Complex Analysis - Maths KU

230 Chapter 4 Elementary Functions
y
φ = 40 φ = 10
–
π
2
φ = 20 φ = 50
Figure 4.27 The isotherms and lines of
heat flux for Example 2
π
2
x
Setting k0 = 40, k1 = 20, k2 = 50, k3 = 10, u1 = −1, u2 = 0, and u3 =1,we
obtain:
Φ(u, v) = 10 + 20
π
Arg (w +1)− 30
π
Arg (w)+40Arg
(w − 1) . (12)
π
Step 4 A solution φ of the Dirichlet problem in the domain D isfound by
replacing the variables u and v in (12) with the real and imaginary partsof
the analytic function f(z) = sin z. Since
sin z =sinx cosh y + i cos x sinh y and w = u + iv,
thisisequivalent to replacing w with sin z in (12). Therefore,
φ(x, y) = 10 + 20
π
Arg (sin(z)+1)− 30
π
Arg (sin z)+40Arg
(sin(z) − 1) (13)
π
isa solution of the Dirichlet problem in D. If desired, the function φ can
be written in termsof x and y, provided that we are careful with our use of
the real arctangent function. In particular, if the values of the arctangent are
chosen to lie between 0 and π, then the function φ in (13) can be written as:
φ(x, y) = 10 + 20
π arctan
� �
cos x sinh y
−
sin x cosh y +1
30
π arctan
� �
cos x sinh y
sin x cosh y
+ 40
π arctan
� �
cos x sin y
.
sin x cosh y − 1
Observe that the function
Ω(z) =10i + 20
π
Ln (sin(z)+1)− 30
π
Ln (sin z)+40Ln
(sin(z) − 1)
π
isanalytic in the domain D given by −π/2