# Entire Solutions for a Quasilinear Problem in the Presence of Sublinear and Super-Linear Terms

- CA Santos
^{1}Email author

**2009**:845946

**DOI: **10.1155/2009/845946

© C. A. Santos 2009

**Received: **31 May 2009

**Accepted: **2 October 2009

**Published: **13 October 2009

## Abstract

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 where are suitable functions and are not identically zero continuous functions. We show that there exists at least one solution for the above-mentioned problem for each for some . Penalty arguments, variational principles, lower-upper solutions, and an approximation procedure will be explored.

## 1. Introduction

In this paper we establish new results concerning existence and behavior at infinity of solutions for the nonlinear quasilinear problem

where , with , denotes the -Laplacian operator; and are continuous functions not identically zero and is a real parameter.

A solution of (1.1) is meant as a positive function with as and

The class of problems (1.1) 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 is the characteristic of the medium. If , the fluids are called pseudoplastics; if Newtonian and if the fluids are called dilatants.

It follows by the nonnegativity of functions of parameter and a strong maximum principle that all non-negative and nontrivial solutions of (1.1) must be strictly positive (see Serrin and Zou [6]). So, again of [6], it follows that (1.1) admits one solution if and only if .

The main objective of this paper is to improve the principal result of Yang and Xu [7] and to complement other works (see, e.g., [8–20] and references therein) for more general nonlinearities in the terms and which include the cases considered by them.

The principal theorem in [7] considered, in problem (1.1), and with . Another important fact is that, in our result, we consider different coefficients, while in [7] problem (1.1) was studied with .

In order to establish our results some notations will be introduced. We set

Additionally, we consider

(*H*_{1}) (i)

(ii)

(*H*_{2}) (i)

(ii)

Concerning the coefficients and ,

(*H*_{3}) (i)

(ii)

Our results will be established below under the hypothesis .

Theorem 1.1.

Remark 1.2.

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 of Theorem 1.1.

We begin with the problem of finding classical solutions for the differential inequality

Our result is.

Theorem 1.3.

Remark 1.4.

Theorems 1.1 and 1.3 are still true with if ( ) hypothesis is replaced by

In fact, ( ) implies ( ) , if . (see sketch of the proof in the appendix).

Remark 1.5.

In Theorem 1.3, it is not necessary to assume that and are continuous up to . It is sufficient to know that are continuous. This includes terms singular in .

The next result improves one result of Goncalves and Santos [21] because it guarantees the existence of radially symmetric solutions in for the problem

where , are continuous and suitable functions and is the ball in centered in the origin with radius .

Theorem 1.6.

The proof of principal theorem (Theorem 1.1) relies mainly on the technics of lower and upper solutions. First, we will prove Theorem 1.3 by defining several auxiliary functions until we get appropriate conditions to define one positive number and a particular upper solution of (1.1) for each .

After this, we will prove Theorem 1.6, 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 Yang [22].

Lemma 1.7.

In the two next sections we will prove Theorems 1.3 and 1.6.

## 2. Proof of Theorem (1.4)

First, inspired by Zhang [20] and Santos [16], we will define functions and by

So, for each , let given by

where

It is easy to check that

and, as a consequence,

Moreover, it is also easy to verify.

Lemma 2.1.

Suppose that and hold. Then, for each ,

(i)

(ii) ,

(iii)

(iv) ,

(v) ,

(vi)

By Lemma 2.1(iii), (iv), and (2.2), the function , given by

is well defined and continuous. Again, by using Lemma 2.1(i) and (ii),

Besides this, , for each , and using Lemma 2.1, it follows that satisfies, for each , the following.

Lemma 2.2.

Suppose that and hold. Then, for each ,

(i) ,

(ii)

(iii)

(iv)

And, in relation to , we have the folowing.

Lemma 2.3.

Suppose that and hold. Then, for each ,

(i) ,

(ii)

Finally, we will define, for each , , by

So, is a continuous function and we have (see proof in the appendix).

Lemma 2.4.

Suppose that and hold. Then,

(i)

(ii)

(iii)

(iv)

(v)

By Lemma 2.4(ii), there exists a such that , where by either ( ) or ( ) , we have

So, by Lemma 2.4(v), there exists a such that . That is,

Let by

where , is given by where is the unique positive and radially symmetric solution of problem

More specifically, by DiBenedetto [23], , for some . In fact, satisfies

So, by (2.10), (2.11), and (2.13), we have for each ,

Hence, after some pattern calculations, we show that there is a such that and

As consequences of (2.9), (2.13) and (2.15), we have and

and hence, by Lemma 2.2 (i), (2.7) and , we obtain

that is, by using (2.2), we have

In particular, making , we get from (2.15), Lemma 2.2(i) and that and satisfies (1.8), for each . That is, is an upper solution to (1.1).

To prove (1.9), first we observe, using Lemma 2.2(i) and (2.15), that

So, by definition of , and hypothesis (1.5), we have

Thus,

Recalling that and using (1.5) again, we obtain

Thus by (2.9), (2.13), and , there is one positive constant such that (1.9) holds. This ends the proof of Theorem 1.3.

## 3. Proof of Theorem (1.5)

To prove Theorem (1.5), we will first show the existence of a solution, say , for each for the auxiliary problem

where In next, to get a solution for problem (1.10), we will use a limit process in .

For this purpose, we observe that

(i) ,

(ii) , by ( ) and by (1.11), it follows that

(iii) is non-increasing, for each

By items (i)–(iii) above, and fulfill the assumptions of Theorem 1.3 in [21]. Thus (3.1) admits one solution , for each Moreover, with satisfying

Adapting the arguments of the proof of Theorem 1.3 in [21], we show

where is the positive first eigenfunction of problem

and , independent of , is chosen (using ( )) such that

with denoting the first eigenvalue of problem (3.4) associated to the .

Hence, by (3.3),

Using ( ), (3.3), the above convergence and Lebesgue's theorem, we have, making in (3.2), that

So, making , after some calculations, we obtain that . This completes the proof of Theorem 1.6.

## 4. 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 problem (1.1), say , such that , where is given by Theorem 1.3. After this, the existence of a solution for the problem (1.1) will be obtained applying Lemma 1.7.

To get a lower solution for (1.1), we will proceed with a limit process in , where is a classical solution of problem (1.10) (given by Theorem 1.6) with , is a suitable function and for and is such that in .

Let

Thus, it is easy to check the following lemma.

Lemma 4.1.

Suppose that and hold. Then,

(i)

(ii) is non-increasing,

(iii) and

Hence, Lemma 4.1 shows that fulfills all assumptions of Theorem 1.6. Thus, for each such that there exists one with and satisfying

equivalently,

Consider extended on by . We claim that

Indeed, first we observe that satisfies Lemma 4.1(ii). So, with similar arguments to those of [21], we show .

To prove , first we will prove that . In fact, if for some , then there is one such that

because and with as .

So, using Lemma A.1 (see the appendix) with , and , we obtain

and from Lemma 4.1(i),

As a consequence of the contradiction hypothesis and the definition of , we get

Recalling that , it follows that

So,

However, this is impossible. To end the proof of claim (4.4), we will suppose that there exist an and such that . Hence, there are with such that , and for all .

Following the same above arguments, we obtain

This is impossible again. Thus, we completed the proof of claim (4.4). Setting

it follows by claim (4.4) that

Moreover, making in (4.3), we use Lebesgue's theorem that

Hence, after some calculations, we obtain and setting it follows, by DiBenedetto [23], that for some . Recalling that and using Lemma 4.1(i), it follows that is a lower solution of (1.1) with

So, by Lemma 1.7, we conclude that problem (1.1) admits a solution. Besides this, the inequality (1.4) is a consequence of a result in [6]. This completes the proof of Theorem 1.1.

## Appendix

Proof of Lemma 2.4.

The proof of item (iv) is an immediate consequence of Lemma 2.3(i). The item (v) follows by Lemma 2.3(i) and (ii) using Lebesgue's Theorem.

Proof.

then the claim (i) of Lemma 2.4 follows from (A.2).

On the other hand, for all , it follows from Lemma 2.1(vi) that

where the last equality is obtained by using ( )-(ii). Hence, using (A.2), the proof of Lemma 2.4(iii) is concluded.

Proof.

This ends the proof of Lemma 2.4.

The next lemma, proved in [21], was used in the proofs of Theorems 1.1 and 1.6. To enunciate it, we will consider , for some , satisfying

and we define the continuous function by

So, we have and

Lemma A.1.

Finally, we will sketch the proof of claim ( ), implies ( ) , if .

Below, will denote several positive constants and , the function

If , by estimating the integral in (A.10), we obtain

Using the assumption in the computation of the first integral above and Jensen's inequality to estimate the last one, we have

Computing the above integral, we obtain

Similar calculations show that

So, by ( ),

On the other hand, if , set

and note that either for all or for some . In the first case, for all . Hence

So has a finite limit as , because . In the second case, for and hence,

Integrating by parts and estimating using , we obtain

Again by ( ), we obtain that is a finite number. This shows the claim.

## Declarations

### Acknowledgment

This research was supported by FEMAT-DF, DPP-UnB.

## Authors’ Affiliations

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