Mathematical and numerical modeling of piezoelectric ... · Mathematical and numerical modeling of piezoelectric transducers e EPI POEMS (UMR CNRS-ENSTA-INRIA) Collaboration CEA (LIST)-INRIA

POems (INRIA) Patrick Joly

Patrick Joly (INRIA)

joint work with Sébastien Impériale (CEA-INRIA)

Mathematical and numerical modeling of piezoelectric transducers

eEPI POEMS (UMR CNRS-ENSTA-INRIA)

Collaboration CEA (LIST)-INRIA

Conference NUMPDE - Heraklion - September 2011

Motivation: Non Destructive Testing

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We want to model the emission and the reception of the signal by the piezoelectric transducer/sensor.


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Motivation: Closer look on the sensor

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Support

Matching layerBacking (viscoelastic)

Piezo-composite

Piezoelectric bars in a polymer

Thin electrodes

Ground electrode


Motivation: Non Destructive Testing

4POems (INRIA) Patrick Joly

ρ∂2

∂t2u−Div C e(u) = f

Linear elastodynamic equations

e(u) = (∇u+∇uT )/2

µ∂

∂tH +∇× E = 0

ε∂

∂tE −∇×H =

Linear Maxwell equations

0− ∂

∂tdT e(u)

Equations of piezoelectricity

ε

H

µ

: magnetic field: electric field

: magnetic permeability: electric permittivity

E

u

ρ

e: displacement field: strain tensor

: density: elasticity tensorC

Piezoelectric effects

5

−Div dE

→d piezoelectric tensor: vector matrix

electric stress

elastic current


ΩaΓint

Equations of piezoelectricity

6

⊂ΩP ΩS

ΩS

ΩS

R3(E,H) ulives in lives in ΩS,

d(x) = 0 ⇔ x ∈ ΩP (support of the coupling terms)

∂ΩP = Γel ∪ Γint

Γel


ρ∂

2

∂t2u−Div Ce(u) = −Div dE + f

µ∂

∂tH +∇× E = 0

ε∂

∂tE −∇×H = − ∂

∂tdT

e(u)

7

The mathematical model

R3

R3

ΩS


Adimensionalization : space and time variables

T ∼

x −→ x/L, t −→ t/T, L = cs T

cs ∼ velocity of elastic waves (speed of sound)

L ∼ typical size of the investigated domain

time scale for elastic waves

ρ∂

2

∂t2u−Div Ce(u) = −Div dE + f

µ∂

∂tH +∇× E = 0

ε∂

∂tE −∇×H = − ∂

∂tdT

e(u)

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The mathematical model

Adimensionalization :

ρr∂

2

∂t2u−Div Cre(u) = −Div

d√C0ε0

E + f

µr∂

∂tH +

1√

µ0ε0

ρ0

C0∇× E = 0

εr∂

∂tE − 1

√µ0ε0

ρ0

C0∇×H = − ∂

∂t

dT

√C0ε0

e(u)

ε(x) = ε0εr(x)µ(x) = µ0µr(x) ρ(x) = ρ0ρr(x) C = C0Cr(x)

ρ0 = maxx

ρ(x) C0 = maxx

maxσ(C(x))

R3

R3

ΩS


A first difficulty

δ := << 1csc∞

=1

δ

= dr

9

c∞ := 1/√µ0ε0 = speed of light

cs :=

C0/ρ0 ~ speed of sound

3× 108~ m/s

< m/s 5× 103

multiscale phenomenon, very hard to handle numerically

ρr∂

2

∂t2u−Div Cre(u) = −Div

d√C0ε0

E + f

µr∂

∂tH +

1√

µ0ε0

ρ0

C0∇× E = 0

εr∂

∂tE − 1

√µ0ε0

ρ0

C0∇×H = − ∂

∂t

dT

√C0ε0

e(u)


ρr∂

2

∂t2u

δ −Div Cre(uδ) = −Div drEδ + f

µr∂

∂tH

δ +1δ∇× E

δ = 0

εr∂

∂tE

δ − 1δ∇×H

δ = − ∂

∂tdT

r e(uδ)

We are going to use an approximate model for smallknown as the quasi-static model

that can be justified rigorously by a limit process

δ

δ → 0

POems (INRIA) 10

The quasi static approach

Patrick Joly

|∂tB| << |∇× E|

−∂tB = ∇× E


Fundamentals of Piezoelectricity.

Takuro Ikeda 1984 !!!


coupled elliptic - hyperbolic system

ρr∂2

∂t2u−Div Cr e(u) = Div dr ∇V + f in ΩS

∇ · εr∇V = ∇ · dTr e(u) in R3

ρr∂

2

∂t2u


µr∂

∂tH

δ +1δ∇× E

δ = 0

εr∂

∂tE

δ − 1δ∇×H

δ = − ∂

∂tdT

r e(uδ)

12


∇× E = 0 =⇒ E = −∇V

∇×H = 0 H = 0=⇒

At the limit , we expect thatδ → 0

div(µrH) = 0

The quasi-static model

We take the divergence

electric potential

and integrate in time


div

ρr∂

2

∂t2u


µr∂

∂tH

δ +1δ∇× E

δ = 0

εr∂

∂tE

δ − 1δ∇×H

δ = − ∂

∂tdT

r e(uδ)

ρ∂2

∂t2u−Div C e(u) = Div d∇V + f in ΩS

∇ · ε∇V = ∇ · dT e(u) in R3

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E = −∇V H = 0

Eδ =

ΩS

ρr|∂tuδ|2 +Cr e(u

δ) : e(uδ)+

R3

εr|Eδ|2 + µr|Hδ|2

E =

ΩS

ρr|∂tu|2 +Cr e(u) : e(u)

+

R3

εr |∇V |2


||uδ − u||L∞

0,T ;H1(ΩS)

+ ||Eδ − E||L∞

0,T ;L2(ΩS)

+ ||Hδ||L∞

0,T ;L2(ΩS)

≤ C T δ

Theorem : Provided that , with f ∈ C3(0, T, L2(ΩS)) E = −∇V

ΩP

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Ingredients of the proof

div(εrAδ) = 0Eδ = −∇V δ +AδHemholtz decomposition with

Energy estimates + Gronwall’s lemma


||uδ − u||L∞

0,T ;H1(ΩS)

+ ||Eδ − E||L∞

0,T ;L2(ΩS)

+ ||Hδ||L∞

0,T ;L2(ΩS)

≤ C T δ


(uδ − u, V δ − V )

with as a source termAδ(uδ − u, V δ − V ) solve quasi-static equations with support inside the piezoelectric domain

≤ C AδL2(ΩP )

Using + stability estimates∇×Aδ ≡ ∇× E

δ = µr δ ∂tHδ

≤ C δα AδL2(ΩP ) ≤

15


Ingredients of the proof

Generalized Hardy-Poincaré-Friedrichs inequality

Aδ

1 + |x|

L2(R3)

≤ C∇ ×AδL2(R3) + div(εr Aδ)L2(R3)


||uδ − u||L∞

0,T ;H1(ΩS)

+ ||Eδ − E||L∞

0,T ;L2(ΩS)

+ ||Hδ||L∞

0,T ;L2(ΩS)

≤ C T δ


div(εrAδ) = 0Eδ = −∇V δ +AδHemholtz decomposition with

∇ ×AδL2(R3) ≤ C δ

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A second difficulty

V


ρ∂2


∇ · ε∇V = ∇ · dT e(u) in R3

d(x) = 0 ⇔ x ∈ ΩP

V

V

u

u

Difficulty : is defined in an unbounded domain (the whole space).

However, we need only inside .

V

V ΩP

εPr = εr|ΩP εextr = εr|R3\ΩP

∂V P

∂n− 1

εPrdTr e(u) · n =

εextr

εPr

∂V ext

∂n

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A second difficulty

From we deduce that ∇ · εr∇V = ∇ · drT e(u)

maxx∈R3\ΩP

εextr (x) << minx∈ΩP

εPr (x)

A good approximate boundary condition is

∂ΩP \ Γelεr∂V

∂n+ dT

r e(u) · n = 0 on

ΩP

n

ΩS

<< 1εPr∂V P

∂n− dT

r e(u) · n = εextr∂V ext

∂n


It is possible to reduce the numerical calculations for to domain without using complex (nonlocal) boundary conditionsthanks to the following observation :

ΩPV

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Piezoelectric sensors: emission / reception


ΩP =N

=1

ΩP

connectedcomponents

RElectrods : perfectly

conducting thin screens

V is constantalong each of them

V = 0

Γ+el

Γ−el

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Ground electrod : the potential will be imposed to 0 .

Boundary conditions

Upper : we apply Ohm’s law.

R

V g = 0Reception

V g = 0Emission

Piezoelectric sensors: emission / reception

V = V g −Rd

dt

Γ+el

(εr∇V − dTr e(u)) · n


Boundary conditions : summary

Electric boundary conditions

V = 0

Elastic boundary conditions

(Cre(u) + dr∇V ) · n = 0 on ∂ΩS

V = V g −Rd

dt

Γ+el


V is constantΓ+elalong each

εr∂V

∂n− dT

r e(u) · n = 0


εr ∂nV − dTr e(u) · n = 0

The retained global model

V = 0

(Cre(u) + dr∇V ) · n = 0 on ∂ΩS

V = V g −Rd

dt

Γ+el



ρ∂2


∇ · ε∇V = ∇ · dT e(u) in R3

u ∈ VS := [H1(ΩS)]3

V0P =

W ∈ H

1(ΩP ) / W = 0 on Γel

VstatP = span

V 1, V 2, ..., V N

supp V = ΩP div(εr∇V ) = 0, (Ω

P )

V = 0 on Γ−,el, V = 1 on Γ+

,el,

∂nV = 0 on ∂ΩP \ (Γ−

,el ∪ Γ+,el),

Variational formulation and discretization

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V = V0 +N

l=1

λV

V ∈ VP := V0P ⊕ Vstat

P


λ : electrod potentials

(V0,λ)−→

u ∈ VS := [H1(ΩS)]3 V ∈ VP := V0

P ⊕ VstatP

V0P =

W ∈ H

1(ΩP ) / W = 0 on Γel

VstatP = span

V 1, V 2, ..., V N


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V 1 V 2 V 3 V 41

0

1

0

1

0

1

0


λ : electrod potentials

(V0,λ)−→V = V0 +

N

l=1

λV

( if is constant )εr

ρ∂2

∂t2u−Div Cr e(u) = Div dr ∇V in ΩS



+ (Neumann like) elastic boundary conditions

∀ v ∈ VSd2

dt2

ΩS

ρu · v +

ΩS

Cr e(u) : e(v) = −

ΩP

dr ∇(V 0 +

i

λiV i) : e(v)

× test functions in VSv

∇ · εr∇V = ∇ · dTr e(u) in ΩP

+ electric boundary conditions

test functions in 1 V0PW0×

∀ W0 ∈ V0P

ΩP

εr ∇V0 +

i

λiV i

·∇W0 =

ΩP

dTr e(u) ·∇W0

∀ v ∈ VSd2

dt2

ΩS

ρu · v +

ΩS

Cr e(u) : e(v) = −

ΩP

dr ∇(V 0 +

i

λiV i) : e(v)

∀ W0 ∈ V0P

ρ∂2


×

1

ΩP

εr ∇V0 ·∇W0 =

ΩP

dTr e(u) ·∇W0

V0P




test functions in VSv



× test functions in W0

test functions

∀ v ∈ VSd2

dt2

ΩS

ρu · v +

ΩS

Cr e(u) : e(v) = −

ΩP

dr ∇(V 0 +

i

λiV i) : e(v)

ρ∂2


×

2 V







×

1 ≤ ≤ Nd

dt

ΩP

εr∇(V0 +

i

λiV i) ·∇V =d

dt

ΩP

dTr e(u) · V −

1

Rλ +

1

RV

g

test functions

∀ v ∈ VSd2

dt2

ΩS

ρu · v +

ΩS

Cr e(u) : e(v) = −

ΩP

dr ∇(V 0 +

i

λiV i) : e(v)

ρ∂2


×

2 V

− 1

Rλ +

1

RV

g







×

ΩP

εr∇V ·∇V

dλ

dt=

d

dt

ΩP

dTr e(u) · V 1 ≤ ≤ N



∀ v ∈ VSd2

dt2

ΩS

ρu · v +

ΩS

Cr e(u) : e(v) = −

ΩP

dr ∇(V 0 +

i

λiV i) : e(v)

ΩP

εr∇V ·∇V

dλ

dt+

1

Rλ =

d

dt

ΩP

dTr e(u) · V +

1

RV

g 1 ≤ ≤ N

ΩP

εr ∇V0 ·∇W0 =

ΩP

dTr e(u) ·∇W0 ∀ W0 ∈ V0

P

Md2u

dt2+Ku = −B V0 −Bstat Λ

AV0 = BTu

MstatdΛ

dt+R−1

stat Λ = BTstat

du

dt+R−1

statVg

After space semi-discretization with finite elements, we get



Md2u


AV0 = BTu

MstatdΛ

dt+R−1

stat Λ = BTstat

du

dt+R−1

statVg

K ←→ −DivCre(·)

B ←→ −Divdr∇

BT ←→ ∇ ·dTr ∇

A ←→ −∇ ·εTr ∇

Mstat,Rstat

diagonal

−→ M

Mass lumping process

diagonal

It is useful to eliminate via to getV0 = A−1BTuV0



Md2u

dt2+K∗ u = −Bstat Λ

MstatdΛ

dt+R−1

stat Λ = BTstat

du

dt+R−1

statVg

Md2u


AV0 = BTu

MstatdΛ

dt+R−1

stat Λ = BTstat

du

dt+R−1

statVg

K∗ = K+BA−1BT augmented stiffness matrix


Sébastien Imperiale 31POems (INRIA)

Md2u

dt2+K∗ u = −Bstat Λ

MstatdΛ

dt+R−1

stat Λ = BTstat

du

dt+R−1

statVg


Time discretization by central finite differences

Mun+1 − 2un + un−1

∆t2+K∗ un = −B∗

statΛn+1 +Λn−1

2

MstatΛn+1 −Λn−1

2∆t+R−1

statΛn+1 +Λn−1

2= BT

statun+1 − un−1

2∆t+R−1

statVng



Mun+1 − 2un + un−1

∆t2+K∗ un = −B∗

statΛn+1 +Λn−1

2


2∆t+R−1

statΛn+1 +Λn−1

2= BT

statun+1 − un−1

2∆t+R−1

statVng

The advantage of the implicit discretization is that the stability condition of the global scheme is not affected by the electric supply process.

K∗unThe discretization in is explicit but the evaluation of requires the solution of a (small, block diagonal) linear system (laplacians in )Ω

p

u


The discretization in is implicit but its calculation is explicit since thematrices are diagonal.

ΛMstat,Rstat



Mun+1 − 2un + un−1

∆t2+K∗ un = −B∗

statΛn+1 +Λn−1

2


2∆t+R−1

statΛn+1 +Λn−1

2= BT

statun+1 − un−1

2∆t+R−1

statVng


En+ 12 :=

1

2M

un+1 − un

∆t· u

n+1 − un

∆t+

1

2K∗ un+1 · un

∆t2

4M−1 K∗ ≤ 1 ∼ ∆t

h≤ Cste

BA−1BT ∼ 2nd order

differential operator

Echos from the bottom

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Numerical results


Piezoelectric sensor (8 bars)

Homogeneous mediumNorm of the displacement

Electric current

time

V g(t) = V g(t− τ)

τ = τ, ∀

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Numerical results


τ

τ

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Numerical results

POems (INRIA) S. Imperiale

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Numerical results

POems (INRIA) S. Imperiale

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Absolute value of the displacement

y-component of the displacement

Electric source :

Gaussian pulseBackingPiezo : homogeneous

That’s all folks !

38Patrick Joly

Thank you for your attention.

POems (INRIA)

Title PresentationVenue 39

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Mathematical and numerical modeling of piezoelectric ... · Mathematical and numerical modeling of piezoelectric transducers e EPI POEMS (UMR CNRS-ENSTA-INRIA) Collaboration CEA (LIST)-INRIA