Abstract

This paper is devoted to the study of a nonlinear Carrier wave equation in an annular membrane associated with Robin-Dirichlet conditions. Existence and uniqueness of a weak solution are proved by using the linearization method for nonlinear terms combined with the Faedo-Galerkin method and the weak compact method. Furthermore, an asymptotic expansion of a weak solution of high order in a small parameter is established.

1. Introduction

In this paper, we consider the following nonlinear Carrier wave equation in the annular membrane:associated with Robin-Dirichlet conditionsand initial conditionswhere , , , are given functions; , are given constants, with . In (1), nonlinear term depends on integral .

Equation (1) herein is the bidimensional nonlinear wave equation describing nonlinear vibrations of annular membrane . In the vibration processing, the area of the annular membrane and the tension at various points change in time. The condition on boundary , that is, , describes elastic constraints where constant has a mechanical signification. And with the boundary condition on requiring , the annular membrane is fixed.

In [1], Carrier established the equation which models vibrations of an elastic string when changes in tension are not small:where is -derivative of the deformation, is the tension in the rest position, is the Young modulus, is the cross section of a string, is the length of a string, and is the density of a material. Clearly, if properties of a material vary with and , then there is a hyperbolic equation of the type [2]:

The Kirchhoff-Carrier equations of form (1) received much attention. We refer the reader to, for example, Cavalcanti et al. [3, 4], Ebihara et al. [5], Miranda and Jutuca [6], Lasiecka and Ong [7], Hosoya and Yamada [8], Larkin [2], Medeiros [9], Menzala [10], Park et al. [11, 12], Rabello et al. [13], and Santos et al. [14], for many interesting results and further references.

The paper consists of four sections. Preliminaries are done in Section 2, with the notations, definitions, list of appropriate spaces, and required lemmas. The main results are presented in Sections 3 and 4.

First, by combining the linearization method for nonlinear terms, the Faedo-Galerkin method, and the weak compact method, we prove that problem (1)–(3) has a unique weak solution.

Next, by using Taylor’s expansion of given functions , , and up to high order , we establish an asymptotic expansion of solution of order in small parameter for, , associated with (1) and (2) with , , , , for all , , . Our results can be regarded as an extension and improvement of the corresponding results of [15, 16].

2. Preliminaries

First, put , , . We omit the definitions of the usual function spaces and denote them by notations , . Let be a scalar product in . Notation stands for the norm in and we denote the norm in Banach space . We call the dual space of We denote , the Banach space of real functions to be measurable, such that , with

With , , we put , , with , and , , , .

On , , we shall use the following norms:respectively.

We remark that , , are the Hilbert spaces with respect to the corresponding scalar products:

The norms in , , and induced by the corresponding scalar products (10) are denoted by , , and .

Consider the following set:

It is obviously that is a closed subspace of and on two norms and are equivalent norms. On the other hand, is continuously and densely embedded in . Identifying with (the dual of ), we have . We note more that the notation is also used for the pairing between and .

We then have the following lemmas.

Lemma 1. The following inequalities are fulfilled: (i) for all .(ii) for all .

Proof of Lemma 1. It is easy to verify the above inequalities via the following inequalities:

Lemma 2. Embedding is compact and for all , we have (i),(ii),(iii),(iv),(v).

Proof of Lemma 2. Embedding is continuous and embedding is compact, so embedding is compact. In what follows, we prove (i)–(v).(i)For all and , (ii)For all and , Integrating over from to , we obtain (iii)For all , (iv)Using integration by part, it leads to for any , so we get (iv).(v)By , we have implying (v).Lemma 2 is proved.

Remark 3. On , two norms and are equivalent. So are two norms and on , and five norms , , , , and on .
Now, we define the following bilinear form:where is a constant.

Lemma 4. Symmetric bilinear form defined by (19) is continuous on and coercive on , that is, (i),(ii),for all , , where and .

Proof of Lemma 4. (i) By , and for all , we have (ii) By inequality , we get Lemma 4 is proved.

Lemma 5. There exists Hilbert orthonormal base of space consisting of eigenfunctions corresponding to eigenvalues such that (i), ,(ii) for all , .Furthermore, sequence is also the Hilbert orthonormal base of with respect to scalar product .
On the other hand, we also have satisfying the following boundary value problem:

Proof. The proof of Lemma 5 can be found in [17, p. 87, Theorem 7.7], with , and as defined by (19).

We also note that operator in (22) is uniquely defined by Lax-Milgram’s lemma; that is,

Lemma 6. On , three norms , , and are equivalent.

Proof of Lemma 6. (i) It is easy to see that, on , two norms , are equivalent, because(ii) For all , and , we have(a) Proof .
It follows from (25) that Hence, This implies By , for all , we have (b) Proof
It follows from (25) that Hence, Thus, This impliesLemma 6 is proved.

Remark 7. The weak formulation of initial-boundary value problem (1)–(3) can be given in the following manner: find , , such that satisfies the following variational equation:for all , a.e., , together with the initial conditions:where is the symmetric bilinear form on defined by (19).

3. The Existence and Uniqueness Theorem

Now, we shall consider problem (1)–(3) with constant and make the following assumptions:(), ;(), with ;() such that and .

Considering fixed and letting and , we put where

Also for each and , we set

We choose first term , suppose thatand associate the following variational problem with problem (1)–(3): find , so thatwhere

Then, we have the following result.

Theorem 8. Let assumptions hold. Then, there exist positive constants , such that the problem (40), (41) has solution .

Proof of Theorem 8. It consists of three steps.
Step 1 (the Faedo-Galerkin approximation (introduced by Lions [18])). Consider basis for as in Lemma 5. Putwhere coefficients satisfy the system of linear differential equations:withThe system of (43) can be rewritten in formin whichNote that, by (39), it is not difficult to prove that system (45), (46) has a unique solution on interval , so let us omit the details.
Step 2 (a priori estimates). We putwhereThen, it follows from (43), (47), and (48) thatWe shall estimate terms on the right-hand side of (49) as follows.
First Term By the following inequalities,we haveSecond Term By the Cauchy-Schwartz inequality, it givesThird Term Similarly, we haveNote thatsoWe also havewhere =
It implies from (39), (56) thatCombining (53), (55), and (57), we obtainFourth Term Equation (43)1 can be rewritten as follows:Hence, it follows after replacing with , thatIntegrate into to getIt follows from (49), (51), (52), (58), and (61) thatwhereBy means of the convergences in (44), we can deduce the existence of constant independent of and such thatfor all ,
Therefore, from (63) and (64), we can choose , such thatFinally, it follows from (62), (64), and (65) thatBy using Gronwall’s Lemma, (67) yieldsfor all , for all and Therefore, we haveStep 3 (limiting process). From (69), there exists a subsequence of , still so denoted, such thatPassing to limit in (43), we have satisfying (40), (41) in On the other hand, it follows from (40)1 and (70)4 that , and hence and the proof of Theorem 8 is complete.

We will use the result obtained in Theorem 8 and the compact imbedding theorems to prove the existence and uniqueness of a weak solution of problem (1)–(3). Hence, we get the main result in this section.

Theorem 9. Let hold. Then, there exist positive constants , satisfying (64)–(66) such that problem (1)–(3) has unique weak solution . Furthermore, the linear recurrent sequence defined by (40), (41) converges to solution strongly in space , with estimate

Proof of Theorem 9. (a) The Existence. First, we note that is a Banach space with respect to norm (see Lions [18]).
We shall prove that is a Cauchy sequence in Let Then, satisfies the variational problem:Taking in (72)1, after integrating into , we getwhereAll integrals on the right-hand side of (73) will be estimated as below.
First Integral By (50) and (74), we haveSecond Integral By , it is clear to see thatHence,Second Integral By , it yieldsHence,Combining (73), (75), (77), and (79), we obtainUsing Gronwall’s lemma, we deduce from (80) thatwhich implies thatIt follows that is a Cauchy sequence in Then, there exists such thatNote that , and then there exists subsequence of such thatWe also note thatHence, from (83) and (85), we obtainOn the other hand, we haveHence, it follows from (83) and (87) thatFinally, passing to limit in (40), (41) as , it implies from (83), (84)1,3, (86), and (88) that there exists satisfyingfor all and the initial conditionsFurthermore, from assumptions , we obtain from (84)4, (86), (88), and (89), thatand thus we have The existence of a weak solution of problem (1)–(3) is proved.
(b) The Uniqueness. Let be two weak solutions of problem (1)–(3). Then, satisfies the variational problem:where ,
We take in (92)1 and integrate in to getwith
Putting , it follows from (93) that Using Gronwall’s lemma, it follows that , that is, .
Therefore, Theorem 9 is proved.

4. Asymptotic Expansion of the Solution with respect to a Small Parameter

In this section, let hold. We make more the following assumptions:(), with .() such that and

Considering the following perturbed problem, where is a small parameter and : with

First, we note that if functions satisfy , , , , then a priori estimates of the Galerkin approximation sequence for problem (1)–(3) corresponding to , , , leads to , where constants , , independent of , are chosen as in (63)–(66), in which , , are replaced with , , , respectively. Hence, limit in suitable function spaces of sequence as , after , is a unique weak solution of problem satisfying

We can prove in a manner similar to the proof of Theorem 9 that limit in suitable function spaces of family as is a unique weak solution of problem (corresponding to ) satisfying .

Next, we shall study the asymptotic expansion of solution with respect to a small parameter . For multi-index , and , we put

We need the following lemma.

Lemma 10. Let and , . Then,where coefficients , , depending on , are defined by the following formulas:

Proof. The proof of Lemma 10 is easy; hence, we omit the details.

Now, we assume that(), with ;() such that .

Let be a unique weak solution of problem ; that is,

Let us consider the sequence of weak solutions , , defined by the following problems: where , , are defined by the following formulas:with , , , are defined by the following formulas:

Then, we have the following theorem.

Theorem 11. Let , , and hold. Then, there exist constants and such that, for every , problem has unique weak solution satisfying the asymptotic estimation up to order as follows:where functions are the weak solutions of problems , , , respectively, and is a constant depending only on , , , , , , , ,

In order to prove Theorem 11, we need the following lemmas.

Lemma 12. Let , be the functions defined by the formulas (100). Put , then we havewith , where is a constant depending only on , , , ,

Proof of Lemma 12. In the case of , the proof of (103) is easy; hence, we omit the details, and we only prove with Put By using Taylor’s expansion of function around point up to order , we obtainwhere, , , ,
By formula (97), we getwhere
Similarly, with , , we also havewhere ,
Hence, we deduce from (106)-(107) thatwhere , are defined by (100).
We deduce from (104), (108) thatwhere , are defined by (100) andBy the boundedness of functions , , in the function space , we obtain from (100), (105), (110) that , where is a constant depending only on , , , Thus, Lemma 12 is proved.

Lemma 13. Let , be the functions defined by formulas (101). Put , and then we havewith , where is a constant depending only on , , ,

Proof of Lemma 13. In the case of , the proof of (111) is easy; hence we omit the details, and we only prove with
Put By using Taylor’s expansion of function around point up to order , we obtainwhereOn the other hand, we also getwhere , are defined by (101).
Using formula (97) again, it follows from (114) thatwhere
We deduce from (112), (115) thatwhere , , are defined by (101) andBy the boundedness of functions , , in function space , we obtain from (101), (113), (117) that , where is a constant depending only on , , , Thus, Lemma 13 is proved.

Remark 14. Lemmas 12 and 13 are a generalization of the formula contained in [19, p. 262, formula ] and it is useful to obtain Lemma 15 below. These lemmas are the key to establish the asymptotic expansion of weak solution of order in small parameter .
Let be the unique weak solution of problem Then, satisfies the problem:where

Lemma 15. Let , , and hold. Then, there exists constant such thatwhere is a constant depending only on , , , , , ,

Proof of Lemma 15. We only prove with
By using formula (103) for function , we obtainwhere , with constant depending only on , , , , ,
By (121), we rewrite as followsHence, we deduce from (103) and (122) thatwhere is bounded in function space by a constant depending only on , , , , ,
On the other hand, we put , , , and we deduce from (111) that whereCombining (99a)-(99b), (119), (123), and (125) leads toBy the boundedness of functions , , , in function space , we obtain from (125), (123), and (126) thatwhere is a constant depending only on , , , , , , , , ,
The proof of Lemma 15 is complete.

Proof of Theorem 11. Consider sequence defined byBy multiplying two sides of (128)1 with and after integrating in , we havewhere Note thatwhere , and Using Lemma 15, (129) gives We estimate the integrals on the right-hand side of (133) as follows.
Estimating We note thatwith
It follows from (134) thatEstimating Similarly,Estimating We have Hence,Estimating Similarly,Combining (133), (135), (136), (138), (139), it leads towhere
By using Gronwall’s lemma, we deduce from (140) thatwhere , ,
We can assume thatWe require the following lemma, and its proof is immediate, so we omit the details.
Lemma  16. Let sequence satisfywhere , are the given constants. Then,Applying Lemma with , , , it follows from (144) thatwhere
On the other hand, linear recurrent sequence defined by (128) converges strongly in space to solution of problem (118). Hence, letting in (145), we get This implies (102). The proof of Theorem 11 is complete.

Competing Interests

The authors declare that they have no competing interests.

Authors’ Contributions

All authors contributed equally in this article. They read and approved the final manuscript.