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UNIVERSITY OF MARIBOR
FACULTY OF NATURAL SCIENCES AND
MATHEMATICS
DEPARTMENT OF MATHEMATICS AND
COMPUTER SCIENCE
GRADUATION THESIS
Tja²a Hrovati£
Maribor, 2013
UNIVERSITY OF MARIBOR
FACULTY OF NATURAL SCIENCES AND
MATHEMATICS
DEPARTMENT OF MATHEMATICS AND
COMPUTER SCIENCE
Graduation Thesis
THE TRUE CATENARY
Advisor: Candidate:
Prof. Dr. Andreas M. Hinz Tja²a Hrovati£
Co-advisor:
Prof. Dr. Uro² Milutinovi¢
Maribor, 2013
UNIVERZA V MARIBORU
FAKULTETA ZA NARAVOSLOVJE IN
MATEMATIKO
ODDELEK ZA MATEMATIKO IN RA�UNALNI�TVO
Diplomsko delo
PRAVA VERI�NICA
Mentor: Kandidatka:
prof. dr. Andreas M. Hinz Tja²a Hrovati£
Somentor:
prof. dr. Uro² Milutinovi¢
Maribor, 2013
Acknowledgements
I reserve the most special thank for my advisor, dr. Andreas M. Hinz. He en-
couraged me to look at the problem of the true catenary and patiently guided
me through.
Additionally I would like to thank my co-advisor dr. Uro² Milutinovi¢, who
helped me with my work after coming back to Slovenia, motivated and en-
couraged me to continue writing in English.
My family and friends deserve the most thanks for supporting me during
my studying and for giving me the chance to explore the thesis abroad.
UNIVERZA V MARIBORU
FAKULTETA ZA NARAVOSLOVJE IN
MATEMATIKO
Izjava
Podpisana Tja²a Hrovati£, rojena 19. maja 1989, ²tudentka Fakultete za nara-
voslovje in matematiko Univerze v Mariboru, ²tudijskega programa Matem-
atika, izjavljam, da je diplomsko delo z naslovom
THE TRUE CATENARY
pri mentorju prof. dr. Andreasu M. Hinzu in somentorju prof. dr. Uro²u Mi-
lutinovi¢u avtorsko delo. V diplomskem delu so uporabljeni viri in literatura
korektno navedeni; teksti niso uporabljeni brez navedbe avtorjev.
Maribor, 4. september 2013
Diploma Thesis Programme
Title: The True Catenary
In your diploma thesis solve the problem of the catenary in the central grav-
itational �eld (i.e. in the −1/r potential).As the main sources use the following articles:
1. J. Denzler, A. M. Hinz, Catenaria vera � the true catenary, Exposition.
Math. 17 (1999), 117�142.
2. M. Razpet, Prava simetri£na veriºnica, Obzornik mat. �z. 57 (2010),
121�133.
Maribor, April 8th, 2013
Advisor: Andreas M. Hinz
Co-advisor: Uro² Milutinovi¢
Program diplomskega dela
Naslov: Prava veriºnica
V diplomskem delu re²ite problem veriºnice v sredi²£nem gravitacijskem polju
(oz. v −1/r potencialu).Kot glavni vir uporabite £lanka:
1. J. Denzler, A. M. Hinz, Catenaria vera � the true catenary, Exposition.
Math. 17 (1999), 117�142.
2. M. Razpet, Prava simetri£na veriºnica, Obzornik mat. �z. 57 (2010),
121�133.
Maribor, 8. april 2013
Mentor: Andreas M. Hinz
Somentor: Uro² Milutinovi¢
HROVATI�, T.: The true catenary
Graduation Thesis, University of Maribor, Faculty of Natural Sci-
ences and Mathematics, Department of Mathematics and Com-
puter Science, 2013
Abstract
In this thesis we introduce the problem of the ideal homogeneous hanging
cable called the catenary. We observe the behaviour of the shape of the
curve. Firstly, we solve the problem of the classical catenary on a �at Earth,
where the gravitational �eld is constant and perpendicular to the ground.
Secondly, we focus on the true symmetric catenary in the central gravita-
tional �eld, which comes from the −1/r potential. In both cases we use the
method of calculus of variations for isoperimetric problems and in particular
the Euler-Lagrange di�erential equation. Lastly, we explain the problem of
the asymmetric case.
Key words: catenary, calculus of variations, Euler-Lagrange equation, curva-
ture, potential energy, di�erential equation.
Math. Subj. Class. (2010): 35A15, 35Q31
HROVATI�, T.: Prava veriºnica
Diplomsko delo, Univerza v Mariboru, Fakulteta za naravoslovje in
matematiko, Oddelek za matematiko in ra£unalni²tvo, 2013
Povzetek
V tem diplomskem delu predstavimo problem idealne vise£e homogene vrvi,
ki jo imenujemo veriºnica in opazujemo obna²anje oblike te krivulje. Najprej
re²imo problem klasi£ne veriºnice na ravni zemlji, kjer je gravitacijsko polje
konstantno in pravokotno na podlago, nato pa se osredoto£imo na pravo veri-
ºnico v sredi²£nem gravitacijskem polju, ki je dobljeno iz −1/r potenciala.
V obeh primerih uporabimo metodo variacijskega ra£una za izoperimetri£ni
problem, ²e posebej Euler-Lagrangeevo diferencialno ena£bo. Nazadnje pa
razloºimo ²e problem asimetri£ne veriºnice.
Klju£ne besede: veriºnica, variacijski ra£un, Euler-Lagrangeeva ena£ba, ukri-
vljenost, potencialna energija, diferencialna ena£ba.
Math. Subj. Class. (2010): 35A15, 35Q31
Raz²irjen povzetek
V diplomskem delu obravnavamo problem prave veriºnice. Glavna razlika
med klasi£no in pravo veriºnico je, da je prva de�nirana na ravni Zemlji,
kjer je gravitacijsko polje konstantno in pravokotno na podlago, slednja pa
je v sredi²£nem gravitacijskem polju, ki je dobljeno iz −1/r potenciala. Pri
izra£unu splo²nih formul obeh je potrebno poznati osnove variacijskega ra£una,
Euler-Lagrangeevo diferencialno ena£bo in izoperimetri£ni problem.
Problem klasi£ne veriºnice je de�niran in opisan s potencialno energijo, ki je
F (y) = gρ∫ x2x1y√
1 + y′2dx in pogoji P (y) =∫ x2x1
√1 + y′2dx = `, y(x1) = y1
in y(x2) = y2, kjer je g teºni pospe²ek, ρ je gostota vrvi, ` pa predstavlja
dolºino. I²£emo funkcijo y(x), ki glede na zgornje pogoje minimizira F (y).
Uporaba Euler-Lagrangeeve ena£be na funkciji f(x, y, y′) = y√
1 + y′2 +
λ√1 + y′2 za izoperimetri£ni problem, kjer je λ parameter, nas pripelje do
splo²ne re²itve y(x) = c ch
(x−Dc
)+E, kjer so c,D in E konstante, ki so v
nadaljevanju poglavja izra£unane.
Pri pravi simetri£ni veriºnici pa zaradi laºjega ra£unanja uvedemo polarne
koordinate in zato i²£emo minimum funkcionala F (r) = −∫ α−α
1
r
√r2 + r′2dϕ
pri pogojih P (r) =∫ α−α
√r2 + r′2dϕ = 2` in r(±α) = r1. Upo²tevamo pa
tudi naravno omejitev r1 sin(α) < ` < r1. Najprej dokaºemo osnovne last-
nosti kot so: lega krivulje v ravnini, konveksnost, kot ϕ = 0 predstavlja
minimum in obstoj maksimalno enega minimuma. Problema se lahko lotimo
na dva razli£na na£ina, kjer pri obeh upo²tevamo predznak ukrivljenosti in s
tem ugotovimo, da za pozitivne c iz Euler-Lagrangeeve ena£be za funkcional
f(r, r′, ϕ) = −1
r
√r2 + r′2 dobimo vise£e krivulje na katere se tudi osredo-
to£imo. V primeru nepozitivnih c pa bi dobili lok ali pa logaritmi£no spi-
ralo. Pri re²evanju po [2] v dobljeno diferencialno ena£bo uvedemo novo
spremnljivko λr =1
uin jo s tem poenostavimo v diferencialno ena£bo drugega
reda s konstantnimi koe�cienti, ki ima obliko u′′ +
(c2 − 1
c2
)u = − 1
c2. Pri
re²evanju upo²tevamo, da je minimum krivulje pri kotu ϕ = 0, saj imamo
simetri£no krivuljo. Da dobimo kon£no re²itvev, vstavimo ²e pogoje dolºine
in robni to£ki. To pa nas privede do r(ϕ) = r1
1− c ch(√
1− c2αc
)1− c ch
(√1− c2ϕc
) , kjer je
pozitiven c dolo£en z`
r1=
c sh
(√1− c2αc
)√1− c2
. Ni pa pomembno ali je vrednost
c ve£ja od 1, enaka 1 ali pa leºi med 0 in 1, saj lahko funkcijo hiperboli£nega
kosinusa v primeru negativnega korena enostavno pretvorimo v kosinus, saj
je hiperboli£ni kosinus de�niran na C. Pri drugem na£inu re²evanja iz [5]
pa najprej uporabimo orientiran kot normalnega vektorja na krivuljo in tako
preoblikujemo r′ v r′ = t tg(ϑ). Tudi v tem primeru uporabimo ukrivljenost
krivulje glede na izhodi²£no to£ko O, nato uvedemo novo neznako τ = tg
(θ
2
)in s tem poenostavimo zapis. Re²iti moramo tri razli£ne integrale s katerimi
izrazimo ϕ, in sicer za 0 < c < 1, c = 1 in c > 1, saj bi v primeru, kjer ne
bi lo£ili integralov glede na vrednost c imeli v integralu korene iz negativnih
²tevil. Z uporabo pogojev pridemo do enake ena£be kot pri prvem re²evanju,
le postopek je malo dalj²i. Pokazali smo, da £e re²itev Euler-Lagrangeeve
ena£be obstaja, potem je to prava veriºnica.
V predzadnjem poglavju pa obravnavamo ²e asimetri£no veriºnico, kjer to£ki,
v katerih krivuljo �ksiramo, nista na enaki vi²ini. S tem se spremenita robni
to£ki integrala. Problem je zato de�niran kot F (r) = −∫ ϕ2
ϕ1
1
r
√r2 + r′2dϕ
pri slede£ih pogojih P (r) =∫ ϕ2
ϕ1
√r2 + r′2dϕ = 2`, r(ϕ1) = r1 in r(ϕ2) = r2.
Naravna �zikalna omejitev je
(r2 − r1
2
)2
+ r1r2(sin(α))2 < `2 <
(r2 + r1
2
)2
,
ki vodi do omejitve za pravo simetri£no veriºnico ko je r1 = r2. Postopek
re²evanja Euler-Lagrangeeve ena£be je enak kot pri simetri£ni veriºnici, raz-
lika je v tem, da ne poznamo kota ϕ0, kjer ima krivulja minimum. Re²itev
u(ϕ) =
1− c ch(√
1− c2(ϕ− ϕ0)
c
)1− c2
dopolnimo z uporabo pogojev, nato
uvedemo nove neznake, ki nam pogoje poenostavijo in tako dobimo determi-
nanto, ki je enaka ni£. Ko preverimo vse moºne re²itve le-te, lahko izra£unamo
ϕ0. Primer, kjer je c = 1, obravnavamo lo£eno.
Contents
1 Introduction 15
2 Basic notation and results 16
3 The classical catenary 22
4 The problem of the true symmetric catenary 29
5 Solving the problem 37
5.1 The �rst way of solving the problem . . . . . . . . . . . . . . 39
5.2 The second way of solving the problem . . . . . . . . . . . . . 45
6 Asymmetric case 57
7 Conclusion and outlook 68
8 References 70
The true catenary
1 Introduction
This thesis consists of seven chapters. In the second chapter we introduce the
basic notation and the meaning of calculus of variations, which is a method
that leads to the Euler-Lagrange di�erential equation, and of the curvature,
which is also used several times in this thesis.
In the third chapter one can �nd the problem and the result of the classical
catenary on the �at Earth. In the forth and �fth chapter there is described a
problem of the true symmetric catenary in the gravitational �eld with −1/rpotential and two di�erent methods of solving it. Firstly, from [2] we use a
trick to make a di�erential equation easier to solve. In the second [5] there
are some di�erential geometric methods which lead to the same result. In
the penultimate chapter we focus on the true asymmetric catenary where we
have to use some other methods to come to the end of calculation. Finally,
the last chapter contains basic conclusions and summaries of the results.
15
The true catenary
2 Basic notation and results
In this chapter we introduce basic terms and formulas which are necessary for
understanding the work in this thesis.
The most important part of it is calculus of variations where we look for
the curve, which minimises the quantity, which depends on the entire curve
such as arc length, surface area or time of descent.
Points A and B have coordinates A(a, y1) and B(b, y2), we consider a fam-
ily of functions y = y(x) on [a, b] which are continuous, have derivatives of
the second order and satisfy boundary conditions y(a) = y1 and y(b) = y2.
We want to �nd a function in this family that minimizes or maximizes an
integral I(y) =∫ baf(x, y(x), y′(x))dx. I is called a functional, de�ned on
F = {y : [a, b] → R | y(a) = y1, y(b) = y2, y has continuous derivatives of
second order}. We present only the case when we try to �nd a minimizing
function, so we seek for a function y0 ∈ F, for which
∀y ∈ F, I(y0) =
∫ b
a
f(x, y0(x), y′0(x))dx ≤
∫ b
a
f(x, y(x), y′(x)) = I(y).
We write I(y) =∫ baf(x, y(x), y′(x))dx shorter as I(y) =
∫ baf(x, y, y′)dx,
where y = y(x) is a function of x. We always assume that the function
given by f(x, y, y′) has continuous partial derivatives of the second order with
respect to x, y and y′. In analysis we would usually write f(x1, x2, x3) where
f has continuous derivatives of the second order with respect to x1, x2 and
x3. For a functional I(y) we can form the Euler-Lagrange equation [3, pp.
99-101], [8, pp. 505-508], which is
∂f
∂y(x, y, y′)− d
dx
(∂f
∂y′(x, y, y′)
)= 0.
16
The true catenary
But we use a shorter notation
∂f
∂y− d
dx
(∂f
∂y′
)= 0. (2.1)
It means that, if y0 ∈ F is a solution of this variational problem, then(∂f
∂y− d
dx
(∂f
∂y′
))(x, y0, y
′0) = 0
for all x ∈ [a, b]. The meaning of this is
∂f
∂x2(x, y0(x), y
′0(x))−
d
dx
(∂f
x3(x, y0(x), y
′0(x))
)= 0.
This is why we search among the solutions of Euler-Lagrange di�erential
equations for the solution of variational problems. These solutions are called
extremals. We have to consider that∂f
∂yand
∂f
∂y′are computed by treating
x, y and y′ as independent variables and that y and y′ are functions of x. In
the next paragraphs we introduce the use of making a derivative by y and
whatd
dx
∂f
∂y′actually means.
During the calculation we use the Chain rule [4, pp. 88-89].
Theorem 2.1 Chain rule
Let f be the function which is de�ned and di�erentiable on an open set U ⊂Rn. Let C be a di�erentiable curve (de�ned for some interval of numbers t)
such that values C(t) = (x1(t), x2(t), ...xn(t)) lie in the open set U . Then the
function f(C(t)) is di�erentiable (as a function of t) and
df(C(t))
dt=
∂f
∂x1
dx1dt
+ ...+∂f
∂xn
dxndt
.
We use C(x) = (x, y, y′), n = 3. Thus
d
dx
∂f
∂y′=
∂2f
∂x∂y′(x, y, y′) +
∂2f
∂y∂y′(x, y, y′)y′ +
∂2f
∂y′2y′′. (2.2)
17
The true catenary
In standard notation it would look like
d
dx
∂f
∂x3(x, y(x), y′(x))
=∂2f
∂x1∂x3(x, y(x), y′(x)) +
∂2f
∂x2∂x3(x, y(x), y′(x))y′(x)+
+∂2f
∂x3∂x3(x, y(x), y′(x))y′′(x),
by using the Chain rule for y(x) ∈ F , n = 3 and C(x) = (x, y(x), y′(x)),
where we make derivatives using variables x1, x2 and x3.
In this thesis there is a situation where x is missing from the function f(x, y, y′) =
g(y, y′). In that case the Euler-Lagrange equation can be integrated into
f − ∂f
∂y′y′ = c, (2.3)
with some c ∈ R. This follows from the identity
d
dx
(∂f
∂y′y′ − f
)= y′
[d
dx
(∂f
∂y′
)− ∂f
∂y
]− ∂f
∂x,
since∂f
∂x= 0 and the expression on the right in brackets is also zero by the
Euler-Lagrange equation. The procedure can be found in [8, pp. 105-110],
[11, pp. 255-258] and [12, pp. 131-139].
In calculus of variation there is an ancient Greek problem called isoperimetric
problem, where we have to �nd a closed planar curve of given length that
encloses the largest area. This can be reduced to �nding extremals for one
functional where a second functional takes a prescribed value. The catenary
problem is an isoperimetric problem. As before we have an integral
I(y) =
∫ b
a
f(x, y, y′)dx
18
The true catenary
and we seek a di�erential equation satis�ed by y(x) that gives a stationary
value of the integral I(y). There are also associated conditions
J(y) =
∫ b
a
g(x, y, y′)dx = c, y(a) = y1, y(b) = y2.
From here we can compose the functional
I + λJ =
∫ b
a
(f(x, y, y′) + λg(x, y, y′))dx, (2.4)
where λ is a parameter called Lagrange multiplier. The method of Lagrange
multipliers is a method which removes the side conditions by introducing a
new variable λ [8, pp. 515-157]. The method of solving the isoperimetrical
problem is the same as the usual extreme using the Euler-Lagrange di�eren-
tial equation (2.1), the derivation is explained in [3, pp. 107-109] and [8, pp.
517-519].
If the function y(x) is the solution of the Euler-Lagrange equation, it does not
mean that the function y(x) is really an extremal. In this thesis we are not
going to solve the problem, whether the solution really minimizes the func-
tional. This either follows from physical considerations or has to be attacked
by some deep mathematics.
More details, derivations and proofs about Euler's di�erential equation, La-
grange multipliers and isoperimetric problems can be found in [3, pp. 95-
114],[8, pp. 502-523], [11, pp. 251-270] and [12, pp. 109-163].
The next important theory is the curvature of a curve, which is represented
by the symbol κ. It measures the rate of the change of the tangent moving
along the curve in the plane. Naturally, the curvature of the straight line is
0 and the curvature of a circle with a radius a is de�ned as κ =1
a, which is
19
The true catenary
a geometric motivation. For a parametrically given curve y(t) = (x(t), y(t)),
the calculations are κ =x′y′′ − y′x′′√(x′2 + y′2)3
from [1, pp. 21]. But the true catenary
problem is de�ned in polar coordinates, where the y-axis represents the angle
0 and therefore we must calculate it. For a planar curve given parametrically
in Cartesian coordinates as (x(ϕ), y(ϕ)) = (r(ϕ) sin(ϕ), r(ϕ) cos(ϕ)), where
we use r = r(ϕ) and then
x = r sin(ϕ)
x′ = r′ sin(ϕ) + r cos(ϕ),
x′′ = 2r′ cos(ϕ)− r sin(ϕ) + r′′ sin(ϕ),
y = r cos(ϕ),
y′ = r′ cos(ϕ)− r sin(ϕ),
y′′ = −2r′ sin(ϕ)− r cos(ϕ) + r′′ cos(ϕ).
In the numerator we have
x′y′′ − y′x′′ = (r′ sin(ϕ) + r cos(ϕ))(−2r′ sin(ϕ)− r cos(ϕ) + r′′ cos(ϕ))−
−(r′ cos(ϕ)− r sin(ϕ))(2r′ cos(ϕ)− r sin(ϕ) + r′′ sin(ϕ))
= r′r′′ cos(ϕ) sin(ϕ)− 2r′2(cos(ϕ))2 − r2(sin(ϕ))2 − rr′ cos(ϕ) sin(ϕ)+
+rr′′(sin(ϕ))2 − 2rr′ sin(ϕ) cos(ϕ)− r′r′′ sin(ϕ) cos(ϕ)− 2r′2(sin(ϕ))2+
+rr′ sin(ϕ) cos(ϕ) + rr′′(cos(ϕ))2 + 2rr′ sin(ϕ) cos(ϕ)− r2(cos(ϕ))2
= rr′′ − 2r′2 − r2,
and in the denominator it stands√(x′2 + y′2)3 =
√((r′ sin(ϕ) + r cos(ϕ))2 + (r′ cos(ϕ)− r sin(ϕ))2)3
=√
(r2 + r′2)3,
20
The true catenary
so the curvature in polar coordinates is
κ =rr′′ − 2r′2 − r2√
(r2 + r′2)3. (2.5)
Additional results about the curvature are in [1] and [9, pp. 13-15].
These are the most important notations for understanding subsequent pages.
21
The true catenary
3 The classical catenary
In this chapter we introduce and solve the problem of the classical catenary,
but at the beginning we assume its existence in the plane, having the lowest
potential energy and being a function of x. We prove all those characteristics
in the next chapter for the true symmetric catenary.
We hang an ideal homogeneous chain with length ` and suspension points
A1(x1, y1) and A2(x2, y2) in an (x, y)-plane perpendicular on the surface of
the Earth considering that x1 < x2. Ideal means that the thickness is zero.
We study its shape when the potential energy is the lowest.
Figure 1: The classical catenary
A small piece of mass dm in the point P (x, y) of the function y = y(x) of the
22
The true catenary
searched curve has a potential energy of
gy dm = gρy ds = gρy√
1 + y′2dx,
where dm = ρ ds and ds is the length of a small piece of the chain. It is
homogeneous and this means that its density ρ is constant. The potential
energy of the chain is
F (y) = gρ
∫ x2
x1
y(x)√
1 + (y′(x))2dx,
but we write it shorter as
F (y) = gρ
∫ x2
x1
y√
1 + y′2dx (3.1)
and it has the length and boundary conditions
P (y) =
∫ x2
x1
√1 + (y′(x))2dx = `, y(x1) = y1, y(x2) = y2
and also in shorter notations are
P (y) =
∫ x2
x1
√1 + y′2dx = `, y(x1) = y1, y(x2) = y2. (3.2)
This is a typical example of an isoperimetric problem. We look for a con-
tinuously di�erentiable function for which the integral (3.1) is minimal with
the condition (3.2). In [10, pp. 76-78] we �nd the de�nition of the problem
but it is solved parametrically. We use standard methods of the calculus of
variations. We form the functional as in (2.4)∫ x2
x1
(y√
1 + y′2 + λ√
1 + y′2)dx, (3.3)
which has to be minimized and where λ is Lagrange's parameter. We can
assume that ρg = 1, because this is a positive constant which does not a�ect
the result. For the function f(x, y, y′) = y√1 + y′2 + λ
√1 + y′2 we use the
23
The true catenary
Euler-Lagrange equation (2.1), but we have a case where x is missing from
the function f , so we use (2.3) where
∂f
∂y′=
yy′√1 + y′2
+λy′√1 + y′2
and (2.3) is
y√
1 + y′2 + λ√
1 + y′2 − y′(
yy′√1 + y′2
+λy′√1 + y′2
)= c,
y + λ = c√
1 + y′2,
where c is a constant. Now we make a derivative of the last equation
y′ = cy′y′′√1 + y′2
and we divide it by y′, where y′ is not identically equal to zero, and we get
cy′′ =√
1 + y′2. (3.4)
We see that c > 0 for chains, because the right-hand side is positive and y′′
is also positive, since the curve is convex. The order of the given di�erential
equation can be reduced by introducing a new variable p = y′. It follows that
p′ = y′′. If we insert the substitutions into the equation (3.4) then we obtain
cp′ =√
1 + p2,
cdp
dx=√1 + p2,
cdp√1 + p2
= dx.
We use a standard integral∫ 1√
1 + x2dx = Arsh(x) for our variables and we
extract x
x = cArsh(p) +D, (3.5)
24
The true catenary
where D is an integration constant. But we know that p = y′ =dy
dxand from
(3.5) we can �nd p
p(x) = sinh
(x−Dc
).
Integration gives us y
y(x) = c cosh
(x−Dc
)+ E, (3.6)
where E is an integration constant. This is the general form of the classi-
cal catenary and we solved it by standard methods using the Euler-Lagrange
equation and �nding its solution.
At this point we start to follow the method from [6] ant then the only thing
that we have to consider next are the given conditions such are length ` and
that the chain is attached to the points A and B. First we insert the coordi-
nates of the point A into (3.6)
y1 = c cosh
(x1 −D
c
)+ E (3.7)
and second the point B into (3.6)
y2 = c cosh
(x2 −D
c
)+ E. (3.8)
The integral (3.2) can be transformed into
` =
∫ x2
x1
√1 +
(sinh
(x−Dc
))2
dx =
∫ x2
x1
√(cosh
(x−Dc
))2
dx
=
∫ x2
x1
cosh
(x−Dc
)dx
=
[c sinh
(x−Dc
)]x2x1
= c
(sinh
(x2 −D
c
)− sinh
(x1 −D
c
)), (3.9)
25
The true catenary
where we have used y′ = sinh
(x−Dc
)and the relation (cosh(x))2−(sinh(x))2 =
1. The task to calculate the c, D and E from (3.7), (3.8) and (3.9) is uniquely
solvable, if
(y2 − y1)2 + (x2 − x1)2 < `2 (3.10)
is true and it means that the length must be longer than the the distance
between the points A and B. If we assume that (3.10) is true, then we
can continue with (3.9). By using the addition formula sinh(x) − sinh(y) =
2 sinh
(x− y2
)cosh
(x+ y
2
)we can transform (3.9) into
2c sinh
(x2 − x1
2c
)cosh
(x2 + x1 − 2D
2c
)= `. (3.11)
From (3.7) and (3.8) we form
y2 − y1 = 2c sinh
(x2 + x1 − 2D
2c
)sinh
(x2 − x1
2c
)(3.12)
by using cosh(x) − cosh(y) = 2 sinh
(x+ y
2
)sinh
(x− y2
). Now we divide
(3.12) by (3.11)
y2 − y1`
=
2c sinh
(x2 − x1
2c
)sinh
(x2 + x1 − 2D
2c
)2c sinh
(x2 − x1
2c
)cosh
(x2 + x1 − 2D
2c
) ,y2 − y1
`= tanh
(x2 + x1 − 2D
2c
). (3.13)
There exists the formula with which we can transform hyperbolic sine into
hyperbolic tangent
sinh
(xy + x2 − 2D
2c
)=
tanh
(x1 + x2 − 2D
2c
)√
1− tanh2
(x1 + x2 − 2D
2c
) ,
26
The true catenary
which is
=
y2 − y1`√
1−(y2 − y1
`
)2,
where we use (3.13). We put the obtained one into (3.12) and we get
y2 − y1 = 2cy2 − y1
`
√1−
(y2 − y1
`
)2sinh
(x2 − x1
2c
),
in case if y2 6= y1, we can divide the upper formula by y2 − y1
sinh
(x2 − x1
2c
)=
`
2c
√1−
(y2 − y1
`
)2
. (3.14)
At this point we introduce new constants
γ =`
x2 − x1
√1−
(y2 − y1
`
)2
(3.15)
and
z =x2 − x1
2c(3.16)
and we transform equation (3.14) into
sinh(z) = zγ. (3.17)
There is a solution z > 0 only if γ > 1 otherwise there is no intersection of
the line y = γz with the function hyperbolic sine. We see that the equation
(3.17) has a unique positive solution which can be numerically calculated and
for z > 0 we can calculate the constants c, D and E,
c =x2 − x1
2z, D =
x1 + x22
− cArth(y2 − y1
`
), E = y1 − c cosh
(x1 −D
c
).
This is the end of calculation except we have to check it for the symmetric
case, when y1 = y2. From the formula (3.13) we extract D
D =x1 + x2
2c
27
The true catenary
and equation (3.11) can be transformed into
`
2c= sinh
(x2 − x1
2c
)and then we introduce the new variable (3.15) which corresponds to γ =
`
x2 − x1and (3.16) and the solutions of (3.17) are the same as stated above.
28
The true catenary
4 The problem of the true symmetric catenary
The aim of this thesis is to describe the true catenary problem. In this chapter
we focus on the de�nition of the symmetric one. The main di�erence between
the classical catenary and the true symmetric catenary is that in the second
case we consider the catenary in a central gravity �eld with −1/r potential,whereas in the �rst case the Earth is assumed to be �at.
Let us de�ne the function F : R3 → R3 of the radius vector r ∈ R3: r 7→ F (r).
Newton's law of universal gravitation says that |F (r)| = GM
|r|2, where G is the
universal gravitational constant and M is the mass of Earth or any other
central body. The gravity �eld would be F (r) = −GMr
|r|3. The minus sign
is necessary, because we have an attractive gravitational �eld. Field F is
conservative, which means that F (r) = − gradU(r), where U : R3 → R is a
scalar potential U de�ned as U(r) = −GM|r|
. For the point O we choose the
gravitational center of Earth or another planet, which corresponds to radius
r = 0 [5, pp. 123].
Figure 2: The true symmetric catenary from [5, pp. 124]
29
The true catenary
Until the end of the chapter we follow the article [2, pp. 120-123]. The length
of the catenary curve, that we seek in this thesis, is �xed at 2`. Homogeneity
is also assumed and it means that the mass density ρ =dm
dsis a constant,
where ds is a di�erential length of the curve. In this chapter we de�ne the
problem of the simplest case, where the catenary has endpoints in di�erent
points A1 and A2, which are on the same height, but the distance in between
and from the attractive center O is large enough. In this situation the cate-
nary is symmetric with respect to A1, A2 and the vertical line which goes
through O.
We de�ne the curve γ : z 7→ γ(z) ∈ R3, which has the same characteristic as
listed above, and it has the lowest center of gravity, otherwise the potential
energy would be converted into kinetic energy, which would lead to the de-
formation of the curve. The curve |γ| can not have more than one minimum,
otherwise there exists z1, z2, where z1 < z2 such that |γ(z1)| = |γ(z2)| ≤ |γ(z)|for all z ∈ [z1, z2]. With re�ection of that curve we would get a lower potential
energy. This means, that there are not two minima.
We assume that the curve γ(z) lies in the plane (A1, A2, 0), otherwise the
projection from R3 onto this plane is shorter than the curve γ(z) and one
may add the extra length at the lowest point as we can see on �gure 3. Then
we obtain a smaller potential energy.
The next task is to eliminate z, so the question is if curves z 7→ γ(z) =
(r(z), ϕ(z)) are graphs of a function ϕ 7→ r(ϕ). This is true, because z 7→ r(ϕ)
is monotonic and convex. Firstly, if it is not monotonic, then a modi�cation
of the curve would lower the potential energy. The arc which does not lie in
the plane A1, A2, O can be replaced with the segment P1P2, where the length
30
The true catenary
Figure 3: Planar curve from [2, pp. 121]
does not change, �gure 4. Secondly, if it is not convex with respect to O,
Figure 4: Not a monotonic curve from [2, pp. 121]
then a re�ection, which is seen on the �gure 5, would also lower the potential
energy.
31
The true catenary
Figure 5: Not convex curve from [2, pp. 121]
The only exceptions left are vertical ends, �gure 6. We can vary c such that
|ac|+ |bc| = constant. The derivative of potential energy of this piece is not 0,
which means that it is not an optimum. It follows that there are no corners.
We assume that there is a single minimum. In a symmetric case the minimum
would lie at the angle ϕ = 0 and we use this from here on, but the justi�cation
of that assumption is given in the chapter of the true asymmetric catenary.
If the x-coordinate of the minimum is not in the middle of A1 and A2, then
Figure 6: Vertialcal ends from [2, pp. 121]
32
The true catenary
we can get another minimum with a re�ection with respect to ϕ = 0, which
is a contradiction to the uniqueness as on �gure 7.
Figure 7: Re�ection of the minimum
To make it easier we introduce a polar coordinate system with pole in point O
and polar axis, which is orthogonal to the line A1A2. The polar coordinates
are:
x = r sin(α),
y = r cos(α) ,
ϕ = arctan(yx
),
r =√x2 + y2.
At this point we change r from vector to scalar, the vector was only needed
for de�ning the problem. The polar angles of points A1 and A2 are α and
−α. A natural limitation is 0 < α <π
2. If α = 0, there would be no curve
but just a straight vertical line. In case α =π
2the catenary would go through
33
The true catenary
the gravitational center, which is physically impossible. The polar radius of
both points is r1.
In polar coordinates we write A1(r1, α) and A2(r1,−α). We seek the equation
of the catenary in polar coordinates r = r(ϕ).
Its potential energy comes from the sum of the arbitrarily small links be-
tween ϕn and ϕn+1, where n ∈ {0, 1, 2, ..., N − 1} ∈ N. It is also known that
ϕ0 = −α and ϕN = α. We de�ne p(ϕ) := (r(ϕ) sin(ϕ), r(ϕ) cos(ϕ)). The
potential energy is
F = limN→+∞
(−
N−1∑n=0
GMρ|p(ϕn+1)− p(ϕn)|
r(ϕn)
)
= −GMρ limN→+∞
(N−1∑n=0
1
r(ϕn)((p1(ϕn+1)− p1(ϕn))2 + (p2(ϕn+1)− p2(ϕn))2)1/2
).
At this point we use the Mean value theorem from [7, pp. 106], because all
the functions are su�ciently smooth.
Theorem 4.1 Mean value theorem
If f is a real continuous function on [a, b], which is di�erentiable in (a, b),
then there is a point x ∈ (a, b) at which
f(b)− f(a)b− a
= f ′(x).
The functions p1 and p2 satisfy the conditions of the Mean value theorem
and therefore for each n there exist ϕ1,n and ϕ2,n from the interval (ϕn, ϕn+1)
such that p1(ϕn+1)− p1(ϕn) = p′1(ϕ1,n)(ϕn+1 − ϕn) and p2(ϕn+1)− p2(ϕn) =p′2(ϕ2,n)(ϕn+1 − ϕn). Now we continue with calculating the potential energy
F = −GMρ limN→+∞
(N−1∑n=0
1
r(ϕn)
√(p′1(ϕ1,n))2 + (p′2(ϕ2,n))2(ϕn+1 − ϕn)
).
34
The true catenary
We can convert the limit of the sum into an integral, replacing ϕn, ϕn+1, ϕ1,n
and ϕ2,n by ϕ which runs through all angles from −α to α
F = −GMρ
∫ α
−α
1
r(ϕ)
√((r(ϕ) sin(ϕ))′)2 + ((r(ϕ) cos(ϕ))′)2.
We use r′(ϕ) =dr
dϕ.
F = −GMρ·
·∫ α
−α
1
r(ϕ)
√(r′(ϕ) sin(ϕ) + r(ϕ) cos(ϕ))2 + (r′(ϕ) cos(ϕ)− r(ϕ) sin(ϕ))2
= −GMρ
∫ α
−α
1
r(ϕ)
√r2(ϕ) + r′2(ϕ)dϕ (4.1)
with condition of length
P (r) =
∫ α
−α
√r2(ϕ) + r′2(ϕ)dϕ = 2`. (4.2)
We also have to consider that an in�nitesimally small mass dm at a distance
r from the point O has a potential energy of −GM dm
r(ϕ). We look for a pos-
itive, di�erentiable function ϕ 7→ r(ϕ) de�ned on the interval [−α, α], whichminimizes the expression in (4.1) with length conditions (4.2). We also have
to consider the additional condition of two suspension points r(±α) = r1.
A natural limit for the length ` is
r1 sin(α) < ` < r1, (4.3)
because the curve lies in the triangle A1A2O, so 2` must be bigger than the
distance between A1 and A2 and smaller than 2r1. Without loss of generality
we can say that GMρ = 1, because G,M and ρ are positive constants and the
result does not depend on them. We look for the minimum of the functional
F (r) = −∫ α
−α
1
r
√r2 + r′2dϕ (4.4)
35
The true catenary
with conditions
P (r) =
∫ α
−α
√r2 + r′2dϕ = 2`, r(±α) = r1. (4.5)
This is a typical problem of the calculus of variations.
Now we write the solution of the de�ned catenary problem which is proved
in the next chapters.
Theorem 4.2 A minimum of the functional in (4.4) under conditions (4.5)
is given by
∀ϕ ∈ [−α, α] : r(ϕ) = r1
1− c cosh(√
1− c2αc
)1− c cosh
(√1− c2ϕc
) , (4.6)
where c > 0 is determined by`
r1=
c sinh
(√1− c2αc
)√1− c2
and α ∈ (0,π
2),
r : [−α, α]→ (0,∞), r1 ∈ (0,∞), ` ∈ (r1 sin(α), r1).
36
The true catenary
5 Solving the problem
In this chapter we prove Theorem 4.2 in two di�erent ways. First, we start
with the part which is the same in both methods from the articles [2] and [5].
Proof of Theorem 4.2
In the �rst step we use the Euler-Lagrange di�erential equation for the isoperi-
metric variational problem (2.1) and we adjust in for our functions and vari-
ablesd
dϕ
(∂f
∂r′
)=∂f
∂r, (5.1)
where we introduce the function
f(r, r′, ϕ) = −1
r
√r2 + r′2 + λ
√r2 + r′2 =
(λ− 1
r
)√r2 + r′2, (5.2)
where λ is the Lagrange parameter. The variable ϕ does not appear explicitly,
so we can reduce the order of the di�erential equation as in (2.3) to
f(r, r′)− r′ ∂f∂r′
(r, r′) = c, (5.3)
where c is a constant. Therefore we get
−1
r
√r2 + r′2 + λ
√r2 + r′2 − r′
(−1
r+ λ
)2r′
2√r2 + r′2
=
=
(−1
r+ λ
)(√r2 + r′2 − r′2√
r2 + r′2
)= c.
The simpli�cation gives us
r(λr − 1) = c√r2 + r′2. (5.4)
If c = 0 then r(rλ− 1) = 0 and because r 6= 0 and if λ > 0 it leads to r = 1λ,
which are arches. If λ = 0, then −r = c√r2 + r′2 and then the equation
37
The true catenary
can be transformed into a di�erential equation cr′ = ±r√1− c2 of which the
solution is a logarithmic spiral, because we have polar coordinates
cdr
dϕ= r√1− c2, dr
r=
√1− c2dϕc
, ln(r) =ϕ√1− c2dϕc
+ lnA,
r = A exp
(1
c
√1− c2ϕ
),
where A and c are constants. This is not a hanging chain.
We focus on the cases where c 6= 0 and λ 6= 0. From here on we continue with
proving the Theorem 4.2 by using two di�erent methods.
38
5.1 The �rst way of solving the problem The true catenary
5.1 The �rst way of solving the problem
In this chapter we continue with proving Theorem 4.2 by following the method
of Hinz and Denzler from [2, pp. 123-128].
We start with the Euler-Lagrange equation (5.1), where the task is to cal-
culate all of its parts for the function (5.2). The right-hand side of (5.1)
is
∂f
∂r=
1
r2(r2 + r′2) +
(λ− 1
r
)r
√r2 + r′2
.
The partd
dϕ
(∂f
∂r′
)is more complicated and therefore we use (2.2) with
variables which are compatible with a problem of the true catenary
d
dϕ
∂f
∂r′=
∂2f
∂ϕ∂r′+
∂2f
∂r∂r′r′ +
∂2f
∂r′2r′′. (5.5)
We calculate the parts of the right-hand side of (5.5)
∂f
∂r′=
(λ− 1
r
)r′
√r2 + r′2
,
∂2f
∂r∂r′=
1
r2r′(r2 + r′2)− rr′
(λ− 1
r
)√
(r2 + r′2)3,
∂2f
∂r′2=
(λ− 1
r
)r2√
(r2 + r′2)3,
∂2f
∂ϕ∂r′= 0.
Having all the parts we can put them back into (5.1)
1
r2(r2 + r′2) + r
(λ− 1
r
)√r2 + r′2
=
1
r2r′2(r2 + r′2)− rr′2
(λ− 1
r
)+ r2r′′
(λ− 1
r
)√
(r2 + r′2)3,
39
5.1 The �rst way of solving the problem The true catenary
which can be transformed into
1
r√r2 + r′2
=
(λ− 1
r
)rr′′ − 2r′2 − r2√
(r2 + r′2)3,
and we see that (5.1) is equivalent to
1
r√r2 + r′2
= κ
(λ− 1
r
), (5.6)
where κ is the curvature (2.5).
If we look at the circle r = a (a > 0), which is concave from the point O
we see that its curvature is
κ =a · 0− a2 − 2 · 0√
(a2 + 02)3= −1
a,
which means that the concave functions have negative curvature, but our
graph is convex so its curvature is positive. By inserting (5.4) into (5.6) we
see that for chains c must be positive
κ =r
c(r2 + r′2).
We continue with the equation
r(λr − 1) = c√r2 + r′2.
In the �rst chapter, we have already eliminated the cases λ = 0 and c = 0.
Here we use the substitution λr =1
u, λr′ = − u
′
u2to �nd the solution in an
easier way. The transformation of our equation is
1
λu
(1
u− 1
)= c
√1
λ2u2+
u′2
λ2u4,
1− u = c sign(λ)√u2 + u′2, (5.7)
40
5.1 The �rst way of solving the problem The true catenary
where the function sign extracts the sign of the real number and is de�ned as
sign(λ) =
1, if λ > 0
0, if λ = 0
−1, if λ < 0.
Now we square the equation (5.7) to eliminate the roots
1− 2u+ u2 = c2(u2 + u′2)
Taking the derivative we get
−2u′ + 2uu′ = c2(2uu′ + 2u′u′′),
and we divide by 2 and u′
u− 1 = c2(u+ u′′)⇔ u′′ +
(c2 − 1
c2
)u = − 1
c2. (5.8)
If u′ = 0, then λr = 0, which means λ = 0, but our condition was that
parameters must be di�erent from 0, otherwise the substitution would be
without sense. Now we have to solve the di�erential equation of second order
with constant coe�cients. The solution is
u(ϕ) = D exp
(√1− c2ϕc
)+ E exp
(−√1− c2ϕc
)+
1
1− c2,
for positve c 6= 1. We want the solution to be expressed with hyperbolic cosine
or sine. The connections between the exponential and hyperbolic functions
are
exp(γx) = sinh(γx) + cosh(γx),
exp(−γx) = − sinh(γx) + cosh(γx).
Thus the general solution in our case is
u(ϕ) = A sinh
(√1− c2ϕc
)+B cosh
(√1− c2ϕc
)+
1
1− c2.
41
5.1 The �rst way of solving the problem The true catenary
At this point we de�ne and �x the unique minimum of this chain at ϕ0, which
follows from convexity, so r′(ϕ0) = 0 = u′(ϕ0). Symmetry is the reason that
our problem has the unique minimum of the chain at ϕ0 = 0.
u′(ϕ) = A cosh
(√1− c2ϕc
) √1− c2c
+B sinh
(√1− c2ϕc
) √1− c2c
.
0 = u′(0) = A
√1− c2c
,
such that A = 0 and
u(ϕ) = B cosh
(√1− c2ϕc
)+
1
1− c2.
With putting this solution back into (5.8) and with ϕ = 0 we determine B =c
c2 − 1. Considering the condition u′(0) = 0 our solution can be expressed
without hyperbolic sine
u(ϕ) =
1− c cosh(√
1− c2ϕc
)1− c2
. (5.9)
The solution is correct for all positive c, except for c = 1. For 0 < c < 1 we
are satis�ed with (5.9), but in case where c > 1, the hyperbolic cosine, which
is de�ned on C, can be easily transformed into cosine by
cosh(ix) =exp(ix) + exp(−ix)
2=
cos(x) + i sin(x) + cos(x)− i sin(x)2
= cos(x),
then
u(ϕ) =
1− c cos(√
c2 − 1ϕ
c
)1− c2
.
The only case we have to discuss separately is c = 1. We insert the value of
c into di�erential equation (5.8)
u′′(ϕ) = −1,
42
5.1 The �rst way of solving the problem The true catenary
u(ϕ) = −ϕ2
2+ Aϕ+B,
0 = u′(0) = A,
u(ϕ) = −ϕ2
2+B.
The solution needs to be inserted back into (5.8) with ϕ = 0 and then B =1
2.
The solution for c = 1 is
u(ϕ) =1− ϕ2
2.
The length ` is given by
` =
∫ α
0
√r2 + r′2dϕ.
From using (5.4) twice we get
√r2 + r′2 =
1
cr(λr − 1) =
(λr − 1)r′√(λr − 1)2 − c2
. (5.10)
At this point we need to �nd the primitive function of the right-hand side of
(5.10) and we see that we �rst need to multiply (5.10) by λ
d
dϕ
√(λr − 1)2 − c2 = λ(λr − 1)r′√
(λr − 1)− c2.
Therefore
` =
[1
λ(√
(λr − 1)2 − c2)]r1r0
,
we change the boundaries, because there is no ϕ in the expression.
` =1
λ
(√(λr1 − 1)2 − c2 −
√(λr0 − 1)2 − c2
)=c
λsign(c)
(√1
c2(λr1 − 1)2 − 1−
√1
c2(λr0 − 1)2 − 1
)
=c
λsign(c)
(√r21 + r′2(α)
r21− 1−
√r20 + r′2(0)
r20− 1
)
43
5.1 The �rst way of solving the problem The true catenary
The part under the root came from equation (5.4), which has provided a
transformation (λr−1)2 = c2
r2(r2+r′2). The second root is 0, because r′2(0) =
0. We have chosen c > 0 so sign(c) > 0 and r′(α) is positive since we consider
the part of the curve where α ∈ (0,π
2). Then
` =r′(α)c
λr1,
`
αr1=
r′(α)c
λα(r(α))2.
Since we made a substitution with u we can also use it here, which leads to
r′(α) = − u′(α)
λu(α)= −u′(α)λ(r(α))2. Then it follows
`
αr1=−u′(α)c
α=
sinh(σ)
σ,
where we use the �rst derivative of (5.9)
u′(α) =
− sinh
(√1− c2c
α
)√1− c2
and a new variable σ :=
√1− c2αc
including c > 0. From (4.3)`
αr1goes from
sin(α)
αto
1
α, this means that we can �nd a unique solution σ for positive c.
The �nal solution is
r(ϕ) = r1
1− c cosh(√
1− c2αc
)1− c cosh
(√1− c2ϕc
) ,where c > 0 is determined by
`
r1=
c sinh
(√1− c2αc
)√1− c2
.
�
44
5.2 The second way of solving the problem The true catenary
5.2 The second way of solving the problem
We continue with proving Theorem 4.2 by the methods from [5, pp. 126-131].
Proof of Theorem (4.2)
Let ϑ be the oriented angle from the normal on a curve to the extended polar
Figure 8: Angles ϑ and µ on catenary picture
radius r as seen on �gure 8. We look for the extremal which is convex from
the point O. Therefore ϑ must continuously rise from negative to positive
values, when ϕ goes from −α to α. It is know that tan(µ) =r
r′if µ is the
angle between the polar radius and the tangent on the curve. The connection
between µ and ϑ comes from µ+ ϑ = 90◦, so
tan(µ) = tan(90◦ − ϑ) = 1
tan(ϑ)=r
r′,
r′ = r tan(ϑ).
45
5.2 The second way of solving the problem The true catenary
Figure 9: The true symmetric catenary with normal vector from [5, pp. 124]
From the formula (5.4) we get
r(λr − 1) = c√r2 + r2 tan2(ϑ),
where we divide the last equation by r, because r 6= 0, so
λr − 1 = c1
cos(ϑ),
r =1
λ
(c
cos(ϑ)+ 1
). (5.11)
From the equalitydr
dϑ=dr
dϕ
dϕ
dϑ,
which is true, because r = r(ϕ), by di�erentiation follows
c sin(ϑ)
λ cos2(ϑ)=dϕ
dϑr tan(ϑ) =
1
λ
dϕ
dϑtan(ϑ)
(1 +
c
cos(ϑ)
),
46
5.2 The second way of solving the problem The true catenary
where we have used r′ = r tan(ϑ) and (5.11). We want to expressdϕ
dϑ
dϕ
dϑ= λ
c sin(ϑ) cos(ϑ)
λ(cos(ϑ))2 tan(ϑ)(cos(ϑ) + c)=
c
c+ cos(ϑ).
We note that ϕ = 0 when ϑ = 0. This is the consequence of the minimality
of the polar radius in the lowest point and equation r′ = r tan(ϑ). The
connection between the angles ϕ and ϑ is
ϕ = c
∫ ϑ
0
dθ
c+ cos(θ). (5.12)
The functions r and ϕ determine a family of extremals in polar parametric
form. But we want more, namely to eliminate the angle ϑ and get an explicit
formula r = r(ϕ). The �rst step is to know the sign of the constant c, where
the use of the curvature κ (2.5) helps us. From the connection between the
angles we know that r′ = r tan(ϑ) anddϕ
dϑ=
c
c+ cos(ϑ). First we calculate
the second derivative of r
r′′ = r′ tan(ϑ) +r
(cos(ϑ))2dϑ
dϕ= r(tan(ϑ))2 +
r
(cos(ϑ))2
(1 +
cos(ϑ)
c
)Now we insert the derivatives into the expression of the curvature
κ =
r2(tan(θ))2 +r2
(cos(θ))2
(1 +
cos(θ)
c
)− r2 − 2r2(tan(θ))2√
(r2 + r2(tan(θ))2)3
=
c(sin(θ))2 + c+ cos(θ)− c(cos(θ))2 − 2c(sin(θ))2
c(cos(θ))2r
(cos(θ))3
=(cos(θ))2
rc.
The extremal, which is still unknown, is convex from the point O and this
means that it must have positive curvature on the whole interval [−α, α].The numerator (cos(θ))2 is always positive on that interval, since 0 < α <
π
2,
47
5.2 The second way of solving the problem The true catenary
which means that c > 0. The case where c = 0 has already been eliminated.
However, we have two constants, c and λ, so when we look back to equation
(5.11), we see that λ > 0. Now we can focus on the integral that will help us
to express ϕ. The easiest way to solve this integral is by using the universal
substitution
τ = tan
(θ
2
),
and then we also need cos(θ), which we can get from(sin
(θ
c
))2
+
(cos
(θ
2
))2
= 1,
and we divide the equation by
(cos
(θ
2
))2
6= 0 and use the substitution
τ 2 + 1 =1(
cos
(θ
2
))2 ,
(cos
(θ
2
))2
=1
τ 2 + 1.
From the connection of the double angle of the cosine we can extract cos(θ)
cos(θ) = 2
(cos
(θ
2
))2
− 1 =1− τ 2
1 + τ 2.
The only missing part now is dθ, which is coming from
dθ =2dτ
1 + τ 2.
With the results obtained we can calculate the integral (5.12)
ϕ(t) = c
∫ t
0
2dτ
(1 + τ 2)
(c+
1− τ 2
1 + τ 2
) = 2c
∫ t
0
dτ
(c− 1)τ 2 + (c+ 1),
where t = tan
(ϑ
2
). At this point we have to solve the integral for di�erent
constants c,
48
5.2 The second way of solving the problem The true catenary
1. 0 < c < 1,
2. c = 1,
3. c > 1.
The question is whether there are any solutions with the condition r1 sin(α) <
l < r1.
1. In the �rst case 0 < c < 1
we make a substitution
β =√1− c τ
and we insert it into the integral
ϕ(t) = 2c
∫ √1−c t0
1√1− c
· dβ
(c− 1)β2
1− c+ c+ 1
= 2c
∫ √1−c t0
dβ√1− c(−β2 + c+ 1)
,
in the tables of integrals we �nd∫dx
ax2 + bx+ c=
1√|4ac− b2|
ln
∣∣∣∣∣2ax+ b−√|4ac− b2|
2ax+ b+√|4ac− b2|
∣∣∣∣∣ ,if 4ac− b2 < 0. In our case a = −
√1− c, b = 0 and c = (c + 1)
√1− c. The
value of the integral is
ϕ(t) =2c
2√1− c2
[ln
∣∣∣∣−2√1− cβ − 2√1− c2
−2√1− cβ + 2
√1− c2
∣∣∣∣]√1−ct
0
=c√
1− c2
ln∣∣∣∣∣∣∣∣
β√1 + c
+ 1
β√1 + c
− 1
∣∣∣∣∣∣∣∣√1−ct
0
.
The solution we want should be expressed with area hyperbolic tangent. The
connection between natural logarithm and area hyperbolic tangent is
Arth(x) =1
2ln
∣∣∣∣1 + x
1− x
∣∣∣∣ .49
5.2 The second way of solving the problem The true catenary
With the transformation we get
ϕ(t) =
[2c√1− c2
Arth
(β√1 + c
)]√1−ct0
=2c√1− c2
Arth
(√1− c√1 + c
t
).
(5.13)
Making the inverse of t = tan
(ϑ
2
)will help us with the calculations later
t =
√1 + c
1− ctanh
(√1− c2ϕ2c
).
2. The second case c = 1
is the simplest, we just insert c = 1 into the integral
ϕ(t) = 2
∫ t
0
dτ
2= τ |t0 = t = tan
(ϑ
2
). (5.14)
3. In the third case, where c > 1,
we make a substitution
β =√c− 1 τ
to simplify the integral
ϕ(t) = 2c
∫ t
0
1√c− 1
· dβ
(c− 1)β2
c− 1+ c+ 1
= 2c
∫ t
0
dβ√c− 1β2 +
√c− 1(c+ 1)
Again in the table of integrals we �nd∫dx
ax2 + bx+ c=
2√4ac− b2
arctan
(2ax+ b√4ac− b2
)+ C,
if 4ac− b2 > 0 and in our case a =√c− 1, b = 0 and c = (c+1)
√c− 1. The
value of the integral is
ϕ(t) =2c√c2 − 1
[arctan
(1√c+ 1
β
)]√c−1t0
=2c√c2 − 1
arctan
(√c− 1√c+ 1
t
),
(5.15)
50
5.2 The second way of solving the problem The true catenary
and the inverse is
t =
√c+ 1
c− 1tan
(√c2 − 1ϕ
2c
).
This is not the end of our calculations, because we want to describe the
catenary in the form r(ϕ). At this point we need the expression for ds. We
use (5.11),(5.12) and the derivative of r, which is r′ =c sin(ϑ)
λ(cos(ϑ))2to get
ds =√r2 + r′2dϕ =
cdϑ
λ(cos(ϑ))2.
By integration we get the length s(ϕ) of the extremal between its apex, the
lowest point, and the point which corresponds to the polar angle ϕ
s(ϕ) =c
λ
∫ ϑ
0
dθ
(cos(θ))2=c
λtan(ϑ) =
cr′(ϕ)
λr(ϕ)= −cr(ϕ)
(1
λr
)′(ϕ). (5.16)
In the third step we have used tan(ϑ) =r′
r. This result is important for
determining the constant c from length ` and angle α.
Again we continue with all three options for c, viz
1. in the �rst case 0 < c < 1
from r =1
λ+
c
λ cos(ϑ)by using the universal substitution from before, where
t = tan
(ϑ
2
), we get
1
λr=
cos(ϑ)
c+ cos(ϑ)=
1− t2
1 + t2
c+1− t2
1 + t2
=1− t2
t2(c− 1) + c+ 1. (5.17)
Calculating the inverse of the angle (5.13) leads to
1
λr(ϕ)=
1− c cosh(√
1− c2ϕc
)1− c2
. (5.18)
51
5.2 The second way of solving the problem The true catenary
If we want the result for s(ϕ), we �rst need a derivative of (5.18).
(1
λr
)′(ϕ) =
−c sinh(√
1− c2ϕc
)√1− c2
c(1− c2)=
− sinh
(√1− c2ϕc
)√1− c2
.
By inserting this into the length (5.16), we get
s(ϕ) = r(ϕ)
c sinh
(√1− c2ϕc
)√1− c2
.
There are some conditions we have to observe in this expression, s(α) = `
and r(α) = r1
` = r1
c sinh
(√1− c2αc
)√1− c2
. (5.19)
We also have to consider all the options of this equation, so we make it easier
by introduction of a new variable
σ =
√1− c2αc
6= 0,
so we get a transcendent equation
` = r1c sinh(σ)
cσ
α
⇔ `
r1α=
sinh(σ)
σ.
The function given by σ 7→ sinh(σ)
σis always bigger than 1, which we can see
from the graph of this function on �gure 10. It is increasing for positive σ. We
get exactly one solution σ. At �rst we know that`
r1α> 1, which means that
r1α < `, but we know from the beginning that ` < r1, therefore 0 < α < 1,
α 6= 0, because we can not divide by 0 and α 6= 1, otherwise r1 < ` < r1,
which is contradiction. Solving the transcendental equation is a numerical
task in concrete examples. When we compute σ, we get the exactly de�ned
constant c =α√
α2 + σ2from the introduction of the variable σ, which is
52
5.2 The second way of solving the problem The true catenary
Figure 10: Functionsinh(x)
x
compatible with condition 0 < c < 1. The curve, which we seek, is described
by the relation
1
λr(ϕ)=
1− c cosh(σϕα
)1− c2
.
For ϕ = α we get r(ϕ) = r1 and
1
λr1=
1− c cosh(σ)1− c2
.
This equation de�nes λ
λ =1− c2
r1(1− c cosh(σ)),
and our �nal solution for the �rst case is
r(ϕ) =(1− c2)r1(1− c cosh(σ))
(1− c2)(1− c cosh
(σϕα
)) = r11− c cosh(σ)
1− c cosh(σϕα
)
= r1
1− c cosh(√
1− c2αc
)1− c cosh
(√1− c2ϕc
)where c is determined by (5.19).
There are two more cases left to solve.
53
5.2 The second way of solving the problem The true catenary
2. The constant c = 1.
In the equation (5.17) we put in c = 1 and we can replace t with ϕ because
of (5.14) and then we get1
λr(ϕ)=
1− ϕ2
2
Di�erentiating the expression for the formula we get(1
λr
)′(ϕ) = −ϕ.
We put it into (5.16) and we get
s(ϕ) = r(ϕ)ϕ,
use the conditions s(α) = ` and r(α) = r1 and then
` = r1α,
which is true only for 0 < α < 1, ` < r1. In this case parameter λ is
λ =2
r1(1− α2),
because ϕ = α and r(ϕ) = r1. The resulting curve is
r(ϕ) = r11− α2
1− ϕ2.
There is just one more case unsolved, namely where
3. constant c > 1.
We solve it with the same method as in the �rst case, so from (5.11) we get
1
λr(ϕ)=
cos(ϑ)
c+ cos(ϑ)=
1− t2
(c+ 1) + (c− 1)t2.
From the length s(ϕ) and the inverse of the integral de�ned by tangents (5.15)
from the second case we produce
1
λr(ϕ)=
1− c cos(√
c2 − 1ϕ
c
)1− c2
.
54
5.2 The second way of solving the problem The true catenary
We put the derivative
(1
λr
)′(ϕ) = c
− sin
(√c2 − 1ϕ
c
)√c2 − 1
,
into the length equation (5.16)
s(ϕ) = r(ϕ)
c sin
(√c2 − 1ϕ
c
)√c2 − 1
.
From the conditions s(α) = ` and r(α) = r1 we get
` = r1
c sin
(√c2 − 1α
c
)√c2 − 1
, (5.20)
into which, to make it easier, we again introduce the new variable
σ =
√c2 − 1α
c6= 0,
and now we get the transcendent equation
`
r1α=
sin(σ)
σ.
The function σ 7→ sin(σ)
σis always smaller than 1 and it is decreasing on the
interval (0, π). We can see that this equation has just one solution σ with
conditions 0 < α <π
2, which is the condition from the beginning and ` < r1α,
but we already know that r1 sin(α) < `, so r1 sin(α) < ` < r1α. It is obvious
thatsin(α)
α<
`
r1α<
sin(σ)
σ, which means that
sin(α)
α<
sin(σ)
σand from
here we can see that σ < α. After we calculate σ we get the exactly de�ned
constant c =α√
α2 − σ2, which satis�es the condition c > 1.
Our curve is de�ned by the relation
1
λr(ϕ)=
1− c cos(σϕα
)1− c2
,
55
5.2 The second way of solving the problem The true catenary
Figure 11: Functionsin(x)
x
We insert the conditions ϕ = α and r(ϕ) = r1 in the previous equation and
with that also λ is de�ned as
1
λr1=
1− c cos(σ)1− c2
,
λ =1− c2
r1(1− c cos(σ)).
At the end the solution for the curve is
r(ϕ) = r11− c cos(σ)
1− c cos(σϕα
) ,
= r1
1− c cos(√
c2 − 1α
c
)1− c cos
(√c2 − 1ϕ
c
)where c is determined from (5.20). �
56
The true catenary
6 Asymmetric case
In this chapter we expand the problem of the true catenary to asymmetric
suspension points. We follow the procedure of calculating from [2, pp. 128-
132].
The true asymmetric catenary is the chain of which suspension points are
Figure 12: True asymmetric catenary from [2, pp. 122]
not on the same height. Because of that the geometric restriction is de�ned
as
0 < α :=ϕ2 − ϕ1
2<π
2, (6.1)
57
The true catenary
which means that the sum of both angles has to be smaller than π, where
A1(r1, ϕ1), ϕ1 < 0 and A2(r2, ϕ2), ϕ2 > 0 meaning
r(ϕ1) = r1, r(ϕ2) = r2. (6.2)
The taut chain would be a straight line going from A1 to A2, where we �rst
use the cosine rule in the triangle A1A2O such that
r21 + r22 − 2r1r2 cos(2α) = (2`)2
and now we use cos(2α) = 1− 2 sin(α) and then it follows
(r2 − r1)2 + 4r1r2(sin(α))2 = (2`)2.
For the maximal chain with the length r1 + r2, which would go through the
gravity center it would be true
(2`)2 = (r2 + r1)2.
Neither the maximal nor the minimal cases can happen, so the physical re-
striction is (r2 − r1
2
)2
+ r1r2(sin(α))2 < `2 <
(r2 + r1
2
)2
. (6.3)
We have used the same restriction in the chapter of the true symmetric cate-
nary for r1 = r2 which gives us (4.3). The problem is de�ned by the functional
F (r) = −∫ ϕ2
ϕ1
1
r
√r2 + r′2dϕ, (6.4)
with conditions
P (r) =
∫ ϕ2
ϕ1
√r2 + r′2dϕ = 2`, r(ϕ1) = r1, r(ϕ2) = r2. (6.5)
We use the Euler-Lagrange equation (5.1) for the function (5.2) and we con-
tinue with (5.4). Again we use the substitution λr =1
uand we are left with
58
The true catenary
solving (5.8), where the solution is (5.9) if c 6= 1. Every chain has a unique
minimum ϕ0, so u′(ϕ0) = 0. We de�ne the function ua as
u(ϕ) = ua(ϕ− ϕ0),
where both of the functions have the same extremal. Calculating the deriva-
tives
u′(ϕ) = u′a(ϕ− ϕ0),
u′′(ϕ) = u′′a(ϕ− ϕ0),
we see that there are no changes, so they both solve the same di�erential
equation and the result is (5.9), so as a consequence we can write
ua(ϕ) = A sinh
(√1− c2(ϕ− ϕ0)
c
)+B cosh
(√1− c2(ϕ− ϕ0)
c
)+
1
1− c2,
and make a derivative of it
u′a(ϕ) = A cosh
(√1− c2(ϕ− ϕ0)
c
) √1− c2c
+B sinh
(√1− c2(ϕ− ϕ0)
c
) √1− c2c
.
From
0 = u′a(ϕ0) = A
√1− c2c
,
such that A = 0 it follows
ua = B cosh
(√1− c2(ϕ− ϕ0)
c
)+
1
1− c2.
Putting this solution back to (5.4) with ϕ−ϕ0 = 0 we determine B =c
c2 − 1.
Considering the condition u′a(ϕ0) = 0 our solution can be expressed without
hyperbolic sine
ua =
1− c cosh(√
1− c2(ϕ− ϕ0)
c
)1− c2
. (6.6)
Here we can �nd the proof that the symmetric catenary has a minimum at
ϕ0 = 0.
59
The true catenary
Proof that the symmetric catenary has a minimum at ϕ0 = 0
There is the symmetric boundary condition u(−α) = u(α) and if follows
uα(−α− ϕ0) = uα(α− ϕ0). We are left with two options:
with either −α− ϕ0 = α− ϕ0 ⇒ −α = α
or −α− ϕ0 = −α + ϕ0 ⇒ ϕ0 = 0.
The �rst case is a contradiction, so we see that symmetric chains really have
a minimum at ϕ0 = 0. �
At this point we can transfer ua back to ra:
ra =1
λua=
1− c2
λ
(1− c cosh
(√1− c2c
(ϕ− ϕ0)
)) ,considering r(ϕ) = ra(ϕ− ϕ0) and r(ϕ0) = ra(0) we can de�ne r0:
r(ϕ0) =1− c2
λ(1− c)=
1 + c
λ=: r0. (6.7)
Without loss of generality we can assume that ϕ1 = 0 with a following second
angle ϕ2 = 2α by considering all of the boundary conditions in (6.2), geometric
restriction in (6.1) and physical restriction in (6.3). We look for a solution for
all ϕ on the interval (ϕ1, ϕ2) with data c > 0, ϕ0 ∈ R and r0 ∈ (0,∞)
u(ϕ0) = ua(0) =1− c1− c2
=1
1 + c,
r(ϕ) =(1 + c)r0
λ
(1− c cosh
(√1− c2c
(ϕ− ϕ0)
)) .Now we use r1 = r(0) and r2 = r(2α), we replace ϕ0 by ψ0 = ϕ0 − α, that ispossible because of the symmetry. Equation (5.9) with (6.7) leads to
r1
1− c cosh(√
1− c2c
(0− α− ψ0)
)1− c2
−r21− c cosh
(√1− c2c
(2α− α− ψ0)
)1− c2
= 0,
60
The true catenary
r1
1− c cosh(√
1− c2c
(α + ψ0)
)1− c2
− r21− c cosh
(√1− c2c
(α− ψ0)
)1− c2
= 0.
(6.8)
We read the length condition`
r1=
c sinh
(√1− c2αc
)√1− c2
for the asymmetric
catenary as
r1
c sinh
(√1− c2c
(α + ψ0)
)√1− c2
+ r2
c sinh
(√1− c2c
(α− ψ0)
)√1− c2
= 2`, (6.9)
where we have used the whole length of the curve. We want to solve equations
(6.8) and (6.9) for positive c and we introduce the new variable by using the
universal substitution t := tanh
(√1− c2c
ψ0
2
)∈ (−1, 1) ∪ iR, where
(tanh
(x2
))2=
cosh(x)− 1
cosh(x) + 1,
(sinh(x))2 = (cosh(x))2 − 1,
therefore
cosh
(√1− c2c
(ψ0)
)=
1 + t2
1− t2,
sinh
(√1− c2c
(ψ0)
)=
2t
1− t2.
Inserting the universal substitution into the �rst term of (6.8) with a note
that σ =
√1− c2c
α, we get
r1
1− c cosh(√
1− c2c
(α + ψ0)
)1− c2
=r1 − t2r1 − r1c cosh(σ)− t2r1c cosh(σ)− 2tr1c sinh(σ)
(1− c2)(1− t2)
61
The true catenary
and from the second term of (6.8)
r2
1− c cosh(√
1− c2c
(α− ψ0)
)1− c2
=r2 − t2r2 − r2c cosh(σ)− t2r2c cosh(σ) + 2tr2c sinh(σ)
(1− c2)(1− t2).
To make it less complicated we de�ne new parameters
a = r2 − r1, b = (r2 − r1)c cosh(σ), d = (r1 + r2)c sinh(σ). (6.10)
A simpli�ed equation without the denominator, which is not important for
solutions, except that t 6= ±1 and c 6= ±1 of (6.8) is
(a+ b)t2 − 2dt− a+ b = 0. (6.11)
Now we do the same for the �rst term of the length condition in (6.9)
r1
c sinh
(√1− c2c
(α + ψ0)
)√1− c2
=r1c sinh(σ) + t2r1c sinh(σ) + 2tr1c cosh(σ)√
1− c2(1− t2),
the second term of (6.9) is
r2
c sinh
(√1− c2c
(α− ψ0)
)√1− c2
=r2c sinh(σ) + t2r2c sinh(σ)− 2tr2c cosh(σ)√
1− c2(1− t2),
and the third term of (6.9) is
2` =2`(1− t2)
√1− c2√
1− c2(1− t)2.
62
The true catenary
We use the substitutions from (6.10) and add a new one
e = 2`√1− c2 (6.12)
and they together give us the simpli�ed equation (6.9)
(d+ e)t2 − 2bt− e+ d = 0. (6.13)
At this point we use a trick and we multiply equations (6.11) and (6.13) by t
and we get
(a+ b)t3 − 2dt2 + (b− a)t = 0,
(d+ e)t3 − 2bt2 + (d− e)t = 0.
The new-found equations, (6.11) and (6.13) together give us the determinant∣∣∣∣∣∣∣∣∣∣∣
a+ b −2d b− a 0
0 a+ b −2d b− ae+ d −2b d− e 0
0 e+ d −2b d− e
∣∣∣∣∣∣∣∣∣∣∣= 0,
where e+ d 6= 0. The result is
(d2 − b2)(d2 − b2 + a2 − e2) = 0. (6.14)
There are two di�erent options. Firstly, if from d2 = b2 we use the �rst option
d = b and we put it back into (6.11) and (6.13), sum them and consequently
we get the equation shown below
(a− e)(t2 − 1) = 0. (6.15)
If a 6= e, then t = ±1, which is a contradiction, because t is de�ned on the
interval (−1, 1), so a = e. It is also possible that d = −b, then
(a+ e)(t2 − 1) = 0, (6.16)
63
The true catenary
and a = −e. The situation where d2 = b2 and a2 6= e2 can not occur. Hence,
d2 − b2 = e2 − a2 is the only case to be solved, which by inserting back into
(6.10) and (6.12) and by using (cosh(σ))2 = 1 + (sinh(σ))2 looks like
−r22c2 + 2r1r2c2(2(sinh(σ))2 + 1)− r21c2 = −r22 + 2r1r2 − r21 + 4`2 − 4`2c2,
and can be transformed into
sinh(σ)
σ=
`2 −
(r2 − r1
2
)2
α2r1r2
1/2
.
The right-hand side runs fromsin(α)
αto
1
α.
We are still left with the problem of �nding ϕ0, where c 6= 1, which gives
us the unique solution for t on the set (−1, 1)∪ iR. Again we consider di�er-
ent options for d2−b2 = e2−a2, which can only be 0 without loss of generality,if e2 − a2 = 0 and inserting (6.12) and a from (6.10) in there, we can extract
c
4`2(1− c2)− (r2 − r1)2 = 0,
c =
√`2 −
(r2 − r1
2
)2
`.
We have eliminated negative values of c, because c has been established as a
positive constant. The t from equation (6.11) is
t1,2 =d∓ ea+ b
, (6.17)
and from equation (6.13) it is
t1,2 =b∓ ad+ e
. (6.18)
64
The true catenary
Now we discuss all the cases to see if there is any solution which ful�ls both
of the equations.
[1]b− ae+ d
=d− ea+ b
⇔ b2 − d2 = a2 − e2 ⇒ a = ±e,
this means (r2 − r1) = 2`√1− c2, with conditions r2 − r1 6= 0 and 0 < c < 1.
This solution is suitable.
[2]b+ a
e+ d=d− ea+ b
⇒ a(a+ b) = 0.
If a = −b, then by putting that back into the (6.17) then we determine that
the denominator is 0. If a = 0, then d2 − b2 = e2 looks like
r22 + r21 + r1r2(2(sinh(σ))2 + 1) = −4`(1− c2),
the left handed side is strictly positive and real, if 0 < c < 1, but then
the right side would be strictly negative which leads to contradiction. The
solutions does not match.
[3]b− ae+ d
=d+ e
a+ b⇒ e(e+ d) = 0,
e = d is not an option, because then the denominator in (6.18) is 0. In case
where e = 0 the equation d2 − b2 + a2 = 0 leads to
(r22 + r21)(1− c2) = −4r1r2(sinh(σ))2,
where the right side is real and strictly negative if 0 < c < 1, but then 1− c2
is positive just like the whole term on the left. Contradiction.
[4]b+ a
e+ d=d+ e
a+ b⇒ e(e+ d) = a(a− b),
⇔ 2`√1− c2(2`
√1− c2 + (r2 + r1)c(sinh(σ))
2) = (r2 − r1)2(1− cosh(σ)).
65
The true catenary
The right-hand side is strictly negative, because cosh(σ) > 1 for 0 < c < 1,
but the left handed side must be also negative, which is impossible, because
all of the factors must strictly be positive.
To summarize, t =b± ed+ e
, a = ±e where 0 < c < 1 is the only non-unique
solution. In all other cases t is unique. By multiplying the equation (6.11)
with (d+e) and the equation (6.13) with −(a+b) and to sum those equations
together we can express
t =eb− adB
,
where we de�nedB := A+de−ab, and A := d2−b2 = e2−a2. From−1 < t < 1
it follows that AB > 0. If A > 0, then B > 0, so ab < ed, whence 0 < A < B.
Otherwise A2 ≤ (ed − ab)2. We know that ab = (r2 − r1)2c cosh(σ) > 0,
therefore there are no other restrictions for the case where ed < 0 , because it
automatically follows that B < 0, but in case ed > 0 we must consider that
ab > ed. Therefore, if A < 0, then ab > ed. Knowing the t, we �rst calculate
ψ0 and then ϕ0.
The only case we have to consider separately is c = 1. We insert the value of
c into (5.8) and get
u′′(ϕ) = −1,
ua(ϕ) = −(ϕ− ϕ0)
2
2+ A(ϕ− ϕ0) +B,
u′a(ϕ0) = A,
ua(ϕ) = −ϕ− ϕ0
2+B.
It needs to be inserted back into (5.8) with ϕ− ϕ0 = 0 to determine B =1
2.
u(ϕ) =1
2(1− ϕ2)
66
The true catenary
is the solution for c = 1 and it leads to
r1(1− (α + ψ0)2)− r2(1− (α− ψ0)
2) = 0, (6.19)
which expresses the same equation as (6.8) for all other positive c. The length
condition is
ψ0(r2 − r1)− α(r1 + r2) = −2l (6.20)
just as (6.9) for positive c 6= 1. For the symmetric curve we can again see
that ψ0 = 0 by putting r1 = r2 back into (6.19), otherwise from (6.20) follows
ψ0 =
r2 + r12
α− `r2 − r1
2
. (6.21)
The calculated results for the true asymmetric catenary can be used to prove
the existence of our solution in Theorem 4.2 and the methods are in [2, pp.
132-141], but we will not prove it in this thesis.
67
The true catenary
7 Conclusion and outlook
In this thesis we explored the behaviour of the hanging chain in a gravita-
tional �eld. The Theorem 4.2 was proven by two di�erent methods.
In both methods we focused only on positive c, which is a constant from
the Euler-Lagrange equation. We used a sign of the curvature for extracting
the sign of the constant c. Moreover we discussed the case c = 1 separately.
The feature of the �rst method was using a trick to introduce a new vari-
able λr =1
uto the di�erential equation and made it solvable by transforming
it into a di�erential equation with constant coe�cients, where we used stan-
dard methods to solve it. We got the same form of solution for 0 < c < 1 and
c > 1 because the hyperbolic cosine could easily be changed into cosine.
In the second method there were introduced several new variables, one of
them was the angle between the radius vector and the tangent on the curve.
We had to solve tree di�erent integrals for 0 < c < 1, c = 1 and c > 1 where
we obtained results with cosine and hyperbolic cosine.
The true asymmetric catenary has had the same formula as the true sym-
metric one, but �rstly we had to consider all the conditions and then we
included a new variable which helped us transform the equations. The prob-
lem was �nally solved by �nding the ϕ0 with which we located the minimum
of the asymmetric curve.
There is a question whether the method from [5] can be used to solve the
asymmetric catenary problem, but since the author in that article used so
many di�erent new variables and had three di�erent cases of c it would be
68
The true catenary
too hard to follow the procedure for the asymmetric case.
Nevertheless, the result is the solution of the Euler-Lagrange equation for
the true catenary but it still has to be proven that this curve really minimizes
the functional.
69
The true catenary
8 References
References
[1] W. Blaschke, H. Reichardt, Einführung in die Di�erentialgeometrie,
Springer, Berlin, 1960.
[2] J. Denzler, A. M. Hinz, Catenaria Vera - The True Catenary , Exposi-
tion. Math. 17 (1999), 117-142.
[3] F. Kriºani£, I. Vidav, Navadne in parcialne diferencialne ena£be.
Variacijski ra£un, Dru²tvo matematikov in �zikov Slovenije, Ljubljana,
1991.
[4] S. Lang, Calculus of several variables , Springer, New York, 1987.
[5] M. Razpet, Prava simetri£na veriºnica, Obzornik za matematiko in �ziko
57 (2010), 121-133.
[6] M. Razpet, Veriºnica, Izleti v matemati£no vesolje, Presentation 2010,
http://http://izleti.famnit.upr.si/201011/slides/FAMNITvesolje2010-
Razpet.pdf, Online, accessed June 6, 2013.
[7] W. Rudin, Principles of mathematical analysis , Third edition, McGraw-
Hill, Auckland, 1976.
[8] G. F. Simmons, Di�erential equations with applications and historical
notes , Second edition, McGraw-Hill, New York, 1991.
[9] J. D. Struik, Lectures on classical di�erential geometry , Addison-Wesley,
London, 1961.
[10] J. L. Troutman, Variational calculus with elementary convexity , Springer,
New York, 1983.
70
REFERENCES The true catenary
[11] I. Vidav, Vi²ja matematika III , Dru²tvo matematikov, �zikov in astro-
nomov SR Slovenije, Ljubljana, 1976.
[12] E. Zakraj²ek, Analiza III , Dru²tvo matematikov, �zikov in astronomov
Slovenije, Ljubljana, 1998.
71
