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volume 4, issue 5, article 106, 2003.

Received 20 February, 2003;

accepted 30 June, 2003.

Communicated by:A. Fiorenza

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Journal of Inequalities in Pure and Applied Mathematics

LINEAR ELLIPTIC EQUATIONS AND GAUSS MEASURE

G. DI BLASIO

Dipartimento di Matematica e Applicazioni “R. Caccioppoli”

Università degli Studi di Napoli “Federico II”

Complesso Monte S. Angelo - Via Cintia. Napoli-ITALY EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 020-03

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Linear Elliptic Equations and Gauss Measure

G. di Blasio

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J. Ineq. Pure and Appl. Math. 4(5) Art. 106, 2003

http://jipam.vu.edu.au

Abstract

In this paper we study a Dirichlet problem relative to a linear elliptic equation with lower-order terms, whose ellipticity condition is given in terms of the functionϕ(x)=(2π)⁻ⁿ²exp(^−|x|²^/2), the density in the Gaussian measure. Using the notion of rearrangement with respect to the Gauss measure, we prove a comparison result with a problem of the same type defined in a half space, with data depending only on the first variable.

2000 Mathematics Subject Classification:35B05, 35J70

Key words: Linear elliptic equation, Comparison theorem, Rearrangements of func- tions, Gauss measure

1. Introduction

The object of this paper is to give comparison results for the solution of the problem

(1.1)







−(a_ij(x)u_x_i)_x

j+b_i(x)u_x_i +c(x)u=f(x)ϕ(x) inΩ

u= 0 on∂Ω

whereϕ(x) = (2π)⁻ⁿ² exp − |x|²/2

is the density in the Gaussian measure, Ωis an open set ofRⁿ(n ≥2)which has Gauss measure less than one,a_ij(x), b_i(x)andc(x), i, j = 1, . . . , n,are measurable functions onΩsuch that

(i) a_ij(x)ξ_iξ_j ≥ϕ(x)|ξ|², ∀ξ∈Rⁿ, a.e. x∈Ω, (ii) (P

b²_i)¹² ≤ϕ(x)B,

(iii) c(x)≥c₀(x)ϕ(x), c₀(x)∈L^∞(Ω),

andfis taken in a suitable weightedL^pspace in order to guarantee the existence of a solutionuof the problem (1.1).

We recall thatu∈H₀¹(ϕ,Ω)is a weak solution of the problem (1.1), if (1.2)

Z

Ω

(a_ij(x)u_x_iψ_x_j +b_i(x)u_x_iψ+c(x)u(x)ψ)dx

= Z

Ω

f(x)ϕ(x)ψdx, ∀ψ ∈H₀¹(ϕ,Ω).

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Let us observe that the operator in (1.1) is uniformly elliptic ifΩis bounded.

It is well known that when Ωis bounded, comparison results for elliptic problems have been obtained via Schwarz symmetrization, for a simpler problem which is defined in a ball and has spherically symmetric data (see for example [1], [2], [4], [3], [6], [17], [20], [19], [18], [21]).

In our case Ω can be not bounded and ellipticity condition (i) is given in terms of the density in the Gaussian measure. We consider solutions of problem (1.1) in the weighted Sobolev space H₀¹(ϕ,Ω) (see § 2) and we compare the solution of the problem (1.1) with the solution of a problem in which the data depend only on the first variable and the domain is a half-space which has the same Gauss measure asΩ.

More precisely let Ω^? = {x= (x₁, x₂, . . . , x_n)∈Rⁿ :x₁ > λ} such that γn(Ω^?) =γn(Ω), and letf^?(x)be the rearrangement with respect to the Gauss measure of the functionf(x)(see §2for the definition). To give an example of the obtained results we consider the casec₀(x) = 0.

Letw(x) =w(x₁)be the solution of

(1.3)







−(w_x₁ϕ(x))_x

1 −Bw_x₁ϕ(x) = f^?(x₁)ϕ(x) in Ω^?,

w= 0 on∂Ω^?.

We prove that the pointwise comparison

(1.4) u^?(x) =u^?(x₁)≤w(x₁) =w^?(x) for a.ex= (x₁, x₂, . . . , x_n)∈Ω^? holds, where u^? andw^? are the rearrangements with respect to the Gauss measure ofuandwrespectively. In this case the comparison also gives an explicit

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estimate of u^? that is also a condition for the existence of the solution for the problem (1.1) in terms of the existence of the solution of (1.3).

If c₀(x) 6= 0, a comparison with a problem which also takes the term

“c(x) u(x)” into account can be found. In this case, depending on the sign ofc₀(x),pointwise comparison (1.4) is false in general, but a comparison between the concentrations can be found (see Theorem3.2).

In the proofs of our results we use tools similar to the classical methods based on the isoperimetric inequality and Schwarz symmetrization (see [2]).

In our case a fundamental rule is played by the isoperimetric inequality with respect to the Gauss measure.

The problem (1.1) has been studied using rearrangement with respect to the Gauss measure in [7], whenb_i(x) =c(x) = 0.

Existence results for weak solutions of problem (1.1) can be obtained for example via the Lax Milgram theorem when (i) and (ii) hold and ^a_ϕ(x)^ij^(x),_ϕ(x)^c(x) ∈ L^∞(Ω)andf(x)∈L²(ϕ,Ω)(see also [15], [22]).

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2. Notations and Preliminary Results

In this section we recall some definitions and results which will be useful in what follows. Let Ω be an open set of Rⁿ and let γn be the n−dimensional Gauss measure onRⁿdefined by

γ_n(dx) = ϕ(x)dx= (2π)⁻ⁿ² exp −|x|² 2

!

dx, x∈Rⁿ

normalized byγ_n(Rⁿ) = 1.

One of the main tools used to prove the comparison result is the isoperimetric inequality with respect to the Gauss measure, to recall this result we define the perimeter with respect to the Gauss measure as (see [13])

P (E) = (2π)⁻ⁿ² Z

∂E

exp −|x|² 2

!

H_n−1(dx),

whereEis a(n−1)−rectificable set andHn−1denotes the(n−1)−dimensional Hausdorff measure. For allλ ∈R, we denote byH(ξ, λ)the half-space defined by

H(ξ, λ) ={x∈Rⁿ : (x, ξ)> λ}

and we setH(ξ, λ) =Rⁿifλ=−∞andH(ξ, λ) =∅ifλ= +∞.

It is well known (see [11]) that among all measurable sets ofRⁿ with pre- scribed Gauss measure, the half-spaces take the smallest perimeter. In other words, the half-spaces are extremal in the isoperimetric problem for the Gauss

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measure. From an appropriate form of the Brunn - Minkowski inequality, the isoperimetric inequality follows (see [8], [11], [10], [12])

P (E)≥P (H(ξ, λ))

for all subsets E ⊂ Rⁿ such that γ_n(E) = γ_n(H(ξ, λ)). For the sake of simplicity we shall considerξ= (1,0, . . . ,0).

Now we consider the notion of rearrangement with respect to the Gauss measure. Ifu is a measurable function inΩ, we define the distribution function of u, denoted byµ, as the Gauss measure of the level set ofu,i.e.

µ:t∈[0,∞[−→µ(t)∈[0,1], where

µ(t) = γ_n({x∈Ω :|u|> t}).

We denote by u^∗ the decreasing rearrangement ofu(with respect to the Gauss measure), i.e.

u^∗(s) = inf{t≥0 :µ(t)≤s}, s ∈]0,1],

the functions µ and u^∗ are decreasing and right - continuous. Moreover the increasing rearrangement ofuis the function defined as follows

u∗(s) =u^∗(γ_n(Ω)−s), s∈[0,1[.

Finally we define the rearrangement ofuwith respect to the Gauss measure, denoted byu^?, as the function whose level sets are half-spaces having the same Gauss measure of the level sets ofu. More precisely we define

Φ (τ) =γ_n({x∈Rⁿ :x₁ > τ}) = 1

√2π Z +∞

τ

exp

−t² 2

dτ

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for allτ ∈R.Thenu^? is a map fromΩ^? into[0,+∞[defined by u^?(x) = u^∗(Φ (x₁)),

whereΩ^? ={x= (x₁, x₂, . . . , x_n)∈Rⁿ:x₁ > λ}such thatγ_n(Ω^?) =γ_n(Ω).

For statements about the properties of rearrangement with respect to a positive measure see, for example, [9], [20], [16].

We recall that iff(x), g(x)are measurable functions, a Hardy type inequality (see [9])

Z

Ω

|f(x)g(x)| γ_n(dx)≤ Z

Ω^?

f^?(x)g^?(x) γ_n(dx) =

Z γn(Ω) 0

f^∗(s)g^∗(s) ds and a Polya-Szëgo inequality for a Lipschitz continuous function u(x) (see [20])

(2.1)

Z

Rⁿ

|∇u^?| γ_n(dx)≤ Z

Rⁿ

|∇u| γ_n(dx) holds. Moreover we have

Z

Ω

|f(x)|^pγ_n(dx) = Z

Ω^?

|f^?(x)|^pγ_n(dx) =

Z γn(Ω) 0

f^∗(s)^pds, that is, theL^p weighted norm is invariant under the rearrangement.

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Now define the weighted Sobolev spaceH₀¹(ϕ,Ω)as the closure ofC₀^∞(Ω) under the norm

(2.2) kuk_H1

0(ϕ,Ω) = Z

Ω

|∇u(x)|²dγ_n(x) ¹₂

.

We remark that the following Poincarè type inequality can be proved.

Proposition 2.1. Let Ω be a open subset of Rⁿ with γ_n(Ω) < 1. For each functionf ∈H₀¹(ϕ,Ω)we have

(2.3) kfk_L2(ϕ,Ω) ≤Ck∇fk_L2(ϕ,Ω), whereCis a constant depending onnandΩ.

Proof. By (2.1) we have the following inequality kfk²_L2

(ϕ,Ω)

k∇fk²_L2 (ϕ,Ω)

≤

kf^?k²_L2 (ϕ,Ω?)

k∇f^?k²_L2 (ϕ,Ω?)

(2.4)

=

R+∞

λ f^?(x₁)²exp

−^x₂²¹ dx₁ R+∞

λ

d

dx1f^?(x₁)2

exp

−^x₂²¹ dx₁

,

whereλis such thatγ_n(Ω^?) =γ_n(Ω)withΩ^? ={x= (x₁, x₂, . . . , x_n)∈Rⁿ : x1 > λ}. The ratio in (2.4) is bounded (see e.g. [14, Theorem 1.3.1./2]).

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The previous Poincaré type inequality ensures the equivalence between the norm (2.2) and the following one

kuk_H1

0(ϕ,Ω)= Z

Ω

|u(x)|²dγ_n(x) ¹₂

+ Z

Ω

|∇u(x)|²dγ_n(x) ¹₂

. Now, we recall the following lemmas which will be useful in the following.

Lemma 2.2. Letf(x), g(x)be measurable, positive functions such that Z α

0

f(x)dx≤ Z α

0

g(x)dx, α∈[0, a]. Ifh(x)≥0is a decreasing function then

Z α 0

f(x)h(x)dx≤ Z α

0

g(x)h(x)dx, α ∈[0, a].

Lemma 2.3. Letzbe a bounded function,K a nonnegative integrable function, andψa function with bounded variation that vanishes at+∞. If

z(t)≤ Z ∞

t

K(s)z(s)ds+ψ(t) for almost everyt >0, then

z(t)≤ Z ∞

t

exp Z s

t

K(τ)dτ

[−dψ(s)]

for almost everyt >0.

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3. Comparison Results

In this section we prove some comparison results for the solution of the problem (1.1). We consider first the casec₀(x) = 0and then the casec₀(x)6= 0.

Theorem 3.1. LetΩbe an open set ofRⁿwithγ_n(Ω)<1and letu∈H₀¹(ϕ,Ω) be the solution of (1.1) with the assumptions (i), (ii) and (iii). Letc₀(x) = 0and (3.1)

Z +∞

λ

exp τ²

2

Z +∞

τ

f^?(σ) exp

B(σ−τ)− σ² 2

dσ

2

dτ <+∞, whereλis such thatΩ^? ={x₁ > λ}.Then we have

(3.2) u^?(x1)≤w(x) for a.e.x∈Ω^? and

(3.3)

Z

Ω

|∇u|^qϕ(x)dx≤ Z

Ω^?

|∇w|^qϕ(x)dx for all0< q ≤2, where

w(x)=w(x1)=

Z x1

λ

exp τ²

2

Z +∞

τ

f^?(σ) exp

B(σ−τ)− σ² 2

dσ

dτ is the solution of the problem (1.3).

Remark 3.1. Condition (3.1) ensures the existence of a solution w(x1) = w^?(x) ∈ H₀¹(ϕ,Ω) of (1.3). It is satisfied for a wide class of functions, for instance for functionsf such that

f^? ≤exp

−B(σ−τ) + σ² 4

(1 +|τ|)¹²⁻ ∀τ ≥λ for some constantsC > 0, >0.

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Remark 3.2. The assumptionγ_n(Ω) <1is made to guarantee that the Poincarè type inequality (2.3) holds.

Proof. If we choose in (1.2), forh >0 andt ∈[0,sup|u|[

ψ(x) =











hsgnu if |u|> t+h, (|u| −t) sgnu if t <|u| ≤t+h,

0 otherwise,

we get 1 h

Z

t<|u|≤t+h

a_ij(x)u_x_iu_x_jdx + 1

h Z

t<|u|≤t+h

b_i(x)u_x_i(|u| −t) sgnudx+ Z

|u|>t+h

b_i(x)u_x_isgnudx + 1

h Z

t<|u|≤t+h

c(x) (|u| −t) sgnudx+ Z

|u|>t+h

c(x)u(x) sgnudx

= 1 h

Z

t<|u|≤t+h

f(x)ϕ(x) (|u| −t) sgnudx+ Z

|u|>t+h

f(x)ϕ(x) sgnudx.

By the ellipticity condition (i), (ii) and (iii) and lettinghgo to 0, we obtain

(3.4) − d

dt Z

|u|>t

|∇u|²ϕ(x)dx

≤B Z

|u|>t

|∇u|ϕ(x)dx+ Z

|u|>t

f(x) sgnuϕ(x)dx

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On the other hand, the coarea formula (see [13]) and the isoperimetric inequality with respect to the Gauss measure give

−d dt

Z

|u|>t

|∇u|ϕ(x)dx≥ Z

∂{|u|>t}^?

ϕ(x)Hn−1(dx) (3.5)

= 1

√2πexp −Φ⁻¹(µ(t))² 2

! ,

where{|u|> t}^? is the half space having Gauss measureµ(t).

Then using (3.5) and the Hölder inequality we obtain (3.6) 1≤(2π)¹² exp Φ⁻¹(µ(t))²

2

!

×(−µ⁰(t))¹²

−d dt

Z

|u|>t

|∇u|²ϕ(x)dx ¹₂

.

Using (3.6) and the Hölder inequality, (3.4) becomes

− d dt

Z

|u|>t

|∇u|²ϕ(x)dx

≤B(2π)¹² Z ∞

t

− d ds

Z

|u|>s

|∇u|²ϕ(x)dx

exp Φ⁻¹(µ(s))² 2

!

×(−µ⁰(s))ds+ Z µ(t)

0

f^∗(s)ds.

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By Gronwall’s Lemma2.3we obtain

−d dt

Z

|u|>t

|∇u|²ϕ(x)dx (3.7)

≤ Z µ(t)

0

exp

"

B(2π)¹² Z µ(t)

r

exp Φ⁻¹(τ)² 2

! dτ

#

f^∗(r)dr

= Z µ(t)

0

exp

B(Φ⁻¹(r)−Φ⁻¹(µ(t)))

f^∗(r)dr.

Using again (3.6) we have (3.8) 1≤2πexp

Φ⁻¹(µ(t))²

(−µ⁰(t))

× Z µ(t)

0

exp

B(Φ⁻¹(r)−Φ⁻¹(µ(t)))

f^∗(r)dr.

Then using (3.8) and integrating between0andt, (3.7) becomes t≤2π

Z γn(Ω) µ(t)

exp

Φ⁻¹(σ)² Z σ

0

exp

B(Φ⁻¹(s)−Φ⁻¹(σ))

f^∗(s)dsdσ, which, puttingµ(t) = sands= Φ (x1),gives

u^?(x) =u^?(x₁) (3.9)

≤ Z x1

λ

exp τ²

2

Z +∞

τ

f^?(σ) exp

B(σ−τ)− σ² 2

dσ

dτ,

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where λ is such that γ_n(Ω^?) = γ_n(Ω) with Ω^? = {x = (x₁, x₂, . . . , x_n) ∈ Rⁿ : x1 > λ}.This completes the proof of (3.2) observing that the right-hand side of (3.9) is the solution of (1.3).

Let us prove now (3.3). Using the Hölder inequality and (3.7) we have (3.10) − d

dt Z

|u|>t

|∇u|^qϕ(x)dx

≤(−µ⁰(t))¹⁻

q 2

Z µ(t) 0

exp

B(Φ⁻¹(r)−Φ⁻¹(µ(t)))

f^∗(r)dr

!^q₂ . By (3.6) we have

(3.11) (−µ⁰(t))⁻

q

2 (2π)⁻^q² exph

−q 2

Φ⁻¹(µ(t))²i

≤

−d dt

Z

|u|>t

|∇u|²ϕ(x)dx ^q₂

. Using (3.11), (3.7) and integrating between0and+∞, (3.10) becomes Z

Ω

|∇u|^qϕ(x)dx≤(2π)^q²

Z γn(Ω) 0

Z s 0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

f^∗(r)dr q

×exphq 2

Φ⁻¹(s)²i ds

= (2π)¹² Z +∞

λ

Z +∞

τ

f^?(σ) exp

B(σ−τ)− σ² 2

dσ

q

×exp q

2τ²− τ² 2

dτ,

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that is, (3.3).

In the next theorem we consider the casec₀(x)6= 0and we compare problem (1.1) with a “symmetrized” problem which also takes account of the influence of the term “c(x)u(x)”.

Theorem 3.2. LetΩbe an open set ofRⁿwithγ_n(Ω)<1and letu∈H₀¹(ϕ,Ω) be a solution of (1.1) with the assumptions (i), (ii) and (iii). Moreover, let

c⁺₀(x) = max{c₀(x),0}, c⁻₀(x) = max{−c₀(x),0}, c⁺_0?(x) =c⁺_0∗(Φ(x₁)).

We will assume that the problem

(3.12)











−(w_x₁ϕ(x))_x

1 −Bw_x₁ϕ(x) +

c⁺_0?(x₁)−c^−?₀ (x₁)

wϕ(x) = f^?(x₁)ϕ(x) inΩ^?,

w= 0 on∂Ω^?,

has a solutionw(x) = w^?(x₁).Then

u^∗(s)≤w^∗(s) holds in[0, s⁰₁],wheres⁰₁ = inf

s:c⁺_0∗(s)>0 ,and Z s

0

u^∗(r)dr≤ Z s

0

w^∗(r)dr holds in]s⁰₁, γn(Ω)].

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Proof. Using (iii), the Schwartz inequality and the Hardy inequality we have,

− Z

|u|>t

c(x)|u|dx≤ − Z

|u|>t

c0(x)|u|ϕ(x)dx (3.13)

≤ − Z µ(t)

0

c⁺_0∗(s)−c^−∗₀ (s)

u^∗(s)ds,

wherec⁺_0∗(s)is the increasing rearrangement ofc⁺₀(x)andµ(t)is the distribution function ofu(x). Proceeding as in Theorem3.1and using (3.13) we obtain (3.14) − d

dt Z

|u|>t

|∇u|²ϕ(x)dx

≤B Z ∞

t

− d ds

Z

|u|>s

|∇u|²ϕ(x)dx ¹₂

(−µ⁰(s))¹²ds

− Z µ(t)

0

c⁺_0∗(s)−c^−∗₀ (s)

u^∗(s)ds+ Z µ(t)

0

f^∗(s)ds.

By (3.6) we have

(3.15) 1

−µ⁰(t) ≤2πexp

Φ⁻¹(s)²

−d dt

Z

|u|>t

|∇u|²ϕ(x)dx

. Following the same steps as in Theorem3.1and using (3.14), (3.15) becomes (3.16) (−u^∗)⁰(s)≤2πexp

Φ⁻¹(s)² Z s

0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

×[f^∗(r) + [c^−∗₀ (r)−c⁺_0∗(r)]u^∗(r)]dr.

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Now if we consider the problem (3.12) we can proceed in the same way except that the inequalities should be replaced by equalities.

Then if we callw(x) =w^?(x)the solution of the problem (3.12), we obtain the following equality

(3.17) (−w^∗)⁰(s) = 2πexp

Φ⁻¹(s)² Z s

0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

×[f^∗(r) + [c^−∗₀ (r)−c⁺_0∗(r)]w^∗(r)]dr.

To prove the comparison result, we putv(s) = u^∗(s)−w^∗(s). By (3.16) and (3.17) we have

(3.18) (−v)⁰(s)≤2πexp

Φ⁻¹(s)² Z s

0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

×[c^−∗₀ (r)−c⁺_0∗(r)]v(r)dr.

Let us supposec^−∗₀ (x), c⁺_0∗(x)6= 0and let us puts⁰₁ = inf

s:c⁺_0∗(s)>0 ands⁰₀ = sup

s:c^−∗₀ (s)>0 .We write V₁(s) =

Z s 0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

c^−∗₀ (r)v(r)dr, s ∈[0, s⁰₀] V₂(s) =

Z s s⁰₁

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

c⁺_0∗(r)v(r)dr, s∈]s⁰₁, γ_n(Ω)].

We assume initially thatc^−∗₀ (s)is continuous ats⁰₀, we have to prove, now, that V₁(s)≤0. Arguing as in [1] we observe that the existence of a solutionw(x) =

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w^?(x)of (3.12), that implies (3.17), guarantees that the problem:











− c^−∗₀ (s)⁻¹Z⁰(s)⁰

= (2π) exp

Φ⁻¹(s)²

Z(s) + 2πexp

Φ⁻¹(s)²

×Rs

0 exp [B(Φ⁻¹(r)−Φ⁻¹(s))]f^∗(r)dr, Z(0) =Z⁰(s⁰₀) = 0,

has the following positive solution Z(s) =

Z s 0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

c^−∗₀ (r)w^∗(r)dr.

This allows us to state (see [5]) that the problem

(3.19)







c^−∗₀ (s)⁻¹ξ⁰(s)⁰

+λ(2π) exp

Φ⁻¹(s)²

ξ(s) = 0, ξ(0) =ξ⁰(s⁰₀) = 0,

has the first eigenvalueλ₁ >1,and consequently in the following problem







c^−∗₀ (s)⁻¹V₁⁰(s)⁰

+ 2πexp

Φ⁻¹(s)²

V₁(s)≥0, V₁(0) =V₁⁰(s⁰₀) = 0,

we have

V₁(s)≤0 and V₁⁰(s)≤0, s∈[0, s⁰₀],

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that is,

(3.20) u^∗(s)≤w^∗(s), s∈[0, s⁰₀]. On the other hand, from (3.18) and (3.20), we have







c⁺_0∗(s)⁻¹V₂⁰(s)⁰

−2πexp

Φ⁻¹(s)²

V₂(s)≥0, V₂(s⁰₁) =V₂⁰(γ_n(Ω)) = 0.

Here we can use the maximum principle to obtain V2(s) ≤ 0. For Lemma2.2 withh(r) = c⁺_0∗(r)⁻¹,we have for eachs∈]s⁰₁, γ_n(Ω)],

(3.21) Z s

s⁰₁

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

u^∗(r)dr

≤ Z s

s⁰₁

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

w^∗(r)dr, that is,

u^∗(s⁰₁)≤w^∗(s⁰₁).

On the other hand, from (3.16), (3.17), (3.20) and the definition ofv(s)we have

−v⁰(s)≤0, s ∈[s⁰₀, s⁰₁].

Now sinceu^∗(s⁰₁)≤w^∗(s⁰₁), integrating betweensands⁰₁ we obtain (3.22) u^∗(s)≤w^∗(s), s∈[s⁰₀, s⁰₁].

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From (3.20) and (3.22) we have

u^∗(s)≤w^∗(s), s ∈[0, s⁰₁]. Moreover, fors∈

s⁰_1,γ_n(Ω)

, (3.21) becomes 0≥

Z s 0

exp

B(Φ⁻¹(r)−Φ⁻¹(s))

[u^∗(r)−w^∗(r)]dr, that is,

Z s 0

u^∗(r)dr≤ Z s

0

w^∗(r)dr, s∈[s⁰₁, γ_n(Ω)].

Finally we can remove the hypothesis about the continuity of c^−∗₀ (s)ats⁰₀ proceeding by approximations.

If c⁺_0∗(x) = 0 or c^−∗₀ (x) = 0 then s⁰₁ = γn(Ω) or s⁰₀ = 0 and the result follows with the obvious modifications.

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