Asymptotic Expansion in the half-space with cavity

# Asymptotic Expansion for Harmonic Functions in the Half-Space with a Pressurized Cavity

A. Aspri, E. Beretta, C. Mascia

Asymptotic Expansion for Harmonic Functionsin the Half-Space with a Pressurized CavityJuly 3, 2019Andrea ASPRI111Dipartimento di Matematica “G. Castelnuovo”, Sapienza – Università di Roma, P.le Aldo Moro, 2 - 00185 Roma (ITALY), aspri@mat.uniroma1.it, Elena BERETTA222Dipartimento di Matematica, Politecnico di Milano, Via Edoardo Bonardi - 20133 Milano (ITALY), elena.beretta@polimi.it, Corrado MASCIA333Dipartimento di Matematica “G. Castelnuovo”, Sapienza – Università di Roma, P.le Aldo Moro, 2 - 00185 Roma (ITALY), mascia@mat.uniroma1.it and Istituto per le Applicazioni del Calcolo, Consiglio Nazionale delle Ricerche (associated in the framework of the program “Intracellular Signalling”)

Abstract. In this paper, we address a simplified version of a problem arising from volcanology. Specifically, as reduced form of the boundary value problem for the Lamé system, we consider a Neumann problem for harmonic functions in the half-space with a cavity . Zero normal derivative is assumed at the boundary of the half-space; differently, at , the normal derivative of the function is required to be given by an external datum , corresponding to a pressure term exerted on the medium at . Under the assumption that the (pressurized) cavity is small with respect to the distance from the boundary of the half-space, we establish an asymptotic formula for the solution of the problem. Main ingredients are integral equation formulations of the harmonic solution of the Neumann problem and a spectral analysis of the integral operators involved in the problem. In the special case of a datum which describes a constant pressure at , we recover a simplified representation based on a polarization tensor.
Keywords. Asymptotic expansions; harmonic functions in the half-space; single and double layer potentials.
2010 AMS subject classifications. 35C20 (31B10, 35J25)

## 1. Introduction

The aim of the paper is to provide a detailed mathematical study of a simplified version of a problem arising from volcanology. The analysis can be considered as a blueprint, useful to address the original problem in a forthcoming research; at the same time, the result has an interest on its own, entering in the stream of the asymptotic analysis for the conductivity equation, see [1, 2, 3, 4, 10] and references therein, with the principal novelties of dealing with a case of an unbounded domain with unbounded boundary and of a different choice of boundary datum (homogeneous on the boundary of the half-space and non-homogeneous on the boundary of the cavity).

The geological problem is the detection of geometrical and physical features of magmatic reservoirs from changes within calderas. The starting evidence is that the magma exerts a force aside the surrounding crust when migrates toward the earth’s surface, producing appreciable horizontal and vertical ground displacements, which can be detected by a variety of modern techniques (see the detailed description provided in [17]). As stated in [6], the main questions that emerge when monitoring volcanoes are how to constrain the source of unrest, that is to estimate the parameters related to its depth, dimension, volume and pressure, and how to better asses hazards associated with the unrest. For this purpose, the measurements of crust deformations are a useful tool to study magmatic processes since they are sensitive to changes in the source pressure and volume.

From a modelling point of view, the displacements are described using the theory of linear elasticity, by replacing caldera with a half-space having a stress-free flat boundary, and the magma chamber by a cavity subjected to an internal pressure; in more detail, let be the displacement vector and the strain tensor, then the elastic model is defined by the linear system of equations

 (1) div(Cˆ∇v)=0in R3−∖C,

with boundary conditions

 (2) (Cˆ∇v)n⋅n=gon ∂C (Cˆ∇v)n⋅τ=0on ∂C (Cˆ∇v)n=0on R2 v(x)→0as |x|→+∞,

where is the pressurized cavity, is a vector-valued function represents the pressure, the unit outer normal vector and the tangential one; is the isotropic elasticity tensor with Lamé parameters and defined as

 C:=λI3⊗I3+2μIsym,

with the identity matrix of order and the fourth order tensor such that . The goal is to determine and geometrical features of (position, volume…) from surface measurements of the displacement field .

In applications, to handle more easily the inverse problem related to (1)–(2), the function is taken constant and equal to a vector . Additionally, from a geological point of view, sometimes it is reasonable to consider the magma chamber small compared to the distance from the boundary of the half-space, see [6, 15, 17]; by adding these hypotheses and fixing the geometry of the cavity, some efforts have been done during the last decades to find some explicit or approximate solutions to the mathematical model. The simplest approximate solution obtained by asymptotic expansions is due to McTigue when the cavity is a sphere, see [15]. The other few solutions available concern ellipsoidal shapes, dike and faults, see [6, 17].

With the ultimate goal to study in detail the elastic problem (1)–(2), establishing an asymptotic expansion in the presence of a pressurized cavity of generic shape, in this paper we analyse a simplified scalar version of this model so as to shed light and mark the path to treat the elastic case. Denoting by the half-space and its boundary, we consider the Laplace equation

 Δu=0in Rd−∖Cε

with boundary conditions

 ∂u∂n=gon ∂Cε,∂u∂xd=0on Rd−1,u(x)→0as |x|→+∞

where is the analogous of the pressurized cavity in the elastic case, with a small parameter controlling its size, is a function defined on and . Obviously, the choice to focus the attention on dimensions greater than two comes from the application we have in mind.

### Presentation of the main result

The main goals of this paper are first to analyse the well-posedness of the scalar problem, find a representation formula of the solution and then determine an asymptotic expansion with respect to the parameter of the solution. For the first two steps we do not need to assume the cavity to be small.

It is worth noticing that in terms of well-posedness the case of the half-space and, in general, of unbounded domain with unbounded boundary, is more difficult to treat compared to bounded or exterior domains since both the control of the solution decay and integrability on the boundary are needed. Indeed, it is typical to treat these problems by means of weighted Sobolev spaces, see for example [5]. In our case, we choose to use the particular symmetry of the half-space to prove the well-posedness in order to mantain a simple mathematical interpretation of the results. Therefore, we bring the problem back to an exterior domain in the whole space for which there is a vast literature on the well-posedness, see [9].

To trackle the issue of the asymptotic expansion, we consider the approach developed by Ammari and Kang based on single and double layer potentials for harmonic functions, see for example [3, 4]. This is the reason why we search an integral representation formula of the solution. To do this, we take advantage of the explicit expression of the Neumann function for the half-space

 N(x,y)=Γ(x−y)+Γ(˜x−y),

where is the fundamental solution of the Laplacian and is the symmetric point of with respect to the -plane, in order to get a representation formula containing only integral contributions on the boundary of . In detail, we find that

 u(x)=∫∂C[N(x,y)g(y)−∂∂nyN(x,y)f(y)]dσ(y),x∈Rd−∖C,

where is the trace of the solution on . From the point of view of the inverse problem we are interested in evaluating the solution on the boundary of the half-space; since , for , the integral formula becomes

 12u(x)=∫∂C[Γ(x−y)g(y)−∂∂nyΓ(x−y)f(y)]dσ(y),x∈Rd−1.

Taking a bounded Lipschitz domain containing the origin and we consider with the assumption that . Therefore, defining , with , and the single layer potential, the main result holds

###### Theorem 1.1.

For any , let such that is independent on . Then, at any , we have

 uε(x) =2εd−1Γ(x−z)∫∂Bˆg(ζ)dσ(ζ)

where denotes a quantity uniformly bounded by with which tends to infinity when goes to zero.

Finally, with the asymptotic expansion in hand, we consider the Neumann boundary datum where is a constant vector in . This particular choice has a double purpose: to reconnect this problem with the constant boundary conditions of the elastic model and to make more explicit the integrals in the asymptotic formula. The result we get contains the same polarization tensor obtained by Friedman and Vogelius in [10] for cavities in a bounded domain.

The organization of the paper is the following. Section 2 is divided into three parts: in the first one we recall some well-known results about harmonic functions and layer potentials; in the second one we examine the well-posedness of the scalar problem; in the third one we get the representation formula of the solution. In Section 3, we state and prove a spectral result crucial for the derivation of our main asymptotic expansion. In Section 4 we present and prove the theorem on the asymptotic expansion and finally we illustrate the result for the particular choice .

### Notation

All the analysis is performed in , with ; denotes the ball with centre and radius , and the area of the -dimensional unit sphere. The scalar product between two vectors is represented by and indicates the unit outward normal vector in on the boundary of some specified domain. The fundamental solution of the Laplace operator in , with , is given by with .

Points , , are decomposed as where . We denote by the half-space and by its boundary. Given a point , its reflected point with respect to the plane is represented by .

## 2. The direct problem

In this Section, we analyse the boundary value problem

 (3) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩\vspace0.2cmΔu=0in Rd−∖C\vspace0.2cm∂u∂n=gon ∂C\vspace0.2cm∂u∂xd=0on Rd−1u→0as |x|→+∞

where is the cavity, with a twofold aim: to establish well-posedness of the problem and to provide a representation formula.

### Preliminaries

The specific symmetry of the half-space permits to show well-posedness by extending the problem to an exterior domain, viz. the complementary set of a bounded set. Hence, it is useful to recall the classical results on the asymptotic behaviour of harmonic functions in exterior domains. Given a bounded domain with , if is harmonic in then is harmonic at infinity if and only if tends to as . Moreover, there exist , such that if , the following estimates hold

 (4) |v(x)|≤M|x|2−d,∣∣vxi(x)∣∣≤M|x|1−d∣∣vxjxk(x)∣∣≤M|x|−d

for . The proof can be found in [9] (see also [8, 14]).

The representation formula makes use of layer potentials whose definitions we now recall; see [3, 9, 14]. Given a bounded Lipschitz domain and a function , we introduce the integral operators

 (5) SΩφ(x) :=∫∂ΩΓ(x−y)φ(y)dσ(y), x∈Rd, DΩφ(x) :=∫∂Ω∂∂nyΓ(x−y)φ(y)dσ(y), x∈Rd∖∂Ω,

which are called, respectively, single and double layer potential relative to the set . By definition, and are well-defined and harmonic in . Further, still for , we have

 SΩφ=O(|x|2−d)andDΩφ=O(|x|1−d)

as . In addition, if has zero mean on , the decay rate of the single layer potential is increased, precisely,

 if∫∂Ωφ(x)dσ(x)=0,% thenSΩφ=O(|x|1−d)

as (such property holds for ).

Next, we introduce the compact operator

 (6) KΩφ(x):=1ωdp.v.∫∂Ω(y−x)⋅ny|x−y|dφ(y)dσ(y)

 (7) K∗Ωφ(x)=1ωdp.v.∫∂Ω(x−y)⋅nx|x−y|dφ(y)dσ(y).

Given a function defined in a neighbourhood of , set

 v(x)∣∣±:=limh→0+v(x±hnx),  x∈∂Ω.

The following relations hold, a.e. in ,

 (8) SΩφ∣∣+=SΩφ∣∣−,∂SΩφ∂nx∣∣±=(±12I+K∗Ω)φ,DΩφ∣∣±=(∓12I+KΩ)φ.

For the proof see [3, 9].

### Well-posedness

Proving existence and uniqueness results for unbounded domains with unbounded boundary is, in general, much more difficult with respect to the case of exterior domains. The main obstacle is the control of both solution decay and integrability on the boundary and usual approach is based on the use of weighted Sobolev spaces [5]. Here, we take advantage of the symmetry property of the half-space to extend the problem to the whole space and to establish well-posedness resorting in a standard Sobolev setting.

Given a bounded Lipschitz domain and the function , we define

 ˜C:={(x′,xd): (x′,−xd)∈C}

and as

 G(x):={g(x)if x∈∂Cg(~x)if x∈∂˜C.
###### Theorem 2.1.

The problem (3) has a unique solution. This solution coincides with the restriction to the half-space of the solution to

 (9) ΔU=0in Rd∖(C∪˜C),∂U∂n=Gon ∂C∪∂˜C,U→0as |x|→+∞.
###### Proof.

The proof is divided into three steps: uniqueness for (9), existence for (9), equivalence between (3) and (9).

1. For , let be such that and set . Given two solutions, and , to problem (9), the difference solves the corresponding homogeneous problem. Multiplying equation by and integrating over the domain , we infer

 0 =∫ΩRW(x)ΔW(x)dx =∫∂BR(0)W(x)∂∂nW(x)dσ(x)−∫ΩR∣∣∇W(x)∣∣2dx,

using integration by parts and boundary conditions. Exploiting the behaviour of the harmonic functions in exterior domains, as described in (4), we get

 ∣∣∫∂BR(0)W(x)∂∂nW(x)dσ(x)∣∣≤CRd−2.

Then, as , we find

 ∫Rd∖Λ∣∣∇W(x)∣∣2dx=0

which implies .

2. We represent the solution of (9) by means of single layer potential

 (10) SΛψ(x)=∫∂ΛΓ(x−y)ψ(y)dσ(y),x∈Rd∖Λ,

with function to be determined. By the properties of single layer potential, is harmonic in , = as and we have

 ∂SΛψ∂n(x)∣∣∣+=12ψ+K∗Λψ,x∈∂Λ.

We now prove that there exists a function such that

 (11) (12I+K∗Λ)ψ(x)=G(x),x∈∂Λ.

To this aim we prove that the operator is injective. Indeed, given , define . Then, from the properties of layer potentials, solves

 ΔV=0inRd∖Λ,∂∂nV=0on∂Λ,V→0as|x|→∞,

hence, from Step 1., the function is identically zero; then it follows that . Finally, observing that is a compact operator and , equation (11) admits a unique solution.

3. To show that , we have to verify that the normal derivative is null on the boundary of the half-space. This is an immediate consequence of the symmetry property

 (12) U(x′,−xd)=U(x′,xd),

which follows from the observation that, by definition of the boundary datum , the function solves (9) and the solution to such problem is unique.

As a consequence of (12), we obtain

 ∂¯u∂xd(x′,xd)=∂U∂xd(x′,xd)=−∂U∂xd(x′,−xd).

Thus the derivative of with respect to computed at any point with is zero. ∎

### Representation formula

Next, we derive an integral representation formula for the solution to problem (3). This makes use of the single and double layer potentials defined in (5) and of contributions relative to the image cavity , given by

 (13) ˜SCφ(x) :=∫∂CΓ(˜x−y)φ(y)dσ(y), x∈Rd, ˜DCφ(x) :=∫∂C∂∂nyΓ(˜x−y)φ(y)dσ(y) x∈Rd∖∂˜C.

These operators, referred to as image layer potentials, can be read as single and double layer potentials on applied to the reflection of the function with respect to coordinate.

###### Theorem 2.2.

The solution to problem (3) is such that

 (14) u(x)=SCg(x)−DCf(x)+˜SCg(x)−˜DCf(x),x∈Rd−∖C,

where are defined in (5), in (13), is the Neumann boundary condition in (3) and is the trace of on .

Using properties of layer potentials, from equation (14), we infer

 f(x)=SCg(x)−(−12I+KC)f(x)−˜DCf(x)+˜SCg(x),x∈∂C,

where is defined in (6). Thus, the trace satisfies the relation

 (12I+KC+˜DC)f=SCg+˜SCg,

which will turn out to be useful in the sequel.

Before proving Theorem 2.2, we first recall the definition of the Neumann function, see [11], that is the solution to

 ΔyN(x,y)=δx(y)in Rd−,∂N∂yd(x,y)=0on Rd−1,

where is the delta function centred in a fixed point and represents the normal derivative on the boundary of the half-space . The classical method of images provides the explicit expression

 N(x,y)=κd|x−y|d−2+κd|˜x−y|d−2.

With the function at hand, the representation formula (14) can be equivalently written as

 (15) u(x) =N(f,g)(x) :=∫∂C[N(x,y)g(y)−∂∂nyN(x,y)f(y)]dσ(y),x∈Rd−∖C,

which we now prove.

###### Proof of Theorem 2.2.

Given such that and , let

 ΩR,ε:=(Rd−∩BR(0))∖(C∪Bε(x)).

We also define as the intersection of the hemisphere with the boundary of the half-space, and with the spherical cap (see Figure 1).

Applying second Green’s identity to and in , we get

 0= ∫ΩR,ε[N(x,y)Δu(y)−u(y)ΔyN(x,y)]dy = ∫∂BhR(0)[N(x,y)∂u∂yd(y)−∂∂ydN(x,y)u(y)]dσ(y) +∫∂BbR(0)[N(x,y)∂u∂ny(y)−∂∂nyN(x,y)u(y)]dσ(y) +∫∂Bε(x)[∂∂nyN(x,y)u(y)−N(x,y)∂u∂ny(y)]dσ(y) −∫∂C[N(x,y)∂u∂ny(y)−∂∂nyN(x,y)u(y)]dσ(y) = I1+I2+I3−N(f,g)(x).

The term is zero since both the normal derivative of the function and are zero above the boundary of the half-space.

Next, taking into account the behaviour of harmonic functions in exterior domains, formulas (4), we deduce

 ∣∣∫∂BbR(0)∂∂nyN(x,y)u(y)dσ(y)∣∣ ≤CR2d−3∫∂BbR(0)dσ(y)=CRd−2, ∣∣∫∂BbR(0)N(x,y)∂u(y)∂nydσ(y)∣∣ ≤CR2d−3∫∂BbR(0)dσ(y)=CRd−2,

where denotes a generic positive constant. As , tends to zero.

Finally, we decompose as

 I3=I31−I32=∫∂Bε(x)∂∂nyN(x,y)u(y)dσ(y)−∫∂Bε(x)N(x,y)∂u∂ny(y)dσ(y).

Using the expression of and the continuity of , we derive

 I31=∫∂Bε(x)∂∂nyN(x,y)u(y)dσ(y)= u(x)∫∂Bε(x)∂∂nyN(x,y)dσ(y) +∫∂Bε(x)[u(y)−u(x)]∂∂nyN(x,y)dσ(y),

which tends to as . Moreover, we infer

 ∣∣I32∣∣ ≤Csupy∈∂Bε(x)∣∣∂u∂ny∣∣∫∂Bε(x)∣∣N(x,y)∣∣dσ(y)

Observing that both the integrals tend to zero when goes to zero because the second one has a continuous kernel while the first one behaves as , we infer that as . Putting together all the results, we obtain (15). ∎

## 3. Spectral analysis

Following the approach of Ammari and Kang, see [3, 4], in this Section, we prove the invertibility of the operator showing that, under suitable assumptions, the following inclusion holds

 σ(KC+˜DC)⊂(−1/2,1/2].

Such task is accomplished by determining the spectrum of the adjoint operator in , relying on the fact that the two spectra are conjugate.

The explicit expression of is in (7). Computing the -adjoint of is straightforward: indeed, given , we have

 ∫∂Cψ(x)˜DCφ(x)dσ(x) =∫∂Cψ(x)(1ωd∫∂C(y−˜x)⋅ny|˜x−y|dφ(y)dσ(y))dσ(x) =∫∂Cφ(y)(1ωd∫∂C(y−˜x)⋅ny|˜x−y|dψ(x)dσ(x))dσ(y)

and thus

 (16) ˜D∗Cφ(x)=1ωd∫∂C(x−˜y)⋅nx|˜y−x|dφ(y)dσ(y).

Note that the kernel of the integral operator is smooth on .

As proved in [13], for smooth domains the eigenvalues of on lie in ; the same result it is also true for Lipschitz domain (see [3, 7]). With the same approach, it can be shown that the same property holds true for .

###### Theorem 3.1.

Let be an open bounded domain with Lipschitz boundary. Then

 σ(K∗C+˜D∗C)⊂(−1/2,1/2].

For completeness, we provide here a complete proof of such fact.

Firstly, we observe that the regular operator on the boundary of the cavity can be seen as the normal derivative of an appropriate single layer potential.

###### Lemma 3.2.

Given we have that

 ˜D∗Cφ(x)=∂∂nx(S˜C˜φ(x)),  x∈∂C,

where is defined by .

###### Proof.

Using the expression (16) of and the identity

 ∇x(1(2−d)|x−y|d−2)=x−y|x−y|d,

we find that

 ˜D∗Cφ(x)=∇x(∫∂Cκdφ(y)|˜y−x|d−2dσ(y))⋅nx.

Given and as previously defined, we have

 ∫∂Cφ(y)|˜y−x|d−2dσ(y) =∫∂˜Cφ(˜z)|˜˜z−x|d−2dσ(z) =∫∂˜Cφ(˜z)|z−x|d−2dσ(z)=∫∂˜C˜φ(z)|z−x|d−2dσ(z),

which gives the conclusion. ∎

We are now ready to prove the main result of this Section.

###### Proof of Theorem (3.1).

Given , let be defined by . By the known properties of single layer potentials, we derive on

 ∂ψ∂n∣∣±=(±12I+K∗C+˜D∗C)φ

and, as a consequence,

 (17) ∂ψ∂n∣∣∣++∂ψ∂n∣∣∣−=2(K∗C+˜D∗C)φ,∂ψ∂n∣∣∣+−∂ψ∂n∣∣∣−=φ.

Taking a linear combination of the two relations in (17), we deduce

 (λI−K∗C−˜D∗C)φ =λ(∂ψ∂n∣∣∣+−∂ψ∂n∣∣∣−)−12(∂ψ∂n∣∣∣++∂ψ∂n∣∣∣−) =(λ−12)∂ψ∂n∣∣∣+−(λ+12)∂ψ∂n∣∣∣−.

If is an eigenvalue of with eigenfunction , then

 (λ−12)∂ψ∂n∣∣∣+−(λ+12)∂ψ∂n∣∣∣−=0,on∂C.

Multiplying such relation by the function and integrating over , we get

 (18) (λ−12)∫∂Cψ(x)∂ψ∂n(x)∣∣∣+dσ(x)−(λ+12)∫∂Cψ(x)∂ψ∂n(x)∣∣∣−dσ(x)=0.

Integrating by parts we have

 (19) =∫C∣∣∇ψ(x)∣∣2dx.

The first integral in (18) can be dealt with as done in the proof of Theorem 2.2. Precisely, given large , applying the Green’s formula in , we get

 =∫∂BhR(0)ψ(x)∂ψ∂xd(x)dσ(x)+∫∂BbR(0)ψ(x)∂ψ∂n(x)∣∣∣+dσ(x) −∫ΩRψ(x)Δψ(x)dx−∫ΩR∣∣∇ψ(x)∣∣2dx,

where is the intersection of the hemisphere with the half-space and is the spherical cap. The quantity is identically zero on the boundary of the half-space since the kernel of the operator is the normal derivative of the Neumann function which, by hypothesis, is null on . Moreover, is harmonic in , so we infer

Recalling the asymptotic behaviour of simple layer potential,

 ∣∣SCφ∣∣+∣∣S˜Cφ∣∣=O(|x|2−d),∣∣∇SCφ∣∣+∣∣∇S˜Cφ∣∣=O(|x|1−d)as  |x|→∞.

we obtain, for some ,

 ∣∣∣∫∂BbR(0)ψ(x)∂ψ∂n(x)∣∣∣+dσ(x)∣∣∣ ≤∫∂BbR(0)∣∣ψ(x)∣∣∣∣∂ψ∂n(x)∣∣∣+∣∣dσ(x) ≤CR2d−3∫∂BbR(0)dσ(x)=1Rd−2.

Passing to the limit , we find

 (20) ∫∂Cψ(x)∂ψ∂n∣∣∣+dσ(x)=−∫Rd−∖¯¯¯¯C∣∣∇ψ(x)∣∣2dx.

Plugging (19) and (20) into (18), we infer the identity

 (λ−12)∫Rd−∖C∣∣∇ψ(x)∣∣2dx+(λ+12)∫C∣∣∇ψ(x)∣∣2dx=0,

that is

 (A+B)λ=12(A−B)

with

 A:=∫Rd−∖C∣∣∇ψ(x)∣∣2dxandB:=∫C∣∣∇ψ(x)∣∣2dx.

The coefficient of is non-zero. On the contrary, if then in which means that , hence, from the second equation in (17), we get in .

Therefore, solving with respect to , we finally get

 (21) λ=12⋅A−BA+B∈[−12,12].

The value is not an eigenvalue for the operator . Indeed, in such a case, we would have

 A=∫Rd−∖C∣∣∇ψ(x)∣∣2dx=0,

and thus in . By definition of , we deduce that on and since is harmonic in