Lecture 21 Revision session Remember Phils Problems and your notes = everything Remember I’m available for questions

Lecture 21Lecture 21

Revision session Revision session

http://www.hep.shef.ac.uk/Phil/PHY226.htmRemember Phils Problems and your notes = everything

Remember I’m available for questions all through Christmas

Revision for the examRevision for the exam

http://www.shef.ac.uk/physics/exampapers/2007-08/phy226-07-08.pdf

Above is a sample exam paper for this course

There are 5 questions. You have to answer Q1 but then choose any 2 others

Previous years maths question papers are up on Phils Problems very soon

Q1: Basic questions to test elementary concepts. Looking at previous years you can expect complex number manipulation, integration, solving ODEs, applying boundary conditions, plotting functions, showing ‘x’ is solution of PDE. Easy stuff.

Q2-5: More detailed questions usually centred about specific topics: InhomoODE, damped SHM equation, Fourier series, Half range Fourier series, Fourier transforms, convolution, partial differential equation solving (including applying an initial condition to general solution for a specific case), Cartesian 3D systems, Spherical polar 3D systems, Spherical harmonics

The notes are the source of examinable material – NOT the lecture presentations

I wont be asking specific questions about Quantum mechanics outside of the notes

http://www.shef.ac.uk/physics/exampapers/2007-08/phy226-07-08.pdf

Revision for the examRevision for the examThe notes are the source of examinable material – NOT the lecture presentations

Things to do now

Read through the notes using the lecture presentations to help where required.

At the end of each section in the notes try Phils problem questions, then try the tutorial questions, then look at your problem and homework questions.

If you can do these questions (they’re fun) then you’re in excellent shape for getting over 80% in the exam.

Any problems – see me in my office or email me

Same applies over holidays. I’ll be in the department most days but email a question or tell me you want to meet up and I’ll make sure I’m in.

Look at the past exam papers for the style of questions and the depth to which you need to know stuff.

You’ll have the standard maths formulae and physical constants sheets (I’ll put a copy of it up on Phils Problems so you are sure what’s on it). You don’t need to know any equations e.g. Fourier series or transforms, wave equation, polars.

Concerned about what you need to know? Look through previous exam questions. 2008/2009 exam will be of very similar style.You don’t need to remember any proofs or solutions (e.g. Parseval, Fourier series, Complex Fourier series) apart from damped SHM which you should be able to do.

You don’t need to remember any equations or trial solutions, eg. Fourier and InhomoODE particular solutions. APART FROM TRIAL FOR COMPLEMENTARY 2ND ORDER EQUATION IS You don’t need to remember solutions to any PDE or for example the Fourier transform of a Gaussian and its key widths, etc. However you should understand how to solve any PDE from start to finish and how to generate a Fourier transform.

Things you need to be able to do:

Everything with complex numbers; solve ODEs and InhomoODEs, apply boundary conditions; integrate and differentiate general stuff; know even and odd functions; understand damped SHM, how to derive its solutions depending on damping coefficient and how to draw them; how to represent an infinitely repeating pattern as a Fourier series, how to represent a pulse as a sine or cosine half range Fourier series; how to calculate a Fourier transform; how to (de)convolve two functions; the steps needed to solve any PDE and apply boundary conditions and initial conditions (usually using Fourier series); how to integrate and manipulate equations in 3D cartesian coordinates; how to do the same in spherical polar coordinates; how to prove an expression is a solution of a spherical polar equation.

mtex

Let’s take a quick look through the courseLet’s take a quick look through the course

and then do the exam from last yearand then do the exam from last year

Binomial and Taylor expansionsBinomial and Taylor expansions

IntegralsIntegralsTry these integrals using the hints provided

2

12cos

2

1cos2 xx )cos(

2

1)cos(

2

1coscos BABABA

2

0

2cos dxnx

2

0

coscos dxmxnx

2

0

2

0 22sin

4

1

2

12cos

2

1 xnx

ndxnx

2

0

2

0

)cos(2

1)cos(

2

1dxxmndxxmn

0)(2

)sin(

)(2

)sin(2

0

2

0

mn

xmn

mn

xmnmn

2

0

cos dxnx 0sin1

2

0

nxn

More integralsMore integrals

Summary

nmallfordxmxnxdxmxnx 0sinsincoscos2

0

2

0

nandmallfordxmxnx 0cossin2

0

Previous page

Remember odd x even function

nallfordxnxdxnx

2

0

22

0

2 cossin Previous page

Even and odd functionsEven and odd functions

So even x even = even even x odd = odd odd x odd = even

x

x

dxodd 0

x

x

x

dxevendxeven0

2

An even function is f(x)=f(-x) and an odd function is f(x)= -f(-x),

Complex numbersComplex numbers

Real

Imaginary

Argand diagram

a

b

Cartesiana + ib

1i

cosra sinrb

r

Polar

222 bar

a

b1tanso

where

ireiribaz )sin(cos

Working with complex numbersWorking with complex numbers

Add / subtract

Multiply / divide

)()()()( dbicaidciba

)(2121

2121 iii errerer

irez Powers ninn erz

Working with complex numbersWorking with complex numbersRootsExample :

Step 1: write down z in polars with the 2πp bit added on to the argument.

Step 2: say how many values of p you’ll need (as many as n) and write out the rooted expression …..

Step 3: Work it out for each value of p….

here n = 2 so I’ll need 2 values of p; p = 0 and p = 1.

If what is z½?39iez

pi

ez

239

602

322

1

39i

i

eez

p

i

ez

2322

1

9

6666

72

322

1

33339

eeeeeezi

iii

i

p = 0

p = 1

11stst order homogeneous ODE order homogeneous ODE

1st method: Separation of variables

)()(

tNdt

tdN

dNdt

N ln N t c t c t c tN e e e Ae

e.g. radioactive decay

gives

2nd method: Trial solution

( ) mtN t AeGuess that trial solution looks like

( )( ) ( )

dN tmN t N t

dt Substitute the trial solution into the ODE

Comparison shows that ( ) tN t Ae m so write

Step 3: General solution is or if m 1=m2

For complex roots solution is which is

same as or

Step 1: Let the trial solution be Now substitute this back into the

ODE remembering that and

This is now called the auxiliary equation

22ndnd order homogeneous ODE order homogeneous ODE

02

2

cxdt

dxb

dt

xda

mtx e

Solving

mxmedt

dx mt xmemdt

xd mt 222

2

Step 2: Solve the auxiliary equation for and

02 cbmam

1m 2m

tmtm BeAex 21 titi BeAex )()(

)][cos()cossin( tEetDtCex tt

im

)( titit BeAeex

Step 4: Particular solution is found now by applying boundary conditions

)( BtAex mt


Step 1: Let the trial solution be So andmtx e mxme

dt

dx mt xmemdt

xd mt 222

2

Example 3: Linear harmonic oscillator with damping

Step 2: The auxiliary is then with roots

Step 3: General solution is then……. HANG ON!!!!!

In the last lecture we determined the relationship between x and t when

02 202

2

xdt

dx

dt

xd

02 20

2 mm 20

2 m

20

2

20

2

20

2 20

2

meaning that will always be real

What if or ???????????????????

(i) Over-damped gives two real roots


Example 3: Damped harmonic oscillator 02 202

2

xdt

dx

dt

xd

Auxiliary is roots are

02 20

2 mm 20

2 m

BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!!

20

2 20

21 m

20

22 m

tmtm BeAex 21

Both m1 and m2 are negative so x(t) is the sum of two exponential decay terms and so tends pretty quickly, to zero. The effect of the spring has been made of secondary importance to the huge damping, e.g. aircraft suspension

(ii) Critically damped gives a single root



2

xdt

dx

dt

xd


02 20

2 mm 20

2 m


)( BtAex t

20

2 21 mm

Here the damping has been reduced a little so the spring can act to change the displacement quicker. However the damping is still high enough that the displacement does not pass through the equilibrium position, e.g. car suspension.



2

xdt

dx

dt

xd


02 20

2 mm 20

2 m


(iii) Under-damped This will yield complex solutions due to presence of square root of a negative number.

Let so thus

20

2

220

2 i220

2

im 1

im 2

The solution is the product of a sinusoidal term and an exponential decay term – so represents sinusoidal oscillations of decreasing amplitude. E.g. bed springs.

As before general solution with complex roots can be written as

)][cos( tEex t

We do this so that is real



2

xdt

dx

dt

xd


02 20

2 mm 20

2 m


20

2 20

2

20

2

Inhomogeneous ordinary differential equations

)(

2

2

tfrxdt

dxq

dt

xdp

Step 4: Apply boundary conditions to find the values of the constants

Step 1: Find the general solution to the related homogeneous equation and call

it the complementary solution . )(txc

Step 2: Find the particular solution of the equation by substituting an

appropriate trial solution into the full original inhomogeneous ODE.

e.g. If f(t) = t2 try xp(t) = at2 + bt + c If f(t) = 5e3t try xp(t) = ae3t If f(t) = 5eiωt try xp(t) =aeiωt If f(t) = sin2t try xp(t) = a(cos2t) + b(sin2t)

If f(t) = cos t try xp(t) =Re[aeiωt] see later for explanationIf f(t) = sin t try xp(t) =Im[aeiωt]

If your trial solution has the correct form, substituting it into the differential equation will yield the values of the constants a, b, c, etc.

)(txp

Step 3: The complete general solution is then .)()()( txtxtx pc

Extra example of inhomo ODE Solve )4sin(2342

2

txdt

dx

dt

xd

Step 1: With trial solution find auxiliary is mtx e 0342 mm

Step 2: So treating it as a homoODE 13 21 mm

Step 3: Complementary solution istt

c BeAetx 3)(

Step 4: Use the trial solution and substitute

it in FULL expression.

tbtatxp 4sin4cos)(

tbtadt

dx4cos44sin4 tbta

dt

xd4sin164cos16

2

2

so

ttbtatbtatbta 4sin2)4sin4cos(3)4cos44sin4(4)4sin164cos16(

ttatbtatb 4sin24sin164sin134cos134cos16 cancelling

ttabtab 4sin24sin)1613(4cos)1316( Comparing sides gives….

2)1613(0)1316( abandab Solving gives425

26

425

32 banda

Step 5: General solution is

ttBeAetxtxtx ttpc 4sin

425

264cos

425

32)()()( 3

Finding partial solution to inhomogeneous ODE using complex formSometimes it’s easier to use complex numbers rather than messy algebra

Since we can write then we can also say

that and where Re and

Im refer to the real and imaginary coefficients of the complex function.

tiFetXtXdt

d )()( 202

2

tiFtFFe ti sincos

tFFe ti cosRe tFFe ti sinIm

Let’s look again at example 4 of lecture 4 notes tFtxtxdt

d cos)()( 2

02

2

Let’s solve the DIFFERENT inhomo ODE

If we solve for X(t) and take only the real coefficient then this will be a

solution for x(t)ti

p AetX )( tip eAidt

dX tip eAdt

Xd 22

2

tititi FeAeAedt

d )()( 20

2202

2

Sustituting so)( 22

0

FA

tip e

FtX

)()(

220

Therefore since take real part )(

cos)(Re)(

220

tFtXtx pp

Finding coefficients of the Fourier Series…Finding coefficients of the Fourier Series…

SummarySummary

1

0

2sin

2cos

2

1)(

nnn L

xnb

L

xnaaxf

L

dxxfL

a00 )(

2L

n dxL

xnxf

La

0

2cos)(

2 L

n dxL

xnxf

Lb

0

2sin)(

2

The Fourier series can be written with period L as

The Fourier series coefficients can be found by:-

Let’s go through example 1 from notes…

Finding coefficients of the Fourier SeriesFinding coefficients of the Fourier SeriesFind Fourier series to represent this repeat pattern.

0 x

1

20

01)(

x

xxf

Steps to calculate coefficients of Fourier series

1. Write down the function f(x) in terms of x. What is period?

2. Use equation to find a0?1][

10

11

1)(

10

2

0

2

00

xdxdxdxxfa

Period is 2

3. Use equation to find an?

0sin1

cos1

cos)0(1

cos)1(1

cos)(1

00

2

0

2

0

n

nxdxnxdxnxdxnxdxnxxfan

4. Use equation to find bn?

00

2

0

2

0

cos1sin

1sin)0(

1sin)1(

1sin)(

1

n

nxdxnxdxnxdxnxdxnxxfbn

nn

n

nn

n

n

nxbn

1cos10coscos1cos1

0

Finding coefficients of the Fourier SeriesFinding coefficients of the Fourier Series

5. Write out values of bn for n = 1, 2, 3, 4, 5, ….

4. Use equation to find bn?

n

n

nnn

n

n

nxbn

cos110coscos1cos1

0

2

1

)1(1

1

1

1cos

1

111

b 0

2

)1(

2

11

2

2cos

2

112

b

3

2

3

)1(

3

11

3

3cos

3

113

b 0

4

)1(

4

11

4

4cos

4

114

b

5

2

5

)1(

5

11

5

5cos

5

115

b

6. Write out Fourier series

with period L, an, bn in the generic form replaced with values for our example

1

0

2sin

2cos

2

1)(

nnn L

xnb

L

xnaaxf

...sinsinsinsincos)(

xxxxn

bxn

aaxfn

nn 55

23

3

22

2

1

2

2

2

2

2

1

10

Fourier Series applied to pulsesFourier Series applied to pulses

If the only condition is that the pretend function be periodic, and since we know that even functions contain only cosine terms and odd functions only sine terms, why don’t we draw it either like this or this?

Odd function (only sine terms)

Even function (only cosine terms)

What is period of repeating pattern now?

Fourier Series applied to pulsesFourier Series applied to pulsesHalf-range sine series

We saw earlier that for a function with period L the Fourier series is:-

1

0

2sin

2cos

2

1)(

nnn L

xnb

L

xnaaxf

where

L

n dxL

xnxf

La

0

2cos)(

2 L

n dxL

xnxf

Lb

0

2sin)(

2

In this case we have a function of period 2d which is odd and so contains only sine terms, so the formulae become:-

11

sin)2(

2sin)(

nn

nn d

xnb

d

xnbxf

dd

d

d

dn dxd

xnxf

ddx

d

xnxf

ddx

d

xnxf

db

0sin)(

2sin)(

1

)2(

2sin)(

2

2

where

Remember, this is all to simplify the Fourier series. We’re still only allowed to look at the function between x = 0 and x = d

I’m looking at top diagram

Fourier Series applied to pulsesFourier Series applied to pulsesHalf-range cosine series

Again, for a function with period L the Fourier series is:-

1

0

2sin

2cos

2

1)(

nnn L

xnb

L

xnaaxf

where

L

n dxL

xnxf

La

0

2cos)(

2 L

n dxL

xnxf

Lb

0

2sin)(

2

Again we have a function of period 2d but this time it is even and so contains only cosine terms, so the formulae become:-

1

01

0 cos2

1

)2(

2cos

2

1)(

nn

nn d

xnaa

d

xnaaxf

dd

d

d

dn dxd

xnxf

ddx

d

xnxf

ddx

d

xnxf

da

0cos)(

2cos)(

1

)2(

2cos)(

2

2

where

Remember, this is all to simplify the Fourier series. We’re still only allowed to look at the function between x = 0 and x = d

I’m looking at top diagram

Fourier Series applied to pulsesFourier Series applied to pulsesSummary of half-range sine and cosine series

The Fourier series for a pulse such as

can be written as either a half range sine or cosine series. However the series is only valid between 0 and d

1

0 cos2

1)(

nn d

xnaaxf

d

n dxd

xnxf

da

0cos)(

2

1

sin)(n

n d

xnbxf

d

n dxd

xnxf

db

0sin)(

2 Half range sine series

Half range cosine series

where

where

Fourier Transforms Fourier Transforms

deFtf ti)(

2

1)(

dtetfF ti

)(

2

1)(

dkekFxf ikx)(

2

1)(

dxexfkF ikx)(2

1)(

where

where

The functions f(x) and F(k) (similarly f(t) and F(w)) are called a pair of Fourier transforms

k is the wavenumber, (compare with ).2

kT

2

Fourier Transforms Fourier Transforms

pxandxp

pxpxf

0

1)(

Example 1: A rectangular (‘top hat’) function

Find the Fourier transform of the function

ikpikpp

p

ikxikx eeik

dxedxexfkF

1

2

1

2

1)(

2

1)(

given that iAiA eei

A 2

1sin

kp

kpp

k

kpee

ikkF ikpikp sin

2

2sin

2

21

2

1)(

This function occurs so often it has a name: it is called a sinc function.

A

AA

sinsinc

Can you plot exponential functions? Can you plot exponential functions?

For any real number a the absolute value or modulus of a is denoted by | a | and is defined as

The ‘one-sided exponential’ function

0

00)(

xe

xxf

x

What does this function look like?

The ‘****’ function

What does this function look like?

xBexf x)(

Fourier Transforms Fourier Transforms Example 4: The ‘one-sided exponential’ function

Show that the function has Fourier transform

0

00)(

xe

xxf

x ikkF

1

2

1)(

0

)(

0 2

1

2

1)(

2

1)( dxedxeedxexfkF ikxikxxikx

ik

eeik

dxekF ikx

1

2

11

2

1

2

1)( 0

0

)(

222

11

2

1)(

k

ik

ik

ik

ikkF

dkekFxf ikx)(

2

1)(

Complexity, Symmetry and the Cosine TransformComplexity, Symmetry and the Cosine Transform

kxdxxf

ikxdxxfkF sin)(

2cos)(

2

1)(

dxexfkF ikx)(2

1)(

We know that the Fourier transform from x space into k space can be written as:-

We also know that we can write sincos iei

So we can say:-

What is the symmetry of an odd function x an even function ?

If f(x) is real and even what can we say about the second integral above ?Will F(k) be real or complex ?

If f(x) is real and odd what can we say about the first integral above ?Will F(k) be real or complex ?

Odd

2nd integral is odd (disappears) and F(k) is real

1st integral is odd (disappears), F(k) is complex

What will happen when f(x) is neither odd nor even ?

Neither 1st nor 2nd integral disappears, and F(k) is usually complex


kxdxxf

ikxdxxfkF sin)(

2cos)(

2

1)(

f(x)f(x) is real and even is real and even

Since we say

As before the 2nd integral is odd, disappears, and F(k) is real

let’s see if we can shorten the amount of maths required for a specific case …

kxdxxfkF e cos)(

2

1)(

so

X

e

X

X e dxxfdxxf0

)(2)(But remember that

So

0

cos)(2

)( kxdxxfkF e

0

cos)(2

)( kxdxkFxf e

Now since F(k) is real and even it must be true that were we to then find the Fourier transform of F(K) , this can be written:-

LET’S GO BACKWARDS


0

cos)(2

)( kxdxxfkF e

0

cos)(2

)( kxdxkFxf e

Fourier cosine transform

Here is the pair of Fourier transforms which may be used when f(x) is real and even only

Example 5: Repeat Example 1 using Fourier cosine transform formula above.

kpk

kxk

dxkxkxdxxfkFp

p

ee sin12

sin12

cos2

cos)(2

)(0

00

pxandxp

pxpxf

0

1)(Find F(k) for this function

Dirac Delta FunctionDirac Delta FunctionThe delta function (x) has the value zero everywhere except at x = 0 where its value is infinitely large in such a way that its total integral is 1.

The Dirac delta function (x) is very useful in many areas of physics. It is not an ordinary function, in fact properly speaking it can only live inside an integral.

(x – x0) is a spike centred at x = x0 (x) is a spike centred at x = 0

1)( 0

dxxx1)(

dxx

Dirac Delta FunctionDirac Delta FunctionThe product of the delta function (x – x0) with any function f(x) is zero except where x ~ x0.

Formally, for any function f(x) )()()( 00 xfdxxxxf

)sin(x )( 0xx

dxxxx )()sin( 0Example: What is ?

0x

)sin()()sin( 00 xdxxxx

0x

)sin( 0x

Dirac Delta FunctionDirac Delta FunctionThe product of the delta function (x – x0) with any function f(x) is zero except where x ~ x0.

Formally, for any function f(x) )()()( 00 xfdxxxxf

Examples

(a) find

dxxxx )(sin 0

(b) find

dxaxx )(2

(c) find the FT of )()( axxf

0sin x

2a

ikaikxikx edxeaxdxexfkF

2

1)(

2

1)(

2

1)(

dxexfkF ikx)(2

1)(

ConvolutionsConvolutions

If the true signal is itself a broad line then what we detect will be a convolution of the signal with the resolution function:

True signal Convolved signal

Resolutionfunction

We see that the convolution is broader then either of the starting functions. Convolutions are involved in almost all measurements. If the resolution function g(t) is similar to the true signal f(t), the output function c(t) can effectively mask the true signal.

http://www.jhu.edu/~signals/convolve/index.html



DeconvolutionsDeconvolutionsWe have a problem! We can measure the resolution function (by studying what we believe to be a point source or a sharp line. We can measure the convolution. What we want to know is the true signal!

This happens so often that there is a word for it – we want to ‘deconvolve’ our signal.

There is however an important result called the ‘Convolution Theorem’ which allows us to gain an insight into the convolution process. The convolution theorem states that:-

( ) 2 ( ) ( )C k F k G ki.e. the FT of a convolution is the product of the FTs of the original functions.

We therefore find the FT of the observed signal, c(x), and of the resolution

function, g(x), and use the result that in order to find f(x). ( )

( )( ) 2

C kF k

G k

( )( )

( ) 2

C kF k

G k

dkkG

kCexf ikx

)(

)(

2

1)(

If then taking the inverse transform,

DeconvolutionsDeconvolutions

Of course the Convolution theorem is valid for any other pair of Fourier transforms so not only does …..

( ) 2 ( ) ( )C k F k G k( )

( )( ) 2

C kF k

G k and therefore

allowing f(x) to be determined from the FT

deFtf ti)(

2

1)(

dkekFxf ikx)(

2

1)(

but also and therefore

allowing f(t) to be determined from the FT

)()(2)( GFC

2)(

)()(

G

CF

Example of convolutionExample of convolution

I have a true signal between 0 < x < ∞ which I detect using a

device with a Gaussian resolution function given by What is the frequency distribution of the detected signal function C(ω) given

that ?

atetf )(

2exp

2)(

2atatg

Let’s find F(ω) first for the true signal …

)()(2)( GFC

dtetfF ti

)(

2

1)(

dtedteeF iattiat )(

2

1

2

1)(

0

)()(

2

1

2

1

ia

edte

iatiat

iaiaia

eF

iat 1

2

110

2

1

2

1)(

0

)(

Let’s find G(ω) now for the resolution signal …

So if then …..

Example of convolutionExample of convolutionWhat is the frequency distribution of the detected signal function C(ω) given

that ?

2exp

2)(

2atatg

)()(2)( GFC

Let’s find G(ω) now for the resolution signal …

dtetgG ti

)(

2

1)(

so

dte

ataG ti

2exp

22

1)(

2

dtti

ataG

2exp

2)(

2We solved this in lecture 10 so let’s go straight to the answer

aG

2exp

2

1)(

2

)()(2)( GFC

ia

F1

2

1)(

aG

2exp

2

1)(

2

and …..

aiaaia

C2

exp2

1

2exp

2

11

2

12)(

22

Poisson’s equation

t

uiVuu

m

2

2

2

2

2

22 1

t

u

cu

Introduction to PDEsIntroduction to PDEs

In many physical situations we encounter quantities which depend on two or more variables, for example the displacement of a string varies with space and time: y(x, t). Handing such functions mathematically involves partial differentiation and partial differential equations (PDEs).

t

u

hu

22 1

02 u

0

2

uAs Laplace but in regions containing mass, charge,

sources of heat, etc.

Electromagnetism, gravitation,

hydrodynamics, heat flow.

Laplace’s equation

Heat flow, chemical diffusion, etc.

Diffusion equation

Quantum mechanicsSchrödinger’s

equation

Elastic waves, sound waves, electromagnetic

waves, etc.Wave equation

For such equations there is a fundamental theorem called the superposition

principle, which states that if and are solutions of the equation then

is also a solution, for any constants c1, c2.

The principle of superpositionThe principle of superpositionThe wave equation (and all PDEs which we will consider) is a linear equation, meaning that the dependent variable only appears to the 1st power.

i.e. In x never appears as x2 or x3 etc.02 202

2

xdt

dx

dt

xd

Waves and Quanta: The net amplitude caused by two or more waves traversing the same space (constructive or destructive interference), is the sum of the amplitudes which would have been produced by the individual waves separately. All are solutions to the wave equation.

1y 2y

2211 ycycy

Can you think when you used this principle last year??

Electricity and Magnetism: Net voltage within a circuit is the sum of all smaller voltages, and both independently and combined they obey V=IR.

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t

Find the solution to the wave equation to predict the displacement of the guitar string at any later time t

The One-Dimensional Wave Equation

A guitarist plucks a string of length L such that it is displaced from the equilibrium position as shown at t = 0 and then released.

Let’s go thorugh the steps to solve the PDE for our specific case …..

Step 1: Separation of the Variables

Since Y(x,t) is a function of both x and t, and x and t are independent of each other then the solutions will be of the form

where the big X and T are functions of

x and t respectively. Substituting this into the wave equation

gives …

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t

Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t

)()(),( tTxXtxY

2

2

22

2 )()()()(

dt

tTd

c

xX

dx

xXdtT

Step 2: Rearrange equation

Rearrange the equation so all the terms in x are on one side and all the terms in t

are on the other:

2

2

22

2 )(

)(

1)(

)(

1

dt

tTd

tTcdx

xXd

xX


(i) (ii)

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 3: Equate to a constant

Since we know that X(x) and T(t) are independent of each other, the only way this can be satisfied for all x and t is if both sides are equal to a constant:

2 2

2 2 2

1 ( ) 1 ( )constant

( ) ( )

d X x d T t

X x dx c T t dt

Suppose we call the constant N. Then we have

Ndx

xXd

xX

2

2 )(

)(

1

)()(

2

2

xXNdx

xXd

Ndt

tTd

tTc

2

2

2

)(

)(

1

)()( 2

2

2

tTNcdt

tTd

which rearrange to …

and

and

(i) (ii)

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 4: Decide based on situation if N is positive or negative

We have ordinary differential equations for X(x) and T(t) which we can solve but the polarity of N affects the solution …..

Linear harmonic oscillator

Unstable equilibrium

Which case we have depends on whether our constant N is positive or negative. We need to make an appropriate choice for N by considering the physical situation, particularly the boundary conditions.

If N is positive

If N is negative xdt

xd 202

2

xdt

xd 22

2

Decide now whether we expect solutions of X(x) and T(t) to be exponential or trigonometric …..

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 4: Continued …..

Remember that if N is negative, solutions will pass through zero displacement many times, whilst if N is positive solutions only tend to zero once.

From this we deduce N must be negative. Let’s write 2kN

22

2

( )( )

d X xk X x

dx

kxBkxAxX sincos)(

So (i) becomes

From lecture 3, this has general solution

and in the same way kctDkctCtT sincos)(

)()(

2

2

xXNdx

xXd )()( 2

2

2

tTNcdt

tTdandFrom before

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 5: Solve for the boundary conditions for X(x)

In our case the boundary conditions are Y(0, t) = Y(L, t) = 0. This means X(0) = X(L) = 0, i.e. X(x) is equal to zero at two different points. (This was crucial in determining the sign of N.)

Now we apply the boundary conditions: X(0) = 0 gives A = 0.

Saying B ≠ 0 then X(L) = 0 requires sin kL = 0, i.e. kL = n.

So k can only take certain values where n is an integer

L

xnBxX nn

sin)( for n = 1, 2, 3, ….

kxBkxAxX sincos)(

L

nkn

So we have

From previous page

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 6: Solve for the boundary conditions for T(t)

kctDkctCtT sincos)(

By standard trigonometric manipulation we can rewrite this as

nnnn L

ctnEctkEtT coscos)(

L

nkn

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 7: Write down the special solution for Y(x,t)

Hence we have special solutions:

nnnnnnnnn L

ctn

L

xnBtxkBtTxXtxY cossincossin)()(),(

We see that each Yn represents harmonic motion with a different wavelength (different frequency). In the diagram below of course time is fixed constant (as it’s a photo not a movie!!):

(Mistake in notes – please correct harmonic numbers in diagram below)

This is the most general answer to the problem. For example, if a skipping rope was oscillated at both its fundamental frequency and its 3rd harmonic, then the rope would look like the dashed line at some specific point in time and its displacement could be described just by the equation :- (mistake in notes at top of page 4, 3rd not 2nd harmonic)

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 8: Constructing the general solution for Y(x,t)

We have special solutions:

nnn L

ctn

L

xnBtxY

cossin),(

Bearing in mind the superposition principle, the general solution of our equation is the sum of all special solutions:

1

cossin),(n

nn L

ctn

L

xnBtxy

3311

3cos

3sincossin),(

L

ct

L

xB

L

ct

L

xBtxy

NB. The Fourier series is a further example of the superposition principle.

Since

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



00

tt

y

0sinsin1

nn

n

L

ctn

L

xn

L

cnB

t

y

1

cossin),(n

n L

ctn

L

xnBtxy

At t = 0 the string is at rest, i.e. , if we differentiate we find

For this to be true for all n and x, and this is only true if

So the general solution becomes

1

cossin),(n

nn L

ctn

L

xnBtxy

Step 8 continued: Constructing the general solution for Y(x,t)

0n0sin

nL

ctn

The guitarist plucked the string of length L such that it was displaced from the equilibrium position as shown and then released at t = 0. This shape can therefore be represented by the half range sine (or cosine) series.

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 9: Use Fourier analysis to find values of Bn

Half-range sine series

1

sin)(n

n L

xnbxy

L

n dxL

xnxy

Lb

0sin)(

2

where

L

xdxy

2)( d

L

xdxy 2

2)(

...5

sin25

83sin

9

8sin

8)(

222

L

xd

L

xd

L

xdxy

It can be shown (see Phils Problems 5.10) that this shape can be represented by

and we see above that the coefficients Bn are the coefficients of the Fourier series

for the given initial configuration at t = 0. Therefore we can write the general solution at t = 0 as …..

Since Fourier series at t = 0 is

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 9 continued: Use Fourier series to find values of Bn

...5

sin25

83sin

9

8sin

8)(

222

L

xd

L

xd

L

xdxy

L

x

L

x

L

x

L

xdxy

7sin

49

15sin

25

13sin

9

1sin

8)0,(

2 Lx 0 for

and the general solution is then at t = 0 the

general solution is

1

sin)0,(n

n L

xnBxy

1

cossin),(n

n L

ctn

L

xnBtxy

Hence, by trusting the superposition principle treating each harmonic as a separate oscillating sinusoidal waveform which is then summed together like a Fourier series to get the resulting shape, we deduce that at later times the configuration of the string will be:-

The solution at t = 0 is

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



Step 10: Finding the full solution for all times

L

x

L

x

L

x

L

xdxy

7sin

49

15sin

25

13sin

9

1sin

8)0,(

2

L

ct

L

x

L

ct

L

x

L

ct

L

x

L

ct

L

xdtxy

7cos

7sin

49

15cos

5sin

25

13cos

3sin

9

1cossin

8),(

2

But we also know that the general solution at all times is

1

cossin),(n

n L

ctn

L

xnBtxy

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t



What does this all mean ????

L

ct

L

x

L

ct

L

x

L

ct

L

x

L

ct

L

xdtxy

7cos

7sin

49

15cos

5sin

25

13cos

3sin

9

1cossin

8),(

2

This means that if you know the initial conditions and the PDE that defines the relationships between all variables, the full solution can be found which describes the shape at any later time.

Want to know how heat passes down a rod, how light waves attenuate and interfere through a prism, how to define time dependent Schrodinger eigenfunctions, or how anything else that is a linear function with multiple variables interacts ???? Then this is what you should use.

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t

)()(),( tTxXtxy

Ndt

tTd

tTcdx

xXd

xX

2

2

22

2 )(

)(

1)(

)(

1

SUMMARY of the procedure used to solve PDEs

9. The Fourier series can be used to find the full solution at all times.

1. We have an equation with supplied boundary conditions

2. We look for a solution of the form

3. We find that the variables ‘separate’

4. We use the boundary conditions to deduce the polarity of N. e.g.

5. We use the boundary conditions further to find allowed values of k and hence X(x).

6. We find the corresponding solution of the equation for T(t).

7. We hence write down the special solutions.

8. By the principle of superposition, the general solution is the sum of all special solutions..

2kN

L

xnBxX nn

sin)( kxBkxAxX sincos)( so

kctDkctCtT sincos)(

nn L

ctnEtT

cos)(

nnn L

ctn

L

xnBtxY

cossin),(

1

cossin),(n

nn L

ctn

L

xnBtxy

L

ct

L

x

L

ct

L

x

L

ct

L

x

L

ct

L

xdtxy

7cos

7sin

49

15cos

5sin

25

13cos

3sin

9

1cossin

8),(

2

3D Coordinate Systems3D Coordinate Systems

2. Integrals in 3D Cartesian Coordinates

We have dV = dx dy dz, and must perform a triple integral over x, y and z. Normally we will only work in Cartesians if the region over which we are to integrate is cuboid.

Example 1 : Find the 3D Fourier transform,

The integral is just the product of three 1D integrals, and is thus easily evaluated:

otherwise

czcbybaxazyxf

0

1 ,,,),,(

spaceall

i dVerfF k.r232

1k )(

)()(

/

z

cikcik

y

bikbik

x

aikaikc

c

zikb

b

yika

a

xikzyx ik

ee

ik

ee

ik

eedzedyedxekkkF

zzyyxx

zyx

2323 2

1

2

1

)()(),,(

This is therefore a product of three sinc functions, i.e.

if

Just integrating over x gives

x

aikaik

x

aikaika

ax

xika

a

xik

ik

ee

ik

ee

ik

edxeI

xxxxx

x

)()()()(

)sin()sin()sin(

)(),,( ckbkak

abc

k

ck

k

bk

k

akkkkF zyx

z

z

y

y

x

xzyx sincsincsinc

2

8

2

82323

kjik zyx kkk and and

kjir zyx

Mistake in notes

Polar Coordinate SystemsPolar Coordinate Systems

1. Spherical Polar Coordinates

Spherical polars are the coordinate system of choice in almost all 3D problems. This is because most 3D objects are shaped more like spheres than cubes, e.g. atoms, nuclei, planets, etc. And many potentials (Coulomb, gravitational, etc.) depend on radius.

Physicists define r, as shown in the figure. They are related to Cartesian coordinates by:

sin cos , sin sin , cosx r y r z r

. 222 zyxr

2. 3D Integrals in Spherical Polars2 sindV r drd d

.20,0,0 r

The volume element is (given on data sheet).

To cover over all space, we take

Example 1 Show that a sphere of radius R has volume 4R3/3.So

R

sphere

drrdddddrrdVV0

2

0

2

0

2

sinsin 3

4

3

3

0

3

020

RrR

cos


3. 2 in Spherical Polars: Spherical Solutions

As given on the data sheet, 2

2 22 2 2 2 2

1 1 1sin

sin sinr

r r r r r

(Spherically symmetric’ means that V is a function of r but not of or .)

Example 3 Find spherically symmetric solutions of Laplace’s Equation 2V(r) = 0.

Therefore we can say 0)(1

)( 22

2

rV

dr

dr

dr

d

rrV

Really useful bit!!!!

If (as in the homework) we were given an expression for V(r) and had to prove that it was a solution to the Laplace equation, then we’d just stick it here and start working outwards until we found the LHS was zero.

If on the other hand we have to find V(r) then we have to integrate out the expression.


0)(1

)( 22

2

rV

dr

dr

dr

d

rrV

0)(2

rV

dr

dr

dr

d

ArVdr

dr )(2

2)(

r

ArV

dr

d

Multiplying both sides by r2 gives

Integrating both sides gives where A is a constant.

This rearranges to and so ….

Integrating we get the general solution: Br

ArV )(

.

We’ve just done Q3(i) of the homework backwards!!! (see earlier note in red)

Extra tips for the examExtra tips for the exam

When we write and say tAty cos)(

500 AAy )(cos)(

50 )(y

We mean that y = 5 when t = 0

When we say prove that is a solution of tAty cos)( ydt

yd 22

2

then to answer the question STICK IT IN BOTH SIDES

When you solve the complementary solution of a 2nd order differential equation, you need to know that the trial solution is ALWAYS

mtex You also need to know the different forms of the complementary solutions

Documents

Lecture 21 Revision session Remember Phils Problems and your notes = everything Remember I’m available for questions