Cubic Splines MATH 375. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Cubic Splines

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1 Cubic Splines MATH 375 J. Robert Buchanan Department of Mathematics Fall 2006

2 Introduction Given data {(x 0, f(x 0 )), (x 1, f(x 1 )),...,(x n, f(x n ))} which we wish to interpolate using a polynomial... Lagrange interpolating polynomial of degree n, Determine n + 1 coefficients, Error term can be difficult to construct and measure. Interpolate piecewise, Piecewise linear result not differentiable at x i, i = 0, 1,...,n, Piecewise quadratic not twice differentiable at x i, i = 0, 1,...,n, Piecewise cubic!

3 Cubic Spline Interpolant Definition Given a function f(x) defined on [a, b] and a set of nodes a = x 0 < x 1 < x 2 < < x n = b, a cubic spline interpolant, S, for f satisfies S is a cubic polynomial, S j on [x j, x j+1 ] for j = 0, 1,...,n 1. S(x j ) = f(x j ) for j = 0, 1,...,n. S j+1 (x j+1 ) = S j (x j+1 ) for j = 0, 1,...,n 2. S j+1 (x j+1) = S j (x j+1) for j = 0, 1,...,n 2. S j+1 (x j+1) = S j (x j+1 ) for j = 0, 1,...,n 2. One of the following boundary conditions is satisfied: S (x 0 ) = S (x n ) = 0 (free or natural BCs). S (x 0 ) = f (x 0 ) and S (x n ) = f (x n ) (clamped BCs).

4 General Form For j = 0, 1,...,n 1 assume S j (x) = a j + b j (x x j ) + c j (x x j ) 2 + d j (x x j ) 3. We must find a j, b j, c j and d j subject to the conditions specified in the definition. Let h j = x j+1 x j then S j (x j ) = a j = f(x j ) S j+1 (x j+1 ) = a j+1 = S j (x j+1 ) = a j + b j h j + c j h 2 j + d j h 3 j So far we know a j for j = 0, 1,...,n 1 and have n equations and 3n unknowns.

5 First Derivative For j = 0, 1,...,n 1 we have Thus S j (x) = b j + 2c j (x x j ) + 3d j (x x j ) 2. S j (x j) = b j S j+1 (x j+1) = b j+1 = S j (x j+1) = b j + 2c j h j + 3d j h 2 j Now we have 2n equations and 3n unknowns.

6 Second Derivative For j = 0, 1,...,n 1 we have Thus S j (x) = 2c j + 6d j (x x j ). S j (x j ) = 2c j S j+1 (x j+1) = 2c j+1 = S j (x j+1 ) = 2c j + 6d j h j Now we have 3n equations and 3n unknowns.

7 Summary of Equations For j = 0, 1,...,n 1 we have a j+1 = a j + b j h j + c j h 2 j + d j h 3 j b j+1 = b j + 2c j h j + 3d j hj 2 c j+1 = c j + 3d j h j. Note: The quantities a j and h j are known. Solve the third equation for d j and substitute into the other two equations.

8 Solving the Equations d j = c j+1 c j 3h j ( a j+1 = a j + b j h j + c j hj 2 cj+1 c j + 3h j = a j + b j h j + h2 j 3 (2c j + c j+1 ) ( ) cj+1 c j b j+1 = b j + 2c j h j + 3 hj 2 3h j = b j + h j (c j + c j+1 ) Solve the second equation for b j. ) h 3 j

9 Solution (continued) b j = 1 h j (a j+1 a j ) h j 3 (2c j + c j+1 ) Replace index j by j 1 to obtain b j 1 = and 1 h j 1 (a j a j 1 ) h j 1 3 (2c j 1 + c j ) b j = b j 1 + h j 1 (c j 1 + c j ). Substitute the first two equations into the last equation.

10 Solution (continued) 1 h j (a j+1 a j ) h j 3 (2c j + c j+1 ) = 1 h j 1 (a j a j 1 ) h j 1 3 (2c j 1 + c j ) + h j 1 (c j 1 + c j ) Collect all terms involving c to one side. h j 1 c j 1 +2c j (h j 1 +h j )+h j c j+1 = 3 h j (a j+1 a j ) 3 h j 1 (a j a j 1 ) We have n 1 equations of this form and n + 1 unknowns.

11 Natural Boundary Conditions If S (x 0 ) = S 0 (x 0) = 2c 0 = 0 then c 0 = 0 and if S (x n ) = S n 1 (x n) = 2c n = 0 then c n = 0. Theorem If f is defined at a = x 0 < x 1 < < x n = b then f has a unique natural cubic spline interpolant.

12 Natural BC Linear System In matrix form the system of n + 1 equations has the form Ac = y where A = h 0 2(h 0 + h 1 ) h h 1 2(h 1 + h 2 ) h h n 2 2(h n 2 + h n 1 ) h n Note: A is a tridiagonal matrix.

13 Natural BC Linear System (continued) A c 0 c 1 c 2. c n 1 c n = 0 3 h 1 (a 2 a 1 ) 3 h 0 (a 1 a 0 ) 3 h 2 (a 3 a 2 ) 3 h 1 (a 2 a 1 ). 3 h n 1 (a n a n 1 ) 3 h n 2 (a n 1 a n 2 ) 0

14 Clamped Boundary Conditions If S (a) = S 0 (a) = f (a) = b 0 then which is equivalent to f (a) = 1 h 0 (a 1 a 0 ) h 0 3 (2c 0 + c 1 ) h 0 (2c 0 + c 1 ) = 3 h 0 (a 1 a 0 ) 3f (a).

15 Likewise if S (b) = S n(b) = f (b) = b n then b n = b n 1 + h n 1 (c n 1 + c n ) 1 = (a n a n 1 ) h n 1 h n 1 3 (2c n 1 + c n ) + h n 1 (c n 1 + c n ) 1 = (a n a n 1 ) + h n 1 h n 1 3 (c n 1 + 2c n ) which is equivalent to h n 1 (c n 1 + 2c n ) = 3f (b) 3 h n 1 (a n a n 1 ). Theorem If f is defined at a = x 0 < x 1 < < x n = b and differentiable at x = a and at x = b, then f has a unique clamped cubic spline interpolant.

16 Clamped BC Linear System In matrix form the system of n + 1 equations has the form Ac = y where A = 2h 0 h h 0 2(h 0 + h 1 ) h h 1 2(h 1 + h 2 ) h h n 2 2(h n 2 + h n 1 ) h n h n 1 2h n 1 Note: A is a tridiagonal matrix.

17 Clamped BC Linear System (continued) A c 0 c 1 c 2. c n 1 c n = 3 h 0 (a 1 a 0 ) 3f (a) 3 h 1 (a 2 a 1 ) 3 h 0 (a 1 a 0 ) 3 h 2 (a 3 a 2 ) 3 h 1 (a 2 a 1 ). 3 h n 1 (a n a n 1 ) 3 h n 2 (a n 1 a n 2 ) 3f (b) 3 h n 1 (a n a n 1 )

18 Coefficients of the Cubic Splines Since a j for j = 0, 1,...,n is known, once we solve the linear system for c j (again for j = 0, 1,...,n) then for j = 0, 1,...,n 1. b j = 1 h j (a j+1 a j ) h j 3 (c j+1 + 2c j ) d j = 1 3h j (c j+1 c j )

19 Natural Cubic Spline Example Consider the following data: x f(x) The coefficients of the natural cubic spline interpolant are a b c d

20 Natural Cubic Spline Example (continued) f x x

21 Clamped Cubic Spline Example Here we will find the clamped cubic spline interpolant to the function f(x) = J 0 ( x) at the nodes x i = 5i for i = 0, 1,...,10. x f(x) Note: f (0) = 0.25 and f (50) =

22 Clamped Cubic Spline Example (continued) The coefficients of the clamped cubic spline interpolant are a b c d E E E 06

23 Clamped Cubic Spline Example (continued) f x x -0.4

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