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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


IMPACT FACTOR 2018: 6.636

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On the moving plane method for boundary blow-up solutions to semilinear elliptic equations

Annamaria Canino / Berardino Sciunzi
  • Corresponding author
  • Dipartimento di Matematica, UNICAL, Ponte Pietro Bucci 31B, 87036 Arcavacata di Rende, Cosenza, Italy
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/ Alessandro Trombetta
Published Online: 2018-07-07 | DOI: https://doi.org/10.1515/anona-2017-0221

Abstract

We consider weak solutions to -Δu=f(u) on Ω1Ω0, with u=c0 in Ω1 and u=+ on Ω0, and we prove monotonicity properties of the solutions via the moving plane method. We also prove the radial symmetry of the solutions in the case of annular domains.

Keywords: Elliptic equations; boundary blow-up; symmetry of solutions; radial symmetry,moving plane method

MSC 2010: 35B01; 35J61; 35J75

1 Introduction

Let Ω0 and Ω1 be two bounded smooth domains of N, N2, such that Ω0Ω1. Moreover, let fC1([0,+)) and c0. We consider weak solutions to the problem

{-Δu=f(u)in Ω1Ω0,u>cin Ω1Ω0,u=con Ω1,u=+on Ω0,(1.1)

i.e., we consider uC1(Ω1¯Ω0¯) such that

Ω1Ω0uφ=Ω1Ω0f(u)φfor all φCc(Ω1Ω0¯)

and

limxx0xΩ1Ω0u(x)=+for all x0Ω0.

When c=0, actually we deal with the case of positive solutions. Necessary and sufficient conditions for the existence of solutions to (1.1) are provided by the classical results of Keller [18] and Osserman [19], under suitable assumptions on the nonlinearity. The literature regarding boundary blow-up solutions is really wide (see, for example, [1, 2, 3, 4, 6, 5, 7, 13, 14, 12, 15, 16, 17, 20, 21]). Here we exploit an adaptation of the celebrated moving plane technique (see [8]) in order to obtain monotonicity properties of the solutions to (1.1). The domain that we consider is not convex and the solutions are not in H01(Ω1Ω0) as in the classical case. This is the same difficulty that occurs when dealing with the study of the uniqueness, symmetry and monotonicity properties of solutions to singular semilinear elliptic equations, see [9, 10, 11]. These problems exhibit in fact some similarities with problem (1.1) although the proofs cannot be adapted to our case.

In the second part of the paper, we prove the radial symmetry of the solutions on annular domains, under suitable assumptions. In our setting, this cannot be done just using the moving plane method, since the domain is not convex, and we prove that the solution is radially symmetric showing directly that the angular derivative is zero. The technique is based on a refined maximum principle for the linearized equation.

Let us introduce some notations. Let ν be a direction in N, with |ν|=1. Given a real number λ, we set

Tλν={xN:xν=λ}

and

xλν=Rλν(x)=x+2(λ-xν)ν,

that is, the reflection of x trough the hyperplane Tλν. We will make the following assumption throughout the paper:

  • (A)

    Ω0 and Ω1 are strictly convex with respect to the ν-direction and symmetric with respect to T0ν.

Moreover, we set

Ωλν={xΩ1:xν<λ}Rλν(Ω0)and(Ωλν)=Rλν(Ωλν).

Also we set

a(ν)=infxΩ1xνandb(ν)=infxΩ0xν.

Observe that, by assumption (A), it follows that Ωλν is nonempty and (Ωλν)Ω1Ω0 for any a(ν)<λ0. We define

uλν(x)=u(xλν)for all xΩλν

and for any a(ν)<λ0.

We are now ready to state our main results.

Theorem 1.1.

Let uC1(Ω1¯Ω0¯) be a weak solution to (1.1). Then

u<uλνin Ωλν for any a(ν)<λ<b(ν).(1.2)

Consequently, it follows that u is strictly increasing with respect to the ν-direction in the set

{xΩ1:a(ν)<xν<b(ν)},with uν>0.

In order to get symmetry results for the solution to (1.1), we restrict our attention to annular domains. We denote by BR the open ball of center 0 and radius R>0 in N. By Theorem 1.1, we immediately deduce the following.

Corollary 1.2.

Let 0<R0<R1 and uC1(BR1¯BR0¯) be a weak solution to (1.1) in BR1BR0. Then ur<0 in BR1BR0.

In the following we state sufficient conditions in order to deduce the radial symmetry of the solution, once we prove the monotonicity. We set

v=-xu,

and we denote by uθ the angular derivative of u.

Theorem 1.3.

Let 0<R0<R1 and let uC1(BR1¯BR0¯) be a weak solution to (1.1) in BR1BR0 that satisfies

uθ=o(v)as |x|R0.(1.3)

If f0, then u is radially symmetric and radially decreasing in BR1BR0.

For the reader’s convenience, let us point out that the condition in (1.3) is inspired by the results in [20, 21].

In Section 2 we give the proof of Theorem 1.1. We prove Theorem 1.3 in Section 3.

2 Proof of Theorem 1.1

Let a(ν)<λ<b(ν). We define

wλν=u-uλν.

We need to prove that wλν<0 in Ωλν. We have

Ωλν(u-uλν)φ=Ωλνf(u)-f(uλν)u-uλν(u-uλν)φfor all φCc(Ωλν),

so that wλν weakly satisfies

-Δwλν=cλ(x)wλν,

where

cλ(x)=f(u)-f(uλν)u-uλνif uuλν  and  cλ(x)=0if u=uλν.

Observe that cλ(x)Lloc(Ωλν) and cλ(x)L(Ωλν) for λ-a(ν) small. Moreover, since wλν=0 on ΩλνTλν, it follows that wλν=c-uλν<0 on ΩλνΩ1 and u<uλν on a neighborhood of ΩλνRλν(Ω0), thanks to the fact that

limxx0xΩλνuλν(x)=+for all x0ΩλνRλν(Ω0).

Then

(wλν)+H01(Ωλν).

Let λ-a(ν) be sufficiently small so that the weak maximum principle in small domains works for the operator Δ+cλ(x) in Ωλν, recalling that cλ(x)L(Ωλν) for λ-a(ν) small. We get wλν0 in Ωλν and, by the strong maximum principle, we obtain

wλν<0in Ωλν.

Set now

μ=sup{λ>a(ν):wtν<0 in Ωtν}for a(ν)<tλ.

We need to prove that μb(ν). On the contrary, suppose μ<b(ν). By continuity, it follows that wμν0 in Ωμν. Moreover, it is possible to apply the same argument of the first part of the proof to obtain, by the strong maximum principle, that

wμν<0in Ωμν.

Take now ε>0 sufficiently small such that μ+ε<b(ν). Given δ>0, we fix a compact set 𝒦Ωμν so that (Ωμν𝒦)<δ2. Since wμν<0 in K, by the continuity of u, it follows that there exists M>0 such that wμνM<0 in K. Therefore, we can find ε0>0 such that

wμ+ενM2<0in K,

whenever 0<εε0. Moreover, choosing ε>0 sufficiently small such that (Ωμ+εν𝒦)<δ and using (wμ+εν)+ as test function in the weak formulation of (1.1), since f is locally Lipschitz continuous, an application of the Poincaré inequality gives

Ωμ+εν𝒦|(wμ+εν)+|2=Ωμ+εν𝒦f(u)-f(uμ+εν)u-uμ+εν(u-uμ+εν)2Cp2(Ωμ+εν𝒦)CΩμ+εν𝒦|(wμ+εν)+|2,

where

C=sups,t[0,uL(Ωμ+ε0ν)]f(t)-f(s)t-s.

Choosing δ sufficiently small, we have Cp2(Ωμ+εν𝒦)C<1 so that (wμ+εν)+=0 in Ωμ+εν𝒦. Hence, wμ+εν0 in Ωμ+εν. By the strong maximum principle, we get

wμ+εν<0in Ωμ+εν.

This gives a contradiction with the definition of μ and shows that μb(ν), namely, (1.2) is proved.

Now, let x and x such that a(ν)<xν<xν<b(ν). Setting λ=(xν+xν)/2, we get xλν=x and

u(x)<uλν(x)=u(x).

Therefore, u is strictly increasing with respect to the ν-direction. To conclude the proof, fix xΩb(ν)ν and let a(ν)<λ<b(ν) be such that xTλνΩ1. We have that -Δwλν=cλ(x)wλν, wλ>0 in Ωλν and wλ=0 in TλνΩ1. By the Hopf lemma, we obtain

0<wλνν=2uν(x).

3 Proof of Theorem 1.3

We start by proving the following proposition.

Proposition 3.1.

Let R0<R1 and uC1(BR1¯BR0¯) be a weak solution to (1.1) in BR1BR0. If f0, then

-Δvf(u)vin BR1BR0.

Proof.

Let x=(x1,,xN)N. Set ui=uei for each i=1,,N. We have

v=-(i=1Nxiui)=-u-i=1Nxiui.

Since -Δu=f(u) in BR1BR0 and f0, we obtain

-Δv=div(u+i=1Nxiui)=2Δu+i=1NxiΔui=-2f(u)-i=1Nxif(u)ui=-2f(u)+f(u)vf(u)v,

so that -Δvf(u)v. ∎

In the following we will also exploit the fact that

-Δuθ=f(u)uθin BR1BR0,

as it follows by direct computation.

Proof of Theorem 1.3.

We shall actually show that uθ0 in BR1BR0. By assumption (1.3), choosing σ>0 sufficiently small, we have uθ+tv0 in BR0+σBR0.

Furthermore, by Corollary 1.2, since v is continuous, we have

v=-x|x||x|u=|x|uν>0,

where ν=-x/|x| with xBR1BR0+σ. Moreover, v>0 on BR1, by the Hopf lemma. Then v>θ(σ)>0 in BR1BR0+σ. Since uθ is bounded in BR1BR0+σ, for t sufficiently large, we have uθ+tv>0 in BR1BR0+σ. Hence,

uθ+tv0in BR1BR0.

Set now

t0=inf{t>0:uθ+tv0 in BR1BR0}.

We need to prove that t0=0. Conversely, suppose t0>0. By the definition of t0 and Proposition 3.1, we obtain

-Δ(uθ+t0v)f(u)(uθ+t0v)in BR1BR0,uθ+t0v0in BR1BR0.

By the strong maximum principle, since uθ=0 and v>0 on BR1, we get

uθ+t0v>0in BR1BR0.

Since (1.3) is in force, there exists δ0>0 such that

uθ+(t0-ε)v>0in BR0+δ0BR0.

Moreover, we have

uθ+t0vm>0in BR1BR0+δ0¯.

By continuity, for ε>0 small, we have that

uθ+(t0-ε)vm2>0in BR1BR0+δ0¯.

Resuming, we have that

uθ+(t0-ε)v>0in BR1BR0.

This contradicts the definition of t0, showing that actually t0=0 and uθ0 in BR1BR0. This is possible only if uθ=0 in BR1BR0, namely, if the solution is radial. We conclude that the solution is also radially decreasing by Theorem 1.1. ∎

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About the article

Received: 2017-10-09

Revised: 2018-04-04

Accepted: 2018-05-14

Published Online: 2018-07-07

Published in Print: 2019-03-01


Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 1–6, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2017-0221.

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