“…4π 5 π 3 .If π 0 = 0 (respectively, {3π3 β ππ 2 = 0, π 1 = 0}; {π 3 β 2π π 2 β ππ 2 = 0, π 1 = 0}), then Lemma 3.1, (73) (respectively, Lemma 3.1, (75); Lemma 3.1, (76)).System{π 1 = 0, π 2 = 0} is incompatible because π
ππ π’ππ‘πππ‘ [π 1 , π 2 , π] = 32π 11 π 8 β 0.Under conditions (65) πΏ 1 = π(2π β π). If π = 0, then Lemma 3.1, (77), and if π = 2π, then Lemma 3.1, (78).Finally, when conditions (65) hold the first Lyapunov quantity isπΏ 1 = (ππ 2 + πππ β 2ππ 2 )(π 2 +π 2 π βπππ).…”