Lecture 21 Continuous Problems Fréchet Derivatives€¦ · Lecture 13 L 1, L ∞ Norm Problems and...
Transcript of Lecture 21 Continuous Problems Fréchet Derivatives€¦ · Lecture 13 L 1, L ∞ Norm Problems and...
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Lecture 21
Continuous Problems
Fréchet Derivatives
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SyllabusLecture 01 Describing Inverse Problems
Lecture 02 Probability and Measurement Error, Part 1
Lecture 03 Probability and Measurement Error, Part 2
Lecture 04 The L2 Norm and Simple Least Squares
Lecture 05 A Priori Information and Weighted Least Squared
Lecture 06 Resolution and Generalized Inverses
Lecture 07 Backus-Gilbert Inverse and the Trade Off of Resolution and Variance
Lecture 08 The Principle of Maximum Likelihood
Lecture 09 Inexact Theories
Lecture 10 Nonuniqueness and Localized Averages
Lecture 11 Vector Spaces and Singular Value Decomposition
Lecture 12 Equality and Inequality Constraints
Lecture 13 L1 , L∞ Norm Problems and Linear Programming
Lecture 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 Analysis
Lecture 18 Varimax Factors, Empircal Orthogonal Functions
Lecture 19 Backus-Gilbert Theory for Continuous Problems; Radon’s Problem
Lecture 20 Linear Operators and Their Adjoints
Lecture 21 Fréchet Derivatives
Lecture 22 Exemplary Inverse Problems, incl. Filter Design
Lecture 23 Exemplary Inverse Problems, incl. Earthquake Location
Lecture 24 Exemplary Inverse Problems, incl. Vibrational Problems
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Purpose of the Lecture
use adjoint methods to compute
data kernels
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Part 1
Review of Last Lecture
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a function
m(x)
is the continuous analog of a vector
m
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a linear operator
ℒ
is the continuous analog of a matrix
L
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a inverse of a linear operator
ℒ-1
is the continuous analog of the inverse
of a matrix
L-1
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a inverse of a linear operator
can be used to solve
a differential equation
if ℒm=f then m=ℒ-1f
just as the inverse of a matrix
can be used to solve
a matrix equation
if Lm=f then m=L-1f
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the inner product
is the continuous analog of dot product
s= aTb
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the adjoint of a linear operator
is the continuous analog of the transpose of
a matrix
LT
ℒ †
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the adjoint can be used to
manipulate an inner product
just as the transpose can be used to
manipulate the dot product
(La) Tb= a T(LTb)
(ℒa, b) =(a, ℒ†b)
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table of adjoints
c(x)
-d/dx
d2/dx2
c(x)
d/dx
d2/dx2
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Part 2
definition of the Fréchet derivatives
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first
rewrite the standard inverse theory
equation in terms of perturbations
a small change in the model
causes a small change in the data
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second
compare with the standard formula for
a derivative
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third
identify the data kernel as
a kind of derivative
this kind of derivative is called a
Fréchet derivative
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definition of a Fréchet derivative
this is mostly lingo
though perhaps it adds a little insight about
what the data kernel is
,
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Part 2
Fréchet derivative of Error
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treat the data as a continuous function d(x)then the standard L2 norm error is
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let the data d(x) be related to the model m(x)by
could be the data kernel integral
=
because integrals are linear operators
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to
a perturbation in the model
causes
a perturbation in the error
now do a little algebra to relate
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if
m(0) implies d(0) with error E(0)
then ...
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if
m(0) implies d(0) with error E(0)
then ...
all this
is just
algebra
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if
m(0) implies d(0) with error E(0)
then ...
use δd = ℒδm
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if
m(0) implies d(0) with error E(0)
then ...
use adjoint
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if
m(0) implies d(0) with error E(0)
then ...
Fréchet derivative of Error
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you can use this derivative to solve and
inverse problem using the
gradient method
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example
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example
this is the relationship between
model and data
d(x) =
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example
construct adjoint
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example
Fréchet derivative of Error
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2
0
2
x
m(x
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
5
x
d(x
)
10 20 30 40 50 60 70 80 90 100-4
-2
0
iteration
log10 e
rror,
Em
(x)
d(x
) x
(A)
(B)
(C)
log 1
0E
/E1
x
iteration
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Part 3
Backprojection
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continuous analog of least squares
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now define the identity operator ℐ
m(x) = ℐ m(x)
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view as a recursion
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view as a recursion
using the adjoint as if it
were theinverse
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example
backprojection
exact
m(x) = ℒ-1 dobs = d dobs / dx
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example
backprojection
exact
m(x) = ℒ-1 dobs = d dobs / dx
crazy!
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0 20 40 60 80 100 120 140 160 180 200
-0.50
0.5
x
m
0 20 40 60 80 100 120 140 160 180 200
-0.5
00.5
x
m
0 20 40 60 80 100 120 140 160 180 200-1
0
1
x
m
(A)
(B)
(C)
m(x
)tru
em
est (
x)m
(1) (
x)
x
x
x
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interpretation as tomography
backprojection
m is slownessd is travel time of a ray from –∞ to x
integrate (=add together) the travel times of all rays that pass through the point x
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discrete analysis
Gm=d
G= UΛVT G-g= VΛ-1UT GT= VΛUT
if Λ-1≈ Λ then G-g≈ GT
backprojection works when the singular
values are all roughly the same size
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suggests scalingGm=d → WGm=Wd
where W is a diagonal matrix chosen to make the singular values more equal in overall size
Traveltime tomography:
Wii = (length of ith ray)-1
so [Wd]i has interpretation of the average slowness
along the ray i.
Backprojection now adds together the average slowness of all rays that interact with the point x.
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0 20 40 60
0
10
20
30
40
50
60
y
x
true model
0 20 40 60
0
10
20
30
40
50
60
y
x
LS estimate
0 20 40 60
0
10
20
30
40
50
60
y
x
BP estimate(A) (B) (C)
y
x
y
x
y
x
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Part 4
Fréchet Derivative
involving a differential equation
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Part 4
Fréchet Derivative
involving a differential equation
seismic wave equationNavier-Stokes equation of fluid flow
etc
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data d is related to field u via an inner product
field u is related to model parameters m via
a differential equation
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pertrubation δd is related to perturbation δu via an
inner product
write in terms of perturbations
perturbation δu is related to perturbation
δm via a differential equation
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what’s the data kernel ?
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easy using adjoints
data inner product with field
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easy using adjoints
data is inner product with field
field satisfies ℒδu= δm
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easy using adjoints
data is inner product with field
field satisfies ℒδu= δm
employ adjoint
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easy using adjoints
data is inner product with field
field satisfies ℒδu= δm
employ adjoint
inverse of adjoint is adjoint ofinverse
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easy using adjoints
data is inner product with field
field satisfies ℒδu= δm
employ adjoint
inverse of adjoint is adjoint ofinverse
data kernel
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easy using adjoints
data is inner product with field
field satisfies ℒδu= δm
employ adjoint
inverse of adjoint is adjoint ofinverse
data kernel
data kernel satisfies “adjoint
differential equation
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most problem involving differential
equations are solved numerically
so instead of just solving
you must solve
and
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so there’s more work
but the same sort of work
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exampletime t instead of position x
field solves a Newtonian-type heat flow equation
where u is temperature and m is heating
data is concentration of chemical whose
production rate is proportional to temperature
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exampletime t instead of position x
field solves a Newtonian-type heat flow equation
where u is temperature
data is concentration of chemical whose
production rate is proportional to temperature
= (bH(ti-t), u)so hi= bH(ti-t)
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we will solve this problem
analytically
using Green functions
in more complicated cases
the differential equation
must be solved numerically
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Newtonian equation
its Green function
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adjoint equation
its Green function
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note that the adjoint Green function
is the original Green function
backward in time
that’s a fairly common patternwhose significance will be pursued in a homework
problem
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we must perform a Green function integral
to compute the data kernel
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time, t
row
, i
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0 50 100 150 200 250 300 350 400 450 5000
0.5
1
t
H(t
,10)
0 50 100 150 200 250 300 350 400 450 5000
1
2
t
m(t
)
0 50 100 150 200 250 300 350 400 450 5000
20
t
u(t
)
0 50 100 150 200 250 300 350 400 450 5000
1000
2000
t
d(t
)(A)
(B)
(C)
(D)
time, t
time, t
time, t
time, t
F(t
,τ=
30
)m
(t)
u(t
)d
(t)
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Part 4
Fréchet Derivative
involving a parameter in
differential equation
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Part 4
Fréchet Derivative
involving a parameter in
differential equation
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previous example
unknown functi on is “forcing”
another possibility
forcing
is knownparameter is unknown
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linearize around a simpler equation
and assume you can solve this equation
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the perturbed equation is
subtracting out the unperturbed equation,ignoring second order terms, and rearranging gives ...
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then approximately
pertubation to parameter acts as an unknown
forcing
so it is back to the form of a forcing
and the previous methodology can be applied