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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 491326, 8 pages

http://dx.doi.org/10.1155/2014/491326

## Geometric Singularities of the Stokes Problem

Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia

Received 11 November 2013; Accepted 12 December 2013; Published 6 January 2014

Academic Editor: Bessem Samet

Copyright © 2014 Nejmeddine Chorfi. 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

When the domain is a polygon of , the solution of a partial differential equation is written as a sum of a regular part and a linear combination of singular functions. The purpose of this paper is to present explicitly the singular functions of Stokes problem. We prove the Kondratiev method in the case of the crack. We finish by giving some regularity results.

#### 1. Introduction

The regularity of the solution of a partial differential equation depends on the geometry of the domain even when the data is smooth. Indeed, for each corner of the polygonal domain a countable family of singular functions can be defined, which depends only on the geometry of the domain. Then the solution of the equation can be written as the sum of a finite number of singular functions multiplied by appropriate coefficients and of a much more regular part. We refer to Kondratiev [1] and Grisvard [2] for their description.

The purpose of our work is to study the singularities of the Stokes equation and the behavior of the solution in the neighborhood of a corner of a polygonal domain of . We are interested to nonconvex domains; we assume that there exists an angle equal either to or to (case of the crack). Handling the singular function is local process, so that there is no restriction to suppose that the nonconvex corner is unique; see Dauge [3]. We deduce the singular function of the velocity from those of the bilaplacian problem with a homogenous boundary conditions by applying the curl operator. We prove the Kondratiev method in the case of the crack. The singularities of the pressure are done by integration from the singular functions of the velocity near the corner.

To approach these problems by a numerical method, we need to take into account the singular functions. Several numeric methods have been proposed in this context; see [4–9]. Since the singular functions are developed for the Stokes problem in this paper, we intend in future work to implement Strang and Fix algorithm, see [10], by the mortar spectral element method. It will be an extension of a work done on an elliptic operator [11, 12].

An outline of this paper is as follows. In Section 2, we present the geometry of the domain and the continuous problem. In Section 3, we give the singular functions and some regularity results. The Kondratiev method is described in Section 4. Section 5 is devoted to the conclusion.

#### 2. The Continuous Problem

We suppose that is a polygonal domain of simply connected and has a connected boundary . is the union of vertex for ; is positive integer. Let be the corner of between and ; is the measure of the angle on . We consider the velocity-pressure formulation of the Stokes problem on the domain .

Find the velocity and the pressure such that where is the viscosity of the fluid that we suppose a positif constant and is the data which represent a density of body forces. Then for in , the functional spaces are for the velocity and for the pressure where The problem (1) is equivalent to the following variational formulation.

For in , find in and in such that for all in and for all in . where where the denotes the duality pairing between and . The bilinear form is continuous on the space and elliptic on ; also the bilinear form is continuous and verifies the following inf-sup condition (see [13, 14]): there exists a nonnull positive constant such that Then we conclude [15] that, for all in the space , the problem (3) has a unique solution in . This solution verifies the following stability condition: where is a positive constant.

We refer to the work of Pironneau [16] for the mathematical modeling of problems resulting from fluids mechanics and Girault and Raviart [17] for the mathematical analysis of the Navier-Stokes equations.

#### 3. Singular Functions and Regularity Results

We recall that, in an open simply connected of , the condition of incompressibility induces the existence of a stream function in the space such that It thus brings to study the regularity of the function , solution of Dirichlet problem for bilaplacian: The regularity of the solution of the problem is related to the geometry of the domain and its behavior is local. Let . We know, see Cattabriga [18] and Ladyzhenskaya [14], the following theorem.

Theorem 1. *For in the space , where , then
**
where is the union of neighborhoods of for .*

To study the function in the neighborhood of a fixed vertex , , it is convenient to introduce the polar coordinates , centered at . We start by enunciating the characteristic equation of bilaplacian: Then we stated the following theorem. We refer to ([17], Chapter 7 Theorem ) and Kondratiev [1] for the proof.

Theorem 2. *Let in . For all in , , the solution of the problem (8) is written as
**
where is in the space .**We set
** is given by
**
where and are real constants, , , and belong to the vectorial space of finite dimension, and are, respectively, the simple and double roots of (10) in the band , excepting if , without exception if .*

We called the unique solution of equation , in the (). We prove, see ([19], chapter 3, 3.3), that is a double root of (10) if and only if or . Hence the sufficient condition on the angle for a double root is Then, if is not a solution of (14), (10) takes the following simplified form:

In the following we suppose that has a unique vertex where is equal to or and the other angles are . We introduce as a neighborhood of . All these angles are different of .

In the case where and , (10) has two real simple roots in the band . We apply the Newton method to approximate those roots and . The functions and are written as follows: with

Proposition 3. *For all , the functions , in , are belonging to the space and are solutions of the problem:
**
where is the boundary of .*

*Proof. *Since is in , we find such that
Then we find in , , and such that:
From the Sobolev injection theorem, we have . If , then . Since , we can take only the value (). Consequently, for . And by construction, we obtain that , for , is the solution of problem (18).

Corollary 4. *For all in , the velocity is written as
**
where is in and there exists a constant such that:
**
where . The functions belong to the space , in , for all .*

#### 4. Kondratiev’s Method Case of the Crack

In the case of the crack there are no known results; we recall the method of kondratiev. We extend this method to the case of crack. We consider the polar coordinates and a truncation function with compact support that does not intersect with the boundary . We define Let ; then, is in the space ; with a compact support included in , then is the solution of the problem:

*Definition 5. *For , one defines the weight Sobolev space:

*Remark 6. *We remark that and .

In the following, we suppose that is in the space , for in . The solution of the problem: is with a compact support; then, there exists a real number such that The Kondratiev’s method consists to change the variable ; then we replace the domain by the domain , and the weight Sobolev space by the ordinary one. We apply the Fourier transformation relative to the first variable of the problem (26).

Proposition 7. *If is in the space , the function belongs to .**For , if belongs to , and let , then the function is in the space .*

*Proof. *Let .

Since , then with , and in .

We denote
We have
and if
then we conclude
since et , then, for . So
We end the proof.

As is the solution of the problem (26), then for the function is a solution of the problem: We have for .

The Fourier's transformation on the variable of the function is is a complex variable; we denote: The function becomes a solution of the problem: For all such that , the operator has as a characteristic polynomial. This polynomial function has two roots and .(i)If and , then the fundamental solutions of the fourth order partial differential equation are (ii)If , the solutions have the form (iii)If , the solutions have the form Then, let the following homogeneous problem We note , solution of the problem (41)}.

Proposition 8. *The set satisfies the following properties:*(1)*if ** and ** where **, then,*(2)*if **, then,*(3)*if *,
* ** and **, which implies that ** is equal to ** where ** and **, then,** *

*Proof. *If and ,
is solution of the problem (41). Since , we obtain the following system:
its determinant is equal to . Then, if , . Otherwise, the system has the following complexes roots: for and , . The space of solutions of the homogeneous equation is of dimension 2. Indeed, for , we obtain

We check that are two independent solutions of the problem (41); then
This ends the proof of 1 and 3.

Now we prove property 2.

If ,
is solution of the problem (41), which gives the following system:
The determinant of this system is zero; then the dimension of is equal to 1. Indeed,
This completed the proof.

*Remark 9. *If
is the set of nonregular values. Since is analytic for , then analytically extends across .

Is then
We verify that and . Hence the multiplicity of is 2 for and . , then, are simple solutions.

Theorem 10. *(1) The solution of the problem (36) can be written in the neighborhood of :
**
where is the solution of problem (41) with and is an analytic function in the neighborhood of with values in .**(2) The solution of the problem (36) is written in the neighborhood of , or , as
**
where and satisfy the same properties as above and satisfies:
*

*Proof. *We verified that any solution to the problem (41) is a linear combination of
In fact, if for or , we have
Therefore, any solution of the problem (41) is linear combination of and . We also verified that and are whole solutions with respect to the variable and if we set
we note that . This implies that is analytical in the neighborhood of for in . is a single root of ; then, admits a simple pole at . On the other side and are roots of multiplicity two of , and poles of multiplicity two of .

In the neighborhood of , , and, for all ,
Then, for , we have
Then
and even
and even for:
Thus, is a solution of the problem (41) at .

The expression of in the neighborhood of for and is
Then, for in the neighborhood of
For , we obtain that and we verify that is solution of the problem (41). And we have:
And if tends to
and verifies the boundary conditions, which completes the proof.

*Remark 11. *We remark that the operator is self-adjoint:
For
This implies that ; thus
where is solution of problem (41).

Let be in such that . We note
We prove by Cauchy’s theorem that
is an analytic function on ; is its residue on the poles . Then,(i)for (ii)for where are solutions of the problem (36).

*Notation.* Using the variables , we introduce the following notations.

For , we define the two singular functions: For , we define four singular functions:

Proposition 12. *For , and belong to the space , and belong to the space and They are solutions to the problem:
**
where is the boundary of .*

*Proof. *It is the same prove that in the case .

For the Stokes problem we handle the case ; thus belongs to the space .

Corollary 13. *For in , the velocity is written in the form
**
where is in and there exist two real constants and such that
**
with
** and belong to the , for all positive.*

For handling the singularities of the pressure we define Indeed for , using the Newton method, we can obtain a good approximation of roots of (10). See [20] for the approximation of . Thus To find the pressure, we note that belongs to , where Then, see [15], there exists a unique function in defined such that: From this equality we determine the singularities of the pressure from the singularities of the velocity.

Then from a regular data where is first singular coefficient. The linearity of (85) gives us and since we explicitly know the singularities of the velocity we can deduce, by simple integration, the singularities of the pressure.

Consider in the case of crack in the case of and .

Proposition 14. *For , belongs to when and belongs to when .*

*Proof. *It is the same proof that in the case of Proposition 3.

#### 5. Conclusion

We summarize below the regularity results for the Stokes problem previously obtained.

(1) If belongs to , where , then belongs to the space , if . This means satisfies the following stability condition:

(2) We know, from Corollaries 4 and 13 and formula (86), that if , is written and if If belongs to , belongs to for , where is the second real solution of (10), in the band ; hence We have the following stability condition:

(3) If belongs to , we can further decompose the regular part of the solution as follows: for where is the third reel solution of (10), in the band ; then is the singular coefficient associated to the second singular function ; then In general we can decompose as where is an integer number.

If belongs to , belongs to for where , -ème reel solution of (10), in the band .

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.

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