THE NEUMANN PROBLEM FOR THE 2-D HELMHOLTZ EQUATION IN A DOMAIN , BOUNDED BY CLOSED AND OPEN CURVES 209

The Neumann problem for the dissipative Helmholtz equation in a connected plane region bounded by closed and open curves is studied. The existence of classical solution is proved by potential theory. The problem is reduced to the Fredholm equation of the second kind, which is uniquely solvable. Our approach holds for both internal and external domains.


INTRODUCTION
The boundary value problems in domains bounded by closed and open curves were not treated in the theory of 2-D PDEs before.Even in the case of Laplace and Helmholtz equations the problems in domains bounded by closed curves [1][2], [5][6][7][8] and problems in the exterior of open arcs [5], [9][10][11] were treated separately, because different methods were used in their analysis.
Previously the Neumann problem in the exterior of an open arc was reduced to the hypersingular integral equation [9][10] or to the infinite algebraic system of equations [11], while the Neumann problem in domains bounded by closed curves was reduced to the Fredholm equation of the second kind [1], [6][7][8].The combination of these methods in case of domains bounded by closed and open curves leads to the integral equation, which is algebraic or hypersingular on open curves and t is an equation of the second kind with compact integral operators on the closed curves.The integral equation on the whole boundary is too complicated and the general theory of similar equations are not constructed currently.The approach suggested in the present paper enables to reduce the Neumann problem in domains bounded by closed and open curves to the Fredholm integral equation on the whole boundary with the help of the nonclassical angular potential.Since the boundary integral equation is Fredholm, the solvability theorem follows from the uniqueness theorem, which is ensured for the Neumann problem in the case of the dissipative Helmholtz equation This approach is based on [3][4], where the problems in the exterior of open curves were reduced to the Fredholm integral equations using the angular potential.

FORMULATION OF THE PROBLEM
By a simple open curve we mean a non-closed smooth arc of finite length without self- intersections [5] In the plane x (xl,x2) 6 R we consider the multiply connected domain bounded by simple open curves r,..., r, c-,, (0, ], n simple closed curves r,...,r, c-,0, o that the curves do not have points in common.We will consider both the case of an external domain and the case of an internal domain, when the curve F encloses all other.We put N N r= Urn, r= Urn, r=rur. The connected domain bonded by F will be called D. We assume that each curve F is para- metricized by the arc lenh s r (. () ((),()), e [,b]}, 1,...,N, k 1,2, so that a < b} < < a, < b < a < b < < a < b and the domn D is to the right when the parameter s increases on F. Therefore points x F and values of the parameter s are in one-to-one correspondence except a, b, wch correspond to the same point x for n 1, N.  The tangent vector to r at the point z(s) we denote by COS (8) X( 8) sinG( 8) X(8).Let , (siG(8)-cosG( 8)) be a ormal vector to 8t x().
The rection of n is chosen such that it will coincide with the rection of T if n is rotated anticlockwise through an angle of r/2.We say, that the nction w(x) belongs to the smoothns class K if ) e (r) c(r), 2) vw e c(rrx), here X is a point-set, consisting of th end-points of r N n=l 3) in the neighborhood of any point x(d) X for some constants C > 0, e > -1 the inequality holds v c (d) (.) wherexx(d) andd=a, ord=b, n=l,...N, 4) there exists a iform for all z(s) e r lit of (n,, V()) as along the normal n.REMARK.In the definition of the class K we consider F Accorng to this definition, w(z) and Vw(z) may have a jump across rx.
Let us formulate the Neumann problem for the ssipative Helmholtz equation in the domn PROBLEM U. To find a fction w(x) of the class K which satisfies the Helmholtz equation ,, () + ,,(z) + Z(z) 0, e r , Z cot, Z > 0, (2.2) and the boundary condition If D is an external domain, then we add the following condition at infinity (2.2c)All conditions of the problem U must be satisfied in the classical sense.By Ow/On on F we mean the limit ensured in the point 4) of the definition of the smoothness class K.The normal derivative 0w/0nz has to be continuous across FI\X and has to take given values on FI\X.At the same time w(x) may have a jump across FI\X.
On the basis of the energy equalities and the technique of equidistant curves [6], we can easily prove the following assertion.THEOREM 1.If F 6 C', A 6 (0, 1], F 6 C2'0, then the problem U has at most one solution. The theorem holds for both internal and external domain 7).

INTEGRAL EQUATIONS AT THE BOUNDARY
Below we assume that f(s) from (2.2b) is an arbitrary function from the Banach space c,(r) c(r), e (0, }.If B1 (F), B2(F2) are Banach spaces of functions given on F and F2, then for functions given on F we introduce the Banach space B(F The kernel V(x, r) is defined on the each curve F, n 1, N by the formula v(, ) / o,( (Z I y()l) d, = e [, b]   where T/(0)(z) is the Hankel function of the first kind (01 Below we suppose that/(a) belongs to the Sanach space C(F1), w 6 (0,1], q e [0,1) and satisfies the following additional conditions / l(a) da O, n l, N. As shown in [3], [4] for such #(a) the angular potential wl[#](x) belongs to the class K.In particular, the inequality (2.1) holds with e -q, if q (0, 1).Moreover, integrating wl[](x) by parts and using (3.2) we express the angular potential in terms of a double layer potential f p(a) ---) (Z lz y(a)l) da, ( []()with the density p(c) ] #(f)d', a e [al, b], n 1, N1. (3.4) Consequently, wl[#](x) satisfies both equation (2.2a) outside F and the conditions at infinity (2.2c).
Let us construct a solution of the problem U This solution can be obtained with the help of potential theory for the Helmholtz equation (2.2a).We seek a solution of the problem in the form of the anlar pontial on F and the single-layer potential on F w](x) w[](x) + w[]( where w[](x)is ven by (3.1), ( ) ) (Z I y(a)l)d.
It follows from the properti of potentials [1], [3-4], [6], that for such p(s) the ction (3.5) belongs to the cls K and satisfies all contions of the problem U except the boundary contion (2.2b).In the ce of the extern domain the fction (3.5) satisfies the contion at inity (.).
To satisfy the bodary contion we pu (3.5) in (2.2b), use the lit form for the anlar potential from [3] d aive the ine equation for the density (s) + f u() where 6(s) 0 if and g(s By 0(z,) we denote the angle between the vector and the Nrection of the normal n.The angle 0(, ) is taken to be positive if it is meured anticlockwise from n and negative if it is measured clockwise om n.Besides, 0(z, ) is continuous in z, F if z .
If s 6 F -, then (3.6) is an equation of the second kind with compact integral operators.If s 6 F , then (3.6) is a singular integral equation [5].
Our further treatment will be aimed to the proof of the solvability of the system (3.2),(3.6) in the Sanach space C[(F 1) q C(F).Moreover, we reduce the system (3.2), (3.6) to a Fredholm equation of the second kind, which can be easily computed by classical methods.
We note A(s,a) (F 2 x F), because F: C'.

THE FREDHOLM INTEGRAL EQUATION AND THE SOLUTION OF THE PROBLEM
Inverting the singular integral operator in (3.8) we arrive at the following integral equation of the second kind [5]: To derive equations for Go, GN-I we substitute #(s) from (4.1) in the conditions (3.2), then we obtain N /tt(a)l,(a)da + Sn,G, Hn, n 1,...,Nx By B we denote the N1 x N1 matrix with the elements B,, from (4.3).As shown in [4], the matrix B is invertible.The elements of the inverse matrix will be called (B-1),.Inverting the matrix B in (4.2) we express the constants Go, GNin terms of tt(s) G.
(B-I).. H.It can be shown using the properties of singular integrals [2], [5], that 0(s), A0(s, a) are Holder functions if s E I'1, o E 1".Therefore, O(s), A(s, a) are also Holder functions if s Fx, a F.
Consequently, any solution of (4.4) belongs to C'/(r 1) and below we look for (s) on '1 in this space.
As noted above, l(s) and A(s,a) are Holder functions if s 6 F , cr 6 F.More precisely (see [4], [5]), (s) e C'(F1), p min{1/2, A} and Al(S,a) belongs to C'(Fl) in s uniformly with respect to a E F. We arrive at the following assertion.
Hence below we will seek/.(s)from C(F).Since A(s, a) C(F x F), the integral operator from (4.5): Ap. f tz.(a)Q-(a)A(s,a)da F is a compact operator mapping C(F) into itself.Therefore, (4.5) is a Fredholm equation of the second kind in the Banach space C(F).
Let us show that homogeneous equation (4.5) has only a trivial solution.Then, according to Fredholm's theorems, the inhomogeneous equation (4.5) has a unique solution for any right-hand side.We will prove this by a contradiction.Let/z.(s)C(F) be a non-trivial solution of the ho- mogeneous equation (4.5).According to the lemma .(s)C'(F1) N C(F), p min{A, 1/2}.
Therefore the function #(s) #.(s)Q-(s) C/=(F ) N C(F2) converts the homogeneous equa- tions (3.7), (4.4) into identities.Using the homogeneous identity (4.4) we check, that/(s) satisfies conditions (3.2).Besides, acting on the homogeneous identity (4.4) with a singular operator with the kernel (s-t) -x we find that #(s) satisfies the homogeneous equation (3.8).Consequently, #(s) satisfies the homogeneous equation (3.6).On the basis of theorem 2, w[#](x) is a solution of the homogeneous problem U.According to theorem 1 w[/l(x) 0, x D\F x.Using the limit formulas for tangent derivatives of an angular potential [3], we obtain 0 0 0 lira wu ](x)-lim w[/](x) #(s) --0, s F -x(s)e(rl)+ x-,(s)e(rl)-OT By (F) + we denote the side of F which is on the left as a parameter s increases and by (F) we denote the other side.
On the basis of the theorem 2 we arrive at the final result.THEOREM 5.If F e C', F e C2', f(s) e C'(F)NC(F), A e (0,1}, then the solution of the problem U exists and is given by (3.5), where #(s) is a solution of equations (3.), (3.6) from C/2(F ) n C(F2), p min{1/2, A} ensured by the theorem .
It can be checked directly that the solution of the problem U satisfies condition (2.1) with -1/2.Explicit expressions for singularities of the solution gradient at the end-points of the open curves can be easily obtained with the help of formulas presented in [4].

(3. 2 )
We say, that/(s) e C(F ) if gl C'(r where C'(F1) is a Holder space with the index and r=l IIcO,w(F1) where Go, GNI-1 are arbitrary constants and 2Q(a)f(a)da Ao(s, a) _lr / Y ( a) Q (I)0(s) l r / a s F F (a)l(a)daWe substitute G, in (4.1) and obtain the integral equation for #(s) on F 1/ (s) + tt(a)A (s, a)da qli's) O1 (s), s e F ,