Lecture 10: Finite Differences for ODEs & Nonlinear Equations

Transcription

1 Lecture 10: Finite Differences for ODEs & Nonlinear Equations J.K. WI3097TU Delft Institute of Applied Mathematics Delft University of Technology 21 November 2012 () Finite Differences & Nonlinear Equations 21 November / 44

2 Outline 1 Review Neumann BC Convection-diffusion equation Stability of Finite Differences 2 Consistency, Stability, Convergence 3 Nonlinear equations Bisection method Fixed Point Iteration () Finite Differences & Nonlinear Equations 21 November / 44

3 Neumann BC y + q(x)y = f (x), x L = 0 < x < x R = 1, y(0) = 0, dy dx (x=1) = 0, (Dirichlet) (Neumann) Similar to the previous case, we have Step 1: Subdivide the domain, h = x R x L N+1 = 1 N+1 Step 2: Define the nodal points, x j = x L + jh = jh, j = 0,..., N + 1. Step 3: Discretize the equation y j w j+1 2w j + w j 1 h 2. () Finite Differences & Nonlinear Equations 21 November / 44

4 Neumann BC Changing the left boundary condition from Dirichlet (y(x R ) = 0) to Neumann ( dy dx (x R ) = 0), changes our approximation through Adding a "ghost" point Approximating the right boundary condition using central differences Vectors increase in length to (N + 1) and matrices increase in size to (N + 1) (N + 1). () Finite Differences & Nonlinear Equations 21 November / 44

5 Convection-diffusion equation y + νy = 0, y(x L ) = 0, y(x R ) = α x L x x R Need to approximate both first and second derivatives. () Finite Differences & Nonlinear Equations 21 November / 44

6 Convection-diffusion equation Define a 1 = ( ) ( ) 1 ν h 2 and b 1 = 1 + ν h 2. Then A = 1 h 2 2 b a 1 2 b a 1 2 b a 1 2 A is no longer symmetric. () Finite Differences & Nonlinear Equations 21 November / 44

7 Condition number Definition (Condition number) The condition number of a matrix A is given by κ(a) = A A 1. This is useful in determining the stability of our system of equations. () Finite Differences & Nonlinear Equations 21 November / 44

8 Condition number For symmetric matrices, the condition number is given by κ(a) = A A 1 = λ max λ min For stability, w w κ(a) f f Small κ(a) affect of perturbations is small. () Finite Differences & Nonlinear Equations 21 November / 44

9 Gerschgorin s Theorem Theorem (Gerschgorin) The eigenvalues of the matrix A lie in the union of circles z a ii where z is a complex number. n j =1 j i a ij = R i, a ii are the diagonal entries a ij are the non-diagonal entries z a ii is a closed disc centered at a ii with radius R. () Finite Differences & Nonlinear Equations 21 November / 44

10 Consistency Recall, for consistency, the local truncation error should go to zero as the step size gets smaller: lim τ = 0 h 0 But, now τ is a vector so we need to look at the individual entries: τ j+1 = Exact ODE Approximating ODE () Finite Differences & Nonlinear Equations 21 November / 44

11 Consistency Consider the equation y + qy = f. At the grid points, this is d 2 y dx 2 x=x j + q j y j = f j. The discretization of this equation is 1 h 2 (y j+1 2y j + y j 1 ) + q j y j = f j. () Finite Differences & Nonlinear Equations 21 November / 44

12 Consistency The componentwise truncation error is then τ j =( d 2 y dx 2 x=x j + q j y j ) ( 1 h 2 (y j+1 2y j + y j 1 ) + q j y j ) =( y j ) ( 1 h 2 (y j+1 2y j + y j 1 )) }{{} Approximation to 2nd derivative Recall that So that d 2 y dx 2 1 h 2 (y j+1 2y j + y j 1 ) = O(h 2 ). τ O(h 2 ) () Finite Differences & Nonlinear Equations 21 November / 44

13 Consistency The local truncation error is second order. This is because of how we chose to approximation y. So, we have consistency lim τ = h 0 O(h2 ) 0 as h 0. () Finite Differences & Nonlinear Equations 21 November / 44

14 Stability Definition (Stability for finite differences) A finite difference scheme is stable if there exists a constant M independent of h such that A 1 M, as h 0. This is equivalent to the system having a unique solution. () Finite Differences & Nonlinear Equations 21 November / 44

15 Stability For the equation y j + q j y j = f j, The approximating system is given by Aw = f, A = 1 h 2 K + M. where w = w 1 w 2 w N 1 w N, f = f 1 f 2 f N 1 f N () Finite Differences & Nonlinear Equations 21 November / 44

16 Stability where A = 1 h 2 K + M M = diag(q 1, q 2,..., q N ) and K is a tri-banded symmetric diagonal matrix K = which means that A is symmetric. () Finite Differences & Nonlinear Equations 21 November / 44

17 Stability Since A is symmetric A has real eigenvalues It also means that we have an exact definition for the matrix norms: A = λ max, A 1 = 1 λ min But, what are the eigenvalues? () Finite Differences & Nonlinear Equations 21 November / 44

18 Stability Case 1: q = 0 for all x. Then A = 1 h 2 K. K = () Finite Differences & Nonlinear Equations 21 November / 44

19 Stability A is a symmetric matrix with eigenvalues λ j = 1 (2 2 cos((n j)hπ)) h2 This means that ) λ min = 1 h 2 (2 2 cos(hπ)) = 4 h 2 sin2 ( hπ 2 = π 2 sin2 (hπ/2) (hπ/2) 2 }{{} 1 λ min π 2 () Finite Differences & Nonlinear Equations 21 November / 44

20 Stability For the maximum eigenvalue of A, we need to look at this in another way: λ max = 1 h 2 (2 2 cos(hπ)) = 4 ( hπ h 2 sin2 2 }{{} 1 ) λ max 4 h 2 () Finite Differences & Nonlinear Equations 21 November / 44

21 Stability Thus, the scheme is stable for q = 0 and the condition number is ( ) ( ) 4 1 κ(a) = A A 1 = h 2 π 2 = Notice, the scheme is stable because A 1 1 π 2, even though the condition number depends on h. ( ) 2 2 hπ () Finite Differences & Nonlinear Equations 21 November / 44

22 aij 4 h 2 Stability Case 2: q(x) 0. We can estimate the eigenvalues using Gerschgorin s theorem. Assume 0 < q min q(x) q max. q min < q min + 4 h 2 λ j q max < q max + 4 h 2, j = 1,..., n because A is symmetric and A 1 1 q min stable scheme () Finite Differences & Nonlinear Equations 21 November / 44

23 Stability What happens with small perturbations? which gives A(w + w) = f + f w = A 1 f 1 λ min f. () Finite Differences & Nonlinear Equations 21 November / 44

24 Stability The relative error is w w 1 f λ min f Using w w 1 f f λ min w f }{{} Effective condition number where λ min π 2 and f w is usually bounded. () Finite Differences & Nonlinear Equations 21 November / 44

25 Convergence For convergence lim y w = 0. h 0 Is this the case? Ay Aw = f + τ f y w A 1 τ 1 π 2 τ But τ O(h 2 ). Therefore y w O(h 2 ) () Finite Differences & Nonlinear Equations 21 November / 44

26 Consistency + Stability = Convergence Consistency τ j+1 = Exact ODE Approximating ODE Stability We also need A 1 M, as h 0. κ(a) = A A 1 to not be too large. Convergence y w τ () Finite Differences & Nonlinear Equations 21 November / 44

27 Nonlinear Equations Given a nonlinear equation, f (x) = 0, we want to determine for what x this equation is satisfied. Methods Bisection method Fixed point iteration Newton-Raphson method () Finite Differences & Nonlinear Equations 21 November / 44

28 Bisection method This idea is based upon the intermediate value theorem. Theorem (Intermediate value theorem) Assume f C[a, b]. Let f (a) f (b) and let F be a number between f (a) and f (b). Then there exists a number c (a, b) such that f (c) = F. () Finite Differences & Nonlinear Equations 21 November / 44

29 Bisection method Given f (x) C[a, b] such that f (a)f (b) < 0, then we know that f (x) changes sign on [a, b]. f (x) = 0 in the inteval [a, b]. () Finite Differences & Nonlinear Equations 21 November / 44

30 Bisection method Idea: Repeatedly half the interval, keeping the half where the change of sign occurs. () Finite Differences & Nonlinear Equations 21 November / 44

31 Bisection method Algorithm Let x = α be the root of f (x) = 0 and ɛ > 0 is the error tolerance. Algorithm 1 Let a 0 = a and b 0 = b 2 Define c j = 1 2 (a j + b j ) 3 If b j c j < ɛ STOP! α = c j. 4 Else if f (b j )f (c j ) 0 then a j+1 = c j, b j+1 = b j. Otherwise a j+1 = a j, b j+1 = c j. 5 j = j Return to step 2. Notice: for each loop, the interval is halved. () Finite Differences & Nonlinear Equations 21 November / 44

32 Bisection method Properties Easy implementation Always converges to a solution. Convergence is slow. Often used to generate a good starting point for other methods. () Finite Differences & Nonlinear Equations 21 November / 44

33 Bisection method Error bounds Let a n, b n, c n be the n th computed values. Then b n+1 a n+1 = 1 2 (b n a n ) = 1 4 (b n 1 a n 1 ) =... = 1 (b a), n 1. 2n 1 () Finite Differences & Nonlinear Equations 21 November / 44

34 Bisection method Error bounds What is the error? The root α is either in [a n, c n ] or [c n, b n ]. α c n c n a n = b n c n = 1 2 (b n a n ) 1 (b a) 2n α c n 1 (b a) 2n () Finite Differences & Nonlinear Equations 21 November / 44

35 Bisection method Error bounds Suppose we want α c n < ɛ. Then what are the number of iterations that we need? α c n 1 (b a) < ɛ 2n 1 2 n < ɛ b a 2 n (b a) > ɛ () Finite Differences & Nonlinear Equations 21 November / 44

36 Bisection method Error bounds ( ) (b a) ln(2 n ) = n ln(2) > ln = ln(b a) ln(ɛ) ɛ n > ln(b a) ln(ɛ) ln(2) # iterations needed to converge to error ɛ. Note: we need to make sure ɛ is larger than the rounding error. () Finite Differences & Nonlinear Equations 21 November / 44

37 Fixed Point Iteration Definition A fixed point of a given function g(x) is a number α such that g(α) = α. Goal: Find the solution, α. Lemma Let g(x) C[a, b] such that a x b. Then a g(x) b and x = g(x) has at least one solution, α [a, b]. () Finite Differences & Nonlinear Equations 21 November / 44

38 Fixed Point Iteration Theorem (Contraction mapping theorem) Assume g(x) C 1 [a, b] such that a x b (which means a g(x) b). Further assume that Then λ max g (x) 1. x [a,b] 1 There exists a unique solution α of x = g(x) in [a, b]. 2 For any initial estimate α 0 [a, b], x n α. 3 α x n λn 1 λ x 0 x 1, n 0. 4 lim n α x n+1 α x n = g (α). () Finite Differences & Nonlinear Equations 21 November / 44

39 Fixed Point Iteration g (α) determines the convergence rate. If g (α) << 1, then we have fast convergence. If g (α) 1 ɛ then we have slow convergence. If g (α) > 1, then we do not have convergence. () Finite Differences & Nonlinear Equations 21 November / 44

40 Fixed Point Iteration Algorithm 1 α 0 = starting value. 2 α n = g(α n 1 ) 3 If lim n α n = α and g(α) C then STOP! α = g(α n ). 4 Else go to Step 2. () Finite Differences & Nonlinear Equations 21 November / 44

41 Fixed Point Iteration Example Let f (x) = x 3 + 3x 4. Find g(x). Root finding says that f (x) = 0. To use fixed point iteration, we must find g(x) = x. () Finite Differences & Nonlinear Equations 21 November / 44

42 Fixed Point Iteration Therefore f (x) = 0 x 3 + 3x 4= 0 x(x 2 + 3) = 4 g(x) = 4 x x = 4 x () Finite Differences & Nonlinear Equations 21 November / 44

43 Fixed Point Iteration Theorem The fixed point iteration always converges. Proof. Recall g(x) C 1 [a, b], a g(x) b, and λ max x [a,b] g (x) 1. Then we have α n α = g(α n 1 ) g(α) = g (ξ) α n 1 α λ α n 1 α lim α n α lim n n λn α 0 α () Finite Differences & Nonlinear Equations 21 November / 44

44 Material addressed 1 Review Neumann BC Convection-diffusion equation Stability of Finite Differences 2 Consistency, Stability, Convergence 3 Nonlinear equations Bisection method Fixed Point Iteration Material in book: Chapter 7, 1-6,8; Chapter 4, 1-3 Useful exercises: Ch7: 1-4; Ch4: 1-3 () Finite Differences & Nonlinear Equations 21 November / 44