The class of problems1.1 appears in many nonlinear phenomena, for instance, in the theoryof quasiregular and quasiconformal mappings 1 3, in the generalized reaction-diffusion theory 4, i
Trang 1Volume 2009, Article ID 845946, 16 pages
doi:10.1155/2009/845946
Research Article
Entire Solutions for a Quasilinear Problem in the Presence of Sublinear and Super-Linear Terms
C A Santos
Department of Mathematics, University of Bras´ ılia, 70910–900 Bras´ılia, DF, Brazil
Correspondence should be addressed to C A Santos,csantos@unb.br
Received 31 May 2009; Revised 13 August 2009; Accepted 2 October 2009
Recommended by Wenming Zou
We establish new results concerning existence and asymptotic behavior of entire, positive, and bounded solutions which converge to zero at infinite for the quasilinear equation −Δp u
a xfu λbxgu, x ∈RN , 1 < p < N, where f, g : 0, ∞ → 0, ∞ are suitable functions and
a x, bx ≥ 0 are not identically zero continuous functions We show that there exists at least one
solution for the above-mentioned problem for each 0≤ λ < λ , for some λ > 0 Penalty arguments,
variational principles, lower-upper solutions, and an approximation procedure will be explored Copyrightq 2009 C A Santos This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
1 Introduction
In this paper we establish new results concerning existence and behavior at infinity of solutions for the nonlinear quasilinear problem
−Δp u axfu λbxgu in R N ,
whereΔp u div|∇u| p−2 ∇u, with 1 < p < N, denotes the p-Laplacian operator; a, b : R N →
0, ∞ and f, g : 0, ∞ → 0, ∞ are continuous functions not identically zero and λ ≥ 0 is a
real parameter
A solution of 1.1 is meant as a positive function u ∈ C1RN with ux → 0 as
|x| → ∞ and
RN |∇u| p−2 ∇u∇ϕdx
RN
a xfu λbxguϕ dx, ∀ϕ ∈ C∞
0
RN
Trang 2The class of problems1.1 appears in many nonlinear phenomena, for instance, in the theory
of quasiregular and quasiconformal mappings 1 3, in the generalized reaction-diffusion theory 4, in the turbulent flow of a gas in porous medium and in the non-Newtonian fluid theory 5 In the non-Newtonian fluid theory, the quantity p is the characteristic of the medium If p < 2, the fluids are called pseudoplastics; if p 2 Newtonian and if p > 2 the
fluids are called dilatants
It follows by the nonnegativity of functions a, b, f, g of parameter λ and a strong
maximum principle that all non-negative and nontrivial solutions of 1.1 must be strictly positivesee Serrin and Zou 6 So, again of 6, it follows that 1.1 admits one solution if
and only if p < N.
The main objective of this paper is to improve the principal result of Yang and Xu7 and to complement other works see, e.g., 8 20 and references therein for more general
nonlinearities in the terms f and g which include the cases considered by them.
The principal theorem in 7 considered, in problem 1.1, fu u m , u > 0, and
g u u n , u > 0 with 0 < m < p − 1 < n Another important fact is that, in our result, we
consider different coefficients, while in 7 problem 1.1 was studied with ax bx, ∀x ∈
RN
In order to establish our results some notations will be introduced We set
ar : min
|x|r a x, br : min
|x|r b x, r ≥ 0,
ar : max
|x|r a x, br : max
|x|r b x, r ≥ 0. 1.3
Additionally, we consider
H1 i lims→ 0fs/s p−1 ∞,
ii lims→ 0fs/s p−1 0,
H2 i lims→ 0gs/s p−1 0,
ii lims→ 0gs/s p−1 ∞.
Concerning the coefficients a and b,
H3 i ∞1 r 1/p−1 a 1/p−1 rdr, ∞1 r 1/p−1 b 1/p−1 rdr < ∞, if 1 < p ≤ 2,
ii ∞1 r p−2N1/p−1 ardr, ∞1r p−2N1/p−1 brdr < ∞, if p ≥ 2.
Our results will be established below under the hypothesis N≥ 3
Theorem 1.1 Consider H1–H3, then there exists one λ > 0 such that for each 0 ≤ λ < λ there exists at least one u u λ ∈ C1RN solution of problem 1.1 Moreover,
C |x| −N−p/p−1 ≤ ux, x ∈ R N , |x| ≥ 1 1.4
for some constant C Cλ > 0 If additionally
f t
t p−1 is nonincreasing and
g t
Trang 3then there is a positive constant D Dλ such that
u2xf u x
4
−1/p−1
≤ D
∞
|x|
t1−N
t
0
as bsds
1/p−1
dt, x∈ RN 1.6
Remark 1.2 If we assume1.5 with ft t m , t > 0, where 0 ≤ m < p − 1, then 1.6 becomes
0 < ux ≤ C
⎛
⎝∞
|x|
t1−N
t
0
as bsds
1/p−1
dt
⎞
⎠
1/2−m/p−1
In the sequel, we will establish some results concerning to quasilinear problems which are relevant in itself and will play a key role in the proof ofTheorem 1.1
We begin with the problem of finding classical solutions for the differential inequality
−Δp v ≥ axfv λbxgv in R N ,
Our result is
Theorem 1.3 Consider H1–H3, then there exists one λ > 0 such that problem1.8 admits, for
each 0 ≤ λ < λ , at least one radially symmetric solution v v λ ∈ C2RN \{0}∩C 1,ν
locRN , for some
ν ∈ 0, 1 Moreover, if in additionally one assumes 1.5, then there is a positive constant D Dλ
such that
v2xf v x
4
−1/p−1
≤ D
∞
|x|
t1−N
t
0
as bsds
1/p−1
dt, x∈ RN 1.9
Remark 1.4 Theorems1.1and1.3are still true with N 2 if H3 hypothesis is replaced by
H3 ∞1 t1−N t
0as bsds 1/p−1 dt < ∞.
In fact,H3 implies H3, if N ≥ 3 see sketch of the proof in the appendix
Remark 1.5 InTheorem 1.3, it is not necessary to assume that f and g are continuous up to
0 It is sufficient to know that f, g : 0, ∞ → 0, ∞ are continuous This includes terms f, g singular in 0
The next result improves one result of Goncalves and Santos21 because it guarantees
the existence of radially symmetric solutions in C2B0, R \ {0} ∩ C1B0, R ∩ CB0, R
for the problem
−Δp u ρxhu in B0, R,
Trang 4where ρ : B0, R → 0, ∞, h : 0, ∞ → 0, ∞ are continuous and suitable functions and
B 0, R ⊂ R Nis the ball inRN centered in the origin with radius R > 0.
Theorem 1.6 Assume ρx ρ|x|, x ∈ R N where ρ : 0, ∞ → 0, ∞, with ρ / 0, is
continuous Suppose that h satisfies (H1 and additionally
h s
then1.10 admits at least one radially symmetric solution u ∈ C2B0, R \ {0} ∩ C1B0, R ∩
C B0, R Besides this, ux u|x|, x ∈ B0, R, and u satisfies
ur u0 −
r
0
t1−N
t
0
s N−1 ρshusds
1/p−1
The proof of principal theoremTheorem 1.1 relies mainly on the technics of lower and upper solutions First, we will proveTheorem 1.3by defining several auxiliary functions
until we get appropriate conditions to define one positive number λ and a particular upper solution of1.1 for each 0 ≤ λ < λ
After this, we will proveTheorem 1.6, motivated by arguments in21, which will permit us to get a lower solution for1.1 Finally, we will obtain a solution of 1.1 applying the lemma below due to Yin and Yang22
Lemma 1.7 Suppose that fx, r is defined on R N1 and is locally H¨older continuous (with γ ∈
0, 1) in x Assume also that there exist functions w, v ∈ C 1,γ
locRN such that
−Δp v ≥ fx, v, x ∈ RN,
−Δp w ≤ fx, w, x ∈ RN,
w x ≤ vx, x ∈ RN,
1.13
and f x, r is locally Lipschitz continuous in r on the set
Then there exists u ∈ C1RN with wx ≤ ux ≤ vx, x ∈ RNsatisfying
RN
|∇u| p−2 ∇u∇ϕ dx
RN
f x, uϕdx, ∀ϕ ∈ C∞
0
RN
In the two next sections we will prove Theorems1.3and1.6
Trang 52 Proof of Theorem 1.4
First, inspired by Zhang20 and Santos 16, we will define functions F : 0, ∞ → 0, ∞ and G : 0, ∞ × 0, ∞ → 0, ∞ by
F s sup
t≥s
f t
t p−1 , s > 0, G τ, s
⎧
⎪
⎨
⎪
⎩
sup
s≤t≤τ
g t
t p−1 , s ≤ τ,
g τ
2.1
So, for each λ ≥ 0, let F λ:0, ∞ × 0, ∞ → 0, ∞ given by
where
F0s s p−1 F s, s > 0, F τ, s s p−1 G τ, s, τ, s > 0. 2.3
It is easy to check that
F0s ≥ fs, s > 0, for each τ > 0, F τ, s ≥ gs, 0 < s ≤ τ 2.4 and, as a consequence,
Moreover, it is also easy to verify
Lemma 2.1 Suppose that H1 and H2 hold Then, for each τ > 0,
i Fτ, s/s p−1 , s > 0 is non-increasing,
ii F0s/s p−1 , s > 0 is non-increasing,
iii lims→ 0Fτ, s/s p−1 sup0<t≤τ gt/t p−1 ,
iv lims→ 0F0s/s p−1 ∞,
v lims→ ∞Fτ, s/s p−1 gτ/τ p−1 ,
vi lims→ ∞F0s/s p−1 0.
ByLemma 2.1iii, iv, and 2.2, the function F λ:0, ∞ × 0, ∞ → 0, ∞, given by
F λ τ, s s s2
0
t/F λ τ, t 1/p−1
is well defined and continuous Again, by usingLemma 2.1i and ii,
F λ τ, s ≥ F λ τ, s 1/p−1 , ∀τ, s > 0. 2.7
Trang 6Besides this, F λ τ, · ∈ C10, ∞, for each τ > 0, and using Lemma 2.1, it follows that F λ
satisfies, for each λ≥ 0, the following
Lemma 2.2 Suppose that H1 and H2 hold Then, for each τ > 0,
i F λ τ, s/s is non-increasing in s > 0,
ii lims→ 0 F λ τ, s/s ∞,
iii lims→ 0 F λ τ, s/s λgτ/τ p−11/p−1 , if λ > 0,
iv lims→ 0 F λ τ, s/s 0, if λ 0.
And, in relation to λ, we have the folowing.
Lemma 2.3 Suppose that H1 and H2 hold Then, for each τ, s > 0,
i F λ1τ, s < F λ2τ, s, if λ1< λ2,
ii F λ τ, s/s → F0s/s, as λ → 0.
Finally, we will define, for each λ ≥ 0, H λ:0, ∞ → 0, ∞, by
H λ τ 1
τ
τ
0
t
So, H λis a continuous function and we havesee proof in the appendix
Lemma 2.4 Suppose that H1 and H2 hold Then,
i limτ→ 0H λ τ 0, for any λ ≥ 0,
ii limτ→ ∞H λ τ ∞, if λ 0,
iii limτ→ ∞H λ τ 0, if λ > 0,
iv H λ1τ, s > H λ2τ, s, if λ1< λ2,
v limλ→ 0H λ τ H0τ, for each τ > 0.
ByLemma 2.4ii, there exists a τ∞> 0 such that H0τ∞ > α 1, where by either H3
orH3, we have
0 < α :
∞
0
t1−N
t
0
as bsds
1/p−1
So, byLemma 2.4v, there exists a λ > 0 such that H λ τ∞ > α That is,
1
τ∞
τ∞
0
t
Let P : 0, ∞ × 0, τ∞ → RNby
P t, s ω t − 1
τ∞
s
0
ς
Trang 7whereω : 0, ∞ → 0, ∞, ω ∈ C20, ∞ ∩ C10, ∞ is given by ωx ω |x|, x ∈ R N
where ω ∈ C2RN \ {0} ∩ C1RN is the unique positive and radially symmetric solution of problem
−Δp ω a|x| b|x| in R N ,
More specifically, by DiBenedetto23, ω ∈ C2RN \ {0} ∩ C 1,ν
locRN , for some ν ∈ 0, 1 In
fact,ω satisfies
ω r α −
r
0
t1−N
t
0
as bsds
1/p−1
So, by2.10, 2.11, and 2.13, we have for each t > 0,
P t, 0 ω t > 0, P t, τ∞ < α − 1
τ∞
τ∞
0
t
F λ τ∞, tdt < 0. 2.14
Hence, after some pattern calculations, we show that there is a ϑ ∈ C20, ∞ ∩ C10, ∞ such that ϑr ≤ τ∞, r≥ 0 and
ω r 1
τ∞
ϑr
0
t
As consequences of2.9, 2.13 and 2.15, we have ϑr → 0, r → ∞ and
r N−1 ω rp−1 ωr 1
τ∞p−1
ϑ r
F λ τ∞, ϑ r
p−1
r N−1ϑ rp−1 ϑ r
p− 1
τ∞p−1
ϑ r
F λ τ∞, ϑ r
p−2
d ds
s
F λ τ∞, s
r N−1ϑ rp
2.16
and hence, byLemma 2.2i, 2.7 and ϑr ≤ τ∞, r ≥ 0, we obtain
−r N−1ϑ rp−1 ϑ r ≥ τ∞
ϑ r
p−1
F λ τ∞, ϑ r p−1
−r N−1 ω rp−1 ωr
r N−1 F λ τ∞, ϑ rar br,
2.17
Trang 8that is, by using2.2, we have
−r N−1ϑ rp−1 ϑ r ≥ r N−1 arF0ϑr λ r N−1 brFτ∞, ϑ r, r ≥ 0. 2.18
In particular, making vx ϑ|x|, x ∈ R N, we get from 2.15, Lemma 2.2i and ω ∈
C2RN \{0}∩C 1,ν
locRN that v ∈ C2RN \{0}∩C 1,ν
locRN and satisfies 1.8, for each 0 ≤ λ ≤ λ
That is, v is an upper solution to1.1
To prove1.9, first we observe, usingLemma 2.2i and 2.15, that
ω r ≥ 1
τ∞
ϑr/2
0
t
F λ τ∞, t1/p−1 dt≥ 1
τ∞
ϑr/2
ϑr/4
t
F λ τ∞, t1/p−1 dt
τ∞
1
F ϑr/4 λ G τ∞, ϑ r/4
1/p−1
ϑr/4, r ≥ 0.
2.19
So, by definition of F, Gτ∞,· and hypothesis 1.5, we have
4
λ G τ∞, ϑ r
4
f ϑr/4
ϑr/4 p−1 λ
g τ∞
τ∞p−1 , r ≥ 0. 2.20 Thus,
ϑr/42p−1
f ϑr/4 ≤ τ
p−1
∞
1 λ
g τ∞
τ∞p−1
ϑr/4 p−1
f ϑr/4 ωr p−1 , r ≥ 0. 2.21 Recalling that ϑr ≤ τ∞, r ≥ 0 and using 1.5 again, we obtain
ϑ r2
4
−1/p−1
≤ 16τ∞
1 λ
g τ∞
τ∞p−1
τ∞/4p−1
f τ∞/4
1/p−1
ω r, r ≥ 0. 2.22
Thus by2.9, 2.13, and vx ϑr, r |x|, for all x ∈ R N, there is one positive constant
D Dλ such that 1.9 holds This ends the proof ofTheorem 1.3
3 Proof of Theorem 1.5
To prove Theorem1.5, we will first show the existence of a solution, say u k ∈ C2B0, R \ {0} ∩ C1B0, R ∩ CB0, R, for each k 1, 2, , for the auxiliary problem
−Δp u ρxh k u in B0, R,
where h k s hs 1/k, s ≥ 0 In next, to get a solution for problem 1.10, we will use a
limit process in k.
Trang 9For this purpose, we observe that
i lim infs→ 0 h k s h1/k > 0,
ii lims→ ∞h k s/s p−1 lims→ ∞hs 1/k/s 1/k p−1 1 1/ks p−1 0, by H1 and by1.11, it follows that
iii h k s/s p−1 hs1/k/s1/k p−1 11/ks p−1 , s > 0 is non-increasing, for each
k 1, 2,
By itemsi–iii above, ρ and h k fulfill the assumptions of Theorem 1.3 in21 Thus 3.1
admits one solution u k ∈ C2B0, R \ {0} ∩ C1B0, R ∩ CB0, R, for each k 1, 2, Moreover, u k x u k |x|, x ∈ R Nwithu k ∈ C20, R ∩ C10, R ∩ C0, R satisfying
u k r u k0 −
r
0
t1−N
t
0
s N−1 ρshu k s 1/kds
1/p−1
dt, 0≤ r ≤ R. 3.2
Adapting the arguments of the proof of Theorem 1.3 in21, we show
cϕ1r ≤ u k1 r 1
k 1 ≤ u k r 1
where ϕ1∈ C2B0, R is the positive first eigenfunction of problem
−r N−1ϕp−2 ϕ
λr N−1 ρrϕp−2 ϕ in B 0, R,
and c > 0, independent of k, is chosen using H1 such that
h
c ϕ1 ∞
with λ1 > 0 denoting the first eigenvalue of problem3.4 associated to the ϕ1
Hence, by3.3,
u k r −→ ur with cϕ1r ≤ ur ≤ |u1|∞ 1, 0 ≤ r ≤ R. 3.6
UsingH1, 3.3, the above convergence and Lebesgue’s theorem, we have, making k → ∞
in3.2, that
ur u0 −
r
0
t1−N
t
0
s N−1 ρshusds
1/p−1
So, making ux u|x|, x ∈ R N , after some calculations, we obtain that u ∈ C2B0, R \ {0} ∩ C1B0, R ∩ CB0, R This completes the proof ofTheorem 1.6
Trang 104 Proof of Main Result: Theorem 1.1
To complete the proof of Theorem 1.1, we will first obtain a classical and positive lower solution for problem1.1, say w, such that w ≤ v, where v is given byTheorem 1.3 After this, the existence of a solution for the problem1.1 will be obtained applyingLemma 1.7
To get a lower solution for1.1, we will proceed with a limit process in u n , where u n
is a classical solution of problem1.10 given byTheorem 1.6 with ρ a, h is a suitable function and R n for n ≥ n0and n0is such thata / 0 in 0, n0
Let
f∞s s p−1 f∞s, s > 0, where f∞s inf
0<t≤s
f t
Thus, it is easy to check the following lemma
Lemma 4.1 Suppose that H1 and H2 hold Then,
i 0 < f∞s ≤ fs ≤ F0s λ F τ∞, s , s > 0,
ii f∞s/s p−1 , s > 0 is non-increasing,
iii lims→ 0f∞s/s p−1 ∞ and lim s→ ∞f∞s/s p−1 0.
Hence,Lemma 4.1 shows that f∞ fulfills all assumptions of Theorem 1.6 Thus, for
each n ∈ N such that n ≥ n0there exists one n ∈ C2B0, n \ {0} ∩ C1B0, n ∩ CB0, n with n x n |x|, x ∈ B0, n and nsatisfying
−r N−1| n|p−2 n
r N−1 arf∞ n r in 0 < r < n,
equivalently,
n r n0 −
r
0
t1−N
t
0
s N−1 asf∞ n sds
1/p−1
Considernextended onn, ∞ by 0 We claim that
Indeed, first we observe that f∞satisfiesLemma 4.1ii So, with similar arguments to those
of21, we show n ≤ n1 , n ≥ n0
To proven ≤ ϑ, first we will prove that n 0 ≤ ϑ0, for all n ∈ N In fact, if n 0 >
ϑ 0 for some n, then there is one T n > 0 such that
ϑ r < n r, r ∈ 0, T n , ϑ T n n T n > 0, 4.5 becausen n 0 and ϑ > 0 with ϑr → 0 as r → ∞.
... establish some results concerning to quasilinear problems which are relevant in itself and will play a key role in the proof ofTheorem 1.1We begin with the problem of finding classical solutions. .. This completes the proof of< /i>Theorem 1.6
Trang 104 Proof of Main Result: Theorem 1.1
To... motivated by arguments in 21, which will permit us to get a lower solution for 1.1 Finally, we will obtain a solution of 1.1 applying the lemma below due to Yin and Yang22
Lemma