Curve Fitting: Splines Chapter 16 Curve Fitting: Splines

Spline Interpolation There are cases where polynomial interpolation is bad Noisy data Sharp Corners (slope discontinuity) Humped or Flat data Overshoot Oscillations

Example: f(x) = sqrt(abs(x)) Interpolation at 4, 3, 2, 1, 0, 1, 2, 3, 4

Spline Interpolation Cubic interpolation 7th-order 5th-order Linear spline

Spline Interpolation Idea behind splines Use lower order polynomials to connect subsets of data points Make connections between adjacent splines smooth Lower order polynomials avoid oscillations and overshoots

Spline Interpolation Use “piecewise” polynomials instead of a single polynomial Spline -- a thin, flexible metal or wooden lath Bent the lath around pegs at the required points Spline curves -- curves of minimum strain energy Piecewise Linear Interpolation Piecewise Quadratic Interpolation Piecewise Cubic Interpolation (cubic splines)

Zero curvatures at end points Drafting Spline Continuous function and derivatives Zero curvatures at end points

Splines There are n-1 intervals and n data points si(x) is a piecewise low-order polynomial

Spline fits of a set of 4 points

Example: Ship Lines waterline

Bow Stern

Spline interpolation for submerged hull (below waterline) Original database Spline interpolation for submerged hull (below waterline)

Linear Splines b is the slope between points Connect each two points with straight line functions connecting each pair of points b is the slope between points

Linear Splines si (x) = ai + bi (x  xi) x1 x2 x3 x4

Identical to Lagrange interpolating polynomials Linear Splines Identical to Lagrange interpolating polynomials

Linear splines Connect each two points with straight line Functions connecting each pair of points are slope

Linear splines are exactly the same as linear interpolation! Example:

Linear Splines Problem with linear splines -- not smooth at data points (or knots) First derivative (slope) is not continuous Use higher-order splines to get continuous derivatives Equate derivatives at neighboring splines Continuous functional values and derivatives

Quadratic Splines Quadratic splines - continuous first derivatives May have discontinuous second and higher derivatives Derive second order polynomial between each pair of points For n points (i=1,…,n): (n-1) intervals & 3(n-1) unknown parameters (a’s, b’s, and c’s) Need 3(n-1) equations

Quadratic Splines 3(n-1) unknowns si(x) = ai + bi (xxi) + ci(xxi)2 f(x5) s2 (x) f(x2) s4 (x) f(x3) s1 (x) s3 (x) f(x4) f(x1) Interval 1 Interval 2 Interval 3 Interval 4 x1 x2 x3 x4 x5 3(n-1) unknowns si(x) = ai + bi (xxi) + ci(xxi)2

Piecewise Quadratic Splines Example with n = 5 The functional must pass through all the points xi (continuity condition) The function values of adjacent polynomials must be equal at all nodes (identical to condition 1.) 2(n-2) + 2 = 2(n-1) The 1st derivatives are continuous at interior nodes xi, (n-2) Assume that the second derivatives is zero at the first points 3(n1) equations for 3(n1) unknowns 2(n1) + (n2) + (1) = 3(n1)

Piecewise Quadratic Splines Function must pass through all the points x = xi (2n  2 eqns) This also ensure the same function values of adjacent polynomials at the interior knots (hi = xi+1 – xi) Apply to every interval (4 intervals, 8 equations)

Piecewise Quadratic Splines Continuous slope at interior knots x = xi (n2 equations) Apply to interior knots x2 , x3 and x4 (3 equations) Zero curvatures at x = x1 (1 equation)

Example: Quadratic Spline

Quadratic Spline Interpolation function [A, b] = quadratic(x, f) % exact solution f(x)=x^3-5x^2+3x+4 % x=[-1 0 2 5 6]; f=[-5 4 -2 19 58]; h1=x(2)-x(1); h2=x(3)-x(2); h3=x(4)-x(3); h4=x(5)-x(4); f1=f(1); f2=f(2); f3=f(3); f4=f(4); f5=f(5); A=[1 0 0 0 0 0 0 0 0 0 0 0 0 h1 h1^2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 h2 h2^2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 h3 h3^2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 h4 h4^2 0 1 2*h1 0 -1 0 0 0 0 0 0 0 0 0 0 0 1 2*h2 0 -1 0 0 0 0 0 0 0 0 0 0 0 1 2*h3 0 -1 0 0 0 1 0 0 0 0 0 0 0 0 0]; b=[f1; f2-f1; f2; f3-f2; f3; f4-f3; f4; f5-f4; 0; 0; 0; 0];

Quadratic Spline Interpolation >> x=[-1 0 2 5 6]; f=[-5 4 -2 19 58]; >> [A,b]=quadratic(x,f) p = A\b -5.0000 9.0000 4.0000 -6.0000 -2.0000 -15.0000 7.3333 19.0000 29.0000 10.0000 dx = 0.1; z1=x(1):dx:x(2); s1=p(1)+p(2)*(z1-x(1))+p(3)*(z1-x(1)).^2; z2=x(2):dx:x(3); s2=p(4)+p(5)*(z2-x(2))+p(6)*(z2-x(2)).^2; z3=x(3):dx:x(4); s3=p(7)+p(8)*(z3-x(3))+p(9)*(z3-x(3)).^2; z4=x(4):dx:x(5); s4=p(10)+p(11)*(z4-x(4))+p(12)*(z4-x(4)).^2; H=plot(z1,s1,'r',z2,s2,'b',z3,s3,'m',z4,s4,'g'); xx=-1:0.1:6; fexact=xx.^3-5*xx.^2+3*xx+4; hold on; G=plot(xx,fexact,'c',x,f,'ro'); set(H,'LineWidth',3); set(G,'LineWidth',3,'MarkerSize',8); H1=xlabel('x'); set(H1,'FontSize',15); H2=ylabel('f(x)'); set(H2,'FontSize',15); H3=title('f(x)=x^3-5x^2+3x+4'); set(H3,'FontSize',15);

Quadratic Spline s4(x) Exact function s2(x) s1(x) s3(x)

Cubic Splines Cubic splines avoid the straight line and the over-swing Can develop method like we did for quadratic – 4(n–1) unknowns – 4(n–1) equations interior knot equality end point fixed interior knot first derivative equality assume derivative value if needed

Piecewise Cubic Splines s2 (x) sn-1 (x) s1 (x) Continuous slopes and curvatures x1 x2 x3 xn-1 xn 4(n1) unknowns

Piecewise Cubic Splines s2”(x) Sn-1”(x) s1”(x) s3”(x) x0 x1 x2 xn-1 xn si (x) - piecewise cubic polynomials si’(x) - piecewise quadratic polynomials (slope) si”(x) - piecewise linear polynomials (curvatures) Reduce to (n-1) unknowns and (n-1) equations for si’’

Cubic Splines Piecewise cubic polynomial with continuous derivatives up to order 2 The function must pass through all the data points gives 2(n-1) equations i = 1,2,…, n-1

Cubic Splines 2. First derivatives at the interior nodes must be equal: (n-2) equations 3. Second derivatives at the interior nodes must be equal:

Cubic Splines The last two equations 4. Two additional conditions are needed (arbitrary) The last two equations Total equations: 2(n-1) + (n-2) + (n-2) +2 = 4(n-1)

Cubic Splines Solve for (ai, bi, ci, di) – see textbook for details Tridiagonal system with boundary conditions c1 = cn= 0

Cubic Splines Tridiagonal matrix

Hand Calculations

Hand Calculations can be further simplified since c1 = c4 = 0 (natural spline)

Cubic Spline Interpolation

Cubic Splines Piecewise cubic splines (cubic polynomials)

Cubic Spline Interpolation The exact solution is a cubic function Why cubic spline interpolation does not give the exact solution for a cubic polynomial? Because the conditions on the end knots are different! In general, f (x0)  0 and f (xn)  0 !!

Cubic Spline Interpolation

>> xx=[-1 0 2 5 6]; f = [-5 4 -2 19 58]; >> Cubic_spline(xx,f); Resulting piecewise function: s1 = (-5.000000)+(11.081218)*(x-(-1.000000)) s2 = (0.000000)*(x-(-1.000000)).^2+(-2.081218)*(x-(-1.000000)).^3 (4.000000)+(4.837563)*(x-(0.000000)) (-6.243655)*(x-(0.000000)).^2+(1.162437)*(x-(0.000000)).^3 (-2.000000)+(-6.187817)*(x-(2.000000)) (0.730964)*(x-(2.000000)).^2+(1.221658)*(x-(2.000000)).^3 (19.000000)+(31.182741)*(x-(5.000000)) (11.725888)*(x-(5.000000)).^2+(-3.908629)*(x-(5.000000)).^3 (i = 1) (i = 2) (i = 3) (i = 4)

Cubic Spline Interpolation Piecewise cubic Continuous slope Continuous curvature zero curvatures at end knots exact solution

End Conditions of Splines Natural spline : zero second derivative (curvature) at end points Clamped spline: prescribed first derivative (clamped) at end points Not-a-Knot spline: continuous third derivative at x2 and xn-1

Continuous third derivatives at x2 and xn-1 >> x = linspace(-1,1,9); y = 1./(1+25*x.^2); >> xx = linspace(-1,1,100); yy = spline(x,y,xx); >> yr=1./(1+25*xx.^2); >> H=plot(x,y,'o',xx,yy,xx,yr,'--'); Continuous third derivatives at x2 and xn-1 Comparison of Runge’s function (dashed red line) with a 9-point not-a-knot spline fit generated with MATLAB (solid green line)

Clamped End Spline - Use 11 values including slopes at end points >> yc = [1 y –4]; % 1 and -4 are the 1st-order derivatives (or slopes)at 1st & last point, respectively. >> yyc = spline(x,yc,xx); >> >> H=plot(x,y,'o',xx,yyc,xx,yr,'--'); Note that first derivatives of 1 and 4 are specified at the left and right boundaries, respectively.

MATLAB Functions One-dimensional interpolations yy = spline (x, y, xx) yi = interp1 (x, y, xi, ‘method’) yi = interp1 (x, y, xi, ‘linear’) yi = interp1 (x, y, xi, ‘spline’) – not-a-knot spline yi = interp1 (x, y,, xi, ‘pchip’) – cubic Hermite yi = interp1 (x, y,, xi, ‘cubic’) – cubic Hermite yi = interp1 (x, y,, xi, ‘nearest’) – nearest neighbor

MATLAB Function: interp1 Piecewise polynomial interpolation on velocity time series for an automobile

Spline Interpolations >> x=[1 2 4 7 9 10] x = 1 2 4 7 9 10 >> y=[3 -2 5 4 9 2] y = 3 -2 5 4 9 2 >> xi=1:0.1:10; >> y1=interp1(x,y,xi,'linear'); >> y2=interp1(x,y,xi,'spline'); >> y3=interp1(x,y,xi,'cubic'); >> plot(x,y,'ro',xi,y1,xi,y2,xi,y3,'LineWidth',2,'MarkerSize',12) >> print -djpeg spline00.jpg

Spline Interpolations Linear Not-a-knot Cubic

MATLAB’s Functions Two-dimensional interpolations zi = interp2 (x, y, z, xi, yi)

MATLAB Function: interp2 >> [x,y,z]=peaks(100); [xi,yi]=meshgrid(-3:0.1:3,-3:0.1:3); >> zi = interp2(x,y,z,xi,yi); surf(xi,yi,zi) >> print -djpeg075 peaks1.jpg >> [x,y,z]=peaks(10); [xi,yi]=meshgrid(-3:0.1:3,-3:0.1:3); >> print -djpeg075 peaks2.jpg 10  10 data base 100 100 data base

CVEN 302-501 Homework No. 11 Chapter 16 Chapter 19 Problem 16.1(a) (30) – Hand Calculation Problem 16.1(b) (20)– MATLAB program not-a-knot spline: yy = spline(x, y, xx) Problem 16.1(c)(20) – MATLAB program Chapter 19 Problem 19.2 (20)– Hand Calculation Problem 19.4 (20)– Hand Calculation Due on Wed. 11/05/08 at the beginning of the period