# Long time evolution of Burgers' equation ($t\to\infty$)

## Problem

Draw the characteristics and describe the evolution for $$t \to \infty$$ of the solution of the problem \begin{align}\begin{cases}u_{t} + u u_{x} = 0 & t > 0 , x \in \mathbb{R} \\ u(x,0) = \phi(x) & x \in \mathbb{R} \end{cases} \end{align} \tag{1}

where $$\phi(x)$$ is given by \phi(x) = \begin{align}\begin{cases} \sin(x) & 0 < x < \pi \\ 0 & x \leq 0 \textrm{ or } x \geq \pi \end{cases} \end{align} \tag{2}

## Attempt

If we take the characteristics of burgers equation

$$u_{t} + u u_{x} = 0 \tag{3}$$

we'll have

$$\frac{dx}{dt} = u \\ \frac{du}{dt} = 0 \tag{4}$$

$$x(t) = ut+x_{0} \\ u = c_{0} \tag{5}$$

then we get that

$$c_{0} = \phi(x_{0}) \implies x(t) =\phi(x_{0})t + x_{0} \tag{6}$$

$$u(x,t) = \phi(x_{0}) = \phi(x-ut) \tag{7}$$ u(x,t) = \begin{align}\begin{cases} \sin(x-c_{0}t) & 0 < x < \pi \\ 0 & x \leq 0 \textrm{ or } x \geq \pi \end{cases} \end{align} \tag{8}

now using trig identities

$$\sin(\alpha-\beta) = \sin(\alpha)\cos(\beta) -\cos(\alpha)\sin(\beta) \tag{9}$$

this gives us

$$\sin(x-c_{0}t) = \sin(x)\cos(c_{0}t) -\cos(x)\sin(c_{0}t) \tag{10}$$

How am I supposed to draw the characteristics? I understand they're between the $$x-t$$ axis. This doesn't look easy. Is there a simple method? Is there a plotting tool?

The characteristic curves issued from the initial data are the curves $$x = x_0 + \phi(x_0) t$$ displayed below: For short times, the solution is given by the method of characteristics, i.e., $$u=\phi(x-ut)$$ is satisfied by $$u$$. Here characteristics intersect at the breaking time $$t_b = {-1}/\inf_x \phi'(x) = 1$$, where a shock wave occurs. The shock-wave abscissa $$x_s(t)$$ satisfies the Rankine-Hugoniot condition x'_s(t) = \frac{1}{2} \big(0 + \sin(x_0(t))\big) \quad\text{where}\quad \left\lbrace \begin{aligned} &x_0(t) + t \sin(x_0(t)) = x_s(t)\\ &0\leq x_0(t)\leq \pi \end{aligned}\right. with initial condition $$x_s(1) = \pi$$. Solving for $$x_0(t)$$, we have $$x'_0(t) = -\frac{1}{2}\frac{\sin x_0(t)}{1 + t \cos x_0(t)}$$ with initial condition $$x_0(1) = \pi$$. Since $$u$$ is constant along characteristics, the maximum value of the solution is $$u_s(t) = u(x_s(t),t) = \sin(x_0(t))$$. As $$t\to {+\infty}$$, we have $$x'_0(t)\to 0$$ and $$x_0(t) \to 0$$. Therefore, $$u_s(t) \to 0$$.