Please answer the following fully, answering all the parts
Posted: Wed May 04, 2022 1:14 pm
Please answer the following fully, answering all the parts
(b) Consider the obstacle K = L(-1+2i, 1 - 2i). (i) Use Theorem 3.3 on page 41 of Book D to prove that the function f(2)= (2+2√1+(3+4i)/z²) is a one-to-one conformal mapping from C-K onto C - K√5/2¹ where K√5/2 = {2:|2| ≤ √5/2}. (ii) Verify that f satisfies the Laurent series condition of the Flow Mapping Theorem. (iii) Use Lemma 4.1 on page 53 of Book D to prove that f(z) = z + 3+4i 4 f(z) and f'(z) = f(z) z√1+(3+4i)/z² (iv) Deduce that the solution to the Obstacle Problem for K with circulation 67 around K is 1 q(z) = (1+ -1+2i 3i 2z f(z) 2 X /1+(3+4i)/z² (v) Verify that lim q(z) = 1 for the function q in part (b)(iv). [2] [3] [2] [7] [2]
Theorem 3.1; once again we omit the proof. Here and later in the unit we use the notation L(a, 3) to denote the closed line segment in C with endpoints a and 3. Theorem 3.3 For a EC-{0}, the function Ja has the following properties. : (a) Ja maps the circle {z |z| = |al} onto the line segment L(-2a, 2a), with J(a) = 2a and J(-a) = -2a. : (b) Ja maps the region {z |z| > |a|} conformally onto the region C-L(-2a, 2a). (c) The restriction of Ja to {z: |z| > |a|} has inverse function J¹ (w) = (w+w√/1-4a²/w²) (w EC-L(-2a, 2a)). (d) Ja has a non-vanishing derivative at all points of C - {0} except z = ta. There is another way to write w = Ja(z) as the composition of three functions which will prove useful in the next section. For this method we
Joukowski function Ja, a 0, introduced in Subsection 3.2, which we will use later in the section. From now on we will usually reverse the roles of the variables z and w when using Joukowski functions and their inverse functions. Lemma 4.1 Let Ja(w) = w+ a²/w, where a 0, and let J¹ be the inverse function of the restriction of Ja to C-Kla. Then, for z EC-L(-2a, 2a), (a)_Jū¹(z) + a²/J¹ (2) = z (b) J¹ (2) - a²/J¹ (2) = 2√√/1-4a²/z² 1 (c) (J¹)'(z) = = Jx¹(z) 1-a²/(Ja¹(z))² 2√/1-4a²/2² Proof Let w= J¹(z). Part (a) holds because z = Ja(w) = w+ a²/w = J¹ (2) + a²/J¹ (2). Part (b) follows from part (a) together with the formula J-1(x) — 1(x + 2/1-40²/2²).
- Please Answer The Following Fully Answering All The Parts 1 (44.04 KiB) Viewed 27 times
Theorem 3.1; once again we omit the proof. Here and later in the unit we use the notation L(a, 3) to denote the closed line segment in C with endpoints a and 3. Theorem 3.3 For a EC-{0}, the function Ja has the following properties. : (a) Ja maps the circle {z |z| = |al} onto the line segment L(-2a, 2a), with J(a) = 2a and J(-a) = -2a. : (b) Ja maps the region {z |z| > |a|} conformally onto the region C-L(-2a, 2a). (c) The restriction of Ja to {z: |z| > |a|} has inverse function J¹ (w) = (w+w√/1-4a²/w²) (w EC-L(-2a, 2a)). (d) Ja has a non-vanishing derivative at all points of C - {0} except z = ta. There is another way to write w = Ja(z) as the composition of three functions which will prove useful in the next section. For this method we
Joukowski function Ja, a 0, introduced in Subsection 3.2, which we will use later in the section. From now on we will usually reverse the roles of the variables z and w when using Joukowski functions and their inverse functions. Lemma 4.1 Let Ja(w) = w+ a²/w, where a 0, and let J¹ be the inverse function of the restriction of Ja to C-Kla. Then, for z EC-L(-2a, 2a), (a)_Jū¹(z) + a²/J¹ (2) = z (b) J¹ (2) - a²/J¹ (2) = 2√√/1-4a²/z² 1 (c) (J¹)'(z) = = Jx¹(z) 1-a²/(Ja¹(z))² 2√/1-4a²/2² Proof Let w= J¹(z). Part (a) holds because z = Ja(w) = w+ a²/w = J¹ (2) + a²/J¹ (2). Part (b) follows from part (a) together with the formula J-1(x) — 1(x + 2/1-40²/2²).