Impedance Transmission Conditions for the Electric Potential across a Highly Conductive Casing

Impedance Transmission Conditions for the Electric Potential across a Highly Conductive Casing Hélène Barucq, Aralar Erdozain, David Pardo, Victor Péron To cite this version: Hélène Barucq, Aralar Erdozain, David Pardo, Victor Péron. Impedance Transmission Conditions for the Electric Potential across a Highly Conductive Casing. V-MAD6 - Valparaíso s Mathematics and its Applications Days, Jan 2016, Valparíso, Chile. <hal-01692286> HAL Id: hal-01692286 https://hal.inria.fr/hal-01692286 Submitted on 24 Jan 2018 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Impedance Transmission Conditions for the Electric Potential across a Highly Conductive Casing Aralar Erdozain 1, Victor Péron 1,2, Hélène Barucq 1,2 and David Pardo 3,4,5 1 INRIA Bordeaux Sud-Ouest, Team magique 3D 2 Université de Pau et des Pays de l Adour 3 University of the Basque Country (UPV/EHU), Leioa, Spain 4 Basque Center for Applied Mathematics (BCAM), Bilbao, Spain 5 Ikerbasque (Basque Foundation for Sciences), Bilbao, Spain 1 / 42

MOTIVATION Main goal: To obtain a better characterization of the Earth s subsurface How: Recording borehole resistivity measurements Procedure: Well Transmitters Receivers Transmitter Receivers 2 / 42

MOTIVATION Practical Difficulties: It is not easy to drill a borehole It may collapse Practical Solutions: Use a metallic casing Surround with a cement layer Problem solved, but... 3 / 42

MOTIVATION Practical Difficulties: It is not easy to drill a borehole It may collapse Practical Solutions: Use a metallic casing Surround with a cement layer Problem solved, but... Numerical problems due to the high conductivity and thinness of the casing 4 / 42

CONFIGURATION OF INTEREST SECTIONED 3D DOMAIN z z MERIDIAN DOMAIN CASING y CASING r x We reduce the 3D problem to a 2D problem due to the axysimmetric configuration. 5 / 42

MAIN IDEA ASYMPTOTIC MODELS REFERENCE MODEL Ω ε i Ω ε e CASING 1st Approach Ω ε i Γ ε i EQUIVALENT CONDITIONS Γ ε e Ω ε e Γ ε i Γ ε e Idea: Replace the casing by equivalent conditions. 2nd Approach Ω i EQUIVALENT CONDITIONS Γ Ω e 6 / 42

REALISTIC SCENARIO 14cm 5cm 1Ωm 2Ωm 5Ωm 20Ωm 5Ωm 5m 2m 5m Conductivity and casing width: { ε = 1.27e 2m σ c = 4.34e6 Ω 1 m 1 σ c ε 3 Classical approach: σ c = α α R High conductive case: 1.27cm, 2.3e 7 Ωm σ c = αε 3 α R 7 / 42

REFERENCES [1] A.A. Kaufman. The electrical field in a borehole with a casing. Geophysics, Vol.55, Issue 1, pp. 29-38, 1990. [2] D.Pardo, C.Torres-Verdín and Z.Zhang. Sensitivity study of borehole-to-surface and crosswell electromagnetic measurements acquired with energized steel casing to water displacement in hydrocarbon-bearing layers. Geophysics, 73 No.6, F261-F268, 2008. [3] M. Duruflé, V. Péron and C. Poignard. Thin Layer Models for Electromagnetism.Communications in Computational Physics 16(1):213-238, 2014. [4] K. Schmidt, A. Chernov, Robust transmission conditions of high order for thin conducting sheets in two dimensions, IEEE Trans. Magn., 50 (2014), pp. 41 44. 8 / 42

OUTLINE 1 Equivalent Conditions 2 Numerical Results 3 Second Approach 4 Application 5 Perspectives 9 / 42

OUTLINE 1 Equivalent Conditions 2 Numerical Results 3 Second Approach 4 Application 5 Perspectives 10 / 42

MODEL PROBLEM - 3D STATIC ELECTRIC POTENTIAL div ((σ iɛω) u) = f in Ω Ω = Ω ε i Ω ε c Ω ε e Γ ε i Γ ε e z x Γ ε e Γ ε i Ω ε i Ω ε c Ω ε e y Where f is a known data, σ is piecewise constant σ i σ = σ c = αε 3 σ e in Ω ε i in Ω ε c in Ω ε e and the solution is expressed as u = u i u c u e in Ω ε i in Ω ε c in Ω ε e 11 / 42

MODEL PROBLEM - 2D Reference problem (P) set in the domain Ω = Ω ε i Ω ε c Ω ε e Γ ε i Γ ε e. We denote the boundary Γ 0 := ( Ω ε i Γ ε i ) ( Ω ε e Γ ε e). y σ i u i = f i in Ω ε i Ω ε i Ω ε c Ω ε e ε σ c u c = 0 in Ω ε c σ e u e = f e in Ω ε e u i = u c on Γ ε i u c = u e on Γ ε e Γ ε i Γ Γ ε e σ i n u i = σ c n u c on Γ ε i x 0 x σ c n u c = σ e n u e on Γ ε e u = 0 on Ω 12 / 42

EXISTENCE AND UNIQUENESS Theorem 1 There exists ε 0 > 0 s.t. for all ε (0, ε 0 ), if f i L 2 (Ω ε i ) and f e L 2 (Ω ε e), then!u H0 1 (Ω) solution of (P) and ) u 1,Ω C ( f i 0,Ω + f εi e 0,Ω. εe Proof: Derive variational formulation, for all v H0 1 (Ω), a(u, v) = b(v). Prove that a coercive and continuous, b continuous in H0 1 (Ω). Apply the Lax-Milgram Lema. 13 / 42

DEFINITIONS Definition 1 We define the jump and mean value of the solution u across the casing as [u] = u Γ ε e u Γ ε i {u} = 1 2 (u Γ εe + u Γ εi ) Definition2 Let u be the reference solution. We say an asymptotic model is of Order k+1, if its solution u [k] satisfies u u [k] H 1 Cε k+1 when ε 0 14 / 42

METHODOLOGY Step1: Derive an Asymptotic Expansion for u when ε 0 In the casing: u c (x, y) = ( ) x ε n U n x0, y ε n N Outside the casing: u(x, y) = n N ε n u n (x, y) Step2: Obtain Equivalent Conditions of order k + 1 by identifying a simpler problem satisfied by the truncated expansion and neglecting the higher order terms in ε u k,ε := u 0 + εu 1 + ε 2 u 2 +... + ε k u k Step3: Prove Convergence rates for asymptotic models 15 / 42

MULTISCALE EXPANSION X 2 Uk + y 2 U k 2 = 0 in Ω ε c σ i ui k = f i in Ω ε i σ e ue k = f e in Ω ε e PROCEDURE U 0 u 0 u k i = U k on Γ ε i U k = u k e on Γ ε e σ i x u k 4 i = α X U k on Γ ε i α X U k = σ e x u k 4 e on Γ ε e U k = 0 on Ω U 1 u 1 U 2 u 2... 16 / 42

MULTISCALE EXPANSION X 2 Uk + y 2 U k 2 = 0 in Ω ε c σ i ui k = f i in Ω ε i σ e ue k = f e in Ω ε e PROCEDURE U 0 u 0 u k i = U k on Γ ε i U k = u k e on Γ ε e σ i x u k 4 i = α X U k on Γ ε i α X U k = σ e x u k 4 e on Γ ε e U k = 0 on Ω U 1 u 1 U 2 u 2... 20 / 42

{ σi u i = f i in Ω ε i EQUIVALENT MODELS FIRST APPROACH (Transmission conditions across the casing) Order 2: u i = 0 on Ω ε i { σe u e = f e in Ω ε e (P 2 ) u e = 0 on Ω ε e σ i u i = f i in Ω ε i Order 4: σ e u e = f e in Ω ε e [u] = 0 [σ n u] = α ε 2 Γ ε {u} u = 0 on Γ 0 (P 4 ) 21 / 42

CONVERGENCE Theorem 2 There exists ε 0 > 0 s.t. for all ε (0, ε 0 ), if f i L 2 (Ω ε i ) and f e L 2 (Ω ε e), then!u [k] V k+1 solution of P k+1, k = 1, 3, and ( ) u [k] 1,Ω ε i Ω ε C f e i 0,Ω ε i + f e 0,Ω ε e u u [k] 1,Ω ε i Ω ε e Cεk+1 Where V 2 = { v : v i H 1 0 (Ω ε i ), v e H 1 0 (Ω ε e) } V 4 = { v : v i H 1 (Ω ε i ), v e H 1 (Ω ε e), Γ ε {v} L 2 (Γ ε ), v Γ ε i = v Γ ε e, v Γ0 = 0 } 22 / 42

OUTLINE 1 Equivalent Conditions 2 Numerical Results 3 Second Approach 4 Application 5 Perspectives 23 / 42

FEM CODE Classic Finite Element Method code Straight triangular elements (h refinement) Lagrange shape functions of any degree (p refinement) Domain Mesh Solution Σ i 5 Σ c Σ e 3 24 / 42

QUALITATIVE COMPARISON Reference Model Order 2 Model Order 4 Model 25 / 42

CONVERGENCE RATES 10 0 ERROR IN LOGARITHMIC SCALE H1 Relative Error (%) 10 5 order 2 order 4 10 10 0.006 0.010 0.016 0.025 0.040 0.063 0.100 0.158 Casing Thickness (m) Casing Thickness (ε) 0.0117 0.0234 0.0469 0.0938 Expected (ε 0) Order 2 Slopes 1.9969 1.9899 1.9638 1.8624 2 Order 4 Slopes 4.0046 4.0039 3.9879 3.8992 4 26 / 42

OUTLINE 1 Equivalent Conditions 2 Numerical Results 3 Second Approach 4 Application 5 Perspectives 27 / 42

EQUIVALENT MODELS SECOND APPROACH (Transmission conditions across the interface Γ) { σi u i = f i in Ω i Order 1: u i = 0 on Ω i { σe u e = f e in Ω e (P 1 ) u e = 0 on Ω e σ i u i = f i in Ω i Order 2: σ e u e = f e in Ω e [u] = ε { n u} [ n u] = 4 ε {u} u = 0 on Ω 28 / 42

Stability Problem Stability problems for the order 2 model of the second approach due to a non coercive term. Γ δ i Γ Γ δ e Solution Use of Artificial Boundaries to move the transmission conditions and recover stability. For δ > 0.5 Γ δ i = {(x, y) : x = x 0 δε, y (0, y 0 )} Γ δ e = {(x, y) : x = x 0 + δε, y (0, y 0 )} 29 / 42

EQUIVALENT MODELS SECOND APPROACH (Transmission conditions across the artificial interfaces) σ i u i = f i in Ω i Order 2: σ e u e = f e in Ω e [u] = ε (1 2δ) { n u} 4 (1 2δ) [ n u] = {u} ε u = 0 on Ω (P 2 ) Stable if δ > 0.5 30 / 42

CONVERGENCE Theorem 3 There exists ε 0 > 0 s.t. for all ε (0, ε 0 ), if f i L 2 (Ω ε i ) and f e L 2 (Ω ε e), then!u [k] V k+1 solution of P k+1, k = 1, 2, and ( ) u [k] 1,Ω ε i Ω ε C f e i 0,Ω ε i + f e 0,Ω ε e u u [k] 1,Ω ε i Ω ε e Cεk+1 Where V 1 = { v : v i H 1 0 (Ω i ), v e H 1 0 (Ω e ) } V 2 = { v : v i H 1 (Ω i ), v e H 1 (Ω e ), v i Γ = v e Γ, v Γ0 = 0 } 31 / 42

QUALITATIVE COMPARISON Reference Model Order 1 Model Order 2 Model 32 / 42

CONVERGENCE RATES L 2 10 1 ERROR IN LOGARITHMIC SCALE L2 Relative Error (%) 10 2 10 3 10 4 order 1 order 2 order 2 displaced 0.006 0.010 0.016 0.025 0.040 Casing Thickness (m) Casing Thickness (ε) 0.008 0.011 0.015 0.020 0.027 Expected (ε 0) Order 1 Slopes 0.972 0.962 0.948 0.928 0.899 1 Order 2 Slopes 2.032 2.011 2.062 2.090 1.9636 2 Order 2 displaced 2.005 2.006 2.007 2.008 2.007 2 33 / 42

CONVERGENCE RATES H 1 10 1 ERROR IN LOGARITHMIC SCALE H1 Relative Error (%) 10 2 10 3 10 4 order 1 order 2 order 2 displaced 0.006 0.010 0.016 0.025 0.040 Casing Thickness (m) Casing Thickness (ε) 0.008 0.011 0.015 0.020 0.027 Expected (ε 0) Order 1 Slopes 0.982 0.975 0.966 0.953 0.935 1 Order 2 Slopes 1.915 1.302 2.463 2.569-0.125 2 Order 2 displaced 2.004 2.005 2.006 2.008 2.007 2 34 / 42

COMPARISON 10 0 ERROR IN LOGARITHMIC SCALE H1 Error 10 5 10 10 10 2 Casing Thickness (m) Order Stability ε-independent Domain Model 1 No Gap 1 Model 2 No Gap 2 Displaced Model 2 Model 1 Gap 2 Model 2 Gap 4 35 / 42

OUTLINE 1 Equivalent Conditions 2 Numerical Results 3 Second Approach 4 Application 5 Perspectives 36 / 42

APPLICATION 1.27cm, 2.3e 7 Ωm Equation: div (σ u) = f z = z 3 z = z 2 z = z 1 Receivers Right hand side: { 1 In the transmitter f = 0 Outside the transmitter z = z 0 Transmitter Objective: Measure the second difference of potential on the Receivers U 2 = u(z 1 ) 2u(z 2 ) + u(z 3 ) Expected Result: (Model of Kaufman) Relation between second difference of potential and rock resistivity of the form U 2 = k ρ 1 2 rock k R 37 / 42

VARYING ROCK CONDUCTIVITY Second Difference of Potential (V) 10 4 10 3 10 2 10 1 10 0 10 1 10 2 10 3 10 4 Rock Resistivity (ohm m) Reference Order 4 Resistivity [1,10] [10,10 2 ] [10 2,10 3 ] [10 3,10 4 ] Expected Reference Model Slopes -0.4914-0.4924-0.4978-0.4991-0.5 Order 4 Slopes -0.4914-0.4924-0.4977-0.4993-0.5 38 / 42

TWO ROCK LAYERS 1.5 1.6 Theoretical Reference Order 4 z 0.15m 0.15m 1.85m σ 1 = 10 σ 1 = 1000 Instruments z Position (m) 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 10 1 10 2 10 3 Second Difference of Potential (V) 39 / 42

OUTLINE 1 Equivalent Conditions 2 Numerical Results 3 Second Approach 4 Application 5 Perspectives 40 / 42

Perspectives Obtain semianalytical solutions to reduce the computational cost. Consider physically more realistic scenarios. Develop 3D electromagnetic models. Study highly deviated boreholes. 41 / 42