int|u|^{2}(1+x_{+})^{2\alpha}\,dx\Bigr{)}^{1/2}\leq
c(T).$

Therefore, since $\alpha-1/2>1/4$, $\alpha\geq 1/4$, Theorem 2.5 can be
applied. In particular, for $t\in[t_{0},T]$, $x\geq x_{0}$

$u_{x}(t,x)=-i\int_{0}^{t}\\!\\!\int G(t-\tau,x-y)f_{1}(\tau,y)\,dyd\tau\\\
-i\int_{0}^{t}\\!\\!\int G_{x}(t-\tau,x-y)f_{2}(\tau,y)\,dyd\tau.$ (5.11)

Note that if $y\in\operatorname*{supp}\eta^{\prime}$ and $x\geq x_{0}$ then
$x-y\geq 1+b_{+}T$. Then according to (A.27)

$|\partial_{x}^{n}G(t-\tau,x-y|)\leq\frac{c(T,n)}{t^{(n+1)/3}}e^{-c_{0}t^{-1/2}}\leq
c_{1}(T,n).$

Therefore,

$\Bigl{|}\int_{0}^{t}\\!\\!\int\partial^{n}_{x}G(t-\tau,x-y)u_{yy}\eta^{\prime}\varphi\,dyd\tau\Bigr{|}\leq
c(T,n)\int_{t_{0}/2}^{T}\int_{x_{0}-2-b_{+}T}^{x_{0}}|u_{yy}|\,dyd\tau\\\ \leq
c_{1}(T,n,t_{0},x_{0}).$

It is easy to see that other terms in the right-hand side of (5.9) belong to
$L_{\infty}(0,T;L_{2}^{\alpha-1/2})$ and can be estimated similarly to (5.4),
other terms in the right-hand side of (5.10) belong to
$L_{\infty}(0,T;L_{1}^{\alpha})$ and can be estimated similarly to (5.6). As a
result,

$\sup\limits_{(t,x)\in[t_{0},T]\times[x_{0},+\infty)}|u_{x}(t,x)|<\infty.$

Moreover the integrals in the right-hand side of (5.11) are continuous with
respect to $(t,x)$. Taking into account Theorem 5.1, we finish the proof. ∎

##  6\. Large-time decay

The prehistory of the methods used in this section see, for example, in [10,
11].

######  Lemma 6.1.

Let the function $d(x)\in H^{1}$ satisfies Condition A for given $\alpha_{0}$
and $R_{0}$. Then for any $T>0$ and $M>0$ there exists a positive constant
$c_{0}=c_{0}(T,M,\alpha_{0},R_{0},\|d\|_{L_{\infty}},\beta)$ such that if
$u_{0}\in H^{1}$ and $\|u_{0}\|_{L_{2}}\leq M$ then for a solution $u(t,x)$ to
problem (1.1), (1.2) from the class $X(\Pi_{T})$ the following inequality
holds:

$\int_{0}^{T}\\!\\!\int_{-R_{0}}^{R_{0}}u^{2}\,dxdt\leq
c_{0}\iint_{\Pi_{T}}d(x)u^{2}\,dxdt.$ (6.1)

###### Proof.

We argue by contradiction. Let us suppose that (6.1) is not verified and let
$\\{u_{0k}(x)\\}_{k\in\mathbb{N}}$, $\\{d_{k}(x)\\}_{k\in\mathbb{N}}$,
$\\{\beta_{k}\\}_{k\in\mathbb{N}}$ be sequences of initial data from $H^{1}$
bounded in $L_{2}$, damping functions from $H^{1}$ verifying Condition A
uniformly with respect to $k$ bounded in $L_{\infty}$ and uniformly bounded
coefficients, where the corresponding solutions
$\\{u_{k}(t,x)\\}_{k\in\mathbb{N}}$ verify

$\displaystyle\lim_{k\to+\infty}\frac{\displaystyle\iint_{\Pi_{T}}d_{k}(x)u_{k}^{2}\,dxdt}{\displaystyle\int_{0}^{T}\\!\\!\int_{-R_{0}}^{R_{0}}u_{k}^{2}\,dxdt}=0.$
(6.2)

Define

$\nu_{k}=\|u_{k}\|_{L_{2}((0,T)\times(-R_{0},R_{0}))},\qquad
v_{k}(t,x)\equiv\frac{u_{k}(t,x)}{\nu_{k}}.$

Then

$\|v_{k}\|_{L_{2}((0,T)\times(-R_{0},R_{0}))}=1\qquad\forall k\in\mathbb{N}$
(6.3)

and $v_{k}$ is a solution from $X(\Pi_{T})$ to a problem

$iv_{kt}+av_{kxx}+ibv_{kx}+iv_{kxxx}+\lambda\nu_{k}|v_{k}|v_{k}+i\beta_{k}\nu_{k}\bigl{(}|v_{k}|v_{k}\bigr{)}_{x}+id_{k}(x)v_{k}=0,$
(6.4) $v_{k}(0,x)=v_{0k}(x)\equiv\frac{u_{0k}(x)}{\nu_{k}}.$ (6.5)

Note that by virtue of (3.22)

$\nu_{k}\leq T^{1/2}M$ (6.6)

and it follows from (6.2) that

$\lim_{k\to+\infty}\iint_{\Pi_{T}}d_{k}(x)v_{k}^{2}\,dxdt=0.$ (6.7)

Next, we show that $\\{v_{0k}(x)\\}_{k\in\mathbb{N}}$ is bounded in $L_{2}$.
Indeed, properties of the functions $d_{k}(x)$ yield that

$\iint_{\Pi_{T}}u_{k}^{2}\,dxdt\leq\frac{1}{\alpha_{0}}\iint_{\Pi_{T}}d_{k}(x)u_{k}^{2}\,dxdt+\int_{0}^{T}\\!\\!\int_{-R_{0}}^{R_{0}}u_{k}^{2}\,dxdt,$

so it succeeds from (3.22) that

$\int
u_{0k}^{2}\,dx\leq\frac{1}{T}\iint_{\Pi_{T}}u_{k}^{2}\,dxdt+2\iint_{\Pi_{T}}d_{k}(x)u_{k}^{2}\,dxdt\\\
\leq\left(2+\frac{1}{\alpha_{0}T}\right)\iint_{\Pi_{T}}d_{k}(x)u_{k}^{2}\,dxdt+\frac{1}{T}\int_{0}^{T}\\!\\!\int_{-R_{0}}^{R_{0}}u_{k}^{2}\,dxdt$

and, therefore,

$\int
v_{0k}^{2}\,dxdt\leq\left(2+\frac{1}{\alpha_{0}T}\right)\iint_{\Pi_{T}}d_{k}(x)v_{k}^{2}\,dxdt+\frac{1}{T},$

whence with the use of (6.7) the desired result on boundedness of
$\\{v_{0k}\\}_{k\in\mathbb{N}}$ follows.

Then according to (3.57) and (6.6) we deduce that

$\\{v_{k}\\}_{k\in\mathbb{N}}\quad\text{is bounded in}\quad
L_{\infty}(0,T;L_{2})\cap L_{2}(0,T;H^{1}(I_{n}))\quad\forall n\in\mathbb{N},\
I_{n}=(-n,n).$ (6.8)

Since $L_{1}(I_{n})\hookrightarrow H^{-1}(I_{n})$ it follows from (6.8) that

$\\{|v_{k}|v_{k}\\}_{k\in\mathbb{N}}\quad\text{is bounded in}\quad
L_{2}(0,T;H^{-1}(I_{n}))\quad\forall n\in\mathbb{N}$

and then with the use of equation (6.4) itself, we obtain that

$\\{v_{kt}\\}_{k\in\mathbb{N}}\quad\text{is bounded in}\quad
L_{2}(0,T;H^{-2}(I_{n}))\quad\forall n\in\mathbb{N}.$

Passing to subsequences $\nu_{k}$, $\beta_{k}$, $\\{v_{0k}\\}$ and
$\\{v_{k}\\}$, still denoted by $\nu_{k}$, $\beta_{k}$, $\\{v_{0k}\\}$,
$\\{v_{k}\\}$ from now on, we get

$\displaystyle\nu_{k}\rightarrow\nu,\quad\beta_{k}\rightarrow\beta,$
$\displaystyle v_{0k}\rightharpoonup v_{0}\quad\text{weakly in}\quad L_{2},$
$\displaystyle v_{k}\rightharpoonup v\quad*-\text{weakly in}\quad
L_{\infty}(0,T;L_{2}),$ (6.9) $\displaystyle v_{k}\rightharpoonup
v\quad\text{weakly in}\quad L_{2}(0,T;H^{1}(I_{n}))\quad\forall
n\in\mathbb{N}$

and since $H^{1}(I_{n})\overset{c}{\hookrightarrow}L_{2}(I_{n})\hookrightarrow
H^{-2}(I_{n})$ by the standard argument that

$v_{k}\rightarrow v\quad\mbox{strongly in}\ \ L_{2}\bigl{(}(0,T)\times
I_{n}\bigr{)}\quad\forall n\in\mathbb{N}.$

Note that it follows from (6.7) and Condition A that $v_{k}\to 0$ in
$L_{2}\bigl{(}(0,T)\times(\mathbb{R}\setminus(-R_{0},R_{0}))\bigr{)}$,
therefore,

$v_{k}\rightarrow v\quad\text{strongly in}\quad L_{2}(\Pi_{T}),$ (6.10)

where

$v(t,x)=0\quad\text{for}\ |x|>R_{0}.$ (6.11)

Let $\phi(t,x)$ be an arbitrary function such that $\phi\in
C^{1}([0,T];H^{1})\cap C([0,T];H^{3})$ and $\phi\bigr{|}_{t=T}=0$. For any $k$

$\iint_{\Pi_{T}}\bigl{[}iv_{k}\phi_{t}-av_{k}\phi_{xx}+ibv_{k}\varphi_{x}+iv_{k}\phi_{xxx}-\lambda\nu_{k}|v_{k}|v_{k}\phi+i\beta_{k}\nu_{k}|v_{k}|v_{k}\phi_{x}\\\
-id_{k}(x)v_{k}\phi\bigr{]}\,dxdt+\int v_{0k}\phi\bigr{|}_{t=0}\,dx=0.$

Passing to the limit as $k\to+\infty$ and taking into account (6.7) and
boundedness of $d_{k}$ in $L_{\infty}$ we obtain an equality

$\iint_{\Pi_{T}}\bigl{[}iv\phi_{t}-av\phi_{xx}+ibv\varphi_{x}+iv\phi_{xxx}-\lambda\nu|v|v\phi+i\beta\nu|v|v\phi_{x}\bigr{]}\,dxdt\\\
+\int v_{0}\phi\bigr{|}_{t=0}\,dx=0.$

that is, $v(t,x)$ is a weak solution to a problem

$iv_{t}+a_{1}v_{xx}+ibv_{x}+iv_{xxx}+\lambda\nu|v|v+i\beta\nu\bigl{(}|v|v\bigr{)}_{x}=0,\quad
v(0,x)=v_{0}(x).$ (6.12)

It follows from (6.9) and (6.11) that $v\in L_{\infty}(0,T;L_{2}^{3/4})$, that
is, it lies in the class of uniqueness. Then it follows from Theorem 5.2 that
$u\in L_{2}(t_{0},T;H^{2}(I))$ and from Theorem 5.3 that $u,u_{x}\in
L_{\infty}((t_{0},T)\times I)$ for any $t_{0}\in(0,T)$ and bounded interval
$I\subset\mathbb{R}$. Note that according to (B.1)

$\bigl{(}|u|u\bigr{)}_{x}=\frac{3}{2}|u|u_{x}+\frac{u^{2}\bar{u}_{x}}{2|u|}=\frac{1}{2}P(u,u_{x}),$

where the function $P$ is defined by formula (B.8). Then according to (B.9)
equation (6.12) satisfies the hypothesis of Theorem 6 from [23] and the unique
continuation property from [23] provides that $v(t,x)\equiv 0$ in $\Pi_{T}$.
Thus, (6.10) yields that $v_{k}\to 0$ in $L_{2}(\Pi_{T})$, which contradicts
(6.3). ∎

######  Theorem 6.2.

Let the function $d(x)\in H^{1}$ satisfies Condition A, $\beta\neq 0$. Then
there exists a constant $\gamma>0$ and for any $M>0$ there exists a constant
$c(M)>0$ such that if $u_{0}\in H^{1}$ and $\|u_{0}\|_{L_{2}}\leq M$ the
corresponding function $u\in X(\Pi_{T})$ $\forall T>0$ which is the solution
to problem (1.1), (1.2) in $\Pi_{T}$, constructed in Theorem 3.5 for
$\alpha=0$, satisfies inequality (1.5).

###### Proof.

Let $T>0$ be fixed. According to (3.22) $\|u(t,\cdot)\|_{L_{2}}$ does not
increase with respect to $t$ and

$\int|u(T,x)|^{2}\,dx+2\iint_{\Pi_{T}}d(x)|u(t,x)|^{2}\,dxdt=\int|u_{0}|^{2}\,dx.$
(6.13)

The properties of the function $d(x)$ provide that

$\int|u(T,x)|^{2}\,dx+2\alpha_{0}\int_{0}^{T}\\!\\!\int_{\mathbb{R}\setminus(-R_{0},R_{0})}|u|^{2}\,dxdt\leq\int|u_{0}|^{2}\,dx$

and, consequently,

$\iint_{\Pi_{T}}|u|^{2}\,dxdt\leq\frac{1}{2\alpha_{0}}\int|u_{0}|^{2}\,dx-\frac{1}{2\alpha_{0}}\int|u(T,x)|^{2}\,dx+\int_{0}^{T}\\!\\!\int_{-R_{0}}^{R_{0}}|u|^{2}\,dxdt.$

By virtue of (6.1) if $\|u_{0}\|_{L_{2}}\leq M$, for $c_{0}=c_{0}(M)$ the last
inequality can be transformed to the following one:

$\iint_{\Pi_{T}}|u|^{2}\,dxdt\leq\frac{1}{2\alpha_{0}}\int|u_{0}|^{2}\,dx-\frac{1}{2\alpha_{0}}\int|u(T,x)|^{2}\,dx+c_{0}\iint_{\Pi_{T}}d(x)|u|^{2}\,dxdt,$

which with the use of (6.13) and monotonicity of $\|u(t,\cdot)\|_{L_{2}}$
yields that

$T\int|u^{2}(T,x)|^{2}\,dx\leq\frac{1}{2\alpha_{0}}\int|u_{0}|^{2}\,dx-\frac{1}{2\alpha_{0}}\int|u(T,x)|^{2}\,dx\\\
+\frac{c_{0}}{2}\Bigl{(}\int|u_{0}|^{2}\,dx-\int|u(T,x)|^{2}\,dx\Bigr{)},$

whence it follows that

$\left(T+\frac{1}{2\alpha_{0}}+\frac{c_{0}}{2}\right)\int|u(T,x)|^{2}\,dx\leq\left(\frac{1}{2\alpha_{0}}+\frac{c_{0}}{2}\right)\int|u_{0}|^{2}\,dx,$

which is equivalent to an inequality

$\int|u(T,x)|^{2}\,dx\leq\frac{1+c_{0}\alpha_{0}}{1+c_{0}\alpha_{0}+2T\alpha_{0}}\int|u_{0}|^{2}\,dx,$
(6.14)

whence with the use of the semigroup property follows that

$\|u(t,\cdot)\|_{L_{2}}^{2}\leq c(M)e^{-\gamma(M)t}\quad\forall\ t\geq 0,$
(6.15)

where

$\gamma(M)=-\frac{1}{T}\ln\frac{1+c_{0}\alpha_{0}}{1+c_{0}\alpha_{0}+2T\alpha_{0}}>0,\quad
c(M)=\left(1+\frac{2T\alpha_{0}}{1+c_{0}\alpha_{0}}\right)M^{2}.$ (6.16)

Define $\gamma=\gamma(1)$, that is the constant from (6.16) corresponding to
$M=1$.

For any $M>0$ according to (6.15) there exists $T_{0}=T_{0}(M)>0$ such that
$\|u(T_{0},\cdot)\|_{L_{2}}\leq 1$ if $\|u_{0}\|_{L_{2}}\leq M$. Then for
$t\geq T_{0}$

$\|u(t,\cdot)\|^{2}_{L_{2}}\leq c(1)e^{\gamma T_{0}}e^{-\gamma t},$

while for $t\in[0,T_{0})$

$\|u(t,\cdot)\|^{2}_{L_{2}}\leq M^{2}e^{\gamma T_{0}}e^{-\gamma t},$

whence (1.5) evidently follows. ∎

######  Theorem 6.3.

Let the function $d(x)\in L_{\infty}$ satisfies Condition A. Then there exists
a constant $\gamma>0$ and for any $M>0$ there exists a constant $c(M)>0$ such
that if $u_{0}\in L_{2}$ and $\|u_{0}\|_{L_{2}}\leq M$ the corresponding
function $u\in C_{w}([0,T];L_{2})$ $\forall T>0$ which is the solution to
problem (1.1), (1.2) in $\Pi_{T}$, constructed in Theorem 3.6 for $\alpha=0$,
satisfies inequality (1.5).

###### Proof.

Approximate the functions $d$ and $u_{0}$ by smooth functions $d_{h}$ and
$u_{0h}$ and introduce the functions $\beta_{h}$ as it is done in the proof of
Theorem 3.6 and consider the corresponding solutions $u_{h}(t,x)\in
X(\Pi_{T})$ $\forall T>0$. Note that if $h\in(0,1]$ Condition A is verified
for functions $d_{h}$ uniformly with respect to $h$. Moreover, the functions
$u_{0h}$ are bounded in $L_{2}$, the functions $d_{h}$ are bounded in
$L_{\infty}$ and the values $\beta_{h}$ are bounded also uniformly with
respect to $h$. Then inequality (6.1) is also uniform and the simple revision
of the proof of Theorem 6.2 shows that functions $u_{h}$ verify estimate (1.5)
uniformly with respect to $h$.

In the proof of Theorem 3.3 the solution $u$ is constructed, in particular, as
the $*$-weak limits of functions $u_{h}$ in $L_{\infty}(0,T;L_{2})$ for any
$T>0$. Then for any $t\in[T,T+1]$

$\|u(t,\cdot)\|^{2}_{L_{2}}\leq c(M)e^{-\gamma T}\leq c(M)e^{\gamma}e^{-\gamma
t},$

which finishes the proof. ∎

Finally, we can present the proof of the main theorem.

###### Proof of Theorem 1.3.

Existence and uniqueness of the weak solution $u\in C_{w}([0,T];L_{2}^{3/4})$
succeeds from Theorems 3.6 and 4.1, while estimate (1.5) under Condition A —
from Theorem refT6.2. ∎

## Appendix A

Let $A>0$ and $a\in\mathbb{R}$, $|a|\leq A$. Introduce a function

$\Phi_{a}(x)\equiv\frac{1}{2\pi}\int e^{i(\xi^{3}-a\xi^{2}+x\xi)}\,d\xi.$
(A.1)

For this function we establish properties similar to the well-known ones for
the classical Airy function.

######  Lemma A.1.

The function $\Phi_{a}$ for any $x\in\mathbb{R}$ exists, is infinitely smooth
and satisfies an equation

$3y^{\prime\prime}-2aiy^{\prime}-xy=0.$ (A.2)

Moreover, for any integer non-negative $n$

$\bigl{|}\Phi_{a}^{(n)}(x)\bigr{|}\leq\begin{cases}\displaystyle
c(A,n)\bigl{(}1+|x|\bigr{)}^{(2n-1)/4},\quad&x\leq 0,\\\
c(A,n)e^{-c_{0}x^{3/2}},\quad&x\geq 0,\end{cases}$ (A.3)

for certain positive constants $c_{0}$, $c(A,n)$. Finally,
$\Phi_{a}=\mathcal{F}^{-1}\left[e^{i(\xi^{3}-a\xi^{2})}\right]$, where the
inverse Fourier transform is understood as an operation in
$\mathcal{S}^{\prime}$.

###### Proof.

Let $n=0$ or $n=1$. Consider functions of the complex variable $z$

$\varphi(z)\equiv z^{3}-az^{2}+xz,\quad g_{n}(z)\equiv z^{n}e^{i\varphi(z)}.$
(A.4)

For $R>0$ define a contour $\gamma$ on $\mathbb{C}$, consisting of the
segments $[0,R]$, $[Re^{i\pi/6},0]$ and the connecting them arc of a
circumference $\gamma_{R}(0,\pi/6)$ of the radius $R$. Then
$\displaystyle\int_{\gamma}g_{n}(z)\,dz=0$. After the change of variables
$z=\zeta^{1/3}$ (the choice of the branch of the root is evident) one obtains
that

$\int_{\gamma_{R}(0,\pi/6)}g_{n}(z)\,dz=\frac{1}{3}\int_{\gamma_{R^{3}}(0,\pi/2)}\zeta^{(n-2)/3}e^{i(\zeta-a\zeta^{2/3}+x\zeta^{1/3})}\,d\zeta.$
(A.5)

Define on the arc $\gamma_{R^{3}}(0,\pi/2)$ the following parameter: let
$\zeta=R^{3}e^{it}$, $t\in[0,\pi/2]$. Then
$\displaystyle\frac{2}{\pi}t\leq\sin t\leq t$ and, therefore,

$\left|e^{i(\zeta-a\zeta^{2/3}+x\zeta^{1/3})}\right|=e^{-R^{3}\sin
t+aR^{2}\sin(2t/3)-xR\sin(t/3)}\leq e^{-R^{3}t/\pi}$

for considerably large $R$ uniformly on any compact set with respect to $x$.
Since $n<2$ it follows from (A.5) that

$\int_{\gamma_{R}(0,\pi/6)}g_{n}(z)\,dz\to 0$ (A.6)

when $R\to+\infty$ uniformly on any compact set with respect to $x$. Next,
after the change of variables $z=re^{i\pi/6}$ we find that

$\int_{0}^{Re^{i\pi/6}}g_{n}(z)\,dz=e^{i\pi(n+1)/6}\int_{0}^{R}r^{n}e^{-r^{3}+ar^{2}\sqrt{3}/2-xr/2}e^{i(-ar^{2}/2+xr\sqrt{3}/2)}\,dr.$
(A.7)

Therefore, when $R\to+\infty$ uniformly on any compact set with respect to $x$

$\int_{0}^{Re^{i\pi/6}}g_{n}(z)\,dz\to
e^{i\pi(n+1)/6}\int_{0}^{+\infty}r^{n}e^{-r^{3}+ar^{2}\sqrt{3}/2-xr/2}e^{i(-ar^{2}/2+xr\sqrt{3}/2)}\,dr.$
(A.8)

As a result, it follows from (A.6) and (A.7) that when $R\to+\infty$ uniformly
on any compact set with respect to $x$

$\int_{0}^{R}g_{n}(\xi)\,d\xi\to\int_{0}^{+\infty}g_{n}(\xi)\,d\xi=\int_{0}^{+\infty\cdot
e^{i\pi/6}}g_{n}(z)\,dz\\\
=e^{i\pi(n+1)/6}\int_{0}^{+\infty}r^{n}e^{-r^{3}+ar^{2}\sqrt{3}/2-xr/2}e^{i(-ar^{2}/2+xr\sqrt{3}/2)}\,dr.$
(A.9)

Similarly, choosing a contour consisting of the segments $[-R,0]$,
$[0,Re^{5i\pi/6}]$ and the arc $\gamma_{R}(5\pi/6,\pi)$ we find that when
$R\to+\infty$ uniformly on any compact set with respect to $x$

$\int_{-R}^{0}g_{n}(\xi)\,d\xi\to\int_{-\infty}^{0}g_{n}(\xi)\,d\xi=\int_{+\infty\cdot
e^{5i\pi/6}}^{0}g_{n}(z)\,dz\\\
=-e^{5i\pi(n+1)/6}\int_{0}^{+\infty}r^{n}e^{-r^{3}-ar^{2}\sqrt{3}/2-xr/2}e^{i(-ar^{2}/2-xr\sqrt{3}/2)}\,dr.$
(A.10)

As a result, it follows from (A.9) and (A.10) that the function
$\displaystyle\Phi_{a}(x)=\frac{1}{2\pi}\int_{\mathbb{R}}g_{0}(\xi)\,d\xi$
exists and

$\Phi_{a}(x)=\frac{1}{2\pi}\Bigl{[}\int_{0}^{+\infty}e^{-r^{3}+ar^{2}\sqrt{3}/2-xr/2}e^{i(-ar^{2}/2+xr\sqrt{3}/2+\pi/6)}\,dr\\\
+\int_{0}^{+\infty}e^{-r^{3}-ar^{2}\sqrt{3}/2-xr/2}e^{i(-ar^{2}/2-xr\sqrt{3}/2-\pi/6)}\,dr\Bigr{]}.$
(A.11)

In particular, this function is infinitely smooth. Differentiating equalities
(A.9) and (A.10) for $n=0$ with respect to $x$ we derive that

$\Phi_{a}^{\prime}(x)=\frac{i}{2\pi}\Bigl{[}\int_{+\infty\cdot
e^{5i\pi/6}}^{0}g_{1}(z)\,dz+\int_{0}^{+\infty\cdot
e^{i\pi/6}}g_{1}(z)\,dz\Bigr{]}\\\ =\frac{i}{2\pi}\int
g_{1}(z)\,dz=\frac{i}{2\pi}\int\xi e^{i(\xi^{3}-a\xi^{2}+x\xi)}\,d\xi.$ (A.12)

Moreover, since $\varphi^{\prime}(z)=3z^{2}-2az+x$

$3\Phi_{a}^{\prime\prime}(x)-2ai\Phi_{a}^{\prime}(x)-x\Phi_{a}(x)\\\
=\frac{1}{2\pi}\Bigl{(}\int_{+\infty\cdot
e^{5i\pi/6}}^{0}+\int_{0}^{+\infty\cdot
e^{i\pi/6}}\Bigr{)}(-3z^{2}+2az-x)e^{i\varphi(z)}\,dz\\\
=\frac{i}{2\pi}e^{i\varphi(z)}\Big{|}_{+\infty\cdot e^{5i\pi/6}}^{+\infty\cdot
e^{i\pi/6}}=0.$

Note that
$\displaystyle\frac{1}{2\pi}\int_{-R}^{R}e^{i(\xi^{3}-a\xi^{2}+x\xi)}\,d\xi\to\Phi_{a}(x)$
when $x\to+\infty$ uniformly on any compact set with respect to $x$.
Therefore, this convergence is valid in the space
$\mathcal{D}^{\prime}(\mathbb{R})$. On the other hand,
$e^{i(\xi^{3}-a\xi^{2})}\chi_{(-R,R)}(\xi)\to e^{i(\xi^{3}-a\xi^{2})}$ in the
space $\mathcal{S}^{\prime}(\mathbb{R})$ when $R\to+\infty$. Since the inverse
Fourier transform is continuous in this space, it follows that
$\displaystyle\frac{1}{2\pi}\int_{-R}^{R}e^{i(\xi^{3}-a\xi^{2}+x\xi)}\,d\xi\to\mathcal{F}^{-1}\bigl{[}e^{i(\xi^{3}-a\xi^{2})}\bigr{]}(x)$
in $\mathcal{S}^{\prime}$. As a result,
$\Phi_{a}=\mathcal{F}^{-1}\bigl{[}e^{i(\xi^{3}-a\xi^{2})}\bigr{]}$ in
$\mathcal{S}^{\prime}$.

It follows from (A.11) that for any integer non-negative $n$

$\bigl{|}\Phi_{a}^{(n)}(x)\Bigr{|}\leq c(A,n),\quad|x|\leq 1.$ (A.13)

In order to obtain estimates (1.3) for $|x|>1$ first consider the case $n\leq
1$. Then according to (A.1) and (A.12) we have to estimate
$\displaystyle\int_{\mathbb{R}}g_{n}(\xi)\,d\xi$.

Let $x>1$. For $R>0$ construct on a complex plane a rectangle
$\widetilde{\gamma}$ with the base $[-R,R]$ and the height $h=(x/3)^{1/2}$.
Then $\displaystyle\int_{\widetilde{\gamma}}g_{n}(z)\,dz=0$. Consider first
the right side of the rectangle. Introduce a parameter $z=R+it$,
$t\in[0,\sqrt{x/3}]$. Then

$\int_{R}^{R+i\sqrt{x/3}}g_{n}(z)\,dz\\\
=i\int_{0}^{\sqrt{x/3}}(R+it)^{n}e^{-3R^{2}t+t^{3}+2aRt-
xt}e^{i(R^{3}-3Rt^{2}-aR^{2}+at^{2}+xR)}\,dt.$

For the fixed $x$ and considerably large $R$

$\Bigl{|}\int_{R}^{R+i\sqrt{x/3}}g_{n}(z)\,dz\Bigr{|}\leq\left(R^{2}+\frac{x}{3}\right)^{n/2}\int_{0}^{\sqrt{x/3}}e^{-R^{2}t}\,dt\\\
\leq\frac{1}{R^{2}}\left(R^{2}+\frac{x}{3}\right)^{n/2}\int_{0}^{+\infty}e^{-\theta}\,d\theta\to
0,\quad R\to+\infty.$

The left side of the rectangle is considered in a similar way. Then passing to
the limit when $R\to+\infty$ we obtain that

$\int
g_{n}(\xi)\,d\xi=\int_{-\infty+i\sqrt{x/3}}^{+\infty+i\sqrt{x/3}}g_{n}(z)\,dz\\\
=\int\bigl{(}\xi+i\sqrt{x/3}\bigr{)}^{n}e^{i[(\xi+i\sqrt{x/3})^{3}-a(\xi+i\sqrt{x/3})^{2}+x(\xi+i\sqrt{x/3})]}\,d\xi\\\
=e^{-2(x/3)^{3/2}}\int\bigl{(}\xi+i\sqrt{x/3}\bigr{)}^{n}e^{-\sqrt{3x}\xi^{2}+2a\sqrt{x/3}\xi}e^{i\operatorname{{Re}}\varphi(\xi+i\sqrt{x/3})}\,d\xi,$

whence it follows that

$\Bigl{|}\int g_{n}(\xi)\,d\xi\Bigr{|}\leq
e^{-2(x/3)^{3/2}}\int\bigl{(}\xi^{2}+\frac{x}{3}\bigr{)}^{n/2}e^{-\sqrt{3x}\xi^{2}+2a\sqrt{x/3}\xi}\,d\xi\\\
=e^{-2(x/3)^{3/2}+a^{2}x^{1/2}3^{-3/2}}\int\bigl{(}\xi^{2}+\frac{x}{3}\bigr{)}^{n/2}e^{-\sqrt{3x}(\xi-a/3)^{2}}\,d\xi\\\
\leq c(A,n)e^{-(x/3)^{3/2}}$

and estimate (A.3) in the case $x>1$, $n\leq 1$ is established.

Now let $x<-1$, $n\leq 1$. Divide the real line into three parts:

$\displaystyle\Omega_{1}=\\{\xi:|x|/2\leq 3\xi^{2}-2a\xi\leq 3|x|/2\\},$
$\displaystyle\Omega_{2}=\\{\xi:3\xi^{2}-2a\xi<|x|/2\\},$
$\displaystyle\Omega_{3}=\\{\xi:3\xi^{2}-2a\xi>3|x|/2\\}.$

Let $\xi\in\Omega_{1}$. Consider first the inequality
$3\xi^{2}-2a\xi\geq|x|/2$. Let

$\xi_{1}=\frac{1}{3}\bigl{(}a-\sqrt{a^{2}+3|x|/2}\bigr{)},\quad\xi_{2}=\frac{1}{3}\bigl{(}a+\sqrt{a^{2}+3|x|/2}\bigr{)}.$
(A.14)

It means that either $\xi\leq\xi_{1}$ or $\xi\geq\xi_{2}$. If $\xi\geq\xi_{2}$
then $\xi\geq|x/6|^{1/2}$ in the case $a\geq 0$, while in the case $a<0$

$\xi\geq\frac{1}{3}\bigl{(}\sqrt{A^{2}+3|x|/2}-A\bigr{)}\geq\frac{1}{3}\bigl{(}\sqrt{A^{2}+3/2}-A\bigr{)}|x|^{1/2}.$

If $\xi\leq\xi_{1}$ then $|\xi|\geq|x/6|^{1/2}$ in the case $a\leq 0$, while
in the case $a>0$

$|\xi|\geq\frac{1}{3}\bigl{(}\sqrt{A^{2}+3|x|/2}-A\bigr{)}\geq\frac{1}{3}\bigl{(}\sqrt{A^{2}+3/2}-A\bigr{)}|x|^{1/2}.$

As a result, if $\xi\in\Omega_{1}$ then $|\xi|\geq c^{-1}(A)|x|^{1/2}$. Next,
consider the inequality $3\xi^{2}-2a\xi\leq 3|x|/2$. Let

$\xi_{3}=\frac{1}{3}\bigl{(}a-\sqrt{a^{2}+9|x|/2}\bigr{)},\quad\xi_{4}=\frac{1}{3}\bigl{(}a+\sqrt{a^{2}+9|x|/2}\bigr{)}.$
(A.15)

It means that $\xi_{3}\leq\xi\leq\xi_{4}$. Then if $a\geq 0$

$|\xi|\leq\xi_{4}\leq\frac{1}{3}\bigl{(}\sqrt{A^{2}+9/2}+A\bigr{)}|x|^{1/2},$

while if $a<0$

$|\xi|\leq|\xi_{3}|\leq\frac{1}{3}\bigl{(}\sqrt{A^{2}+9/2}+A\bigr{)}|x|^{1/2}.$

As a result, if $\xi\in\Omega_{1}$ then

$c^{-1}(A)|x|^{1/2}\leq|\xi|\leq c(A)|x|^{1/2}.$ (A.16)

Note that $\varphi^{\prime\prime}(\xi)=6\xi-2a$. Then for $\xi\in\Omega_{1}$

$|\varphi^{\prime\prime}(\xi)|=\frac{1}{|\xi|}|3\xi^{2}-2a\xi+3\xi^{2}|\geq
3|\xi|\geq c^{-1}(A)|x|^{1/2}$ (A.17)

and application of the Van-der-Corput lemma yields that

$\Bigl{|}\int_{\Omega_{1}}g_{n}(\xi)\,d\xi\Bigr{|}\\\ \leq
c\Bigl{(}\inf\limits_{\xi\in\Omega_{1}}|\varphi^{\prime\prime}(\xi)|\Bigr{)}^{-1/2}\Bigl{(}\sup\limits_{\xi\in\Omega_{1}}|\xi|^{n}+\int_{\Omega_{1}}\bigl{|}(\xi^{n})^{\prime}\bigr{|}\,d\xi\Bigr{)}\leq
c(A)|x|^{(2n-1)/4}.$ (A.18)

Let $\xi\in\Omega_{2}$. Then $\xi_{1}\leq\xi\leq\xi_{2}$. It means that

$|\xi|\leq\frac{1}{3}\bigl{(}\sqrt{A^{2}+3/2}+A\bigr{)}|x|^{1/2}=c(A)|x|^{1/2}.$

Therefore,

$|\varphi^{\prime}(\xi)|=|3\xi^{2}-2a\xi+x|\geq|x|/2\geq
c^{-1}(A)(1+\xi^{2}).$ (A.19)

Then

$\int_{\Omega_{2}}g_{n}(\xi)\,d\xi=\frac{\xi^{n}}{i\varphi^{\prime}(\xi)}e^{i\varphi(\xi)}\Big{|}_{\xi_{1}}^{\xi_{2}}-\frac{1}{i}\int_{\Omega_{2}}\Bigl{(}\frac{\xi^{n}}{\varphi^{\prime}(\xi)}\Bigr{)}^{\prime}e^{i\varphi(\xi)}\,d\xi.$
(A.20)

Here

$\Bigl{(}\frac{\xi^{n}}{\varphi^{\prime}(\xi)}\Bigr{)}^{\prime}=\frac{(\xi^{n})^{\prime}\varphi^{\prime}(\xi)-\xi^{n}\varphi^{\prime\prime}(\xi)}{(\varphi^{\prime}(\xi))^{2}}$

and since $|\xi^{n}\varphi^{\prime\prime}(\xi)|\leq c(A)(\xi^{2}+1)$

$\Bigl{|}\int_{\Omega_{2}}g_{n}(\xi)\,d\xi\Bigr{|}\leq
c(A)|x|^{n/2-1}+c(A)\int_{\Omega_{2}}\frac{d\xi}{1+\xi^{2}}\leq
c(A)|x|^{n/2-1}\\\
+\frac{c_{1}(A)}{|x|^{1/4}}\int\frac{d\xi}{(1+\xi^{2})^{3/4}}\leq
c_{2}(A)|x|^{-1/4}\leq c_{2}(A)|x|^{(2n-1)/4}.$ (A.21)

Let $\xi\in\Omega_{3}$. Then either $\xi\leq\xi_{3}$ or $\xi\geq\xi_{4}$ and

$|\varphi^{\prime}(\xi)|\geq|3\xi^{2}-2a\xi|-|x|\geq|x|/2\geq 1/2.$ (A.22)

Since $|x|\leq 2(3\xi^{2}-2a\xi)/3$ it follows that

$|\varphi^{\prime}(\xi)|\geq
3\xi^{2}-2a\xi-|x|\geq\frac{1}{3}(3\xi^{2}-2a\xi).$

If $a\xi\leq 0$, then with the use of (A.22) we obtain that

$|\varphi^{\prime}(\xi)|\geq(1+\xi^{2})/4.$ (A.23)

Let $a\xi>0$. If $a>0$ then $\xi\geq\xi_{4}$, if $a<0$ then $\xi\leq\xi_{3}$.
In both cases we obtain that

$|\xi|\geq\frac{1}{3}\bigl{(}|a|+\sqrt{a^{2}+9|x|/2}\bigr{)}.$

Since $|x|\geq a^{2}/A^{2}$

$|\xi|\geq\frac{|a|}{3}\bigl{(}1+\sqrt{1+\frac{9}{2A^{2}}}\bigr{)}=\frac{2}{1-\widetilde{c}}\frac{|a|}{3},$

where

$\widetilde{c}=\frac{\sqrt{A^{2}+9/2}-A}{\sqrt{A^{2}+9/2}+A}=c^{-1}(A),\quad\widetilde{c}\in(0,1),$

that is $(1-\widetilde{c})|\xi|\geq 2|a|/3$, and, therefore,

$|\varphi^{\prime}(\xi)|\geq\widetilde{c}\xi^{2}+|\xi|\bigl{(}(1-\widetilde{c})|\xi|-2|a|/3\bigr{)}\geq\widetilde{c}\xi^{2}\geq
c^{-1}(A)(1+\xi^{2}).$ (A.24)

Inequalities (A.22), (A.23), (A.24) mean that in the case $\xi\in\Omega_{3}$
the function $|\varphi^{\prime}(\xi)|$ verifies the same properties as in the
case $\xi\in\Omega_{2}$ in (A.19). Then we can apply the same argument as in
(A.20) and obtain the same estimate as in (A.21).

As a result, estimate (A.3) is established for $n\leq 1$. For other values of
$n$ it follows from equality (A.2). ∎

Now consider the fundamental solution of the differential operator
$\mathcal{L}(\partial_{t},\partial_{x})=\partial_{t}+\partial_{x}^{3}-ia\partial_{x}^{2}+b\partial_{x}$
(see (2.5)). Then it follows from (2.6) and (A.1) that

$G(t,x)=\frac{\theta(t)}{t^{1/3}}\Phi_{at^{1/3}}\left(\frac{x-bt}{t^{1/3}}\right).$
(A.25)

The properties of the function $\Phi_{a}$ imply that the function $G(t,x)$ is
infinitely smooth for $t>0$.

######  Lemma A.2.

For any $T>0$, $0<t\leq T$ and integer non-negative $n$

$\bigl{|}\partial_{x}^{n}G(t,x)\bigr{|}\leq\begin{cases}\displaystyle
c(T,n,a,b)t^{-q(n)}\bigl{(}1+|x|\bigr{)}^{(2n-1)/4},\quad&x\leq 0,\\\
c(T,n,a,b)t^{-(n+1)/3}e^{-c_{0}x^{3/2}T^{-1/2}},\quad&x\geq 0,\end{cases}$
(A.26)

for certain positive constants $c(T,n,a,b)$, $c_{0}$, where $q(0)=1/3$,
$q(n)=(2n+1)/4$ for $n\geq 1$. Moreover, if $b\leq 0$, $x\geq 0$ or $b\geq 0$,
$x\geq bt$, then

$\bigl{|}\partial_{x}^{n}G(t,x)\bigr{|}\leq
c(T,n,a,b)t^{-(n+1)/3}e^{-c_{0}x^{3/2}t^{-1/2}}.$ (A.27)

###### Proof.

First consider the case $b\geq 0$. Let $x\geq bt$, then since

$\Bigl{(}\frac{x-bt}{t^{1/3}}\Bigr{)}^{3/2}\geq\frac{1}{\sqrt{2}}\Bigl{(}\frac{x}{t^{1/3}}\Bigr{)}^{3/2}-b^{3/2}T$

inequality (A.3) in the case $x\geq 0$ yields that

$|\partial_{x}^{n}G(t,x)|\leq\frac{c(|a|T^{1/3},n)}{t^{(n+1)/3}}e^{-c_{0}\bigl{(}\frac{x-bt}{t^{1/3}}\bigr{)}^{3/2}}\leq\frac{c(T,n,a,b)}{t^{(n+1)/3}}e^{-c_{0}x^{3/2}(2t)^{-1/2}}.$

Let $0\leq x\leq bt$, then since

$1\leq 1+\frac{|x-bt|}{t^{1/3}}\leq 1+bT^{2/3}$

inequality (A.3) in the case $x\leq 0$ yields that

$|\partial_{x}^{n}G(t,x)|\leq\frac{c(|a|T^{1/3},n,bT^{2/3})}{t^{(n+1)/3}}\leq\frac{c(T,n,a,b)}{t^{(n+1)/3}}e^{-c_{0}(bT)^{3/2}T^{-1/2}}\\\
\leq\frac{c(T,n,a,b)}{t^{(n+1)/3}}e^{-c_{0}x^{3/2}T^{-1/2}}.$

Finally, let $x\leq 0$, then since

$1+\frac{|x|}{T^{1/3}}\leq 1+\frac{|x-bt|}{t^{1/3}}\leq
1+bT^{2/3}+\frac{|x|}{t^{1/3}}$

inequality (A.3) in the case $x\leq 0$ yields that

$|G(t,x)|\leq c(T,a,b)t^{-1/3}(1+|x|)^{-1/4}$ (A.28)

and if $n\geq 1$

$|\partial_{x}^{n}G(t,x)|\leq\frac{c(T,n,a,b)}{t^{(n+1)/3}}(1+\frac{|x|}{t^{1/3}})^{(2n-1)/4}\leq
c_{1}(T,n,a,b)t^{-q(n)}(1+|x|)^{(2n-1)/4}.$ (A.29)

Now consider the case $b<0$. If $x\geq 0$ then inequality (A.3) in the case
$x\geq 0$ yields that

$|\partial_{x}^{n}G(t,x)|\leq\frac{c(|a|T^{1/3},n)}{t^{(n+1)/3}}e^{-c_{0}\bigl{(}\frac{x-bt}{t^{1/3}}\bigr{)}^{3/2}}\leq\frac{c(T,n,a)}{t^{(n+1)/3}}e^{-c_{0}x^{3/2}t^{-1/2}}.$

If $bt\leq x\leq 0$, then $|x|\leq|b|T$ and inequality (A.3) in the case
$x\geq 0$ yields that

$|\partial_{x}^{n}G(t,x)|\leq\frac{c(|a|T^{1/3},n)}{t^{(n+1)/3}}\leq\frac{c(T,n,a,b)}{t^{(n+1)/3}}(1+|x|)^{(2n-1)/4}\\\
\leq\frac{c_{1}(T,n,a,b)}{t^{q(n)}}(1+|x|)^{(2n-1)/4}.$

Finally, if $x\leq bt$ then since

$\frac{1}{1+|b|T^{2/3}}\Bigl{(}1+\frac{|x|}{T^{1/3}}\Bigr{)}\leq
1+\frac{|x-bt|}{t^{1/3}}\leq 1+\frac{|x|}{t^{1/3}}$

inequality (A.3) in the case $x\leq 0$ similarly to (A.28), (A.29) yields
estimates (A.26) also in this case. ∎

######  Lemma A.3.

For any $\alpha\geq 0$, $\epsilon>0$, $T>0$, $t\in(0,T]$, $n=0$ or $n=1$, and
$y\in\mathbb{R}$

$\|\partial_{x}^{n}G(t,\cdot-y)\rho^{1/2}_{\alpha,\epsilon}\|_{L_{2}(\mathbb{R})}\leq
c(\alpha,\epsilon,T,a,b)t^{-(5n+4)/12}\rho^{1/2}_{\alpha,\epsilon}(y)(1+y_{+})^{(2n+1)/4}.$
(A.30)

###### Proof.

Note that $(5n+4)/12=q(n)$ for $n=0$ and $n=1$, $q(n)\geq(n+1)/3$, where
$q(n)$ is a number from (A.26).

First let $y\geq 0$, then ($\rho\equiv\rho_{\alpha,\epsilon)}$)

$\int\rho(x)|\partial_{x}^{n}G(t,x-y)|^{2}\,dx=\int_{-\infty}^{0}\dots
dx+\int_{0}^{y}\dots dx+\int_{y}^{+\infty}\dots dx.$

Inequalities (A.26) imply that

$t^{2q(n)}\int_{-\infty}^{0}\dots dx\leq
c\int_{-\infty}^{0}(1+y-x)^{(2n-1)/2}e^{2\epsilon x}\,dx\\\ \leq
c(1+y)^{1/2}\int_{-\infty}^{0}(1+|x|)^{1/2}e^{2\epsilon
x}\,dx=c(\epsilon)(1+y)^{1/2},$ $t^{2q(n)}\int_{0}^{y}\dots dx\leq
c\int_{0}^{y}(1+y-x)^{(2n-1)/2}\rho(x)\,dx\\\ \leq
c\rho(y)\int_{0}^{y}(1+y-x)^{(2n-1)/2}\,dx\leq c_{1}\rho(y)(1+y)^{(2n+1)/2},$
$t^{2(n+1)/3}\int_{y}^{+\infty}\dots dx\leq
c\int_{y}^{+\infty}e^{-2c_{0}(x-y)^{3/2}T^{-1/2}}(1+x)^{2\alpha}\,dx\\\
=c\int_{0}^{+\infty}e^{-2c_{0}\theta^{3/2}T^{-1/2}}(1+\theta+y)^{2\alpha}\,d\theta\\\
\leq
c(1+y)^{2\alpha}\int_{0}^{+\infty}e^{-2c_{0}\theta^{3/2}T^{-1/2}}(1+\theta)^{2\alpha}\,d\theta\leq
c(\alpha)\rho(y).$

If $y<0$ then

$\int\rho(x)|G_{x}(t,x-y)|^{2}\,dx=\int_{-\infty}^{y}\dots
dx+\int_{y}^{0}\dots dx+\int_{0}^{+\infty}\dots dx.$

Here

$t^{2q(n)}\int_{-\infty}^{y}\dots dx\leq
c\int_{-\infty}^{y}(1+y-x)^{(2n-1)/2}e^{2\epsilon x}\,dx\\\ \leq ce^{2\epsilon
y}\int_{0}^{+\infty}(1+\theta)^{1/2}e^{-2\epsilon\theta}\,d\theta\leq
c(\epsilon)\rho(y),$ $t^{2(n+1)/3}\int_{y}^{0}\dots dx\leq
c\int_{y}^{0}e^{-2c_{0}(x-y)^{3/2}T^{-1/2}}e^{2\epsilon x}\,dx\\\ \leq
ce^{2\epsilon
y}\int_{0}^{+\infty}e^{-2c_{0}\theta^{3/2}T^{-1/2}}e^{2\epsilon\theta}\,d\theta\leq
c(\epsilon)\rho(y),$ $t^{2(n+1)/3}\int_{0}^{+\infty}\dots dx\leq
c\int_{0}^{+\infty}e^{-2c_{0}(x-y)^{3/2}T^{-1/2}}(1+x)^{2\alpha}\,dx\\\ \leq
c(T)e^{2\epsilon
y}\int_{0}^{+\infty}e^{-2c_{0}\theta^{3/2}T^{-1/2}}(1+\theta)^{2\alpha}e^{2\epsilon\theta}\,d\theta\leq
c(\alpha,\epsilon)\rho(y).$

∎

## Appendix B

######  Lemma B.1.

Let $\varphi\in W_{1}^{1}(I)$ for certain interval $I\subset\mathbb{R}$, then
for a.e. $x\in I$ there exists $|\varphi(x)|^{\prime}$,

$|\varphi|^{\prime}=\begin{cases}\displaystyle\frac{\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime}}{2|\varphi|},\quad&\varphi(x)\neq
0,\\\ 0,\quad&\varphi(x)=0,\end{cases}$ (B.1)

in particular,

$\bigl{|}|\varphi|^{\prime}\bigr{|}\leq|\varphi^{\prime}|.$ (B.2)

###### Proof.

This formula is obvious if $\varphi(x)\neq 0$. The function $|\varphi(x)|$ is
absolutely continuous on every bounded segment in $I$ and, thus,
differentiable a.e. on $I$. Let $x_{0}\in I$ be such point, where
$|\varphi(x)|$ is differentiable, $\varphi(x_{0})=0$ and there exists a
sequence $x_{n}\to x_{0}$ such that $\varphi(x_{n})=0$. Then

$|\varphi(x_{0})|^{\prime}=\lim\limits_{x_{n}\to
x_{0}}\frac{|\varphi(x_{n})|-|\varphi(x_{0})|}{x_{n}-x_{0}}=0.$

It remains to note that the set of isolated zeros of the function $\varphi$
has a zero measure. ∎

######  Lemma B.2.

Let $\varphi\in W^{2}_{1}(I)$ for certain interval $I\subset\mathbb{R}$, then
for a.e. $x\in I$ there exists
$\bigl{(}|\varphi(x)|^{\prime}\varphi(x)\bigr{)}^{\prime}$ and

$\bigl{|}\bigl{(}|\varphi|^{\prime}\varphi\bigr{)}^{\prime}\bigr{|}\leq
3|\varphi^{\prime}|^{2}+|\varphi\varphi^{\prime\prime}|.$ (B.3)

###### Proof.

At the point $x\in I$, where exist derivatives $\varphi^{\prime}(x)$,
$\varphi^{\prime\prime}(x)$ and $\varphi(x)\neq 0$, it is easy to see that
there exists

$\bigl{|}\bigl{(}|\varphi|^{\prime}\varphi\bigr{)}^{\prime}\bigr{|}=\frac{\varphi^{\prime}}{2|\varphi|}(\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime})\\\
+\frac{\varphi}{2|\varphi|}(\varphi^{\prime\prime}\bar{\varphi}+2|\varphi^{\prime}|^{2}+\varphi\bar{\varphi}^{\prime\prime})-\frac{\varphi}{4|\varphi|^{3}}(\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime})^{2}\equiv\Phi(x).$
(B.4)

Set $\Phi(x)=0$ when $\varphi(x)=0$.

For $\delta>0$ consider the functions $g_{\delta}$ from (1.7), then for a.e.
$x\in I$

$\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}=\frac{(\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime})}{2(|\varphi|^{2}+\delta)^{1/2}}$
(B.5)

and

$\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}\varphi\to\frac{\varphi(\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime})}{2|\varphi|}$

in $L_{1}(I)$ when $\delta\to+0$. Moreover, for a.e. $x\in I$ there exists
$\bigl{(}\bigl{(}g_{\delta}(|\varphi(x)|^{2})\bigr{)}^{\prime}\varphi(x)\bigr{)}^{\prime}$
and

$\bigl{|}\bigl{(}\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}\varphi\bigr{)}^{\prime}\bigr{|}=\frac{\varphi^{\prime}}{2(|\varphi|^{2}+\delta)^{1/2}}(\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime})\\\
+\frac{\varphi}{2(|\varphi|^{2}+\delta)^{1/2}}(\varphi^{\prime\prime}\bar{\varphi}+2|\varphi^{\prime}|^{2}+\varphi\bar{\varphi}^{\prime\prime})-\frac{\varphi}{4(|\varphi|^{2}+\delta)^{3/2}}(\varphi^{\prime}\bar{\varphi}+\varphi\bar{\varphi}^{\prime})^{2}.$
(B.6)

Let $\mathcal{E}$ be s set of points $x\in I$ where either the function
$\varphi$ is not twice differentiable or $x$ is an isolated zero of the
function $\varphi$. Then this set has a zero measure. Moreover, similarly to
the proof of Lemma B.1 if $x\not\in\mathcal{E}$ and $\varphi(x)=0$ then
$\varphi^{\prime}(x)=0$. Then for $x\in I\setminus\mathcal{E}$

$\bigl{(}\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}\varphi\bigr{)}^{\prime}\to\Phi(x)$

and

$\bigl{(}\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}\varphi\bigr{)}^{\prime}\to\Phi$

in $L_{1}(I)$ when $\delta\to+0$. Therefore, there exists
$\bigl{(}|\varphi|^{\prime}\varphi\bigr{)}^{\prime}=\Phi$ and estimate (B.3)
holds. ∎

######  Remark B.3.

It follows from (B.2), (B.3), (B.5) and (B.6) that uniformly with respect to
$\delta\geq 0$

$\bigl{|}\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}\bigr{|}\leq|\varphi^{\prime}|,\quad\bigl{|}\bigl{(}\bigl{(}g_{\delta}(|\varphi|^{2})\bigr{)}^{\prime}\varphi\bigr{)}^{\prime}\bigr{|}\leq
3|\varphi^{\prime}|^{2}+|\varphi\varphi^{\prime\prime}|.$ (B.7)

######  Lemma B.4.

Consider a function of two complex variables $u$ and $v$

$P(u,v)\equiv 3|u|v+\frac{u^{2}\bar{v}}{|u|}.$ (B.8)

If $|u|,|v|,|\widetilde{u}|,|\widetilde{v}|\leq M$, then

$|P(u,v)-P(\widetilde{u},\widetilde{v})|\leq
6M|u-\widetilde{u}|+4M|v-\widetilde{v}|.$ (B.9)

###### Proof.

In fact,

$P(u,v)-P(\widetilde{u},\widetilde{v})=3\bigl{(}|u|-|\widetilde{u}|\bigr{)}v+3|\widetilde{u}|(v-\widetilde{v})+\left(\frac{u^{2}}{|u|}-\frac{\widetilde{u}^{2}}{|\widetilde{u}|}\right)\bar{v}+\frac{\widetilde{u}^{2}}{|\widetilde{u}|}\overline{(v-\widetilde{v})}.$

Let, for example, $|\widetilde{u}|\leq|u|$, then

$\frac{u^{2}}{|u|}-\frac{\widetilde{u}^{2}}{|\widetilde{u}|}=\frac{u^{2}|\widetilde{u}|-|u|\widetilde{u}^{2}}{|u\widetilde{u}|}=\frac{(u^{2}-\widetilde{u}^{2})|\widetilde{u}|+\widetilde{u}^{2}\bigl{(}|\widetilde{u}|-|u|\bigr{)}}{|u\widetilde{u}|}$

and so

$\left|\frac{u^{2}}{|u|}-\frac{\widetilde{u}^{2}}{|\widetilde{u}|}\right|\leq\left(2+\frac{|\widetilde{u}|}{|u|}\right)|u-\widetilde{u}|\leq
3|u-\widetilde{u}|.$

∎

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