+ h4. + h5. 6! f (5) i. (C.3) Since., it yields

Transcription

1 Appendix C. Derivation of the Numerical Integration Formulae C.1. Derivation of the Numerical Integration of dy(x) / dx = f (x) For a given analytical or taulated function f (x), the left column in Tale 3.1 shows how to determine y(x) numerically, where dy(x) / dx = f (x). In this section, we will show how to otain the integration rules listed in the left column of Tale 3.1 for the given accuracy. Based on Taylor s series expansion, we have + hf i + h2 2! C.1.1. The first-order numerical integration f i + h3 3! f (2) i + h4 4! f (3) i + h5 5! f (4) i + h6 6! f (5) i +... (C.1) Oviously, the first-order numerical integration should e + hf i + O(h 2 ) (C.2) C.1.2. The second-order numerical integration (Trapezoidal rule) Let the second-order numerical integration e + h(af i + f i+1 ) + O(h 3 ) (C.3) Since f i+1 + h f i + O(h 2 ) (C.4) Sustituting Eq. (C.4) into Eq. (C.3) to eliminate f i+1, it yields or + h{af i + [ f i + h f i + O(h 2 )]} + O(h 3 ) + h(a + ) f i + h 2 f i + O(h 3 ) (C.5) Comparing the coefficients in Eqs. (C.1) and (C.5), it yields a + = 1 and = 1/ 2. Hence, a = 1/ 2. As a result, we otain the well-known Trapezoidal rule, + h( f i + f i+1 ) + O(h 3 ) (C.6) 2 C-1

2 C.1.3. The fourth-order numerical integration (Simpson s rule) Let a fourth-order numerical integration e + h(a f i + f i+(1/2) + c f i+1 ) + O(h 5 ) (C.7) Since f i+(1/2) + h 2 f (h / 2)2 i + 2! f i (2) + (h / 2)3 3! f i (3) + O(h 4 ) (C.8) f i+1 + h f i + h2 2! f (2) i + h3 3! f (3) i + O(h 4 ) (C.9) Sustituting Eqs. (C.8) and (C.9) into Eq. (C.7) to eliminate f i+(1/2) and f i+1, respectively, it yields or + h{af i + [ f i + h 2 f (h / 2)2 i + 2! f i (2) + (h / 2)3 3! f i (3) + O(h 4 )] + c[ f i + h f i + h2 2! f (2) i + h3 3! f (3) i + O(h 4 )]} + O(h 5 ) + h(a + + c) f i + h 2 ( c) f + h3 ( ! + 1 2! c) f (2) i + h 4 ( ! + 1 3! c) f (3) i + O(h 5 ) (C.10) Comparing the coefficients in Eqs. (C.1) and (C.10), it yields a + + c = c = 1 2! ! + 1 2! c = 1 3! (C.11) (C.12) (C.13) where Eqs. (C.12) and (C.13) can e rewritten as c = 1!2 1 / 2! 2!2 2 / 3! (C.14) Solving Eq. (C.14), it yields = 4 / 6, c = 1/ 6. Sustituting these results into Eq. (C.11), it yields a = 1/ 6. It is interesting to note that the following equation is satisfied C-2

3 ! + 1 3! c = 1 4! Thus, the following Simpson rule is indeed a fourth-order-accuracy integration form + h( 1 6 f i f i+(1/2) f i+1 ) + O(h5 ) (C.15) C.1.4. The Simpson s 3/8 rule (a fourth-order numerical integration) Let a fourth-order numerical integration e + h(a f i + f i+(1/3) + c f i+(2/3) + d f i+1 ) + O(h 5 ) (C.16) Since f i+(1/3) + h 3 f (h / 3)2 i + 2! f i (2) + (h / 3)3 3! f i (3) + O(h 4 ) (C.17) f i+(2/3) + 2h 3 (2h / 3)2 f i + 2! f i (2) + (2h / 3)3 3! f i (3) + O(h 4 ) (C.18) f i+1 + h f i + h2 2! f (2) i + h3 3! f (3) i + O(h 4 ) (C.19) Sustituting Eqs. (C.17), (C.18), and (C.19) into Eq. (C.16), then comparing the coefficients in the resulting equation with the coefficients in equation (C.1), it yields a + + c + d = 1 (C.20) c d = 1!3 1 / 2! 2!3 2 / 3! 3!3 3 / 4! (C.21) Solving Eq. (C.21), it yields = 3 / 8, c = 3 / 8, d = 1/ 8. Sustituting these results into Eq. (C.20), it yields a = 1/ 8. We have otained the Simpson s 3/8 law + h( 1 8 f i f i+(1/3) f i+(2/3) f i+1) + O(h 5 ) (C.22) C-3

4 C.1.5. The sixth-order numerical integration Let a sixth-order numerical integration e + h(a f i + f i+(1/4) + c f i+(1/2) + d f i+(3/4) + e f i+1 ) + O(h 7 ) (C.23) Since f i+(1/4) + h 2 f (h / 4)2 i + 2! f i (2) + (h / 4)3 3! f i (3) + (h / 4)4 4! f i (4) + (h / 4)5 5! f i (5) + O(h 6 ) (C.24) f i+(1/2) + h 2 f (h / 2)2 i + 2! f i (2) + (h / 2)3 3! f i (3) + (h / 2)4 4! f i (4) + (h / 2)5 5! f i (5) + O(h 6 ) (C.25) f i+(1/2) + 3h 4 (3h / 4)2 f i + 2! f i (2) + (3h / 4)3 3! f i (3) + (3h / 4)4 4! f i (4) + (3h / 4)5 5! f i (5) + O(h 6 ) (C.26) f i+1 + h f i + h2 2! f (2) i + h3 3! f (3) i + h4 4! f (4) i + h5 5! f (5) i + O(h 6 ) (C.27) Sustituting Eqs. (C.24), (C.25), and (C.26) into Eq. (C.23), then comparing the coefficients in the resulting equation with the coefficients in equation (C.1), it yields a + + c + d + e = 1 (C.28) c d e = 1!4 1 / 2! 2!4 2 / 3! 3!4 3 / 4! 4!4 4 / 5! (C.29) Solving Eq. (C.29), it yields = 32 / 90, c = 12 / 90, d = 32 / 90, e = 7 / 90. Sustituting these results into Eq. (C.28), it yields a = 7 / 90. Note that the following equation is also satisfied c d e = 5!4 5 / 6! Thus, we have otained the sixth-order integration formula C-4

5 + h( 7 90 f i f i+(1/4) f i+(1/2) f i+(3/4) f i+1 ) + O(h7 ) (C.30) C.1.6. Summary of the higher-order numerical integration In summary, we can otain the higher-order numerical integration of dy(x) / dx = f (x) ased on the procedures discussed in this Section C.1. C.2. Derivation of the Runge-Kutta Method (This section has een revised on ) For dy(t) / dt = f [t, y(t)], the right column in Tale 3.1 shows numerical method to otain y(t), for a given initial value of y 0 = y(t = 0). Based on the Taylor s series expansion, we have y n+1 = y n + Δt 1! ( f ) + (Δt)2 ( df t=t n 2! dt ) + (Δt)3 ( d 2 f t=t n 3! dt ) + (Δt)4 ( d 3 f 2 t=t n 4! dt ) + O((Δt) 5 f (4) ) 3 t=t n (C.31) Since d dt f = ( t + f y ) f Eq. (C.31) can e rewritten as y n+1 = y n + Δt( f ) t=t n + (Δt)2 2 ( f y ) t=t n + (Δt)3 6 [ t ( f y ) + f y ( f y )] t=t n + (Δt)4 24 { t [ t ( f y ) + f y ( f y )] + f y [ t ( f y ) + f y ( f y )]} + O((Δt) 5 f (4) ) t=t n (C.32) C.2.1. The Second-Order Runge-Kutta Method Let us assume a second-order expression of y n+1 y n+1 = y n + Δt[a f 1 + ]+O((Δt) 3 f (2) ) (C.33) where f 1 = f (t n, y n ), y n + Δt f 1 ) (C.34) (C.35) C-5

6 Since the 2-dimensional Taylor s series expansion of f (t + Δt, y + Δy) can e written as f (t + Δt, y + Δy) = f (t, y) + (Δt t + Δy y ) f (t, y)+ 1 2! (Δt t + Δy y )2 f (t, y) k! (Δt t + Δy y )k f (t, y)+... (C.36) Let ( f ) t=t n = f (t n, y n ). Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + Δt, y n + Δt( f ) t=t n ) into Eq. (C.35), it yields = ( f ) t=t n + Δt( f y ) t=t n +O[(Δt) 2 f (2) ] (C.37) Sustituting Eqs. (C.34) and (C.37) into Eq. (C.33), it yields y n+1 = y n + Δt{a( f ) t=t n + [( f ) t=t n + Δt( f y ) t=t n ]} + O[(Δt) 3 f (2) ] (C.38) Comparing the coefficients in Eqs. (C.32) and (C.38), it yields a = = 1/ 2. Thus, we have otained the expression of the second-order Runge-Kutta method y n+1 = y n + Δt[ 1 2 ( f ) t=t n f (t n + Δt, y n + Δt( f ) t=t n )]+ O((Δt) 3 f (2) ) (C.39) The second-order Runge-Kutta method shown in Eq. (C.39) is similar to the Trapezoidal rule of the second-order integration shown in the first column of Tale 3.1, ut is different from the second-order Runge-Kutta method shown in the second column of Tale 3.1. Apparently, the expression of the second-order Runge-Kutta method is not unique. Let us assume another second-order expression of y n+1 y n+1 = y n + Δt[a f 1 + ]+O((Δt) 3 f (2) ) (C.40) where f 1 = f (t n, y n ) (C.41) 2, yn + Δt 2 f 1) (C.42) Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + (Δt / 2), y n + (Δt / 2)( f ) t=t n ) into Eq. (C.42), it yields = ( f ) t=t n + Δt 2 ( f y ) t=t n +O[(Δt) 2 f (2) ] (C.43) Sustituting Eqs. (C.41) and (C.43) into Eq. (C.40), it yields C-6

7 y n+1 = y n + Δt{a( f ) t=t n + [( f ) t=t n + Δt 2 ( f y ) t=t n ]}+O[(Δt) 3 f (2) ] (C.44) Comparing the coefficients in Eqs. (C.32) and (C.44), it yields a = 0 and =1. Thus, we have otained another expression of the second-order Runge-Kutta method y n+1 = y n + Δt f (t n + Δt 2, yn + Δt 2 ( f ) t=t n )+O((Δt) 3 f (2) ) (C.45) The second-order Runge-Kutta method shown in Eq. (C.45) is the same as the one shown in the second column of Tale 3.1. The expression given in Eq. (C.45) is also similar to the second-order Lax-Wendroff scheme discussed in Section 3.1. C.2.2. The Fourth-Order Runge-Kutta Method Let us assume a fourth-order expression of y n+1 y n+1 = y n + Δt[a f c + d f 4 ]+ O((Δt) 5 f (4) ) (C.46) where f 1 = f (t n, y n ) (C.47) 2, yn + Δt 2 f 1) (C.48) 2, yn + Δt 2 ) (C.49) f 4, y n + Δt ) (C.50) Let ( f ) t=t n = f (t n, y n ). Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + (Δt / 2), y n + (Δt / 2)( f ) t=t n ) into Eq. (C.48), it yields = ( f ) t=t n + Δt 2 ( f y ) + 1 t=t n 2 (Δt 2 )2 ( 2 f t + 2 f 2 f 2 t y + f 2 y ) 2 t=t n (Δt 2 )3 ( 3 f t + 3 f 3 f 3 t 2 y + 3 f 3 t y + f 3 2 y ) 3 t=t n (Δt 2 )4 ( 4 f t + 4 f 4 f 4 t 3 y f t 2 y + 4 f 4 2 t y + f 4 f 4 3 y ) +O[(Δt) 5 f (5) ] 4 t=t n (C.51) Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + (Δt / 2), y n + (Δt / 2) ) into Eq. (C.49), it yields C-7

8 = ( f ) t=t n + Δt 2 ( f 2 y ) + 1 t=t n 2 (Δt 2 )2 ( 2 f t + 2 f 2 f 2 2 t y + ( f 2 f 2 )2 y ) 2 t=t n (Δt 2 )3 ( 3 f t f t 2 y + 3( ) 2 3 f t y + ( f 3 f 2) 3 2 y ) 3 t=t n (Δt 2 )4 ( 4 f t + 4 f 4 f 4 2 t 3 y + 6( f 4 f 2) 2 t 2 y + 4( f 4 f 2) 3 2 t y + ( f 4 f 2) 4 3 y ) +O[(Δt) 5 f (5) ] 4 t=t n (C.52) Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + Δt, y n + Δt ) into Eq. (C.50), it yields f 4 = ( f ) t=t n + Δt( f 3 y ) + 1 t=t n 2 (Δt)2 ( 2 f t (Δt)3 ( 3 f t + 3 f 3 f 3 3 t 2 y + 3( f 3 f 3) 2 t y + ( f 3 f 3) 3 2 y ) 3 t=t n (Δt)4 ( 4 f t + 4 f f t y + ( f 2 f 3 )2 y ) 2 t=t n 4 f t 3 y + 6( f 4 f 3) 2 t 2 y + 4( f 4 f 3) 3 2 t y + ( f 4 f 3) 4 3 y ) +O[(Δt) 5 f (5) ] 4 t=t n (C.53) Sustituting Eqs. (C.47), (C.51)-(C.53) into Eq. (C.46), it yields y n+1 = y n + Δt[(a + + c + d)( f ) t=t n ] +(Δt) 2 [( 2 + c 2 + d)( f y ) t=t n ] + (Δt)3 2 [( 4 + c 4 + f d)( 2 t + 2 f 2 f 2 t y + f 2 y ) + ( c 2 t=t n 2 + d)( f y ) ( f t=t n y ) ] t=t n + (Δt)4 6 [( 8 + c 8 + f d)( 3 t + 3 f 3 f 3 t 2 y + 3 f 3 t y + f 3 2 y ) 3 t=t n + ( 3c 8 + 3d f 4 )( 2 t + 2 f 2 f 2 t y + f 2 y ) ( f 2 t=t n y ) t=t n + ( 3d 2 )( f y ) ( f t=t n y ) 2 t=t n + ( 3c 4 + 3d)( f ) t=t n ( f y ) t=t n ( 2 f y 2 ) t=t n + ( 3c 4 + 3d)( f y ) t=t n ( 2 f t y ) t=t n ] + (Δt)5 24 [( 16 + c 16 + f d)( 4 t + 4 f 4 f 4 t 3 y + 6 +O((Δt) 6 f (5) ) 4 f t 2 y + 4 f 4 2 t y + f 4 f 4 3 y ) +...] 4 t=t n (C.54) C-8

9 For easy comparison, we rewrite Eq. (C.32) into the following form y n+1 = y n + Δt( f ) t=t n + (Δt)2 2 ( f y ) t=t n + (Δt)3 f 6 [( 2 t + 2 f 2 f 2 t y + f 2 f f 2 )+ ( 2 y y )( f y )] t=t n + (Δt)4 f 24 [( 3 t + 3 f 3 f 3 t 2 y + 3 f 3 t y + f 3 2 y ) 3 + ( 2 f t f 2 f t y + 2 f y 2 )( f y ) + ( f y )( f y )2 + 3( f )( f y )( 2 f y 2 ) + 3( f y )( 2 f t y )] t=t n + (Δt)5 f 120 [( 4 t + 4 f 4 f 4 t 3 y + 6 +O((Δt) 6 f (5) ) 4 f t 2 y + 4 f 4 2 t y + f 4 f 4 3 y ) +...] 4 t=t n (C.55) If Eq. (C.54) is equal to Eq. (C.55), the coefficients in Eqs. (C.54) and (C.55) should e identical to each other. Namely, a + + c + d = c 2 + d = c 4 + d = 1 3 c 2 + d = c 8 + d = 1 4 3c 8 + 3d 4 = 1 4 C-9 3d 2 = 1 4 3c 4 + 3d = c 16 + d = 1 5 (C.56) (C.57) (C.58) (C.59) (C.60) (C.61) (C.62) (C.63) (C.64)

10 Solving Eqs. (C.56)-(C.59), we can find a set of solutions a = d = 1/ 6 and = c = 1/ 3. These solutions also satisfy Eqs. (C.60)-(C.63), ut do not satisfy Eq. (C.64). Thus, we have otained the expression of the fourth-order Runge-Kutta method y n+1 = y n + Δt[ 1 6 f f 4 ]+O((Δt)5 f (4) ) (C.65) where f 1 = f (t n, y n ) 2, yn + Δt 2 f 1 ) 2, yn + Δt 2 ) f 4, y n + Δt ) C.2.3. Looking for the Third-Order and Higher-Order Expressions of y n+1 Let us assume a third-order expression of y n+1 y n+1 = y n + Δt[a f c ]+O((Δt) 4 f (3) ) (C.66) where f 1 = f (t n, y n ) k, yn + Δt k f 1) l, yn + Δt ) l (C.67) (C.68) (C.69) Let ( f ) t=t n = f (t n, y n ). Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + (Δt / k), y n + (Δt/ k)( f ) t=t n ) into Eq. (C.68), it yields = ( f ) t=t n + Δt k ( f y ) + 1 t=t n 2 (Δt k )2 ( 2 f t + 2 f 2 f 2 t y + f 2 y ) +O[(Δt) 3 f (3) ] 2 t=t n (C.70) Sustituting the 2-dimensional Taylor s series expansion of the function f (t n + (Δt / l), y n + (Δt / l) ) into Eq. (C.69) yields = ( f ) t=t n + Δt l ( f 2 y ) + 1 t=t n 2 (Δt l )2 ( 2 f t + 2 f 2 f 2 2 t y + ( f 2 f 2) 2 y ) +O[(Δt) 3 f (3) ] 2 t=t n Sustituting Eqs. (C.67), (C.70) and (C.71) into Eq. (C.66), it yields C-10 (C.71)

11 y n+1 = y n + Δt[(a + + c)( f ) t=t n ] +(Δt) 2 [( k + c l )( f y ) t=t n ] + (Δt)3 2 [( k + c f 2 l 2 )( 2 t + 2 f 2 f 2 t y + f 2 y ) ] 2 t=t n +(Δt) 3 [ c kl ( f y ) t=t n ( f y ) t=t n ]+O((Δt) 4 f (3) ) (C.72) For easy comparison, we cast Eq. (C.32) into the following form y n+1 = y n + Δt( f ) t=t n + (Δt)2 2 ( f y ) t=t n + (Δt)3 f 6 [( 2 t + 2 f 2 f 2 t y + f 2 f f 2 )+ ( 2 y y )( f y )] +O((Δt) 4 f (3) ) t=t n (C.73) If Eq. (C.72) is equal to Eq. (C.73), the coefficients in Eqs. (C.72) and (C.73) should e identical to each other. Namely, a + + c = 1 k + c l = 1 2 k + c 2 l = c kl = 1 6 (C.74) (C.75) (C.76) (C.77) Solving Eqs. (C.56)-(C.59), we can find a set of solutions a =1/ 4, = 0, c = 3 / 4, k = 3, and l = 3 / 2. Thus, we have otained an expression of the third-order Runge-Kutta method y n+1 = y n + Δt[ 1 4 f ]+O((Δt) 4 f (3) ) (C.78) where f 1 = f (t n, y n ) 3, yn + Δt 3 f 1) = f (t n Δt, yn Δt ) C-11

12 Exercise C.1 Find a general expression of the second-order expression of y n+1 Let us assume a second-order expression of y n+1 y n+1 = y n + Δt[a f 1 + ]+ O((Δt) 3 f (2) ) (C.79) where f 1 = f (t n, y n ) k, yn + Δt k f 1 ) Please find possile solutions of a,, and k. (C.80) (C.81) Answer: Sustituting Eqs. (C.80) and (C.70) into Eq. (C.79), it yields Possile solutions include: a + = 1 k = 1 2 Type 1 solutions:{ a = = 1/ 2, k = 1}, which resemle the Trapezoidal rule. Type 2 solutions:{ a = 0, =1, k = 2 }, which resemle the second-order Lax-Wendroff scheme. General solutions:{ = k / 2, a = 1 (k / 2), for all real numer k }. (C.82) (C.83) Exercise C.2 Find a fifth-order expression of y n+1 Let us assume a sixth-order expression of y n+1 y n+1 = y n + Δt[a f c + d f 4 + g f 5 ]+ O((Δt) 6 f (5) ) (C.84) where f 1 = f (t n, y n ) j, yn + Δt f 1 ) j k, yn + Δt k ) f 4 l, yn + Δt ) l f 5 m, yn + Δt m f 4 ) Please find possile solutions of a,,c,d,g, and j,k,l,m. (C.85) (C.86) (C.87) (C.88) (C.89) C-12

13 Exercise C.3 Find a sixth-order expression of y n+1 Let us assume a sixth-order expression of y n+1 y n+1 = y n + Δt[a f c + d f 4 + g f 5 + h f 6 ]+O((Δt) 7 f (6) ) (C.90) where f 1 = f (t n, y n ) j, yn + Δt f 1 ) j k, yn + Δt k ) f 4 l, yn + Δt ) l f 5 m, yn + Δt m f 4 ) f 6 p, yn + Δt p f 5 ) Please find possile solutions of a,,c,d,g,h, and j,k,l,m, p. (C.91) (C.92) (C.93) (C.94) (C.95) (C.96) C-13

14 C.3. Derivation of the Adams' Formulae Let dy / dt = f (y,t), y n = y(t = nδt), and [ f (k ) ] n = [d k f / dt k ] t=nδt. The Taylor series expansion of y n+1 is given y y n+1 = y n + hf n + h2 2! [ f (1) ] n + h3 3! [ f (2) ] n + h4 4! [ f (3) ] n hm m! [ f (m 1) ] n + O(h m+1 f (m) ) (C.97) For Adams' open formula, the m -th order expression of y n+1 can e written as y n+1 = y n + h[a 0 f n + a 1 f n 1 + +a 2 f n 2 + a 3 f n a (m 1) f n (m 1) ]+ O(h m+1 ) (C.98) For Adams close formula, the m -th order expression of y n+1 can e written as y n+1 = y n + h[ +1 f n f n + 1 f n f n (m 2) f n (m 2) ]+ O(h m+1 ) (C.99) The Taylor series expansions of f n+1, f n 1,.., f n (m 1) are given elow: f n+1 = f n + h[ f (1) ] n + h2 2! [ f (2) ] n + h3 3! [ f (3) ] n + h4 4! [ f (4) ] n hm 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.100) f n 1 = f n h[ f (1) ] n + h2 2! [ f (2) ] n h3 3! [ f (3) ] n + h4 4! [ f (4) ] n ( h)m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.101) f n 2 = f n 2h[ f (1) ] n h 2 2! [ f (2) ] n 2 3 h 3 3! [ f (3) ] n ( 2h)m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.102) f n 3 = f n 3h[ f (1) ] n h 2 2! [ f (2) ] n 3 3 h 3 3! [ f (3) ] n ( 3h)m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.103) h 2 f n (m 2) = f n (m 2)h[ f (1) ] n + (m 2) 2 2! [ f h 3 (2) ] n (m 2) 3 3! [ f (3) ] n ( h) m (m 2) m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.104) C-14

15 h 2 f n (m 1) = f n (m 1)h[ f (1) ] n + (m 1) 2 2! [ f h 3 (2) ] n (m 1) 3 3! [ f (3) ] n ( h) m (m 1) m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.105) Sustituting Eqs. (C.101)~(C.105) into Eq. (C.98), and then comparing the coefficients of [ f (k ) ] n with the Eq. (C.97) it yields a 0 + a 1 + a 2 + a a (m 1) = 1 (C.106) (m 1) (m 1) (m 1) m 1 3 m 1 (m 1) m 1 a 1 a 2 a 3 a (m 1) = 1 / 2 +1 / 3 1 / 4 ( 1) m 1 / m (C.107) Sustituting Eqs. (C.100)~(C.104) into Eq. (C.99), and then comparing the coefficients of [ f (k ) ] n with the Eq. (C.97) it yields (m 2) = 1 (C.108) (m 2) (m 2) (m 2) 3 1 ( 1) m 1 ( 2) m 1 ( m + 2) m (m 2) 1/ 2 1/ 3 = 1/ 4 1/ m (C.109) Solutions of Eqs. (C.106) and (C.107) yield the Adams' open formulae as the ones shown in Tale C.2. Solutions of Eqs. (C.108) and (C.109) yield the Adams' close formulae as the ones shown in Tale C.3. Note that a different way to show the derivations of the Adams open and close formulae can e found in the textook y Hilderand (1976). C-15

16 Tale C.2. The nth order Adams' Open Formulae (also called Adams-Bashforth Formulae) Order of Solving dy /d t = f or y / t = f explicitly with h = Δt Accuracy 1 st y n +1 = y n + h[ f n ] + O(h 2 f (1) ) 2 nd y n +1 = y n + h[ 3 2 f n 1 2 f n 1 ] + O(h 3 f (2) ) 3 rd y n +1 = y n + h[ f n f n f n 2 ] + O(h 4 f (3) ) 4 th y n +1 = y n + h[ f n f n f n f n 3 ] + O(h 5 f (4) ) 5 th y n +1 = y n 6 th y n +1 = y n + h[ f n f n f n f n f n 4 ] + O(h 6 f (5) ) + h[ f n f n f n f n f n f n 5 ] + O(h 7 f (6) ) Tale C.3. The nth order Adams' Close Formulae (also called Adams-Moulton Formulae) Order of Solving dy /d t = f or y / t mplicitly with h = Δt Accuracy 1 st y n +1 = y n + h[ f n +1 ] + O(h 2 f (1) ) 2 nd y n +1 = y n + h[ 1 2 f n f n ] + O(h 3 f (2) ) 3 rd y n +1 = y n + h[ 5 12 f n f n 1 12 f n 1 ] + O(h 4 f (3) ) 4 th y n +1 = y n + h[ 9 24 f n f n 5 24 f n f n 2 ] + O(h 5 f (4) ) 5 th y n +1 = y n 6 th y n +1 = y n + h[ f n f n f n f n f n 3 ] + O(h 6 f (5) ) + h[ f n f n f n f n f n f n 4 ] + O(h 7 f (6) ) C-16

17 C.4. Derivation of the Open Formulae at half time steps For open formula, the m -th order expression of y n+1 can e written as y n+1 = y n + h[a 0.5 f n a 0.5 f n a 1.5 f n a 2.5 f n a (m 0.5) f n (m 0.5) ]+ O(h m+1 ) (C.110) The Taylor series expansions of f n+0.5, f n 0.5,.., f n (m 0.5) are given elow: f n+0.5 = f n + h 2 [ f (1) ] n h 2 2! [ f (2) ] n h m 1 h m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) 3! [ f (3) ] n h 4 4! [ f (4) ] n (C.111) f n 0.5 = f n h 2 [ f (1) ] n h 2 2! [ f (2) ] n h m 1 ( h) m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) 3! [ f (3) ] n h 4 4! [ f (4) ] n (C.112) f n 1.5 = f n 3 2 h[ f (1) ] n h 2 2! [ f (2) ] n h 3 3! [ f (3) ] n h 4 4! [ f (4) ] n m 1 2 m 1 ( h) m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.113) f n 2.5 = f n 5 2 h[ f (1) ] n h 2 2! [ f (2) ] n h 3 3! [ f (3) ] n h 4 4! [ f (4) ] n m 1 2 m 1 ( h) m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.114) h 2 f n (m 0.5) = f n (m 0.5)h[ f (1) ] n + (m 0.5) 2 2! [ f h 3 (2) ] n (m 0.5) 3 3! [ f (3) ] n ( h) m (m 0.5) m 1 (m 1)! [ f (m 1) ] n + O(h m f (m) ) (C.115) Sustituting Eqs. (C.111)~(C.115) into Eq. (C.110), and then comparing the coefficients of [ f (k ) ] n C-17

18 with the Eq. (C.97) it yields ! 1 1/ 2 1/ 2 3 / 2 5 / 2! (m 0.5) 1/ 2 2 1/ / / 2 2! (m 0.5) 2 1/ 2 3 1/ / / 2 3! (m 0.5) 3 "!!!! " 1/ 2 m 1 ( 1/ 2) m 1 ( 3 / 2) m 1 ( 5 / 2) m 1! [ (m 0.5)] m 1 a 0.5 a 0.5 a 1.5 a 2.5 " a (m 0.5) 1 1/ 2 = 1/ 3 1/ 4 " 1/ m (C.116) Solutions of Eqs. (C.116) are shown in Tale C.4. Order of Accuracy Tale C.4. The nth order Open Formulae at half time steps Solving dy /d t = f or y / t = f explicitly with h = Δt 2 nd y n+1 = y n + h[1 f n f n 0.5 ]+ O(h 3 f (2) ) 3 rd y n+1 = y n + h[ f n f n f n 1.5 ]+ O(h 4 f (3) ) 4 th y n+1 = y n + h[ f n f n f n f n 2.5 ]+ O(h 5 f (4) ) References Hilderand, F. B., Advanced Calculus for Applications, 2 nd edition, Prentice-Hall, Inc., Englewood, Cliffs, New Jersey, Press, W. H., B. P. Flannery, S. A. Teukolsky, and W. T. Vetterling, Numerical Recipes (in C or in FORTRAN and Pascal), Camridge University Press, Camridge, Shampine, L. F., and M. K. Gordon, Computer Solution of Ordinary Differential Equation: the Initial Value Prolem, W. H. Freeman and Company, San Francisco, C-18