12.1. Burgers equation

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

# Chapter 12. Nonlinear equations

## 12.1. Burgers equation

### Two problems

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 $f$.   For $f(u)=u$ as $u_- < u_+$ characteristics and solution consecutive profiles (slightly shifted up)

Case 2. $f(u_-) > f(u_+)$. In this case characteristics collide For $f(u)=u$ as $u_- > u_+$ characteristics

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_+))$: For$f(u)=u$as$u_- < u_+$characteristics and a line of jump (profiles are just shifted steps) ### Shock waves Huston, we have a problem However allowing discontinuous solutions to (\ref{eq-12.1.6}) we opened a floodgate to many discontinuous solution and broke unicity. Indeed, let us return to Case 1 and construct solution in the same manner as in Case 2. Then we get (\ref{eq-12.1.8})--(\ref{eq-12.1.7}) solution with$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$\varphiU(\varphi)\ge 0$. It was proven Theorem 1. 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. ### Examples Example 1. The truly interesting example is Burgers equation ($f(u)=u) with initial conditions which are not monotone. So we take \begin{equation} u(x,0)=\phi(x):=\left\{\begin{aligned} 1& &&x<-1,\\ -1& &&&-1< x< 0,\\ 1& &&x > 1. \end{aligned}\right. \label{eq-12.1.14} \end{equation} Then obviously the solution is first provided by a combination of Case 1 and Case 2: \begin{equation} u(x,t)=\left\{\begin{aligned} 1& &&x<-1,\\ -1& &&&-1< x< -t,\\ \frac{x}{t}& &&& -t< x <t,\\ 1& && x > t. \end{aligned}\right. \label{eq-12.1.15} \end{equation} This holds as0< t<1$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$u\mapsto -u. Example 2. Consider now \begin{equation} u(x,0)=\phi(x):=\left\{\begin{aligned} 2& &&x<-1,\\ 0& &&&-1< x< 1,\\ -2& &&x > 1. \end{aligned}\right. \label{eq-12.1.17} \end{equation} Then for0< t< 1, we have two shock waves: \begin{equation} u(x,t)=\left\{\begin{aligned} u_-& &&x<-1+t,\\ u_0& &&&-1+ t< x < 1- t,\\ u_+& &&x > 1-t \end{aligned}\right. \label{eq-12.1.18} \end{equation} and fort=1$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} 