Answers from STACKEXCHANGE

1 Is it possible to solve the following ordinary differential equation: (t ) = x (t )sin( (t )) y (t )cos( (t )), t I , I interval from R,where x , y : I R are two given continuous functions?Christian Blatter, Switzerland: A simplification can be obtained in the following

way: Write x (t ), y (t )

= r (t )

cos(t ), sin(t )

,

where the functions t 7 r (t ) > 0 and t 7 (t ) can be considered as given. The ODEthen becomes

(t ) = r (t ) sin (t )(t ) (1)

Now introduce a new independent variable by means of

:=(t ) :=

t0

r (s )d s ,

and denote differentiation with respect to by a dot. The ODE (1) then becomes

() = sin () () ,

where () :=1()

can again be considered as given.

2 Let f ,g : I R be two function in C k (I ), with the property that f 2(t ) + g 2(t ) =1, t I . Is there a function : I R, C k (I ), such that:

cos( (t )) = f (t )sin( (t )) = g (t )

?If so, i wonder if an explicit formula can be found.Moreover, if f and g are real analytic function, is it true that has the same prop-

erty?Note that, C k (I ) is the class of all continuous function, having derivative of order k

that is continuous on I .

WWW (Mathlinks): Heres a chaining argument: Consider arcs A on the unit circleof length less thanpi. For any such A there is a real-valued, real analytic function h ona neighborhood of A such that (x , y ) = (cosh (x , y ), sinh (x , y )) for all (x , y ) A. Thish is just a piece of the argument "function". Furthermore, if h1,h2 are two functionswith this property, then h2 = h1+2npi for some n Z. It follows that if h1 = h2 at somepoint of A, then h1 = h2 on all of A. We are using the connectedness of A here.

Another connectedness fact: if A1,A2 are two such arcs with non-empty intersec-tion, h1,h2 are argument functions on A1,A2 respectively, and h1 = h2 at some point

1

of A1A2, then h1 = h2 on A1A2. This follows from the above and the connectednessof A1A2. Note that A1A2 need not be connected for any pair of arcs A1,A2, but herewere saved because of the length restriction.

Let F = ( f ,g ). Partition I = [0, 1] with points 0 = x0 < x1 < < xn = 1 and de-fine overlapping open subintervals I1 = [x0, x2), In1 = (xn2, xn ] and in between Ik =(xk1, xk+1). We can choose this partition so that F (Ik ) has length less than pi for eachk . Then we can choose "arguments"hk on neighbrohoods of F (Ik ) as follows: Choosean h1 that works on F (I1) as described above. Now choose y1 I1 I2. We can chooseh2 that works on F (I2) *and* such that h2(F (y1)) = h1(F (y1)). We then get h1 = h2 onF (I1) F (I2) from the results above. Because F (I1 I2) F (I1) F (I2) we get h1 F =h2 F on I1 I2.

We can continue chaining maps together in this way with agreement on the over-laps. We then define = hk F on each Ik . This is well defined on [0, 1] since there isno overlap problem. Because each hk is real analytic, the map preserves whateversmoothness F has, including real analyticity.John: Yes, by a very slight generalization of the path lifting theorem (PLT). Treat

your functions f and g together as a path

: I S1 : t 7 ( f (t ),g (t ))This lifts, by the PLT, to a map to the universal cover, R, of S1,

: I Rwith the property that

p = where p is the universal covering map,

p :R S1 : t 7 (cos(t ), sin(t ))You can then define (t ) = (t ).Now all thats left is to prove differentiability, which is straightforward.Comentarii:-I wonder if it is posibile in that particular case to find an explicit formula for (t ). I

need this in solving some differential equations. Anyway, thanks a lot for your answer.It helped me. - Bogdan Oct 17 at 17:16

- Sure, sort of: (t ) atan2(g (t ), f (t )), with the "approx" because you might needto add multiples of 2pi to make it continuous at points where g (t ) = 1. Alterna-tively, you could write (t ) = arg( f (t ) + i g (t )), where "arg" is the multi-valued com-plex "argument" function. Either one of these is probably sufficient, in the sensethat even if (t ) is hard to write down, (t ) is relatively easy, because locally (t ) =C +arctan(g (t )/ f (t )), where C is some multiple of 2pi, and for your DE, may be allthats needed. - John Oct 17 at 19:16

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-I came across another problem. If the functions f and g are analytic on I , then sois , too? -Bogdan

-@Bogdan In this situation yes, is analytic if f and g are, because the coveringmap is analytic and locally invertible (the derivative of t (cos t , sin t ) is nonzeroeverywhere) - Daniel Fischer Oct 20 at 22:06

3 Let f ,g : I R be two real, non-constant, analytic functions on an open intervalI R. If there is a smooth function h : I R (i.e. h C(I )) such that: ( f h )(t ) =g (t ), t I , is it true that h is also analytic on I ?Yiorgos S. Smyrlis, Cyprus: If we add that f does not vanish nowhere, then the

answer is yes.It can be checked locally. As f is real-analytic, it extends analytically in an neigh-

bourhood of h (I ) inC - We assume that f is real analytic on an interval J , with h (I )J .

Let now h (t0) =w0 h (I ).As f (w0) 6= 0, then f possesses an analytic inverse F in a neighbourhoodD (w0, r )

C of t0, and hence h = F g , for t I , such that h (t ) (w0 r,w0 + r ), and thereforeh is analytic in a neighboorhood of t0.

However, if f vanishes, then it DOES NOT hold. For example h (x ) = |x | and f (x ) =x 2.

-Thanks for your answer! I put that question trying to solve this:.... There f = cos. Doyou think that maybe its not true what is asked at that topic? Bogdan Oct 20 at 21:38

-@Bogdan There you have two functions, cos and sin, and their derivatives dontvanishat the samepoints, so everywhere oneof the two is locally invertible.Daniel Fischer

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