NREV, IREV

If NREV=1 then it is assumed that the system is reversible under the transformation $t \to -t$ and $U(i) \to -U(i)$ for all $i$ with IREV(i)>0. Then only half the homoclinic solution is solved for with right-hand boundary conditions specifying that the solution is symmetric under the reversibility (see ChSp:93). The number of free parameters is then reduced by one. Otherwise IREV=0.