SPLINES 135(ii) Built-in(clamped) end: Where the first derivatives at t 0 and tn are specified as Φ 00 ′()t = g 0 or
Φn′–1() = tgnn, that isΦ 00 ′()t =–s 0 h 0 /2 – (y 0 /h 0 – s 0 h 0 /6) + (y 1 /h 0 – s 1 h 0 /6) = g 0and Φnn′–1()t–1 = –sn–1hn–1/2– (yn–1/hn–1– sn–1hn–1/6) + (yn/hn–1– snhn–1/6) = gn (5.11)(iii) Quadratic end spans: Where the end spans are quadratic, the end curvatures are constant, that
is,s 0 =s 1 and sn–1=sn.
We may use different combinations of end conditions from the above. Note that Eq. (5.10) form
a tri-diagonal system that can be solved efficiently to get the piecewise composite spline
Φ(t) = {Φi(t),i= 0,... , n–1}. We can set the values of yi,i = 0,... , n to shape the polynomial spline
as a basis function shown in Figure 5.2. As is, a polynomial spline is a two-dimensional composite
curve, however, with few disadvantages. Relocation of one or more data points requires computing
the entire spline again. Also, cubic polynomial splines are curvature continuous everywhere implying
that it may not be possible to model real life curves with slope or curvature discontinuities.
Example 5.1. Compute a cubic polynomial spline to fit the data points (0, 0), (1, 3) and (2, 0) with
free end conditions.
The three knots t 0 = 0, t 1 = 1 and t 2 = 2 are uniformly placed so that h 0 = h 1 = 1. From Eq. (5.10),
the following equation is to be solved for unknown slopes, s 0 ,s 1 and s 2.
s
h
s
hhs
hy
hy
hhy
h0
0
1
012
12
12 1
02
120
0+ 4 21
+1
+ =3
+ 3^1 –^1 –
⎛^3
⎝⎞
⎠⎛
⎝⎜⎞
⎠⎟For free end conditions, s 0 = s 2 = 0. Further using y 0 ,y 1 and y 2 as 0, 3 and 0 respectively,
4 s 1 = 0 ⇒s 1 = 0Using Eq. (5.7)
Φ 0 23–1( ) = [1 ]^231000
1111
0100
01230
3
0
0tttt = 9tt– 6⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥and
Φ 1 23–1( ) = [1 ]^22111 1
124 8
012 3
014123
0
0
0tttt = 6ttt– 27 + 36 – 12⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥⎡⎣⎢
⎢
⎢
⎢
⎢⎤⎦⎥
⎥
⎥
⎥
⎥Note that Φ 0 (1) = Φ 1 (1) = 3. Further, ΦΦ 01 ′′(1) = (1) = 0 and ΦΦ 01 ′′(1) = ′′(1)= – 18. A plot of the two
cubic spline segments is shown in Figure 5.4.