The shape of a hanging chain. a project in the calculus of variations

Transcription

1 The shape of a hanging hain a projet in the alulus of variations April 15, 218

2 2

3 Contents 1 Introdution 5 2 Analysis Model Extremal graphs Minimality of extremals Computation Approximating the onstant The Lambert W funtion Alternative reursive approximation Numerial Code(s) Experiment Relation to physial parameters Some speifi shapes Error Generalizations

4 4 CONTENTS

5 Chapter 1 Introdution If a hain or able has its ends fixed at two different points and hangs under the influene of gravity, it takes the shape of a hyperboli osine urve. We now desribe this shape preisely and explain how it arises as a minimizer of potential energy among many possible shapes Figure 1.1: the shape of a hain hanging from its endpoints in gravity 5

6 6 CHAPTER 1. INTRODUCTION

7 Chapter 2 Analysis 2.1 Model Let l > be the length of the hain and let ρ denote the linear density of mass along the length of hain. Choose x, y-oordinates with the left end of the hain fixed at (, ) and the right end at (1, u 1 ). We have made a hoie of units here so that the horizontal distane between the fixed endpoints is 2 units. This is equivalent to saling the system given in some partiular initial units. We ould also assume u 1 has a speifi sign, say u 1 >, but this is not neessary. Given the length onstraint on the hain, we must have 1 + u 2 1 < l 2. (2.1) There are many urves of length l onneting (, ) to (1, u 1 ). Among these onsider C 1 urves given by the graph of a funtion u : [, 1] R. The length onstraint may then be written as 1 + [u (x)] 2 dx = l. Assuming a onstant gravitational field G = g(, 1) and zero potential at y =, we may integrate to approximate the potential energy of a portion of the hain having mass m j = ρ 1 + [u (x j )]2 x j : approximate potential energy V j = 7 u(x j ) ρg 1 + u (x j )2 x j dy.

8 8 CHAPTER 2. ANALYSIS Figure 2.1: an alternative hain shape and the assoiated potential energy The potential energy assoiated with a point mass is given by the work required to move the mass from a position of zero potential to another position, that is, F T where F is the fore field, γ is a path onneting a position γ of zero potential to the position of the mass, and T is the unit tangent vetor along the path. In this ase the fore F = m jg = mj g(, 1) is assumed onstant, and the integral amounts to the fore multiplied by the vertial distane to equilibrium: V j = ρgu(x j ) 1 + u (x j )2 x j. Summing over all model portions of hain and taking the limit as the maximum portion length tends to, we find an expression for the total potential energy as a funtion of the hain shape determined by u: potential energy V = lim ρgu(x j ) 1 + u (x j )2 x j = ρgu(x) 1 + u (x) 2 dx. By the Leibniz /Maupertuis priniple of virtual work, or Hamilton s ation priniple, the observable shape u should be a ritial point for subjet to the onstraint V [u] = L[u] = ρgu(x) 1 + u (x) 2 dx 1 + u (x) 2 dx = l.

9 2.2. EXTREMAL GRAPHS 9 Under the assumption that ρ and g are positive onstants, we may replae the expression for V above with V [u] = u(x) 1 + u (x) 2 dx Introduing a Lagrange multiplier λ assoiated with the onstraint and assuming the existene of the model shape within the admissible lass A = {u C 2 [, 1] : u() =, u(1) = u 1 }, we set F = V + λl and obtain the neessary ondition δf u [φ] = d dǫ (u + ǫφ + λ) 1 + (u + ǫφ ) 2 dx ǫ= = for all φ C (, 1). Differentiating under the integral and evaluating, we find [ φ ] u φ 1 + u 2 + (u + λ) dx =. 1 + u 2 We may integrate by parts in the seond term to obtain [ ( (u + λ)u 1 + u 2 ) u 2 ] φ = for all φ C (, 1). Finally, we may apply the fundamental lemma of the alulus of variations to obtain a two point boundary value problem for a seond order nonlinear ordinary differential equation for the observed shape u: ( (u + λ)u 1 + u 2 ) = 1 + u 2, u() =, u(1) = u 1. We know this equation is satisfied even under the assumption u C 1 [, 1]. 2.2 Extremal graphs Using the assumed regularity of the observed shape u, we an also write u (u + λ) (1 + u 2 ) + u 2 = 1 + u 2 3/2 1 + u 2

10 1 CHAPTER 2. ANALYSIS or (u + λ)u = 1 + u 2. Under the assumption u () >, whih (based on observation of the shape of atual physial hanging hains) seems rather reasonable, we an solve for the Lagrange multiplier and find λ = 1 + u () 2 u () >. More generally, whenever u + λ, we an write u 1 + u 2 u = 1 u + λ u. In partiular, integrating from x = to x, u u () t 1 + t 2 dt = u u() 1 t + λ dt or 1 [ ln(1 + u 2 ) ln(1 + u () 2 ) ] = ln(u + λ) ln λ. 2 It follows that 1 + u 2 ( u ) u () = 2 λ + 1. (2.2) Let us pause at this point to onsider the first integral equation u F p (u, u ) F(u, u ) = (2.3) where is some onstant and F(z, p) = (z + λ) 1 + p 2 is the Lagrangian assoiated with F. We have used instead of here to simplify things later. After a omputation, we find That is, u u 2 = 1 + u 2 u + λ. 1 + u 2 = 1 (u + λ). Taking the onstant = λ/ 1 + u () 2, whih it must be, we see several things. First of all, any solution of the first integral equation with

11 2.2. EXTREMAL GRAPHS 11 will give a solution of (2.2). It is possible to get a solution of (2.3) with the hoie =, but in this ase, we must take u λ =, and we must therefore have u 1 =. This is, indeed, not a solution of the Euler-Lagrange equation for F = V +λl, but this possibility represents the exeptional ase of Proposition 1.17 in the book of Buttazzo, Giaquinta, and Hildebrandt [1] in whih the onstraint is degenerate. In this ase, the solution u gives the shortest path between (, ) and (1, u 1 ) = (1, ) and is, therefore, a ritial point for the length funtional L providing the onstraint. When, we obtain from the first integral equation a global justifiation for our assumption u + λ. This is beause every solution of the Euler-Lagrange equation must be a solution of the first integral equation. Only the solution u in the ase u 1 = and l = 2 is exeptional. Finally, the first integral equation tells us something about the sign of u + λ beause 1 + u u () = 2 (u + λ). λ It follows that u+λ and λ must share the same sign, and under our, seemingly justified, assumption u () >, that sign is positive. Thus, we may proeed to solve either the Euler-Lagrange equation or the first integral equation under this assumption. Making the substitution v = (u+λ) 1 + u () 2 /λ, we find u = ± v 2 1 or λ 1 + u () 2v = ± v 2 1. It follows that 1 + osh v osh u () v() = ± 2 (x + 1), λ [ ] 1 + u () v = u () (u+λ) = osh ± 2 (x + 1) + osh v(), λ λ or [ λ 1 + u = λ+ 1 + u () osh u () 2 (x + 1) ± osh ] 1 + u 2 λ () 2.

12 12 CHAPTER 2. ANALYSIS This looks rather ompliated, but it does tell us that the extremals have the form of hyperboli osine urves. This also onfirms that the onstant from the first integral equation should be positive with < extremals orresponding to maximizers of the energy. Substituting the value of from the first integral equation and differentiating, we also see u = sinh ((x + 1)/ ± osh ) 1 + u () 2. This allows us to nominally loate the vertex or lowest point on the hyperboli osine urve whih ours for x = µ = osh (λ/). In terms of this parameter, the extremals may be written as ( ) x µ u = λ + osh. There are now three unknown parameters λ, µ, and, but the initial ondition u() = implies ( ) 1 + µ λ = osh and ( ) ( ) x µ 1 + µ u = osh osh. The other endpoint ondition takes the symmetri form ( ) ( ) 1 µ 1 + µ osh osh = u 1. Another equation we an use to determine the parameters and µ is given by the length onstraint L[u] = l. ( ) ( ) x µ x µ u = sinh and 1 + u 2 = osh 2. Therefore, L[u] = ( ) 1 + u 2 x µ dx = osh dx,

13 2.2. EXTREMAL GRAPHS 13 and writing down L[u] = l we are led to the fundamental symmetri system: ( ) ( ) 1 µ 1 + µ osh osh = u 1. (2.4) and ( ) ( ) 1 µ 1 + µ sinh + sinh = l. (2.5) In this symmetri form, it is possible to eliminate µ as follows: Square both equations and subtrat the first from the seond, noting l 2 u We get ( ) ( ) ( ) ( )] 1 µ 1 + µ 1 µ 1 + µ [ osh osh + 2 sinh sinh = l 2 u 2 1. That is, + osh ( ) ( ) ( ) = 1 + osh 2 + sinh 2 = l2 u That is, sinh ( ) 1 = 1 l 2 2 u 2 1 > 1. (2.6) In this way, we obtain a single transendental equation for. One an show sinh(1/) is monotone dereasing in for > and takes every value greater than 1. Let us verify the equivalent assertions for the funtion f(z) = sinh z/z. First of all if z ց, we have by L Hopital s rule sinh z lim zց z = lim zց osh z = 1 and sinh z lim zր z = lim zր osh z =. Also, f z osh z sinh z (z) =. z 2 Setting f 1 (z) = z osh z sinh z we see f 1 () = and f 1 (z) = z sinh z > for z >. In partiular, f 1 (z) > for z >, so f (z) > for z >. Also, lim f f 1 (z) = lim (z) zց zց 2z =. We have shown that f takes every value on [1, ) uniquely and has a welldefined inverse on that interval. Thus, we have a unique solution

14 14 CHAPTER 2. ANALYSIS Figure 2.2: sinh z/z and its inverse = 1 ). (1 f 2 l2 u 2 1 One we know >, we an expand (2.4) to see ( ) 1 ( µ ) 2 sinh sinh = u 1. Therefore, substituting from (2.6), ( µ = sinh u 1 l2 u 1 2 ) (2.7) This ompletes our initial analysis of the extremals for the problem of determining the shape of a hanging hain. We summarize our disussion as follows: Theorem 1. For eah length l > 2 and eah value u 1 R with u 2 1 < l 2 4, there are unique values > and µ satisfying ( ) 1 sinh = 1 ( ) l 2 2 u 2 1 and µ = sinh u 1 2 l2 u 1

15 2.3. MINIMALITY OF EXTREMALS 15 suh that ( ) ( ) x µ 1 + µ u(x) = osh osh, x 1 is the unique onvex extremal for the funtional F[u] = [u(x) + λ] 1 + u (x) 2 dx in A = {u C 1 [, 1] : u() =, u(1) = u 1 } subjet to the onstraint and with L[u] = 1 + u (x) 2 dx = l ( ) 1 + µ λ = osh. The uniqueness extends generally to the funtional F (or more preisely to λ) in the sense that for other values of λ, there is no extremal of the resulting funtional for the onstrained problem. In pratie, some analysis is required to set up the omputation approximating solutions of the transendental equation for. We inlude this analysis in the setion on omputation and now address the minimality of these extremals. Note 1. If and λ are allowed to be negative negative, one finds onave extremals (satisfying u < ) whose gravitational potential energy is greater than the onvex solutions. 2.3 Minimality of extremals We have established the existene of a unique (onvex) atenary extremal given by the graph of a funtion u C [, 1] and satisfying u() =, u(1) = u 1, and 1 + u 2 dx = l.

16 16 CHAPTER 2. ANALYSIS The funtion u satisfies ( ) ( ) x µ 1 + µ u(x) = osh osh where > is the unique solution of sinh(1/) = l 2 u 2 1 µ = sinh ( u 1 l2 u 1 2 ). (2.8) / 2 >, and We now wish to establish the following result. Theorem 2. The funtion u given in (2.8) is the unique minimizer of on subjet to V [u] = u 1 + u 2 dx A = {u C 1 [, 1] : u() =, u(1) = u 1 } L[u] = 1 + u 2 dx = l. A fundamental diffiulty in establishing this result is that the Lagrangian F(z, p) = (z + λ) 1 + p 2 assoiated with the augmented funtional F = V + λl where ( ) 1 + µ λ = osh > is not (always) onvex. Showing this is Problem 3 of Chapter 3 in the book [2] of Troutman. Following Troutman, we take the speial ase u 1 =. In this ase µ =, and the extremal is given by ( x u(x) = osh λ with λ = osh ) ( ) 1. On the other hand, the funtion u satisfies u A, and δf u [v]. Taking v = u, we have u + v = u and showing F is not onvex amounts to showing F[u ] F[u] <

17 2.3. MINIMALITY OF EXTREMALS 17 (under some irumstanes). In fat, F[u ] F[u] = λ dx (u + λ) 1 + u 2 dx ( ) 1 ( x ) = 2 osh osh 2 dx ( ) 1 = 2 osh [ ( ) ] 2x osh + 1 dx 2 ( ) ( ) 1 = 2 osh sinh [ ( ) ( ) ( ) ] osh sinh osh 1. Sine x sinh x as x ր, we see that for > small enough ( ) 1 sinh > 2, and F[u ] F[u] <. Realling that is determined by ( ) 1 sinh = 1 l 2 2 u 2 1 = l 2, we find nononvexity for hains of any length l > 4. In spite of this nononvexity, Troutman suggests a rephrasing of the problem whih leads to a muh stronger result than Theorem 2 above. The funtion u determines a parametri urve parameterized by arlength. This is given by the funtion x C 1 ([, l] R 2 ) by x(s) = (ξ(s), η(s)) where { [ ξ(s) = µ + sinh s sinh ( 1+µ η(s) = u(ξ(s)) = osh ( sinh [ s sinh ( 1+µ This parametri map x also satisfies x 1 and 2 = )] ξ (s) ds = )]) ( osh 1+µ ) (2.9). 1 η 2 ds. Now if we let x = (ξ, η) C 1 ([, l] R 2 ) be any parametri urve parameterized by arlength ( x 1) with x() = (, ) and x(l) = (1, u 1 ), then the potential energy expression V [u] = u 1 + u 2 dx

18 18 CHAPTER 2. ANALYSIS generalizes to V 1 [x] = η ds. To see this, we may again onsider a portion of hain of mass m j = ρ s j loated at a point x(s j ). The potential energy of this partiular setion of hain is approximately η ρg s j dy = ρgη s j. Summing over a partition of suh portions and taking the limit as the maximum length s j tends to zero (and dividing out by the onstant ρg as usual), we arrive at the expression for V 1 above. The following result treats these Figure 2.3: parametri hain shape: These shapes are also not required to satisfy ξ 1 though the one illustrated does. (Atually, this shape has length a little longer than the original atenary hain shape.) general parametri urves of length l onneting (, ) to (1, u 1 ) and asserts that the atenary graph extremal is the unique minimizer among suh urves. Theorem 3. The atenary graph satisfying (2.9) is the unique minimizer of on V 1 [x] = η ds B = {x C 1 ([, l] R 2 ) : x() = (, ), x(l) = (1, u 1 ), x 1}

19 2.3. MINIMALITY OF EXTREMALS 19 subjet to L 1 [x] = 1 η 2 ds = 2. Finally, we simplify the previous result slightly and prove something even more general. It will be noted that the funtionals appearing above only depend on the seond oordinate funtion of x, namely, η C 1 [, l]. Thus, it makes sense to extend their domains and rename them: and V 1 : C 1 [, l] R by V 1 [η] = η ds L 1 : {η C 1 [, l] : η (s) 1 for s l} R by L 1 [x] = We now state the main result. Theorem 4. The seond omponent of the parametri map defined in (2.9) is the unique minimizer of on subjet to V 1 [η] = η ds B = {x C 1 ([, l] R 2 ) : η() =, η(l) = u 1 } L 1 [η] = 1 η 2 ds = 2. Notie the absene of the ondition x 1 in the definition of B. Notie, furthermore, that the funtional L 1 is not (even) defined on all of B, but only on B 1 = {η B : η (s) 1 for s l}. Proof of Theorem 4: We show first that η from (2.9) is the unique minimizer of G[η] = (V 1 L 1 )[η] = [η ] 1 η 2 ds on B 1 (without onstraint). This follows from two fats 1 η 2 ds.

20 2 CHAPTER 2. ANALYSIS 1. The augmented funtional G = V 1 L 1 is stritly onvex on B 1 in the sense that G[η+v] G[η] δg η [v] whenever η, η+v B 1 with equality only if v. 2. The funtion η from (2.9) is an extremal for G, that is δg η [v] = whenever η + v B 1. If we an establish these two assertions, we may apply the following (easy) theorem on minimizing onvex funtionals: Theorem 5. If G is stritly onvex on B 1 and for every v C 1 [, 1] suh that η + v B 1 we have δg η [v] = then we have G[η ] G[η] for all η B 1. The strit onvexity does not follow immediately beause the augmented Lagrangian G(z, p) = z 1 p 2 is not stritly seond order onvex. We do have D 2 G = ( (1 p 2 ) 3/2 ). Therefore, for eah v C 1 [, l] suh that η + v B 1, we have G(η + v, η + v ) G(η, η ) = G z (η, η )v + G p (η, η )v + 2(1 p 2 )3/2v 2 G z (η, η )v + G p (η, η )v with equality only if v = (pointwise). Integrating this inequality G[η +v] G[η] = [G z (η, η )v+g p (η, η )v ] ds+ 2 v 2 (1 p 2 )3/2 ds δg η[v] with equality only if v. But if η + v B 1, then v() = v(l) =, so equality implies v. This establishes the strit onvexity of G. On the other hand, the Euler-Lagrange equation for G is ( η = η 2)

21 2.3. MINIMALITY OF EXTREMALS 21 where the derivatives are with respet to the arlength s. To ompute this for the funtion η from the arlength parameterization of the atenary we observe first that Therefore, s = ξ ( ) 1 + u 2 ξ µ dx = sinh + sinh dξ ds = 1 osh ( ). ξ µ ( 1 + µ Having made this observation/alulation we have from (2.9) η (s) = du dx (ξ)dξ ds = sinh ( ξ µ ) osh ( ). ξ µ Therefore, ( ) d η ds 1 η 2 = d ( sinh dx ( )) ξ µ x=ξ dξ ds = 1, and η is a C 2 lassial extremal for G. In partiular, δg η [v], and G[η + v] G[η] whenever η + v B 1 with equality only if v. The usual argument of Theorem 5 now applies. That is, it happens that L 1 [η] = 1 η 2 ds = 2, so for any v C 1 [, l] suh that η + v B and for whih L 1 [η + v] = 2, we have V 1 [η + v] L 1 [η + v] = G[η + v] G[η] = V 1 [η] L 1 [η] with equality only if v. Sine L 1 [η + v] = L 1 [η] = 2, we have V 1 [η + v] V 1 [η] with equality only if v. This establishes Theorem 4. Proof of Theorem 3: If x = ( ξ, η) B satisfies 1 η 2 ds = 2 ).

22 22 CHAPTER 2. ANALYSIS and x is the parametri atenary, then η B 1 B and satisfies L 1 [ η] = 2. Thus, by Theorem 4 V 1 [ x] = V 1 [ η] V 1 [η] = V 1 [x] with equality only if η η. We have, in partiular, V 1 [ x] V 1 [x] for all x B satisfying the onstraint L 1 [ x] = 2. In the ase of equality we have ξ = ± 1 η 2 and 2 = ξ ds = 1 η 2 ds. Sine η (s) = 1 for at most one arlength s, we onlude ξ = 1 η 2 and x x. Finally we prove the initial (and weakest) assertion. Proof of Theorem 2: If ũ A and L[ũ] = 1 ũ 2 dx = l, then the graph of ũ may be parameterized by arlength to give a parameterized urve x B satisfying the onstraint L 1 [ x] = 1 η 2 ds = 2. By Theorem 3, we know V 1 [ x] V 1 [x] with equality only if x = x. Changing variables, we find and V 1 [ x] = V 1 [x] = The result evidently follows. η ds = η ds = ũ 1 + ũ 2 dx = V [ũ] u 1 + u 2 dx = V [u]. Note 2. The onave extremals are loal maximizers.

23 Chapter 3 Computation 3.1 Approximating the onstant The key pratial diffiulty remaining in determining the shape of a hanging hain is finding an approximation for the solution of ( ) 1 sinh = 1 l u2 1. We reall the funtion f(z) = sinh z/z restrited to z > has a well defined inverse as indiated in Figure 2.2, and if we an obtain a simple (and adequately aurate) approximation for the funtion g(v) = f (v) defined for v <, then a root find algorithm may be used to quikly determine the value of. Let us briefly larify the meaning of and need for a simple and adequately aurate approximation. A standard root find algorithm, like Newton s method or (to be more speifi) Mathematia s FindRoot requires an initial guess. In order for the algorithm to return a value quikly and, in the presene of multiple solutions, a orret value, it is required that the initial guess be simple that is easy to alulate based on the value of v in terms of standard funtions and adequately aurate so that one does not find extraneous roots. In the ase at hand, in fat, the ommand FindRoot[ Sinh[x]/x == v, {x, startingx(v)}] (3.1) where {x, startingx(v)} represents the diretive to solve for x using the initial approximation startingx(v), ould produe a negative value if some are is not taken with the value of startingx(v); this is beause f(x) = 23

24 24 CHAPTER 3. COMPUTATION sinh x/x is an even funtion. We now desribe how to obtain a simple and adequately aurate approximation for the value g(v) where g = f. We will use the following notation to distinguish among deimal approximation values. We write f(a) b to mean b is a rough approximation for f(a), usually a simple and adequately aurate approximation. We use f(z) b to mean b is a numerially generated approximation aurate to a large (at least 5) deimal plaes. We start with the observation that lim vց1 g (v) = +. In order to get an aurate approximation for g = f near v = 1, we ompute the next derivative of f: f (z) = z2 f 1 2zf 1 z 4 Again, using L Hopital s rule = z2 sinh z 2z osh z + 2 sinh z z 3. lim f osh z (z) = lim zց zց 3 = 1 3. Thus, we have to leading order v 1+g(v) 2 /6 or g(v) g (v) = 6(v 1). This provides an approximation with tolerane deaying to.1 around v = 1.1. See Figure Figure 3.1: an initial approximation of f (v): The vertial line near v = 1.1 gives a tolerane for the approximation around.1. This approah may be applied with enter of expansion any z > for a sequene of suh values z 1, z 2,... In this way, we obtain a olletion

25 3.2. THE LAMBERT W FUNCTION 25 of approximations g j eah adequately aurate on some interval ontaining sinh(z j )/z j. The Taylor approximation for f at z = z j gives v = f(z) f(z j ) + f (z j )(z z j ) + f (z j ) 2 Solving for z g(v) we have (z z j ) 2. g(v) z j + f (z j ) + f (z j ) 2 2f (z j )(f(z j ) v). (3.2) f (z j ) Taking the first suh approximation entered at v 1 = 1.1 with z 1 = sinh(1.1)/ , we obtain an approximation g 1 (v) aurate to within.1 for 1.1 < v v 2 = 1.57; see Figure Figure 3.2: a seond approximation of f (v): The vertial line near v = 1.57 gives a tolerane for the approximation around.1. Before proeeding to further approximations in our sequene, we onsider approximation of g(v) when v is large. 3.2 The Lambert W funtion There is a standard speial funtion we an use to approximate g(v) = f (v) for large v. This is Lambert s W funtion known as ProdutLog in Mathematia. Roughly speaking, W is the inverse of the omplex valued funtion h(z) = ze z defined on C. More preisely, W is defined on the Riemann surfae W of h whih we now partially desribe. The basi formula for h(z) is h(x + iy) = e x [x os y y sin y + i(y osy + x sin y)].

26 26 CHAPTER 3. COMPUTATION The funtion ξ(y) = y oty is even and smooth when restrited to π < y < π (or any interval kπ < y < (k + 1)π). Also, ξ is inreasing when restrited to y < π and has an inreasing inverse η : [, ) [, π) whose graph gives the upper boundary of the region Ω on the left in Figure 3.3. The region Figure 3.3: a fundamental domain for ze z and its Riemann surfae (initial sheet) Ω = {z = x + iy : η (x) < y η (x)} is a fundamental domain for h with image W indiated on the right with a branh ut extending from /e along the negative real axis. The inverse mapping W : W C is the primary branh of the Lambert funtion. In partiular, the funtion W restrited to the real interval [/e, ) gives the inverse of h(x) = xe x for x. The graph of the restrition of h to the real line is indiated in Figure 3.4 Unfortunately, we need the inverse on the omplementary interval (, ] whih is not given by W. If we exit W along the image of the graph of η, whih is the top of the branh ut along the negative real axis, we enter the region W of the Riemann surfae on whih the inverse W is defined. A onvenient urve for referene within W is obtained by onsidering again the funtion ξ(y) = y ot y restrited to π < y < 2π. Sine ξ (y) = oty + y s 2 y = 1 ( osy y ), sin y sin y

27 3.2. THE LAMBERT W FUNCTION Figure 3.4: The restritions to the intervals (, ] and [, ) are invertible. and < sin y < 1 while y > π > 1 on the interval π < y < 2π, we see ξ is inreasing and takes all values in R. Thus, this funtion has in inverse η : R (π, 2π) whose graph is indiated above the graph of η in Figure 3.5. The region Figure 3.5: a portion of the fundamental region assoiated with W Ω = {z = x + iy : < y η(x)} {x R : x /e} again overs the plane. The upper half plane in the image is part of W as before, but the lower half plane is now part of W. The image of the segment

28 28 CHAPTER 3. COMPUTATION < x along the negative real axis is the segment /e w < in W. Thus, the inverse of the restrition of h(x) = xe x to (, ] is given by h (ξ) = W (ξ) for /e ξ <. Reall our funtion g(v) = f (v) where f(z) = sinh z/z. When z is real and signifiantly greater than z =, the value of f(z) is approximated by e z /(2z). Setting ξ = z, this beomes f(z) 1 2ξe ξ = 1 2h(ξ). That is, if v = f(z), then ( ξ h 1 ) ( = W 1 ), 2v 2v and ( g(v) = z h 1 ) ( = W 1 ). 2v 2v The approximation ( g (v) = W 1 ) 2v gives an approximation of g(v) within.1 for v 2.5; see Figure Figure 3.6: approximation of f (v) for v large: The vertial line near v = 2.5 gives a tolerane for the approximation around.1. Thus, we see our sequene of approximations g j (v) has a termination point when we an approximate on the interval [, 2.5], or (given the approximations g and g 1, when we an approximate for 1.57 < v < 2.5). In fat, It

29 3.3. ALTERNATIVE RECURSIVE APPROXIMATION 29 is adequate to obtain approximations g 2,...,g 6 orresponding to v 2 = 1.57, v 3 = 1.78, v 4 = 1.93, v 5 = 2.7, and v 6 = With the eight approximations, g,...,g, we obtain a fast and reliable implementation of (3.1). 3.3 Alternative reursive approximation The following was not used in our numerial solution of the equation sinh z/z = v, but it may be of independent interest. For v > 1, we also have reourse to a reursive approximation sheme. Piking an initial value g, for example, we ould take g = sinh (v) whih will be smaller than g(v) as long as v > sinh(1). We see the atual value g(v) satisfies sinh g(v) g(v) = v or g(v) = sinh (vg(v)). This suggests setting g 1 = sinh (vg ) and g j+1 = sinh (vg j ) in general for j =, 1, 2,.... Conjeture 1. The sequene g j tends (upward) to g(v) with the estimate g(v) g j+1 g j+1 g j. For example, if we take v = sinh[6]/6, then g = sinh (v) g 1 = sinh (g v) g 2 = sinh (g 1 v) g 3 = sinh (g 2 v) Numerial Code(s) We begin with a first setion in Mathematia obtaining a reliable numerial representation of the funtion g. As indiated in (3.2), expressions for f, f, and f are used in our initial approximations of g. With this in mind, we define

30 3 CHAPTER 3. COMPUTATION f[z ] = Sinh[z]/z; fp[z ] = D[Sinh[z]/z, z]; fpp[z ] = D[Sinh[z]/z, z, 2]; The undersore appearing in z indiates the delaration of a variable in Mathematia. As suggested in (3.1) we begin with a general form of the funtion whih remains unevaluted: ggeneral[vinput, vguess ] := FindRoot[Sinh[findz]/findz == vinput, {findz, vguess}][[1, 2]] The rypti [[1,2]] is required to extrat the numerial value of the Mathematia output from FindRoot whih has the form { findz value}. Literally, Take the first objet within the vetor (urly brakets) and take the seond entry value. We define our initial approximation near v = 1: g[w ] = Sqrt[6 (v-1)]; Setting our requirement for deviation from the atual inverse funtion g at g =.1, we determine the first of six approximation points as desribed in onnetion with Figure 3.1 above. The first value at whih a new approximation is required is v 1 = 1.1. The others are determined iteratively and are given by v 2 = 1.57, v 3 = 1.78, v 4 = 1.93, v 5 = 2.7, and v 6 = 2.24 as mentioned above. The ode determining v 1 is maxvalidity = FindRoot[ Sqrt[6(wfind - 1)] - ggeneral[wfind,sqrt[6(wfind-1)]] ==.1, {wfind, 1.2}][[1, 2]] > where we have indiated output by >. It may also be noted that we have taken the initial guess in ggeneral to be the value of the first approximation g. The guess forwfind in this evaluation is made from the plot in Figure 3.1. We then have six short subsetions in whih we determine the next six loal approximations. These have the following form: v1 = maxvalidity z1 = Sinh[1.1]/1.1 > g1[v ] = z1 + (-fp[z1] + Sqrt[fp[z1]^2-2 fpp[z1] (f[z1] - v)])/fpp[z1]

31 3.4. NUMERICAL CODE(S) 31 maxvalidity1 = FindRoot[g1[wfind] - ggeneral[wfind,g1[wfind]] ==.1, {wfind,1.5}][[1, 2]] > In eah ase, a plot similar to the one appearing in Figure 3.2 is required to make a rough guess for the last implementation of RootFind to determine the interval of validity (and the next point of evaluation). When these six subsetions are omplete, we define ginfinity[v ] = -ProdutLog[-1, -1/(2 v)] from whih we obtained minvalidity = FindRoot[gGeneral[wfind,g6[wfind]] - ginfinity[wfind] ==.1, {wfind, 2.3}][[1, 2]] > Finally, we define a global approximation using the Mathematia funtion Pieewise used to define pieewise funtions: gglobal[v ] = Pieewise[ Sqrt[6 (v - 1)], 1 < v < 1.1, g1[v], < v < 1.57, g2[v], < v < 1.78, g3[v], < v < 1.93, g4[v], < v < 2.7, g5[v], < v < 2.24, g6[v], < v < 2.5, ginf[v], v > ] Finally, we omplete this (main) setion by setting g[v ] := ggeneral[ v, gglobal[v] ] This ompletes the numerial approximation of f(z) = sinh(z)/z. When we have done this, equation (2.6) an be solved numerially and the relations (2.7) and (2.8) may be implemented to obtain a normalized hain shape defined for x 1: = 1/g[Sqrt[ell^2 - u1^2]/2]; mu = - ArSinh[u1/Sqrt[ell^2 - u1^2]]; unormalized[x ]= Cosh[(x - mu)/] - Cosh[(1 + mu)/])

32 32 CHAPTER 3. COMPUTATION It remains to take speifi physial oordinates, sale them to determine an appropriate l and u 1, and then sale the shape determined above to fit the physial problem. We inlude the oding for this in the disussion of the experiment below.

33 Chapter 4 Experiment 4.1 Relation to physial parameters Let us say we are given a hain of length L > and two points (A, B) and (C, D) in the plane with C > A and L 2 > (C A) 2 + (D B) 2. In our ode, we implement a test: The following numbers should be positive: C - A L^2 - (C - A)^2 - (D - B)^2 In normalized oordinates we obtain l = 2 C A L and u 1 = 2 (D B). C A Thus, if the test returns a positive result, we define u1 = 2 (D - B)/(C - A); ell = 2 L/( - a); We use these values to determine the parameters and µ aording to the analysis above. We have also the relation X = A + C A 2 (x + 1) or x = + 2 (X A). C A 33

34 34 CHAPTER 4. EXPERIMENT Substituting into the solution above, we obtain the physial hain shape is modeled by the graph of U(X) = B+ C A [ osh 1 ( ) ( )] µ (X A) 1 µ osh. 2 C A with the obvious implementation. 4.2 Some speifi shapes On the next page we produe several hain shapes (produed for a hain of length 2 feet) whih are saled down to fit the page. One an take any small piee of hain and test the results by omparing to the shapes on this page.

35 4.2. SOME SPECIFIC SHAPES Figure 4.1: several hain shapes to test with physial hains

36 36 CHAPTER 4. EXPERIMENT 4.3 Error Normally, we would inlude a setion on error analysis here, but the physial hains agree with the predited shapes so losely, that it is not lear how to measure any error. 4.4 Generalizations There are at least three obvious generalizations for this problem and two more that I will list as well. Some have been done as projets before. I will list them in (roughly) the order of inreasing diffiulty. Eah has something to do with hanging the density of the hain or the flexibility of the hain. 1. If two lengths of hain of different densities are joined and the resulting single hain with a pieewise onstant density hangs from two fixed points, then one an model the shape with a straightforward modifiation of the disussion above. 2. One ould onsider finitely many links, assumed to be straight line segments, hinged at the endpoints, and having presribed lengths l 1, l 2,...,l k. This leads to a very interesting disrete minimization. Questions onerning approximation of shapes for the other problems is an obvious related question. 3. A diret generalization of the first problem above is to onsider a general density ρ = ρ(s) as a smooth funtion of arlength. This is a muh more diffiult problem. It is also somewhat diffiult to easily obtain a large variety of subjet hains for experimentation and modeling, though the next two problems suggest some possibilities. 4. One an onsider attahing weights to various points on a hanging hain. If there are many links and the weights vary ontinuously, this an approximate some hains of hains of variable density. 5. One an inlude an elasti energy, whih is an essentially different kind of problem. Nevertheless, a thin beam supported at both sides and supporting its own weight in gravity will sag/hang in an interesting shape whih an be modeled.

37 Bibliography [1] G. Buttazzo, M. Giaquinta, and S. Hildebrandt. One-dimensional Variational Problems. Oxford, [2] J. Troutman. Variational Calulus and Optimal Control, Seond Edition. Springer,