Answers to Homework 6: Interpolation: Spline Interpolation

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1 Mth 128A Spring 2002 Hndout # 17 Sergey Fomel Mrch 14, 2002 Answers to Homework 6: Interpoltion: Spline Interpoltion 1 In clss, we interpolted the function f (x) = 1 x stisfied the nturl boundry conditions t the points x = 2,4,5 with the cubic spline tht S () = 0 ; (1) S (b) = 0 (2) for = 2 nd b = 5 () Chnge conditions (1-2) to the clmped boundry conditions S () = f () ; (3) S (b) = f (b), (4) find the corresponding cubic spline nd evlute it t x = 3 Is the result more ccurte thn the one of the nturl cubic spline interpoltion? Note: No progrmming is necessry, but clcultor might help Solution: Let the cubic spline in the intervl from x = 2 to x = 4 be the polynomil S 1 (x) = 05 + b 1 (x 2) + c 1 (x 2) 2 + d 1 (x 2) 3 nd the spline in the intervl from x = 4 to x = 5 be the polynomil S 2 (x) = b 2 (x 4) + c 2 (x 4) 2 + d 2 (x 4) 3 The six coefficients b 1,c 1,d 1,b 2,c 2,d 2 re the unknowns tht we need to determine From the interpoltion conditions, we get S 1 (4) = b 1 + 4c 1 + 8d 1 = f (4) = 025 ; S 2 (5) = b 2 + c 2 + d 2 = f (5) = 02 From the smoothness conditions t the internl point, we get S 1 (4) = b 1 + 2c 1 (4 2) + 3d 1 (4 2) 2 = S 2 (4) = b 2 ; S d 1 (4 2) = S 2 2 Finlly, from the boundry conditions, we get S 1 (2) = b 1 = f (2) = 025 ; S 2 (5) = b 2 + 2c 2 + 3d 2 = f (5) = 004 1

2 Thus, we hve six liner equtions to determine the six unknowns In the mtrix form, the equtions re b c d b 2 = c d The equtions cn be solved, for exmple, by successive elimintion of unknowns We get b 1 = 025, then c d b c 2 = d Tke c 1 = d 1, then Tke d 1 = 025b 2, then b 2 c 2 d 2 d 1 b 2 c 2 d 2 = = Tke b 2 = c 2, then [ ] [ c2 d 2 ] = [ ] Finlly, tke c 2 = d 2 nd get d 2 = The finl nswer is Evluting the spline t x = 3, we get d 2 = c 2 = 0005 b 2 = d 1 = c 1 = b 1 = 025 S(3) = S 1 (3) = 05 + b1 + c 1 + d 1 = This is closer to the exct result f (3) = thn the result of the nturl spline interpoltion (S(3) = ) 2

3 (b) Prove tht if S(x) is cubic spline tht interpoltes function f (x) C 2 [,b] t the knots = x 1 < x 2 < < x n = b nd stisfies the clmped boundry conditions (3-4), then [ S (x) ] b 2 dx [ f (x) ] 2 dx (5) Hint: Divide the intervl [, b] into subintervls nd use integrtion by prts in ech subintervl Solution: Let us form the difference D(x) = f (x) S(x) From the integrl equlity [ f (x) ] b 2 dx = [ S (x) ] b 2 dx + [ D (x) ] b 2 dx + 2 we cn see tht the theorem will be proved if we cn prove tht S (x) D (x)dx, Indeed, the integrl S (x) D (x)dx = 0 [ f (x) ] 2 dx in this cse will be equl to the integrl [ S (x) ] 2 dx plus some non-negtive quntity [ D (x) ] 2 dx Applying integrtion by prts, we get S (x) D (x)dx = S (x)d (x) b b S (x) D (x)dx The first term is zero becuse of the clmped boundry conditions: D () = f () S () = 0 ; D (b) = f (b) S (b) = 0 The integrl in the second term cn be divided into subintervls, s follows: n 1 S (x) D (x)dx = 3 k=1 x k+1 x k S (x) D (x)dx

4 Integrtion by prts in ech subintervl produces x k+1 x k S (x) D (x)dx = S (x) D(x) xk+1 x k x k+1 x k S (4) (x) D(x)dx The first term in the expression bove is zero becuse of the interpoltion condition D(x k ) = f (x k ) S(x k ) = 0, k = 1,2,,n The second term is zero becuse the spline S(x) in ech subintervl is cubic polynomil nd hs zero fourth derivtive We hve proved tht which proves the theorem S (x) D (x)dx = 0, 2 The nturl boundry conditions for cubic spline led to system of liner equtions with the tridigonl mtrix 2(h 1 + h 2 ) h h 2 2(h 2 + h 3 ) h 3 0 h 3 0, (6) hn h n 2 2(h n 2 + h n 1 ) where h k = x k+1 x k The textbook shows tht the clmped boundry conditions led to the mtrix 2h 1 h h 1 2(h 1 + h 2 ) h h 2 2(h 2 + h 3 ) 0 (7) hn 2 0 hn 2 2(h n 2 + h n 1 ) h n h n 1 2h n 1 Find the form of mtrices tht correspond to two other populr types of boundry conditions: () not knot conditions: (b) periodic conditions: S 1 2) = S 2 2) ; (8) S n 2 n 1) = S n 1 n 1) (9) S 1 (x 1) = S n 1 (x n) ; (10) S 1 1) = S n 1 n) (11) 4

5 Here S k (x) represent the spline function on the intervl from x k to x k+1, k = 1,2,,n 1 The periodic conditions re pplied when S(x 1 ) = S(x n ) Solution: The centrl prt of the mtrix will lwys hve the sme tridigonl structure, which results from the recursive reltionship c k 1 h k 1 + 2c k (h k 1 + h k ) + c k+1 h k = 3 ( f [x k, x k+1 ] f [x k 1, x k ]), where c k is the second-order coefficient in the spline expression S k (x) = f k + b k (x x k ) + c k (x x k ) 2 + d k (x x k ) 3, x k x x k+1,k = 1,2,,n 1 The boundry conditions will only ffect the first nd the lst rows of the mtrix () The not knot conditions trnsform into the equtions Using the recursive reltionship d 1 = d 2 ; d n 2 = d n 1 d k = c k+1 c k 3h k, where h k = x k+1 x k, the conditions further trnsform to c 2 c 1 = c 3 c 2 ; h 1 h 2 c n 1 c n 2 = c n c n 1, h n 2 h n 1 where c n = S n 1 (x n)/2 Using this two conditions, we cn eliminte c 1 nd c n from the system with the help of the expressions ( c 1 = c h ) 1 h 2 c n = c n 1 ( 1 + h n 1 h n 2 The first eqution in the system is then h 1 c 3 ; h 2 ) c n 2 h n 1 h n 2 ( ( ) c 1 h 1 +2c 2 (h 1 +h 2 )+c 3 h 2 = c 2 3h 1 + 2h 2 + h2 1 )+c 3 h 2 h2 1 = 3 ( f [x 2, x 3 ] f [x 1, x 2 ]), h 2 h 2 nd the lst eqution is c n 2 ( c n 2 h n 2 + 2c n 1 (h n 2 + h n 1 ) + c n h n 1 = ) ( ) + c n 1 h n 2 h2 n 1 h n 2 3h n 1 + 2h n 2 + h2 n 1 h n 2 = 3 ( f [x n 1, x n ] f [x n 2, x n 1 ]) 5

6 The mtrix tkes the form 3h 1 + 2h 2 + h2 1 h 2 h 2 h2 1 h h 2 2(h 2 + h 3 ) h 3 0 h 3 0 hn h n 2 h2 n 1 h n 2 3h n 1 + 2h n 2 + h2 n 1 h n 2 Alterntive forms re possible (b) The periodic boundry conditions led to the equtions b 1 = b n 1 + 2c n 1 h n 1 + 3d n 1 h 2 n 1 c 1 = c n After eliminting c n from the system, the first eqution trnsforms to or f [x 1, x 2 ] 2c 1 + c 2 3 h 1 = f [ x n 1, x n ] + 2c n 1 + c 1 3 h n 1 2c 1 (h 1 + h n 1 ) + c 2 h 1 + c n 1 h n 1 = 3 ( f [x 1, x 2 ] f [ x n 1, x n ]) The system mtrix is 2(h 1 + h n 1 ) h h n 1 h 1 2(h 1 + h 2 ) h h hn 2 h n h n 2 2(h n 2 + h n 1 ) Alterntive forms re possible 3 The lgorithm for solving tridigonl symmetric systems, presented in clss, decomposes symmetric tridigonl mtrix into product of lower nd upper bidigonl mtrices, s follows: 1 b α β b 1 2 b 2 b 0 b 2 1 α β 0 = 0 b bn 1 0 βn b n 1 n 0 0 b n 1 α n The lgorithm for solving the liner system 1 b b 1 2 b 2 0 b 2 0 bn b n 1 n c 1 c 2 c n = g 1 g 2 g n 6

7 is summrized below TRIDIAGONAL( 1, 2,, n,b 1,b 2,,b n 1, g 1, g 2,, g n ) 1 α for k 1,2,,n 1 3 do 4 β k b k /α k 5 α k+1 k+1 b k β k 6 c 1 g 1 7 for k 2,3,,n 8 do 9 c k g k β k 1 c k 1 10 c n c n /α n 11 for k n 1,n 2,,1 12 do 13 c k c k /α k β k c k+1 14 return c 1,c 2,,c n () The lgorithm will fil (with division by zero) if ny α k is zero Prove tht, in the cse of cubic spline interpoltion with the nturl boundry conditions, α k > b k > 0, k = 1,2,,n Hint: Strt with k = 1 nd use the method of mthemticl induction Solution: In the cse of the nturl boundry conditions, nd Let us first check the cse k = 1: k = 2 (h k + h k+1 ), k = 1,2,,n 2 b k = h k+1 > 0, k = 1,2,,n 3 α 1 = 1 = 2(h 1 + h 2 ) > h 2 = b 1 The theorem is stisfied Using the method of mthemticl induction, let us ssume tht α k > b k for some k nd prove tht the nlogous inequlity is true for k + 1 Indeed, the lgorithm shows tht α k+1 = k+1 b2 k α k The ssumed inequlity implies tht Therefore, b 2 k α k < b k α k+1 > k+1 b k = 2 (h k + h k+1 ) h k+1 = h k+1 + 2h k > 0 QED 7

8 (b) Design n lterntive lgorithm, where the tridigonl mtrix is fctored into the product of upper nd lower bidigonl mtrices, s follows: 1 b b 1 2 b 2 0 b 2 0 = bn b n 1 n ˆα 1 b ˆα 2 b 2 ˆβ ˆβ 2 bn ˆα n 0 0 ˆβ n 1 1 Solution: Mtching the digonl elements, we rrive t the system of equtions ˆα 1 + b 1 ˆβ 1 = 1 ˆα n 1 + b n 1 ˆβ n 1 = n 1 ˆα n = n Mtching the off-digonl elements leds to the system ˆα 2 ˆβ 1 = b 1 ˆα n ˆβ n 1 = b n 1 Together, the two systems define the bckwrd recursion ˆα n = n ; { ˆβ k = b k / ˆα k+1 ˆα k = k b k ˆβ k, k = n 1,n 2,,1 After the decomposition, the upper nd lower bidigonl mtrices re inverted using recursion in the opposite directions The finl lgorithm is TRIDIAGONAL2( 1, 2,, n,b 1,b 2,,b n 1, g 1, g 2,, g n ) 1 ˆα n n 2 for k n 1,n 2,,1 3 do 4 ˆβ k b k / ˆα k+1 5 ˆα k k b k ˆβ k 6 c n g n 7 for k n 1,n 2,,1 8 do 9 c k g k ˆβ k c k+1 10 c 1 c 1 / ˆα 1 11 for k 2,3,,n 12 do 13 c k c k / ˆα k ˆβ k 1 c k 1 14 return c 1,c 2,,c n 8

9 4 (Progrmming) In this ssignment, you cn use your own implementtion of the nturl cubic spline lgorithm or librry function For your convenience, here is the lgorithm summry: NATURAL SPLINE COEFFICIENTS(x 1, x 2,, x n, f 1, f 2,, f n ) 1 for k 1,2,,n 1 2 do 3 h k x k+1 x k 4 b k ( f k+1 f k )/h k 5 for k 2,3,,n 1 6 do 7 k 2 (h k + h k 1 ) 8 g k b k b k 1 9 c c n 0 11 c 2,c 3,,c n 1 TRIDIAGONAL( 2, 3,, n 1,h 2,h 3,,h n 2, g 2, g 3,, g n 1 ) 12 for k 1,2,,n 1 13 do 14 d k (c k+1 c k )/h k 15 b k b k (2c k + c k+1 ) h k 16 c k 3c k 17 return b 1,b 2,,b n 1,c 1,c 2,,c n 1,d 1,d 2,,d n 1 SPLINE EVALUATION(x, x 1, x 2,, x n, f 1, f 2,, f n,b 1,b 2,,b n 1,c 1,c 2,,c n 1,d 1,d 2,,d n 1 ) 1 for k n 1,n 2,,1 2 do 3 h x x k 4 if h > 0 5 then exit loop 6 S f k + h (b k + h (c k + h d k )) 7 return S Using your progrm, interpolte Runge s function f (x) = 1 on set of n regulrly spced 1+25 x 2 spline knots 2(k 1) x k = 1 + n 1, k = 1,2,,n Tke n = 5,11,21 nd compute the interpoltion spline S(x) nd the error f (x) S(x) t 41 regulrly spced points You cn either plot the error or output it in tble Does the interpoltion ccurcy increse with the number of knots? Answer: 9

10 x The ccurcy does increse with the number of knots 10

11 Solution: C progrm: #include <stdlibh> /* for lloction */ #include <sserth> /* for ssertion */ #include "splineh" /* Function: tridigonl Symmetric tridigonl system solver n - dt length dig[n] - digonl offd[n-1] - off-digonl x[n] - in: right-hnd side, out: solution */ void tridigonl(int n, const double* dig, const double* offd, double* x) { double *, *b; int k; /* llocte storge */ = (double*) mlloc(n*sizeof(double)); b = (double*) mlloc((n-1)*sizeof(double)); ssert (!= NULL && b!= NULL); /* LU decomposition */ [0] = dig[0]; for (k=0; k < n-1; k++) { b[k] = offd[k]/[k]; [k+1] = dig[k+1] - b[k]*offd[k]; /* inverting L */ for (k=1; k < n; k++) { x[k] = x[k] - b[k-1]*x[k-1]; /* inverting U */ x[n-1] /= [n-1]; for (k=n-2; k >=0; k--) { x[k] = x[k]/[k] - b[k]*x[k+1]; free (); free (b); /* Function: tridigonl Alterntive form of symmetric tridigonl system solver n - dt length dig[n] - digonl offd[n-1] - off-digonl x[n] - in: right-hnd side, out: solution */ void tridigonl2(int n, const double* dig, const double* offd, double* x) { double *, *b; int k; 11

12 /* llocte storge */ = (double*) mlloc(n*sizeof(double)); b = (double*) mlloc((n-1)*sizeof(double)); ssert (!= NULL && b!= NULL); /* UL decomposition */ [n-1] = dig[n-1]; for (k=n-2; k >= 0; k--) { b[k] = offd[k]/[k+1]; [k] = dig[k] - b[k]*offd[k]; /* inverting U */ for (k=n-2; k >= 0; k--) { x[k] = x[k] - b[k]*x[k+1]; /* inverting L */ x[0] /= [0]; for (k=1; k < n-1; k++) { x[k] = x[k]/[k] - b[k-1]*x[k-1]; free (); free (b); /* Function: spline_coeffs Compute spline coefficients for interpolting nturl cubic spline n - number of knots x[n] - knots f[n] - function vlues coeff[4][n] - coefficients */ void spline_coeffs(int n, const double* x, const double* f, double* coeff[]) { double *, *h, *b, *c, *d; int k; h = (double*) mlloc((n-1)*sizeof(double)); = (double*) mlloc((n-2)*sizeof(double)); ssert (h!= NULL); /* renme for convenience */ b = coeff[1]; c = coeff[2]; d = coeff[3]; for (k=0; k < n-1; k++) { h[k] = x[k+1] - x[k]; /* intervl length */ coeff[0][k] = f[k]; b[k] = (f[k+1]-f[k])/h[k]; /* divided difference */ for (k=0; k < n-2; k++) { [k] = 2*(h[k+1] + h[k]); /* digonl */ c[k+1] = b[k+1] - b[k]; /* right-hnd side */ c[0] = 0; /* solve the tridigonl system */ 12

13 tridigonl(n-2,, h, c+1); for (k=0; k < n-1; k++) { if (k < n-2) { d[k] = (c[k+1]-c[k])/h[k]; b[k] -= (c[k+1]+2*c[k])*h[k]; else { d[k] = -c[k]/h[k]; b[k] -= 2*c[k]*h[k]; c[k] *= 3; /* Function: spline_evl Evlute cubic spline n - number of knots y - where to evlute x[n] - knots coeff[4][n] - spline coefficients */ double spline_evl(int n, double y, const double* x, double* coeff[]) { double h, s; int i, k; /* find the intervl for x */ for (k=n-2; k >=0; k--) { h = y - x[k]; if (h >= 0) brek; if (k < 0) k = 0; /* evlute cubic by Horner s rule */ s = coeff[3][k]; for (i=2; i >=0; i--) { s = s*h + coeff[i][k]; return s; #include <stdlibh> /* for lloction */ #include <stdioh> /* for output */ #include <sserth> /* for ssertion */ #include "splineh" /* Runge s function */ double runge (double x) { return (1/(1+25*x*x)); int min (void) { int i, k, n[]={5,11,21, nx, ny=41; double *x, *f, *coeff[4], *y, xk, s, e; y = (double*) mlloc (ny*sizeof(double)); 13

14 ssert (y!= NULL); /* regulr grid for plotting */ for (k=0; k < ny; k++) { y[k] = *k/(ny-1); /* three cses */ for (i=0; i < 3; i++) { nx = n[i]; /* llocte spce for tble */ x = (double*) mlloc (nx*sizeof(double)); f = (double*) mlloc (nx*sizeof(double)); ssert (x!= NULL && f!= NULL); /* llocte coefficients */ for (k=0; k < 4; k++) { coeff[k] = (double*) mlloc ((nx-1)*sizeof(double)); ssert (coeff[k]!= NULL); /* build the tble */ for (k=0; k < nx; k++) { xk = *k/(nx-1); f[k] = runge(xk); x[k] = xk; /* compute coefficients */ spline_coeffs(nx, x, f, coeff); /* evlute the spline function */ for (k=0; k < ny; k++) { xk = y[k]; s = spline_evl(nx, xk, x, coeff); /* spline */ e = runge(xk)-s; /* error */ /* print out the tble */ printf("%d %f %f %g\n", k, xk, s, e); free (x); free (f); for (k=0; k < 4; k++) { free (coeff[k]); exit(0); 14

15 5 (Progrmming) The vlues in the tble specify {x, y points on curve {x(t), y(t) x y In this ssignment, you will reconstruct the curve using cubic splines nd interpolting independently x(t) nd y(t) We don t know the vlues of t t the spline knots but cn pproximte them For exmple, we cn tke t to represent the length long the curve nd pproximte it by the length of the liner segments: t 1 = 0 ; t k = t k 1 + (x k x k 1 ) 2 + (y k y k 1 ) 2, k = 2,3,,n Clculte spline coefficients for the nturl cubic splines interpolting x(t) nd y(t), then evlute the splines t 100 regulrly spced points in the intervl between t 1 nd t n nd plot the curve Wht other boundry conditions would be pproprite in this exmple? Answer:

16 The periodic boundry conditions would be more pproprite in this cse Solution: C progrm: #include <stdlibh> /* for lloction */ #include <stdioh> /* for output */ #include <mthh> /* for mth functions */ #include <sserth> /* for ssertion */ #include "splineh" int min (void) { const int nt=13, nt1=100; int k; double x[] = {25,13,-025,0,025,-13,-25,-13,025,0,-025,13,25; double y[] = {0,-025,13,25,13,-025,0,025,-13,-25,-13,025,0; double t[13], *xc[4], *yc[4], t1[100], tk, x1, y1; for (k=0; k < 4; k++) { xc[k] = (double*) mlloc ((nt-1)*sizeof(double)); yc[k] = (double*) mlloc ((nt-1)*sizeof(double)); ssert (xc[k]!= NULL && yc[k]!= NULL); /* find the knots */ t[0] = 0; for (k=1; k < nt; k++) { t[k] = t[k-1] + hypot(x[k]-x[k-1],y[k]-y[k-1]); /* regulr grid for plotting */ for (k=0; k < nt1; k++) { t1[k] = k*t[nt-1]/(nt1-1); /* spline coefficients for x(t) nd y(t) */ spline_coeffs(nt, t, x, xc); spline_coeffs(nt, t, y, yc); /* evlute the spline function */ for (k=0; k < nt1; k++) { tk = t1[k]; x1 = spline_evl(nt, tk, t, xc); y1 = spline_evl(nt, tk, t, yc); /* print out the tble */ printf("%d %f %f %g\n", k, tk, x1, y1); exit(0); 16