Maria Cameron Theoretical foundations. Let. be a partition of the interval [a, b].

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1 Maria Cameron 1 Interpolation by spline functions Spline functions yield smooth interpolation curves that are less likely to exhibit the large oscillations characteristic for high degree polynomials Splines are used eg, in applications in graphics, numerical methods (eg, for BVP), signal processing The simplest splines are the cubic splines We will restrict ourselves to their considerations 11 Theoretical foundations Let be a partition of the interval [a, b] := {a = x 0 < x 1 < < x n = b} Definition 1 A cubic spline S on is a real function S : [a, b] R with the properties: (1) S C 2 [a, b], (2) S coincides on every subinterval [x j, x j+1 ], j = 0, 1,, n 1 with a polynomial of degree at most 3 Thus a cubic spline consists of cubic polynomials glued in such a manner that their values together with the values of their first two derivatives coincide at the interior nodes x j, j = 1, 2,, n 1 Suppose a function f is given at the nodes x j, j = 0, 1,, n, ie, f(x j ) = f j, j = 0, 1,, n 1 We denote the vector {f 0,, f n } by F We will denote a spline function that interpolates f at these points by S (F ; ) Note that S (F, ) is not uniquely defined by the sequence F Indeed, we have 4n coefficients of the cubic polynomials and n (n 1) = 4n 2 conditions to satisfy Hence we a free to assign two additional conditions that would allow us to determine S (F ; ) uniquely The typical conditions are: (1) zero second derivatives at the ends: S (F ; a) = S (F ; b) = 0, (2) periodicity: S (F ; a) = S (F ; b), S (F ; a) = S (F ; b), (3) assigned first derivatives at the ends: S (F ; a) = f 0, S (F ; b) = f n A prerequisite for the second condition is that f 0 = f n There is a theorem proven eg, in [1] that the spline function is unique for each of these cases 1

2 2 12 Setting up a system of equations for a cubic spline Throughout this section we will stick with a fixed partition := {a = x 0 < x 1 < < x n = b} and fixed vector of f s values F = {f 0,, f n } We will use the following notations for the intervals between the nodes and their lengths: I j := [x j, x j+1 ], j = 0, 1,, n 1, and +1 := x j+1 x j The second derivatives of the spline functions at the nodes are called moments and denoted M j := S (F, x j ), J = 0, 1,, n We will show that the spline function S (F ; x) is completely characterized by its moments, and the moments are found by solving a system of linear equations The second derivative of a spline function coincides with a linear function on each subinterval, and these linear functions can be described in terms of the moments M j : (1) S (F ; x) = M j x j+1 x +1 Integrating Eq (1) we obtain: + M j+1 x x j +1 x I j [x j, x j+1 ] (2) S (F ; x) = M j (x j+1 x) M j+1 (x x j ) A j, (3) S (F ; x) = M j (x j+1 x) M j+1 (x x j ) A j (x x j ) + B j, for x I j, j = 0, 1,, n, where A j and B j are constants of integration From S (F ; x j ) = f j and S (F ; x j+1 ) = f j+1 we obtain the following equations for A j and B j : M j h 2 j B j = f j, Hence M j+1 h 2 j A j +1 + B j = f j+1 (4) (5) B j = f j M j h 2 j+1 6, A j = f j+1 f j (M j+1 M j ) This yields the following representation of the spline function in terms of its moments: (6) S (F ; x) = α j + β j (x x j ) + γ j (x x j ) 2 + δ j (x x j ) 3 for x I j,

3 3 where (7) α j = f j, (8) (9) (10) γ j = M j 2, β j = S (F ; x j ) = M j +1 2 δ j = S (F, x+ j ) = M j+1 M j A j = f j+1 f j (M j+1 + 2M j ), Now we need to calculate the moments M j The continuity of S (F ; x) at the interior nodes yields n 1 equations for the moments M j Using Eqs (3) and (5) we obtain Therefore, for j = 1, 2,, n 1 we have S (F ; x) = M j (x j+1 x) M j+1 (x x j ) f j+1 f j (M j+1 M j ) S (F ; x j ) = f j f j 1 Since S (F ; x+ j ) = S (F ; x j ), we have (11) + 3 M j + 6 M j 1, S (F ; x + j ) = f j+1 f j M j +1 6 M j+1 6 M j M j M j+1 = f j+1 f j +1 f j f j 1 for j = 1, 2,, n 1 these are n 1 equations for n + 1 unknown moments The remaining two equations can be gained from the boundary conditions In the first case, S (F ; a) = S (F ; b) = 0 we set M 0 = M n = 0 In the periodic case we set M 0 = M n and h n 6 M n 1 + h n + h 1 M n + h M 1 = f 1 f n f n f n 1 h 1 h n Exercise Obtain additional two conditions for the case of assigned first derivatives at the end points: (12) (13) h 1 3 M 0 + h 1 6 M 1 = f 1 f 0 f h 0, 1 h n 6 M n 1 + h n 3 M n = f n f n f n 1 h n Eqs (11), (12) and (13) can be written in the common format µ j M j 1 + 2M j + λ j M j+1 = d j, j = 1, 2,, n 1,

4 4 where (14) (15) (16) λ j := , µ j := 1 λ j =, + +1 { 6 fj+1 f j d j := f } j f j 1 In the case M 0 = M n = 0 we set λ 0 = 0, d 0 = 0, µ n = 0, d n = 0 In result, we arrive at the following system of equations (17) 2 λ 0 0 µ 1 2 λ 1 µ 2 2 λ n 1 0 µ n 2 M 0 M 1 M n = d 0 d 1 d n Exercise Write out the system of equations for the momenta for the periodic case and for the case of assigned first derivatives at the endpoints Theorem 1 The matrix in Eq (17) is nonsingular for any partition of [a, b] Proof We will denote the Matrix in Eq (17) by A If follows from the definitions of λ j and µ j that µ j + λ j = 1, µ j 0, λ j 0 Therefore, the matrix A is strictly diagonal dominant Therefore, for any vector z max j z j max Az j j (Check this!) Hence Az = 0 if and only if z = 0

5 13 A fast solver for tridiagonal matrices The matrix in Eq (17) is tridiagonal There is a fast solver for such kind of systems that involves O(n) flops q 0 = λ 0 2 ; u 0 = d 0 2 ; λ n = 0; for k = 1 : n p k = µ k q k 1 + 2; q k = λ k p k ; u k = d k µ k u k 1 p k ; end for M n = u n ; for k = n 1 : 0 end for M k = q k M k+1 + u k ; 5 14 Convergence properties of cubic spline functions Definition 2 The fineness of the given partition is := max j As we know, the interpol ating polynomials do not necessarily converge to f no matter how smooth f is as the fineness of the partition tends to zero The spline interpolants do converge to f nicely under mild conditions First we will introduce some notations We will denote by F the vector of the second derivatives of f at the nodes The vector of moments M satisfies Eq (17) We will write The residual r is defined as AM = d r := d AF = A(M F ) Theorem 2 Suppose the first derivatives of f are assigned If f C 4 [a, b] and f (4) (x) L for x [a, b], then M F r 3 4 L 2

6 6 Proof By definition of the residual we have r j = d j µ j f (x j 1 ) 2f (x j ) λ j f (x j+1 ) { 6 fj+1 f j = f } j f j f (x j 1 ) 2f (x j ) +1 f (x j+1 ) Using Taylor s expansion around x j we obtain r j = (f f + h2 j+1 6 f + h3 j+1 24 f (τ 1 ) f + 2 f h2 j 6 f + h3 j 24 f (τ 2 )) [ f f + h2 j f (τ 3 ) [ ] 2f +1 f + +1 f + h2 j f (τ 4 ) [ ] 1 h 3 j+1 = f (τ 1 ) + h3 j 4 f (τ 2 ) h3 j 2 f (τ 3 ) +1 2 f (τ 4 ) Here all omitted arguments are x j and τ j [x j 1, x j+1 ] Therefore, for j = 1, 2,, n 1 r j 3 4 Lh3 j+1 + h3 j +1 + = 3 4 L(h2 j +1 + h 2 j+1) 3 4 L 2 The first and the last intervals are handled in a similar manner For them we have Since r = A(M F ) and A 1 we get r L 2 and r n 3 4 L 2 M F r 3 4 L 2 Theorem 3 Suppose f C 4 [a, b] and f (4) (x) L for x [a, b] Let be a partition of the interval [a, b] and K be a constant such that +1 K, j = 0, 1,, n 1 If S is the spline function which interpolates the values of f at the nodes of the partition and satisfies S (a) = f (a) and S (b) = f (b), ]

7 7 then there exist constants c k 2 independent of the partition, such that for x [a, b] (18) (19) (20) (21) f(x) S (x) c 0 L 4, f (x) S (x) c 1 L 3, f (x) S (x) c 2 L 2, f (x) S (x) c 3 LK Proof First we prove Eq (21) For x [x j 1, x j ], S (x) f (x) = M j M j 1 f (x) = M j f (x j ) M j 1 f (x j 1 ) + f (x j ) f (x) [f (x j 1 ) f (x)] f (x) Using the previous theorem and the Taylor expansion at x we get S (x) f (x) 3 2 L (x j x)f + (x j x) 2 f (η 1 ) 2 (x j 1 x)f (x j 1 x) f (η 2 ) f 3 2 L 2 + L 2 2 2LK 2 = 2L 2 Here η 1, η 2 [x j 1, x j ] To prove Eq (20) we observe that for every x (a, b) there is a closes node x j = x j (x) Assume without loss of generality that x x j (x) so that x [x j 1, x j ] and x j (x) x /2 /2 Then f (x) S (x) = f (x j ) S (x j ) + x x j (x) (f (t) S (t))dt 3 4 L L L 2, x (a, b) Next we prove Eq (19) In addition to the boundary points ξ 0 := a and ξ n+1 := b there exist by Rolle s theorem n points ξ j (x j 1, x j ) such that f (ξ j ) = S (ξ j ) for any x (a, b) there exists a closest one ξ j = ξ j (x), and x ξ j (x) <

8 8 Therefore, x f (x) S (x) = (f (t) S (t))dt ξ(x) Finally we prove Eq (18) We have f(x) S (x) = x 7 4 L 2 = 7 4 L 3 x j (x) (f (t) S (t))dt L 3 = 7 8 L 4, x [a, b] 15 Spline interpolation in Matlab The Matlab command interp1 is capable of finding the cubic spline The default mode is the linear interpolation, while adding spline in the argument results in finding the cubic spline Figure 1 is generated by the following sequence of commands >> t=linspace(0,5,1000); >> p=linspace(0,5,20); >> flin=interp1(p,sin(p^2),t); >> fspl=interp1(p,sin(p^2),t, spline ); >> fig=figure; >> hold on; >> grid; >> plot(t,sin(t^2), LineWidth,1) >> plot(t,flin, LineWidth,1, color, k ) >> plot(t,fspl, LineWidth,1, color, r ) >> plot(p,sin(p^2), b, MarkerSize,20) >> axis tight >> set(gca, DataAspectRatio,[1 1 1]) References [1] J Stoer and R Bulirsch, Introduction to Numerical Analysis, Third Edition, Springer, 2002

9 Figure 1 Linear interpolation and cubic spline interpolation 9