Solving the inviscid Burgers' equation $u_t + uu_x = 0$

Question

Here, we are given the initial condition: $$u(0, x) = \begin{cases} \, 1 & \text{when } x \ge 0 \\ \, -1 & \text{when } x < 0. \end{cases}$$ I am aware that then solution has the general form $u(t, x) = u(u(x)t + x_0)$. My question is how to find where the solutions exist? I think it's clear that it will not be the entire $t$-$x$ plane, but I don't know how to think about these recursive situations.

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$

$u(0,x)=\begin{cases}1&\text{when}~x\geq0\\-1&\text{when}~x<0\end{cases}$ :

$\therefore u=\begin{cases}1&\text{when}~x-ut\geq0\\-1&\text{when}~x-ut<0\end{cases}=\begin{cases}1&\text{when}~x-t\geq0\\-1&\text{when}~x+t<0\end{cases}$

Hence $u(t,x)=\begin{cases}1&\text{when}~x-t\geq0\\-1&\text{when}~x+t<0\\c&\text{otherwise}\end{cases}$

Answer 2

Hint: Here is a weak 1-parameter solution $$u(t,x)~=~\theta (x\!-\!t\!+\!t_0)\theta(x) - \theta(-x\!-\!t\!+\!t_0)\theta(-x)+\frac{x}{t\!-\!t_0}\theta(t\!-\!t_0\!-\!|x|) , $$ $$ (t,x)~\in~\mathbb{R}^2\backslash (]-\infty,t_0]\times \{0\} ), \qquad t_0~\in~[0,\infty[, \tag{1}$$ to the inviscid Burgers' equation

$$ \frac{\partial u}{\partial t} +\frac{1}{2}\frac{\partial (u^2)}{\partial x}~=~0,\qquad u(t\!=\!0,x)~=~{\rm sgn}(x).\tag{2}$$

The solution (1) is defined and continuous everywhere in the plane $\mathbb{R}^2$ except for a halfline, where it has a kink. Physically, the solution (1) describes the dispersion of a kink at some future time $t=t_0$. If we take the parameter $t_0\to \infty$, we get the constant kink solution $$ u(t,x)~=~{\rm sgn}(x), \qquad x~\neq~ 0.\tag{3}$$ The solutions (1) & (3) are not the only solutions, cf. above comments by user Mattos & Hans Lundmark. This is due to rarefaction in the future region $t>|x|$ and shock in the past region $-t>|x|$, cf. e.g. Scott A. Sarra's website.

Sketched check of eq. (1): $$u^2~=~\theta (|x|\!-\!t\!+\!t_0)+\frac{x^2}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|) .\tag{4}$$ $$ \frac{\partial (u^2)}{\partial x}~=~\left({\rm sgn}(x)-\frac{x|x|}{(t\!-\!t_0)^2} \right)\delta (|x|\!-\!t\!+\!t_0)+\frac{2x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|)$$ $$~=~ \frac{2x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|).\tag{5}$$ $$ \frac{\partial u}{\partial t} ~=~ -\delta (x\!-\!t\!+\!t_0)\theta(x)+ \delta (x\!+\!t\!-\!t_0)\theta(-x)+\frac{x}{t\!-\!t_0}\delta (|x|\!-\!t\!+\!t_0)-\frac{x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|)$$ $$~=~-\frac{x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|).\tag{6}$$