arXiv:0804.1513v1 [math.DG] 9 Apr 2008
GEOMETRY OF WHIPS AND CHAINS STEPHEN C. PRESTON
1. Introduction The purpose of this note is to explore the geometry of the inextensible string (or whip) which is fixed at one end and free at the other end. In the absence of gravity, the inextensible string is a geodesic motion in the space of curves in R2 parametrized by length and passing through a fixed point. This is a submanifold of the space of all curves in R2 , and the Lagrange multiplier from the constraint is the tension in the string. Hence the string satisfies a wave equation, where the tension is generated by the velocity implicitly by solving a one-dimensional Laplace equation. The situation seems closely analogous to the motion of an incompressible fluid, where the configuration space is the space of volumepreserving diffeomorphisms, a submanifold of the space of all diffeomorphisms, and the Lagrange multiplier is the pressure. This is also a geodesic motion, but the Riemannian geometry is still not well-understood. It is our hope that understanding the one-dimensional situation better will help in understanding the two- and three-dimensional versions. Part of the problem with the inextensible string is that the geodesic equation is not an ordinary differential equation on an infinitedimensional manifold (unlike the situation for incompressible fluids). The derivative loss cannot be avoided, and thus one must treat it as a hyperbolic partial differential equation. However, since the tension must be zero at the free end, the equation degenerates at the free end, and the standard approach to symmetric hyperbolic equations must be modified to deal with this. Although these issues are not discussed here, we will study them in a forthcoming paper. The other serious issue is that since the tension is not given by a positive function of the speed, position, and time, but rather given implicitly by the solution of an ordinary differential equation, it is possible for tension to become negative. This corresponds to the differential equation changing type from hyperbolic to elliptic, and drastically complicates the problem. We show that in the absence of gravity, the 1
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STEPHEN C. PRESTON
tension is always nonnegative, while with gravity, the tension is nonnegative as long as the string hangs below the fixed point. To better understand the necessary conditions on weak solutions (which has remained controversial in recent papers on falling strings with kinks), we consider the corresponding finite-dimensional model: a system of N particles linked by rigid massless rods. In the absence of gravity, this is also a geodesic motion on an N-torus with a certain Riemannian metric. The geodesic equation ends up being a sort of discretization of the wave equation for the inextensible string, which still conserves energy. Furthermore, the tension in the chain converges to the tension in the corresponding whip as N → ∞. As a result, the Riemannian curvature of TN converges to the Riemannian curvature of the space of curves, and one can therefore understand the geometry of the infinite-dimensional space quite well by studying the finite-dimensional approximation. This suggests that one may be able to use the same approach to study incompressible fluids. Namely, take a collection of particles on a torus T2 in a grid, and constrain them so that the areas of quadrilaterals are constant. (One could consider constraining triangle areas, but this seems too rigid; it does not appear possible to approximate a general area-preserving diffeomorphisms by such maps.) The curvature of the finite-dimensional manifold should approach that of the full area-preserving diffeomorphism group, and hence one expects one could understand the geometry of fluid mechanics well by a finite-dimensional approximation. 2. Discrete string Let us compute the geometry of a discrete string (a chain), which we model by n + 1 point masses joined in R2 by rigid rods of length n1 whose mass is negligible. We assume the 0th point is fixed at the origin while the nth point is free. The unconstrained Lagrangian is n
n
1X 2 X |x˙ i | − ghxi, e2 i. L= 2 i=1 i=1
In addition the constraints are given by 1 |xi − xi−1 |2 = 2 , 1 ≤ i ≤ n. n Therefore the equations of motion are (1)
¨ i = −ge2 + λi+1 (xi+1 − xi ) + λi (xi−1 − xi ) x
GEOMETRY OF WHIPS AND CHAINS
3
for 1 ≤ i ≤ n, where we are fixing x0 = 0 and λn+1 = 0 to make the equations satisfied also at the endpoints. The constraint equations determine the Lagrange multipliers λ, which are obviously just the discrete tensions. We get 2 λi −λi−1 hxi −xi−1 , xi−1 −xi−2 i n2 = 0), while for i = 0 we get
|vi −vi−1 |2 = −λi+1 hxi −xi−1 , xi+1 −xi i+ for 1 ≤ i < n (again using λn+1
1 λ1 . n2 We can simplify these λ-equations somewhat if we change variables. For each 1 ≤ i ≤ n, let xi = xi−1 + n1 (cos θi , sin θi ), where θi ∈ S 1 . Then 1 hxi − xi−1 , xi+1 − xi i = 2 cos (θi+1 − θi ) n while 1 |vi − vi−1 |2 = 2 |θ˙i |2 . n So the constraint equation for λ becomes |v1 |2 − ghx1, e2 i = −λ2 hx1 , x2 − x1 i +
(2)
− cos (θi+1 − θi )λi+1 + 2λi − cos (θi − θi−1 )λi−1 = |θ˙i |2
for 2 ≤ i ≤ n and − cos (θ2 − θ1 )λ2 + λ1 = |θ˙1 |2 − ng sin θ1 .
(3)
For the evolution equation, we have (4)
θ¨i = λi+1 sin (θi+1 − θi ) − λi−1 sin (θi − θi−1 )
for 2 ≤ i ≤ n and θ¨1 = λ2 sin (θ2 − θ1 ) − ng cos θ1 .
(5)
Our first question is whether the tension is positive. We can write equations (3) and (2) as n X
Mij λj = |θ˙i |2 − ngδi1 sin θ1 ,
j=1
where the matrix M is 1
− cos (θ2 −θ1 ) 0 ... − cos (θ2 −θ1 ) 2 − cos (θ3 −θ2 ) ... 0 − cos (θ3 −θ2 ) 2 ...
M =
.. .
0
.. .
0
.. .
0
0 0 0
.. .
0 0 0
.. .
... − cos (θn −θn−1 ) 2
.
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STEPHEN C. PRESTON
If we let ai = cos (θi+1 − θi ) for 1 ≤ i ≤ n−1, then standard Gaussian elimination yields b1 −a1 0 ... 0 | 1 0 0 ... 0 0
b
2 0 0 .. .. . .
0
−a2 ... 0 | b3 ... 0 |
.. .
0
0
.. .
... bn
a1 b1 a1 a2 b1 b2
.
.. | Qn−1 | i=1
ai bi
1
0
... 0
a2 b2
1
... 0
.. .
ai i=2 bi
Qn−1
.. .
ai i=3 bi
Qn−1
.. .
... 1
where the sequence bi is defined by b1 = 1, bi+1 = 2 − for every i, we see that every bi satisfies 1 ≤ bi ≤ 2. Reducing further, we obtain the inverse matrix (6)
ij
M =
n X
m=max i,j
Thus the tensions are
a2i . bi
Since |ai | ≤ 1
m−1 m−1 1 Y ak Y al . bm k=i bk l=j bl
λi = M (|θ˙1 |2 − ng sin θ1 ) + i1
n X
M ij |θ˙j |2 .
j=2
If every ai ≥ 0, then every component of M −1 is nonnegative. So as long as |θi+1 − θi | ≤ π for every 1 ≤ i ≤ n − 1 and sin θ1 ≤ 0, we will have all tensions λi ≥ 0, regardless of |θ˙i |. If any ai < 0, then some M ij < 0, and thus if θ˙k = δjk , then λi < 0. Additionally, we always have M 11 > 0, so that if sin θ1 > 0, then when all θ˙k = 0, we have λ1 = −ngM 11 sin θ1 . Let us now look at the curvature; we will assume for the moment that g = 0 to get a purely geometric problem. We consider Mn = (S 1 )n as a submanifold of R2n , with induced Riemannian metric given by the kinetic energy formula above. We know that the equation of a constrained geodesic is always d2 xk dxj dxi ∂k = Bk ∂i , ∂j ∂k . dt2 dt dt For any vector u ∈ Tθ Mn , we have huk − uk−1 , xk − xk−1 i = 0, for every 1 ≤ k ≤ n. So uk − uk−1 = n1 ηk (− sin θk , cos θk ) for some ηk ∈ R. In terms of the η’s, we can write (using (4)) (7)
Bk (u, u) = λ(u, u)k+1(xk+1 − xk ) + λ(u, u)k (xk−1 − xk ),
for 1 ≤ k ≤ n, where (8)
λ(u, u)i =
n X j=1
M ij |ηj |2 .
GEOMETRY OF WHIPS AND CHAINS
5
Since the ambient space is flat, the curvature is, by the GaussCodazzi formula, hR(u, v)v, ui = hB(u, u), B(v, v)i − hB(u, v), B(u, v)i. To simplify this, we first compute n 1 X hB(u, u), B(v, v)i = 2 λ(u, u)k+1λ(v, v)k+1 + λ(u, u)k λ(v, v)k n k=1 − λ(u, u)k+1λ(v, v)k cos (θk+1 − θk )
− λ(u, u)k λ(v, v)k+1 cos (θk+1 − θk ) 1 = 2 λ(v, v)1 λ(u, u)1 − λ(u, u)2 cos (θ2 − θ1 ) n n 1 X 2λ(u, u)k λ(v, v)k + 2 n k=2 − λ(u, u)k−1λ(v, v)k cos (θk − θk−1 )
− λ(u, u)k+1λ(v, v)k cos (θk+1 − θk ) = = Similarly we have
1 n2 1 n2
n X n X k=1 l=1 n X n X
Mkl (θ)λ(u, u)lλ(v, v)k M ij (θ)|ηi |2 |ξj |2 .
i=1 j=1
hB(u, v), B(u, v)i =
n n 1 X X ij M (θ)ηi ξi ηj ξj . n2 i=1 j=1
Therefore the curvature is n 1 X ij 2 2 (9) hR(u, v)v, ui = 2 M (θ) ηi ξj − ηi ξi ηj ξj n i,j=1 =
n 2 1 X ij M (θ) η ξ − η ξ . i j j i 2n2 i,j=1
Thus the curvature is nonnegative in every section if and only if every M ij (θ) ≥ 0, which also happens if and only if every tension is nonnegative regardless of the initial velocity. To get a bound on the curvature, we want a bound on the components M ij . We see from equation (6) that M ij ≤ max(n + 1 − i, n + 1 − j). The extreme case is when all θ’s are the same (so that ai = 1 for every i), which corresponds to a straight string. Then we have M ij =
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STEPHEN C. PRESTON
max(n + 1 − i, n + 1 − j). Therefore the sectional curvature satisfies 0 < K < n. 3. Smooth string Now let us consider the continuum model of an smooth inextensible string. The configuration space is the set CL,0 of all C ∞ maps η : [0, 1] → R2 such that η(0) = 0 and hη ′ (s), η ′(s)i ≡ 1. This is a closed submanifold of C0 = {η ∈ C ∞ ([0, 1], Rd ) | η(0) = 0}. The tangent spaces are Tη C0 = {X ∈ C ∞ ([0, 1], R2 ) | X(s) ∈ Tη(s) R2 ∀s,
X(0) = 0}
and Tη CL,0 = {X ∈ Tη C0 | hX ′(s), η ′(s)i = 0 ∀s}. The Riemannian metric is defined by Z 1 hhX, Y ii = hX(s), Y (s)iη(s) ds. 0
In this metric, the orthogonal space to Tη CL,0 is ⊥ n d σ(s)η ′ (s) Tη CL,0 = Y ∈ C ∞ ([0, 1], R2) Y (s) = ds
o for some σ ∈ C ([0, 1], R) with σ(1) = 0 . ∞
The orthogonal projection of a vector X ∈ Tη C0 should be PL (X) = X −
d (ση ′), ds
where σ satisfies (10)
σ ′′ (s) − hη ′′ (s), η ′′(s)iσ(s) = hX ′ (s), η ′(s)i,
σ(1) = 0,
but we have a problem with the boundary condition at s = 0. We want PL to map into Tη CL,0 , but PL (X)(0) = X(0) − σ ′ (0)η ′ (0) − σ(0)η ′′ (0) = −σ ′ (0)η ′ (0) − σ(0)η ′′(0). Now η ′′ (0) is orthogonal to η ′ (0), so if η ′′ (0) 6= 0, we need both σ(0) = 0 and σ ′ (0) = 0, which will generally be impossible. So we do not have a nice orthogonal projection. There are two ways to get around this. (1) Consider only curves with η ′′ (0) = 0. This forces X ′′ (0) = 0 as well, and then we have to worry about PL (X) also satisfying (PL (X))′′ (0) = 0. This forces η iv (0) = 0 and X iv (0) = 0, and so on... We end up being forced to require that all derivatives of η exist at s = 0 (which is why I assumed η ∈ C ∞ ([0, 1], R2 )) and that all even derivatives of η vanish at s = 0. Alternatively,
GEOMETRY OF WHIPS AND CHAINS
7
η extends to a C ∞ curve on [−1, 1] such that η(−s) = −η(s). This observation leads to the second idea. (2) We could assume from the start that η is the restriction of an odd curve on [−1, 1]. Then the tangent space will automatically require X odd. We then have two boundary conditions for (10): σ(−1) = 0 and σ(1) = 0. Finally, the solution of (10) will automatically be even in s by symmetry, and then we can just restrict to [0, 1]. The advantage of this approach is that we don’t need to assume any more smoothness than is absolutely necessary for (10) to make sense. This is of course the same kind of trick that’s used to get simple solutions to the constant-coefficient linear wave equation with one free and one fixed boundary. Either of these two tricks will force σ ′ (0) = 0, so we may as well assume this as our other boundary condition. These difficulties do not arise if both ends of the curve are free. If both ends of the curve are fixed, we need oddness through both endpoints, again as with the simplest wave equation. Now, the first thing we do with the orthogonal projection is to com⊥ pute the second fundamental form B : Tη CL,0 × Tη CL,0 → Tη CL,0 defined by B(X, Y ) = ∇X Y − PL (∇X Y ). We obtain d (11) B(X, Y ) = σXY (s)η ′ (s) ds where (12)
′′ σXY (s) − κ2 (s)σXY (s) = −hX ′ (s), Y ′ (s)i.
This formula implies the geodesic equation d2 η dη dη , =B , dt2 dt dt
or more explicitly,
ηtt (s, t) = ∂s σXX (s, t)ηs (s, t) ,
∂s2 σXX (s, t) − |ηss (s, t)|2 σXX (s, t) = −|ηst (s, t)|2 , with boundary conditions as discussed. If we want to incorporate gravity, then we have ηtt (s, t) = ∂s σ(s, t)ηs (s, t) − ge2 (13) (14)
σss (s, t) − |ηss (s, t)|2σ(s, t) = −|ηst (s, t)|2 ,
with boundary condition σ(1, 0) = 0. To determine the boundary condition at s = 1, we assume ηss (0, t) = 0 as before. If we want
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STEPHEN C. PRESTON
ηtt (0, t) = 0, we must have σs (0, t)ηs (0, t) = ge2 , which cannot be satisfied unless e2 k ηs (0, t). Instead, motivated in part by the finite case above, we set σs (0, t) = ghe2 , ηs (0, t)i, which is equivalent to forcing the acceleration parallel to the curve to be zero. If, as before, we use the fact that |ηs | ≡ 1 to write ηs (s, t) = cos θ(s, t), sin θ(s, t) , then differentiating the first of equations (13) with respect to s yields (15)
θtt (s, t) = 2σs (s, t)θs (s, t) + σ(s, t)θss (s, t)
(16)
σss (s, t) − θs (s, t)2 σ(s, t) = −θt (s, t)2
with boundary conditions σs (0, t) = g cos θ(0, t) and σ(1, t) = 0. Now we want to compute the curvature; again we assume g = 0 to get a purely geometric formula. Formula (11) gives, via the Gauss-Codazzi equations, the curvature formula: hhR(X, Y )Y, Xii = hhB(X, X), B(Y, Y )ii − hhB(X, Y ), B(X, Y )ii Z 1 d d ′ ′ σXX (s)η (s) , σY Y (s)η (s) = ds ds 0 d d ′ ′ − σXY (s)η (s) , σXY (s)η (s) ds ds ds Z 1 = σXX (s) − σY′′ Y (s) + κ2 (s)σY Y (s) ds 0 Z 1 ′′ − σXY (s) − σXY (s) + κ2 (s)σXY (s) ds. 0
If we denote the Green function of equation (12) by G(s, q), so that Z 1 σXY (s) = G(s, q)hX ′(q), Y ′ (q)i dq, 0
then the curvature formula becomes Z 1Z 1 hhR(X, Y )Y, Xii = G(s, q) |X ′ (q)|2 |Y ′ (s)|2 0 0 ′ − hX (q), Y ′ (q)i hX ′(s), Y ′ (s)i ds dq Z Z d X 2 1 1 1 G(s, q) Xi′ (s)Yj′ (q) − Xj′ (q)Yi′ (s) ds dq, = 2 0 0 i,j=1 using the symmetry of the Green function. Thus if G(s, q) ≥ 0 for all s and q, we get nonnegative curvature.
GEOMETRY OF WHIPS AND CHAINS
9
To prove positivity of G(s, q), we first compute: G(q, q) =
Z
1
G(s, q) δ(s − q) ds Z 1 =− G(s, q) Gss (s, q) − κ2 (s)G(s, q) ds 0 Z 1 = Gs (s, q)2 + κ2 (s)G(s, q)2 ds > 0.
0
0
So G(q, q) > 0. We also know that Gs (s = q + , q) − Gs (s = q − , q) = −1. Since the boundary conditions are Gs (0, q) = 0 and G(1, q) = 0, we can say the following. (1) If G(0, q) < 0, then G(s, q) < 0 for s near 0. As long as G(s, q) < 0, we have Gss (s, q) ≤ 0, which implies Gs is nonincreasing. Since Gs (0, q) = 0, this means Gs (s, q) ≤ 0 as long as G(s, q) < 0. Thus there is no way G(s, q) can increase to 0 or any positive number on 0 < s < q. This makes G(q, q) > 0 impossible, a contradiction. Therefore G(0, q) > 0, and furthermore G(s, q) > 0 for all s ∈ (0, q). (2) If Gs (1, q) > 0 then G(s, q) < 0 for s close to 1. As long as G(s, q) < 0, we have Gss (s, q) ≤ 0, and so Gs (s, q) is nonincreasing. So Gs (s, q) > 0 as long as G(s, q) < 0, so G cannot have a turning point on q < s < 1. This again makes G(q, q) > 0 impossible, implying a contradiction. So Gs (1, q) < 0, which implies G(s, q) > 0 for all s ∈ (q, 1). This proves that the curvature is nonnegative as long as η is smooth. In fact the sectional curvature must be strictly positive in all nontrivial two-planes, since it can only vanish if there is a c such that X ′ (s) = cY ′ (s) for every s. Since X(0) = Y (0), we must have X(s) = cY (s) for all s, which means X and Y are linearly dependent in Tη CL,0 . Polarizing the formula for curvature, we obtain Z 1Z 1 hhR(Y, X)X, W ii = G(s, q) |X ′ (q)|2 hY ′ (s), W ′ (s)i 0 0 ′ − hX (q), Y ′ (q)i hX ′(s), W ′ (s)i ds dq Z 1Z 1 ∂ ′ 2 ′ =− |X (q)| G(s, q)Y (s) , W (s) ∂s 0 0 ∂ ′ ′ ′ G(s, q)X (s) , W (s) ds dq. + hX (q), Y (q)i ∂s
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STEPHEN C. PRESTON
Therefore the curvature operator is ∂ hZ 1 i Y 7→ PL G(s, q) − |X ′(q)|2 Y ′ (s) + hX ′ (q), Y ′ (q)iX ′(s) dq . ∂s 0
Even for smooth X, this operator is not bounded in L2 . (By contrast, the curvature operator in the volumorphism group for a fixed C 1 vector field X is bounded in L2 and in any other Sobolev space.) 4. Convergence We don’t have a good existence and uniqueness result for the continuous equation; the closest we have is Reeken [R2], who proved short-time existence for a chain in R3 in something like the Sobolev space H 17 with initial conditions close to a chain hanging straight down (for technical reasons, Reeken assumes the chain is hanging from (0, 0, ∞) with nonconstant gravity). This result is already quite difficult. It is conceivable that we could get a simpler result using the two-dimensional equation derived above. Of course, the finite-dimensional approximation is the flow of a vector field on a compact manifold (a bounded subset of the tangent space T (S 1 )n , since the kinetic energy is bounded). Thus the finite-dimensional approximation has a unique solution existing for all time and for any n. We can thus ask whether solutions of the PDE can be constructed as limits of the ODE as n → ∞. It is easy to see that if we knew a smooth solution of the PDE existed, then the discrete model would converge to it. Let us write 1 2 1 2 1 θn : , , . . . , 1 × R → S and σn : , . . . , 1 × R → R. n n n n We will set θn ( nk , t) = θk (t) and σn ( nk , t) = Then equations (2) and (4) become (17)
d2 θn dt2
k ,t n
= n2 σn
k+1 ,t n
− cos θn
− n2 σn
k+1 ,t n
− cos θn
− θn
k ,t n
−
k ,t n
for 1 ≤ k ≤ n.
− θn nk , t k−1 , t sin θn nk , t − θn n
sin θn
and (18)
1 λ (t) n2 k
σn
θn k−1 ,t n
k+1 ,t n
k+1 ,t n
σn
+ 2σn
k−1 ,t n
k−1 ,t n
k ,t n
1 = 2 n
d θn dt
k ,t n
2
The operators appearing in equations (17) and (18) are symmetric discretizations of derivative operators; if σ and θ are sufficiently
GEOMETRY OF WHIPS AND CHAINS
11
smooth, we can write n2 σ x + n1 , t sin θ x + n1 , t − θ x, t − n2 σ x − n1 , t sin θ x, t − θ x − n1 , t = σ(x)θ′′ (x) + 2σ ′ (x)θ′ (x) + O n12 and 2 1 − n cos θ x + n , t − θ x, t σ x + n1 , t + 2n2 σ x, t − n2 cos θ x, t − θ x − n1 , t σ x − n1 , t
= θ′ (x)2 σ(x) − σ ′′ (x) + O
1 n2
where the constant in O n12 depends on the C 4 norms of both θ and σ. Therefore, we can say that if σ and θ form a sufficiently solution of (13), then the motion of the discrete chain with n elements converges to the motion of the smooth chain as n → ∞. We have convergence not only of the position and velocity, but also of the acceleration and the tension, which is not necessarily typical (for example, a system with strong returning force to a submanifold has position and velocity converging to the geodesic motion on the submanifold, but the acceleration does not converge). Because the convergence is so strong, we can speculate about convergence of the motion in the nonsmooth case. Suppose the continuous string has a kink in it. There has been some serious analysis of the possible jump conditions one should impose on the equations in this case. See for example Reeken [R1], who used energy conservation and momentum conservation to classify possible motions of a kink. Also D. Serre used an approach similar to yours, relaxing the constraint to |x′ (s)| ≤ 1 and requiring that the tension always be nonnegative, and derived jump conditions. But this only sets up the differential equations, and there’s no guarantee that they have solutions. (Serre found blowup phenomena in a couple of examples.) Indeed, since we expect the motion of n constrained particles to converge to the motion of a chain, the fact that the particles can (ideally) support negative tension in certain motions (when there is an acuteangle kink or when the chain starts with the first link above the fixed point) seems to suggest that the motion of the continuous string should also support negative tension. Of course this destroys the actual equations, which suggests that the motion of n constrained particles has no limit in some cases.
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STEPHEN C. PRESTON
We will look at the limit as n → ∞ by examining the inverse matrix M ij given by (6). We expect the components of this matrix to converge to the Green function as n1 M ij ≈ G ni , nj . To understand
what happens to M ij , we first look at the sequence bi appearing in its definition. We have bi+1
a2i =2− , bi
where ai = cos (θi+1 − θi ).
Now if the discrete curve approaches something smooth, then we will have 1 θi+1 − θi ≈ κi , n where κi = θ˙ i+1/2 , the approximate curvature. Thus as long as the n curve is smooth, we will have ai ≈ 1 and thus also bi ≈ 1. Numerically experimenting, we find that the sequence bi is approximated very well by (19)
bi = 1 +
1 f n
i n
where f satisfies the differential equation
,
f ′ (s) = κ2 (s) − f 2 (s),
(20)
with initial condition f (0) = 0. This equation of course comes from the Riccati trick f (s) = y ′(s)/y(s), where y ′′ (s) − κ2 (s)y(s) = 0. Now suppose there is a kink in the curve, say of angle α at position so with so ≈ k/n for some integer k. Then θk+1 −θk = α, and so all ai ’s are close to 1 except for ak = cos α. We then have bk+1 being rather far from 1, which means for equation (19) to be valid, we must have lim f (s) = +∞. In fact this is precisely what happens numerically: f s→so
satisfies equation (20) to the left of so with initial condition f (0) = 0; f also satisfies (20) to the right of so with condition lim f (s) = +∞. In s→so
particular, f does not depend on the size of the kink, only the fact that there is one. If there are multiple kinks, we get the same condition: f approaches infinity from the right side of the kink. Knowing ai and bi , we know M ij . Recall from (6) the formula ij
M =
n X
m=max i,j
m−1 m−1 1 Y ak Y al . bm k=i bk l=j bl
GEOMETRY OF WHIPS AND CHAINS
13
Pick any x and y in [0, 1] and s ≥ max(x, y). For large n, choose i = nx, κ2 j = ny, and m = ns. Since ak ≈ 1 − 2nk2 except at the kink, we have ( −1 ns Y 1 if so ∈ / (x, s) ak → as n → ∞ cos α if s ∈ (x, s) o k=nx for any x and s. On the other hand, we have ns −1 −1 ns i h Y X 1 k bk = exp ln 1 + f n n k=nx k=nx Z s ns −1 X k 1 1 f = exp +O → exp f (σ) dσ . n k=nx n n x Thus we have
G(x, y) =
Z
1
φ(x, s)φ(y, s)e−
Rs x
f (σ) dσ −
e
Rs y
f (τ ) dτ
ds
max(x,y)
where
φ(x, s) =
(
cos α x < so < s . 1 otherwise
Thus we see G(x, y) ≥ 0 even if the curve has one or more kinks. So in the absence of gravity, the curvature is nonnegative in all sections. If there is gravity, then G(x, y) may be negative, if the string is pointing upwards at the fixed point. If we use a generalized curvature that incorporates the Hessian of the potential energy, then we will also get negative generalized curvature in this case. References R1. M. Reeken, The equation of motion of a chain, Math. Z. 155 no. 3, 219–237 (1977). R2. M. Reeken, Classical solutions of the chain equation I, Math. Z. 165 143–169 (1979). S. D. Serre, Un mod`ele relax´e pour les cˆ ables inextensibles. RAIRO Modl. Math. Anal. Numr. 25, no. 4, 465–481 (1991). Department of Mathematics, University of Colorado, Boulder, CO 80309-0395 E-mail address:
[email protected]
