arXiv:1710.04265v1 [cs.NA] 11 Oct 2017

Solutions of Quadratic First-Order ODEs applied to
Computer Vision Problems
Plane Curve Reconstruction
UAH
David Casillas-Perez and Daniel Pizarro
October 13, 2017

Supervisor:

Daniel Pizarro

Abstract
The article proves the existence of a maximum of two possible solutions to the initial value
problem composed by the planar-perspective equation and an initial condition. This initial
value problem has a geometric interpretation. Solutions are curves than pass trough the initial
condition which is point of the plane.

1

Description of the problem

Let C be a regular curve in the plane R2 parameterized by the perspective parameterization X(t) =
ρ(t)(t, 1), where t ∈ I ⊂ R is an interval and ρ : I 7−→ R is the objective function. A complete
description of parametric curves is available in any generic differential geometric book as Kreyszig
(1991). Let U (t) be a function of class U (t) ∈ C 1 (I, R) defined in the interval I except for a
finite number of points A where the function is of class U (t) ∈ C ∞ (A, R) and where the first-order
derivative of the ρ function is null. The function U (t) is positive function in I and it has the
geometric interpretation of being the square of the modulus of the velocity vector of the curve C.
It means that U (t) = ||~v (t)||2 . This function U (t) is known. Consider the next non-linear firstorder ordinary differential equation (ODE ) (strictly speaking, the equation is a differential algebraic
equation DAE):
dρ 2
ε
+ ρ2 = U,
(1)
dt
where ρ(t) is the function that we have to find which is positive or null ρ(t) ≥ 0. The function ε = (1 + t2 )2 is a polynomial of degree 4. The variable t is the independent variable. We
consider the Initial Value Problem (IVP ) (also called the Cauchy problem) composed by the equation 1 (also called planar-perspective reconstruction equation) with the initial condition ρ(t0 ) = ρ0
in a neighborhood t0 ∈ J ⊂ I. The formal description of an initial value problem appear in the
book Tenenbaum and Pollard (1985).

1

2

Hypotheses

We want to prove that the IVP composed by the equation 1 and any Initial Condition (t0 , ρ0 ),
inside the domain of the equation, has a maximum of two possible solutions ρ1 (t) and ρ2 (t). In
other words, given a point in the plane p = (t0 , ρ0 ) compatible with the equation, there are only a
maimum of two curves that fulfill the equation 1 and pass through the point p = (t0 , ρ0 ).
Figure 1 shows empirical evidences that there are only two possible curves which passes trough
different initial conditions. The red stright line represent the curve we want to reconstruct. We

Figure 1: SfT of a stright line.
only know the square modulus of the velocity vector ~v . The magenta curve is called maximal curve.
This curve is obtained vanishing the term ρt = 0 in the original equation and solving ρ. It has the
property of contain all the points of solutions with derivative null. Any extrema of this curve is a
point with null derivative. In 1, the red stright line is tangent to the magenta maximal curve in a
point p with null derivative (it will be critical points during the proof). All this kind of points have
the normal of the curve parallel to the optical ray (the line that join p with the origin). The blue
and green curves are the two solutions that we obtain with different non-null initial conditions. We
can see that there exist solutions which pass trough the origin and go to the y < 1 semi-plane.
Figure 2 shows the reconstructed curves of the red straight line in a initial condition with null

2

derivative. We observe empirically that there exist only two possible analytic solutions or 4 C 1
function if we mix all the branches.

Figure 2: SfT of a stright line in a null initial condition.
Remark 1 It is possible to have curves without any non-null initial condition. For instance, the
red straight line in figure 2 with a positive grow has non-null initial conditions.

2.1

Case 1: The derivative in the initial condition is not null

Solving the equation 1 taking ρ(t0 ) = ρ0 , we obtain two different values of the derivatives in t0 . We
call both values ρ̇(t0 ) = ρ11) and ρ̇(t0 ) = ρ21) = −ρ11) . By the hypothesis, the derivative in the initial
condition is not null, therefore, both values have equal modulus and different sign.
The hypothesis ensures that both solutions are different and non-equal to 0 in t0 . Consequently, we
have two different ODEs (one per sign):
r
dρ
ρ2
=± U− ,
(2)
dt
ε
Calling the Picard-Lindelöf theorem for each ODE with the initial condition ρ0 , it is guaranteed
the existence and the uniqueness of both two solutions ρ1 (t) y ρ2 (t). Of course, we have to fulfill
3

the conditions of Picard-Lindelöf theorem. Picard-Lindelöf theorem is formally described in the
book Tenenbaum and Pollard (1985).
Now we wonder what happend when the initial condition is a critical point, that means, a point
where the derivative vanishes. Given a critical point, there exist two and only two curves that pass
through this point?
If the initial condition ρ(t0 ) = ρ0 is a critical point, the equation 1 becomes algebraic. To show that
there exist only two solutions we cannot call directly Picard-Lindelöf theorem. To prove that there
exist only two solutions we follow the next steps:

2.2

Case 2: The derivative in the initial condition is null

Firstly, we derivate both sides of the equation 1. We will use the next notation: ρji) = "ith derivative
of ρ raised to the j th power".
ε1) ρ21) + 2ερ1) ρ2) + 2ρ0) ρ1) = U1)

(3)

The obtained equation is second-order. Besides, observe how the highest-order of the equation is
not quadratic. Therefore, any bifurcation appears. Since we supposed that the first-order derivative
is null in the initial condition, the equation 3 is an identity and necessary it must fulfill U1) = 0.
Contrary, the equation would not have solution. We cannot solve the IVP with this equation and
the initial condition (t0 , ρ0 ) with null derivative. We repeat the process and derivate both sides of
the previous equation again:
ε2) ρ21) + 4ε1) ρ1) ρ2) + 2ε(ρ22) + ρ1) ρ3) ) + 2(ρ21) + ρ0) ρ2) ) = U2)

(4)

Now, the equation obtained is third-order. The coefficient of the higest-order term is the same in
both equation and equal to 2ερ1) . As we supposed that first-order derivative in the initial condition
is null, the equation 4 becomes a quadratic in t = t0 :
2ερ22) + ρ0) ρ2) = U2)

(5)

The previous equation 5 produce a bifurcation. Solving the equation, we obtain two different values
for the second-order derivatives in the initial condition due to its quadratic character. In this point,
we assume that the discriminant ρ20) + 8U2) ε > 0 in t = t0 to get the two different values. We will
refer to this pair of values of the second-order derivative as θ1 and θ2 . Again, we repeat the process
and derivate both side of the equation:
ε3) ρ21) + 6ε2) ρ1) ρ2) + 6ε1) (ρ22) + ρ1) ρ3) ) + 2ε(3ρ2) ρ3) + ρ1) ρ4) ) + 2(3ρ1) ρ2) + ρ0) ρ3) ) = U3)

(6)

The obtained equation is fourth-order. At this point, we prove the existence of only two possible
solutions. We can calculate the value of the third-order derivative for each value of the initial
condition (there are two because of the bifurcation) (ρ0 , 0, θ1 ) and (ρ0 , 0, θ2 ). Vanishing the firstorder derivative, we obtain the next third-order equation:
6ε1) ρ22) + 6ερ2) ρ3) + 2ρ0) ρ3) = U3)

(7)

For each initial condition (ρ0 , 0, θ1 ) y (ρ0 , 0, θ2 ) we obtain an unique value of the third-order derivative
(due to the fact that it is not going to appear a quadratic term in the last equation, at least in
the highest-order term, in this case ρ3) ). We will prove at the end of the article. If we follow
this procedure, we can build a Taylor serie for each initial condition calculates a = (ρ0 , 0, θ1 , ...) y
4

b = (ρ0 , 0, θ2 , ...) in t0 . Assuming that the function U (t) is analytic in the critical points, that means,
the Taylor serie of the function U (t) converge in a neighborhood (an interval) t0 ∈ K around θ0 . As
the operations that involves the equation (squares, sums, derivatives) are friendly with the analytic
property of the functions, then, Taylor series converge. That means that, the solutions derived are
analytic and particularly ρ1 (t) and ρ2 (t). Consequently, the solutions converge in a neighbouthood
of t0 .

ρ1 (t) =
ρ2 (t) =

∞
X
ai
i=0
∞
X
i=0

i!

(t − t0 )i

bi
(t − t0 )i
i!

t∈J
(8)
t∈J

Now, we prove the existence of only two solutions. As the obtained function analytic in t0 , the
Taylor serie converges in a neighbourhood J arround t0 . For any point of the t ∈ J, t 6= t0 , the
first-order derivative is not null and, consequently, que can apply the Picard-Lindellöf theorem to
guarantee the existence of only 2 solutions that pass through the critical point.
There exist a degenerate case where all the coefficients of the Taylor series are null except ρ0 . In
this case, the function ρ(t) = ρ0 is a constant. It is the only case where there is only one solution
that matches with the circumference.
Next points are important to generalize the proof for all order of the derivative. In the previous
proof, we assumed that the bifurcation appeared in the second-order derivative, but, in general, the
bifurcation may appear in the nth derivative.
a. Whatever the number of times we derivate the equation, the coeficient of the highest-order of
the derivative is always 2ερ1) . We can prove it easily using the Leibniz rule:
(n)

(f · g)

n  
X
n (k) (n−k)
=
f g
k

(9)

k=0

Substituting f = g = ρ(1) , we find that:
x(n) = (ρ(1) · ρ(1) )(n) =

n  
X
n (k+1) (n−k+1)
ρ
ρ
k

(10)

n  
X
n (k) (n−k)
ρ ρ
k

(11)

k=0

We do the same for the non-derivative term:
y (n) = (ρ · ρ)(n) =

k=0

Now, the original equation 1 can be rewritten as follow:
(12)

εx + y = U
The successive derivatives that it will be appear can be expressed as follow:
n  
X
n
k=0

k

ε(k) x(n−k) + y (n) = U (n)

5

n∈N

(13)

Fixing the iteration n, we always find that the highest-order derivative is obtained from the
term k = 0 of the summatory of the equation 13 and the terms k = 0 and k = n of the
summatory of the equation x(n) in 10 yielding to:

 
 
n (1) (n+1)
n (n+1) (1)
= 2ερ(1) ρ(n+1)
(14)
ε
ρ ρ
+
ρ
ρ
0
n
Consequently, this term vanishes in critical points reducing the order of the differential equation
in one.
b. Before continuing the proof, we show some first iterations of the expressions x(n) , y (n) and the
equation 13.
c. We show x(n) :
n=0
n=1
n=2
n=3
n=4
n=5
n=6
..
.
n=i−1
n=i
n=i+1
..
.

ρ(1) ρ(1)

2 10 ρ(1) ρ(2)


2 20 ρ(1) ρ(3) + 21 ρ(2) ρ(2)


2 30 ρ(1) ρ(4) + 2 31 ρ(2) ρ(3)



2 40 ρ(1) ρ(5) + 2 41 ρ(2) ρ(4) + 42 ρ(3) ρ(3)



2 50 ρ(1) ρ(6) + 2 51 ρ(2) ρ(5) + 2 52 ρ(3) ρ(4)




2 60 ρ(1) ρ(7) + 2 61 ρ(2) ρ(6) + 2 62 ρ(3) ρ(5) + 63 ρ(4) ρ(4)
..
.
 (1) (i)
 (2) (i−1)
 ( i−2 ) ( i+4 )
 ( i ) ( i+2 )
2
2
2 i−1
ρ + 2 i−1
ρ
+ · · · + 2 i−1
ρ 2 + 2 i−1
i−4 ρ
i−2 ρ 2 ρ
0 ρ
1 ρ
2
2


 ( i ) ( i+4 )
 i+2 i+2
i
2 0i ρ(1) ρ(i+1) + 2 1i ρ(2) ρ(i) + · · · + 2 i−2
ρ 2 ρ 2 + ii ρ( 2 ) ρ( 2 )
2
2
 (1) (i+2)

 ( i+2 ) ( i+4 )
i+1  ( 2i ) ( i+6
i+1 (2) (i+1)
)
2
ρ
2 i+1
ρ
+ 2 i+1
ρ
ρ
+
2
ρ
ρ
+
·
·
·
+
2
ρ 2 ρ 2
i−2
i
0
1
2
2
..
.
(15)

6

d. We show y (n) :
n=0
n=1
n=2
n=3
n=4
n=5
n=6
..
.
n=i−1
n=i
n=i+1
..
.

ρ(0) ρ(0)

2 10 ρ(0) ρ(1)


2 20 ρ(0) ρ(2) + 21 ρ(1) ρ(1)


2 30 ρ(0) ρ(3) + 2 31 ρ(1) ρ(2)



2 40 ρ(0) ρ(4) + 2 41 ρ(1) ρ(3) + 42 ρ(2) ρ(2)



2 50 ρ(0) ρ(5) + 2 51 ρ(1) ρ(4) + 2 52 ρ(2) ρ(3)




2 60 ρ(0) ρ(6) + 2 61 ρ(1) ρ(5) + 2 62 ρ(2) ρ(4) + 63 ρ(3) ρ(3)
..
.
 (0) (i−1)
 (1) (i−2)
 ( i−4 ) ( i+2 )
 ( i−2 ) ( i )
2
2
2 i−1
ρ 2 + 2 i−1
ρ 2
ρ
+ 2 i−1
ρ
+ · · · + 2 i−1
i−2 ρ
i−4 ρ
0 ρ
1 ρ
2
2
 i−2 i+2
 i


i
2 0i ρ(0) ρ(i) + 2 1i ρ(1) ρ(i−1) + · · · + 2 ii ρ( 2 ) ρ( 2 ) + ii ρ( 2 ) ρ( 2 )
2
2
 ( i ) ( i+2 )
 (0) (i+1)
 (1) (i)
 ( i−2 ) ( i+4 )
2
ρ 2 + 2 i+1
ρ 2 ρ 2
2 i+1
ρ
+ 2 i+1
ρ + · · · + 2 i+1
i−2 ρ
i
0 ρ
1 ρ
2
2
..
.
(16)

e. Now, we show the next iteration:
n=1

ε(1) x(0) + ε(0) x(1) + y (1) = U (1)

n=2

ε(2) x(0) + 2ε(1) x(1) + ε(0) x(2) + y (2) = U (2)

n=3

ε(3) x(0) + 3ε(2) x(1) + 3ε(1) x(2) + ε(0) x(3) + y (3) = U (3)

n=4

ε(4) x(0) + 4ε(3) x(1) + 6ε(2) x(2) + 4ε(1) x(3) + ε(0) x(4) + y (4) = U (4)

n=5

ε(5) x(0) + 5ε(4) x(1) + 10ε(3) x(2) + 10ε(2) x(3) + 5ε(1) x(4) + ε(0) x(5) + y (5) = U (5)

n=6
..
.

ε(6) x(0) + 6ε(5) x(1) + 15ε(4) x(2) + 20ε(3) x(3) + 15ε(2) x(4) + 6ε(1) x(5) + ε(0) x(6) + y (6) = U (6)
..
.

 (i−2) (1)
 (1) (i−2)
 (0) (i−1)
i−1 (i−1) (0)
x + i−1
x + · · · + i−1
x
+ i−1
x
+ y (i−1) = U (i−1)
0 ε
1 ε
i−2 ε
i−1 ε


 (1) (i−1)

i (i) (0)
i
+ 1i ε(i−1) x(1) + · · · + i−1
ε x
+ ii ε(0) x(i) + y (i) = U (i)
0 ε x

 (i) (1)
 (1) (i)
 (0) (i+1)
i+1 (i+1) (0)
x + i+1
+ · · · + i+1
x + i+1
x
+ y (i+1) = U (i+1)
0 ε
1 ε x
i ε
i+1 ε
..
.
(17)

n=i−1
n=i
n=i+1
..
.

7

f. The first time that we derivate the equation 1 and evaluate the initial condition (ρ0 , 0), we
obtain an identity and we cannot solve the second-order derivate. If we derivate again, we
obtain a second-order equation 5 in ρ(2) variable. We can obtain two different roots or one
double root. Derivating again and substituting the value of ρ(1) = 0 yields to the equation 7.
It is possible to solve the the value of ρ(3) if we know all the previous derivatives. The only
constrain is that the coefficient of this derivative has to be non-null 2(ρ(0) + 3ερ(2) 6= 0 in order
to solve it. But, there is a value of ρ(2) that vanishes the coefficient of this derivative and obeys
2(ρ(0) + 3ερ(2) ) = 0. We can prove that after the second order derivative, for each iteration of
the equation, the highest-order of the equation has the next coefficient:
2(ρ(0) + (i + 1)ρ(2) )

(18)

If this coefficient does not vanish for any iteration, then, we will achieve all the coefficients of
Taylor Series. We can prove it also that if for an iteration, this coefficient vanishes, then, in
the rest of iteration this coefficient is not vanished and we obtain a a system of more variables
than equations. But they are singular cases. For instance, if ρ(2) ≥ 0 then this coefficient will
never vanishes for any iteration.
The relation that has to obey ρ(0) and ρ(2) is for becoming in a special case is not common.
It is related with a countable infinite set.
g. Assuming that we can obtain all the coefficients, we can build a Taylor serie and impossing
analiticity in singular points. Taylor series of functions in analytical points converges in a
neighbourhood of the point to the function (see the book Can (2008)). Calling Picard-Lindelöf
to any point of the neighbourhood of the point proves the existence and the uniqueness as we
show in case of derivative non-null.
Quod erat demonstrandum

References
(2008). Taylor expansions and applications, pages 223–255. Springer Milan, Milano.
Kreyszig, E. (1991). Differential Geometry. Dover Childrens Books.
Tenenbaum, M. and Pollard, H. (1985). Ordinary Differential Equations. DOVER PUBN INC.

8

