Existence results for a quasilinear elliptic problem with a gradient term via shooting method

Applied Mathematics and Computation 218 (2011) 4161–4168

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Existence results for a quasilinear elliptic problem with a gradient term via shooting method Dragos-Patru Covei Constantin Brancusi University of Tg-Jiu, Str. Grivitei, Nr. 1, Targu-Jiu, Gorj, Romania

a r t i c l e

i n f o

Keywords: Quasilinear problem Radial solution Entire solution

a b s t r a c t The main purpose of our paper is to complete and improve a theorem of Dupaigne, Ghergu and Radulescu [Lane-Emden-Fowler equations with convection and singular potential, J. Math. Pures Appliquees (Journal de Liouville, 87(2007), 563–581).] showing the existence of solution for quasilinear elliptic equations where the nonlinearity depends on x, u and gradient term. The proofs combine O.D.E. techniques and shooting arguments. Previous developments require a monotonicity of the nonlinearity, while our main result is applied to a larger class of nonlinearities. Ó 2011 Elsevier Inc. All rights reserved.

1. Introduction The aim of this paper is to prove an existence theorem for positive radial solutions of the nonlinear elliptic problems involving both singular nonlinearity and convection (gradient) terms of this type

8 > Dp u þ bðxÞjrujp1 ¼ aðxÞf ðuÞ in RN ; > < u > 0 in RN ; > > : lim ðxÞ ¼ 0;

ð1:1Þ

jxj!1

where 1 < p < N, N P 3 and Dp :¼ r  (jr  jp2r) stands for the p-Laplacian operator, a; b : RN ! ð0; 1Þ are radial continuous while f : (0, 1) ? (0, 1) is a C1-function, singular at zero, for instance, in the sense that lims?0+f(s) = 1. In recent years equations of these types have been subject to rather deep investigations (see [10] for a detailed discussion). They are usually known in the literature as the Lane, Emden and Fowler problems. Such equations arise in different mathematical models occurring in: generalized reaction–diffusion theory, non-Newtonian filtration process, the turbulent flow of a gas in a porous medium, the equilibrium configurations of spherically symmetric gaseous stellar objects etc. We refer in particular to the books [1,3,7] and articles [4,13,15] in which explicit models are given. Notice that many works have been devoted to study this kind of problems. Our research is in part motivated by the study of the non-radially symmetric ground states solutions for the equation

Du  bðxÞjrujq ¼ aðxÞf ðuÞ in RN ; u > 0 in RN ;

lim uðxÞ ¼ 0;

jxj!1

intensively discussed in the works [5,9,10,12,14] and their references. We also quote the paper [17] where the case of a nonlinearity f that decreases on (0,1) have been considered while b(x) = 0 and the work [16] where the case b(x) = 0, p = 2 and f(s) = sk + sc (where k P 0, 0 6 c < 1) are also treated. These study obviously give more relevance to the problem of existence of radial solutions. E-mail address: [email protected] 0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.09.047

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The main objective here is to extend the existence theory developed in [11] to the more general problem (1.1). The main difficulties are that the operator is degenerate (resp. singular) for p > 2 (resp. p < 2) (see [7] for details), the nonlinearity f is singular at zero and all of these are combined with the nonlinear gradient term jrujp1. The approach we use is based on O.D.E. techniques and shooting arguments. In the present paper, we shall consider problem (1.1) for the potential functions a, b which are radially symmetric and the nonlinearity f that satisfies the following conditions

f ðsÞ sp1

is nonincreasing in ð0; 1Þ;

lim inf f ðsÞ > 0; s!0

lim

s!1

ð1:2Þ

f ðsÞ ¼ 0: sp1

ð1:3Þ

A typical model of f is

f ðsÞ ¼ sk þ sc ;

0 6 c < p  1:

where k P 0;

The main results in this article are the following theorems. Theorem 1.1. (see also [10, Theorem 3.3, pp. 575]). Let BR be the ball of radius R. Assume (1.2), (1.3). Then for each sufficiently large R,

(

Dp u þ bðxÞjrujp1 ¼ aðxÞf ðuÞ in BR ; u > 0 in BR ;

u ¼ 0 in@BR ;

ð1:4Þ

has a radially symmetric solution in CðBR Þ \ C 1 ðBR Þ \ C 2 ðBR n f0gÞ. Theorem 1.2. Assume (1.2), (1.3) and

R1 0

R1 0

1

1

rp1 aðrÞp1 dr < 1; r

ðp2ÞNþ1 p1

if 1 < p 6 2;

ð1:5Þ

aðrÞdr < 1; if 2 6 p < N:

Then there is a radially symmetric function

v 2 C 1 ðRN Þ \ C 2 ðRN n f0gÞ such that

8 < Dp v þ bðxÞjrv jp1 P aðxÞf ðv Þ in RN n f0g; N : v > 0 in R ;

lim

jxj!1

v ðxÞ ¼ 0:

ð1:6Þ

Combining Theorem 1.1 with Theorem 1.2 yields the following existence theorem for (1.1). Theorem 1.3. Assume (1.2), (1.3) and (1.5). Then (1.1) has a radially symmetric solution u in C 1 ðRN Þ \ C 2 ðRN n f0gÞ. Remark 1. Regarding Theorem 1.3, it will be shown that u 2 C 2 ðRN Þ if and only if p 6 2. Additionally, such a solution is uniquely determined if f(s)/(s + b)p1 is nonincreasing for some b > 0. See Section 3. We stress that these results are the main contributions in the literature of the problem (1.1), since radially symmetric solutions for these families of nonlinearities have not been detected before, except for the monotone class of functions considered in the article [10]. 2. Proof of the main theorems 2.1. A sketch of the proof of the Theorem 1.1 The proof of this theorem is analogous to that of Theorem 1.3 in [11]. Since we just consider radially symmetric solutions of (1.4) we will in fact study the following singular initial value problem where we have set r = jxj,

 0 Rr 1 bðtÞdt 0 p2 0 Rr r N1 e 0 ju j u ¼ aðrÞf ðuÞ in ð0; 1Þ; bðtÞdt r N1 e 0 uð0Þ ¼ d > 0; u0 ð0Þ ¼ 0: 

ð2:1Þ

Rr bðtÞdt Assuming that u0 is non-positive, multiplying Eq. (2.1) by r N1 e 0 and integrating it with the initial conditions, we find that problem (2.1) may be equivalently written in terms of the nonlinear integral equation

uðrÞ ¼ d 

Z r 0

t1N e



Rt 0

bðzÞdz

Z

t 0

1 p1 Rs bðtÞdt sN1 e 0 aðsÞf ðuðsÞÞds dt:

ð2:2Þ

D.-P. Covei / Applied Mathematics and Computation 218 (2011) 4161–4168

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Our idea is to regard this as an operator equation Su(r) = u(r) with S defined by

SuðrÞ ¼ d 

Z r

t 1N e



Rt 0

bðzÞdz

0

Z

t

1 p1 Rs bðtÞdt sN1 e 0 aðsÞf ðuðsÞÞds dt;

ð2:3Þ

0

and so a solution of (2.1) will be obtained as a fixed point of the operator (2.3). To show this we give an inequality that will be used all over in the paper and is motivated by Goncalves and Santos [11] and Diaz-Saà [8]. Given T, h > 0 we consider the set

X :¼ fw 2 C 1 ð½0; TÞjw P hg; and for w1, w2 2 X let H : ½0; T ! R be the continuous function

HðsÞ :¼ sN1 e

Rs 0

bðzÞdz

        1=p 0 p2 1=p 0 1p  1=p 0 p2 1=p 0 1p w1 w1p ðw1  w2 ÞðsÞ:  w2  ðw2 Þ w2p   w1 

All of these give the following expected inequality, which is the key tool of the proofs. Lemma 2.1. If w1, w2 2 X and 0 6 S 6 U 6 T, then

HðUÞ  HðSÞ 6

Z S

2

U

N1 6 r e 6 6 4

Rr 0

0  0 p2 1=p  1=p 0  w2  ðw2 Þ

bðzÞdz 

p1

w2p

 0 3 Rr  0 p2 bðzÞdz  1=p  1=p 0 r N1 e 0  w1  ðw1 Þ 7 7  7ðw1  w2 Þdr: p1 5 w1p

ð2:4Þ

Sketch of proof. Motivated by [11], [2, Proposition 1] let J : L1 ð½0; TÞ ! R [ f1g,

JðwÞ :¼

( R 1 U p

S

Rs bðzÞdz sN1 e 0 jðw1=p Þ0 jp ds; w 2 X

1;

w R X;

where 0 6 S 6 U 6 T. A standard straightforward computation shows that X and J are both convex. Since J is convex, hJ0 (w1)  J0 (w2), w1  w2i P 0 and Lemma 2.1 follows. The two lemmas below are about solving (2.1) and exploring properties of its solutions regarding distinct values of d. Lemma 2.2. Let assumption (1.2) be satisfied. For any number d > 0 there is some T(d) 2 (0, 1] and a unique solution, u :¼ u(, d) 2 C1([0, T(d))) \ C2((0, T(d))) of problem (2.1) which possesses the following properties (i) u(r) ? 0 as r ? T(a) provided T(a) < 1; (ii) u(, d1) < u(, d2) in [0, T(d1)) if d1 < d2 and moreover T(d1) < T(d2). Sketch of proof. (i) First, it is proved the existence of a unique local solution of problem (2.1) on C([0, e]). This is fairly standard and is based on the Banach contraction mapping principle. Now, letting

TðdÞ :¼ supfr > 0jð2:1Þ has an only solution in ½0; rg; and repeat the proof in [11] to see that the function u 2 C([0, T(d)]) \ C1([0, T(d))) \ C2((0, T(d))) for which u(, T(d)) = 0 is the desired solution. (ii) Assume, on the contrary u(r, d1) < u(r, d2) for r 2 [0, T) and u(T, d1) = u(T, d2) for some T < T(d1). Lemma 2.1 with w1 = up(r, d1) and w2 = up(r, d2) implies the following

" # ju0 ðr; d1 Þjp2 u0 ðr; d1 Þ ju0 ðr; d2 Þjp2 u0 ðr; d2 Þ  6 0 for r 2 ð0; TÞ: uðr; d1 Þ uðr; d2 Þ Hence

uðr;d2 Þ uðr;d1 Þ

1<

is nondecreasing for r 2 [0, T], and so

uð0; d2 Þ uðT; d2 Þ 6 ¼ 1; uð0; d1 Þ uðT; d1 Þ

is a contradiction. Thus u(r, d1) < u(r, d2) for r 2 [0, T(d1)) which gives T(d1) < T(d2). This finishes the proof of Lemma 2.2. Proceeding as in the proof of Lemma 2.4 in [11] we can prove.

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Lemma 2.3. Suppose assumption (1.2) is satisfied. Let {an} be a sequence in (0, 1) such that an % a or an & a for some a > 0 and let u(, an), u(, a) be the solutions given by Lemma 2.2. If K 2 (0, min{T(a), supnT(an)}) then n!1

(i) uð; an Þ ! uð; aÞ in C([0, K]); n!1 (ii) u0 ðr; an Þ ! u0 ðr; aÞ pointwise in [0, K]. The continuation of the Proof of Theorem 1.1. Consider the set C :¼ {d > 0jT(d) P R} and let D :¼ inf C > 0. Using Lemmas 2.2 and 2.3 it is sufficient to apply the technique used in the proof of Theorem 1.3 of [11] to see that the solution u(r, D) corresponding to the problem (2.1) satisfies u(r, D) > 0 for 0 6 r < T(D) and u(T(D), D) = 0, where is proved easily and that the following useful equality holds T(D) = R. In that case, u(x) :¼ u(jxj) is the desired solution of (1.4). 2.2. Proof of Theorem 1.2 Let

K ¼:

Z

 Z n1N

1 0

n

rN1 aðrÞdr

1 p1

dn:

0

We can easy see that the spherically symmetric function

Z r Z wðrÞ ¼ K  n1N 0

n N1

r

aðrÞdr

1 p1

dn;

ð2:5Þ

0

is a solution to the initial value problem

8   <  r N1 jwjp2 w0 0 ¼ rN1 aðrÞ in ð0; 1Þ; : wð0Þ ¼ K; w0 ð0Þ ¼ 0; w > 0 in ½0; 1Þ:

ð2:6Þ

We wish to stress that

Z

wðrÞ :¼

1



n1N

Z

r

1 p1

n

rN1 aðrÞdr

dn;

ð2:7Þ

0

is bounded. For this it is convenient to distinguish the following two cases: 1 < p 6 2 and 2 6 p < N. The case where 1 < p 6 2. Since in this case

16

1 < 1; p1

by the Hölder inequality for integrals, we have

wðrÞ ¼

Z

1

1N

1

n p1 np1 r

 Z n 1=ðp1Þ Z n  Z 1 2N 1 N1 1 1 rN1 aðrÞdr dn 6 n p1 r p1 ap1 ðrÞdr dn: n 0 n 0 r

Using an integration by parts in the left hand side and L’ Hôpital rule, we have

Z

1

2N

n p1 r

1 n

Z

n

N1



1

 Z n N1 1 d  2N n p1 r p1 ap1 ðrÞdr dn dn r 0 Z R

Z R Z r 1 2N N1 1 2N N1 1 p1 ¼ lim ½naðnÞp1 dn  R p1 r p1 ap1 ðrÞdr þ r p1 r p1 ap1 ðrÞdr N  2 R!1 r 0 0 h i R N1 1 N2 R R 1 1 2N R r N1 1 R p1 p1 ap1 ðnÞdn þ r p1 p1 ap1 ðnÞdn  R n n n p1 ap1 ðnÞdn r 0 0 p1 lim ¼ N2 N  2 R!1 R p1 Z 1  Z r 1 1 2N N1 1 p1 ¼ np1 ap1 ðnÞdn þ r p1 n p1 ap1 ðnÞdn ; R > r: N2 r 0

r p1 ap1 ðrÞdr dn ¼ 

0

p1 N2

Z

1

ð2:8Þ

Now, by the second mean value theorem for integrals, there is r1 2 (0, r) such that

Z

r

N1

1

n p1 ap1 ðnÞdn ¼

0

Z

r

N2

1

1

0

since N > 2. From (2.8)–(2.9) it follows finally

wðrÞ 6

p1 N2

Z 0

1

N2

n p1 np1 ap1 ðnÞdn ¼ r p1

n1=ðp1Þ a1=ðp1Þ ðnÞdn:

Z

r

r1

1

1

N2

np1 ap1 ðnÞdn 6 r p1

Z 0

r

1

1

np1 ap1 ðnÞdn

ð2:9Þ

D.-P. Covei / Applied Mathematics and Computation 218 (2011) 4161–4168

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1 The case where 2 6 p < N. In this case 0 < p1 6 1 if

Z

n

rN1 aðrÞdr < 1 for n > 0;

0

then

Z

R



1N

Z

r

n r

1 p1

n N1

aðrÞdr

dn <

Z

0

R

 pN p  1  pN R p1  r p1 : pN

1N

n p1 dn ¼

r

ð2:10Þ

Using (2.10), we have

wðrÞ 6

 pN  p  1 pN pN p1 lim R p1  r p1 ¼ r p1 ; p  N R!1 Np

since p < N. Rn If there exists n0 > 0 such that 0 0 rN1 aðrÞdr ¼ 1 then

Z

n

r

N1

aðrÞdr

1 p1

Z

n

rN1 aðrÞdr for all n P n0

6

0

0

which implies

Z

R

1N

Z

r

1 p1

n

rN1 aðrÞdr

n p1

dn 6

0

Z

R

1N

Z

n p1 r

n

rN1 aðrÞdrdn:

ð2:11Þ

0

Moreover, integrating by parts

Z

R

1N

Z

r

Z Z n p  1 1 d  2N n p1 rN1 aðrÞdrdn N  p r dn 0 Z R  Z R Z r Nðp2Þþ1 pN pN p1 ¼ n p1 aðnÞdn  R p1 rN1 aðrÞdr þ r p1 rN1 aðrÞdr Np r 0 0 h i R Np R pN R R Nðp2Þþ1 r N1 R N1 p1 n p1 aðnÞdn þ r p1 0 r aðrÞdr  0 r aðrÞdr r p1 R ¼ : Np Np R p1

n

rN1 aðrÞdrdn ¼ 

n p1

0

ð2:12Þ

From (2.11), (2.12) it follows, by the L’Hopital rule

h i R Np R pN R R Nðp2Þþ1 r R R p1 r n p1 aðnÞdn þ r p1 0 rN1 aðrÞdr  0 rN1 aðrÞdr p1 lim wðrÞ ¼ n r aðrÞdr dn 6 Np N  p R!1 r 0 R p1 Z R  Z r Nðp2Þþ1 pN p1 lim ¼ n p1 aðnÞdn þ r p1 rN1 aðrÞdr : N  p R!1 r 0 Z

1

1N p1

Z

1 p1

n

N1

Again, by the second mean value theorem for integrals, there is r2 2 (0, r) such that

Z

r

nN1 aðnÞdn ¼

0

Z

r

Np

Nðp2Þþ1 p1

np1 n

Np

aðnÞdn ¼ r p1

0

Z

r

Nðp2Þþ1 p1

n

Np

aðnÞdn 6 r p1

Z

r2

r

n

Nðp2Þþ1 p1

aðnÞdn;

0

and so w is bounded. Clearly w(r) ? 0 as r ? 1. We want to prove now the existence of an upper-solution to (1.1). Consider the function 1 ~f p ðtÞ :¼ ðf ðtÞ þ 1Þp1 ;

t > 0;

ð2:13Þ

and note that the items below hold true, (i) 1 ~f ðtÞ P f ðtÞp1 > 0; ðiiÞ p

~f ðtÞ p t

is decreasing; ðiiiÞ

lim

t!1

~f ðtÞ p ¼ 0: t

ð2:14Þ

Now set, for s > 0

F p ðsÞ :¼

1 Cp

Z 0

s

t dt; ~f p ðtÞ

where Cp is a positive constant which will be chosen later. Clearly, Fp(0) = 0 and Fp is increasing. Using (2.14)(iii) it follows s!1 that, F p ðsÞ ! 1. Applying the Implicit Function Theorem,

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D.-P. Covei / Applied Mathematics and Computation 218 (2011) 4161–4168

Z v ðrÞ

1 Cp

wðrÞ :¼

t dt; ~f p ðtÞ

0

ð2:15Þ

for some C2((0, 1)) \ C1([0, 1))-function v. On the other hand using (2.14)(ii) we obtain

Z v ðrÞ

t ~f p ðtÞ

0

dt 6

Z v ð0Þ 0

t dt ¼ C p wð0Þ ¼ C p K < 1; ~f p ðtÞ

we deduce that there exists a positive constant M, such that v(r) 6 v(0) 6 M. Finally, since ~f p ðtÞ > 1 we have

Z v ð0Þ

Cp K ¼

t ~f ðtÞ p

0

dt 6

Z v ð0Þ

tdt ¼

0

ðr

0

jw ðrÞj

p2



1 w ðrÞÞ ¼ Cp 0

0

p1

s!1

If not, this contradicts with FðsÞ ! 1, so

1 1 v ð0Þ2 6 M2 ; 2 2

and then Cp :¼ max{M, 2K}. We want to prove that respect to r we have N1

v(0) 6 M.

v

is an upper-solution to (1.1). Indeed, differentiating in (2.15) with

!p1

v ~f p ðv Þ

ðrN1 jv 0 ðrÞjp2 v 0 ðrÞÞ0

!p2 !!  p1 1 v d v rN1 jv 0 jp : þ ðp  1Þ ~f p ðv Þ Cp dv ~f p ðv Þ

ð2:16Þ

Now, using (2.14)(iii), (2.16) and (2.14)(i), it follows that

 p1 Cp ~f p ðv Þp1 rN1 aðrÞ 6 r N1 aðrÞf ðv ðrÞÞ: ðr N1 jv 0 ðrÞjp2 v 0 ðrÞÞ0 6 

v

Remarking that by (2.15) v0 (0) = 0 and limr?1v(r) = 0 it follows that Theorem 1.2 is now complete.

v is a radially symmetric solution of (1.6). The proof of

2.3. Proof of Theorem 1.3 We take an integer j sufficiently large such that (1.4) with R = j + k has, by Theorem 1.1, a radially symmetric solution, say uk 2 C1([0, j + k)) \ C([0, j + k]) for each integer k P 1. Consider uk(r) extended on [j + k, 1) by 0, i. e. uk(r) = 0 for all r P j + k. We claim that,

0 6 u1 6 u2 6    6 uk 6    6 v :

ð2:17Þ

We will show first that uk 6 uk+1. Indeed, we claim that uk(0) 6 uk+1(0). Otherwise, both uk(r) > uk+1(r) for r 2 [0, T) and uk(T) = uk+1(T) for some T 2 (0, j + k). Taking r 2 (0, T) and using Lemma 2.1 and (1.2) we have

r

N1

Rr

"  u0 p2 u0 kþ1 kþ1

 0 p2 0 # u  u 

 k p1 k upk  upkþ1 up1 u kþ1 k 0  0 3 2 Rt Rt    p2 0 qðzÞdz qðzÞdz  0 N1 N1 u0 ðtÞp2 u0 ðtÞ  u ðtÞ 0 0 Z r t e t e u ðtÞ kþ1 kþ1 k k 6 7 p 6 7 u  up dt 6  kþ1 4 5 k p1 p1 ukþ1 uk 0

e

0

¼

qðzÞdz

Z

r

0

" # Rt f ðuk Þ f ðukþ1 Þ  p qðzÞdz uk  upkþ1 dt 6 0: tN1 e 0 aðtÞ p1  p1 uk ukþ1

As a consequence,

 0 p2 0 u  u k k up1 k which gives,

1<



uk ukþ1

 0 p2 0 u  u kþ1 kþ1 up1 kþ1

P 0;

is nondecreasing in (0, T), and then

uk ð0Þ uk ðTÞ 6 ¼ 1; ukþ1 ð0Þ ukþ1 ðTÞ

but this is a contradiction and so uk(0) 6 uk+1(0). To end the proof, we will suppose that there exist an k and r > 0 such that uk(r) > uk+1(r). Hence there are S, U 2 (0, j + k) with r 2 (S, U), such that uk(S) = uk+1(S), uk(U) = uk+1(U) and uk(r) > uk+1(r) for all r 2 (S, U).

D.-P. Covei / Applied Mathematics and Computation 218 (2011) 4161–4168

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Following the same above arguments, we find,

1¼

uk ðSÞ uk ðrÞ uk ðUÞ 6 6 ¼ 1; ukþ1 ðSÞ ukþ1 ðrÞ ukþ1 ðUÞ

r 2 ½S; U;

so that, uk(r) = uk+1(r), r 2 [S, U] which is impossible again. This shows that uk 6 uk+1. To complete to proof of (2.17), it remains to show that uk 6 v. Reasoning as in the proof of uk 6 uk+1 we get the desired conclusion. The proof of (2.17) is complete. Setting

lim uk ðrÞ ¼ uðrÞ;

r P 0;

k!1

it follows by the above arguments that

0 < uðrÞ 6 v ðrÞ for all r P 0: By the proof of Theorem 1.1,

uk ðrÞ ¼ uk ð0Þ 

Z r



s1N e

Rs 0

bðzÞdz

Z

0

s

1 p1 Rt bðzÞdz t N1 e 0 aðtÞf ðuk ðtÞÞdt ds;

r P 0:

ð2:18Þ

0

Set r > 0, pick k0 such that j + k0 P r + 1 and notice that by (2.17), uk P uk0 for k P k0. Recalling that u0k and v0 are nonpositive and using (1.2) and (2.17),

tN1 e

Rt 0

bðzÞdz

f ðuk0 ðsÞÞ

aðtÞf ðuk ðtÞÞ 6 v ð0Þp1

uk0 ðsÞ

p1

Rt bðzÞdz t N1 e 0 aðtÞ;

t 2 ½0; s:

We remark that the last function above belongs to L1((0, s)). According Lebegue’s theorem, we deduce

Z

s

Z s Rt Rt bðzÞdz bðzÞdz t N1 e 0 aðtÞf ðuk ðtÞÞdt ! tN1 e 0 aðtÞf ðuðtÞÞdt;

0

s 2 ½0; r;

0

and employing, once more, arguments as above, we obtain

Z r 0 k!1

!

s1N e

Z r



Rs 0

Z

bðzÞdz

s

tN1 e

0

s1N e



Rs 0

bðzÞdz

Z

0

Rt 0

bðzÞdz

1 p1

aðtÞf ðuk ðtÞÞdt

ds

1 p1 Rt bðzÞdz tN1 e 0 aðtÞf ðuðtÞÞdt ds:

s

0

Passing to the limit in (2.18) we infer that,

uðrÞ ¼ uð0Þ 

Z r



s1N e

Rs 0

bðzÞdz

Z

0

s

1 p1 Rt bðzÞdz t N1 e 0 aðtÞf ðuðtÞÞdt ds:

0

Since 1  p1 Z r Rr Rt  bðzÞdz bðzÞdz u0 ðrÞ ¼  r 1N e 0 tN1 e 0 aðtÞf ðuðtÞÞdt ds:

0

u00 ðrÞ ¼ 

 2p Z R R p1 hðrÞ 1N  r bðzÞdz r N1 t bðzÞdz r e 0 t e 0 aðtÞf ðuðtÞÞdt ; p1 0

ð2:19Þ

where

 Z r Rr Rr Rt  bðzÞdz  bðzÞdz bðzÞdz hðrÞ :¼ aðrÞf ðuðrÞÞ þ ð1  NÞr N e 0  r1N bðrÞe 0 t N1 e 0 aðtÞf ðuðtÞÞdt;

ð2:20Þ

0

we obtain that u 2 C1([0, 1)) \ C2((0, 1)). This together with the fact that u 6 v, showed that u is radially symetric solution of (1.1), thereby completing the proof of Theorem 1.3. 3. Comments on remark We regard to Remark 1. By (2.20), we get

lim hðrÞ ¼ að0Þf ðuð0ÞÞ þ ð1  NÞ r!0

að0Þf ðuð0ÞÞ að0Þf ðuð0ÞÞ ¼ : N N

4168

D.-P. Covei / Applied Mathematics and Computation 218 (2011) 4161–4168

On the other hand,

  Z r Rt Rr  bðzÞdz bðzÞdz lim r 1N e 0 t N1 e 0 aðtÞf ðuðtÞÞdt ¼ 0; r!0 0   Z r Rr Rt  bðzÞdz bðzÞdz 1N 0 lim r bðrÞe tN1 e 0 aðtÞf ðuðtÞÞdt ¼ 0: r!0

0

Hence, by (2.19) limr?0u00 (r) exists if and only if p 6 2, that is u 2 C2([0, 1)) if and only if p 6 2. Now let u, v be such solutions of (1.1). By Lemma 2.2 we can assume u P v. Let w1 :¼ (v + b)p and w2 :¼ (u + b)p and notice that w1, w2 2 X. Taking r > 0 and using Lemma 2.1 and (1.2), as in the proof of Lemma 2.2, we find,

ju0 jp2 u0 p1

ðu þ bÞ



jv 0 jp2 v 0 ðv þ bÞp1

P 0;

and since u0 , v0 6 0, we infer that, vuþb is nondecreasing in (0, 1), so that, þb

Z r



t1N e

Rt 0

bðzÞdz

Z

0

t

sN1 e

Rs 0

bðzÞdz

1 p1

aðsÞf ðuðsÞÞds

0

dt 6

uðrÞ þ b v ðrÞ þ b

1 p1 Z t Z r Rs Rt  bðzÞdz bðzÞdz t 1N e 0 sN1 e 0 aðsÞf ðv ðsÞÞds dt

0

0

ð3:1Þ By (3.1), the above inequality and the fact that limr?1u(r) = limr?1u(r) = 0, we find

Rr 16



0 uð0Þ ¼ lim  v ð0Þ r!1 R r 0



t1N e

t

1N 

e

Rt 0

Rt 0

bðzÞdz

Rt

bðzÞdz

Rt

sN1 e 0

0

sN1 e

Rs 0

Rs 0

bðzÞdz

bðzÞdz

1 p1 aðsÞf ðuðsÞÞds dt

aðsÞf ðv ðsÞÞds

1 p1

6 1; dt

so that, by Lemma 2.2, u = v. This proves Remark 1. 4. Open problem It seems interesting to ask if the reader can obtain an analogous result for the problem with different gradient power

8 < Dp u  bðxÞjrujq ¼ aðxÞf ðuÞ in RN ; N : u > 0 in R ;

lim uðxÞ ¼ 0;

ð4:1Þ

jxj!1

where q 2 (0, p], 1 < p < N, N P 3 while b; a : RN ! ½0; 1Þ are continuous radially symmetric functions satisfying (1.5) and f : (0, 1) ? (0, 1) is a C1-function, singular at zero satisfying (1.2), (1.3). In the particular case q = p  1, we have the above main results. Even so, for such a result to be established for the problem (4.1), the proof would require significant innovation. The author consider that the reader can be start in the proving such a result with the article [6]. References [1] R. Aris, The mathematical theory of diffusion and reaction in catalysts, Volumes I (The Theory of the Steady State 444 pp.) and II (Questions of Uniqueness, Stability, and Transient Behavior 217 pp.), Clarendon Press, Oxford, 1975. [2] H. Brezis, Equations et inequations non lineaires dans les espaces vectoriels en dualite, (French) Ann. Inst. Fourier (Grenoble) 18 (fasc. 1) (1968) 115– 175. [3] S. Chandrasekhar, An Introduction to the Study of Stellar Structure, Dover, New York, 1957, corrected republication of original (1939) edition, ISBN: 0486604136 v. [4] S. Chandrasekhar, On stars, their evolution and their stability, Nobel lecture (1983). [5] D.P. Covei, Existence and uniqueness of solutions for the Lane, Emden and Fowler type problem, Nonlinear Anal. 72 (5) (2010) 2684–2693. [6] M. Chipot, F.B. Weissler, Some blowup results for a nonlinear parabolic problem with a gradient term, SIAM J. Math. Anal. 20 (4) (1989) 886–907. [7] J.I. Diaz, Nonlinear partial differential equations and free boundaries, Pitman Research Notes in Mathematics 106 (1985). [8] J.I. Díaz, J.E. Saa, Existence et unicité de solutions positives pour certaines equations elliptiques quasilinéaires, C. R. Acad. Sci. Paris Sér. I Math. 305 (12) (1987) 521–524. [9] T.-L. Dinu, Entire positive solutions of the singular Emden–Fowler equation with nonlinear gradient term, Results Math. 43 (1–2) (2003) 96–100. [10] L. Dupaigne, M. Ghergu, V. Radulescu, Lane-Emden-Fowler equations with convection and singular potential, J. Math. Pures Appl. 87 (6) (2007) 563– 581. [11] J.V. Goncalves, C.A.P. Santos, Positive solutions for a class of quasilinear singular equations, Electron. J. Differential Equations (56) (2004) 15 (online). [12] J.V. Goncalves, F.K. Silva, Existence and nonexistence of ground state solutions for elliptic equations with a convection term, Nonlinear Anal. 72 (2) (2010) 904–915. [13] E. Momoniat, C. Harley, An implicit series solution for a boundary value problem modelling a thermal explosion, Math. Comput. Modelling 53 (1–2) (2011) 249–260. [14] H. Xue, X. Shao, Existence of positive entire solutions of a semilinear elliptic problem with a gradient term, Nonlinear Anal. 71 (7–8) (2009) 3113–3118. [15] P.J.E. Peebles, Star distribution near a collapsed object, Astrophys. J. 178 (1972) 371–376. [16] S. Yijing, L. Shujie, Structure of ground state solutions of singular semilinear elliptic equations, Nonlinear Anal. 55 (4) (2003) 399–417. [17] J. Yuan, Z. Yang, Existence and asymptotic behavior of radially symmetric ground states of quasi-linear singular elliptic equations, Appl. Math. Comput. 216 (1) (2010) 213–220.