About an inverse problem for a free boundary compressible problem in hydrodynamic lubrication

Khalid Ait Hadi; Guy Bayada; Mohamed El Alaoui Talibi

doi:10.1515/jiip-2014-0087

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About an inverse problem for a free boundary compressible problem in hydrodynamic lubrication

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Published/Copyright: November 17, 2015

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From the journal Journal of Inverse and Ill-posed Problems Volume 24 Issue 5

Abstract

In this paper an inverse problem is considered for a non-coercive partial differential equation, issued from a mass conservation cavitation model for a slightly compressible fluid. The cavitation phenomenon and compressibility take place and are described by the Elrod model. The existence of an optimal solution is proven. Optimality conditions are derived and some numerical results are given.

Keywords: Inverse problem; Elrod–Adams model; optimality system; numerical results

MSC 2010: 35J70; 35Q35; 49K20; 65N21; 65N30

Funding statement: This work was partially supported by the French–Morocco cooperation CNRS-CNRST (SPM02/12) and the CNRST project URAC-01.

A Some properties of BV spaces

See [26].

Lemma 1

The following statements hold.

(Lower semicontinuity) Suppose fk∈BV⁢(Ω), k=1,2,…, and fk→f in L1⁢(Ω). Then
∥D⁢f∥⁢(Ω)≤lim⁡infk→+∞⁡∥D⁢fk∥⁢(Ω).
(Compactness) Let fk∈BV⁢(Ω), k=1,2,…, satisfying
supk⁡∥fk∥BV⁢(Ω)<+∞.
Then there exist a subsequence (fkj)1≤j≤+∞ and f∈BV⁢(Ω) such that
fkj→f in ⁢L1⁢(Ω).

Using the previous lemma, we can prove the following result.

Proposition 2

The following statements hold.

𝒰ad is a non-empty convex closed subset of BV⁢(Ω)∩L∞⁢(Ω).
Let (hn) be a sequence of 𝒰ad. Then there exists h∈𝒰ad such that hn→h in Lλ⁢(Ω) (for a subsequence) for all λ∈[1,+∞[.

Proof.

(i) Let (hn)n∈ℕ⊂𝒰ad. Assertion (ii) of Lemma 1 implies that there exists h∈BV⁢(Ω) such that hn→h in L1⁢(Ω) (for a subsequence). However, as 0<a≤hn≤b a.e. in Ω, it follows that 0<a≤h≤b a.e. in Ω. According to Lemma 1 (i), we have

∥D⁢h∥⁢(Ω)≤lim⁡infn→+∞⁡∥D⁢hn∥⁢(Ω)≤C∗,

therefore, h∈𝒰ad.

(ii) Thanks to the previous assertion, there exists h∈𝒰ad such that hn→h in L1⁢(Ω) (for a subsequence), in addition

∥hn-h∥Lλ⁢(Ω)λ≤(b-a)λ-1⁢∥hn-h∥L1⁢(Ω)

for all λ∈[1,+∞[, from which the result follows. ∎

B Some technical lemmas

Here, we give the general theorem:

Theorem 1

Let uϵ be a solution of the following problem:

(B.1)uϵ∈H1⁢(Ω)⁢ such that ⁢uϵ⁢(1)=0⁢, ⁢d⁢uϵd⁢x⁢(0)=0⁢ and -dd⁢x⁢((aϵ⁢(x))⁢d⁢uϵd⁢x)-gϵ⁢d⁢uϵd⁢x=Lϵ.

We suppose that there exist constants C1,C2,C3 and C4 independent of ϵ such that

(B.2)0<C1≤aϵ⁢(x)≤C2 for all ⁢x∈[0,1],

(B.3)Lϵ∈L2(]0,1[),gϵ∈L∞(]0,1[),

(B.4)∥uϵ∥L∞(]0,1[)≤C3,∥Lϵ∥L2(]0,1[)≤C4,

(B.5)gϵ≥0.

Then

(B.6)∥uϵ∥H1(]0,1[)+∥gϵ⁢d⁢uϵd⁢x∥L1(]0,1[)+∥gϵ⁢d⁢uϵd⁢x∥H-1(]0,1[)≤C.

Let us first prove the following technical lemma in which V=φ∈H1(]0,1[),φ(1)=0.

Lemma 2

We have

∫]0,1[|gϵ⁢d⁢uϵd⁢x|⁢ζ=∫]0,1[sign0⁡(d⁢uϵd⁢x)⁡(aϵ⁢d⁢uϵd⁢x⁢d⁢ζd⁢x-Lϵ⁢ζ) for all ⁢ζ∈V,

where

sign0⁡(z)={1if ⁢z>0,0if ⁢z=0,-1if ⁢z<0.

Proof.

Let sign0σ be the regularization of the function sign0 such that

sign0σ⁡(z)={1if ⁢z>σ,zσif -σ≤z≤σ,-1if ⁢z<-σ,

with σ>0. From (B.3) and the fact that uϵ belongs to V, we get that aϵ⁢d⁢uϵd⁢x belongs to H1(]0,1[). It follows that for any ζ∈V, then ζ⁢sign0σ⁡(aϵ⁢d⁢uϵd⁢x) also lies in V and can be chosen as a test function in (B.1). So we have

(B.7)∫]0,1[gϵ⁢ζ⁢sign0σ⁡(aϵ⁢d⁢uϵd⁢x)⁡d⁢uϵd⁢x=∫]0,1[sign0σ⁡(aϵ⁢d⁢uϵd⁢x)⁡(aϵ⁢d⁢uϵd⁢x⁢d⁢ζd⁢x-Lϵ⁢ζ)+∫]0,1[aϵ⁢d⁢uϵd⁢x⁢(sign0σ)′⁢(aϵ⁢d⁢uϵd⁢x)⁢ζ⁢dd⁢x⁢(aϵ⁢d⁢uϵd⁢x).

|(sign0σ)′⁢(d⁢qη⁢ϵd⁢x)⁢d⁢qη⁢ϵd⁢x|≤1,

if we set

I=∫]0,1[aϵ⁢d⁢uϵd⁢x⁢(sign0σ)′⁢(aϵ⁢d⁢uϵd⁢x)⁢ζ⁢dd⁢x⁢(aϵ⁢d⁢uϵd⁢x),

we get

I≤C2⁢∫(|d⁢uϵd⁢x|≤σ)|d⁢uϵd⁢x⁢(sign0σ)′⁢(d⁢uϵd⁢x)|⁢|ζ⁢dd⁢x⁢(aϵ⁢d⁢uϵd⁢x)|≤C2⁢∫(|d⁢uϵd⁢x|≤σ)|ζ⁢dd⁢x⁢(aϵ⁢d⁢uϵd⁢x)|.

However

and the right-hand side is equal to 0 (by an application of [32, Lemma A.4]). Then

∫]0,1[aϵ⁢d⁢uϵd⁢x⁢(sign0σ)′⁢(d⁢uϵd⁢x)⁢ζ⁢dd⁢x⁢(aϵ⁢d⁢uϵd⁢x)→0.

Passing to the limit on σ in (B.7) and using (B.5), we obtain the result of the lemma. ∎

Proof of Theorem 1.

Let us define

λϵ=gϵ⁢d⁢uϵd⁢x,γϵ=11+∥λϵ∥L1(]0,1[)+∥λϵ∥H-1(]0,1[),u~ϵ=γϵ⁢uϵ,λ~ϵ=γϵ⁢λϵ.

Then

(B.8)γϵ+∥λ~ϵ∥L1(]0,1[)+∥λ~ϵ∥H-1(]0,1[)=1,

and from (B.1) we have

(B.9)-dd⁢x⁢(aϵ⁢d⁢u~ϵd⁢x)=λ~ϵ+γϵ⁢Lϵ.

Taking u~ϵ as a test function in (B.9), we get

∫]0,1[aϵ⁢(d⁢u~ϵd⁢x)2=∫]0,1[λ~ϵ⁢u~ϵ+∫]0,1[γϵ⁢Lϵ⁢u~ϵ=γϵ⁢∫]0,1[(λ~ϵ+γϵ⁢Lϵ)⁢uϵ.

From (B.4), uϵ is bounded in L∞(]0,1[). Moreover, λ~ϵ+γϵ⁢Lϵ is bounded in L1(]0,1[). From (B.8) and (B.4), using (B.2), we get

(B.10)∥u~ϵ∥H1(]0,1[)2≤C⁢γϵ.

Now choosing ϕ∈H01(]0,1[) as a test function in (B.9), we get

∫]0,1[aϵ⁢d⁢u~ϵd⁢x⁢d⁢ϕd⁢x=∫]0,1[λ~ϵ⁢ϕ+∫]0,1[γϵ⁢Lϵ⁢ϕ,

which is rewritten as

∫]0,1[λ~ϵ⁢ϕ=∫]0,1[aϵ⁢d⁢u~ϵd⁢x⁢d⁢ϕd⁢x-∫]0,1[γϵ⁢Lϵ⁢ϕ,

so that

|∫λ~ϵϕ|≤C2∥u~ϵ∥∥H1(]0,1[)ϕ∥H01(]0,1[)+γϵ∥Lϵ∥L2(]0,1[)∥ϕ∥H01(]0,1[).

From (B.4) we get

(B.11)∥λ~ϵ∥H-1(]0,1[)≤C⁢(γϵ+∥u~ϵ∥H1(]0,1[)).

Now, let us suppose:

Assumption (H)

λϵ is not bounded in L1(]0,1[)∪H-1(]0,1[).

Then γϵ tends (for a subsequence) to 0, then u~ϵ and λ~ϵ converge to 0 respectively in H1(]0,1[) and H-1(]0,1[) and according to (B.8) we get

(B.12)∥λ~ϵ∥L1(]0,1[)→1.

We aim to compute ∥λ~ϵ∥L1(]0,1[) by way of the characterization

(B.13)∥λ~ϵ∥L1(]0,1[)=supw∈𝔹⁢(0,1)⁡∫01|λ~ϵ|⁢w⁢𝑑x,

in which 𝔹⁢(0,1) denotes the unit ball of C0⁢([0,1]). Let us choose ϕ in H1(]0,1[) such that ϕ≥1 so that for any w in 𝔹⁢(0,1),

(B.14)w≤ϕ.

Such a function ϕ cannot belong to V as it cannot be 0 for x=1. So we locally modify ϕ by introducing for any small non-negative parameter σ:

vσ⁢(z)={σif ⁢z≤1-σ,ϕ⁢(1)-σσ⁢z+ϕ⁢(1)⁢(σ-1)+σσif ⁢z≥1-σ.

Let us write from (B.14)

(B.15)∫01|λ~ϵ|⁢w≤∫01|λ~ϵ|⁢ϕ=∫01|λ~ϵ|⁢(ϕ-vσ)+∫01|λ~ϵ|⁢vσ.

We can use ζ=γϵ⁢(ϕ-vσ) as a test function in 2 so that (B.15) leads to

∫01|λ~ϵ|⁢w≤∫01sign0⁡(d⁢u~ϵd⁢x)⁡aϵ⁢d⁢u~ϵd⁢x⁢d⁢ϕd⁢x-∫01sign0⁡(d⁢u~ϵd⁢x)⁡aϵ⁢d⁢u~ϵd⁢x⁢d⁢vσd⁢x-γϵ⁢∫01sign0⁡(d⁢u~ϵd⁢x)⁡Lϵ⁢(ϕ-vσ)+∫01|λ~ϵ|⁢vσ,

which is rewritten as

(B.16)∫01|λ~ϵ|⁢w≤I1-I2-I3+I4

From (B.2) and (B.10), we have

|I1|≤C⁢γϵ⁢∥d⁢ϕd⁢x∥L2(]0,1[).

Using (B.2) and the definition of vσ,

I2≥0.

From (B.4) and the definition of vσ,

I3≥-C⁢γϵ⁢(1+∥vσ∥L2(]0,1[)).

As λ~ϵ belongs to L∞(]0,1[), it follows from (B.3), (B.4) and vσ≥0 that

I4≤∥λ~ϵ∥L∞(]0,1[⁢∥vσ∥L1(]0,1[).

As the constant C does not depend on σ,w or ϵ, we can first let σ tend to zero so that ∥vσ∥L2(]0,1[)→0 and we get

∫01|λ~ϵ|⁢w≤C⁢γϵ⁢∥d⁢ϕd⁢x∥L2(]0,1[)+C⁢γϵ.

As ϕ does not depend on w and as γϵ tends to zero, we get

∥λ~ϵ∥L1(]0,1[)≤C⁢γϵ,

which is impossible from (B.12) and Assumption (H).

So Assumption (H) is not true and this implies that there exists a constant C>0 such that

(B.17)∥λ~ϵ∥L1(]0,1[)+∥λ~ϵ∥H-1(]0,1[)≤C.

According to (B.10) and the definition of u~ϵ, we have

(B.18)∥uϵ∥H1(]0,1[)2≤Cγϵ=C⁢(1+∥λ~ϵ∥L1(]0,1[)+∥λ~ϵ∥H-1(]0,1[)).

Then from (B.17) we get

(B.19)∥uϵ∥H1(]0,1[)≤C,

so estimate (B.6) is obtained from (B.17) and (B.19). ∎

C About the computation of the state equation for the one-dimensional problem

The following procedure is a generalization of the one described in [8] for the incompressible problem. This procedure is described in detail for a particular gap h⁢(x) corresponding to the numerical example in Section 5:

h∈C0,1(]0,1[),d⁢hd⁢x<0 on ]0,E[,h(x)=hmin on [E,F],d⁢hd⁢x>0 on ]F,1].

It shows how to compute the solution of (𝒫h) by a quasi-analytical method. The main idea is that it is equivalent to solve (𝒫h) or to compute the solution of

(C.1)-h3d⁢u+d⁢x+(1+1βu)h=Θ0 in 𝒟′([0,1[),

which can be considered as a mass flow conservation equation.

Lemma 1

For a sufficiently large β, the variational inequality (C.2) has a unique solution us which belongs to H2(]0,1[,

(C.2)∫01h3⁢d⁢usd⁢x⁢d⁢(ϕ-us)d⁢x-1β⁢us⁢h⁢d⁢(ϕ-us)d⁢x≤∫01h⁢d⁢(ϕ-us)d⁢x for all ⁢ϕ∈K.

Proof.

It is obvious using the Poincaré inequality to prove the coercivity of the quadratic term and the regularity properties of the solutions of the variational inequality (C.2) (see [39, p. 148]). ∎

Let Ω+={x∈]0,1[:us(x)>0 a.e.} and Ω0={x∈]0,1[:us(x)=0 a.e.}.

Corollary 2

The set Ω0 lies in the divergent region of the gap (i.e. where d⁢hd⁢x≥0) and there exists a value Qs such that

(C.3)-h3⁢d⁢usd⁢x+(1+1β⁢us)⁢h=Qs in ⁢Ω+.

Moreover, for any free boundary xs in Ω+¯∩Ω0, we have

(C.4)us⁢(xs)=d⁢usd⁢x⁢(x⁢s)=0.

Proof.

The two first assertions are obtained by using as test functions in (C.2) respectively us+ϕ with ϕ∈𝒟⁢(Ω0),ϕ>0 and us+-α⁢ϕ with ϕ∈𝒟⁢(Ω+) for small constant α. The last assertion is a direct consequence of the C1-regularity of us. ∎

Corollary 3

The set Ω+ has only one connected component ]0,xs[ with xs in ]F,1].

Proof.

In each connected component of Ω+ there is at least a point xm where us is maximal and d⁢usd⁢x⁢(xm)=0. If this component is not identical to ]0,1[, one of its boundary at least is a free boundary xs which satisfies (C.4) so that from (C.3) we get

(1+us⁢(xm))β⁢h⁢(xm)=(1+us⁢(xs))β⁢h⁢(xs).

As us⁢(xm)>us⁢(xs)=0, this implies h⁢(xm)<h⁢(xs), and as xs cannot belong to [0,E[, this inequality prevents any other configuration of Ω+ than ]0,xs[ with xs is in ]F,1]. ∎

Lemma 4

Let ui be the solutions of the differential equations

-h3⁢d⁢uid⁢x+(1+1β⁢ui)⁢h=Qi,i=1,2,

such that Q1>Q2 and u1⁢(x*)≥u2⁢(x*) for some x*∈]0,1[. Then u1⁢(x)≥u2⁢(x) for any x<x*.

Proof.

The lemma is obtained by direct computation of the solution of the differential equation satisfied by w⁢(x)=u1-u2. ∎

Lemma 5

There exists a continuous solution Us of Ph corresponding to the data θ0⁢s=Qsh⁢(0) in which Qs is defined in (C.3).

Proof.

Let Us⁢(x)=us⁢(x) for x∈Ω+=[0,xs] and Us⁢(x)=(Qs-h⁢(x))⁢βh⁢(x) for x∈Ω0=[xs,1]. By definition, U⁢s=us+ in Ω+ and U⁢s<0 in Ω0 as a consequence of the fact that Qs=h⁢(xs) from Corollary 2 and that h⁢(x)>h⁢(xs) in Ω0. Moreover, the mass flow conservation

[-h3⁢d⁢Usd⁢x+(1+Usβ)⁢h⁢(x)]x=xs-=[-h3⁢d⁢Usd⁢x+(1+Usβ)⁢h⁢(x)]x=xs+

is satisfied on xs. ∎

Figure 5

Schematic of the gap (h(x), the solution of the variational inequality.

The computational procedure.

For any Θ0 with

(C.5)hminh⁢(0)<Θ0<Θ0⁢s

it will be proved that the cavitation area corresponding to the solution ud of (𝒫h) is made of two parts namely (0,A) and (B,1) (Figure 5), the first one is located in the convergent part of the bearing and the second one in the divergent part. The solution ud is computed by solving a differential equation:

Let B the point in ]F,1[ such that h⁢(B)=Θ0⁢h⁢(0). From (C.5) this point B is unique and located between F and xs.
A backward ordinary differential procedure iteration procedure is carried out from x=B to solve the equation
(C.6)-h3⁢d⁢udd⁢x+(1+1β⁢ud)⁢h=Θ0⁢h⁢(0).

By construction, d⁢udd⁢x⁢(B)=0. Moreover, d2⁢udd⁢x2⁢(B)>0 so that ud is non-negative in some domain (A,B) in which A is the nearest point of B such that ud⁢(A)=0.

As Θ0<Θ0⁢s, then us⁢(B)<0, and applying Lemma 4 with u1=us and u2=ud, we get ud⁢(0)≤us⁢(0)=0 so that A≥0. Moreover, d⁢udd⁢x must be non-negative in A so that A lies in {x:h⁢(x)>h⁢(B)}. The solution ud is defined by

ud⁢(x)={(Θ0⁢h⁢(0)-h⁢(x))⁢βh⁢(x)on [B,1],the solution of (C.6)on [A,B],(Θ0⁢h⁢(0)-h⁢(x))⁢βh⁢(x)on [0,A[.

By construction, h⁢(x)>Θ0⁢h⁢(0) on [0,A]∪[B,1], so that ud<0 in this area and ud>0 on [A,B[. Moreover, the mass flow equation is satisfied in all points of ]0,1[ and the boundary condition on x=0 is fulfilled. So ud is a solution of (C.1) and in turn of (𝒫h).

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Received: 2013-11-25

Revised: 2015-6-15

Accepted: 2015-9-15

Published Online: 2015-11-17

Published in Print: 2016-10-1

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Articles in the same Issue

https://doi.org/10.1515/jiip-2014-0087

Keywords for this article

Inverse problem; Elrod–Adams model; optimality system; numerical results