Lecture 19 Continuous Problems: Backus-Gilbert Theory and Radon’s Problem
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Transcript of Lecture 19 Continuous Problems: Backus-Gilbert Theory and Radon’s Problem
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Lecture 19
Continuous Problems:Backus-Gilbert Theory
andRadon’s Problem
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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
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Purpose of the Lecture
Extend Backus-Gilbert theory to continuous problems
Discuss the conversion ofcontinuous inverse problems to discrete problems
Solve Radon’s Problemthe simplest tomography problem
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Part 1
Backus-Gilbert Theory
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Continuous Inverse Theory
the data are discretebut
the model parameter is a continuous function
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One or several dimensions
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model functiondata
One or several dimensions
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hopeless to try to determine estimates of model function at a particular depth
m(z0) = ?localized average is the only way to go
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hopeless to try to determine estimates of model function at a particular depth
m(z0) = ?localized average is the only way to go
the problem is that an integral, such as the data kernel integral, does not depend upon the value of m(z) at a “single point” z0
continuous version of resolution matrix
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let’s retain the idea that the“solution”
depends linearly on the data
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let’s retain the idea that the“solution”
depends linearly on the data
continuous version of generalized inverse
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implies a formula for R
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<m>=G-gdcomparison to discrete case
d=Gm<m>=Rm
R=G-gG
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implies a formula for R
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Now define the spread of resolution as
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fine generalized inverse
that minimizes the spread J with the constraint that
= 1
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J has exactly the same form as the discrete case
only the definition of S is different
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Hence the solution is thesame as in the discrete case
where
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furthermore, just as we did in the discrete case, we can add the size of
the covariance
where
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and leads to a trade-off of resolution and variance
as beforethis just changes the definition of S
spread of resolution
size
of v
aria
nce α=1
α=0
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Part 2
Approximating aContinuous Problemas a Discrete Problem
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approximation using finite number of known functions
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approximation using finite number of known functions
known functions
unknown coefficients= discrete model parameters
continuous function
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posssible fj(x)’s
voxels (and their lower dimension equivalents)
polynomials
splines
Fourier (and similar) series
and many others
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does the choice of fj(x) matter?
Yes!
The choice implements prior informationabout the properties of the solution
The solution will be different depending upon the choice
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conversion to discrete Gm=d
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special case of voxels
fi(x) =1 if x inside Vi0 otherwise
integral over voxel j
size controlled by the scale of variation of m(x)
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center of voxel j
approximation when Gi(x) slowly varying
size controlled by the scale of variation of Gi(x)
more stringent condition than scale of variation of m(x)
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Part 3
Tomography
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Greek Root
tomosa cut, cutting, slice, section
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“tomography”as it is used in geophysics
data are line integrals of the model function
curve i
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you can force this into the form
if you want
Gi(x)but the Dirac delta function is not square-
integrable, which leads to problems
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Radon’s Problem
straight line raysdata d treated as a continuous variable
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xθ
yu s
(u,θ) coordinate system forRadon Transform
integrate over this line
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Radon Transformm(x,y) → d(u,θ)
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-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
true image
y
x
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
Radon transform
theta
u
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
reconstructed image
y
x
(A) m(x,y) (B) d(u,θ)
yθ
x
u
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Inverse Problem
find m(x,y) given d(u,θ)
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Solution via Fourier Transforms
x→kx kx → x
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now Fourier transformu→ku
now change variables (u,θ) →(x,y)
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now Fourier transformu→ku
Fourier transform of m(x,y)evaluated on a line of slope θ
now change variables(s,u) →(x,y)Fourier transform
of d(u,θ)
J=1, by the way
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xθ0
yu d(u,θ0 )
ky
kxθ0
m(x,y) m(kx,ky)
FT
^̂
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Learned two things
1. Proof that solution exists and unique, based on “well-known” properties of Fourier Transform
2. Recipe how to invert a Radon transform using Fourier transforms
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-1 -0.5 0 0.5 1
-1
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true image
y
x
-1 -0.5 0 0.5 1
-1
-0.5
0
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Radon transform
theta
u
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
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reconstructed image
y
x
(A) (B) (C)
y yθ
x xu