Computer Aided Engineering Design

136 COMPUTER AIDED ENGINEERING DESIGN

5.4 B-Splines (Basis-Splines)

In Section 5.3, generation of a cubic polynomial spline Φ(t) was discussed with Φi(ti) = yi and
Φi(ti+ 1 ) = yi+ 1 i = 0,... , n– 1 with slope and curvature continuity at each knot, t. There were n+ 1
unknownss 0 ,... , sn, the first derivatives of the spline at each knot, with n–1 equations. Two
additional conditions were required to match the number of unknowns. In all, data points and two
conditions, that is, n+ 3 conditions were needed to completely determine the spline with n knot
spans. Construction of cubic polynomial spline is performed next such that its form appears like a
bell-shaped basis function much like the one in Figure 5.4. Consider a cubic spline Φ(t) with Φ(t),
Φ′(t) and Φ′′(t) all zero at each end of the knot vector leading to 6 conditions. From above, we can
observe that the number of spans n, required to determine a unique cubic spline is 3 (n+3 = 6). Let
the four knots be denoted by ti–3,ti–2,ti–1 and ti. The solution obtained over these knots, that is, Φ(t)
≡ 0 is trivial however and, therefore, we would need to increase the number of spans by 1 or introduce
a new knot, say ti–4. Thus, for n = 4 or in the knot span ti–4≤t≤ti, an additional condition is required.
(This is because from among the required n+ 3 = 7 conditions, 6 are already known). We can specify
a non-zero value of the spline at an internal knot, or alternatively, can standardize the spline. A way
suggested by Cox (1972) and de Boor (1972) is

t

t

i

i

tdt

–4 m

() =^1

∫

Φ (5.12)

wherem is the order (degree + 1) of the spline. For a cubic spline, m = 4. We can realize that
computing the cubic spline as above is an arduous procedure. Example 5.2 provides an insight even
though it is simplified for a uniform knot span.

Example 5.2.Construct a standard cubic spline over the knot span ti = i,i = 0,... , 4.
We may use the fact here that the knot placement being uniform and the boundary conditions being
symmetric, the standardized spline will be symmetric about t = 2. It is thus required to compute the
spline only in two segments, Φ 0 (t) in 0≤t≤ 1 and Φ 1 (t) in 1≤t≤ 2. Since the spline is cubic,

4

3.5

3

2.5

2

1.5

1

0.5

0

y Φ 0 (t) Φ 1 (t)

0 0.5 1 1.5 2

t

Figure 5.4 Plot of Splines ΦΦΦΦΦ 0 (t) = 9t^2 – 6t^3 , 0 ≤t≤ 1 and ΦΦΦΦΦ 1 (t) = 6t^3 – 27t^2 + 36t – 12, 1 ≤t≤ 2