Pure and Applied Mathematics Journal
Volume 4, Issue 5, October 2015, Pages: 189-215

Existence of Time Periodic Solutions of New Classes of Nonlinear Problems

Luisa Toscano1, Speranza Toscano2

1Dep. of Math. and Appl."R. Caccioppoli",Univ. of Naples "Federico II", via Cintia, Monte S. Angelo, Italy

2Dep. of Civil Ing,. Second Univ. of Naples, fac. of Ing., Aversa (CE), Italy

Email address:

(L. Toscano)
(S. Toscano)

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 RN+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 (c8))). 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 >q1),(i21) 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 c8) 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 u0 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 (c2)) and Theor.2.3 (case (c9) with m1=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 (c3)) and Theor.2.3 (case (c9) with m1=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 (c3)), Theor.2.3 (case (c9) with m1=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 (c4) with m1=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 (c4) with m1=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 (c6))). 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 (c6))).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).            

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