Southeast Asian Bulletin of Mathematics (2002) 26: 235–244 Southeast Asian Bulletin of Mathematics : Springer-Verlag 2002 On Solvability of a Boundary Value Problem for Singular Integral Equations Nguyen Van Mau and Nguyen Tan Hoa Hanoi University of Science, Vietnam, 334 Nguyen Trai, Dong Da, Hanoi, Vietnam AMS Subject Classification: 47G05, 45GO5, 45E05 Abstract This paper deals with the solvability of boundary value problems for singular integral equations of the form (i)–(ii) By an algebraic method we reduce the problem (i)–(ii) to a system of linear algebraic equations which gives all solutions in a closed form Keywords: initial and co-initial operators, singular integral equations, boundary value problem Introduction The theory of general boundary value problem induced by right invertible operators were investigated by Przeworska-Rolewicz and has been developed by many other mathematicians (c.f [2], [4]) In this paper, we give an application of this theory to solve the following boundary value problem: ẵK n ỵ K n1 Kc ịjtị ẳ f tị; Fj K j jịtị ẳ jj tị; G0 f ÞðtÞ ¼ 0; j ¼ 0; ; n À 1; ðGj RjÀ1 R0 Þ f tị ẳ 0; jj tị A Ker K; iị if k > 0; j ¼ 1; ; n À if k < 0; ðiiÞ where Kc is an operator of multiplication by the function cðtÞ; Fj ; Gj ð j ¼ 0; ; n À 1Þ are initial and co-initial operators, respectively, and btị Kjịtị ẳ atịjtị ỵ pi jtị dt; t G Àt k ¼ Ind K: Preliminaries We recall some notations and results which are used in the sequel (see [2], [4]) 236 Ng.V Mau and Ng.T Hoa Let G be a simple regular closed arc in complex plane and let X ẳ H m Gị < m < 1ị Denote by Dỵ the domain bounded by G (assume that A Dỵ ) and by D -its complement including the point at infinity The set of all linear operators with domains and ranges contained in X will be denoted by LðX Þ Write L ðX Þ :¼ fA A LðX Þ : dom A ¼ X g Let RðX Þ be the set of all right invertible operators belonging to LðX Þ For D A RðX Þ, we denote by RD the set of all its right inverses An operator F A LðX Þ is said to be an initial operator for an operator D A RðX Þ corresponding to a right inverse R of D if F ẳ F; F dom Dị ẳ Ker D; FR ¼ 0: Denote by FD the set of all initial operators for D A RðX Þ The following fact is well-known: F A LðX Þ is an initial operator for D A RðX Þ corresponding to R A RD if and only if F ¼ I À RD on dom D Let LðX Þ be the set of all left invertible operators belonging to L ðX Þ For V A LðX Þ, we denote by LV the set of all its left inverser If V A LðX Þ and L A LV then the operator G :¼ I À VL is called the co-initial operators for V corresponding to L A LV Denote by GV the set of all co-initial operators for V A LðX Þ Lemma (see [2]) Let A; B A LðX Þ, Im A H dom B, Im B H dom A Then equation I ABịx ẳ y has solutions if and only if I BAịu ẳ By does and there is one-to-one correspondence between the two sets of solutions, given by u ¼ Bx $ x ¼ y ỵ Au: The Results Let Kjịtị :ẳ atịjtị ỵ bðtÞ pi ð jðtÞ dt; t G Àt where aðtÞ; bðtÞ A X , a ðtÞ À b tị ẳ 1, 0 k ẳ Ind K Denote btịZtị jtị dt; R0 jịtị :ẳ atịjtị pi Ztịt tị G where Ztị ẳ e Gtị tk=2 ; ln tk atịbtị atịỵbtị dt: Gtị ẳ tt 2pi G On Solvability of a Boundary Value Problem for Singular Integral Equations 237 It is known that (see [2], [4]) i) If k > then K is right invertible and RK ¼ fR ¼ R0 þ ðI À R0 KÞT : T A L ðX Þg: Let F0 ; ; FnÀ1 be given initial operators for K corresponding to R0 ; ; RnÀ1 A RK , where Rj ẳ R0 ỵ I R0 KịTj j ẳ 1; ; n À 1Þ; T1 ; ; TnÀ1 A L ðX Þ: ð1Þ ii) If k < then K is left inverbible and LK ẳ fR ẳ R0 ỵ TI KR0 ị : T A LX ị; dom T ẳ ðI À KR0 ÞX g: In this case, let G0 ; ; GnÀ1 be given co-initial operators for K corresponding to R0 ; ; Rn1 A LK , where Rj ẳ R0 ỵ Tj I KR0 ị j ẳ 1; ; n À 1Þ; T1 ; ; TnÀ1 A LðX Þ: ð2Þ Lemma If k > 0, then F0 jịtị ẳ k1 X uk jịck tị on X ; kẳ0 where ck tị ẳ btịZtịt k ðk ¼ 0; ; k À 1ị and uk jị k ẳ 0; ; k À 1Þ are linear functionals which are defined by uk jị ẳ 2pi t k1k jtị À À G ðtÞ pi G e ð ! jðt1 Þ dt1 dt; G t1 À t ð where GÀ ðtÞ is a boundary value of the function GðzÞ in D Proof We have F0 jịtị ẳ ẵI R0 Kịjtị atịbtị jtị dt ẳ jtị a ðtÞjðtÞ À pi GtÀt ! ð ð bðtÞZðtÞ bðtÞ jt1 ị atịjtị ỵ dt1 dt ỵ pi pi G t1 À t G ZðtÞðt À tÞ ð aðtÞbðtÞ jðtÞ dt ẳ jtị a tịjtị pi Gtt ! jỵ tị dt j tị dt ỵ btịZtị ; pi G X ỵ tịt À tÞ pi G X À ðtÞðt À tÞ 3ị 238 Ng.V Mau and Ng.T Hoa ỵ where jỵ tị ẳ 12 ẵI ỵ Sịjtị, j tị ẳ 12 ẵI ỵ Sịjtị, X ỵ tị ẳ e G tị , X tị ẳ tk e G tị (Gỵ tị; G tị are boundary values of the function Gzị in Dỵ ; D , respectively) On the other hand pi jỵ tị dt j tị dt ỵ pi G X tịt tị G X tịt tị jỵ tị j tịt k dt ẳ ỵ X tị pi G e G ðtÞ ðt À tÞ ð kÀ1 k X jỵ tị t k j tị dt t t k1k j tị ẳ ỵ dt À X ðtÞ pi G e G ðtÞ ðt À tị kẳ0 pi G e G tị ẳ ẳ k1 k jỵ tị j tị X t ỵ ỵ X tị X tị kẳ0 pi btị atị jtị ỵ Ztị Ztịpi t k1k j tị dt e G ðtÞ G ð kÀ1 k X jðtÞ t dt À pi GtÀt k¼0 ð ð t kÀ1Àk jÀ ðtÞ dt: À e G ðtÞ G Hence ! ð ð kÀ1 X bðtÞZðtÞt k t kÀ1Àk jðt1 Þ F0 jịtị ẳ dt1 dt: jtị G tị pi G t1 À t 2pi G e k¼0 The lemma is proved By similar arguments, we obtain the following result: Lemma If k < 0, then ðG0 jÞðtÞ ¼ jkjÀ1 X vk ðjÞck ðtÞ on X ; k¼0 where ck tị ẳ btịt k k ẳ 0; ; jkj À 1Þ and vk ðjÞ ðk ¼ 0; jkj À 1Þ are linear functionals defined by vk jị ẳ 2pi G tjkjÀ1Àk e G À ðtÞ jðtÞ À ZðtÞ pi ð ! jðt1 Þ dt1 dt G Zðt1 Þðt1 À tÞ where GÀ ðtÞ is a boundary value of the function GðzÞ in DÀ In the sequel, for every function ctị A X , we write Kc jịtị ẳ cðtÞjðtÞ: ð4Þ On Solvability of a Boundary Value Problem for Singular Integral Equations 239 Consider singular integral equation of the form ẵK n ỵ K n1 Kc ịjtị ẳ f ðtÞ; ð5Þ with mixed boundary conditions iÞ ðFj K j jịtị ẳ jj tị; G0 f ịtị ẳ 0; iiị jj tị A Ker K; j ẳ 0; ; n À ðGj R0 Rj1 f ịtị ẳ 0; if k > 0; j ¼ 1; ; n À if k < 0; ð6Þ where f ðtÞ; cðtÞ A X ; Fj ; Gj ð j ¼ 0; ; n À 1Þ are defined by (1) and (2), respectively, < n A N Theorem Suppose that ỵ ctịatị G ctịbtị 0 for all t A G Then every solution of the problem (5)–(6) can be found in a closed form Proof Let k > 0, we have K A RðX Þ Hence, the equation (5) is equivalent to the equation jtị ẳ R0 Rn1 K n1 Kc jịtị ỵ R0 Rn1 f ịtị ỵ R0 Rn2 zn1 ịtị ỵ ỵ R0 z1 ịtị ỵ z0 tị; where z0 tị; ; znÀ1 ðtÞ A Ker K are arbitrary Thus, the problem (5)(6) is equivalent to the equation jtị ẳ R0 Rn1 K n1 Kc jịtị ỵ R0 Rn1 f ịtị ỵ R0 Rn2 jn1 ịtị ỵ ỵ R0 j1 ịtị ỵ j0 tị; i.e ẵI ỵ R0 Rn1 K n1 Kc ịjtị ẳ f1 tị; 7ị where f1 tị ẳ R0 Rn1 f ịtị ỵ R0 Rn2 jn1 ịtị ỵ ỵ R0 j1 ịtị ỵ j0 ðtÞ: By the Taylor-Gontcharov formula for right invertible operators (see [4]), (7) is iquivalent to the equation 20 4@I þ R0 I À F1 À nÀ1 X k¼2 ! R1 RkÀ1 Fk K kÀ1 Kc Aj5tị ẳ f1 tị: 8ị By lemma 1, in order to solve the equation (8) it is enough to solve the equation 240 Ng.V Mau and Ng.T Hoa " I ỵ Kc R0 F1 Kc R0 n1 X ! # R1 RkÀ1 Fk K k1 Kc R0 c tị ẳ gtị; kẳ2 where gtị ¼ " nÀ1 X I À F1 À R1 RkÀ1 Fk K k¼2 kÀ1 ! # Kc f1 ðtÞ: Rewrite this equation in the form " I þ Kc R0 À F1 Kc R0 À nÀ1 X # ! R1 RkÀ1 Fk K kÀ1 Kc R0 KZ f tị ẳ gtị; kẳ2 9ị where ftị ẳ ctị=Ztị From lemma 2, we have Fk K k1 Kc R0 KZ fịtị ẳ k1 X ujk fịcj tị; k ẳ 1; ; n 1; jẳ0 where ujk fị ẳ uj ẵI Tk KÞK kÀ1 Kc R0 KZ f; uj ðjÞ; cj ðtÞ ð j ¼ 0; ; k À 1Þ are defined by (3) Hence, (9) is of the form ðMfÞðtÞ À nÀ1 X kÀ1 X ujk ðfÞcjk ðtÞ ¼ gðtÞ; ð10Þ k¼1 j¼0 where cj1 ðtÞ :¼ cj tị, cjk tị ẳ R1 Rk1 cj ÞðtÞ ðk ¼ 2; ; n À 1ị, j ẳ 0; ; k and ctịbtịZtị Mfịtị :ẳ ẵ1 ỵ ctịatịZtịftị pi ð fðtÞ dt: GtÀt ð11Þ Write this equation in the form Mfịtị q X u~k fịc~k tị ẳ gtị; 12ị kẳ1 where q ẳ kn 1ị, f~ u1 ðfÞ; ; u~q ðfÞg is a permutation of fujk fị; j ẳ 0; k À 1; k ¼ 1; ; n À 1g and fc~1 ðtÞ; ; c~q ðtÞg is obtained by this permutation from the set of functions fcjk tị; j ẳ 0; ; k À 1; k ¼ 1; ; n À 1g On Solvability of a Boundary Value Problem for Singular Integral Equations 241 Denote where cðtÞbðtÞZ1 ðtÞ Ð ftị ẵ1 ỵ atịctịftị ỵ G Z1 tịttị dt pi Nfịtị :ẳ ; ỵ atịctị c ðtÞb ðtÞ ZðtÞ Ã 1=2 ÂÀ Á2 Z1 tị ẳ Ztịe G1 tị tk1 =2 ỵ aðtÞcðtÞ À b ðtÞc ðtÞ ; Á À k ln t Gtị G1 tị ẳ dt; 2pi G tt Gtị ẳ ỵ atịctị ỵ btịctị ; ỵ atịctị btịctị k1 ẳ Ind Gtị: If k1 ẳ 0, then M is invertible and M À1 ¼ N Hence, the equation (12) is equivalent to the equation fðtÞ À q X u~k ðfÞðN c~k ÞðtÞ ¼ ðNgÞðtÞ: ð13Þ k¼1 Without loss of generality, we can assume that fN c~k ịtịgkẳ1; q is a linearly independent system Then every solution of (13) can be found in a closed form by means of the system of linear algebraic equations u~j ðfÞ À q X ajk u~k fị ẳ u~k Ngị; j ẳ 0; ; q; kẳ1 where ajk ẳ u~j N c~k ị; k; j ¼ 1; ; q If k1 > 0, then M is right invertible and N is a right inverse of M Hence, the equation (12) is equivalent to the equation fðtÞ À q X u~k fịN c~k ịtị ẳ Ngịtị ỵ ytị; kẳ1 where ytị A Ker M is arbitrary We now can solve this equation by the same method as for the equation (13), i.e every its solution can be found in a closed form If k1 < 0, then M is left invertible and N is a left inverse of M Hence, the equation (12) is equivalent to the system q X > > > ftị u~k fịN c~k ịtị ẳ Ngịtị; > > > kẳ1 < q P > > u~k fịc~k tị gtị ỵ > > > kẳ1 > : t nÀ1 dt ¼ 0; n ¼ 1; ; jk1 j; Z1 ðtÞ G 242 Ng.V Mau and Ng.T Hoa i.e q X > > > ftị u~k fịN c~k ịtị ẳ Ngịtị; > > < k¼1 q > X > > > bnk u~k fị ẳ fn ; > : 14ị n ¼ 1; ; jk1 j; k¼1 where fn ẳ gtịt n1 dt ; Z1 tị G bnk ẳ ~ ck tịt n1 dt: Z1 ðtÞ G Without loss of generality, we can assume that fN c~k ịtịgkẳ1; q is a linearly independent system Every solution of (14) can be found in a closed form by means of the system of linear algebraic equations q X > > > ~ u ajk u~k ðfÞ ¼ u~j ðNgÞ; j ¼ 1; ; q; fị > j > < kẳ1 q > X > > > bnk u~k fị ẳ fn ; > : n ¼ 1; ; jk1 j; kẳ1 where akj ẳ u~k N c~j ị, k; j ¼ 1; ; q Thus, every solution of the equation (12) can be found in a closed form Due to the result of Lemma 1, every solution of the problem (5)–(6) is defined by the formula jtị ẳ R0 KZ fịtị ỵ f1 tị; where ftị is a solution of the equation (12), i.e every solution of the problem (5)– (6) can be found in a closed form Let k < 0, we have K A LðX Þ The Taylor-Gontcharov formula for left invertible operators (see [4]) and (6) together imply n ½ðI À K RnÀ1 R0 ị f tị ẳ G0 f ịtị ỵ " n1 X k ! # K Gk Rk1 R0 f tị ẳ 0; kẳ1 i.e f tị ẳ ẵK n Rn1 R0 Þ f ðtÞ: Hence, the problem (5)–(6) is equivalent to the equation ẵK n ỵ K n1 Kc ịjtị ¼ ðK n RnÀ1 R0 f ÞðtÞ; On Solvability of a Boundary Value Problem for Singular Integral Equations 243 i.e ẵK ỵ Kc ịjtị ẳ KRn1 R0 f ÞðtÞ: ð15Þ If jðtÞ is a solution of (15) then G0 Kc jịtị ẳ G0 KRn1 R0 f ịtị G0 Kjịtị ẳ 0; Thus, (15) is equivalent to the system & ½ðI ỵ R0 Kc ịjtị ẳ f2 tị; G0 Kc jịtị ¼ 0; ð16Þ where f2 ðtÞ ¼ ðRnÀ1 R0 f ÞðtÞ From Lemma 3, ðG0 Kc jÞðtÞ ¼ if and only if vk ðKc jÞ ¼ 0; k ¼ 0; ; jkj À 1; where vk jị k ẳ 0; ; jkj À 1Þ are defined by (4) Hence, the system (16) is equivalent to the system & ẵI ỵ R0 Kc ịjtị ẳ f2 tị; vk Kc jị ẳ 0; k ¼ 0; ; jkj À 1: Consider the system of equations & ẵI ỵ Kc R0 ịctị ẳ Kc f2 ịtị; vk cị ẳ 0; k ¼ 0; ; jkj À 1: ð17Þ It is easy to check that the system (16) has solutions if and only if the system (17) does Moreover, if jðtÞ is a solution of (16) then cðtÞ ¼ cðtÞjðtÞ is a solution of (17) Conversely, if cðtÞ is a solution of (17) then jtị ẳ btịctị ỵ btịZtị é ctị G Ztịttị dt pi ỵ f2 tị ỵ ctịatị ỵ ctịbtị 18ị is a solution of (16) Hence, in order to solve the system (16), it is enough to solve the system (17) Rewrite (17) in the form & Mfịtị ẳ Kc f2 ịtị; v~k fị ¼ 0; k ¼ 0; ; jkj 1: where ftị ẳ ctị=Ztị, v~k fị ẳ vk ðKZ fÞ and M is defined by (11) ð19Þ 244 Ng.V Mau and Ng.T Hoa By the same method as for the equation (12), every solution of this system can be found in a closed form So every solution of the problem (5)(6) is defined by the formula jtị ẳ btịZtịftị ỵ btịZtị é ftị G ttị dt pi þ cðtÞaðtÞ þ cðtÞbðtÞ þ f2 ðtÞ ; where fðtÞ is a solution of the system (19), i.e every solution of the problem (5)– (6) can be found in a closed form The theorem is proved References Gakhov, F.D.: Boundary value problems, Oxford, 1966 (3rd Russian complemented and corrected edition, Moscow, 1977) Mau, Ng.V.: Boundary value problems and controllability of linear system with right invertible operators, Warszawa, 1992 Mau, Ng.V.: Generalized algebraic elements and linear singular integral equations with transformed arguments, WPW, Warszawa, 1989 Przeworska-Rolewicz, D.: Algebraic Analysis, PWN and Reidel, Warszawa-Dordrecht, 1988 Przeworska-Rolewicz, D.: Equations with transformed argument, An algebraic approach, Amsterdam-Warszawa, 1973 ... sequel, for every function cðtÞ A X , we write Kc jịtị ẳ ctịjtị: 4ị On Solvability of a Boundary Value Problem for Singular Integral Equations 239 Consider singular integral equation of the form... WPW, Warszawa, 1989 Przeworska-Rolewicz, D.: Algebraic Analysis, PWN and Reidel, Warszawa-Dordrecht, 1988 Przeworska-Rolewicz, D.: Equations with transformed argument, An algebraic approach, Amsterdam-Warszawa,... problems and controllability of linear system with right invertible operators, Warszawa, 1992 Mau, Ng.V.: Generalized algebraic elements and linear singular integral equations with transformed arguments,