Lecture 24 Exemplary Inverse Problems including Vibrational Problems
description
Transcript of Lecture 24 Exemplary Inverse Problems including Vibrational Problems
Lecture 24
Exemplary Inverse Problemsincluding
Vibrational Problems
SyllabusLecture 01 Describing Inverse ProblemsLecture 02 Probability and Measurement Error, Part 1Lecture 03 Probability and Measurement Error, Part 2 Lecture 04 The L2 Norm and Simple Least SquaresLecture 05 A Priori Information and Weighted Least SquaredLecture 06 Resolution and Generalized InversesLecture 07 Backus-Gilbert Inverse and the Trade Off of Resolution and VarianceLecture 08 The Principle of Maximum LikelihoodLecture 09 Inexact TheoriesLecture 10 Nonuniqueness and Localized AveragesLecture 11 Vector Spaces and Singular Value DecompositionLecture 12 Equality and Inequality ConstraintsLecture 13 L1 , L∞ Norm Problems and Linear ProgrammingLecture 14 Nonlinear Problems: Grid and Monte Carlo Searches Lecture 15 Nonlinear Problems: Newton’s Method Lecture 16 Nonlinear Problems: Simulated Annealing and Bootstrap Confidence Intervals Lecture 17 Factor AnalysisLecture 18 Varimax Factors, Empircal Orthogonal FunctionsLecture 19 Backus-Gilbert Theory for Continuous Problems; Radon’s ProblemLecture 20 Linear Operators and Their AdjointsLecture 21 Fréchet DerivativesLecture 22 Exemplary Inverse Problems, incl. Filter DesignLecture 23 Exemplary Inverse Problems, incl. Earthquake LocationLecture 24 Exemplary Inverse Problems, incl. Vibrational Problems
Purpose of the Lecture
solve a few exemplary inverse problems
tomographyvibrational problems
determining mean directions
Part 1
tomography
di = ∫ray i m(x(s), y(s)) dsray i
tomography:data is line integral of model function
assume ray path is knownx
y
discretization:model function divided up into M pixels mj
data kernel
Gij = length of ray i in pixel j
data kernel
Gij = length of ray i in pixel jhere’s an easy,
approximate way to calculate it
ray i
start with G set to zero
then consider each ray in sequence
∆s
divide each ray into segments of arc length ∆s
and step from segment to segment
determine the pixel index, say j, that the center of each line segment falls within
add ∆s to Gijrepeat for every segment of every ray
You can make this approximation indefinitely accurate simply by
decreasing the size of ∆s
(albeit at the expense of increase the computation time)
Suppose that there are M=L2 voxels
A ray passes through about L voxels
G has NL2 elementsNL of which are non-zero
so the fraction of non-zero elements is1/Lhence G is very sparse
In a typical tomographic experiment
some pixels will be missed entirely
and some groups of pixels will be sampled by only one ray
In a typical tomographic experiment
some pixels will be missed entirely
and some groups of pixels will be sampled by only one ray
the value of these pixels is completely undetermined
only the average value of these pixels is determined
hence the problem is mixed-determined(and usually M>N as well)
soyou must introduce some sort of a priori
information to achieve a solutionsay
a priori information that the solution is small
or
a priori information that the solution is smooth
Solution Possibilities1. Damped Least Squares (implements smallness): Matrix G is sparse and very large
use bicg() with damped least squares function
2. Weighted Least Squares (implements smoothness): Matrix F consists of G plus
second derivative smoothinguse bicg()with weighted least squares function
Solution Possibilities1. Damped Least Squares: Matrix G is sparse and very large
use bicg() with damped least squares function
2. Weighted Least Squares: Matrix F consists of G plus
second derivative smoothinguse bicg()with weighted least squares function
test case has very good ray coverage,
so smoothing probably
unnecessary
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
xtrue model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated model
True model
x
y sources and receivers
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
xtrue model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated model
x
yRay Coverage
x
y just a “few” rays shown
else image is black
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated modelData, plotted in Radon-style coordinates
angle θ of ray
dist
ance
r of ray
to
cent
er o
f im
age
Lesson from Radon’s Problem:Full data coverage need to achieve exact solution
minor data gaps
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
yx
estimated modelEstimated model
x
y
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
yx
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated model
True model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
yx
estimated modelEstimated model
x
y
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
yx
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated model
Estimated model
streaks due to minor data gapsthey disappear if ray density is doubled
but what if the observational geometry is poor
so that broads swaths of rays are missing ?
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
xtrue model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated model
(A) (B)
(C) (D)
x
y
x
y
x
y
θ
r
complete angular coverage
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
xtrue model
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
true model with rays
-150 -100 -50 0 50 100 150
0
10
20
30
40
theta
R
traveltimes
0 10 20 30 40 50 60
0
10
20
30
40
50
60
y
x
estimated model
(A) (B)
(C) (D)
x
y
x
y
x
y
θ
r
incomplete angular coverage
Part 2
vibrational problems
statement of the problem
Can you determine the structure of an objectjust knowing the
characteristic frequencies at which it vibrates?
frequency
the Fréchet derivativeof frequency with respect to velocity
is usually computed using perturbation theory
hence a quick discussion of what that is ...
perturbation theory
a technique for computing an approximate solution to a complicated problem, when
1. The complicated problem is related to a simple problem by a small perturbation
2. The solution of the simple problem must be known
simple example
we know the solution to this equation: x0=±c
Here’s the actual vibrational problem
acoustic equation withspatially variable sound velocity v
acoustic equation withspatially variable sound velocity v
frequencies of vibrationor
eigenfrequencies
patterns of vibrationor
eigenfunctionsor
modes
v(x) = v(0)(x) + εv(1)(x) + ...
assume velocity can be written as a perturbation
around some simple structurev(0)(x)
eigenfunctions known to obey orthonormality relationship
now represent eigenfrequencies and eigenfunctions as power series in ε
represent first-order perturbed shapes as sum of
unperturbed shapes
now represent eigenfrequencies and eigenfunctions as power series in ε
plug series into original differential equation
group terms of equal power of εsolve for first-order perturbation
in eigenfrequencies ωn(1)and eigenfunction coefficients bnm(use orthonormality in process)
result
result for eigenfrequencies
write as standard inverse problem
standard continuous inverse problem
standard continuous inverse problem
perturbation in the eigenfrequencies are the
data
perturbation in the velocity structure is the
model function
standard continuous inverse problem
depends upon theunperturbed velocity structure,the unperturbed eigenfrequency
and the unperturbed mode
data kernel or Fréchet derivative
1D organ pipe
unperturbed problem has constant velocity
0
h x
open end,p=0
closed enddp/dx=0perturbed problem has
variable velocity
00.1
0.20.3
0.40.5
0.60.7
0.80.9
1-1 0 10
h x
p=0
dp/dx=0
p1
x
00.1
0.20.3
0.40.5
0.60.7
0.80.9
1-1 0 1
x
p2 p300.1
0.20.3
0.40.5
0.60.7
0.80.9
1-1 0 1
x𝜔1
modes
frequencies 𝜔2 𝜔3 𝜔 0
solution to unperturbed problem
position , x
velocity, v
perturbedunperturbed
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
2
4
6
8
10
velocity structure
How to discretize the model function?
m is veloctity function evaluated at sequence of points equally spaced in x
our choice is very simple
0 10 20 30 40 50 60 700
0.5
1
the dataa list of frequencies of vibration
true, unperturbedtrue, perturbedobserved = true, perturbed + noise
frequency
ωi
mjthe data kernel
Solution Possibilities1. Damped Least Squares (implements smallness): Matrix G is not sparse
use bicg() with damped least squares function
2. Weighted Least Squares (implements smoothness): Matrix F consists of G plus
second derivative smoothinguse bicg()with weighted least squares function
Solution Possibilities1. Damped Least Squares (implements smallness): Matrix G is not sparse
use bicg() with damped least squares function
2. Weighted Least Squares (implements smoothness): Matrix F consists of G plus
second derivative smoothinguse bicg()with weighted least squares function
our choice
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
2
4
6
8
10
position , x
velocity, v
the solution
true estimated
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
2
4
6
8
10
position , x
velocity, v
the solution
true estimated
mi
mjthe model resolution matrix
mi
mjthe model resolution matrix
what is this?
This problem has a type of nonuniqueness
that arises from its symmetry
a positive velocity anomaly at one end of the organ pipe
trades off with a negative anomaly at the other end
this behavior is very commonand is why eigenfrequency data
are usually supplemented with other data
e.g. travel times along rays
that are not subject to this nonuniqueness
Part 3
determining mean directions
statement of the problemyou measure a bunch of directions (unit vectors)
what’s their mean?
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x1
x2
x 3
x y
zdata
mean
what’s a reasonableprobability density function
for directional data?
Gaussian doesn’t quite workbecause
its defined on the wrong interval(-∞, +∞)
θ ϕcentral vector
datum
coordinate system
distribution should be symmetric in ϕ
-3 -2 -1 0 1 2 30
0.2
0.4
0.6
0.8
1
theta
p(th
eta)
angle, θ π-π
p(θ)Fisher distribution
similar in shape to a Gaussian but on a sphere
-3 -2 -1 0 1 2 30
0.2
0.4
0.6
0.8
1
theta
p(th
eta)
angle, θ π-π
p(θ)Fisher distribution
similar in shape to a Gaussian but on a sphere
“precision parameter”quantifies width of p.d.f.
κ=5
κ=1
solve by
direct application of
principle of maximum likelihood
maximize joint p.d.f. of data
with respect toκ and cos(θ)
x: Cartesian components of observed unit vectors
m: Cartesian components of central unit vector; must constrain |m|=1
likelihood function
constraint
unknownsm, κC = = 0
Lagrange multiplier equations
Results
valid when κ>5
Results
central vector is parallel to the vector that you get by putting all the observed unit vectors end-to-end
Solution PossibilitiesDetermine m by evaluating simple formula
1. Determine κ using simple but approximate formula
2. Determine κ using bootstrap method
our choice
only valid when κ>5
Application to Subduction Zone Stresses
Determine the mean direction ofP-axes
of deep (300-600 km) earthquakesin the Kurile-Kamchatka subduction zone
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1N
E
N
E
data central direction bootstrap