Solvability of a system of dual integral equations of a mixed boundary value problem for the laplace equation - Nguyen Thi Ngan

SOLVABILITY OF A SYSTEM OF DUAL INTEGRAL EQUATIONS OF A MIXED BOUNDARY VALUE PROBLEM FOR THE LAPLACE EQUATION Nguyen Thi Ngan1 and Nguyen Thi Minh, College of Education -TNU Summary. The aim of the present paper is to consider solvability and solution of a system of dual integral equations involving Fourier transforms occurring in mixed boundary value problems for Laplace equation. The uniqueness and existence theorems are proved. A method for reducing system of dual equations to a system of Fredholm integral equations of second kind is also proposed. Key words: Fourier transform, dual integral equations, Laplace equation, mixed boundary value problems. 1 Introduction Dual integral equations arise when integral trans- forms are used to solve mixed boundary value prob- lems of mathematical physics and mechanics. For- mal technique for solving such equations have been developed extensively, but their solvability prob- lems have been considered comparatively weakly [1, 2]. The solvabilities of dual integral equations involving Fourier transforms and dual series equa- tions involving orthogonal expansions of general- ized functions were considered in [3, 4]. The solvability problems for systems of dual equa- tions so far we know have been not considered. The aim of the present work is to consider existence and uniqueness problems for a system of dual integral equations involving Fourier transforms occurring in one mixed boundary value problem for the Laplace equation. A method for reducing system of dual equations to a system of Fredholm integral equa- tions of second kind is also proposed. Consider the following problem: find a solution of the Laplace equation ∂2Φ ∂x2 + ∂2Φ ∂y2 = 0, (−∞ < x <∞, 0 < y < h) (1.1) subject to the boundary conditions−Φ(x, 0) = f1(x), x ∈ (a, b),∂Φ ∂y (x, 0) = 0, x ∈ R \ (a, b), (1.2) ∂Φ ∂y (x, h) = f2(x), x ∈ (a, b), Φ(x, h) = 0, x ∈ R \ (a, b), (1.3) where f1, f2 are given functions. We will solve the problem (1.1)-(1.3) by the method of Fourier transformation. For a suitable func- tion f(x), x ∈ R = (−∞,∞)(for example, f(x) ∈ L1(R)), direct and inverse Fourier transforms are defined by the formulas fˆ(ξ) = F [f ](ξ) = ∫ ∞ −∞ f(x)eixξdx, (1.4) f˘(ξ) = F−1[f ](ξ) = 1 2pi ∫ ∞ −∞ f(x)e−ixξdx. (1.5) The Fourier transforms of tempered generalized functions can be found, for example, in [5,6]. The formulated problem (1.1)-(1.3) is reduced to the following system of dual integral equations F−1 [ sinh(ξh) ξ cosh(ξh) uˆ1(ξ)− uˆ2(ξ) cosh(ξh) ] (x) = f1(x), x ∈ (a, b), F−1 [ uˆ1(ξ) cosh(ξh) + ξ sinh(ξh) cosh(ξh) uˆ2(ξ) ] (x) = f2(x), x ∈ (a, b), u1(x) = 0, x ∈ R \ (a, b), u2(x) = 0, x ∈ R \ (a, b), (1.6) where u1(x) = F −1[uˆ1(ξ)](x) = ∂Φ(x, 0) ∂y , u2(x) = F −1[uˆ2(ξ)](x) = Φ(x, h). 2 Solvability of the system of dual equations Let R be a real axis, S = S(R) and S ′ = S ′(R) be the Schwartz spaces of basic and generalized func- tions, respectively [5, 6]. Denote by F and F−1 the Fourier transform and inverse Fourier transform de- fined on S ′. It is known that these operators are au- tomorphisms on S and S ′. For a suitable ordinary function f(x) (for example, f ∈ L1(R)), the direct 1Tel: 0979791223 Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên and inverse Fourier transforms are defined by for- mulas in (1.4) and (1.5), respectively. The symbol denotes a value of the generalized func- tion f ∈ S ′ on the basic function ϕ ∈ S, besides, (f, ϕ) := . Let Hs := Hs(R)(s ∈ R) be the Sobolev space de- fined as a closure of the set C∞o (R) of infinitely dif- ferentiable functions with a compact support with respect to the norm [6] ||u||s := [ ∫ ∞ −∞ (1 + |ξ|)2s|uˆ(ξ)|2dξ ]1/2 <∞. (2.1) The space Hs is Hilbert with the following scalar product (u, v)s := ∫ ∞ −∞ (1 + |ξ|)2sû(ξ)v̂(ξ)dξ. (2.2) Let Ω = (a, b) be a certain interval in R. The sub- space of Hs(R) consisting of functions u(x) with suppu ⊂ Ω is denoted by Hso(Ω)[6], while the space of functions v(x) = pu(x), where u ∈ Hs(R) and p is the restriction operator to Ω is denoted byHs(Ω). The norm in Hs(Ω) is defined by ||v||Hs(Ω) = inf l ||lv||s, where the infimum is taken over all possible exten- sions lv ∈ Hs(R). Let X be a linear topological space. We denote the direct product of two elements X by X2. A topol- ogy in X2 is given by the usual topology of the di- rect product. We shall use bold letters for denoting vector-values and matrices Denote by u a vector of the form (u1, u2), and S2 = S×S, (S′)2 = S ′×S ′. For the vectors u ∈ (S ′)2, ϕ ∈ S2 and put = 2∑ j=1 . The Fouruier transform and inverse Fourier trans- form of a vector u ∈ (S′)2 are the vectors uˆ = F±1[u] = (F±1[u1], F±1[u2]), defined by the equa- tions [5]: = . (2.3) Let Hsj , H sj o (Ω), Hsj (Ω) be the Sobolev spaces, where j = 1, 2; Ω is a certain set of intervals in R. We put ~s = (s1, s2), H~s = Hs1 ×Hs2 , H~so(Ω) = Hs1o (Ω)×Hs2o (Ω), H~s(Ω) = Hs1(Ω)×Hs2(Ω). A scalar product and a norm in H~s and H~so(Ω) are given by the formulas (u,v)~s = 2∑ j=1 (uj , vj)sj , ||u||~s = ( 2∑ j=1 ||uj ||2sj )1/2 , where ||uj ||sj and (uj , vj)sj are given by the for- mula (2.1) and (2.2), respectively. A norm in H~s(Ω) is defined by the equation ||u||H~s(Ω) := ( 2∑ j=1 inf lj ||ljuj ||2sj )1/2 , where lj - extension operator of the uj ∈ Hsj (Ω) from Ω to R. Theorem 2.1. Let Ω ⊂ R,u = (u1, u2) ∈ H~s(Ω), f ∈ H−~s(Ω) and lf = (l1f1, l2f2) be an extension of f from Ω to R belonging H−~s(R), then the integrals [f ,u] := 2∑ j=1 ∫ ∞ −∞ l̂jfj(t)ûj(t)dt (2.4) do not depend on the choice of the extension lf . Therefore, this formula defines a linear continuous functional on H~so(Ω). Conversely, for every linear continuous functional Φ(u) on H~so(Ω) there exists an element f ∈ H−~s(Ω) such that Φ(u) = [u, f ] and ||Φ|| = ||f ||H−~s(Ω). Proof. The proof is based on the fact, that the set (C∞o (Ω)) 2 is dense in H~so(Ω), ~s = (s1, s2), and on the Riesz theorem. Consider the Fourier integral operators of the form (Au)(x) := F−1[A(t)û(t)](x), where A(t) = ||aij(t)||2×2 is a square matrix of order two , u = (u1, u2) T is a vector, transposed to the line vector (u1, u2), and û(t) := F [u] = (F [u1], F [u2]) T . We introduce following classes. Definition 2.1. Let α ∈ R. We say that the func- tion a(t) belongs to the class σα(R), if |a(t)| 6 C1(1 + |t|)α, ∀t ∈ R, and belongs to the class σα+(R), if C2(1 + |t|)α 6 a(t) 6 C1(1 + |t|)α, ∀t ∈ R, where C1 and C2 are certain posi- tive constants. Lemma 2.2. [4]. Let a(t) > 0 and such that (1 + |t|)−αa(t) is a bounded continuous function on R. Suppose moreover that, there are the positive limits of the function (1+ |t|)−αa(t) when t→ ±∞. Then a(t) ∈ σα+(R). Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên Definition 2.2. Let A(t) = ||aij(t)||2×2, t ∈ R be a square matrix of second order, where aij(t) are continuous functions on R, αi ∈ R, (i = 1, 2), ~α = (α1, α2). Denote by Σ ~α(R) the class of square ma- trices A(t) = ||aij(t)||2×2, such that aii(t) ∈ σαi(R), aij(t) ∈ σβij (R), βij 6 1 2 (αi + αj). We shall say that the matrix A(t) belongs to the class Σ~α+(R), if wTAw ≥ C1 2∑ j=1 (1 + |t|)αj |wj |2, ∀w = (w1, w2)T , where C1 is a positive constant. Finally, the matrix A(t) belongs to class Σ~αo (R), if A(t) ∈ Σ~α and Re wTAw ≥ 0, ∀w = (w1, w2)T . Lemma 2.3. Let the matrix A(t) = A+(t) belongs to the class Σ~α+(R). Then scalar product and norm in H~α/2(R) can be defined by the formulas (u,v)A+,~α/2 = ∫ ∞ −∞ F [vT ](t)A+(t)F [u](t)dt, (2.5) ||u||A+,~α/2 = (∫ ∞ −∞ F [uT ](t)A+(t)F [u](t)dt )1/2 , (2.6) respectively. Lemma 2.4. Let A(t) ∈ Σα(R). Then the Fourier integral operator Au defined by the formula F−1[A(t)uˆ(t)](x) is bounded from H~α/2(R) into H−~α/2(R). Lemma 2.5. Let Ω be a bounded subset of intervals in R. Then immersion of H~s(Ω) into H~s−~ε(Ω) is a completely continuous, where ~ε = (ε1, ε2) > 0 ⇔ εj > 0, j = 1, 2. Proof. The proof is based on the fact that the im- mersion Hsj (Ω) into Hsj−εj (Ω), εj > 0 is com- pletely continuous if Ω is bounded in R (see, [3, 6]). The system (1.6) can be rewritten in the form{ pF−1[A(ξ)û(ξ)](x) = f(x), x ∈ Ω := (a, b), p′F−1[û(ξ)](x) = 0, x ∈ Ω′ := R \ Ω, (2.7) where f̂(x) = (f̂1(x), f̂2(x)) T , û(ξ) = F [u] = (û1(ξ), û2(ξ)) T , and A(ξ) =  tanh(ξh) ξ − 1 cosh(ξh) 1 cosh(ξh) ξ tanh(ξh)  , p and p′ denote restriction operators to Ω and Ω′ respectively. Denote ~α = (α1, α2) T = (−1, 1)T . It is not difficult to show that A(ξ) ∈ Σα˜o (R). We make the assump- tions f(x) ∈ H−~α/2(Ω) and shall find the solution u(x) = F−1[uˆ](x) in the space H~α/2(R). Theorem 2.6. (Uniqueness) The system (2.7) has at most one solution u ∈ H~α/2o (a, b). Proof. Let u ∈ H~α/2o be a solution of the homoge- neous system of the system (2.7). Using the formu- las (2.4)-(2.6) and Lemma 2.4, we can show that [Au,u] = ∫ ∞ −∞ uˆT(ξ)A(ξ)uˆ(ξ)dξ = 0, from which it follows that u ≡ 0. Denote (Au)(x) = pF−1[A(ξ)uˆ(ξ)](x) (2.8) and rewrite (2.7) in the form A(u)(x) = f(x), x ∈ Ω. (2.9) Our purpose now is to establish an existence of the solution of the system (2.9) in the space H~α/2o (Ω), ~α = (−1, 1)T . Let us introduce the matrices A+(ξ) =  tanh(ξh)ξ 0 0 ξ coth(ξh)  , B(ξ) =  0 −1 cosh(ξh) 1 cosh(ξh) −ξ cosh(ξh). sinh(ξh)  . It is clear that A+(ξ) ∈ Σ~α+, B(ξ) = A(ξ)−A+(ξ) ∈ Σ−~β , ~β = (β1, β2), βj >> 1. Theorem 2.7. (Existence) . If f ∈ H−~α/2(Ω), then the system (2.9) has a unique solution u ∈ H~α/2o (Ω). Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên Proof. We represent the operator A defined by the equation (2.8) in the form A = A+ +B, where A+u = pF −1[A+uˆ], Bu = pF−1[Buˆ], uˆ = F [u]. Note that, the system of equations A+u(x) = k(x), u(x) ∈ H~α/2o (Ω) is disintegrated to two distinct equations. These equations were investigated in [4]. Due to [4], the operator A−1+ is bounded from H−~α/2(Ω) into H~α/2o (Ω) and due to of Lemma 2.5, the operator Bu defined by (2.8) is completely continuous from H~α/2o (Ω) into H−~α/2(Ω). In this case, we represent the system (2.9) in the form A+u +Bu = f . Hence, we have u +A−1+ Bu = A −1 + f . (2.10) Since the operator A−1+ B is completely continu- ous, the system (2.10) is Fredholm , and from the uniqueness of the solution it follows that this sys- tem has a unique solution. Therefore, in this case, the system (2.9) has a unique solution u ∈ H~α/2(R). The proof is complete. 3 Reduction to a system of Fredholm integral equations In this section we propose a method for solving the system of dual integral equations (1.6). We intro- duce the following definitions. Definition 3.1. Denote by Cˆmo (a, b) the class of continuous functions u(x) ∈ S′(R), such that u(x) ∈ Cm−1[a, b], u(k)(x) = 0 (k = 0, 1, ...,m − 1), x 6∈ (a, b), u(m)(x) ∈ L2(a, b). Definition 3.2. Let ρ(x) = √ (x− a)(b− x) (a < x < b). We denote by L2ρ±1(a, b) Hilbert spaces of functions with respect to the scalar product and the norm (u, v)Lρ±1 = ∫ b a ρ±1(x)u(x)v(x)dx, ||u||Lρ±1 = √ (u, u)Lρ±1 < +∞. The following lemma holds. Lemma 3.1. Let ϕ ∈ L2ρ(a, b). Denote by ϕ0 the zero-extension of the function ϕ on R. Then, ϕ0 ∈ H−1/2o (a, b). In the spaces L2ρ±1(a, b) we consider the singular integral operator SΩ[ϕ](x) = 1 pii ∫ b a ϕ(t) x− tdt, x ∈ Ω = (a, b), where the integral is taken in the sense of Cauchy principal value. Theorem 3.2. The operator SΩ is bounded in the spaces L2ρ±1(a, b). In the sequel we shall need the following inverse formula for the Cauchy integral. Theorem 3.3. Under the assumption that f(x) ∈ L2ρ(a, b) ∩H1/2(a, b) the integral equation 1 pii ∫ b a ϕ(t) x− tdt = f(x) (3.1) in the L2ρ(a, b) has the solution ϕ(x) = 1 piiρ(x) ∫ b a f(t)ρ(t) x− t dt+ C ρ(x) , (3.2) where C is an arbitrary constant. Now we turn to the system (2.7). We shall find the function u1(x) = F −1[uˆ1](x) and u2(x) = F−1[uˆ2](x) in the form u1(x) = dv1(x) dx , v1(x) ∈ H1/2o (a, b) ∩ L2ρ−1(a, b), (3.3) u2(x) = 1 2 ∫ b a v2(t)sign(x− t)dt, v2 ∈ L2ρ(a, b) ⊂ H−1/2o (a, b) (3.4) where v2 ∈ O1(a, b), i.e.∫ b a v2(x)dx = 0. (3.5) Taking the Fourier transforms of the functions u1, u2 and v1, v2 in virtue of (3.3) and (3.4) we get uˆ1(ξ) = (−iξ)vˆ1(ξ), uˆ2(ξ) = 1 (−iξ) vˆ2(ξ). (3.6) Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên Substituting equations (3.6) into (2.7), after some transforms we get the system F−1 [ signξ. tanh(ξh).vˆ1(ξ) + vˆ2(ξ) ξ cosh(ξh) ] (x) = if1(x), x ∈ (a, b), F−1 [ ξ.vˆ1(ξ) cosh(ξh) − signξ. tanh(ξh).vˆ2(ξ) ] (x) = if2(x), x ∈ (a, b). (3.7) Using following formula F−1[sign(ξ).F [v]](x) = 1 pii ∫ b a v(t)dt x− t , we can transform the system (3.7) to the following system of singular integral equations: 1 pii ∫ b a v1(t)dt x− t + ∫ b a v1(t)k11(x− t)dt + ∫ b a v2(t)k12(x− t)dt = if1(x), 1 pii ∫ b a v2(t)dt x− t + ∫ b a v1(t)k21(x− t)dt + ∫ b a v2(t)k22(x− t)dt = −if2(x), v1(x) ∈ L2ρ−1(a, b) ∩H1/2o (a, b), v2 ∈ L2ρ(a, b) ⊂ H−1/2o (a, b), (3.8) where k11(x) = k22(x) = 2i pi ∫ ∞ 0 e−2ξh 1 + e−2ξh sin ξxdξ, (3.9) k12(x) = −i pi ∫ ∞ 0 sin ξx ξ cosh(ξh) dξ, k21(x) = i pi ∫ ∞ 0 ξ sin ξx cosh(ξh) dξ. (3.10) Consider the system of singular integral equations (3.8). Due to the inverse formula (3.2) we have vm(x) + 1 piiρ(x) 2∑ k=1 ∫ b a vk(τ)Lmk(x, τ)dτ = 1 piiρ(x) ∫ b a ρ(t)(−ifm(t)) x− t dt+ Cm ρ(x) , (3.11) where Cm are arbitrary constants and Lmk(x, τ) = ∫ b a ρ(t)kmk(t− τ) x− t dt. For determining the constant Cm we use the con- dition (3.5) and we have Cm = 0(m = 1, 2). Hence, we have the system of Fredholm integral equations ϕm(x) + 2∑ k=1 ∫ b a ϕk(τ)Kmk(x, τ)dτ = hm(x), a < x < b, (m = 1, 2), (3.12) where ϕm(x) = √ ρ(x)vm(x), Kmk(x, τ) = 1 pii Lmk(x, τ)√ ρ(x) √ ρ(τ) , hm(x) = 1 pii √ ρ(x) ∫ b a ρ(t)(−ifm(t)) x− t dt. Thus, we have proved the following theorem. Theorem 3.4. The system of dual integral equa- tions (1.6) with respect to u(x) = F−1[û](x) ∈ H~α/2o (a, b), ~α = (−1, 1)T , is equivalent to the system of Fredholm integral equations (3.12) with respect to ϕm(x) = √ ρ(x)vm(x) ∈ L2(a, b)(m = 1, 2). If fm(x) ∈ L2ρ(a, b)(m = 1, 2), then the system of sin- gular integral equations (3.8) has a unique solution v ∈ (L2ρ(a, b))2. In this case, the solution with re- spect to u(x) = F−1[û](x) ∈ H~α/2o (a, b) of the sys- tem of dual equations (1.6) is given by the formula (3.3) and (3.4). TÝnh gi¶i ®­îc cña mét hÖ ph­¬ng tr×nh cÆp tÝch ph©n trong bµi to¸n biªn hçn hîp cña ph­¬ng tr×nh Laplace ®èi víi miÒn h×nh d¶i Tãm t¾t. Trong bµi b¸o nµy chóng t«i xÐt tÝnh gi¶i ®­îc cña mét hÖ ph­¬ng tr×nh cÆp tÝch ph©n Fourier gÆp trong bµi to¸n biªn hçn hîp cña ph­¬ng tr×nh Laplace ®èi víi miÒn h×nh d¶i. §· chøng minh c¸c ®Þnh lý tån t¹i vµ duy nhÊt nghiÖm cña hÖ trong kh«ng gian Sobolev cña c¸c hµm suy réng. Tr×nh bµy ph­¬ng ph¸p ®­a hÖ ph­¬ng tr×nh cÆp nãi trªn vÒ hÖ ph­¬ng tr×nh tÝch ph©n Fredholm lo¹i hai trong kh«ng gian c¸c hµm b×nh ph­¬ng kh¶ tÝch. Tõ kho¸: BiÕn ®æi Fourier, hÖ ph­¬ng tr×nh cÆp tÝch ph©n, ph­¬ng tr×nh Laplace, bµi to¸n biªn hçn hîp. Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên References [1] Ufliand Ia.S., Method of Dual Equations in Problems of Mathematical Physics - Leningrad, Nauka, 1977 (in Russian). [2] Mandal B.N. , Advances in Dual Integral Equa- tions - Chapman & Hall / CRC Press, Boca Raton, 1998. [3] Nguyen Van Ngok and Popov G.Ya, Dual in- tegral equations associated with Fourier trans- forms Ukrainskii matematicheskii Zhurnal.- 1986,-38, N2.- p. 188-195. (in Russian). 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