12.1. Burgers equation

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# 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