Geology 5670/6670Inverse Theory

27 Feb 2015

© A.R. Lowry 2015Read for Wed 25 Feb: Menke Ch 9 (163-188)

Last time: The Sensitivity Matrix (Revisited)• The Sensitivity Matrix, or Kernel Matrix, G can be thought of as the matrix of derivatives: and this holds true regardless of the nature of the model equation/operator F!• If F is linear, the derivatives are independent of the model parameter space m and the second derivatives are zero, so the minimum error solution is found in a single step.• If F is nonlinear but the derivatives can be evaluated analytically, derivatives depend on m and gradient search methods iteratively search for minimum error.• If analytical derivatives can’t be found, can evaluate derivatives numerically using:

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Gij = ∂Fi

∂m j

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Gij = Fia − Fi

b

m ja − m j

b

For inequality constraints on an L2 problem, use Quadratic Programming. General statement of the QP problem is something like:

Minimize subject to So want to express our problem in these forms. Note that:

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ϕ x( ) = 12

xT

Qx +ζT

x

€

F x ≥ h, E x = r

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eT

e = Gm − d( )T

Gm − d( ) = mT

GT

− dT ⎛

⎝ ⎜ ⎞ ⎠ ⎟Gm − d( )

= mT

GT

Gm −2dT

Gm + dT

d

€

12

mT

GT

Gm − dT

Gm

So our objective function to minimize is

(i.e. let ; )

If our constraints are of the form

Like with linear programming, can treat quadratic programming as a black box and find a suitable algorithm (e.g. in Matlab) to solve.

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Q = GT

G

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ζ =−dT

G

€

l ≤ m ≤ u

€

I−I

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥m ≥ l

−u

⎡

⎣ ⎢

⎤

⎦ ⎥

€

F = I−I

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥; h = l

−u

⎡

⎣ ⎢

⎤

⎦ ⎥

€

G m − m x( ) = d − d x ⇒ Gm r = d r

Stochastic Inversion:Suppose we know (or expect) something about the model parameters m before we begin the inversion, i.e., we know the expected value and an a priori covariance matrix

We can express this problem as

in which the remainder part of the model has

We seek to find a generalized inverse G+ that minimizes the mean square error:

of:

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C ma

= mmT

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m x = m

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m r = 0

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L = ˜ m − m t( )T

˜ m − m t( )

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m r = G+d r

Expanding and re-writing using a few of our math tricks,

Minimizing this is equivalent to minimizing:

You can convince yourself (if so inclined) that the solution to this minimax problem is:

or:

(Does this remind you of anything we’ve seen before?)

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L = Tr C ma

−C md C d−1

C dm + G+

−C md C d−1 ⎡

⎣ ⎢ ⎤ ⎦ ⎥C d G

+−Cmd C d

−1 ⎡ ⎣ ⎢

⎤ ⎦ ⎥

T ⎧ ⎨ ⎪

⎩ ⎪ ⎫ ⎬ ⎪

⎭ ⎪

€

Tr G+C d G

+T

+ G+

−C md C d−1 ⎡

⎣ ⎢ ⎤ ⎦ ⎥C d G

+−C md C d

−1 ⎡ ⎣ ⎢

⎤ ⎦ ⎥

T ⎧ ⎨ ⎪

⎩ ⎪ ⎫ ⎬ ⎪

⎭ ⎪

€

G+

= C md C d−1

= C ma

GT

GC ma

GT

+Cε ⎛ ⎝ ⎜

⎞ ⎠ ⎟

−1

€

⇒ G+

= GT

Cε−1

G +C ma −1 ⎛

⎝ ⎜

⎞

⎠ ⎟

−1

GT

Cε−1