Nonlinear Algebraic Equations. Lectures INF2320 p. 1/88

Transcription

1 Nonlinear Algebraic Equations Lectures INF2320 p. 1/88

2 Lectures INF2320 p. 2/88 Nonlinear algebraic equations When solving the system u (t) = g(u), u(0) = u 0, (1) with an implicit Euler scheme we have to solve the nonlinear algebraic equation u n+1 t g(u n+1 ) = u n, (2) at each time step. Here u n is known and u n+1 is unknown. If we let c denote u n and v denote u n+1, we want to find v such that where c is given. v t g(v) = c, (3)

3 Nonlinear algebraic equations First consider the case of g(u) = u, which corresponds to the differential equation u = u, u(0) = u 0. (4) The equation (3) for each time step, is now v t v = c, (5) which has the solution v = 1 1 t c. (6) The time stepping in the Euler scheme for (4) is written u n+1 = 1 1 t u n. (7) Lectures INF2320 p. 3/88

4 Lectures INF2320 p. 4/88 Nonlinear algebraic equations Similarly, for any linear function g, i.e., functions on the form g(v) = α+βv (8) with constants α and β, we can solve equation (3) directly and get v = c+α t 1 β t. (9)

5 Lectures INF2320 p. 5/88 Nonlinear algebraic equations Next we study the nonlinear differential equation u = u 2, (10) which means that g(v) = v 2. (11) Now (3) reads v t v 2 = c. (12)

6 Lectures INF2320 p. 6/88 Nonlinear algebraic equations This second order equation has two possible solutions v + = t c 2 t (13) and v = t c 2 t. (14) Note that lim t t c 2 t =. Since t is supposed to be small and the solution is not expected to blow up, we conclude that v + is not correct.

7 Lectures INF2320 p. 7/88 Nonlinear algebraic equations Therefore the correct solution of (12) to use in the Euler scheme is v = t c. (15) 2 t We can now conclude that the implicit scheme can be written on computational form u n+1 t u 2 n+1 = u n (16) u n+1 = t u n 2 t. (17)

8 Lectures INF2320 p. 8/88 Nonlinear algebraic equations We have seen that the equation can be solved analytically when or v t g(v) = c (18) g(v) = v (19) g(v) = v 2. (20) Generally it can be seen that we can solve (18) when g is on the form g(v) = α+βv+γv 2. (21)

9 Lectures INF2320 p. 9/88 Nonlinear algebraic equations For most cases of nonlinear functions g, (18) can not be solved analytically A couple of examples of this is g(v) = e v or g(v) = sin(v)

10 Lectures INF2320 p. 10/88 Nonlinear algebraic equations Since we work with nonlinear equations on the form u n+1 u n = t g(u n+1 ) (22) where t is a small number, we know that u n+1 is close to u n. This will be a useful property later. In the rest of this lecture we will write nonlinear equations on the form f(x) = 0, (23) where f is nonlinear. We assume that we have available a value x 0 close to the true solution x (, i.e. f(x ) = 0). We also assume that f has no other zeros in a small region around x.

11 Lectures INF2320 p. 11/88 The bisection method Consider the function f(x) = 2+x e x (24) for x ranging from 0 to 3, see the graph in Figure 1. We want to find x = x such that f(x ) = 0

12 Lectures INF2320 p. 12/ x* x=3 5 f(x) y x Figure 1: The graph of f(x) = 2+x e x.

13 Lectures INF2320 p. 13/88 The bisection method An iterative method is to create a series {x i } of approximations of x, which hopefully converges towards x For the Bisection Method we choose the two first guesses x 0 and x 1 as the endpoints of the definition domain, i.e. x 0 = 0 and x 1 = 3 Note that f(x 0 ) = f(0) > 0 and f(x 1 ) = f(3) < 0, and therefore x 0 < x < x 1, provided that f is continuous We now define the mean value of x 0 and x 1 x 2 = 1 2 (x 0 + x 1 ) = 3 2

14 Lectures INF2320 p. 14/ x =0 0 x 2 =1.5 x 1 =3 5 y x Figure 2: The graph of f(x) = 2 + x e x and three values of f : f(x 0 ), f(x 1 ) and f(x 2 ).

15 Lectures INF2320 p. 15/88 The bisection method We see that f(x 2 ) = f( 3 2 ) = 2+3/2 e3/2 < 0, Since f(x 0 ) > 0 and f(x 2 ) < 0, we know that x 0 < x < x 2 Therefore we define x 3 = 1 2 (x 0 + x 2 ) = 3 4 Since f(x 3 ) > 0, we know that x 3 < x < x 2 (see Figure 3) This can be continued until f(x n ) is sufficiently small

16 Lectures INF2320 p. 16/ x 3 =0.75 x 2 =1.5 y x Figure 3: The graph of f(x) = 2+x e x and two values of f : f(x 2 ) and f(x 3 ).

17 Lectures INF2320 p. 17/88 The bisection method Written in algorithmic form the Bisection method reads: Algorithm 1. Given a, b such that f(a) f(b) < 0 and given a tolerance ε. Define c = 1 2 (a+b). while f(c) > ε do if f(a) f(c) < 0 then b = c else a = c c := 1 2 (a+b) end

18 Lectures INF2320 p. 18/88 Example 11 Find the zeros for f(x) = 2+x e x using Algorithm 1 and choose a = 0, b = 3 and ε = In Table 1 we show the number of iterations i, c and f(c) The number of iterations, i, refers to the number of times we pass through the while-loop of the algorithm

19 Lectures INF2320 p. 19/88 i c f(c) Table 1: Solving the nonlinear equation f(x) = 2 + x e x = 0 by using the bisection method; the number of iterations i, c and f(c).

20 Lectures INF2320 p. 20/88 Example 11 We see that we get sufficient accuracy after 21 iterations The next slide show the C program that is used to solve this problem The entire computation uses seconds on a Pentium III 1GHz processor Even if this quite fast, we need even faster algorithms in actual computations In practical applications you might need to solve billions of nonlinear equations, and then every micro second counts

21 Lectures INF2320 p. 21/88 #include <stdio.h> #include <math.h> double f (double x) { return 2.+x-exp(x); } inline double fabs (double r) { return ( (r >= 0.0)? r : -r ); } int main (int nargs, const char** args) { double epsilon = 1.0e-6; double a, b, c, fa, fc; a = 0.; b = 3.; fa = f(a); c = 0.5*(a+b); while (fabs(fc=(f(c))) > epsilon) { if ((fa*fc) < 0) { b = c; } else { a = c; fa = fc; } c = 0.5*(a+b); } printf("final c=%g, f(c)=%g\n",c,fc); return 0; }

22 Lectures INF2320 p. 22/88 Newton s method Recall that we have assumed that we have a good initial guess x 0 close to x (where f(x ) = 0) We will also assume that we have a small region around x where f has only one zero, and that f (x) 0 Taylor series expansion around x = x 0 yields f(x 0 + h) = f(x 0 )+hf (x 0 )+O(h 2 ) (25) Thus, for small h we have f(x 0 + h) f(x 0 )+hf (x 0 ) (26)

23 Lectures INF2320 p. 23/88 Newton s method We want to choose the step h such that f(x 0 + h) 0 By (26) this can be done by choosing h such that Solving this gives We therefore define f(x 0 )+hf (x 0 ) = 0 h = f(x 0) f (x 0 ) x 1 def = x 0 + h = x 0 f(x 0) f (x 0 ) (27)

24 Lectures INF2320 p. 24/88 Newton s method We test this on the example studied above with f(x) = 2+x e x and x 0 = 3 We have that Therefore We see that f (x) = 1 e x x 1 = x 0 f(x 0) f (x 0 ) = 3 5 e3 1 e 3 = f(x 0 ) = f(3) and f(x 1 ) = f(2.2096) 4.902, i.e, the value of f is significantly reduced

25 Lectures INF2320 p. 25/88 Newton s method We can now repeat the above procedure and define x 2 def = x 1 f(x 1) f (x 1 ), (28) and in algorithmic form Newton s method reads: Algorithm 2. Given an initial approximation x 0 and a tolerance ε. k = 0 while f(x k ) > ε do x k+1 = x k f(x k) f (x k ) k = k+1 end

26 Lectures INF2320 p. 26/88 Newton s method In Table 2 we show the results generated by Newton s method on the above example. k x k f(x k ) Table 2: Solving the nonlinear equation f(x) = 2 + x e x = 0 by using Algorithm 25 and ε = 10 6 ; the number of iterations k, x k and f(x k ).

27 Lectures INF2320 p. 27/88 Newton s method We observe that the convergence is much faster for Newton s method than for the Bisection method Generally, Newton s method converges faster than the Bisection method This will be studied in more detail in Project 1

28 Lectures INF2320 p. 28/88 Example 12 Let f(x) = x 2 2, and find x such that f(x ) = 0. Note that one of the exact solutions is x = 2 Newton s method for this problem reads or x k+1 = x k x2 k 2 2x k x k+1 = x2 k + 2 2x k

29 Lectures INF2320 p. 29/88 Example 12 If we choose x 0 = 1, we get x 1 = 1.5, x 2 = , x 3 = Comparing this with the exact value x = , we see that a very accurate approximation is obtained in only 3 iterations.

30 Lectures INF2320 p. 30/88 An alternative derivation The Taylor series expansion of f around x 0 is given by f(x) = f(x 0 )+(x x 0 ) f (x 0 )+O((x x 0 ) 2 ) Let F 0 (x) be a linear approximation of f around x 0 : F 0 (x) = f(x 0 )+(x x 0 ) f (x 0 ) F 0 (x) approximates f around x 0 since F 0 (x 0 ) = f(x 0 ) and F 0(x 0 ) = f (x 0 ) We now define x 1 to be such that F(x 1 ) = 0, i.e. f(x 0 )+(x 1 x 0 ) f (x 0 ) = 0

31 Lectures INF2320 p. 31/88 An alternative derivation Then we get x 1 = x 0 f(x 0) f (x 0 ), which is identical to the iteration obtained above We repeat this process, and define a linear approximation of f around x 1 F 1 (x) = f(x 1 )+(x x 1 ) f (x 1 ) x 2 is defined such that F 1 (x 2 ) = 0, i.e. x 2 = x 1 f(x 1) f (x 1 )

32 Lectures INF2320 p. 32/88 An alternative derivation Generally we get x k+1 = x k f(x k) f (x k ) This process is illustrated in Figure 4

33 Lectures INF2320 p. 33/88 f(x) x 3 x 2 x 1 x 0 Figure 4: Graphical illustration of Newton s method.

34 Lectures INF2320 p. 34/88 The Secant method The secant method is similar to Newton s method, but the linear approximation of f is defined differently Now we assume that we have two values x 0 and x 1 close to x, and define the linear function F 0 (x) such that F 0 (x 0 ) = f(x 0 ) and F 0 (x 1 ) = f(x 1 ) The function F 0 (x) is therefore given by F 0 (x) = f(x 1 )+ f(x 1) f(x 0 ) x 1 x 0 (x x 1 ) F 0 (x) is called the linear interpolant of f

35 Lectures INF2320 p. 35/88 The Secant method Since F 0 (x) f(x), we can compute a new approximation x 2 to x by solving the linear equation F(x 2 ) = 0 This means that we must solve f(x 1 )+ f(x 1) f(x 0 ) x 1 x 0 (x 2 x 1 ) = 0, with respect to x 2 (see Figure 5) This gives x 2 = x 1 f(x 1)(x 1 x 0 ) f(x 1 ) f(x 0 )

36 x f x x x f x Figure 5: The figure shows a function f = f(x) and its linear interpolant F between x 0 and x 1. Lectures INF2320 p. 36/88

37 Lectures INF2320 p. 37/88 The Secant method Following the same procedure as above we get the iteration x k+1 = x k f(x k)(x k x k 1 ) f(x k ) f(x k 1 ), and the associated algorithm reads Algorithm 3. Given two initial approximations x 0 and x 1 and a tolerance ε. k = 1 while f(x k ) > ε do x k+1 = x k f(x k ) k = k+1 end (x k x k 1 ) f(x k ) f(x k 1 )

38 Lectures INF2320 p. 38/88 Example 13 Let us apply the Secant method to the equation f(x) = 2+x e x = 0, studied above. The two initial values are x 0 = 0, x 1 = 3, and the stopping criteria is specified by ε = Table 3 show the number of iterations k, x k and f(x k ) as computed by Algorithm 3 Note that the convergence for the Secant method is slower than for Newton s method, but faster than for the Bisection method

39 k x k f(x k ) Lectures INF2320 p. 39/88 Table 3: The Secant method applied with f(x) = 2+x e x =0.

40 Example 14 Find a zero of which has a solution x = 2. f(x) = x 2 2, The general step of the secant method is in this case x k+1 =x k f(x k ) x k x k 1 f(x k ) f(x k 1 ) =x k (x 2 k 2) x k x k 1 x 2 k x2 k 1 =x k x2 k 2 x k + x k 1 = x kx k x k + x k 1 Lectures INF2320 p. 40/88

41 Lectures INF2320 p. 41/88 Example 14 By choosing x 0 = 1 and x 1 = 2 we get x 2 = x 3 = x 4 = This is quite good compared to the exact value x = Recall that Newton s method produced the approximation in three iterations, which is slightly more accurate

42 Fixed-Point iterations Above we studied implicit schemes for the differential equation u = g(u), which lead to the nonlinear equation u n+1 t g(u n+1 ) = u n, where u n is known, u n+1 is unknown and t > 0 is small. We defined v = u n+1 and c = u n, and wrote the equation v t g(v) = c. We can rewrite this equation on the form where v = h(v), (29) h(v) = c+ t g(v). Lectures INF2320 p. 42/88

43 Lectures INF2320 p. 43/88 Fixed-Point iterations The exact solution, v, must fulfill v = h(v ). This fact motivates the Fixed Point Iteration: with an initial guess v 0. v k+1 = h(v k ), Since h leaves v unchanged; h(v ) = v, the value v is referred to as a fixed-point of h

44 Lectures INF2320 p. 44/88 Fixed-Point iterations We try this method to solve x = sin(x/10), which has only one solution x = 0 (see Figure 6) The iteration is x k+1 = sin(x k /10). (30) Choosing x 0 = 1.0, we get the following results x 1 = , x 2 = , x 3 = , which seems to converge fast towards x = 0.

45 Lectures INF2320 p. 45/88 y y = x 10π y = sin(x/10) 10π x Figure 6: The graph of y = x and y = sin(x/10).

46 Lectures INF2320 p. 46/88 Fixed-Point iterations We now try to understand the behavior of the iteration. From calculus we recall for small x we have Using this fact in (30), we get and therefore sin(x/10) x/10. x k+1 x k /10, x k (1/10) k. We see that this iteration converges towards zero.

47 Lectures INF2320 p. 47/88 Convergence of Fixed-Point iterations We have seen that h(v) = v can be solved with the Fixed-Point iteration v k+1 = h(v k ) We now analyze under what conditions the values {v k } generated by the Fixed-Point iterations converge towards a solution v of the equation. Definition: h = h(v) is called a contractive mapping on a closed interval I if (i) h(v) h(w) δ v w for any v,w I, where 0 < δ < 1, and (ii) v I h(v) I.

48 Lectures INF2320 p. 48/88 Convergence of Fixed-Point iterations The Mean Value Theorem of Calculus states that if f is a differentiable function defined on an interval [a, b], then there is a c [a,b] such that f(b) f(a) = f (c)(b a). It follows from this theorem that h in is a contractive mapping defined on an interval I if and h(v) I for all v I h (ξ) < δ < 1 for all ξ I, (31)

49 Lectures INF2320 p. 49/88 Convergence of Fixed-Point iterations Let us check the above example x = sin(x/10) We see that h(x) = sin(x/10) is contractive on I = [ 1,1] since h (x) = 1 10 cos(x/10) 1 10 and x [ 1,1] sin(x/10) [ 1,1].

50 Lectures INF2320 p. 50/88 Convergence of Fixed-Point iterations For a contractive mapping h, we assume that for any v,w in a closed interval I we have h(v) h(w) δ v w, where 0 < δ < 1, The error, e k = v k v, fulfills v I h(v) I e k+1 = v k+1 v = h(v k ) h(v ) δ v k v =δe k.

51 Lectures INF2320 p. 51/88 Convergence of Fixed-Point iterations It now follows by induction on k, that e k δ k e 0. Since 0 < δ < 1, we know that e k 0 as k. This means that we have convergence lim v k = v. k We can now conclude that the Fixed-Point iteration will converge when h is a contractive mapping.

52 Speed of convergence We have seen that the Fixed-Point iterations fulfill e k e 0 δ k. Assume we want to solve this equation to the accuracy e k e 0 ε. We need to have δ k ε, which gives k ln(δ) ln(ε) Therefore the number of iterations needs to satisfy k ln(ε) ln(δ) Lectures INF2320 p. 52/88

53 Lectures INF2320 p. 53/88 Existence and Uniqueness of a Solution For the equations on the form v = h(v), we want to answer the following questions a) Does there exist a value v such that b) If so, is v unique? c) How can we compute v? v = h(v )? We assume that h is a contractive mapping on a closed interval I such that for all v, w. h(v) h(w) δ v w, where 0 < δ < 1, (32) v I h(v) I (33)

54 Lectures INF2320 p. 54/88 Uniqueness Assume that we have two solutions v and w of the problem, i.e. v = h(v ) and w = h(w ) (34) From the assumption (32) we have where δ < 1. But (34) gives h(v ) h(w ) δ v w, v w δ v w which can only hold when v = w, and consequently the solution is unique.

55 Lectures INF2320 p. 55/88 Existence We have seen that if h is a contractive mapping, the equation can only have one solution. h(v) = v (35) If we now can show that there exists a solution of (35) we have answered (a), (b) and (c) above Below we show that assumptions (32) and (33) imply existence

56 Lectures INF2320 p. 56/88 Cauchy sequences First we recall the definition of Cauchy sequences. A sequence of real numbers, {v k }, is called a Cauchy sequence if, for any ε > 0, there is an integer M such that for any m,n M we have v m v n < ε (36) Theorem: A sequence {v k } converges if and only if it is a Cauchy sequence Under we shall show that the sequence, {v k }, produced by the Fixed-Point iteration, is a Cauchy series when assumptions (32) and (33) hold

57 Lectures INF2320 p. 57/88 Existence Since v n+1 = h(v n ), we have v n+1 v n = h(v n ) h(v n 1 ) δ v n v n 1 By induction, we have v n+1 v n δ n v 1 v 0 In order to show that {v n } is a Cauchy sequence, we need to bind v m v n We may assume that m > n, and we see that v m v n = (v m v m 1 )+(v m 1 v m 2 )+...+(v n+1 v n )

58 Existence By the triangle-inequality, we have v m v n v m v m 1 + v m 1 v m v n+1 v n (37) gives consequently v m v m 1 δ m 1 v 1 v 0 v m 1 v m 2 δ m 2 v 1 v 0. v n+1 v n δ n v 1 v 0 v m v n v m v m 1 + v m 1 v m v n+1 v n ( δ m 1 + δ m δ n) v 1 v 0 Lectures INF2320 p. 58/88

59 Lectures INF2320 p. 59/88 Existence We can now estimate the power series δ m 1 + δ m δ n = δ n 1( δ+δ δ m n) δ n 1 δ k k=1 = δ n δ So v m v n δn 1 1 δ v 1 v 0 δ n 1 can be as small as you like, if you choose n big enough

60 Lectures INF2320 p. 60/88 Existence This means that for any ε > 0, we can find an integer M such that v m v n < ε provided that m,n M, and consequently {v k } is a Cauchy sequence. The sequence is therefore convergent, and we call the limit v Since v = lim k v k = lim k h(v k ) = h(v ) by continuity of h, we have that the limit satisfies the equation

61 Lectures INF2320 p. 61/88 Systems of nonlinear equations We start our study of nonlinear equations, by considering a linear system that arises from the discretization of a linear 2 2 system of ordinary differential equations, u (t) = v(t), u(0) = u 0, v (t) = u(t), v(0) = v 0. (37) An implicit Euler scheme for this system reads u n+1 u n t = v n+1, v n+1 v n t = u n+1, (38) and can be rewritten on the form u n+1 + t v n+1 = u n, t u n+1 + v n+1 = v n. (39)

62 Lectures INF2320 p. 62/88 Systems of linear equations We can write this system on the form Aw n+1 = w n, (40) where A = ( 1 t t 1 ) and w n = ( u n v n ). (41) In order to compute w n+1 = (u n+1,v n+1 ) T from w n = (u n,v n ), we have to solve the linear system (40). The system has a unique solution since det(a) = 1+ t 2 > 0. (42)

63 Lectures INF2320 p. 63/88 Systems of linear equations And the solution is given by w n+1 = A 1 w n, where ( ) A t = 1+ t 2. (43) t 1 Therefore we get ( ) u n+1 v n+1 = = 1 1+ t t 2 ( ( 1 t t 1 u n t v n t u n + v n )( ) u n v n ) (44). (45)

64 Lectures INF2320 p. 64/88 Systems of linear equations We write this as u n+1 = 1 1+ t (u 2 n t v n ), v n+1 = 1 1+ t (v 2 n + t u n ). (46) By choosing u 0 = 1 and v 0 = 0, we have the analytical solutions u(t) = cos(t), v(t) = sin(t). (47) In Figure 7 we have plotted (u,v) and (u n,v n ) for 0 t 2π, t = π/500. We see that the scheme provides good approximations.

65 t Figure 7: The analytical solution (u = cos(t),v = sin(t)) and the numerical solution (u n,v n ), in dashed lines, produced by the implicit Euler scheme. Lectures INF2320 p. 65/88

66 Lectures INF2320 p. 66/88 A nonlinear system Now we study a nonlinear system of ordinary differential equations u = v 3, u(0) = u 0, v = u 3, v(0) = v 0. (48) An implicit Euler scheme for this system reads u n+1 u n t = v 3 n+1, v n+1 v n t = u 3 n+1, (49) which can be rewritten on the form u n+1 + t v 3 n+1 u n = 0, v n+1 t u 3 n+1 v n = 0. (50)

67 Lectures INF2320 p. 67/88 A nonlinear system Observe that in order to compute (u n+1,v n+1 ) based on (u n,v n ), we need to solve a nonlinear system of equations We would like to write the system on the generic form f(x,y) = 0, g(x, y) = 0. (51) This is done by setting f(x,y) = x+ t y 3 α, g(x,y) = y t x 3 β, (52) α = u n and β = v n.

68 Lectures INF2320 p. 68/88 Newton s method When deriving Newton s method for solving a scalar equation we exploited Taylor series expansion p(x) = 0 (53) p(x 0 + h) = p(x 0 )+hp (x 0 )+O(h 2 ), (54) to make a linear approximation of the function p, and solve the linear approximation of (53). This lead to the iteration x k+1 = x k p(x k) p (x k ). (55)

69 Lectures INF2320 p. 69/88 Newton s method We shall try to extend Newton s method to systems of equations on the form f(x,y) = 0, g(x, y) = 0. (56) The Taylor-series expansion of a smooth function of two variables F(x, y), reads F(x+ x,y+ y) = F(x,y)+ x F x (x,y)+ y F y (x,y) +O( x 2, x y, y 2 ). (57)

70 Lectures INF2320 p. 70/88 Newton s method Using Taylor expansion on (56) we get f(x 0 + x,y 0 + y) = f(x 0,y 0 )+ x f x (x 0,y 0 )+ y f y (x 0,y 0 ) and +O( x 2, x y, y 2 ), (58) g(x 0 + x,y 0 + y) = g(x 0,y 0 )+ x g x (x 0,y 0 )+ y g y (x 0,y 0 ) +O( x 2, x y, y 2 ). (59)

71 Lectures INF2320 p. 71/88 Newton s method Since we want x and y to be such that f(x 0 + x,y 0 + y) 0, g(x 0 + x,y 0 + y) 0, (60) we define x and y to be the solution of the linear system f(x 0,y 0 )+ x f x (x 0,y 0 )+ y f y (x 0,y 0 ) = 0, g(x 0,y 0 )+ x g x (x 0,y 0 )+ y g y (x 0,y 0 ) = 0. (61) Remember here that x 0 and y 0 are known numbers, and therefore f(x 0,y 0 ), f x (x 0,y 0 ) and f y (x 0,y 0 ) are known numbers as well. x and y are the unknowns.

72 Newton s method (61) can be written on the form )( ) x y ( f0 x g 0 x f 0 y g 0 y = ( f 0 g 0 ). (62) where f 0 = f(x 0,y 0 ), g 0 = g(x 0,y 0 ), f 0 x = f x (x 0,y 0 ), etc. If the matrix ) A = ( f0 x g 0 x f 0 y g 0 y (63) is nonsingular. Then ( ) x = y ( f0 x g 0 x f 0 y g 0 y ) 1 ( f 0 g 0 ). (64) Lectures INF2320 p. 72/88

73 Lectures INF2320 p. 73/88 Newton s method We can now define ( ) ( ) ( x 1 x 0 = + y 1 y 0 x y ) = ( x 0 y 0 ) ( f0 x g 0 x f 0 y g 0 y ) 1 ( f 0 g 0 ). And by repeating this argument we get ( x k+1 y k+1 ) = ( x k y k ) ( fk x g k x f k y g k y ) 1 ( f k g k ), (65) where f k = f(x k,y k ), g k = g(x k,y k ) and f k x = f x (x k,y k ) etc. The scheme (65) is Newton s method for the system (56).

74 Lectures INF2320 p. 74/88 A Nonlinear example We test Newton s method on the system e x e y = 0, ln(1+x+y) = 0. (66) The system have analytical solution x = y = 0. Define f(x,y) = e x e y, g(x,y) = ln(1+x+y). x k+1 y k+1 The iteration in Newton s method (65) reads ) = ( x k y k ) ( e x k e y k x k +y k 1+x k +y k ) 1 ( e x k e y k ln(1+x k + y k ) ).(67)

75 Lectures INF2320 p. 75/88 A Nonlinear example The table below shows the computed results when x 0 = y 0 = 1 2. k x k y k We observe that, as in the scalar case, Newton s method gives very rapid convergence towards the analytical solution x = y = 0.

76 Lectures INF2320 p. 76/88 The Nonlinear System Revisited We now go back to nonlinear system of ordinary differential equations (48), presented above. For each time step we had to solve f(x,y) = 0, g(x, y) = 0, (68) where f(x,y) = x+ t y 3 α, g(x,y) = y t x 3 β. (69) We shall now solve this system using Newton s method.

77 Lectures INF2320 p. 77/88 The Nonlinear System Revisited We put x 0 = α, y 0 = β and iterate as follows where ( x k+1 y k+1 ) = ( x k y k ) ( fk x g k x f k y g k y ) 1 ( f k g k ), (70) f k = f(x k,y k ), g k = g(x k,y k ), f k x = f x (x k,y k ) = 1, g k x = g x (x k,y k ) = 3 t xk, 2 f k y = f y (x k,y k ) = 3 t y 2 k, g k y = g y (x k,y k ) = 1.

78 Lectures INF2320 p. 78/88 The Nonlinear System Revisited The matrix A = ( fk x g k x f k y g k y ) = ( 1 3 t y 2 k 3 t xk 2 1 ) (71) has its determinant given by: det(a) = 1+9 t 2 xk 2 y2 k > 0. So A 1 is well defined and is given by ( ) A t y 2 k = 1+9 t 2 xk 2 y2 k 3 t xk 2. (72) 1 For each time-level we can e.g. iterate until f(x k,y k ) + g(x k,y k ) < ε = (73)

79 Lectures INF2320 p. 79/88 The Nonlinear System Revisited We have tested this method with t = 1/100 and t [0,1] In Figure 8 the numerical solutions of u and v are plotted as functions of time, and in Figure 9 the numerical solution is plotted in the (u, v) coordinate system In Figure 10 we have plotted the number of Newton s iterations needed to reach the stopping criterion (73) at each time-level Observe that we need no more than two iterations at all time-levels

80 Lectures INF2320 p. 80/ t Figure 8: The numerical solutions u(t) and v(t) (in dashed line) of (48) produced by the implicit Euler scheme (49) using u 0 = 1, v 0 = 0 and t = 1/100.

81 Lectures INF2320 p. 81/ v u Figure 9: The numerical solutions of (48) in the (u, v)-coordinate system, arising from the implicit Euler scheme (49) using u 0 = 1, v 0 = 0 and t = 1/100.

82 Lectures INF2320 p. 82/ t Figure 10: The graph shows the number of iterations used by Newton s method to solve the system (50) at each time-level.

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