$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

Consider equation \begin{equation} u_t + f(u)u_x =0,\qquad t >0 \label{eq-12.1.1} \end{equation} Then we consider a problem \begin{equation} u(x,0)= \left\{\begin{aligned} u_{-} &&x <0,\\\\ u_{+} &&x >0 \end{aligned}\right. \label{eq-12.1.2} \end{equation} There are two cases:

**Case 1.**
$f(u_-) < f(u_+)$.

In this case characteristics \begin{equation} \frac{dt}{1}=\frac{dx}{f(u)}=\frac{du}{0} \label{eq-12.1.3} \end{equation} originated at $\{(x,0):\, 0< x<\infty\}$ fill $\{x< f(u_-)t,\, t>0\}$ where $u=u_-$ and $\{x > f(u_+)t,\, t>0\}$ where $u=u_+$ and leave sector $\{ f(u_-)t < x< f(u_+)t,\, t>0\}$ empty. In this sector we can construct continuous self-similar solution $u= g(x/t)$ and this construction is unique provided $f$ is monotone function (say increasing) \begin{equation} f'(u)>0 \label{eq-12.1.4} \end{equation} (and then $f(u_-) < f(u_+)$ is equivalent to $u_< u_+$). Namely \begin{equation} u(x,t)= \left\{\begin{aligned} u_-& &&x < f(u_-)t,\\ g\left(\frac{x}{t}\right)& &&f(u_-)t < x < f(u_+)t,\\ u_+& &&x >f(u_+)t \end{aligned}\right. \label{eq-12.1.5} \end{equation} provides solution for (\ref{eq-12.1.1})-(\ref{eq-12.1.2}) where $g$ is an inverse function to $f$.

**Case 2.**
$f(u_-) > f(u_+)$. In this case characteristics collide

and to provide solution we need to reformulate our equation (\ref{eq-12.1.1}) as \begin{equation} u_t + (F(u))_x =0,\qquad t >0 \label{eq-12.1.6} \end{equation} where $F(u)$ is a primitive of $f$: $F'(u)=f(u)$. Now we can understand equation in a weak sense and allow discontinuous (albeit bounded) solutions. So, let us look for \begin{equation} u=\left\{\begin{aligned} u_-& &&x < st,\\ u_+& &&x >st \end{aligned}\right. \label{eq-12.1.7} \end{equation} where so far $s$ is unknown. Then $u_t= -s[u_+-u_-]\delta (x-st)$, $u_x= [F(u_+)-F(u_-)]\delta (x-st)$ where brackets contain jumps of $u$ and $F(u)$ respectively and (\ref{eq-12.1.6}) means exactly that \begin{equation} s= \frac{[F(u_+)-F(u_-)]}{u_+-u_-}$ \label{eq-12.1.8} \end{equation} which is equal to $F'(v)=f(v)$ at some $v\in (u_-, u_+)$ and due to (\ref{eq-12.1.4}) $s\in (f(u_-), f(u_+))$:

**Huston, we have a problem**

However allowing discontinuous solutions to (\ref{eq-12.1.6}) we opened a floodgate to many discontinuous solution and broke unicity. Indeed, let us return to Case 1 and construct solution in the same manner as in Case 2. Then we get (\ref{eq-12.1.8})--(\ref{eq-12.1.7}) solution with $s=F'(v)$ at some $v\in (u_+, u_-)$ and due to (\ref{eq-12.1.4}) $s\in (f(u_+), f(u_-))$:

So, we got two solutions: new discontinuous and old continuous (\ref{eq-12.1.5}). In fact, situation is much worse since there are many hybrid solutions in the form (\ref{eq-12.1.8}) albeit with discontinuous $g$. To provide a uniqueness we need to weed out all such solutions.

**Remark 1.**
Equation (\ref{eq-12.1.6}) is considered to be a *toy-model* for gas dynamics. Discontinuous solutions are interpreted as *shock waves* and solution in Case 1 are considered *rarefaction waves*. Because of this was formulated principle *there are no shock rarefaction waves* which mathematically means $u(x-0,t)\ge u(x+0,t)$ where $u(x\pm 0,t)$ are limits from the right and left respectively.

However it is not a good condition in the long run: for more general solutions these limits may not exist. To do it we multiply equation (\ref{eq-12.1.1}) by $u$ and write this new equation \begin{equation} uu_t + f(u)uu_x =0,\qquad t >0 \label{eq-12.1.9} \end{equation} in the divergent form \begin{equation} (\frac{1}{2}u^2)_t + (\Phi (u))_x =0,\qquad t >0 \label{eq-12.1.10} \end{equation} where $\Phi(u)$ is a primitive of $uf(u)$.

**Remark 2.**
Let us observe that while equations (\ref{eq-12.1.1}) and (\ref{eq-12.1.9}) are equivalent for continuous solutions, equations (\ref{eq-12.1.6}) and (\ref{eq-12.1.10}) are not equivalent for discontinuous ones. Indeed, for solution (\ref{eq-12.1.7}) the left-hand expression in (\ref{eq-12.1.10}) is
\begin{equation}
K \delta (x-st)
\label{eq-12.1.11}
\end{equation}
with
\begin{multline}
K(u_-,u_+) = -s\frac{1}{2}[u_+^2-u_-^2] + [\Phi(u_+)-\Phi(u_-)] =\\
-\frac{1}{2}(u_+ + u_-)[F(u_+)-F(u_-)] + [\Phi(u_+)-\Phi(u_-)]\qquad
\label{eq-12.1.12}
\end{multline}

**Exercise 1.**
Prove that $K(u_-,u_+)\gtrless 0$ as $u_+ \gtrless u_-$. To do it consider
$K(u-v,u+v)$, observe that it is $0$ as $v=0$ and $\partial_v K(u-v,u+v)=
v (f(u+v)-f(u-v))>0$ for $v\ne 0$ due to (\ref{eq-12.1.4}).

So for "good" solutions \begin{equation} (\frac{1}{2}u^2)_t + (\Phi (u))_x \le 0,\qquad t >0 \label{eq-12.1.13} \end{equation} where we use the following

**Definition 1.**
Distribution $U\ge 0$ if for all non-negative test functions $\varphi$
$U(\varphi)\ge 0$.

It was proven

**Theorem 1.**
Assume that $f'(u)>0$. Then solution to the problem (\ref{eq-12.1.6}), (\ref{eq-12.1.2}) with additional restriction (\ref{eq-12.1.13}) exists and is unique as $\phi$ is *bounded total variation function*.

**Remark 3.**
Restriction (\ref{eq-12.1.13}) is interpreted as *entropy cannot decrease.*

**Example 1.**
The truly interesting example is Burgers equation ($f(u)=u$) with initial conditions which are not monotone. So we take
\begin{equation}
u(x,0)=\phi(x):=\left\{\begin{aligned}
1& &&x1,\\
-1& &&&-1< x< 0,\\
1& &&x > 1.
\end{aligned}\right.
\label{eq-12.1.14}
\end{equation}
Then obviously the solution is first provided by a combination of
Case 1 and Case 2:
\begin{equation}
u(x,t)=\left\{
\begin{aligned}
1& &&x > 1,\\
-1& && -1< x< -t,\\
\frac{x}{t}& && -t< x < t,\\
1& && x > t.
\end{aligned}\right.
\label{eq-12.1.15}
\end{equation}
This holds as $0< t<1$ because at $t>1$ rarefaction and shock waves collide.

Now there will be a shock wave at $x=\xi (t)$, $t>1$. On its left $u=1$, on its right $u=\xi t^{-1}$ and therefore slope is a half-sum of those: \begin{equation} \frac{d\xi}{dt}=\frac{1}{2} + \frac{\xi}{2t}; \label{eq-12.1.16} \end{equation} this ODE should be solved with initial condition $\xi(1)=-1$ and we have $\xi(t)=t-2t^{\frac{1}{2}}$;

Observe that while $\max_{-\infty < x < \infty} u(x,t)=1$, $\min _{-\infty < x < \infty} u(x,t)=\xi(t)t^{-1}=1-2t^{-\frac{1}{2}}$ and \begin{equation*} \bigl(\max_{-\infty < x < \infty} u(x,t)- \min_{-\infty < x < \infty} u(x,t)\bigr)\to 0\qquad \text{as } t\to +\infty. \end{equation*}

We can consider example with $u(x,0)=-\phi(x,0)$ by changing $x\mapsto -x$ and $u\mapsto -u$.

**Example 2.**
Consider now
\begin{equation}
u(x,0)=\phi(x):=\left\{\begin{aligned}
2& &&x1,\\
0& &&&-1< x< 1,\\
-2& &&x > 1.
\end{aligned}\right.
\label{eq-12.1.17}
\end{equation}
Then for $0< t< 1$, we have two shock waves:
\begin{equation}
u(x,t)=\left\{\begin{aligned}
&u_- &&x1+t,\\
&u_0 &&&-1+ t< x < 1- t,\\
&u_+ &&x > 1-t
\end{aligned}\right.
\label{eq-12.1.18}
\end{equation}
and for $t=1$ both shock waves collide at $x=0$ and then for $t>1$
\begin{equation}
u(x,t)=\left\{\begin{aligned}
2& &&x<0,\\\\
-2& &&x > 0.
\end{aligned}\right.
\label{eq-12.1.19}
\end{equation}

One can prove that

**Theorem 2.**
Assume that $f'(u)>0$. Then solution $v=v(x,t,\varepsilon)$ of the problem
\begin{align}
&v_t+uu_x=\varepsilon v_{xx},\qquad t>0, -\infty < x<\infty
\label{eq-12.1.20}\\
&v|_{t=0}=f(x).
\label{eq-12.1.21}
\end{align}
converges (in the sense of distributions) to solution $u(x,t)$ of (\ref{eq-12.1.1}) with the same initial $u(x,0)=f(x)$ condition as $\varepsilon\to +0$.

We claim that this solution satisfies (\ref{eq-12.1.13}). Indeed, from (\ref{eq-12.1.1}) we conclude that \begin{equation*} (\frac{1}{2}u^2)_t+(\Phi(u))_x= \varepsilon uu_{xx}= \varepsilon (\frac{1}{2}u^2)_{xx} - \varepsilon u_x^2; \end{equation*} applying it to test function $\varphi \ge 0$ we pass to the limit in the left (requires justification) and in the first term on the right (also requires justification, results in $0$) but the last term on the right is $-\varepsilon u_x^2 (\varphi)$ which results in $\le 0$ (but not necessarily equal).

It turns out that the convergence is uniform away from shockwave, but around shockwave appears a layer of the width of magnitude $\varepsilon$ where convergence cannot be uniform (because there continuous and even infinitely smooth solution to the viscous Burgers equation (\ref{eq-12.1.20}) converges to the discontinuous one.