QUANTITATIVE JACOBIAN DETERMINANT BOUNDS FOR THE CONDUCTIVITY EQUATION IN HIGH CONTRAST COMPOSITE MEDIA

. We consider the conductivity equation in a bounded domain in R d with d ≥ 3. In this study, the medium corresponds to a very contrasted two phase homogeneous and isotropic material, consisting of a unit matrix phase, and an inclusion with high conductivity. The geometry of the inclusion phase is so that the resulting Jacobian determinant of the gradients of solutions DU takes both positive and negatives values. In this work, we construct a class of inclusions Q and boundary conditions φ such that the determinant of the solution of the boundary value problem satisﬁes this sign-changing constraint. We provide lower bounds for the measure of the sets where the Jacobian determinant is greater than a positive constant (or lower than a negative constant). Diﬀerent sign changing structures where introduced in [9], where the existence of such media was ﬁrst established. The quantitative estimates provided here are new.


YVES CAPDEBOSCQ AND SHAUN CHEN YANG ONG
It is known [16] that in this setting J is piecewise continuous, so in particular measurable, and the sets appearing in (3) are indeed measurable. Unlike the two dimensional case where such inclusions do not exist [4], when d ≥ 3 examples have been provided in [9]. The purpose of our study is to establish lower bounds for the measures of J − = J −1 ((−∞, 0)) and J + = J −1 ((0, ∞)) , by exhibiting open subsets included in both sets. The bounds we derive depend on some geometrical characteristics of Ω and Q.
The open domain Ω is either convex and polygonal, or convex with a C 2 boundary.
• The inclusion Q ⊂ Ω is closed with non-empty interior and Q is connected.
• The (hyper)surface ∂Q is C 2 . Not all boundary conditions lead to sign changing properties: a constant boundary condition leads to a null Jacobian determinant everywhere, for example. The following definition details sufficient assumptions, as we will see. Definition 1.1. Given α > 1, we say that φ ∈ C 1,σ R d ; R d for some σ ∈ (0, 1) is an α-admissible boundary condition if for all x in ∂Ω, and every i in {1, . . . , d} there holds e i · φ (x) = −e i · φ (x − 2 (x · e i ) e i ) , e i · φ (x) = e i · φ (x − 2 (x · e j ) e j ) for any j = i, These assumptions are sufficient to show that the Jacobian determinant changes sign in Ω, see Proposition 2. Our result is the following.
Theorem 1.2. Suppose that Assumption 1 holds. Then there exists two positive constants τ and C depending on Ω and Q only such that for any α > 0, any k ≥ C max 1, 1 α d τ and any α-admissible φ in C 1,σ R d the solution of (1) satisfies where the points p i ∈ Ω \ Q, i = 1, . . . , 2d are all located on different coordinate half-axes.
For a more detailed version of this result see Proposition 8. The motivation for this investigation stems from several applications in the field of the so-called hybrid imaging inverse problems. In that context, a non-vanishing Jacobian determinant is key in the assessment of the parameters of the PDE from the knowledge of its solution. Sign changing structures typically renders such approach unsuccessful. Whether the Jacobian determinant vanishes is also an issue in numerical homogenisation: in [8] a Cordes type conditions is imposed, namely for some C Ω small enough, in order to avoid sign changes in the Jacobian determinant. Establishing bounds for (det DU ) −1 (−∞, −τ 0 ) and (det DU ) −1 (τ 0 , ∞) is a first step towards an optimisation scheme aimed at constructing topologically simple structures in which the sign of the Jacobian determinant can be prescribed in most of the domain.
The ambient space dimension plays an important role in this problem. In two dimensions, the Radó-Choquet-Kneser Theorem, and its extension to γ-harmonic maps [3,2,4] states that given a bi-continuous map from the boundary ∂Ω of a two dimensional domain Ω onto the boundary ∂Σ of a convex two dimensional domain Σ generates a harmonic extension of that map (which is defined as the solution to (1)) within the domain which is a diffeomorphism from Ω to Σ. In particular, for any γ ∈ L ∞ (Ω) bounded below by a positive constant, the logarithm of the Jacobian determinant belongs to the space of functions of bounded mean oscillation log |det DU | ∈ BM O Ω , therefore either J + or J − is null. The situation is completely different in higher dimensions (d ≥ 3). Counter-examples to the positivity of the Jacobian determinant for harmonic extensions have been known for several decades see [15,19,17]. Allowing for two phase materials, one can establish [10] that no choice of boundary data φ ∈ H 1 2 ∂Ω) 3 whose harmonic extension satisfies a strong positive determinant constraint could enforce a local positive determinant constraint for all two phase isotropic conductivity matrices. Considering a domain Ω Ω, ρ > 0, x 0 ∈ Ω so that B ρ (x 0 ) ⊂ Ω and the harmonic system and denoting A(x 0 , λ, ρ) the set containing all boundary data whose harmonic extension satisfy the strong positive Jacobian determinant bound in the ball of radius ρ centred at the point x 0 the set A(x 0 , λ, ρ) is non-empty for λ sufficiently small as the identity map I d is in A(x 0 , λ, ρ). Considering a two phase isotropic [0, 1] d periodic conductivity γ, which satisfies (3) and some symmetry properties, the following result is established. [10] and [1]). Given ρ > 0, x 0 ∈ Ω , such that B ρ (x 0 ) ⊂ Ω , and λ > 0, then there exist n > 0, depending on ρ, Ω, Ω and λ only, a universal constant τ > 0 and two open subsets B + and B − of B ρ (x 0 ) such that and for any φ in A(x 0 , ρ, λ), where U n is the γ(n·)-harmonic extension of φ.
The proof of Theorem 1.3 relied on a key result in [9] which showed that for a specific inclusion Q and γ = 1 + (k − 1)χ Q I d defined on the three dimensional unit cube Y = [0, 1] 3 , the periodic corrector matrix P = Dζ associated with the two phase homogenisation problem which is the solution ζ to for some open neighbourhoods Y + and Y − . While τ is established to be positive, its size is not quantified. Theorem 1.2 provides a lower bound to the measure of Y + , Y − and to τ in estimates (5) in the case of Dirichlet boundary conditions. Our paper is devoted to the proof of Theorem 1.2. In Section 2 we show that Assumption 1 is sufficient to guarantee that both J + and J − are non empty provided k is large enough. We adapt a method introduced in [9], and show that it is sufficient to focus on the study of the determinant on particular lines. The next two sections are devoted to quantitative estimates. In Section 3 we turn to the case k = ∞, and establish a counterpart to Theorem 1.2 in that case. In Section 4 we use layer potential estimates to establish our main result by means of an explicit convergence rate of the Jacobian determinant when the contrast k tends to infinity.
2. Symmetrical inclusions with sign-changing Jacobian determinants. The first step, adapted from [9], is to derive a simplified form for the determinant.
Lemma 2.1. Let γ = 1 + (k − 1)χ Q I d be as above, and suppose Assumption 1 holds and that φ is α-admissible. If U satisfies the Dirichlet problem −div(γDU ) = 0 in Ω, then the Jacobian matrix DU is diagonal along the line L := Ω ∩ R × {0} d−1 , and the Jacobian determinant is given by Proof. We first show that the solution U i for i = 1, . . . , d corresponding to the scalar equation is odd with respect to x i and even with respect to We write H 1 φi (Ω) as the affine space φ i + H 1 0 (Ω). Notice that from the symmetry of Ω and φ, V i ∈ H 1 φi (Ω). Moreover, Using the symmetry of γ and denoting T i : Ω → Ω the change of variables T i (x i , z) = (−x i , z), a straightforward computation (and a change of variable) yields Considering the variational formulation for U i , we note that U i is the unique minimizer of the Dirichlet energy, namely U i satisfies Thus, by the uniqueness of minimisers, we deduce that The evenness with respect to x j is established by similar arguments.
In particular, we may apply the same argument to the function ∈ Ω, a straightforward computation shows that for i = j, and for i = j, altogether, this implies that for y = (x 1, − x 2 , . . . , −x d ), In particular, for s = (x 1 , 0), Definition 2.2. Given k > 0, we denote by U k ∈ H 1 Ω; R d the solution to the Dirichlet problem We referred to this solution as U previously; the k dependence is underlined in order to compare U k to its perfectly conducting limit. We write and where U ∞ is defined in (12).
Let us first establish the limit problem for (1) as k → ∞. We provide its proof for the reader's convenience. A similar proof can be found in [7]. This can also be deduced from the layer potential approach followed in Section 4. Lemma 2.3. Suppose Assumption 1 holds and that φ is α-admissible. Let U k denote the solution to (9) with conductivity Furthermore, U ∞ = 0 on Q.
Proof. Consider the variational formulation for U k given by  and, on the other hand, as M ⊂ H 1 φ (Ω; R d ). From the weak lower semi-continuity of norms, we deduce hence, the weak limit U ∞ minimizes the functional I in M. Finally, by (13), we have Combining this fact with (14), the sequence of energies I k (U k ) converge to I(U ∞ ), hence by the uniform convexity of L 2 norms, the sequence DU k converges strongly in L 2 (Ω) to the limit DU ∞ where U ∞ satisfies the equation (12) as claimed.
Finally, for any i ∈ {1, . . . , d}, since (U k ) i is odd with respect to the variable x i , as shown in Lemma 2.1, and U k → U ∞ in H 1 (Ω), the limit (U ∞ ) i is also odd in x i . Thus (U ∞ ) i ({x i = 0} ∩ Ω) = 0. Since Q is a connected component of Ω, we infer that (U ∞ ) i must be identically 0 on the whole of Q.
We now proceed to show that Assumption 1 and α-admissible boundary conditions (1.1) are sufficient to obtain sign changing structures.

Proposition 1. Under the assumptions of Lemma 2.3, there holds
Proof. In this proof, we write U ∞ = (U i ) 1≤i≤d . Thanks to Lemma 2.1, and because of the continuity of DU , it is sufficient to focus on the line Thus, thanks to weak maximum principle, this implies that U i > 0 in the interior of D i . Observe that since U 1 is odd with respect to x 1 and even with respect to the other variables, we have U 1 (0) = 0. By assumption, B(0, ) ∩ Q = ∅ and Re 1 ∩ Q = ∅, therefore there exists s 1 > 0 such that x * = s 1 e 1 ∈ R + × {0} d−1 ∩ ∂Q, and U 1 (se 1 ) > 0 for any 0 < s < s 1 . In particular, we have F : s → U 1 (se 1 ) satisfies Since Q is a C 2 inclusion, elliptic regularity asserts that the solution U is smooth in the interior of Ω \ Q and is continuous up to the boundary of Q, hence, we infer that F (s) = ∂U1 ∂x1 changes sign for some s ∈ (0, s 1 ). Next, we show that for i = 1, ∂Ui ∂xi preserves a constant sign along the line = (0, s 1 ) × {0} d−1 . To see this, we note that ⊂ Ω \ Q . For any point x ∈ , we may find an open ball centred at x with radius r Since U i is strictly positive in the open ball B i,x , Hopf's lemma ensures that Since x is chosen arbitrary on the line , together with the continuity of DU , we can conclude Proposition 2. Suppose that Assumption 1 holds and that φ is α-admissible. There exists k 0 > 0 and δ 0 > 0 such that for k ≥ k 0 , the Jacobian determinant J k given by (10) satisfies Proof. Observing that the solution U k is harmonic and hence smooth away from the inclusion Q, we may infer from standard regularity estimates that in any compact subset K ⊂ Ω \ Q, the sequence of functions {U k } k≥k0 is uniformly bounded in C 2 norm. By Ascoli-Arzela's theorem, and the limiting argument of Lemma 2.3, U k converges to U ∞ in C 1 norm as k → ∞ in any fixed set K. Thus, in some compact which is our thesis.
3. Volume estimates for the sign changing set of det DU ∞ . In this section, we present a method which enables us to derive quantitative bounds for the Jacobian determinant of DU ∞ in the perfectly conducting case. We will make use of the result of Lemma 2.1, namely that the Jacobian determinant is diagonal along the line Obtaining a pointwise bound for det DU ∞ along this line corresponds to estimating each of the partial derivatives ∂ i (U ∞ ) i along the line L. There are several methods available to do so. One approach is to use Harnack's inequality which states that if H is a positive harmonic function in some ball B(x 0, , R), then it follows from the Poisson formula that In particular, if H vanishes at some point y 0 ∈ ∂B (x 0 , R), this implies the quantitative Hopf's inequality ∂H ∂ν To obtain a quantitative result, we will use that the domain is (for lack of a better word) tube connected, which roughly means that Ω \ Q is path connected with 'thick' paths.
open set of a hyperplane of R d containing the origin, and v a vector normal to S, we denote C (x, S, v) the cylinder given by We name its boundaries The first cylinder cuts the exterior boundary: The start of the next cylinder is within the previous one.
Remark 1. For explicit constructions, one can consider cylinders whose cross sections are rectangles or discs, see Appendix B.

Proposition 3.
If Ω and Q satisfy Assumption 1, then Ω \ Q is tube connected.
Proof. Consider first the case when a ∈ (Ω \ Q) satisfies the initial constraint. The construction is straightforward afterwards. To address also the case when a · e i = 0 for some i, apply the above construction withã given bỹ and add a final cylinder to connect to a. Lemma 3.3. Suppose Assumption 1 holds and that φ is α-admissible. Then, there exists p 1 ∈ L, where L is the set given by (16), and two positive constants M 1 and M 0 such that Lower bounds for M 0 and M 1 can be computed depending on geometric invariants of the surface ∂Q, and depending on the tubes connecting p 1 , another point (p 0 ∈ L) and the origin, to the exterior boundary of Ω. Possible locations for p 1 , p 0 ∈ L can be determined based on an exterior ball condition satisfied at y 0 , the point of contact between L and Q closest to the origin.
The following quantitative lower bound for the measure of the domains where the Jacobian determinant is positive or negative follows.
Proof. Thanks to a variant of Hadamard's inequality [12] there holds for any x, y ∈ Ω, Furthermore, thanks to Lemma 3.4, on Ω \ Q only, and the result follows.
Proof of Lemma 3.3. In this proof, we write U for U ∞ .
First step: construction of the point p 1 . Since ∂Q is C 2 , it satisfies the exterior ball condition, and we may place a ball B (p 0 , r 0 ) centred at some (intermediate) point p 0 ∈ L, tangential to the boundary of the inclusion ∂Q at the point of contact connecting p 0 to the exterior boundary. Consider the following Dirichlet problems 1 2 , and in turn, by the maximum principle Iterating this argument, we obtain that Thanks to the quantitative variant of Hopf's Lemma, we may apply (17) for U 1 in the ball B (p 0 , r 0 ) and at the point y 0 to obtain Since U 1 ∈ C 1,σ Ω \ Q , see Lemma 3.4 below, we can choose p 1 between p 0 and y 0 such that Indeed, for every θ ∈ (0, 1) Second step: lower bound for ∂ i U i (p 1 ) for i = 2, . . . , d. At the point p 1 , for each i = 2, . . . , d let C x i m , S i m , v i m 1≤i≤N be the tubes connecting p 1 to the exterior boundary of Ω (for a possibly different N from the one above). We note that due to the evenness of χ Q in the direction e i , U i = 0 on L. Consider the following Dirichlet problems for m = 1, . . . N − 1, and arguing as above we have Third step: upper bound for J ∞ (p 1 ) and lower bound for J ∞ (0). In the first two steps, we have obtained that which is the announced upper bound. By the same argument as in the second step, applied at the origin, we have Turning to ∂ 1 U 1 (0), we notice that because of the symmetry of χ Q , U 1 (0) = 0. We thus repeat once again the second step method, to obtain and the conclusion follows from (21) and (22).

Remark 2.
Using explicitly cylinders to connect the points is certainly not necessary. An alternative construction would be to make use auxiliary problems defined in Ω \ Q ∩ (0, ∞) d . The motivation for this detailed approach is that it allows explicit bounds (as closed form solutions can be written down) in practical cases, see Proposition 4.
Proof. See [18], and [13] for the case when Ω is a convex polygon.
Example 1. Consider the case d = 3, Suppose that Q is a torus centred at the origin with minor radii a, and major radii + a, with 0 < < 1 − 2a. We write down the coordinates of p 1 , and provide lower bounds for M 0 and M 1 as defined in Lemma 3.3.

Proposition 4.
Suppose that Q is a proper torus centred at the origin with minor radii a, and major radii + a, with 0 < < 1 − 2a.
Remark 3. Example 1 satisfies most of the requirements listed in Assumption 1, except that Q ∩ Re 3 = ∅. The reader will notice that Q ∩ Re 1 = ∅ is enough for Lemma 3.3 to hold.
4. Continuous dependence of the gradient field in the conductivity contrast k. In this section, we investigate the continuous dependence of DU k , where U k is the solution of (9), when k varies. Several authors have investigated this question including [16,5,14,7]. We follow an integral equation approach. We show that U k admits a representation formula similar to that in [6], and we use the classical theory of weakly singular boundary integral operators to show the continuous dependence of DU k with respect to k as k → ∞.
In this section we prove the following result.
Proposition 5. Under the assumptions of Theorem 1.2, suppose U k ∈ H 1 Ω; R d solves the conductivity equation (9). Then U k converges to U ∞ in C 1,σ Ω \ Q . Furthermore there holds where K SL is a constant independent of the conductivity parameter k and the boundary value φ.
We remind the readers of classical definitions of layer potential operators. Given ψ ∈ C 0,σ (∂Q), the single layer potential S Q ψ and double layer potential D Q ψ are defined as where ω d denotes the d − 1 dimensional surface area of the unit ball. For a function u defined on R d \ ∂Q, we denote: , ν x , x ∈ ∂Q where ν x is the outward unit normal to ∂Q at the point x. Note that by assumption Q is orientable and the outer and inner limits ∂ ∂νx u | ± (x) are well defined.
The single layer and double layer potential also satisfy the well known jump formulas given by denotes the Poincaré-Neumann operator and K Q (ψ) is the L 2 -adjoint of K * Q given by We refer the reader to [11] for proofs of the jump formulas. Denoting u k to be the solution of the following scalar Dirichlet boundary value problem with the Dirichlet to Neumann map, Λ k : it is proven for example in [6] that the solution u k admits a unique layer potential representation given by for k ∈ (0, ∞], where the function H is harmonic of the form and ψ k satisfies the integral equation: Observe that when Q is perfectly conducting, the jump formulas (25) assert that Lemma 4.1. The Dirichlet to Neumann map (27) is uniformly bounded in H − 1 2 ∂Ω with respect to k. In particular, the following estimates hold Proof. Since f ∈ H 1 2 ∂Ω , its harmonic extensionf on Ω, which belongs in H 1 Ω , is such that f and trace( f ) = f . We define ξ ∈ W 1,∞ (Ω; R) as the cut-off given by If we callg the harmonic extension of g, W being the minimiser of a Dirichlet problem, Testing the equation (27) Since u k is a minimiser of a Dirichlet problem, we have .

YVES CAPDEBOSCQ AND SHAUN CHEN YANG ONG
Altogether, we have obtained which implies (30). The second inequality follows from the same argument, with g = 1. In that case W Lemma 4.2. Let u k denote the solution to (27). Given f ∈ H 1 2 ∂Ω , the function H k given by (28) where C 0 depends only on dist (Q, ∂Ω), |Ω \ Q|, |∂Ω| and d, and is given by (31).
Proof. Let x be an arbitrary point in Q. A straightforward computation of ∂ αβ H k (x) gives Write δ 0 = dist (Q, ∂Ω). Using that δ 0 ≤ |x − y| for all y ∈ ∂Ω, and the bounds derived in Lemma 4.1 we obtain Remark 4. Since dist (Q, ∂Ω) > 0, the function H k = (−S Ω • Λ k + D Ω ) f which is defined as a boundary layer potential on ∂Ω is trivially smooth on Q. However, in the spirit of tracking down constants, the purpose of the lemma above is to provide a quantitative version for the second order derivatives of H k on the set Q.
Proposition 6. Suppose that ψ, ψ k ∈ C 0,σ ∂Q solve the integral equations respectively, with H k be the harmonic function defined in (28). Then for any k > 1, there holds where the constant C Q is a geometrical constant which depends on the spectral radius of the operator K * Q .
Proof. Thanks to Proposition 10 from the appendix, a straightforward computation shows that for any φ ∈ C 0,σ (∂Q), λ n , λ ∈ R such that λ n > λ and λ is greater than the spectral radius of K * Q we have The triangle inequality then shows that where ρ = lim sup n K * Q n 1 n and M (ρ, λ) ≤ C(Q) |λ|−ρ with C(Q) independent of λ. Using Proposition 10, we know that the spectral radius of K * Q on the set C 0,σ 0 ∂Q := f ∈ C 0,σ ∂Q : ∂Q f = 0 is less than 1 2 . Applying the above argument to λ = 1 2 and λ n = k+1 2(k−1) , we obtain , as announced. Proof. Let ψ be an arbitrary odd function belonging to H 1 2 ∂Ω . Denoting the map z(y) = −y for y ∈ R d , observe that if ψ is odd and Ω has cubic symmetry, a straightforward computation shows that for any x ∈ R d , and similarly, as ν(−y) = −ν(y) for all y ∈ ∂Ω, thus establishing our thesis.

Corollary 2.
Let Ω be as the above and let K * Ω be the Poincaré-Neumann operator given in (26). Denoting the integral equation given by is uniquely solvable for any f ∈ C 0,σ odd (∂Ω), moreover, the solution ψ k belongs to C 0,σ odd ∂Ω . Proof. Since ∂Ω has cubic symmetry, it is clear that the set C 0,σ odd (∂Ω) is non-empty. Thanks to Lemma 4.3, moreover, as C 0,σ odd is a closed subspace of Proposition 9 from the appendix ensures that the spectral radius of K * Ω is less than Proposition 7. Under the assumptions of Theorem 1.2, the solution U k to (9) converges to U ∞ , the solution of the perfect conductivity equation (12) in C 1,σ Ω \ Q . Furthermore, there holds where K SL is independent of the conductivity parameter k.
Proof. Let U i,k denote the i-th component of the vector U k . By the representation formula (28), U i,k = H i,k + S Q ψ k i where ψ k satisfies the integral equation: Observe from the proof of Lemma 2.1 that U i,k is an odd function in Ω and therefore its gradient ∇U i,k is even for all x ∈ Ω. The symmetry property of ∂Ω then implies that ∇U i,k · ν Ω is an odd function so that as shown in Lemma 4.3, the function is also odd. Since Q is a symmetrical inclusion by Assumption 1, we may reiterate the argument to show that the function (H i,k · ν) | ∂Q is also odd. Since H k is in C 2 Q by Lemma 4.2 , (H i,k · ν) | ∂Q is in C 0,σ odd for any σ ∈ (0, 1). Then thanks to [11,Theorem 2.30] together with Corollary 2 we deduce the existence of the a unique ψ k in C 0,σ odd . Next, write V i,k := H i,k + S Q ψ i where ψ i solves the integral equation

YVES CAPDEBOSCQ AND SHAUN CHEN YANG ONG
The function V i,k is odd thanks to Lemma 4.3. Thus V i,k = 0 on Q. Consider the function W i,k = U i,k − V i,k . By construction, we have moreover, as ∂Q is C 2 and ψ k , ψ ∈ C 0,σ ∂Q , the following estimate holds, see [11,Theorem 2.17], Lemma 3.4 implies that Thus, we have for any i ∈ {1, . . . , d}, which is our statement.
We are now ready to prove our main result, which follows from the following proposition.
Proposition 8. Assume that Assumption 1 holds and that φ is α-admissible. Let M 0 and M 1 denote the values given by Lemma 3.3, and K SL be the constant given by Proposition 7. Write There exists non-empty sets A 1 − , . . . , A d − and A + such that for all k > k 0 , for all α-admissible φ, the solution U k of (1) satisfies . The constant C(Ω, Q) depends on the regularity of Ω and Q only. Furthermore , and where the point p i is constructed on R + e i ∩(Ω \ Q).
Proof. Using Corollary 1, we observe that, writing and there holds and Thanks to Proposition 7, we have therefore arguing as in the proof of Corollary 1, and for any positive k such that the triangle inequality and (33) shows for all x ∈ B 1 , and for all x ∈ B 0 there holds Combining these two statements, and writing τ 0 = 1 4 α d min {M 0 , M 1 }, we have the set containments which holds uniformly for all k > k 0 := 1 + 4 · K SL · max 1, Finally, because Assumption 1 imposes axial symmetry and because φ is αadmissible, all of the above developments are valid for every canonical direction e 1 , . . . , e d and both in the positive and negative half lines, thus, where p i is constructed similarly to p 1 on the half-axes Re i .
5. Additional remarks. The method presented here does not provide optimal bounds for J −1 k ((−∞, 0)) and J −1 k ((0, ∞)), as the restriction to tubes we have used is not necessary. Its advantage is that it delivers explicit bounds in practical cases. It does give a rule of thumb to obtain larger bounds: the larger the cross section of the connecting tubes, the better, by comparison.
The amplitude of the boundary condition does not play a role in the definition of the sets where the sign of the Jacobian determinant is controlled. It appears as a multiplicative factor for the (positive or negative) value of the determinant on these sets.
The fact that γ is piecewise constant is not required for Proposition 2 to hold. We can argue by a perturbation argument. Consider a regularised γ using the standard mollifier η ∈ C ∞ (R), where η (x) = c η exp −1 1−|x/ | 2 for |x| < and 0 otherwise, in which, the constant c η is a normalizing constant so that η L 1 R d = 1. Observing that the regularised conductivity matrixγ = γ η is smooth and yet satisfies Assumption 1, we observe the Jacobian determinant of the solution DŨ with coefficient matrixγ would still satisfy the sign changing phenomenon. The main result, however, makes use of layer potential techniques which are ill-suited for variable coefficients.
It is worth noting that for the perfectly conducting case, the measure of the sets where the Jacobian determinant is positive or negative is not controlled by the Lebesgue measure of the inclusion Q; in particular it may happen when Q has zero measure. Consider the case d = 3 where Σ is a crack inclusion, or a surface of co-dimension 1 which has a C 2 boundary edge and satisfies Assumption 1, for example lying in the plane x 3 = 0. The perfect conductivity equation in this case reads −∆U = 0 in Ω \ Σ, The solution to (34) is non-trivial and a classical solution of C 2 Ω \ Σ ∩ C Ω \ Σ may be obtained by Perron's method of sub-solutions, since surfaces of co-dimension 1 with a C 2 boundary edge consists only of regular points to the Laplace operator. By the continuity of the solution and the fact that the weak maximum principle still holds here, the proof of Proposition 1 still applies even though the gradient DU is likely to be unbounded as it approaches the boundary edge of Σ.
Proposition 9. Let ∂Ω be a C 1,α domain, consider the space then H 0 is a Hilbert space in which its norm is equivalent to the usual space H 1 2 (∂Ω). Moreover, the operator K * Ω : H 0 → H 0 is compact, self-adjoint, and only has real discrete spectrum accumulating to 0. Then the spectrum of the operator K * Q : C 0,α 0 (∂Q) → C 0,α 0 (∂Q) consists only of discrete eigenvalues σ n where σ n ∈ (− 1 2 , 1 2 ), moreover, the Neumann series converges in operator norm for all λ such that |λ| > sup n σ n .
Proof. Suppose that φ ∈ C 0,α 0 (∂Q) and φ satisfies σ n I − K * Q (φ) = 0, where φ = 0. Consider the following quantities since we may apply the divergence theorem and the jump formula of S Q (φ) to find Thus, if A and B are not identically 0, this shows that |σ n | < 1 2 and we are done. Now consider the case that σ n = − 1 2 which corresponds to the case B = 0. Since B = 0, we deduce from (35) that S Q (φ) is constant on R d \ Q. Moreover, by the usual decay estimates thus S Q (φ) ≡ 0 on R d \ Q, which further implies that φ is identically 0 by the injectivity of S Q . Since we have assumed that φ = 0, this implies a contradiction and thanks to (36), this also implies that σ n = − 1 2 is not an eigenvalue.
When σ n = 1 2 , we have that A = 0, so we deduce again from (35) that S Q (φ) = m, a constant in Q. This, in turn, yields Since by our assumption φ ∈ C 0 0 (∂Q), it must be that B = 0, which further implies that S Q (φ) is a constant in R 3 \ Q. By the same argument as before S Q (φ) must be 0 in R 3 \ Q, which implies m = 0 by the continuity of S Q (φ), therefore implying φ = 0 and hence a contradiction. Combining these two facts, we deduce that the spectrum of K * Q in C 0 (∂Q) lies in the open interval σ n ∈ (− 1 2 , 1 2 ). Lastly, consider the space By Proposition 9 in the appendix, the spectrum of K * Q in H 0 consists only of real discrete eigenvalues converging to 0 with spectral radius sup n σ n < 1 2 . Since we deduce that the spectrum of K * ∂Q in the space C 0 0 (∂Q) also consists only of discrete eigenvalues with spectral radius sup n σ n < 1 2 . Thus, we may apply Theorem 5.1 Appendix B.
Then y 0 = inf (Ω \ Q ∩ (R + e 1 )) = , 0, 0) and there holds Proof. We parametrize Q as follows By the weak maximum principle, we have U 1 | ∂D ≥ V 1 | ∂D so that V 1 is a sub-solution with respect to U 1 . We construct the solution to V 1 by the standard method of separation of variables in D and obtain where k m,n = π √ 2 m 2 + n 2 C m,n = 8 where we used Lemma 5.2 to derive a lower bound. Subsequently, we infer by the comparison principle that so that by (17) (and by symmetry), which is our thesis. There holds det DU (p 1 ) ≤ −M 1 , where M 1 is given in Proposition 4.
Proof. Thanks to Lemma 3.4 there holds which together with 38 yields The point p 1 lies on the line (s, 0, 0) and at a distance r 2 = κ away from the boundary of ∂Q. In the Introducing the cube D 2 := p 1 + (−τ, +τ ) × (0, 1 2 τ ) × (−1, 1) and taking V 2 to be the solution to the Dirichlet problem we have 0 ≤ V 2 | ∂D2 ≤ U 2 | ∂D2 and V 2 is a sub-solution with respect to U 2 . We compute the solution V 2 as