= Solution
Write $h=\log\psi$ and $q=2\epsilon h_\theta$, with $\epsilon>0$. Direct differentiation gives
$$
q_Z-qq_\theta-\epsilon q_{\theta\theta}
=2\epsilon\partial_\theta[h_Z-\epsilon(h_{\theta\theta}+h_\theta^2)]
=2\epsilon\partial_\theta\left(\frac{\psi_Z-\epsilon\psi_{\theta\theta}}{\psi}\right).
$$
Thus the <Cole-Hopf transformation> with this negative-flux convention maps the <heat equation> to the stated <viscous Burgers equation>. Conversely its residual is independent of $\theta$ and can be removed by multiplying $\psi$ by a function of $Z$.
Convolving each initial exponential with the <Gaussian heat kernel> gives
$$
\psi(\theta,Z)=\sum_{j=1}^N e^{A_j},
\qquad A_j=\frac{q_j(\theta-\theta_j)+q_j^2Z/2}{2\epsilon}.
$$
The resulting <finite-exponential Cole-Hopf solution> is
$$
\boxed{q(\theta,Z)=\frac{\sum_jq_je^{A_j}}{\sum_je^{A_j}}.}
$$
It is a <softmax function> average of the constant states, so $q_1\leq q\leq q_N$ and $q_\theta=\operatorname{Var}_w(q_j)/(2\epsilon)\geq0$.
For $N=2$, put $\Delta=q_2-q_1>0$ and $\overline q=(q_1+q_2)/2$. Equality of the two exponents occurs at
$$
\theta_s(Z)=\frac{q_2\theta_2-q_1\theta_1}{\Delta}-\overline q Z.
$$
Therefore
$$
\boxed{q=\overline q+\frac\Delta2\tanh\frac{\Delta(\theta-\theta_s)}{4\epsilon}.}
$$
This is an increasing viscous shock from $q_1$ to $q_2$, moving at speed $-\overline q$ and with transition width of order $\epsilon/\Delta$. That speed agrees with the <Rankine-Hugoniot condition> for flux $-q^2/2$; its concavity makes the increasing jump the entropy shock.
For general $N$, use the shock frame $\theta=-(q_1+q_N)Z/2+\xi$. The leading $Z$ coefficient in $A_j$ is $q_j(q_j-q_1-q_N)/(4\epsilon)$. The endpoint coefficients coincide, and every interior coefficient is smaller by $(q_j-q_1)(q_N-q_j)/(4\epsilon)>0$. The <extremal-state asymptotics of exponential Burgers solutions> therefore reduce the large-$Z$ profile to the same tanh shock with states $q_1,q_N$ and centre
$$
\theta_s=\frac{q_N\theta_N-q_1\theta_1}{q_N-q_1}-\frac{q_1+q_N}{2}Z.
$$
At a fixed $\theta$, rather than in the moving frame, the endpoint of largest $|q_j|$ dominates. If $q_N=-q_1$, both endpoint rates coincide and the limiting shock is stationary. Intermediate plateaus disappear in the moving-frame limit.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-88-burgers-states.png]
{title=Intermediate states disappear from the finite-exponential Burgers solution in the moving shock frame}
Back to article page
