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Open Access Research Article

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

CA Santos

Author Affiliations

Department of Mathematics, University of Brasília, 70910–900 Brasília, DF, Brazil

Boundary Value Problems 2009, 2009:845946  doi:10.1155/2009/845946

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2009/1/845946


Received:31 May 2009
Revisions received:13 August 2009
Accepted:2 October 2009
Published:13 October 2009

© 2009 The Author(s)

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.

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

(11)

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

(12)

The class of problems (1.1) appears in many nonlinear phenomena, for instance, in the theory of quasiregular and quasiconformal mappings [13], 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., [820] 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

(13)

Additionally, we consider

()(i)

(ii)

() (i)

(ii)

Concerning the coefficients and ,

() (i)

(ii)

Our results will be established below under the hypothesis .

Theorem 1.1.

Consider , then there exists one such that for each there exists at least one solution of problem (1.1). Moreover,

(14)

for some constant . If additionally

(15)

then there is a positive constant such that

(16)

Remark 1.2.

If we assume (1.5) with , , where , then (1.6) becomes

(17)

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

(18)

Our result is.

Theorem 1.3.

Consider , then there exists one such that problem (1.8) admits, for each , at least one radially symmetric solution , for some . Moreover, if in additionally one assumes (1.5), then there is a positive constant such that

(19)

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

(110)

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

Theorem 1.6.

Assume where , with , is continuous. Suppose that satisfies ( and additionally

(111)

then (1.10) admits at least one radially symmetric solution . Besides this, and satisfies

(112)

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.

Suppose that is defined on and is locally Hölder continuous (with ) in . Assume also that there exist functions such that

(113)

and is locally Lipschitz continuous in on the set

(114)

Then there exists with satisfying

(115)

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

(21)

So, for each , let given by

(22)

where

(23)

It is easy to check that

(24)

and, as a consequence,

(25)

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

(26)

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

(27)

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

(28)

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

(29)

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

(210)

Let by

(211)

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

(212)

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

(213)

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

(214)

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

(215)

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

(216)

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

(217)

that is, by using (2.2), we have

(218)

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

(219)

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

(220)

Thus,

(221)

Recalling that and using (1.5) again, we obtain

(222)

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

(31)

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

(32)

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

(33)

where is the positive first eigenfunction of problem

(34)

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

(35)

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

Hence, by (3.3),

(36)

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

(37)

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

(41)

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

(42)

equivalently,

(43)

Consider extended on by . We claim that

(44)

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

(45)

because and with as .

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

(46)

and from Lemma 4.1(i),

(47)

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

(48)

Recalling that , it follows that

(49)

So,

(410)

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

(411)

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

(412)

it follows by claim (4.4) that

(413)

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

(414)

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

(415)

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.

Acknowledgment

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

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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.

By Lemma 2.2(i),

(A1)

So, using (2.2), (2.5), and Lemma 2.1(i) and (ii), we get

(A2)

Since, by Lemma 2.1(iv),

(A3)

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

(A4)

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

Proof.

In this case (),

(A5)

That is, does not depend on . So, by L'Hopital and Lemma 2.2(iv),

(A6)

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

(A7)

and we define the continuous function by

(A8)

So, we have and

Lemma A.1.

If , then

(A9)

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

Below, will denote several positive constants and , the function

(A10)

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

(A11)

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

(A12)

Computing the above integral, we obtain

(A13)

Similar calculations show that

(A14)

So, by (),

(A15)

On the other hand, if , set

(A16)

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

(A17)

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

(A18)

Integrating by parts and estimating using , we obtain

(A19)

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