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The distribution span{∂x,∂y+x∂z} is not tangent to any regular level sets

Statement

Let O⊆R3 be open and nonempty, and consider the derivations A:=∂x and B:=∂y+x ∂z acting on C2 functions on O. If G:O→R is of class C2 with

AG=0andBG=0on O,

then DG=0 on O. Consequently, if G:O→R is C2 and DG(q)≠0 for every q∈O (so that every level set of G is a regular level set with tangent space ker⁡DG(q), Regular and critical points, regular and critical values, and level sets, The tangent space to a regular level set), then the distribution D(q)=span⁡{A(q),B(q)} of The velocity constraint z′=xy′: virtual displacements, the d'Alembert law and the vakonomic law cannot equal ker⁡DG(q) for every q∈O; that is, no such G has level sets whose tangent planes are everywhere the virtual displacements of the constraint z′=xy′.

Facts & Assumptions

Given: The open set O, a C2 function G:O→R, and the two derivations A=∂x, B=∂y+x∂z.

[F1]

Partial derivatives, the gradient and the Jacobian (Directional derivatives and partial derivatives of a map U⊆Rm→Rn, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Ck maps and multi-index derivative notation in Euclidean space, Ck Euclidean maps and diffeomorphisms): AG=∂xG and BG=∂yG+x∂zG are the displayed C1 functions, and DG=(∂xG,∂yG,∂zG).

[F2]

Equality of mixed partial derivatives for a C2 function (Clairaut--Schwarz theorem for continuous second partial derivatives): ∂x∂yG=∂y∂xG and ∂x∂zG=∂z∂xG on O.

Proof

technique · direct; bracket the two defining vector fields and use the equality of mixed partials
1.1F1F3givenalgebra

The hypotheses AG=0 and BG=0 on O give, by the product rule and linearity of the derivations, A(BG)−B(AG)=0−0=0 on O; that is, [A,B]G=0 where [A,B]G:=ABG−BAG.

1.2F2F3algebra

Compute [A,B]G from [F3]: ABG−BAG=(∂x∂yG+∂zG+x∂x∂zG)−(∂y∂xG+x∂z∂xG)=∂zG, because [F2] cancels the first pair and the pair multiplied by x.

2.1F1step 1.1step 1.2algebra

Steps 1.1 and 1.2 give ∂zG=0 on O. Then BG=∂yG+x∂zG=∂yG=0, and AG=∂xG=0; hence all three partial derivatives vanish and DG=0 on O.

3.1step 2.1F1∎

Consequently, if DG(q)≠0 at every point of O, no C2 function satisfies AG=BG=0; in particular no such G has D(q)=ker⁡DG(q) for every q, since equality with the kernel would force A and B to annihilate G.

Remarks

  • The computation shows that the constraint distribution of z′=xy′ is not tangent to a two-dimensional regular level foliation: steps 1.1-1.2 identify the bracket [A,B] on C2 functions as the derivation ∂z, and the hypothesis that A and B annihilate a defining function forces ∂zG=0 and hence a vanishing gradient. The conclusion needed by the classical-mechanics development is drawn directly, without invoking a converse of the Frobenius theorem.

  • The proposition is the precise sense in which z′=xy′ is a non-holonomic constraint: it is not equivalent to G(q)=constant for any C2 function with nonvanishing differential. Nothing is asserted about constraints that merely look similar or about kinematic mechanisms; those belong to the consuming physical models.

Depends on

Dependency tree · 2 levels

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