Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 305, pp. 1–16.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
NONLINEAR ELLIPTIC EQUATIONS WITH GENERAL
GROWTH IN THE GRADIENT RELATED TO GAUSS
MEASURE
YUJUAN TIAN, CHAO MA, FENGQUAN LI
Abstract. In this article, we establish a comparison result through symmetrization for solutions to some problems with general growth in the gradient.
This allows to get sharp estimates for the solutions, obtained by comparing
them with solutions of simpler problems whose data depend only on the first
variable. Furthermore, we use such result to prove the existence of bounded
solutions. All the above results are based on the study of a class of nonlinear
integral operator of Volterra type.
1. Introduction
Let Ω be an open subset of Rn . We consider the Dirichlet problem whose prototype is
− div(ϕ(x)|∇u|p−2 ∇u) = H(x, u, ∇u) in Ω,
(1.1)
u = 0 on ∂Ω,
where |H(x, s, ξ)| ≤ ϕ(x) (f (x) + θ|ξ|q ) with p − 1 < q ≤ p, 1 < p < 2, θ > 0 and
2
ϕ(x) = (2π)−n/2 exp − |x|2 is the density of Gauss measure. Problem (1.1) is
related to the generator of Ornstein-Uhlenbeck semigroup.
As Ω is bounded, the operator in (1.1) is uniformly elliptic. In this case, it is well
known that one can use Schwarz symmetrization to estimate the solutions of elliptic
equations in terms of the solutions of radially symmetric problems. This kind of
issue has been faced in [8, 24, 19, 2] for linear equations. As regards nonlinear
equations, for the case of p − 1 growth in the gradient, comparison results are
obtained in [3, 18]. Optimal summability of solutions are discussed in [1]. For
the case of p growth in the gradient, using Schwarz symmetrization, the existence
of bounded solutions are obtained in [9, 15, 16, 18]. For the case of q(p − 1 <
q ≤ p) growth in the gradient, similar results can be found in [14, 26]. Recently,
symmetrization techniques have also been applied to equations involving fractional
Laplacian operators (see [28, 29, 13]).
In our case, since Ω maybe unbounded, the degeneracy of the operator does not
allow to use the classical approach via Schwarz symmetrization. This leads us to
2010 Mathematics Subject Classification. 35J60, 35J70, 35J15.
Key words and phrases. Comparison results; symmetrization; gauss measure;
nonlinear elliptic equation.
c
2015
Texas State University.
Submitted July 22, 2015. Published December 16, 2015.
1
2
Y. TIAN, C. MA, F. LI
EJDE-2015/305
consider Gauss symmetrization based on the structure of the problem. By Gauss
symmetrization, comparison results for linear equations have been obtained, with
a simpler problem which is defined in a half space and has data depending only
on the first variable (see [6, 10, 12]). Nonlinear equations with p − 1 growth in
the gradient have also been discussed in [11]. However, the case with other growth
in the gradient has not been studied until now. In this paper, we deal with such
problem (with q(p − 1 < q ≤ p) growth in the gradient).
Our aim is to prove a sharp comparison result which allows us to estimate the
solutions of (1.1) in terms of the symmetric solutions of the following “symmetrized”
problem
−D1 (ϕ|D1 v|p−2 D1 v) = ϕf ] + θϕ|D1 v|q in Ω] ,
(1.2)
v = 0 on ∂Ω] ,
where Ω] is a half space with the same Gauss measure as Ω and f ] is Gauss
symmetrization of f . To this end, we first discuss the existence of symmetric
solutions to (1.2) and give the regularity results of such solutions, which is a key
step for the comparison results. Moreover, by the comparison results, we are able
to prove the existence of bounded solutions to (1.1) in weighted Sobolev space
W01,p (ϕ, Ω). Note that the assumption 1 < p < 2 is necessary to the existence of
bounded solutions. We will give an example in the Appendix to show that there
may be no solution to (1.1) if p ≥ 2.
There are two main difficulties in studying (1.1). One is due to the fact that
the operator is in general not uniformly elliptic, for instance when Ω is an unbounded domain. Another is due to the presence of general growth in the gradient.
Therefore, the present approaches can not be applied to our case. To overcome
the above difficulties, based on the properties of the weighted rearrangement, we
convert the problems into the study of a class of Volterra integral operator. This
class of Volterra integral operator was introduced in [20, 17]. By discussing the
existence of fixed point to the Volterra integral operator, we obtained the existence
and non-existence of symmetric solution to the “symmetrized” problem (1.2). The
sharp comparison results are obtained by proving a new type of comparison principle for the Volterra integral operator. The methods developed in this article can
also be used to study the corresponding variational inequalities.
This article is organized as follows: In Section 2 we give some preliminary results.
In Section 3, the main results of this paper are stated. In Section 4, three results
for a class of Volterra integral operator are proved. In Section 5, 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.
Let γn be the n-dimensional normalized Gauss measure on Rn defined as
|x|2 dx, x ∈ Rn .
2
We denote by Φ(τ ) the measure of the half space {x ∈ Rn : x1 > τ }, i.e.
Z +∞
1
t2 n
Φ(τ ) = γn ({x ∈ R : x1 > τ }) = √
exp −
dt, τ ∈ R.
2
2π τ
dγn = ϕ(x)dx = (2π)−n/2 exp −
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NONLINEAR ELLIPTIC EQUATIONS
3
Observe that (see [27])
Φ−1 (t)2 exp −
≤ αt(1 − log t)1/2 ,
2
Φ−1 (t)2 exp −
≥ βt(1 − log t)1/2 ,
2
t ∈ (0, γn (Ω)),
(2.1)
t ∈ (0, γn (Ω)),
(2.2)
where α and β are positive constants depending on γn (Ω). Now we give the notion
of rearrangement.
Definition 2.1. 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)].
If Ω] = {x = (x1 , x2 , · · · , xn ) ∈ Rn : x1 > λ} is 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 is
u] (x) = u? (Φ(x1 )),
x ∈ Ω] ,
with
u? (s) = u? (γn (Ω) − s),
s ∈ (0, γn (Ω)).
The properties of rearrangement with respect to Gauss measure or a positive measure have been widely considered, see [7, 5, 21, 22, 23] for instance. Here we just
recall that
(a) (Hardy-Little inequality)
Z
γn (Ω)
u? (s)v ? (s)ds =
Z
u] (x)v ] (x)dγn ≤
Ω]
0
Z
≤
u] (x)v ] (x)dγn =
Ω]
Z
|u(x)v(x)|dγn
Ω
Z γn (Ω)
u? (s)v ? (s)ds,
0
where u and v are measurable functions.
(b) (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.
Finally, we recall that the weighted Sobolev space W01,p (ϕ, Ω) is the closure of
∞
C0 (Ω) under the norm
Z
Z
1/p
p
kukW 1,p (ϕ,Ω) =
|∇u(x)| ϕ dx +
|u(x)|p ϕ dx
.
Ω
Ω
4
Y. TIAN, C. MA, F. LI
EJDE-2015/305
3. Statement of main results
In this article, we consider the problem
− div(a(x, u, ∇u)) = H(x, u, ∇u)
u∈
W01,p (ϕ, Ω)
in Ω,
(3.1)
∩ L∞ (Ω),
where Ω is an open subset of Rn (n ≥ 2) with Gauss measure less than one, a :
Ω × R × Rn → Rn and H : Ω × R × Rn → R are Carathéodory functions such that
for all (s, ξ) ∈ R × Rn and for almost every x ∈ Ω,
(A1) a(x, s, ξ)ξ ≥ ϕ(x)|ξ|p , with 1 < p < 2;
(A2) |a(x, s, ξ)| ≤ c1 ϕ(x) k(x) + |s|p−1 + |ξ|p−1 , with c1 > 0, k(x) ≥ 0 and
0
k ∈ Lp (ϕ, Ω);
(A3) |H(x, s, ξ)| ≤ ϕ(x) (f (x) + θ|ξ|q ), with θ > 0, p − 1 < q ≤ p, f (x) ≥ 0 and
f ∈ L∞ (Ω).
Definition 3.1. We say that u ∈ W01,p (ϕ, Ω) ∩ L∞ (Ω) is a solution of (1.1), if
Z
Z
a(x, u, ∇u)∇ψ dx =
H(x, u, ∇u)ψ dx, ∀ψ ∈ W01,p (ϕ, Ω) ∩ L∞ (Ω). (3.2)
Ω
Ω
Remark 3.2. The assumption 1 < p < 2 is necessary. As p ≥ 2, there may be no
solution to (3.1) (see an example in the Appendix).
First, let us turn our attention to the “symmetrized” problem (1.2), discussing
the existence and regularity of a unique symmetric solution, which is a key step for
the comparison results.
√
γ
γ
q
, γ 0 = γ−1
and M0 = θ 2πβ −1 (1 − log γn (Ω))−1/2 .
Theorem 3.3. Let γ = p−1
If
1
1
(3.3)
kf kL∞ (Ω) ≤ 0 (γM0 ) 1−γ ,
γ
then there exists a unique solution to (1.2) such that v(x) = v ] (x). Moreover,
v ∈ C 1 (Ω] \x1 = +∞) ∩ W 1,∞ (Ω] ) with provides the estimates
k∇vkL∞ (Ω] ) ≤ C1 (β, γ, γn (Ω)),
kvkL∞ (Ω] ) ≤ C2 (β, γ, γn (Ω)),
where C1 , C2 are constants depending only on β, γ and γn (Ω).
In the case f (x) ≡ f0 , we have the following nonexistence result for (1.2).
Theorem 3.4. Let
f0 ≥ A(γn (Ω)),
(3.4)
where
Z s
0
1
γ
γγ
α γ s i γ−1
√
A(s) =
, F1 (s) =
(1 − ln τ )− 2 dτ,
θ(γ − 1)
F
(s)
2π
1
0
for s ∈ [0, γn (Ω)]. Then (1.2) has no symmetric solution.
h
1
Remark 3.5. By computations, it follows that A(γn (Ω)) > γ10 (γM0 ) 1−γ . Thus, the
above theorem proposes an example to show that (1.2) has no symmetric solution
without the assumption (3.3).
Now, the comparison results can be stated by the following theorem.
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NONLINEAR ELLIPTIC EQUATIONS
5
Theorem 3.6. Assume that (A1)–(A3) hold. Let u be a solution to (3.1) and v be
a solution to (1.2) such that v(x) = v ] (x). Then
Z
u] (x) ≤ v(x), x ∈ Ω] ,
Z
p
η(|∇u| )ϕ dx ≤
η(|∇v|p )ϕ dx,
(3.5)
(3.6)
Ω]
Ω
where η is a concave and nondecreasing function on [0, +∞). Moreover, if f (x) 6≡ 0,
the equality in (3.5) holds if and only if
Ω = Ω] ,
u(x) = δu] (x),
a.e. x ∈ Ω] ,
a1 (x, u, ∇u) = δϕ|D1 u] |p−2 D1 u] ,
n
X
Di ai (x, u, ∇u) = 0
a.e. x ∈ Ω] ,
0
(3.7)
]
in D (Ω ),
i=2
H(x, u, ∇u) = δϕ(f ] (x) + θ|D1 u] |q ),
a.e. x ∈ Ω]
modulo a rotation with δ = ±1.
Remark 3.7. Equality (3.7) implies that the comparison result (3.5) is sharp in the
sense that as the equality holds, problem (3.1) is equivalent to its “symmetrized”
problem (1.2) modulo a rotation.
The estimates we have found can be applied to prove the existence result by
using the well known approximation techniques [4].
Theorem 3.8. Let (A1)–(A3) and (3.3) hold. Assume that
[a(x, s, ξ1 ) − a(x, s, ξ2 )] · (ξ1 − ξ2 ) > 0,
for ξ1 6= ξ2 .
Then there exists at least one solution to (3.1).
4. Results for Volterra integral operators
This section is devoted to study a class of Volterra integral operator. We prove
three results, i.e. the comparison principle, the existence of fixed point and the
nonexistence of fixed point for the Volterra integral operator, which will be useful
in proving the main results of this paper.
Assume h(s, ξ) : [0, T ] × R → R is a Carathéodory function. Consider the
following Volterra integral operator (see [20, 17])
Z t
K : D(K) ⊆ C([0, T ]) → C([0, T ]), Kψ(t) =
h(τ, ψ(τ ))dτ, ∀ψ ∈ D(K).
0
Definition 4.1. We say that the operator K has property (m) if for all ψ1 , ψ2 ∈
D(K) and a ∈ [0, T ), there exist constants b ∈ (a, T ] and m(a, b, ψ1 , ψ2 ) ∈ [0, 1)
such that for any t ∈ (a, b],
ψ1 − ψ2 ∞
.
(4.1)
kh(·, ψ1 (·)) − h(·, ψ2 (·))kL1 (a,t) ≤ tm(a, b, ψ1 , ψ2 )
L (a,t)
s
Lemma 4.2. Let the operator K satisfy property (m) and h(t, ·) be nondecreasing
for a.e. t ∈ [0, T ]. If u, v ∈ D(K) are such that u ≤ Ku, v ≥ Kv, then we have
u ≤ v.
In particular, the equation w = Kw possesses at most one solution in D(K).
(4.2)
6
Y. TIAN, C. MA, F. LI
EJDE-2015/305
Proof. We argue by contradiction. If (4.2) does not hold, then there must exist
a ∈ [0, T ) and b1 ∈ (a, T ] such that u(t) ≤ v(t) for t ∈ [0, a] and u(t) > v(t)
for t ∈ (a, b1 ]. Set b2 = min{b1 , b} with b is the constant of property (m). Then
u(t) > v(t) in (a, b2 ]. Since u ≤ Ku, v ≥ Kv and h(t, ·) is nondecreasing for a.e.
t ∈ [0, T ], it follows that for ∀t ∈ (a, b2 ],
Z t
|u(t) − v(t)| = u(t) − v(t) ≤
(h(τ, u(τ )) − h(τ, v(τ ))) dτ
0
Z a
Z t
=
(h(τ, u(τ )) − h(τ, v(τ ))) dτ +
(h(τ, u(τ )) − h(τ, v(τ ))) dτ
0
a
Z t
≤
(h(τ, u(τ )) − h(τ, v(τ ))) dτ
a
≤ kh (·, u(·)) − h (·, v(·)) kL1 (a,t) .
(4.3)
Using property (m), we obtain
u−v
kL∞ (a,t) , ∀t ∈ (a, b2 ].
(4.4)
s
Taking the maximum over t ∈ (a, b2 ] and noting m(a, b, u, v) ∈ [0, 1), we have
u−v
u−v
u−v
k
kL∞ (a,b2 ) ≤ m(a, b, u, v)k
kL∞ (a,b2 ) < k
kL∞ (a,b2 ) ,
s
s
s
which is a contradiction. Thus
|u(t) − v(t)| ≤ tm(a, b, u, v)k
u ≤ v, ∀u, v ∈ D(K),
and the uniqueness claim easily follows. Then the lemma is proved.
Let
s
√
Φ−1 (τ )2 γ γ
ψ (τ )dτ,
2
0
We shall deal with two types of domains
Z
Kψ(s) =
f ? (τ ) + θ
2π exp
s ∈ [0, γn (Ω)].
D1 (K) = {ψ ∈ C([0, γn (Ω)]) : M ≥ 0, 0 ≤ ψ(s) ≤ M s},
D2 (K) = {ψ ∈ C([0, γn (Ω)]) : Mψ ≥ 0, 0 ≤ ψ(s) ≤ Mψ s},
and let Ri (K) be the range of K on Di (K), i = 1, 2.
1
Lemma 4.3. Let M = (γM0 ) 1−γ in D1 (K). If (3.3) holds, then the equation
w = Kw has a unique solution in D1 (K). Furthermore, the solution is also unique
in D2 (K).
Proof. First, we prove that R1 (K) ⊆ D1 (K). For ∀ψ ∈ D1 (K) and s ∈ [0, γn (Ω)],
we have by (2.2) that
Z s
√
Φ−1 (τ )2 γ γ
Kψ(s) =
f ? (τ ) + θ 2π exp
ψ (τ )dτ
2
0
Z
γ
√
γ s 1
≤ skf kL∞ (Ω) + θ 2πβ −1
(M τ )γ dτ
(4.5)
τ (1 − log τ )1/2
0
√
γ 2πβ −1 M (1 − log γn (Ω))−1/2
s
≤ kf kL∞ (Ω) + θ
= kf kL∞ (Ω) + M0 M γ s.
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NONLINEAR ELLIPTIC EQUATIONS
7
1
Under the assumption (3.3), noting that M = (γM0 ) 1−γ , we obtain
kf kL∞ (Ω) + M0 M γ ≤ M.
Thus,
0 ≤ Kψ(s) ≤ M s,
s ∈ [0, γn (Ω)].
This implies that R1 (K) ⊆ D1 (K).
Next, we verify that the operator K is compact with respect to the uniform
topology of C([0, γn (Ω)]). To prove this, let us first show the equicontinuity of
R1 (K) in C([0, γn (Ω)]). Take any Kψ ∈ R1 (K). For any 0 ≤ a < b ≤ γn (Ω), it
follows
|Kψ(b) − Kψ(a)|
Z b
Φ−1 (τ )2 γ
√
?
ψ γ (τ )dτ
≤
f (τ ) + θ 2π exp
2
a
Z
γ
√
γ b 1
(M τ )γ dτ
≤ kf kL∞ (Ω) (b − a) + θ 2πβ −1
τ (1 − log τ )1/2
a
≤ (kf kL∞ (Ω) + M0 M γ )(b − a).
(4.6)
Then, R1 (K) is equicontinuous in C([0, γn (Ω)]). Also, the family of functions from
R1 (K) is uniformly bounded (see (4.5)), i.e.
0 ≤ Kψ ≤ kf kL∞ (Ω) + M0 M γ γn (Ω).
(4.7)
By Ascoli-Arzela theorem, R1 (K) is relatively compact in C([0, γn (Ω)]), and hence
the operator K is compact.
Since the domain D1 (K) is bounded, closed and convex, by Schauder’s fixed point
theorem there exists at least one solution w ∈ D1 (K) to the equation w = Kw.
Now, we study the uniqueness of the solution w. By D1 (K) ⊆ D2 (K), it suffices
to show that the solution w is unique in D2 (K).
Next we check that K satisfies property (m) in D2 (K). For all ψ1 , ψ2 ∈ D2 (K),
0 ≤ a < b ≤ γn (Ω) and a < t ≤ b, we have
kh(·, ψ1 (·)) − h(·, ψ2 (·))kL1 (a,t)
Z t √
Φ−1 (τ )2 γ
≤
θ 2π exp
|ψ1γ (τ ) − ψ2γ (τ )|dτ
2
a
γ
Z √
γ t
1
≤ θγ 2πβ −1
max{ψ1γ−1 (τ ), ψ2γ−1 (τ )}
1/2
τ
(1
−
log
τ
)
a
× |ψ1 (τ ) − ψ2 (τ )|dτ
Z t
√
γ
γ
1
≤ θγ 2πβ −1 max{Mψ1 , Mψ2 }γ−1
τ γ−1
1/2
τ
(1
−
log
τ
)
a
× |ψ1 (τ ) − ψ2 (τ )|dτ
γ
√
fγ−1 (t − a)k ψ1 − ψ2 kL∞ (a,t)
≤ θγ
2πβ −1 (1 − log γn (Ω))−1/2 M
τ
ψ1 − ψ2
a
γ−1
f
≤ tγM0 M
1− k
kL∞ (a,t)
b
τ
ψ1 − ψ2
= tm(a, b, ψ1 , ψ2 )k
kL∞ (a,t) ,
τ
(4.8)
8
Y. TIAN, C. MA, F. LI
EJDE-2015/305
where
fγ−1 (1 − a )
m(a, b, ψ1 , ψ2 ) = γM0 M
b
f
with M = max{Mψ1 , Mψ2 }.
From limb→a+ m(a, b, ψ1 , ψ2 ) = 0, it follows that there exists b0 > a such that
0 ≤ m(a, b0 , ψ1 , ψ2 ) < 1. Thus (4.8) implies that K satisfies property (m) in D2 (K).
Since h(t, ψ) is nondecreasing for t ∈ [0, γn (Ω)], we have by Lemma 4.2 that the
solution of w = Kw is unique in D2 (K). Thus the proof is complete.
For f (x) ≡ f0 , we have the following nonexistence result.
Lemma 4.4. Let (3.4) hold. Then the equation z = Kz has no solution in D1 (K).
Proof. We argue by contradiction. Assume that there exists a solution z ∈ D1 (K)
of z = Kz. Let z0 (s) = f0 s and
Z s √
Φ−1 (τ )2 γ
γ
2π exp
zm−1
(τ )dτ, s ∈ [0, γn (Ω)].
(4.9)
zm (s) = θ
2
0
First, we claim that
f γ m m
0
γ γ−1 ,
zm (s) ≥ sA(s)
A(s)
m≥1
(4.10)
where
0
γγ
A(s) =
θ(γ − 1)
h
α
√
2π
γ
1
s i γ−1
,
F1 (s)
Z
F1 (s) =
s
γ
(1 − ln τ )− 2 dτ.
0
To prove (4.10), set
Z
s
−γ/2
(1 − ln τ )
Fm+1 (s) =
γ
Fm
(τ )dτ.
0
Thus,
zm (s) ≥ θ
γ m −1
γ−1
−γ
√2π γ m+1
γ−1
m
f0γ s−
γ m −γ
γ−1
Fm (s).
(4.11)
α
Indeed, by (2.1) and (4.9), we have
Z s √
Φ−1 (τ )2 γ
z1 (s) = θ
2π exp
z0γ (τ )dτ
2
0
√2π γ Z s γ
1
≥θ
f0γ
τ γ dτ
1/2
α
τ
(1
−
ln
τ
)
0
√2π γ
γ
=θ
f0 F1 (s),
α
which proves (4.11) for m = 1. Assume it holds for m − 1. Then (4.11) can be
established by induction on m. Next, treat the term Fm (s) in (4.11). Actually,
m m−δ
γ m −1
Y
γ − 1 γ
γ−1
Fm (s) ≥
F
(s), m ≥ 1.
(4.12)
1
γδ − 1
δ=1
The case m = 1 is obvious. Suppose (4.12) holds for some m. Then
Z s
−γ/2 γ
Fm+1 (s) =
(1 − ln τ )
Fm (τ )dτ
0
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NONLINEAR ELLIPTIC EQUATIONS
9
m m+1−δ Z s
γ m+1 −γ
Y
γ − 1 γ
−γ/2
γ−1
≥
(τ )dτ
(1
−
ln
τ
)
F
1
γδ − 1
0
δ=1
m m+1−δ Z s
γ m+1 −γ
Y
γ − 1 γ
γ−1
=
F
(τ )dF1 (τ )
1
γδ − 1
0
δ=1
=
m+1
Y
δ=1
γ − 1 γ
γδ − 1
m+1−δ
γ m+1 −1
γ−1
F1
(s),
which proves (4.12). Moreover,
m m m−δ
m−δ
Pm
Y
Y
P
m−δ
m−δ
γ − 1 γ
γ − 1 γ
− m
δ=1 γ
δ=1 δγ
≥
=
(γ
−
1)
γ
γδ − 1
γδ
δ=1
δ=1
= (γ − 1)
γ m −1
γ−1
γ
γ(γ m −1)
m
γ−1 − (γ−1)2
(4.13)
.
Combining (4.11), (4.12) and (4.13), we know that (4.10) holds.
On the other hand, A(s) is a decreasing continuous function on (0, γn (Ω)] and
lims→0+ A(s) = +∞. Then the range of A is [A(γn (Ω)), +∞). Recalling that f0 ≥
A(γn (Ω))), thus there must exist a constant s∗ ∈ (0, γn (Ω)] such that f0 = A(s∗ ).
f0
As s ≥ s∗ , it follows that A(s)
≥ 1 and
m
zm (s) ≥ sA(s)γ γ−1 .
P∞
m
Note that limm→∞ γ γ−1 = +∞. We conclude that m=0 zm (s) = +∞ for s ∈
[s∗ , γn (Ω)]. On the other hand, by inducing on k, it is easy to prove that
k
X
z(s) ≥
∀k ∈ N.
zm (s),
(4.14)
m=0
Indeed, since z(s) = Kz(s) ≥ f0 s = z0 (s), we obtain
Z s √
Φ−1 (τ )2 γ
z(s) = f0 s + θ
2π exp
z γ (τ )dτ ≥ z0 (s) + z1 (s),
2
0
which proves the claim for k = 1. Now assume (4.14) holds for some k ∈ N . Then
Z
√
s
z(s) ≥ z0 (s) + θ
2π exp
k
Φ−1 (τ )2 γ X
2
0
≥ z0 (s) + θ
k
X
Z
m=0
= z0 (s) +
k
X
s
√
2π exp
zm+1 (s) =
m=0
k+1
X
m=0
Φ−1 (τ )2 γ
2
0
γ
zm (τ ) dτ
γ
zm
(τ )dτ
zm (s).
m=0
Thus for k ∈ N, (4.14) holds. However, recalling that z ∈ D1 (K), we obtain
M s ≥ z(s) ≥
∞
X
zm (s) = +∞,
s ∈ [s∗ , γn (Ω)],
m=0
which is a contradiction. Thus the lemma is proved.
10
Y. TIAN, C. MA, F. LI
EJDE-2015/305
5. Proofs of the main results
First, let us enunciate the following lemma, the proof of which is not supplied
here since it follows the same lines as in [14, 18].
Lemma 5.1. Let u be a solution of (1.1) and v be a solution of (1.2) such that
v(x) = v ] (x). Then
Φ−1 (s)2 p
p−1 1
0
√ exp −
− u? (s)
2
2π
Z s
h Z s √
Φ−1 (σ)2 p−q q−p+1 i
0
≤
f ? (τ ) exp θ
2π exp
−u? (σ)
dσ dτ
2
0
τ
(5.1)
and
Φ−1 (s)2 p
p−1 1
0
√ exp −
− v ? (s)
2
2π
Z s
h Z s √
Φ−1 (σ)2 p−q
q−p+1 i
0
=
f ? (τ ) exp θ
2π exp
− v ? (σ)
dσ dτ,
2
0
τ
(5.2)
a.e. s ∈ (0, γn (Ω)).
Proof of Theorem 3.3. Let
Z γn (Ω) √
Φ−1 (τ )2 p0 1
v(x) = V (Φ(x1 )) =
2π exp
w p−1 (τ )dτ, x ∈ Ω] ,
2
Φ(x1 )
(5.3)
where w is the unique solution of w = Kw obtained in Lemma 4.3. Clearly,
v(x) = v ] (x), x ∈ Ω] . By (5.3) and (2.2),
1
x2 |∇v(x)| = D1 v(x) = − √ V 0 (Φ(x1 )) exp − 1
2
2π
√
2 1
1
x
p−1
w p−1 (Φ(x1 ))
=
2π exp 1
2
1
− p−1
√ 1 1
≤
2π p−1 βΦ(x1 )(1 − log Φ(x1 ))1/2
(M Φ(x1 )) p−1
1
√
p−1
≤
2πβ −1 M (1 − log γn (Ω))−1/2
.
Moreover, since 1 < p < 2,
γn (Ω)
Z
kvkL∞ (Ω] ) = V (0) =
√
0
≤
=
√
√
2πβ −1
p0
M p−1
2πβ −1
p0
M p−1
1
1
0
1
Φ−1 (τ )2 p p−1
w
(τ )dτ
2
Z γn (Ω) p0 1
1
τ p−1 dτ
τ (1 − log τ )1/2
0
Z γn (Ω)
p0
(1 − log τ )− 2 τ −1 dτ
2π exp
0
√
p0
p0
1
2
= 0
2πβ −1 M p−1 (1 − log γn (Ω))− 2 +1 .
p −2
Then v ∈ W01,p (ϕ, Ω] ) ∩ L∞ (Ω] ).
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NONLINEAR ELLIPTIC EQUATIONS
11
On the other hand, for all ψ ∈ W01,p (ϕ, Ω] ) ∩ L∞ (Ω] ), by (5.3) and the fact that
w = Kw in D1 (K), we have
Z
ϕ(D1 v)p−1 D1 ψ dx
Ω]
Z
1
x2 p−1
p−1
=
ϕ √ exp − 1
(−V 0 (Φ(x1 )))
D1 ψ dx
2
]
2π
ZΩ
=
w(Φ(x1 ))D1 ψ dx
Ω]
Z
Z
=
Ω]
Φ(x1 )
h
f ? (τ ) + θ
√
2π exp
0
Φ−1 (τ )2 γ
2
i
wγ (τ ) dτ D1 ψ dx.
Integrating by part on the right-hand side of the third equality to obtain
Z
p−1
ϕ (D1 v)
D1 ψ dx
]
Ω
Z h
√
i
x 2 γ γ
=
f ] (x) + θ 2π exp 1
w (Φ(x1 )) ψϕ dx
2
Ω]
Z
Z
x2 q
1
ψϕ dx
=
f ] ψϕ dx +
θ − V 0 (Φ(x1 )) √ exp − 1
2
]
]
2π
ZΩ
ZΩ
=
f ] ψϕ dx +
θ|D1 v|q ψϕ dx.
Ω]
Ω]
Hence, v is a symmetric solution to (1.2).
Next, we show that v is the unique symmetric solutions to (1.2). Indeed, assume
that there exists another symmetric solution v1 . Let
Z s
h Z s √
Φ−1 (σ)2 p−q
?
w1 (s) =
f (τ ) exp θ
2π exp
2
0
τ
(5.4)
i
0
q−p+1
?
× − v1 (σ)
dσ dτ.
It follows from (5.2) that
√
Φ−1 (s)2 p0 1
0
− v1? (s) =
w1p−1 (s), a.e. s ∈ (0, γn (Ω)).
2π exp
(5.5)
2
Furthermore, a simple computation shows that
√
Φ−1 (s)2 γ
w1γ , a.e. s ∈ (0, γn (Ω)),
w10 (s) = f ? (s) + θ 2π exp
2
which gives w1 = Kw1 . By (5.4), (2.2) and Hölder inequality,
Z s
Z
h √
p−q
s
1
?
−1 p−q
w1 (s) ≤
f (τ ) exp θ 2πβ
dσ
1/2
0
τ σ(1 − log σ)
Z s 0
q−p+1 i
× −
v1? (σ)dσ
dτ
τ
Z s
h √
Z s
p−q i
p−q
1
q−p+1
≤
f ? (τ ) exp θ 2πβ −1
kv1 kL
dσ
dτ
∞ (Ω] )
1/2
0
τ σ(1 − log σ)
Z s
h Z s
p−q
Z s
2(p−q)
i
3
3
?
−1
− 23
σ −1 dσ
dτ,
≤
f (τ ) exp C1
σ (1 − log σ) dσ
0
τ
τ
12
Y. TIAN, C. MA, F. LI
where C1 = θ
√
EJDE-2015/305
p−q
q−p+1
2πβ −1
kv1 kL
∞ (Ω] ) . Since
Z s
3
σ −1 (1 − log σ)− 2 dσ ≤ 2(1 − log γn (Ω))−1/2 ,
τ
by Young’s inequality with ε, the above inequality becomes
Z s
i
h Z s
2(p−q)
3
dτ
f ? (τ ) exp C2
σ −1 dσ
w1 (s) ≤
τ
Z0 s
Z
s
≤
f ? (τ ) exp C1ε C2 + εC2
σ −1 dσ dτ
0
τ
Z s
s
= C2ε
f ? (τ )( )εC2 dτ,
τ
0
where
C2 = θ
√
2πβ −1
p−q
−1/2
q−p+1
kv1 kL
∞ (Ω] ) (2(1 − log γn (Ω))
)
p−q
3
,
C1ε and C2ε are positive constants depending on ε. Take ε small enough such that
εC2 < 1. Then
0 ≤ w1 (s) ≤ C3ε kf kL∞ (Ω) s.
Thus, w1 ∈ D2 (K) satisfying w1 = Kw1 . By Lemma 4.3, we know w1 = w. Noting
v1? (γn (Ω)) = 0, from (5.5) we have
Z γn (Ω) √
Φ−1 (τ )2 p0 1
?
w p−1 (τ )dτ.
2π exp
v1 (s) =
2
s
Hence v1 = v and then the uniqueness is proved.
Remark 5.2. From the above proof, we find that v ? and w can be expressed by
each other via the following equations:
Z γn (Ω) √
Φ−1 (τ )2 p0 1
?
v (s) =
(5.6)
2π exp
w p−1 (τ )dτ
2
s
and
Z
w(s) =
0
s
h Z
f ? (τ ) exp θ
s
√
2π exp
τ
Φ−1 (σ)2 p−q
2
q−p+1 i
0
− v ? (σ)
dσ dτ.
(5.7)
Proof of Theorem 3.4. Assume that there exists a symmetric solution v to (1.2).
From Remark 5.2, we see that w defined by (5.7) is a solution of the equation
z = Kz,
z ∈ D1 (K),
which contradict with Lemma 4.4. Thus the proof is complete.
Proof of Theorem 3.6.
Step 1. First we verify (3.5). Take
Z s
Φ−1 (σ)2 p−q
h Z s √
q−p+1 i
0
ρ(s) =
f ? (τ ) exp θ
2π exp
− u? (σ)
dσ dτ.
2
0
τ
Then ρ ∈ D2 (K) and ρ(s) ≤ Kρ(s) for s ∈ [0, γn (Ω)]. Now remembering that K
has property (m) (see the proof of Lemma 4.3) and w = Kw in D2 (K), by Lemma
4.2 we obtain
ρ(s) ≤ w(s), s ∈ [0, γn (Ω)].
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NONLINEAR ELLIPTIC EQUATIONS
13
Moreover, (5.1) and (5.2) imply that
Φ−1 (s)2 p
p−1 1
0
√ exp −
− u? (s)
≤ ρ(s) a.e. s ∈ (0, γn (Ω))
2
2π
and
Φ−1 (s)2 p
p−1 1
0
√ exp −
= w(s) a.e. s ∈ (0, γn (Ω)).
− v ? (s)
2
2π
0
0
Thus, −u? (s) ≤ −v ? (s), a.e. s ∈ (0, γn (Ω)). Note that u? (γn (Ω)) = v ? (γn (Ω)) =
0. We have
u? (s) ≤ v ? (s), s ∈ [0, γn (Ω)].
The proof of (3.6) can be done by repeating the proof of [14, Theorem 4.1].
Step 2. The equality case of (3.5) will be studied. Sufficiency is obvious. Let us
prove the necessity. Assume that u? (s) = v ? (s) for s ∈ [0, γn (Ω)]. By Polya-Szëgo
principle and (3.6), we have
Z
Z
Z
Z
p
] p
p
|∇v| ϕ dx =
|∇u | ϕ dx ≤
|∇u| ϕ dx ≤
|∇v|p ϕ dx.
Ω]
Ω
Ω]
Ω
Thus,
Z
|∇u] |p ϕ dx =
Ω]
Z
|∇u|p ϕ dx
Ω
and then Ω = Ω] and |u| = u] modulo a rotation, which implies u = δu] and Ω]
modulo a rotation with δ = ±1. Now for h > 0, we take


if |u(x)| > t + h,
sign u(x)
ψ(x) = (|u(x)|−t)h sign u(x) if t < |u(x)| ≤ t + h,


0
otherwise
as the test function in (3.2) and let h → 0. Since δu = |u| = u] = v ] = v, we have
by Hardy-Littlewood inequality that
Z
d
−
|∇u] |p ϕ dx
dt u] >t
Z
Z
d
d
p
|∇u| ϕ dx ≤ −
a(x, u, ∇u)∇u dx
=−
dt |u|>t
dt |u|>t
Z
Z
Z
=
H(x, u, ∇u)udx ≤
f uϕ dx + θ
|∇u|q uϕ dx
|u|>t
|u|>t
|u|>t
(5.8)
Z
Z
] ]
] q ]
≤
f u ϕ dx + θ
|∇u | u ϕ dx
]
]
Zu >t
Z u >t
=
f ] vϕ dx + θ
|∇v|q vϕ dx
v>t
v>t
Z
Z
d
d
=−
|∇v|p ϕ dx = −
|∇u] |p ϕ dx.
dt v>t
dt u] >t
Thus, the above equalities hold. In particular,
Z
Z
d
d
−
a(x, u, ∇u)∇u dx = −
|∇u] |p ϕ dx,
dt |u|>t
dt u] >t
Z
Z
Z
H(x, u, ∇u)udx =
f ] u] ϕ dx + θ
|∇u] |q u] ϕ dx,
|u|>t
u] >t
u] >t
(5.9)
(5.10)
14
Y. TIAN, C. MA, F. LI
Z
f u] ϕ dx =
u] >t
EJDE-2015/305
Z
f ] u] ϕ dx.
(5.11)
u] >t
By [25, Lemma 4.3], from (5.11) on has f = f ] a.e. x ∈ Ω] . Then combine (5.10)
and assumption (iii) to discover that
H(x, u, ∇u) = δ(f ] ϕ + θ|D1 u] |q ϕ),
a.e. x ∈ Ω]
Similarly, from (5.9) and assumption (i) we have
δa(x, u, ∇u)∇u] = |∇u] |p ϕ.
Recalling (5.6), we have
D1 u] (x) = D1 v ] (x) =
√
1
1
x21 p−1
w p−1 (Φ(x1 )) > 0,
2
Di u] (x) = 0, i = 2, 3, . . . , n.
2π exp
a.e. x ∈ Ω] ,
Hence,
a1 (x, u, ∇u) = δ|D1 u] |p−2 D1 u] ϕ,
a.e. x ∈ Ω] .
From the definition of solution it follows that for all ψ ∈ W01,p (ϕ, Ω] ) ∩ L∞ (Ω] ),
Z
Z X
n
ai (x, u, ∇u)Di ψ dx
δ|D1 u] |p−2 D1 u] D1 ψϕ dx +
Ω]
Ω] i=2
Z
Z
a(x, u, ∇u)∇ψ dx =
H(x, u, ∇u)ψ dx
Z
]
] q
=
δ(f ϕ + θ|D1 u | ϕ)ψ dx =
δ(f ] ϕ + θ|D1 v|q ϕ)ψ dx
Ω]
Ω]
Z
Z
=
δ|D1 v|p−2 D1 vD1 ψϕ dx =
δ|D1 u] |p−2 D1 u] D1 ψϕ dx.
=
ZΩ
Ω
Ω]
Then
Z
Ω]
n
X
Ω]
ai (x, u, ∇u)Di φ dx = 0,
∀ψ ∈ W01,p (ϕ, Ω] ) ∩ L∞ (Ω] ),
i=2
which completes the proof.
6. Appendix
In this section give an example to Remark 3.2. Assume that v is a solution of
the problem
−D1 (ϕ|D1 v|p−2 D1 v) = ϕ + θ|D1 v|q ϕ in Ω] ,
(6.1)
v ∈ W01,p (ϕ, Ω] ) ∩ L∞ (Ω] )
and Y is the solution of the problem
−D1 (ϕ|D1 Y |p−2 D1 Y ) = ϕ in Ω] ,
Y =0
on ∂Ω] .
For any given k > 0, let


k
Tk (s) = s


−k
if s > k,
if |s| ≤ k,
if s < −k.
(6.2)
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NONLINEAR ELLIPTIC EQUATIONS
15
Take Tk ((v − Y )− ) as the test function in (6.1) and (6.2), and subtract the two
results. We obtain
Z
0≥
ϕ(|D1 v|p−2 D1 v − |D1 Y |p−2 D1 Y )(Y − v)dx ≥ 0.
{0<(v−Y )− ≤k}
Thus, γn ({0 < (v − Y )− ≤ k}) = 0. Let k → +∞. Then γn ({(v − Y )− > 0}) = 0,
which implies v ≥ Y a.e. in Ω] . However, if p ≥ 2,
kvkL∞ (Ω] ) ≥ kY kL∞ (Ω] ) = Y ? (0)
Z γn (Ω) √
Φ−1 (τ )2 p0 1
2π exp
τ p−1 dτ
=
2
0
Z γn (Ω) p0 1
1
≥C
τ p−1 dτ
τ (1 − log τ )1/2
0
Z γn (Ω)
p0
=C
(1 − log τ )− 2 τ −1 dτ = +∞.
0
∞
This contradicts with v ∈ L (Ω] ). As p ≥ 2, problem (6.1) has no solution.
Acknowledgements. This research was supported by the National Natural Science Foundation of China (Grant No. 11501333), 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), by the Natural Science Foundation
of Shandong Province (Grant No. ZR2015AL015) and by the Doctoral Fund of
University of Jinan (Grant No. XBS1337).
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Yujuan Tian (corresponding author)
School of Mathematical Sciences, Shandong Normal University, Jinan 250014, China
E-mail address: tianyujuan0302@126.com
Chao Ma
School of Mathematical Sciences, University of Jinan, Jinan 250022, China
E-mail address: chaos ma@163.com
Fengquan Li
School of Mathematical Sciences, Dalian University of Technology, Dalian 116024,
China
E-mail address: fqli@dlut.edu.cn