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coudy5,692 19 49

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-1Let x, y: [a, b] → R, a < b, a, b ∈ R be two smooth functions (x, y ∈ C∞([a, b])). How can I prove thatthere is a unique function θ : [a, b] → R, θ ∈ C∞([a, b]) such that:

x(t)sinθ(t) = y(t)cosθ(t), ∀ t ∈ [a, b].(x(t0), y(t0)) = (x0, y0) ≠ (0, 0) is given

θ(t0) = θ0 is given too such that x(t0)sinθ(t0) = y(t0)cosθ(t0).

dg.differential-geometry smooth-manifolds

share improve this question asked May 3 at 11:51

closed as off-topic by Deane Yang, Benoît Kloeckner, Igor Belegradek, Alex Degtyarev, HughThomas May 3 at 14:33

This question appears to be off-topic. The users who voted to close gave these specific reasons:

"MathOverflow is for mathematicians to ask each other questions about their research. SeeMath.StackExchange to ask general questions in mathematics." – Benoît Kloeckner, Hugh Thomas"This question does not appear to be about research level mathematics within the scope defined in the helpcenter." – Deane Yang, Igor Belegradek, Alex Degtyarev

If this question can be reworded to fit the rules in the help center, please edit the question.

1 There cannot be uniqueness because your assumptions do not prevent x, y from vanishing on a subinterval,and on that subinterval θ can be chosen arbitrarily. – Igor Belegradek May 3 at 12:20

1You need (x(t), y(t)) to be non-zero for all t. If the curve t ↦ (x(t), y(t)) spends some time at the origin,you can choose θ as you wish there.

Assuming that (x, y) is non-zero, your question boils down to representing the curve in polarcoordinates. Writing x = rcos(η), y = rcos(η), we get

r(cosηsinθ − sinηcosθ) = rsin(θ − η) = 0

Given some initial condition, it is classical that we can choose η(t) uniquely if r is non-zero everywhere,by the fact that the standard projection R ↦ S1 given by η ↦ (cos(η), sin(η)) is a covering map. Then

η(t) − θ(t) must belong to π2 mod π and thus by continuity, this difference is constant, the constant being

given by the initial condition.

EDIT: here is a counterexample to the existence of such θ when (x, y) is allowed to vanish. Choose a C∞

function ϕ on R such that this function together with all its derivatives vanish at the origin. We take

x(t) = ϕ(t)cos(1/t), y(t) = ϕ(t)sin(1/t)

It is not difficult to show that x and y are smooth but the angle has no limit at t = 0. What is happening isthat the vector (x(t), y(t)) is spinning very fast near the origin.

share improve this answer edited May 3 at 15:13 answered May 3 at 12:22

It remains true (the uniqueness) when it is given that: (x(t), y(t)) = (0, 0) ⟺ t ∈ {t1 < t2 < … < tn}? – Student May 3 at 12:29

Yes if the angle has a limit at each ti, in which case you can glue together the solutions on each of theintervals. But this may not be the case, see my answer. – coudy May 3 at 12:53

Asked 1 Month AgoViewed 115 TimesActive 1 Month Ago

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intervals. But this may not be the case, see my answer. – coudy May 3 at 12:53

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