Poroelasticity Zhuoran Wang Colorado State University Zhuoran Wang - - PowerPoint PPT Presentation

Poroelasticity

Zhuoran Wang

Colorado State University

Zhuoran Wang Poroelasticity

Linear poroelasticity

Poroelasticity equation:

• −∇ · (2µε(u) + λ(∇ · u)I) + α∇p = f

∂t(c0p + 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: u = − 1 4π sin(2πt) cos(2πx) sin(2πy) sin(2πx) cos(2πy)

• .

The strain tensor is: ε(u) = 1 2(∇u + (∇u)T) 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). ∇p = 2π sin(2πt) cos(2πx) sin(2πy) sin(2πx) cos(2πy)

• Zhuoran Wang

Poroelasticity

Numerical experiments for linear poroelasticity

So right hand side of linear poroelasticity: f = (−2µ − λ + α)2π sin(2πt) cos(2πx) sin(2πy) sin(2πx) cos(2πy)

• ,

s = (sin(2πx) sin(2πy))(2π cos(2πt)(c0 + α) + 8π2κ 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.

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical displacement elementwise at final time

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical pressure elementwise at fime time

• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8

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

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical displacement elementwise at final time

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical pressure elementwise at fime time

• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8

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

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical displacement elementwise at final time

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical pressure elementwise at fime time

• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8

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

κ = 10−6 κ = 1 κ = 103 error max(ErrDsplT) max(ErrPresT) max(ErrDsplT) max(ErrPresT) max(ErrDsplT) max(ErrPresT) n = 16 2.6488E-03 2.4404E-01 8.1006E-03 1.6961E-02 8.3049E-03 1.2560E-02 n = 32 1.6019E-03 1.3404E-01 3.9255E-03 5.4317E-03 4.0192E-03 3.1964E-03 n = 128 8.6522E-04 7.0095E-02 1.9366E-03 1.9185E-03 1.9811E-03 8.0311E-04 Zhuoran Wang Poroelasticity

• 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 value u(n)

h (·) and then calculate the L2 error on the domain Ω.

• Ω
• u(·, tn) − u(n)

h (·)

• 2

, 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.

• E∈εh
• E
• u(·, tn) − u(n)

h (·)

• 2

.

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 L2(L2)err =

• NT
• n=1

∆tn

• Ω
• u(·, tn) − u(n)

h (·)

• 2

.

Zhuoran Wang Poroelasticity

• Numer. Exp.: Rectangular Meshes:

Profiles of numerical displacement & pressure

Table: Convergence rates of errors in the numerical displacement with time steps,Q1.

n L2L2ErrDispl.

• conv. rate

n = 8 2.1187E-03 – n = 16 6.8777E-04 1.6232 n = 32 2.6531E-04 1.3742 n = 64 1.1697E-04 1.1815 n = 128 5.5224E-05 1.0828

Zhuoran Wang 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 Lam´ e 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

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Numerical pressure elementwise at final time

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

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