lecture 20 continuous problems linear operators and their adjoints
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Lecture 20 Continuous Problems Linear Operators and Their Adjoints. Syllabus. - PowerPoint PPT PresentationTRANSCRIPT
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Lecture 20
Continuous Problems
Linear Operators and Their Adjoints
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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
Teach you a tiny bit of analysis
enough for you to understand
Linear Operators and their Adjoints
because they are the core techniqueused in the so-called
adjoint method of computing data kernels
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everything we do today
is based on the idea of
generalizingdiscrete problems to continuous problems
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a function
m(x)
is the continuous analog of a vector
m
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a function
m(x)
is the continuous analog of a vector
m
simplification: one spatial dimension x
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comparison
m is of length Mm(x) is infinite dimensional
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What is the continuous analogof a matrix L ?
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We’ll give it a symbol, ℒand a name, a linear operator
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Matrix times a vector is another vector
b = L aso we’ll want
linear operator on a functionis another function
b(x) = ℒ a(x)
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Matrix arithmetic is not communative
L(1) L(2) a ≠ L(2) L(1) a so we’ll not expect that property
for linear operators, either
ℒ (1) ℒ(2) a(x) ≠ ℒ (2) ℒ(1) a(x)
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Matrix arithmetic is associative
(L(1) L(2)) L(3) a = L(1) (L(2) L(3) ) a so well want
that property for linear operators, too
(ℒ (1) ℒ(2) )ℒ(3) a(x) = ℒ (1) (ℒ(2) ℒ(3)) a(x)
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Matrix arithmetic is distributive
L [a+b] = La + Lbso well want
that property for linear operators, too
ℒ [a(x)+ b(x)] = ℒ a(x)+ ℒ b(x)
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Hint to the identity of ℒmatrices can approximate
derivativesand integrals
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B
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LAa ≈ da/dx Laa ≈B
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Linear Operatorℒany combination of functions,
derivativesand integrals
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ℒ a(x)= c(x) a(x) ℒ a(x)= da/dx
ℒa(x) = b(x)da/dx + c(x) d2a/dx 2 ℒ a(x)= ∫0x a(ξ)dξ
ℒ a(x)= f(x)∫0∞ a(ξ)g(x,ξ ) dξ
all perfectly good ℒ a(x)’s
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What is the continuous analogof the inverse L-1 of a matrix L ?
call it ℒ -1
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ProblemLA not square, so has no inverse
derivative determines a function only up to an additive constant
Patch by adding top row that sets the constant
Now LB LC = I
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lesson 1: ℒ may need to include boundary conditions
lesson 2:
if
ℒ = d/dx ℒ -1 =
then
since
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the analogy to the matrix equation
L m = fand its solution m = L-1 f
is the differential equationℒ m = fand its Green function solution
ℒ
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so the inverse to a differential operator ℒ
is the Green function integral
ℒ -1 a(x) =
where F solves
a(ξ)
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What is the continuous analogyto a dot product ?
aTb = Σi ai bi
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The continuous analogyto a dot product
s = aTb = Σi ai biis the inner product
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squared length of a vector
|a|2 = aTasquared length of a function
|a|2 = (a,a)
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important property of a dot product
(La)T b = aT (LTb)
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important property of a dot product
(La)T b = aT (LTb) what is the continuous analogy ?
(ℒa, b )= (a, ?b )
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in other words ...
what is the continuous analogy of the transpose of a matrix?
(ℒa, b )= (a, ?b ) by analogy , it must be another linear operatorsince transpose of a matrix is another matrix
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in other words ...
what is the continuous analogy of the transpose of a matrix?
(ℒa, b )= (a, ?b ) give it a name “adjoint “ and a symbol ℒ †
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so ...
(ℒa, b )= (a, ℒ † b )
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so, given ℒ, how do you determine ℒ † ?
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so, given ℒ, how do you determine ℒ † ?
various ways ,,,
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the adjoint of a function is itself
if ℒ=c(x) then ℒ † =c(x)
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the adjoint of a function is itself
if ℒ=c(x) then ℒ † =c(x) a function is self-adjoint
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the adjoint of a function is itself
if ℒ=c(x) then ℒ † =c(x) self-adjoint operator anagous to a symmetric matrixx
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the adjoint of d/dx(with zero boundary consitions)
is –d/dx
if ℒ=d/dx then ℒ † =-d/dx
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the adjoint of d/dx(with zero boundary consitions)
is –d/dx
if ℒ=d/dx then ℒ † =-d/dxintegration by parts
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the adjoint of d2/dx 2 is itself
a function is self-adjoint
apply integration by parts twice
if ℒ=d2/dx 2 then ℒ † =d2/dx 2
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trick using Heaviside
step function
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properties of adjoints
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table of adjoints
c(x)-d/dxd2/dx2
c(x)d/dxd2/dx2
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analogiesmLLm=fL-1f=L-1ms=aTb(La) Tb= a T(LTb)LT
m(x)ℒℒm(x)=f(x)ℒ-1f(x) =ℒ-1f(x)s=(a(x), b(x))(ℒa, b) =(a, ℒ†b)ℒ†
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how is all this going to help us?
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step 1
recognize thatstandard equation of inverse theory
is an inner productdi = (Gi, m)
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step 2
suppose that we can show thatdi = (hi, ℒm)then do thisdi = (ℒ†hi, m)soGi = ℒ†hi
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step 2
suppose that we can show thatdi = (hi, ℒm)then do thisdi = (ℒ†hi, m)soGi = ℒ†hi
formula for the data kernel