Abstract: - In this paper, we are interested in the singular positive solutions of the inhomogeneous radial equation

|u0|p−2u00(r) + N−1

r|u0|p−2u0(r) + uq(r) + f(r) = 0, r > 0,

where N≥1,p > 2,q > 1and fis a continuous radial and strictly positive function on (0,+∞).

More precisely, we study the solutions uthat cannot be extended by continuity at zero, that is, lim

r→0u(r) = +∞.

We give existence and nonexistence results and we describe the behavior of entire solutions near infinity. The

study needs some assumptions on p,q,Nand explicit conditions on the inhomogeneous term f.

Key-Words: - Inhomogeneous elliptic equation; entire solutions; strictly positive solutions; energy function;

asymptotic behavior near infinity.

1 Introduction

The purpose of this paper is to study the following

equation

(|u0|p−2u0)0+N−1

r|u0|p−2u0+uq+f(r) = 0, r > 0,

(1)

where N≥1,p > 2,q > 1and fis a continuous

radial and strictly positive function on (0,+∞).

We are interested in the singular positive solutions

of (1) that satisfy lim

r→0u(r)=+∞. The study

is a continuation of the work carried out by [6],

where the authors proved the existence of a maxi-

mal solution udefined on ]0, rmax[such that u∈

C0(]0, rmax[) ∩C1(]0, rmax[) and |u0|p−2u00∈

C1(]0, rmax[) where 0< rmax ≤+∞and satisfy-

ing

(P)((|u0|p−2u0)0+N−1

r|u0|p−2u0+uq+f= 0, r > 0,

lim

r→0u(r) = +∞,lim

r→0r(N−1)/(p−1)u0(r) = 0,

where N≥1,p > 2,q > 1and fis a strictly pos-

itive, continuous radial function on (0,+∞). They

presented also the behavior of singular solutions near

the origin. In this work, we give the existence of en-

tire solutions of problem (P), present their behavior

near infinity, and prove the nonexistence results.

Equation (1) can be considered as a natural gener-

alization of pure Laplacian case p= 2 studied in the

papers, [2], [3]. It is presented as follows

u00(r)+N−1

ru0(r)+uq(r)+f(r) = 0, r > 0,(2)

where p > 1,N≥3and fis a strictly positive, con-

tinuous radial function on (0,+∞). Equation (2) ap-

peared in probability theory in the study of stochas-

tic processes. It plays a central role in establish-

ing some limit theorems for super-Brownian motion,

[15]. Therefore, it has been extensively studied in

much literature. [3], studied the existence and nonex-

istence of solutions of equation (2). He proved that

if q≤N

N−2or if q > N

N−2and f≥Cr

q−1for

some constant C > 0, equation (2) does not have any

solution. But when q > N

N−2and fis dominated by

a function of the form C/(1+r)N−2

qfor C > 0suffi-

ciently small, equation (2) has a solution. [2], studied

the existence of global positive solutions of equation

(2), and in the paper, [1], presented the asymptotic

behavior near the origin and infinity of positive radial

solutions. We also refer the readers to see, [14], [8],

[17] for more details about the equation (2) and the

references therein.

When f≡0, equation (2) becomes the classic

Emden-Fowler equation. In [9], [10], [11], Emden-

Fowler gave the existence results and a classification

of global radial solutions on RNand RN\{0}. In the

Behavior of Entire Solutions of a Nonlinear Elliptic Equation with An

Inhomogeneous Singular Term

ARIJ BOUZELMATE, HIKMAT EL BAGHOURI

LaR2A Laboratory, Department of Mathematics,

Faculty of Sciences, Abdelmalek Essaadi University

Tetouan, MOROCCO

Received: July 26, 2022. Revised: February 21, 2023. Accepted: March 17, 2023. Published: April 13, 2023.

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case N > 2, two critical values N

N−2and N+ 2

N−2

appear. [12], presented local and global results in the

non-radial case when q < N+ 2

N−2. [7], have just ex-

tended them to the critical case q=N+ 2

N−2.

The motivation to study the equation (1) grew

from earlier work of [16], in case f≡0and p > 2

for the equation

|u0|p−2u00+N−1

r|u0|p−2u0+uq(r) = 0, r > 0.

(3)

They have shown the existence of two critical val-

ues N(p−1)

N−pand N(p−1) + p

N−p. [13], studied

the existence of global solutions and asymptotic be-

havior near the origin of radial solutions when q <

N(p−1)

N−p. The non-radial case was proved by [5].

In this paper, we shall further extend the analysis

of the equation (3) by adding an inhomogeneous sin-

gular term fwhich has an impact on the existence and

the asymptotic behavior of solutions of equation (1).

We show under some assumptions that if the term f

behaves like r−pq/(q+1−p)near infinity, then the solu-

tion uof problem (P)behaves like r−p/(q+1−p)near

infinity. In particular, we have the following results

in the case N > p and q > N(p−1)

N−p,

(i)If

f(r) = q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

r−pq/(q+1−p),

then problem (P)possesses a positive explicit solu-

tion given by

u(r) = p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!1/(q+1−p)

r−p/(q+1−p).(4)

(ii)If f(r)∼

+∞lr−pq/(q+1−p)for some lsuch that

0< l ≤q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

then there exists a solution uof problem (P)such that

u(r)∼

+∞br−p/(q+1−p)for some b > 0.

(iii)If f(r)∼

+∞lr−pq/(q+1−p)for some lsuch that

l > q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

then problem (P)does not possess any entire solu-

tion.

The rest of the paper is divided in three sections.

In section 2, we give the existence of entire solutions

of problem (P). In section 3, we present the asymp-

totic behavior of solutions near infinity in the cases

q6=N(p−1) + p

N−pand q=N(p−1) + p

N−p. The last

section is devoted to the nonexistence results.

2 Existence of Entire Solutions

In this section, we study the existence of entire so-

lutions of problem (P)while recalling that the exis-

tence of maximal solutions of problem (P)defined

on (0, rmax),0< rmax ≤+∞has been established

if N > p and q > N(p−1)

N−pin the paper, [6].

Theorem 2.1. Assume that N > p and q >

N(p−1)

N−p. Then problem (P)possesses an entire so-

lution u.

Proof. Let ube a maximal solution of problem (P)

defined on (0, rmax), where 0< rmax ≤+∞. We

will proceed in four steps to prove that uis entire.

Step 1.u0<0on the set {r∈(0, rmax) : u(r)>

0}.

First, we prove that uis strictly decreasing near 0. We

argue by contradiction. Suppose that there exists a

small rsuch that u0(r) = 0, then by equation (1), we

have |u|p−2u00(r) = −uq(r)−f(r)<0(because

f > 0and u > 0near 0). Therefore u06= 0 near

0. Moreover, since lim

r→0u(r) = +∞, then umust be

decreasing near 0. If there exists a first zero r0∈

(0, rmax)of u0such that u0(r0) = 0 and u(r0)>

0, then |u|p−2u00(r0)≥0, but by equation (1) we

have |u0|p−2u00(r0) = −uq(r0)−f(r0)<0. We

deduce that u0<0on the set {r∈(0, rmax) : u(r)>

0}.

Step 2.u > 0and u0<0on (0, rmax).

Let r1∈(0, rmax)the first zero of u. Since u≥

0on (0, rmax), then necessarily u0(r1) = 0. Using

equation (1), we have

rN−1|u0|p−2u00(r) = −rN−1uq(r)−rN−1f(r).

(5)

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We integrate (5) on (r, r1)for r∈(0, r1)and since

(uq+f)>0on (0, r1)and u0(r1) = 0, we obtain

u0(r)>0on (0, r1). This contradicts the first step.

Consequently u > 0on (0, rmax)and by the first step

u0<0on (0, rmax).

Step 3.u(r) = Or−p

q+1−pfor any r∈(0, rmax).

Since fis positive, then by equation (5), we have for

any r∈(0, rmax)

rN−1|u0|p−2u00(r)≤ −rN−1uq(r).(6)

Integrating this last inequality on r

2, rfor r∈

(0, rmax)and using the fact that u0(r)<0on

(0, rmax), we obtain for any r∈(0, rmax)

|u0|p−2u0(r)<−2N−1

N2Nruq(r).

Since u > 0and u0<0on (0, rmax), then for any

r∈(0, rmax)we have

u0(r)u−q/(p−1)(r)<−2N−1

N2N1/(p−1)

r1/(p−1).

Since q > N(p−1)

N−p> p −1, then for any r∈

(0, rmax)we have

u(p−1−q)/(p−1)0(r)>q+ 1 −p

p−12N−1

N2N1/(p−1)

r1/(p−1).

Integrating this last inequality on (0, r)for r∈

(0, rmax)and using the fact that lim

r→0u(r) = +∞and

q > p −1, we obtain

0< u(r)≤Mr−p/(q+1−p)for any r∈(0, rmax),

(7)

where

M= N2N

2N−1p

q+ 1 −pp−1!1/(q+1−p)

.(8)

Step 4.rmax = +∞.

Suppose by contradiction that rmax <+∞. Then

lim

r→rmax

u(r) = +∞. But by letting r→rmax in (7),

we get a contradiction. Consequently the solution u

is entire.

3 Asymptotic Behavior Near Infinity

In this section, we investigate the asymptotic be-

havior near the infinity of solutions of problem (P).

Under some additional assumptions on f, we prove

that any solution of (P)must behave like r−p/(q+1−p)

at infinity. The study requires some ideas from [4].

First, let us define for any c6= 0 the following

function

(rcu(r))0=rc−1Ec(r),(9)

where

Ec(r) = cu(r) + ru0(r).(10)

By equation (1), for any r > 0such that u0(r)6= 0

we have

(p−1) u0

p−2(r)E0

c(r) =(p−1) c−N−p

p−1×

u0

p−2u0(r)−ruq(r)

−rf(r).(11)

Therefore, if Ec(r0) = 0 for some r0>0, equation

(1) implies that

(p−1) rp−1

0u0

p−2(r0)E0

c(r0) = −(p−1) c−N−p

p−1×

|c|p−2c up−1(r0)−rp

0uq(r0)

−rp

0f(r0)(12)

We start with the following preliminary results.

Proposition 3.1. Let ube a solution of problem (P).

Then

u(r)>0and u0(r)<0,for any r > 0,(13)

Moreover, if q > p −1, then

0< u(r)≤M r−p/(q+1−p),(14)

where Mis given by (8).

Proof. We follow the same reasoning of the first three

steps of the proof of Theorem 2.1 by taking rmax =

+∞.

Proposition 3.2. Assume that N > p. Let ube a

solution of problem (P). Then E(N−p)/(p−1)(r)>0

for large r.

Proof. Taking c=N−p

p−1in (11), we see that

(N−p)/(p−1)(r)<0for any r > 0. Therefore,

E(N−p)/(p−1)(r)6= 0 for large r. Suppose by contra-

diction that E(N−p)/(p−1)(r)<0for large r. Then,

lim

r→+∞E(N−p)/(p−1)(r)∈[−∞,0[. We distinguish

two cases.

Case 1.lim

r→+∞E(N−p)/(p−1)(r) = −∞.

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Using Proposition 3.1, we have u(r)>0and u0(r)<

0on (0,+∞), which implies that lim

r→+∞u(r)exists

and is finite. Now, by relation (10) we have necessar-

ily lim

r→+∞ru0(r) = −∞, which is impossible since u

is positive.

Case 2.−∞ <lim

r→+∞E(N−p)/(p−1)(r)<0.

Then ru0(r)converges necessarily to 0when r→

+∞. This implies that lim

r→+∞u(r)<0(because

N > p). Which is impossible.

We conclude that E(N−p)/(p−1)(r)>0for large

Proposition 3.3. Assume that N > p and q > p −1.

Let ube a solution of problem (P). Then the function

rp/(q+1−p)+1u0(r)is bounded for large r.

Proof. Using Propositions 3.1 and 3.2, we have uis

strictly decreasing and E(N−p)/(p−1) >0, for large r.

This implies that

0< r|u0(r)|<N−p

p−1u(r)for large r. (15)

Recall by Proposition 3.1, that rp/(q+1−p)u(r)is

bounded for any r > 0. Therefore, by (15), we easily

get that rp/(q+1−p)+1u0(r)is bounded for large r.

Now, to prove the next Theorems, we introduce

the following change of variable which will be very

useful. So let us define the function

υ(t) = rp

q+1−pu(r)where t=ln r. (16)

So υsatisfies the following equation

y0(t) + N−pq

q+ 1 −py(t) + υq(t) + j(t) = 0,

(17)

where

j(t) = epq

q+1−ptf(et),(18)

y(t) = |k|p−2k(t),(19)

k(t) = υ0(t)−p

q+ 1 −pυ(t).(20)

It’s easy to see that

k(t) = rp

q+1−p+1u0(r).(21)

Proposition 3.4. Assume that N > p and q >

N(p−1)

N−p. Let ube a solution of problem (P). Sup-

pose that rpq/(q+1−p)f(r)and rp/(q+1−p)u(r)con-

verge when r→+∞. Then rp/(q+1−p)+1u0(r)con-

verges also when r→+∞and

lim

r→+∞rp/(q+1−p)+1u0(r) =

−p

q+ 1 −plim

r→+∞rp/(q+1−p)u(r).(22)

Proof. According to Proposition 3.1 and logarithmic

change (16), we have υis bounded for large t, which

gives by Proposition 3.3 that kis bounded for large t.

Therefore by expression (19), we have yis bounded

for large t. We show that yis monotone for large

t. Suppose by contradiction that there exist two se-

quences {γi}and {ρi}going to +∞as i→+∞such

that {γi}and {ρi}are respectively local minimum

and local maximum of y, satisfying γi< ρi< γi+1

and

− ∞ <lim inf

t→+∞y(t) = lim

i→+∞y(γi)

<lim sup

t→+∞

y(t) = lim

i→+∞y(ρi)≤0.(23)

Using equation (17) and the fact that y0(γi) =

y0(ρi) = 0, we obtain

N−pq

q+ 1 −py(γi) + υq(γi) + j(γi) =

N−pq

q+ 1 −py(ρi) + υq(ρi) + j(ρi) = 0.

Since υ(t)and j(t)converge when ttends to +∞and

N > pq

q+ 1 −p, then lim

i→+∞y(γi) = lim

i→+∞y(ρi),

which contradicts the estimate (23). Therefore k(t)

converges when t→+∞. So, by (20), υ0(t)

converges necessarily to 0when t→+∞. Con-

sequently, rp/(q+1−p)+1u0(r)converges when r→

+∞and (22) is satisfied.

Proposition 3.5. Assume that N > p and q >

N(p−1)

N−p. Let ube a solution of problem (P). Sup-

pose that lim

r→+∞rpq/(q+1−p)f(r) = l > 0.

If lim

r→+∞rp/(q+1−p)u(r) = b. Then bis one of the two

roots λ1and λ2of the equation

sq−N−pq

q+ 1 −p p

q+ 1 −pp−1

sp−1+l= 0,

(24)

such that 0< λ1≤λ2.

In particular, if

l=q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

(25)

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then

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!1/(q+1−p)

(26)

Proof. Using logarithmic change (16), we have υ(t)

converges to b≥0when t→+∞. Hence,

combining with Proposition 3.4, k(t)converges also

and lim

t→+∞k(t) = −p

q+ 1 −pb. Therefore, by

(19), lim

t→+∞y(t) = −p

q+ 1 −pp−1

bp−1. Since

lim

t→+∞j(t) = l, then by equation (17), y0(t)converges

necessarily to 0. Therefore, by letting t→+∞in the

same equation, we obtain

bq−Λq+1−pbp−1+l=l−ψ(b) = 0,(27)

where

Λ = N−pq

q+ 1 −p p

q+ 1 −pp−1!1/(q+1−p)

(28)

and

ψ(s) = Λq+1−psp−1−sq, s ≥0.(29)

A simple study gives

max

s≥0ψ(s) = ψ p−1

q1/(q+1−p)

Λ!=L, (30)

where

L=q+ 1 −p

p−1p−1

qq/(q+1−p)

Λq

=q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

(31)

Since l > 0, we can easily see that the equation

l−ψ(s) = 0 has two roots λ1and λ2such that 0<

λ1≤λ2. Then, by (27), b=λ1>0or b=λ2>0.

Moreover if l=L, this last equation has only one ex-

plicit root, that is, λ1=λ2=p−1

q1/(q+1−p)

Λ.

Hence by (27) and (30), we have explicitly

b=p−1

q1/(q+1−p)

= p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!1/(q+1−p)

(32)

Proposition 3.6. Assume that N > p and q >

N(p−1)

N−p. Let ube a solution of problem (P). Sup-

pose that lim

r→+∞rpq/(q+1−p)f(r) = l > 0. Then

lim inf

r→+∞rp/(q+1−p)u(r)>0(33)

and

lim sup

r→+∞

rp/(q+1−p)+1u0(r)<0.(34)

Proof. The proof will be done in two steps.

Step 1.lim inf

r→+∞rp/(q+1−p)u(r)>0.

Assume by contradiction that

lim inf

r→+∞rp/(q+1−p)u(r) = 0. This means that

lim inf

t→+∞υ(t)=0. Since υ(t)is bounded for large t,

we distinguish two cases.

•υ(t)is monotone for large t.

Then lim

t→+∞υ(t) = 0. Since u0(r)<0for any r > 0

and E(N−p)/(p−1)(r)>0for large r(by Proposition

3.2), then we have for large t,

0<|k(t)|<N−p

p−1υ(t).(35)

It follows that lim

t→+∞k(t) = 0 and by relation

(19), lim

t→+∞y(t)=0. Therefore, by equation (17),

lim

t→+∞y0(t) = −l < 0. But this is a contradiction

with lim

t→+∞y(t) = 0.

•υ(t)is oscillating for large t.

Then there exists a sequence {µi}going to +∞

as i→+∞such that υhas a local minimum

in µi. Therefore using our hypotheses, we have

lim

i→+∞υ(µi) = 0,υ0(µi) = 0 and υ00(µi)≥0(υ00 ex-

ists because u0<0). Which gives lim

i→+∞k(µi)=0

and k0(µi)≥0and so lim

i→+∞y(µi) = 0 and

y0(µi)≥0. Therefore, by equation (17), we have

lim

i→+∞y0(µi) = −l < 0. This is a contradiction.

It follows from both cases that

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lim inf

r→+∞rp/(q+1−p)u(r)>0.

Step 2.lim sup

r→+∞

rp/(q+1−p)+1u0(r)<0.

Since u0<0, we assume by contradiction that

lim sup

r→+∞

rp/(q+1−p)+1u0(r)=0. This means that

lim sup

t→+∞

k(t) = 0. In the same way as the first step,

since k(t)is bounded for large t(by Proposition 3.3),

we distinguish two cases.

•k(t)is monotone for large t.

Then lim

t→+∞k(t) = 0. That is,

lim

r→+∞rp/(q+1−p)+1u0(r) = 0. This yields by

L’Hôpital’s rule that lim

r→+∞rp/(q+1−p)u(r) = 0. But

this contradicts the fact that lim inf

r→+∞rp/(q+1−p)u(r)>

0by the first step.

•k(t)oscillates for large t.

Then there exists a sequence {ρi}going to +∞as

i→+∞such that khas a local maximum in ρi.

Therefore, lim

i→+∞k(γi)=0and k0(γi)=0and

so lim

i→+∞y(γi)=0and y0(γi)=0. Therefore, by

equation (17), we have lim

i→+∞υq(γi) = −l < 0. This

is impossible since υis positive.

We deduce that lim sup

r→+∞

rp/(q+1−p)+1u0(r)<0. The

proof is complete.

The following theorem gives a sufficient condition

to obtain explicitly lim inf

r→+∞rp/(q+1−p)u(r).

Theorem 3.7. Assume that N > p and q >

N(p−1)

N−p. Let ube a solution of problem (P). Sup-

pose that

lim

r→+∞rpq/(q+1−p)f(r) = q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

(36)

Then

lim inf

r→+∞rp/(q+1−p)u(r) =

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!1/(q+1−p)

Proof. Note that if υconverges, we obtain the result

directly by using Proposition 3.5. Otherwise, since υ

is bounded, it must oscillate. Then there exists a se-

quence {µi}going to +∞as i→+∞such that υ

has a local minimum in µi. Therefore υ0(µi) = 0 and

υ00(µi)≥0. Hence, using (20), we have k(µi) =

−p

q+ 1 −pυ(µi)and k0(µi) = υ00(µi)≥0. This

yields

y(µi) = −p

q+ 1 −pp−1

υp−1(µi)

and

y0(µi) = (p−1) |k(µi)|p−2k0(µi)≥0.

Using equation (17) with t=γi, we obtain

0≤y0(µi) = Λq+1−pυp−1(µi)−υq(µi)−j(µi)

=ψ(υ(µi)) −j(µi)

≤L−j(µi),

where ψand Lare respectively given by (29) and

(31). Since lim

t→+∞j(t) = L, then lim

i→+∞ψ(υ(µi)) =

lim

i→+∞j(µi) = L. Hence, according to (30),

lim

i→+∞υ(µi) = lim inf

t→+∞υ(t) = p−1

q1/(q+1−p)

Λ,

where Λis given by (28). This completes the proof.

Now, we study the convergence of rp/(q+1−p)u(r)

at infinity to improve the previous result which gives

only lim inf

r→+∞rp/(q+1−p)u(r). For this, we assume that

fis differentiable and satisfies the following condi-

tions:

(K1)Z+∞

1rpq/(q+1−p)f+

rdr < +∞,

(K2)Z+∞

1rpq/(q+1−p)f−

rdr < +∞.

The study depends on the comparison of qwith

N(p−1)

N−pand N(p−1) + p

N−p. We start with the case

N(p−1)

N−p< q 6=N(p−1) + p

N−p.

Theorem 3.8. Assume that N > p and q >

N(p−1)

N−p. Let ube a solution of problem (P). Sup-

pose that lim

r→+∞rpq/(q+1−p)f(r) = l > 0and fsat-

isfies

(i) (K1)if q > N(p−1) + p

N−p,

(ii) (K2)if N(p−1)

N−p< q < N(p−1) + p

N−p.

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Then

l≤q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

(37)

and lim

r→+∞rp/(q+1−p)u(r) = bwhere bis one of the

two roots λ1or λ2of equation (24) such that 0<

λ1≤λ2.

To demonstrate Theorem 3.8, we need the classic

result of [12], of which we recall the proof.

Lemma 3.9. Let gbe a positive differentiable func-

tion satisfying

(i)Z+∞

g(t)dt < +∞for large t0.

(ii)g0(t)is bounded for large t.

Then lim

t→+∞g(t) = 0.

Proof. We argue by contradiction and we suppose

that lim

t→+∞g(t)6= 0. Then there exist ε > 0and a

sequence {ti}going to +∞as i→+∞satisfying

g(ti)≥2ε. Since g0(t)is bounded for large t, then

there exists a constant C > 0such that |g0(t)| ≤ Cfor

large t. By the mean value Theorem for g, we have

g(t)≥εfor |t−ti|<ε

Choose a subsequence t0

isuch that t0

0> t0and t0

i−1+2ε

Ct0

0for i > 1. Therefore

i=1 Zt0

i−1

g(t)dt >

i=1 Zt0

i−1+ε/C

i−1

g(t)dt

≥ε2

Cn→+∞as n→+∞.

This implies that

Z+∞

g(t)dt = +∞.

This contradiction completes the proof.

Now, we can prove Theorem 3.8.

Proof. Define the following energy function associ-

ated with equation (17),

I(t)= p−1

p|k(t)|p+p

q+ 1 −py(t)υ(t)

−A

pp

q+ 1 −pp−1

υp(t)

+υq+1(t)

q+ 1 +lυ(t),(38)

where

A=q(N−p)−(N(p−1) + p)

q+ 1 −p.(39)

Note that the energy function Iplays a central role

in the study of the convergence of rp/(q+1−p)u(r).

First, using Proposition 3.3 we have k(t)is bounded

for large t, which yields that y(t)is bounded for large

t. Hence I(t)is bounded for large t.

The rest of the proof will be done in three steps.

Step 1.The function I(t)converges when t→+∞.

A simple computation gives

I0(t) = −AZ(t)−(j(t)−l)υ0(t),(40)

where

Z(t) =|k(t)|p−1−p

q+ 1 −pp−1υp−1(t)×

|k(t)| − p

q+ 1 −pυ(t).(41)

Integrating (40) on (T, t)for large T, we get

I(t) = C(T)−AR(t)−(j(t)−l)υ(t)+Zt

j0(s)υ(s)ds,

(42)

where

C(T) = I(T)+(j(T)−l)υ(T)(43)

and

R(t) = Zt

Z(s)ds. (44)

Since the function s→sp−1is monotone, then

Z(t)≥0, which in turn implies that the function

R(t)is positive and increasing. By our hypotheses,

we have A6= 0, then relation (42) can be expressed

as follows:

R(t) = −I(t)

A−1

A(j(t)−l)υ(t) + 1

AZt

j0(s)υ(s)ds

+C(T)

A.(45)

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Using the fact that υ(t)and I(t)are bounded for large

t, lim

t→+∞j(t) = l,−(j0(s))−≤j0(s)≤(j0(s))+

and Z+∞

Tj0(s)+ds < +∞from (K1)if A > 0

or Z+∞

Tj0(s)−ds < +∞from (K2)if A < 0,

we obtain R(t)is bounded for large t. Consequently,

R(t)converges when t→+∞, which yields that

Z+∞

Z(s)ds exists. By letting t→+∞in (42), we

deduce that I(t)converges to a real number noted d

when t→+∞.

Step 2.lim

t→+∞υ0(t) = 0.

By expressions (41) and (20) and the fact that k(t)<

0for large t, we have just to prove that lim

t→+∞Z(t) =

0. Since Z+∞

Z(s)ds < +∞, then by Lemma 3.9,

it suffices to prove that Z0(t)is bounded for large t.

Using expression (41), Z(t)can be written as follows:

Z(t) =|y(t)|p/(p−1) +p

q+ 1 −pυ(t)y(t)

+p

q+ 1 −pp−1

υp−1(t)υ0(t).(46)

Therefore

Z0(t) = p

p−1k(t)y0(t) + p

q+ 1 −py(t)υ0(t)

q+ 1 −pυ(t)y0(t)+(p−1) p

q+ 1 −pp−1

υp−2(t)υ02(t) + p

q+ 1 −pp−1

υp−1(t)υ00(t).

(47)

Since υ(t), k(t)and j(t)are bounded for large t,

then combining with (20) and (17), υ0(t)and y0(t)are

bounded also for large t. Hence, it remains to prove

that υ00(t)is bounded for large t. According to (20),

we have

υ00(t) = k0(t) + p

q+ 1 −pυ0(t).(48)

Then, it suffices to prove that k0(t)is bounded for

large t.

Since k(t)<0for large t, we have by (19)

k0(t) = 1

p−1|k(t)|2−py0(t).(49)

According to Proposition 3.6, we have lim sup

t→+∞

k(t)<

0. Therefore, there exists a constant M > 0such that

k(t)≤ −Mfor large t. Which implies that, |k(t)|2−p

is bounded for large t. Therefore k0(t)is bounded

for large tand by relations (48) and (47), Z0(t)is

bounded for large t. Hence, by Lemma 3.9, we get

lim

t→+∞Z(t) = 0 and therefore lim

t→+∞υ0(t) = 0.

Step 3.υ(t)converges when t→+∞.

Suppose by contradiction that υ(t)is oscillating for

large t. Then there exist two sequences {µi}and {νi}

tend to +∞when i→+∞such that {µi}and {νi}

are respectively local minimum and local maximum

of υ, satisfying µi< νi< µi+1 and

0<lim inf

t→+∞υ(t) = lim

i→+∞υ(µi) = α

<lim sup

t→+∞

υ(t) = lim

i→+∞υ(νi) = β < +∞.(50)

Since υ0(µi) = υ0(νi)=0, then by relations (38),

(50), (19) and (20), we obtain

lim

i→+∞I(µi) = ϕ(α)and lim

i→+∞I(νi) = ϕ(β),

(51)

where

ϕ(s) = ls −Λq+1−p

psp+sq+1

q+ 1 =ls −Zs

ψ(r)dr,

(52)

for any s≥0and ψis given by (29). Since

lim

t→+∞I(t) = dby the first step, then

ϕ(α) = ϕ(β) = d. (53)

Then, there exist δ∈(α, β)and xi∈(µi, νi)such

that υ(xi) = δ, ϕ0(δ) = 0 and ϕ(δ)6=d. On the other

hand, using step 2, we have lim

i→+∞υ0(xi)=0, so by

(20), we obtain lim

i→+∞k(xi) = −p

q+ 1 −pδ. There-

fore lim

i→+∞I(xi) = ϕ(δ) = d. Which gives a con-

tradiction. Hence υ(t)converges when t→+∞.

Let lim

t→+∞υ(t) = b. So, by Proposition 3.5, bis one

of the two roots λ1and λ2of equation (24) such that

0< λ1≤λ2. Finally, by (30), l=ψ(b)≤Lwhere

Lis given by (31). The proof is complete.

Now we give the following result which deals with

the critical case q=N(p−1) + p

N−p.

Theorem 3.10. Assume that N > p and q=

N(p−1) + p

N−p. Let ube a solution of problem (P).

Suppose that lim

r→+∞rpq/(q+1−p)f(r) = l > 0and f

satisfies (K1)or (K2). Then usatisfies one of the fol-

lowing cases:

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(i)lim

r→+∞rp/(q+1−p)u(r) = λ1or λ2given in The-

orem 3.8.

(ii)rp/(q+1−p)u(r)is oscillating and

λ1≤α=lim inf

r→+∞rp/(q+1−p)u(r)< λ2

< β =lim sup

r→+∞

rp/(q+1−p)u(r)≤Γ,(54)

where Γ6=λ1is the root of the equation

lΓ + Γq+1

q+ 1 −Λq+1−p

pΓp

=p−1

pN−p

pp

λp

1−N(p−1) + p

Np λNp/(N−p)

(55)

Moreover, αand βsatisfy the following estimates

l=1

pN−p

ppβp−αp

β−α

−N−p

βNp/(N−p)−αN p/(N−p)

β−α(56)

and

N(N−p)p−1<βp−αp

βNp/(N−p)−αN p/(N−p)

≤1

p−1p

N−ppN(p−1) + p

N.(57)

In both cases, we have the estimate (37).

Proof. According to the notations in the proof of The-

orem 3.8, we have A= 0, where Ais given by (39).

Using the fact that υ(t)is bounded, lim

t→+∞j(t) = l,

j0(s)≤(j0(s))+for (K1)and j0(s)≥ − (j0(s))−

for (K2), we deduce that the energy function Icon-

verges when t→+∞.

Since υ(t)is bounded, we have two possibilities:

•υ(t)converges when t→+∞, then similarly

to Theorem 3.8, lim

t→+∞υ(t) = λ1or λ2, where λ1

and λ2are two roots of the equation (24) such that

0< λ1≤λ2. Moreover, the estimate (37) is satis-

fied.

•υ(t)oscillates, then there exist two sequences {µi}

and {νi}tend to +∞when i→+∞such that {µi}

and {νi}are respectively local minimum and local

maximum of υ, satisfying µi< νi< µi+1 and re-

lation (50). Therefore k0(µi) = υ00(µi)≥0and

k0(νi) = υ00(νi)≤0and so by equation (17), we

have

0≤y0(µi) = ψ(υ(µi)) −j(µi)(58)

and

0≥y0(νi) = ψ(υ(νi)) −j(νi),(59)

where ψis given by (29). On the other hand, accord-

ing to the proof of Theorem 3.8, we have

lim

t→+∞I(t) = ϕ(α) = ϕ(β),

where ϕis given by (52). Since q=N(p−1) + p

N−p

and α < β, then

l=Λq+1−p

βp−αp

β−α−1

q+ 1

βq+1 −αq+1

β−α

pN−p

ppβp−αp

β−α

−N−p

βNp/(N−p)−αN p/(N−p)

β−α.

This proves (56). Now, a simple study of the func-

tion ϕgives ϕ0(λ1) = ϕ0(λ2) = 0,ϕ0(s)>0

for 0< s < λ1,ϕ0(s)<0for λ1< s < λ2,

ϕ0(s)>0for s > λ2and lim

s→+∞ϕ(s) = +∞. There-

fore, there exists Γ> λ2such that ϕ(Γ) = ϕ(λ1).

Since ψ(λ1) = land q=N(p−1) + p

N−p, then

ϕ(Γ) = p−1

pN−p

pp

λp

1−N(p−1) + p

Np λNp/(N−p)

(60)

which gives (55). To prove estimate (54), we let i→

+∞in (58) and (59), we obtain

ψ(β)≤l≤ψ(α),(61)

that is by (52)

ϕ0(α)≤0≤ϕ0(β).(62)

Combining this with the study of ϕand the fact that

ϕ(α) = ϕ(β), we deduce that λ1≤α < λ2<

β. Moreover, we have β≤Γ, otherwise ϕ(β)>

ϕ(Γ) = ϕ(λ1)≥ϕ(α), which contradicts ϕ(α) =

ϕ(β). Consequently, (54) is satisfied. Concerning

(57), we use the fact that l > 0and (56) to obtain

the left inequality. To prove the right inequality of

(57), we have β > α > 0, then by (61), we have

βψ(β)≤lβ =ϕ(β) + Λq+1−p

pβp−βq+1

q+ 1

and

αψ(α)≥lα =ϕ(α) + Λq+1−p

pαp−αq+1

q+ 1,

which in turn implies that

βψ(β)−Λq+1−p

pβp+βq+1

q+ 1 ≤ϕ(β) = ϕ(α)≤

αψ(α)−Λq+1−p

pαp+αq+1

q+ 1.

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With the expression of ψ, we get

p−1

pΛq+1−pβp−q

q+ 1βq+1

≤p−1

pΛq+1−pαp−q

q+ 1αq+1.

The right inequality of (57) can be easily obtained

since q=N(p−1) + p

N−p. The proof is complete.

4 Nonexistence Results

This section concerns the nonexistence theorems of

entire solutions of problem (P). The study depends

on the parameters N, p, q and the comparison of f(r)

with r−pq/(q+1−p).

Theorem 4.1. Assume that N≤por N > p and

p−1≤q≤N(p−1)

N−por N > p and q < p−1. Then

problem (P)does not possess any entire solution.

Proof. Let ube a solution of problem (P). We dis-

tinguish five cases.

Case 1. N≤p.

It’s easy to see by Proposition 3.1 and equation

(5) that the function r→rN−1|u0|p−2u0(r)is de-

creasing and strictly negative on (0,+∞). Then

lim

r→+∞rN−1|u0|p−2u0(r)∈[−∞,0[. Therefore, there

exists a constant M0>0such that

rN−1|u0|p−2u0(r)<−M0,(63)

for large r. So

|u0(r)|> M1/(p−1)

0r(1−N)/(p−1),

for large r. We get a contradiction by integrating

this last inequality for large rand using the fact that

N≤p.

Case 2. N > p and q=p−1.

Recall, by Proposition 3.2, that E(N−p)/(p−1)(r)>0

for large r. Then estimation (15) is satisfied. Combin-

ing this with equation (1) and the fact that q=p−1,

we obtain

|u0|p−2u00(r)<up−1(r)×

"(N−1) N−p

p−1p−1

r−p−1#,

for large r. In particular, we get |u0|p−2u00(r)<0

for large r. Since u0(r)<0, then

lim

r→+∞|u0|p−2u0(r)∈[−∞,0[. Therefore

lim

r→+∞u(r) = −∞, which is impossible.

Case 3. N > p and p−1< q < N(p−1)

N−p.

In this case we have necessarily N−p

p−1<

q+ 1 −p, which implies by (14) that

lim

r→+∞rN−p

p−1u(r)=0. But this contradicts the

fact that rN−p

p−1u(r)is positive and strictly increasing

by relation (9) and Proposition 3.2.

Case 4. N > p and q=N(p−1)

N−p.

Using relations (15) and (5), then for large rwe

obtain

rN−1|u0|p−2u00(r)≤ − N−p

p−1−q

rN+q−1|u0|q.

(64)

Which can be written as

−φ0(r)≤ − N−p

p−1−q

r−1φq/(p−1)(r)for large r,

(65)

where

φ(r) = rN−1|u0|p−1.(66)

Integrating relation (65) on (R, r)for large Rand us-

ing the fact that q > p −1, we obtain for large r,

φ(p−1−q)/(p−1)(r)≤φ(p−1−q)/(p−1)(R)

+p−1−q

p−1N−p

p−1−q

ln r

R.

(67)

By letting r→+∞in the last inequality, we get a

contradiction with the fact that φis positive.

Case 5. N > p and q < p −1.

Since fis positive, then by equation (5), we have

rN−1|u0|p−2u00(r)≤ −rN−1uq(r)for any r > 0.

(68)

Integrating this last inequality on r

2, rfor r > 0

and using the fact that u0(r)<0on (0,+∞), we

obtain

|u0|p−2u0(r)<−2N−1

N2Nruq(r)for any r > 0.

Since u > 0and u0<0on (0,+∞), then for any

r > 0

u0(r)u−q/(p−1)(r)<−2N−1

N2N1/(p−1)

r1/(p−1).

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Integrating this last inequality on (R, r)for large R

and using the fact that q < p −1, we obtain

u(p−1−q)/(p−1)(r)≤u(p−1−q)/(p−1)(R)

−p−1−q

p2N−1

N2N1/(p−1)

rp/(p−1) −Rp/(p−1).(69)

By letting r→+∞in the last inequality, we get a

contradiction with the fact that uis positive.

we deduce that problem (P)does not possess any en-

tire solution in all cases.

Theorem 4.2. Assume that N > p and q >

N(p−1)

N−p. Suppose that

lim inf

r→+∞rpq/(q+1−p)f(r)>q+ 1 −p

p−1×

p−1

qN−pq

q+ 1 −p p

q+ 1 −pp−1!q/(q+1−p)

Then problem (P)does not possess any entire solu-

tion.

Proof. Let ube a solution of problem (P). First

we show that rp/(q+1−p)u(r) is strictly monotone

for large r. This amounts to proving by (9) that

Ep/(q+1−p)(r)6= 0 for large r.

Suppose that there exists a large rsuch that

Ep/(q+1−p)(r)=0. Taking c=p

q+ 1 −pin (12)

and multiplying by rpq/(q+1−p)−1, we obtain

(p−1) rpq/(q+1−p)−1u0

p−2E0

p/(q+1−p)(r) =

Λq+1−prp(p−1)/(q+1−p)up−1(r)−rpq/(q+1−p)uq(r)

−rpq/(q+1−p)f(r),(70)

where Λis given in relation (28).

Using the change of variable (16), we see that the re-

lation (70) is equivalent to

(p−1) rpq/(q+1−p)−1u0

p−2(r)E0

p/(q+1−p)(r)

=ψ(υ(t)) −j(t),(71)

where ψis given respectively by (29). Since

lim inf

r→+∞rpq/(q+1−p)f(r)> L, then there exists ε > 0

such that

j(t)≥L+εfor large t. (72)

Recall, by (30), that max

s≥0ψ(s) = L, where Lis given

by (31).Then by (71), E0

p/(q+1−p)(r)<0and so

Ep/(q+1−p)(r)6= 0 for large r.

Now, we have υ(t)is strictly monotone for large t

and bounded by Proposition 3.1, then υ(t)converges

when t→+∞. Let lim

t→+∞υ(t) = b≥0.

We distinguish two cases according to the monotonic-

ity of υ0(t)for large t.

Case 1. υ0(t)is monotone for large t.

Since k(t)is bounded for large tby Proposition 3.3,

then v0(t)is bounded for large t. Therefore v0(t)con-

verges and necessarily lim

t→+∞υ0(t) = 0 (because υis

bounded). Therefore lim

t→+∞k(t) = −p

q+ 1 −pband

so lim

t→+∞y(t) = −p

q+ 1 −pp−1

bp−1. Accord-

ing to equation (17) and estimate (72), we have

y0(t)≤ − N−pq

q+ 1 −py(t)−υq(t)−L−ε,

for large t. But

lim

t→+∞−N−pq

q+ 1 −py(t)−υq(t)−L−ε

=ψ(b)−L−ε≤ −ε,

Hence there exists a constant M > 0such that y0(t)≤

−Mfor large t. Integrating this last inequality on

(T, t)for large T, we get lim

t→+∞y(t) = −∞. Which

gives a contradiction.

Case 2. υ0(t)is not monotone for large t.

Since υ(t)is strictly monotone for large t, then we

have two possibilities.

•υ0(t)>0for large t. Then lim inf

t→+∞υ0(t) = 0.

Otherwise, there exits C > 0such that υ0(t)≥C

for large t, then integrating this last inequality near

+∞, we obtain lim

t→+∞υ(t)=+∞, which is im-

possible. Hence there exists ζitends to +∞when

i→+∞such that ζiis a local minimum of υ0sat-

isfying lim

i→+∞υ0(ζi)=0. Consequently, we have

lim

i→+∞y(ζi) = −p

q+ 1 −pp−1

bp−1.On the other

hand, by deriving relation (20) and taking account

that υ00(ζi)=0, we get lim

i→+∞k0(ζi)=0. There-

fore lim

i→+∞y0(ζi)=0. Taking t=ζiin equation

(17) and tending i→+∞, we obtain lim

i→+∞j(ζi) =

ψ(b)≤L=max

s≥0ψ(s). But this contradicts the fact

that lim inf

t→+∞j(t)> L.

•υ0(t)<0for large t. In the same way, using the fact

that υis bounded, we deduce that lim

t→+∞sup υ0(t) =

0. Then there exists τitends to +∞when i→

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+∞such that τiis a local minimum of υ0satisfy-

ing lim

i→+∞υ0(τi) = 0 and υ00(τi) = 0. Therefore

lim

i→+∞y0(τi) = 0 and lim

i→+∞j(τi) = ψ(b)≤L.

Which is impossible like the previous case.

We conclude that problem (P)does not possess any

entire solution.

5 Conclusion

In this paper, we have studied the existence, the

nonexistence and the asymptotic behavior near

infinity of global singular solutions of problem (P).

The difficulty of this work lies in the influence of

the inhomogeneous term fwhich is positive and is

equivalent to the function r−pq/(q+1−p)near infinity.

Under some conditions, we prove that the singular

solution of problem (P)is equivalent to the function

r−p/(q+1−p)near infinity. The cases where fis

not positive or negligible in front to the function

r−pq/(q+1−p)near infinity are not yet treated and will

be the subject of a future study.

Acknowledgment:

The authors express their gratitude to the editor and

reviewers for their comments and suggestions which

have improved the quality of this paper.

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Arij Bouzelmate and Abdelilah Gmira. It brings

together the techniques of Nonlinear Analysis. All

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Scientific Article or Scientific Article Itself

No funding was received for conducting this study.

Conflicts of Interest

The authors have no conflicts of interest to declare

that are relevant to the content of this article.

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Arij Bouzelmate proposed the subject of the article

to his Ph.D. student Hikmat El Baghouri. This work

is an extension of a paper published in the Nonlinear

Functional Analysis and Applications journal by

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