1. Brachistochrone Problem
An Exploration of the Calculus of Variation
Xining Li
Introduction
In mathematicsq, a brachistochrone curve, or
curve of fastest descent, is the curve that would
carry an idealized point-like body, starting at
rest and moving along the curve, without fric-
tion, under constant gravity, to a given end
point in the shortest time. Given two points A
and B, with A not lower than B, only one up-
side down cycloid passes through both points,
has a vertical tangent line at A, and has no
maximum points between A and B: the brachis-
tochrone curve. The curve does not depend on
the body’s mass or on the strength of the grav-
itational constant. The problem can be solved
with the tools from the calculus of variations
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
x
Frictionless Tracks
x**2
x
x**(0.5)
tan(0.785398*x)
sin(1.5708*x)
Figure 1: Frictionless Tracks—GNUplot
The following equation is called the conser-
vation of energy:
mgy − mgy0 =
1
2
mv2 (1)
Based on Equation (1), we can get the follow-
ing equations.
v = 2gy (2)
dt =
ds
v
(3)
T =
dx2 + dy2
√
2gy
(4)
T =
1 + y
2
√
2gy
dx (5)
To calculate the Equation (5), we need to know
a basic theorem.
Fundamental Lemma
Lemma 1 (Fundamental Lemma of Calculus
of Variation).
b
a
M(x)h(x)dx = 0
With M(x) ∈ C2[a, b], for any arbitrary
h(x) ∈ C2[a, b] such that h(a) = 0 and h(b) =
0, we can conclude that M(x) = 0.
b
a
M(x)h(x)dx = 0 (6)
We are given that for any arbitrary h(x) such
that h(a) = 0 and h(b) = 0. Because the h(x) is
arbitrary, we let h(x) = −M(x)(x − a)(x − b)
∴ M(x)h(x) = M2(x)(−(x − a)(x − b)), for
a ≤ x ≤ b
Its easy to see that for a ≤ x ≤ b, (x − a) > 0
and (x − b) < 0.
∴ (−(x − a)(x − b)) > 0. Also, M2(x) ≥ 0
Equation (6) can be rewritten as this.
b
a
M2(x)(−(x − a)(x − b))dx = 0. Suppose
that M(x) > 0 at x = ξ for a ≤ ξ ≤ b, then
there would be a neighborhood in following
fashion because of M(x) ∈ C2[a, b].



[a, ξ0) when x = a
(ξ1, b] when x = b
(ξ2, ξ3) when a < x < b
In any of these cases, we know that
b
a
M2(x)(−(x − a)(x − b))dx > 0, which
contradicts Equation (6).
∴ M(x) = 0 is true when x ∈ [a, b]
Euler-Lagrange Equation
¯F(x) = F(x) + D(x) (7)
¯F(x) : the collection of possible functions
F(x) : the solution
D(x) : the difference between ¯F(x) and F(x)
We can re-write the equation (7) in equation
(8).
¯F(x) = F(x) + η(x) (8)
We can define that F(x, y + η, y + η ) =
1+¯y
2
2g¯y
.
¯F(x) : the collection of possible functions
F(x) : the solution
D(x) : the difference between ¯F(x) and D(x)
η(x) :
D(x)
: an variable
With the concept of Basic Methods of Calcu-
lus of Variation, we can re-write the Equation
(5) in this following way.
I( ) = T(¯y) =
x2
x1
F(x, y + η, y + η )dx
(9)
As approaches to 0, I( ) would reach its
minimum value because we are looking for the
minimum time. At the same time ¯y = y.
Because I( ) would reach its minimum value
as approaches to 0. We can say that
dI
dt
= 0 as
= 0.
Now we let u = y + η, and v = y + η , then
F(x, y + η, y + η ) = F(x, u, v)
Now we take the partial derivative of F with
respect to . We have.
∂F
∂
=
∂F
∂x
∂x
∂
+
∂F
∂u
∂u
∂
+
∂F
∂v
∂v
∂
(10)
u : y + η
v : y + η , which is the derivative of y
∂x
∂
: since x and are not related, this term is
equal to 0
∂u
∂
: η
∂v
∂
: η
∂F
∂
=
∂F
∂u
η +
∂F
∂v
η (11)
We know that x is not related to , and is ap-
proaching to 0. Based on the previous knowl-
edge, the following equation can be written.
0 =
dI( )
d




=0
=
x2
x1
∂F
∂y
η +
∂F
∂y
η dx (12)
The following equation is the alternative of
Equation (12)
0 =
x2
x1
η
∂F
∂y
−
d
dx
∂F
∂y
dx (13)
η is arbitrary.η(x1) = 0, η(x2) = 0. ∴
∂F
∂y
−
d
dx
∂F
∂y
= 0
The following Equation is called the Euler-
Lagrange Equation:
∂F
∂y
−
d
dx
∂F
∂y
= 0 (14)
To get the alternative form of Euler-Lagrange
Equation, let’s consider the following equa-
tions.
dF
dx
(x, y, y ) =
∂F
∂x
dx
dx
+
∂F
∂y
dy
dx
+
∂F
∂y
dy
dx
(15)
d
dx
y
∂F
∂y
=
∂F
∂y
dy
dx
+ y
d
dx
∂F
∂y
(16)
∂F
∂y
dy
dx
=
dF
dx
−
∂F
∂x
−
∂F
∂y
dy
dx
(17)
d
dx
y
∂F
∂y
=
dF
dx
−
∂F
∂x
−y
∂F
∂y
−
d
dx
∂F
∂y
(18)
d
dx
y
∂F
∂y
=
dF
dx
−
∂F
∂x
(19)
Here is the alternative form of Euler-
Lagrange Equation.
∂F
∂x
−
d
dx
F − y
∂F
∂y
= 0 (20)
The alternative form of Euler-Lagrange Equa-
tion can be very useful when dealing with
equation without x.
∂F
∂x
= 0
d
dx
F − y
∂F
∂y
= 0 (21)
F − y
∂F
∂y
= C (22)
[1]
Brachistochrone’s Solution
T =
x2
x1
1 + y
2
√
2gy
dx (23)
F = (1 + y
2
)1/2(2gy)−1/2 (24)
∂F
∂y
= (2gy)−1/2
1
2
(1 + y
2
)−1/22y (25)
1 + y
2
2gy
−
y y
√
2gy 1 + y
2
= C (26)
1
√
2gy 1 + y
2
= C (27)
(2gy) 1 + y
2
= C1 (28)
y 1 + y
2
= K (29)
y
K − y
dy = dx (30)
Here y is re-parameterized.
y =
K
2
(1 − cos(θ)) + V (31)
y = y − V =
K
2
(1 − cos(θ)) (32)
dy = dy =
K
2
sin(θ)dθ (33)
For simplicity, we can solve the relationship
between y and x instead.
The relationship between y and x can be re-
written as another differential equation.
K
2
(1 − cos(θ))
K −
K
2
(1 − cos(θ))
K
2
sin(θ)dθ = dx (34)
K
2
(1 − cos(θ))
K
2
+
K
2
cos(θ)
K
2
sin(θ)dθ = dx (35)



x =
K
2
(θ − sin(θ)) + H
y =
K
2
(1 − cos(θ)) + V
(36)
We can assume that the trajectory started at
point (H,V) when θ is equal to 0.



x =
K
2
(θ − sin(θ))
y =
K
2
(1 − cos(θ))
(37)
0
0.5
1
1.5
2
0 0.5 1 1.5 2
t-sin(t), 1-cos(t)
Figure 2: The Parametric Equation—GNUplot
Cycloid’s Geometry
−1.2−0.8−0.4 0.2 0.6 1. 1.4 1.8 2.2
0.2
0.4
0.6
0.8
1.
1.2
1.4
1.6
1.8
2.
A
B
c
a C
D
d
e
E
r α = 1.34rad
g
h
FG j k
H
m
Figure 3: Cycloid
As we already know, ◦A is rotating without
slipping and getting to ◦C.
∴
DE = BD (38)
αr = DE = BD (39)
x = BH = GF − EF = rα − r sin(α) (40)
y = BG = DC − FC = r − r cos(α) (41)
x = r(α − sin(α))
y = r(1 − cos(α))
(42)
As we can see, the K in Equation (37) is the di-
ameter of the rotating circle.
Cycloid’s Tautochronic
A tautochrone curve is the curve that the time
take by an object sliding without friction in uni-
form gravity to its lowest point is independent
of its starting point. The derivative of Equation
(42) is Equation (43)
dx
dθ = r(1 − cos(θ))
dy
dθ = r sin(θ)
(43)
ds = r 2 − 2 cos(θ)dθ (44)
For the easiest case, the object is sliding from
the highest point of the inverted cycloid to the
lowest point of the inverted cycloid. Based on
Equation (4) and Equation (43), the following
equation can be achieved.
T =
π
θ0
r 2 − 2 cos(θ)
2gr(cos(θ0) − cos(θ))
dθ (45)
sin
x
2
=
1 − cos(x)
2
(46)
cos
x
2
=
1 + cos(x)
2
(47)
cos(x) = 2 cos
x
2
2
− 1 (48)
The Equation (45) can be written as the fol-
lowing equation.
T =
r
g
π
θ0
sin 1
2θ
cos2(1
2θ0) − cos2(1
2θ)
dθ (49)
Set the variable u in the following fashion.
u =
cos(1
2θ)
cos(1
2θ0)
, du =
− sin(1
2θ)
2 cos(1
2θ0)
dθ (50)
T = 2
r
g
1
0
du
1 − u2
(51)
It is a arcsin integration.
T = π
r
g
(52)
The time taken by an object sliding without
friction in uniform gravity to its lowest point is
independent of its starting point.
Acknowledgement
I hereby acknowledge the accomplishment
of all the alumni of UC-Berkeley who have
worked on the LATEX and the Linux OS. Without
your dedication, it would be impossible for me
to finish my academic poster freely and effec-
tively.
References
[1] Hans Sagan. Introduction to the Calculus of
Variations (Kindle Edition). Dover Publica-
tions, 1992.