It was conjectured by M. Glasser and S. Davison and later proved by A.
Eremenko that the certain animals should gather close to each other in order to
decrease the total heat loss. In this paper we show that it is not always true
for the individual heat loss. This gives a negative answer to a question posed
by A. Eremenko

Ivanisvili, Paata

English

arXiv

eScholarship - University of California

UC IrvineUC Irvine Previously Published WorksTitleA bundling problem revisitedPermalinkhttps://escholarship.org/uc/item/1js5q2rkAuthorIvanisvili, PaataPublication Date2016-02-02Copyright InformationThis work is made available under the terms of a Creative Commons Attribution License, available at https://creativecommons.org/licenses/by/4.0/ Peer reviewedeScholarship.org Powered by the California Digital LibraryUniversity of CaliforniaA BUNDLING PROBLEM REVISITEDPAATA IVANISVILIABSTRACT. It was conjectured by M. Glasser and S. Davison and later proved by A. Eremenko that the certainanimals should gather close to each other in order to decrease the total heat loss. In this paper we show that it isnot always true for the individual heat loss. This gives a negative answer to a question posed by A. Eremenko.1. MOTIVATION, MODEL AND RESULTSIn [2], [1] Glasser and Davison consider the following problem: let B1 and B2 are two disjoint balls in R3with equal radii. Let d ≥ 0 be a distance between the balls. Consider a harmonic function in the complementof the balls, i.e., R3 \B1∪B2, such that u|∂B j = 1 and lim|x|→∞ u(x) = 0. LetQ(d) =∫∂B1∂u∂ndσ +∫∂B2∂u∂ndσbe the heat flux where n is the outward unit normal vector to a sphere. Is it true that the quantity Q(d) isincreasing as the function of distance between the balls?The problem arises from the question why certain warm blooded animals like armadillos can keep eachother warm by huddling together. In this simple model the balls B1 and B2 represent uniform spherical animalsin R3 with body temperature 1 and medium temperature 0. The harmonic function u(x) represents timeindependent temperature in R3 \B1∪B2, and the quantity Q(d) represents the total heat loss of both animals:the total amount of heat given off by the animals as a function of distance between the balls. Presumablymoving animals closer together decreases the heat loss Q(d), and it was confirmed numerically in [2] that thequantity Q(d) is increasing. However no mathematical proof was given until A. Eremenko [3] gave a rigorousproof in more general setting. In Eremenko’s argument it was noticed that4πQ= inf{I(µ) : supp µ ⊂ ∂B1∪∂B2, µ(∂B1∪∂B2) = 1, µ ≥ 0}(1)whereI(µ) =∫ ∫ dµ(x)dµ(y)|x− y|.The monotonicity of Q follows from the fact that if ϕ : R3 → R3 is continuous, one-to-one, and |ϕ(x)−ϕ(y)| ≥ |x− y| then I(ϕ∗µ)≤ I(µ) where ϕ∗µ(A) = µ(ϕ−1(A)) for any Borel measurable A⊂ R3.Notice that each individual animal B j feels only his own heat loss Q j(d)Q j(d) =∫∂B j∂u∂ndσ ,but not the total Q = Q1 +Q2. Therefore the behavior of the animals we have discussed could be driven byindividual feelings but not the abstract “common goal”. In case of equal balls the individual heat loss Q jis monotonically increasing because Q1 = Q2 = Q2 . It is natural to think that the individual heat loss Q j ismonotonically increasing for the balls of different radii. In [3, 4] the following question was asked:Question. Are the quantities Q j(d), j = 1,2 monotonically increasing if the balls have different radii?2010 Mathematics Subject Classification. 31B05, 31B15, 31B20, 42B37, 26E05.Key words and phrases. Bundling problem, heat loss, bispherical coordinates, harmonic functions, capacity, hypergeometricfunctions, digamma function.1arXiv:1602.00983v1  [math.AP]  2 Feb 20162 PAATA IVANISVILIIn Section 2 we will show that if r j denotes the radius of the ball B j for j = 1,2, then the of heat loss Q1 isnot monotonically increasing provided that r1r2 > ` where `≈ 1.95 is the positive solution of the equation2(1+ x3)(γ +ψ(11+ x))+ x2 + x+2(x2− x)ψ ′(11+ x)= 0.(2)In the above notation γ is the Euler’s constant, and ψ is the digamma function.In other words, if armadillo A is at least twice as big as armadillo B, then A should keep some nonzerodistance from B in order to minimize the heat loss while B should try to be as close as possible to A.In Section 2 we obtain the following asymptotic expression for Q1(d) which implies the conclusion (2).Let r1 > r2 > 0, d ≥ 0, and let the temperature of the balls be constant and equal to T0 > 0. ThenQ1(d)4πT0r1=−r2(γ +ψ(r2r1+r2))r1 + r2−d6r1(r1 + r2)3[2(r31 + r32)(γ +ψ(r2r1 + r2))+ r21r2 + r22r1 +2(r21r2− r22r1)ψ ′(r2r1 + r2)]+o(d).as d→ 0.2. TWO BALLS OF UNEQUAL RADIIWe consider a bispherical coordinate system(x,y,z) =(asinη cosφcosh µ− cosη,asinη sinφcosh µ cosη,asin µcosh µ− cosη),where−∞≤ µ ≤∞, 0≤η ≤ π , 0≤ϕ ≤ 2π and a> 0 so that the foci F1 and F2 coincide with the centers of B1and B2. A general solution of the Laplace equation in bispherical coordinate system (under the assumptionsthat the solution does not depend on ϕ which is true in our case) is given by the expression (see page 1298,[6])T (µ,η) =√cosh µ− cosη∞∑n=0[Ane(n+12 )µ +Bne−(n+12 )µ]Pn(cosη),where Pn are Legandre polynomials. Let r1 > 0 and r2 > 0 be the radii of B1 and B2 correspondingly. Thecorresponding values of the µ coordinate µ = µ1 > 0 on ∂B1 and µ = µ2 < 0 on ∂B2 will be determined byr j = a|sinh µ j| for j = 1,2. Notice also that the distance d between the balls can be obtained as followsd + r1 + r2 = a(coth µ1− coth µ2) = r1 cosh µ1 + r2 cosh µ2.Assume that the temperature on ∂B j is a constant Tj, j = 1,2. Then by using the generating function forthe Legandre polynomials1√cosh µ− cosη=√2∞∑n=0e−(n+12 )|µ|Pn(cosη),(3)the boundary condition T (µ j,η) = Tj impliesAne(n+12 )µ j +Bne−(n+12 )µ j = Tj√2e−(n+12 )|µ j|, j = 1,2 n≥ 0.(4)Let us compute the surface element dσ in the bispherical coordinates:|(x,y,z)η × (x,y,z)φ |=a2 sinη(cosh µ− cosη)2.The heat loss Q1 can be computed as followsQ1 =∫∫∂B1∂T∂n∣∣∣∣µ=µ1a2 sinηdηdφ(cosh µ1− cosη)2.A BUNDLING PROBLEM REVISITED 3Notice that on ∂B1 we have∂T∂n=(cosh µ1− cosηa)∂T∂ µ=sinh µ12a√cosh µ1− cosη∞∑n=0[Ane(n+12 )µ1 +Bne−(n+12 )µ1]Pn(cosη)+(cosh µ1− cosη)3/2a∞∑n=0[(n+12)Ane(n+12 )µ1− (n+ 12)Bne−(n+12 )µ1]Pn(cosη) =sinh µ12aT (µ1,η)+(cosh µ1− cosη)3/2a∞∑n=0(n+12)[Ane(n+12 )µ1−Bne−(n+12 )µ1]Pn(cosη).(5)The boundary conditions (4) imply thatAne(n+12 )µ1−Bne−(n+12 )µ1 =√2e−(n+12 )µ1T1 + e−(n+12 )(3µ1−2µ2)T1−2T2e−(n+12 )(µ1−2µ2)1− e−(n+ 12 )(2µ1−2µ2)=∞∑k=0(T1√2e−(n+12 )(µ1+k(2µ1−2µ2))+T1√2e−(n+12 )(3µ1−2µ2+k(2µ1−2µ2))−T22√2e−(n+12 )(µ1−2µ2+k(2µ1−2µ2)))=T1√2e−(n+12 )µ1 +2√2∞∑k=0(T1e−(n+12 )(3µ1−2µ2+k(2µ1−2µ2))−T2e−(n+12 )(µ1−2µ2+k(2µ1−2µ2))).Thus (5) takes the following form∂T∂n=sinh µ12aT (µ1,η)+(cosh µ1− cosη)3/2aT1√2∞∑n=0(n+12)e−(n+12)µ1Pn(cosη)+(cosh µ1− cosη)3/2a·2√2T1∞∑n=0∞∑k=0(n+12)e−(n+12 )(3µ1−2µ2+k(2µ1−2µ2))Pn(cosη)− (cosh µ1− cosη)3/2a·2√2T2∞∑n=0∞∑k=0(n+12)e−(n+12 )(µ1−2µ2+k(2µ1−2µ2))Pn(cosη).If we differentiate (3) with respect to µ we obtain the following identity12sinh µ(cosh µ− cosη)−3/2 =√2∞∑n=0(n+12)e−(n+12 )µPn(cosη) for µ > 0.(6)Using (6) we further simplify the expression for ∂T∂n :∂T∂n=sinh µ12aT (µ1,η)+sinh µ12aT1+∞∑k=0T1asinh(3µ1−2µ2 + k(2µ1−2µ2)) ·(cosh µ1− cosηcosh(3µ1−2µ2 + k(2µ1−2µ2))− cosη)3/2−∞∑k=0T2asinh(µ1−2µ2 + k(2µ1−2µ2))(cosh µ1− cosηcosh(µ1−2µ2 + k(2µ1−2µ2))− cosη)3/24 PAATA IVANISVILIWe notice that T (µ1,η) = T1. Thensinh µ12a T (µ1,η)+sinh µ12a T1 =sinh µ1a T1 and finally we haveQ1 =∫π0[2πaT1 sinh µ1sinη(cosh µ1− cosη)2+∞∑k=0(2πaT1 sinh(3µ1−2µ2 + k(2µ1−2µ2)) ·(cosh µ1− cosη)−1/2 sinη(cosh(3µ1−2µ2 + k(2µ1−2µ2))− cosη)3/2−2πaT2 sinh(µ1−2µ2 + k(2µ1−2µ2)) ·(cosh µ1− cosη)−1/2 sinη(cosh(µ1−2µ2 + k(2µ1−2µ2))− cosη)3/2)]dη .By substituting cosη = x we obtainQ1 =∫ 1−1[2πaT1 sinh µ11(cosh µ1− x)2+∞∑k=0(2πaT1 sinh(3µ1−2µ2 + k(2µ1−2µ2)) ·(cosh µ1− x)−1/2(cosh(3µ1−2µ2 + k(2µ1−2µ2))− x)3/2−2πaT2 sinh(µ1−2µ2 + k(2µ1−2µ2)) ·(cosh µ1− x)−1/2(cosh(µ1−2µ2 + k(2µ1−2µ2))− x)3/2)]dx =I1 +∞∑k=0(I2,k + I3,k).Let us calculate each term separately. We remind that sinh µ12a T1 =T12r1. Therefore we haveI1 =4πaT1 sinh µ1sinh2 µ1= 4πT1r1.We notice the following subtle identitysinh(B)∫ 1−1(cosh(A)− x)−1/2dx(cosh(B)− x)3/2=2sinh(A+B2)(7)for all real numbers A and B whenever the both sides of (7) make sense. ThenI2,k =4πaT1sinh(2µ1−µ2 + k(µ1−µ2)),I3,k =−4πaT2sinh(µ1−µ2 + k(µ1−µ2)).Taking into account that r1 = asin µ1 we obtainQ1 = 4πT1r1 +4πa∞∑k=0T1sinh(2µ1−µ2 + k(µ1−µ2))− T2sinh(µ1−µ2 + k(µ1−µ2))=4πT1r1 +4πr1 sinh(µ1)∞∑k=1T1sinh(µ1 + k(µ1−µ2))− T2sinh(k(µ1−µ2)).Further we consider the case when T1 = T2 = T0 > 0. In order to investigate the monotonicity of Q1 withrespect to d, it is enough to investigate the monotonicity of the following functionf (d) := sinh(µ1)∞∑k=11sinh(µ1 + k(µ1−µ2))− 1sinh(k(µ1−µ2)).A BUNDLING PROBLEM REVISITED 5We notice thatcosh µ1 =(d + r1 + r2)2 + r21− r222(d + r1 + r2)r1and cosh µ2 =(d + r1 + r2)2 + r22− r212(d + r1 + r2)r2.Let x = eµ1 > 1 and y = e−µ2 > 1. Then f (d) takes the formf (d) =(x− 1x)∞∑k=1(x(xy)kx2(xy)2k−1− (xy)k(xy)2k−1)(8)wherex = 1+d(d +2r2)+√(d +2r1 +2r2)(d +2r2)(d +2r1)d2(d + r1 + r2)r1;(9)y = 1+d(d +2r1)+√(d +2r1 +2r2)(d +2r2)(d +2r1)d2(d + r1 + r2)r2.(10)By using the identity 1t−1 =1t ∑∞k=0 t−k two times for the terms inside the summation (8), and by Fubini’stheorem the expression for f (d) can be simplified as followsf (d) =(x− 1x)∞∑j=0x−2 j−1−1(xy)2 j+1−1.If r1 = r2 then, as we already mentioned in (1), it is known that f is monotonically increasing (see also aproof in [5] without resorting to (1)). Therefore it is enough to study the sign of limd→0+ f ′(d) for differentradii r1 and r2. Further we assume that r1 > r2. Let z = xy. Setf (d) =(x− 1x)∞∑j=0x−2 j−1−1z2 j+1−1=(x− 1x)∞∑k=0g(k),where g(s) = x−2s−1−1z2s+1−1 . Since x,z > 1 it is easy to see that g(s) ∈C∞([0,∞)) and all its derivatives tend to zeroas k→ ∞. Therefore By Euler–Maclaurin formula we have(x− 1x)∞∑k=0g(k) =(x− 1x)(∫∞0g(s)ds+g(0)2− g′(0)12)−(x− 1x)∫∞0B2({1− s})2g(2)(s)ds(11)where B2(x) = x2− x+ 12 is the Bernoulli polynomial, and {x} represents the fractional part of x.We will compute the asymptotic behavior of each term in (11) separately as t → 0. First notice that (9)impliesx− 1x= 2√2r2(r1 + r2)r1d1/2 +1√2r21− r1r2 + r22(r1 + r2)3/2√r2r3/21d3/2 +O(d2).(12)We have ∫∞0g(t)dt =∫∞0x−1e−2t lnx−1ze2t lnz−1dt =1lnz∫∞0x−1e−slnxlnz −1zes−1ds.We notice that ∫∞01zes−1ds = ln(zz−1).Therefore1lnz∫∞0x−1e−slnxlnz −1zes−1ds =−ln( zz−1)2lnz+12x lnz∫∞0e−slnxlnzzes−1ds.6 PAATA IVANISVILISet α = lnxlnz . Then∫∞0e−sαzes−1ds =∫∞1y−α[−1y+zzy−1]dy =− 1α+ z∫∞1y−αzy−1dy =− 1α+ x∫ 1/z0sα−11− sds =1z∞∑k=0(z−1)kα +1+ k=z−11+α 2F1(1,1+α;2+α;z−1),where2F1(a,b;c;z) =∞∑n=0(a)n(b)n(c)nznn!denotes the hypergeometric function where (a)n = a(a+1) · · ·(a+n−1) if n≥ 1 and (a)0 = 1. On the otherhand it is known that (see [7])2F1(a,b;a+b;v) =Γ(a+b)Γ(a)Γ(b)(∞∑k=0(a)k(b)kk!2(− ln(1− v)+2ψ(k+1)−ψ(a+ k)−ψ(b+ k))(1− v)k)for all 0 < 1− v < 1, where ψ is the digamma function. Therefore we obtain12x lnz∫∞0e−slnxlnzzes−1ds =12xz lnz2F1(1,1+α;2+α;z−1)1+α=12xz lnz∞∑k=0((1+α)kk!(− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k).Note that when d→ 0+ we have α = r2r1+r2 +O(√d), 1− z−1 =√2(r1+r2)r1r2d1/2 +O(d) and ln(1− z−1)(1−z−1)3 = O(d3/2 lnd). It is known that ψ(t) = ln(t)+O(1/t) for t→∞. Therefore if d is sufficiently small wehave α < 1 and thus for k ≥ 3 we obtain∣∣∣∣((1+α)kk! (− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k)∣∣∣∣≤ 10kb(r1,r2)| lnd|(c(r1,r2)d1/2)k,where b(r1,r2)> 0 and c(r1,r2)> 0 are some finite numbers depending on r1 and r2. Therefore for sufficientlysmall d we have∣∣∣∣∣ ∞∑k=3((1+α)kk!(− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k)∣∣∣∣∣≤10b(r1,r2)| lnd|∞∑k=3k(c(r1,r2)d1/2)k ≤ 100b(r1,r2)| lnd|c(r1,r2)3d3/2 = O(d3/2 lnd).We obtain∞∑k=0((1+α)kk!(− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k)=2∑k=0((1+α)kk!(− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k)+O(d3/2 lnd).A BUNDLING PROBLEM REVISITED 7Notice that (x− 1x )12xz lnz =r2r1+r2+O(√d). Thus we obtain(x− 1x)[∫∞0g(s)ds+g(0)2− g′(0)12]=(x− 1x)[g(0)2− g′(0)12−ln( zz−1)2lnz+12xz lnz(O(d3/2 lnd)+2∑k=0((1+α)kk!(− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k))]=(x− 1x)[x−1−12(z−1)+z lnx− lnx+ z lnz− zx lnz6x(z−1)2−ln( zz−1)2lnz+12xz lnz2∑k=0(1+α)kk!(− ln(1− z−1)+ψ(k+1)−ψ(1+ k+α))(1− z−1)k]+O(d3/2 lnd)We note that ψ(1) = −γ , then one can check that when d→ 0, after some routine computations, using (9),(10) and the identity ψ(x+1) = ψ(x)+ 1x several times, the above expression takes the following form−r2(γ +ψ(2r2+r1r1+r2))r1 + r2−d6r1(r1 + r2)3[(r31 + r32)(2γ +2ψ(r2r1 + r2))+ r21r2 + r22r1 +2(r21r2− r22r1)ψ ′(r2r1 + r2)]+o(d).We are left with showing that the term(x− 1x)∫∞0B2({1−s})2 g(2)(s)ds in (11) behaves as o(d) for sufficientlysmall d. Since x− 1x = O(√d) it is enough to show that∫∞0B2({1−s})2 g(2)(s)ds = o(√d). We have∣∣∣∣∫ ∞0 B2({1− s})g(2)(s)ds∣∣∣∣= 13∣∣∣∣∫ ∞0 B3({1− s})g(3)(s)∣∣∣∣≤ ∫ ∞0 |g(3)|ds(13)where B3(x) = x3− 32 x2 + 12 x is Bernoulli polynomial. Consider a function f (s) =esp−1es−1 where p = −lnxlnxy .Clearly f ((2s+1) lnxy) = g(s). We need the following technical lemma:Lemma 1. Let −1≤ p≤ 0. Then f (3)(s)≥ 0 for all s≥ 0.Before we proceed to the proof of the lemma we will show how the desired estimate follows from thelemma. First notice that p = − lnxlnxy ∈ [−1,0] because x,y > 1. Therefore sgn(g(3)) = sgn( f (3)) > 0 for alls > 0, and we have ∫∞0|g(3)|ds =−g′′(0).After some straightforward computations one can show that g′′(0) = O(d) as d→ 0. We will omit the detailsof the unnecessary computations.Finally, we obtain thatQ1(d)4πT0r1= 1−r2(γ +ψ(2r2+r1r1+r2))r1 + r2−d6r1(r1 + r2)3[(r31 + r32)(2γ +2ψ(r2r1 + r2))+ r21r2 + r22r1 +2(r21r2− r22r1)ψ ′(r2r1 + r2)]+o(d) =−r2(γ +ψ(r2r1+r2))r1 + r2−d6r1(r1 + r2)3[(r31 + r32)(2γ +2ψ(r2r1 + r2))+ r21r2 + r22r1 +2(r21r2− r22r1)ψ ′(r2r1 + r2)]+o(d)8 PAATA IVANISVILIwhere T0 = T1 = T2 is the temperature of the balls.It remains to prove the technical lemma.Proof. Notice thatf (3)(s) =es(p+3)(p−1)3 + es(p+2)(−3p3 +6p2−4)+ es(p+1)(3p3−3p2−3p−1)− p3esp + e3s +4e2s + es(es−1)4(14)It is enough to show that the coefficient ak of skk! for k ≥ 0 of numerator in (14) is nonnegative. Indeedak = (p+3)k(p−1)3 +(p+2)k(−3p3 +6p2−4)+(p+1)k(3p3−3p2−3p−1)− pk+3 +3k +4 ·2k +1 =J1,k(p)+ J2,k(p)+ J3,k(p),whereJ1,k = (p+1)k(3p3−3p2−3p−1)+1;J2,k = (p+2)k(−3p3 +6p2)− pk+3J3,k = (p+3)k(p−1)3−4(p+2)k +3k +4 ·2k;We notice that a0 = a1 = a2 = a3 = 0, a4 = 6p2(p−1)2 ≥ 0 and a5 = 4p(p−1)(6p2(p+1)−14p+1)≥ 0because of the assumptions on p. It is enough to show that J1,k,J2,k and J3,k are nonnegative for all k ≥ 6.Indeed, since p ∈ [−1,0] we haveJ1,k(p) = (p+1)k(3p3−3p2−3p−1)+1≥ (p+1)(3p3−3p2−3p−1)+1 = p(3p3−6p−4)≥ 0For J2,k if k is even then−pk+3 ≥ 0 and there is nothing to prove. Assume that k = 2m+1 where m≥ 3. ThenJ2,k = (p+2)2m+1(−3p3 +6p2)− p2m+4 ≥ (p+2)(−3p3 +6p2)− p4 = 4p2(3− p2)≥ 0It remains to show that J3,k(p)≥ 0. We have J3,k(0) = 0, andJ′3,k(p) = k(p+3)k−1(p−1)3 +3(p+3)k(p−1)2−4k(p+2)k−1 =(k+3)(p+2)k−1[(1+1p+2)k−1(p−1)2(p+ 9− k3+ k)− 4kk+3].If k≥ 9 there is nothing to prove. We assume k = 6,7 and 8. In this case the only interesting situation is whenk−93+k ≤ p≤ 0. Then(1+1p+2)k−1(p−1)2(p+ 9− k3+ k)− 4kk+3≤(32)k−1(p−1)2(p+ 9− k3+ k)− 4kk+3=(32)k−1[p3− 3(k−1)k+3p2 +3(k−5)k+3p+1k+3(9− k−4k(23)k−1)]≤ 0.The last inequality follows because the signs of the coefficients of the polynomial alternate, and p≤ 0.3. CONCLUSIONSIt follows from the previous section that if r1 > `r2 (where ` ≈ 1.95 is a positive solution of (2)) then theheat loss of the big ball, i.e., Q1(d) is decreasing when d ∈ [0,ε) where ε > 0 is sufficiently small. Noticethat when d→ ∞ we have x≈ dr1 , and y≈dr2therefore limd→∞ f (d) = 0 and thusQ1(∞) = 4πT0r1 >−4πT0r1(r2(γ +ψ( r2r1+r2 ))r1 + r2)= Q1(0).A BUNDLING PROBLEM REVISITED 9FIGURE 1. The heat loss Q1(d) of the big ball. T0 = 1,r1 = 20,r2 = 1, 0≤ d ≤ 80The last inequality is justified because 1>−x(γ+ψ(x)) for x∈ (0,1) follows from the fact that ψ(x+1)>−γfor x ∈ (0,∞). So there exists a minimal value of Q1(d) on [0,∞), i.e., a nonzero distance when the heat lossof the big ball is minimal.The numerical computations show that, in fact Q1(d) is decreasing on the interval [0,c) and then it isincreasing on (c,∞) where c ≈ r1. The heat loss of the small ball Q2(d) is always increasing for d ≥ 0.Figure 1 represents the graph of Q1(d) where T0 = 1,r1 = 20,r2 = 1 and 0≤ t ≤ 80.Acknowledgments. The author thanks Alexandre Eremenko, Benjamin Jaye and Fedor Nazarov for helpfuldiscussions and suggestions that led to simplifications of some proofs.REFERENCES[1] M. L. Glasser, Problem 77-5, SIAM review, Vol. 19, no. 1 (1977), 148.[2] M. L. Glasser and S. .G. Davison, A bundling problem, SIAM review, Vol. 20, no. 1 (1978), 178–180.[3] A. Eremenko, Sleeping armadillos and Dirichlet’s Principle, preprint,http://www.math.purdue.edu/~eremenko/dvi/armadillo.pdf[4] A. Eremenko (http://mathoverflow.net/users/25510/alexandre-eremenko), A problem of potentialtheory arising in biology, URL (version: 2015-09-06): http://mathoverflow.net/q/217530[5] F. Petrov (http://mathoverflow.net/users/4312/fedor-petrov), Curious inequality, URL(version: 2015-09-09): http://mathoverflow.net/q/217784[6] P. H. Morse, H. Feshbach, Methods of Theoretical Physics, McGraw-Hill, New York, 1953.[7] A. Gil, J. Segura, and N. M. Temme, Numerical Methods for Special Functions, SIAM, Philadelphia, 2007.DEPARTMENT OF MATHEMATICS, KENT STATE UNIVERSITY, KENT, OH 44240, USAE-mail address: ivanishvili.paata@gmail.com

A bundling problem revisited

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