1. Introduction
Let
be a bounded domain with a
-boundary
. In this paper we study the following anisotropic
-equation
Given
, we define
Let
and for
we introduce the anisotropic
r-Laplacian differential operator defined by
In (
1), the left hand side (the differential operator) is the sum of two such operators with different exponents. Equations driven by the sum of two differential operators of different nature, such as the anisotropic
-equations of the present work, arise in the mathematical models of many physical processes. We refer to the survey papers of Marano–Mosconi [
1] and Rădulescu [
2] and the references therein. In particular, anisotropic equations arise in elasticity (see Zhikov [
3,
4]) and in the study of electrorheological and magnetorheological fluids (see Rǔžička [
5] and Versaci–Palumbo [
6]). For other papers dealing with the sum of two differential operators of different nature (mostly
-Laplacian) we refer to Candito–Gasiński–Livrea [
7], Gasiński–Klimczak–Papageorgiou [
8], Gasiński–Papageorgiou [
9,
10,
11,
12], Gasiński–Winkert [
13,
14], and for anisotropic problems governed by the
-Laplacian we refer to Gasiński–Papageorgiou [
15,
16]. Finally for the use of the eigenproblem to molecules we refer to Jäntsch [
17], and Teng–Lu [
18].
In the reaction (right hand side of (
1)), the function
is a Carathéodory function (that is, for all
,
is measurable and for almost all
,
is continuous) and asymptotically as
, we can have resonance with respect to the principal eigenvalue of
. Using variational tools from the critical point theory, together with suitable truncation techniques and Morse theory (critical groups), we show that problem (
1) has at least three nontrivial smooth solutions, one positive, one negative and the third nodal (sign-changing). For isotropic problems, such three solutions theorem was proved for Dirichlet problems driven by the
p-Laplacian by Liu [
19] (Theorem 1.2). In that paper, the reaction
asymptotically as
is uniformly nonresonant with respect to the principal eigenvalue of
and no nodal solutions are obtained. For the same problem, the resonant case was examined by Liu–Su [
20], who obtained two nontrivial solutions, but without providing sign information for them.
The study of anisotropic equations is lagging behind and there is only the work of Fan–Zhao [
21], who produced nodal solutions for a class of radially symmetric equations driven by the anisotropic
p-Laplacian. Our work here appears to be the first one producing three nontrivial smooth solutions with sign information for resonant anisotropic
-equations.
2. Mathematical Background—Hypotheses
The study of problem (
1) requires the use of Lebesgue and Sobolev spaces with variable exponents. For a comprehensive presentation of such spaces, we refer to the book of Diening–Harjulehto–Hästö-Rǔžička [
22].
By
we denote the space of all functions
which are measurable. As usual, we identify two such functions which differ only on a Lebesgue-null set. Given
, the variable exponent Lebesgue space
is defined by
We equip this space with the so-called “Luxemburg norm”, defined by
The space
is separable and uniformly convex (hence reflexive too, by the Milman–Pettis theorem; see Papageorgiou–Winkert [
23] (Theorem 3.4.28, p. 225) or Gasiński–Papageorgiou [
24] (Theorem 5.89, p. 853)). If
then
and we have
. In addition the following Hölder-type inequality holds
These function spaces have many properties similar to the classical Lebesgue -space. So, if and for all , then embeds continuously into .
Using the variable exponent Lebesgue spaces, given
, we can define the variable exponent Sobolev space
as follows
In this definition, the gradient
is understand in the weak sense. The space
is equipped with the following norm
For the sake of notational simplicity, we will write .
When
(that is,
is Lipschitz continuous on
), then we can also define
The spaces
and
are separable and uniformly convex (thus reflexive). For the space
, the well-known Poincaré inequality is still valid, namely there exists
such that
This inequality implies that on
we can consider the equivalent norm
The Sobolev embedding theorem can be extended to the present setting. More precisely, let
and set
for all
(the critical Sobolev exponent corresponding to
). Suppose that
,
,
and also
Then the following embeddings are true
The study of the anisotropic Lebesgue and Sobolev spaces uses the following modular function
for
. This modular function is closely related to the Luxemburg norm.
Proposition 1. If , and , then
(a) .
(b) (resp. , ) (resp. , ).
(c) .
(d) .
(e) .
(f) .
We know that for
, we have
So, to every
, correspond functions
such that
with
. Then we introduce the operator
defined by
This operator has the following properties (see Gasiński–Papageorgiou [
15] (Proposition 2.5)).
Proposition 2. The operator is bounded (that is, maps bounded sets into bounded sets), continuous, strictly monotone (hence maximal monotone too) and of type , which means that
“if in and , than in .”
In addition to the anisotropic spaces, we will also use the space
This is an ordered Banach space with positive (order) cone
This cone has a nonempty interior given by
with
n being the outward unit normal on
.
We will need also some information about the spectrum of
. So, with
, we consider the following eigenvalue problem
We say that
is an “eigenvalue”, if problem (
2) admits a nontrivial solution
, known as a corresponding “eigenfunction”. We introduce the set
In contrast to the isotropic case, in the anisotropic case it can happen that
(see Fan–Zhang–Zao [
25] (Theorem 3.1)). However, if there exists
(
) such that for all
, the function
is monotone on
and if
, then there exists a principal eigenvalue
with corresponding eigenfunction
(see Fan–Zhang–Zao [
25] (Theorem 3.3)). Moreover, we have
Let
X be a Banach space,
and
. We introduce the following sets
Suppose that
is a topological pair such that
. For
, by
we denote the
k-th relative singular homology group for the pair
with integer coefficients. Let
be isolated and
. Then the critical groups of
at
u are defined by
with
U being a neighbourhood of
u such that
. The excision property of singular homology implies that this definition of critical groups is independent of the choice of the neighbourhood
U.
Next let us fix our basic notation. If
, then
. So, given
, we set
. We have
Furthermore, for
, we write
and
. Recall that by the Poincaré inequality we have
We say that a set is “downward directed” if, for every pair , we can find such that , ; similarly, we say that S is “upward directed”, if for every pair , we can find such that , . Finally as for the Luxemburg norm, we write for all .
Now we are ready to introduce the hypotheses on the data of problem (
1).
Hypothesis 1. and there exists such that for all , the function is monotone on , and for all .
Hypothesis 2. is a Carathéodory function such that for a.a. and
- (i)
for a.a. , all , with ;
- (ii)
and uniformly for a.a. , with ;
- (iii)
there exist such that - (iv)
there exist , and such that and
Remark 1. Hypothesis 2(ii) incorporates in our framework problems which are resonant with respect to the principal eigenvalue . Hypothesis Hypothesis 2(iv) implies the presence of a local concave term near zero.
Example 1. The following function satisfies hypotheses :where , and (note that ); see Figure 1. We will need the following lemma.
Lemma 1. If , for a.a. and , then there exists such that Proof. We proceed indirectly. So, suppose that the lemma is not true. Then we can find a sequence
such that
Evidently we may assume that
for all
. We have
Suppose that the sequence
is not bounded. Then by passing to a subsequence if necessary we may assume that
so, by Proposition 1, also
We set
,
. Differentiating, we have
so
with
as
. Hence
where
denotes the Lebesgue measure on
.
Then Proposition 1 and the Poincaré inequality, imply that the sequence is bounded.
Passing to a suitable subsequence if necessary, we may assume that
The modular function
is continuous, convex on
. Therefore it is sequentially weakly lower semicontinuous. Hence we have
See (
7). Furthermore, (
8) implies that
See (
9) and recall the hypothesis on
, thus
Suppose that
. then from (
4), passing to the limit as
, we have
, a contradiction.
Next suppose that
for all
. Using this in (
10), from first inequality there, we obtain
which contradicts (
3). This proves the boundedness of the sequence
. Then we may assume that
From (
10) and reasoning as above (replacing
with
), we reach again a contradiction. So, the assertion of the lemma is true. □
3. Solutions of Constant Sign
We introduce the energy (Euler) functional for problem (
1) and the positive and negative truncations of it. So, we consider the following three functionals
:
for all
.
Proposition 3. If hypotheses and hold, then the functionals φ, are coercive.
Proof. We do the proof for the functional , the proof for the functionals and being similar.
Hypotheses
imply that there exists
such that
Suppose that
is not coercive. Then we can find a sequence
and
such that
From the inequality in (
12) and since
for all
(see hypotheses
), we have
so
See (
3), thus
hence
with some
(see (
11)). Using also Proposition 1 and Poincaré’s inequality, we conclude that
From (
12) and (
13), it follows that
so by Proposition 1, we have
We set
,
. From the proof of Lemma 1 (see (
6) and (
7)), we have
with
and for some
, so the sequence
is bounded.
From (
12) and (
13), we have
for some
, so
thus
Using Hypothesis H2(ii), we obtain
with
,
for a.a.
(see Aizicovici–Papageorgiou–Staicu [
26] (proof of Proposition 16)). So, if in (
16) we pass to the limit as
and use (
14) and (
15) and the fact that the modular function
is sequentially weakly lower semicontinuous (being continuous convex), we obtain
If
, then from Lemma 1 and (
17), we have
so
(see Proposition 1). Then from (
16) we see that
a contradiction, since
for all
.
If
for a.a.
, then from (
17) and (
3) we have
so
If , then as above we have a contradiction.
If
for all
(see (
18)), we have
From (
12) and (
13), we have
for some
, so
See (
3), thus
for some
(see (
11)) and hence
But from (
19) and Fatou’s lemma, we have that
a contradiction.
Therefore we have that the sequence
is bounded, and thus the sequence
is bounded (see (
13)), a contradiction (see (
12)). This proves the coercivity of
.
In a similar fashion, we show that the functionals and are coercive too. □
Now that we have the coercivity of the functionals , we can use the direct method of the calculus of variations to produce two constant sign solutions.
Proposition 4. If hypotheses and hold, then problem (1) has at least two constant sign solutionsboth local minimizers of the energy functional φ. Proof. From Proposition 3 we know that
is coercive. Furthermore, using the Sobolev embedding theorem, we see that
is sequentially weakly lower semicontinuous. Hence by the Weierstrass–Tonelli theorem, we can find
such that
so
and hence
We test (
21) with
and obtain
so
.
Let
and choose
small such that
Here
is as postulated in Hypothesis H2(iv). Using the fact that
and
, we can write that
So, choosing
even smaller, we see that
so
Then from (
21) it follows that
is a positive solution of (
1) and we have
From Theorem 4.1 of Fan–Zhao [
27], we know that
. Then Lemma 3.3 of Fukagai–Narukawa [
28] (see also Tan–Fang [
29] (Corollary 3.1)), we infer that
. Note that Hypothese H2(i),(iv) imply that
So, from the anisotropic maximum principle of Papageorgiou–Qui–Rădulescu [
30] (Proposition 4) (see also Zhang [
31] (Theorem 1.2)), we have that
.
Note that
. Therefore we see that
so by Proposition 3.3 of Gasiński–Papageorgiou [
15] and Theorem 3.2 of Tan–Fang [
29],
Similarly, working with the functional , we produce a negative solution which is a local minimizer of . □
In fact we can show that there exist extremal constant sign solutions, that is, a smallest positive solution and a biggest negative solution. In
Section 4, we will use these extremal constant sign solutions in order to generate a nodal (sign-changing) solution.
To obtain the extremal constant sign solutions, we need to do some preliminary work.
Let
(resp.
) be the set of positive (resp. negative) solutions of Problem (
1). From Proposition 4 and its proof, we know that
We will produce a lower bound for the set
and an upper bound for the set
. To this end, note that on account of hypotheses
, we have
for some
.
This unilateral growth condition on
leads to the consideration of the following auxiliary anisotropic Dirichlet problem
For this problem, we have the following existence and uniqueness result.
Proposition 5. If hypotheses hold, then problem (23) has a unique positive solution . Moreover, since problem is odd is the unique negative solution of (23). Proof. First we prove the existence of a positive solution. For this purpose we introduce the
-functional
defined by
for all
.
Since
, we see that
is coercive. Furthermore, it is sequentially weakly lower semicontinuous. So, we can find
such that
Given
, since
, as in the proof of Proposition 4, we see that for
small we have
so
From (
24), we have
so
for all
. Choosing
in (
25), we obtain
so
and
.
As before (see the proof of Proposition 4), using the anisotropic regularity theory and the anisotropic maximum principle, we obtain that .
Next we show the uniqueness of this positive solution. For this purpose, we consider the integral functional
defined by
From Theorem 2.2 of Takáč-Giacomoni [
32], we know that
j is convex. Suppose that
is another positive solution of (
23). Again we have that
. Then using Proposition 4.1.22 of Papageorgiou–Rădulescu–Repovš [
33] (p. 274), we infer that
So, if
(the effective domain of
j) and
, then for
small we have
Hence the convexity of
j implies the Gâteaux differentiability of
j at
and at
in the direction
h. Moreover, using the chain rule and Green’s identity, we obtain
The convexity of
j implies the monotonicity of
. Therefore
so
. This proves the uniqueness of the positive solution
of (
23). Since problem (
23) is odd, it follows that
is the unique negative solution of problem (
23). □
These solutions will serve as bounds of and respectively.
Proposition 6. If hypotheses and hold, then for all and for all .
Proof. We do the proof for the set , the proof for the set being similar.
So, let
and introduce the Carathéodory function
defined by
We set
and consider the
-functional
defined by
for all
. Evidently
is coercive (see (
26)) and sequentially weakly lower semicontinuous. Therefore we can find
such that
If
, then we can find
small such that
(recall that
and use Proposition 4.1.22 of Papageorgiou–Rădulescu–Repovš [
33] (p. 274)). Using (
26) and the fact that
, we see that by taking
even smaller if necessary, we will have
so
See (
27), thus
.
In (
28) first we use the test function
. We obtain
so
,
.
Next, in (
28) we use
. We have
See (
26), (
22) and since
, so
(see Proposition 2).
Thus we have proved that
where
.
From (
29), (
26), (
28) and Proposition 5, it follows that
so
for all
(see (
29)).
In a similar fashion, we show that
□
Next following some ideas of Filippakis–Papageorgiou [
34], we show that
is downward directed and
is upward directed.
Proposition 7. If hypotheses and hold, then is downward directed and is upward directed.
Proof. We do the proof for , the proof for being similar.
For
, we consider the function
for
. The function
is Lipschitz continuous. Let
. Since
, from the chain rule for isotropic Sobolev spaces (see Papageorgiou–Rădulescu–Repovš [
33] (Proposition 1.4.2, p. 22)), we have
(recall that by Rademacher’s theorem; see Gasiński–Papageorgiou [
35] (Theorem 1.5.8, p. 56)
is differentiable almost everywhere). It follows that
Let
and consider
,
. We introduce the test functions
Evidently
(see (
30)). We have
We add these two equations, divide with
and then let
. Note that
and
So, in the limit as
, we obtain
(recall that
).
Let
. From (
31) we infer that
is an upper solution for Problem (
1). Then, by a standard truncation technique (see for example the proof of Proposition 6), we produce
satisfying
. hence
,
and this proves that
is downward directed.
Similarly we show that is upward directed. □
Now we are ready to produce the extremal constant sign solutions.
Proposition 8. If hypotheses and hold, then has a smallest element ( for all ), has a biggest element ( for all ).
Proof. From Proposition 7 we know that
is downward directed. Using Lemma 3.10 of Hu–Papageorgiou [
36] (p. 178), we can find a decreasing sequence
such that
We have
and
(see Proposition 6).
Testing (
32) with
and using (
33) and Hypothesis H2(i), we see that the sequence
is bounded. So, we may assume that
In (
32) we use
, pass to the limit as
and use (
33) and Hypothesis H2(i). We obtain
so
since
is monotone, thus
(see Proposition 2).
So, if in (
32) we pass to the limit as
and use (
35), then
Furthermore, from (
33), we have
Then (
36) and (
37) imply that
,
.
Similarly working with the set , we produce such that . Note that since is upward directed, we can find an increasing sequence such that . □
4. Nodal Solution
In this section we produce a nodal solution for problem (
1). The idea is to use truncations in order to focus on the order interval
Then on account of the extremality of the solutions
and
, any nontrivial solution of (
1) located in
and distinct from
and
will be nodal. To produce such a solution, we will combine tools from critical point theory and from Morse theory (critical groups).
We start with a result which provides the critical groups of the energy functional
at the origin. The result is a consequence of Hypothesis H2(iv) and follows from Proposition 6 of Leonardi–Papageorgiou [
37].
Proposition 9. If hypotheses and hold, then As mentioned above, to concentrate on the order interval
, we will use truncations. For this purpose, we introduce the function
g defined by
This is a Carathéodory function. We will also use the positive and negative truncations of
, namely the Carathéodory functions
We set
and consider the
-functionals
defined by
for all
.
Since
and
, using Proposition 9 and a simple homotopy invariance argument as in the proof of Proposition 4.4 of Papageorgiou–Rădulescu–Repovš [
38], we obtain the following result.
Proposition 10. If hypotheses and hold, then Now we are ready to produce nodal solutions.
Proposition 11. If hypotheses and hold, then problem (1) has a nodal solution Proof. Using (
38) and (
39), we can easily see that
On account of the extremality of
and
, we have
Claim. and are local minimizers of .
From (
38) and (
39) it is clear that the functionals
are coercive. Furthermore, they are sequentially weakly lower semicontinuous. So, we can find
such that
Let
and choose
small so that
(recall that
and use Proposition 4.1.22 of Papageorgiou–Rădulescu–Repovš [
33] (p. 274)). Using Hypothesis H2(iv) and recalling that
, by choosing
even smaller, we will have
so, by (
41) also
thus
and hence
From (
38) and (
39) it is clear that
Since
, it follows that
thus by Proposition 3.3 of Gasiński–Papageorgiou [
15] and Theorem 3.2 of Tan–Fang [
29], also
Similarly for using this time the functional .
This proves the Claim.
Without any loss of generality, we may assume that
The reasoning is similar if the opposite inequality holds.
From (
40) we see that we may assume that
is finite. Otherwise we already have an infinity of nodal solutions of (
1) and so we are done. By Theorem 5.7.6 of Papageorgiou–Rădulescu–Repovš [
33] (p. 449), we can find
small such that
From (
38) it is clear that
is coercive. So, using Proposition 5.1.15 of Papageorgiou–Rădulescu–Repovš [
33] (p. 369), we have that
Then (
42) and (
43) permit the use of the mountain pass theorem. So, we can find
such that
(see (
40) and (
42)). From (
42) and (
44), it follows that
Since
is a critical point of
of the mountain pass type, using Theorem 6.5.8 of Papageorgiou–Rădulescu–Repovš [
33] (p. 527), we have
From (
45) and Proposition 10, we infer that
. Therefore
□
Finally we can state the following multiplicity theorem for Problem (
1).
Theorem 1. If hypotheses and hold, then Problem (1) has at least three nontrivial solutions Remark 2. In this paper we examined resonant anisotropic problems in which the resonance occurs from the left of (see Hypothesis H2(ii)). This made the relevant energy functionals coercive (see Proposition 3). It is an interesting open problem what can be said if the resonance if from the right of . In this case the functionals fail to be coercive.