The Hanging Chain. John McCuan. January 19, 2006

Transcription

1 The Hanging Chain John MCuan January 19, Introdution We onsider a hain of length L attahed to two points (a, u a and (b, u b in the plane. It is assumed that the hain hangs in the plane under a onstant gravitational aeleration g and has shape determined by the graph of a funtion u : [a, b] R. Evidently, we must also assume L 2 (b a 2 + (u b u a 2. After translating and saling, we may assume that a =, b = 1 and u a = 0. An expression for the potential energy of the system is given by E 0 = ρg 1 + u (x 2 (u(x u 0 dx where u 0 is some referene height whih, as we shall see, plays no signifiant role. We assume initially, however, that the shape of the hain determined by u = u minimizes the energy E 0 among all funtions u C 1 [, 1] suh that u( = 0, u(1 = u 1 and 1 + u (x 2 dx = L. Simplifiations In onsidering suh a minimization problem, one notes that the positive physial onstants ρ and g appear in the expression for the energy simply 1

2 as multipliative onstants and an be negleted (i.e., assumed to be 1. Furthermore, we an write the resulting energy as E 0 = = 1 + u (x 2 u(xdx 1 + u (x 2 u 0 dx 1 + u (x 2 u(xdx Lu 0. Sine the last term Lu 0 onstitutes simply an additive onstant in the energy, it plays no role in the minimization either. Thus, we onsider the following simplified problem: Minimize E = over the lass of admissible funtions A 0 = {u C 1 [, 1] : u( = 0, u(1 = u 1, Compliations 1 + u (x 2 u(x dx (1 1 + u (x 2 dx = L}. It turns out to be advantageous to onsider lasses of funtions that are larger than A 0. In priniple, this makes the problem harder. For the time being, the most important modifiation to our disussion above involves the lass of funtions A with presribed endpoint values but arbitrary length; in this lass, we introdue the ondition 1 + u (x 2 dx = L as a separate onstraint. This leads to onsideration of the following modified problem: Minimize over E λ = 1 + u (x 2 (u(x λdx A = {u C 1 [, 1] : u(0 = 0, u(1 = u 1 }. 2

3 This modifiation of the problem may appear suspet. First of all, we already threw away the onstant referene value u 0 whih appeared in preisely the same position where λ appears now. Furthermore, we seem to have simply thrown away the onstraint. Notie, however, that we are onsidering here a fundamentally different problem. The relation with the original problem is not so lear. What is lear is that if we an solve this problem for every λ (finding some minimizing funtion u = u λ and it happens to turn out that the length orresponding to one of these solutions is L, then we have solved the original problem. In that ase, u λ A 0 and E λ (u λ = E(u λ λl E λ (u for every u A A 0 ; speializing this inequality to u A 0, we find E(u λ E(u. One might also ask at this point how we know suh a value λ will exist. We will give a justifiation of existene for a reasonably large lass of funtionals later. For now we simply point out the analogy with the method of Lagrange multipliers from elementary alulus (whih we will in fat use in our justifiation. In that situation, one wishes to minimize f : Ω R subjet to a onstraint g = onstant where Ω is a domain in R n. Assuming the onstraint is nondegenerate, i.e., Dg(x min 0, there is some λ for whih Df(x min = λdg(x min. That is to say, the point x min is a ritial point for the funtion f λg. Taking one step bak, we ould summarize the method by saying: In order to minimize f subjet to g = onstant, try to minimize f λg instead with no onstraint; then find λ so that the onstraint is satisfied. There is another modifiation of our minimization problem that will eventually have to be faed. It turns out that for many problems of this sort, the orret set of funtions to onsider is not C 1, but the set of ontinuous funtions (C 0 with pieewise ontinuous derivatives. This will beome an important point later, but we are going to ignore it for now. We note, however, that if we were able to solve this problem in the generality suggested, that is, if we were able to find a funtion that minimized the energy above over the lass of all pieewise differentiable ontinuous funtions with the presribed endpoint values and length, and it just happened to turn out that the minimizer was in C 1 [, 1], then we would be ompletely justified in ignoring the more general lass of funtions for this problem. Similarly, if it happened that the minimizer was in C 2 [, 1], then we ould make that assumption from the outset as well. In the ase of a hanging hain, both these things turn out to be true, and we will make this assumption in our disussion here. 3

4 2 The First Variation and the First Integral Through proedures whih will be explained later, we arrive at the expression [ 1 δeu λ (φ = + u 2 φ + u λ ] 1 + u 2 u φ dx. For a fixed u A we an think of δeu λ as a funtional on C funtional is alled the first variation of E λ at u. [, 1]. This Key Fat: If u = u λ minimizes E λ over A, then for every φ C [, 1]. δe λ u (φ = 0 This fat is ompletely analogous to the fat that the first derivative vanishes at minimum points for funtions on R n. In fat, it follows from appliation of that fat, as we will explain when we derive the expression for the first variation. This is the point at whih we will use the assumption that u C 2 [, 1]. Suh an assumption allows us to integrate by parts in the integral involving φ to obtain [ ( 1 ] δeu λ (φ = + u 2 u λ 1 + u 2 u φ dx. The vanishing of this expression for every φ implies ( u λ 1 + u 2 u 1 + u 2 = 0, whih is an ordinary differential equation. It is natural to make the substitution v = u λ, and one finds ( v = v 2 We an integrate one to find v = λ. (2 1 + v u ( 2 4

5 At this point, it is onvenient to onsider two speial ases. If λ = 0, then we have u 0. This is ertainly a plausible solution in the ase u 1 = 0 and L = 2. Another speial ase to onsider is that when u ( = 0 (and of ourse, λ 0. Evidently, we have v ( = v( 2 λ 2 = 0, and it follows from the original equation that v ( = /λ 0. Therefore, v (x 0 for x in some interval (, + ɛ. No matter what the sign of λ, we find that v/λ = v/ > 1 on the same interval. Thus, we may solve (2 for v to obtain v / (v/2 1 for x < + ɛ and integrate from to x to get or osh (v/ = (x + 1/, = 1/ (3 ( x + 1 u(x = osh. (4 This funtion and all its derivatives happen to be defined and finite for all x R and for x [, 1] in partiular. Moreover, by the uniqueness properties of ordinary differential equations (again under the assumption that u C 2 this expression must give the unique solution for all x. More generally, assuming λ 0, we find that (2 implies 2 v 2 = v 2 2. Under the further assumption that u ( 0, we have that v( = λ >. Solving for v, we find that (v/ (v/2 1 = ±1. (5 Noting that v/ > 1 at least in some interval to the right of, the same integration as before yields ( u(x = osh ± x osh ( λ/ + λ [ ( ( ] x µ µ + 1 = osh osh, 5

6 where µ = osh ( λ/. Again, this is a niely behaved funtion for all values of x and must be the unique solution of the ordinary differential equation for a given λ,, and hoie of sign (± in the definition of µ. Note also, that the expression just derived for u redues to that given in (4 for the ase u ( = 0 on substitution of this relation. Thus, aside from the situation u 0, all solutions of the ordinary differential equation have the form u(x = [ osh ( ( x µ 1 + µ osh We wish to rule out the possibility that < 0. Lemma 1 If u(x is given by the formula (6 with < 0, then [ ( ( ] x + µ 1 µ ũ(x = osh osh ]. (6 has the same endpoint values, the same length, and stritly lower energy. Proof: Clearly, ũ( = 0. It is also easily heked that ũ(1 = u(1. In order to alulate the lengths, we note that ( ( x + µ x µ ũ (x = sinh and u (x = sinh. Therefore, the length assoiated with ũ is ( [ ( ( ] x + µ µ + 1 µ 1 osh dx = sinh sinh ( x µ = osh dx, whih is the length assoiated with u. This establishes that the lengths are equal and shows also that [ ( ( ] 1 + µ 1 µ L = sinh + sinh > 0 (7 for all values of and µ. This latter fat an also be obtained with an estimate by expanding the expression using the addition formula for sinh: ( 1 ( µ ( 1 L = 2 sinh osh > 2 sinh > 0. (8 6

7 Finally, we onsider the energies. Sine we have shown that the lengths are the same, it suffies to show E(u > E(ũ where E is the energy given in (1 without the Lagrange multiplier term. E(u = = [ osh ( ( x µ 1 + µ osh ( 1 + µ dx osh osh 2 ( x µ ] osh L ( x µ dx where L here denotes the ommon length alulated above. Sine ũ is obtained from u simply by swithing the signs of and µ, we obtain a similar expression for E(ũ. Making a hange of variables ξ = x in the expression yields Thus, [ ( ( ] ( x + µ 1 µ x + µ osh osh osh [ ( ( ] ( µ ξ 1 µ µ ξ = osh osh osh [ ( ( ] 1 µ 1 + µ = E(u + osh osh L E(ũ = [ E(u E(ũ = 2E(u osh = 2 ( ( 1 µ 1 + µ osh ( x µ osh 2 dx [ ( ( 1 µ 1 + µ osh + osh dx dξ ] L ] L. The integral appearing on the right may be evaluated with the help of an addition formula for osh: ( x µ 1 [ ( 2 osh 2 dx = osh 2 x µ ] + 1 dx ( ( ( ( ] 1 µ 1 µ 1 + µ 1 + µ = [osh 2 sinh + osh sinh + 2 7

8 Finally, if we expand the seond term using the expression (7 for L, we find ( x µ E(u E(ũ = 2 osh 2 dx [ ( ( ] 1 µ 1 + µ osh + osh L ( ( ( ( ] 1 µ 1 + µ 1 + µ 1 µ = [osh 2 sinh + osh sinh + 2 ( 2 = 2 sinh + 2 [ ( ] 2 = sinh 2. The funtion f( = sinh(2/ is inreasing for < 0, and lim f( = 2. Thus we see that E(u E(ũ > 0 for < 0, and this ompletes the proof of the lemma. In light of Lemma 1, we need only onsider funtions of the form u(x = [ osh ( ( x µ 1 + µ osh ]. (9 with > 0. Notie that µ gives the x-oordinate of the vertex of the hyperboli osine urve under onsideration, and is a saling fator for the entire urve. Theorem 1 Given L > 0 and u 1 satisfying L 2 > u , there exists exatly one pair (µ, R (0, suh that the funtion defined in (9 satisfies u(1 = u 1 and 1 + u 2 dx = L. Proof: Sine sinh is invertible, we may solve the equation [ ( ( ] 1 µ 1 + µ u(1 = osh osh = u 1 8

9 for µ in terms of (and u 1. In fat, using the addition formula for osh, this equation beomes ( 1 ( µ u(1 = 2 sinh sinh = u 1, so that ( µ = sinh u1 2f( where f( = sinh(1/. Substituting this into the equation ( 1 ( µ ( µ L = 2 sinh osh = 2f( osh (10 from (8, we obtain L = 2f( 1 + u2 1 4f( 2. (11 The funtion f is easily seen to be dereasing and onvex for > 0 with lim f( = + and lim f( = In partiular, f( > 1 for > 0. With this in mind, equation (11 beomes L = 4f( 2 + u 2 1 or L2 u 2 1 f( =. (12 2 Notie that the value on the right is a fixed onstant greater than 1 sine L2 u 2 1 > 2. Therefore, aording to the properties of f, equation (12 has exatly one positive solution. This solution and the µ obtained from substituting it into (10 evidently give the unique pair asserted to exist in the theorem. 3 Numerial Calulations The equation f( = L 2 u 2 1/2 is fairly easy to solve by Newton s method, if one keeps in mind that the positive solution is required and f( is even. 9

10 It does not seem easy to determine an analyti initial value 0 to start Newton s method whih always leads to the positive root. One reasonable initial value, however, seems to be obtained by replaing, initially, f( with f 0 ( 0 = 1/ Notie that f 0 has the monotonoity properties and limiting behavior of f for 0 > 0. This leads to the initial value 0 = 2 L2 u