$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\sgn}{\operatorname{sgn}}$ $\newcommand{\rank}{\operatorname{rank}}$
Recall that we consider Cauchy problem for Burgers equation \begin{align} &u_t+uu_x=\varepsilon u_{xx},\qquad t>0, -\infty < x<\infty \tag{9.1.1}\\ &u|_{t=0}=f(x). \tag{9.1.2} \end{align}
Example 1.
Let $\ f(x)=\left\{\begin{aligned} &f_- && x < 0,\\ &f_+ &&x > 0 \end{aligned}\right.$ with $\ f_- > f_+$.
Then as $\varepsilon=0$ the proper solution is $u_0(t)= \left\{\begin{aligned} &f_- && x < X (t):=vt,\\ &f_+ &&x > X (t) \end{aligned}\right.\ $ with $v:=\frac{1}{2}(f_++f_-)$.
As $\varepsilon>0$ there is a stationary solution of the Burgers equation \begin{gather} U_\varepsilon (x,t)= \frac{1}{2}(f_+ +f_-) \pm \frac{1}{2}(f_+ -f_-) \bigl(1 - e^{-g\varepsilon ^{-1}|x-X (t)|}\Bigr), \qquad g= \frac{1}{2}(f_- - f_+). \label{eq-9.2.1} \end{gather}
Remark 1.
Now consider the general case. Assume that
Condition 1. $f(x)$ is a smooth function as $x\le \bar{x}$ and as $x\ge \bar{x}$ but has a jump at $x=\bar{x}$: \begin{gather} f_- := f(\bar{x}-0)>f(\bar{x}+0)=:f_+. \label{eq-9.2.3} \end{gather}
Then for $0< t < T$ (with small constant $T>0$) a proper solution $u_0(x,t)$ has a jump at $x= X (t)$ which is the solution to \begin{gather} \frac{dX }{dt}=\frac{1}{2}\bigl(f(z_-(X,t)) + f(z_+(X ,t))\bigr), \label{eq-9.2.4}\\ X (0)=\bar{x} \label{eq-9.2.5} \end{gather} where $z_\pm (x,t)$ is a solution of \begin{gather} x= z +tf(z), \label{eq-9.2.9}\\ z_\pm \gtrless \bar{x} \label{eq-9.2.6} \end{gather} and one can prove that this equation has exactly one solution in $(\bar{x}-CT,\bar{x})$ and one solution in $(\bar{x},\bar{x}+CT)$.
Then \begin{gather} u_0(x,t)= f (z_\pm (t)) \qquad \text{as } x\gtrless X (t). \label{eq-9.2.7} \end{gather}
Remark 2.
Consider kind of stationary wave from Example 1 \begin{gather} U_\varepsilon (x,t) \sim \sum_{n\ge 0} U_n(x,t) \varepsilon^n. \label{eq-9.2.8} \end{gather} The outer problem (that is, as $|x-X (t)|\ge \varepsilon^{1-\delta}$) could be solved by the metod of characteristics: \begin{align} &\left\{ \begin{aligned} &\frac{\partial U_0} {\partial t} + U_0 \frac{\partial U_0} {\partial x}=0,\\ &U_0(x,0,\varepsilon)=f(x), \end{aligned}\right.\label{eq-9.2.10}\\[2pt] &\left\{\begin{aligned} &\frac{\partial U_1} {\partial t} + U_0 \frac{\partial U_1} {\partial x} + U_1 \frac{\partial U_0} {\partial x} = \frac{\partial^2U_0}{\partial x^2},\\ &U_1(x,0)=0, \end{aligned}\right.\label{eq-9.2.11} \end{align} Then as before $U_0 = f(z_\pm (x,t))$ as $x\gtrless X (t)$.
Consider inner zone $\{ (x,t) \colon |x-X (t)| \le \varepsilon^{1-\delta}\}$ and introduce there variable $\xi:= \varepsilon^{-1} (x-X(t))$. Then \begin{gather*} \frac{\partial u} {\partial t}= -\varepsilon^{-1} X'(t)\frac{\partial w} {\partial \xi } +\frac{\partial w} {\partial t} \end{gather*} and \begin{gather*} \frac{\partial u} {\partial x}=\varepsilon^{-1}\frac{\partial w} {\partial \xi}. \end{gather*} Then equation in the inner zone is \begin{gather*} -\varepsilon^{-1} X' \frac{\partial w} {\partial \xi} + \frac{\partial w} {\partial t}+ \varepsilon^{-1}w \frac{\partial w} {\partial \xi} =\varepsilon^{-1}\frac{\partial^2 w} {\partial \xi^2} \end{gather*} or \begin{gather*} \frac{\partial^2 w} {\partial \xi^2} = -\varepsilon \frac{\partial w} {\partial t} + (w-X')\frac{\partial w} {\partial \xi }. \end{gather*} Then looking for \begin{gather*} w(\xi,t,\varepsilon)\sim \sum_{n\ge 0} W_n (\xi,t)\varepsilon^n \end{gather*} we arrive to the inner problem \begin{align*} &\frac{\partial^2 W_0} {\partial \xi^2}=(W_0-X') \frac{\partial W_0} {\partial \xi},\\ &\frac{\partial^2 W_1}{\partial \xi^2}= \frac{\partial W_0}{\partial t} + (W_0-X')\frac{\partial W_1}{\partial \xi } +W_1 \frac{\partial W_0}{\partial \xi}. \end{align*} Integrating by $\xi$ we get \begin{align*} &\frac{\partial W_0} {\partial \xi }=\frac{1}{2}W_0^2 -X'W_0 +A(t) \end{align*} and then \begin{gather*} \frac{dW_0}{\frac{1}{2}W_0^2 -X'W_0 +A(t)}=d\xi. \end{gather*}
Let us compare inner and outer solutions: \begin{gather*} W_0 (\mp \infty)= u_\mp (t):= u(X(t)\mp 0)=f(z_\mp (X(t),t). \end{gather*} In particular, $\lim_{\xi\to \pm \infty}\frac{\partial W_0}{\partial \xi}=0$. Then, as $\xi\to \mp \infty$ we get \begin{align*} &0=\frac{1}{2}u_- ^2 - X' u_- +A(t),\\ &0=\frac{1}{2}u_+ ^2 - X' u_+ +A(t). \end{align*} Then \begin{gather*} 0=\frac{1}{2}\bigl(u_+^2 - u_-^2\bigr) - X'\bigl(u_+ - u_-\bigr) \end{gather*} and then \begin{gather*} X'=\frac{1}{2}\bigl(u_+ + u_- \bigr), \qquad A(t)=u_-(t)u_+(t). \end{gather*} Here the first equality is fulfilled by definition of $X(t)$.
We can now find an inner solution \begin{align*} \frac{1}{2}\xi =&\int \frac{dW_0}{(W_0-u_+)(W_0-u_-)}\\ =&(u_--u_+)^{-1}\Bigl[ \ln \frac{u_- - W_0}{W_0- u_+}+ \ln B(t)\Bigr] \end{align*} and \begin{gather*} W_0(\xi,t)= \frac {u_- (t)-u_+(t)E(\xi,t)}{1-E(\xi,t)},\qquad E(\xi,t):=B(t)e^{\frac{1}{2}\bigl(u_-``(t)-u_+(t)\bigr)\xi}. \end{gather*}
To determine the constant of integration, $B(t)$, one must perform the matching to higher order.