Equilibrium points Linearization at an equilibrium

190 G. De Marco

3. Equilibrium points

The equilibria of a nonholonomic system correspond to constant solutions, and are to be found at the points q ∈  such that for some λ ∈ 3 we have −Sq = B ∗ λ . Notice that when the system is written in the form 10 the equilibria are exactly the zeroes of f ; and λ = hq is found from the second equation in 7: the set of equilibria is a subset of the graph of the function h :  → 3, an n−dimensional submanifold of X × 3 diffeomorphic to . If Sq = 0, then the preceding equation is satisfied by λ = 0; an unconstrained equilibrium remains of course an equilibrium when velocity constraints are added, but there are other equilibria, as remarked in the introduction. In the generic case, the set of equilibria {q, λ ∈ X × 3 : Qq, 0 + B ∗ qλ = 0} will be an m−dimensional submanifold of X × 3, contained in the graph of h, which then projects onto an m−dimensional submanifold of X . This is certainly the case if the solution set is non–empty, and the linear operator Q ′ q, 0[ · ] + B ∗ ′ [ · , λ] ∈ LX has rank n for every q, λ in the solution set; λ acts then as a system of parameters for the manifold. This is the generic situation, but exceptions are not hard to find sections 7, 8.

4. Linearization at an equilibrium

If 8q, p = p, f q − Rq[ p] + gq[ p, p], the differential of 8 at an equilibrium point q , 0 is 11 8 ′ q , 0 = 1 X f ′ q −Rq . We want to study f ′ q see section 2 for the definition of f , R, etc.. For this, the following is a crucial result: P ROPOSITION 3.1. Pq is a projector onto the space K qkerBq and K q −1 Pq has kerBq as image, kerBq ⊥ as kernel. Dimostrazione. For simplicity, omit q from the operators; P is a projector iff it is idempotent, and this is true iff B ∗ D −1 B K −1 is idempotent, which is immediate to check: B ∗ D −1 B K −1 B ∗ D −1 B K −1 = B ∗ D −1 B K −1 B ∗ D −1 B K −1 = B ∗ D −1 D D −1 B K −1 = B ∗ D −1 B K −1 ; B ∗ D −1 B K −1 is a projector onto the space B ∗ 3 , with kernel K kerB: all this is im- mediate. This implies that 1 X − B ∗ D −1 B K −1 has kernel B ∗ 3 = kerB ⊥ , and K kerB as image. Linearization of nonholonomic systems 191 From f q = K −1 q PqSq we get f ′ q[ · ] = −K −1 qK ′ [ · , K −1 q PqSq] + K −1 q P ′ q[ · , Sq] + PqS ′ q[ · ] The idempotence of P implies that P ′ q[ · , Sq] = P ′ q[ · , PqSq] + Pq P ′ [ · , Sq]; substituting in the above we get f ′ q[ · ] = −K −1 qK ′ [ · , f q] K −1 q P ′ q[ · , PqSq] + Pq P ′ [ · , Sq] + PqS ′ q[ · ] , and at an equilibrium point q we have Pq Sq = K q f q = 0 = f q ; at equilibria we are then able to write: f ′ q [ · ] = K −1 q Pq P ′ [ · , Sq ] + S ′ q [ · ] . C OROLLARY 3.1. At an equilibrium point q the images of f ′ q and of Rq are con- tained in kerBq . Dimostrazione. See above for f ′ q ; for R, simply recall that R = K −1 P F. It will also be useful to differentiate f q written in the form f q = K −1 qSq + B ∗ qhq; we get f ′ q[ · ] = −K −1 qK ′ [ · , K −1 qSq + B ∗ qhq] + K −1 qS ′ q[ · ] + B ∗′ q[ · , hq] + B ∗ qh ′ q[ · ], which at equilibria becomes 12 f ′ q [ · ] = K −1 q S ′ q [ · ] + B ∗′ q [ · , hq ] + B ∗ q h ′ q [ · ].

5. Linear algebra for the characteristic equations