Hanging cables with small bending stiffness

Nonlinear Analysis, Theory, Methods & Applicolions, Printed in Great Britain.

HANGING

Vol. 20, No. 10. pp. 1193-1204.

1993. 0

CABLES WITH SMALL BENDING

0362-546X/93 $6.00+ .OO 1993 Pergamon Press Ltd

STIFFNESS

PETER WOLFE Department of Mathematics, University of Maryland, College Park, MD 20742, U.S.A. (Received 10 June 1991; received for publication 26 November Key words and phrases:

1992)

Nonlinear elasticity, nonlinear two point boundary value problems, singular

perturbations of 0.D.E.s. 1. INTRODUCTION IN THIS paper we study the problem of the equilibrium configuration of a cable suspended between two fixed supports. The only load on the cable is its own weight. We model the cable as a Cosserut rod, and assume the cable may undergo extension and flexure but not shear. We wish to study cables having small bending stiffness, which we do by introducing a small parameter E into the constituative equations (stress-strain laws) for the cable. When E = 0 the equilibrium equations reduce to those which arise when the cable is modeled as a string with no resistance to bending. These equations have been analyzed by Antman [l]. He shows that there is always exactly one solution of this problem in which the cable is in tension. The problem for small positive E is a singular perturbation of the string problem. For our problem, based on the rod model, in addition to stipulating the position of the ends of the cable, we must impose additional boundary conditions. We impose the condition that the cable is clamped at both ends. The solution to the string model will not satisfy these additional conditions. Thus we would expect that as E tends to zero the solution of the rod model will tend to the solution of the string model in the interior of the cable. However, in order to satisfy the additional boundary conditions boundary layer corrections must be included to obtain an approximate solution to the rod problem for small E. In a previous paper [2] we considered a similar problem, that of a conducting wire in a magnetic field. This problem was chosen as a starting point for a theory of rods with small bending stiffness because the corresponding reduced (string) problem can be solved explicitly and the solution has a simple form. The main point of that paper was to show how to compute the boundary layer corrections starting from an ad hoc ansatz. There was no claim that the solution constructed there was actually asymptotic to an exact solution as E tends to zero. After completing this paper we became aware of the work of Schmeiser [3] and Schmeiser and Weiss [4]. These papers (and the references cited therein) deal with a general theory of singular perturbations for systems of nonlinear ordinary differential equations. In these papers it is proven that, in fact, the approximate solutions constructed are asymptotic to an exact solution. It became clear that our work could easily be put into the framework of this theory. We also realized that our technique was far more general than originally thought and could be applied to problems more complicated than our model problem addressed in [2]. In Section 2 we will describe the rod model and our concept of small bending stiffness. In Section 3 we will describe the string model and its solution. In Section 4 we will state the general theorem on singular perturbations for nonlinear systems of ordinary differential equations and in Section 5 we will apply this theorem to our problem. Section 6 consists of some concluding remarks. 1193

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P. WOLFE

2. THE

ROD

MODEL

We consider a cable suspended between two fixed supports. The only load on the cable is its own weight. We model the cable as a rod. By a rod we understand a slender three dimensional body. Let {i, j, k) be a fixed orthonormal basis in Euclidean three-space, E3. The unit vector j is interpreted as pointing down. We define a position vector function r E E3 of the real variable s E [0, 11. Here s is interpreted as a scaled arc length parameter of the line of centroids of the rod in a reference configuration so s identifies material cross-sections of the rod. Thus, in a deformed configuration r(s) is the position of the material point at the centroid of the section s. The cable (rod) is assumed to be welded to fixed supports at 0 and ai + bj, a > 0. The force on the cable is its own weight

f(s) = -W)(sM

(2.1)

where (PA)(S) is the mass density per unit unstretched length at s and g is the acceleration of gravity. We assume here that f is a continuously differentiable function of s. Let n(s) denote the contact force and m(s) the contact couple or bending moment exerted by the material of (s, l] on the material of [0, s]. We denote by a prime, differentiation with respect to s. Then the equations of equilibrium for the cable are n’ + f = 0, m’ + r’

X

n = 0.

(2.2) (2.3)

We look for solutions of the problem for which r(s) lies in span(i, j), i.e. we consider planar deformations of the cable so that r = xi + yj. Thus the equilibrium equations become (2.2) and M’ + (r’ x n) * k = 0,

(2.4)

where M=m.k.

(2.5)

We assume that the cable is unshearable so that the material section perpendicular to r’(s) in the deformed configuration. We now set x’ = vcose,

y’ = v sin 6,

of the cable at s remains

(2.6a, b)

e, = cos Bi + sin t9j,

(2.6~)

e2 = -sin

Bi + cos dj,

(2.6d)

r’ = ve,.

(2.6e)

so that

We decompose

n as

n=Ne,+He,,

(2.7)

and set 8’ = U. Then u and v are the strains of the problem.

cw

Hanging

1195

cables

We set n, = n . i = Ncos 8 - H sin 8,

(2.9a)

nY = n-j

(2.9b)

F(s) = so

=

NsinB + Hcos8,

(2.9~)

‘(pA)(o)gdo, s0

that (2.9d)

f(s) = -F’(s)j. We assume F’(s) > 0, 0 5 s I 1. The equilibrium equations then become n’x = 0 7

(2.10a)

n; - F’ = 0,

(2. lob) (2.1Oc)

M’ + VH =O.

The elastic properties of the cable are embodied in the constituative equations relating the stresses M and N to the strains u and u. Thus, we assume that there are twice continuously differentiable functions R x (O,m) x [O, l] 3 (U, u, s) --f (A?@, u, s), *(U, u, s)) E R2, such that M(s) = ti(u(s),

N(s) = fi(u(s), u(s), s).

u(s), s),

(2.11)

We further assume that the cable is hyperelastic, meaning that there is a stored energy function Q(u, u, s) such that for (u, u, s) E R x (0, m) x [0, l] %(U, u, s) = (D”(U,u, s).

A(& u, s) = @u(u, u, s),

(2.12)

Here subscripts denote partial derivatives. We assume that Q, is three times continuously differentiable, convex and coercive in the sense that @(U, u, s)/(u2 + u2)U --f 00

as u2 + u2 + 00

iif, fiu fiu1 l

cD(U,u, s) + 43

if”

[

Equations (2.13a, b) imply

for some (Y> l/2,

(2.13a)

as u + 0,

(2.13b)

is positive definite.

(2.13c)

fi(U, u,s) -+

--co 00

A&u, u,s)

-+ +w

fi(O,l,s)

= 0,

as u + 0, asu+oo,

(2.14a)

as u -+ 200.

(2.14b)

A?(O, u, s) = 0.

(2.15)

Finally we assume See [5] for a discussion of the physical significance of these conditions. In this model the stress The boundary conditions reflecting the condition that the

H is an unknown of the problem.

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1196

cable is clamped at both ends are x(0) = 0,

x(1) = a > 0,

Y(O) = 0, e(0) = 0,

Y(1) = 6,

0(l) = 0.

(2.16a) (2.16b)

Thus, the boundary value problem for the rod model of the cable consists of (2.6a, b), (2.8), (2.10) along with the constituative relations (2.11) and the boundary conditions (2.16). We can prove existence of a solution to this problem by an argument following the lines of that in [6]. This solution is found by minimizing the total energy functional over the set of all possible configurations of the cable which satisfy the boundary conditions (2.16). We now embed this problem in a continuum of problems by setting (2.17)

@&’ u, s) = @(EU,u, s),

and considering the problem with stored energy function Qe for E > 0. Then instead of (2.11) we have M(s) = &fi(&U, u, s).

(2.18)

Thus, if e is small we see that the bending moment is small, even if the corresponding strain is not. This is what we mean by the phrase “small bending stiffness”. Formally, in the limit as E -+ 0 we have M = 0 and from (2.10~) H = 0. Thus, n is parallel to r’. In fact n = N&u, s)e, ,

(2.19)

N,(?J, s) = R(O, u, s).

(2.20)

where

From (2.14a) and (2.15) we have NfJ(l,s) = 0,

N&o,@ -+ --o;) as u -+ 0,

N,(u,s)

‘00

as ~+a.

(2.21)

The convexity of @ implies

;N,(u,s) >

0.

(2.22)

Equations (2.2), (2.19), (2.21), (2.22) describe a string model. In the next section we will study this model and obtain a solution. In Section 5 we will show how to use this solution to construct an approximate solution to the rod model with CD= QE which, as E -+ 0, is asymptotic to an exact solution of the problem. 3. THE

STRING

MODEL

In this section we describe Antman’s solution for the problem in which the cable is modeled as a string [l]. Thus, the problem consists of the equilibrium equations (2.10a, b) with the force given by (2.9c, d), the stress-strain law (2.19) and the boundary conditions (2.16a). In this model H = 0, so that integrating (2.10a, b) and using (2.9a, b) with N = N,(v, s) we find Ncos 0 = c!,

Nsin 8 = fl + F,

(3.1)

Hanging cables

where CYand p are constants and F is defined in (2.9c). Thus, tanO=PtF,

(Y

N = +x$x~ + (fi + F)‘) = +a.

(3.2b)

It is shown that a > 0 implies that N cannot change sign on [O, 11. We will consider the case in which the plus sign is taken in (3.2b), i.e. the case in which the cable is always in tension. Now from (2.6) and (2.16a, b) 1

u(.s)(cos 8(.s)i+ sin O(.s)j]ds.

ai + bj = r(1) - r(0) =

(3.3)

10 Equations (2.21) and (2.22) imply we can solve the equation N = N,(v, s) for u. Hence we may write u = t(N, s).

(3.4)

Thus, (3.3) may be rewritten

s“‘$ii 1*

ai + bj =

‘) (ai

o

+ (/I + F(s))j) ds

P; F)i + Qb,P; F)j,

= P(a,

(3.5)

where 6 is given by (3.2b). We define N WY

~1

=

%rl,

4

drl,

s0

(3.6)

and ‘1 y(&s), s) d.s - c2a- /lb.

I(a, /3) = I0

(3.7)

Then (3.5) is equivalent to the vanishing of the gradient of I. It follows from (2.21), (3.4), and (3.6) that T(a,j3) -+ 00 as a2 + /I2 -+ co. Therefore, I, which is continuous, has an absolute minimum at a point (cx*,p*) where the gradient of I must vanish. Since a > 0,CY*> 0.Moreover, it follows from the positivity of u and CNthat the matrix

c

pa ps

Q,

I

QB

(3.8)

is positive definite (this fact will be used in Section 5) so that I is strictly convex. Hence (3.5) has at most one solution. Thus, we have the following theorem. THEOREM 3.1 [l]. There is exactly one tensile (N > 0) solution of the string model of the hanging cable for a > 0.

Actually, more can be said about the shape of the cable. THEOREM 3.2 [l]. The configuration r = xi + yj of the cable admits the parameterization y = y(x), 0 5 x I a with the function y a strictly convex function of x.

P. WOLFE

1198

In Section 5 we will show how this solution forms the basis for a singular perturbation scheme which is used to compute an approximate solution of the rod model with small bending stiffness. 4. SINGULAR

PERTURBATIONS

In this section we state the basic result for regular singularly perturbed boundary value problems for nonlinear systems of ordinary differential equations. The material in this section is taken from [3]. We will, for the most part, use the notation of that paper. The functions b and f and the variable y in this section have nothing to do with the force f, the number b and the coordinate y introduced in Section 2. We consider a problem of the form

&Y’ = f(y, z, s, E),

z’ = dY, z, s, E),

(4.1)

@y(O), z(o), y(l), Z(I)) = o, where y E R”, 7; E lRmand f, g and b are smooth mappings. We suppose that the reduced equation 0 =f(y*,Z*,&O)

(4.2)

has an isolated solution y* = +(z*, s). We assume the n x n matrix fy(4(z*, s), z*, s, 0) has n_ eigenvalues with negative real parts and n, eigenvalues with positive real parts with n- + n, = n. We look for a solution of (4.1) of the form Y(& E) = m*(s), 4 + Ly(a> + RY(4 + O(E),

(4.3)

z(s, E) = z*(s) + O(E), where o = S/E, r = (1 - s)/E, Ly(o0) = Ry(o0) = 0. We substitute (4.3) into (4.1) and let E tend to zero. The result is (4.4a) Z *’ = s(+(Z*> s), s, O), dLy/da

= f($(z*(O), 0) +

dRy/dt = -f(~(z*(l), b(4(z*(O), 0) +

LY(O),

z*(O),

LY, Z*(O),

(4.4b)

0, (9,

(4.4c)

I) + Ry, z*(l), 1, O), G(z*(l),

1) +

Ry(O), z*(l)) = 0,

(4.4d) (4.4e)

Ly(m) = Ry(c0) = 0. THEOREM 4.1 [3]. (z*,

Let f, g

Ly, Ry) be a solution

and

b be continuously

of (4.4)

differentiable

such that the linearization

with respect of (4.4)

at (z*,

to all variables.

Let

Ly, Ry) has only the

trivial solution. Then there are constants .sO,6 > 0 such that for 0 < E 5 .sOa solution of (4.1) (y(s, E), z(.s, a)) exists which satisfies (4.3) and is unique in a ball with radius 6 and center (+(-z*(s), s) + Ly(a) + Ry(r), z*(s)) in the space C’[O, 11.

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Hanging cables

5. THE

ASYMPTOTIC

SOLUTION

In this section we apply the theory of Section 4 to the rod model of the cable. This model consists of equations (2.6a, b), (2.8), (2. lo), constituative relations (2.11) with stored energy function aE given by (2.17) and the boundary conditions (2.16). Now (2.11) becomes M(s) = &A(&& u, s),

N(s) = R(&U, u, s).

(5.1)

We let w = EM.

(5.2)

As in Section 3 we can integrate (2.10a) and (2.10b) to obtain Ncose

- Hsine

Nsin8

+ Hcostl = p + F,

(5.3b)

N = a cos e + (p + F) sin 8,

(5.4a)

= (Y,

(5.3a)

where CYand j? are constants. Hence,

H = -CY sin 0 + (p + F)

cos 8.

(5.4b)

Equation (2.10~) becomes EM; + vH = 0,

(5.5)

M, = iG(w, v, s).

(5.6)

N = fi(w, U,s).

(5.7)

where By (5.1) and (5.2)

The convexity and coercivity of the energy functional
(5.8a)

u = l’(M, , N, s).

(5.8b)

Thus, we may write our system as

Eel = W, EM; = ~(a sin e -- (p + F) cos e),

(5.9a) (5.9b)

cos 8,

(5.9c)

y’ = u sin e,

(5.9d)

0!l = 0,

(5.9e)

p’ =o.

(5.9f)

X’

=

v

In (5.9) we use (5.8) and (5.4a) to express w and v as functions of M, and 0, the so-called “fast variables’ ’ . We will apply the results of Section 4 to the system consisting of (5.9) along with

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P. WOLFE

the boundary conditions (2.16). The reduced equations (cf. (4.2)) are W(M,, N s) = 0,

(5.10a)

V(M, , N, ~)(a sin 8 - (/I + F) cos 19) = 0,

(5.10b)

with N given by (5.4a). By (2.15) and (5.6) a solution of (5.10a) is given by M, = 0,

(5.11a)

a! sin 0 - (b + F) cos 8 = 0,

(5.11b)

while (5.10b) is satisfied if

that is H = 0. We may rewrite (5.11 b) as tanB=P+

(5.11c)

(Y

(cf. (3.2b)). We will check the matrix condition mentioned after (4.2) after using (5.11) to obtain a solution of the equations corresponding to (4.4a). These equations are (5.9c)-(5.9f) with 0 given by (5.1 lc) and v = V(0, N, s). The relevant boundary conditions are (2.16a). But this is precisely the problem considered in Section 3 with ir(N, s) = I/(0, N, s), (cf. (3.4)). By theorem 3.1 there exists a unique tensile (N > 0) solution of this problem. We let (x(s),y(s), e(s), (Y,/3) be the tensile solution to this problem. (In the notation of Section 3, (CY, /3) = (c-w*, /3*) but we prefer to omit the superscripts for now.) We now consider the Jacobian of the left-hand side of (5.10) at the solution (5.11). By (2.12), (2.15) and the convexity of @ I&(0, N, s) = W,(O, N, s) = 0,

I&(0, N, s) > 0.

Thus, the relevant Jacobian is IV&O, N, s) 0 i

0 V(0, N, s)((Ycos

e +

(p +

F)

sin 0)I .

But by (5.4a) cxcos 0 + (p + F) sin B = N > 0. So this matrix has two positive eigenvalues and the matrix condition is satisfied. We now turn to the boundary layer at s = 0. We denote the boundary layer corrections to the variables M1 and 8 by m and I$ (instead of the awkward LM, and Lo). We use a dot to denote differentiation with respect to o. The equations corresponding to (4.4b) are (5.12a)

i = W(m, n, O), ti = v(m, n, O)(CXsin(0, + 4) - p

c0s(e,

+

$)I,

(5.12b)

where &, = e(O) and n = 01cos(0, + 4) + p sin(0, + 4). But since CY cos B,, + p sin 0, = N(0) = N, > 0 and CY sin B0 - /3 cos 0, = 0, equation (5.12) can be rewritten as 4 = W(m, No cos 4, 0),

(5.13a)

ti = k’(m, N, cos 4,O)N, sin 4.

(5.13b)

Hanging

cables

1201

The boundary conditions are 00 + 9(O) = 0,

(5.14a)

qqco) = m(a) = 0.

(5.14b)

In regard to the boundary condition (5.14a), we observe that it is precisely the failure of the string model to satisfy the boundary condition (2.16b) which makes it necessary to append the boundary layer correction terms to this solution in order to obtain an approximate solution to the rod model for E > 0. The origin is a stationary point for system (5.13). In order to understand the nature of this stationary point we linearize (5.13) about (0,O). We note that W,(O, N,, 0) = wg > 0, V(0, N,, 0) = ug > 0. Thus, the linearized system is 4 = worn,

(5.15a)

ti = u,N,9.

(5.15b)

Clearly system (5.15) has one positive and one negative eigenvalue. Thus, there is exactly one pair of orbits of (5.13) entering the origin as o + co. Along these orbits m and $ decay exponentially. We wish to show that there is some m, such that the point (m,, , - 19,)lies on one of these orbits. It will then follow that the solution of (5.13) passing through this point at o = 0 will decay exponentially as cr -+ 00. This solution will then satisfy the boundary conditions (5.14). In order to analyze (5.13) we recall that the functions W and I/ arise from inverting (5.6) and (5.7) where Ii?= a’,,

iQ= @“.

Thus, if we set Y*(m, n) = m W(m, n, 0) + nV(m, n, 0) - @( W(m, n, 0), V(m, n, 0), 0),

(the Legendre-Fenchel

transformation)

(5.16)

we have

YXm, n) = w(m, n, O),

Y,*(m, n) = V(m, n, 0).

(5.17)

The function Y* is convex and coercive in m, i.e. Y*(m, n)/m + 00 as Irnl -+ co uniformly for n in a bounded interval. Hence, we see that if we set Y(m, 4) = Y*(m, % cos +),

(5.18)

then (5.13) can be written m = -Y’,, (5.19) i=YVV i.e. the system is Hamiltonian. We must show that, given a value 0,) we can find m, such that the orbit starting at (m,, - f?,,) as o = 0 approaches (0,O) as o + co. Since Y is conserved along orbits, this orbit is identified by the condition that along it Y(m, 9) = Y(0, 0) = Y*(O, No).

(5.20)

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Actually we have -n/2 < 8,, < 0 (cf. theorem 3.2). We will show that the orbit entering the origin from 4 > 0, if traced backward, approaches the stationary point at (0,271) as o + --oo. This will prove the result. We show this by ruling out all other possibilities. We first observe that, since Y* is coercive in m, m must stay bounded along the orbit. Furthermore, the orbit cannot cross any line C#J= constant more than twice and cannot cross the lines 4 = 0 or 4 = 2n at all. This is because ‘Pm = W which is positive for m > 0 and negative for m < 0. Thus, the orbit, on which Y is constant, cannot cross the line 4 = constant twice on either side of the 4 axis and cannot cross the lines $ = 0 or 4 = 271 at all. Thus, by the Poincare-Bendixson theorem the orbit must approach a stationary point as o + --00. But the only stationary points in the strip 0 I 4 5 27~ are at (0,0), (0, rc), and (0,2n). The orbit cannot approach (0, n) because Y*(O, No) # Y*(O, -NJ. It cannot approach (0,O) because to do so it would have to cross the line 4 = 0, which, as we have seen, it cannot do. The only remaining possibility is that the orbit approaches (0,271) as o + --co. This completes the construction of the boundary correction terms at s = 0. The boundary correction terms at s = 1 can be handled in a similar manner. (If the rod is homogeneous and it happens that b = 0, the solution will be symmetric about s = l/2 and the boundary correction terms at s = 1 can be obtained using this symmetry.) In order to apply theorem 4.1 we must show that the linearization of our equations about the solution we have constructed has only the trivial solution. We start with the system (5.9c)-(5.9f) which corresponds to (4.4a). These are the equations of Section 3 with u = I/(0, N, s) = ir(N, s) x,

w49 4

=

6(s) y' =

W(4,s) 6(s)

ay

(5.21a)

(P + m))>

(5.21b)

CY’ = 0,

(5.21~)

p’ = 0.

(5.21d)

In (5.21) 6 = N is given by (3.2a). The linearized

equations

(for the variables

xi, yl, aI, PI) are (5.22a)

(5.22b)

The boundary

a; = 0,

(5.22~)

p; = 0.

(5.22d)

conditions

are x,(O) = y,(O) = x,(l)

If we integrate

= y,(l)

= 0.

(5.23)

(5.22a) and (5.22b) from 0 to 1 and use (5.23) we obtain = 0,

(5.24a)

Qola~ + Qs& = 0,

(5.24b)

P,a,

+ P&

Hanging cables

1203

where P and Q are defined by (3.5). But we know that the matrix (3.8) is follows that CY~= /3i = 0 and hence xi = yi = 0, i.e. (5.22), (5.23) has only We now turn to the linearization of (5.13) about the solution we have we denote by (m(a), ~$(a)). We note that by construction &a) # 0 for equations are i,

= Y,#,&

+ Ylmmmir

(5.25a)

- ‘y,+m,.

(5.25b)

m1 = -Y’,6& In (5.25) the derivatives

of Y are evaluated

positive definite. It the trivial solution. constructed, which 0 I (T < 00. These

at (m, 4). The boundary

4,(O) = &(=J) = ml(m)

conditions

are (5.26)

= 0.

We observe that (ti, 4) is a solution of (5.25). By the method that the general solution of (5.25) is

of reduction

of order [7] we find

(T

‘Pm, I#-”da’

(5.27a)

,

0

41 =

0

c,i+czii

But in order to satisfy the boundary &(uz). By L’Hopital’s rule

Y,,&’

do’.

(5.27b)

0

condition

4,(O) = 0, we must have c, = 0. Let us consider

CT

i-2ym,

lim 4 o+m

da,

=

lim

J~~-“J’mm

0-m

0

da’

=

lim -ylmm=+a IT’D=

i-’

*

f#l

(Recall that Ym, is evaluated at (m, C#J) and, thus, has a limit as o -+ 00. This limit is positive by the convexity of Y*.) Thus, we must have c2 = 0 and we conclude that (5.25), (5.26) has only the trivial solution. We may therefore apply theorem 4.5 to conclude. THEOREM 5.1. If @ satisfies the conditions stated in Section 2 then there exists a solution of the problem consisting of (5.9) with the boundary conditions (2.16) having the following form x(s) =

a*

s

suo, W),

Y(S) =

sV(O,&O, t) (8*

i0

s(t)

where (a*, p*) is the value of ((Y,p) found s(t) =

t)

s(t)

0

J(cY*)2+

0(s) = arctan

dt

+ O(E),

(5.28a)

+ F(t)) dt + O(E),

in Section

3 and

(p* + F(t))Z.

P* + m) (y*

+ 4(o) + 9(r)

Ml(s) = m(4 + ~(4 + O(E),

(5.28~) + O(e),

(5.29a) (5.29b)

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P. WOLFE

where CJ = S/E, T = (1 - s)/E, m and 4 are obtained the corresponding boundary layer terms at s = 1.

by solving

(5.13), (5.14) and m and _ C#J are (5.30)

M(s) = &MI(S), CY= CY*+ O(E),

(5.31a)

P = P* + O(E),

(5.31b)

N(s) = 01cos 19(s) + (/? + F(s)) sin B(s).

(5.32)

In (5.32) (Yand B are given by (5.31) and I!?by (5.29a). U =;

W(M,,N,s),

(5.33a)

U = V(M, ) 3, s)

(5.33b)

where M, is given by (5.29b) and N by (5.32). If we assume further smoothness of the strain energy function @ we can compute higher order correction terms, both in the interior and in the boundary layers. In fact the equations satisfied by these correction terms are linear. They are inhomogeneous versions of (5.22) and (5.25). 6. CONCLUSIONS

In this paper we studied singular perturbations of tensile (N > 0) solutions of the string model (no resistance to bending). We used the fact that we were dealing with the (unique) tensile solution very strongly, both in the analysis of (5.13) and also of (5.24). In [2] the author shows that under certain conditions there can exist compressive (N < 0) solutions of the string model. For these solutions, our construction, i.e. the solution of (5.13), must fail. In this case the origin is a center for (5.13) and no orbits can enter or leave it. It would be interesting to determine whether in fact these compressive solutions “disappear ” when we introduce a small amount of bending stiffness. It is clear that the techniques of this paper may be applied to problems in which the cable is subjected to other types of loading. Acknowledgement-The and some suggestions

author would like to thank Christian which led directly to this paper.

Schmeiser

for some extremely

helpful

comments

on [2]

REFERENCES 1. ANTMANS. S., Multiple equilibrium states of nonlinearly elastic strings, SIAM J. uppl. Math. 37, 588-604 (1979). 2. WOLFE P., Asymptotic analysis of a rod with small bending stiffness, Q. appl. Math. 49, 53-65 (1991). 3. SCHMEISER C., Finite deformations of thin beams. Asymptotic analysis by singular perturbation methods, IMA J. appl. Math. 34, 155-164 (1985). 4. SCHMEISER C. & WEISS R., Asymptotic analysis of singular singularly perturbed boundary value problems, SIAM J. math. Analysis 17, 560-579 (1986). 5. ANTMAN S. S. & DUNN J. E., Qualitative behavior of buckled nonlinear elastic arches, J. E1usticit.y 10, 225-239 (1980). 6. SEIDMAN T. & WOLFE P., Equilibrium states of an elastic conducting rod in a magnetic field, Archs ration. Mech. Analysis 102, 307-329 (1988). 7. HURWITZ W., Lectures on Ordinary Differential Equations. Wiley, New York (1958).