partial differential equations -...

1

© Dr. Nidal M. Ershaidat

Phys. 601: Mathematical Physics

Physics Department

Yarmouk University

Chapter 3 Partial Differential Equations

in Physics

© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

2

Overview

1.Partial Differential Equations

2.First-Order Differential Equations

(Review)

3.Separation of Variables

4.Singular Points

5.Series Solutions – Forbenius' Method

6.A Second Solution

7.Nonhomogeneous Equations –

Green's Function

8.Heat Flow, or Diffusion, PDE


3

Must Read

Chapter 1 in Mathematical Methods for Physics, H. W. Wyld.

For basic concepts please review:Introduction to Mathematical Physics, 2nd Edition 2004, Nabil M. Laham and Nabil Y. Ayoub

http://ctaps.yu.edu.jo/physics/Phys201

Partial Differential Equations


5

IntroductionIn physics the knowledge of the force in an

equation of motion usually leads to a

differential equation.

Thus, almost all the elementary and numerous

advanced parts of theoretical physics are

formulated in terms of differential equations.

Sometimes these are ordinary differential

equations in one variable (abbreviated ODEs).

More often the equations are partial differential

equations (PDEs) in two or more variables.


6

Basic CalculusLet us recall from calculus that the operation of taking an ordinary or partial derivative is a linear operation (LLLL),

( ) ( )( )dx

db

dx

da

dx

yxbyxad ψ+

ϕ=

ψ+ϕ ,,1

( ) ( )( ) ( ) ( )dx

yxb

dx

yxa

x

yxbyxa ,,,, ψ∂+

ϕ∂=

∂

ψ+ϕ∂2

for ODEs involving derivatives in one variable x

only and no quadratic, (dψ/dx)2, or higher

powers.

Similarly, for partial derivations,

2


7

Thus, ODEs and PDEs appear as linear operator equations,

Linear OperatorsIn general

( ) ( ) ψ+ϕ=ψ+ϕ LLL baba

4

where F is a known (source) function of one

(for ODEs) or more variables (for PDEs), LLLL is a

linear combination of derivatives, and ψ is the

unknown function or solution.

( ) F=ψL

3


8

Superposition Principle

Since the dynamics of many physical systems involve

just two derivatives, for example, acceleration in

classical mechanics and the kinetic energy operator,

∼∇∼∇∼∇∼∇2, in quantum mechanics, differential equations of

second-order occur most frequently in physics.

The superposition principle for homogeneous PDEs

states that any linear combination of solutions is again

a solution if F = 0.

If F = 0 and ψψψψ1 and ψψψψ2 are solutions of eq. 4 then the

linear combination a ψψψψ1 + b ψψψψ2 (a and b ∈∈∈∈ CCCC) is also a

solution.

*Maxwell’s and Dirac’s equations are first-order but involve

two unknown functions. Eliminating one unknown yields a

second-order differential equation for the other.

Examples of PDEs


10

1. Laplace’s Equation

Laplace’s Equation 02 =ψ∇

This very common and very important equation occurs in studies of

a. electromagnetic phenomena, including electrostatics, dielectrics, steady currents, and magnetostatics,

b. hydrodynamics (irrotational flow of perfect fluid and surface waves),

c. heat flow,

d. gravitation.

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11

2. Poisson’s Equation

Poisson’s Equation0

2 ερ−=ψ∇

In contrast to the homogeneous Laplace equation, Poisson’s equation is

nonhomogeneous with a source term −ρρρρ/εεεε0.

In its most general form, Poisson's equation is written

vu =∇2

where u(r) is some scalar potential which is to

be determined, and v(r) is a known “source

function”.

6

7


12

The solutions to Poisson's equation are completely

superposable. Thus, if u1 is the potential generated by

the source function v1 and u2 is the potential generated

by the source function v1 so that

Poisson’s Equation - Details

The most common boundary condition applied to this

equation is that the potential u is zero at infinity.

22

2

11

2, vuvu =∇=∇

( )2121

2

2

2

1

2 vvuuuu +=+∇=∇+∇

then the potential generated v1+v2 is u1+u2, since

Poisson's equation has this property because it is linear in both the potential and the source term.

3

13

Poisson’s Equation - Electrostatics

The electric field generated by a set of stationary charges (density ρρρρ) can be written as the gradient of a scalar potential, so that

φφφφ∇∇∇∇−−−−====E

0

2

εεεε

ρρρρ−−−−====φφφφ∇∇∇∇

Combining the previous two equations we obtain a partial differential equation for the scalar potential:

The same electric field verifies the first of four field equations (Maxwell’s Equations) which describe all electromagnetic phenomena.

0εεεε

ρρρρ====••••∇∇∇∇==== EEdiv

10

9

8


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Other Examples


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Other Examples

Solution Solution

TechniquesTechniques


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General Features• Separation of variables The PDE is split into ODEs that are related by common constants that appear as

eigenvalues of linear operators, Lψ = lψ, usually in one variable. Helmholtz equation, and Schrödinger equation are examples of this.

• Conversion of a PDE into an integral equation using Green’s functions applies to inhomogeneous PDEs

• Other analytical methods, such as the use of integral transforms (This course Chapter 6 = Arfken Chapter 15)

• Frobenius’ power-series method may be used for ODEs It does not always work but is often the simplest method when it does. Classes of Classes of PDEPDE’’ss

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Second-order PDE’s

• Elliptical PDE’s involve the Laplacian ∇ ∇ ∇ ∇2

or c-2 ∂ ∂ ∂ ∂2/∂∂∂∂t2 + ∇ ∇ ∇ ∇2

•Parabolic PDE’s a ∂∂∂∂/∂∂∂∂t+ ∇ ∇ ∇ ∇2

•Hyperbolic PDE’s c-2 ∂ ∂ ∂ ∂2/∂∂∂∂t2 - ∇ ∇ ∇ ∇2 Ordinary Differential Ordinary Differential

EquationsEquations

IntroductionIntroduction


22

The order of the highest derivative is called

the order of the differential equations.

Definition:

An ordinary differential equation is any

equation which contains a function of one

independent variable and its derivatives.

The solution of an ODE is any relationship

between the unknown function with its

variable which verifies the ODE.

ODE’s


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Examples 1

ODE’s- Examples

2yxyxxd

yd====++++ This is a 1st-order ODE.

Finding the solution of these ODE’s is

finding the function y(x).

We shall see in this section how to solve

these ODE’s, i.e. how to find y(x).

2

4

4

yxyxxd

yd

xd

yd====++++−−−− This is a 4th-order ODE. Special FirstSpecial First--order ODEorder ODE’’ss

5

Separable EquationsSeparable Equations


26

Any equation which can be written in the

form

General Form of Separable 1st-order Differential Equations

The solution to this equation, called general

solution, is obtained by directly integrating

it, i.e.(((( )))) (((( )))) CdyyGxdxF ====++++ ∫∫∫∫∫∫∫∫

C is an integration constant.

(((( )))) (((( )))) 0====++++ dyyGxdxF SDE1

SDE2

is called separable 1st-order DE.


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If we have particular conditions that impose

a certain value for the constant C then the

solution is said to be a particular solution.

Boundary Conditions

The particular conditions are called

boundary conditions (BC) Linear EquationsLinear Equations


29

A linear 1st-order differential equation is

defined as:

General Form of Linear 1st-order Differential Equations

( ) ( )xQyxPx

y=+

d

d

where P(x), Q(x) are general functions of the

independent variable x,

If Q(x) = 0 then the linear equation LDE1

becomes a separable equation.

LDE1


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Homogeneous Linear 1st-order Differential Equations

(((( )))) 0====++++ yxPxd

yd

(((( )))) xdxPy

yd−−−−====⇒⇒⇒⇒

Integrating both sides we have:

(((( )))) CxdxPyln ++++−−−−==== ∫∫∫∫(((( )))) ∫∫∫∫ ++++−−−−====⇒⇒⇒⇒ CxdxPexp)x(y (((( )))) ∫∫∫∫−−−−==== xdxPexpA

Where A=eC =constant.

LDE2

LDE3

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Inhomogeneous Linear 1st-order Differential Equations

(((( )))) (((( ))))xQyxPxd

yd====++++

Q(x)#0 for any x

The solution of Eq. LDE4 is obtained by

making the its left hand side an exact derivative of some function.

Solution using the integrating factorSolution using the integrating factor

This is done by multiplying both sides by

the function f(x) which we call the

integrating factor.

LDE4


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A differential of the form

y

fh

x

fg

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂==== ,

is called exact differential if is path independent.

Exact Differential

∫∫∫∫ df

(((( )))) (((( ))))dyyxhdxyxgdf ,, ++++====

This is verified if and only if: dyy

fdx

x

fdf

∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂====

so g and h must verify

But andxy

f

y

g

∂∂∂∂∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂ 2

yx

f

x

h

∂∂∂∂∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂2

x

h

y

g

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂⇒⇒⇒⇒


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Solution of IL- 1st- order DE

First we multiply by f(x)

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))xQxfyxPxfxd

ydxf ====++++

If the lhs of this equation is required to be

an exact differential of some function then we must have:

(((( )))) (((( ))))xPxfxd

df====

This last equation is a separable linear 1st-order DE and its solution is easy:

(((( )))) (((( )))) IedxxPxf ======== ∫∫∫∫exp


34

Thus we have

Solution of IL-1st-order DE

(((( )))) (((( ))))xQeyxPexd

yde III ====++++

Which we can rewrite as follows:

Integrating both sides we get:

Therefore the solution y(x) is given by:

(((( )))) (((( ))))xQexd

eyd II

====

(((( )))) CdxxQeey II ++++==== ∫∫∫∫

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))xIxIxIeCdxxQeexy

−−−−−−−− ++++==== ∫∫∫∫


35

Solve the equation:

Example 1

Solution: Here we have

xyxxd

yd====++++

(((( )))) xxP ==== (((( )))) xxQ ====

(((( )))) (((( )))) 22

expexpx

edxxedxxPxf I ================ ∫∫∫∫∫∫∫∫(((( )))) 222 222 xxx

eCdxxeexy−−−−−−−− ++++==== ∫∫∫∫

22 22 xuxeduedxxe ======== ∫∫∫∫∫∫∫∫

(((( )))) 222 222 xxx eCeexy −−−−−−−− ++++==== (((( )))) 22

1 xeCxy −−−−++++====⇒⇒⇒⇒

Exact EquationsExact Equations

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In this case the solution of the above equation is:

An equation of the form:

Where M(x,y) and N(x,y) are functions of the

independent variables x and yis called an exact first-order ordinary

differential equation if the lhs can be expressed

as an exact differential dU of a function U(x,y)

such that:

General Form of Exact Equations

(((( )))) (((( )))) 0,, ====++++ ydyxNxdyxM E1

(((( )))) (((( )))) ydyxNxdyxMdU ,, ++++====

(((( )))) CyxU ====, = constant


38

Sometimes Eq. E1 may not be exact but we

can make it exact by multiplying it by a

proper function (called the integration

factor) as the one we have seen in the

inhomogeneous LDE.

This is verified if and only of:

The Integration Factor

yxx

N

y

M

∂∂∂∂

∂∂∂∂====

∂∂∂∂

∂∂∂∂


39

Solve the equation:

Example 2

Solution: This equation can be put in the

form of an exact equation.

(((( )))) (((( ))))yxyxyxd

yd−−−−++++++++==== sinsintan

(((( )))) (((( ))))[[[[ ]]]] 0tansinsin ====−−−−−−−−++++++++ ydyxdyxyx Homogeneous Homogeneous

Differential EquationsDifferential Equations

41

A homogeneous differential equation is a first-

order differential equation which can be

written in the form:

To solve such an equation we use the change

of variable

Thus we have:

Substituting in Eq. H1 we get:

General Form of Homogeneous DE’s

====

x

yF

xd

ydH1

(((( ))))xvxy ====

(((( ))))xd

dvxxv

xd

yd++++====

(((( )))) (((( ))))vFxd

dvxxv ====++++ H2


42

General Form of Homogeneous DE’s

The previous equation (H2) can be written in

the form:

(((( )))) vvF

dv

x

dx

−−−−====

Variables are separated and the general solution of H3 is obtained by integrating

both sides

H3

(((( ))))∫∫∫∫∫∫∫∫ ++++−−−−

==== CvvF

dv

x

dxH4

Where C is a constant

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Solve the equation:

Example 31

Solution: This equation can be put in the

form of an exact equation.

++++====

x

y

x

y

xd

ydsin

(((( ))))xvxy ====

(((( ))))xd

dvxxv

xd

yd++++====⇒⇒⇒⇒

(((( )))) (((( )))) (((( ))))vxvxd

dvxxv sin++++====++++

(((( )))) x

xd

v

dv====

sin© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

44

Example 32

(((( ))))

−−−−

====++++==== ∫∫∫∫ 2

cosln2

sinlnsin

lnvv

Cv

dvx

====

2tanlnln

vcx

====

2tan

vcx

x

y

c

xv ====

==== −−−−1tan2 (((( ))))

====⇒⇒⇒⇒ −−−−

c

xxxy 1tan2

Second-order Linear

Differential Equations


46

Any equation which can be written in the

form

Various methods are used to solve such

equations. These methods depend on the

nature of P(x), Q(x) and R(x).

General Form of 2nd-order Linear Differential Equations

( ) ( ) ( )xRyxQx

yxP

x

y=++

d

d

d

d2

2

Where P(x), Q(x) and R(x) are general

functions of the independent variable x,

is called second-order Linear Differential

equation.

2nd-order Linear Differential

Equations with Constant

Coefficients


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We shall focus here on the 2nd-order LDE with

constant coefficients, i.e. the cases where the

coefficients P(x) and Q(x) are constant.

Homogeneous and Inhomogeneous 2nd-order Linear ODE

( )xRybx

ya

x

y=++

d

d

d

d2

2

This particular form of 2nd-order LDE is said to

be homogeneous if the rhs = 0, i.e. R(x) = 0.

If R(x) # 0 then the 2nd-order LDE is said to

be inhomogeneous.

( )xRybyay =+′+′′or

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49

The 2nd-order ODE

Homogeneous 2nd-order Linear ODE

02

2

=++ ybx

ya

x

y

d

d

d

d

is homogeneous because each term contains

y(x) or a derivative;

0=+′+′′ ybyayor

It is linear because each y, dy/dx or d2y/dx2

appears at the first power. No products (i.e. y

dy/dx or y d2y/dx2 appear)

Homogeneous 2nd-order

Linear Differential Equations

51

0====++++′′′′++++′′′′′′′′ ybyay02

2

====++++++++ ybxd

yda

xd

yd

As we said these equations are of the form:

Homogeneous 2nd-order Linear Differential Equations

We discuss here the method used to solve

such an equation.

02 =++ ybyay DD

We can rewrite Eq. LDE1 as:

First let’s use symbol D to stand for and D2

to stand for ( and ). xd

d

2

2

xd

dyyD ′′′′==== yyD ′′′′′′′′====2

LDE1

LDE2

or


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2

42 baa −−−−−−−−−−−−====ββββ

Where αααα and ββββ are the roots of the quadratic

equation Q. They are given by:

Factorizing we have:

Finding a solution

The quadratic equation (also called the

auxiliary equation)

D2 + a D + b

(((( )))) 02 ====++++++++ ybDaD

2

42 baa −−−−++++−−−−====αααα

LDE3

(D - αααα)(D - ββββ)can be written in the form:

Q


53

Let u = (D - ββββ) y. Then we have

Recipe1

(D - ββββ) y = A e αααα x

This 1st-order separable diff. eq. has for

solution (see previous lecture):

Substituting u in Eq. LDE3 we get:

u’ - αααα u = 0

xeAu α=

LDE4(D - αααα) u = 0

LDE5

or

y’ - ββββ y = A e αααα xor

This is a linear 1st-order diff. eq. Its solution is:


54

Recipe2

See previous lecture.

BxdeeAexpey xxx ++++==== ∫∫∫∫ββββ−−−−ααααββββ++++

The solution depends on the roots αααα and ββββ.

(((( )))) xxx eBxdeeAy ββββββββ−−−−ααααββββ ++++==== ∫∫∫∫

Case 1: αααα#ββββ(((( ))))

(((( ))))x

xx eB

eeAy ββββ

ββββ−−−−ααααββββ ++++

ββββ−−−−αααα××××====

(((( ))))x

x

eBeA

yββββ

αααα

++++ββββ−−−−αααα

====⇒⇒⇒⇒

where

xxeBeC

ββββαααα ++++====

(((( ))))ββββ−−−−αααα====

AC

LDE6

LDE7

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Recipe3Case 2: αααα=ββββ

With

The solution is similar for the case αααα#ββββ where

we replace αααα and ββββ by their complex

expressions, i.e.

Case 3: αααα and ββββ are complex conjugates

(((( )))) xxx eBxAeBxdeAy ββββββββββββ ++++====++++==== ∫∫∫∫

(((( )))) (((( )))) (((( ))))xixi eBeCxy δδδδ−−−−γγγγδδδδ++++γγγγ ++++====

which can be rewritten as:

δδδδ++++γγγγ====αααα i δδδδ−−−−γγγγ====ββββ i

In this case the determinant is

negative and the roots αααα and ββββ are of the form:2

4 2ab −−−−====δδδδ

2a====γγγγ

(((( )))) [[[[ ]]]]xixix eBeCexy δδδδ−−−−δδδδ++++γγγγ ++++====© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

56

Recipe4or

Where K and ψψψψ are arbitrary constants.

or

Where C1 and C2 are arbitrary constants related to the (original) integration constants A and B.

(((( )))) (((( )))) xxxeBxAeBxdeAxy

ββββββββββββ ++++====++++==== ∫∫∫∫(((( )))) (((( )))) (((( ))))[[[[ ]]]]xBCixBCexy

x δδδδ−−−−++++δδδδ++++====⇒⇒⇒⇒ γγγγ sincos

(((( )))) [[[[ ]]]]xCixCexyx δδδδ++++δδδδ==== γγγγ sincos 21

or

(((( )))) (((( ))))ψψψψ++++δδδδ==== γγγγ xeKxyx

sin


58

We notice that while we only need to determine

one arbitrary (integration) constant in the case of

first-order ODE’s, two arbitrary (integration)

constants appear in the case of 2nd-order ODE’s.

This observation can be generalized for differential

equations of any order. The number of arbitrary

constants equals strictly the order of the equation.

Finally, the integration constants are obtained

knowing the particular conditions which we called

Boundary conditions.

In physics we always have an idea of such

conditions and the general solution is, in general,

perfectly defined (known).

Integration constants

Inhomogeneous 2nd-order Linear

Differential Equations with

Constant Coefficients


60

1- The solution obtained in setting R(x) =0 which

is called the complementary solution (yc)

As we said these equations are of the form:

The general solution of Eq. ILDE1 is the sum

of two solutions:

2- A particular solution which satisfies Eq. ILDE1.

This solution is called the particular solution (yP)

The general solution is the sum:

( )xRybx

ya

x

y=++

d

d

d

d

2

2

Inhomogeneous 2nd-order Linear DE

ILDE1

y(x) = yc(x) + yP(x)


61

In this case we have:

( )xRybyay =++ 222 ppp'''

y(x) is the general solution if and only if

there is only one distinct particular solution

One Distinct Particular Solution

Let’s suppose that there are 2 distinct

particular solutions yp1 and yp2

( )xRybyay =++ 111 ppp'''

Subtracting Eq. ILDE2 from Eq. ILDE3 we get

( ) ( ) ( ) ( )xRyybyyayy =−+−+− 121212 pppppp''''''

ILDE2

ILDE3

11


62

Which means that the function yp2 – yp1 must

be a solution of Eq. ILDE1.


But yc is the solution of Eq. ILDE2. Therefore

there is only one distinct particular solution

yp which when added to yc gives the general

solution of Eq. ILDE1


63


Eq. ILDE1 can be written as:

Let u = (D - ββββ) y. Then Eq. ILDE2 becomes:

where αααα and ββββ are the roots of the auxiliary

equation

(((( ))))(((( )))) (((( ))))xRyDD ====ββββ−−−−αααα−−−− ILDE4

This is a 1st-order linear diff. eq. in u and x,

whose solution is (see previous paragraph):

u’ - αααα u = R(x)

(((( )))) 1CdxexReuxdxI xx ′′′′++++====⇒⇒⇒⇒αααα====αααα−−−−==== ∫∫∫∫∫∫∫∫ αααα−−−−αααα−−−−

ILDE5(D - αααα) u = R(x)


64


Let (((( )))) (((( ))))∫∫∫∫ αααα−−−−==== dxexRxF x

(((( )))) xx eCxFeu αααααααα ′′′′++++==== 1

ILDE6

We thus have:

This is a first-order linear diff. eq. such that:

Eq. ILDE1 can be rewritten as:

(((( )))) xx eCxFeyy αααααααα ′′′′++++====ββββ−−−−′′′′ 1ILDE7

(((( ))))[[[[ ]]]] 21 CdxeCxFeeey xxxx ++++′′′′++++==== ∫∫∫∫ ααααααααββββ−−−−ββββ−−−−

(((( )))) (((( ))))21 CdxeCdxexFe xxx ++++′′′′++++==== ∫∫∫∫∫∫∫∫

ββββ−−−−ααααααααββββ−−−−

Suppose αααα#ββββ(((( )))) (((( ))))

(((( ))))xxxx eCe

CdxxFeey ββββααααββββ−−−−ααααββββ ++++

ββββ−−−−αααα

′′′′++++==== ∫∫∫∫ 2

1ILDE8


65

Put then

where

which is y = yC + yP

We recognize in Eq. ILDE9

Therefore the particular solution yyPP is

contained in the expression

General Solution

(((( ))))∫∫∫∫ααααββββ−−−−ββββββββαααα ++++++++==== dxxFeeeeCeCy xxxxx

21

(((( ))))x

CC

ββββ−−−−αααα

′′′′==== 1

1

(((( )))) (((( ))))∫∫∫∫ ββββ−−−−ααααββββ==== dxxFeey xxP

(((( )))) (((( ))))∫∫∫∫ αααα−−−−==== dxexRxF x

ILDE9

xxC eCeCy ββββαααα ++++==== 21

Cy

ILDE10

Py


66

The general solution of an inhomogeneous

2nd-order differential equation of the form:

with

Summary

xxC eCeCy ββββαααα ++++==== 21

is y = yC + yP

(((( )))) (((( ))))∫∫∫∫ββββ−−−−ααααββββ==== dxxFeey xx

P

(((( )))) (((( ))))∫∫∫∫αααα−−−−==== dxexRxF x

(((( ))))xRybxd

yda

xd

yd====++++++++

2

2

ILDE1

Particular Solutions yp for

Various Functions R(x).

Method of Undetermined

Coefficients

12


69

We shall consider here the most frequent

cases for R(x) in physics.

Various Forms of R(x)

1) R(x) = A = constant

2) R(x) = K ecx , where K and c are constants

3) R(x) = K sin(ωωωωx+δδδδ),

where K, ωωωω and δδδδ are constants.

4) , i.e. R(x) = Polynomial of

degree n

(((( )))) ∑∑∑∑====

====n

i

ii xaxR

0

(((( )))) ∑∑∑∑====

====

n

i

ii

xc xaexR

0

5)


70

Let’s consider the equation:

Case 1: R(x) = Constant

Where A is a constant.

Aybyay ====++++′′′′++++′′′′′′′′ LDE2

(((( )))) xx eA

xdeAxF αααα−−−−αααα−−−−

αααα−−−−======== ∫∫∫∫

(((( ))))

ββββαααα====

αααα−−−−==== ∫∫∫∫

αααα−−−−ββββ−−−−ααααββββ Axde

Aeey xxx

P

b====ββββααααb

AyP ====⇒⇒⇒⇒

2

42

baa −−−−−−−−−−−−====ββββ,

baa

2

42 −−−−++++−−−−

====αααα

71

Example 4

Solve the equation:

Case 2: αααα=ββββ

General solution:

444 ====++++′′′′−−−−′′′′′′′′ yyy

14

4============

b

AyP

044 ====++++′′′′−−−−′′′′′′′′ yyy

αααα========××××−−−−−−−−

====−−−−−−−−−−−−

====ββββ 22

44164

2

42 baa

22

44164

2

42

====××××−−−−++++

====−−−−++++−−−−

====ααααbaa

(((( )))) xC eBxAy ββββ++++====

(((( )))) 1++++++++==== ββββ xeBxAy© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

72

which equals the rhs of the equation.

[[[[ ]]]] 4448844442 ++++++++++++−−−−−−−−−−−−++++++++ BxABxAABxAAe x

Checking:

Checking the solution

444 ====++++′′′′−−−−′′′′′′′′ yyy(((( )))) 1++++++++==== ββββ xeBxAy

(((( )))) (((( ))))(((( ))))BxAAeeBxAeA'y xxx ++++ββββ++++====++++ββββ++++==== ββββββββββββ

(((( ))))(((( )))) xx eABxAAe''y ββββββββ ββββ++++++++ββββ++++ββββ====

[[[[ ]]]] 4444442 22 ++++++++++++ββββ−−−−ββββ−−−−−−−−ββββ++++ββββ++++ββββββββ BxABxAABxAAe x

[[[[ ]]]] xeABxAA''y ββββββββ++++ββββ++++ββββ++++ββββ==== 22

4====

Using the value ββββ = 2, we get

73

( )( ) ( )

( )( )( )xx

xxx xy

ββ

βααβ

βα

α

−

−−

−−=

−= ∫

c

c

P

ecc

Ke

deec

Ke

The linear 2nd-order differential equation is:

Case 2: R(x) = K ecx

where K and c are constants.

xceKybyay ====++++′′′′++++′′′′′′′′ LDE2

i) c # αααα and c # ββββ.

( )( )

( )xxx xx αα

α−−

−== ∫

cce

c

KdeeKF

xcP eGy ====⇒⇒⇒⇒

G can be determined using the method of

undetermined coefficients.© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

74

Case 2: R(x) = K ecx

(((( ))))∫∫∫∫ββββ−−−−ααααββββ==== xdexKey xx

P

ii) c = αααα and c # ββββ.

(((( )))) xKxdKxdeeKxF xcx ============ ∫∫∫∫∫∫∫∫αααα−−−−

The term Deααααx can be “absorbed” into yc, the

complementary solution and the general form for the particular solution in this case is:

(((( ))))

(((( ))))

(((( ))))

(((( ))))

ββββ−−−−αααα−−−−

ββββ−−−−αααα====

ββββ−−−−ααααββββ−−−−ααααββββ

2

xxx

P

eexKey

xxP eDexGy αααααααα ++++====⇒⇒⇒⇒

xxcP exGexGy αααα========

13


75

Solve the equation

y = yC + yP

First yc

We are in the case αααα # ββββ

Example 51

xeyyy ====−−−−′′′′++++′′′′′′′′ 2

(((( ))))2

2

2411

2

42

−−−−====−−−−××××−−−−−−−−−−−−

====−−−−−−−−−−−−

====ββββbaa

(((( ))))1

2

2411

2

42

====−−−−××××−−−−++++−−−−

====−−−−++++−−−−

====ααααbaa

02 ====−−−−′′′′++++′′′′′′′′ yyy

xxC eBeCy ββββαααα ++++====⇒⇒⇒⇒


76

Example 52

xxP eDexGy αααααααα ++++====⇒⇒⇒⇒

x

D

x

G

P eexy

−−−−++++====⇒⇒⇒⇒9

1

3

1

(((( ))))

(((( ))))

(((( ))))

(((( ))))

ββββ−−−−αααα−−−−

ββββ−−−−αααα====

ββββ−−−−ααααββββ−−−−ααααββββ

2

xxx

P

eexKey

−−−−==== −−−−

93

332

xxx

P

eexey

The general solution is finally:

P

Cy

xx

y

xx eexeBeAy

−−−−++++++++++++==== −−−−

9

1

3

12

xxx exeBeAy3

1

9

12 ++++++++

−−−−==== −−−− xx eBeAx 2

9

1

3

1 −−−−++++

−−−−++++====

Finding yP: We are in the case αααα#ββββ, αααα = 1 and ββββ = -2


77

xxx exeBeAy3

1

9

1 2 ++++++++

−−−−==== −−−−

Checking the solution

xxxx eexeBeAy3

2

3

14

9

1 2 ++++++++++++

−−−−====′′′′′′′′ −−−−

xeyyy ====−−−−′′′′++++′′′′′′′′ 2

xxxx eexeBeAy3

1

3

12

9

1 2 ++++++++−−−−

−−−−====′′′′ −−−−

Exercise: Check the calculations© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

78

xx exdedv αααα−−−−αααα−−−−

αααα−−−−⇒⇒⇒⇒====

1

Case 3: R(x) = K sin(ωωωωx+δδδδ)The linear 2nd-order differential equation is:

Where K, ωωωω and δδδδ are constants.

(((( ))))δδδδ++++ωωωω====++++′′′′++++′′′′′′′′ xKybyay sin LDELDE22

(((( )))) (((( ))))dxxduxu δδδδ++++ωωωωωωωω====⇒⇒⇒⇒δδδδ++++ωωωω==== cossin

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

δδδδ++++ωωωω

αααα

ωωωω++++δδδδ++++ωωωω

αααα−−−−==== ∫∫∫∫

αααα−−−−αααα−−−− xdxcosexsineKxF xx1

(((( )))) (((( )))) xdxeKxF x δδδδ++++ωωωω==== ∫∫∫∫ αααα−−−− sin

Integrating by parts:

∫∫∫∫∫∫∫∫ −−−−==== duvvudvu


79

Case 3: Solution1

Integrating by parts the integral on the r.h.s

of the previous equation:

(((( )))) (((( )))) (((( ))))dxxexexdxe xxx ∫∫∫∫∫∫∫∫ δδδδ++++ωωωωαααα

ωωωω−−−−δδδδ++++ωωωω

αααα−−−−====δδδδ++++ωωωω αααα−−−−αααα−−−−αααα−−−− sincoscos1

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

(((( ))))

(((( ))))∫∫∫∫

∫∫∫∫

δδδδ++++ωωωωαααα

ωωωω−−−−




−−−−====δδδδ++++ωωωω

αααα−−−−


αααα−−−−αααα−−−−

xdxsine

xdxcose

xsinexdxsine

x

x

xx

2

2

2

1


80

Case 3: Solution2

Rearranging we get:

(((( ))))

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


αααα


αααα−−−−====


αααα

ωωωω++++


αααα−−−−∫∫∫∫

xxe

xdxe

x

x

cossin

sin12

2

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


αααα


αααα

ωωωω++++αααα

====αααα−−−−

xxe

KxFx

cossin

12

2

14


81

Case 4

Case 6

(((( )))) ∑∑∑∑====

====n

i

ii xaxR

0

(((( )))) ∑∑∑∑====

====n

i

ii

xc xaexR

0Case 5

(((( )))) (((( ))))∑∑∑∑====

====n

i

ixc xfexR

0

See Laham-Ayoub pages 267-275

Method of Variation of

Parameters


83

A general lengthy method, essentially

used when the integral

is not straightforward.

Method of Variation of Parameters

(((( )))) (((( ))))∫∫∫∫ αααα−−−−==== dxexRxF x


84

(We are in the case of αααα and ββββ complex

conjugates). yC is given by:

Find the general solution of

Solution: Here we have:

Example illustrating the method

xyy

2sin

14 ====++++′′′′′′′′ V1

V2

V3

i2−−−−====ββββ⇒⇒⇒⇒i22

4400====

××××−−−−++++====αααα

xCxCyC 2sin2cos 21 ++++====

Complementary solution yC

(((( ))))x

xR2sin

1====


85

We, now, assume that the general solution is of the form:

General Solution1

V4

The technique now is to impose a relation between the functions f1 and f2 since we have only one equation to be satisfied.

A smart choice is to impose that:

(((( )))) (((( )))) (((( )))) (((( )))) xxfxxfxxfxxfdx

dy2sin2cos2cos22sin2 2121

′′′′++++′′′′++++++++−−−−====

Deriving Eq. V4 gives:

Where f1 and f2 are suitable functions of x.

V5

V6

(((( )))) (((( )))) xxfxxfy 2sin2cos 21 ++++====

(((( )))) (((( )))) 02sin2cos 21 ====′′′′++++′′′′ xxfxxf

86

General Solution2

V7

Thus the diff. Eq. V1 can be written as:

This leaves us with

Deriving Eq. V7 gives:

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

(((( )))) (((( ))))(((( ))))x

xxfxxf

xxfxxfxxfxxf

2sin

12sin2cos4

2cos22sin22sin42cos4

21

2121

====++++++++

′′′′++++′′′′−−−−++++−−−−

(((( )))) (((( )))) xxfxxfdx

dy2cos22sin2 21 ++++−−−−====

(((( )))) (((( )))) (((( )))) (((( )))) xxfxxfxxfxxfdx

ydsin 2cos2222sin42cos4 21212

2

′′′′++++′′′′−−−−++++−−−−====

V8

V9

(((( )))) (((( ))))xsin

xcosxfxsinxf2

12222 21 ====′′′′++++′′′′−−−−⇒⇒⇒⇒ V10

15


87

V10

General Solution3

Multiplying Eq. V6 by 2 cos 2x we have:

Multiplying Eq. V10 by sin 2x we have:

Adding Eq. V11 and Eq. V12 we get:

Which gives

(((( )))) (((( )))) 02sin2cos22cos2 22

1 ====′′′′++++′′′′ xxxfxxf V11

V6(((( )))) (((( )))) 02sin2cos 21 ====′′′′++++′′′′ xxfxxf

(((( )))) (((( )))) 12sin2cos22sin2 22

1 −−−−====′′′′−−−−′′′′ xxxfxxf V12

(((( )))) (((( ))))x

xxfxxf2sin

12cos22sin2 21 ====′′′′++++′′′′−−−−

12 1 −−−−====′′′′f V13

112

1Cxf ++++−−−−==== V14

88

Integrating Eq. V6 (change of variable u = sin 2x)we get:

The general solution is thus:

or

General solution4

V6(((( ))))

xx

xxff

2tan2

1

2sin

2cos12 ====′′′′−−−−====′′′′

V15[[[[ ]]]] 22 2sinln4

1

4

1

2sin2

2cosCx

u

du

x

dxxf ++++============ ∫∫∫∫∫∫∫∫

V16[[[[ ]]]] xCxxCxy 2sin2sinln4

12cos

2

121

++++++++

++++−−−−====

V4[[[[ ]]]]xxxxCxCy 2sinln

4

12cos

2

12sin2cos 21 ++++−−−−++++====

Cy

Py

Other Second-order

Differential Equation


90

The differential equation reads:

Solution

Differential Equations with y missing

Where F is some functions of x and y’,

====

xd

yd,xF

xd

yd2

2

We make the substitution 2

2

xd

yd

xd

du

xd

ydu ====⇒⇒⇒⇒====

(((( ))))uxFxd

du,====

This is a first-order diff. eq. which does not

need to be linear. Its solution depends on F.

O1

O2

O3


91

Differential Equations with y missing

where the other arbitrary constant is

included in U(x).

Suppose we find a solution u = U(x), then

(((( )))) (((( )))) 2CdxxUyxUxd

dy++++====⇒⇒⇒⇒==== ∫∫∫∫


92

This is a separable first-order diff. eq. We

have:

Solve

Example 6

(((( )))) 12

2

1Cexdxexu xx ++++−−−−====++++==== −−−−−−−−∫∫∫∫

xex

xd

yd −−−−++++====2

2

We make the substitution 2

2

xd

yd

xd

du

xd

ydu ====⇒⇒⇒⇒====

xexxd

du −−−−++++====

∫∫∫∫

++++−−−−====⇒⇒⇒⇒==== −−−− dxCexyu

xd

yd x1

2

2

1

213

6

1CxCex

x ++++++++++++==== −−−−

16


93

The differential equation reads:

Solution

Differential Equations with x missing

Where F is some functions of y and y’.

====

xd

ydyF

xd

yd,

2

2

We make the substitution xd

dy

dy

du

xd

du

xd

ydu ====⇒⇒⇒⇒====

(((( ))))u,xFxd

duu ====

This is a first-order Diff. eq. which does not

need to be linear. Its solution depends on F.

dy

duu

xd

du

xd

yd========

2

2

In Eq. O4

O4

O5

O6


94

Differential Equations with x and y' missing

See Laham-Ayoub page 280


95

This is a differential equation of the form:

To solve such an equation we first reduce it to

a linear second-order diff. eq. with constant

coefficient.

Make the substitution:

Euler or Cauchy Differential Equation1

Where a2, a1, and a0 are constant coefficients.

(((( ))))xfyaxd

ydxa

xd

ydxa ====++++++++ 012

22

2

uex ====

O7

xedu

dx u ======== uedxduxdx

du −−−−============11

O8

and


96


And

i.e.

Substituting O8, O9 and O10 in O7 we get:

du

yde

xd

du

du

yd

xd

yd u−−−−======== O9

du

de

xd

d u−−−−====

du

yde

du

yde

du

yde

du

ydee

du

yde

du

de

xd

dy

dx

d

xd

yd

uu

uuuuu

2

2

22

2

2

2

2

−−−−−−−−

−−−−−−−−−−−−−−−−−−−−

−−−−====

++++−−−−============

O10

(((( ))))uuuuuu efyadu

ydeea

du

yde

du

ydeea ====++++++++

−−−− −−−−−−−−−−−−

012

2

222

2


97


Substituting O8, O9 and O10 in O7 we get:

du

yde

xd

du

du

yd

xd

yd u−−−−======== O9

du

yde

du

yde

du

yde

du

ydee

du

yde

du

de

xd

dy

dx

d

xd

yd

uu

uuuuu

2

2

22

2

2

2

2

−−−−−−−−

−−−−−−−−−−−−−−−−−−−−

−−−−====

++++−−−−============

O10

(((( ))))uuuuuu efyadu

ydeea

du

yde

du

ydeea ====++++++++

−−−− −−−−−−−−−−−−

012

2

222

2

uex ==== O8


98


Which gives

Rearranging we get:

(((( ))))uefyadu

yda

du

yd

du

yda ====++++++++

−−−− 012

2

2

( ) ( )uefyadu

daa

du

da =+−+ 0212

2

2

yyO11

Eq. O11 is a linear second-order differential

equation with constant coefficients. And we

know how to solve it.

1



Physics Department

Yarmouk University

Chapter 3 Partial Differential Equations in

Physics

©Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

2

Overview

1. Partial Differential Equations

2. First-Order Differential Equations (Review)

3. Separation of Variables

4. Singular Points

5. Series Solutions – Forbenius' Method

6. A Second Solution

7. Nonhomogeneous Equations – Green's

Function

8. Heat Flow, or Diffusion, PDE

Separation of Variables


4

First technique for the solution of partial differential

equations splits the partial differential equation of n

variables into n ordinary differential equations.

Introduction

Each separation introduces an arbitrary constant of

separation. If we have n variables, we have to introduce

n−1 constants, determined by the conditions imposed in

the problem being solved.

Cartesian Coordinates


6

Helmholtz EquationUsing the definition of the Laplacian in Cartesian coordinates :

11

022

2

2

2

2

2

====ψψψψ++++∂∂∂∂

ψψψψ∂∂∂∂++++

∂∂∂∂

ψψψψ∂∂∂∂++++

∂∂∂∂

ψψψψ∂∂∂∂k

zyx13becomes

022 ====ψψψψ++++ψψψψ∇∇∇∇ kthe Helmholtz equation

2

2

2

2

2

22

zyx ∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂++++

∂∂∂∂

∂∂∂∂====∇∇∇∇⋅⋅⋅⋅∇∇∇∇====∇∇∇∇

12

(((( )))) (((( )))) (((( )))) (((( ))))zZyYxXz,y,x ====ψψψψ

If we consider that ψψψψ(x,y,z) is of the form

14

2

7

Separation of Variables – Step 1

15

17

then we have:

16

022

2

2

2

2

2

====++++ψψψψ

++++ψψψψ

++++ψψψψ

ZYXkdz

dYX

dy

dZX

dx

dZY

Dividing Eq. 15 by XYZ we obtain:

0111 2

2

2

2

2

2

2

====++++++++++++ kdz

Zd

Zdy

Yd

Ydx

Xd

X

or2

2

2

22

2

2 111

dz

Zd

Zdy

Yd

Yk

dx

Xd

X−−−−−−−−−−−−====

Eq. 17 exhibits one separation of variables. The lhs is a

function of x alone, whereas the right-hand side depends only

on y and z and not on x. But x, y, and z are all independent coordinates.


8

The equality of both sides depending on different

variables means that the behavior of x as an

independent variable is not determined by y and z.

(1st) Separation of Variables

19

1822

21l

dx

Xd

X−−−−====

22

2

2

22 11

ldz

Zd

Zdy

Yd

Yk −−−−====−−−−−−−−−−−−

Therefore, each side must be equal to a constant, a constant of separation. The choice in general is:


9

Now we rewrite Eq. 19 as:

We have:

(2nd & 3rd) Separation of Variables

2122

21m

dy

Yd

Y−−−−====

22222

21nmlk

dz

Zd

Z−−−−====++++++++−−−−====

A second separation is achieved. The same procedure is now applied to Eq. 20. We choose the constant of

separation here = -m2

202

222

2

2 11

dz

Zd

Zlk

dy

Yd

Y−−−−++++−−−−====

and

This is a third separation (n2 + l2 +m2 = k2).

22


10

Three ODE's

- n2 = -k2+l2+m2 is introduced in order to produce a symmetric set of equations.

Solving the PDE (Eq. 12) is now equivalent to solving a set of three ODE's, namely equations 18, 21 and 22.

The solution depends on the three constants n, l and m and we normally write:

(((( )))) (((( )))) (((( )))) (((( ))))nnmlmln zZyYxXz,y,x ====ψψψψ 23


11

The most general solution of Eq. 13 is a linear combination of

solutions ψψψψm

Or since n2 = l2 + m2 – k2

The coefficients alm are chosen to permit ψψψψ to satisfy the boundary conditions of the problem.

As a rule, these boundary conditions lead to a discrete values

for l and m.

The Solution

(((( )))) (((( )))) (((( )))) (((( ))))zZyYxXz,y,x nmlml ====ψψψψ 24

∑∑∑∑ ψψψψ====ψψψψm,l

mlmla 25

Circular Cylindrical Coordinates

3


13

First Separation

26

27

28

Using the definition of the Laplacian in circular cylindrical coordinates, Eq. 12 is written as:

In a first separation Eq. 26 can be rewritten as:

011 2

2

2

2

2

2

2

====ψψψψ++++ψψψψ

∂∂∂∂

∂∂∂∂++++

ϕϕϕϕ∂∂∂∂

∂∂∂∂

ΦΦΦΦρρρρ++++

ρρρρ∂∂∂∂

∂∂∂∂ρρρρ

ρρρρ∂∂∂∂

∂∂∂∂

ρρρρ

∇∇∇∇

kz

(((( )))) (((( )))) (((( )))) (((( ))))zZz,, ϕϕϕϕΦΦΦΦρρρρΡΡΡΡ====ϕϕϕϕρρρρψψψψ

Here the solution is of the form:

2

22

2

2

2111

dz

Zd

Zk

d

d

d

d

d

d−−−−====++++

ϕϕϕϕ

ΦΦΦΦ


ρρρρ

ΡΡΡΡρρρρ

ρρρρΡΡΡΡρρρρ

14

22

21l

dz

Zd

Z−−−−====−−−−

A 2nd order ODE of a function of z on the rhs of Eq. 28 appears

to depend on a function of ρρρρ and ϕϕϕϕ on its lhs. We resolve this by setting each side of Eq. 28 equal to the same constant. We

choose this separation constant to be --l l 22.

Setting k2+l2=n2, multiplying by ρρρρ2, and rearranging terms, we obtain

2nd Separation

and

29Zldz

Zd 22

2

====

30222

2

2

11lk

d

d

d

d

d

d−−−−====++++

ϕϕϕϕ

ΦΦΦΦ


ρρρρ

ΡΡΡΡρρρρ

ρρρρΡΡΡΡρρρρ

312

222 1

ϕϕϕϕ

ΦΦΦΦ

ΦΦΦΦ−−−−====ρρρρ++++

ρρρρ

ΡΡΡΡρρρρ

ρρρρΡΡΡΡ

ρρρρ

d

dn

d

d

d

d

Then


15

3rd Separation – Back to Helmholtz

A 2nd order ODE of a function of ϕϕϕϕ on the rhs of Eq. 31

appears to depend on a function of ρ ρ ρ ρ on its lhs. We resolve this by setting each side equal to the same

constant. We choose this separation constant to be -m2.

33

For the ρρρρ dependence we have:

32

and we have:

ΦΦΦΦ−−−−====ϕϕϕϕ

ΦΦΦΦ 22

2

md

d

(((( )))) 0222 ====ΡΡΡΡ−−−−ρρρρ++++

ρρρρ

ΡΡΡΡρρρρ

ρρρρρρρρ mn

d

d

d

d

16

The separation of variables of several known PDE's gives rise to Bessel's DE. We have seen this in the case of Helmholtz equation. It can also be seen in the case of separation of variables of Laplace's equation in parabolic coordinates.

Eq. 33 is Bessel's differential equation.

The solution of Bessel's DE introduces Bessel functions of the first and the second type. See Arfken Chapter 11)

Bessel Equation

30

31(((( )))) 0222 ====ΡΡΡΡ−−−−ρρρρ++++

ρρρρ

ΡΡΡΡρρρρ

ρρρρρρρρ mn

d

d

d

d


17

Solution

Similarly to the case of the Cartesian case, the general solution is of the form:

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))∑∑∑∑∑∑∑∑ ϕϕϕϕΦΦΦΦρρρρΡΡΡΡ====ϕϕϕϕρρρρψψψψ====ϕϕϕϕρρρρψψψψm,n

nmnmnm

n,m

nm .zZaz,,z,, 30

The solution of the original PD Helmholtz equation is now equivalent to solving three ODE's!

Spherical Polar Coordinates

4

19

Same Procedure

32

33

Using the definition of the Laplacian in spherical coordinates, Eq. 12 is written as:

(((( )))) (((( )))) (((( )))) (((( ))))ϕϕϕϕΦΦΦΦθθθθΘΘΘΘ====ϕϕϕϕθθθθψψψψ rR,,r

Here the solution is of the form:

0sin

1sin

sin

11 22

2

2222

2 ====ΦΦΦΦΘΘΘΘ++++ϕϕϕϕ

ΦΦΦΦ

θθθθΘΘΘΘ++++

θθθθ

ΘΘΘΘθθθθ

θθθθΦΦΦΦ

θθθθ++++

ΦΦΦΦΘΘΘΘ Rk

d

d

rR

d

d

d

dR

rdr

dRr

dr

d

r

0sin

1sinsin

sin

1 222 ====ψψψψ++++


ψψψψ∂∂∂∂

θθθθϕϕϕϕ∂∂∂∂

∂∂∂∂++++

θθθθ∂∂∂∂

ψψψψ∂∂∂∂θθθθ


∂∂∂∂++++

∂∂∂∂


∂∂∂∂

∂∂∂∂θθθθ

θθθθk

rr

rr31

Substituting Eq. 32 into Eq. 31, we get:

In general, and in quantum mechanics in particular, the

separation starts by the r dependence split off first. Here we

start with ϕϕϕϕ.©Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

20

A 2nd order ODE of a function of ϕϕϕϕ on the rhs of Eq. 35 appears

to depend on a function of r and θθθθ on its lhs. We resolve this by setting each side equal to the same constant. Again, we choose this separation constant to be -m2.

First Separation

34

Dividing by RΘΦΘΦΘΦΘΦ and rearranging we have:

22

2

2222

2 sin

1sin

sin

11k

d

d

rd

d

d

d

rdr

dRr

dr

d

rR−−−−====

ϕϕϕϕ

ΦΦΦΦ

θθθθΦΦΦΦ++++

θθθθ

ΘΘΘΘθθθθ

θθθθθθθθΘΘΘΘ++++

Multiplying by r2 sin2θθθθ we can isolate the term containing

the ODE in ϕϕϕϕ, i.e. (1st separation)

2

2

22

2222 1

sinsin

11sin

ϕϕϕϕ

ΦΦΦΦ

ΦΦΦΦ====

θθθθ

ΘΘΘΘθθθθ

θθθθΘΘΘΘθθθθ−−−−

−−−−−−−−θθθθ

d

d

d

d

d

d

rdr

dRr

dr

d

Rrkr 35

33


21

Again the variables are separated. Each side must be equal

to the same constant, which we call Q. We thus have:

2nd and 3rd Separations

And we are left with:

37θθθθ

++++

θθθθ

ΘΘΘΘθθθθ

θθθθΘΘΘΘθθθθ−−−−====++++

2

2222

sinsin

sin11 m

d

d

d

dkr

dr

dRr

dr

d

R

38,r

RQRk

dr

dRr

dr

d

r0

12

222

====−−−−++++

39.Qm

d

d

d

d0

sinsin

sin1 2

====ΘΘΘΘ++++ΘΘΘΘθθθθ

−−−−

θθθθ

ΘΘΘΘθθθθ

θθθθΘΘΘΘθθθθ

36221

md

d−−−−====

ϕϕϕϕ

ΦΦΦΦ

ΦΦΦΦ


22

Eq. 39

is identified as the associated Legendre equation, in which

the constant Q = l(l+1) where l is a non-negative number.

Associated Legendre Equation

AL.Qm

d

d

d

d0

sinsin

sin1 2


−−−−

θθθθ

ΘΘΘΘθθθθ

θθθθΘΘΘΘθθθθ

If k2 is positive then Eq. 38

becomes the spherical Bessel equation.

SB,r

RQRk

dr

dRr

dr

d

r0

12

222

====−−−−++++


23

Solution

Similarly to the case of the Cartesian case, the general solution is of the form:

(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))∑∑∑∑∑∑∑∑ ϕϕϕϕΦΦΦΦθθθθΘΘΘΘ====ϕϕϕϕθθθθψψψψ====ϕϕϕϕθθθθψψψψ

m,Q

mmQQmQ

m,Q

mQmQ rRa,,r,,r 40

The solution of the original PD Helmholtz equation is now equivalent to solving three ODE's!

Restrictions on k

5


25

k2 defines the type of equation

Obviously the restriction on k2 to be a constant is rather

severe. The previous procedure is also valid if k2 is not a

constant but a function of r for example (as in the case of the Schrödinger equation for the hydrogen atom and many other areas in physics).

We discussed the separation of variables in Helmholtz

equation without defining k.

It is even still valid if k2(r,θθθθ,ϕϕϕϕ) is of the form:

46(((( )))) (((( )))) (((( )))) 2222

2 11kh

sinrg

rrfk ′′′′++++ϕϕϕϕ

θθθθ++++θθθθ++++====

as we will see in the next paragraph.

26

The Most Famous Application

The time independent SWE for the hydrogen atom is given by:

41

The Schrödinger wave equation for the hydrogen atom

has the form of Helmholtz equation (Eq. 12) with k being

a function of r only (it does not depend on the angular coordinates).

02

22

====ψψψψ

−−−−

χχχχ−−−−++++ψψψψ∇∇∇∇

µµµµ−−−− E

r

where µµµµ and E are, respectively, the reduced mass and the total

energy of the (e, p) system and

45(((( )))) (((( )))) ÅeV414106110944

199

0

2

0

2

.e.ee

====××××××××====εεεεππππ

====εεεεππππ

====χχχχ −−−−


27

Other Important Examples

Hermite's equation

44

The Laguerre and associated Laguerre equations

(((( )))) 012

22 ====αααα++++−−−−++++ y

dx

dyx

dx

dyx 43

0222

2

====αααα++++−−−− ydx

dyx

dx

dy


28

First we write the Laplacian in spherical coordinates, i.e.

Verify that

Example 7 – Arfken Exercise 9.3.4

(((( )))) (((( )))) (((( )))) 0sin

11

sin1

sinsinsin

1

2222

22

====ψψψψ

ϕϕϕϕ

θθθθ++++θθθθ++++++++++++



θθθθϕϕϕϕ∂∂∂∂

∂∂∂∂++++


ψψψψ∂∂∂∂θθθθ


∂∂∂∂++++

∂∂∂∂


∂∂∂∂

∂∂∂∂θθθθ

θθθθ

hr

gr

rfk

rr

rr

(((( )))) (((( )))) (((( )))) (((( )))) (((( )))) 011

22222 ====ϕϕϕϕθθθθψψψψ

ϕϕϕϕ

θθθθ++++θθθθ++++++++++++ϕϕϕϕθθθθψψψψ∇∇∇∇ ,,rh

sinrg

rrfk,,r 47

Solution: We follow exactly the procedure.

48


29

Assuming a solution of the form

Example 7 – Solution

In a first step we separate the variable ϕϕϕϕ from r and θθθθ

(((( )))) (((( )))) (((( )))) (((( ))))ϕϕϕϕΦΦΦΦθθθθΘΘΘΘ====ϕϕϕϕθθθθψψψψ rR,,r

(((( )))) (((( )))) (((( )))) 0sin

11sin

1sin

sin

11

2222

2

2

2222

2

====ΦΦΦΦΘΘΘΘ

ϕϕϕϕ

θθθθ++++θθθθ++++++++++++

ϕϕϕϕ

ΦΦΦΦ

θθθθΘΘΘΘ++++

θθθθ

ΘΘΘΘθθθθ

θθθθθθθθΦΦΦΦ++++

ΦΦΦΦΘΘΘΘ

Rhr

gr

rfk

d

d

rR

d

d

d

d

rR

dr

dRr

dr

d

r

49


30

Multiplying by r2 sin2θθθθ we can isolate the term containing

the ODE in ϕϕϕϕ, i.e. (1st separation)

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

(((( ))))ϕϕϕϕ++++ϕϕϕϕ

ΦΦΦΦ

ΦΦΦΦ====

++++

θθθθ++++

θθθθ

ΘΘΘΘθθθθ


++++

θθθθ−−−−

hd

d

kgrd

d

d

d

rrf

dr

dRr

dr

d

rRr

2

2

222

22

22

1

1sin

sin

11sin

First Separation

50

Dividing by RRΘΦΘΦΘΦΘΦΘΦΘΦΘΦΘΦ and rearranging we have:

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

(((( )))) 22

2

22

22

2

1

sin

1

sinsin111

khd

d

r

gd

d

d

d

rrf

dr

dRr

dr

d

rR

−−−−====

ϕϕϕϕ++++

ϕϕϕϕ

ΦΦΦΦ

ΦΦΦΦθθθθ++++

θθθθ++++

θθθθ

ΘΘΘΘθθθθ


++++

51

6


31

(((( )))) (((( )))) 2222

22

22 1sin

sin

11sin mkg

rd

d

d

d

rrf

dr

dRr

dr

d

rRr −−−−====

++++

θθθθ++++

θθθθ

ΘΘΘΘθθθθ


++++

θθθθ−−−− (((( )))) (((( ))))

θθθθ++++

θθθθ++++

θθθθ

ΘΘΘΘθθθθ

θθθθΘΘΘΘθθθθ−−−−====

++++++++

2

2222

sinsin

sin11 m

gd

d

d

dkrrf

dr

dRr

dr

d

R53

Again the variables are separated. Each side must be equal to the same constant, which we call Q. We thus have:

2nd and 3rd Separations

And we are left with:

(((( )))) 221

mhd

d−−−−====ϕϕϕϕ++++

ϕϕϕϕ

ΦΦΦΦ

ΦΦΦΦ

54

52

(((( )))) 0sin

sinsin

12

2


−−−−ΘΘΘΘθθθθ++++

θθθθ

ΘΘΘΘθθθθ

θθθθθθθθQ

mg

d

d

d

d

(((( ))))0

12

22

22 ====−−−−++++++++

r

RQRk

r

Rrf

dr

dRr

dr

d

r55


32

Overview




4. Singular Points




Function


Singular PointsSingular Points


34

Ordinary Point

A general homogeneous 2nd order ODE is written as:

Here we introduce the concept of a singular point, or singularity (as applied to a differential equation).

(((( )))) (((( )))) 0====++++′′′′++++′′′′′′′′ yxQyxPy

This is useful in1- classifying ODEs and 2- investigating the feasibility of a series solution.

Ordinary Points

If at x = x0, P(x) and Q(x) remain finite then x0 is said be an ordinary point.

55

35

Singular Point

If at x = x0, P(x) and/or Q(x) diverge then x0 is said be a singular point.

There are two types of singular points:

1) If either P(x) or Q(x) diverges as x→→→→x0 but (x−x0)P(x)and (x − x0)2 Q(x) remain finite as x→→→→x0, then x = x0 is called a regular, or nonessential, singular point.

If P(x) diverges faster than 1/(x−x0) so that (x−x0)P(x)goes to infinity as x→→→→x0, or Q(x) diverges faster than 1/(x− x0)2 so that

(x − x0)2Q(x) goes to infinity as x→→→→x0, then point

x = x0 is labeled an irregular, or essential, singularity.


36

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

(((( )))) (((( ))))2

124

13

22

122

1

2

2

2

2

dz

zydz

dz

zdyz

zdz

zydz

dz

zdyz

dx

dz

dx

xdy

dz

d

dx

xyd

−−−−−−−−

−−−−−−−−

++++====

−−−−

−−−−−−−−====

====

The case x →→→→ ∞∞∞∞The analysis of point x→∞→∞→∞→∞ is similar to that in the case of functions of a complex variable.

The technique is to set x = 1/z and substitute into the ODE and then let z→→→→0 (here point z is finite). We use the chain rule and write:

(((( )))) (((( )))) (((( )))) (((( ))))dz

zdyz

dz

zdy

xdx

dz

dz

zdy

dx

xdy 12

1

2

1 1 −−−−−−−−−−−−

−−−−====−−−−========56

7


37

(((( )))) (((( )))) 0112 14

122

2

====++++

−−−−++++ −−−−−−−− yzQ

zdz

dyzP

zzdz

yd

The case x →→→→ ∞∞∞∞Using eqs. 56, Eq. 53 is transformed into:

as z → → → → 0 and we apply the previous rules, i.e.

58(((( )))) (((( ))))

4

1

2

1

and2

z

zQQ

z

zPzP

−−−−−−−−

====′′′′−−−−

====′′′′

At x = ∞∞∞∞ (z = 0), the behavior of Eq. 57 depends on the behavior of the new coefficients:

57

If P' and Q' remain finite then x=∞∞∞∞ is a regular point.

If P' diverges more rapidly than 1/z and Q' diverges more

rapidly than 1/z2 then xx==∞∞∞∞∞∞∞∞ is a regular singular pointis a regular singular point. In the

other cases x=∞∞∞∞ is an essential singularity.


38

Bessel's equation is

which we rewrite as:

Both P(x) and Q(x) diverge at point x=0. This point is a regular singularity

(((( )))) 0222 ====−−−−++++′′′′++++′′′′′′′′ ynxyxyx 59

Example 8 – Bessel Arfken Example 9.4.1

(((( ))))

(((( ))))

011

2

2

====

−−−−++++′′′′++++′′′′′′′′ y

x

ny

xy

xQxP

60


39

61

Let z = 1/x, then we have:

Example 8 – Investigate x = ∞∞∞∞

from which we can see that Q' diverges as 1/z4 then x=∞∞∞∞ is

an irregular or essential, singularity.

(((( )))) 011

2 2242

2

====−−−−++++++++

′′′′′′′′

yznzdz

dy

dz

yd

QP

40

Table 1 shows the singularities of some ODE we shall study later in this course.

Other Examples


41

Overview




4. Singular Points




Function


Series Solutions Series Solutions ––ForbeniusForbenius' Method' Method

8


43

Ferdiand Georg FrobeniusFrom Wikipedia, the free encyclopedia

Ferdinand Georg Frobenius (October 26, 1849 – August 3, 1917) was a German mathematician, best known for his contributions to the theory of elliptic functions, differential equations and to group theory.

http://http://en.wikipedia.org/wiki/Ferdinand_Georg_Frobeniusen.wikipedia.org/wiki/Ferdinand_Georg_Frobenius©Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations

44

IntroductionNot every differential equation can be solved — a solution may not exist. There may be no function that satisfies the differential equation.

If a solution does exist it may not be possible to express it in closed form in terms of the elementary functions.

There are differential equations of great importance in physics that cannot be solved in terms of elementary functions.

In such cases one must turn to approximate methods such as power series. The solutions to many differential equations are expressible in terms of a power series.


45

Thus one method of solving a linear 2nd-order homogeneous ODE is a series expansion.

There is a second independent solution but no third solution exists!

This second solution is developed by two methods: an integral method and a power series containing a logarithmic term

A basic question rises: when the method of series substitution can be expected to work?

The answer is given by The answer is given by Fuchs’ theoremFuchs’ theorem..

Series Solutions


46

Fuchs’ theorem asserts that "we can always obtain at least one power-series solution, provided we are expanding about a point that is an ordinary point or at worst a regular singular point.

In physics this very gentle condition is almost always In physics this very gentle condition is almost always satisfied.satisfied.

In other words, the power expansion solution will always work provided the point of expansion is no worse than a regular singular point.

Fuch's Theorem


47

The 2nd-order ODE


(((( )))) (((( )))) 02

2

====++++++++ yx

y

x

yxQ

d

dxP

d

d

is homogeneous because each term contains y(x) or a

derivative;

(((( )))) (((( )))) 0====++++′′′′++++′′′′′′′′ yyy xQxPor

It is linear because each y, dy/dx or d2y/dx2 appears at the

first power. No products (i.e. y dy/dx or y d2y/dx2 ) appear.

62


48

The solution we shall try is of the form of a power series,

i.e.

Undetermined Coefficients

(((( )))) 0, 00

≠≠≠≠==== ∑∑∑∑∞∞∞∞

====λλλλ

λλλλ++++λλλλ axaxy

k

The condition a0 = 0 means that the power of the lowest nonvanishing term of the series is available as a

parameter. k need not be an integer.

The coefficients aλλλλ are undetermined and we shall see through an example how we find them.

63

9


49

The equation of motion of a classical oscillator is the

solution of the 2nd-order linear ODE:

where ωωωω is the angular frequency (ωωωω2 = k/m)

We know the solutions are sine functions, i.e.

022

2

====ωωωω++++ ydx

yd64

Example 9 – Linear Classical Oscillator

(((( )))) (((( )))) xAxyorxAxy ωωωω====ωωωω==== cos,sin 65

We try: (((( ))))

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

≠≠≠≠==== ∑∑∑∑∞∞∞∞

====λλλλ

λλλλ++++λλλλ

33

2210

00

0,

xaxaxaax

axaxy

k

k

66


50

(((( ))))(((( )))) 010

2

0

2 ====ωωωω++++−−−−λλλλ++++λλλλ++++ ∑∑∑∑∑∑∑∑∞∞∞∞

====λλλλ


∞∞∞∞

====λλλλ

−−−−λλλλ++++λλλλ

kk xaxkka

Substituting in Eq. 62 we obtain:

We differentiate Eq. 66 twice and plug the result in Eq. 62

(LDE1)), as follows:

67


68

(((( )))) ,0

1∑∑∑∑∞∞∞∞

====λλλλ

−−−−λλλλ++++λλλλ λλλλ++++==== kxka

dx

dy

69

(((( ))))(((( )))) ,10

22

2

∑∑∑∑∞∞∞∞

====λλλλ

−−−−λλλλ++++λλλλ −−−−λλλλ++++λλλλ++++==== kxkka

dx

yd


51

(((( ))))(((( )))) 010

2

0

2 ====ωωωω++++−−−−λλλλ++++λλλλ++++ ∑∑∑∑∑∑∑∑∞∞∞∞

====λλλλ


∞∞∞∞

====λλλλ

−−−−λλλλ++++λλλλ

kk xaxkka

We write here the first 3 terms of the series in Eq. 69


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

(((( ))))

++++++++++++++++ωωωω

++++++++++++++++++++

++++++++++++++++++++−−−−

++++++++++++

++++

−−−−−−−−

33

22

110

2

13

21

12

0

23

1211

kkkk

k

kkk

xaxaxaxa

xkka

xkkaxkkaxkka

70


52


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

(((( ))))(((( ))))(((( )))) 012

23

121

1

22

11

23

02

2

11

20

====ωωωω++++++++++++++++++++++++

ωωωω++++++++++++++++

ωωωω++++++++++++++++

++++++++

−−−−

++++++++

++++

−−−−

−−−−

jkjj

k

k

k

k

xajkjka

xakka

xakka

xkka

xkka

71

53

The Indicial Equation

The lowest power being xk-2 (for λλλλ= 0), the coefficient a0(k(k-1)) must vanish and a0 ≠ ≠ ≠ ≠ 0 implies that k(k-1) = 0.

The uniqueness of power series implies that the

coefficients of each power of x on the lhs must vanish individually.

This is satisfied in the case where k is either equal to 0 or

1.

The condition k(k-1) = 0 is called the indicial equation. This equation and its roots is important for our analysis!

72(((( )))) 01 ====−−−−kk


54

Recurrence Relation

Before considering these two possibilities for k, we return to Eq. 63 and Eq. 71 and demand that the remaining net

coefficients, say, the coefficient of xk+j (j ≥ 0), vanish. We

set λλλλ = j +2 in the first summation and λλλλ = j in the second.

(They are independent summations and λλλλ is a dummy index.) This results in

73

(((( ))))(((( )))) 012 22 ====ωωωω++++++++++++++++++++++++ jj ajkjka

or(((( ))))(((( ))))12

2

2++++++++++++++++

ωωωω−−−−====++++

jkjkaa jj

This is a two-term recurrence relation.

10


55

For this example, if we start with a0, Eq. 73 leads to the

even coefficients a2, a4, and so on, and ignores a1, a3,

a5, and so on.

Given aj, we may compute aj+2 and then aj+4, aj+6, and so on up as far as desired.

Even Coefficients

The previous recurrence relation is valid for any

integer j ≥ ≥ ≥ ≥0.


56

if k = 1, a1(k+1)k=0 means that a1 is necessarily zero.

In the case k = 1, it is exactly the inverse where the even coefficients vanish (except a0 of course which we assumed # 0from the beginning!)

if k = 0 then the coefficient a1(k+1)k=0 means that the coefficient a1 is arbitrary.

Odd Coefficients

0753 ================ aaa

(((( )))) (((( )))) (((( ))))(((( ))))(((( ))))++++++++++++ωωωω

++++++++++++++++++++−−−−++++++++

−−−−−−−−

22

110

22

11

20 1211

kkk

kkk

xaxaxa

xkkaxkkaxkka

Let us set it equal to zero. In which case all the odd

coefficients will vanish, i.e.

57

The Indicial Equation – Case k = 0

which leads to

The recurrence relation becomes

In a compact form we have:

74(((( ))))(((( ))))12

2

2++++++++

ωωωω−−−−====++++

jjaa jj

75

etc!

aaa

!aaa

!aa

,665

,434

,2

6

0

2

46

4

0

2

24

2

02

ωωωω−−−−====

⋅⋅⋅⋅

ωωωω−−−−====

ωωωω++++====

⋅⋅⋅⋅

ωωωω−−−−====

ωωωω−−−−====

76(((( ))))(((( ))))

,an

an

nn 0

2

2 !21

ωωωω−−−−====


58

Solution – Case k = 0The solution is thus:

77(((( ))))

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

(((( ))))xa

xx

xx

xxaxxy k

ωωωω====

++++

ωωωω++++++++

ωωωω++++++++

ωωωω−−−−++++========

cos

!60

!40

!201

0

65

43

2

00

0


59

The Indicial Equation – Case k = 1The solution in this case is of the form:

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

(((( )))) (((( )))) (((( )))) (((( ))))xsina

!

x

!

x

!

xx

a

!

x

!

x

!

x

!axxy k

ωωωωωωωω

====

++++


ωωωω++++

ωωωω−−−−ωωωω

ωωωω====

++++


ωωωω++++

ωωωω−−−−========

0543

0

642

01

1

753

75311

78

We leave the proof of Eq. 78 as an exercise

60

To summarize this approach, we may write Eq. 63 schematically asshown in Fig. 2.

Summary

From the uniqueness of power series (Section 5.7), the total

coefficient of each power of x must vanish—all by itself.

The requirement that the first coefficient (1) vanish leads to the indicial equation, Eq. 69. The second coefficient is handled by

setting a1 = 0. The vanishing of the coefficient of xk (and higher powers, taken one at a time) leads to the recurrence relation, Eq. 73.

11

61

This series substitution, known as Frobenius’ method, has given us two series solutions of the linear oscillator equation. However, there are two points about such series solutions that must be stronglyemphasized:

Frobenius Method

1. The series solution should always be substituted back into the The series solution should always be substituted back into the differential equation, to see if it works, as a precaution againdifferential equation, to see if it works, as a precaution against st algebraic and logical errors. If it works, it is a solution.algebraic and logical errors. If it works, it is a solution.

2. The acceptability of a series solution depends on its converg2. The acceptability of a series solution depends on its convergence ence (including asymptotic convergence). It is quite possible for (including asymptotic convergence). It is quite possible for

FrobeniusFrobenius’’ method to give a series solution that satisfies the original method to give a series solution that satisfies the original differential equation when substituted in the equation but that differential equation when substituted in the equation but that does does not converge over the region of interest. Legendrenot converge over the region of interest. Legendre’’s differential s differential equation illustrates this situationequation illustrates this situation..

Symmetry of SolutionsSymmetry of Solutions


66

The 2nd-order ODE (Eq. 62)


(((( )))) (((( )))) 02

2

====++++++++ yx

y

x

yxQ

d

dxP

d

d

can be written as:

83(((( )))) (((( )))) ,xxL 0====y

where (((( )))) (((( )))) (((( )))).xQd

dxP

d

dxL ++++++++====

xx2

2

is a differential operator

84


67

Going back to the simple harmonic equation, we have

Parity

(((( )))) 22

2

ωωωω++++====xd

dxL

We can see that:

(((( )))) (((( ))))xLxL −−−−====

Whenever the differential operator has a specific parity or

symmetry, either even or odd, we may interchange +x and

−x, and Eq. 83 becomes

85

86

(((( )))) (((( )))) 0====−−−−±±±± xyxL 87

+ +if if LL((xx)) is even, is even, −− if if LL((xx)) is odd. is odd.


68

Parity

Clearly, if Clearly, if yy((xx)) is a solution of the differential equation, is a solution of the differential equation,

yy((−−xx)) is also a solution. is also a solution.

(((( )))) (((( )))) (((( ))))[[[[ ]]]] (((( )))) (((( ))))[[[[ ]]]] ,2

1

2

1xyxyxyxyxy −−−−−−−−++++−−−−++++==== 88

gives an even solutiongives an even solution gives an odd solutiongives an odd solution

Then any solution may be resolved into even and odd Then any solution may be resolved into even and odd parts,parts,

69

equations (or differential operators) all exhibit this even parity; that is, their P(x) in Eq. 63 is odd and Q(x) is even. Solutions of all of them may be presented as series of even powers of x and separate

series of odd powers of x.

Parity vs. Solutions

Legendre, 80(((( )))) (((( )))) 0121 2

22 ====++++++++−−−−−−−− yll

dx

dyx

dx

ydx

(((( )))) ,yndx

dyx

dx

ydx 01 2

2

22 ====++++−−−−−−−− 81

(((( )))) ,yadx

dyxc

dx

ydx 02

2

====−−−−−−−−++++

,ydx

dyx

dx

yd0222

2

====αααα++++−−−−

89

Chebyshev,

Bessel,

simple harmonic oscillator,

and Hermite 90

12


70

Parity

The Laguerre differential operator has neither even nor odd symmetry; hence its solutions cannot be expected to exhibit even or odd parity.

91(((( )))) 012

2

====++++−−−−++++ yadx

dyx

dx

ydx

Parity is important in quantum mechanics. This is mainly due to the importance of parity conservation or non-conservation by the fundamental interactions.

Usually, we find that wave functions are either even or odd, meaning that they have a definite parity.

Most interactions (beta decay is the big exception) are also even or odd, and the result is that parity is conserved.


71

Overview




4. Singular Points




Function


72

A Second Solution

SelfSelf--Reading Reading Arfken 6Arfken 6thth Edition Section 9.6 (pp 578Edition Section 9.6 (pp 578--592)592)Arfken 7Arfken 7thth Edition Section 7.6 (pp 358Edition Section 7.6 (pp 358--375)375)


73

SummarySummary

TwoTwo solutions form the complete solution of a linear, solutions form the complete solution of a linear, homogeneous, secondhomogeneous, second--order ODEorder ODE——assuming that the assuming that the

point of expansion is no worse than a regular point of expansion is no worse than a regular singularity (singularity (Fuch'sFuch's theorem). theorem).

At least one solution can always be obtained by series At least one solution can always be obtained by series

substitution.substitution.

A second, linearly independent solution can be A second, linearly independent solution can be

constructed by the constructed by the WronskianWronskian double integral. double integral.

No third, linearly independent solution exists. No third, linearly independent solution exists.

1



Physics Department

Yarmouk University

Chapter 3 Partial Differential Equations in

Physics – Part 3


2

Overview




4. Singular Points




Function


Nonhomogeneous Equations


4

2nd-Order Linear nonhomgeneous ODE

The 2nd-order ODE

(((( )))) (((( )))) (((( ))))xFxQd

dxP

d

d====++++++++ y

x

y

x

y2

2

is nonhomogeneous because its rhs does not

contain y(x) or a derivative;

It is linear because each y, dy/dx or d2y/dx2 appears

at the first power. No products (i.e. y dy/dx or y

d2y/dx2 ) appear.

92

F(x) is called the source term.


5

Non Homogeneous 2nd-order ODE

The general solution in this case is the sum of the general solution of the corresponding homogeneous equation and a particular solution, as we said earlier.

The particular solution can be obtained by:

1- The method of variation of parameters, or

2- Techniques such as Green's function.

6

Here we discuss the method using Green’s

function in order to find the particular solution, yp,

linearly dependent on the source term F(x).

Green's Function*

For a brief introduction to Green’s function method, as applied to the solution of a nonhomogeneous PDE, it is helpful to use the electrostatic analog.

See also Arfken 7th Edition: Chapter 10And http://mathworld.wolfram.com/GreensFunction.html

2

The Electrostatic Analog

©©©© Dr. Nidal M. Ershaidat Dr. Nidal M. Ershaidat Dr. Nidal M. Ershaidat Dr. Nidal M. Ershaidat ---- Mathematical Physics Mathematical Physics Mathematical Physics Mathematical Physics ---- Phys. 601 Phys. 601 Phys. 601 Phys. 601 ---- Chapter 3 Partial Differential EquationsChapter 3 Partial Differential EquationsChapter 3 Partial Differential EquationsChapter 3 Partial Differential Equations

8

In the absence of charges, ρρρρ = 0, this equation becomes Laplace's equation:

The electrostatic potential of a charge distribution ρ ρ ρ ρ in free space at some point in space is defined by Poisson's equation:

The Electrostatic Potential

0

2

εεεε

ρρρρ−−−−====ψψψψ∇∇∇∇ 93

02 ====ψψψψ∇∇∇∇ 94


9

For a single point charge and using Coulomb's force, the

electrostatic field at point P (defined by the position

vector ) is the force per unit positive charge, i.e.

The Electrostatic Field

(((( )))) rr

qrEr

r

qqF q

ˆ4

1ˆ

4

12

02

0

0 εεεεππππ====⇒⇒⇒⇒

εεεεππππ====

where is the unit vector in the direction of .r

rr

====ˆ r

If q>0, is parallel to and they point in the same direction.

E

r

If q>0, is anti=parallel to .E

r

95

r

10

For a discrete distribution of point charges The

electrostatic field at point P, which we consider, now, placed at the origin, is obtained using the superposition principle and is defined by:

Electric Potential

(((( )))) i

n

i i

iq r

r

qrE ˆ

4

1

1

20∑∑∑∑

====εεεεππππ

====

96

We know that the corresponding electrostatic potential is given by:

(((( )))) ∑∑∑∑====

εεεεππππ========ψψψψ

n

i i

i

r

qr

104

10

For a continuous distribution of charge of density ρρρρ(r) we have:

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

∫∫∫∫ ττττρρρρ

εεεεππππ========ψψψψ d

r

rr

04

10

97

98


11

For the same distribution ρρρρ, defined by the

position and point P, placed at distance from the origin we have:

Point P not at the origin

99

satisfies Poisson's equation.

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

∫∫∫∫ ττττ−−−−

ρρρρ

εεεεππππ====ψψψψ 2

21

2

01

4

1d

rr

rr

1rr

====2rr

====

(((( ))))1r

ψψψψ


12

Defining Position Vectors

andFigure 1-2: Definition of

O

21 rr

−−−−====r21

21ˆrr

rr

−−−−

−−−−====r

3

Green's FunctionGreen's Function

14

Green's FunctionGreen's Function (symbol G) is required to satisfy Poisson's equation with a point source placed at

the point defined by the vector r2, i.e.

Where δδδδ is the Dirac Delta function (distribution) defined in 1-D by:

(((( ))))212 rrG

−−−−δδδδ−−−−====∇∇∇∇ 100

(((( ))))

≠≠≠≠====∞∞∞∞++++

====δδδδ00

0

x

xx 101

(((( ))))∫∫∫∫∞∞∞∞−−−−

∞∞∞∞−−−−

====δδδδ 1dxx 102

G is physically the potential at point r1

corresponding to a unit source at r2.

15

Now we make use of Eq. 93 and Eq. 100 and we have:

The problem can be simplified;

104

Green's Theorem states that*

Green's Theorem

(((( )))) (((( ))))∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫ ⋅⋅⋅⋅∇∇∇∇−−−−∇∇∇∇====∇∇∇∇−−−−∇∇∇∇ σσσσψψψψψψψψττττψψψψψψψψ

dGGdGG 222

103

,22

22 ∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫ ττττψψψψ∇∇∇∇====ττττ∇∇∇∇ψψψψ dGdG

105(((( )))) (((( ))))(((( )))) (((( ))))

.,

20

2212212 ∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫ ττττ

εεεε

ρρρρ−−−−====ττττ−−−−δδδδψψψψ−−−− d

rrrGdrrr

* See Doc 7

1) Assuming that the integrand falls off faster than r−2 and taking

the volume so large that the surface integral on the rhs vanishes, leaving

16

We know that:

Comparing Eq. 100 with Eq. 107 we get the value of G, i.e.

Plugging this result into Eq. 106 we obtain Eq. 99, i.e.

The lhs of Eq. 102 is simply which gives:

Dirac Delta Function

106

107(((( )))) (((( ))))2121

22

4

1,

4

1rr

rrorr

r

−−−−δδδδ−−−−====

−−−−ππππ∇∇∇∇δδδδ−−−−====

ππππ∇∇∇∇

(((( )))) (((( )))) (((( )))) .,1

22210

1 ∫∫∫∫∫∫∫∫∫∫∫∫ ττττρρρρεεεε

====ψψψψ drrrGr

(((( ))))1r

ψψψψ

108(((( )))) .4

1,

211

rrrrG

−−−−ππππ====

109(((( ))))(((( ))))

.4

12

21

2

01 ∫∫∫∫∫∫∫∫∫∫∫∫ ττττ

−−−−

ρρρρ

εεεεππππ====ψψψψ d

rr

rr

17

where is a linear differential operator.

The Green's function is taken to be a solution of:

The Green's function depends on boundary conditions and the particular solution is thus:

We generalize the previous result for any nonhomogeneous 2nd-order linear ODE of the form

Generalization

110(((( )))) (((( ))))11 rfry

−−−−====L

112(((( )))) (((( )))) (((( )))) ., 22211 ∫∫∫∫∫∫∫∫∫∫∫∫ ττττ==== drfrrGry

L

111(((( )))) (((( ))))2121 , rrrrG

−−−−δδδδ−−−−====L


18

It appears as a weighting functionweighting function or propagator functionpropagator function

that enhances or reduces the effect of the charge element

ρρρρ(r2)dττττ2 according to its distance from the field point r1.

For the simple, but important, electrostatic case we obtain Green’s function, by Gauss’ law.

Looking carefully at Eq. 109. one can give a physical interpretation to Green's function.

Physical Meaning

Green’s function, represents the effect of a unit

point source at in producing a potential at . 2r

1r

(((( ))))21 ,rrG

4

19

In summary, Green’s function, , is a solution of Eq.

100 or Eq. 111 more generally.

Summary

(((( ))))212 rrG

−−−−δδδδ−−−−====∇∇∇∇ 100

(((( ))))21 ,rrG

111(((( )))) (((( ))))2121 , rrrrG

−−−−δδδδ−−−−====L

It enters in an integral solution of our differential equation, as in Eq. 109.

109(((( ))))(((( ))))

.4

12

21

2

01 ∫∫∫∫∫∫∫∫∫∫∫∫ ττττ

−−−−

ρρρρ

εεεεππππ−−−−====ψψψψ d

rr

rr

Symmetry of Green's Function

21

In order to prove it under more general conditions we require that:

Green’s function is symmetric, i.e.

Symmetry

(((( )))) (((( )))) .,, 1221 rrGrrG

==== 113

(((( )))) (((( ))))[[[[ ]]]] (((( )))) (((( )))) (((( ))))111 ,, rrrrGrqrrGrp

−−−−δδδδ−−−−====λλλλ++++∇∇∇∇⋅⋅⋅⋅∇∇∇∇

This property is obvious in the electrostatic case.

114

corresponding to a mathematical point source at

. 1rr

====

and are arbitrary well-behaved functions

of the distance .

(((( ))))rp (((( ))))rq

r


22

The Green’s function, G(r,r2), satisfies the same

equation, but the subscript 1 is replaced by

subscript 2.

Symmetry – Proof1

The Green’s functions, G(r,r1) and G(r,r2), have the

same values over a given surface S of some volume of finite or infinite extent, and their normal derivatives have the same values over the surface

S,

Then G(r,r2) is a sort of potential at r, created by a

unit point source at r2.

23

We multiply the equation for G(r,r1) by G(r,r2) and the equation for G(r,r2) by G(r,r1) and then subtract the two:

Symmetry – Proof2

The first term in Eq. 115,The first term in Eq. 115,

may be replaced bymay be replaced by

(((( )))) (((( )))) (((( ))))[[[[ ]]]] (((( )))) (((( )))) (((( ))))[[[[ ]]]](((( )))) (((( )))) (((( )))) (((( ))))2112

2112

rrr,rGrrr,rG

r,rGrpr,rGr,rGrpr,rG

−−−−δδδδ++++−−−−δδδδ−−−−====

∇∇∇∇⋅⋅⋅⋅∇∇∇∇−−−−∇∇∇∇⋅⋅⋅⋅∇∇∇∇

(((( )))) (((( )))) (((( ))))[[[[ ]]]] ,,, 12 rrGrprrG

∇∇∇∇⋅⋅⋅⋅∇∇∇∇

(((( )))) (((( )))) (((( ))))[[[[ ]]]] (((( )))) (((( )))) (((( )))) .,,,, 1212 rrGrprrGrrGrprrG

∇∇∇∇⋅⋅⋅⋅∇∇∇∇−−−−∇∇∇∇⋅⋅⋅⋅∇∇∇∇

115

24

Symmetry – Proof3 A similar transformation is carried out on the second term. Then

integrating over the volume whose surface is S and using Green’s theorem, we obtain a surface integral:

The terms on the The terms on the rhsrhs appear when we use the Dirac delta functions appear when we use the Dirac delta functions in Eq. 115 and when calculating the volume integration. in Eq. 115 and when calculating the volume integration.

More generally, in the complex plane we have the selfMore generally, in the complex plane we have the self--adjoint form: adjoint form:

(((( )))) (((( )))) (((( )))) (((( )))) (((( )))) (((( ))))(((( )))) (((( ))))1221

1112

,,

,,,,

rrGrrG

rrGrprrGrrGrprrG

++++−−−−====

∇∇∇∇−−−−∇∇∇∇∫∫∫∫

(((( )))) (((( )))) .,, 1221 rrGrrG

==== 117

(((( )))) (((( )))) .,, 12*

21 rrGrrG

==== 118

116

With the boundary conditions imposed on the Green’s function, thWith the boundary conditions imposed on the Green’s function, the e surface integral vanishes and surface integral vanishes and

5

25

Example of Use of Green's Function26

Example1

27

Example3

Fundamental Green's Functions

29

Table 10.1 (Arfken 7th Edition)

Forms of Green's Function

6

Spherical Polar Coordinate Expansion

Circular Cylindrical Coordinate Expansion


Next Lecture

Chapter 4Sturm-Liouville Eigenvalue

Problem

1



Physics Department

Yarmouk University

Chapter 3 Partial Differential Equations

in Physics

: : : :

Chapter 1’

Vector Calculushttp://ctaps.yu.edu.jo/physics/Courses/Phys201/Chapter1P

Appendix 3-1

Solid Angle

© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3 Partial Differential Equations - Appendices

4

Surface integrals become simpler when

using solid angles.

Solid Angle - Definition

The notion of solid angle is a “kind of”

generalization of plane (1D) angle in a 2D

space.

Solid angle can be interpreted as the angle

with which a surface is seen by a point.

This geometrical concept is very common

and important in physics.


5

Consider the element of area da. Let C be

the center of coordinates and the vector

representing the position of da.

Solid Angle – Geometrical Definition

The (element of) solid angle

is defined as:

2

cosd

r

da θ=Ω

Where θθθθ is the angle between

the direction of and that of

the normal to the element of

area .n

r

r

→r


6

In a vector notation, dΩΩΩΩ can be written as:

Solid Angle –Definition using vectors

2

0ˆˆ

r

dar nd

⋅⋅⋅⋅====ΩΩΩΩ

Where is the unit vector in

the direction of .r

0r→r

3r

danr

⋅⋅⋅⋅====

• Solid angle is dimensionless

• dΩΩΩΩ can be positive or negative

(following the sign of )nr

⋅⋅⋅⋅

• The unit of solid angles is

called the steradian (symbol sr)

2


7

The Meaning of Solid AngleConsider a sphere, center C and radius R

∫∫∫∫∫∫∫∫⋅⋅⋅⋅

====ΩΩΩΩ====ΩΩΩΩ2

0

r

danrd

The whole solid angle is given by adding all

the elements of solid angle, i.e.

ππππ====ππππ××××========ΩΩΩΩ⇒⇒⇒⇒ ∫∫∫∫ 4411 2

22R

RRda

10 ====⋅⋅⋅⋅n

r

All elements of area are “seen” by C as being at distance r = R. And the normal to element of area is parallel to the vector .

Thus we have:

r

∫∫∫∫⋅⋅⋅⋅

====3r

danr

© Dr. Nidal M. Ershaidat - Mathematical Physics - Phys. 601 - Chapter 3

8

The previous result is generalized to any closed surface and we always have:

ΩΩΩΩ = 4ππππ

ππππ====⋅⋅⋅⋅

====ΩΩΩΩ====ΩΩΩΩ ∫∫∫∫∫∫∫∫ 4ˆˆ

2

0

r

dar nd

The solid angle subtended by any simple closed surface at any interior point is 4ππππ sr.

If C lies outside the closed surface then

ΩΩΩΩ = 0.

9

222

01

0cosˆ

r

da

r

da

r

dar========

⋅⋅⋅⋅====ΩΩΩΩ

nd

As we can see from the figure, the element

of solid angle dΩΩΩΩ1 = dΩΩΩΩ2 in magnitude.

ΩΩΩΩ = 0 if C lies outside the surface

But while dΩΩΩΩ1 is positive, dΩΩΩΩ2 is negative.

is anti-parallel to n

01r

2222

022

cosˆˆ

r

da

r

da

r

dar−−−−====

ππππ====

⋅⋅⋅⋅====ΩΩΩΩ

nd

Integrating over all the dΩΩΩΩ’s gives zero.

Appendix 3-2

Green's Theorem

Potential Theorem


12

Green’s theorem is a useful corollary of

Gauss’s divergence theorem.

Consider two scalar functions φφφφ1 and φφφφ2

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.

Green’s Theorem1

13

We start from the identity:

Green’s Theorem2

(((( )))) 22

12121 φφφφ∇∇∇∇φφφφ++++φφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====φφφφ∇∇∇∇φφφφ⋅⋅⋅⋅∇∇∇∇

Now Let’s integrate this equation over the

volume V:

(((( )))) ∫∫∫∫∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ++++ττττφφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====ττττφφφφ∇∇∇∇φφφφ⋅⋅⋅⋅∇∇∇∇VVV

ddd 22

12121

The integral on the lhs of the previous equation can

be transformed to a surface integral using Gauss’s

divergence theorem and we can, thus, write:

(((( )))) ∫∫∫∫∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ++++ττττφφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====φφφφ∇∇∇∇φφφφVVS

ddd 22

12121

a

3

14

Green’s Theorem3

And this is the Green’s Theorem in the first form:

(((( )))) ∫∫∫∫∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ++++ττττφφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====φφφφ∇∇∇∇φφφφVVS

ddd 22

12121

a

By exchanging φφφφ1 and φφφφ2 we obtain the following equation G2

∫∫∫∫∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ++++ττττφφφφ∇∇∇∇⋅⋅⋅⋅φφφφ∇∇∇∇====

φφφφ∇∇∇∇φφφφ

→→→→→→→→→→→→

VVS

ddad 12

21212

G1

G2

Subtracting equation G2 from equation G1 we get:

(((( )))) (((( ))))∫∫∫∫∫∫∫∫ ττττφφφφ∇∇∇∇φφφφ−−−−φφφφ∇∇∇∇φφφφ====φφφφ∇∇∇∇φφφφ−−−−φφφφ∇∇∇∇φφφφVS

dd 12

222

11221 a

G3

G3 is known as the second formsecond form of of Green’s

Theorem.

Potential Theorem


16

Consider a force which satisfies the equation:

which means that is irrotational.

This suggests the existence of a scalar function

φφφφ such that:

In terms of gradient, the magnitude of the force

equals the magnitude of the gradient of the scalar

function φφφφ and its direction is the opposite

direction of this gradient. (See lecture on Gradient

and the negative sign)

Scalar Potential1

0====××××∇∇∇∇→→→→→→→→F

We know that , φφφφ being a scalar function,0====φφφφ∇∇∇∇××××∇∇∇∇→→→→→→→→

→→→→F

→→→→F

φφφφ∇∇∇∇−−−−====→→→→→→→→

F


17

And the scalar function φφφφ that satisfies the

equation:

We have seen that such a force is said to be

conservative. According to the previous

discussion a force is conservative if it

satisfies the irrotational relation:

Alternatively we can associate to any force

satisfying the irrotational relation a scalar

function (which depends on space coordinates)

is called a scalar potential.

Scalar Potential2

0====××××∇∇∇∇→→→→→→→→F

→→→→F

φφφφ∇∇∇∇−−−−====→→→→→→→→

F


18

Using Stokes’ theorem

Which means that the work done by a

conservative force over a closed path is zero.

We also equivalently say that the work done by

a conservative force does not depend on the

nature of the path but only on its initial and

final points.

gives in the case of a conservative force

Conservative Forces and Stokes’ Theorem

∫∫∫∫∫∫∫∫→→→→→→→→→→→→→→→→→→→→

⋅⋅⋅⋅××××∇∇∇∇====⋅⋅⋅⋅SC

adFrdF

0====⋅⋅⋅⋅∫∫∫∫→→→→→→→→

CrdF

)( 0====××××∇∇∇∇→→→→→→→→F


19

Tow kinds of potential appear in physics:

scalar potential and vector potential

In general the term potential includes the

effects of any external force (action) on a

given system.

We often express this influence in terms of

“potential energy”.

This energy is “stored” in the system and

when the external agent “leaves” the

system, the system uses the “potential

energy” to perform actions itself.

Scalar and Vector Potentials

4


20

Consider a vector which satisfies the equation:→→→→A

We can create a vector potential in a similar manner to the case of the scalar potential

Vector Potential

For a vector satisfying the relation:→→→→B

0====

××××∇∇∇∇⋅⋅⋅⋅∇∇∇∇

→→→→→→→→→→→→A

0====⋅⋅⋅⋅∇∇∇∇→→→→→→→→B

We can define a vector such that:→A

→→→→→→→→→→→→××××∇∇∇∇==== AB

Theoretically there are an infinite number of

potential vectors satisfying the equation VP.

VP

is called the vector potential.→→→→A

Poisson’s Equation and

Laplace’s Equation


22

From Gauss’s law we have

If ρρρρ=0 then the resulting equation is called

Laplace’s equation

Poisson’s Equation

We can write:

And from the potential theory we have:0εεεε

ρρρρ====⋅⋅⋅⋅∇∇∇∇

→→→→→→→→E

φφφφ∇∇∇∇−−−−====→→→→→→→→

E

P

This is Poisson’s equation.

0εεεε

ρρρρ====φφφφ∇∇∇∇⋅⋅⋅⋅∇∇∇∇

→→→→→→→→

0

2

εεεε

ρρρρ====φφφφ∇∇∇∇

L02 ====φφφφ∇∇∇∇


25

Problem 1. 61

( ) 22 2F

xzyxyy

z =+∂

∂=

∂

∂( ) 222 3

F, xxyzx

zz

y=−

∂

∂=

∂

∂

( ) yxyzyxzz

x 22F 3 =−

∂

∂=

∂

∂ ( ) yxzyxxx

z 22F

, 2 =+∂

∂=

∂

∂

( ) 222 323F

yzxxyzxxx

y−=−

∂

∂=

∂

∂

( ) 23 322F

yzxyzyxyy

x −=−∂

∂=

∂

∂

0F =×∇⇒→→


26

⇒⇒⇒⇒

b) Find a potential φφφφ(x,y,z) such that

Problem 1. 61 (b)

φ∇−=→→

F

Solution:

φ−=

∂

φ∂+

∂

φ∂+

∂

φ∂−=

→

→

rd

dkj

yi

xˆˆˆ

zF

32 yxyx +−=φ z 32 yxyx +− z 2zz −− yx2

23223 zyxzyx −+−=φ

∫→→

⋅−=φ rdF ( )dyxyzx∫ −− 223

( )dzzyx∫ +− 22

(((( ))))dxyzyx∫∫∫∫ −−−−−−−−==== 32

Appendix 3-3

Dirac Delta Function

5

28

The starting point is the definition:

where f(x) is any well-behaved function and the

integration includes the origin.

The Dirac delta function is defined by its assigned properties

Definition

(((( ))))rr

δδδδππππ−−−−====

∇∇∇∇ 4

12

(((( ))))

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

0,

0,0

fdxxxf

x

xx

====δδδδ

====∞∞∞∞≠≠≠≠

====δδδδ

∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

which can be generalized to

(((( )))) (((( )))) (((( )))) .00 xfdxxxxf ====−−−−δδδδ∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

3-2-1

3-2-1

3-2-3

3-2-4


29

Mathematically, the delta function is not a

function, because it is too singular.

can be regarded as an "operator" which pulls

the value of a function at zero.

Instead it is said to be a distribution. It is a

generalized idea of functions but can be used

only in integrals. In fact the integral

δδδδ Distribution

(((( ))))∫∫∫∫ δδδδ dxx 3-2-5

30

PropertiesAn important property, which is a special case of

Eq. 3-2-3 (taking f(x) = 1) is:

(((( )))) .1====δδδδ∫∫∫∫∞∞∞∞++++

∞∞∞∞−−−−

dxx

δδδδ(x) must be an infinitely high, infinitely thin spike at

x = 0, as in the description of the charge density for a point charge.

3-2-6

Delta function can be seen as the limit of a Gaussian

(((( ))))22

2

0 2

1lim

σσσσ−−−−

→→→→σσσσ σσσσππππ====δδδδ x

ex

or a Lorentzian (((( ))))220

1lim

εεεε++++

εεεε

ππππ====δδδδ

→→→→εεεε xx 3-2-8

3-2-7

partial differential equations -...

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