Sensitivity analysis for a class of semi-coercive...

Sensitivity analysis for a class ofsemi-coercive variational inequalities

using recession tools.

Samir ADLYUniversite de Limoges

France

Workshop Internationalsur les mathematiques et l’environnement

Essaouira, 23-24 novembre 2012.

References

The talk is based on the following papers :

I K. ADDI, S. ADLY, D. GOELEVEN, H. SAOUD,A sensitivity analysis of a class of semi-coercivevariational inequalities using recession tools, JOGO 2007.

I K. ADDI, S. ADLY, B. BROGLIATO, D. GOELEVEN,A method using the approach of Moreau andPanagiotopoulos for the mathematical formulation ofnon-regular circuits in electronics, Nonlinear Analysis,Hybrid Systems 2007.

I S. ADLY, E. ERNST, M. THERA, Stability of Non-coerciveVariational Inequalities, Communications in ContemporaryMathematics Vol. 4, 1, 145-160 (2002).

Outline

1. Non-coercive Variational inequalities : a state of art2. Stability of semi-coercive variational inequalities3. Ideal diode model : a complementarity formulation4. Set-Valued Ampere-Volt characteristics in

electronics : the convex case5. Some applications in electronics and mechanics

Outline

Non-coercive Variational inequalities : a state of art

Stability of semi-coercive variational inequalities

Ideal diode Model

Set-Valued Ampere-Volt characteristic in electronics

Some applications in electronics and mechanics

Position of the problem

Let us consider the problem :

V .I.(A, f ,Φ,K )

Find u ∈ K such that

〈Au − f , v − u〉+ Φ(v)− Φ(u) ≥ 0,

∀v ∈ KI X is a reflexive Banach space ,I K ⊂ X is a closed convex subset (nonemty),I A : X → X ∗ is an operator,I f ∈ X ∗,I Φ ∈ Γ0(X ) is a convex l.s.c. and proper function.

The coercive case

Several existence results for V .I.(A, f ,Φ,K ) are knownwhen the operator A

(i) is linear and coercive , i.e. ∃α > 0 such that :

〈Au,u〉 ≥ α‖u‖2, ∀u ∈ X,

(ii) is non-linear and coercive i.e. :

lim‖u‖→+∞

〈Au,u〉‖u‖ = +∞.

See the contributions of Brezis, Browder, J.L. Lions,Mosco, Stampacchia, Fichera etc . . .

The non-coercive case

For the Non-coercive case, we refer to the works of :I Fichera (1964)I J.L. Lions & G. Stampacchia (1967)I C. Baiocchi, F. Gastaldi, F. Tomarelli (1985)I C. Baiocchi, G. Buttazzo, F. Gastaldi, F. Tomarelli

(1988)I F. Tomarelli (1993)I D. Goeleven (1994)I S. Adly, D. Goeleven, M. Thera (1996) etc ...I A. Auslender (1996).

MotivationLet us consider the following classical Neumann problem :

N (f )

find u ∈ H1(Ω) such that−∆u = f , in Ω∂u∂n

= 0, on ∂Ω

It is well-known that N (f ) has a solution if and only if∫Ω

f (x)dx = 0.

If we replace f by fε = f + ε (with ε > 0), then N (fε) hasno solutions.

The problem N (f ) is unstable.

Obstacle problem without friction.Let Ω be an open bounded subset of R2 (representing athin elastic membrane).For f ∈ L2(Ω) and Ψ a given obstacle, we consider thefollowing problem :

(P)

−∆u ≥ f , in Ω(−∆u − f )(u −Ψ) = 0, in Ωu = 0, on ∂Ω

fdx

x1

x2

xu(x)

The variational formulation of the problem (P) is :

(P)

Find u ∈ K such that〈Au − f , v − u〉 ≥ 0, ∀v ∈ K

I X = H10 (Ω)

I K = v ∈ X | v ≥ ΨI 〈Au, v〉 =

∫Ω

∇u · ∇v dx

I 〈f , v〉 =

∫Ω

f vdx

If the condition u = 0 on ∂Ω is replaced by∂u∂n

= 0 on ∂Ω

then the operator A : X := H1(Ω)→ X ′ is no morecoercive.

If the problem (P) has a solution, then∫

Ω

fdx ≤ 0.

If∫

Ω

fdx < 0, then the problem has a solution.

Equivalently :

Necessary Condition : f ∈[

ker A ∩ K∞]

.

Suffisante Condition : f ∈ Int(

[ker A ∩ K∞])

.

The problem is unstable on the boundary.

(ker(A) ∩ K∞

)

Stable

instableinstable

The puffed-up membrane

Membrane

Obstacle

The puffed-up membrane

Find u ∈ K = v ∈W 1,2(Ω) : v ≥ Ψ on Ω∫

Ω

∇u · ∇(v − u)dx +

∫∂Ω

g(|v | − |u|)dσ ≥∫

Ω

f (v − u)dx ,

∀v ∈ K .

Necessary condition :∫∂Ω

gdσ ≥∫

Ω

fdx .

Suffisante condition :∫∂Ω

gdσ >∫

Ω

fdx .

Outline



Ideal diode Model



We consider the following finite dimensional variationalinequality

VI(A, f , ϕ,K )

Find u ∈ K such that〈Mu − q, v − u〉+ Φ(v)− Φ(u) ≥ 0, ∀v ∈ K

I M ∈ Rn×n is a symmetric and positive semidefinitematrix ;

I Φ : Rn → R ∪ +∞ is a proper, convex, lowersemicontinuous and bounded from below ;

I K ⊂ Rn is a closed and convex set such that0 ∈ Dom Φ ∩ K .

I The recession function :

Φ∞(x) = limt→+∞

Φ(x0 + tx)− Φ(x0)

t, x0 ∈ dom(Φ).

(epi (Φ)

)∞

= epi (Φ∞).

I The polar cone :

C = f ∈ X ∗ : 〈f , x〉 ≤ 0, ∀x ∈ C.

C

C

I The support function :

σC(f ) = supx∈C〈f , x〉.

σC (f

2 )

.

f2

f1

C

σC (f1 )

I The Barrier cone : B(C) = dom σC.

I The Fenchel-Legendre conjugate :

Φ∗(f ) = supx∈X〈f , x〉 − Φ(x).

Φ(x)

−Φ∗(f)

f

〈f,x〉

=0

Let us finally recall the following proposition

PropositionLet Ψ ∈ Γ0(Rn) and p ∈ Rn be given. We have :

(i) p ∈ Dom Ψ∗ ⇐⇒ Ψ∞(w) ≥ 〈p,w〉, ∀w ∈ Rn ;

(ii) p ∈ Int (Dom Ψ∗)⇐⇒ Ψ∞(w) > 〈p,w〉, ∀w ∈ Rn\0.

The solutions set of VI(A, f , ϕ,K ) will be denoted bySol(M,q,Φ,K ).The following resolvent set will also play an important role

R(A,Φ,K ) = q ∈ Rn : Sol(M,q,Φ,K ) 6= ∅.

Let us introduce the following functionΨ : Rn → R ∪ +∞ defined by

Ψ(u) =12‖Qu‖2 + Φ(u) + IK (u), (1)

where Q = I − Pker(M) and Pker(M) denotes the orthogonalprojector from Rn to ker(M).

LetΨ(u) =

12‖u − Pker(M)(u)‖2 + Φ(u) + IK (u),

We have the following lemma :

LemmaSuppose that the assumptions (H) hold. We have

I Dom Ψ∗ = R(M) + Dom(

Φ + IK)∗.

I Ψ∞(w) = Iker(M)(w) + Φ∞(w) + IK∞(w),

PropositionA necessary condition for the existence of a solution ofVI(A, f , ϕ,K ) is that

Φ∞(w) ≥ 〈q,w〉, ∀w ∈ ker(M) ∩ K∞. (2)

PropositionSuppose that assumptions (H) hold. We have

IntR(M,Φ,K ) =

q ∈ Rn : Φ∞(w) > 〈q,w〉,∀w ∈ ker(M) ∩ K∞ \ 0

DefinitionThe variational inequality VI(A, f , ϕ,K ) is stable if thereexists ε > 0 such that for any Mε ∈ S+

n (R), any vectorqε ∈ q + εBn, any Φε ∈ Γ0(Rn) bounded from below, andany nonempty closed convex set Kε satisfying thefollowing conditions

0 ∈ Dom Φε ∩ Kε

ker(M) ∩ ker(Φ∞) ∩ K∞ = ker(Mε) ∩ ker((Φε)∞) ∩ (Kε)∞

the perturbed problem VI(Mε,qε,Φε,Kε) has at least onesolution.

TheoremAssume that assumptions (H) are satisfied. Then thevariational inequality VI(A, f , ϕ,K ) is stable in the senseof Definition 2 if and only if

Φ∞(w) > 〈q,w〉, ∀w ∈ ker(M) ∩ K∞, w 6= 0.

Ideal diode Model

THE APPROACH OF MOREAU AND PANAGIOTOPOULOS:

USE IT IN ELECTRONICS

K. ADDI and D. GOELEVEN

IREMIA, University of La Reunion

97400 Saint-Denis, France

[email protected], [email protected]

S. ADLY

DMI-XLIM, University of Limoges

87060 Limoges, France

[email protected]

B. BROGLIATO

INRIA Rhones-Alpes

38334 Saint-Ismier, France

[email protected]

Ideal Diode

i (mA)

V (Volts)

+ !

i

V

• V < 0, i = 0 =⇒ diode is blocking ;• i > 0,V = 0 =⇒ diode is conducting ;

Complementarity formulation

V ≤ 0, i ≥ 0, Vi = 0 ⇐⇒ min−V , i = 0

V ∈ ∂ΨR+(i) ⇐⇒ i ∈ ∂Ψ∗R+(V ) = ∂ΨR−(V )

ΨR+(x) :=

0 if x ≥ 0+∞ if x < 0 ∂ΨR+(x) :=

R− if x = 00 if x > 0∅ if x < 0


u =

Ri︷︸︸︷UR +

∈ ∂ΨR+ (i)︷︸︸︷V +E ⇐⇒ E + Ri − u ∈ −∂ΨR+(i)


ER

+ i − uR∈ −∂ΨR+(i)⇐⇒ −E

R+

uR∈ i + ∂ΨR+(i)

i = (id + ∂ΨR+)−1(u − E

R) =

1R

max0,u − E

u < E =⇒ diode is blocking

u ≥ E =⇒ diode is conducting

Numerical simulation : Ideal diode

INPUT SIGNAL t 7→ u(t)

D. Goeleven - Variational inequalities in electronics - 16

Numerical simulation: Ideal diode

INPUT SIGNAL t !" u(t)

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

input vo

ltage

OUTPUT SIGNAL t !" Vo(t) := V (t) + E = u(t) # Ri(t)

= u(t) # R max0,u(t) # E

R = u(t) + min0, E # u(t)

= minu(t), E

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

outp

ut vo

ltage

Numerical simulation : Ideal diode

OUTPUT SIGNAL t 7→ Vo(t) := V (t) + E = u(t)− Ri(t)

= u(t)− R max0, u(t)− ER

= u(t) + min0,E − u(t)

= minu(t),E


Numerical simulation: Ideal diode

INPUT SIGNAL t !" u(t)

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

input vo

ltage

OUTPUT SIGNAL t !" Vo(t) := V (t) + E = u(t) # Ri(t)

= u(t) # R max0,u(t) # E

R = u(t) + min0, E # u(t)

= minu(t), E

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

outp

ut vo

ltage

Outline



Ideal diode Model



Set-Valued Ampere-Volt characteristic inelectronics

• Diode• Zener Diode• Varactor• Triode• Tetrode• Transistor• Diac• Triac• SiliconControlledRectifier

Set-Valued Ampere-Volt characteristic : theDiode model

V

i

+ -

V1

V2

1

-100

V (Volts)

i (mA)

There is a voltage point, called the knee voltage V1, at whichthe diode begins to conduct and a maximum reverse voltage,called the peak reverse voltage V2,that will not force the diodeto conduct.

Set-Valued Ampere-Volt characteristic : thecomplete diode model

V1

V2

IR

i (mA)

-100

100

1

-0.001

V (Volts)

Illustrates a complete diode model which includes theeffects of the natural resistance of the diode

Set-Valued Ampere-Volt characteristic : theZener diode model

V

i

+ -

V1

V2

I1I2

-10

-10

1

V (Volts)

i (mA)

Zener diode is a good voltage regulator to maintain aconstant voltage regardless of minor variations in loadcurrent or input voltage.


V

V

i

+ -

V1

V2

i

VV1

V2

i

(a) (b)

I1

I2

Illustrates a typical voltage current characteristics of adiac.


V

I

IG

i

V

i

(a) (b)

V

i

(c)

V

IG3

IG2

IG1

<IG3<IG2=IG10

+

_

Illustrates the AV-characteristic of a three-terminal siliconcontrolled rectifier which is used for start/stop controlcircuit for a direct current motor, lamp...

Outline



Ideal diode Model



Clipping circuit

u(t)

R

V+

-

+-

E

i

V

i

+ -

V1

V2

1

-100

V (Volts)

i (mA)

The electrical superpotential of the practical diode is

ϕPD(x) =

V1x if x ≥ 0

V2x if x < 0, (x ∈ R).

Thenϕ∗PD(z) = I[V2,V1](z), (z ∈ R)

We see that

∂ϕPD(x) =

V2 if x < 0

[V2,V1] if x = 0

V1 if x > 0

, (x ∈ R)

∂ϕ∗PD(z) =

R− if z = V2

0 if z ∈]V2,V1[

R+ if z = V1

∅ if z ∈ R\ [V2,V1]

, (z ∈ R).

recovers the volt-ampere characteristic (V , i). Theampere-volt characteristic of the practical diode can thusbe written as

V ∈ ∂ϕPD(i)⇐⇒ i ∈ ∂ϕ∗PD(V )⇐⇒ ϕPD(i) + ϕ∗PD(V ) = iV .

Using Kirchhoff’s law the problem is equivalent toVI(R,E− u, ϕPD,R), i.e.

i ∈ K := R : (Ri+E−u)(v−i)+ϕPD(v)−ϕPD(i) ≥ 0,∀v ∈ R.

Here R > 0 and for each E ,u ∈ R.

Moreover :

i(t) = (idR + ∂ϕPD)−1(u(t)− E

R)

= argminx∈R12|x − (

u(t)− ER

)|2 + ϕPD(x).

andVo(t) = u(t)− Ri(t). (3)

A static model

E1 < E2

D2D1

u(t)

R

V1 V2

i2

i1

i

KIRCHOFF’S LAWS

E1 + R(i1 + i2)− u = +V1 ∈ −∂ΨR−(i1)

E2 + R(i1 + i2)− u = −V2 ∈ −∂ΨR+(i2)

Non-coercive complementarity formalism

A︷︸︸︷(R RR R

) I︷︸︸︷(i1i2

)+

U︷︸︸︷(E1 − uE2 − u

)∈ −∂ΨR−×R+(I).

Here the matrix A is positive semidefinite and symmetric.A sufficient condition for the existence of at least onesolution is :

〈q, v〉 > 0, ∀v ∈ kerA ∩ (R− × R+) \ 0.

Non-coercive complementarity formalism

〈q, v〉 > 0, ∀v ∈ kerA ∩ (R− × R+) \ 0.We have

kerA∩(R−×R+) = v = (v1, v2) ∈ R2 : v1 ≤ 0, v2 = −v1.

Then, for all v ∈ kerA ∩ R− × R+, v 6= 0, we get

qT v = (E1 − u)v1 + (E2 − u)v2 = v2(E2 − E1) > 0.

Then I is determined as the unique solution of thecomplementarity system :

I ∈ K , AI + U ∈ K ∗, IT (AI + U) = 0

m

VARIATIONAL INCLUSION SYSTEM

AI + U ∈ −∂ΨR+×R+(I)

Moreover

i1 + i2 =

u − E1

Rif u < E1

0 if E1 ≤ u ≤ E1u − E2

Rif u > E2

.

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

inpu

t vol

tage


INPUT t !" u(t)

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

input voltage

OUTPUT t !" Vo(t) = Ri(t)

0 1 2 3 4 5 6 7 8 9 10!1.5

!1

!0.5

0

0.5

1

1.5

time

outp

ut voltage

Non-regular electronic circuits

• A ∈ Rn×n, B ∈ Rn×m, C ∈ Rm×n, D ∈ Rn×p,• j : R×Rm → R, x 7→ j(t , x) locally Lipschitz for all t ≥ 0,• u ∈ L1

loc(0,+∞;Rp), x0 ∈ Rn.Problem P(x0) : Find x : [0,+∞[→ Rn; andyL : [0,+∞[→ Rm such that :

x ∈ C0([0,+∞[;Rn), ByL ∈ L1loc(0,+∞;Rn),

dxdt∈ L1

loc(0,+∞;Rn), x(0) = x0,

dxdt

(t) = Ax(t)− ByL(t) + Du(t), a.e. t ≥ 0,

y(t) = Cx(t), ∀ t ≥ 0, and yL(t) ∈ ∂2j(t , y(t)), a.e. t ≥ 0.

Non-regular circuit with VAP-admissibledevice

(H) There exists a symmetric and invertible matrixR ∈ Rn×n such that

R−2CT = B.

Problem Q(x0) : Find z : [0,+∞[→ Rn; t 7→ z(t) suchthat :

z ∈ C0([0,+∞[;Rn),dzdt∈ L1

loc(0,+∞;Rn), z(0) = Rx0,

dzdt

(t) ∈ RAR−1z(t) + RDu(t)− R−1CT∂2j(t ,CR−1z(t)).


Suppose that assumption (H) is satisfied.

If (x , yL) is solution of Problem P(x0) then z = Rx issolution of Problem Q(x0).

Reciprocally, if z is solution of Problem Q(x0) then thereexists a function yL such that (R−1z, yL) is solution ofProblem P(x0).


SetL = CR−1.

andJ(t ,X ) = j(t ,LX ), (X ∈ Rn, t ≥ 0)

then

∂2J(t ,X ) ⊂ R−1CT∂2j(t ,CR−1X ), (X ∈ Rn, t ≥ 0)

⇓

DIFFERENTIAL INCLUSION

dzdt

(t) ∈ RAR−1z(t) + RDu(t)− ∂2J(t , z(t)), a.e. t ≥ 0


DIFFERENTIAL INCLUSION

dzdt

(t) ∈ RAR−1z(t) + RDu(t)− ∂2J(t , z(t)), a.e. t ≥ 0

⇓

VARIATIONAL INEQUALITY

〈dzdt

(t)− RAR−1z(t)− RDu(t), v − z(t)〉+

+J(t , v)− J(t , z(t)) ≥ 0,∀v ∈ Rn, a.e. t ≥ 0.

Example

IG

u(t)

R

L

C0

Example

dx1

dtdx2

dt

=

A︷︸︸︷ 0 1

− 1LC0

−RL

( x1

x2

)−

B︷︸︸︷(0

−1L

)yL+

D︷︸︸︷(01L

)u,

y =

C︷︸︸︷(0 −1

)( x1

x2

)and

yL ∈ ∂C,2jSCR(., y).

(H) There exists a symmetric and invertible matrixR ∈ Rn×n such that

R−2CT = B.

R =

1√C0

0

0√

L

Hence assumption (H) is satisfied.

Example 2

u

L2

L3

R1

R2

R3

C4

dx1

dtdx2

dtdx3

dt

=

A︷︸︸︷0 1 0

− 1L3C4

−(R1 + R3)

L3

R1

L3

0R1

L2−(R1 + R2)

L2

x1

x2

x3

−

B︷︸︸︷0 0

1L3

1L3

− 1L2

0

(

yL,1

yL,2

)+

D︷︸︸︷00

1L2

u,

yL,1 ∈ ∂jD(−x3 + x2)yL,2 ∈ ∂jZ (x2)

(4)

Setting

y =

C︷︸︸︷(0 1 −10 1 0

) x1

x2

x3

jZD(X ) = jZ (X1) + jD(X2), (X ∈ R2)

yL,1 ∈ ∂jD(−x3 + x2)yL,2 ∈ ∂jZ (x2)

⇐⇒ yL ∈ ∂jZD(Cx).

A simple computation shows that the matrix

R =

1√C4

0 0

0√

L3 0

0 0√

L2

is convenient.Hence assumption (H) is satisfied.

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