Cubic splines are preferred for many applications. Consider the spline composed ofcubic polynomials of the form
(1)
so that,(2) and
(3) for
The interpolating function must pass through all the data points, be continuous,
and have continuous derivative at the interior points. Furthermore, there are enough degrees
of freedom to impose continuous curvature as well (continuity of second derivative) at the interior points.
(4)
(5)
(6)
(7)
We haveunknowns and
equations (
from each of (1) and (2) and
from each of (3) and (4). Hence, we have the freedom to impose two extra conditions.
We can saythe width of the
interval, and
Next, we shall write all the equations in terms of(given) and C-k (unknown) only.
From equations (3) and (7)
From equations (1) and (4),
From equations (1) and (5)
So far we have obtained expressions for the coefficientsand
in terms of
and
All that remains is to match the slopes at the interior points.
From equations (2) and (6):
Substitute into this(8) and
(9) to give
After some simple algebra, we obtain a system ofequations to solve for
The problem as been reduced from findingcoefficients
and
to finding
values of the curvature
We only haveequations for
but we can obtain two additional equations by specifying
end conditions on the curvature, i.e. conditions involvingand
-
Natural or free boundaries: take
i.e. the end splines have no curvature at the end points.
-
Clamped boundaries: fix the slopes at each end to specified values.
-
Periodic boundaries: take
The last equations can be written in matrix form,
In the case of natural splines (i.e. for), the matrix is non-singular and has a tri-diagonal structure. The solution for
is unique and can easily be obtained using rapid tri-diagonal matrix solvers.
We can solvefor
and from this construct the splines
Example: Find the natural spline for the following data:
|
1 |
2 |
3 |
4 |
|
0 |
1 |
3 |
4 |
|
2 |
4 |
3 |
4 |
From this
MC=F becomes
Hence
Thus, using equations (8), (9), andwe obtain