Mathematical Problems in Engineering
Volume 2016 (2016), Article ID 8031638, 18 pages
http://dx.doi.org/10.1155/2016/8031638

Research Article

Linear Approximation and Asymptotic Expansion of Solutions for a Nonlinear Carrier Wave Equation in an Annular Membrane with Robin-Dirichlet Conditions

Le Thi Phuong Ngoc,¹ Le Huu Ky Son,2,3 Tran Minh Thuyet,⁴ and Nguyen Thanh Long³

¹University of Khanh Hoa, 01 Nguyen Chanh Str., Nha Trang City, Vietnam
²Department of Fundamental sciences, Ho Chi Minh City University of Food Industry, 140 Le Trong Tan Str., Tan Phu Dist., Ho Chi Minh City, Vietnam
³Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University-Ho Chi Minh City, 227 Nguyen Van Cu Str., Dist. 5, Ho Chi Minh City, Vietnam
⁴Department of Mathematics, University of Economics of Ho Chi Minh City, 59C Nguyen Dinh Chieu Str., Dist. 3, Ho Chi Minh City, Vietnam

Received 23 April 2016; Accepted 8 September 2016

Academic Editor: Yuri Vladimirovich Mikhlin

Copyright © 2016 Le Thi Phuong Ngoc et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 conditions

and initial conditions

where , , , 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 implies

Lemma 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 that

and associate the following variational problem with problem (1)–(3): find , so that

where

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. Put

where coefficients satisfy the system of linear differential equations:

with

The system of (43) can be rewritten in form

in which

Note 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 put

where

Then, it follows from (43), (47), and (48) that

We shall estimate terms on the right-hand side of (49) as follows.
First Term By the following inequalities,

we have

Second Term By the Cauchy-Schwartz inequality, it gives

Third Term Similarly, we have

Note that

so

We also have

where =
It implies from (39), (56) that

Combining (53), (55), and (57), we obtain

Fourth Term Equation (43)₁ can be rewritten as follows:

Hence, it follows after replacing with , that

Integrate into to get

It follows from (49), (51), (52), (58), and (61) that

where

By means of the convergences in (44), we can deduce the existence of constant independent of and such that

for all ,
Therefore, from (63) and (64), we can choose , such that

Finally, it follows from (62), (64), and (65) that

By using Gronwall’s Lemma, (67) yields

for all , for all and Therefore, we have

Step 3 (limiting process). From (69), there exists a subsequence of , still so denoted, such that

Passing 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 get

where

All integrals on the right-hand side of (73) will be estimated as below.
First Integral By (50) and (74), we have

Second Integral By , it is clear to see that

Hence,

Second Integral By , it yields

Hence,

Combining (73), (75), (77), and (79), we obtain

Using Gronwall’s lemma, we deduce from (80) that

which implies that

It follows that is a Cauchy sequence in Then, there exists such that

Note that , and then there exists subsequence of such that

We also note that

Hence, from (83) and (85), we obtain

On the other hand, we have

Hence, it follows from (83) and (87) that

Finally, passing to limit in (40), (41) as , it implies from (83), (84)1,3, (86), and (88) that there exists satisfying

for all and the initial conditions

Furthermore, from assumptions , we obtain from (84)₄, (86), (88), and (89), that

and 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 get

with
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 have

with , 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 obtain

where

, , , ,
By formula (97), we get

where
Similarly, with , , we also have

where ,
Hence, we deduce from (106)-(107) that

where , are defined by (100).
We deduce from (104), (108) that

where , are defined by (100) and

By 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 have

with , 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 obtain

where

On the other hand, we also get

where , are defined by (101).
Using formula (97) again, it follows from (114) that

where
We deduce from (112), (115) that

where , , are defined by (101) and

By 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 that

where is a constant depending only on , , , , , ,

Proof of Lemma 15. We only prove with
By using formula (103) for function , we obtain

where , with constant depending only on , , , , ,
By (121), we rewrite as follows

Hence, we deduce from (103) and (122) that

where is bounded in function space by a constant depending only on , , , , ,
On the other hand, we put , , , and we deduce from (111) that

where

Combining (99a)-(99b), (119), (123), and (125) leads to

By the boundedness of functions , , , in function space , we obtain from (125), (123), and (126) that

where is a constant depending only on , , , , , , , , ,
The proof of Lemma 15 is complete.

Proof of Theorem 11. Consider sequence defined by

By multiplying two sides of (128)₁ with and after integrating in , we have

where

Note that

where , and

Using Lemma 15, (129) gives

We estimate the integrals on the right-hand side of (133) as follows.
Estimating We note that

with
It follows from (134) that

Estimating Similarly,

Estimating We have

Hence,

Estimating Similarly,

Combining (133), (135), (136), (138), (139), it leads to

where
By using Gronwall’s lemma, we deduce from (140) that

where , ,
We can assume that

We require the following lemma, and its proof is immediate, so we omit the details.
Lemma  16. Let sequence satisfy

where , are the given constants. Then,

Applying Lemma with , , , it follows from (144) that

where
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.

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