Problem 7 of Chapter 4 in Evans PDE, viscous conservation law 
Consider the viscous conservation law
  \begin{equation}
u_t+F(u)_x-au_{xx}=0,\quad\text{in }\mathbb{R}\times (0,\infty),\quad(1)
\label{3-1}
\end{equation}
  where $a>0$ and $F$ is uniformly convex. 
(a). Show $u$ solves above PDE if $u(x,t)=v(x-\sigma t)$ and $v$ is defined implicitly by the formula 
  $$
s=\int_{c}^{v(s)} \frac{a}{F(z)-\sigma z+b} d z \quad(s \in \mathbb{R}),
$$
  where $b$ and $c$ are constants. 
(b). Demonstrate that we can find a traveling wave satisfying 
  $$
\lim\limits_{s\rightarrow-\infty}v(s)=u_l,\lim\limits_{s\rightarrow \infty}v(s)=u_r
$$
  for $u_l>u_r$, if and only if 
  $$
\sigma=\frac{F(u_l)-F(u_r)}{u_l-u_r}.
$$
(c). Let $u^\varepsilon$ denote the above traveling wave solution of (1) for $a=\varepsilon$, with $u^\varepsilon(0,0)=\frac{u_l+u_r}{2}$. Compute $\lim\limits_{\varepsilon\rightarrow 0}u^\varepsilon$ and explain your answer. 

My try:
(a). It is easy to check $u$ solve the PDE.
(b). By definition of $v$, we have 
$$
F(v(s))-\sigma v(s)+b=av'(s).
$$
If we have $v'(\pm\infty)=0$, then 
$$
F(u_l)-\sigma u_l+b=0=F(u_r)-\sigma u_r+b
$$
which implies $\sigma=\frac{F(u_l)-F(u_r)}{u_l-u_r}$. But I don't know how to show that $v'(\pm\infty)=0$.
(c). I have no much idea about it.
 A: This post is related.
(b) Let us inject the traveling wave Ansatz $u(x,t) = v(s)$ with $s = x-\sigma t$ in the PDE. Then, we integrate the second-order ODE so-obtained once, so that
$$
F(v) - \sigma v + b = a v'  ,
$$
where $b$ is an integration constant*. One notes that if $v$ has a finite limit at $\pm \infty$, then $v'$ too. By virtue of the mean value theorem, $v'$ vanishes at $\pm \infty$. Thus, the limits of $v$ must satisfy
$$
\sigma u_l - F(u_l) = b = \sigma u_r - F(u_r) \, ,
$$
i.e. $\sigma\, (u_l - u_r) = F(u_l) - F(u_r)$. Injecting the expression of $\sigma$ so-obtained in the differential equation, we have
$$
a v' = F(v) - F(u_r) - \sigma (v - u_r) \leq 0 , \qquad \forall\; v \in [u_r, u_l]
$$
due to the convexity of $F$ (the graph of a convex function lies below its chords). It is therefore possible to connect $u_l$ and $u_r$ with a smooth decreasing function $v$ if $u_l > u_r$, and if $\sigma$ represents the rate of increase of $F$ over $[u_r, u_l]$.
(c) Now, we are left with $u^\varepsilon(x,t) = v(s)$ and $s=x - \sigma t$, where
$$
s = \int_{c}^{v(s)} \frac{\varepsilon}{F(z)-\sigma z + b} dz \, ,
\qquad
\sigma = \frac{F(u_l) - F(u_r)}{u_l - u_r} .
$$
The constant $b$ equals simultaneously $\sigma u_l - F(u_l)$ and $\sigma u_r - F(u_r)$. Evaluating the implicit integral equation at $s=0$, we may choose $c = \frac12(u_l + u_r)$.
On the one hand, the numerator of the integrand vanishes as $\varepsilon \to 0$. On the other hand, we know from (b) that the denominator of the integrand has a constant (negative) sign for $z \in ]u_l, u_r[$, and that it vanishes as $z\to u_l$ or $z\to u_r$. Therefore, to keep the integral equal to a given $s\neq 0$ as $\varepsilon \to 0$, we must either increase the value of $v(s)\to u_l$ if $s<0$, or decrease the value of $v(s)\to u_r$ if $s>0$. Finally, we obtain a piecewise constant solution, with a single discontinuity located at $x=\sigma t$. It is a shock wave that satisfies the Rankine-Hugoniot condition. This solution corresponds to the vanishing viscosity solution by definition.

*This differential equation is separable as
$$
\text{d}s = \frac{a\, \text{d}v}{F(v)-\sigma v + b}
$$
which provides the implicit formula in OP.
