Institut de Physique du Globe de Paris & Université Pierre et Marie Curie (Paris VI) Course on Inverse Problems Albert Tarantola Lesson XII: Optimization and Linear Problems
Recall: We have seen that the posterior volumetric probability for the model parameters, whose general expression is f post ( m ) = 1 ν f prior ( m ) g obs ( o ( m ) ) . becomes, using the Gaussian model for all uncertainties, f post ( m ) = k exp ( − S ( m ) ) , where the misfit function S ( m ) is the sum of squares -1 ( m − m prior ) 2 S ( m ) = ( m − m prior ) t C prior -1 ( o ( m ) − o obs ) + ( o ( m ) − o obs ) t C obs . What happens if the forward modeling relation o = o ( m ) is, in fact, linear?
Linear problems: We have assumed the prior information cor- responds to a Gaussian distribution with mean m prior and co- variance C prior , and that the observations are described by a Gaussian distribution with mean o obs and covariance C obs . If the forward relation is linear, o = o ( m ) = O m , then, the posterior distribution in the model space is also a Gaussian distribution, with the following mean and covariance: prior ) -1 ( O t C -1 m post = ( O t C -1 obs O + C -1 obs o obs + C -1 prior m prior ) = m prior + ( O t C -1 prior ) -1 O t C -1 obs O + C -1 obs ( o obs − O m prior ) = m prior + C prior O t ( O C prior O t + C obs ) -1 ( o obs − O m prior ) C post = ( O t C -1 obs O + C -1 prior ) -1 = C prior − C prior O t ( O C prior O t + C obs ) -1 O C prior
The posterior distribution in the observable parameter space is also Gaussian, with mean O m post and covariance O C post O t .
Remark: The five elements of a linear inverse problem are m prior , C prior , o obs , C obs , and O . The expressions for m post and C post have been given in the slides above. What these expressions become if, in fact, we don’t have any a priori in- formation (and the forward relation o = O m is redundant)? C prior → ∞ I ; ( then m prior disappears ) m post = ( O t C -1 obs O ) -1 O t C -1 obs o obs C post = ( O t C -1 obs O ) -1 And if the matrix O is square and invertible? m post = O -1 o obs
4.2 Ray Tomography Using Blocks ❊①❡❝✉t❛❜❧❡ ♥♦t❡❜♦♦❦ ❛t ❤tt♣✿✴✴✇✇✇✳✐♣❣♣✳❥✉ss✐❡✉✳❢r✴⑦t❛r❛♥t♦❧❛✴❡①❡r❝✐❝❡s✴❝❤❛♣t❡r❴✵✹✴❘❛②❚♦♠♦❣r❛♣❤②✳♥❜ This exercise corresponds to a highly idealized version of an X-ray tomography experi- ment. A 2D medium is characterized by a parameter m ( x , y ) whose physical dimension is the inverse of a length. This parameter may take any real value (including negative values). Assume that when a ray R i is materialized in the medium (using a source and a receiver), we are able to measure the observable parameter o i = � R i d ℓ m ( x , y ) , (4.1) where ℓ is the length along the ray. The goal of the exercise is to use some observed values o i obs to infer the values of the function m ( x , y ) .
❊①❡❝✉t❛❜❧❡ ♥♦t❡❜♦♦❦ ❛t ❤tt♣✿✴✴✇✇✇✳✐♣❣♣✳❥✉ss✐❡✉✳❢r✴⑦t❛r❛♥t♦❧❛✴❡①❡r❝✐❝❡s✴❝❤❛♣t❡r❴✵✹✴❘❛②❚♦♠♦❣r❛♣❤②✳♥❜ 11 10 09 08 12 07 13 06 14 05 15 04 16 03 17 02 18 01 01020304 19 05060708 20 09101112 21 13141516 22 Figure 4.1: Geometry of the ‘X-ray experiment’. This drawing is at scale 1:1 (the length of the side of the blocks is 1 cm ). To simplify the problem, the medium is divided into blocks, numbered from 1 to 16 (see figure 4.1), so, instead of evaluating the function m ( x , y ) , we only need to evaluate
the discrete values { m 1 , m 2 , . . . , m 16 } . With this discretization, the relation (4.1) becomes α = 16 o i = O i α m α ∑ , (4.2) α = 1 where O i α is the length of the ray i in the block α . We shall use 22 rays, so the index i runs from 1 to 22. For short, equation (4.2) shall be written o = O m . (4.3) With the geometry of the 16 blocks and the 22 rays represented in figure 4.1, it is easy to see that the matrix O is (only the nonzero elements are indicated) − − − − − − − − − − − − − − − a − − − − − − − − − − − a − − a − − − − − − − − a − − a − − a − − − − − − − − − − − − − − a a a a − − − − − − − − − − − − − a a a − − − − − − − − − − − − − − a a a − − − − − − − − − − − − − − − − − − b − − − b − − − b − − − b − − b − − − b − − − b − − − b − − b − − − b − − − b − − − b − − − − − − − − − − − − − − O = b b b b , (4.4) − − − − − − − − − − − − − − − a − − − − − − − − − − − − − − a a − a − − − − a − − − − a − − − − a − − − − a − − − − a − − − − a − − − − a − − − − a − − − − a − − − − − − − − − a − − − − a − − − − − − − − − − − − − − − − − a − − − − − − − − − − − − b b b b − − − − − − − − − − − − b b b b − − − − − − − − b b b b − − − − − − − − − − − − − − − − b b b b √ with a = 2 cm and b = 1 cm . In a real-life application, the components of the matrix O should be automatically com- ✭✯ ❚❤❡ ♠❛tr✐① ❖ ✐♥ t❤❡ ♦ ❂ ❖ ♠ r❡❧❛t✐♦♥ ✯✮ ❯♥♣r♦t❡❝t❬❖❪❀ ❖ ❂ ❚❛❜❧❡ ❬✵ ✱ ④✐✱✶ ✱✷✷⑥ ✱ ④❥ ✱✶ ✱✶✻⑥❪❀ sq✷ ❂ ❙qrt ❬✷❪❀ ❖❬❬✵✶ ✱✶✻❪❪ ❂ sq✷❀ ❖❬❬✵✷ ✱✶✷❪❪ ❂ sq✷❀❖❬❬✵✷ ✱✶✺❪❪ ❂ sq✷❀ ❖❬❬✵✸ ✱✵✽❪❪ ❂ sq✷❀❖❬❬✵✸ ✱✶✶❪❪ ❂ sq✷❀❖❬❬✵✸ ✱✶✹❪❪ ❂ sq✷❀ ❖❬❬✵✹ ✱✵✹❪❪ ❂ sq✷❀❖❬❬✵✹ ✱✵✼❪❪ ❂ sq✷❀❖❬❬✵✹ ✱✶✵❪❪ ❂ sq✷❀❖❬❬✵✹ ✱✶✸❪❪ ❂ sq✷❀ ❖❬❬✵✺ ✱✵✸❪❪ ❂ sq✷❀❖❬❬✵✺ ✱✵✻❪❪ ❂ sq✷❀❖❬❬✵✺ ✱✵✾❪❪ ❂ sq✷❀ ❖❬❬✵✻ ✱✵✷❪❪ ❂ sq✷❀❖❬❬✵✻ ✱✵✺❪❪ ❂ sq✷❀ ❖❬❬✵✼ ✱✵✶❪❪ ❂ sq✷❀ ❖❬❬✵✽ ✱✵✹❪❪ ❂ ✶❀❖❬❬✵✽ ✱✵✽❪❪ ❂ ✶❀❖❬❬✵✽ ✱✶✷❪❪ ❂ ✶❀❖❬❬✵✽ ✱✶✻❪❪ ❂ ✶❀ ❖❬❬✵✾ ✱✵✸❪❪ ❂ ✶❀❖❬❬✵✾ ✱✵✼❪❪ ❂ ✶❀❖❬❬✵✾ ✱✶✶❪❪ ❂ ✶❀❖❬❬✵✾ ✱✶✺❪❪ ❂ ✶❀
with a = 2 cm and b = 1 cm . In a real-life application, the components of the matrix O should be automatically com- puted using the positions of the end-points of the ray, but —as writing the associated code is tedious— in this simple exercise, we can enter the components “by hand” as follows 1 ✭✯ ❚❤❡ ♠❛tr✐① ❖ ✐♥ t❤❡ ♦ ❂ ❖ ♠ r❡❧❛t✐♦♥ ✯✮ ❯♥♣r♦t❡❝t❬❖❪❀ ❖ ❂ ❚❛❜❧❡ ❬✵ ✱ ④✐✱✶ ✱✷✷⑥ ✱ ④❥ ✱✶ ✱✶✻⑥❪❀ sq✷ ❂ ❙qrt ❬✷❪❀ ❖❬❬✵✶ ✱✶✻❪❪ ❂ sq✷❀ ❖❬❬✵✷ ✱✶✷❪❪ ❂ sq✷❀❖❬❬✵✷ ✱✶✺❪❪ ❂ sq✷❀ ❖❬❬✵✸ ✱✵✽❪❪ ❂ sq✷❀❖❬❬✵✸ ✱✶✶❪❪ ❂ sq✷❀❖❬❬✵✸ ✱✶✹❪❪ ❂ sq✷❀ ❖❬❬✵✹ ✱✵✹❪❪ ❂ sq✷❀❖❬❬✵✹ ✱✵✼❪❪ ❂ sq✷❀❖❬❬✵✹ ✱✶✵❪❪ ❂ sq✷❀❖❬❬✵✹ ✱✶✸❪❪ ❂ sq✷❀ ❖❬❬✵✺ ✱✵✸❪❪ ❂ sq✷❀❖❬❬✵✺ ✱✵✻❪❪ ❂ sq✷❀❖❬❬✵✺ ✱✵✾❪❪ ❂ sq✷❀ ❖❬❬✵✻ ✱✵✷❪❪ ❂ sq✷❀❖❬❬✵✻ ✱✵✺❪❪ ❂ sq✷❀ ❖❬❬✵✼ ✱✵✶❪❪ ❂ sq✷❀ ❖❬❬✵✽ ✱✵✹❪❪ ❂ ✶❀❖❬❬✵✽ ✱✵✽❪❪ ❂ ✶❀❖❬❬✵✽ ✱✶✷❪❪ ❂ ✶❀❖❬❬✵✽ ✱✶✻❪❪ ❂ ✶❀ ❖❬❬✵✾ ✱✵✸❪❪ ❂ ✶❀❖❬❬✵✾ ✱✵✼❪❪ ❂ ✶❀❖❬❬✵✾ ✱✶✶❪❪ ❂ ✶❀❖❬❬✵✾ ✱✶✺❪❪ ❂ ✶❀
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