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1 et e a domain in the c omplex plane A mapping is said to b onformal atap oint if is analytic at every p oint and 0 Theorem 132 Supp ose that a tr ansformation x y iv x y is c onformal on a smo oth ar If along a function u v satis es either of t ID: 23120

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## Presentations text content in LECTURE ConformalMappingT ec hniques Definition

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LECTURE 13 ConformalMappingT ec hniques Definition 13.1 et e a domain in the c omplex plane. A mapping is said to b onformal atap oint if is analytic at every p oint and =0 Theorem 13.2 Supp ose that a tr ansformation )= x; y )+ iv x; y is c onformal on a smo oth ar . If along , a function u; v satis es either of the c onditions u; v (13.1) dh dn (13.2) wher isar al c onstant and dn denotes the derivative along the dir ction normal to , then the function x; y )= x; y ;v x; y )) satis es the c orr esp onding c ondition x; y (13.3) dH dN (13.4) wher dh dN denotes derivatives normal to Theorem 13.3 Supp ose that the image of an analytic function )= x; y )+ iv x; y de ne d on a domain is another domain .If u; v is a harmonic function de ne don , then the function x; y )= x; y ;v x; y )) is harmonic in Application Find the electrostatic p oten tial in the space enclosed b y the half circle =1, and the line = 0 when = 0 on the circular b oundary and = 1 on the line segmen t [-1,1]. Consider the transformation )= 1+ (13.5) maps the upp er half of the unit circle on to the rst quadran t of the plane and the in terv al [-1,1] on to the p ositiv axis. e can determine the image of the region describ ed ab o eb y guring out ho w the b oundaries are mapp ed. The circular part of the b oundary can b e parameterized b )= i 49

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13.CONF ORMALMAPPINGTECHNIQUES 50 and so the image of this b oundary b is the curv )= i 1+ i = tan ( = 2) whic h coincides with the p ositiv e real -axis. The line segmen t [-1,1] can b e parameterized b )= 1] and the image of [-1,1] b is 1+ 1] It is clear that this corresp onds to a line running along the p ositiv e imaginary axis. T o see ho w the in terior of the semi-circular region is mapp ed, w ec ho ose an arbitrary p oin t, sa , and compute its image. 1+ 1+ 3+ (1 + )(3 10 4+2 10 This is eviden tly a p oin t lying in the rst quadran t. By con tin uit y argumen ts w e can conclude that all p oin ts of the original semi-circular region m ust b e mapp ed in to the rst quadran t. The next step is to nd a solution of Laplace's equation that satis es the b oundaries conditions u; 0) = 1 (0 ;v )= 0 (13.6) No w the imaginary part of the analytic function Log )= (ln( )+ i is a harmonic function that satis es the b oundary conditions (13.2). In terms of the co ordinates and this function is u; v )= Im Log iv arctan No w all w eha etodono w is pull bac k this function to the plane. F rom (13.5) w eha iv iy 1+ iy (13.7) (1 iy )(1 + iy (1 + iy )(1 + iy (13.8) 1+2 1+2 (13.9) And so 1+ 2 (13.10) 1+ 2 (13.11) Th us, x; y )= arctan (13.12) Homew ork: 4.8.4, 4.8.5