Positive solutions for nonlinear problems involving the one-dimensional {\phi}-Laplacian
Uriel Kaufmann, Leandro Milne

TL;DR
This paper investigates the existence of positive solutions for a class of nonlinear differential equations involving the one-dimensional {\
Contribution
It introduces new existence results for positive solutions of {\
Findings
Existence of positive solutions under sublinear conditions.
Results extend to cases where r is zero or m is non-negative.
Method combines sub and supersolution techniques with nonlinear estimates.
Abstract
Let , and be a real parameter. Let be the differential operator given by , where is an odd increasing homeomorphism and . We study the existence of positive solutions for problems of the form in on , where is a continuos function which is, roughly speaking, sublinear with respect to . Our approach combines the sub and supersolution method with some estimates on related nonlinear problems. We point out that our results are new even in the cases and/or…
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Taxonomy
TopicsAdvanced Mathematical Modeling in Engineering · Nonlinear Partial Differential Equations · Nonlinear Differential Equations Analysis
Positive solutions for nonlinear problems involving the one-dimensional -Laplacian††thanks: 2000 Mathematics Subject Clasification. 34B15;
34B18; 35J25. ††thanks: Key words and phrases. Elliptic one-dimensional problems, -Laplacian, positive solutions. ††thanks: Partially supported by Secyt-UNC 30720150100019CB.
U. Kaufmann, L. Milne *E-mail addresses. *[email protected] (U. Kaufmann, Corresponding Author), [email protected] (L. Milne).
FaMAF, Universidad Nacional de Córdoba, (5000) Córdoba, Argentina
Abstract
Let , and be a real parameter. Let be the differential operator given by , where is an odd increasing homeomorphism and . We study the existence of positive solutions for problems of the form
[TABLE]
where is a continuous function which is, roughly speaking, sublinear with respect to . Our approach combines the sub and supersolution method with some estimates on related nonlinear problems. We point out that our results are new even in the cases and/or .
1 Introduction
Let , and be a real parameter. Let us consider problems of the form
[TABLE]
where is an odd increasing homeomorphism and is a continuous function. The existence of positive solutions for problems as (1.1) involving the so-called -Laplacian have been widely studied in the literature (see e.g. [1, 2, 4, 5, 11, 14, 15, 16, 23] and the references therein) and appear in diverse applications such as reaction-diffusion systems, nonlinear elasticity, glaciology, population biology, combustion theory, and non-Newtonian fluids, see for instance [8, 10, 12, 17]. We mention also that these kind of problems arise naturally in the study of radial solutions for nonlinear equations in annular domains (see e.g. [21] and its references).
When and with and , the existence of positive solutions for (1.1) was considered in [13], even for sign-changing weights (see also [9, 6] for the analogous -dimensional problem). We note, however, that for the computations in [13] it was crucial the homogeneity of both and , which of course is no longer true here.
Let us now introduce the following assumptions on and :
with in and on any subinterval of 2.
with 3.
There exist increasing homeomorphisms such that for all 4.
There exist such that for and all
Under some standard growth conditions on (which allow both sublinear and superlinear nonlinearities) and assuming and , it was proved that (1.1) possesses a positive solution for all (see [20, Theorem 1.1]), and recently in [22, Theorem 2] the authors extended this result to certain and not requiring that . We point out that these hypothesis impose, in particular, rather strong restrictions on
[TABLE]
Indeed, the existence of as above implies that for all and , while the existence of entails that for all . Let us note that the first and third of these conditions are not satisfied for instance by exponential-like nonlinearities, and the remaining one does not hold for example for logarithmic-like functions.
On the other side, a similar result was established in [3, Corollary 3.4] assuming and . We observe that the first inequality in also implies that for all , while the second one requires that (and this does not occur, for instance, with logarithmic-like nonlinearities). Let us add that in all these works the main tool utilized was some kind of Krasnoselskii-type fixed point theorem in cones.
Following a different approach, in Theorem 3.2 below we shall improve substantially the aforementioned results in the sublinear case, under much weaker conditions on both and . In fact, regarding the assumptions on , we shall only require that in . Furthermore, we shall see that the solutions in as . In order to derive our theorems, we shall rely on the well-known sub and supersolution method, combined with upper and lower estimates on some related nonlinear problems.
Also, under some additional hypothesis on and , we shall prove in Theorem 3.4 similar results for the differential operator
[TABLE]
where . Moreover, as a consequence of Theorems 3.2 and 3.4, we shall deduce the existence of (nontrivial) nonnegative solutions for sign-changing weights , see Corollary 3.5.
The rest of the article is organized as follows. In the next section we collect some auxiliary results, while in Section 3 we shall state and prove our main theorems. Finally, at the end of the paper we present several examples illustrating our conditions and their relations with the ones already mentioned (see also Remarks 3.1 and 3.3).
2 Preliminaries
Let be an odd increasing homeomorphism and . We start compiling some necessary facts about the problem
[TABLE]
Remark 2.1**.**
For every , (2.1) admits a unique solution such that is absolutely continuous and that the equation holds pointwise . In fact, one can see that
[TABLE]
where is the unique constant such that . Furthermore, the solution operator is continuous (see e.g. [7, Lemma 2.1]).
The following lemma shows that is a nondecreasing operator. Although this result should probably be well-known, we have not been able to find a proof in the literature.
Lemma 2.2**.**
Let with . Then in .
Proof. Let , , and suppose by contradiction that . Let be a connected component of . Note that, either by the continuity of and or by the boundary condition in (2.1), on . Taking into account this, multiplying (2.1) (with in place of ) by and integrating by parts we get
[TABLE]
Since we can argue in the same way with the equation involving and in , recalling that is increasing we infer that
[TABLE]
and thus in . Furthermore, in because on . Contradiction.
For with we define
[TABLE]
and
[TABLE]
Observe that, since , is well defined and (and so, ). Let us also set
[TABLE]
The next lemma provides some useful upper and lower bounds for when is nonnegative.
Lemma 2.3**.**
Let with . Then in it holds that
[TABLE]
Proof. Let . Since is increasing and in , using (2.2) we see that is nonincreasing and so is concave in . Hence, since on and we deduce that and therefore
[TABLE]
Employing again the fact that is increasing and (2.10) we find that
[TABLE]
and thus from the concavity of we obtain the second inequality in (2.9).
Let us prove the first inequality in (2.9). We first claim that
[TABLE]
In order to verify this, let be some point where reaches its maximum (and so ). We note that . Indeed, when this is obvious. If , then by (2.2) we have for all , with by (2.10). In particular, is increasing for such and thus as asserted. Hence, recalling the concavity of we get that for all
[TABLE]
Analogously, for
[TABLE]
and the claim is proved.
Suppose now that . Taking into account that is an homeomorphism with , that and , we derive that . Then, recalling (2.2), that is increasing and ,
[TABLE]
Assume now that . In this case we rewrite as
[TABLE]
where is the unique constant such that . Moreover, reasoning as in the previous paragraph we see that . Therefore,
[TABLE]
Taking into account (2.11), (2.12) and (2.13) we may infer the first inequality in (2.9) and this concludes the proof.
Remark 2.4**.**
Let with .
- (i)
Observe that, since , the constant that appears in the first term of the inequalities in (2.9) is strictly positive. 2. (ii)
For any with in , note that and . Therefore, by the above lemma we have that
[TABLE]
Let be a Carathéodory function (that is, is measurable for all and is continuous for ). Let be as in (1.2), and let us now consider problems of the form
[TABLE]
We say that is a *subsolution *of (2.14) if there exists a finite set such that , for each and
[TABLE]
If the inequalities in (2.15) are inverted, we say that is a supersolution of (2.14).
For the reader’s convenience we state the following existence theorem in the presence of well-ordered sub and supersolutions (for a proof, see for instance [18, Theorem 7.16]).
Theorem 2.5**.**
Let and be sub and supersolutions respectively of (2.14) such that for all . Suppose there exists such that
[TABLE]
Then there exists solution of (2.14) with in .
3 Main results
Before proving our main results, let us introduce the following conditions on and .
H1. There exist and an increasing homeomorphism defined in such that and
[TABLE]
H1’. There exists such that
[TABLE]
H2. There exist such that
[TABLE]
F1. There exist such that
[TABLE]
F1’. There exist such that
[TABLE]
We notice that . Let us also mention that the inequality in (3.4) appears (but for large values of and ) in the so-called condition referred to Young functions (see e.g. [19]).
Remark 3.1**.**
- (i)
Note that if the condition (3.3) holds automatically since is increasing and thus in that case H1’ reduces to (3.2). On the other hand, if H1 is true with for some , fixing in (3.1) we see that H1 implies (3.2). In other words, in this particular case, in “small” domains H1 is stronger than H1’. However, in general, these hypothesis are independent (see examples (a2) and (d) at the end of the paper). 2. (ii)
Suppose that fulfills H1’ or H1 with for some . Then the condition
[TABLE]
is sufficient in order for H2 to hold. Indeed, in any case we may assume (3.2) (see (i)). Hence, given any , there exists such that for all . Also, (3.7) implies that for every there exists such that for all . It follows that for all and ,
[TABLE]
and thus H2 is valid. We observe however that (3.7) is not necessary for H2 to be true (see examples (a4), (b) and (c) below). 3. (iii)
Let us point out that if is differentiable in and
[TABLE]
then one can readily verify that H2 holds with . 4. (iv)
It is not difficult to check that the hypothesis H1 and H2 are independent, and that the same is true for H1’ and H2, see examples (a), (a2) and (d).
Our results shall provide us with solutions that lie in the interior of the positive cone of , which is denoted by
[TABLE]
Theorem 3.2**.**
Let with .
(i) Assume H1 and F1 with
[TABLE]
Then for all there exists solution of (1.1).
(ii) Assume H1’ and F1’ with
[TABLE]
Then for all there exists solution of (1.1).
Moreover, in both (i) and (ii) it holds that
[TABLE]
Remark 3.3**.**
When is the -Laplacian, i.e. with , clearly H1 (with ) and H1’ (with in place of in (3.2)) hold. Furthermore, (3.8) is valid if and only if , so in this case we have the usual growth condition that characterizes the sublinear problems. Observe also that, since for the -Laplacian in (ii) we can take any and , Theorem 3.2 (i) and (ii) provide here the same result.
Proof. Let . We start proving (i). Let , be given by H1 and F1 accordingly. By the the continuity of and the fact that , there exists such that
[TABLE]
for all , where is given by (3.3). Also, let and be as in (2.8) and set
[TABLE]
It follows from the definition of that . Let us also write
[TABLE]
We now observe that by (3.8) there exists such that for all . Hence,
[TABLE]
for . We notice next that H1 says that for all and , and therefore
[TABLE]
for all and .
Let us choose
[TABLE]
and for such define . Since and in , the second inequality in (2.9) and (3.11) tell us that . Consequently, taking into account (3.12), (3.13) and (3.14), employing F1 and Remark 2.4 (ii) we deduce that
[TABLE]
In other words, is a subsolution of (1.1).
On the other side, we see that H1 yields that for all and . Thus, for such and and so,
[TABLE]
for all and . Let now . Recalling (3.12), (3.14) and (3.16) and utilizing again F1 and Lemma 2.3, we get that
[TABLE]
and hence is a supersolution of (1.1). Moreover, since and is nondecreasing (see Lemma 2.2) we infer that in . Then, we may apply Theorem 2.5 to obtain a solution of (1.1) with in , and since it also holds that .
Let us prove (ii). Let be given by H1’. We note that (3.2) implies that for all and some . Hence, we have that for , or equivalently,
[TABLE]
for such . We now set
[TABLE]
and similarly to (i) we define , picking
[TABLE]
where is such that . As in the proof of (i) we have that . Thus, taking into account (3.9), (3.17) and (3.18) and arguing as in (3.15) we derive that
[TABLE]
On the other hand, let (by (3.3)). For all we have and so
[TABLE]
for all . Let with
[TABLE]
Recalling F1’, the upper bound given by Lemma 2.3 and that and in , employing (3.20) and (3.21) we infer that
[TABLE]
Furthermore, enlarging if necessary so that and utilizing Lemma 2.2 we can achieve that in and thus we obtain a solution of (1.1).
Finally, let us prove (3.10). Let be fixed, and consider . We first observe that the solutions obtained in either (i) or (ii) can be chosen such that with independent of . Indeed, since is a supersolution of (1.1) for any , and since the above part of the proof provides arbitrary small subsolutions of (1.1) (that converge to [math] in as , by the second inequality in (2.9)), it follows from Theorem 2.5 that there exist solutions of (1.1) such that for all . So, as claimed. Taking into account this, the upper estimate in Lemma 2.3 yields that
[TABLE]
uniformly in as and so .
We choose next such that . Integrating (1.1) over we get that and hence by the above paragraph we see that as . Now, for any , we integrate (1.1) over to find that
[TABLE]
uniformly when . Thus, the proof of (3.10) is complete.
We next consider the case with , that is, the problem
[TABLE]
Theorem 3.4**.**
Let with . Assume that fulfills H2, and suppose and satisfy the hypothesis of Theorem 3.2 (i) or (ii), with for some in case (i). If either in or and , then for all there exists solution of (3.22). Moreover, these satisfy (3.10).
Proof. The proof follows the lines of the proof of Theorem 3.2 and hence we only indicate the minor changes that are needed.
Let and suppose the hypothesis of Theorem 3.2 (i) hold. Let be given by H2 and pick such that
[TABLE]
For such define . Taking in (3.1) (and recalling that here for some ) we get that there exists such that
[TABLE]
where is given by (3.3). Taking into account that in , using Lemma 2.3 and H2 we derive that
[TABLE]
Now, assume first that in . By (3.8) we have that . Thus, making smaller if necessary, since , from (3.15) and (3.24) we get that
[TABLE]
On the other hand, if and , for all sufficiently small, also from (3.15) and (3.24) we deduce that
[TABLE]
Hence, in any case we obtain a subsolution of (3.22) which belongs to . Furthermore, these subsolutions tend uniformly to zero (by Lemma 2.3) as . Therefore, since the solutions given by Theorem 3.2 (which also lie in ) are supersolutions of (3.22), Theorem 2.5 yields the desired solution . Moreover, it also follows that , and similar computations to those in the last part of the proof of Theorem 3.2 show that satisfy (3.10).
Suppose now the assumptions of Theorem 3.2 (ii) hold. Then we set , where is given by H1’. Since (3.23) is true by (3.2), proceeding as in (3.24) we have that
[TABLE]
Therefore, employing (3.19) in place of (3.15) and arguing as in the above two paragraphs we can construct arbitrarily small subsolutions and thus the proof can be completed as before.
As a direct consequence of Theorems 3.2 and 3.4 we are able to provide an existence result also for
[TABLE]
where or accordingly, in the case where changes sign in . As usual, we write with .
Corollary 3.5**.**
Let such that there exists an open interval with in . Suppose the hypothesis of one of the above theorems are satisfied, with in place of . Then for all there exists nonnegative (and nontrivial) solution of (3.25). Moreover, these satisfy (3.10).
Proof. Let , and let be the solution of (3.25) with in place of , provided by some of the above theorems. It is clear that is a supersolution of (3.25).
On the other side, since in , an inspection of the proofs of the aforementioned theorems show that we can find some with in and such that
[TABLE]
Define now by in and in . Then is a subsolution of (3.25) and this yields the existence assertion.
To conclude the proof we note that the last assertion follows similarly to the previous theorems, having in mind that in and that uniformly as .
Examples. We assume that since we may extend oddly.
(a) Let be continuous and nondecreasing, with increasing in for some if . Define
[TABLE]
Then fulfills H1 with and because
[TABLE]
for all and . Furthermore, in this case the condition (3.8) of Theorem 3.2 is true if and only if .
Let us note that here satisfies H2 if and only if does. In particular, taking some which does not fulfill H2 we obtain a function that satisfies H1 but not H2 (one such is for instance for and ).
We finally point out that if , the above paragraph together with Remark 3.1 (i) imply the existence of some which satisfies H1’ but not H2.
Let us exhibit next some interesting particular cases:
(a1) Let
[TABLE]
Since is increasing, by the first paragraph in (a) we get that H1 holds, and it is also clear that H2 is true with and any .
(a2) Let
[TABLE]
A brief computation shows that is increasing and so fulfills H1. Moreover, taking into account Remark 3.1 (ii) we see that satisfies H2. Note also that if , then (3.3) is not valid. In particular, does not fulfill H1’ in this case (and for any , *neither *satisfies the conditions and at the introduction nor the one in [22]).
(a3) Let
[TABLE]
One can check that is increasing and therefore H1 holds. Also, recalling Remark 3.1 (ii) we deduce that satisfies H2. We observe that *does not *satisfy the assumptions and or the one in [22].
(a4) Let
[TABLE]
Since is increasing we infer that H1 is valid. Although (3.7) *does not *hold with , one can verify that fulfills H2 with and .
(b) Let
[TABLE]
It can be proved that cannot be written as in (3.26) with and nondecreasing. Let us demonstrate, however, that satisfies H1 and H2. Let . We choose such that for all . Then, for such and all we have that
[TABLE]
and H1 holds. Also, employing the first inequality in (3.27) it is easy to see that H2 is true with and any . Let us finally note that (3.7) *is not *valid with any .
(c) Let
[TABLE]
Then fulfills (3.2) with and also (3.3). In other words, H1’ holds (let us remark that, despite it is less direct, one can prove that also satisfies H1 with and for some large enough). Although (3.7) *does not *hold with , from Remark 3.1 (iii) we deduce that H2 is valid since
[TABLE]
for all and all .
(d) Let
[TABLE]
One can readily check (3.2) and (3.3), and thus H1’ is true. Also, utilizing again Remark 3.1 (ii) we get that H2 holds. Let us observe that *does not *satisfy H1 (and hence *neither *fulfills the conditions or at the introduction) because
[TABLE]
and so there is not a continuous such that (3.1) is valid and . Furthermore, this tell us that neither meets the assumptions in [22].
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