= Solution
For the negative-flux <Inviscid Burgers equation>, the <method of characteristics> gives
$$
\frac{d\theta}{dz}=-f,\qquad \frac{df}{dz}=0.
$$
The <characteristic curve> starting at $\theta_0$ therefore has $\theta=\theta_0-f_0(\theta_0)z$, and hence
$$
\boxed{f\big(z,\theta_0-f_0(\theta_0)z\big)=f_0(\theta_0).}
$$
For smooth data this describes a single-valued classical solution as long as the <characteristic flow map> is invertible. Its Jacobian is $1-zf_0'(\theta_0)$; after <characteristic crossing>, one must instead select a <weak solution> with the appropriate <entropy solution> condition.
The conservation form is $f_z+G(f)_\theta=0$, with <conservation law flux> $G(f)=-f^2/2$. Integrating this <scalar conservation law> across a moving discontinuity, or differentiating its Heaviside representation as a <distribution>, yields the <Rankine-Hugoniot condition>
$$
\theta_s'(f_R-f_L)=G(f_R)-G(f_L).
$$
For distinct one-sided limits it simplifies to
$$
\boxed{\theta_s'=-\frac{f_R+f_L}{2}.}
$$
This jump-speed relation is exact for the Burgers conservation law; an additional entropy condition is needed to distinguish a physical compressive <shock wave> from an expansion discontinuity.
The step gives the <Burgers Riemann problem with negative flux>. If $U>0$, <characteristic curves> from the left have speed zero, while those from the right have speed $-U$: they converge. The entropy <shock wave> has speed $-U/2$ and thus
$$
\boxed{f(z,\theta)=\begin{cases}0,&\theta<-Uz/2,\\ U,&\theta>-Uz/2,\end{cases}\qquad U>0.}
$$
Characteristics enter this <shock> from both sides, since $0>-U/2>-U$.
If $U<0$, the right-hand speed $-U$ is positive and the two families separate. A smooth steep approximation to the initial step spreads into a <rarefaction wave>. In the fan, the self-similar <characteristic curve> relation is $-f=\theta/z$, giving
$$
\boxed{f(z,\theta)=\begin{cases}0,&\theta<0,\\-\theta/z,&0\leq\theta\leq-Uz,\\U,&\theta>-Uz,\end{cases}\qquad U<0.}
$$
The endpoint values match continuously. The discontinuity moving at $-U/2$ would satisfy the jump condition even for $U<0$, but its <characteristic curves> leave the discontinuity and it fails entropy admissibility. For $U=0$ the solution is identically zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-70-burgers-characteristics.png]
{title=Characteristics of the negative-flux Burgers step: compression gives a shock for positive U, while negative U gives a rarefaction fan}
{height=420}
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