Rosenbrock Methods€¦ · 3 = 5:000000000000e 01 ^b 3 = 1:388888888888e 01 Table:ROS3w (with a...

OutlineIntroduction

StabilityOrder conditions

Computational resultsEnd

Rosenbrock Methods

Yue Yu

December 9, 2009

Yue Yu Rosenbrock Methods

OutlineIntroduction



IntroductionInspirationspecial cases

StabilityA-stability and L-stability

Order conditionsRooted treesγii = γW-methods

Computational resultsComputational efficiency

EndConclusionReferences


OutlineIntroduction



Inspirationspecial cases

We start with a diagonally IRK method

kni = hf (yn +

i−1∑j=1

αijknj + αiiki ), i = 1, · · · , s

yn+1 = yn +s∑

j=1

bjkj

applied to the autonomous differential equation

y ′ = f (y).

Linearizing the first formula yields

kni = hf (yn +

i−1∑j=1

αijknj ) + hf ′(yn +

i−1∑j=1

αijknj )αiik

ni , i = 1, · · · , s.


OutlineIntroduction




kni = hf (yn +

i−1∑j=1

αijknj ) + hf ′(yn +

i−1∑j=1

αijknj )αiik

ni , i = 1, · · · , s.

Replacing the Jacobians with J = f ′(yn) for the computationalcost:

kni = hf (yn +

i−1∑j=1

αijknj ) + hJαiik

ni , i = 1, · · · , s.

and introducing additional linear combinations of terms to gainfurther freedom:

kni = hf (yn +

i−1∑j=1

αijknj ) + hJ

i∑j=1

γijknj , i = 1, · · · , s


OutlineIntroduction




Definition:An s-stage Rosenbrock method is given by the formulas

kni = hf (yn +

i−1∑j=1

αijknj ) + hJ

i∑j=1

γijknj , i = 1, · · · , s

yn+1 = yn +s∑

j=1

bjkj

where αij , γij , bi are the determining coefficients and J = f ′(yn).


OutlineIntroduction




Non-autonomous problem:The equation

y ′ = f (x , y)

can be converted to autonomous form by adding x ′ = 1. So thes-stage Rosenbrock method for non-autonomous case could bewritten as

kni = hf (xn+αih, y

n+i−1∑j=1

αijknj )+γih

2 ∂f

∂x(xn, yn)+h

∂f

∂y(xn, yn)

i∑j=1

γijknj

yn+1 = yn +s∑

j=1

bjkj

where the additional coefficients are given by

αi =i−1∑j=1

αij , γi =i∑

j=1

γij .


OutlineIntroduction




Implicit differential equations:Suppose the problem is of the form

My ′ = f (x , y)

where M is a constant, nonsingular matrix. Then applying ans-stage Rosenbrock method, we can get

Mkni = hf (yn +

i−1∑j=1

αijknj ) + hJ

i∑j=1

γijknj , i = 1, · · · , s

yn+1 = yn +s∑

j=1

bjkj .

Then the inversion of M is advoided.


OutlineIntroduction



A-stability and L-stability

Applying the Rosenbrock method to the test equation y ′ = λy ,then the numerical solution becomes yn+1 = R(hλ)yn with

R(z) = 1 + zbT (I − zB)−1en

wherebT = (b1, · · · , bs), B = (αij + γij)

si ,j=1.

Since B is a lower triangular matrix, the stability function can beinvestigated similarly as for DIRK methods.


OutlineIntroduction




Examples of L-stable Rosenbrock methods:

γ = 2.928932188134e − 01

α21 = 1.000000000000e + 00 γ21 = −5.857864376269e − 01

b1 = 5.000000000000e − 01 b2 = 5.000000000000e − 01

Table: ROS2


OutlineIntroduction




−15 −10 −5 0 5 10 15−15

−10

−5

0

5

10

15stable region for ROS2 when \delta=1−\sqrt{2}/2

Figure: stablity region for ROS2 scheme


OutlineIntroduction




And you can also do the time step control with embeded methods[6]:

γ = 4.358665215084e − 01

α21 = 6.666666666666e − 01 γ21 = 3.635068368900e − 01α31 = 6.666666666666e − 01 γ31 = −8.996866791992e − 01α32 = 6.666666666666e − 01 γ32 = −1.537997822626e − 01

b1 = 2.500000000000e − 01 b̂1 = 7.467047032740e − 01

b2 = 2.500000000000e − 01 b̂2 = 1.144064078371e − 01

b3 = 5.000000000000e − 01 b̂3 = 1.388888888888e − 01

Table: ROS3w (with a strongly A-stable 2rd order method embeded)


OutlineIntroduction



Rooted treesγii = γW-methods

s-stage Rosenbrock method:

kni = hf (yn +

i−1∑j=1

αijknj ) + hJ

i∑j=1

γijknj , i = 1, · · · , s

yn+1 = yn +s∑

j=1

bjkj .

Theorem 7.4 [5]:A Rosenbrock method with J = f ′(y0) is of order p iff∑

j

bjΦj(t) =1

γ(t), ∀t ∈ T (p).

Therefore, we can apply the Rooted trees method to derive theorder conditions just as what we do for Runge-Kutta methods.


OutlineIntroduction




The only difference is that at singly-branched vertices of thecorresponding trees αjk is replaced by αjk + γjk . The conditions upto order 4 are showed as in Table 3.

order RHS(=1/γ(t)) Φj(t)

1 1 1

2 12

∑k(αjk + γjk)

3 13

∑k,l αjkαjl

3 16

∑k,l(αjk + γjk)(αkl + γkl)

4 14

∑k,l ,m αjkαjlαjm

4 18

∑k,l ,m αjk(αkl + γkl)αjm

4 112

∑k,l ,m(αjk + γjk)αklαkm

4 124

∑k,l ,m(αjk + γjk)(αkl + γkl)(αlm + γlm)

Table: order conditions for original RosenbrockYue Yu Rosenbrock Methods

OutlineIntroduction





kni = hf (yn +

i−1∑j=1

αijknj ) + hJ

i∑j=1

γijknj , i = 1, · · · , s

yn+1 = yn +s∑

j=1

bjkj .

When γii = γ for all i , we only need to do the LU decompositiononce. Then the order conditions can be derived similarly as IRKmethods, except that these diagonal γ appear only forsingly-branched vertices.


OutlineIntroduction




The conditions up to order 4 are showed as in Table 4.

order RHS Φj(t)

1 1 1

2 12 − γ

∑′k(αjk + γjk)

3 13

∑′k,l αjkαjl

3 16 − γ + γ2

∑′k,l(αjk + γjk)(αkl + γkl)

4 14

∑′k,l ,m αjkαjlαjm

4 18 −

γ3

∑′k,l ,m αjk(αkl + γkl)αjm

4 112 −

γ3

∑′k,l ,m(αjk + γjk)αklαkm

4 124 −

γ2 + 3γ2

2 − γ3

∑′k,l ,m(αjk + γjk)(αkl + γkl)(αlm + γlm)

Table: order conditions for simplified Rosenbrock 1


OutlineIntroduction





kni = hf (yn +

i−1∑j=1

αijknj ) + hJ

i∑j=1

γijknj , i = 1, · · · , s

yn+1 = yn +s∑

j=1

bjkj .

In Rosenbrock, the Jacobian matrix J = ∂f∂y must be evaluated at

every step, which can make the computations costly. So peoplesearched for Rosenbrock methods which assure classical order forall approximations A of ∂f

∂y , and this kind of methods are calledMethods with Inexact Jacobian or W-Methods for short.


OutlineIntroduction




The order conditions up to order 3 are showed as in Table 5.

order RHS Φj(t)

1 1 1

2 12

∑k αjk

2 0∑

k γjk

3 13

∑k,l αjkαjl

3 16

∑k,l αjkαkl

3 0∑

k,l αjkγkl

3 0∑

k,l γjkαkl

3 0∑

k,l γjkγkl

Table: order conditions for W-Rosenbrock


OutlineIntroduction



Computational efficiency

Here we compare ROS2 and Crank-Nicolson method for a Burger’sequation with the following form

ut + (u2

2)x = εuxx + g(u, x , t)

where

g(u, x , t) = (cos(x + t)− cos2(x + t) + sin(x + t))u + cos(x + t)u2

with the initial condition

u(x , 0) = esin(x).

The accurate solution is computted by setting ∆t = 0.0001.Here we discretize u in space with 1000 points. The computingregion is [−π, π], with end time T = 1.0.


OutlineIntroduction



Computational efficiency

When ε = 1.0, the L-2 errors are listed in table 6 and the time costare listed in table 7.

Table: L-2 error at T = 1.0, ∆x = 2π/999, ∆t = T/N

N 10 20 40 80 160

ROS2 1.357e-03 2.951e-04 6.574e-05 1.279e-05 4.413e-06

CN 7.354e-04 1.806e-04 4.203e-05 8.590e-06 5.221e-06

Table: time cost (seconds) at T = 1.0, ∆x = 2π/999, ∆t = T/N

N 10 20 40 80 160

ROS2 7 13 28 55 106

CN 30 48 85 148 245


OutlineIntroduction



ConclusionReferences

To sum up, Rosenbrock methods have the following advantages

I Do not need iterations for each step.

I Can be made L-stable, suitable for stiff problems.

I Easily derived order conditions with rooted trees.

and disadvantages

I Problems at dealing with boundary conditions for interiorstages.


OutlineIntroduction




I H.H. Rosenbrock, Some general implicit processes for thenumerical solution of differential equations, Comput. J. , 5,329330, 1963.

I J.G. Verwer, An analysis of Rosenbrock methods for nonlinearstiff initial value problems, SIAM J. Numer. Anal. 19,155-170, 1982.

I J.G. Verwer, Instructive experiments with someRunge-Kutta-Rosenbrock methods, Comp. and Math. withAppls. 7, 217-229, 1982.

I J.G. Verwer, S. Scholz, J.G. Blom and M. Louter-Nool, Aclass of Runge-Kutta-Rosenbrock methods for stiff differentialequations, ZAMM 63, 13-20, 1983.


OutlineIntroduction




I E. Hairer and G. Wanner, Solving Ordinary DifferentialEquations II: Stiff and Differential-Algebraic Equations,Springer Series in Computational Mathematics 14, SpringerVerlag, 2nd edition, 1996.

I J.Rang and L.Angermann, New Rosenbrock W-Methods ofOrder 3 for Partial Differentail Algebraic Equations of Index 1,BIT Numerical Mathematics 45, 761-787, 2005.


Rosenbrock Methods€¦ · 3 = 5:000000000000e 01 ^b 3 = 1:388888888888e 01 Table:ROS3w (with a...

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Transcript of Rosenbrock Methods€¦ · 3 = 5:000000000000e 01 ^b 3 = 1:388888888888e 01 Table:ROS3w (with a...