Lecture 10
Nonuniqueness
and
Localized Averages
Syllabus
Lecture 01
Lecture 02
Lecture 03
Lecture 04
Lecture 05
Lecture 06
Lecture 07
Lecture 08
Lecture 09
Lecture 10
Lecture 11
Lecture 12
Lecture 13
Lecture 14
Lecture 15
Lecture 16
Lecture 17
Lecture 18
Lecture 19
Lecture 20
Lecture 21
Lecture 22
Lecture 23
Lecture 24
Describing Inverse Problems
Probability and Measurement Error, Part 1
Probability and Measurement Error, Part 2
The L2 Norm and Simple Least Squares
A Priori Information and Weighted Least Squared
Resolution and Generalized Inverses
Backus-Gilbert Inverse and the Trade Off of Resolution and Variance
The Principle of Maximum Likelihood
Inexact Theories
Nonuniqueness and Localized Averages
Vector Spaces and Singular Value Decomposition
Equality and Inequality Constraints
L1 , L∞ Norm Problems and Linear Programming
Nonlinear Problems: Grid and Monte Carlo Searches
Nonlinear Problems: Newton’s Method
Nonlinear Problems: Simulated Annealing and Bootstrap Confidence Intervals
Factor Analysis
Varimax Factors, Empirical Orthogonal Functions
Backus-Gilbert Theory for Continuous Problems; Radon’s Problem
Linear Operators and Their Adjoints
Fréchet Derivatives
Exemplary Inverse Problems, incl. Filter Design
Exemplary Inverse Problems, incl. Earthquake Location
Exemplary Inverse Problems, incl. Vibrational Problems
Purpose of the Lecture
Show that null vectors are the source of nonuniqueness
Show why some localized averages of model parameters are
unique while others aren’t
Show how nonunique averages can be bounded using prior
information on the bounds of the underlying model parameters
Introduce the Linear Programming Problem
Part 1
null vectors as the source of
nonuniqueness
in linear inverse problems
suppose two different solutions
exactly satisfy the same data
since there are two
the solution is nonunique
then the difference between the
solutions satisfies
the quantity
mnull = m(1) – m(2)
is called a null vector
it satisfies
G mnull = 0
an inverse problem can have more than
one null vector
mnull(1) mnull(2) mnull(3)...
any linear combination of null vectors is a
null vector
αmnull(1) + βmnull(2) +γmnull(3)
is a null vector for any α, β, γ
suppose that a particular choice of
model parameters
mpar
satisfies
G mpar=dobs
with error E
then
has the same error E
for any choice of αi
since
e = dobs-Gmgen = dobs-Gmpar + Σi αi 0
since
since αi is arbitrary
the solution is nonunique
hence
an inverse problem is
nonunique
if it has null vectors
example
consider the inverse problem
Gm
a solution with zero error is
mpar=[d1, d1, d1, d1]T
the null vectors are easy to work out
note that
of these vectors is zero
times any
the general solution to the inverse
problem is
Part 2
Why some localized averages are
unique
while others aren’t
let’s denote a weighted average of
the model parameters as
<m> = aT m
where a is the vector of weights
let’s denote a weighted average of
the model parameters as
<m> = aT m
where a is the vector of weights
a may or may not be “localized”
examples
a = [0.25, 0.25, 0.25, 0.25]T
not localized
a = [0. 90, 0.07, 0.02, 0.01]T
localized near m1
now compute the average of the
general solution
now compute the average of the
general solution
if this term is zero for all i,
then <m> does not depend on αi,
so average is unique
an average
T
<m>=a m
is unique
if the average of all the null vectors
is zero
if we just pick an average
out of the hat
because we like it ... its nicely localized
chances are that it will not zero all the null
vectors
so the average will not be unique
relationship to model resolution R
relationship to model resolution R
aT is a linear combination of the rows
of the data kernel G
if we just pick an average
out of the hat
because we like it ... its nicely localized
its not likely that it can be built out of the rows
of G
so it will not be unique
suppose we pick a
average that is not unique
is it of any use?
Part 3
bounding localized averages
even though they are nonunique
we will now show
if we can put weak bounds on m
they may translate into stronger
bounds on <m>
example
with
so
example
with
so
nonunique
but suppose mi is bounded
0 > mi > 2d1
smallest α3= -d1
largest α3= +d1
smallest α3= -d1
largest α3= +d1
(2/3) d1 >
<m>
> (4/3)d1
smallest α3= -d1
largest α3= +d1
(2/3) d1 >
<m>
> (4/3)d1
bounds on <m> tighter than bounds on mi
the question is how to do this in more
complicated cases
Part 4
The Linear Programming Problem
the Linear Programming problem
the Linear Programming problem
flipping sign
switches
minimization to
maximization
flipping signs of A and b
switches to ≥
in Business
quantity of each
product
profit
unit profit
maximizes
physical limitations of factory
government regulations
etc
no negative
production
care about both profit z and product quantities x
in our case
first minimize
then
maximize
m
not needed
<m>
Gm=d
a
bounds on m
care only about <m>, not m
In MatLab
Example 1
simple data kernel
one datum
sum of mi is zero
bounds
|mi| ≤ 1
average
unweighted average of K model parameters
bounds on absolute value of
weighted
bound onaverage
|<m>|
1.5
1
0.5
0
2
4
6
8
10
12
width
K
14
16
18
20
bounds on absolute value of
weighted
bound onaverage
|<m>|
1.5
1
0.5
0
2
4
6
8
10
12
14
16
18
20
width
K
if you know that the sum of 20 things is zero
and
if you know that the things are bounded by ±1
then you know
the sum of 19 of the things is bounded by about ±0.1
bounds on absolute value of
weighted
bound onaverage
|<m>|
1.5
1
0.5
0
2
4
6
8
10
12
14
16
18
20
width
K
for K>10
<m> has tigher bounds
than mi
Example 2
more complicated data kernel
dk weighted average of first 5k/2 m’s
bounds
0 ≤ mi ≤ 1
average
localized average of 5 neighboring
model parameters
1.5
1
0
2
width, w
4
depth,
z zi
6
8
10
j
0.5
G
j
mtrue
dobs
=
i
i
-0.5
0
=
≈=
mi (zm
i)
(B)
(A)
(A)
2
width, w
4
i
depth,
z zi
i
=
6
8
10
complicated G
but
reminiscent of Laplace
Transform kernel
j
1.5
0
=
≈=
1
0
j
0.5
G
mi (zm
i)
(B)
mtrue
-0.5
dobs
(A)
2
width, w
4
i
depth,
z zi
i
=
6
8
10
j
true mi increased
with depth zi
1.5
0
=
≈=
1
0
j
0.5
G
mi (zm
i)
(B)
mtrue
-0.5
dobs
(A)
1.5
1
0
j
0.5
G
mi (zm
i)
(B)
mtrue
-0.5
dobs
0
2
width, w
=
≈=
4
i
depth,
z zi
i
=
6
8
10
j
minimum
length solution
(A)
1.5
1
0
j
0.5
G
mi (zm
i)
(B)
mtrue
-0.5
dobs
0
2
width, w
=
≈=
4
i
depth,
z zi
i
=
6
8
10
lower bound j
on solution
upper
bound
on solution
(A)
1.5
1
0
j
0.5
G
mi (zm
i)
(B)
mtrue
-0.5
dobs
0
2
width, w
=
≈=
4
i
depth,
z zi
i
=
6
8
10
lower bound j
on average
upper
bound
on
average