Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 05, pp. 1–17.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
COMPARISON RESULTS FOR ELLIPTIC VARIATIONAL
INEQUALITIES RELATED TO GAUSS MEASURE
YUJUAN TIAN, CHAO MA
Abstract. In this article, we study linear elliptic variational inequalities that
are defined on a possibly unbounded domain and whose ellipticity condition is
given in terms of the density of Gauss 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.
1. Introduction
This article concerns the problem
Z
a(u, ψ − u) ≥
f (ψ − u)ϕ dx,
∀ψ ∈ H01 (ϕ, Ω), ψ ≥ 0,
Ω
u ∈ H01 (ϕ, Ω),
(1.1)
u ≥ 0,
where
Z
a(u, ψ − u) =
n
X
aij Di uDj (ψ − u) dx −
Ω i,j=1
Z X
n
+
bi uDi (ψ − u) dx
Ω i=1
Z
di Di u(ψ − u) dx +
Ω i=1
n
Z X
n
cu(ψ − u) dx,
Ω
|x|2
ϕ(x) = (2π)− 2 e− 2 is the density of Gauss measure, Ω is an open subset of
Rn (n ≥ 2) with Gauss measure less than one, aij , bi , di , c and f are measurable
functions on Ω that satisfy the following assumptions:
(A1) P
aij /ϕ, c/ϕ ∈ L∞ (Ω), f ∈ L2 (ϕ, Ω);
n
2
n
(A2)
i,j=1 aij (x)ξi ξj ≥ ϕ(x)|ξ| , a.e. x ∈ Ω, ∀ξ ∈ R ;
1/2
Pn
2
(A3)
≤ Bϕ(x), a.e. x ∈ Ω, B > 0;
i=1 |bi (x) + di (x)|
Pn
0
∞
(A4)
i=1 Di bi (x) + c(x) ≥ c0 (x)ϕ(x) in D (Ω), c0 ∈ L (Ω);
We obtain a priori estimates for the solutions of (1.1) using rearrangement techniques. As Ω is bounded, the operator is uniformly elliptic. This issue has been
2000 Mathematics Subject Classification. 35J86, 35J70, 35B45.
Key words and phrases. Comparison results; rearrangements; gauss measure;
Elliptic variational inequality.
c
2015
Texas State University - San Marcos.
Submitted November 19, 2014. Published Janaury 5, 2015.
1
2
Y. TIAN, C. MA
EJDE-2015/05
studied by many authors, firstly by Weinberger [28] and Talenti [24]. Actually it
is well known that, one can use Schwarz symmetrization to estimate the solutions
of elliptic and parabolic equations in terms of the solutions of (one dimentional)
radially symmetric problems (for a comprehensive bibliography see the paper [10]
and [27]). For variational inequalities, similar comparison results in terms of linear
elliptic variational inequalities can be found, for example, in [1, 3, 18, 19]; while
nonlinear elliptic variational inequalities are discussed, for example, in [5, 20]. In
the same case, comparison results for parabolic variational inequalities can be found
in [11, 14].
In the elliptic variational inequalities (1.1), since Ω maybe unbounded, the degeneracy of the operator does not allow to use the classical approach via Schwarz
symmetrization. Based on the structure of the problem, it is more appropriate to
use the Gauss symmetrization as has been done for elliptic and parabolic equations
in [7, 8, 12, 13]. Our aim is to compare the solutions of problem (1.1) with the symmetric solutions of a problem in which the data depend only on the first variable
and the domain is a half-space, i.e. the following “symmetrized” problem
Z
a] (v, ψ − v) ≥
f ] ϕ(ψ − v)dx, ∀ψ ∈ H01 (ϕ, Ω] ), ψ ≥ 0,
(1.2)
Ω]
v ∈ H01 (ϕ, Ω] ), v ≥ 0,
where
a] (v, ψ − v) =
Z
Z
ϕD1 vD1 (ψ − v) dx −
Ω]
Z
BϕD1 v(ψ − v) dx +
Ω]
c0] ϕv(ψ − v) dx,
Ω]
where Ω] is a half space with the same Gauss measure as Ω, f ] is the Gauss symmetrization of f and c0] is the decreasing Gauss symmetrization of c0 . To this end,
by following arguments in [1, 19], we first discuss the existence of symmetric solutions to the “symmetrized” problem (1.2), which is a key step for the comparison
results. However, in the equation case, the papers [12, 13] always assume that the
“symmetrized” problem has a symmetric solution instead of studying the existence
conditions for such solutions. Our results (Theorem 3.1) make up for that in large
extent. In addition, as an application of the comparison results, we prove an estimates of the Lorentz-Zygmund norm of u in terms of the norm of the symmetric
solutions v.
The main tools we use are Gauss symmetrization and the properties of the
weighted rearrangement. It is worth noting that the method used in equation
case for obtaining the comparison results can not be applied to the variational
inequalities (1.1). In this paper, we combine the property of the first eigenvalue
(Lemma 4.3) with the maximum principle to overcome the difficulties and get the
desired results.
This article is organized as follows: Section 2 is devoted to give some notation
and preliminary results; in Section 3, the main results of this paper are stated; in
Section 4, we finish the proof of the main results.
2. Notation and preliminary results
In this section, we recall some definitions and results which will be useful in what
follows. First, we recall that the wieghted Sobolev space W01,p (ϕ, Ω) is the closure
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COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
3
of C0∞ (Ω) under the norm
kukW 1,p (ϕ,Ω) =
Z
Z
p
|∇u(x)| ϕ dx +
Ω
1,2
W0 (ϕ, Ω)
p1
|u(x)|p ϕ dx .
Ω
When p = 2, the space
is also denoted by H01 (ϕ, Ω).
Let γn be the n-dimensional normalized Gauss measure on Rn defined as
n
|x|2 dγn = ϕ(x)dx = (2π)− 2 exp −
dx, x ∈ Rn .
2
Set
Φ(τ ) = γn ({x ∈ Rn : x1 > τ })
Z +∞
t2
exp(− )dt,
= (2π)−1/2
2
τ
∀τ ∈ R ∪ {−∞, +∞}.
In [16] we observe that
Φ−1 (t)2
)
2
1 1/2
t(2 log t )
−1/2 exp(−
lim (2π)
t→0+ ,1−
√
t(2 log 1 )1/2
Remark 2.1 ([26]). By limt→0+ t(1−logtt)1/2 = 2
and note that (2.1) and the fact γn (Ω) < 1, we have
Φ−1 (t)2 exp −
≤ αt(1 − log t)1/2 ,
2
Φ−1 (t)2 exp −
≥ βt(1 − log t)1/2 ,
2
where α and β are two positive constants depending
= 1.
and limt→1−
(2.1)
t(2 log 1t )1/2
t(1−log t)1/2
= 0
t ∈ (0, γn (Ω)),
(2.2)
t ∈ (0, γn (Ω)),
(2.3)
on γn (Ω).
Now we give the notion of rearrangement.
Definition 2.2. If u is a measurable function in Ω and µ(t) = γn ({x ∈ Ω : |u| > t})
is the distribution function of u, then we define the decreasing rearrangement of u
with respect to Gauss measure as
u? (s) = inf{t ≥ 0 : µ(t) ≤ s},
]
s ∈ [0, γn (Ω)].
n
Let Ω = {x = (x1 , x2 , . . . , xn ) ∈ R : x1 > λ} be the half-space such that γn (Ω) =
γn (Ω] ). Then
u] (x) = u? (Φ(x1 )), x ∈ Ω]
denote the increasing Gauss symmetrization of u (or Gauss symmetrization of u).
Similarly, the decreasing Gauss symmetrization of u will be
u] (x) = u? (Φ(x1 )),
x ∈ Ω] ,
with
u? (s) = u? (γn (Ω) − s),
s ∈ (0, γn (Ω)).
Properties of rearrangement with respect to Gauss measure or a positive measure
have been widely considered in [9, 22, 23, 25], for instance. Here we just recall the
following:
Hardy-Little inequality:
Z γn (Ω)
Z
Z
?
]
u? (s)v (s)ds =
u] (x)v (x)dγn ≤
|u(x)v(x)|dγn
0
Ω]
Ω
4
Y. TIAN, C. MA
Z
≤
]
EJDE-2015/05
Z
]
γn (Ω)
u (x)v (x)dγn =
Ω]
u? (s)v ? (s)ds,
0
where u and v are measurable functions.
Polya-Szëgo principle: Let u ∈ W01,p (ϕ, Ω) with 1 < p < +∞. Then
k∇u] kLp (ϕ,Ω] ) ≤ k∇ukLp (ϕ,Ω) ,
and equality holds if and only if Ω = Ω] and |u| = u] modulo a rotation.
Now we recall the definition and main properties of the Lorentz-Zygmund space
(see [6]).
Definition 2.3. For any measurable function u, 0 < q, p ≤ +∞ and −∞ < α <
+∞, set
 R
 γn (Ω) t p1 (1 − log t)α u? (t)q dt 1/q if 0 < q < +∞,
t
0
kukLp,q (log L)α (ϕ,Ω) =
1
sup
α ?
p
if q = +∞.
t∈(0,γn (Ω)) t (1 − log t) u (t)
(2.4)
We say that u belongs to the Lorentz-Zygmund space Lp,q (log L)α (ϕ, Ω) if
kukLp,q (log L)α (ϕ,Ω) < +∞.
Remark 2.4. It is clear that the space Lp,q (log L)0 (ϕ, Ω) is just the Lorentz space
Lp,q (ϕ, Ω). As p = q, the space Lp,p (log L)0 (ϕ, Ω) is the Lebesgue space Lp (ϕ, Ω).
Remark 2.5. For 1 < p ≤ +∞, 1 ≤ q ≤ +∞ and −∞ < α < +∞, (2.4) is a
quasinorm. Replacing u? (t) with
Z
1 t ?
u?? (t) =
u (s)ds,
t 0
we obtain an equivalent norm.
Remark 2.6. If 0 < r < p ≤ +∞, 0 < q, s ≤ +∞ and −∞ < α, β < +∞, then
Lp,q (log L)α (ϕ, Ω) ⊆ Lr,s (log L)β (ϕ, Ω)
and when the first exponents are the same,
Lp,q (log L)α (ϕ, Ω) ⊆ Lp,s (log L)β (ϕ, Ω),
whenever either
q ≤ s and α ≥ β
or
q > s and α +
1
1
>β+ .
q
s
Remark 2.7. The space Lp,q (log L)α (ϕ, Ω) is nontrivial if and only if one of the
following conditions holds
p < +∞,
1
< 0,
q
p = +∞, q = +∞ and α = 0.
p = +∞ and α +
The following imbedding theorem in Lorentz-Zygmund space is a straight consequence of the Sobolev logarithmic inequalities. It has been proved in [13] by using
the properties of rearrangement.
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COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
5
Proposition 2.8. Let Ω be an open subset of Rn with γn (Ω) < 1. If f ∈ W01,p (ϕ, Ω)
with 1 ≤ p < +∞, then f ∈ Lp (log L)1/2 (ϕ, Ω) and
kf kLp (log L)1/2 (ϕ,Ω) ≤ Ck∇f kLp (ϕ,Ω) .
The following Hardy inequalities are also needed in this article [6].
Proposition 2.9. Suppose that r > 0, 1 ≤ q ≤ +∞ and −∞ < α < +∞. Let Ψ
be a nonegative measurable function on (0, 1). If 1 ≤ q < +∞, then
Z t
q dt 1/q
Z 1
Z 1
q dt 1/q
Ψ(s)ds
t−r (1 − log t)α
t1−r (1 − log t)α Ψ(t)
≤C
t
t
0
0
0
(2.5)
holds. Moreover if q = +∞, then
Z t
sup t−r (1 − log t)α
Ψ(s)ds ≤ C sup t1−r (1 − log t)α Ψ(t) ,
(2.6)
0<t<1
0
0<t<1
where the positive constant C = C(r, q, α) is independent of Ψ.
3. Statement of main results
The main results of this article are the following three theorems. First, the
existence result of symmetric solutions to the “symmetrized” variational problem
are presented, which is a key step for the comparison results. Let
c+
0 (x) = max{c0 (x), 0},
c−
0 (x) = max{−c0 (x), 0},
−]
+
−
]
c+
0] (x) = (c0 (x))] , c0 (x) = (c0 (x)) .
Theorem 3.1. Set
A] v = −D1 (ϕD1 v) − BϕD1 v + c0] ϕv
in Ω] .
Suppose that: (1) c0 (x) ≥ 0, or (2) c−
0 (x) 6≡ 0. Let one of the following equivalent
conditions be satisfied:
(a) The first eigenvalue λ1 of the problem
B|x |
1
+ c−]
ϕΨ in Ω] ,
A] Ψ = λ 1
0
2
(3.1)
Ψ ∈ H01 (ϕ, Ω] )
is positive.
(b) There exists a constant ξ > 0 such that
Z
Z
Z
ϕ|D1 Z|2 dx −
BϕZD1 Zdx +
c0] ϕZ 2 dx
Ω]
Ω]
Ω]
Z
≥ξ
ϕ|D1 Z|2 dx, ∀Z ∈ H01 (ϕ, Ω] ).
(3.2)
Ω]
Then the maximum principle holds for A] ; i.e.,
A] v ≥ 0 in Ω] and v ≥ 0 on ∂Ω] imply v ≥ 0 in Ω] ,
(3.3)
and problem (1.2) has unique symmetric solution.
Now, the comparison result between problems (1.1) and (1.2) can be stated in
the following theorem.
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Y. TIAN, C. MA
EJDE-2015/05
Theorem 3.2. Let (A1)–(A4) and one of the conditions in Theorem 3.1 hold.
Suppose that u is the solution of problem (1.1) and v = v ] is the solution of problem
(1.2). Then
(1) As c0 (x) ≤ 0, we have
u? (s) ≤ v ? (s),
(2) As
c+
0 (x)
Z
s
s ∈ [0, γn (Ω)].
(3.4)
6≡ 0, it follows that
u? (s) ≤ v ? (s), s ∈ [0, s1 ],
Z s
−1
−1
eBΦ (σ) v ? (σ)dσ,
eBΦ (σ) u? (σ)dσ ≤
(3.5)
s ∈ [s1 , γn (Ω)],
(3.6)
s1
s1
where s1 = inf{s ∈ [0, γn (Ω)] : c0? (s) > 0}.
Using the above comparison results, it is possible to obtain estimates of the
solutions to problem (1.1) in terms of the solutions to “symmetrized” problem
(1.2).
Theorem 3.3. Suppose that Lorentz-Zygmund spaces Lp,q (log L)α (ϕ, Ω) are nontrivial. Under the same assumptions of Theorem 3.2, we have
(1) If c0 (x) ≤ 0, then
kukLp,q (log L)α (ϕ,Ω) ≤ kvkLp,q (log L)α (ϕ,Ω] )
(3.7)
with 0 < p, q ≤ +∞ and −∞ < α < +∞.
(2) If c0 (x) > 0, then for all 0 < ε < p1 ,
B2
−1
kukLp,q (log L)α (ϕ,Ω) ≤ Ce[ 2ε −BΦ
(γn (Ω))]
kvk
p
L 1−pε
,q
ε
(log L)α− 2 (ϕ,Ω] )
,
(3.8)
where C is a positive constant depending on p, q and γn (Ω), 1 < p < +∞, 1 ≤ q ≤
+∞ and −∞ < α < +∞.
(3) Otherwise,
−1
kukLp,q (log L)α (ϕ,Ω) ≤ e[BΦ
(s1 )−BΦ−1 (γn (Ω))]
kvkLp,q (log L)α (ϕ,Ω] ) ,
(3.9)
with s1 defined in Theorem 3.2, 1 < p ≤ +∞, 1 ≤ q ≤ +∞ and −∞ < α < +∞.
Remark 3.4. As Ω is bounded, by Schwarz symmetrization, it is possible to estimate Lp,q (log L)α norm of u by the same norm of v, the solutions to the “symmetrized” problems defined on balls with coefficients depending only on the radial
(see Corollary 4.1 in [14]). However, in our case that Ω maybe unbounded, it is
impossible to get the same result as before. In the above theorem, we estimate the
Lp,q (log L)α norm of u in terms of a little stronger Lorentz-Zygmund norm of v.
4. Proof of main results
Proof of Theorem 3.1. First, we can see that if |x1 | ∈ L∞ (log L)−1/2 (ϕ, Ω] ), every
term in the weak form of (3.1) makes sense. In fact, for any Y ∈ H01 (ϕ, Ω] ), using
Hölder inequality and Hardy-Littlewood inequality, we obtain
Z
Z
1/2 Z
1/2
|x1 |ΨY ϕ dx ≤
(|x1 |Ψ)2 ϕ dx
Y 2 ϕ dx
Ω]
Ω]
γn (Ω)
≤
Z
0
Ω]
1/2
|x1 |?2 (s)|Ψ|?2 (s)ds
kY kL2 (ϕ,Ω] )
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COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
≤
|x1 |?2 (s)(1 − log s)−1
sup
7
1/2
0<s<γn (Ω)
×
hZ
γn (Ω)
|Ψ|? (s)(1 − log s)1/2
2
i1/2
ds
kY kL2 (ϕ,Ω] )
0
≤ k|x1 |kL∞ (log L)−1/2 (ϕ,Ω] ) k|Ψ|kL2 (log L)1/2 (ϕ,Ω] ) kY kL2 (ϕ,Ω] ) .
R
By Proposition 2.8, |Ψ| ∈ L2 (log L)1/2 (ϕ, Ω] ). Thus, Ω] |x1 |ΨY ϕ dx is finite. The
remainder terms are similarly considered. Hence, it remains us to prove that |x1 | ∈
L∞ (log L)−1/2 (ϕ, Ω] ).
If λ ≥ 0, by Ω] = {x = (x1 , x2 , . . . , xn ) ∈ Rn : x1 > λ}, we have x1 > λ ≥ 0.
Therefore,
|x1 |? (s) = x?1 (s) = Φ−1 (s).
(4.1)
Moreover, from Remark 2.1 it follows that
√
Φ−1 (s)2
Φ−1 (s)
− 2πe 2
lim
= lim+ 1
−1/2 (− 1 )
s→0+ (1 − log s)1/2
s→0
2 (1 − log s)
s
√ s(1 − log s)1/2
= lim 2 2π
Φ−1 (s)2
s→0+
e− 2
√ s(1 − log s)1/2 s(2 log 1s )1/2
= lim 2 2π
Φ−1 (s)2
s→0+
s(2 log 1s )1/2
e− 2
√
√
1
1
= 2 2π √ √
= 2.
2 2π
Then there exist M1 > 0 and δ0 ∈ (0, γn (Ω)) such that
Φ−1 (s)(1 − log s)−1/2 ≤ M1 ,
s ∈ (0, δ0 ).
Since Φ−1 (s)(1 − log s)−1/2 is continuous on [δ0 , γn (Ω)], there exists a constant
M2 > 0 such that
Φ−1 (s)(1 − log s)−1/2 ≤ M2 ,
s ∈ [δ0 , γn (Ω)).
Thus,
Φ−1 (s)(1 − log s)−1/2 ≤ max{M1 , M2 },
s ∈ (0, γn (Ω)).
Recalling (4.1), we have
k|x1 |kL∞ (log L)−1/2 (ϕ,Ω] ) =
sup
−1
Φ (s)(1 − log s)−1/2 ≤ max{M1 , M2 }.
0<s<γn (Ω)
If λ < 0, setting
µ(t) = γn ({x ∈ Ω] : |x1 | > t}),
ν(t) = γn ({x ∈ Ω] : x1 > t}),
we have


ν(t)
µ(t) = 2ν(t) − (1 − γn (Ω))


γn (Ω)
and
|x1 |? (s) = inf{t ≥ 0 : µ(t) ≤ s}
t ≥ −λ,
0 ≤ t < −λ,
t<0
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Y. TIAN, C. MA
EJDE-2015/05
(
inf{t ≥ 0 : ν(t) ≤ s}
s ∈ [0, 1 − γn (Ω)],
=
s+1−γn (Ω)
inf{t ≥ 0 : ν(t) ≤
} s ∈ (1 − γn (Ω), γn (Ω)]
2
(
x?1 (s)
s ∈ [0, 1 − γn (Ω)],
=
n (Ω)
x?1 s+1−γ
s
∈ 1 − γn (Ω), γn (Ω)
2
(
Φ−1 (s)
s ∈ [0, 1 − γn (Ω)],
=
−1 s+1−γn (Ω)
Φ
s ∈ 1 − γn (Ω), γn (Ω) .
2
Using the same method as in the case λ > 0, we obtain that there exists M > 0
such that
|x1 |? (s)(1 − log s)−1/2 ≤ M,
which implies |x1 | ∈ L∞ (log L)−1/2 (ϕ, Ω] ).
Now we give the proof of the equivalence of (a) and (b).
(a)⇒ (b) The first eigenvalue of (3.1) can be characterized by the Rayleigh
principle as
R
R
R
ϕ|D1 Q|2 dx − B Ω] ϕQD1 Q dx + Ω] c0] ϕQ2 dx
Ω]
λ1 =
min
. (4.2)
R B|x1 |
Q∈H01 (ϕ,Ω] ),Q6=0
2
+ c−]
0 (x) ϕQ dx
2
Ω]
−]
In view of (4.2) and the fact that c0] = c+
0] − c0 , integrating by parts, we have
Z
Z
Z
2
ϕ|D1 Z| dx − B
ϕZD1 Z dx +
c0] ϕZ 2 dx
Ω]
Ω]
Ω]
Z Z
B|x1 |
−]
2
≥ (1 − ξ)λ1
+ c0 ϕZ dx + ξ
ϕ|D1 Z|2 dx
2
Ω]
Ω]
Z
Z
−]
2
− ξB
ϕZD1 Z dx + ξ
(c+
0] − c0 )ϕZ
Ω]
Ω]
Z
Z
2
B
B|x1 |
+ c−]
x1 ϕZ 2 dx
ϕZ
dx
−
ξ
= (1 − ξ)λ1
0
2
Ω] 2
Ω]
Z
Z
−]
+
2
+ξ
(c0] − c0 )ϕZ dx + ξ
ϕ|D1 Z|2 dx
Ω]
Ω]
Z
Z
B|x1 |
−] 2
≥ ((1 − ξ)λ1 − ξ)
+ c0 ϕZ dx + ξ
ϕ|D1 Z|2 dx,
2
]
]
Ω
Ω
1
for all Z ∈ H01 (ϕ, Ω] ). Then we can choose 0 < ξ < λ1λ+1
such that (3.2) holds.
(b) ⇒ (a) Assume that Ψ is the eigenfunction corresponding to λ1 . Then
R
R
R
|D1 Ψ|2 ϕ dx − B Ω] ϕΨD1 Ψ dx + Ω] c0] Ψ2 ϕ dx
Ω]
.
λ1 =
R
B|x1 |
2
+ c−]
0 (x) Ψ ϕ dx
2
Ω]
Recalling (b) and Proposition 2.8 and noting that Ψ 6≡ 0, we have
R
R
Cξ Ω] |Ψ|2 ϕ dx
ξ Ω] |D1 Ψ|2 ϕ dx
λ1 ≥ R
≥R
> 0,
B|x1 |
B|x1 |
Ψ2 ϕ dx
Ψ2 ϕ dx
+ c−]
+ c−]
0
0
2
2
Ω]
Ω]
where C is a positive constant depending on γn (Ω). Thus (a) holds.
Now, we prove that the maximum principle holds for A] . As c0 (x) ≥ 0, it is
−
obvious. As c−
as a test function of (3.3) and using (b), we
0 (x) 6≡ 0, taking v
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COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
obtain
Z
0≤ξ
Z
− 2
v − A] v − dx ≤ 0,
ϕ|D1 v | dx ≤
Ω]
9
Ω]
which implies v − = 0. Thus v ≥ 0 on ∂Ω] .
On the other hand, from [17] we see that (1.2) has unique solution v. Thus
the solution depends only on the first variable. To prove v = v ] , for the sake of
simplicity, we suppose that v is sufficiently smooth. Set E = {x ∈ Ω] : v > 0}.
Then v satisfies
A] v = f ] on E,
which gives
− ϕD11 v + (x1 − B)ϕD1 v + c0] ϕv = f ] ϕ on E.
(4.3)
Differentiating (4.3) with respect to x1 and taking K = D1 v, we obtain
(A] + ϕ)K = −ϕD11 K + (x1 − B)ϕD1 K + c0] ϕK + ϕK
= ϕD1 f ] − vϕD1 c0] ≥ 0
on E.
Observing E = {x ∈ Ω : x1 > ξ0 }, where ξ0 = Φ−1 (γn ({x ∈ Ω] : v > 0})), it
follows v(ξ0 ) = 0. Thus K ≥ 0 on ∂E.
Since A] +ϕ satisfies the property (b), we obtain K = D1 v ≥ 0 on E by applying
the maximum principle to the operator A] + ϕ. Thus v = v ] on Ω] . The proof is
complete.
]
Before proving Theorem 3.2, we need the following lemmas.
Lemma 4.1. Assume that (A1)–(A4) hold. Let u be the solution to problem (1.1)
and v = v ] be the solution to problem (1.2). Then
Z s
0
−1
2
−1
−1
− u? (s) ≤ 2πe[Φ (s) −BΦ (s)]
eBΦ (σ) [f ? (σ) − c0? (σ)u? (σ)]dσ,
(4.4)
0
for s ∈ [0, γn ({u > 0})], and
?0
− v (s) = 2πe
[Φ−1 (s)2 −BΦ−1 (s)]
Z
s
−1
eBΦ
(σ)
[f ? (σ) − c0? (σ)v ? (σ)]dσ,
(4.5)
0
for s ∈ [0, γn ({v > 0})].
The proof of the above lemma follows the same lines as in [12, 13]; we omit it.
Lemma 4.2. Assume that (A1)–(A4) hold. Let u and v = v ] be the solutions to
problem (1.1) and (1.2) respectively. If c0? is continuous at s1 , then w = u? − v ?
satisfies
Z s
−1
2
−1
−1
− w0 (s) ≤ −2πe[Φ (s) −BΦ (s)]
eBΦ (σ) c0? (σ)w(σ)dσ,
(4.6)
0
for s ∈ [0, γn ({u > 0})].
Proof. As γn ({u > 0}) ≤ γn ({v > 0}), it is obvious that (4.6) holds. As γn ({u >
0}) > γn ({v > 0}), (4.6) holds on [0, γn ({v > 0})]. Moreover, using the regularity theory (see [15]), v belongs to H 2 (ϕ, Ω] ) and then v ? ∈ C 1 (0, γn (Ω)]. Thus
0
v ? (γn ({v > 0})) = 0. It follows from (4.5) that
Z γn ({v>0})
−1
eBΦ (σ) [f ? (σ) − c0? (σ)v ? (σ)]dσ = 0.
(4.7)
0
10
Y. TIAN, C. MA
EJDE-2015/05
Noting that w(s) = u? (s) on [γn ({v > 0}), γn ({u > 0})], we combine (4.7) and
(4.4) to discover that
nZ s
−1
0
[Φ−1 (s)2 −BΦ−1 (s)]
eBΦ (σ) [f ? (σ) − c0? (σ)w(σ)]dσ
−w (s) ≤ 2πe
0
Z
γn ({v>0})
o
c0? (σ)v ? (σ)dσ
0
n Z s
−1
−1
2
−1
eBΦ (σ) c0? (σ)w(σ)dσ
= 2πe[Φ (s) −BΦ (s)] −
0
Z s
o
−1
+
eBΦ (σ) f ? (σ)dσ , s ∈ [γn ({v > 0}), γn ({u > 0})].
−1
eBΦ
−
(σ)
γn ({v>0})
If we show
f ? (s) < 0
on [γn ({v > 0}), γn (Ω)],
(4.8)
then (4.6) is proved.
Now it remains to prove (4.8). In fact, as γn ({v > 0}) ≤ s1 , (4.7) yields
Z γn ({v>0})
Z γn ({v>0})
−1
BΦ−1 (σ) ?
e
f (σ) =
eBΦ (σ) c0? (σ)v ? (σ)dσ ≤ 0.
0
0
Therefore f ? can not be nonnegative on [0, γn ({v > 0})].
Rs
−1
As γn ({v > 0}) > s1 , taking V (s) = s1 eBΦ (σ) c0? (σ)v ? (σ)dσ, from (4.5) we
obtain
0
−1
−1
2
−1
− e−BΦ (s) (c0? (s))−1 V 0 (s) + 2πe[Φ (s) −BΦ (s)] V
Z s1
hZ s
i
−1
2
−1
−1
−1
= 2πe[Φ (s) −BΦ (s)]
eBΦ (σ) f ? (σ)dσ −
eBΦ (σ) c0? (σ)v ? (σ)dσ
0
Z s0
−1
2
−1
−1
≥ 2πe[Φ (s) −BΦ (s)]
eBΦ (σ) f ? (σ)dσ, s ∈ (s1 , γn ({v > 0})).
0
?
If f ≥ 0 on [0, γn ({v > 0})], observing c0? (s1 ) = 0, V satisfies
0
−1
−1
2
−1
− e−BΦ (s) (c0? (s))−1 V 0 (s) + 2πe[Φ (s) −BΦ (s)] V ≥ 0 on (s1 , γn ({v > 0})),
V (s1 ) = 0,
V 0 (s1 ) = 0.
(4.9)
By the maximum principle (see [21]), we obtain V ≤ 0 in (s1 , γn ({v > 0})) which
contradict with the fact that V > 0 in (s1 , γn ({v > 0})). Thus f ? can not be
nonnegative in [0, γn ({v > 0})] and we obtain the desired result.
Lemma 4.3 (P.163 in [4]). Let
n
X
L=−
Di (aij (x)Dj ) + c(x).
i,j=1
Consider the eigenvalue problem
LΨ = λ1 P Ψ
Ψ=0
in D,
on Γ0 ,
∂Ψ
+ ηΨ = 0
∂ν
on Γ1 ,
(4.10)
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COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
11
where D ⊆ Rn , Γ0 is a subset of ∂D, Γ1 = ∂D − Γ0 and P is a positive function
in D. If λ1 is the first eigenvalue of (4.10), then
λ1 ≤ sup
x∈D
Lh
,
Ph
where h is any positive function in D satisfying the same boundary conditions as
Ψ.
Lh
,
λ1 ≥ inf
x∈D P h
where h is positive in D, h ≥ 0 on Γ0 and
∂h
∂ν
+ ηh ≥ 0 on Γ1 .
Lemma 4.4 ([2]). Let f, g be measurable positive functions such that
Z r
Z r
f (σ)dσ ≤
g(σ)dσ, r ∈ [0, ρ].
0
0
If h ≥ 0 is a decreasing function in [0, ρ], then
Z r
Z r
f (σ)h(σ)dσ ≤
g(σ)h(σ)dσ,
0
r ∈ [0, ρ].
0
Proof of Theorem 3.2. Suppose that c0 (x) is smooth in Ω.
(1) As c (x) ≤ 0, we assume c0 (x) < 0. In this case, c−?
0 = −c0? . Set W (s) =
R s BΦ−1 (σ)0 −?
e
c
(σ)w(σ)dσ.
By
(4.6),
we
have
0
0
−BΦ−1 (s) −?
0
−1
2
−1
− e
(c0 (s))−1 W 0 ≤ 2πe[Φ (s) −BΦ (s)] W in (0, γn ({u > 0})),
W (0) = 0,
W 0 (γn ({u > 0})) ≤ 0.
(4.11)
Thus, w ≤ 0 on [0, γn ({u > 0})]. In fact, let E be a half space whose Gauss measure
is γn ({u > 0}). Consider the eigenvalue problem
e −] ϕΨ
A] Ψ = λc
0
Ψ∈
in E,
H01 (ϕ, E).
The first eigenvalue of (4.12) can be characterized as
R
R
R
ϕ|D1 Q|2 dx − B E ϕQD1 Q dx + E c0] ϕQ2 dx
E
e=
.
λ
min
R −]
Q∈H01 (ϕ,E),Q6=0
c ϕQ2 dx
E 0
(4.12)
(4.13)
By (b) of Theorem 3.1, we have
Z
Z
Z
ϕ|D1 Q|2 dx − B
ϕQD1 Q dx +
c0] ϕQ2 dx
E
E
E
Z
≥β
ϕ|D1 Q|2 dx > 0, Q ∈ H01 (ϕ, E) and Q 6≡ 0 in E
E
e > 0.
Hence, λ
e then both Ψ and |Ψ| minimize
Now if Ψ is an eigenfunction corresponding to λ,
(4.13). |Ψ| is also an eigenfunction. Then we can take Ψ ≥ 0. Moreover, Ψ satisfies
R
R
R
ϕ|D1 Ψ|2 dx − B E ϕΨD1 Ψ dx + E c0] ϕΨ2 dx
E
e
λ=
.
(4.14)
R −]
c ϕΨ2 dx
E 0
12
Y. TIAN, C. MA
EJDE-2015/05
Integrating by parts and using Hardy-Littlewood inequality and Polya-Szëgo principle, we obtain
R
R
R
|D1 Ψ|2 ϕ dx − B E x21 Ψ2 ϕ dx + E c0] Ψ2 ϕ dx
E
e
λ=
R −]
c Ψ2 ϕ dx
E 0
R
R
R
(4.15)
|D1 Ψ] |2 ϕ dx − B E ( x21 )] Ψ]2 ϕ dx + E c0] Ψ]2 ϕ dx
E
≥
.
R −]
c Ψ]2 ϕ dx
E 0
Noting that x]1 = x1 , we conclude from (4.15) that
R
R
R
|D1 Ψ] |2 ϕ dx − B E x21 Ψ]2 ϕ dx + E c0] Ψ]2 ϕ dx
E
e
λ≥
.
R −]
c Ψ]2 ϕ dx
E 0
Thus, the above inequality, (4.13) and (4.14) imply Ψ = Ψ] . In addition,
Z s
0
−1
?
e + 1)e[Φ−1 (s)2 −BΦ−1 (s)]
− Ψ? (s) = 2π(λ
eBΦ (σ) c−?
0 (σ)Ψ (σ)dσ,
(4.16)
0
Rs
−1
?
for s ∈ [0, γn ({u > 0})]. Let Θ(s) = 0 eBΦ (σ) c−?
0 (σ)Ψ (σ)dσ. Then (4.16) can
be written as
−1
−1 0 0
e + 1)e[Φ−1 (s)2 −BΦ−1 (s)] Θ in (0, γn ({u > 0})),
− e−BΦ (s) (c−?
Θ = 2π(λ
0 (s))
Θ(0) = 0,
Θ0 (γn ({u > 0})) = 0.
(4.17)
If λ is the first eigenvalue of the problem
−1
2
−1
−1
−1 0 0
− e−BΦ (s) (c−?
U = λ2πe[Φ (s) −BΦ (s)] U
0 (s))
U (0) = 0,
in (0, γn ({u > 0})),
0
U (γn ({u > 0})) = 0,
(4.18)
it follows from Lemma 4.3 that
−1
λ≥
−1 0 0
−[e−BΦ (s) (c−?
Θ]
0 (s))
e + 1.
=λ
−1
2
−1
[Φ
(s)
−BΦ
(s)]
s∈(0,γn ({u>0}))
2πe
Θ
inf
e + 1 > 1. By [4, Lemma 4.7], we have W 0 ≤ 0 on [0, γn ({u > 0})].
Therefore, λ ≥ λ
That is w ≤ 0 on [0, γn ({u > 0})]. The result in the case c(x) ≤ 0 can be proved
by approximation techniques (see [2]).
(2) As c+
0 (x) 6≡ 0, we let
s2 = inf{s ∈ [0, γn (Ω)] : c0? (s) ≥ 0}.
Then s2 ≤ s1 and c0? (s2 ) = 0. Moreover, noting that c0? (s) < 0 on (0, s2 ), we
obtain c−?
0 (s) = −c0? (s) on (0, s2 ). Take
Z s
−1
W1 (s) =
eBΦ (σ) c−?
0 (σ)w(σ)dσ.
0
By (4.6), we obtain
0
−1
−1
2
−1
−1
− e−BΦ (s) (c−?
W10 (s) ≤ 2πe[Φ (s) −BΦ (s)] W1 (s)
0 (s))
W1 (0) = 0,
W10 (s2 )
in (0, s2 ),
= 0.
Proceeding as in case (1), it follows that
u? (s) ≤ v ? (s),
s ∈ [0, s2 ].
(4.19)
EJDE-2015/05
COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
13
On the other hand, letting
Z
s
W2 (s) =
−1
eBΦ
(σ)
c0? (σ)w(σ)dσ,
s1
Inequalities (4.6) and (4.19) yield
0
−1
−1
2
−1
− e−BΦ (s) (c0? (s))−1 W20 (s) ≤ −2πe[Φ (s) −BΦ (s)] W2 (s)
in (s1 , γn ({u > 0})],
W2 (s1 ) = 0,
W20 (γn ({u > 0})) ≤ 0.
By the maximum principle, we have W2 (s) ≤ 0 on [s1 , γn ({u > 0})]. That is,
Z s
Z s
−1
BΦ−1 (σ)
?
eBΦ (σ) c0? (σ)v ? (σ)dσ, s ∈ [s1 , γn ({u > 0})].
e
c0? (σ)u (σ)dσ ≤
s1
s1
According to Lemma 4.4, we obtain
Z s
Z s
−1
−1
eBΦ (σ) u? (σ)dσ ≤
eBΦ (σ) v ? (σ)dσ,
s1
s ∈ [s1 , γn ({u > 0})],
s1
which implies u? (s1 ) ≤ v ? (s1 ). Finally applying (4.19) to (4.6), it follows
0
0
− u? (s) ≤ −v ? (s),
s ∈ [s2 , s1 ]
(4.20)
Integrating (4.20) from s to s1 , we obtain
u? (s) ≤ v ? (s)
in [s2 , s1 ],
which completes the proof.
At last, we can remove the smooth assumption on c0 (x) by approximations.
Remark 4.5. For the variational problem, since f ? maybe negative in [0, γn ({u >
0})], the method in the equation case are failed to obtain (3.4)–(3.6). Here, we use
the properties of the first eigenvalue (Lemma 4.3) and maximal principle to obtain
the desired results.
Proof of Theorem 3.3. (1) If c0 (x) ≤ 0, inequality (3.7) follows from (3.4).
(2) If c0 (x) > 0, we have
Z s
Z s
−1
BΦ−1 (σ) ?
e
u (σ)dσ ≤
eBΦ (σ) v ? (σ)dσ, s ∈ [0, γn (Ω)],
(4.21)
0
0
Set
kuk?Lp,q (log L)α (ϕ,Ω)
( R
=
q 1/q
1
t p (1 − log t)α u?? (t) dt
t 1
supt∈(0,γn (Ω)) t p (1 − log t)α u?? (t)
γn (Ω)
0
if 0 < q < +∞,
if q = +∞.
Remark 2.5 implies that the quasinorm k·kLp,q (log L)α (ϕ,Ω) is equivalent to the norm
k · k?Lp,q (log L)α (ϕ,Ω) when p > 1 and q ≥ 1.
As q < +∞, using (4.21), (2.3) and (2.5), we obtain
kuk?q
Lp,q (log L)α (ϕ,Ω)
Z γn (Ω) Z
q dt
1
1 t ?
=
t p (1 − log t)α
u (σ)dσ
t 0
t
0
q
Z γn (Ω) Z t
1
dt
α −BΦ−1 (t) 1
BΦ−1 (σ) ?
p
≤
t (1 − log t) e
e
u (σ)dσ
t
t
0
0
14
Y. TIAN, C. MA
Z
γn (Ω)
≤
α −BΦ−1 (t) 1
1
p
t (1 − log t) e
γn (Ω)
=
0
×
≤e
1
t
t
Z
e
−1
1
t p (1 − log t)α e−BΦ
−1
eBΦ
(σ) ?
v (σ)dσ
2
−1
(σ)− √B2ε )2 + B
2Φ
2ε
−BqΦ−1 (γn (Ω)]
γn (Ω)
Z
]
εΦ−1 (σ)2
2
e
v ? (σ)dσ
1
t p (1 − log t)α
0
Z
γn (Ω)
t
0
q dt
t
0
2
[q B
2ε
q dt
(t)
√ε
[−(
t
Z
t
0
Z
EJDE-2015/05
1
t
t
Z
e
εΦ−1 (σ)2
2
v ? (σ)dσ
q dt
t
0
Z
t
q dt
1
?
ε v (σ)dσ
ε
t
0
0 σ (1 − log σ) 2
Z
γ
(Ω)
n
q dt
−1
1
B2
ε
t p −ε (1 − log t)α− 2 v ? (t)
≤ Ce[q 2ε −BqΦ (γn (Ω)]
t
0
B2
−1
(γn (Ω)]
B2
−1
(γn (Ω)]
≤ Ce[q 2ε −BqΦ
= Ce[q 2ε −BqΦ
kvkq
p
1
t p (1 − log t)α
,q
L 1−pε
1
t
ε
(log L)α− 2 (ϕ,Ω] )
,
where C is a positive constant depending on p, q and γn (Ω).
As q = ∞, (3.8) can be obtained by the same method as before with (2.5)
replaced by (2.6).
(3) It follows from Theorem 3.2 that
Z
u? (s) ≤ v ? (s), s ∈ [0, s1 ],
Z s
s
−1
−1
eBΦ (σ) u? (σ)dσ ≤
eBΦ (σ) v ? (σ)dσ,
s1
s ∈ [s1 , γn (Ω)],
s1
where 0 < s1 < γn (Ω).
If q < +∞,
kuk?q
Lp,q (log L)α (ϕ,Ω)
γn (Ω)
Z
=
1
p
t (1 − log t)
α1
t
0
s1
Z
=
1
t p (1 − log t)α
0
Z
γn (Ω)
+
s1
s1 Z
≤
1
γn (Ω)
+
s1
u? (σ)dσ
0
1
1
t
Z
q dt
t
0
t (1 − log t)α
t p (1 − log t)α
u? (σ)dσ
t
1
p
0
Z
1
t
Z
t
Z
1
t
Z
u? (σ)dσ
0
v ? (σ)dσ
0
1
t
t
t
t
t p (1 − log t)α
q dt
Z
0
q dt
t
q dt
t
t
u? (σ)dσ
q dt
t
Denote by I2 the second term on the right-hand side of the above inequality. Then
Z γn (Ω) h
Z
Z t
iq dt
1
1 s1 ?
I2 =
t p (1 − log t)α
u (σ)dσ +
u? (σ)dσ
t
t
s1
0
s1
Z γn (Ω) h
Z s1
Z t
iq dt
−1
−1
1
1
≤
t p (1 − log t)α
v ? (σ)dσ + e−BΦ (t)
eBΦ (σ) v ? (σ)dσ
t
t
s1
0
s1
Z γn (Ω) h
Z s1
Z t
i
q
−1
−1
1
1
dt
≤
t p (1 − log t)α
v ? (σ)dσ + e[BΦ (s1 )−BΦ (t)]
v ? (σ)dσ
t
t
s1
0
s1
EJDE-2015/05
≤e
COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
[qBΦ−1 (s1 )−qBΦ−1 (γn (Ω))]
Z
γn (Ω)
1
p
t (1 − log t)
α1
t
s1
Z
t
v ? (σ)dσ
q dt
t
0
15
.
Hence,
−1
[qBΦ
kuk?q
Lp,q (log L)α (ϕ,Ω) ≤ e
(s1 )−qBΦ−1 (γn (Ω))]
kvk?q
Lp,q (log L)α (ϕ,Ω) .
If q = +∞,
kuk?Lp,∞ (log L)α (ϕ,Ω)
1
t p (1 − log t)α u?? (t)
=
sup
t∈(0,γn (Ω))
= max
n
≤ max
n
1
sup [t p (1 − log t)α u?? (t)],
t∈(0,s1 )
o
1
[t p (1 − log t)α u?? (t)]
sup
t∈(s1 ,γn (Ω))
1
p
sup [t (1 − log t)α v ?? (t)],
sup
1
[t p (1 − log t)α
t∈(s1 ,γn (Ω))
t∈(0,s1 )
1
t
Z
t
o
u? (σ)dσ] .
0
By the same method as the case q < +∞, we have
Z t
h 1
i
α1
p
sup
t (1 − log t)
u? (σ)dσ
t 0
t∈(s1 ,γn (Ω))
Z t
h 1
i
[BΦ−1 (s1 )−BΦ−1 (γn (Ω))]
α1
p
t (1 − log t)
≤e
sup
v ? (σ)dσ .
t 0
t∈(s1 ,γn (Ω))
Thus,
−1
kuk?Lp,∞ (log L)α (ϕ,Ω) ≤ e[BΦ
(s1 )−BΦ−1 (γn (Ω))]
kvk?Lp,∞ (log L)α (ϕ,Ω) .
Conclusions. This article studies a class of linear elliptic variational inequalities
which are defined on a possibly unbounded domain and whose ellipticity condition is
given in terms of the density of Gauss measure. Using the notion of rearrangement
with respect to the Gauss measure, we prove a comparison result with the symmetric
solutions of a “symmetrized” problem in which the data depend only on the first
variable and the domain is a half-space. To this end, we first discuss the existence of
symmetric solutions to the “symmetrized” problem, which make up for the previous
results. In addition, as an application of the comparison results, we prove an
estimates of the Lorentz-Zygmund norm of u in terms of the norm of the symmetric
solutions v.
Acknowledgements. This research was supported by the Tianyuan Special Funds
of the National Natural Science Foundation of China (Grant no. 11426146), by the
Promotive Research Fund for Excellent Young and Middle-Aged Scientists of Shandong Province (Grant no. BS2012SF026), and by the Doctoral Fund of University
of Jinan (Grant no. XBS1337).
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1990.
16
Y. TIAN, C. MA
EJDE-2015/05
[3] A. Alvino, G. Trombetti, P. L. Lions, S. Matarasso; Comparison results for solutions of
elliptic problems via symmetrization, Annales de l’I. H. P., section C, vol. 16, no. 2, pp.
167-188, 1999.
[4] C. Bandle; Isopermetric Inequalities and Application, Monographs and Studies in Math., No.
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EJDE-2015/05
COMPARISON RESULTS FOR ELLIPTI INEQUALITIES
17
Yujuan Tian (corresponding author)
School of Mathematical Sciences, Shandong Normal University, Jinan, Shandong 250014,
China
E-mail address: tianyujuan0302@126.com
Chao Ma
School of Mathematical Sciences, University of Jinan, Jinan, Shandong 250022, China
E-mail address: chaos ma@163.com