Numerical Methods for Differential Equations
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1 Numerical Methods for Differential Equations Chapter 2: Runge Kutta and Multistep Methods Gustaf Söderlind Numerical Analysis, Lund University
2 Contents V Runge Kutta methods 2. Embedded RK methods and adaptivity 3. Implicit Runge Kutta methods 4. Stability and the stability function 5. Linear multistep methods 6. Difference operators 2 / 61
3 1. Runge Kutta methods Given an IVP y = f (t, y), y(0) = y 0 use numerical integration to approximate integrals f (τ, y(τ)) dτ tn+1 y(t n+1 ) = y(t n ) + t n s y(t n+1 ) y(t n ) + h b j f (t n + c j h, y(t n + c j h)) j=1 Let {Y j } s j=1 denote numerical approximations to {y(t n + c j h)} s j=1 A Runge Kutta method then has the form y n+1 = y n + s b j hf (t n + c j h, Y j ) j=1 3 / 61
4 The explicit Runge Kutta computational process Sample vector field to obtain stage derivatives hy j = hf (t n + c j h, Y j ) at stage values i 1 Y i = y n + a i,j hy j j=1 and advance solution one step by a linear combination y n+1 = y n + s j=1 b j hy j 4 / 61
5 The Butcher tableau Weights, nodes and stages An s-stage RK method has nodes {c i } s i=1 and weights {b j} s i=1 The Butcher tableau of an explicit RK method is c 2 a 2, c s a s,1 a s,2 0 b 1 b 2 b s or c A b T Simplifying assumption c i = s j=1 a i,j (row sums of RK matrix) 5 / 61
6 Simplest case A two-stage ERK A two-stage explicit RK method has three free coefficients The simplifying assumption determines the nodes hy 1 = hf (t n, y n ) hy 2 = hf (t n + c 2 h, y n + h a 21 Y 1) y n+1 = y n + [ b 1 hy 1 + b 2 hy 2 ] Butcher tableau c 1 = 0, c 2 = a c 2 a 21 0 b 1 b 2 6 / 61
7 Derivation of two-stage ERK s Using hy 1 = hf (t n, y n ), expand hy 2 in Taylor series around t n, y n hy 2 = hf (t n + c 2 h, y n + h a 21 f (t n, y n )) = hf + h 2 [c 2 f t + a 21 f y f ] + O(h 3 ) Insert into y n+1 = y n + b 1 hy 1 + b 2 hy 2 and use c 2 = a 21 to obtain y n+1 = y n + (b 1 + b 2 )hf + h 2 b 2 c 2 (f t + f y f ) + O(h 3 ) Expand exact solution in Taylor series and match terms y = f y = f t + f y y = f t + f y f y(t + h) = y + hf + h2 2 (f t + f y f ) + O(h 3 ) 7 / 61
8 One-parameter family of 2nd order two-stage ERK s Match terms to get conditions for order 2 b 1 + b 2 = 1 b 2 c 2 = 1/2 (consistency) Note Consistent RK methods are always convergent Two equations, three unknowns there is a one-parameter family of 2nd order two-stage ERK methods with Butcher tableau b 0 1 b b 1 2b 8 / 61
9 Example 1 The modified Euler method Put b = 1 to get the Butcher tableau /2 1/ hy 1 = hf (t n, y n ) hy 2 = hf (t n + h/2, y n + hy 1/2) y n+1 = y n + hy 2 Second order two-stage explicit Runge Kutta (ERK) method 9 / 61
10 Example 2 Heun s method Put b = 1/2 to get /2 1/2 hy 1 = hf (t n, y n ) hy 2 = hf (t n + h, y n + hy 1) y n+1 = y n + (hy 1 + hy 2)/2 Second order two-stage ERK, compare to the trapezoidal rule 10 / 61
11 Third order three-stage ERK Conditions for 3rd order ( c 2 = a 21 ; c 3 = a 31 + a 32 ) b 1 + b 2 + b 3 = 1 b 2 c 2 + b 3 c 3 = 1/2 b 2 c2 2 + b 3 c3 2 = 1/3 b 3 a 32 c 2 = 1/6 Classical RK /2 1/ /6 2/3 1/6 Nyström scheme /3 2/ /3 0 2/3 0 1/4 3/8 3/8 11 / 61
12 Exercise Construct the Butcher tableau for the 3-stage Heun method. hy 1 = hf (t n, y n ) hy 2 = hf (t n + h/3, y n + hy 1/3) hy 3 = hf (t n + 2h/3, y n + 2hY 2/3) y n+1 = y n + (hy 1 + 3hY 3)/4 Is the method of order 3? 12 / 61
13 Classical RK4 4th order, 4-stage ERK The original RK method (1895) hy 1 = hf (t n, y n ) hy 2 = hf (t n + h/2, y n + hy 1/2) hy 3 = hf (t n + h/2, y n + hy 2/2) hy 4 = hf (t n + h, y n + hy 3) y n+1 = y n (hy 1 + 2hY 2 + 2hY 3 + hy 4) 13 / 61
14 Classical RK4... Butcher tableau /2 1/ /2 0 1/ /6 1/3 1/3 1/6 Note s-stage ERK methods of order p = s exist only for s 4 There is no 5-stage ERK of order 5 14 / 61
15 Order conditions... leave it to the professionals An s-stage ERK method has s + s(s 1)/2 coefficients to choose, but there are overwhelmingly many order conditions # of available coefficients stages s coefficients # of order conditions and min # of stages to achieve order p order p conditions min stages ?? 15 / 61
16 2. Embedded RK methods Two methods in a single Butcher tableau (RK34) hy 1 = hf (t n, y n ) hy 2 = hf (t n + h/2, y n + hy 1/2) hy 3 = hf (t n + h/2, y n + hy 2/2) hz 3 = hf (t n + h, y n hy 1 + 2hY 2) hy 4 = hf (t n + h, y n + hy 3) y n+1 = y n (hy 1 + 2hY 2 + 2hY 3 + hy 4) order 4 z n+1 = y n (hy 1 + 4hY 2 + hz 3) order 3 The difference y n+1 z n+1 can be used as an error estimate 16 / 61
17 Adaptive RK methods Example Use an embedded pair, e.g. RK34 Local error estimate r n+1 := y n+1 z n+1 = O(h 4 ) Adjust the step size h to make local error estimate equal to a prescribed error tolerance tol Simplest step size updating scheme makes r n tol ( ) tol 1/p h n+1 = h n r n+1 Time step adaptivity using local error control 17 / 61
18 Advanced adaptive RK methods Very high precision There are many state-of-the-art embedded ERK methods, e.g. Dormand Prince DOPRI45 Dormand Prince DOPRI78 Cash Karp CK5 Advanced adaptivity uses discrete control theory and digital filters h n+1 = ρ n h n ( ) tol β1 /p ( tol ρ n = r n+1 r n ) β2 /p ρ α n 1 PI control, ARMA filters &c., via control parameters (β 1, β 2, α) 18 / 61
19 3. Implicit Runge Kutta methods (IRK) In ERK, the matrix A in the tableau is strictly lower triangular In IRK, A may have nonzero diagonal elements or even be full hy i = hf (t n + c i h, y n + y n+1 = y n + s i=1 b i hy i s j=1 a i,j hy j ) The method is implicit and requires equation solving to compute the stage derivatives {Y i }s i=1 19 / 61
20 Implicit Runge Kutta methods... In stage value stage derivative form Y i = y n + s j=1 a i,j hy j hy i = hf (t n + c i h, Y i ) y n+1 = y n + s i=1 b i hy i Method coefficients (A, b, c) are represented in Butcher tableau 20 / 61
21 One-stage IRK methods Implicit Euler (order 1) Implicit midpoint method (order 2) /2 1/2 1 hy 1 = hf (t n + c 1 h, y n + a 11 hy 1) y n+1 = y n + b 1 hy 1 Taylor expansion of y n+1 = y n + b 1 hf (t n + c 1 h, y n + a 11 hy 1 ) y n+1 = y n + b 1 hf (t n + c 1 h, y n + a 11 hy 1) 21 / 61
22 Taylor expansions for one-stage IRK Match terms in y n+1 = y + h b 1 f + h 2 (b 1 c 1 f t + a 11 f y f ) + O(h 3 ) y(t n+1 ) = y + h f + h2 2 (f t + f y f ) + O(h 3 ) Condition for order 1 (consistency) b 1 = 1 Condition for order 2 c 1 = a 11 = 1/2 Conclusion Implicit Euler is of order 1 and the implicit midpoint method is the only one-stage 2nd order IRK 22 / 61
23 4. Stability Applying an IRK to the linear test equation y = λy, we get hy i = hλ (y n + s j=1 a i,j hy j ) Introduce hy = [hy 1 hy s ] T and 1 = [1 1 1] T R s Then (I hλa)hy = hλ1y n so hy = hλ(i hλa) 1 1y n and y n+1 = y n + s j=1 b j hy j = [1 + hλb T (I hλa) 1 1]y n 23 / 61
24 The stability function Theorem For every Runge-Kutta method applied to the linear test equation y = λy we have where the rational function y n+1 = R(hλ)y n R(z) = 1 + zb T (I za) 1 1 is called the method s stability function. If the method is explicit, then R(z) is a polynomial of degree s 24 / 61
25 RK4 stability region Contours of R(z) 25 / 61
26 A-stability of RK methods Definition The method s stability region is the set D = {z C : R(z) 1} Theorem If R(z) maps all of C into the unit circle, then the method is A-stable Corollary No explicit RK method is A-stable (For ERK R(z) is a polynomial, and P(z) as z ) 26 / 61
27 A-stability and the Maximum Principle Theorem A Runge Kutta method with stability function R(z) is A-stable if and only if all poles of R have positive real parts, and R(iω) 1 for all ω R This is the Maximum Principle in complex analysis Example 0 1/4 1/4 Y 1 = f (y n + hy 1 /4 hy 2 /4) 2/3 1/4 5/12 Y 2 = f (y n + hy 1 /4 + 5hY 2 /12) 1/4 3/4 y n+1 = y n + h(y 1 + 3Y 2 )/4 27 / 61
28 Example... Applied to the test equation, we get the stability function y n+1 = with poles 2 ± i 2 C +, and hλ hλ (hλ)2 y n R(iω) 2 = ω ω ω4 Conclusion R(hλ) 1 hλ C. The method is A-stable 28 / 61
29 5. Linear Multistep Methods A multistep method is a method of the type y n+1 = Φ(f, h, y 0, y 1,..., y n ) using values from several previous steps Explicit Euler y n+1 = y n + h f (t n, y n ) Trapezoidal rule y n+1 = y n + h ( f (tn,yn)+f (t n+1,y n+1 ) 2 ) Implicit Euler y n+1 = y n + h f (t n+1, y n+1 ) are all one-step (RK) methods, but also LM methods 29 / 61
30 Multistep methods and difference equations A k-step multistep method replaces the ODE y = f (t, y) by a difference equation k a j y n+j = h j=0 k b j f (t n+j, y n+j ) j=0 Generating polynomials k k ρ(w) = a j w j σ(w) = b j w j j=0 j=0 Coefficients are normalized either by a k = 1 or σ(1) = 1 b k 0 implicit ; b k = 0 explicit 30 / 61
31 Trivial (one-step) examples Explicit Euler y n+1 y n = hf (t n, y n ) ρ(w) = w 1 σ(w) = 1 Implicit Euler y n+1 y n = hf (t n+1, y n+1 ) ρ(w) = w 1 σ(w) = w Trapezoidal rule y n+1 y n = h 2 (f (t n+1, y n+1 ) + f (t n, y, n)) ρ(w) = w 1 σ(w) = (w + 1)/2 31 / 61
32 Adams methods (J.C. Adams, 1880s) Suppose we have the first n + k approximations y m = y(t m ), m = 0, 1,..., n + k 1 Rewrite y = f (t, y) by integration y(t n+k ) y(t n+k 1 ) = tn+k t n+k 1 f (τ, y(τ)) dτ Approximate by an interpolation polynomial on t n, t n 1,... f (τ, y(τ)) P(τ) 32 / 61
33 Neptune (1846) As seen from Voyager 2 33 / 61
34 Voyager 2 34 / 61
35 ... and Adams got the final word 130 years later Voyager orbit ( ) computed using Adams Moulton methods 35 / 61
36 Triton, Neptune s moon 36 / 61
37 Adams Bashforth methods (explicit) Approximate P(τ) = f (τ, y(τ)) + O(h k ), degree k 1 polynomial P(t n+j ) = f (t n+j, y(t n+j )) j = 0,..., k 1 tn+k Then y(t n+k ) = y(t n+k 1 ) + P(τ) dτ + O(h k+1 ) t n+k 1 Adams-Bashforth method (k-step, order p = k) k 1 y n+k = y n+k 1 + b j hf (t n+j, y n+j ) j=0 where b j = h 1 t n+k t n+k 1 ϕ j (τ) dτ from Lagrange basis polynomials 37 / 61
38 Coefficients of AB1 For k = 1 y n+1 = y n + b 0 hf (t n, y n ) the coefficient is determined by tn+1 tn+1 b 0 = h 1 ϕ 0 (τ) dτ = h 1 1 dτ = 1 t n t n y n+1 = y n + hf (t n, y n ) Conclusion AB1 is the explicit Euler method 38 / 61
39 Coefficients of AB2 For k = 2 y n+2 = y n+1 + h [b 1 f (t n+1, y n+1 ) + b 0 f (t n, y n )] with coefficients tn+2 b 0 = h 1 τ t n+1 dτ = 1 t n+1 t n t n+1 2 tn+2 b 1 = h 1 τ t n dτ = 3 t n+1 t n 2 t n+1 y n+2 = y n hf (t n+1, y n+1 ) 1 2 hf (t n, y n ) 39 / 61
40 Initializing an Adams method The first step of AB2 is y 2 = y 1 + h [ 3 2 f (t 1, y 1 ) 1 ] 2 f (t 0, y 0 ) While y 0 is obtained from the initial value, y 1 must be computed with a one-step method, e.g. AB1 y 1 = y 0 + h f (t 0, y 0 ) [ 3 y n+2 = y n+1 + h 2 f (t n+1, y n+1 ) 1 ] 2 f (t n, y n ), n 0 Multistep software is generally self-starting (with gearbox ) 40 / 61
41 A computational comparison AB1 vs AB2 Solve y = y 2 cos t, y 0 = 1/2, t [0, 8π] using 48 and 480 steps AB1: Solutions Errors AB2: Solutions Errors 41 / 61
42 How do we check the order of a multistep method? The order of consistency is p if the local error is k k a j y(t n+j ) h b j y (t n+j ) = O(h p+1 ) j=0 j=0 Try if the formula holds exactly for polynomials y(t) = 1, t, t 2, t 3,... Insert y = t m and y = mt m 1 into the formula, taking t n+j = jh k k k a j (jh) m h b j m(jh) m 1 = h m (a j j m b j mj m 1 ) j=0 j=0 j=0 42 / 61
43 Order conditions Multistep methods Theorem A k-step method is of consistency order p if and only if it satisfies the following conditions k j m a j j=0 k = m j m 1 b j, j=0 m = 0, 1,..., p k k j p+1 a j (p + 1) j p b j j=0 j=0 A multistep method of consistency order p is exact for polynomials of degree p. Problems with solutions y = P(t) are solved exactly 43 / 61
44 Stability Unlike RK methods there are two distinct kinds of stability notions Finite step stability This is concerned with for what nonzero step sizes h the method can solve the linear test equation y = λy without going unstable. It determines for what problem classes the method is useful. Same idea as for RK Zero stability This is concerned with whether a multistep method can solve the trivial problem y = 0 without going unstable. If not, the method is useless: zero stability is necessary for convergence. Multsitep methods only 44 / 61
45 Zero stability The root condition Definition A polynomial ρ(w) satisfies the root condition if all its zeros are on or inside the unit circle, and the zeros of unit modulus are simple Definition A multistep method whose generating polynomial ρ(w) satisfies the root condition is called zero stable Examples ρ(w) = (w 1)(w 0.5) ρ(w) = (w 1)(w + 1) ρ(w) = (w 1) 2 (w 0.5) ρ(w) = (w 1)(w ) Adams methods have ρ(w) = w k 1 (w 1) and are zero stable 45 / 61
46 The Dahlquist equivalence theorem Theorem A multistep method is convergent if and only if it is zero-stable and consistent of order p 1 (without proof) Example k-step Adams-Bashforth methods are explicit methods of consistency order p = k and have ρ(w) = w k 1 (w 1) they are convergent of convergence order p = k Example k-step Adams-Moulton methods are implicit methods of consistency order p = k + 1 and have ρ(w) = w k 1 (w 1) they are convergent of convergence order p = k / 61
47 Dahlquist s first barrier theorem Theorem The maximal order of a zero-stable k-step method is { k for explicit methods p = k + 1 if k is odd for implicit methods k + 2 if k is even 47 / 61
48 Backward differentiation methods BDF2 Construct a two-step 2nd order method of the form α 2 y n+2 + α 1 y n+1 + α 0 y n = hf (t n+2, y n+2 ) Order conditions for p = 2 α 2 + α 1 + α 0 = 0; 2α 2 + α 1 = 1; 4α 2 + α 1 = y n+2 2y n y n = hf (t n+2, y n+2 ) ρ(w) = 3 2 (w 1)(w 1 ) BDF2 is convergent of order / 61
49 Backward differentiation formulas (BDF) Backward difference operator y n+k = y n+k y n+k 1 with j y n+k = j 1 y n+k j 1 y n+k 1, j > 1 Theorem (without proof) The k-step BDF method k j=1 j j y n+k = hf (t n+k, y n+k ) is convergent of order p = k if and only if 1 k 6 Note BDF methods are designed for stiff problems 49 / 61
50 BDF1-6 stability regions y = hλ The methods are stable outside the indicated areas 20 Stability regions of BDF1 6 methods / 61
51 A-stability of multistep methods y = hλ Applying a method to y = λy produces a difference equation k k a j y n+j = hλ b j y n+j j=0 j=0 The characteristic equation (with z = hλ) ρ(w) zσ(w) = 0 has k roots w j (z). The method is A-stable if and only if Re z 0 w j (z) 1, with simple unit modulus roots (root condition) 51 / 61
52 Dahlquist s second barrier theorem Theorem (without proof) The highest order of an A-stable multistep method is p = 2. Of all 2nd order A-stable multistep methods, the trapezoidal rule has the smallest error Note There is no order restriction for Runge Kutta methods, which can be A-stable for arbitrarily high orders A multistep method can be useful although it isn t A-stable 52 / 61
53 6. Difference operators Differentiation D : y ẏ, where D = d/dt Forward shift E : y(t) y(t + h) Forward shift applied to sequences y = {y n } n=0 Ey = {y n+1 } n=0 Shorthand notation (Ey) n = y n+1 Ey n = y n+1 53 / 61
54 Taylor s theorem Operator version Expand in Taylor series y(t + h) = y(t) + hẏ(t) + h2 2 ÿ(t) +... = (1 + hd + (hd)2 + (hd)3 2! 3! = e hd y(t) +... ) y(t) Taylor s theorem E = e hd 54 / 61
55 Forward difference operator Using short-hand notation y(t) = y(t + h) y(t) y n = y n+1 y n Note = E 1 55 / 61
56 Forward differences of higher order Recursive definition y n = y n+1 y n k y n = ( k 1 y n ) In particular, 2nd order difference 2 y n = (y n+1 y n ) = (y n+2 y n+1 ) (y n+1 y n ) = y n+2 2y n+1 + y n 56 / 61
57 Finite difference approximation of derivatives Approximation of derivatives dy dx y x d 2 y dx 2 y n/ x y n 1 / x x y n+1 2y n + y n 1 x 2 2 y x 2 57 / 61
58 Backward difference operator Backward difference y(t) = y(t) y(t h) y n = y n y n 1 Bwd Difference = 1 E 1 Backward shift E 1 = e hd 58 / 61
59 Linear operators All operators under consideration are linear L(αu + βv) = αlu + βlv Allows addition and multiplication (assoc + dist laws) The operators are commutative (L 1 L 2 ) u = (L 2 L 1 ) u There is a zero and a unit operator, 0 and 1 The operators form an operator algebra 59 / 61
60 Operator series and operator calculus Taylor s theorem e hd = 1 Formal inversion and power series expansion hd = log(1 ) = Apply to differential equation y = f (y) 60 / 61
61 Derivation of BDF methods Differential equation y = f (y) implies hdy = hf (y) Replace with operator series hd = j /j. Truncate at k terms ) ( k y n = hf (y n ) k 1 This is the k-step BDF method The formula is exact for polynomials of degree k, but zero stable only for k 6. BDF1 6 are covergent of order p = k 61 / 61
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