= Solution
Set $q=\log\psi$ and use the sign-appropriate <Cole-Hopf transformation> $f=2\varepsilon q_\theta$. Direct differentiation gives
$$
f_z-ff_\theta-\varepsilon f_{\theta\theta}
=2\varepsilon\partial_\theta\left(q_z-\varepsilon q_{\theta\theta}-\varepsilon q_\theta^2\right)
=2\varepsilon\partial_\theta\left(\frac{\psi_z-\varepsilon\psi_{\theta\theta}}\psi\right).
$$
Thus the <heat equation> $\psi_z=\varepsilon\psi_{\theta\theta}$ implies the <viscous Burgers equation> with the required negative nonlinear sign. Conversely, vanishing of the Burgers residual makes the final quotient depend only on $z$, and a multiplicative time-dependent factor in $\psi$ removes it. The algebraic substitution works for any nonzero $\varepsilon$ where $\psi\ne0$.
For a forward dissipative initial-value solution, take \b[$\varepsilon>0$ and $z>0$]. This is essential for the supplied <heat kernel>: at $\varepsilon<0$, $z>0$, its real <Gaussian integral> diverges even for $\psi(0,\phi)=1$. More generally that kernel requires $\varepsilon z>0$. The physical viscosity convention selects the positive case; the mere condition $\varepsilon\ne0$ does not justify a forward <Gaussian function> formula.
Integrating $\partial_\theta\log\psi(0,\theta)=f_0(\theta)/(2\varepsilon)$ gives a positive continuous initial factor, normalized as
$$
\psi(0,\phi)=\begin{cases}1,&\phi\leq0,\\ e^{U\phi/(2\varepsilon)},&\phi\geq0.\end{cases}
$$
Its <heat kernel> <convolution> converges for both signs of $U$, since a <Gaussian function> dominates the one-sided exponential. Define
$$
J(b)=\int_b^\infty e^{-y^2/(4\varepsilon z)}dy,\qquad
S=\frac{U(2\theta+Uz)}{4\varepsilon}.
$$
The left half-line contributes $J(\theta)$; completing the square on the right half-line gives $e^S J(-\theta-Uz)$. Thus
$$
\psi(z,\theta)=\frac{J(\theta)+e^S J(-\theta-Uz)}{\sqrt{4\pi\varepsilon z}}.
$$
When differentiating, the two moving-endpoint <Gaussian function> terms cancel, because
$$
e^S e^{-(\theta+Uz)^2/(4\varepsilon z)}=e^{-\theta^2/(4\varepsilon z)}.
$$
Only $\partial_\theta e^S=(U/(2\varepsilon))e^S$ remains in the logarithmic <derivative>. The <viscous Burgers step solution with negative flux> is therefore
$$
\boxed{f(z,\theta)=\frac{U}{1+\alpha e^{-S}},\qquad
\alpha=\frac{J(\theta)}{J(-\theta-Uz)}.}
$$
The denominator is strictly positive, so $f$ lies between $0$ and $U$, irrespective of the sign of $U$. Its initial limits away from $\theta=0$ are the required step.
For the spatial limits, the <Gaussian function> tails must be compared with $e^S$; inspecting that exponential alone gives the wrong inference when $U<0$. As $b\to+\infty$, the <complementary error function> asymptotic gives
$$
J(b)\sim\frac{2\varepsilon z}{b}e^{-b^2/(4\varepsilon z)},\qquad
J(b)\to\sqrt{4\pi\varepsilon z}\quad(b\to-\infty).
$$
At $\theta\to+\infty$, $J(\theta)e^{-S}$ tends to zero even for $U<0$, because its quadratic decay dominates any exponential linear in $\theta$. Hence $\alpha e^{-S}\to0$. At $\theta\to-\infty$, use the identity above to obtain
$$
e^S J(-\theta-Uz)\sim\frac{2\varepsilon z}{-\theta-Uz}e^{-\theta^2/(4\varepsilon z)}\to0,
$$
while $J(\theta)$ approaches its full <Gaussian integral>. Thus
$$
\boxed{f(z,\theta)\to U\ (\theta\to+\infty),\qquad f(z,\theta)\to0\ (\theta\to-\infty),}
$$
for both signs of $U$.
Since $J$ is strictly decreasing, $\alpha=1$ holds exactly when $\theta=-\theta-Uz$. At this point $S=0$, so the <midpoint symmetry of a viscous Burgers step> gives
$$
\boxed{\alpha=1\quad\Longleftrightarrow\quad\theta=-Uz/2,\qquad f=U/2.}
$$
More generally the exact symmetry is $f(z,-Uz-\theta)=U-f(z,\theta)$. For $U>0$, the midpoint follows the inviscid <shock> trajectory. For $U<0$, it is instead the center of an expanding fan, not a <shock>.
The <vanishing viscosity approximation> makes the comparison precise. For fixed $z>0$ and $U>0$, near $\theta=-Uz/2$ both tail integrals tend to the full <Gaussian integral>, so $\alpha\to1$ and
$$
f\sim\frac{U}{1+\exp\{-U(\theta+Uz/2)/(2\varepsilon)\}}.
$$
The layer has width $O(\varepsilon/U)$ and tends to the entropy <shock>, with value $U/2$ at its center. Outside it the limits are $0$ and $U$.
For $U<0$, inside $0<\theta<-Uz$ both <Gaussian function> tails have large positive lower limits. Their asymptotics give
$$
\alpha e^{-S}\sim\frac{-\theta-Uz}{\theta},\qquad
f\longrightarrow-\frac{\theta}{z}.
$$
Outside this interval the limits are $0$ on the left and $U$ on the right. Thus positive viscosity selects the <rarefaction wave>, including the same midpoint value as the inviscid fan. The two sign cases agree with the preceding <entropy solution> construction, while a negative diffusivity would not furnish this dissipative selection.
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