1. Sea f una función analítica que transforma un dominio D en un dominio D. Si U es armonica en D, entonces la función real u(x, y) = U(f(z)) es armonica en D. Teorema de transformación de funciones armónicas Proof We will give a special proof for the special case in which D is simply connected. If U has a harmonic conjugate V in D, then H = U + iV is analytic in D, and so the composite function H(f(z)) = U(f(z)) + iV(f(z)) is analytic in D. It follow that the real part U(f(z)) is harmonic in D.

2. Solving Dirichlet Problems Using Conformal Mapping 1.Find a conformal mapping w = f(z) that transform s the original region R onto the image R. The region R may be a region for which many explicit solutions to Dirichlet problems are known. 2.Transfer the boundary conditions from the R to the boundary conditions of R. The value of u at a boundary point  of R is assigned as the value of U at the corresponding boundary point f(  ).

3. 3.Solve the Dirichlet problem in R. The solution may be apparent from the simplicity of the problem in R or may be found using Fourier or integral transform methods. 4.The solution to the original Dirichlet problems is u(x, y) = U(f(z)).

4. Ejemplo: The function U(u, v) = (1/  ) Arg w is harmonic in the upper half- plane v > 0 since it is the imaginary part of the analytic function g(w) = (1/  ) Ln w. Use this function to solve the Dirichlet problem en la figura superior (a). Solution The analytic function f(z) = sin z maps the original region to the upper half-plane v  0 and maps the boundary segments to the segments shown in Fig 20.14(b). The harmonic function U(u, v) = (1/  ) Arg w satisfies the transferred boundary conditions U(u, 0) = 0 for u > 0 and U(u, 0) = 1 for u < 0.

5. Example 7 From C-1 in Appendix IV, the analytic function maps the region outside the two open disks |z| < 1 and |z – 5/2| < ½ onto the circular region r 0  |w|  1,. See Fig 20.15(a) and (b).

6. Fig 20.15

7. Example 7 (2) In problem 10 of Exercise 14.1, we found is the solution to the new problem. From Theorem 20.2 we conclude that the solution to the original problem is

8. A favorite image region R for a simply connected region R is the upper half-plane y  0. For any real number a, the complex function Ln(z – a) = log e |z – a| + i Arg (z – a) is analytic in R and is a solution to the Dirichlet problem shown in Fig 20.16.

9. Fig 20.16

10. It follows that the solution in R to the Dirichlet problem with is the harmonic function U(x, y) = (c 0 /  )(Arg(z – b) – Arg(z – a))

11. Fig 20.35(a)

12. Example 6 Solve the Dirichlet problem in Fig 20.35(a) using conformal mapping by constructing a linear fractional transformation that maps the given region into the upper half-plane.

13. Example 6 (2) Solution The boundary circles |z| = 1 and |z – ½ | = ½ each pass through z = 1. We can map each boundary circle to a line by selecting a linear fractional transformation that has a pole at z = 1. If we require T(i) = 0 and T(-1) = 1, then Since, T maps the interior of |z| = 1 onto the upper half- plane and maps |z – ½ | = ½ onto the line v = 1. See Fig 20.35(b).

14. Example 6 (3) The harmonic function U(u, v) = v is the solution to the simplified Dirichlet problem in the w-plane, and so u(x, y) = U(T(z)) is the solution to the original Dirichlet problem in the z-plane.

15. Example 6 (4) The level curves u(x, y) = c can be written as and are therefore circles that pass through z = 1. See Fig 20.36.

16. Fig 20.36

17. 20.4 Schwarz-Christoffel Transformations Special Cases First we examine the effect of f(z) = (z – x 1 )  / , 0 <  < 2 , on the upper half-plane y  0 shown in Fig 20.40(a).

18. The mapping is the composite of  = z – x 1 and w =   / . Since w =   /  changes the angle in a wedge by a factor of  / , the interior angle in the image is (  /  )  = . See Fig 20.40(b).

19. Next assume that f(z) is a function that is analytic in the upper half-plane and that jas the derivative (1) where x 1 < x 2. We use the fact that a curve w = w(t) is a line segment when the argument of its tangent vector w(t) is constant. From (1) we get (2)

20. Since Arg(t – x) =  for t < x, we can find the variation of f (t) along the x-axis. They are shown in the following table. See Fig 20.41.

21. Fig 20.41

22. Let f(z) be a function that is analytic in the upper half-plane y > 0 and that has the derivative (3) where x 1 < x 2 < … < x n and each  i satisfies 0 <  i < 2 . Then f(z) maps the upper half-plane y  0 to a polygonal region with interior angles  1,  2, …,  n. THEOREM 20.4 Schwarz-Christoffel Formula

23. Comments (i) One can select the location of three of the points x k on the x-axis. (ii) A general form for f(z) is (iii) If the polygonal region is bounded only n – 1 of the n interior angles should be included in the Schwarz-Christoffel formula.

24. Example 1 Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the strip |v|  1, u  0. See Fig 20.42.

25. Example 1 (2) Solution We may select x 1 = −1, x 2 = 1 on the x-axis, and we will construct a mapping f with f(−1) = −i, f(1) = i. Since  1 =  2 =  /2, (3) gives

26. Example 1 (3)

27. Example 2 Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the region shown in Fig 20.43(b).

28. Example 2 (2) Solution We may select x 1 = −1, x 2 = 1 on the x-axis, and we will require f(−1) = ai, f(1) = 0. Since  1 = 3  /2  2 =  /2, (3) gives

29. Example 2 (3)

30. Example 3 Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the region shown in Fig 20.44(b).

31. Example 3 (2) Solution Since the region is bounded, only two of the 60  interior angles should be included. If x 1 = 0, x 2 = 1, we obtain

32. Example 3 (3)

33. Example 4 Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the upper half-plane with the horizontal line v = , u  0, deleted. See Fig 20.45.

34. Example 4 (2) Solution The nonpolygonal target region can be approximated by a polygonal region by adjoining a line segment from w =  i to a pint u 0 on the negative u-axis. See Fig 20.45(b). If we require f(−1) =  i, f(0) = u 0, then Note that as u 0 approaches − ,  1 and  2 approach 2  and 0, respectively.

35. Example 4 (3) This suggests we examine the mappings that satisfy w = A(z + 1) 1 z -1 = A(1 + 1/z) or w = A(z + Ln z) + B. First we determine the image of the upper half-plane under g(z) = z + Ln z and then translate the image region if needed. For t real g(t) = t + log e |t| + i Arg t If t < 0, Arg t =  and u(t) = t + log e |t| varies from −  to −1. It follows that w = g(t) moves along the line v =  from −  to −1.

36. Example 4 (4) When t > 0, Arg t = 0 and u(t) varies from −  to . Therefore g maps the positive x-axis onto the u-axis. We can conclude that g(z) = z + Ln z maps the upper half-plane onto the upper half-plane with the horizontal line v = , u  −1, deleted. Therefore w = z + Ln z + 1 maps the upper half-plane onto the original target region.

37. 20.5 Poisson Integral Formulas Introduction It would be helpful if a general solution could be found for Dirichlet problem in either the upper half-plane y  0 or the unit disk |z| = 1. The Poisson formula fro the upper half-plane provides such a solution expressing the value of a harmonic function u(x, y) at a point in the interior of the upper half-plane in terms of its values on the boundary y = 0.

38. Formula for the Upper Half- Plane Assume that the boundary function is given by u(x, 0) = f(x), where f(x) is the step function indicated in Fig 20.55.

39. The solution of the corresponding Dirichlet problem in the upper half-plane is (1) Since Arg(z – b) is an exterior angle formed by z, a and b, Arg(z – b) =  (z) + Arg(z – a), where 0 <  < , and we can write (2)

40. The superposition principle can be used to solve the more general Dirichlet problem in Fig 20.56.

41. If u(x, 0) = u i for x i-1  x  x i, and u(x, 0) = 0 outside the interval [a, b], then from (1) (3) Note that Arg(z – t) = tan -1 (y/(x – t)), where tan -1 is selected between 0 and , and therefore (d/dt) Arg(z – t) = y/((x – t) 2 + y 2 ).

43. Let u(x, 0) be a piecewise-continuous function on every finite interval and bounded on －  < x < . Then the function defined by is the solution of the corresponding Dirichlet problem on the upper half-plane y  0. THEOREM 20.5 Poisson Integral Formula for the Upper Half-Plane

44. Example 1 Find the solution of the Dirichlet problem in the upper half-plane that satisfies the boundary condition u(x, 0) = x, where |x| < 1, and u(x, 0) = 0 otherwise. Solution By the Poisson integral formula

45. Example 1 (2)

46. Example 2 The conformal mapping f(z) = z + 1/z maps the region in the upper half-plane and outside the circle |z| = 1 onto the upper half- plane v  0. Use the mapping and the Poisson integral formula to solve the Dirichlet problem shown in Fig 20.57(a).

47. Fig 20.57

48. Example 2 (2) Solution Using the result of Example 4 of Sec 20.2, we can transfer the boundary conditions to the w-plane. See Fig 20.57(b). Since U(u, 0) is a step function, we will use the integrated solution (3) rather than the Poisson integral. The solution to the new Dirichlet problem is

49. Example 2 (3)

50. Let u(e i  ) be a bounded and piecewise continuous for －     . Then the solution to the corresponding Dirichlet Problem on the open units disk |z| < 1 is given by (5) THEOREM 20.6 Poisson Integral Formula for the Unit Disk

51. Geometric Interpretation Fig 20.58 shows a thin membrane (as a soap film) that has been stretched across a frame defined by u = u(e i  ).

52. Fig 20.58

53. The displacement u in the direction perpendicular to the z-plane satisfies the two-dimensional wave equation and so at equilibrium, the displacement function u = u(x, y) is harmonic. Formula (5) provides an explicit solution for u and has the advantage that the integral is over the finite interval [− ,  ].

54. Example 3 A frame for a membrane is defined by u(e i  ) = |  | for −     . Estimate the equilibrium displacement of the membrane at (−0.5, 0), (0, 0) and (0.5, 0). Solution Using (5),

55. Example 3 (2) When (x, y) = (0, 0), we get For the other two values of (x, y), the integral is not elementary and must be estimated using a numerical solver. We have u(−0.5, 0) = 2.2269, u(0.5, 0) = 0.9147.

56. Fourier Series Form Note that u n (r,  ) = r n cos n  and v n (r,  ) = r n sin n  are each harmonic, since these functions are the real and imaginary parts of z n. If a 0, a n, b n are chosen to be the Fourier coefficients of u(e i  ) for −  <  < , then

58. Example 4 Find the solution of the Dirichlet problem in the unit disk satisfying the boundary condition u(e i  ) = sin 4 . Sketch the level curve u = 0. Solution Rather than using (5), we use (6) which reduces to u(r,  ) = r 4 sin 4 . Therefore u = 0 if and only if sin 4  = 0. This implies u = 0 on the lines x = 0, y = 0 and y =  x.

59. Example 4 (2) If we switch to rectangular coordinates, u(x, y) = 4xy(x 2 – y 2 ). The surface of u(x, y) = 4xy(x 2 – y 2 ), the frame u(e i  ) = sin 4 , and the system of level curves were sketched in Fig 20.59.

60. Fig 20.59

61. 20.6 Applications Vector Fields A vector field F(x, y) = P(x, y)i + Q(x, y)j in a domain D can also be expressed in the complex form F(x, y) = P(x, y) + iQ(x, y) Recall that div F =  P/  x +  Q/  y and curl F = (  Q/  x −  P/  y)k. If we require both of them are zeros, then (1)

62. (i) Suppose that F(x, y) = P(x, y) + Q(x, y) is a vector field in a domain D and P(x, y) and Q(x, y) are continuous and have continuous first partial derivatives in D. If div F = 0 and curl F = 0, then complex function is analytic in D. (ii) Conversely, if g(z) is analytic in D, then F(x, y) = defined a vector field in D for which div F = 0 and curl F = 0. THEOREM 20.7 Vector Fields and Analyticity

63. THEOREM 20.7 Proof If u(x, y) and v(x, y) denote the real and imaginary parts of g(z), then u = P and v = −Q. Then Equations in (2) are the Cauchy-Riemann equations for analyticity.

64. Example 1 The vector field F(x, y) = (−kq/|z − z 0 | 2 )(z − z 0 ) may be interpreted as the electric field by a wire that is perpendicular to the z-plane at z = z 0 and carries a charge of q coulombs per unit length. The corresponding complex function is

65. Example 2 The complex function g(z) = Az, A > 0, is analytic in the first quadrant and therefore gives rise to the vector field

66. Potential Functions Suppose that F(x, y) is a vector field in a simply connected domain D with div F = 0 and curl F = 0. By Theorem 18.8, the analytic function g(z) = P(x, y) − iQ(x, y) has an antiderivative (4) in D, which is called a complex potential for the vector filed F.

67. Therefore F = , and the harmonic function  is called a (real) potential function of F.

68. Example 3 The potential  in the half-plane x  0 satisfies the boundary conditions  (0, y) = 0 and  (x, 0) = 1 for x  1. See Fig. 20.68(a). Determine a complex potential, the equipotential lines, and the field F.

69. Fig. 20.68

70. Example 3 (2) Solution We knew the analytic function z = sin w maps the strip 0  u   /2 in the w-plane to the region R in question. Therefore f(z) = sin - 1 z maps R onto the strip, and Fig 20.68(b) shows the transferred boundary conditions. The simplified Dirichlet problem has the solution U(u, v) = (2/  )u, and so  (x, y) = U(sin -1 z) = Re((2/  ) sin -1 z) is the potential function on D, and G(z) = (2/  )u sin -1 z is a complex potential for F.

71. Example 3 (3) Note that the equipotential lines  = c are the images of the equipotential lines U = c in the w-plane under the inverse mapping z = sin w. We found that the vertical lines u = a is mapped onto a branch of the hyperbola

72. Example 3 (4) Since the equipotential lines U = c, 0 < c < 1 is the vertical line u =  /2c, it follows that the equipotential lines  = c is the right branch of the hyperbola

73. Steady-State Fluid Flow The vector V(x, y) = P(x, y) + iQ(x, y) may also be expressed as the velocity vector of a two-dimensional steady-state fluid flow at a point (x, y) in a domain D. If div V = 0 and curl V = 0, V has a complex velocity potential G(z) =  (x, y) +  (x, y) that satisfies

74. Here special importance is placed on the level curves  (x, y) = c. If z(t) = x(t) + iy(t) is the path of a particle, then

75. The function  (x, y) is called a stream function and the level curves  (x, y) = c are streamlines for the flow.

76. Example 4 The uniform flow in the upper half-plane is defined by V(x, y) = A(1, 0), where A is a fixed positive constant. Note that |V| = A, and so a particle in the fluid moves at a constant speed. A complex potential for the vector field is G(z) = Az = Ax + iAy, and so the streamlines are the horizontal lines Ay = c. See Fig 20.69(a). Note that the boundary y = 0 of the region is itself streamline.

77. Fig 20.69(a)

78. Example 5 The analytic function G(z) = z 2 gives rise to the vector field in the first quadrant. Since z 2 = x 2 − y 2 + i(2xy), the stream function is  (x, y) = 2xy and the streamlines are the hyperbolas 2xy = c. See Fig 20.69(b).

79. Fig 20.69(b)

80. Suppose that G(x) =  (x, y) + i  (x, y) is analytic in a region R and  (x, y) is continuous on the boundary of R. Then V(x, y) = defined an irrotational and incompressible fluid flow in R. Moreover, if a particle is placed is placed inside R, its path z = z(t) remains in R. THEOREM 20.8 Streamline

81. Example 6 The analytic function G(z) = z + 1/z maps the region R in the upper half-plane and outside the circle |z| = 1onto the upper half- plane v  0. The boundary of R is mapped onto the u-axis, and so v =  (x, y) = y – y/(x 2 + y 2 ) is zero on the boundary of R. Fig 20.70 shows the streamlines. The velocity field is given by

82. Example 6 (2) It follows that V  (1, 0) for large values of r, and so the flow is approximately uniform at large distance from the circle |z| = 1. The resulting flow in R is called flow around a cylinder.

83. Fig 20.70

84. Example 7 The analytic function f(w) = w + Ln w + 1 maps the upper half-plane v  0 to the upper half-plane y  0, with the horizontal line y = , x  0, deleted. See Example 4 in Sec 20.4. If G(z) = f -1 (z) =  (x, y) + i  (x, y), then G(z) maps R onto the upper half-plane and maps the boundary of R onto the u-axis. Therefore  (x, y) = 0 on the boundary of R.

85. Example 7 (2) It is not easy to find an explicit formula for  (x, y). The streamlines are the images of the horizontal lines v = c under z = f(w). If we write w = t + ic, c > 0, then the streamlines can be See Fig 20.71.

86. Fig 20.71

87. Example 8 The analytic function f(w) = w + e w + 1 maps the strip 0  v   onto the region R shown in Fig 20.71. Therefore G(z) = f -1 (z) =  (x, y) + i  (x, y) maps R back to the strip and from M-1 in the Appendix IV, maps the boundary line y = 0 onto the u-axis and maps the boundary line y =  onto the horizontal line v = . Therefore  (x, y) is constant on the boundary of R.

88. Example 8 (2) The streamlines are the images of the horizontal lines v = c, 0 < c < , under z = f(w). If we write w = t + ic, then the streamlines can be See Fig 20.72.

89. Fig 20.72