A NUMERICAL APPROACH MAHMOUDREZA BAZARGANZADEH
Abstract. Through this paper we shall study the existence and the uniqueness of the quadrature domains in a sub-domain ofRNby potential theory techniques. We also derive a partial differential equation for the problem and study the prop-erties of the corresponding free boundary. An application of the problem is also given. As a by-product, a numerical scheme based on the level set method is presented and it is followed by a numerical simulation.
1.
Problem Setting
Suppose that K ⊂ Rn, n≥ 2, has a smooth boundary and λ denotes the Lebesgue
measure. Let c > 0 and µ be a positive finite measure with compact support in K. In the present work, we are looking for a function u and the set
Ω :={x ∈ K : u(x) > 0},
containing supp(µ) where u solves the following free boundary problem:
(1.1) ∆u = cλ⌊Ω−µ, in K, u = 0, in K\ Ω, u≥ 0, in K, see Figure (1).
Our aim is to investigate this problem by employing the concept of the partial balayage and study the existence and uniqueness of its solution. To do this, we will establish a link between the theory of the quadrature domain, QD and (1.1) by introducing a special quadrature domain which is called quadrature domain, K-QD. We show that Ω which satisfies (1.1), is a K- K-QD.
The organization of this paper is as follows:
• In Section 2 we study the partial balayage and the properties of the
satura-tion set.
• In Section 3 we define K- QD and K- sub quadrature domain, K- SQD
and prove the existence and uniqueness of the K- SQD.
• In section 4 we extract a PDE formulation for the K- SQD. • An application of the K- QD is given in section 5.
• In the last section we construct a numerical scheme to approximate the
solution of (1.1).
Key words and phrases. Free boundary problem, Quadrature domain, Level set method.
u = 0
supp(µ) Ω :={u > 0}
K
Γ0
Figure 1. We consider Γ0 ̸= ∅ and look for the domain Ω.
2.
Introduction and Background
Free boundary problems similar to (1.1) have been studied by numerous authors for several decades. For instance, in the case of Γ0 = ∅ the problem was studied
by Shahgholian and Gustafsson in [11]. Moreover, if|∇u| = 0 on Γ0 and K =
Rnthen the problem could be similar to the one phase quadrature domain, see for
instance [8] and references therein.
In [9] the authors have investigated the existence and the geometrical properties of the classical QD by partial balayage techniques. The existence of the solution to (1.1) can not be expected unless the measure µ is concentrated enough.
In this paper we assume that Γ0 ̸= ∅ and we note that ∇u could be non-zero on
Γ0.
2.1. Notations
We denote the characteristic function of Ω by χΩand G always presents the
fun-damental solution of the Laplace operator inRN, N ≥ 2. It is verified that if Ω is
open and bounded then Gy(x) := G(x− y), as a function of x ∈ Ω, is harmonic
for all y∈ Ωcand−Gy(x) is subharmonic function for all y ∈ Ω. The Newtonian
potential of the measure µ is denoted by Uµand is defined by
Uµ(x) := (G ⋆ µ)(x) = ∫
RN
G(x− y)dµ(y), for x ∈ RN,
and it satisfies the Poisson equation−∆Uµ= µ in the distribution sense. It is easy also to see that if supp(µ)⊂ Ω then Uµis harmonic inRN\ Ω. For simplicity we write UχΩ= UΩ.
Definition 2.1. Suppose that u is superharmonic on the open set K ⊂ RN and h is a harmonic function on K and µ is a finite measure with compact support in K.
• If h ≤ u on K then h is called a harmonic minorant of u. The function h is
the greatest harmonic minorant of u if h is harmonic minorant and v≤ h whenever v is a harmonic minorant of u.
• The Green function of K is denoted by
GK(x, y) := Gy(x)− hy(x),
where hy is the greatest harmonic minorant of Gy.
• A superharmonic function of the form UKµ(x) := (GK⋆ µ)(x) =
∫
GK(x, y) dµ(y) ∀x ∈ K,
is called the potential of µ on K.
We also use the following notations in this paper,
Uµ:= G ⋆ µ Newtonian potential of µ,
GK Green function of K,
UKµ := GK⋆ µ Green Potential of µ on K,
µKc Classical balayage of µ with respect to K,
VKµ See (2.1),
BKµ :=−∆VKµ Partial balayage of µ with respect to K,
Sµ
K Saturated set for B µ
K, see (2.3),
ωKµ {VKµ< UKµ}.
2.2. Classical and Partial Balayage
Roughly speaking, the classical balayage of a measure µ with respect to a bounded domain K ⊂ RN is the cleaning process from any mass of the measure µ in K such that the potential does not change in Kc. We denote the classical balayage of
µ on K by ν = Bal(µ, Kc) or ν = µKc where we need the following conditions
for the measure ν:
{
ν = 0 in K,
Uν = Uµ, in Kc.
One way to construct the balayage of µ is solving the following Dirichlet problem where u = Uµ− Uν,
{
−∆u = µ in K, u = 0, in RN\ K,
therefore ν = µ + ∆u inRN. We note that there are other different ways to find
ν. For instance Uν is the largest function V satisfying
{
V ≤ Uµ in RN, −∆V ≤ 0 in K.
We refer to [7] and references therein for more information and details of the clas-sical balayage.
2.2.1. Partial balayage
We mentioned that in the classical balayage process, one cleans the domain K com-pletely from any mass sitting there. If we make some partial cleaning in K then we will deal with partial balayage. Roughly speaking, partial balayage of µ is a measure which in some sense is maximal along a constraint that is smaller than Lebesgue measure and gravi-equivalent to µ outside the set K. The concept of partial balayage was developed by Gustafsson and Sakai [9].
Suppose that λ denotes Lebesgue measure and let µ be a measure with compact support in K. The partial balayage of µ to λ, is denoted by BKµ and is defined by
BKµ :=−∆VKµ where VKµis the largest function V satisfying (2.1)
{
V ≤ UKµ in RN, −∆V ≤ λ in K.
Gustafsson and Sakai in [9] proved that VKµexists and V satisfying (2.1) is upper semi continuous. We also note that V +|x|2n2 is a subharmonic function. Let us recall the proof of the next lemma which may be found in [9]. See also [18] Lemma 2.1. Lemma 2.2. Let µ, λ and K be as above and VKµdenotes the largest element sat-isfies in (2.1) then (2.2) VKµ≤ UKµ inRN, BKµ =−∆VKµ≤ λ in K, VKµ= UKµ in Kc, BKµ =−∆VKµ= λ in ωKµ :={VKµ< UKµ}.
Proof. The proof of first and second part is directly obtained from (2.1). The third
one is a conclusion of balayage definition. Suppose that−∆VKµ ̸= λ in ωµK then for a x∈ ωKµ one can find a small ball Bε = Bε(x)⊂ ωKµ such that
(−∆VKµ)⌊Bε< λ⌊Bε,
where we have considered−∆VKµas a measure. If V solves {
−∆V = λ in Bε
V = VKµ on ∂Bε,
then the maximum principle states that VKµ≤ V in Bε. We also can choose ε such
that V ≤ UKµ in Bε. Now if we consider
˜
V :=
{
V in Bε
VKµ in K\ Bε
then ˜V belongs to the test class (2.1) which contradicts the maximality of VKµ. Thus we conclude that −∆Vµ K = λ in ω µ K={V µ K < U µ K}.
□ Definition 2.3. The saturated set for BKµ is defined by
(2.3) SKµ =∪{E; E ⊂ K is open and BKµ = λ in E}.
By (2.2) it is straightforward to obtain ωµK ={VKµ < UKµ} ⊆ SKµ. We note that this
inclusion may be strict, even when µ has a compact support inSKµ, see Remark 2
below. Clearly if µ increases thenSKµ and ωKµ also increase.
Proposition 2.4. There exist a measure ηK ≥ 0 with supp(ηK)⊂ ∂ω µ K∩ ∂K such that BKµ = λ Sµ K + µ (SKµ)c∩K + ηK (2.4) = λ ωµ K + µ (ωµ K)c∩K + ηK. In particular case, when K =RN then ηK = 0.
Remark 1. The result (2.4) is called structure formula of partial balayage and states
the classical balayage of a measure is always singular with respect to the Lebesgue measure. This result has been proved by Gardiner and Sjödin in [6].
By considering (2.4) and Lemma 2.2 the following lemma is obtained, see [17]. Lemma 2.5. If λ(supp(µ)) = 0 then by considering the above notations
Sµ K = (ω µ K) ◦ ∩ K. Moreover, if K is a half space then
Sµ K = ω
µ K.
Remark 2. The assumption λ(supp(µ)) = 0 in the previous lemma is crucial. For
example if for an appropriate constant c and a bounded domain D far from the origin we set µ = δ0+ cλ|Dthen it is not hard to see that
ωµK = Br(0), but SKµ = Br(0)∪ D.
Lemma 2.6. Suppose that for an open set D ⊂ K we have µ ≥ CχDfor a constant
C > 1. If supp(µ)⊂ D then supp(µ) ⊂ ωµK.
Proof. For x∈ ∂D one can find a small ball B ⊂ D such that λ≥ λ ωµ K + µ (ωµ K)c∩K + ηK = BKµ ≥ BCχD K ≥ B CχB K = χB′,
where B′ is a small ball that is contained in K. The last equality is obtained by the mean value theorem. Since x ∈ B′ ⊂ K and supp(µ) ⊂ D, it follows that
D⊂ ωKµ and therefore supp(µ)⊂ ωµK.
3.
K-quadrature domain
Let Ω be a subset of K and HK(Ω) denotes the set of all Green potentials, UKν,
where ν is a signed Radon measure with compact support in K\ Ω. It is clear that all functions in HK(Ω) are harmonic in Ω.
The following proposition is well known in potential theory and its proof may be found in [1].
Proposition 3.1. Suppose that h is harmonic in a bounded open set K such that Ω⊂⊂ K. There exists a measure ν with compact support in K \ Ω and h = UKν
in Ω.
Definition 3.2. Suppose that K ⊂ RN and µ is a measure with compact support. We say Ω⊆ K is a K-QD if (3.1) ∫ Ω h dx = ∫ h dµ, ∀h ∈ HK(Ω), and supp(µ)⊂ Ω.
Then we call (3.1) a K-quadrature identity and we write Ω∈ Q(µ, K).
It is clear that if ∂Ω does not meet the boundary of K then a K- QD coincides with the ordinary quadrature domain which has been studied extensively. For in-stance, see [13] and [14].
Example 3.3. Suppose that x0 ∈ Ω ⊂ K, α > 0. If
∫
Ωh(x)dx = αh(x0) for
every h∈ HK(Ω) then Ω is a K- QD which is so called harmonic ball, with center
at x0 and radius α, relative to K and we write Ω(x0, α), see [17]. Obviously if
Ω(x0, α) ⊂ K then it coincides with the standard ball with center x0 and radius
α. This is derived by the well known mean value property for harmonic functions
over the balls. In fact, the balls are the only domain with this property, see [4].
Remark 3. We note that if Ω = K then Q(µ, K) ={K}, it means that K is always
a K−quadrature domain by the definition. In this case we call it a trivial K- QD. Now let SK(Ω) be the set of all UKµ where µ is a signed Radon measure with
compact support in K and µ≤ 0 in Ω.
Definition 3.4. We say that a bounded open set Ω ⊂ K is a K-sub quadrature
domain, K- SQD, if supp(µ)⊂ Ω and
(3.2) ∫ Ω s dx≥ ∫ s dµ, ∀s ∈ SK(Ω).
Then we call (3.2) a K-sub quadrature identity and we write Ω∈ SQ(µ, K). Ob-viously, SQ(µ, K)⊂ Q(µ, K).
For example, let x0 ∈ Ω and α > 0. If
∫
Ωs(x)dx≥ αs(x0) for all s∈ SK(Ω)
then Ω is called subharmonic ball, see [17]. For the fundamental properties and related results of the classical quadrature domains, see for instance, [9] and [15].
3.1. An equivalent statement
Now suppose that Ω∈ Q(µ, K) and choose x ∈ K \Ω, then GK(x−·) ∈ HK(Ω).
Therefore (3.1) implies Uλ|Ω K = ∫ GK(x, y)d(λ|Ω) = ∫ Ω GK(x, y)dλ = ∫ GK(x, y)dµ(y) = UKµ.
Now to have simple notation let Uλ|Ω
K = UKΩ, and define
vµK,Ω:= UKµ − UKΩ.
From now on, whenever it is clear from the context, we will drop the indication K and µ and use vΩinstead of vK,Ωµ .
Theorem 3.5. Suppose that K ⊂ RN and µ is a measure with compact support and supp(µ)⊂ Ω ⊂ K. For vΩ, the following holds:
• Ω is a K- QD if and only if vΩ = 0 in K\ Ω.
• Ω is a K- SQD if and only if vΩ≥ 0 in K and vΩ = 0 in K\ Ω.
Proof. For every h ∈ HK(Ω), Proposition 3.1 implies that there is a measure ν
with compact support such that supp(ν) ⊂ K \ Ω and h = UKν in Ω. If vΩ = 0
then we have by Fubini's theorem ∫ Ω h dλ− ∫ h dµ = ∫ Ω UKν dλ− ∫ UKν dµ = ∫ UKΩdν− ∫ UKµ dν = ∫ vΩdν = 0.
This shows that Ω is a K- QD. Similarly if x ∈ K, then −GK(x− ·) ∈ SK(Ω)
and by similar discussion we get vΩ≥ 0 in K and vΩ= 0 in Ωc. The proof of the
second part is similar to the first one. □
3.2. Existence and Uniqueness
In general, if one does not consider appropriate conditions on the measure then the existence of the corresponding QD is not necessarily achieved. Sakai has proved that if (3.3) sup r>0 µ(Br(x)) λ(Br(x)) ≥ 2 N,
then one obtains a unique SQD, see [16]. We use a stronger version of Sakai's con-dition (3.3) which have recently stated by Gardiner and Sjödin to get the existence of the two phase QDs, see [5].
According to (2.1) if we set Aµ:={u = UKµ − V > 0 and ∆u ≤ λ − µ}, then
Now if µ1 and µ2are two positive measures with disjoint compact supports in K
and µ1≤ µ2 then
(3.4) Aµ2 ⊆ Aµ1.
Lemma 3.6. If µ is a measure with compact support in K and satisfies
(3.5) lim sup
r→0
µ(Br(x))
λ(Br(x)) ≥ 2
n, for all x∈ supp(µ),
then supp(µ)⊂ ωµK.
Proof. Let x0 ∈ supp(µ) \ ωKµ and define uµ := UKµ − VKµ then uµ(x0) = 0
because of the definition of ωµK. We can choose a small r0such that ωµK˜ exists and
ωKµ˜ ⊂ K where ˜µ := µ⌊Br0(x0). By (3.5) the Sakai's condition is fulfilled hence
uµ˜(x0) > 0. On the other hand ˜µ(x0)≤ µ(x0) and (3.4) gives that
0 < uµ(x˜ 0)≤ uµ(x
0) = 0,
which is a contradiction. This proves that supp(µ)⊂ ωKµ. □ Proposition 3.7. Suppose that K⊂ RN and µ is a measure with compact support, supp(µ) ⊂ K. Then SQ(µ, K) ̸= ∅ provided µ is concentrated enough in the sense of Lemma 2.6 or Lemma 3.6.
Proof. We claim that Ω := ωµK ={VKµ< UKµ} is a K- SQD. According to
Theo-rem 3.5 it is enough to show that uµΩ ≥ 0 in K and uµΩ = 0 in K\ Ω. By Lemma 2.6 or Lemma 3.6 we get that supp(µ)⊂ ωKµ. The definition of the partial balayage states that
0 < UKµ − VKµ< UKµ − UKΩ = uµΩ, on K,
and (2.2) implies that uµΩ = 0 in K\ Ω. □
Remark 4. We also can show that for every Ω∈ SQ(µ, K) we have ωKµ ⊆ Ω ⊆ Sµ
K, see [17].
Proposition 3.8. Suppose that Ω1∈ Q(µ, K), Ω2∈ SQ(µ, K). Then
∂Ω1∩ K ⊂ Ω2.
Proof. According to Theorem 3.5 we have UKµ ≥ UΩ2 K in K and U µ K= U Ω1 K in K\ Ω1, then for v := UΩ1 K − U Ω2
K the following is derived,
(3.6)
{
v≥ 0 in K\ Ω1, (especially on ∂Ω1),
∆v = χΩ2 − χΩ1 ≤ 0 in K \ Ω1,
and consequently v ≥ 0 in K. Now assume that x ∈ (∂Ω1 \ Ω2)∩ K. Then
v is a nonzero superharmonic function in a small neighborhood of x, say Br(x).
Moreover, v vanishes on Br(x)∩ ∂Ω1 then by maximum principle it should be
The above discussion implies that the K- SQD is unique up to a Lebesgue null set.
Corollary 3.9. If Ω1, Ω2 ∈ SQ(µ, K) then Ω1 ≡ Ω2up to a Lebesgue null set.
Proposition 3.10. Suppose that Ω1 ∈ Q(µ, K), Ω2 ∈ SQ(µ, K). If K \ Ω2 is
connected and Ω1∪ Ω2 ̸= K then
Ω1 ⊂ SKµ.
In particular, if K is a half space then the only K−quadrature domain is ωµK. Proof. According to Lemma 2.5 and Remark 4, one needs to prove that Ω1⊆ Ω2.
By connectedness of K \ Ω2, for x ∈ Ω1\ Ω2 either (∂Ω1 \ Ω2)∩ K ̸= ∅ that
is impossible due to Proposition 3.8 or K\ Ω2 ⊂ Ω1which is not the case by the
assumption. The second part is clear by Lemma 2.5. □
4.
PDE formulation
In this section we extract a PDE formulation for the K- SQD. By Lemma 2.2 we can show that
u = uµ= UKµ − VKµ,
solves the following free boundary problem
(4.1) −∆u ≥ µ − λ in K, −∆u = µ − λ in ωµ K ={u > 0}, u≥ 0 inRN, u = 0 in (ωKµ)c,
which is understood in the distribution sense.
Let K⊂ RNbe an open and bounded set and µ be a finite measure with compact support in K. Now suppose that a function u and an open set Ω ={u > 0} ⊂ K solve the problem (1.1), then we have the following lemma. The proof is a direct consequence of (4.1) and Theorem 3.5.
Lemma 4.1. If u and Ω ={u > 0} satisfy the problem (1.1) then Ω ∈ SQ(µ, K). To be more precise, we can prove that
Ω∈ SQ(µ, K) ≡ UKΩ = UKµ in K∩ Ωc, ∇UΩ K =∇U µ K in K∩ Ωc, UKΩ ≥ UKµ in Ω. ≡ ∆u = χΩ− µ in K, u = 0 on ∂Ω∪ ∂K, ∇u = 0 on ∂Ω∩ K, u > 0 in Ω.
Remark 5. In general, without any assumption on µ and∇u one can not expect the
existence of the solution. More precisely, we can write ∫ Ω ∆u dx = ∫ Ω (1− µ) dx, and by Green formula it turns to
∫ Γ0 ∂u ∂ν ds = ∫ Ω (1− µ) dx.
In other words, if the measure is not concentrated enough, i.e., 1− µ is very small or∇u is too large then we lose the existence. For more details, we refer the reader to [9].
Example 4.2. Let µ = cδa+ where a+ ∈ K = Rn+ := {x ∈ Rn : x1 > 0} and
c > 0. Suppose that (u+, Ω+) is the corresponding solution of the problem (1.1). Now one can oddly reflect Ω+ with respect to Π := {x ∈ Rn : x1 = 0}. If
we denote the reflected domain by Ω− and the reflected function by u− then the problem turns to the following two phase free boundary problem
(4.2) { ∆u = χ Ω+ − δa+− (χΩ− − δa−) in R n, u =|∇u| = 0 on ∂Ω\ Π,
where a−is the reflected point of a+with respect to Π and Ω = Ω+∪ Ω−. For simplicity, let n = 1 then for some constants A1, A2, B1, B2we have the following
form for u+and u−
u+(x) = x
2
2 + A1x + B1 and u
−(x) =−x2
2 + A2x + B2.
Consider that Ω+ touches Π at points α and β. Now taking into account the fol-lowing conditions
{
u+(α) = u−(α) = u+(β) = u−(β) = 0,
∇u+(α) =∇u−(α) and∇u+(β) =∇u−(β),
one can seek for the unknowns A1, A2, B1, B2and α, β and consequently ωµK =
Ω+={u+> 0} is determined.
5.
A moving boundary problem in rock mechanics
The author owes much of the material in this section to B. Gustafsson [10]. We establish a link between a moving boundary problem in rock mechanics and K-SQDs. Consider K as a porous medium (rock) and supp(µ), as the source of a fluid (water) that is located in K, see Figure 1. Concerning the continuity equation we can write
(5.1) ∂θ
∂t =∇ · (θV) + µ,
• θ : is the fluid weight per unit rock volume, i.e, degree of saturation. • V : is the velocity of water.
Moreover, by Darcy's law we have
(5.2) θV =−κ∇p.
Here, p is the pressure of the water and κ > 0 is the permeability of the rock. Then we can remove the velocity from (5.1) and (5.2) and obtain
(5.3) ∂θ
∂t =∇ · (κ∇p) + µ,
where both p and θ are unknown. In [10] and references therein, one is able to find a relation between these two unknowns. Briefly, this relation can be described as follows, (5.4) θ∈ f(p) := ∅ for p < 0, {θ ∈ R : 0 ≤ θ ≤ θ0} for p = 0, {θ0} for p > 0,
where θ0is the value of θ corresponding to the complete saturation.
This expresses that at any time the rock region consists of two parts, the dry and the saturated parts. Let us define the saturated part
Ωt={x ∈ K : θ(x, t) = θ0},
and consequently the dry part is K\ Ωt. In other words, if x∈ Ωtthen there exists
a neighborhood of x such that θ = θ0. We also assume that Ωtis an open set in K.
In addition, it is clear that p≥ 0 in Ωtand p = 0 in K\ Ωt. In fact, the saturated
part is unknown and it depends on the time, so we deal with a moving boundary problem.
From now on, we consider (5.3) and (5.4) in space-time region K× [0, T ] with
T > 0. In order to have a unique solution, we need appropriate boundary
condi-tions. Suppose that ∂K is split into two disjoint parts Γ0, Γ1 := ∂K\ Γ0 with the
following boundary conditions (5.5)
{
p = 0 on Γ0, ∂p
∂n = 0 on Γ1.
The second boundary condition states that Γ1is that part of ∂K which is
imperme-able, see Figure 1. Set
u(x, t) :=
∫ t 0
p(x, ζ)dζ
and integrate (5.3) with respect to t by considering θ(x, 0) = 0. Then we obtain
The new unknown function u is called Baiocchi transformation. The boundary conditions (5.5) can be rewrite
(5.7) u = 0 on Γ0, ∂u ∂n = 0 on Γ1, u≥ 0 for all t∈ [0, T ]. In addition, if p > 0 then u > 0. From (5.6) and
(5.8) θ∈ f(p = ∂u
∂t),
we can eliminate θ and get
(5.9) ∇ · (κ∇u) + tµ ∈ f(∂u
∂t).
Lemma 5.1. With the above notations we have θ∈ f(u).
Proof. Suppose that (x0, t0)∈ K ×(0, T ) and u(x0, t0) > 0, then by the definition
of Baiocchi transformation we have p(x0, τ ) > 0 for some τ ∈ (0, t0). Therefore
p is positive in that neighborhood of (x0, τ ) and consequently θ = θ0 in a
neigh-borhood of (x0, τ ). On the other hand θ is an increasing function of t, i.e., ∂θ∂t ≥ 0
hence θ = θ0a.e. in a neighborhood of (x0, t0). It means that u > 0 implies θ = θ0
a.e. and one obtains θ∈ f(u). □
Remark 6. By using θ∈ f(u) instead of (5.8) we will remove time derivative and
get
(5.10) ∇ · (κ∇u) + tµ ∈ f(u),
which is an elliptic problem with time as a parameter.
Remark 7. (Hele-Shaw problem in the half plane) If K =Rn+then the problem is similar to the Hele-Shaw problem in the half plane. This problem could be used to model a free boundary problem arising from a fluid injection between two parallel plates when the fluid is constrained inRn+ and it is allowed to leave through the boundary Γ0. For the Hele-Shaw problem in the half plane see [2] and [3].
Here we will prove that the saturated part of the rock is a K− quadrature domain. Theorem 5.2. Suppose that K is bounded (as a rock) and for simplicity let θ0 =
1, κ = 1. Moreover, let µ be a positive finite measure µ, such that supp(µ)⊂ K.
If Ω is the saturated part of the rock at the time t0, then Ω is a K- SQD.
Proof. In the saturated part we know that the pressure p and u are positive then due
to (5.10) one obtains ∆u = 1− t0µ in Ω where Ω is the saturated part. Putting all
the boundary conditions (5.7) and the governing equation together, we obtain
(5.11) ∆u = χΩ− t0µ in K, ∂u ∂n = 0 on Γ1, u≥ 0 in K, u = 0 in K\ Ω.
It is clear that (5.11) is a special case of (4.1) hence the solution of this problem is
a K− quadrature domain. □
6.
Numerical Simulations
Roughly speaking, one can tackle with the problem by two numerical approaches. First, a variational formulation is considered and the corresponding cost functional is minimized which requires the calculus of shape optimization. Second, a sequence of elliptic problems, defined on a sequence of converging domains, may be consid-ered. Then by using an appropriate updating process in every iteration, one tracks the interface. In this paper, we apply the second scenario which is based on the level set method, see for instance [12].
First let us recall the level set formulation. The key point of this approach is to represent domains and their boundaries as level sets of a continuous function ϕ without considering boundary parametrization. For tracking the motion of an open set Ω(t), t∈ R+in the set K, one can define a function ϕt : K× R+ → R such
that
Ω(t) ={ϕ(x, t) < 0 : x ∈ K}, and the zero level set will be represented by
Γt= ∂Ω(t) ={ϕ(x, t) = 0 : x ∈ K}.
If the evolution of the shape is determined by a velocity field, i.e.,
dx
dt(t) = V(x(t), t),
then the corresponding level set function ϕ is determined by the first-order Hamilton-Jacobi equation
∂ϕ
∂t + V· ∇ϕ = 0 in K × R
+.
Now let F = V· n where n is the outward normal vector on Γ and n = ∇ϕ
|∇ϕ|.
Therefore we are able to compute the level set function by
(6.1) ∂ϕ
∂t + F|∇ϕ| = 0 in K × R
+.
Note that we have to extend the velocity field in the whole K and solve the equation. If ϕ is considered as the sign distance function then, the level set equation (6.1) turns to
(6.2) ∂ϕ
∂t + F = 0 in K× R
Consider Ωtis an approximation of Ω such that ∂Ωt∩ ∂K ̸= ∅. Now let utbe the
time parameterized solution of ∆ut= χΩt − µ in K, ut= 0 on ∂K, |∇ut| = 0 on ∂Ωt∩ K.
Let nt be the outward normal vector on ∂Ωt. We would like to have utnt as a
velocity field which moves ∂Ωt∩ K in the normal direction and decreases when
Ωt approaches to Ω, the solution of problem (1.1). In other words, the updated
domain Ωt+1is found by moving the boundary of Ωtin the normal direction such
that utvanishes there. Let xt∈ ∂Ωtthen for the corresponding point xt+1∈ Ωt+1
we get
∂Ωt+1= ∂Ωt+ utnt.
Hence by the level set formulation (6.2), the evolution of Ωtcould be obtained by
(6.3) ∂ϕ
∂t + ut= 0 on ∂Ωt.
To proceed the level set method we have to extend the equation (6.3) to whole K. For this purpose, we rewrite the level set formulation in K
(6.4) ∂ϕ
∂t + ut= 0 in K.
The numerical algorithm is summarized as follows. Choose a tolerance, T OL << 1.
• Let Ω0 ⊂ K be an initial guess such that supp(µ) ⊂ Ω0.
• Find utby solving (6.5) ∆ut= χΩt − µ in K, ut= 0 on ∂K, ut≥ 0 in K.
• Solve (6.4) to get ϕ and consequently obtain Ωt+1.
• Solve the PDE formulation (6.5) for Ωt+1and find ut+1.
• If sup |ut+1(x)| ≤ T OL for x ∈ ∂Ωt+1∩ ∂K, then stop otherwise iterate
the previous steps.
Example 6.1. We simulate a numerical example of a K- SQD where K = B1 is
a unit ball and µ = 30χB2. Here B2 is a ball of radius 0.2, centered at (0,−
1 2).
We consider P as the initial guess, see Figure 2 which shows the initial guess and the first iteration, Ω1. Figures 3 and 4 depict the numerical approximation of the
corresponding B1- SQD and Figure 5 states the value of u on the free boundary.
Acknowledgments. The author thanks Prof. H. Shahgholian to suggest this problem and also Prof. T. Sjödin for carefully reading this article and for their valuable comments.
P
Ω1
Figure 2. This figure shows the support of the measure, the inside ball, and the initial guess, P and the first numerical approximation, i.e., Ω1.
Figure 4. The surface of the numerical approximation.
Figure 5. The value of function corresponding to the numerical approximation illustrated in Figure 3.
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Department of Mathematics, Uppsala University, Uppsala, Sweden E-mail address: Reza@math.uu.se
