= Solution
Let $h=\log\psi$, with $\psi>0$, and set $f=2\alpha h_\theta$. Direct differentiation gives
$$
f_Z-ff_\theta-\alpha f_{\theta\theta}
=2\alpha\partial_\theta\left\{h_Z-\alpha h_{\theta\theta}-\alpha h_\theta^2\right\}.
$$
The <heat equation> $\psi_Z=\alpha\psi_{\theta\theta}$ implies $h_Z=\alpha(h_{\theta\theta}+h_\theta^2)$, so the bracket vanishes. Conversely its being independent of $\theta$ can be absorbed into a $Z$-dependent multiplicative normalization of $\psi$, leaving $f$ unchanged. This proves the <Cole-Hopf transformation> with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero $\alpha$, the given forward <Gaussian function> diffusion kernel and a physical vanishing-<viscosity> limit require $\alpha>0$.
Take $U,L>0$ for the stated <Burgers N-wave>. Integrating $\partial_\theta\log\psi(0,\theta)=f_0(\theta)/(2\alpha)$ and normalizing the exterior value to one gives
$$
\psi_0(\theta)=\begin{cases}
\exp\{U(L^2-\theta^2)/(4\alpha)\},&|\theta|<L,\\
1,&|\theta|\geq L.
\end{cases}
$$
This function is continuous at $\pm L$; its logarithmic derivative has the specified jumps. Convolution with the <heat kernel> is positive and solves the <heat equation> for $Z>0$. Splitting the integral into the exterior baseline and the interior correction yields
$$
\psi=1-I_\alpha(\theta,L,Z)
+\frac{e^{UL^2/(4\alpha)}}{\sqrt{4\pi\alpha Z}}
\int_{-L}^{L}\exp\left\{-\frac{U\varphi^2}{4\alpha}-\frac{(\varphi-\theta)^2}{4\alpha Z}\right\}d\varphi.
$$
Put $a=1+UZ$. Completing the square gives
$$
U\varphi^2+\frac{(\varphi-\theta)^2}{Z}
=\frac aZ\left(\varphi-\frac\theta a\right)^2+\frac{U\theta^2}{a}.
$$
Changing the interior integration variable to $\eta=a\varphi$ changes its limits to $\pm La$ and the <Gaussian function> width to $Za$. The Jacobian and normalization leave the factor $a^{-1/2}$. Thus the <Cole-Hopf solution for a Burgers N-wave> is
$$
\boxed{\psi=1-I_\alpha(\theta,L,Z)+I_\alpha(\theta,La,Za)\widehat\psi,\qquad
\widehat\psi=a^{-1/2}\exp\left\{\frac U{4\alpha}\left(L^2-\frac{\theta^2}{a}\right)\right\},\qquad
f=2\alpha\partial_\theta\log\psi.}
$$
Here $I_\alpha$ is the normalized <Gaussian function> mass of its indicated interval. Its explicit <error function> representation is
$$
I_\alpha(\theta,b,w)=\frac12\left\{
\operatorname{erf}\left(\frac{b-\theta}{\sqrt{4\alpha w}}\right)
+\operatorname{erf}\left(\frac{b+\theta}{\sqrt{4\alpha w}}\right)\right\}.
$$
For fixed $w>0$, the <Gaussian function> concentrates at $\varphi=\theta$ as $\alpha\downarrow0$. Consequently
$$
\boxed{I_\alpha(\theta,L,w)\longrightarrow H(L^2-\theta^2)}
$$
away from the endpoints; at $\theta=\pm L$ the limit is $1/2$. The transition layer has width $O(\sqrt{\alpha w})$. This is an <approximate identity> argument, not a uniform step approximation across the endpoints.
For fixed $Z>0$ and $|\theta|\ll L$, both interval masses tend to one, and the exponentially large positive $\widehat\psi$ dominates the exterior correction. Its logarithmic derivative therefore gives
$$
\boxed{f(Z,\theta)\simeq-\frac{U\theta}{1+UZ}\qquad(|\theta|\ll L,\ \alpha\downarrow0).}
$$
For $|\theta|\gg La$, both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore $\boxed{f(Z,\theta)\simeq0}$ in the specified far exterior.
An exponentially weighted <Gaussian function> tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with $\widehat\psi$ gives the sharper inviscid <Burgers N-wave> fronts $|\theta|=L\sqrt{1+UZ}$, not $La$. Away from these fronts, the <vanishing-viscosity limit> is
$$
f(Z,\theta)\longrightarrow\begin{cases}
-U\theta/(1+UZ),&|\theta|<L\sqrt{1+UZ},\\
0,&|\theta|>L\sqrt{1+UZ}.
\end{cases}
$$
Inside $|\theta|<La$, this follows from the sign of $L^2-\theta^2/a$ in the exponential. Outside $La$, the constrained <Gaussian function> maximum lies at an interval endpoint and has negative exponent. The right <shock wave>'s speed is $UL/(2\sqrt a)$, equal both to the derivative of $L\sqrt a$ and to minus half the sum of its two limiting states; the left <shock wave> is its reflection. This checks consistency with the <Rankine-Hugoniot condition> and shows that the requested near-center and far-exterior approximations are compatible with the full <entropy> limit.
Back to article page
