Comparison results for solutions to p-Laplace equations with Robin boundary conditions

In the last decades comparison results of Talenti type for Elliptic Problems with Dirichlet boundary conditions have been widely investigated. In this paper, we generalize the results obtained in arXiv:1909.11950 to the case of p-Laplace operator with Robin boundary conditions. The point-wise comparison, obtained in arXiv:1909.11950 only in the planar case, holds true in any dimension if p is sufficiently small.


Introduction
Let β be a positive parameter and let Ω be a bounded open set of R n , n ≥ 2, with Lipschitz boundary. Let f ∈ L p ′ (Ω) be a non-negative function. We consider the following problem A function u ∈ W 1,p (Ω) is a weak solution to (1) if We want to establish a comparison principle with the solution to the following symmetrized problem where Ω ♯ is the ball centered in the origin with the same measure of Ω and f ♯ is the Schwarz rearrangement of f (see next section for its definition).

2
This kind of problems has been widely investigated in the last decades. The first step is contained in [10], where Talenti proved a pointwise comparison result between u ♯ and v in the case of Dirichlet Laplacian. After this, several papers generalized the result of Talenti: for instance, the one by Talenti himself [11], in which the operator is a generic non-linear operator in divergence form, or the one by Alvino, Lions and Trombetti [2] in which the authors deal with both elliptic and parabolic cases: in both papers, Dirichlet boundary conditions are considered. Different kind of boundary conditions are considered by Alvino, Nitsch and Trombetti in [3], where they establish a comparison between a suitable norm of u and v, respectively solution to They found out that if f is a non-negative function in L 2 (Ω), then where · L k,q is the so called Lorentz norm, whose definition can be found in next section. Moreover, the authors in [3] were able to establish a comparison á la Talenti, in the case f ≡ 1 and n = 2. This will be the starting point of our work: it will be clear that our results coincide with the one in [3] in the case p = 2. For completeness sake, we cite that this wasn't the first result in this sense, indeed in [6] the authors study a comparison result for the p-torsion, that is the case f ≡ 1, with a completely different argument, obtaining Another work that is worth to be mentioned is [1], where the authors obtained similar results to [3] in the case of mixed Dirichlet and Robin boundary conditions. This paper is organized as follows. In the next section, we give some basic notions about rearrangements of functions and Lorentz spaces. Moreover, we list some properties of the solutions to problems (1) and (3). In section 3, we prove the main results about comparison of the two solutions in terms of the Lorentz norm.
In particular, we prove We observe that from Theorem 1.1, we have that, if p ≥ n Theorem 1.2. Assume that f ≡ 1 and let u and v be the solutions to (1) and (3) respectively.
Then we explicitly observe that in the case f ≡ 1 from Theorem 1.2 we have that While we have the point-wise comparison only for p ≤ n n − 1 .
In section 4, using tools from section 3, we give a new proof of the Faber-Krahn inequality with Robin boundary conditions in the case p ≥ n. This topic was already studied in the papers by Bucur, Giacomini, Daners and Trebeschi, [4], [5] , [6] and [7] where the authors proved the Faber-Krahn inequality for the eigenvalues of the Laplacian, or of the p-Laplacian, with Robin boundary conditions, for every p > 1. Actually, the results in [4] are more general, since they hold for every p > 1, but they are obtained with completely different tools than the ones contained in our paper. Finally, in section 5, we provide some examples and open problems and we discuss the optimality of our results.
It is easily checked that u, u * e u ♯ are equi-distributed, so it follows that An important propriety of the decreasing rearrangement is the Hardy-Littlewood inequaliy, that isˆΩ Definition 2.3. Let 0 < p < +∞ and 0 < q ≤ +∞. The Lorentz space L p,q (Ω) is the space of those functions such that the quantity: Let us observe that for p = q the Lorentz space coincides with the L p space, as a consequence of the well known Cavalieri's Principlê See [12] for more details on Lorentz space. Let us consider the functional . This functional is well defined and its Euler-Lagrange equation is exactly (1). If we show that the functional admits a minimum, our problem will always have a solution.
1) Let us show that the functional is bounded from below, indeed using the parametric Young inequality, we have In the last inequality we used the Sobolev inequality with trace term In general, the quantity λ 1,β (Ω) denotes the first eigenvalue of the p-Laplacian with Robin boundary conditions, whose definition is given in (28), which can be also seen as a trace constant of the set Ω.
If ε is small enough, then the quantity is non negative, and then

2) Compactness and lower semicontinuity.
Let {u i } be a minimizing sequence. We can assume that F(u i ) ≤ m + 1, ∀i. Using again the Young inequality, we have Then, the minimizing sequence {u i } is bounded in W 1,p (Ω), so there exists a subsequence { u i k } weakly converging in W 1,p (Ω) and strongly in L p (Ω) to a function u. Let us show that u is the minimum.
The function t p is strictly convex for p > 1, so Passing to the limit for k → ∞, by the weak convergence of {u i k } the integral over Ω on the right-hand side goes to 0. The integral over ∂Ω goes to 0 as well. Indeed, the space This ensures us that u is the minimum of the functional.
The uniqueness of the minimum follows from the fact that F(u) is the sum of a strictly convex part and a linear part.
We observe that the solutions u and v to (1) and (3) respectively are both p-superharmonic and then, by the strong maximum principle in [13], it follows that they achieve their minima on the boundary. Denoting by u m and v m the minimum of u and v respectively, thanks to the positiveness of β and the Robin boundary conditions, we have that u m ≥ 0 and v m ≥ 0. Hence u and v are strictly positive in the interior of Ω. Moreover we can observe that in fact, if we consider . A consequence of (10) that will be used in what follows is that

Useful lemmas
Let u be the solution to (1). For t ≥ 0, we denote by and by where |·| is the Lebesgue measure on R n and P er(·) is the perimeter.
If v is the solution to (3), using the same notations, we set Because of the invariance of the p-Laplacian and of the Schwarz rearrangement of f by rotation, there exists a radial solution to (3) and, by uniqueness of solutions, this solution is v.
Since v is radial, positive and decreasing along the radius then, for to Ω ♯ and strictly contained in it.
be a continuous and differentiable function satisfying, for some non negative constant C, the following differential inequality Then we have Proof. Dividing both sides of the differential inequality by τ p , we obtain Now, we integrate from τ 0 to τ and we obtain which gives (i).
In order to obtain (ii), we just take into account (i) in the differential inequality. (1) and (3) respectively. Then for almost every t > 0 we have

Lemma 2.2. Let u and v be solutions to
where γ n = nω 1/n n p p−1 . And Proof. Let t > 0 e h > 0, we choose the test function Dividing by h, using coarea formula and letting h go to 0, we have that for a. e. t > 0 So, using the isoperimetric inequality, for a. e. t > 0 we have Then (12) follows. We notice that if v is the solution to (3), than all the inequalities are verified as equalities, so we have (13). Moreover,ˆτ Proof. If we integrate the quantity from 0 to +∞, by Fubini theorem, we obtain where the last equality follows from the fact that u solves (1). Analogouslyˆ∞ Since u is positive, we obtain, ∀t ≥ 0, and the proof of lemma 2.3 is complete.
Remark 2.1. It can be observed that, since ∂V t ∩ ∂Ω ♯ is empty for t ≥ v m and φ(t) = |Ω| for t ≤ v m , for all δ > 0 and for all t, we havê

Main results
Now we prove Theorem 1.1 and Theorem 1.2 .
Proof of Theorem 1.1.
Multiplying (12) by t p−1 µ(t) δ and integrating from 0 to τ ≥ v m , by the previous Lemma, we obtainˆτ dω, we can integrate by parts both sides of the last inequality, getting are in the hypothesis of Lemma 2.1 (Gronwall), namely The previous inequality becomes an equality if we replace Passing to the limit as τ → ∞, we get and hence To prove the inequality (5), it is enough to show that Let us consider equation (16), let us integrate by parts the first term on the right-hand side from 0 to τ and then let us pass to the limit as τ → ∞, we havê we obtain the (17). To this aim, we multiply (12) by t p−1 F (µ(t))µ(t) − (n−1)p n(p−1) and integrate. First, we observe that, by the choice k ≤ n(p − 1) (n − 2)p + n , it follows that the function h(l) = F (l)l − (n−1)p

n(p−1)
is non decreasing. Hence, we obtain

If we integrate by parts both sides of the last expression and set
where Of course, the inequality holds as an equality if we replace µ(t) with φ(t), so we get, keeping in mind that as H µ (τ ), H φ (τ ) → 0. This proves (18), and hence (5). The fact that both H µ and H φ go to 0 as τ goes to infinity can be easily deduced distinguishing the cases.
and analogously for H φ , which concludes the proof.
(i) Firstly, we observe thatˆµ Let us multiply both sides by Taking into account remark 2.1, if we replace µ(t) with φ(t) the previous inequality holds as equality.
Hence, we getˆτ Then an integration by parts gives Finally, using Gronwall's Lemma with the function ξ(τ ) =ˆτ The quantity on the right-hand side is non-positive, thanks to (10), so and, remembering that, we get the point-wise inequality of the functions.
(ii) Now we want to show that We multiply (20) by t p−1 µ(t) and equality holds if we replace µ with φ. In order to be shorter, we set We point out that (23) follows by (21) if η ≥ 0, namely With these notations and keeping in mind that µ is a non increasing function, we have from (23) tahtˆτ Let us set G(ℓ) =ˆℓ 0 w η dw = ℓ η+1 η + 1 , let us integrate by parts both sides of (24) in order to obtain Setting (25) reads as follows τ ξ ′ (τ ) ≤ (p − 1)ξ(τ ) + C Hence, using Gronwall's Lemma 2.1 with τ 0 = v m , we get Again, if we replace µ with φ, the previous inequality holds as an equality and ξ(v m ) is less or equal than the same quantity obtained by replacing µ with φ, as (10) holds. Keeping in mind (11), we have Passing to the limit as τ → +∞, we get To conclude the proof, we have to show that We consider (24), pass to the limit as τ → +∞ and integrate by parts the first term on the right-hand sidê Hence it is enough to show that To this aim, we multiply (20) by Since k ≤ n(p − 1) n(p − 1) − p , using Lemma 2.3 and the fact that the function G(ℓ)ℓ −(1− 1 If we replace µ with φ the previous inequality holds as an equality, thanks to (2.1). Now, let us integrate by parts both sides of (26), obtaining the (27) reads as follows τ ξ ′ (τ ) ≤ ξ(τ ) + C Again, using Gronwall's Lemma 2.1, we get Keeping in mind that for φ the previous inequalities hold as equality and the fact that G is not decreasing, ξ(v m ) is less or equal to the same quantity obtained by replacing µ with φ. Hence, we obtain and passing to the limit as τ → +∞, we finally get indeed, as in the proof of Theorem 1.1, H µ (τ ) and H φ (τ ) go to 0 as τ → ∞. That concludes the proof.  (1) and (3) respectively. Then, if p ≥ n, we have Moreover in the case f ≡ 1, Theorem 1.2 gives and the point-wise comparison for p ≤ n n − 1 .
Proof. If p ≥ n the upper bounds of k, in both cases (4) e (5), are greater than 1 and so we can choose k = 1. The assertion follows from the fact that .

Faber-Krahn inequality
We recall that the first eigenvalue of p-Laplace operator with Robin boundary conditions is obtained as the minimum of the Rayleigh quotients, i.e., λ 1,β (Ω) = min We can observe that if u achieves the minimum of Rayleigh quotients, so does |u|. From this we have that u in non-negative. Furthermore, we have if u λ 1 ≥ 0, as a consequence of Harnack inequality, u λ 1 > 0. Another important thing is that the eigenvalue is simple. Indeed, as shown in [8], if Ω is smooth enough and u and v are to eigenfunctions referred to the first eigenvalue, we can choose as test Summing the two equations, we have Now, using the well known inequalities, which hold true for each w 1 and w 2 ∈ R n , if we consider the case p ≥ 2, we choose w 2 = ∇ log u and w 1 = ∇ log v, we obtain Hence, we obtain that v∇u = u∇v a.e. in Ω, and so there exists a constant K for which u = Kv. This means that λ 1 is simple. For the proof of (29), we refer to [8].
The following corollary of Theorem 1.1 holds true, that is Faber-Krahn inequality.  (1) and (3), respectively. Then , if p ≥ n, we have Proof. Let u an eigenfunction referred to the first eigenvalue of (1), then it solves Now, let z be a solution to the following problem In that case, corollary 3.1 giveŝ and hence, by Hölder inequalitŷ Therefore, observing that we can write the eigenvalue λ 1,β (Ω) in the following way, we obtain

Conclusions
We have been able to extend the results obtained for the Laplacian to the p-Laplacian. Many problems remain open, such as Open Problem In the assumptions of Theorem 1.2, does the point-wise comparison hold also for p > n n−1 ? We have already observed in the corollary 3.1 that if p ≥ n we have an estimate on the L p norms of u and v. Can we generalize this estimate also for q = p? We know for sure that for q = ∞ this can't be done, as it can be seen in the following example. Proof. Let us consider the Taylor expansion of (1 + y) p , we get If we choose β < n − 1 p − 1 p−1 we have − 1 (nβ) α n − 1 n(p − 1) + 1 n α+1 < 0. In order to have (30), we need .