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q 2 Integrating the equation in (1) over Ω, we see that Ω u − u = 0 for every solution of (1), hence every nontrivial solution is sign changing. Let us consider the| fu| nctional R 1 1 ϕ : W 1,2(Ω) R, ϕ(u)= ∇u 2 dx u q dx. → 2 Ω | | − q Ω | | Z Z If 1 < q < 2, ϕ is of class C1, and critical points of ϕ are precisely the weak solutions of (1). Moreover, 2,α Ω since the nonlinearity in (1) is Hölder continuous, weak solutions u of (1) are in Cloc ( ) by elliptic regularity, and the restriction of u to the open set u = 0 is of class C∞. If q = 1, then ϕ fails to be differentiable and weak solutions of (1) are in general{ 6 not} of class C2, but they are still strong solutions 2,p Ω ∞ 1,α Ω α contained in Wloc ( ) for every p < and thus contained in Cloc ( ) for every (0,1). The purpose of this paper is to derive the existence of solutions of (1) with minimal∈ energy and to characterize these solutions both variationally and in terms of their qualitative properties. We first con- sider the case 1 < q < 2. In order to obtain least energy nodal solutions, we minimize the functional ϕ on the set

1,2 q 2 1,2 (4) N := u W (Ω) : u − udx = 0 W (Ω). ∈ Ω | | ⊂ n Z o We shall see that minimizers of ϕ N solve (1), so these minimizers are precisely the least energy nodal solutions of (1). Note that this property| does not follow from the Lagrange multiplier rules since N is not a C1-manifold if q < 2. Our main result for the case 1 < q < 2 is the following.

Theorem 1.1. Suppose that 1 < q < 2, and let m := inf ϕ(u). Then we have: u N ∈ (i) ϕ attains the value m < 0 on N . 1 (ii) Every minimizer u of ϕ on N is a sign changing solution of (1) such that u− (0) Ω has vanishing Lebesgue measure. ⊂ (iii) If Ω is a bounded radial domain, then every minimizer u of ϕ on N is foliated Schwarz symmet- ric. (iv) If Ω = B1(0) is the unit ball, then every minimizer u of ϕ on N is a nonradial function. We add some comments on these results. The property (i) follows by standard arguments based on the 2 weak lower continuity of the Dirichlet integral u Ω ∇u dx. To show that every minimizer of ϕ on N is a solution of (1), we use a saddle point characterization7→ | | of N (see Lemma 2.1 below). The most R difficult part is the unique continuation property of minimizers of ϕ on N , i.e., the fact that their zero q 2 sets have vanishing Lebesgue measure. Note that, due to the fact that the nonlinearity u u − u is not locally Lipschitz, the linear theory on unique contination does not apply. Moreover, as can7→ | be| seen from very simple ODE examples already, nontrivial solutions of semilinear equations of the type ∆u = f (u) with non-Lipschitz f may have very large zero sets. It is an interesting open problem whether− every nontrivial solution of (1) has the unique continuation property;we conjecture that this is true. The proofof (iii) is again quite short and essentially follows the arguments in [3]. In contrast, the nonradiality property for least energy nodal solutions stated in (iv) is not immediate. The idea is to use properties of directional derivatives of u. For problems with C1-nonlinearities, nonradiality properties have successfully been derived via directional derivatives in the case of Dirichlet problems [1] and Neumann problems [7], while the methods in these papers differ significantly due to the impact of the boundary conditions. A particular difficulty of the present problem is to analyze for which solutions the problem (1) has a meaningful linearization, see Proposition 2.3 for a first result on this question. Let us now consider the case q = 1. In this case, the functional ϕ is not differentiable, so that the techniques used when 1 < q < 2 can not be applied. Moreover, the saddle point characterization in 1,2 Lemma 2.1 fails in the case q = 1, i.e., for the set N = u W (Ω) : Ω sgn(u)dx = 0 . Nevertheless, we derive the same conclusions as in Theorem 1.1 by adjusting{ ∈ the variational principle.} More precisely, R LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 3 we consider minimizers of the restriction of ϕ to the set

M := u W 1,2(Ω) : u > 0 u < 0 u = 0 . ∈ |{ }| − |{ }| ≤ |{ }| Note that M is strictly larger than N . Our main results for the case q 1 are collected in the following = Theorem.

Theorem 1.2. Suppose that q = 1 in (1) and the definition of ϕ, andlet m := inf ϕ(u). Then we have: u M ∈ (i) ϕ attains the value m < 0 on M . 1 (ii) Every minimizer u of ϕ on M is a sign changing solution of (1) such that u− (0) Ω has vanishing Lebesgue measure. ⊂ (iii) If Ω is a bounded radial domain, then every minimizer u of ϕ on M is foliated Schwarz symmet- ric. (iv) If Ω = B1(0) is the unit ball, then every minimizer u of ϕ on M is a nonradial function. The general strategy for the proofs of (i)-(iii) is the same as in Theorem 1.1, but the details are quite different due to the geometry of M and the fact that ϕ fails to be differentiable in the case q = 1. The proof of (iv) is completely different, since there seems to be no way to use directional derivatives to prove nonradiality. Instead, our proof of (iv) is based on inequalities comparing the value m with the least energy of radial nodal solutions of 1 (in the case q = 1). In fact, the latter value can be computed explicitly once we have shown that least energy radial nodal solutions are strictly monotone in the radial variable and therefore have exactly two nodal domains. We then compare this value with upper estimates x1 for the value m obtained by using the test functions v(x)= x1 (if n = 2) or v(x)= (if n 3). x ≥ The paper is organized as follows. After proving some fundamental properties| of| the functional ϕ in the case 1 < q < 2 (Section 2), we show that least energy nodal solutions satisfy a unique continuation property (Section 3), and we then deal with symmetry results in radially symmetric domains (Section 4). In particular, Theorem 1.1 will readily follow from Lemma 2.2 and Theorems 3.4, 4.1 and 4.2 below. In Section 5, we turn to the case q = 1 and prove Theorem 1.2. Finally, we mention that it is not straightforward to obtain similar results for the Dirichlet problem corresponding to (1). Indeed, least energy solutions of the Dirichlet problem might have different varia- tional characterizations on different domains, so the situation is more complicated than in the Neumann case. The Dirichlet problem is considered in a paper in preparation by the authors.

Acknowledgement. Part of this paperwas written duringseveral visits of E.P.to the Goethe-Universität in Frankfurt. He would like to thank the institution for its kind hospitality.

2. THE VARIATIONAL FRAMEWORK IN THE CASE q > 1. For fixed q [1,2) and u W 1,2(Ω) with ∇u 0, we put ∈ ∈ 6≡ 1 q 2 q Ω u − ∞ (5) t∗(u)= | | 2 (0, ). Ω ∇u ∈  R | |  It is easy to see that t∗(u) is the unique minimizerR of the function (0,∞) R, t ϕ(tu). → 7→ Suppose that 1 < q < 2 from now on, and consider the set N defined in (4). For any u N 0 , the value t (u) is well defined and satisfies t (u)u N and ϕ(t (u)u) < 0. This in particular∈ implies\{ that} the ∗ ∗ ∈ ∗ infimum m of ϕ N is negative. The next lemma highlights the saddle point structure given by ϕ and the set N . | 4 ENEA PARINI, TOBIAS WETH

Lemma 2.1. 1 q 2 (i) For every u L (Ω) there exists precisely one c = c(u) R such that Ω u+c (u+c)dx = 0. ∈ ∈ | | − Moreover, the map L1(Ω) R,u c(u) is continuous. → 7→ R (ii) If u W 1,2(Ω), then ∈ ∂ ∂ (6) ϕ(u + c) > 0 for c < c(u) and ϕ(u + c) < 0 for c > c(u). ∂c ∂c In particular, ϕ(u + c(u)) = maxϕ(u + c). c R ∈ 1 q 2 Proof. (i) Let u L (Ω). Since the map R R, t t − t is strictly increasing, there exists at most ∈ q 2 → 7→ | | one c = c(u) such that Ω u + c (u + c)dx = 0. Moreover, since | | − R q 2 q 2 lim u + c − (u + c)dx = ∞ and lim u + c − (u + c)dx = ∞, c ∞ Ω | | − c ∞ Ω | | →− Z → Z 1 1 there exists precisely one such c = c(u). Now consider u,un L (Ω), n N such that un u in L (Ω). ∈ ∈ → We first show that c(un) remains bounded as n ∞. Suppose by that, after passing to a → subsequence, c(un) +∞ as n ∞. Passing again to a subsequence, we may assume that c(un) > 0 for → → all n and un u pointwise a.e. in Ω. Moreover, by [12, Lemma A.1] we may assume that there exists 1 → u˜ L (Ω) with un u˜ a.e. in Ω for all n N. Since u˜ un + c(un) a.e. in Ω for all n N, we also have∈ | |≤ ∈ − ≤ ∈ q 2 q 2 u˜ − u˜ un + c(un) − (un + c(un)) a.e. in Ω for all n N. −| | ≤ | | ∈ q 2 Hence, since un + c(un) (un + c(un)) ∞ pointwise a.e. in Ω, Fatou’s Lemma implies that | | − → q 2 un + c(un) − (un + c(un)) ∞ as n ∞, Ω | | → → Z which contradicts the definition of the map c. In the same way, we obtain a contradiction when assuming that c(un) ∞ for a subsequence. Consequently, c(un) remains bounded as n ∞. We now argue →− → by contradiction, supposing that c(un) c(u) as n ∞. Then we may pass to a subsequence such that 6→ → c(un) c = c(u) as n ∞. Since the map → 6 → 1 q 2 L (Ω) R, u u − u → 7→ Ω | | Z 1 is continuous and un + c(un) u + c in L (Ω), we have that → q 2 u + c − (u + c)dx = 0. Ω | | Z By the uniqueness property noted above, we then deduce that c = c(u), a contradiction. We thus conclude 1 that c(un) c(u) as n ∞, and this shows the continuity of the map c : L (Ω) R. (ii) We have→ → → ∂ > 0 for c < c u ; q 2 ( ) ϕ(u + c)= u + c − (u + c)dx ∂c − Ω | | < 0 for c > c(u), Z ( as claimed. 

Lemma 2.2. The functional ϕ attains the value m < 0 on N . Moreover, every minimizer u of ϕ on N is a sign changing solution of (1) Proof. We first note that, as a consequence of Lemma 2.1(ii), we have 2 2 Ω 2 q 2 Ω 2 qµ 1 ∇ 2 N u Lq Ω = min u + c Lq Ω − min u + c 2 Ω − 2− u dx for u , k k ( ) c R k k ( ) ≤ | | c R k kL ( ) ≤ | | Ω | | ∈ ∈ ∈ Z LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 5 where µ2 > 0 is the first nontrivial eigenvalue of the Neumann Laplacian on Ω. As a consequence, the functional ϕ is coercive on N . Let (un)n N be a minimizing sequence for ϕ. Then (un) is bounded, ⊂ 1,2 N q and we may pass to a subsequence such that un ⇀ u W (R ). Then un u in L (Ω), ∈ → 2 2 ∇u dx liminf ∇un dx Ω | | ≤ n ∞ | | Z → and q 2 q 2 un − un dx u − udx as n ∞. Ω | | → Ω | | → Z Z Consequently, we have u N , and u satisfies ϕ(u) liminfϕ(un). Hence u is a minimizer for ϕ on N . ∈ ≤ n ∞ Next, we let u N be an arbitrary minimizer for ϕ on→ N . We show that u is a critical point of ϕ. ∈ 1,2 Arguing again by contradiction, we assume that there exists v W (Ω) such that ϕ′(u)v < 0. Since ϕ is a C1-functional on W 1,2(Ω), there exists ε > 0 with the following∈ property: 1,2 For every w W (Ω) with w 1,2 Ω < ε and every t (0,ε) we have ∈ k kW ( ) ∈ ϕ(u + w +tv) ϕ(u + w) εt. ≤ −

Since the map c is continuous and c(u)= 0 by definition of c, there exists t (0,ε) such that c(u + ∈ k tv) 1,2 Ω < ε, and thus kW ( ) ϕ(u +tv + c(u +tv)) ϕ(u + c(u +tv)) εt ϕ(u) εt < ϕ(u). ≤ − ≤ − Since u + tv + c(u + tv) N , this contradicts the definition of m. Finally, since m < 0 by the remarks in the beginning of this∈ section, every minimizer u N of ϕ is a nonzero function and therefore sign changing by the definition of N . ∈  We close this section with a result on the existence of second derivatives of ϕ which we will need in Section 4.1 below. Proposition 2.3. Let (7) W := v C1(Ω) : ∇v(x) = 0 for every x Ω with v(x)= 0 . { ∈ 6 ∈ } Then W C1(Ω) is an open (with respect to the C1-topology) having the following properties: ⊂ 2 2 (i) If ∂Ω is of class C , then the restriction ϕ W is of class C with | q 2 1 ϕ′′(u)(v,w)= ∇v∇wdx (q 1) u − vw foreveryv,w C (Ω). Ω − − Ω | | ∈ Z Z ∂Ω 2,1 W 3,p Ω 1 (ii) If is of class C , and u is a weak solution of (1), then u W ( ) for p (1, 2 q ), ∈ 2,p ∈ ∈ − and the partial derivatives ux W (Ω) are strong solutions of the problem i ∈ q 2 (8) ∆ux = (q 1) u − ux in Ω. − i − | | i Proof. It is easy to see that W is open in C1(Ω). We first show Claim 1: If s (0,1) and 1 p < 1 , then the map ∈ ≤ s p s γs : W L (Ω), γs(u) u − → 7→ | | is well defined and continuous. To see this, let K W be a compact subset (with respect to the C1-norm). We claim that exists κ > 0 such that ⊂ (9) u δ min κδ, Ω for every δ > 0, u K . |{| |≤ }|≤ { | |} ∈ 1 Ω 1 RN In order to provethis estimate, we consider a bounded linear extension map ext : C ( ) Cb ( ), where 1 RN 1 RN → Cb ( ) denotes the Banach space of bounded C -functions on with bounded gradient. Such a map 6 ENEA PARINI, TOBIAS WETH

∂Ω C 2 L K 1 RN Ω RN Ω RN exists since is of class . We put := ext( ) Cb ( ) and b := x : dist(x, ) < b for b > 0. Since L is compact, there exists a constant⊂ b > 0 such that { ∈ }⊂

∂v (10) max (x) 2b for every v L and x Ωb with v(x) b. j=1,...,N ∂x j ≥ ∈ ∈ | |≤

We now fix v0 L . Then there exists a positive integer d = d(v0) and a finite number of cubes W1,...,Wd of equal length∈l > 0 such that N (I) every cube has the form [x1,x1 + l] [xN,xN + l] with some x R ; d ×···× ∈ (II) Ω Wi Ωb; ⊂ i=1 ⊂ ∂ (III) osc vS0 < b for i = 1,...,d, j = 1,...,N. ∂ x j 2 Wi

Moreover, there exists a neighborhood U (v0) of v0 in L such that ∂v (11) osc < b for i = 1,...,d, j = 1,...,N and v U (v0). Wi ∂x j ∈

For every v U (v0), i 1,...,d and δ (0,b) we then have ∈ ∈{ } ∈ N 1 l − (12) x Wi : v(x) δ δ. |{ ∈ | |≤ }|≤ b

∂ v Indeed, if there exists x Wi with v(x) δ b, then ∂ b on Wi for some j = j(i) by (10) and (11); ∈ | |≤ ≤ x j ≥ in particular, v is strictly monotone in the j-th coordinate direction on Wi. Hence (12) easily follows by

Fubini’s theorem. As a consequence, we have the estimate

N 1 dl − x Ω : v(x) δ δ for every v U (v0), δ (0,b). |{ ∈ | |≤ }|≤ b ∈ ∈ Since L is compact, it can be covered by finitely many neighborhoods constructed as above, and hence there exists d > 0 such that ∗ x Ω : v(x) δ d δ for every v L , δ (0,b). |{ ∈ | |≤ }|≤ ∗ ∈ ∈ Ω By the construction of L , (9) follows with κ := max d , | | . As a consequence of (9), we have { ∗ b } ∞ ∞ ∞ sp sp 1 1 u − dx = u − τ dτ = u τ− ps dτ min κτ− ps , Ω dτ < ∞, Ω | | 0 |{| | ≥ }| 0 |{| |≤ }| ≤ 0 { | |} Z Z Z Z K 1 γ γ for every u , since ps > 1. In particular, the map s is well defined. To see the continuity of s, let ∈ 1 (un)n W be a sequence such that un u as n ∞ with respect to the C -norm. We then consider the ⊂ → → compact set K := un,u : n N and κ > 0 such that (9) holds. For given ε > 0, we then fix c > 0 sufficiently small such{ that ∈ } ∞ p 1 (13) 2 min κτ− ps , Ω dτ < ε. 2c ps { | |} Z( )− By Lebesgue’s theorem, it is easy to see that

s s p (14) ( un − u − ) dx 0 as n ∞. u >c | | − | | → → Z| | LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 7

Moreover, there exists n0 N be such that u c un 2c for n n0. Consequently, ∈ {| |≤ } ⊂ {| |≤ } ≥ p s s p 1 ps ps un − u − dx 2 − un − + u − dx u c | | − | | ≤ u c | | | | Z| |≤ Z| |≤   p 1 ps ps 2 − un − dx + u − dx ≤ un 2c | | u c | | Z| ∞|≤ Z| |≤ ∞  p 1 1 1 = 2 − un τ− ps dτ + u τ− ps dτ ps ps (2c)− |{| |≤ }| c− |{| |≤ }| ∞Z Z  p 1 2 min κτ− ps , Ω dτ < ε for n n0. ≤ 2c ps { | |} ≥ Z( )− Combining this with (14), we conclude that p s s limsup un − u − dx ε. n ∞ Ω | | − | | ≤ → Z Since ε > 0 was given arbitrarily, we conclude that

p γ γ p s s ∞ s(un) s(u) p Ω = un − u − dx 0 as n . k − kL ( ) Ω | | − | | → → Z Hence Claim 1 follows. 2 We now turn to the proof of (i). To show that ϕ W is of class C , it suffices to show that | 1 ψ : W R, ψ(u)= u q dx → q Ω | | Z is of class C2 with q 2 1 (15) ψ′′(u)(v,w) = (q 1) u − vw for every v,w C (Ω). − Ω | | ∈ Z By standard arguments, ψ is of class C1 with

q 2 1 ψ′(u)v = u − uv for every v C (Ω). Ω | | ∈ Z 1 Let u W and v,w C (Ω) with v ∞ Ω , w ∞ Ω < 1. For t R 0 we have ∈ ∈ k kL ( ) k kL ( ) ∈ \{ } 1 ψ′(u +tw)v ψ′(u)v = It + Jt t − with   q 2 q 2 q 2 q 2 u +tw − (u +tw) u − u 1 u +tw − (u +tw) u − u It = | | − | | vdx, Jt = | | − | | vdx. u > t t t u t t Z| | | | Z| |≤| | Note that, with s := 2 q (0,1), − ∈ 1 s It = (q 1) u + τtw − vwdxdτ − 0 u > t | | Z Z| | | | 1 1 s = (q 1) γs(u + τtw)vwdxdτ u + τtw − vwdxdτ − 0 Ω − 0 u t | | Z Z Z Z| |≤| |  with 1 q 2 γs(u + τtw)vwdxdτ γs(u)vwdx = u − vwdx as t 0 0 Ω → Ω Ω | | → Z Z Z Z and, by applying Hölder’s inequality with some p (1, 1 ), ∈ s 1 1 1 τ s τ 1 p γ τ τ u + tw − vwdxd u t − s(u + tw) Lp(Ω) d 0 as t 0. 0 u t | | ≤ |{| | ≤ | |}| 0 k k → → Z Z| |≤| | Z

8 ENEA PARINI, TOBIAS WETH

Moreover, by choosing κ > 0 such that (9) holds for u, we find that q 1 q 1 1 q 1 q 1 q 2 u − u − Jt u +tw − + u − dx = t − + w + dx | |≤ t u t | | | | | | u t t t | | Z| |≤| | Z| |≤| | q 2 q 1  q 1 q 1 t − (2 − + 1) u t κ t − (2 − + 1) 0 as t 0. ≤ | | |{| |≤ }|≤ | | → → Combining these estimates, we conclude that

1 q 2 ψ′(u +tw)v ψ′(u)v (q 1) u − vwdx as t 0. t − → − Ω | | →   Z Hence the second directional derivatives of ϕ exists at u W and satisfy (15). By Claim 1 above, it also follows that the second derivatives depend continuously∈ on u W , so that ψ C2(W ). The proof of (i) is thus complete. ∈ ∈ q 2 ∞ Ω ε To prove (ii), put v := u − u. Then for f Cc ( ) and > 0 sufficiently small we have, by the diver- gence theorem, | | ∈ ∂ ∂ σ ε v xi f dx xi vfdx v f d 0 as 0. u ε − u ε ≤ u =ε | || | → → Z| |≥ Z| |≥ Z| |

Hence ∂ ∂ ∂ q 2 v xi f dx = lim v xi f dx = lim xi vfdx = (q 1) u − uxi f dx Ω ε 0+ u ε ε 0+ u ε − Ω | | Z → Z| |≥ → Z| |≥ Z C ∞ Ω q 2 for f c ( ), which shows that vxi = (q 1) u − uxi in distributional sense. Considering s = 2 q ∈ − | p| Ω 1 ∆ −Ω again, we deduce from Claim 1 above that vxi L ( ) for p (1, 2 q ). Since moreover u = v in , ∈ ∈ − − it follows from standard regularity theory that u W 3,p(Ω). Moreover, ∈ ∆ux = ∂x ( ∆u)= vx in Ω − i i − i in strong sense for i = 1,...,N, which shows (8). 

3. THE UNIQUE CONTINUATION PROPERTY IN THE CASE q > 1 1 In this section we still consider the case 1 < q < 2, and we show that the set u− (0) Ω has zero Lebesgue measure for every minimizer of ϕ on N . For this we need some preliminaries. We⊂ recall that, for a measurable subset A RN , a point x RN is called a point of density one for A if ⊂ ∈ A Br(x) lim | ∩ | = 1. r 0 Br(0) → | | If A RN is measurable, then, by a classical result (see e.g. [5, p.45]), a.e. x A is a point of density one for A⊂. We also need the following simple observation. ∈ Lemma 3.1. Let α > 0,andlet f : (0,∞) R be a nonnegative function with the following properties: → (i) f is bounded on [ε,∞) for every ε > 0; (ii) f (r)= o(r α ) as r ∞. − → Then for any r > 0 there exists s > 0 with α s f (s) f (r) and f (t) 2 f (s) for t ,2s . ≥ ≤ ∈ 2 h i Proof. We argue by contradiction. If the assertion was , we would find r > 0 and a sequence (sn)n ⊂ (0,∞) such that s0 = r and, for every n N, ∈ sn α (16) sn 1 ,2sn and f (sn 1) > 2 f (sn). + ∈ 2 + h i LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 9

Without loss of generality, we may also assume that f0 := f (r)= f (s0) > 0 (otherwise we replace sn by sn+1 for every n). From (16) we deduce n nα (17) sn 2− r and f (sn) 2 f0. ≥ ≥ ∞ α α N Assumption (i) then implies that sn 0 as n , whereas (17) implies that f (sn)sn r f0 for all n . This contradicts the assumption f (r→)= o(r →α ) as r 0. ≥ ∈  − → Proposition 3.2. Let u be a solution of q 2 (18) ∆u = u − u in Ω, − | | 2 1 and suppose that x0 Ω is a pointof density one for the set u (0) Ω. Thenu(x)= o( x 2 q ) asx x0. ∈ − ⊂ | | − → Here we recall that, by elliptic regularity theory, a distributional solutions u of (18) contained in 1,2 Ω 2,α Ω α Wloc ( ) is in fact a classical solution in Cloc ( ) for some > 0.

Proof of Proposition 3.2. Without loss, we may assume that x0 = 0 Ω and that u is bounded in Ω ∈ N (otherwise we replace Ω by a compactly contained subdomain containing x0). We extend u to all of R N 1 by setting u 0 on R Ω. Since 0 is a point of density one for the set u− (0) and u is continuous in 0 and bounded≡ on Rn, the\ function 2 ∞ R 2 q f : (0, ) , f (r)= r− − sup u(x) → x =r | | | | α 2 satisfies the assumptions of Lemma 3.1 with = 2 q . We first show the following − Claim: The function f is bounded on (0,∞).

Arguing by contradiction, we assume that there exists a sequence of radii rn > 0, n N such that ∈ f (rn) ∞ as n ∞. By Lemma 3.1, we then find sn > 0, n N such that, for all n N, → → ∈ ∈ sn (19) f (sn) f (rn) and f (t) C f (sn) for t ,2sn , ≥ ≤ ∈ 2 q h i 2 q ∞ ∞ where C := 2 − . In particular, f (sn) as n . By the definition of f , this implies that sn 0 as ∞ → → Ω N Ω → n , so without loss we may assume that B2sn (0) for all n . We now put 0 := B2(0) B 1 (0) → ⊂ ∈ \ 2 and define, for n N, the functions vn : Ω0 R as ∈ → 2 2 q s−n − u(snx) vn(x)= . f (sn)

The functions vn solve the equations q 2 q 2 (20) ∆vn = f (sn) − vn − vn in Ω0, − | | whereas 2 2 2 x 2 q ( x sn)− 2 q u(snx) 2 2 q f ( x sn) 2 − − − 2 q Ω N vn(x) = | | | | | | | | 2 − C for x 0, n . | | f (sn) ≤ f (sn) ≤ ∈ ∈ 1 RN Moreover,there exists a sequence of points xn S := y : y = 1 such that u(snxn) = sup x =sn u(x) = 2 ∈ { ∈ | | } | | | | | | 2 q sn− f (sn) and hence vn(xn) = 1 for n N. Using (20), elliptic regularity theory and the fact that q 2 | | ∈ f (sn) 0 as n ∞, we may pass to a subsequence such that − → → 1 xn x¯ S , → ∈ 1 vn v in C (Ω0), → loc 10 ENEA PARINI, TOBIAS WETH where v is a harmonic function in Ω0 such that v(x¯)= 1. In particular, v 0. On the other hand, since 0 1 6≡ is a point of density one for the set u− (0), the sets An := x B2(0) B 1 (0) : vn(x) = 0 satisfy { ∈ \ 2 6 } An x B2s : u(x) = 0 | | |{ ∈ n 6 }| 0 as n ∞ B2(0) ≤ B2s (0) → → | | | n | and thus v 0 a.e. in B2(0) B 1 (0). This is a contradiction, and thus the above claim is true. ≡ \ 2 To finish the proof of the proposition, it thus remains to show that f (r) 0 as r 0. Arguing again by → → contradiction, we assume that there exists ε > 0 and a sequence of radii rn > 0, n N such that rn 0 ∞ ε N Ω ∈N → as n and f (rn) for all n . We may assume that B2rn (0) for all n . We then consider → ≥ 2 ∈ ⊂ ∈ 2 q wn : Ω0 R, wn(x)= rn− − u(rnx) for n N. It follows from the claim above that the functions wn are → ∈ uniformly bounded in Ω0. Moreover, wn solves q 2 ∆wn = wn − wn in B2(0) B 1 (0), − | | \ 2 1 and there exists a sequence of points xn S , n N such that w(xn)= f (rn) ε for all n N. Using elliptic regularity theory again, we may pass∈ to a∈ subsequence such that ≥ ∈ 1 xn x¯ S , → ∈ 1 wn w in C (Ω0), → loc 2,α q 2 where w C (Ω0) is a weak solution of ∆w = w w in Ω0 such that w(x¯) ε. In particular, ∈ loc − | | − ≥ w 0. However, by the same argument as above, the sets An := x B2(0) B 1 (0) : wn(x) = 0 satisfy 6≡ { ∈ \ 2 6 } lim An = 0 and therefore w 0 a.e. in B2(0) B 1 (0). This yields a contradiction, and thus we conclude n ∞ | | ≡ \ 2 that→ f (r) 0 as r 0, as required.  → → Next, we consider the family of energy functionals

1,2 1 2 1 q (21) ϕr : W (Br(0)) R, ϕr(v)= ∇v dx v dx → 2 B 0 | | − q B 0 | | Z r( ) Z r( ) for r > 0. We also consider the scaling map

1,2 1,2 2 x (22) Tr : W (B1(0)) W (Br(0)), [Trv](x)= r 2 q v . → − r We then have the following.   1,2 Proposition 3.3. Let u W (Ω) be a solution of (18) in Ω, andlet x0 Ω be a point of density one for 1 ∈ 1,2 ∈ u (0) Ω. Moreover, let K W (B1(0)) be a compact set such that − ⊂ ⊂ 0 cK := supϕ1(v) < 0. v K ∈ Then there exists r0 > 0 with the following property: Ω ϕ ϕ Br0 (x0) is contained in , and for every r (0,r0) and every v K we have (u + vr) < (u), where 1,2 ∈ ∈ vr W (Ω) is defined by ∈ 0 [Trv](x x0) x Br(x0); vr(x)= − ∈ 0 x Ω Br(x0). ( ∈ \ Proof. Without loss, we may assume that x0 = 0 Ω. Let v K. We then have ∈ ∈ 1 q q q ϕ(u + vr)= ϕ(u)+ ϕr(vr)+ ∇u∇vr dx u + vr u vr )dx Br(0) − q Br(0) | | − | | − | | Z Z   1 q q 2 (23) ϕ(u)+ ϕr(vr)+ ∇u∇vr dx + u q vr − vru )dx ≤ Br(0) q Br(0) | | − | | Z Z   LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 11 where in the last step we used that

q q q 2 u + vr vr q vr − vru dx 0 Br(0) | | − | | − | | ≥ Z   by the convexity of the function t t q. We note that 7→ | | q 2 ∇u∇vr dx = u − uvr dx B 0 B 0 | | Z r( ) Z r( ) and therefore 2(q 1) 2 2q q 1 − N+ N+ ∇u∇v dx u −∞ v dx = o(r 2 q )r 2 q v dx = o(r 2 q ) as r 0. r L (Br(0)) r − − − Br(0) ≤k k Br(0) | | B1(0) | | → Z Z Z Moreover, 2q 2q q 2 q N+ 2 q u dx = o(r − ) Br(0) = o(r − ) B 0 | | | | Z r( ) and 2 N+ 2(q 1) N+ 2q q 2 ∞ q 1 2 q 2 −q q 1 2 q vr − vrudx u L (Br(0)) vr − dx o(r − )r − v − dx = o(r − ) B 0 | | ≤k k B 0 | | ≤ B 0 | | Z r( ) Z r( ) Z 1( ) q x as r 0. Finally, since ∇vr(x)= r 2 q ∇v( ) for x Br(0), we find that → − r ∈ 2q 2q 2 q 2 2 q q r − x r − x ϕr(vr)= ∇v dx v dx 2 Br(0) r − q Br(0) r Z Z 2q 1   1   N+ 2 q ∇ 2 q = r − v (x) dx v(x) dx 2 B 0 | | − q B 0 | | Z 1( ) Z 1( ) 2q   N+ 2 q ϕ = r − 1(v). Inserting these estimates in (23), we obtain

N+ 2q N+ 2q ϕ(u + vr) ϕ(u)+ r 2 q ϕ1(v)+ o(1) ϕ(u)+ r 2 q cK + o(1) ≤ − ≤ −     Since K is compact, it is easy to see that these estimates are uniform in v K. Since moreover cK < 0, the claim follows. ∈ 

1 Theorem 3.4. Let u be a minimizer of ϕ on N . Then u− (0) has vanishing Lebesgue measure. 1 Proof. Suppose by contradiction that u− (0) > 0. Then there exists a point x0 of density one for the set 1 | | u− (0). Without loss of generality, we can suppose x0 = 0. We fix two arbitrary nonnegative nontrivial 1,2 functions v1,v2 W (B1(0)) with disjoint support, and consider the path ∈ 0 1,2 γ : [0,1] W (B1(0)), γ(s)= t∗((1 s)v1 sv2) ((1 s)v1 sv2), → 0 − − · − − where the function t∗ is defined in (5). It is clear that ϕ(γ(t)) < 0 for t [0,1]. Applying Proposition 3.3 γ 1,2 ∈ ϕ to the compact set K := ([0,1]) W0 (B1(0)), wemayfix r > 0 sufficiently small such that (u+vr) < ϕ(u) for every v K. Since K is⊂ connected and ∈ > 0 for v γ 0 K, q 2 = ( ) u + vr − (u + vr)dx ∈ Ω | | < 0 for v = γ(1) K, Z ( ∈ there exists v K such that u + vr N . This however contradicts the assumption that u is a minimizer of ϕ in N . ∈ ∈ 

Corollary 3.5. Let u be a minimizer of ϕ on N ,andletx0 Ω be a pointwith u(x0)= 0. Then u changes ∈ sign in every neighborhood of x0. 12 ENEA PARINI, TOBIAS WETH

Proof. Suppose by contradiction that there is a ball Br(x0) Ω such that u does not change sign in Br(x0). ⊂ Without loss, we may assume that u 0 on Br(x0). By Theorem 3.4, u 0 in Br(x0). Since u solves ≥ 6≡ (1) and is therefore superharmonic in Br(x0), the strong maximum principle implies that u > 0 in Br(x0), contrary to the assumption that u(x0)= 0. 

4. SYMMETRY RESULTS

We add a result on minimizers of ϕ N in the case where the underlying domain is radial, i.e., a ball or an annulus in RN centered at zero. | Theorem 4.1. Suppose that Ω RN is a radial bounded domain. Then every minimizer u of ϕ on N is foliated Schwarz symmetric. ⊂ Here we recall that a function u defined on a radial domain is said to be foliated Schwarz symmetric if RN θ x there is a unit vector p , p = 1 such that u(x) only depends on r = x and = arccos x p and ∈ | | | | | | · u is nonincreasing in θ.  

Proof. Let u N be a minimizer of ϕ N , and pick x0 Ω 0 with u(x0)= max u(x) : x = x0 . ∈x0 | ∈ \{ } N { | | | |} We put p := , and we let Hp denote the family of all open halfspaces H in R such that p H and x0 | | N N ∈ 0 ∂H. For H Hp we consider the reflection σH : R R with respect to the the hyperplane ∂H. We∈ claim the following:∈ →

(24) Forevery H Hp, we have u u σH on H Ω. ∈ ≥ ◦ ∩ To prove this, we fix H Hp and recall a simple rearrangement, namely the polarization of u with respect to H defined by ∈

max u(x),u(σH (x)) , x Ω H uH (x)= { } ∈ ∩ min u(x),u(σH (x)) , x Ω H. ( { } ∈ \ It is well known and fairly easy to prove (see e.g. [13]) that

2 2 q q q 2 q 2 ∇uH dx = ∇u dx, uH dx = u dx and uH − uH dx = u − udx. Ω | | Ω | | Ω | | Ω | | Ω | | Ω | | Z Z Z Z Z Z Consequently, uH N and ϕ(uH )= ϕ(u), so that uH is also a minimizer of ϕ on N . Hence, by ∈ Theorem 1.1, both u and uH are solutions of (1). Therefore w := uH u is a nonnegative function in Ω H satisfying − ∩ q 2 q 2 ∆w = uH − uH u − u 0 in H Ω. − | | − | | ≥ ∩ The strong maximum principle then implies that either w 0 or w > 0 in H Ω. The latter case is ruled ≡ ∩ out since x0 H Ω and w(x0)= uH (x0) u(x0)= 0 by the choice of x0. We therefore obtain w 0, ∈ ∩ − ≡ hence u = uH and (24) holds. By continuity, it follows from (24) that u is symmetric with respect to every hyperplane containing p, so it is axially symmetric with respect to the axis pR. Hence u(x) only depends on r = x and θ = | | x θ arccos x p . Moreover, it also follows from (24) that u is nonincreasing in the polar angle . We thus | | · conclude thatu is foliated Schwarz symmetric.  N Theorem 4.2. Let Ω R be the unit ball, and let u N be a minimizer of ϕ N . Then u is not radially symmetric. ⊂ ∈ | Proof. By Lemma 2.2, u is a solution of (1), so u C2,α (Ω) by elliptic regularity. We suppose by contradiction that u is radially symmetric, and we write∈ u(r) := u( x ) for simplicity. Then u solves | | N 1 q 2 (25) u′′ + − u′ + u − u = 0 in (0,1], u′(0)= u′(1)= 0, r | | LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 13 where the prime denotes the radial derivative. We first prove Claim 1: u(0) = 0, u(1) = 0, and u has only finitely many zeros in (0,1). Indeed, with the6 transformation6 2 N r − , N 3; y(t)= u(r) with t = ≥ 1 logr, N = 2, ( − (25) is transformed into the problem q 2 (26)y ¨+ p(t) y − y = 0, t [1,∞), y˙(1)= 0 | | ∈ with 1 2(N 1) N −2 t− − , N 3, p t (N 2)2 ≥ ( )=  − 2(t 1) e− − , N = 2. Since u 0 we have y 0 in [1,∞). Hence y(1) = 0 by local uniqueness and continuability of solutions  to (26)6≡ (see e.g. [14]),6≡ and thus u(1) = 0. Moreover,6 by the non-oscillation criterion for (26) given in [8, Theorem 6], y has only finitely many6 zeros in [1,∞), so u only has finitely many zeros in (0,1). Finally, since u does not change sign in some neighborhood of 0, the strong maximum principle implies that u(0) = 0. Thus Claim 1 is proved. It now follows6 from Hopf’s boundary lemma that u W , where W is defined in (7). Consequently, u 3,p 1 ∈ 2,p 1 ∈ W (Ω) for p (1, ) by Proposition 2.3, and ux W (Ω) C (Ω) solves the linearized Dirichlet ∈ 2 q 1 ∈ ∩ problem − q 2 ∆ux = (q 1) u ux in Ω, (27) − 1 − | | − 1 ux = 0 on ∂Ω.  1 The boundary condition follows from the fact that ∇u 0 on ∂Ω since u is radial and satisfies Neumann ≡ boundary conditions. Let H be the hyperplane x1 = 0 . We first prove the following 1 { } Claim 2: If w C (Ω) is antisymmetric with respect to H and such that ϕ′′(u)(w,w) < 0, then ϕ(u + tw + c(u +tw))∈< ϕ(u) for t > 0 sufficiently small. Indeed, by Proposition 2.3(i) we have, for every c R, the Taylor expansion ∈ t2 ϕ(u + c +tw)= ϕ(u + c)+tϕ (u + c)w + ϕ (u + c)(w,w)+ o(t2), ′ 2 ′′ where the quantity o(t2) is locally uniform in c. Since u is radially symmetric and w is antisymmetric with respect to H , we have

q 2 ϕ′(u + c)w = ∇u∇wdx u + c − (u + c)wdx = 0. Ω − Ω | | Z Z Hence there exist M > 0 and δ > 0 such that ϕ(u +tw + c) ϕ(u + c) Mt2 ϕ(u) Mt2 for t , c < δ. ≤ − ≤ − | | | | Since c(u +tw) c(u)= 0 as t 0 as a consequence of Lemma 2.1, we deduce that → → ϕ(u +tw + c(u +tw)) ϕ(u) Mt2 < ϕ(u) for t > 0 sufficiently small. ≤ − Hence Claim 2 is proved. Next, we consider an arbitrary function w C1(Ω) which is antisymmetric with respect to H . By Claim 2 and the minimizing property of u, we have∈ 2 0 ϕ′′(u)(w +tux ,w +tux )= ϕ′′(u)(w,w)+ 2tϕ′′(u)(ux ,w)+t ϕ′′(u)(ux ,ux ) for every t R ≤ 1 1 1 1 1 ∈ and also ϕ ∇ 2 q 2 2 ′′(u)(ux ,ux )= ux dx (q 1) u − ux dx = 0, 1 1 Ω | 1 | − − Ω | | 1 Z Z 14 ENEA PARINI, TOBIAS WETH

since ux1 is a solution of (27). These relations imply that q 2 (28) 0 = ϕ′′(u)(ux ,w)= ∇ux ∇wdx (q 1) u − ux wdx = (ux )ν wdσ, 1 Ω 1 − − Ω | | 1 ∂ Ω 1 Z Z Z where the last equality follows again from (27). Since (ux )ν C(∂Ω) is antisymmetric with respect to 1 ∈ H and (28) holds for every w C1(Ω) which is antisymmetric with respect to H , we conclude that ∂Ω ∈ (ux1 )ν = 0 on , and in particular ux1,x1 (e1)= 0. In the radial variable, we thus have q 2 0 = u′′(1)= u(1) − u(1) −| | by (25) and therefore u(1)= 0, contrary to Claim 1. The proof is finished. 

5. THECASE q = 1 In this section we are concerned with the case q = 1, i.e., with the boundary value problem ∆u = sgn(u), in Ω (29) − uν = 0 on ∂Ω.  We will suppose that the boundary of Ω is of class C1,1. As already noted in the introduction, the varia- tional framework of Section 2 does not extend in a straightforward way to the case q = 1. In particular, the functional 1 ϕ : W 1,2(Ω) R, ϕ(u)= ∇u 2 dx u dx → 2 Ω | | − Ω | | Z Z is not differentiable. Moreover, while every (weak) solution of (29) is contained in the set

u W 1,2(Ω) sgn(u)dx = 0 , { ∈ | Ω } Z this set does not have the nice intersection property given by Lemma 2.1(i). Indeed, any function u W 1,2(Ω) satisfying ∈ 0 < u > 0 u < 0 < u = 0 |{ }| − |{ }| |{ }| has the property that

sgn(u + c)dx = 0 forevery c R. Ω 6 ∈ Z As a consequence, many of the arguments used in the previous sections do not apply in the case q = 1. Instead, we will consider the larger set M := u W 1,2(Ω) : u > 0 u < 0 u = 0 . ∈ |{ }| − |{ }| ≤ |{ }| We collect useful properties of M . First, we may rewrite the defining property for u M as ∈ (30) sgn (u)dx 0 sgn (u)dx, Ω ≤ ≤ Ω + Z − Z where R sgn+(t) := 1t 0 1t<0 and sgn (t) := 1t>0 1t 0 for t . ≥ − − − ≤ ∈ We also point out that, in contrast to the definition in the case q > 1, the set M also contains nonzero functions which do not change sign. We also need the following facts. Lemma 5.1. (i) If u M , then ∈ u + c dx > u dx foreveryc R 0 . Ω | | Ω | | ∈ \{ } Z Z (ii) For u W 1,2(Ω), we have u M ∈ ∈ sgn(u c)dx 0 sgn(u + c)dx foreveryc > 0. Ω − ≤ ≤ Ω Z Z LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 15

(iii) If u W 1,2(Ω) and γ : [0,1] W 1,2(Ω) L∞(Ω) is a continuous curve such that ∈ → ∩ sgn(u + γ(0))dx > 0 and sgn(u + γ(1))dx < 0, Ω Ω Z Z then there exists s0 [0,1] with u + γ(s0) M . (iv) For every u W 1,2∈(Ω) there exists a unique∈ c(u) R such that u + c(u) M . Moreover, the map u c(u∈) is continuous. ∈ ∈ 7→ (v) If u,v W 1,2(Ω) satisfy u v, then c(u) c(v). ∈ ≤ ≥ + Proof. (i) Let c < 0. If u 0, then obviously Ω u+c dx > Ω u dx. Suppose now that u = 0. We then claim that ≤ | | | | 6 R R (31) 0 < u < c > 0. |{ | |}| Indeed, we have v := min(u+, c ) 0, since v > 0 on the set where u is positive. Moreover, if we suppose by contradiction that 0 < u <| |c 6≡ = 0, then ∇v 0a.e. on Ω (see e.g. [6, Lemma 7.7]), so that v equals a positive constant a.e.|{ in Ω. This| |}| however implies≡ that u > 0a.e. in Ω, contrary to the assumption u M . Hence we conclude that (31) holds. As a consequence, we estimate, for c < 0, ∈

u + c dx u dx = c u 0 + ( u + c u)dx + ( u + c u)dx Ω | | − Ω | | | ||{ ≤ }| 0 c u 0 c 0 < u < c c u c | ||{ ≤ }| − | ||{ | |}| − | ||{ ≥ | |}| = c u 0 u > 0 0, | | |{ ≤ }| − |{ }| ≥ as claimed. If c > 0, a similar argument yields  u + c dx u dx > 0. Ω | | − Ω | | Z Z (ii) This simply follows from the fact that u M is equivalent to (30), whereas ∈ sgn(u + c) sgn+(u) sgn (u) sgn(u c) for every c > 0 ≥ ≥ − ≥ − and Ω + sgn(u + c) sgn+(u), sgn(u c) sgn (u) pointwise in as c 0 . → − → − → (iii) Consider

s0 := sup s [0,1) : sgn(u + γ(s))dx > 0 , ∈ Ω Z and let w := u + γ(s0). We use (ii) to show that w M . Let (sn)n [0,s 0] be a sequence with sn s0 and ∈ ⊂ →

(32) sgn(u + γ(sn))dx > 0 forevery n N. Ω ∈ Z For given c > 0, there exists n N with ∈ u + γ(sn) > 0 w + c > 0 and w + c < 0 u + γ(sn) < 0 { }⊂{ } { }⊂{ } and thus Ω sgn(w + c)dx > 0 by (32). Now if s0 = 1, the assumption implies that

R 0 > sgn(w)dx sgn(w c)dx for all c > 0 Ω ≥ Ω − Z Z and thus w M by (ii). Suppose finally that s0 < 1, and suppose by contradiction that ∈ sgn(w c)dx > 0 forsome c > 0. Ω − Z 16 ENEA PARINI, TOBIAS WETH

By the continuity of γ, there exists ε > 0 such that u + γ(s) w c for s [s0,s0 + ε) and therefore ≥ − ∈

sgn(u + γ(s))dx > 0 for s [s0,so + ε). Ω ∈ Z This contradicts the definition of s0. Hence sgn(w c)dx 0 forevery c > 0, Ω − ≤ Z and by (ii) we conclude that w = u + γ(s0) M . (iv) Let u W 1,2(Ω). The uniqueness of c∈= c(u) R with u + c M is an immediate consequence of (i). To see∈ the existence, we note that sgn(u c) ∈ 1 as c +∈∞ a.e. in Ω. Hence, by Lebesgue’s ± →± → theorem, there exists c0 > 0 with

sgn(u c0)dx > 0. ± Ω ± Z Applying (iii) to the path s (1 2s)c0 now yields the existence of c [ c0,c0] such that u + c M . The continuity follows similarly7→ as− (but more easily than) in Lemma 2.1.∈ − ∈ (v) Suppose by contradiction that c(u) < c(v)=: c. We then have u u + c v + c in Ω and therefore ≤ ≤ sgn (u + c)dx sgn (v + c)dx 0 Ω ≤ Ω ≤ Z − Z − and sgn (u + c)dx sgn (u)dx 0. Ω + ≥ Ω + ≥ Z Z By (30), we then have u + c M , with contradicts the uniqueness statement in (iv).  ∈ We now consider the variational problem related to the minimax value m := infϕ = inf supϕ(u + c) M u W 1,2(Ω) c R ∈ ∈ Note that the second equality is an immediate consequence of Lemma 5.1(i), (iv). Note also that m < 0, since for every u M 0 we have ∈ \{ } u 2 u 2 ϕ 1 k kL1(Ω) k kL1(Ω) (33) (t∗(u)u)= 2 < 0 with t∗(u)= 2 , −2 Ω ∇u dx Ω ∇u dx | | | | whereas t∗(u)u M . R R The main result∈ of this section is the following. Theorem 5.2. The value m < 0 is attained by ϕ on M . Moreover, every u M with ϕ(u)= m is a nontrivial solution of (29) such that its zero set u = 0 Ω has vanishing Lebesgue∈ measure. { }⊂ The proof of this Theorem is split in two steps. We first show the following. Lemma 5.3. The functional ϕ attains the value m < 0 on M . Moreover, every minimizer u of ϕ on M is a sign changing solution of (29). Proof. We first note that for all u M we have, by Lemma 5.1(i) ∈ 2 2 Ω 2 Ω µ 1 ∇ 2 u 1 Ω = min u + c 1 Ω min u + c 2 Ω 2− u dx, k kL ( ) c R k kL ( ) ≤ | | c R k kL ( ) ≤ | | Ω | | ∈ ∈ Z where µ2 > 0 is the first nontrivial Neumann eigenvalue of ∆ on Ω. As a consequence, the functional ϕ − is coercive on M . Let (un)n M be a minimizing sequence for ϕ. Then (un) is bounded, and we may ⊂ 1,2 1 pass to a subsequence such that un ⇀ u˜ W (Ω). Then un u˜ in L (Ω), ∈ → 2 2 ∇u˜ dx liminf ∇un dx Ω | | ≤ n ∞ Ω | | Z → Z LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 17 and

u˜ + c dx = lim un + c dx lim un dx = u˜ dx for all c R. Ω | | n ∞ Ω | | ≤ n ∞ Ω | | Ω | | ∈ Z → Z → Z Z Consequently,u ˜ M by Lemma 5.1(i) and (iv), and ∈ ϕ(u˜) liminfϕ(un)= m. n ∞ ≤ → By definition of m, equality holds and thus ϕ attains its minimum on M . Next, we let u M be an arbitrary minimizer for ϕ on M , and we show that u is a solution of (29). We first show that ∈ (34) sgn (u)vdx ∇u∇vdx sgn (u)vdx for every v W 1,2(Ω), v 0. Ω ≤ Ω ≤ Ω + ∈ ≥ Z − Z Z Arguing by contradiction, we assume that there exists v W 1,2(Ω), v 0 such that ∈ ≥ (35) sgn (u)vdx > ∇u∇vdx. Ω Ω Z − Z Note that for a,c R we have a + c a + sgn (a)c. Hence for every c 0, t 0 we have ∈ | | ≥ | | − ≤ ≥

u + c +tv 1 Ω u 1 Ω + sgn (u)(c +tv)dx u 1 Ω +t sgn (u)vdx, k kL ( ) ≥k kL ( ) Ω ≥k kL ( ) Ω Z − Z − where the last inequality follows from the fact that Ω sgn (u)dx 0 since u M . Since v 0 implies that c(u +tv) 0 for t > 0 by Lemma 5.1(v), we find that− ≤ ∈ ≥ ≤ R 1 2 ϕ(u +tv + c(u +tv)) = ∇(u +tv) dx u +tv + c(u +tv) 1 Ω 2 Ω | | −k kL ( ) Z 2 1 2 t 2 ∇u dx u 1 Ω +t ∇u∇vdx sgn (u)vdx + ∇v dx ≤ 2 Ω | | −k kL ( ) Ω − Ω 2 Ω | | Z Z Z − Z  t2  = m +t ∇u∇vdx sgn (u)vdx + ∇v 2 dx Ω − Ω 2 Ω | | Z Z − Z < m for t > 0 sufficiently small  by (35). This contradicts the definition of m. Hence the first inequality in (34) holds, and the second inequality is proved by a similar argument. As a consequence, we have

+ 1,2 ∇u∇vdx sgn (u)v sgn (u)v− dx v 1 Ω for every v W (Ω). Ω ≤ Ω + − ≤k kL ( ) ∈ Z Z −  ∆ ∞ Ω R 1 Ω Consequently, the distributional Laplacian u : C0 ( ) is continuous with respect to the L ( )-norm and is therefore represented by a function w L∞(Ω)→satisfying − ∈ (36) sgn (u) w sgn+(u). − ≤ ≤ 2,p Ω ∞ ∆ Then, by elliptic regularity theory, it follows that u Wloc ( ) for all p (1, ) with u = w. Moreover, we have ∇u 0 and w = ∆u 0 a.e. on the set ∈u = 0 (see e.g. [6,∈ Lemma 7.7]).− Hence, by (36) we may assume≡ that w = sgn(−u), and≡ thus u is a solution{ of (29).} Finally, to show that u is sign changing, we first note that u = 0 since ϕ(u)= m < 0. Suppose by contradiction that u 0, then u is also superharmonic by (29), and hence6 u > 0 in Ω by the strong maximum principle, which≥ contradicts the fact that u M . Similarly, we get a contradiction assuming that u 0. Hence u changes sign in Ω. ∈  ≤ In order to complete the proof of Theorem 5.2, we need to show that every minimizer u M of ϕ has the unique continuation property, that is, u = 0 has measure zero. The argument is similar∈ as in the case q > 1, but changes are required at some{ points.} We start with the following.

Proposition 5.4. Let u be a solution of (29), and suppose that x0 Ω is a point of density one for the set 1 2 ∈ u (0) Ω. Then u(x)= o( x ) as x x0. − ⊂ | | → 18 ENEA PARINI, TOBIAS WETH

Proof. The argument is similar as the proof of Proposition 3.2. Without loss, we assume that x0 = 0, and we extend u to all of RN by setting u 0 on RN Ω. Applying Lemma 3.1 as in the proof of Proposition 3.2 with α = 1, β = 2, we see that≡ the function\

2 g : (0,∞) R, g(r) := r− sup u(x) → x =r | | | | is bounded. To show that g(r) 0 as r 0, we argue by contradiction, assuming that there exists ε > 0 → → and a sequence of radii rn > 0, n N such that rn 0 as n ∞ and g(rn) ε for all n N. We Ω ∈ N → →Ω ≥ ∈ may assume that B2rn (0) for all n . We then consider 0 := B2(0) B 1 (0) and the functions ⊂ ∈ \ 2 2 wn : Ω0 R, wn(x)= r u(rnx) for n N which are uniformly bounded in Ω0. For n N, wn solves → n− ∈ ∈ ∆wn = sgn(wn) in B2(0) B 1 (0), − \ 2 1 and there exists a sequence of points xn S , n N such that w(xn)= g(rn) ε for all n N. Using elliptic regularity theory again, we may pass∈ to a∈ subsequence such that ≥ ∈

1 xn x¯ S ; → ∈ 1 wn w in C (Ω0), → loc ∞ 1 sgn(wn) ⇀∗ f in L (Ω0) = (L (Ω0))∗,

1,α where w C (Ω0) is a weak solution of ∆w = f in Ω0 such that w(x¯) ε. In particular, w 0. ∈ loc − ≥ 6≡ However, by the same argument as in the proof of Proposition 3.2, the sets An := x B2(0) B 1 (0) : { ∈ \ 2 wn(x) = 0 satisfy lim An = 0 and therefore w 0 a.e. in B2(0) B 1 (0). This yields a contradiction, 6 } n ∞ | | ≡ \ 2 and thus we conclude→ that g(r) 0 as r 0, as required.  → → 1,2 Next, for r > 0 and q = 1, we consider the functional ϕr : W (Br(0)) R and the rescaling map 1,2 1,2 → Tr : W (B1(0)) W (Br(0)) defined in (21), (22), respectively. We then have the following. → 1,2 Proposition 5.5. Let u W (Ω) be a solution of (29) in Ω, andlet x0 Ω be a point of density one for 1 ∈ 2,2 ∈ u (0) Ω. Moreover, let K W (B1(0)) be a compact set such that − ⊂ ⊂ 0 cK := supϕ1(v) < 0, v K ∈ Then there exists r0 > 0 with the following property: Ω ϕ ϕ Br0 (x0) is contained in , and for every r (0,r0) and every v K we have (u + vr) < (u), where 2,2 2,2 ∈ ∈ vr W (Br(x0)) W (Ω) is defined by ∈ 0 ⊂ 0 [Trv](x x0) x Br(x0); vr(x)= − ∈ 0 x Ω Br(x0). ( ∈ \ The proof is somewhat different than the proof of Proposition 3.3. Note that we need the stronger 2,2 assumption K W (B1(0)) here. This assumption is not optimal but suffices for our purposes. ⊂ 0

Proof. Without loss, we may assume that x0 = 0 Ω. Let v K. We then have ∈ ∈

ϕ(u + vr)= ϕ(u)+ ϕr(vr)+ ∇u∇vr dx u + vr u vr )dx Br(0) − Br(0) | | − | | − | | Z Z   (37) ϕ(u)+ ϕr(vr)+ ∇u∇vr dx + 2 u dx. ≤ B 0 B 0 | | Z r( ) Z r( ) LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 19

2,2 Note that, since vr W (Br(0)), we have ∈ 0 N ∇u∇vr dx = u∆vr dx r u(rx)[∆v](x)dx B 0 B 0 ≤ B 0 Z r( ) Z r( ) Z 1( ) N+2 N+2 = o(r ) ∆v 1 = o(r ) as r 0. k kL (B1(0)) → Moreover, 2 N+2 u dx = o(r ) Br(0) = o(r ) as r 0. B 0 | | | | → Z r( ) x Finally, since ∇vr(x)= r∇v( ) for x Br(0), we find that r ∈ 2 2 1 x x N+2 ϕr(vr)= r ∇v dx v dx = r ϕ1(v). 2 Br(0) r − Br(0) r  Z   Z   

Inserting these estimates in (37), we obtain N+2 N+2 ϕ(u + vr) ϕ(u)+ r ϕ1(v)+ o(1) ϕ(u)+ r cK + o(1) . ≤ ≤     Since K is compact, it is easy to see that these estimates are uniform in v K. Since moreover cK < 0, the claim follows. ∈ 

1 Theorem 5.6. Let u be a minimizer of ϕ on M . Then u− (0) has vanishing Lebesgue measure. 1 Proof. Suppose by contradiction that u− (0) > 0. Then there exists a point x0 of density one for the set 1 | | u− (0). Without loss of generality, we can suppose x0 = 0. We fix two arbitrary nonnegative nontrivial 2 functions v1,v2 C (B1(0)) with disjoint support, and consider the path ∈ c 2 γ : [0,1] C (B1(0)), γ(s)= t∗((1 s)v1 sv2) ((1 s)v1 sv2), → c − − · − − where the map t is defined in (33). It is clear that ϕ(γ(s)) < 0 for s [0,1]. We also define ∗ ∈ 2 γr := Tr γ : [0,1] C (Br(0)). ◦ → c 2,2 Applying Proposition 5.5 to the compact set K := γ([0,1]) W (B1(0)), we may fix r > 0 sufficiently ⊂ 0 small such that ϕ(u + γr(s)) < ϕ(u) for every s [0,1]. Moreover, by making r smaller if necessary and ∈ 1 using again the fact that x0 = 0 is a point of density one for the set u− (0)= 0, we may assume that

u = 0 γr(0) > 0 > 0 and u = 0 γr(1) < 0 > 0. |{ }∩{ }| |{ }∩{ }| As a consequence of these inequalities and the fact that u is a solution of (29), we find that

sgn(u + γr(0))dx > sgn(u)dx = 0 > sgn(u + γr(1))dx < 0. Ω Ω Ω Z Z Z By Lemma 5.1(iii), there exists s0 [0,1] such that u + γr(s0) M . This however contradicts the as- sumption that u is a minimizer of ϕ∈in M . ∈ 

In the following, we restrict our attention to the case where Ω is a radial bounded domain in RN . In this case we also consider

Mr := u M : u radial and mr := infϕ, { ∈ } Mr so that Mr M and mr m. ⊂ ≥ Theorem 5.7. Let Ω RN be a radial bounded domain. Then we have: ⊂ (i) If u M satisfies ϕ(u)= m, then u is foliated Schwarz symmetric. ∈ (ii) If u Mr satisfies ϕ(ur)= mr, then u is strictly monotone in the radial variable. ∈ 20 ENEA PARINI, TOBIAS WETH

(iii) If Ω = B is the unit ball in RN , we have π 1 1 (38) m < mr = π + ln2 , ≤−18 − −16 8   if N = 2 and, for N 3, ≥ 2 1 2 N 2 ωN (2− N 1)N + 2 − N (39) m ωN − < mr = − , ≤− 2N2(N 1) − 2 (N 2)(N + 2) − − where, as usual, ωN denotes the measure of B . Hence every minimizer u M of ϕ M is a nonradial function. | | ∈ | Proof. (i) The proof is very similar to the proof of Theorem 4.1, and we use the notation introduced there. Ω x0 H Pick x0 0 with u(x0)= max u(x) : x = x0 . We put p := x , and we let H p. As in the ∈ \{ } { | | | |} | 0| ∈ proof of Theorem 4.1, it suffices to show that u uH on H Ω. Since ≡ ∩ 2 2 ∇uH dx = ∇u dx, uH dx = u dx and sgn (uh) dx = sgn (u) dx, Ω | | Ω | | Ω | | Ω | | Ω Ω Z Z Z Z Z ± Z ± we find that uH M and ϕ(uH )= ϕ(u), so that uH is also a minimizer of ϕ on M . Hence, by Theo- ∈ rem 1.1, both u and uH are solutions of (29). Therefore w := uH u is a nonnegative function in Ω H satisfying − ∩ ∆w = sgn(uH ) sgn(u) 0 in H Ω. − − ≥ ∩ The strong maximum principle then implies that either w 0 or w > 0 in H Ω. The latter case is ruled ≡ ∩ out since x0 H Ω and w(x0)= uH (x0) u(x0)= 0 by the choice of x0. We therefore obtain w 0 and ∈ ∩ − ≡ hence u uH on H Ω, as required. (ii) We only≡ consider∩ the case where Ω = B is the unit ball in RN ; the proof in the case of an annulus is similar. Let u Mr satisfy ϕ(ur)= mr. Then, by similar arguments as in the proof of Theorem 2.2, u is ∈ a radial solution of (29). Suppose by contradiction that there exists r0 (0,1) such that u (r0)= 0. We ∈ ′ claim that we can choose r0 minimally, i.e., such that

(40) u′(r) = 0 for r (0,r0). 6 ∈ Indeed, suppose by contradiction that there exists a sequence of rn (0,1), n N such that u (rn)= 0 for ∈ ∈ ′ all n and rn 0 as n ∞. Without loss, we may assume that rn > rn+1 for every n. Since the function N 1 → → r r − u′(r) is strictly monotone on every interval on which u has no zero, we conclude that there exists 7→ 1 sn (rn+1,rn) such that u(sn)= 0. Since u is of class C , we therefore conclude that u(0)= u′(0)= 0. ∈ 2 u′(r) This however implies that u 0, since the absolutely continuous function r h(r)= 2 + u(r) is decreasing on [0,1], which follows≡ from the fact that 7→ | |

(N 1) 2 (41) h′(r)= u′(r)[u′′(r)+ sgn(u(r))] = − u′(r) 0 fora.e. r (0,1). − r ≤ ∈ We thus conclude that we can choose r0 (0,1) such that (40) holds. It is then easy to see (e.g. by using ∈ the Hopf boundary lemma) that u(r0) = 0. Let Ω1 := Br (0) and Ω2 := B Br (0). Then u solves (29) 6 0 \ 0 both on Ω1 and Ω2, so that sgn(u)dx = 0 = sgn(u)dx. Ω Ω Z 1 Z 2 We now define 2u(r0) u(x), x Ω1; v W 1,2(B), v(x)= − ∈ ∈ u(x), x Ω2. ( ∈ Moreover, we let c = c(v) R be given by Lemma 5.1(iv), so that w := v + c Mr. We then have ∈ ∈ (42) ∇w 2 dx = ∇v 2 dx = ∇u 2 dx. B | | B | | B | | Z Z Z LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 21

Moreover,

(43) w dx = u 2u(r0) c dx u dx Ω | | Ω | − − | ≥ Ω | | Z 1 Z 1 Z 1 by Lemma 5.1(i) applied to the domain Ω1. Similarly, (44) w dx = u + c dx u dx Ω | | Ω | | ≥ Ω | | Z 2 Z 2 Z 2 by Lemma 5.1(i) applied to the domain Ω2. Moreover, if equality holds in both (43) and (44), then Lemma 5.1(i) implies that 2u(r0)+ c = 0 and c = 0, hence u(r0)= 0 contrary to what we have seen earlier. Hence at least one of the inequalities (43) and (44) is strict, so that

(45) w dx > u dx Ω | | Ω | | Z Z and therefore ϕ(w) < ϕ(u)= mr as a consequence of (42) and (45). This contradicts the definition of mr, and thus the proof is finished. (iii) Similar arguments as in the proof of Lemma 5.3 show that mr is attained by a radial solution u Mr of (29). By (ii), we know that u is strictly monotone in the radial variable, and therefore it is up to∈ sign uniquely given by 1 1 r2 + for 0 r a := 1 u(r)= −4 8 ≤ ≤ √2  1 1 2 1 1  lnr + r ln2 for a r 1  −2 4 − 8 − 4 ≤ ≤ for N = 2, and by  1 2 2 1 (a r ) for 0 r a := 2 N 2N − ≤ ≤ − u(r)=  2 1 2 2 N 2 2− N (N + 2)  r − + r for a r 1  2N N 2 − N 2 ≤ ≤ − − !  for N 3. Note here that the value a is determined by the condition that u > 0 = u < 0 . For N = 2,≥ we have |{ }| |{ }| 1 1 ∇u 2 = u = 2π + ln2 B | | B | | −16 8 Z Z   and therefore 1 1 mr = ϕ(u)= π + ln2 , − −16 8   thus showing the right inequality in (38). For N 3, ≥ ω 2 1 2 2 N 1 N 2 1 1 (2− N 1)N + 2 − N ∇u = u = 2 N− = ωN − B | | B | | 2 N 2 − N + 2 − N 2 (N 2)(N + 2) Z Z  − −  − and therefore 2 1 2 ωN (2− N 1)N + 2 − N mr = ϕ(u)= − , − 2 (N 2)(N + 2) − which is the right equality in (39). To see the left inequalities in (38) and (39), we consider the functions s N x us(x)= x1 x for s > . We then have 7→ | | − 2 ω 2 s 1 1 s+1 N us dx x1 x − dx = x dx = B | | ≥ B | | N B | | N + s + 1 Z Z Z and 2 ∇ 2 2s 2 2 2(s 1) ω N + s + 2s us dx = x + (s + 2s)x1 x − dx = N . B | | B | | | | N + 2s Z Z   22 ENEA PARINI, TOBIAS WETH

M B us dx We now have vs for the function vs = csus with cs := | | 2 and therefore ∇us dx ∈ BR| | 2 R us dx ϕ 1 B | | ω N + 2s m (vs)= 2 N 2 2 . ≤ −2 R ∇us dx ≤− 2(N + s + 2s)(N + s + 1) B | | Thus the left inequalities in (38) andR (39) follow by choosing s = 0 if N = 2, and s = 1 if N 3. Finally, for the middle inequality in (39) we need to prove: − ≥ 2 2 (2 N 1)N + 21 N N 2 (46) − − − < − (N 2)(N + 2) N2(N 1) − − for N 3, which is equivalent to ≥ 2 4 2 3 2 2 (2− N 1)N + 2− N N + 2(1 2−N )N + 4N 8 < 0 − − − 3 2 2 2 2 N (2− N 1)N + 2− N + (1 2− N ) + 4N 8 < 0. ⇔ − N − −   To this aim, we will prove that the function h : [3,+∞) R defined by → 2 2 2 2 h(t) := h1(t)+ h2(t)+ h3(t) = (2− t 1)t + 2− t + (1 2− t ) − t − is maximized for t = 3, and that h(3) < 1. Inequality (46) will be implied by N3 + 4N 8 < 0, which − − − is true for N 3. The function h1 is such that ≥ 1 1 t t t 1 4− 4 t +t + ln4 4 −t (ln2)2 h (t)= − , h (t)= . 1′  t  1′′ t3 Since h (t) 0 as t +∞ and h (t) > 0 for t > 0, h1 is monotone decreasing. h2 is clearly strictly 1′ → → 1′′ decreasing, while h3 satisfies t 2 1 2 −t (4 t 1)t + ln4 − h′ (t)= < 0 3 −  t3  for t > 0. It is then easily verified that h(3)= 1 (5√3 2 7) < 1.  3 − REFERENCES 1. A. Aftalion and F. Pacella, Qualitative properties of nodal solutions of semilinear elliptic equations in radially symmetric domains, C. R. Math. Acad. Sci. Paris 339 (2004), no. 5, 339–344. 2. J. M. Arrieta, A. Rodriguez-Bernal, and J. Valero, Dynamics of a reaction-diffusion equation with a discontinuous nonlinearity, I. J. Bifurcation and Chaos 16 (2006), 2965–2984. 3. T. Bartsch, T. Weth, and M. Willem, Partial symmetry of least energy nodal solutions to some variational problems, J. Anal. Math. 96 (2005), 1–18. MR MR2177179 4. K. C. Chang, The obstacle problem and partial differential equations with discontinuous nonlinearities, Comm. Pure Appl. Math. 33 (1980), no. 2, 117–146. 5. L. C. Evans and R. F. Gariepy, Measure theory and fine properties of functions, Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, 1992. 6. D. Gilbarg and N. S. Trudinger, Elliptic partial differential equations of second order, Classics in Mathematics, Springer-Verlag, Berlin, 2001, Reprint of the 1998 edition. 7. P. Girão and T. Weth, The shape of extremal functions for Poincaré-Sobolev-type inequalities in a ball, J. Funct. Anal. 237 (2006), no. 1, 194–223. MR MR2239264 8. H. E. Gollwitzer, Nonoscillation theorems for a nonlinear differential equation, Proc. Amer. Math. Soc. 26 (1970), 78–84. MR 0259243 (41 #3885) 9. Danielle Hilhorst and José-Francisco Rodrigues, On a nonlocal diffusion equation with discontinuous reaction, Adv. Differen- tial Equations 5 (2000), no. 4-6, 657–680. 10. J. Rauch, Discontinuous semilinear differential equations and multiple valued maps, Proc. Amer. Math. Soc. 64 (1977), 277– 282. 11. J. L. Vazquez, The porous medium equation, Oxford Science Publications, 2007. LEAST ENERGY NODAL SOLUTIONS TO SUBLINEAR NEUMANN PROBLEMS 23

12. M. Willem, Minimax theorems, Progress in Nonlinear Differential Equations and their Applications, 24, Birkhäuser Boston Inc., Boston, MA, 1996. 13. , Principes d’analyse fonctionnelle, Nouvelle Bibliothèque Mathématique [New Mathematics Library], 9, Cassini, Paris, 2007. MR 2567317 (2010g:46001) 14. J. S. W. Wong, On the generalized Emden-Fowler equation, SIAM Rev. 17 (1975), 339–360. MR 0367368 (51 #3610)

AIX MARSEILLE UNIVERSITÉ,CNRS, CENTRALE MARSEILLE,I2M, UMR 7373, 13453 MARSEILLE,FRANCE E-mail address: [email protected]

INSTITUT FÜR MATHEMATIK,GOETHE-UNIVERSITÄT FRANKFURT,D-60054 FRANKFURT AM MAIN,GERMANY E-mail address: [email protected]