Runge-Kutta Methods
(fun for everyone)
Dave Goulet
goulet@rose-hulman.edu
Mathematics, RHIT
May 15, 2013
1 / 55
Outline
Introduction
Calculus and Differential Equations
Trees and Combinatorics
Linear Algebra, Complex Variables, & Geometry
Continuous Optimization
Algorithms
Beyond Runge-Kutta
2 / 55
Runge-Kutta and Runge-Kutta-like Methods
High Quality Solvers:
I
Matlab
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Fortran
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I
I
ode45
ode23
ode23s∗
DOPRI
ROS∗
RODAS∗
many others
Mathematica (NDSolve)
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I
I
ExplicitRungeKutta
ImplicitRungeKutta
SymplecticPartitionedRungeKutta
Maple (dsolve, type=numeric)
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I
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rkf45
ck45
dverk78
rosenbrock∗
Pedagogical Solvers:
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Maple’s classical uses six low quality textbook RK methods.
3 / 55
What’s wrong with ode45 or dsolve?
Consider the probelm y’=-50(1-y) with y(0)=0.
1
y
0.8
0.6
0.4
0.2
0
exact
ode45
0
2
4
6
8
10
t
12
14
16
18
20
ode45 is adaptive. What values for the time steps did it choose?
4 / 55
Explicit RK Methods are Not Suitable for All Problems
ode45 used 325 steps and 2077 functional evaluations.
error
y
1
0.5
0
1
h
0
0.1
0
0
2
4
6
8
10
t
12
14
16
18
20
Multiply work by 106 or 109 for many realistic problems.
5 / 55
Calculus &
Differential Equations
6 / 55
Euler’s Method: Approximation of Derivatives
Begin with a first order ODE.
y 0 (x) = f (x, y (x))
Approximate slopes in a simple way.
y 0 (xn ) ≈
y (xn+1 ) − y (xn )
yn+1 − yn
yn+1 − yn
≈
=
xn+1 − xn
xn+1 − xn
h
Replace the ODE with a difference equation.
yn+1 − yn
= f (xn , yn )
h
The Forward Euler Method (1768):
yn+1 = yn + hf (xn , yn )
xn+1 = xn + h
7 / 55
Euler’s Method: Approximation of Integrals (skip)
Begin with a first order ODE, integrating both sides.
y 0 (x) = f (x, y (x))
=⇒
y (xn + h) − y (xn ) =
Z
xn +h
f (x, y (x))dx
xn
Approximate the integral in a simple way.
Z
xn +h
xn
f (x, y (x))dx ≈ hf (xn , y (xn ))
Replace the ODE with a difference equation.
yn+1 − yn = hf (xn , yn )
The Forward Euler Method:
yn+1 = yn + hf (xn , yn )
xn+1 = xn + h
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How Good is Forward Euler?
Accuracy If FWE is applied to the exact solution, y (xn ), how large
will the error, |yn+1 − y (xn+1 )|, be?
Stability Does FWE amplify small errors in an undesirable way? Do
decaying solutions decay as they should?
Quality As FWE iterates, does it provide information on and
control over accuracy and stability? For a given accuracy,
how computationally efficient is FWE?
9 / 55
Accuracy of Forward Euler: Taylor Polynomials
Apply FWE to an autonomous ODE, y 0 = f (y ). Compute the error.
h2
yn+1 − y (xn+1 ) = yn + hf (yn ) − y (xn ) + hy 0 (xn ) + y 00 (c)
2
2
h
= h(f (yn ) − y 0 (xn )) − y 00 (c)
2
h2 00
= − y (c)
2
= O(h2 )
Forward Euler is a first order method.
Method is p th order if truncation error is O(hp+1 ).
10 / 55
Stability of Forward Euler: The Dahlquist Test Problem
The ODE y 0 = λy has exponentially decaying solutions when <(λ) < 0.
y (x + h) = e hλ y (x) = e z y (x)
Apply FWE to the ODE.
yn+1 = yn + hλyn = (1 + hλ)yn ≡ (1 + z)yn
Numerical solution decays iff |1 + z| < 1. A circle is not a half plane!
11 / 55
Stabilty Region for Forward Euler
Exact
FWE
1
|1 + z| < 1
−1
−2
0
12 / 55
Numerical Solutions Using Forward Euler
dy/dt=λ y
4
Exact
z=-2.5
z=-1.5
z=-0.5
2
0
−2
−4
0
0.2
0.4
0.6
0.8
1
t
13 / 55
Quality of Forward Euler
I
Without the exact solution, how can you gauge accuracy?
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Without the exact solution, how can you detect instability?
I
How can you control step size to achieve accuracy and stability?
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If your implementation efficient?
14 / 55
The Butcher Table: 2-stage examples
Idea: estimate ∆y using slope estimates in multiple locations.
Heun’s Method (1900)
Generic 2-Stage Method
k1 = hf (xn , yn )
k1 = hf (xn + c1 h, yn + a11 k1 + a12 k2 )
k2 = hf (xn + h, yn + k1 )
1
1
yn+1 = yn + k1 + k2
2
2
k2 = hf (xn + c2 h, yn + a21 k1 + a22 k2 )
yn+1 = yn + b1 k1 + b2 k2
Butcher tables organize the coefficients.
0
1
0
1
0
0
1
2
1
2
c1
c2
a11
a21
a12
a22
b1
b2
Where ci = ai1 + ai2 , the ith row sum of A.
xn + ci h is the location where the ith slope estimate is being estimated.
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The Butcher Table: General Case
An s-stage method.
Where ci =
c
P
A
b
T
j
c1
c2
..
.
a11
a21
..
.
a12
a22
..
.
···
···
..
.
a1s
a2s
..
.
cs
as1
as2
···
ass
b1
b2
···
bs
aij = (A1)i .
k = hf (xn + hc, yn 1 + Ak)
yn+1 = yn + b T k
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Accuracy: Computing Truncation Errors with Calculus
Apply RK method to y 0 = f (y ). Compute the truncation error.
yn+1 − y (xn + h) = yn +
X
j
bj kj −
X hj
j
j!
y (j) (xn )
Derivatives can be computed successively:
y 0 = f (y )
y 00 = f 0 (y )y 0 = f 0 f
y 000 = f 00 ff + f 0 f 0 f
y (4) = f 000 fff + f 00 f 0 ff + f 00 ff 0 f + f 00 ff 0 f + f 0 f 00 ff + f 0 f 0 f 0 f
Can be automated, but still tedious.
Expansion of kj in Taylor series is hard.
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Accuracy: Tedious Taylor Series Expansions. (skip)
Example: 2-stage explicit method (easiest case).
yn+1 = yn + b1 k1 + b2 k2
= yn + b1 hf + b2 hf (y + a21 hf )
= yn + h(b1 + b2 )f + h2 b2 a21 f 0 f +
y (x + h) = yn + hf +
h3
2 00 2
b2 a21
f f + ...
2
h2 0
h3
f f − (f 00 f 2 + (f 0 )2 f ) + . . .
2
6
Order 1 requires b1 + b2 = 1. Order 2 requires b2 a21 = 1/2. Order 3
isn’t possible.
yn+1 − y (x + h) =
h3
(3a21 − 2) f 00 f 2 − 2(f 0 )2 f + . . .
12
Deriving order conditions this way is hard.
Order 14 methods with 35 stages have been constructed.
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Graph Theory &
Combinatorics
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Order Conditions: Rooted Trees
y0 = f
y 00 = f 0 f
y 000 = f 00 ff + f 0 f 0 f
y (4) = f 000 fff + 3f 00 ff 0 f + f 0 f 00 ff + f 0 f 0 f 0 f
Butcher (1963) utilized correspondences between rooted trees and
elementary differentials.
q
f
q
q
q
q
q
q q
A q
q
q
q
q
q q
A q
q
q
q q
A q
ff 0 f 0 f 0 f f 00 ff f 0 f 0 f 0 f f 0 f 00 ff f 00 ff 0 f
q q q
@q
f 000 fff
Butcher’s idea: Use combinatoric techniques to compute tree data and
Taylor series expansions. Prove equivalences.
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Order Conditions: Butcher’s Theorem
The tree densities, γ, and elementary weights, Φ, satisfy Φ(t) =
q q
A q
q
q
q
q
q q
A q
q
q
q q
A q
q q q
@q
6
3
24
12
8
4
f 0f 0f
f 00 ff
f 0f 0f 0f
f 0 f 00 ff
f 00 ff 0 f
f 000 fff
q
q
q
q
q
q
1
2
f
ff 0
P
bi
P
bi ci
1
γ(t) .
P
bi aij cj
P
bi ci2
P
bi aij ajk ck
P
bi aij cj2
P
bi aij ci cj
P
bi ci3
I
Conditions easily derivable by hand or by CAS (Sofroniu 1994).
I
Solving multivariable polynomial equations is hard.
I
Solving algorithms exist, e.g., using a Grobner basis (Sofroniu 2004).
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Example: An 8th Order Condition. (skip)
u1
u2
u1
A
A Au
4
@
@
t=
@u8
u1
u1
A
A Au3
γ(t) = 8 ∗ 4 ∗ 3 ∗ 2 ∗ 1 ∗ 1 ∗ 1 ∗ 1 = 192
X
Φ(t) =
bi (aij cj2 )(aik (ck (akl cl )))
i,j,k,l
X
bi aij aik akl cj2 ck cl =
i,j,k,l
1
192
Only 114 more 8th order conditions to go!
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Example: 3-stage 3rd Order Explicit Method
General case and an example.
0
c2
c3
0
a21
a31
0
0
a32
0
0
0
b1
b2
b3
0
1
1
Order 0:
c2 = a21 ,
Order 1:
b1 + b2 + b3 = 1
Order 2:
b2 c2 + b3 c3 = 1/2
Order 3:
b2 c22 + b3 c32 = 1/3,
0
1
0
0
1
2
1
2
0
0
0
2
3
0
1
3
c3 = a31 + a32
b3 (a31 c1 + a32 c2 ) = 1/6
4 equations in 6 unknowns: use free variables to improve quality.
The example satisfies several important additional criteria.
23 / 55
How much accuracy is possible? (skip)
Butcher’s approach reveals limitations on s-stage order p methods.
Figure summarizes many important theorems.
exp.
imp.
exp.
imp.
100
10
free parameters
maximum order
12
8
6
4
60
40
20
2
0
80
0
5
stages
10
0
0
5
stages
10
24 / 55
Linear Algebra,
Complex Variables,
& Geometry
25 / 55
Stability Functions
RK methods are of the form
k = hf (xn + hc, yn 1 + Ak),
yn+1 = yn + b T k
Apply to test problem, y 0 = λy with <(λ) < 0.
k = hλ(yn 1 + Ak),
yn+1 = yn + b T k
Define z = hλ.
yn+1 = 1 + zb T (I − zA)−1 1 yn
= Φ(z)yn
Φ(z) is the Stability Function.
Exact update would be yn+1 = e z yn . How does Φ(z) compare to e z ?
26 / 55
Stability Functions: Padé Approximations
The stability function is rational (proof by Cramer’s Rule).
Φ(z) = 1 + zb T (I − zA)−1 1 =
det(I − zA + z1b T )
det(I − zA)
Idea: Choose Φ(z) to be a Padé approximation of e z :
Approx
1+z
Method
Forward Euler
Order
1
Stages
1
Type
explicit
1 + z + z 2 /2
1
1−z
2+z
2−z
6 + 2z
6 − 4z + z 2
12 + 6z + z 2
12 − 6z + z 2
Heun
2
2
explicit
Backward Euler
1
1
implicit
Midpoint Rule
2
1
implicit
Radau IIa (1969)
3
2
implicit
Gauss
4
2
implicit
27 / 55
Padé Approximations
1
0.5
F. Euler
Heun
B. Euler
Midpoint
Radau IIa
Gauss
0
−0.5
−1
−20
−15
−10
−5
0
28 / 55
Stability Functions
When applied to y 0 = λy , how alike are the exact and numerical solution?
y (xn + h) = e z y (xn )
yn+1 = Φ(z)yn
Geometric Method 1: Stability Regions
I
S = {z ∈ C : |Φ(z)| < 1} is the Stability Region, i.e., the region
where the numerical solution will decay.
I
Compare this to the left half plane, |e z | < 1 for <(z) < 0.
Geometric Method 2: Order Stars
I
Ideally, |e z | ≈ |Φ(z)|, so that numerical solutions and exact solutions
decay at similar rates.
I
Φ(z)e −z is the Order Star. Study the set {z ∈ C : |Φ(z)e −z | > 1}.
29 / 55
Stability Regions and Order Stars: An Example
Consider RK4 (Kutta, 1901), Φ = 1 + z + 12 z 2 + 16 z 3 +
k1 = hf (xn , yn )
0
1
2
1
2
1
1 4
24 z .
1
2
k2 = hf (xn + h/2, yn + k1 /2)
1
2
0
0
0
1
1
6
1
3
1
3
k3 = hf (xn + h/2, yn + k2 /2)
k4 = hf (xn + h, yn + k3 )
1
6
yn+1 = yn + (k1 + 2k2 + 2k3 + k4 )/6
6
6
3
3
0
0
−3
−3
−6
−6 −3 0
3
6
−6
−6 −3 0
3
6
30 / 55
Geometry of Padé Approximations: Stability Regions
Midpoint
F. Euler
B. Euler
6
6
6
3
3
3
0
−3
0
−3
0
−3
−6
−6 −3 0 3
Heun
6
−6
−6 −3 0 3
Gauss
6
−6
6
6
6
3
3
3
0
−3
0
−3
0
−3
−6
−6 −3 0
3
6
−6
−6 −3 0
3
6
−6
−6 −3 0 3
Radau
6
−6 −3 0
6
3
31 / 55
Geometry of Padé Approximations: Order Stars
Midpoint
F. Euler
B. Euler
6
6
6
3
3
3
0
−3
0
−3
0
−3
−6
−6 −3 0 3
Heun
6
−6
−6 −3 0 3
Gauss
6
−6
6
6
6
3
3
3
0
−3
0
−3
0
−3
−6
−6 −3 0
3
6
−6
−6 −3 0
3
6
−6
−6 −3 0 3
Radau
6
−6 −3 0
6
3
32 / 55
The Geometric Perspective
accuracy and stability ←→ geometry of approximations to e z .
Theorem (Wanner et al., 1978)
I
The number of lobes in the order star meeting at the origin is equal
to the order of the method.
I
If the poles of Φ are in the right half plane and if the order star
doesn’t intersect the imaginary axis, then the stability region
contains the entire left half plane.
6
6
3
3
0
0
−3
−3
−6
−6 −3 0
3
6
−6
−6 −3 0
Dormand & Prince, 1980 (Matlab’s ode45).
3
6
33 / 55
Accuracy and Stability are not Independent
Define B =diag(b) and C =diag(c) with


 
 
a11 · a1s
b1
1
·  b =  ·  1 = ·
A= ·
as1 · ass
bs
1
 
c1
c =  ·  = A1
cs
The order conditions up to 4th :
bT 1 = 1
b T A1 = 1/2
b T C 2 1 = 1/3
b T A2 1 = 1/6
b T C 3 1 = 1/4
b T CA2 1 = 1/8
b T AC 2 1 = 1/12
b T A3 1 = 1/24
Theorem
Φ = 1 + zb T 1 + z 2 b T A1 + z 3 b T A2 1 + z 4 b T A3 1 + . . .
Order conditions cause Φ(z) to resemble e z , for small z.
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Implicit Methods can be Very Stable. (skip)
Theorem
If A is singular, then
lim
<(z)→−∞
|Φ(z)| =
6 0
If A is nonsingular, then the Laurent series expansion of Φ(z) is
Φ(z) = 1 − b T A−1 1 +
∞
X
(−1)n
n=1
zn
b T (A−1 )n+1 1
and
lim
<(z)→−∞
|Φ(z)| = 0 if and only if
b T A−1 1 = 1
35 / 55
Explicit Methods Lack Stability. (skip)
Theorem
When the method is explicit, A is singular, and the stability function is a
polynomial.
Φ(z) =
det(I − zA + z1b T )
= det(I − zA + z1b T )
det(I − zA)
so
lim
<(z)→−∞
|Φ(z)| = ∞
Instability is guaranteed.
How far can the stability boundary be extended?
36 / 55
Continuous Optimization
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Optimizing the Stability Region: Adding Stages. (skip)
s=p
(s,p)=(2,2),(2,3),(3,3)
s=p+1
4
0
−4
−8
−4
0 −8
−4
0
−8
−4
0
4
0
−4
−32
−16
0
38 / 55
Optimizing Stability Regions. (skip)
Fehlberg, 1970
6 stages, 5th order, explicit
(Maple’s default)
Bogacki & Shampine, 1996
8 stages, 5th order, explicit
6
6
3
3
0
0
−3
−3
−6
−6
−6
−3
0
3
6
−6
−3
0
3
6
39 / 55
Optimizing the CFL Condition. (skip)
Nonlinear Optimization Problem
Find an explicit RK method with maximal positive CFL number given:
1. order = p and stages = s.
2. The method satisfies certain efficiency constraints.
3. The method is Total Variation Diminishing.
Theorem (Ruuth and Spiter, 2004)
I
Heun is the optimal 2nd order 2-stage method.
I
RK3TVD is the optimal 3rd order 3-stage method.
I
There is no optimal nth order n-stage method for
n ≥ 4.
Solution: Add more stages!
0
1
0
1
0
0
1
2
1
2
0
1
0
1
0
0
1
2
1
4
1
4
0
0
0
1
6
1
6
2
3
40 / 55
Optimizing the CFL Condition: Explicit SSP
Ruuth and Spiter’s SSP(9,5) with CLF number 2.695.
0
0.234
0.285
0.297
0.611
0.546
0.760
0.714
0.914
0
0.234
0.110
0.050
0.143
0.045
0.058
0.114
0.096
0
0
0.174
0.079
0.227
0.071
0.093
0.180
0.151
0
0
0
0.167
0.240
0.143
0.109
0.132
0.111
0
0
0
0
0
0.298
0.227
0.107
0.090
0
0
0
0
0
0.013
−0.010
−0.129
−0.108
0
0
0
0
0
0
0.281
0.133
0.112
0
0
0
0
0
0
0
0.175
0.147
0
0
0
0
0
0
0
0
0.312
0
0
0
0
0
0
0
0
0
0.088
0.102
0.111
0.158
−0.060
0.197
0.071
0.151
0.179
4
0
−4
−8
−4
0
41 / 55
Choosing the Right Method is Important
ode45 struggles because of stability problems.
6
1
3
0
0.5
−3
−6
6
−6 −3 0
3
6
0
0
5
10
0
0.5
−3
−6 −3 0
3
6
0
vs.
similar to ode23s
Implicit Rosenbrock
4th order
528 function evals.
h chosen for accuracy
1
3
−6
ode45
Explicit RK
5th order
2077 function evals.
h chosen for stability
0
5
10
42 / 55
Algorithms
43 / 55
High Quality RK Methods Require High Quality Algorithms
“The numerical solution of ordinary differential equations is an old topic
and, perhaps surprisingly, methods discovered around
the turn of the century are still the basis of
the most effective, widely used codes for this
purpose. Great improvements in efficiency have been made, but it is
probably fair to say that the most significant
achievements have been in reliability,
convenience, and diagnostic capabilities
.”-
Shampine and Gear, 1979
44 / 55
Efficiency: Embedding and the FSAL Property
Embedded: two different reconstructions from the same k values.
First Stage As Last: s-stage method is effectively s-1 stages.
Sofroniu, 2004
Bogaki-Shampine, 1989
Mathematica: ExplicitRungeKutta 2,1
Matlab: ode23
0
1
1
0
0
1
0
0
1
2
1
2
0
0
0
1
2
1
2
0
2nd order
1
− 61
1
6
1st order
1
2
3
4
1
0
1/2
0
0
0
4
9
0
0
0
0
1
3
4
9
0
3rd order
1
4
1
3
1
8
2nd order
2
9
3
4
1
3
2
9
7
24
0
0
0
ks = f (yn+1 ) so that k1 of the next time step is already computed.
The lower order method helps with error estimation and control.
45 / 55
Closer Look at FSAL
Suppose the nth set of k values is computed, and the solution updated.
k1 = hf (yn )
k2 = hf (yn + k1 )
k3 = hf (yn + k1 /2 + k2 /2)
yn+1 = yn + k1 /2 + k2 /2
When the next set of k values is computed, the new k1 will be the old k3 .
k1 = hf (yn+1 )
= hf (yn + k1 /2 + k2 /2)
= k3
A k value is reused, so an n stage method is effectively n − 1 stages.
46 / 55
Reliability: Adjusting Step Size for Accuracy
Use an order p + 1 method with an embedded order p method.
yn+1 − y (xn + h) = ahp+1 + O(hp+2 )
ȳn+1 − y (xn + h) = O(hp+2 )
Subtract
yn+1 − ȳn+1 = ahp+1 + O(hp+2 )
Suppose ĥ achieves the optimal error tolerance.
Etol = aĥp+1 + O(ĥp+2 )
Divide
Etol
=
yn+1 − ȳn+1
ĥ
h
!p+1
+ O(h + ĥ)
Adjust
1
p+1
Etol
ĥ = h yn+1 − ȳn+1 47 / 55
Step Size Adjustment is an I Controller
Consider the step size control formula.
1
p+1
Etol
ĥ = h yn+1 − ȳn+1 Compute logarithms and rearrange.
1
(log |Etol | − log |yn+1 − ȳn+1 |)
p+1
Controller Change = (Integral Gain) (Set Point − Process Variable)
log(ĥ) − log(h) =
u[n] − u[n − 1] = Ki e[n]
Making it a PI Controler can provide better convergence.
u[n] − u[n − 1] = Ki e[n] + Kp (e[n] − e[n − 1])
Gustafsson, 1991
ĥ = h 0.7
p+1
Etol
yn+1 − ȳn+1 0.4
yn − ȳn p+1
Etol 48 / 55
PI Control of DOPRI54
ode45
dopri54, i control
dopri54, pi control
49 / 55
Diagnostic: Detecting the Stability Border
Hairer & Wanner, 1996: For DOPRI54, the quantity
−0.08536k1 + 0.088k2 − 0.0096k3 + 0.0052k4 + 0.00576k5 − 0.004k6 h
yn+1 − ȳn+1
is < 1 near the stability border and > 1 away from it.
z hovers the border of the
smaller stability region
switching → 223 function evals.
50 / 55
Summary
I
Software packages provide general purpose solvers.
I
Pick the method which is right for your problem.
I
Custom methods can be superior.
I
There are tradeoffs between accuracy and stability.
I
Efficiency, reliability, diagnostics, and control are key.
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Parallel Peer Schemes
RK schemes do this:
kn = hf (yn 1 + Akn )
yn+1 = yn + b T kn
The kn,i are computed in serial.
Parallel Peer schemes do something like this:
kn = hf (yn 1 + Akn−1 )
yn+1 = yn + b T kn
kn,i is computed independently of kn,j , so parallelization is natural.
kn,i = hf (yn + ai1 kn−1,1 + . . . + ai,s kn−1,s )
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Potential for high number of stages, s= number of nodes.
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Can be made very stable with no implicit equations to solve.
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Can be made superconvergent, p > s. (Weiner et al. 2008).
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Rosenbrock Methods: How Implicit is Implicit Enough?
Implicit methods require too much computation. One stage example:
k = hf (yn + a11 k)
If f is costly to evaluate, iterative methods of solution,e.g., Newton’s
method, are expensive.
Rosenbrock (1963): k is usually small, so use a Taylor polynomial.
k = hf (yn ) + hf 0 (yn )a11 k
I
Easy to generalize, easy to derive order conditions.
I
I
Implicit, but linear. Linear solves are cheap.
Stability regions are the same as implicit methods.
I
Parallel Peer versions exist (Weiner et al. 2005).
I
Generalizations approximate f 0 , e.g., Matlab’s ode23s (1997).
53 / 55
Multistage Rosenbrock Methods
Start with
k1 = hf (yn ) + hf 0 (yn )γ1 k1
Collect k1 on the left.
(1 − hγ1 f 0 (yn ))k1 = hf (yn )
Add more stages.
(1 − hγ2 f 0 (yn ))k2 = hf (yn + a21 k1 ) + d21 k1
(1 − hγ3 f 0 (yn ))k3 = hf (yn + a31 k1 + a32 k2 ) + d31 k1 + d32 k2
General case:
(1 − hγi f 0 (yn ))ki = hf (yn + aiT k) + hdiT k
For systems, each stage requires one linear solve and each time step one
Jacobian computation.
(I − hγi J)ki =hf (yn + aiT k) + hdiT k
Choosing all γi equal means only one LU decomposition per time step.
54 / 55
Rosenbrock Example
Example used in the talk. (4th/3rd order embedded)
k1 = hf (yn ) + hf 0 (yn )k1 /2
k2 = f (yn + k1 ) − 2k1 + hf 0 (yn )k2 /2
k3 = f (yn + (3k1 + k2 )/8) − k1 /4 + hf 0 (yn )k3 /2
k4 = f (yn + (3k1 + k2 )/8) + 3k1 /4 − k2 /8 − k3 /2 + hf 0 (yn )k3 /2
0
1
0
1
0
0
1
2
1
2
3
8
3
8
1
8
1
8
0
0
0
0
0
0
0
0
1
6
1
6
0
2
3
1
6
1
6
2
3
0
1
2
0
−2
− 14
0
3
8
1
2
− 18
0
0
1
2
1
−2
0
0
0
1
2
55 / 55