Symmetric double bubbles in the Grushin plane

We address the double bubble problem for the anisotropic Grushin perimeter $P_\alpha$, $\alpha\geq 0$, and the Lebesgue measure in $\mathbb R^2$, in the case of two equal volumes. We assume that the contact interface between the bubbles lays on either the vertical or the horizontal axis. Since no regularity theory is available in this setting, in both cases we first prove existence of minimizers via the direct method by symmetrization arguments and then characterize them in terms of the given area by first variation techniques. Angles at which minimal boundaries intersect satisfy the standard 120-degree rule up to a suitable change of coordinates. While for $\alpha=0$ the Grushin perimeter reduces to the Euclidean one and both minimizers coincide with the symmetric double bubble found in [Foisy et al., Pacific J. Math. (1993)], for $\alpha=1$ vertical interface minimizers have Grushin perimeter strictly greater than horizontal interface minimizers. As the latter ones are obtained by translating and dilating the Grushin isoperimetric set found in [Monti Morbidelli, J. Geom. Anal. (2004)], we conjecture that they solve the double bubble problem with no assumptions on the contact interface.

where {E i : i = 1, . . . , m} is an m-bubble cluster and P P is given by For m = 1, (P) is the isoperimetric problem. When M is either the Euclidean space, the n-dimensional sphere S n or the hyperbolic space, endowed with the Riemannian perimeter and volume, minimizers are known to be metric balls, see [27].
Regarding the case m ≥ 2, Plateau experimentally established in [25] that soap films are made of constant mean curvature smooth surfaces meeting in threes along an edge, the so-called Plateau border, at an angle of 120 degrees. These Plateau borders, in turn, meet in fours at a vertex at an angle of arccos(− 1 3 ) ≃ 109.47 degrees (the tetrahedral angle). Existence and regularity of minimizers of (P) in the Euclidean setting (R n , P, L n ) were proved by Almgren in his celebrated work [1]. Here, P denotes the standard De Giorgi perimeter and L n the n-dimensional Lebesgue measure. Plateau's observations were rigorously confirmed by Taylor in [28] (a proof of the same result in higher dimensions n ≥ 4 was announced in [29]). The case n = 2 was treated separately in [22].
When m = 2, (P) is the double bubble problem. This is the type of problem that we address in this paper for an anisotropic perimeter, called the Grushin perimeter, see Section 1.2. In the Euclidean setting, the natural candidate solution is the so-called standard double bubble given by three (n − 1)-dimensional spherical cups intersecting in an (n − 2)-dimensional sphere at an angle of 120 degrees (for equal volumes v 1 = v 2 , the central cup is indeed a flat disc). The first proof of this result for n = 2 was given in [10] exploiting the analysis carried out in [22]. A second proof appeared in [8]. The case n = 3 was established first in [13] for equal volumes and then in [14] with no restrictions. The case n ≥ 4 was finally solved in [26].
The double bubble problem has been addressed also in other spaces. For M = S n the problem was completely solved for n = 2 in [17], while for n ≥ 3 only partial results are available, see [5,7]. The double bubble problem was completely solved on the 2dimensional boundary of the cone in R 3 , where minimizers are either two concentric circles or a circle lens, see [15], and on the flat 2-torus, where five types of minimizers occur, see [6].
For m ≥ 3, problem (P) is still unsolved even in the Euclidean case and presents several interesting open questions, see [16,Part IV]. The case M = R 2 , m = 3 was solved in [30]. For a detailed review on minimal partition problems, see [16,23].
1.2. Our setting. In this paper, we address problem (P) for m = 2, where M = R 2 , V = L 2 is the 2-dimensional Lebesgue measure and, for α ≥ 0, P = P α is the Grushin α-perimeter given by P α (E) = sup E (∂ x ϕ 1 + |x| α ∂ y ϕ 2 ) dxdy : ϕ 1 , ϕ 2 ∈ C 1 c (R 2 ), sup the Grushin isoperimetric set. This is E α = {(x, y) ∈ R 2 : |y| ≤ ϕ α (|x|), |x| ≤ 1}, where the profile function ϕ α : [0, 1] → [0, +∞) is given by See also [12] for a generalization to higher dimensional Grushin structures. We remark that a regularity theory for almost minimizers of the α-perimeter is not yet available. We refer to [18,20,21] for some partial results in the strictly related setting of Heisenberg group. For this reason, the case m = 2 in problem (P) cannot be addressed in full generality for the Grushin perimeter following the approach of [10,22]. As a matter of fact, a candidate solution of the double bubble problem in the sub-Riemannian setting has not been proposed yet. In this paper we study possible configurations and formulate a "standard double bubble conjecture" in this context. To this purpose, we study the case of two equal volumes and we assume that the contact interface between the two bubbles is contained in one of the two coordinate axes.

The problem for vertical interfaces. For any
For a given v > 0, we define the class of admissible sets as The first problem that we treat is where, for any E ∈ A x (v), we let P x α (E) = 1 2 P α (E +x ) + P α (E −x ) + P α (E) ∈ [0, +∞]. (1.6) When α = 0, we simply write P x 0 = P x .

Main results.
Existence of minimizers to problems (1.5) and (1.9) is proved in Theorems 4.2 and 5.2 respectively. Our approach to characterize them is based on a rearrangement technique which has its own interest (see Section 3 and Theorems 4.1, 5.1) which exploits the existence of a change of coordinates that transforms the Grushin plane (R 2 , P α , L 2 ) in the transformed plane (R 2 , P, M α ). Here, P is the standard Euclidean perimeter and M α is a weighted area, see Section 2.2. Thanks to this rearrangement we deduce symmetry of minimizers, which yields a complete characterization of the minimal double bubbles with constrained vertical and horizontal interface by a first variation argument (see Theorems 4.4,5.4). Our main results are resumed in the following theorem.  (1.9), then, up to vertical translations, we have Both double bubbles with constrained interface in (R 2 , P α , L 2 ) consist of three smooth curves with constant mean α-curvature (see Figures 3,4,6,7). Even though they do not satisfy the 120-degree Plateau's rule in the Grushin plane, they do satisfy it in the transformed plane (R 2 , P, M α ), see Sections 4.2.3, 5.2.3. This new phenomenon gives some insights about the possible structure of the singular set of minimizers in this sub-Riemannian context.
In conclusion, by comparing the two minimal double bubbles in the case α = 1, see Remark 5.7, we obtain a candidate solution to the general double bubble problem for the Grushin perimeter: the configuration with vertical interface has perimeter strictly greater than the one with horizontal interface. This establishes a connection with the standard double bubble in the Euclidean setting: in fact, the minimal double bubble with horizontal interface is obtained by translating and dilating the Grushin isoperimetric set found in [19], similarly to the Euclidean case. This leads us to conjecture that this configuration may solve the double bubble problem with no assumptions on the contact interface.
2. Preliminaries on the Grushin perimeter 2.1. Representation formulas. We start recalling some representation formulas for the α-perimeter that we will use in the sequel. Let E ⊂ R 2 be a bounded open set with Lipschitz boundary. Then The sets E ϕ and E ψ have finite α-perimeter in the cylinders D ×R and R×D respectively. Moreover, formula (2.1) implies that and 2.2. Transformed plane. As observed in [19], there exists a change of coordinates that allows us to identify the Grushin plane (R 2 , P α , L 2 ) with the transformed (Grushin) plane (R 2 , P, M α ), where P is the Euclidean perimeter and M α is a weighted 2-dimensional Lebesgue measure. Precisely, consider the functions Φ, Ψ : Clearly, the functions Φ and Ψ are homeomorphisms with Φ −1 = Ψ and, for any ξ = 0, | det JΦ(ξ, η)| = |(α + 1)ξ| − α α+1 . By [19,Proposition 2.3], for any L 2 -measurable set E ⊂ R 2 , the transformed set F = Ψ(E) satisfies where c(α) > 0 is a constant depending only on α ≥ 0. The equality in (2.6) is achieved on the Grushin isoperimetric set where the isoperimetric profile ϕ α : [0, 1] → [0, r α ] is given by (1.2) and we let r α = ϕ α (0). The isoperimetric set is unique up to dilations and vertical translations. Observe that the isoperimetric profile satisfies for all x ∈]0, 1[ and that We remark that the boundary of the isoperimetric set E α is not smooth. Precisely, if α ∈ N, then ∂E α is C α+1 but not C α+2 around the y-axis.
2.4. Additional terminology. We conclude this section introducing some additional terminology. We say that a set is an open interval for every t ∈ R. Moreover, the set E is y-Schwarz symmetric (respectively, x-Schwarz symmetric) if it is both y-symmetric and y-convex (respectively, x-symmetric and x-convex). We denote by S x the class of L 2 -measurable, x-symmetric sets in R 2 and by S * y the class of L 2 -measurable and y-Schwarz symmetric sets in R 2 . The classes S y and S * x are analogously defined.

Proof.
Since E is measurable, the horizontal section E y t is measurable for a.e. t ∈ R, so that the functions λ E , ϕ E : R → [0, +∞] are well-defined a.e. Moreover, since E ∈ S * y , we have E y t = E y −t for every t ∈ R and E y t ⊂ E y s for every 0 < s < t. Thus λ E (respectively, ϕ E ) is equivalent to a function even on R and non-increasing (respectively, non-decreasing) on ]0, +∞[. Since a monotone function can only have countably many discontinuities in its domain, the conclusion follows.

Rearrangement and first properties.
Let E ⊂ H be a measurable set such that E ∈ S * y . Let λ E , ϕ E : R → [0, +∞] be the functions given by Lemma 3.2. Let dom ϕ E be the set of points where ϕ E is finite. By Definition 3.1, it is enough to work on dom ϕ E . Let D = {d k : k ∈ N} ⊂]0, +∞[ be the set of discontinuity points of the function ϕ E in dom ϕ E . For all k ∈ N, we set We define the function τ E : R → [0, +∞] as Note that τ is even on R, non-decreasing and left-continuous on ]0, +∞[, and such that τ E (t) ≤ ϕ E (t) for t ∈ dom ϕ E . We thus define the horizontal rearrangement of E as It is easy to see that E ⋆ ⊂ H is measurable and such that E ⋆ ∈ S y . The following result shows that the rearrangement defined in (3.3) does not modify x-convex sets.
Proof. Up to a modification of E on a negligible set, we can directly assume that see [16,Lemma 14.6]. Therefore, we just need to prove that τ E (y) = 0 for every y ∈ R.
To do so, let 0 < s < t and note that E y t ⊂ E y s because E ∈ S * y . By (3.4), this means that Recalling (3.1) and the definition of τ E in (3.2), this concludes the proof.

Approximation lemma and elementary inequalities.
In the proof of the main result of this section, Theorem 3.6 below, we will need the following approximation result. See also [ Proof. Since L 2 (E) < +∞, it is not restrictive to assume that E is bounded, see [16,Remarks 13.12]. Let us setẼ = (x, y) ∈ R 2 : (|x|, y) ∈ E , the symmetrization of E with respect to the y-axis. It is immediate to see thatẼ ∈ S x ∩ S * y is bounded with P (Ẽ) < +∞ and P (Ẽ; ∂H) = 0. By [16,Theorem 13.8], there exists a sequence (Ẽ k ) k∈N of bounded open sets with smooth boundary such that as k → +∞. Arguing as in the proof of [11, Lemma 2.1], one can prove that (Ẽ k ) k∈N ⊂ S x ∩ S * y . Now let us set E k =Ẽ k ∩ H + for each k ∈ N. Since P (Ẽ) = 2P (E; H + ) and P (Ẽ k ) = 2P (E k ; H + ) by symmetry, from (3.5) we deduce that as k → +∞. We claim that P (E k ) → P (E) as k → +∞. Indeed, by (3.6), it is enough to show that P (E k ; ∂H) → P (E; ∂H) (3.7) as k → +∞. Recalling (3.6), the limit in In the proof of Theorem 3.6 below, we will need the following elementary inequalities, which we prove here for the reader's convenience.
and thus, in particular, Therefore, by Cauchy-Schwarz inequality, we get we deduce (3.9).

Rearrangement theorem.
We are now ready to prove the main result of this section. The argument follows the strategy outlined in the proof of [16,Theorem 14.4].
Moreover, equality holds in (3.10) if and only if E y t is equivalent to an interval for a.e. t ∈ R, in which case E ⋆ = E up to negligible sets.
Proof. We divide the proof in two steps.
Step 1. Let us assume that E is a bounded open set in R 2 with polyhedral boundary and that the outer unit normal to E (that is elementarily defined at H 1 -a.e. point of ∂E) is never orthogonal to e 1 . By this assumption and by the implicit function theorem, we get that Here {I ± h } h=1,...,M is a finite family of non-overlapping bounded intervals such that Figure 1. If E is a bounded open set with polyhedral boundary, with outer unit normal never orthogonal to e 1 , then ∂E is parametrized as in (3.11). In particular, These two properties, guaranteed by (3.11), imply the continuity of λ E ,ũ andṽ.
Note that λ E is a continuous function. The function ϕ E :

symmetric interval centered at the origin and
The functionsũ,ṽ are affine, even, non-negative and continuous functions on I such thatũ ≤ṽ andũ (respectively,ṽ) is non-decreasing (respectively, non-increasing) on In particular, we get that orh = 0 if the condition in brackets is empty. Then, we geth ≥h, and hence see Figure 2. We now prove that P (E ⋆ ) ≤ P (E). We have By inequality (3.9) in Lemma 3.5, we obtain that The inequality above allows us to deduce some more precise information. Let D ⊂ I be the set of those t ∈ I such that Since clearly it holds by Cauchy-Schwarz inequality we deduce that In addition, from (3.13) we deduce that We can also obtain a stronger inequality. Indeed, by (3.12) and (3.14), we have Then, sinceh ≥h, arguing as before we get that Step 2. Let E ⊂ H be a measurable set such that L 2 (E) < +∞, P (E) < +∞ and E ∈ S * y . By Lemma 3.4, there exists a sequence (E k ) k∈N of bounded open sets with polyhedral boundary such that (3.16) and, as k → +∞, Without loss of generality, we can assume that for every k ∈ N the outer unit normal to E k is never orthogonal to e 1 while keeping (3.16) and (3.17). Let us set and D k ⊂ I k the set of those t ∈ I k such that (E k ) y t is not an interval. Applying step one to each E k , we get and, by the lower semicontinuity of the perimeter, Moreover, by Fatou's lemma, we have By the one-dimensional isoperimetric inequality, P (E y t ) ≥ 2 for a.e. t ∈ R. Therefore, P (E y t ) = 2 for a.e. t ∈ I. By [16,Proposition 12.13], for every such t, E y t is equivalent to an interval. Thanks to Lemma 3.3, this concludes the proof.
3.5. Boundary regularity. We conclude this section showing that x-convex and y-Schwarz symmetric sets are bounded and have Lipschitz boundary.
Thus, by (3.22) and [3, ThenF ∈ S * x ∩ S * y . Arguing as in the proof of [19, Theorem 3.1] (see also [12, Section 5.1]), we conclude that ∂G and ∂F are both union of the image of four Lipschitz curves. As a consequence, ∂F is piecewise Lipschitz. This proves that ∂E is piecewise Lipschitz and the proof is complete.

Existence of minimizers with vertical interface.
4.1.1. Reduction to more symmetric sets. In this section we prove the existence of solutions to problem (1.5). The following result restricts the class of admissible sets to more symmetric ones. For the notation we refer to Section 2.4 Proof. We split the proof in several steps. To avoid heavy notation, during the proof we will omit the x-superscript for the sets E ±x defined in (1.3).
Since any open convex set with finite perimeter is bounded, the set G 2 is bounded and thus L 2 (G 2 ) < +∞, which immediately implies that L 2 (F 2,+ ) < +∞ as claimed.
We can thus apply Theorem 3.6 to the set F 2,+ . We define with equality if and only if F 2,+ is equivalent to a ξ-convex set. Therefore P ξ (F 3 ) ≤ P ξ (F 2 ) with equality if and only if F 2,+ is equivalent to a ξ-convex set. We claim that M α (F 3,+ ) ≥ M α (F 2,+ ). Indeed, by the definition of the horizontal rearrangement in (3.3), we have that On the other hand, since L 1 (A) = L 1 (B) and ess inf(A \ B) ≥ ess sup(B \ A), we have that Thus, by Tonelli's theorem and (4.2), we conclude that M α (F 3,+ ) ≥ M α (F 2,+ ) as claimed.
In conclusion, setting , with strict inequality if and only if E 2,+ is not equivalent to a x-convex set.

Existence of minimal double bubbles with vertical interface.
We are now ready to prove the existence of minimizers to problem (1.5).

Theorem 4.2 (Existence of minimizers with vertical interface). Let α ≥ 0 and fix
Proof. We study the existence of solutions by the direct method of the calculus of variation. By Theorem 4.1, the class of admissible sets can be restricted to x-transformed-convex, as in (4.1) . By the isoperimetric inequality (2.6), for any E ∈ B x (v) we have that P x α (E) ≤ k for some constant k = k(α, v) ∈]0, +∞[ depending only on α ≥ 0 and v > 0.
We claim that any Thus, by the representation formula (2.2), we get α + 1 and thus 2a α+1 ≤ (α + 1)k. In particular, the convex set F +ξ = Φ(E +x ) is bounded and contained in [0,ā] × R for someā depending on a and α. Let Then, by convexity, we get k ≥ P (F +ξ ) ≥ ā 2 + 2b 2 , which immediately implies that b depends only on α, v > 0. Now, let (E h ) h∈N ⊂ B x (v) be a minimizing sequence for the problem (1.5), namely The space of function of bounded variation BV (R 2 ) is compactly embedded in L 1 loc (R 2 ) and therefore, possibly extracting a subsequence, there exists a measurable set a.e. and in L 1 (R 2 ). Letting E = Φ(F ), it follows that χ E h → χ E a.e. and in L 1 (R 2 ). Up to negligible sets, we have that E ∈ B x (v). Thus there exist r ∈]0, +∞[ and Moreover, by the lower semicontinuity of the α-perimeter, we have which implies that E is a minimum. This concludes the proof.

Regular minimizers.
In this section, we solve the minimal partition problem (1.5). In Theorem 4.2, we proved the existence of a minimizer E ∈ A x (v) ∩ S x ∩ S * y that is bounded, x-transformed-convex and of the form for some x-profile function f ∈ C([0, r]) ∩ Lip loc (]0, r[) with r ∈]0, +∞[. We call such a minimizer a x-regular minimizer. If E is a x-regular minimizer, then where P x α is defined in (1.  Proof. Let E ∈ A (v) be a minimizer of problem (1.5). Consider the set E * = (x, y) ∈ R 2 : (|x|, y) ∈ E +x .
. By the minimality of E, we must have P x α (E * ) = P x α (E). Thus E * is also a minimizer of problem (1.5) and it is x-symmetric, so E * = E 0 up to vertical translations. In particular, E +x = E +x 0 up to vertical translations. With a similar argument, we also get E −x = E −x 0 up to vertical translations. Since E 0 is x-symmetric, this implies that E = E 0 up to vertical translations.

Characterization and examples.
We are now ready for the main result of this section. In Theorem 4.4 below, we prove that, given α ≥ 0 and v > 0, the x-regular minimizer of problem (1.5) is unique and has smooth boundary far from the y-axis. By Lemma 4.3 and Section 4.2.1, this is the unique minimizer of problem (1.5) up to vertical translations.
Step 1: differential equation for the profile. We perform a first variation argument. Let Then, by (4.8) and the minimality of E, after an integration by parts we find Thus, by the fundamental lemma of the Calculus of Variations, there exists a constant k ∈ R such that d dx for all x ∈]0, r[. Integrating, we get for all x ∈]0, r[, which implies |kx + d| < 1 for all x ∈]0, r[. In particular, we deduce that |d| ≤ 1 and for all x ∈]0, r[. As a consequence, f ∈ C ∞ (]0, r[). By the regularity theory of Λ-minimizer of perimeter, the boundary ∂E is smooth far from the y-axis. Therefore, we must have By the expression of f ′ in (4.14), this implies that k = 0 and kr + d = −1. As a consequence, since r > 0 and |d| ≤ 1, we get k < 0 and d ∈] − 1, 1].
Step 2: proof of f (0) > 0. Passing to the limit in (4.14) as x → 0 we must have that Assume that f (0) = 0 by contradiction. We distinguish two cases.
We then consider the set We thus define and the claim follows from the Taylor's expansion of the function ε → P x α (F ε ). But this contradicts the minimality of E.
Step 4: characterization of the profile. By the expression of f ′ in (4.14), for x ∈ [0, r] we can compute using the information kr + 1 2 = −1 found in step 1.
by (4.10). This is the profile function of a circle of radius 1 |k| and center (0, 1 2|k| ). In particular, we have f ′ (0) = 1 √ 3 and the angle γ = arctan f ′ (0) is given by see Figure 3. Thanks to Theorem 4.4, up to Euclidean translations, the unique minimizer of problem (1.5) for α = 0 is the symmetric double bubble found in [10]. Example 4.6 (The Grushin case). In the Grushin case α = 1, the profile function defined in (4.9) can be explicitly computed and we find by (4.10), see Figure 4. Thanks to Theorem 4.4, up to vertical translations, this is the x-profile function of the unique minimizer of problem (1.5) for α = 1.  Figure 5. In the Euclidean case α = 0, we have ϑ = π 6 , as we found in Example 4.5 accordingly to the well-known regularity theory. In the Grushin case α > 0, instead, we have ϑ = 0, which means that the minimizer of problem (1.5) has a C 1 -boundary consisting of two symmetric curves joining the vertical interface at two triple points with right angles. However, if we transform the Grushin plane (R 2 , P α , L 2 ) into the Euclidean plane with weighted volume (R 2 , P, M α ) using the maps defined in (2.4), then the set F = Ψ(E) has ξ-profile functionf : [0,r] → [0, +∞[ given bŷ An elementary computation shows that the profile angle at the interface in the transformed plane is given byγ = π 2 +θ, wherê In other words, the problem (1.5) reformulated in the transformed plane (R 2 , P, M α ) has a unique minimizer consisting of two symmetric curves joining the vertical interface at two triple points with angles 2π 3 . 5. The double bubble problem for horizontal interfaces 5.1. Existence of minimizers with horizontal interface.
5.1.1. Reduction to more symmetric sets. In this section we prove the existence of solutions to problem (1.9). The following result restricts the class of admissible sets to more symmetric ones.
x ∩ S y , bounded and y-transformed-convex, such that P y α (Ẽ) ≤ P y α (E). Moreover, in the case α > 0, E ∈ S y is equivalent to a y-transformed-convex set if and only if P y α (Ẽ) = P y α (E). In addition, there exist r ∈]0, +∞[ and g ∈ C([0, r]) ∩ Lip loc (]0, r[) such thatẼ = (x, y) ∈ R 2 : |x| ≤ g(|y|), |y| ≤ r . (5.1) Proof. The proof is similar to the one of Theorem 4.1 and we just sketch it. To avoid heavy notation, during the proof we will omit the y-superscript for the two sets E ±y defined in (1.7).
Step 1: y-symmetrization. It is not restrictive to assume that P α (E + ) ≤ P α (E − ) (otherwise, we can reflect E with respect to the x-axis). We thus define Clearly, E 1 ∈ A y (v) is y-symmetric. As in step 1 of the proof of Theorem 4.1, one can prove that P x α (E 1 ) ≤ P x α (E).
We can now conclude as in step 4 and step 5 of the proof of Theorem 4.1 with minor modifications. We leave the details to the reader.

Existence of minimal double bubbles with horizontal interface.
We are now ready to prove the existence of minimizers to problem (1.9).
Thus, by the representation formula (2.3), we get and thus 2b ≤ k. In particular, the convex set F +η = Φ(E +y ) is bounded and contained in R × [0,b] for someb depending on b and α. Let Then, by convexity, we get k ≥ P (F +η ) ≥ 2ā 2 +b 2 , which immediately implies that a depends only on α, v > 0. The proof can now be concluded similarly to the one of Theorem (4.2), with minor modifications. We leave the details to the reader.

Regular minimizers.
In this section, we solve the minimal partition problem (1.9). In Theorem 5.2, we proved the existence of a minimizer E ∈ A y (v) ∩ S x ∩ S * y that is bounded, y-transformed-convex and of the form for some y-profile function g ∈ C([0, r]) ∩ Lip loc (]0, r[) with r ∈]0, +∞[. We call such a minimizer a y-regular minimizer. In case of a y-regular minimizer E, the functional (1.10) takes the form

Characterization and examples.
We are now ready for the main result of this section. In Theorem 5.4 below, we prove that, given α ≥ 0 and v > 0, the y-regular minimizer of problem (1.9) is unique and has smooth boundary far from the x-axis. By Lemma 5.3 and and Section 5.2.1, this is the unique minimizer of problem (1.9) up to vertical translations.
Note that E ε ∈ A y (v) for all ε ∈ R small, since g(y) > 0 for every y ∈]0, r[ by definition. Then, by (5.3) and the minimality of E, after an integration by parts we find Thus, by the fundamental lemma of the Calculus of Variations, there exists a constant c ∈ R such that for all y ∈]0, r[. In addition, by the regularity theory of Λ-minimizers of perimeter, the boundary ∂E is smooth far from the y-axis. Therefore we must have g ∈ C ∞ (]0, r[).
Step 2: Performing a first variation argument and arguing as in the proof of [19, Theorem 3.2], we find a constant b < 0 such that for all y ∈]f (δ), r[. Inserting (5.10) in (5.7), we get b = c.
Step 3: g has a strict maximum in ]0, r[. Define G : ]0, r[→ R by setting for all y ∈]0, r[. The differential equation in (5.7) can be rewritten as for all y ∈]0, r[. Combining (5.11) and (5.13), we get g(y) α g ′ (y) for all y ∈]0, r[. Thus g ′ can change sign at most one time on the interval ]0, r[, because the function appearing on the right-hand side of (5.14) is strictly monotone since g(y) > 0 for all y ∈]0, r[. By contradiction, assume that g ′ does not change sign on ]0, r[, so that, by (5.9), g ′ (y) < 0 for all y ∈]0, r[. Therefore, by the implicit function theorem, we can extend the function f : [−δ, δ] → [0, +∞[ to the interval [−g(0), g (0)]. Note that we must have that g(0) > 0, since g(r) = 0 by (5.9) and since we are assuming that g ′ < 0 on ]0, r[. We can thus repeat the argument contained in step 2 for all x ∈ [0, g(0)] and y ∈ [0, r] and deduce that passing to the limit in (5.10) as y → 0, because c < 0 as we found in step 2.
Note that E ε ∈ A y (v) for all ε ∈ R small, since g(y) > 0 for every y ∈ [0, r[. Then, by (5.3) and the minimality of E, we find By the arbitrariness of ψ, we deduce that Therefore we must have g ′ (0) > 0, contradicting (5.15). We conclude that g ′ must change sign exactly one time on the interval ]0, r[. So there must be a pointŷ ∈]0, r[ such that g ′ (ŷ) = 0, g ′ (y) > 0 for y ∈]0,ŷ[ and g ′ (y) < 0 for y ∈]ŷ, r[. In particular, g has a strict maximum point at y =ŷ.
Note that We thus define Then F ε ∈ A y (v) and P y α (F ε ) = λ α+1 ε P y α (E ε ). We claim that P y α (F ε ) < P y α (E) for any ε > 0 sufficiently small. A direct computation gives and the claim follows from the Taylor's expansion of the function ε → P y α (F ε ). But this contradicts the minimality of E.
In other words, the problem (1.5) reformulated in the transformed plane (R 2 , P, M α ) has a unique minimizer consisting of two symmetric curves joining the vertical interface at two triple points with angles 2π 3 . Remark 5.7 (Comparison of the two minimal bubbles for α = 1). Having in mind the general problem (P) for m = 2, we want to understand which of the two double bubbles characterized in Theorems 4.4 and 5.4 may be a candidate solution to the double bubble problem for equal areas in the Grushin plane (R 2 , P α , L 2 ). Since the expressions of the values of problems (1.5) and (1.9) given in (4.11) and (5.6) are not easily comparable for an arbitrary α > 0, we restrict our analysis to the case α = 1.
Besides its computational manageability, the case α = 1 is of particular interest since it is connected with the Heisenberg group H 1 . This is the framework of the famous Pansu's conjecture about the shape of isoperimetric sets, see [24], which is still unsolved. Pansu's set can be obtained by rotating the set E α in (2.7) for α = 1 around the vertical axis in R 3 . Our analysis might give some insights on the candidate solutions to the double bubble problem in H 1 .
Now let E x and E y be the minimal bubbles given by Theorems 4.4 and 5.4 respectively. From (4.11) and Example 4.6 we get