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Sea f una función analítica que transforma un dominio D en un dominio D  . Si U es armonica en D  , entonces

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.

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Sea f una función analítica que transforma un dominio D en un dominio D  . Si U es armonica en D  , entonces

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  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 • 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. • 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. 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. • 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 r0 |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 foundis 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) = loge|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) = (c0/)(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 asand are therefore circles that pass through z = 1. See Fig 20.36.

  16. Fig 20.36

  17. 20.4 Schwarz-Christoffel Transformations • Special CasesFirst we examine the effect of f(z) = (z–x1)/, 0 <  < 2, on the upper half-plane y  0 shown in Fig 20.40(a).

  18. The mapping is the composite of  = z–x1 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 x1 < x2. 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. THEOREM 20.4 Let f(z) be a function that is analytic in the upper half-plane y > 0 and that has the derivative (3)where x1 < x2 < … < xnand each i satisfies 0 < i < 2. Then f(z) maps the upper half-plane y  0to a polygonal region with interior angles 1, 2, …, n. Schwarz-Christoffel Formula

  23. Comments (i) One can select the location of three of the points xk 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 x1 = −1, x2 = 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) SolutionWe may select x1 = −1, x2 = 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 x1 = 0, x2 = 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 u0 on the negative u-axis. See Fig 20.45(b). If we require f(−1) = i, f(0) = u0, thenNote that as u0 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)1z-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 realg(t) = t + loge |t| + i Arg tIf t < 0, Arg t =  and u(t) = t + loge |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) = uifor xi-1 x xi, 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 + y2).

  42. THEOREM 20.5 Let u(x, 0) be a piecewise-continuous function on everyfinite 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. Poisson Integral Formula for the Upper Half-Plane

  43. 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

  44. Example 1 (2)

  45. 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).

  46. Fig 20.57

  47. Example 2 (2) SolutionUsing 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

  48. Example 2 (3)

  49. THEOREM 20.6 Let u(ei) 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) Poisson Integral Formula for the Unit Disk

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