qr code reader java app The Schwarz-Christoffel Transformation in Java

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The Schwarz-Christoffel Transformation
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The Schwarz-Christoffel transformation maps The interior of a polygon to the upper half plane The boundary of the polygon to the real axis Here we give a heuristic explanation of the transform (not a formal derivation) and state the result Our discussion is out of Levinson and Redheffer (see the bibliography list at the end of the book) Consider the polygon shown in Fig 101 In what follows, we assume the polygon is in the w plane and that it is mapped to the upper half of the z plane As noted in the gure, the curve enclosing the region is traversed in the positive or counterclockwise sense We de ne the interior angles at the vertices by
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Now we assume the existence of a function w = f ( z ) that maps the interior of the polygon to the upper half plane (the existence of the function is implied by the Riemann mapping theorem) If this mapping is one-to-one and conformal (angle preserving) then it follows that f ( z ) is analytic for y > 0 and continuous for y 0
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Figure 101 A polygon to be mapped to the upper half plane
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Schwarz-Christoffel
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Figure 102 dz is a vector on the x axis of the z plane
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We can also assume the existence of an inverse mapping, which we denote by z = g( w) The inverse mapping is analytic on the polygon s interior Moreover, it is continuous in the interior of the polygon and on it s boundary The boundary of the polygon is mapped onto the real axis Suppose that the vertices of the polygon are mapped onto the points of the real x axis denoted by x1 , x 2 , , x n Now, since w = f ( z ), it follows that dw = f ( z ) dz (101) Next we assume that the mapping and its inverse w = f ( z ), z = g( w) are analytic on the sides of the polygon in addition to its interior Picking a point w on the polygon which is not a vertex, the image of w is a point z The point dz is a positive vector on the real axis x This is shown in Fig 102 It should follow that dw is a vector on the edge of the polygon pointing in the positive sense (in the counterclockwise direction) This is shown in Fig 103 The arguments of f and w are then related in the following way: dw arg[ f ( z )] = arg dz (102)
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Figure 103 If we take dz to be a positive pointing vector on the real axis, then dw is a vector on the edge of the polygon pointing in the positive sense
Complex Variables Demysti ed
Now imagine moving the point w around the polygon and moving the image point z in the positive direction along the x axis Each time a point w moves past a vertex of angle 1 , the argument changes as (1 1 ) But arg dz = 0 Therefore, to the left of the point x1 arg[ f ( z )] is a constant, but at the point x1 it will change by (1 1 ) to maintain Eq (102) As z moves from x < x1 to x > x1, arg[( z x1 )] decreases from to 0 This means that the argument of ( z x1 ) 1 1 is changed by (1 1 ) This change will occur at every vertex So we choose f ( z ) = A( z x1 ) 1 / 1 ( z x 2 ) 2 / 1 Then, using Eq (101) dw = A( z x1 ) 1 / 1 ( z x 2 ) 2 / 1 dz Integrating, we obtain the form w = f ( z ): w = A ( z x1 ) 1 / 1 ( z x 2 ) 2 / 1 ( z x n ) n / 1 dz + B (105) ( z x n ) n / 1 (104) ( z x n ) n / 1 (103)
The constants A and B are in general complex that indicate the orientation, size, and location of the pentagon in the w plane To obtain the mapping in Eq (105), three points out of x1 , x 2 ,, x n can be chosen If a point x j is taken at in nity then the / 1 factor ( z x j ) j is not included in Eq (105) EXAMPLE 101 Find the image in the upper half plane using the transformation w= when 0 < k < 1 SOLUTION Looking at Eq (105), we see that this mapping is a Schwarz-Christoffel transformation Let s rewrite the integral in a more suggestive form: w = (1 t 2 ) 1/ 2 (1 k 2t 2 ) 1/ 2 dt
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