12.1. Burgers equation


# Chapter 12. Nonlinear equations

## 12.1. Burgers equation

### Two problems

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

Case 1. $f(u_-) < f(u_+)$. In this case characteristics $$\frac{dt}{1}=\frac{dx}{f(u)}=\frac{du}{0} \label{eq-12.1.3}$$ 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) $$f'(u)>0 \label{eq-12.1.4}$$ (and then $f(u_-) < f(u_+)$ is equivalent to $u_< u_+$). Namely 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} 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 $$u_t + (F(u))_x =0,\qquad t >0 \label{eq-12.1.6}$$ 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 u=\left\{\begin{aligned} u_-& &&x < st,\\ u_+& &&x >st \end{aligned}\right. \label{eq-12.1.7} 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 $$s= \frac{[F(u_+)-F(u_-)]}{u_+-u_-} \label{eq-12.1.8}$$ 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 $$uu_t + f(u)uu_x =0,\qquad t >0 \label{eq-12.1.9}$$ in the divergent form $$(\frac{1}{2}u^2)_t + (\Phi (u))_x =0,\qquad t >0 \label{eq-12.1.10}$$ 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 $$K \delta (x-st) \label{eq-12.1.11}$$ 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 $$(\frac{1}{2}u^2)_t + (\Phi (u))_x \le 0,\qquad t >0 \label{eq-12.1.13}$$ 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. 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 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} Then obviously the solution is first provided by a combination of Case 1 and Case 2: 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} 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: $$\frac{d\xi}{dt}=\frac{1}{2} + \frac{\xi}{2t}; \label{eq-12.1.16}$$ 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 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} Then for $0< t< 1$, we have two shock waves: 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} and for $t=1$ both shock waves collide at $x=0$ and then for $t>1$ u(x,t)=\left\{\begin{aligned} 2& &&x<0,\\ -2& &&x > 0. \end{aligned}\right. \label{eq-12.1.19}