Linear Interpolating Splines

Transcription

1 Jim Lambers MAT 772 Fall Semester Lecture 17 Notes Tese notes correspond to Sections 112, 11, and 114 in te text Linear Interpolating Splines We ave seen tat ig-degree polynomial interpolation can be problematic However, if te fitting function is only required to ave a few continuous derivatives, ten one can construct a piecewise polynomial to fit te data We now precisely define wat we mean by a piecewise polynomial Definition (Piecewise polynomial) Let [a, b] be an interval tat is divided into subintervals [x i, x i+1 ], were i = 0,, n 1, x 0 = a and x n = b A piecewise polynomial is a function p(x) defined on [a, b] by p(x) = p i (x), x i 1 x x i, i = 1, 2,, n, were, for i = 1, 2,, n, eac function p i (x) is a polynomial defined on [x i 1, x i ] Te degree of p(x) is te maximum degree of eac polynomial p i (x), for i = 1, 2,, n It is essential to note tat by tis definition, a piecewise polynomial defined on [a, b] is equal to some polynomial on eac subinterval [x i 1, x i ] of [a, b], for i = 1, 2,, n, but a different polynomial may be used for eac subinterval We first consider one of te simplest types of piecewise polynomials, a piecewise linear polynomial Let f C[a, b] Given te points x 0, x 1,, x n defined as above, te linear spline s L (x) tat interpolates f at tese points is defined by s L (x) = f(x i 1 ) x x i + f(x i ) x x i 1, x [x i 1, x i ], i = 1, 2,, n x i 1 x i x i x i 1 Te points x 0, x 1,, x n are te knots of te spline Before we study te accuracy of linear splines, we introduce some terminology and notation First, we say tat a function f is absolutely continouous on [a, b] if its derivative is finite almost everywere in [a, b] (meaning tat it is not finite on at most a subset of [a, b] tat as measure zero), is integrable on [a, b], and satisfies x a v (s) dx = v(x) v(a), a x b Any continuously differentiable function is absolutely continuous, but te converse is not necessarily true 1

2 Example For example, f(x) = x is absolutely continuous on any interval of te form [ a, a], but it is not continuously differentiable on suc an interval Next, we define te Sobolev spaces H k (a, b) as follows Te space H 1 (a, b) is te set of all absolutely continuous functions on [a, b] wose derivatives belong to L 2 (a, b) Ten, for k > 1, H k (a, b) is te subset of H k 1 (a, b) consisting of functions wose (k 1)st derivatives are absolutely continuous, and wose kt derivatives belong to L 2 (a, b) If we denote by C k [a, b] te set of all functions defined on [a, b] tat are k times continuously differentiable, ten C k [a, b] is a proper subset of H k (a, b) For example, any linear spline belongs to H 1 (a, b), but does not generally belong to C 1 [a, b] Example Te function f(x) = x /4 belongs to H 1 (0, 1) because f (x) = 4 x 1/4 is integrable on [0, 1], and also square-integrable on [0, 1], since 1 1 f (x) 2 9 dx = 16 x 1/2 = x1/2 = However, f / C 1 [a, b], because f (x) is singular at x = 0 0 Now, if f C 2 [a, b], ten by te error in Lagrange interpolation, on eac subinterval [x i 1, x i ], for i = 1, 2,, n, we ave f(x) s L (x) = f (ξ) (x x i 1 )(x x i ) 2 If we let i = x i x i 1, ten te function (x x i 1 )(x x i ) acieves its maximum absolute value at x = (x i 1 + x i )/2, wit a maximum value of 2 i /4 If we define = max 1 i n i, ten we ave f s L f, were denotes te -norm over [a, b] One of te most useful properties of te linear spline s L (x) is tat among all functions in H 1 (a, b) tat interpolate f(x) at te knots x 0, x 1,, x n, it is te flattest Tat is, for any function v H 1 (a, b) tat interpolates f at te knots, To prove tis, we first write Ten, applying integration by parts, we obtain s L 2 v 2 v 2 2 = v s L v s L, s L + s L v s L, s L = b a [v (x) s L(x)]s L(x) dx 2

3 = = n i=1 i=1 xi [v (x) s L(x)]s L(x) dx x i 1 { n [v(x) s L (x)]s L(x) } x xi i [v(x) s x i 1 L (x)]s L(x) dx x i 1 However, s L is a linear function on eac subinterval [x i 1, x i ], so s L (x) 0 on eac subinterval Furtermore, because bot v(x) and s L (x) interpolate f(x) at te knots, te bounday terms vanis, and terefore v s L, s L = 0, wic establises te result Basis Functions for Linear Splines Lagrange interpolation allows te unique polynomial p n (x) of degree n tat interpolates f(x) at te knots x 0, x 1,, x n to be expressed in te convenient form p n (x) = n f(x i )L n,i (x) i=0 A similar form can be obtained for te linear spline s L (x) using linear basis splines, wic are piecewise linear functions tat are equal to one at one of te knots, and equal to zero at all oter knots Tese functions, known as at functions due to te sapes of teir graps, are defined as follows: { (x1 x)/ φ 0 (x) = 1 x 0 x < x 1,, 0 x 1 x x n 0 x 0 x < x i 1, (x x φ i (x) = i 1 )/ i x i 1 x < x i,, i = 1, 2,, n 1, (x i+1 x)/ i+1 x i x < x i+1, 0 x i+1 x x n { 0 x0 x < x φ n (x) = n 1, (x x n 1 )/ n x n 1 x x n Ten, te linear spline can be expressed as s L (x) = n f(x i )φ i (x) i=0

4 Cubic Splines Typically, piecewise polynomials are used to fit smoot functions, and terefore are required to ave a certain number of continuous derivatives Tis requirement imposes additional constraints on te piecewise polynomial, and terefore te degree of te polynomials used on eac subinterval must be cosen sufficiently ig to ensure tat tese constraints can be satisfied Cubic Spline Interpolation A spline is a piecewise polynomial of degree k tat as k 1 continuous derivatives Te most commonly used spline is a cubic spline, wic we now define Definition (Cubic Spline) Let f(x) be function defined on an interval [a, b], and let x 0, x 1,, x n be n + 1 distinct points in [a, b], were a = x 0 < x 1 < < x n = b A cubic spline, or cubic spline interpolant, is a piecewise polynomial s(x) tat satisifes te following conditions: 1 On eac interval [x i 1, x i ], for i = 1,, n, s(x) = s i (x), were s i (x) is a cubic polynomial 2 s(x i ) = f(x i ) for i = 0, 1,, n s(x) is twice continuously differentiable on (a, b) 4 Eiter of te following boundary conditions are satisfied: (a) s (a) = s (b) = 0, wic is called free or natural boundary conditions, and (b) s (a) = f (a) and s (b) = f (b), wic is called clamped boundary conditions If s(x) satisfies free boundary conditions, we say tat s(x) is a natural spline x 0, x 1,, x n are called te nodes of s(x) Te points Clamped boundary conditions are often preferable because tey use more information about f(x), wic yields a spline tat better approximates f(x) on [a, b] However, if information about f (x) is not available, ten free boundary conditions must be used instead Constructing Cubic Splines Suppose tat we wis to construct a cubic spline interpolant s(x) tat fits te given data (x 0, y 0 ), (x 1, y 1 ),, (x n, y n ), were a = x 0 < x 1 < < x n = b, and y i = f(x i ), for some known function f(x) defined on [a, b] From te preceding discussion, tis spline is a piecewise polynomial of te form s(x) = s i (x) = d i (x x i 1 ) + c i (x x i 1 ) 2 + b i (x x i 1 ) + a i, i = 1, 2,, n, x i 1 x x i Tat is, te value of s(x) is obtained by evaluating a different cubic polynomial for eac subinterval [x i 1, x i ], for i = 1, 2,, n 4

5 We now use te definition of a cubic spline to construct a system of equations tat must be satisfied by te coefficients a i, b i, c i and d i for i = 1, 2,, n We can ten compute tese coefficients by solving te system Because s(x) must fit te given data, we ave a i = y i 1, i = 1, 2,, n If we define i = x i x i 1, for i = 1, 2,, n, and define a n+1 = y n, ten te requirement tat s(x) is continuous at te interior nodes implies tat we must ave s i (x i ) = s i+1 (x i ) for i = 1, 2,, n 1 Furtermore, because s(x) must fit te given data, we must also ave s(x n ) = s n (x n ) = y n Tese conditions lead to te constraints d i i + c i 2 i + b i i + a i = a i+1, i = 1, 2,, n To ensure tat s(x) as a continuous first derivative at te interior nodes, we require tat s i (x i) = s i+1 (x i) for i = 1, 2, n 1, wic imposes te constraints d i 2 i + 2c i i + b i = b i+1, i = 1, 2,, n 1 Similarly, to enforce continuity of te second derivative at te interior nodes, we require tat s i (x i) = s i+1 (x i) for i = 1, 2,, n 1, wic leads to te constraints d i i + c i = c i+1, i = 1, 2,, n 1 Tere are 4n coefficients to determine, since tere are n cubic polynomials, wit 4 coefficients eac However, we ave only prescribed 4n 2 constraints, so we must specify 2 more in order to determine a unique solution If we use free boundary conditions, ten tese constraints are c 0 = 0, d n n + c n = 0 On te oter and, if we use clamped boundary conditions, ten our additional constraints are b 0 = z 0, d n 2 n + 2c n n + b n = z n, were z i = f (x i ) for i = 0, 1,, n Having determined our constraints tat must be satisfied by s(x), we can set up a system of linear equations Ax = b based on tese constraints, and ten solve tis system to determine te coefficients a i, b i, c i, d i for i = 1, 2, n In te case of free boundary conditions, A is an 5

6 (n + 1) (n + 1) matrix is defined by ( ) 2 A = 0 2 2( 2 + ) 0 n 1 2( n 1 + n ) n and te (n + 1)-vectors x and b are c 1 c 2 x =, b = c n (a a 2 ) 1 (a 2 a 1 ) n (a n+1 a n ) n 1 (a n a n 1 ) 0, were c n+1 = s (x n )/2 In te case of clamped boundary conditions, we ave ( ) 2 A = 0 2 2( 2 + ) 0 n 1 2( n 1 + n ) n 0 0 n 2 n and x = c 1 c 2 c n+1, b = 1 (a 2 a 1 ) z 0 2 (a a 2 ) 1 (a 2 a 1 ) n (a n+1 a n ) z n n 1 (a n a n 1 ) n (a n+1 a n ) Once te coefficients c 1, c 2,, c n+1 ave been determined, te remaining coefficients can be computed as follows: 1 Te coefficients a 1, a 2,, a n+1 ave already been defined by te relations a i = y i 1, for i = 0, 1,, n 6

7 2 Te coefficients b 1, b 2,, b n are given by b i = 1 i (a i+1 a i ) i (2c i + c i+1 ), i = 1, 2,, n Te coefficients d 1, d 2,, d n can be obtained using te constraints d i i + c i = c i+1, i = 1, 2,, n Example We will construct a cubic spline interpolant for te following data on te interval [0, 2] j x j y j / / Te spline, s(x), will consist of four pieces {s j (x)} 4 j=1, eac of wic is a cubic polynomial of te form s j (x) = a j + b j (x x j 1 ) + c j (x x j 1 ) 2 + d j (x x j 1 ), j = 1, 2,, 4 We will impose free, or natural, boundary conditions on tis spline, so it will satisfy te conditions s (0) = s (2) = 0, in addition to te essential conditions imposed on a spline: it must fit te given data and ave continuous first and second derivatives on te interval [0, 2] Tese conditions lead to te following system of equations tat must be solved for te coefficients c 1, c 2, c, c 4, and c 5, were c j = s (x j 1 )/2 for j = 1, 2,, 5 We define = (2 0)/4 = 1/2 to be te spacing between te interpolation points c 1 = 0 (c 1 + 4c 2 + c ) = y 2 2y 1 + y 0 (c 2 + 4c + c 4 ) = y 2y 2 + y 1 (c + 4c 4 + c 5 ) = y 4 2y + y 2 c 5 = 0 Substituting = 1/2 and te values of y j, and also taking into account te boundary conditions, we obtain 1 6 (4c 2 + c ) = 2 7

8 Tis system as te solutions 1 6 (c 2 + 4c + c 4 ) = (c + 4c 4 ) = 48 c 1 = 516/7, c 2 = 720/7, c = 684/7 Using te relation a j+1 = y j, for j = 0, 1, 2,, and te formula we obtain Finally, using te formula we obtain b j = a j+1 a j (2c j + c j+1 ), j = 1, 2,, 4, b 1 = 184/7, b 2 = 74/7, b = 4, b 4 = 46/7 d j = c j+1 c j, j = 1, 2,, 4, d 1 = 44/7, d 2 = 824/7, d = 96/7, d 4 = 456/7 We conclude tat te spline s(x) tat fits te given data, as two continuous derivatives on [0, 2], and satisfies natural boundary conditions is 44 7 x x2 + if x [0, 05] 824 s(x) = 7 (x 1/2) (x 1/2) (x 1/2) 4 if x [05, 1] 96 7 (x 1) (x 1)2 4(x 1) + 5 if x [1, 15] (x /2) (x /2) (x /2) 6 if x [15, 2] Te grap of te spline is sown in Figure 1 Well-Posedness and Accuracy For bot boundary conditions, te system Ax = b as a unique solution, wic leads to te following results Teorem Let x 0, x 1,, x n be n+1 distinct points in te interval [a, b], were a = x 0 < x 1 < < x n = b, and let f(x) be a function defined on [a, b] Ten f as a unique cubic spline interpolant s(x) tat is defined on te nodes x 0, x 1,, x n tat satisfies te natural boundary conditions s (a) = s (b) = 0 Teorem Let x 0, x 1,, x n be n + 1 distinct points in te interval [a, b], were a = x 0 < x 1 < < x n = b, and let f(x) be a function defined on [a, b] tat is differentiable at a and b Ten f 8

9 Figure 1: Cubic spline tat passing troug te points (0, ), (1/2, 4), (1, 5), (2, 6), and (, 7) 9

10 as a unique cubic spline interpolant s(x) tat is defined on te nodes x 0, x 1,, x n tat satisfies te clamped boundary conditions s (a) = f (a) and s (b) = f (b) Just as te linear spline is te flattest interpolant, in an average sense, te natural cubic spline wit te least average curvature Specifically, if s 2 (x) is te natural cubic spline for f C[a, b] on [a, b] wit knots a = x 0 < x 1 < < x n = b, and v H 2 (a, b) is any interpolant of f wit tese knots, ten s 2 2 v 2 Tis can be proved in te same way as te corresponding result for te linear spline It is tis property of te natural cubic spline, called te smootest interpolation property, from wic splines were named A spline is a flexible curve-drawing aid tat is designed to produce a curve y = v(x), x [a, b], troug prescribed points in suc a way tat te strain energy E(v) = b a v (x) 2 (1 + v (x) 2 ) dx is minimized over all functions tat pass troug te same points, wic is te case if te curvature is small on [a, b] 10