We begin by computing towards with the option IEQUIB=-2 which means that both equilibria are solved for as part of the continuation process.
BR PT TY LAB PAR(3) L2-NORM ... PAR(1) 1 5 2 4.528332E-01 3.726787E-01 ... 1.364973E-01 1 10 3 3.943370E-01 3.303798E-01 ... 1.044119E-01 1 15 4 3.358942E-01 2.873213E-01 ... 7.515570E-02 1 20 5 2.772726E-01 2.433403E-01 ... 4.952636E-02 1 25 6 2.181955E-01 1.981358E-01 ... 2.845849E-02 1 30 EP 7 1.581633E-01 1.512340E-01 ... 1.292975E-02Alternatively, for this problem there exists an analytic expression for the two equilibria. This is specified in the subroutine PVLS of she.f. Re-running with IEQUIB=-1
BR PT TY LAB PAR(3) L2-NORM ... PAR(1) 1 5 2 4.432015E-01 3.657716E-01 ... 1.310559E-01 1 10 3 3.723085E-01 3.142439E-01 ... 9.300982E-02 1 15 4 3.008842E-01 2.611556E-01 ... 5.933966E-02 1 20 5 2.286652E-01 2.062194E-01 ... 3.179939E-02 1 25 6 1.555409E-01 1.491652E-01 ... 1.239897E-02 1 30 EP 7 8.107462E-02 9.143108E-02 ... 2.386616E-03This output is similar to that above, but note that it is obtained slightly more efficiently because the extra parameters PAR(12-21) representing the coordinates of the equilibria are no longer part of the continuation problem. Also note that AUTO has chosen to take slightly larger steps along the family. Finally, we can continue in the opposite direction along the family from the original starting point (again with IEQUIB=-1).
BR PT TY LAB PAR(3) L2-NORM ... PAR(1) 1 5 8 4.997590E-01 4.060153E-01 ... 1.637322E-01 1 10 9 5.705299E-01 4.551872E-01 ... 2.065264E-01 1 15 10 6.416439E-01 5.031844E-01 ... 2.507829E-01 1 20 11 7.133301E-01 5.500668E-01 ... 2.959336E-01 1 25 12 7.857688E-01 5.958712E-01 ... 3.415492E-01 1 30 13 8.590970E-01 6.406182E-01 ... 3.872997E-01 1 35 EP 14 9.334159E-01 6.843173E-01 ... 4.329270E-01The results of both computations are presented in Figure 25.1, from which we see that the orbit shrinks to zero as PAR(1)=.