Bifurcation Analysis for Time Steppers
Laurette Tuckerman
[email protected]
T HE T HREE T OOLS OF
C OMPUTATIONAL F LUID DYNAMICS
Time stepping:
∂tU = LU + N (U )
Steady state solving:
0 = LU + N (U )
Linear stability analysis:
λu = Lu + NU u
Heat Equation
2
∂tu = ∂xx
u
kX
max
uk (t) sin kx
u(x, t) =
k=1
2
∂tuk = −k uk
E XACT SOLUTION
uk (t + ∆t) = e−k
2
∆t
uk (t)
E XPLICIT E ULER
uk (t + ∆t) = uk (t) − k2 ∆tuk (t)
= (1 − k2 ∆t)uk (t)
As kmax → ∞, ∆tmax =
2
2
kmax
→0
I MPLICIT E ULER
uk (t + ∆t) =
−
2
(1 + k ∆t)uk (t + ∆t) =
uk (t + ∆t) =
Matrix version:
u(t + ∆t) =
uk (t)
k2 ∆tuk (t + ∆t)
uk (t)
(1 + k2 ∆t)−1 uk (t)
(I − ∆tL)−1 u(t)
NAVIER -S TOKES E QUATIONS
∂t U =
−(U · ∇)U − ∇P
+ ν∇2U
= −(I − ∇∇−2∇·)(U · ∇)U + ν∇2U
=
N (U )
+ LU
• Time stepping (Direct Numerical Simulation)
∂tU = N (U ) + L U ≡ A(U )
Explicit/Implicit Euler:
U (t + ∆t) = U (t) + ∆t[N (U (t)) + LU (t + ∆t)]
= (I − ∆tL)−1 (I + ∆tN )U (t) ≡ BU (t)
• Steady state solving
0 = N (U ) + L U
Newton’s method:
AU u = A(U )
U ←U −u
• Linear stability analysis:
λu = NU u + L u ≡ AU u
Arnoldi/block power method:
AU ∆t
un+1 = A−1
un
U un or un+1 = e
NU u ≡ −(U · ∇)u − (u · ∇)U
AU u = NU u + Lu
S TEADY STATE SOLVING
0 = F (U )
Newton’s method
0 = F (U − u) ≈ F (U ) − DF (U )u
DF (U )u = F (U )
U ←U −u
How to solve linear systems?
1) Direct: Gaussian Elimination = LU + Backsolve
Storage: M 2
Time: M 3
For 3D case with Mx = My = Mz = 102 , we have M = 106
M 2 = 1012
M 3 = 1018
2) Iterative: Conjugate Gradient methods
Use only matrix-vector products u → Au
For an arbitrary matrix,
Each product u → Au requires M 2 operations
Convergence requires M iterations
Can gain:
If A is structured or sparse, then u → Au takes ∼ M ops
If A is well-conditioned, convergence takes few iterations.
M3
C ONDITIONING
A is well conditioned if its eigenvalues lie close together.
The best conditioned matrix is a multiple of the identity.
The condition number is, roughly,
max eig of A κ(A) ∼ min eig of A P RECONDITIONING
Au = v
P Au = P v
where:
P is easy to act with
P A is better conditioned than A
Extreme cases:
P = I (easy to act with but no improvement)
P = A−1 (perfect preconditioner but impossible)
Our case:
AU u = Lu + NU u
= ∇2 u − (U · ∇)u − (u · ∇)U
For 3D case with Mx = My = Mz = 100,
eigs of L range from ∼ −1 to −(Mx2 + My2 + Mz2 ) = −30000.
Idea: L is the main source of difficulty =⇒ Use P = L−1
Question: Where do we get L−1?
Answer: Already present in a timestepping code!
U (t + ∆t) − U (t) = (I − ∆tL)−1 (I + ∆tN ) − I U (t)
= (I − ∆tL)−1 [I + ∆tN − (I − ∆tL)] U (t)
= (I − ∆tL)−1 ∆t(N + L) U (t)
(B − I)U (t) ≡ U (t + ∆t) − U (t) has same roots as (N + L)!
•In time-stepping, ∆t must be small (∼ 10−2 ) to insure
(I − ∆tL)−1 (1 + ∆tNU ) ≈ exp((L + NU )∆t).
•Here, ∆t plays algebraic role, and can (should) be large (& 102 ).
•∆t interpolates between P = I and P = L−1 .
• Called Stokes preconditioning
O NE N EWTON S TEP
(I − ∆tL)−1 ∆t(NU + L) u = (I − ∆tL)−1 ∆t(N + L) U
(I − ∆tL)−1 (I + ∆tNU ) − I u = (I − ∆tL)−1 (I + ∆tN ) − I U
{z
}
{z
}
|
|
difference between two
difference between two
widely spaced consecutive
widely spaced consecutive
linearized timesteps
timesteps
Solve linear system with B I -CGSTAB
H.A. van der Vorst, Bi-CGSTAB: A fast and smoothly converging variant of
Bi-CG for the solution of nonsymmetric linear systems, SIAM J. Sci. Stat.
Comput. 13, 631 (1992)
∼ 30 lines of code
C ONTINUATION
Goal
0 = RN (U ) + LU
0 = p(U, R) − p̄ where
Ui some component
R
Newton step
(U, R) not solution, so try (U − u, R − r)
0 = (R − r)N (U − u) + L(U − u)
= RN (U ) + LU − RNU u − rN (U ) − Lu + O(r, u)2
Ui − p̄ − ui
0 = p(U − u, R − r) − p̄ =
R − p̄ − r
RNU + L
N (U )
0 0...0 1 0...0
1
|
{z
}
or
u
r

RN
(U
)
+
LU

Ui − p̄
=
R − p̄

If p(U, R) = R (i.e. set Reynolds number),
then set R = p̄, r = 0 and get previous case:
L−1 [RNU + L] [u] = L−1 [RN (U ) + LU ]
If p(U, R) = Ui , then must solve extended system for (u, r).
L−1 (RNu + L) L−1 N (U )
0 0...0 1 0...0
0
u
r
=
L−1 (RN (U ) + LU )
Ui − p̄

Set ui = Ui − p̄

Use only vectors and
−1
Calculate L (RNU + L)u
 operators of length M
Add L−1 N (U )r
T RAVELING WAVES : U (x − Ct, y, z)
Goal
0 = C∂xU + N (U ) + LU
0 = p(U ) − p̄
C∂x + NU + L ∂xU
0 0...0 1 0...0 0
u
c
=
C∂xU + N (U ) + LU
Ui − p̄
A XISYMMETRIC S PHERICAL C OUETTE F LOW WITH
σ = 0.18
C.K. Mamun & L.S. Tuckerman (1995)
A XISYMMETRIC S PHERICAL C OUETTE F LOW
WITH σ = 0.18
A XISYMMETRIC S PHERICAL C OUETTE F LOW
WITH σ = 0.18
L INEAR S TABILITY A NALYSIS
λu = Lu + NU u
How to calculate eigenpairs (λ, u)?
1) Direct: Diagonalisation = QR decomposition
Storage: M 2
Time: M 3
For 3D case with Mx = My = Mz = 102 , we have M = 106
M 2 = 1012
M 3 = 1018
2) Iterative: Calculate a few desired eigenpairs.
Use only matrix-vector products u → Au
To diagonalise an arbitrary matrix,
Each product u → Au requires M 2 operations
Generating M eigenpairs requires M iterations
Can gain:
If A is structured or sparse, then u → Au takes ∼ M ops.
Aim method at desired eigenvalues.
M3
M ATRIX T RANSFORMATIONS
If A u = λ u
then f (A) u = f (λ) u
f (A) =
fj
P
chosen
ically
desired
to
j
fj Aj
dynamextract
eigenvalues:
principle of A RPACK
f (A) = eA∆t
f (A) = A−1
(Sorensen et al.)
E XPONENTIAL P OWER M ETHOD
un+1 = (I−∆tL)−1(I+∆tNU )un ≈ e∆t(L+NU )un
Approximation valid for ∆t ≪ 1
Time-stepping linearized evolution equation
Enhancement factor at each iteration is
e∆tλ1 where λ1 > λ2 > · · ·
∆tλ & 1
e 2 I NVERSE P OWER M ETHOD
un+1 = (L + NU )−1un
Stokes preconditioning:
(L + NU )un+1
−1
(I − ∆tL) ∆t(L + NU )un+1
(I − ∆tL)−1 [I + ∆tNU − (I − ∆tL)] un+1
(I − ∆tL)−1 (I + ∆tNU ) − I un+1
|
{z
}
difference between two widely spaced
consecutive linearized timesteps
=
=
=
=
un
(I − ∆tL)−1 ∆tun
(I − ∆tL)−1 ∆tun
(I − ∆tL)−1 ∆tun
{z
}
|
one Stokes timestep
Solve with Conjugate Gradient (Bi-CGSTAB) method.
Enhancement factor at each iteration is λλ12 ≫ 1
for λ1 ≈ 0
Can shift to find eigenvalues closest to s
λ2 − s for λ1 ≈ s
λ − s ≫ 1
1
A XISYMMETRIC S PHERICAL C OUETTE F LOW WITH
σ = 0.18
Basic flow at
Re = 650
Leading
eigenvector
Inverse Power Method on Spherical Couette Flow
∆t =100, 10, 1, 0.1, 0.01
CGcrit = 10−7 (•, •), 10−9 (△, △)
s = −0.1, 0
s = −0.152, −0.15, −0.1, 0
M = 4096, 16384
S UMMARY
Time stepping
Steady-state solving Linear stability analysis
∂t U = (N + L)U
0 = (N + L)U
λu = (NU + L)u
Implicit/explicit Euler
Newton
Inverse power/Arnoldi
U (t + ∆t) = BU (t)
= (I − ∆tL)−1
(I + ∆tN )U (t)
(NU + L)u
= (N + L)U
U ←U −u
(NU + L)un+1 = un
≡ P (I + ∆tN )U (t)
AU u = AU
P AU u = P AU
AU un+1 = un
P AU un+1 = P un
3−4
Newton steps
3−4
Inverse Arnoldi steps
200 BiCGSTAB
iters/step
200 BiCGSTAB
iters/step
B OSE -E INSTEIN C ONDENSATION
Ultra-cold coherent state of matter
Predicted by Bose (1924) and Einstein (1925)
Realized experimentally by Cornell, Ketterle, Wieman (1995)
Nobel prize (2001)
Gross-Pitaevskii / Nonlinear Schrödinger Equation
1
∂tΨ = i [ ∇2 + µ − V (r) − a|Ψ|2]Ψ
|
{z
}
|2{z }
N
L
V (x) =
1
2
|ω · x|2 =
1
2
(ωr r 2 + ωz z 2 )
(cylindrical trap)
Spatial discretisation up to M = 102 × 102 × 102 = 106
Eigenvalues, energies determine decay rates of condensate.
Hamiltonian Systems
f (A) = A
f (A) = eA∆t
f (A) = A−1
f (A) = A−2
S TEADY STATE SOLVING :
0 = LΨ + N (Ψ)
Newton’s method + Stokes preconditioning + B I CGSTAB
L INEAR STABILITY OF STEADY STATE Ψ:
1
∂t ψ = i [( ∇2 + µ − V (r))ψ − aΨ2 (2ψ + ψ ∗ )]
2
A
R
ψ
ψI
≡
I
0
−(L + DN )
L + DN R
0
DN R ≡ µ − V (x) − 3aΨ2
DN I ≡ µ − V (x) − aΨ2
R
ψ
ψI
ψR
ψI
=
A2
R −(L + DN I )(L + DN R )
0
ψ
R
I
0
−(L + DN )(L + DN )
ψI
Inverse square power method
with Stokes preconditioning and shift:
(A2 − s2 I)ψn+1 = ψn
L−2 (A2 − s2 I)ψn+1 = L−2 ψn
Solve with B I CGSTAB
Hamiltonian saddle-node bifurcation
of hyperbolic and elliptic fixed points
p
A
B
0.075
0.075
0.05
0.05
0.025
0.025
0
-0.025
p
Q-
Q+
0
-0.05
-0.05
-0.075
-0.075
-0.2
-0.1
0
0.1
0.2
Q-
Q+
-0.025
-0.2
-0.1
0
C
ø
0.05
0.002
0.025
0.001
Q+
0
-0.4
-0.2
0.2
-0.025
-0.001
-0.05
-0.002
-0.075
-0.2
0.2
D
0.075
p
0.1
q
q
-0.1
0
q
0.1
0.2
A
B
C
Q-
0.4
q
2000
1300
1600
1200
1200
E+
E
E
E+
1100
1000
E
E
900
800
8.0
2.0
2
4.0
2
2+
2+
1.0
0.0
0.0
2
−4.0
800
1000
1200
N
1400
ωz = ωr /5 (cigar)
2
NG
NE
1600
1800
−1.0
1400
1600
N
1800
NG
NE
2000
2200
ωr = ωz /5 (pancake)
S UMMARY
Time stepping
Steady-state solving Linear stability analysis
∂t U = (N + L)U
0 = (N + L)U
λu = (NU + L)u
Implicit/explicit Euler
Newton
Inverse power/Arnoldi
U (t + ∆t) = BU (t)
= (I − ∆tL)−1
(I + ∆tN )U (t)
(NU + L)u
= (N + L)U
U ←U −u
(NU + L)un+1 = un
≡ P (I + ∆tN )U (t)
AU u = AU
P AU u = P AU
AU un+1 = un
P AU un+1 = P un
3−4
Newton steps
3−4
Inverse Arnoldi steps
200 BiCGSTAB
iters/step
200 BiCGSTAB
iters/step