Runge-Kutta methods. With orders of Taylor methods yet without derivatives of f (t, y(t))
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1 Runge-Kutta metods Wit orders of Taylor metods yet witout derivatives of f (t, y(t))
2 First order Taylor expansion in two variables Teorem: Suppose tat f (t, y) and all its partial derivatives are continuous on D def = {(t, y) a t b, c y d.} Let (t 0, y 0 ), (t, y) D, t = t t 0, y = y y 0. Ten f (t, y) = P 1 (t, y) + R 1 (t, y), were
3 First order Taylor expansion in two variables Teorem: Suppose tat f (t, y) and all its partial derivatives are continuous on D def = {(t, y) a t b, c y d.} Let (t 0, y 0 ), (t, y) D, t = t t 0, y = y y 0. Ten f (t, y) = P 1 (t, y) + R 1 (t, y), were P 1 (t, y) = f (t 0, y 0 ) + t f t (t 0, y 0 ) + y f y (t 0, y 0 ), R 1 (t, y) = 2 t 2 for some point (ξ, µ) D. 2 f t 2 (ξ, µ) + t y 2 f t y (ξ, µ) + 2 y 2 2 f (ξ, µ), y 2
4 Second order Taylor expansion in two variables Teorem: Suppose tat f (t, y) and all its partial derivatives are continuous on D def = {(t, y) a t b, c y d.} Let (t 0, y 0 ), (t, y) D, t = t t 0, y = y y 0. Ten f (t, y) = P 2 (t, y) + R 2 (t, y), were
5 Second order Taylor expansion in two variables Teorem: Suppose tat f (t, y) and all its partial derivatives are continuous on D def = {(t, y) a t b, c y d.} Let (t 0, y 0 ), (t, y) D, t = t t 0, y = y y 0. Ten ( P 2 (t, y) = f (t 0, y 0 ) + + f (t, y) = P 2 (t, y) + R 2 (t, y), were ) t f t (t 0, y 0 ) + y f y (t 0, y 0 ) ( 2 t 2 f 2 t 2 (t 0, y 0 ) + t y 2 f t y (t 0, y 0 ) + 2 y 2 f 2 y 2 (t 0, y 0 ) ),
6 Second order Taylor expansion in two variables Teorem: Suppose tat f (t, y) and all its partial derivatives are continuous on D def = {(t, y) a t b, c y d.} Let (t 0, y 0 ), (t, y) D, t = t t 0, y = y y 0. Ten ( P 2 (t, y) = f (t 0, y 0 ) + + ( 2 t 2 R 2 (t, y) = 1 3 3! f (t, y) = P 2 (t, y) + R 2 (t, y), j=0 were ) t f t (t 0, y 0 ) + y f y (t 0, y 0 ) 2 f t 2 (t 0, y 0 ) + t y ( 3 j ) 3 j t j y 2 f t y (t 0, y 0 ) + 2 y 2 3 f t 3 j y (ξ, µ) j ) 2 f y 2 (t 0, y 0 ),
7 Runge-Kutta Metod of Order Two (III) Midpoint Metod w 0 = α, w j+1 = w j + f ( t j + 2, w j + ) 2 f (t j, w j ), j = 0, 1,, N 1. Two function evaluations for eac j, Second order accuracy. No need for derivative calculations
8 General 2nd order Runge-Kutta Metods w 0 = α; for j = 0, 1,, N 1, w j+1 = w j + (a 1 f (t j, w j ) + a 2 f (t j + α 2, w j + δ 2 f (t j, w j ))).
9 General 2nd order Runge-Kutta Metods w 0 = α; for j = 0, 1,, N 1, w j+1 = w j + (a 1 f (t j, w j ) + a 2 f (t j + α 2, w j + δ 2 f (t j, w j ))). Two function evaluations for eac j, Want to coose a 1, a 2, α 2, δ 2 for igest possible order of accuracy.
10 General 2nd order Runge-Kutta Metods w 0 = α; for j = 0, 1,, N 1, w j+1 = w j + (a 1 f (t j, w j ) + a 2 f (t j + α 2, w j + δ 2 f (t j, w j ))). Two function evaluations for eac j, Want to coose a 1, a 2, α 2, δ 2 for igest possible order of accuracy. local truncation error τ j+1 () = y(t j+1) y(t j ) (a 1 f (t j, y(t j ))+a 2 f (t j +α 2, y(t j )+δ 2 f (t j, y(t j )))) = y (t j ) + 2 y (t j ) + O( 2 ) ( f (a 1 + a 2 ) f (t j, y(t j )) + a 2 α 2 t (t j, y(t j )) + a 2 δ 2 f (t j, y(t j )) f ) y (t j, y(t j )) + O( 2 ).
11 local truncation error τ j+1 () = y (t j ) + 2 y (t j ) + O( 2 ) ( f (a 1 + a 2 ) f (t j, y(t j )) + a 2 α 2 t (t j, y(t j )) + a 2 δ 2 f (t j, y(t j )) f ) y (t j, y(t j )) + O( 2 ). were y (t j ) = f (t j, y(t j )),
12 local truncation error τ j+1 () = y (t j ) + 2 y (t j ) + O( 2 ) ( f (a 1 + a 2 ) f (t j, y(t j )) + a 2 α 2 t (t j, y(t j )) + a 2 δ 2 f (t j, y(t j )) f ) y (t j, y(t j )) + O( 2 ). were y (t j ) = f (t j, y(t j )), y (t j ) = d f d t (t j, y(t j )) = f t (t j, y(t j )) + f (t j, y(t j )) f y (t j, y(t j )). For any coice wit we ave a second order metod a 1 + a 2 = 1, a 2 α 2 = a 2 δ 2 = 2, τ j+1 () = O( 2 ).
13 local truncation error τ j+1 () = y (t j ) + 2 y (t j ) + O( 2 ) ( f (a 1 + a 2 ) f (t j, y(t j )) + a 2 α 2 t (t j, y(t j )) + a 2 δ 2 f (t j, y(t j )) f ) y (t j, y(t j )) + O( 2 ). were y (t j ) = f (t j, y(t j )), y (t j ) = d f d t (t j, y(t j )) = f t (t j, y(t j )) + f (t j, y(t j )) f y (t j, y(t j )). For any coice wit we ave a second order metod a 1 + a 2 = 1, a 2 α 2 = a 2 δ 2 = 2, τ j+1 () = O( 2 ). Four parameters, tree equations
14 General 2nd order Runge-Kutta Metods w 0 = α; for j = 0, 1,, N 1, w j+1 = w j + (a 1 f (t j, w j ) + a 2 f (t j + α 2, w j + δ 2 f (t j, w j ))). a 1 + a 2 = 1, a 2 α 2 = a 2 δ 2 = 2, Midpoint metod: a 1 = 0, a 2 = 1, α 2 = δ 2 = 2, w j+1 = w j + f ( t j + 2, w j + ) 2 f (t j, w j ). Modified Euler metod: a 1 = a 2 = 1 2, α 2 = δ 2 =, w j+1 = w j + 2 (f (t j, w j ) + f (t j+1, w j + f (t j, w j ))).
15 3rd order Runge-Kutta Metod (rarely used in practice) w 0 = α; for j = 0, 1,, N 1, w j+1 = w j + ( ( f (t j, w j ) + 3f t j , w j f (t j + 3, w j + )) 3 f (t j, w j )) def = w j + φ(t j, w j ).
16 3rd order Runge-Kutta Metod (rarely used in practice) w 0 = α; for j = 0, 1,, N 1, w j+1 = w j + ( ( f (t j, w j ) + 3f t j , w j f (t j + 3, w j + )) 3 f (t j, w j )) def = w j + φ(t j, w j ). local truncation error τ j+1 () = y(t j+1)y(t j ) φ(t j, y(t j )) = O( 3 ).
17 4t order Runge-Kutta Metod w 0 = α; for j = 0, 1,, N 1, k 1 = f (t j, w j ), ( k 2 = f t j + 2, w j + 1 ) 2 k 1, ( k 3 = f t j + 2, w j + 1 ) 2 k 2, k 4 = f (t j+1, w j + k 3 ), w j+1 = w j (k 1 + 2k 2 + 2k 3 + k 4 ). 4 function evaluations per step
18 Example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
19 Example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
20 Example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
21 Example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
22 Adaptive Error Control (I) dy = f (t, y(t)), a t b, y(a) = α. dt Consider a variable-step metod wit a well-cosen function φ(t, w, ): w 0 = α. for j = 0, 1,, coose step-size j = t j+1 t j, set wj+1 = w j + j φ (t j, w j, j ).
23 Adaptive Error Control (I) dy = f (t, y(t)), a t b, y(a) = α. dt Consider a variable-step metod wit a well-cosen function φ(t, w, ): w 0 = α. for j = 0, 1,, coose step-size j = t j+1 t j, set wj+1 = w j + j φ (t j, w j, j ). Adaptively coose step-size to satisfy given tolerance
24 Adaptive Error Control (II) Given an order-n metod w0 = α. for j = 0, 1,, w j+1 = w j + φ (t j, w j, ), = t j+1 t j, local truncation error (LTE) τ j+1 () = y(t j+1) y(t j ) φ (t j, y(t j ), ) = O ( n ). Given tolerance τ > 0, we would like to estimate largest step-size for wic τ j+1 () τ.
25 Adaptive Error Control (II) Given an order-n metod w0 = α. for j = 0, 1,, w j+1 = w j + φ (t j, w j, ), = t j+1 t j, local truncation error (LTE) τ j+1 () = y(t j+1) y(t j ) φ (t j, y(t j ), ) = O ( n ). Given tolerance τ > 0, we would like to estimate largest step-size for wic τ j+1 () τ. Approac: Estimate τ j+1 () wit order-(n + 1) metod w j+1 = w j + φ (t j, w j, ), for j 0.
26 LTE Estimation (I) Assume w j y(t j ), w j y(t j ) (only estimating LTE).
27 LTE Estimation (I) Assume w j y(t j ), w j y(t j ) (only estimating LTE). φ(t, w, ) is order-n metod τ j+1 () def = w j y(t j ) = y(t j+1 ) y(t j ) φ (t j, y(t j ), ) y(t j+1 ) (w j + φ (t j, w j, )) y(t j+1 ) w j+1 = O ( n ).
28 LTE Estimation (I) Assume w j y(t j ), w j y(t j ) (only estimating LTE). φ(t, w, ) is order-n metod τ j+1 () def = w j y(t j ) = φ(t, w, ) is order-(n + 1) metod, y(t j+1 ) y(t j ) φ (t j, y(t j ), ) y(t j+1 ) (w j + φ (t j, w j, )) y(t j+1 ) w j+1 = O ( n ). τ j+1 () def = w j y(t j ) = y(t j+1 ) y(t j ) φ (t j, y(t j ), ) ( y(t j+1 ) w j + φ ) (t j, w j, ) y(t j+1 ) w j+1 = O ( n+1).
29 LTE Estimation (III) Assume w j y(t j ), w j y(t j ) (only estimating LTE).
30 LTE Estimation (III) Assume w j y(t j ), w j y(t j ) (only estimating LTE). φ(t, w, ) is order-n metod τ j+1 () y(t j+1) w j+1 = O ( n ).
31 LTE Estimation (III) Assume w j y(t j ), w j y(t j ) (only estimating LTE). φ(t, w, ) is order-n metod τ j+1 () y(t j+1) w j+1 φ(t, w, ) is order-(n + 1) metod, = O ( n ). τ j+1 () y(t j+1) w j+1 = O ( n+1).
32 LTE Estimation (III) Assume w j y(t j ), w j y(t j ) (only estimating LTE). φ(t, w, ) is order-n metod τ j+1 () y(t j+1) w j+1 φ(t, w, ) is order-(n + 1) metod, = O ( n ). τ j+1 () y(t j+1) w j+1 = O ( n+1). terefore τ j+1 () y(t j+1) w j+1 + w j+1 w j+1 = O ( n+1) + w j+1 w j+1 LTE estimate: τ j+1 () w j+1 w j+1 = O ( n ).
33 step-size selection (I) LTE estimate: τ j+1 () w j+1 w j+1
34 step-size selection (I) LTE estimate: τ j+1 () w j+1 w j+1 Since τ j+1 () = O ( n ), assume τ j+1 () K n were K is independent of.
35 step-size selection (I) LTE estimate: τ j+1 () w j+1 w j+1 Since τ j+1 () = O ( n ), assume τ j+1 () K n were K is independent of. K sould satisfy K n w j+1 w j+1. Assume LTE for new step-size q satisfies given tolerance ɛ: τ j+1 (q ) ɛ, need to estimate q.
36 step-size selection (II) LTE estimate: K n τ j+1 () w j+1 w j+1
37 step-size selection (II) LTE estimate: K n τ j+1 () w j+1 w j+1 Assume LTE for q satisfies given tolerance ɛ: τ j+1 (q ) K (q ) n = q n K n q n w j+1 w j+1 ɛ.
38 step-size selection (II) LTE estimate: K n τ j+1 () w j+1 w j+1 Assume LTE for q satisfies given tolerance ɛ: τ j+1 (q ) K (q ) n = q n K n q n w j+1 w j+1 ɛ. 1 new step-size estimate: q ɛ n w j+1 w j+1
39 Summary Given order-n metod w j+1 = w j + φ (t j, w j, ), = t j+1 t j, j 0, and given order-(n + 1) metod w j+1 = w j + φ (t j, w j, ), for j 0. for eac j, compute w j+1 = w j + φ (t j, w j, ), w j+1 = w j + φ (t j, w j, ), new step-size q sould satisfy 1 1 q ɛ n ɛ n w j+1 w j+1 =. φ (t j, w j, ) φ (t j, w j, )
40 Runge-Kutta-Felberg: 4 t order metod, 5 t order estimate w j+1 = w j k k k k 5, w j+1 = w j k k k k k 6, were
41 Runge-Kutta-Felberg: 4 t order metod, 5 t order estimate w j+1 = w j k k k k 5, w j+1 = w j k k k k k 6, were k 1 = f (t j, w j ),
42 Runge-Kutta-Felberg: 4 t order metod, 5 t order estimate w j+1 = w j k k k k 5, w j+1 = w j k k k k k 6, were k 1 = f (t j, w j ), ( k 2 = f t j + 4, w j + 1 ) 4 k 1, ( k 3 = f t j + 3 8, w j k ) 32 k 2, ( k 4 = f t j , w j k k ) 2197 k 3, ( k 5 = f t j +, w j k 1 8 k k ) 4104 k 4, ( k 6 = f t j + 2, w j 8 27 k k k k 4 11 ) 40 k 5.
43 Runge-Kutta-Felberg: step-size selection procedure compute a conservative value for q: 1 q = ɛ 4 2 ( w j+1 w j+1 ).
44 Runge-Kutta-Felberg: step-size selection procedure compute a conservative value for q: 1 q = ɛ 4 2 ( w j+1 w j+1 ). make restricted step-size cange: { 0.1, if q 0.1, = 4, if q 4.
45 Runge-Kutta-Felberg: step-size selection procedure compute a conservative value for q: 1 q = ɛ 4 2 ( w j+1 w j+1 ). make restricted step-size cange: { 0.1, if q 0.1, = 4, if q 4. step-size can t be too big: = min (, max ).
46 Runge-Kutta-Felberg: step-size selection procedure compute a conservative value for q: 1 q = ɛ 4 2 ( w j+1 w j+1 ). make restricted step-size cange: { 0.1, if q 0.1, = 4, if q 4. step-size can t be too big: = min (, max ). step-size can t be too small: if < min ten declare failure.
47 Runge-Kutta-Felberg: example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
48 Runge-Kutta-Felberg: example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
49 Runge-Kutta-Felberg: example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t.
50 Runge-Kutta-Felberg: example dy Initial Value ODE = y t 2 + 1, 0 t 2, y(0) = 0.5, dt exact solution y(t) = (1 + t) e t. t j j y(t j ) w j w j
51 Runge-Kutta-Felberg: solution plots Initial Value ODE dy dt = y t 2 + 1, 0 t 2, y(0) = 0.5.
52 Runge-Kutta-Felberg: solution plots Initial Value ODE dy dt = y t 2 + 1, 0 t 2, y(0) = 0.5.
53 Runge-Kutta-Felberg: solution errors Initial Value ODE dy dt = y t 2 + 1, 0 t 2, y(0) = 0.5.
54 Runge-Kutta-Felberg: solution errors Initial Value ODE dy dt = y t 2 + 1, 0 t 2, y(0) = 0.5.
55 Runge-Kutta-Felberg: truncation errors Initial Value ODE actual def = dy dt y(t j ) w j j = y t 2 + 1, 0 t 2, y(0) = 0.5., def estimate = w j w j j.
56 Runge-Kutta-Felberg: truncation errors Initial Value ODE actual def = dy dt y(t j ) w j j = y t 2 + 1, 0 t 2, y(0) = 0.5., def estimate = w j w j j.
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