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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?

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The characteristic curves issued from the initial data are the curves $x = x_0 + \phi(x_0) t$ displayed below:

char

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, $u_s(t) \to 0$.

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