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Constrained radial symmetry for the infinity-Laplacian
Nonlinear Analysis: Real World Applications 37 (2017) 239–248
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Nonlinear Analysis: Real World Applications www.elsevier.com/locate/nonrwa
Constrained radial symmetry for the infinity-Laplacian Antonio Greco Department of Mathematics and Informatics, via Ospedale 72, 09124 Cagliari, Italy
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Article history: Received 14 June 2016 Received in revised form 5 October 2016 Accepted 23 February 2017 Available online 19 March 2017 Keywords: Infinity-Laplacian Overdetermined problems Radial symmetry
abstract Three main results concerning the infinity-Laplacian are proved. Theorem 1.1 shows that some overdetermined problems associated to an inhomogeneous infinityLaplace equation are solvable only if the domain is a ball centered at the origin: this is the reason why we speak of constrained radial symmetry. Theorem 1.2 deals with a Dirichlet problem for infinity-harmonic functions in a domain possessing a spherical cavity. The result shows that under suitable control on the boundary data the unknown part of the boundary is relatively close to a sphere. Finally, Theorem 1.4 gives boundary conditions implying that the unknown part of the boundary is exactly a sphere concentric to the cavity. Incidentally, a boundary-point lemma of Hopf’s type for the inhomogeneous infinity-Laplace equation is obtained. © 2017 Elsevier Ltd. All rights reserved.
1. Introduction A celebrated result of Serrin [1] started an interesting research branch in the theory of PDE’s and raised attention to the power of the moving plane method (see [2]). The method had been previously used by Alexandrov [3] for his soap bubble theorem. To be more precise, Theorem 1 in [1] asserts that if there exists a solution u ∈ C 2 (Ω ) of the problem { ∆u = −1 in Ω ; (1) ∂u u = 0, = constant on ∂Ω , ∂n where Ω is a bounded domain (=open, connected set) of class C 2 in RN with outer normal n, then Ω is a ball and u has the specific form u(x) =
R2 − |x − x0 | 2N
2
where R is the radius of the ball Ω = BR and x0 denotes its center. Weinberger [4] described an alternative method of proof based on Green’s formula and less demanding in terms of regularity. Problem (1) is called E-mail address: [email protected]. http://dx.doi.org/10.1016/j.nonrwa.2017.02.016 1468-1218/© 2017 Elsevier Ltd. All rights reserved.
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overdetermined because the torsion problem, i.e. the Dirichlet problem { ∆u = −1 in Ω ; u=0 on ∂Ω admits a unique (=determined) solution: existence fails, in general, if we prescribe the further boundary condition ∂u = constant. ∂n
(2)
Serrin discovered that the ball is the only domain making problem (1) solvable. Garofalo and Lewis ([5], see also [6] and the references therein) extended the result to the p-Laplacian: the operator ∆p u = p−2 div(|∇u| ∇u) with p ∈ (1, +∞). In [5,6], more general operators than the p-Laplacian are taken into account, and solutions are intended in the weak sense. When p = +∞, instead, the situation is different. 2 Let us concentrate on the classical ∞-Laplacian, i.e. the operator ∆∞ u = uνν |uν | where the subscript ν denotes differentiation in the direction of ν = −∇u/|∇u|. A motivation for the name of infinity-Laplacian, based on a relation with the p-Laplacian, is explained in [7]: however, wrong expectations often arise when trying to devise properties of the first operator by letting p → +∞ in the second (see below). Solutions of ∆∞ u = −1 are intended in the viscosity sense. More precisely, a solution of the ∞-torsion problem { ∆∞ u = −1 in Ω ; (3) u=0 on ∂Ω is a function u ∈ C 0 (Ω ) vanishing on ∂Ω and satisfying both of the following conditions at each interior point x0 ∈ Ω : (a) if there exists a function φ0 of class C 2 in a neighborhood of x0 such that the difference φ0 (x) − u(x) has a local minimum at x0 , then for every such function φ the inequality φij (x0 ) φi (x0 ) φj (x0 ) ≥ −1 is satisfied; (b) if there exists a function ψ0 of class C 2 in a neighborhood of x0 such that the difference ψ0 (x)−u(x) has a local maximum at x0 , then for every such function ψ the inequality ψij (x0 ) ψi (x0 ) ψj (x0 ) ≤ −1 holds. Here the subscripts i, j denote differentiation with respect to xi , xj , and the summation from 1 to N over repeated indices is understood. It is worth noticing that if u is not smooth enough at x0 to let some C 2 -function φ0 have the required property, then condition (a) is trivially satisfied, and a similar remark holds for (b). An example is given below, concerning the function uR in (4). A standard reference on viscosity solutions is [8] (see also [9, pp. 238–239] and [10, Section 2]). It is known that if Ω is a bounded open subset of RN then there exists a unique viscosity solution of problem (3) (see [11, Theorems 2 and 5]). The solution is differentiable in Ω [12], and if the boundary ∂Ω is also differentiable then u is differentiable up to ∂Ω [13]. Finally, the solution uR in the ball Ω = BR centered at x0 is the radial function uR (x) =
34/3 ( 4/3 4/3 ) R − |x − x0 | . 4 2
(4)
Indeed uR (x) is smooth for x ̸= x0 and satisfies uνν |uν | = −1 in the classical, as well as in the viscosity sense in the punctured ball BR \ {x0 }. Concerning the center x0 , due to the exponent 4/3 no function ψ0 of class C 2 in a neighborhood of x0 can touch uR from below there, i.e., the difference ψ0 (x) − uR (x) cannot have a local maximum at x0 . Consequently, condition (b) given before is satisfied. Moreover, every function φ of class C 2 in a neighborhood of x0 such that the difference φ(x)−u(x) has a local minimum at x0 must be stationary at x0 , and therefore the inequality φij (x0 ) φi (x0 ) φj (x0 ) ≥ −1 in condition (a) holds. In conclusion, uR solves problem (3) in the ball Ω = BR , as claimed (see also [9, p. 239] and [10, Example 2.1]). Unlike the case when p is finite, adding the boundary condition (2) to the Dirichlet problem (3) does not force the domain Ω to be a ball: thus, Serrin–Weinberger’s theorem fails for the ∞-Laplacian. Indeed,
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1
Fig. 1. The solution u ∈ C 1, 3 (Ω ) in a stadium-like planar domain.
Buttazzo and Kawohl [9, p. 241] gave a geometric condition on the (supposedly C 2 ) domain Ω sufficient to ensure the solvability of the overdetermined problem { ∆∞ u = −1 in Ω ; (5) ∂u u = 0, = constant on ∂Ω . ∂n The result was extended to C 1 -domains by Crasta and Fragal`a [7, Theorem 2]. In order to state the geometric condition we need to refer to the cut locus Σ (Ω ) = Ω \ G, also denoted by R(Ω ), which is the complement in Ω of the largest open subset G ⊂ Ω such that every point x ∈ G has a unique closest point on ∂Ω . We also need the set M (Ω ) ⊂ Σ (Ω ) of those points where the distance function d(x, ∂Ω ) attains its maximum. In [7, Theorem 2] the authors show that if Ω is of class C 1 and if Σ (Ω ) = M (Ω )
(6)
then there exists a solution u of the overdetermined problem (5). Furthermore u is a web-function in the sense that u(x) depends on the distance dist(x, ∂Ω ) only. In fact, the solution u is given by ( )4/3 a4 − a3 − 3 dist(x, ∂Ω ) u(x) = (7) 4 and therefore ∂u/∂n = −a on ∂Ω . The parameter a has to be chosen such that a3 /3 equals the inner radius of Ω , i.e. the distance ρΩ from the set Σ (Ω ) = M (Ω ) to the boundary ∂Ω . Which domains satisfy the geometric condition (6)? If Ω is a ball then clearly Σ (Ω ) = M (Ω ) =the center of Ω , and the solution (7) equals the radial function (4). It is worth mentioning that the ball is the only convex domain of class C 2 satisfying (6) (see [14, Theorem 12]). An example of a non-radially symmetric domain satisfying (6) is the convex envelope of two congruent (distinct) balls: see Fig. 1. A general procedure to construct such kind of domains was indicated on p. 241 of [9] and then developed in [14] (see also the survey [10] with several pictures). Although the additional condition (2), alone, does not force the domain Ω to be a ball when the overdetermined problem (5) is solvable, Serrin–Weinberger’s theorem has been recovered by adding some restrictions on the domain Ω and under the assumption that the solution u is C 1 near ∂Ω and up to there (see [15, Corollary 23]). In Theorem 1.1 we address our attention to domains containing the origin, and in place of (2) we consider the following non-autonomous condition: ∂u − = q(|x|) (8) ∂n (where the minus sign is just to deal with non-negative functions). We show that if the prescribed function q(r) grows fast enough with respect to r then the overdetermined problem { ∆∞ u = −1 in Ω ; (9) ∂u u = 0, − = q(|x|) on ∂Ω ∂n
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is solvable only if Ω is a ball centered at the origin. This is the reason why we speak of a constrained radial symmetry. Note that a solution of (9) is sought in the whole space C 0 (Ω ), not only among web-functions. Before proceeding further let us show that without restrictions on q problem (9) is solvable in a proper ellipse. Example. Let Ω ⊂ R2 be a proper ellipse in canonical position, and let u be the solution (differentiable up ∂u to the boundary) of the ∞-torsion problem (3). Define the function g : ∂Ω → R by letting g(x) = − ∂n (x) at each point x = (x1 , x2 ) ∈ ∂Ω . Since problem (3) is invariant under reflections with respect to the coordinate axes, and by uniqueness, the function g satisfies g(x1 , x2 ) = g(±x1 , ±x2 ) for every x ∈ ∂Ω and for every combination of the signs in front of the coordinates x1 , x2 . Consequently we can define q(r) as follows. Let R1 = min |x| and R2 = max |x|. x∈∂Ω
x∈∂Ω
(10)
For every r ∈ [R1 , R2 ] define q(r) = g(x) where x is any point on the boundary ∂Ω satisfying |x| = r. Since g is symmetric as mentioned above, the definition of q(r) is independent of the choice of x. Thus we end up with a solvable instance of problem (9) although Ω is neither a disc nor a stadium-like domain. The solvability of the overdetermined problem (9) is related to the solvability of Eq. (11). Theorem 1.1. Let Ω be a bounded domain in RN , N ≥ 2, containing the origin and with a differentiable boundary, and let q(r) be any real-valued function defined in the closed interval [R1 , R2 ], where R1 , R2 are as in (10). (i) Suppose that the equation q(r) − (3r)1/3 = 0
(11)
possesses a unique solution R ∈ [R1 , R2 ], and the difference q(r) − (3r)1/3 has the same sign as r − R. Then problem (9) is solvable only in the special case when R1 = R2 = R, i.e. when Ω = BR (0). If, instead, R1 < R2 , then problem (9) is unsolvable. (ii) Suppose that the ratio q(r)/r1/3 is strictly increasing. Then problem (9) is solvable only if Ω is a ball centered at the origin. (iii) If q is continuous, and if Eq. (11) does not possess any solution, then problem (9) is unsolvable. The proof is given in Section 2 and is based on comparison with radial solutions. The procedure was applied in [16] to the classical torsion problem and to corresponding problems involving the p-Laplacian, as well as some equations of constant curvature. Recent progress on the classical torsion problem was done in [17]. The p-Laplacian has also been considered in [18,19]. In particular, it is interesting to compare problem (9) with { ∆p u = −1 in Ω ; (12) ∂u u = 0, − = q(|x|) on ∂Ω . ∂n It is known that if Ω is a bounded domain containing the origin, and if the ratio q(r)/rp−1 is monotone non-decreasing,
(13)
then problem (12) is solvable only if Ω is a ball centered at the origin [19, Theorem 1.1]. Thus, the monotonicity assumption on the ratio q(r)/r1/3 in Claim (ii) of Theorem 1.1 is not predictable just letting p → +∞ in (13). It is worth noticing that, contrary to what one may expect, the weak solution up ∈ W01,p (Ω )
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of the equation ∆p u = −1 does not converge to a viscosity solution u of ∆∞ u = −1: see [9, Section 4] for details. Let us now turn our attention to constrained radial symmetry for domains with cavities (rings). In Section 3 we prove the following result: Theorem 1.2 (Approximate Radial Symmetry). Let Ω ⊂ RN be a bounded domain containing the closure of a given ball B0 = BR0 (0) and with a differentiable boundary, and let R1 , R2 > R0 be as in (10). Let u be the solution of the Dirichlet problem ⎧ in Ω \ B 0 ; ⎨∆∞ u = 0 (14) u = a(x) on ∂B0 ; ⎩ u=b on ∂Ω , where a(x) is a continuous function and b a given constant. Define a1 = max|x|=R0 a(x), a2 = min|x|=R0 a(x), q1 = supx∈∂Ω |∇u(x)|, q2 = inf x∈∂Ω |∇u(x)|, and suppose that a2 < b. Then b − a1 R1 − R0 q2 ≤ q1 . b − a2 R2 − R0
(15)
Note that the Dirichlet problem for the infinity-Laplace equation with continuous boundary values is uniquely solvable [20, Corollary 3.14]. Furthermore, since the boundary data are constant along ∂Ω , which is differentiable by assumption, the solution of (14) is differentiable up to there [21] hence the gradient ∇u is well defined for x ∈ ∂Ω . An inequality corresponding to (15) for the Laplace operator was proved in [22, Corollary 1] and then refined in [19, Theorem 5.1]. Its purpose is to show that when the boundary values a(x), x ∈ ∂B0 , and the boundary gradient |∇u(x)|, x ∈ ∂Ω , are close to two constants then the ratio (R1 − R0 )/(R2 − R0 ) is close to 1 and therefore the domain Ω is relatively close to a ball centered at the origin (of course, oscillations of the boundary are neglected). Theorem 1.2 implies as a special case the following known result: Corollary 1.3. Let Ω be as in Theorem 1.2, and let a < b be constants. If there exists a solution of the overdetermined problem ⎧ in Ω \ B 0 ; ⎪ ⎨∆∞ u = 0 u=a on ∂B0 ; (16) ⎪ ⎩u = b, ∂u = constant on ∂Ω , ∂n then Ω is a ball centered at the origin. The more general case when the closed ball B 0 in problem (16) is replaced by a compact convex subset K ⊂ RN , and Ω is a domain containing K, is considered in [23, Theorem 3.3]. In the present paper the assumption that ∂u/∂n = constant on ∂Ω is relaxed as follows. Suppose that the overdetermined problem ⎧ in Ω \ B 0 ; ⎪ ⎨∆∞ u = 0 u=a on ∂B0 ; (17) ⎪ ⎩u = b, ∂u = q(|x|) on ∂Ω , ∂n is solvable. In order to prove that Ω is a ball centered at the origin it is enough that the product (r −R0 ) q(r) is strictly increasing. Indeed, we have: Theorem 1.4. Let Ω , R1 , R2 be as in Theorem 1.2, and let a < b be constants. Let q be a real-valued function defined in the closed interval [R1 , R2 ].
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(i) Suppose that the equation (r − R0 ) q(r) = b − a
(18)
possesses a unique solution R ∈ [R1 , R2 ], and the difference (r − R0 ) q(r) − (b − a) has the same sign as r − R. Then problem (17) is solvable only in the special case when R1 = R2 = R, i.e. when Ω = BR (0). If, instead, R1 < R2 , then problem (17) is unsolvable. (ii) Suppose that the product (r − R0 ) q(r) is strictly increasing. Then problem (17) is solvable only if Ω is a ball centered at the origin. (iii) If q is continuous, and if Eq. (18) does not possess any solution, then problem (17) is unsolvable. Problems of type (17) associated to the Laplace operator and to the p-Laplacian are considered in [19,22,24,25] as well as in [26], which is the paper that started the series. Consider for instance a positive, differentiable function q(r) such that ) d ( N −1 r (q(r))p−1 ≥ −rN −1 for some p ∈ (1, +∞). dr
(19)
Theorem 1.3 in [19] implies, in particular, that if the overdetermined problem ⎧ in Ω \ B 0 ; ⎪ ⎨∆p u = 0 u=a on ∂B0 ; ⎪ ⎩u = b, ∂u = q(|x|) on ∂Ω ∂n admits a weak solution u ∈ C 1 (Ω \ B 0 ) then Ω is a ball centered at 0. The result was previously obtained as a special case of [22, Theorem 4] under the assumption that rN −1 (q(r))p−1 is increasing instead of (19). Observe that the assumption on the product (r − R0 ) q(r) in Theorem 1.4 does not follow just letting p → +∞ in (19) . The theorem is proved in Section 3 by comparison with the solution in the annulus. Let us mention, in conclusion, that constrained radial symmetry for the equation ∆u = δ (=Dirac’s delta) has been investigated in [27]. Other kinds of constrained symmetry have also been considered: for instance, the domain of the problem turns out to be a cylinder in [25,28] and an ellipsoid in [29]. Overdetermined problems for the fractional Laplacian have also been investigated: see [30] for the free radial symmetry, and [31] for constrained symmetry. 2. Proof of Theorem 1.1 and Hopf’s Lemma Proof of Theorem 1.1. Claim (i). Let us check that the overdetermined problem (9) is solvable in the ball Ω = BR (0) whose radius R satisfies Eq. (11). By computation, from (4) we immediately deduce that the radial solution uR of the Dirichlet problem (3) satisfies (cf. [9, p. 238]) −
∂uR = (3 R)1/3 , ∂n
(20)
hence the additional condition (8) reads as (3 R)1/3 = q(R). Since this holds by assumption, problem (9) is solvable, as claimed. To prove that there is no other possibility, let Ω be now any bounded domain. Before proceeding further, let us verify that the solution u of the Dirichlet problem (3) is positive in Ω . Observe that the null function u0 ≡ 0 trivially satisfies the inequality ∆∞ u0 ≥ −1 as well as the boundary condition u0 = 0 on ∂Ω , hence by the comparison principle [11, Theorem 4] we have u ≥ 0 in Ω . Now let Br (x0 ) be a ball of radius r > 0 centered anywhere and contained in Ω , and consider the solution ur of problem (3) in Br (x0 ), which is obtained by replacing R with r in (4). Since u ≥ 0 on ∂Br (x0 ), the comparison principle implies u ≥ ur > 0 in Br (x0 ). Since x0 ∈ Ω is arbitrary, we have u > 0 in all of Ω , as claimed.
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Now assume that u is a solution of the overdetermined problem (9). Denote by Bi = BRi (0) the ball of radius Ri , i = 1, 2, centered at the origin, so that B1 ⊂ Ω ⊂ B2 . Define the function ui as follows: ui (x) =
34/3 ( 4/3 4/3 ) Ri − |x| . 4
Thus, ui is the solution of problem (3) in the ball Bi . Arguing as above we find u1 ≤ u in B1 .
(21)
Concerning u2 , we have u = 0 ≤ u2 along the boundary ∂Ω : hence by the comparison principle [11, Theorem 4] we obtain u ≤ u2 in all of Ω . Let us consider a point P1 ∈ ∂B1 ∩ ∂Ω , i.e., a point on ∂Ω such that |P1 | = R1 . Observe that the outer normal n to ∂Ω at P1 equals P1 /R1 = the outer normal to B1 . By (20) and (21), and since u1 = u at P1 , we deduce (3 R1 )1/3 = −
∂u ∂u1 (P1 ) ≤ − (P1 ) = q(R1 ), ∂n ∂n
(22)
where the last equality comes from (8). Since Eq. (11) has a unique solution R ∈ [R1 , R2 ] by assumption, and the difference q(r) − (3r)1/3 is negative for any r < R, inequality (22) implies R1 = R (the case R1 > R being excluded). Arguing at a point P2 ∈ ∂B2 ∩ ∂Ω in a similar manner we find q(R2 ) = −
∂u ∂u2 (P2 ) ≤ − (P2 ) = (3 R2 )1/3 ∂n ∂n
(23)
and consequently R2 = R. But then we deduce R1 = R2 and Claim (i) follows. Claim (ii). If problem (9) is solvable, then arguing as before we arrive at (22) and (23), hence the ratio ρ(r) = q(r)/r1/3 satisfies ρ(R2 ) ≤ 31/3 ≤ ρ(R1 ). But since ρ(r) is strictly increasing by assumption, we must have R1 = R2 and the conclusion follows. To prove Claim (iii) we assume the solvability of problem (9) and show that Eq. (11) also has a solution. Using again (22) and (23) we see that the difference q(r) − (3r)1/3 is non-negative at R1 and non-positive at R2 . The claim follows because q is continuous. □ As a by-product of the proof above (see in particular (22)) it turns out that the difference q(r) − (3r)1/3 must be non-negative at r = R1 in order that problem (9) is solvable. Since the Dirichlet problem (3) is translation-invariant, the argument proves the following boundary-point lemma: Lemma 2.1 (Hopf). Let Ω be a bounded open subset of RN satisfying the interior sphere condition at some x1 ∈ ∂Ω , i.e. there exists BR (x0 ) ⊂ Ω such that x1 ∈ ∂BR (x0 ). Then the solution u of the Dirichlet problem (3) satisfies lim inf h→0+
u(x1 + (x0 − x1 ) h/R) ≥ (3 R)1/3 > 0. h
Hopf’s lemma for infinity-harmonic functions (solutions of ∆∞ u = 0) is found in [32, Lemma 1]. For the fractional Laplacian see for instance [31]. 3. Proofs of Theorems 1.2 and 1.4 The proofs of Theorems 1.2 and 1.4 are based on the comparison principle for infinity-harmonic functions [20, Corollary 3.11]. More precisely, the solution u is compared to the solutions u1 , u2 in suitable annuli. Let R > R0 > 0 and define BR = BR (0) and B0 = BR0 (0), for shortness. We have:
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Lemma 3.1. Let a, b be two constants. The unique viscosity solution of the Dirichlet problem ⎧ in BR \ B 0 ; ⎨∆∞ u = 0 u=a on ∂B0 ; ⎩ u=b on ∂BR ,
(24)
is the radial function uR0 ,R (x) given by uR0 ,R (x) =
b−a (|x| − R0 ) + a. R − R0
(25)
Proof . If a = b then the function uR0 ,R reduces to the constant uR0 ,R ≡ a which trivially satisfies (24). x b−a in the closed annulus If, instead, a ̸= b then uR0 ,R has a non-vanishing gradient DuR0 ,R (x) = R−R 0 |x| B R \ B0 , hence the unit vector ν = −DuR0 ,R /|DuR0 ,R | = sign(a − b) x/|x| is well defined. Since uνν ≡ 0 2 it turns out that uR0 ,R is a classical solution of the equation uνν |uν | = 0 and a fortiori it is a viscosity solution of (24). Uniqueness follows from [20, Corollary 3.14]. □ Proof of Theorem 1.2. We concentrate on the case when a1 < b, otherwise (15) holds trivially. Observe that the solution u of (14) satisfies a2 ≤ u ≤ b in Ω \ B0 .
(26)
Indeed, the constant functions u(x) ≡ a2 and u(x) = b are infinity-harmonic. Furthermore, we have a2 ≤ a(x) = u(x) ≤ a1 for |x| = R0 and we are assuming a1 < b. Finally, a2 ≤ a1 < b = u(x) on ∂Ω . Hence (26) follows from the comparison principle [20, Theorem 3.11]. As a consequence of (26) we have ∂u/∂n ≥ 0 on ∂Ω and therefore we may write ∂u/∂n = |∇u| there. From this we get the inequalities q2 ≤
∂u ≤ q1 ∂n
on ∂Ω ,
which enter into (27) and (28). Now denote by Bi = BRi (0), i = 1, 2, the ball of radius Ri centered at the origin, as in Section 2, and let ui (x) be the radial solution of the Dirichlet problem (24) in the annulus Bi \ B 0 with boundary value a = ai . In other terms, define ui = uR0 ,Ri where uR0 ,R is given in (25). Let us compare ui to the solution u of problem (14). When |x| = R0 we have u2 (x) ≤ u(x) ≤ u1 (x) by definition. When |x| = R1 , instead, we have u(x) ≤ b = u1 (x) by (26), hence u(x) ≤ u1 (x) in B 1 \ B0 . Concerning u2 , since u2 (x) ≤ b = u(x) for x ∈ ∂Ω , by the comparison principle we get u2 (x) ≤ u(x) in Ω \ B0 . Since u = u1 at every point P1 ∈ ∂B1 ∩ ∂Ω , and since the outer normals of ∂B1 and ∂Ω coincide there, we get b − a1 ∂u1 ∂u = (P1 ) ≤ (P1 ) ≤ q1 . R1 − R0 ∂n ∂n
(27)
Furthermore, since u2 = u at every P2 ∈ ∂B2 ∩ ∂Ω , we obtain q2 ≤
∂u ∂u2 b − a2 (P2 ) ≤ (P2 ) = . ∂n ∂n R2 − R0
By (27) and (28) we arrive at (15), and the proof is complete.
□
(28)
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To prove Theorem 1.4 we now let a1 = a2 = a in (27) and (28). Proof of Theorem 1.4. Claim (i). Let us check that the overdetermined problem (17) is solvable in the annulus BR \ B 0 whose outer radius R satisfies Eq. (18). From (25) we see that the radial solution uR0 ,R of the Dirichlet problem (24) satisfies b−a ∂ uR0 ,R = on ∂BR ∂n R − R0 hence the additional boundary condition in the overdetermined problem (17), which reads as ∂ uR ,R = q(R) on ∂BR ∂n 0 reduces to (R−R0 ) q(R) = b−a, which holds true by hypothesis. Therefore the overdetermined problem (17) is solvable in BR \ B 0 , as claimed. To proceed further, assume now that the overdetermined problem (17) has a solution u in the domain Ω \ B 0 . Arguing as in the proof of Theorem 1.2, and letting a1 = a in (27), we obtain b−a ∂u ≤ (P1 ) = q(R1 ), R1 − R0 ∂n
(29)
where the last equality follows from the additional boundary condition in (17). This and the assumption that (r − R0 ) q(r) < b − a for all r < R imply R1 = R, the case R1 > R being impossible because R ∈ [R1 , R0 ]. Similarly, letting a2 = a in (28) we obtain q(R2 ) =
b−a ∂u (P2 ) ≤ . ∂n R2 − R0
(30)
But since (r − R0 ) q(r) > b − a for all r > R we deduce R2 = R = R1 and Claim (i) follows. Claim (ii). If problem (17) is solvable, then by (29) and (30) we see that the product p(r) = (r − R0 ) q(r) satisfies p(R2 ) ≤ b−a ≤ p(R1 ). But since p(r) is strictly increasing by assumption we conclude that R1 = R2 , which proves the claim. To prove Claim (iii) suppose again that there exists a solution u of the overdetermined problem (17). From (29) and (30) it follows that the difference (r −R0 ) q(r)−(b−a) is non-negative at R1 and non-positive at R2 . Since q(r) is continuous, Eq. (18) must have at least a solution, contrary to the assumption. Hence Claim (iii) must hold, and the proof is complete. □ Acknowledgment The author is a member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). References
[1] J. Serrin, A symmetry problem in potential theory, Arch. Ration. Mech. Anal. 43 (1971) 304–318. [2] H. Berestycki, L. Nirenberg, On the method of moving planes and the sliding method, Bol. Soc. Brasil. Mat. (N.S.) 22 (1991) 1–37. [3] A.D. Alexandrov, A characteristic property of spheres, Ann. Mat. Pura Appl. 58 (1962) 303–315. [4] H.F. Weinberger, Remark on the preceding paper of Serrin, Arch. Ration. Mech. Anal. 43 (1971) 319–320. [5] N. Garofalo, J.L. Lewis, A symmetry result related to some overdetermined boundary value problems, Amer. J. Math. 111 (1989) 9–33. [6] A. Farina, B. Kawohl, Remarks on an overdetermined boundary value problem, Calc. Var. Partial Differential Equations 31 (2008) 351–357.
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[7] G. Crasta, I. Fragal` a, A symmetry problem for the infinity Laplacian, Int. Math. Res. Not. IMRN 2015 (2015) 8411–8436. [8] M.G. Crandall, H. Ishii, P.-L. Lions, User’s guide to viscosity solutions of second order partial differential equations, Bull. Amer. Math. Soc. (N.S.) 27 (1992) 1–67. [9] G. Buttazzo, B. Kawohl, Overdetermined boundary value problems for the ∞-laplacian, Int. Math. Res. Not. IMRN 2 (2011) 237–247. [10] G. Crasta, I. Fragal` a, Geometric issues in pde problems related to the infinity Laplace operator, Radon Ser. Comput. Appl. Math. (in press). [11] G. Lu, P. Wang, Inhomogeneous infinity Laplace equation, Adv. Math. 217 (2008) 1838–1868. [12] E. Lindgren, On the regularity of solutions of the inhomogeneous infinity Laplace equation, Proc. Amer. Math. Soc. 142 (2014) 277–288. [13] G. Hong, Boundary differentiability for inhomogeneous infinity Laplace equations, Electron. J. Differential Equations 72 (2014) 6 pp. [14] G. Crasta, I. Fragal` a, On the characterization of some classes of proximally smooth sets, ESAIM Control Optim. Calc. Var. 22 (2016) 710–727. [15] G. Crasta, I. Fragal` a, On the Dirichlet and Serrin problems for the inhomogeneous infinity Laplacian in convex domains: Regularity and geometric results, Arch. Ration. Mech. Anal. 218 (2015) 1577–1607. [16] A. Greco, Symmetry around the origin for some overdetermined problems, Adv. Math. Sci. Appl. 13 (2003) 387–399. [17] C. Bianchini, A. Henrot, P. Salani, An overdetermined problem with non-constant boundary condition, Interfaces Free Bound. 16 (2014) 215–241. [18] A. Greco, Comparison principle and constrained radial symmetry for the subdiffusive p-laplacian, Publ. Mat. 58 (2014) 485–498. [19] A. Greco, Constrained radial symmetry for monotone elliptic quasilinear operators, J. Anal. Math. 121 (2013) 223–234. [20] R. Jensen, Uniqueness of lipschitz extensions: Minimizing the sup norm of the gradient, Arch. Ration. Mech. Anal. 123 (1993) 51–74. [21] G. Hong, Boundary differentiability of infinity harmonic functions, Nonlinear Anal. 93 (2013) 15–20. [22] A. Henrot, G.A. Philippin, Approximate radial symmetry for solutions of a class of boundary value problems in ring-shaped domains, J. Appl. Math. Phys. (ZAMP) 54 (2003) 784–796. [23] J. Manfredi, A. Petrosyan, H. Shahgholian, Symmetry in multi-phase overdetermined problems, Calc. Var. Partial Differential Equations 14 (2002) 359–384. [24] A. Greco, Radial symmetry and uniqueness for an overdetermined problem, Math. Methods Appl. Sci. 24 (2001) 103–115. [25] L.E. Payne, G.A. Philippin, On two free boundary problems in potential theory, J. Math. Anal. Appl. 161 (1991) 332–342. [26] A. Henrot, G.A. Philippin, H. Pr´ ebet, Overdetermined problems on ring shaped domains, Adv. Math. Sci. Appl. 9 (1999) 737–747. [27] C. Babaoglu, H. Shahgholian, Symmetry in multi-phase overdetermined problems, J. Convex Anal. 18 (2011) 1013–1024. [28] A. Greco, Boundary point lemmas and overdetermined problems, J. Math. Anal. Appl. 278 (2003) 214–224. [29] A. Greco, A characterization of the ellipsoid through the torsion problem, J. Appl. Math. Phys. (ZAMP) 59 (2008) 753–765. [30] M.M. Fall, S. Jarohs, Overdetermined problems with fractional laplacian, ESAIM Control Optim. Calc. Var. 21 (2015) 924–938. [31] A. Greco, R. Servadei, Hopf’s lemma and constrained radial symmetry for the fractional laplacian, Math. Res. Lett. 23 (2016) 863–885. [32] G. Hong, Counterexample to c1 boundary regularity of infinity harmonic functions, Nonlinear Anal. 104 (2014) 120–123.
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Nonlinear Analysis: Real World Applications www.elsevier.com/locate/nonrwa
Constrained radial symmetry for the infinity-Laplacian Antonio Greco Department of Mathematics and Informatics, via Ospedale 72, 09124 Cagliari, Italy
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Article history: Received 14 June 2016 Received in revised form 5 October 2016 Accepted 23 February 2017 Available online 19 March 2017 Keywords: Infinity-Laplacian Overdetermined problems Radial symmetry
abstract Three main results concerning the infinity-Laplacian are proved. Theorem 1.1 shows that some overdetermined problems associated to an inhomogeneous infinityLaplace equation are solvable only if the domain is a ball centered at the origin: this is the reason why we speak of constrained radial symmetry. Theorem 1.2 deals with a Dirichlet problem for infinity-harmonic functions in a domain possessing a spherical cavity. The result shows that under suitable control on the boundary data the unknown part of the boundary is relatively close to a sphere. Finally, Theorem 1.4 gives boundary conditions implying that the unknown part of the boundary is exactly a sphere concentric to the cavity. Incidentally, a boundary-point lemma of Hopf’s type for the inhomogeneous infinity-Laplace equation is obtained. © 2017 Elsevier Ltd. All rights reserved.
1. Introduction A celebrated result of Serrin [1] started an interesting research branch in the theory of PDE’s and raised attention to the power of the moving plane method (see [2]). The method had been previously used by Alexandrov [3] for his soap bubble theorem. To be more precise, Theorem 1 in [1] asserts that if there exists a solution u ∈ C 2 (Ω ) of the problem { ∆u = −1 in Ω ; (1) ∂u u = 0, = constant on ∂Ω , ∂n where Ω is a bounded domain (=open, connected set) of class C 2 in RN with outer normal n, then Ω is a ball and u has the specific form u(x) =
R2 − |x − x0 | 2N
2
where R is the radius of the ball Ω = BR and x0 denotes its center. Weinberger [4] described an alternative method of proof based on Green’s formula and less demanding in terms of regularity. Problem (1) is called E-mail address: [email protected]. http://dx.doi.org/10.1016/j.nonrwa.2017.02.016 1468-1218/© 2017 Elsevier Ltd. All rights reserved.
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overdetermined because the torsion problem, i.e. the Dirichlet problem { ∆u = −1 in Ω ; u=0 on ∂Ω admits a unique (=determined) solution: existence fails, in general, if we prescribe the further boundary condition ∂u = constant. ∂n
(2)
Serrin discovered that the ball is the only domain making problem (1) solvable. Garofalo and Lewis ([5], see also [6] and the references therein) extended the result to the p-Laplacian: the operator ∆p u = p−2 div(|∇u| ∇u) with p ∈ (1, +∞). In [5,6], more general operators than the p-Laplacian are taken into account, and solutions are intended in the weak sense. When p = +∞, instead, the situation is different. 2 Let us concentrate on the classical ∞-Laplacian, i.e. the operator ∆∞ u = uνν |uν | where the subscript ν denotes differentiation in the direction of ν = −∇u/|∇u|. A motivation for the name of infinity-Laplacian, based on a relation with the p-Laplacian, is explained in [7]: however, wrong expectations often arise when trying to devise properties of the first operator by letting p → +∞ in the second (see below). Solutions of ∆∞ u = −1 are intended in the viscosity sense. More precisely, a solution of the ∞-torsion problem { ∆∞ u = −1 in Ω ; (3) u=0 on ∂Ω is a function u ∈ C 0 (Ω ) vanishing on ∂Ω and satisfying both of the following conditions at each interior point x0 ∈ Ω : (a) if there exists a function φ0 of class C 2 in a neighborhood of x0 such that the difference φ0 (x) − u(x) has a local minimum at x0 , then for every such function φ the inequality φij (x0 ) φi (x0 ) φj (x0 ) ≥ −1 is satisfied; (b) if there exists a function ψ0 of class C 2 in a neighborhood of x0 such that the difference ψ0 (x)−u(x) has a local maximum at x0 , then for every such function ψ the inequality ψij (x0 ) ψi (x0 ) ψj (x0 ) ≤ −1 holds. Here the subscripts i, j denote differentiation with respect to xi , xj , and the summation from 1 to N over repeated indices is understood. It is worth noticing that if u is not smooth enough at x0 to let some C 2 -function φ0 have the required property, then condition (a) is trivially satisfied, and a similar remark holds for (b). An example is given below, concerning the function uR in (4). A standard reference on viscosity solutions is [8] (see also [9, pp. 238–239] and [10, Section 2]). It is known that if Ω is a bounded open subset of RN then there exists a unique viscosity solution of problem (3) (see [11, Theorems 2 and 5]). The solution is differentiable in Ω [12], and if the boundary ∂Ω is also differentiable then u is differentiable up to ∂Ω [13]. Finally, the solution uR in the ball Ω = BR centered at x0 is the radial function uR (x) =
34/3 ( 4/3 4/3 ) R − |x − x0 | . 4 2
(4)
Indeed uR (x) is smooth for x ̸= x0 and satisfies uνν |uν | = −1 in the classical, as well as in the viscosity sense in the punctured ball BR \ {x0 }. Concerning the center x0 , due to the exponent 4/3 no function ψ0 of class C 2 in a neighborhood of x0 can touch uR from below there, i.e., the difference ψ0 (x) − uR (x) cannot have a local maximum at x0 . Consequently, condition (b) given before is satisfied. Moreover, every function φ of class C 2 in a neighborhood of x0 such that the difference φ(x)−u(x) has a local minimum at x0 must be stationary at x0 , and therefore the inequality φij (x0 ) φi (x0 ) φj (x0 ) ≥ −1 in condition (a) holds. In conclusion, uR solves problem (3) in the ball Ω = BR , as claimed (see also [9, p. 239] and [10, Example 2.1]). Unlike the case when p is finite, adding the boundary condition (2) to the Dirichlet problem (3) does not force the domain Ω to be a ball: thus, Serrin–Weinberger’s theorem fails for the ∞-Laplacian. Indeed,
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241
1
Fig. 1. The solution u ∈ C 1, 3 (Ω ) in a stadium-like planar domain.
Buttazzo and Kawohl [9, p. 241] gave a geometric condition on the (supposedly C 2 ) domain Ω sufficient to ensure the solvability of the overdetermined problem { ∆∞ u = −1 in Ω ; (5) ∂u u = 0, = constant on ∂Ω . ∂n The result was extended to C 1 -domains by Crasta and Fragal`a [7, Theorem 2]. In order to state the geometric condition we need to refer to the cut locus Σ (Ω ) = Ω \ G, also denoted by R(Ω ), which is the complement in Ω of the largest open subset G ⊂ Ω such that every point x ∈ G has a unique closest point on ∂Ω . We also need the set M (Ω ) ⊂ Σ (Ω ) of those points where the distance function d(x, ∂Ω ) attains its maximum. In [7, Theorem 2] the authors show that if Ω is of class C 1 and if Σ (Ω ) = M (Ω )
(6)
then there exists a solution u of the overdetermined problem (5). Furthermore u is a web-function in the sense that u(x) depends on the distance dist(x, ∂Ω ) only. In fact, the solution u is given by ( )4/3 a4 − a3 − 3 dist(x, ∂Ω ) u(x) = (7) 4 and therefore ∂u/∂n = −a on ∂Ω . The parameter a has to be chosen such that a3 /3 equals the inner radius of Ω , i.e. the distance ρΩ from the set Σ (Ω ) = M (Ω ) to the boundary ∂Ω . Which domains satisfy the geometric condition (6)? If Ω is a ball then clearly Σ (Ω ) = M (Ω ) =the center of Ω , and the solution (7) equals the radial function (4). It is worth mentioning that the ball is the only convex domain of class C 2 satisfying (6) (see [14, Theorem 12]). An example of a non-radially symmetric domain satisfying (6) is the convex envelope of two congruent (distinct) balls: see Fig. 1. A general procedure to construct such kind of domains was indicated on p. 241 of [9] and then developed in [14] (see also the survey [10] with several pictures). Although the additional condition (2), alone, does not force the domain Ω to be a ball when the overdetermined problem (5) is solvable, Serrin–Weinberger’s theorem has been recovered by adding some restrictions on the domain Ω and under the assumption that the solution u is C 1 near ∂Ω and up to there (see [15, Corollary 23]). In Theorem 1.1 we address our attention to domains containing the origin, and in place of (2) we consider the following non-autonomous condition: ∂u − = q(|x|) (8) ∂n (where the minus sign is just to deal with non-negative functions). We show that if the prescribed function q(r) grows fast enough with respect to r then the overdetermined problem { ∆∞ u = −1 in Ω ; (9) ∂u u = 0, − = q(|x|) on ∂Ω ∂n
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is solvable only if Ω is a ball centered at the origin. This is the reason why we speak of a constrained radial symmetry. Note that a solution of (9) is sought in the whole space C 0 (Ω ), not only among web-functions. Before proceeding further let us show that without restrictions on q problem (9) is solvable in a proper ellipse. Example. Let Ω ⊂ R2 be a proper ellipse in canonical position, and let u be the solution (differentiable up ∂u to the boundary) of the ∞-torsion problem (3). Define the function g : ∂Ω → R by letting g(x) = − ∂n (x) at each point x = (x1 , x2 ) ∈ ∂Ω . Since problem (3) is invariant under reflections with respect to the coordinate axes, and by uniqueness, the function g satisfies g(x1 , x2 ) = g(±x1 , ±x2 ) for every x ∈ ∂Ω and for every combination of the signs in front of the coordinates x1 , x2 . Consequently we can define q(r) as follows. Let R1 = min |x| and R2 = max |x|. x∈∂Ω
x∈∂Ω
(10)
For every r ∈ [R1 , R2 ] define q(r) = g(x) where x is any point on the boundary ∂Ω satisfying |x| = r. Since g is symmetric as mentioned above, the definition of q(r) is independent of the choice of x. Thus we end up with a solvable instance of problem (9) although Ω is neither a disc nor a stadium-like domain. The solvability of the overdetermined problem (9) is related to the solvability of Eq. (11). Theorem 1.1. Let Ω be a bounded domain in RN , N ≥ 2, containing the origin and with a differentiable boundary, and let q(r) be any real-valued function defined in the closed interval [R1 , R2 ], where R1 , R2 are as in (10). (i) Suppose that the equation q(r) − (3r)1/3 = 0
(11)
possesses a unique solution R ∈ [R1 , R2 ], and the difference q(r) − (3r)1/3 has the same sign as r − R. Then problem (9) is solvable only in the special case when R1 = R2 = R, i.e. when Ω = BR (0). If, instead, R1 < R2 , then problem (9) is unsolvable. (ii) Suppose that the ratio q(r)/r1/3 is strictly increasing. Then problem (9) is solvable only if Ω is a ball centered at the origin. (iii) If q is continuous, and if Eq. (11) does not possess any solution, then problem (9) is unsolvable. The proof is given in Section 2 and is based on comparison with radial solutions. The procedure was applied in [16] to the classical torsion problem and to corresponding problems involving the p-Laplacian, as well as some equations of constant curvature. Recent progress on the classical torsion problem was done in [17]. The p-Laplacian has also been considered in [18,19]. In particular, it is interesting to compare problem (9) with { ∆p u = −1 in Ω ; (12) ∂u u = 0, − = q(|x|) on ∂Ω . ∂n It is known that if Ω is a bounded domain containing the origin, and if the ratio q(r)/rp−1 is monotone non-decreasing,
(13)
then problem (12) is solvable only if Ω is a ball centered at the origin [19, Theorem 1.1]. Thus, the monotonicity assumption on the ratio q(r)/r1/3 in Claim (ii) of Theorem 1.1 is not predictable just letting p → +∞ in (13). It is worth noticing that, contrary to what one may expect, the weak solution up ∈ W01,p (Ω )
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of the equation ∆p u = −1 does not converge to a viscosity solution u of ∆∞ u = −1: see [9, Section 4] for details. Let us now turn our attention to constrained radial symmetry for domains with cavities (rings). In Section 3 we prove the following result: Theorem 1.2 (Approximate Radial Symmetry). Let Ω ⊂ RN be a bounded domain containing the closure of a given ball B0 = BR0 (0) and with a differentiable boundary, and let R1 , R2 > R0 be as in (10). Let u be the solution of the Dirichlet problem ⎧ in Ω \ B 0 ; ⎨∆∞ u = 0 (14) u = a(x) on ∂B0 ; ⎩ u=b on ∂Ω , where a(x) is a continuous function and b a given constant. Define a1 = max|x|=R0 a(x), a2 = min|x|=R0 a(x), q1 = supx∈∂Ω |∇u(x)|, q2 = inf x∈∂Ω |∇u(x)|, and suppose that a2 < b. Then b − a1 R1 − R0 q2 ≤ q1 . b − a2 R2 − R0
(15)
Note that the Dirichlet problem for the infinity-Laplace equation with continuous boundary values is uniquely solvable [20, Corollary 3.14]. Furthermore, since the boundary data are constant along ∂Ω , which is differentiable by assumption, the solution of (14) is differentiable up to there [21] hence the gradient ∇u is well defined for x ∈ ∂Ω . An inequality corresponding to (15) for the Laplace operator was proved in [22, Corollary 1] and then refined in [19, Theorem 5.1]. Its purpose is to show that when the boundary values a(x), x ∈ ∂B0 , and the boundary gradient |∇u(x)|, x ∈ ∂Ω , are close to two constants then the ratio (R1 − R0 )/(R2 − R0 ) is close to 1 and therefore the domain Ω is relatively close to a ball centered at the origin (of course, oscillations of the boundary are neglected). Theorem 1.2 implies as a special case the following known result: Corollary 1.3. Let Ω be as in Theorem 1.2, and let a < b be constants. If there exists a solution of the overdetermined problem ⎧ in Ω \ B 0 ; ⎪ ⎨∆∞ u = 0 u=a on ∂B0 ; (16) ⎪ ⎩u = b, ∂u = constant on ∂Ω , ∂n then Ω is a ball centered at the origin. The more general case when the closed ball B 0 in problem (16) is replaced by a compact convex subset K ⊂ RN , and Ω is a domain containing K, is considered in [23, Theorem 3.3]. In the present paper the assumption that ∂u/∂n = constant on ∂Ω is relaxed as follows. Suppose that the overdetermined problem ⎧ in Ω \ B 0 ; ⎪ ⎨∆∞ u = 0 u=a on ∂B0 ; (17) ⎪ ⎩u = b, ∂u = q(|x|) on ∂Ω , ∂n is solvable. In order to prove that Ω is a ball centered at the origin it is enough that the product (r −R0 ) q(r) is strictly increasing. Indeed, we have: Theorem 1.4. Let Ω , R1 , R2 be as in Theorem 1.2, and let a < b be constants. Let q be a real-valued function defined in the closed interval [R1 , R2 ].
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(i) Suppose that the equation (r − R0 ) q(r) = b − a
(18)
possesses a unique solution R ∈ [R1 , R2 ], and the difference (r − R0 ) q(r) − (b − a) has the same sign as r − R. Then problem (17) is solvable only in the special case when R1 = R2 = R, i.e. when Ω = BR (0). If, instead, R1 < R2 , then problem (17) is unsolvable. (ii) Suppose that the product (r − R0 ) q(r) is strictly increasing. Then problem (17) is solvable only if Ω is a ball centered at the origin. (iii) If q is continuous, and if Eq. (18) does not possess any solution, then problem (17) is unsolvable. Problems of type (17) associated to the Laplace operator and to the p-Laplacian are considered in [19,22,24,25] as well as in [26], which is the paper that started the series. Consider for instance a positive, differentiable function q(r) such that ) d ( N −1 r (q(r))p−1 ≥ −rN −1 for some p ∈ (1, +∞). dr
(19)
Theorem 1.3 in [19] implies, in particular, that if the overdetermined problem ⎧ in Ω \ B 0 ; ⎪ ⎨∆p u = 0 u=a on ∂B0 ; ⎪ ⎩u = b, ∂u = q(|x|) on ∂Ω ∂n admits a weak solution u ∈ C 1 (Ω \ B 0 ) then Ω is a ball centered at 0. The result was previously obtained as a special case of [22, Theorem 4] under the assumption that rN −1 (q(r))p−1 is increasing instead of (19). Observe that the assumption on the product (r − R0 ) q(r) in Theorem 1.4 does not follow just letting p → +∞ in (19) . The theorem is proved in Section 3 by comparison with the solution in the annulus. Let us mention, in conclusion, that constrained radial symmetry for the equation ∆u = δ (=Dirac’s delta) has been investigated in [27]. Other kinds of constrained symmetry have also been considered: for instance, the domain of the problem turns out to be a cylinder in [25,28] and an ellipsoid in [29]. Overdetermined problems for the fractional Laplacian have also been investigated: see [30] for the free radial symmetry, and [31] for constrained symmetry. 2. Proof of Theorem 1.1 and Hopf’s Lemma Proof of Theorem 1.1. Claim (i). Let us check that the overdetermined problem (9) is solvable in the ball Ω = BR (0) whose radius R satisfies Eq. (11). By computation, from (4) we immediately deduce that the radial solution uR of the Dirichlet problem (3) satisfies (cf. [9, p. 238]) −
∂uR = (3 R)1/3 , ∂n
(20)
hence the additional condition (8) reads as (3 R)1/3 = q(R). Since this holds by assumption, problem (9) is solvable, as claimed. To prove that there is no other possibility, let Ω be now any bounded domain. Before proceeding further, let us verify that the solution u of the Dirichlet problem (3) is positive in Ω . Observe that the null function u0 ≡ 0 trivially satisfies the inequality ∆∞ u0 ≥ −1 as well as the boundary condition u0 = 0 on ∂Ω , hence by the comparison principle [11, Theorem 4] we have u ≥ 0 in Ω . Now let Br (x0 ) be a ball of radius r > 0 centered anywhere and contained in Ω , and consider the solution ur of problem (3) in Br (x0 ), which is obtained by replacing R with r in (4). Since u ≥ 0 on ∂Br (x0 ), the comparison principle implies u ≥ ur > 0 in Br (x0 ). Since x0 ∈ Ω is arbitrary, we have u > 0 in all of Ω , as claimed.
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Now assume that u is a solution of the overdetermined problem (9). Denote by Bi = BRi (0) the ball of radius Ri , i = 1, 2, centered at the origin, so that B1 ⊂ Ω ⊂ B2 . Define the function ui as follows: ui (x) =
34/3 ( 4/3 4/3 ) Ri − |x| . 4
Thus, ui is the solution of problem (3) in the ball Bi . Arguing as above we find u1 ≤ u in B1 .
(21)
Concerning u2 , we have u = 0 ≤ u2 along the boundary ∂Ω : hence by the comparison principle [11, Theorem 4] we obtain u ≤ u2 in all of Ω . Let us consider a point P1 ∈ ∂B1 ∩ ∂Ω , i.e., a point on ∂Ω such that |P1 | = R1 . Observe that the outer normal n to ∂Ω at P1 equals P1 /R1 = the outer normal to B1 . By (20) and (21), and since u1 = u at P1 , we deduce (3 R1 )1/3 = −
∂u ∂u1 (P1 ) ≤ − (P1 ) = q(R1 ), ∂n ∂n
(22)
where the last equality comes from (8). Since Eq. (11) has a unique solution R ∈ [R1 , R2 ] by assumption, and the difference q(r) − (3r)1/3 is negative for any r < R, inequality (22) implies R1 = R (the case R1 > R being excluded). Arguing at a point P2 ∈ ∂B2 ∩ ∂Ω in a similar manner we find q(R2 ) = −
∂u ∂u2 (P2 ) ≤ − (P2 ) = (3 R2 )1/3 ∂n ∂n
(23)
and consequently R2 = R. But then we deduce R1 = R2 and Claim (i) follows. Claim (ii). If problem (9) is solvable, then arguing as before we arrive at (22) and (23), hence the ratio ρ(r) = q(r)/r1/3 satisfies ρ(R2 ) ≤ 31/3 ≤ ρ(R1 ). But since ρ(r) is strictly increasing by assumption, we must have R1 = R2 and the conclusion follows. To prove Claim (iii) we assume the solvability of problem (9) and show that Eq. (11) also has a solution. Using again (22) and (23) we see that the difference q(r) − (3r)1/3 is non-negative at R1 and non-positive at R2 . The claim follows because q is continuous. □ As a by-product of the proof above (see in particular (22)) it turns out that the difference q(r) − (3r)1/3 must be non-negative at r = R1 in order that problem (9) is solvable. Since the Dirichlet problem (3) is translation-invariant, the argument proves the following boundary-point lemma: Lemma 2.1 (Hopf). Let Ω be a bounded open subset of RN satisfying the interior sphere condition at some x1 ∈ ∂Ω , i.e. there exists BR (x0 ) ⊂ Ω such that x1 ∈ ∂BR (x0 ). Then the solution u of the Dirichlet problem (3) satisfies lim inf h→0+
u(x1 + (x0 − x1 ) h/R) ≥ (3 R)1/3 > 0. h
Hopf’s lemma for infinity-harmonic functions (solutions of ∆∞ u = 0) is found in [32, Lemma 1]. For the fractional Laplacian see for instance [31]. 3. Proofs of Theorems 1.2 and 1.4 The proofs of Theorems 1.2 and 1.4 are based on the comparison principle for infinity-harmonic functions [20, Corollary 3.11]. More precisely, the solution u is compared to the solutions u1 , u2 in suitable annuli. Let R > R0 > 0 and define BR = BR (0) and B0 = BR0 (0), for shortness. We have:
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Lemma 3.1. Let a, b be two constants. The unique viscosity solution of the Dirichlet problem ⎧ in BR \ B 0 ; ⎨∆∞ u = 0 u=a on ∂B0 ; ⎩ u=b on ∂BR ,
(24)
is the radial function uR0 ,R (x) given by uR0 ,R (x) =
b−a (|x| − R0 ) + a. R − R0
(25)
Proof . If a = b then the function uR0 ,R reduces to the constant uR0 ,R ≡ a which trivially satisfies (24). x b−a in the closed annulus If, instead, a ̸= b then uR0 ,R has a non-vanishing gradient DuR0 ,R (x) = R−R 0 |x| B R \ B0 , hence the unit vector ν = −DuR0 ,R /|DuR0 ,R | = sign(a − b) x/|x| is well defined. Since uνν ≡ 0 2 it turns out that uR0 ,R is a classical solution of the equation uνν |uν | = 0 and a fortiori it is a viscosity solution of (24). Uniqueness follows from [20, Corollary 3.14]. □ Proof of Theorem 1.2. We concentrate on the case when a1 < b, otherwise (15) holds trivially. Observe that the solution u of (14) satisfies a2 ≤ u ≤ b in Ω \ B0 .
(26)
Indeed, the constant functions u(x) ≡ a2 and u(x) = b are infinity-harmonic. Furthermore, we have a2 ≤ a(x) = u(x) ≤ a1 for |x| = R0 and we are assuming a1 < b. Finally, a2 ≤ a1 < b = u(x) on ∂Ω . Hence (26) follows from the comparison principle [20, Theorem 3.11]. As a consequence of (26) we have ∂u/∂n ≥ 0 on ∂Ω and therefore we may write ∂u/∂n = |∇u| there. From this we get the inequalities q2 ≤
∂u ≤ q1 ∂n
on ∂Ω ,
which enter into (27) and (28). Now denote by Bi = BRi (0), i = 1, 2, the ball of radius Ri centered at the origin, as in Section 2, and let ui (x) be the radial solution of the Dirichlet problem (24) in the annulus Bi \ B 0 with boundary value a = ai . In other terms, define ui = uR0 ,Ri where uR0 ,R is given in (25). Let us compare ui to the solution u of problem (14). When |x| = R0 we have u2 (x) ≤ u(x) ≤ u1 (x) by definition. When |x| = R1 , instead, we have u(x) ≤ b = u1 (x) by (26), hence u(x) ≤ u1 (x) in B 1 \ B0 . Concerning u2 , since u2 (x) ≤ b = u(x) for x ∈ ∂Ω , by the comparison principle we get u2 (x) ≤ u(x) in Ω \ B0 . Since u = u1 at every point P1 ∈ ∂B1 ∩ ∂Ω , and since the outer normals of ∂B1 and ∂Ω coincide there, we get b − a1 ∂u1 ∂u = (P1 ) ≤ (P1 ) ≤ q1 . R1 − R0 ∂n ∂n
(27)
Furthermore, since u2 = u at every P2 ∈ ∂B2 ∩ ∂Ω , we obtain q2 ≤
∂u ∂u2 b − a2 (P2 ) ≤ (P2 ) = . ∂n ∂n R2 − R0
By (27) and (28) we arrive at (15), and the proof is complete.
□
(28)
A. Greco / Nonlinear Analysis: Real World Applications 37 (2017) 239–248
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To prove Theorem 1.4 we now let a1 = a2 = a in (27) and (28). Proof of Theorem 1.4. Claim (i). Let us check that the overdetermined problem (17) is solvable in the annulus BR \ B 0 whose outer radius R satisfies Eq. (18). From (25) we see that the radial solution uR0 ,R of the Dirichlet problem (24) satisfies b−a ∂ uR0 ,R = on ∂BR ∂n R − R0 hence the additional boundary condition in the overdetermined problem (17), which reads as ∂ uR ,R = q(R) on ∂BR ∂n 0 reduces to (R−R0 ) q(R) = b−a, which holds true by hypothesis. Therefore the overdetermined problem (17) is solvable in BR \ B 0 , as claimed. To proceed further, assume now that the overdetermined problem (17) has a solution u in the domain Ω \ B 0 . Arguing as in the proof of Theorem 1.2, and letting a1 = a in (27), we obtain b−a ∂u ≤ (P1 ) = q(R1 ), R1 − R0 ∂n
(29)
where the last equality follows from the additional boundary condition in (17). This and the assumption that (r − R0 ) q(r) < b − a for all r < R imply R1 = R, the case R1 > R being impossible because R ∈ [R1 , R0 ]. Similarly, letting a2 = a in (28) we obtain q(R2 ) =
b−a ∂u (P2 ) ≤ . ∂n R2 − R0
(30)
But since (r − R0 ) q(r) > b − a for all r > R we deduce R2 = R = R1 and Claim (i) follows. Claim (ii). If problem (17) is solvable, then by (29) and (30) we see that the product p(r) = (r − R0 ) q(r) satisfies p(R2 ) ≤ b−a ≤ p(R1 ). But since p(r) is strictly increasing by assumption we conclude that R1 = R2 , which proves the claim. To prove Claim (iii) suppose again that there exists a solution u of the overdetermined problem (17). From (29) and (30) it follows that the difference (r −R0 ) q(r)−(b−a) is non-negative at R1 and non-positive at R2 . Since q(r) is continuous, Eq. (18) must have at least a solution, contrary to the assumption. Hence Claim (iii) must hold, and the proof is complete. □ Acknowledgment The author is a member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilit`a e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). References
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