D bản đồ khu vực này vào H qua w = z + 1 / z, và H, § một lần nữa có thể được thực hiện để § = Ay cho chúng ta một tiềm năng phức tạp È 1 ˘ F (z) = AÍz + ˙ Î z ˚ Đểxem phần thực và ảo, sử dụng hình thức cực: z = lại. Điều này cho phép Vì vậy, bây giờ chúng ta có thể nhận được tiềm năng và sắp xếp hợp lý theo hình thức cực | This region D maps into H via w z 1 z and on H can again be taken to be Ay Giving us a complex potential F z A z z To see the real and imaginary i0 . This gives parts use polar form z re F z A rei0 e i0 A . 1 r rJ cos 0 i A 1 r - TJ sin 0 So we can now get the potentials and streamlines in polar form Equipotentials . 1 r -r cos 0 const Kind of complicated to draw these piggies - wait until we do things parametrically in the next section. Streamlines r - r sin 0 const Again these are not standard curves. However at large distances 1 r 0 and so the streamlines are 1r sin 0 const or y const horizontal lines. Velocity Field F z A 1 - 4 z . whence F z A 1 - J2 z . We get stagnation points when the velocity equals zero so we see that this gives z 1. Exercise Set 12 Hand In 1. Compute all the details for the flow around a corner of 60 . 2. Flow through an Aperture Use a conformal map to model the following flow. 41 The width of the aperture is set to 2a. To model this Suggestion Consider what the inverse sine function does to this region. 13. Parametrically Representing Streamlines and Using Technology Sometimes it is hard to draw the streamlines from the implicit equation that defines them. An analytic approach short of seeing directly what the curves are as we have done up to now is to find an equation for dy dx using implicit differentiation and then drawing the integral curves using technology. However a more direct way is the following which hinges on inverting the conformal mappings we have been using up to now. Proposition Streamlines and Equipotentials go to Streamlines and Equipotential S pose H is a conformal invertible map with P i is a complex potential on H P H C . Then the image under F 1 of the streamlines and equipotentials on H are the streamlines and equipotentials on D. Proof The associated complex potential on D is as we have seen given by composition Q P F Therefore its streamlines are specified by setting imaginary part equal to a constant Im P
