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1. 3. Riemann-Hilbert Problems : ,gnC~{ c>O c=O if if if -1 c < o. 1. Let Zo, Zl, ... , Zn denote points on r which one meets consecutively as one traverses r in a counterclockwise manner, such that Zn = Zo, and such that uv changes sign at most once on each closed sub-arc Zj_1Zj of r, j = 1,2, ... ,n. 9) Proof. Let us define a vector function (U, V) on r by (U, V) = (sgn u, sgn v). 12) valid, provided UV changes sign at most once on the segment Zj-l, Zj. 13) 1. 9). 10), we have I = 0. In this case (Uj - 1 , Vj-d = (Uj , Vj) for j = 1,2, ...

9). 10), we have I = 0. In this case (Uj - 1 , Vj-d = (Uj , Vj) for j = 1,2, ... 13), we have J = 0. Hence if uv does not change sign on r, then I = J. Next, let (0, (1, ... , (m denote the distinct consecutive points located on r in a counterclockwise manner, at which uv changes sign, and such that (m = (0. 14) 3j = ~ i:~l I Ju Jv I for j = 1,2, ... , m, with the integration taken along r. As t traverses r from (j-1 to (j, the vector (u, v) changes according to one of the following: (i) (ii) (iii) (iv) (0, a) (0, a) (a, 0) (a,O) (0, b) (b,O) (b,O) (O,b), -t -t -t -t with a and b non-zero, since (u,v) =I (0,0) on r.

We remark also that if we multiply Z as defined by r. 3. 20). 4 SOLUTION OF THE NON-HOMOGENEOUS PROBLEM In the previous subsection we constructed a fundamental function Z, which played a role in the solution ofthe homogeneous Riemann-Hilbert problem. We now use this same fundamental function to solve the non-homogeneous Riemann-Hilbert problem in the cases of either a contour or an open arc, via the same simple technique. Let us now define the non-homogeneous problem. Non-Homogeneous Problem. Let r denote either a smooth contour or a smooth open arc, and let G and 9 be some given functions belonging ° r.

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Continued Fractions - Analytic Theory and Applns by W. Jones, W. Thron


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