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Perturbation formula for the two-phase membrane problem

Advances in Difference Equations20112011:19

https://doi.org/10.1186/1687-1847-2011-19

  • Received: 11 January 2011
  • Accepted: 29 June 2011
  • Published:

Abstract

A perturbation formula for the two-phase membrane problem is considered. We perturb the data in the right-hand side of the two-phase equation. The stability of the solution and the free boundary with respect to perturbation in the coefficients and boundary value is shown. Furthermore, continuity and differentiability of the solution with respect to the coefficients are proved.

Keywords

  • Free boundary problems
  • Two-phase membrane
  • Perturbation

Introduction

Let λ ± be non-negative Lipschitz continuous functions, where Ω is a bounded open subset of n with smooth boundary. Assume further that g W 1,2(Ω)∩ L (Ω) and g changes sign on ∂Ω. Let . Consider the functional
(1.1)
which is convex, weakly lower semi-continuous and hence attains its infimum at some point u K. The Euler-Lagrange equation corresponding to the minimizer u is given by Weiss [1] and is called the two-phase membrane problem:
(1.2)
where χ A denotes the characteristic function of the set A, and
is called the free boundary. The free boundary consists of two parts:
and

By Ω+(u) and Ω - (u) we denote the sets {x Ω: u(x) > 0} and {x Ω: u(x) < 0}, respectively. Also, Λ(u) denotes the set {x Ω: u(x) = 0}.

The regularity of the solution, the Hausdorff dimension and the regularity of the free boundary are discussed in [25]. In [5], on the basis of the monotonicity formula due to Alt, Caffarelli, and Friedman, the boundedness of the second-order derivatives D 2 u of solutions to the two-phase membrane problem is proved. Moreover, in [3], a complete characterization of the global two-phase solution satisfying a quadratic growth at a two-phase free boundary point and at infinity is given. In [4] it has been shown that if λ + and λ - are Lipschitz, then, in two dimensions, the free boundary in a neighborhood of each branch point is the union of two C 1-graphs. Also, in higher dimensions, the free boundary has finite (n - 1)-dimensional Hausdorff measure. Numerical approximation for the two-phase problem is discussed in [6].

In this article, by perturbation we mean the perturbation of the coefficients λ + and λ - and the perturbation of the boundary values g. The case of the one phase obstacle problem has been studied in [7].

For given (λ +,λ - ) C 0,1(Ω) × C 0,1(Ω), Equation 1.2 has a unique solution for
  1. 1
    < p < ∞ (see [8]). Define the map
    (1.3)
     
where u is the solution of (1.2) corresponding to the coefficients λ + and λ - . The main results in this paper are the following:
  1. 1.

    The stability of solution with respect to boundary value and coefficients is shown.

     
  2. 2.
    Let . By , we mean the solution of problem (1.2) with coefficients (λ + + εh 1) and (λ - + εh 2). If we Consider the map T : (λ +, λ - ) u, for given parameters λ + and λ + and a fixed Dirichlet condition, then the Gateaux derivative of this map is characterized in . More precisely, it is shown in Theorem 3.4 that
     
where
  1. 3.
    (Theorem 3.5) Assuming that all free boundary points are one-phase points (points such that u = 0), a stability result for the free boundary in the flavor of [7] is proved which says that
     

Were Γ± = u(x) > 0} ∩ Ω. The function δ is constructed as a solution of certain Dirichlet problem in . The vector v 1 stands for the exterior unit normal vector to .

The structure of article is organized as follows. In the next section, stability of solution with respect to boundary value and coefficients is studied. In Section 3, we prove that the map T is Lipschitz continuous (Theorem 3.1) and differentiable (Theorem 3.4).

Preliminary analysis and stability results

In this section, we state some lemmas which have been proved in the case of one-phase obstacle problem (see [9]). The following proposition shows the stability in L -norm. In what follows, we will denote by B r (x 0) the ball of radius r centered at x 0 and, for simplicity, we use the notation B r = B r (0).

Proposition 2.1. Let u i for i = 1, 2 be the solution of the following problem
(1.4)
If g 1g 2g 1 + ε, then u 1u 2u 1 + ε. In particular,
Proof. First, we show that u 1u 2. Denote ; then, for all the following inequalities hold.
and
These inequalities imply that
which shows that
One can see that on the boundary of , the following holds:
Note that by assumptions on g 1 and g 2, the inequality u 1 - u 2 ≤ 0 will hold on the . Thus, we have,
(1.5)
By maximum principle, we obtain that

which is impossible. Therefore, .

Let u 3 be the solution to the following problem:
(1.6)
An analysis similar to the one above shows that if v = u 1 + ε - u 3, then v ≥ 0, which implies

Lemma 2.2. Assume that . Let u be a solution to
and let u ε solve
with u = u ε = g on ∂ B 1. Then

Proof. Let ε > 0,; we will show that u ε u. Set D = {x B 1 : u ε (x) > u(x)}. If u ε ≤ 0, on D, then u < 0 on D and Δu = - λ - ≤ - (λ - - ε) ≤ Δu ε : On the other hand, if u ε > 0; then Δu ε = λ + + ε ≥ Δu. Therefore, Δu ε ≥ Δu and, by maximum principle, D = .

Now we claim that also u + εvu ε in B 1, where v is the solution to Δv = 1 with zero Dirichlet boundary data in B 1. Assume that
Note that v(x) ≤ 0 in B 1, and so we have
Then, for all , the following inequalities hold:
and
In , we have
Therefore, we have
This shows that u + εvu ε in , which is impossible. Since
this implies that u ε ≥ - + u. Note that in the case when ε < 0, with the assumption one can prove that

Remark 1. An analysis similar to Lemma 2.2 shows that if the coefficients λ ± be perturbed by ±ε, then |u ε - u| ≤ .

Remark 2. The proofs of Proposition 2.1 and Lemma 2.2 show that if u and v solve the following problems, respectively:
and
with , then u ≥ v. In particular,
Theorem 2.3. Let u k be a sequence of minimizer to (1.1), respectively with data g k and , such that
and
Then,

where u is the minimizer of (1.1) with data g and potential λ ±.

Proof. First, one can see that g is an admissible boundary data, i.e., g changes sign on the boundary by the strong convergence of g k in . We denote by u* the solution to minimization problem (1.1) with data g and λ ±. Consider the minimum levels c k = I k (u k ) and c* = I(u*). Also the convergence of the boundary traces g k and of the , ensures a bound on the sequence c k . Since the sequence of functionals {I k } is uniformly coercive, from the fact that I k (u k ) ≤ C, we infer a bound on the sequence ; therefore, we can assume, up to a subsequence, that
Furthermore, by the weak continuity of the trace operator, we obtain
The weak lower semi-continuity of the norm implies
and we also have
Note that the level
is not necessarily a minimum, but, by the previous discussion it satisfies the inequalities
We shall prove that c 0 = c*. Suppose, by contradiction, that c* < c 0. Consider the harmonic extensions (denoted with the same notations) on Ω of g i 's and of g and introduce
Then, by construction
(1.7)
We define w k = u* + h k , and observe that w k |∂Ω = g k . Moreover, by (1.7),
(1.8)
Hence, it follows from the definition of c k that
On the other hand, (1.8) gives

which implies that c* ≥ c 0. Finally, from the equality of the minima c 0 = c = c*, we also deduce the strong convergence of u k in H 1(Ω). □

Perturbation formula for the free boundary

In this section, we prove the continuity and differentiability of the map T. The case of one-phase obstacle problem was studied by Stojanovic [7].

Theorem 3.1. Assume λ +, λ - L p (Ω) for . The map (λ +, λ - ) u is Lipschitz continuous in the following sense. If u i for i = 1, 2 solves
(1.9)
then
and for

We first prove the following lemma:

Lemma 3.2. If
(1.10)
then
where δ > 0, solves
(1.11)
Moreover, the same argument can be applied with
(1.12)
Proof. Let
(1.13)
(1.14)
Then, by the same proof as in the first part of Lemma 2.2, one gets
where u 3 and u 4 solve Equation 1.2 with coefficients , , respectively. Relation (1.10) gives
(1.15)
Also, by the choice of , we have
(1.16)
We will show that
First, note that
Therefore,
Rearranging the above terms gives
Multiplying by (u 4 -(u 3 + δ))+ and integrating by parts gives
(1.17)
Then,
It follows that
Note that
Then, we have
However,

In the last equation, we have used (1.16).

Thus we completed the proof of Theorem 3.1.

Proof of Theorem 3.1. By elliptic regularity and Lemma 3.2, we have
and, consequently, the Sobolev embedding for , implies
Therefore,
Now if we assume , then it will follows that |u 2 - u 1| < δ. To complete the proof, assume that
Set . Then, we have
and
By Equation 1.11, we obtain

The proof of Theorem 3.4 uses the following theorem, proved by I. Blank in [9].

Theorem 3.3. (Linear Stability of the Free Boundary in the one phase case). Suppose that the free boundary is locally uniformly C 1, α regular in B 1. Let w, w ε be the solutions of the following one-phase problems, respectively,
and
Then, for ε small enough, we have
(1.18)
Remark 3. The analogue of Theorem 3.3 can be proved for the two-phase membrane problem in the following cases:
  1. (1)

    When all the points are regular one-phase points (cf. Theorem 3.3).

     
  2. (2)

    When all the points are two-phase points with |u| = 0 (branching points).

     
  3. (3)

    When |u| is uniformly bounded from below (cf. Estimate 1.19).

     

Although we could not prove this theorem for the two-phase case in general, there are grounds, however, to suggest that it holds true in this case as well.

The proof of part (3) is as follows. Suppose ε > 0, h 1 > 0, h 2 < 0 and . Then Lemma 2.2 implies that
Also, for x Ω+B r where r is small enough, which gives
Thus, is positive provided that , which shows
(1.19)

Now we shall prove that the map is differentiable in the following sense:

Theorem 3.4. The mapping
defined by u = T(λ +, λ -) is differentiable. Furthermore, if . Then, there exists , such that
where
(1.20)

In Equation 1.20, denotes the (n - 1)-dimensional Hausdorff measure.

Proof. We have
and
Therefor,
(1.21)
We multiply both sides of (1.21) by and integrate by parts and we obtain
Note that
and
Therefore,
The Hölder inequality implies
Moreover, by the Poincaré inequality, we have
(1.22)
From (1.22), the weak convergence to a limit, denoted by , follows (for a subsequence). Here, we show that satisfies (1.20). Multiply (1.21) by a test function ϕ, where ϕ has compact support in , and then divide by ε,
(1.23)
Assume that d is the distance between supp(ϕ) and . If , then, (since ) for ε small enough, we have
and so . This means that, for each ϕ, one can chose ε small enough such that
In particular, passing to the limit in (1.23), we obtain that in the set , equation
holds. Similarly, in the set , one has

Now let x 0 be a one-phase regular point for and where x ε has minimal distance to x 0.

Assumption In what follows, we assume that the estimate (1.18) in Theorem 3.3 also holds for one-phase points in our case. A straightforward calculation gives

which shows that at one-phase regular points.

To complete the proof, let us assume that . Let ν denote the normal to the free boundary at x 0, that is . Assume that B r (x 0) is a ball centered at x 0 where r is small enough. Since u(x 0) ≠ 0, then can be represented as (x', f(x')) where f is a C 1, α graph. We have
(1.24)
Let Ω ε be the region between and . From (1.21) we obtain
The term converges weakly as ε → 0, to a measure μ with support on Γ"(u). For any ball B r (x 0) with x 0 Γ"(u), set
Estimate (1.19) shows that μ is a finite measure, since
We want to prove that
(1.25)
Then, μ can be written as (see [10], Chapter I)
Let d be the distance of x 0 to in direction of v, using Taylor expansion, we get
(1.26)
In order to show (1.25), we have
where is the measure of . In addition, we have
Therefore,

We deduce that, satisfies (1.20).

Remark 4. If for all free boundary points u = 0, which means that Γ(u) = Γ'(u), then
where is the unique solution of the elliptic equation
Remark 5. Consider the following two-phase problem in dimension one (n = 1), where λ 1, λ 2 are constants.

Straightforward calculations show that if , then the set {x Ω: u(x) = 0} has a positive measure. In this setting, an interesting question is which conditions in higher dimensions will imply that the zero set has positive measure in B 1.

Example 1 Let , . Consider the equation
One can obtain
Consequently, one computes

By Weiss [1], we know that the Hausdorff dimension of Γ = ∂{u > 0} ∂{u < 0} is less than or equal to n - 1 and by Edquist et al. [2] the regularity of the free boundary is C 1. Let d Γ denote the measure ; the restriction of the (n - 1)-dimensional Hausdorff measure on the set Γ. Moreover, let v 1 be the unit normal exterior to ∂{u > 0} and v 2 be the unit normal to ∂{u < 0} exterior to {u < 0}.

Theorem 3.5. Assume that the free boundary points are one-phase points, and let δ be the same as defined in Remark 4. Then, we have
weakly in H -1(Ω) as ε → 0. In addition
Proof. To begin with, observe that
Then, for a test function one obtains
(1.27)
The left-hand side of Equation 1.27 is
Let ε → 0, in (1.27); then, by the notations introduced in Remark 4, one has
Integrating by parts gives
In the view of Remark 4, we have
Finally, we conclude that
and

Declarations

Acknowledgements

The author thanks Henrik Shahgholian for initiating this work and for useful suggestions. Moreover, the author would like to express his great sense of gratitude to the referees for carefully reading the article and coming with many helpful suggestions.

Authors’ Affiliations

(1)
Faculty of Sciences, Persian Gulf University, Bushehr, 75168, Iran

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Copyright

© Bozorgnia; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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