Math. Ann. 268, 59-84 (1984) A m �9 Springer-Verlag 1984

The Rotating String

M i c h a e l R e e k e n

Fachbereich Mathematik, Gesamthochschule Wuppertal, Gaussstrasse 20, D-5600 Wuppertal 1, Federal Republic of Germany

Contents

1. Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 2. The Equations of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . 60 3. General Properties of the Associated Functional Map . . . . . . . . . . . . . . 63 4. Properties of the Set of Solutions of ~i = 0 . . . . . . . . . . . . . . . . . . . 64 5. Asymptotic Behaviour of the Functional Map . . . . . . . . . . . . . . . . . 66 6. The Derivative of (~ at # = 0 . . . . . . . . . . . . . . . . . . . . . . . . . 69 7. Bifurcation Analysis at Infinity . . . . . . . . . . . . . . . . . . . . . . . . 71 8. Connecting the Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 73 9. Solving the Full-Set of Equations . . . . . . . . . . . . . . . . . . . . . . . 82

10. Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

1. Nota t ion

Fig. 1

x 3

Xl + i x 2

/ / J (z(s),u(s))

/ ,,/L ~ p l (s=L)

lg

60 M. Reeken

[0, L] (z(s), u(s))

re(s)

N

Tt

9~

o)

A , A

01 02 Oo g

X Y w

S

represents the reference configuration of the string is the location in space of the point parametrized by s. (For notational convenience the Xl -xz -p lane is represented by ~) is the mass distribution on the string, re(s) is the mass of the string contained in the interval [0, s], In the most general case it will be a function a) of bounded total variation b) strictly increasing is the tension in the string. It is a given function with the following properties: a) N:(O, ~ ) ~ R b) strictly increasing and onto. (Further restrictions will be imposed when they are needed) is the inverse of N. It has the following properties: a) 91:R-~(O, oo) b) strictly increasing and onto

9~(t) is given by 93/(0 :-- .9X has a singularity at t - -0

t is the angular velocity of the rotating frame is given by fl(s, c) : = fl = re(s) - c is given by A(k, s) : -- A = ~ + fiE

A ( k , ~ , s ) : = A = l ]/]kl2 32 V 7 §

will be a measurable subset of [0, L]

~ + 1 , s c I , is given by ~((s) : = ( _ 1, s c I c

(QI,0) is the position of Po (02 eick, UO) is the position of P1 is given by 0o:=01--Q2 eir the constant of gravity. By incorporating it into m we can treat g as being one r := ~lz'(s)l 2 +u'(s) 2 measures the stretching of the string at s X : = C ([O,L]) x ~ . 2 Y : = F x 6 s

w : = y - ~ o92zdm(a), new dependent variable replacing z o

S : = {(x, y) ~ X x YIA(w, s) has a zero on [0, L]} - {(x, y) ~ X x Ylc ~ [0, L] and w(c) = 0}

the singular set, where differentiability of the functional maps breaks down

2. The Equations of Equilibrium

(E 1)

(E 2)

Rotating String 61

These equations have to be completed by the boundary conditions:

zO) = 01, z ( L ) = Q 2 e i* , (BC 1)

u(O) = O, u (L) = u o . (BC 2)

The prime denotes differentiation (in the distributional sense) with respect to s. By a well-known theorem on distributions of one variable there are constants y e and c e R such that

N(~) z ' - - 7 -- i 0)2zdm(rr) (E 1")

N(~) u '= m - c. (E 2*)

A straightforward computation shows that

N(~(s)) 2 = 7 - 0 ) 2 ! zdm(a)l 2 + ( m ( s ) - - c ) 2 .

This implies that N(~(s ) ) is of bounded total variation. Using the shorthand s

introduced earlier we write this with w(s) : = y - o) 2 S zdm(a) as 0

N(~(s))2 = iw12 + fie = A(w(s ) , s) z .

We do not have any information on the sign of N. Instead of making ad-hoc assumptions on the continuity of N we assume that N is non-negative on I and negative on I c. I is just any measurable subset of [0, L]. I is so to speak an additional constraint imposed.

Now we can determine N:

~ + 1 , s e I , N(~(s ) ) = X(s )A(w, s) = z A , )~(s) = ( _ 1, s ~ I c .

Using the inverse 9l of N we determine 3:

~(s) = 91(zA) �9

Returning to (E 1") and (E 2*) we get

This is turned into

N ~ w'

91(z A ) z" - - w = 9~(x A )w ,

)~A

u' = 91(zA~) fl = ~(;(A)fl. zA

62 M. Reeken

Using the boundary conditions for the point Po leads to

z(s) = ~ + i 931(zA)wda, 0

$

u(s) = ~ 9~(ZA)/~d~. 0

w /~ Remark. Division by xA does not lead to any problems because ~ and ~ are

bounded functions since A = ]/@2 + w z. z and u have now been expressed in terms of w. It remains to find the equation

which determines w. To find it we only have to insert the above formula for z into the definition of w:

w(s) = 7 - o32 i zdm(a) , 0

w(s)+ 032 QI+ 9-Jl(zA)wdo d i n ( a ) - 7 = 0 . 0

This has to be completed by the remaining boundary conditions for P1- Thus, we are led to the following system of equations replacing (E 1), (E 2), (BC 1), (BC 2):

w + 092 i (m(s) - m(a))gJl(xA)wda + ~ 1 o32m- 7 = 0, (1) 0

L

~o + ~ ?O~(zA)wda = 0, (2) 0

L

Uo -- ~ ~O~(zA)flda = O. (3) 0

Given any solution of(l), (2), (3) it is a straightforward exercise to compute that $

z(s) = o, + f ~ t (ZA)wd~, 0

s

u(s) = ~ ~J~(zA)fldr 0

is a solution of (E 1), (E 2), (BC 1), (BC 2). That any solution (z, u, I) of the original set of equations leads to a solution

(w, ~, c) of (1), (2), (3) has just been derived. So, the two problems are equivalent. It is remarkable that the system of(E 1), (E 2), (BC 1), (BC 2) is broken up into an

uncountable number of equations parametrized by the measurable subsets of [0, L]. Each of these problems is essentially an elliptic one as one recognizes if(l) is turned into a second order differential equation and boundary conditions are determined from (I) and (2). The original problem was nonelliptic and second derivatives are in general singular distributions. The new unknown w (the tensile force in the string) has bounded second order derivatives.

The idea of characterizing distributional solutions by introducing the set I has been developed in the context of nonrotating strings in [4] and extended in [3].

Rotating String 63

3. General Properties of the Associated Functional Map

For the sake of simplicity we restrict attention to the case m(a) = a. Some of the properties established can be carried over to the general case others seem to fail or become excessively difficult to demonstrate.

We define maps

( 6 [C( [0 , L]) • F. 2] • Cc'([0, L]) x

sS: [Q( [o , L]) x R 2] x

given by:

w + (s -- a)~J~(zA)wda + ~ 1 e)2s - ~ ei~ , (6(w, q, o , c, ~p, Qo, ql, Uo) =

[ ~o + ! 9Jl(zA)wda,

L

O(w, ~, o.~, c, ~p, Qo, ~ , Uo) = Uo -- S 9J~(zA)fida. 0

C~ is the space of complex valued continuously differentiable functions. We have written the two factors ~,. in the image space as a single complex factor ~. This is just for computational convenience. It has no deeper significance. In our shorthand notation (6 is a map from X • Y into X and .~ from X • Y into R.

(6 and ~ are defined everywhere.

This is a consequence of the obvious estimate given below:

[gJ~(zA)wI=9~(zA)~[

19J~(zA)31 J

They ensure that all the integrations can be carried out. The integral term in the first component has a first derivative which is itself absolutely continuous.

(6 and ~ are everywhere continuous.

Consider a sequence (xi, yi)~(Xo, Yo) in X • Y. All the terms not involving integrals are obviously continuous. Concerning the integrals we remark that we have the integrand uniformly bounded by a constant. Furthermore, the integrands converge pointwise except for the point, where w 2 +fi2 takes the value zero. Because fl is a monotone function there is at most one such point. Therefore, all the integrals converge to the proper limits.

(6 is of the form "identity minus compact operator".

We only have to check on the first component. But as we already remarked the integral term admits a second generalized derivative which is a bounded function. From the estimate

d_~ i ( s - a)gJlo( a )wda = [!IJ2(z A )w[ < 9~Oi A ) < O(Iwlco, c) 0

64 M. Reeken

we conclude that this part of the map is compact on CI([0, L]). The remainder of the first component only involves parameters and scalar variables and is, therefore, also compact.

ffJ is continuously differentiable on the complement of

2; = {(x, y)~ X x Y[A(w, s) has a zero on [0, L]}.

Just formally computing we get an additional power of A into the denominator. It destroys integrability of the integrands on S. If x r U then A can be bounded away from zero. By standard arguments differentiation can then be carried out under the integral sign.

4. Properties of the Set of Solutions of (~ = 0

For convenience we introduce the following notation:

S:= {w ~ Ca([O,L])]Rew and I m w have simple zeroes only},

X s : = X c ~ ( S x R2),

~ : = { ( x , y ) e X x Ylffffx, y)=O},

~(P) : = {(x, y) c ~[P(x, y) is true},

3 : = z = ~ l + I ~ ( z A ) w d ~ l ( x , y ) e ~ , 0

3(Q) : = {z c 3}Q(z) is true}.

We are going to study properties of ~ which are important for the analysis of the problem

~(eo #O)cX~ x Y.

For r = 0 we get w = 1' but 5o + 0 prevents ? from being zero. If ~o is non-zero we study the integral equation (1) locally near any point s o c [O,L]. We have to distinguish between two cases. First, if c 4: So then we can rewrite the equation as

w(s) + o~ 2 ~ (s o -- a)92~(zA)wdtz + o)2Q1 So - - 0

-4-(S--So)[032s!~Jf~(zA)wdff'4-0)2Q11

+ i (s-a)gJl(zA)wda SO

= w(s) + W(So) + (s-- So)W'(So) + i (s-- a)~(zA)wda = O. $o

If W(So) = w'(So) = 0 then the integral equation defines a contraction near s = So and w = 0 is the only solution.

Rotating String 65

V ~ s ) : -

Second, if c = So and W(So)= 0 the argument is more complicated. We define w(s)

- - - for s + So. Wean be extended by continuity because S - - S 0

lim W(s) = lim w(s) - W(So) = w'(So) = : W(so) . s-~so S~So S - - S o

Dividing the equation we just have studied by s - S o we get an equation for W:

0 = W(s) + W(so) + i s - ~ 9l(X[a- Soll//lw(a)l 2 + I) s g n ( a - So)W(a) da . 3o S - S o ]//1 + [w(o-)[ 2

This again defines a contraction locally. If W(so)= 0 then W ( s ) = 0 is the only solution. Both results together imply that any solution which has a zero of order higher than one is identically zero. But zero cannot occur as a solution under the restrictions made. If only Rew or Imw has a zero then the first argument applies to the equation for Rew or Imw. This completes the proof.

The next proposit ion is crucial because it will allow us to distinguish between different continua in ~.

I f an equilibrium configuration (z, u) for co + 0 intersects the axis o f rotation then it lies entirely in a plane through the axis o f rotation.

Remark. This implies that such an intersection only occurs if either ~b = 0 or n or if QI or Q2 is zero or written compactly if A : = ~1Q2 sin~b = 0. So we can condense the result into the formula 3(V sz(s) + O) 3 3 (A +- 0).

For the proof we consider the Wronskian W(s)" = w~(s), Wz(S) As w satisfies w 2 ( s ) "

the O D E of second order

w" + co2?Ol(zA)w = 0

we know that W is a constant. If (z, u) intersects the axis of rotat ion z must have a zero. But w ' = - ~o2z. It follows that W=O. We determine the values ofw and w' at 0 and L:

w(O) = ?, w'(O) = - ~x~o 2 , w'(L) = - Q2ei%92 .

From W - 0 we deduce

0 = W(0) = Im ff(0)w'(0) = - ~ 1 ~02 Im f = + ? 2Q ~02,

0 = W(L) = - Im ~(L)Q2eiO~o 2 = - Q2~o 2 Im ~(L)e ie~ .

If ~1+0 it follows that 72=0. But then w2 has zero initial conditions [w2(0) -= w~(0) = 0-] so that w2 =- O. Here the integral equation is used for which we have already proved unique solvability independently of possible singularities of ~J/(zA). Consequently, the entire solution lies in a plane.

If Q l = 0 but Q2+0 then Im~(L)e i~=0. ~ : = w e - ~ satisfies the terminal conditions k2 (L)= k~(L)= 0. So, w2 = 0 and w is planar.

Ife~ and ~2 are non-zero then we have 72 = 0 and Im(we-~*) = 0. Looking again at ~ we have terminal conditions ~2(L)= ~ ( L ) = 0 which imply that w 2 - 0. But k(0) = ? e -~6 and k2(0)= 0 implies that sin~b =0. This ends the proof.

66 M. Reeken

We continue with another assertion about planar solutions:

I f 01 = 02 = 0 all equilibrium configurations for co + 0 are planar.

In that case any solution w of(l) and (2) gives rise to a solution ~ = we -'"~g~ of

+ co2 i (s - a)?iJlOcA)#da - 17[ = 0. 0

Since #2(0)= 0, ~ ( 0 ) = 0 we conclude that w 2 - 0. We conclude with the following useful result:

~(A =I=0 v c(~[O,L])c~Z=O.

For cr [0,L] this is obvious because fl+0. For 0102 sin~b+0 the function w does not have a zero, because otherwise W(s) would be identically zero. We just have shown above that this implies 0xP2 sin~b=0. This proves the assertion.

5. Asymptotic Behaviour of the Functional Map

We have separated ~ because it has not the same asymptotic properties as (5. Just from looking at 15 it seems appropriate to study the following map

6 : [c~([O, L]) x ~ ] x [S ~ x ~5]__,[c ~([0, L]) x ~2],

, co, c, ,oo, ol,uo) or explicitly

v + coz i ( s - a)?Ol(zA)vda + #co201s- e i~ , 0

Ig(v, #, co; c, lp, Oo, Ol,Uo)= L

#00 + 1 9J~(Z,'t)vda, 0

where --1 2 V ,2 +B z-- 1/Iv12+,2# 2.

Coming from (5 with # = 171- i the map ~i is only defined for #>0. I f wejust admit arbitrary # # 0 in the definition we see by inspection that the map ~ has a certain symmetry:

i $ ( - v, - # , co; c, ~p + re, . . . )= - tg(v, ~, co; c,q~, . . .).

Therefore, solutions with p < 0 are just mirror images of the solutions of the original map.

We are now going to show that ~i can be extended by continuity to # = 0. Topologically it means that we glue together two copies of X along a cylinder at infinity and extend the map to the whole space. To be precise we shall denote this new space by X. Thus, ~ becomes a map from X x Y into )( x Y.

has a unique continuous extension to # = O.

Rotating String 67

The only trouble arises in the integrands, where A contains inverse powers of#. At this point we have to be specific about the asymptotic nature of 9l. For simplicity we assume that 9l is asymptotically linear: l i m ~ ( t ) = n o .

t ~ o o

lim 9~(t)= 0 follows from the general form of N we had assumed from the t ~ ct~

beginning. From these assumptions it follows that there is a constant C > 0 such that

and

I~( t ) [~C for I t l~ l

9 l ( 0 = C for I t [< l .

This yields a uniform estimate for the integrands:

[gYl(ZA)v r <__ C[v[ if A >__ 1,

Ivl I~(z2)vl=~(zA)!~ =~(zA)pVlvl2+p2B 2 <c~ if A< 1.

Therefore, if we consider any sequence 2i--.2 o in X then the integrands are uniformly bounded. Let s be a point at which v o has no zero. Then

~ ( Z l / % ( ) l /#~ ( )2) ( ) ( ) ( ) lim ~ s 2 -.[- s - c ~ v~ s = c~ s nov o s i--* oo

because the square root tends to infinity so that 9)l either goes to no if X(s) = + 1 or to zero if X(s)= - 1.

If Vo(S)= 0 we cannot say anything about vi(s). But we have the a priori bound #i

which implies that for #i-~0 and vi(s)~O we get lim Jg(xfti)vi =0 = noQ(S)Vo(S). i ~ o o

That is the formula is generally true:

lim J/l (z A( vi, #i) )vi = noC tVo . i -~ oo

The pointwise limit and the uniform boundedness of the integrands allows us to interchange the integration and the limit. Thus, we have shown that:

- $

v + o)2 ~ ( s - a ) n o Q v d a - e i~ ,

~(v ,O,~;~P,C, Oo, Ol,Uo)= L 0

S noclvda. 0

This proves the following proposition:

~i[,= o is independent o f the parameters c, 00, 01, Uo.

~[u = o = 0 has a countable number o f continua o f solutions in X~ x Y. They have the form ( v f i~, O, +_ ~j ; c, ~p, 0o, 01, Uo), j = I, 2 . . . . . coj and vj are solutions of

$

v i + w ] ~ ( s - a ) n o Q v i d a - 1 = O, vj = Oj, 0

L

noClVjda = O . 0

68 M. Reeken

2 (D j2 is an increasing sequence, toj # 0 and v i hasj nodes, vj is concave, where v i_>_ 0 and convex, where vj < O.

That to = 0 does not lead to a solution is clear. The first equation implies v = const, the second implies then v j--0 which is definitely not a solution. If we differentiate the integral equation twice we get:

V" + to2noClV = 0.

From the integral equation we also rcad off initial conditions v(0) -- 1, v'(0)= 0. The second equation of (5 = 0 implies v'(L)= 0.

The boundary value problem

V" + to2noClV = 0 ,

v'(O) = v'(L) = 0

d 2 is of a well-known type. Let L = ds 2 +1 with boundary condition

v'(O) = v ' (L)= O. Applying standard results to the symmetric compact operator

L- 1 / 2 ( n o c I - - 1)L- 1/2

we get an infinite sequence of positive eigenvalues to} tending to infinity. They are simple eigenvalues of the boundary value problem with eigenfunctions v s having j nodes. As the problem is linear v~ can be chosen in such a way that vs(0 ) = 1. The concavi ty (convexity) follows from

~ < 0 , v>__0, v" = - toinoclv l > O, v < 0

ending the proof. The map ~ does not behave as nicely. We define

L

~(v, #, to; c, ~o, 0o, 01, Uo) :=u0 -- I ~ ( Z J ) f l da . 0

.~ has the same kind of mirror symmetry as (5. If (v, g, to, c, ~ . . . . ) is a solution of ~ - - 0 then ( - v , - # , t o , c , ~ + ~ . . . . ) is one.

From our earlier estimates it is obvious that 9Jl(ZA) does not have a unique well defined limit for #--+0 because the presence ofv in the intcgrand was crucial. But we can restrict our attention to )(, x ~z.

The map ~ has a unique continuous extension to # = 0 on J(~ x Y.

As before we need a uniform estimate on the integrand:

0J/(zA)fl[ = 9~(ZA) Aft--; ~ C if .4 __< 1,

[gJl(ZJ)fl[<=C.(L+[c[) if J > l .

Assume (xi, y i )~ (xo , Yo) in )(~ x Y.

Rotating String 69

Ifs does not coincide with one of the zeroes Of Vo then ,~!im ~ / ~ . + fl~ = o0,

and therefore, lim 9J~(ZAi) = cfno. Therefore, i--~ oo

L

$(v, 0, o); c,~v . . . . ) = U o - I c,noflda. 0

This proves our assertion. The map ~ has another important property:

lim ~(v,p,o); ~D,C,~o ,~ I ,Ho) : -~C~

uniformly on uo-bounded subsets of the remaining variables provided 2(1)> 0.

This follows from our estimates:

L L = U o - I ~(zd)sd~ + c I ~(zd)d~

0 0 L

The integrands are all bounded above because A = ] [//[v[2 V ] A2 "[- f12 ~ [C[- L for big c.

The second coefficient of c in ~ is negative but goes to zero as Icl--* oe because lim ~ ( t ) = 0. ~ ~Jl(A)da > no2(I) because gJl(t) > no for t > 0. This concludes the

proof.

6. The Derivative of ~i at p = 0

In order to continue our analysis we want to differentiate ~[,=o. From the properties of ff~ we know that (~ will be continuously differentiable for/~ ~= 0 if we avoid the singular set.

(~ will denote the partial Frechet derivative of (~ with respect to the i-th variable.

The partial Frechet derivatives of Cb for #~: 0 are given by: s $ P

h + )2 1 (s-a)9~(zA)hda + 0)21 ( s -a ) 9J~ (~A) Regh o o Z A v ~ - d a ,

@lh = L ~ 9~'(ZA) Re~h . 9~(Zft)hda + o zA o v ~ - d a ,

[ 2 ~- - ~ ' ( Z A) [vl 2v- . l--O) J ( S - - a ) ~ z~- aa+o)20a s, ^, ~ 0 2)~A # ~2--1 ~ ( z A ) lvl v . .

2o) I ( s - a)fg~()jl)vda + 2o)#Qls, , 0

63= 0.

70 M. Reeken

L

The presence of a term like S 9JlOcA)hda shows that there is no well defined limit for o

#--,0. The existence of the limit of I~i for #--*0 depended precisely upon the fact that 9J/(xA ) was multiplied by v.

The situation can be saved though by taking into account that the solutions are in general confined to the open subset )?s x Y = S x R 2 x Y. On that subset the derivative has a unique limit.

Let us take a closer look at the Frechet derivatives we computed earlier. There are essentially three functions occurring in the integrands which have to be handled when taking the limit #-+0. One of them, 9J/(zA), has been treated before. The other two are:

, ~ v R e v h 9JI'(Z,~) v Revh _ zA~IR (zA) Ilv~ ~ - f l 2 ] , #2ZA

We are looking for a uniform bound for these integrands as (xi, y i )~ (Xo, Yo) in (g~ x ~)n.#.

At this point we have to impose further conditions on 9l:

lira 9J/'(t)t 2 = nl .

t g l ' - gl Since 9Jl'(t)= t ~ this amounts to saying that 9l'(t)t differs asymptotically

only by a constant from 9l. This also implies lim fOl'(t)t-= O. t - * o O

From our earlier assumptions it follows that lim 9Jl'(t)t 2 =0. This also implies lim 9J/'(t)t=O. t - , - |

t ~ - - 0 0

The functions we have to bound have already been rearranged in such a manner that the factors in square brackets are uniformly bounded by 1.9Jl(t) and 9Jl'(t)t are both singular at t = 0 and are bounded as rtl~oe. 9Jl'(t)t 2 is even bounded everywhere. So, it suffices to bound / t / away from zero to get the desired uniform estimate. It is at this point that the restriction of being in ()~s x Y)m/~ c enters. The limit function v 0 has finitely many zeroes all different from c. So, we choose intervals around them which do not contain c. Then Ai> 60 > 0 for all sufficiently big i because on these intervals

#2 +/~2 >1/~1__>6o>0"

On the complement of these intervals Ivol has a lower limit 6o > 0. So,

~/lvil 2 > Ivil 60 7 > _ - # - 2 #

for all sufficiently big i.

Rotating String 71

What remains to be done is to determine the pointwise limits of these functions. These limits are

V0 O, nlcl Ivol

for all s except the zeroes of v o because for any s with Vo(S ) =i= 0 the expression -~i tends to infinity. This completes the proof of the following proposition:

As # tends to zero the Frechet derivatives of ~J with respect to (v, It, 00) converge on (Xs • y ) ~ c to the followin9 expressions:

�9 S

h + 002 ~ ( s - a)noclhda, lim ff~h = L o ~-~o ~ noclhda ,

0

lira ~' ( s -~)n lcI ~ d ~ + 002 Qls '

~ o ~ 2 = ~ ! v - n~Cl~da+Oo,

200 i ( s - a)noc,vd~ , lim (~ = ~0. o tt~O

Because the first and the third limit is equal to the corresponding derivative of ~lu=o the map ~ is continuously differentiable on (f(~ x Y)c~S ~.

7. Bifurcation Analysis at Infinity

We study the solvability of the following linear equation:

lh at- (~2~ + (~3 TM ,

where the derivatives are taken at one of the continua of solutions of ~lu=o =0. (v j, 00~) satisfy the equations

vj + i Cs - O)noC,V,a - 1 = 0 , = 0

L noCxV flt~ = O .

0 The system of equations for (h, 3, v) takes the following form:

h+00Z i (s-a)n~ + 6 ( 2 o ~176176 i (s--tT)nlc'eiWdt7) o

+ 2co jr i (s-- r = k, 0

! noclhdtT + 6 (Qo - ! nl ctei~~ = ~ .

72 M. Reeken

We deduce a necessary condition for the solvability of this system by differentiating the first equation and multiplying it by v~. Then we integrate over [0, L] and perform some integrations by part.

( ) 0 = ' ' 2 z L 2 vj nlctei~da'da vjh drr + o9j ~ v) ~ nocthda'da + 6 (.oj Q l V j [ o - - O)j 0 0 0 0

L a L

+ 2m jr ~ vj ~ noclvjei~da'dtr - y vjk'da 0 0 0

L L _ _ ~ ~ / 2 - (vjh - ~j vjnocih)da + o~]vj(L) ~. nocthda

0 0

+6 2 L ~oj Qlvjlo + ~o 2 S vjnlcte i~d~- o~]vj(L) nxct ei~&r J O

L L L

- 2o~iv ~ noc,vZei~da + 2~ojvvj(L) ~ noctvjeiV&r- ~ v~k'da. 0 0 0

Using the second equation of the linear system for (h, 6, v) and all the information on vj we have we come up with:

~]vj(L) ~ - 6 Qo- n~ct e"~d +6og~'01vjl~6-fo~2vj(L)~.nlc, e''~da 0

L L

-2~ojv ! noC,v}ei'~da - ! v;k'd(r=O.

Rearranging this we get an important result:

L

6[e~(v,(L)Qzei4'-Q1)]-v[2mjei~no!v}da ] = !v)k'da-co2vj(L)o~.

The right side is a complex number determined by the given element (k, ~). The two real numbers 6 and v are uniquely determined iff the two complex numbers in square brackets are linearly independent over R.

Because vj is a solution of a linear second order ODE with v'(L) = 0 it follows that vj(L)#O. So, the coefficient of 6 is zero iff Q1 =~2 =0. We exclude the case al = P2 = 0. If ~p # arg _+ (vj(L)Q2 ei4~ - ~ 1) then 6 and v are uniquely determined by k and a. Inserting them into the first equation we can solve uniquely for h because the integral equation defines a contraction on C1([0, L]) if a suitable norm is used. If we then repeat the calculation we did before it appears that h just satisfies the second equation, as well. Thus, we have shown"

For ~p+arg--k(vj(L)o2ei4~--~l ) the trivial solutions are the only solutions of I~i[u= o in a neiohbourhood of (el%j, O, o)j; c, ip, Qo, Q1, Uo).

We are going to demonstrate that the index of these solutions is plus or minus one and changes sign when we cross t p f : = arge(v~(L)o2ei*-oO, where e = +_ 1 is chosen in such a way that sin~pj+ > 0. This is possible because sin ~of # 0. We do this by showing that the Frechet derivative which is invertible for ~p. ~pf has a simple eigenvalue going through zero as ~p goes through ~pf. The Frechet derivative ^ / ! 151h + Ig26 + I~i;v is of the form "identity minus compact". It depends analytically

Rotating String 73

on ~o. The necessary and sufficient condition to be satisfied is

a6 + bei~v = 2AOp),

where a : = 092(v~(L)Qze i ~ Q,), b : = 2o~jn 1 1 v2da, I

L

I ' ' = vjh d a - o~j vj(L) (6 + iv). 0

Differentiating and setting ~p = ~Pf we get

a6" + be i~; v" + ibe itp~ v = 2"A(lpf ).

But ae-i~; =real number. So, it follows

by = 2' Im (A (~of)e-iv;)

and since b 4:0 we conclude that 2'4: 0. If we project the branch bifurcating from this point onto the / . t - o~-plane it is

straightforward to compute the direction of that curve at the bifurcation point. In fact,

de) v j ( L ) o 2 ei4~ - - Q 1 - ,~

= ~ J 2noSv2da e J # 0

I

O)j = 2n o S vyda (v~(L)Q2 cos~--Q, + iv;(L)e2 sin~)(costp? -- i s in~?)

I

O)j

= 2no I v2da (vj(L)Q2 cosq~ c o s ~ p / - 01 cos~p/+ vj(L)Qz simp simp/) I

O)j

= 2no I v2da (vj(L)Q2 cos(~ - t p f ) - r cos~pf). I

The imaginary part vanishes by the definition of ~pf.

8. Connecting the Solutions

In order to see how the continua of solutions anchored at the trivial solutions at infinity connect with each other and how they extend in different directions of the underlying space we have to find appropriate a priori estimates and if possible invariants which distinguish between certain continua.

We want to demonstrate that the winding number of z is such an invariant. We define the winding number in the usual way as

d argz(a) da F ( z ) ' = 1 ! da

for all z with the appropriate boundary conditions and z(s)4:0 for all s ~ z.

74 M. Reeken

We have shown in Sect. 4 that for A = ~1~2 sin~b + 0 the function z has no zeroes. From

z(0)=Q~ and z(L)=p2 e~~

we conclude immediately that

F(z)= 2~ + n , n ~ Z .

We shall consider F a function of (x, y)~ X s x Y:

1 ! d a r g w ' ( a ) d a = l !w~w'~-w'zw'~ r(x):= ~ d~ ~ wl 2u d~

=

2~ o 9~(zA) w]2--~ wi: d~ = ~ s W(0) o wl ~ + wl ~ d~

= -~- 7 2 q l ( D d" L ~2il(zA) �9

2o9 Jo w'12 + w~ 2 aa "

,b We set F ' : = F - 2~ = [F]. We call it the reduced winding number. It is the winding

number of the closed curve we get when we extend the given curve by the straight line segment from the endpoint to the initial point. From this formula we conclude

F'(x) is constant on any connected component of ~(A # 0).

The proof is based on the fact that the integrand has no singularity because neither w] 2 + w~ 2 nor w 2 + w2 2 has a zero on the continuum considered. Since the integrand is continuous as a function from the continuum into L~176 L]) the winding number is continuous. Taking only discrete values it must be constant.

We want to determine the winding number for the trivial solutions a t /~=0. These solutions are contained in a line through the origin. So, the winding number is not defined. But we can establish a relation between the winding number on a continuum and the number of nodes of the trivial solution which is the limit of that continuum as #--.0.

We had established that there are asymptotic solutions (vjeit~, O, +_ ogj, ~pj,... ). As ~P7 =~PJ+ + n the function just changes sign: vje i~j = - v i e it~. Both are contained in a line through the origin and have opposite sign. They havej interior nodes. The zeros of v~ need not be isolated but because of the concavity (convexity) of vj they occur on closed intervals I , and v~ changes sign exactly once on a

P~ r sufficiently narrow open interval covering anyone of the I .s.The e are exactly j + 1 such intervals I . and two of them contain one endpoint each.

We have sketched a possible graph for v2 and v~. There are rectilinear parts in the graphs. They occur when s e Y.

Rota t i ng St r ing 75

v 2

i i

i I

I

I I

i

i t i i I

i

I

llj ~z i lh__ 0 - - bO,,1 +

O 0 Fig, 2 sg n v2 ~ = -

Because the integral for F cannot be evaluated explicitly we have to look for a different way of computing the winding number. Assume z is a continuously differentiable curve with z + 0 and x 1 = Rez, x 2 = Imz having simple zeroes only. Then the winding number can be computed by counting the intersections with either axis.

Because the curve we are studying is flattened out for p ~ 0 we would like to rescale things transversally in such a way that it does not collapse. Let us try therefore the following scheme: first, we rotate the frame so that the x 1-axis contains the asymptotic solutions. The equations in this frame are

I) § (0 2 i (S - - ~7)~(Z . f f l )vda § p(02 Q~se-iwo _ ei~ = O, 0

L Oo e-iwo + ~ ~ ( X d ) v d o = O.

0

Now we try to normalize v2 by setting gz := v2 and rewriting the equations in P

terms of vl and f2. g2 is actually just the original w2 as it appeared in 15.

U 1 § (02 ~ (S - - a)gJl(zA)vlda + ~/(02 o 1S c o s I~o - - c o s 1~ = 0 ,

/~2 § (021 (S - - O')gJ~O(/l)/~2do" - - (02Q i s s i n ~o o - - -

L # Re0oe-i~o + S fIJl(ZA)vlda = O,

0

L Imooe-i~o + S 9-y~(ZA)g2d 6 = O,

0

where now A : - V # 2

sintp = 0 ,

76 M. Reeken

If we move along the continuum for #--,0 then we can repeat earlier arguments from Sect. 5 to determine the limit. It comes out as

V 1 "~- (D 2 ~ (S - - a ) n o c l v l d a - 1 = O,

V2 + (D2 ~ (S - - a ) n o c i ~ 2 d o - o92~o 1S sin~po - - •o = O,

L [. nocivlda = O, 0

L imOo e - i~o + ~ nocl6Eda = O.

0

We have used the fact that ~p-oO and that t%" = lira s m ~ exists. This follows from ~-~0 ]2

the fact that the continuum is a differentiable branch a t / t = O. But then we get

dip d~p lim sinw = lim cos~p - u-.0 # ~-~o d# d-~ ~=o

The solutions of this system can be determined. (vl, co) is just (u, COo) and the second component is a solution of the same homogeneous second order differential equation as u but with non homogeneous boundary conditions:

tY' + r = 0,

if(0) = ~02~1 sin~po, if(L) = ~OoZ(ql sin ~vo + Im00 e-i~~

with the additional condition tT(0)= s: o. The necessary and sufficient condition for solvability is

o~2(u(L) (01 sin ~Po + Im Qo e - iwo) _ Q 1 sin ~Po)

= - 0~2(0~ sin~po + 02 sin(~b- Ipo)u(L)) = O.

This is the defining relation for ~Po in a disguised form. So, the solution u exists. Obviously, it is not uniquely defined because ~ + au is also a solution. But a is then uniquely determined by ~(O)+au(O)=~(O)+a=~c o. This proves that we have obtained a vertically rescaled version of the asymptotic solution.

When we now determine the winding number we have to keep in mind that the rescaling changes the angle between the endpoints in the x~-xz-p lane . The winding number changes accordingly. The endpoints of (v], ~ ) for # ~ 0 are

Vtl ( 0 ) = --j[/O)201 COSlPo-'-~0 ,

t~(0) = cn2ql sin~Po ~ 0 2 q l sin~Po,

v'l(L) = v](0) + #~02 ReQoe-iwo = _ #c0202 cos (~b - ~po)~0,

t~(L) = ~(0) + co 2 Ira00 e-i~~ - ~2 ~1 sin~po _., ~02 01 sin~po - o YLS

For # = 1 the curve (v'~, tT~) has the same endpoints as (w], w~). Let ~(x) be the angle between the endpoints of (v], 6~). The analytical formula for q~(x) is

Rotating String 77

simPo --~/t~02~2 COS(~--/PO) -I- i602~1 u(L)

~(x)=arg _~tco20~ cOSWo+iCo201 simp ~

= arg [#2 ~ cos(~b- lpo) cos~po

cos~po simpo~ sin2~po~. J+u J It follows that on the continuum

limn~(x)={Orc if s g n u ( L ) = l , , o if s g n u ( L ) = - l .

From the monotonicity of v~(O) and v'l(L ) it follows also that

0 < q~(x) < rr for all/~.

implies that the winding number 2~ + n on the continuum we are considering This

1 - sgn u(L) is changed to n +

4 It remains to determine the winding number of the limit of(v], f~) for/~-o0. The

winding number of that curve can be determined by counting properly the number of intersections of the curve with the x2-axis. The appropriate formula is

F = - �89 ~ ~(si) sgn u"(s~) sgn if(Sl), s i~Z

where Z = set of zeroes of u' and e(s) = )�89 S 0, L,

[ 1, s e (0, L). To be precise this formula

does not apply to the curve (u', tT') because u" is not continuous and the zeroes ofu ' are not necessarily isolated. But that formula has a variant which is sufficient for our purpose. We know that the zeroes of u' might be intervals and we know also that u' changes sign exactly once on a sufficiently small open interval covering the zero-interval. If we can make sure that tT' has constant sign on an open interval covering the zero-set then the formula can be adapted. But u and t~ satisfy the same second order ODE:

u" + oO2noclU = O,

~" + toZnocrft = O .

Therefore, the Wronskian ..u", _~' is constant and equal to u(O)/~/(O) ~ U ~ ( O ) u ( O )

=cogr sin~0 o. From the defining relation of ~o o it is obvious that sinhoo40. So, wherever u' equals zero the sign of ff is determined by the sign of u:

sgnf f= sgn sin ~Oo sgnu.

But we also know that u + O, wherever u ' = O. Thus, we come to the following conclusion: If I~ . . . . , In are the zero-intervals of u' we define

78 M. Reeken

ai: = sgn sin~po sgnu(Ii) and

S - 1 if u is concave on a neighbourhood of Ii, 6i

+ 1 if u is convex on a neighbourhood of Ii.

l n - 1 Then F = -�88 + a.6.)+ - ~ ai6 i which is a rather obvious generalization of

2 1=2

the formula originally presented. Our last step is to recognize that in fact the sign of u is determined by its being convex or concave, sgnu is positive on the concave parts and negative on the convex parts. Therefore, the numbers ai6 i are all equal to - 1 . The number of concave/convex parts is one more than the number v(u) of (interior) nodes of u and coincides with the number of zero-intervals of u'. So, F

equals + sgn sin~po (v (u ) - 1 ~) + = + sgnsin~o o. We can also determine

sgnu(L) from the number of nodes: sgnu(L) = ( - 1) *(u). This concludes the proof of the following proposition:

The continua emanating from (v3 ei~';, +_icoj, ~p; . . . . ) have positive winding

number equal to + , those emanating from (vje '~'j, +_icoj,~[ . . . . ) have

negative winding number equal to 2~ + - "

Remark. The difference of the winding numbers of the continua emanating from the same "cercle" at infinity is the number of nodes of the asymptotic solution vj.

de) ~It a bifurcation point (uei~~ the sign of ~fi~ is given by

sgne) o sgn sinlpo.

The proof is contained in the following diagram (see Fig. 3) and the formula

do) ~Oo d# = -2no ~. u2da Re [(u(L)q2 ei~ - 00e iwo]

I

which we have derived in Sect. 7. In order to connect the bifurcating solutions to each other we derive the

following results which give barriers excluding certain connections.

All solutions in ~(o)= O, A t-O)have reduced winding number O.

This follows from the fact that the z E 3(0) = 0) lies in the straight line through the endpoints. In terms of the physical configuration: all solutions of the catenary problem lie in a vertical plane through the endpoints.

The canonical projection of ~(Io)l, I#[, [y[ bounded) onto C 1 is precompact.

From the equation v"+o92~(ZA)v=O and V / ( 0 ) = 0 1 ( O 2 ~ w e get suplv"l s

<= o)2C([p[ + Iv[co). This implies [V'[co_-< ~o2(01[p[ + C([p[ + [rico)). Assume IV[co is

Rotating String 79

Fig. 3 P2eir < O)

u(L)P2eir �9 O)

e140

,o(U(L) < O)

IJ unbounded on the continuum and consider g : - Then the set of g is

Ivlco" precompact because of the estimate obtained. Let # ' = IV[co x. Then g satisfies the following set of equations:

g + co2 i (s - tr)gJl(xA)gda + #'#Q ~ o . ) 2 s - ei~#" = O, 0

L

#'#Qo + I ~(z/T)~d~ = O, 0

where ~ : = 1 / Igl2 _~f12. g(## ' ) ~

We can take the limit along a suitable subsequence (x i, Yi) and get

L Vo + e)~ ~ ( s - ~r)noclg&r = O,

0

L

[. noqg&r = O. 0

But this equation has only g - o as solution. This is a contradiction because a sequence of functions on the unit ball in C o cannot converge to zero. Therefore, the assumption IVfc ~ unbounded has to be abandoned and precompactness of the set of v itself is established.

The canonical projection of ~(]to I, 101, lYl bounded) onto C 1 is precompact.

Translating the estimate for v into one for w we get sup Iw"l < ~o2c(1 + Iwlco) and Iw'lco<~O2(ox+lWlco). By the same argument as before we establish that the assumption IWlco unbounded is contradictory. Then precompactness follows from the estimates.

80 M. Reeken

Now we come to the last step establishing the full picture for the case A + 0. We want to exclude the possibility that one of the branches coming from the asymptotic solutions goes off to oo in the m-direction. The solution v satisfies the equation

V" + foz~fJl(zA)v = 0 ,

or(O) = - O 1 0 ) 2 ~ , v'(L) = - - e20)2ei (~

d 2 I f 0) 2 happens to be such that the linear operator ~sZs 2 + 0)29Jl(zA) with zero

boundary conditions on the first derivative has a nontrivial nullspace then this nullspace is one-dimensional. The necessary condition for existence of a solution v of the inhomogeneous boundary value problem is that the nullvectors u and v satisfy the following relation:

~,o(L)v'(L)- t/o,(0)v'(0) = - e z0)Zl,~ei4~fG(L ) + ~ 1/,10)2/~(0) = 0 .

Since uo can be chosen real this cannot hold for A +0. Therefore, along a continuum starting at one of the asymptotic solutions 0)2 is never a critical value

d 2 for the operator ~ + 0)2~I(zA ). We can turn this around and say that 0)2 is

always sandwiched between two neighbouring critical values. If t! 2 t t u~, + 0) ?olO~A)u,o = 0 and uo,(0) = u~,(L) = 0

L then obviously ~ ?iJlO~A)u,oda = 0. The problem can be recast in the following form:

o d2 Let L be the selfadjoint operator - ~ with zero boundary conditions on the first

derivative L has a unique inverse on

K = {u ~ L2 t ! udcr =O } .

1 1/2 1 This inverse L- has a unique positive square root. Then (E u ~ , ~ - ) are the

eigensolutions of the compact operator U ~/2~OI(zA)L- 1/2 on K. Starting at one of the asymptotic solutions this operator is L-1/2nociL-1/2 and 0)2 is one of its positive eigenvalues say the n th from the top. Going along the continuum the operator changes continuously in the uniform norm because the function gJI(xA) has no singularity as there are no zeroes of M and g)10~A) changes continuously in C~ Therefore, the (n - 1)th, nth, and (n + 1)~t eigenvalue changes continuously

1 and ~ - is sandwiched between the first two or the last two, depending on the

bifurcating branch. If we can find a fixed (n+ 1) dimensional subspace of K on which the operators L-1/2~iI(zA)L-1/2 are positive definite with a fixed positive lower bound 2o then the (n + 1) th eigenvalue is bounded below by this bound and

consequently we have > ;to or 0)2 < ~-0" Suppose i is not empty. Consider the

set K1 = L1/e {u e C~lsuppu ~ I}.

Rotating String 81

K~ is contained in K and dimK1 = ~ because the set ofu defined above has infinite dimension, so applying L 1/2 to it certainly does not change that. But on finite dimensional subspaces of K1 the operator [7 ~/zgJI(zA)L-1/2 is bounded below:

(LW2u, L- x/zgJlL- 1/2LWZu) (u, 9J/, u) lul z > Vono, (ZWZu, ZWZu) -]ZV2u}2 >n~ =

where Vo depends in some complicated fashion upon the set I. This concludes the proof of the following important proposition:

I f I has non-empty interior the continua emanatin9 from the asymptotic solutions are bounded in co.

of nodes of v, J ~d winding

schemat ic s o l u t i o n d iagram f o r t~e

case A # 0 Fig. 4

82 M. Reeken

9. Solving the Full-Set of Equations

Until now we have treated the equation ~ = 0 treating c as an independent parameter. The full-set of equations is ~ = 0 and ~ = 0 which fixes a value ofc. For p = 0 the equation ~(v, 0, o9; c, 02 . . . . ) = 0 is independent of v, 09, ~p, Qo, 01 and has a unique solution

S noada- Uo Co = Co(Uo) =

2(I)no

provided 2(I) 4= 0. It follows that for all ~p, 00, 01, uo,j (vie i~, O, +mj; Co, ~p, 00, QI, Uo) is a solution

of ~ = O. We introduce new variables

v=vjei~+Av, /~=O+A#, m= +_mj+Am, c=co(uo)+Ac.

Now we consider the family of maps:

6 , [(cJ x ~,~) x ~1] x ~,?--,(cJ x ~ ) ,

.~, : [(Co 1 X ~.~.2) X ~-~1- I X ~X.5--+~. 1

given by

6t(Av , A~A, Ao, Ac; Ip, 00, 01, b/0) : ~--- 6 1 1 . = o A r + 6~l.=oA~+ 6~l.=oAm+ t 6 r,

~t(Av, A#, Am, Ac; ~p, 0o, 01, Uo) : = 2(i)noAc + t~, .

6 , denotes the remainder term in the first order Taylor development of ~ at (v f~,O, +_mj;Co, W, Oo, Qt,Uo) with respect to the variables (v, /z, m). .~, is the remainder of the zero order Taylor development of .~. (.~ is not differentiable but continuous.) Comparing ~(v, 0,_+mj;c, lp . . . . ) with the first order Taylor polynomial at (vie ~, O, + mj; Co, ~p . . . . ) shows that

6,(v, 0, +mj; c,~;, . . . ) - 0 .

A simular comparison shows that

.~,(v, 0, m; c,~o, . . . ) - 0 .

From our earlier computations it follows that ( ~ ) = 0 has solutions

(vie i'~, O, +_ mj; Co, ~p, 00, 01, Uo) which are isolated in the (v, p, m, c)-space if ~p + ~pf. We want to show that the index of these solutions is not zero and changes when we cross ~pf. We have proved this already in Sect. 7 for ~ for arbitrary parameter

value c. We want to extend this result to the map ( ~ ) which corresponds in an

obvious way to ~ . If we can prove that ~t has no zeroes on a suitable small

neighbourhood of (Av, A~, Am, Ac)=O then we can deform to -~o without

Rotating String 83

changing the index. But ~o is a cartesian product of two maps. I~i o was treated

before while ~0 is linear. The well-known product formula for the index proves then the desired result.

We prove the admissibility of the deformation. For t = 0 we get

~o(Av, A~t, A~,, Ac,...) = 03'1 I,= oa r + ~ 1 , = oA/z + ~ ; I , = oAr

From the results in Sect. 7 we know that for ~p 4: ~p+ the only solution of ~i o = 0 is (Av, A#,A~o)=O. From ~ o = 0 we get Ac=O. For t4:0 we get from

6i '1 . = o( Av, A ~, A~o) = - t ~ r

by solving the linear system determining A# and Ar (see Sect. 7) that A/z and A~o are of order o(e) if (Av, Ala, Aco, Ac) is confined to an e-ball around 0. If e is small enough this implies that (A#, Aco)=0. But then ~i,_=0 and we are left with

A !

C~t(Av , O, O, Ac; lp . . . . ) = 6)11, = oar = 0.

For ~p4:~pf^this has the solution Av=O. But for A p = 0 we have ~t=2(1)noAc. Therefore, ~ t = 0 implies Ac=O.

The computation of the index of ~o is trivial, that of 6o has been done in

( ) changes from (~ Sect. 7. The index of 0 for the map ~ - 1 to + i or vice versa when ~ passes through ~ f .

We have now established that we are in the situation described in [1]. The

discussion of Sect. 8 can now be carried over to the map and, therefore,

with no essential change and leads to the following theorem:

Assume that

lira 9l(t) = no, lim tg~'(t)- 9~(t) = n~, E--* o0 t t ~ O 0

I4:r and A=olo2sin~b4:0.

((fi)=Oconsistingofsolutions Then there are continua ~ , (n ~ JE ) of solutions of ~9

with reduced winding number n which have topological dimension five at every point and which intersect the trivial solutions at infinity in the way described by Fig. 4. (The drawing is actually a section for fixed 0o, ~1, and Uo.)

10. Conclusions

The previous result is valid for any fixed I with 14: ~b but we don't say any thing about how the continua ~ . behave as I varies. The conjecture is that they form continua which are topologically at least as complicated as ~ S x s p a c e of admissible sets I. For the nonrotating case (the catenary problem) this has been proved in [3] but there no bifurcation occurs. For the bifurcation problem technical obstacles remain for the time being.

84 M. Reeken

The problem of rotating strings has been treated before. In [5] the author has used the idea of continuing solutions starting at infinity for the case of an inextensible string with endpoints on the axis of rotation. Only one continuum was constructed. The apparently more natural idea of bifurcating from the trivial solutions in the axis leads to serious problems because those trivial solutions do in general have a kink leading to a weird singularity. This problem was solved in [2]. In our approach we should be able to reproduce the bifurcation diagram by studying what happens when ~1Q2~0. (This relates to a subset of the excluded set A = 0.) We plan to give the details in a separate paper.

We think there are several interesting points. One is the strange way this nonlinear "nonelliptic" second order equation can be reduced to a family of nonlinear "elliptic" equations parametrized by the measurable subsets of an interval. The other is the topological invariant characterizing the continua. For the rotating string with endpoints on the axis of rotation the invariant is the number of nodes of the unknown function. This is simply a consequence of the fact that we deal with a second-order ODE, a situation recurring in so many physical examples. If the string is not fixed on the axis of rotation but unsymmetricalty (not in a plane through the axis of rotation) we have to deal with a system of ODE's having no natural nodal properties which are preserved along continua. But we have shown in Sect. 8 that those nodal properties in the symmetric case are actually a degenerate version of the property of having a certain winding number. The nodes develop when a string wound n times around the axis becomes flattened out until it lies in a plane through the axis of rotation. Thus, the nodal properties are just the shadow of something quite different but physically manifest. In other words enlarging the problem by studying asymmetrically suspended strings turned out not to be a complication but rather to shed new light on the structure of the problem which is somewhat hidden in the symmetric case.

References

I. Alexander, J.C., Antman, S.S.: Global and local behavior of bifurcating multidimensional continua of solutions for multiparameter nonlinear eigenvalue problems. Arch. Rational Mech. Anal. 76, 339-354 (1981)

2. Alexander, J.C., Antman, S.S., Deng, S.T.: Nonlinear eigenvalue problems for the whirling of heavy elastic strings. II. New methods in global bifurcation theory. Proc. Roy. Soc. Edinburgh

3. Alexander, J.C., Reeken, M.: On the topological structure of the set of generalized solutions of the catenary problem. Proc. Roy. Soc. Edinburgh

4. Reeken, M.: Exotic equilibrium states of the elastic string. Proc. Roy. Soc. Edinburgh 5. Reeken, M.: Rotating chain fixed at two points vertically above each other. Rocky Mountain, J.

Math. 10, (1980)

Received November 30, 1983