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International Journal of Mathematics and Mathematical Sciences
Volume 2012 (2012), Article ID 269607, 20 pages
http://dx.doi.org/10.1155/2012/269607
Research Article

The Dirichlet Problem for the 2D Laplace Equation in a Domain with Cracks without Compatibility Conditions at Tips of the Cracks

KIAM, Miusskaya Square 4, Moscow 125047, Russia

Received 21 March 2012; Accepted 11 June 2012

Academic Editor: Vladimir Mityushev

Copyright © 2012 P. A. Krutitskii. 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

We study the Dirichlet problem for the 2D Laplace equation in a domain bounded by smooth closed curves and smooth cracks. In the formulation of the problem, we do not require compatibility conditions for Dirichlet's boundary data at the tips of the cracks. However, if boundary data satisfies the compatibility conditions at the tips of the cracks, then this is a particular case of our problem. The cases of both interior and exterior domains are considered. The well-posed formulation of the problem is given, theorems on existence and uniqueness of a classical solution are proved, and the integral representation for a solution is obtained. It is shown that weak solution of the problem does not typically exist, though the classical solution exists. The asymptotic formulae for singularities of a solution gradient at the tips of the cracks are presented.

1. Introduction

Boundary value problems in 2D domains with cracks (double-sided open arcs) are very important for applications [1].

It is known that if the Dirichlet problem for the Laplace equation is considered in a 2D domain bounded by sufficiently smooth closed curves, and if the function specified in the boundary condition is smooth enough, then existence of a classical solution follows from existence of a weak solution. In the present paper, we consider the Dirichlet problem for the Laplace equation in both interior and exterior 2D domain bounded by closed curves and double-sided open arcs (cracks) of an arbitrary shape. The Dirichlet condition is specified on the whole boundary, that is, on both closed curves and on the cracks, so that different functions may be specified on opposite sides of the cracks. The case of this problem, when the Dirichlet boundary data satisfies compatibility conditions at the tips of the cracks, has been previously studied in [26], where theorems on existence and uniqueness of a classical solution have been proved, and the integral representation for a classical solution has been obtained. In the present paper, this problem is considered in case when the Dirichlet boundary data may not satisfy the compatibility conditions at the tips of the cracks. However if boundary data satisfies the compatibility conditions at the tips of the cracks, then this is a particular case of our problem. We prove that there exists a unique classical solution to our problem and obtain an integral representation for the classical solution. In addition, we prove that a weak solution to our problem may not exist even if both cracks and functions in the boundary conditions are sufficiently smooth. This result follows from the fact that the square of the gradient of a classical solution, basically, is not integrable near the ends of the cracks, since singularities of the gradient are rather strong there. This result is very important for numerical analysis, it shows that finite elements and finite difference methods cannot be applied to numerical treatment of the Dirichlet problem in question directly, since all these methods imply existence of a weak solution. To use difference methods for numerical analysis, one has to localize all strong singularities first and next to use difference method in a domain excluding the neighbourhoods of the singularities. The asymptotic formulae for singularities of a solution gradient at the tips of the cracks are presented.

2. Formulation of the Problem

By an open curve we mean a simple smooth nonclosed arc of finite length without self-intersections [7].

Let be a set of curves, which may be both closed and open. We say that (or ) if curves belong to the class (or ) with the Hölder exponent .

In a plane in Cartesian coordinates , we consider a multiply connected domain bounded by simple open curves and simple closed curves , , in such a way that all curves do not have common points, in particular, endpoints. We will consider both the case of an exterior domain and the case of an interior domain when the curve encloses all others. Set The connected domain bounded by and containing curves will be called , so that . We assume that each curve is parametrized by the arc length : so that, and the domain is placed to the right when the parameter increases on . The points and values of the parameter are in one-to-one correspondence except the points , , which correspond to the same point for . Further on, the sets of the intervals on the -axis will be denoted by , , and also.

Set ,  , and . The tangent vector to in the point , in the direction of the increment of , will be denoted by , while the normal vector coinciding with after rotation through an angle of in the counterclockwise direction will be denoted by . According to the chosen parametrization ,   . Thus, is an inward normal to on . By we denote the point set consisting of the endpoints of .

We consider as a set of cracks (or double-sided open arcs). The side of the crack , which is situated on the left when the parameter increases, will be denoted by , while the opposite side will be denoted by .

We say that the function belongs to the smoothness class , if (1),(2)in the neighbourhood of any point , both the inequality and the equality hold, where the curvilinear integral of the first kind is taken over a circumference of a radius with the center in the point , in addition, is a normal in the point , directed to the center of the circumference and or ,  .

Remark 2.1. By we denote the class of continuous in functions, which are continuously extensible to the sides of the cracks from the left and from the right, but their limiting values on can be different from the left and from the right, so that these functions may have a jump on . To obtain the definition of the class we have to replace by and by in the previous sentence.

Let us formulate the Dirichlet problem for the Laplace equation in a domain (interior or exterior, see Figures 1 and 2).

269607.fig.001
Figure 1: An example of an interior domain.
269607.fig.002
Figure 2: An example of an exterior domain.

Problem
Find a function from the class , so that obeys the Laplace equation in and satisfies the boundary conditions If is an exterior domain, then we add the following condition at infinity:
All conditions of the problem must be satisfied in a classical sense. The boundary conditions (2.8) on must be satisfied in the interior points of , their validity at the ends of is not required.

Theorem 2.2. If , then there is no more than one solution to the problem .

Proof. It is sufficient to prove that the homogeneous problem admits the trivial solution only. Let be a solution to the homogeneous problem with . Let be a disc of a small enough radius with the center in the point ( or ,  ). Let be a set consisting of such points of the curve which do not belong to discs and . We choose a number so small that the following conditions are satisfied:(1)for any the set of points is a unique non-closed arc for each , (2)the points belonging to are placed outside the discs , for any ,(3)discs of radius with centers in different ends of do not intersect.
Set ,  ,  . If is an exterior domain, then we set , where is a disc with a center in the origin and with sufficiently large radius .
Since ,  (remind that ), and since , and owing to the lemma on regularity of solutions of elliptic equations near the boundary [8, lemma 6.18], we obtain . Since , we observe that for any . By we mean . Analogously, in case of exterior domain for . Let be an interior domain. Since the boundary of a domain is piecewise smooth, we write out Green's formula [9, page 328] for the function : By the exterior (with respect to ) normal on at the point is denoted. By the superscripts + and − we denote the limiting values of functions on and on , respectively. Since satisfies the homogeneous boundary condition (2.8) on , we observe that and for any . Therefore, identity (2.10) takes the form Setting in (2.11), taking into account that and using the relationships (2.6), (2.5) we obtain From the homogeneous boundary conditions (2.8) we conclude that in , where is an interior domain.
Let be an exterior domain. Since the boundary of a domain is piecewise smooth and since for any , we may write Green's formula in a domain for a harmonic function [9, c.328]: By on we denote an outward (with respect to ) normal in the point . It follows from condition (2.9) and from the theorem on behaviour of a gradient of a harmonic function at infinity [9, page 373] that Consequently, and formula (2.13) transforms to the formula (2.10) as . Repeating all arguments, presented above for the case of an interior domain , we arrive to formula (2.12). Taking into account homogeneous boundary conditions (2.8), we obtain from (2.12) that in , where is an exterior domain. Thus, in all cases in . Theorem is proved.

Remark 2.3. The maximum principle cannot be used for the proof of the theorem even in the case of an interior domain , since the solution to the problem may not satisfy the boundary condition (2.8) at the ends of the cracks, and it may not be continuous at the ends of the cracks.

3. Properties of the Double Layer Potential on the Open Curve

Let be an open curve of class . Assume that is parametrized by the arc length . The points and values of the parameter are in one-to-one correspondence, so the segment will be also denoted by . The tangent vector to in the point , in the direction of the increment of , will be denoted by , while the normal vector to in the point will be denoted by . According to the chosen parametrization ,  . The side of the crack , which is situated on the left when the parameter increases, will be denoted by , while the opposite side will be denoted by . Let be a set of the ends of .

Set , and consider the double layer potential for Laplace equation in a plane Set , , . If , then , since where and are constants, ,  . We took into account in deriving the latter inequality that (see lemma 1 in [10]), whence

Consider the integral of the Cauchy type with the real density : then . It follows from properties of the Cauchy type integral that if , then . This means that the potential is continuously extensible to from the left and from the right in interior points (though its values on from the left and from the right may be different). If, in addition, , then the potential is continuously extensible to the end , where or (see [7, Section   15.2]). Set then is a polar angle of the coordinate system with the origin in the point . Formulae for define the angle with indeterminacy up to ( is an integer number). Let be a disc of a sufficiently small radius with the center in ( or ). From asymptotic formulae describing behavior of at the ends of [7, Section   22], we may derive the asymptotic formulae for at the ends of . Namely, for any and , the formula holds: Here by we mean some fixed branch of this function, so that the branch varies continuously in in a neighbourhood of the point , cut along . The upper sign is taken if , while the lower sign is taken if . The function is continuous as . Moreover, is continuous in outside and is continuously extensible from the left and from the right to the part of lying in . It follows from formula (3.7) that for any and the inequality holds

Now we will study properties of derivatives of the double layer potential. It follows from Cauchy-Riemann relations that On the other hand, if , then for : where Since , then , so one can show that (the proof repeats the given above proof of the fact that , if ). From (3.9) and (3.10) and from properties of the Cauchy type integral [7, Section   15], it follows that if , then , that is, is continuously extensible to from the left and from the right in interior points, though the limiting values of on from the left and from the right can be different. We can write (3.10) in the form where It follows from [7, 22] that for all ( or ), such that , the inequality holds where and are constants.

Comparing formulae (3.9) and (3.12), we obtain that for and , the formulae hold where are constants. The upper sign in formulae is taken if , while the lower sign is taken if . It follows from [7, Section   15.2] that if , then the functions and are continuously extensible to the end . Moreover, if and , then for the functions and , the formulae hold: which can be derived, using the asymptotics for from [7, 22]. The upper sign in formulae is taken if , while the lower sign is taken if . Functions and are continuously extensible to the end . By we mean some fixed branch of this function, which varies continuously in in a neighbourhood of a point , cut along .

Let , and let be a normal in the point , directed to the center of the circumference , that is, , then we obtain from (3.15) for Therefore, according to (3.16): since on . From here we obtain that if , so Now let be a fixed positive number (sufficiently small). Using formulae (3.15) and setting ,  , we consider the integral over the disc : The integral converges according to estimates (3.16): Hence, if integral (3.22) converges, then the integral converges as well (as a difference of two convergent integrals), but the integral converges if and only if , while in other cases diverges. Thus, the integral (3.22) converges if and only if . Consequently belongs to with small if and only if . Let us formulate obtained results in the form of the theorem.

Theorem 3.1. Let be an open curve of class , . Let be a disc of a sufficiently small radius with the center in the point ( or ). (i) If , then and for any , such that , the relationships (3.7) and (3.8) hold.(ii) If , then(1);(2)for any , such that , the formulae (3.15) hold, in which the functions and satisfy relationships (3.16) and (3.17);(3)for the property (3.21) holds;(4) belongs to for sufficiently small if and only if .

Remark 3.2. Each function of class is continuous in , is continuously extensible to from the left and from the right, but limiting values of such a function on from the left and from the right can be different, that is, the function may have a jump on .
Let us study smoothness of the direct value of the double layer potential on the open curve.

Lemma 3.3. Let be an open curve of class , , and let . Let be the direct value of the double layer potential on . Then

Proof. Let us prove that . Taking into account that , we find Note that is a point on corresponding to . So, we may put . For , we have [10, 3] where Note that the function does not equal zero anywhere on and ; therefore, Further, Consequently, . Similarly, Consider , where Then where and . According to [7, Section   5.7], the following representation holds . Using properties of Hölder functions [7], we obtain the representation where ,  . By formal differentiation under the integral, we find The validity of differentiation under the integral can be proved in the same way as at the end of [9, Section   1.6] (Fubini theorem on change of integration order is used). Taking into account the obtained representation for and applying results of [7, Section   51.1], we obtain that . Lemma is proved.

4. Existence of A Classical Solution

We will construct the solution to the problem in assumption that , , . Note that we do not require compatibility conditions at the tips of the cracks, that is, we we do not require that for any . We will look for a solution to the problem in the form where is the double layer potential. The potential satisfies the Laplace equation (2.7) in and belongs to the class according to Theorem 3.1. Limiting values of the potential on are given by the formula where is the direct value of the potential on .

The function in (4.1) must be a solution to the following problem.

Problem D
Find a function , which obeys the Laplace equation (2a) in the domain and satisfies the boundary conditions If , then is the direct value of the potential on .
If is an exterior domain, then we add the following condition at infinity:
All conditions of the problem have to be satisfied in a classical sense. Obviously, . It follows from Lemma 3.3 that (here by we mean the direct value of the potential on ). So, and .
We will look for the function in the smoothness class .
We say that the function belongs to the smoothness class if (1), where is a pointset consisting of the endpoints of . (2)in a neighbourhood of any point for some constants , the inequality holds, where and or , .

The definition of the functional class is given in the remark to the definition of the smoothness class . Clearly, , that is, if , then .

It can be verified directly that if is a solution to the problem D in the class , then the function (4.1) is a solution to the problem .

Theorem 4.1. Let , , , . Then the solution to the problem in the smoothness class exists and is unique.

Theorem 4.1 has been proved in the following papers: (1)in [2, 3], if is an interior domain; (2)in [4], if is an exterior domain and ;(3)in [5, 6], if and so is an exterior domain.

In all these papers, the integral representations for the solution to the problem D in the class are obtained in the form of potentials, densities in which are defined from the uniquely solvable Fredholm integroalgebraic equations of the second kind and index zero. Uniqueness of a solution to the problem D is proved either by the maximum principle or by the method of energy (integral) identities. In the latter case, we take into account that a solution to the problem belongs to the class . Note that the problem D is a particular case of more general boundary value problems studied in [36].

Note that conditions of Theorem 4.1 hold if ,  ,  ,  ,  . From Theorems 3.1 and 4.1 we obtain the solvability of the problem .

Theorem 4.2. Let ,  ,  ,  ,  . Then a solution to the problem exists and is given by the formula (4.1), where is a unique solution to the problem in the class   ensured by Theorem 4.1.

Uniqueness of a solution to the problem follows from Theorem 2.2. In fact, the solution to the problem found in Theorem 4.2 is a classical solution. Let us discuss, under which conditions this solution to the problem is not a weak solution.

5. Nonexistence of a Weak Solution

Let be a solution to the problem defined in Theorem 4.2 by the formula (4.1). Consider a disc with the center in the point and of radius ( or ,  ). In doing so, is a fixed positive number, which can be taken small enough. Since , we have and (this follows from the definition of the smoothness class ). Let and . It follows from (4.1) that , whence since . Assume that belongs to , then, integrating this inequality over , we obtain . Consequently, if , then . However, according to Theorem 3.1, if , then does not belong to . Therefore, if , then our assumption that does not hold, that is, . Thus, if among numbers there exists such a number that , then for some , we have , so , where is a Sobolev space of functions from , which have generalized derivatives from . We have proved the following theorem.

Theorem 5.1. Let conditions of Theorem 4.2 hold and among numbers there exists such a number , that (i.e., compatibility condition does not hold at the tip ). Then the solution to the problem , ensured by Theorem 4.2, does not belong to for some , whence it follows that it does not belong to . Here is a disc of a radius with the center in the point .

By we denote a class of functions, which belong to on any bounded subdomain of . If conditions of Theorem 5.1 hold, then the unique solution to the problem , constructed in Theorem 4.2, does not belong to , and so it is not a weak solution. We arrive to the following corollary.

Corollary 5.2. Let conditions of Theorem 5.1 hold, then a weak solution to the problem in the class of functions does not exist.

Remark 5.3. It should be stressed that even if closed curves and cracks are very smooth and if boundary data is very smooth as well, but if there exists a tip of the crack, where the compatibility condition does not hold, then a weak solution of the problem in the class of functions does not exist.

Remark 5.4. Even if the number , mentioned in Theorem 5.1, does not exist, then the solution to the problem , ensured by Theorem 4.2, may not be a weak solution to the problem . The Hadamard example of a nonexistence of a weak solution to a harmonic Dirichlet problem in a disc with continuous boundary data is given in [11, Section   12.5] (the classical solution exists in this example).

Clearly, , since is a set of zero measure.

6. Singularities of the Gradient of the Solution at the Endpoints of the Cracks

It follows from Theorems 4.2 and 4.1 that the gradient of the solution of problem given by formula (4.1) can be unbounded at the endpoints of the cracks .

Let be a solution of the Problem ensured by Theorem 4.1. Let be a solution of the Problem ensured by Theorem 4.2 and given by formula (4.1). Let be one of the endpoints of . In the neighbourhood of , we introduce the system of polar coordinates We assume that if , and if . We recall that is the angle between the direction of the axis and the tangent vector to at the point .

Hence, if , and if .

Thus, the angle varies continuously in the neighbourhood of the endpoint , cut along .

Let be a solution of the integral equation ensured by Theorem 4 in [2] or by Theorem 4.4 in [3] or by Theorem 4 in [4]. The integral representation for the solution of the Problem is constructed in [24] on the basis of the function that is a solution of the certain integral equation.

Alternatively, one can assume that is an element of a solution to equations ensured by Corollary 3.2 in [5] or by Theorem 4 in [6]. The solution of the Problem is constructed in [5, 6] with the help of the function , which is an element of a solution to certain equations.

We will use the notation , and put if if .

At first, we study the behaviour of the gradient of a solution of the problem at the tips of the cracks. Using the representation of the derivatives of harmonic potentials in terms of Cauchy type integrals (see [10]) and using the properties of these integrals near the endpoints of the integration line, presented in [7], we can prove the following assertion.

Theorem 6.1. Let be a solution of the problem ensured by Theorem 4.1. Let be an arbitrary endpoint of the cracks , that is, and or for some . Then the derivatives of the solution of the problem in the neighbourhood of have the following asymptotic behaviour.
If , then
If , then By we denote functions which are continuous at the endpoint . Moreover, the functions denoted as are continuous in the neighbourhood of the endpoint cut along and are continuously extensible to and to from this neighbourhood.

The formulas of the theorem demonstrate the following curious fact. In the general case, the derivatives of the solution of problem near the endpoint of cracks behave as . However, if , then is bounded and even continuous at the endpoint .

On the basis of Theorem 3.1, Theorem 6.1 and formula (4.1), we may study the behaviour of the gradient of a solution of the problem at the tips of the cracks. Using notations introduced above and notations from Section 3, we arrive at the following assertion.

Theorem 6.2. Let be a solution of the problem ensured by Theorem 4.2. Let be an arbitrary endpoint of the cracks , that is, and or for some . Then the derivatives of the solution of the problem in the neighbourhood of have the following asymptotic behaviour.
If , then
If , then