Complex Analysis - Maths KU

226 Chapter 4 Elementary Functions
Φ = –2
D′
Φ = 3
–1
1
∇ 2 Φ = 0
v
Figure 4.21 The transformed Dirichlet
problem for Example 1
u
Step 1 Inspection of the domain D in Figure 4.20 suggests that we take D ′
to be a domain bounded by the lines u = −1 and u = 1 in which a solution
of the associated Dirichlet problem is given by (2).
Our first step is to find an analytic mapping from D onto D ′ . In order
to do so, we first rotate the region D through π/4 radianscounterclockwise
about the origin. Under thisrotation, the boundary linesy = x + 2 and y = x
are mapped onto the vertical lines u = − √ 2 and u = 0, respectively. If we
next magnify thisdomain by a factor of √ 2, we obtain a domain bounded by
the lines u = −2 and u = 0. Finally, we translate this image by 1 in order to
obtain a domain bounded by the lines u = 1 and u = −1 asdesired. Recall
from Section 2.3 that rotation through π/4 radiansabout the origin isgiven
by the mapping R(z) =e iπ/4 , magnification by √ 2 isgiven by M(z) = √ 2z,
and translation by 1 is given by mapping T (z) =z +1. Therefore, the domain
D ismapped onto the domain D ′ by the composition
f(z) =T (M(R(z))) = √ 2e iπ/4 z +1=(1+i)z +1.
Since the function f isa linear function, it isentire, and so we have completed
Step 1.
Step 2 We now transform the boundary conditions on D to boundary conditionson
D ′ . In order to do so, we must find the images under w = f(z) of
the boundary lines y = x and y = x +2 of D. By replacing the symbol z with
x + iy, we can express the mapping w =(1+i)z + 1 as:
w =(1+i)(x + iy)+1=x − y +1+(x + y)i. (8)
From (8) we find that the image of the boundary line y = x + 2 isthe set of
points:
w = u + iv = x − (x +2)+1+(x +(x + 2)) i = −1+2(x +1)i
which isthe line u = −1. In a similar manner, we also find that the image of
the boundary line y = x is the set of points:
w = u + iv = x − (x)+1+(x +(x)) i =1+2xi
which isthe line u = 1. Therefore, the boundary condition φ(x, x +2)=−2
istransformed to the boundary condition Φ(−1, v)=−2, and the boundary
condition φ(x, x) = 3 istransformed to the boundary condition Φ(1, v)=3.
See Figure 4.21.
Step 3 A solution of the Dirichlet problem in D ′ isgiven by (2) with x and
y replaced by u and v, and with k0 = −2 and k1 =3:
Φ(u, v) =
3 − (−2)
u +
2
−2+3
2
= 5 1
u +
2 2 .
Step 4 The final step in our solution is to substitute the real and imaginary
partsof f into Φ for the variables u and v to obtain the desired solution φ.
From (8) we see that the real and imaginary parts of f are:
u(x, y) =x − y + 1 and v(x, y) =x + y,