Cubic Hermite spline

Unit interval [0, 1]

On the unit interval ${\displaystyle [0,1]}$, given a starting point ${\displaystyle {\boldsymbol {p}}_{0}}$ at ${\displaystyle t=0}$ and an ending point ${\displaystyle {\boldsymbol {p}}_{1}}$ at ${\displaystyle t=1}$ with starting tangent ${\displaystyle {\boldsymbol {m}}_{0}}$ at ${\displaystyle t=0}$ and ending tangent ${\displaystyle {\boldsymbol {m}}_{1}}$ at ${\displaystyle t=1}$, the polynomial can be defined by

$${\displaystyle {\boldsymbol {p}}(t)=\left(2t^{3}-3t^{2}+1\right){\boldsymbol {p}}_{0}+\left(t^{3}-2t^{2}+t\right){\boldsymbol {m}}_{0}+\left(-2t^{3}+3t^{2}\right){\boldsymbol {p}}_{1}+\left(t^{3}-t^{2}\right){\boldsymbol {m}}_{1},}$$

where t ∈ [0, 1].

Interpolation on an arbitrary interval

Interpolating ${\displaystyle x}$ in an arbitrary interval ${\displaystyle (x_{k},x_{k+1})}$ is done by mapping the latter to ${\displaystyle [0,1]}$ through an affine (degree-1) change of variable. The formula is

$${\displaystyle {\boldsymbol {p}}(x)=h_{00}(t){\boldsymbol {p}}_{k}+h_{10}(t)(x_{k+1}-x_{k}){\boldsymbol {m}}_{k}+h_{01}(t){\boldsymbol {p}}_{k+1}+h_{11}(t)(x_{k+1}-x_{k}){\boldsymbol {m}}_{k+1},}$$

where ${\displaystyle t=(x-x_{k})/(x_{k+1}-x_{k})}$, and ${\displaystyle h}$ refers to the basis functions, defined below. Note that the tangent values have been scaled by ${\displaystyle x_{k+1}-x_{k}}$ compared to the equation on the unit interval.

Uniqueness

The formula specified above provides the unique third-degree polynomial path between the two points with the given tangents.

Proof. Let ${\displaystyle P,Q}$ be two third-degree polynomials satisfying the given boundary conditions. Define ${\displaystyle R=Q-P,}$ then:

${\displaystyle R(0)=Q(0)-P(0)=0,}$

${\displaystyle R(1)=Q(1)-P(1)=0.}$

Since both ${\displaystyle Q}$ and ${\displaystyle P}$ are third-degree polynomials, ${\displaystyle R}$ is at most a third-degree polynomial. So ${\displaystyle R}$ must be of the form

$${\displaystyle R(x)=ax(x-1)(x-r).}$$

Calculating the derivative gives

$${\displaystyle R'(x)=ax(x-1)+ax(x-r)+a(x-1)(x-r).}$$

We know furthermore that

${\displaystyle R'(0)=Q'(0)-P'(0)=0,}$

${\displaystyle R'(0)=0=ar,}$

(1)

${\displaystyle R'(1)=Q'(1)-P'(1)=0,}$

${\displaystyle R'(1)=0=a(1-r).}$

(2)

Putting ( 1 ) and ( 2 ) together, we deduce that ${\displaystyle a=0}$, and therefore ${\displaystyle R=0,}$ thus ${\displaystyle P=Q.}$

Representations

We can write the interpolation polynomial as

$${\displaystyle {\boldsymbol {p}}(t)=h_{00}(t){\boldsymbol {p}}_{0}+h_{10}(t)(x_{k+1}-x_{k}){\boldsymbol {m}}_{0}+h_{01}(t){\boldsymbol {p}}_{1}+h_{11}(t)(x_{k+1}-x_{k}){\boldsymbol {m}}_{1}}$$

where ${\displaystyle h_{00}}$, ${\displaystyle h_{10}}$, ${\displaystyle h_{01}}$, ${\displaystyle h_{11}}$ are Hermite basis functions. These can be written in different ways, each way revealing different properties:

	expanded	factorized	Bernstein
${\displaystyle h_{00}(t)}$	${\displaystyle 2t^{3}-3t^{2}+1}$	${\displaystyle (1+2t)(1-t)^{2}}$	${\displaystyle B_{0}(t)+B_{1}(t)}$
${\displaystyle h_{10}(t)}$	${\displaystyle t^{3}-2t^{2}+t}$	${\displaystyle t(1-t)^{2}}$	${\displaystyle {\tfrac {1}{3}}B_{1}(t)}$
${\displaystyle h_{01}(t)}$	${\displaystyle -2t^{3}+3t^{2}}$	${\displaystyle t^{2}(3-2t)}$	${\displaystyle B_{3}(t)+B_{2}(t)}$
${\displaystyle h_{11}(t)}$	${\displaystyle t^{3}-t^{2}}$	${\displaystyle t^{2}(t-1)}$	${\displaystyle -{\tfrac {1}{3}}B_{2}(t)}$

The "expanded" column shows the representation used in the definition above. The "factorized" column shows immediately that ${\displaystyle h_{10}}$ and ${\displaystyle h_{11}}$ are zero at the boundaries. You can further conclude that ${\displaystyle h_{01}}$ and ${\displaystyle h_{11}}$ have a zero of multiplicity 2 at 0, and ${\displaystyle h_{00}}$ and ${\displaystyle h_{10}}$ have such a zero at 1, thus they have slope 0 at those boundaries. The "Bernstein" column shows the decomposition of the Hermite basis functions into Bernstein polynomials of order 3:

$${\displaystyle B_{k}(t)={\binom {3}{k}}\cdot t^{k}\cdot (1-t)^{3-k}.}$$

Using this connection you can express cubic Hermite interpolation in terms of cubic Bézier curves with respect to the four values ${\textstyle {\boldsymbol {p}}_{0},{\boldsymbol {p}}_{0}+{\frac {1}{3}}{\boldsymbol {m}}_{0},{\boldsymbol {p}}_{1}-{\frac {1}{3}}{\boldsymbol {m}}_{1},{\boldsymbol {p}}_{1}}$ and do Hermite interpolation using the de Casteljau algorithm. It shows that in a cubic Bézier patch the two control points in the middle determine the tangents of the interpolation curve at the respective outer points.

We can also write the polynomial in standard form as

$${\displaystyle {\boldsymbol {p}}(t)=\left(2{\boldsymbol {p}}_{0}+{\boldsymbol {m}}_{0}-2{\boldsymbol {p}}_{1}+{\boldsymbol {m}}_{1}\right)t^{3}+\left(-3{\boldsymbol {p}}_{0}+3{\boldsymbol {p}}_{1}-2{\boldsymbol {m}}_{0}-{\boldsymbol {m}}_{1}\right)t^{2}+{\boldsymbol {m}}_{0}t+{\boldsymbol {p}}_{0}}$$

where the control points and tangents are coefficients. This permits efficient evaluation of the polynomial at various values of t since the constant coefficients can be computed once and reused.

Finite difference

The simplest choice is the three-point difference, not requiring constant interval lengths:

${\displaystyle {\boldsymbol {m}}_{k}={\frac {1}{2}}\left({\frac {{\boldsymbol {p}}_{k+1}-{\boldsymbol {p}}_{k}}{x_{k+1}-x_{k}}}+{\frac {{\boldsymbol {p}}_{k}-{\boldsymbol {p}}_{k-1}}{x_{k}-x_{k-1}}}\right)}$

for internal points ${\displaystyle k=2,\dots ,n-1}$, and one-sided difference at the endpoints of the data set.

Cardinal spline

A cardinal spline, sometimes called a canonical spline, ^[4] is obtained ^[5] if

${\displaystyle {\boldsymbol {m}}_{k}=(1-c){\frac {{\boldsymbol {p}}_{k+1}-{\boldsymbol {p}}_{k-1}}{x_{k+1}-x_{k-1}}}}$

is used to calculate the tangents. The parameter c is a tension parameter that must be in the interval [0, 1]. In some sense, this can be interpreted as the "length" of the tangent. Choosing c = 1 yields all zero tangents, and choosing c = 0 yields a Catmull–Rom spline in the uniform parameterization case.

Catmull–Rom spline

For tangents chosen to be

${\displaystyle {\boldsymbol {m}}_{k}={\frac {{\boldsymbol {p}}_{k+1}-{\boldsymbol {p}}_{k-1}}{2}}}$

a Catmull–Rom spline is obtained, being a special case of a cardinal spline. This assumes uniform parameter spacing.

The curve is named after Edwin Catmull and Raphael Rom. The principal advantage of this technique is that the points along the original set of points also make up the control points for the spline curve. ^[7] Two additional points are required on either end of the curve. The uniform Catmull–Rom implementation can produce loops and self-intersections. The chordal and centripetal Catmull–Rom implementations ^[8] solve this problem, but use a slightly different calculation. ^[9] In computer graphics, Catmull–Rom splines are frequently used to get smooth interpolated motion between key frames. For example, most camera path animations generated from discrete key-frames are handled using Catmull–Rom splines. They are popular mainly for being relatively easy to compute, guaranteeing that each key frame position will be hit exactly, and also guaranteeing that the tangents of the generated curve are continuous over multiple segments.

Kochanek–Bartels spline

A Kochanek–Bartels spline is a further generalization on how to choose the tangents given the data points ${\displaystyle {\boldsymbol {p}}_{k-1}}$, ${\displaystyle {\boldsymbol {p}}_{k}}$ and ${\displaystyle {\boldsymbol {p}}_{k+1}}$, with three parameters possible: tension, bias and a continuity parameter.

Monotone cubic interpolation

If a cubic Hermite spline of any of the above listed types is used for interpolation of a monotonic data set, the interpolated function will not necessarily be monotonic, but monotonicity can be preserved by adjusting the tangents.