\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 01, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/01\hfil Weak asymptotic solution]
{Weak asymptotic solution for a non-strictly hyperbolic system of conservation
laws}

\author[H. Singh, M. R. Sahoo, O. P. Singh \hfil EJDE-2015/01\hfilneg]
{Harendra Singh, Manas Ranjan Sahoo, Om Prakash Singh}  % in alphabetical order

\address{Harendra Singh \newline
Department of Mathematical Sciences, IIT (BHU)\\
Varanasi 221005, India}
\email{harendrasingh.rs.apm12@iitbhu.ac.in}

\address{Manas Ranjan Sahoo \newline
Department of Mathematical Sciences, IIT (BHU)\\
Varanasi 221005, India}
\email{sahoo@math.tifrbng.res.in}

\address{Om Prakash Singh \newline
Department of Mathematical Sciences, IIT (BHU)\\
Varanasi 221005, India}
\email{opsingh.apm@iitbhu.ac.in}

\thanks{Submitted December 5, 2014. Published January 5, 2015.}
\subjclass[2000]{35A20, 35F25, 35R05}
\keywords{System of PDEs; initial conditions; weak asymptotic solutions}

\begin{abstract}
 In this article, we construct the weak asymptotic solution developed
 by Panov and Shelkovich for piecewise known solutions to a prolonged
 system of conservation laws. This is done by introducing four singular waves
 along a discontinuity curve, which in turn implies the existence of
 weak asymptotic solutions for the Riemann type initial data.
 By piecing together the Riemann problems, we construct weak asymptotic solution
 for general type initial data.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

Systems of conservation laws arise in many physical contexts are not
strictly hyperbolic. For such systems classical theories of
Glimm \cite{G1} and Lax \cite{la1} do not apply.
Because of the appearance of product of distributions, it is difficult
to define the notion of solutions for these problems.
One way to avoid this is to work with the generalized space of Colmbeau.
For details see \cite{ob1} and \cite{co2}.

A system of this kind was introduced by Joseph and Vasudeva Murthy\cite{j2}, namely,
\begin{equation}
(u_j)_t+\sum_{i=1}^{j}(\frac{u_i u_{j-i+1}}{2})_x=0,\quad j=1,2,\dots, n.
\label{*}
\end{equation}
For $n=1$, system \eqref{*} is Burger's equation, which is well studied
by Hopf \cite{h1}. For $n=2$ case is an one dimensional model for the large
scale structure formation of universe, see,  \cite{w1}.
Using vanshing viscosity approach it is observed by Joseph \cite{j1}
that the second component contain $\delta$ measure concentrated along
the line of discontinuity. The case $n=3$ is studied in \cite{j3}.
Solution is constructed in the Colombeau setting. If $u_1=u, u_2=v, u_3=w$,
\eqref{*} becomes
\begin{equation}
u_t +(\frac{u^2}{2})_x=0,\quad v_t+(uv)_x=0,\quad w_t+(\frac{v^2}{2} +uw)_x=0.
\label{*2}
\end{equation}
A similar system,
\begin{equation}
u_t +(u^2)_x=0,\quad v_t+(2uv)_x=0,\quad w_t+2(v^2 +uw)_x=0,
\label{*3}
\end{equation}
is studied by Panov and Shelkovich \cite{s1}. In \cite{s1} a concept of
weak asymptotic solution is introduced and a solution is constructed under
this consideration and generalized integral formulation is introduced
for piecewise continuous data. Note that the system
\eqref{*3} can be obtained from \eqref{*2} using the transformation
$(u,v,w)\to (2u,v,\frac{w}{2})$. The case $n=4$ is studied
by joseph and Sahoo \cite{jm1}. In \cite{jm1}, using vanshing viscosity
approach a solution is constructed for Riemann type initial data and based on
this a weak integral formulation is given.

In this paper we use the weak asymptotic method introduced by Panov
and Shelkovich \cite{s1} to study the case $n=4$.
Putting $u_1=u$, $u_2=v$, $u_3=w$, $u_4=z$ and followed by a linear transformation,
the system \eqref{*} leads to the  system
\begin{equation}
\begin{gathered}
u_t +(u^2)_x=0,\quad v_t+(2uv)_x=0\\
w_t+2(v^2 +uw)_x=0,\quad z_t+ 2((3vw+uz)_x)=0.
\end{gathered} \label{e1.2}
\end{equation}

The aim of this paper is to study the above system \eqref{e1.2} with initial
conditions
\begin{equation}
u(x,0)=u_0 (x),\quad v(x,0)=v_0 (x), \quad w(x,0)=w_0 (x), \quad z(x,0)=z_0 (x).
\label{e1.3}
\end{equation}
The content of the paper is as follows. We construct weak asymptotic solution
by connecting two known solutions from the left and right.
As a special case we derive weak asymptotic solution for the Riemann type initial
data. Then we construct a weak asymptotic solution when
the initial data for $u$ is a monotonic increasing function and initial
data for $v,w$ and $z$ are locally integrable functions.

\section{weak asymptotic solution for Riemann type initial data}

In this section we connect two classical solutions by introducing a
discontinuity curve in asymptote level. First of all we recall the
definition of \emph{weak asymptotic solution} as introduced in \cite{s1, s2}.

\begin{definition} \rm
Let us define
\begin{gather*}
L_1 (u)=u_t +(u^2)_x,\quad
L_2 (u,v)= v_t+(2uv)_x\\
L_3 (u,v,w)=w_t+2(v^2 +uw)_x,\quad
L_4 (u,v,w,z)=z_t+ 2((3vw+uz)_x).
\end{gather*} %\label{e2.1}
$(u^{\epsilon}, v^{\epsilon}, w^{\epsilon}, z^{\epsilon})$ is said to be
weak asymptotic solution to  problem \eqref{e1.2} with
initial data \eqref{e1.3} if
\begin{equation}
\begin{gathered}
\int L_1[u(x,t,\epsilon)]\psi(x)dx=o (1),\\
\int L_2[u(x,t,\epsilon),v(x,t,\epsilon)]\psi(x)dx=o (1),\\
\int L_3[u(x,t,\epsilon),v(x,t,\epsilon),w(x,t,\epsilon)]\psi(x)dx=o (1),\\
\int L_4[u(x,t,\epsilon),v(x,t,\epsilon),w(x,t,\epsilon),z(x,t,\epsilon)]
 \psi(x)dx=o(1),
\end{gathered}\label{e2.2}
\end{equation}
and initial conditions  satisfy
\begin{equation}
\begin{gathered}
\int\Big(u(x,0,\epsilon)-u_0 (x)\Big)\psi(x)dx=o(1),\\
\int\Big(v(x,0,\epsilon)-v_0 (x)\Big)\psi(x)dx=o(1),\\
\int\Big(w(x,0,\epsilon)-w_0 (x)\Big)\psi(x)dx=o(1),\\
\int\Big(z(x,0,\epsilon)-z_0 (x)\Big)\psi(x)dx=o(1),
\end{gathered}\label{e2.3}
\end{equation}
for all $\psi \in D(R)$.
\end{definition}

To study weak asymptotic analysis first we need the following
Lemma as in \cite{s1}, regarding the superpositions of the singular
waves $\delta,\delta',\delta''$ and $ \delta''' $.

\begin{lemma} \label{lem2.2}
Let $\{w_i \}_{i\in I}$ be an indexed set of Friedrich mollifiers satisfying
$$
w_i (x)=w_i (-x), \quad \int w_i =1.
$$
Define
\[
 H_i(x,\epsilon)=w_{0i}(\frac{x}{\epsilon})=\int_{-
\infty}^{\frac{x}{\epsilon}}w_i(y)d y, \quad
\delta_i (x,\epsilon)=\frac{1}{\epsilon}w_i(\frac{x}{\epsilon}), \quad
\delta^{k}_i(x,\epsilon)=\frac{1}{\epsilon^{k+1}}w_i ^{k}(\frac{x}{\epsilon}).
\]
The above assumptions implies the following asymptotic
expansions, in the sense of distributions,
\begin{gather*}
(H_i(x,\epsilon))^{r}=H(x)+O_{D'}(\epsilon),\quad
(H_i(x,\epsilon)(H_j(x,\epsilon))=H(x)+O_{D'}(\epsilon)\\
(H_i(x,\epsilon))^{r}\delta_j(x,\epsilon)=\delta(x)\int{w_{0i}^{r}(y)w_j(y)d
y}+O_{D'}(\epsilon)\\
(\delta_i(x,\epsilon))^{2}=\frac{1}{\epsilon}\delta(x)\int{w_i ^{2}(y)d
y}+O_{D'}(\epsilon)\\
H_i(x,\epsilon)\delta_j '(x,\epsilon)=-\frac{1}{\epsilon}\delta(x)\int{w_i(y)w_j (y)d
y}+\delta{'}(x)\int{w_{0i}(y)w_j(y)d y}+O_{D'}(\epsilon)\\
H_i(x,\epsilon)\epsilon^{2}\delta_j'''(x,\epsilon)=\frac{1}{\epsilon}\delta(x)\int{w_i'(y)\delta_j'
(y)d y}+O_{D'}(\epsilon),\\
\delta_i (x,\epsilon).\delta_j
(x,\epsilon)=\frac{1}{\epsilon}\delta(x)\int{w_i(y)w_j(y)dy}+O_{D'}(\epsilon)\\
\delta_i (x,\epsilon)\delta_j'(x,\epsilon)=\frac{1}{\epsilon}\delta'(x)\int
{y w_i(y)w_j'(y)d y}+O_{D'}(\epsilon),\\
H_i(x,\epsilon)\delta_j''(x,\epsilon)=\frac{1}{\epsilon}\delta(x)\int
{w_{0i}(y)w_j(y)d y}+\frac{1}{2}\delta''(x)\int {y^{2}w_{0i}(y)w_j(y)d
y}+O_{D'}(\epsilon)\\
\delta_i(x,\epsilon))\epsilon^{2}\delta_j'''(x,\epsilon)=\frac{1}{\epsilon}\delta'(x)
\int y w_i(y) w_j'''(y) dy+O_{D'}(\epsilon)\\
H_i(x,\epsilon)\epsilon^{2}\delta_j''''(x,\epsilon)=\frac{1}{\epsilon}\delta'(x)
\int y w_{0i}(y) w_j''''(y) dy+O_{D'}(\epsilon)
%\label{e2.4}
\end{gather*}
where $ \langle O_{D'}(\epsilon),\psi(x)\rangle\to 0 $ for every test function
$ \psi $.
\end{lemma}

\begin{proof}
Let $\psi \in D(\mathbb{R})$ be any test function.
The first six relations can be found in \cite{s1}; so
wee prove from the seventh onward.

Now we prove seventh asymptotic expansion.
Using change of variable formula $(x=\epsilon y)$,
employing third order Taylor expansion
$\psi(\epsilon y)=\psi(0)+\epsilon y\psi'(0) 
+\frac{1}{2} \epsilon^2y^2 \psi''(0) + \epsilon^3 y^3 O(1)$,
and the fact that $\int y w_i(y)w_{j}(y)dy=0$, we have
\begin{align*}
\langle\delta_i(x,\epsilon)\delta_j(x,\epsilon),\psi(x)\rangle
&=\int\frac{1}{\epsilon}w_i(\frac{x}{\epsilon})
 \frac{1}{\epsilon}w_{j}(\frac{x}{\epsilon})\psi(x)dx \\
& =\frac{1}{\epsilon}\int w_i(y)w_{j}(y)\psi(\epsilon y)d y\\
&=\frac{1}{\epsilon}\psi(0)\int w_i(y)w_{j}(y)d
y+\psi'(0)\int y w_i(y)w_{j}(y)d
y+O(\epsilon)\\
&=\frac{1}{\epsilon}\delta(x)\int { w_i(y)w_j(y)d
y}+O(\epsilon).
\end{align*}

Now we prove eighth asymptotic expansion. Using change of variable formula
$(x=\epsilon y)$, employing third order Taylor expansion,
$\psi(\epsilon y)=\psi(0)+\epsilon y\psi' (0)+\frac{1}{2}\epsilon^2 y^2 \psi''(0)
+\epsilon^3 y^3 O(1)$,
 and the fact that $\int y w_i(y)w_{j}(y)d
y=0$, we have
\begin{align*}
\langle\delta_i(x,\epsilon)\delta_j'(x,\epsilon),\psi(x)\rangle
&=\frac{1}{\epsilon^2}\int w_i(y)w_{j}'(y)\psi(\epsilon y)d y \\
&=\frac{1}{\epsilon^2}\psi(0)\int w_i(y)w_{j}'(y)dy
+\frac{1}{\epsilon}\psi'(0)\int y w_i(y)w_{j}'(y)dy\\
&\quad +\frac{1}{2}\psi''(0)\int y^{2} w_i(y)w_{j}'(y)d y +O(\epsilon)\\
&=\frac{1}{\epsilon}\delta'(x)\int {y w_i(y)w_j'(y)dy}+O(\epsilon).
\end{align*}
In the above calculation we also used the identity
$$
\int w_i(y)w_{j}'(y)dy=\int {y^2 w_i(y)w_j'(y)d y}=0
$$.

Following an analysis similar as above, we prove the remaining identities.
 Details are as follows:
\begin{align*}
&\langle H_i(x,\epsilon)\delta_j''(x,\epsilon),\psi(x)\rangle\\
&=\int w_{0i}(y)\frac{1}{\epsilon^2}w_{j}''(y)\psi(\epsilon y)d y \\
&=\int w_{0i}(y)\frac{1}{\epsilon^2}w_{j}''(y)(\psi(0)+\epsilon y
\psi'(0)+\frac{\epsilon^{2} y^2}{2} \psi''(0))d y+O(\epsilon) \\
&=\frac{1}{\epsilon}\delta'(x)\int {y w_{0i}(y)w_j''(y)d
y}+\frac{1}{2}\delta''(x)\int {y^{2}w_{0i}(y)w_j''(y)d
y}+O(\epsilon),
\end{align*}

\begin{align*}
&\langle \delta_i(x,\epsilon))\epsilon^{2}
 w_j'''(x,\epsilon),\psi(x)\rangle\\
&=\frac{1}{\epsilon^{2}}\int w_i(y) w_j'''(y)\psi(\epsilon y)d y \\
&=\frac{1}{\epsilon^{2}}\int
w_i(y) w_j'''(y)(\psi(0)+\epsilon y
\psi'(0)+\frac{\epsilon^{2} y^2}{2} \psi''(0))d y+O(\epsilon) \\
&=\frac{1}{\epsilon^{2}}\delta(x)\int
w_i(y) w_j'''(y)dy+{\frac{1}{\epsilon}\delta'(x) \int y
w_i(y) w_j'''(y) dy }\\&+\frac{1}{2}\delta''(x)\int  y^2
w_i(y) w_j'''(y) dy +O(\epsilon)\\
&={\frac{1}{\epsilon}\delta'(x) \int y w_i(y) w_j'''(y) dy
}+O(\epsilon)\\
\end{align*}

\begin{align*}
&\langle H_i(x,\epsilon)\epsilon^2\delta_j''''(x,\epsilon),\psi(x)\rangle\\
&=\int w_{0i}(y)\frac{1}{\epsilon^2}w_{j}''''(y)\psi(\epsilon y)(\psi(0)+\epsilon y
\psi'(0)+\frac{\epsilon^{2} y^2}{2} \psi''(0))d y +O(\epsilon)\\
&=\frac{1}{\epsilon}\delta'(x)
\int y w_{0i}(y) w_j''''(y) dy+O(\epsilon)
\end{align*}
\end{proof}

It is observed in \cite{jm1}, that the vanishing viscosity limit for
the component $z$ admits combinations of $ \delta, \delta',\delta'' $ waves.
So we choose ansatz as the combination of the above singular waves along
the discontinuity curve. But this is not enough as it is clear in the
construction of $w$, see \cite{s1}. In \cite{s1}, a correction term is
added in the component $w$ to construct weak asymptotic solution.
As the solution for the component is more complicated, extra care has to be
taken to accomplish this.This is done by choosing the correction term
carefully in the component $z$.

\begin{theorem} \label{thm2.3}
The following ansatz
\begin{equation}
\begin{gathered}
u(x,t,\epsilon)=u_2(x,t)+[u]H_{u}(-x+\phi(t),\epsilon), \\
v(x,t,\epsilon)=v_2(x,t)+[v]H_{v}(-x+\phi(t),\epsilon)
 +e(t)\delta_e(-x+\phi(t),\epsilon),\\
\begin{aligned}
w(x,t,\epsilon)
&=w_2(x,t)+[w]H_{w}(-x+\phi(t),\epsilon)+g(t)\delta_g(-x+\phi(t),\epsilon)\\
&\quad +h(t)\delta'_h(-x+\phi(t),\epsilon)+R_w(-x+\phi(t),\epsilon),
\end{aligned}\\
\begin{aligned}
z(x,t,\epsilon)
&=z_2(x,t)+[z]H_{z}(-x+\phi(t),\epsilon)+l(t)\delta_l(-x+\phi(t),\epsilon)\\
&\quad +m(t)\delta'_{m}(-x+\phi(t),\epsilon)
+n(t)\delta''_{n}(-x+\phi(t),\epsilon)\\
&\quad +R_{z} (-x+\phi(t),\epsilon),
\end{aligned}
\end{gathered} \label{e2.5}
\end{equation}
where
\begin{gather*}
R_w(x,t, \epsilon)=\epsilon^2 P(t) \delta_P'''(-x+\phi(t),\epsilon),\\
R_z(x,t, \epsilon)=\epsilon^2 (Q(t) \delta_R
'''(-x+\phi(t),\epsilon)+R(t) \delta_R ''''(-x+\phi(t),\epsilon).
\end{gather*}
is weak asymptotic solution to the problem \eqref{e1.2} if the following
relations hold:
\begin{gather*}
L_1[u_1]=0,\quad L_1[u_2]=0, \\
L_2[u_1,v_1]=0, \quad  L_2[u_2,v_2]=0, \\
L_3[u_1,v_1,w_1]=0, \quad
L_3[u_2,v_2,w_2]=0,\\
\dot{\phi}(t)=(u_1+u_2)\big|_{x=\phi(t)},\quad
\dot{e}(t)=[u](v_1+v_2)\big|_{x=\phi(t)} \\
\dot{g}(t)=(2[v](v_1+v_2)+[u](w_1+w_2)\big|_{x=\phi(t)}, \quad
\frac{d}{dt}(h(t)[u(\phi(t),t)])=\frac{d}{dt}e^2 (t),\\
\int {w_{0u}(y)w_j(y)d y}=\int {y^2w_{0v}(y)w_e(y)d y}=\frac{1}{2},
 \quad j=e,g,h ,\\
\int{w_{u}(y)w_h(y)d y}=\int{w_{e}^2(y)d y} ,\quad
 P(t)=\frac{A}{u_1(\phi(t),t)},\quad \text{where A is a constant},\\
L_4[u_1,v_1,w_1,z_1]=0,\quad L_4[u_2,v_2,w_2,z_2]=0, \\
\dot{l}(t)=-[z]\dot{\phi}(t)+2[3vw+uz],\\
\int {w_{0u}(y)w_l(y)d y}=\frac{1}{2}\int {y^2w_{0u}(y)w_n(y)d y}
=\int {w_{0u}(y)w_m(y)d y}=\frac{1}{2},\\
\begin{aligned}
\dot{m}(t)&=2[3\{(v_2+[v]\int {w_{0v}(y)w_g(y)d y}) g(t)+(w_2+[w]\int
{w_{0w}(y)w_e(y)d y})e(t)\}\\
&\quad +3\{(v_{2x} +[v_x]\int {w_{0v}(y)w_h(y)dy})h(t)\}\\
&\quad +(u_{2x}+\frac{[u_x]}{2})m(t)+(u_{1xx}
+\frac{[u_{xx}]}{2})n(t)],
\end{aligned}\\
\dot{n}(t)=2[3\{(v_2 +[v]\int {w_{0v}(y)w_h(y)d y})h(t)\}-(2u_{2x}
+[u_x])n(t)],\\
\begin{aligned}
R(t)&=\frac{1}{[u]\int {w_{0u}(y) w_R''''(y)d y}}
\Big[3e(t)h(t)\int {y w_{e}(y)w_h'(y)d y}\\
&\quad +3e(t)p(t)\int {y w_{e}(y) w_P'''(y)d y}\Big]
\end{aligned}\\
\begin{aligned}
Q(t)&=\frac{1}{[u]\int {w_{u}'(y) w_Q'(y)d y}}
\Big[3e(t)g(t)\int {w_{e}(y)w_g(y)d y}\\
&\quad -3[v]h(t)\int { w_{v}(y)w_h(y)d y}
-[u]m(t)\int {w_{u}(y)w_m(y)d y}\\
&\quad +\frac{[u]n(t)}{2}+[u_x]R(t)\int{w_{0u}(y) w_R''''(y)d y}\Big]
\end{aligned} %\label{e2.6}
\end{gather*}
\end{theorem}

\begin{proof}
If the  first thirteen relations above hold,
then the expression for $ u, v $ and
$ w $ in \eqref{e2.5} is a weak asymptotic solution, is shown in \cite{s1}.
So, we only prove that the expression for the component 
$z$ in equation \eqref{e2.5} is a weak asymptotic solution.


Multiplying the ansatz given for $v$ and $w$ in the equation \eqref{e2.5}
and using lemma \ref{lem2.2}, we obtain
\begin{align*}
&v(x,t,\epsilon)w(x,t,\epsilon)\\
&=v_2w_2+[vw]H(-x+\phi(t))
+\Big\{(v_2+[v]\int {w_{0v}(y)w_g(y)d y}) g(t)\\
&\quad +(w_2+[w]\int {w_{0w}(y)w_e(y)d y})e(t)\Big\}\delta(-x+\phi(t))\\
&\quad +(v_2 +[v]\int {w_{0v}(y)w_h(y)d
y})h(t)\delta'(-x+\phi(t))\\
&\quad +(e(t)g(t)\int { w_{e}(y)w_g(y)dy}-[v]h(t)\int { w_{v}(y)w_h(y)d
y})\frac{1}{\epsilon}\delta(-x+\phi(t))\\
&\quad +(e(t)h(t)\int {y
w_{e}(y)w_h'(y)d y}+e(t)p(t)\int {y w_{e}(y) w_P'''(y)d
y})\frac{1}{\epsilon}\delta'(-x+\phi(t))\\
&\quad +O_{D'}(\epsilon).
\end{align*}
Similarly,
\begin{align*}
&u(x,t,\epsilon)z(x,t,\epsilon)\\
&=u_2z_2+[uz]H(-x+\phi(t))\\
&\quad +[u_2+[u]\int
{w_{0u}(y)w_l(y)d y}]l(t)\delta(-x+\phi(t))\\
&\quad +[u_2+[u]\int
{w_{0u}(y)w_m(y)d y}]m(t)\delta'(-x+\phi(t))\\
&\quad +[u_2 +\frac{[u]}{2}\int
{y^2 w_{0u}(y)w_n(y)d y}]n(t)\delta''(-x+\phi(t))\\
&\quad +[-[u]m(t)\int {w_{u}(y)w_m(y)d y}+[u]n(t)\int {w_{0u}(y)w_n(y)d
y}\\
&\quad +[u]Q(t)\int {w_{u}'(y) w_Q'(y)d
y}]\frac{1}{\epsilon}\delta(-x+\phi(t))\\
&\quad +[u]R(t)\int{w_{0u}(y) w_R''''(y)d
y}\frac{1}{\epsilon}\delta'(-x+\phi(t))+O_{D'}(\epsilon).
\end{align*}
Arranging the coefficient of $ \delta $ and the derivatives,
 $ \frac{1}{\epsilon}\delta $ and $\frac{1}{\epsilon} \delta'$ of
$ 3v(x,t,\epsilon)w(x,t,\epsilon)+u(x,t,\epsilon)z(x,t,\epsilon) $, we obtain
\begin{align*}
& 3v(x,t,\epsilon)w(x,t,\epsilon)+u(x,t,\epsilon)z(x,t,\epsilon)\\
&=(3v_2w_2+u_2z_2)+[3vw+uz]H(-x+\phi(t))\\
&\quad +[3\{(v_2+[v]\int {w_{0v}(y)w_g(y)d y}) g(t)
 +(w_2+[w]\int {w_{0w}(y)w_e(y)dy})e(t)\}\\
&\quad +(u_2+[u]\int {w_{0u}(y)w_l(y)d y})l(t)+3\{(v_{2x} +[v_x]\int
{w_{0v}(y)w_h(y)d y})h(t)\}\\
&\quad +(u_{2x}+[u_x]\int {w_{0u}(y)w_m(y)dy})m(t)\\
&\quad +(u_{2xx} +\frac{[u_{xx}]}{2}\int {y^2 w_{0u}(y)w_n(y)dy})n(t)]
 \big|_{x=\phi(t)}\delta(-x+\phi(t))\\
&\quad +[3\{(v_2 +[v]\int{w_{0v}(y)w_h(y)d y})h(t)\}
 +(u_2+[u]\int {w_{0u}(y)w_m(y)dy})m(t)\\
&\quad -2(u_{2x} +\frac{[u_x]}{2}\int {y^2 w_{0u}(y)w_n(y)dy})n(t)]
 \big|_{x=\phi(t)}\delta'(-x+\phi(t))\\
&\quad +[u_2+\frac{[u]}{2}\int {y^2 w_{0u}(y)w_n(y)dy}]n(t)
 \big|_{x=\phi(t)}\delta''(-x+\phi(t))\\
&\quad +\Big[3e(t)g(t)\int {w_{e}(y)w_g(y)d y}-3[v]h(t)
 \int { w_{v}(y)w_h(y)d y}\\
&\quad -[u]m(t) \int {w_{u}(y)w_m(y)d y} +[u]n(t)\int {w_{0u}(y)w_n(y)d y}\\
&\quad +[u]Q(t)\int{w_{u}'(y) w_Q'(y)d y}\\
&\quad +[u_x]R(t)\int {w_{0u}(y) w_R''''(y)dy}\Big]\big|_{x=\phi(t)}
 \frac{1}{\epsilon}\delta(-x+\phi(t))\\
&\quad +\Big[3e(t)h(t)\int {y w_{e}(y)w_h'(y)d y}+3e(t)p(t)
 \int {y w_{e}(y) w_P'''(y)dy}\\
&\quad +[u]R(t)\int {w_{0u}(y) w_R''''(y)d y}\Big]\big|_{x=\phi(t)}
\frac{1}{\epsilon}\delta'(-x+\phi(t))+O_{D'}(\epsilon).
\end{align*}

\begin{equation}
\begin{aligned}
&( 3v(x,t,\epsilon)w(x,t,\epsilon)+u(x,t,\epsilon)z(x,t,\epsilon))_x \\
&=(3v_2w_2+u_2z_2)_x+[(3vw+uz)_x]H(-x+\phi(t))-[3vw+uz]\delta(-x+\phi(t))\\
&\quad -[3\{(v_2+[v]\int{w_{0v}(y)w_g(y)d y}) g(t)+(w_2+[w]\int {w_{0w}(y)w_e(y)d
y})e(t)\}\\
&\quad +(u_2+[u]\int {w_{0u}(y)w_l(y)d y})l(t)+3\{(v_{2x} +[v_x]\int
{w_{0v}(y)w_h(y)d y})h(t)\}\\
&\quad +(u_{2x}+[u_x]\int {w_{0u}(y)w_m(y)dy})m(t)\\
&\quad +(u_{2xx} +\frac{[u_{xx}]}{2}\int {y^2 w_{0u}(y)w_n(y)d
y}) n(t)]\big|_{x=\phi(t)}\delta'(-x+\phi(t))\\
&\quad -[3\{(v_2 +[v]\int {w_{0v}(y)w_h(y)d y})h(t)\}+(u_2+[u]
\int {w_{0u}(y)w_m(y)d y})m(t)\\
&\quad -2(u_{2x} +\frac{[u_x]}{2}\int {y^2 w_{0u}(y)w_n(y)d
y})n(t)]\big|_{x=\phi(t)}\delta''(-x+\phi(t))\\
&\quad -[u_2+\frac{[u]}{2}\int {y^2 w_{0u}(y)w_n(y)d
y}]n(t)\big|_{x=\phi(t)}\delta'''(-x+\phi(t))\\
&\quad -\Big[3e(t)g(t)\int {w_{e}(y)w_g(y)d y}-3[v]h(t)
 \int { w_{v}(y)w_h(y)d y}\\
&\quad -[u]m(t)\int{w_{u}(y)w_m(y)d y}\\
&\quad +[u]n(t)\int {w_{0u}(y)w_n(y)d y}+[u]Q(t)\int
{w_{u}'(y) w_Q'(y)d y}\\
&\quad +[u_x]R(t)\int {w_{0u}(y) w_R''''(y)d y}\Big]\big|_{x=\phi(t)}
 \frac{1}{\epsilon}\delta'(-x+\phi(t))\\
&\quad -\Big[3e(t)h(t)\int {y w_{e}(y)w_h'(y)d y}+3e(t)p(t)\int {y w_{e}(y)
 w_P'''(y)d y}\\
&\quad +[u]R(t)\int {w_{0u}(y) w_R''''(y)d y}\Big]
 \big|_{x=\phi(t)}\frac{1}{\epsilon}\delta''(-x+\phi(t))+O_{D'}(\epsilon).
\end{aligned}\label{e2.6*}
\end{equation}

Differentiating $z$ with respect to $t$,
\begin{equation}
\begin{aligned}
&z_t(x,t,\epsilon)\\
&=z_{1t}+[z_t]H(-x+\phi(t))+\Big[[z]\dot{\phi}(t)+\dot{l}(t)\Big]
 \delta(-x+\phi(t))\\
&\quad +\big[l(t)\dot{\phi}(t)+\dot{m}(t)\big]\delta'(-x+\phi(t))
 +\big[m(t)\dot{\phi}(t)+\dot{n}(t)\big]\delta''(-x+\phi(t))\\
&\quad +n(t)\dot{\phi}(t)\delta'''(-x+\phi(t))+O_{D'}(\epsilon).
\end{aligned} \label{e2.7}
\end{equation}
Putting the value of $ z_t(x,t,\epsilon)$ from the equations \eqref{e2.7}
 and $ ( 3v(x,t,\epsilon)w(x,t,\epsilon)+u(x,t,\epsilon)z(x,t,\epsilon))_x $
from the equations \eqref{e2.6*} in the fourth equation of \eqref{e1.2}, we obtain
\begin{align*}
&z_t+ 2 ((3vw+uz)_x)\\
&=z_{1t}+2(3v_2w_2+u_2z_2)_x+\Big[[z_t]+2[(3vw+uz)_x]\Big]H(-x+\phi(t))\\
&\quad +\Big[[z]\dot{\phi}(t)+\dot{l}(t)-2[3vw+uz]\Big]\delta(-x+\phi(t))\\
&\quad +\Big[l(t)\dot{\phi}(t)+\dot{m}(t)-2[3\{(v_2+[v]\int {w_{0v}(y)w_g(y)d
y}) g(t)\\
&\quad +(w_2+[w]\int {w_{0w}(y)w_e(y)d y})e(t)\}\\
&\quad +(u_2+[u]\int
{w_{0u}(y)w_l(y)d y})l(t)+3\{(v_{2x} +[v_x]\int {w_{0v}(y)w_h(y)d
y})h(t)\}\\
&\quad +(u_{2x}+[u_x]\int {w_{0u}(y)w_m(y)d y})m(t)\\
&\quad +(u_{2xx}+\frac{[u_{xx}]}{2}\int {y^2 w_{0u}(y)w_n(y)d
y})n(t)]\Big]\delta'(-x+\phi(t))\\
&\quad +\Big[m(t)\dot{\phi}(t)+\dot{n}(t)-2[3\{(v_2
+[v]\int {w_{0v}(y)w_h(y)d y})h(t)\}\\
&\quad +(u_2+[u]\int {w_{0u}(y)w_m(y)dy})m(t)\\
&\quad -(2u_{2x} +[u_x]\int {y^2 w_{0u}(y)w_n(y)d
y})n(t)]\Big]\delta''(-x+\phi(t))
\\
&\quad +\Big[n(t)\dot{\phi}(t)-[2u_2 +[u]\int {y^2 w_{0u}(y)w_n(y)d
 y}]n(t)\Big]\delta'''(-x+\phi(t))\\
&\quad -2\Big[3e(t)g(t)\int {w_{e}(y)w_g(y)d y}-3[v]h(t)
 \int { w_{v}(y)w_h(y)d y}\\
&\quad -[u]m(t)\int{w_{u}(y)w_m(y)d y}\\
&\quad +[u]n(t)\int {w_{0u}(y)w_n(y)d y}+[u]Q(t)\int
 {w_{u}'(y) w_Q'(y)d y}\\
&\quad +[u_x]R(t)\int {w_{0u}(y) w_R''''(y)dy}\Big]\big|_{x=\phi(t)}
 \frac{1}{\epsilon}\delta'(-x+\phi(t))\\
&\quad -2\Big[3e(t)h(t)\int
{y w_{e}(y)w_h'(y)d y}+3e(t)p(t)\int {y w_{e}(y) w_P'''(y)d
y}\\
&\quad +[u]R(t)\int {w_{0u}(y) w_R''''(y)d
y}\Big]\big|_{x=\phi(t)}\frac{1}{\epsilon}\delta''(-x+\phi(t))+O_{D'}(\epsilon).
\end{align*}
So if the relations 14-21 holds then the coefficients of $ \delta $
and their derivatives, $ \frac{1}{\epsilon}\delta $ and
 $\frac{1}{\epsilon} \delta'$ vanishes. The proof is complete.
\end{proof}

For Riemann type data the above expression is simple, and it is described in
the following corollary.

\begin{corollary} \label{coro2.4}
If $u_i,v_i,w_i$ $z_i$ for $i=1,2$ are constants then
expression \eqref{e2.5} is a weak asymptotic solution provided the
following equalities hold.
\begin{gather*}
\dot{\phi}(t)=(u_1+u_2)\big|_{x=\phi(t)},\quad
\dot{e}(t)=[u](v_1+v_2)\big|_{x=\phi(t)}, \\
\dot{g}(t)=(2[v](v_1+v_2)+[u](w_1+w_2)\big|_{x=\phi(t)} , quad
\frac{d}{dt}(h(t)[u(\phi(t),t)])=\frac{d}{dt}e^2 (t)\\
\int {w_{0u}(y)w_j(y)d y}=\int {y^2w_{0v}(y)w_e(y)d y}=\frac{1}{2},\quad j=e,g,h ,\\
\int{w_{u}(y)w_h(y)d y}=\int{w_{e}^2(y)d y} ,\quad
P(t)=\frac{A}{u_1(\phi(t),t)},\quad \text{where $A$ is a constant},\\
\dot{l}(t)=-[z]\dot{\phi}(t)+2[3vw+uz],\\
\int {w_{0u}(y)w_l(y)d y}=\frac{1}{2}\int {y^2w_{0u}(y)w_n(y)d y}=\int
{w_{0u}(y)w_m(y)d y}=\frac{1}{2},\\
\dot{m}(t)=2[3\{(v_2+[v]\int {w_{0v}(y)w_g(y)d y}) g(t)+(w_2+[w]\int
{w_{0w}(y)w_e(y)d y})e(t)\},\\
\dot{n}(t)=2[3\{(v_2 +[v]\int {w_{0v}(y)w_h(y)d y})h(t)\},\\
\begin{aligned}
R(t)&=\frac{1}{[u]\int {w_{0u}(y) w_R''''(y)d y}}[3e(t)h(t)\int {y
w_{e}(y)w_h'(y)d y}\\
&\quad +3e(t)p(t)\int {y w_{e}(y) w_P'''(y)d y}],
\end{aligned}
\end{gather*}
\begin{equation}
\begin{aligned}
Q(t)&=\frac{1}{[u]\int {w_{u}'(y) w_Q'(y)d y}}\Big[3e(t)g(t)\int {
w_{e}(y)w_g(y)d y}\\
&\quad -3[v]h(t)\int { w_{v}(y)w_h(y)d y}
 -[u]m(t)\int {w_{u}(y)w_m(y)d y}+\frac{[u]n(t)}{2}\Big]
\end{aligned} \label{e2.8}
\end{equation}
\end{corollary}

Piecing together the Riemann problems we construct a weak asymptotic
solution for general type initial data under the assumption that $u$ is a
monotonic increasing function.

\begin{theorem} \label{thm2.5}
If $u_0,v_0,w_0$ and $z_0$ are locally integrable functions on
$\mathbb{R}$, and $u_0$ is monotonic increasing, then there exists 
weak asymptotic solution $ (u,v,w,z) $ to the system \eqref{e1.2}
with initial data \eqref{e1.3}.
\end{theorem}

\begin{proof}
Let $\phi$ be a test function on $\mathbb{R}$ having support in $[-K, K]$.
Given $\epsilon>0$, there exist piecewise constant functions
$(u_{0\epsilon},v_{0\epsilon},w_{0\epsilon},z_{0\epsilon})$ such that
\begin{gather*}
\int_{[-K,K]}|u_0(x)-u_{0\epsilon}(x)|dx
<\epsilon,\quad \int_{[-K,K]}|v_0(x)-v_{0\epsilon}(x)|dx <\epsilon,\\
\int_{[-K,K]}|w_0(x)-w_{0\epsilon}(x)|dx
<\epsilon,\quad \int_{[-K,K]}|z_0(x)-z_{0\epsilon}(x)|dx <\epsilon.
\end{gather*}
In addition to this we can take $u_{0\epsilon}$ monotonic increasing and
all functions have same points of discontinuities.
$(u_{0\epsilon},v_{0\epsilon},w_{0\epsilon},z_{0\epsilon})$ in $[-K,K]$
can be represented as
\begin{gather*}
u_{0\epsilon}=\sum_{i=1}^{n}{u_{0i}(H(x-a_{i-1})-H(x-a_i))},\\
v_{0\epsilon}=\sum_{i=1}^{n}{v_{0i}(H(x-a_{i-1})-H(x-a_i))},\\
w_{0\epsilon}=\sum_{i=1}^{n}{w_{0i}(H(x-a_{i-1})-H(x-a_i))},\\
z_{0\epsilon}=\sum_{i=1}^{n}{z_{0i}(H(x-a_{i-1})-H(x-a_i))}.
\end{gather*}
Since $ u_{0\epsilon} $ is a monotonic increasing function, discontinuity
curve arising in the solution of $ (u,v,w,z)$ do not intersect for any
time. So the following functions are weak asymptotic solutions
\begin{gather*}
\begin{aligned}
u(x,t,\eta)
&=u_{01}H_u(-x+c_1 t+a_1,\eta)+\sum_{i=2}^{n-1} u_{0i} \Big(H_u(x-c_{i-1}t
 -a_{i-1},\eta)\\
&\quad -H_u(x-c_it-a_i,\eta)\Big)+u_{0n}(H_u(x-c_{n-1}t-a_{n-1},\eta),
\end{aligned}\\
\begin{aligned}
v(x,t,\eta)
&=v_{01}H_v(-x+c_1 t+a_1,\eta)+\sum_{i=2}^{n-1} v_{0i} \Big(H_v(x-c_{i-1}t-a_{i-1},
\eta) \\
&\quad -H_v(x-c_it-a_i,\eta)\Big)  +v_{0n} H_v(x-c_{n-1}t-a_{n-1},\eta)\\
&\quad +\sum_{i=1}^{n-1} e_i(t)\delta_e(-x+c_it,\eta),
\end{aligned}\\
\begin{aligned}
w(x,t,\eta)
&=w_{01}H_w(-x+c_1 t+a_1,\eta)+\sum_{i=2}^{n-1} w_{0i}\Big(H_w(x-c_{i-1}t-a_{i-1},\eta)\\
&\quad -H_w(x-c_it-a_i,\eta)\Big) +w_{0n} H_w(x-c_{n-1}t-a_{n-1},\eta)\\
&\quad +\sum_{i=1}^{n-1} g_i(t)\delta_g(-x+c_i t,\eta)
 +\sum_{i=1}^{n-1} h_i(t)\delta_h'(-x+c_i t,\eta)\\
&\quad +\sum_{i=1}^{n-1} R_{wi}(-x+c_i t,\eta),
\end{aligned}\\
\begin{aligned}
z(x,t,\eta)
&=z_{01}H_z(-x+c_1 t+a_1,\eta)+\sum_{i=2}^{n-1} z_{0i}
 \Big(H_z(x-c_{i-1}t-a_{i-1},\eta)\\
&\quad -H_z(x-c_it-a_i,\eta)\Big)
 +z_{0n}(H_z(x-c_{n-1}t-a_{n-1},\eta))\\
&\quad +\sum_{i=1}^{n-1} l_i(t)\delta_l(-x+c_i t,\eta)
 +\sum_{i=1}^{n-1}{m_i(t)\delta_m'(-x+c_it,\eta)}\\
&\quad +\sum_{i=1}^{n-1}{n_i(t)\delta_n''(-x+c_i t,\eta)}
 +\sum_{i=1}^{n-1}{R_{zi}(-x+c_i t,\eta)},
\end{aligned}
\end{gather*}
where $ e_i$, $g_i$, $h_i$, $l_i$, $m_i$, $n_i$, $R_{wi}$ and $R_{zi}$ satisfy
\eqref{e2.8} with $u_1$, $u_2$, $v_1$, $v_2$, $w_1$, $w_2$, $z_1$, $z_2$, $e$,
$g$, $h$, $l$, $m$, $n$, $R_w$ and $R_z$
replaced by $ u_{i-1}$, $u_i$, $v_{i-1}$, $v_i$, $w_{i-1}$, $w_i$, $z_{i-1}$,
$z_i$, $e_i$, $g_i$, $h_i$, $l_i$, $m_i$, $n_i$, $R_{wi}$ and $R_{zi}$.
Given $\epsilon >0$ choose $\eta(\epsilon)$ small enough such that the
following estimates hold.
\begin{gather*}
\big|\int L_1[u(x,t,\eta(\epsilon))]\psi(x)dx\big|<\eta(\epsilon),\quad
\big|\int L_2[u(x,t,\eta(\epsilon)),v(x,t,\eta(\epsilon))]\psi(x)\big|<\epsilon,\\
\big|\int L_3[u(x,t,\eta(\epsilon)),v(x,t,\eta(\epsilon)),w(x,t,\eta(\epsilon))]
 \psi(x)dx\big|<\epsilon,\\
\big|\int L_4[u(x,t,\eta(\epsilon)),v(x,t,\eta(\epsilon)),w(x,t,\eta(\epsilon)),
z(x,t,\eta(\epsilon))]\psi(x)dx\big|<\epsilon,\\
\big|\int\Big(u(x,0,\eta(\epsilon))-u_0 (x)\Big)\psi(x)dx\big|<
2\epsilon,\\
\big|\int\Big(v(x,0,\eta(\epsilon))-v_0 (x)\Big)\psi(x)dx\big|<2\epsilon,\\
\big|\int\Big(w(x,0,\eta(\epsilon))-w_0 (x)\Big)\psi(x)dx\big|<
2\epsilon,\\
 \big|\int\Big(z(x,0,\eta(\epsilon))-z_0 (x)\Big)\psi(x)dx\big|<2\epsilon.
\end{gather*}
Define
\begin{align*}
&(\bar{u}(x,t,\epsilon),\bar{v}(x,t,\epsilon),\bar{w}(x,t,\epsilon),
\bar{z}(x,t,\epsilon))\\
&=(u(x,t,\eta(\epsilon),v(x,t,\eta(\epsilon),w(x,t,\eta(\epsilon),
z(x,t,\eta(\epsilon)).
\end{align*}
Then $(\bar{u},\bar{v},\bar{w},\bar{z})$ is a weak asymptotic solution
of system \eqref{e1.2}-\eqref{e1.3}.
\end{proof}


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\end{document}