On the initial-boundary value problem for a non-local elliptic-hyperbolic system related to the short pulse equation

Abstract

In this paper, we prove the well-posedness of the initial-boundary value problem for a non-local elliptic-hyperbolic system related to the short pulse equation. Our arguments are based on energy estimates and passing to the limit in a vanishing viscosity approximation of the problem.

1 Introduction

This paper is dedicated to the well-posedness analysis of the following initial boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu -q^2\partial _x (uv)=bP, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x P=u, &{}\qquad t>0,\,x>0,\\ \displaystyle \alpha \partial _{x}^2v +\beta \partial _x v +\gamma v=\kappa u^2, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x u(t,0)=g(t), &{}\qquad t>0,\\ \displaystyle P(t,0)=0, &{}\qquad t>0,\\ \displaystyle v(t,0)=h(t), &{}\qquad t>0,\\ \displaystyle u(0,x)=u_0(x)&{}\qquad x>0. \end{array}\right. } \end{aligned}$$

(1.1)

The equations in (1.1) belonging to the large class of systems of the form

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu+g\partial _x u^3 - a\partial _{x}^3u +q\partial _x (uv)=bP, \\ \displaystyle \partial _x P=u, \\ \displaystyle \alpha \partial _{x}^2v +\beta \partial _x v +\gamma v=\kappa u^2, \end{array}\right. } \end{aligned}$$

(1.2)

are termed continuum spectrum pulse equations [7,8,9, 55, 60, 61, 68, 79]. They describe the dynamics of the electrical field u of linearly polarized continuum spectrum pulses in optical waveguides, including fused-silica telecommunication-type or photonic-crystal fibers, as well as hollow capillaries filled with transparent gases or liquids.

The constants $a,\,b,\,g,\,q,\,\alpha ,\,\kappa ,\,\beta ,\,\gamma$, in (1.2), take into account the frequency dispersion of the effective linear refractive index and the nonlinear polarization response, the excitation efficiency of the vibrations, the frequency and the decay time (see [8, 9, 79]).

From a mathematical point of view, in [29], the well-posedness of the classical solutions of the Cauchy associated with (1.2) is proven.

Taking $b=\alpha =\beta =0$, (1.2) reads

$$\begin{aligned} \partial _tu +\left( g+\frac{q\kappa }{\gamma }\right) \partial _x u^3 -a\partial _{x}^3u=0, \end{aligned}$$

(1.3)

which is known as modified Korteweg–de Vries equation (see [23, 45, 59, 75, 81]).

In [6, 7, 10, 63,64,65], it is proven that (1.3) is a non-slowly-varying envelope approximation model that describes the physics of few-cycle-pulse optical solitons. In [34, 59], the Cauchy problem for (1.3) is studied, while, in [23, 75], the convergence of the solution of (1.3) as $a\rightarrow 0$ to the unique entropy solution of the following scalar conservation law

$$\begin{aligned} \partial _tu +\left( g+\frac{q\kappa }{\gamma }\right) \partial _x u^3 =0 \end{aligned}$$

(1.4)

is proven.

On the other hand, taking $a=\alpha =\beta =0$ in (1.2), we have the following equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu+\left( g+\frac{q\kappa }{\gamma }\right) \partial _x u^3 =bP, \\ \displaystyle \partial _x P=u. \end{array}\right. } \end{aligned}$$

(1.5)

It was introduced by Kozlov and Sazonov [61] as a model equation describing the nonlinear propagation of optical pulses of a few oscillations duration in dielectric media, and Schäfer and Wayne [76] as a model equation describing the propagation of ultra-short light pulses in silica optical fibers.

In [3, 4, 30, 63,64,65], the authors show that (1.5) is also a non-slowly-varying envelope approximation model that describes the physics of few-cycle-pulse optical solitons. Meanwhile, [5, 24, 72, 74] show that (1.5) is a particular Rabelo equation which describes pseudospherical surfaces.

System (1.5) is also deduced in [82] to describe the short pulse propagation in nonlinear metamaterials characterized by a weak Kerr-type nonlinearity in their dielectric response.

It also is interesting to remind that equation (1.5) was proposed earlier in [69] in the context of plasma physic and that similar equations describe the dynamics of radiating gases [62, 77]. Moreover, [31, 52,53,54] show that (1.5) is also a model for ultrafast pulse propagation in a mode-locked laser cavity in the few-femtosecond pulse regime. Finally, an interpretation of (1.5) in the context of Maxwell equations is given in [71].

From a mathematical point of view, wellposedness results for the Cauchy problem of (1.5) are proven in the context of energy spaces (see [50, 70, 80]). Similar results are proven in [20, 27, 41, 51] in the context of entropy solutions, while, in [19, 33, 43, 73], the wellposedness of the homogeneous initial boundary value problem is studied. Finally, the convergence of a finite difference scheme is studied in [42].

Observe that, taking $\alpha =\beta =0$ and $a\not =0$, (1.2) reads

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu+\left( g+\frac{q\kappa }{\gamma }\right) \partial _x u^3-a\partial _{x}^3u =bP, \\ \displaystyle \partial _x P=u. \end{array}\right. } \end{aligned}$$

(1.6)

It was derived by Costanzino, Manukian and Jones [48] in the context of the nonlinear Maxwell equations with high-frequency dispersion. Kozlov and Sazonov [61] show that (1.6) is an more general equation than (1.5) to describe the nonlinear propagation of optical pulses of a few oscillations duration in dielectric media.

Mathematical properties of (1.6) are studied in many different contexts, including the local and global well-posedness in energy spaces [48, 70] and stability of solitary waves [48, 67], while, in [34], the well-posedness of the classical solutions is proven.

In analogy with the convergence result of (1.3) to (1.4), in [27, 28], the convergence of the solution of (1.6) as $a\rightarrow 0$ to the unique entropy solution of (1.5) was proved.

Taking $g=a=0$ in (1.2), we have the following system:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu+q\partial _x (uv)=bP, \\ \displaystyle \partial _x P=u, \\ \displaystyle \alpha \partial _{x}^2v +\beta \partial _x v +\gamma v=\kappa u^2, \end{array}\right. } \end{aligned}$$

(1.7)

which represents a non-local formulation of (1.5) (see also [37]).

Conservation laws with non-local flux can be found in the context of traffic flow modeling [1, 12, 14, 15, 39, 44, 46, 47, 49, 56,57,58], in the context of sedimentation dynamic modeling [11] and in the context of slow erosion modeling [2, 16, 78].

In [32, 37], the well-posedness of the classical solution of the Cauchy associated with (1.7) is proven. Here we continue the analysis started in [37], studying the initial boundary value problem associated to (1.7).

Since our argument does not depend on the sign of the coefficient q here we assume

$$\begin{aligned} q<0 \end{aligned}$$

and for the sake of notational simplicity from now on we write ${-q^2}$ instead of q

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu -q^2\partial _x (uv)=bP, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x P=u, &{}\qquad t>0,\,x>0,\\ \displaystyle \alpha \partial _{x}^2v +\beta \partial _x v +\gamma v=\kappa u^2, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x u(t,0)=g(t), &{}\qquad t>0,\\ \displaystyle P(t,0)=0, &{}\qquad t>0,\\ \displaystyle v(t,0)=h(t), &{}\qquad t>0,\\ \displaystyle u(0,x)=u_0(x),&{}\qquad x>0, \end{array}\right. } \end{aligned}$$

(1.8)

where $\partial _x u(t,0)$ is the trace of $\partial _x u(t,x)$ at $x=0$.

We can rewrite the problem as a boundary value problem for a single integro-differential equation.

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu -q^2\partial _x (u\mathcal {V}_t[u(t,\cdot )])=b\int _0^x u(t,y)dy, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x u(t,0)=g(t), &{}\qquad t>0,\\ \displaystyle u(0,x)=u_0(x),&{}\qquad x>0, \end{array}\right. } \end{aligned}$$

where $v(t,x)=\mathcal {V}_t[u(t,\cdot )](x)$ is the solution of the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \alpha \partial _{x}^2v +\beta \partial _x v +\gamma v=\kappa u^2, &{}\qquad t>0,\,x>0,\\ \displaystyle v(t,0)=h(t), &{}\qquad t>0,\\ \end{array}\right. } \end{aligned}$$

On the function g, we assume

$$\begin{aligned} g\in W^{1,\infty }(0,\infty ), \end{aligned}$$

(1.9)

while on the function h, we assume

$$\begin{aligned} 0<\kappa ^2_1\le h(t), \quad h\in L^{\infty }(0,\infty ), \end{aligned}$$

(1.10)

for some constant $\kappa _1>0$. On the initial datum, we assume that

$$\begin{aligned}&u_0\in L^1(0,\infty )\cap H^1(0,\infty ), \quad \int _{0}^{\infty }u_0(x) dx=0, \end{aligned}$$

(1.11)

$$\begin{aligned}&\left\| P_0 \right\| ^2_{L^2(0,\infty )}=\int _{0}^{\infty }\left( \int _{0}^xu_0(y)dy\right) ^2 dx <\infty , \end{aligned}$$

(1.12)

$$\begin{aligned}&\text {where}\>P_0(x)=\int _{0}^{x}u_0(y) dy,\qquad x>0. \end{aligned}$$

(1.13)

The zero mean requirement in (1.11) and the $L^2$ one in (1.13) imply that the solution u of (1.8) satisfies the same conditions at every time $t>0$ (see [17, 22, 35]), i.e.

$$\begin{aligned} \int _0^\infty u(t,x)dx=0,\qquad \left\| P(t,\cdot ) \right\| ^2_{L^2(0,\infty )}<\infty ,\qquad t\ge 0. \end{aligned}$$

Those properties will play a key role in the estimates of the next sections.

In addition, on the constants $\alpha ,\,\beta ,\,\kappa$, we assume

$$\begin{aligned} \frac{\alpha }{\kappa }>0, \quad \frac{\beta }{\kappa }\le 0, \end{aligned}$$

(1.14)

and either (1.15) or (1.16)

$$\begin{aligned}&b,\,\beta ,\,\gamma \ne 0, \quad \alpha \beta <0, \quad \text {or}, \end{aligned}$$

(1.15)

$$\begin{aligned}&b\ne 0, \quad \beta =0, \quad \alpha \gamma <0. \end{aligned}$$

(1.16)

Observe that, in all cases, $\alpha \ne 0$. Therefore, we may set it equal to 1 and work with only three constants.

We use the following definition fo solutions.

Definition 1.1

A triplet of real valued functions $(u,\,v,\, P)$ defined on $[0,\infty )\times [0,\infty )$ is a distributional solution of (1.8) if

• $u\in L^2(0,\infty ;H^2(0,\infty )), \,P\in L^2(0,\infty ;H^1(0,\infty )), \,v\in L^2(0,\infty ;H^1(0,\infty ))$;

• $\partial _x u(\cdot ,0)=g,\,P(\cdot ,0)=0, \, v(\cdot ,0)=h$ in the sense of traces;

• for every test function $\varphi \in C^\infty (\mathbb {R}\times (0,\infty ))$ with compact support

  $$\begin{aligned} \begin{aligned} \int _0^\infty \int _0^\infty \left( u\partial _t\varphi -q^2uv\partial _x \varphi +bP\varphi \right) dtdx+\int _0^\infty u_0(x)\varphi (0,x)dx&=0\\ \int _0^\infty \int _0^\infty \left( P \partial _x \varphi +u\varphi \right) dtdx&=0,\\ \int _0^\infty \int _0^\infty \left( \alpha v\partial _{x}^2\varphi -\beta v\partial _x \varphi +\gamma v\varphi -\kappa u^2\varphi \right) dtdx&=0. \end{aligned} \end{aligned}$$

  (1.17)

The assumptions (1.14), (1.15) and (1.16) on the constants guarantee the boundedness of the $L^2$ norm of u in time (see Lemma 2.3 below).

The main results of this paper are the following theorem.

Theorem 1.1

Assume (1.9), (1.10), (1.11), (1.13), (1.12), (1.14) and either (1.15) or (1.16). Given $T>0$, there exists a unique distributional solution $(u,\,v,\, P)$ of (1.8) in the sense of Definition 1.1 such that

$$\begin{aligned}&u \in H^1((0,T)\times (0,\infty ))\cap L^{\infty }(0,T;H^1(0,\infty )), \end{aligned}$$

(1.18)

$$\begin{aligned}&v\in L^{\infty }(0,T;H^3(0,\infty )), \end{aligned}$$

(1.19)

$$\begin{aligned}&P\in L^{\infty }(0,T;H^2(0,\infty )), \end{aligned}$$

(1.20)

$$\begin{aligned}&\int _{0}^{\infty }u(t,x)dx=0, \quad t\ge 0. \end{aligned}$$

(1.21)

Moreover, if $(u_1,\,v_1,\,P_1)$ and $(u_2,\,v_2,\,P_2)$ are two solutions of (1.8), we have that

$$\begin{aligned} \begin{aligned} \left\| u_1(t,\cdot )-u_2(t,\cdot ) \right\| _{L^2(0,\infty )}\le&e^{C(T)t}\left\| u_{1,0}-u_{2,0} \right\| _{L^2(0,\infty )},\\ \left\| v_1(t,\cdot )-v_2(t,\cdot ) \right\| _{H^2(0,\infty )}\le&e^{C(T)t}\left\| u_{1,0}-u_{2,0} \right\| _{L^2(0,\infty )}, \end{aligned} \end{aligned}$$

(1.22)

for some suitable $C(T)>0$, and every $0\le t\le T$.

Finally, we decided to study only the stability with respect to the initial datum in order to shorten the arguments.

In the next theorem we give a necessary condition on the constants that gives some additional regularity on the solutions.

Theorem 1.2

Given $T>0$ assuming either (1.14) or (1.15), if

$$\begin{aligned} \beta \gamma < 0, \end{aligned}$$

(1.23)

there exists an unique distributional solution $(u,\,v,\, P)$ of (1.8) in the sense of Definition 1.1 such that (1.18), (1.20) and (1.21) hold, while

$$\begin{aligned} \begin{aligned}&v\in H^1((0,T)\times \mathbb {R})\cap L^{\infty }(0,T;H^3(\mathbb {R}))\cap W^{1,\infty }((0,T)\times \mathbb {R})\\&\partial _{tx}^2v\in L^{\infty }(0,T;L^2(0,\infty ))\cap L^{\infty }((0,T)\times (0,\infty )), \quad \partial _t\partial _{x}^2v \in L^{\infty }(0,T;L^2(0,\infty )), \end{aligned} \end{aligned}$$

(1.24)

for every $0\le t\le T$. Moreover, assuming (1.14), one between (1.15) or (1.16), and

$$\begin{aligned} 0<\kappa ^2_1\le h(t), \quad h\in W^{1,\infty }(0,\infty ), \end{aligned}$$

(1.25)

there exists an unique distributional solution $(u,\,v,\, P)$ of (1.8) such that (1.18), (1.20), (1.21), (1.24) hold. Finally, if $(u_1,\,v_1,\,P_1)$ and $(u_2,\,v_2,\,P_2)$ are two solutions of (1.8), (1.22) holds.

The paper is organized as follows. In Sect. 2, we prove several a priori estimates on a vanishing viscosity approximation of (1.8). Those play a key role in the proof of our main result, that is given in Sect. 3. In Sects. 4 and 5, we prove Theorem 1.2, under Assumptions (1.23) and (1.25), respectively.

2 Vanishing viscosity approximation

Our existence argument is based on passing to the limit in a vanishing viscosity approximation of (1.8)

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu_{\varepsilon }-q^2\partial _x (u_{\varepsilon }v_\varepsilon )=bP_\varepsilon +\varepsilon \partial _{x}^2u_{\varepsilon }, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x P_\varepsilon =u_{\varepsilon }, &{}\qquad t>0,\,x>0,\\ \displaystyle \alpha \partial _{x}^2v_\varepsilon +\beta \partial _x v_\varepsilon +\gamma v_\varepsilon =\kappa u_{\varepsilon }^2, &{}\qquad t>0,\,x>0,\\ \displaystyle \partial _x u_{\varepsilon }(t,0)=\ge (t), &{}\qquad t>0,\\ \displaystyle P_\varepsilon (t,0)=0, &{}\qquad t>0,\\ \displaystyle v_\varepsilon (t,0)=h_{\varepsilon }(t), &{}\qquad t>0,\\ \displaystyle u_{\varepsilon }(0,x)=u_{\varepsilon ,\,0}(x),&{}\qquad x>0, \end{array}\right. } \end{aligned}$$

(2.1)

where $0<\varepsilon <1$ and $u_{\varepsilon ,0}, g_{\varepsilon }\,\ge ,\,h_{\varepsilon }$ are $C^{\infty }$ approximations of $u_0,\,g,\, h$ such that

$$\begin{aligned} \begin{aligned}&\left\| u_{\varepsilon ,0} \right\| _{H^1(0,\infty )}\le \left\| u_0 \right\| _{H^1(0,\infty )},\quad \int _{0}^{\infty }u_{\varepsilon ,0}(x) dx=0, \quad \varepsilon \left\| \partial _{x}^2u_{\varepsilon , 0} \right\| _{L^2(0,\infty )}\le C_0,\\&\left\| g_{\varepsilon } \right\| _{W^{1,\infty }(0,\infty )}\le C_0, \quad 0<\kappa _1^2\le h_{\varepsilon }(t), \quad \left\| h_{\varepsilon } \right\| _{L^{\infty }(0,\infty )}\le C_0,\\&\left\| P_{\varepsilon ,0} \right\| _{L^2(0,\infty )}\le \left\| P_0 \right\| _{L^2(0,\infty )}, \>\text {with}\>P_{\varepsilon ,0}(x)=\int _{0}^{x}u_{\varepsilon ,0}(y) dy, \end{aligned} \end{aligned}$$

(2.2)

and $C_0$ is a constant independent on $\varepsilon$. The existence of a unique smooth solution

$$\begin{aligned} u_{\varepsilon }\in C^\infty ([0,\infty )\times [0,\infty ))\cap H^2((0,\infty )\times (0,\infty )) \end{aligned}$$

can be proved using the same arguments as in [25, 36, 38]. Finally, we want to point put that the requirement $\varepsilon \left\| \partial _{x}^2u_{\varepsilon , 0} \right\| _{L^2(0,\infty )}\le C_0$ is quite common in the arguments based on vanishing viscosity and it states the fact that the $H^2$ norm of $u_{\varepsilon , 0}$ is not uniformly bounded with respect to $\varepsilon$ but it blows-up as $\varepsilon \rightarrow 0$.

Let us prove some a priori estimates on $u_{\varepsilon }$, $P_\varepsilon$ and $v_\varepsilon$. We denote with C all the the constants which depend only on the initial data, and with C(T), the constants which depend also on T. Moreover, we always assume that $\varepsilon \in (0,1)$ is given,l $(u_{\varepsilon },\,v_\varepsilon ,\,P_\varepsilon )$ is a solution of (2.1), and that (2.2) holds.

We begin by proving the following lemma.

Lemma 2.1

For each $t>0$, we have that

$$\begin{aligned} P_\varepsilon (t,\infty )&=0, \end{aligned}$$

(2.3)

$$\begin{aligned} \int _{0}^{\infty }u_{\varepsilon }(t,x) dx&=0. \end{aligned}$$

(2.4)

Proof

Arguing as in [32, Lemma 2.2], or [18, Lemma 2.1], we have (2.3).

We prove (2.4). Integrating the second equation of (2.1) on (0, x) and using the boundary conditions, we have that

$$\begin{aligned} P_\varepsilon (t,x)=\int _{0}^{x}u_{\varepsilon }(t,y) dy. \end{aligned}$$

(2.5)

By (2.3) and (2.5), we get

$$\begin{aligned} \int _{0}^{\infty }u_\varepsilon (t,x) dx=P_\varepsilon (t,\infty )=0, \end{aligned}$$

which gives (2.4). $\square$

Lemma 2.2

For each $t>0$, we have that,

$$\begin{aligned} \int _{0}^{\infty }u_{\varepsilon }^2\partial _x v_\varepsilon dx=-\frac{\alpha }{2\kappa }(\partial _x v_\varepsilon (t,0))^2+\frac{\beta }{\kappa }\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\frac{\gamma }{2\kappa }h_{\varepsilon }^2(t). \end{aligned}$$

(2.6)

Proof

Multiplying the third equation of (2.1) by $\partial _x v_\varepsilon$, thanks to (2.1), integrating on $(0,\infty )$, we get

$$\begin{aligned} \kappa \int _{0}^{\infty }u_{\varepsilon }^2\partial _x v_\varepsilon dx=&\, \alpha \int _{0}^{\infty }\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon dx +\beta \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \gamma \int _{0}^{\infty }v_\varepsilon \partial _x v_\varepsilon dx\\ =&\, -\frac{\alpha }{2}(\partial _x v_\varepsilon (t,0))^2 +\beta \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\frac{\gamma }{2}v_\varepsilon ^2(t,0)\\ =&\, -\frac{\alpha }{2}(\partial _x v_\varepsilon (t,0))^2 +\beta \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\frac{\gamma }{2}h_{\varepsilon }^2(t), \end{aligned}$$

which gives (2.6). $\square$

We continue by proving an $L^2$-estimate unform in $\varepsilon$.

Lemma 2.3

Fix $T>0$ and assume (1.15), or (1.16). Then, there exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \begin{aligned} \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+2\varepsilon \int _{0}^{\infty }\left\| \partial _x u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )} ds +\frac{q^2\kappa _1^2}{2}\int _{0}^{t}u_{\varepsilon }^2(s,0)ds\\&+\frac{q^2\alpha }{2\kappa }\int _{0}^{t}(\partial _x v_\varepsilon (s,0))^2ds-\frac{q^2\beta }{\kappa }\int _{0}^{t}\left\| \partial _x v_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le C(T), \end{aligned} \end{aligned}$$

(2.7)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. Multiplying the first equation of (2.1) by $2u_{\varepsilon }$, thanks to (2.1), an integration on $(0,\infty )$ gives

$$\begin{aligned} \frac{d}{dt}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}=&\, 2\int _{0}^{\infty }u_{\varepsilon }\partial _tu_{\varepsilon }dx\\ =&\, 2q^2\int _{0}^{\infty }u_{\varepsilon }\partial _x (u_{\varepsilon }v_\varepsilon ) dx +2b\int _{0}^{\infty }P_\varepsilon u_{\varepsilon }dx +2\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _{x}^2u_{\varepsilon }dx\\ =&\, 2q^2\int _{0}^{\infty }u_{\varepsilon }\partial _x (u_{\varepsilon }v_\varepsilon ) dx +2b\int _{0}^{\infty }P_\varepsilon u_{\varepsilon }dx \\&-2\varepsilon u_{\varepsilon }(t,0)\partial _x u_{\varepsilon }(t,0)-2\varepsilon \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ =&\, 2q^2\int _{0}^{\infty }u_{\varepsilon }\partial _x (u_{\varepsilon }v_\varepsilon ) dx +2b\int _{0}^{\infty }P_\varepsilon u_{\varepsilon }dx\\&-2\varepsilon g_{\varepsilon }(t)u_{\varepsilon }(t,0)-2\varepsilon \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Therefore, we have that

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad =2q^2\int _{0}^{\infty }u_{\varepsilon }\partial _x (u_{\varepsilon }v_\varepsilon ) dx +2b\int _{0}^{\infty }P_\varepsilon u_{\varepsilon }dx-2\varepsilon \ge (t)u_{\varepsilon }(t,0). \end{aligned} \end{aligned}$$

(2.8)

Observe that, by the boundary condition on $P_\varepsilon$ and (2.3),

$$\begin{aligned} 2b\int _{0}^{\infty }P_\varepsilon u_{\varepsilon }dx=2b\int _{0}^{\infty }P_\varepsilon \partial _x P_\varepsilon dx=0. \end{aligned}$$

(2.9)

Moreover, by (2.1) and (2.6),

$$\begin{aligned} 2q^2\int _{0}^{\infty }&u_{\varepsilon }\partial _x (u_{\varepsilon }v_\varepsilon ) dx=-2q^2u_{\varepsilon }^2(t,0)v_\varepsilon (t,0)-2q^2\int _{0}^{\infty }u_{\varepsilon }\partial _x u_{\varepsilon }v_\varepsilon dx \nonumber \\ =&\, -q^2u_{\varepsilon }^2(t,0)v_\varepsilon (t,0)+q^2\int _{0}^{\infty }\partial _x v_\varepsilon u_{\varepsilon }^2 dx\nonumber \\ =&\, -q^2u_{\varepsilon }^2(t,0)v_\varepsilon (t,0)-\frac{q^2\alpha }{2\kappa }(\partial _x v_\varepsilon (t,0))^2 +\frac{q^2\beta }{\kappa }\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\frac{q^2\gamma }{2\kappa }h_{\varepsilon }^2(t)\nonumber \\ =&\, -q^2h_{\varepsilon }(t)u_{\varepsilon }^2(t,0) -\frac{q^2\alpha }{2\kappa }(\partial _x v_\varepsilon (t,0))^2 +\frac{q^2\beta }{\kappa }\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\frac{q^2\gamma }{2\kappa }h_{\varepsilon }^2(t)\nonumber \\ \le&-q^2\kappa _1^2u_{\varepsilon }^2(t,0) -\frac{q^2\alpha }{2\kappa }(\partial _x v_\varepsilon (t,0))^2 +\frac{q^2\beta }{\kappa }\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\frac{q^2\gamma }{2\kappa }h_{\varepsilon }^2(t).\nonumber \\ \end{aligned}$$

(2.10)

Using (2.9) and (2.10) in (2.8)

$$\begin{aligned}{} & {} \frac{d}{dt}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+q^2\kappa _1^2u_{\varepsilon }^2(t,0)+\frac{q^2\alpha }{2\kappa }(\partial _x v_\varepsilon (t,0))^2\nonumber \\{} & {} \qquad -\frac{q^2\beta }{\kappa }\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le -2\varepsilon g_{\varepsilon }(t)u_{\varepsilon }(t,0)-\frac{q^2\gamma }{2\kappa }h_{\varepsilon }^2(t). \end{aligned}$$

(2.11)

Since $0<\varepsilon <1$, thanks to (1.10) and the Young’s inequality,

$$\begin{aligned} 2\varepsilon \vert g_{\varepsilon }(t)\vert \vert u_{\varepsilon }(t,0)\vert \le&~ C\vert u_{\varepsilon }(t,0)\vert \le C+\frac{q^2\kappa _1^2}{2}u_{\varepsilon }^2(t,0),\qquad \left| \frac{q^2\gamma }{2\kappa }\right| h_{\varepsilon }^2(t)\le ~ C. \end{aligned}$$

Consequently, (2.11) becomes,

$$\begin{aligned} \frac{d}{dt}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+2\varepsilon \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\frac{q^2\kappa _1^2}{2}u_{\varepsilon }^2(t,0)\nonumber \\&+\frac{q^2\alpha }{2\kappa }(\partial _x v_\varepsilon (t,0))^2-\frac{q^2\beta }{\kappa }\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C. \end{aligned}$$

Integrating on (0, t), by (2.2) we get

$$\begin{aligned}&\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon \int _{0}^{t}\left\| \partial _x u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds+\frac{q^2\kappa _1^2}{2}\int _{0}^{t}u_{\varepsilon }^2(s,0)ds\nonumber \\&\qquad +\frac{q^2\alpha }{2\kappa }\int _{0}^{t}(\partial _x v_\varepsilon (s,0))^2ds-\frac{q^2\beta }{\kappa }\int _{0}^{t}\left\| \partial _x v_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\&\quad \le \left\| u_{\varepsilon ,0} \right\| ^2_{L^2(0,\infty )}C +Ct\le C +Ct\le C(T), \end{aligned}$$

which gives (2.7). $\square$

We continue by proving the Lipschitz continuity of $v_\varepsilon$

Lemma 2.4

Given $T>0$. Assume either (1.15) or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le&C(T), \end{aligned}$$

(2.12)

$$\begin{aligned} \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le&C(T), \end{aligned}$$

(2.13)

$$\begin{aligned} \left\| v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le&C(T), \end{aligned}$$

(2.14)

$$\begin{aligned} \left\| v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le&C(T), \end{aligned}$$

(2.15)

for every $0\le t\le T$.

Proof

(Proof assuming (1.15)) Let $0\le t\le T$. We begin by proving that

$$\begin{aligned} 2\beta ^2\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\alpha \beta (\partial _x v_\varepsilon (t,0))^2\le C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) . \end{aligned}$$

(2.16)

Multiplying the third equation of (2.1) by $2\beta \partial _x v_\varepsilon$, we have that

$$\begin{aligned} 2\beta \alpha \partial _x v_\varepsilon \partial _{x}^2v_\varepsilon +2\beta ^2(\partial _x v_\varepsilon )^2+2\beta \gamma v_\varepsilon \partial _x v_\varepsilon =2\beta \kappa u_{\varepsilon }^2\partial _x v_\varepsilon . \end{aligned}$$

(2.17)

Observe that

$$\begin{aligned} \begin{aligned} 2\beta \alpha \int _{0}^{\infty }\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon dx =&\, -\beta \alpha (\partial _x v_\varepsilon (t,0))^2,\\ 2\beta \gamma \int _{0}^{\infty }v_\varepsilon \partial _x v_\varepsilon dx=&\, -\beta \gamma v_\varepsilon ^2(t,0)=-\beta \gamma h_{\varepsilon }^2(t). \end{aligned} \end{aligned}$$

(2.18)

Thanks to (1.15) and (2.18), an integration of (2.17) on $(0,\infty )$ gives

$$\begin{aligned} 2\beta ^2\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\alpha \beta (\partial _x v_\varepsilon (t,0))^2=2\beta \kappa \int _{0}^{\infty }u_{\varepsilon }^2\partial _x v_\varepsilon dx +\beta \gamma h_{\varepsilon }^2(t). \end{aligned}$$

(2.19)

Since, using Lemma 2.3,

$$\begin{aligned}&2\vert \beta \kappa \vert \int _{0}^{\infty }u_{\varepsilon }^2\vert \partial _x v_\varepsilon \vert dx +\vert \beta \gamma \vert h_{\varepsilon }^2(t)\nonumber \\&\qquad \le 2\vert \beta \kappa \vert \left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \vert \beta \gamma \vert h_{\varepsilon }^2(t)\\&\qquad \le C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) , \nonumber \end{aligned}$$

(2.20)

(2.16) follows from (2.19) and (2.20).

We prove that for every $t\in [0,T]$

$$\begin{aligned} \left\| v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le&C(T)\sqrt{1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}}, \end{aligned}$$

(2.21)

$$\begin{aligned} \left\| v_\varepsilon (t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le&C(T)\sqrt{1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}}. \end{aligned}$$

(2.22)

Multiplying the third equation of (2.1) by $2\gamma v_\varepsilon$, we get

$$\begin{aligned} 2\gamma \alpha v_\varepsilon \partial _{x}^2v_\varepsilon +2\gamma \beta v_\varepsilon \partial _x v_\varepsilon + 2\gamma ^2v_\varepsilon ^2=2\gamma \kappa u_{\varepsilon }^2v_\varepsilon . \end{aligned}$$

(2.23)

Observe that, thanks to (2.1),

$$\begin{aligned} \begin{aligned} 2\gamma \alpha \int _{0}^{\infty }v_\varepsilon \partial _{x}^2v_\varepsilon dx=&\, -2\gamma \alpha v_\varepsilon (t,0)\partial _x v_\varepsilon (t,0) -2\gamma \alpha \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ =&\, -2\gamma \alpha h_{\varepsilon }(t)\partial _x v_\varepsilon (t,0) -2\gamma \alpha \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\ 2\gamma \beta \int _{0}^{\infty }v_\varepsilon \partial _x v_\varepsilon dx=&\, -\gamma \beta v_\varepsilon ^2(t,0)=-\gamma \beta h_{\varepsilon }^2(t). \end{aligned} \end{aligned}$$

(2.24)

Therefore, integrating (2.23) on $(0,\infty )$, thanks to (2.24),

$$\begin{aligned} \begin{aligned} 2\gamma ^2\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}=&\, 2\gamma \kappa \int _{0}^{\infty }u_{\varepsilon }^2v_\varepsilon dx +\gamma \beta h_{\varepsilon }^2(t)\\&+2\gamma \alpha h_{\varepsilon }(t)\partial _x v_\varepsilon (t,0)+2\gamma \alpha \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned} \end{aligned}$$

(2.25)

Due to (2.2), (2.7) and the Young’s inequality,

$$\begin{aligned} 2\vert \gamma \kappa \vert \int _{0}^{\infty }u_{\varepsilon }^2\vert v_\varepsilon \vert dx\le&2\vert \gamma \kappa \vert \left\| v_\varepsilon (t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ \le&C(T)\left\| v_\varepsilon (t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le \frac{C(T)}{D_1}+D_1\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )},\\ \gamma \beta h_{\varepsilon }^2(t)+2\vert \gamma \alpha \vert \vert h_{\varepsilon }(t)\vert \vert \partial _x v_\varepsilon (t,0)\vert \le&C+2C|\partial _x v_\varepsilon (t,0)|\le C +(\partial _x v_\varepsilon (t,0))^2, \end{aligned}$$

where $D_1$ is an arbitrary positive constant, which will be specified later. It follows from (2.25) that

$$\begin{aligned} \begin{aligned} 2\gamma ^2\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le&\frac{C(T)}{D_1}+D_1\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\\&+C +(\partial _x v_\varepsilon (t,0))^2 +2\vert \gamma \alpha \vert \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned} \end{aligned}$$

(2.26)

Due to (2.2) and the Young’s inequality,

$$\begin{aligned} v_\varepsilon ^2(t,x)=&\, 2\int _{0}^{x}v_\varepsilon \partial _x v_\varepsilon dy +h_{\varepsilon }^2(t)\le 2\int _{0}^{\infty }\vert v_\varepsilon \vert \vert \partial _x v_\varepsilon \vert dx +C\\ \le&\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ C. \end{aligned}$$

Hence,

$$\begin{aligned} \left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\le \left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ C. \end{aligned}$$

(2.27)

Since $D_1>0$, (2.26) implies

$$\begin{aligned} 2\gamma ^2\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le&\frac{C(T)}{D_1}+D_1\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+\left( 1+\frac{1}{D_1}\right) C +(\partial _x v_\varepsilon (t,0))^2+\left( 2\vert \gamma \alpha \vert +\frac{1}{D_1}\right) \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

that is

$$\begin{aligned} \left( 2\gamma ^2-D_1\right) \left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le&\; \frac{C(T)}{D_1}+\left( 1+\frac{1}{D_1}\right) C \\&+\left( 1+\frac{1}{D_1}\right) C +(\partial _x v_\varepsilon (t,0))^2\\&+\left( 2\vert \gamma \alpha \vert +\frac{1}{D_1}\right) \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Choosing $D_1=\gamma ^2$, we have that

$$\begin{aligned} \gamma ^2\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T)\left( 1+(\partial _x v_\varepsilon (t,0))^2+\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) . \end{aligned}$$

(2.28)

(2.21) follows from (2.16) and (2.28), while (2.16), (2.21) and (2.27) give (2.22).

We prove (2.12). Multiplying the third equation of (2.1) by $2\alpha \partial _x v_\varepsilon$, we have

$$\begin{aligned} 2\alpha ^2\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon +2\alpha \beta (\partial _x v_\varepsilon )^2 +2\alpha \gamma v_\varepsilon \partial _x v_\varepsilon =2\alpha \kappa u_{\varepsilon }^2\partial _x v_\varepsilon . \end{aligned}$$

(2.29)

Observe that,

$$\begin{aligned} \begin{aligned} 2\alpha ^2\int _{0}^{x}\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon dy=&\, \alpha ^2(\partial _x v_\varepsilon (t,x))^2 -\alpha ^2(\partial _x v_\varepsilon (t,0))^2,\\ 2\alpha \gamma \int _{0}^{x}v_\varepsilon \partial _x v_\varepsilon dy=&\, \alpha \gamma v_\varepsilon ^2(t,x)-\alpha \gamma v_\varepsilon ^2(t,0)=\alpha \gamma v_\varepsilon ^2(t,x)-\alpha \gamma h_{\varepsilon }^2(t). \end{aligned} \end{aligned}$$

(2.30)

Integrating (2.29) on (0, x), thanks to (2.2), (2.7) and (2.30), we have

$$\begin{aligned} \alpha ^2(\partial _x v_\varepsilon (t,x))^2=&\; 2\alpha \kappa \int _{0}^{x}u_{\varepsilon }^2\partial _x v_\varepsilon dx +\alpha ^2(\partial _x v_\varepsilon (t,0))^2-2\alpha \beta \int _{0}^{x}(\partial _x v_\varepsilon )^2 dy \\&-\alpha \gamma v_\varepsilon ^2(t,x)-\alpha \gamma h_{\varepsilon }^2(t)\\ \le&\;2\vert \alpha \kappa \vert \int _{0}^{\infty }u_{\varepsilon }^2\vert \partial _x v_\varepsilon \vert dx +\alpha ^2(\partial _x v_\varepsilon (t,0))^2+2\vert \alpha \beta \vert \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+\vert \alpha \gamma \vert \left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )} +\vert \alpha \gamma \vert h_{\varepsilon }^2(t)\\ \le&\;2\vert \alpha \kappa \vert \left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +2\alpha ^2(\partial _x v_\varepsilon (t,0))^2\\&+2\vert \alpha \beta \vert \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\vert \alpha \gamma \vert \left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )} +C\\ \le&\;C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) +2\alpha ^2(\partial _x v_\varepsilon (t,0))^2\\&+2\vert \alpha \beta \vert \left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\vert \alpha \gamma \vert \left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}. \end{aligned}$$

Consequently, by (2.16) and (2.22), we have that

$$\begin{aligned} \alpha ^2(\partial _x v_\varepsilon (t,x))^2\le C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) . \end{aligned}$$

(2.31)

Hence,

$$\begin{aligned} \alpha ^2\left\| \partial _x v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}-C(T)\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}-C(T)\le 0, \end{aligned}$$

(2.32)

which gives (2.12).

Finally, (2.13), (2.14) and (2.15) follows from (2.12), (2.13), (2.14) and (2.15), respectively. $\square$

Proof

(Proof assuming (1.16)) Let $0\le t\le T$. We begin by observing that, thanks to (1.16), the third equation of (2.1) reads

$$\begin{aligned} \alpha \partial _{x}^2v_\varepsilon +\gamma v_\varepsilon =\kappa ^2u_{\varepsilon }^2. \end{aligned}$$

(2.33)

Following [18], or [33], in order to work with homogeneous boundary conditions we define

$$\begin{aligned} W_1(t,x)=v_\varepsilon (t,x)-h_{\varepsilon }(t)e^{-x}, \quad x\ge 0. \end{aligned}$$

(2.34)

We have that

$$\begin{aligned} \begin{aligned} \partial _x W_1(t,x)=&\, \partial _x v_\varepsilon (t,x) +h_{\varepsilon }(t)e^{-x},\\ \partial _{x}^2W_1(t,x)=&\, \partial _{x}^2v_\varepsilon (t,x)-h_{\varepsilon }(t)e^{-x}, \end{aligned} \end{aligned}$$

(2.35)

and

$$\begin{aligned} W_1(t,0)=v_\varepsilon (t,0)-h_{\varepsilon }(t)=0. \end{aligned}$$

(2.36)

Due to (2.34) and (2.35), (2.33) is equivalent to the following one:

$$\begin{aligned} \alpha \partial _{x}^2W_1 +\gamma W_1=\kappa ^2u_{\varepsilon }^2 +\alpha h_{\varepsilon }(t)e^{-x}+\gamma h_{\varepsilon }(t)e^{-x}. \end{aligned}$$

(2.37)

Multiplying (2.37) by $2\gamma W_1$, we have that

$$\begin{aligned} 2\alpha \gamma W_1\partial _{x}^2W_1 +2\gamma ^2 W_1^2=2\gamma \kappa ^{{2}}u_{\varepsilon }^2 W_1 +2\alpha \gamma h_{\varepsilon }(t)e^{-x} W_1 +2\gamma ^2h_{\varepsilon }(t)e^{-x} W_1. \end{aligned}$$

(2.38)

Observe that, thanks to (2.36),

$$\begin{aligned} 2\alpha \gamma \int _{0}^{\infty } W_1\partial _{x}^2W_1 dx=-2\alpha \gamma \left\| \partial _x W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

(2.39)

Consequently, integrating (2.38) on $(0,\infty )$, by (1.16) and (2.39), we have that

$$\begin{aligned} \begin{aligned}&-2\alpha \gamma \left\| \partial _x W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\gamma ^2\left\| W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad =2\gamma \kappa \int _{0}^{\infty }u_{\varepsilon }^2 W_1 dx + 2\alpha \gamma h_{\varepsilon }(t)\int _{0}^{\infty }e^{-x} W_1 dx+2\gamma ^2h_{\varepsilon }(t)\int _{0}^{\infty }e^{-x} W_1 dx. \end{aligned} \end{aligned}$$

(2.40)

Observe that

$$\begin{aligned} \int _{0}^{\infty }e^{-2x} dx=\frac{1}{2}. \end{aligned}$$

(2.41)

Thanks to (2.2), (2.41), the fact that $\gamma \not =0$, and the Young’s inequality,

$$\begin{aligned} 2\vert \alpha \gamma \vert \vert h_{\varepsilon }(t)\vert \int _{0}^{\infty }e^{-x}\vert W_1\vert dx\le&C\int _{0}^{\infty }e^{-x}\vert W_1\vert dx\\ =&\int _{0}^{\infty }\left| \frac{Ce^{-x}}{\gamma }\right| \left| \gamma W_1\right| dx\le C+\frac{\gamma ^2}{2}\left\| W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\ 2\gamma ^2\vert h_{\varepsilon }(t)\vert \int _{0}^{\infty }e^{-x} \vert W_1\vert dx\le&C\gamma ^2\int _{0}^{\infty }e^{-x} \vert W_1\vert dx\\ \le&C+\frac{\gamma ^2}{2}\left\| W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

The $L^2$ estimate stated in (2.7) and (2.40) gives

$$\begin{aligned} \begin{aligned} -\alpha \gamma \left\| \partial _x W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\gamma ^2\left\| W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ \le&\; \vert \gamma \kappa \vert \int _{0}^{\infty }u_{\varepsilon }^2 \vert W_1\vert dx+C\\ \le&\;\vert \gamma \kappa \vert \left\| W_1 \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C\\ \le&\;C(T)\left( 1+\left\| W_1 \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) . \end{aligned} \end{aligned}$$

(2.42)

We prove that

$$\begin{aligned} \left\| W_1 \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le C(T). \end{aligned}$$

(2.43)

Due to (2.36), (2.42) and the Hölder inequality,

$$\begin{aligned} W_1^2(t,x)=&\, 2\int _{0}^{x}W_1\partial _x W_1 dy\le 2\int _{0}^{\infty }\vert W_1\vert \vert \partial _x W_1\vert dx\\ \le&\;2\left\| W_1(t,\cdot ) \right\| _{L^2(0,\infty )}\left\| \partial _x W_1(t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T)\left( 1+\left\| W_1 \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) , \end{aligned}$$

for every $x\ge 0$. Hence,

$$\begin{aligned} \left\| W_1 \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}-C(T)\left\| W_1 \right\| _{L^{\infty }((0,T)\times (0,\infty ))}-C(T)\le 0, \end{aligned}$$

which gives (2.43).

Observe that, by (2.42) and (2.43), we have that

$$\begin{aligned} -\alpha \gamma \left\| \partial _x W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| W_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T). \end{aligned}$$

(2.44)

We prove (2.15). Thanks to (2.2) and (2.34),

$$\begin{aligned} \vert v_\varepsilon (t,x)\vert \le&\left| W_1(t,x) +h_{\varepsilon }(t)e^{-x}\right| \le \vert W_1(t,x)\vert +\vert h_{\varepsilon }(t) e^{-x}\vert \\ \le&\left\| W_1 \right\| _{L^{\infty }((0,T)\times (0,\infty ))}+C\le C(T). \end{aligned}$$

Therefore,

$$\begin{aligned} \left\| v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le C(T), \end{aligned}$$

that is (2.15).

We prove (2.14). Thanks to (2.2), (2.34) and the Young’s inequality,

$$\begin{aligned} v_\varepsilon ^2(t,x)=\left( W_1(t,x)+h_{\varepsilon }(t)e^{-x}\right) ^2\le 2W_1^2(t,x)+2h_{\varepsilon }^2(t)e^{-2x}\le 2W_1^2(t,x) +Ce^{-2x}. \end{aligned}$$

(2.45)

Integrating (2.45) on $(0,\infty )$, by (2.41) and (2.44), we have (2.14).

In a similar way, thanks to (2.2), (2.35), (2.44) and the Young’s inequality, we have (2.13).

Finally, we prove (2.12). We begin by proving that

$$\begin{aligned} \alpha ^2(\partial _x v_\varepsilon (t,0)^2)\le C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) . \end{aligned}$$

(2.46)

Multiplying (2.33) by $-2\alpha \partial _x v_\varepsilon$, we have that

$$\begin{aligned} -2\alpha ^2\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon -2\alpha \gamma v_\varepsilon \partial _x v_\varepsilon =-2\alpha \kappa ^2u_{\varepsilon }^2\partial _x v_\varepsilon . \end{aligned}$$

(2.47)

Observe that, thanks to (1.8), we have that

$$\begin{aligned} \begin{aligned} -2\alpha ^2\int _{0}^{\infty }\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon dx=&\, \alpha ^2(\partial _x v_\varepsilon (t,0))^2,\\ -2\alpha \gamma \int _{0}^{\infty }v_\varepsilon \partial _x v_\varepsilon dx=&\, \alpha \gamma v_\varepsilon ^2(t,0)=\alpha \gamma h_{\varepsilon }^2(t,0). \end{aligned} \end{aligned}$$

(2.48)

Therefore, integrating (2.47) on $(0,\infty )$, by (1.16), (2.2), (2.7) and (2.48), we get

$$\begin{aligned} \alpha ^2(\partial _x v_\varepsilon (t,0))^2=&\, -\alpha \kappa \int _{0}^{\infty }u_{\varepsilon }^2\partial _x v_\varepsilon dx -\alpha \gamma h_{\varepsilon }^2(t,0)\\ \le&\;\vert \alpha \kappa \vert \int _{0}^{\infty }u_{\varepsilon }^2\vert \partial _x v_\varepsilon \vert dx+C\\ \le&\;\vert \alpha \kappa \vert \left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +C\\ \le&\;C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) , \end{aligned}$$

which gives (2.46).

Now, we prove (2.12). Multiplying (2.33) by $2\alpha \partial _x v_\varepsilon$, thanks to (2.1), an integration on (0, x) gives

$$\begin{aligned} \alpha ^2(\partial _x v_\varepsilon (t,x))^2=2\alpha \kappa ^2\int _{0}^{x}u_{\varepsilon }^2\partial _x v_\varepsilon dx +\alpha ^2(\partial _x v_\varepsilon (t,0))^2-\alpha \gamma v_\varepsilon ^2(t,x)+\alpha \gamma h_{\varepsilon }^2(t,0). \end{aligned}$$

Therefore, by (2.2), (2.7), (2.14) and (2.46),

$$\begin{aligned} \alpha ^2(\partial _x v_\varepsilon (t,x))^2\le&\; 2\vert \alpha \kappa \vert \int _{0}^{x}u_{\varepsilon }^2\vert \partial _x v_\varepsilon \vert dx+\alpha ^2(\partial _x v_\varepsilon (t,0))^2\\&+\vert \alpha \gamma \vert v_\varepsilon ^2(t,x)+\vert \alpha \gamma \vert h_{\varepsilon }^2(t,0)\\ \le&\;2\vert \alpha \kappa \vert \int _{0}^{\infty }u_{\varepsilon }^2\vert \partial _x v_\varepsilon \vert dx+\alpha ^2(\partial _x v_\varepsilon (t,0))^2\\&+\vert \alpha \gamma \vert v_\varepsilon ^2(t,x)+\vert \alpha \gamma \vert h_{\varepsilon }^2(t,0)\\ \le&\; 2\vert \alpha \kappa \vert \left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) +\vert \alpha \gamma \vert \left\| v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\\ \le&\; C(T)\left( 1+\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\right) . \end{aligned}$$

Therefore, we have (2.32), which gives (2.12). $\square$

Lemma 2.5

Assume (1.15), or (1.16). Define

$$\begin{aligned} F_\varepsilon (t,x)=\int _{0}^{x}P_\varepsilon (t,y) dy,\qquad t,\,x\ge 0. \end{aligned}$$

(2.49)

Then for every $t\ge 0$ we have

$$\begin{aligned} bF_\varepsilon (t,\infty )&=b\int _{0}^{\infty }P_\varepsilon (t,x) dx=q^2u_{\varepsilon }(t,0)h_{\varepsilon }(t)+\varepsilon g_{\varepsilon }(t),\end{aligned}$$

(2.50)

$$\begin{aligned} F_\varepsilon ^2(t,\infty )&\le Cu_{\varepsilon }^2(t,0) +C, \end{aligned}$$

(2.51)

where

$$\begin{aligned} \qquad F_\varepsilon (t,\infty )=\lim _{x\rightarrow \infty }F_\varepsilon (t,x). \end{aligned}$$

Proof

Integrating the first equation of (2.1) on (0, x), thanks to (2.1) and (2.49), we have that

$$\begin{aligned} \begin{aligned} bF_\varepsilon (t,x)=&\, \int _{0}^{x}\partial _tu_{\varepsilon }(t,y)dy-q^2u_{\varepsilon }(t,x)v_\varepsilon (t,x)+q^2u_{\varepsilon }(t,0)h_{\varepsilon }(t)\\&-\varepsilon \partial _x u_{\varepsilon }(t,x)+\varepsilon g_{\varepsilon }(t), \end{aligned} \end{aligned}$$

(2.52)

for every $x\ge 0$. Observe that, differentiating (2.4) with respect to t, we have that

$$\begin{aligned} \frac{d}{dt}\int _{0}^{\infty }u_{\varepsilon }(t,x) dx=\int _{0}^{\infty }\partial _tu_{\varepsilon }(t,0)=0, \end{aligned}$$

(2.53)

therefore (2.50) follows from (2.52) and the fact that $u_{\varepsilon },\,v_\varepsilon ,\,\partial _x v_\varepsilon$ vanish at infinity.

Finally, we prove (2.51). Since $0<\varepsilon <1$, by (2.2) and the Young’s inequality,

$$\begin{aligned} b^2F_\varepsilon ^2(t,\infty )=&\, \left( q^2u_{\varepsilon }(t,0)h_{\varepsilon }(t)+\varepsilon \ge (t)\right) ^2\le 2q^4u_{\varepsilon }^2(t,0)h_{\varepsilon }^2(t)+2\varepsilon ^2g_{\varepsilon }^2(t)\\ \le&Cu_{\varepsilon }^2(t,0)+2g_{\varepsilon }^2(t)\le Cu_{\varepsilon }^2(t,0)+C, \end{aligned}$$

which gives (2.51). $\square$

Lemma 2.6

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2e^{C(T)t}\varepsilon \int _{0}^{t}e^{-C(T)s}\left\| u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le C(T), \end{aligned}$$

(2.54)

for every $0\le t\le T$. Moreover, we have that

$$\begin{aligned} \left\| P_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le C(T). \end{aligned}$$

(2.55)

Proof

Let $0\le t\le T$ and $x\ge 0$. We begin by observing that, differentiating (2.5) with respect to t, we have that

$$\begin{aligned} \partial _tP_\varepsilon (t,x)=\frac{d}{dt}\int _{0}^{x}u_{\varepsilon }(t,y)dy=\int _{0}^{x}\partial _tu_{\varepsilon }(t,y) dy. \end{aligned}$$

(2.56)

It follows from (2.52) and (2.56) that

$$\begin{aligned} \partial _tP_\varepsilon -q^2u_{\varepsilon }v_\varepsilon +q^2u_{\varepsilon }(t,0)h_{\varepsilon }(t)-\varepsilon \partial _x u_{\varepsilon }(t,x)+\varepsilon \ge (t)=bF_\varepsilon . \end{aligned}$$

(2.57)

Observe that, thanks to (2.1) and (2.49),

$$\begin{aligned} \begin{aligned} -2q^2\int _{0}^{\infty }P_\varepsilon u_{\varepsilon }v_\varepsilon dx=&\, -2q^2\int _{0}^{\infty }P_\varepsilon \partial _x P_\varepsilon v_\varepsilon dx=q^2\int _{0}^{\infty }\partial _x v_\varepsilon P_\varepsilon ^2 dx,\\ -2\varepsilon \int _{0}^{\infty }P_\varepsilon \partial _x u_{\varepsilon }(t,x)=&\, 2\varepsilon \int _{0}^{\infty }\partial _x P_\varepsilon u_{\varepsilon }dx=2\varepsilon \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\ 2b\int _{0}^{\infty }F_\varepsilon P_\varepsilon dx=&\, 2b\int _{0}^{\infty }F_\varepsilon \partial _x F_\varepsilon dx=bF_\varepsilon ^2(t,\infty ). \end{aligned} \end{aligned}$$

(2.58)

Therefore, thanks to (2.49), (2.50) and (2.58), multiplying (2.57) by $2P_\varepsilon$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad =-q^2\int _{0}^{\infty }\partial _x v_\varepsilon P_\varepsilon ^2 dx+bF_\varepsilon ^2(t,\infty )\\&\qquad \quad -q^2u_{\varepsilon }(t,0)h_{\varepsilon }(t)F_\varepsilon (t,\infty ) -\varepsilon g_{\varepsilon }(t)F_\varepsilon (t,\infty ). \end{aligned} \end{aligned}$$

(2.59)

Since $0<\varepsilon <1$, due to Lemma 2.4, (2.2), (2.51) and the Young’s inequality,

$$\begin{aligned} q^2\int _{0}^{\infty }\vert \partial _x v_\varepsilon \vert P_\varepsilon ^2 dx\le&q^2\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\le C(T)\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\ \vert b\vert F_\varepsilon ^2(t,\infty )\le&Cu_{\varepsilon }^2(t,0) +C,\\ q^2\vert u_{\varepsilon }(t,0)\vert \vert h_{\varepsilon }(t)\vert \vert F_\varepsilon (t,\infty )\vert \le&C\vert u_{\varepsilon }(t,0)\vert \vert F_\varepsilon (t,\infty )\vert \le Cu_{\varepsilon }^2(t,0) +CF_\varepsilon ^2(t,\infty )\\ \le&Cu_{\varepsilon }^2(t,0) +C,\\ \varepsilon \vert \ge (t)\vert \vert F_\varepsilon (t,\infty )\vert \le&C\vert F_\varepsilon (t,\infty )\vert \le Cu_{\varepsilon }^2(t,0) +C. \end{aligned}$$

Consequently, by (2.59),

$$\begin{aligned}&\frac{d}{dt}\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T)\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ Cu_{\varepsilon }^2(t,0) +C. \end{aligned}$$

The Gronwall Lemma and (2.7) give

$$\begin{aligned} \left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+2e^{C(T)t}\varepsilon \int _{0}^{t}e^{-C(T)s}\left\| u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\ \le&C +Ce^{C(T)t}\int _{0}^{t}e^{-C(T)s}u_{\varepsilon }^2(s,0)ds \\ \le&C(T) + C(T)\int _{0}^{t}u_{\varepsilon }^2(s,0)ds\le C(T), \end{aligned}$$

that is (2.54).

Finally, we prove (2.55). Due to (2.1), (2.7), (2.54) and the Hölder inequality, for every $x\ge 0$

$$\begin{aligned} P_\varepsilon ^2(t,x)=2\int _{0}^{x}P_\varepsilon \partial _x P_\varepsilon dy=&\, 2\int _{0}^{x}P_\varepsilon u_{\varepsilon }dy\le 2\int _{0}^{\infty }\vert P_\varepsilon \vert \vert u_{\varepsilon }\vert dx\\ \le&\; 2\left\| P_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\left\| u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T). \end{aligned}$$

Hence,

$$\begin{aligned} \left\| P_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\le C(T), \end{aligned}$$

which gives (2.55). $\square$

Lemma 2.7

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| u_{\varepsilon }(t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le&\; C(T)\sqrt{\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}}, \end{aligned}$$

(2.60)

$$\begin{aligned} \left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le&\; C(T)\left( 1+\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\right) , \end{aligned}$$

(2.61)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. We prove (2.60). We begin by observing that, by (2.7) and the Hölder inequality,

$$\begin{aligned} u_{\varepsilon }^2(t,x)=&\, 2\int _{0}^{x}u_{\varepsilon }\partial _x u_{\varepsilon }dy\le 2\int _{0}^{\infty }\vert u_{\varepsilon }\vert \vert \partial _x u_{\varepsilon }\vert dx\\ \le&\;2\left\| u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T)\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}. \end{aligned}$$

Hence,

$$\begin{aligned} \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\le C(T)\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}, \end{aligned}$$

which gives (2.60).

Finally, we prove (2.61). By the third equation of (2.1) and Lemma 2.4, we have that

$$\begin{aligned} \vert \alpha \vert \vert \partial _{x}^2v_\varepsilon \vert =&\, \left| \kappa u_{\varepsilon }^2 -\beta \partial _x v_\varepsilon -\gamma v_\varepsilon \right| \\ \le&\;\vert \kappa \vert u_{\varepsilon }^2 +\vert \beta \vert \vert \partial _x v_\varepsilon \vert +\vert \gamma \vert \vert v_\varepsilon \vert \\ \le&\;\vert \kappa \vert \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )} +\vert \beta \vert \left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times \mathbb {R})}+\gamma \left\| v_\varepsilon \right\| _{L^{\infty }((0,T)\times \mathbb {R})}\\ \le&\;C(T)\left( 1+\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\right) . \end{aligned}$$

Therefore, since $\alpha \not =0$ (see (1.14)),

$$\begin{aligned} \left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le C(T)\left( 1+\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\right) . \end{aligned}$$

(2.62)

(2.61) follows from (2.60) and (2.62). $\square$

Following [18], or [33], in order to work with homogeneous boundary conditions we define

$$\begin{aligned} W_2(t,x)=u_{\varepsilon }(t,x)+g_{\varepsilon }(t)e^{-x}, \quad t,\,x\ge 0. \end{aligned}$$

(2.63)

Thanks to (2.63), we have that

$$\begin{aligned} \partial _tW_2(t,x)=&\, \partial _tu_{\varepsilon }(t,x)+g_{\varepsilon }'(t)e^{-x},\nonumber \\ \partial _x W_2(t,x)=&\, \partial _x u_{\varepsilon }(t,x)-g_{\varepsilon }(t)e^{-x},\\ \partial _{x}^2W_2(t,x)=&\, \partial _{x}^2u_{\varepsilon }(t,x)+g_{\varepsilon }(t)e^{-x}.\nonumber \end{aligned}$$

(2.64)

Moreover, by (2.1), (2.2) and (2.64),

$$\begin{aligned} \partial _x W_2(t,0)=\partial _x u_{\varepsilon }(t,0)-g_{\varepsilon }(t)=0, \quad \left\| W_{2}(0,\cdot ) \right\| _{L^2(0,\infty )}\le C. \end{aligned}$$

(2.65)

Thanks to (2.63) and (2.64), the first equation of (2.1) reads

$$\begin{aligned} \begin{aligned}&\partial _tW_2 -q^2v_\varepsilon \partial _x W_2-q^2W_2\partial _x v_\varepsilon \\&\qquad =bP_\varepsilon +\varepsilon \partial _{x}^2W_2-\varepsilon g_{\varepsilon }(t)e^{-x} +g_{\varepsilon }'(t)e^{-x}+q^2g_{\varepsilon }(t)e^{-x}v_\varepsilon -q^2g_{\varepsilon }(t)e^{-x}\partial _x v_\varepsilon . \end{aligned} \end{aligned}$$

(2.66)

We prove the following lemma.

Lemma 2.8

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| W_2(t,\cdot ) \right\| _{L^2(0,\infty )}\le&\;C(T), \end{aligned}$$

(2.67)

$$\begin{aligned} \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le&\; C\left( \left\| \partial _x W_2(t,\cdot ) \right\| _{L^2(0,\infty )}+1\right) , \end{aligned}$$

(2.68)

$$\begin{aligned} \left\| u_{\varepsilon }(t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le&\;C(T)\sqrt{1+\left\| \partial _x W_2(t,\cdot ) \right\| _{L^2(0,\infty )}}, \end{aligned}$$

(2.69)

$$\begin{aligned} \left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| _{L^{\infty }(0,\infty )}\le&\; C(T)\left( 1+\left\| \partial _x W_2(t,\cdot ) \right\| _{L^2(0,\infty )}\right) , \end{aligned}$$

(2.70)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$ and $x\ge 0$. We prove (2.67). We begin by observing that, by (2.2), (2.63) and the Young’s inequality,

$$\begin{aligned} W_2^2(t,x)=\left( u_{\varepsilon }(t,x)+g_{\varepsilon }(t)e^{-x}\right) ^2\le u_{\varepsilon }^2(t,x)+4g_{\varepsilon }^2(t)e^{-2x}\le u_{\varepsilon }^2(t,x)+Ce^{-2x}. \end{aligned}$$

Integrating on $(0,\infty )$, by (2.7), we get

$$\begin{aligned} \left\| W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le \left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C\int _{0}^{\infty }e^{-2x} dx\le C(T)+C\le C(T), \end{aligned}$$

which gives (2.67).

We prove (2.68). By (2.2), (2.64) and the Young’s inequality,

$$\begin{aligned} (\partial _x u_{\varepsilon }(t,x))^2=&\, \left( \partial _x W_2(t,x)+g_{\varepsilon }(t)e^{-x}\right) ^2\le (\partial _x W_2(t,x))^2+g_{\varepsilon }^2(t)e^{-2x}\\ \le&\; (\partial _x W_2(t,x))^2+Ce^{-2x}. \end{aligned}$$

(2.41) and an integration on $(0,\infty )$ gives

$$\begin{aligned} \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C\left( \left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+1\right) \end{aligned}$$

Therefore, we get

$$\begin{aligned} \left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le C\left( \left\| \partial _x W_2(t,\cdot ) \right\| _{L^2(0,\infty )}+1\right) , \end{aligned}$$

which gives (2.68).

Finally, (2.69) follows from (2.60) and (2.68), while (2.61) and (2.68) give (2.70). $\square$

Lemma 2.9

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned}&\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^2W_2(s,\cdot ) \right\| ^2_{L^2(0,\infty )} ds\le C(T), \end{aligned}$$

(2.71)

$$\begin{aligned}&\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )},\,\left\| u_{\varepsilon } \right\| _{L^{\infty }((0,T)\times (0,\infty ))},\,\left\| \partial _{x}^2v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le C(T), \end{aligned}$$

(2.72)

$$\begin{aligned}&\varepsilon \int _{0}^{t}\left\| \partial _{x}^2u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le C(T), \end{aligned}$$

(2.73)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. We begin by proving (2.71). Multiplying (2.66) by $-2\partial _{x}^2W_2$, thanks to (2.1) and (2.65), an integration on $(0,\infty )$ gives

$$\begin{aligned} -2\int _{0}^{\infty }\partial _{x}^2W_2\partial _tW_2dx=&\, \frac{d}{dt}\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ =&\, -2q^2\int _{0}^{\infty }v_\varepsilon \partial _x W_2\partial _{x}^2W_2 dx -2q^2\int _{0}^{\infty }W_2\partial _{x}^2W_2\partial _x v_\varepsilon dx\\&-2b\int _{0}^{\infty }P_\varepsilon \partial _{x}^2W_2 dx-2\varepsilon \left\| \partial _{x}^2W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+2\varepsilon g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _{x}^2W_2 dx-2g_{\varepsilon }'(t)\int _{0}^{\infty }e^{-x}\partial _{x}^2W_2 dx\\&-2q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}v_\varepsilon \partial _{x}^2W_2 dx+2q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _x v_\varepsilon \partial _{x}^2W_2 dx\\ =&\, 3q^2\int _{0}^{\infty }\partial _x v_\varepsilon (\partial _x W_2)^2 dx +2q^2\int _{0}^{\infty }W_2\partial _x W_2\partial _{x}^2v_\varepsilon dx \\&+2b\int _{0}^{\infty }u_{\varepsilon }\partial _x W_2 dx -2\varepsilon \left\| \partial _{x}^2W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&{+}2\varepsilon g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _x W_2 dx {-}2g_{\varepsilon }'(t)\int _{0}^{\infty }e^{-x}\partial _x W_2 dx\\&{+}2q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}v_\varepsilon \partial _x W_2 dx+4q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _x v_\varepsilon \partial _x W_2 dx\\&-2q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _{x}^2v_\varepsilon \partial _x W_2 dx. \end{aligned}$$

Therefore, we have that

$$\begin{aligned}&\frac{d}{dt}\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+2\varepsilon \left\| \partial _{x}^2W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\nonumber \\&\qquad = 3q^2\int _{0}^{\infty }\partial _x v_\varepsilon (\partial _x W_2)^2 dx +2q^2\int _{0}^{\infty }W_2\partial _x W_2\partial _{x}^2v_\varepsilon dx\\&\qquad \quad +2b\int _{0}^{\infty }u_{\varepsilon }\partial _x W_2 dx-2\varepsilon g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _x W_2 dx\nonumber \\&\qquad \quad -2g_{\varepsilon }'(t)\int _{0}^{\infty }e^{-x}\partial _x W_2 dx-2q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}v_\varepsilon \partial _x W_2 dx\nonumber \\&\qquad \quad +4q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _x v_\varepsilon \partial _x W_2 dx-2q^2g_{\varepsilon }(t)\int _{0}^{\infty }e^{-x}\partial _{x}^2v_\varepsilon \partial _x W_2 dx.\nonumber \end{aligned}$$

(2.74)

Since $0<\varepsilon <1$, using Lemma 2.4, (2.2), (2.7), (2.41), (2.67), (2.70) and the Young’s inequality, we estimate separaltely each term in (2.74)

$$\begin{aligned}&3q^2\int _{0}^{\infty }\vert \partial _x v_\varepsilon \vert (\partial _x W_2)^2 dx\le 3q^2\left\| \partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2\int _{0}^{\infty }\vert W_2\partial _{x}^2v_\varepsilon \vert \vert \partial _x W_2\vert dx\le \int _{0}^{\infty }W_2^2(\partial _{x}^2v_\varepsilon )^2 dx+q^4\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\left\| W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+q^4\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}+q^4\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C(T),\\&2\vert b\vert \int _{0}^{\infty }\vert u_{\varepsilon }\vert \vert \partial _x W_2\vert dx\le b^2\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+ \left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2(\varepsilon \vert g_{\varepsilon }(t)\vert + \vert g_{\varepsilon }'(t)\vert )\int _{0}^{\infty }e^{-x}\vert \partial _x W_2\vert dx\le 2C \int _{0}^{\infty }e^{-x}\vert \partial _x W_2\vert dx\\&\qquad \le C+C\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2\vert g_{\varepsilon }(t)\vert \int _{0}^{\infty }e^{-x}\vert v_\varepsilon \vert \vert \partial _x W_2\vert dx\le 2C\int _{0}^{\infty }e^{-x}\vert v_\varepsilon \vert \vert \partial _x W_2\vert dx\\&\qquad \le C\int _{0}^{\infty }e^{-2x}v_\varepsilon ^2 dx +C\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C\left\| v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\int _{0}^{\infty }e^{-2x} dx+ C\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+ C\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&4q^2\vert g_{\varepsilon }(t)\vert \int _{0}^{\infty }e^{-x}\vert \partial _x v_\varepsilon \vert \vert \partial _x W_2\vert dx\le 2C\int _{0}^{\infty }e^{-x}\vert \partial _x v_\varepsilon \vert \vert \partial _x W_2\vert \\&\qquad \le C\int _{0}^{\infty }e^{-2x}dx+\int _{0}^{\infty }(\partial _x v_\varepsilon )^2(\partial _x W_2)^2 dx\\&\qquad \le C+\left\| \partial _x v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad \le C+C(T)\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\&2q^2\vert g_{\varepsilon }(t)\vert \int _{0}^{\infty }e^{-x}\vert \partial _{x}^2v_\varepsilon \vert \vert \partial _x W_2\vert dx\le 2C\int _{0}^{\infty }e^{-x}\vert \partial _{x}^2v_\varepsilon \vert \vert \partial _x W_2\vert dx\\&\qquad \le C\int _{0}^{\infty }e^{-2x}(\partial _{x}^2v_\varepsilon )^2 dx +C\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}+C\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C(T). \end{aligned}$$

It follows from (2.74) that

$$\begin{aligned} \frac{d}{dt}\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+2\varepsilon \left\| \partial _{x}^2W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )} \le C(T)\left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C(T). \end{aligned}$$

The Gronwall Lemma and (2.65) give

$$\begin{aligned} \left\| \partial _x W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+2\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^2W_2(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\ \le&C + C(T)e^{C(T)t}\int _{0}^{t}e^{-C(T)s}ds\le C(T), \end{aligned}$$

that is (2.71).

(2.72), follows from (2.68), (2.69), (2.70) and (2.71).

Finally, we prove (2.73). By (2.2), (2.64) and the Young’s inequality, we have that

$$\begin{aligned} \begin{aligned} (\partial _{x}^2u_{\varepsilon }(t,x))^2=&\, \left( \partial _{x}^2W_2(t,x)-g_{\varepsilon }(t)e^{-x}\right) ^2\le 2(\partial _{x}^2W_2(t,x))^2+2g_{\varepsilon }^2(t)e^{-2x}\\ \le&2(\partial _{x}^2W_2(t,x))^2+Ce^{-2x}. \end{aligned} \end{aligned}$$

(2.75)

Multiplying (2.75) by $\varepsilon$ and integrating on $(0,\infty )$, thanks to (2.41), we have that

$$\begin{aligned} \varepsilon \left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le&\; 2\varepsilon \left\| \partial _{x}^2W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +2C\varepsilon \int _{0}^{\infty }e^{-2x} dx\\ \le&\;2\varepsilon \left\| \partial _{x}^2W_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C\varepsilon . \end{aligned}$$

Being $0<\varepsilon <1$, integrating on (0, t), by (2.71), we get

$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _{x}^2u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le&\; 2\varepsilon \int _{0}^{t}\left\| \partial _{x}^2W_2(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds+\varepsilon Ct\\ \le&\;2\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^2W_2(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds+ Ct\\ \le&\;C(T), \end{aligned}$$

which gives (2.73). $\square$

Lemma 2.10

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le&\; C(T), \end{aligned}$$

(2.76)

$$\begin{aligned} \left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le&\; C(T), \end{aligned}$$

(2.77)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. Multiplying the third equation of (2.1) by $2\alpha \partial _{x}^2v_\varepsilon$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned} 2\alpha ^2\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}=&\, 2\kappa \alpha \int _{0}^{\infty }u_{\varepsilon }^2\partial _{x}^2v_\varepsilon dx -2\beta \alpha \int _{0}^{\infty }\partial _x v_\varepsilon \partial _{x}^2v_\varepsilon dx\\&-2\gamma \alpha \int _{0}^{\infty }v_\varepsilon \partial _{x}^2v_\varepsilon dx. \end{aligned} \end{aligned}$$

(2.78)

Due to Lemma 2.4, (2.7), (2.72) and the Young’s inequality, we obtain

$$\begin{aligned}&2\vert \kappa \vert \vert \alpha \vert \int _{0}^{\infty }u_{\varepsilon }^2\vert \partial _{x}^2v_\varepsilon \vert dx\le 2\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^4 dx +\frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le 2\kappa ^2\left\| u_{\varepsilon } \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +\frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T) + \frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert \beta \vert \vert \alpha \vert \int _{0}^{\infty }\vert \partial _x v_\varepsilon \vert \vert \partial _{x}^2v_\varepsilon \vert dx\le 2\beta ^2\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+\frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert \gamma \vert \vert \alpha \vert \int _{0}^{\infty }\vert v_\varepsilon \vert \vert \partial _{x}^2v_\varepsilon \vert dx\le 2\gamma ^2\left\| v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+\frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Consequently, by (2.78),

$$\begin{aligned} \frac{\alpha ^2}{2}\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T), \end{aligned}$$

(2.79)

which gives (2.76).

Finally, we prove (2.77). Differentiating the third equation of (2.1) with respect to x, we have that

$$\begin{aligned} \alpha \partial _{x}^3v_\varepsilon +\beta \partial _{x}^2v_\varepsilon +\gamma \partial _x v_\varepsilon =2\kappa u_{\varepsilon }\partial _x u_{\varepsilon }. \end{aligned}$$

(2.80)

Multiplying (2.80) by $2\alpha \partial _{x}^3v_\varepsilon$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned} 2\alpha ^2\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}=&\, 4\alpha \kappa \int _{0}^{\infty }u_{\varepsilon }\partial _x u_{\varepsilon }\partial _{x}^3v_\varepsilon dx -2\alpha \beta \int _{0}^{\infty }\partial _{x}^2v_\varepsilon \partial _{x}^3v_\varepsilon dx\\&-2\alpha \gamma \int _{0}^{\infty }\partial _x v_\varepsilon \partial _{x}^3v_\varepsilon dx. \end{aligned} \end{aligned}$$

(2.81)

Due to Lemma 2.4, (2.72) and the Young’s inequality,

$$\begin{aligned}&4\vert \alpha \vert \vert \kappa \vert \int _{0}^{\infty }\vert u_{\varepsilon }\partial _x u_{\varepsilon }\vert \vert \partial _{x}^3v_\varepsilon \vert dx\le 8\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _x u_{\varepsilon })^2 dx +\frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le 8\kappa ^2\left\| u_{\varepsilon } \right\| ^2_{L^{\infty }((0,\infty )\times (0,\infty ))}\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+\frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert \alpha \vert \vert \beta \vert \int _{0}^{\infty }\vert \partial _{x}^2v_\varepsilon \vert \vert \partial _{x}^3v_\varepsilon \vert dx\le 4\beta ^2\left\| \partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+\frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert \alpha \vert \vert \gamma \vert \int _{0}^{\infty }\vert \partial _x v_\varepsilon \vert \vert \partial _{x}^3v_\varepsilon \vert dx\le 2\gamma ^2\left\| \partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\frac{\alpha ^2}{2}\left\| \partial _{x}^3u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+\frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Therefore, by (2.81),

$$\begin{aligned} \frac{\alpha ^2}{2}\left\| \partial _{x}^3v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T), \end{aligned}$$

which gives (2.77). $\square$

Following [26, Lemma 3.2], or [40, Lemma 2.2], we prove the following $H^2$ estimate on $u_{\varepsilon }$.

Lemma 2.11

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \begin{aligned}&\varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +3\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \quad + \frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _t\partial _x u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds +\varepsilon \int _{0}^{t}\left\| \partial _tu_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le C(T), \end{aligned} \end{aligned}$$

(2.82)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$ and $x\ge 0$. Let A be a positive constant, which will be specified later. Multiplying the first equation of (2.1) by

$$\begin{aligned} -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\end{aligned}$$

we have that

$$\begin{aligned}&\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) \partial _tu_{\varepsilon }-q^2\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) v_\varepsilon \partial _x u_{\varepsilon }\nonumber \\&\qquad \quad -q^2\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) u_{\varepsilon }\partial _x v_\varepsilon -b\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) P_\varepsilon \\&\qquad =\varepsilon \left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) \partial _{x}^2u_{\varepsilon }.\nonumber \end{aligned}$$

(2.83)

Observe that, since the traces of $\partial _t\partial _x u_{\varepsilon }$ and P at $x=0$ are $g_{\varepsilon }'$ and 0 respectively, by (2.1), we have that

$$\begin{aligned}&\int _{0}^{\infty }\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+A\varepsilon \partial _tu_{\varepsilon }\right) \partial _tu_{\varepsilon }dx\nonumber \\&\qquad =2\varepsilon g_{\varepsilon }'(t)\partial _tu_{\varepsilon }(t,0)+2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +2A\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&-q^2\int _{0}^{\infty }\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) v_\varepsilon \partial _x u_{\varepsilon }dx\nonumber \\&\qquad =-q^2\varepsilon h_{\varepsilon }(t)g_{\varepsilon }'(t)g_{\varepsilon }(t)-2q^2\varepsilon \int _{0}^{\infty }\partial _x v_\varepsilon \partial _x u_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx\nonumber \\&\qquad \quad -2q^2\varepsilon \int _{0}^{\infty }v_\varepsilon \partial _{x}^2u_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx -2q^2A\varepsilon \int _{0}^{\infty }v_\varepsilon \partial _x u_{\varepsilon }\partial _tu_{\varepsilon }dx,\nonumber \\&-q^2\int _{0}^{\infty }\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) u_{\varepsilon }\partial _x v_\varepsilon dx\nonumber \\&\qquad =-2q^2\varepsilon u_{\varepsilon }(t,0)\partial _x v_\varepsilon (t,0)g_{\varepsilon }'(t)-2q^2\varepsilon \int _{0}^{\infty }\partial _x u_{\varepsilon }\partial _x v_\varepsilon \partial _t\partial _x u_{\varepsilon }dx\nonumber \\&\qquad \quad -2q^2\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _{x}^2v_\varepsilon \partial _t\partial _x u_{\varepsilon }dx -2q^2A\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _x v_\varepsilon \partial _tu_{\varepsilon }dx,\nonumber \\&-b\int _{0}^{\infty }\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) P_\varepsilon dx\nonumber \\&\qquad =-2b\varepsilon \int _{0}^{\infty }\partial _x P_\varepsilon \partial _t\partial _x u_{\varepsilon }dx -2bA\varepsilon \int _{0}^{\infty }P_\varepsilon \partial _tu_{\varepsilon }dx\nonumber \\&\qquad =-2b\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx-2bA\varepsilon \int _{0}^{\infty }P_\varepsilon \partial _tu_{\varepsilon }dx,\nonumber \\&\varepsilon \int _{0}^{\infty }\left( -2\varepsilon \partial _t\partial _{x}^2u_{\varepsilon }+2A\varepsilon \partial _tu_{\varepsilon }\right) \partial _{x}^2u_{\varepsilon }dx\nonumber \\&\qquad =-\frac{d}{dt}\left( \varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \nonumber \\&\qquad \qquad -2A\varepsilon ^2g_{\varepsilon }(t)\partial _tu_{\varepsilon }(t,0).\nonumber \end{aligned}$$

(2.84)

Therefore, integrating (2.83) on $(0,\infty )$, thanks to (2.84), we have

$$\begin{aligned}&\frac{d}{dt}\left( \varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \nonumber \\&\qquad \quad +2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +2A\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad ={-}2\varepsilon g_{\varepsilon }'(t)\partial _tu_{\varepsilon }(t,0)+{2}q^2\varepsilon h_{\varepsilon }(t)g_{\varepsilon }'(t)g_{\varepsilon }(t)+4q^2\varepsilon \int _{0}^{\infty }\partial _x v_\varepsilon \partial _x u_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx\nonumber \\&\qquad \quad +2q^2\varepsilon \int _{0}^{\infty }v_\varepsilon \partial _{x}^2u_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx +2q^2A\varepsilon \int _{0}^{\infty }v_\varepsilon \partial _x u_{\varepsilon }\partial _tu_{\varepsilon }dx\nonumber \\&\qquad \quad +2q^2\varepsilon u_{\varepsilon }(t,0)\partial _x v_\varepsilon (t,0)g_{\varepsilon }'(t)+2q^2\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _{x}^2v_\varepsilon \partial _t\partial _x u_{\varepsilon }dx\nonumber \\&\qquad \quad +2q^2A\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _x v_\varepsilon \partial _tu_{\varepsilon }dx+2b\varepsilon \int _{0}^{\infty }u_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx\nonumber \\&\qquad \quad +2bA\varepsilon \int _{0}^{\infty }P_\varepsilon \partial _tu_{\varepsilon }dx-2A\varepsilon ^2g_{\varepsilon }(t)\partial _tu_{\varepsilon }(t,0).\nonumber \end{aligned}$$

(2.85)

Since $0< \varepsilon <1$, due to Lemma 2.4, (2.2), (2.7), (2.54), (2.72), and the Young’s inequality, we can estimate all the previous terms in (2.85) as follows.

$$\begin{aligned}&2\varepsilon \vert g_{\varepsilon }'(t)\vert \vert \partial _tu_{\varepsilon }(t,0)\vert \le 2\varepsilon C\vert \partial _tu_{\varepsilon }(t,0)\vert \le C+\varepsilon ^2(\partial _tu_{\varepsilon }(t,0))^2\le C+\varepsilon (\partial _tu_{\varepsilon }(t,0))^2,\\&q^2\varepsilon \vert h_{\varepsilon }(t)\vert \vert g_{\varepsilon }'(t)\vert \vert g_{\varepsilon }(t)\vert \le q^2\vert h_{\varepsilon }(t)\vert \vert g_{\varepsilon }'(t)\vert \vert g_{\varepsilon }(t)\vert \le C,\\&4q^2\varepsilon \int _{0}^{\infty }\vert \partial _x v_\varepsilon \partial _x u_{\varepsilon }\vert \vert \partial _t\partial _x u_{\varepsilon }\vert dx=2\varepsilon \int _{0}^{\infty }\left| \frac{2q^2\partial _x v_\varepsilon \partial _x u_{\varepsilon }}{\sqrt{D_2}}\right| \left| \sqrt{D_2}\partial _t\partial _x u_{\varepsilon }dx\right| dx\\&\qquad \le \frac{4q^4\varepsilon }{D_2}\int _{0}^{\infty }(\partial _x v_\varepsilon )^2(\partial _x u_{\varepsilon })^2 dx +D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{4q^4}{D_2}\left\| \partial _x v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{C(T)}{D_2}+ D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2\varepsilon \int _{0}^{\infty }\vert v_\varepsilon \partial _{x}^2u_{\varepsilon }\vert \vert \partial _t\partial _x u_{\varepsilon }\vert dx=2\varepsilon \int _{0}^{\infty }\left| \frac{q^2v_\varepsilon \partial _{x}^2u_{\varepsilon }}{\sqrt{D_2}}\right| \left| \sqrt{D_2}\partial _t\partial _x u_{\varepsilon }\right| dx\\&\qquad \le \frac{q^4\varepsilon }{D_2}\int _{0}^{\infty }v_\varepsilon ^2(\partial _{x}^2u_{\varepsilon })^2 dx + D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{q^4\varepsilon }{D_2}\left\| v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{C(T)\varepsilon }{D_2}\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2A\varepsilon \int _{0}^{\infty }\vert v_\varepsilon \partial _x u_{\varepsilon }\vert \vert \partial _tu_{\varepsilon }\vert dx=2A\varepsilon \int _{0}^{\infty }\left| \frac{q^2v_\varepsilon \partial _x u_{\varepsilon }}{\sqrt{D_3}}\right| \left| \sqrt{D_3}\partial _tu_{\varepsilon }\right| dx\\&\qquad \le \frac{q^4A\varepsilon }{D_3}\int _{0}^{\infty }v_\varepsilon ^2(\partial _x u_{\varepsilon })^2 dx + AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{q^4A}{D_3}\left\| v_\varepsilon \right\| ^2_{L^{\infty }((0,\infty )\times (0,\infty ))}\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{C(T)A}{D_3} + AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2\varepsilon \vert u_{\varepsilon }(t,0)\vert \vert \partial _x v_\varepsilon (t,0)\vert \vert g_{\varepsilon }'(t)\vert \le C \vert u_{\varepsilon }(t,0)\vert \vert \partial _x v_\varepsilon (t,0)\vert \le Cu_{\varepsilon }^2(t,0) +C(\partial _x v_\varepsilon (t,0))^2\\&\qquad \le C u_{\varepsilon }^2(t,0) +C\left\| \partial _x v_\varepsilon (\cdot ,0) \right\| ^2_{L^{\infty }(0,T)}\le C u_{\varepsilon }^2(t,0)+C(T), \\&2q^2\varepsilon \int _{0}^{\infty }\vert u_{\varepsilon }\partial _{x}^2v_\varepsilon \vert \vert \partial _t\partial _x u_{\varepsilon }\vert dx=2\varepsilon \int _{0}^{\infty }\left| \frac{q^2u_{\varepsilon }\partial _{x}^2v_\varepsilon }{\sqrt{D_2}}\right| \left| \sqrt{D_2}\partial _t\partial _x u_{\varepsilon }\right| dx\\&\qquad \le \frac{q^4\varepsilon }{D_2}\int _{0}^{\infty }u_{\varepsilon }^2(\partial _{x}^2v_\varepsilon )^2 dx +D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{q^4}{D_2}\left\| \partial _{x}^2v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{C(T)}{D_2}+ D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2A\varepsilon \int _{0}^{\infty }\vert u_{\varepsilon }\partial _x v_\varepsilon \vert \vert \partial _tu_{\varepsilon }\vert dx=2A\varepsilon \int _{0}^{\infty }\left| \frac{q^2u_{\varepsilon }\partial _x v_\varepsilon }{\sqrt{D_3}}\right| \left| \sqrt{D_3}\partial _tu_{\varepsilon }\right| dx\\&\qquad \le \frac{q^4A\varepsilon }{D_3}\int _{0}^{\infty }u_{\varepsilon }^2(\partial _x v_\varepsilon )^2 dx + AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{q^4A}{D_3}\left\| \partial _x v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{AC(T)}{D_3} + AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert b\vert \varepsilon \int _{0}^{\infty }\vert u_{\varepsilon }\vert \vert \partial _t\partial _x u_{\varepsilon }\vert dx=2\varepsilon \int _{0}^{\infty }\left| \frac{bu_{\varepsilon }}{\sqrt{D_2}}\right| \left| \sqrt{D_2}\partial _t\partial _x u_{\varepsilon }\right| dx\\&\qquad \le \frac{b^2\varepsilon }{D_2}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{C(T)}{D_2}+D_2\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert b\vert A\varepsilon \int _{0}^{\infty }\vert P_\varepsilon \vert \vert \partial _tu_{\varepsilon }\vert dx=2A\varepsilon \int _{0}^{\infty }\left| \frac{bP_\varepsilon }{\sqrt{D_3}}\right| \left| \sqrt{D_3}\partial _tu_{\varepsilon }\right| dx\\&\qquad \le \frac{Ab^2\varepsilon }{D_3}\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{AC(T)}{D_3}+ AD_3\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2A\varepsilon ^2\vert \ge (t)\vert \vert \partial _tu_{\varepsilon }(t,0)\vert \le 2AC\varepsilon \vert \partial _tu_{\varepsilon }(t,0)\vert \le A^2C\varepsilon +\varepsilon (\partial _tu_{\varepsilon }(t,0))^2\\&\qquad \le AC+\varepsilon (\partial _tu_{\varepsilon }(t,0))^2, \end{aligned}$$

where $D_2,\,D_3$ are two positive constants, which will be specified later. It follows from (2.85) that

$$\begin{aligned}&\frac{d}{dt}\left( \varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \\&\qquad \quad +2\left( 1-2D_3\right) \varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\left( 2-3D_3\right) \varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left( 1+A+\frac{1}{D_2}+\frac{A}{D_3}\right) +Cu_{\varepsilon }^2(t,0) +2\varepsilon (\partial _tu_{\varepsilon }(t,0))^2\\&\qquad \quad + \frac{C(T)\varepsilon }{D_2}\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Choosing $D_2=\frac{1}{4}$ and $D_3=\frac{1}{3}$, we have that

$$\begin{aligned}&\frac{d}{dt}\left( \varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \nonumber \\&\qquad \quad +\varepsilon \left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left( 1+A\right) +Cu_{\varepsilon }^2(t,0) +2\varepsilon (\partial _tu_{\varepsilon }(t,0))^2+ C(T)\varepsilon \left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \nonumber \end{aligned}$$

(2.86)

Observe that, thanks to the Hölder inequality and the Young’s inequality,

$$\begin{aligned} 2\varepsilon (\partial _tu_{\varepsilon }(t,0))^2=&\, -2\varepsilon \int _{0}^{\infty }\partial _tu_{\varepsilon }\partial _t\partial _x u_{\varepsilon }dx \le 2\varepsilon \int _{0}^{\infty }\vert \partial _tu_{\varepsilon }\vert \vert \partial _t\partial _x u_{\varepsilon }\vert dx\\ \le&\;2\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\\ \le&\;2\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\frac{\varepsilon }{2}\left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Consequently, by (2.86),

$$\begin{aligned}&\frac{d}{dt}\left( \varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +A\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \\&\qquad \quad +\frac{\varepsilon }{2}\left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +\left( A-2\right) \varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)\left( 1+A\right) +Cu_{\varepsilon }^2(t,0) + C(T)\varepsilon \left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Taking $A=3$, we have that

$$\begin{aligned}&\frac{d}{dt}\left( \varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +3\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \\&\qquad \quad +\frac{\varepsilon }{2}\left\| \partial _t\partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +\varepsilon \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+Cu_{\varepsilon }^2(t,0) + C(T)\varepsilon \left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Integrating on $(0,\infty )$, by (2.2), (2.7) and (2.73), we get

$$\begin{aligned}&\varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +3\varepsilon ^2\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \quad + \frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _t\partial _x u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds +\varepsilon \int _{0}^{t}\left\| \partial _tu_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\&\qquad \le C +C(T)t +C\int _{0}^{t}u_{\varepsilon }^2(s,0)ds +2C(T)\varepsilon \int _{0}^{t}\left\| \partial _{x}^2u_{\varepsilon }(s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\&\qquad \le C(T), \end{aligned}$$

which gives (2.82). $\square$

In order to prove the compactness of the family $u_{\varepsilon }$ we prove the following estimate on the time derivative of $u_{\varepsilon }$.

Lemma 2.12

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T), \end{aligned}$$

(2.87)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. Multiplying the first equation in (2.1) by $2\partial _tu_{\varepsilon }$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned} 2\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}=&\, 2q^2\int _{0}^{\infty }\partial _x u_{\varepsilon }v_\varepsilon \partial _tu_{\varepsilon }dx +2q^2\int _{0}^{\infty }u_{\varepsilon }\partial _x v_\varepsilon \partial _tu_{\varepsilon }dx\\&+2b\int _{0}^{\infty }P_\varepsilon \partial _tu_{\varepsilon }dx +2\varepsilon \int _{0}^{\infty }\partial _{x}^2u_{\varepsilon }\partial _tu_{\varepsilon }dx. \end{aligned} \end{aligned}$$

(2.88)

Due to Lemma 2.4, (2.7), (2.54), (2.72), (2.82) and the Young’s inequality, we can estimate the previous terms as follows

$$\begin{aligned}&2q^2\int _{0}^{\infty }\vert \partial _x u_{\varepsilon }v_\varepsilon \vert \vert \partial _tu_{\varepsilon }\vert dx\le 2q^4\int _{0}^{\infty }(\partial _x u_{\varepsilon })^2v_\varepsilon ^2 dx +\frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le 2q^4 \left\| v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _x u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+ \frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2\int _{0}^{\infty }\vert u_{\varepsilon }\partial _x v_\varepsilon \vert \vert \partial _tu_{\varepsilon }\vert dx\le 2q^4\int _{0}^{\infty }u_{\varepsilon }^2(\partial _x v_\varepsilon )^2 dx + \frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le 2q^4\left\| \partial _x v_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+ \frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert b\vert \int _{0}^{\infty }\vert P_\varepsilon \vert \vert \partial _tu_{\varepsilon }\vert dx\le 2b^2\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T) + \frac{1}{2}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\varepsilon \int _{0}^{\infty }\vert \partial _{x}^2u_{\varepsilon }\vert \vert \partial _tu_{\varepsilon }\vert dx=2\int _{0}^{\infty }\left| \sqrt{3}\varepsilon \partial _{x}^2u_{\varepsilon }\right| \left| \frac{\partial _tu_{\varepsilon }}{\sqrt{3}}\right| dx\\&\qquad \le 3\varepsilon ^2\left\| \partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \frac{1}{3}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+\frac{1}{3}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

It follows from (2.88) that

$$\begin{aligned} \frac{1}{6}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T), \end{aligned}$$

which gives (2.87). $\square$

Lemma 2.13

Fix $T>0$ and assume (1.15), or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| \partial _tv_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T), \end{aligned}$$

(2.89)

for every $0\le t\le T$.

Proof

Thanks to the estimates (2.60) and (2.87), the claim follows differentiating with respect to t the third equation in (2.1) and using the same argument developed for (2.14). $\square$

3 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1.

Using the Sobolev Immersion Theorem [13], we begin by proving the following result.

Lemma 3.1

Fix $T>0$. There exist a subsequence $\{(u_{\varepsilon _k},\,v_{\varepsilon _k},\,P_{\varepsilon _k}) \}_{k\in \mathbb {N}}$ of $\{(u_{\varepsilon },\,v_\varepsilon ,\,P_\varepsilon )\}_{\varepsilon >0}$ and a limit triplet $(u,\,v,\,P)$ which satisfies (1.18), (1.19) and (1.20) such that

$$\begin{aligned} \begin{aligned}&u_{\varepsilon _k}\rightarrow u\text { a.e.~and in }L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty ,\\&u_{\varepsilon _k}\rightharpoonup u\text { in }H^1((0,T)\times (0,\infty )),\\&v_{\varepsilon _k}\rightharpoonup v\text { in }H^1((0,T)\times (0,\infty )),\\&P_{\varepsilon _k}\rightarrow \int _{0}^{x} u(t,y)dy\text { a.e.~and in } L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty . \end{aligned} \end{aligned}$$

(3.1)

Moreover, $(u,\,v,\,P)$ is solution of (1.8)satisfying (1.21).

Proof

Let $0\le t\le T$. We begin by observing that, thanks to Lemmas 2.3, 2.8, and 2.12

$$\begin{aligned} \{u_{\varepsilon }\}_{\varepsilon >0}\text { is uniformly bounded in } H^1((0,T)\times (0,\infty )). \end{aligned}$$

(3.2)

Therefore, there exists a subsequence $\{u_{\varepsilon _k}\}_{k\in \mathbb {N}}$ of $\{u_{\varepsilon }\}_{\varepsilon >0}$ and an function u such that

$$\begin{aligned}&u_{\varepsilon _k}\rightarrow u\text { a.e.~and in }L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty , \end{aligned}$$

(3.3)

$$\begin{aligned}&u_{\varepsilon _k}\rightharpoonup u\text { in } H^1((0,T)\times (0,\infty )). \end{aligned}$$

(3.4)

Observe that by (2.1), we have that

$$\begin{aligned} P_{\varepsilon _k}(t,x)=\int _{0}^{x}u_{\varepsilon _k}(t,y) dy. \end{aligned}$$

(3.5)

Therefore, by (3.3), (3.5) and the Hölder inequality, we have that

$$\begin{aligned} P_{\varepsilon _k}\rightarrow \int _{0}^{x} u(t,y)dy\text { a.e.~and in } L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty . \end{aligned}$$

(3.6)

Moreover, by Lemmas 2.4 and 2.13, we have that

$$\begin{aligned} v_{\varepsilon _k}\rightharpoonup v\text { in }H^1((0,T)\times (0,\infty )). \end{aligned}$$

(3.7)

(3.3), (3.6), (3.7) and (3.4) give (3.1).

Observe that, by Lemmas 2.7, 2.8 and 2.9 , we have that

$$\begin{aligned} u\in L^{\infty }(0,T;H^1(0,\infty )), \end{aligned}$$

(3.8)

while by Lemmas 2.7, 2.6, 2.8 and 2.9,

$$\begin{aligned} P\in L^{\infty }(0,T;H^2(0,\infty )). \end{aligned}$$

Additionally, by Lemmas 2.4 and 2.10,

$$\begin{aligned} v\in L^{\infty }(0,T;H^3(\mathbb {R})). \end{aligned}$$

(3.9)

Therefore, (1.17), (1.18), (1.19) and (1.20) holds and $(u,\,v,\,P)$ is solution of (1.8).

Finally, (1.21) follows from (3.1) and Lemma 2.4. $\square$

Following [25, Theorem 1.1], we prove Theorem 1.1.

Proof of Theorem 1.1

Lemma 3.1 gives the existence of a solution of (1.8) such that (1.18), (1.19) and (1.20) hold.

We prove (1.22). Given $t\ge 0,\,x\ge 0$. Let $(u_1,\,v_1,\,P_1)$ and $(u_2,\,v_2,\,P_2)$ be two solutions of (1.8), which satisfy (1.18), that is

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu_i -q^2\partial _x (u_iv_i)=bP_i, \quad &{}t>0,\, x>0,\\ \displaystyle \partial _x P_i=u_i, \quad &{}t>0,\, x>0,\\ \displaystyle \alpha \partial _{x}^2v_i +\beta \partial _x v_i +\gamma v_i=\kappa u_i^2, \quad &{}t>0,\, x>0,\\ \displaystyle \partial _x u_i(t,0)=g(t), \quad &{}t>0,\\ \displaystyle P_i(t,0)=0, \quad &{}t>0,\\ \displaystyle v_i(t,0)=h(t), \quad &{}t>0,\\ \displaystyle u_i(0,x)=u_{i,\,0}(x),\quad &{}x>0, \end{array}\right. } \qquad i=1,2. \end{aligned}$$

Then, the triplet $(U,\, V,\, U)$ defined by

$$\begin{aligned} \begin{aligned}&U(t,x)=u_1(t,x)-u_2(t,x), \quad V(t,x)=v_1(t,x)-v_2(t,x), \\&\Omega (t,x)=\int _{0}^{x}U(t,y)dy =\int _{0}^{x}u_1(t,y)dy-\int _{0}^{x}u_2(t,y)dy, \end{aligned} \end{aligned}$$

(3.10)

is solution of the following Cauchy problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tU -q^2\partial _x (u_1v_1-u_2v_2)=b\Omega , \quad &{}t>0,\, x>0,\\ \displaystyle \partial _x \Omega =U, \quad &{}t>0,\, x>0,\\ \displaystyle \alpha \partial _{x}^2V +\beta \partial _x V +\gamma V=\kappa (u_1^2- u_2^2), \quad &{}t>0,\, x>0,\\ \displaystyle \partial _x U(t,0)=0, \quad &{}t>0,\\ \displaystyle \Omega (t,0)=0, \quad &{}t>0,\\ \displaystyle V(t,0)=0, \quad &{}t>0,\\ \displaystyle U(0,x)=u_{1,\,0}(x)-u_{2,\,0}(x),\quad &{}x>0. \end{array}\right. } \end{aligned}$$

(3.11)

Observe that, thanks to (1.21) and (3.10),

$$\begin{aligned} \Omega (t,\infty )=\int _{0}^{\infty }U(t,x)dx =\int _{0}^{\infty }u_1(t,x)dx-\int _{0}^{\infty }u_2(t,x)dy=0. \end{aligned}$$

(3.12)

Moreover, since $u_1,\,u_2\in L^\infty (0,T;H^1(0,\infty ))$, for very $0\le t\le T$, we can define

$$\begin{aligned} C(T)=\mathop {\mathrm {ess\,sup}}\limits _{(0,T)\times \mathbb {R}}\Big \{\vert u_1\vert +\vert u_2\vert \Big \}. \end{aligned}$$

(3.13)

We prove that

$$\begin{aligned} \left\| V(t,\cdot ) \right\| ^2_{H^2(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

(3.14)

We need to distinguish two cases. We begin by assuming (1.15). Multiplying the third equation of (3.11) by $2\beta \partial _x V$, an integration on $(0,\infty )$ and (3.10) give

$$\begin{aligned} \begin{aligned} 2\beta \alpha \int _{0}^{\infty }\partial _{x}^2V\partial _x V dx&+2\beta ^2\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )} \\&+2\gamma \beta \int _{0}^{\infty }V\partial _x V dx=2\beta \kappa \int _{0}^{\infty }\partial _x V(u_1+u_2)U dx. \end{aligned} \end{aligned}$$

(3.15)

Since,

$$\begin{aligned} 2\beta \alpha \int _{0}^{\infty }\partial _{x}^2V\partial _x V=-\alpha \beta (\partial _x V(t,0))^2,\quad 2\gamma \beta \int _{0}^{\infty }V\partial _x V dx=0, \end{aligned}$$

it follows from (1.15) and (3.15) that

$$\begin{aligned} 2\beta ^2\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\alpha \beta (\partial _x V(t,0))^2=2\beta \kappa \int _{0}^{\infty }(u_1+u_2)U\partial _x V dx. \end{aligned}$$

(3.16)

Due to (3.13) and the Young’s inequality,

$$\begin{aligned}&2\vert \beta \vert \vert \kappa \vert \int _{0}^{\infty }\vert u_1+u_2\vert \vert U\vert \vert \partial _x V\vert dx\le 2\vert \beta \vert C(T)\int _{0}^{\infty }\vert U\vert \vert \partial _x V\vert dx\\&\qquad \le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \beta ^2\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Consequently, by (3.16),

$$\begin{aligned} \beta ^2\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\alpha \beta (\partial _x V(t,0))^2\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

(3.17)

Squaring the equation for V in (3.11), by (3.10), we get

$$\begin{aligned} \begin{aligned} \alpha ^2(\partial _{x}^2V)^2&+\beta ^2(\partial _x V)^2 +\gamma ^2 V^2 +2\alpha \beta \partial _x V\partial _{x}^2V\\&+2\alpha \gamma V\partial _{x}^2V+ 2\beta \gamma V\partial _x V=\kappa ^2(u_1+u_2)^2U^2. \end{aligned} \end{aligned}$$

(3.18)

Observe that, by (3.11),

$$\begin{aligned} 2\alpha \gamma \int _{0}^{\infty } V\partial _{x}^2V dx=&\, -2\alpha \gamma \left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned}&\alpha ^2\left\| \partial _{x}^2V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\beta ^2\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \quad = \kappa ^2\int _{0}^{\infty }(u_1+u_2)^2U^2dx+2\alpha \gamma \left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\alpha \beta (\partial _x V(t,0))^2. \end{aligned} \end{aligned}$$

(3.19)

Due to (1.15), (3.13) and (3.17),

$$\begin{aligned} \kappa ^2\int _{0}^{\infty }(u_1+u_2)^2U^2dx&+2\vert \alpha \gamma \vert \left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+\vert \alpha \beta \vert (\partial _x V(t,0))^2\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Therefore, by (3.19),

$$\begin{aligned} \alpha ^2\left\| \partial _{x}^2V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\beta ^2\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Defining

$$\begin{aligned} \tau ^2=\min \{\alpha ^2,\,\beta ^2,\, \gamma ^2\}, \end{aligned}$$

we have that

$$\begin{aligned} \tau ^2\left\| V(t,\cdot ) \right\| ^2_{H^2(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

which gives (3.14).

We continue by assuming (1.16). Since $\beta =0$, by (3.10), the third equation of (3.11) reads

$$\begin{aligned} \alpha \partial _{x}^2V+\gamma V=\kappa (u_1+u_2)U. \end{aligned}$$

(3.20)

Squaring (3.20), we have that

$$\begin{aligned} \alpha ^2(\partial _{x}^2V)^2 +\gamma ^2 V^2 +2\alpha \gamma V\partial _{x}^2V= \kappa ^2(u_1+u_2)^2U^2. \end{aligned}$$

(3.21)

Since, by (3.11)

$$\begin{aligned} 2\alpha \int _{0}^{\infty }\gamma V\partial _{x}^2V dx=-2\alpha \gamma \left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

(3.22)

it follows from an integration of (3.21) on $(0,\infty )$, (1.16) and (3.13) that

$$\begin{aligned} \alpha ^2\left\| \partial _{x}^2V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}-2\alpha \gamma \left\| \partial _x V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| V(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Reminding that $\alpha \gamma <0$ we define

$$\begin{aligned} \tau _1^2=\min \{\alpha ^2,\,-2\alpha \gamma ,\,\gamma ^2\}, \end{aligned}$$

we have that

$$\begin{aligned} \tau ^2_1\left\| V(t,\cdot ) \right\| ^2_{H^2(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

which gives (3.14).

We prove that

$$\begin{aligned} \left\| V(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

(3.23)

for every $0\le t\le T$.

Due to (3.11) and the Hölder inequality,

$$\begin{aligned} V^2(t,x)=2\int _{0}^{x}V\partial _x V dy\le 2\int _{0}^{\infty }\vert V\vert \vert \partial _x v\vert dx\le 2\left\| V(t,\cdot ) \right\| _{L^2(0,\infty )}\left\| \partial _x V(t,\cdot ) \right\| _{L^2(0,\infty )}. \end{aligned}$$

Consequently, thanks to (3.14),

$$\begin{aligned} V^2(t,x)\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \end{aligned}$$

which gives (3.23).

In a similar way, we have that

$$\begin{aligned} \left\| \partial _x V(t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

(3.24)

Observe that, by (3.10),

$$\begin{aligned} \partial _x (u_1v_1-u_2v_2)=\partial _x (u_1v_1-u_2v_1+u_2v_1-u_2v_2)= \partial _x (v_1U)+\partial _x (u_2V). \end{aligned}$$

Therefore, the first equation of (3.11) is equivalent to the following one:

$$\begin{aligned} \partial _tU=b\Omega +q^2\partial _x (v_1U)+q^2\partial _x (u_2V). \end{aligned}$$

(3.25)

Multiplying (3.25) by 2U, an integration on $(0,\infty )$ gives

$$\begin{aligned} \frac{d}{dt}\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}=2b\int _{0}^{\infty }\Omega U dx +2q^2\int _{\mathbb {R}}U\partial _x (v_1U) dx +2q^2\int _{0}^{\infty }U\partial _x (u_2V) dx. \end{aligned}$$

(3.26)

Observe that by (3.12) and the second equation of (3.11),

$$\begin{aligned} 2b\int _{0}^{\infty }\Omega U dx=2b\int _{0}^{\infty }\Omega \partial _x \Omega dx=b\Omega ^2(t,\infty )=0. \end{aligned}$$

(3.27)

Moreover, since $v_1(t,0)=h(t)$,

$$\begin{aligned} \begin{aligned} 2q^2\int _{0}^{\infty }U\partial _x (v_1U) dx=&\, -2q^2h(t)U^2(t,0) -2q^2\int _{0}^{\infty }U\partial _x U v_1 dx\\ =&\, -q^2h(t)U^2(t,0)+q^2\int _{0}^{\infty }\partial _x v_1U^2 dx. \end{aligned} \end{aligned}$$

(3.28)

It follows from (1.10), (3.26), (3.27) and (3.28) that

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+q^2\kappa _1^2U^2(t,0)\\&\qquad \le q^2\int _{0}^{\infty }\partial _x v_1U^2 dx+ 2q^2\int _{0}^{\infty }\partial _x u_2U V dx+ 2q^2\int _{0}^{\infty } u_2U\partial _x V dx. \end{aligned} \end{aligned}$$

(3.29)

Fix $T>0$. By (3.8) and (3.9), there exists a constant $C(T)>0$ such that

$$\begin{aligned} \left\| u_2(t,\cdot ) \right\| _{L^2(0,\infty )},\, \left\| \partial _x u_2(t,\cdot ) \right\| _{L^2(0,\infty )},\, \left\| \partial _x v_1(t,\cdot ) \right\| _{L^{\infty }(\mathbb {R})}\le C(T), \end{aligned}$$

(3.30)

Due to (3.23), (3.24), (3.30) and the Young’s inequality,

$$\begin{aligned}&q^2\int _{0}^{\infty }\vert \partial _x v_1\vert U^2 dx\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}, \\&2q^2\int _{0}^{\infty }\vert \partial _x u_2\vert \vert U\vert \vert V\vert dx\le q^4\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\int _{0}^{\infty }(\partial _x u_2)^2V^2 dx\\&\qquad \le q^4\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \left\| V(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\left\| \partial _x u_1(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le q^4\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +C(T)\left\| V(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2q^2\int _{0}^{\infty } \vert u_2\vert \vert U\vert \vert \partial _x V\vert dx\le q^4\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \int _{0}^{\infty }u_2^2(\partial _x V)^2 dx\\&\qquad \le q^4\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )} +\left\| u_2(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\\&\qquad \le q^4\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C(T)\left\| \partial _x V(t,\cdot ) \right\| ^2_{L^{\infty }(0,\infty )}\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

It follows from (3.29) that

$$\begin{aligned} \frac{d}{dt}\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+q^2\kappa _1^2U^2(t,0)\le C(T)\left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

The Gronwall Lemma and (3.11) give

$$\begin{aligned} \left\| U(t,\cdot ) \right\| ^2_{L^2(0,\infty )}+q^2\kappa _1^2e^{C(T)t}\int _{0}^{t}e^{-C(T)s}U^2(s,0)ds\le e^{C(T)t}\left\| U_0 \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

(3.31)

(1.22) follows from (3.10), (3.14) and (3.31). $\square$

4 Proof of Theorem 1.2 assuming (1.23)

In this section, we prove Theorem 1.2 assuming (1.23). We consider (2.1), which is an approximation of (1.8), such that (2.2) holds.

Arguing as in Section 2, we have Lemmas 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 2.10, 2.11 and 2.12.

We prove the following result.

Lemma 4.1

Fix $T>0$. Assume (1.15) and (1.23). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \begin{aligned} \frac{\alpha ^2}{2}\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+\frac{\gamma ^2}{2}\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T), \end{aligned} \end{aligned}$$

(4.1)

for every $0\le t\le T$. In particular, we have that

$$\begin{aligned} \left\| \partial _tv_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le&C(T), \end{aligned}$$

(4.2)

$$\begin{aligned} \left\| \partial _t\partial _x v_\varepsilon \right\| _{L^{\infty }((0,T)\times (0,\infty ))}\le&C(T). \end{aligned}$$

(4.3)

Proof

Let $0\le t\le T$. We begin by proving that

$$\begin{aligned} \beta ^2\left\| \partial _x \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )} -\beta \alpha (\partial _t\partial _x v_\varepsilon (t,0))^2 -\gamma \beta (\partial _tv_\varepsilon (t,0))^2\le C(T). \end{aligned}$$

(4.4)

Differentiating the third equation of (2.1) with respect to t, we have that

$$\begin{aligned} \alpha \partial _t\partial _{x}^2v_\varepsilon +\beta \partial _t\partial _x v_\varepsilon +\gamma \partial _tv_\varepsilon =2\kappa u_{\varepsilon }\partial _tu_{\varepsilon }. \end{aligned}$$

(4.5)

Multiplying (4.5) by $2\beta \partial _t\partial _x v_\varepsilon$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned} 2\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+ 2\beta \alpha \int _{0}^{\infty }\partial _t\partial _x v_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx\\&+ 2\beta \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _x v_\varepsilon dx=4\kappa \beta \int _{0}^{\infty }u_{\varepsilon }\partial _tu_{\varepsilon }\partial _t\partial _x v_\varepsilon dx. \end{aligned} \end{aligned}$$

(4.6)

Observe that

$$\begin{aligned} 2\beta \alpha \int _{0}^{\infty }\partial _t\partial _x v_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx=&\, -\beta \alpha (\partial _t\partial _x v_\varepsilon (t,0))^2 , \\ 2\beta \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _x v_\varepsilon dx=&\, -\beta \gamma (\partial _tv_\varepsilon (t,0))^2. \end{aligned}$$

Consequently, by (1.15), (1.23) and (4.6), we have that

$$\begin{aligned} \begin{aligned} 2\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&-\beta \alpha (\partial _t\partial _x v_\varepsilon (t,0))^2 \\&-\beta \gamma (\partial _tv_\varepsilon (t,0))^2=4\kappa \beta \int _{0}^{\infty }u_{\varepsilon }\partial _tu_{\varepsilon }\partial _t\partial _x v_\varepsilon dx. \end{aligned} \end{aligned}$$

(4.7)

Due to (2.72), (2.87) and the Young’s inequality,

$$\begin{aligned}&2\vert \kappa \vert \vert \beta \vert \int _{0}^{\infty }\vert u_{\varepsilon }\partial _tu_{\varepsilon }\vert \vert \partial _t\partial _x v_\varepsilon \vert dx\le \kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2 dx +\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\nonumber \\&\qquad \le \kappa ^2\left\| u_{\varepsilon } \right\| ^2_{L^{\infty }((0,t)\times (0,\infty )}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )} + \beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T) + \beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}.\nonumber \end{aligned}$$

(4.8)

Therefore, by (4.7),

$$\begin{aligned} \beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\beta \alpha (\partial _t\partial _x v_\varepsilon (t,0))^2-\beta \gamma (\partial _tv_\varepsilon (t,0))^2\le C(T), \end{aligned}$$

which gives (4.4).

We prove (4.1). Squaring (4.5), an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ \le&\; 4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2 dx-2\alpha \beta \int _{0}^{\infty }\partial _t\partial _x v_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx\\&-2\alpha \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx -2\beta \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _x v_\varepsilon dx. \end{aligned} \end{aligned}$$

(4.9)

Observe that

$$\begin{aligned} -2\alpha \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx=2\alpha \gamma \partial _tv_\varepsilon (t,0)\partial _t\partial _x v_\varepsilon (t,0)+2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Consequently, by (1.16), (1.23) and (4.9),

$$\begin{aligned} \begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\gamma ^2\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ \le&\;4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2 dx-2\alpha \beta \int _{0}^{\infty }\partial _t\partial _x v_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx\\&-2\beta \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _x v_\varepsilon dx +2\alpha \gamma \partial _tv_\varepsilon (t,0)\partial _t\partial _x v_\varepsilon (t,0)\\&+2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned} \end{aligned}$$

(4.10)

Due to (2.72), (2.87), (4.4) and the Young’s inequality,

$$\begin{aligned}&4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2 dx\le 4\kappa ^2\left\| u_{\varepsilon } \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T),\\&2\vert \alpha \vert \vert \beta \vert \int _{0}^{\infty }\vert \partial _t\partial _x v_\varepsilon \vert \vert \partial _t\partial _{x}^2v_\varepsilon \vert dx\le 2\beta ^2\left\| \partial _x \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ \frac{\alpha ^2}{2}\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le C(T)+ \frac{\alpha ^2}{2}\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\&2\vert \beta \vert \vert \gamma \vert \int _{0}^{\infty }\vert \partial _tv_\varepsilon \vert \vert \partial _t\partial _x v_\varepsilon \vert dx\le \frac{\gamma ^2}{2}\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )} +2\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&\qquad \le \frac{\gamma ^2}{2}\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+C(T),\\&2\vert \alpha \vert \vert \gamma \vert \vert \partial _tv_\varepsilon (t,0)\vert \vert \partial _t\partial _x v_\varepsilon (t,0)\vert \le \alpha ^2(\partial _tv_\varepsilon (t,0))^2 +\gamma ^2(\partial _x \partial _tv_\varepsilon (t,0))^2 \le C(T),\\&2\vert \alpha \gamma \vert \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T). \end{aligned}$$

Therefore, by (1.15), (1.23) and (4.10), we get

$$\begin{aligned} \frac{\alpha ^2}{2}\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+\frac{\gamma ^2}{2}\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T), \end{aligned}$$

which gives (4.1).

We prove (4.2). Thanks to (4.1), (4.4) and the Hölder inequality, we obtain for $t\ge 0,\,x\ge 0$

$$\begin{aligned} (\partial _tv_\varepsilon (t,x))^2=&\, 2\int _{0}^{x}\partial _tv_\varepsilon \partial _t\partial _x v_\varepsilon dy +(\partial _tv_\varepsilon (t,0))^2\le 2\int _{0}^{\infty }\vert \partial _tv_\varepsilon \vert \vert \partial _t\partial _x v_\varepsilon \vert dx +C(T)\\ \le&2\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}+C(T)\le C(T). \end{aligned}$$

Hence,

$$\begin{aligned} \left\| \partial _tv_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\le C(T), \end{aligned}$$

which gives (4.2).

In a similar way, we can prove (4.3). $\square$

Using the Sobolev immersion Theorem, we begin by proving the following result.

Lemma 4.2

Fix $T>0$. There exist a subsequence $\{u_{\varepsilon _k},\,v_{\varepsilon _k},\,P_{\varepsilon _k}\}_{k\in \mathbb {N}}$ of $\{u_{\varepsilon },\,v_\varepsilon ,\,P_\varepsilon \}_{\varepsilon >0}$ and a triplet $(u,\,v,\,P)$ which satisfies (1.18), (1.19) and (1.24) such that

$$\begin{aligned} \begin{aligned}&u_{\varepsilon _k}\rightarrow u\text { a.e.~and in }L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty ,\\&u_{\varepsilon _k}\rightharpoonup u\text { in }H^1((0,T)\times (0,\infty )),\\&v_{\varepsilon _k}\rightarrow v\text { a.e.~and in }L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty ,\\&v_{\varepsilon _k}\rightharpoonup v\text { in }H^1((0,T)\times (0,\infty )),\\&P_{\varepsilon _k}\rightarrow \int _{0}^{x} u(t,y)dy\text { a.e.~and in } L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty . \end{aligned} \end{aligned}$$

(4.11)

Moreover, $(u,\,v,\,P)$ is solution of (1.8) satisfying (1.21).

Proof

Let $0\le t\le T$. Arguing as in Lemma 3.1, we have (3.2), (3.6), (3.7) and (3.4). Moreover, thanks to (3.4) and Lemma 2.1, (1.21) holds.

Observe that, thanks to Lemmas 2.4, 2.10 and 4.1, we have that

$$\begin{aligned} v_{\varepsilon _k}\rightarrow v\text { a.e. and in } L^{p}_{loc}((0,\infty )\times (0,\infty )), 1\le p<\infty . \end{aligned}$$

(4.12)

Therefore, (4.11) is proven.

Observe that, again by Lemmas 2.4, 2.10 and 4.1, we have that

$$\begin{aligned} v\in L^{\infty }(0,T;H^3(\mathbb {R}))\cap W^{1,\infty }((0,T)\times \mathbb {R}), \end{aligned}$$

while, by Lemma 4.1,

$$\begin{aligned} \partial _{tx}^2v\in L^\infty (0,T;L^2(0,\infty ))\cap L^{\infty }((0,T)\times (0,\infty )), \quad \partial _t\partial _{x}^2v \in L^\infty (0,T;L^2(0,\infty )). \end{aligned}$$

Arguing as in Lemma 3.1, the proof is conclued. $\square$

Proof

(Proof of Theorem 1.2 assuming (1.23)) Lemma 4.2 gives the existence of a solution of (1.8), such that (1.18), (1.20) and (1.24) hold. Arguing as in Theorem 1.1, we have (1.22). $\square$

5 Proof of Theorem 1.2 assuming (1.25)

In this section, we prove Theorem 1.2 assuming (1.25). We consider (2.1), which is an approximation of (1.8), such that (2.2) holds, and

$$\begin{aligned} \left\| h_{\varepsilon }' \right\| _{L^{\infty }(0,\infty )}\le C, \end{aligned}$$

(5.1)

where C is a constant independent on $\varepsilon$.

Arguing as in Section 2, we have Lemmas 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 2.10, 2.11 and 2.12.

We prove the following result.

Lemma 5.1

Fix $T>0$ and assume (1.15) or (1.16). There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )},\,\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )},\,\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T), \end{aligned}$$

(5.2)

for every $0\le t\le T$. Moreover, we have (4.2) and (4.3).

Proof

Let $0\le t\le T$. Assume (1.15). We begin by proving that

$$\begin{aligned} \beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\alpha \beta (\partial _t\partial _x v_\varepsilon (t,0))^2\le C(T). \end{aligned}$$

(5.3)

Thanks to (1.8), we have that

$$\begin{aligned} \partial _tv_\varepsilon (t,0)=h_{\varepsilon }'(t). \end{aligned}$$

(5.4)

Consequently, by (4.7), we have that

$$\begin{aligned} 2\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&-\beta \alpha (\partial _t\partial _x v_\varepsilon (t,0))^2 \\ =&\, 2\kappa \beta \int _{0}^{\infty }u_{\varepsilon }\partial _tu_{\varepsilon }\partial _t\partial _x v_\varepsilon dx -\beta \gamma h_{\varepsilon }^2(t). \end{aligned}$$

Therefore, by (4.8) and (5.1), we get

$$\begin{aligned} 2\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}-\beta \alpha (\partial _t\partial _x v_\varepsilon (t,0))^2\le C(T), \end{aligned}$$

which gives (5.3).

Now, we prove (5.2). Arguing as in Lemma 4.1, we have (4.9). Moreover, thanks to (5.4),

$$\begin{aligned} -2\alpha \beta \int _{0}^{\infty }\partial _t\partial _x v_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx=&\, \alpha \beta (\partial _t\partial _x v_\varepsilon (t,0)^2dx,\\ -2\alpha \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _{x}^2\partial _t\partial _{x}^2v_\varepsilon dx =&\, 2\alpha \gamma h_{\varepsilon }(t)\partial _x v_\varepsilon (t,0)+2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )},\\ -2\beta \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _x v_\varepsilon dx=&\, \beta \gamma h_{\varepsilon }^2(t). \end{aligned}$$

Therefore, by (1.15) and (4.9), we have that

$$\begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\nonumber +\gamma ^2\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )} \\ \le&\; 4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2 dx+\alpha \beta (\partial _t\partial _x v_\varepsilon (t,0))^2+2\alpha \gamma h_{\varepsilon }(t)\partial _x v_\varepsilon (t,0) \\&+2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\beta \gamma h_{\varepsilon }^2(t).\nonumber \end{aligned}$$

(5.5)

Due to (2.72), (2.87), (5.3) and the Young’s inequality,

$$\begin{aligned}&4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2 dx\le 4\kappa ^2\left\| u_{\varepsilon } \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T), \\&\vert \alpha \beta \vert (\partial _t\partial _x v_\varepsilon (t,0))^2\le C(T), \\&2\vert \alpha \gamma \vert \vert h_{\varepsilon }(t)\vert \vert \partial _x v_\varepsilon (t,0)\vert \le 2C\vert \partial _t\partial _x v_\varepsilon \vert \le C+ (\partial _t\partial _x v_\varepsilon (t,0))^2\le C(T),\\&2\vert \alpha \gamma \vert \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\vert \beta \gamma \vert h_{\varepsilon }^2(t)\le C(T)+C\le C(T). \end{aligned}$$

Consequently, by (5.5), we have that

$$\begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\beta ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\nonumber +\gamma ^2\left\| \partial _tv_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T), \end{aligned}$$

which gives (5.2).

Arguing as in Lemma 4.1, we have (4.2) and (4.3).

Assume (1.16). Differentiating (2.33) with respect to t, we have that

$$\begin{aligned} \alpha \partial _t\partial _{x}^2v_\varepsilon +\gamma \partial _tv_\varepsilon =2\kappa u_{\varepsilon }\partial _tu_{\varepsilon }. \end{aligned}$$

(5.6)

Squaring (5.6), an integration on $(0,\infty )$ gives

$$\begin{aligned} \begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+\gamma ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\&+2\alpha \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx=4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2dx. \end{aligned} \end{aligned}$$

(5.7)

Observe that, by (5.4), we have that

$$\begin{aligned} 2\alpha \gamma \int _{0}^{\infty }\partial _tv_\varepsilon \partial _t\partial _{x}^2v_\varepsilon dx=-2\alpha \gamma h_{\varepsilon }(t)\partial _t\partial _x v_\varepsilon (t,0) -2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}. \end{aligned}$$

Therefore, by (1.16), (5.7), we have that

$$\begin{aligned} \begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&-2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ =&\, 4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2dx+2\alpha \gamma h_{\varepsilon }(t)\partial _t\partial _x v_\varepsilon (t,0). \end{aligned} \end{aligned}$$

(5.8)

Due to (2.72), (2.87) and (5.1),

$$\begin{aligned}&4\kappa ^2\int _{0}^{\infty }u_{\varepsilon }^2(\partial _tu_{\varepsilon })^2dx\le 4\kappa ^2\left\| u_{\varepsilon } \right\| ^2_{L^{\infty }((0,T)\times (0,\infty ))}\left\| \partial _tu_{\varepsilon }(t,\cdot ) \right\| ^2_{L^2(0,\infty )}\le C(T),\\&2\vert \alpha \gamma \vert \vert h_{\varepsilon }(t)\vert \vert \partial _t\partial _x v_\varepsilon (t,0)\vert \le C\vert \partial _t\partial _x v_\varepsilon (t,0)\vert . \end{aligned}$$

Therefore, by (5.8),

$$\begin{aligned} \begin{aligned} \alpha ^2\left\| \partial _t\partial _{x}^2v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&-2\alpha \gamma \left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+\gamma ^2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\\ \le&\;C(T)\left( 1+\vert \partial _t\partial _x v_\varepsilon (t,0)\vert \right) . \end{aligned} \end{aligned}$$

(5.9)

We prove that

$$\begin{aligned} \vert \partial _t\partial _x v_\varepsilon (t,0)\vert \le C(T), \end{aligned}$$

(5.10)

for every $0\le t\le T$. Due to (5.9) and the Hölder inequality,

$$\begin{aligned} (\partial _t\partial _x v_\varepsilon (t,0))^2=&\, -2\int _{0}^{\infty }\partial _t\partial _x u_{\varepsilon }\partial _t\partial _{x}^2u_{\varepsilon }dx\le 2\int _{0}^{\infty }\vert \partial _t\partial _x v_\varepsilon \vert \vert \partial _t\partial _{x}^2v_\varepsilon \vert dx\\ \le&\;2\left\| \partial _t\partial _x v_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}\left\| \partial _t\partial _{x}^2u_{\varepsilon }(t,\cdot ) \right\| _{L^2(0,\infty )}\le C(T)\left( 1+\vert \partial _t\partial _x v_\varepsilon (t,0)\vert \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} (\partial _t\partial _x v_\varepsilon (t,0))^2-C(T)\vert \partial _t\partial _x v_\varepsilon (t,0)\vert -C(T)\le 0, \end{aligned}$$

which gives (5.10).

(5.2) follows from (5.9) and (5.10).

Finally, arguing as in Lemma 4.1, we have (4.2) and (4.3). $\square$

Arguing as in Sect. 4, we have Theorem 1.2 assuming (1.25).

6 Conclusions

In this paper we proved the well-posedness of a non local approximation of the short pulse equation. Since in this problem the evolutive equation is a transport equation with smooth coefficients several approaches for the well-posedness could be used (e.g. the fixed point one). We used the vanishing viscosity one because we plan to design finite difference numerical schemes for the problem. Indeed such schemes have an intrinsic diffusion similar to the one produced by vanishing viscosity. Moreover, we plan to study the boundary controllability of this nonlocal approximation of the short pulse equation.