math 529 – mathematical methods for physical sciences ii
TRANSCRIPT
MATH 529 – Mathematical Methodsfor Physical Sciences II
Christoph Kirsch
April 27, 2011
Contents
0 Overview of the lecture 30.1 Partial differential equations . . . . . . . . . . . . . . . . . . . 40.2 Complex analysis . . . . . . . . . . . . . . . . . . . . . . . . . 6
1 Review of Fourier analysis 81.1 Fourier transform: definition . . . . . . . . . . . . . . . . . . . 81.2 Fourier transform: properties . . . . . . . . . . . . . . . . . . 81.3 Periodic functions: Fourier series . . . . . . . . . . . . . . . . 91.4 Application: solution of ODEs . . . . . . . . . . . . . . . . . . 10
12 Partial Differential Equations (PDEs) 1212.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.2 1D Wave Equation: Vibrating String . . . . . . . . . . . . . . 1612.3 1D Wave Equation: Separation of Variables, Use of Fourier
Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1812.4 D’Alembert’s Solution of the Wave Equation. Characteristics. 2612.5 Heat Equation: Solution by Fourier Series . . . . . . . . . . . 3112.6 1D Heat Equation: Solution by Fourier Integrals and Transforms 3812.7 2D Wave Equation: Vibrating Membrane . . . . . . . . . . . . 4612.8 Rectangular Membrane. Double Fourier Series . . . . . . . . . 4812.9 Laplacian in Polar Coordinates. Circular Membrane. Fourier-
Bessel Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
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12.10Laplace’s Equation in Cylindrical and Spherical Coordinates.Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
12.11Solution of PDEs by Laplace Transforms . . . . . . . . . . . . 7212.12Application: hydrogen-like atomic orbitals . . . . . . . . . . . 7312.13Application: Unbounded domains and artificial boundaries.
Dirichlet-to-Neumann map. . . . . . . . . . . . . . . . . . . . 7612.14Review of Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . 79
12.14.1 Basic Concepts (12.1) . . . . . . . . . . . . . . . . . . . 7912.14.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 8012.14.3 Separation of Variables (12.3, 12.5, 12.8, 12.9, 12.10) . 8012.14.4 Method of Characteristics (12.4) . . . . . . . . . . . . . 8812.14.5 Fourier Integrals (12.6) . . . . . . . . . . . . . . . . . . 90
13 Complex Numbers and Functions 9213.1 Complex Numbers. Complex Plane. . . . . . . . . . . . . . . . 9213.2 Polar Form of Complex Numbers. Powers and Roots . . . . . 9713.3 Derivative. Holomorphic Function. . . . . . . . . . . . . . . . 10013.4 Cauchy-Riemann Equations. Laplace’s Equation. . . . . . . . 10313.5 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . 10813.6 Trigonometric and Hyperbolic Functions . . . . . . . . . . . . 10913.7 Logarithm. General Power . . . . . . . . . . . . . . . . . . . . 110
14 Complex Integration 11314.1 Line Integral in the Complex Plane . . . . . . . . . . . . . . . 11314.2 Cauchy’s Integral Theorem . . . . . . . . . . . . . . . . . . . . 12114.3 Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . 12514.4 Derivatives of Holomorphic Functions . . . . . . . . . . . . . . 128
15 Power Series, Taylor Series 13215.1 Sequences, Series, Convergence Tests . . . . . . . . . . . . . . 13215.2 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13815.3 Functions Given by Power Series . . . . . . . . . . . . . . . . 14115.4 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . 147
16 Laurent Series. Residue Integration 15416.1 Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 15416.2 Singularities and Zeros. Infinity . . . . . . . . . . . . . . . . . 15916.3 Residue Integration Method . . . . . . . . . . . . . . . . . . . 163
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16.4 Residue Integration of Real Integrals . . . . . . . . . . . . . . 168
17 Conformal Mapping 17417.1 Geometry of Holomorphic Functions: Conformal Mapping . . 17417.2 Linear Fractional Transformations . . . . . . . . . . . . . . . . 17817.4 Conformal Mapping by Other Functions . . . . . . . . . . . . 181
18 Complex Analysis and Potential Theory 18418.1 Electrostatic Fields . . . . . . . . . . . . . . . . . . . . . . . . 18418.2 Use of Conformal Mapping. Modeling . . . . . . . . . . . . . . 18818.3 Heat Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 19118.4 Fluid Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19318.5 Poisson’s Integral Formula for Potentials . . . . . . . . . . . . 19618.6 Review of the Second Part of the Lecture . . . . . . . . . . . . 198
18.6.1 Final Exam . . . . . . . . . . . . . . . . . . . . . . . . 19818.6.2 Complex Numbers and Functions (13) . . . . . . . . . 19918.6.3 Complex Integration (14, 16) . . . . . . . . . . . . . . 20318.6.4 Power Series, Taylor Series, Laurent Series (15, 16) . . 20518.6.5 Conformal Mapping (17) . . . . . . . . . . . . . . . . . 208
0 Overview of the lecture
This is the second half of a one year course on mathematical methods forphysical sciences. The course web site is http://www.unc.edu/~ckirsch/
MATH529, which contains the course syllabus, course material such as theselecture notes, as well as the homework assignments. Please check this website regularly for updates.
The two topics covered in the lecture are
• partial differential equations
• complex analysis
The course is based on the textbook
Erwin Kreyszig: Advanced Engineering Mathematics. 9th Edi-tion; Wiley, 2006
3
We will basically go through chapters 12–18 of this textbook.It is presumed that students are familiar with multivariable calculus.
Fourier series, integrals and transforms will be reviewed as needed. Thislecture will connect naturally to the first half, MATH 528. Students whohave not attended that lecture may notice a larger gap in the first few weeks.
In addition to the weekly homework assignments, there will be two exams,a mid-term exam on March 3 (75 minutes, during regular class time), and thefinal exam on May 3, 8:00 AM (as set by the University Registrar). Classes ofMarch 1 and April 26 will be review sessions in preparation of the respectiveexam. For the mid-term exam, you are allowed to use a summary of thelecture notes on three pages, and six pages for the final exam. Homeworkassignments and exams are graded and each counts for 1/3 of the total coursegrade.
Please ask questions as they occur, and take advantage of the office hours,too!
0.1 Partial differential equations
A differential equation is a mathematical equation for an unknown functionof one or several variables that relates the values of the function itself andits derivatives of various orders. If the unknown function is of one variable,say u(x), we have an ordinary differential equation (ODE) of the form
F (x, u, u′, u′′, . . . ) = 0. (1)
Solution techniques for these equations (series, Laplace transform, . . . ) havebeen discussed in MATH 528.
For an unknown function of several variables, u(x), x = (x1, . . . , xd) ∈ Rd,we have a partial differential equation (PDE) of the form
F (x, u,∇u,Hu, . . . ) = 0, (2)
(with a possibly different function F than in (1), of course) with the partialderivatives
∇u =
∂u∂x1...∂u∂xd
, Hu =
∂2u∂x21
. . . ∂2u∂x1∂xd
.... . .
...∂2u
∂xd∂x1. . . ∂2u
∂x2d
, (3)
4
and so on for higher order derivatives. Often, one of the variables is time, t.PDEs often come up in the mathematical modeling of physical phenomena
in continuous media, such as the propagation of sound, heat conduction,electrostatics, electrodynamics, fluid flow, and elasticity. In this context, weusually deal with (initial-)boundary value problems, where a PDE is to besolved in a region Ω ⊂ Rd (typically d ∈ 1, 2, 3), together with boundaryconditions on ∂Ω and possibly initial conditions at time t = 0.
In this course, we will discuss analytical techniques for solving boundaryvalue problems which involve three important linear, second order PDEs,namely the
• wave equation, utt = ∆u,
• heat equation, ut = ∆u,
• Laplace equation, ∆u = 0,
with the time derivatives
ut :=∂u
∂t, utt :=
∂2u
∂t2, (4)
and with the Laplacian
∆u :=∂2u
∂x21
+ · · ·+ ∂2u
∂x2d
(5)
in d space dimensions. Numerical methods for PDEs, such as the finiteelement, finite volume, and finite difference methods, are covered in othercourses, like MATH 566.
We will present techniques such as the method of separation of variables,which leads to a representation of the solution as a Fourier series.Example: The solution of the one-dimensional initial-boundary value problem
ut = c2uxx, x ∈ (0, L), t > 0, (6)
u(0, t) = u(L, t) = 0, t > 0, (7)
u(x, 0) = f(x), x ∈ [0, L], (8)
is given by the Fourier series
u(x, t) =∞∑n=1
Bn sin(nπxL
)e−λ
2nt, λn :=
cnπ
L, (9)
5
where the coefficients Bn are Fourier sine coefficients of the initial data, f :
Bn =2
L
L∫0
f(y) sin(nπyL
)dy. (10)
Other techniques presented in this course are the method of character-istics, and integral transform methods, in particular Fourier and Laplacetransforms. If the boundary value problem is given in a circular (2D) orspherical (3D) domain, a transformation to polar or spherical coordinates isuseful. Solutions then involve special functions, such as Bessel and Legendrefunctions.
0.2 Complex analysis
We will introduce complex numbers z ∈ C. Every non-constant polynomialin one variable with coefficients in C has a root in C (fundamental theoremof algebra), thus C is the algebraic closure of the field of real numbers, R.Complex numbers may be represented geometrically in the complex plane,z = (x, y), x, y ∈ R. With the imaginary unit, i ∈ C, i2 = −1, we may write
z = x+ iy = r (cosϕ+ i sinϕ) = reiϕ, (11)
withx = Re(z), y = Im(z), r = |z|, ϕ = arg(z). (12)
We will discuss the elementary operations in the field of complex numbers(addition, subtraction, multiplication, division, complex conjugation, . . . ).
For complex functions, f : D → C, D ⊂ C, we will discuss what itmeans to be differentiable, leading to the notion of holomorphic functions.A function
z 7→ f(z) := u(x, y) + iv(x, y), x = Re(z), y = Im(z), (13)
is holomorphic if and only if the partial derivatives of u and v satisfy theCauchy-Riemann equations,
∂u
∂x=
∂v
∂y, (14)
∂u
∂y= −∂v
∂x. (15)
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We will look at line integrals of holomorphic functions, and study their be-havior near singularities.
In potential theory, we study harmonic functions (solutions to Laplace’sequation with continuous second-order derivatives). In two space dimensions,potential theory coincides with the investigation of holomorphic functions,because of the Cauchy-Riemann equations. Thus we establish a link betweencomplex analysis and partial differential equations.
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1 Review of Fourier analysis
Fourier analysis will be an important tool in our analysis of partial differentialequations. Therefore it is reviewed here, although it has already been coveredin MATH 528.
1.1 Fourier transform: definition
The Fourier transform of a function f : Rd → C is defined by
f(k) ≡ F [f ](k) :=1
(2π)d/2
∫Rd
f(x)e−ik·x dx, k ∈ Rd, (16)
if the integral exists (!). The Fourier transform is a function f : Rd → C,and the inverse Fourier transform is defined by
f(x) ≡ F−1[f ](x) :=1
(2π)d/2
∫Rd
f(k)eik·x dk, x ∈ Rd. (17)
Here, the complex number i ∈ C denotes the imaginary unit, i2 = −1.Remark: The letter ξ is also used instead of k to denote the “Fourier space”variable.
1.2 Fourier transform: properties
Important properties of the Fourier transform include the linearity,
F [af + bg] = aF [f ] + bF [g], a, b ∈ C, (18)
and the convolution theorem
F [f ∗ g] = (2π)d/2F [f ]F [g], (19)
where the d-dimensional convolution of the functions f, g : Rd → C is definedby
(f ∗ g)(x) :=
∫Rd
f(x− y)g(y) dy =
∫Rd
f(y)g(x− y) dy, x ∈ Rd. (20)
8
Writing f ≡ F [f ], g ≡ F [g], and using the inverse Fourier transform (17)on both sides of (19), we obtain a different representation of the convolution(20):
(f ∗ g)(x) =
∫Rd
f(k)g(k)eik·x dk, x ∈ Rd. (21)
The formula (21) will help us in solving partial differential equations.We continue in one space dimension (d = 1), where the Fourier transform
f may be interpreted as the spectral density of a function f .Remark: In one space dimension, ω is commonly used instead of k to denotethe “Fourier space” variable.
If the n-th order derivative of the function f ,
f (n) ≡ dnf
dxn, (22)
exists and has a Fourier transform (this requires f to decay sufficiently fastas |x| → ∞), then we have the relation
F [f (n)](ω) = (iω)nF [f ](ω). (23)
The property (23) may be used to transform an ordinary differential equationinto an algebraic equation (similar to the Laplace transform, cf. MATH 528),and will also useful to reduce the number of variables in a partial differentialequation.
1.3 Periodic functions: Fourier series
If f : R→ C is periodic, i. e. if
∃L > 0 : f(x+ L) = f(x) ∀x ∈ R, (24)
then the spectrum f is discrete. This can be seen by comparing the inverseFourier transform of f with its shifted version,
f(x) = F−1[f ](x) =1√2π
∞∫−∞
f(ω)eiωx dω, (25)
f(x+ L) = F−1[f ](x+ L) =1√2π
∞∫−∞
f(ω)eiω(x+L) dω. (26)
9
For the two expressions on the right-hand side of (25), (26) to be equal forall x ∈ R, we require that
f(ω) = 0 ∀ω : eiωL 6= 1 ⇔ f(ω) = 0 ∀ω :ωL
2π6∈ Z. (27)
With (27), the inverse Fourier transform reduces to a series
f(x) =∞∑
n=−∞
cnei 2πnLx, cn =
1
L
L∫0
f(x)e−i2πnLx dx, n ∈ Z. (28)
Using Euler’s formulaeit = cos t+ i sin t, (29)
we may also write the Fourier series using sine and cosine functions:
f(x) = a0 +∞∑n=1
(an cos
(2πn
Lx
)+ bn sin
(2πn
Lx
)), (30)
with
a0 =1
L
L∫0
f(x) dx, an =2
L
L∫0
f(x) cos
(2πn
Lx
)dx, n ≥ 1, (31)
bn =2
L
L∫0
f(x) sin
(2πn
Lx
)dx, n ≥ 1. (32)
Remark: Problem 1 in Problem Set 1.
1.4 Application: solution of ODEs
Consider the forced damped harmonic oscillator (mass-spring-dashpot sys-tem, RLC circuit)
y(t) + 2κy(t) + Ω2y(t) = f(t), Ω, κ > 0, (33)
for the unknown time-variant function y(t). Taking the Fourier transform F(16) on both sides of (33), and using properties (18) and (23), we obtain thealgebraic equation (
−ω2 + 2κiω + Ω2)y(ω) = f(ω). (34)
10
We define the function g : R→ C,
g(ω) :=1√2π
1
−ω2 + 2κiω + Ω2= − 1√
2π
1
(ω − ω1)(ω − ω2), (35)
which has poles atω1,2 = iκ±
√Ω2 − κ2, (36)
and solve (34) for y(ω):
y(ω) =√
2πf(ω)g(ω). (37)
Now we apply the inverse Fourier transform and use the convolution propertyto obtain
y(t) = F−1[y](t)(17)=
1√2π
∞∫−∞
y(ω)eiωt dω(37)=
∞∫−∞
f(ω)g(ω)eiωt dω (38)
(21)= (f ∗ g)(t)
(20)=
∞∫−∞
f(τ)g(t− τ) dτ. (39)
It remains to determine the function g = F−1[g] (which is also called aGreen’s function for this problem):
g(t) = F−1[g](t)(17),(35)
= − 1
2π
∞∫−∞
eiωt
(ω − ω1)(ω − ω2)dω (40)
The integral in (40) may be evaluated as a contour integral in the complexplane, using Cauchy’s Residue Theorem. We will learn about this in theComplex Analysis part of this course. The result is
g(t) =
sin(√
Ω2 − κ2t)
e−κt√Ω2−κ2 , κ < Ω (underdamped)
te−κt, κ = Ω (critically damped)
sinh(√
κ2 − Ω2t)
e−κt√κ2−Ω2 , κ > Ω (overdamped)
. (41)
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Figure 1: Qualitative behavior of a damped harmonic oscillator (33), forκ < Ω (solid line) κ = Ω (dashed line), and κ > Ω (dash-dot line)
12 Partial Differential Equations (PDEs)
As mentioned in the overview, PDEs often come up in the mathematicalmodeling of physical phenomena in continuous media, such as the propaga-tion of sound, heat conduction, electrostatics, electrodynamics, fluid flow,and elasticity. There are several mathematical textbooks on the subject; Iwould recommend
Lawrence C. Evans: Partial Differential Equations. AMS, 1998
to students who like to learn more about the theory of PDEs.
12.1 Basic Concepts
In the Kreyszig textbook, this section is in words only. We shall follow amore formal approach here to describe the general concept. The specialPDEs treated later in the lecture will not take advantage of the notationintroduced here, but it is useful to be acquainted with it.
We consider a function u : Ω → R, where Ω ⊂ Rd is an open subset. Asmentioned in the overview, one of the variables is often time, t. For the partialderivatives of u, we use the multiindex notation: α = (α1, . . . , αd) ∈ Nd
0 is amultiindex of order
|α| = α1 + · · ·+ αd. (42)
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Given a multiindex α ∈ Nd0, we define the partial derivative
Dαu :=∂|α|u
∂xα11 · · · ∂x
αdd
≡ ∂α1x1· · · ∂αdxd u. (43)
Example: d = 3, α = (2, 0, 1), |α| = 3. We obtain the partial derivative
D(2,0,1)u =∂3u
∂x21∂x3
. (44)
If the partial derivative exists, Dαu : Ω → R is again a function. Partialderivatives of order k ∈ N0 (if they exist) are collected in a k-th order tensor,
Dku := Dαu |α ∈ Nd0, |α| = k, k ∈ N0. (45)
The tensor Dku may be represented by a k-dimensional array.Note: For d > 1, there should be no confusion between Dku, k ∈ N0, andDαu, α ∈ Nd
0, because α is a multiindex and k is a scalar.Examples: D2u is a second-order tensor, which may be represented by a d×dmatrix:
D2u =
∂2u∂x21
· · · ∂2u∂x1∂xd
. . .∂2u∂xdx1
· · · ∂2u∂x2d
. (46)
This is the Hessian matrix, Hu, of u. Similarly, D1u ≡ Du may be repre-sented by the gradient vector, ∇u. Finally, D0u ≡ u.
Definition 1 (partial differential equation) For k ∈ N (k 6= 0), an expres-sion of the form
F((Dku)(x), (Dk−1u)(x), . . . , (Du)(x), u(x), x
)= 0, x ∈ Ω, (47)
is called a k-th order partial differential equation (PDE), where
F : Rdk × Rdk−1 × · · · × Rd × R× Ω→ R (48)
is given, and whereu : Ω→ R (49)
is unknown.
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Remarks:
1. x ∈ Ω is called the independent variable in the equation, whereas u :Ω→ R is called the dependent variable (because it depends on x ∈ Ω).
2. A typical mathematical model is given by a PDE (47), together withauxiliary conditions on some part of Γ ⊂ ∂Ω, such as boundary condi-tions or initial conditions (if time is involved). The problem is thencalled an (initial-)boundary value problem ((I)BVP).
3. A (classical) solution u of a k-th order PDE must, of course, have allpartial derivatives up to order k. There is the notion of weak solutionstoo, but we will not go into that in this lecture.
Definition 2 (classification of PDEs)
1. The PDE (47) is called linear if it has the form∑|α|≤k
aα(x) (Dαu) (x) = f(x), (50)
for given functions f and aα, |α| ≤ k. A linear PDE is called homo-geneous if f ≡ 0.
2. The PDE (47) is called quasilinear if it has the form∑|α|=k
aα(x) (Dαu) (x)+a0
((Dk−1u)(x), . . . , (Du)(x), u(x), x
)= 0. (51)
Remarks:
1. A linear PDE depends linearly on u and on all partial derivatives,i. e. the function F in (47) is linear in Dku,Dk−1u, . . . , Du, u, but isnot necessarily linear in x. A linear PDE is homogeneous if each ofits terms contains either u or one of its partial derivatives. Otherwise,(47) is called nonhomogeneous.
2. In a quasilinear PDE, the coefficient of the highest order derivative doesnot depend on the unknown function. Some authors call a PDE of theform (51) semilinear, and they use a weaker condition for quasilinearPDEs.
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Examples of PDEs:
• linear, first order equation:
– linear transport equation: ut + b · ∇u = 0
• linear, second order equations:
– wave equation: utt = ∆u
– heat (diffusion) equation: ut = ∆u
– Laplace’s equation: ∆u = 0
– general wave equation: utt − div (A∇u) + b · ∇u = 0
• linear, higher order equations:
– Airy’s equation: ut + uxxx = 0
– beam equation: ut + uxxxx = 0
• nonlinear equations:
– minimal surface equation: div
(∇u√
1 + |∇u|2
)= 0
– scalar conservation law: ut + div (F (u)) = 0
– scalar reaction-diffusion equation: ut −∆u = f(u)
• linear systems (the unknown u is a vector-valued function):
– evolution equations of linear elasticity:
utt − µ∆u− (λ+ µ)∇ (divu) = 0,
where the vector Laplacian is defined by ∆ := ∇div − curlcurl
– Maxwell’s equations:
Ett = curlB,
Btt = −curlE,
divB = 0,
divE = 0.
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• nonlinear systems:
– reaction-diffusion system:
ut −∆u = f(u),
– Navier-Stokes equations for incompressible, viscous flow:
ut + (u · ∇)u−∆u = −∇p,divu = 0.
and many more!A boundary value problem may not have a solution at all, or it may have
more than one solutions. For a well-posed problem, we require both existenceand uniqueness of the solution, and additionally that the solution dependscontinuously on the data given in the problem (such as boundary conditions).
Theorem 1 (Superposition) If u1, u2 are solutions of a homogeneous linearPDE in some region Ω, then
u = c1u1 + c2u2 (52)
with any constants c1, c2 is also a solution of that PDE in Ω.
Proof: Use the linearity of both the PDE and of the partial derivatives (Prob-lem Set 2).
12.2 1D Wave Equation: Vibrating String
We want to derive a mathematical model of small transverse vibrations ofan elastic string, such as a violin string. The string is assumed to be fixed atthe points x = 0, L. The deflection of the string at point x ∈ [0, L] and timet > 0 is given by u(x, t) [m]. For a fixed t > 0, the function u(·, t) describesthe shape of the string at time t, whereas for a fixed x ∈ [0, L], the functionu(x, ·) describes the vertical motion of this point on the string over time. Wemake the following simplifying physical assumptions (idealization!):
1. The mass per unit length of the string, ρ [kgm−1], is constant (homoge-neous material). The string is perfectly elastic and does not offer anyresistance to bending.
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2. The tension T [N] caused by stretching the string before fastening itat the ends is so large that the action of the gravitational force on thestring can be neglected.
3. The string performs small transverse motions in a vertical plane.
To derive the model, we consider the forces acting on a small portion of thestring (a line element [x, x + ∆x] of length 0 < ∆x 1). Because of themodel assumptions, these will be tensile forces, i. e. tangential to the curveof the string.
After restriction to the vertical plane where the transverse motion takesplane, we denote the tensile forces at the two endpoints of the portion byF 1,F 2 ∈ R2. The resultant F 1 + F 2 is acting on the line element. Becauseof Newton’s second law of motion and due to the assumption that there isno horizontal motion, we have
F 1 + F 2 =
(0
ρ∆x∂2u∂t2
(x, t)
). (53)
We write
F 1 =
(−T1 cosα−T1 sinα
), F 2 =
(T2 cos βT2 sin β
), (54)
and conclude that
T2 cos β − T1 cosα = 0, (55)
T2 sin β − T1 sinα = ρ∆x∂2u
∂t2(x, t). (56)
Thus the magnitude of the horizontal component of each force is constantand must be equal to T : T1 cosα = T2 cos β = T . After division of (56) byT we may write
T2 sin β
T2 cos β− T1 sinα
T1 cosα= tan β − tanα =
ρ
T∆x
∂2u
∂t2(x, t). (57)
Now the tangent of the angles is given by the slope of the curve u(·, t), andcan be written using spatial derivatives of u:
tanα =∂u
∂x(x, t), tan β =
∂u
∂x(x+ ∆x, t). (58)
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Equation (57), after division by ∆x, becomes
1
∆x
(∂u
∂x(x+ ∆x, t)− ∂u
∂x(x, t)
)=ρ
T
∂2u
∂t2(x, t). (59)
As ∆x→ 0, the left-hand side of (59) converges to the second spatial deriva-tive of u evaluated at x, and because our considerations were independentof the horizontal position of the line element, we finally obtain the linear,second-order PDE
∂2u
∂t2= c2∂
2u
∂x2in (0, L), c :=
√T
ρ[ms−1]. (60)
Equation (60) is called the one-dimensional wave equation.
12.3 1D Wave Equation: Separation of Variables, Useof Fourier Series
In the previous section, we have derived the one-dimensional wave equation(60), which governs small transverse vibrations of an elastic string of lengthL > 0. This PDE is given in Ω := (0, L) × (0,∞) ⊂ R2, and according toour Definitions 1 and 2, it is a second-order, linear and homogeneous partialdifferential equation. In particular, Theorem 1 applies to the solutions of(60).
We complete the one-dimensional wave equation (60) by boundary andinitial conditions, so that we obtain the following (well-posed) initial-boundaryvalue problem:
∂2u
∂t2= c2∂
2u
∂x2, x ∈ (0, L), t > 0, c :=
√T
ρ, (61)
u(0, t) = u(L, t) = 0, t > 0, (62)
u(x, 0) = f(x),∂u
∂t(x, 0) = g(x), x ∈ [0, L]. (63)
For compatibility of initial and boundary conditions, we require that f(0) =f(L) = 0. The initial conditions (63) specify the initial deflection, f(x), andthe initial velocity, g(x).
The method presented here to solve the initial-boundary value problem(61)–(63) consists of three steps:
18
1. separation of variables (product method): write the unknown functionas a product of functions with fewer variables. From the PDE, deriveseparate differential equations for each factor.
2. Find solutions of these differential equations which satisfy the auxiliaryconditions.
3. Using the superposition principle (Thm. 1), combine these solutions ina series ( Fourier series) and determine coefficients from the data.
Separation of Variables We are looking for solutions u 6≡ 0 of the one-dimensional wave equation (61) of the form
u(x, t) = F (x)G(t), x ∈ (0, L), t > 0. (64)
If we insert (64) into (61), we obtain
FG = c2F ′′G, x ∈ (0, L), t > 0. (65)
where ˙ denotes time derivatives and where ′ denotes spatial derivatives. Di-vision by c2FG leads to
G
c2G=F ′′
F, x ∈ (0, L), t > 0. (66)
The left-hand side of (66) depends only on t, whereas the right-hand side of(66) depends only on x, thus the variables have been separated. We concludethat both sides must be equal to a constant, the separation constant k ∈ R.Then we obtain the following set of separate ODEs:
F ′′ = kF, x ∈ (0, L), (67)
G = c2kG, t > 0. (68)
Remark: A PDE which can be broken down into a set of separate equations oflower dimensionality by a method of separating variables is called a separablePDE.
Auxiliary Conditions and Solution of Separate ODEs We solve theODE (67) for F first.
19
Proposition 1 The general solution of the ODE (67) is given by
F (x; k) =
Aeωx +Be−ωx, k = ω2, ω > 0Ax+B, k = 0A cos(ωx) +B sin(ωx), k = −ω2, ω > 0
, A,B ∈ R. (69)
From the boundary conditions (62) we infer with (64) that
F (0)G(t) = F (L)G(t) = 0, t > 0. (70)
Because G 6≡ 0 (otherwise u ≡ FG ≡ 0), we obtain the following boundaryconditions for F in (67):
F (0) = F (L) = 0. (71)
Using Proposition 1, we verify that for k ≥ 0, F ≡ 0 is the only solutionof (67) which satisfies the boundary conditions (71). The case k < 0, ork = −ω2, ω > 0, is more interesting: With Proposition 1, we obtain fromthe boundary conditions (71):
A = 0, B sin(ωL) = 0. (72)
F 6≡ 0 requires
B 6= 0(72)⇒ sin(ωL) = 0
ω,L>0⇔ ωL = nπ, n ∈ N. (73)
Setting B = 1 in Proposition 1 (move any constant factor over to the functionG) leads to infinitely many solutions
Fn(x) = sin(nπLx), n ∈ N, (74)
of (67). The value of the separation constant is thus given by
k = −ω2 = −n2π2
L2, n ∈ N. (75)
We use (75) in the ODE (68) for G, which then becomes
G = −λ2nG, t > 0, λn := cω =
cnπ
L> 0, n ∈ N. (76)
20
We use Proposition 1 again, and conclude that the general solutions for Gn
are given by
Gn(t) = Bn cos(λnt) +B∗n sin(λnt), n ∈ N, Bn, B∗n ∈ R. (77)
Using the addition theorem for the cosine function, we may also write
Gn(t) = An cos (λnt+ ϕn) , Bn = An cosϕn, B∗n = −An sinϕn, (78)
for n ∈ N. An and ϕn then have the interpretation of amplitude and phase,respectively. Now, solutions of the one-dimensional wave equation (61) sat-isfying the boundary conditions (62) are given by
un(x, t) = Fn(x)Gn(t) = (Bn cos(λnt) +B∗n sin(λnt)) sin(nπLx), (79)
for n ∈ N. The functions un are called the eigenfunctions of the vibrat-ing string, and the values λn are called the eigenvalues, n ∈ N. The setλ1, λ2, . . . is called the spectrum.Remark: Notice that the functions un in (79) satisfy
∂2un∂t2
= −λ2nun, and
∂2un∂x2
= −n2π2
L2un, n ∈ N. (80)
Compare this with the eigenvalue problem Av = λv, from linear algebra,where solutions (v, λ) consist of eigenvectors and eigenvalues. Instead of amatrix, we now have more general linear operators, ∂2
t and ∂2x. Spectra of
operators are investigated in functional analysis.Each un represents a harmonic motion with frequency λn/(2π) = cn/(2L)
[Hz]. This motion is called the n-th normal mode (or n-th harmonic) of thestring. For n = 1, this is the fundamental mode ( pitch), whereas highermodes (n > 1) are perceived as overtones. The n-th normal mode has n− 1nodes, that is, points on the string that do not move (in addition to the fixedendpoints):
x =k
nL, k = 1, 2, . . . , n− 1, ⇒ sin
(nπLx)
= 0. (81)
The amplitudes of the normal modes determine the timbre of the sound.Tuning of the string means changing its fundamental frequency,
c
2L=
1
2L
√T
ρ[Hz]. (82)
For given values of ρ and L, this is achieved by varying the tension, T , in acertain range.
21
Superposition and Determination of Coefficients With Theorem 1,we conclude that the superposition,
u(x, t) :=∞∑n=1
un(x, t) =∞∑n=1
(Bn cos(λnt) +B∗n sin(λnt)) sin(nπLx), (83)
is also a solution of (61) (again, we moved all scaling constants to the coef-ficients Bn, B
∗n). It is easily verified that the function u in (83) satisfies the
boundary conditions (62). The remaining coefficients Bn, B∗n ∈ R, n ∈ N,
now need to be computed from the initial data in (63). We evaluate u in(83) as t→ 0 to obtain an expression for the initial displacement:
u(x, 0) =∞∑n=1
Bn sin(nπLx)
!= f(x), x ∈ [0, L]. (84)
By differentiation of (83) with respect to t and evaluating as t→ 0, we alsoobtain an expression for the initial velocity:
∂u
∂t(x, 0) =
∞∑n=1
B∗nλn sin(nπLx)
!= g(x), x ∈ [0, L]. (85)
We conclude that Bn, B∗n must be the coefficients of the Fourier sine series
of f and g, respectively:
Bn =2
L
L∫0
f(x) sin(nπLx)dx, (86)
B∗nλn =2
L
L∫0
g(x) sin(nπLx)dx. (87)
Summary The formal solution of the boundary value problem (61)–(63)may be written as a series,
u(x, t) =∞∑n=1
(Bn cos(λnt) +B∗n sin(λnt)) sin(nπLx), x ∈ [0, L], t ≥ 0,
(88)
22
where, for n ∈ N, λn, Bn, and B∗n are given by
λn =cnπ
L, (89)
Bn =2
L
L∫0
f(x) sin(nπLx)dx, (90)
B∗n =2
cnπ
L∫0
g(x) sin(nπLx)dx. (91)
The function u is a classical solution of the boundary value problem (61)–(63), if the series in (88) converges, and if that is also true for all partialderivatives up to order 2 (differentiated termwise). This requires the dataf, g to be sufficiently smooth.Example: We consider a triangular initial deflection and zero initial velocity:
f(x) := 1− 2
L
∣∣∣∣x− L
2
∣∣∣∣ , g(x) := 0, x ∈ [0, L]. (92)
Then we obtain the coefficients
Bn =8
π2n2sin(nπ
2
), B∗n = 0, n ∈ N. (93)
We compute
sin(nπ
2
)=
0, n even(−1)k, n = 2k + 1, k ∈ N0 (⇒ n odd)
. (94)
So we obtain the solution
u(x, t) =8
π2
∞∑k=0
(−1)k
(2k + 1)2cos
((2k + 1)πc
Lt
)sin
((2k + 1)π
Lx
), (95)
for x ∈ [0, L], t ≥ 0. The amplitudes in the alternative representation (78)are given by
An =
0, n even
8π2n2 , n odd
. (96)
Therefore, only odd harmonics are present in this wave, and their amplitudedecays like n−2 as n → ∞. There is a theorem from Fourier analysis which
23
says that the smoother the function, the faster the decay of the Fourier coef-ficients with the index. Therefore, the decay is slower, for example O(n−1),n → ∞, for the sawtooth and square waves (Problem Set 1), which are noteven continuous functions.
For c = L = 1, we plot the formal solution u (series (95) truncated atk = 200) and the amplitude spectrum up to λn = 100π:
24
Fourier series ansatz From what we have seen previously in this section,we may try to write the solution of (61)–(63) as a Fourier series in space withunknown time-variable coefficients. In other words, we make the ansatz
u(x, t) = a0(t) +∞∑n=1
(an(t) cos
(nπLx)
+ bn(t) sin(nπLx))
. (97)
From the boundary conditions (62) we conclude that an ≡ 0, n ≥ 0, so thatthe series becomes
u(x, t) =∞∑n=1
bn(t) sin(nπLx). (98)
With formal differentiation, and with the PDE (61), we obtain
utt(x, t)− c2uxx(x, t) =∞∑n=1
(bn(t) + c2n
2π2
L2bn(t)
)sin(nπLx)
= 0, (99)
for x ∈ (0, L), t > 0. This yields an ODE for each Fourier coefficient bn:
bn + λ2nbn = 0, t > 0, n ≥ 1. (100)
Again with Proposition 1, we obtain the solution
bn(t) = bn(0) cos(λnt) +bn(0)
λnsin(λnt), n ≥ 1. (101)
25
From the initial conditions (63) we obtain, with termwise differentiation,
u(x, 0) =∞∑n=1
bn(0) sin(nπLx)
!= f(x), x ∈ [0, L], (102)
ut(x, 0) =∞∑n=1
bn(0) sin(nπLx)
!= g(x), x ∈ [0, L]. (103)
So the initial values of bn are given as Fourier sine coefficients of the initialdata:
bn(0) =2
L
L∫0
f(x) sin(nπLx)dx, n ≥ 1, (104)
bn(0) =2
L
L∫0
g(x) sin(nπLx)dx, n ≥ 1. (105)
By comparison with the solution (83) derived before, we note that bn(0) =Bn, bn(0) = B∗nλn, as expected.
12.4 D’Alembert’s Solution of the Wave Equation. Char-acteristics.
Recall the series solution of the wave equation (61) with homogeneous Dirich-let boundary conditions (62) (value of u prescribed as 0) and initial data f ,g:
u(x, t) =∞∑n=1
(Bn cos
(nπLct)
+B∗n sin(nπLct))
sin(nπLx), x ∈ [0, L], t ≥ 0,
(106)where, for n ∈ N, λn, Bn, and B∗n are given by
Bn =2
L
L∫0
f(x) sin(nπLx)dx, (107)
B∗n =2
cnπ
L∫0
g(x) sin(nπLx)dx. (108)
26
We assume for simplicity that g ≡ 0, i. e. B∗n = 0, n ∈ N. With the additiontheorem for the sine function, we write
sin(nπLx)
cos(nπLct)
=1
2
(sin(nπL
(x+ ct))
+ sin(nπL
(x− ct)))
, (109)
for n ∈ N. Therefore,
u(x, t) =1
2
∞∑n=1
Bn sin(nπL
(x+ ct))
+1
2
∞∑n=1
Bn sin(nπL
(x− ct))(110)
=1
2(f ∗(x+ ct) + f ∗(x− ct)) , (111)
where f ∗ is defined as the odd periodic extension of the initial deflection fwith period 2L, i. e.
f ∗(x) :=
f(x), x ∈ [0, L)−f(2L− x), x ∈ [L, 2L)
, x ∈ [0, 2L), (112)
and extended periodically. Note that because of the compatibility conditionf(0) = f(L) = 0, the function f ∗ is continuous.Remark: In Problem Set 3, you will verify that the Fourier series of f ∗ isindeed given by
f ∗(x) =∞∑n=1
Bn sin(nπLx), Bn =
2
L
L∫0
f(x) sin(nπLx)dx, n ∈ N.
(113)Equation (111) motivates the ansatz
u(x, t) = ϕ(x+ ct) + ψ(x− ct) (114)
for the solution of the wave equation (61), with two sufficiently smooth func-tions ϕ and ψ (Problem 4 of Problem Set 2). Equation (114) is d’Alembert’ssolution of the wave equation. It states that the solution is the average oftwo waves, u`(x, t) := ϕ(x+ ct) traveling to the left, and ur(x, t) := ψ(x− ct)traveling to the right. From the initial conditions (63) we conclude that
u(x, 0) = ϕ(x) + ψ(x)!
= f(x), (115)
ut(x, 0) = cϕ′(x)− cψ′(x)!
= g(x). (116)
27
Solutions are found to be
ϕ(x) =1
2(ϕ− ψ) (x0) +
1
2f(x) +
1
2c
x∫x0
g(y) dy, (117)
ψ(x) = −1
2(ϕ− ψ) (x0) +
1
2f(x) +
1
2c
x0∫x
g(y) dy. (118)
Then we obtain the following representation of the solution u:
u(x, t) = ϕ(x+ ct) + ψ(x− ct) =1
2(f(x+ ct) + f(x− ct)) +
1
2c
x+ct∫x−ct
g(y) dy.
(119)In the special case g ≡ 0, we conclude from the boundary conditions (62)that f must be an odd, 2L-periodic function (Problem Set 3), and thus weobtain again (111).
Let’s have a closer look at the left and right traveling waves u`,r: appar-ently, they satisfy the first-order (!) transport equations
(u`,r)t ∓ c (u`,r)x = 0, (120)
and thus they are constant along the characteristics
x± ct = const. (121)
The characteristics are solutions x(t) of the characteristic equation
(x+ c)(x− c) = x2 − c2 = 0. (122)
We define the transformed coordinates v(x, t) := x+ ct, w(x, t) := x− ct andwrite
u(x, t) = U(v(x, t), w(x, t)). (123)
By the chain rule, we obtain the following expressions for the partial deriva-tives of u up to second order:
ux = Uvvx + Uwwx, (124)
uxx = Uvvv2x + Uvwvxwx + Uvvxx + Uwvvxwx + Uwww
2x + Uwwxx, (125)
uxt = Uvvvxvt + Uvwvxwt + Uvvxt + Uwvvtwx + Uwwwxwt + Uwwxt,(126)
ut = Uvvt + Uwwt, (127)
utx = Uvvvxvt + Uvwvtwx + Uvvtx + Uwvvxwt + Uwwwxwt + Uwwtx,(128)
utt = Uvvv2t + Uvwvtwt + Uvvtt + Uwvvtwt + Uwww
2t + Uwwtt. (129)
28
With our definitions of v, w, vx ≡ wx ≡ 1, vt = c, wt = −c, these expressionssimplify considerably and the one-dimensional wave equation transforms to
utt − c2uxx = −2c2 (Uvw + Uwv) = 0. (130)
If the second derivatives of u are continuous, we have Uwv ≡ Uvw and weobtain the wave equation in normal form:
Uvw = 0. (131)
Now the PDE (131) may be integrated in two steps to yield
Uv(v, w) = h(v), U(v, w) =
∫h(v) dv︸ ︷︷ ︸=:ϕ(v)
+ψ(w). (132)
Finally, we obtain
u(x, t) = U(v(x, t), w(x, t)) = ϕ(x+ ct) + ψ(x− ct), (133)
which is d’Alembert’s solution, (114).In the previous example for the wave equation, we have guessed the char-
acteristics in the beginning, but the method of characteristics is more general.We shall illustrate it here for second-order quasilinear PDEs in two variables,which are of the form
A(x, y)uxx(x, y) + 2B(x, y)uxy(x, y) + C(x, y)uyy(x, y)
= F (x, y, u(x, y), ux(x, y), uy(x, y)) (134)
(compare with Def. 2), with real-valued functions A, B, C, and F . Here weassume that second derivatives of u are continuous. Using new independentvariables v(x, y), w(x, y) and writing
u(x, y) = U(v(x, y), w(x, y)), (135)
we obtain expressions for the partial derivatives of u as before. Inserting intothe PDE (134), we obtain(
Av2x + 2Bvxvy + Cv2
y
)Uvv +
(Aw2
x + 2Bwxwy + Cw2y
)Uww +
+2 (Avxwx +B(vxwy + vywx) + Cvywy)Uvw = F (v, w, U, Uv, Uw). (136)
29
We now write y as a function of x: y = y(x). Because characteristic curvesare of the form v(x, y) = const. or w(x, y) = const., we know that the totalderivatives must vanish:
dv
dx(x, y(x)) = vx(x, y(x)) + vy(x, y(x))y′(x) = 0, (137)
dw
dx(x, y(x)) = wx(x, y(x)) + wy(x, y(x))y′(x) = 0. (138)
This allows us to eliminate vx and wx in the PDE for U above:(Ay′
2 − 2By′ + C) (v2yUvv + 2vywyUvw + w2
yUww)
= F (v, w, U, Uv, Uw).
(139)Now if y(x) is a solution of the nonlinear ODE (the characteristic equationassociated to PDE (134))
Ay′2 − 2By′ + C = 0, (140)
then we obtain the first-order PDE
F (v, w, U, Uv, Uw) = 0, (141)
to be solved for U .We analyze the characteristic equation (140) in more detail: it is obviously
quadratic in y′, so that we may factorize(y′ − B −
√B2 − ACA
)(y′ − B +
√B2 − ACA
)= 0. (142)
There are three cases to distinguish, depending on the discriminant B2−AC:
1. If B2 −AC > 0, then there are two real-valued solutions y1,2(x) of thecharacteristic ODE, corresponding to two characteristics. The associ-ated PDE (134) is called hyperbolic.Example: wave equation, uxx− c2uyy = 0. We have B2−AC = c2 > 0,and the characteristic ODE is given by (y′ + c)(y′ − c) = 0, as we haveseen earlier.
2. If B2 − AC = 0, then there is one solution y(x) of the characteristicODE, corresponding to one characteristic. The associated PDE (134)is called parabolic.Example: heat equation, −c2uxx = uy. We have A = −c2 < 0 andB = C = 0, so that B2 −AC = 0. The characteristic ODE is given byy′ = 0.
30
3. If B2 − AC < 0, then there are two complex-valued solutions y1,2(x).The associated PDE (134) is called elliptic.Example: Laplace equation, uxx+uyy = 0. We have B2−AC = −1 < 0.
Because A, B, C are functions of x, y in general, the PDE (134) may be ofmixed type, i. e. it may have different type depending on the location in thexy plane.Example: The Euler-Tricomi equation (used in the study of transonic flow)
uxx − xuyy = 0 (143)
is hyperbolic in the halfspace x > 0, parabolic at x = 0 and elliptic in thehalf space x < 0.
12.5 Heat Equation: Solution by Fourier Series
Mathematical model Consider a body of homogeneous material and de-note its temperature by u(x, t) [K], x ∈ Ω ⊂ R3, t ≥ 0. Let V ⊂ Ω denote abounded control volume with surface S := ∂V . The amount of heat trans-ferred per unit time from the control volume to the neighboring material isgiven by the integral of the heat flux q [Wm−2] over the boundary,
∂Q
∂t= −
∫S
q · n ds = −∫V
divq dx, (144)
if no work is done and if there are no heat sources or sinks inside of V . In(144), n denotes the outward unit normal vector field on S. We used thedivergence theorem in the second equation of (144). Quantities on both sidesof (144) have the physical unit of power [W]. Fourier’s law states that theheat flux q is proportional to the negative temperature gradient,
q = −k∇u, (145)
with the thermal conductivity k [Wm−1K−1], i. e. heat flows from warmer tocolder regions. Assuming a constant k, the right-hand side of (144) becomes
−∫V
divq dx =
∫V
k∆u dx (146)
31
For the left-hand side of (144) we write
∂Q
∂t=∂Q
∂u
∂u
∂t=
∫V
cpρ∂u
∂tdx, (147)
with the specific heat capacity cp [Jkg−1K−1] and with the mass density ρ[kgm−3]. Therefore we obtain∫
V
(cpρut − k∆u) dx = 0, (148)
and because the control volume V ⊂ Ω was chosen arbitrarily, we obtain thePDE
ut − c2∆u = 0, in Ω× (0,∞), c2 :=k
cpρ[m2s−1]. (149)
Equation (149) is called the heat equation, with the thermal diffusivity c2. Itis a second-order, linear, parabolic PDE. The modeling approach used here isvery common. You might have seen it already in a lecture on mathematicalmodeling, such as MATH 564.
We use separation of variables to obtain solutions in the one-dimensionalcase. For that purpose, we consider the initial-boundary value problem
ut = c2uxx, in (0, L)× (0,∞), (150)
u(0, t) = u(L, t) = 0, t ≥ 0, (151)
u(x, 0) = f(x), x ∈ [0, L]. (152)
A physical interpretation of the solution is the temperature distribution ina thin bar of length L > 0, which is insulated in the lateral direction, keptat a constant temperature at the endpoints, and has a prescribed initialtemperature distribution.
Separation of Variables Again we write the solution as a product of twofunctions, u(x, t) = F (x)G(t) (64). Then we obtain from (150):
FG = c2F ′′G, in (0, L)× (0,∞), (153)
and after division by c2FG we have separated the variables:
G
c2G=F ′′
F= k ∈ R, (154)
32
because the left-hand side of (154) depends only on t, whereas the right-handside of (154) depends only on x. We obtain the separate ODEs
F ′′ = kF, in (0, L), (155)
G = c2kG, in (0,∞). (156)
Auxiliary Conditions and Solution of Separate ODEs From theboundary conditions (151), we conclude, with the assumption that u 6≡ 0,that F (0) = F (L) = 0. Using Proposition 1, we infer that only the casek < 0 is interesting. As before, we obtain the following solutions for F :
Fn(x) = sin(nπLx), n ∈ N, (157)
and the values of the separation constant are given by −(nπ/L)2, n ∈ N.The ODEs for G then become
G+ λ2nG = 0, λn =
cnπ
L, n ∈ N. (158)
The general solution of (158) is given by
Gn(t) = Bne−λ2nt, n ∈ N, Bn ∈ R. (159)
Then we obtain the eigenfunctions
un(x, t) = Fn(x)Gn(t) = Bn sin(nπLx)e−λ
2nt, n ∈ N, (160)
with associated eigenvalues λn.Remark: Notice that
∂un∂t
= −λ2nun,
∂2un∂x2
= −(λnc
)2
un, n ∈ N. (161)
The n-th eigenfunction has n−1 nodes like for the wave equation. Unlike thenormal modes of a vibrating string, however, these eigenfunctions decreaseexponentially in time, and the rate of decrease increases with n.
33
Superposition and Determination of Coefficients With the superpo-sition principle, Theorem 1, we may write the solution u(x, t) as a Fourierseries,
u(x, t) =∞∑n=1
un(x, t) =∞∑n=1
Bn sin(nπLx)e−λ
2nt. (162)
As t→ 0, we obtain with the initial conditions (152)
u(x, 0) =∞∑n=1
Bn sin(nπLx)
= f(x), (163)
so that the constants Bn must be the Fourier sine coefficients of f :
Bn =2
L
L∫0
f(x) sin(nπLx)dx, n ∈ N. (164)
Summary The formal solution of the boundary value problem (150)–(152)may be written as a series,
u(x, t) =∞∑n=1
un(x, t) =∞∑n=1
Bn sin(nπLx)e−λ
2nt, x ∈ [0, L], t ≥ 0, (165)
where, for n ∈ N, λn and Bn are given by
λn =cnπ
L, (166)
Bn =2
L
L∫0
f(x) sin(nπLx)dx, n ∈ N. (167)
Example: We consider a triangular initial temperature distribution:
f(x) := 1− 2
L
∣∣∣∣x− L
2
∣∣∣∣ . (168)
We already know from the example in Section 12.3 that the coefficients Bn
are given by
Bn =
8π2
(−1)k
(2k+1)2, n = 2k + 1, k ∈ N0
0, n even, (169)
34
so that the solution of (150)–(152) with initial data f is given by
u(x, t) =8
π2
∞∑k=0
(−1)k
(2k + 1)2sin
((2k + 1)π
Lx
)exp
(−c
2(2k + 1)2π2
L2t
),
(170)for x ∈ [0, L], t ≥ 0. We plot the solution for L = c = 1 in the xt plane inthe following figure:
Steady 2D heat problems. Laplace equation. At steady state, we haveut ≡ 0, so that the heat equation (149) simplifies to the Laplace equation,
∆u = 0, in Ω. (171)
In two space dimensions, this yields again a PDE in two variables (noticehow we try to avoid problems with more than two variables):
uxx + uyy = 0, in Ω ⊂ R2. (172)
This second-order, linear, elliptic PDE needs to be completed with boundaryconditions on ∂Ω, so that we obtain a boundary value problem (BVP). Wedistinguish three different types of BVPs:
35
1. In a Dirichlet problem, the value of u is prescribed on ∂Ω. This is calleda Dirichlet boundary condition.
2. In a Neumann problem, the normal derivative of ∂nu is prescribed on∂Ω. This is called a Neumann boundary condition.
3. In a mixed BVP, the value of u is prescribed on ΓD and the value of∂nu is prescribed on ΓN , where ∂Ω = ΓD ∪ ΓN , ΓD ∩ ΓN = ∅. This isa (special case of a) Robin boundary condition.
We consider a rectangle, Ω = (0, a) × (0, b) ⊂ R2, a, b > 0, and Dirichletboundary conditions
u =
f(x), x ∈ [0, a], y = b0, otherwise
, on ∂Ω = ([0, a]× 0, b)∪(0, a × [0, b]) .
(173)We need f(0) = f(a) = 0 for compatibility. Separation of variables, u(x, y) =F (x)G(y), leads to
F ′′
F= −G
′′
G= k ∈ R, (174)
so that we obtain the separate ODEs
F ′′ = kF, in (0, a), (175)
G′′ = −kG, in (0, b). (176)
We solve for F first. From the boundary conditions we conclude that F (0) =F (a) = 0. With Proposition 1 we obtain (once again) the solutions
Fn(x) = sin(nπax), n ∈ N, (177)
and the values of the separation constant are given by −(nπ/a)2 < 0. There-fore, the ODEs for the functions Gn, n ∈ N, are given by
G′′n =(nπa
)2
Gn, in (0, b). (178)
With Proposition 1, we obtain the general solutions
Gn(y) = Anenπay +Bne
−nπay, An, Bn ∈ R, n ∈ N. (179)
36
The boundary condition at y = 0 implies that Gn(0) = An +Bn = 0, n ∈ N,so that we may eliminate the constants Bn, n ∈ N, in (179). The functionsGn are then of the form
Gn(y) = An(enπay − e−
nπay)
= 2An sinh(nπay). (180)
Now the eigenfunctions of our Dirichlet problem are given by
un(x, y) = Fn(x)Gn(y) = A∗n sinh(nπay)
sin(nπax), n ∈ N. (181)
With Theorem 1, we obtain a series representation for the solution u. Eval-uated as y → b this becomes
u(x, b) =∞∑n=1
un(x, b) =∞∑n=1
A∗n sinh(nπab)
sin(nπax)
!= f(x), (182)
because of the boundary conditions (173). We conclude that
A∗n sinh(nπab)
=2
a
a∫0
f(x) sin(nπax)dx. (183)
So we finally obtain the formal solution to the BVP (172), (173):
u(x, y) =∞∑n=1
A∗n sinh(nπay)
sin(nπax), x ∈ [0, a], y ∈ [0, b], (184)
where the coefficients are determined from the boundary data f by
A∗n =2
a sinh(nπab) a∫
0
f(x) sin(nπax)dx, n ∈ N. (185)
Example: We consider triangular boundary data
f(x) := 1− 2
a
∣∣∣x− a
2
∣∣∣ . (186)
From the previous examples we know that
A∗n =
1
sinh(nπa b)8π2
(−1)k
(2k+1)2, n = 2k + 1, k ∈ N0
0, n even, (187)
37
so that the solution is given by
u(x, y) =8
π2
∞∑k=0
(−1)k
(2k + 1)2
sinh(
(2k+1)πa
y)
sinh(
(2k+1)πa
b) sin
((2k + 1)π
ax
). (188)
In the following figure, we plot the solution for a = 1, b = 2:
0
0.5
1
1.5
2 0
0.5
1
0
0.2
0.4
0.6
0.8
1
yx
12.6 1D Heat Equation: Solution by Fourier Integralsand Transforms
In the previous section, we stated the one-dimensional heat equation as amodel for the heat conduction in a thin bar of length L > 0, which is laterallyinsulated. We may now consider the equation for an “infinitely long” bar.This leads to an initial-value (or Cauchy) problem
ut = c2uxx, in R× (0,∞), (189)
u = f, on R× 0. (190)
38
The Cauchy problem (189), (190) is a fairly good model for the heat conduc-tion in a very long, thin bar, where the temperature values on the boundarybecome unimportant.
Separation of Variables We may still use separation of variables to solvethe Cauchy problem (189), (190), so we write
u(x, t) = F (x)G(t), x ∈ R, t > 0. (191)
The separate ODEs are the same as before (155), (156):
F ′′ = kF, in R, (192)
G = c2kG, in (0,∞). (193)
Solution of Separate ODEs Although we do not impose any boundaryconditions on u, we still require a physically meaningful solution, such asthat it remains bounded:
∃M > 0 : |u(x, t)| ≤M, ∀x ∈ R, t > 0. (194)
This implies of course that both F and G must be bounded functions. WithProposition 1, we conclude that the only bounded solution of (192) for k ≥ 0is the zero function F ≡ 0, so that we may again restrict our considerationsto the case of a negative separation constant, k = −ω2, ω > 0. In that case,solutions of (192), (192) are given by
F (x;ω) = A(ω) cos(ωx) +B(ω) sin(ωx), x ∈ R, (195)
G(t;ω) = e−c2ω2t, t > 0. (196)
Note that any scaling factor in G may be moved over to A and B. Now theproduct
u(x, t;ω) := F (x;ω)G(t;ω) = (A(ω) cos(ωx) +B(ω) sin(ωx)) e−c2ω2t (197)
satisfies
ut = FG = −c2ω2FG = −c2ω2u, (198)
c2uxx = c2F ′′G = −c2ω2FG = −c2ω2u, (199)
so that u(·, ·;ω) is a solution of the heat equation (189), for any ω > 0.
39
Superposition Unlike in the case of a bounded domain, the eigenvalues−c2ω2, ω > 0, do not form a countable set here. Therefore, we need asuperposition of the functions u(·, ·;ω) in the form of an integral instead ofa series: the general solution of (189) is given by
u(x, t) =
∞∫0
u(x, t;ω) dω =
∞∫0
(A(ω) cos(ωx) +B(ω) sin(ωx)) e−c2ω2t dω
(200)(any scaling factor in front of u(x, t;ω) may be moved over to A,B).
Determination of Coefficients The functions A and B now need to bedetermined from the initial data f (190). For that purpose, we evaluate (200)as t→ 0:
u(x, 0) =
∞∫0
A(ω) cos(ωx) +B(ω) sin(ωx) dω!
= f(x), x ∈ R. (201)
In complete analogy with Fourier series, we may also show that any functionf may be written as a real Fourier integral of the form
f(x) =
∞∫0
a(ω) cos(ωx) + b(ω) sin(ωx) dω, (202)
where the coefficient functions a, b are given by
a(ω) =1
π
∞∫−∞
f(x) cos(ωx) dx, b(ω) =1
π
∞∫−∞
f(x) sin(ωx) dx, ω > 0.
(203)You will verify this in Problem Set 4. Therefore, we have determined ourcoefficients A,B to be
A(ω) =1
π
∞∫−∞
f(y) cos(ωy) dy, B(ω) =1
π
∞∫−∞
f(y) sin(ωy) dy, ω > 0.
(204)
40
We have chosen the integration variable y instead of x because we want toinsert (204) into (200):
u(x, t) =
∞∫0
1
π
∞∫−∞
f(y) cos(ωy) dy cos(ωx) +1
π
∞∫−∞
f(y) sin(ωy) dy sin(ωx)
e−c2ω2t dω
=1
π
∞∫0
∞∫−∞
f(y) (cos(ωx) cos(ωy) + sin(ωx) sin(ωy)) e−c2ω2t dy dω (205)
=1
π
∞∫0
∞∫−∞
f(y) cos (ω(x− y)) e−c2ω2t dy dω (206)
=
∞∫−∞
f(y)1
π
∞∫0
cos (ω(x− y)) e−c2ω2t dω dy, (207)
where we have used an addition theorem for the cosine function as wellas changed the order of integration. The inner integral of (207) may beevaluated with the formula
∞∫0
e−ax2
cos(bx) dx =1
2
√π
aexp
(− b
2
4a
). (208)
We obtain
1
π
∞∫0
cos (ω(x− y)) e−c2ω2t dω =
1√4πc2t
exp
(−(x− y)2
4c2t
)= K(c2t, x, y),
(209)where the heat kernel K is defined by
K(t, x, y) :=1√4πt
exp
(−(x− y)2
4t
), t > 0, x, y ∈ R. (210)
So we may write the solution to the Cauchy problem (189), (190) in the form
u(x, t) =
∞∫−∞
f(y)K(c2t, x, y) dy =: (Tf)(x, t), (211)
41
where T is an integral transform. Because K depends only on the differencex− y, it is actually a convolution kernel, and we may write (211) in the form
(Tf)(x, t) =
∞∫−∞
f(y)ϕ(x− y, c2t) dy =(f ∗ ϕ(·, c2t)
)(x), (212)
with
ϕ(x, t) :=1√4πt
exp
(−x
2
4t
), x ∈ R, t > 0. (213)
Therefore, the solution u(·, t) to the Cauchy problem (189), (190) is givenat any time t > 0 by the convolution of the initial data with the Gauss-Weierstrass kernel ϕ(·, c2t). Compare the kernel ϕ (213) with the probabil-ity density function of a normal distribution with mean µ = 0 and varianceσ2 = 2t. This kernel is also used in the (generalized) Weierstrass transform,Wt: u(x, t) =Wc2t[f ](x) (t > 0 is interpreted as a parameter in this context).The transformation Wt may also be inverted (this requires complex analy-sis), so that we can reconstruct the initial data from the solution (cf. imagesharpening).Example: We consider a triangular initial distribution
f(x) :=
1− 2|x|, x ∈ [−1/2, 1/2]0, otherwise
, x ∈ R. (214)
The solution u(x, t) of the Cauchy problem (189), (190) may be expressedusing the error function erf, which is defined by
erf(x) :=2√π
x∫0
e−z2
dz, x ∈ R. (215)
42
We obtain
u(x, t) =1√
4πc2t
∞∫−∞
f(y) exp
(−(x− y)2
4c2t
)dy (216)
=1√
4πc2t
1/2∫−1/2
(1− 2|y|) exp
(−(x− y)2
4c2t
)dy (217)
= (x+ 1/2)
(erf
(x+ 1/2
2c√t
)− erf
(x
2c√t
))+ (218)
−(x− 1/2)
(erf
(x
2c√t
)− erf
(x− 1/2
2c√t
))+ (219)
+
√4c2t
πexp
(− x2
4c2t
)(exp
(−x+ 1/4
4c2t
)− 1
)+ (220)
−√
4c2t
πexp
(− x2
4c2t
)(1− exp
(x− 1/4
4c2t
)), (221)
for x ∈ R, t ≥ 0. Notice that although the initial data is compactly sup-ported, f(x) = 0, |x| ≥ 1/2, this is not true for the solution u(x, t), t > 0.In fact, we have |u(x, t)| > 0 ∀x ∈ R, for any t > 0!
43
Fourier transform We consider the spatial Fourier transform of the solu-tion u,
u(ω, t) := F [u(·, t)](ω) =1√2π
∞∫−∞
u(x, t)e−iωx dx, ω ∈ R, t > 0. (222)
We apply the Fourier transform F to both sides of the PDE (189) to obtain
F [ut(·, t)] = c2F [uxx(·, t)], in R, t > 0. (223)
For the terms in (223) we obtain
F [ut(·, t)](ω) =1√2π
∞∫−∞
ut(x, t)e−iωx dx = ut(ω, t), (224)
F [uxx(·, t)](ω) =1√2π
∞∫−∞
uxx(x, t)e−iωx dx = −ω2u(ω, t), (225)
if we assume sufficiently fast decay of u(x, t) and ux(x, t) as |x| → ∞. Ap-plying the Fourier transform to the initial condition (190), we obtain
F [u(·, 0)] = F [f ], in R. (226)
We have
F [u(·, 0)](ω) = u(ω, 0), (227)
F [f ](ω) = f(ω), (228)
where f denotes the Fourier transform of the initial data f . Now we havetransformed the Cauchy problem (189), (190) into a new initial-value problemfor u which involves only time derivatives:
ut = −c2ω2u, in R× (0,∞), (229)
u = f , on R× 0. (230)
The general solution of (229) is given by
u(ω, t) = C(ω)e−c2ω2t, ω ∈ R, t > 0, (231)
44
and from the initial condition (230) we obtain for the coefficient C:
u(ω, 0) = C(ω) = f(ω), ω ∈ R. (232)
Now the solution u of the Cauchy problem (189), (190) is given by the inversespatial Fourier transform of u:
u(x, t) = F−1[u(·, t)](x) =1√2π
∞∫−∞
f(ω)e−c2ω2teiωx dω (233)
We would like to use the convolution formula (21), and write
u(x, t) =
∞∫−∞
f(ω)g(ω, t)eiωx dω, g(ω, t) :=1√2πe−c
2ω2t. (234)
The Fourier transform of a Gaussian function is another Gaussian function,and we obtain
F−1[g(·, t)](x) =1√
4πc2texp
(− x2
4c2t
)= ϕ(x, c2t), (235)
with the Gauss-Weierstrass kernel ϕ (213). With the convolution formula(21), we obtain again the form (212) for the solution of the Cauchy problem(189), (190). We have used the Fourier transform method already in Sec-tion 1.4 (review part of this lecture) to solve an ODE. As we have seen there,the method also works for nonhomogeneous equations.
A related method is to assume that the solution u may be represented asan inverse Fourier transform,
u(x, t) =1√2π
∞∫−∞
u(ω, t)eiωx dω, (236)
and to insert this ansatz into the problem (189), (190). This will lead to anew problem (229), (230) for u, which is hopefully easier to solve.
Notice that for problems given in bounded domains, the integral in (236)reduces to a series. That leads to the Fourier series ansatz of the solution,which we have used several times in the previous sections. You have solvedseveral initial-boundary value problems already with this ansatz in the Prob-lem Sets, some of them involving nonhomogeneous PDEs.
45
12.7 2D Wave Equation: Vibrating Membrane
In a similar way as for the vibrating string (Section 12.2) we now wantto derive a mathematical model of small transverse vibrations of an elasticmembrane, such as a drumhead. The membrane at rest covers a boundeddomain Ω ⊂ R2 and is fixed on the boundary ∂Ω. The deflection of themembrane at point (x, y) ∈ Ω and time t > 0 is given by u(x, y, t) [m]. For afixed t > 0, the function u(·, t) describes the shape of the membrane at timet, whereas for a fixed (x, y) ∈ Ω, the function u(x, y, ·) describes the verticalmotion of this point on the membrane over time. We make the followingsimplifying physical assumptions (idealization!):
1. The mass per unit area of the membrane, ρ [kgm−2] is constant (ho-mogeneous membrane). The membrane is perfectly flexible and offersno resistance to bending.
2. The membrane is stretched and then fixed along its entire boundary∂Ω. The tension per unit length T [Nm−1] caused by stretching themembrane is the same at all points and in all directions and does notchange during the motion. The tension is so large that the action ofthe gravitational force on the membrane can be neglected.
3. The deflection u(x, y, t) of the membrane during the motion is smallcompared to the size of the membrane, and all angles of inclination aresmall.
To derive the model, we consider the forces acting on a small portion of themembrane (an area element [x, x + ∆x] × [y, y + ∆y] of area ∆x∆y, with0 < ∆x,∆y 1). Because of the model assumptions, these will be tensileforces, i. e. tangential to the membrane.
We consider the four forces acting on the edges of the area element, F 1,2x ,
F 1,2y . With the assumption of small deflections we conclude that the magni-
tudes of these forces are given by
|F 1,2x | = T∆y, |F 1,2
y | = T∆x. (237)
We write F 1,2y in polar coordinates in the yz plane:
F 1y =
0−T∆x cosα−T∆x sinα
, F 2y =
0T∆x cos βT∆x sin β
. (238)
46
The angles of inclination α, β are small by assumption, so that cosα 'cos β ' 1. The y-component of the resulting force in y-direction, F 1
y + F 2y,
is thus negligible and we have
T∆x cosα ' T∆x cos β ' T∆x. (239)
The z-component of the resulting force in y-direction is given by
T∆x (sin β − sinα) ' T∆x (tan β − tanα) (240)
= T∆x (uy(x2, y + ∆y, t)− uy(x1, y, t)) , (241)
for any values x1, x2 ∈ (x, x+ ∆x). In the same way, we write F 1,2x in polar
coordinates in the xz plane and obtain that the x-component of the resultingforce in x-direction, F 1
x + F 2x, is negligible and that the z-component of the
resulting force in x-direction is given by
T∆y (ux(x+ ∆x, y2, t)− ux(x, y1, t)) , (242)
for any values y1, y2 ∈ (y, y+∆y). The only non-zero component of the totalresulting force F 1
x + F 2x + F 1
y + F 2y is thus the z-component. It is given by
T∆x∆y
(ux(x+ ∆x, y2, t)− ux(x, y1, t)
∆x+uy(x2, y + ∆y, t)− uy(x1, y, t)
∆y
).
(243)By Newton’s second law of motion, (243) must be equal to the mass of thearea element times its vertical acceleration,
ρ∆x∆y∂2u
∂t2(x, y, t). (244)
If we let ∆x,∆y → 0, (243) converges to the sum of the second partialderivatives, so that we obtain
∂2u
∂t2(x, y, t) =
T
ρ
(∂2u
∂x2(x, y, t) +
∂2u
∂y2(x, y, t)
). (245)
The right-hand side of (245) may be written as the 2D Laplacian applied tou, and because the location of the area element within Ω was arbitrary, wefinally obtain the two-dimensional wave equation
∂2u
∂t2= c2∆u in Ω× (0,∞), c :=
√T
ρ[ms−1]. (246)
47
12.8 Rectangular Membrane. Double Fourier Series
In the previous section, we have derived the two-dimensional wave equation(246), which governs small vibrations of an elastic membrane. This PDEis given in Ω × (0,∞) ⊂ R3, and according to our Definitions 1 and 2, itis a second-order, linear and homogeneous partial differential equation. Inparticular, Theorem 1 applies to the solutions of (246), which will allow usto write formal solutions as superpositions of eigenfunctions, just as we didfor the one-dimensional PDEs in the previous sections.
We proceed as in Section 12.3: We complete the two-dimensional waveequation (246) by boundary and initial conditions, so that we obtain thefollowing initial-boundary value problem:
∂2u
∂t2= c2∆u, in Ω× (0,∞), c :=
√T
ρ, (247)
u = 0, on ∂Ω× (0,∞), (248)
u = f,∂u
∂t= g, on Ω× 0, (249)
with initial data f, g : Ω → R. For compatibility of initial and boundaryconditions, we require that f = 0 on ∂Ω. The initial conditions (249) specifythe initial deflection f and the initial velocity g.
The method presented here to solve the initial-boundary value problem(247)–(249) consists of three steps:
1. separation of variables (product method): write the unknown functionas a product of functions with fewer variables. From the PDE, deriveseparate differential equations for each factor.
2. Find solutions of these differential equations which satisfy the boundaryconditions.
3. Using the superposition principle (Thm. 1), combine these solutions ina series ( Fourier series) and determine coefficients from the initialdata.
Separation of Variables We are looking for solutions u 6≡ 0 of the two-dimensional wave equation (247) of the form
u(x, y, t) = F (x, y)G(t), (x, y) ∈ Ω, t > 0. (250)
48
If we insert (250) into (247), we obtain
FG = c2 (Fxx + Fyy)G, in Ω× (0,∞), (251)
and after division by c2FG we have separated the time variable from thespatial variables:
G
c2G=Fxx + Fyy
F= k ∈ R. (252)
The separate differential equations are thus given by
Fxx + Fyy = kF, in Ω, (253)
G = c2kG, in (0,∞). (254)
Notice that the differential equation for F (253) is still a PDE. For a negativek = −λ2, (253) becomes a two-dimensional Helmholtz equation.
We consider a rectangular membrane, Ω := (0, a) × (0, b), with a, b > 0.Then we use the method of separating variables again and write F (x, y) =H(x)Q(y), (x, y) ∈ Ω. We obtain
H ′′Q+HQ′′ − kHQ = 0, (255)
which we can separate after division by HQ as
H ′′
H= −Q
′′
Q+ k = p ∈ R. (256)
The separate ODEs for H and Q are then given by
H ′′ = pH, in (0, a), (257)
Q′′ = (k − p)Q, in (0, b). (258)
We summarize our findings: after separation of variables we obtain threeseparate ODEs with two (!) separation constants k, p ∈ R:
H ′′ = pH, in (0, a), (259)
Q′′ = (k − p)Q, in (0, b), (260)
G = c2kG, in (0,∞). (261)
After solutions forH, Q, G are found, the function u(x, y, t) := H(x)Q(y)G(t)is a solution to the two-dimensional wave equation (247).
49
Auxiliary Conditions and Solution of Separate ODEs From theboundary condition (248) we conclude that
H(0) = H(a) = Q(0) = Q(b) = 0, (262)
for a rectangular domain Ω = (0, a)× (0, b). With Proposition 1 we find thatsolutions H,Q 6≡ 0 exist for p < 0 and k − p < 0 only. For p = −ω2, ω > 0,we have the general solution
H(x) = A cos(ωx) +B sin(ωx), (263)
and with the boundary conditions, we obtain
A = 0, B sin(ωa) = 0 ⇒ B = 0 ∀ω 6= mπ
a, m ∈ N. (264)
Therefore, solutions H 6≡ 0 are given by
Hm(x) = sin(mπax), m ∈ N, (265)
where any scaling factor has been moved to the function G. For k−p = −ν2,ν > 0, and with the same arguments as before, we have
Qn(y) = sin(nπby), n ∈ N. (266)
Notice that the separation constant k is given by k = p−ν2 = − (ω2 + ν2) <0. The functions Fmn := HmQn, m,n ∈ N,
Fmn(x, y) = sin(mπax)
sin(nπby)
(267)
are solutions of two-dimensional Helmholtz equations:
∆Fmn + π2
(m2
a2+n2
b2
)Fmn = 0. (268)
The ODEs for the time-dependent functions Gmn are given by
Gmn = −λ2mnGmn, in (0,∞), m, n ∈ N, (269)
with the eigenvalues
λmn := c√ω2 + ν2 = cπ
√m2
a2+n2
b2, m, n ∈ N, (270)
50
so that the general solutions are given by (Proposition 1)
Gmn(t) = Bmn cos(λmnt) +B∗mn sin(λmnt), Bmn, B∗mn ∈ R, (271)
for m,n ∈ N. The functions umn := FmnGmn,
umn(x, y, t) = (Bmn cos(λmnt) +B∗mn sin(λmnt)) sin(mπax)
sin(nπby),
(272)m,n ∈ N, are eigenfunctions of the second-order differential operators inboth space and time:
∂2t umn = FmnGmn = −λ2
mnFmnGmn = −λ2mnumn, (273)
c2∆umn = c2∆FmnGmn = −λ2mnFmnGmn = −λ2
mnumn, (274)
and therefore they are solutions of the two-dimensional wave equation (247)which satisfy the boundary condition (248). The frequency of umn is
λmn2π
=c
2
√m2
a2+n2
b2=c√m2b2 + n2a2
2ab[Hz], m, n ∈ N. (275)
Compare this value with the frequency cn/(2L) of the n-th normal mode ofa vibrating string (Section 12.3). In the 2D case, it is possible that there areseveral modes with the same frequency, depending on the values of a and b!Example: Consider a square membrane of area 1, i. e. a = b = 1. We obtainthe modes
umn(x, y, t) = (Bmn cos(λmnt) +B∗mn sin(λmnt)) sin (mπx) sin (nπy) , (276)
with eigenvaluesλmn = cπ
√m2 + n2, (277)
for m,n ∈ N. Because λmn = λnm, The modes umn and unm have the samefrequencies. However, these are two different functions if m 6= n. This canbe seen from the location of the nodal lines, i. e. points (x, y) ∈ Ω withFmn(x, y) = sin (mπx) sin (nπy) = 0:
x =k
m, k = 1, . . . ,m− 1 or y =
`
n, ` = 1, . . . , n− 1. (278)
For example, the modes u12 and u21 both have the same frequency c√
5/2.Their nodal lines are located at y = 1/2 and x = 1/2, respectively. Any
51
linear combination of u12 and u21 is again a solution of the two-dimensionalwave equation (247) which satisfies the boundary condition (248), and it hasthe same frequency c
√5/2. For example, by choosing B12 = 1, B21 = p ∈ R,
B∗12 = B∗21 = 0, we obtain the vibration
cos(cπ√
5t) (F12(x, y) + pF21(x, y)) , p ∈ R. (279)
The nodal lines coincide with the zeros of F12 +pF21, and so they are definedby the equation
sin(πx) sin(2πy) + p sin(2πx) sin(πy) = 0. (280)
We find that solutions (x, y) of (280) satisfy
cos(πy) + p cos(πx) = 0. (281)
For every value of p ∈ R, we obtain different nodal lines.
52
There may also be more than two pairs (m,n) which lead to vibrations ofthe same frequency. It is not trivial to determine how many modes existfor a given frequency. In the following figure, we plot the lattice points(m,n) ∈ 1, . . . , 102 together with the level sets of the function x2 + y2.Each quarter circle corresponds to one frequency, and lattice points whichlie on the same curve correspond to vibration modes which have the samefrequency. Notice that there are curves with three or four lattice points onthem.
53
In the following table, we write all possible values of m2 + n2, m,n ∈1, . . . , 102 together with the pairs (m,n) which correspond to these val-ues.
54
m2 + n2 2 5 8 10 13 17 18 20 25(m,n) (1, 1) (1, 2) (2, 2) (1, 3) (2, 3) (1, 4) (3, 3) (2, 4) (3, 4)(m,n) (2, 1) (3, 1) (3, 2) (4, 1) (4, 2) (4, 3)
m2 + n2 26 29 32 34 37 40 41 45 50(m,n) (1, 5) (2, 5) (4, 4) (3, 5) (1, 6) (2, 6) (4, 5) (3, 6) (1, 7)(m,n) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) (5, 4) (6, 3) (5, 5)(m,n) (7, 1)
m2 + n2 52 53 58 61 65 68 72 73 74(m,n) (4, 6) (2, 7) (3, 7) (5, 6) (1, 8) (2, 8) (3, 8) (6, 6) (5, 7)(m,n) (6, 4) (7, 2) (7, 3) (6, 5) (4, 7) (8, 2) (8, 3) (7, 5)(m,n) (7, 4)(m,n) (8, 1)
m2 + n2 80 82 85 89 90 97 98 100 101(m,n) (4, 8) (1, 9) (2, 9) (5, 8) (3, 9) (4, 9) (7, 7) (6, 8) (1, 10)(m,n) (8, 4) (9, 1) (6, 7) (8, 5) (9, 3) (9, 4) (8, 6) (10, 1)(m,n) (7, 6)(m,n) (9, 2)
m2 + n2 104 106 109 113 116 117 125 128 130(m,n) (2, 10) (5, 9) (3, 10) (7, 8) (4, 10) (6, 9) (5, 10) (8, 8) (7, 9)(m,n) (10, 2) (9, 5) (10, 3) (8, 7) (10, 4) (9, 6) (10, 5) (9, 7)
m2 + n2 136 145 149 162 164 181 200(m,n) (6, 10) (8, 9) (7, 10) (9, 9) (8, 10) (9, 10) (10, 10)(m,n) (10, 6) (9, 8) (10, 7) (10, 8) (10, 9)
Also notice that the frequency ratios
λmnλ11
=
√m2 + n2
2, m, n ∈ N, (282)
are not integer values in general: the sound of most drums is characterizedby strongly inharmonic spectra.
Superposition and Determination of Coefficients The superpositionof the modes umn leads to the representation of the formal solution of (247),
55
(248) as a double Fourier series:
u(x, y, t) =∞∑
m,n=1
umn(x, y, t) (283)
=∞∑
m,n=1
(Bmn cos(λmnt) +B∗mn sin(λmnt)) sin(mπax)
sin(nπby).
Evaluated as t→ 0 we obtain
u(x, y, 0) =∞∑m=1
∞∑n=1
Bmn sin(nπby)
sin(mπax)
(284)
=∞∑m=1
bm(y) sin(mπax)
!= f(x, y), (285)
ut(x, y, 0) =∞∑m=1
∞∑n=1
B∗mnλmn sin(nπby)
sin(mπax)
(286)
=∞∑m=1
b∗m(y) sin(mπax)
!= g(x, y), (287)
because of the initial conditions (249). We solve for the coefficients Bmn intwo steps: first, we find that bm(y), m ∈ N, are the Fourier sine coefficientsof f(·, y), for any y ∈ [0, b]:
bm(y) =∞∑n=1
Bmn sin(nπby)
=2
a
a∫0
f(x, y) sin(mπax)dx, (288)
and second we observe that Bmn, m,n ∈ N, are the Fourier sine coefficientsof bm, m ∈ N:
Bmn =2
b
b∫0
bm(y) sin(nπby)dy =
2
b
b∫0
2
a
a∫0
f(x, y) sin(mπax)dx sin
(nπby)dy,
(289)so that the coefficients Bmn are given by
Bmn =4
ab
a∫0
b∫0
f(x, y) sin(mπax)
sin(nπby)dy dx, m, n ∈ N. (290)
56
In exactly the same way, we obtain for the coefficients B∗mn:
B∗mn =4
abλmn
a∫0
b∫0
g(x, y) sin(mπax)
sin(nπby)dy dx, m, n ∈ N. (291)
Summary The formal solution of the initial-boundary value problem (247)–(249) with a rectangular domain Ω = (0, a)× (0, b), a, b > 0, may be writtenas a double Fourier series,
u(x, y, t) =∞∑
m,n=1
(Bmn cos(λmnt) +B∗mn sin(λmnt)) sin(mπax)
sin(nπby),
(292)where, for m,n ∈ N, λmn, Bmn, and B∗mn are given by
λmn = cπ
√m2
a2+n2
b2, (293)
Bmn =4
ab
a∫0
b∫0
f(x, y) sin(mπax)
sin(nπby)dy dx, (294)
B∗mn =4
abλmn
a∫0
b∫0
g(x, y) sin(mπax)
sin(nπby)dy dx. (295)
Example: We consider a rectangular membrane with a = 4, b = 2, propaga-
tion speed c =√
5, and initial data g ≡ 0 and
f(x, y) = 0.1(4x− x2)(2y − y2). (296)
We have B∗mn = 0, m,n ∈ N, and for Bmn we obtain
Bmn =1
20
4∫0
(4x− x2) sin(mπ
4x)dx
2∫0
(2y − y2) sin(nπ
2y)dy. (297)
The integrals over x and y may be evaluated using integration by parts. We
57
obtain
4∫0
(4x− x2) sin(mπ
4x)dx =
8
mπ
4∫0
(2− x) cos(mπ
4x)dx (298)
=32
m2π2
4∫0
sin(mπ
4x)dx (299)
=128
m3π3(1− (−1)m) (300)
=
256m3π3 , m odd0, m even
, (301)
and
2∫0
(2y − y2) sin(nπ
2y)dy =
4
nπ
2∫0
(1− y) cos(nπ
2y)dy (302)
=8
n2π2
2∫0
sin(nπ
2y)dy (303)
=16
n3π3(1− (−1)n) (304)
=
32n3π3 , n odd0, n even
. (305)
Therefore we obtain
Bmn =
2048
5m3n3π6 , m, n odd0, otherwise
, m, n ∈ N, (306)
and the solution of (247)–(249) is given by
u(x, y, t) =2048
5π6
∞∑m,n=1m,n odd
1
m3n3cos
(√5π
4
√m2 + 4n2t
)sin(mπ
4x)
sin(nπ
2y),
(307)for x, y ∈ [0, 4]× [0, 2], t ≥ 0.
58
12.9 Laplacian in Polar Coordinates. Circular Mem-brane. Fourier-Bessel Series.
Laplacian in Polar Coordinates We consider polar coordinates (r, ϕ),which are related to the cartesian coordinates (x, y) via
x = r cosϕ, y = r sinϕ, r ≥ 0, ϕ ∈ [0, 2π). (308)
These relations can be inverted and we obtain
r =√x2 + y2, ϕ = arctan
(yx
), (x, y) ∈ R2. (309)
We write u(x, y) = U(r(x, y), ϕ(x, y)). With the chain rule we can expressthe partial derivatives of u using the partial derivatives of U as
ux = Urrx + Uϕϕx, (310)
uxx = Urrr2x + Urϕrxϕx + Urrxx + Uϕrϕxrx + Uϕϕϕ
2x + Uϕϕxx, (311)
uy = Urry + Uϕϕy, (312)
uyy = Urrr2y + Urϕryϕy + Urryy + Uϕrϕyry + Uϕϕϕ
2y + Uϕϕyy. (313)
With Urϕ ≡ Uϕr, ∆u may be written as
uxx + uyy = (r2x + r2
y)Urr + 2(rxϕx + ryϕy)Urϕ + (ϕ2x + ϕ2
y)Uϕϕ +
+(rxx + ryy)Ur + (ϕxx + ϕyy)Uϕ. (314)
The partial derivatives of r(x, y) and ϕ(x, y) are given by
rx =x
r, rxx =
y2
r3, ry =
y
r, ryy =
x2
r3, (315)
ϕx = − y
r2, ϕxx =
2xy
r4, ϕy =
x
r2, ϕyy = −2xy
r4, (316)
where we have used the quotient rule and
arctan′(x) =1
1 + x2, x ∈ R. (317)
So we obtain
uxx + uyy = Urr +1
rUr +
1
r2Uϕϕ. (318)
59
Therefore, the Laplacian in polar coordinates is given by
∆ =∂2
∂r2+
1
r
∂
∂r+
1
r2
∂2
∂ϕ2. (319)
For a circular membrane with radius R > 0,
Ω := (r cosϕ, r sinϕ) ∈ R2 | r ∈ (0, R), ϕ ∈ [0, 2π), (320)
we obtain the following Dirichlet boundary value problem for u = u(r, ϕ, t):
utt = c2
(urr +
1
rur +
1
r2uϕϕ
), in Ω× (0,∞), (321)
u = 0, on ∂Ω× (0,∞), (322)
u = f, ut = g, on Ω× 0. (323)
Remark: Notice that in (r, ϕ)-coordinates, Ω is again a rectangle.
Separation of Variables We look for a solution of the form
u(r, ϕ, t) = F (r, ϕ)G(t), in Ω× (0,∞). (324)
We already know the first separation step from Section 12.8:
∆F = Frr +1
rFr +
1
r2Fϕϕ = −k2F, in Ω, (325)
G = −c2k2G, in (0,∞), (326)
with k > 0, where the Laplacian ∆ is defined by (319).Remark: Notice that we restrict ourselves to the case of a negative separationconstant −k2 here. This will turn out to be the right choice.We separate the spatial PDE by assuming F (r, ϕ) = H(r)Q(ϕ), which leadsto
r2H′′
H+ r
H ′
H+ k2r2 = −Q
′′
Q= ν2, (327)
for ν ≥ 0.Remark: Notice the restriction to a non-negative separation constant ν2.The separate ODEs for H and Q are given by
r2H ′′ + rH ′ + k2r2H = ν2H, in (0, R), (328)
Q′′ = −ν2Q, in [0, 2π). (329)
60
Auxiliary Conditions and Solution of Separate ODEs The angularfunctions Q must be continuous and 2π-periodic, and by Proposition 1, suchsolutions are found for ν = n ∈ N0 (this also justifies our choice of a non-negative second separation constant):
Q1n(ϕ) = cos(nϕ), Q2
n(ϕ) = sin(nϕ), n ∈ N0. (330)
For the radial equation (328), we use the transformation z := kr and writew(z) := H(r). Then the ODE (328) transforms to
z2w′′ + zw′ +(z2 − ν2
)w = 0, z > 0. (331)
Equation (331) is Bessel’s differential equation with parameter ν > 0. Forinteger values of the parameter, ν = n ∈ N0, a fundamental system of solu-tions is given by the Bessel functions of the first and second kind, Jn, Yn,n ∈ N0. Only the Bessel functions of the first kind are bounded as z → 0.
The boundary conditions (322) require that Jn(kR) = 0, n ∈ N0. We obtaininfinitely many solutions which involve the values kmn > 0, m ∈ N, n ∈ N0.They satisfy
Jn(kmnR) = 0. (332)
Notice that the values kmn need to be approximated numerically (or foundin a table) in practice.Remark: If we choose the first separation constant to be positive, we findthe modified Bessel functions of the first and second kind, In, Kn, n ∈ N0.
61
Only the functions In are bounded as z → 0. But they do not have anypositive zeros, so they cannot satisfy the boundary condition at r = R. Thecase of a zero separation constant is left as an exercise (Problem Set 5).We obtain the following solutions for the radial equation (328):
Hmn(r) = Jn(kmnr), m ∈ N, n ∈ N0. (333)
Therefore, Hmn is the n-th order Bessel function of the first kind, scaled suchthat its m-th zero is located at r = R, m ∈ N, n ∈ N0. This implies thatHmn has m−1 zeros inside the domain r < R. We illustrate this for n = 0, 1,m = 1, . . . , 4, R = 10, in the following figures:
62
The functions F 1,2mn := HmnQ
1,2n are solutions of the two-dimensional Helmholtz
equation in polar coordinates (325) which vanish at r = R:
∆F 1,2mn + k2
mnF1,2mn = 0, m ∈ N, n ∈ N0. (334)
With λmn := ckmn, the ODE in time (326) becomes
G+ λ2mnG = 0, (335)
with solutions
Gmn(t) = Amn cos(λmnt) +Bmn sin(λmnt), m ∈ N, n ∈ N0. (336)
We obtain the following eigenfunctions for the Dirichlet boundary value prob-lem (321)–(323):
umn(r, ϕ, t) = Gmn(t)F 1mn(r, ϕ) = Gmn(t)Jn(kmnr) cos(nϕ), (337)
u∗mn(r, ϕ, t) = G∗mn(t)F 2mn(r, ϕ) = G∗mn(t)Jn(kmnr) sin(nϕ), (338)
for m ∈ N, n ∈ N0, with frequencies λmn/(2π) and constants Amn, Bmn ∈ Rand A∗mn, B
∗mn ∈ R, respectively.
63
Superposition and Determination of Coefficients We use the super-position principle (Thm. 1) to obtain an expression for the general solutionu(r, ϕ, t). At t = 0, the solution has the form of a Fourier series in ϕ:
u(r, ϕ, 0) = a0(r) +∞∑n=1
(an(r) cos(nϕ) + a∗n(r) sin(nϕ))!
= f(r, ϕ), (339)
where the coefficients a0 and an, a∗n, n ∈ N, are given by Fourier-Bessel series
in r:
a0(r) =∞∑m=1
Am0J0(km0r), (340)
an(r) =∞∑m=1
AmnJn(kmnr), a∗n(r) =∞∑m=1
A∗mnJn(kmnr). (341)
As in Section 12.8, we determine the unknown coefficients Am0 and Amn, A∗mn,
m,n ∈ N, in two steps. First, the coefficients a0, an, a∗n, n ∈ N, are deter-
mined as the Fourier coefficients of f(r, ·), for any r ∈ [0, R]:
a0(r) =1
2π
2π∫0
f(r, ϕ) dϕ, an(r) =1
π
2π∫0
f(r, ϕ) cos(nϕ) dϕ, n ∈ N,
a∗n(r) =1
π
2π∫0
f(r, ϕ) sin(nϕ) dϕ, n ∈ N. (342)
Second, the coefficients Am0 and Amn, A∗mn, m ∈ N, are the Fourier-Bessel
coefficients of a0, an, a∗n, n ∈ N, respectively.
Remark: The working principle behind the classical Fourier series are theorthogonality relations for the functions sin(nx) and cos(nx), n ∈ N0:
2π∫0
cos(mx) cos(nx) dx =
0, m 6= n,π, m = n ≥ 12π, m = n = 0
, m, n ∈ N0, (343)
2π∫0
sin(mx) sin(nx) dx =
0, m 6= n,π, m = n ≥ 10, m = n = 0
, m, n ∈ N0, (344)
2π∫0
sin(mx) cos(nx) dx = 0, m, n ∈ N0. (345)
64
Fourier-Bessel coefficients are found in the same way as the classical Fouriercoefficients, but by using the following orthogonality relation for the Besselfunctions instead:
R∫0
rJn(k`nr)Jn(kmnr) dr =R2
2Jn+1(kmnR)2δ`m, n ∈ N0, (346)
with the Kronecker delta
δ`m :=
1, ` = m0, ` 6= m
. (347)
12.10 Laplace’s Equation in Cylindrical and SphericalCoordinates. Potential
Consider two point masses located at x0 ∈ R3 and x ∈ R3. Define
r(x) := x− x0, (348)
then the distance between the two particles is given by r(x) := |r(x)|. Thegravitational force on the particle at x ∈ R3 is given by Newton’s law ofuniversal gravitation, and it is of the form
F (x) =c
r(x)2
x0 − x|x0 − x|
= −cr(x)
r(x)3∈ R3, (349)
or, in short, F = −cr/r3.Remark: In this example, we have c = Gm0m, whereG ' 6.67·10−11 Nm2kg−2
denotes the gravitational constant and where m0 and m [kg] denote the massof the particle at x0 and x, respectively. We obtain the same form for F ifwe consider charged particles and use Coulomb’s law instead.With ∇r = r/r we may also write
F = −c∇rr2
= ∇(cr
). (350)
Therefore the scalar field f := −c/r [J] is called the potential of F . So wehave F = −∇f , and we compute
∆f = div (∇f) = −divF(349)= c div
( rr3
)= c
(∇(
1
r3
)· r +
1
r3divr
).
(351)
65
Furthermore, we have
∇(
1
r3
)= −3r2∇r
r6= −3r
r5⇒ ∇
(1
r3
)· r = − 3
r3, (352)
divr = 3 ⇒ 1
r3divr =
3
r3, (353)
so that the potential f satisfies the three-dimensional Laplace’s equation
∆f =∂2f
∂x2+∂2f
∂y2+∂2f
∂z2= 0. (354)
The theory of solutions of (354) is called potential theory. Solutions of (354)that have continuous second partial derivatives are known as harmonic func-tions. In Section 12.5, we had solved the two-dimensional Laplace’s equationin cartesian coordinates with the method of separating variables. In thissection, we shall see how the technique is used in three space dimensions.
Cylindrical Coordinates Cylindrical coordinates (r, ϕ, z) are related tothe cartesian coordinates (x, y, z) via
x = r cosϕ, y = r sinϕ, r ≥ 0, ϕ ∈ [0, 2π). (355)
These are just polar coordinates used in the xy plane; the height is un-changed. Therefore, the Laplacian is very easy to compute in these coordi-nates, and we obtain a similar expression as in Section 12.9 (319):
∆ =∂2
∂r2+
1
r
∂
∂r+
1
r2
∂2
∂ϕ2+
∂2
∂z2. (356)
The unknown function u may be written as u(r, ϕ, z) = F (r, ϕ)G(z) and wemay proceed exactly as in Section 12.9.
Spherical Coordinates The spherical coordinates (r, ϕ, ϑ) are related tothe cartesian coordinates (x, y, z) via
x = r cosϕ sinϑ, y = r sinϕ sinϑ, z = r cosϑ, (357)
with r ≥ 0, ϕ ∈ [0, 2π) and ϑ ∈ [0, π] (⇒ sinϑ ≥ 0). These relations can beinverted as
r =√x2 + y2 + z2, ϕ = arctan
(yx
), ϑ = arccos
(zr
), (358)
66
for (x, y, z) ∈ R3. With the chain rule and using the partial derivativesof r, ϕ, ϑ with respect to x, y, z, we obtain for the Laplacian in sphericalcoordinates:
∆ =∂2
∂r2+
2
r
∂
∂r+
1
r2∆S2 , (359)
where the Laplace-Beltrami operator (spherical Laplacian) ∆S2 is given by
∆S2 =1
sin2 ϑ
∂2
∂ϕ2+
1
sinϑ
∂
∂ϑ
(sinϑ
∂
∂ϑ
). (360)
We consider Dirichlet boundary value problems which are posed either insideor outside of a sphere with radius R > 0: our domain Ω is either
Ω− := x ∈ R3 | |x| < R ( “interior problem”), or (361)
Ω+ := x ∈ R3 | |x| > R ( “exterior problem”). (362)
Notice that Ω− is a bounded domain, whereas Ω+ is not. In both cases, theboundary value problem has the form
∆u = 0 in Ω, (363)
u(R,ϕ, ϑ) = f(ϕ, ϑ), ϕ ∈ [0, 2π), ϑ ∈ [0, π], (364)
limr→∞
u(r, ϕ, ϑ) = 0, ϕ ∈ [0, 2π), ϑ ∈ [0, π]. (365)
Separation of Variables We write the unknown function in the formu(r, ϕ, ϑ) = F (r)G(ϕ, ϑ). We separate the radial variable from the angularvariables by
r2F′′
F+ 2r
F ′
F= − 1
G∆S2G = k ∈ R, (366)
and the separate differential equations are given by
r2F ′′ + 2rF ′ − kF = 0, r < R or r > R, (367)
∆S2G+ kG = 0, ϕ ∈ [0, 2π), ϑ ∈ [0, π]. (368)
We solve forG first. For that purpose, we separate again, G(ϕ, ϑ) = H(ϕ)Q(ϑ),and we obtain
−H′′
H=
sinϑ
Q(sinϑQ′)
′+ k sin2 ϑ = µ2 ∈ R. (369)
67
The separate ODEs are now given by
H ′′ + µ2H = 0, ϕ ∈ [0, 2π), (370)
sin2 ϑQ′′ + sinϑ cosϑQ′ +(k sin2 ϑ− µ2
)Q = 0, ϑ ∈ [0, π]. (371)
Auxiliary Conditions and Solution of Separate ODEs 2π-periodicsolutions of (370) are found for µ = m ∈ Z, and they are given by
Hm(ϕ) = eimϕ. (372)
Remark: We prefer to use the complex exponentials here, instead of cos(mϕ)and sin(mϕ), m ∈ Z.In (371) we use z := cosϑ ∈ [−1, 1] to obtain
d
dϑ= − sinϑ
d
dz,
d2
dϑ2= (1− z2)
d2
dz2− z d
dz. (373)
Then the ODE (371) transforms to
(1− z2)w′′ − 2zw′ +
(k − µ2
1− z2
)w = 0, (374)
for the function w(z) = Q(ϑ). For k = ν(ν + 1), this becomes Legendre’sdifferential equation with parameters ν and µ.Remark: Every number k ≥ −1/4 can be written in the form k = ν(ν + 1),with ν ∈ R. For k < 1/4, we have ν ∈ C.Periodic and continuous solutions are found for ν = n ∈ N0, 0 ≤ |m| ≤ n,and they are given by the associated Legendre functions
Qmn(ϑ) = P |m|n (cosϑ), 0 ≤ |m| ≤ n. (375)
For integer values of the parameters, these functions are polynomials (of de-gree n). For m = 0, they reduce to the Legendre polynomials, P 0
n ≡ Pn.
Remark: The polynomial P|m|n has n− |m| real zeros in [−1, 1], 0 ≤ |m| ≤ n.
We may normalize the product Gmn := HmQmn, and so we define the spher-ical harmonics Ynm : [0, 2π)× [0, π]→ C (!)
Ynm(ϕ, ϑ) :=
√2n+ 1
4π
(n− |m|)!(n+ |m|)!
P |m|n (cosϑ)eimϕ, 0 ≤ |m| ≤ n. (376)
68
Remark: For any n ∈ N0, there are 2n+1 spherical harmonics Ynm. The firstindex is often denoted by ` instead of n. Y m
` is also a common notation.The spherical harmonics Ynm are eigenfunctions of the spherical Laplacian,
∆S2Ynm + n(n+ 1)Ynm = 0, (377)
and they satisfy the orthogonality relations
2π∫0
π∫0
Ynm(ϕ, ϑ)Yn′m′(ϕ, ϑ) sinϑ dϑ dϕ = δnn′δmm′ , (378)
where the bar denotes complex conjugation.Remark: Notice that Ynm = Yn,−m, so that the real part of Ynm is given byRe(Ynm) = (Ynm+Yn,−m)/2, and therefore Re(Yn,−m) = Re(Yn,m), 0 ≤ |m| ≤n. Spherical harmonics Yn0 are called zonal, and Yn,±n are called sectoral.All other spherical harmonics are called tesseral.Nodal lines of the real and imaginary parts of Ynm are located at the zerosof the associated Legendre polynomials and of the cos and sin functions.Visualization on the unit sphere S2 (red: positive, blue: negative), for n = 5:
69
70
With the separation constant k = n(n+ 1), the ODE for the radial functionsF becomes
r2F ′′ + 2rF ′ − n(n+ 1)F = 0. (379)
This is a Cauchy-Euler differential equation, in which powers of r are equalto the order of the derivative in each term. Solutions are given by
Fn(r) = rn, F ∗n(r) =1
rn+1, n ∈ N0. (380)
Solutions of the three-dimensional Laplace equation (171) in spherical coor-dinates are given by
umn(r, ϕ, ϑ) = AmnrnYnm(ϕ, ϑ), u∗mn(r, ϕ, ϑ) =
A∗mnrn+1
Ynm(ϕ, ϑ), (381)
for n ∈ N0, 0 ≤ |m| ≤ n.
Superposition and Determination of Coefficients Interior Problem:We must have A∗mn = 0, because the solution must be bounded as r → 0.Therefore, the solution in this case is given by
u−(r, ϕ, ϑ) =∞∑n=0
n∑m=−n
AmnrnYnm(ϕ, ϑ), r ≤ R. (382)
Evaluated as r → R, we obtain
u−(R,ϕ, ϑ) =∞∑n=0
n∑m=−n
AmnRnYnm(ϕ, ϑ)
!= f(ϕ, ϑ). (383)
Using the orthogonality relations for the spherical harmonics, the coefficientsAmn may be computed as
Amn =1
Rn
2π∫0
π∫0
f(ϕ, ϑ)Ynm(ϕ, ϑ) sinϑ dϑ dϕ. (384)
Exterior Problem: We must have Amn = 0, because the solution must vanishin the limit as r →∞. Therefore, the solution for this case is given by
u+(r, ϕ, ϑ) =∞∑n=0
n∑m=−n
A∗mnrn+1
Ynm(ϕ, ϑ), r ≥ R. (385)
71
Evaluated as r → R, we obtain
u+(R,ϕ, ϑ) =∞∑n=0
n∑m=−n
A∗mnRn+1
Ynm(ϕ, ϑ)!
= f(ϕ, ϑ). (386)
The coefficients A∗mn are given by
A∗mn = Rn+1
2π∫0
π∫0
f(ϕ, ϑ)Ynm(ϕ, ϑ) sinϑ dϑ dϕ. (387)
Notice that A∗mn = R2n+1Amn. If we want to study the behavior of thesolution u as r →∞, it is useful to write
u+(r, ϕ, ϑ) =∞∑n=0
(R
r
)n+1
fn(ϕ, ϑ), fn(ϕ, ϑ) :=n∑
m=−n
A∗mnRn+1
Ynm(ϕ, ϑ).
(388)A Taylor expansion in 1/r of u(·, ϕ, ϑ) yields
u+(r, ϕ, ϑ) =R
rf0(ϕ, ϑ) +O(r−2), r →∞, ∀ϕ, ϑ. (389)
It turns out that f0(ϕ, ϑ) is actually constant, and equal to the average ofthe boundary data f over the unit sphere S2:
f0 =1
4π
2π∫0
π∫0
f(ϕ, ϑ) sinϑ dϑ dϕ. (390)
12.11 Solution of PDEs by Laplace Transforms
We may transform a PDE involving both space and time into a family ofPDEs involving spatial variables only. This is accomplished by the Laplacetransform, which for a function f : (0,∞)→ R is defined by
f#(s) ≡ L[f ](s) :=
∞∫0
f(t)e−st dt, s ∈ C. (391)
72
if the integral exists (it does for a large class of functions, for Re(s) largeenough). L is a linear transform and has an inverse, L−1. Using integrationby parts, we find that the Laplace transform of the n-th derivative of F isgiven by
L[f (n)](s) = sL[f (n−1)](s)− f (n−1)(0). (392)
By induction we conclude that
L[f (n)](s) = snL[f ](s)− sn−1f(0)− · · · − sf (n−2)(0)− f (n−1)(0). (393)
Remark: Because of property (393), the Laplace transform is particularlysuited to initial value problems, which model physical systems that are atrest for all times t < 0 and suddenly change at t = 0. If, on the other hand,the system is time-invariant, it is better to use the Fourier transform instead.We may use (393) to eliminate the time derivatives in a PDE. Consider thefollowing initial value problem for the wave equation:
Utt = c2∆U, in Ω× (0,∞), c > 0, (394)
U = 0, Ut = 0, in Ω× 0. (395)
This problem is not well posed, but we will take care of the boundary condi-tions later. The Laplace transform u(x; s) := L[U(x, ·)](s) of the unknownfunction U(x, t) satisfies the Helmholtz equation
∆u+ k2u = 0, in Ω, k :=is
c∈ C. (396)
Any boundary condition imposed on U on ∂Ω × (0,∞) also needs to beLaplace transformed, which yields a new boundary condition for u on ∂Ω.Thus we end up with a boundary value problem in Ω. For simple shapes ofΩ, we can solve this problem analytically, as we have seen for a rectanglein 2D (Section 12.8 double Fourier series), a disk in 2D (Section 12.9 Fourier-Bessel series), as well as in the case of a spherical boundary in 3D(Section 12.10 spherical harmonics).
12.12 Application: hydrogen-like atomic orbitals
A hydrogen-like atom is one with a single electron (charge −e, where e '1.60·10−19 C denotes the elementary charge) and a nucleus of charge Ze, withthe atomic number Z ∈ N. The quantum state of the electron is described by
73
the wave function Ψ : R3 × (0,∞)→ C: the function |Ψ|2 is the probabilitydensity for the position of the electron at time t > 0. The time evolution ofthe wave function Ψ is described by the Schrodinger equation
i~∂Ψ
∂t= − ~2
2me
∆Ψ + VΨ, (397)
where i2 = −1 and where ~ ' 1.05 · 10−34 Js denotes the reduced Planckconstant, me ' 9.11·10−31 kg denotes the electron mass and V (x) [J] denotesthe potential energy at position x ∈ R3. A Fourier transform in time leadsto an expansion into standing waves of the form
Ψ(x, t;ω) = ψ(x;ω)e−iωt, (398)
with ψ : R3 → C, |Ψ|2 = |ψ|2. The functions ψ satisfy the time independentSchrodinger equation
Eψ = − ~2
2me
∆ψ + V ψ, (399)
where E = E(ω) = ~ω denotes the total energy of the electron. We wish tosolve the eigenvalue problem associated to (399), i. e. find all energy levels Efor which (399) has a solution, as well as the solution ψ itself. An analyticsolution is possible for simple cases, such as the one considered here, wherethe potential is given by
V (x) = − 1
4πε0
Ze2
|x|, (400)
with the vacuum permittivity ε0 ' 8.85 · 10−12 Fm−1. Here we have assumedwithout loss of generality that the nucleus is located at the origin 0 ∈ R3.We write the stationary Schrodinger equation (399) in spherical coordinates:(
− ~2
2me
(∂2
∂r2+
2
r
∂
∂r+
1
r2∆S2
)− 1
4πε0
Ze2
r
)ψ = Eψ. (401)
Expanding the solution ψ into spherical harmonics,
ψ(r, ϕ, ϑ;E) =∞∑`=0
∑m=−`
R`m(r;E)Y`m(ϕ, ϑ), (402)
74
we obtain, with ∆S2Y`m + `(`+ 1)Y`m = 0, for 0 ≤ |m| ≤ `, ` ∈ N0:(− ~2
2me
(d2
dr2+
2
r
d
dr− `(`+ 1)
r2
)− 1
4πε0
Ze2
r
)R`(r;E) = ER`(r;E),
(403)for r > 0. We have dropped the index m from R` because the radial ODE(403) is independent of that index. We keep ` ∈ N0 fixed for the momentand define the function u(r) := rR`(r;E).
R`(r;E) =u(r)
r, R′`(r;E) =
ru′(r)− u(r)
r2, (404)
R′′` (r;E) =r2u′′(r)− 2ru′(r) + 2u(r)
r3. (405)
− ~2
2me
u′′(r) +
(~2
2me
`(`+ 1)
r2− Ze2
4πε0r
)u(r) = Eu(r), (406)
for r > 0. We consider the ODE (406) in the limit as r →∞:
− ~2
2me
u′′(r) = Eu(r), r →∞. (407)
Bound states are characterized by a negative energy E < 0, and accordingto Proposition 1, the asymptotic behavior of u is given by
u ∼ e−κr, r →∞, κ :=
√−2meE
~2> 0 [m−1]. (408)
We transform variables: ρ := κr (notice that this scaling depends on E) andobtain the following ODE for v(ρ(r)) = u(r), from (406):
ρ2v′′(ρ) =(ρ2 − 2νρ+ `(`+ 1)
)v(ρ), ρ > 0, ν := − Ze2κ
8πε0E> 0. (409)
We consider the ODE (409) as ρ→ 0:
ρ2v′′(ρ) ∼ `(`+ 1)v(ρ), ρ→ 0. (410)
This is a Cauchy-Euler equation, with solutions ρ−` and ρ`+1. Because vmust be bounded as ρ→ 0, its asymptotic behavior must be given by
v ∼ ρ`+1, ρ→ 0. (411)
75
We factor out both asymptotics and write v(ρ) := e−ρρ`+1w(ρ), with a newunknown function w. It satisfies the second-order ODE
ρw′′(ρ) + ((2`+ 1) + 1− 2ρ)w′(ρ) + 2 (ν − `− 1)w(ρ) = 0, ρ > 0. (412)
with z := 2ρ, w(z) = w(ρ), we obtain
zw′′(z) + ((2`+ 1) + 1− z) w′(z) + (ν − `− 1) w(z) = 0. (413)
Solutions are found for ν = n ∈ N, ` < n, and they are given by the associatedLaguerre polynomials
w(z) = L2`+1n−`−1(z) =
n−`−1∑k=0
(−1)k(
n+ `n− `− 1− k
)zk
k!. (414)
We can now also express the admissible energy states as
En = −Z2
n2
mee4
32π2ε20~2
, n ∈ N. (415)
The radial solutions are then given by
Rn`(r) := R`(r;En) = κe−κr(κr)`L2`+1n−`−1(2κr) (416)
The solution can now be given as a superposition of the products
ψn`m(r, ϕ, ϑ) := cn`Rn`(r)Y`m(ϕ, ϑ), 0 ≤ |m| ≤ ` < n, (417)
with some normalization constants cn` > 0. So the wave function for then-th energy state is of the form
ψ(r, ϕ, ϑ;En) =n−1∑`=0
∑m=−`
cn`Rn`(r)Y`m(ϕ, ϑ), n ∈ N. (418)
12.13 Application: Unbounded domains and artificialboundaries. Dirichlet-to-Neumann map.
Many practical problems, such as scattering problems, are posed in un-bounded or at least very large domains. In these problems, the scatterer
76
is typically a region of space with complicated properties, such as inhomo-geneities, nonlinearities, obstacles and sources. Analytical methods are thusnot applicable to the problem, and we need to use numerical methods for itssolution.
However, with numerical methods we immediately run into a conflict withthe unbounded domain. Typical grid-based numerical methods (such as thefinite difference, finite element, and finite volume method) cannot handle in-finite domains, so the problem is not (yet) solvable. One approach (there areothers!) is to introduce an artificial boundary B, which separates space intoa bounded domain Ω− and an unbounded region Ω+. The domain Ω− shouldbe chosen such that it contains all the complicated features of the scatterer.Ω+ will contain the “uninteresting region”, where the medium is supposedto be simple. The idea is then to solve the problem numerically in Ω− only.We still need to impose a boundary condition on B ⊆ Ω− in order to obtaina well-posed problem. In the context of wave propagation, this boundarymust be such that waves impinging on B can pass that interface withoutany spurious reflection. Such reflection would be non-physical (because theboundary itself is only artificial), but it would spoil the numerical solutionthroughout Ω−. A boundary condition on B with the desired property istherefore called a non-reflecting boundary condition (NRBC, NBC).
The idea is that because the problem is supposed to be simple in Ω+, wecan solve it analytically and represent the solution in Ω+ using the boundaryvalues on B = ∂Ω+. We can then get an analytic expression for the Dirichlet-to-Neumann (DtN) map M ,
M : u|B 7→∂u
∂n
∣∣∣∣B
. (419)
Then the boundary condition
∂nu = Mu, on B, (420)
(Robin type) is an exact non-reflecting boundary condition for the compli-cated problem in Ω−. To find an analytic expression forM , we usually assumea simple shape for B, such as a sphere with radius R > 0 (then ∂n ≡ ∂r).
After Laplace transform in time, if necessary, an acoustic scattering prob-lem satisfies a Helmholtz equation in Ω+ ⊂ R3, together with the Rellich-
77
Sommerfeld condition (which ensures uniqueness of the solution):
∆u+ k2u = 0 in Ω+, (421)
lima→∞
∫|x|=a
∣∣∣∣∂u∂r − iku∣∣∣∣2 ds = 0. (422)
We may solve this problem using separation of variables. It is natural for thisgeometry to use spherical coordinates. We expand the unknown function uinto spherical harmonics,
u(r, ϕ, ϑ) =∞∑n=0
n∑m=−n
umn(r)Ynm(ϕ, ϑ), r ≥ R. (423)
Then we obtain the following ODE for the coefficients umn, 0 ≤ |m| ≤ n:(d2
dr2+
2
r
d
dr− n(n+ 1)
r2+ k2
)umn(r) = 0, r ≥ R, (424)
limr→∞
(d
dr− ik
)umn(r) = 0. (425)
A transformation z := kr leads to a Bessel differential equation for umn(r) =w(z)/
√z. Together with the radiation condition, we find that solutions are
given by the Hankel functions of the first kind of order n+ 1/2, n ∈ N0:
umn(r) = AmnH
(1)n+1/2(kr)√r
, r ≥ R, (426)
The coefficients Amn ∈ C can be determined from the boundary values ofumn at r = R:
umn(R) = AmnH
(1)n+1/2(kR)√R
⇒ Amn =
√R
H(1)n+1/2(kR)
umn(R), (427)
so that the coefficients umn may also be written as
umn(r) =
√R
r
H(1)n+1/2(kr)
H(1)n+1/2(kR)
umn(R). (428)
78
We compute the derivative with respect to r and evaluate at r = R to obtain
u′mn(R) =1
REn(kR)umn(R), (429)
with the DtN kernel
En(z) =zH
(1)n+1/2
′(z)
H(1)n+1/2(z)
− 1
2, n ∈ N0. (430)
Now we may write the DtN map M as a spherical harmonic expansion
(Mu)(ϕ, ϑ) =∂u
∂n
∣∣∣∣B
(ϕ, ϑ) =∂u
∂r(R,ϕ, ϑ)
=1
R
∞∑n=0
n∑m=−n
En(kR)umn(R)Ynm(ϕ, ϑ) (431)
=1
R
∞∑n=0
n∑m=−n
En(kR)
∫B
u|B Ynm dsYnm(ϕ, ϑ).
In a numerical scheme, the exact non-reflecting boundary condition will needto be approximated by truncating the series in M . In 3D, the efficient eval-uation of M is another issue: the application of the non-reflecting boundarycondition must not be more costly than the numerical method used inside ofΩ−.
12.14 Review of Chapter 12
12.14.1 Basic Concepts (12.1)
A partial differential equation of order k ∈ N for an unknown function u :Ω→ R, Ω ⊆ Rd open, d > 1 (the total number of variables), is an expressionof the form
F((Dku)(x), (Dk−1u)(x), . . . , (Du)(x), u(x), x
)= 0, x ∈ Ω, (432)
with a function F : Rdk×Rdk−1×· · ·×Rd×R×Ω→ R. A typical mathemat-ical model is given by (432), together with auxiliary conditions on Γ ⊆ ∂Ω,
79
such as initial and/or boundary conditions. We have mainly looked at ho-mogeneous, linear PDEs, which are of the form∑
|α|≤k
aα(x)(Dαu)(x) = 0, (433)
with coefficients aα : Ω → R, |α| ≤ k. For these PDEs, the superpositionprinciple (Theorem 1) is valid.
12.14.2 Examples
We have considered
• problems in two variables (d = 2):
– 1 spatial variable + time: 1D wave equation, utt = c2uxx, c > 0(12.3, 12.4), 1D heat equation ut = c2uxx, c > 0 (12.5, 12.6)
– 2 spatial variables: 2D Laplace equation, ∆u = 0 (steady heatproblems, 12.5)
• problems in three variables (d = 3):
– 2 spatial variables + time: 2D wave equation, utt = c2∆u, c > 0(12.8, 12.9)
– 3 spatial variables: 3D Laplace equation, ∆u = 0 (12.10)
12.14.3 Separation of Variables (12.3, 12.5, 12.8, 12.9, 12.10)
The method of separating variables aims at transforming a PDE into a systemof differential equations involving fewer independent variables, which may besolved more easily than the original PDE.
1. Separation of Variables For problems involving time, we write thedependent variable in the form u(x, t) = F (x)G(t). In the case of wave andheat equations, respectively, this leads to
utt − c2∆u = FG− c2∆FG = 0, in Ω× (0,∞), (434)
ut − c2∆u = FG− c2∆FG = 0, in Ω× (0,∞), (435)
80
where Ω ⊆ Rd−1 denotes the spatial domain only, from now on. After divisionby c2FG we obtain
G
c2G=
∆F
F, in Ω× (0,∞), (436)
G
c2G=
∆F
F, in Ω× (0,∞). (437)
Because the left-hand side of (436), (437) depends only on the time variablet > 0, and the right-hand side of (436), (437) depends only on the spatialvariable(s) x ∈ Ω, both sides must be equal to a constant, the separationconstant k ∈ R. So we obtain two separate equations, one in space and onein time (which is an ODE):
∆F = kF, in Ω, (+ boundary conditions), (438)
and either G = c2kG, in (0,∞), (+ initial conditions), (439)
or G = c2kG, in (0,∞), (+ initial conditions). (440)
The separation constant k ∈ R is part of the solution of the spatial eigenvalueproblem (438), which involves d−1 variables. The Laplace equation, ∆u = 0,is just a special case of (438), with k = 0. Therefore, the same solutionstrategies apply.
2. Solution of the spatial eigenvalue problem
d = 2: In this case, the spatial eigenvalue problem is given by
F ′′ = kF, in Ω ⊆ R, (441)
together with boundary conditions. The general solution of (441) wasgiven in Proposition 1:
F (x; k) =
Aeωx +Be−ωx, k = ω2, ω > 0Ax+B, k = 0A cos(ωx) +B sin(ωx), k = −ω2, ω > 0
, A,B ∈ R.
(442)Let the domain be given by Ω = (0, L), L > 0. The boundary condi-tions on ∂Ω = 0, L will determine which of these solutions are feasi-ble; in general we need to check each case for the separation constantk separately.
81
With Dirichlet boundary conditions F (0) = F (L) = 0, we find only thetrivial solution F ≡ 0 for k ≥ 0. However, we find non-trivial solutionsfor k = −(nπ/L)2, n ∈ N:
Fn(x) = sin(nπLx), n ≥ 1, (443)
where we have moved any constant factor over to the time-dependentfunctions Gn(t).
With Neumann boundary conditions F ′(0) = F ′(L) = 0, we find onlythe trivial solution F ≡ 0 for k > 0. However, we find non-trivialsolutions for k = −(nπ/L)2, n ∈ N:
Fn(x) = cos(nπLx), n ≥ 0. (444)
Notice the constant solution, F0!
With mixed boundary conditions F (0) = F ′(L) = 0, we find only thetrivial solution F ≡ 0 for k ≥ 0. However, we find non-trivial solutionsfor k = −((n+ 1/2)π/L)2, n ≥ 0:
Fn(x) = sin
((n+
1
2
)π
Lx
), n ≥ 0. (445)
d = 3: The spatial differential equation for F will be a PDE in two spacedimensions. We choose coordinates such that the boundary of Ω becomesas simple as possible. So, if Ω = (0, a)× (0, b), a, b > 0, is a rectangle,we use cartesian coordinates (x, y), and if Ω is a disk of radius R > 0(or an annulus), we use polar coordinates (r, ϕ). The Laplacian in thesetwo coordinate systems is given by
∆ =∂2
∂x2+
∂2
∂y2=
∂2
∂r2+
1
r
∂
∂r+
1
r2
∂2
∂ϕ2. (446)
We separate the spatial variables by writing u(x, y) = H(x)Q(y), oru(r, ϕ) = H(r)Q(ϕ), and we obtain
H ′′
H= −Q
′′
Q+ k = p ∈ R, (447)
r2H′′
H+ r
H ′
H− kr2 = −Q
′′
Q= p ∈ R. (448)
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The two separate ODEs are thus given by either
H ′′ = pH, in (0, a), (449)
Q′′ = (k − p)Q, in (0, b), (450)
or r2H ′′ + rH ′ − (kr2 + p)H = 0, in [0, R), (451)
Q′′ = −pQ, in [0, 2π). (452)
Again, to find feasible values of the separation constants as well as thesolutions themselves, we require the boundary conditions for u on ∂Ω.We assume homogeneous Dirichlet boundary conditions here.
(x, y): As in the 1D case, we find solutions for p = −(mπ/a)2, m ∈ N,k − p = −(nπ/b)2, n ∈ N. They are given by
Hm(x) = sin(mπax), m ∈ N, (453)
Qn(y) = sin(nπby), n ∈ N. (454)
The spatial eigenfunctions are thus given by
Fmn(x, y) = sin(mπax)
sin(nπby), m, n ∈ N, (455)
with
∆Fmn = kFmn = −π2
(m2
a2+n2
b2
)Fmn, m, n ∈ N. (456)
(r, ϕ): With the condition that the angular functions Q be 2π-periodic,we obtain the following solutions for p = n2, n ∈ N0:
Q(1)n (ϕ) = cos(nϕ), Q(2)
n (ϕ) = sin(nϕ), n ∈ N0. (457)
For the radial functions H, we need to go through the possiblecases for the first separation constant k ∈ R.
k = ω2, ω > 0: The radial equation becomes
r2H ′′ + rH ′ −(ω2r2 + n2
)H = 0, in [0, R). (458)
After transformation z := ωr, w(z) := H(r), we obtain themodified Bessel’s differential equation:
z2w′′ + zw′ −(z2 + n2
)w = 0, in [0, ωR). (459)
83
The general solutions are given by the modified Bessel func-tions In, Knn≥0. If we require the solution to be boundedas r → 0, and with the homogeneous Dirichlet boundary con-ditions at r = R, we find only the trivial solution H ≡ 0 inthis case.
k = 0: The radial equation becomes a Cauchy-Euler ODE:
r2H ′′ + rH ′ − n2H = 0, in [0, R), (460)
with solutions Hn(r) = Anrn + Bnr
−n, n ≥ 0 (Problem Set5). Again, boundedness of the solution as r → 0 and thehomogeneous Dirichlet boundary condition at r = R yieldsonly the trivial solution H ≡ 0.
k = −ω2, ω > 0: The radial equation becomes
r2H ′′ + rH ′ +(ω2r2 − n2
)H = 0, in [0, R), (461)
and after transformation z := ωr, w(z) := H(r), we obtain
z2w′′ + zw′ +(z2 − n2
)H = 0, in [0, kR). (462)
General solutions are given by the Bessel functions Jn, Ynn≥0.Only the Bessel functions of the first kind are bounded asz → 0. From the homogeneous Dirichlet boundary condi-tions, we obtain the equations
Jn(ωR) = 0, n ∈ N0. (463)
There are infinitely many positive solutions ωmn, m ∈ N,for every n ∈ N0. So the radial eigenfunctions are given byHmn(r) = Jn(ωmnr), r ∈ [0, R). Now the spatial eigenfunc-tions are given by
F (1)mn(r, ϕ) = Jn(ωmnr) cos(nϕ), (464)
F (2)mn(r, ϕ) = Jn(ωmnr) sin(nϕ), (465)
for m ∈ N, n ∈ N0, with
∆F (1,2)mn = −ω2
mnF(1,2)mn , m ∈ N, n ∈ N0. (466)
84
d = 4: For this case, we did not consider a time variable in the lecture, sowe will only consider the case k = 0, for which we obtain the Laplaceequation in 3D. We have considered three coordinate systems, depend-ing on the shape of the spatial domain Ω: cartesian, cylindrical, andspherical.
(x, y, z): The Laplacian in cartesian coordinates is given by
∆ =∂2
∂x2+
∂2
∂y2+
∂2
∂z2. (467)
The separation of variables is like in cartesian coordinates in 2D,and leads to a triple Fourier series in x, y, and z.
(r, ϕ, z): The Laplacian in cylindrical coordinates is given by
∆ =∂2
∂r2+
1
r
∂
∂r+
1
r2
∂2
∂ϕ2+
∂2
∂z2. (468)
The separation of variables leads to a Fourier series in z and aFourier-Bessel series in r and ϕ, similar to polar coordinates in2D.
(r, ϕ, ϑ): The Laplacian in spherical coordinates is given by
∆ =∂2
∂r2+
2
r
∂
∂r+
1
r2∆S2 , (469)
where the Laplace-Beltrami operator on the unit sphere S2 ⊂ R3
is given by
∆S2 =1
sin2 ϑ
∂2
∂ϕ2+
1
sinϑ
∂
∂ϑ
(sinϑ
∂
∂ϑ
). (470)
The eigenfunctions of ∆S2 are given by the spherical harmonics
Ynm(ϕ, ϑ) :=
√2n+ 1
4π
(n− |m|)!(n+ |m|)!
P |m|n (cosϑ)eimϕ, 0 ≤ |m| ≤ n.
(471)
The functions P|m|n are the associated Legendre polynomials, which
reduce to the Legendre polynomials for m = 0. The spherical har-monics satisfy
∆S2Ynm = −n(n+ 1)Ynm, 0 ≤ |m| ≤ n, (472)
85
and the orthogonality relations for the spherical harmonics aregiven by ∫
S2
YnmYn′m′ ds = δnn′δmm′ , 0 ≤ |m| ≤ n. (473)
The radial functions Fn(r) then satisfy the ODE
r2F ′′n + 2rF ′n − n(n+ 1)Fn = 0, n ∈ N0. (474)
This is a Cauchy-Euler ODE, with solutions given by
Fn(r) = rn, F ∗n(r) = r−(n+1), n ∈ N0. (475)
Eigenfunctions are now given by
umn(r, ϕ, ϑ) = Fn(r)Ynm(ϕ, ϑ), u∗mn(r, ϕ, ϑ) = F ∗n(r)Ynm(ϕ, ϑ),(476)
for 0 ≤ |m| ≤ n.
3. Superposition and Determination of Coefficients
d = 2: The general solution of the one-dimensional heat or wave equation isgiven by a superposition of the products un(x, t) = Fn(x)Gn(t), whereGn satisfy the appropriate ODE. Evaluated as t→ 0, we obtain Fourierseries in x, with coefficients to be determined from the initial data.
If the boundary conditions are not mixed, it is faster to use the Fourierseries ansatz
u(x, t) = a0(t) +∞∑n=1
(an(t) cos
(nπLx)
+ bn(t) sin(nπLx))
. (477)
Depending on the boundary conditions, either the coefficients an or bnwill vanish, and we obtain ODEs in time for the remaining coefficients,by plugging into the PDE.
For the wave equation with Dirichlet boundary conditions, we obtainan ≡ 0, n ≥ 0, and
utt(x, t)− c2uxx(x, t) =∞∑n=1
(bn(t) +
(cnπL
)2
bn(t)
)sin(nπLx)
= 0.
(478)
86
For the heat equation with Neumann boundary conditions, we obtainbn ≡ 0, n ≥ 1, and
ut(x, t)− c2uxx(x, t) = a0(t) +∞∑n=1
(an(t) +
(cnπL
)2
an(t)
)cos(nπLx)
= 0. (479)
The initial values of the coefficients an, bn are given as the Fouriercoefficients of the initial data:
u(x, 0) = a0(0) +∞∑n=1
(an(0) cos
(nπLx)
+ bn(0) sin(nπLx))
,(480)
ut(x, 0) = a0(0) +∞∑n=1
(an(0) cos
(nπLx)
+ bn(0) sin(nπLx))
,(481)
where again either the coefficients an or bn vanish due to the boundaryconditions.
d = 3: Again, we consider both cartesian and polar coordinates.
(x, y): The general solution is given by a superposition of products
umn(x, y, t) = Fmn(x, y)Gmn(t). (482)
Evaluated as t → 0, we obtain double Fourier series in x and y,with coefficients determined from the initial data, similar to thed = 2 case.
Again, if the boundary conditions in, say, the x-variable are notmixed, it is faster to use a Fourier series ansatz of the form
u(x, y, t) = a0(y, t)+∞∑m=1
(an(y, t) cos
(mπax)
+ bn(y, t) sin(mπax))
.
(483)The same ansatz may be used for steady problems, i. e. the 2DLaplace equation; in that case of course, the coefficients an, bn willdepend only on y.
87
(r, ϕ): The general solution is given by a superposition of the products
umn(r, ϕ, t) = F(1)mn(r, ϕ)Gmn(t) and u∗mn(r, ϕ, t) = F
(2)mn(r, ϕ)G∗mn(t).
Evaluated as t → 0, we obtain Fourier-Bessel series in r and ϕ,the coefficients of which can again be evaluated from the initialdata.
d = 4: The general solution of the Laplace equation in 3D is given by thesuperposition
u(r, ϕ, ϑ) =∞∑n=0
n∑m=−n
(Amnr
n +Bmn
rn+1
)Ymn(ϕ, ϑ), (484)
where the coefficients are determined from the boundary data. Thesemay also be given by conditions as r → 0 or r → ∞. For example,for the interior problem in the sphere r < R, we must have Bmn = 0,0 ≤ |m| ≤ n, so that the solution is bounded as r → 0. In that case,we have
u−(r, ϕ, ϑ) =∞∑n=0
n∑m=−n
AmnrnYnm(ϕ, ϑ), r ≤ R, ϕ ∈ [0, 2π), ϑ ∈ [0, π],
(485)and the coefficients Amn are determined from the boundary data:
Amn =1
Rn
∫S2
u|r=R Ynm ds. (486)
12.14.4 Method of Characteristics (12.4)
A second-order quasilinear PDE in two variables is of the form
A(x, y)uxx(x, y) + 2B(x, y)uxy(x, y) + C(x, y)uyy(x, y) (487)
= F (x, y, u(x, y), ux(x, y), uy(x, y)). (488)
The type of this PDE is determined by the discriminant B2 − AC:
B2 − AC > 0 : hyperbolic (489)
B2 − AC = 0 : parabolic (490)
B2 − AC < 0 : elliptic (491)
88
we write down the characteristic ODE
Ay′2 − 2By′ + C = 0. (492)
With the solutions y(x), we define new variables v(x, y) and w(x, y). If thereis only one characteristic, we typically set v(x, y) := x and define w usingthe characteristic.
For the new variables v(x, y) and w(x, y), the second partial derivativesof u(x, y) = U(v(x, y), w(x, y)) are obtained from the chain rule as
uxx = Uvvv2x + 2Uvwvxwx + Uvvxx + Uwww
2x + Uwwxx, (493)
uxy = Uvvvxvy + Uvw (vxwy + vywx) + Uvvxy + Uwwwxwy + Uwwxy,(494)
uyy = Uvvv2y + 2Uvwvywy + Uvvyy + Uwww
2y + Uwwyy (495)
Plugging this into the PDE will lead to its normal form, which should beeasier to solve than the original PDE.Example: Consider the PDE
uxx + 2uxy − uyy = 0. (496)
We have A = B = 1, C = −1, which implies B2 −AC = 2 > 0, so this PDEis hyperbolic. The characteristic ODE is given by
y′2 − 2y′ − 1 =
(y′ − (1 +
√2))(
y′ − (1−√
2))
= 0, (497)
with solutions y1,2(x) = (1 ±√
2)x + c1,2. We define the new variables v, was
v(x, y) = −(1 +√
2)x+ y, w(x, y) = −(1−√
2)x+ y, (498)
with partial derivatives
vx = −(1 +√
2), vy = 1, vxx = vxy = vyy = 0, (499)
wx = −(1−√
2), wy = 1, wxx = wxy = wyy = 0. (500)
We obtain from the PDE:
uxx + 2uxy − uyy = (3 + 2√
2)Uvv − 2Uvw + (3− 2√
2)Uww + (501)
−(2 + 2√
2)Uvv − 4Uvw − (2− 2√
2)Uww + (502)
−Uvv − 2Uvw − Uww = −8Uvw, (503)
89
so that the normal form of the PDE is given by Uvw = 0. We integrate twiceto obtain
Uv(v, w) = f(v), U(v, w) = F (v) +G(w). (504)
The general solution of the PDE is now given by
u(x, y) = U(v(x, y), w(x, y)) = F (v(x, y)) +G(w(x, y)) (505)
= F (−(1 +√
2)x+ y) +G(−(1−√
2)x+ y). (506)
12.14.5 Fourier Integrals (12.6)
For unbounded domains, we may need Fourier integrals instead of Fourierseries, because the eigenvalues may not form a countable set anymore. Wehad looked at the Cauchy problem for the one-dimensional heat equation:
ut = c2uxx, in R× (0,∞), (507)
u = f, on R× 0. (508)
Separation of variables leads to the separate ODEs
F ′′ = kF, in R, (509)
G = c2kG, in (0,∞). (510)
In this case, we require a bounded solution as |x| → ∞, which excludes allbut the trivial solution for k ≥ 0. For k = −ω2, ω > 0, on the other hand,we find uncountably many oscillatory solutions in space:
F (x;ω) = A(ω) cos(ωx) +B(ω) sin(ωx), x ∈ R, (511)
G(t;ω) = e−c2ωt, t > 0. (512)
Eigenfunctions are now given by
u(x, t;ω) = F (x;ω)G(t;ω), ω > 0, (513)
and the superposition is an integral:
u(x, t) =
∞∫0
(A(ω) cos(ωx) +B(ω) sin(ωx)) e−c2ωt dω. (514)
90
The coefficients A, B are determined from the initial data, f :
u(x, 0) =
∞∫0
(A(ω) cos(ωx) +B(ω) sin(ωx)) dω!
= f(x), x ∈ R, (515)
from which we conclude that
A(ω) =1
π
∞∫−∞
f(y) cos(ωy) dy, B(ω) =1
π
∞∫−∞
f(y) sin(ωy) dy. (516)
After some manipulation we were able to write the solution u as an integraltransform of the initial data f ,
u(x, t) =
∞∫−∞
f(y)K(c2t, x, y) dy, (517)
with the heat kernel
K(t, x, y) :=1√4πt
exp
(−(x− y)2
4t
), t > 0, x, y ∈ R. (518)
It turns out that K(t, ·, ·) is a convolution kernel, K(t, x, y) = ϕ(x − y, t),where ϕ denotes the Gauss-Weierstrass kernel.
91
13 Complex Numbers and Functions
Complex analysis is the theory of functions of a complex variable. Holomor-phic functions are of particular interest; these are complex differentiable inevery point z ∈ U in a domain U ⊆ C. If f = u + iv is holomorphic, thenthe real and imaginary parts u, v, understood as functions of two real vari-ables (x, y) ∈ R2, satisfy the Cauchy-Riemann equations, which is a systemof two first-order partial differential equations. If, furthermore, u and v havecontinuous second partial derivatives, they are harmonic functions and thussolutions to Laplace’s differential equation (cf. Chapter 12) in two space di-mensions. Therefore, potential theory in 2D (to be treated in Chapter 18)is related to complex analysis, and this shall be the main motivation for itsstudy in this lecture.
There are other useful results from complex analysis, such as that Fourierseries may be written down in a somewhat simpler way by using complexexponentials, or that certain complicated integrals can be evaluated by theelegant method of complex integration, the calculus of residues (Chapter 16).
13.1 Complex Numbers. Complex Plane.
We follow the Cayley-Dickson construction to obtain the field of complexnumbers, C, from the field of real numbers, R.
Ordered pairs of real numbers We consider ordered pairs of real num-bers,
(x, y) ∈ R× R. (519)
Addition We define the addition + of ordered pairs componentwise, i. e. by
(x1, y1) + (x2, y2) := (x1 + x2, y1 + y2). (520)
The set of ordered pairs together with the addition just defined forms acommutative group:
1. Closure is obvious from the definition.
92
2. Associativity:
((x1, y1) + (x2, y2)) + (x3, y3) = (x1 + x2, y1 + y2) + (x3, y3)(521)
= (x1 + x2 + x3, y1 + y2 + y3)(522)
= (x1, y1) + (x2 + x3, y2 + y3)(523)
= (x1, y1) + ((x2, y2) + (x3, y3))(524)
3. Identity element: the pair (0, 0) ∈ R× R is the additive identity:
(0, 0) + (x, y) = (x, y) = (x, y) + (0, 0). (525)
4. Inverse element: for a pair (x, y) ∈ R×R, the additive inverse is givenby (−x,−y) ∈ R× R:
(x, y) + (−x,−y) = (0, 0) = (−x,−y) + (x, y). (526)
5. Commutativity:
(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) = (x2 + x1, y2 + y1)(527)
= (x2, y2) + (x1, y1). (528)
Remark: If we add a scalar product λ(x, y) := (λx, λy), λ ∈ R, we obtaina vector space, which is of course R2 with the standard basis (1, 0), (0, 1).Addition and subtraction of ordered pairs can thus be interpreted as additionand subtraction of vectors in R2.
Multiplication We define the multiplication · of ordered pairs by
(x1, y1) · (x2, y2) ≡ (x1, y1)(x2, y2) := (x1x2 − y1y2, x1y2 + y1x2). (529)
Field of Complex Numbers The set of ordered pairs together with themultiplication and addition defined above form a field:
1. Closure with respect to both addition and multiplication is obviousfrom the definitions.
93
2. Associativity: proven already for the addition. For the multiplicationwe obtain
((x1, y1)(x2, y2)) (x3, y3) = (x1x2 − y1y2, x1y2 + y1x2)(x3, y3) (530)
= ((x1x2 − y1y2)x3 − (x1y2 + y1x2)y3, (x1x2 − y1y2)y3 + (x1y2 + y1x2)x3)
= (x1x2x3 − y1y2x3 − x1y2y3 − y1x2y3, x1x2y3 − y1y2y3 + x1y2x3 + y1x2x3)
= (x1(x2x3 − y2y3)− y1(x2y3 + y2x3), x1(x2y3 + y2x3) + y1(x2x3 − y2y3))
= (x1, y1)(x2x3 − y2y3, x2y3 + y2x3) = (x1, y1) ((x2, y2)(x3, y3)) (531)
3. Commutativity: already shown for the addition. For the multiplicationwe obtain
(x1, y1)(x2, y2) = (x1x2 − y1y2, x1y2 + y1x2) (532)
= (x2x1 − y2y1, x2y1 + y2x1) (533)
= (x2, y2)(x1, y1) (534)
4. Identity elements: we already know the additive identity (0, 0) ∈ R×R.The multiplicative identity is given by the pair (1, 0) ∈ R× R:
(1, 0)(x, y) = (x, y) = (x, y)(1, 0). (535)
5. Inverse elements: we already know the additive inverse (−x,−y) ∈ R×R for a pair (x, y) ∈ R×R. For (x, y) 6= (0, 0) (“we can’t divide by 0!”),the multiplicative inverse is given by (x, y)−1 = (x/(x2 + y2),−y/(x2 + y2)):
(x, y)(x/(x2 + y2),−y/(x2 + y2)
)= (1, 0) (536)
=(x/(x2 + y2),−y/(x2 + y2)
)(x, y).
6. Distributivity of multiplication over addition:
(x1, y1) ((x2, y2) + (x3, y3)) = (x1, y1)(x2 + x3, y2 + y3) (537)
= (x1(x2 + x3)− y1(y2 + y3), x1(y2 + y3) + y1(x2 + x3)) (538)
= (x1x2 + x1x3 − y1y2 − y1y3, x1y2 + x1y3 + y1x2 + y1x3)(539)
= (x1x2 − y1y2, x1y2 + y1x2) + (x1x3 − y1y3, x1y3 + y1x3)(540)
= (x1, y1)(x2, y2) + (x1, y1)(x3, y3) (541)
94
The field of complex numbers is denoted by C. The fundamental theoremof algebra states that every non-constant polynomial in one variable withcoefficients in C has a root in C, i. e. C is algebraically closed.Remark: The field of real numbers, R, is not algebraically closed, becausethe polynomial x2 + 1 (which is non-constant and has real coefficients) hasno root in R.
By construction, any complex number z ∈ C may be written as an orderedpair, z = (x, y), and we call x := Rez ∈ R the real part of z and y := Imz ∈ Rthe imaginary part of z.
Complex Plane We go back to the vector space notion with the scalarproduct: if we identify the standard basis vectors with complex numbers,
(1, 0) ≡ 1, (0, 1) ≡ i, (542)
with the imaginary unit i ∈ C, we may also write
z = (Rez, Imz) = (x, y) = x(1, 0)+y(0, 1) = x ·1+y ·i = x+iy = Rez+iImz.(543)
We can visualize complex numbers as points in the complex plane, which isisomorphic to R2. By the definition of multiplication, we have
i2 = i · i = (0, 1)(0, 1) = (−1, 0) = −(1, 0) = −1. (544)
Now we may compute sums and products of complex numbers just as wewould for real numbers:
z1 + z2 = (Rez1 + iImz1) + (Rez2 + iImz2) (545)
= (x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2), (546)
= (Rez1 + Rez2) + i(Imz1 + Imz2), (547)
z1z2 = (Rez1 + iImz1)(Rez2 + iImz2) (548)
= (x1 + iy1)(x2 + iy2) = (x1x2 + ix1y2 + iy1x2 + i2y1y2)
= (x1x2 − y1y2) + i(x1y2 + y1x2) (549)
= (Rez1Rez2 − Imz1Imz2) + i(Rez1Imz2 + Imz1Rez2). (550)
We proceed in a similar way for the additive identity and inverse:
(0, 0) = 0(1, 0) + 0(0, 1) = 0 · 1 + 0 · i = 0, (551)
−z = −(Rez, Imz) = (−Rez,−Imz) = −Rez − iImz. (552)
95
Complex conjugate For a complex number z ∈ C, we define the complexconjugate z∗ ≡ z ∈ C by
z = (x, y) := (x,−y) = x− iy = Rez − iImz. (553)
Now we may express the real and imaginary parts of a complex number z ∈ Cby
1
2(z + z) =
1
2(Rez + iImz + Rez − iIm) = Rez, (554)
1
2i(z − z) =
1
2i(Rez + iImz − Rez + iImz) = Imz. (555)
Complex conjugation has the properties
z1 + z2 = (Rez1 + Rez2)− i(Imz1 + Imz2) (556)
= Rez1 − iImz1 + Rez2 − iImz2 = z1 + z2, (557)
z1z2 = (Rez1Rez2 − Imz1Imz2)− i(Rez1Imz2 + Imz1Rez2) (558)
= (Rez1 − iImz1)(Rez2 − iImz2) = z1 z2, (559)
z = Rez − iImz = Rez + iImz = z. (560)
Complex conjugation is simply identity for real numbers, which satisfy Imz =0. The following product is a non-negative number:
zz = (x− iy)(x+ iy) = x2 − i2y2 = x2 + y2 = (Rez)2 + (Imz)2 ≥ 0. (561)
The number|z| :=
√zz =
√(Rez)2 + (Imz)2 (562)
is the absolute value or modulus of z ∈ C.Remark: The vector space mentioned before becomes a normed vector spacewith the norm |(x, y)| =
√x2 + y2. This is, of course, the Euclidean norm of
the vector (x, y) ∈ R2.With the complex conjugate, we may write the multiplicative inverse in asimpler way:
z−1 = (x, y)−1 =(x/(x2 + y2),−y/(x2 + y2)
)=
(x,−y)
|z|2=
z
|z|2.(563)
As with any group, the inverse operations subtraction and division of complexnumbers are defined using addition and multiplication by
z1 − z2 := z1 + (−z2), z1, z2 ∈ C (564)z1
z2
:= z1z−12 =
z1z2
|z2|2, z1, z2 ∈ C, z2 6= 0. (565)
96
13.2 Polar Form of Complex Numbers. Powers andRoots
The polar form of a complex number z ∈ C is motivated from using polarcoordinates in the complex plane:
x = r cosϕ, y = r sinϕ, r ≥ 0, ϕ ∈ [0, 2π). (566)
Then we havez = x+ iy = r (cosϕ+ i sinϕ) , (567)
with|z| =
√x2 + y2 = r, (568)
so the absolute value |z| is just the distance of the point z from the originin the complex plane. The argument of a complex number z ∈ C \ 0,arg z = ϕ, is determined by the equations:
cosϕ =x
r, sinϕ =
y
r. (569)
Notice that the argument ϕ is not uniquely determined by the equationtanϕ = y/x.
Triangle Inequality Complex numbers cannot be ordered, but we havethe triangle inequality
|z1 + z2| ≤ |z1|+ |z2|, z1, z2 ∈ C. (570)
By induction, we may also show that
|z1 + z2 + · · ·+ zn| ≤ |z1|+ |z2|+ · · ·+ |zn|, zk ∈ C, k = 1, . . . , n. (571)
These equations are, of course, the same as the triangle inequalities in R2.
97
Multiplication and Division in Polar Form The product of two com-plex numbers z1, z2 ∈ C in polar form is given by
z1z2 = r1(cosϕ1 + i sinϕ1)r2(cosϕ2 + i sinϕ2) (572)
= r1r2(cosϕ1 cosϕ2 − sinϕ1 sinϕ2) +
+ir1r2(sinϕ1 cosϕ2 + cosϕ1 sinϕ2) (573)
= r1r2 (cos(ϕ1 + ϕ2) + i sin(ϕ1 + ϕ2)) , (574)
where we have used addition theorems for trigonometric functions. Therefore,the absolute value and angle of the product z1z2 are given by
|z1z2| = r1r2 = |z1||z2|, (575)
arg(z1z2) = ϕ1 + ϕ2 = arg z1 + arg z2. (576)
With z1 = z2 = z, and by induction, we conclude that
|zn| = |z|n, (577)
arg (zn) = n arg z. (578)
By comparing the two formulas for zn, we obtain De Moivre’s formula:
(cosϕ+ i sinϕ)n = cos(nϕ) + i sin(nϕ). (579)
To obtain the absolute value and argument for the quotient, we write
z1 =z1
z2
z2 ⇒ |z1| =∣∣∣∣z1
z2
∣∣∣∣ |z2| (580)
and therefore we have ∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
. (581)
In a similar way we obtain for the argument:
arg z1 = arg
(z1
z2
)+ arg z2, (582)
and thus
arg
(z1
z2
)= arg z1 − arg z2. (583)
Therefore, we may write the quotient z1/z2 in polar form as
z1
z2
=r1
r2
(cos(ϕ1 − ϕ2) + i sin(ϕ1 − ϕ2)) . (584)
98
Roots The n-th root of z ∈ C, n ∈ N, is the solution of the polynomialequation wn = z. We write the numbers w, z ∈ C in polar form as
z = r (cosϕ+ i sinϕ) , w = R (cosϑ+ i sinϑ) . (585)
Then, with De Moivre’s formula, we obtain
wn = Rn (cos(nϑ) + i sin(nϑ)) = r (cosϕ+ i sinϕ) , (586)
and thusRn = r, cos(nϑ) = cosϕ, sin(nϑ) = sinϕ. (587)
There are n solutions, given by
R = r1/n = n√r, ϑk =
ϕ+ 2kπ
n, k = 0, . . . , n− 1. (588)
Therefore, the n-th root of z is n-valued:
n√z = wk = n
√|z|(
cos
(arg z + 2kπ
n
)+ i sin
(arg z + 2kπ
n
))(589)
= n√|z|(
cos(arg z
n
)+ i sin
(arg z
n
))(cos
(2kπ
n
)+ i sin
(2kπ
n
)),
= n√|z|(
cos(arg z
n
)+ i sin
(arg z
n
))(cos
(2π
n
)+ i sin
(2π
n
))k,
for k = 0, . . . , n−1. In particular, for the n-th roots of unity, n√
1, we obtain,with |1| = 1, arg 1 = 0:
n√
1 = cos
(2kπ
n
)+ i sin
(2kπ
n
)=
(cos
(2π
n
)+ i sin
(2π
n
))k, (590)
for k = 0, . . . , n−1. These points lie on the unit circle in the complex plane,and they form the vertices of a regular n-sided polygon.Remark: It is a common mistake to just write down the solution for k = 0,and to neglect the other values of k, for example to give the solution of
x2 = 5 (n = 2) as x =√
5 instead of x = ±√
5. (591)
99
13.3 Derivative. Holomorphic Function.
We want to consider functions f : C→ C. In order to analyze these functions,we need a few topological definitions. First of all, we notice that the function
d : C× C→ R, d(z1, z2) := |z1 − z2|, (592)
with the absolute value | · | of a complex number defined as in (562), is ametric on C, namely the metric induced by the norm | · | (Problem Set 7).Therefore, (C, d) is a metric space. We define the open balls (disks) withcenter a ∈ C and radius ρ > 0 by
Bρ(a) := z ∈ C | d(z, a) < ρ. (593)
The open balls generate a topology on C, making it a topological space. Anysubset U ⊆ C inherits the topology from C and becomes a topological spaceitself. Standard topological definitions apply:
Definition 3 Let U ⊆ C denote a subset with the topology inherited from C.
1. U is open if every point z ∈ U has an open neighborhood contained inU :
∀ z ∈ U ∃ ε > 0 : Bε(z) ⊆ U. (594)
Remark: ε will depend on z in general.
2. U is closed if its complement C \ U is open.
3. U is disconnected if it is the union of two disjoint nonempty open sets:
∃ ∅ 6= V,W ⊆ C, V,W open, V ∩W = ∅ : U = V ∪W. (595)
Otherwise, U is connected.
4. If U is both open and connected, it is called a domain.
5. The closure U of U consists of the points z ∈ C which are “close” toU :
z ∈ U :⇔ ∀ ε > 0 ∃w ∈ U : w ∈ Bε(z). (596)
Remark: w will depend on ε in general. If z ∈ U , we may have w = z.
100
6. The interiorU of U consists of the points z ∈ U which are “not on the
edge” of U :
z ∈U :⇔ ∃ ε > 0 : Bε(z) ⊆ U. (597)
Remark: U open ⇔ U =U (compare Definitions 1 and 6). We also
have the dualityU= C \ C \ U and U = C \ (
C \ U).
7. The boundary of U , ∂U , is the set of points which are in the closure of
U , but not in the interior: ∂U := U\U= U ∩ C \ U . In other words,
every open ball around a boundary point z ∈ ∂U contains points fromboth U and C \ U .
Complex function For a function f : C → C, z 7→ w = f(z), we maysplit both the argument and the value into real and imaginary parts:
f(x+ iy) = u(x, y) + iv(x, y), (598)
with real-valued functions u, v : R2 → R.Examples:
1. f(z) := z2 + 3z. Then we have
f(x+ iy) = (x+ iy)2 + 3 (x+ iy) = x2− y2 + 3x+ i (2xy + 3y) , (599)
so that we find u(x, y) = x2 − y2 + 3x, v(x, y) = 2xy + 3y.
2. f(z) := 2iz + 6z. Then we have
f(z) = 2i(x+ iy) + 6(x− iy) = 6x− 2y + i (2x− 6y) , (600)
and so we find u(x, y) = 6x− 2y, v(x, y) = 2x− 6y.
Limit, Continuity
Definition 4 Let f : C→ C, z 7→ f(z), denote a complex function.
1. The limit of f as z approaches z0 is L if f(z) is “close” to L for all z“close” to z0:
∀ ε > 0 ∃ δ > 0 : f(z) ∈ Bε(L) ∀ z ∈ Bδ(z0), z 6= z0. (601)
101
In this case we writelimz→z0
f(z) = L. (602)
Remark: z0 need not be in the domain of f , nor does L need to be inthe range of f . z may approach z0 from any direction in the complexplane.
2. f is continuous at z0 ∈ C if f(z0) is defined and
limz→z0
f(z) = f(z0). (603)
3. f is continuous in the domain U ⊆ C if it is continuous at all pointsz0 ∈ U .
Derivative
Definition 5 Let f denote a complex function. The derivative of f at z0 ∈ Cis defined by
f ′(z0) := limz→z0
f(z)− f(z0)
z − z0
, (604)
provided this limit exists. In this case, f is said to be differentiable at z0.
Complex differentiability is a strong condition, because it means that onwhatever path in the complex plane the value z approaches z0, the quotient(f(z)− f(z0))/(z − z0) always approaches the same value.Examples:
1. The function f(z) := z2 is differentiable for all z0 ∈ C with derivativef ′(z0) = 2z0:
limz→z0
z2 − z20
z − z0
= limz→z0
(z + z0)(z − z0)
z − z0
= limz→z0
(z + z0) = 2z0. (605)
2. The function f(z) := z is not differentiable at any point z0 ∈ C: forany two points z, z0 ∈ C, we have
f(z)− f(z0)
z − z0
=z − z0
z − z0
=z − z0
z − z0
=Re(z − z0)− iIm(z − z0)
Re(z − z0) + iIm(z − z0)=: w.
(606)
102
If Re(z−z0) = 0, we have w = −1, and if Im(z−z0) = 0, we have w = 1.Therefore, we may construct a sequence znn∈N with limn→∞ zn → z0,so that the sequence wnn∈N oscillates between 1 and −1, i. e. is notconvergent.
If f is differentiable at z0, it must be continuous at z0, but the opposite is nottrue. The rules for differentiation are the same as for real functions (productrule, quotient rule, chain rule).
Holomorphic functions
Definition 6 A function f which is defined and differentiable at every pointz ∈ U ⊆ C is called holomorphic in U . A holomorphic function is a functionthat is holomorphic in some domain U ⊆ C. A function which is holomorphicin all of C is an entire function.
Example:
1. Monomials zn, n ∈ N0, are entire functions and so are polynomials,i. e. functions of the form
p(z) = c0 + c1z + c2z2 + · · ·+ cnz
n, c0, . . . , cn ∈ C. (607)
2. Rational functions
f(z) :=p(z)
q(z), (608)
where p, q are polynomials, are differentiable everywhere except at thezeros of q, assuming that common factors have been canceled.
13.4 Cauchy-Riemann Equations. Laplace’s Equation.
The Cauchy-Riemann equations are a system of two linear first-order par-tial differential equations which provide a criterion for differentiability of acomplex function:
Theorem 2 If a complex function f(x+ iy) = u(x, y) + iv(x, y) is holomor-phic in U ⊆ C, then the first partial derivatives of u and v exist and theysatisfy the Cauchy-Riemann equations
ux = vy, uy = −vx, in U. (609)
103
Theorem 3 If two real-valued continuous functions u, v : R2 → R havecontinuous first partial derivatives that satisfy the Cauchy-Riemann equations(609) in some domain U ⊆ C, then the complex function f(x+iy) := u(x, y)+iv(x, y) is holomorphic in U .
Remark: valid under weaker conditions (Looman, 1923; Menchoff, 1936)Sketch of Proofs:
• Theorem 2: For a point z0 ∈ U , we know that the derivative of f isgiven by
f ′(z0) = limz→z0
f(z)− f(z0)
z − z0
(610)
(and that this limit exists). We choose two particular paths z → z0
such that
f ′(x0 + iy0) = limx→x0
u(x, y0)− u(x0, y0)
x− x0
+ i limx→x0
v(x, y0)− v(x0, y0)
x− x0
,
f ′(x0 + iy0) = −i limy→y0
u(x0, y)− u(x0, y0)
y − y0
+ limy→y0
v(x0, y)− v(x0, y0)
y − y0
.
From these equations, we conclude that the first partial derivatives ofu, v exist at (x0, y0) and also that
ux(x0, y0) = vy(x0, y0), uy(x0, y0) = −vx(x0, y0). (611)
This is true for any point z0 = x0 + iy0 ∈ U .Remark: We also conclude from this proof that
f ′(x0 +iy0) = ux(x0, y0)+ivx(x0, y0) = −iuy(x0, y0)+vy(x0, y0). (612)
• Theorem 3: Choose z0 ∈ U . Because U is open, it contains an openball around z0 = (x0, y0), and therefore we may find a second pointz = (x, y) ∈ U such that the line segment which connects the twopoints lies entirely in U . Because of the differentiability of u and v, wemay apply the mean value theorem in R2. It guarantees the existence ofpoints m1,m2 ∈ R2 on the line segment connecting the points (x0, y0)and (x, y), such that
u(x, y)− u(x0, y0) = (x− x0)ux(m1) + (y − y0)uy(m1), (613)
v(x, y)− v(x0, y0) = (x− x0)vx(m2) + (y − y0)vy(m2). (614)
104
Then we have, with the Cauchy-Riemann equations (609),
f(z)− f(z0) = u(x, y)− u(x0, y0) + i (v(x, y)− v(x0, y0))(615)
= (x− x0)ux(m1) + (y − y0)uy(m1) +
+i ((x− x0)vx(m2) + (y − y0)vy(m2)) (616)(609)= (x− x0)ux(m1)− (y − y0)vx(m1) +
+i ((x− x0)vx(m2) + (y − y0)ux(m2)) . (617)
With z − z0 = x− x0 + i(y − y0) we obtain
f(z)− f(z0)
z − z0
= ux(m1) + ivx(m1) +
−iy − y0
z − z0
(ux(m1)− ux(m2)) +
−ix− x0
z − z0
(vx(m1)− vx(m2)) . (618)
We let z → z0 and with the continuity of the partial derivatives of uand v we obtain
limz→z0
f(z)− f(z0)
z − z0
= ux(x0, y0) + ivx(x0, y0) (619)
= limx→x0
u(x, y0)− u(x0, y0)
x− x0
+ i limx→x0
v(x, y0)− v(x0, y0)
x− x0
.
We have shown existence of the limit for a path along the line connect-ing z and z0. Because z was arbitrarily chosen in an open ball aroundz0, f is differentiable at z0. This is true for any z0 ∈ U , and thereforef is holomorphic in U .
Examples:
1. f(z) = z2 is an entire function. We have u(x, y) = x2 − y2, v(x, y) =2xy. The partial derivatives are given by
ux = 2x = vy, uy = −2y = −vx, (620)
so u, v satisfy the Cauchy-Riemann equations (609) everywhere in R2
105
2. f(z) = z = x− iy, so that u(x, y) = x, v(x, y) = −y. We obtain
ux = 1 = −vy, uy = 0 = −vx, (621)
so the first Cauchy-Riemann equation is not satisfied. By Theorem 2, fcannot be differentiable. We have verified that before, in Section 13.3.
3. f(z) = ex(cos y + i sin y). We find u(x, y) = ex cos y, v(x, y) = ex sin yand therefore
ux = ex cos y = vy, uy = −ex sin y = −vx. (622)
This is true in any subset U ⊆ C. By Theorem 3, f is an entire function.
4. Assume that a function f is holomorphic in U ⊆ C and satisfies |f(z)| =k = const., z ∈ U . We want to show that f must be a constant function.
We know that |f(x+ iy)|2 = u(x, y)2 + v(x, y)2 = k2. We differentiatewith respect to x and y to obtain
uux + vvx = 0, uuy + vvy = 0. (623)
By Theorem 2, u and v must satisfy the Cauchy-Riemann equations(609), so that
uux − vuy = 0, uuy + vux = 0. (624)
We may eliminate uy from the first equation and ux from the secondand obtain
(u2 + v2)ux = 0, (u2 + v2)uy = 0. (625)
Now if u2 + v2 = k2 = 0, we have u ≡ v ≡ 0 and thus f ≡ 0. If k2 6= 0,we conclude that ux ≡ uy ≡ 0. With the Cauchy-Riemann equations,we also obtain that vx ≡ vy ≡ 0. Therefore u ≡ const., v ≡ const. andhence f = u+ iv ≡ const..
In polar form, f(r(cosϕ + i sinϕ)) = u(r, ϕ) + iv(r, ϕ), and the Cauchy-Riemann equations are given by (Problem Set 8)
ur =1
rvϕ, vr = −1
ruϕ. (626)
106
Laplace’s Equation. Harmonic Functions The real and imaginaryparts of a holomorphic function satisfy Laplace’s equation:
Theorem 4 If f(x + iy) = u(x, y) + iv(x, y) is holomorphic in a domainU ⊆ C, then both u and v satisfy Laplace’s equation:
∆u = uxx + uyy = 0, ∆v = vxx + vyy = 0, in U, (627)
and they have continuous second partial derivatives in U .
Sketch of proof: Taking second partial derivatives of u, we obtain
uxx = (vy)x = vyx, uyy = −(vx)y = −vxy. (628)
We shall prove in Chapter 14 that the derivative of a holomorphic functionis again holomorphic, which implies that u and v have continuous partialderivatives of any order. This means that we can exchange the order ofdifferentiation: vyx = vxy, and therefore uxx = −uyy. Similar for v.
A harmonic function is a twice continuously differentiable function whichsatisfies Laplace’s equation. Harmonic functions are studied in potential the-ory. Because the real and imaginary parts of a holomorphic function in Care harmonic functions in R2, complex analysis is useful for potential theoryin two space dimensions (to be treated in Chapter 18).
Let a function u be defined in some set Ω ⊂ R2. The function v is aharmonic conjugate of u if f(x+ iy) := u(x, y) + iv(x, y) is holomorphic.Example: u(x, y) := x2 − y2 − y. We have ux = 2x, uy = −2y − 1, anduxx = 2 = −uyy, so that u is a harmonic function. We want to solve theCauchy-Riemann equations (609) for the partial derivatives of v:
vx!
= −uy = 2y + 1, vy!
= ux = 2x. (629)
We obtain by integration of the second equation:
v(x, y) = 2xy + h(x) ⇒ vx(x, y) = 2y + h′(x), (630)
and from the first equation we conclude that h′(x)!
= 1. Therefore, h(x) =x + c, c ∈ R, and v(x, y) = 2xy + x + c. The corresponding holomorphicfunction is given by
f(x+ iy) = x2 − y2 − y + i (2xy + x+ c) = (x+ iy)2 + i(x+ iy + c), (631)
or f(z) = z2 + i(z + c), z ∈ C. The conjugate of a given harmonic functionis uniquely determined up to an arbitrary additive constant.
107
13.5 Exponential Function
The exponential function exp : C→ C is defined by
exp(x+ iy) ≡ ex+iy := ex(cos y + i sin y) = ex cos y + iex sin y. (632)
We shall now list several important properties of this function:
• extension of real exponential function: z = x ∈ R ⇒ ez = ex
• exp is an entire function (Section 13.4, Example 3)
• derivative: exp′(z) = exp(z)
exp′(x+ iy) = (ex cos y)x + i (ex sin y)x = ex cos y + iex sin y
= ex(cos y + i sin y) = exp(x+ iy) (633)
• exp(z1 + z2) = exp(z1) exp(z2), z1, z2 ∈ C
– Special case z1 := x ∈ R, z2 := iy, y ∈ R, then ex+iy = exeiy.Compare with the definition (632) ⇒ Euler’s formula
eiy = cos y + i sin y, y ∈ R. (634)
Then the polar form of z ∈ C may also be written as
z = r(cosϕ+ i sinϕ) = reiϕ. (635)
– Absolute value
|eiy| =√
cos2 y + sin2 y = 1,∣∣ex+iy
∣∣ = ex > 0. (636)
This implies that the function exp never vanishes.
– Set r = 1 and choose specific values for ϕ:
eiπ2 = i, eiπ = −1, ei
3π2 = −i, e2πi = 1. (637)
– Periodicity with period 2πi:
ez+2πi = eze2πi = ez, z ∈ C. (638)
All the values that exp(z) can assume are already assumed in thefundamental region Imz ∈ (−π, π] (or any other horizontal stripof width 2π in the complex plane). This implies that exp is notan injective (one-to-one) function.
108
13.6 Trigonometric and Hyperbolic Functions
With the complex exponential function, trigonometric and hyperbolic func-tions may also be generalized to the complex numbers: for z ∈ C, we define
cos z :=1
2
(eiz + e−iz
), sin z :=
1
2i
(eiz − e−iz
), (639)
as well as
cosh z :=1
2
(ez + e−z
), sinh z :=
1
2
(ez − e−z
). (640)
Since the complex exponential is entire, so are the functions cos, sin, cosh,and sinh. We may also consider quotients such as
tan z :=sin z
cos z, tanh z :=
sinh z
cosh z. (641)
The derivatives of these functions are given like for the real functions, namelyby
cos′ z = − sin z, sin′ z = cos z, (642)
cosh′ z = sinh z, sinh′ z = cosh z. (643)
We determine the real and imaginary parts from the definitions:
cos(x+ iy) =1
2
(ei(x+iy) + e−i(x+iy)
)=
1
2
(eixe−y + e−ixey
)(644)
=1
2e−y (cosx+ i sinx) +
1
2ey (cosx− i sinx) (645)
= cosx1
2
(ey + e−y
)− i sinx
1
2
(ey − e−y
)(646)
= cosx cosh y − i sinx sinh y. (647)
In the same way, we find
sin(x+ iy) = sinx cosh y + i cosx sinh y, (648)
cosh(x+ iy) = coshx cos y + i sinhx sin y, (649)
sinh(x+ iy) = sinhx cos y + i coshx sin y. (650)
109
Absolute values are given by
| cos z|2 = cos2 x cosh2 y + sin2 x sinh2 y (651)
cosh2 y−sinh2 y=1= cos2 x(1 + sinh2 y) + sin2 x sinh2 y (652)
= cos2 x+ (cos2 x+ sin2 x) sinh2 y (653)
= cos2 x+ sinh2 y, (654)
and in the same way
| sin z|2 = sin2 x+ sinh2 y (655)
| cosh z|2 = cosh2 x− sin2 y (656)
| sinh z|2 = sinh2 x+ sin2 y (657)
Notice that the complex trigonometric functions sin, cos are not bounded.All zeros of sin, cos lie on the real axis, because the only solution of sinh y = 0is y = 0. The following formulas are also true for the complex trigonometricfunctions:
cos(z1 ± z2) = cos z1 cos z2 ∓ sin z1 sin z2, (658)
sin(z1 ± z2) = sin z1 cos z2 ± cos z1 sin z2, (659)
and
cos2 z + sin2 z = 1. (660)
The following relations between the complex trigonometric and hyperbolicfunctions do not exist for their real counterparts:
cosh(iz) = cos z, sinh(iz) = i sin z, (661)
cos(iz) = cosh z, sin(iz) = i sinh z. (662)
13.7 Logarithm. General Power
The natural logarithm w = log z (sometimes denoted by ln z) is defined as theinverse of the exponential function: ew = z. We already know that ew 6= 0
110
∀w ∈ C, and therefore log is undefined at z = 0. For w = u + iv, z = reiϕ,r > 0, we obtain the equations
eueiv = reiϕ. (663)
The solutions u, v are given by
u = log r, v = ϕ+ 2πn, n ∈ Z. (664)
The natural logarithm of z is multivalued: w = log r + i(ϕ + 2πn), n ∈ Z.The principal value, Logz, is the one which lies in the fundamental regionof the exponential function: Im(Logz) ∈ (−π, π]. Then the other values oflog z are given by
log z = Logz + 2πin, n ∈ Z. (665)
Theorem 5 The natural logarithm log (665) is holomorphic except for z ≤0. Its derivative is given by
log′ z =1
z, z 6= 0, arg z 6= π. (666)
The negative real axis z < 0 is called the branch cut of log.
Proof: We know that the logarithm cannot be defined for z = 0. For all othervalues z = reiϕ ∈ C, r > 0, ϕ ∈ [0, 2π), the principal value of the logarithmis given by
Logz =
log r + iϕ, ϕ ∈ [0, π]log r + i(ϕ− 2π), ϕ ∈ (π, 2π)
. (667)
Therefore, log is not continuous at ϕ = π, and thus not holomorphic forz < 0. For values z ∈ C \ z ≤ 0 we have, in polar form
log(reiϕ
)= log r + i (ϕ+ 2πn) = u(r, ϕ) + iv(r, ϕ), n ∈ Z, (668)
with u(r, ϕ) := log r, v(r, ϕ) := ϕ + 2πn. We verify the Cauchy-Riemannequations
ur =1
r=
1
rvϕ, vr = 0 = −1
ruϕ, (669)
111
so log is holomorphic in C \ z ≤ 0, by Theorem 3. For the derivative, wefind
log′(reiϕ
)= cosϕur(r, ϕ)− 1
rsinϕuϕ(r, ϕ) + i cosϕvr(r, ϕ)− i1
rsinϕvϕ(r, ϕ)
=1
rcosϕ− i1
rsinϕ =
1
re−iϕ, (670)
so that log′ z = 1/z. The general power of z ∈ C is defined by
zc := ec log z, z, c ∈ C, z 6= 0. (671)
For c = n and c = 1/n, n ∈ N, we obtain the n-th power and n-th root ofz ∈ C, respectively (Section 13.2).
112
14 Complex Integration
For functions of real numbers f : R → R, we know indefinite integrals (orantiderivatives)
F (x) =
∫f(x) dx, F ′(x) = f(x), (672)
and definite integrals over an interval [a, b] ⊂ R:
b∫a
f(x) dx = F (b)− F (a) (673)
(fundamental theorem of calculus). For complex functions f : C→ C, indef-inite integrals are defined in the same way, and definite integrals generalizeto line integrals in the complex plane.
14.1 Line Integral in the Complex Plane
Curves in C A curve C in C is represented by a continuous mapping (path)γ : [a, b]→ C, a, b ∈ R, as
C = γ(t) | t ∈ [a, b]. (674)
Examples:
• γ(t) := t+3it, t ∈ [0, 2], parametrizes a segment of the line Imz = 3Rez.
• γ(t) := 4 cos t+ 4i sin t, t ∈ [0, 2π), parametrizes the circle |z| = 4.
Notice that the parametrization for a given curve C is not unique. We denoteby −C the curve oriented in the opposite direction:
−C := γ(a+ b− t) | t ∈ [a, b]. (675)
The arc length of a curve C is defined by
L(C) := supa=t0<t1<···<tn−1<tn=b
n∈N
n∑m=1
|γ(tm)− γ(tm−1)| . (676)
The arc length is an intrinsic property of the curve C, i. e. it does notdepend on the choice of the parametrization γ. The curve C is rectifiable ifL(C) <∞. This is true for Lipschitz-continuous functions γ, for example.
113
Definition of the line integral For a given n ∈ N, we choose n+1 pointsin the interval [a, b],
a = t0 < t1 < · · · < tn = b. (677)
With the mapping γ we obtain n+ 1 points zm := γ(tm) ∈ C, m = 0, . . . , n.Now we can divide the curve C into n segments:
C =n⋃
m=1
Cm, Cm = γ(t) | t ∈ [tm−1, tm], m = 1, . . . , n. (678)
Consider a rectifiabe curve C. For a continuous complex function f which isdefined (at least) at each point z ∈ C, and for any choice of points ζm ∈ Cm,m = 1, . . . , n, we define the sums
Sn :=n∑
m=1
f(ζm) (zm − zm−1) . (679)
Remark: Sn depends on the partition of the interval [a, b] and on the choicesof ζ1, . . . , ζn.The line integral of f over the path of integration C is defined by∫
C
f(z) dz := limn→∞
Sn. (680)
Under the assumptions that f is continuous and C is rectifiable, the limiton the right-hand side of (680) exists. It turns out that it is independentof the choice of points ζm, m = 1, . . . , n, and also that the line integral isindependent of the choice of the parametrization γ of C. If the curve C isclosed, γ(b) = γ(a), then we also write∮
C
f(z) dz. (681)
Basic Properties
1. Linearity:∫C
(k1f1(z) + k2f2(z)) dz = k1
∫C
f1(z) dz + k2
∫C
f2(z) dz. (682)
114
2. Sense reversal: ∫−C
f(z) dz = −∫C
f(z) dz. (683)
3. Partitioning of path:∫C
f(z) dz =
∫C1
f(z) dz +
∫C2
f(z) dz (684)
Evaluation via antiderivative This evaluation method requires a com-plex function which is holomorphic on a simply connected domain.
Definition 7 1. A subset U ⊆ C is path-connected (0-connected) if forany two points z0, z1 ∈ U there is a (continuous) path γ : [0, 1] → Uwith γ(0) = z0, γ(1) = z1.
2. A path-connected subset U ⊆ C is simply connected (1-connected) ifany path from z0 ∈ U to z1 ∈ U can be continuously transformed intoany other such path, while staying within U, for any z0, z1 ∈ U .
Remarks:
• Path-connected sets are connected, but not every connected set is path-connected.
• Simply connected sets are necessarily path-connected, and thereforealso connected.
• Any subset of C with a hole is not simply connected.
A complex function which is holomorphic on a simply connected, open setU ⊆ C has an antiderivative:
Theorem 6 Let f : C → C be holomorphic in a simply connected domainU . Then the antiderivative F of f exists (F ′(z) = f(z) ∀ z ∈ U) and is alsoholomorphic. Furthermore, for any path in U joining two points z0, z1 ∈ U ,we have
z1∫z0
f(z) dz = F (z1)− F (z0). (685)
115
Proof: Section 14.2 (Cauchy’s Integral Theorem)
Remark: We writez1∫z0
instead of∫C
in this case, because the value of the integral
is the same for any curve C from z0 to z1.Examples:
1. U = C, f(z) = z2:
1+i∫0
z2 dz =z3
3
∣∣∣∣1+i
0
=1
3(i+ 1)3 = −2
3+
2
3i. (686)
2. U = C \ z ≤ 0, f(z) = 1/z:
i∫−i
1
zdz = Logi− Log(−i) =
iπ
2−(−iπ
2
)= iπ. (687)
Evaluation via parametrization This method is not restricted to holo-morphic functions, nor does it require special topological properties.
Theorem 7 Let C = γ(t) | t ∈ [a, b] be a regular piecewise C1-curve (i. e. γcontinuous in [a, b], piecewise continuously differentiable in (a, b) and γ(t) 6=0 ∀ t ∈ (a, b)), and let f : C→ C be a continuous function on C. Then
∫C
f(z) dz =
b∫a
f (γ(t)) γ(t) dt. (688)
116
Proof: We write f(x+ iy) = u(x, y) + iv(x, y), zm = xm + iym, m = 0, . . . , n,ζm = ξm + iηm, m = 1, . . . , n. Then, for any n ∈ N:
Sn =n∑
m=1
(u(ξm, ηm) + iv(ξm, ηm)) (xm − xm−1 + i(ym − ym−1)) (689)
=n∑
m=1
(u(ξm, ηm)(xm − xm−1)− v(ξm, ηm)(ym − ym−1)) +
+in∑
m=1
(v(ξm, ηm)(xm − xm−1) + u(ξm, ηm)(ym − ym−1)) (690)
=n∑
m=1
(u(ξm, ηm)−v(ξm, ηm)
)·(xm − xm−1
ym − ym−1
)+
+in∑
m=1
(v(ξm, ηm)u(ξm, ηm)
)·(xm − xm−1
ym − ym−1
). (691)
With γ = λ+iµ, we define r(t) := (λ(t), µ(t))>. The function r parametrizesa curve in R2, which contains the points (ξm, ηm)>, m = 1, . . . , n, and(xm, ym)>, m = 0, . . . , n. We also denote this curve by C. We also define thevector fields
F 1(x, y) :=
(u(x, y)−v(x, y)
), F 2(x, y) :=
(v(x, y)u(x, y)
)(692)
Then the sum Sn (691) converges to a line integral in R2, as n→∞:
limn→∞
Sn =
∫C
F 1(r) · dr + i
∫C
F 2(r) · dr. (693)
Because r is also a regular piecewise C1 function, we may write
limn→∞
Sn =
∫C
f(z) dz =
b∫a
F 1(r(t)) · r(t) dt+ i
b∫a
F 2(r(t)) · r(t) dt. (694)
117
Using the definitions, we write
∫C
f(z) dz =
b∫a
u(λ(t), µ(t))λ(t) dt−b∫
a
v(λ(t), µ(t))µ(t) dt+
+i
b∫a
v(λ(t), µ(t))λ(t) dt+ i
b∫a
u(λ(t), µ(t))µ(t) dt (695)
=
b∫a
(u(λ(t), µ(t)) + iv(λ(t), µ(t)))(λ(t) + iµ(t)
)dt(696)
=
b∫a
f(γ(t))γ(t) dt (697)
Examples:
1. f(z) = 1/z, γ(t) = eit, t ∈ [0, 2π] (C is the unit circle in the complexplane, oriented counterclockwise). Then γ(t) = ieit, and therefore
∮C
1
zdz =
2π∫0
e−itieit dt = i
2π∫0
1 dt = 2πi. (698)
Remark: 1/z is holomorphic in D := C \ 0, but D is not simplyconnected. Therefore, Theorem 6 (which yields zero for the integralover any closed path) is not applicable!
2. z0 ∈ C, f(z) = (z − z0)m, m ∈ Z, γ(t) = z0 + ρeit, ρ > 0, t ∈ [0, 2π].That is, C = z ∈ C | |z − z0| = ρ is the circle around z0 with radiusρ. We obtain
f(γ(t)) =(z0 + ρeit − z0
)m= ρmeimt, (699)
γ(t) = iρeit. (700)
118
Then∮C
(z − z0)m dz =
2π∫0
ρmeimtiρeit dt = iρm+1
2π∫0
ei(m+1)t dt (701)
= iρm+1
2π∫0
cos ((m+ 1)t) dt+ i
2π∫0
sin ((m+ 1)t) dt
=
2πi, m = −10, m ∈ Z \ −1 . (702)
3. If Theorem 6 is applicable, then the value of the integral is indepen-dent of the path connecting the points z0, z1 ∈ C. This is not true ingeneral: Consider the function f(z) = Rez, which is continuous butnot holomorphic (Thm. 2), and two curves C1, C2 represented by
γ1(t) := t+ 2it, t ∈ [0, 1], (703)
γ2(t) :=
t, t ∈ [0, 1)1 + 2i(t− 1), t ∈ [1, 2]
. (704)
Then
f(γ1(t)) = t, t ∈ [0, 1], f(γ2(t)) =
t, t ∈ [0, 1)1, t ∈ [1, 2]
. (705)
Both curves connect the points 0 and 1 + 2i in the complex plane, andboth are regular and piecewise C1:
γ1(t) = 1 + 2i, t ∈ [0, 1], γ2(t) =
1, t ∈ [0, 1)2i, t ∈ [1, 2]
. (706)
119
Then, by Theorem 7:
∫C1
f(z) dz =
1∫0
f(γ1(t))γ1(t) dt =
1∫0
t(1 + 2i) dt, (707)
= (1 + 2i)t2
2
∣∣∣∣10
=1
2+ i, (708)
∫C2
f(z) dz =
2∫0
f(γ2(t))γ2(t) dt =
1∫0
t dt+
2∫1
2i dt (709)
=t2
2
∣∣∣∣10
+ 2i t|21 =1
2+ 2i, (710)
so we have ∫C1
f(z) dz 6=∫C2
f(z) dz. (711)
Bounds for Integrals. ML-Inequality
Proposition 2 If f is bounded on C, i. e. ∃M > 0: |f(z)| ≤ M ∀ z ∈ C,then ∣∣∣∣∣∣
∫C
f(z) dz
∣∣∣∣∣∣ ≤ML, (712)
where L = L(C) denotes the arc length of the curve C.
Proof: For the sums (679), we obtain (zm = γ(tm))
|Sn| ≤n∑
m=1
|f(ζm)| |γ(tm)− γ(tm−1)| ≤Mn∑
m=1
|γ(tm)− γ(tm−1)| (713)
≤ ML(C), (714)
by the definition of the arc length (676). Example: f(z) = z2, γ(t) = t+ it, t ∈ [0, 1]. L(C) = |1 + i| =
√2, and
|f(γ(t))| = |(1 + i)t|2 = 2t2 ≤ 2, t ∈ [0, 1]. (715)
120
Therefore the line integral of f over C can be bounded as∣∣∣∣∣∣∫C
z2 dz
∣∣∣∣∣∣ ≤ 2√
2. (716)
With Theorem 6, we obtain the exact value∫C
z2 dz =z3
3
∣∣∣∣1+i
0
=(1 + i)3
3= −2
3+
2
3i, (717)
and therefore ∣∣∣∣∣∣∫C
z2 dz
∣∣∣∣∣∣ =2√
2
3. (718)
14.2 Cauchy’s Integral Theorem
This is the main theorem in this Chapter, and it will allow us to prove Thm. 6.
Theorem 8 Let U ⊆ C be open and simply connected, let the complex func-tion f be holomorphic in U , and let C be a rectifiable, closed curve in U .Then ∮
C
f(z) dz = 0. (719)
Proof: We write f(x + iy) = u(x, y) + iv(x, y). This proof uses Green’stheorem, which requires that the partial derivatives of u and v are continuous.Note that Thm. 2 implies only existence of these partial derivatives, but notcontinuity (Goursat’s proof works without assuming continuity of f ′, but it’smore complicated)! We know from the proof of Thm. 7 that∮
C
f(z) dz =
∫C
(u dx− v dy) + i
∫C
(v dx+ u dy) (720)
121
By Green’s theorem∫C
(u dx− v dy) =
∫U
(−vx − uy) dx dy, (721)
∫C
(v dx+ u dy) =
∫U
(ux − vy) dx dy. (722)
From the Cauchy-Riemann equations (Thm. 2) ux = vy, uy = −vx, weconclude that (721), (722) vanish. Examples
1. If f is an entire function, the integral over any closed curve C vanishes:∮C
ez dz = 0,
∮C
cos z dz = 0,
∮C
zn dz = 0, n ∈ N0. (723)
2. Unit circle C = z ∈ C | |z| = 1. Functions which are not entirebut holomorphic on and inside of C: tan z is holomorphic except at(2k+1)π/2, k ∈ Z, (z2 +4)−1 is holomorphic except at z = ±2i. Thesepoints lie outside of C, and therefore∮
C
tan z dz = 0,
∮C
1
z2 + 4dz = 0. (724)
3. γ(t) = eit, f(z) = z is not holomorphic, and therefore Thm. 8 is notapplicable. We compute, with Thm. 7:
∮C
z dz =
2π∫0
e−itieit dt = 2πi 6= 0. (725)
4. Domains which are not simply connected: f(z) := z−1 is holomorphic inC\0, but this domain is not simply connected. We have (Example 1after Thm. 7, Section 14.1), for γ(t) = eit:∮
C
1
zdz = 2πi 6= 0. (726)
122
Independence of Path
Theorem 9 If the complex function f is holomorphic in a simply connecteddomain U ⊆ C, then the integral of f is independent of path in U , i. e. itsvalue depends only on the two endpoints of C.
Proof: Let z1, z2 be any points in U . Consider two curves C1, C2 from z1 toz2 which do not intersect each other (so z1, z2 are the only common points ofC1 and C2). The path from z1 to z2 on C1 and back from z2 to z1 on −C2 isclosed, and from Thm. 8 we know that∫
C1
f dz +
∫−C2
f dz = 0 ⇒∫C1
f dz = −∫−C2
f dz =
∫C2
f dz. (727)
For paths with a finite number of common points, we may integrate overloops between these points.
Principle of Deformation of Path From the path independence (Thm. 9),we conclude that the integral of f over any continuous transformation of acurve C is the same as the integral of f over C, as long as the deformingpath contains only points at which f is holomorphic.Example: We know from Example 2 after Thm. 7 (Section 14.1) that∮
C
(z − z0)m dz =
2πi, m = −10, m ∈ Z \ −1 , (728)
where C is a circle in the complex plane with center z0 and radius ρ > 0.This is, in fact, true for any closed curve enclosing z0.
Existence of Indefinite Integral We may now prove Thm. 6: The con-ditions of Thm. 8 are satisfied. The line integral of f from any z0 ∈ U to anyz ∈ U is independent of path in U (Thm. 9). For a fixed z0 ∈ U , we define
F (z) :=
z∫z0
f(ζ) dζ. (729)
123
We show that F is holomorphic in U and that F ′(z) = f(z). We form thedifference quotient
F (z + ∆z)− F (z)
∆z=
1
∆z
z+∆z∫z0
f(ζ) dζ −z∫
z0
f(ζ) dζ
=1
∆z
z+∆z∫z
f(ζ) dζ.
(730)Because U is open, we may choose ∆z such that the curve segment from zto z + ∆z lies inside of U . With
1
∆z
z+∆z∫z
f(z) dζ = f(z)1
∆z
z+∆z∫z
dζ =z + ∆z − z
∆zf(z) = f(z), (731)
we obtain
F (z + ∆z)− F (z)
∆z− f(z) =
1
∆z
z+∆z∫z
(f(ζ)− f(z)) dζ. (732)
f is holomorphic, and therefore continuous, at z (Problem Set 7). Let ε > 0be given, then
∃ δ > 0 : |f(ζ)− f(z)| < ε ∀ |ζ − z| < δ. (733)
We choose ∆z such that |∆z| < δ, and with the ML-inequality (Prop. 2), weobtain∣∣∣∣F (z + ∆z)− F (z)
∆z− f(z)
∣∣∣∣ =1
|∆z|
∣∣∣∣∣∣z+∆z∫z
(f(ζ)− f(z)) dζ
∣∣∣∣∣∣ ≤ 1
|∆z|ε|∆z| = ε.
(734)By the definition of the limit and of the derivative, this proves that
F ′(z) = lim∆z→0
F (z + ∆z)− F (z)
∆z= f(z). (735)
z ∈ U was arbitrary, and therefore F is holomorphic in U and is an an-tiderivative of f . It remains to show that we may use any antiderivativeto evaluate (685). Indeed, if we have another holomorphic function G withG′(z) = f(z) ∀ z ∈ U , then F ′(z)−G′(z) = 0 in U , hence F −G is constantin U . Now
F (z1)− F (z0)− (G(z1)−G(z0)) = F (z1)−G(z1)− (F (z0)−G(z0)) = 0,(736)
for two points z0, z1 ∈ U , and so (685) holds for any antiderivative of f .
124
Cauchy’s Integral Theorem for Multiply Connected Domains Ifa complex function f is holomorphic in a domain U∗ containing a doublyconnected domain U with boundary curves C1, C2 (both with the same ori-entation), then U can be cut into two simply connected domains U1 and U2,in each of which Thm. 8 is applicable:∮
∂U1
f(z) dz =
∮∂U2
f(z) dz = 0, (737)
where both integrals are taken counterclockwise, for example. If we take thesum of these two integrals, the integrals over the cuts will cancel, and weobtain
0 =
∮C1
f(z) dz +
∮−C2
f(z) dz =
∮C1
f(z) dz −∮C2
f(z) dz, (738)
so that ∮C1
f(z) dz =
∮C2
f(z) dz. (739)
This procedure can be generalized to any multiply connected domain.
14.3 Cauchy’s Integral Formula
This is the most important consequence from Cauchy’s Integral Theorem(Thm. 8), and it is useful for evaluating integrals. Furthermore, it will helpus prove that holomorphic functions have derivatives of all orders (Section14.4), and that they are analytic (Section 15.4).
Theorem 10 Let the complex function f be holomorphic in a simply con-nected domain U . Then for any point z0 in U and any closed curve C in Uthat encloses z0, ∮
C
f(z)
z − z0
dz = 2πif(z0), (740)
the integration being taken counterclockwise.
Remark: The domain U does not need to be simply connected, as long asthe curve C can be continuously transformed into an arbitrarily small circle
125
around z0, while staying inside of U .Proof: We write∮
C
f(z)
z − z0
dz = f(z0)
∮C
1
z − z0
dz +
∮C
f(z)− f(z0)
z − z0
dz (741)
= 2πif(z0) +
∮C
f(z)− f(z0)
z − z0
dz. (742)
It remains to prove that ∮C
f(z)− f(z0)
z − z0
dz = 0. (743)
The integrand is holomorphic in U \ z0. We define the closed curve Kρ :=z ∈ C | |z − z0| = ρ, ρ > 0, oriented counterclockwise. For ρ > 0 smallenough, we have, with the principle of deformation of path:∮
C
f(z)− f(z0)
z − z0
dz =
∮Kρ
f(z)− f(z0)
z − z0
dz. (744)
The function f is holomorphic, and therefore continuous, at z0. Let ε > 0 begiven. Then
∃ δ > 0 : |f(z)− f(z0)| < ε ∀ |z − z0| < δ, z 6= z0. (745)
If we choose 0 < ρ < δ, we obtain∣∣∣∣f(z)− f(z0)
z − z0
∣∣∣∣ < ε
ρ, on Kρ. (746)
From the ML-Inequality (Prop. 2), we obtain∣∣∣∣∣∣∣∮Kρ
f(z)− f(z0)
z − z0
dz
∣∣∣∣∣∣∣ <ε
ρ2πρ = 2πε. (747)
ε > 0 was arbitrary, so the integral must vanish. Remark: We already know from Cauchy’s Integral Theorem (Thm. 8) thatthe integral in (740) vanishes if C does not enclose z0.Examples:
126
1. z0 = 2, f(z) = ez:∮C
ez
z − 2dz = 2πiez|z=2 = 2πie2 ' 46.4i, (748)
for any closed curve C enclosing z = 2.
2. z0 = i/2, f(z) = (z3/2− 3)/(z − i/2):∮C
z3 − 6
2z − idz =
∮C
12z3 − 3
z − i2
dz = 2πi
(1
2z3 − 3
)∣∣∣∣z=i/2
=π
8−6πi, (749)
for any closed curve C enclosing z = i/2.
3. ∮C
z2 + 1
z2 − 1dz =
∮C
z2 + 1
(z + 1)(z − 1)dz (750)
Four cases:
a) C encloses z = 1 but not z = −1:∮C
z2 + 1
z2 − 1dz =
∮C
z2+1z+1
z − 1dz = 2πi
z2 + 1
z + 1
∣∣∣∣z=1
= 2πi, (751)
because f(z) := (z2 + 1)/(z+ 1) is holomorphic in any set U with−1 6∈ U .
b) C encloses z = −1 but not z = 1:∮C
z2 + 1
z2 − 1dz =
∮C
z2+1z−1
z + 1dz = 2πi
z2 + 1
z − 1
∣∣∣∣z=−1
= −2πi, (752)
because f(z) := (z2 + 1)/(z− 1) is holomorphic in any set U with1 6∈ U .
c) Both points z = 1 and z = −1 lie outside of C:∮C
z2 + 1
z2 − 1dz = 0, (753)
because f(z) := (z2 +1)/(z2−1) is holomorphic in any set U with−1, 1 6⊆ U .
127
d) Both points z = 1 and z = −1 lie inside of C: We draw two closedcurves C2, C3 around z = 1 and z = −1, respectively. WithCauchy’s integral theorem we obtain∮
C
z2 + 1
z2 − 1dz =
∮C2
z2 + 1
z2 − 1dz +
∮C3
z2 + 1
z2 − 1dz (754)
=
∮C2
z2+1z+1
z − 1dz +
∮C3
z2+1z−1
z + 1dz (755)
= 2πi− 2πi = 0, (756)
from cases a) and b).
14.4 Derivatives of Holomorphic Functions
Cauchy’s Integral Formula (Thm. 10) may be used to show that holomorphicfunctions have derivatives of any order, which is different from the situationin real calculus. This is Cauchy’s differentiation formula:
Theorem 11 If the complex function f is holomorphic in a domain U ⊆ C,then it has derivatives of all orders in U , which are then also holomorphicfunctions in U . The values of these derivatives at a point z0 ∈ U are givenby
f (n)(z0) =n!
2πi
∮C
f(z)
(z − z0)n+1dz, n ∈ N. (757)
Here, C is any closed curve in U that encloses z0 and whose full interiorbelongs to U ; and we integrate counterclockwise around C.
Remark: For n = 0, this is Cauchy’s Integral Formula (Thm. 10). Derivativesfor n ≥ 1 are obtained by formally differentiating under the integral sign,with respect to z0.Proof: We know from Thm. 10 that we have a contour integral representationof f at any point z0 ∈ U :
f(z0) =1
2πi
∮C
f(z)
z − z0
dz. (758)
128
Choose a point z0 ∈ U . The derivative of f at z0 is given by
f ′(z0) = lim∆z→0
f(z0 + ∆z)− f(z0)
∆z, (759)
if the limit exists. With (758) we have
f(z0 + ∆z)− f(z0)
∆z=
1
2πi∆z
∮C
f(z)
z − (z0 + ∆z)dz −
∮C
f(z)
z − z0
dz
=
1
2πi∆z
∮C
(z − z0)f(z)− (z − z0 −∆z)f(z)
(z − z0 −∆z)(z − z0)dz
=1
2πi
∮C
f(z)
(z − z0 −∆z)(z − z0)dz. (760)
Now we compute∮C
f(z)
(z − z0 −∆z)(z − z0)dz−
∮C
f(z)
(z − z0)2dz =
∮C
∆zf(z)
(z − z0 −∆z)(z − z0)2dz.
(761)We now show that this integral approaches 0 as ∆z → 0. f is continuouson C, and therefore bounded: ∃K > 0 : |f(z)| ≤ K ∀ z ∈ C. Let d be themiminum distance from z0 to any point on C:
d := minz∈C|z − z0| ⇒ 1
|z − z0|2≤ 1
d2∀ z ∈ C (762)
Furthermore, with the triangle inequality,
d ≤ |z − z0| ≤ |z − z0 −∆z|+ |∆z|, ∀ z ∈ C. (763)
For |∆z| ≤ d/2 we have
d
2≤ d− |∆z| ≤ |z − z0 −∆z| ⇒ 1
|z − z0 −∆z|≤ 2
d, ∀ z ∈ C. (764)
Let L := L(C) denote the arc length of the curve C. For |∆z| ≤ d/2, we usethe ML-inequality to obtain∣∣∣∣∣∣∮C
∆zf(z)
(z − z0 −∆z)(z − z0)2dz
∣∣∣∣∣∣ ≤ |∆z|K 2
d
1
d2=
2K
d3|∆z| → 0, ∆z → 0.
(765)
129
This proves that f ′(z0) exists, and is given by
f ′(z0) =1
2πi
∮C
f(z)
(z − z0)2dz. (766)
Therefore f ′ may be represented at z0 as a contour integral. This is true forany z0 ∈ U . We repeat the previous argument with f replaced by f ′ and(758) replaced by (766) to obtain a contour integral representation of f ′′ atany z0 ∈ U . Thus we can prove Cauchy’s differentiation formula for anyn ∈ N. Examples:
1. For any contour enclosing the point πi (counterclockwise):∮C
cos z
(z − πi)2dz = 2πi (cos z)′
∣∣z=πi
= −2πi sin(πi) = 2π sinhπ. (767)
2. For any countour which encloses 1 but ±2i lie outside (counterclock-wise):∮C
ez
(z − 1)2(z2 + 4)dz = 2πi
(ez
z2 + 4
)′∣∣∣∣z=1
= 2πiez(z2 + 4)− ez2z
(z2 + 4)2
∣∣∣∣z=1
=6eπ
25i ' 2.05i. (768)
Cauchy’s Inequality. Liouville’s and Morera’s Theorems Cauchy’sdifferentiation formula (Thm. 11) has several interesting corollaries. Wechoose C := z ∈ C | |z − z0| = r, for an r > 0. With |f(z)| ≤ M ,z ∈ C, we obtain using the ML-inequality
∣∣f (n)(z0)∣∣ =
n!
2π
∣∣∣∣∣∣∮C
f(z)
(z − z0)n+1dz
∣∣∣∣∣∣ ≤ n!
2πM
1
rn+12πr =
n!M
rn. (769)
Equation (769) is Cauchy’s inequality.
Theorem 12 (Liouville) If an entire function is bounded in absolute valuein the whole complex plane, then this function must be constant.
130
Proof: Assume that |f(z)| ≤ K for all z ∈ C. For a z0 ∈ C, we concludefrom Cauchy’s inequality that |f ′(z0)| ≤ K/r. This holds for every r > 0,because f is entire. It follows that f ′(z0) = 0. Since z0 is arbitrary, we havef ′(z) = 0 for all z ∈ C, hence ux = vx = 0, and uy = vy = 0, by Thm. 2.Thus u ≡ const. and v ≡ const., and f = u+ iv ≡ const..
Theorem 13 (Morera) If a complex function f is continuous in a simplyconnected domain U and if ∮
C
f(z) dz = 0 (770)
for every closed curve C in U , then f is holomorphic in U .
Remark: This is the converse of Cauchy’s Integral Theorem (Thm. 8).Proof: In Section 14.2, we had derived the antiderivative
F (z) =
z∫z0
f(ζ) dζ, z ∈ U, (771)
which is holomorphic in U and satisfies F ′(z) = f(z). By Thm. 11, F ′ isholomorphic in U , which implies that f is holomorphic in U .
131
15 Power Series, Taylor Series
Chapters 15 and 16 deviate from the “main road”, which leads us from PDEsvia Complex Analysis to Potential Theory in 2D (Chapter 18). This separatepath will lead to the technique of Residue Integration (Section 16.3), whichis useful, for example, to evaluate certain complicated real integrals (Section16.4).
In this chapter, we will introduce series and (complex) analytic functions,and we will prove that every holomorphic function is analytic.
15.1 Sequences, Series, Convergence Tests
These concepts are very similar as in real calculus. We will go through themrather quickly and not present all of the proofs in detail.
Definition 8 1. A sequence znn≥1 is a function from N to C. Thenumbers zn ∈ C, n ∈ N, are the terms of the sequence.
2. A sequence znn≥1 is convergent if it has a limit, i. e. there is a c ∈ Cso that limn→∞ zn = c, or
∀ ε > 0 ∃N ∈ N : |zn − c| < ε ∀n > N. (772)
Otherwise the sequence is called divergent.
3. A Cauchy sequence is a sequence znn≥1 with the property that theterms are getting closer and closer together:
∀ ε > 0 ∃N ∈ N : |zm − zn| < ε ∀m,n > N. (773)
Remark: Every convergent sequence is a Cauchy sequence. In fact, the space(C, d), with the metric d induced by the absolute value | · |, is a completemetric space, so that the opposite is also true!Examples:
1. The sequence in/nn≥1 is convergent with limit 0.
2. The sequences inn≥1 and (1 + i)nn≥1 are divergent.
132
Definition 9 1. Given a sequence znn≥1, we call the numbers
sn :=n∑
m=1
zm = z1 + z2 + · · ·+ zn, n ∈ N, (774)
the n-th partial sums of the series∞∑m=1
zm = z1 + z2 + · · · (775)
The complex numbers z1, z2, . . . are the terms of the series. The partialsums snn≥1 form a new sequence.
2. A series is convergent if its sequence of partial sums snn≥1 converges,say limn→∞ sn = s ∈ C. In that case, we call s the sum of the series,and we write
s =∞∑m=1
zm = z1 + z2 + · · · (776)
Otherwise the series is called divergent.
3. A series z1 + z2 + · · · is called absolutely convergent if the series∞∑m=1
|zm| (777)
is convergent. A series which is convergent but not absolutely conver-gent is called conditionally convergent.
4. The series
Rn :=∞∑
m=n+1
zm = zn+1 + zn+2 + · · · , n ∈ N, (778)
is called the remainder of the series (775) after the term zn.
Remarks:
1. For convergent series, we have Rn = s− sn → 0, n→ 0.
2. Due to completeness, every absolutely convergent series is convergent(triangle inequality and Cauchy criterion).
Example: The alternating harmonic series∑
n≥1 ((−1)n+1/n) converges, butnot absolutely.
133
Tests for Convergence and Divergence of Series The following crite-rion is commonly used to establish divergence of a series:
Theorem 14 If a series z1 + z2 + · · · converges, then limn→∞ zn = 0.
Proof: Since zn = sn − sn−1, n ≥ 2, we have
limn→∞
zn = limn→∞
(sn − sn−1) = 0, (779)
because the sequence snn≥1 is convergent by assumption, and therefore aCauchy sequence. Remark: The harmonic series
∑n≥1(1/n), satisfies limn→∞(1/n) = 0, but is
divergent. So limn→∞ zn = 0 is necessary, but not sufficient, for convergence.The Cauchy convergence principle allows us to establish convergence of a
series without knowing its sum:
Theorem 15 A series z1 + z2 + · · · is convergent if and only if the sequenceof partial sums snn≥1 is a Cauchy sequence:
∀ ε > 0 ∃N ∈ N : |sn+p − sn| =
∣∣∣∣∣n+p∑
m=n+1
zm
∣∣∣∣∣ < ε ∀n > N, p ∈ N. (780)
From Cauchy’s convergence principle, we may derive several convergencetests.
Theorem 16 (Comparison Test) If a series z1 + z2 + · · · is given and wecan find a convergent (real) series b1 + b2 + · · · with bn ≥ 0, n ∈ N, such that|zn| ≤ bn, n ∈ N, then the series z1 + z2 + · · · converges absolutely.
Proof: Let ε > 0 be given. We can find an N ∈ N such that
bn+1 + · · ·+ bn+p < ε ∀n > N, p ∈ N. (781)
For those values of n and p, we have
|zn+1|+ · · ·+ |zn+p| ≤ bn+1 + · · ·+ bn+p < ε. (782)
Then by Thm. 15, the series |z1| + |z2| + · · · converges, i. e. the series z1 +z2 + · · · converges absolutely.
A series often used in a comparison test is the geometric series.
134
Proposition 3 The geometric series
∞∑m=0
qm = 1 + q + q2 + · · · (783)
converges with the sum 1/(1− q) if |q| < 1 and diverges if |q| ≥ 1.
Proof: For |q| ≥ 1 we have |qm| ≥ 1, m ≥ 0, and Thm. 14 implies divergence.Let |q| < 1. The n-th partial sum is given by
sn = 1 + q + · · ·+ qn, n ∈ N, (784)
so thatqsn = q + · · ·+ qn + qn+1. (785)
After subtraction, we are left with
sn − qsn = (1− q)sn = 1− qn+1. (786)
Since q 6= 1, we have 1− q 6= 0, and we may solve (786) for sn:
sn =1− qn+1
1− q=
1
1− q− qn+1
1− q→ 1
1− q, n→∞, (787)
since |q| < 1. In the following sections, we will mostly use the ratio test to establish
convergence of a series.
Theorem 17 If a series z1 + z2 + · · · with zn 6= 0, n ∈ N, has the propertythat for some N ∈ N and for some q < 1∣∣∣∣zn+1
zn
∣∣∣∣ ≤ q, ∀n > N, (788)
then this series converges absolutely. If∣∣∣∣zn+1
zn
∣∣∣∣ ≥ 1, ∀n > N, (789)
then the series z1 + z2 + · · · diverges.
135
Remark: q = 1 is not sufficient (cf. harmonic series)!Proof: From (789), we have that |zn+1| ≥ |zn|, n > N , so divergence of theseries follows from Thm. 14. If (788) holds, then |zn+1| ≤ q|zn|, n > N . Inparticular, |zN+p| ≤ q|zN+p−1| ≤ q2|zN+p−2| ≤ · · · ≤ qp−1|zN+1|, p ∈ N. WithProp. 3, we find that
|zN+1|+ |zN+2|+ · · · ≤ |zN+1|(1 + q + q2 + · · ·
)≤ |zN+1|
1
1− q. (790)
The comparison test (Thm. 16) yields absolute convergence of the seriesz1 + z2 + · · · .
If the sequence of ratios in Thm. 17 converges, we get the more convenient
Theorem 18 If a series z1 + z2 + · · · with zn 6= 0, n ∈ N, is such that
limn→∞
∣∣∣∣zn+1
zn
∣∣∣∣ = L, (791)
then
a) If L < 1, the series converges absolutely.
b) If L > 1, the series diverges.
c) If L = 1, the series may converge or diverge, i. e. the test fails andpermits no conclusion.
Proof:
a) We define kn := |zn+1/zn| and let L = 1 − b < 1. By definition of thelimit, kn must eventually get close to 1− b, say, kn ≤ q := 1− b/2 < 1,n > N , for some N . Convergence of z1 + z2 + · · · now follows fromThm. 17.
b) For L = 1 + c > 1 we have kn ≥ 1 + c/2 > 1, n > N∗, for some N∗
sufficiently largte, which implies divergence of z1 +z2 + · · · by Thm. 17.
c) The (divergent) harmonic series∑
n≥1(1/n) satisfies
limn→∞
∣∣∣∣zn+1
zn
∣∣∣∣ = limn→∞
n
n+ 1= 1. (792)
136
On the other hand, the series∑
n≥1(1/n2) also satisfies
limn→∞
∣∣∣∣zn+1
zn
∣∣∣∣ = limn→∞
n2
(n+ 1)2= 1. (793)
This series is convergent, however, as can be seen from the inequality
sn =n∑
m=1
1
m2≤ 1 +
n∫1
1
x2dx = 2− 1
n. (794)
The (real) sequence snn≥1 is bounded, and it is also monotone in-creasing, and thus convergent (integral test for a real series).
Examples:
1. The series∞∑n=0
(100 + 75i)n
n!(795)
is convergent: We have∣∣∣∣zn+1
zn
∣∣∣∣ =|100 + 75i|n+1n!
|100 + 75i|n(n+ 1)!=|100 + 75i|n+ 1
=125
n+ 1→ 0, n→∞.
(796)Absolute convergence of the series now follows from Thm. 18 with L =0. The sum of this series is exp(100 + 75i).
2. Consider the two sequences ann≥0 with an := 2−3ni, n ≥ 0, andbnn≥0 with bn := 2−(3n+1), n ≥ 0. Consider the series
a0 + b0 + a1 + b1 + · · · = i+1
2+i
8+
1
16+
i
64+
1
128+ · · · (797)
The sequence of ratios of absolute values of successive terms are givenby
1
2,1
4,1
2,1
4, . . . (798)
so that Thm. 18 is not applicable. From Thm. 17, however, we inferabsolute convergence of the series.
Another important convergence test is the root test, and there are severalothers.
137
15.2 Power Series
Power series are the most important series in complex analysis because weshall see that their sums are holomorphic functions, and every holomorphicfunction can be represented by power series.
Definition 10 The power series with center z0 ∈ C and coefficients an ∈ C,n ∈ N0, is given by
∞∑n=0
an(z − z0)n = a0 + a1(z − z0) + a2(z − z0)2 + · · · . (799)
Remark: Notice that a power series is a function of z ∈ C. For z fixed, allthe concepts for series with constant terms in the last section apply. Usuallya series with variable terms will converge for some z and diverge for others.Examples:
1. Convergence in a disk: the geometric series
∞∑n=0
zn = 1 + z + z2 + · · · (800)
converges absolutely if |z| < 1 and diverges if |z| ≥ 1 (Prop. 3).
2. Convergence for every z: the power series
∞∑n=0
zn
n!= 1 + z +
z2
2!+z3
3!+ · · · (801)
is absolutely convergent for every z ∈ C. In fact, by the ratio test(Thm. 18) ∣∣∣∣∣
zn+1
(n+1)!
zn
n!
∣∣∣∣∣ =|z|n+ 1
→ 0, n→∞. (802)
3. Convergence only at z0: the power series
∞∑n=0
n!zn = 1 + z + 2z2 + 6z3 + · · · (803)
138
converges only at z = 0, but diverges for every z 6= 0. In fact, from theratio test (Thm. 17)∣∣∣∣(n+ 1)!zn+1
n!zn
∣∣∣∣ = (n+ 1)|z| ≥ 1 ∀n ≥ 1
|z|− 1. (804)
Theorem 19 a) Every power series (799) converges at the center z0.
b) If (799) converges at z = z1 6= z0, it converges absolutely for every zcloser to z0 than z1, |z − z0| < |z1 − z0|.
c) If (799) diverges at z = z2, it diverges for every z farther away from z0
than z2, |z − z0| > |z2 − z0|.
Proof:
a) For z = z0, the series reduces to the single term a0.
b) Convergence at z = z1 implies by Thm. 14 that an(z1 − z0)n → 0,n→∞. Therefore, the sequence an(z1− z0)nn≥0 must be bounded:
|an(z1 − z0)n| < M ∀n ∈ N0. (805)
Then we have
|an(z − z0)n| =∣∣∣∣an(z1 − z0)n
(z − z0
z1 − z0
)n∣∣∣∣ ≤M
∣∣∣∣ z − z0
z1 − z0
∣∣∣∣n , n ∈ N0.
(806)Summation over n gives
∞∑n=0
|an(z − z0)n| ≤M∞∑n=1
∣∣∣∣ z − z0
z1 − z0
∣∣∣∣n . (807)
For |z− z0| < |z1− z0|, the series on the right-hand side is a converginggeometric series (Prop. 3). Absolute convergence of (799) now followsfrom the comparison test (Thm. 16).
c) If we had convergence at a point z3 with |z3 − z0| > |z2 − z0|, then wewould have convergence at z2 by b), a contradiction.
139
Radius of Convergence of a Power Series
Definition 11 Consider circles with center z0, which include all the pointsat which a given power series (799) converges. The smallest such circle,
z ∈ C | |z − z0| = R, (808)
is called the circle of convergence and its radius R the radius of convergenceof (799). We write R =∞ if the series (799) converges for all z, and R = 0if it converges only at the center.
Remark: From Thm. 19, we conclude that (799) converges for every |z−z0| <R. Because R is as small as possible, we also have that the series (799)diverges for all z with |z − z0| > R. No general statements can be made forthe points on the circle, |z − z0| = R.Example: The following three series all have convergence radius R = 1. Onthe circle of convergence,
•∑n
zn
n2converges:
∑n
|z|n
n2=∑n
1
n2,
•∑n
zn
nconverges at z = −1, but diverges at z = 1 (harmonic series),
•∑n
zn diverges:∑n
|z|n =∑n
1n
The Cauchy-Hadamard formula allows us to determine the radius of conver-gence from the coefficients of the power series:
Theorem 20 Suppose that the sequence |an+1/an|, n ∈ N, converges withlimit L∗. If L∗ = 0, then R =∞ (convergence everywhere). If L∗ > 0, thenthe convergence radius is given by
R =1
L∗= lim
n→∞
∣∣∣∣ anan+1
∣∣∣∣ . (809)
If |an+1/an| → ∞, then R = 0 (convergence only at the center).
140
Proof: We consider the absolute values of the ratios∣∣∣∣an+1(z − z0)n+1
an(z − z0)n
∣∣∣∣ =
∣∣∣∣an+1
an
∣∣∣∣ |z − z0| → L∗|z − z0| =: L, n→∞. (810)
We want to use the ratio test, Thm. 18. For L∗ > 0, we have convergenceif L = L∗|z − z0| < 1, i. e. if |z − z0| < 1/L∗, and divergence if |z − z0| >1/L∗. Therefore, R = 1/L∗ is the convergence radius of the power series.L∗ = 0 implies L = 0 which gives convergence for all z by the ratio test. If|an+1/an| → ∞, then |an+1/an||z − z0| > 1 for any z 6= z0 and all sufficientlylarge n. This implies divergence for all z 6= z0 by the ratio test, Thm. 17. Example: For the power series
∞∑n=0
(2n)!
(n!)2(z − 3i)n (811)
with center z0 = 3i, we compute the convergence radius
R = limn→∞
( (2n)!(n!)2
(2n+2)!((n+1)!)2
)= lim
n→∞
(n+ 1)2
(2n+ 2)(2n+ 1)=
1
4. (812)
Remark: Thm. 20 can be extended to cases where the sequence of ratios doesnot converge: we have
R =1
L, L = lim
n→∞|an|1/n, (813)
and even
R =1˜, ˜ the greatest limit point of the sequence
|an|1/n
n≥0
. (814)
15.3 Functions Given by Power Series
For a power series with center z0 ∈ C and with positive convergence radiusR > 0, we consider the function
f(z) :=∞∑n=0
anzn = a0 + a1z + a2z
2 + · · · , z ∈ BR(z0) (815)
141
(we use the notation with open balls again, cf. Section 13.3). We say that fis represented by the power series of that it is developed in the power seriesnear z0.Example: The function f(z) := 1/(1 − z) is represented by the geometricseries in the interior of the unit circle:
1
1− z=∞∑n=0
zn, z ∈ B1(0), (816)
by Prop. 3.
Definition 12 A complex function f is analytic at z0 ∈ C if it can be rep-resented as a power series centered at z0 with positive convergence radius:
∃R > 0, ann≥0 : f(z) =∞∑n=0
an(z − z0)n, ∀ z ∈ BR(z0). (817)
We want to prove uniqueness of this representation first.
Theorem 21 If a complex function f is analytic at z0, then f is continuousat z0.
Proof: From (817), we know that f(z0) = a0, so we must prove that limz→z0 f(z) =a0. For any r ∈ (0, R), the power series (817) converges absolutely if|z − z0| ≤ r, by Thm. 19. Hence the series
S :=∞∑n=1
|an|rn−1 =1
r
∞∑n=1
|an|rn (818)
converges. For 0 < |z − z0| ≤ r we obtain
|f(z)− a0| =
∣∣∣∣∣∞∑n=1
an(z − z0)n
∣∣∣∣∣ ≤ |z − z0|∞∑n=1
|an||z − z0|n−1 (819)
≤ |z − z0|∞∑n=1
|an|rn−1 = |z − z0|S. (820)
If S = 0, we are done. Otherwise, let ε > 0 be given, and set
δ := minr,ε
S
. (821)
Then we have |z − z0|S < ε, for all z ∈ Bδ(z0), and therefore f(z) → a0,z → z0.
142
Theorem 22 Let the power series∞∑m=0
am(z − z0)m,∞∑m=0
bm(z − z0)m (822)
both be convergent in BR(z0), for an R > 0, and let them both have the samesum in BR(z0). Then the series are identical, i. e. an = bn, n ∈ N0.
Remark: This implies that if a complex function f is analytic at z0, then itspower series representation near z0 is unique.Proof: We proceed by induction with respect to n. By assumption,
∞∑m=0
am(z − z0)m =∞∑m=0
bm(z − z0)m, z ∈ BR(z0). (823)
By Theorem 21, the sums of these power series are continuous at z0. Considerz0 6= z ∈ BR(z0) and let z → z0 on both sides; this yields a0 = b0, i. e. theassertion is true for n = 0. Now assume that am = bm for all m ≤ n. Wesubtract equal terms from both sides of (823) and divide by (z− z0)n+1 6= 0:
∞∑m=n+1
am(z − z0)m−(n+1) =∞∑
m=n+1
bm(z − z0)m−(n+1). (824)
Like before we let z → z0 and conclude that an+1 = bn+1.
Operations on Power Series This discussion will prepare for our maingoal in this section, namely, to show that a function which is analytic at z0
is holomorphic in BR(z0).
• Termwise addition or subtraction of two power series with the samecenter z0 ∈ R and radii of convergence R1, R2 yields a power serieswith center z0 and radius of convergence R ≥ minR1, R2. This isbecause we may add/subtract the partial sums sn and s∗n, n ∈ N, termby term and sue limn→∞ (sn ± s∗n) = limn→∞ sn ± limn→∞ s
∗n.
• Termwise multiplication of two power series with the same center z0 ∈C,
f(z) :=∞∑k=0
ak(z − z0)k, z ∈ BR1(z0), (825)
g(z) :=∞∑m=0
bm(z − z0)m, z ∈ BR2(z0), (826)
143
leads to the Cauchy product
s(z) = f(z)g(z) =∞∑n=0
cn(z − z0)n, z ∈ BR(z0), (827)
with R = minR1, R2 and
cn =n∑k=0
akbn−k, n ∈ N0. (828)
• Termwise differentiation and integration of power series is permissible:
Theorem 23 The power series
∞∑n=1
nan(z − z0)n−1 (829)
obtained by differentiating the power series∑
n≥0 an(z − z0)n term byterm has the same radius of convergence as the original series.
Example: Differentiating the geometric series (R = 1) twice term byterm and multiplying with z2/2 yields
∞∑n=2
(n
2
)zn, (830)
which also has radius of convergence R = 1, by Thm. 23.
Theorem 24 The power series
∞∑n=0
ann+ 1
(z − z0)n+1 (831)
obtained by integrating the power series∑
n≥0 an(z− z0)n term by termhas the same radius of convergence as the original series.
Both theorems can be proved with the Cauchy-Hadamard formula(Thm. 20) or with the generalizations (813) or (814).
144
Power Series Represent Holomorphic Functions
Theorem 25 A power series with center z0 ∈ C and radius of convergenceR > 0 represents a holomorphic function in BR(z0). The derivatives of thisfunction are obtained by differentiating the original series term by term. Allthe series thus obtained have the same radius of convergence R. Hence bythe first statement, each of them represents a holomorphic function.
Corollary 1 A complex function f which is analytic at z0 ∈ C is holomor-phic in BR(z0), where R > 0 denotes the radius of convergence of the powerseries representation of f near z0.
Remark: In the next section, we will show that the opposite is also true(Taylor’s theorem).Proof of the Theorem: The proof is in three steps:
1. We write the power series representation of f and the derived series by
f(z) =∞∑n=0
an(z − z0)n, f1(z) =∞∑n=1
nan(z − z0)n−1, z ∈ BR(z0).
(832)We want to show that f is holomorphic in BR(z0) and f ′ = f1. Withtermwise addition, we obtain
f(z + ∆z)− f(z)
∆z−f1(z) =
∞∑n=2
an
((z + ∆z − z0)n − (z − z0)n
∆z− n(z − z0)n−1
),
(833)for z ∈ BR(z0).
2. With b := z + ∆z − z0, a := z − z0, ∆z = b− a, we have
(z + ∆z − z0)n − (z − z0)n
∆z−n(z− z0)n−1 =
bn − an
b− a−nan−1, n ≥ 2.
(834)We prove by induction that
bn − an
b− a− nan−1 = (b− a)
n−2∑k=0
(k + 1)akbn−2−k
︸ ︷︷ ︸=:An
, n ≥ 2. (835)
145
• n = 2:
b2 − a2
b− a− 2a =
(b+ a)(b− a)
b− a− 2a = b− a = (b− a)A2 (836)
• m 7→ m+ 1: Assume the formula is true for n = m. We write
bm+1 − am+1
b− a=
bm+1 − bam + bam − am+1
b− a(837)
=b(bm − am) + (b− a)am
b− a= b
bm − am
b− a+ am
= b((b− a)Am +mam−1
)+ am (838)
= (b− a)(bAm +mam−1
)+ (m+ 1)am. (839)
From the definition of Am we obtain
bAm+mam−1 =m−2∑k=0
(k+1)akbm−1−k+mam−1 =m−1∑k=0
(k+1)akbm−1−k = Am+1,
(840)so that
bm+1 − am+1
b− a− (m+ 1)am = Am+1. (841)
Therefore we have
(z + ∆z − z0)n − (z − z0)n
∆z−n(z−z0)n−1 = ∆z
n−2∑k=0
(k+1)(z−z0)k(z+∆z−z0)n−2−k,
(842)for n ≥ 2.
3. We use this to obtain
f(z + ∆z)− f(z)
∆z−f1(z) =
∞∑n=2
an∆zn−2∑k=0
(k+1)(z−z0)k(z+∆z−z0)n−2−k.
(843)For |z| ≤ R0 and |z + ∆z| ≤ R0, R0 < R, we estimate∣∣∣∣∣n−2∑k=0
(k + 1)(z − z0)k(z + ∆z − z0)n−2−k
∣∣∣∣∣ ≤n−2∑k=0
(k+1)Rn−20 =
n(n− 1)
2Rn−2
0 .
(844)
146
Therefore we have∣∣∣∣f(z + ∆z)− f(z)
∆z− f1(z)
∣∣∣∣ ≤ |∆z|12∞∑n=2
|an|n(n− 1)Rn−20 . (845)
The second derived series of f at z = z0 +R0 is given by
∞∑n=2
ann(n− 1)Rn−20 , (846)
and because of R0 < R it converges absolutely by Thms. 23 and 19.Thus, finally∣∣∣∣f(z + ∆z)− f(z)
∆z− f1(z)
∣∣∣∣ ≤ |∆z|K(R0), K(R0) :=1
2
∞∑n=2
|an|n(n−1)Rn−20 .
(847)Because R0 < R is arbitrary, and letting ∆z → 0, we conclude that f isdifferentiable at any point z ∈ BR(z0) (i. e. f holomorphic in BR(z0)),and that its derivative is represented by the derived series.
The statements about the higher derivatives follow by induction.
15.4 Taylor and Maclaurin Series
The Taylor series of a complex function f is given by the power series
f(z) =∞∑n=0
an(z − z0)n, (848)
where the coefficients ann≥0 are given by
an =1
n!f (n)(z0)
Thm. 11=
1
2πi
∮C
f(ζ)
(ζ − z0)n+1dζ, n ∈ N0. (849)
A Maclaurin series is a Taylor series with center z0 = 0. The remainder ofthe Taylor series (848) after the term an(z − z0)n is
Rn(z) =∞∑
k=n+1
ak(z − z0)k =(z − z0)n+1
2πi
∮C
f(ζ)
(ζ − z0)n+1(ζ − z)dζ, (850)
147
n ≥ 0. Thus we obtain Taylor’s formula with remainder:
f(z) = f(z0) + f ′(z0)(z − z0) + · · ·+ f (n)(z0)
n!(z − z0)n +Rn(z), (851)
n ≥ 0. Taylor’s theorem states that every holomorphic function can berepresented by Taylor series (with various centers):
Theorem 26 Let the complex function f be holomorphic in a domain U ⊆C, and let z0 ∈ U . Then there exists precisely one Taylor series (848) withcenter z0 that represents f . This representation is valid in the largest opendisk with center z0 in which f is holomorphic. The remainders Rn, n ≥ 0, of(848) can be represented in the form (850). The coefficients ann≥0 satisfythe inequality
|an| ≤M
rn, n ≥ 0, M := max
|z−z0|=r|f(z)|, (852)
where r > 0 is such that Br(z0) ⊆ U .
Corollary 2 A complex function f which is holomorphic in U ⊆ C is ana-lytic at every point z0 ∈ U .
Remark: Now we understand why holomorphic and analytic are often usedsynonymously. Keep in mind, however, that this equivalence did not followimmediately from the definition of a holomorphic function (Section 13.3)!Proof of the theorem: From Cauchy’s Integral Formula (Thm. 10), we knowthat
f(z) =1
2πi
∮C
f(ζ)
ζ − zdζ, (853)
if z is enclosed by C. We choose C = z ∈ C | |z − z0| = r, and we develop1/(ζ − z) in powers of z − z0:
1
ζ − z=
1
ζ − z0 − (z − z0)=
1
(ζ − z0)(
1− z−z0ζ−z0
) (854)
Because ζ ∈ C and |z − z0| < r, we have∣∣∣∣z − z0
ζ − z0
∣∣∣∣ < 1. (855)
148
From the finite geometric sum
1 + q + · · ·+ qn =1− qn+1
1− q=
1
1− q− qn+1
1− q, q 6= 1, (856)
we conclude that
1
1− q= 1 + q + · · ·+ qn +
qn+1
1− q, (857)
for any n ∈ N0. Using this with q := z−z0ζ−z0 we obtain
1
ζ − z=
1
ζ − z0
(1 +
z − z0
ζ − z0
+ · · ·+(z − z0
ζ − z0
)n)+
1
ζ − z
(z − z0
ζ − z0
)n+1
.
(858)We insert this into (853):
f(z) =1
2πi
∮C
f(ζ)
ζ − z0
dζ +z − z0
2πi
∮C
f(ζ)
(ζ − z0)2dζ + · · ·
+(z − z0)n
2πi
∮C
f(ζ)
(ζ − z0)n+1dζ +Rn(z), (859)
with Rn given by (850). The integrals are those in (849) related to thederivatives, so that we have proved the Taylor formula (851). Because theholomorphic function f has derivatives of all orders (Thm. 11), we may taken in (851) arbitrarily large. For n → ∞, we obtain (848). This series willconverge and represent f if and only if limn→∞Rn(z) = 0, which we prove asfollows. Because z lies inside of C and ζ ∈ C, we have |ζ − z| > 0. Becausef is holomorphic inside and on C (by the choice of r), f is bounded on C,and so is the function f(ζ)/(ζ − z), say,∣∣∣∣ f(ζ)
ζ − z
∣∣∣∣ ≤ M, ∀ ζ ∈ C. (860)
Also, C has the radius r = |ζ−z0| and the length 2πr. By the ML-inequality(Prop. 2), we obtain from (850):
|Rn(z)| =|z − z0|n+1
2π
∣∣∣∣∣∣∮C
f(ζ)
(ζ − z0)n+1(ζ − z)dζ
∣∣∣∣∣∣ (861)
≤ |z − z0|n+1
2πM
1
rn+12πr = Mr
∣∣∣∣z − z0
r
∣∣∣∣n+1
. (862)
149
Because z lies inside of C we have |z−z0| < r, or |(z−z0)/r| < 1, and thereforeRn(z) → 0, n → ∞. This proves that the Taylor series converges and hasthe sum f(z). Uniqueness follows from Thm. 22. Finally, the inequality forthe coefficients (852) follows from (849) and from Cauchy’s inequality (769).Remarks:
1. We may achieve any prescribed accuracy in approximating f(z) by apartial sum of (848) by choosing n large enough.
2. On the circle of convergence of the Taylor series (848) there is at leastone singular point of f , i. e. a point c at which f is not holomorphic(otherwise the radius of convergence would be larger). We also saythat f is singular at c or has a singularity at c. Hence the radius ofconvergence R of (848) is usually equal to the distance from z0 to thenearest singular point of f .
Power Series as Taylor Series The relation to Section 15.3 is given by
Theorem 27 A power series with a nonzero radius of convergence is theTaylor series of its sum.
Proof: Given the power series
f(z) =∞∑m=0
am(z − z0)m, z ∈ BR(z0), (863)
we have f(z0) = a0. By Thm. 25, we obtain by termwise differentiation
f (n)(z) =∞∑m=n
m!
(m− n)!am(z − z0)m−n, z ∈ BR(z0), n ∈ N, (864)
and therefore f (n)(z0) = n!an. Therefore we have
f(z) =∞∑m=0
am(z − z0)m =∞∑m=0
1
m!f (m)(z0)(z − z0)m, (865)
which is the Taylor series of f with center z0. Remark: We know that holomorphic functions have derivatives of all or-ders (Thm. 11), and that they can always be represented by Taylor series
150
(Thm. 26). This is not true in general for real functions. As an example,consider the function
f(x) :=
0, x = 0
e−1/x2 , otherwise, x ∈ R. (866)
This function has derivatives of all orders, with f (n)(0) = 0, n ∈ N, andtherefore it cannot be represented by a Maclaurin series in an open disk withcenter 0.
Important Special Taylor Series These are as in calculus, with x re-placed by a complex number z.
1. f(z) = 1/(1 − z). We have f (n)(z) = n!/(1 − z)n+1, n ∈ N0. Forz0 = 0, we obtain f (n)(0) = n!, n ∈ N0 and thus the coefficients of theMaclaurin series are given by an = f (n)(0)/n! = 1, n ∈ N0. Hence theMaclaurin series of f near 0 is
1
1− z=∞∑n=0
zn, z ∈ B1(0). (867)
f is singular at z = 1; this point lies on the circle of convergence.
2. f(z) = exp z. We have f (n)(z) = exp(z), n ∈ N0 (Section 13.5). Forz0 = 0, we obtain f (n)(0) = 1, n ∈ N0 and thus the coefficients of theMaclaurin series are given by an = 1/n!, n ∈ N0. Hence the Maclaurinseries of f near 0 is
exp z =∞∑n=0
zn
n!, z ∈ C. (868)
3. Trigonometric and Hyperbolic functions: We use (868) in the defini-tions of cos, sin, cosh, sinh (Section 13.6) to obtain Maclaurin series
cos z =∞∑n=0
(−1)nz2n
(2n)!, sin z =
∞∑n=0
(−1)nz2n+1
(2n+ 1)!, (869)
cosh z =∞∑n=0
z2n
(2n)!, sinh z =
∞∑n=0
z2n+1
(2n+ 1)!, (870)
for z ∈ C.
151
4. The function f(z) = Log(1 + z) has derivatives f (n)(z) = (−1)n−1(n−1)!/(1 + z)n, n ∈ N (Thm. 5). For z0 = 0 we obtain f(0) = 0 andf (n)(0) = (−1)n−1(n − 1)!, n ∈ N and thus the coefficients of theMaclaurin series are given by a0 = 0, an = (−1)n−1/n, n ∈ N. Hencethe Maclaurin series of f near 0 is
Log(1 + z) =∞∑n=0
(−1)n
n+ 1zn+1, z ∈ B1(0). (871)
We may also write the Maclaurin series of g(z) := −f(−z) = −Log(1−z) as
Log
(1
1− z
)= −Log(1− z) =
∞∑n=0
zn+1
n+ 1, z ∈ B1(0). (872)
By adding both series, we obtain
Log
(1 + z
1− z
)= 2
∞∑n=0
z2n+1
2n+ 1, z ∈ B1(0). (873)
Practical Methods There are several methods which allow us to obtainTaylor series more quickly than by the use of coefficient formulas. Becauseof uniqueness, they will be the same regardless of the method used.
1. Substitution:
1
1 + z2=
1
1− (−z2)
(867)=
∞∑n=0
(−z2)n =∞∑n=0
(−1)nz2n, z ∈ B1(0).
(874)
2. Integration: For f(z) = arctan z we have f ′(z) = 1/(1 + z2). We mayintegrate (874) termwise to obtain
arctan z =∞∑n=0
(−1)n
2n+ 1z2n+1, z ∈ B1(0). (875)
arctan is a multi-valued function. This series represents the valueswhich satisfy |Re(arctan z)| < π/2.
152
3. Development by Using the Geometric Series: we had used this in theproof of Thm. 26. For R := |c− z0| > 0 we have
1
c− z=
1
c− z0
∞∑n=0
(z − z0
c− z0
)n, z ∈ BR(z0). (876)
4. Binomial Series, Reduction by Partial Fractions: For the rational func-tion
f(z) =2z2 + 9z + 5
z3 + z2 − 8z − 12=
1
(z + 2)2+
2
z − 3, (877)
we use the binomial series
1
(1− z)m=∞∑n=0
(n+m− 1
n
)zn, z ∈ B1(0), m ∈ N (878)
(for m = 1, we obtain the geometric series, as expected). We look fora development at z0 = 1 and therefore write
1
(z + 2)2=
1
(3 + (z − 1))2 =1
9
1(1−
(− z−1
3
))2 (879)
=1
9
∞∑n=0
(−1)n(n+ 1)
(z − 1
3
)n, z ∈ B3(1), (880)
2
z − 3= − 2
2− (z − 1)= − 1
1− z−12
(881)
= −∞∑n=0
(z − 1
2
)n, z ∈ B2(1). (882)
Addition yields the Taylor series of f near 1:
f(z) =∞∑n=0
((−1)n(n+ 1)
3n+2− 1
2n
)(z − 1)n, z ∈ B2(1). (883)
153
16 Laurent Series. Residue Integration
Laurent series generalize Taylor series, because they also allow for negativeinteger powers of z − z0. Therefore, they will converge in an annulus withcenter z0. Hence by a Laurent series we can represent a given function f thatis holomorphic in an annulus and may have singularities outside the ring aswell as in the “hole” of the annulus. Laurent series will help us in classifyingsingularities of complex functions (16.2), and will also lead to a powerful andelegant integration method, called residue integration (16.3, 16.4).
16.1 Laurent Series
If in an application we want to develop a function f in powers of z−z0 whenf is singular at z0, we cannot use a Taylor series. Instead we may use a newkind of series, called Laurent series.
Theorem 28 Let the complex function f be holomorphic in a domain con-taining two concentric circles C1, C2 with center z0 and the annulus betweenthem. Then f can be represented by the Laurent series
f(z) =∞∑
n=−∞
an(z − z0)n =−1∑
n=−∞
an(z − z0)n +∞∑n=0
an(z − z0)n, (884)
consisting of all integer powers of z − z0. The coefficients of this Laurentseries are given by the integrals
an =1
2πi
∮C
f(ζ)
(ζ − z0)n+1dζ, n ∈ Z, (885)
taken counterclockwise around any closed curve C that lies in the annulusand encloses the inner circle C2.
This series converges and represents f in the enlarged open annulus ob-tained from the given annulus by continuously increasing the outer circle C1
and decreasing C2 until each of the two circles reaches a point where f issingular.
In the important special case that z0 is the only singular point of f insideC2, this circle can be shrunk to the point z0, giving convergence in a disk
154
except at the center: z ∈ C | 0 < |z − z0| < R. In this case (!) the series(or finite sum) of the negative powers of (884),
−1∑n=−∞
an(z − z0)n =∞∑n=1
a−n(z − z0)n
(886)
(the notation bn instead of a−n, n ∈ N, is also used), is called the principalpart of f at z0 (or of that Laurent series (884)).
Proof: The proof is in several steps:
1. The nonnegative powers are those of a Taylor series. We use Cauchy’sIntegral Formula (Thm. 10), which in our annulus (doubly connected)is given by
f(z) =1
2πi
∮C1
f(ζ)
ζ − zdζ
︸ ︷︷ ︸=:g(z)
+
− 1
2πi
∮C2
f(ζ)
ζ − zdζ
︸ ︷︷ ︸
=:h(z)
, (887)
where the integration is taken counterclockwise over both C1 and C2.The first integral is transformed exactly as in the proof of Taylor’stheorem (Thm. 26), which yields
g(z) =∞∑n=0
an(z − z0)n, an =1
2πi
∮C1
f(ζ)
(ζ − z0)n+1dζ, (888)
for n ∈ N0. By the principle of deformation of path we may replace C1
by C, which proves the formula for the coefficients (885) for n ∈ N0.
2. For the negative powers, we consider the second term in (887). Becausez lies outside of the circle C2, we have |z−z0| > |ζ−z0|, for ζ ∈ C2. Wedo a similar development as in the proof of Taylor’s theorem (Thm. 26):
1
ζ − z=
1
ζ − z0 − (z − z0)=
−1
(z − z0)(
1− ζ−z0z−z0
) . (889)
We use (857) to obtain
1
ζ − z= − 1
z − z0
(1 +
ζ − z0
z − z0
+ · · ·+(ζ − z0
z − z0
)n)− 1
z − ζ
(ζ − z0
z − z0
)n+1
,
(890)
155
n ∈ N. Multiplication by −f(ζ)/(2πi) and integration over C2 on bothsides now yield
h(z) =1
(z − z0)
1
2πi
∮C2
f(ζ) dζ +1
(z − z0)2
1
2πi
∮C2
(ζ − z0)f(ζ) dζ + · · ·
+1
(z − z0)n+1
1
2πi
∮C2
(ζ − z0)nf(ζ) dζ +R∗n(z),
with
R∗n(z) :=1
(z − z0)n+1
1
2πi
∮C2
(ζ − z0)n+1 f(ζ)
z − ζdζ, (891)
for n ∈ N. We may again replace C2 by C. Therefore
h(z) =n∑
m=1
a−m(z − z0)m
+R∗n(z), n ∈ N, (892)
with coefficients as given in (885), and it remains to show that R∗n(z)→0, n→∞.
3. Because f is bounded on C2 and z 6∈ C2 we have∣∣∣∣ f(ζ)
z − ζ
∣∣∣∣ < M, ∀ ζ ∈ C2. (893)
With the radius of the circle C2, |ζ − z0| = r2 > 0, and L = 2πr2, andwe obtain from the ML-inequality (Prop. 2)
|R∗n(z)| ≤ 1
2π|z − z0|n+1rn+1
2 ML = Mr2
(r2
|z − z0|
)n+1
, n ∈ N.
(894)Because z lies outside of the circle C2 we have |z − z0| > r2, so thatthe right-hand side approaches 0 as n→∞. Therefore, we have estab-lished convergence of (884) with coefficients (885) in the given annulusbounded by C1 and C2.
4. It remains to show that the annulus can be enlarged as stated in thetheorem. The part of the Laurent series with the nonnegative powers(representing g) is a Taylor series; hence it converges in the disk D
156
with center z0 whose radius equals the distance from z0 to the closestsingularity of g. Also g must be singular at all points outside C1 wheref is singular.
The part with the negative powers (representing h) is a power seriesin 1/(z − z0). Let the given annulus be r2 < |z − z0| < r1, where r1
and r2 are the radii of C1 and C2, respectively. Then we have 1/r2 >1/|z− z0| > 1/r1, so that this power series in 1/(z− z0) must convergeat least in the disk 1/|z−z0| < 1/r2. This implies that h is holomorphicoutside of C2 (Thm. 25). Also h must be singular inside C2 where fis singular, and the series of the negative powers of (884) convergesin a region E exterior to the circle with center z0 and radius equalto the maximum distance from z0 to the singularities of f inside C2.The intersection D ∩ E is the enlarged open annulus characterized inLaurent’s theorem.
Remark: The Laurent series of a given holomorphic function f in its annulusof convergence is unique. However, f may have different Laurent series intwo annuli with the same center; see the examples below.
As for Taylor series, we may use other methods than the integral formula(885) to obtain the Laurent series of f . Due to uniqueness, they will be thesame regardless of the method used.Examples:
1. Use of Maclaurin series: f(z) = z−5 sin z, z0 = 0. With (869) we obtain
f(z) =∞∑n=0
(−1)nz2n−4
(2n+ 1)!=
∞∑n=−2
(−1)nz2n
(2n+ 5)!, |z| > 0. (895)
This is the Laurent series of f at 0, with principal part z−4 − (6z2)−1.
2. Substitution: f(z) = z2 exp(1/z), z0 = 0. With (868) with z replacedby 1/z we obtain
f(z) =∞∑n=0
z2−n
n!=
∞∑n=−2
z−n
(n+ 2)!=
2∑n=−∞
zn
(2− n)!, |z| > 0. (896)
This is the Laurent series of f at 0, and the principal part is an infiniteseries.
157
3. Development of 1/(1− z)
a) in nonnegative powers of z: z0 = 0. We have
1
1− z=∞∑n=0
zn, |z| < 1, (897)
b) in negative powers of z: z0 = 0. We write
1
1− z=
−1
z(1− 1
z
) (867)= −1
z
∞∑n=0
(1
z
)n= −
∞∑n=0
z−(n+1)(898)
= −∞∑n=1
z−n = −−1∑
n=−∞
zn, |z| > 1. (899)
4. Laurent Expansions in Different Concentric Annuli: f(z) = 1/(z3−z4),z0 = 0. With the previous example, we obtain
• 1
z3 − z4=∞∑n=0
zn−3 =∞∑
n=−3
zn, 0 < |z| < 1,
• 1
z3 − z4= −
−1∑n=−∞
zn−3 = −−4∑
n=−∞
zn, |z| > 1.
5. Use of Partial Fractions: z0 = 0,
f(z) =−2z + 3
z2 − 3z + 2=
1
1− z− 1
z − 2. (900)
We already know the Laurent series for the first fraction from Example3. For the second fraction, we obtain
− 1
z − 2=
1
2(1− z
2
) =1
2
∞∑n=0
(z2
)n=∞∑n=0
zn
2n+1, |z| < 2,(901)
− 1
z − 2= − 1
z(1− 2
z
) = −1
z
∞∑n=0
(2
z
)n= −
∞∑n=0
2n
zn+1(902)
= −−1∑
n=−∞
zn
2n+1|z| > 2. (903)
Now we may add the series:
158
a) f(z) =∞∑n=0
zn +∞∑n=0
zn
2n+1=∞∑n=0
(1 +
1
2n+1
)zn, |z| < 1
b) f(z) = −−1∑
n=−∞
zn +∞∑n=0
zn
2n+1, 1 < |z| < 2,
c) f(z) = −−1∑
n=−∞
zn −−1∑
n=−∞
zn
2n+1= −
−1∑n=−∞
(1 +
1
2n+1
)zn, |z| > 2.
Remark: If f in Thm. 28 is holomorphic inside C2, then the coefficients infront of the negative powers in (884) are zero by Cauchy’s Integral Theorem(Thm. 8), so that the Laurent series reduces to a Taylor series. Examples3a) and 5a) illustrate this.
16.2 Singularities and Zeros. Infinity
Definition 13 A complex function f is singular at z0 ∈ C or has a singu-larity at z0 if f is not differentiable (perhaps not even defined) at z0 ∈ C, butevery open ball Br(z0) ⊆ C, r > 0 contains points at which f is differentiable.Such a point z0 ∈ C is called a singular point of f . z0 ∈ C is an isolatedsingularity if there is an open ball Br(z0) which contains no further singular-ities of f , i. e. if f is holomorphic on Br(z0) \ z0. An isolated singularityz0 of f is
1. removable, if f can be made differentiable at z0 by assigning a suitablevalue f(z0).
2. a pole of order m ∈ N, if the principal part of the Laurent series of fat z0 (886) is of the form
−1∑n=−m
an(z − z0)n =m∑n=1
a−n(z − z0)n
, a−m 6= 0. (904)
If m = 1, z0 is called a simple pole.
3. an (isolated) essential singularity, if it is neither removable nor a pole.
Remarks:
159
• The function f(z) = tan(1/z) has a nonisolated singularity at z0 = 0.We will not discuss this type of singularity here.
• f has an essential singularity z0 ∈ C if and only if the principal part ofthe Laurent series of f at z0 contains infinitely many terms.
• Be careful to consider the right Laurent series when analyzing singular-ities! It must be the one which is valid in the immediate neighborhoodof z0.
Examples:
1. The function f(z) = sin z/z has a removable singularity at z = 0. Itis removed by assigning f(0) = 1. This can be seen from the Laurentseries of f at 0, which we obtain from (869):
f(z) =∞∑n=0
(−1)nz2n
(2n+ 1)!= 1− z2
3!+z4
5!−+ · · · , z ∈ C. (905)
2. The function
f(z) =1
z(z − 2)5+
3
(z − 2)2(906)
has a simple pole at z = 0 and a pole of order 5 at z = 2. From theexamples in Section 16.1 we know that f(z) = z−5 sin z has a fourth-order pole at z = 0 and that f(z) = 1/(z3 − z4) has a third order poleat z = 0
3. The functions
exp
(1
z
)=
0∑n=−∞
zn
(−n)!, |z| > 0, (907)
sin
(1
z
)=
0∑n=−∞
(−1)−nz2n−1
(−2n+ 1)!, |z| > 0, (908)
have an (isolated) essential singularity at z = 0.
The behavior of complex functions near singularities is characterized by thefollowing theorems.
160
Theorem 29 If a holomorphic function f has a pole at z = z0, then |f(z)| →∞ as z → z0 in any manner.
Example: The function f(z) = 1/z2 hs a second-order pole at z = 0 We have|f(z)| = |z|−2 →∞ as z → 0 in any manner.
Theorem 30 (Picard) A holomorphic function f with an isolated singularityat a point z0 takes on every value, with at most one exceptional value, inBε(z0), for any ε > 0.
Example: Consider the function f(z) = exp(1/z), which has an essentialsingularity at z = 0. For a given complex number c = c0 exp(iα) 6= 0, wewant to find z ∈ C with f(z) = c. We set z = r exp(iϕ) and obtain theequation
exp(1/z) = exp(cosϕ
r
)exp
(−isinϕ
r
)= c0 exp(iα), (909)
so that we need to solve
cosϕ = r log c0, sinϕ = −rα. (910)
We have 1 = cos2 ϕ+ sin2 ϕ = r2(log2 c0 + α2
), so that
r =1√
log2 c0 + α2. (911)
Now we may replace α by α+2πn, n ∈ Z, without changing the given numberc. This allows us to make r > 0 arbitrarily small, and so a complex numberz with f(z) = c can be found in Bε(0), for any ε > 0.
Zeros of Holomorphic Functions
Definition 14 A complex function f which is holomorphic in a domain U ⊆C has a zero at z0 ∈ U if f(z0) = 0. A zero of f has order n ∈ N iff (m)(z0) = 0, m < n and f (n)(z0) 6= 0. If n = 1, z0 is called a simple zero.
Remark: With Taylor’s theorem (Thm. 26) we conclude that the first n− 1terms of the Taylor series of f at z0 vanish, if z0 is a zero of order n of f ,i. e. the Taylor series of f at z0 has the form
f(z) =∞∑m=n
am(z−z0)m = (z−z0)n∞∑m=0
am+n(z − z0)m︸ ︷︷ ︸=:g(z)
, z ∈ BR(z0), (912)
161
with an 6= 0. Unlike singularities, zeros are always isolated:
Theorem 31 The zeros of a holomorphic function f 6≡ 0 are isolated; thatis, each of them has a neighborhood that contains no further zeros of f .
Proof: The factor (z − z0)n in (912) is zero only at z = z0. Because an 6= 0for a zero of order n ∈ N, we have g(z0) 6= 0. Because g is holomorphic inBR(z0) (Thm. 25), it must be continuous in BR(z0) and therefore g(z) 6= 0in some open neighborhood Bε(z0). Hence the same holds for f .
Poles are often caused by zeros in the denominator. For example, thepoles of tan are located at the zeros of cos.
Theorem 32 Let the complex function f be holomorphic in a domain U ⊆ Cand assume that it has a zero of order n ∈ N at z = z0 ∈ U . Then the function1/f(z) has a pole of order n at z0; and so does the function h(z)/f(z), if his analytic at z0 and h(z0) 6= 0.
Riemann Sphere. Point at Infinity For the study of the behavior ofcomplex functions as |z| → ∞, it is useful to use a representation of complexnumbers on the so-called Riemann sphere. This is a sphere S ⊂ R3 withdiameter 1. We let its “north pole” be located at (0, 0, 1). Then the “southpole” of the sphere is located at (0, 0, 0). Now we represent any complexnumber z ∈ C by (Rez, Imz, 0) ∈ R3. The stereographic projection of z on Sis given by the intersection point of the line from (Rez, Imz, 0) to (0, 0, 1) withS. There is exactly one such point on S for any z ∈ C. Conversely, each pointon S represents a complex number, except for the north pole (0, 0, 1). Thissuggests that we introduce an additional point, called the point at infinityand denoted by ∞ (“infinity”), and let its image be (0, 0, 1). The complexplane together with ∞ is called the extended complex plane.
Analytic or Singular at Infinity Instead of investigating f for large |z|,we may set z = 1/w and investigate f(z) = f(1/w) =: g(w) in a neighbor-hood of w = 0. We say that f is analytic or singular, respectively, at infinityif g is analytic or singular, respectively, at 0. We also define
g(0) = limw→0
g(w), (913)
if this limit exists. We say that f has a zero of order n ∈ N at inifinity ifg has a zero of order n at 0. Similarly for poles and essential singularities.Examples:
162
1. f(z) = 1/z2. Then we have g(w) = w2, and so we say that f is analyticat ∞ and has a second-order zero at ∞.
2. f(z) = z3. We have g(w) = 1/w3, and so we say that f is singular at∞ and has a third-order pole at ∞.
3. f(z) = exp(z). We have g(w) = exp(1/w), and so we say that f hasan essential singularity at ∞. Similarly, cos and sin have an essentialsingularity at ∞.
We know from Liouville’s theorem (Thm. 12) that a nonconstant entire func-tion must be unbounded. Hence it has a singularity at ∞, which is a pole ifthe function is a polynomial or an essential singularity if it is not.
A holomorphic function whose only singularities in the complex plane(without ∞) are poles is called a meromorphic function. Examples are ra-tional functions with nonconstant denominator, tan, cot, sec, and csc.
16.3 Residue Integration Method
The purpose of Cauchy’s residual integration method is the evaluation ofintegrals ∮
C
f(z) dz (914)
taken around a closed curve C. There are two cases:
• if f is holomorphic everywhere on C and inside C, such an integral(914) is zero by Cauchy’s Integral Theorem (Thm. 8), and we are done.
• if f is singular at a point z0 inside of C, but is otherwise holomorphicon C and inside of C, then f has a Laurent series representation nearz0 which converges in z ∈ C | 0 < |z − z0| < R, for some R > 0(Thm. 28). By the coefficient formula (885), we have
a−1 =1
2πi
∮C
f(z) dz. (915)
If we can find the coefficient a−1 of the Laurent series representation off at z0 without using the integral formula (885), then we may use (915)to evaluate the integral. The coefficient with index −1 of the Laurent
163
series of f at a singular point z0 is called the residue of f at z0, and wedenote it by
a−1 = Resz=z0f(z) ≡ Res(f, z0) (916)
Example: f(z) = z−4 sin z, C = z ∈ C | |z| = 1 (counterclockwise). f issingular at 0 and 0 lies inside of C. From (869) we obtain the Laurent seriesof f at 0:
f(z) =∞∑n=0
(−1)nz2n−3
(2n+ 1)!=
∞∑n=−2
(−1)nz2n+1
(2n+ 5)!, |z| > 0, (917)
so that the residue of f at 0 is given by Res(f, 0) = a−1 = −1/3! = −1/6.Then we obtain ∮
C
sin z
z4dz = 2πiRes(f, 0) = −π
3i. (918)
Again, be careful to consider the right Laurent series: it must be the onewhich is valid in the immediate neighborhood of z0.
Formulas for Residues If the singularity of f at z0 is a pole of orderm ∈ N, then the residue of f at z0 is given by
Res(f, z0) =1
(m− 1)!limz→z0
(dm−1
dzm−1((z − z0)mf(z))
). (919)
Notice that if the function in the large brackets is continuous at z0, then wemay just evaluate it at z0. To prove this, we consider the Laurent series of fvalid near z0:
f(z) =∞∑
n=−m
an(z − z0)n, z ∈ BR(z0) \ z0, (920)
with a−m 6= 0. We multiply both sides by (z − z0)m to obtain
g(z) := (z − z0)mf(z) =∞∑n=0
an−m(z − z0)n, z ∈ BR(z0), (921)
which is the Taylor series of g with center z0. From Taylor’s theorem (Thm. 26)we obtain for the coefficient a−1 = am−1−m:
a−1 =1
(m− 1)!g(m−1)(z0). (922)
164
If f(z) = p(z)/q(z) with p(z0) 6= 0 and where q has a simple zero at z0,then f has a simple pole at z0 (Thm. 32). The Taylor series of q at z0 isgiven by
q(z) = (z − z0)∞∑n=1
1
n!q(n)(z0)(z − z0)n−1. (923)
We also have, using the formula derived before for m = 1:
Res(f, z0) = limz→z0
((z − z0)f(z)) (924)
= limz→z0
(p(z)∑∞
n=11n!q(n)(z0)(z − z0)n−1
)=p(z0)
q′(z0). (925)
Example: The function
f(z) =50z
z3 + 2z2 − 7z + 4=
50z
(z + 4)(z − 1)2(926)
has a second-order pole at z = 1. With (919) we obtain the residue
Res(f, 1) = limz→1
d
dz
((z − 1)2f(z)
)= lim
z→1
d
dz
(50z
z + 4
)= lim
z→1
200
(z + 4)2= 8.
(927)
Several Singularities Inside the Contour. Residue Theorem Residueintegration can be extended from the case of a single singularity to the caseof several singularities within the curve C. This is the purpose of the residuetheorem.
Theorem 33 Let the complex function f be holmorphic on a closed curve Cand inside of C, except for finitely many singular points z1, . . . , zk, k ∈ N,inside of C. Then the integral of f taken counterclockwise around C is givenby ∮
C
f(z) dz = 2πik∑j=1
Res(f, zj). (928)
Proof: We enclose each of the singular points zj in a circle Cj with radius smallenough that those k circles and C are all separated. Then f is holomorphic
165
in the multiply connected domain bounded by C,C1, . . . , Ck, as well as onthe boundary of this domain. With Cauchy’s Integral Theorem (Thm. 8) wehave ∮
C
f(z) dz =k∑j=1
∮Cj
f(z) dz, (929)
where all integrals are taken counterclockwise. Now we use (915) and (916)to obtain ∮
Cj
f(z) dz = 2πiRes(f, zj), j = 1, . . . , k. (930)
Now just take the sum over j. Remark: For k = 1 and f(z) = g(z)/(z− z0)n+1, n ∈ N0, with g holomorphicon and inside of C, we obtain Cauchy’s Integral Formula and Cauchy’s Dif-ferentiation Formula (Thms. 10 & 11) for g. So Thm. 33 is a generalizationof these.Examples:
1. We consider the function
f(z) :=4− 3z
z2 − z=
4− 3z
z(z − 1), (931)
which is singular at z ∈ 0, 1. Each singularity is a simple pole. Wecompute the residues
Res(f, 0) = limz→z0
(zf(z)) =4− 3z
z − 1
∣∣∣∣z=0
= −4, (932)
Res(f, 1) = limz→z0
((z − 1)f(z)) =4− 3z
z
∣∣∣∣z=1
= 1. (933)
Now we may consider four different curves C:
• both 0 and 1 lie inside of C. Then we obtain with Thm. 33∮C
f(z) dz = 2πi (Res(f, 0) + Res(f, 1)) = −6πi. (934)
• 0 lies inside of C, 1 lies outside. Then we obtain∮C
f(z) dz = 2πiRes(f, 0) = −8πi. (935)
166
• 1 lies inside of C, 0 lies outside. We obtain∮C
f(z) dz = 2πiRes(f, 1) = 2πi. (936)
• both 0 and 1 lie outside of C. Then we obtain with Thm. 8∮C
f(z) dz = 0. (937)
2. f(z) = (tan z)/(z2 − 1), C = z ∈ C | |z| = 3/2. tan is holomorphicexcept at the zeros of cos, which lie at (2k+ 1)π/2, k ∈ Z, i. e. outsideof C. The zeros of the denominator lie at z = ±1, i. e. inside of C. fhas simple poles at z = ±1. We compute the residues of f at ±1:
Res(f,±1) =tan z
2z
∣∣∣∣z=±1
= ±1
2tan(±1) =
1
2tan 1. (938)
With the residue theorem (Thm. 33) we obtain∮C
f(z) dz = 2πi (Res(f,−1) + Res(f, 1)) = 2πi tan 1. (939)
3. C = z ∈ C | 9(Rez)2 + (Imz)2 = 9 (counterclockwise),
f(z) =zeπz
z4 − 16+ zeπ/z. (940)
By the linearity of the integral, we may integrate each term of f sepa-rately. The first term of f has simple poles at z ∈ ±2,±2i, but onlythe points ±2i lie inside of C. We compute the residuals
Res
(zeπz
z4 − 16,±2i
)=zeπz
4z3
∣∣∣∣z=±2i
=±2ie±2πi
4(±2i)3=±2i
∓32i= − 1
16. (941)
The second term of f has an essential singularity at 0, which lies insideof C. Its Laurent series at 0 is given by
zeπ/z = z
∞∑n=0
1
n!
(πz
)n=∞∑n=0
πn
n!z−n+1 =
−1∑n=−∞
π1−n
(1− n)!zn (942)
167
so that
Res(zeπ/z, 0
)=π2
2. (943)
Now from the residue theorem (Thm. 33), we obtain∮C
f(z) dz = 2πiRes
(zeπz
z4 − 16,−2i
)+ 2πiRes
(zeπz
z4 − 16, 2i
)+
+2πiRes(zeπ/z, 0
)= π
(π2 − 1
4
)i ' 30.2i. (944)
16.4 Residue Integration of Real Integrals
Integrals of Rational Functions of cosϑ and sinϑ We first considerintegrals of the type
J =
2π∫0
F (cosϑ, sinϑ) dϑ, (945)
where F is a real rational function of cosϑ and sinϑ with finite values on theinterval of integration.
We set z = eiϑ and obtain
cosϑ =1
2
(eiϑ + e−iϑ
)=
1
2
(z +
1
z
), (946)
sinϑ =1
2i
(eiϑ − e−iϑ
)=
1
2i
(z − 1
z
). (947)
So we obtain a complex rational function f with f(eiϑ) = F (cosϑ, sinϑ). Wealso find that dϑ = dz/(iz), and therefore the integral (945) can be evaluatedby the complex line integral
J =
∮C
f(z)
izdz, (948)
with C = z ∈ C | |z| = 1, oriented counterclockwise.Example: F (cosϑ, sinϑ) = 1/(
√2− cosϑ). We use z = eiϑ to obtain
F (cosϑ, sinϑ) =1√
2− 12
(eiϑ + e−iϑ)=
−2eiϑ
e2iϑ − 2√
2eiϑ + 1
!= f(eiϑ), (949)
168
so that we obtain
f(z) =−2z
z2 − 2√
2z + 1=
−2z
(z −√
2− 1)(z −√
2 + 1). (950)
Now
2π∫0
1√2− cosϑ
dϑ = −2
i
∮C
1
(z −√
2− 1)(z −√
2 + 1)dz. (951)
The integrand has simple poles at z =√
2± 1; only the pole at z =√
2− 1lies inside the unit circle. By the residue theorem (Thm. 33) we have
− 2
i
∮C
1
(z −√
2− 1)(z −√
2 + 1)dz = −4πRes
(1
(z −√
2− 1)(z −√
2 + 1),√
2− 1
)
= −4π1
z −√
2− 1
∣∣∣∣z=√
2−1
= 2π. (952)
Improper Integral of the First Kind We consider real integrals overthe whole real axis:
∞∫−∞
f(x) dx = lima→−∞
c∫a
f(x) dx+ limb→∞
b∫c
f(x) dx, (953)
for any c ∈ R. The Cauchy principal value of the integral (953) is given by
pr. v.
∞∫−∞
f(x) dx := limR→∞
R∫−R
f(x) dx. (954)
The principal value of the integral may exist even if the integral does not. Wenow consider a closed curve C = [−R,R] ∪ SR (oriented counterclockwise),which coincides with the interval [−R,R] on the real axis, and where SR is acurve in the complex plane such that C is closed and encloses all singularities
169
of f in the upper half-plane of the complex plane (we could also choose thelower half-plane). By the residue theorem (Thm. 33) we know that
∮C
f(z) dz =
∫SR
f(z) dz +
R∫−R
f(x) dx = 2πi∑
Imzj>0
Res(f, zj). (955)
We want to show that the integral over SR vanishes under the followingassumptions: We assume that f is a rational function f(z) = p(z)/q(z)(meromorphic), with q(x) 6= 0, x ∈ R (i. e. no poles on the real axis), anddeg q ≥ deg p+ 2. We also assume that SR is a semicircle in the upper half-plane with center 0 and radius R, where R > 0 has to be chosen large enoughso that C = [−R,R] ∪ SR encloses all poles of f (zeros of q) in the upperhalf-plane. The length of SR = z ∈ C | |z| = R, Imz ≥ 0 is πR. With theassumption on the degrees of p and q, we may find two constants k,R0 > 0so that
|f(z)| < k
|z|2, |z| > R0. (956)
With the ML-inequality (Prop. 2) we obtain∣∣∣∣∣∣∫SR
f(z) dz
∣∣∣∣∣∣ < k
R2πR =
kπ
R, R > R0, (957)
and so the integral over SR approaches 0 as R→∞. We obtain
∞∫−∞
f(x) dx = 2πi∑
Imzj>0
Res(f, zj). (958)
Example: The function f(x) := 1/(1 + x4) satisfies the degree requirement.
(Simple) Poles of f are located at the zeros of z4 + 1, i. e. in ei(2k+1)π/43k=0;
only the first two of these poles lie in the upper half-plane. With Res(f, z0) =p(z0)/q(z0) (Section 16.3) we compute
Res(f, eiπ/4
)=
1
4z3
∣∣∣∣z=eiπ/4
=1
4e−i3π/4 = −1
4eiπ/4, (959)
Res(f, e3iπ/4
)=
1
4z3
∣∣∣∣z=e3iπ/4
=1
4e−9iπ/4 =
1
4e−iπ/4. (960)
170
With the formula derived above, we obtain
∞∫−∞
f(x) dx = −2πi
4
(eiπ/4 − e−iπ/4
)= −πi
22i sin
(π4
)=
π√2. (961)
Because f(x) = f(−x) is an even function, we also obtain
∞∫0
f(x) dx =1
2
∞∫−∞
f(x) dx =π
2√
2. (962)
Fourier Integrals We may extend the method described before to inte-grals of the form
∞∫−∞
f(x) cos(sx) dx,
∞∫−∞
f(x) sin(sx) dx, s ∈ R. (963)
We consider the corresponding integral∮C
f(z)eisz dz, s > 0, (964)
over the same contour C as before (if s < 0 we would go via the lowerhalf-plane). Now the integrand is the function f(z)eisz, where we assumeagain that f = p/q is a rational function with q(x) 6= 0, x ∈ R, and thatdeg q ≥ deg p+ 2. The estimate used before works again because
|eisz| = e−sImz ≤ 1, s > 0, Imz ≥ 0, (965)
which implies that |f(z)eisz| ≤ |f(z)|, and we may proceed as before. So wehave
∞∫−∞
f(x)eisx dx = 2πi∑
Imzj>0
Res(f(z)eisz, zj), (966)
171
and if we take the real and imaginary parts on both sides, we obtain
∞∫−∞
f(x) cos(sx) dx = −2π∑
Imzj>0
Im(Res(f(z)eisz, zj)
), (967)
∞∫−∞
f(x) sin(sx) dx = 2π∑
Imzj>0
Re(Res(f(z)eisz, zj)
). (968)
Example: f(x) = 1/(k2 + x2), k > 0. The integrand eisz/(k2 + z2) has twosimple poles at z = ±ik, and only z = ik lies in the upper half-plane. Wecompute
Res(f(z)eisz, ik
)=eisz
2z
∣∣∣∣z=ik
=e−sk
2ik, (969)
and∞∫
−∞
eisx
k2 + x2dx = 2πi
e−sk
2ik=π
ke−ks ∈ R. (970)
The imaginary part of this number is 0, so that the Fourier sine coefficientof f is also 0.
Improper Integral from Section 12.6 In Section 12.6 we had used theintegral (208)
∞∫0
e−ax2
cos(bx) dx, a, b > 0. (971)
We will now show how to compute its value. The function f(z) := e−z2
isentire. We define the closed curve
Cs := Rez ∈ [−R,R], Imz = 0 ∪ Rez = R, Imz ∈ [0, s] ∪ (972)
∪Rez ∈ [−R,R], Imz = s ∪ Rez = −R, Imz = [0, s],
172
for s > 0 (oriented counterclockwise). With z = x + iy we obtain, withThm. 8
0 =
∮Cs
e−z2
dz =
R∫−R
e−x2
dx+ ie−R2
s∫0
ey2−2iRy dy +
+es2
−R∫R
e−x2−2isx dx− ie−R2
0∫s
ey2−2iRy dy (973)
As R→∞, the terms with e−R2
vanish. We obtain for the real part
∞∫−∞
e−x2
cos (2sx) dx = e−s2
∞∫−∞
e−x2
dx (974)
Because the integrand is an even function of x we obtain
∞∫0
e−x2
cos (2sx) dx =e−s
2
2
∞∫−∞
e−x2
dx. (975)
After transformation of the original integral, we have with s := b/(2√a)
∞∫0
e−ax2
cos(bx) dx =1√a
∞∫0
e−x2
cos
(2
b
2√ax
)dx (976)
=1
2exp
(− b
2
4a
)1√a
∞∫−∞
e−x2
dx (977)
The remaining integral can be evaluated in various ways, and yields√π, for
example like this: ∞∫−∞
e−x2
dx
2
=
∞∫−∞
∞∫−∞
e−(x2+y2) dx dy =
2π∫0
∞∫0
e−r2
r dr dϕ = 2π
∞∫0
re−r2
dr
= −π e−r2∣∣∣∞0
= π. (978)
So we obtain (208).
173
17 Conformal Mapping
We now go back to the “main road”, which leads from Chapter 12 (PDEs),via Chapters 13, 14 & 17, to Chapter 18 (Potential Theory in 2D).
In this chapter, we interpret complex functions f : C → C as mappingsfrom R2 to R2, and thus consider a geometric approach to complex anal-ysis. This new approach gives new insight on holomorphic functions; itsimportance is similar to the study of curves (x, f(x)) ∈ R2 |x ∈ R for realfunctions.
Conformal mapping will also yield a standard method for solving bound-ary value problems in two-dimensional potential theory, by transforming acomplicated region into a simple one.
17.1 Geometry of Holomorphic Functions: ConformalMapping
We interpret a complex function f as a mapping of its domain of definition(in C ' R2) onto its image (in C ' R2):
w = f(z) = u(x, y) + iv(x, y), z = x+ iy. (979)
We shall refer to such a mapping as “the mapping w = f(z)”, and we refer tothe “z-plane” and the “w-plane” to distinguish between the spaces containingthe domain and image of f . We use cartesian coordinates (x, y) and (u, v) orpolar coordinates (r, ϑ) and (R,ϕ) to represent points in the z- and w-planes,respectively.Example: The mapping w = z2. Using polar forms z = reiϑ in the z-plane
and w = Reiϕ in the w-plane, we have w = z2 = r2e2iϑ, and thereforeR = r2, ϕ = 2ϑ (cf. Section 13.2). Hence circles |z| = r = r0 in the z-planeare mapped onto circles |w| = R = r2
0 in the w-plane and rays argz = ϑ = ϑ0
in the z-plane onto rays argw = ϕ = 2ϑ0 in the w-plane.In Cartesian coordinates we have z = x+ iy and
u = Rew = Re(z2) = x2 − y2, v = Imw = Im(z2) = 2xy. (980)
Hence vertical lines x = c = const. in the z-plane are mapped onto u = c2−y2,v = 2cy. So we obtain
v2 = 4c2y2 = 4c2(c2 − u
), (981)
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which defines for each c ∈ R a parabola in the w-plane that opens to the left.Similarly, horizontal lines y = k = const. in the z-plane are mapped ontoparabolas in the w-plane,
v2 = 4k2(k2 + u
), (982)
which are parabolas opening to the right, for each k ∈ R.
Conformal Mapping A mapping w = f(z) is called conformal if it pre-serves angles between oriented curves, in magnitude as well as in sense.
Theorem 34 The mapping w = f(z) by a holomorphic function f is con-formal, except at critical points, that is, points at which the derivative of fvanishes.
Proof: We consider a curve C = γ(t) | t ∈ [a, b] in the z-plane, with γ :[a, b]→ C (cf. Section 14.1). The tangent to C at z0 = γ(t0) ∈ C is given byγ(t0) ∈ C, which is to be understood as a vector in the z-plane, attached toz0. The image of C under f is given by the curve C∗ = f(γ(t)) | t ∈ [a, b]in the w-plane, and the tangent vector to C∗ at w0 = f(z0) = f(γ(t0)) ∈ C∗is given by the chain rule: f ′(z0)γ(t0) ∈ C (because f is holomorphic, itis differentiable at z0). Now the angle between the two tangent vectors isgiven by the argument of the quotient, arg (f ′(z0)) (cf. Section 13.2), whichis well-defined if f ′(z0) 6= 0. We have found that the tangent vector at z0 toany curve C with z0 ∈ C is rotated by the angle arg (f ′(z0)) under f . Thisimplies conformality of the mapping, if f ′(z0) 6= 0. Examples:
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1. The mapping w = zn, n = 2, 3, . . . , is conformal, except at z = 0,where f ′(0) = 0.
2. The mapping w = f(z) = z + 1/z. We obtain
w = u+ iv = r (cosϑ+ i sinϑ) +1
r(cosϑ− i sinϑ) , (983)
and therefore
u = a cosϑ, v = b sinϑ, a := r +1
r, b := r − 1
r. (984)
For r = const., r 6= 1, we obtain
u2
a2+v2
b2= cos2 ϑ+ sin2 ϑ = 1, (985)
and therefore circles |z| = const., |z| 6= 1, in the z-plane are mapped toellipses in the w-plane. Circles in the z-plane with center slightly awayfrom the origin, however, may be mapped to very different shapes inthe w-plane, such as a Joukowski airfoil.
The derivative of f is given by
f ′(z) = 1− 1
z2=z2 − 1
z2=
(z + 1)(z − 1)
z2, (986)
and therefore the mapping is not conformal at z = ±1.
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3. The mapping w = f(z) = exp z. We find w = exp(x+iy) = exp x exp(iy),so that |w| = ex and argw = y (mod 2π). The derivative of f vanishesnowhere, and so the fundamental region y ∈ (−π, π] in the z-planeis mapped bijectively and conformally onto the w-plane without theorigin, C \ 0 (Section 13.5).
4. The mapping w = f(z) = Logz. To obtain the mapping by the inversez = f−1(w) of w = f(z), we interchange the roles of the z and the wvariables. For this example we obtain z = expw. We know from theprevious example that exp maps the fundamental region v ∈ (−π, π] inthe w-plane onto the z-plane except for the origin. Hence w = f(z) =Logz maps the z-plane without the origin and cut along the negativereal axis (where Im(Logz) jumps by 2π and thus is not differentiable,cf. Thm. 5) conformally onto the horizontal strip v ∈ (−π, π] in thew-plane.
By definition of the derivative we have
limz→z0
∣∣∣∣f(z)− f(z0)
z − z0
∣∣∣∣ = |f ′(z0)|. (987)
Therefore, the mapping w = f(z) magnifies (or shortens) the lengths of shortlines in the z-plane by approximately the factor |f ′(z0)|. The image (in thew-plane) of a small figure in the z-plane conforms to the original figure inthe sense that it has approximately the same shape. Since f ′(z0) varies frompoint to point, however, a large figure in the z-plane may have an image inthe w-plane whose shape is quite different from that of the original figure.
More on the condition f ′(z0) 6= 0: From Section 13.4 and from Thm. 2 inparticular, we know that
|f ′(z0)|2 = |ux + ivx|2 = (ux)2 + (vx)
2 = uxvy − uyvx. (988)
If we write the mapping in R2, we have F : R2 → R2, F := (u, v)>. TheJacobian of F is given by
DF =
(ux uyvx vy
)⇒ detDF = uxvy − uyvx (989)
(we have omitted the arguments), so that we have |f ′(z0)|2 = det (DF (x0, y0)).Therefore f ′(z0) 6= 0 implies that the mapping w = f(z) is invertible in asmall neighborhood of z0 (Inverse Function Theorem).
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17.2 Linear Fractional Transformations
An important class of mappings are linear fraction transformations (or Mobiustransformations), which are of the form
w = f(z) =az + b
cz + d, (990)
a, b, c, d ∈ C, ad− bc 6= 0. Then we have
f ′(z) =a(cz + d)− (az + b)c
(cz + d)2=
ad− bc(cz + d)2
6= 0, z ∈ C, (991)
and by Thm. 34, the mapping (990) is conformal everywhere in C. Specialcases of (990) include
• a = d = 1, c = 0: w = z + b (translations)
• |a| = d = 1, b = c = 0: w = az, |a| = 1 (rotations)
• c = 0, d = 1: w = az + b, a 6= 0 (linear transformations)
• a = d = 0, b = c = 1: w = 1/z (inversion in the unit circle)
Example: The inversion w = 1/z. We use polar forms z = reiϑ, w = Reiϕ toobtain
Reiϕ =1
reiϑ=
1
re−iϑ ⇒ R =
1
r, ϕ = −ϑ (mod 2π). (992)
Hence the unit circle |z| = r = 1 in the z-plane is mapped onto the unitcircle |w| = R = 1 in the w-plane (oriented in the opposite direction). Welook at points (x, y) in the z-plane which satisfy the equation
A(x2 + y2) +Bx+ Cy +D = 0, A,B,C,D ∈ R. (993)
For A = 0, the points lie on a straight line (Bx + Cy = −D), whereas forA 6= 0, the points lie on a circle:(
x+B
2A
)2
+
(y +
C
2A
)2
=B2 + C2 − 4AD
4A2(994)
With z = x+ iy (993) becomes
Azz +B
2(z + z) +
C
2i(z − z) +D = 0. (995)
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Replacing z = 1/w and multiplying by ww we have
A+B
2(w + w)− C
2i(w − w) +Dww = 0. (996)
With w = u+ iv, we obtain
D(u2 + v2) +Bu− Cv + A = 0, (997)
which is of the same form as (993). We conclude that circles and straightlines in the z-plane are mapped to circles and straight lines in the w-planeby the mapping w = 1/z.
This is not only true for the inversion, but for every linear fractionaltransformation. Before we prove that, it is useful to notice that every linearfractional transformation can be written as a composition of inversions andlinear transformations:
• If c = 0, we must have d 6= 0. Then we have w =a
dz +
b
d.
• If c 6= 0, we write
w =az + b
cz + d=
acz + bc
c(cz + d)=a(cz + d)− ad+ bc
c(cz + d)(998)
=a
c− ad− bcc(cz + d)
= −ad− bcc
1
cz + d+a
c. (999)
This is a composition w = (f1 f2 f3)(z) = f1 (f2 (f3(z))), where
f1(z) = −ad− bcc
z +a
c, f2(z) =
1
z, f3(z) = cz + d. (1000)
f2 is the inversion, and f1, f3 are linear transformations.
Therefore, if we can prove a theorem for both the inversion and for lineartransformation, then this theorem is valid for every linear fractional trans-formation.
Theorem 35 Every linear fractional transformation (990) maps circles andlines in the z-plane to circles and lines in the w-plane.
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Proof: We already know that his is true for the inversion w = 1/z. We nowprove the same property for linear transformations w = az + b, a 6= 0. Wereplace z by (w − b)/a in (995) and use w = u+ iv to obtain
0 = Aw − ba
w − ba
+B
2
(w − ba
+w − ba
)+C
2i
(w − ba− w − b
a
)+D
=A
|a|2ww +
1
|a|2Re((
(B − iC)a− 2Ab)w)
+ A
∣∣∣∣ ba∣∣∣∣2 −BRe
(b
a
)− CIm
(b
a
)+D
=A
|a|2(u2 + v2) +
BRea− CIma− 2AReb
|a|2u+
BIma+ CRea− 2AImb
|a|2v +
+A
∣∣∣∣ ba∣∣∣∣2 −BRe
(b
a
)− CIm
(b
a
)+D, (1001)
which is again of the save form as (993). Now we use that every linearfractional transformation (990) is a composition of two linear transformationsand an inversion.
Extended Complex Plane Linear fractional transformations may be in-verted, which yields another linear fractional transformation:
w = f(z) =az + b
cz + d, z = f−1(w) =
dw − b−cw + a
. (1002)
For c 6= 0, the point z = −d/c in the z-plane is mapped to∞ in the w-plane.In the same way, w = a/c is the image of z =∞ in the w-plane. Therefore,every linear fractional transformation is defined on the extended complexplane C ∪ ∞, and is a bijective conformal map onto itself.
Fixed Points Fixed points of a mapping w = f(z) are points in the z-planethat are mapped to the same points in the w-plane: They are the solutionsof the nonlinear equation f(z) = z.Examples:
• The identity mapping w = z has every point as a fixed point.
• The mapping w = z has infinitely many fixed points (the points on thereal axis Imz = 0).
• The mapping w = 1/z has two fixed points, z = ±1
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• A rotation w = az, |a| = 1, has one fixed point, z = 0
• A translation w = z + b, b 6= 0, has no fixed point
For a linear fractional transformation, we obtain the fixed points
az + b
cz + d= z ⇔ cz2 − (a− d)z − b = 0. (1003)
From this we can prove
Theorem 36 A linear fractional transformation, not the identity, has atmost two fixed points. If a linear fractional transformation is known to havethree or more fixed points, it must be the identity mapping, w = z.
Proof: For c 6= 0, we have a quadratic equation for z, which has two solutionsin C. For c = 0, a 6= d, we have a linear equation with one solution in C. Wehave infinitely many solutions if and only if a = d 6= 0, b = c = 0, in whichcase we have the identity mapping.
17.4 Conformal Mapping by Other Functions
So far, we have discussed the mapping by zn, z + 1/z, exp z, and Logz,and linear fractional transformations. In this section we shall turn to themapping by trigonometric and hyperbolic holomorphic functions. Formulasderived for these functions in Section 13.6 will be helpful.
1. Sine Function: The mapping w = sin z satisfies (648)
u = sinx cosh y, v = cosx sinh y. (1004)
Because sin is 2π-periodic, it will not be one-to-one if we consider it inthe full z-plane. Therefore, we restrict z to the vertical strip S := z ∈C | |Rez| ≤ π/2. The derivative f ′(z) = cos z vanishes at z = ±π/2, sothe mapping is not conformal on the boundary of S. This maps verticallines in the z-plane (x = const.) onto hyperbolas in the w-plane, andhorizontal lines in the z-plane (y = const.) onto ellipses in the w-plane:
x = const. :u2
sin2 x− v2
cos2 x= cosh2 y − sinh2 y = 1 (hyperbolas),
y = const. :u2
cosh2 y+
v2
sinh2 y= sin2 x+ cos2 x = 1 (ellipses).
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Exceptions are the lines x = ±π/2, which are mapped onto [1,∞) and(−∞,−1], respectively, the line y = 0, which is mapped onto [−1, 1],and the line x = 0, which is mapped onto the line u = 0.
2. Cosine Function: we write
w = cos z = sin(z +
π
2
), (1005)
so that this mapping is a composition of a translation to the right byπ/2 and of the mapping by sin, which was discussed before.
3. Hyperbolic Sine: Here we use the relation
w = sinh z = −i sin(iz) (1006)
to see that this mapping is a composition of a counterclockwise rotationby 90, followed by the sine mapping, and followed by a clockwiserotation by 90.
4. Hyperbolic Cosine: We use the relation
w = cosh z = cos(iz), (1007)
which means that this mapping is a composition of a counterclockwiserotation by 90, followed by the cosine mapping. The semi-infinite stripz ∈ C |x ≥ 0, y ∈ [0, π] in the z-plane is mapped onto the upper halfof the w-plane (v ≥ 0) by the cosh mapping. We have
u = coshx cos y, v = sinhx sin y. (1008)
y ∈ [0, π] implies that sin y takes on every non-negative value. Fromx ≥ 0 we have sinhx ≥ 0, and therefore v ≥ 0. Furthermore, cos ytakes on all values in [−1, 1] and coshx ≥ 1. Therefore u takes onevery real number.
5. Tangent Function: We write
w = tan z =sin z
cos z=
12i
(eiz − e−iz)12
(eiz + e−iz)= −ie
2iz − 1
e2iz + 1. (1009)
With Z := e2iz this becomes
w = −iZ − 1
Z + 1= −iW, W :=
Z − 1
Z + 1. (1010)
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Therefore, the tangent mapping is a linear fraction transformation, pre-ceded by an exponential mapping and followed by a clockwise rotationby 90.
With Z = e−2ye2ix, we see that the strip S := z ∈ C |x ∈ (−π/4, π/4)in the z-plane is mapped onto the right half of the Z-plane, X > 0.The linear fractional transformation maps the right half of the Z-planeonto the interior of the unit disk in the W -plane. A clockwise rotationby π/2 follows, so that w = tan z maps the strip S in the z-plane ontothe interior of the unit disk in the w-plane.
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18 Complex Analysis and Potential Theory
The theory of solutions of Laplace’s equation (cf. Sec. 12.10) is called poten-tial theory. Laplace’s equations occurs in many fields such as gravitation,electrostatics, heat conduction, fluid flow etc. In two space dimensions, andin cartesian coordinates, we are interested in solutions of the second-orderPDE
∆Φ = Φxx + Φyy = 0, in Ω ⊆ R2, (1011)
for the (real-valued) potential Φ : Ω → R. We know from Section 13.4 thatsolutions of (1011) with continuous second derivatives (harmonic functions)are closely related to holomorphic functions Φ + iΨ where Ψ is a harmonicconjugate of Φ.
The restriction to two space dimensions is often justified when the poten-tial Φ is independent of one of the space coordinates. In this last chapter ofthe lecture, we shall consider this connection and its consequences in detailand illustrate it by typical boundary value problems from electrostatics, heatconduction, and hydrodynamics.
18.1 Electrostatic Fields
The electric force F [N ] between charged particles is given by Coulomb’slaw. This force is the gradient of a function Φ: F = qE, E = −∇Φ, withthe electric field E [Vm−1] = [NC−1] and the electrostatic potential Φ [V](cf. Sec. 12.10; there we looked at Newton’s law of universal gravitation, butCoulomb’s law has a similar form). Equipotential surfaces are level sets of Φ(Φ ≡ const.; lines in 2D). The electric field is perpendicular to these surfaces.Examples:
1. Two parallel plates kept at potentials Φ1, Φ2, respectively, extendingto infinity. Choose a cartesian coordinate system (x, y, z) such that theplates are parallel to the yz-plane, and located at x = −1 and x = 1.Due to translation invariance of the problem in both y- and z-direction,we may restrict our consideration to the x-direction, where we have asecond-order ODE with Dirichlet boundary conditions:
Φ′′ = 0, in (−1, 1), Φ(−1) = Φ1, Φ(1) = Φ2. (1012)
The general solution is given by Φ(x) = ax + b, and the constants a, b
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are determined from the boundary conditions:
Φ(−1) = −a+ b = Φ1
Φ(1) = a+ b = Φ2⇒ a = 1
2(Φ2 − Φ1)
b = 12
(Φ1 + Φ2)(1013)
Therefore, the potential between two parallel plates is given by
Φ(x) =1
2(Φ2 − Φ1)x+
1
2(Φ1 + Φ2) . (1014)
The equipotential surfaces are parallel planes, x = const.
2. Two coaxial cylinders extending to infinity on both ends, and keptat potentials Φ1, Φ2, respectively. We choose a cylindrical coordinatesystem (r, ϑ, z) where the z-axis coincides with the axis of the cylinders.Because the problem is translation invariant in z-direction, we considera cross-section of the domain between the cylinders r ∈ [R1, R2], ϑ ∈[0, 2π). With the Laplacian in polar coordinates (Section 12.9), weobtain the PDE r2Φrr + rΦr + Φϑϑ = 0. Because the problem is alsoinvariant with respect to rotations around the origin, it turns out thatΦ actually depends only on r, which yields an ODE with boundaryconditions:
rΦ′′ + Φ′ = 0, in (R1, R2), Φ(R1) = Φ1, Φ(R2) = Φ2. (1015)
The ODE can be solved by separation of variables and integrating:
(log Φ′)′=
Φ′′
Φ′= −1
r⇒ log Φ′ = − log r + a ⇒ Φ′ =
a
r,
(1016)and a second integration yields
Φ(r) = a log r + b. (1017)
The constants a, b are again determined from the boundary conditions(cf. Problem Set 6): a = (Φ2 − Φ1)/(logR2 − logR1), b = (Φ1 logR2 −Φ2 logR1)/(logR2 − logR1). The equipotential surfaces are concentriccylinders r = const.
3. Two non-parallel plates extending to infinity and intersecting at anangle α ∈ (0, π], which are kept at potentials Φ1 and Φ2. We choose a
185
cartesian coordinate system such that the z-axis is coincides with theline of intersection between the plates. Due to translation invariance inz-direction of the problem, we may restrict ourselves to a cross-section,which we now interpret as the complex plane, z = x+ iy. The domainbetween the two plates can be characterized by ϑ := Im (Logz) ∈(−α/2, α/2), which is a circular sector in the right half of the complexplane.Remark: The imaginary part of the principal value of the logarithmof a complex number z is sometimes called the principal value of theargument of z, Im (Logz) ≡ Argz ∈ (−π, π].We exclude z = 0, so that r > 0. The function Log is holomorphicin this domain (Thm. 5), and so its real and imaginary parts are bothharmonic functions (Thm. 4). Only the imaginary part is constantalong the rays ϑ = const. (Re (Logz) = log r is not), and therefore thefunction
Φ(r, ϑ) = a+ bϑ (1018)
is the general solution of the Laplace equation in this domain. From theboundary conditions, we determine a = (Φ1 + Φ2)/2, b = (Φ2−Φ1)/α.The equipotential surfaces are half-planes ϑ = const. ∈ (−α/2, α/2),r > 0.
Complex Potential Let Φ be harmonic in some domain Ω ⊆ R2 and Ψ aharmonic conjugate of Φ in Ω (Section 13.4). Then
F (z) = Φ(x, y) + iΨ(x, y), z = x+ iy, (1019)
is a holomorphic function in Ω (by definition of the harmonic conjugate,Section 13.4). This complex function F is called the complex potential cor-responding to the real potential Φ. The harmonic conjugate and thus thecomplex potential for a given harmonic function Φ are uniquely determinedup to a real additive constant.
Technically, F is easier to handle than Φ and Ψ. Physically, the harmonicconjugate Ψ of Φ is also meaningful: the complex potential F defines aconformal mapping, except where F ′(z) = 0. Therefore, the curves Ψ =const. intersect the equipotential lines Φ = const. at right angles. Hencethey have the direction of the electric force and, therefore, are called lines offorce. They are the field lines of the electric field E. Charged particles (such
186
as electrons in an electron microscope, etc.) will travel along these curves.Examples:
1. For the potential Φ(x, y) = ax+b from Example 1 before, Ψ(x, y) := ayis a harmonic conjugate, so that the complex potential is given byF (z) = ax+ b+ iay = a(x+ iy) + b = az + b. The lines of force (levelsets of Ψ) are given by horizontal straight lines y = const. parallel tothe x-axis.
2. For the potential Φ(r, ϑ) = a log r + b = aRe (Logz) + b from Exam-ple 2 before, a conjugate is given by Ψ(r, ϑ) = aIm (Logz), so that thecomplex potential is given by F (z) = aLogz + b. The lines of force arestraight lines through the origin.Remark: With R1 → 0 and R2 → ∞, we obtain the solution of anexterior problem, namely for the electric potential of a line source, per-pendicular to the complex plane and through the origin. This could bea charged wire, for example. In that case, F (z) = KLogz.
3. For the potential Φ(r, ϑ) = a + bϑ = a + bIm (Logz) from Example 3before, we can get the complex potential
F (z) = a− ibLogz = a+ bIm (Logz)− ibRe (Logz) , (1020)
that is, Ψ(r, ϑ) = −bRe (Logz) = −b log r. The lines of force are con-centric circles, r = const.
Superposition More complicated potentials can often be obtained by su-perposition (cf. Thm. 1). As an example, we consider the potential of a pairof oppositely charged wires (line sources). Consider charged wires perpen-dicular to the z-plane and cutting the plane at z = ±c. From the previousexamples, we know that the complex potentials are given by
F+(z) = KLog (z − c) , F−(z) = −KLog (z + c) . (1021)
The complex potential of the combination of the two line sources is thesuperposition
F (z) = F+(z) + F−(z) = K (Log (z − c)− Log (z + c)) = KLog
(z − cz + c
).
(1022)Equipotential lines are circles with variable centers and radii, and the linesof force cross these at a right angle:
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18.2 Use of Conformal Mapping. Modeling
When we solve boundary value problems in complicated domains, the idea isto use a conformal mapping to map the given domain onto one for which thesolution is known or can be found more easily. This solution is then mappedback to the given domain. This works because harmonic functions remainharmonic under conformal mapping:
Theorem 37 Let Φ∗ be harmonic in a domain Ω∗ in the w-plane. Supposethat w = u + iv = f(z) is holomorphic in a domain Ω in the z-plane andmaps Ω conformally onto Ω∗. Then the function
Φ(x, y) := Φ∗ (u(x, y), v(x, y)) , (1023)
defined in the z-plane, is harmonic in Ω.
Remarks:
1. (Alternative proof) With a harmonic conjugate Ψ∗ of Φ∗, we obtainthe complex potential F ∗(w) := Φ∗(u, v) + iΨ∗(u, v) in the w-plane,corresponding to Φ∗. We define the complex function F (z) := F ∗(f(z))in the z-plane, which is holomorphic (chain rule). By Thm. 4, thefunction Φ(x, y) := Re(F (z)) = Re(F ∗(f(z))) = Φ∗(u(x, y), v(x, y)) isharmonic.
188
2. F = F ∗ f is called the pullback of F ∗ by f . The theorem states thatthe pullback of a complex potential by a conformal mapping is also acomplex potential.
Proof: (without using the harmonic conjugate) We have to prove that ∆Φ = 0in Ω. By the chain rule
Φx = Φ∗uux + Φ∗vvx, Φy = Φ∗uuy + Φ∗vvy (1024)
and
Φxx = Φ∗uuu2x + Φ∗uvuxvx + Φ∗uuxx + Φ∗vuuxvx + Φ∗vvv
2x + Φ∗vvxx,(1025)
Φyy = Φ∗uuu2y + Φ∗uvuyvy + Φ∗uuyy + Φ∗vuuyvy + Φ∗vvv
2y + Φ∗vvyy. (1026)
We take the sum and collect terms:
∆Φ = Φxx + Φyy = Φ∗uu(u2x + u2
y
)+ (Φ∗uv + Φ∗vu) (uxvx + uyvy) +
+Φ∗u (uxx + uyy) + Φ∗vv(v2x + v2
y
)+ Φ∗v (vxx + vyy) (1027)
Because f is holomorphic, we have uxvx + uyvy = 0 by the Cauchy-Riemannequations (Thm. 2). With the same theorem, we may also eliminate thederivatives with respect to y: u2
y = (−vx)2 = v2x and v2
y = u2x. Because
f(z) = u(x, y)+iv(x, y) is holomorphic, we also have uxx+uyy = vxx+vyy = 0(Thm. 4). The remaining terms in the sum are
∆Φ = Φ∗uu(u2x + u2
y
)+ Φ∗vv
(v2x + v2
y
)= (Φ∗uu + Φ∗vv)
(u2x + v2
x
),(1028)
which is 0 because Φ∗ was assumed to be harmonic. Example: Potential between noncoaxial cylinders. In the z-plane, we considerthe domain between the circles C1 := z ∈ C | |z| = 1 and C2 := z ∈C | |z − 2/5| = 2/5. We look for the electrostatic potential in this domain,with prescribed values on the circles. This leads to a boundary value probleminvolving Laplace’s equation and Dirichlet boundary conditions. We firstshow that the linear fractional transformation
w =z − bbz − 1
, |b| < 1, (1029)
maps the unit disk onto the unit disk. Indeed, for |z| = 1 we have
|z − b| = |z − b| = |z||z − b| = |zz − bz| = |1− bz| = |bz − 1|, (1030)
189
and therefore |w| = 1. Furthermore, z = 0 maps to w = −b with |w| = |b| <1, and so any point on the unit disk in the z-plane is mapped to the unitdisk in the w-plane. For |z − 2/5| = 2/5, we proceed in a similar way:
|z − b| = |z − b| = 5
2
∣∣∣∣z − 2
5
∣∣∣∣ |z − b| = ∣∣∣∣52(z − 2
5
)(z − b
)∣∣∣∣ (1031)
=
∣∣∣∣52(z − 2
5
)(z − 2
5+
2
5− b)∣∣∣∣ (1032)
=
∣∣∣∣52(z − 2
5
)(z − 2
5
)+
5
2
(z − 2
5
)(2
5− b)∣∣∣∣ (1033)
=
∣∣∣∣25 + z − 5
2bz − 2
5+ b
∣∣∣∣ =
∣∣∣∣(5
2b− 1
)z − b
∣∣∣∣ (1034)
= |b|∣∣∣∣(5
2− 1
b
)z − 1
∣∣∣∣ != α
∣∣bz − 1∣∣ . (1035)
This is true if
1 > |b| = α ∧ b2 − 5
2b+ 1 = 0. (1036)
The quadratic equation is satisfied if b ∈ 1/2, 2, and b = α = 1/2 satisfies|b| < 1. With this choice of b, we have |w| = 1/2, i. e. the circle C2 in thez-plane is mapped onto w ∈ C | |w| = 1/2 in the w-plane by the mapping
w = f(z) =2z − 1
z − 2, (1037)
so that we have two concentric circles in the w-plane. From Example 2before, we know that the general complex potential in the w-plane is givenby F ∗(w) = aLogw + b, where the coefficients a, b are determined from theboundary conditions on Φ∗. The pullback of F ∗ under the mapping w = f(z)is given by
F (z) = F ∗(f(z)) = aLog
(2z − 1
z − 2
)+ b, (1038)
and it is also a complex potential, by Thm. 37, corresponding to the harmonicfunction
Φ(x, y) = Re
(aLog
(2z − 1
z − 2
)+ b
)= a log
∣∣∣∣2z − 1
z − 2
∣∣∣∣+ b. (1039)
190
18.3 Heat Problems
Laplace’s equation also governs steady heat problems (cf. Sections 12.5, 12.6):the heat equation is given by Tt = c2∆T , where T [K] denotes the tempera-ture and where c [m2s−1] denotes the thermal diffusivity. We have Tt ≡ 0 insteady state, i. e. Laplace’s equation. Therefore in two space dimensions, wemay again use methods from complex analysis. In this application, T is alsocalled the heat potential, and it is the real part of the complex heat potential
F (z) = T (x, y) + iΨ(x, y). (1040)
191
The curves T ≡ const., are called isotherms and the curves Ψ ≡ const. arecalled heat flow lines. Heat flows along these lines from higher to lower tem-peratures.Examples: The examples considered in the previous section can now be rein-terpreted as problems on heat flow: the electrostatic potential lines nowbecome isotherms, and the lines of electrical force become lines of heat flow.So we can immediately write down the heat potential between two plates, twocylinders and so on. We consider two new examples with different boundaryconditions:
1. We consider the first quadrant of the unit disk. The temperature onthe straight line segments is prescribed (Dirichlet boundary conditions),but on the arc, the domain is assumed to be insulated, so that ∂nT = 0(Neumann boundary condition). We use polar coordinates (r, ϑ), sothat ∂n ≡ ∂r on the arc. The angular solution from Example 3 beforesatisfies the Neumann boundary condition on the arc. The complexpotential is thus given by F (z) = a−ibLogz with real part T (x, y) = a+bϑ, ϑ = Im (Logz). Coefficients a, b are determined from the Dirichletboundary values.
2. We consider the upper half-plane. On the boundary (real axis), thetemperature is prescribed for x < −1 and x > 1 (two different values)and the domain is insulated on (−1, 1). The upper half of the z-planeis mapped onto a semi-infinite vertical strip in the w-plane, S := w ∈C |u ∈ (−π/2, π/2), v ≥ 0, by the inverse of the sine mapping:
w = f(z) = arcsin z. (1041)
The Dirichlet boundary segments are mapped onto the semi-infinitevertical lines u = ±π/2 (v > 0) and the Neumann boundary segmentis mapped to v = (−π/2, π/2), u = 0. Thus we have ∂n ≡ −∂v. Thisis related to Example 1 before (the parallel plates), so that we obtainF ∗(w) = aw + b, where the constants a, b are determined from theDirichlet boundary values. The pullback by f is given by
F (z) = F ∗(f(z)) = a arcsin z + b. (1042)
192
18.4 Fluid Flow
The Navier-Stokes equations for an incompressible, Newtonian fluid are givenby the following system of nonlinear PDEs:
ρ
(∂v
∂t+ (v · ∇)v
)= −∇p+ µ∆v + f , (1043)
divv = 0, (1044)
where ρ [kgm−3] denotes the mass density of the fluid, v [ms−1] the velocity,p [Pa] the pressure, µ [Pas] the (dynamic) viscosity and f [Nm−3] denotesthe volume force density (such as from gravitational forces). The vectorLaplacian is given by ∆ = ∇div − curlcurl.
These equations are difficult to solve. A simplification can be made whenadvective inertial forces are small compared to viscous forces. This assump-tion is valid in the case of small Reynolds numbers, Re 1, i. e. for flowat small length scales, small velocities or high viscosity. In these cases, thefluid exhibits Stokes flow or creeping flow, which is steady, i. e. vt ≡ 0:
µ∆v + f = ∇p, divv = 0. (1045)
This is a system for four linear PDEs. Next we assume irrotational flow,i. e. curlv = 0. In this case, we may write v = ∇Φ with the velocity potentialΦ (Helmholtz decomposition). The equations simplify to a system of twodecoupled PDEs for the pressure p and for the velocity potential Φ:
∆p = divf , ∆Φ = 0. (1046)
193
The PDE for the velocity potential Φ is Laplace’s equation. We restrict our-selves again to two space dimension, so that we can use complex analysis. Inthe corresponding complex potential F (z) = Φ(x, y) + iΨ(x, y), the functionΨ is called the stream function. Lines Ψ ≡ const. are called streamlines ofthe fluid motion. In this context, the velocity vector v = (v1, v2)> is usuallywritten as a complex number:
V = v1 + iv2 = Φx + iΦy = Φx − iΨx, (1047)
where we have used the second Cauchy-Riemann equation. Therefore wehave V (x, y) = F ′(z).
Similar to what we did for the electrostatic and heat problems before,we may interpret any holomorphic function F as a complex potential, whosereal and imaginary part will be the velocity potential and stream function ofa certain flow problem.Examples:
1. The complex potential F (z) = z2 = x2 − y2 + 2ixy models a flow with
Equipotential lines Φ(x, y) = x2 − y2 = const., (1048)
Streamlines Ψ(x, y) = 2xy = const., (1049)
which are hyperbolas. The velocity vector is given by V (x, y) = F ′(z) =2z = 2(x−iy), that is v1 = 2x, v2 = −2y, with speed |V | = 2
√x2 + y2.
This flow may be interpreted as flow around a corner. The flow speedalong the streamline 2xy = c is minimal at x = y =
√c/2, where the
cross-section of the channel is large.
194
2. The complex potential F (z) = z + 1/z gives, in polar coordinates,
F (z) = reiϑ +1
re−iϑ =
(r +
1
r
)cosϑ+ i
(r − 1
r
)sinϑ (1050)
(cf. Section 17.1). Therefore, the streamlines are given by
Ψ(x, y) =
(r − 1
r
)sinϑ = const. (1051)
195
18.5 Poisson’s Integral Formula for Potentials
We have seen in the previous sections that boundary value problems involv-ing the Laplace equation in complicated domains in two space dimensionscan be transformed to problems in standard domains, such as a disk, by asuitable conformal mapping. In this section, we shall investigate the solutionof Laplace’s equation in a disk in more detail.
Consider a circle with radius R > 0, CR := γ(α;R) |α ∈ [0, 2π], γ(t) :=Reiα, and a complex function F which is holomorphic in a simply connecteddomain Ω that contains CR.
1
2πi
∮CR
F (ζ)
ζ − zdζ =
1
2π
2π∫0
F (Reiα)Reiα
Reiα − zdα =
F (z), |z| < R0, |z| > R
, z ∈ Ω,
(1052)from Thms. 7, 8, and 10. We choose z ∈ C with |z| < R and define Z :=R2/z, |Z| = R2/|z| > R. Therefore
0 =1
2π
2π∫0
F (Reiα)Reiα
Reiα − Zdt =
1
2π
2π∫0
F (Reiα)z
z −Re−iαdα (1053)
Subtracting the integrals, we obtain
F (z) =1
2π
2π∫0
F (Reiα)
(Reiα
Reiα − z− z
z −Re−iα
)dα (1054)
=1
2π
2π∫0
F (Reiα)R2 − |z|2
|Reiα − z|2dα. (1055)
Finally, with z = reiϑ, we have
F (reiϑ) =1
2π
2π∫0
F (Reiα)R2 − r2
R2 − 2Rr cos(ϑ− α) + r2dα. (1056)
This formula is valid for any z ∈ C, |z| < R. Now we write F (reiϑ) =
196
Φ(r, ϑ) + iΨ(r, ϑ) and take the real part on both sides:
Φ(r, ϑ) =1
2π
2π∫0
Φ(R,α)R2 − r2
R2 − 2Rr cos(ϑ− t) + r2dα, r < R, ϑ ∈ [0, 2π).
(1057)Poisson’s integral formula (1057) expresses the potential Φ inside of the diskwith radius R > 0 centered at the origin using the boundary values, Φ(R, ·).
Series for Potentials in Disks We may also write this formula as a series.For that purpose, we note first that the quotient under the integral is thereal part of the complex function
ζ + z
ζ − z=
(ζ + z)(ζ − z)
(ζ − z)(ζ − z)=|ζ|2 + 2iIm(zζ)− |z|2
|ζ − z|2, (1058)
where ζ = Reiα and z = reiϑ. Now we use the geometric series to write
ζ + z
ζ − z=
1 + zζ
1− zζ
=
(1 +
z
ζ
) ∞∑n=0
(z
ζ
)n=∞∑n=0
(z
ζ
)n+∞∑n=0
(z
ζ
)n+1
(1059)
= 1 + 2∞∑n=1
(z
ζ
)n= 1 + 2
∞∑n=1
( rR
)nein(ϑ−α). (1060)
For the real part, we obtain
R2 − r2
R2 − 2Rr cos(ϑ− α) + r2= Re
(ζ + z
ζ − z
)= 1 + 2
∞∑n=1
( rR
)ncos (n(ϑ− α))
= 1 + 2∞∑n=1
( rR
)n(cos(nϑ) cos(nα) + sin(nϑ) sin(nα)) .
Now Poisson’s integral formula (1057) becomes a Fourier series:
Φ(r, ϑ) = a0 +∞∑n=1
( rR
)n(an cos(nϑ) + bn sin(nϑ)) , (1061)
197
which involves the Fourier coefficients of the boundary values Φ(R, ·):
a0 =1
2π
2π∫0
Φ(R,α) dα, an =1
π
2π∫0
Φ(R,α) cos(nα) dα, n = 1, 2, . . . ,
bn =1
π
2π∫0
Φ(R,α) sin(nα) dα, n = 1, 2, . . . . (1062)
Comparte this with Section 12.10 where we had derived the analogous for-mula for the interior problem of the 3D Laplace equation. In 3D, this involvedspherical harmonics Ynm(ϕ, ϑ), 0 ≤ |m| ≤ n (2n + 1 linearly independentfunctions for each n ∈ N0), whereas in 2D, we have trigonometric functionscos(nϑ), sin(nϑ) (two linearly independent functions for each n ∈ N, one forn = 0).Example: For R = 1, with the boundary values
Φ(1, α) =
−α/π, α ∈ (−π, 0]α/π, α ∈ (0, π]
, (1063)
we obtain bn = 0, n ∈ N, because Φ(1, ·) is an even function. We also have
a0 =1
2π
− 0∫−π
α
πdα +
π∫0
α
πdα
=1
2, (1064)
an =1
π
− 0∫−π
α
πcos(nα) dα +
π∫0
α
πcos(nα) dα
=2 ((−1)n − 1)
n2π2,(1065)
so that
Φ(r, ϑ) =1
2− 4
π2
∞∑k=0
r2k+1
(2k + 1)2cos((2k + 1)ϑ). (1066)
18.6 Review of the Second Part of the Lecture
18.6.1 Final Exam
The final exam of MATH 529 will take place on Tuesday, May 3, 2011, 8:00– 11:00 AM, in Phillips Hall, Room 332 (the regular lecture hall). Please
198
read the paragraph on “Final Examinations” in the Academic Proceduressection of the 2010–2011 Undergraduate Bulletin(http://www.unc.edu/ugradbulletin/procedures1.html), to be informed aboutwhat to do in case you are unable to take the exam at the scheduled time.As stated in the course syllabus, you are allowed to use a summary of thelecture notes on 6 (six) pages (US letter format, double-sided) for the finalexam. This is meant as 6 sheets of paper, with text on both sides.
The exam will cover topics from Chapter 12 (up to and including 12.10)and 13–17 (not 16.4). The problems will focus on the application of themethods treated in this course; I want to see that students are able to recog-nize which method should be used for a given problem, and that they applyit correctly. I will not ask for proofs, and I will try to avoid complicated aux-iliary calculations (such as partial fraction decompositions, finding roots ofhigh order polynomials, or computing high-order derivatives of complicatedfunctions). There will be ∼ 9 problems in the final exam, which means ∼ 20minutes to solve each problem.
18.6.2 Complex Numbers and Functions (13)
Complex Numbers. Complex Plane We had constructed the field ofcomplex numbers C by considering ordered pairs (x, y) of real numbers. Acomplex number z ∈ C may be represented in various ways in the complexplane C ' R2, such as
z = x+ iy = r(cosϕ+ i sinϕ) = reiϕ, i2 = −1, (1067)
where x := Rez denotes the real part of z, y := Imz the imaginary part,r = |z| =
√zz =
√(Rez)2 + (Imz)2 the absolute value, and ϕ = arg z the
argument of z. It is defined by the equations
cosϕ =x
r, sinϕ =
y
r, (1068)
and therefore multi-valued, because of the periodicity of sin and cos. Wetypically restrict the argument to either arg z ∈ [0, 2π) or Argz ∈ (−π, π].The latter is convenient when working with complex logarithms, becauseArgz = Im(Logz), and the branch cut of Arg coincides with the branch cutof Log. We have
arg z =
Argz, Imz ≥ 02π + Argz, Imz < 0
, z 6= 0. (1069)
199
The complex conjugate z ∈ C of a complex number z ∈ C is defined byz = Rez − iImz; geometrically, this corresponds to a mirroring on the realaxis in the complex plane, and therefore z = z ⇔ Imz = 0. The absolutevalue |z| =
√zz induces a metric on C, d : C×C→ C, d(z1, z2) := |z1 − z2|.
Thus we can measure distances between complex numbers (Problem Set 7).In particular, we have the triangle inequality |z1+z2| ≤ |z1|+|z2|, z1, z2 ∈ C.
Polar Form of Complex Numbers. Powers and Roots. The polarform of a complex number is useful to represent products, quotients, powersand roots of complex numbers:
z1z2 = |z1||z2| exp (i(arg z1 + arg z2)) , (1070)
z1
z2
=|z1||z2|
exp (i(arg z1 − arg z2)) , (1071)
zn = |z|n exp (in arg z) , (1072)
n√z = n
√|z| exp
(iarg z
n
)exp
(i2πk
n
), k = 0, . . . , n− 1. (1073)
Derivative. Holomorphic Function. With the metric d(z1, z2) := |z1 −z2| on the complex numbers induced by the absolute value (which defines anorm on C), we may introduce open balls Bρ(a) := z ∈ C | d(z, a) < ρ,a ∈ C, ρ > 0. The open balls generate a topology on C, making it atopological space. A subset U ⊆ C is open if every point in U has an openneighborhood contained in U . U is connected if it cannot be written as theunion of two disjoint nonempty open sets.
For a complex function f : z 7→ w = f(z), both the argument and thevalue may be split into real and imaginary parts: f(x+iy) = u(x, y)+iv(x, y),with real-valued functions u, v : R2 → R. The limit of f as z approaches z0
is L if
∀ ε > 0 ∃ δ > 0 : f(z) ∈ Bε(L) ∀ z ∈ Bδ(z0), z 6= z0. (1074)
in this case we write limz→z0 f(z) = L. f is continuous at z0 ∈ C if f(z0) isdefined and limz→z0 = f(z0). f is differentiable at z0 ∈ C, if the limit
f ′(z0) := limz→z0
f(z)− f(z0)
z − z0
(1075)
200
exists. In this case, the limit value is the derivative of f at z0.Examples:
• f(z) := z2 is differentiable at all z0 ∈ C,
• f(z) := z is not differentiable at any z0 ∈ C.
The rules for differentiation are the same as for real functions (product rule,quotient rule, chain rule).
A complex function f which is differentiable at every point z ∈ U ⊆ C (Uis typically assumed to be a domain, i. e. open and connected), is holomorphicin U . A function which is holomorphic in all of C is an entire function. Aholomorphic function is necessarily continuous (Problem Set 7).
Cauchy-Riemann Equations. Laplace’s Equation. The following twotheorems provide a criterion for differentiability of a complex function:Thm. 2: f(x + iy) = u(x, y) + iv(x, y) holomorphic in U ⊆ C ⇒ the firstpartial derivatives of u and v exist, and they satisfy the Cauchy-Riemannequations (system of linear first-order PDEs)
ux = vy, uy = −vx in U. (1076)
Thm. 3: u, v : R2 → R with continuous first partial derivatives satisfythe Cauchy-Riemann equations in some domain U ⊆ C ⇒ f(x + iy) :=u(x, y) + iv(x, y) is holomorphic in U .From the proof of Thm. 2, we also concluded that the derivative of a holo-morphic function f can be written as
f ′(x0 + iy0) = ux(x0, y0) + ivx(x0, y0) = −iuy(x0, y0) + vy(x0, y0). (1077)
In polar form, we write f(reiϕ) = u(r, ϕ)+iv(r, ϕ), and the Cauchy-Riemannequations are given by (Problem Set 8)
ur =1
rvϕ, vr = −1
ruϕ. (1078)
The following theorem was useful in potential theory in 2D (Chapter 18):Thm. 4: f holomorphic in some domain U ⊆ C ⇒ both the real andimaginary part of f are harmonic functions in U .A harmonic function is a twice continuously differentiable function u which
201
satisfies Laplace’s equation ∆u = 0 in some domain Ω ⊆ R2. The conjugateof a harmonic function u is a function v such that f(x+iy) := u(x, y)+iv(x, y)is holomorphic. The harmonic conjugate of a given function u is uniquelydetermined up to some real additive constant. It can be found by solvingthe Cauchy-Riemann equations. In potential theory, u is typically somereal potential (electric potential, temperature, . . . ), whereas its harmonicconjugate gives the lines of force, heat flow lines, . . . , which are perpendicularto the equipotential lines. We had seen this in Chapter 18.
Exponential Function The function f(z) = ez, z ∈ C is defined by
exp(x+ iy) := ex cos y + iex sin y. (1079)
It satisfies exp(z1 + z2) = exp z1 exp z2, from which we infer Euler’s formulaeiy = cos y+ i sin y, |eiy| = 1. We also know that exp′ z = exp z, and that expnever vanishes. The function is periodic with period 2πi: ez+2πi = ez, z ∈ C.It maps the fundamental region Imz ∈ (−π, π] (a horizontal strip) onto thewhole complex plane without the origin, C \ 0.
Trigonometric and Hyperbolic functions With the complex exponen-tial function, we may define trigonometric and hyperbolic functions for com-plex numbers in exactly the same way as for real numbers:
cos z :=1
2(eiz + e−iz), sin z =
1
2i(eiz − e−iz) (1080)
cosh z :=1
2(ez + e−z), sinh z =
1
2(ez − e−z). (1081)
Addition theorems for these functions remain valid in the complex. In thecomplex only, trigonometric and hyperbolic functions are related via
cosh(iz) = cos z, sinh(iz) = i sin z, (1082)
cos(iz) = cosh z, sin(iz) = i sinh z. (1083)
Logarithm. General Power. The natural logarithm is the inverse of theexponential function, and therefore multi-valued (because exp is not one-to-one in all of C). For z = reiϕ, r > 0, we obtain
w = log z = log r + i(ϕ+ 2πn), n ∈ Z. (1084)
202
The principal value of log is the one which lies in the fundamental region ofexp: Im(Logz) ∈ (−π, π]:
Logz =
log r + iϕ, ϕ ∈ [0, π]log r + i(ϕ− 2π), ϕ ∈ (π, 2π)
. (1085)
All values of log are then given by log z = Logz + 2πin, n ∈ Z.Thm. 5: The natural logarithm is holomorphic except for z ≤ 0. Its deriva-tive is given by
log′ z =1
z, z 6= 0, arg z 6= π. (1086)
The negative real axis z < 0 is called the branch cut of log.With the complex exponential and logarithm, we may define the generalpower of two complex numbers by
zc := exp (c log z) , z, c ∈ C. (1087)
This is multi-valued, because log is multi-valued. The principal value of zc
uses the principal value Log instead of log.
18.6.3 Complex Integration (14, 16)
Line Integral in the Complex Plane A curve C ⊂ C is defined by
C := γ(t) | t ∈ [a, b], a, b ∈ R, γ : [a, b]→ C. (1088)
We usually assume a regular curve, i. e. γ is continuous and piecewise con-tinuously differentiable, and that γ(t) 6= 0, t ∈ (a, b). The line integral of acomplex function f along the curve C can then be defined in a similar way asthe Riemann integral in R, with a partition a = t0 < t1 · · · < tn = b, n ∈ N:
Sn :=n∑
m=1
f(ζm)(zm − zm−1), zm = γ(tm), ζm = γ(τm), τm ∈ [tm−1, tm].
(1089)Then ∫
C
f(z) dz := limn→∞
Sn. (1090)
From this definition, we infer properties such as the linearity of the integral.It is not, however, used to evaluate integrals in practice. For that purpose,
203
we have two other theorems:Thm. 6: f holomorphic in U ⊆ C, U a simply connected domain (!). Thenthe antiderivative F of f exists (F ′(z) = f(z) ∀ z ∈ U) and is also holo-morphic in U . For any path in U joining the points z0, z1 ∈ U , we have
z1∫z0
f(z) dz = F (z1)− F (z0). (1091)
Notice that the domain U does not appear explicitly in the integral. U ⊆ Cis typically not prescribed, but it has to be chosen such that it is open andsimply connected, contains C and such that f is holomorphic in U . Thisdomain U ⊆ C occurs in several other integral theorems. In general, it maybe impossible to find such a domain, in which case these theorems are notapplicable.Example: f(z) = z2, then we can choose U = C.
1+i∫0
z2 dz =z3
3
∣∣∣∣1+i
0
= −2
3+
2
3i. (1092)
When Thm. 6 is not applicable (such as for functions with singularities insideof a closed curve C), we may need to use a parametrization of C to write theintegral as a real integral:Thm. 7: C regular, piecewise C1 curve, f continuous complex function.∫
C
f(z) dz =
b∫a
f(γ(t))γ(t) dt. (1093)
Example: z0 ∈ C, f(z) = (z − z0)m, m ∈ Z, C := z ∈ C | |z − z0| = ρ,ρ > 0. For m < 0, f is not differentiable at z0 and C encloses z0: therefore,there is no simply connected domain U ⊆ C which contains C and in whichf is holomorphic, so that Thm. 6 cannot be applied (we could use Thm. 6for m ≥ 0, though, in which case f is entire). With Thm. 7, we obtain∮
C
(z − z0)m dz = · · · =
2πi, m = −10, m ∈ Z \ −1 . (1094)
We have seen several theorems which are usually easier to use than Thm. 7,but require less strong conditions than Thm. 6:
204
CIT, CIF, CDF, Residue theorem The integral theorems by Cauchy(Thms. 8, 10, 11) are all special cases of the residue theorem (Chapter 16):Thm. 33: f holomorphic on a closed curve C and inside of C, except forfinitely many singular points z1, . . . , zk, k ∈ N inside of C (!). Then theintegral of f taken counterclockwise around C is given by∮
C
f(z) dz = 2πik∑j=1
Res(f, zj). (1095)
In the special cases of CIT, CIF, CDF, the integrand has at most one singularpoint inside of C (k = 1), and it is either holomorphic in a simply connecteddomain which contains C (CIT), or it has a pole of order n + 1 at z0 ∈ C,which lies inside of C (CDF; for n = 0 (simple pole), this is CIF).
Here, the residue of f at the singular point z0 is defined as the coefficienta−1 of the Laurent series of f at z0:
f(z) =∞∑
m=−∞
an(z − z0)n, 0 < |z − z0| < R, ⇒ Res(f, z0) = a−1.
(1096)If f has an essential singularity at z0, we typically need to compute theLaurent series of f explicitly in order to determine the value of a−1 (bysubstitution in a Maclaurin series, for example). For poles, however, thereis an easier way to compute the residue, which involves differentiation: theresidue of f at a pole of order m ∈ N, z0 ∈ C, is given by
Res(f, z0) =1
(m− 1)!limz→z0
(dm−1
dzm−1((z − z0)mf(z))
). (1097)
For a function of the form f(z) = p(z)/q(z) (p, q not necessarily polynomials!)with p(z0) 6= 0 and where q has a simple zero at z0 (⇒ f has a simple poleat z0), this simplifies further to
Res(f, z0) =p(z0)
q′(z0). (1098)
18.6.4 Power Series, Taylor Series, Laurent Series (15, 16)
In order to get to the residue theorem in Chapter 16, we had to consider afew more theoretical topics, concerning power series. We had seen that every
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power series of the form
∞∑n=0
an(z − z0)n, z0 ∈ C, an ∈ C, n ∈ N0 (1099)
(no negative powers!) converges at the center z0. Furthermore, if a powerseries converges at z = z1 6= z0, it converges absolutely for all z which arecloser to z0 than z1, |z − z0| < |z1 − z0|. On the other hand, if a powerseries diverges at z = z2, it diverges for every z farther away from z0 than z2,|z− z0| > |z2− z0| (Thm. 19). We had defined the convergence radius R of apower series as the radius of the smallest circle with center z0 containing allpoints z ∈ C for which the power series converges; R = 0 and R =∞ are alsopossible. We have also found that termwise integration and differentiationdo not change the radius of convergence (Thms. 23 & 24). The radius ofconvergence can be computed with the Cauchy-Hadamard formula:Thm. 22: Suppose that the sequence |an+1/an|, n ∈ N, converges with limitL∗. If L∗ = 0, then R =∞ (convergence everywhere). If L∗ > 0, then
R =1
L∗= lim
n→∞
∣∣∣∣ anan+1
∣∣∣∣ . (1100)
If |an+1/an| → ∞, n→∞, then R = 0 (convergence only at the center).We may need to bring a given series into the form (1099) before we can applythe formula.Example: We consider the following function from Problem Set 11:
z cos1
z=∞∑n=0
(−1)nz−2n+1
(2n)!=∞∑n=0
(−1)n
(2n)!
(1
z
)2n−1
. (1101)
We use termwise integration to eliminate the constant in the power of w :=1/z: The power series
∞∑n=1
(−1)n
(2n)!
w2n
2n=∞∑n=1
(−1)n
(2n)!2n(w2)n (1102)
thus obtained is a power series in w2 in the form (1099) with the sameradius of convergence as the original power series (it is not a power series
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in z, because it contains negative powers of z; the function has an essentialsingularity at z = 0). We look at the quotient
∣∣∣∣an+1
an
∣∣∣∣ =
∣∣∣∣∣∣(−1)n+1
(2(n+1))!2(n+1)
(−1)n
(2n)!2n
∣∣∣∣∣∣ =(2n)!2n
(2n+ 2)!(2n+ 2)=
2n
(2n+ 1)(2n+ 2)2(1103)
With de l’Hopital’s rule, for example, we verify that the limit satisfies L∗ = 0.Therefore, the power series converges for
|w|2 =1
|z|2<∞ ⇔ |z| > 0. (1104)
We call a function f analytic at z0 ∈ C if it can be represented as a powerseries in z of the form (1099) which converges in BR(z0), R > 0. Thm. 25says that such a function is holomorphic in BR(z0), and Thm. 26 (Taylor’stheorem) states that every holomorphic function f can be locally representedat z0 as a power series, namely the Taylor series of f at z0, where the co-efficients an = f (n)(z0)/n! are given by the values of the derivatives of f atz0. This representation is valid in the largest open disk with center z0 inwhich f is holomorphic. From these two theorems we concluded that holo-morphic functions are analytic and vice versa. Practical methods to obtainTaylor series without using the coefficient formulas include substitution intoor termwise integration of already known Taylor series. The geometric seriesis also often used as a tool.
In the Laurent series of a complex function f at z0 we also allow fornegative powers of (z − z0). These are therefore not power series in (z −z0) (of the form (1099)) anymore, but may be brought into that form witha transformation of variables, as we had just seen in the example for theLaurent series of z cos(1/z). We are usually interested in the Laurent serieswhich converges for 0 < |z − z0| < R, where R = ∞ is also possible. Thisseries allows us to analyze an (isolated) singularity of f at z0, by looking atthe principal part
−1∑n=−∞
an(z − z0)n =∞∑n=1
a−n(z − z0)n
. (1105)
Depending on the highest order non-vanishing coefficient, we identify thesingularity as removable, a pole, or an essential singularity. Poles and zeros
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of functions are related in the sense that if a function f has a zero of ordern ∈ N at z0, then the function 1/f has a pole of order n at z0.
We had defined the complex “infinity” via stereographic projection ontothe Riemann sphere. ∞ gets mapped onto the north pole of this sphere understereographic projection. This motivates the extended complex plane, whichincludes∞. The behavior of a function f at∞ is analyzed by looking at thebehavior of the function g(w) := f(1/w) at w = 0.
18.6.5 Conformal Mapping (17)
Holomorophic functions f : C→ C can be interpreted as conformal mappings(from R2 to R2), conformality meaning that angles between oriented curvesare preserved, in magnitude as well as in sense. Thm. 34 states that themapping w = f(z) by a holomorphic function is conformal except at criticalpoints, where f ′(z0) = 0.
We looked at linear fractional transformations (LFTs) in particular, whichare of the form
w = f(z) :=az + b
cz + d, a, b, c, d ∈ C, ad− bc 6= 0. (1106)
Such a mapping is conformal everywhere in R2; special cases of LFTs includetranslations, rotations, linear transformations and the inversion on the unitcircle. An LFT maps circles and lines in the z-plane to circles and lines inthe w-plane (Thm. 35). It is a bijective conformal mapping of the extendedcomplex plane C ∪∞, and the inverse is given by
z = f−1(w) =dw − b−cw + a
. (1107)
For c 6= 0, the image of z = −d/c is∞ (in the w-plane), and the image of∞(in the z-plane) is w = a/c. An LFT has at most two fixed points, f(z) = z,unless it is the identity mapping (Thm. 36). The fixed points of an LFT arethe solutions of the polynomial equation
cz2 − (a− d)z − b = 0. (1108)
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References
[1] L. C. Evans: Partial Differential Equations. AMS, 1998
[2] E. Kreyszig: Advanced Engineering Mathematics. 9th Edition; Wiley,2006
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