complex analysis - ctaps.yu.edu.jo

Physics Department, Yarmouk University, Irbid Jordan

Phys. 601 Mathematical Physics

Dr. Nidal M. Ershaidat Doc. 1

1

Complex Analysis

Definition

A complex function is a one in which the independent variable and the

dependent variable are both complex numbers. More precisely, a complex

function is a function whose domain and range are subsets of the complex

plane.

The study of functions of a complex variable is known as complex analysis

and has enormous practical use in applied mathematics as well as in physics.

Often, the most natural proofs for statements in real analysis or even

number theory employ techniques from complex analysis

Functions of a complex variable provide us some powerful and widely useful

tools in theoretical physics. For example:

◊ Some operators in Quantum mechanics are complex entities.

◊ Some important physical quantities are complex variables (the wave-

function Ψ)

Many physical quantities that were originally real become complex as a

simple physical theory is made more general.

Examples

1. The most famous example is real energy associated with an energy

level which becomes complex when the finite lifetime of the level is

considered. The energy has the form En = En + i Γ (where h/Γ is the finite

life time of this state).

2. The real index of refraction of light becomes a complex quantity when

absorption is included.

2

Importance in Physics

1. Mapping

For many pairs of functions u and v, both u and v satisfy Laplace’s

equation

2222=,0=),(=)( yxuru +ρθρ∇∇

r

2222=,0=),(=)( yxvrv +ρθρ∇∇

r

Either u or v may be used to describe a two-dimensional electrostatic

potential. The other function, which gives a family of curves orthogonal to

those of the first function, may then be used to describe the electric field

Ev.

Transforming such a real problem in the complex plane, we also say

domain, is called mapping. A similar situation holds for the

hydrodynamics of an ideal fluid in irrotational motion. The function u

might describe the velocity potential, whereas the function v would then

be the stream function.

In many cases in which the functions u and v are unknown, mapping

permits us to create a coordinate system tailored to the particular

problem.

2. Analytic Continuation

Second-order differential equations of interest in physics may be solved

by power series. The same power series may be used in the complex

plane to replace x by the complex variable z.

The dependence of the solution f(z) at a given z0 on the behavior of f(z)

elsewhere gives us greater insight into the behavior of our solution and a

powerful tool (analytic continuation) for extending the region in which the

solution is valid.

3. Real to Imaginary Change

◊ The change of a parameter k from real to imaginary, k→ik, transforms

the Helmholtz equation into the diffusion equation.

3

◊ The same change transforms the Helmholtz equation solutions (Bessel

and spherical Bessel functions) into the diffusion equation solutions

(modified Bessel and modified spherical Bessel functions).

4. Integrals in the Complex Plane

◊ Evaluating definite integrals;

◊ Inverting power series;

◊ Forming infinite products;

◊ Obtaining solutions of differential equations for large values of the

variable (asymptotic solutions);

◊ Investigating the stability of potentially oscillatory systems;

◊ Inverting integral transforms.


Phys. 601 Mathematical Physics 1


1

Functions of a Complex Variable - Summary

DEFINITION

A function of a complex, or a complex function, is a one in which the

independent variable and the dependent variable are both complex

numbers. More precisely, a complex function is a function whose domain and

range are subsets of the complex plane. A function of a complex variable f(z)

has a real part and an imaginary part as complex numbers, i.e.

(((( )))) (((( )))) (((( ))))zvizuzf ++++====

where u(z) = u(x,y) and v(z) = v(x,y) are real valued functions.

CALCULUS IN THE COMPLEX SPACE

m The derivative of a complex function f(z) of a complex variable is

defined, as for a real function, by:

( ) ( )( )

( )zfdz

df

z

f

zzz

zfzzf

zz′==

δ

δ=

−δ+

−δ+

→δ→δ 00limlim

CAUCHY-RIEMANN CONDITIONS

f(z) has a derivative df/dz if and only if the Cauchy-Riemann conditions:

dy

dv

dx

du==== and

dx

dv

dy

du−−−−====

are satisfied.

These Cauchy–Riemann conditions are necessary for the existence of a

derivative of f(z); that is, if df/dz exists, the Cauchy–Riemann conditions must

hold.

Conversely, if the Cauchy–Riemann conditions are satisfied and the partial

derivatives of u(x, y) and v(x, y) are continuous, the derivative df/dz exists.

2

ANALYTIC (HOLOMORPHIC) FUNCTIONS

If f(z) has a derivative df/dz at z=z0 and in some region around z0, we say that

f(z) is analytic (holomorphic) at z=z0.

If f(z) is analytic everywhere in the (finite) complex plane the function is said

to be entire.

If f’(z) does not exist at z=z0 then we say that z0 is a singularity point.

Holomorphic functions are the central objects of study in complex

analysis. A holomorphic function is a complex-valued function of one or

more complex variables that is complex differentiable in a neighborhood of

every point in its domain. The existence of a complex derivative is a very

strong condition, for it implies that any holomorphic function is actually

infinitely differentiable and equal to its own Taylor series.

While The derivative of a function of a real variable is rather a local property,

the derivative of a function of a complex variable not only governs the local

behavior of the complex function, but controls the distant behavior as well.

The phrase "holomorphic at a point z0" means not just differentiable at z0,

but differentiable everywhere within some neighborhood of z0 in the complex

plane.

CONTOUR INTEGRALS

The integral of a function of a complex variable over a contour in the

complex plane may be defined in close analogy to the (Riemann) integral of

a real function integrated along the real x-axis. This means that the integral

of a function of a complex variable between two points is simply the area

enclosed by this function between these two points. The integral of f(z) along

a specified contour C (from z=z0 to z=z’0) is called the contour integral of f(z).

CAUCHY INTEGRAL FORMULA (2ND BASIC THEOREM)

For an analytic function f(z) on a closed contour C, in the mathematical sense

(counterclockwise) and within the interior region bounded by C. We have

(((( )))) (((( ))))

====−−−−∫∫∫∫ contourtheo0

contourthei

2

1

0

00

0utsidez

nsidezzfdz

zz

zf

i Cπ

3

z0 is any point in the interior region bounded by C, i.e. z # z0.

Although f(z) is assumed analytic, the integrand f(z)/(z - z0) is not analytic at

z = z0 unless f(z0) = 0.

nth DERIVATIVE OF A FUNCTION OF A COMPLEX VARIABLE

( )( )

( )dz

zz

zf

i

nzf

Cn

n ∫ +−π

=1

0

)(

2

!

TAYLOR EXPANSION

Taylor expansion of the function f(z) about z = z0

(((( )))) (((( ))))(((( ))))(((( ))))

!n

zfzzzf

n

n

n 0

0

0∑∑∑∑∞∞∞∞

====

−−−−====

LAURENT EXPANSION

Laurent expansion of a complex function f(z) is a generalization of Taylor

expansion in the presence of singularities. It is a representation of that

function as a power series which includes terms of negative degree.

It may be used to express complex functions in cases where a Taylor series

expansion cannot be applied. It is defined by:

(((( )))) (((( ))))∑∑∑∑∞∞∞∞

∞∞∞∞−−−−====

−−−−====

n

nn zzazf 0

Where The series coefficients are defined by:

(((( ))))

(((( )))).

zz

zdzf

ia

Cnn ∫∫∫∫ ++++

−−−−′′′′

′′′′′′′′====

10

2

1

π

ANALYTICITY AND TAYLOR EXPANSION

◊ An analytic function is a function that is locally given by a convergent

power series.

◊ A function is analytic if and only if it is equal to its Taylor series in some

neighborhood of every point.


Phys. 601 Mathematical Physics 1


1

Finding an analytic function given its real part.

Find the analytic function (((( )))) (((( )))) (((( ))))y,xviy,xuzw ++++==== If (((( )))) 23

3 yxxy,xu −−−−==== (Arfken

Problem 6.2.5 (a))

SOLUTION

w(z) being analytic, its real part and its imaginary part should satisfy

Cauchy-Riemann Conditions, i.e.

y

v

x

u

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂ (1)

and x

v

y

u

∂∂∂∂

∂∂∂∂−−−−====

∂∂∂∂

∂∂∂∂

(2)

Knowing u(x,y) we can use eq. 1(y

v

x

u

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂) and deduce v(x,y), as follows:

y

vyx

x

u

∂∂∂∂

∂∂∂∂====−−−−====

∂∂∂∂

∂∂∂∂ 2233

(((( )))) dyyxdv22

33 −−−−==== (3)

Integrating eq. 3 we get:

(((( )))) (((( )))) (((( ))))xHyyxdyyxy,xv ++++−−−−====−−−−==== ∫∫∫∫3222

333 (4)

where H(x) (should be) is a real valued function. Now we plug this function in

eq. 2 and we obtain

(((( ))))xHyxyx ′′′′++++−−−−====−−−− 66 (((( )))) (((( )))) CxHxH ====⇒⇒⇒⇒====′′′′⇒⇒⇒⇒ 0 (5)

where C is a constant (C ∈ R) because v(x,y) is real valued. Finally we have:

(((( )))) (((( )))) (((( ))))Cyyxiyxxzw ++++−−−−++++−−−−====3223

33 (6)

or (((( )))) (((( )))) CiyixCiyiyxyxixzw ++++++++====++++−−−−−−−−++++====33223

33

Hence (((( )))) Cizzw ++++====3

(7)

One needs to know the value of the function w(z) at one point in order to fix

the value of the constant C.

2

Finding an analytic function given its imaginary part.

Find the analytic function (((( )))) (((( )))) (((( ))))y,xviy,xuzw ++++==== If (((( )))) y

exy,xv−−−−==== sin (Arfken

Problem 6.2.5 (b))

SOLUTION

w(z) being analytic, its real part and its imaginary part should satisfy

Cauchy-Riemann Conditions (eq. 1 and eq. 2).

Knowing v(x,y) we can use eq. 2(y

v

x

u

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂) and deduce u(x,y), as follows:

x

uex

y

v y

∂∂∂∂

∂∂∂∂====−−−−====

∂∂∂∂

∂∂∂∂ −−−−sin (8)

(((( )))) dxexduy−−−−−−−−==== sin (9)

Integrating eq. 11 we get:

(((( )))) (((( )))) (((( ))))yHexdxexy,xu yy ++++====−−−−====−−−−−−−−∫∫∫∫ cossin (10)

where H(y) is a real valued function. Now we plug this function in eq. 2 and

we obtain

(((( ))))yHexexx

v

y

u yy ′′′′++++−−−−====−−−−⇒⇒⇒⇒∂∂∂∂

∂∂∂∂−−−−====

∂∂∂∂

∂∂∂∂ −−−−−−−−coscos (11)

From which we obtain:

(((( )))) (((( )))) CyHyH ====⇒⇒⇒⇒====′′′′ 0 (12)

where C is a constant (C ∈ R) because u(x,y) is real valued. Finally we have:

(((( )))) Cexy,xuy ++++====

−−−−cos (13)

and (((( )))) (((( )))) (((( ))))yyexiCexzw

−−−−−−−− ++++++++==== sincos

Or (((( )))) (((( )))) (((( )))) CieCieCiexixzw yixiyxiy ++++====++++====++++++++====++++−−−−−−−−

sincos (14)

Hence (((( )))) Ciezwzi ++++====

One needs to know the value of the function w(z) at one point in order to fix

the value of the constant C.




1

Gauss Probability Integral

A) Calculating the Integral:

( ) dxexxn

n ∫+∞

∞−

α−=α2

I (1)

For n = 2p (even, n and p ∈ N) we have the following difference equation:

( )

( )

α−==

α−=

α−

α−

α−−α−−α−

2

222

0

422

22222

x

n

nn

xpxpxp

exd

d...

exd

dex

d

dex =

(2)

And for n = 2p+1 (odd, n and p ∈ N) we have the following relation

( )

( )

α−==

α−=

α−

α−

α−−α−−α−+

2

222

1

322

21212

x

p

pn

xppxpxp

exd

d

exd

dex

d

dex

K

=

(3)

Integral (1) can be rewritten in the form:

(even values of n): ( ) ( ) ( ) ( ) ( )αα

−=αα

−=α − 0222 IIIp

pp

pp

p

d

d

d

dp

(odd values of n): ( ) ( ) ( ) ( ) ( )αα

−=αα

−=α −+ 11212 IIIp

pp

pp

p

d

d

d

dp

(4)

for p = 0, 1, 2, 3, …

Thus we only need to compute ( )α0I and ( )α1I and use the previous relation

to get all the other integrals

2

1. Computing ( ) dxex∫

+∞

∞−

α−=α2

0I , Gauss’s probability integral

For this we rather calculate 20I

( ) ( )∫∫∫

+∞

∞−

+α−

+∞

∞−

α−

+∞

∞−

α− =

=α dydxedyedxe yxyx

222220I (5)

Using the polar cylindrical coordinates

φ=φ= inry,rx scos (6)

we have:

φ==+ ddrrdydxryx and222 (7)

the integral can be rewritten as:

( ) ∫∫ ∫+∞

α−

+∞ π

α− π=φ=α

00

2

0

20

222 drreddrre rr

I (9)

Using a simple change of variable: r2 = u we have:

α=

α==

∞

α−

+∞

α−

+∞

α− ∫∫ 21

21

21 0

00

2 uureduedrre (10)

Thus we get:

( )α

π=α2

0I

( )α

π==α⇒ ∫

+∞

∞−

α−dxe

x2

0I (11)

2. Computing ( ) dxexx∫

+∞

∞−

α−=α2

1I

We have (see eq. 9):

( )α

==α ∫+∞

∞−

α−

212

1 dxexx

I (12)

3

3. Finding the other integrals ( )αnI , n > 1

( ) ( ) ( )30

22 2

12

α

π=

α

π

α−=α

α−==α ∫

∞

∞−

α−

d

d

d

ddxex

xII (13)

( ) ( ) ( )21

33

2

121

α=

αα−=α

α−==α α−

∞

∞−

∫ d

d

d

ddxex

wxII (14)

( ) ( ) ( ) ( ) ( )502

22

24

4 43

α

π=α

α−=α

α−==α α−

∞

∞−

∫ IIId

d

d

ddxex

wx (15)

( ) ( ) ( ) ( )3

112

2

35

5α

=αα

−=αα

−==α α−

∞

∞−

∫ IIId

d

d

ddxex

wx (16)

Table 1 gives the results for n = 6, 7, 8, 9 and 10.

( )76 8

15

α

π=αI

( )47

3

α=αI

( )98 16

105

α

π=αI

( )59

12

α=αI

( )1110 32

945

α

π=αI

Table 1: Value of the integral 1 for n = 6, 7, 8, 9 and 10

4

B) Application

Second excited state of a 1D Simple Harmonic Oscillator

Consider a 1D-SHO, mass m = 0.1 kg and frequency ω = 3 ×1014 rad s-1, in its

second excited stated defined by the wavefunction:

( ) ( ) 25222

224 tix

eexAt,xω−α−−α×=Ψ

where h

ω=α

m (Note that [ ] 2−=α L )

1. Computing the energy:

eV49250103Å103

ÅeV197025

25

25

21

2 141182 .

sc

cE =××

×≈ω=ω=ω

+=

−

hhh (12)

2. Computing the normalization constant

( ) ( ) ( )1

2222

2241

−

α−

∞+

∞−

∞+

∞−

−α=⇒=ΨΨ ∫∫ dxexAdxt,xt,x

x** (12)

( )

∫∫∫

∫∞+

∞−

α−

∞+

∞−

α−

∞+

∞−

α−

α−

∞+

∞−

+α−α=

−α=

dxedxexdxex

dxex

xxx

x

222

2

41616

24

242

22J

α×+

α×α−

α×α=

224

2216

2216 024

2IIIJ

α

π×+

α

π××α−

α

π××α=

221

82

241

322

483

32 352

J

α

π=

α

π=

α

π+

α

πα−

α

πα= 8

232

24

216

248 3

2

5

4

J

( )4121

21

6464

π

α=

α

π==⇒

−

−JA

Computing the expectation values of the potential energy and the kinematic

energy 2Ψ

V and 2Ψ

T respectively, in this state.

5

a) 2ΨΨΨΨ

V

( ) ( )

( )

( ) ( ) ( )[ ]

[ ]

22

32

322

322

357222

246222

222426222

222222

222

2

21

451

452

19621

8

221

1962

44824021

22

21

42

443

162

88

1516

21

4161621

4161621

2421

21

222

2

2

Emm

mAmA

mA

IIImA

dxexdxexdxexmA

dxexxmA

dxt,xxmt,xV

xxx

x

*

≡ω=ωα

=α

π××ω×

π

α=

α

π

ω××=

α

π+−

ω×=

α

π××+

α

π××α−

α

π××α

ω×=

α+αα−αα

ω×=

+α−α

ω×=

−α

ω×=

Ψ

ωΨ=

∫∫∫

∫

∫

∞+

∞−

α−

∞+

∞−

α−

∞+

∞−

α−

∞+

∞−

α−

∞+

∞−

Ψ

h

b) 2Ψ

T

( )( )

( )dxt,xdx

t,xd

mt,xT

**

∫∞+

∞−

ΨΨ

Ψ−Ψ= 22

222

2 22

h

For this one needs to compute ( )2

22

dx

t,xd Ψ

( ) ( )[ ]( ) ( )( ) ( )

( ) ( )2532

252

2522

2

2

2

84

2224

22

tix

tix

tix

eexxA

eexxxA

eexAdx

d

dx

t,xd

ω−α−

ω−α−

ω−α−

α+α−=

α−−α+α=

−α=Ψ

( ) ( )( )[ ] ( )

[ ] ( )252243

2532222

22

2

2

8288

284812

tix

tix

eexxA

eexxxxAdx

t,xd

ω−α−

ω−α−

α+α−α=

α−α+α−+α+α−=Ψ

221

45

45

25

2EVET =ω=ω−ω=><−><=

Ψhhh

We leave as an exercise the explicit calculation of 2Ψ

T




1

Gamma Function (Factorial Function)

A) DEFINITION(S)

1) Euler’s Infinite Limit

( )( )( ) ( )

LL

L,,,,z,n

nzzzz

nlimz

z

n3210

21

321−−−≠

+++

⋅⋅≡Γ

∞→ (1)

2) Euler’s Definite Integral

(((( )))) (((( )))) 0,0

1 >>>>ℜℜℜℜ≡≡≡≡ΓΓΓΓ ∫∫∫∫∞∞∞∞

−−−−−−−−zdttez

zt (2)

3) Weierstraββββ’s Infinite Product

( ),e

n

zez

z

nz

n

z −∞

=

γ ∏

+≡

Γ1

11

(3)

where γ is defined as the limiting difference between the harmonic series and

the natural logarithm, i.e.

( ) 577215661901

1

.nlnk

lim

n

kn

=

−=γ ∑

=∞→

(4)

B) PROPERTIES

Its importance stems from its usefulness in developing other functions (Beta

function) and integrals that have direct physical application.

2) Recurrence relation (difference equation)

( ) ( )zzz Γ≡+Γ 1 (5)

3) Definition of the factorial:

( ) ( ) ( ) ( ) ( )!1132111 −=−⋅⋅=−Γ−≡Γ nnnnn L (6)

2

4) The Weierstraβ definition shows that Γ(z) has poles at z = 0, -1, -2, -3 … and

that [Γ(z)]-1 has poles in the finite complex plane, which means that Γ(z) has

no zeros.

5) Duplication Formula (Legendre)

( ) ( )zzzz

22

12

12 Γπ=

+ΓΓ−

(7)

6) Reflection Formula

( ) ( )z

zzπ

π=−ΓΓ

sin1 (8)

6) Complex conjugate

( ) ( )**zz Γ=Γ (9)

C) THE FACTORIAL FUNCTION

Generalization of the factorial to the complex plane

( ) .zR,zdttezt

1!0

−>≡∫∞

− (10)

Fig. 1 shows the factorial function Γ(x+1) extended to negative values, where

the values of z are restrained to be purely real (z = x ∈ R),

Figure 1: ( ) !1 xx =+Γ

C) THE DOUBLE FACTORIAL

( ) ( ) ( )!2

!1212531!!12

n

nn...n

n

+=+=+ L (11)

( ) !22642!!2 nn...nn== L (12)

( ) !!1!!! -nnn = (13)

3

D) THE DIGAMMA AND POLYGAMMA FUNCTIONS

Definition of the digamma function (((( ))))1++++ψψψψ z

( ) ( ) ,nzzz

nlimzdz

dz

n

+−−

+−

+−=+Γ≡+ψ

∞→

1

2

1

1

1ln1ln1 L (14)

Definition of the polygamma function (((( ))))(((( ))))1++++ψψψψ zm

( )( ) ( )

( )( )

L,,,m,nz

m

zdz

dz

n

m

m

m

mm

3211

!1

!ln1

11

1

1

1

=+

−−=

=+ψ

∑∞

=+

+

+

+

(15)

D) MACLAURIN EXPANSION

Maclaurin series is a polynomial series representation of an infinitely

differentiable function f(z), whose value and the values of all its derivatives,

exist at z = 0 and is given by:

( )( ) n

n

n

zn

f∑

∞

= 1!

0 (16)

which we can write as:

( ) ( )( ) ( ) ( )nn

zz

n

zzln

n

nnn

n

n

ξ−+γ−=ψ=+Γ ∑∑∞

=

−∞

= 1

1

1

11!

1 (17)

where ( ) ,n

m

n

m∑∞

=

=ξ1

1is the Riemann zeta function.

E) STIRLING'S SERIES - ASYMPTOTIC EXPANSIONS

Striling's series is an asymptotic expansion of Γ(z) for large |z|.

( ) L−+−+−

++π=+Γ

53 1260

1

360

1

12

1ln

2

12ln

2

11ln

zzzzzzz (18)

Taking the exponential of both side of the previous equation we obtain:

−+−π= −L

53 1260

1

360

1

12

1exp2!

zzzezzz

zz (19)

4

BETA FUNCTION

Definition

Also known as Euler function of the first kind, Beta function is frequently

encountered in the evaluation of many definite integrals.

( )

( )!1

!!

sincos2112

0

1212

++=

θθθ=++Β ∫π

++

nm

nm

dn,mnm

(20)

Equivalently the Beta function in terms of Gamma function is given by:

( ) ( ) ( )( )qp

qpq,p

+Γ

ΓΓ=Β (21)

It is easy to see the symmetric property of the Beta function, i.e.

( ) ( )p,qq,p Β=Β (22)

INTEGRAL FORM OF BETA FUNCTION

( ) ( ) ,dttty,xyx∫

−− −=Β

1

0

11 1 (23)




1

2ND ORDER ORDINARY DIFFERENTIAL EQUATIONS

FINDING THE PARTICULAR SOLUTION

Consider the inhomogeneous 2nd order differential equation

( ) .2

2

tfdt

d

dt

d=++ yb

ya

y (1)

The general solution of Eq. (1) is the sum of two solutions (functions):

( ) ( ) ( ) ,ttt PC yyy += (2)

where ( )tCy is the complementary solution obtained by solving the

corresponding homogenous ODE, i.e. 02

2

=++ ybx

ya

x

y

d

d

d

d and ( )tPy is the so-

called particular solution.

The particular solution of Eq. 1 takes the term ( )tf on the right hand side

into account. The complementary solution is transient in nature, so from a

physical point of view, the particular solution will usually dominate the

response of the system at large times. The method of determining ( )tPy is to

guess a suitable functional form containing arbitrary constants, and then to

choose the constants to ensure it is indeed the solution. If our guess is

incorrect, then no values of these constants will satisfy the differential

equation, and so we have to try a different form. Clearly this procedure

could take a long time; fortunately, there are some guiding rules on what to

try for the common examples of ( )tf :

(1) ( )tf is a polynomial in t, ( )tf is of the form ( ) nttf = .

The trial particular solution is also polynomial in t, with terms up to the

same power. Note that the trial particular solution is a power series in t,

even if ( )tf contains only a single terms n

tA .

2

(2) ( ) C∈= keAtftk

, .

The trial particular solution is ( ) tkP eBt =y .

(3) ( ) ktAktAtf cosorsin= .

The trial particular solution is ( ) ktCktAtyP cossin += , i.e. even though ( )tf

contains only a sine or cosine term, we need both sine and cosine terms

for the particular solution.

(4) ( ) teAteAtftt ββ= ααcosorsin .

The trial particular solution is ( ) ( )tCtBetyx

P β+β= αcossin .

(5) ( )tf is a polynomial of order n in t, multiplied by ekt.

The trial particular solution is a polynomial in t with coefficients to be

determined, multiplied by ekt.

(6) ( )tf is a polynomial of order n in t, multiplied by ktsin .

The trial particular solution is (((( )))) (((( ))))∑∑∑∑====

++++====

n

j

jjjP tktcosDktsinCty

0

If the trial particular solution or part of it is identical to one of the terms of

the complementary function, then the trial particular solution must be

multiplied by an extra power of t. Therefore, we need to find the

complementary solution before we try to work out the particular one. What

is meant by “identical in form”? It means that the ratio of their t

dependences is a constant. Thus -2e-t and Ae

-t are identical in form, but e

-t

and e-2t are not.




GREEN’S THEOREM

A) GREEN’S THEOREM

Green’s theorem is a useful corollary of Gauss’s divergence theorem.

Consider two scalar functions φ1 and φ2 defined over a region of space of

volume V surrounded by a surface S. We shall assume that the 2 functions

and their first derivatives are finite and continuous over V.

We start from the identity:

( )2

2

12121φ∇φ+φ∇⋅φ∇=φ∇φ⋅∇

��

(1)

Now Let’s integrate this equation over the volume V:

( ) ∫∫∫ τφ∇φ+τφ∇⋅φ∇=τφ∇φ⋅∇VVV

ddd2

2

12121

��

(2)

The integral on the lhs of the previous equation can be transformed to a

surface integral using Gauss’s divergence theorem and thus, one can write:

∫∫∫ τφ∇φ+τφ∇⋅φ∇=⋅

φ∇φ

VVSddad 2

212121

��

�

�

(3)

Eq. (3) is the Green’s Theorem in the first form.

By exchanging φ1 and φ2 we obtain the following equation

∫∫∫ τφ∇φ+τφ∇⋅φ∇=⋅

φ∇φ

VVSddad 1

221212

��

�

�

(4)

Subtracting Eq. (4) from Eq. (3) we get:

∫∫ τ

φ∇φ−φ∇φ=⋅

φ∇φ−φ∇φ

VSdad 1

222

211221

�

��

(5)

Eq. (5) is known as the Green’s Theorem in the second form.




1

POTENTIAL THEOREM

Consider a vector (force) F

r which satisfies the equation:

0=×∇ F

rr (1)

which means that F

r is irrotational.

We know that 0=φ∇×∇rr

for any scalar function φ. This suggests the

existence of a scalar function φ such that

φ∇−=rr

F (2)

i.e., the magnitude of the force equals the magnitude of the gradient of the

scalar function φ and its direction is in the opposite direction of this gradient.

Such a force (vector) is said to be conservative. According to the previous

discussion a force F

r is conservative if it satisfies the irrotational relation (1).

The work of such a force is path independent.

The scalar function φ that satisfies Eq. (2) is called a scalar potential.

Alternatively we can associate to any force satisfying the irrotational relation

a scalar function (which depends on space coordinates).

Now consider a vector A

r

which satisfies the equation:

( ) 0=×∇⋅∇ A

rr (3)

For a vector B

r

satisfying the relation:

0=⋅∇ B

rr (4)

We can define a vector A

r

such that:

AB

rrr×∇= (5)

A

r

is called the vector potential.

2

SCALAR AND VECTOR POTENTIAL – ELECTROMAGNETIC THEORY

Theoretically there are an infinite number of scalar potentials satisfying Eq.

(2) and potential vectors satisfying the Eq. (5). Tow kinds of potential appear

in physics: scalar potential and vector potential.

The electric field of a distribution of charges is given by:

φ∇−=rr

E (6)

The scalar potential φ is undetermined to an arbitrary additive constant since

the transformation

φ = φ + c (7)

leaves the electric field invariant.

The magnetic field of a distribution of currents is given by:

AB

rrr×∇= (8)

However, the divergence of A

r

has no physical significance. In fact, we are

completely free to choose A

rr⋅∇ to be whatever we like. Note that, according

to Eq. (11), the magnetic field is invariant under the transformation

ψ∇−→rrr

AA (9)

In other words, the vector potential is undetermined to the gradient of a

scalar field (ψ). This is just another way of saying that we are free to choose

A

rr⋅∇ .

GAUGE TRANSFORMATIONS

The transformations Eq. (7) and Eq. (9) are examples of what mathematicians

call gauge transformations. These transformations are used to determine a

unique scalar potential or a unique vector potential.

The choice of a particular function ψ or a particular constant c is referred to

as a choice of the gauge. We are free to fix the gauge to be whatever we

like. The most sensible choice is the one which makes our equations as

simple as possible.

The usual gauge for the scalar potential φ , known as the Coulomb gauge, is

such that φ → 0 at infinity. The usual gauge for A

r is such that

0=⋅∇ A

rr (10)

complex analysis - ctaps.yu.edu.jo

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