$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\sgn}{\operatorname{sgn}}$ $\newcommand{\rank}{\operatorname{rank}}$
In previous Section 9.1 we presented solution to Cauchy problem for Burgers equation in the form: \begin{equation} u(x,t;\varepsilon)= \frac{\int_{-\infty}^{\infty} e^{-\frac{1}{2\varepsilon} \Phi(x,z,t)} \frac{1}{t} (x-z)\,dz} {\int_{-\infty}^{\infty} e^{-\frac{1}{2\varepsilon} \Phi(x,z,t)}\,dz} \label{eq-9.2.1} \end{equation} with \begin{equation} \Phi(x,z,t)=F(z)+\frac{1}{2 t} (x-z)^2,\qquad F'(z)=f(z). \label{eq-9.2.2} \end{equation}
We look for asymptotics as $\varepsilon\to +0$. Let us apply Laplace method Section 3.1 we need to find absolute minima (with respect to $z$) of $\Phi(x,z,t)$. Each of them is a stationary point of $\Phi(x,z,t)$: $f(z)-\frac{1}{t}(x-z)=0$ which is equivalent to \begin{equation} x=z+tf(z) \label{eq-9.2.3} \end{equation} which is the first equation of (9.1.5).
Then if $z$ is a single non-degenerate absolute minimum of $\Phi(x,z,t)$ then cancelling the same factor in the numerator and denominator we arrive to \begin{equation} u(x,t,\varepsilon)= \frac{1}{t}(x-z)+O(\varepsilon)=f(z)+O(\varepsilon). \label{eq-9.2.4} \end{equation} This is the case as $0< t< t^*$ with \begin{equation} t^*:=1/\max (-f'(z)) \label{eq-9.2.5} \end{equation} is the time before discontinuities appear.
But what happens after this? Assume that $\max (-f'(z))$ is achieved in a single point $z^*$. Further, we assume that this is non-degenerate maximum. Then as $t$ passes through $t^*$ a single non-degenerate minimum of $\Phi (x,z,t)$ morphs into two non-degenerate local minima and a non-degenerate local maxima in between. We are not interested in the maxima and we look for an absolute minima. As long as $x\ne \xi(t)$ where $\xi(t)$ is a position of the shock wave described by ODE (9.1.8) with $t> t^*$ and $\xi(t^*)= z^* + t^*f(z^*)$ only one of them is an absolute. Namely, as $x>\xi(t)$ it is the right one, and as $x<\xi(t)$ it is the left one.
Example 9.2.1 Let $f$ be an odd function, then $u(x,t,\varepsilon)$ is an odd function with respect to $x$. Assume that $f'(z)<0$, $f''(z)>0$ as $z>0$. Then $z^*= 0$, $t^*=-1/f'(0)$, $\xi(t)=0$ (as $t\ge t^*$ and $v (\pm 0,t)= \pm f(z(t)$ where $z(t)>0$ is a solution to $z+tf'(z)=0$.
In particular, as $f(z)=-\arctan (x)$ we have $z(t)=\sqrt{t-1}$ and $v (\pm 0,t)= \mp \arctan(\sqrt{t-1})$.
Again we arrive to $u(x,t,\varepsilon)=v(x,t)+O(\varepsilon)$ as long as we are in the situation we described (in future some shock waves may collide or dissipate).
Assume that $t >t^*$ and $t$ is fixed but $t-t^*$ is small enough.
Then we still have two non-degenerate minima. As $x=\xi(t)$ we denote the left minimum by $z^-(t)$ and the right minimum by $z^+(t)$. Then we know that both of minimal values coincide \begin{equation} \Phi (\xi(t),z^+(t),t)=\Phi (\xi(t),z^-(t),t). \label{eq-9.2.6} \end{equation} Consider formula (\ref{eq-9.2.1}). Then as $|x-\xi(t)|\ll 1$, $|z-z^\pm (t)|\ll 1$ we have \begin{multline} \Phi (x,z,t) \approx \Phi (\xi(t) , z^\pm (t),t) \\ +\bigl(\frac{1}{2} f'(z^\pm(t))+\frac{1}{t}\bigr)(z-z^\pm(t))^2 -\frac{1}{t}(x-\xi(t)) z^\pm(t) + \label{eq-9.2.7} \end{multline} where we neglect terms which are $o\bigl(|x-\xi(t)|+|z-z^\pm|^2\bigr)$ and observe that \begin{equation} k^\pm (t):=\Phi_{zz} (\xi(t) , z^\pm (t),t) = F''(z^\pm (t)) +\frac{1}{t}= f'(z^\pm (t)) +\frac{1}{t}>0. \label{eq-9.2.8} \end{equation} Then \begin{multline} u(x,t;\varepsilon)\approx\\ \frac{\frac{1}{\sqrt{k^+}} \exp \bigl( \frac{1} {2\varepsilon t} z^+ (t)(x-\xi(t))\bigr) v^+(t) + \frac{1}{\sqrt{k^-}} \exp \bigl( \frac{1} {2\varepsilon t} z^- (t)(x-\xi(t))\bigr) v^-(t)} {\frac{1}{\sqrt{k^+}} \exp \bigl( \frac{1} {2\varepsilon t} z^+ (t)(x-\xi(t))\bigr) + \frac{1}{\sqrt{k^-}} \exp \bigl( \frac{1} {2\varepsilon t} z^- (t)(x-\xi(t))\bigr) }\\ \shoveleft{=\frac{1}{2}\bigl(v^+(t)+v^-(t)\bigr)}\\ + \frac{1}{2}\bigl(v^+(t)-v^-(t)\bigr) \frac{\sqrt{k^-/k^+} \exp \bigl( \frac{1} {2\varepsilon t} (z^+ (t)-z^-(t))(x-\xi(t))\bigr) -1 } {\sqrt{k^-/k^+} \exp \bigl( \frac{1} {2\varepsilon t} (z^+ (t)-z^-(t))(x-\xi(t))\bigr) +1 } \label{eq-9.2.9} \end{multline} where $v^\pm (t)=v(\xi(t)\pm 0,t)$.
In particular $u(x,t;\varepsilon)\approx v^\pm$ as $|x-\xi(t)|\gg \varepsilon$ and $\pm (x-\xi(t))>0$.
Sure, finding asymptotics as $|x-\xi(t)|\ll 1$ and $t-t^*|\ll 1$ is much more difficult.