Poroelasticity
Zhuoran Wang
Colorado State University
Zhuoran Wang
Poroelasticity
Linear poroelasticity
Poroelasticity equation:
(
−∇ · (2µε(u) + λ(∇ · u)I) + α∇p = f
∂t (c0 p + α∇ · u) + ∇ · (−K∇p) = s
,
(1)
where µ = 1, λ = 1, α = 1, c0 = 0.1, K = κI. It is a coupled PDEs
for poroelasticity.
Zhuoran Wang
Poroelasticity
Numerical experiments for linear poroelasticity
We test the example which is on (0, 1)2 for linear poroelasicity.
Dirichlet boundary condition for displacement is uD = u and for
pressure is pD = p.
Zhuoran Wang
Poroelasticity
Numerical experiments for linear poroelasticity
u is the known vector valued displacement function:
1
cos(2πx) sin(2πy )
.
u = − sin(2πt)
sin(2πx) cos(2πy )
4π
The strain tensor is:
1
ε(u) = (∇u + (∇u)T )
2
The stress tensor is:
σ(u) = 2µε(u) + λ(∇ · u) · I
Zhuoran Wang
Poroelasticity
Numerical experiments for linear poroelasticity
p is the known scalar valued pressure function:
p = sin(2πt) sin(2πx) sin(2πy ).
cos(2πx) sin(2πy )
∇p = 2π sin(2πt)
sin(2πx) cos(2πy )
Zhuoran Wang
Poroelasticity
Numerical experiments for linear poroelasticity
So right hand side of linear poroelasticity:
cos(2πx) sin(2πy )
f = (−2µ − λ + α)2π sin(2πt)
,
sin(2πx) cos(2πy )
s = (sin(2πx) sin(2πy ))(2π cos(2πt)(c0 + α) + 8π 2 κ sin(2πt)).
Zhuoran Wang
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Following figures are numerical displacement and pressure based on
different κ when n = 32.
Numerical displacement elementwise at final time
1
Numerical pressure elementwise at fime time
1
0.8
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.2
0.5
0.5
0
0.4
0.4
-0.2
0.3
0.3
0.2
0.2
0.1
0.1
0.6
0.4
-0.4
-0.6
0
0
0.2
0.4
0.6
0.8
1
-0.8
0
0
0.2
0.4
0.6
0.8
1
Figure: Left: Displacement with n = 32, κ = 10−6 . Right: Pressure with n =
32, κ = 10−6 .
Zhuoran Wang
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Numerical pressure elementwise at fime time
Numerical displacement elementwise at final time
1
1
0.9
0.8
0.9
0.8
0.8
0.6
0.7
0.7
0.4
0.6
0.6
0.2
0.5
0.5
0
0.4
0.4
-0.2
0.3
0.3
-0.4
0.2
0.2
-0.6
0.1
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.8
0
0
0.2
0.4
0.6
0.8
1
Figure: Left: Displacement with n = 32, κ = 1. Right: Pressure with n = 32,
κ = 1.
Zhuoran Wang
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Numerical displacement elementwise at final time
1
Numerical pressure elementwise at fime time
1
0.9
0.8
0.9
0.8
0.8
0.6
0.7
0.7
0.4
0.6
0.6
0.2
0.5
0.5
0
0.4
0.4
-0.2
0.3
0.3
-0.4
0.2
0.2
-0.6
0.1
0.1
0
0
0.2
0.4
0.6
0.8
1
-0.8
0
0
0.2
0.4
0.6
0.8
1
Figure: Left: Displacement with n = 32, κ = 103 . Right: Pressure with n =
32, κ = 103 .
Zhuoran Wang
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
The table shows the maximum of differences of the exactly
displacement and numerical displacement, and the maximum of
differences of the exactly pressure and numerical pressure with n =
16, 32, 64, κ = 10−6 , 1, 103 .
Table: Errors of numerical value with different κ
error
n = 16
n = 32
n = 128
κ = 10−6
max(ErrDsplT) max(ErrPresT)
2.6488E-03
2.4404E-01
1.6019E-03
1.3404E-01
8.6522E-04
7.0095E-02
κ=1
max(ErrDsplT) max(ErrPresT)
8.1006E-03
1.6961E-02
3.9255E-03
5.4317E-03
1.9366E-03
1.9185E-03
Zhuoran Wang
Poroelasticity
κ = 103
max(ErrDsplT) max(ErrPresT)
8.3049E-03
1.2560E-02
4.0192E-03
3.1964E-03
1.9811E-03
8.0311E-04
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
For calculating L2 error in space in one time step, find the
difference between exact displacement u(·, tn ) and the numerical
(n)
value uh (·) and then calculate the L2 error on the domain Ω.
Z 2
(n)
u(·, tn ) − uh (·) ,
Ω
where tn means the time step. On the unit square domain, we
have a mesh. And we calculate the L2 error on each element
simultaneously.
2
X Z (n)
u(·, tn ) − uh (·) .
E ∈εh
E
Zhuoran Wang
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Here, we have the same time step ∆t with NT time steps totally.
So the L2 error in displacement and time is
v
u NT
Z 2
uX
(n)
t
L2(L2)err =
∆tn
u(·, tn ) − uh (·) .
n=1
Zhuoran Wang
Ω
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Table: Convergence rates of errors in the numerical displacement with
time steps,Q1.
n
n
n
n
n
n
=
=
=
=
=
8
16
32
64
128
L2L2ErrDispl.
2.1187E-03
6.8777E-04
2.6531E-04
1.1697E-04
5.5224E-05
Zhuoran Wang
conv. rate
–
1.6232
1.3742
1.1815
1.0828
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Another example is on a square domain. The final time as
T = 10−3 . The value of permeability is κ = 10−6 . The Lamé
coefficients are λ = 12500 and µ = 8333. On the top edge of the
domain, p = 0, σn = (0, −1)T . The boundary conditions of other
sides are: ∇p · n = 0, u = 0.
Zhuoran Wang
Poroelasticity
Numer. Exp.: Rectangular Meshes:
Profiles of numerical displacement & pressure
Numerical pressure elementwise at final time
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0.2
0.4
0.6
0.8
1
Figure: Left: Numerical Pressure with n = 40. Right: Contours of numerical
pressure with n = 40.
The figure shows the numerical pressure at final time. From the
contour of the numerical pressure, we can see the pressure value.
Zhuoran Wang
Poroelasticity