Natural cubic splines

Transcription

1 Natural cubic splines Arne Morten Kvarving Department of Mathematical Sciences Norwegian University of Science and Technology October

2 Motivation We are given a large dataset, i.e. a function sampled in many points. We want to find an approximation in-between these points. Until now we have seen one way to do this, namely high order interpolation - we express the solution over the whole domain as one polynomial of degree N for N + 1 data points. a = t 0 t 1 t 2 t 3 t n 1 b = t n x

3 Motivation Let us consider the function Known as Runge s example. f (x) = x 2. While what we illustrate with this function is valid in general, this particular function is constructed to really amplify the problem.

4 Motivation Figure: Runge s example plotted on a grid with 100 equidistantly spaced grid points.

5 Motivation 8 7 exact interpolated Figure: Runge s example interpolated using a 15th order polynomial based on equidistant sample points.

6 Motivation It turns out that high order interpolation using a global polynomial often exhibit these oscillations hence it is dangerous to use (in particular on equidistant grids). Another strategy is to use piecewise interpolation. For instance, piecewise linear interpolation. y 0 s 0 (x) y 1 s 1 (x) s n 1 (x) x 0 x 1 x 2 x n 1 x n x

7 Motivation Figure: Runge s example interpolated using piecewise linear interpolation. We have used 7 points to interpolate the function in order to ensure that we can actually see the discontinuities on the plot.

8 A better strategy - spline interpolation We would like to avoid the Runge phenomenon for large datasets we cannot do higher order interpolation. The solution to this is using piecewise polynomial interpolation. However piecewise linear is not a good choice as the regularity of the solution is only C 0. These desires lead to splines and spline interpolation. s 0 (x) s 1 (x) s 2 (x) s n 1 (x) a = t 0 t 1 t 2 t 3 t n 1 b = t n x

9 Splines - definition A function S(x) is a spline of degree k on [a, b] if S C k 1 [a, b]. a = t 0 < t 1 < < t n = b and S 0 (x), t 0 x t 1 S 1 (x), t 1 x t 2 S(x) =. S n 1 (x), t n 1 x t n where S i (x) P k.

10 Cubic spline S 0 (x) = a 0 x 3 + b 0 x 2 + c 0 x + d 0, t 0 x t 1 S(x) =. S n 1 (x) = a n 1 x 3 + b n 1 x 2 + c n 1 x + d n 1, t n 1 x t n. which satisfies S(x) C 2 [t 0, t n ] : S i 1 (x i ) = S i (x i ) S i 1(x i ) = S i (x i ) S i 1(x i ) = S i (x i ), i = 1, 2,, n 1.

11 Cubic spline - interpolation Given (x i, y i ) n i=0. Task: Find S(x) such that it is a cubic spline interpolant. The requirement that it is to be a cubic spline gives us 3(n 1) equations. In addition we require that S(x i ) = y i, i = 0,, n which gives n + 1 equations. This means we have 4n 2 equations in total. We have 4n degrees of freedom (a i, b i, c i, d i ) n 1 i=0. Thus we have 2 degrees of freedom left.

12 Cubic spline - interpolation We can use these to define different subtypes of cubic splines: S (t 0 ) = S (t n ) = 0 - natural cubic spline. S (t 0 ), S (t n ) given - clamped cubic spline. } S 0 (t 1 ) = S 1 (t 1 ) S n 2 (t n 1 ) = S n 1 (t n 1 ) - Not a knot condition (MATLAB)

13 Natural cubic splines Task: Find S(x) such that it is a natural cubic spline. Let t i = x i, i = 0,, n. Let z i = S (x i ), i = 0,, n. This means the condition that it is a natural cubic spline is simply expressed as z 0 = z n = 0. Now, since S(x) is a third order polynomial we know that S (x) is a linear spline which interpolates (t i, z i ). Hence one strategy is to first construct the linear spline interpolant S (x), and then integrate that twice to obtain S(x).

14 Natural cubic splines The linear spline is simply expressed as S i x t i+1 x t i (x) = z i + z i+1. t i t i+1 t i+1 t i We introduce h i = t i+1 t i, i = 0,, n which leads to S (x) = z i+1 x t i h i + z i t i+1 x h i. We now integrate twice S i (x) = z i+1 (x t i ) 3 + z i (t i+1 x) 3 6h i 6h i + C i (x t i ) + D i (t i+1 x).

15 Natural cubic splines Interpolation gives: Continuity yields: S i (t i ) = y i z i 6 h2 i + D i h i = y i, i = 0,, n. S i (t i+1 ) = y i+1 z i+1 6 h2 i + C i h i = y i+1.

16 Natural cubic splines We insert these expressions to find the following form of the system S i (x) = z i+1 (x t i ) 3 + z i (t i+1 x) 3 6h i 6h ( i yi+1 + z ) i+1 h i 6 h i (x t i ) ( yi + h ) i h i 6 z i (t i+1 x). We then take the derivative.

17 The derivative reads Natural cubic splines S i (x) = z i+1 2h i In our abscissas this gives (x t i ) 2 z i 2h i (t i+1 x) (y i+1 y i ) h i h } i 6 (z i+1 z i ). {{ } b i S i (t i ) = 1 2 z ih i + b i h i 6 z i h iz i S i (t i+1 ) = z i+1 2 h i + b i h i 6 z i h iz i S i 1 (t i ) = 1 3 z ih i h i 1z i 1 + b i 1 S i (t i ) = S i 1 (t i ) 6 (b i b i 1 ) = h i 1 z (h i 1 + h i ) z i + h i z i+1.

18 Natural cubic splines - algorithm This means that we can find our solution using the following procedure: First do some precalculations h i = t i+1 t i, i = 0,, n 1 b i = 1 h i (y i+1 y i ), i = 0,, n 1 v i = 2 (h i 1 + h i ), i = 1,..., n 1 u i = 6 (b i b i 1 ), i = 1,, n 1 z 0 = z n = 0

19 Natural cubic splines - algorithm Then solve the tridiagonal system v 1 h 1 h 1 v 2 h 2 h 2 v 3 h hn 2 h n 2 v n 1 z 1 z 2 z 3. z n 2 z n 1 = u 1 u 2 u 3. u n 2 u n 1.

20 Natural cubic splines - example Given the dataset i x i y i h i = x i+1 x i b i = 1 h i (y i+1 y i ) v i = 2 (h i 1 + h i ) u i = 6 (b i b i 1 ) The linear system reads [ ] [ ] z1 = z 2 [ ]

21 Natural cubic splines - example We find z 0 = 0.5, z 1 = This gives us our spline functions S 0 (x) = (x 0.9) (x 0.9) (1.3 x) S 1 (x) = (x 1.3) (1.9 x) x S 2 (x) = (x 1.9) (x 1.9) (2.1 x)

22 Gaussian elimination of tridiagonal systems Assume we are given a general tridiagonal system d 1 c 1 b 1 a 1 d 2 c 2 b ,.. cn 1 b n 1 b n a n 1 d n First elimination (second row) yields d 1 c 1 0 d2 c , cn 1 a n 1 d n b 1 b 2. b n 1 b n, d 2 = d 2 a 1 d 1 c 1 b 2 = b 2 a 1 d 1 b 1

23 Gaussian elimination of tridiagonal systems This means that the elimination stage is for i =2,, n end m = a i 1 /d i 1 d i = d i mc i 1 b i = b i mb i 1 And the backward substitution reads where b 1 = b 1. x n = b n /d n for i =n 1,, 1 ) x i = ( b i c i x i+1 / d i end

24 Gaussian elimination of tridiagonal systems This will work out fine as long as d i 0. Assume that d i > a i 1 + c i - i.e. diagonal dominance. For the eliminated system diagonal domiance means that d i < c i. We now want to show that diagonal domiance of the original system implies that the eliminated system is also diagonal dominant.

25 Gaussian elimination of tridiagonal systems We now assume that d i 1 > c i 1. This is obviously satisfied for d 1 = d 1. d i = d i a i 1 d i 1 c i 1 d i a i 1 d i 1 c i 1 > a i 1 c i a i 1 = c i. Hence the diagonal domiance is preserved which means that d i 0. The algorithm produces a unique solution.

26 Why cubic splines? Now to motivate why we use cubic splines. First, let us introduce a measure for the smoothness of a function: µ(f ) = b We then have the following theorem Theorem a (f (x)) 2 dx. (1) Given interpolation data (t i, y i ) n i=0. Among all functions f C 2 [a, b] which interpolates (t i, y i ), the natural cubic spline is the smoothest, where smoothness is measured through (1).

27 We need to prove that Introduce Why cubic splines? µ(f ) µ(s) f C 2 [a, b]. g(x) = S(x) f (x), Inserting this yields µ(f ) = b a g(x) C 2 [a, b] g (t i ) = 0, i = 0,, n. ( S (x) g (x) ) 2 dx = µ(s) + µ(g) 2 b a S (x)g (x) dx Now since µ(g) > 0, we have proved our result if we can show that b a S (x)g (x) dx = 0.

28 Why cubic splines? We have that b a S (x)g (x) dx = g (x)s (x) b b a g (x)s (x) dx First part on the right hand side is zero since z 0 = z n = 0. Second part we split in an integral over each subdomain b a n 1 g (x)s (x) dx = i=0 i=0 t i+1 t i a g (x)s (x) dx n 1 ti+1 = 6a i g (x) dx n 1 = 6a i g(x) t i+1 t i = 0. i=0 t i

29 Cubic spline result Figure: Runge s example interpolated using cubic spline interpolation based on 15 equidistant samples.