rk tutorial

8/3/2019 RK Tutorial

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Introduction Formulation Taylor series: exact solution Approximation Order conditions

John Butchers tutorialsIntroduction to RungeKutta methods

( t ) = 1 ( t )

Introduction to RungeKutta methods


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Introduction

It will be convenient to consider only autonomous initial valueproblems

y (x ) = f (y(x )) , y(x 0) = y0,

f : R N R N .

The Euler method is the simplest way of obtaining numericalapproximations at

x 1 = x0 + h, x 2 = x1 + h , . . .

using the formula

yn = yn 1 + hf (yn 1), h = x n x n 1, n = 1 , 2, . . . .



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Introduction

It will be convenient to consider only autonomous initial valueproblems

y (x ) = f (y(x )) , y(x 0) = y0,

f : R N R N .

The Euler method is the simplest way of obtaining numericalapproximations at

x 1 = x0 + h, x 2 = x1 + h , . . .

using the formula

yn = yn 1 + hf (yn 1), h = x n x n 1, n = 1 , 2, . . . .



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This method can be made more accurate by using either themid-point quadrature formula

yn = yn 1 + hf yn 1 + 12 hf (yn 1) .

or the trapezoidal rule quadrature formula:

yn = yn 1 + 12 hf (yn 1) +12 hf yn 1 + hf (yn 1) .

These methods from Runges 1895 paper are second orderbecause the error in a single step behaves like O (h3).

This is in contrast to the rst order Euler method where theorder behaviour is O (h 2).



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yn = yn 1 + hf yn 1 + 12 hf (yn 1) .






d i l i l i l i A i i O d di i


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yn = yn 1 + hf yn 1 + 12 hf (yn 1) .






I t d ti F l ti T l i t l ti A i ti O d diti


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yn = yn 1 + hf yn 1 + 12 hf (yn 1) .








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A few years later, Heun gave a full explanation of order 3methods.

Shortly afterwards Kutta gave a detailed analysis of order 4methods.

In the early days of RungeKutta methods the aim seemed tobe to nd explicit methods of higher and higher order.

Later the aim shifted to nding methods that seemed to beoptimal in terms of local truncation error and to nding built-in

error estimators.With the emergence of stiff problems as an importantapplication area, attention moved to implicit methods.




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y pp




Later the aim shifted to nding methods that seemed to beoptimal in terms of local truncation error and to nding built-inerror estimators.

With the emergence of stiff problems as an importantapplication area, attention moved to implicit methods.




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y pp

Formulation of RungeKutta methods

In carrying out a step we evaluate s stage valuesY 1, Y 2, . . . , Y s

and s stage derivatives

F 1 , F 2, . . . , F s ,

using the formula F i = f (Y i ).Each Y i is dened as a linear combination of the F j added on toy0:

Y i = y0 + hs

j =1

a ij F j , i = 1 , 2, . . . , s ,

and the approximation at x 1 = x 0 + h is found from

y1 = y0 + hs

i=1

bi F i .




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Formulation of RungeKutta methods

In carrying out a step we evaluate s stage valuesY 1, Y 2, . . . , Y s

and s stage derivatives

F 1 , F 2, . . . , F s ,

using the formula F i = f (Y i ).Each Y i is dened as a linear combination of the F j added on toy0:

Y i = y0 + hs

j =1

a ij F j , i = 1 , 2, . . . , s ,

and the approximation at x 1 = x 0 + h is found from

y1 = y0 + hs

i=1

bi F i .




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We represent the method by a tableau:c1 a 11 a 12 a 1sc

2a

21a

22 a

2s......

......

cs a s 1 a s 2 a ssb1 b2 bs

or, if the method is explicit, by the simplied tableau0c2 a 21...

...... . . .

cs a s 1 a s 2 a s,s 1b1 b2 bs 1 bs

In each case, ci (i = 1 , 2, . . . ) is dened as s j =1 a ij . The valueof ci indicates the point X i = x 0 + hc i for which Y i is a goodapproximation to y(X i ).




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2a

21a

22 a

2s......

......



...... . . .






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2a

21a

22 a

2s......

......



...... . . .






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Examples:

y1 = y0 + 0 hf (y0) + 1 hf y0 +1

2 hf (y0)

012

1

2

0 1




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Examples:

y1 = y0 + 0 hf (y0) + 1 hf y0 +1

2 hf (y0)

012

1

2

0 1

Y 1 Y 2




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Examples:

y1 = y0 + 0 hf (y0) + 1 hf y0 +1

2 hf (y0)

012

1

2

0 1

Y 1 Y 2

y1 = y0 +1

2 hf (y0) +1

2 hf y0 + 1 hf (y0)

01 11

212




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Examples:

y1 = y0 + 0 hf (y0) + 1 hf y0 +12 hf (y0)

012

1

2

0 1

Y 1 Y 2

y1 = y0 +1

2 hf (y0) +1

2 hf y0 + 1 hf (y0)

01 11

212

Y 1 Y 2




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Taylor series of exact solution

We need formulae for the second, third, . . . , derivatives.y (x ) = f (y(x ))y (x ) = f (y(x ))y (x )

= f (y(x )) f (y(x ))

y (x ) = f (y(x ))(f (y(x )) , y (x ))+ f (y(x )) f (y(x ))y (x )= f (y(x ))( f (y(x )) , f (y(x )))

+ f (y(x )) f (y(x )) f (y(x ))

This will become increasingly complicated as we evaluate higherderivatives.Hence we look for a systematic pattern.Write f = f (y(x )), f = f (y(x )), f = f (y(x )), . . . .



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= f (y(x )) f (y(x ))


+ f (y(x )) f (y(x )) f (y(x ))





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= f (y(x )) f (y(x ))


+ f (y(x )) f (y(x )) f (y(x ))





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= f (y(x )) f (y(x ))


+ f (y(x )) f (y(x )) f (y(x ))





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= f (y(x )) f (y(x ))


+ f (y(x )) f (y(x )) f (y(x ))





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= f (y(x )) f (y(x ))


+ f (y(x )) f (y(x )) f (y(x ))





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y (x ) = f f

y (x ) = f f f f

y (x ) = f (f , f )f

f f

+ f f f f f f

The various terms have a structure related to rooted-trees.Hence, we introduce the set of all rooted trees and somefunctions on this set.




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y (x ) = f f

y (x ) = f f f f

y (x ) = f (f , f ) f f f

+ f f f f f f





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y (x ) = f f

y (x ) = f f f f

y (x ) = f (f , f ) f f f

+ f f f f f f





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Let T denote the set of rooted trees:

T = , , , , , , , , . . .

We identify the following functions on T .




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T = , , , , , , , , . . .


In this table, t will denote a typical tree




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T = , , , , , , , , . . .


In this table, t will denote a typical treer (t ) order of t = number of vertices




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T = , , , , , , , , . . .


In this table, t will denote a typical treer (t ) order of t = number of vertices (t ) symmetry of t = order of automorphism group




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T = , , , , , , , , . . .


In this table, t will denote a typical treer (t ) order of t = number of vertices (t ) symmetry of t = order of automorphism group (t ) density of t




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T = , , , , , , , , . . .


In this table, t will denote a typical treer (t ) order of t = number of vertices (t ) symmetry of t = order of automorphism group (t ) density of t (t ) number of ways of labelling with an ordered set




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T = , , , , , , , , . . .


In this table, t will denote a typical treer (t ) order of t = number of vertices (t ) symmetry of t = order of automorphism group (t ) density of t (t ) number of ways of labelling with an ordered set (t ) number of ways of labelling with an unordered set




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T = , , , , , , , , . . .



F (t )(y0) elementary differential




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T = , , , , , , , , . . .



F (t )(y0) elementary differentialWe will give examples of these functions based on t =




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t =




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t =

r (t ) = 7 7 65

1 2 43




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t =

r (t ) = 7 7 65

1 2 43

(t ) = 8




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t =

r (t ) = 7 7 65

1 2 43

(t ) = 8

(t ) = 63 7 331 1 11




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t =

r (t ) = 7 7 65

1 2 43

(t ) = 8

(t ) = 63 7 331 1 11

(t ) = r ( t )! (t ) (t ) = 10




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t =

r (t ) = 7 7 65

1 2 43

(t ) = 8

(t ) = 63 7 331 1 11

(t ) = r ( t )! (t ) (t ) = 10

(t ) = r ( t )! (t ) = 630




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t =

r (t ) = 7 765

1 2 43

(t ) = 8

(t ) = 63 7 331 1 11

(t ) = r ( t )! (t ) (t ) = 10

(t ) = r ( t )! (t ) = 630

F (t ) = f f (f , f ), f (f , f ) f f f

f f f f




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These functions are easy to compute up to order-4 trees:

t

r (t ) 1 2 3 3 4 4 4 4

(t ) 1 1 2 1 6 1 2 1

(t ) 1 2 3 6 4 8 12 24

(t ) 1 1 1 1 1 3 1 1

(t ) 1 2 3 6 4 24 12 24

F (t ) f f f f (f , f ) f f f f (3)(f , f , f ) f (f , f f ) f f (f , f ) f f f f




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The formal Taylor expansion of the solution at x 0 + h is

y(x 0 + h ) = y0 +tT

(t )hr ( t )

r (t )!F (t )(y0)

Using the known formula for (t ), we can write this as

y(x 0 + h ) = y0 +tT

hr (t )

(t ) (t )F (t )(y0)

Our aim will now be to nd a corresponding formula for theresult computed by one step of a RungeKutta method.

By comparing these formulae term by term, we will be able toobtain conditions for a specic order of accuracy.




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y(x 0 + h ) = y0 +tT

(t )hr ( t )

r (t )!F (t )(y0)


y(x 0 + h ) = y0 +tT

hr (t )

(t ) (t )F (t )(y0)





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y(x 0 + h ) = y0 +tT

(t )hr ( t )

r (t )!F (t )(y0)


y(x 0 + h ) = y0 +tT

hr (t )

(t ) (t ) F (t )(y0)





Taylor series of approximation


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We need to evaluate various expressions which depend on thetableau for a particular method.





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We need to evaluate various expressions which depend on thetableau for a particular method.These are known as elementary weights.





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We need to evaluate various expressions which depend on thetableau for a particular method.These are known as elementary weights.We use the example tree we have already considered toillustrate the construction of the elementary weight ( t ).

t =i

kj

l m on



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We need to evaluate various expressions which depend on thetableau for a particular method.These are known as elementary weights.We use the example tree we have already considered toillustrate the construction of the elementary weight ( t ).

t =i

kj

l m on

(t ) =s

i,j,k,l,m,n,o =1

bi a ij a ik a jl a jm a kn a ko

Simplify by summing over l ,m,n,o :

(t ) =s

i,j,k =1

bi a ij c2 j a ik c2k



Now add ( t ) to the table of functions:


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Now add ( t ) to the table of functions:

t

r (t ) 1 2 3 3 (t ) 1 1 1 1 (t ) 1 2 3 6(t ) bi bi ci bi c2i bi a ij c j

t

r (t ) 4 4 4 4 (t ) 1 3 1 1 (t ) 4 24 12 24(t ) bi c3i bi ci a ij c j bi a ij c2 j bi a ij a jk ck




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The formal Taylor expansion of the numerical approximation tothe solution at x0 + h is

y1 = y0 +tT

(t )h r (t )

r (t )!(t )F (t )(y0)


y1 = y0 +tT

h r (t )

(t )(t )F (t )(y0)




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The formal Taylor expansion of the numerical approximation tothe solution at x0 + h is

y1 = y0 +tT

(t )h r (t )

r (t )!(t )F (t )(y0)


y1 = y0 +tT

h r (t )

(t )(t )F (t )(y0)



Order conditions


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To match the Taylor series

y(x 0 + h ) = y0 +tT

h r ( t )

(t ) (t )F (t )(y0)

y1 = y0 +tT

h r (t )

(t )(t )F (t )(y0)

up to h p terms we need to ensure that

(t ) =1

(t ),

for all trees such thatr (t ) p.

These are the order conditions.



Order conditions


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To match the Taylor series

y(x 0 + h ) = y0 +tT

h r ( t )

(t ) (t )F (t )(y0)

y1 = y0 +tT

h r (t )

(t )(t )F (t )(y0)

up to h p terms we need to ensure that

(t ) =1

(t ),

for all trees such thatr (t ) p.

These are the order conditions.



The order conditions will be illustrated in the case of explicit 4t g th d ith d 4


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stage methods with order 4.

t (t ) = 1 (t)

b1 + b2 + b3 + b4 = 1

b2c2 + b3c3 + b4c4 = 12b2c22 + b3c23 + b4c24 =

1

3b3a 32c2 + b4a 42c2 + b4a 43c3 = 16

b2c32 + b3c33 + b4c34 =14

b3c3a 32c2 + b4c4a 42c2 + b4c4a 43c3 = 18b3a 32c22 + b4a 42c22 + b4a 43c23 =

112

b4a 43a 32c2 = 124


Review order conditions Quadrature connections Order 2 Order 3 Order 4

John Butchers tutorials


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Low order RungeKutta methods

01

2

1

2

12

0 12

1 0 0 1

1

6

1

3

1

3

1

6



Review of order conditions


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Recall the order-4 conditions for a 4-stage method:

t (t) = 1 (t )b1 + b2 + b3 + b4 = 1

b2c2 + b3c3 + b4c4 = 12

b2c22 + b3c23 + b4c24 = 13b3a32c2 + b4a42c2 + b4a43c3 = 16

b2c32 + b3c33 + b4c34 =14

b3c3a32c2 + b4c4a42c2 + b4c4a43c3 =18

b3a32c22 + b4a42c22 + b4a43c23 =112

b4a43a32c2 = 124



For order p ( p 4), no more than p stages are required.


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For order p ( p 4), no more than p stages are required.Th di i f h b f d h 4 b


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The conditions for these can be found, when p < 4, byomitting conditions corresponding to trees with greaterthan p vertices



For order p ( p 4), no more than p stages are required.Th diti f th b f d h < 4 b


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The conditions for these can be found, when p < 4, byomitting conditions corresponding to trees with greaterthan p vertices(that is by omitting trees with order greater than p)



For order p ( p 4), no more than p stages are required.Th diti f th b f d h < 4 b


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The conditions for these can be found, when p < 4, byomitting conditions corresponding to trees with greater

than p vertices(that is by omitting trees with order greater than p)omitting all terms in ( t) with subscripts greater than p.



For order p ( p 4), no more than p stages are required.The conditions for these can be found when p < 4 by


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Order 2:b1 + b2 = 1

b2c2 = 12



For order p ( p 4), no more than p stages are required.The conditions for these can be found when p < 4 by


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Order 2:b1 + b2 = 1

b2c2 = 12Order 3:

b1 + b2 + b3 = 1

b2c2 + b3c3 =12

b2c22 + b3c23 =13

b3a32c2 = 16



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Connection with quadrature


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In the special case of a differential equation of the form:dydx = (x),

integration over a single step using an s-stage RungeKuttamethod is equivalent to the approximation

x 1x 0 (x)dx h si=1 bi (x0 + hc i ) (*)We will examine how well this approximation works for the sspecial choices of given by

(x) = ( x x0)k 1, k = 1 , 2, . . . , s .

The error in (*) is equal to

si=1 bi ck 1i

1k h

k ()

To obtain order s, the coefficient of hk in () must be zero forall k = 1 , 2, . . . , s .





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(x) = ( x x0)k 1, k = 1 , 2, . . . , s .


si=1 bi ck 1i

1k h

k ()






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(x) = ( x x0)k 1, k = 1 , 2, . . . , s .


si=1 bi ck 1i

1k h

k ()




From a consideration of special quadrature problems, we haveseen that necessary conditions for order s are that


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s

i=1 bi ck 1i =

1k , k = 1 , 2, . . . , s . (*)





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s

i=1 bi ck 1i =

1k , k = 1 , 2, . . . , s . (*)

These are equivalent to those Runge-Kutta order conditionswhich correspond to the trees





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s

i=1 bi ck 1i =

1k , k = 1 , 2, . . . , s . (*)


It will usually be convenient to choose a quadrature formula asthe rst step in deriving a RungeKutta method.





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s

i=1b

ick 1

i= 1

k, k = 1 , 2, . . . , s . (*)


It will usually be convenient to choose a quadrature formula asthe rst step in deriving a RungeKutta method.Once this is done, the bi and the ci are known and the nal task

will be to satisfy the remaining order conditions by choosingsuitable values of the a ij .





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s

i=1b

ick 1

i= 1

k, k = 1 , 2, . . . , s . (*)


It will usually be convenient to choose a quadrature formula asthe rst step in deriving a RungeKutta method.Once this is done, the bi and the ci are known and the nal task

will be to satisfy the remaining order conditions by choosingsuitable values of the a ij .In the remaining sections of this tutorial, these ideas will beapplied to nding methods up to order 4.

Low order RungeKutta methods Review order conditions Quadrature connections Order 2 Order 3 Order 4

Methods with order 2


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The two quadrature conditions are

b1 + b2 = 1 ,b2c2 = 12 ,

corresponding to the trees and .

There are no additional trees (or additional order conditions) tosatisfy, so all we have to do is choose c2 = 0 and immediatelywe nd that b2 = 1 / 2c2 , b1 = 1 1/ 2c2 .

Here are three methods based on convenient choices of c2. Notethat the rst two methods are due to Runge.

012

120 1

01 1

12

12

023

2314

34




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b1 + b2 = 1 ,b2c2 = 12 ,




012

120 1

01 1

12

12

023

2314

34




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b1 + b2 = 1 ,b2c2 = 12 ,




012

120 1

01 1

12

12

023

2314

34




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There are now three quadrature conditions corresponding to the

trees , and .



h h d d d h


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trees , and .These are

b1 + b2 + b3 = 1 , (a)b2c2 + b3c3 = 12 , (b)

b2c22 + b3c

23 =

13 . (c)



Th h d di i di h


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b1 + b2 + b3 = 1 , (a)b2c2 + b3c3 = 12 , (b)

b2c22 + b3c

23 =

13 . (c)

There is also an additional condition corresponding to :

b3a32c2 = 16 . (d)



Th th d t diti di t th


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b1 + b2 + b3 = 1 , (a)b2c2 + b3c3 = 12 , (b)

b2c22 + b3c

23 =

13 . (c)


b3a32c2 = 16 . (d)The steps in nding a method are to choose suitable ci , solve

(a), (b) and (c) for the bi and nally solve (d) for a32 .



Th th d t diti di g t th


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b1 + b2 + b3 = 1 , (a)b2c2 + b3c3 = 12 , (b)

b2c22 + b3c

23 =

13 . (c)


b3a32c2 = 16 . (d)The steps in nding a method are to choose suitable ci , solve

(a), (b) and (c) for the bi and nally solve (d) for a32 .Note that in the choice of the ci and the evaluation of the bi ,the value b3 = 0 must be avoided, otherwise the solution of (d)becomes impossible.

Low order Runge Kutta methods


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Recall the conditions for order 4 but ordered differently:


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Recall the conditions for order 4, but ordered differently:

b1 + b2 + b3 + b4 = 1 , (1)b2c2 + b3c3 + b4c4 = 12 , (2)b2c22 + b3c

23 + b4c

24 = 13 , (3)

b2c32 + b3c33 + b4c

34 = 14 , (4)

b3a32c2 + b4a42c2 + b4a43c3 = 16 , (5)b3c3a32c2 + b4c4a42c2 + b4c4a43c3 = 18 , (6)

b3a32c22 + b4a42c22 + b4a43c

23 =

112 , (7)

b4a43a32c2 = 124 . (8)Given c2, c3, c4, carry out the three steps:

1 solve for b1, b2, b3, b4 from (1), (2), (3), (4),2 solve for a32 , a42 , a43 from (5), (6), (7),3 substitute the results found in steps 1 and 2 into (8) and

check for consistency.

Low order Runge Kutta methods Review order conditions Quadrature connections Order 2 Order 3 Order 4




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b1 + b2 + b3 + b4 = 1 , (1)b2c2 + b3c3 + b4c4 = 12 , (2)b2c22 + b3c

23 + b4c

24 = 13 , (3)

b2c32 + b3c33 + b4c

34 = 14 , (4)


b3a32c22 + b4a42c22 + b4a43c

23 =

112 , (7)








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b1 + b2 + b3 + b4 = 1 , (1)b2c2 + b3c3 + b4c4 = 12 , (2)b2c22 + b3c

23 + b4c

24 = 13 , (3)

b2c32 + b3c33 + b4c

34 = 14 , (4)


b3a32c22 + b4a42c22 + b4a43c

23 =

112 , (7)








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b1 + b2 + b3 + b4 = 1 , (1)b2c2 + b3c3 + b4c4 = 12 , (2)b2c22 + b3c

23 + b4c

24 = 13 , (3)

b2c32 + b3c33 + b4c

34 = 14 , (4)


b3a32c22 + b4a42c22 + b4a43c

23 =

112 , (7)








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b1 + b2 + b3 + b4 = 1 , (1)b2c2 + b3c3 + b4c4 = 12 , (2)b2c22 + b3c

23 + b4c

24 = 13 , (3)

b2c32 + b3c33 + b4c

34 = 14 , (4)


b3a32c22 + b4a42c22 + b4a43c

23 =

112 , (7)





If c2, c3 and c4 are treated as parameters, and these steps are


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p pcarried out, it is found that the consistency condition yielded inStep 3 is surprisingly simple. This condition is:

c4 = 1.

For specic choices of c2 and c3 to be used with c4 = 1, it

sometimes happens that some step of the process cannot becarried out, for example because of a vanishing denominator,but fortunately, many cases exist when there is no trouble.

We conclude this tutorial by presenting a number of examples

of order 4 methods.

L d R g K tt th d Review order conditions Quadrature connections Order 2 Order 3 Order 4



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carried out, it is found that the consistency condition yielded inStep 3 is surprisingly simple. This condition is:

c4 = 1.




of order 4 methods.

L d R K tt th d Review order conditions Quadrature connections Order 2 Order 3 Order 4



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c4 = 1.




of order 4 methods.

L d R K tt th d Review order conditions Quadrature connections Order 2 Order 3 Order 4



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c4 = 1.




of order 4 methods.

L d R K h d


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John Butchers tutorials

Implicit RungeKutta methods


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12

36

14

14

36

12 + 36 14 + 36 14

12

12


Since we have an order barrier, which says that order p RKmethods require more than p stages if p > 4, we might ask howto get around this barrier.


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For explicit methods, solving the order conditions becomesincreasingly difficult as the order increases but everythingbecomes simpler for implicit methods.

For example the following method has order 5:

014

18

18

710

1100

1425

320

1 27 0 571

143281

250567

554




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014

18

18

710

1100

1425

320

1 27 0 571

143281

250567

554




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014

18

18

710

1100

1425

320

1 27 0 571

143281

250567

554




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014

18

18

710

1100

1425

320

1 27 0 571

143281

250567

554


We could check the order of this method by verifying the 17order conditions but there is an easier way.A method has order 5 if it satises the B(5), C(2) and D(2)conditions


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conditions.A method satises B( k), C( k), D( k) and E( k, ) if

s

i=1bi c j 1i =

1 j , j = 1 , 2, . . . , k , B(k)

s

j =1a ij c1 j = 1 ci , i = 1 , 2, . . . , s , = 1 , 2, . . . , k , C(k)

s

i=1bi c1i a ij =

1 b j (1 c

j ), j = 1 , 2, . . . , s , = 1 , 2, . . . , k , D(k)

s

i,j =1bi cm 1i a ij cn 1 j = 1(m + n )n , m, n = 1 , 2, . . . , s , E(k, )

and B(5), C(2) and D(2) are easy to check for this method.




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s

i=1bi c j 1i =

1 j , j = 1 , 2, . . . , k , B(k)

s

j =1a ij c1 j = 1 ci , i = 1 , 2, . . . , s , = 1 , 2, . . . , k , C(k)

s

i=1bi c1i a ij =

1 b j (1 c

j ), j = 1 , 2, . . . , s , = 1 , 2, . . . , k , D(k)

s






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s

i=1bi c j 1i =

1 j , j = 1 , 2, . . . , k , B(k)

s

j =1a ij c1 j = 1 ci , i = 1 , 2, . . . , s , = 1 , 2, . . . , k , C(k)

s

i=1bi c1i a ij =

1 b j (1 c

j ), j = 1 , 2, . . . , s , = 1 , 2, . . . , k , D(k)

s






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s

i=1bi c j 1i =

1 j , j = 1 , 2, . . . , k , B(k)

s

j =1a ij c1 j = 1 ci , i = 1 , 2, . . . , s , = 1 , 2, . . . , k , C(k)

s

i=1bi c1i a ij =

1 b j (1 c

j ), j = 1 , 2, . . . , s , = 1 , 2, . . . , k , D(k)

s




The most important types of fully implicit methods (that isA can have any structure) are


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Gauss methods of order 2 s, characterized by B(2 s) andC(s). To satisfy B(2 s), the ci must be zeros of P s (2x 1) = 0, where P s is the Legendre polynomial of degree s.

Radau IIA methods of order 2 s 1, characterized bycs = 1, B(2 s 1) and C( s). The ci are zeros of P s (2x 1) P s1(2x 1) = 0.

Both these families of methods are A-stable.

But both are very expensive to implement and both can sufferfrom order reduction.




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Outline proof that Gauss methods have order 2 s

C (s)


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B (2s)

D (s)

E (s, s ) p=2 s

AND

AND

AND

AND

AND



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This idea of choosing A as a lower triangular matrix can betaken further by avoiding diagonal zeros.

If all the diagonal elements are equal, we get the


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Diagonally-Implicit methods of R. Alexander and theSemi-Explicit methods of S. P. Nrsett (referred to assemi-implicit by J.C. Butcher in 1965).

The following third order L-stable method illustrates what ispossible for DIRK methods

12 (1 + )

12 (1 )

1 14 ( 62 + 16 1) 14 (6

2 20 + 5) 14 ( 6

2+ 16 1)

14 (6

2 20 + 5)

where 0.4358665215 satises 16 32 +3

2 3 = 0.





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12 (1 + )

12 (1 )

1 14 ( 62 + 16 1) 14 (6

2 20 + 5) 14 ( 6

2+ 16 1)

14 (6

2 20 + 5)

where 0.4358665215 satises 16 32 +3

2 3 = 0.





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12 (1 + )

12 (1 )

1 14 ( 62 + 16 1) 14 (6

2 20 + 5) 14 ( 6

2+ 16 1)

14 (6

2 20 + 5)

where 0.4358665215 satises 16 32 +3

2 3 = 0.





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12 (1 + )

12 (1 )

1 14 ( 62 + 16 1) 14 (6

2 20 + 5) 14 ( 6

2+ 16 1)

14 (6

2 20 + 5)

where 0.4358665215 satises 16 32 +3

2 3 = 0.


A SIRK method is characterised by the equation(A) = {}.That is A has a one-point spectrum.

For DIRK methods the stages can be computed independentlyd ll f f h f


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and sequentially from equations of the form

Y i hf (Y i ) = a known quantity.

Each stage requires the same factorised matrix I h J to

permit solution by a modied Newton iteration process (whereJ f/y ).

How then is it possible to implement SIRK methods in asimilarly efficient manner?

The answer lies in the inclusion of a transformation to Jordancanonical form into the computation.



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For DIRK methods the stages can be computed independentlyd i ll f i f h f


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For DIRK methods the stages can be computed independentlyd ti ll f ti f th f


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For DIRK methods the stages can be computed independentlyd ti ll f ti f th f


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For DIRK methods the stages can be computed independentlyand sequentially from equations of the form


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Suppose the matrix T transforms A to canonical form as follows

T 1AT = A


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where

A = (I J ) = 1 0 0 0 0 1 1 0 0 0

0 1 1 0 0...

......

......

0 0 0 1 00 0 0 1 1



T 1AT = A


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where

A = (I J ) = 1 0 0 0 0 1 1 0 0 0

0 1 1 0 0...

......

......

0 0 0 1 00 0 0 1 1



T 1AT = A


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where

A = (I J ) = 1 0 0 0 0 1 1 0 0 0

0 1 1 0 0...

......

......

0 0 0 1 00 0 0 1 1


Consider a single Newton iteration, simplied by the use of thesame approximate Jacobian J for each stage.

Assume the incoming approximation is y0 and that we areattempting to evaluate


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attempting to evaluate

y1 = y0 + h(bT I )F

where F is made up from the s subvectors F i = f (Y i ),

i = 1 , 2, . . . , s .The implicit equations to be solved are

Y = ey0 + h(A I )F

where e is the vector in R n with every component equal to 1and Y has subvectors Y i , i = 1 , 2, . . . , s





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y1 = y0 + h(bT I )F



Y = ey0 + h(A I )F






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y1 = y0 + h(bT I )F



Y = ey0 + h(A I )F






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y1 = y0 + h(bT I )F



Y = ey0 + h(A I )F






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y1 = y0 + h(bT I )F



Y = ey0 + h(A I )F



The Newton process consists of solving the linear system

(I s I hA J )D = Y ey0 h(A I )F

and updating


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and updatingY Y D

To benet from the SI property, write

Y = ( T 1 I )Y, F = ( T 1 I )F, D = ( T 1 I )D,

so that


The following table summarises the costs




and updating


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p gY Y D


Y = ( T 1 I )Y, F = ( T 1 I )F, D = ( T 1 I )D,

so that






and updating


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p gY Y D


Y = ( T 1 I )Y, F = ( T 1 I )F, D = ( T 1 I )D,

so that






and updating


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p gY Y D


Y = ( T 1 I )Y, F = ( T 1 I )F, D = ( T 1 I )D,

so that






and updating


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p gY Y D


Y = ( T 1 I )Y, F = ( T 1 I )F, D = ( T 1 I )D,

so that




without withtransformation transformation

LU factorisation s 3 N 3 N 3

Transformation s 2 N Backsolves s 2 N 2 sN 2


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Transformation s 2 N

In summary, we reduce the very high LU factorisation cost to alevel comparable to BDF methods.

Also we reduce the back substitution cost to the same work perstage as for DIRK or BDF methods.

By comparison, the additional transformation costs areinsignicant for large problems.






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Stage order s means thats

j =1a ij (ci ) = c i0 (t)dt,

for any polynomial of degree s 1 This implies that


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for any polynomial of degree s 1. This implies that

Ack1 = 1k ck , k = 1 , 2, . . . , s ,

where the vector powers are interpreted component bycomponent.

This is equivalent to

Akc

0=

1k!c

k, k = 1 , 2, . . . , s ()



j =1a ij (ci ) = c i0 (t)dt,



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Ack1 = 1k ck , k = 1 , 2, . . . , s ,



Akc

0=

1k!c

k, k = 1 , 2, . . . , s ()



j =1a ij (ci ) = c i0 (t)dt,



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Ack1 = 1k ck , k = 1 , 2, . . . , s ,



Ak

c0

=1k!c

k

, k = 1 , 2, . . . , s ()



j =1a ij (ci ) = c i0 (t)dt,



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Ack1 = 1k ck , k = 1 , 2, . . . , s ,



Ak

c0

=1k!c

k

, k = 1 , 2, . . . , s ()



j =1a ij (ci ) = c i0 (t)dt,



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Ack1 = 1k ck , k = 1 , 2, . . . , s ,



Ak

c0

=1k!c

k

, k = 1 , 2, . . . , s ()


From the Cayley-Hamilton theorem

(A I )s c0 = 0

and hences


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i=0

si

( )s i Ai c0 = 0 .

Substitute from ( ) and it is found that

s

i=0

1i!

si

( )s i ci = 0 .



(A I )s c0 = 0

and hences


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i=0

si

( )s i Ai c0 = 0 .


s

i=0

1i!

si

( )s i ci = 0 .



(A I )s c0 = 0

and hences s


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i=0

si

( )s i Ai c0 = 0 .


s

i=0

1i!

si

( )s i ci = 0 .


Hence each component of c satisess

i=0

1i!

si

x

i= 0

That is x


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Lsx

= 0

where LS denotes the Laguerre polynomial of degree s.

Let 1, 2, . . . , s denote the zeros of Ls so that

ci = i , i = 1 , 2, . . . , s

The question now is, how should be chosen?



i=0

1i!

si

x

i= 0

That is x


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Lsx

= 0



ci = i , i = 1 , 2, . . . , s




i=0

1i!

si

x

i= 0

That is x


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Lsx

= 0



ci = i , i = 1 , 2, . . . , s




i=0

1i!

si

x

i= 0

That is x


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Lsx

= 0



ci = i , i = 1 , 2, . . . , s



Unfortunately, to obtain A-stability, at least for orders p > 2, has to be chosen so that some of the ci are outside the interval[0, 1].

This effect becomes more severe for increasingly high orders and


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This effect becomes more severe for increasingly high orders andcan be seen as a major disadvantage of these methods.

We will look at two approaches for overcoming thisdisadvantage.

However, we rst look at the transformation matrix T forefficient implementation.





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However, we rst look at the transformation matrix T forefficient implementation.





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However, we rst look at the transformation matrix T for

efficient implementation.





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However, we rst look at the transformation matrix T for

efficient implementation.


Dene the matrix T as follows:

T =

L0(1) L1(1) L2(1) Ls

1(1)L0(2) L1(2) L2(2) Ls1(2)L0(3) L1(3) L2(3) Ls1(3)


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......

......

L0(s ) L1(s ) L2(s ) Ls1(s )

It can be shown that for a SIRK method

T 1AT = (I J )



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There are two ways in which SIRK methods can be generalized

In the rst of these we add extra diagonally implicit stages so

that the coefficient matrix looks like this:A 0


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A 0W I

,

where the spectrum of the p p submatrix A is( A) = {}For s p = 1 , 2, 3, . . . we get improvements to the behaviour of

the methods


A second generalization is to replace order by effectiveorder.

This allows us to locate the abscissae where we wish.

In DESIRE methods:


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In DESIRE methods:

Diagonally Extended Singly Implicit Runge-Kutta methodsusing Effective order

these two generalizations are combined.

This seems to be as far as we can go in constructing efficientand accurate singly-implicit Runge-Kutta methods.




In DESIRE methods:


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In DESIRE methods:







In DESIRE methods:


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In DESIRE methods:







In DESIRE methods:


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In DESIRE methods:





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