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 ...
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CHAPTER 2. <strong>AVO</strong> MODELING 12<br />
and strain components. According to <strong>the</strong> usual geometry depicted in Figure 2.1.<br />
σij = λ∆δij + 2µeij, (2.6)<br />
where σij is <strong>the</strong> stress component, eij is <strong>the</strong> strain component and ∆ is <strong>the</strong> dilation which<br />
is <strong>the</strong> sum <strong>of</strong> <strong>the</strong> normal strain components. <strong>the</strong> indices ij refers to <strong>the</strong> various combination<br />
x, y and z directions. <strong>the</strong> constants λ and µ (shear modulus) are <strong>the</strong> Lamé parameters.<br />
The following boundary conditions holds at z = 0.<br />
Boundary Condition 1: The tangential displacement component is continuous across <strong>the</strong><br />
boundary,<br />
ux1 = ux2. (2.7)<br />
Using <strong>the</strong> first two equations <strong>of</strong> equation (2.5) and setting z=0, we can easily see that<br />
ω<br />
A0 cos θi − ω<br />
A1 cos θi + ω<br />
B1 sin ϕr = ω<br />
A2 cos θt − ω<br />
B2 sin ϕt. (2.8)<br />
α1<br />
α1<br />
β1<br />
Boundary Condition 2: The normal displacement component is continuous across <strong>the</strong><br />
boundary,<br />
α2<br />
uz1 = uz2. (2.9)<br />
Using <strong>the</strong> third and <strong>the</strong> fourth equations <strong>of</strong> equation (2.5) and setting z=0, we have<br />
ω<br />
A0 sin θi + ω<br />
A1 sin θr + ω<br />
B1 cos ϕr = ω<br />
A2 sin θt + ω<br />
B2 cos ϕt. (2.10)<br />
α1<br />
α1<br />
β1<br />
Boundary Condition 3: The normal stress component is continuous across <strong>the</strong> boundary.<br />
According to equation (2.6), <strong>the</strong> normal stress component can be written as<br />
where<br />
We can rewrite equation (2.11) as<br />
α2<br />
σzz = λ(exx + ezz) + 2µezz, (2.11)<br />
exx = ∂ux<br />
∂x ,<br />
ezz = ∂uz<br />
. (2.12)<br />
∂z<br />
σzz = (λ + 2µ) ∂uz<br />
∂z<br />
β2<br />
β2<br />
+ λ∂ux . (2.13)<br />
∂x