= Solution
The integrated <travelling wave> equation is $v'=Q(v)/\varepsilon$. A finite limiting value $a$ at either end must satisfy $Q(a)=0$. Otherwise continuity of $Q$ makes $v'$ eventually have a fixed sign and an absolute value bounded below, which is incompatible with convergence to $a$. Therefore
$$
F(u_l)-\sigma u_l+b=0,\qquad F(u_r)-\sigma u_r+b=0.
$$
Subtracting yields the <Rankine-Hugoniot condition>
$$
\sigma(u_l-u_r)=F(u_l)-F(u_r).
$$
For distinct end states this gives
$$
\boxed{\sigma=\frac{F(u_l)-F(u_r)}{u_l-u_r}.}
$$
\b[Distinctness 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 $\sigma$.
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