Alain Bensoussan
email: alain.bensoussan@utdallas.edu

Laurent Mertz
email: mertz@ann.jussieu.fr

Degenerate Dirichlet Problems Related to the Ergodic Property of an Elasto-Plastic Oscillator Excited by a Filtered White Noise *

Keywords: Random vibration, ergodic diffusion, stochastic variational inequalities

A stochastic variational inequality is proposed to model an elasto-plastic oscillator excited by a filtered white noise. We prove the ergodic properties of the process and characterize the corresponding invariant measure. This extends Bensoussan-Turi's method (Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators, AMO, 2008) with a significant additional difficulty of increasing the dimension. Two points boundary value problem in dimension 1 is replaced by elliptic equations in dimension 2. In the present context, Khasminskii's method (Stochastic Stability of Differential Equations, Sijthoff and Noordhof,1980) leads to the study of degenerate Dirichlet problems with partial differential equations and nonlocal boundary conditions.

Introduction

Nonlinear oscillators subjected to vibrations represent useful models for predicting the response of mechanical structures when stressed beyond the elastic limit. When the excitation is a white noise, it has received considerable interest over the past decades. In a previous work [START_REF] Bensoussan | Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators[END_REF], the authors have considered the response of a white noise excited elasto-plastic oscillator using a stochastic variational inequality formulation. The results in [START_REF] Bensoussan | Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators[END_REF] provide a framework to assess the accuracy of calculations made in the literature (see e.g., (4; 5; 6), and the references therein). In this paper, instead of considering white noise input signal whose power spectral density (PSD) is constant, we consider an excitation with a non-constant PSD which could be a more realistic framework. We consider the excitation as the velocity of a "reflected" Ornstein-Uhlenbeck process. Therefore, comparing with the elasto plastic oscillator excited by white noise, a third process occurs in the variational inequality. Consider w(t) and w(t), two independent Wiener processes and x(t) an Ornstein-Ulhenbeck reflected process dx(t) = -αx(t)dt + dw(t) + 1 {x(t)=-L} dξ 1 t -1 {x(t)=L} dξ 2 t .

(1..1)

We have -L x(t) L. In this model, the stochastic excitation is given by -β x(t)dt + d w(t).

The stochastic variational inequality model is given by: When β = 0, x(t) is involved in the dynamic of y(t) and then this model will be referred as the 2d case. Whereas if β = 0, x(t) is not involved in the dynamic of y(t) and then (y(t), z(t)) satisfy the elastoplastic oscillator problem of (1) which will be referred as the 1d case. The infinitesimal generator of the process (x(t), y(t), z(t)), denoted by Λ is given by: • {m(x, y, z), (x, y, z) ∈ O} is the elastic component,

                 dx(t) = -
Λ : φ → Aφ if z ∈] -Y,Y [, B ± φ if z = ±Y, ±y > 0.
• {m(x, y,Y ), (x, y) ∈ O + } is the positive plastic component,

• and {m(x, y, -Y ), (x, y) ∈ O -} is the negative plastic component.

In addition, m satisfies in the sense of distributions the following equation in O,

α ∂ ∂ x [xm] + ∂ ∂ y [(β x + c 0 y + kz)m] -y ∂ m ∂ z + 1 2 ∂ 2 m ∂ x 2 + 1 2 ∂ 2 m ∂ y 2 = 0 in O,
and on the boundary

ym + ∂ ∂ x [xm] + ∂ ∂ y [(β x + c 0 y + kY )m] + 1 2 ∂ 2 m ∂ x 2 + 1 2 ∂ 2 m ∂ y 2 = 0, in O + -ym + ∂ ∂ x [xm] + ∂ ∂ y [(β x + c 0 y -kY )m] + 1 2 ∂ 2 m ∂ x 2 + 1 2 ∂ 2 m ∂ y 2 = 0, in O - m = 0, in (-L, L) × (-∞, 0) × {Y} ∪ (-L, L) × (0, ∞) × {-Y}.
The proof will be based on solving a sequence of interior and exterior Dirichlet problems, which are interesting in themselves. We will put in parallel the 1d and 2d cases, in order to facilitate the reader's work. In the 1d case, the variable x disappears (β = 0), we will still use the notation A, B + , B -for the operators defined above without x.

Let us mention that our study presents a mathematical interest because it generalizes the method proposed by the first author and J. Turi [START_REF] Bensoussan | Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators[END_REF] in the case of higher dimension. Non-local boundary conditions expressed in the form of differential equations in dimension 1 are replaced by elliptic partial differential equations (PDEs) in dimension 2. In the first case, there are two semi-explicit solutions. Thus, the non-local boundary conditions are reduced to two unknown numbers. In the second case, we do not know explicit formulas for solutions of the both elliptic on the boundary. In this context, these two solutions depend on two unknown functions respectively defined on the set (-L, L).

In addition, the choice of the excitation (1..1) is also motivated by two technical considerations:

• the first is to force x(t) (through the processes ξ 1 , ξ 2 ) to evolve in the compact set [-L, L]. Thus, as part of our proof, a compactness argument allows to show the ergodic property of the triple (x(t), y(t), z(t)). Note that this is not a problem in terms of applications, because if we choose L large enough, then the process x(t) is similar to an Ornstein-Ulhenbeck process.

• the second is the uncorrelation of w(t) and w(t). In our approach, based on PDEs associated to the triple (x(t), y(t), z(t)), we avoid the appearance of cross-derivative terms in the infinitesimal generator Λ . In the case where w(t) and w(t) are correlated, these cross-derivative terms yield more technical difficulties.

The interior Dirichlet problem

In this section, we prove existence and uniqueness to the homogeneous interior Dirichlet problem.

Some background on the interior Dirichlet problem in the 1d case

Let us recall the interior Dirichlet problem from [START_REF] Bensoussan | Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators[END_REF]. Let ȳ1 > 0, Notation 2..1

D 1 := (-ȳ1 , ȳ1 ) × (-Y,Y ); D + 1 := (0, ȳ1 ) × {Y }; D - 1 := (-ȳ1 , 0) × {-Y} and
D ε 1 := (-ȳ1 + ε, ȳ1 -ε) × (-Y,Y), ε > 0.
Denote τ1 := inf{t > 0, |y(t)| = ȳ1 } and consider φ ∈ L ∞ (-Y,Y ). We will use the following notation where η Y and η -Y are constants, and

E p (•) := E{• |(y(0), z(0)) = p}. It is shown that E (y,z) (φ (z( τ1 ))) solves a nonlocal Dirichlet problem: Find η ∈ L ∞ (D 1 ) ∩ C 0 (D ε 1 ), ∀ε > 0 such that Aη = 0 in D 1 , B + η = 0 in D + 1 , B -η = 0 in D - 1 (2..1) with η( ȳ1 , z) = φ (z), η(-ȳ1 , z) = 0, if z ∈ (-Y,Y ).
I(a, b) := b a exp(c 0 λ 2 + 2kY λ )dλ ȳ1 0 exp(c 0 λ 2 + 2kY λ )dλ
.

The nonlocal condition is restricted to the value of these two constants. Based on these semi-explicit expressions a subset K of H 1 (D 1 ) is defined for proving existence of the solution to (2..1) , by

K :=                  v ∈ H 1 (D 1 ), v(-ȳ1 , z) = φ (z), v(-ȳ1 , z) = 0, v(y,Y ) = v Y I(y, ȳ1 ) + φ (Y )I(0, y), 0 y ȳ1 v(y, -Y ) = v -Y I(-ȳ1 , y), -ȳ1 y 0 v Y , v -Y are constant with |v ±Y | φ L ∞ .
The set K is convex and not empty if φ (z) ∈ H 1 (-Y,Y ). We take v(y, z) = φ (z)I(0, y)1 {y>0} .

∆ 1 := (-L, L) × (-ȳ1 , ȳ1 ) × (-Y,Y ), ∆ + 1 := (-L, L) × (0, ȳ1 ) × {Y }, ∆ - 1 := (-L, L) × (-ȳ1 , 0) × {-Y} and ∆ ε 1 := (-L, L) × (-ȳ1 + ε, ȳ1 -ε) × (-Y,Y ), ∀ε > 0. Denote τ1 := inf{t > 0, |y(t)| = ȳ1 } and consider φ ∈ L ∞ ((-L, L) × (-Y,Y )).
Similarly as before we use the notation E p (•) := E{• |(x(0), y(0), z(0)) = p}. We want to define E (x,y,z) (φ (x( τ1 ), z( τ1 ))) as the solution of the interior Dirichlet problem stated below.

Problem 1 Find η ∈ L ∞ (∆ 1 ) ∩ C 0 (∆ ε 1 ), ∀ε > 0 such that Aη = 0 in ∆ 1 , B + η = 0 in ∆ + 1 , B -η = 0 in ∆ - 1 and η x (±L, y, z) = 0 in (y, z) ∈ (-ȳ1 , ȳ1 ) × (-Y,Y ), η(x, ȳ1 , z) = φ (x, z) in (x, z) ∈ (-L, L) × (-Y,Y ), η(x, -ȳ1 , z) = 0 in (x, z) ∈ (-L, L) × (-Y,Y ).
This is formal. We should consider the case of φ smooth first and precise the functional space, then proceed with the regularization.

As in the 1d-case, this problem is a nonlocal problem but the boundary condition are in two dimensions. Thus we need to solve partial differential equations for η on the boundary at z = Y , and z = -Y respectively. Here we do not have semi-explicit solution, indeed η(x, y,Y ) solves

B + η = 0 on (-L, L) × (0, ȳ1 ) with η x (±L, y,Y ) = 0, η(x, ȳ1 ,Y ) = φ (x, z) (2..2)
with η(x, 0,Y ) = η Y (x), and η(x, y, -Y ) solves

B -η = 0 on (-L, L) × (-ȳ1 , 0) with η x (±L, y, -Y ) = 0, η(x, -ȳ1 , -Y ) = 0 (2..3) with η(x, 0, -Y ) = η -Y (x) where η Y (x) and η Y (x) are unknown function with η ±Y L ∞ φ L ∞ . Next,
we give a convenient formulation of the boundary condition (2..2)-(2..3).

2.2..1 Boundary conditions

In order to reformulate Problem 1, we consider first the equation (2..2) on the boundary ∆ + 1 . Define β + (x, y) the solution of the mixed Dirichlet-Neuman problem

               B + β + = 0, in (-L, L) × (0, ȳ1 ), β + x (±L, y) = 0, in (0, ȳ1 ), β + (x, ȳ1 ) = φ (x,Y ), in (-L, L), β + (x, 0) = 0, in (-L, L). (2..4) PROPOSITION 2..3 If φ (x,Y ) ∈ H 1 (-L, L)
then there exists a unique solution to the equation (2..4)

β + ∈ H 1 ((-L, L) × (0, ȳ1 )) satisfying β + L ∞ φ L ∞ . Proof. Define D + Y := (-L, L) × (0, ȳ1 ) and consider on H 1 (D + Y ) the bilinear form b Y (ξ , χ) = 1 2 D + Y ξ x χ x + ξ y χ y dxdy + D + Y αxξ x + (c 0 y + kY + β x)ξ y χdxdy. For λ sufficiently large b Y (ξ , χ) + λ (ξ , χ) is coercive. Now, define the convex set K Y = {ξ ∈ H 1 (D + Y ), ξ (x, ȳ1 ) = φ (x,Y ) and ξ (x, 0) = 0} which is not empty since φ (x,Y ) ∈ H 1 2 (-L, L). Take z ∈ K Y with z L ∞ φ L ∞ , we define ξ λ to be the unique solution of b Y (ξ λ , χ -ξ λ ) + λ (ξ λ , χ -ξ λ ) λ (z, χ -ξ λ ), ξ λ ∈ K Y , ∀χ ∈ K Y (2..5)
We have defined a map

T λ (z) = ξ λ from K Y → K Y . Let us check that ξ λ L ∞ φ L ∞ = C. (2..6) Indeed, in (2..5) we take χ = ξ λ -(ξ λ -C) + ∈ K Y . Hence -b Y (ξ λ , (ξ λ -C) + ) -λ (ξ λ , (ξ λ -C) + ) λ (z, (ξ λ -C) + ) and b Y ((ξ λ -C) + , (ξ λ -C) + ) + λ |(ξ λ -C) + | L 2 -λ (z + C, (ξ λ -C) + ).
Since z +C 0 it follows that (ξ λ -C) + = 0, hence ξ λ C. Similarly, we check that (-ξ λ -C) + = 0, hence we have (2..6). Consider then the sequence ξ n λ defined by

b Y (ξ n+1 λ , χ -ξ n+1 λ ) + λ (ξ n+1 λ , χ -ξ n+1 λ ) λ (ξ n λ , χ -ξ n+1 λ ) (2..7) with ξ 0 λ ∈ K Y , ξ 0 λ L ∞ φ L ∞ .
We can take ξ 0

λ (x, y) = y ȳ1 φ (x,Y ). From (2..6), we have ξ n λ L ∞ (D + Y )
C and from (2..7), we have

ξ n λ H 1 (D + Y )
C ′ . Then, we can consider a subsequence, also denoted by ξ n λ such that

ξ n λ → ξ in H 1 (D + Y ) weakly and in L ∞ (D + Y ) weakly ⋆ also, ξ n λ → ξ in L 2 (D + Y ) strongly. From (2..7) we obtain b Y (ξ , χ -ξ ) 0, ξ ∈ K Y , ∀χ ∈ K Y .
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise7 of 46

We conclude easily that ξ is a solution of (2..4) also ξ L ∞ φ L ∞ . Now, in order to prove uniqueness, we must prove that a solution ξ ∈ H 1 (D + Y ) satisfying

B + ξ = 0 in (-L, L) × (0, ȳ1 ) ξ x (-L, y) = ξ x (L, y) = 0 ξ (x, ȳ1 ) = ξ (x, 0) = 0 (2..8)
is identically zero. Consider χ := ξ x then we have

B + χ + α χ + β ξ y = 0 in D + Y , χ(-L, y) = χ(L, y) = 0, χ(x, ȳ1 ) = χ(x, 0) = 0, hence χ ∈ H 1 0 (D + Y ). This implies ξ xx ∈ L 2 (D + Y ), ξ yx ∈ L 2 (D + Y ). From equation (2..8) we deduce ξ yy ∈ L 2 (D + Y ). Hence ξ ∈ H 2 (D + Y ).
In particular, ξ is continuous on D+ Y . We have ξ L ∞ = 0, so ξ = 0. Now, consider the following convex sets,

K Y,M := {χ + ∈ H 1 ((-L, L) × (0, ȳ1 )) satisfying (2..10); χ + (x, ȳ1 ) = 0; χ + L ∞ M} (2..9)
where

L -L ȳ1 0 { 1 2 (χ + x ψ x + χ + y ψ y ) + (αxχ + x + (β x + c 0 y + kY )χ + y )ψ}dxdy = 0, ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )) with ψ(x, 0) = ψ(x, ȳ1 ) = 0. (2..10)
and

K -Y,M := {χ -∈ H 1 ((-L, L) × (-ȳ1 , 0)) satisfying (2..12); χ -(x, -ȳ1 ) = 0; χ - L ∞ M} (2..11)
where

L -L 0 -ȳ1 { 1 2 (χ - x ψ x + χ - y ψ y ) + (αxχ - x + (β x + c 0 y -kY )χ - y )ψ}dxdy = 0, ∀ψ ∈ H 1 ((-L, L) × (-ȳ1 , 0)) with ψ(x, 0) = ψ(x, -ȳ1 ) = 0.
(2..12)

These sets are not empty since they contain 0.

REMARK 2..1 • ∀π(y) function of y such that π(0) = 0 , {χπ; χ ∈ K Y,M } and {χπ; χ ∈ K -Y,M } are H 1 -bounded. • If χ ∈ K Y, φ L ∞ , denoting ω := β + + χ, we have B + ω = 0 and max(|ω(x, 0)|, |ω(x, ȳ1 )|) φ L ∞ , so a maximum principle implies ω L ∞ φ L ∞ .
Using the sets K Y, φ and K -Y, φ , the following result gives a convenient formulation of the boundary conditions in Problem 1. PROPOSITION 2.. [START_REF] Feau | Les méthodes probabilistes en mécanique sismique. Applications aux calculs de tuyauteries fissurées[END_REF] The Problem 1 can be reformulated in the following way:

find η ∈ L ∞ (∆ 1 ) ∩ C 0 (∆ ε 1 ), ∀ε > 0 such that Aη = 0 in ∆ 1 , η(x, y,Y ) -β + (x, y) ∈ K Y, φ , η(x, y, -Y ) ∈ K -Y, φ and η x (±L, y, z) = 0, in (y, z) ∈ (-ȳ1 , ȳ1 ) × (-Y,Y ), η(x, ȳ1 , z) = φ (x, z), in (x, z) ∈ (-L, L) × (-Y,Y ), η(x, -ȳ1 , z) = 0, in (x, z) ∈ (-L, L) × (-Y,Y ).
Proof. First, we can obtain the generic solution of (2..2) by considering any function χ + which satisfies

χ + ∈ H 1 ((-L, L) × (0, ȳ1 ))
and

B + χ + = 0, χ + x (±L, y) = 0, χ + (x, ȳ1 ) = 0.
(2..13) Note that we have not defined the value of χ + for y = 0, hence χ + is certainly not unique. We add the condition that χ + is bounded by φ L ∞ . We define in a similar way the function χ -such that

χ -∈ H 1 ((-L, L) × (-ȳ1 , 0)) and B -χ -= 0, χ - x (±L, y) = 0, χ -(x, -ȳ1 ) = 0. (2..14)
Then, interpreting (2..13) as (2..10) and (2..14) as (2..12) repectively, we obtain χ + ∈ K Y, φ and χ -∈ K -Y, φ . Hence, the set of solutions of (2..2) and (2..3) can be written as follows

η(x, y,Y ) -β + (x, y) ∈ K Y, φ , y > 0; η(x, y, -Y ) ∈ K -Y, φ , y < 0.

2.2..2 Approximation (part 1)

We study Problem 1 by a regularization method in the next proposition. Define

A ε := A + ε 2 ∂ 2 ∂ z 2 .
PROPOSITION 2.. [START_REF] Feau | Probabilistic response of an elastic perfectly plastic oscillator under Gaussian white noise[END_REF] The following problem:

find η ε ∈ L ∞ (∆ 1 ) ∩ H 1 (∆ 1 ) such that η ε L ∞ φ L ∞ , A ε η ε = 0 in ∆ 1 , η ε (x, y,Y ) -β + (x, y) ∈ K Y, φ , η ε (x, y, -Y ) ∈ K -Y, φ (2.
.15)

and η ε x (±L, y, z) = 0, in (y, z) ∈ (-ȳ1 , ȳ1 ) × (-Y,Y ), η ε z (x, y, ±Y ) = 0, in (x, ∓y) ∈ (-L, L) × (0, ȳ1 ), η ε (x, ȳ1 , z) = φ (x, z), in (x, z) ∈ (-L, L) × (-Y,Y ), η ε (x, -ȳ1 , z) = 0, in (x, z) ∈ (-L, L) × (-Y,Y ),
has a unique solution.

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise9 of 46

As in the 1d case, we formulate a variational inequality to prove existence of solutions. In the present context, we consider a convex subset of H 1 (∆ 1 ) which is adapted to the two dimensional boundary condition by

K =            ψ ∈ H 1 (∆ 1 ), ψ ∞ φ ∞ , ψ(x, ȳ1 , z) = φ (x, z), for (x, z) ∈ (-L, L) × (-Y,Y ), ψ(x, -ȳ1 , z) = 0, for (x, z) ∈ (-L, L) × (-Y,Y ), ψ(., .,Y ) -β + (x, y) ∈ K Y, φ L ∞ , ψ(., ., -Y ) ∈ K -Y, φ L ∞ . PROPOSITION 2..6 The set K is a closed non-empty subset of H 1 (∆ 1 ).
Proof. The fact that K is closed follows from the continuity of the trace operator. Now, pick the function, L,L), then the function ψ defined by (2..14) belongs to K. REMARK 2..2 If u ∈ K and w ∈ H 1 (∆ 1 ) with w(x, ± ȳ1 , z) = 0 and w(x, y, ±Y ) = 0, for 0 < ±y < ȳ1 then u + w ∈ K.

ψ(x, y, z) = y ȳ1 φ (x, z) - z 2Y φ (x,Y ) - 1 2 φ (x,Y ) 1 {y 0} + ( z 2Y + 1 2 )β + (x, y)1 {y 0} . (2..16) We have ψ(x, ȳ1 , z) = φ (x, z), ψ(x, -ȳ1 , z) = 0, ψ(x, y,Y ) = β + (x, y), y > 0, ψ(x, y, -Y ) = 0, y < 0. So, if φ (x, z) ∈ H 1 (∂ ∆ 1 ), φ (x,Y ) ∈ H 1 (-
Consider the bilinear form

a(u, v) = 1 2 ∆ 1 {εu z v z + u y v y + u x v x }dxdydz + ∆ 1 (β x + c 0 y + kz)u y vdxdydz - ∆ 1 yu z vdxdydz + ∆ 1 αxu x vdxdydz.
Equation (2..15) is formulated as follows a(u, vu) 0, ∀v ∈ K, u ∈ K. Proof of Proposition 2..5. First, existence is proved by variational argument. For λ sufficiently large a(u, v) + λ (u, v) is coercive on H 1 (∆ 1 ) and for f ∈ L 2 (∆ 1 ) we can solve the variational inequality

a(u, v -u) + λ (u, v -u) ( f , v -u), ∀v ∈ K, u ∈ K.
We define the map u = T λ w where u is the unique solution of

a(u, v -u) + λ (u, v -u) λ (w, v -u), ∀v ∈ K, u ∈ K.
The following lemma shows that T λ is a contracting map. (see proof in Appendix)

LEMMA 2..1 If w L ∞ φ L ∞ then u L ∞ φ L ∞ . Moreover, taking v = u 0 ∈ K we deduce a(u, u) + λ |u| 2 L 2 a(u, u 0 ) + λ (u, u 0 ) + λ γ|u 0 -u| L 1 . That implies u H 1 (∆ 1 ) M,
for a constant M which depends only of λ , ε, H 1 norm of u 0 and γ. Now, we define

K = {w ∈ K : w L ∞ φ L ∞ , w H 1 M}.
We have

T λ : L 2 (∆ 1 ) → L 2 (∆ 1
) is continuous, maps K into itself and K is a compact subset of L 2 . Then Schauder's theorem implies T λ has a fixed point u ∈ K which satisfies

a(u, v -u) 0, ∀v ∈ K.
Let us check that u is solution of (2..15). As u ∈ K, we have

B + u = 0 in ∆ + B -u = 0 in ∆ - u(x, ȳ1 , z) = φ (x, z) in (-L, L) × (-Y,Y ) u(x, -ȳ1 , z) = 0 in (-L, L) × (-Y,Y ) Moreover, ∀v ∈ H := v ∈ H 1 (∆ 1 ) such that v(x, ± ȳ1 , z) = 0 v(x, y, ±Y ) = 0, 0 < ±y < ȳ1 ,
we have u + v ∈ K and a(u, v) 0. Then,

A ε u = 0 in the sense H ′
and integration by parts gives

ε 2 L -L 0 -ȳ1 u z (x, y,Y )v(x, y,Y )dxdy - ε 2 L -L ȳ1 0 u z (x, y, -Y )v(x, y, -Y )dxdy + ȳ1 -ȳ1 Y -Y u x (L, y, z)v(L, y, z)dydz - ȳ1 -ȳ1 Y -Y u x (-L, y, z)v(-L, y, z)dydz = 0.
Uniqueness of the solution to problem (2..15

) comes from u L ∞ φ L ∞ . 2.2..3 Approximation (part 2)
When φ is smooth, we can exhibit a solution to the problem (2..17) by extracting a converging subsequence of η ε . Let θ a smooth fonction such that θ (± ȳ1 ) = 0. Denote π(y) := y p θ (y) q for some p, q.

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise11 of 46 PROPOSITION 2.. [START_REF] Khasminkii | Stochastic Stability of Differential Equations[END_REF] The following problem: find η ∈ L ∞ (∆ 1 ) such that ηπ ∈ H 1 (∆ 1 ),

Aη = 0 in ∆ 1 , η(x, y,Y ) -β + (x, y) ∈ K Y, φ , η(x, y, -Y ) ∈ K -Y, φ (2.
.17) and

η x (±L, y, z) = 0 in (y, z) ∈ (-ȳ1 , ȳ1 ) × (-Y,Y ), η(x, ȳ1 , z) = φ (x, z) in (x, z) ∈ (-L, L) × (-Y,Y ), η(x, -ȳ1 , z) = 0 in (x, z) ∈ (-L, L) × (-Y,Y ),
has a unique solution.

As in the 1d case, the key ingredient of the proof is to bound uniformly the norm of first derivative w.r.t. z using the auxiliary function π. Proof. From the previous section, we have η ε ∞ φ ∞ , hence η ε → η in L ∞ ⋆. We also have a(η ε , u 0 -η ε ) 0, for some u 0 ∈ K. So, we deduce estimates in the following lemma:

(see proof in Appendix) LEMMA 2..2 We have ε ∆ 1 (η ε z ) 2 dxdydz C; ∆ 1 (η ε x ) 2 dxdydz C; ∆ 1 (η ε y ) 2 dxdydz C.
It is licit to test (2..15) with η ε z y 2p-1 θ 2q . We have

∆ 1 ε 2 η ε zz + 1 2 η ε yy + 1 2 η ε xx + yη ε z -αxη ε x -(β x + c 0 y + kz)η ε y η ε z y 2p-1 θ 2q = 0.
So, we obtain

∆ 1 (η ε z π) 2 1 4 L 0 ȳ1 0 {(η ε y (x, y,Y )) 2 + (η ε x (x, y,Y )) 2 }y 2p-1 θ 2q dxdy + 1 4 L 0 0 -ȳ1 {(η ε y (x, y, -Y )) 2 + (η ε x (x, y, -Y )) 2 }|y| 2p-1 θ 2q dxdy + 1 2 ∆ 1 η ε y η ε z ((2p -1)y 2p-2 θ 2q + y 2p-1 2qθ ′ θ 2q-1 ) + 2(β x + c 0 y + kz)y 2p-1 θ 2q dxdydz + ∆ 1 η ε x η ε z (αxy 2p-1 θ 2q )dxdydz.
(2..18) Moreover, Remark 2..1 allows to bound the two integrals on the boundary that yields the following estimate:

η ε z π L 2 C π . Denoting v ε := η ε π, we have v ε H 1 (∆ 1 )
Cπ so we can extract a weakly converging subsequence

v ε → v in H 1 (∆ 1 ) and v = ηπ ∈ H 1 (∆ 1 )
. We can check that η satisfies the boundary condition of Problem 1. First, let us check that

π(y)Aη = 0 in H -1 (∆ 1 ), π(y)η x (±L, y, z) = 0 in (H 1 2 ((-L, L) × (-ȳ1 , ȳ1 ))) ′ .
As v ε ∈ H 2 (∆ 1 ) and πA ε η ε = 0, we have

-A ε v ε = f (η, η y ) in strong sense with f (η, η y ) := -1 2 {π ′′ η ε + 2π ′ η ε y } + (c 0 y + kz + αx)π ′ η ε . We obtain that ∀φ ∈ H 1 (∆ 1 ), φ (x, y, ±Y ) = 0 and φ (x, ± ȳ1 , z) = 0, ε 2 ∆ 1 v ε z φ z + 1 2 ∆ 1 v ε y φ y + 1 2 ∆ 1 v ε x φ x + ∆ 1 αxv ε x + (β x + c 0 y + kz)v ε y -yv ε z φ = ∆ 1 f (η ε , η ε y )φ .
Now, when ε goes to 0, we have

1 2 ∆ 1 v y φ y + 1 2 ∆ 1 v x φ x + ∆ 1 αxv x + (β x + c 0 y + kz)v y -yv z φ = ∆ 1 f (η, η y )φ .
We deduce we have in H -1 (∆ 1 ), firstly -Av = f (η, η y ) which is equivalent to πAη = 0 and secondly that choice of test function implies πη x (±L, y, z) = 0 in (H

1 2 ((-L, L) × (-ȳ1 , ȳ1 )) ′ . Then, we check that η(x, y,Y ) -β + (x, y) ∈ K Y, φ ; η(x, y, -Y ) ∈ K -Y, φ .
We know that η ε ∈ H 1 (∆ 1 ), its trace is well defined and satisfies

γ(η ε )(x, y,Y ) = χ +,ε + β + ; χ +,ε ∈ K Y, φ ; y > 0 γ(η ε )(x, y, -Y ) = χ -,ε ; χ -,ε ∈ K -Y, φ ; y < 0 with χ ±,ε L ∞ φ L ∞ . (2..19)
We also have χ -,ε , χ +,ε satisfy respectively (2..20) and (2..21)

L -L ȳ1 0 { 1 2 (χ +,ε x ψ x + χ +,ε y ψ y ) + (αxχ +,ε x + (β x + c 0 y + kY )χ +,ε y )ψ}dxdy = 0, ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )) with ψ(x, 0) = ψ(x, ȳ1 ) = 0. (2..20)
and

L -L 0 -ȳ1 { 1 2 (χ -,ε x ψ x + χ -,ε y ψ y ) + (αxχ -,ε x + (β x + c 0 y -kY )χ -,ε y )ψ}dxdy = 0, ∀ψ ∈ H 1 ((-L, L) × (-ȳ1 , 0)) with ψ(x, 0) = ψ(x, -ȳ1 ) = 0. (2..21)
First, we study convergence of the sequence χ ±,ε and we deduce the PDEs satisfied by lim ε→0 χ ±,ε . In particular (2..19) implies

χ +,ε → χ + in L 2 ((-L, L) × (0, ȳ1 )) weakly, χ -,ε → χ -in L 2 ((-L, L) × (-ȳ1 , 0)) weakly.
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise13 of 46 And (2..20) and (2..21) imply

χ +,ε π H 1 C and χ +,ε π → χ + π in H 1 weakly, χ -,ε π H 1 C and χ -,ε π → χ -π in H 1 weakly. Denote ξ ±,ε := χ ±,ε y 2 . From (2..20), we obtain B + ξ +,ε = 0 in (-L, L) × (-ȳ1 , ȳ1 ), in H -1 ((-L, L) × (0, ȳ1 )). Since the operator B + is strictly elliptic then ξ +,ε ∈ H 2 ((-L, L) × (0, ȳ1 )). We also have ξ +,ε x (±L, y) = 0 in (H 1 2 (0, ȳ1 )) ′ . As ξ +,ε ∈ H 2 (∆ 1 )
and πB + χ +,ε = 0, we obtain

-B + ξ +,ε = g(χ +,ε , χ +,ε y ) in a strong sense with g(χ +,ε , χ +,ε y ) := -χ +,ε {1 -2y(αx + c 0 y + kY )} -2yχ +,ε y . We obtain that ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )), ψ(x, 0) = ψ(x, ȳ1 ) = 0, L -L ȳ1 0 1 2 (ξ +,ε x ψ x + ξ +,ε y ψ y ) + αxξ +,ε x + (β x + c 0 y + kY )ξ +,ε y -yξ +,ε z ψ = L -L ȳ1 0 g(χ ε , χ ε y )ψ.
Now, when ε goes to 0, we have

L -L ȳ1 0 1 2 (ξ + x ψ x + ξ + y ψ y ) + αxξ + x + (β x + c 0 y + kY )ξ + y -yξ + z ψ = L -L ȳ1 0 g(χ, χ y )ψ.
We deduce that in H -1 ((-L, L) × (0, ȳ1 )), we firstly have -B + ξ + = g(χ + , χ + y ), which is equivalent to y 2 B + χ + = 0 and secondly that choice of test functions implies

y 2 χ + x (±L, y) = 0 in (H 1 2 ((0, ȳ1 ))) ′ . To summarize, we have ξ + ∈ H 1 ((-L, L) × (0, ȳ1 )) and          -B + ξ + = g(χ + , χ + y ) in (-L, L) × (0, ȳ1 ), ξ + x (±L, y) = 0 in (0, ȳ1 ), ξ + (x, ȳ1 ) = 0 in (-L, L). Hence π χ + ∈ H 1 ((-L, L) × (0, ȳ1 )), χ + L ∞ φ L ∞ and          πB + χ + = 0 in (-L, L) × (0, ȳ1 ), π χ + x (±L, y) = 0 in (0, ȳ1 ), χ + (x, ȳ1 ) = 0 in (-L, L). (2..22) Similarly, we have π χ -∈ H 1 ((-L, L) × (-ȳ1 , 0)), χ - L ∞ φ L ∞ and          πB + χ -= 0 in (-L, L) × (-ȳ1 , 0), π χ - x (±L, y) = 0 in (-ȳ1 , 0), π χ -(x, -ȳ1 ) = 0 in (-L, L). (2..23) First, γ(πη ε ) → π(χ + + β + ) in H 1 ((-L, L) × (0, ȳ1 )) weakly. Secondly, the weak convergence of πη ε → πη in H 1 (∆ 1 ) implies the weak convergence of γ(πη ε ) → γ(πη) in H 1 2 (∂ ∆ 1 )
. By uniqueness of the limit, we deduce γ(πη) = π(χ + + β + ). Finally, we verify that η(x, ȳ1 , z) = φ (x, z); η(x, ȳ1 , z) = 0.

Using Green formula, we obtain

∀ψ ∈ H 1 (∆ 1 ), ∆ 1 η ε y ψ + ∆ 1 η ε ψ y = ∂ ∆ 1 φ ψ n(y)dσ .
Now, we can let ε tend to 0, we obtain

∀ψ ∈ H 1 (∆ 1 ), ∆ 1 η y ψ + ∆ 1 ηψ y = ∂ ∆ 1 φ ψ n(y)dσ . 2.2..4 Approximation (part 3) Now, φ ∈ L ∞ (∂ ∆ 1 ), we introduce a sequence of function {φ k , k 0} ⊂ H 1 (∂ ∆ 1 ) such that φ k → φ in L 2 (∂ ∆ 1 )
. We denote η k the solution of the Problem 2..17 with φ k as boundary condition. From the previous section we have

η k ∈ L ∞ (∆ 1 ) satisfies π(y)η k ∈ H 1 (∆ 1 ), Aη k = 0 in ∆ 1 , η k (x, y,Y ) -β k,+ (x, y) ∈ K Y, φ , η k (x, y, -Y ) ∈ K -Y, φ and η k x (±L, y, z) = 0 in (y, z) ∈ (-ȳ1 , ȳ1 ) × (-Y,Y ), η k (x, ȳ1 , z) = φ k (x, z) in (x, z) ∈ (-L, L) × (-Y,Y ), η k (x, -ȳ1 , z) = 0 in (x, z) ∈ (-L, L) × (-Y,Y ).
where

β k,+ (x, y) ∈ H 1 ((-L, L) × (0, ȳ1 )) solves the problem (2..4) with φ k (x,Y ) as boundary condition. Moreover η k ∞ φ ∞ and β k ∞ φ ∞ .
Let us check that the sequence β k has a limit. PROPOSITION 2..8 We have

β k,+ (x, y) → β + in L 2 ((-L, L) × (0, ȳ1 )) weakly, πβ k,+ (x, y) → πβ + in H 1 ((-L, L) × (0, ȳ1 ))weakly
and the limit β + solves the problem (2..4) with φ (x,Y ) as boundary condition.

Proof. We have ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )) with ψ(x, 0) = ψ(x, ȳ1 ) = 0, 1 2 β k,+ x ψ x + β k,+ y ψ y + αxβ k,+ x + (c 0 y + kY + β x)β k,+ y ψ = 0. (2..24)
In particular, the choice of ψ = π(y)β k,+ with π(0

) = π( ȳ1 ) = 0 gives 1 2 (πβ k,+ x ) 2 +(πβ k,+ y ) 2 + π(y)π ′ (y)β k,+ y β k,+ + αxβ k,+ x +(c 0 y+kY +β x)β k,+ y π 2 β k,+ = 0.
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise15 of 46

This implies

(πβ k,+ x ) 2 C π and (πβ k,+ y ) 2 C π .
We deduce that we can extract a subsequence such that we have

β k,+ → β + in L 2 weakly and πβ k,+ → πβ + in H 1 weakly. Now, denote γ k := π(y)β k,+ . We have B + γ k = -β k,+ ( π ′′ 2 -(β x + c 0 y + kY )π ′ ) -β k,+ y π ′ . Also ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )) with ψ(x, 0) = ψ(x, ȳ1 ) = 0, 1 2 γ k x ψ x + γ k y ψ y + αxγ k x + (c 0 y + kY + β x)γ k y ψ (2..25) = {-β k,+ ( π ′′ 2 -(β x + c 0 y + kY )π ′ ) -β k,+ y π ′ }ψ.
When k goes to +∞ in (2..25), we obtain

B + β + = 0; β + x (±L, y) = 0. Then, from (2..26), ∀ψ ∈ H 2 ∩ H 1 0 , β k,+ ψ yy = - β k,+ y ψ y + φ k ψ y n(x)dσ , (2.
.26) we deduce taking limit when k goes to

+∞ ∀ψ ∈ H 2 ∩ H 1 0 , β + ψ yy = - β + y ψ y + φ ψ y n(x)dσ ,
and we obtain the Dirichlet boundary condition

β + (x, ȳ1 ) = φ (x,Y ), β + (x, 0) = 0.
THEOREM 2..9 The Problem 1 has a unique solution.

Proof. Testing Aη k with ηθ 2 , we obtain

1 2 (η k y θ ) 2 + 1 2 (η k x θ ) 2 1 2 (η k ) 2 (θ θ ′ ) ′ + 1 2 (c 0 y + kz + β x)(η k ) 2 (θ 2 ) ′ + 1 2 αθ 2 (η k ) 2 - 1 2 αx(η k θ ) 2 n(x)dσ + 1 2 yθ 2 (η k ) 2 n(z)dσ .
So, we have

(η k x θ ) 2 C; (η k y θ ) 2 C.
Testing (Aη k ) with η k z y 2p-1 θ 2q , we deduce the following lemma LEMMA 2.. [START_REF] Doob | Stochastic Processes[END_REF] We have

∆ 1 (η k z π) 2 1 4 L 0 ȳ1 0 {(η k y (x, y,Y )) 2 + (η k x (x, y,Y )) 2 }y 2p-1 θ 2q dxdy + 1 4 L 0 0 -ȳ1 {(η k y (x, y, -Y )) 2 + (η k x (x, y, -Y )) 2 }|y| 2p-1 θ 2q dxdy + (p -1 2 ) ∆ 1 (η k y θ )(η k z π)y p-2 θ q-1 + q ∆ 1 (η k y θ )(η k z π)y p-1 θ q-2 θ ′ + ∆ 1 (c 0 y + kz + β x)(η k y θ )(η k z π)y p-1 θ q-1 + ∆ 1 αx(η k x θ )(η k z π)y p-1 θ q .
(2..27)

Moreover, η k z π L 2 C π . Denoting v k := η k π, we have v k H 1 (∆ 1 )
Cπ so we can extract a weakly converging subsequence

v k → v in H 1 (∆ 1 ) and v = ηπ ∈ H 1 (∆ 1 )
. Similarly as before, we can check that η satisfies the boundary condition of Problem 1 which is summarized in the following lemma. (see proof in Appendix). LEMMA 2.. [START_REF] Feau | Les méthodes probabilistes en mécanique sismique. Applications aux calculs de tuyauteries fissurées[END_REF] We have π(y

)Aη = 0 in H -1 (∆ 1 ), π(y)η x (±L, y, z) = 0 in (H 1 2 ((-L, L) × (-ȳ1 , ȳ1 ))) ′ , η(x, y,Y ) -β + (x, y) ∈ K Y, φ , η(x, y, -Y ) ∈ K -Y, φ , and η(x, ȳ1 , z) = φ (x, z); η(x, ȳ1 , z) = 0.

2.2..5 Local regularity in the interior Dirichlet problem

In this section, we derive local regularity properties of the function v related to the interior problem. We recall that v := πη ∈ H 1 (∆ 1 ) and we have

- 1 2 v xx - 1 2 v yy + ρ 0 (x, y, z)v y + αxv x -yv z = ηρ 1 (x, y, z) + η y ρ 2 (y) (2..28)
where we denote

ρ 0 (x, y, z) := β x + c 0 y + kz, ρ 1 (x, y, z) := π ′′ 2 -ρ 0 (x, y, z)π ′ , ρ 2 (y) = π ′ (y). Let ȳ < ȳ1 , Notation 2..10 ∆ 1 (δ ) := {(x, y, z) ∈ ∆ 1 , |y -ȳ1 | > δ , |y + ȳ1 | > δ } ∆ 1 (δ , γ) := {(x, y, z) ∈ ∆ 1 (δ ), |z -Y | > γ} H 1 (δ ) := H 1 (-Y,Y ; H 1 (-L, L; H 1 (-ȳ + δ , ȳ -δ ))) Recall that η ∈ H 1 (δ ) means η, η x , η y , η z , η xy , η xz , η zy , η xyz ∈ L 2 (∆ 1 (δ )). PROPOSITION 2..11 We have ∀δ , γ > 0, η ∈ H 1 (δ ); η ∈ C 0 (∆ 1 (δ )); η ∈ H 2 (∆ 1 (δ , γ)), (2..29) and η H 1 (δ ) M 1 ; η H 2 (∆ 1 (δ ,γ)) M 2 .
(2..30)

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise17 of 46

Moreover, the trace of η at y = ȳ, denoted by h(x, z) := η(x, ȳ, z) satisfies

h ∈ H 1 ((-L, L) × (-Y,Y )) ∩ C 0 ((-L, L) × (-Y,Y )). (2.
.31)

Proof.

The proof relies on the following estimates (see proof in Appendix). We have

η xx π 2 ∈ L 2 (∆ 1 ); η xy π 2 ∈ L 2 (∆ 1 ); η yy π 2 ∈ L 2 (∆ 1 ), η xz y 2 π 4 ∈ L 2 (∆ 1 ); η yz y 2 π 3 ∈ L 2 (∆ 1 ).
For p and q large enough, we have η xyz y p π q ∈ L 2 (∆ 1 ); η yyz y p π q ∈ L 2 (∆ 1 ); ρ(z)η xxz y p π q ∈ L 2 (∆ 1 ); ρ(z)η zz y p π q ∈ L 2 (∆ 1 ).

The exterior Dirichlet problem

In this section, we prove existence and uniqueness to the homogeneous exterior Dirichlet problem. Let us recall the exterior Dirichlet problem from [START_REF] Bensoussan | Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators[END_REF]. Denote τ := inf{t > 0, |y(t)| = ȳ} and consider

h ± ∈ L ∞ (-Y,Y ). It is shown that E (y,z) (h ± (z( τ ))) solves a Dirichlet problem: Find ζ ∈ L ∞ (D) ∩ C 0 (D ε ), ∀ε > 0 such that Aζ = 0 in D, B ± ζ = 0, in D ± (3..1) with ζ (± ȳ, z) = h ± (z), -Y < z < Y.
By solving the ordinary differential equation on the boundary, there is

ζ y (y, ±Y ) = K exp(c 0 y 2 ± 2kY y).
As a bounded solution is sought, K must be 0. Hence, ζ (y, ±Y ) = h ± (±Y ) and then problem 3..1 was recast in (1) by

Aζ = 0 in D, ζ (y, ±Y ) = h ± (±Y ), in D ± (3..2) with ζ (± ȳ, z) = h ± (z), -Y < z < Y.
∆ d := (-L, L) × (-∞, -ȳ) × (-Y,Y ), ∆ u := (-L, L) × ( ȳ, +∞) × (-Y,Y ) ∆ := ∆ d ∪ ∆ u , ∆ + := (-L, L) × ( ȳ, +∞) × {Y }, ∆ -:= (-L, L) × (-∞, -ȳ) × {-Y} and ∆ ε u := ∆ u ∩ {|y| > ȳ + ε}, ε > 0.
Due to the symmetry in the exterior Dirichlet problem, we can consider positive values of y only. Denote τ := inf{t > 0, y(t) = ȳ} and consider h ∈ L ∞ ((-L, L) × (-Y,Y )), we define E (x,y,z) (h(x( τ), z( τ ))) as the solution of the exterior Dirichlet Problem 2:

Problem 2 Find ζ ∈ L ∞ (∆ u ) ∩ C 0 (∆ ε u ), ∀ε > 0 such that Aζ = 0 in ∆ u , B + ζ = 0 in ∆ + and ζ x (±L, y, z) = 0, in (y, z) ∈ ( ȳ, +∞) × (-Y,Y ), ζ (x, ȳ, z) = h(x, z), in (x, z) ∈ (-L, L) × (-Y,Y ). 3.2..1 Boundary condition Find ζ + ∈ L ∞ (∆ + ) such that B + ζ + = 0 in ∆ + , (3..3)
and

ζ + x (±L, y, z) = 0 in (y, z) ∈ ( ȳ, +∞) × (-Y,Y ), ζ + (x, ȳ, z) = h(x, z) in (x, z) ∈ (-L, L) × (-Y,Y ). Define H 1 2 (∆ + ) := u : ∆ + → R, ∆ + u 2 + u 2 x + u 2 y 1 + y 2 dxdy < ∞ . PROPOSITION 3..3 Assume that there exists H ∈ H 1 2 (∆ + ) such that H(x, ȳ) = h(x,Y ).
Then there exists one and only one solution to the problem (3..3) with

ζ + ∈ H 1 2 (∆ + ), ζ + L ∞ h(.,Y ) L ∞ .
Proof. First, we prove uniqueness. It is sufficient to prove that if we have

B + ζ = 0, ζ x (±L, y) = 0 and ζ (x, ȳ) = 0 with ζ bounded then we obtain ζ = 0. Set u(x, y) := ζ (x, y) exp(-c 0 2 y 2 ) then 1 2 u xx + 1 2 u yy -αxu x -u y (β x + kY ) + uc 0 (- c 0 2 y 2 -y(β x + kY) + 1 2 ) = 0, (x, y) ∈ ∆ + , u x (±L, y) = 0, y > ȳ, u(x, ȳ) = 0, x ∈ (-L, L).
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise19 of 46

We can assume that -c 0 2 y 2y(β x + kY ) + 1 2 < 0, y ȳ ( ȳ sufficiently large). Let us prove that u = 0. Indeed if u has a positive maximum if cannot be at y = ∞. But then this contradicts maximum principle from (3..3). Similarily, we cannot have a negative minimum. Now, we adress existence. Let

W 1 2 (∆ + ) :=    u ∈ H 1 2 (∆ + ), u W 1 2 (∆ + ) := u L ∞ + ∆ + u 2 x + u 2 y (1 + y 2 ) 2 dxdy 1 2 < ∞    .
We define a bilinear form on

H 1 2 (∆ + ) × W 1 2 (∆ + ) by a(u, v) := 1 2 ∆ + u x v x (1 + y 2 ) 2 dxdy + 1 2 ∆ + u y v y (1 + y 2 ) 2 dxdy -2 ∆ + u y vy (1 + y 2 ) 3 dxdy +α ∆ + xu x v (1 + y 2 ) 2 dxdy + ∆ + (β x + c 0 y + kY )u y v (1 + y 2 ) 2 dxdy.
We next define

a γ (u, v) := a(u, v) + γ ∆ + uv (1 + y 2 ) 2 dxdy.
We finally define a bilinear form on

H 1 2 (∆ + ) × H 1 2 (∆ + ) by a γ,δ (u, v) := 1 2 ∆ + u x v x (1 + y 2 ) 2 dxdy + 1 2 ∆ + u y v y (1 + y 2 ) 2 dxdy -2 ∆ + u y vy (1 + y 2 ) 3 dxdy +α ∆ + xu x v (1 + y 2 ) 2 dxdy + ∆ + (β x + c 0 y - ȳ 1 + δ y + c 0 ȳ + kY ) u y v (1 + y 2 ) 2 dxdy +γ ∆ + uv (1 + y 2 ) 2 dxdy. If v ∈ W 1 2 (∆ + ), a γ,δ (u, v) = a γ (u, v) - ∆ + c 0 (y -ȳ)δ y 1 + δ y u y v (1 + y 2 ) 2 dxdy. If u ∈ W 1 2 (∆ + ), we can compute a γ,δ (u, u) = 1 2 ∆ + u 2 x (1 + y 2 ) 2 dxdy + 1 2 ∆ + u 2 y (1 + y 2 ) 2 dxdy -2 ∆ + u y uy (1 + y 2 ) 3 dxdy +α ∆ + xu x u (1 + y 2 ) 2 dxdy + ∆ + (β x + c 0 ȳ + kY ) (1 + y 2 ) 2 u y udxdy - c 0 2 ∆ + u 2 (1 + y 2 ) 2 1 1 + δ y - δ (y -ȳ) (1 + δ y) 2 - 4(y -ȳ)y (1 + δ y)(1 + y 2 ) dxdy +γ ∆ + u 2 (1 + y 2 ) 2 dxdy.
And see that

a γ,δ (u, u) 1 2 ∆ + u 2 x (1 + y 2 ) 2 dxdy + 1 2 ∆ + u 2 y (1 + y 2 ) 2 dxdy -2 ∆ + u y uy (1 + y 2 ) 3 dxdy +α ∆ + xu x u (1 + y 2 ) 2 dxdy + ∆ + (β x + c 0 ȳ + kY ) (1 + y 2 ) 2 u y udxdy +(γ - c 0 2 ) ∆ + u 2 (1 + y 2 ) 2 dxdy
and the right hand does not depend on δ . Moreover, we can define a constant γ depending only on the constants α, β , c 0 , ȳ, k,Y but not on δ such that

a γ,δ (v, v) a 0 v 2 H 1 2 (∆ + ) , a 0 > 0. (3..4)
The constant a 0 depends only on α, β , c 0 , ȳ, k,Y . If f (x, y) is bounded, we consider the problem

a γ,δ (u, v -u) γ ∆ + f (v -u) (1 + y 2 ) 2 dxdy, ∀v ∈ K, u ∈ K (3..5)
where

K := {v ∈ H 1 2 (∆ + ), v(x, ȳ) = h(x,Y
)} which is not empty from the assumption. Then from the coercivity (3..4) and results of the theory of Variational Inequalities (3..5) has one and only one solution u γδ ( f ) (writing u for u γδ ( f ) to simplify notation). If w ∈ H 1 2 (∆ + ) satisfies w(x, ȳ) = 0 then u + w ∈ K, hence

a γ,δ (u, w) = γ ∆ + f w (1 + y 2 ) 2 dxdy
and thus also

- 1 2 u xx - 1 2 u yy + αxu x + (β x + c 0 y - ȳ 1 + δ y + c 0 ȳ + kY )u y + γu = γ f , (x, y) ∈ ∆ + , u x (±L, y) = 0, y ∈ ( ȳ, +∞), u(x, ȳ) = h(x,Y ), x ∈ (-L, +L). Also if M f := max{ h(.,Y ) ∞ , f L ∞ } then u γδ ( f ) L ∞ M f .
Moreover,

a γδ (u, H -u) γ ∆ + f (H -u) (1 + y 2 ) 2 dxdy hence a γδ (u, u) a γδ (u, H) -γ ∆ + f (H -u) (1 + y 2 ) 2 dxdy
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise21 of 46 and

a γδ (u, H) = a γ (u, H) - ∆ + c 0 (y -ȳ)δ y (1 + δ y) u y H (1 + y 2 ) 2 dxdy C u H 1 2 (∆ + ) H W 1 2 (∆ + ) + γCM f h(.,Y ) ∞
where C depends only on constants α, β , c 0 ȳ, k,Y . From (3..4), we deduce easily that

u γδ ( f ) H 1 2 (∆ + ) C γ ( f ). (3..6)
Letting δ → 0, we obtain

u γδ → u γ ( f ) in H 1 2 (∆ + ) weakly and in L ∞ (∆ + ) weakly-⋆, u γ ( f ) ∈ K
We deduce easily from (3..5) that u γ ( f ) satisfies

a γ (u, v -u) γ ∆ + f (v -u) (1 + y 2 ) 2 dxdy (3..7) ∀v ∈ W 1 2 (∆ + ), v ∈ K, u ∈ K, u ∞ M f , u H 1 2 (∆ + ) C γ ( f )
where again we write u for u γ ( f ). The solution of (3..7) is unique. Indeed, we first claim that

a(u, w) = γ ∆ + f w (1 + y 2 ) 2 dxdy, ∀w ∈ W 1 2 (∆ + ), w(x, ȳ) = 0.
But then if u 1 , u 2 are two solutions

a γ (u 1 , u 1 -u 2 ) = γ ∆ + f u 1 -u 2 (1 + y 2 ) 2 dxdy, a γ (u 2 , u 1 -u 2 ) = γ ∆ + f u 1 -u 2 (1 + y 2 ) 2 dxdy,
hence a γ (u 1u 2 , u 1u 2 ) = 0. However, from (3..4) we also have

a γ (v, v) a 0 v 2 H 1 2 (∆ + ) , ∀v ∈ H 1 2 (∆ + )
Therefore u 1u 2 = 0. We next consider a sequence ζ n with ζ 0 = h(.,Y ) L ∞ given by the solution of

a γ (ζ n+1 , v -ζ n+1 ) γ ∆ + ζ n (v -ζ n+1 ) (1 + y 2 ) 2 dxdy, (3..8) where ζ n+1 ∈ K, ∀v ∈ W 1 2 (∆ + ), v ∈ K, ζ n+1 L ∞ max ( h(.,Y ) L ∞ , ζ n L ∞ ) . Considering ζ 1 , we have a γ (ζ 1 , v -ζ 1 ) γ ∆ + ζ 0 (v -ζ 1 ) (1 + y 2 ) 2 dxdy. Take v = ζ 1 -(ζ 1 -ζ 0 ) + , which is admissible, then a γ (ζ 1 , (ζ 1 -ζ 0 ) + ) γ ∆ + ζ 0 (ζ 1 -ζ 0 ) + (1 + y 2 ) 2 dxdy and a γ (ζ 1 -ζ 0 , (ζ 1 -ζ 0 ) + ) = a γ (ζ 1 , (ζ 1 -ζ 0 ) + ) -γ ∆ + ζ 0 (ζ 1 -H) +
(1 + y 2 ) 2 dxdy 0 from the assumptions. Therefore (ζ 1 -ζ 0 ) + = 0 which implies ζ 1 ζ 0 and also

ζ 1 L ∞ h(.,Y ) L ∞ . Then by induction if ζ n ζ n-1 , ζ n L ∞ h(.,Y ) L ∞ . We obtain ζ n+1 ζ n , ζ n+1 L ∞ h(.,Y ) L ∞ . Also a γ (ζ n+1 , ζ 1 -ζ n+1 ) γ ∆ + ζ n (ζ 1 -ζ n+1 ) (1 + y 2 ) 2 dxdy a γ (ζ n+1 , ζ n+1 ) a γ (ζ n+1 , ζ 1 ) -γ ∆ + ζ n (ζ 1 -ζ n+1 ) (1 + y 2 ) 2 dxdy a(ζ n+1 , ζ 1 ) + γ ∆ + ζ 1 ζ n+1 (1 + y 2 ) 2 dxdy -γ ∆ + ζ n (ζ 1 -ζ n+1 ) (1 + y 2 ) 2 dxdy Hence ζ n+1 H 1 2 (∆ + ) C γ ζ 1 H 1 2 (∆ + ) + h(.,Y ) ∞ . The sequence ζ n → ζ is monotone decreasing and ζ n → ζ in H 1 2 (∆ + ) weakly, ζ ∈ K, ζ L ∞ h(.,Y ) L ∞ .
Hence in the limit,

a(ζ , v -ζ ) 0, ∀v ∈ W 1 2 (∆ + ), v ∈ K, ζ ∈ K, ζ L ∞ h(.,Y ) L ∞ (3..9)
and ζ is solution of equation (3..3).

3.2..2 The Cauchy problem

The exterior Dirichlet Problem 2 is equivalent to

Aζ = 0, (x, y, z) ∈ (-L, L) × ( ȳ, ∞) × (-Y,Y ), ζ (x, ȳ, z) = h(x, z), (x, z) ∈ (-L, L) × (-Y,Y ), ζ (x, y,Y ) = ζ + (x, y), (x, y) ∈ (-L, L) × ( ȳ, ∞), ζ x (±L, y, z) = 0, (y, z) ∈ ( ȳ, ∞) × (-Y,Y ) (3..10) with B + ζ + = 0, (x, y) ∈ (-L, L) × ( ȳ, ∞), ζ + (x, ȳ) = h(x,Y ), x ∈ (-L, L), ζ + x (±L, y) = 0, y ∈ ( ȳ, ∞) and ζ + L ∞ h(.,Y ) L ∞ .
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise23 of 46 This is a Cauchy problem, with z taking place of time and (x, y) as the space variable. We write it as

ζ z + 1 y 1 2 ζ yy + 1 2 ζ xx -αxζ x -(β x + c 0 y + kz)ζ y = 0 (3..11)
and using the notation

A (z)u(x, y) := - 1 y 1 2 u yy + 1 2 u xx -αxu x -(β x + c 0 y + kz)u y we obtain -∂ ζ ∂ z + A (z)ζ = 0, (x, y) ∈ ∆ + , z < Y, ζ (x, y,Y ) = ζ + (x, y), (x, y) ∈ (-L, L) × ( ȳ, ∞), ζ x (±L, y, z) = 0, (y, z) ∈ ( ȳ, ∞) × (-Y,Y ), ζ (x, ȳ, z) = h(x, z), (x, z) ∈ (-L, L) × (-Y,Y ). (3..12)
So it is a Cauchy problem with mixed Dirichlet-Neuman boundary conditions. We consider the space H 1 2 (∆ + ), with a new norm

u 2 ⋆ = ∆ + 1 y u 2 x + u 2 y (1 + y 2 ) 2 dxdy + ∆ + u 2 (1 + y 2 ) 2 dxdy
which is not equivalent to

u 2 = ∆ + u 2 x + u 2 y (1 + y 2 ) 2 dxdy + ∆ + u 2 (1 + y 2 ) 2 dxdy.
The norm in L 2 2 (∆ + ) is defined by

|u| 2 ⋆ := ∆ + u 2 (1 + y 2 ) 2 dxdy u 2 ⋆ .
We define on H 1 2 (∆ + ) the bilinear continuous form

A (z)(u, v) := 1 2 ∆ + u x v x + u y v y y(1 + y 2 ) 2 dxdy - 1 2 ∆ + u y v(1 + 3y 2 ) y 2 (1 + y 2 ) 3 dxdy + ∆ + (αxu x + (β x + c 0 y + kz)u y ) v y(1 + y 2 ) 2 dxdy.
Moreover,

A (z)(u, u) = 1 2 ∆ + u 2 x + u 2 y y(1 + y 2 ) 2 dxdy - 1 2 ∆ + u y u(1 + 3y 2 ) y 2 (1 + y 2 ) 3 dxdy + ∆ + (αxu x + (β x + c 0 ȳ + kz)u y ) u y(1 + y 2 ) 2 dxdy - 1 2 ∆ + u 2 c 0 d dy ( y - ȳ y(1 + y 2 ) 2 )dxdy and for |z| < Y , a 0 ∆ + u 2 x + u 2 y y(1 + y 2 ) 2 dxdy -b 0 ∆ + u 2 (1 + y 2 ) 2 dxdy
where the constants a 0 , b 0 depend only on α, β , c 0 , k, ȳ,Y . Therefore also,

< A (z)u, u > a 0 u 2 H 1 2 (∆ + ) -b ∆ + u 2 (1 + y 2 ) 2 dxdy.
The problem (3..12) is equivalent to

-( ∂ ζ ∂ z , w) + A (z)(ζ , w) = 0, ∀w ∈ H 1 2 (∆ + ), w(x, ȳ) = 0 (3..13) and ζ (x, y,Y ) = ζ + (x, y) , ζ (x, ȳ, z) = h(x, z).
We assume that there exists

H(z) ∈ H 1 (-Y,Y ; H 1 2 (∆ + )) with H(z)(x, ȳ) = h(x, z), ∀x, z. (3..14) Writing ζ (x, y, z) = ζ (x, y, z) -H(z)(x, y), we deduce -( ∂ ζ ∂ z , w) + A (z)( ζ , w) = ( ∂ H ∂ z (z)(x, y), w) -A (z)(H(z), w), ∀w ∈ H 1 2 (∆ + ), w(x, ȳ) = 0. ζ (x, y,Y ) = ζ + (x, y) -H(Y )(x, y) ζ (x, ȳ, z) = 0
Under this form, we obtain one and only one solution

ζ (x, y, z) ∈ L 2 -Y,Y ; H 1 2,0 (∆ + ) , ∂ ζ ∂ z (x, y, z) ∈ L 2 -Y,Y ; H 1 2,0 (∆ + ) ′
where H 1 2,0 (∆ + ) denotes the subspace of H 1 2 (∆ + ) of function which vanish at ȳ. We now prove that

ζ L ∞ h(.,Y ) L ∞ (3..15)
We will consider ∆ + R := {-L < x < L, ȳ < y < R} for R large. We begin with an approximation of the boundary condition ζ + with ζ + R solution of (we delete +)

1 2 ζ R,xx + 1 2 ζ R,yy -αxζ R,x -(β x + c 0 y + kY )ζ R,y = 0, (x, y) ∈ (-L, L) × ( ȳ,Y ), ζ R,x (±L, y) = 0, y ∈ ( ȳ, ∞), ζ R (x, ȳ) = h(x,Y ), x ∈ (-L, L), ζ R (x, R) = 0,
x ∈ (-L, L).

(3..16)

We can assume h 0, otherwise we decompose

h = h + -h -. We extend ζ R (x, y) by 0 for y > R. The sequence of functions ζ R (x, y) is increasing and ζ R L ∞ h(.,Y ) L ∞ . Let θ (y) = 1 if 0 < y < 1 2 and
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise25 of 46 0 if y > 1 be a smooth function. We may assume ȳ < R 2 . Let v ∈ W 1 2 (∆ + ), v ∈ K and test (3..16) with

vθ ( y R )-ζ R ( 1+y
2 ) 2 which vanishes at y = ȳ and y = R. Setting θ R (y) = θ ( y R ), we obtain,

1 2 ∆ + R ζ R,x (v x θ R -ζ R,x ) (1 + y 2 ) 2 dxdy + 1 2 ∆ + R ζ R,y (v y θ R -ζ R,y )
(1 + y 2 ) 2 dxdy (3..17)

+ 1 2R ∆ + R ζ R,y vθ ′ R (1 + y 2 ) 2 dxdy - ∆ + R 2ζ R,y (vθ R -ζ R )y (1 + y 2 ) 3 y dxdy + ∆ + R αxζ R (vθ R -ζ R ) (1 + y 2 ) 2 dxdy + ∆ + R (β x + c 0 y + kY ) (1 + y 2 ) 2 ζ R,y (vθ R -ζ R )dxdy = 0.
and thus also a

(ζ R , vθ R -ζ R ) = 0. Recalling that a(u, u) + γ ∆ + u 2 (1 + y 2 ) 2 dxdy a 0 u 2 H 1 2 (∆ + ) , we get a 0 ζ R 2 H 1 2 (∆ + ) a(ζ R , vθ R ) + C h(.,Y ) 2 L ∞ , from which we deduce easily, ζ R H 1 2 (∆ + )
C.

We then consider the limit

ζ R → ζ monotone increasing , ζ R → ζ in H 1 2 (∆ + ) weakly Hence, a(ζ , v) a(ζ , ζ )
which implies that ζ is the solution ζ + of (3..3). We next consider the approximation of (3..10) for ȳ < y < R. We write

Aζ R = 0, (x, y) ∈ ∆ R × (-Y,Y ), ζ R (x, y,Y ) = ζ + R (x, y), (x, y) ∈ ∆ R , ζ R (x, ȳ, z) = h(x, z), (x, z) ∈ (-L, L) × (-Y,Y ), ζ R (x, R, z) = 0, (x, z) ∈ (-L, L) × (-Y,Y ), ζ R,x (±L, y, z) = 0, (y, z) ∈ ( ȳ, R) × (-Y,Y ). (3..18)
We write (3..18) as a Cauchy problem

∂ ζ R ∂ z + A (z)ζ R = 0, in ∆ R , ζ R (x, y, R) = ζ + R (x, y), (x, y) ∈ ∆ + R , ζ R,x (±L, y, z) = 0, (y, z) ∈ ( ȳ, R) × (-Y,Y ), ζ R (x, ȳ, z) = h(x, z), (x, z) ∈ (-L, L) × (-Y,Y ), ζ R (x, R, z) = 0, (x, z) ∈ (-L, L) × (-Y,Y ) (3..19)
and extend ζ R by 0 for y > R. If w ∈ H 1 2 (∆ + ), w(x, ȳ) = 0, we can write

-( ∂ ζ R ∂ z , wθ R ) + A (z)(ζ R , θ R w) = 0, ζ R (x, y,Y ) = ζ + R (x, y), (x, y) ∈ ∆ + R , ζ R (x, ȳ, z) = h(x, z), (x, z) ∈ (-L, L) × (-Y,Y ).
(3..20)

However, from (3..19), we deduce ζ R L ∞ h(.,Y ) L ∞ . (3..21)
We can pass to the limit in (3..21) and check that ζ R → ζ solution of (3..13). This proves (3..15).

Local regularity in the exterior Dirichlet problem

In this section, we derive local regularity properties of the solution ζ to the exterior Dirichlet problem. Recall that ȳ < ȳ1 ,

Notation 3..4 ∆ d (δ ) := {(x, y, z) ∈ ∆ d , y < -ȳ -δ }, ∆ u (δ ) := {(x, y, z) ∈ ∆ u , y > ȳ + δ }, ∆ -(δ ) := ∆ d (δ ) ∩ ∆ -, ∆ + (δ ) := ∆ u (δ ) ∩ ∆ + , ∆ d (δ , γ) := {(x, y, z) ∈ ∆ d (δ ), -Y + γ < z}, ∆ u (δ , γ) := {(x, y, z) ∈ ∆ u (δ ), z < Y -γ}, H d (δ ) := H 1 (-Y,Y ; H 1 (-L, L; H 1 (-ȳ1 -δ , -ȳ1 + δ ))), H u (δ ) := H 1 (-Y,Y ; H 1 (-L, L; H 1 ( ȳ1 -δ , ȳ1 + δ )))
and

∆ (δ ) := ∆ d (δ ) ∪ ∆ u (δ ), ∆ (δ , γ) := ∆ d (δ , γ) ∪ ∆ u (δ , γ), H(δ ) := H d (δ ) ∩ H u (δ ).
Again, thanks to the symmetry in the exterior Dirichlet problem, we consider only positive values of y.

PROPOSITION 3..5 [Local regularity of the exterior Dirichlet problem] We have

∀δ , γ > 0, ζ ∈ H u (δ ); ζ ∈ C 0 (∆ u (δ )); ζ ∈ H 2 (∆ u (δ , γ)), (3..22) and ζ H u (δ ) M 1, φ ; ζ H 2 (∆ u (δ ,γ)) M 2, φ . (3..23) Moreover, the trace of ζ at y = ȳ1 , denoted g(x, z) := ζ (x, ȳ, z) satisfies g ∈ H 1 ((-L, L) × (-Y,Y )) ∩ C 0 ((-L, L) × (-Y,Y )).
Γ ± 1 := (-L, L) × {± ȳ1 } × (-Y,Y ); Γ ± := (-L, L) × {± ȳ} × (-Y,Y ) and Γ1 := Γ - 1 ∪ Γ + 1 ; Γ := Γ -∪ Γ + 4.1. Construction of the operator P Consider φ := (φ -, φ + ) ∈ L ∞ ( Γ1 ).
Following the same procedure of the 1d case, we first solve the interior Dirichlet problem for η with the boundary condition

   η = φ + in Γ + 1 , η = φ -in Γ - 1 . Then η L ∞ max( φ + , φ -).
Then, we solve the exterior Dirichlet problem for ζ with the boundary condition

   ζ = η in Γ + , ζ = η in Γ -. Then ζ L ∞ η L ∞ φ L ∞ .
For p1 ∈ Γ1 , let us define Pφ ( p1 ) := ζ ( p1 ).

Probabilistic interpretation of the operator P

Let us recall from Khasminskii [START_REF] Khasminkii | Stochastic Stability of Differential Equations[END_REF] the probabilistic interpretation of the operator P. For p ∈ Γ and for p1 ∈ Γ1 , we denote by γ1 ( p; .) the distribution of z( τ1 ) starting from p and by γ( p1 ; .) the distribution of z( τ) starting from p1 . We showed

η( p) = Γ1 φ (u) γ1 ( p; du) and ζ ( p1 ) = Γ η(v) γ( p1 ; dv).
Also, using Fubini theorem we can write

Pφ ( p1 ) = Γ1 φ (u) γ γ1 ( p1 , du),
where γ γ1 ( p1 , du) := Γ γ( p1 ; dv) γ1 (v; du).

The nonhomogeneous interior Dirichlet problem

Consider f a bounded function on ∆ 1 , we want to define E (x,y,z) τ1 0 f (x(s), y(s), z(s))ds as the solution of the non-homogeneous interior Dirichlet Problem 3:

Problem 3 Find χ ∈ L ∞ (∆ 1 ) ∩ C 0 (∆ ε 1 ), ∀ε > 0 such that Aχ + f = 0 in ∆ 1 , B ± χ + f = 0 in ∆ ± 1 , χ(x, ± ȳ1 , z) = 0 in (-L, L) × (-Y,Y ), χ x (±L, y, z) = 0 in (-ȳ1 , ȳ1 ) × (-Y,Y ). Consider Φ(x, y, z) := exp λ (c 0 kz 2 + c 0 y 2 ); λ 1. THEOREM 5..1
The Problem 3 has a unique solution. Moreover, we have

χ L ∞ expλ (c 0 kY 2 + c 0 ȳ2 1 ),
where λ depends on f .

Proof. The uniqueness of a solution of Problem 3 is argued as the uniqueness for the homogeneous Dirichlet interior problem. The existence can be proven by the regularization technique used previously. Now, we give a L ∞ bound for χ. We have

φ x = 0; φ xx = 0 φ y = 2yc 0 λ φ ; φ yy = 2c 0 λ φ + (2c 0 λ y) 2 φ φ z = 2zc 0 kλ φ and yφ z -αxφ x -φ y (β x + c 0 y + kz) + 1 2 φ yy + 1 2 φ xx = -2β c 0 λ xyφ -2c 2 0 λ y 2 φ + c 0 λ φ + 2(c 0 λ y) 2 = c 0 λ φ -c 0 λ 2β xyφ + 2(c 0 y) 2 λ (λ -1)φ c 0 λ (1 -2β L ȳ1 )φ + 2(c 0 y) 2 λ (λ -1)φ f
where ȳ1 can be chosen as 2 ȳ1 < 1 2β L , and λ max(1,

f c 0 (1-2β L ȳ1 ) ).
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise31 of 46

Using that φ (x, y, ±Y ) = exp λ (c 0 kY 2 ) exp λ (c 2 0 y 2 ) we obtain for z = Y, 0 < y < ȳ1 .

-αxφ x -φ y (β x + c 0 y + kY ) +

1 2 φ yy + 1 2 φ xx = -φ y (x, y,Y )(c 0 y + kY + β x) + 1 2 φ yy (x, y,Y ) = λ c 0 φ (x, y,Y ) 1 -2(kY + β x)y + 2c 0 (λ -1)y 2 λ c 0 φ (x, y,Y ) 1 -2(kY + β L)y + 2c 0 (λ -1)y 2 λ c 0 1 - (kY + β L) 2 2c 0 (λ -1) f , if λ max(2 f c 0 , 1 + (kY +β L) 2 c 0
). Similar estimates hold for z = -Y, -ȳ1 < y < 0. Consider u := φχ, then u satisfies the following inequalities

Au 0 in ∆ 1 ; B ± u 0 in ∆ ± 1 (5..1)
and the following boundary conditions

u(x, ± ȳ1 , z) = exp(λ c 0 kz 2 + c 0 ȳ2 1 ) in (-L, L) × (-Y,Y ), u x (±L, y, z) = 0 in (-ȳ1 , ȳ1 ) × (-Y,Y ). (5..2)
The maximum of u cannot be attained in ∆ 1 or on the boundaries z = ±Y, x = ±L. It can only be attained for y = ȳ1 or y = ȳ1 . Hence, u(x, y, z) exp(λ

(c 0 kY 2 + c 0 ȳ2 1 )) which implies χ(x, y, z) -exp(λ (c 0 kY 2 + c 0 ȳ2 1 )). Now, consider v = -(φ + χ), then v satisfies inequalities (5..3) Av 0 in ∆ 1 ; B ± v 0 in ∆ ± 1 (5..3)
and the boundary condition (5..4)

v(x, ± ȳ1 , z) = -exp(λ c 0 kz 2 + c 0 ȳ2 1 ) in (-L, L) × (-Y,Y ), v x (±L, y, z) = 0 in (-ȳ1 , ȳ1 ) × (-Y,Y ). (5..4)
Again, the minimum of v cannot be attained in ∆ 1 or on the boundaries z = ±Y, x = ±L. Hence

v(x, y, z) -exp(λ (c 0 kY 2 + c 0 ȳ2 1 )) which implies χ(x, y, z) exp(λ (c 0 kY 2 + c 0 ȳ2 1 )).
Now, let us derive some estimates on derivatives. Let M := exp(λ (c 0 kY 2 + c 0 ȳ1 2 )). Using the test function m 0 (x, y, z) = exp(-c 0 (y 2 + kz 2 )) exp(-αx 2 ) we obtain now a priori estimates on the partial derivatives of χ. We test [START_REF] Doob | Stochastic Processes[END_REF]

with m 0 χ ∆ 1 χ 2 y m 0 + ∆ 1 χ 2 x m 0 = ∂ ∆ 1 ,z=Y yχ 2 m 0 - ∂ ∆ 1 ,z=-Y yχ 2 m 0 + 2 ∆ 1 f χm 0 - ∆ 1 c 0 β xχ 2 m 0 ∂ ∆ 1 ,z=Y,y>0 yχ 2 m 0 - ∂ ∆ 1 ,z=-Y,y<0 yχ 2 m 0 + 2 ∆ 1 f χ C(M).
Hence,

∆ 1 χ 2 y m 0 C(M) ; ∆ 1 χ 2 x m 0 C(M).
We have then,

∆ 1 (y 2 χ z ) 2 m 0 C(M) and ∆ 1 (y 2 χ yy ) 2 m 0 C(M).
The function χ is smooth outside a neighborhood of y = 0.

The nonhomogeneous exterior Dirichlet problem

Consider the function Ψ (y) := log(y) + K, K such that log( ȳ1 ) + K > 0 and the space of functions X ∞ Ψ := {uψ, u ∈ L ∞ (∆ u )}. We want to define E (x,y,z) τ 0 f (x(s), y(s), z(s))ds as the solution of the nonhomogeneous exterior Dirichlet Problem 4:

Problem 4 Find ξ ∈ X ∞ Ψ ∩ C 0 (∆ ε u ), ∀ε > 0 such as Aξ + f = 0 in ∆ u , B + ξ + f = 0 in ∆ + , ξ (x, ȳ, z) = 0 in (-L, L) × (-Y,Y ), ξ x (±L, y, z) = 0 in ( ȳ, +∞) × (-Y,Y ).
THEOREM 6.. [START_REF] Bensoussan | Degenerate Dirichlet Problems Related to the Invariant Measure of Elasto-Plastic Oscillators[END_REF] The Problem 4 has a unique solution. Moreover, we have -ψ(y) ξ (x, y, z) ψ(y).

Proof. We justify uniqueness of the solution taking f = 0 and setting ξ ξ = wψ α , α > 1.

We show w = 0. In particular, w satisfies ξ = wψ α ξ x = w x ψ α ; ξ y = w y ψ α + wαψ y ψ α-1 ; ξ z = w z ψ α ξ xx = w xx ψ α ; ξ yy = w yy ψ α + 2w y αψ y ψ α-1 + wα ψ yy ψ α-1 + (ψ y ) 2 ψ α-2 (α -1)

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise33 of 46 then 1 2

w xx ψ α + 1 2 {w yy ψ α + 2w y αψ y ψ α-1 wα ψ yy ψ α-1 + (ψ y ) 2 ψ α-2 (α -1)} -(c 0 y + kz + β x){w y ψ α + wαψ y ψ α-1 } -αxw x ψ α + yw z ψ α = 0.
Collecting terms related to ψ α , we obtain

1 2 w xx + 1 2 {w yy + 2w y α ψ y ψ + wα ψ yy ψ + (ψ y ) 2 ψ 2 (α -1)} +(-c 0 y -kz -β x){w y + wα ψ y ψ } -αxw x + yw z = 0,
which implies for z ∈ (-Y,Y ) and y > ȳ,

1 2 w xx + 1 2 w yy + w y {α ψ y ψ -c 0 -kz -β x} + wα ψ { 1 2 ψ yy + (α -1) 2
(ψ y ) 2 ψ -(c 0 y + kz + β x)αψ y } -αxw x + yw z = 0 and for z = Y ; y > ȳ, 1 2 w xx + 1 2 w yy + w y {α ψ y ψ -c 0 -kY -β x} + wα ψ { 1 2 ψ yy + (α -1) 2
(ψ y ) 2 ψ -(c 0 y + kY + β x)αψ y } -αxw x + yw z = 0 and w(x, ȳ, z) = 0.
We can find α > 1 with α -1 small so that

1 2 ψ yy -(c 0 y + kz + β x) + (α -1) (ψ y ) 2 ψ -γ c 0 2 + (α -1)γ y 2 (log(y) + K) < 0.
Since w → 0 as y → ∞, a positive maximum can be attained only for a finite y ⋆ . But this is impossible from the equation. So w is negative,w 0. But, w is also positive, w 0. Hence ξ = 0 and uniqueness follows. Now, existence is demonstrated by the following approximation

Aξ R + f = 0, in ∆ R , B + ξ R + f = 0, in ∆ + R , ξ (x, ȳ, z) = 0, ξ (x, R, z) = 0, ξ x (±L, y, z) = 0. Using u R = ξ R -ψ, we obtain Au R 0, in ∆ R , B + u R 0, in ∆ + R , u R (x, ȳ, z) = -ψ( ȳ), u R
x (±L, y, z) = 0.

Necessarily, u R 0. The sequence ξ R is monotone increasing and converges towards a solution. Let us show estimates on u. Suppose f 0, we will show that 0 ξ ψ. Then,

ψ y (-c 0 y -kz -β x) + 1 2 ψ yy = γ[ 1 y (-c 0 y -kz -β x) - 1 2y 2 ] γ[ 1 y (-c 0 y -kz) - 1 2y 2 ] γ[ 1 y (-c 0 y -kz) - 1 2y 2 ] -γ c 0 2 , if ȳ > 2kY c 0 .
Define γ such that γc 0 2 f . We then have, setting u = ξψ, Au 0, in ∆

B + u 0, in ∆ + u(x, ȳ, z) = -ψ( ȳ) u x (±L, y, z) = 0
It follows that u 0. We can show similarly that ξ + φ 0. Hence, we have -ψ ξ ψ. This defines a linear operator from L ∞ Γ1 in L ∞ ( Γ1 ).

Probabilistic interpretation of the operator T

For any A a Borel subset of O, consider the two following measures of occupation of A by the process (x(t), y(t), z(t)) starting at p ∈ O, namely ν χ (p; A) := E p τ1 0 1 A (x(s), y(s), z(s))ds ν ξ (p; A) := E p τ 0 1 A (x(s), y(s), z(s))ds .

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise35 of 46

We have shown

χ(p) = O f (q)dν χ (p; dq) and ξ ( p) = O f ( q)dν ξ ( p; d q).
For any p1 ∈ Γ1 , we have

T( f )( p1 ) := E p1 τ1 0 f (x(s), y(s), z(s))ds + Γ E q τ 0
f (x(s), y(s), z(s))ds dγ( p1 ; dq) T( f ) integrates f over a time cycle of the Markov chain stopped on Γ1 , that is (x( τ1,k ), y( τ1,k ), z( τ1,k )).

Construction of the invariant measure ν

Now, define

ν( f ) := L -L Y -Y T f (r, ȳ1 , s)γ + ⋆ (dr, ds) + L -L Y -Y T f (r, -ȳ1 , s)γ - ⋆ (dr, ds) L -L Y -Y T 1(r, ȳ1 , s)γ + ⋆ (dr, ds) + L -L Y -Y T 1(r, -ȳ1 , s)γ - ⋆ (dr, ds)
.

The denominator is well defined and is stricly positive. Now, we want to solve the complete problem.

Problem 5 Find u such that uψ -1 is bounded for |y| > | ȳ| and

Au + f = 0, in O B + u + f = 0, in O + B -u + f = 0, in O - u x (±L, y, z) = 0, in {x = ±L} ∩ O
Considering Problems 1, 2, 3 and 4, in the following result we use a functional analysis method to prove that ν is the unique invariant distribution associated to the solution (x(t), y(t), z(t)) of the stochastic variational inequality (1..2).

THEOREM 7..1 The Problem 5 has a unique solution if and only if ν( f ) = 0.

Proof. Similar arguments as the 1d case are considered. Uniqueness is guaranteed by the ergodic property of the operator P. Suppose ν( f ) = 0. We define χ the solution of the Interior non-homogeneous Dirichlet Problem 3 and ξ the solution of the Exterior non-homogeneous Dirichlet Problem 4. We set χ 0 := χ and ξ 0 := ξ , and for k 0, we define χ k+1 by

Aχ k+1 = 0, in ∆ 1 , B + χ k+1 = 0, in ∆ + 1 , B -χ k+1 = 0, in ∆ - 1 , χ k+1 x (±L, y, z) = 0, in (-ȳ1 , ȳ1 ) × (-Y,Y ) with χ k+1 (x, ȳ1 , z) = ξ k (x, ȳ1 , z); χ k+1 (x, -ȳ1 , z) = ξ k (x, -ȳ1 , z).
Afterwards, we define ξ k+1 by

Aξ k+1 = 0, in ∆ u , B + ξ k+1 = 0, in ∆ + u , ξ k+1 x (±L, y, z) = 0, in ( ȳ, +∞) × (-Y,Y ) with ξ k+1 (x, ȳ, z) = χ k+1 (x, ȳ, z).
Similarly, we define ξ k+1 by

Aξ k+1 = 0, in ∆ d , B -ξ k+1 = 0, in ∆ - d , ξ k+1 x (±L, y, z) = 0, in (-∞, -ȳ) × (-Y,Y ) with ξ k+1 (x, -ȳ, z) = χ k+1 (x, -ȳ, z).
That means

ξ 0 | Γ1 = T f ; ξ k+1 | Γ1 = P(ξ k | Γ1 ).
Next, we set ξ k = ξ 0 + ξ 1 + ... + ξ k ; χk = χ 0 + χ 1 + ... + χ k then ξ k | Γ1 = T f + P(T f ) + ... + P k (T f ).

Now, let us remark that χk satisfies the following equation:

A χk + f = 0, in ∆ 1 , B + χk + f = 0, in ∆ + 1 , B - χk + f = 0, in ∆ - 1 , χk x (±L, y, z) = 0, in (-∞, -ȳ) × (-Y,Y ) (7..1) with χk (x, ȳ1 , z) = ξ k-1 (x, ȳ1 , z); χk (x, -ȳ1 , z) = ξ k-1 (x, -ȳ1 , z)
ξ k satisfies the following equations:

A ξ k + f = 0, in ∆ u , B + ξ k + f = 0, in ∆ + u , ξ k x (±L, y, z) = 0, in ( ȳ, +∞) × (-Y,Y ) (7..2) ξ k (x, ȳ, z) = χk (x, ȳ, z)
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise37 of 46 and

A ξ k + f = 0, in ∆ d , B + ξ k + f = 0, in ∆ - d , ξ k x (±L, y, z) = 0, in (-∞, -ȳ) × (-Y,Y ) (7..3) ξ k (x, -ȳ, z) = χk (x, -ȳ, z). The condition ν( f ) = 0 means Γ1 T f (x)dπ(x) = Γ1 T f (r, -ȳ1 , s)dπ 1 (r, s) + Γ1 T f (r, ȳ1 , s)dπ 2 (r, s) = 0.
From the estimate (4..3), we obtained ξ k converges in L ∞ ( Γ1 ). Now, we notice that χk -χ is a solution of the interior homogeneous Dirichlet problem with ( χk - χ)| Γ1 = ξ k-1 | Γ1 and ξ k -ξ is a solution of the Exterior homogeneous Dirichlet problem with

( ξ k - ξ )| Γ = ( χk -χ)| Γ . Then, we obtain ξ k -ξ L ∞ χk -χ L ∞ ξ k-1 L ∞ C.
We can extract a subsequence such that ξ k → ξ ; χk → χ.

Moreover, χ satisfies equation:

A χ + f = 0, in ∆ 1 , B + χ + f = 0, in ∆ + 1 , B -χ + f = 0, in ∆ - 1 , χx (±L, y, z) = 0, in (-ȳ1 , ȳ1 ) × (-Y,Y ) (7..4) χ(x, ȳ1 , z) = ξ (x, ȳ1 , z); χ(x, -ȳ1 , z) = ξ (x, -ȳ1 , z)
and ξ satisfies equations:

A ξ + f = 0, in ∆ u , B + ξ + f = 0, in ∆ + u , ξx (±L, y, z) + f = 0, in ( ȳ, +∞) × (-Y,Y ) (7..5) ξ (x, ȳ, z) = χ(x, ȳ, z) A ξ + f = 0, in ∆ d , B -ξ + f = 0, in ∆ - d , ξx (±L, y, z) = 0, in (-∞, -ȳ) × (-Y,Y ) (7..6) ξ (x, -ȳ, z) = χ(x, -ȳ, z).
Then, we must have χ = ξ in ȳ < ±y < ȳ1

Thus, the function

u = χ in ∆ 1 , ξ in ∆ c
(u -ξ k )| Γ = (u -χk )| Γ . Hence, u -ξ k L ∞ ( Γ1 ) u -χk L ∞ ( Γ1 ) u -ξ k-1 L ∞ ( Γ1 ) u L ∞ ( Γ1 ) and so, ξ k L ∞ ( Γ1 ) C.
We have

ξ k = (k + 1) Γ1 T f (x)dπ(x) + k ∑ j=0 P j (T( f -ν( f )))
and since the sum is bounded, we obtain

(k + 1) Γ1 T f (x)dπ(x) is bounded. That leads to Γ1 T f (x)dπ(x) = 0 That implies ν( f ) = 0. Consider now ϕ a smooth function in [-L, L] × R × [-Y,Y] with compact support. If we take f = -Aϕ, f (x, y,Y ) = -B + ϕ, f (x, y, -Y ) = -B -ϕ,
then ϕ is solution of (5) for this f . From Theorem 7..1, we have

L -L ∞ -∞ Y -Y 1 2 ϕ yy + 1 2 ϕ xx -αxϕ x -(β x + c 0 y + kz)ϕ y + yϕ z dν(x, y, z) + L -L ∞ 0 1 2 ϕ yy + 1 2 ϕ xx -αxϕ x -(β x + c 0 y + kY )ϕ y dν(x, y,Y ) + L -L 0 -∞ 1 2 ϕ yy + 1 2 ϕ xx -αxϕ x -(β x + c 0 y -kY )ϕ y dν(x, y, -Y ) = 0.
If ν has a density m, then we deduce that

α ∂ ∂ x [xm] + ∂ ∂ y [(β x + c 0 y + kz)m] -y ∂ m ∂ z + 1 2 ∂ 2 m ∂ x 2 + 1 2 ∂ 2 m ∂ y 2 = 0 in (0, L) × R × (-Y,Y ) (7..7)
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise39 of 46 in the sense of distributions. If we test (7..7) with ϕ and integrate by parts, we obtain

- L -L +∞ -∞ ym(x, y,Y )ϕ(x, y,Y )dxdy + L 0 +∞ -∞ ym(x, y, -Y )ϕ(x, y, -Y )dxdy + L -L ∞ -∞ Y -Y m(x, y, z) 1 2 ϕ yy + 1 2 ϕ xx -αxϕ x -(β x + c 0 y + kz)ϕ y + yϕ z dxdydz = 0
and comparing with (7..7)

- L -L 0 -∞ ym(x, y,Y )ϕ(x, y,Y )dxdy + L 0 +∞ 0 ym(x, y, -Y )ϕ(x, y, -Y )dxdy - L -L ∞ 0 m(x, y,Y ) 1 2 ϕ yy + 1 2 ϕ xx -αxϕ x -(β x + c 0 y + kY )ϕ y + yϕ dxdy - L -L 0 -∞ m(x, y, -Y ) 1 2 ϕ yy + 1 2 ϕ xx -αxϕ x -(β x + c 0 y -kY )ϕ y -yϕ dxdy = 0 we finally deduce ym + ∂ ∂ x [xm] + ∂ ∂ y [(β x + c 0 y + kY )m] + 1 2 ∂ 2 m ∂ x 2 + 1 2 ∂ 2 m ∂ y 2 = 0, on O + -ym + ∂ ∂ x [xm] + ∂ ∂ y [(β x + c 0 y -kY )m] + 1 2 ∂ 2 m ∂ x 2 + 1 2 ∂ 2 m ∂ y 2 = 0, on O - m = 0, on (-L, L) × (-∞, 0) × {Y} ∪ (-L, L) × (0, ∞) × {-Y}
The proof of the main result is complete.

Appendix

(u -γ) + (x, ± ȳ1 , z) = 0 if x ∈ (-L, L), z ∈ (-Y,Y ) (u -γ) + (x, y, ±Y ) = 0 if x ∈ (-L, L), 0 < ±y < ȳ1 and (u + γ) -(x, ± ȳ1 , z) = 0 if x ∈ (-L, L), z ∈ (-Y,Y ) (u + γ) -(x, y, ±Y ) = 0 if x ∈ (-L, L), 0 < ±y < ȳ1 . Choose v = u -(u -γ) + ∈ K and we obtain -a(u, (u -γ) + ) -λ (u, (u -γ) + ) -λ (w, (u -γ) + ) a(u, (u -γ) + ) + λ (u, (u -γ) + ) λ (w, (u -γ) + ).
Switching the first argument by (u -γ) in the previous inequality, we obtain

a((u -γ) + , (u -γ) + ) + λ |(u -γ) + | 2 L 2 λ (w -γ, (u -γ) + )
and w γ implies (u -γ) + = 0. Taking v = u + (u + γ) -similar arguments yields (u + γ) -= 0. Proof of Lemma 2..2. We have the following expressions,

a(η ε , η ε ) = ∆ 1 ε 2 (η ε z ) 2 + 1 2 (η ε y ) 2 + 1 2 (η ε x ) 2 + (β x + kz + c 0 y)η ε η ε y -yη ε η ε z + αxη ε η ε x and a(η ε , u 0 ) = ∆ 1 ε 2 η ε z u 0z + 1 2 η ε y u 0y + 1 2 η ε x u 0x + (β x + kz + c 0 y)u 0 η ε y -yu 0 η ε z + αxu 0 η ε x . Inequality a(η ε , η ε ) a(η ε , u 0 ) means ε 2 ∆ 1 (η ε z ) 2 + 1 2 ∆ 1 (η ε x ) 2 + 1 2 ∆ 1 (η ε y ) 2 ε 2 ∆ 1 η ε z u 0z + 1 2 ∆ 1 η ε y u 0y - ∆ 1 (β x + c 0 y + kz)η ε η ε y + 1 2 ∆ 1 (β x + c 0 y + kz)u 0 η ε y + 1 2 ∆ 1 η ε x u 0x + ∆ 1 αxη ε η ε x - ∆ 1 αxu 0 η ε x + ∆ 1 yη ε η ε z - ∆ 1 yu 0 η ε z
We apply Cauchy-Schwartz inequality to the first two terms, then we apply Green formula to the last one and we get

ε 2 ∆ 1 (η ε z ) 2 + 1 2 ∆ 1 (η ε x ) 2 + 1 2 ∆ 1 (η ε y ) 2 ε ∆ 1 (η ε z ) 2 1 2 ( ∆ 1 (u ε 0z ) 2 ) 1 2 + ∆ 1 (η ε y ) 2 { 1 2 ( ∆ 1 (u ε 0y ) 2 ) 1 2 + φ ( ∆ 1 (β x + c 0 y + kz) 2 ) 1 2 } + ∆ 1 (η ε x ) 2 { 1 2 ( ∆ 1 (u ε 0x ) 2 ) 1 2 + α φ ( ∆ 1 x 2 ) 1 2 + α( ∆ 1 x 2 u 0 ) 1 2 } + φ 2 L -L ȳ1 -ȳ1 |y|dxdy + φ ( ∆ 1 |yu 0z |) + L -L ȳ1 -ȳ1 |y||u 0 (x, y,Y )| + L -L ȳ1 -ȳ1 |y||u 0 (x, y, -Y )|
with c 1 ,c 2 ,c 3 and c 4 are positive constant, we get

√ εη ε z 2 L 2 + η ε y 2 L 2 + η ε x 2 L 2 √ εc 1 √ εη ε z L 2 + c 2 η ε y L 2 + c 3 η ε x L 2 + c 4 .
Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise41 of 46

Proof of Lemma 2..3. We have -(2p -1) Testing (Aη k ) with η k z y 2p-1 θ 2q , we deduce

∆ 1 (η k y θ )(η k z π)y p-2 θ q-1 -2q ∆ 1 (η k y θ )(η k z π)y p-1 θ q-2 θ ′ , ∆ 1 η k xx η k z y 2p-1 θ 2q = - 1 2 ∆ 1 ((η k x ) 2 ) z y 2p-1 θ 2q = - 1
+(p - 1 2 ) ∆ 1 (η k y θ )(η k z π)y p-2 θ q-1 + q ∆ 1 (η k y θ )(η k z π)y p-1 θ q-2 θ ′ + ∆ 1 (c 0 y + kz + β x)(η k y θ )(η k z π)y p-1 θ q-1 + ∆ 1 αx(η k x θ )(η k z π)y p-1 θ q .
Proof of Lemma 2..4.

1.

With

f (η k , η k y ) := -η k ( π ′′ 2 -(c 0 y + kz + αx)π ′ ) -η k y π ′ ,
we have ∀ψ ∈ H 1 (∆ 1 ), ψ(x, y, ±Y ) = 0 and ψ(x, ± ȳ1 , z) = 0,

1 2 ∆ 1 v k y ψ y + 1 2 ∆ 1 v k x ψ x + ∆ 1 αxv k x + (β x + c 0 y + kz)v k y -yv k z ψ = ∆ 1 f (η k , η k y )ψ.
Now, when k goes to +∞, we get

1 2 ∆ 1 v y ψ y + 1 2 ∆ 1 v x ψ x + ∆ 1 αxv x + (β x + c 0 y + kz)v y -yv z ψ = ∆ 1 f (η, η y )ψ.
We deduce that in H -1 (∆ 1 ) we firstly have -Av = f (η, η y ) which is equivalent to πAη = 0 and secondly that choice of test function implies πη x (±L, y, z) = 0 in (H First, we study convergence of the sequence χ ±,k and we deduce PDEs satisfied by lim k→0 χ ±,k . In particular (8..1) implies χ +,k → χ + in L 2 ((-L, L) × (0, ȳ1 )) weakly, χ -,k → χ -in L 2 ((-L, L) × (-ȳ1 , 0)) weakly.

and equalities (8..2), (8..3) imply χ +,k π H 1 C and χ +,k π → χ + π in H 1 ((-L, L) × (0, ȳ1 )) weakly, χ -,k π H 1 C and χ -,k π → χ -π in H 1 ((-L, L) × (-ȳ1 , 0)) weakly.

Denote ξ ±,k := χ ±,k y 2 and g(χ +,k , χ +,k y ) := -χ +,k {1 -2y(αx + c 0 y + kY )} -2yχ +,k y . From (8..2), we have B Y ξ +,k = g(χ +,k , χ +,k y ) in (-L, L) × (-ȳ1 , ȳ1 ), in H -1 ((-L, L) × (0, ȳ1 )).

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise43 of 46

The operator B + is strictly elliptic then ξ +,k ∈ H 2 ((-L, L)×(0, ȳ1 )). We have ξ +,k x (±L, y) = 0 in (H 1 2 (0, ȳ1 )) ′ . As ξ +,k ∈ H 2 (∆ 1 ) and πB + χ +,k = 0, we have -B + ξ +,k = g(χ +,k , χ +,k y ) in a strong sense

We obtain that ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )), ψ(x, 0) = ψ(x, ȳ1 ) = 0, We deduce that in H -1 ((-L, L) × (0, ȳ1 )) we have -B + ξ + = g(χ + , χ + y ) which is equivalent to y 2 B + χ + = 0 and that choice of test function implies y 2 χ +

x (±L, y) = 0 in (H 1 2 ((0, ȳ1 ))) ′ . (b) Firstly, γ(πη k ) → π(χ + + β + ) in H 1 ((-L, L) × (0, ȳ1 )) weakly. Secondly, weak convergence of πη k → πη in H 1 (∆ 1 ) implies weak convergence of γ(πη k ) → γ(πη) in H 1 2 (∂ ∆ 1 ). By uniqueness of the limit, we have γ(πη) = π(χ + + β + ).

Using

  -αxu x -(β x + c 0 y + kz)u y + yu z , -αxu x -(β x + c 0 y + kY )u y , B -u := 1 2 u yy + 1 2 u xx -αxu x -(β x + c 0 y -kY )u y .

Notation 1.. 2 O

 2 := (-L, L) × R × (-Y,Y ); O + := (-L, L) × (0, +∞) × {Y}; O -:= (-L, L) × (-∞, 0) × {-Y}.As the main result of the paper we prove the following: THEOREM 1..3 There exists one and only one probability measure ν on O ∪ O -∪ O + satisfying O Aφ dν + O - Bφ dν + O + B + φ dν = 0, ∀φ smooth. Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise3 of 46 Moreover, ν has a probability density function m such that O m(x, y, z)dxdydz + O + m(x, y,Y )dxdy + O - m(x, y, -Y )dxdy = 1, where

  Since η( ȳ1 ,Y ) = φ (Y ) and η(-ȳ1 , -Y ) = 0, there are semi-explicit solutions by solving ordinary differential equations for η on the boundary at z = Y , and z = -Y respectively, η(y,Y ) = η Y I(y, ȳ1 ) + φ (Y )I(0, y), 0 < y ȳ1 ; η(y, -Y ) = η -Y I(-ȳ1 , y), -ȳ1 y < 0,

3. 1 .

 1 Some background on the exterior Dirichlet problem in the 1d case Notation 3..1 D d := (-∞,ȳ) × (-Y,Y ), D u := ( ȳ, +∞) × (-Y,Y ), D := D d ∪ D u , D + := ( ȳ, +∞) × {Y }, D -:= (-∞,ȳ) × {-Y } and D ε := D ∩ {|y| > ȳ + ε}, ε > 0.

3. 2 .

 2 The exterior Dirichlet problem in the 2d case Notation 3..2

PROPOSITION 3 .. 6 [ 4 .

 364 Local regularity of the exterior Dirichlet problem on the boundary z= Y ] We have ∀δ > 0, ζ ∈ H 2 (∆ + (δ )); ζ H 2 (∆ + (δ )) M 3, φ . (3..24) Similar results hold for negative values of y. Proofs of Proposition 3..5 and Proposition 3..6 rely on similar estimates related to the local regularity of the interior Dirichlet problem. Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise27 of 46 The ergodic operator P Notation 4..1

7. The operator T and the invariant measure ν 7 . 1 .

 71 Construction of the operator TConsider f ∈ L ∞ (O), following the same procedure of the 1d case, we first solve the interior nonhomogeneous Dirichlet problem for χ with f at the right hand side, then we solve the exterior nonhomogeneous Dirichlet problem for ξ with f at the right hand side and χ as a boundary condition. Also, for any p1 ∈ Γ1 , we define the operator T f ( p1 ) := ξ ( p1 ).

1 is the solution of Problem 5 .

 15 Now, suppose that Problem 5 has a unique solution u. Considering the same sequences χk and ξ k , we have that uχk is a solution of the interior homogeneous Dirichlet problem with (uχk )| Γ1 = (uξ k-1 )| Γ1 and u -ξ k is a solution of the exterior homogeneous Dirichlet problem

) 2 )) 2 ) 2

 222 z y 2p-1 θ 2q -(2p -1) (x, y,Y )y 2p-1 θ 2q + (x, y, -Y )y 2p-1 θ 2q

) 2 ) 2 1 ( 1 ( 1 ∆ 1 αxη k x η k z y 2p-1 θ 2q = ∆ 1 αx

 2211111 (x, y,Y )y 2p-1 θ 2q + (x, y, -Y )y 2p-1 θ 2q and ∆ c 0 y + kz + β x)η k y η k z y 2p-1 θ 2q = ∆ c 0 y + kz + β x)(η k y θ )(η k z π)y p-1 θ q-(η k x θ )(η k z π)y p-1 θ q .

  y ) 2 (x, y,Y )y 2p-1 θ 2q - y ) 2 (x, y, -Y )y 2p-1 θ 2q x ) 2 (x, y,Y )y 2p-1 θ 2q - x ) 2 (x, y, -Y )y 2p-1 θ 2q

1 2 (-

 2 (-L, L) × (-ȳ1 , ȳ1 )) ′ .2. (a)We know that πη k ∈ H 1 (∆ 1 ), its trace is well defined and satisfies,γ(πη k )(x, y,Y ) = π(χ +,k + β k,+ ); y > 0 γ(πη k )(x, y, -Y ) = π χ -,k ; y < 0 with χ ±,k L ∞ φ L ∞ (8..1)and χ -,k , χ +,k satisfy respectively (8..2) and (8..3) x ψ x + χ +,k y ψ y ) + (αxχ +,k x + (β x + c 0 y + kY )χ +,k y )ψ}dxdy = 0, ∀ψ ∈ H 1 ((-L, L) × (0, ȳ1 )) with ψ(x, 0) = ψ(x, ȳ1 ) ȳ1 { 1 2 (χ -,k x ψ x + χ -,k y ψ y ) + (αxχ -,k x + (β x + c 0 y -kY )χ -,k y )ψ}dxdy = 0,∀ψ ∈ H 1 ((-L, L) × (-ȳ1 , 0)) with ψ(x, 0) = ψ(x, -ȳ1 ) = 0. (8..3)

  x ψ x + ξ +,k y ψ y ) + αxξ +,k x + (β x + c 0 y + kY )ξ +,k yyξ +,k k , χ k y )ψ.Now, when k goes to 0, we haveL x ψ x + ξ + y ψ y ) + αxξ + x + (β x + c 0 y + kY )ξ + yyξ + z ψ = , χ y )ψ.

  αx(t)dt + dw(t) + 1 {x(t)=-L} dξ 1 t -1 {x(t)=L} dξ 2 t , dy(t) = -(β x(t) + c 0 y(t) + kz(t))dt + d w(t),

	(dz(t) -y(t)dt)(ζ -z(t)) 0,	(1..2)
	|ζ | Y,
	|z(t)| Y.

  Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise5 of 46 2.2. The interior Dirichlet problem in the 2d case

	Notation 2..2

: Proofs of lemmas 2..1,2..2,2..3,2..4 Proof

of Lemma 2..1. Denote γ = φ L ∞ . Notice we have

  Green formula ∀ψ ∈ H 1 0 (∆ 1 ) ∩ H 2 (∆ 1 ), LEMMA 8..1 We have η xx π 2 ∈ L 2 (∆ 1 ); η xy π 2 ∈ L 2 (∆ 1 ); η yy π 2 ∈ L 2 (∆ 1 )Proof. Denote w := v x . Deriving equation (2..28) with respect to x, we obtain equality (8..4), Λ (x, y, z)w y + αxw xyw z = -β v y -αv x + ρ 1 η x + (ρ 1 ) x η + ρ 2 (y)η xy (8..4)

						∆ 1	η k ψ yy +	∆ 1	η k y ψ y =
						∆ 1	ηψ yy +	∆ 1	η y ψ y =
	-w yy + Testing equality (8..4) with π 2 , we obtain 1 2 w xx -1 2
	1 2 ∆ 1	(w x π) 2 +	1 2 ∆ 1	(w y π) 2 = -	∆ 1	w y ππ ′ w -	∆ 1	Λ (x, y, z)w y wπ 2 -	∆ 1	αxw x wπ 2
				+	∆ 1	yw z wπ 2 -β	∆ 1	v y wπ 2 -α	∆ 1	v x wπ 2 +	∆ 1	ρ 1 η x wπ 2
				+	∆ 1	(ρ 1 ) x ηwπ 2 +	∆ 1	η xy ρ 2 (y)wπ 2

∂ ∆ 1 φ k ψ y dσ . Now, when k goes to ∞, we obtain ∀ψ ∈ H 1 0 (∆ 1 ) ∩ H 2 (∆ 1 ), ∂ ∆ 1 φ ψ y dσ .

* This research was partially supported by a grant from CEA, Commissariat à l'énergie atomique and by the National Science Foundation under grant [DMS-0705247] † This research in the paper was supported by WCU (World Class University) program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology [R31 -20007]. c Institute of Mathematics and its Applications 2011; all rights reserved.

By construction, the operator P is the transition probability associated to the Markov chain {(x( τ1,k ), y( τ1,k ), z( τ1,k ))} k 0 where τ1,0 = 0 and τk+1 := inf{t > τ1,k , |y(t)| = ȳ}; τ1,k+1 := inf{t > τk+1 , |y(t)| = ȳ1 }.

Note that τk+1 = τ • θ τ1,k and τ1,k+1 = τ1 • θ τk+1 where θ t is the shift operator. Proof. A Borel subset of Γ1 can be written as B := B -× {y 1 } ∪ B + × {-y 1 } with B + , B -are Borel subsets of (-L, L) × (-Y,Y ). We have

Consider φ + = 1 B + and φ -= 1 B -in the interior Dirichlet problem, then

We will prove the ergodic property of the operator P in the four following steps.

Doob's criterion from (3).

The operateur P is ergodic if we prove the following Doob's criterion sup λ p, p(B) < 1, ∀p, p ∈ Γ1 and ∀B.

Denote η k and ζ k the solution of the interior and exterior Dirichlet problems with φ + = 1 B +k and φ -= 1 B -k , where B +k and B -k are deduced from B k as previously. We have

Now, extracting a subsequence of p k and pk , we deduce p k → p ⋆ and pk → p⋆ in Γ1 .

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise29 of 46 Also,

From (3..22), we know that

and

Proof. We must have lim

Contradiction.

Suppose p ⋆ = (x ⋆ , ȳ1 , z ⋆ ) and set

Maximum principle for parabolic operator applied to ζ * implies p ⋆ ∈ Ξ 1 ∪ Ξ 2 . Then p ⋆ cannot be in Ξ 1 because of the Neuman condition and p ⋆ cannot be in Ξ 2 because of the boundary condition z = Y . A similar argument yields a contradiction with p ⋆ = (x ⋆ , -ȳ1 , z ⋆ ).

Conclusion.

From ergodic theory, there exists a unique probability mesure γ ⋆ = (γ - ⋆ , γ + ⋆ ) on Γ1 and K, ρ > 0 such that ∀n 0, P n φ -

It is easy to verify that v xx π, v xy π ∈ L 2 (∆ 1 ).

Denote w := v y . Deriving equation (2..28), with respect to x, we obtain equality (8..5),

Then, we test equality (8..5) with π 2 w, we obtain

Again, it is easy to verify that v yx π, v yy π ∈ L 2 (∆ 1 ).

LEMMA 8..2 We have

Proof.

Degenerate Dirichlet problems Related to the Ergodic property of the elasto-plastic oscillator excited by a filtered white noise45 of 46

We test equality (8..4) with y 2p-1 π 2q w z , we obtain

∆ 1

(w y π)(w z y p π q )(y p-2 π q-1 ) + q ∆ 1 (w y π)(w z y p π q )(y p-1 π q-1 π ′ ).

We have ∆ 1

(y p π q w z ) 2 1 4 {w 2 x (x, y,Y ) + w 2 y (x, y,Y )}y 2p-1 π 2q dxdy

∆ 1

(w y π)(w z y p π q )(y p-2 π q-1 ) + q ∆ 1 (w y π)(w z y p π q )(y p-1 π q-2 π ′ )

For p 2 and q 3, some calculations give v xz y 2 π 3 ∈ L 2 (∆ 1 ).

∆ 1

( wy π)( wz y p π q )(y p-2 π q-1 ) + q ∆ 1 ( wy π)( wz y p π q )(y p-1 π q-2 π ′ ) + ∆ 1 c 0 w(y p π q wz )y p-1 π q + ∆ 1 Λ (x, y, z)( wy π)( wz y p π q )y p-1 π q-1 + ∆ 1 αx( wx π)( wz y p π q )y p-1 π q-1 -∆ 1 v z (y p π q wz )y p-1 π q - ∆ 1 ρ 1 (v y -ηπ ′ )(y p π q wz )y p-1 π q-1 -∆ 1

(ρ 1 ) y η(y p π q wz )y p-1 π q -∆ 1

(ρ 1 ) x η(y p π q wz )y p-1 π q .

-∆ 1 ρ 2 (y)η yy π 2 (y p π q wz )y p-1 π q-2 For p 2 and q 2, some calculations gives v yz y 2 π 2 ∈ L 2 (∆ 1 ).

LEMMA 8..3 For p and q large enough, we have η xyz y p π q ∈ L 2 (∆ 1 ); η yyz y p π q ∈ L 2 (∆ 1 ); ρ(z)η xxz y p π q ∈ L 2 (∆ 1 ); ρ(z)η zz y p π q ∈ L 2 (∆ 1 ).


Emmanuel Vincent
email: emmanuel.vincent@inria.fr

Improved perceptual metrics for the evaluation of audio source separation

Keywords: audio source separation, objective evaluation, PEASS

We aim to predict the perceived quality of estimated source signals in the context of audio source separation. Recently, we proposed a set of metrics called PEASS that consist of three computation steps: decomposition of the estimation error into three components, measurement of the salience of each component via the PEMO-Q auditory-motivated measure, and combination of these saliences via a nonlinear mapping trained on subjective opinion scores. The parameters of the decomposition were shown to have little influence on the prediction performance. In this paper, we evaluate the impact of the parameters of PEMO-Q and the nonlinear mapping on the prediction performance. By selecting the optimal parameters, we improve the average correlation with mean opinion scores (MOS) from 0.738 to 0.909 in a cross-validation setting. The resulting improved metrics are used in the context of the 2011 Signal Separation Evaluation Campaign (SiSEC).

Introduction

Audio source separation is the task of extracting the signal of each sound source from a given mixture. In a number of applications such as speech enhancement for hearing aids or denoising of old music recordings, the separation performance amounts to the subjective judgment of listeners.

A popular set of performance metrics can be obtained by decomposing the estimation error into three components, namely target distortion, interference and artifacts, and measuring the salience of these components via energy ratios termed signal to distortion ratio (SDR), source image to spatial distortion ratio (ISR), signal to interference ratio (SIR) and signal to artifacts ratio (SAR) [START_REF] Vincent | Achievements and remaining challenges[END_REF]. Despite the wide use of the associated BSS Eval toolbox1 , e.g. within the annual Signal Separation Evaluation Campaign (SiSEC) [START_REF] Vincent | Achievements and remaining challenges[END_REF], these metrics are known to poorly correlate with subjective performance for certain mixtures involving e.g. low-frequency sounds or time-varying distortion. Two different routes have been taken to increase correlation: assessing the overall distortion via auditorymotivated measures such as PEAQ [START_REF]Method for objective measurements of perceived audio quality[END_REF] or PEMO-Q [START_REF] Huber | PEMO-Q-A new method for objective audio quality assessment using a model of auditory perception[END_REF], or combining energy ratios via linear or nonlinear mappings trained on subjective opinion scores [START_REF] Fox | Towards a model of perceived quality of blind audio source separation[END_REF].

In [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF], we combined these two routes via a three-step procedure consisting of 1. decomposing the estimation error into target distortion, interference and artifacts components, 2. assessing the salience of each component via PEMO-Q, 3. combining these saliences via trained nonlinear mappings.

We distributed the resulting metrics termed overall perceptual score (OPS), target-related perceptual score (TPS), interference-related perceptual score (IPS) and artifacts-related perceptual score (APS), as the version 1.0 of a toolkit called PEASS 2 . Each of the above three steps involves one or more design parameters. In [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF], we showed that the parameters of the first step have little influence on the prediction performance. In this paper, we evaluate the impact of the parameters of the two latter steps and select the optimal parameters maximizing the correlation with mean opinion scores (MOS). The resulting improved metrics are distributed as the version 2.0 of PEASS and used among others for the evaluation of the algorithms submitted to SiSEC 2011.

The structure of the rest of the paper is as follows. In Section 2, we summarize the computation of the PEASS metrics and highlight the parameters involved in each step. In Section 3, we describe the evaluation protocol and show the effect of each parameter on the prediction performance. We conclude in Section 4.

The PEASS metrics

For a given set of separated sources, we aim to predict the perceived quality of the estimated multichannel spatial image s j (t) of each source j, i.e. its contribution to all mixture channels, relatively to the true spatial image s j (t) [START_REF] Vincent | Achievements and remaining challenges[END_REF]. The PEASS metrics [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF] involve three computation steps outlined in the introduction. In the following, we summarize each step with a focus on the two latter steps, including the internal computations of PEMO-Q which were not detailed in [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF].

Distortion decomposition

In the first step, the estimation error s j (t)s j (t) is split into three components: target distortion e target j (t), interference e interf j (t) and artifacts e artif j (t) such that

s j (t) -s ij (t) = e target j (t) + e interf j (t) + e artif j (t). (1)
This is achieved by passing the signals through a bank of gammatone filters [START_REF] Hohmann | Frequency analysis and synthesis using a gammatone filterbank[END_REF], partitioning the output into overlapping time frames, performing decomposition (1) in each subband and each time frame by least-squares projection onto the subspaces spanned by delayed versions of the true source spatial image signals, and reconstructing time-domain signals by filterbank inversion. Compared with BSS Eval, this step aims to improve the handling of time-varying distortion.

PEMO-Q component saliences

In the second step, the perceptual salience of these components is assessed as

q o j = PEMO-Q( s j , s j ) (2)
q t j = PEMO-Q( s j , s j -e target j ) (3)
q i j = PEMO-Q( s j , s j -e interf j ) (4)
q a j = PEMO-Q( s j , s j -e artif j ) (5)
where PEMO-Q( x, x) ∈ [-1, 1] is the perceptual similarity measured by PEMO-Q between a test signal x and a reference signal x. Compared with BSS Eval, this step accounts for auditory masking and dynamic compression phenomena. PEMO-Q first computes internal auditory representations X i and X i of each channel i of x and x via the computational auditory model in [START_REF] Dau | A quantitative model of the "effective" signal processing in the auditory system: I. Model structure[END_REF][START_REF] Dau | Modeling auditory processing of amplitude modulation. I. Detection and masking with narrow-band carriers[END_REF]. This model comes in two versions and consists of: R1 subband decomposition via a bank of gammatone filters linearly spaced on the equivalent rectangular bandwidth (ERB) scale between f min and f max , R2 for each subband, halfwave rectification, first-order autoregressive (AR) lowpass filtering with 1 kHz cutoff, and summation with a threshold a thresh , R3 amplitude compression by five consecutive nonlinear feedback loops emphasizing rapid changes up to a maximum amplitude ratio of r max for each loop, R4 either first-order AR lowpass filtering with 8 Hz cutoff (lowpass version [START_REF] Dau | A quantitative model of the "effective" signal processing in the auditory system: I. Model structure[END_REF])

or decomposition via a bank of eight first-order AR bandpass filters with center frequencies ranging from 0 to 129 Hz (modulation version [START_REF] Dau | Modeling auditory processing of amplitude modulation. I. Detection and masking with narrow-band carriers[END_REF]).

This mimics the effect of haircells in the inner ear and modulation processing in the auditory cortex. The outputs X i and X i are either two-dimensional timefrequency representations for the lowpass version or three-dimensional timefrequency-rate representations for the modulation version. The perceptual similarity between X i and X i is then measured by [START_REF] Huber | Objective assessment of audio quality using an auditory processing model[END_REF][START_REF] Huber | PEMO-Q-A new method for objective audio quality assessment using a model of auditory perception[END_REF] S1 partial assimilation of the two representations in each time-frequency-rate bin (t, f, m) as

X itf m ← αX itf m + (1 -α) X itf m if | X itf m | < |X itf m |, S2
computation of the time-varying linear cross-correlation between X i and X i over time frames of length l corr3 , S3 computation of the time-varying root mean square (RMS) amplitude of X i over time frames of length l amp , S4 computation of the p-th percentile of the cross-correlation series weighted by the RMS amplitude.

This attempts to model the perception of global similarity based on the local similarities between the signals. Finally, the overall scalar similarity PEMO-Q( x, x) is selected as the minimum of the channel-wise similarity over all channels i.

Trained nonlinear mapping

In the third step, the saliences in (2)-( 5) are combined by [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF] M1 optional log-mapping from [-1, 1] to R via q k j ← log((1 + q k j )/(1q k j )) [START_REF] Huber | Objective assessment of audio quality using an auditory processing model[END_REF], M2 selection of one or more saliences forming a feature vector q j , M3 transformation into a scalar objective score via a feedforward neural network (NN) [START_REF] Haykin | Neural Networks[END_REF] composed of n lay layers of n neur neurons trained on subjective scores.

Compared with BSS Eval, this accounts for the different perceptual importance of each distortion component by which artifacts may be heard as more disturbing than interference for instance. Four different perceptual assessment tasks were considered in [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF]: global quality, preservation of the target source, suppression of other sources, and absence of additional artificial noise. For each task, a different feature vector was selected and a different NN was trained by minimizing the RMS error between the predicted and the actual subjective opinion scores. This resulted in four metrics called OPS, TPS, IPS and APS, respectively.

3 Effect of the design parameters

Data and evaluation procedure

Each processing block from R1 to M3 involves some design parameters listed above. In order to evaluate their effect on the prediction performance, we consider the set of 6400 subjective scores collected in [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF] using the MUltiple Stimuli with Hidden Reference and Anchor (MUSHRA) protocol [START_REF]Method for the subjective assessment of intermediate quality levels of coding systems[END_REF]. For each of 10 mixtures and each of the four tasks listed in Section 2.3, 20 subjects were asked to score 8 test sounds, including 4 real-world sounds produced by actual source separation algorithms, one hidden reference and 3 anchors. The scoring scale ranges from 0 to 100, where larger means better. The anchors are artificial sounds with low quality ensuring that the whole scale is used. For information about the variance of subjective scores and outliers, see [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF]. In order to avoid overfitting, a 200-fold cross-validation procedure is used. For each fold, the scores of 19 subjects over 9 mixtures are used for training while testing is performed on the scores of the remaining subject over the remaining mixture. The prediction accuracy is assessed via the linear correlation between the predicted scores and the MOS.

Main results

The version 1.0 of PEASS relies on the following default parameters of PEMO-Q: modulation version, f min = 235 Hz, f max = 14500 Hz, a thresh = 10 -5 , r max = +∞, α = 0.5, l corr = +∞ and l amp = +∞ 4 . The mapping consists of a 1.5-layer NN 5 without input log-mapping. For each mixture and subject, all 8 test sounds were used for training but only the 4 real-world sounds for testing. The best feature vector among 3 or 4 candidates and the best number of neurons were then selected so as to maximize accuracy over the test set [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF].

In subsequent experiments, we found this approach to be unsuitable for two reasons. First, the absence of references and anchors in the test set resulted in objective metrics that do not span the whole range from 0 to 100 and thus fail to handle better or poorer sounds than those in that set. Second, the 10 references in the training set drew the NN to better fit scores close to 100 instead of uniformly fitting all scores. In order to avoid these drawbacks, we adopt a consistent approach from now on, whereby all real-world sounds and anchors but only one reference are employed in each training and testing fold. The resulting baseline performance of version 1.0 is displayed in the top row of Table 1.

Due to the large number of design parameters, we optimize these parameters in three successive stages, from higher-level to lower-level ones. For simplicity and computational efficiency, the same parameters are used for all four metrics, except the optimal feature vector and number of neurons which depend on the metric. The resulting performance after each stage is shown in the bottom three lines of Table 1. On average, the accuracy improves from 0.738 to 0.909 when combining all three stages. This huge improvement is mostly due to the optimization of higher-level parameters in the first stage, while the two other stages have less impact. We analyze each stage in more details in the following.

Detailed impact of the mapping and the version of PEMO-Q

The top half of Table 2 describes the effect of the number of neurons n neur and the feature vectors. By simply selecting the optimal n neur (first row) and features (second row), we greatly improve the performance of the TPS and significantly improve that of the three other metrics, resulting in an average accuracy of 0.868. This is a direct consequence of the consistent training approach discussed above, but also of the fact that all possible feature vectors are tested here. Indeed, none of the optimal feature vectors belongs to the list of candidate vectors previously tested in [START_REF] Emiya | Subjective and objective quality assessment of audio source separation[END_REF].

Table 3 describes the effect of the other parameters of the nonlinear mapping and the version of PEMO-Q. The use of a 2-layer NN with input log-mapping along with the lowpass version of PEMO-Q appears optimal for all metrics except the APS and yields an optimal average accuracy of 0.882. The corresponding feature vectors are shown in the bottom line of Table 2.

Table 2. Accuracy as a function of the feature vectors and of the baseline or the optimal mapping and version of PEMO-Q, assuming optimal number of neurons and default PEMO-Q parameters.

Mapping and version Feature vector OPS TPS IPS APS Average baseline baseline 0.799 0.710 0.860 0.905 0.819 optimal [q o j q a j ] [q t j q i j ] [q o j q i j q a j ] [q o j q t j q a j ] 0.868 0.871 0.747 0.935 0.920 optimal baseline 0.901 0.801 0.865 0.834 0.850 optimal [q o j q a j ] [q t j q i j q a j ] [q o j q i j q a j ] [q t j q a j ] 0.882 0.909 0.815 0.934 0.870 After fixing the optimal mapping and version of PEMO-Q, we consider the parameters of the PEMO-Q similarity measure in a second stage. The effect of each parameter is illustrated in Figure 1. Among the tested values, the average accuracy appears to increase with p and decrease with α and l corr . This effect is particularly significant for the APS, which may be due to the nonstationary nature of artifacts calling for local rather than global correlation between the reference and the test representation. The optimal values are α = 0.25, l corr = 100 ms, l amp = 1 s and p = 0.5, yielding an average accuracy of 0.898.

Detailed impact of the PEMO-Q internal representation

After fixing the optimal parameters of the similarity measure, we consider the parameters of the internal representation in a last stage. The effect of each parameter is illustrated in Figure 2. Among the tested values, the average accuracy appears to increase with f min and r max and decrease with f max and a thresh . This effect is significant for all metrics except the IPS. The optimal parameters are the default f min , f max and r max along with a thresh = 10 -6 , yielding an average accuracy of 0.909.

Conclusion and perspectives

We examined the impact of various design parameters over the accuracy of the PEASS metrics. By adopting a consistent training approach together with unconstrained feature selection, we improved the accuracy from 0.738 to 0.868 in a cross-validation setting. By optimizing the parameters of PEMO-Q and the nonlinear mapping, we further increased it to 0.909. These results show that the mapping from the error component saliences to the metrics is crucial, while fine tuning of auditory parameters has smaller impact. The resulting improved metrics have been released as version 2.0 of PEASS and used within SiSEC 2011.

Fig. 1 .

 1 Fig.1. Accuracy as a function of one of the three parameters (lcorr, lamp), α and p given the optimal values of the two other parameters, assuming optimal mapping and version of PEMO-Q and default PEMO-Q internal representation. Note that infinite durations are equivalent to 10 s here, since the duration of the test signals is 5 s.

Fig. 2 .

 2 Fig.2. Accuracy as a function of one of the three parameters (fmin, fmax), a thresh and rmax given the optimal values of the two other parameters, assuming optimal mapping and version of PEMO-Q and PEMO-Q similarity metric.

Table 1 .

 1 Accuracy after successive parameter optimization stages.

	Optimization stage	OPS TPS IPS APS Average
	Baseline (version 1.0)	0.799 0.396 0.860 0.896 0.738
	Optimal mapping and PEMO-Q version	0.909 0.815 0.934 0.870 0.882
	Optimal PEMO-Q similarity measure	0.925 0.812 0.931 0.924 0.898
	Optimal PEMO-Q internal representation 0.922 0.864 0.926 0.925 0.909

Table 3 .

 3 Accuracy as a function of the version of PEMO-Q, the optional log-mapping and the number of NN layers n lay , assuming optimal feature vectors and numbers of neurons in each case and default PEMO-Q parameters.

	Version Log-mapping n lay OPS TPS IPS APS Average
	filterbank	no yes	1.5 0.871 0.747 0.935 0.920 0.868 2 0.877 0.759 0.912 0.924 0.868 1.5 0.884 0.784 0.928 0.916 0.878 2 0.884 0.761 0.926 0.909 0.870
	lowpass	no yes	1.5 0.886 0.794 0.940 0.869 0.872 2 0.877 0.788 0.919 0.878 0.866 1.5 0.903 0.775 0.939 0.839 0.864 2 0.909 0.815 0.934 0.870 0.882
	3.4 Detailed impact of the PEMO-Q similarity measure

http://bass-db.gforge.inria.fr/bss eval/

http://bass-db.gforge.inria.fr/peass/

A slightly distinct processing is applied in the modulation version. See[START_REF] Huber | PEMO-Q-A new method for objective audio quality assessment using a model of auditory perception[END_REF] for details.

p is irrelevant here due to the use of global correlation (lcorr = +∞).

This term refers to a 2-layer NN with linear output layer.

Acknowledgment

This work was supported by the EUREKA Eurostars i3Dmusic project funded by Oseo. We would like to thank Rainer Huber, Volker Hohmann and Valentin Emiya for discussions about this work.