30.07.2013 Views

Regularization of the AVO inverse problem by means of a ...

Regularization of the AVO inverse problem by means of a ...

Regularization of the AVO inverse problem by means of a ...

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

CHAPTER 2. <strong>AVO</strong> MODELING 13<br />

At this point, we need to express <strong>the</strong> Lamé parameters in terms <strong>of</strong> density, P-wave and<br />

S-wave velocities in order to express <strong>the</strong> final result solely in terms <strong>of</strong> density and velocities<br />

<strong>of</strong> <strong>the</strong> media. To this end, <strong>the</strong> following relationships are useful<br />

α = (<br />

Using <strong>the</strong>se relations in equation (2.13), we have<br />

From which follows,<br />

λ + 2µ<br />

)<br />

ρ<br />

1<br />

2 ,<br />

β = ( µ 1<br />

) 2 . (2.14)<br />

ρ<br />

2 ∂uz<br />

σzz = ρα<br />

∂z<br />

ρ1α 2 ∂uz1<br />

1<br />

∂z + ρ1β 2 ∂ux1<br />

1<br />

∂x<br />

(σzz)1 = (σzz)2<br />

∂ux<br />

+ ρβ2 . (2.15)<br />

∂x<br />

= ρ2α 2 ∂uz2<br />

2<br />

∂z + ρ2β 2 2<br />

∂ux2<br />

. (2.16)<br />

∂x<br />

Substituting equations (2.5) into equation (2.13) and applying differentiation, one can easily<br />

show that<br />

ρ1α1(1 − 2β 2 1p 2 ) ω<br />

A0 + ρ1α1(1 − 2β 2 1p 2 ) ω<br />

α1<br />

= ρ2α2(1 − 2β 2 2p 2 ) ω<br />

ω<br />

A1 − 2ρ1β<br />

α1<br />

2 1p cos ϕr B1<br />

β1<br />

A2 − 2ρ2β<br />

α2<br />

2 2p cos ϕt<br />

β2<br />

ω<br />

B2. (2.17)<br />

Boundary Condition 4: The tangential stress component is continuous across <strong>the</strong> bound-<br />

ary i.e According to <strong>the</strong> relation given <strong>by</strong> equation (2.6), <strong>the</strong> tangential stress component<br />

can be written as<br />

But exz can be expressed as<br />

σxz = 2µexz. (2.18)<br />

exz = 1<br />

2 (∂uz<br />

∂x<br />

∂ux<br />

+ ). (2.19)<br />

∂z<br />

Using this expression and equation (2.14) in equation (2.18), we can write <strong>the</strong> boundary<br />

condition as<br />

ρ1β 2 1( ∂uz1<br />

∂x<br />

∂ux1<br />

+<br />

∂z ) = ρ2β 2 2( ∂uz2<br />

∂x<br />

∂ux2<br />

+ . (2.20)<br />

∂z

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!