Advection Through Irregular Grids - University of …dfg/dsloan.pdf · Advection Through Irregular...
Transcript of Advection Through Irregular Grids - University of …dfg/dsloan.pdf · Advection Through Irregular...
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Advection Through Irregular Grids
David F Griffiths
&
Mohammed Babatin
with thanks to Phil Gresho
Mathematics Division
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 1/28
Outline
• Regular grids, dispersion & Group velocity.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 2/28
Outline
• Regular grids, dispersion & Group velocity.
• Smoothly varying gridsMapped grids: xj = G(j/N), (G ∈ C ∩ PD)Kreiss & Oliger (73), Giles & Thompkins (83, 85), Vichnevetsky (87, 89).Lighthill (78), Whitham (74)
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 2/28
Outline
• Regular grids, dispersion & Group velocity.
• Smoothly varying gridsMapped grids: xj = G(j/N), (G ∈ C ∩ PD)Kreiss & Oliger (73), Giles & Thompkins (83, 85), Vichnevetsky (87, 89).Lighthill (78), Whitham (74)
• Sensitivity to grid perturbations, Gresho & Sani (98).1D analogue of unstructured grids.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 2/28
The Problem
The PDE:
ut + ux = 0, 0 < x ≤ 1, t > 0
with periodic BCs.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 3/28
The Problem
The PDE:
ut + ux = 0, 0 < x ≤ 1, t > 0
with periodic BCs.
The Methods:Finite Difference/ Finite element discretizations in space such that theylead to systems of ODEs
Mu + Au = 0,
where:M is symmetric p.d.A is skew–symmetric.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 3/28
The Problem
The PDE:
ut + ux = 0, 0 < x ≤ 1, t > 0
with periodic BCs.
The Methods:Finite Difference/ Finite element discretizations in space such that theylead to systems of ODEs
Mu + Au = 0,
where:M is symmetric p.d.A is skew–symmetric.
We then have the energy conservation:
d
dxu
T Mu = 0.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 3/28
The Simplest Method
We use a grid of N points:
0 = x0 ≤ x1 ≤ · · · ≤ xN = 1
and define the grid sizes: hj = xj − xj−1.
FDM
(xj+1 − xj−1)uj + uj+1 − uj−1 = 0,
j = 0 : N with periodic BCs.
• 2nd order FD approximation to G′(s)ut + us = 0, i.e. ut + ux = 0 with x = G(s).
• Lumped linear FEM.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 4/28
Trick 1
On a uniform grid of size h:
ut + ux = 0 ⇒ 2huj + uj+1 − uj−1 = 0
⇓
Define : vj(t) = (−1)juj(t)
⇓
vt − vx = 0 ⇐ 2hvj − (vj+1 − vj−1) = 0
Hence, vj approximates the pde with phase speed −1. The pde is a 1–way waveequation but the method allows waves in both directions.We can perform a similar trick with most methods—we have to identify the eigenvectorcorresponding to stationary solutions (Mathai, 95).
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 5/28
Trick 1
On a uniform grid of size h:
ut + ux = 0 ⇒ 2huj + uj+1 − uj−1 = 0
⇓
Define : vj(t) = (−1)juj(t)
⇓
vt − vx = 0 ⇐ 2hvj − (vj+1 − vj−1) = 0
Hence, vj approximates the pde with phase speed −1. The pde is a 1–way waveequation but the method allows waves in both directions.
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Solution u
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Solution v
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 5/28
Uniform Grid: Modified equation
The FDM 2huj + uj+1 − uj−1 = 0 is a 4th order approximation of
ut + ux +1
6h2uxxx = 0.
This has fundamental solutions given by Airy functions (Hedstrom (75))
Ai(x) =1
2π
∞
−∞
exp i(kx −1
3k3)dk.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 6/28
Uniform Grid: Modified equation
When u(x, 0) is a δ–function at x = x0,
u(x, t) =
�
2
h2t
� 1/3
Ai
�
2
h2t
� 1/3
(x − t − x0)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−40
−20
0
20
40Modified equation: Airy function and its image
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−20
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uj for x > 1/2, (−1)juj for x < 1/2.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 6/28
Uniform Grid: Modified equation
When w(x, t) =
�
∞
xu(x, t)dt (IC is a step function), wj(t) = h
�Ji=j ui(t):
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
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1.5Modified equation: step response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
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1.5Numerical solution at t = 0.4
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 6/28
The Dispersion Relation
On a uniform grid (h), this method has fundamental solutions of the form
uj(t) = eiΘj(t), Phase Function : Θj(t) = 2πkxj − ωt,
where ω (the frequency) is related to k (the wavenumber)by the dispersion relation
ω =sin 2πkh
h, |k| ≤ N/2
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 7/28
The Dispersion Relation
On a uniform grid (h), this method has fundamental solutions of the form
uj(t) = eiΘj(t), Phase Function : Θj(t) = 2πkxj − ωt,
where ω (the frequency) is related to k (the wavenumber)by the dispersion relation
ω =sin 2πkh
h, |k| ≤ N/2
Note that ω is a solution of the eigenvalue problem (Mu + Au = 0)
ωMv = (−iA)v
(iA is Hermitian, so ω is real).
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 7/28
The Dispersion Relation
On a uniform grid (h), this method has fundamental solutions of the form
uj(t) = eiΘj(t), Phase Function : Θj(t) = 2πkxj − ωt,
where ω (the frequency) is related to k (the wavenumber)by the dispersion relation
ω =sin 2πkh
h, |k| ≤ N/2
Note that ω is a solution of the eigenvalue problem (Mu + Au = 0)
ωMv = (−iA)v
(iA is Hermitian, so ω is real).
The pde has corresponding solutions u(xj , t) = ei(2πkxj−ωt) when
ω = 2πk
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 7/28
Dispersion Relations: Examples
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k h
ω h k = N/4
k = N/3Lumped linear FElinear FE
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 8/28
Dispersion Relations: Examples
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ω h k = N/4
k = N/3Lumped linear FElinear FE
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k h
ω h
FD methods of orders 2, 4, ..., 10, 20, 40
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 8/28
Dispersion Relations: Examples
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ω h k = N/4
k = N/3Lumped linear FElinear FE
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k h
ω h
FD methods of orders 2, 4, ..., 10, 20, 40
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ω h
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1
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k h
ω h
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10Cubic FEM
For FEM of degree n, the dispersion relation is pn = 0, where (Mathai (95))
p1(r) = (2 + cos θ)r − 3 sin θ, θ = 2πkh
p2(r) = (3 − cos θ)r2 + 8 sin θ − 20(1 − cos θ),
p3(r) = (4 + cos θ)r3 − 15 sin θr2 − 30(4 + 3 cos θ)r + 210 sin θ,
As n → ∞, FDM −→ pseudo–spectral method (Fornberg (87))
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 8/28
Wavepackets
A wavepacket is defined by
u(x, t) = A(x, t)eiΘ(x,t),
where A is a slowly varying function of x and t, and
• Θ(x, t) = 2πkx − ωt, for constant coefficients / uniform grid,
• ∂tΘ = −ω and ∂xΘ = 2πk for varying coefficients / irregular grid
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 9/28
Wavepackets
A wavepacket is defined by
u(x, t) = A(x, t)eiΘ(x,t),
where A is a slowly varying function of x and t, and
• Θ(x, t) = 2πkx − ωt, for constant coefficients / uniform grid,
• ∂tΘ = −ω and ∂xΘ = 2πk for varying coefficients / irregular grid
0 0.2 0.4 0.6 0.8 1−1
−0.5
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20 40 60 80 1000
5
10
15|FFT|, N = 100 grid points
The IC is a wavepacket, its FFT is localized to a few wavenumbers, so the solution isconfined to a low dimensional subspace spanned by the corresponding eigenvectors.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 9/28
Wavepacket Analysis
With grid mapping: xj = G(sj), sj = jh, we substitute
uj(t) = Aj(t)eiΘj (t),
into the FDM and use
∂tΘ = −ω, ∂sΘ = 2πk.
The leading term in the result gives
ω =2 sin(Θj+1 − Θj−1)/2
hj+1 + hj=
2 sin 2πhkj
hj+1 + hj,
where kj is the mean wavenumber on (sj−1, sj+1).k varies to compensate for variation in h so that ω is constant.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 10/28
Wavepacket Analysis II
First term:
ω =2 sin 2πhk(sj)
hj+1 + hj.
The second asymptotic term in the wavepacket analysis leads to
At + cgAx = 0,
cg = cos 2πhk(x) =1
2π
∂ω
∂k(x)
is the Group velocity.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 11/28
Wavepacket Analysis II
First term:
ω =2 sin 2πhk(sj)
hj+1 + hj.
The second asymptotic term in the wavepacket analysis leads to
At + cgAx = 0,
cg = cos 2πhk(x) =1
2π
∂ω
∂k(x)
is the Group velocity.ExampleWe take a wavepacket with N = 200, k = 28 and superimpose on thenumerical solutions the curves defined by
x = cos 2πhk(x)
with starting wavenumbers 23,28, 33 and starting values x(0) chosen to 10, 0,−10 gridpoints from the packet centre, respectively.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 11/28
Numerical Results: linearly varying grid
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N = 200, k = 28: linearly increasing grid
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ContoursPredicted paths
Right figure shows contours of |A|.The analysis predicts the location and spread of the wavepacket but accuracy falls off ast increases.As the packet moves to the right, hj increases ⇒ kj must increase to keep ω constant.
However, the group velocity → 0 as kjhj → 14
(the peak of the dispersion relation),i.e., it predicts that the packet comes to rest (where it remains).We need a more sophisticated approach - our model only allows “one–way” waves.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 12/28
Staggered Grid Approach
We saw earlier that (−1)juj(t) is a solution whenever uj(t) is a solution. This suggeststhat we relabel the grid variables in an odd–even pattern
u0, v1, u2, . . . , u2j , v2j+1, . . .
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 13/28
Staggered Grid Approach
We saw earlier that (−1)juj(t) is a solution whenever uj(t) is a solution. This suggeststhat we relabel the grid variables in an odd–even pattern
u0, v1, u2, . . . , u2j , v2j+1, . . .
We the suppose that
u2j(t) = A2j(t) exp(iΘ2j(t))
v2j+1(t) = B2j+1(t) exp(iΘ2j+1(t)).
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 13/28
Staggered Grid Approach
We saw earlier that (−1)juj(t) is a solution whenever uj(t) is a solution. This suggeststhat we relabel the grid variables in an odd–even pattern
u0, v1, u2, . . . , u2j , v2j+1, . . .
We the suppose that
u2j(t) = A2j(t) exp(iΘ2j(t))
v2j+1(t) = B2j+1(t) exp(iΘ2j+1(t)).
These lead to the same dispersion relation as before together with the PDEs
At + cos 2πhk(x)Bx = 0
Bt + cos 2πhk(x)Ax = 0
A + B is constant along x = cos 2πhk(x)
A − B is constant along x = − cos 2πhk(x)
(h = 1/N ). We solve these with the BC v(x∗) = 0 where x∗ is where k(x∗) = N/4.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 13/28
Piecewise constant grid example
t = 0
Coarse
A−B = constantFine
A+B = constant
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 14/28
Piecewise constant grid example
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N = 200, k = 28
0 20 40 60 80 100 120 140 160 180 2000
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Grid sizes
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 14/28
Piecewise constant grid: Analysis
Grid: hj =h j ≤ M
H M < jh = 1/n, H = 1/N .
The corresponding eigenvalue problem gives
j < M : 2ihωuj + uj−1 − uj−1 = 0,⇒ ω =sin 2πkh
h
j > M : 2iHωuj + uj−1 − uj−1 = 0,⇒ ω =sin 2πKH
H
Given k, we can only solve for K when |ωH| < 1.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 15/28
Piecewise constant grid: Analysis
Grid: hj =h j ≤ M
H M < jh = 1/n, H = 1/N .
The corresponding eigenvalue problem gives
j < M : 2ihωuj + uj−1 − uj−1 = 0,⇒ ω =sin 2πkh
h
j > M : 2iHωuj + uj−1 − uj−1 = 0,⇒ ω =sin 2πKH
H
Given k, we can only solve for K when |ωH| < 1.
N/2 n/2
N
n
|ω H|>1
|ω H|<1
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 15/28
Low Frequencies
200 grid points, h = 1/n, H = 1/N , N = 125, n = 250.
uj =exp(2πikh) j < M
exp(2πiKH) j > M
N/2 n/2
N
n
|ω H|>1
|ω H|<1
−1
−0.5
0
0.5
1ω
181=31.2345
−1
−0.5
0
0.5
1ω
123=108.5047
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0.01
Grid sizes
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 16/28
High Frequencies, Evanescent Modes
h = 1/n, H = 1/N , N < ω < n
uj =exp(2πikh) j < M
ijr±j j > M
r = ωH +
�
ω2H2 − 1, (r > H/h)
The support of each eigenvector is essentially confined to the fine grid.Thus, a wavepacket initiated on the fine grid and carrying a sufficiently highwavenumber, will be confined to the grid for all time.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 17/28
High Frequencies, Evanescent Modes
h = 1/n, H = 1/N , N < ω < n
uj =exp(2πikh) j < M
ijr±j j > M
r = ωH +
�
ω2H2 − 1, (r > H/h)
The support of each eigenvector is essentially confined to the fine grid.Thus, a wavepacket initiated on the fine grid and carrying a sufficiently highwavenumber, will be confined to the grid for all time.
N/2 n/2
N
n
|ω H|>1
|ω H|<1
−1
−0.5
0
0.5
1ω
3=249.784
N = 125, n = 250
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0.01
Grid sizes
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 17/28
High Frequencies, Evanescent Modes
h = 1/n, H = 1/N , N < ω < n
uj =exp(2πikh) j < M
ijr±j j > M
r = ωH +
�
ω2H2 − 1, (r > H/h)
The support of each eigenvector is essentially confined to the fine grid.Thus, a wavepacket initiated on the fine grid and carrying a sufficiently highwavenumber, will be confined to the grid for all time.
N/2 n/2
N
n
|ω H|>1
|ω H|<1
−1
−0.5
0
0.5
1ω
77=172.7069
N = 125, n = 250
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0.01
Grid sizes
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 17/28
Partial Reflection/Transmission
With a wavepacket, those frequencies above cutoff will be reflected, while those belowwill be transmitted:
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200
250
cuto
ff
Dispersion relations
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N = 200, k = 28
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David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 18/28
Gradual Transition
h ≤ hj ≤ H, linear variation.
�
= 1/N
Trick 2: A similarity transformation with the diagonal matrix diag(1, i,−1,−i, 1, . . . )
converts the Hermitian matrix iA to a symmetric matrix:
iD
��
��
�
0 1 0 −1
−1 0 1 0
0 −1 0 1
1 0 −1 0
��
��
�
DH =
��
��
�
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
��
��
�
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 19/28
Gradual Transition
h ≤ hj ≤ H, linear variation.
�
= 1/N
Trick 2: Define uj(t) = e−iωtijvj(t) so that (hj + hj+1)uj + uj−1 − uj−1 = 0
becomes
2ω(hj + hj+1)vj = vj+1 + vj−1
which is a 2nd order FD approximation of
v′′(s) =a(s − s0)
a = 4(H − h)ω/
�2, s0 =(1 − ωh)/
�
H′ω, H(s) = Nh(s)
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 19/28
Gradual Transition
h ≤ hj ≤ H, linear variation.
�
= 1/N
Trick 2: This has solution v(s) = Ai(a1/3(s − s0)),where Ai(·) is the Airy function(y′′ = xy).
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 19/28
Gradual Transition
h ≤ hj ≤ H, linear variation.
�
= 1/N
Trick 2: This has solution v(s) = Ai(a1/3(s − s0)),where Ai(·) is the Airy function(y′′ = xy).
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Grid size hj−1
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Eigenvector V19
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 19/28
Gradual Transition
h ≤ hj ≤ H, linear variation.
�
= 1/N
Trick 2: This has solution v(s) = Ai(a1/3(s − s0)),where Ai(·) is the Airy function(y′′ = xy).The eigenvector penetrates further into the coarse grid as ω decreases:
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Grid size hj
−1
−0.5
0
V1
−1
0
1 V19
−1
0
1 V61
−1
0
1 V121
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 19/28
Grid Sensitivity
We now turn our attention to the sensitivity of solutions to perturbations in the grid:
xj = jh + εhgj , j = 0 : N
with g0 = gN = 0, ε < 1/2 and |gj | < 1. The perturbations gj may come from amapping : gj = g(jh), or they might be “random” numbers from [−1, 1].The approximation of ut + ux = 0 then leads to the eigenvalue problem
ωM(ε)v = A(ε)v
where,M is symmetric p.d.A is Hermitian (we have replaced −iA by A).
We assume that
M(ε) = M + εM + . . . , A(ε) = A + εA + . . . ,
Henceforth, M , A, etc will denote M(0), A(0), resp.In FEM, M is linear in ε and A is independent of ε.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 20/28
Examples: Lumped LFEM
N = 80, t = 20. Both examples below use two grids, a uniform grid and the sameperturbed grid.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 21/28
Examples: Lumped LFEM
N = 80, t = 20. Both examples below use two grids, a uniform grid and the sameperturbed grid. In the first, the grid perturbation has no effect
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1
1.5u
0(t)
ueps
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 21/28
Examples: Lumped LFEM
N = 80, t = 20. Both examples below use two grids, a uniform grid and the sameperturbed grid.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1.5
−1
−0.5
0
0.5
1
1.5
u0(t)
ueps
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 21/28
Perturbation I
Define the matrices
Λ(ε) = diag(ω1, . . . , ωN ), X(ε) = [v1, . . . , vN ],
containing the eigenvalues & eigenvectors so
XH (ε)A(ε)X(ε) = Λ(ε)
and we suppose that the eigenvectors are normalized so that
XH(ε)M(ε)X(ε) = I.
We develop series
Λ(ε) = Λ + εΛ + · · ·
X(ε) = X(I + εR + · · · )
We write
�
M = XHMX, etc.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 22/28
Perturbation II
We find
λj =
�
Ajj −
�
Mjjλj
Rjk =λk
�
Mjk −
�
Ajk
λj − λk, λj 6= λk
The solution of the system Mu + Au = 0 is then
u(t; ε) = u(t; 0) + εX(−e−iΛtΛt + Re−iΛt + e−iΛtRH)u(0) + O(ε2)
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 23/28
Perturbation II
We find
λj =
�
Ajj −
�
Mjjλj
Rjk =λk
�
Mjk −
�
Ajk
λj − λk, λj 6= λk
The solution of the system Mu + Au = 0 is then
u(t; ε) = u(t; 0) + εX(−e−iΛtΛt + Re−iΛt + e−iΛtRH)u(0) + O(ε2)
• We will generally get the largest perturbations when λj 6= 0,
• Difficulty defining Rjk when there are multiple eigenvalues. Suppose that the
corresponding eigenspace is spanned by a subset X of columns of X. We have
to choose the basis in such a way that the diagonal of X(
�
MΛ −
�
A)X is zero.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 23/28
Unimodal Perturbations
We now suppose that the grid perturbations are generated by a Fourier mode ofwavenumber r:
gj = sin 2πrj/N.
This means that we can evaluate the terms
�
M and
�
A: in the following, we will always
have A = 0.Linear FEM
We find
�
M = 0 and so Λ(ε) = Λ + O(e2).
The critical terms in perturbing the solutions are due to R: this has nonzero terms only inthe two diagonals distance ±r from the main diagonal (akin to the sidebands of Sloan &
Mitchell (86) in nonlinear instability): X(ε) = X(I + εR + . . . )
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 24/28
Unimodal Perturbations
We now suppose that the grid perturbations are generated by a Fourier mode ofwavenumber r:
gj = sin 2πrj/N.
This means that we can evaluate the terms
�
M and
�
A: in the following, we will always
have A = 0.Linear FEM
We find
�
M = 0 and so Λ(ε) = Λ + O(e2).
The critical terms in perturbing the solutions are due to R: this has nonzero terms only inthe two diagonals distance ±r from the main diagonal (akin to the sidebands of Sloan &
Mitchell (86) in nonlinear instability): X(ε) = X(I + εR + . . . )
0 0 ∗ 0 0 0
0 0 0 ∗ 0 0
∗ 0 0 0 ∗ 0
0 ∗ 0 0 0 ∗
0 0 ∗ 0 0 0
0 0 0 ∗ 0 0
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 24/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
• the locations of these maxima will specify which modes (columns of X) in theinitial data are affected. The magnitudes depend on the separation ofeigenvalues:
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
• the locations of these maxima will specify which modes (columns of X) in theinitial data are affected. The magnitudes depend on the separation ofeigenvalues:
Linear FEM j
ω
3.5 N
− 3.5 N
N/2 N
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
• the locations of these maxima will specify which modes (columns of X) in theinitial data are affected. The magnitudes depend on the separation ofeigenvalues:
0 5 10 15 20 25 30 35 400
10
20
30
40
5 10 15 20 25 30 350
50
100
150
200
max(max(|ℜ (R)|))
r
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
• the locations of these maxima will specify which modes (columns of X) in theinitial data are affected. The magnitudes depend on the separation ofeigenvalues:
0 5 10 15 20 25 30 35 400
10
20
30
40
5 10 15 20 25 30 350
50
100
150
200
max(max(|ℜ (R)|))
r
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
U
x
N=80, LFEM
u0=F
1u
0(t)
ueps
−0.01
0
0.01Term 1
−0.05
0
0.05
Term 2 + Term 3
10 20 30 40 50 60 70 80−5
0
5
j
FF
T(g
)
Fourier transform of random grid
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
• the locations of these maxima will specify which modes (columns of X) in theinitial data are affected. The magnitudes depend on the separation ofeigenvalues:
0 5 10 15 20 25 30 35 400
10
20
30
40
5 10 15 20 25 30 350
50
100
150
200
max(max(|ℜ (R)|))
r
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
U
Random grid modified by removing offending mode 36
x
u0=F
1u
0(t)
ueps
−2
0
2
x 10−3 Term 1
−5
0
5
x 10−3 Term 2 + Term 3
10 20 30 40 50 60 70 80−5
0
5
j
FF
T(g
)
Fourier transform of random grid
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Application: Linear FEM
To apply these results,
• for each grid mode r ∈ [1, N/2] we compute the two largest elements of |R|
• the locations of these maxima will specify which modes (columns of X) in theinitial data are affected. The magnitudes depend on the separation ofeigenvalues:
0 5 10 15 20 25 30 35 400
10
20
30
40
5 10 15 20 25 30 350
50
100
150
200
max(max(|ℜ (R)|))
r We conclude that the linear FEM isprone to wiggles when the grid has particular high frequency components
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 25/28
Lumped Linear FEM
This is more complicated since the results depend on the parity of N
• N odd: simple eigenvalues
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 26/28
Lumped Linear FEM
This is more complicated since the results depend on the parity of N
• N odd: simple eigenvalues
• N a multiple of 4: λN/4 and λ3N/4 are simple, the remainder double.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 26/28
Lumped Linear FEM
This is more complicated since the results depend on the parity of N
• N odd: simple eigenvalues
• N a multiple of 4: λN/4 and λ3N/4 are simple, the remainder double.
• N not a multiple of 4: all eigenvalues double.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 26/28
Lumped Linear FEM
This is more complicated since the results depend on the parity of N
• N odd: simple eigenvalues
• N a multiple of 4: λN/4 and λ3N/4 are simple, the remainder double.
• N not a multiple of 4: all eigenvalues double.
0 5 10 15 20 25 30 35 400
10
20
30
40N = 79, FDM
k
5 10 15 20 25 30 350
10
20
30
40
50
max(max(|ℜ (R)|))
rDavid F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 26/28
Lumped Linear FEM
This is more complicated since the results depend on the parity of N
• N odd: simple eigenvalues
• N a multiple of 4: λN/4 and λ3N/4 are simple, the remainder double.
• N not a multiple of 4: all eigenvalues double.
5 10 15 20 25 30 35
5
10
15
20
25
30
35
40N = 80, FDM
k
5 10 15 20 25 30 350
10
20
30
40max(max(|ℜ (R)|))
rDavid F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 26/28
Lumped Linear FEM
This is more complicated since the results depend on the parity of N
• N odd: simple eigenvalues
• N a multiple of 4: λN/4 and λ3N/4 are simple, the remainder double.
• N not a multiple of 4: all eigenvalues double.
−1.5
−1
−0.5
0
0.5
1
1.5DFM, N = 80, t=20, r = 36, ε = 2−6.6
U
u0(t)
ueps
−0.5
0
0.5
Term 1
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−4−2
024
x 10−3 Term 2 + Term 3
x
10 20 30 40 50 60 70 800
2
4
6
j
FF
T(u
0)
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 26/28
Quadratic FEM
5
10
15
20
25
30
35
40N = 79, QFEM (consistent mass)
k
5 10 15 20 25 30 350
50
100
150
max(max(|ℜ (R)|))
r
This shows that wiggles will only be stimulated if the initial data is sufficiently rich inwavenumbers > 14 and there are high frequency components in the grid perturbation.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 27/28
Quadratic FEM
5 10 15 20 25 30 35
5
10
15
20
25
30
35
40N = 80, QFEM (lumped mass)
k
5 10 15 20 25 30 350
100
200
300
400
500
600max(max(|ℜ (R)|))
r
For Lumped Quadratic FEMs, the situation is not quite so favourable. The wavenumbersinvolved from the IC are lower.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 27/28
Summary
• Fourier Analysis and the ideas of dispersion and group velocity can begeneralized to smoothly varying grids via wavepackets.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 28/28
Summary
• Fourier Analysis and the ideas of dispersion and group velocity can begeneralized to smoothly varying grids via wavepackets.
• Waves can get trapped in regions of high grid refinement unless they can beresolved by the coarse grids. This applies also to advection–diffusion with smalldiffusion (Shishkin grids) (G, Gresho & Silvester, 04).
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 28/28
Summary
• Fourier Analysis and the ideas of dispersion and group velocity can begeneralized to smoothly varying grids via wavepackets.
• Waves can get trapped in regions of high grid refinement unless they can beresolved by the coarse grids. This applies also to advection–diffusion with smalldiffusion (Shishkin grids) (G, Gresho & Silvester, 04).
• Methods that have multiple eigenvalues on uniform grids are particularly sensitiveto grid perturbations.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 28/28
Summary
• Fourier Analysis and the ideas of dispersion and group velocity can begeneralized to smoothly varying grids via wavepackets.
• Waves can get trapped in regions of high grid refinement unless they can beresolved by the coarse grids. This applies also to advection–diffusion with smalldiffusion (Shishkin grids) (G, Gresho & Silvester, 04).
• Methods that have multiple eigenvalues on uniform grids are particularly sensitiveto grid perturbations.
• Quadratic FEMs appear not to be sensitive to perturbations.
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 28/28
Summary
• Fourier Analysis and the ideas of dispersion and group velocity can begeneralized to smoothly varying grids via wavepackets.
• Waves can get trapped in regions of high grid refinement unless they can beresolved by the coarse grids. This applies also to advection–diffusion with smalldiffusion (Shishkin grids) (G, Gresho & Silvester, 04).
• Methods that have multiple eigenvalues on uniform grids are particularly sensitiveto grid perturbations.
• Quadratic FEMs appear not to be sensitive to perturbations.
• Thank you for yourattention!
David F Griffiths, Dundee University http://www.maths.dundee.ac.uk/dfg – p. 28/28