The 2D Dirichlet Problem for the Propagative Helmholtz Equation in an Exterior Domain with Cracks and Singularities at the Edges

The Dirichlet problem for the 2D Helmholtz equation in an exterior domain with cracks is studied. The compatibility conditions at the tips of the cracks are assumed. The existence of a unique classical solution is proved by potential theory. The integral representation for a solution in the form of potentials is obtained. The problem is reduced to the Fredholm equation of the second kind and of index zero, which is uniquely solvable. The asymptotic formulae describing singularities of a solution gradient at the edges endpoints of the cracks are presented. The weak solution to the problem may not exist, since the problem is studied under such conditions that do not ensure existence of a weak solution.


Introduction
The 2D Dirichlet boundary value problem for the Helmholtz equation in an exterior multiply connected domain bounded by closed curves is considered in monographs on mathematical physics, for instance, in 1-3 .The review on studies of the Dirichlet problem for this equation in the exterior of cracks is given in 4 .The present paper is an attempt to combine these problems and to consider exterior domains containing cracks.From a practical stand point such domains have great significance, because cracks model both cracks in solids and wings or double-sided screens in fluids.
So, we study Dirichlet problem in an exterior domain bounded by closed curves and cracks.The theorems on existence and uniqueness of a classical solution are proved.The integral representation for a solution to a problem in the form of potentials is obtained.The problem is reduced to the uniquely solvable Fredholm integral equation of the second kind and index zero.To derive uniquely solvable integral equation on the whole boundary we use International Journal of Mathematics and Mathematical Sciences the modified integral equation approach 5, 6 on the closed curves.Since the derived integral equation of the 2nd kind is uniquely solvable, we may obtain its numerical solution in a very simple way, just by discretization and inversion of the matrix.Substituting numerical solution of the integral equation into potentials, we obtain numerical solution to the exterior Dirichlet problem in a very simple way as well.The integral representation for a solution presented in the present paper enables us to derive asymptotic formulae for singularities of a gradient of the solution at the tips of the cracks.
The Dirichlet problem for the Helmholtz equation in the exterior of several closed curves in a plane and the Dirichlet problem for the Helmholtz equation in the exterior of several curvilinear cracks in a plane are particular cases of our problem.
It is important to stress that the boundary data on the closed curves in the present paper is assumed to be only continuous.This means that weak solution may not exist in our problem, though classical solution exists.In other words, the problem in this paper is studied under conditions, which are not sufficient for existence of a weak solution in H 1 loc and weak solution may not exist, but these conditions are sufficient for existence of a classical solution.This curious fact follows from the Hadamard example of a nonexistence of a weak solution to the Dirichlet problem for Laplacian in the unit disc with continuous boundary data classical solution exists in this example .Roughly speaking, continuity of a Dirichlet data on smooth closed curves does not ensure existence of a weak solution in the Dirichlet problem.The Hadamard example of existence of a classical solution and nonexistence of a weak solution is presented and is discussed in the book 7, section 12.5 by Sobolev himself, who invented Sobolev's spaces for analysis of weak solvability of boundary value problems.
Numerical methods for the Dirichlet and Neumann problems for the Laplace and Helmholtz equations in the exterior of cracks in a plane have been developed in 8, 9 on the basis of boundary integral equations.Numerical simulation for engineering problems with cracks is presented in 10-12 using boundary element method.Problems with a crack in electromagnetoelasticity have been reduced to integral equations in 13 , and numerical solutions for some model problems have been obtained.The Dirichlet problem for elasticity equations in an exterior of several arbitrary curvilinear cracks in a plane has been reduced to the uniquely solvable integral equations in 14 .

Formulation of the Problem
By an open curve we mean a simple smooth nonclosed arc of finite length without selfintersections 15 .In the plane x x 1 , x 2 ∈ R 2 we consider an exterior multiply connected domain bounded by simple open curves , λ ∈ 0, 1 , so that the curves do not have common points; in particular, they do not have common endpoints.Suppose that The exterior connected domain bounded by Γ 2 will be denoted by D. We assume that each curve Γ k n is parametrized by the arc length s: and the domain D is to the right when the parameter s increases on Γ 2 n .Therefore points x ∈ Γ and values of the parameter s are in one-to-one correspondence except for a 2 n and b 2 n , which correspond to the same point x for n 1, . . ., N 2 .Below the sets of the intervals on the Os axis

2.2
The tangent vector to Γ at the point x s is denoted by τ x cos α s , sin α s , where cos α s x 1 s , and sin α s x 2 s .Let n x sin α s , − cos α s be the normal vector to Γ at x s .The direction of n x is chosen such that it will coincide with the direction of τ x if n x is rotated anticlockwise through an angle of π/2.
We consider Γ 1 as a set of cracks.The side of Γ 1 which is on the left, when the parameter s increases, will be denoted by Γ 1 and the opposite side will be denoted by Γ 1 − .We say that the function w x belongs to the smoothness class K if we denote the class of continuous in D \ Γ 1 functions, which are continuously extensible to the sides of the cracks Γ 1 \ X from the left and from the right, but their limiting values on Γ 1 \ X can be different from the left and from the right, so that these functions may have a jump on Γ 1 \ X.The functions of class C 0 D \ Γ 1 \ Γ 2 \ X belong to the class C 0 D \ Γ 1 if they are continuously extensible to Γ 2 from D and if they are continuously extensible to the tips of the cracks Γ 1 .
Let us formulate the Dirichlet problem for the Helmholtz equation in the exterior domain D \ Γ 1 see Figure 1 .

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Problem U. Find a function u x of the class K which satisfies the Helmholtz equation: the boundary conditions: and the radiating conditions at infinity:

2.4c
All conditions of the Problem U must be satisfied in the classical sense.Problem U includes two particular cases.In the first particular case, there are no cracks Γ 1 i.e., Γ 1 ∅ , then we get the Dirichlet problem for the Helmholtz equation in the exterior of several closed curves Γ 2 in a plane see 1-3 .In another particular case, there are no closed curves Γ 2 i.e., Γ 2  ∅ , and we obtain the Dirichlet problem for the Helmholtz equation in the exterior of several curvilinear cracks Γ 1 in a plane 4 .
On the basis of the Rellich lemma 1 , energy equalities 2 , and the regularity of the solution to the homogeneous Dirichlet problem near the boundary Γ 2 see 16, lemma 6.18 , we can easily prove the following assertion.Proof.It is sufficiently to prove that the homogeneous Problem U admits the trivial solution only.Let u 0 x be a solution to the homogeneous Problem U with F s ≡ F − s ≡ 0 and F s ≡ 0. Let S r be an open disc with a center in the origin and with sufficiently large radius r.Assume that Γ ⊂ S r 0 for some r 0 and assume that r > r 0 . Since and owing to the lemma on regularity of solutions of elliptic equations near the boundary 16, lemma 6.18 , we obtain n n 1, . . ., N 1 by a closed contour and write first Green's formula for u 0 x in a domain, bounded by these contours, by Γ 2 and by ∂S r .Then we allow to shrink closed contours onto cracks Γ 1 and use smoothness of the function u 0 x .In this way we arrive at the identity for u 0 x in the domain which is true for any r > r 0 , where r 0 is some constant.The curviliniear integral of the 1st kind is taken over ∂S r .By u 0 x the complex conjugate function to u 0 x is denoted.Clearly, u 0 x belongs to class K and satisfies homogeneous boundary conditions.
By the superscripts and − we denote the limiting values of functions on Γ 1 and on Γ 1 − , respectively.Since u 0 x satisfies the homogeneous boundary condition 2.4b on Γ, we rewrite identity * in the following form: Taking the imaginary part, we obtain the identity which is true for any r > r 0 .Tending r → ∞ in this identity and taking into account conditions 2.4c at infinity, we obtain lim whence u 0 x ≡ 0 in D \ Γ 1 on the basis of the Rellich lemma 1, section 229 .Thus, u 0 x is a trivial solution to the homogeneous Problem U. Consequently, the homogeneous Problem U has only the trivial solution, and the theorem is proved owing to the linearity of the Problem U.

Integral Equations at the Boundary
To prove existence of a solution to the Problem U, we assume that

3.1b
The conditions 3.1b are compatibility conditions for functions F s and F − s at the tips of the cracks.To solve Problem U we discuss some preliminary matter.
If B 1 Γ 1 and B 2 Γ 2 are Banach spaces of functions given on Γ 1 and Γ 2 , then for functions given on Γ we introduce the Banach space B 1 Γ 1 ∩ B 2 Γ 2 with the norm

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We consider the angular potential from 4 for 2.4a on Γ 1 : The kernel V x, σ is defined on each curve Γ 1 n , n 1, . . ., N 1 , by where H 1 0 z is the Hankel function of the first kind 3 :

3.4
Here in after we suppose that ν σ belongs to C 0,λ Γ 1 and satisfies the following additional conditions:

3.5
As shown in 4 , for such ν σ the angular potential v 1 ν x belongs to the class K.In particular, the condition 2.3 is satisfied for any ∈ 0, 1 .Moreover, integrating v 1 ν x by parts and using 3.5 we express the angular potential in terms of a double-layer potential: with the density Consequently, v 1 ν x satisfies both equation 2.4a outside Γ 1 and the conditions at infinity 2.4c .
Let us construct a solution to the Problem U.This solution can be obtained with the help of potential theory for the Helmholtz equation 2.4a .We look for a solution to the problem in the following form: where v 1 ν x is given by 3.2 , 3.6 , and 3.9 The density ν σ must belong to C 0,λ Γ 1 and must satisfy conditions 3.5 .
We will look for μ s in the Banach space We say that μ s belongs to the Banach space C ω q Γ 1 with some ω ∈ 0, 1 and q ∈ 0, 1 if where C 0,ω Γ 1 is a H ölder space with the exponent ω.The norm in the Banach space C ω q Γ 1 is defined by

3.11
It can be checked directly with the help of 4 that for such μ s the function w 1 μ x satisfies 2.4a and belongs to the class K.In particular, the inequality 2.3 holds with −q if q ∈ 0, 1 .The potential w 2 μ x satisfies 2.4a and belongs to C 0 D ∩C 2 D .It is clear that the function 3.8 satisfies conditions at infinity 2.4c .So, the function 3.8 with densities μ s and ν s subject to requirements described before satisfies all conditions of the Problem U except for the boundary conditions 2.4b .
To satisfy the boundary conditions we substitute 3.8 in 2.4b and arrive at the system of integral equations for the densities μ s and ν s :

3.12a
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3.12b
where ρ s is defined in terms of ν s in 3.7 .
To derive limit formulas for the angular potential, we used its expression in the form of a double-layer potential 3.6 .
Equation 3.12a is obtained as x → x s ∈ Γ 1 ± and comprises two integral equations.The upper sign denotes the integral equation on Γ 1 , and the lower sign denotes the integral equation on Γ 1 − .
In addition to the integral equations written before we have the conditions 3.5 .Subtracting the integral equations 3.12a and using 3.7 , we find

3.13
We note that ν s is found completely and satisfies all required conditions, in particular, conditions 3.5 .Hence, the angular potential of 3.2 and 3.6 is found completely as well.
We introduce the function f s on Γ by where F s is specified on Γ 2 in 2.4b and We set

3.15
Adding the integral equations 3.12a and taking into account 3.12b we obtain the integral equation for μ s on Γ:

3.16
where f s is given in 3.14 .

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Thus, if μ s is a solution of 3.16 in the space C ω q Γ 1 ∩ C 0 Γ 2 , ω ∈ 0, 1 , q ∈ 0, 1 , then the potential 3.8 with ν s from 3.13 satisfies all conditions of the Problem U. We arrive at the following statement.Theorem 3.1.If Γ ∈ C 2,λ , if conditions 3.1a and 3.1b hold, and if equation 3.16 has a solution μ s from the Banach space C ω q Γ 1 ∩ C 0 Γ 2 , ω ∈ 0, 1 , q ∈ 0, 1 , then a solution to the Problem U exists and is given by formula 3.8 , where ν s is defined in 3.13 .
If s ∈ Γ 2 , then 3.16 is an equation of the second kind.If s ∈ Γ 1 , then 3.16 is an equation of the first kind and its kernel has a logarithmic singularity, because where h z is a smooth function.Indeed, as z → 0 0,

3.18
Our further treatment will be aimed to the proof of the solvability of equation 3.16 in the Banach space C ω q Γ 1 ∩ C 0 Γ 2 .Moreover, we reduce equation 3.16 to a Fredholm equation of the second kind and of index zero, which can be easily computed by classical methods.
By differentiating equation 3.16 on Γ 1 we reduce it to the following singular integral equation on Γ 1 :

3.19a
where the function h z is defined by 3.17 , and ϕ 0 x, y is the angle between the vector xy and the direction of the normal n x .The angle ϕ 0 x, y is taken to be positive if it is measured anticlockwise from n x and negative if it is measured clockwise from n x .Besides, ϕ 0 x, y is continuous in x, y ∈ Γ if x / y.Equation 3.16 on Γ 2 we rewrite in the following form: where

3.21
We note that 3.19a is equivalent to 3.16 on Γ 1 if and only if 3.19a is accompanied by the following additional conditions:

3.22
The system of 3.19a , 3.19b , and 3.22 is equivalent to 3.16 .
It can be easily proved see 4 for details that sin ϕ 0 x s , y σ

The Fredholm Integral Equation and the Solution to the Problem
Inverting the singular integral operator in 3.24 , we arrive at the following integral equation of the second kind 4, 15 : where and G 0 , . . ., G N 1 −1 are arbitrary constants.We set sign s − a 1 n 1 as s a 1 n ; then the function sign s − a 1 n belongs to C ∞ Γ 1 in s variable for n 1, . . ., N 1 .It can be shown using the properties of singular integrals 15 that Φ 1 s and A 1 s, σ are H ölder continuous functions if s ∈ Γ 1 and σ ∈ Γ.Consequently, any solution of 4.1 belongs to C ω 1/2 Γ 1 with some ω ∈ 0, 1 , and here in after we look for μ s on Γ 1 in this space.
We set and rewrite 4. 1 and 3.19b in the form of one equation: where To derive equations for G 0 , . . ., G N 1 −1 we substitute μ s from 4.1 and 3.19b in the conditions 3.22 ; then in terms of μ * s we obtain where

4.7
By • we denote the variable of integration in the potential w μ x in 3.9 .Thus, the system of 3.19a , 3.19b , and 3.22 for μ s has been reduced to the system of 4.4 and 4.6 for the function μ * s and constants G 0 , . . ., G N 1 −1 .It is clear from our consideration that any solution of system of 4.4 and 4.6 generates a solution to the system of 3.19a , 3.19b , and 3.22 .As noted before, Φ 1 s and A 1 s, σ are H ölder continuous functions if s ∈ Γ 1 , and σ ∈ Γ.More precisely see 4 , Φ 1 s ∈ C 0,p Γ 1 , p min{1/2, λ}, and A 1 s, σ belongs to C 0,p Γ 1 in s uniformly with respect to σ ∈ Γ.Using these properties we can prove the following.
The condition Φ s ∈ C 0,p Γ 1 ∩ C 0 Γ 2 holds if conditions 3.1a and 3.1b hold.Hence here in after we will look for μ * s from C 0 Γ .Since A 1 s, σ ∈ C 0 Γ 1 × Γ , and due to the special representation for A 2 s, σ from 3.19b , the integral operator from 4.4 is a compact operator mapping C 0 Γ into itself.Indeed, one can check using Arzela theorem 17 that the integral operator with the kernel A 1 s, σ is a compact operator mapping C 0 Γ into C 0 Γ 1 , while the integral operator with the kernel A 2 s, σ is a compact operator mapping C 0 Γ 1 into C 0 Γ 2 .Moreover, it can be verified directly with the help of the Arzela theorem that the integral operator To verify equicontinuity in the Arzela theorem, we may use the property of uniform continuity in x on Γ 2 for the functions |x − y| 1/3 and |x − y| 1/3 ln |x − y|.In doing so, we may use the Cauchy-Bunyakovski inequality for estimates.
We rewrite 4.4 in the following operator form: where P is the operator of multiplication of the row P 1 − δ s s 0 , . . ., s N 1 −1 by the column G G 0 , . . ., G N 1 −1 T .The operator P is finite-dimensional from E N 1 into C 0 Γ and therefore compact 18 .

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Now we rewrite 4.6 in the following form: B is a N 1 × N 1 matrix consisting of the elements B nm from 4.7 .The operator L acts from where The operators B − I N 1 and L are finite-dimensional and therefore compact 19 .
We consider the columns . We write the system of 4.9 and 4.10 in the form of one equation: where I is the identity operator in the space C 0 Γ ×E N 1 .It is clear that R is a compact operator mapping C 0 Γ × E N 1 into itself, since it consists of compact operators.Therefore, 4.13 is a Fredholm equation of the second kind and of index zero in the space C 0 Γ × E N 1 see 17, 18, 20 .
Let us show that homogeneous equation 4.13 has only a trivial solution.Then, according to Fredholm's alternative 17, 18, 20 , the inhomogeneous equation 4.13 has a unique solution for any right-hand side.Let be an arbitrary solution of the homogeneous equation 4.13 .According to Lemma 4.1, Therefore the function the column G 0 convert the homogeneous equations 4.1 , 3.19b and 4.6 into identities.For instance, 3.19b takes the following form:

4.16a International Journal of Mathematics and Mathematical Sciences
Using the homogeneous identities 4.1 and 3.19b , we check that the homogeneous identities 4.6 are equivalent to w μ 0 a 1 n 0, n 1, . . ., N 1 .

4.16b
Besides, acting on the homogeneous identity 4.1 with a singular operator with the kernel s − t −1 we observe that μ 0 s satisfies the homogeneous equation 3.24 : It follows from 4.16a , 4.16b and 4.16c that μ 0 s satisfies the homogeneous equation 3.16 .On the basis of Theorem 3.1, u 0, μ 0 x ≡ w μ 0 x is a solution to the homogeneous Problem U.According to Theorem 2.2, w μ 0 x ≡ 0 for x ∈ D \ Γ 1 .Using the limit formulas for normal derivatives of a single-layer potential on Γ 1 , we have Hence, w μ 0 x w 2 μ 0 x ≡ 0 for x ∈ D, and μ 0 s satisfies 4.16a , which can be written as
We have proved the following assertion.
We recall that Φ belongs to the class of smoothness required in Corollary 4.3 if conditions 3.1a and 3.1b hold.Besides, 4.13 is equivalent to the system of 4.4 and 4.6 .As mentioned before, if {μ * s , G 0 , . . ., G N 1 −1 } is a solution of the system of 4.4 and 4.6 , and a solution of the system of 3.19a , 3.19b and 3.22 , and therefore μ s satisfies 3.16 .We obtain the following statement.It can be checked directly that the solution to the Problem U satisfies condition 2.3 with −1/2.Explicit expressions for singularities of the solution gradient at the endpoints of the cracks will be presented in the next section.Theorem 4.5 ensures existence of a classical solution to the Problem U when Γ ∈ C 2,λ , λ ∈ 0, 1 , and conditions 3.1a and 3.1b hold.The uniqueness of the classical solution follows from Theorem 2.2.On the basis of our consideration we suggest the following scheme for solving the Problem U. First, we find the unique solution of the Fredholm equation 4.13 from C 0 Γ × E N 1 .This solution automatically belongs to C Finnaly, substituting ν s from 3.13 and μ s in 3.8 we obtain the solution to the Problem U. Remark 4.6.It is important to stress that the solution u x to the Problem U, ensured by Theorem 4.5, is a classical solution, but it may be not a weak solution to the Problem U. In other words, classical solution to the Problem U exists and is ensured by Theorem 4.5, but weak solution to the Problem U may not exist in H 1 loc D \ Γ 1 space.This follows from the fact that Dirichlet data on the closed curves Γ 2 is assumed to be continuous only.Continuity of a Dirichlet boundary data on closed curves is not sufficient for existence of a weak solution International Journal of Mathematics and Mathematical Sciences in H 1 loc D \ Γ 1 space.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 the book 7, section 12.5 by Sobolev himself the classical solution exists in this example .

Singularities of the Gradient of the Solution at the Endpoints of the Cracks
As noted at the end of Section 4, the gradient of the solution to Problem U can be unbounded at the endpoints of the cracks Γ 1 , so that the gradient of the solution to the Problem U satisfies estimate 2.3 with the exponent −1/2.We will now make a detailed analysis of the behaviour of ∇u x at the endpoints of Γ 1 .
Let u x be a solution to the Problem U ensured by Theorem 4.5 and given by 3.8 .Let x d ∈ X be one of the endpoints of Γ 1 .In the neighbourhood of x d , we introduce the system of polar coordinates: n .Using the representation of the derivatives of harmonic potentials in terms of Cauchy type integrals see 4 and using the properties of these integrals near the endpoints of the integration line, presented in 15 , we can prove the following assertion.

5.4
By O 1 one denotes functions which are continuous at the endpoint x d .Moreover, the functions denoted by O 1 are continuous in the neighbourhood of the endpoint x d cut along Γ 1 and are continuously extensible to Γ 1 and to Γ 1 − from this neighbourhood.
The formulas of the theorem demonstrate the following curious fact.In the general case, the derivatives of the solution to the Problem U near the endpoint x d of cracks Γ 1 behave as However, if μ 1 d ν d 0, then ∇u x will be bounded and even continuous at the endpoint x d ∈ X.

Figure 1 :
Figure 1: An example of an exterior domain.

x 1 xα d α a 1 n0 if d a 1 n and α d α b 1 nd μ 1 a 1 n μ 1 a 1 n0 if d a 1 n , and μ 1 d μ 1 b 1 n μ 1 b 1 n − 0 if d b 1
1 d |x − x d | cos ϕ, x 2 x 2 d |x − x d | sin ϕ.5.1We will assume thatϕ ∈ α d , α d 2π if d a 1 n and ϕ ∈ α d − π, α d π if d b 1 n .We recall that α s is the angle between the direction of the Ox 1 axis and the tangent vector τ x to Γ 1 at the point x s .Hence, − 0 if d b 1 n .Thus, the angle ϕ varies continuously in the neighbourhood of the endpoint x d , cut along Γ 1 .We will use the notation μ 1 s μ s |s − d| 1/2 Q −1 s μ * s |s − d| 1/2 and put μ 1

Theorem 5 . 1 .a 1 n , then ∂ ∂x 1 u x μ 1 a 1 n 2 x − x a 1 n 1 /2 sin ϕ α a 1 n 2 − ν a 1 n 2π − sin α a 1 n ln x − x a 1 n ϕ cos α a 1 nμ 1 a 1 n 2 x − x a 1 n 1 /2 cos ϕ α a 1 n 2 − ν a 1 n 2π cos α a 1 n ln x − x a 1 n ϕ sin α a 1 n O 1 .
Let u x be a solution to the Problem U ensured by Theorem 4.5.Let x d be an arbitrary endpoint of the cracks Γ 1 , that is, x d ∈ X and d a 1 n or d b 1 n for some n 1, . . ., N 1 .Then the derivatives of the solution to the Problem U in the neighbourhood of x d have the following asymptotic behaviour.If d