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Reduced d'Alembert equation for an immersion: generalized forces and the quadratic correction

Statement

Let n,m≥1, let Ω⊆Rn be open and nonempty, and let X:Ω→Rm be of class C3 with DX(q) injective for every q∈Ω; fix an immersion in the sense of Submersions and immersions between Euclidean open sets (this forces n≤m). Let M∈Mm(R) be symmetric positive definite, meaning MT=M and uTMu>0 for every 0≠u∈Rm, let D⊆R×Rm×Rm be open and let F:D→Rm be continuous, let I⊆R be an open interval (Intervals of R: the nine order-convex forms, nondegeneracy, and length), and let q:I→Ω be a C2 curve with x(t):=X(q(t)) and (t,x(t),x′(t))∈D for all t∈I. Suppose there is a continuous R:I→Rm with

Mx′′(t)=F(t,x(t),x′(t))+R(t),R(t)⋅DX(q(t))δ=0for all δ∈Rn, t∈I.

Then, writing DX(q(t))T for the transpose and (Xqq[q′,q′])k:=∑i,j∂i∂jXk(q)qi′qj′, one has

DX(q)TMDX(q) q′′=DX(q)TF(t,x,x′)−DX(q)TM Xqq[q′,q′]

on I. The vector DX(q)TF(t,x,x′) is the generalized force of the ambient force F in the coordinates q, and DXTMXqq[q′,q′] is the acceleration correction of the parametrisation; when X is affine, the correction vanishes.

Facts & Assumptions

Given: The open set Ω, the immersion X, the constant matrix M, the field F, the curve q with x=X∘q, and the reaction R satisfying the two displayed relations.

[F1]

Differentiating x=X∘q twice by the chain rule (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Coordinate covariance of the Euler-Lagrange expression under time-dependent point transformations, Ck Euclidean maps and diffeomorphisms, Ck maps and multi-index derivative notation in Euclidean space): x′=DX(q)q′ and x′′=ddt(DX(q))q′+DX(q)q′′=(Xqq[q′,q′])+DX(q)q′′.

[F2]

R(t) annihilates the tangent space of the immersion: R(t)⋅DX(q(t))δ=0 for all δ; equivalently DX(q(t))TR(t)=0, because (DXTR)⋅δ=R⋅DXδ for every δ (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, Linear map between vector spaces over the same field).

Proof

technique · direct; pull the ambient equation back by the transpose of the derivative
1.1F2F3given

Multiplying the ambient equation Mx′′=F+R on the left by DX(q(t))T gives DXTMx′′=DXTF+DXTR=DXTF, because DXTR=0 by [F2].

2.1F1step 1.1algebra∎

Substituting the second-derivative chain rule of [F1] into step 1.1 gives DXTM(Xqq[q′,q′]+DX q′′)=DXTF. By associativity of matrix products this is DXTMDX q′′=DXTF−DXTMXqq[q′,q′], which is the displayed equation.

Remarks

Depends on

Dependency tree · 2 levels

Sources