Part IB Numerical Analysis

Transcription

1 Prt IB Numericl Anlysis Theorems with proof Bsed on lectures by G. Moore Notes tken by Dexter Chu Lent 2016 These notes re not endorsed by the lecturers, nd I hve modified them (often significntly) fter lectures. They re nowhere ner ccurte representtions of wht ws ctully lectured, nd in prticulr, ll errors re lmost surely mine. Polynomil pproximtion Interpoltion by polynomils. Divided differences of functions nd reltions to derivtives. Orthogonl polynomils nd their recurrence reltions. Lest squres pproximtion by polynomils. Gussin qudrture formule. Peno kernel theorem nd pplictions. [6] Computtion of ordinry differentil equtions Euler s method nd proof of convergence. Multistep methods, including order, the root condition nd the concept of convergence. Runge-Kutt schemes. Stiff equtions nd A-stbility. [5] Systems of equtions nd lest squres clcultions LU tringulr fctoriztion of mtrices. Reltion to Gussin elimintion. Column pivoting. Fctoriztions of symmetric nd bnd mtrices. The Newton-Rphson method for systems of non-liner lgebric equtions. QR fctoriztion of rectngulr mtrices by Grm-Schmidt, Givens nd Householder techniques. Appliction to liner lest squres clcultions. [5] 1

2 Contents IB Numericl Anlysis (Theorems with proof) Contents 0 Introduction 3 1 Polynomil interpoltion The interpoltion problem The Lgrnge formul The Newton formul A useful property of divided differences Error bounds for polynomil interpoltion Orthogonl polynomils Sclr product Orthogonl polynomils Three-term recurrence reltion Exmples Lest-squres polynomil pproximtion Approximtion of liner functionls Liner functionls Gussin qudrture Expressing errors in terms of derivtives 12 5 Ordinry differentil equtions Introduction One-step methods Multi-step methods Runge-Kutt methods Stiff equtions Introduction Liner stbility Implementtion of ODE methods Locl error estimtion Solving for implicit methods Numericl liner lgebr Tringulr mtrices LU fctoriztion A = LU for specil A Liner lest squres 20 2

3 0 Introduction IB Numericl Anlysis (Theorems with proof) 0 Introduction 3

4 1 Polynomil interpoltion IB Numericl Anlysis (Theorems with proof) 1 Polynomil interpoltion 1.1 The interpoltion problem 1.2 The Lgrnge formul Theorem. The interpoltion problem hs exctly one solution. Proof. We define p P n [x] by Evluting t x i gives p(x j ) = p(x) = f k l k (x). f k l k (x j ) = f k δ jk = f j. So we get existence. For uniqueness, suppose p, q P n [x] re solutions. Then the difference r = p q P n [x] stisfies r(x j ) = 0 for ll j, i.e. it hs n + 1 roots. However, non-zero polynomil of degree n cn hve t most n roots. So in fct p q is zero, i.e. p = q. 1.3 The Newton formul Theorem (Recurrence reltion for Newton divided differences). For 0 j < k n, we hve f[x j,, x k ] = f[x j+1,, x k ] f[x j,, x k 1 ] x k x j. Proof. The key to proving this is to relte the interpolting polynomils. Let q 0, q 1 P k j 1 [x] nd q 2 P k j stisfy q 0 (x i ) = f i i = j,, k 1 q 1 (x i ) = f i q 2 (x i ) = f i i = j + 1,, k i = j,, k We now clim tht q 2 (x) = x x j x k x j q 1 (x) + x k x x k x j q 0 (x). We cn check directly tht the expression on the right correctly interpoltes the points x i for i = j,, k. By uniqueness, the two expressions gree. Since f[x j,, x k ], f[x j+1,, x k ] nd f[x j,, x k 1 ] re the leding coefficients of q 2, q 1, q 0 respectively, the result follows. 4

5 1 Polynomil interpoltion IB Numericl Anlysis (Theorems with proof) 1.4 A useful property of divided differences Lemm. Let g C m [, b] hve continuous mth derivtive. Suppose g is zero t m + l distinct points. Then g (m) hs t lest l distinct zeros in [, b]. Proof. This is repeted ppliction of Rolle s theorem. We know tht between every two zeros of g, there is t lest one zero of g C m 1 [, b]. So by differentiting once, we hve lost t most 1 zeros. So fter differentiting m times, g (m) hs lost t most m zeros. So it still hs t lest l zeros. Theorem. Let {x i } n i=0 [, b] nd f Cn [, b]. Then there exists some ξ (, b) such tht f[x 0,, x n ] = 1 n! f (n) (ξ). Proof. Consider e = f p n C n [, b]. This hs t lest n + 1 distinct zeros in [, b]. So by the lemm, e (n) = f (n) p (n) n must vnish t some ξ (, b). But then p (n) n = n!f[x 0,, x n ] constntly. So the result follows. 1.5 Error bounds for polynomil interpoltion Theorem. Assume {x i } n i=0 [, b] nd f C[, b]. Let x [, b] be noninterpoltion point. Then where e n ( x) = f[x 0, x 1,, x n, x]ω( x), ω(x) = n (x x i ). Proof. We think of x = x n+1 s new interpoltion point so tht i=0 p n+1 (x) p n (x) = f[x 0,, x n, x]ω(x) for ll x R. In prticulr, putting x = x, we hve p n+1 ( x) = f( x), nd we get the result. Theorem. If in ddition f C n+1 [, b], then for ech x [, b], we cn find ξ x (, b) such tht e n (x) = 1 (n + 1)! f (n+1) (ξ x )ω(x) Proof. The sttement is trivil if x is n interpoltion point pick rbitrry ξ x, nd both sides re zero. Otherwise, this follows directly from the lst two theorems. Corollry. For ll x [, b], we hve f(x) p n (x) 1 (n + 1)! f (n+1) ω(x) 5

6 1 Polynomil interpoltion IB Numericl Anlysis (Theorems with proof) Lemm (3-term recurrence reltion). The Chebyshev polynomils stisfy the recurrence reltions T n+1 (x) = 2xT n (x) T n 1 (x) with initil conditions T 0 (x) = 1, T 1 (x) = x. Proof. cos((n + 1)θ) + cos((n 1)θ) = 2 cos θ cos(nθ). Theorem (Miniml property for n 1). On [ 1, 1], mong ll polynomils 1 p P n [x] with leding coefficient 1, 2 T n 1 n minimizes p. Thus, the 1 minimum vlue is 2. n 1 Proof. We proceed by contrdiction. Suppose there is polynomil q n P n with leding coefficient 1 such tht q n < 1 2 n 1. Define new polynomil r = 1 2 n 1 T n q n. This is, by ssumption, non-zero. Since both the polynomils hve leding coefficient 1, the difference must 1 hve degree t most n 1, i.e. r P n 1 [x]. Since 2 T n 1 n (X k ) = ± 1 2, nd n 1 q n (X n ) < 1 2 by ssumption, r lterntes in sign between these n + 1 points. n 1 But then by the intermedite vlue theorem, r hs to hve t lest n zeros. This is contrdiction, since r hs degree n 1, nd cnnot be zero. Corollry. Consider w = n (x x i ) P n+1 [x] i=0 for ny distinct points = {x i } n i=0 [ 1, 1]. Then min ω = 1 2 n. This minimum is chieved by picking the interpoltion points to be the zeros of T n+1, nmely ( ) 2k + 1 x k = cos 2n + 2 π, k = 0,, n. Theorem. For f C n+1 [ 1, 1], the Chebyshev choice of interpoltion points gives f p n n (n + 1)! f (n+1). 6

7 2 Orthogonl polynomils IB Numericl Anlysis (Theorems with proof) 2 Orthogonl polynomils 2.1 Sclr product 2.2 Orthogonl polynomils Theorem. Given vector spce V of functions nd n inner product,, there exists unique monic orthogonl polynomil for ech degree n 0. In ddition, {p k } n form bsis for P n[x]. Proof. This is big induction proof over both prts of the theorem. We induct over n. For the bse cse, we pick p 0 (x) = 1, which is the only degree-zero monic polynomil. Now suppose we lredy hve {p n } n stisfying the induction hypothesis. Now pick ny monic q n+1 P n+1 [x], e.g. x n+1. We now construct p n+1 from q n+1 by the Grm-Schmidt process. We define p n+1 = q n+1 This is gin monic since q n+1 is, nd we hve p n+1, p m = 0 q n+1, p k p k, p k p k. for ll m n, nd hence p n+1, p = 0 for ll p P n [x] = p 0,, p n. To obtin uniqueness, ssume both p n+1, ˆp n+1 P n+1 [x] re both monic orthogonl polynomils. Then r = p n+1 ˆp n+1 P n [x]. So r, r = r, p n+1 ˆp n+1 = r, p n+1 r, ˆp n+1 = 0 0 = 0. So r = 0. So p n+1 = ˆp n 1. Finlly, we hve to show tht p 0,, p n+1 form bsis for P n+1 [x]. Now note tht every p P n+1 [x] cn be written uniquely s p = cp n+1 + q, where q P n [x]. But {p k } n is bsis for P n[x]. So q cn be uniquely decomposed s liner combintion of p 0,, p n. Alterntively, this follows from the fct tht ny set of orthogonl vectors must be linerly independent, nd since there re n + 2 of these vectors nd P n+1 [x] hs dimension n + 2, they must be bsis. 2.3 Three-term recurrence reltion Theorem. Monic orthogonl polynomils re generted by with initil conditions p k+1 (x) = (x α k )p k (x) β k p k 1 (x) p 0 = 1, p 1 (x) = (x α 0 )p 0, where α k = xp k, p k p k, p k, β k = p k, p k p k 1, p k 1. 7

8 2 Orthogonl polynomils IB Numericl Anlysis (Theorems with proof) Proof. By inspection, the p 1 given is monic nd stisfies p 1, p 0 = 0. Using q n+1 = xp n in the Grm-Schmidt process gives p n+1 = xp n p n+1 = xp n xp n, p k p k, p k p k p n, xp k p k, p k p k We notice tht p n, xp k nd vnishes whenever xp k hs degree less thn n. So we re left with = xp n xp n, p n p n, p n p n p n, xp n 1 p n 1, p n 1 p n 1 = (x α n )p n p n, xp n 1 p n 1, p n 1 p n 1. Now we notice tht xp n 1 is monic polynomil of degree n so we cn write this s xp n 1 = p n + q. Thus p n, xp n 1 = p n, p n + q = p n, p n. Hence the coefficient of p n 1 is indeed the β we defined. 2.4 Exmples 2.5 Lest-squres polynomil pproximtion Theorem. If {p n } n re orthogonl polynomils with respect to,, then the choice of c k such tht p = c k p k minimizes f p 2 is given by nd the formul for the error is c k = f, p k p k 2, f p 2 = f 2 Proof. We consider generl polynomil We substitute this in to obtin p = f p, f p = f, f 2 c k p k. f, p k 2 p k 2. c k f, p k + 8 c 2 k p k 2.

9 2 Orthogonl polynomils IB Numericl Anlysis (Theorems with proof) Note tht there re no cross terms between the different coefficients. We minimize this qudrtic by setting the prtil derivtives to zero: 0 = c k f p, f p = 2 f, p k + 2c k p k 2. To check this is indeed minimum, note tht the Hessin mtrix is simply 2I, which is positive definite. So this is relly minimum. So we get the formul for the c k s s climed, nd putting the formul for c k gives the error formul. 9

10 3 Approximtion of liner functionls IB Numericl Anlysis (Theorems with proof) 3 Approximtion of liner functionls 3.1 Liner functionls 3.2 Gussin qudrture Proposition. There is no choice of ν weights nd nodes such tht the pproximtion of w(x)f(x) dx is exct for ll f P 2ν[x]. Proof. Define Then we know q(x) = ν (x c k ) P ν [x]. w(x)q 2 (x) dx > 0, since q 2 is lwys non-negtive nd hs finitely mny zeros. However, So this cnnot be exct for q 2. ν b k q 2 (c n ) = 0. Theorem (Ordinry qudrture). For ny distinct {c k } ν [, b], let {l k} ν be the Lgrnge crdinl polynomils with respect to {c k } ν. Then by choosing the pproximtion L(f) = b k = w(x)l k (x) dx, w(x)f(x) dx ν b k f(c k ) is exct for f P ν 1 [x]. We cll this method ordinry qudrture. Theorem. For ν 1, the zeros of the orthogonl polynomil p ν re rel, distinct nd lie in (, b). Proof. First we show there is t lest one root. Notice tht p 0 = 1. Thus for ν 1, by orthogonlity, we know w(x)p ν (x)p 1 (x) dx = w(x)p ν (x) dx = 0. So there is t lest one sign chnge in (, b). We hve lredy got the result we need for ν = 1, since we only need one zero in (, b). Now for ν > 1, suppose {ξ j } m j=1 re the plces where the sign of p ν chnges in (, b) (which is subset of the roots of p ν ). We define q(x) = m (x ξ j ) P m [x]. j=1 10

11 3 Approximtion of liner functionls IB Numericl Anlysis (Theorems with proof) Since this chnges sign t the sme plce s p ν, we know qp ν mintins the sme sign in (, b). Now if we hd m < ν, then orthogonlity gives q, p ν = w(x)q(x)p ν (x) dx = 0, which is impossible, since qp ν does not chnge sign. Hence we must hve m = ν. Theorem. In the ordinry qudrture, if we pick {c k } ν to be the roots of p ν (x), then get we exctness for f P 2ν 1 [x]. In ddition, {b n } ν re ll positive. Proof. Let f P 2ν 1 [x]. Then by polynomil division, we get f = qp ν + r, where q, r re polynomils of degree t most ν 1. We pply orthogonlity to get w(x)f(x) dx = Also, since ech c k is root of p ν, we get ν b k f(c k ) = w(x)(q(x)p ν (x) + r(x)) dx = ν b k (q(c k )p ν (c k ) + r(c k )) = w(x)r(x) dx. ν b k r(c k ). But r hs degree t most ν 1, nd this formul is exct for polynomils in P ν 1 [x]. Hence we know w(x)f(x) dx = w(x)r(x) dx = ν b k r(c k ) = ν b k f(c k ). To show the weights re positive, we simply pick s specil f. Consider f {l 2 k }ν P 2ν 2[x], for l k the Lgrnge crdinl polynomils for {c k } ν. Since the qudrture is exct for these, we get 0 < w(x)l 2 k(x) dx = ν b j l 2 k(c j ) = j=1 ν b j δ jk = b k. j=1 Since this is true for ll k = 1,, ν, we get the desired result. 11

12 4 Expressing errors in terms of IBderivtives Numericl Anlysis (Theorems with proof) 4 Expressing errors in terms of derivtives Theorem (Peno kernel theorem). If λ nnihiltes polynomils of degree k or less, then λ(f) = 1 k! for ll f C k+1 [, b], where K(θ)f (k+1) (θ) dθ 12

13 5 Ordinry differentil equtions IB Numericl Anlysis (Theorems with proof) 5 Ordinry differentil equtions 5.1 Introduction 5.2 One-step methods Theorem (Convergence of Euler s method). (i) For ll t [0, T ], we hve lim y n y(t) = 0. h 0 nh t (ii) Let λ be the Lipschitz constnt of f. Then there exists c 0 such tht e n ch eλt 1 λ for ll 0 n [T/h], where e n = y n y(t n ). Proof. There re two prts to proving this. We first look t the locl trunction error. This is the error we would get t ech step ssuming we got the previous steps right. More precisely, we write y(t n+1 ) = y(t n ) + h(f, t n, y(t n )) + R n, nd R n is the locl trunction error. For the Euler s method, it is esy to get R n, since f(t n, y(t n )) = y (t n ), by definition. So this is just the Tylor series expnsion of y. We cn write R n s the integrl reminder of the Tylor series, R n = By some creful nlysis, we get tn+1 t n (t n+1 θ)y (θ) dθ. R n ch 2, where c = 1 2 y. This is the esy prt, nd tends to go rther smoothly even for more complicted methods. Once we hve bounded the locl trunction error, we ptch them together to get the ctul error. We cn write e n+1 = y n+1 y(t n+1 ) ) = y n + hf(t n, y n ) (y(t n ) + hf(t n, y(t n )) + R n ( ) = (y n y(t n )) + h f(t n, y n ) f(t n, y(t n )) R n Tking the infinity norm, we get e n+1 y n y(t n ) + h f(t n, y n ) f(t n, y(t n )) + R n e n + hλ e n + ch 2 = (1 + λh) e n + ch 2. 13

14 5 Ordinry differentil equtions IB Numericl Anlysis (Theorems with proof) This is vlid for ll n 0. We lso know e 0 = 0. Doing some lgebr, we get Finlly, we hve n 1 e n ch 2 (1 + hλ) j ch λ ((1 + hλ)n 1). j=0 (1 + hλ) e λh, since 1 + λh is the first two terms of the Tylor series, nd the other terms re positive. So (1 + hλ) n e λhn e λt. So we obtin the bound e n ch eλt 1. λ Then this tends to 0 s we tke h 0. So the method converges. 5.3 Multi-step methods Theorem. An s-step method hs order p (p 1) if nd only if nd s ρ l = 0 s ρ l l k = k for k = 1,, p, where 0 0 = 1. Proof. The locl trunction error is s ρ l y(t n+l ) h s σ l l k 1 s σ l y (t n+l ). We now expnd the y nd y bout t n, nd obtin ( s ) ( s ) h k s ρ l y(t n ) + ρ l l k k σ l l k 1 y (k) (t n ). k! This is O(h p+1 ) under the given conditions. Theorem. A multi-step method hs order p (with p 1) if nd only if s x 0. Proof. We expnd ρ(e x ) xσ(e x ) = O(x p+1 ) ρ(e x ) xσ(e x ) = s ρ l e lx x 14 s σ l e lx.

15 5 Ordinry differentil equtions IB Numericl Anlysis (Theorems with proof) We now expnd the e lx in Tylor series bout x = 0. This comes out s ( s s ) 1 s ρ l + ρ l l k k σ l l k 1 x k. k! So the result follows. Theorem (Dhlquist equivlence theorem). A multi-step method is convergent if nd only if (i) The order p is t lest 1; nd (ii) The root condition holds. Lemm. An s-step bckwrd differentition method of order s is obtined by choosing s 1 ρ(w) = σ s l ws l (w 1) l, l=1 with σ s chosen such tht ρ s = 1, nmely σ s = Proof. We need to construct ρ so tht ( s l=1 ) 1 1. l ρ(w) = σ s w s log w + O( w 1 s+1 ). This is esy, if we write ( ) 1 log w = log w ( = log 1 w 1 ) w ( ) l 1 w 1 =. l w l=1 Multiplying by σ s w s gives the desired result. 5.4 Runge-Kutt methods 15

16 6 Stiff equtions IB Numericl Anlysis (Theorems with proof) 6 Stiff equtions 6.1 Introduction 6.2 Liner stbility Theorem (Mximum principle). Let g be nlytic nd non-constnt in n open set Ω C. Then g hs no mximum in Ω. 16

17 7 Implementtion of ODE methods IB Numericl Anlysis (Theorems with proof) 7 Implementtion of ODE methods 7.1 Locl error estimtion 7.2 Solving for implicit methods 17

18 8 Numericl liner lgebr IB Numericl Anlysis (Theorems with proof) 8 Numericl liner lgebr 8.1 Tringulr mtrices 8.2 LU fctoriztion 8.3 A = LU for specil A Theorem. A sufficient condition for the existence for both the existence nd uniqueness of A = LU is tht det(a k ) 0 for k = 1,, n 1. Proof. Strightforwrd induction. Theorem. If det(a k ) 0 for ll k = 1,, n, then A R n n hs unique fctoriztion of the form A = LDÛ, where D is non-singulr digonl mtrix, nd both L nd Û re unit tringulr. Proof. From the previous theorem, A = LU exists. Since A is non-singulr, U is non-singulr. So we cn write this s U = DÛ, where D consists of the digonls of U nd Û = D 1 U is unit. Theorem. Let A R n n be non-singulr nd det(a k ) 0 for ll k = 1,, n. Then there is unique symmetric fctoriztion A = LDL T, with L unit lower tringulr nd D digonl nd non-singulr. Proof. From the previous theorem, we cn fctorize A uniquely s We tke the trnspose to obtin A = LDÛ. A = A T = Û T DL T. This is fctoriztion of the form unit lower - digonl - unit upper. By uniqueness, we must hve Û = LT. So done. Theorem. Let A R n n be positive-definite mtrix. Then det(a k ) 0 for ll k = 1,, n. Proof. First consider k = n. To show A is non-singulr, it suffices to show tht Ax = 0 implies x = 0. But we cn multiply the eqution by x T to obtin x T Ax = 0. By positive-definiteness, we must hve x = 0. So done. Now suppose A k y = 0 for k < n nd y = R k. Then y T A k y = 0. We invent new x R n by tking y nd pd it with zeros. Then x T Ax = 0. By positive-definiteness, we know x = 0. Then in prticulr y = 0. 18

19 8 Numericl liner lgebr IB Numericl Anlysis (Theorems with proof) Theorem. A symmetric mtrix A R n n is positive-definite iff we cn fctor it s A = LDL T, where L is unit lower tringulr, D is digonl nd D kk > 0. Proof. First suppose such fctoriztion exists, then x T Ax = x T LDL T x = (L T x) T D(L T x). We let y = L T x. Note tht y = 0 if nd only if x = 0, since L is invertible. So x T Ax = y T Dy = ykd 2 kk > 0 if y 0. Now if A is positive definite, it hs n LU fctoriztion, nd since A is symmetric, we cn write it s A = LDL T, where L is unit lower tringulr nd D is digonl. Now we hve to show D kk > 0. We define y k such tht L T y k = e k, which exist, since L is invertible. Then clerly y k 0. Then we hve So done. D kk = e T k De k = y T k LDL T y k = y T k Ay k > 0. Proposition. If bnd mtrix A hs bnd width r nd n LU fctoriztion A = LU, then L nd U re both bnd mtrices of width r. Proof. Strightforwrd verifiction. 19

20 9 Liner lest squres IB Numericl Anlysis (Theorems with proof) 9 Liner lest squres Theorem. A vector x R n minimizes Ax b 2 if nd only if A T (Ax b) = 0. Proof. A solution, by definition, minimizes f(x) = Ax b, Ax b = x T AAx 2x T A T b + b T b. Then s function of x, the prtil derivtives of this must vnish. We hve So necessry condition is f(x) = 2A T (Ax b). A T (Ax b). Now suppose our x stisfies A T (Ax b) = 0. Then for ll x R n, we write x = x + y, nd then we hve Ax b 2 = A(x + y) b = Ax b + 2y T A T (Ax b) + Ay 2 = Ax b + Ay 2 Ax b. So x must minimize the Eucliden norm. Corollry. If A R m n is full-rnk mtrix, then there is unique solution to the lest squres problem. Proof. We know ll minimizers re solutions to (A T A)x = A T b. The mtrix A being full rnk mens y 0 R n implies Ay 0 R m. Hence A T A R n n is positive definite (nd in prticulr non-singulr), since x T A T Ax = (Ax) T (Ax) = Ax 2 > 0 for x 0. So we cn invert A T A nd find unique solution x. Proposition. A mtrix A R m n cn be trnsformed into upper-tringulr form by pplying n Householder reflections, nmely H n H 1 A = R, where ech H n introduces zero into column k nd leves the other zeroes lone. Lemm. Let, b R m, with b, but = b. Then if we pick u = b, then H u = b. 20

21 9 Liner lest squres IB Numericl Anlysis (Theorems with proof) Proof. We just do it: H u = where we used the fct tht = b. 2( T b) 2 2 T ( b) = ( b) = b, b + b 2 Lemm. If the first k 1 components of u re zero, then (i) For every x R m, H u x does not lter the first k 1 components of x. (ii) If the lst (m k + 1) components of y R m re zero, then H u y = y. Lemm. Let, b R m, with k. m b ḳ.. b m, but Suppose we pick m m 2 j = b 2 j. j=k j=k u = (0, 0,, 0, k b k,, m b m ) T. Then we hve H u = ( 1,, k 1 b k,, b m ). 21