Existence of Time Periodic Solutions of New Classes of Nonlinear Problems
Luisa Toscano^{1}, Speranza Toscano^{2}
^{1}Dep. of Math. and Appl."R. Caccioppoli",Univ. of Naples "Federico II", via Cintia, Monte S. Angelo, Italy
^{2}Dep. of Civil Ing,. Second Univ. of Naples, fac. of Ing., Aversa (CE), Italy
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To cite this article:
Luisa Toscano, Speranza Toscano. Existence of Time Periodic Solutions of New Classes of Nonlinear Problems.Pure and Applied Mathematics Journal.Vol.4, No. 5, 2015, pp. 189-215. doi: 10.11648/j.pamj.20150405.11
Abstract: We study the existence of one or more weak periodic solutions of nonlinear evolution PDEs in a cylinder of R^{N+1 }with conditions on lateral surface by using the results connected to a general evolution variational equation depending on a parameter.
Keywords: Time Periodic, Evolution PDEs, Nonstationarity, Weak Periodic Solutions
1. Introduction
Let . Let be an open, bounded and connected set, with boundary .
For N=1, the condition ""means that is a bounded open interval.
Let us set
the class of the real functions defined a.e. in , measurable and T-periodic with respect to t; with .
We denote by F the linear map .
Let and V be a closed subspace of such that We do not exclude . Let be a norm equivalent to the one of on V.
Set , we consider the normed spaces:
(1.1)
Remark 1.1. When we assume
We recall ([16], Chap.23) are reflexive and separable Banach spaces. It is not difficult to prove that are Banach spaces; the restriction of F to is a norm-preserving linear transformation into .
Consequently are reflexive and separable spaces.
Remark 1.2. These conclusions hold even if .In this case =and .
Let us denote by the duality between (dual space of ) and , and by Fréchet differential operator. Let A0, 0 (j=1,…,m; ) and B be real functionals defined in satisfying the conditions
Let us consider the following problem.
Problem .Find such that
where is a real parameter,
Problem is a particular case of Problem(P) studied in ([10], [12]) by using the Lagrange multipliers and the "algebraic" approach which is based on the fibering method [14]. In ([11], [12]) many applications of the results connected to Problem (P) related to nonlinear elliptic systems are present.
In section 2 we considered convenient about Problem to state existence theorems (Theorems 2.1-2.3) included in the results of ([4], [7] [8], [10], [12]) whose validity, by the way, depends on . Furthermore we added Propositions 2.1 and 2.2 useful in some concrete cases in order to establish the nonstationarity of the found solutions.
In section 3 we study some evolution PDEs in the cylinder Q, with also nonlocal nonlinearities and with different conditions on . About these problems, whose variational formulation is included in Problem , we find the existence of one or more weak periodic solutions, giving also some sufficient conditions to their nonstationarity.
It is known that the search for periodic solutions of nonlinear problems has attracted the attention of many researchers. In particular Pohozaev in ([13], [15]) introduced "the separation variables method" for nonlinear equations in which it is possible to find weak periodic solutions in the form As far as we know, the methods developed in literature are not applicable in this article.
2. Existence Theorems Preliminary Results
As we set:
We find the solvability of Problem in the cases based on the following assumptions
is nonempty and bounded in .
At first we consider the cases in which one of the assumptions is present:
Let us introduce the open set A of the space :
Theorem 2.1 ( [10], Section 2 ). Under assumptions , in case we have:
.
When is a vector lattice, if and , then is solution of Problem . Consequently we can assume .
When m>1, for any the equation has only one positive root and we have
Evidently
functionals
belong to. (2.1)
Theorem 2.2 ([10],Section 2;[12],Section 2).Under assumptions , in cases we have:
.
When is a vector lattice, if and then are solutions of Problem . Consequently we can assume .
Proposition 2.1. In cases let be as in Theorems 2.1 and 2.2. If is such that , then
(2.2)
Proof. Let us set . We note that ,and . Since and , there exist and only one function such that and ; moreover = .
Let such that ; then
(2.3)
In . Then , from which (2.2) since
.
In ,.
Consequently , i.e. (2.2). In fact, since
we have
□
Let us pass to the cases in which is present:
In for any the equation has only one positive root with . Let us set .
Theorem 2.3 ([10], Section 4 ).Let hold. In case
.
In case
.
When is a vector lattice, if and then in and are solutions of Problem . Consequently we can assume .
Proposition 2.2. Let be as in Theor. 2.3. In case if is such that , then
(2.4)
In case if is such that , then
(2.5)
Proof. As let us consider the function of . Since and , there exist and only one function such that , and we have
In , where , with such that we have
;
then
from which (2.4) follows.
In , where , we note that relations imply that there exist an open ball centered in included in and only one functional such that for any . Since , set such that , we have
.
Then functional
belongs to .
Taking into account that
,
we have
from which (2.5) since . □
3. Some Applications
In this section we suppose and set
the Lebesgue measure on R;
the outward orthogonal unitary vector to ;
the class of the real functions defined a.e.in R, measurable and T-periodic;
the support of the restriction of .
We warn that the weak continuity in of the functionals present in the applications can be easily proved by using embedding Sobolev theorems [1],a compactness lemma ([6], Theor.5.1 page 58) and the isomorphism F (section 1).
We add the following clarification:
Application 3.1 (connected to Theor.2.1). Let us assume in the definition (1.1) of , then
,
and let us set as any
where
(3.1)
Problem becomes:
Find such that
(3.2)
Each solution u of (3.2) is for definition a weak solution of the problem:
(3.3)
Evidently
(3.4)
Let be the first eigenvalue and the first eigenfunction of the problem:
We remember that [3] and
(3.5)
(3.5) implies that
;
then as
(3.6)
from which
Then, since
we have
(3.7)
Relations (3.4),(3.7) and Theor.2.1 let us to state the following proposition.
Proposition 3.1. Under conditions (3.1), with as in (3.7) problem (3.3) has at least two weak solutions and (,,).
Remark 3.1. If , we have . Then, since , it results in . Additionally, when , since , holds if and only if .
Proposition 3.2. Let if . If there exist a measurable set with , a limit point such that a.e. in and a.e. in, then is nonstationary.
Proof. The additional assumption on implies that
with continuous embedding. (3.8)
Reasoning by contradiction, let . Set a.e. in, a Lebesgue theorem assures that
(3.9)
Let . Since by (3.8) , from (3.2) withand we get
(3.10)
Let us add that
the left side of (3.10) is (3.11)
where
Since , from (3.10),(3.11) we get
(3.12)
Then (3.12) contradicts (3.9). □
Application 3.2 (connected to Theor.2.1 and Theor.2.3 (case (c_{8}))). Let us assume in the definition (1.1) of , then
,
and let us set as any
where
(3.13)
Problem becomes:
Find such that
(3.14)
Each solution u of (3.14) is for definition a weak solution of the problem:
(3.15)
Let us introduce the conditions
0 in Q, (3.16)
(3.17)
Evidently
Proposition 3.3.Under conditions (3.13) (with p>1 and not necessarily >q_{1}),(i_{21}) holds if l<0.
Proof. Let l<0. Reasoning by contradiction, as any there exists such that
Then with we have
moreover there exists such that (within a subsequence)
Consequently
,
from which and the contradiction □
Proposition 3.4. Under conditions (3.13), (3.16) and (3.17), there exists satisfying the condition
Consequently holds if .
Proof. Reasoning by contradiction, as any such that
Then with such that (within a subsequence)
from which since (3.17) we get . Then and the contradiction □
Proposition 3.5. Under conditions (3.13) and (3.17), there exists such that holds if .
Proof. Reasoning by contradiction, as any such that
(3.18)
(3.19)
(3.20)
Relation (3.20) implies there exists strictly increasing such that
.
Set , from (3.18), (3.19) we get
(3.21)
(3.22)
Let such that (within a subsequence)
From (3.21), (3.22) we get
from which since (3.17) and the contradiction □
Propositions 3.3-3.5, Theorems 2.1 and 2.3 (case c_{8}) allow the following proposition.
Proposition 3.6. Under assumptions (3.13) we have:
when (3.16) and (3.17) hold, with problem (3.15) has at least two weak solutions and (,,);
when (3.17) holds, with problem (3.15) has at least two weak solutionsand (,).
Consequently, when (3.16) and (3.17) hold, with problem (3.15) has at least four different weak solutions.
Proposition 3.7.Let (3.13), (3.16) and (3.17) hold. Let . If there exists a measurable set with such that a.e. in , then u_{0} is nonstationary.
Proof. Reasoning by contradiction let . With a.e. in we have
(3.23)
Let . Setting in (3.14) , we have
from which
where
Then a.e.in, and this contradicts (3.23). □
Proposition 3.8. Let (3.13) and (3.17) hold. Let . Let one of the following conditions holds:
There exists a measurable set with such that a.e. in ; (3.24)
There exist a measurable set with , and a limit point of I such that a.e. in , ; (3.25)
There exist [resp.there exists ] as in , and , with and , such thatand . (3.26)
Then is nonstationary.
Proof. Under condition (3.24) [resp. (3.25)], reasoning by contradiction let
Let . Relation (3.14) with becomes
from which
and then the contradiction
Under condition (3.26), according to Prop.2.2 it is sufficiently to prove that
If , then there exists such that and
(3.27)
Let be as in . Since , we have
Then
(3.28)
Let us add
where
Since as
from (3.28) we get that it is possible to choose such that (3.27) holds with . □
Application 3.3 (connected to Theor.2.2 (case (c_{2})) and Theor.2.3 (case (c_{9}) with m_{1}=1)). We premise some clarifications. Let
(3.29)
We note
with continuous embedding . (3.30)
Let .We note the norm on X
is equivalent to the natural one on X. It is also equivalent to the norm of . In fact, there exist such that for any
( [2], Theor.8.2 page 444),
(from (3.30)).
Therefore is a closed subspace of . It is easy to verify that
then, set ,we have
(3.31)
(3.32)
Let us assume in the definition (1.1) of and V=X, then
,
and let us set for any
where
(3.33)
Problem becomes:
Find such that
(3.34)
Each solution u of (3.34) is for definition a weak solution of the problem:
(3.35)
Let us introduce the conditions
There exist a compact set with and an open
set such that , (3.36)
(3.37)
We note
0 in Q
besides
since as we have
Taking into account from (3.32)
we have
Proposition 3.9. Under conditions (3.36) and (3.37), there exists such that
Proof. Reasoning by contradiction, for any there exist such that
Then with such that (within a subsequence), from the relations
passing to limit as we get
(3.38)
(3.39)
Since from (3.31) according to (3.38) we get in Q. Consequently
from which the contradiction □
Proposition 3.10. Under conditions (3.37), there exists such that
Proof. Reasoning by contradiction, as in Prop.3.5 there exist ,with ,and such that
Then, set , we have
(3.40)
(3.41)
Let such that (within a subsequence). Relations (3.40), (3.41) imply
from which, taking into account (3.37), we deduce that in Q and the contradiction □
How established far allows the following result.
Proposition 3.11 (Theor.2.2(case ); Theor.(2.3) (case )). Under assumptions (3.29) and (3.33) we have:
when (3.36) and (3.37) hold, with problem (3.35) has at least two nonstationary weak solutions and (,,);
when (3.37) holds, with problem (3.35) has at least two weak solutionsand (,).
Consequently, when (3.36) and (3.37) hold, with problem (3.35) has at least four different weak solutions.
Proposition 3.12. Let . If there exist a measurable set with and a limit point of I such that and as j=2,…,m a.e. in , then is nonstationary.
Proof. In fact, if , we get the contradiction
□
Let us suppose has the following structure (according to (3.36) and (3.37)):
°0 ,
there exist as in such that
and (3.42)
In addition let us suppose:
(3.43)
Proposition 3.13. Let (3.42) and (3.43) hold. Let .If , then is nonstationary.
Proof. It is sufficient (Prop.2.2) to prove that
If , then there exists such that and
(3.44)
Let be as in . We note that
Then
and as j=2,…,m
Let us note that
besides, set , we get
Since
,
with a suitable fulfills (3.44). □
Remark 3.2. It is not difficult to set functions as in (3.42) such that .
Application 3.4 (connected to Theor.2.2 (case (c_{3})) and Theor.2.3 (case (c_{9}) with m_{1}=m)). Let us assume in the definition (1.1) of , n=2 and , then
and let us set as any
where
(3.45)
Problem becomes:
Find such that
(3.46)
Each solution u of (3.46) is for definition a weak solution of the problem:
(3.47)
Let us introduce the conditions
0 in Q, (3.48)
(3.49)
Evidently
0 in Q
Reasoning as in Prop.3.9 and Prop.3.10, we prove that
under conditions (3.48) and (3.49), there exists such that holds if
under condition (3.49), there exists such that holds with
Then
Proposition 3.14. (Theor.2.2 (case (c_{3})), Theor.2.3 (case (c_{9}) with m_{1}=m)).Under conditions (3.45) we have:
when (3.48) and (3.49) hold, with problem (3.47) has at least one nonstationary weak solution (,,);
when (3.49) holds, with problem (3.47) has at least one weak solution ().
Consequently, when (3.48) and (3.49) hold, with problem (3.47) has at least two different weak solutions.
Proposition 3.15. Let . Let be true one of the following conditions:
There exist a measurable set and a limit point of I such that
(3.50)
There exist a measurable set ,a limit point of I and such that
(3.51)
Then is nonstationary.
Proof. If , then with we have
from which the contradictions
□
Relations (3.48), (3.49) in particular fulfill when
(3.52)
Proposition 3.16. Let (3.52) holds. Let . Then is nonstationary.
Proof. Reasoning by contradiction let Since °0 in Ω, set in (3.46) we have
Then
(3.53)
and moreover
(3.54)
since Condition °0 implies there exists a compact set with and.Let where .
Since we choose such that Then taking into account (3.53), (3.54), from (3.46) with we get the contradiction
□
Application 3.5 (connected to Theor.2.2 (case (c_{4}) with m_{1}=m-1)). Let us assume in the definition (1.1) of , then
and let us set as any
where
(3.55)
Problem becomes:
Find such that
(3.56)
Each solution u of (3.56) is for definition a weak solution of the problem:
(3.57)
Evidently. Let us add that set , there exists satisfying the condition:
(3.58)
In fact, otherwise, for each there exist
Then with we have
from which, passing to limit as , we get .
(3.58) implies holds even if .
Proposition 3.17. (Theor.2.2 (case (c_{4}) with m_{1}=m-1)).Under conditions (3.55), with
problem (3.57) has at least one weak solution (,,).
About the nonstationarity of ,let us introduce the conditions:
There exist a measurable set and a limit point of I such that
for almost any and as j=1,…,m-1; (3.59)
There exist an open interval ,and ° in I , with such that as almost every and as each
(3.60)
There exist such that for almost any and
(3.61)
Remark 3.3. It is easy to find assumptions on and a such that the inequality in (3.61) holds.
Proposition 3.18. Let . If one of the conditions (3.59) - (3.61) holds, then is nonstationary.
Proof. Reasoning by contradiction, let in Q.
When (3.59) holds, we have the contradiction
where
(from (3.58)) if .
When (3.60) holds, we have with
and this contradicts hypothesis.
When (3.61) holds, it is sufficient(Prop.2.1) to prove that
There exists such that and
(3.62)
With we have
Then
Since as j=1,…,m
,
with a suitable , fulfills (3.62). □
Remark 3.4. It is easy to prove that Propositions 3.17 and 3.18 also hold when
with and a as in (3.55).
Application 3.6 (connected to Theor.2.2 (case (c_{6}))). Let us assume in the definition (1.1) of , n=2 and , then
,
and let us set as any
where
(3.63)
Problem becomes:
Find such that
(3.64)
Each solution u of (3.64) is for definition a weak solution of the problem:
(3.65)
We note that
Then
Proposition 3.19. (Theor.2.2 (case (c_{6}))).Under conditions (3.63), with problem (3.65) has at least one weak solution (,,).
Proposition 3.20. Let . Let one of the following conditions be fulfilled:
There exist a measurable set and a limit point of I such that
for almost any and as j=1,…,m;
There exist an open interval ,and ° in I , with such that as almost every and as each ;
There exist such that for almost any as j=.and ,as almost each and for each where
Then is nonstationary.
Proof. We reason as in Prop. (3.18).
References