8.1 Introduction. Consider the initial value problem (IVP):

Transcription

1 8.1 Introduction Consider the initial value problem (IVP): y dy dt = f(t, y), y(t 0)=y 0, t 0 t T. Geometrically: solutions are a one parameter family of curves y = y(t) in(t, y)-plane. Assume solution y = Y (t) exists & is unique exactly one solution curve passes through each point (t 0,y 0 ). Example 8.1 dy dt = ty1/3, y(1) = 1. General solution: ( t 2 ) 3/2 + C y = ±. 3 Different C give different solution curves. C = 2 for solution curve through (1, 1): Y = ( t ) 3/2. E.g. Y (1.1) = (1.07) 3/2 = If family of solutions diverge as t increases from t 0 IVP is ill-conditioned (unstable), otherwise well-conditioned stable.

2 y t Figure 8.1: Solution curves of Example 8.1 for C = 8, 7,...,12. The circle is the initial point (1, 1). IVP (linearly) well-conditioned if J = f/ y < 0int 0 t T (ode unperturbed). Let y = Y (t) be perturbed at t = τ to Y (τ)+δ. If no further perturbation, y = Y (t)+η(t), where η = f(t, Y + η) Y = f(t, Y + η) f(t, Y )=f y (t, ξ)η = Jη, η(τ) =δ, τ t T, where ξ lies between Y & Y + η, bymvt.hence t η(t) =δe Jdt τ. J<0 η(t) < δ, t > 0,

3 i.e. disturbance does not grow. J J 0 < 0, J 0 constant, η(t) 0ast. J 0 perturbations grow IVP ill-conditioned. IVP is stiff on [t 0,T] if solution slowly varies over most of [t 0,T], J<0&J 1. E.g. y = 1000(y sin t)+cost, y(0) = 1, 0 t π. Solution: y = e 1000t +sint. Varies slowly like sin t over most of [0,π] but drops rapidly to zero near t = 0. Not stiff on [0, 0.005]. Expansion methods: determine constants c i where Y (t) N i=1 c i φ i (t) in terms of known basis functions {φ i (t)}, e.g. polynomial, trigonometric, rational, orthogonal functions. Discretisation methods: approximate Y (t) att 1,t 2,...,t n by y 1,y 2,...,y n,wherey i Y (t i ) & usually y i y(t i ), i =1, 2,...,n. Ignore roundoff error for the present. f i f(t i,y i ), Y i Y (t i ), Y i Y (t i ). In general, y k+1 is calculated from y 1,y 2,...,y k and f 1,f 2,...,f k, which have already been computed, and y 0.Methodism-step if y k+1 is calculated from y k,y k 1,...,y k m+1 & f k,f k 1,...,f k m+1.notself-starting if m>1. In general, (t 1,y 1 ) does not lie on solution curve. Even if no further errors are made y 2,y 3,...,y n will lie on the wrong solution curve. The error in the first step has propagated to the succeeding steps. In practice, errors occur at every step; e.g. (t 2,y 2 ) does not lie on the solution curve through (t 1,y 1 ). However, the numerical

4 solution will be acceptable provided all the points (t i,y i ), i =1, 2,...,n, lie closer than some given tolerance to the solution curve through (t 0,y 0 ). The computed solution ỹ k includes roundoff error, usually a secondary consideration since y k ỹ k Y k y k. Can derive discretisation methods using various techniques. E.g. integration of the differential equation, t y(t) =y 0 + f(t, y) dt. t 0 Unknown y occurs in integrand; compare numerical quadrature. y y = Y (t) t 0 t Figure 8.2: Geometric interpretation of quadrature methods.

5 Example 8.2 t1 f(t, y) dt (t 1 t 0 )f(t 0,y 0 ), y(t 1 ) y 0 +(t 1 t 0 )f(t 0,y 0 ). t 0 Repeat to get Euler s method, y k+1 = y k + hf(t k,y k ). h = t k+1 t k is the step-size. Backward Euler method: t1 f(t, y) dt (t 1 t 0 )f(t 1,y 1 ), y k+1 = y k + hf(t k+1,y k+1 ). t 0 The trapezoidal rule gives the trapezoidal rule, Simpson s 1 -rule gives Milne s method, 3 y k+1 = y k + h 2 {f(t k,y k )+f(t k+1,y k+1 )}. y k+1 = y k 1 + h 3 {f(t k 1,y k 1 )+4f(t k,y k )+f(t k+1,y k+1 )}. Not all good methods. E.g. Milne s method: more accurate than others above but (weakly) unstable. Discretisation methods must be convergent & consistent. 1. Method is convergent if y k (h) Y (t), as h max 0, k, t k (h) t, where h max =max i h i. Smaller steps give approximate solutions closer to exact solution. Larger number of steps to reach fixed value of t. Round-off error ignored.

6 2. Method is consistent if y k+1 (h) y k (h) Y (t), as h max 0, k, t k (h) t. h k+1 3. Local truncation error or local error of method at t k is e T k := ȳ k Y k, where ȳ k is determined using method assuming preceding values of y i are exact, i.e. y i = Y i for i =0, 1,...,k Method is order p if local truncation error = O(h p+1 ). 5. Propagated error at t k is e P k := y k ȳ k. Propagated error vanishes if no error at preceding steps t i, i =0, 1, 2,...,k Global truncation error or global error at t k is e k := y k Y k =(y k ȳ k )+(ȳ k Y k )=e P k + e T k. 7. Stability: method may amplify local truncation error at each step to eventually swamp true solution. If propagated error is not amplified at each step, i.e. e P k < e k 1,methodisstable; otherwise unstable. Linear stability: approximate ode y = f(t, y) locally by (y Y ) = f(t, y) f(t, Y )=f y (t, ξ)(y Y ).

7 Locally means for h L,whereL = length scale of Y (t); f y (t, ξ) treated as a constant. (For systems replace f y by Jacobian of system.) Thus stability is usually only considered for f(t, y) =λy, whereλ is a constant (complex for systems). Methods in this course are convergent when stable.

8 8.2 Systems and Higher-Order Equations Methods mostly generalise to first-order systems, with initial conditions In vector notation, y 1 = f 1 (t, y 1,y 2,...,y n ) y 2 = f 2 (t, y 1,y 2,...,y n ). y n = f n (t, y 1,y 2,...,y n ). y i (t 0 )=y 0 i, i =1, 2,...,n. y = f(t, y), y(t 0 )=y 0. A first-order system of n differential equations requires n scalar conditions to determine the n independent constants of integration. Conditions at one point give IVP. Conditions at two or more (up to n) different points give boundary-value problems. Higher-order ode s & systems of higher-order ode s subject to initial conditions can be transformed to first-order systems. E.g. convert where u (j) = d j u/dt j (j =1,...,n), to u (n) = f(t, u, u (1),u (2),...,u (n 1) ), y 1 = y 2 y 2 = y 3

9 . y n 1 = y n y n = f(t, y 1,y 2,...,y n 1 ) by taking y 1 = u, y 2 = u (1), y 3 = u (2),..., y n = u (n 1). First-order system is (linearly) well-conditioned (or stable) ifjacobian matrix, J(y) = f 1 f 1 y 1 y 2 f 2 f 2 y 1 y 2. f n y 1. f n y 2 has only eigenvalues with negative real part. If J has eigenvalues with positive real part then system is usually ill-conditioned. If η is a perturbation in y, then f 1 y n f 2 y n. f n y n η = f(t, y + η) f(t, y) Jη, using a Taylor expansion. Perturbations η will be damped out if eigenvalues all have negative real parts. If J has eigenvalues with widely differing negative real parts system is stiff.,

10 8.3 Truncated Taylor Series Expand exact solution in 3-term Taylor series about t 0, Y (t) =y(t 0 )+Y (t 0 )(t t 0 )+ 1 2 Y (t 0 )(t t 0 ) Y (ξ)(t t 0 ) 3. Determine Y (t 0 )&Y (t 0 ) using ode: Y (t 0 )=f(t 0,y 0 ). To find Y (t 0 ) differentiate y = f(t, y): y = f t + f y dy dt = f t + f y f. Thus Y 0 = f t (t 0,y 0 )+f y (t 0,y 0 )Y 0. Example 8.3 dy dt = ty1/3, y(1) = 1. Four term Taylor expansion about t =1. Solution y = ty 1/3 y = y 1/ ty 2/3 y Y (1) = 1, Y (1) = 4/3, Y (1) = 8/9. Hence y = 2 3 y 2/3 y 2 9 ty 5/3 (y ) ty 2/3 y. Y (t) =1+(t 1) (t 1) (t 1)

11 Y (1.1) , 6 significant figures. Accuracy of Taylor series worsens the further t is from t 0. So expand in new Taylor series about t 1 near t 0 using y 1 from old series to approximate Y (t 1 ). Repeat. Thus, Y k+1 = Y k + hy k h2 Y k h3 Y (ξ). Express Y k and Y k in terms of f and its derivatives, Y k+1 = Y k + hf(t k,y k )+ 1 2 h2 {f t (t k,y k )+f y (t k,y k )f(t k,y k )} + O(h 3 ). The one-step three-term Taylor series discretisation method y k+1 = y k + hy k h2 y k, = y k + hf(t k,y k )+ 1 2 h2 {f t (t k,y k )+f y (t k,y k )f(t k,y k )}. More terms in Taylor series can usually achieve greater accuracy but disadvantages: f(t, y) may be too complicated to differentiate; differentiations increase chance of blunders; Taylor series may not exist; e.g. y = t 1/2 y, y(0) = 0, y (0) does not exist. Taylor series are not a general purpose method.

12 8.4 Euler s Method No differentiation in Taylor series method if only two terms: Euler s method: y k+1 = y k + hy k = y k + hf(t k,y k ). Poor accuracy requires very small step-size h & large number of steps. Use more accurate methods unless storage limitations demand Euler. Example 8.4 dy dt = ty1/3, y(1) = 1. Find Y (1.1) (= exactly) using Euler s method with (a) 1 step & (b) 10 steps. Solution t 0 =1,y 0 =1,f(t, y) =ty 1/3. (a) h =0.1, y 1 =1+0.1f(1, 1) = =1.100 Y (1.1) (b) h =0.01, Y (1.1) Taylor series of Y (t) aboutt k : Y k+1 = Y k + hf(t k,y k )+ 1 2 h2 Y (ξ k ), t k <ξ k <t k+1. Subtract Euler s method to get the global error at t k+1, e k+1 = Y k+1 y k+1 = Y k y k + h{f(t k,y k ) f(t k,y k )} h2 Y (ξ k ). Global error consists of two parts (ignore round-off error):

13 1. Local error at t k+1 : e T k+1 = 1 2 h2 Y (ξ k ). Thus 2. Propagated error at t k+1 : Y k = y k global error = local truncation error. e P k+1 = Y k y k + h{f(t k,y k ) f(t k,y k )} =(1+hJ(ξ))(Y k y k )=(1+hJ)e k, where ξ lies between Y k & y k. e k+1 (1 + hj)e k + e T k+1. Error decays if amplification factor 1 + hj satisfies 1+hJ < 1, i.e. if 2 <hj<0. Well-conditioned IVP J<0 Euler s method stable only if 0 <h< 2/J. Stiff IVP J<0& J 1 very small step-size for stability. Essential difficulty of stiff problems: very small steps to maintain stability. Local error control: y y k 1,k := y k y k 1. t k t k 1 Local error from t k to t k+1 <ɛif h chosen such that 1 2 Y h y k 1,k h2 ɛ or h 2ɛ y k 1,k.

14 Local relative error control: h min 0.9 2ɛ y k y k 1,k,h max h max = maximum step-size; safety factor = 0.9; change step only every few steps. Check derivative approximation using computed value of y k+1 y y k,k+1 := y k+1 y k t k+1 t k. Accept step change only if 1 2 y k,k+1 h2 small enough, otherwise reject & reduce h further. Euler s method for systems: stable if 1+hλ < 1 for any eigenvalue λ of system Jacobian matrix J. Geometrically: interior of circle, centre 1, radius 1, in complex plane..

15 8.5 Runge-Kutta Methods Explicit Two-Stage Runge-Kutta Methods Achieve accuracy of any truncated Taylor series method without differentiation by extra evaluations of f(t, y) ateachstep.e.g. Taylor expansion for function of two variables: k 1 = hf(t k,y k ) k 2 = hf(t k + αh, y k + βk 1 ) y k+1 = y k + ak 1 + bk 2. f(x 0 + x, y 0 + y) =f(x 0,y 0 )+f x (x 0,y 0 )x + f y (x 0,y 0 )y Replace x 0, x, y 0 & y by t k, αh, y k and βh: k 2 = hf(t k,y k )+h{f t (t k,y k )αh + f y (t k,y k )βk 1 } + O(h 3 ).

16 Substitute into method: y k+1 = y k +(a + b)hf(t k,y k )+h 2 {bαf t (t k,y k )+bβf y (t k,y k )f(t k,y k )} + O(h 3 ). Subtract from Taylor expansion of Y (t) aboutt k to obtain global error : where propagated error: & local error: e k+1 := Y k+1 y k+1 = e P k+1 + et k+1, e P k+1 = Y k y k +(a + b)h{f(t k,y k ) f(t k,y k )} + bαh 2 {f t (t k,y k ) f t (t k,y k )} + bβh 2 {f y (t k,y k )f(t k,y k ) f y (t k,y k )f(t k,y k )} + O(h 3 ) e T k+1 =(1 a b)hf(t k,y k )+ ( ) ( ) bα h 2 f t (t k,y k )+ 2 bβ h 2 f y (t k,y k )f(t k,y k )+O(h 3 ). Propagated error vanishes if y k = Y k ; usually non-zero due to errors at previous steps. Local error contains Y k but not y k ; due to approximate nature of numerical method. 2nd-order 2-stage RK methods: generally e T k+1 = O(h), but e T k+1 = O(h 3 )if: a + b =1, bα = 1 2, bβ = 1 2.

17 Modified Euler Method: b =1/2 a =1/2, α =1=β: k 1 = hf(t k,y k ) k 2 = hf(t k+1,y k + k 1 ) y k+1 = y k (k 1 + k 2 ). Example 8.5 dy dt = ty1/3, y(1) = 1. Approximate Y(1.1) using the modified Euler method. Solution t 0 =1,y 0 =1,h =0.1, f(t, y) =ty 1/3. k 1 = 0.1f(1, 1) = 0.1 1=0.1 k 2 = 0.1f(1.1, 1.1) = y 1 = 1+ 1 ( ) =

18 For 2nd-order 2-stage methods global error reduces to e k+1 = Y k y k + h{f(t k,y k ) f(t k,y k )} h2 {f t (t k,y k ) f t (t k,y k )} = h2 {f y (t k,y k )f(t k,y k ) f y (t k,y k )f(t k,y k )} + O(h 3 ) { ( y + hf + 1 ) } y 2 h2 (f t + ff y ) + O(h 3 ) e k + e T k+1, (t k,ξ) where ξ lies between y k & Y k. To O(h 2 ) amplification factor is G = y { y + hf h2 (f t + ff y )} (tk,ξ). Stability of two-stage methods requires G < 1. Linear stability requires G < 1forf(t, y) =λy, i.e. 1+hλ h2 λ 2 < 1.

19 8.5.2 s-stage Runge-Kutta Methods (s >2) Explicit s-stage Runge-Kutta method: k 1 = hf(t k,y k ) k 2 = hf(t k + α 2 h, y k + β 21 k 1 ) k 3 = hf(t k + α 3 h, y k + β 31 k 1 + β 32 k 2 ). k s = hf(t k + α s h, y k + β s1 k 1 + β s2 k β s,s 1 k s 1 ) y k+1 = y k + a 1 k 1 + a 2 k a s k s. α 2 β 21 α 3 β 31 β 32. α s β s1 β s2... β s,s 1 a 1 a 2... a s 1 a s Table 8.2: Tabular form of explicit Runge-Kutta methods. Determine α i, β ij, a j by agreement with Taylor expansion of solution to terms O(h p ). Local error = O(h p+1 ); order of method = p. Highest order methods explicitly constructed of order 10 with 18 & 17 stages.

20 Classical Fourth-Order Four-Stage Method: k 1 = hf(t k,y k ) k 2 = hf(t k h, y k k 1) k 3 = hf(t k h, y k k 2) k 4 = hf(t k+1,y k + k 3 ) y k+1 = y k (k 1 +2k 2 +2k 3 + k 4 ). Example 8.6 dy dt = ty1/3, y(1) = 1. Approximate Y(1.1) using the classical Runge-Kutta method. Solution t 0 =1,y 0 =1,h =0.1, f(t, y) =ty 1/3. k 1 = 0.1f(1, 1) = 0.1 k 2 = 0.1f(1.05, 1.05) = k 3 = 0.1f(1.05, ) = k 4 = 0.1f(1.1, ) = y 1 = 1+ 1 ( ) =

21 Runge-Kutta methods are easily applied to systems. E.g. the modified Euler method, k 1 = hf(t k, y k ) k 2 = hf(t k+1, k k + k 1 ) k k+1 = y k (k 1 + k 2 ) where k 1 & k 2 are vectors. For system of two equations: (k 1 ) 1 = hf 1 (t k, (y 1 ) k, (y 2 ) k ) (k 1 ) 2 = hf 2 (t k, (y 1 ) k, (y 2 ) k ) (k 2 ) 1 = hf 1 (t k+1, (y 1 ) k +(k 1 ) 1, (y 2 ) k +(k 1 ) 2 ) (k 2 ) 2 = hf 2 (t k+1, (y 1 ) k +(k 1 ) 1, (y 2 ) k +(k 1 ) 2 ) (y 1 ) k+1 = (y 1 ) k {(k 1) 1 +(k 2 ) 1 } (y 2 ) k+1 = (y 2 ) k {(k 1) 2 +(k 2 ) 2 }. If method has order p for a single equation then it has order p for a system, if p 4. If p>4 the order of method for systems may be <p.

22 8.5.3 Accuracy and Stability of Runge-Kutta Methods (A) The global error of an explicit Runge-Kutta method of order p is e k+1 = Ge k + e T k+1, where the amplification factor to O(h p )is G = {y + hy (1) + 12 y h2 y (2) p! } hp y (p). The y (j) here must be expressed in terms of t and y using the differential equation. The conditions for stability and linear stability are, respectively, G < 1and 1+hλ h2 λ p! hp λ p < 1. For Runge-Kutta methods the region of linear stability increases with the order of the method. The method of estimating the local truncation error from a higher derivative (y for the Euler method) relies on the form of the truncation error being known and reasonably simple. This is not true for Runge-Kutta methods. Instead, two other methods are commonly used for local error control, extrapolation and embedding. Extrapolation for a p-th order method uses the fact that the local error for a single step from t k to t k+1 of size h is of the form e T k+1 = Ch p+1 + O(h p+2 ), where C is a constant. The local error for two steps of size h/2 fromt k to t k+1 is thus approximately 2C(h/2) p+1. If the corresponding estimates of Y (t k+1 )arey k+1 (h) andy k+1 (h/2), then, neglecting other errors, Y (t k+1 ) y k+1 (h)+ch p+1 y k+1 (h/2) + 2C(h/2) p+1. (t k,ξ)

23 This gives the following estimate for the local truncation error, Ch p+1 y k+1(h) y k+1 (h/2) 1 2 p. If this is larger or much smaller than some specified error tolerance the stepsize h can be adjusted. To get a local error ɛ requires a new stepsize H such that CH p+1 ɛ. Eliminating C, or ( H h H h ) p+1 { (1 2 p )ɛ y k+1 (h) y k+1 (h/2) (1 2 p } 1 )ɛ p +1. y k+1 (h) y k+1 (h/2) Various safeguards are needed, such as those in 8.4, to make a robust local error control. The major drawback of the method is the extra work required at each step.

24 Embedding methods use the stages of the method to calculate two approximations to the solution of different orders y k+1 = y k + a 1 k 1 + a 2 k a s k s ŷ k+1 = y k +â 1 k 1 +â 2 k â s k s. y k+1 ŷ k+1 then provides an estimate of the local error. six-stage method For example, England s fourth-order k 1 = hf(t k,y k ) k 2 = hf(t k h, y k k 1) k 3 = hf(t k h, y k k k 2) k 4 = hf(t k+1,y k k 2 +2k 3 ) k 5 = hf(t k h, y k (7k 1 +10k 2 + k 4 )) k 6 = hf(t k h, y k (28k 1 125k k 3 +54k 4 378k 5 )) y k+1 = y k (k 1 +4k 3 + k 4 ) with the local error estimate e T k+1 = y k+1 ŷ k+1 = ( 42k 1 224k 3 21k k k 6 ). Another example is the Dormand-Prince 5(4) method see Table 8.5.

25 8.6 Multistep Methods General linear m-step method of fixed step-size ( α m + β m 0): y k+1 = α 1 y k +α 2 y k α m y k m+1 +h{β 0 f(t k+1,y k+1 )+β 1 f(t k,y k )+...+β m f(t k m+1,y k m+1 )}. Explicit if β 0 =0. Implicit if β 0 0: unknown y k+1 occurs on both sides solve non-linear equation for y k+1 at each step (if f non-linear in y). Multistep methods for m>1moreefficient. Adams Methods α 1 =1, α 2 = α 3 =...= α m =0. Adams-Bashforth Methods Explicit β 0 =0. y i = f(t i,y i ). Polynomial p AB m 1(t) ofdegreem 1interpolatesm points (t i,y i)(i = k m +1,...,k): y(t k+1 )=y(t k )+ tk+1 t k y dt y k+1 = y k + tk+1 t k p AB m 1 (t) dt. Adams-Moulton Methods Implicit β 0 0. Polynomial p AM m (t) ofdegreem interpolates m +1points(t i,y i )(i = k m +1,...,k+1): y(t k+1 )=y(t k )+ tk+1 t k y dt y k+1 = y k + tk+1 t k p AM m (t) dt.

26 p β 1 β 2 β 3 β 4 β 5 β 6 C p Table 8.3: Parameters of pth-order Adams-Bashforth method (p = 1,...,6). p β 0 β 1 β 2 β 3 β 4 β 5 C p Table 8.4: Parameters of pth order Adams-Moulton method (p = 1,...,6).

27 p β 0 α 1 α 2 α 3 α 4 α 5 α 6 C p Table 8.5: Parameters of pth order Gear (closed) method (p = 1,...,6). k y y y k+1 = α 1 y k + β 0 y k+1 + β 1 y k + β 2 y k 1 + β 3 y k = α 1 1 t 1 1 = β 0 + β 1 + β 2 + β 3 2 t 2 2t 1 = 2( β 0 β 2 2β 3 ) 3 t 3 3t 2 1 = 3( β 0 + β 2 + 4β 3 ) 4 t 4 4t 3 1 = 4( β 0 β 2 8β 3 ) Table 8.6: Equations determining disposable parameters of 3-step Adams-Moulton method.

28 Example 8.7 Solution p AM 2 (t) = Integrating 2-step Adams-Moulton method. Quadratic polynomial interpolating (t k 1,y k 1) (t k,y k)&(t k+1,y k+1): (t t k )(t t k+1 ) (t k 1 t k )(t k 1 t k+1 ) y k 1 + (t t k 1)(t t k+1 ) (t k t k 1 )(t k t k+1 ) y k + (t t k 1 )(t t k ) (t k+1 t k )(t k+1 t k 1 ) y k+1 = 1 2h 2 (t t k)(t t k+1 )y k 1 1 h 2 (t t k 1)(t t k+1 )y k + 1 2h 2 (t t k 1)(t t k )y k+1. tk+1 t k p AM 2 (t) dt = I 1 y k 1 I 2y k + I 3y k+1, where I 1 = 1 tk+1 (t t 2h 2 k )(t t k+1 ) dt = 1 h τ(τ h) dτ = 1 t k 2h h3, letting t = t k + τ. Similarly, I 2 = 2 3 h3, I 3 = 5 6 h3. tk+1 t k p AM 2 (t) dt = h 12 y k 1 + 2h 3 y k + 5h 12 y k+1 y k+1 = y k + tk+1 replacing y by appropriate f. t k p AM 2 (t) dt = y k + h 12 {5f(t k+1,y k+1 )+8f(t k,y k ) f(t k 1,y k 1 )},

29 Gear s Methods β 0 0, β 1 = β 2 =...= β m =0. Polynomial p G m(t) ofdegreem interpolates m +1points(t i,y i )(i = k m +1,...,k+1): Example step Gear method. dp G m (t k+1) dt = f(t k+1,y k+1 ). Solution Quadratic polynomial interpolating (t k 1,y k 1 )(t k,y k )&(t k+1,y k+1 ): p G 2 (t) = 1 2h 2 (t t k)(t t k+1 )y k 1 1 h 2 (t t k 1)(t t k+1 )y k + 1 2h 2 (t t k 1)(t t k )y k+1. Differentiating f(t k+1,y k+1 ) = dpg 2 dt Rearranging tk+1 = 1 2h (t 2 k+1 t k )y k 1 1 h (t 2 k+1 t k 1 )y k + 1 2h (2t 2 k+1 t k 1 t k )y k+1 = 1 2h y k 1 2 h y k + 3 2h y k+1. y k+1 = 4 3 y k 1 3 y k 1 + 2h 3 f(t k+1,y k+1 ).

30 m-step Adams-Moulton method exact for problems with polynomial solutions of degree m + 1. Since the method is linear in f, itmustbeexactfory = t k, k =0, 1,...,m+1, and this determines the disposable parameters of the method, α 1,β 0,β 1,...,β m. Similarly, the m-step Adams-Bashforth method, with disposable parameters α 1,β 1,...,β m, and Gear s method, with disposable parameters α 1,...,α m,β 0, are exact for problems with polynomial solutions of degree m, in particular for y = t k, k =0, 1,...,m. Example step Adams-Moulton method. Determine α 1,β 0,β 1,β 2,β 3 by requiring method to be exact for y =1,t,t 2,t 3,t 4.The factor h ensures parameters do not depend on h or k. Hence take h =1, t k =0 t k+1 =1, t k 1 = 1, t k 2 = 2. Solution α 1 =1, β 0 = 9 24, β 1 = 19 24, β 2 = 5 24, β 3 = 1 24.

31 8.6.1 Accuracy and Stability of Multistep Methods The m-step Adams-Moulton, Adams-Bashforth and Gear s methods are of orders m +1, m and m, and the local errors are e T = C p h p+1 y (p+1) (ξ k ), where p is the order. These methods are convergent within their regions of stability. The propagated error of multistep methods at t k+1 is e P k+1 = y k+1 ȳ k+1 = The global error is m i=1 e k+1 = e P k+1 + e T k+1 = α i y k i+1 + h m i=1 Setting α 0 = 1 thiscanbewrittenas m i=0 m i=0 α i e k i+1 + h β i y k i+1 m i=0 m i=1 α i Y k i+1 h m i=0 β i Y k i+1. β i f y (t k i+1,ξ k i+1 )e k i+1 + e T k+1. {α i + hβ i f y (t k i+1,ξ k i+1 )}e k i+1 = e T k+1. For linear stability, f(t, y) =λy, and the errors are related by m i=0 {α i + hλβ i }e k i+1 = e T k+1. This is an inhomogeneous finite-difference equation with constant coefficients. homogeneous equation are of the form e k = C 1 ξ k 1 + C 2 ξ k C m ξ k m Solutions of the

32 if the associated characteristic equation has m distinct roots ξ 1,ξ 2,...,ξ m,where ρ(ξ)+hλσ(ξ) =0 ρ(ξ) =α 0 ξ m + α 1 ξ m α m, σ(ξ) =β 0 ξ m + β 1 ξ m β m. Secular terms must be added if there are repeated roots. The error will not grow and the method is absolutely stable if the roots of the characteristic polynomial satisfy ξ i 1, i =1, 2,... and there is no repeated root on the circle ξ = 1. There is a region of the complex hλ plane for which the roots of the characteristc equation satisfy the absolute stability condition ξ 1. For hλ = 0 the characteristic equation reduces to ρ(ξ) =0. Forsmallhλ the roots of ρ(ξ) provide a first approximation to the roots of ρ(ξ) +hλσ(ξ). It can be shown (Gear 1971) that a necessary and sufficient condition for a method of order p 1 to be stable is that the roots of ρ(ξ) satisfy ξ 1 and that the roots on ξ = 1 be simple. A stable method is only weakly stable if there is more than one distinct root of ρ(ξ) on ξ =1anditisstrongly stable if there is only one root of ρ(ξ) on ξ = 1. The Adams and Gear methods are strongly stable. With weakly stable methods such as Milne s method there is the possibility of parasitic instability: for hλ small one of the roots of ρ(ξ) on ξ = 1 behaves like the true solution, which is y = Ce λt, while the other root(s) on ξ = 1 contribute terms which may swamp the true solution.

33 1 0.8 p=1 0.6 p=2 0.4 p=3 0.2 p=4 p=5 Im hλ 0 p= Re hλ Figure 8.3: The pth order Adams-Bashforth method is stable within the contour labelled p = 1, 2, 3, 4, 5, 6.

34 3 p=3 p=2 2 p=4 1 p=5 p=1 p=6 Im hλ Re hλ Figure 8.4: The pth order Adams-Moulton method is stable outside the contour labelled p = 1, to the left of the contour labelled p = 2 and within the contours labelled p =3, 4, 5, 6.

35 p=6 15 p=5 10 p=4 5 p=3 p=2 Im hλ 0 p= Re hλ Figure 8.5: The pth order Gear method is stable outside the contours labelled p =1, 2, 3, 4, 5, 6. Note the stability regions are infinite in the left half-plane.

36 8.6.2 Solution of Implicit Methods Implicit methods are more stable than explicit methods. Implicit methods require solution of non-linear equations at every step (β 0 β): Functional Iteration y k+1 = y k + βhf(t k+1,y k+1 )+g(y k,y k 1,...,y k m+1 ) 0=F (y) :=y y k βhf(t k+1,y) g. y (j+1) k+1 = y k + βhf(t k+1,y (j) k+1 )+g. Initial estimate y (0) k+1 of y k+1 predicted (P) by explicit method; e.g. Adams-Bashforth method of order m or m + 1 predictor for Adams-Moulton method of order m + 1. Function f is evaluated is found. For better stability another E is done giving a PECE method. 1 or 2 iterations give y k+1 to an accuracy of same order as local error of implicit method. Adams methods generally used for these predictor-corrector methods. (E) & correction (C) y (1) k+1

37 Functional iteration methods only converge if βhj < 1: y (j+1) k+1 y (j) k+1 = { y k + βhf(t k+1,y (j) k+1 )+g} { y k + βhf(t k+1,y (j 1) k+1 )+g}. = βh { f(t k+1,y (j) k+1 ) f(t k+1,y (j 1) = βhj(ξ (j) k+1 ) ( y (j) k+1 y(j 1) k+1 ) k+1 )} For stiff problems, where J 1, this forces such a small step-size, h<(β J ) 1 predictor-corrector methods are practically useless for stiff problems. 1, that Newton s Method y (j+1) k+1 = y (j) k+1 { F (y (j) k+1 )} 1 F (y (j) k+1 ), F =1 βhj. Newton s method suitable for stiff problems, since it converges regardless of h if y (0) k+1 accurate. is sufficiently Non-stiff: use Adams predictor-corrector methods: more accurate than Gear s methods for same number of function evaluations. Stiff: use Gear s methods; Adams methods unstable without very small step-size.