Consider $u_t + uu_x + \alpha u = 0$ for $t > 0$, all $x$ where $\alpha > 0$ is a
constant. Find the characteristic equations for the equation with
initial data $u(x, 0) = f(x)$ given. Show that a shock cannot form if $\alpha
\geq \max_{r \in H}|f'(r)|$ where $H = \{r : f'(r) < 0\}$ or if $H$ is empty.
So far, I've found the characteristics by parametrizing
$$\begin {cases} x_s=u, x(0,r)=r \\ t_s=1,t(0,r)=0 \\ u_s = -\alpha u, u(0,r)=f(r)\end {cases}$$
Then
$\frac{du}{ds}=-\alpha u \Rightarrow u = C_1 e^{-\alpha s}$. Considering the initial condition, $u = f(r)e^{-\alpha s}.$
$\frac{dt}{ds} = 1 \Rightarrow t = s$ (since $t(0,r)=0$)
$\frac{dx}{ds}=u=f(r)e^{-\alpha s} \Rightarrow x = -\frac{1}{\alpha}f(r)e^{-\alpha s}+\frac{1}{\alpha}f(r)+r$ (since $x(0,r)=r$),
i.e. $x = -\frac{1}{\alpha}f(r)e^{-\alpha t}+\frac{1}{\alpha}f(r)+r$
So how do we show that a shock cannot form?
Best Answer
This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves $$ x = -f(x_0)\frac{e^{-\alpha t} - 1}{\alpha} + x_0 $$ along which the solution satisfies $$ u = f\left(x - \frac{e^{\alpha t}-1}{\alpha} u\right)e^{-\alpha t} . $$ Computing $\frac{\text d x}{\text d x_0}$, we find that this derivative vanishes at a given positive time $t$ -i.e. a shock wave forms- if $$ -\frac{\ln(1+\alpha/f'(x_0))}{\alpha} = t >0 . $$ For this to be possible, the logarithm should be negative, and thus, $f'(x_0)<0$ (in other words, $x_0\in H$). However, if $-\alpha < f'(x_0) < 0$ for $x_0$ in $H$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $\max_{x_0 \in H}|f'(x_0)| \leq \alpha$ or if $H = \emptyset$. Alternatively, this condition may be written $\inf_{x_0 \in\Bbb R} f'(x_0) \geqslant -\alpha$.