Download MAT389 Fall 2014, Problem Set 5 (due Oct 23) Holomorphic functions

MAT389 Fall 2014, Problem Set 5 (due Oct 23)
Holomorphic functions
5.1 For each of the functions below, determine the largest domain over which they are
holomorphic.
(i) f (z) =
eiz
,
z 2 − 2z + 1
(ii) f (z) = log |z| + i Arg z,
(iii) f (z) = (z 3 − 1)¯
z.
Note: the function eiz = ei(x+iy) = e−y+ix = e−y (cos x + i sin x) behaves as follows with
respect to the Wirtinger operators:
∂eiz
= 0,
∂ z¯
∂eiz
= ieiz
∂z
Hint: in (ii), use the Cauchy–Riemann equations in polar form (see Problem 4.6).
∗
5.2 Prove that the composition of two entire functions is again an entire function.
Note: I forgot to introduce this terminology in class: an entire function is a function
that is holomorphic on the whole of the complex plane. Nothing more, nothing less.
For example, any polynomial is an entire function. Another example is the function
f (z) = ex (cos y + i sin y): in question 8 of the first midterm, I asked you to find out where
this function is differentiable; the answer was that it is everywhere differentiable, hence
everywhere holomorphic, i.e., entire.
Hint: remember the chain rule!
5.3 Check that f (z) = 3x + y + i(3y − x) is an entire function. Can you write it in terms of z
in some simple form?
Harmonic conjugates
5.4 Check that each of the functions u(x, y) below is harmonic at every (x, y) ∈ R2 , and
find the unique harmonic conjugate, v(x, y), satisfying v(0, 0) = v0 . Express the resulting
holomorphic (entire, in fact) functions, f (z) = u(x, y) + iv(x, y), in terms of z.
(i) ∗ u(x, y) = ax + by + c (where a, b, c ∈ R) and v0 = 0,
(ii) u(x, y) = x2 − y 2 − 2x and v0 = 1,
(iii) u(x, y) = y 3 − 3x2 y and v0 = 0,
(iv) u(x, y) = x4 − 6x2 y 2 + y 4 and v0 = 0,
(v) ∗ u(x, y) = e2y cos 2x and v0 = 1.
5.5 Let u, v : Ω ∈ R2 → R be harmonic function. Show that v is harmonic conjugate to u if
and only if −u is a harmonic conjugate to v.
Steady temperatures
The variation over time of temperature in a given (homogeneous, isotropic) material is controlled by
the so-called heat equation,
∂T
= α ∆T,
∂t
where the temperature T is a function of space and time, ∆ is the Laplacian, and α is a positive
constant known as the thermal diffusivity of the material in question. Time-independent temperature
distributions are then solutions to the Laplace equation ∆T = 0.
Complex Analysis can be used to find solutions of the above in two-dimensional contexts —e.g., a
thin layer of the material that is insulated in the vertical direction).
∗
5.6 Let Ω ⊂ R2 be the infinite strip {(x, y) ∈ R2 | 0 < y < 1}, and notice that its boundary ∂Ω
consists of two pieces —namely, the lines y = 0 and y = 1. Consider the boundary-value
problem given by
∆T = 0
on Ω,
T (x, 0) = T0 ,
T (x, 1) = T1
(1)
Check that a solution —the unique solution, in fact— to (1) is given by the real part of
the function
f (z) = T0 − i(T1 − T0 )z
(which is holomorphic on Ω).
Comment: there is no need to appeal to Complex Analysis to find the solution to (1). By
symmetry, it cannot depend on the coordinate x along the strip; only on the coordinate y
across the strip. Then ∆T = 0 is reduced to Tyy = 0, so T must be a linear function of y.
Making it satisfy the boundary conditions fixes which linear function it must be.
∗
5.7 Show that the transformation
z = g(w) =
1
log |w| + i Arg w
π
defines a holomorphic function on the upper half-plane H of the w-plane, and that it maps
said upper half-plane to the infinite strip Ω in the z-plane. Where are the two pieces of ∂Ω
coming from?
∗
5.8 Show that the real part of f ◦ g provides a solution to the boundary-value problem
(
T0 if x > 0
∆T = 0 on H,
T (x, 0) =
T1 if x < 0
(2)
Comment: (2) is not as easy as (1), though you could still make a symmetry argument to
conclude that the solution should be independent of the radial coordinate and depend only
on the angular one. Notice that the radial (resp., angular) coordinate in (2) corresponds to
the x (resp. y) coordinate in (1) through the transformation z = g(w).
Electrostatic potentials
Electric and magnetic fields (in three-dimensional vacuum) obey Maxwell’s equations:
∇·E=
ρ
0
∂B
∂t
∂E
∇ × B = µ0 J + 0
∂t
∇×E=−
∇·B=0
where
− E and B denote the electric and magnetic fields, respectively;
− ρ and J are the electric charge density (charge per unit volume) and electric current density
(current per unit area), respectively; and
− 0 and µ0 are universal constants known as the permittivity and permeability of free space.
In static field configurations with zero electric charge and density, the electric and magnetic fields
decouple: the electric field only enters into the first two equations; the magnetic field, only into the
last two. For E, we have
∇ · E = 0,
∇×E=0
In this situation, we can show that E is (minus) the gradient of some function V , which we call the
electrostatic potential 1 . It satisfies the Laplace equation, for
∆V = ∇ · (∇V ) = −∇ · E = 0
In problems with translational symmetry along the vertical axis, V only depends on the x and y
coordinates. Since it is a harmonic function, we might be able to use Complex Analysis!
5.9 If you have taken a course on electromagnetism, you have probably seen (even if you don’t
remember right now) that the electrostatic potential created by an infinite wire is given by
V =
λ
r0
log ,
2π0
r
where λ is the linear charge density of the wire, r is the distance to it, and r0 is an arbitrary
constant. In terms of the three-dimensional picture, we are placing the wire along the
z-axis, and r is the radial coordinate in a cylindrical coordinate system.
Now think of that same wire inside a cylinder of unit radius and parallel to it. If we keep
the surface of the cylinder at a constant value of the potential (physically, this means
that the electric field is perpendicular to it), the potential function above also provides a
solution to the Laplace equation ∆V = 0 on the inside of the cylinder (except at the center
of it where the wire is located), and satisfies a Dirichlet boundary condition: V |r=1 = V0 .
(i) Calculate the value of V0 in terms of r0 .
(ii) Show that the equipotential lines (the curves where the potential is constant) are
circles, and find an expression for their radius in terms of the value of the potential.
1
This is only true on simply-connected regions of three-dimensional space, but let’s obviate this finer point here.
5.10 Let us forget now about the direction parallel to the wire. The interior of the cylinder then
corresponds to the punctured unit disc D× . It is not simply-connected, so we might not be
able to find a harmonic conjugate to the potential V on the whole of it. In fact, this is
precisely the case. But —again, as in class— we can find a harmonic conjugate (let us call
it U ) on the simply-connected domain consisting of the unit disc take away the origin and
the negative x-axis. Give an explicit expression for it.
5.11 Consider the M¨
obius transformation
z = T (w) = −
w−i
w+i
It maps the upper half-plane H to the unit disc D, and the point w = i to the origin of the
z-plane (see Problem 2.12).
(i) Show that the composition (V + iU ) ◦ T defines a holomorphic function on the set
H − {w = 0 + iv | v ≥ 1}.
(ii) Show that the real part of (V + iU ) ◦ T extends across {w = 0 + iv | v > 1}, i.e., that
it doesn’t have any discontinuity there.
(iii) Conclude that W = Re[(V + iU ) ◦ T ] provides a solution to the Laplace equation on
H − {w = i} satisfying W |∂H = V0 .
(iv) Find equations for the equipotential lines.
Fluid flow
Under certain conditions, the velocity field V of a fluid can be treated much in the same way as the
electric field of the last few problems. We need to assume that the flow is stationary, irrotational and
incompressible. In other words, that the following equations are satisfied:
∂V
= 0,
∂t
∇ × V = 0,
∇ · V = 0.
As in the case of the electric field, we can write2 V as the gradient of some function φ —usually
referred to as the velocity potential — which satisfies the Laplace equation:
∆φ = ∇ · (∇φ) = ∇ · V = 0.
We can now attack problems with translational symmetry along an axis with Complex Analysis
techniques.
ˆ (here x
ˆ is unit vector pointing in the positive
5.12 Consider the stationary flow V = v0 x
x-direction) in the upper-half plane H = {(x, y) ∈ R2 | y > 0}.
(i) Find a velocity potential φ that gives this velocity field.
(ii) For your choice of φ, find a harmonic conjugate ψ. Show that the velocity field above
is tangent to the level curves ψ = const. In particular ψ takes a constant value along
the boundary of H.
2
Once again, on simply-connected domains; once again, don’t worry about it for now.
Figure 1: Fluid flow around a cylinder
The situation in this last problem is paradigmatical. From a physical point of view, the velocity field
should be parallel to the boundary —a component perpendicular to it would signal that there is a
flux of fluid through that boundary. A potential for such a velocity field will then never be constant
along the boundary; a harmonic conjugate to it, though, always will.
A strategy for finding the velocity field of a fluid near an impermeable boundary is then the following.
1) Find a solution of the Laplace equation ∆ψ = 0 on a domain Ω, subject to the boundary
condition ψ|∂Ω = const (if the boundary consists of different connected components, the
constant value of ψ may be different in different components).
2) Find a harmonic conjugate to ψ on Ω, and let the velocity potential φ be minus that function
(recall Problem 5.5).
3) Take the gradient of that velocity potential.
The easiest way of visualizing this velocity field, though, is to simply look at the level curves of ψ,
which coincide with the integral lines of the vector field V = ∇φ (try it in the problem above!).
We will now find the velocity field of a fluid around a cylinder (see Figure 1). The first observation is
that the picture is symmetric with respect to reflection across the x-axis. In particular, the velocity
field is parallel to the x-axis (when x < −1 and x > 1, of course), and so it is enough to solve the
problem in the domain
n
o
Ω = H ∩ (x, y) ∈ R2 |x2 + y 2 > 1 ;
that is, the upper-half of Figure 1.
5.13 Show that the transformation z = g(w) = w + 1/w takes the boundary of Ω to the real
axis, and Ω to the upper half-plane H.
5.14 Consider the function f (z) = ψ − iφ, where φ and ψ are the functions you constructed in
Problem 5.12. According to Problem 5.5, f (z) is a holomorphic function on the upper-half
plane. Composing g ◦ f then gives you a holomorphic function on Ω. The velocity field
we are after is then −∇ Im(g ◦ f ), and its integral lines are given by the level curves of
Re(g ◦ f ). Find equations for the latter, and plot a few of them (probably aided by some
computer algebra system; e.g., Wolfram Alpha).
Poisson’s equation
In the above problems, we have been working with harmonic functions. The reason why Complex
Analysis is so helpful with those is because holomorphic functions transform harmonic functions to
harmonic functions —through the process of lifting a harmonic function to a holomorphic function
by finding a harmonic conjugate, composing and then taking the real part.
There are other cases in which holomorphic functions might help transform some problems into
easier ones. For example, electrostatic potentials in the presence of nonzero charge densities (recall
Maxwell’s equations above) satisfy Poisson’s equation
∆V = ∇ · (∇V ) = −∇ · E = −
ρ
0
The following two problems explore how functions satisfying a two-dimensional Poisson equation are
transformed under composition with a holomorphic function.
5.15 Suppose that a holomorphic function w = f (z) = u(x, y) + iv(x, y) maps a domain Dz in
the z plane onto a domain Dw in the w plane; and let a function h(u, v), with continuous
partial derivatives of the first and second order, be defined on Dw . Use the chain rule for
partial derivatives to show that if H(x, y) = h u(x, y), v(x, y) , then
h
i
Hxx (x, y) + Hyy (x, y) = huu (u, v) + hvv (u, v) |f 0 (z)|2
Hint: in the simplifications you will need to use the Cauchy–Riemann equations for f , and
the fact that u and v are harmonic on Dz .
5.16 Let p(u, v) be a function that has continuous partial derivatives of the first and second
orders and satisfies Poisson’s equation
puu (u, v) + pvv (u, v) = Φ(u, v)
in a domain Dw of the w plane, where Φ is a prescribed function. Show how it follows from
the previous problem that if a holomorphic function w = f (z) = u(x, y) + iv(x, y) maps a
domain Dz onto the domain Dw , then the function
P (x, y) = p u(x, y), v(x, y)
satisfies the Poisson equation
Pxx (x, y) + Pyy (x, y) = Φ u(x, y), v(x, y) |f 0 (z)|2
on Dz .