THE SOLUTION OF THE THIRD PROBLEM FOR THE LAPLACE EQUATION ON PLANAR DOMAINS WITH SMOOTH BOUNDARY AND INSIDE CRACKS AND MODIFIED JUMP CONDITIONS ON CRACKS

This paper studies the third problem for the Laplace equation on a bounded planar domain with inside cracks. The third condition ∂u/∂n+hu= f is given on the boundary of the domain. The skip of the function u+ − u− = g and the modified skip of the normal derivatives (∂u/∂n)+− (∂u/∂n)− +hu+ = f are given on cracks. The solution is looked for in the form of the sum of a modified single-layer potential and a double-layer potential. The solution of the corresponding integral equation is constructed.


Introduction
Krutitskii studied in [6] the boundary value problem for the Helmholtz equation outside several cuts in the plane.Two boundary conditions were given on the cuts.One of them specified the jump of the unknown function.Another one of the type of the Robin condition contained the jump of the normal derivative of an unknown function and the one-side limit of this function on the cuts.He looked for a solution of the problem in the form of the sum of a single-layer potential and an angular potential.He reduced the problem to the solution of a uniquely solvable integral equation.So, he proved the unique solvability of the problem.
This paper studies the boundary value problem for the Laplace equation in a bounded planar domain with several cuts inside.The cuts in [6] are smooth open arcs.The cracks in this paper are arbitrary closed subsets of such sets.The Robin condition is given on the boundary of the domain.The same conditions as that in [6] are on the cuts.The case when the jumps of the solution and its normal derivatives are given on cracks is included.So, this problem is a generalization of the problem studied in [4,5].I looked for a solution in a similar form like in [6] but instead of a single-layer potential, I used a modified single-layer potential.I reduced the problem to the solution of an integral equation and constructed the solution of this equation.So, I constructed a solution of the problem which is a new result even if there are no cuts inside the domain.

Formulation of the problem
Let M ⊂ R m .We denote by C 0 (M) the set of all continuous functions on M. If k is a positive integer, denote by C k (M) the set of all functions f such that there is continuous D α f on M for each multi-index α with the length at most k.If 0 < β < 1, denote by C β (M) the set of β-Hölder functions on M, that is, the set of continuous functions f on M such that If k is a positive integer and 0 < β < 1, denote by C k+β (M) the set of all functions f ∈ C k (M) such that D α f ∈ C β (M) for each multi-index α with the length k.
We say that a bounded open set H ⊂ R 2 has C α boundary ∂H if there exist a finite number of "local" coordinate systems (x k , y k ) (k = 1,...,m) and a finite number of functions denote by u(x) (u + (x), u − (x)) the limit of u at x with respect to G (H + , H − ), respectively.We will study the Robin problem for the Laplace equation on the cracked open set G.
Let h ∈ C 0 (∂G), h ≥ 0, f ∈ C 0 (∂G), g ∈ C 0 (Γ).We say that a function u in G is a solution of the problem (2) there is an extension of u onto the function from C 1 (clH + ); (3) there is an extension of u onto the function from 0, we will talk about the Neumann problem.In the opposite case, we will talk about the Robin problem.

Uniqueness
Denote by Ᏼ k the k-dimensional Hausdorff measure normalized so that Ᏼ k is the Lebesgue measure in R k .Theorem 3.1.Let h ∈ C 0 (∂G), h ≥ 0, f ≡ 0, g ≡ 0, and let u be a solution of the problem (2.2).Then u is constant in G.If there is x ∈ ∂G so that h(x) > 0, then u = 0 in G.

Necessary conditions for the solvability
Let K be a closed subset of ∂H − .For a function f defined on K, denote where τ(x) = (−n 2 (x),n 1 (x)) is the unit tangential vector of H − at x.
, and let u be a solution of the problem (2.2).Then Proof.Since g Γ = u + − u − in ∂H + and u + ,u − ∈ C 1 (∂H + ), we deduce that g Γ ∈ C 1 (∂H + ).If h ≡ 0, we get, using Green's theorem and the fact that 4 Third problem on cracked domains Suppose now that h ∈ C 0 0 (∂H + ).Since

Single-layer potentials
Fix R > 0. For f ∈ C 0 (∂G), define as the modified single-layer potential with density f .In the case of several sets, we will write G R f .(For R = 1, we get the usual single-layer potential.)Note that R f differs by constants.The function R f is harmonic in R 2 \ ∂G.
where M depends on G, R, and β.
Easy calculation yields that this operator is closed.According to the closed graph theorem (see [13, Chapter II, Section 6, Theorem 1]), it is bounded, similarly for H + .

Double-layer potentials
as the double-layer potential with density g.If V is a bounded open set with C 1 boundary, g ∈ C(∂V ), and n V (y) denotes the exterior unit normal of V at y, define on as the double-layer potential corresponding to V with density g.
where M depends on G, R, and β.
The boundary of H + is formed by finitely many Jordan curves.Fix one of these curves T. Denote g T = g on T, g T = 0 elsewhere.Let T be parametrized by the arc length s : T = {ϕ(s); s ∈ [a,b]}, and H + is to the right when the parameter s increases on T.
], denote by v(x,ϕ(s)) the increment of the argument of y − x along the curve {y = ϕ(t); t ∈ [a,s]}, and is the angular potential corresponding to f .Define Using boundary properties of single-layer potentials (see [2, Theorem 2.2.13]), we can deduce that n + (x) we have the function Ᏸ g is constant in the interior of T and in the exterior of T as well.Thus , it is bounded by the closed graph theorem (see [13, Chapter II, Section 6, Theorem 1]), similarly for H − .

Reduction of the problem
Let H, H + have boundary of class C 1+γ , where 0 < γ < 1.Let 0 We will look for a solution u of the problem (2.2) in the form According to Lemma 6.1 and [9, Theorem 1], the function u is a solution of the problem (2.2) if and only if the function v is a solution of the problem where ) Dagmar Medková 7 We will look for a solution of the problem (7.2) in the form of a modified single-layer potential R w, where w ∈ C β 0 (∂G) and R > diamG.Define for According to [9, Theorem 2], ) Since the modified single layer potential R w is a harmonic function in G, we get using Lemma 5.1 that R w is a solution of the problem (7.2) if and only if where   8 Third problem on cracked domains Notation 8.2.Let X be a real Banach space.Denote by complX = {x + iy; x, y ∈ X} the complexification of X with the norm x + iy = x + y .If T is a bounded linear operator on X, we define T(x + iy) = Tx + iT y as the bounded linear extension of T onto compl X. Definition 8.3.The bounded linear operator T on the Banach space X is called Fredholm if α(T), the dimension of the kernel of T, is finite, the range T(X) of T is a closed subspace of X, and β(T), the codimension of T(X), is finite.The number i(T According to Section 7, the modified single-layer potential R T 0,R ϕ is a solution of the problem (2.2) with h ≡ 0, g ≡ 0, and f = T 2 0,R ϕ = 0.According to Theorem 3.1, there is a constant c such that R T 0,R ϕ = c on G.
According to Section 7, the modified single-layer potential R ϕ is a solution of the problem (2.2) with h ≡ 0, g ≡ 0, and f = T 0,R ϕ.
Proposition 8.6.Let H, H + have boundary of class C 1+γ , where According to Lemma 8.5, the operator T 0,R is injective in T 0,R (C Then there is a unique solution of the problem (2.2).
Proof.Fix R > diamG.According to Proposition 8.6, there is Let F be given by (7.3).Then is a solution of the problem (2.2) by Section 7.This solution is unique by Theorem 3.1.
Then there is a solution of the problem (2.2) if and only if (4.2) holds.This solution is unique up to an additive constant.
Proof.If there is a solution of the problem (2.2), the relation (4.2) holds by Proposition 4.1.Suppose now that (4.2) holds.Fix R > diamG.Denote by T the restriction of T 0,R onto C β 0,0 (∂G).Then there is T −1 by Proposition 8.6.Let F be given by (7.4).Then u = Ᏸg + R T −1 F is a solution of the problem (2.2) by Section 7.This solution is unique up to an additive constant by Theorem 3.1.

Solution of the problem
by Lemma 5.3 and h ≥ 0, we get Proof.We can suppose that λ = 0. Lemma 9.1 yields λ > 0 and we thus can suppose that f ∈ C β 0 (∂G).Since R λ −1 f is a solution of the problem (2.2) with h ≡ 0, g ≡ 0 (see Section 7), Proposition 4.1 gives that f fulfills (4.2).Since T 0,R f = f on Γ, we deduce from T 0,R f = λ f that λ = 1 or f = 0 on Γ.We can restrict ourselves to the case when f = 0 on Γ.

Dagmar Medková 11
Fix r > 0 such that clG ⊂ Ω r (0) ≡ {y ∈ R 2 ;|x| < r} and put Using Lemma 5.3, we get Notation 9.3.Let X be a complex Banach space and let T be a bounded linear operator on X. Denote by σ(T) the spectrum of T and by r(T) = sup{|λ|; λ ∈ σ(T)} the spectral radius of T.

β 0 (
∂G) to C α (∂H).Since the identity operator is a compact operator from C α (∂H) to C β (∂H), the operator K * G is a compact linear operator from C β 0 (∂G) to C β (∂H) by [13, Chapter X, Section 2].Since K * G is a bounded linear operator from C β 0 (∂G) to C β (∂H), the operator T 0,R is a bounded linear operator in C β 0 (∂G).Since T h,R − T 0,R is a bounded linear operator in C β 0 (∂G) by Lemma 5.2, the operator T h,R is a bounded linear operator in C β 0 (∂G).