Complex Analysis - Maths KU

438 Chapter 7 Conformal Mappings
ψ being constant on C. Put yet another way, the boundary of D must be a
streamline of the flow. The following summarizes this discussion.
Streamlining
Suppose that the complex velocitypotential Ω(z) =φ(x, y) +iψ(x, y)
is analytic in a domain D and that ψ is constant on the boundaryof D.
Then f(z) =Ω ′ (z) is a complex representation of the velocityfield of a
flow of an ideal fluid in D. Moreover, if a particle is placed in D and
allowed to flow with the fluid, then its path z(t) remains in D.
Manystreamlining problems can be solved using conformal mappings in
a manner similar to that presented for solving Dirichlet and Neumann problems.
In order to do so, we consider the complex velocitypotential as an
conformal mapping of the z-plane to the w-plane. If z(t) =x(t)+iy(t) isa
parametrization of a streamline ψ(x, y) =c2 in the z-plane, then
w(t) =Ω(z(t)) = φ(x(t),y(t)) + iψ(x(t),y(t)) = φ(x(t),y(t)) + ic2.
Thus, the image of a streamline under the conformal mapping w = Ω(z) is
a horizontal line in the w-plane. Since the boundary C is required to be a
streamline, the image of C under w = Ω(z) must be a horizontal line. That
is, we can determine the complex velocitypotential byfinding a conformal
mapping from D onto a domain in the w-plane that maps the boundary C of
D onto a horizontal line. It is often the case, however, that it is easier to find a
conformal mapping z = Ω −1 (w) from, say, the upper half-plane v>0ontoD
that takes the boundary v = 0 onto the boundary C of D. Ifz = Ω −1 (z) isa
one-to-one function, then its inverse w = Ω(z) is the desired complex velocity
potential. In summary, we have the following method for solving streamlining
problems.
Solving a Streamlining Problem
If w = Ω(z) =φ(x, y)+iψ(x, y) is a one-to-one conformal mapping of
the domain D in the z-plane onto a domain D ′ in the w-plane such that
the image of the boundary C of D is a horizontal line in the w-plane, then
f(z) =Ω ′ (z) is a complex representation of a flow of an ideal fluid in D.
EXAMPLE 4 Flow arounda Corner
Construct a flow of an ideal fluid in the first quadrant.
Solution Let D denote the first quadrant x>0, y>0. From EntryE-4
of Appendix III with the identification α = 2, we see that w = Ω(z) =z 2 is
a one-to-one conformal mapping of the domain D onto the upper half-plane
v>0 and that the image of the boundaryof D under this mapping is the real
axis v = 0. Therefore, f(z) =Ω ′ (z) =2¯z is a complex representation of the