Research Article | Open Access
Linear Approximation and Asymptotic Expansion of Solutions for a Nonlinear Carrier Wave Equation in an Annular Membrane with Robin-Dirichlet Conditions
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.
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 , 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 :
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. , Miranda and Jutuca , Lasiecka and Ong , Hosoya and Yamada , Larkin , Medeiros , Menzala , Park et al. [11, 12], Rabello et al. , and Santos et al. , 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.
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].
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 .
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:
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
Considering fixed and letting and , we put where
Also for each and , we set
Then, we have the following result.
Proof of Theorem 8. It consists of three steps.
Step 1 (the Faedo-Galerkin approximation (introduced by Lions )). 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