Electronic Journal of Differential Equations, Vol. 2005(2005), No. 139, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE AND UNIQUENESS OF POSITIVE SOLUTIONS TO
A QUASILINEAR ELLIPTIC PROBLEM IN RN
DRAGOS-PATRU COVEI
Abstract. We prove the existence of a unique positive solution to the problem
−∆p u = a(x)f (u)
in RN , N > 2. Our result extended previous works by Cirstea-Radulescu and
Dinu, while the proofs are based on two theorems on bounded domains, due
to Diaz-Saà and Goncalves-Santos.
1. Introduction
Our purpose in this paper is to study the problem
−∆p u = a(x)f (u)
in RN
u>0
u(x) → 0
in RN
(1.1)
as |x| → ∞ ,
where N > 2, ∆p u, (1 < p < ∞) is the p-Laplacian operator and the function a(x)
satisfies the following hypotheses:
0,α
(A1) a(x) ∈ Cloc
(RN ) for some α ∈ (0, 1);
(A2) a(x) > 0 in RN ;
(A3) For Φ(r) = max|x|=r a(x) and p < N ,
Z ∞
r1/(p−1) Φ1/(p−1) (r)dr < ∞ if 1 < p ≤ 2
0
Z ∞
(p−2)N +1
r p−1 Φ(r)dr < ∞ if 2 ≤ p < ∞ .
0
This problem has been studied extensively in the case p = 2 and f (u) = u−γ ,
with γ > 0. Lazer and McKenna [12] studied the special case when Ω ⊂ RN
(N ≥ 1) is a bounded domain with smooth boundary. They proved the existence
and the uniqueness of a positive solution u ∈ C 2+α (Ω) ∩ C(Ω) with homogeneous
Dirichlet boundary condition, provided that a(x) ∈ C α (Ω) and a(x) > 0 for all
x ∈ Ω.
2000 Mathematics Subject Classification. 35J60, 35J70.
Key words and phrases. Quasilinear elliptic problem; uniqueness; existence; nonexistence;
lower-upper solutions.
c
2005
Texas State University - San Marcos.
Submitted June 8, 2005. Published December 5, 2005.
Supported by grant ET 65/2005 from ANCS-MEDC .
1
2
D. P. COVEI
EJDE-2005/139
The existence of entire positive solutions on RN for γ ∈ (0, 1) and under certain
additional hypotheses has been established by Edelson [7] and Kusano-Swanson
[10].
Kusano-Swanson proved that the problem (1.1) has an entire positive solution
in R2 with logarithmic growth at ∞ if a(x) > 0, x > 0, a(x) ∈ C(0, ∞) and
Z ∞
t(Logt)−γ max a(x) dt < ∞.
e
|x|=t
Edelson proved the existence of a solution provided that
Z ∞
rN −1+γ(N −2) (max a(x))dt < ∞,
1
|x|=t
for some γ ∈ (0, 1). This result is generalized for any γ > 0 via the sub- and super
solutions method in Shaker [13] and by other methods by Dalmasso [4].
Shaker proved that problem (1.1) with p = 2 and f (u) = u−γ , γ > 0 has an
entire positive solution u(x) such that c1 ≤ u(x)|x|q|N −2| ≤ c2 for some c1 , c2 and
0 < q < 1 as x → ∞ if
α
(1) a(x) ∈ Cloc
(RN ), a(x) > 0 for x ∈ RN \{0};
(2) There exists 0 < c < 1 such that cΦ(|x|) ≤ a(x) ≤ Φ(|x|) where Φ(r) :=
max|x|=r a(x) , r ∈ [0, ∞);
R∞
(3) 1 rN −1+γ(N −2) (max|x|=t a(x))dt < ∞.
Lair and Shaker continued in [11] the study of (1.1) for p = 2 and f (u) = u−γ ,
γ > 0. Under the above conditions the authors proved the existence of a unique
2,α
positive solution u ∈ Cloc
(RN ) vanishing at infinity to this special problem.
Zhang [14], imposed the following condition to guarantee the existence of positive
solutions to problem (1.1):
(A4) f ∈ C 1 ((0, ∞), (0, ∞)), lims&0+ limf (s) = ∞, and f 0 (s) < 0, for all s ∈
(0, ∞), namely, f is strictly decreasing in (0, ∞).
Under the above condition Zhang’s proved that problem (1.1) has a unique pos2+α
itive solution, u ∈ Cloc
(RN ), vanishing at infinity.
Cirstea-Radulescu [2] and Dinu [6] extended the results of Lair, Shaker and
Zhang for the case of a nonlinearity that is not necessarily decreasing on (0, ∞).
Our aim is to extend the results of Cirstea-Radulescu and Dinu in the sense that
1 < p < ∞. More exactly, let f : (0, ∞) → (0, ∞) be a C 1 function that satisfies
the following assumptions:
(F1) There exists β > 0 such that the mapping u 7→ f (u)/(u+β)p−1 is decreasing
on (0, ∞)
(F2) limu&0 f (u)/up−1 = +∞ and f is bounded in a neighbourhood of +∞.
Our main results are the following:
Theorem 1.1. Under hypotheses (F1), (F2), (A1), (A2), (A3), problem (1.1) has
a unique positive global solution vanishing at infinity.
Theorem 1.2. Suppose a(r) is a positive radial function which is continuous on
RN and fulfills
Z ∞
r1/(p−1) a1/(p−1) (r)dr = ∞ if 2 ≤ p < ∞
0
Then (1.1) has no positive radial solution.
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
3
Theorem 1.3. Problem (1.1) has no positive radial solution if p ≥ N .
Theorem 1.4. Suppose a(r) is a positive radial function which is continuous on
RN and
Z ∞
(p−2)N +1
r p−1 a(r)dr = ∞ if 1 < p ≤ 2.
0
Then (1.1) has no positive radial solution.
2. Uniqueness
Suppose u and v are arbitrary solutions of problem (1.1). Let us show that u ≤ v
or, equivalently, ln(u(x) + β) ≤ ln(v(x) + β), for any x ∈ RN . Assume the contrary.
Since, we have
lim (ln(u(x) + β) − ln(v(x) + β)) = 0,
|x|→∞
we deduce that
max(ln(u(x) + β) − ln(v(x) + β))
RN
exists and is positive. At that point, say x0 , we have
∇(ln(u(x0 ) + β) − ln(v(x0 ) + β)) = 0,
so
1
1
· ∇u(x0 ) =
· ∇v(x0 ),
u(x0 ) + β
v(x0 ) + β
and
1
1
· |∇u(x0 )|p−2 =
· |∇v(x0 )|p−2 .
p−2
(u(x0 ) + β)
(v(x0 ) + β)p−2
By (f1) we obtain
f (v(x0 ))
f (u(x0 ))
<
.
p−1
(u(x0 ) + β)
(v(x0 ) + β)p−1
Since 0 ≥ ∆(ln(u(x0 ) + β) − ln(v(x0 ) + β)), it follows that
(2.1)
(2.2)
∆v(x0 )
∆u(x0 )
≤
,
u(x0 ) + β
v(x0 ) + β
so
1
1
· |∇u(x0 )|p−2 ∆u(x0 ) ≤
· |∇v(x0 )|p−2 ∆v(x0 ) . (2.3)
p−1
(u(x0 ) + β)
(v(x0 ) + β)p−1
Since
|∇ ln(u(x0 ) + β)|p−2 =
1
· |∇u(x0 )|p−2 ,
(u(x0 ) + β)p−2
it follows that
∇(|∇ ln(u(x0 ) + β)|p−2 )
= −(p − 2)
|∇u(x0 )|p−2 (u(x0 ) + β)p−3
∇(|∇u(x0 )|p−2 )
·
∇u(x
)
+
.
0
(u(x0 ) + β)p−2
(u(x0 ) + β)2(p−2)
Then
∇(|∇ ln(u(x0 ) + β)|p−2 ) · ∇(ln(u(x0 ) + β))
= −(p − 2)
|∇u(x0 )|p−2 |∇u(x0 )|2
∇(|∇u(x0 )|p−2 ) · ∇u(x0 )
+
(u(x0 ) + β)p
(u(x0 ) + β)p−1
(2.4)
4
D. P. COVEI
EJDE-2005/139
and
|∇ ln(u(x0 ) + β)|p−2 ∆(ln(u(x0 ) + β)) =
|∇u(x0 )|p−2 ∆u(x0 )
|∇u(x0 )|p
−
.
p−1
(u(x0 ) + β)
(u(x0 ) + β)p
So, by (2.1), (2.2), (2.3) and (2.4) we have
0 ≥ ∆p (ln(u(x0 ) + β)) − ∆p (ln(v(x0 ) + β))
∆p u(x0 )
|∇u(x0 )|p
∆p v(x0 )
− (p − 1)
−
p−1
(u(x0 ) + β)
(u(x0 ) + β)p
(v(x0 ) + β)p−1
|∇v(x0 )|p
+ (p − 1)
(v(x0 ) + β)p
∆p v(x0 )
∆p u(x0 )
−
=
(u(x0 ) + β)p−1
(v(x0 ) + β)p−1
f (u(x0 ))
f (v(x0 ))
= −a(x0 )(
−
)>0
p−1
(u(x0 ) + β)
(v(x0 ) + β)p−1
=
which is a contradiction. Hence u ≤ v. By symmetry we also have v ≤ u, and the
proof is complete.
3. Existence of a solution
We first show that our hypothesis (F1) implies limu&0 f (u) exists, finite or +∞.
f (u)
f (u)
Indeed, since (u+β)
p−1 is decreasing, there exists L := limu&0 (u+β)p−1 ∈ (0, +∞].
It follows that limu&0 f (u) = Lβ p−1 .
To prove the existence of a solution to Problem (1.1), we need to employ a
corresponding result by Diaz-Saà [5] for bounded domains. They considered the
problem
−∆p u = g(x, u) in Ω
u ≥ 0 in Ω
(3.1)
u(x) = 0
on ∂Ω,
N
where Ω ⊂ R is a bounded domain with smooth boundary and g(x, u) : Ω ×
[0, ∞) → R.
Assume that
-for a.e. x ∈ Ω the function u → g(x, u) is continuous on [0, ∞)
and the function u → g(x, u)/up−1 is decreasing on (0, ∞);
∞
-for each u ≥ 0 the function x → g(x, u) belongs to L (Ω);
-there exists C > 0 such that g(x, u) ≤ C(u
p−1
(3.2)
(3.3)
+ 1) a.e. x ∈ Ω, for all u ≥ 0.
(3.4)
Under these hypotheses on g, Diaz-Saà [5] proved that there is at most one
solution of (1.1).
Let us consider the problem
−∆p uk = a(x)f (uk ),
uk (x) = 0,
if |x| < k,
if |x| = k.
(3.5)
The following two distinct situations may occur:
Case 1: f is bounded on (0, +∞). In this case, as we have initially observed,
limu&0 f (u) exists and is finite, so f can be extended by continuity at the origin.
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
5
To obtain a solution to (3.5), it is sufficinet to verify that the hypotheses of the
Diaz-Saà theorem are fulfilled.
* Since f ∈ C 1 ((0, ∞), (0, ∞)) it follows that the mapping u → a(x)f (u) is
continuous in [0, ∞).
(u)
f (u)
(u+β)p−1
* From a(x) ufp−1
= a(x) (u+β)
p−1 ·
up−1 , using positivity of a and (F1) we
(u)
deduce that the function u → a(x) ufp−1
is decreasing on (0, ∞).
0,α
* For all u ≥ 0, since a(x) ∈ Cloc
(RN ), we obtain x → a(x)f (u) belongs to
L∞ (Ω).
f (u)
f (u)
up−1
1
* By limu→∞ up−1
+1 = limu→∞ up−1 · up−1 +1 = 0 and f ∈ C ((0, ∞), (0, ∞)),
p−1
there exists C > 0 such that f (u) ≤ C(u +1) for all u ≥ 0. Therefore, a(x)f (u) ≤
C(up−1 + 1) for all u ≥ 0.
(u)
(u)
* Observe that a0 (x) = limu&0 p(x)f
= +∞ and a∞ (x) = limu→+∞ p(x)f
=
up−1
up−1
0. Thus by Diaz-Saa, problem (3.5) has a unique solution uk which, by the maximum principle, is positive in |x| < k.
Case 2. limu&0 f (u) = +∞. We will apply the method of sub- and supersolutions
in order to find a solution to the problem (3.5). We first observe that 0 is a
subsolution for this problem.
We construct in what follows a positive supersolution. By the boundedness of
f in a neighbourhood of +∞, there exists A > 0 such that f (u) ≤ A, for any
u ∈ (1, +∞). Let f0 : (0, 1] → (0, +∞) be a continuous nonincreasing function such
that f0 ≥ f on (0, 1]. We can assume without loss of generality that f0 (1) = A. Set
(
f0 (u), if 0 < u ≤ 1,
g(u) =
A,
if u > 1.
Then g is a continuous nonincreasing function on (0, +∞). Let h : (0, +∞) →
(0, +∞) be a C 1 nonincreasing function such that h ≥ g. Thus by in [8, Theorem
1.3] the problem
−∆p U = p(x)h(U ),
U = 0,
if |x| < k,
if |x| = k.
has a positive solution. Now, since h ≥ f on (0, +∞), it follows that U is supersolution of (3.5).
In both cases studied above we define uk = 0 for |x| > k. Using a comparision
principle argument as already done above for proving the uniqueness, we can show
that uk ≤ uk+1 on RN .
We now justify the existence of a continuous function v : RN → R such that
uk ≤ v in RN . We first construct a positive radially symmetric function w such
that −∆p w = Φ(r), (r = |x|) on RN and limr→∞ w(r) = 0. A straightforward
computation shows that
Z rh
Z ξ
i1/(p−1)
1−N
w(r) := K −
ξ
σ N −1 Φ(σ)dσ
dξ,
0
0
where
Z
K=
0
∞
h
ξ
1−N
Z
0
ξ
σ N −1 Φ(σ)dσ
i1/(p−1)
dξ.
6
D. P. COVEI
EJDE-2005/139
We first show that (A3) implies that
Z ξ
Z +∞ h
i1/(p−1)
σ N −1 Φ(σ)dσ
ξ 1−N
dξ,
0
0
is finite.
Theorem 3.1. If j : I ⊆ R → R is a locally integrable nonnegative function, then
Z b
1 Z b
h
1
j(x)dx ≤ (resp. ≥)
j h (x)dx
b−a a
b−a a
for all a, b ∈ I, a < b and 1 ≤ h < +∞ (resp 0 < h ≤ 1)
Case 1: Let 1 < p ≤ 2, so 0 < p − 1 ≤ 1, follows that 1 ≤
3.1 for any r > 0, we have
Z r
hξ Z ξ
i1/(p−1)
1−N
p−1
σ N −1 Φ(σ)dσ
ξ
dξ
ξ 0
0
Z
Z r
h1 ξ
i1/(p−1)
1−N
dξ
=
σ N −1 Φ(σ)dσ
ξ p−1 ξ 1/(p−1)
ξ 0
0
Z r
Z ξ
2−N 1
N −1
≤
ξ p−1
σ p−1 Φ1/(p−1) (σ)dσdξ
ξ 0
0
Z r
Z ξ
2−N
N −1
=
ξ p−1 −1
σ p−1 Φ1/(p−1) (σ)dσdξ
0
1
p−1
< +∞. By Theorem
0
Z
Z ξ
N −1
p − 1 r d 2−N
p−1
=−
ξ
σ p−1 Φ1/(p−1) (σ)dσdξ
N − 2 0 dξ
Z r 0
Z r
i
2−N
N −1
p−1 h
=
− r p−1
σ p−1 Φ1/(p−1) (σ)dσ +
ξ 1/(p−1) Φ1/(p−1) (ξ)dξ .
N −2
0
0
Now, by L’Hôpital’s rule, we have
Z r
Z
h
2−N
N −1
1/(p−1)
p−1
p−1
σ
Φ
(σ)dσ +
lim − r
r→∞
= lim
0
−
Rr
0
σ
Φ
1/(p−1)
(σ)dσ + r
r
r
1
p−1
ξ 1/(p−1) Φ1/(p−1) (ξ)dξ
0
N −1
p−1
r→∞
Z
r
N −2
p−1
Rr
0
ξ 1/(p−1) Φ1/(p−1) (ξ)dξ
N −2
p−1
1/(p−1)
Φ
(ξ)dξ
= lim
ξ
r→∞ 0
Z ∞
=
ξ 1/(p−1) Φ1/(p−1) (ξ)dξ < ∞,
0
1
Case 2: Let 2 ≤ p < +∞, so 1 ≤ p − 1, it follows that 1 ≥ p−1
> 0. Set
Z ξ
σ N −1 Φ(σ)dσ ≤ 1 for ξ > 0, or
0
Z
ξ
σ N −1 Φ(σ)dσ > 1
for ξ > 0,
0
In the first case
hZ
0
ξ
σ N −1 Φ(σ)dσ
i1/(p−1)
≤ 1,
i
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
so
Z
r
1−N
ξ p−1
hZ
0
ξ
σ N −1 Φ(σ)dσ
i1/(p−1)
r
Z
dξ ≤
0
7
1−N
ξ p−1 dξ
0
is finite as r → ∞ and N > p. In the second case,
hZ ξ
i1/(p−1) Z ξ
σ N −1 Φ(σ)dσ
≤
σ N −1 Φ(σ)dσ
0
for ξ ≥ 0, so
Z r
hZ
1−N
p−1
ξ
0
0
ξ
σ
N −1
Φ(σ)dσ
i1/(p−1)
r
Z
dξ ≤
0
ξ
1−N
p−1
0
Z
ξ
σ N −1 Φ(σ)dσdξ .
0
Integration by parts gives
Z r
Z ξ
1−N
ξ p−1
σ N −1 Φ(σ)dσdξ
0
0
Z
Z ξ
p − 1 r d p−N
ξ p−1
σ N −1 Φ(σ)dσdξ
N − p 0 dξ
0
Z r
Z r
(p−2)N +1
p−N
p−1
N
−1
(−r p−1
σ
Φ(σ)dσ +
ξ p−1 Φ(ξ)dξ) .
=
N −p
0
0
=−
Now, by L’ Hôpital’s rule, we have
Z r
Z
h
p−N
lim − r p−1
σ N −1 Φ(σ)dσ +
r→∞
= lim
0
−
Rr
0
σ
N −1
Φ(σ)dσ + r
r
r
ξ
(p−2)N +1
p−1
Φ(ξ)dξ
i
0
r→∞
Z
r
N −p
p−1
Rr
0
ξ
(p−2)N +1
p−1
Φ(ξ)dξ
N −p
p−1
(p−2)N +1
p−1
= lim
ξ
Φ(ξ)dξ
r→∞ 0
Z ∞
(p−2)N +1
=
ξ p−1 Φ(ξ)dξ < ∞,
0
From cases 1 and 2 above, it follows that
Z ∞
1
p−1
·
ξ p−1 Φ1/(p−1) (ξ)dξ if 1 < p < 2, or
K=
N −2 0
Z ∞
(p−2)N +1
p−1
K=
·
ξ p−1 Φ(ξ)dξ if 2 ≤ p < +∞.
N −p 0
Clearly, for all r > 0,
Z ∞
p−1
·
ξ 1/(p−1) Φ1/(p−1) (ξ)dξ if 1 < p ≤ 2, or
w(r) <
N −2 0
Z ∞
(p−2)N +1
p−1
w(r) <
·
ξ p−1 Φ(ξ)dξ if 2 ≤ p < +∞,
N −p 0
An upper-solution to (1.1) will be constructed. Consider the function f (u) =
(f (u) + 1)1/(p−1) , for u > 0.
Note that the hypothesis u → f (u)/(u + β)p−1 is a decreasing function on (0, ∞)
implies that u → f (u)/up−1 is a decreasing function on (0, ∞), because v+β
≤
v
u+β
⇔
vu
+
βu
≤
vu
+
vβ
⇔
β(u
−
v)
≤
0,
is
true
∀u
≤
v
and
β
>
0.
We
have
u
(F1’) f (u) ≥ f (u)1/(p−1)
8
D. P. COVEI
EJDE-2005/139
(F2’) limu&0 f (u)/u = ∞ and u 7→ f (u))/up−1 is decreasing on (0, ∞).
R v(r)
Let v be a positive function such that w(r) = C1 0 tp−1 /f (t) dt, where C
R C 1/(p−1) p−1
is a positive constant such that KC ≤ 0
t /f (t) dt. We prove that we
can find C
R x> 0 with this property. From our hypothesis (F2’) we obtain that
limx→+∞ 0 tp−1 /f (t) dt = +∞. Now using L’Hôpital’s rule we have
Z x p−1
1
t
x
lim
dt = lim
= +∞.
x→∞ xp−1 0 f (t)
x→∞ (p − 1)f (x)
Rx
This means that there exists x1 > 0 such that 0 tp−1 /f (t) dt ≥ Kxp−1 , for all
x ≥ x1 . It follows that for any C ≥ x1 ,
Z C 1/(p−1) p−1
t
KC ≤
dt.
f (t)
0
But w is a decreasing function, and this implies that v is a decreasing function too.
Then
Z v(r) p−1
Z v(0) p−1
Z C 1/(p−1) p−1
t
t
t
dt ≤
dt = C · w(0) = C · K ≤
dt.
f (t)
f (t)
f (t)
0
0
0
It follows that v(r) ≤ C 1/(p−1) for all r > 0. From w(r) → 0 as r → +∞ we deduce
v(r) → 0 as r → +∞. By the choice of v we have
∇w =
1 v p−1
·
∇v
C f (v)
and ∆w =
so
|∇w|p−2 =
1
C p−2
1 v p−1
1 v p−1 0
·
∆v +
|∇v|2 .
C f (v)
C f (v)
v p−1 p−2
|∇v|p−2 .
f (v)
It follows that
1 v p−1
v p−1 p−2
1 v p−1 0
) |∇v|p−2
∆v + (
) |∇v|2
C
C f (v)
C f (v)
f (v)
p−1
p−1
v
1
v
v p−1 0
1
)p−1 |∇v|p−2 ∆v + p−1 (
)p−2 (
) |∇v|p ,
= p−1 (
C
C
f (v)
f (v)
f (v)
|∇w|p−2 ∆w =
1
(
p−2
so
∇(|∇w|p−2 ) · ∇w
n 1
o 1 v p−1
v p−1 p−2
1 h v p−1 p−2 i0
p−2
p−2
=
(
)
∇(|∇v|
)
+
(
)
|∇v|
∇v
·[ ·
∇v]
C p−2 f (v)
C p−2 f (v)
C f (v)
1
v p−1 p−2
1 h v p−1 p−2 i0 v p−1
= p−1 (
) ∇(|∇v|p−2 ) · ∇v + p−1 (
)
|∇v|p ,
C
C
f (v)
f (v)
f (v)
so that
∆p w =
=
i
v p−1 p−1 h
p−2
p−2
|∇v|
∆v
+
∇(|∇v|
)
·
∇v
C p−1 f (v)
1
v p−1 p−2
v p−1 0
1 h v p−1 p−2 i0 v p−1
+ p−1
|∇v|p (
) + p−1
|∇v|p
C
C
f (v)
f (v)
f (v)
f (v)
1
1
C p−1
(
1
v p−1 p−2 v p−1 0
v p−1 p−1
).
) (
) ∆p v + (p − 1) p−1 |∇v|p (
C
f (v)
f (v)
f (v)
(3.6)
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
9
From (3.6) we deduce that
∆p w =
1
C p−1
(
1
v p−1 p−1
v p−1 p−2 v p−1 0
) ∆p v + (p − 1) p−1 |∇v|p (
) (
).
C
f (v)
f (v)
f (v)
From (3.7) and the fact that u →
deduce that
∆p v ≤ C p−1
f (u)
up−1
(3.7)
is a decreasing function on (0, +∞), we
f (v) p−1
f (v) p−1
∆p w = −C p−1 p−1
Φ(r) ≤ −f (v)Φ(r).
v p−1
v
(3.8)
By (3.7) and (3.8) and using in an essential manner the hypothesis (F1), as already
done for proving the uniqueness, we obtain that uk ≤ v for |x| ≤ k and, hence, for all
RN . Now we have a bounded increasing sequence u1 ≤ u2 ≤ · · · ≤ uk ≤ dots ≤ v
with v vanishing at infinity. Thus there exists a function, say u ≤ v such that
uk → u pointwise in RN . Using
Z r
h
i1/(p−1)
u0 (r) = r1−N
σ N −1 p(σ)f (u(σ))dσ
,
0
Rr
p(r)f (u(r)) + (1 − N )r−N 0 σ N −1 p(σ)f (u(σ))dσ
u00 (r) = −
p−1
Z r
2−p
h
i p−1
1−N
N −1
× r
,
σ
p(σ)f (u(σ))dσ
0
2−p
≥ 0 ⇐⇒ 1 < p ≤ 2
p−1
R r N −1
σ
p(σ)f (u(σ))dσ
=0
lim 0
N
r→0
R r N −1 r
σ
p(σ)f (u(σ))dσ
=0
lim 0
r→0
rN −1
it is easy to prove that u(r) ∈ C 2 (RN ) if 1 < p ≤ 2 because limr→∞ u00 (r) is finite
and u(r) ∈ C 1 (RN ) if 2 < p < ∞ because limr→∞ u0 (r) is finite.
4. Proof of Theorem 1.2
Suppose (1.1) has a solution u(r), then
(rN −1 |u0 (r)|p−2 u0 (r))0 = −rN −1 f (u(r))a(r),
integrating from 0 to r, we have
|u0 (r)|p−2 u0 (r) = −r1−N
Z
r
σ N −1 f (u(σ))a(σ)dσ,
0
hence u0 (r) < 0. We put ln(u(r) + 1) := u(r) > 0 for all r > 0. Then we have
∆p u(r) =
∆p u(r)
|∇u(r)|p
− (p − 1)
.
p−1
(u(r) + 1)
(u(r) + 1)p
Then u(r) satisfies
0
1 N −1
|∇u(r)|p
f (u(r))a(r)
0
p−2 0
r
(−u
(r))
u
(r)
+
(p
−
1)
=−
.
rN −1
(u(r) + 1)p
(u(r) + 1)p−1
(4.1)
10
D. P. COVEI
EJDE-2005/139
Multiplying (4.1) by rN −1 and integrating on (0, ξ) yield
ξ
Z
0
ξ
Z
ξ
0
σ N −1 |∇u(σ)|p
dσ
(u(σ) + 1)p
Z
ξ
Z
0
(−u0 (σ))p−1 σ N −1 dσ − (p − 1)
N −1
f (u(σ))a(σ)σ
(u(σ) + 1)p−1
=
0
dσ,
equivalently
0
p−1 N −1
(−u (ξ))
ξ
ξ
Z
−
0
σ N −1 |∇u(σ)|p
dσ =
(p − 1)
(u(σ) + 1)p
0
f (u(σ))a(σ)σ N −1
dσ . (4.2)
(u(σ) + 1)p−1
Multiplying equation (4.2) by ξ 1−N , we deduce
0
p−1
−ξ
(−u (ξ))
1−N
ξ
Z
σ N −1 |∇u(σ)|p
dσ = ξ 1−N
(u(σ) + 1)p
(p − 1)
0
ξ
Z
0
f (u(σ))a(σ)σ N −1
dσ.
(u(σ) + 1)p−1
(4.3)
From (4.3), we have
p−1
− u0 (ξ)
≥ ξ 1−N
Z
0
ξ
f (u(σ))a(σ)σ N −1
dσ,
(u(σ) + 1)p−1
so
−u0 (ξ) ≥ ξ
1−N
p−1
ξ
hZ
0
f (u(σ))a(σ)σ N −1 i1/(p−1)
dσ
,
(u(σ) + 1)p−1
(4.4)
integrating (4.4) on (0, r), we have
Z
u(0) − u(r) ≥
r
ξ
1−N
p−1
hZ
0
0
ξ
f (u(σ))a(σ)σ N −1 i1/(p−1)
dξ.
dσ
(u(σ) + 1)p−1
We observe that u(r) < u(0), for all r > 0 implies u(r) < u(0), for all r > 0.
f (u)
If β ≥ 1, then the function u 7→ (u+β)
p−1 is decreasing on (0, +∞). This implies
f (u(σ))
f (u(0))
>
.
(u(σ) + 1)p−1
(u(0) + 1)p−1
(4.5)
Since u is positive, we have
r
Z
ξ
1−N
p−1
0
ξ
hZ
0
f (u(σ))a(σ)σ N −1 i1/(p−1)
dσ
dξ ≤ u(0),
(u(σ) + 1)p−1
∀r > 0,
substituting (4.5) into this expression, we obtain
Z
r
ξ
0
1−N
p−1
hZ
0
ξ
a(σ)σ N −1 dσ
i1/(p−1)
dξ ≤
u(0) + 1
1
f (u(0)) p−1
u(0) < ∞.
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
11
1
Let 2 ≤ p < +∞, so 1 ≤ p − 1, follows that 1 ≥ p−1
> 0. We have
Z r
Z
hξ ξ
i1/(p−1)
1−N
ξ p−1
σ N −1 a(σ)dσ
dξ
ξ 0
0
Z r
h1 Z ξ
i1/(p−1)
1−N
σ N −1 a(σ)dσ
ξ p−1 ξ 1/(p−1)
dξ
=
ξ 0
0
Z r
Z ξ
2−N 1
N −1
≥
ξ p−1
σ p−1 a1/(p−1) (σ)dσdξ
ξ 0
0
Z ξ
Z r
N −1
2−N
σ p−1 a1/(p−1) (σ)dσdξ
ξ p−1 −1
=
0
0
Z
Z ξ
N −1
p − 1 r d 2−N
p−1
=−
σ p−1 a1/(p−1) (σ)dσdξ
ξ
N − 2 0 dξ
0
Z r
Z r
2−N
N −1
p−1
(−r p−1
σ p−1 a1/(p−1) (σ)dσ +
ξ 1/(p−1) a(ξ)1/(p−1) dξ)
=
N −2
0
Z r 0h
i
N −2
N −2
p−1 1
1/(p−1)
≥
r p−1 − (t) p−1 t
a1/(p−1) (t)dt
−2
N − 2 r Np−1
0
Z r/2
−2
−2
p − 1 1 Np−1
r Np−1
≥
r
−( )
t1/(p−1) a1/(p−1) (t)dt
−2
N − 2 r Np−1
2
0
Z r/2
1 N −2
p−1
t1/(p−1) a1/(p−1) (t)dt → ∞ as r → ∞.
(1 − ( ) p−1 )
=
N −2
2
0
So
∞>
u(0) + 1
u(0) ≥ ∞,
f (u(0))1/(p−1)
which is a contradiction.
p−1
If β < 1 then the function u 7→ (u+β)
(u+1)p−1 is increasing on (0, +∞). In this case
Z r 1−N h Z ξ
f (u(σ))a(σ)σ N −1 i1/(p−1)
u(0) ≥
ξ p−1
dξ
dσ
(u(σ) + 1)p−1
0
0
Z r 1−N h Z ξ
f (u(σ))a(σ)(u(σ) + β)p−1 σ N −1 i1/(p−1)
=
ξ p−1
dξ
dσ
(u(σ) + β)p−1 (u(σ) + 1)p−1
0
0
Z r
hZ ξ
i1/(p−1)
1−N
f (u(0))1/(p−1)
≥
β
ξ p−1
σ N −1 a(σ)dσ
dξ,
u(0) + β
0
0
which implies
u(0) + β
∞>
u(0) ≥
f (u(0))1/(p−1) β
Z
r
ξ
1−N
p−1
hZ
0
ξ
σ N −1 a(σ)dσ
i1/(p−1)
0
which is a contradiction.
5. Proof of Theorem 1.3
Assume u is positive for r > 0 and satisfies
(rN −1 |u0 (r)|p−2 u0 (r))0 = −rN −1 f (u(r))a(r).
Since f (u(r))a(r) is positive for r > 0, follows that
(rN −1 |u0 (r)|p−2 u0 (r))0 < 0,
for r > 0,
dξ ≥ ∞,
12
D. P. COVEI
EJDE-2005/139
and that rN −1 |u0 (r)|p−2 u0 (r) is a decreasing function. Because this function is
decreasing and u0 < 0,
rN −1 |u0 (r)|p−2 u0 (r) ≤ −C,
for r ≥ R,
where C is positive constant. As a consequence
1−N
−u0 (r) ≥ C1 r− p−1 ,
with C1 > 0.
Integrating this inequality from R to r we have
Z r
1−N
u(R) − u(r) ≥ C1
r− p−1 dr,
for r ≥ R.
R
Letting r → ∞, we arrive at a contradiction.
6. Proof of Theorem 1.4
As in proof of Theorem 1.2, we have
Z r 1−N h Z ξ
f (u(σ))a(σ)σ N −1 i1/(p−1)
u(0) − u(r) ≥
ξ p−1
dσ
dξ,
(u(σ) + 1)p−1
0
0
We observe that u(r) < u(0), for all r > 0 implies u(r) < u(0), for all r > 0. If
f (u)
β ≥ 1 then the function u 7→ (u+β)
p−1 is decreasing on (0, +∞). This implies
f (u(σ))
f (u(0))
>
,
p−1
(u(σ) + 1)
(u(0) + 1)p−1
(6.1)
Since u is positive, we have
Z r 1−N h Z ξ
f (u(σ))a(σ)σ N −1 i1/(p−1)
dσ
ξ p−1
dξ ≤ u(0),
(u(σ) + 1)p−1
0
0
substituting 6.1 into this expression we obtain
Z r 1−N h Z ξ
i1/(p−1)
a(σ)σ N −1 dσ
dξ ≤
ξ p−1
u(0) + 1
u(0) < ∞.
f (u(0))1/(p−1)
0
0
Let 1 < p < 2, so 0 < p − 1 < 1, it follows that 1 <
Z
∀r > 0
1
p−1
< +∞. Set
ξ
rN −1 a(r)dr < 1
for ξ > 0,
or
0
Z
ξ
rN −1 a(r)dr ≥ 1
for ξ > 0,
0
In the second case, we have
hZ ξ
i1/(p−1) Z
σ N −1 a(σ)dσ
≥
0
ξ
σ N −1 a(σ)dσ,
0
so
Z
r
ξ
0
1−N
p−1
hZ
0
ξ
σ
N −1
a(σ)dσ
i1/(p−1)
Z
dξ ≥
r
ξ
0
1−N
p−1
Z
0
ξ
σ N −1 a(σ)dσdξ .
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
13
Integration by parts gives
Z r
Z ξ
1−N
ξ p−1
σ N −1 a(σ)dσdξ
0
0
Z
Z ξ
p − 1 r d p−N
p−1
σ N −1 a(σ)dσdξ
=−
ξ
N − p 0 dξ
0
Z r
Z r
(p−2)N +1
p−N
p−1
=
σ N −1 a(σ)dσ +
ξ p−1 a(ξ)dξ)
(−r p−1
N −p
0
Z r 0h
i (p−2)N +1
N −p
N −p
p−1 1
r p−1 − (t) p−1 t p−1 p(t)dt
≥
−p
N − p r Np−1
0
Z r/2
(p−2)N +1
N −p
−p
p−1 1
r Np−1
p−1
≥
t p−1 a(t)dt
−( )
)
−p (r
N − p r Np−1
2
0
Z r/2
(p−2)N
+1
N
−p
1
p−1
(1 − ( ) p−1 )
t p−1 a(t)dt
=
N −p
2
0
= ∞ as r → ∞.
Then
∞>
u(0) + 1
u(0) ≥ ∞,
f (u(0))1/(p−1)
which is a contradiction.
If β < 1 we have u+β
u+1 > β ⇐⇒ u + β > βu + β ⇐⇒ (1 − β)u > 0 is true. In this
case we have
Z r 1−N h Z ξ
f (u(σ))a(σ)σ N −1 i1/(p−1)
u(0) ≥
dσ
ξ p−1
dξ
(u(σ) + 1)p−1
0
0
Z r 1−N h Z ξ
f (u(σ))a(σ)(u(σ) + β)p−1 σ N −1 i1/(p−1)
dσ
=
ξ p−1
dξ
(u(σ) + β)p−1 (u(σ) + 1)p−1
0
0
Z r
hZ ξ
i1/(p−1)
1−N
f (u(0))1/(p−1)
β
ξ p−1
σ N −1 a(σ)dσ
≥
dξ,
u(0) + β
0
0
which implies
∞>
u(0) + β
u(0) ≥
f (u(0))1/(p−1) β
Z
r
1−N
ξ p−1
ξ
hZ
0
σ N −1 a(σ)dσ
i1/(p−1)
dξ ≥ ∞,
0
which is a contradiction.
R ξ (p−2)N +1
In the first case we observe that we can not have 0 r p−1 a(r)dr = ∞ because
Z r
Z r
Z ξ
1−N
1−N
p−1
p−1
ξ
dξ >
ξ
σ N −1 a(σ)dσdξ
0
0
≥
0
p−1
1 N −p
(1 − ( ) p−1 )
N −p
2
Z
r/2
t
(p−2)N +1
p−1
a(t)dt → ∞ as r → ∞
0
which is a contradiction.
1
Remark 6.1. Let 2 ≤ p < +∞. Then 1 ≥ p−1
> 0. From the above proofs we
observe if that
Z ξ
1/(p−1)
σ N −1 a(σ)dσ
≤1
0
14
D. P. COVEI
EJDE-2005/139
then
Z r/2
−2
p−1
1 Np−1
(1 − ( )
)
t1/(p−1) a1/(p−1) (t)dt
N −2
2
0
Z r
hZ ξ
i1/(p−1)
1−N
σ N −1 a(σ)dσ
ξ p−1
dξ
≤
0
Z0 r
1−N
≤
ξ p−1 dξ.
0
As r → ∞, we have
R∞
0
1
t p−1 a1/(p−1) (t)dt 6= ∞.
On the other hand, if
ξ
Z
σ N −1 a(σ)dσ
1/(p−1)
> 1,
0
then
Z r/2
1 N −2
p−1
(1 − ( ) p−1 )
t1/(p−1) a1/(p−1) (t)dt
N −2
2
0
Z r
hZ ξ
i1/(p−1)
1−N
ξ p−1
σ N −1 a(σ)dσ
dξ
≤
0
p−1
≤
N −p
Then if
R∞
0
Z
0
r/2
t
(p−2)N +1
p−1
a(t)dt.
0
t1/(p−1) a1/(p−1) (t)dt = ∞ we have
R∞
0
t
(p−2)N +1
p−1
a(t)dt = ∞.
1
p−1
Remark 6.2. Let1 < p ≤ 2. Then 1 ≤
< +∞. From the above proofs we
observe that
Z r
Z
hZ ξ
i1/(p−1)
1−N
p − 1 ∞ 1/(p−1) 1/(p−1)
dξ ≤
ξ p−1
σ N −1 a(σ)dσ
t
a
(t)dt.
N −2 0
0
0
Rξ
If 0 σ N −1 a(σ)dσ ≥ 1, then
Z r/2
(p−2)N +1
−p
1 Np−1
N −1
(1 − ( )
)
t p−1 a(t)dt
N −p
2
0
Z r
hZ ξ
i1/(p−1)
1−N
≤
ξ p−1
dξ .
σ N −1 a(σ)dσ
0
Then if
R∞
0
t
(p−2)N +1
p−1
0
a(t)dt = ∞ we have
R∞
0
t1/(p−1) a1/(p−1) (t)dt = ∞.
Acknowledgments. The author thanks Professor V. Radulescu for proposing this
problem, as well as for his valuable suggestions on this subject.
References
[1] H. Brezis, L. Oswald, Remarks on sublinear elliptic equations, Nonlinear Anal., T.M.A. 10
(1986), 55-64.
[2] F. Cirstea and V. Radulescu, Existence and uniqueness of positive solutions to a semilinear
elliptic problem in RN , J. Math. Anal. Appl. 229 (1999), 417-425.
[3] M. G. Crandall, P. H. Rabinowitz and L. Tartar; On a Dirichlet problem with a singular
nonlinearity, Comm. Partial Diff. Equations 2 (1977), 193-222.
[4] R. Dalmasso, Solutions d’equations elliptiques semi-lineaires singulieres, Ann. Mat. Pura
Appl. 153 (1988), 191-201.
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QUASILINEAR ELLIPTIC PROBLEMS
15
[5] J. I. Diaz, J. E. Saa, Existence et unicite de solutions positives pour certaines equations
elliptiques quasilineaires, CRAS 305 Serie I (1987), 521-524;
[6] T. L. Dinu, Entire solutions of sublinear elliptic equations in anisotropic media, J. Math.
Anal. Appl., in press (doi:10.1016/j.jmaa.2005.09.015).
[7] A. Edelson, Entire solutions of singular elliptic equations, J. Math. Anal. Appl. 139 (1989),
523-532.
[8] J. V. Goncalves and C. A. Santos, Positive solutions for a class of quasilinear singular
equations, Journal of Differential Equations, Vol. 2004(2004), No. 56, pp 1-15.
[9] N. Hirano, N. Shioji, Existence of positive solutions for singular Dirichlet problems, Diff. Int.
Eqns., 14 (2001) 1531-1540.
[10] T. Kusano and C. A. Swanson, Entire positive solutions of singular semilinear elliptic equations, Japan J. Math. 11 (1985), 145-155.
[11] A .V. Lair and A. W. Shaker, Entire solutions of a singular semilinear elliptic problem, J.
Math. Anal. Appl. 200 (1996), 498-505.
[12] A. C. Lazer and P. J. McKenna, On a singular nonlinear elliptic boundary value problem,
Proc. Amer. Math. Soc. 111 (1991), 721-730.
[13] A. W. Shaker, On singular semilinear elliptic equations, J. Math. Anal. Appl. 173 (1993),
222-228.
[14] Z. Zhang, A remark on the existence of entire solutions of a singular semilinear elliptic
problem, J. Math. Anal. Appl. 215 (1997), 579-582.
Dragos-Patru Covei
Constantin Brâncuşi’ University of Târgu-Jiu, Bulevardul Republicii, nr. 1, 210152
Târgu-Jiu, Romania
Email: dragoscovei77@yahoo.com
Corrigendum posted on October 8, 2007
The author wants to correct some misprints and to clarify the existence and the
non-existence of solutions to the problem considered in this article.
Page 2, line 8. In the formula
Z ∞
rN −1+γ(N −2) (max a(x))dt < ∞,
|x|=t
1
N −1+γ(N −2)
N −1+γ(N −2)
replace the factor r
with t
.
Page 2, line 17. In the formula
Z ∞
rN −1+γ(N −2) (max a(x))dt < ∞,
|x|=t
1
N −1+γ(N −2)
N −1+γ(N −2)
replace the factor r
with t
.
Page 3, line 1. Replace “Problem (1.1) has no positive radial solution if p ≥ N .”
with “Suppose a(r) is a positive radial function which is continuous on RN . Then
problem (1.1) has no positive radial solution if p ≥ N .”
Page 4, line 16. Replace “bounded domain” with “smooth open bounded domain”.
Page 4, line 21. Replace “bounded domain with smooth boundary” with “smooth
open bounded domain”.
Page 5, line 25-26. Replace “[8, Theorem 1.3]” with “[8, Theorem 1.3 and lowerupper solutions method]”
Page 5, line 27. In the equation
−∆p U = p(x)h(U ),
replace p(x) with a(x).
if |x| < k,
16
D. P. COVEI
EJDE-2005/139
Page 7, line 17. In the formula
K=
p−1
·
N −2
Z
∞
1
ξ p−1 Φ1/(p−1) (ξ)dξ
0
replace the symbol “ = ” with “ ≤ ”.
Page 7, line 18. Replace
Z ∞
(p−2)N +1
p−1
ξ p−1 Φ(ξ)dξ
K=
·
N −p 0
if 2 ≤ p < +∞.
with “
K≤
p−1
·
N −p
Z
∞
ξ
(p−2)N +1
p−1
Z
Φ(ξ)dξ
0
∞
or K ≤ Const. +
1−N
ξ p−1 dξ,
1
if 2 ≤ p < +∞ and for the above considered cases.”
Page 7, line 21. Replace
Z ∞
(p−2)N +1
p−1
w(r) <
·
ξ p−1 Φ(ξ)dξ if 2 ≤ p < +∞.
N −p 0
with “
p−1
w(r) ≤
·
N −p
Z
∞
ξ
(p−2)N +1
p−1
0
Z
Φ(ξ)dξ
or w(r) ≤ Const. +
∞
1−N
ξ p−1 dξ.
1
if 2 ≤ p < +∞ and for the above considered cases.”
Page 8, line 1. “(F2’)” must be replaced by “(F2’) limu&0 f (u)/u = ∞, limu%∞ f (u)/u =
0 and u 7→ f (u)/up−1 is decreasing on (0, ∞).”
Page 9. line 10. Replace “Using” with “If u(x) is radially symmetric solution
(see [8] for conditions to a(x) and f (u(x))) then, using”
Page 9. line 15. Replace
R r N −1
σ
p(σ)f (u(σ))dσ
lim 0
=0
r→0
rN
with
Rr
σ N −1 a(σ)f (u(σ))dσ
a(0)f (u(0))
=
r→0
rN
N
Page 9. line 16. Replace p(σ) with a(σ).
Page 9. After line 18, insert “If u(x) is a weak solution, then applying the regularity theory for quasilinear elliptic equations (see for example [16] or [15, Theorem
1.3]) we find that u ∈ C 1,α (RN ).”
1−N
1−N
Page 12. line 5. Replace r− p−1 with r p−1 .
1−N
1−N
Page 12. line 7. Replace r− p−1 with r p−1 .
Page 13. line 5. Replace p(t) with a(t).
Page 14. After line 18 insert “On the other hand, if
Z ξ
1/(p−1)
σ N −1 a(σ)dσ
< 1,
lim
0
0
R∞
(p−2)N +1
p−1
a(t)dt 6= ∞ as in Case 2, Theorem 1.1.”
we observe that 0 t
Also the author wants to present an alternative proof for the existence of solutions
(see section 3. Existence of a solution).
EJDE-2005/139
QUASILINEAR ELLIPTIC PROBLEMS
17
Proof. The existence of solutions will be established by solving the approximate
problems
−∆p u = a(x)f (u + ε), if |x| < k,
(6.2)
u(x) = 0, if |x| = k,
for ε > 0 and then showing the convergence of uε as ε → +0 to a solution u. It is
clear that the problems (6.2) has a unique solution which is due to Diaz-Saà. In the
next steps we established some properties for such solution. For this, let ε := εn be
a decreasing sequence converging to 0 and set un := uεn with n > k ≥ 1 in (6.2).
By [15] we see that un ≥ c0,Bk ϕ1,Bk and there exists some function uk ∈ C(B k )
such that
((i) un → uk a.e. in Bk as n → ∞,
(ii) uk ≥ c0,k ϕ1 a.e. in Bk ,
where ϕ1 := ϕ1,Bk is the first eigenfunction for the eigenvalue λ1 of (−∆p ) in
W01,p (Bk ) and Bk := {x ∈ RN : |x|}. Moreover using Diaz-Saà’s comparison
lemma we have a sequence {uk } (which is 0 for |x| > k), as in the present paper,
such that
u1 ≤ u2 ≤ · · · ≤ uk ≤ · · · ≤ v in RN ,
where v is the same function as above and so the existence of solution u to the
problem (1.1) is proved.
This alternative proof is treated more generally in [15]. With this alternative
proof we observe that it is sufficient to apply the rest the reference [5]. This technique is inspired by [3] and by another results due to Goncalves and Santos, which
are treated more generally in [15].
References
[15] D. P. Covei, Existence and asymptotic behavior of positive solution to a quasilinear elliptic
problem in RN , Nonlinear Analysis: TMA, DOI 10.1016/j.na.2007.08.039.
[16] E. DiBenedetto, C 1,α - local regularity of weak solutions of degenerate elliptic equations,
Nonlinear Anal. 7 (1983), 827-850.