= Solution
For zero viscosity the <Inviscid Burgers equation> is $q_X-q q_\theta=0$. Along a <characteristic curve>, $dq/dX=0$ and $d\theta/dX=-q$. Labelling a characteristic by its initial position $\eta$ gives
$$
\boxed{q=f(\eta),\qquad\theta=\eta-Xf(\eta),\qquad q(\theta,X)=f(\theta+Xq).}
$$
Where $f$ is differentiable, $q_\theta=f'(\eta)/(1-Xf'(\eta))$. The first characteristic crossing occurs at $X=1/\max f'$ if the maximum positive slope exists. A multivalued continuation is not a physical wave: rewrite the equation as the <conservation law> $q_X+\partial_\theta(-q^2/2)=0$ and replace the overturning part by an entropy-admissible <shock wave>. The <Rankine-Hugoniot conditions> give its trajectory
$$
\frac{d\theta_s}{dX}=\frac{-q_R^2/2+q_L^2/2}{q_R-q_L}=-\frac{q_L+q_R}{2}.
$$
The states on either side are supplied by the characteristics meeting the shock. Conservation supplies the equal-area construction, and the entropy condition requires characteristics to enter the shock. Since the flux is concave, a jump with $q_L<q_R$ is compressive.
For the rising step, $q_L=0$ and $q_R=U>0$, the left characteristics have slope zero and the right ones slope $-U$. They overlap immediately. The <Burgers Riemann problem with negative flux> is solved by
$$
\boxed{q(\theta,X)=\begin{cases}0,&\theta<-UX/2,\\U,&\theta>-UX/2,\end{cases}\qquad\theta_s(X)=-UX/2.}
$$
It is a shock, not a <rarefaction wave>; the minus sign in the nonlinear flux is decisive.
For positive viscosity, substitute $q=2\epsilon\partial_\theta\log\psi$ into the <viscous Burgers equation>. Writing $w=\log\psi$ gives
$$
q_X-qq_\theta-\epsilon q_{\theta\theta}
=2\epsilon\partial_\theta\left[w_X-\epsilon(w_{\theta\theta}+w_\theta^2)\right].
$$
An $X$-dependent multiplicative factor in $\psi$ removes the possible integration function. The <Cole-Hopf transformation> therefore gives $\psi_X=\epsilon\psi_{\theta\theta}$. Normalize its initial value continuously at zero:
$$
\psi(\theta,0)=\begin{cases}1,&\theta<0,\\e^{U\theta/(2\epsilon)},&\theta>0.\end{cases}
$$
Split the <heat kernel> integral at zero. Completing the square in the positive-half integral yields
$$
\boxed{\psi(\theta,X)=\frac12\operatorname{erfc}\left(\frac{\theta}{2\sqrt{\epsilon X}}\right)
+\frac12e^{U\theta/(2\epsilon)+U^2X/(4\epsilon)}\operatorname{erfc}\left(-\frac{\theta+UX}{2\sqrt{\epsilon X}}\right).}
$$
When differentiating, the Gaussian terms from the two <complementary error functions> cancel, because
$$
\frac{U\theta}{2\epsilon}+\frac{U^2X}{4\epsilon}-\frac{(\theta+UX)^2}{4\epsilon X}=-\frac{\theta^2}{4\epsilon X}.
$$
Thus the exact <viscous Burgers step solution with negative flux> is
$$
\boxed{q(\theta,X)=\frac{U e^{U\theta/(2\epsilon)+U^2X/(4\epsilon)}\operatorname{erfc}\!\left(-\frac{\theta+UX}{2\sqrt{\epsilon X}}\right)}
{\operatorname{erfc}\!\left(\frac{\theta}{2\sqrt{\epsilon X}}\right)+e^{U\theta/(2\epsilon)+U^2X/(4\epsilon)}\operatorname{erfc}\!\left(-\frac{\theta+UX}{2\sqrt{\epsilon X}}\right)}.}
$$
For $X>0$ its denominator is positive, its tails tend to $0$ and $U$, and it approaches the initial step as $X\downarrow0$. At $\theta=-UX/2$ it equals $U/2$. Its <vanishing-viscosity limit> gives the entropy shock above. At large $U\sqrt{X/\epsilon}$ near the moving shock center it approaches the travelling profile $U/[1+e^{-U(\theta+UX/2)/(2\epsilon)}]$, with thickness of order $\epsilon/U$.
Back to article page
