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

Poroelasticity Zhuoran Wang Colorado State University Zhuoran Wang Poroelasticity

Linear poroelasticity Poroelasticity equation: � −∇ · (2 µε ( u ) + λ ( ∇ · u ) I ) + α ∇ p = f , (1) ∂ t ( c 0 p + α ∇ · u ) + ∇ · ( − K ∇ p ) = s where µ = 1 , λ = 1 , α = 1 , c 0 = 0 . 1 , K = κ I . 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 u D = u and for pressure is p D = p . Zhuoran Wang Poroelasticity

Numerical experiments for linear poroelasticity u is the known vector valued displacement function: u = − 1 � cos(2 π x ) sin(2 π y ) � 4 π sin(2 π t ) . sin(2 π x ) cos(2 π y ) The strain tensor is: ε ( u ) = 1 2( ∇ u + ( ∇ u ) T ) The stress tensor is: σ ( u ) = 2 µε ( u ) + λ ( ∇ · u ) · I Zhuoran Wang Poroelasticity

Numerical experiments for linear poroelasticity p is the known scalar valued pressure function: p = sin(2 π t ) sin(2 π x ) sin(2 π y ) . � cos(2 π x ) sin(2 π y ) � ∇ p = 2 π sin(2 π t ) sin(2 π x ) cos(2 π y ) Zhuoran Wang Poroelasticity

Numerical experiments for linear poroelasticity So right hand side of linear poroelasticity: � cos(2 π x ) sin(2 π y ) � f = ( − 2 µ − λ + α )2 π sin(2 π t ) , sin(2 π x ) cos(2 π y ) s = (sin(2 π x ) sin(2 π y ))(2 π cos(2 π t )( c 0 + α ) + 8 π 2 κ sin(2 π t )) . 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 pressure elementwise at fime time Numerical displacement elementwise at final time 1 1 0.8 0.9 0.9 0.6 0.8 0.8 0.4 0.7 0.7 0.2 0.6 0.6 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 -0.8 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Left : Displacement with n = 32, κ = 10 − 6 . Right : Pressure with n = Figure: 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 0.6 0.8 0.7 0.4 0.7 0.6 0.2 0.6 0.5 0 0.5 0.4 0.4 -0.2 0.3 0.3 -0.4 0.2 0.2 -0.6 0.1 0.1 -0.8 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 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 pressure elementwise at fime time Numerical displacement elementwise at final time 1 1 0.9 0.8 0.9 0.8 0.6 0.8 0.7 0.4 0.7 0.6 0.2 0.6 0.5 0 0.5 0.4 0.4 -0.2 0.3 0.3 -0.4 0.2 0.2 -0.6 0.1 0.1 -0.8 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Left : Displacement with n = 32, κ = 10 3 . Right : Pressure with n = Figure: 32, κ = 10 3 . 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 , 10 3 . Table: Errors of numerical value with different κ κ = 10 − 6 κ = 10 3 κ = 1 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 ( · , t n ) and the numerical value u ( n ) h ( · ) and then calculate the L2 error on the domain Ω. � 2 � � � u ( · , t n ) − u ( n ) h ( · ) , � � � Ω where t n means the time step. On the unit square domain, we have a mesh. And we calculate the L2 error on each element simultaneously. � 2 � � � u ( · , t n ) − u ( n ) � h ( · ) . � � � E E ∈ ε h Zhuoran Wang Poroelasticity

Numer. Exp.: Rectangular Meshes: Profiles of numerical displacement & pressure Here, we have the same time step ∆ t with N T time steps totally. So the L2 error in displacement and time is � N T � � 2 � � � � u ( · , t n ) − u ( n ) � L2(L2)err = ∆ t n h ( · ) . � � � � Ω n =1 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, σ n = (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