Cubic Splines. Antony Jameson. Department of Aeronautics and Astronautics, Stanford University, Stanford, California, 94305

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1 Cubic Splines Antony Jameson Department of Aeronautics and Astronautics, Stanford University, Stanford, California, References on splines 1. J. H. Ahlberg, E. N. Nilson, J. H. Walsh. Theory of splines and their applications, Academic Press, I. J. Schoenberg (ed.). Spline functions, (Symposium at U. Wisconsin, 1969), Academic Press, I. J. Schoenberg. Quart. Appl. Math. 4, 1946, pp , pp C. H. Reinsch. Numer. Math. 10, 1967, pp Smoothing by spline functions. 5. Schulz, Spline Analysis. 1

2 2 Definition of spline A spline is a piecewise polynomial in which the coefficients of each polynomial are fixed between knots or joints. Figure 1: Typically cubics are used. Then the coefficients are chosen to match the function and its first and second derivatives at eacoint. There remain one free condition at each end, or two conditions at one end. However, using only starting conditions the spline is unstable. In general with n th degree polynomials one can obtain continuity up to the n 1 derivative. The most common spline is a cubic spline. Then the spline function y(x) satisfies y (4) (x) = 0, y(3)(x) = const, y (x) = a(x) + h. But for a beam between simple supports y (x) = M(x) EI where M(x) varies linearly. Thus a spline is the curve obtained from a draughtsman s spline. 2

3 Figure 2: Draughtsman s spline 3 Equations of cubic spline Let data be given at x 0, x 1, x n with values y 0, y 1, y n. Let S(x) bet the spline. Let M j = S (x j ), = x j x j 1 be moment at j th point. Then between x j 1 and x j S (x) = M j 1 x j x + M j x x j 1 But S (x) = M j 1 (x j x) M j (x x j 1 ) A (x j x) 3 (x x j 1 ) 3 S(x) = M j 1 + M j + Ax + B 6 6 S(x j 1 ) = y j 1, S(x j ) = y j. We obtain ( (x j x) 3 (x x j 1 ) 3 h 2 ) ( j xj x h 2 ) j x xj S(x) = M j 1 +M j + y j 1 M j 1 + y j + M j

4 S (x) = M j 1 (x j x) M j (x x j 1 ) y j y j 1 M j M j 1 6 We now have continuity of S (x) and S(x) at x j. We also require continuity of S (x). Now S (x j ) = 6 M j M j + y j y j 1 S (x + j ) = +1 M j M j+1 + y j+1 y j +1 Equating these gives equations for the M j at every interior node: 6 M j M j M j+1 = y j+1 y j +1 y j y j 1 To complete the system we need 2 additional equations. Setting S (x 0 ) = M 0 = 0, S (x n ) = M n = 0, corresponds to simple supports at the ends of the beam. Better accuracy is obtained if S (x 0 ) and S (x n ) are known. Then from the equation for S (x n ) we have h 1 3 M 0 + h 1 6 M 1 = y 1 y 0 h 1 y 0 h n 6 M n 1 + h n 3 M n = y n y n y n 1 h n If neither S (x) nor S (x) are known at the ends, one may set S (x) = 0, or M 1 M 0 = 0 M n 1 M n = 0 Alternatively one may use some linear combination of these end conditions. In any 4

5 case one obtains a tridiagonal set of equations which may be written 2 λ 0 µ 1 2 λ 1 µ n 1 2 λ n 1 µ n 2 M 0 M 1 M n 1 M n = d 0 d 1 d n 1 d n where at interior points λ j = , µ j = 1 λ j d j = 6 (y j+1 y j )/+1 (y j y j 1 )/ + +1 and λ 0, µ n, d 0, d n depend on the end conditions. These equations are diagonally dominant, leading to stable behaviour. Spline equations for first derivatives Let m j = S (x j ) Then S(x) = m j 1 (x j x) 2 (x x j 1 ) h 2 j +y j 1 (x j x) 2 [2(x x j 1 ) + ] h 3 j 5 m j (x x j 1 ) 2 (x j x) h 2 j + y j (x x j 1 ) 2 [2(x j x) + ] h 3 j

6 S (x j x)(2x j 1 + x j 3x) (x x j 1 )(2x j x j 1 3x) (x) = m j 1 m h 2 j j h 2 j +6 y j y j 1 (x h 3 j x)(x x j 1 ) j S (x) = 2m j 1 2x j + x j 1 3x h 2 j 2m j 2x j 1 + x j 3x) h 2 j +6 y j y j 1 (x h 3 j + x j 1 2x) j S (x j ) = 2m j 1 + 4m j 6 y j y j 1 h 2 j S (x + j ) = 4m j 2m j y j+1 y j h 2 j+1 Equating these gives ( 1 1 m j ) m j + 1 m j+1 = 3 y j y j y j+1 y j h 2 j h 2 j+1 Then 2 µ 0 λ 1 2 µ 1 λ N 1 2 µ N 1 λ N 2 m 0 m 1 m N 1 m N = c 0 c 1 c N 1 c N where at interior points c j = 3λ j y j y j 1 + 3µ j y j+1 y j +1 6

7 The end condition S = 0 gives µ 0 = 1, c 0 = 3 y j y 0 λ N = 1, c N = 3 y N y N 1 h N 1 4 Minimum curvature property of splines Consider all functions f(x) such that f(x i ) = y i Then the spline S(x) with S (x 0 ) = S (x n ) = 0 is that function f(x) which minimize: xn x 0 f (x) 2 dx (Ahlberg, p. 76) Proof = xn xn x 0 f (x) 2 dx 2 x 0 (f (x) S (x)) 2 dx (f (x) S (x)) S (x)dx S (x) 2 dx The middle term is n i=1 xi Integrating by parts it becomes x i 1 (f (x) S (x)) S (x)dx n n [(f (x) S (x)) S (x)] i i 1 i=1 i=1 xi x i 1 (f (x) S (x)) S (x)dx 7

8 S (x) is constant in each segment, and can be taken out of each integral in the second term. But xi which vanishes because x i 1 (f (x) S (x)) dx = [f(x) S(x)] i i 1 f(x i ) = S(x i ) = y i, i = 0, 1,... n Also the contributions to the sum in the first term vanish at every interior node, leaving (f (x n ) S (x n )) S (x n ) (f (x 0 ) S (x 0 )) S (x 0 ) Thus xn x 0 (f (x) S (x)) S (x)dx = 0 if either S (x) = 0 at each end or f (x) = S (x). Consequently xn x 0 f (x) 2 dx xn xn S (x) 2 dx = (f (x) S (x)) 2 dx > 0 x 0 x 0 unless f(x) = S(x). 8

9 5 Errors in spline approximation If the maximum interval is h then f(x) S(x) h 4 f (x) S (x) h 3 f (x) S (x) h 2 f (x) S (x) h This is proved by Ahlberg, Nilson and Walsh, p. 29, provided that f (4) is continuous. To study convergence properties of splines we need an estimate of the norm of the inverse of the matrix B of the equations for the spline. Taking x = max x i the induced norm of B is Bx B = sup x 0 x = max i n b ij j=1 9

10 With a diagonally dominant matrix we have n y = Bx = max b ij x j i j=1 n b ij x j j=1 where x = x k, or y { b kk j k { min i b kj b ii j i } b ij x } x Thus B 1 B 1 y = sup y 0 y [ min i { b i i j i b ij }] 1 or for the spline matrix B 1 max [ (2 λ 0 ) 1, (2 λ n ) 1, 1 ] Now let δ = max and let max δ β <. Then the spline has the following convergence propety as the maximum mesh spacing 10

11 δ is reduced: Theorem A: (Ahlberg, p. 22) Let f(x) be continuous on [a, b] and let max [ λ 0, µ n ] < 2. Then if S(x) is a spline fit to f(x) with x 0 = a, x n = b, then f(x) S(x) 0 as δ 0. Also if f(x) f(x ) κ x x α, 0 α 1 then f(x) S(x) = O (δ α ). Proof: On [x j 1, x j ] the spline equations can be arranged as S(x) f(x) = M j 1(x j x) [ ] (x j x) 2 h 2 j + M [ j(x x j 1 ) (x xj 1 ) 2 hj] f j + f j 1 2 f(x) + (f j f j 1 ) x j + x j 1 2x 2 By maximizing the coefficients of M j and M j 1 we find that they do not exceed h 2 j/3 5/2. Let A i j be the elements of B 1 where B is the coefficient matrix. Then n 1 (f i+1 f i )/h i+1 (f i f i 1 )/h i M j = 6 A ji + A j0 d 0 + A jn d n. h i + h i+1 i=1 11

12 Also let µ(f, δ) be the modulus of continuity of f(x) on [a, b] defined as µ(f, δ) = sup f(x) f(x ) x x <δ Then remembering the condition that δ β, we find that h 2 j (f i+1 f i )/h i+1 (f i f i 1 )/h i h i + h i+1 β2 µ(f, δ) Thus h 2 j ( M j 1 + M j ) 2 B 1 { 6β 2 µ(f, δ) + δ 2 ( d 0 + d n ) } and finally S(x) f(x) B 1 { 6β 2 µ(f, δ) + δ 2 ( d 0 + d n ) } ( + µ(f, δ) ) 2 If the function satisfies f(x) f(x ) κ(x x ) α, 0 < α 1 then µ(f, δ) κδ α establishing the required result, in light of the inequality satisfied by B 1, provided that λ < 2, µ n < 2. If f is in C 2 we can prove Theorem B: (Ahlberg, p. 27) 12

13 f (p) x S (p) (x) = O(δ 2 p ) where S(x) satisfies end conditions 2M 0 + M 1 = 6 ( ) f1 f 0 f 0 h 1 h 1 or M 0 = f 0. Proof: If B is the matrix of the spline we have BM = d whence B ( M d ) ( = I B ) d 3 3 The right side is 1 3 d 0 d 1 µ 1 (d 1 d 0 ) λ 1 (d 2 d 1 ) d N d N 1 Now at interior points d j 6 = (f j+1 f j )/+1 (f j f j 1 )/ + +1 = f[x j 1, x j, x j+1 ] = 1 2 f (ξ j ), x j 1 ξ j x j+1 13

14 Also by the Taylor theorem d 0 6 = (f 1 f 0 )/h 1 f 0 h 1 = 1 2 f (ξ 0 ), x 0 ξ 0 x 1 Thus if µ is the modulus of continuity of f, f (x) f (x ) µ(f, δ), x x δ then here since λ j + µ j = 1, ( I B ) d 3 3µ(f, δ) Also with the given end conditions B 1 1 so M d ( 3 B 1 I B ) d 3 3µ(f, δ) Now since d j 3 = f (ξ j ) it also follows that f d 3 µ(f, δ) whence M f 4µ(f, δ) and since S is linear on each interval S f 5µ(f, δ) 14

15 Now if f (x) f (x ) κ x x α, 0 < α 1 we have S f 5κδ α Since S(x j ) = f(x j ), by Rolle s theorem, there is a point ξ j in the interval [x j 1, x j ] at which S (ξ j ) = f (ξ j ). Then on this interval f (x) S (x) = x ξ j (f (x) S (x)) 5δµ(f, δ) and integrating a second time f(x) S(x) 5 2 δ2 µ(f, δ) Even stronger convergence theorems can be proved. Ahlberg (p. 29) gives: Theorem C: Let f (4) (x) be continuous then f (p) (x) S (p) (x) = O ( δ 3 p), p = 0, 1, 2, 3 B spline 15

16 6 Smoothing splines Figure 3: B spline Instead of requiring a spline to coincide with the function at specidied points one may simply require it to pass closer and permit some smoothing. To do this let S(x) be chosen to minimize R 1 xn x 0 f (x) 2 dx + R 2 n (f(x i ) y i ) 2 Q i=0 i Then by the calculus of variations the solution can be shown to be a spline witoints at x 1, x 2, x n 1 which does not necessarily pass through thte points. Reinsch (Numer. Math 10, 1967, pp ) shows that the equations now reduce to a five diagonal system, which can still be solved with a number of operations proportional to n. 16

17 7 Change of independent variable With any type of polynomial fitting or spline fitting difficulties occur if the function y(x) has too large a slope. In extreme cases the given points may represent a closed curve or other curve that bends back on itself. Figure 4: In such a case the x i do not form a monotone sequence, and also dy is inifinite at dx the nose, while dx is infinite at the top and bottom. To overcome this difficulty we dy need to parameterize the curve and set x = x(t) y = y(t) where t is a new independent variable that increases monotonely. Then form separate splines for x(t) and y(t). 17

18 8 Integration Splines are useful for integration when function is defined by a table of coordinates at unequal intervals. Conclusions Acknowledgements References 18