Entire Bounded Solutions for a Class of Quasilinear Elliptic Equations

Volume 2007, Article ID 16407,8pages doi:10.1155/2007/16407

Research Article

Entire Bounded Solutions for a Class of Quasilinear Elliptic Equations

Zuodong Yang and Bing Xu

Received 29 June 2006; Accepted 17 October 2006 Recommended by Shujie Li

We consider the problem−div(|∇u|^p−2∇u)=a(x)(u^m+λuⁿ),x∈R^N,N≥3, where 0<

m < p−1< n,a(x)≥0,a(x) is not identically zero. Under the condition thata(x) satisfies (H), we show that there existsλ0>0 such that the above-mentioned equation admits at least one solution for allλ∈(0,λ0). This extends the results of Laplace equation to the case ofp-Laplace equation.

Copyright © 2007 Z. Yang and B. Xu. 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.

In this work, we are interested in studying the existence of solutions to the following quasilinear equation:

−div|∇u|^p−2∇u=a(x)u^m+λuⁿ, x∈R^N,N≥3, (1)

where 0< m < p−1< n,a(x)≥0,a(x) is not identically zero. We will assume through- out the paper thata(x)∈C(R^N). Equations of the above form are mathematical models occuring in studies of thep-Laplace equation, generalized reaction-diffusion theory [1], non-Newtonian fluid theory, and the turbulent flow of a gas in porous medium [2]. In the non-Newtonian fluid theory, the quantitypis characteristic of the medium. Media with p >2 are called dilatant fluids and those with p <2 are called pseudoplastics. Ifp=2, they are Newtonian fluids.

Problem (1) for bounded domains with zero Dirichlet condition has been extensively studied (even for more general sublinear functions). We refer in particular to [3–10] (see also the references therein). Whenp=2, the related results have been obtained by [11–

16] (including bounded domains with zero Dirichlet condition orR^N). Our existence

results extend that of Brezis and Kamin (see [11, Theorem 1]) for semilinear problem, and complement results in [3–10].

u∈W^1,p(R^N)∩C¹(R^N) is called a entire weak solution to (1) if

R^N|∇u|^p−2∇u· ∇ψ dx=

R^Na(x)u^m+λuⁿψ dx ∀ψ∈C0^∞

R^N

(2) andu >0 inR^N.

Definition 1. u∈W^1,p(R^N)∩C¹(R^N) is called a supersolution to problem

div|∇u|^p−2∇u+f(x,u)=0 (3)

if

R^N|∇u|^p−2∇u· ∇ψ dx≥

R^N f(x,u)ψ dx ∀ψ∈C0^∞

R^N

(4) andu >0 inR^N. As always, a subsolutionuis defined by reversing the inequalities.

From [3], we have the following lemma.

Lemma 1. Suppose that f(x,u) is defined onR^N+1and is locally H¨older continuous (with exponentλ∈(0, 1)) inx.uis a subsolution anduis a supersolution to (3) withu≤uon R^N, and suppose that f(x,u) is locally Lipschitz continuous inuon the set

(x,u) :x∈R^N,w(x)≤u≤v(x). (5) Then, (3) possesses an entire solutionu(x) satisfying

w(x)≤u(x)≤v(x), x∈R^N. (6)

Definition 2. Say that a functiona(x)∈C(R^N), a(x)≥0, has the property (H) if the linear problem

−div|∇u|^p−2∇u=a(x), inR^N, (7) has a bounded solution.

Remark 1. Ifa(x) satisfies

H∞= _∞

0

s^1−N

_s

0t^N−1ψ(t)dt 1/(p−1)

ds <∞, (8)

whereψ(r)=max_|x|=ra(x), thena(x) has the property (H).

In fact, because

V(x)= _∞

|x|

1 s^N−1

_s

0σ^N−1ψ(σ)dσ 1/ p−1

ds (9)

which is a solution for the−div(|∇V|^p−2∇V)=ψ(r) inR^Nand lim_|x|→∞V(x)=0, soV is a supersolution for (7). On the other hand, 0 is a subsolution for (7), then (7) exists bounded entire solution.

Remark 2. IfN≥3,N > p, then condition (8) ofRemark 1is replaced by 0<

_∞

1 r^1/(p−1)ψ(r)^1/(p−1)dr <∞ if 1< p≤2, (A) 0<

_∞

1 r^{((p−2)N+1)/(p−1)}ψ(r)dr <∞ ifp≥2. (B) Let

J(r)= _r

0

t^1−N

_t

0s^N−1ψ(s)ds 1/(p−1)

dt. (10)

In fact, if 1< p≤2, by estimating the above integral, J(r)≤C1+

_r

1t^{(1−N)/(p−1) t}

0s^N−1ψ(s)ds 1/(p−1)

dt. (11)

Using the assumptionN≥3 in the computation of the first integral above and Jensen’s inequality to estimate the last one,

J(r)≤C2+C3

_r

1t^{(3−N−p)/(p−1)} _t

1s^{(N−1)/(p−1)}ψ(s)^1/(p−1)dsdt. (12) Computing the above integral, we obtain

J(r)≤C2+C4

_r

1t^1/(p−1)ψ(t)^1/(p−1)dt. (13) Applying (A) in the above integral, we infer thatH∞=lim_r→∞J(r)<∞. On the other hand, ifp≥2, set

H(t)= _t

0s^N−1ψ(s)ds (14)

and note that either H(t)≤1 fort >0 orH(t0)=1 for some t0>0. In the first case, H^1/(p−1)≤1, and hence,

J(r)= _r

0t^{(1−N)/(p−1)}H(t)^1/(p−1)dt≤C5+ _r

1t^{(1−N)/(p−1)}dt (15) so thatJ(r) has a finite limit because p < N. In the second case,H(s)^1/(p−1)≤H(s) for s≥s0and hence,

J(r)≤C6+ _r

1t^{(1−N)/(p−1)} _t

0s^N−1ψ(s)dsdt. (16)

Estimating and integrating by parts, we obtain J(r)≤C6+ p−1

N−p 1

0t^N−1ψ(t)dt + p−1

N−p

r

1t^{((p−2)N+1)/(p−1)}ψ(t)dt−r^{(p−N)/(p−1)} _r

0t^N−1ψ(t)dt

≤C7+C8

_r

1t^{((p−2)N+1)/(p−1)}ψ(t)dt.

(17)

By (B),H∞=lim_r→∞J(r)<∞. Lemma 2. Problem

−div|∇u|^p−2∇v=a(x)u^m, inR^N,N≥3, (18) has a bounded solution if and only ifa(x) satisfies (H). Moreover, there is a minimal positive solution of (18).

Proof

Sufficient condition. Let

BR=

x∈R^N:|x|< R (19)

and letu_Rbe the solution of

−div|∇u|^p−2∇u=a(x)u^m inB_R,

u=0 on∂BR. (20)

It is well known thatuRexists and is unique (see [5]). The sequenceuRis increasing with R. Indeed, letR> R. ThenuRis a supersolution for (20). We now construct a subsolution ufor (20) andu≤u_R. FromLemma 1, we will imply that there is a solutionufor (20) betweenuanduR. Since the unique solution isuR, it follows thatuR≤uRinBR. Foru, we may takeεψ1whereψ1satisfies

−div∇ψ1^p−2∇ψ1

=λ1a(x)ψ1^p−2ψ1 inB_R,

ψ1=0 on∂BR. (21)

We now prove that the sequenceu_Rremains bounded asR→ ∞. In fact,

uR≤CU (22)

for some appropriate constantC. Indeed,CUis a supersolution for the (20) since

−div∇(CU)^p−2∇(CU)=C^p−1a(x)≥a(x)(CU)^m, (23)

provided that

C^p−1−m≥ U^m_∞. (24)

Thereforeu=lim_R→∞u_Rexists anduis a solution of (18) satisfying

u≤CU. (25)

Clearly,uis the minimal solution. In fact, ifuis another solution of (18) thenuR≤uon BRby the above argument and thusu≤u.

Necessary condition. Supposeuis bounded positive solution of (18) and set v= p−1

p−1−mu^{(p−1−m)/(p−1)}. (26) Then

−div|∇v|^p−2∇v=mu^−m−1|∇u|^p+a(x)≥a(x). (27) The solutionwRof the problem

−div∇wR^p−2∇wR

=a(x), x∈BR,

w_R=0, x∈∂B_R (28)

satisfiesw_R≤v. Thusw_Rincreases asR→ ∞to a bounded solution of (7).

Theorem 1. Suppose thata(x) satisfies (H), then there exists

λ0= p−1−m

n−p+ 1E^{(p−1−n)/(p−1−m)−n}

n−p+ 1 n−m

(n−m)/(p−1₋m)

, (29)

hereE=ess sup_x∈RNe(x),e(x) is a bounded solution of (18), such that forλ∈(0,λ0), (1) has an entire bounded solution. If (1) has an entire bounded solution, then (7) has an entire bounded solution.

Proof. Firstly, we prove that there existsλ0>0 such that for all λ∈(0,λ0), (1) has a bounded solution. Sincea(x) satisfies (H), we have that

−div|∇u|^p−2∇u=a(x) (30) has a bounded solutione(x), letE=ess sup_x∈RNe(x), we consider the following function:

λ(t)=t^p−1−E^mt^m tⁿE^{n =}

1 Eⁿ

t^p−1−n−E^mt^m−n, t >0, (31)

forλ(t) first derivation, we have λ(t)= 1

Eⁿ

(p−1−n)t^p−2−n−(m−n)E^mt^m−n−1 (32)

letλ(t)=0, it follows that t0=

E^m(n−m) n−p+ 1

1/(p−1₋m)

. (33)

By simple calculation, we obtain thatt0 is maximal value point ofλ(t), it is clear that λ(t0)=λ0. Then for allλ∈[0,λ0],∃T=T(λ)>0 satisfies (T^p−1−E^mT^m)/TⁿEⁿ≥λ, it follows that for allλ∈[0,λ0], such thatT^p−1≥T^mE^m+λTⁿEⁿ,Teis a supersolution of (1), in fact

−div∇(Te)^p−2∇(Te)= −T^p−1div|∇e|^p−2∇e=T^p−1a(x)

≥a(x)T^mE^m+λTⁿEⁿ≥a(x)(Te)^m+λ(Te)ⁿ. (34) FromLemma 2, problem (18) has a positive solutionu0, thenεu0is a subsolution of (1), in fact , for allλand sufficiently small, we haveε(0< ε <1),

−div∇

ε^1/(p−1)u0^p−2∇

ε^1/(p−1)u0

= −εdiv∇u0^p−2∇u0

=εa(x)u^m0 ≤a(x)εu0

_m +λεu0

_n

. (35) Setεsufficiently small, such thatε^1/(p−1)u0< Te, then for 0< λ < λ0,ε^1/(p−1)u0< u <

Te, therefore (1) has a bounded solution.

Secondly, if (1) has a positive solution, then (3) has a positive solution. Let us define λ^∗=supλ >0|(1) has at least one bounded positive solution. (36) Apparently, 0< λ < λ^∗. Supposeuis a bounded positive solution of (1) and for allλ∈ (0,λ^∗), setv=((p−1)/(p−1−m))u^{(p−1−m)/(p−1)}. Then

−div|∇v|^p−2∇v=

p−1 p−1−m

_p−1

−div∇

u^{(p−1−m)/(p−1)p−2}∇

u^{(p−1−m)/(p−1)}

= −

p−1 p−1−m

_p−1

div

p−1−m p−1

_p−1

u^−m|∇u|^p−2∇u

= −divu^−m|∇u|^p−2∇u=mu^−m−1|∇u|^p−div|∇u|^p−2∇uu^−m

=mu^−m−1|∇u|^p+a(x)1 +λu^n−m≥a(x).

(37)

The solutionwRof the problem

−div∇wR^p−2∇wR

=a(x), x∈BR, wR=0, x∈∂BR

(38) satisfieswR≤v. ThuswRincreases asR→ ∞to a bounded solution of (3).

Acknowledgments

This project is supported by the National Natural Science Foundation of China (no.

10571022); the Natural Science Foundation of Jiangsu Province Educational Department (no. 04KJB110062; no. 06KJB110056), and the Science Foundation of Nanjing Normal University (no. 2003SXXXGQ2B37).

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Zuodong Yang: Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, Jiangsu Nanjing 210097, China

Email address:zdyang [email protected]

Bing Xu: Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University, Jiangsu Nanjing 210097, China

Email address:[email protected]