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The quaternion lift of a prescribed continuous angular velocity: $q^{'} = \frac{1}{2} q φ (ω (t))$ keeps $S^{3}$ invariant, its rotation path has body angular velocity $ω$ , and the initial value problem has a unique solution on the whole interval

Statement

Let $I \subseteq R$ be a nondegenerate open interval, let $ω : I \to R^{3}$ be continuous, and consider on the open time-state domain $D : = I \times H \subseteq R \times H$ the initial value problem

$q^{'} (t) = \frac{1}{2} q (t) φ (ω (t)), q (t_{0}) = q_{0},$

read as the first-order system of First-order systems, initial value problems, and solutions on intervals with time-state domain $D$ , state space $H$ , and right-hand side $F : D \to H$ , $F (t, q) = \frac{1}{2} q φ (ω (t))$ , where $φ$ , $S^{3}$ and $ρ$ are those of The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts.

If $q : I \to H$ is a $C^{1}$ solution of $q^{'} = \frac{1}{2} q φ (ω (t))$ , then $N \circ q$ is constant on $I$ ; if $q (t_{0}) \in S^{3}$ for one $t_{0} \in I$ , then $q (I) \subseteq S^{3}$ and the curve

$R : = ρ \circ q : I ⟶ S O (3), R (t) : = ρ (q (t)),$

is of class $C^{1}$ with $R^{'} (t) = R (t) Ω_{ω (t)}$ for every $t \in I$ ; consequently $ω$ is the body angular velocity of $R$ in the sense of The body angular velocity of a $C^{1}$ rotation path: $Ω = R^{T} R^{'}$ is skew, $Ω z = ω \times z$ determines a unique continuous $ω$ , and $R^{'} = R Ω$ . 2. For every $t_{0} \in I$ and $q_{0} \in S^{3}$ the problem has exactly one maximal solution, and its domain is all of $I$ ; that is, there is exactly one $C^{1}$ curve $q : I \to S^{3}$ with $q (t_{0}) = q_{0}$ and $q^{'} = \frac{1}{2} q φ (ω (t))$ on $I$ , and clause 1 applies to it.

Facts & Assumptions

Given: The interval $I$ , the continuous curve $ω$ , the initial data $(t_{0}, q_{0}) \in I \times H$ , and the objects $H, φ, S^{3}, ρ$ of The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts.

[F1]

$H$ is $R^{4}$ with its coordinate norm $∥ \cdot ∥_{2}$ , the quaternion product, and the identification $φ : R^{3} \to Im H$ ; $N (x) = ∥ x ∥_{2}^{2}$ ; $S^{3} = N^{- 1} (1)$ ; and $ρ (q) (x) = q x q^{- 1}$ for $q \neq 0$ (The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F2]

For $p = s + φ (u)$ , $q = t + φ (w)$ one has $p q = (s t - ⟨ u, w ⟩_{2}) + φ (s w + t u + u \times w)$ ; for $C^{1}$ curves $p, q : I \to H$ , $(p q)^{'} = p^{'} q + p q^{'}$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 1).

[F3]

$N (p q) = N (p) N (q)$ , and for a $C^{1}$ curve $q : I \to H$ one has $(N \circ q)^{'} = 2 Re (q^{'} \overset{ˉ}{q}) = 2 ⟨ q^{'}, q ⟩_{2}$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 2).

[F4]

$q φ (ω) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥_{2}^{2}) ω + 2 ⟨ u, ω ⟩_{2} u + 2 a (u \times ω))$ for $q = a + φ (u)$ , in particular its real part is $0$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 3).

[F5]

For $q_{0} \in S^{3}$ one has $ρ (q_{0}) \in S O (3)$ , and $R \mapsto ρ (R)$ is a group homomorphism on $S^{3}$ (The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts).

[F6]

For a $C^{1}$ curve $R : I \to S O (3)$ the body angular velocity $ω_{R}$ exists uniquely, with $R^{'} = R Ω_{ω_{R}}$ (The body angular velocity of a $C^{1}$ rotation path: $Ω = R^{T} R^{'}$ is skew, $Ω z = ω \times z$ determines a unique continuous $ω$ , and $R^{'} = R Ω$ ).

[F7]

The system is an initial value problem in the sense of First-order systems, initial value problems, and solutions on intervals with open domain $D = I \times H$ . In applying Local Lipschitz continuity in the state variable, locally uniform in time and parameters, a compact time-state cylinder here is a product $J \times K \subseteq D$ with $J \subseteq I$ a compact interval and $K \subseteq H$ compact; the condition requires one finite state-Lipschitz constant on each such product. If $F$ is continuous and satisfies this condition, then Picard-Lindelof local existence and uniqueness holds, and there is a unique maximal solution (Picard-Lindelöf local existence and uniqueness for first-order systems, Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, Extensions, maximal solutions, maximal intervals, and global solutions); any two solutions through the same initial data agree on every common interval containing the initial time (Picard-Lindelöf local existence and uniqueness for first-order systems).

[F8]

Compact continuation: if a solution whose graph lies in a compact subset $K$ of the domain approaches a finite endpoint, then it extends past that endpoint (Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it).

[F9]

For every finite $n \geq 1$ , a subset of $R^{n}$ is compact if and only if it is closed and bounded; continuous preimages of closed sets are closed, singletons in a metric space are closed, and polynomial maps such as $N$ are continuous (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open cover, subcover, compact metric space, and compact subset of a metric space, Metric continuity characterisations, with countable choice for the sequential converse, A space is $T_{1}$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology, $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, $C^{k}$ Euclidean maps and diffeomorphisms).

[F10]

The Euclidean norm is a norm on $R^{3}$ and satisfies $∣ ∥ u ∥_{2} - ∥ v ∥_{2} ∣ \leq ∥ u - v ∥_{2}$ , so $t \mapsto ∥ ω (t) ∥_{2}$ is continuous; a continuous real function on a nonempty compact metric space is bounded (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The reverse triangle inequality in a normed space, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct; the field is Lipschitz with the continuous bound $\|\omega\|$, the norm is conserved by the conjugation formula, and compact continuation keeps the solution on the compact sphere

1.1F1F3F7F10algebra

The map $F (t, q) = \frac{1}{2} q φ (ω (t))$ on $D = I \times H$ is continuous, since the product is bilinear and continuous and $ω$ is continuous. For $q, q^{'} \in H$ and $t \in I$ one has $F (t, q) - F (t, q^{'}) = \frac{1}{2} (q - q^{'}) φ (ω (t))$ , so by [F3] and [F1] $∥ F (t, q) - F (t, q^{'}) ∥_{2} = \frac{1}{2} ∥ q - q^{'} ∥_{2} ∥ φ (ω (t)) ∥_{2} = \frac{1}{2} ∥ q - q^{'} ∥_{2} ∥ ω (t) ∥_{2}$ . The norm continuity and compact boundedness in [F10] give a finite bound $M_{J} : = \max_{t \in J} ∥ ω (t) ∥_{2}$ on every nonempty compact time interval $J \subset I$ . For a compact time-state cylinder $J \times K$ as specified in [F7], $L = M_{J} / 2$ is one finite state-Lipschitz constant, since the estimate holds for every $q, q^{'} \in H$ at each $t \in J$ . Thus $F$ satisfies the hypotheses of [F7].

1.2F3F4algebra

Let $q : I \to H$ be a $C^{1}$ solution. By [F3] and the differential equation, $(N \circ q)^{'} = 2 Re (q^{'} \overset{ˉ}{q}) = Re (q φ (ω (t)) \overset{ˉ}{q})$ , and this is $0$ by [F4]. Hence $N \circ q$ is constant on $I$ ; if $N (q (t_{0})) = 1$ , this constant is $1$ and $q (I) \subseteq S^{3}$ .

1.3F1F9

The set $S^{3} = N^{- 1} ({1})$ is closed and bounded in $R^{4}$ : it is the preimage of the closed singleton ${1}$ under the continuous map $N$ , and every $q \in S^{3}$ has $∥ q ∥_{2}^{2} = N (q) = 1$ . By [F9] it is compact.

2.1step 1.1F7

By [F7] and step 1.1 the problem has a unique maximal solution $x : I_{x} \to H$ through $(t_{0}, q_{0})$ , and its domain $I_{x}$ is an open interval containing $t_{0}$ .

2.2step 1.2F2F5F6algebra

Let $q : I \to H$ be any $C^{1}$ solution with $q (t_{0}) \in S^{3}$ , so $q (I) \subseteq S^{3}$ and $ρ (q (t)) \in S O (3)$ for all $t$ by [F5]; fix $v \in R^{3}$ and put $f (t) : = q (t) φ (v) \overset{ˉ}{q} (t)$ . By [F2] the curve $f$ is $C^{1}$ with $f^{'} = q^{'} φ (v) \overset{ˉ}{q} + q φ (v) \overline{q^{'}}$ , and $q^{'} = \frac{1}{2} q φ (ω (t))$ , $\overline{q^{'}} = \frac{1}{2} φ (- ω) \overset{ˉ}{q}$ by the anti-automorphism identity $\overline{p q} = \overset{ˉ}{q} \overset{ˉ}{p}$ of [F3] and the linearity of conjugation ( $\overline{φ (ω)} = φ (- ω)$ ). The finite product formula of [F2] gives $φ (ω) φ (v) + φ (v) φ (- ω) = 2 φ (ω \times v)$ , so $f^{'} (t) = \frac{1}{2} q (φ (ω) φ (v) + φ (v) φ (- ω)) \overset{ˉ}{q} = q (t) φ (ω (t) \times v) \overset{ˉ}{q} (t)$ . The $a$ -th entry of the matrix $R (t) = ρ (q (t))$ is the component of $R (t) e_{a} = f (e_{a})$ read through $φ$ , so each entry of $R$ is $C^{1}$ and, by the displayed identity with $v = e_{a}$ , $R^{'} (t) e_{a} = ρ (q (t)) (φ (ω (t) \times e_{a})) = R (t) Ω_{ω (t)} e_{a}$ for $a = 1, 2, 3$ . Both sides are linear in the vector they act on, so $R^{'} (t) = R (t) Ω_{ω (t)}$ , and [F6] identifies $ω$ as the body angular velocity of $R$ . This proves clause 1.

3.1step 1.2step 2.1step 1.3F8F9algebra

Assume $q_{0} \in S^{3}$ , so that the maximal solution $x$ of step 2.1 satisfies $x (I_{x}) \subseteq S^{3}$ by step 1.2. If $I_{x} \neq I$ , then since both are intervals containing $t_{0}$ either $β : = \sup I_{x} < \sup I$ or $α : = \inf I_{x} > \inf I$ . In the first case $t_{0} < β < \sup I$ ; choose $b \in I$ with $β < b < \sup I$ and choose $δ > 0$ with $β - δ > t_{0}$ . The set $K : = [t_{0}, b] \times S^{3}$ lies in $D = I \times H$ , since $I$ is an interval containing $t_{0}$ and $b$ . It is closed and bounded in $R^{5} ≅ R \times H$ : $[t_{0}, b]$ is closed and bounded in $R$ , and $S^{3}$ is closed and bounded in $R^{4}$ by step 1.3. Hence $K$ is compact by [F9]. For every $t \in (β - δ, β)$ the graph point $(t, x (t))$ lies in $K$ , so [F8], applied to $x ∣_{[t_{0}, β)}$ with finite endpoint $β$ , extends $x$ past $β$ , contradicting maximality. If instead $α : = \inf I_{x} > \inf I$ , define the reflected intervals $I^{*} : = {s : 2 t_{0} - s \in I}$ and $I_{x}^{*} : = {s : 2 t_{0} - s \in I_{x}} \subset I^{*}$ , and define $\tilde{x} (s) : = x (2 t_{0} - s)$ . Direct differentiation shows that $\tilde{x}$ solves the same type of equation on $I_{x}^{*}$ with continuous rate $s \mapsto - ω (2 t_{0} - s)$ ; reflection carries any proper extension of $\tilde{x}$ to a proper extension of $x$ , so $\tilde{x}$ is maximal. Its finite right endpoint $2 t_{0} - α$ lies strictly below the right endpoint $2 t_{0} - \inf I$ of $I^{*}$ , so the first-case argument applied to $\tilde{x}$ gives a contradiction. Therefore $I_{x} = I$ , and the problem with $q_{0} \in S^{3}$ has a solution on all of $I$ .

4.1step 3.1step 2.2F7∎

Finally, let $q, q^{''} : I \to H$ be two solutions with the same value at $t_{0}$ . By [F7] any two solutions through the same initial data agree on every common interval containing $t_{0}$ , and $I$ is such an interval; hence $q = q^{''}$ . Combined with step 3.1 (existence on $I$ for $q_{0} \in S^{3}$ for the maximal solution, whose restrictions are the solutions) and step 2.2, this proves clause 2 and completes the proof.

Remarks

The hypotheses are minimal in the usual sense: $ω$ is only continuous, the equation is linear in $q$ with coefficient $\frac{1}{2} φ (ω (t))$ , and no regularity of $ω$ beyond continuity is used; the solution is $C^{1}$ but not claimed to be more regular unless $ω$ is.
The invariance argument uses the identity $(N \circ q)^{'} = Re (q φ (ω) \overset{ˉ}{q})$ and the scalar-part computation of the conjugation formula; it does not use the unit condition at intermediate times, so it applies along the whole maximal solution.
If $I$ is replaced by a non-open interval, the statement is read on its interior; the maximal-domain conclusion is a statement about open domains. Nothing here asserts completeness of the field in any other sense.

The quaternion lift of a prescribed continuous angular velocity: $q^{'} = \frac{1}{2} q φ (ω (t))$ keeps $S^{3}$ invariant, its rotation path has body angular velocity $ω$ , and the initial value problem has a unique solution on the whole interval

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The quaternion lift of a prescribed continuous angular velocity: q′=12q φ(ω(t))q′=21​qφ(ω(t)) keeps S3S3 invariant, its rotation path has body angular velocity ωω, and the initial value problem has a unique solution on the whole interval

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The quaternion lift of a prescribed continuous angular velocity: $q^{'} = \frac{1}{2} q φ (ω (t))$ keeps $S^{3}$ invariant, its rotation path has body angular velocity $ω$ , and the initial value problem has a unique solution on the whole interval