1 Introduction
In this document we are going to prove that a cubic Bézier curve
(1) |
with the special control point configuration
(2) |
has at most one local extremum of curvature when and .
2 Special cases
We need to treat two special cases first. When , the curve degenerates to a line segment, which has a kink (a point with infinite curvature) at , so it clearly has exactly one local curvature extremum.
With this handled, let us set, without loss of generality,
(3) |
where . The second special case is when . Once again, the curve degenerates to a line segment. When , its curvature is always , otherwise it has a single kink.
In the following we will assume .
3 Curvature extrema
The signed curvature of a planar polynomial curve is given [1] as
(4) |
where is the first derivative of with respect to etc. The derivative of its square is
(5) |
so
(6) |
Here means that the curvature is equal to , and it corresponds to inflection points. Consequently,
(7) |
corresponds to curvature extrema.
4 Main proof
Let denote the left-hand side of Eq. (7), applied to our curve, with control points given as in (3). We need to show that has at most one 0-crossing when and .
First note that
(8) |
which is positive for . Let us denote the expression in brackets with . We have , and
(9) |
from which we can see that is negative at , and increases with . We can conclude that as goes from to , first decreases, and then it may increase, but since , it always remains negative, so
(10) |
The derivative of with respect to is given as
(11) |
where
(12) | ||||
(13) |
It is easy to see that , since its derivative with respect to is negative, and is positive. Consequently, the signs of and are the same.
As for , note that it is a quadratic function of , and the coefficient of is , which is negative for . We can also prove
(14) |
by ascertaining that , and , as this means that decreases as goes from to , starting from a negative value, and thus itself is always negative, as well.
4.1 Case I:
The maximum of for a fixed value is found by solving , giving
(15) |
When , this will be in the interval, and
(16) | ||||
(17) | ||||
(18) |
Since is quadratic in , and also , and , we can see that as goes from to , the value of decreases, starting from a negative value, and while it may start to increase, it remains negative.
In summary, we have shown that in this case the maximum of is negative, so , i.e., the curvature decreases monotonically.
4.2 Case II:
In this case takes its maximum over the interval at . We can also state the following (these will be proved below):
(19) | ||||
(20) | ||||
(21) |
Equation (21) shows that decreases monotonically as goes from to , while by Eq. (20) it starts from a positive value, so it may have at most one 0-crossing. Consequently first increases, starting from a negative value (Eq. 14), and then it may decrease.
When , since this is its maximal value, it means that is always negative, so the curvature decreases monotonically.
Otherwise has exactly one 0-crossing, so first decreases, starting from a positive value, and then increases, as goes from to . Since in this case (Eq. 19), there is exactly one curvature extremum.
4.2.1 Proof of Eq. (19)
Since
(22) |
it can be seen that can only be positive when . Under these constraints it holds that , so
(23) |
Consequently, the assumption implies .
Restructuring
(24) | ||||
(25) |
shows that is negative when is outside the circle with center and radius . But will be outside for any , which proves Eq. (19).
4.2.2 Proof of Eq. (20)
4.2.3 Proof of Eq. (21)
Derivating twice,
(27) |
which is negative for all .
References
- [1] M. P. doCarmo, Differential geometry of curves and surfaces. Prentice-Hall, 1976.
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