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: giuseppina.diblasio@dma.unina.it
volume 4, issue 5, article 106,
2003.
Received 20 February, 2003;
accepted 30 June, 2003.
Communicated by: A. Fiorenza
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
020-03
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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
n
function ϕ(x)=(2π)− 2 exp(−|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 functions, Gauss measure
Contents
Linear Elliptic Equations and
Gauss Measure
G. di Blasio
Title Page
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Notations and Preliminary Results . . . . . . . . . . . . . . . . . . . . . . 6
3
Comparison Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
References
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1.
Introduction
The object of this paper is to give comparison results for the solution of the
problem

in Ω
 − (aij (x)uxi )xj + bi (x)uxi + c(x)u = f (x)ϕ(x)
(1.1)

u=0
on ∂Ω
n
where ϕ(x) = (2π)− 2 exp − |x|2 /2 is the density in the Gaussian measure,
Ω is an open set of Rn (n ≥ 2) which has Gauss measure less than one, aij (x),
bi (x) and c(x), i, j = 1, . . . , n, are measurable functions on Ω such that
(i) aij (x)ξi ξj ≥ ϕ(x) |ξ|2 ,
P 1
(ii) ( b2i ) 2 ≤ ϕ(x)B,
(iii) c (x) ≥ c0 (x) ϕ(x),
∀ξ ∈ Rn , a.e. x ∈ Ω,
G. di Blasio
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c0 (x) ∈ L∞ (Ω),
and f is taken in a suitable weighted Lp space in order to guarantee the existence
of a solution u of the problem (1.1).
We recall that u ∈ H01 (ϕ, Ω) is a weak solution of the problem (1.1), if
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Z
(1.2)
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(aij (x)uxi ψxj + bi (x)uxi ψ + c(x)u(x)ψ)dx
Ω
Z
=
f (x)ϕ(x)ψdx, ∀ψ ∈ H01 (ϕ, Ω).
J. Ineq. Pure and Appl. Math. 4(5) Art. 106, 2003
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Let us observe that the operator in (1.1) is uniformly elliptic if Ω is bounded.
It is well known that when Ω is 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 as Ω.
More precisely let Ω? = {x = (x1 , x2 , . . . , xn ) ∈ Rn : x1 > λ} such that
γn (Ω? ) = γn (Ω), and let f ? (x) be the rearrangement with respect to the Gauss
measure of the function f (x) (see § 2 for the definition). To give an example of
the obtained results we consider the case c0 (x) = 0.
Let w (x) = w (x1 ) be the solution of

in Ω? ,
 − (wx1 ϕ (x))x1 − Bwx1 ϕ (x) = f ? (x1 )ϕ (x)
(1.3)

w=0
on ∂Ω? .
We prove that the pointwise comparison
(1.4) u? (x) = u? (x1 ) ≤ w (x1 ) = w? (x) for a.e x = (x1 , x2 , . . . , xn ) ∈ Ω?
holds, where u? and w? are the rearrangements with respect to the Gauss measure of u and w respectively. In this case the comparison also gives an explicit
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G. di Blasio
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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
of c0 (x) , pointwise comparison (1.4) is false in general, but a comparison between the concentrations can be found (see Theorem 3.2).
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], when bi (x) = c(x) = 0.
Existence results for weak solutions of problem (1.1) can be obtained for
aij (x) c(x)
example via the Lax Milgram theorem when (i) and (ii) hold and ϕ(x)
, ϕ(x) ∈
∞
2
L (Ω) and f (x) ∈ L (ϕ, Ω) (see also [15], [22]).
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G. di Blasio
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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 on Rn defined by
!
2
n
|x|
dx,
x ∈ Rn
γn (dx) = ϕ (x) dx = (2π)− 2 exp −
2
normalized by γn (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])
!
Z
2
|x|
−n
exp −
Hn−1 (dx) ,
P (E) = (2π) 2
2
∂E
where E is a (n−1)−rectificable set and Hn−1 denotes the (n−1)−dimensional
Hausdorff measure. For all λ ∈ R, we denote by H (ξ, λ) the half-space defined
by
H (ξ, λ) = {x ∈ Rn : (x, ξ) > λ}
and we set H (ξ, λ) = Rn if λ = −∞ and H (ξ, λ) = ∅ if λ = +∞.
It is well known (see [11]) that among all measurable sets of Rn with prescribed 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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G. di Blasio
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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 γn (E) = γn (H (ξ, λ)). For the sake of
simplicity we shall consider ξ = (1, 0, . . . , 0).
Now we consider the notion of rearrangement with respect to the Gauss measure. If u is a measurable function in Ω, we define the distribution function of
u, denoted by µ, as the Gauss measure of the level set of u, i.e.
µ : t ∈ [0, ∞[ −→ µ (t) ∈ [0, 1] ,
Linear Elliptic Equations and
Gauss Measure
G. di Blasio
where
µ (t) = γn ({x ∈ Ω : |u| > t}) .
∗
We denote by u the decreasing rearrangement of u (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 of u is the function defined as follows
∗
u∗ (s) = u (γn (Ω) − s) ,
s ∈ [0, 1[ .
Finally we define the rearrangement of u with respect to the Gauss measure,
denoted by u? , as the function whose level sets are half-spaces having the same
Gauss measure of the level sets of u. More precisely we define
2
Z +∞
t
1
n
exp −
dτ
Φ (τ ) = γn ({x ∈ R : x1 > τ }) = √
2
2π τ
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for all τ ∈ R. Then u? is a map from Ω? into [0, +∞[ defined by
u? (x) = u∗ (Φ (x1 )) ,
where Ω? = {x = (x1 , x2 , . . . , xn ) ∈ Rn : x1 > λ} such that γn (Ω? ) = γn (Ω).
For statements about the properties of rearrangement with respect to a positive measure see, for example, [9], [20], [16].
We recall that if f (x) , g (x) are measurable functions, a Hardy type inequality (see [9])
Z
Z
Z
Linear Elliptic Equations and
Gauss Measure
γn (Ω)
f ∗ (s) g ∗ (s) ds
G. di Blasio
and a Polya-Szëgo inequality for a Lipschitz continuous function u (x) (see
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?
|f (x) g (x)| γn (dx) ≤
?
f (x) g (x) γn (dx) =
Ω?
Ω
0
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[20])
Z
|∇u | γn (dx) ≤
(2.1)
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?
Rn
|∇u| γn (dx)
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holds. Moreover we have
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Z
Ω
|f (x)|p γn (dx) =
Z
Ω?
|f ? (x)|p γn (dx) =
Z
γn (Ω)
f ∗ (s)p ds,
0
that is, the Lp weighted norm is invariant under the rearrangement.
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Now define the weighted Sobolev space H01 (ϕ, Ω) as the closure of C0∞ (Ω)
under the norm
Z
12
2
(2.2)
kukH 1 (ϕ,Ω) =
|∇u (x)| dγn (x) .
0
Ω
We remark that the following Poincarè type inequality can be proved.
Proposition 2.1. Let Ω be a open subset of Rn with γn (Ω) < 1. For each
function f ∈ H01 (ϕ, Ω) we have
Linear Elliptic Equations and
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kf kL2 (ϕ,Ω) ≤ C k∇f kL2 (ϕ,Ω) ,
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(2.3)
where C is a constant depending on n and Ω.
Proof. By (2.1) we have the following inequality
(2.4)
kf k2L2
(ϕ,Ω)
k∇f k2L2
(ϕ,Ω)
≤
kf ? k2L2
(ϕ,Ω? )
k∇f ? k2L2 ?
(ϕ,Ω )
2
R +∞ ?
x
2
f (x1 ) exp − 21 dx1
λ
,
=R
2
2
+∞
x1
d
? (x )
f
exp
−
dx
1
1
dx1
2
λ
where λ is such that γn (Ω? ) = γn (Ω) with Ω? = {x = (x1 , x2 , . . . , xn ) ∈ 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
12 Z
Z
12
2
2
|u (x)| dγn (x)
kukH 1 (ϕ,Ω) =
|∇u (x)| dγn (x) .
+
0
Ω
Ω
Now, we recall the following lemmas which will be useful in the following.
Lemma 2.2. Let f (x) , g (x) be measurable, positive functions such that
Z α
Z α
f (x) dx ≤
g (x) dx,
α ∈ [0, a] .
0
0
G. di Blasio
If h(x) ≥ 0 is a decreasing function then
Z α
Z α
f (x) h(x)dx ≤
g (x) h(x)dx,
0
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α ∈ [0, a] .
0
Lemma 2.3. Let z be a bounded function, K a nonnegative integrable function,
and ψ a function with bounded variation that vanishes at +∞. If
Z ∞
z (t) ≤
K (s) z (s) ds + ψ (t)
t
for almost every t > 0, then
Z
z (t) ≤
t
for almost every t > 0.
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∞
Z
exp
s
K (τ ) dτ [−dψ (s)]
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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 case c0 (x) = 0 and then the case c0 (x) 6= 0.
Theorem 3.1. Let Ω be an open set of Rn with γn (Ω) < 1 and let u ∈ H01 (ϕ, Ω)
be the solution of (1.1) with the assumptions (i), (ii) and (iii). Let c0 (x) = 0 and
2
Z +∞ 2 Z +∞
σ2
τ
?
(3.1)
exp
f (σ) exp B (σ − τ ) −
dσ dτ < +∞,
2
2
λ
τ
where λ is such that Ω? = {x1 > λ} . Then we have
u? (x1 ) ≤ w(x)
(3.2)
for a.e. x ∈ Ω?
Linear Elliptic Equations and
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and
Z
Z
q
|∇u| ϕ (x) dx ≤
(3.3)
|∇w|q ϕ (x) dx
for all 0 < q ≤ 2,
Ω?
Ω
where
2 Z +∞
τ
σ2
?
w (x)=w (x1 )=
exp
f (σ) exp B (σ − τ ) −
dσ dτ
2
2
λ
τ
is the solution of the problem (1.3).
Z
x1
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 functions f such that
1
σ2
?
f ≤ exp −B (σ − τ ) +
(1 + |τ |) 2 −
∀τ ≥ λ
4
for some constants C > 0, > 0.
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Remark 3.2. The assumption γn (Ω) < 1 is made to guarantee that the Poincarè
type inequality (2.3) holds.
Proof. If we choose in (1.2), for h > 0 and t ∈ [0, sup |u|[


h sgn u
if |u| > t + h,



(|u| − t) sgn u if t < |u| ≤ t + h,
ψ(x) =



 0
otherwise,
we get
Z
1
aij (x)uxi uxj dx
h t<|u|≤t+h
Z
Z
1
+
bi (x)uxi (|u| − t) sgn udx +
bi (x)uxi sgn udx
h t<|u|≤t+h
|u|>t+h
Z
Z
1
+
c(x) (|u| − t) sgn udx +
c(x)u (x) sgn udx
h t<|u|≤t+h
|u|>t+h
Z
Z
1
=
f (x)ϕ(x) (|u| − t) sgn udx +
f (x)ϕ(x) sgn udx.
h t<|u|≤t+h
|u|>t+h
By the ellipticity condition (i), (ii) and (iii) and letting h go to 0, we obtain
Z
d
(3.4)
−
|∇u|2 ϕ(x)dx
dt |u|>t
Z
Z
≤B
|∇u| ϕ (x) dx +
f (x) sgn uϕ(x)dx
|u|>t
|u|>t
Linear Elliptic Equations and
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On the other hand, the coarea formula (see [13]) and the isoperimetric inequality
with respect to the Gauss measure give
Z
Z
d
(3.5)
−
|∇u| ϕ(x)dx ≥
ϕ(x)Hn−1 (dx)
dt |u|>t
∂{|u|>t}?
!
Φ−1 (µ (t))2
1
,
= √ exp −
2
2π
where {|u| > t}? is the half space having Gauss measure µ(t).
Then using (3.5) and the Hölder inequality we obtain
!
2
−1
1
Φ
(µ
(t))
(3.6)
1 ≤ (2π) 2 exp
2
12
Z
1
d
2
0
2
× (−µ (t)) −
|∇u| ϕ(x) dx .
dt |u|>t
Using (3.6) and the Hölder inequality, (3.4) becomes
Z
d
−
|∇u|2 ϕ(x)dx
dt |u|>t
Z ∞
Z
1
d
2
2
≤ B (2π)
−
|∇u| ϕ(x)dx exp
ds |u|>s
t
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!
Φ−1 (µ (s))2
2
Z µ(t)
× (−µ0 (s)) ds +
f ∗ (s) ds.
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By Gronwall’s Lemma 2.3 we obtain
Z
d
(3.7) −
|∇u|2 ϕ(x)dx
dt |u|>t
"
Z
µ(t)
≤
exp B (2π)
0
Z
=
1
2
Z
µ(t)
exp
r
Φ−1 (τ )2
2
!
#
dτ f ∗ (r) dr
µ(t)
exp B(Φ−1 (r) − Φ−1 (µ (t))) f ∗ (r) dr.
Linear Elliptic Equations and
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0
Using again (3.6) we have
G. di Blasio
(3.8) 1 ≤ 2π exp Φ−1 (µ (t))2 (−µ0 (t))
Z µ(t)
×
exp B(Φ−1 (r) − Φ−1 (µ (t))) f ∗ (r) dr.
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Then using (3.8) and integrating between 0 and t, (3.7) becomes
Z
γn (Ω)
t ≤ 2π
µ(t)
exp Φ−1 (σ)2
Z
σ
exp B(Φ−1 (s) − Φ−1 (σ)) f ∗ (s) dsdσ,
0
which, putting µ (t) = s and s = Φ (x1 ) , gives
(3.9)
u? (x) = u? (x1 )
Z x1 2 Z +∞
τ
σ2
?
≤
exp
f (σ) exp B (σ − τ ) −
dσ dτ,
2
2
λ
τ
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where λ is such that γn (Ω? ) = γn (Ω) with Ω? = {x = (x1 , x2 , . . . , xn ) ∈
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
Z
d
(3.10) −
|∇u|q ϕ (x) dx
dt |u|>t
! 2q
Z µ(t)
q
1−
≤ (−µ0 (t)) 2
.
exp B(Φ−1 (r) − Φ−1 (µ (t))) f ∗ (r) dr
0
By (3.6) we have
(3.11)
Linear Elliptic Equations and
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G. di Blasio
− 2q
(−µ0 (t))
h q i
q
(2π)− 2 exp −
Φ−1 (µ (t))2
2
2q
Z
d
2
≤ −
|∇u| ϕ (x) dx .
dt |u|>t
Using (3.11), (3.7) and integrating between 0 and +∞, (3.10) becomes
q
Z
Z γn (Ω) Z s
∗
q
q
−1
−1
|∇u| ϕ (x) dx ≤ (2π) 2
exp B(Φ (r) − Φ (s)) f (r) dr
Ω
0
0
h q i
2
−1
× exp
Φ (s) ds
q
Z +∞ Z 2+∞
2
1
σ
?
= (2π) 2
f (σ) exp B (σ − τ ) −
dσ
2
λ
τ
q 2 τ2
× exp τ −
dτ,
2
2
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that is, (3.3).
In the next theorem we consider the case c0 (x) 6= 0 and 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. Let Ω be an open set of Rn with γn (Ω) < 1 and let u ∈ 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} , c0 (x) = max {−c0 (x), 0} , c0? (x) = c0∗ (Φ(x1 )).
We will assume that the problem

− (wx1 ϕ (x))x1 − Bwx1 ϕ (x)



−?
?
?
+ c+
(3.12)
0? (x1 ) − c0 (x1 ) wϕ(x) = f (x1 )ϕ (x) in Ω ,



w=0
on ∂Ω? ,
has a solution w(x) = w? (x1 ). Then
u∗ (s) ≤ w∗ (s)
holds in [0, s01 ] , where s01 = inf s : c+
(s)
>
0
, and
0∗
Z s
Z s
∗
u (r) dr ≤
w∗ (r) dr
0
holds in ]s01 , γn (Ω)].
Linear Elliptic Equations and
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Proof. Using (iii), the Schwartz inequality and the Hardy inequality we have,
Z
Z
(3.13)
c(x) |u| dx ≤ −
c0 (x) |u| ϕ(x)dx
−
|u|>t
|u|>t
µ(t)
Z
+
∗
c0∗ (s) − c−∗
0 (s) u (s)ds,
≤−
0
c+
0∗ (s)
where
is the increasing rearrangement of c+
0 (x) and µ(t) is the distribution function of u(x). Proceeding as in Theorem 3.1 and using (3.13) we obtain
Z
d
(3.14) −
|∇u|2 ϕ(x)dx
dt |u|>t
21
Z
Z ∞
1
d
2
|∇u| ϕ(x)dx (−µ0 (s)) 2 ds
≤B
−
ds |u|>s
t
Z µ(t)
Z µ(t)
+
∗
−∗
−
c0∗ (s) − c0 (s) u (s)ds +
f ∗ (s)ds.
0
0
By (3.6) we have
(3.15)
Z
−1 2 1
d
2
≤ 2π exp Φ (s)
−
|∇u| ϕ(x)dx .
−µ0 (t)
dt |u|>t
Following the same steps as in Theorem 3.1 and using (3.14), (3.15) becomes
Z
−1 2 s
∗ 0
(3.16) (−u ) (s) ≤ 2π exp Φ (s)
exp B(Φ−1 (r) − Φ−1 (s))
0
+
∗
× [f ∗ (r) + [c−∗
0 (r) − c0∗ (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 call w(x) = w? (x) the solution of the problem (3.12), we obtain
the following equality
(3.17)
Z
∗ 0
(−w ) (s) = 2π exp Φ−1 (s)2
s
exp B(Φ−1 (r) − Φ−1 (s))
0
+
∗
× [f ∗ (r) + [c−∗
0 (r) − c0∗ (r)]w (r)]dr.
∗
∗
To prove the comparison result, we put v(s) = u (s) − w (s). By (3.16) and
(3.17) we have
(3.18)
0
(−v) (s) ≤ 2π exp Φ−1 (s)2
Z
exp B(Φ−1 (r) − Φ−1 (s))
+
× [c−∗
0 (r) − c0∗ (r)]v (r) dr.
+
Let us suppose
c−∗
0 (x) , c0∗(x) 6= 0 and
and s00 = sup s : c−∗
0 (s) > 0 . We write
c+
0∗
exp B(Φ−1 (r) − Φ−1 (s)) c−∗
0 (r) v(r)dr,
s∈
[0, s00 ]
exp B(Φ−1 (r) − Φ−1 (s)) c+
0∗ (r) v(r)dr,
s ∈ ]s01 , γn (Ω)] .
let us put
s01
= inf s :
(s) > 0
Z0 s
V2 (s) =
s01
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s
V1 (s) =
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s
0
Z
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0
We assume initially that c−∗
0 (s) is continuous at s0 , we have to prove, now, that
V1 (s) ≤ 0. Arguing as in [1] we observe that the existence of a solution w (x) =
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w? (x) of (3.12), that implies (3.17), guarantees that the problem:

0
−1 2 −1 2 −1 0
−∗

−
c
(s)
Z
(s)
=
(2π)
exp
Φ
(s)
Z
(s)
+
2π
exp
Φ (s)

0



Rs
× 0 exp [B(Φ−1 (r) − Φ−1 (s))] f ∗ (r) dr,




 Z (0) = Z 0 (s0 ) = 0,
0
has the following positive solution
Z s
∗
Z (s) =
exp B(Φ−1 (r) − Φ−1 (s)) c−∗
0 (r) w (r)dr.
0
This allows us to state (see [5]) that the problem

0
−1 0
 c−∗
ξ (s) + λ (2π) exp Φ−1 (s)2 ξ (s) = 0,
0 (s)
(3.19)

ξ (0) = ξ 0 (s00 ) = 0,
has the first eigenvalue λ1 > 1, and consequently in the following problem

0
−1 0
 c−∗
V1 (s) + 2π exp Φ−1 (s)2 V1 (s) ≥ 0,
0 (s)

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V1 (0) = V10 (s00 ) = 0,
Page 19 of 24
we have
V1 (s) ≤ 0
and
V10 (s) ≤ 0,
s ∈ [0, s00 ] ,
J. Ineq. Pure and Appl. Math. 4(5) Art. 106, 2003
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that is,
u∗ (s) ≤ w∗ (s),
(3.20)
s ∈ [0, s00 ] .
On the other hand, from (3.18) and (3.20), we have

0
−1 2 −1 0
 c+
(s)
V
(s)
−
2π
exp
Φ (s) V2 (s) ≥ 0,
0∗
2

V2 (s01 ) = V20 (γn (Ω)) = 0.
Here we can use the maximum principle to obtain V2 (s) ≤ 0. For Lemma 2.2
−1
0
with h(r) = c+
0∗ (r) , we have for each s ∈ ]s1 , γn (Ω)] ,
Z
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s
(3.21)
s01
exp B(Φ−1 (r) − Φ−1 (s)) u∗ (r)dr
Z s
≤
exp B(Φ−1 (r) − Φ−1 (s)) w∗ (r)dr,
s01
that is,
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u∗ (s01 ) ≤ w∗ (s01 ).
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On the other hand, from (3.16), (3.17), (3.20) and the definition of v (s) we have
Close
−v 0 (s) ≤ 0,
Now since u
∗
(s01 )
≤w
∗
(s01 ),
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s ∈ [s00 , s01 ] .
integrating between s and
s01
Page 20 of 24
we obtain
J. Ineq. Pure and Appl. Math. 4(5) Art. 106, 2003
(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, for s ∈ s01, γn (Ω) , (3.21) becomes
Z s
0≥
exp B(Φ−1 (r) − Φ−1 (s)) [u∗ (r) − w∗ (r)]dr,
0
that is,
Z
0
s
u∗ (r)dr ≤
Z
s
w∗ (r)dr,
s ∈ [s01 , γn (Ω)] .
0
0
Finally we can remove the hypothesis about the continuity of c−∗
0 (s) at s0 proceeding by approximations.
−∗
0
0
If c+
0∗ (x) = 0 or c0 (x) = 0 then s1 = γn (Ω) or s0 = 0 and the result
follows with the obvious modifications.
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References
[1] A. ALVINO, P.L. LIONS AND G. TROMBETTI, Comparison results for
elliptic and parabolic equations via Schwarz symmetrization, Ann. Inst. H.
Poincaré Anal. Non Linéaire, 7 (1990), 37–65.
[2] A. ALVINO AND G. TROMBETTI, Sulle migliori costanti di maggiorazione per una classe di equazioni ellittiche degeneri, Ricerche Mat., 27
(1978), 413–428.
[3] A. ALVINO AND G. TROMBETTI, Equazioni ellittiche con termini di
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[5] C. BANDLE, Isoperimetric Inequalities and applications, Pitman London,
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[8] C. BORELL, The Brunn-Minkowski inequality in the Gauss space, Invent.
Math., 30 (1979), 194–200.
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[9] K.M. CHONG AND N.M. RICE, Equimisurable rearrangements of
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[10] A. EHRHARD, Symmétrisation dans l’space de Gauss, Math. scand., 53
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[11] A. EHRHARD, Inégalités isoperimétriques et intégrales de Dirichlet
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[12] A. EHRHARD, Éléments extremaux pour les inégalitès de BrunnMinkowski gaussennes, Ann. Inst. H. Poincarè Anal. non linèaire,
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[15] M.K.W. MURTY AND G. STAMPACCHIA, Boundary value problems for
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[18] G. TALENTI, Nonlinear elliptic equation, rearrangements of functions and
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