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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?

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
$$
u = fleft(x - frac{e^{alpha t}-1}{alpha} uright)e^{-alpha t} .
$$
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, $x_0in H$). 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 $inf_{x_0 inBbb R} f'(x_0) geqslant -alpha$.

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 :
$$u+alpha x=Phileft(ue^{-alpha t}right) tag 2$$
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.

$$f(x)+alpha x=Phileft(f(x)e^{0}right)=Phileft(f(x)right)$$
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.$(2)$.
$$u+alpha x=ue^{-alpha t}+alpha f^{-1}left(ue^{-alpha t}right)$$
$$f^{-1}left(ue^{-alpha t}right)=x+frac{u}{alpha}left(1-e^{-alpha t}right)$$
$$ue^{-alpha t}=fleft(x+frac{u}{alpha}left(1-e^{-alpha t}right)right) tag 3$$
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.