5 The mapping f 1 z z 4 takes the shaded region in Figure 756 onto the upper

# 5 the mapping f 1 z z 4 takes the shaded region in

• Homework Help
• 186
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 169 - 171 out of 186 pages.

5. The mapping f 1 ( z ) = z 4 takes the shaded region in Figure 7.56 onto the upper semi-disk of radius 1. The upper semi-disk is mapped onto the first quadrant by f 2 ( w 1 ) = i 1 - w 1 1 + w 1 . Thus the mapping w = φ ( z ) = i 1 - z 4 1 + z 4 takes the shaded region in Figure 7.56 onto the first quadrant. It is also a conformal mapping, being the composition of such mappings. We now determine the image of the boundary. The rays at angle 0 and π/ 4 are mapped onto the interval [ - 1 , 1] by f 1 . The quarter of a circle is mapped to the upper semi-circle by f 1 . The mapping f 2 takes the interval [ - 1 , 1] onto the upper half of the imaginary axis, and the semi-circle onto the right half of the real axis. Thus the problem in the first quadrant of the w -plane becomes: Δ U = 0 , U ( α ) = 100 if α > 0 , U ( ) = 0 if α > 0 . The solution in the w -plane is U ( w ) = a Arg ( w ) + b , where a and b must be chosen in according to the boundary conditions: a Arg ( w )+ b = 100, when Arg = 0 and a Arg ( w )+ b = 0, when Arg = π 2 . Thus b = 100 and a = - 200 π . Hence U ( w ) = - 200 π Arg w + 100 and so u ( z ) = U ( φ ( z )) = - 200 π Arg i 1 - z 4 1 + z 4 + 100 . 9. To solve the given problem, we can proceed as in Example 7.3.8 and make the necessary changes. A much quicker way is based is to use the solution in Example 7.3.8 and superposition, as follows. Let u 1 ( z ) denote the solution in Example 7.3.8. Let u 2 ( z ) = 100. It is clear that u 2 is harmonic for all z . Thus u 2 - u 1 is harmonic in the shaded region of Fig. 18. On the real axis, we have u 2 ( z ) - u 1 ( z ) = 100 - 0 = 100. On the upper semi-circle, we have u 2 ( z ) - u 1 ( z ) = 100 - 100 = 0. Thus 100 - u 1 is the solution, where u 1 is the solution in Example 7.3.8. 13. This problem is very similar to that in Example 7.3.3. The first step is to map the region to an annular region bounded by concentric circles. This can be done by using the linear fractional transformation φ ( z ) = 4 z - 1 4 - z .
Section 7.3 Solving Dirichlet Problems with Conformal Mappings 169 Then φ ( z ) = 4 z - 1 4 - z = z - 1 4 1 - z 4 = φ 1 4 ( z ) . Using φ , we map the outer circle to the unit circle and the inner circle to the circle of radius 1 4 and center at 0. As in Example 7.3.3, the solution is U ( w ) = 100 + 100 ln | w | + ln 4 ln(1 / 4) = 100 - 100 ln | w | + ln 4 ln 4 = - 100 ln | w | ln 4 . Thus the solution is u ( z ) = - 100 ln 4 ln 4 z - 1 4 - z . 17. (a) The solution of the Dirichlet Problem in Figure 7.6.7 is obtained by applying the Poisson formula (7.3.5), with f ( s ) = s if - 1 < s < 1, f ( s ) = - 1 if s < - 1 and f ( s ) = 1 if s > 1: u ( x + iy ) = y π Z 1 - 1 s ( x - s ) 2 + y 2 ds + y π Z - 1 -∞ - 1 ( x - s ) 2 + y 2 ds + y π Z 1 ds ( x - s ) 2 + y 2 = I 1 + I 2 + I 3 . We compute each integral separately: I 1 = - y π Z 1 - 1 x - s ( x - s ) 2 + y 2 ds + y π Z 1 - 1 x ( x - s ) 2 + y 2 ds = y π 1 2 ln ( ( x - s ) 2 + y 2 ) 1 - 1 + xy π Z 1 - 1 ds ( s - x ) 2 + y 2 = - y 2 π ln ( ( x + 1) 2 + y 2 ) - ln ( ( x - 1) 2 + y 2 ) + x Z 1 - 1 ds ( s - x y ) 2 + 1 = - y 2 π ln ( x + 1) 2 + y 2 ( x - 1) 2 + y 2 + x π Z 1 s = - 1 du u 2 + 1 (Let u = s - x y . ) = - y 2 π ln ( x + 1) 2 + y 2 ( x - 1) 2 + y 2 + x π tan - 1 s - x y 1 s = - 1 = - y 2 π ln ( x + 1) 2 + y 2 ( x - 1) 2 + y 2 + x π h tan - 1 1 + x y + tan - 1 1 - x y i .