Wellposedness of bounded solutions of the non-homogeneous initial boundary for the short pulse equation

Abstract

The short pulse equation provides a model for the propagation of ultra-short light pulses in silica optical fibers. It is a nonlinear evolution equation. In this paper the wellposedness of bounded solutions for the inhomogeneous initial boundary value problem associated to this equation is studied.

1 Introduction

The short pulse equation has the form

$$\begin{aligned} 2\partial _{x_1}\partial _{\phi }A_{0}+\chi ^{(3)}\partial _{\phi \phi }^2A_{0}^3 + \frac{1}{c_2^2}A_{0} =0, \end{aligned}$$

(1.1)

where $A_0$ is the light wave amplitude, $\phi = \frac{t-x}{\varepsilon }$, $x_1= \varepsilon x$, $\varepsilon$ is a small scale parameter, and $\chi ^{(3)}$ is the third order magnetic susceptibility (1.1) was introduced recently by Schäfer and Wayne [22] as a model equation describing the propagation of ultra-short light pulses in silica optical fibers. It provides also an approximation of nonlinear wave packets in dispersive media in the limit of few cycles on the ultra-short pulse scale. Numerical simulations [5] show that the short pulse equation approximation to Maxwell’s equations in the case when the pulse spectrum is not narrowly localized around the carrier frequency is better than the one obtained from the nonlinear Schrödinger equation, which models the evolution of slowly varying wave trains. Such ultra-short plays a key role in the development of future technologies of ultra-fast optical transmission of informations.

In [4] the author studied a new hierarchy of equations containing the short pulse equation (1.1) and the elastic beam equation, which describes nonlinear transverse oscillations of elastic beams under tension. He showed that the hierarchy of equations is integrable. He obtained the two compatible Hamiltonian structures and constructed an infinite series of both local and nonlocal conserved charges. Moreover, he gave the Lax description for both systems. The integrability and the existence of solitary wave solutions have been studied in [20, 21].

Well-posedness and wave breaking for the short pulse equation have been studied in [17, 22], respectively.

Boyd [3] (Table 4.1.2, p 212) shows that, for some polymers, $\chi ^{(3)}$ is a negative constant. Therefore, (1.1) reads

$$\begin{aligned} 2\partial _{x_1}\partial _{\phi }A_{0}-k^2\partial _{\phi \phi }^2A_{0}^3 + \frac{1}{c_2^2} A_{0} =0, \quad \chi ^{(3)}=-k^2. \end{aligned}$$

(1.2)

Following [1, 12, 13, 15], we consider the admensional form of (1.2)

$$\begin{aligned} \partial _x \left( \partial _tu + 3u^2 \partial _x u\right) =u. \end{aligned}$$

(1.3)

Indeed, multiplying (1.2) by $-c_2^2$, we have

$$\begin{aligned} -2c_2^2\partial _{x_1}\partial _{\phi }A_{0}+c_2^2k^2\partial _{\phi \phi }^2A_{0}^3 = A_{0} \end{aligned}$$

(1.4)

Consider the following Robelo transformation (see [1, 13, 15]):

$$\begin{aligned} x_1=D_1 t , \quad \phi = D_2 x , \end{aligned}$$

(1.5)

where $D_1$ and $D_2$ are two constants that will be specified later. Therefore,

$$\begin{aligned} \partial _{x_1}= D_1\partial _t, \quad \partial _{\phi }=D_2 \partial _x . \end{aligned}$$

(1.6)

Taking $A_{0}(x_1,\phi )=u(t,x)$, it follows from (1.1) and (1.6) that

$$\begin{aligned} -2c_2^2 D_1D_2\partial _x (\partial _tu) + 3c_2^2k^2 D_2^2\partial _x \left( u^2\partial _x u\right) =u. \end{aligned}$$

(1.7)

We choose $D_1$, $D_2$ so that

$$\begin{aligned} 2c_2^2 D_{1}D_{2}=-1,\qquad c_2^2k^2 D_2^2=1, \end{aligned}$$

that is

$$\begin{aligned} D_{1}=-\frac{k}{2c_2}, \quad D_{2}=\frac{1}{c_2 k}. \end{aligned}$$

(1.8)

Therefore, (1.3) follows from (1.7) and (1.8).

It is interesting to remind that equation (1.3) was proposed earlier in [19] in the context of plasma physic. Moreover, similar equations describe the dynamics of radiating gases [16, 23].

We are interested in the initial-boundary value problem for this equation, so we augment (1.3) with the boundary condition

$$\begin{aligned} u(t,0)=g(t), \qquad t>0, \end{aligned}$$

(1.9)

and the initial datum

$$\begin{aligned} u(0,x)=u_0(x), \qquad x>0, \end{aligned}$$

(1.10)

on which we assume that

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

(1.11)

On the function

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

(1.12)

we assume that

$$\begin{aligned} \left\| P_{0} \right\| ^2_{L^2(0,\infty )}=\int _{0}^{\infty }\left( \int _{0}^{x} u_{0}(y)dy\right) ^2 dx < \infty . \end{aligned}$$

(1.13)

On the boundary datum $g$, we assume that

$$\begin{aligned} g(t)\in L^{\infty }(0,\infty ). \end{aligned}$$

(1.14)

Integrating (1.3) in $(0,x)$ we gain the integro-differential formulation of (1.3) (see [20])

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu+3u^2\partial _x u = \int ^x_0 u(t,y) dy,&{}\qquad t>0, \ x>0,\\ u(t,0)=g(t),&{} \qquad t>0,\\ u(0,x)=u_0(x), &{}\qquad x>0, \end{array}\right. } \end{aligned}$$

(1.15)

that is equivalent to

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu+ 3 u^2\partial _x u= P, &{}\qquad t>0, \ x>0,\\ \partial _x P=u, &{}\qquad t>0, \ x>0,\\ u(t,0)=g(t),&{} \qquad t>0,\\ P(t,0)=0,&{} \qquad t>0,\\ u(0,x)=u_0(x), &{}\qquad x>0. \end{array}\right. } \end{aligned}$$

(1.16)

One of the main issues in the analysis of (1.16) is that the equation is not preserving the $L^1$ norm, as a consequence the nonlocal source term $P$ and the solution u are a priori only locally bounded. Indeed, from (1.15) and (1.16) is clear that we cannot have any $L^\infty$ bound without an $L^1$ bound. Since we are interested in the bounded solutions of (1.3), some assumptions on the decay at infinity of the initial condition $u_0$ are needed. The unique useful conserved quantities are

$$\begin{aligned} t\longmapsto \int u(t,x)dx=0,\qquad t\longmapsto \int u^2(t,x)dx. \end{aligned}$$

In the sense that if $u(t,\cdot )$ has zero mean at time $t=0$, then it will have zero mean at any time $t>0$. In addition, the $L^2$ norm of $u(t,\cdot )$ is constant with respect to t. Therefore, we require that initial condition $u_0$ belongs to $L^2\cap L^\infty$ and has zero mean.

Due to the regularizing effect of the P equation in (1.16) we have that

$$\begin{aligned} u\in L^{\infty }((0,T)\times (0,\infty ))\Longrightarrow P\in L^{\infty }(0,T;W^{1,\infty }(0,\infty )), \quad T>0. \end{aligned}$$

(1.17)

Therefore, if a map $u\in L^{\infty }((0,T)\times (0,\infty )),\,T>0,$ satisfies, for every convex map $\eta \in C^2(\mathbb {R})$,

$$\begin{aligned} \partial _t\eta (u)+ \partial _x q(u)-\eta '(u) P\le 0, \qquad q(u)=\int ^u 3\xi ^2 \eta '(\xi )\, d\xi , \end{aligned}$$

(1.18)

in the sense of distributions, then [11], Theorem 1.1] provides the existence of strong trace $u^\tau _0$ on the boundary $x=0$.

We give the following definition of solution (see [2]):

Definition 1.1

We say that $u\in L^{\infty }((0,T)\times (0,\infty ))$, $T>0$, is an entropy solution of the initial-boundary value problem (1.3), (1.9), and (1.10) if for every nonnegative test function $\phi \in C^2(\mathbb {R}^2)$ with compact support, and $c\in \mathbb {R}$

$$\begin{aligned}&\int _{0}^{\infty }\int _{0}^{\infty }\Big (\vert u - c\vert \partial _t\phi +\mathrm {sign}\left( u-c\right) \left( u^3- c^3\right) \partial _x \phi \Big )dtdx\nonumber \\&\quad +\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\mathrm {sign}\left( u-c\right) P\phi dtdx\nonumber \\&\quad +\int _{0}^{\infty }\mathrm {sign}\left( g(t)-c\right) \left( (u^\tau _0(t))^3-c^3\right) \phi (t,0)dt\nonumber \\&\quad +\int _{0}^{\infty }\vert u_{0}(x)-c\vert \phi (0,x)dx\ge 0, \end{aligned}$$

(1.19)

where $u^\tau _0(t)$ is the trace of u on the boundary $x=0$.

The main result of this paper is the following theorem.

Theorem 1.1

Assume (1.11), (1.13), (1.14). The initial-boundary value problem (1.3), (1.9) and (1.10) possesses an unique entropy solution u in the sense of Definition 1.1. Moreover, if u and v are two entropy solutions of (1.3), (1.9), (1.10) in the sense of Definition 1.1 the following inequality holds

$$\begin{aligned} \left\| u(t,\cdot )-v(t,\cdot ) \right\| _{L^1(0,R)}\le e^{C(T) t}\left\| u(0,\cdot )-v(0,\cdot ) \right\| _{L^1(0,R+C(T)t)}, \end{aligned}$$

(1.20)

for almost every $0<t<T$, $R>0$, and some suitable constant $C(T)>0$.

The paper is organized as follows. In Sect. 2 we prove several a priori estimates on a vanishing viscosity approximation of (1.16). Those play a key role in the proof of our main result, that is given in Sect. 3

2 Vanishing viscosity approximation

Our existence argument is based on passing to the limit in a vanishing viscosity approximation of (1.16).

Fix a small number $\varepsilon >0$, and let $u_\varepsilon =u_\varepsilon (t,x)$ be the unique classical solution of the following mixed problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu_\varepsilon +3u_\varepsilon ^2\partial _x u_\varepsilon =P_\varepsilon + \varepsilon \partial _{xx}^2u_\varepsilon ,&{}\quad t>0,\ x>0,\\ \partial _x P_\varepsilon =u_\varepsilon ,&{}\quad t>0,\ x>0,\\ u_\varepsilon (t,0)=g_\varepsilon (t),&{}\quad t>0,\\ P_\varepsilon (t,0)=0,&{}\quad t>0,\\ u_\varepsilon (0,x)=u_{0,\varepsilon }(x),&{}\quad x>0, \end{array}\right. } \end{aligned}$$

(2.1)

where $u_{\varepsilon ,0}$ and $g_\varepsilon$ are $C^\infty (0,\infty )$ approximations of $u_{0}$ and $g$ such that

$$\begin{aligned}&u_{0,\varepsilon }\rightarrow u_{0},\quad \text {a.e. and in } L^p(0,\infty ),\, 1\le p< \infty ,,\nonumber \\&P_{0,\varepsilon }\rightarrow P_{0},\quad \text {in } L^2(0,\infty ),\nonumber \\&g_\varepsilon \rightarrow g,\quad \text {a.e. and in } L^p_{loc}(0,\infty ),\, 1\le p< \infty ,\nonumber \\&\left\| u_{\varepsilon ,0} \right\| _{L^{\infty }(0,\infty )}\le \left\| u_0 \right\| _{L^{\infty }(0,\infty )}, \quad \left\| u_{\varepsilon ,0} \right\| _{L^2(0,\infty )}\le \left\| u_0 \right\| _{L^2(0,\infty )},\nonumber \\&\left\| u_{\varepsilon ,0} \right\| _{L^{4}(0,\infty )}\le \left\| u_0 \right\| _{L^{4}(0,\infty )}, \quad \int _{0}^{\infty } u_{\varepsilon ,0}(x) dx =0,\nonumber \\&\left\| P_{\varepsilon ,0} \right\| _{L^2(0,\infty )}\le \left\| P_0 \right\| _{L^2(0,\infty )}, \quad \left\| g_\varepsilon \right\| _{L^{\infty }(0,\infty )}\le C_0, \end{aligned}$$

(2.2)

and $C_0$ is a constant independent on $\varepsilon$.

Clearly, (2.1) is equivalent to the integro-differential problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu_\varepsilon +3u_\varepsilon ^2\partial _x u_\varepsilon =\int _0^x u_\varepsilon (t,y)dy+ \varepsilon \partial _{xx}^2u_\varepsilon ,&{}\quad t>0,\ x>0 ,\\ u_\varepsilon (t,0)=g_\varepsilon (t),&{}\quad t>0,\\ u_\varepsilon (0,x)=u_{\varepsilon ,0}(x),&{}\quad x>0. \end{array}\right. } \end{aligned}$$

(2.3)

Let us prove some a priori estimates on $u_\varepsilon$ and $P_\varepsilon$, denoting with $C_0$ the constants which depend only on the initial data, and $C(T)$ the constants which depend also on $T$.

Arguing as [9], Lemma 1], or [12], Lemma 2.2.1], we have the following result.

Lemma 2.1

The following statements are equivalent

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

(2.4)

$$\begin{aligned} \frac{d}{dt}\int _0^{\infty } u_\varepsilon ^2dx + 2\varepsilon \int _{0}^{\infty }(\partial _x u_\varepsilon )^2 dx&=\frac{3}{2}g_\varepsilon ^4(t)+2\varepsilon g_\varepsilon (t)\partial _x u_\varepsilon (t,0),\quad t>0. \end{aligned}$$

(2.5)

Proof

Let $t>0$. We begin by proving that (2.4) implies (2.5). Multiplying (2.3) by $u_\varepsilon$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \frac{d}{dt}\int _{0}^{\infty }u_\varepsilon ^2 dx= & {} 2\int _{0}^{\infty }u_\varepsilon \partial _tu_\varepsilon dx\\= & {} 2\varepsilon \int _{0}^{\infty }u_\varepsilon \partial _{xx}^2u_\varepsilon dx - 6\int _{0}^{\infty }u_\varepsilon ^3 \partial _x u_\varepsilon dx+2\int _{0}^{\infty }u_\varepsilon \left( \int _{0}^xu_\varepsilon dy \right) dx\\= & {} 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon (t) - 2\varepsilon \int _{0}^{\infty }(\partial _x u_\varepsilon )^2 dx +\frac{3}{2}g_\varepsilon ^4(t)\\&+2\int _{0}^{\infty }u_\varepsilon \left( \int _{0}^xu_\varepsilon dy \right) dx. \end{aligned}$$

By (2.1),

$$\begin{aligned} 2\int _{0}^{\infty }u_\varepsilon \left( \int _{0}^xu_\varepsilon dy \right) dx= 2\int _{0}^{\infty }P_\varepsilon \partial _x P_\varepsilon dx=P_\varepsilon ^2(t,\infty ). \end{aligned}$$

Then,

$$\begin{aligned} \frac{d}{dt}\int _{0}^{\infty }u_\varepsilon ^2 dx +2\varepsilon \int _{0}^{\infty }(\partial _x u_\varepsilon )^2 dx= P_\varepsilon ^2(t,\infty )+2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon (t) +\frac{3}{2}g_\varepsilon ^4(t). \end{aligned}$$

(2.6)

Thanks to (2.4),

$$\begin{aligned} \lim _{x\rightarrow \infty }P_\varepsilon ^2(t,x) =\left( \int _{0}^{\infty } u_\varepsilon (t,x) dx\right) ^2=0. \end{aligned}$$

(2.7)

(2.6) and (2.7) give (2.5).

Let us show that (2.5) implies (2.4). We assume by contradiction that (2.4) does not hold, namely:

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

By (1.16),

$$\begin{aligned} P_\varepsilon ^2(t,\infty )= \left( \int _{0}^{\infty }u_\varepsilon (t,x) dx\right) ^2 \ne 0. \end{aligned}$$

Therefore, (2.6) gives

$$\begin{aligned} \frac{d}{dt}\int _{0}^{\infty }u_\varepsilon ^2 dx +2\varepsilon \int _{0}^{\infty }(\partial _x u_\varepsilon )^2 dx\ne 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon (t) +\frac{3}{2}g_\varepsilon ^4(t), \end{aligned}$$

which is in contradiction with (2.5).

Lemma 2.2

For each $t \ge 0$, (2.4) holds true. In particular, we have that

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2+2\varepsilon \int _0^t \left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le C_{0}(t+1) +2\varepsilon \int _{0}^{t}g_\varepsilon (t)\partial _x u_\varepsilon (t,0) ds.\nonumber \\ \end{aligned}$$

(2.8)

Proof

We begin by observing that $\partial _tu_\varepsilon (t,0)=g_\varepsilon '(t)$, being $u_\varepsilon (t,0)=g_\varepsilon (t)$. It follows from (2.3) that

$$\begin{aligned} \varepsilon \partial _{xx}^2u_\varepsilon (t,0)&= \partial _tu_\varepsilon (t,0) + 3u_\varepsilon ^2(t,0)\partial _x u_\varepsilon (t,0)-\int _{0}^{0}u_\varepsilon (t,x)dx \nonumber \\&= g_\varepsilon '(t) + 3g_\varepsilon ^2(t)\partial _x u_\varepsilon (t,0). \end{aligned}$$

(2.9)

Differentiating (2.3) with respect to $x$, we have

$$\begin{aligned} \partial _x (\partial _tu_\varepsilon + 3u_\varepsilon ^2\partial _x u_\varepsilon - \varepsilon \partial _{xx}^2u_\varepsilon )=u_\varepsilon . \end{aligned}$$

From (2.9), and being $u_\varepsilon$ a smooth solution of (2.3), an integration over $(0,\infty )$ gives (2.4). Lemma 2.1 says that also (2.5) holds true. Therefore, integrating (2.5) on $(0,t)$, for (2.2), we have

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2&+2\varepsilon \int _0^t \left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\ \le&\left\| u_0 \right\| ^2_{L^2(0,\infty )}+ \frac{3}{2}\int _{0}^{t}g_\varepsilon ^4(s)ds+2\varepsilon \int _{0}^{t}g_\varepsilon (s)\partial _x u_\varepsilon (s,0)ds\\ \le&\left\| u_0 \right\| ^2_{L^2(0,\infty )}+ \frac{3}{2}\left\| g_\varepsilon \right\| ^4_{L^{\infty }(0,\infty )}t+ 2\varepsilon \int _{0}^{t}g_\varepsilon (s)\partial _x u_\varepsilon (s,0)ds\\ \le&\left\| u_0 \right\| ^2_{L^2(0,\infty )}+C_{0}t + 2\varepsilon \int _{0}^{t}g_\varepsilon (s)\partial _x u_\varepsilon (s,0)ds, \end{aligned}$$

which gives (2.8).

Lemma 2.3

We have that

$$\begin{aligned} \lim _{x\rightarrow \infty }F_\varepsilon (t,x)=\int _{0}^{\infty }P_\varepsilon (t,x)dx= \varepsilon \partial _x u_\varepsilon (t,0)-g_\varepsilon ^3(t), \end{aligned}$$

(2.10)

where

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

(2.11)

Proof

We begin by observing that, integrating on $(0,x)$ the second equation of (2.1), we get

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

(2.12)

Differentiating (2.12) with respect to $t$, we have

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

(2.13)

It follows from (2.4) and (2.13) that

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

(2.14)

Integrating on $(0,x)$ the first equation of (2.1), thanks to (2.13), we have

$$\begin{aligned} \partial _tP_\varepsilon (t,x) +u_\varepsilon ^3(t,x)- g_\varepsilon ^3(t)-\varepsilon \partial _x u_\varepsilon (t,x) + \varepsilon \partial _x u_\varepsilon (t,0)=\int _{0}^{x}P_\varepsilon (t,y)dy. \end{aligned}$$

(2.15)

It follows from the regularity of $u_\varepsilon$ that

$$\begin{aligned} \lim _{x\rightarrow \infty }\left( u_\varepsilon ^3(t,x))-\varepsilon \partial _x u_\varepsilon (t,x)\right) =0. \end{aligned}$$

(2.16)

(2.14) and (2.16) give (2.10).

Arguing as in [8], Lemma$2.3$], we prove the following lemma.

Lemma 2.4

Let $T>0$. There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{4}(0,\infty )}&+2\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\nonumber \\&+12\varepsilon \int _{0}^{t}\left\| u_\varepsilon (s,\cdot )\partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\nonumber \\&+4\varepsilon \int _{0}^{t}\left\| u_\varepsilon (s,\cdot ) \right\| _{L^2(0,\infty )}^2 ds +\varepsilon ^2\int _{0}^{t}\left( \partial _x u_\varepsilon (s,0)\right) ^2 ds \le C(T), \end{aligned}$$

(2.17)

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

Proof

Let $0\le t\le T$. We begin by observing that (2.11) and (2.15) imply

$$\begin{aligned} \partial _tP_\varepsilon (t,x)= F_\varepsilon (t,x)-u_\varepsilon ^3(t,x)+g_\varepsilon ^3(t)+\varepsilon u_\varepsilon (t,x)-\varepsilon \partial _x u_\varepsilon (t,0). \end{aligned}$$

(2.18)

Multiplying (2.18) by $P_\varepsilon$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \frac{d}{dt}\int _{0}^{\infty }P_\varepsilon ^2 dx= & {} 2\int _{0}^{\infty }P_\varepsilon \partial _tP_\varepsilon dx\nonumber \\= & {} 2\int _{0}^{\infty }P_\varepsilon F_\varepsilon dx -2\int _{0}^{\infty }u_\varepsilon ^3P_\varepsilon dx+2g_\varepsilon ^3(t)\int _{0}^{\infty }P_\varepsilon dx\nonumber \\&+2\varepsilon \int _{0}^{\infty }\partial _x u_\varepsilon P_\varepsilon dx -2\varepsilon \partial _x u_\varepsilon (t,0)\int _{0}^{\infty }P_\varepsilon dx. \end{aligned}$$

(2.19)

By (2.1),

$$\begin{aligned} 2\int _{0}^{\infty }\partial _x u_\varepsilon P_\varepsilon dx = -2\varepsilon \int _{0}^{\infty }u_\varepsilon \partial _x P_\varepsilon dx= -2\varepsilon \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2, \end{aligned}$$

(2.20)

while, in light of (2.11) and (2.10),

$$\begin{aligned} 2\int _{0}^{\infty }P_\varepsilon F_\varepsilon dx&= 2\int _{0}^{\infty }F_\varepsilon \partial _x F_\varepsilon dx \nonumber \\&=F_\varepsilon ^2(t,\infty )= \left( \varepsilon \partial _x u_\varepsilon (t,0)-g_\varepsilon ^3(t)\right) ^2\nonumber \\&=\varepsilon ^2\left( \partial _x u_\varepsilon (t,0)\right) ^2- 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t)+ g_\varepsilon ^6(t). \end{aligned}$$

(2.21)

Using again (2.10),

$$\begin{aligned} -2\varepsilon \partial _x u_\varepsilon (t,0)\int _{0}^{\infty }P_\varepsilon dx&= -2\varepsilon ^2\left( \partial _x u_\varepsilon (t,0)\right) ^2 + 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t),\nonumber \\ 2g_\varepsilon ^3(t)\int _{0}^{\infty }P_\varepsilon dx&= 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t)-2g_\varepsilon ^6(t). \end{aligned}$$

(2.22)

(2.19), (2.20), (2.21) and (2.22) give

$$\begin{aligned} \frac{d}{dt}\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}=&-2\varepsilon \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2 -2\int _{0}^{\infty }u_\varepsilon ^3P_\varepsilon dx\\&-\varepsilon ^2\left( \partial _x u_\varepsilon (t,0)\right) ^2 - g_\varepsilon ^6(t)+ 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t), \end{aligned}$$

that is,

$$\begin{aligned} \frac{d}{dt}\left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}&+2\varepsilon \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2+\varepsilon ^2\left( \partial _x u_\varepsilon (t,0)\right) ^2\nonumber \\ =&-2\int _{0}^{\infty }u_\varepsilon ^3P_\varepsilon dx - g_\varepsilon ^6(t)+ 2\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t). \end{aligned}$$

(2.23)

Multiplying (2.1) by $2u_\varepsilon ^3$, an integration on $(0,\infty )$ gives

$$\begin{aligned} \frac{d}{dt}\left( \frac{1}{2}\int _{0}^{\infty }u_\varepsilon ^4 dx\right)&=2\int _{0}^{\infty }u_\varepsilon ^3\partial _tu_\varepsilon dx\\&=-6\int _{0}^{\infty }u_\varepsilon ^5 \partial _x u_\varepsilon dx +2\int _{0}^{\infty }u_\varepsilon ^3P_\varepsilon dx + 2\varepsilon \int _{0}^{\infty }u_\varepsilon ^3\partial _{xx}^2u_\varepsilon dx\\&=g_\varepsilon ^6(t)+2\int _{0}^{\infty }u_\varepsilon ^3P_\varepsilon dx+ 2 \varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t)-6\varepsilon \int _{0}^{\infty }u_\varepsilon ^2(\partial _x u_\varepsilon )^2dx, \end{aligned}$$

that is

$$\begin{aligned} \frac{d}{dt}\left( \frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{4}(0,\infty )}\right)&+ 6\varepsilon \left\| u_\varepsilon (t,\cdot )\partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\nonumber \\ =&g_\varepsilon ^6(t)+2\int _{0}^{\infty }u_\varepsilon ^3P_\varepsilon dx+ 2 \varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t). \end{aligned}$$

(2.24)

Adding (2.23), (2.24), we get

$$\begin{aligned}&\frac{d}{dt}\left( \frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{4}(0,\infty )}+ \left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \nonumber \\&\quad + 6\varepsilon \left\| u_\varepsilon (t,\cdot )\partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ 2\varepsilon \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2\nonumber \\&\quad +\varepsilon ^2\left( \partial _x u_\varepsilon (t,0)\right) ^2 = 4\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t). \end{aligned}$$

(2.25)

Due to the Young inequality,

$$\begin{aligned} 4\varepsilon \partial _x u_\varepsilon (t,0)g_\varepsilon ^3(t)\le \left| \varepsilon \partial _x u_\varepsilon (t,0)\right| \left| 4g_\varepsilon ^3(t)\right| \le \frac{\varepsilon ^2}{2}\left( \partial _x u_\varepsilon (t,0)\right) ^2+8g_\varepsilon ^6(t). \end{aligned}$$

(2.26)

It follows from (2.25), (2.26) that

$$\begin{aligned}&\frac{d}{dt}\left( \frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{4}(0,\infty )}+ \left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}\right) \nonumber \\&\quad + 6\varepsilon \left\| u_\varepsilon (t,\cdot )\partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+ 2\varepsilon \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2\nonumber \\&\quad +\frac{\varepsilon ^2}{2}\left( \partial _x u_\varepsilon (t,0)\right) ^2 \le 8g_\varepsilon ^6(t). \end{aligned}$$

(2.27)

Integrating (2.27) on (0,t), by (2.2), we have

$$\begin{aligned}&\frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{4}(0,\infty )}+ \left\| P_\varepsilon (t,\cdot ) \right\| ^2_{L^2(0,\infty )}+6\varepsilon \int _{0}^{t}\left\| u_\varepsilon (s,\cdot )\partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\&\quad +2\varepsilon \int _{0}^{t}\left\| u_\varepsilon (s,\cdot ) \right\| _{L^2(0,\infty )}^2 ds + \frac{\varepsilon ^2}{2}\int _{0}^{t}\left( \partial _x u_\varepsilon (s,0)\right) ^2 ds \\&\qquad \le \left\| u_0 \right\| ^4_{L^{4}(0,\infty )}+ \left\| P_0 \right\| ^2_{L^2(0,\infty )}+8\int _{0}^{t}g_\varepsilon ^6(s)ds\\&\qquad \le C_{0} + 8\left\| g_\varepsilon \right\| ^6_{L^{\infty }(0,\infty )}t\le C_{0}\left( 1+8t\right) , \end{aligned}$$

which gives (2.17).

Lemma 2.5

Let $T>0$. There exists a constant $C(T)>0$, independent on $\varepsilon$, such that

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(0,\infty )}^2+2\varepsilon \int _0^t \left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\le C(T), \end{aligned}$$

(2.28)

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

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

(2.29)

Proof

We begin by observing that, using the Young inequality,

$$\begin{aligned} 2\varepsilon g_\varepsilon (t)\partial _x u_\varepsilon (t,0)\le 2 \left| g_\varepsilon (t) \right| \left| \varepsilon \partial _x u_\varepsilon (t,0)\right| \le g_\varepsilon ^2(t) + \varepsilon ^2 \left( \partial _x u_\varepsilon (t,0)\right) ^2. \end{aligned}$$

Therefore, in light of (2.2) and (2.17),

$$\begin{aligned} 2\varepsilon \int _{0}^{t}g_\varepsilon (s)\partial _x u_\varepsilon (s,0)ds&\le 2\int _{0}^{t}\left| g_\varepsilon (t)\right| \left| \varepsilon \partial _x u_\varepsilon (t,0)\right| dx\nonumber \\&\le \int _{0}^{t}g_\varepsilon ^2(s)ds + \varepsilon ^2 \int _{0}^{t}\left( \partial _x u_\varepsilon (s,0)\right) ^2 ds\nonumber \\&\le \left\| g_\varepsilon \right\| ^2_{L^{\infty }(0,\infty )} t + \varepsilon ^2 \int _{0}^{t}\left( \partial _x u_\varepsilon (s,0)\right) ^2 ds\nonumber \\&\le C_{0}t + \varepsilon ^2 \int _{0}^{t}\left( \partial _x u_\varepsilon (s,0)\right) ^2 ds \le C(T). \end{aligned}$$

(2.30)

(2.28) follows from (2.8) and (2.30).

Finally, we prove (2.29). Due to (2.1), (2.17), (2.28) and the Hölder inequality,

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

Therefore,

$$\begin{aligned} \vert P_\varepsilon (t,x)\vert \le C(T), \end{aligned}$$

which gives (2.29).

Lemma 2.6

Let $T>0$. We have

$$\begin{aligned} \left\| u_\varepsilon \right\| _{L^\infty ((0,T)\times (0,\infty ))}\le \left\| u_0 \right\| _{L^\infty (0,\infty )}+C(T). \end{aligned}$$

(2.31)

Proof

Due to (2.1) and (2.29),

$$\begin{aligned} \partial _tu_\varepsilon +3u_\varepsilon ^2\partial _x u_\varepsilon -\varepsilon \partial _{xx}^2u_\varepsilon \le C(T). \end{aligned}$$

Since the map

$$\begin{aligned} {\mathcal F}(t):=\left\| u_0 \right\| _{L^\infty (0,\infty )}+ C(T)t, \end{aligned}$$

solves the equation

$$\begin{aligned} \frac{d{\mathcal F}}{dt}= C(T) \end{aligned}$$

and

$$\begin{aligned} \max \{u_\varepsilon (0,x),0\}\le {\mathcal F}(t),\qquad (t,x)\in (0,T)\times (0,\infty ), \end{aligned}$$

the comparison principle for parabolic equations implies that

$$\begin{aligned} u_\varepsilon (t,x)\le {\mathcal F}(t),\qquad (t,x)\in (0,T)\times (0,\infty ). \end{aligned}$$

In a similar way we can prove that

$$\begin{aligned} u_\varepsilon (t,x)\ge -{\mathcal F}(t),\qquad (t,x)\in (0,T)\times (0,\infty ). \end{aligned}$$

Therefore,

$$\begin{aligned} \vert u_\varepsilon (t,x)\vert \le \left\| u_0 \right\| _{L^\infty (0,\infty )}+ C(T)t\le \left\| u_0 \right\| _{L^\infty (0,\infty )}+ C(T)T, \end{aligned}$$

which gives (2.31).

3 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1.

Let us begin by proving the existence of a distributional solution to (1.3), (1.9), (1.10) satisfying (1.19).

Lemma 3.1

Let $T>0$. There exists a function $u\in L^{\infty }((0,T)\times (0,\infty ))$ that is a distributional solution of (1.16) and satisfies (1.19).

We construct a solution by passing to the limit in a sequence $\left\{ u_{\varepsilon }\right\} _{\varepsilon >0}$ of viscosity approximations (2.1). We use the compensated compactness method [24].

Lemma 3.2

Let $T>0$. There exists a subsequence $\{u_{\varepsilon _k}\}_{k\in \mathbb {N}}$ of $\{u_\varepsilon \}_{\varepsilon >0}$ and a limit function $u\in L^{\infty }((0,T)\times (0,\infty ))$ such that

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

(3.1)

Moreover, we have

$$\begin{aligned} P_{\varepsilon _k}\rightarrow P \hbox { a.e. and in } L^{p}_{loc}(0,T;W^{1,p}_{loc}(0,\infty )), 1\le p<\infty , \end{aligned}$$

(3.2)

where

$$\begin{aligned} P(t,x)=\int _0^x u(t,y)dy,\qquad t\ge 0,\quad x\ge 0, \end{aligned}$$

(3.3)

and (1.19) holds true.

Proof

Let $\eta :\mathbb {R}\rightarrow \mathbb {R}$ be any convex $C^2$ entropy function, and $q:\mathbb {R}\rightarrow \mathbb {R}$ be the corresponding entropy flux defined by $q'(u)=3u^2\eta '(u)$. By multiplying the first equation in (2.1) with $\eta '(u_\varepsilon )$ and using the chain rule, we get

$$\begin{aligned} \partial _t\eta (u_\varepsilon )+\partial _x q(u_\varepsilon ) =\underbrace{\varepsilon \partial _{xx}^2\eta (u_\varepsilon )}_{=:\mathcal {L}_{1,\varepsilon }} \, \underbrace{-\varepsilon \eta ''(u_\varepsilon )\left( \partial _x u_\varepsilon \right) ^2}_{=: \mathcal {L}_{2,\varepsilon }} \, \underbrace{+\eta '(u_\varepsilon ) P_\varepsilon }_{=: \mathcal {L}_{3,\varepsilon }}, \end{aligned}$$

where $\mathcal {L}_{1,\varepsilon }$, $\mathcal {L}_{2,\varepsilon }$, $\mathcal {L}_{3,\varepsilon }$ are distributions. Let us show that

$$\begin{aligned} \mathcal {L}_{1,\varepsilon }\rightarrow 0\quad in\ H^{-1}((0,T)\times (0,\infty )), T>0. \end{aligned}$$

Since

$$\begin{aligned} \varepsilon \partial _{xx}^2\eta (u_\varepsilon )=\partial _x (\varepsilon \eta '(u_\varepsilon )\partial _x u_\varepsilon ), \end{aligned}$$

from (2.28) and Lemma 2.6,

$$\begin{aligned} \left\| \varepsilon \eta '(u_\varepsilon )\partial _x u_\varepsilon \right\| ^2_{L^2((0,T)\times (0,\infty ))}&\le \varepsilon ^2\left\| \eta ' \right\| ^2_{L^{\infty }(J_T)}\int _{0}^{T}\left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\&\le \varepsilon \left\| \eta ' \right\| ^2_{L^{\infty }(J_T)}C(T)\rightarrow 0, \end{aligned}$$

where

$$\begin{aligned} J_T=\left( -\left\| u_0 \right\| _{L^\infty (0,\infty )}- C(T), \left\| u_0 \right\| _{L^\infty (0,\infty )}+C(T)\right) . \end{aligned}$$

We claim that

$$\begin{aligned} \{\mathcal {L}_{2,\varepsilon }\}_{\varepsilon >0} \hbox { is uniformly bounded in } L^1((0,T)\times (0,\infty )), T>0. \end{aligned}$$

Again by (2.28) and Lemma 2.6,

$$\begin{aligned} \left\| \varepsilon \eta ''(u_\varepsilon )(\partial _x u_\varepsilon )^2 \right\| _{L^1((0,T)\times (0,\infty ))}&\le \left\| \eta '' \right\| _{L^{\infty }(J_T)}\varepsilon \int _{0}^{T}\left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(0,\infty )}ds\\&\le \left\| \eta '' \right\| _{L^{\infty }(J_T)}C(T). \end{aligned}$$

We have that

$$\begin{aligned} \{\mathcal {L}_{3,\varepsilon }\}_{\varepsilon >0} \hbox { is uniformly bounded in } L^1_{loc}((0,T)\times (0,\infty )), T>0. \end{aligned}$$

Let $K$ be a compact subset of $(0,T)\times (0,\infty )$. Using (2.29) and Lemma 2.6,

$$\begin{aligned} \left\| \eta '(u_\varepsilon )P_\varepsilon \right\| _{L^1(K)}&=\int _{K}\vert \eta '(u_\varepsilon )\vert \vert P_\varepsilon \vert dtdx\\&\le \left\| \eta ' \right\| _{L^{\infty }(J_T)}\left\| P_\varepsilon \right\| _{L^{\infty }(I_{T})}\vert K \vert . \end{aligned}$$

Therefore, Murat’s lemma [18] implies that

$$\begin{aligned} \left\{ \partial _t\eta (u_\varepsilon )+\partial _x q(u_\varepsilon )\right\} _{\varepsilon >0} \hbox { lies in a compact subset of } H_{\mathrm {loc}}^{-1}((0,T)\times (0,\infty )). \end{aligned}$$

(3.4)

The $L^{\infty }$ bound stated in Lemma 2.6, (3.4), and the Tartar’s compensated compactness method [24] give the existence of a subsequence $\{u_{\varepsilon _k}\}_{k\in \mathbb {N}}$ and a limit function $u\in L^{\infty }((0,T)\times (0,\infty )),\,T>0,$ such that (3.1) holds.

(3.2) follows from (3.1), the Hölder inequality and the identity

$$\begin{aligned} P_{\varepsilon _{k}}=\int _{0}^{x}u_{\varepsilon _{k}} dy, \quad \partial _x P_{\varepsilon _{k}}=u_{\varepsilon _{k}}. \end{aligned}$$

Finally, we prove (1.19).

Let $k\in \mathbb {N}$, $c\in \mathbb {R}$ be a constant, and $\phi \in C^{\infty }(\mathbb {R}^2)$ be a nonnegative test function with compact support. Multiplying the first equation of(2.1) by $\mathrm {sign}\left( u_\varepsilon -c\right)$, we have

$$\begin{aligned} \partial _t\vert u_{\varepsilon _k}-c\vert&+\partial _x \left( \mathrm {sign}\left( u_{\varepsilon _k}-c\right) \left( u_{\varepsilon _k}^3-c^3\right) \right) \\&-\mathrm {sign}\left( u_{\varepsilon _k}-c\right) P_{\varepsilon _k}-\varepsilon _k\partial _{xx}^2\vert u_{\varepsilon _k}-c\vert \le 0. \end{aligned}$$

Multiplying by $\phi$ and integrating over $(0,\infty )^2$, we get

$$\begin{aligned}&\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\left( \vert u_{\varepsilon _k}-c\vert \partial _t\phi +\left( \mathrm {sign}\left( u_{\varepsilon _k}-c\right) \left( u_{\varepsilon _k}^3-c^3\right) \right) \partial _x \phi \right) dtdx\\&\qquad +\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\mathrm {sign}\left( u_{\varepsilon _k}-c\right) P_{\varepsilon _k}dtdx -\varepsilon _k\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\partial _x \vert u_{\varepsilon _k}-c\vert \partial _x \phi dtdx\\&\qquad + \int _{0}^{\infty }\vert u_{0}(x) -c\vert \phi (0,x) dx + \int _{0}^{\infty }\mathrm {sign}\left( g_{\varepsilon _{k}}(t)-c\right) \left( g^3_{\varepsilon _{k}}(t)-c^3\right) \phi (t,0)dt\\&\qquad -\varepsilon _k\int _{0}^{\infty }\partial _x \vert u_{\varepsilon _k}(t,0) -c\vert \phi (t,0)dt \ge 0. \end{aligned}$$

Thanks to (2.2) and Lemmas 2.5 and 2.6, when $k\rightarrow \infty$, we have

$$\begin{aligned}&\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\left( \vert u -c\vert \partial _t\phi +\left( \mathrm {sign}\left( u-c\right) \left( u^3-c^3\right) \right) \partial _x \phi \right) dtdx\\&\qquad +\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\mathrm {sign}\left( u-c\right) P dtdx + \int _{0}^{\infty }\vert u_{0}(x) -c\vert \phi (0,x) dx\\&\qquad +\int _{0}^{\infty }\mathrm {sign}\left( g(t)-c\right) \left( g^3(t)-c^3\right) \phi (t,0)dt\\&\qquad -\lim _{\varepsilon _k}\varepsilon _k\int _{0}^{\infty }\partial _x \vert u_{\varepsilon _k}(t,0) -c\vert \phi (t,0)dt \ge 0. \end{aligned}$$

We have to prove that (see [2])

$$\begin{aligned}&\lim _{\varepsilon _k}\varepsilon _k\int _{0}^{\infty }\partial _x \vert u_{\varepsilon _k}(t,0) -c\vert \phi (t,0)dt\nonumber \\&\qquad = \int _{0}^{\infty }\mathrm {sign}\left( g(t)-c\right) \left( g^3(t)-(u^\tau _0(t))^3\right) \phi (t,0)dt. \end{aligned}$$

(3.5)

Let $\{\rho _{\nu }\}_{\nu \in \mathbb {N}}\subset C^{\infty }(\mathbb {R})$ be such that

$$\begin{aligned} 0\le \rho _{\nu } \le 1, \quad \rho _{\nu }(0)=1, \quad \vert \rho '_{\nu }\vert \le 1, \quad x\ge \frac{1}{\nu } \quad \Longrightarrow \quad \rho _{\nu }(x)=0. \end{aligned}$$

(3.6)

Using $(t,x) \mapsto \rho _{\nu }(x)\phi (t,x)$ as test function for the first equation of (2.1) we get

$$\begin{aligned}&\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\left( u_{\varepsilon _k}\partial _t\phi \rho _{\nu }+u_{\varepsilon _k}^3\partial _x \phi \rho _{\nu } + u_{\varepsilon _k}^3\phi \rho '_{\nu }\right) dtdx +\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }P_{\varepsilon _k}\phi \rho _{\nu } dtdx\\&\qquad -\varepsilon _k\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\partial _x u_{\varepsilon _k}\left( \partial _x \phi \rho _{\nu }+\phi \rho '_{\nu }\right) dtdx + \int _{0}^{\infty } u_{0}(x)\phi (0,x)\rho _{\nu }(x)dx\\&\qquad +\int _{0}^{\infty } g^3_{\varepsilon _k}(t)\phi (t,0) dt -\varepsilon _k\int _{0}^{\infty }\partial _x u_{\varepsilon _k}(t,0)\phi (t,0) dt=0. \end{aligned}$$

As $k\rightarrow \infty$, we obtain that

$$\begin{aligned}&\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }\left( u\partial _t\phi \rho _{\nu }+u^3\partial _x \phi \rho _{\nu } + u^3\phi \rho '_{\nu }\right) dtdx +\int _{0}^{\infty }\!\!\!\!\int _{0}^{\infty }P\phi \rho _{\nu }dtdx\\&\qquad + \int _{0}^{\infty } u_{0}(x)\phi (0,x)\rho _{\nu }dx +\int _{0}^{\infty } g^3(t)\phi (t,0) dt\\&\quad =\lim _{\varepsilon _k}\varepsilon _k\int _{0}^{\infty }\partial _x u_{\varepsilon _k}(t,0)\phi (t,0) dt. \end{aligned}$$

Sending $\nu \rightarrow \infty$, we get

$$\begin{aligned} \lim _{\varepsilon _k}\varepsilon _k\int _{0}^{\infty }\partial _x u_{\varepsilon _k}(t,0)\phi (t,0) dt= \int _{0}^{\infty }\left( g^3(t)- (u^\tau _0(t))^3\right) \phi (t,0) dt. \end{aligned}$$

Therefore, due to the strong convergence of $g_{\varepsilon _k}$ and the continuity of $g$ we have

$$\begin{aligned}&\lim _{\varepsilon _k}\varepsilon _k\int _{0}^{\infty }\partial _x \vert u_{\varepsilon _k}(t,0)-c\vert \phi (t,0)dt\\&\qquad = \lim _{\varepsilon _k} \int _{0}^{\infty }\partial _x u_{\varepsilon _k}(t,0)\mathrm {sign}\left( u_{\varepsilon _k}(t,0)-c\right) \phi (t,0)dt\\&\qquad = \lim _{\varepsilon _k} \int _{0}^{\infty }\partial _x u_{\varepsilon _k}(t,0)\mathrm {sign}\left( g_{\varepsilon _k}(t)-c\right) \phi (t,0)dt\\&\qquad = \int _{0}^{\infty }\mathrm {sign}\left( g(t)-c\right) \left( g^3(t)- (u^\tau _0(t))^3\right) \phi (t,0) dt, \end{aligned}$$

that is (3.5).

Proof of Theorem 1.1 Lemma (3.2) gives the existence of an entropy solution $u$ for (1.15), or equivalently (1.16).

We observe that, fixed $T>0$, the solutions of (1.15), or equivalently (1.16), are bounded in $(0,T)\times \mathbb {R}$. Therefore, using [6], Theorem 1.1], $u$ is unique and (1.20) holds true. $\square$