Sign changing solutions of Poisson's equation

Let Ω be an open, possibly unbounded, set in Euclidean space Rm with boundary ∂Ω , let A be a measurable subset of Ω with measure |A| and let γ∈(0,1) . We investigate whether the solution vΩ,A,γ of −Δv=γ1Ω∖A−(1−γ)1A with v=0 on ∂Ω changes sign. Bounds are obtained for |A| in terms of geometric characteristics of Ω (bottom of the spectrum of the Dirichlet Laplacian, torsion, measure or R ‐smoothness of the boundary) such that essinfvΩ,A,γ⩾0 . We show that essinfvΩ,A,γ<0 for any measurable set A , provided |A|>γ|Ω| . This value is sharp. We also study the shape optimisation problem of the optimal location of A (with prescribed measure) which minimises the essential infimum of vΩ,A,γ . Surprisingly, if Ω is a ball, a symmetry breaking phenomenon occurs.


Introduction
Let Ω be an open, possibly unbounded, set in Euclidean space R m with boundary ∂Ω, and with, possibly infinite, measure |Ω|. It is well known [4] that if the bottom of the Dirichlet Laplacian defined by is bounded away from 0, then −Δv = 1, v = 0 on ∂Ω, has a unique weak solution denoted by v Ω , which is non-negative, and which satisfies, The m-dependent constant in the right-hand side of (2) has been improved in [13], and subsequently in [24]. If |Ω| < ∞ then, by the Faber-Krahn inequality, λ(Ω) > 0, and by (2), v Ω ∈ H 1 0 (Ω), and v Ω ∈ L 1 (Ω). For an arbitrary open set Ω, we define the torsion, or torsional rigidity, by Note that, under the assumption λ(Ω) > 0, by (2) the solution of an equation like in (1) with a right-hand side f ∈ L ∞ (Ω) can be defined by approximation on balls for the positive and negative parts of f . For a measurable subset A ⊂ Ω, with λ(Ω) > 0, and 0 < γ < 1, we denote by v Ω,A,γ the solution of These hypotheses on A, Ω and γ will not be repeated in the statements of all lemmas and theorems below. This paper investigates whether the solution of (3) satisfies essinf v Ω,A,γ < 0. Whether this holds depends on the geometry of Ω, and on the size and the location of the set A ⊂ Ω. This question shows up in a variety of situations. We refer, for instance, to [17], where v is a scalar potential and the right-hand side stands for a magnetic field which changes sign. The influence of the magnetic field on the asymptotic behaviour of the bottom of the spectrum of the Pauli operator is effective, provided that the scalar potential has constant sign, that is, essinf v Ω,A,γ = 0. In fluid mechanics, the function v can be interpreted as a vorticity stream function, for a vorticity taking the values γ and − (1 − γ). If v Ω,A,γ changes sign, then there exist at least two stagnation points. More situations where the sign question of the state function is put in relationship with sign changing data can be found in [7,11,21] and, in some biological models, [19].
It follows immediately from the definition that for a homothety tΩ, t > 0 of Ω, we have the scaling relations and C + (tΩ, γ) = t m C + (Ω, γ).
Furthermore, if Ω 1 , Ω 2 are disjoint open sets, then C − (Ω 1 ∪ Ω 2 , γ) = min{C − (Ω 1 , γ), C − (Ω 2 , γ)}, This paper concerns the analysis of these quantities and their dependence on Ω. It turns out that C + (Ω, γ) = γ|Ω| for arbitrary open sets Ω with finite measure. On the contrary, C − (Ω, γ) is very sensitive to the geometry. We find its main properties, give basic estimates, establish isoperimetric and isotorsional inequalities, and we discuss the shape optimisation problem related to the optimal location of the set A in order to minimise the essential infimum. Below we show that, in general, we have to assume some regularity of Ω in order to have In Theorem 3 below, we show that if Ω is bounded, and ∂Ω is of class C 2 , then C − (Ω) > 0. In order to quantify this assertion, we introduce some notation. For a non-empty open set Ω, we denote by diam(Ω) = sup{|x − y| : x ∈ Ω, y ∈ Ω}. We denote the complement R m \ E of E by E c , and the closure of E by E. Furthermore, B r (x) := {y ∈ R m : |x − y| < r} denotes the open ball centred at x of radius r. If x = 0, we simply write B r . We set ω m = |B 1 |. For x ∈ Ω, we letx ∈ ∂Ω be a point such that |x −x| = min{|x − z| : z ∈ ∂Ω}. We recall the following from [2, p. 280].
We also recall that a bounded Ω with C 2 boundary ∂Ω is R-smooth for some R > 0.
If Ω is an open, bounded set in R m with a C 2 and R-smooth boundary, then and The following inequality gives an upper bound for C − (Ω, γ) in terms of λ(Ω).
and where C 2 (m) is the constant in the Kohler-Jobin inequality (37) below.
This implies that if Ω is an open set with T (Ω) < ∞, then The optimal coefficient of T (Ω) m/(m+2) in (9) is not known. However, the Kohler-Jobin inequality suggests to prove (or disprove) optimality for balls.
In particular, if Ω is unbounded, then C − (Ω, γ) = 0. The value of C 3 (m) can be read off from the proof in Section 6.
We see from Theorems 1 and 3 that C − (B R , γ) < C + (B R , γ). The isoperimetric inequality below generalises this to arbitrary open sets with finite measure.
The theorem above implies that C − (Ω, γ) C(m, γ)|Ω| for every open set of finite measure with C(m, γ) < γ. The proof of Theorem 6 relies on the relaxation of the shape optimisation problem (11) to the larger class of quasi-open sets. We shall prove that the supremum is attained at some quasi-open set Ω * for which C − (Ω * , γ) < γ|Ω * |.
The optimal value C(m, γ) = C−(Ω * ,γ) is not known, nor whether Ω * is open. The symmetry breaking phenomenon for balls stated in Theorem 7 below does not support the ball to be a maximiser.
Given a constant c ∈ (C − (Ω, γ), |Ω|), there exists at least one set A ⊆ Ω, |A| = c such that essinf v Ω,A,γ < 0. A natural question is to find the best location of the set A of measure c, which minimises essinf v Ω,A,γ . This question is of particular interest for values of c close to C − (Ω, γ), as this gives information on where the geometry of Ω is most sensitive to negative values. We prove the following shape optimisation result for the optimal location.
has a solution. Moreover, if Ω is a ball B, then, depending on the value of c, the optimal locations may be radial or not.
The existence of an optimal set relies partly on a concavity property of the shape functional A → essinf v Ω,A,γ . We point out that the proof relies on both the concavity and the analysis of optimality conditions in relationship with the partial differential equation (3) (see [9]). If Ω is a ball B and c is close to |B|, then the optimal location is a ball. If c is close to C − (B, γ), then the optimal location is no longer radial. This symmetry breaking phenomenon occurs at a value c ∈ (C − (B, γ), γ|B|), and is supported by analytical, and numerical computations.
Theorem 7 can be interpreted both as a (rather non-standard) shape optimisation problem or as an optimisation problem in a prescribed class of rearrangements, see, for example, [1]. We also refer to the paper of Burton and Toland [8] for models of steady waves with vorticity, where the distribution of the vorticity is prescribed, but we point out that our problem is essentially of different nature since the functional to be minimised is not an energy of the problem.
The proofs of Theorems 1-7 are deferred to Sections 2-8 below.

Proof of Theorem 1
In order to simplify notation, throughout the paper, if Ω is an open set and A ⊂ R m is measurable, not necessarily contained in Ω, by v Ω,A,γ we mean v Ω,Ω∩A,γ .
Proof. Firstly, assume that Ω ⊂ R m is an open set with finite measure. Assume that A ⊂ Ω is a measurable set such that v Ω,A,γ 0. In a first step, we shall prove that |A| γ|Ω|. As a consequence, C + (Ω, γ) γ|Ω|.
Indeed, since v Ω,A,γ 0, one can use Talenti's theorem (see, for instance, [18, Theorem 3.1.1]) in the following way. We denote by v * the Schwarz rearrangement of v Ω,A,γ , and by f * the rearrangement of γ1 Ω\A − (1 − γ)1 A . There exist two positive values 0 < r 1 < r 2 such that f * = γ1 Br 1 − (1 − γ)1 Br 2 \Br 1 , where r 1 is such that |B r1 | = |Ω \ A| and |B r2 | = |Ω|. By Talenti's theorem, we get By elementary computations, one gets the expression for v Br 2 ,B c r 1 ,γ . Indeed, the solution v Br 2 ,B c r 1 ,γ is radially symmetric and satisfies the equation with initial condition v (0) = 0, and v(r 2 ) = 0. Moreover, the solution is C 1,α regular, for some α > 0. We integrate separately on [0, r 1 ], and on [r 1 , r 2 ], and write the equality of the left and right derivatives in r 1 , namely v − (r 1 ) = v + (r 1 ). Hence, we get In general, from the positivity of v Br 2 ,B c r 1 ,γ , one gets that v (r 2 ) 0. Hence, As a by-product of the computation, we observe that the constant γ in Theorem 1 is sharp, and that equality holds for the ball. As soon as, This means that as v(r 2 ) = 0, the solution is not positive near the boundary of the ball.
The construction is based on the following observation. There exists a finite family of In every ball, we display the set A i of measure γ|B i | in an annulus centred at the centre of B i and having ∂B i as external boundary. Hence v Bi,Ai,γ 0. Moreover, since the sets B i are mutually disjoint, we get that v ∪iBi,∪iAi,γ 0.
We have the following.
Proof. As a consequence of the hypotheses, we get Hence, by the maximum principle, Coming back to the proof of Theorem 1, using the additivity and monotonicity property of C + , we get that The theorem follows by letting ε → 0.

Proof of Theorem 2
We first introduce some basic notation and properties. For a non-empty open set Ω ⊂ R m , we denote by G Ω (x, y), x ∈ Ω, y ∈ Ω, x = y, the kernel of the resolvent of the Dirichlet Laplacian acting in L 2 (Ω). This function exists and is well defined for all x = y, provided m 3. It also exists for m = 2, for example, under the hypothesis that the torsion function v Ω defined by approximation on balls is locally finite. The resolvent kernel is non-negative, symmetric in x and y and is monotone increasing in Ω. That is, if If The monotonicity in (13) implies that both the torsion function v Ω and torsion T (Ω) are monotone increasing in Ω.
We have also that Formula (14) implies that Proof of Theorem 2. Let Ω = ΔOAB be a triangle, with α := ∠BOA π 3 at the origin, and oriented such that the positive x-axis is the bisectrix of that angle. Let W α be the infinite wedge with vertex at O, and edges at angles ± 1 2 α with the positive x-axis, which contain the two sides OA and OB of Ω. Let W α,c be the radial sector with area c and edges at angles ± 1 2 α. Then W α,c ⊂ Ω for all c sufficiently small. We have by monotonicity that In Cartesian coordinates where s = tan(α/2). In polar coordinates x = (r; θ), we have by [22, p. 279] for the sector with We observe that for θ = 0 the terms in the series in the right-hand side of (17) are alternating and decreasing in absolute value. Hence By (15) which is negative for all x 1 sufficiently small. We see from the proof above that we could have chosen any angle of the triangle, provided that angle is strictly less than π/2. The proof above also shows that the infinite wedge W α , α < π/2 with radial sector W α,c , c > 0 has a sign changing solution v Wα,Wα,c,γ .

Proof of Theorem 3
Proof of Theorem 3. Let us start by observing that the following covering property holds: for every x ∈ Ω, there exists a ball B of radius R such that x ∈ B ⊂ Ω. Indeed, letx ∈ ∂Ω be a point which realises the distance to the boundary. Since the boundary of Ω is of class C 2 , then x −x is normal to the boundary ∂Ω atx. If |x −x| R, then B R (x) ⊂ Ω. If |x 0 −x| < R, then x belongs to the ball of radius R tangent to ∂Ω atx.
Assume for a contradiction that the infimum being attained at x ε . Taking a sequence ε → 0, we may assume (up to extracting suitable subsequences) that Then uniformly on Ω. This is a consequence of the elliptic regularity of the solutions, which are uniformly bounded in C 1,α (Ω) for some α > 0. Consequently, v Ω,g,γ 0 in Ω. Indeed, for x * a minimum point of v Ω,g,γ with v Ω,g,γ (x * ) < 0, we can modify g slightly to find a new functioñ g, such that 0 g 1, From the density of the characteristic functions, we can find a sequence of setsÃ δ such that Consequently, v Ω,g,γ (x * ) = 0. There are two possibilities: either x * ∈ Ω, or x * ∈ ∂Ω. Assume first that x * ∈ Ω. As a consequence of the covering property, there exists a ball B of radius R such that x 0 ∈ B ⊂ Ω. In particular, this implies that v Ω,g,γ 0 on ∂B. The maximum principle gives Case 1. In case v B,g,γ (x * ) < 0, we immediately get a contradiction since, as above, we can build a sequence of setsÃ δ ⊂ B such that 1Ã This contradicts (19).
Case 2. In case v B,g,γ (x * ) = 0, we claim that either g is itself a characteristic function, or we can find another functiong such that 0 g 1, Assume that g is a characteristic function. Then g = 1 A . Taking a new set A ⊂Ã ⊂ B, such that |A| < |Ã| < C − (B R , γ) we get by the maximum principle that v B,Ã,γ (x * ) < 0, in contradiction with the definition of C − (B R , γ).
Assume that g is not a characteristic function on B. Then, for some value δ > 0, the set U δ = {x ∈ B : δ g(x) 1 − δ} has positive Lebesgue measure. We putg = g + s1 U δ , where s > 0 is small enough such that Bg < C − (B R , γ). By the maximum principle, we get v B,g,γ (x * ) < 0. In this case, we are back to Case 1.
Assume now that x * ∈ ∂Ω. Let n x * be the outward normal vector at x * . Let x ε be the projection on ∂Ω of x ε . Since Ω is of class C 2 , we get x ε → x * and that there exists a point y ε on the segment [x ε , x ε ] such that ∇v Ω,Aε,γ (y ε ) · n xε 0. Passing to the limit, we get ∇v Ω,g,γ (x * ) · n x * 0.
Using the R-smoothness at x * , the ball B ⊂ Ω of radius R tangent to ∂Ω at x * stays in Ω.
Since v Ω,g,γ 0, by the maximum principle, we get v Ω,g,γ v B,g,γ . By the Hopf maximum principle, applied to v Ω,g,γ − v B,g,γ on B at the minimum point x * ∈ ∂B, we have either that In the first situation, which means that v B,g,γ takes negative values close to x * . Then, we conclude as in Case 1 above. In the second situation, if we find a point x ∈ ∂B ∩ Ω, we can conclude as in Case 2 since v B,g,γ (x) = 0. The alternative is that ∂B ⊂ ∂Ω so that Ω = B, and we have a contradiction.
To prove (5) we let m 3, and let H be an open half-space. Then where x * is the reflection of x with respect to ∂H, and By (14), and monotonicity we have that where Hx is the half-space tangent to Let By (20)- (22) and radial rearrangement of A about x, we have The following will be used in the proof of (5), and in the proof of (55) and (56) in Remark 6 below.
We first consider the case |x −x| > R. Then, by domain monotonicity of the torsion function and (24), We next consider the case |x −x| R.
By (24) and (25) The right-hand side of (26) is non-negative for r A γR/(4m). This is, by (23), equivalent to (5). Consider the case m = 2. Then By the triangle inequality, Hence, we have that The remaining arguments follow those of the case m 3, as the right-hand side above equals the right-hand side of (24). To prove (6), we let m 3. By scaling it suffices to prove (6) for R = 1. Let a ∈ (0, 1). We obtain an upper bound for a such that v B1,Ba,γ (0) < 0. Note that Hence, by (27) The right-hand side of (28) is negative for a > γ 1/2 . This implies (6).

Proof of Theorem 4
Proof of Theorem 4. The proof of Theorem 4 relies on some basic facts on the connection between torsion function, Green function and heat kernel. These have been exploited elsewhere in the literature. See, for example, [3]. We recall that (see [10,14,15]) the heat equation has a unique, minimal, positive fundamental solution p Ω (x, y; t), where x ∈ Ω, y ∈ Ω, t > 0. This solution, the heat kernel for Ω, is symmetric in x, y, strictly positive, jointly smooth in x, y ∈ Ω and t > 0, and it satisfies the semigroup property for all x, y ∈ Ω and t, s > 0. If Ω is an open subset of R m , then, by minimality, It is a standard fact that for Ω open in R m , whenever the integral with respect to t converges. We have By the heat semigroup property, we have that for x ∈ Ω, y ∈ Ω, t > 0, Furthermore, for all s ∈ (0, t), So choosing s = t/2 in (34), and subsequently using (33) gives that By (31), both diagonal heat kernels in the right-hand side of (35) are bounded by (2πt) −m/2 , and p Ω (x, y; t) 1/3 (4πt) −m/6 e −|x−y| 2 /(12t) . Hence, by (35), The Kohler-Jobin inequality asserts that (see, for instance, [5]) there exists C 2 (m) > 0 such that for every open set Ω, This, together with the Lieb inequality, implies where We estimate the integral of v Ω,A,γ on the set A as follows: By monotonicity, we have that For all x ∈ A , and for all y ∈ Ω \ A, we have that

By (32) and (36) and the preceding inequality,
By (38), (40), (39) and (41), we find In order to bound the right-hand side of (42) from above, we have where we have used the scaling λ(B r1 ) = r −2 1 λ(B 1 ). In order to bound the first term in the right-hand side of (43) from above, we use the inequality e −x By (43) and (44), we obtain that the right-hand side of (42) is bounded from above by 0, provided with C 1 (m) given by (8). This implies the bound for C − (Ω, γ) in (9).

Proof of Theorem 5
We start with the following.
Lemma 10. There exists ε = ε(m, γ) > 0 such that for every open set Ω ⊆ R m with finite torsion and for every x 0 ∈ Ω, the following holds: Note that a consequence of the lemma above, for every δ > 0 Proof. Assume for the moment that Ω is bounded and smooth. Let for some value ε > 0 that will be specified later in the proof. We observe that v Ω,B1(x0),γ is Lipschitz so that for every r ∈ (0, 1), one can define The function M : (0, 1) → R is Lipschitz and bounded from above by ε. If there exists some r ∈ (0, 1) such that M (r) = 0, then the assertion of the theorem is proved since one gets by the maximum principle that v Ω,B1(x0),γ 0 on B r (x 0 ). So, we can assume that M > 0 on (0,1). Then, the supremum above is achieved at a point x r ∈ ∂B r (x 0 ) ∩ Ω. Moreover, in the viscosity sense on (0, 1). For every 0 < ε < R 1, we introduce the equation By the comparison principle (see, for instance, [23, Theorem 1.1]), we get that M φ ε,R on (R, d). In particular, this implies that φ is non-negative. If M is differentiable at R, then φ ε,R (R) M (R). Multiplying the equation for φ ε,R by r m−1 and integrating between r and R gives Dividing by r m−1 and integrating over (ε, R) yields Since M is Lipschitz and lim ε→0 Finally, Integrating over (0,1) gives Since M 0, Taking into account that M γv Ω , and putting concludes the proof. Assume now that Ω is open and with finite torsion. Assume that Let (Ω n ) n be an increasing sequence of open, smooth sets such that Ω = ∪ n Ω n . For all n sufficiently large, x 0 ∈ Ω n . Moreover, by the maximum principle, v Ωn (x) ε for a.e. x ∈ B 1 (x 0 ).
Proof of Theorem 5.
The key observation is that the class of quasi-open sets is the largest class of sets where the Dirichlet-Laplacian problem is well defined in the Sobolev space H 1 0 , and satisfies a strong maximum principle (see [12]). Of course, any open set is also quasi-open. Although the reader may only be interested in open sets, we are forced to work with quasi-open ones since the crucial step of the proof is the existence of a quasi-open set Ω * which maximises the left-hand side of (11).
The strategy of the proof is as follows. We analyse the shape optimisation problem and prove in Step 1 below the existence of a maximiser Ω * . Denoting C (m, γ) = C − (Ω * , γ)/|Ω * |, we then prove in Step 2 that C (m, γ) < γ by a direct estimate on Ω * . We start with the following observation. Assume that (Ω n ) n is a sequence of quasi-open sets of R m , |Ω n | 1, such that v Ωn converges strongly in L 2 (R m ), and pointwise almost everywhere to some function v. Let us denote Ω := {v > 0}. We then have Indeed, in order to prove this assertion, let us consider a set A ⊆ Ω such that essinf v Ω,A,γ < 0. We have and hence Following [6,Lemma 4.3.15], there exists larger setsΩ n ⊃ Ω n , |Ω n | 2, such that for a subsequence (still denoted with the same index) lim n→+∞ vΩ n ,A∩Ωn,γ (x) = v Ω,A,γ (x), for a.e. x ∈ R m . Since essinf v Ω,A,γ < 0, we get for n large enough that essinf vΩ n,A∩Ωn γ < 0 for n large enough. Lemma 8 (which also holds in the class of quasi-open sets) implies that essinf v Ωn,A∩Ωn,γ < 0, since the right-hand side equals to γ, γ > 0 onΩ n \ Ω n . Consequently, C − (Ω n , γ) |Ω n ∩ A|. Passing to the limit, which implies the assertion.
Let us prove now that the shape optimisation problem (51) has a solution. In order to prove this result, it is enough to consider a maximising sequence (Ω n ) of quasi-open, quasi-connected subsets of R m , with |Ω n | = 1. We first notice that the diameters of Ω n are uniformly bounded, so that up to a translation, all of them are subsets of the same ball B. This is a consequence of Theorem 5 which by approximation holds as well on quasi-open, quasi-connected sets. Indeed, this is essentially a consequence of (45) which passes to the limit by approximation.
Then, the existence result is immediate from the compact embedding of H 1 0 (B) L 2 (B) and the observation above: there exists a subsequence such that v Ωn converges strongly in L 2 (R m ) and pointwise almost everywhere to some function v. Taking Ω * := {v > 0}, and using the upper semi-continuity result (52) together with the lower semicontinuity of the Lebesgue measures coming from (53), we conclude that Ω * is optimal. Note that the smoothness of ∂Ω implies that v Ω,A,γ ∈ C 1,α (Ω).
Firstly, we extend the shape functional m on the closure of the convex hull of

Denote by
One naturally extends the functional m to the set F by defining v Ω,f,γ as the solution of −Δv = f in H 1 0 (Ω). We shall prove in the sequel that the relaxation of the shape optimisation problem (12) on the set F has a solution in F. Precisely, we solve min{m(f ) : f ∈ F}. (54) Clearly, F is compact for the weak-L ∞ -topology, so that we can assume that (f n ) n is a minimising sequence which converges in weak-L ∞ to f . We know, by the Calderon-Zygmund inequality, that (v Ω,fn,γ ) n are uniformly bounded in W 2,p (Ω), for every p < ∞. In particular, for p large enough, this implies that v Ω,fn,γ converges uniformly to v Ω,f,γ . Consequently, this implies that m(f n ) converges to m(f ) so that f is a solution to the optimisation problem (54). Secondly, we prove that there exists some set A such that f = γ1 Ω\A − (1 − γ)1 A . To prove this, we exploit both the concavity property of the map f → m(f ), and the structure of the partial differential equation. Assume for contradiction that the set has non-zero measure, for some ε > 0. Let A 1 , A 2 ⊂ A ε be two disjoint sets, such that |A 1 | = |A 2 |. We consider the functions , ε). Then, f 1 , f 2 ∈ F, and by linearity, we have Consequently, . We distinguish between two situations: v Ω,f,γ (x * ) = 0, and v Ω,f,γ (x * ) < 0. If we are in the first situation, then x * could belong to ∂Ω. In this case, for all admissible sets A, we have v Ω,A,γ 0, the minimal value, which is 0 being attained on ∂Ω. In this case, every admissible set A is a solution to the shape optimisation problem. If we are in the second situation, then necessarily x * ∈ Ω. By linearity, In particular, for every pair of points x, y ∈ A ε \ {x * } with density 1 in A ε , we get Since G Ω is harmonic on Ω \ {x * }, we get that G Ω is constant in Ω \ {x * }, in contradiction with the fact that it is a fundamental solution.
Finally, this implies that |A ε | = 0 for every ε > 0. Hence, f is a characteristic function.
Remark 1. Clearly, the solution of the shape optimisation problem above is, in general, not unique. If the minimal value is 0, then any admissible set A is a solution. If the minimal value is strictly negative, then there are geometries with non-uniqueness. For example, if Ω is the union of two disjoint balls with the same radius, then A is a subset of one of the two balls.
Remark 2. Assume Ω = B R , and |B R | c γ|B R |. The solution to the shape optimisation problem (12) is given by the (concentric) ball B rc , of mass c, c = |B rc |. Indeed, there are two possibilities. This follows directly from Talenti's theorem applied to −v BR,A,γ in case A ⊂ B R has measure c, and v BR,A,γ 0.
Assume now that v BR,A,γ changes sign on B R . We define the sets Ω + = {v BR,A,γ > 0} and Ω − = {v BR,A,γ < 0}. In view of Theorem 1, we have that |A ∩ Ω + | γ|Ω + | and |A ∩ Ω − | γ|Ω − |. We use Talenti's theorem on Ω − , and get that the essential infimum of the function v B R ,B r ,γ is not larger than the infimum of v BR,A,γ , where B R , B r are the balls centred at the origin of measures |Ω − |, |Ω − ∩ A|, respectively. We claim that v BR,Br c ,γ v B R ,B r ,γ . Indeed, making a suitable rescaling by a factor t 1 such that |t(Ω − ∩ A)| = c γ|B R |, the function v tB R ,tB r ,γ has an essential infimum lower than that of v B R ,B r ,γ . We finally notice  set A which gives a lower essential infimum. This fact is observed numerically, if, for instance, the set A is a disc, centred at (0.52,0) of radius r c = 0.432. Of course, the fact that in this case the essential infimum is strictly negative can be directly deduced from estimates of the Poisson formula. In Figures 1 and 2 below, we display the (rescaled) numerical solutions computed with MATLAB. If c is less than the critical value, the infimum is equal to 0, and is attained for an infinite number of solutions to the shape optimisation problem.
Hence, the position of the set A is a suitable lower/upper level set of v Ω . Remark 6. If |A| C − (B R , γ), then v Ω,A,γ 0, and Ω v Ω,A,γ 0. Below we improve the bound |A| ( γ 4m ) m ω m R m in (5) for Ω v Ω,A,γ 0 to hold. The Proceedings of the London Mathematical Society is wholly owned and managed by the London Mathematical Society, a not-for-profit Charity registered with the UK Charity Commission. All surplus income from its publishing programme is used to support mathematicians and mathematics research in the form of research grants, conference grants, prizes, initiatives for early career researchers and the promotion of mathematics.