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Symmetry results for Serrin-type problems in doubly connected domains

In this work, we employ the technique developed in arXiv:2109.11255 to prove rotational symmetry for a class of Serrin-type problems for the standard laplacian. We also discuss in some length how our strategy compares with the classical moving plane method.


Introduction and statement of the main results
In his pioneering paper [35], Serrin refined Alexandrov's method of moving planes [6] and employed it to prove that a function defined in a bounded domain having constant nonzero laplacian and satisfying constant Dirichlet and Neumann boundary condition is necessarily rotationally symmetric. In fact, in [35] it was shown that similar results hold more generally for a vast class of elliptic PDEs. Serrin's result and the method of moving planes have given rise to a prolific field of research on overdetermined boundary value problems. Nowadays Serrin's technique has been employed to study analogous problems in a number of work, among which we mention [8,14,23,27]. An alternative proof of Serrin's result was given by Weinberger [38] using a completely different method, essentially based on a comparison argument with the model solutions. Weinberger's method and in particular its use of a P-function has also experienced quite some success [16,18,19,22,34]. For further insights on these methods and their applications, see also [28] and references therein.
The above mentioned papers all gave rise to interesting characterizations of solutions supported in a ball. In this work, we are instead interested in characterizations of solutions supported in an annulus. Let us explain our setting in more details. Let E i ⋐ E o ⊂ R 2 be two nonempty simply connected bounded domains, whose boundaries ∂E o = Γ o and ∂E i = Γ i are simple closed smooth curves, and let Ω = E o \ E i . By contruction, Ω is a bounded domain that is not simply connected, with boundary ∂Ω = Γ i ⊔ Γ o .
We are interested in studying pairs (Ω, u), where Ω has the above structure and u : Ω → R is a solution to the following problem where a, b, α, β are constants and ν is the unit-normal pointing outside Ω. This overdetermined boundary value problem has been well studied, especially in the case where a > b and α ≥ 0. Under this assumption, the method of moving planes shows its strength and allows to prove some powerful rigidity results. A particularly nice one is the following, due to Reichel.
Then Ω is an annulus and u is rotationally symmetric with ∂u/∂|x| < 0.
It is worth mentioning that Reichel's result is slightly stronger, as it also works for more general PDEs and it allows E i (and thus Γ i ) to be disconnected. In [36], Sirakov generalized previous results [1,5,40] and proved that Reichel's thesis remains true if one allows different values of a and α on different connected components of Γ i . He was also able to replace the hypothesis u < a with the weaker α ≥ 0. We also remark that similar symmetry results have been proven to hold for more general families of elliptic operators. In [36] it is shown that the same result works for quasi-linear regular strongly elliptic operator. Similar results also hold for quasi-linear possibly degenerate elliptic operators [5], fully nonlinear operators [32] and for the fractional laplacian [37]. For further symmetry results on this and related problems, see also [7,16,17,29,33].
On the other hand, it seems that less attention has been brought to the case a < b. One possible explaination is that the moving plane method works well when the model solution are monotonically decreasing, but seems to be harder to employ when the model solutions are monotonically increasing. In this work, our aim is to discuss this case and to prove the analogue of Theorem 1.1, namely: Then Ω is an annulus and u is rotationally symmetric with ∂u/∂|x| > 0.
The strategy of the proof is based on the method developed in [2]. There, problem (1.1) has been studied in the case a = b = 0, using a new comparison strategy. Of course, under that hypothesis, monotonicity is lost and there are some complications, especially concerning the behaviour of the solution near the set of the maxima of the function u. However, we will show that the core of the method employed in [2] is perfectly suited to study the case discussed in this paper. In fact, we can adapt most of the argument in [2] to prove Theorem 1.2.
Perhaps surprisingly, our method proves to be less effective in the more studied case b < a. We will try to explain along the work why this is the case. Nevertheless, we will be able to employ our method to prove a slightly weaker version of Theorem 1.1 (namely, we will need to assume the additional condition 2a + α 2 ≤ 2b + β 2 , see Theorem 4.1). As we will see, the proof is also less elementary than the one for Theorem 1.2. It does seem then that the moving plane method is still the one best suited to characterize rotationally symmetric solutions that are radially decreasing. Nevertheless, our alternative proof may be of interest: it is novel and it expands on our method in ways that may find further applications in the future.
Let us conclude with some final comments on the above results and on future applications. First of all, one natural question would be whether similar characterizations can be achieved for model solutions that are not monotonic along the radial coordinate. In this case, the situation is less clear as there are negative results: in [25] it is proven that there exist solutions to (1.1) with α = β that are not rotationally symmetric. In the case a = b = 0, there are both positive ([2, Theorem B and Theorem C]) and negative results ([2, Theorem A]). Another natural question concerns whether our method can be applied to more general elliptic equations (in the same way as the Weinberger's comparison method [38] was then developed in [18,19,22]) or to other problems in Riemannian geometry. In fact, the most part of our arguments (with the notable exception of the Pohozaev identity) do not rely on the structure of the Euclidean space at all. As a couple of examples of further applications, this method has proven to be quite successful to characterize static spacetimes in General Relativity [10,11,12,13] and somewhat similar techniques have been employed for other problems in General Relativity [4,21] and for p-harmonic functions in manifolds with nonnegative Ricci curvature in [3,20].
The paper is structured as follows. In Section 2 we show that for any acceptable choice of a, b, α, β there is a rotationally symmetric solution solving problem (1.1). This is necessary in order to start our method, as we need to select a model solution to compare with. Section 3 is the heart of the paper: a crucial gradient estimate (Theorem 3.1) is introduced and exploited to obtain some important area bounds for the boundary components (Proposition 3.2). In Subsection 3.3 these two results are then used, in combination with a divergence theorem argument, to prove Theorem 1.2. Unfortunately, the method described in Subsection 3.3 does not work to prove Theorem 1.1. In Section 4 we then develop a new alternative argument, based on a combination of the Pohozaev identity and the isoperimetric inequality, leading to the proof of Theorem 4.1, which is a weaker version of Theorem 1.1. Finally, in Section 5 we further comment on the complications that one has to deal with when studying the case a > b, trying to analyze where they come from and how one may try to overcome them. Ultimately, the purpose of this final section is that of understanding what are the limits of our method and how far they can be pushed.

Setting up the comparison argument
Our first aim is that of understanding whether there are some relations that are always in place between the constants a, α, b, β. The main tool that will help us in this regard is the well known Pohozaev identity [30]. In our framework, this formula has the following form is the position vector and ν is the exterior normal to ∂Ω. Recalling that div X = 2 and using the divergence theorem, we deducê From this formula we deduce some relations between a, b, α, β, summarized in the following result.
As a consequence, we can rewrite (2.1) in the following two formŝ If b < u < a in Ω, then the left hand side of the first equation is negative, whereas if a < u < b in Ω, then the left hand side of the second equation is negative. In both cases, we deduce 4a+α 2 > 4b+β 2 . The signs for α and β follow immediately from the hypothesis that a is the maximum (resp. minimum) of u and b is the minimum (resp. maximum) of u. The fact that β = 0 when b < u < a and that α = 0 when a < u < b is a consequence of the Hopf Lemma (recall ∆u = −2 < 0).
In order to start our comparison argument, we first need to take a close look at the model solutions. In particular, we first have to make sure that we always have a model solution to compare with, for any acceptable value of a, b, α, β.
It can be easily checked that the rotationally symmetric solutions to ∆u = −2 have the form with L, M ∈ R. When M ≤ 0, these solutions are monotonically decreasing with respect to |x|. When M > 0, these solutions are monotonically increasing for |x| < √ M and they are monotonically decreasing for |x| > √ M .
Remark 1. As in [2], one can fix the value of L by means of a rescaling of the function and of the domain, ending up with the one-parameter family However, for our purposes in this paper this is not necessary, as it is actually easier to work directly with (2.2).
We expect that for any acceptable choice (according to Proposition 2.1) of a, b, α, β, there should be a solution of the form (2.2) solving our problem (1.1). This is true in the case a < b, as guaranteed by the following lemma: Proof. The proof of the lemma is more of a long exercise. We start by imposing the conditions we want on L, M, r o , r i , namely and we employ the last two equations to obtain Since we are assuming α < 0 and β ≥ 0, in order for r o , r i to be positive, we need to choose the plus sign in (2.4) and we need M > 0. Substituting in the first two equations, we then deduce that we have a solution if and only if we can find a zero for the function Taking the limit at zero and infinity, we compute It follows by continuity of F that there is a value 0 < M < ∞ such that F (M ) = 0, as wished.
The case a > b appears to be more complicated. In the next lemma we will show that we can still find a model to compare with as long as we assume the additional condition 2a+α 2 ≤ 2b+β 2 . This is the first instance in which we notice complications in the case a > b. More specifically, it seems that negative values of M are harder to deal with. In this respect, the hypothesis 2a Proof. Proceeding as in the proof of Lemma 2.2, we find again the expressions (2.4) for r o and r i . Notice that, in contrast with the previous case, both signs for ± in (2.4) are possible. However, since we are ultimately interested in having M ≥ 0, this again essentially forces us to choose the plus sign in the formulae (2.4) for r i and r o . Therefore, as in the previous lemma, we reduce ourselves to finding a zero for the function Then we compute Therefore, since we are assuming 4a + 2α 2 − 4b − 2β 2 ≤ 0, we conclude again that there is a value 0 ≤ M < ∞ such that F (M ) = 0.

Exploiting the comparison argument
In the previous section we have developed a way to compare a general solution of (1.1) with a model solution (2.2). Building on that, in this section we develop our comparison technique. In Subsection 3.1 we prove a crucial gradient estimate, that is then exploited in Subsection 3.2 to prove area bounds for the boundary components Γ i and Γ o . The results in both these subsection work in both the case a < b and b < a. In Subsection 3.3 we will then specialize to the case a < b and we will prove Theorem 1.2. From now on, the constants L ∈ R, M ≥ 0, r o > 0, r i > 0 will always be the ones prescribed by Lemmata 2.2 and 2.3. We are now ready to set up our comparison argument, in the spirit of [2]. We start by defining the pseudo-radial function Ψ : Ω → R implicitly via the following identity From the Implicit Function Theorem, it is easily seen that Ψ is well defined as long as Ψ 2 = M inside Ω. From now on we will always assume that a < u < b or b < u < a, and from this it easily follows that the function Ψ actually takes values between r i (attained at Γ i ) and r o (attained at Γ o ). On the other hand, Lemmata 2.2 and 2.3 tell us that either r i < r o ≤ √ M or √ M ≤ r i < r o . Therefore, for all our purposes, the function Ψ : Ω → [r i , r o ] will always be well defined.
We also introduce the two functions The function W 0 corresponds to the value that |∇u| 2 would have on the model solution. In fact, it is easily checked that, for the model solutions (2.2), it holds W 0 ≡ W pointwise. Our strategy is then that of comparing W and W 0 in Ω, ultimately trying to prove that they have to coincide. First of all, notice that, by construction, W and W 0 coincide on the boundary of Ω, namely when Ψ = r o and when Ψ = r i . We will now show that, under our hypotheses, W is actually controlled by W 0 on the whole Ω.
Theorem 3.1 (Gradient Estimates). Let (Ω, u) be a solution to problem (1.1) with respect to some choice of a, α, b, β. Suppose that a < u < b or b < u < a. In the case b < u < a, assume further that 2a + α 2 ≤ 2b + β 2 . Let W , W 0 be defined as above. Then it holds  [11,12,13] for static spacetimes). This function is really helpful as it allows to have an explicit formula for W 0 . This is crucial in the computations that follows: we will see that Ψ will appear in all the important computations in this section.
Proof. The proof is essentially the same as the one in [2], but let us give some comments. Let L ∈ R, M ≥ 0, r o > 0, r i > 0 be the constants prescribed by Lemmata 2.2 and 2.3. The crucial estimate is deduced starting from the following quantity The heuristic behind the choice of the quantity above is essentially the fact that it can be computed to be zero on the model solution. Computing explicitly its square norm and using the fact that it is nonnegative, we obtain the following estimate for the hessian of u: Starting now from the Bochner formula and plugging in this estimate, we obtain an elliptic inequality for the quantity W − W 0 where we have used ′ to denote the differentiation with respect to Ψ. We now need to find a function γ such that the coefficients of the zero order terms have the right sign. A good choice is to set which corresponds to choosing We know that Ψ 2 = M inside Ω, hence γ is well defined in the interior of Ω. However, Ψ 2 may attain the value M on the boundary ∂Ω: this happens when α = 0 (which implies from (2.4) that r i = √ M ) or when β = 0 (which implies from (2.4) that r o = √ M ). Let us for the moment assume that this does not happen, so that F γ is well defined on the whole Ω. With this choice of γ, it is easily seen that F γ satisfies It remains to discuss the case where Ψ 2 = M on one of the boundary components, say Γ i (for Γ o , the same argument apply). In this case, we consider ε > 0 and a small neighborhood U ε = {x ∈ Ω : d(x, Γ i ) < ε} of Γ i , and we work on the domain Ω ε = Ω − U ε . Clearly F γ is well defined and satisfies the Maximum Principle in Ω ε . We will then obtain the desired result by taking ε → 0, provided we can prove that F γ → 0 as we approach Γ i . It is easily seen that F γ goes to zero if and only if W/ √ W 0 goes to zero, which in turn is equivalent to  Suppose that a < u < b or b < u < a. In the case b < u < a, assume further that 2a+ α 2 ≤ 2b+ β 2 . Then Proof. We will prove the area bound for Γ i only, as the one for Γ o is proven in the same way (up to some sign differences coming from the fact that ∇Ψ is pointing outside Ω on Γ o ). Let us also suppose for the moment that β = 0, so that the gradient of u does not vanish on Γ i . Notice that has norm |∇Ψ| = W/W 0 equal to 1 on Γ i and less than 1 inside Ω thanks to (3.3). In particular, the inequality ∇|∇Ψ| 2 | ∇Ψ ≤ 0 holds on Γ i and the curvature κ of Γ i at any point of Γ i satisfies Integrating κ ≥ r i on Γ i , recalling that the total curvature of a simple curve is equal to 2π, we get This concludes the proof of the area bound for Γ i . Let us now discuss the case β = 0. In this case, one has W = W 0 = 0 on Γ i , so |∇Ψ| = W/W 0 is not well defined. Let p ∈ Γ i and let us extend u to a smooth function in a neighborhood U of p. We can do it by means of Whitney Extension Theorem [39], that clearly applies to our case since u is smooth up to the boundary (see for instance [24,Theorem 6.19]). With a slight abuse of notation, let us still call u the smooth function defined on U that coincides with the original u in Ω ∩ U . In the spirit of [11, Theorem 3.1], we consider the signed distance function r to Γ i . If U is taken to be small enough, the function r is known to be smooth and can be used as a coordinate. We can then proceed exactly as in [11,Theorem 3.1] to show that inside U . From this, it easily follows that W = 4r 2 + o(r 2 ) and W 0 = 4r 2 + o(r 2 ), hence |∇Ψ| goes to 1 as we approach Γ i . We can then mimik the argument given for the case β = 0: the first equality in (3.7) still holds for the curvature of the level sets of u close to Γ i , therefore at the limit we obtain again that the curvature of Γ i is greater than or equal to 1/r i , from which one concludes.
3.3. Proof of Theorem 1.2. In this subsection, we focus on the case a < b and we prove Theorem 1.2. For simplicity, let us start by discussing the case where ∇u does not vanish on the boundary of Ω, namely for the moment let us assume β = 0 (notice that we know that α is strictly positive in the case a < b, so α cannot vanish). Using the divergence theorem and recalling that ∆u = −2, we computê where ν is the outward unit normal to ∂Ω = Γ i ⊔ Γ o . Since we are in the case a < b, then ν = −∇u/|∇u| on Γ i and ν = ∇u/|∇u| on Γ o . Setting W = |∇u| 2 as usual, and recalling the expression (3.2) of W 0 in terms of Ψ, the identity above can be written aŝ (3.10) Notice that the left hand side is nonnegative because of (3.3). Furthermore, since we are focusing on the case a < b, we have Ψ Since by construction W = W 0 on ∂Ω, from (3.10) we immediately deduce On the other hand, the area bounds in Proposition 3.2 give us the opposite inequality. Therefore the equality must hold in (3.11), and from (3.10) in particular we obtain that W ≡ W 0 in Ω. The result now follows from the rigidity statement in Theorem 3.1. It remains to discuss the case where β is equal to zero. In this case, the boundary term in (3.9) is ill-defined, as one has Ψ 2 = r 2 o = M on Γ o . To avoid this problem, we just need to apply the divergence theorem in Ω \ {u > b − ε}, and then take the limit as ε → 0. From (3.8) it follows easily that the quantity ∇u | ν /(M − Ψ 2 ) goes to 1/Ψ = 1/r o as we approach Γ o . The proof then proceeds exactly as in the case β = 0.

Refined Pohozaev
In this section we focus on the case a > b and we prove Theorem 1.1 under the additional hypothesis that 2a + α 2 ≤ 2b + β 2 . Let us give the precise statement for the convenience of the reader.
As discussed, all the arguments in Subsections 3.1 and 3.2 still work in this case. Unfortunately however, the strategy employed in Subsection 3.3 to complete the proof for a < b, does not work when a > b. In fact, one can check that in this case the inequality that one gets at the end has the opposite sign of (3.11), hence it does not combine with the area bounds from Proposition 3.2 anymore. We will then need to exploit a more delicate argument based on sharp estimates coming from the Pohozaev identity and the isoperimetric inequality.
Let us start by writing where ϕ is a function of u to be chosen later. We then use the divergence theorem to write the following sequence of identities (we denote byφ the derivative of ϕ with respect to u) where k ∈ R is a constant, then one can check that it holds (2ϕ − 4u)/φ = W 0 . Combining (4.1) with the Pohozaev identity (2.1), with this choice of ϕ we obtain Since |Ω| = |E o | − |E i |, we can write the above equation in the following way: Recall from Proposition 2.1 that 4a + α 2 > 4b + β 2 . We can then combine the isoperimetric inequality, the area bounds in Proposition 3.2 and the identity to obtain the following inequalitŷ Recall that the constants appearing in the above formula are related as follows: Using these identities and formula (4.2) for ϕ, with some computations we can rewrite the above inequality as follows: Let us now analyzeφ. Differentiating (4.2), we geṫ One can check that the quantity in square brakets is monotonically decreasing in Ψ. Since we know from Lemma 2.3 that Ψ 2 ≥ r 2 i ≥ M , in order forφ to be positive in Ω it is then sufficient to choose Choosing k exactly equal to the above value, recalling from (3.3) that W ≤ W 0 , we then get from (4.3): But we have the opposite inequality from Proposition 3.2, therefore everything must be an equality. As a consequence, W = W 0 on the whole Ω and we conclude using the rigidity statement in Theorem 3.1.

Further comments
In the previous sections we have shown that we are able to deal with the case a > b and prove Theorem 1.1 if we assume the additional hypothesis 2a + α 2 ≤ 2b + β 2 . While this hypothesis is somewhat restrictive, it still allows to get some nontrivial applications. As a particularly relevant example, we now argue that our weaker version of Theorem 1.1, namely Theorem 4.1, is still strong enough to deal with the cases of interest in [2]. There, Theorem 1.1 was invoked in the proof of [2, Theorem B] on a domain Ω o (the outer domain, in the terminology of [2]) for a function u satisfying problem (1.1) with α = 0, b = 0 and β 2 /2a ≥ 1 (the latter inequality followed from [2, Theorem 2.1]). Under those hypotheses, the inequality 2a + α 2 ≤ 2b + β 2 is trivially satisfied. In other words, Theorem 4.1 is enough for the intended applications in [2].
The rest of this section is devoted to further comment on the hypothesis 2a + α 2 ≤ 2b + β 2 . Specifically, we discuss what can go wrong when this hypothesis is not in place and what are the complications that one would need to overcome in order to prove Theorem 1.1 in full generality using our approach.
The first basic step that one needs in order to start our comparison argument is of course to have a model to compare with. In other words, given a solution (Ω, u) of problem (1.1) for some a, b, α, β with a > b, we need to have a model solution (2.2) solving the same problem. This is granted by Lemma 2.3 when 2a + α 2 ≤ 2b + β 2 . Unfortunately, an analogous result seems harder to prove when 2a + α 2 > 2b + β 2 .
In case we are able to show that there is a model solution to compare with, there is still another crucial complication, namely the proof of Theorem 3.1. In fact, in order to prove the gradient estimate W ≤ W 0 , we relied on the Maximum Principle applied to the elliptic inequality (3.5). Unfortunately, when M < 0 (this can happen when 2a + α 2 > 2b + β 2 ), the zeroth order term of (3.5) has the wrong sign, hence we cannot apply the Maximum Principle. It is possible with some work to find a workaround, at least under the additional hypothesis that r 2 i ≥ −M . Theorem 5.1 (Gradient Estimates, M < 0). Let (Ω, u) be a solution to problem (1.1) and suppose that b < u < a. Suppose that there exist L, M, r o , r i such that the corresponding model solution (2.2) solves the problem for the same a, b, α, β. Let W , W 0 be defined as in (3.2). If r 2 i ≥ −M > 0, then it holds W ≤ W 0 in Ω ,

1)
Moreover, if W = W 0 at some point in the interior of Ω, then (Ω, u) is rotationally symmetric.
Proof. We start by observing that we can write (3.4) as If we ask the quantity in the latter square bracket to be equal to zero, we obtain the following formula for γ where k ∈ R is a constant. In particular we get Since Ψ 2 ≥ r 2 i ≥ −M > 0, choosing k to be big enough we have that the right hand side of (5.3) is negative at any point in Ω. As a consequence, for such k, from (5.2) we get We can then apply the Maximum Principle to find out that F γ is nonpositive on the whole Ω. This concludes the proof.
The question still remains open of whether it is possible to remove the hypothesis r 2 i ≥ −M in the above theorem. Looking at the proof, it is clear that the freedom in the choice of k should allow to weaken that hypothesis. However, it does seem hard to remove it entirely.
The rest of our arguments does not seem to depend heavily on the hypothesis 2a + α 2 ≤ 2b + β 2 . In other words, even in the case 2a+ α 2 > 2b+ β 2 , our method should allow to conclude the desired rigidity, provided we have a model solution to compare with and the gradient estimate W ≤ W 0 is in force. In order to prove Theorem 1.1 in full generality using our approach, these seem to be the two issues that are left to address.