Shock formation condition in IVP of $u_t + uu_x + \alpha u = 0$

Question

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?

Answer 1

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 $\inf_{x_0 \in\Bbb R} f'(x_0) \geqslant -\alpha$.

Answer 2

Harry49 already answered to the question.

I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.

The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below. $$u_t+uu_x=-\alpha u \tag 1$$ The Charpit-Lagrange equations are : $$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{(-\alpha u)}$$ A first family of characteristic curves comes from $\frac{dx}{u}=\frac{du}{(-\alpha u)}$ : $$u+\alpha x =c_1$$ A second family of characteristic curves comes from $\frac{dt}{1}=\frac{du}{(-\alpha u)}$ : $$ue^{-\alpha t}=c_2$$ The general solution of the PDE Eq.$(1)$ expressed on the form of implicit equation is : $$u+\alpha x=\Phi\left(ue^{-\alpha t}\right) \tag 2$$ where $\Phi$ is an arbitrary function (to be determined according to boundary condition).

Condition : $u(x,0)=f(x)$ with $f(x)$ a known (given) function.

$$f(x)+\alpha x=\Phi\left(f(x)e^{0}\right)=\Phi\left(f(x)\right)$$ Let $X=f(x)$ and $x=f^{-1}(X)$

$f^{-1}$ denotes the inverse function of $f$. $$\Phi(X)=X+\alpha f^{-1}(X)$$ So, the function $\Phi$ is determined. We put it into Eq.$(2)$. $$u+\alpha x=ue^{-\alpha t}+\alpha f^{-1}\left(ue^{-\alpha t}\right)$$ $$f^{-1}\left(ue^{-\alpha t}\right)=x+\frac{u}{\alpha}\left(1-e^{-\alpha t}\right)$$ $$ue^{-\alpha t}=f\left(x+\frac{u}{\alpha}\left(1-e^{-\alpha t}\right)\right) \tag 3$$ Eq.$(3)$ is the implicite form of the analytic solution of $u_t+uu_x+\alpha u=0$ with condition $u(x,0)=f(x)$.

The explicite form of $u(x,t)$ requires to solve Eq.$(3)$ for $u$. The possibility to do it analytically depends on the kind of function $f$.

Of course, this is only for information without answering to the OP question as mentioned at first place.