 Solution

The integrated travelling wave equation is $v^{\prime }=Q(v)/\epsilon$. A finite limiting value $a$ at either end must satisfy $Q(a)=0$. Otherwise continuity of $Q$ makes $v^{\prime }$ eventually have a fixed sign and an absolute value bounded below, which is incompatible with convergence to $a$. ThereforeSubtracting yields the Rankine-Hugoniot conditionFor distinct end states this givesDistinctness is needed for the printed quotient. If $u_l=u_r$, the identity is $0=0$. A nonconstant global profile is strictly monotone by the scalar ordinary differential equation, so cannot have equal finite end states. The equal-state profiles here are constant and their representation allows any $\sigma$.