LECTURE NOTES - SPLINE INTERPOLATION. 1. Introduction. Problems can arise when a single high-degree polynomial is fit to a large number

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1 LECTURE NOTES - SPLINE INTERPOLATION DR MAZHAR IQBAL 1 Introduction Problems can arise when a single high-degree polynomial is fit to a large number of points High-degree polynomials would obviously pass through all the data points themselves, but they can oscillate wildly between data points due to round-off errors and overshoot In such cases, lower-degree polynomials can be fit to subsets of the data points If the lower-degree polynomials are independent of each other, a piecewise approximation is obtained An alternate approach is to fit a lowerdegree polynomial to connect each pair of data points and to require the set of lower-degree polynomials to be consistent with each other in some sense This type of polynomial is called a spline function, or simply a spline Splines can be of any degree Linear splines are simply straight line segments connecting each pair of data points Linear splines are independent of each other from interval to interval Linear splines yield first-order approximating polynomials The slopes (ie, first derivatives) and curvature (ie, second derivatives) are discontinuous at every data point Quadratic splines yield second-order approximating polynomials The slopes of the quadratic splines can be forced to be continuous at each data point, but the curvatures (ie, the second derivatives) are still discontinuous A cubic spline yields a third-degree polynomial connecting each pair of data points The slopes and curvatures of the cubic splines can be forced to be continuous at each data point In fact, these requirements are necessary to obtain the additional conditions required to fit a cubic polynomial to two data points Higher-degree splines can be defined in a similar manner However, cubic splines have proven to be a good compromise between accuracy and complexity Consequently, the concept 1

2 DR MAZHAR IQBAL of cubic splines is developed in this section The name spline comes from the thin flexible rod, called a spline, used by draftsmen to draw smooth curves through a series of discrete points The spline is placed over the points and either weighted or pinned at each point Due to the flexure properties of a flexible rod (typically of rectangular cross section), the slope and curvature of the rod are continuous at each point A smooth curve is then traced along the rod, yielding a spline curve Construction Definition For a given set of data points (x k, y k ), k = 0 : N, the cubic spline s(x) consists of N cubic polynomial s k (x), s assigned to each subinterval satisfying certain constraints We define cubic spline as a function S(x) = S j (x) on interval [x j, x j+1 ] for n = 0, 1,, n 1 defined as (1) S j (x) = a j + b j (x x j ) + c j (x x j ) + d j (x x j ) 3 and the requirements () (3) (4) (5) () S j (x j ) = y j, j = 0, 1,, n 1, S n 1 (x n ) = y n S j (x j+1 ) = S j+1 (x j+1 ) S j(x j+1 ) = S j+1(x j+1 ) j (x j+1 ) = j+1(x j+1 ) Construction From condition we get S j (x j ) = a j y j = a j for i = 0, 1, n 1 The equation 1 for the node j + 1 can be written as (7) S j+1 (x) = a j+1 + b j+1 (x x j+1 ) + c j+1 (x x j+1 ) + d j+1 (x x j+1 ) 3

3 LECTURE NOTES - SPLINE INTERPOLATION 3 If we evaluate this on node x j+1 we get (8) S j+1 (x j+1 ) = a j+1 and (9) S j (x j+1 ) = a j + b j (x j+1 x j ) + c j (x j+1 x j ) + d j (x j+1 x j ) 3 if we write h j = x j+1 x j the above equation can be written as (10) S j (x j+1 ) = a j + b j h j + c j h j + d j h 3 j By invoking the condition 4 we get (11) a j+1 = a j + b j h j + c j h j + d j h 3 j or (1) y j+1 = y j + b j h j + c j h j + d j h 3 j Taking derivatives of the equation 1 we get (13) (14) S j(x) = b j + c j (x x j ) + 3d j (x x j ) j (x) = c j + d j (x x j ) similarly (15) j+1(x) = c j+1 + d j+1 (x x j+1 ) Evaluating the above equations at x j and x j+1 we get j (x j ) = c j & j+1(x j+1 ) = c j+1 Now invoking we get c j + d j h j = c j+1

4 4 DR MAZHAR IQBAL Let d j = S j+1 (x j+1) j (x j) h j j (x j ) = m j for j = 0, 1,, n 1 and n 1(x n ) = m n Therefore, (1) d j = m j+1 m j h j and (17) c j = m j Replacing the values of c and d in equation 11 we get y j+1 = y j + b j h j + m j h j + m j+1 m j h j h 3 j b j h j = (y j+1 y j ) m jh j m j+1h j + m jh j (18) b j = (y j+1 y j ) h j m jh j 3 m j+1h j or b j = f[x j, x j+1 ] c j h j d j h j Now we are left with m j s as unknowns We now invoke condition 5 on s j 1 and s j splines S j(x) = b j + c j (x x j ) + 3d j (x x j )

5 LECTURE NOTES - SPLINE INTERPOLATION 5 S j 1(x) = b j 1 + c j 1 (x x j 1 ) + 3d j 1 (x x j 1 ) By condition 5 at the common node x j we get b j = b j 1 + c j 1 h j 1 + 3d j 1 h j 1 Replacing the values of b j s, c j & d j, from equations 18,17 and 1, we get the relationship between different m j s as y j+1 y j m j + m j+1 h j h j = y j y j 1 h j 1 m j 1 + m j h j 1 +m j 1 h j 1 +3 m j mj 1 h j 1 f[x j, x j+1 ] f[x j 1, x j ] = m j + m j+1 h j m j 1 + m j h j 1 +m j 1 h j 1 +3 m j mj 1 h j 1 m j h j +m j+1 h j m j 1 h j 1 m j h j 1 +m j 1 h j 1 +3m j h j 1 3m j 1 h j 1 = [f[x j, x j+1 ] f[x j 1, x j ]] (19) m j 1 h j 1 + (h j 1 + h j )m j + m j+1 h j = [f[x j, x j+1 ] f[x j 1, x j ]] j = 1,,, n 1 Now we will employ the two missing conditions Depending upon the conditions we will have different splines These are discussed in the succeeding paragraphs 1 Natural Spline If the second derivatives at the end points are set equal to zero then the resulting splines are called Natural Splines It is the curve obtained by forcing a flexible elastic rod through the data points but letting the slope at the

6 DR MAZHAR IQBAL ends be free to equilibrate to the position that minimizes the oscillatory behavior of the curve It is useful for fitting a curve to experimental data that are significant to several significant digits The two boundary conditions that we apply are m 0 = 0 & m n = 0 From equation 19 we have For j = 1 m 0 h 0 + (h 0 + h 1 )m 1 + h 1 m = [f[x 1, x ] f[x 0, x 1 ]] For j = n m j 1 h j 1 + (h j 1 + h j )m j + m j+1 h j = [f[x j, x j+1 ] f[x j 1, x j ]] For j = n 1 m n h n + (h n + h n 1 )m n 1 + h n 1 m n = [f[x n 1, x n ] f[x n, x n 1 ]] On applying m 0 = 0 & m n = 0 in the above equations resulting system of equations will be (h 0 + h 1 ) h 1 0 h 1 (h 1 + h ) h 0 h n (h n + h n 1 ) m 1 m m n 1 = (f[x 1 x ] f[x 0 x 1 ]) (f[x n 1 x n ] f[x n x n 1 ])

7 LECTURE NOTES - SPLINE INTERPOLATION 7 Clamped Spline The clamped spline involves the slope at the ends This spline can be visualized as the curve obtained when a flexible elastic rod is forced to pass through the data points, and the rod is clamped at each end with a fixed slope This spline is useful to a draftsman for drawing a smooth curve through several points In this type of cubic spline the slopes at the end points are specified As per the formulation of the linear system to calculate m j s We are required to find the values of m 0 & m n These are zero in the natural spline case but here wee will have to calculate these using the given information of the first derivative We know that (0) b j = f[x j x j+1 ] c j h j d j h j also S j(x) = b j + c j (x x j ) + 3d j (x x j ) (1) S j(x j ) = b j Particularly () S 0(x 0 ) = b 0 from 0 we get b 0 = f[x 0 x 1 ] c 0 h 0 d 0 h 0 = f[x 0 x 1 ] m 0 h 0 m 1 m 0 = f[x 0 x 1 ] m 0 3 h 0 m 1 h 0 h 0

8 8 DR MAZHAR IQBAL m 0 = 3 h 0 [f[x 0 x 1 ] b 0 ] m 1 = 3 h 0 [f[x 0 x 1 ] S 0(x 0 )] m 1 Here S (x 0 ) ie 1 st derivative at the node x 0 is same as the S 0(x 0 ) Because, S 0 (x 0 ) is the spline with center at x 0 In order to calculate m n we must use S (x n ) = given As we have n 1 splines Therefore, S n 1 (x) will be the last spline defined in the last interval [x n 1, x n ] defined as S n 1 (x) = a n 1 + b n 1 (x x n 1 ) + c n 1 (x x n 1 ) + d n 1 (x x n 1 ) 3 S n 1(x) = b n 1 + c n 1 (x x n 1 ) + 3d n 1 (x x n 1 ) (3) S n 1(x n ) = b n 1 + c n 1 h n 1 + 3d n 1 h n 1 Replacing the value of b n 1 in equation 3 we get S n 1(x n ) = f[x n 1 x n ] c n 1 h n 1 d n 1 h n 1 + c n 1 h n 1 + 3d n 1 h n 1 = f[x n 1 x n ] + c n 1 h n 1 + d n 1 h n 1 Now replacing the values of C n 1 & d n 1 we get S n 1(x n ) = f[x n 1 x n ] + m n 1 h n 1 + m n m n 1 h n 1 h n 1 = f[x n 1 x n ] + m n 1 h n 1 + m n 3 h n 1 m n = 3 [ ] S h n 1(x n ) f[x n 1xn ] m n 1 n 1

9 LECTURE NOTES - SPLINE INTERPOLATION 9 From equation 19 we have For j = 1 m 0 h 0 + (h 0 + h 1 )m 1 + h 1 m = [f[x 1, x ] f[x 0, x 1 ]] replacing the value of m 0 we get (h 0 + h 1 )m 1 + h 1 m = [f[x 1, x ] f[x 0, x 1 ]] 3[f[x 0 x 1 ] S 0(x 0 )] + m 1 h 0 ( 3 h 0 + h 1 )m 1 + h 1 m = [f[x 1, x ] f[x 0, x 1 ]] 3[f[x 0 x 1 ] S 0(x 0 )] For j = n 1 m n h n + (h n + h n 1 )m n 1 + h n 1 m n = [f[x n 1, x n ] f[x n, x n 1 ]] Replacing the values of m n we get m n h n +(h n +h n 1 )m n 1 = [f[x n 1, x n ] f[x n, x n 1 ]] 3[S (x n ) f[x n 1, x n ]+ m n 1 h n 1 m n h n +(h n + 3 h n 1)m n 1 = [f[x n 1, x n ] f[x n, x n 1 ]] 3[S (x n ) f[x n 1, x n ] The resulting system of equations will be ( 3 h 0 + h 1 ) h 1 0 h 1 (h 1 + h ) h 0 h n (h n + 3 h n 1) m 1 m m n 1 =

10 10 DR MAZHAR IQBAL [f[x 1, x ] f[x 0, x 1 ]] 3[f[x 0 x 1 ] S 0(x 0 )] [f[x, x 3 ] f[x 1, x ]] [f[x n 1, x n ] f[x n, x n 1 ]] 3[S (x n ) f[x n 1, x n ] 3 Endpoint Curvature Adjusted Spline Imposing the values of the second derivatives at the endpoints permits the practitioner to adjust the curvature at the endpoints The construction of linear system for this spline is left as an exercise References [1] Joe D Hoffman, Numerical Methods for Engineers and Scientists,nd ed,001, [] Wan Y Yang, Applied Numerical Methods Using Matlab,Wiely 005 [3] Jhon H Mathews and Kurtis D Fink, Numerical Methods Using Matlab, Pearson,004 [4] Curtis F Gerald and Patrick O Wheatley, Applied Numerical Analysis, Pearson,004 [5] Steven C Cahpra and Raymond P Canale, Numerical Methods for Engineers, 5th Ed, Macraw-Hill,00 [] Burden and Faires, Numerical Analysis, 8th Ed