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Perturbation formula for the two-phase membrane problem
Advances in Difference Equations volume 2011, Article number: 19 (2011)
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.
Introduction
Let λ
± :Ω → ℝ be non-negative Lipschitz continuous functions, where Ω is a bounded open subset of ℝ n with smooth boundary. Assume further that g ∈ W
1,2(Ω)∩ L
∞(Ω) and g changes sign on ∂Ω. Let . Consider the functional
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ1_HTML.gif)
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:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ2_HTML.gif)
where χ A denotes the characteristic function of the set A, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equa_HTML.gif)
is called the free boundary. The free boundary consists of two parts:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equb_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equc_HTML.gif)
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 [2–5]. 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
< 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.
The stability of solution with respect to boundary value and coefficients is shown.
-
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Eque_HTML.gif)
-
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ4_HTML.gif)
If g 1 ≤ g 2 ≤ g 1 + ε, then u 1 ≤ u 2 ≤ u 1 + ε. In particular,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equg_HTML.gif)
Proof. First, we show that u
1 ≤ u
2. Denote ; then, for all
the following inequalities hold.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equh_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equi_HTML.gif)
These inequalities imply that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equj_HTML.gif)
which shows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equk_HTML.gif)
One can see that on the boundary of , the following holds:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equl_HTML.gif)
Note that by assumptions on g
1 and g
2, the inequality u
1 - u
2 ≤ 0 will hold on the . Thus, we have,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ5_HTML.gif)
By maximum principle, we obtain that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equm_HTML.gif)
which is impossible. Therefore, .
Let u 3 be the solution to the following problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ6_HTML.gif)
An analysis similar to the one above shows that if v = u 1 + ε - u 3, then v ≥ 0, which implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equn_HTML.gif)
□
Lemma 2.2. Assume that
. Let u be a solution to
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equo_HTML.gif)
and let u ε solve
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equp_HTML.gif)
with u = u ε = g on ∂ B 1. Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equq_HTML.gif)
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 + εv ≤ u ε in B 1, where v is the solution to Δv = 1 with zero Dirichlet boundary data in B 1. Assume that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equr_HTML.gif)
Note that v(x) ≤ 0 in B 1, and so we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equs_HTML.gif)
Then, for all , the following inequalities hold:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equt_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equu_HTML.gif)
In , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equv_HTML.gif)
Therefore, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equw_HTML.gif)
This shows that u + εv ≤ u
ε in , which is impossible. Since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equx_HTML.gif)
this implies that u
ε ≥ -Cε + u. Note that in the case when ε < 0, with the assumption one can prove that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equy_HTML.gif)
□
Remark 1. An analysis similar to Lemma 2.2 shows that if the coefficients λ ± be perturbed by ±ε, then |u ε - u| ≤ Cε.
Remark 2. The proofs of Proposition 2.1 and Lemma 2.2 show that if u and v solve the following problems, respectively:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equz_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equaa_HTML.gif)
with , then u ≥ v. In particular,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equab_HTML.gif)
Theorem 2.3. Let u
k
be a sequence of minimizer to (1.1), respectively with data g
k
and
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equac_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equad_HTML.gif)
Then,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equae_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equaf_HTML.gif)
Furthermore, by the weak continuity of the trace operator, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equag_HTML.gif)
The weak lower semi-continuity of the norm implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equah_HTML.gif)
and we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equai_HTML.gif)
Note that the level
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equaj_HTML.gif)
is not necessarily a minimum, but, by the previous discussion it satisfies the inequalities
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equak_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equal_HTML.gif)
Then, by construction
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ7_HTML.gif)
We define w k = u* + h k , and observe that w k |∂Ω = g k . Moreover, by (1.7),
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ8_HTML.gif)
Hence, it follows from the definition of c k that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equam_HTML.gif)
On the other hand, (1.8) gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equan_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ9_HTML.gif)
then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equao_HTML.gif)
and for
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equap_HTML.gif)
We first prove the following lemma:
Lemma 3.2. If
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ10_HTML.gif)
then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equaq_HTML.gif)
where δ > 0,
solves
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ11_HTML.gif)
Moreover, the same argument can be applied with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ12_HTML.gif)
Proof. Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ13_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ14_HTML.gif)
Then, by the same proof as in the first part of Lemma 2.2, one gets
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equar_HTML.gif)
where u
3 and u
4 solve Equation 1.2 with coefficients ,
, respectively. Relation (1.10) gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ15_HTML.gif)
Also, by the choice of , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ16_HTML.gif)
We will show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equas_HTML.gif)
First, note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equat_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equau_HTML.gif)
Rearranging the above terms gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equav_HTML.gif)
Multiplying by (u 4 -(u 3 + δ))+ and integrating by parts gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ17_HTML.gif)
Then,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equaw_HTML.gif)
It follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equax_HTML.gif)
Note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equay_HTML.gif)
Then, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equaz_HTML.gif)
However,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equba_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbb_HTML.gif)
and, consequently, the Sobolev embedding for
, implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbc_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbd_HTML.gif)
Now if we assume , then it will follows that |u
2 - u
1| < δ. To complete the proof, assume that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Eqube_HTML.gif)
Set . Then, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbf_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbg_HTML.gif)
By Equation 1.11, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbh_HTML.gif)
□
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,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbi_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbj_HTML.gif)
Then, for ε small enough, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ18_HTML.gif)
Remark 3. The analogue of Theorem 3.3 can be proved for the two-phase membrane problem in the following cases:
-
(1)
When all the points are regular one-phase points (cf. Theorem 3.3).
-
(2)
When all the points are two-phase points with |∇u| = 0 (branching points).
-
(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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbk_HTML.gif)
Also, for x ∈ Ω+ ∩ B
r
where r is small enough, which gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbl_HTML.gif)
Thus, is positive provided that
, which shows
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ19_HTML.gif)
Now we shall prove that the map is differentiable in the following sense:
Theorem 3.4. The mapping
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbm_HTML.gif)
defined by u = T(λ
+, λ
-) is differentiable. Furthermore, if
. Then, there exists
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbn_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ20_HTML.gif)
In Equation
1.20,
denotes the (n - 1)-dimensional Hausdorff measure.
Proof. We have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbo_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbp_HTML.gif)
Therefor,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ21_HTML.gif)
We multiply both sides of (1.21) by and integrate by parts and we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbq_HTML.gif)
Note that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbr_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbs_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbt_HTML.gif)
The Hölder inequality implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbu_HTML.gif)
Moreover, by the Poincaré inequality, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ22_HTML.gif)
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,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ23_HTML.gif)
Assume that d is the distance between supp(ϕ) and . If
, then, (since
) for ε small enough, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbv_HTML.gif)
and so . This means that, for each ϕ, one can chose ε small enough such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbw_HTML.gif)
In particular, passing to the limit in (1.23), we obtain that in the set , equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbx_HTML.gif)
holds. Similarly, in the set , one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equby_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equbz_HTML.gif)
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, αgraph. We have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ24_HTML.gif)
Let Ω
ε
be the region between and
. From (1.21) we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equca_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcb_HTML.gif)
Estimate (1.19) shows that μ is a finite measure, since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcc_HTML.gif)
We want to prove that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ25_HTML.gif)
Then, μ can be written as (see [10], Chapter I)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcd_HTML.gif)
Let d be the distance of x
0 to in direction of v, using Taylor expansion, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ26_HTML.gif)
In order to show (1.25), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equce_HTML.gif)
where is the measure of
. In addition, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcf_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcg_HTML.gif)
We deduce that, satisfies (1.20).
□
Remark 4. If for all free boundary points ∇u = 0, which means that Γ(u) = Γ'(u), then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equch_HTML.gif)
where is the unique solution of the elliptic equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equci_HTML.gif)
Remark 5. Consider the following two-phase problem in dimension one (n = 1), where λ 1, λ 2 are constants.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcj_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equck_HTML.gif)
One can obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcl_HTML.gif)
Consequently, one computes
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcm_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcn_HTML.gif)
weakly in H -1(Ω) as ε → 0. In addition
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equco_HTML.gif)
Proof. To begin with, observe that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcp_HTML.gif)
Then, for a test function one obtains
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equ27_HTML.gif)
The left-hand side of Equation 1.27 is
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcq_HTML.gif)
Let ε → 0, in (1.27); then, by the notations introduced in Remark 4, one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcr_HTML.gif)
Integrating by parts gives
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcs_HTML.gif)
In the view of Remark 4, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equct_HTML.gif)
Finally, we conclude that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcu_HTML.gif)
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1847-2011-19/MediaObjects/13662_2011_Article_13_Equcv_HTML.gif)
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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.
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Bozorgnia, F. Perturbation formula for the two-phase membrane problem. Adv Differ Equ 2011, 19 (2011). https://doi.org/10.1186/1687-1847-2011-19
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DOI: https://doi.org/10.1186/1687-1847-2011-19