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This article was downloaded by:[Tuan, Nguyen Minh] On: 23 January 2008 Access Details: [subscription number 789785478] Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK Complex Variables and Elliptic Equations An International Journal Publication details, including instructions for authors and subscription information: http://www.informaworld.com/smpp/title~content=t713455999 On a class of singular integral equations with the linear fractional Carleman shift and the degenerate kernel Le Huy Chuan a; Nguyen Van Mau b; Nguyen Minh Tuan b a Department of Applied Physics, Graduate School of Engineering, Osaka University, Japan b Faculty of Mathematics Mechanics and Informatics, Department of Analysis, University of Hanoi, Hanoi, Vietnam Online Publication Date: 01 February 2008 To cite this Article: Chuan, Le Huy, Van Mau, Nguyen and Tuan, Nguyen Minh (2008) 'On a class of singular integral equations with the linear fractional Carleman shift and the degenerate kernel', Complex Variables and Elliptic Equations, 53:2, 117 - 137 To link to this article: DOI: 10.1080/17476930701619782 URL: http://dx.doi.org/10.1080/17476930701619782 PLEASE SCROLL DOWN FOR ARTICLE Full terms and conditions of use: http://www.informaworld.com/terms-and-conditions-of-access.pdf This article maybe used for research, teaching and private study purposes Any substantial or systematic reproduction, re-distribution, re-selling, loan or sub-licensing, systematic supply or distribution in any form to anyone is expressly forbidden The publisher does not give any warranty express or implied or make any representation that the contents will be complete or accurate or up to date The accuracy of any instructions, formulae and drug doses should be independently verified with primary sources The publisher shall not be liable for any loss, actions, claims, proceedings, demand or costs or damages whatsoever or howsoever caused arising directly or indirectly in connection with or arising out of the use of this material Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Complex Variables and Elliptic Equations Vol 53, No 2, February 2008, 117–137 On a class of singular integral equations with the linear fractional Carleman shift and the degenerate kernel LE HUY CHUANy, NGUYEN VAN MAUz and NGUYEN MINH TUAN*z yDepartment of Applied Physics, Graduate School of Engineering, Osaka University, Japan zFaculty of Mathematics Mechanics and Informatics, Department of Analysis, University of Hanoi, 334, Nguyen Trai Str., Hanoi, Vietnam Communicated by R P Gilbert (Received March 2007; in final form June 2007) This article deals with the solvability, the explicit solutions of a class of singular integral equations with a linear-fractional Carleman shift and the degenerate kernel on the unit circle by means of the Riemann boundary value problem and of a system of linear algebraic equations All cases about index of the coefficients in the equations are considered in detail Keywords: Integral operators; Singular Integral equations; Riemann boundary value problems AMS Subject Classifications: 47G05; 45G05; 45E05 Introduction Singular integral equations with a shift (SIES) have been studied for a long time (see [1,2] and references therein) Many papers devoted to singular integral operators with a shift (SIOS) are given to the construction of the Fredholm theory Once M G Krein called the Fredholm theory of linear operators a rough theory, and the theory describing its defect subspaces a delicate theory [3] However, the Fredholm theory of these operators brings about only one thing, the defect of dimensions of kernel of the operator and its dual operator In other words, it is only the defect of the numbers of linear independent solutions of the homogeneous equations reduced by the operator and the corresponding dual operator So the question of solving (and even of estimating the numbers of solutions) of the corresponding equations actually remains open [4] There are only a few special types of SIES for which it is possible to answer this question to some extent [2,5] Among the SIES of this type not reducible to two-term boundary value problems, the most general and important is the class of singular integral equations with a *Corresponding author Email: nguyentuan@vnu.edu.vn Complex Variables and Elliptic Equations ISSN 1747-6933 print/ISSN 1747-6941 online ß 2008 Taylor & Francis http://www.tandf.co.uk/journals DOI: 10.1080/17476930701619782 Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 118 L H Chuan et al linear-fractional Carleman shift In our view, the singular integral equations with a linear-fractional Carleman shift in the unit circle, in addition, deserve the interest Factorization is the main method used by some authors to investigate Fredholm and solvability theory for SIES (see [3,4] and references therein) In [6], two of us gave a general formula of linear-fractional Carleman shifts on the unit circle and solved by means of Riemann boundary value problem for a class of singular integral equations with a linear-fractional Carleman shift on the unit circle In this article, we study the solvability for a class of singular integral equations with a linear-fractional Carleman shift and with the degenerate kernels on the unit circle In genaral, one knows that the singular integral operator of Cauchy’s type (denoted by S) not commute with Carleman shift opetaor (denoted by W), but the difference beetwen them WS À SW is a compact operator [2]) In section 2, we obtain some identities relating to those operators The scheme of our investigation is divided into two parts: first, we move the degenerate kernels to the right-side hand of the equation Based on the identity W n ¼ I, we construct the orthogonal projectors and reduce the equation to a system of singular integral equations without shift and solve this system by means of Riemann boundary value problem Second, we reconstruct the solution of the orginal equation from the solutions of system of equations that can be solved, but its solution depends on some unknown parameters As indicated below, the equations of the type (1.1) can be solved by means of Riemann boundary value problem and by of a system of linear algebraic equations Let À ¼ {t C, jtj ¼ 1} be the unit circle on the complex plane C and let X :ẳ H(), 0551 Let ! tị ẳ t þ , t þ ð À ¼ 1; 6¼ 0Þ be a Carleman linear-fractional function shift of order of n on À, i.e., ! : À ! À; > > > < !n tị ẳ t; for every t !k tị :ẳ !!k1 tịị; !0 tị tị; > for some t ; k ẳ 1, 2, , n À 1, ! k ðtÞ 6 t; > > : ! positive orientation of À The such functions ! (t) might be of the form tÀ > , if n ¼ 2; < " ! tị ẳ t > : ei t À , if n > 2, " À1 t " cos2 ðk=nÞ, for some k {1, 2, , n À 1}, where j j51; cos ¼ À 2ð1 À Þ (k, n) ¼ [6] In this article, we study the solvability of the equations of the form Z nÀ1 bðtÞ X ’ ð Þ nk d " a tị tị ỵ n kẳ0 i !k tị Z m X ỵ aj tịbj ị ịd ẳ ftị, i À j¼1 where ‘ n À 1, "1 ¼ e2i=nị, " ẳ "1 : 1:1ị Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations 119 Some identities of singular integral operator of Cauchy’s type and linear-fractional shift operator on unit circle Consider the following operators in X: Z ’ ð Þ d , ðS’Þ ðtÞ ¼ i À À t ðW’Þ ðtÞ ¼ ’ð! tịị, n 1X Pk ẳ "n1j W jỵ1 , n jẳ1 k 2:1ị k ẳ 1, 2, : In the sequel, we shall need the following identities [7,8]: n P > k > W ¼ "kj Pj ; k ¼ 1; 2, , n, > > > j¼1 > < Pk Pj ¼ kj Pj ; k; j ¼ 1; 2, , n, > n > X > > > Pj ẳ I; > : 2:2ị jẳ1 where kj is the Kronecker symbol For every a X we write (K a ’) (t) ¼ a (t) ’ (t) Lemma 2.1 Let a X be fixed Then for every (k, j ), k, j {1, 2, , n} there exists an element b X such that Kb Xj & Xk and Pk Ka Pj ¼ Kb Pj , where xk :¼ P( X ) The Function b(t) will be denoted by akj (t) and determined as follows akj tị ẳ n 1X " jk a!ỵ1 tịị: n ẳ1 ỵ1 2:3ị By using (2.2) we get Proof Pk Ka Pj ¼ n n 1X 1X ỵ1 "n1 W K P ẳ "n1 a!ỵ1 tịịW ỵ1 Pj a j n ¼1 k n ¼1 k n n n X 1X 1X ỵ1 "ỵ1ị a! tịị " P P ẳ "ỵ1ị a!ỵ1 tịị"ỵ1 Pj ỵ1 j j k n ¼1 k n ¼1 ¼1 ! n n 1X 1X jÀk Àk j ¼ " " a!ỵ1 tịịPj ẳ " a!ỵ1 tịị Pj ẳ akj tịPj ; n ẳ1 ỵ1 ỵ1 n ẳ1 ỵ1 ẳ where ak j tị ẳ n 1X " jk a!ỵ1 tịị: n ẳ1 ỵ1 Putting akj(t) :ẳ b(t), we obtain b X and Pk Ka Pj ¼ Kb Pj Lemma 2.2 Let a X be fixed Then for any k, j {1, 2, , n}, we have Pk Kak j ¼ Kak j Pj , ð2:4Þ g Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 120 L H Chuan et al where akj(t) are determined by (2.4) Proof For any ’ X we have n 1X "n1 Wỵ1 akj tịtị n ẳ1 k ! ! n n 1X 1X ỵ1ị ỵ1 jk " W " a!ỵ1 tịị tị ẳ n ẳ1 k n ẳ1 ỵ1 " # n n 1X 1X ỵ1ị jk jk "ỵ1 "ỵ1 a!ỵ1ỵỵ1 tịị "kj W ỵ1 tị ẳ ỵ1 "k n ẳ1 n ẳ1 Pk Kakj ịtị ẳ Pk akj tịtị ẳ n ỵ1 1X k akj tị"kj tị ỵ1 "ỵ1 W n ẳ1 " # n 1X n1 ỵ1 " W tị ẳ akj tịPj ịtị ẳ Kakj Pj ịtị: ¼ akj ðtÞ n ¼1 j ¼ g Thus Pk Kakj Kakj Pj : Lemma 2.3 Let ’ X Then for every z C we have (1) SW ịzị ẳ W k Sịzị W k1 Sị = ị; k ẳ 1; 2, , W ẳ I: (2) Pk Sịzị ẳ SPk ịzị ỵ 1="k 1ịSPk ị = ị; k ẳ 1; 2, , n À 1, k where and are !zị ẳ z ỵ ị= z ỵ ịị the coefficients of the linear-fractional function Proof (1) By induction on k For k ẳ we have Z ! ịị d : SWịzị ¼ i À À z Put ¼ !À1 xị ẳ x ; x ỵ then d ẳ dx: x ỵ ị2 Therefore, Z xị1= x ị2 ị dx x = x ỵ ị z Z xị dx ẳ i x þ Þðx À À zðÀ x þ ÞÞ Z 1 ẳ xịdx i x z ỵ = z ỵ ị x = ị Z Z xị xị dx dx ẳ WSịzị Sị : ẳ i x !zị i x = ị SWịzị ẳ i 121 Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations So, we obtain ; ðSW’ÞðzÞ ¼ ðWS’ÞðzÞ À ðS’Þ for any z C: ð2:5Þ Suppose that (i) is true for k ¼ m For k ẳ m ỵ we find SW mỵ1 ịzị ẳ ẵSWW m ịzị m m ẳ W ẵSW ịzị SW ị ẳ ẵWSW ịzị ẵSW ị ! ! À ðW m S’Þ À ðW mÀ1 S’Þ ¼ W ðW m S’ÞðzÞ À ðW mÀ1 S’Þ ! À ðW m Sị ỵ W m1 Sị : ẳ W mỵ1 Sịzị À W ðW mÀ1 S’Þ m m Hence W ẵW m1 Sị = ị ẳ W m1 Sị = ị; provided W m1 Sị = ị is a constant Therefore mỵ1 mỵ1 m : SW ịzị ẳ W Sịzị W Sị The first part of the lemma is proved (2) Rewrite the equality in (1) in the form W k Sịzị ẳ SW k ịzị ỵ W k1 Sị : We find ! n n 1X 1X ni1 iỵ1 ni1 iỵ1 i " W Sịzị ẳ " SW ịzị ỵ W Sị Pk Sịzị ẳ n iẳ1 k n i¼1 k " ! # " ! # n n 1X 1X ni1 iỵ1 ni i "k W "k W S ẳ S zị ỵ n iẳ1 "k n iẳ1 ẳ SPk ịzị ỵ Pk Sị : "k Substituting z ẳ = ị in formula (2.6), we get Pk Sị ẳ SPk ị : "k ð2:6Þ ð2:7Þ If k {1, 2, , n À 1} then "k 6¼ In this case, substituting (2.7) in (2.6) we receive SPk ị : g Pk Sịzị ẳ SPk ịzị ỵ "k À Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 122 L H Chuan et al Comment From the identity (2.5), one can say that the operators S and W not commute to each other, but the difference of WS and SW at the a function ’(t) always equals ðS’Þð = Þ: Reducing equation (1.1) to a system of singular integral equations We now represent the equation (1.1) in the following form a tị tị ỵ btịP Sịtị ỵ Z m X aj tịbj ị ịd ẳ f tị; i jẳ1 3:1ị where a, b, f, a1, , am, b1, , bm X are given, and S, P‘ (1 ‘ n À 1) are the operators defined by (2.1) Suppose that a(t) is a non-vanishing function on À Denote by Mbj ; j ¼ 1, , m, the linear functionals on X defined as follows Mbj ị ẳ i Z bj ị ịd ; for any X: À Put j ¼Mbj(’), j=1, , m We reduce equation (3.1) to the following problem: Find solutions of equation atịtị ỵ btịP Sịtị ẳ ftị m X j aj tị 3:2ị jẳ1 depended on the parameters 1, , m, and fulfilled the following conditions Mbj ị ẳ j ; j ¼ 1, : ð3:3Þ Lemma 3.1 Let ’ X Then ’ is a solution of (3.2) if and only if {’k ¼ Pk’, k ¼ 1, 2, , n} is a solution of the following system a tị k tị ỵ bk tịS ịtị bk tị S ị ỵ ẳ fk tị, " k ẳ 1, 2, , n, 3:4ị where a tị ẳ n Y a ! j ỵ1 tịị; jẳ1 bk tị ẳ n n Y 1X jỵ1 "k b! tịị a! ỵ1 tịị; n jẳ1 jỵ1 ẳ1 3:5ị 6ẳj f k tị " # n m n Y X 1X n1j jỵ1 jỵ1 ẳ "k f ! tịị a ! tịị a! ỵ1 tịị: n jẳ1 ẳ1 ẳ1 6ẳj 123 Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations Proof Suppose that ’ X is a solution of (3.2) We then have n Y að! ỵ1 tịịtị ỵ btị ẳ1 " ẳ ftị m X # j aj tị jẳ1 n Y a!ỵ1 tịịP Sịtị ẳ1 6ẳn1 n Y a!ỵ1 tịị: ẳ1 6ẳn1 Applying the projections Pk, k ¼ 1, 2, , n to both sides of above equation and using the Lemmas 2.1 and 2.2, we obtain n n X Y 61 "k b! jỵ1 tịị a! ỵjỵ2 tịị5P Sịtị a tịPk ịtị ỵ n jẳ1 jỵ1 ẳ1 6ẳn1 " # n m n Y X 1X nj1 jỵ1 jỵ1 ẳ "k f ! tịị a ! tịị a! ỵjỵ2 tịị: n jẳ1 ẳ1 ẳ1 3:6ị 6ẳn1 It is easy to see that n Y a! ỵjỵ2 tịị ẳ1 6ẳn1 n Y a!ỵ1 ðtÞÞ for any j f1; 2, , ng: ¼1 6¼j Hence, (3.6) is equivalent to the following system a tịPk ịtị ỵ bk tịP Sịtị ẳ f k tị; k ẳ 1; 2, , n: ð3:7Þ Using Lemma 2.3, we rewrite the system (3.7) in the form bà ðtÞ aà ðtÞðPk ịtị ỵ bk tịSP ịtị ỵ k SP ị ẳ f k tị, k ẳ 1, 2, , n: "‘ À Thus (P1’, P2’ , , Pn’) is a solution of (3.4) Conversely, suppose that there exists X such that (P1’, P2’ , , Pn’) is a solution of (3.4) Summing by k from to n, we obtain ! X n n X à à SP ị bk tị SP ịtị ỵ f k tị: 3:8ị ẳ a tị tị ỵ " ‘ k¼1 k¼1 From (3.5), we get n X k¼1 n n X 1X jỵ1 "k tịị jỵ1 b! n Y a! ỵ1 tịị n ẳ1 jẳ1 kẳ1 6ẳj " # n n n X 1X Y jỵ1 ẳ "k tịị a! ỵ1 tịị b! jỵ1 n ẳ1 jẳ1 kẳ1 bk tị ẳ 6ẳj ẳ btị n Y ẳ1 6ẳn1 a! ỵ1 tịị: 3:9ị Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 124 L H Chuan et al Similarly, n X " f k tị ẳ f tị m X # jẳ1 kẳ1 n Y j aj tị a! ỵ1 tịị: 3:10ị ẳ1 6ẳn1 Therefore, (3.8) is equivalent to the following equality a tịtị ỵ btị n Y a! ỵ1 tịị SP ịtị ỵ ẳ1 6ẳn1 ¼ fðtÞ À m X j¼1 ! ðSP‘ ’Þ "‘ À !Y n j aj ðtÞ að! ỵ1 tịị: ẳ1 6ẳn1 This implies atịtị ỵ btịP Sịtị ¼ fðtÞ À m X g j aj ðtÞ: j¼1 Lemma 3.2 If (’1, ’2, , ’n) is a solution of system (3.4) then (P1’1, P2’2 , , Pn’n) is also its solution Proof Suppose ( 1, 2, , n) is a solution of the system (3.4) Applying the projections Pk to both sides of k-th equation of (3.4) we get aà ðtÞðPk k ịtị ỵ Pk bk tịS ịtị ỵ ! bk tị S ị ẳ Pk f k tịị: " À ð3:11Þ It is easy to see that Pk f k tịị ẳ f k tị and Pk bk tị ẳ bk tịP : 3:12ị Substituting (3.12) into (3.11), we obtain a tịPk k ịtị ỵ bk tịP S’‘ ÞðtÞ bÃk‘ ðtÞ P‘ ðS’‘ Þ ỵ ẳ f k tị: " 3:13ị Provided ðS’‘ Þð = Þ is a constant function, then P‘ n 1X nj1 jỵ1 S Þ " W ðS’‘ Þ ¼ n j¼1 ‘ X n nj1 ẳ S ị " ẳ 0: n jẳ1 3:14ị Using Lemma 2.3, (3.13) is equivalent to the following equation a tịPk k ịtị ỵ bk tịSP ịtị ỵ bk tị SP Þ ¼ f Ãk ðtÞ; "‘ À Thus, (P1’1, P2’2, , Pn’n) is a solution of (3.4) k ¼ 1; 2, , n: Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations Theorem 3.1 125 The equation (3.2) has solutions in X if and only if the following equation a tị tị ỵ b tịS ịtị ỵ b tị S ị ẳ f tị " ð3:15Þ has solutions Moreover, if ’‘(t) is a solution of equation (3.15) then equation (3.2) has a solution given by formula tị ẳ f tị Pm jẳ1 j aj ðtÞ À bðtÞðP‘ S’‘ ÞðtÞ : aðtÞ ð3:16Þ Proof Suppose that ’ X is a solution of equation (3.2) By Lemma 3.1, (P1’, P2’, , Pn’) is a solution of system (3.4) Hence, P‘’ is a solution of (3.15) Conversely, suppose that ’‘(t) is a solution of (3.15) In this case, system (3.4) has solution (’1, ’2, , ’n) determined by the formula f Ãk ðtÞ À bÃk‘ ðtÞðS’‘ ÞðtÞ À ððbÃk‘ tịị=" 1ịịS ị = ị ; a tị k ẳ 1, 2, , n; k 6¼ ‘: k tị ẳ 3:17ị By Lemma 3.2, we have (P11, P2’2, , Pn’n) is also a solution of (3.4) Put ẳ n X Pk k : 3:18ị k¼1 It is clear that Pk’ ¼ Pk’k This means that (P1’, P2’ , , Pn’) is a solution of (3.4) From Lemma 3.1 it follows that is a solution of (3.2) Moreover, from (3.17) and (3.18) we get tị ẳ n X kẳ1 ẳ Pk k ẳ n X kẳ1 Pk f k tị bÃk‘ ðtÞðS’‘ ÞðtÞ À ðbÃk‘ ðtÞÞ=ð"‘ À 1ÞðS’‘ Þð = Þ aà ðtÞ ! n bÃk‘ ðtÞ X à à P f ðtÞ À b ðtÞðP S’ ÞðtÞ À ðS’ Þ : ‘ ‘ ‘ ‘ k‘ aà tị kẳ1 k " 3:19ị Substituting (3.9), (3.10), (3.14) into (3.19) we obtain P f ðtÞ À m jẳ1 j aj tị btịP S ịtị tị ẳ : aðtÞ g The proof is complete The solvability of equation (3.15) We set Dỵ ẳ fz C : jzj51g; DÀ ¼ fz C : jzj41g Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 126 L H Chuan et al Denote by H(Dỵ),H(D) the sets of the analytic functions in Dỵ and D respectively Consider the equation (3.15) bà ðtÞ aà ðtÞ’‘ ðtÞ ỵ b tịS ịtị ỵ S ị ẳ f tị: " Put ẩ zị ẳ 2i Z ’‘ ð Þ d ; À Àz z CnÀ: According to SokhotskiPlemelij formula, we have [1] tị ẳ ẩỵ tị ẩ tị, 4:1ị S ịtị ẳ ẩỵ tị ỵ ẩ tị: 4:2ị Moreover, S Þð = Þ ¼ 2È‘ ð = Þ: Put 0 ¼ È‘ ð = Þ We reduce equation (3.15) to the following boundary problem: find a sectionally analytic function ẩ(z) (ẩ(z) ẳ ẩỵ(z) for z Dỵ, ẩ(z) ẳ ẩ(z) for z D) vanishes at infinity, ẩ = ị ẳ 0 and satisfies the following linear relation on ẩỵ tị ẳ Gtịẩ tị ỵ gtị; t ; 4:3ị where Gtị ẳ a tị b tị f tị 0 2b tịị=" 1ịị ; gtị ẳ : a tị ỵ b tị a tị ỵ b tị 4:4ị Suppose that a tị ặ bÑ‘ ðtÞ are the non-vanishing functions on À Then G(t), g(t) X and G(t) 6¼ for any t Put Z d ln Gtị, { ẳ Gtị ẳ 2i Z lnẵ { G ị d , zị ẳ 2i z ỵ Xỵ zị ẳ e zị , X zị ẳ z{ e zị : ð4:5Þ Using the results in [2] (p 16–20) we get the following cases: (1) { ! The equation (4.3) has general solution is given by formula ! Z g ị d ẩ zị ẳ Xzị ỵ P{1 zị 2i Xỵ ị z ẳ Xzị ẫzị 0 Bzị ỵ P{1 zị , 4:6ị where ẫzị ẳ 2i Z f ị d , þ à À X ð Þða ð Þ þ b‘‘ ð ÞÞ À z ð4:7Þ 127 Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations Bzị ẳ 2i Z 2b ị=" 1ị d , ỵ ịa ị ỵ b ịị z X P{1 zị 0, 4:8ị if { ẳ 0, and P{1 zị ẳ p1 ỵ p2 z ỵ ỵ p{ z{1 , if { ! 1, 4:9ị which is a polynomial of degree { À with arbitrary complex coefficients The function È‘(z) determined in (4.6) is a solution of problem (4.3) if ẩ = ị ẳ 0 , that is ! ẫ 0 B ỵ P{1 ¼ 0 : X This implies ! ! 0 ỵ X B ẳX ẫ ỵ P{1 : 4:10ị (i) If ỵ X = ịB = ị 6ẳ : from (4.10), we get 0 ẳ X = ịẵẫ = ị þ P{À1 ð = Þ : þ Xð = ÞBð = Þ ð4:11Þ In this case, the general solution of problem (4.3) is given by formula ! X = ịẵẫ = ị ỵ P{1 = ị Bzị ỵ P{1 zị , ẩ zị ẳ Xzị ẫzị ỵ X = ịB = ị 4:12ị where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) and P{À1(z) is a polynomial of degree { À with arbitrary complex coefficients (ii) If ỵ X = ịB = ị ¼ : from (4.10) we get ỵ P{1 ẳ 0: 4:13ị ẫ Then, the general solution of problem (4.3) is given by formula ẩ zị ẳ Xzịẵẫzị 0 Bzị ỵ P{1 zị, ð4:14Þ where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8), 0 is arbitrary and P{À1(z) is a polynomial of degree { À with complex coefficients satisfying condition (4.13) (2) {50 The necessary condition for the problem (4.3) to be solvable is that Z g ị {1 d ẳ 0, k ẳ 1, , {: ỵ À X ð Þ Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 128 L H Chuan et al This condition can be written as follows Z f Ñ ð Þ {1 d ẳ 0 ỵ X ịa ị ỵ b ịị Z 2b ịị=" 1ịị k1 d ; ỵ X ịa ị ỵ b ịị k ẳ 1, , {: 4:15ị In this case, we have P{À1(z) So we receive (i) If ỵ X = ịB = ị 6ẳ : from (4.11) we get 0 ẳ X = ịẫ = ị : þ Xð = ÞBð = Þ Hence, (4.15) becomes the following condition Z f ị k1 X = ịẫ = ị d ẳ ỵ ịa ị ỵ b ịị ỵ X = ịB = ị X Z 2b ị=" 1ị k1 d ; ỵ X ịa ị ỵ b ịị 4:16ị k ẳ 1, , À {: If the condition (4.16) is satisfied then the solution of the problem (4.3) is given by formula ! Xð = ÞÉð = Þ BðzÞ ; È‘ ðzÞ ẳ Xzị ẫzị ỵ X = ịB = ị 4:17ị where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) (ii) If ỵ X = ịB = ị ẳ : from (4.13) we get ẫ ẳ 0: 4:18ị If the conditions (4.15) and (4.18) are satisfied then the solution of the problem (4.3) is given by formula  ẩ zị ẳ Xzị ẫzị 0 Bzị ; ð4:19Þ where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) and 0 is determined from condition (4.15) Now we can formulate the main results about solutions of the equation (3.15) in the following form Theorem 4.1 Suppose that the functions a tị ặ b tị does not vanish on (1) If ỵ X = ịB = ị 6ẳ and { ! then equation (3.15) has solutions ’‘ which satisfy the following formula ! X = ịẵẫ = ị ỵ P{1 = ị ỵ B tị ỵ P{1 tị S tị ẳ Xỵ tị ẫỵ tị ỵ X = ịB = ị ! X = ịẵẫ = ị ỵ P{1 = ị ỵ X tị ẫ tị B tị ỵ P{1 tị , 4:20ị ỵ X = ịB = ị Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations 129 where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) and P{ À 1(z) is a polynomial of degree { À with arbitrary complex coefficients (2) If ỵ X = ịB = ị 6ẳ and {50 then equation (3.15) is solvable if the condition (4.16) is satisfied In this case, equation (3.15) has unique solution which satisfies the formula (4.20), where P{À1(z) (3) If ỵ X = ịB = ị ẳ and { ! then equation (3.15) has solutions ’‘ which satisfy the following formula  S tị ẳ Xỵ tị ẫỵ tị 0 Bỵ tị ỵ P{1 tị ỵ X tịẵẫ tị 0 B tị ỵ P{1 tị, 4:21ị where X(z), ẫ(z), B(z) are determined by (4.5), (4.7), (4.8), 0 is arbitrary and P{À1(z) is a polynomial of degree { À with complex coefficients satisfying the condition (4.13) (4) If ỵ X = ịB = ị ẳ and {50 then the equation (3.15) is solvable if the condition (4.15) and (4.18) are satisfied In this case, equation (3.15) has unique solution which satisfies the formula (4.21), where P{À1(z) and 0 is determined from the condition (4.15) Proof (1) From assumption it follows that the problem (4.3) has a solution È‘(z) determined by (4.12) Therefore, equation (3.15) has a solution ’‘(t) determined by (4.1) Moreover, from (4.2) we get À S’‘ ðtÞ ẳ ẩỵ tị ỵ ẩ tị ! X = ị ẵẫ = ị þ P{À1 ð = Þ þ B ðtÞ þ P{À1 ðtÞ ỵ X = ịB = ị ! X = ị ẵẫ = ị ỵ P{1 = ị þ XÀ ðtÞ ÉÀ ðtÞ À B ðtÞ þ P{À1 tị : ỵ X = ịB = ị ẳ Xỵ tị ẫỵ tị À Similarly, the cases (2), (3), (4) can be proved g The solvability of equation (3.1) Theorems 3.1 and 4.1 show that if a tị ặ b tị 6ẳ on À then equation (3.2) is solvable in a closed form In this section, we study which solutions of (3.2) will be the solution of (3.1), i.e., the solutions of (3.2) need to satisfy the condition (3.3) Consider the following cases: (1) ỵ X = ịB = ị 6ẳ 0; { ! 0: By using Theorems 3.1 and 4.1, we have solutions of (3.2) given by the following formula tị ẳ f tị Pm jẳ1 j aj tị btịP S’‘ ÞðtÞ ; aðtÞ ð5:1Þ Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 130 L H Chuan et al where S’‘(t) is determined by (4.20) From (3.5) and (4.7) we get  à Z ð1=nÞ Pn "nÀ1Àj f ! jỵ1 ịị Pm a !jỵ1 ịị Qnẳ1 a! ỵ1 ịị jẳ1 ẳ1 d 6ẳj ẫzị ẳ ỵ 2i z X ịa ị ỵ b ịị m X ẳ ẫ1 zị A zị, 5:2ị ẳ1 where Z 1=nị Pn "n1j f ! jỵ1 ịị Qnẳ1 a!ỵ1 ịị jẳ1 d 6ẳj , ỵ z X ịa ị ỵ b ịị Z 1=nị Pn "n1j a ! jỵ1 ịị Qnẳ1 a! ỵ1 ịị jẳ1 d 6ẳj : A zị ẳ 2i z Xỵ ịa ị ỵ b ịị ẫ1 zị ẳ 2i 5:3ị 5:4ị Substituting (4.9), (5.2) into (4.20) we find À À S’‘ ðtÞ ẳ ẵXỵ tịẫỵ tị ỵ X tịẫ1 tị m X j ẵXỵ tịAỵ j tị ỵ X tịAj tị jẳ1 Pm X = ị ẫ1 = ị jẳ1 j Aj = ị ỵ ỵ X = ịB = ị ỵ { X P{ jẳ1 pj = ị j1 ẵX ỵ tịB ỵ tị ỵ X tịB tị pj t j1 ẵX ỵ tị ỵ X tị: jẳ1 Since we can rewrite (5.1) in the form tị ẳ f tị b tịP ẵX ỵ tịẫ1 ỵ tị þ X À ðtÞÉÀ ðtÞ aðtÞ m X À aj tị btịP ẵXỵ tịAỵ j tị ỵ X tịAj tị atị jẳ1 P P{ j1 X = ị ẫ1 = ị m jẳ1 j Aj = ị ỵ jẳ1 pj = ị ỵ ỵ X = ịB = ị j btịP ẵX ỵ tịB ỵ tị ỵ X tịB tị atị j1 ỵ { X btịP t ẵX tị ỵ X tị , pj atị jẳ1 5:5ị where X(z), B(z), ẫ1(z), A1(z), , Am(z) are determined by (4.5), (4.8), (5.3), (5.4), and p1, , p{ are arbitrary The function ’ is a solution of the equation (3.1) if it satisfies the condition (3.3), that is Mbk ’ ¼ k ; k ¼ 1, , m: Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations 131 Substituting (5.5) into the last condition, we obtain " X jÀ1 # m m { { X X X k ẳ dk ekj j ỵ ẫ1 j Aj pj gkj pj fk ỵ jẳ1 jẳ1 jẳ1 j¼1 " ! X ! jÀ1 # m { X ẳ dk ỵ ẫ1 ekj þ fk Aj gkj À fk pj ; fk À j À j¼1 j¼1 k ¼ 1, 2, , m, 5:6ị where dk ẳ Mbk ekj ẳ Mbk f tị btịP ẵXỵ tịẫỵ tị ỵ X tịẫ1 tị ; atị ! aj tị btịP ẵXỵ tịAỵ j tị ỵ X tịAj tị ; atị X = ị btịP ẵX ỵ tịB ỵ tị ỵ X tịB tị ; fk ẳ Mbk ỵ X = ịB = ị atị btịP tj1 ẵXỵ tị ỵ X tị gkj ẳ Mbk : atị 5:7ị Put 1 B d1 ỵ ẫ f C p1 1 C B C B B C B C C , B B C ¼@ A , P¼@ A , D¼B C C B A @ m mÂ1 p{ {1 dm ỵ ẫ fm m1 B e11 ỵ f1 A1 e1m ỵ f1 Am C C B C B C , E¼B C B C B @ A em1 ỵ fm A1 emm ỵ fm Am mÂm 0 {À1 C B g11 À f1 Á Á Á g1{ À f1 C B C B C B G¼B C C B @ {À1 A gm1 À fm Á Á Á gm{ À fm mÂ{: ð5:8Þ Now we write (5.6) in the form of matrix condition I ỵ Eị ẳ D À GP, ð5:9Þ where I is the unit matrix So we can formulate that the function ’ determined by (5.5) is a solution of (3.1) if and only if (1 , , m) satisfy the condition (5.9) Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 132 L H Chuan et al (2) ỵ X = ịB = ị 6ẳ 0; {50: From Theorems 3.1 and 4.1 it follows that the equation (3.2) has solutions if and only if the condition (4.16) satisfied If this is in case, then P{À1 So, the solutions of (3.2) are given by as follows tị ẳ ftị btịP ẵXỵ tịẫỵ tị ỵ X tịẫ1 tị atị m X aj tị btịP ẵXỵ tịAỵ j tị ỵ X tịAj tị j atị jẳ1 Pm X = ị ẫ1 = ị jẳ1 j Aj = ị btịP ẵXỵ tịBỵ tị ỵ X tịB tị : ỵ atị ỵ X = ịB = ị 5:10ị Therefore, the function determined by (5.10) is a solution of the equation (3.1) if and only if (1, , m) satisfy the following matrix condition I ỵ E Þ ¼ D, ð5:11Þ where E and D are determined by (5.8) On the other hand, substituting (3.5), (5.2) into (4.16) we get m m X X 0 dk À ek ¼ É1 fk Aj fk À j , k ¼ 1, 2, , { , 5:12ị ẳ1 jẳ1 where Z 1=nị Pn "n1j f ! jỵ1 ịị Qnẳ1 a! ỵ1 ịị jẳ1 6ẳj k1 d þ ð Þðaà ð Þ þ bà ð ÞÞ X À ‘‘ Z 1=nị Pn "n1j a ! jỵ1 ịị Qnẳ1 a! ỵ1 ịị jẳ1 6ẳj e0k ẳ k1 d Xỵ ịa ị ỵ b ịị Z X = ị 2b ịị=" 1ịị k1 d : fk ẳ ỵ X = ịB = ị Xỵ ịa ị ỵ b ịị dk0 ẳ ð5:13Þ Put 0 À É f d B 1 C C B C B C B D ¼B C C B C B A @ 0 dÀ{ À É1 fÀ{ À{Â1, 0 0 À f A À f A Á Á Á e e 1 1m m C B 11 C B C B C B E ¼B : C C B B C @ A 0 0 eÀ{1 À fÀ{ A1 À fÀ{ Am Á Á Á eÀ{m À{Âm ð5:14Þ 133 Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations We write (5.12) in the form of matrix condition E 0 ẳ D 0: 5:15ị Combining (5.11) and (5.15) we can say that the function determined by (5.10) is a solution of (3.1) if and only if (1 , , m) satisfy the following matrix condition 0 D IỵE ẳ : 5:16ị D m{ị1 E m{ịm (3) ỵ X = ịB = ị ¼ 0, { ! 0: Then the solutions of the equation (3.2) are given by the following formula tị ẳ ftị btịP ẵXỵ tịẫỵ tị ỵ X ðtÞÉ1 ðtÞ aðtÞ À À m X aj ðtÞ btịP ẵXỵ tịAỵ j tị ỵ X tịAj tị j atị jẳ1 btịP ẵXỵ tịBỵ tị ỵ X tịB tị atị j1 ỵ { X btịP t ẵX tị ỵ X tị , pj atị jẳ1 ỵ 0 5:17ị where X(z), B(z), ẫ1(z), A1(z), , Am(z) are determined by (4.5), (4.8), (5.3), (5.4), 0 is an arbitrary complex number and p1, , p{ satisfy the condition (4.13) Substituting (5.2) in (4.13) we obtain X X jÀ1 m { É1 j Aj pj ¼ 0: ỵ 5:18ị jẳ1 jẳ1 The function is a solution of the equation (3.1) if it satisfies the condition (3.3) Substituting (5.17) into (3.3) we get k ẳ dk m X ekj j ỵ 0 fk jẳ1 ẳ dk ỵ 0 fk { X gkj pj j¼1 m X ekj j À j¼1 { X gkj pj , k ¼ 1, 2, , m, where dk, ekj, fk, gkj are determined by (5.7) Put 1 f1 1 d1 B C B C B C C ¼@ A , D¼@ A , F¼B @ A , m mÂ1 dm mÂ1 fm mÂ1 1 g11 Á Á Á g1m e11 Á Á Á e1m B C C B C E ¼ @ , G¼B A A @ em1 emm mm 5:19ị jẳ1 gmj Á Á Á gm{ mÂ{ p1 B C C PẳB @ A , p{ {1 : 5:20ị Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 134 L H Chuan et al We write (5.19) in the form of matrix condition I ỵ Eị ẳ D þ 0 F À GP: ð5:21Þ Combining (5.18) and (5.21) we can say that the function ’ determined by (5.17) is a solution of (3.1) if and only if (1, , m) satisfy the following matrix condition I ỵ Eị ẳ D ỵ 0 F GP; ð5:22Þ where 0 1 G B B I ỵ E C {1 C IỵEẳ@ ; G ¼ @ 0 , A A A1 Á Am mỵ1ịm mỵ1ị{ F C B D D¼@ , Fẳ : A mỵ1ị1 ẫ1 mỵ1ị1 5:23ị (4) ỵ X = ịB = ị ẳ 0, {50: By Theorems 3.1 and 4.1 the equation (3.2) has solutions if and only if the condition (4.15) and the conditions (4.18) are satisfied Then we have { > with P{À1 So the solutions of (3.2) are given by the following formula tị ẳ ftị btịP ẵXỵ tịẫỵ tị ỵ X tịẫ1 tị atị m X aj tị btịP ẵXỵ tịAỵ j tị ỵ X tịAj tị j atị jẳ1 ỵ 0 btịP ẵX þ ðtÞB þ ðtÞ þ X À ðtÞB À ðtÞ : aðtÞ ð5:24Þ The function ’ determined by (5.24) is a solution of the equation (3.1) if and only if (1, , m) satisfy the following matrix condition I ỵ Eị ẳ D ỵ 0 F; 5:25ị where I ỵ E; D; F are determined by (5.23) On the other hand, (4.15) is equivalent to the condition m X e0kj j ¼ 0 f 0k ; k ¼ 1; 2, , À{; d 0k 5:26ị jẳ1 dK0 , ekj0 , where (5) Put fk0 , are determined by (5.13) d10 B C C D0 ¼ B , @ A dÀ{ À{Â1 Àf10 B C C F0 ¼ B , @ A ÀfÀ{ À{Â1 e011 B E0 ¼ B @ eÀ{1 Á Á Á e1m C C : A Á Á Á eÀ{m À{Âm ð5:27Þ 135 Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations We write (5.26) in the form of matrix condition E ẳ D ỵ 0 F : ð5:28Þ Combining (5.25) and (5.28) we can say that the function ’ determined by (5.24) is a solution of (3.1) if and only if (1, , m) satisfy the following matrix condition IỵE D F ẳ ỵ 0 E D mỵ1{ị1 F mỵ1{ị1: mỵ1{ịm 5:29ị Remark 5.1 Among the matrices fD, E, G, F, D , E , F , D, I ỵ EÞ, Fg, there are three matrices D, D , D which depending on f(t), the remaining ones are completely determined by a(t), b(t), a1(t), , am(t), b1(t), , bm(t) Theorem 5.1 Suppose aÃ2 tị b2 tị 6ẳ for any t (1) ỵ X = ịB = ị 6ẳ 0, { ! 0: Put r ẳ rank I ỵ E ị G ịmmỵ{ị , where E, G are determined by (5.8) Then the equation (3.1) is solvable if and only if the matrix D determined by (5.8) satisfies the matrix condition rankððI ỵ E ị GDịmmỵ{ỵ1ị ẳ r: If this is in case, the solutions of the equation (3.1) are given by the formula (5.5), where (1, , m, p1, , p{ ) satisfy (5.9) Moreover, we can choose m ỵ { r coefficients in {1 , , m, p1, , p{ } which are arbitrary so that (t) is uniquely determined by these coefficients In particular, if r ¼ m then the equation (3.1) is solvable for any function f(t) (2) ỵ X = ịB = ị 6ẳ 0; {50: Put IỵE , r ẳ rank E0 m{ịm where E, E are determined by (5.8), (5.14) Then the equation (3.1) is solvable if and only if the function f(t) determines D and D by the formulas (5.8), (5.14) which satisfy the following matrix condition IỵE D rank ẳ r: E D m{ịmỵ1ị 5:30ị If this is in case, the solutions of the equation (3.1) are given by the formula (5.10), where (1, , m) satisfy (5.16) In particular, if r ¼ m and the condition (5.30) is satisfied then the equation (3.1) has unique solution (3) ỵ X = ịB = ị ẳ 0; { ! 0: Put r ¼ rank I ỵ E F G mỵ1ịmỵ1ỵ{ị , Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 136 L H Chuan et al where I ỵ E; F; G are determined by (5.23) The equation (3.1) is solvable if and only if the matrix D determined by formulas (5.23) satisfies the matrix condition rank I ỵ E F G D ẳ r: mỵ1ịmỵ2ỵ{ị If the above condition satisfied then solutions of the equation (3.1) are given by the formula (5.17), where (0, , m, p1, , p{ ) satisfy (5.22) Moreover, we can choose m ỵ ỵ { r coefficients in {0, , m, p1, , p{ } which are arbitrary so that ’(t) is uniquely determined by these coefficients In particular, if r ẳ m ỵ then the equation (3.1) is solvable for any function f (t) (4) ỵ X = ịB = ị ẳ 0; {50: Put r ẳ rank IỵE F E0 F0 , mỵ1{ịmỵ1ị where I ỵ E, F, E , F are determined by (5.23) and (5.27) Then the equation (3.1) is solvable if and only if the matrix D determined by the formulas (5.23), (5.27) satisfies the matrix condition IỵE F D ẳ r: rank E0 F0 D0 mỵ1{ịmỵ2ị 5:31ị If the above condition satisfied then solutions of the equation (3.1) are given by the formula (5.24), where (0, 1, , m) satisfy (5.29) In particular, if r ¼ m þ and the condition (5.31) is satisfied then the equation (3.1) has unique solution Proof (1) From assumption it follows that the equation (3.1) has solutions if and only if there exists (1, , m) and (p1, , p{ ) which satisfy (5.9) We can rewrite (5.9) in the form ðI ỵ Eị G mmỵ{ị P ẳ D: mỵ{ị1 Therefore P ị is a solution of the following equation I ỵ Eị G X ẳ D: ð5:32Þ It follows that the necessary and sufficient condition for which the equation (3.1) has solutions is that the equation (5.32) has solutions in Cmỵ{ Since rankI ỵ E ịGDị ẳ rankI ỵ E ịGị ẳ r: If this is in case, then by using (5.32) we can express r coefficients in {1, , m , p1, , p{ } by m ỵ { À r remaining ones In particular, if r ¼ m then the equation (5.32) has solutions with any D Therefore the equation (3.1) is solvable with any f(t) Similarly, the cases (2), (3), (4) can be proved g Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008 Singular integral equations 137 Acknowledgements This work is partially supported by National basic Research Program in Natural Science, Vietnam References [1] Gakhov, F.D., 1990, Boundary Value Problems (New York: Dover Publications) [2] Litvinchuc, G.S., 2000, Solvability Theory of Boundary Value Problems and Singular Integral Equations with Shift (Dordrecht, Boston, London: Kluwer Academic Publisers) [3] Kravchenko, V.G and Litvinchuk, G.S., 1994, Singular integral equations with Carleman linear fractional shift Complex Variables: Theory and Applications, 26, 69–78 [4] Kravchenko, V.G and Shaev, A.K., 1991, The theory of solvability of singular integral equations with a linear-fractional Carleman shift Soviet Mathematics Doklady, 43(1), 73–77 [5] Vasilevski, N.L and Litvinchuk, G.S., 1975, Doklady Akademii Nauk SSSR, 221, 269–271 The solvability theory of one class of singular integral equations with an involution English translation in Soviet Mathematics Doklady, 16(2), 318–321 [6] Chuan, L.H and Tuan, N.M., 2003, On the singular integral equations with Carleman shift in the case of the vanishing coefficient Acta Mathematica Vietnamica, 28(5), 13–27 [7] Mau, Ng.V and Tuan, Ng.M., 1996, On solutions of integral equations with analytic kernels and rotations Annale Polonici Mathematici, LXIII, 3, 293–300 [8] Mau, Ng.V., 1989, On the solvability in closed form of the class of the complete singular integral equations Differential Equations, USSR, T., 25(2), 307–311 ... for a class of singular integral equations with a linear- fractional Carleman shift and with the degenerate kernels on the unit circle In genaral, one knows that the singular integral operator of. .. final form June 2007) This article deals with the solvability, the explicit solutions of a class of singular integral equations with a linear- fractional Carleman shift and the degenerate kernel. .. Singular integral equations with Carleman linear fractional shift Complex Variables: Theory and Applications, 26, 69–78 [4] Kravchenko, V.G and Shaev, A. K., 1991, The theory of solvability of singular