Nontrivial solutions of quasilinear elliptic equations with natural growth term
Marco Degiovanni, Alessandra Pluda

TL;DR
This paper establishes the existence of multiple solutions for a class of quasilinear elliptic equations with natural growth terms, using an adapted degree theory approach under invariant assumptions.
Contribution
It introduces a novel adaptation of degree theory to prove multiple solutions for quasilinear elliptic equations with natural growth, invariant under diffeomorphisms.
Findings
Proved existence of multiple solutions.
Developed an adapted degree theory method.
Applicable under invariant assumptions.
Abstract
We prove the existence of multiple solutions for a quasilinear elliptic equation containing a term with natural growth, under assumptions that are invariant by diffeomorphism. To this purpose we develop an adaptation of degree theory.
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Nontrivial
solutions of quasilinear elliptic
equations with natural growth term
Marco Degiovanni
Dipartimento di Matematica e Fisica
Università Cattolica del Sacro Cuore
Via dei Musei 41
25121 Brescia, Italy
and
Alessandra Pluda
Fakultät für Mathematik
Universität Regensburg
Universitätstrasse 31
93053 Regensburg, Germany
Dedicated to Gianni Gilardi
Abstract.
We prove the existence of multiple solutions for a quasilinear elliptic equation containing a term with natural growth, under assumptions that are invariant by diffeomorphism. To this purpose we develop an adaptation of degree theory.
Key words and phrases:
Quasilinear elliptic equations, divergence form, natural growth conditions, multiple solutions, degree theory, invariance by diffeomorphism.
2010 Mathematics Subject Classification:
35J66, 47H11
The authors are member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM)
1. Introduction
Consider the quasilinear elliptic problem
[TABLE]
where is a bounded, connected and open subset of and
[TABLE]
are Carathéodory functions such that: **
for every there exist and satisfying
[TABLE]
for a.e. and every with .
The existence of a weak solution of (1.1), namely a solution of
[TABLE]
follows from the results of [6, 7], provided that a suitable a priori -estimate holds, possibly related to the existence of a pair of sub-/super-solutions (see e.g. [6, Théorème 2.1] and [7, Theorems 1 and 2]). Moreover, each weak solution is locally Hölder continuous in (see e.g. [10, Theorem VII.1.1]).
The existence of multiple solutions, in the semilinear case and under suitable regularity assumptions, has also been proved for instance in [2]. Here we are interested in the existence of multiple nontrivial solutions, as implies that , under assumptions that do not imply an a priori -estimate. Let us state our main result.
Theorem 1.1**.**
there exist such that
[TABLE] 2.
the function is differentiable at for a.e. .
Consider the eigenvalue problem
[TABLE]
and denote by , , the sequence of the eigenvalues repeated according to multiplicity.
If there exists with and even, then problem (1.2) admits at least three nontrivial solutions in with
[TABLE]
If is constant and , the result is essentially contained in [3, 15], which in turn developed previous results of [4]. Actually, in those papers it is enough to assume that , because in that case (1.2) is the Euler-Lagrange equation of a suitable functional and variational methods, e.g. Morse theory, can be applied.
In our case there is no functional and also degree theory arguments cannot be applied in a standard way, because of the presence of the term . Let us point out that our assumptions do not imply that the solutions of (1.2) belong to , so that the natural growth term plays a true role.
Let us also mention that our statement has an invariance property.
Remark 1.2**.**
Let be an increasing -diffeomorphism with . Then the following facts hold:
the functions , and satisfy the assumptions of Theorem 1.1 if and only , and , defined as
[TABLE]
do the same; in particular, we have
[TABLE] 2.
a function is a solution of (1.2) if and only if is a solution of (1.2) with , and replaced by , and , respectively.
From the definition of we see that the term cannot be omitted in the equation, if we want to ensure this kind of invariance.
When (1.2) is the Euler-Lagrange equation associated to a functional, the question of invariance under suitable classes of diffeomorphisms has been already treated in [14], where it is shown that problems with degenerate coercivity can be reduced, in some cases, to coercive problems.
In the next sections we develop an adaptation of degree theory suited for our setting and then we prove Theorem 1.1 by a degree argument. Under assumptions that are not diffeomorphism-invariant, a degree theory for quasilinear elliptic equations with natural growth conditions has been already developed in [1]. Here we find it more convenient to reduce the equation (1.2) to a variational inequality possessing as obstacles a pair of sub-/super-solutions, according to an approach already considered for instance in [12].
2. Topological degree in reflexive Banach spaces
Let be a reflexive real Banach space.
Definition 2.1**.**
A map , with , is said to be of class if, for every sequence in weakly convergent to some in with
[TABLE]
it holds .
More generally, if is a metrizable topological space, a map , with , is said to be of class if, for every sequence in with weakly convergent to in , convergent to in and
[TABLE]
it holds (we write instead of ).
Assume now that is a bounded and open subset of , a bounded and continuous map of class , a closed and convex subset of and . We aim to consider the variational inequality
[TABLE]
Remark 2.2**.**
It is easily seen that the set
[TABLE]
is compact (possibly empty).
According to [8, 11, 13], if the variational inequality (2.1) has no solution , one can define the topological degree
[TABLE]
Let us recall some basic properties.
Proposition 2.3**.**
If (2.1) has no solution , then
[TABLE]
Proposition 2.4**.**
If (2.1) has no solution , and we set
[TABLE]
then
[TABLE]
Theorem 2.5**.**
If (2.1) has no solution , then .
Theorem 2.6**.**
If (2.1) has no solution and there exists such that
[TABLE]
then .
Theorem 2.7**.**
If and are two disjoint open subsets of and (2.1) has no solution , then
[TABLE]
Definition 2.8**.**
Let , be closed and convex subsets of . The sequence is said to be Mosco-convergent to if the following facts hold:
if , for any and is weakly convergent to in , then ; 2.
for every there exist and a sequence strongly convergent to in with for any .
Theorem 2.9**.**
Let be a bounded and open subset of , be a bounded and continuous map of class and , , be a family of closed and convex subsets of such that, for every sequence convergent to in , the sequence is Mosco-convergent to .
Then the following facts hold:
the set of pairs , satisfying
[TABLE]
is compact (possibly empty); 2.
if the problem (2.2) has no solution and we set
[TABLE]
then is independent of .
Proof.
If is a sequence in constituted by solutions of (2.2), then up to a subsequence is weakly convergent to some in and is convergent to some in . Then and there exists a sequence strongly convergent to in with . It follows
[TABLE]
whence
[TABLE]
Then and . For every there exists a sequence strongly convergent to in with . From
[TABLE]
it follows
[TABLE]
so that the set introduced in assertion is compact.
Assume now that the problem (2.2) has no solution . It is enough to prove that is locally constant.
Suppose first that for any . By Michael selection theorem (see e.g. [5, Theorem 1.11.1]) there exists a continuous map such that for any .
If we set
[TABLE]
then , and satisfy the same assumptions and
[TABLE]
by Proposition 2.4. Moreover for any . Therefore we may assume, without loss of generality, that for any .
Given , there exist a bounded and open subset of and such that
[TABLE]
and such that (2.2) has no solution in
[TABLE]
with . From Theorem 2.7 we infer that
[TABLE]
From [11, Theorem 4.53 and Proposition 4.61] we deduce that is constant on .
In general, given , let us distinguish the cases and .
If , by the Mosco-convergence there exists such that for any . By the previous step we infer that is constant on .
If , from Theorem 2.5 we infer that . Assume, for a contradiction, that there exists a sequence convergent to with for any . Again from Theorem 2.5 we infer that the problem (2.2) has a solution , in particular , for any . Up to a subsequence, is weakly convegent to some , whence by the Mosco-convergence, and a contradiction follows. ∎
Now let be a bounded and open subset of , let be a metrizable topological space and let
[TABLE]
be two Carathéodory functions. We will denote by the usual norm in and write , instead of , .
In this section, we assume that and satisfy the controllable growth conditions in the sense of [10], uniformly with respect to . In a simplified form enough for our purposes, this means that: **
there exist , , , and such that
[TABLE]
for a.e. and every , , ; such a is clearly unique.
It follows
[TABLE]
and the map defined by
[TABLE]
is continuous and bounded on , whenever is bounded in .
Theorem 2.10**.**
we have
[TABLE]
for a.e. and every , , , with .
Then is of class .
Proof.
See e.g. [13, Theorem 1.2.1]. ∎
3. Quasilinear elliptic variational inequalities with
natural growth conditions
Again, let be a bounded and open subset of and let now
[TABLE]
be two Carathéodory functions. In this paper we are interested in the case in which and satisfy the natural growth conditions in the sense of [10]. More precisely, we assume that: **
there exist and, for every , , , and such that
[TABLE]
for a.e. and every , with ; such a is clearly unique; 2.
we have
[TABLE]
for a.e. and every , with .
Then we can define a map
[TABLE]
by
[TABLE]
Remark 3.1**.**
Assume that and also satisfy and . An easy density argument shows that, if
[TABLE]
then
[TABLE]
Consider also a -quasi upper semicontinuous function and a -quasi lower semicontinuous function , and set
[TABLE]
where is any -quasi continuous representative of (see e.g. [9]).
We aim to consider the solutions of the variational inequality
[TABLE]
We denote by the set of solutions of (). We will simply write , if no confusion can arise.
For every , we also set
[TABLE]
Proposition 3.2**.**
A function satisfies () if and only if
[TABLE]
Proof.
Assume that is a solution of () and let with a.e. in . Since is a nonincreasing sequence of nonpositive -quasi upper semicontinuous functions converging to -q.e. in , by [9, Lemma 1.6] there exists a sequence in converging to in with -q.e. in . Without loss of generality, we may assume that -q.e. in .
Then it follows that , whence
[TABLE]
Going to the limit as , we get
[TABLE]
If with a.e. in , the argument is similar. Since every can be written as with , it follows
[TABLE]
Since , the converse is obvious. ∎
We are also interested in the invariance of the problem with respect to suitable transformations.
Let us denote by the set of increasing -diffeomorphisms such that and by the set of -functions .
For any and , we define
[TABLE]
by
[TABLE]
If we define the Carathéodory functions
[TABLE]
by
[TABLE]
it easily follows that
[TABLE]
We also set and, given a set of real valued functions, .
It is easily seen that
[TABLE]
for every and .
We also say that is of Euler-Lagrange type, if there exists a function
[TABLE]
such that
is measurable for every ; 2.
is of class for a.e. ; 3.
we have
[TABLE]
for a.e. and every .
Taking into account Proposition 3.2, the next two results are easy to prove.
Proposition 3.3**.**
For every , the following facts hold:
the functions satisfy and with the same ; 2.
we have
[TABLE]
and is -quasi upper semicontinuous, while is -quasi lower semicontinuous (here we agree that and ); 3.
* is a solution of () if and only if is a solution of the corresponding variational inequality with , and replaced by , and , respectively;* 4.
the pair is of Euler-Lagrange type if and only if the pair does the same; moreover, if is associated with , then
[TABLE]
is associated with .
Proposition 3.4**.**
For every , the following facts hold:
the functions satisfy and with the same ; 2.
* is a solution of () if and only if the same is a solution of the corresponding variational inequality with and replaced by and , respectively.*
Remark 3.5**.**
Let with nonconstant, let and let , .
Then is not of Euler-Lagrange type. However, if we take , then , which is given by
[TABLE]
is of Euler-Lagrange type with
[TABLE]
Therefore the property of being of Euler-Lagrange type is not invariant under the transformation induced by , which plays in fact the role of “integrating factor”. By the way, if then we take , we get
[TABLE]
which are simply related to
[TABLE]
Proposition 3.6**.**
For every there exist and , depending only on , and , such that
[TABLE]
for a.e. and every , with .
Proof.
If we set with , we have
[TABLE]
whence the existence of . The existence of can be proved in a similar way. ∎
4. Quasilinear elliptic variational inequalities with
natural growth conditions depending on a parameter
Again, let be a bounded and open subset of and let now be a metrizable topological space and
[TABLE]
be two Carathéodory functions satisfying and uniformly with respect to .
More precisely, we assume that and satisfy and: **
there exist and, for every , , , and such that
[TABLE]
for a.e. and every , and with ; again, such a is clearly unique.
Then we can define a map
[TABLE]
by
[TABLE]
Consider also, for each , a -quasi upper semicontinuous function and a -quasi lower semicontinuous function , set
[TABLE]
and assume the following form of continuity related to the Mosco-convergence: **
for every sequence convergent to in , the following facts hold:
- •
if is a sequence weakly convergent to in , with for any and bounded in , then ;
- •
for every there exist and a sequence in which is bounded in and strongly convergent to in , with for any .
Then consider the parametric variational inequality
[TABLE]
Theorem 4.1**.**
Let be a sequence of solutions of () with bounded in and convergent to some in with .
Then admits a subsequence strongly convergent in to some and is a solution of ().
Proof.
Let and let be a sequence strongly convergent to in , with for any and bounded in . If we set
[TABLE]
and define , accordingly, it is easily seen that all the assumptions are still satisfied and now . Therefore, we may assume without loss of generality that for any .
Let be such that for any . We claim that is bounded in . Actually, by Propositions 3.4 and 3.6, we may assume, without loss of generality, that
[TABLE]
[TABLE]
which implies that is bounded in .
In a similar way one finds that is bounded in , so that is weakly convergent, up to a subsequence, to some in and . Let be a sequence strongly convergent to in , with for any and bounded in .
Let be a continuous function such that for and for . Then, let
[TABLE]
It is easily seen that each is also a solution of () with and replaced by and . Moreover, and satisfy both the assumptions , and the assumptions of [1, Theorem 4.2]. In particular, there exist , and such that
[TABLE]
for a.e. and every , and .
Now, if , the proof of [1, Theorem 4.2] can be repeated in a simplified form, as is bounded in . We have only to observe that, if is the solution of
[TABLE]
then there exists such that
[TABLE]
so that
[TABLE]
It follows
[TABLE]
and now the proof of [1, Theorem 4.2] can be repeated with minor modifications, showing that is strongly convergent to in . If , the argument is similar and simpler.
It is easily seen that is a solution of (). ∎
Remark 4.2**.**
In the previous theorem the assumption is crucial to ensure that is bounded in .
Consider , , , and , where .
If with and are the solutions of (), then the sequence is unbounded in .
5. Topological degree for quasilinear elliptic variational
inequalities with natural growth conditions
Consider again the setting of Section 3. Throughout this section, we also assume that: **
the functions and are bounded.
It follows that is automatically bounded in .
Theorem 5.1**.**
The set is (strongly) compact in (possibly empty).
Proof.
If , we have . Otherwise the assertion follows from Theorem 4.1. ∎
Definition 5.2**.**
We denote by the family of the subsets of which are both open and closed in with respect to the -topology. We will simply write , if no confusion can arise.
Fix a continuous function such that
[TABLE]
then set, for any and ,
[TABLE]
If we consider and define
[TABLE]
it is easily seen that satisfy and . Moreover, for every , they satisfy , if is restricted to . In particular, we can define a continuous map by
[TABLE]
and, by Theorem 2.10, this map is of class . We will simply write , if no confusion can arise.
Proposition 5.3**.**
For every , the following facts hold:
there exist a bounded and open subset of and such that
[TABLE]
and such that the variational inequality () has no solution ; in particular, the degree is defined whenever ; 2.
if have the same properties of and , and , are as in , then
[TABLE]
for every and ; 3.
if and also satisfy , and
[TABLE]
then we have
[TABLE]
for every and , provided that , , , are as in with respect to and , respectively.
Proof.
By definition of , there exists a bounded and open subset of such that
[TABLE]
If is a sequence of solutions of () with and , then is bounded in by . By Theorem 4.1, up to a subsequence is convergent to some in and is a solution of (). Then and a contradiction follows. Therefore, there exists such that () has no solution and assertion is proved.
To prove , define
[TABLE]
and consider . Arguing as before, we find with , , such that
[TABLE]
has no solution with , and .
From Theorem 2.7 we infer that
[TABLE]
On the other hand, we have
[TABLE]
by Theorem 2.9. Then assertion also follows.
The proof of is quite similar. ∎
Definition 5.4**.**
For every , we set
[TABLE]
where , , are as in and . We will simply write , if no confusion can arise.
Proposition 5.5**.**
Assume that and satisfy, instead of , the more specific controllable growth condition:
there exist , , , and such that
[TABLE]
for a.e. and every , .
Then the map is continuous, bounded on bounded subsets and of class . Moreover, if is a bounded and open subset of such that () has no solution and
[TABLE]
then and
[TABLE]
Proof.
It is easily seen that this time and satisfy and , for belonging to all .
Then the assertions follow from Theorems 2.10 and 2.9. ∎
Theorem 5.6**.**
Let with . Then .
Proof.
Let and be as in of Proposition 5.3. If , from Theorem 4.1 we infer that there exists such that () has no solution with . From Theorem 2.5 we deduce that
[TABLE]
and a contradiction follows. ∎
Along the same line, the additivity property can be proved taking advantage of Theorems 2.7 and 4.1.
Theorem 5.7**.**
Let with . Then and
[TABLE]
Theorem 5.8**.**
Let
[TABLE]
be two Carathéodory functions satisfying and with respect to and set
[TABLE]
Let also, for each , be a -quasi upper semicontinuous function and a -quasi lower semicontinuous function, define as in section 4 and assume that:
- •
the functions are bounded uniformly with respect to , we have for any and assumption is satisfied.
Then the following facts hold:
the set
[TABLE]
is (strongly) compact in (possibly empty); 2.
if is open and closed in with respect to the topology of and
[TABLE]
then for any and is independent of .
Proof.
First of all, the set is compact by Theorem 4.1. To prove assertion , let be a bounded and open subset of such that
[TABLE]
In particular, we have for any .
Let , let
[TABLE]
for and define
[TABLE]
It is easily seen that and satisfy and with respect to , so that we can consider the problem
[TABLE]
Since is compact, by Theorem 4.1 there exists such that (5.1) has no solution with and .
By Definition 5.4 we infer that
[TABLE]
and the assertion follows from Theorem 2.9. ∎
Remark 5.9**.**
By Theorem 5.8 and Proposition 5.5, can be calculated also by other approximation techniques, with respect to the one used in Definition 5.4.
Theorem 5.10**.**
If , then .
Proof.
Define, for ,
[TABLE]
It is easily seen that and satisfy assumptions and , so that Theorem 5.8 can be applied. If we take , we get
[TABLE]
Let and let
[TABLE]
with large enough to guarantee that , and
[TABLE]
From Proposition 5.5 and Theorem 2.6 we infer that
[TABLE]
and the assertion follows. ∎
Proposition 5.11**.**
Let and let and . Then , and
[TABLE]
Proof.
If we set
[TABLE]
the assertion follows from Theorem 5.8. ∎
6. Proof of Theorem 1.1
We aim to apply the results of the previous sections to
[TABLE]
By hypothesis , assumptions and are satisfied with . Moreover, if and are as in hypothesis , then and satisfy assumption .
Denote by , , the sequence of the eigenvalues of (1.3), repeated according to multiplicity, and set, for a matter of convenience, .
Finally, define , , , and as before, observe that and set
[TABLE]
It is easily seen that
[TABLE]
Let us also set
[TABLE]
so that
[TABLE]
Proposition 6.1**.**
For every there exist , depending only on and , such that
[TABLE]
Proof.
It is a simple variant of Proposition 3.6. ∎
Proposition 6.2**.**
Let
[TABLE]
be a Carathéodory function such that for every there exists satisfying
[TABLE]
and such that
[TABLE]
If is a solution of the variational inequality () with
[TABLE]
then satisfies the equation
[TABLE]
Proof.
Let with a.e. in , let and let
[TABLE]
Since , it follows
[TABLE]
On the other hand, we have
[TABLE]
whence
[TABLE]
Since , we can go to the limit as , obtaining
[TABLE]
Arguing on , one can prove in a similar way that
[TABLE]
whence
[TABLE]
and the assertion follows. ∎
Proposition 6.3**.**
Let be connected and assume that satisfies a.e. in , on a set of positive measure and
[TABLE]
Then we have
[TABLE]
Proof.
By Propositions 3.4 and 6.1, we may assume without loss of generality that
[TABLE]
Then for every with a.e. in , we have
[TABLE]
with .
From [11, Theorem 8.15 and Remark 8.16] the assertion follows. ∎
Lemma 6.4**.**
Assume that is connected and that . Moreover, according to , let be such that
[TABLE]
Let also be a continuous function, with and outside , and consider the problem
[TABLE]
where
[TABLE]
Denote by the set of solutions of (6.1) and let
[TABLE]
Then there exist such that
[TABLE]
Proof.
If we set
[TABLE]
it is easily seen that assumptions and are satisfied. From Theorem 5.8 we infer that is compact in .
First of all, we claim that there exists such that
[TABLE]
By Propositions 3.4 and 6.1, we may assume without loss of generality that
[TABLE]
Assume, for a contradiction, that is a sequence in with . Then we may suppose, without loss of generality, that is convergent to [math] in and a.e. in and that is convergent to some . Let with and, up to a subsequence, weakly convergent to some in .
If with a.e. in , we have
[TABLE]
which implies that is bounded hence convergent, up to a subsequence, to some .
Then we also get
[TABLE]
whence
[TABLE]
If we choose , where is a positive eigenfunction of (1.3) associated with , we get
[TABLE]
whence .
Finally, the choice in (6.1) yields
[TABLE]
We infer that and a contradiction follows.
With this choice of , we also have
[TABLE]
for every . Now we claim that there exists such that
[TABLE]
By Propositions 3.4 and 6.1, now we may assume without loss of generality that
[TABLE]
Assume, for a contradiction, that is a sequence in the set at the left hand side with and . Then, up to a subsequence, is convergent to some in and a.e. in , while is convergent to some . By Propositions 6.2 and 6.3 we have a.e. in . If we write with , we have that is weakly convergent, up to a subsequence, to some in and, on the other hand, is convergent to [math] a.e. in , as . The choice in (6.1) yields
[TABLE]
whence
[TABLE]
We infer that and a contradiction follows. ∎
Proposition 6.5**.**
If is connected and , then we have and
[TABLE]
Proof.
Let , , , and be as in Lemma 6.4. We aim to apply again Theorem 5.8 with
[TABLE]
By Lemma 6.4 the set is open and closed in . From Theorem 5.8 we infer that
[TABLE]
and that
[TABLE]
Now we claim that . By Propositions 3.4 and 6.1 we may assume without loss of generality that
[TABLE]
If we take with a.e. in in (6.1), we see that . Moreover, if , the choice yields , hence
[TABLE]
Therefore , whence and the claim is proved.
From Theorem 5.10 we infer that
[TABLE]
and the assertion concerning follows.
The assertion concerning can be proved in a similar way. ∎
Proposition 6.6**.**
If there exists with , then and
[TABLE]
Proof.
If we set
[TABLE]
[TABLE]
[TABLE]
it is easily seen that assumptions and are satisfied. We aim to apply Theorem 5.8.
We claim that there exists such that, if and , then . Assume, for a contradiction, that is a sequence in with and . Let with and, up to a subsequence, weakly convergent to some in and convergent to some in .
Given , if we have
[TABLE]
whence
[TABLE]
Going to the limit as , we get
[TABLE]
If , we have
[TABLE]
and the same conclusion easily follows.
Then we infer that
[TABLE]
whence , as [math] is not in the sequence .
By Propositions 3.4 and 6.1, we may assume without loss of generality that
[TABLE]
If , the choice in (6.2) yields
[TABLE]
which implies that . If the argument is analogous and simpler. In a similar way one can show that and a contradiction follows. Therefore, there exists with the required property.
In particular, we can apply Theorem 5.8 with
[TABLE]
obtaining
[TABLE]
and
[TABLE]
On the other hand, if we set
[TABLE]
from Proposition 5.5 we infer that
[TABLE]
Now, if we set
[TABLE]
from [11, Theorem 4.53 and Proposition 4.61] we deduce that
[TABLE]
Finally, from [13, Theorem 2.5.2] it follows that
[TABLE]
∎
Proof of Theorem 1.1.
From Proposition 6.5 we know that
[TABLE]
By Theorem 5.6 and Propositions 6.2 and 6.3 we infer that there exist at least two solutions and of (1.2) with
[TABLE]
Assume, for a contradiction, that
[TABLE]
with by Proposition 6.6. From Theorems 5.10 and 5.7 we infer that
[TABLE]
and a contradiction follows. Therefore there exists
[TABLE]
By Proposition 6.2 is a sign-changing solution of (1.2). According to [10, Theorem VII.1.1], each is locally Hölder continuous in . ∎
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