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Sequential Methods for Coupled Geomechanics and Multiphase Flow

Sequential Methods for Coupled Geomechanics and Multiphase Flow

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126 CHAPTER 5. FIXED-STRAIN AND FIXED-STRESS SPLITS<br />

From Equation 5.28, the fixed-strain split does not inherit the contractivity relative to<br />

the norm · N at the stage of A p sn. As a result, the fixed-strain split is not contractive with<br />

respect to the full problem.<br />

When solving the mechanical problem after the flow problem, A u sn, we obtain<br />

d dχ 2<br />

N<br />

dt<br />

=<br />

=<br />

<br />

<br />

Ω<br />

Ω<br />

<br />

dσ ′ : ˙<br />

dεe − dκ · ˙<br />

dξ + 1<br />

dσ ′ <br />

: dεdΩ ˙ −<br />

<br />

dpd ˙p dΩ<br />

M<br />

<br />

dσ ′ <br />

: dεp<br />

˙ + dκ · dξ ˙ dΩ<br />

Ω <br />

D<br />

<br />

d p≥0<br />

(from d ˙p = 0)<br />

= −D d p ≤ 0, (5.29)<br />

<br />

dσ<br />

Ω<br />

′ <br />

: dεdΩ ˙ = 0 from Equation 5.27 .<br />

5.3.2 Contractivity of the fixed-stress split<br />

Again we consider two arbitrary initial conditions <strong>and</strong> study the contractivity of the fixed-<br />

stress split. In the fixed-stress split, the original operator A is decomposed as follows:<br />

⎡<br />

⎣ dun<br />

dpn ⎤<br />

⎦ Ap ss<br />

⎡<br />

−→ ⎣ du∗<br />

dpn+1 ⎤<br />

⎦ Au ss<br />

−→<br />

⎡<br />

⎣ dun+1<br />

⎤<br />

dpn+1 ⎦, where<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

A p ss : ˙<br />

dm + Div dv = 0, δdσv ˙ = 0,<br />

A u ss : Div dσ = 0, dp = 0,<br />

⇒ Div dσ ′ = 0,<br />

(5.30)<br />

which has homogeneous boundary conditions with no source terms. Note that δ ˙<br />

dσv = 0 is<br />

equivalent to δ ˙<br />

dσ = 0 in A p ss, since pressure variations affect the volumetric stress, not the<br />

shear stress. Using Equation 5.3, the initial conditions of A p ss in Equation 5.30 become<br />

Div d ˙σt=0 = 0, Div dσ(t) t=0 = 0. (5.31)<br />

First, we show the contractivity of the fixed-stress split when solving the flow problem<br />

A p ss. In A p ss of Equation 5.30, δ ˙<br />

dσ = 0 yields ˙<br />

dσ(t) − ˙<br />

dσt=0 = 0. Combining this result

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