Lecture 21
Revision session
Remember I’m available for questions all through Christmas
Remember Phils Problems and your notes = everything
http://www.hep.shef.ac.uk/Phil/PHY226.htm
Revision 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
Revision for the exam
The 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.
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.
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.
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
mt
xe
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.
Let’s take a quick look through the course
and then do the exam from last year
Binomial and Taylor expansions
Integrals
Try these integrals using the hints provided
cos 2 x 
1
1
cos 2 x 
2
2
2
 cos nx dx
0
2
2
cos
 nx dx
0
2
nm
1
1
cos( A  B)  cos( A  B)
2
2
2
1

  sin nx   0
n
0
2
2
1
x
1
1
   cos 2nx   dx   sin 2nx    
2
2
2 0
 4n
0
 cos nx cos mx dx
0
cos A cos B 
2
2
1
1
  cos(n  m) xdx   cos(n  m) x dx
2
2
0
0
2
2
 sin( n  m) x 
 sin( n  m) x 


0


 2(n  m)  0  2(n  m)  0
More integrals
Summary
2
2
 cos nx cos mx dx   sin nx sin mx dx  0
0
for all m  n
Previous page
0
2
 sin nx cos mx dx  0
for all m and n
Remember odd x even function
0
2
2
2
2
sin
nx
dx

cos

 nx dx   for all n
0
0
Previous page
Even and odd functions
So even x even = even
even x odd = odd
odd x odd = even
x
x
x
0
An even function is f(x)=f(-x) and
an odd function is f(x)= -f(-x),
 even dx  2 even dx
x
 odd dx  0
x
Complex numbers
i  1
Argand diagram
Cartesian
a + ib
Imaginary
r
b

Polar
a  r cos
b  r sin 
Real
a
z  a  ib  r (cos   i sin  )  re
so
where
r  a b
2
2
2
i
b
  tan  
a
1
Working with complex numbers
Add / subtract
Multiply / divide
Powers
(a  ib )  (c  id )  (a  c)  i (b  d )
i1
r1e  r2e
z  re
i
i 2
 r1r2e
n ni
z r e
n
i (1  2 )
Working with complex numbers
Roots
Example : If
z  9e
i

what is z½?
3
Step 1: write down z in polars with the 2πp bit added on to the
argument.
z  9e


i   2p 
3

Step 2: say how many values of p you’ll need (as many as n) and
write out the rooted expression …..
i

1
  2p 
here n = 2 so I’ll need 2 values of p; p = 0 and p = 1. 2
2 3

z  9e
Step 3: Work it out for each value of p….
1
2
p = 0 z  9e
p=1
1
2
i

  2 0 
2 3

z  9e
 3e
i

  2 
2 3

i

6
 3e
 7 
i

 6 
 3e
 
i   
6

i
 3e e
i
6

 3e 6
1st order homogeneous ODE
e.g. radioactive decay
dN (t )
 N (t )
dt
1st method: Separation of variables
dN
 N    dt gives
ln N  t  c
N  et c  et ec  Aet
2nd method: Trial solution
Guess that trial solution looks like
N (t )  Aemt
Substitute the trial solution into the ODE
Comparison shows that
m  
dN (t )
 mN (t )   N (t )
dt
so write
N (t )  Ae
 t
2nd order homogeneous ODE
d 2x
dx
Solving a 2  b  cx  0
dt
dt
Step 1: Let the trial solution be
x  emt
Now substitute this back into the
dx
d 2x
mt
2 mt
2
 me  mx and
ODE remembering that

m
e

m
x
2
dt
dt
This is now called the auxiliary equation
Step 2: Solve the auxiliary equation for
Step 3: General solution is
For complex roots m  
same as
x  Ae
m1t
am 2  bm  c  0
m1
 Be
and
m2
m2t or
x  emt ( A  Bt )
if m 1=m2
 i solution is x  Ae ( i )t  Be ( i ) t which is
x  e t ( Aei t  Be i t ) or x  e t (C sin  t  D cos  t )  Ee t [cos(  t   )]
Step 4: Particular solution is found now by applying boundary conditions
2nd order homogeneous ODE
Example 3: Linear harmonic oscillator with damping
Step 1: Let the trial solution be
Step 2: The auxiliary is then
x
d 2x
dx
2

2



x0
0
2
dt
dt
d 2x
mt So dx
mt
 me  mx and 2  m 2e mt  m 2 x
e
dt
dt
m 2  2m   02  0
with roots m   
 2  02
Step 3: General solution is then……. HANG ON!!!!!
In the last lecture we determined the relationship between x and t when    0
2
 2  02
meaning that
What if
 2  02
or
 2  02
2
will always be real
???????????????????
2nd order homogeneous ODE
2
d
x
dx
2
Example 3: Damped harmonic oscillator

2



x0
0
2
dt
dt
Auxiliary is
m 2  2m   02  0
roots are m   
 2  02
BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!!
(i) Over-damped
 2   02 gives two real roots m1     2  02
m2     2  02
x  Ae
m1t
 Be
m2 t
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
2nd order homogeneous ODE
2
d
x
dx
2
Example 3: Damped harmonic oscillator

2



x0
0
2
dt
dt
Auxiliary is
m 2  2m   02  0
roots are m   
 2  02
BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!!
(ii) Critically damped
 2  02 gives a single root m1  m2  
t
x  e ( A  Bt )
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.
2nd order homogeneous ODE
2
d
x
dx
2
Example 3: Damped harmonic oscillator

2



x0
0
2
dt
dt
Auxiliary is
m 2  2m   02  0
roots are m   
 2  02
BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!!
(iii) Under-damped
 2  02 This will yield complex solutions due to
presence of square root of a negative number.
Let
 2  02   2 so  2  02    2  i thus
We do this so that  is real
As before general solution with complex
roots can be written as
 t
x  Ee [cos(  t   )]
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.
m2    i
m1    i
2nd order homogeneous ODE
2
d
x
dx
2
Example 3: Damped harmonic oscillator

2



x0
0
2
dt
dt
Auxiliary is
m 2  2m   02  0
roots are m   
 2  02
BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!!
 
2
2
0
 2  02
 2  02
Inhomogeneous ordinary differential equations
d 2x
dx
p 2  q  rx  f (t )
dt
dt
Step 1: Find the general solution to the related homogeneous equation and call
it the complementary solution
xc (t. )
Step 2: Find the particular solution
x p (t ) of the equation by substituting an
appropriate trial solution into the full original inhomogeneous ODE.
e.g.
If
If
If
If
If
If
f(t) = t2
f(t) = 5e3t
f(t) = 5eiωt
f(t) = sin2t
f(t) = cos t
f(t) = sin t
try
try
try
try
try
try
xp(t) = at2 + bt + c
xp(t) = ae3t
xp(t) =aeiωt
xp(t) = a(cos2t) + b(sin2t)
xp(t) =Re[aeiωt] see later for explanation
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.
Step 3: The complete general solution is then
x(t )  xc (t )  x p (t ) .
Step 4: Apply boundary conditions to find the values of the constants
Extra example of inhomo ODE
Step 1: With trial solution x  emt
d 2x
dx

4
 3x  2 sin( 4t )
Solve
2
dt
dt
find auxiliary is m 2  4m  3  0
Step 2: So treating it as a homoODE m1  3 m2  1
3t
t
Step 3: Complementary solution is xc (t )  Ae  Be
Step 4: Use the trial solution x p (t )  a cos 4t  b sin 4t
it in FULL expression.
so
and substitute
dx
d 2x
 4a sin 4t  4b cos 4t
 16a cos 4t  16b sin 4t
2
dt
dt
(16a cos 4t  16b sin 4t )  4(4a sin 4t  4b cos 4t )  3(a cos 4t  b sin 4t )  2 sin 4t
cancelling
16b cos 4t 13a cos 4t 13b sin 4t 16a sin 4t  2 sin 4t
(16b  13a) cos 4t  (13b  16a) sin 4t  2 sin 4t Comparing sides gives….
32
26
(16b  13a)  0 and (13b  16a)  2 Solving gives a  
and b  
425
425
Step 5: General solution is
x(t )  xc (t )  x p (t )  Ae 3t  Be t 
32
26
cos 4t 
sin 4t
425
425
Finding partial solution to inhomogeneous ODE using complex form
Sometimes it’s easier to use complex numbers rather than messy algebra
Since we can write
that

Feit  F cos t  iF sin t then we can also say

Re Feit  F cos t
and


Im Feit  F sin t
where Re and
Im refer to the real and imaginary coefficients of the complex function.
Let’s look again at example 4 of lecture 4 notes
Let’s solve the DIFFERENT inhomo ODE
d2
2
x
(
t
)


0 x (t )  F cos t
2
dt
d2
2
it
X
(
t
)


X
(
t
)

Fe
0
dt 2
If we solve for X(t) and take only the real coefficient then this will be a
2
d
Xp
it dX p
it
2 it
solution for x(t) X p (t )  Ae
 Aie


A

e
2
dt
dt
F
d2
Sustituting ( 2  02 ) Aeit  ( 2  02 ) Aeit  Feit so A  2
2
(



)
dt
0
Therefore since X p (t ) 
F
t
e it take real part x p (t )  ReX p (t )  F cos
2
2
(0   )
(02   2 )
Finding coefficients of the Fourier Series…
Summary
The Fourier series can be
written with period L as

1
2nx
2nx
f ( x)  a0   an cos
 bn sin
2
L
L
n 1
The Fourier series coefficients can be found by:2 L
2 L
2nx
2 L
2nx
a0   f ( x)dx an   f ( x) cos
dx
dx bn   f ( x) sin
0
0
0
L
L
L
L
L
Let’s go through example 1 from notes…
Finding coefficients of the Fourier Series
Find Fourier series to represent this repeat pattern.
1
Steps to calculate coefficients of Fourier series

0
2
1
f ( x)  
1. Write down the function f(x) in terms of x. What is period?
0
2. Use equation to find a0?
a0 
1


2
0
3
x
0 x 
  x  2
Period is 2
1 
1 2
1
f ( x)dx   1dx   0dx  [ x]0  1
 0
 

3. Use equation to find an?
an 
1


2
0
f ( x) cos nx dx 
1



0
(1) cos nx dx 
1

2

(0) cos nx dx 
1


1  sin nx 
cos nx dx  
0
  n  0


0
4. Use equation to find bn?
bn 
1

2
0
f ( x) sin nx dx 

1


0
(1) sin nx dx 
1
2
 
(0) sin nx dx 
 1  cos nx 
 1  cos n cos 0   1  cos n 1 
bn 


 





  n 0   n
n    n
n
1



0

 1  cos nx 
sin nx dx 
  n  0
Finding coefficients of the Fourier Series

4. Use equation to find bn?
 1  cos nx 
 1  cos n cos 0  1  1 cos n 
bn 



  



  n 0   n
n   n
n 
5. Write out values of bn for n = 1, 2, 3, 4, 5, ….
b1 
1  1 cos1
 
 1
1
b3 
1  1 cos 3  1  1 (1)  2
 
  

 3
3   3
3  3
b5 
1  1 cos 5
 
 5
5
 1  (1)  2
  1 

1  
 
b2 
1  1 cos 2
 
 2
2
b4 
1  1 cos 4  1  1 (1) 
 
   0
 4
4   4 4 
 1  1 (1) 
   0
  2 2 
 1  1 (1)  2
  

5  5
  5

1
2nx
2nx
 bn sin
6. Write out Fourier series f ( x)  a0   an cos
2
L
L
n 1
with period L, an, bn in the generic form replaced with values for our example

1
2nx
2nx 1 2
2
2
f ( x)  a0   an cos
 bn sin
  sin x 
sin 3 x 
sin 5 x  ...
2
2
2
2 
3
5
n 1
Fourier 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 pulses
Half-range sine series
We saw earlier that for a function with period L the Fourier series is:
1
2nx
2nx
f ( x)  a0   an cos
 bn sin
2
L
L
n 1
where
an 
2 L
2nx
f
(
x
)
cos
dx
L 0
L
bn 
2 L
2nx
f
(
x
)
sin
dx
L 0
L
In this case we have a function of period 2d which is odd and so contains
only sine terms, so the formulae become:-
2nx 
nx
f ( x)   bn sin
  bn sin
where
(2d ) n 1
d
n 1

2
bn 
2d

d
d
2nx
1 d
nx
2 d
nx
f ( x) sin
dx   f ( x) sin
dx   f ( x) sin
dx

d
0
( 2d )
d
d
d
d
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 pulses
Half-range cosine series
Again, for a function with period L the Fourier series is:
1
2nx
2nx
f ( x)  a0   a n cos
 bn sin
2
L
L
n 1
where
an 
2 L
2nx
f
(
x
)
cos
dx
L 0
L
bn 
2 L
2nx
f
(
x
)
sin
dx
L 0
L
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
2nx 1
nx
f ( x)  a0   an cos
 a0   an cos
2
( 2d ) 2
d
n 1
n 1
2
an 
2d

d
d
where
2nx
1 d
nx
2 d
nx
f ( x) cos
dx   f ( x) cos
dx   f ( x) cos
dx

d
0
( 2d )
d
d
d
d
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 pulses
Summary 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
Half range
sine series
nx
f ( x)   bn sin
d
n 1

where

1
nx
Half range
f ( x)  a0   an cos
cosine series
2
d
n 1
where
2 d
nx
bn   f ( x) sin
dx
0
d
d
2 d
nx
an   f ( x) cos
dx
0
d
d
Fourier Transforms
f (t ) 
f ( x) 
1
2
1
2




it
F ( )e d


ikx
F (k )e dk
where
where
F ( ) 
F (k ) 
1
2
1
2






f (t )e it dt
f ( x)e ikx dx
The functions f(x) and F(k) (similarly f(t) and F(w)) are called a pair of
Fourier transforms
2
2
k is the wavenumber, k 
(compare with  
).

T
Fourier Transforms
Example 1: A rectangular (‘top hat’) function
Find the Fourier transform of the function
1  p  x  p
f ( x)  
0  p  x and x  p
1
F (k ) 
2
F (k ) 



f ( x )e

ikx
1
dx 
2
given that sin A 

p
e
p
ikx

1 iA iA
e e
2i

 1 1 ikp ikp
dx 
e e
2 ik



1 1 ikp ikp
2 sin kp
2 p sin kp
e e


2 ik
2 k
2 kp
This function occurs so often it has a name: it is called a sinc function.
sin A
sinc A 
A
Can you plot exponential functions?
The ‘one-sided exponential’ function
What does this function look like?
0
f ( x)   x
e
x0
x0
The ‘****’ function
For any real number a the absolute value or modulus of a is denoted by | a |
and is defined as
What does this function look like?
f ( x)  Be
 x
  x  
Fourier Transforms
1
f ( x) 
2
Example 4: The ‘one-sided exponential’ function
0
f ( x)   x
e
Show that the function
1
F (k ) 
2

1
F (k ) 
2



f ( x )e

0
e
ikx
 x (   ik )
x0
x0



has Fourier transform F (k ) 
1
dx 
2


0
e
x ikx
e
1
dx 
2



0
F (k )e ikx dk
1
1
2   ik
e  x (  ik ) dx

1
1
1
1

0
dx 
e e 
2   ik
2   ik
1
1    ik 
1
F (k ) 


2   ik    ik 
2
   ik 
 2
2 
 k 
Complexity, Symmetry and the Cosine Transform
We know that the Fourier transform from x space
into k space can be written as:-
F (k ) 
1
2



f ( x)e ikx dx
We also know that we can write ei  cos   i sin 
So we can say:-
1
F (k ) 
2



i
f ( x) cos kxdx 
2

What is the symmetry of an odd function x an even function ?


f ( x) sin kxdx
Odd
If f(x) is real and even what can we say about the second integral above ?
Will F(k) be real or complex ? 2nd integral is odd (disappears) and F(k) is real
If f(x) is real and odd what can we say about the first integral above ?
Will F(k) be real or complex ? 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
Complexity, Symmetry and the Cosine Transform
1
F (k ) 
2
Since we say



i
f ( x) cos kxdx 
2



f ( x) sin kxdx
let’s see if we can shorten the amount of maths required for a specific case …
f(x) is real and even
As before the 2nd integral is odd, disappears, and F(k) is real
1
so F (k ) 
2
So F (k ) 
2





0
f e ( x) cos kxdx But remember that

X
X
X
f e ( x)dx  2 f e ( x)dx
0
f e ( x) cos kxdx
LET’S GO BACKWARDS
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:
f ( x) 
2

0
Fe (k ) cos kxdx
Complexity, Symmetry and the Cosine Transform
Fourier cosine transform
2

F (k ) 

0
f ( x) 
f e ( x) cos kxdx
2


0
Fe (k ) cos kxdx
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.
1  p  x  p
Find F(k) for this function f ( x)  
0  p  x and x  p
Fe (k ) 
2



0
f e ( x) cos kxdx 
2


p
0
p
cos kx dx 
2 1
2 1

sin
kx

sin kp


 k
 k
0
Dirac Delta Function
The delta function d(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 d(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.
d(x) is a spike centred at x = 0



d ( x) dx  1
d(x – x0) is a spike centred at x = x0



d ( x  x0 ) dx  1
Dirac Delta Function
The product of the delta function d(x – x0) with any function f(x) is zero
except where x ~ x0.
Formally, for any function f(x)
Example: What is






f ( x)d ( x  x0 ) dx  f ( x0 )
sin( x)d ( x  x0 ) dx ?
d ( x  x0 )
sin( x )
x0
sin( x0 )
x0



sin( x)d ( x  x0 ) dx  sin( x0 )
Dirac Delta Function
The product of the delta function d(x – x0) with any function f(x) is zero
except where x ~ x0.

Formally, for any function f(x)


f ( x)d ( x  x0 ) dx  f ( x0 )
Examples
(a) find



sin x d ( x  x0 ) dx
sin x0

(b) find
2
2
x
d
(
x

a
)
dx
a

F (k ) 
(c) find the FT of f ( x)  d ( x  a )
1
F (k ) 
2



f ( x )e
ikx
1
dx 
2



1
2



f ( x)e ikx dx
d ( x  a)e ikx dx 
1 ika
e
2
Convolutions
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
Resolution
function
Convolved
signal
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
Deconvolutions
We 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:-
C (k )  2 F (k )G (k )
i.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 F (k ) 
C (k )
in order to find f(x).
G (k ) 2
1
C (k )
F
(
k
)

f
(
x
)

If
then taking the inverse transform,
2
G (k ) 2

ikx
e


C (k )
dk
G (k )
Deconvolutions
Of course the Convolution theorem is valid for any other pair of Fourier
transforms so not only does …..
C (k )  2 F (k )G (k )
C (k )
and therefore F (k ) 
G (k ) 2
allowing f(x) to be determined from the FT
but also
C ( )  2 F ( )G ( )
f ( x) 
1
2



F (k )e ikx dk
C ( )
and therefore F ( ) 
G ( ) 2
allowing f(t) to be determined from the FT
f (t ) 
1
2



F ( )e it d
Example of convolution
I have a true signal
f (t )  e  at between 0 < x < ∞ which I detect using a
  at 2 
a

exp 
device with a Gaussian resolution function given by g (t ) 
2
 2 
What is the frequency distribution of the detected signal function C(ω) given
that C ( ) 
2 F ( )G ( ) ?
Let’s find F(ω) first for the true signal …
1
F ( ) 
2

1
 at it
e
e
dt


2

 1  e  t ( a  i ) 
1
F ( ) 



2  a  i  0
2



e t ( a i ) dt
F ( ) 
1
2



2
e t ( a i ) dt 
1 
1  1 

0






a

i

a

i

2 



Let’s find G(ω) now for the resolution signal …
1



f (t )e it dt
 t ( a  i )


1  e


2  a  i  0
Example of convolution
What is the frequency distribution of the detected signal function C(ω) given
that C ( ) 
2 F ( )G ( ) ?
Let’s find G(ω) now for the resolution signal …
  at 2 
a
 so G( ) 
g (t ) 
exp 
2
 2 
1
2



1
G ( ) 
2



g (t )e it dt
  at 2  it
a
e dt
exp 
2
 2 
  at 2

We solved this in lecture 10 so let’s go


G( ) 
exp

i

t
dt
 2

straight to the answer
2 


 2 
1

G( ) 
exp 
2 
 2a 
So if C ( )  2 F ( )G ( ) then …..
a
F ( ) 

 2 
1
1  1 
 and …..
exp 

 G( ) 
2  a  i 
2 
 2a 
 2 
 2 
1  1
1
 1
 

C ( )  2 
exp 
exp 

 2 a  i  2 
 2a  2  a  i 
 2a 
Introduction 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).
2
1

u
2
 u 2 2
c t
Wave equation
Elastic waves, sound
waves, electromagnetic
waves, etc.
2 2
u

 u  Vu  i
2m
t
1 u
 2u  2
h t
Schrödinger’s
equation
Quantum mechanics
Diffusion
equation
Heat flow, chemical
diffusion, etc.
 u0
Laplace’s
equation
Electromagnetism,
gravitation,
hydrodynamics, heat flow.

2
 u
0
Poisson’s
equation
As Laplace but in regions
containing mass, charge,
sources of heat, etc.
2
The principle of superposition
The wave equation (and all PDEs which we will consider) is a linear equation,
meaning that the dependent variable only appears to the 1st power.
2
d
x
dx
2
i.e. In

2



0x 0
2
dt
dt
x never appears as x2 or x3 etc.
For such equations there is a fundamental theorem called the superposition
principle, which states that if
y1 and y2 are solutions of the equation then
y  c1 y1  c2 y2 is also a solution, for any constants c1, c2.
Can you think when you used this principle last year??
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.
Electricity and Magnetism: Net voltage within a circuit is the
sum of all smaller voltages, and both independently and
combined they obey V=IR.
The One-Dimensional Wave Equation
 y ( x , t ) 1  y ( x, t )
 2
2
2
x
c
t
2
2
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.
Find the solution to the wave equation to predict the
displacement of the guitar string at any later time t
Let’s go thorugh the steps to solve the PDE
for our specific case …..
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
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
Y ( x, t )  X ( x)T (t )
where the big X and T are functions of
x and t respectively. Substituting this into the wave equation
2
2
d
X
(
x
)
X
(
x
)
d
T (t )
gives … T (t )

dx 2
c2
dt 2
Step 2: Rearrange equation
Rearrange the equation so all the terms in x are on one side and all the terms in t
2
2
1
d
X
(
x
)
1
d
T (t )
are on the other:

X ( x) dx 2
c 2T (t ) dt 2
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time 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: 1 d 2 X ( x)
1 d 2T (t )
X ( x)
dx
2

2
c T (t ) dt
2
 constant
Suppose we call the constant N. Then we have
1 d 2 X ( x)
N
2
X ( x) dx
and
1 d 2T (t )
N
c 2T (t ) dt 2
(i)
which rearrange to …
d 2 X ( x)
 N X ( x)
dx 2
(i)
and
(ii)
d 2T (t )
2

Nc
T (t )
2
dt
(ii)
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
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 …..
If N is negative
d 2x
 02 x
2
dt
If N is positive
d 2x
  2x
2
dt
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.
Decide now whether we expect
solutions of X(x) and T(t) to be
exponential or trigonometric …..
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
d 2 X ( x)
From before
 N X ( x)
2
dx
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
d 2T (t )
2

Nc
T (t )
and
2
dt
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 N  k 2
2
d
X ( x)
So (i) becomes
 k 2 X ( x)
2
dx
From lecture 3, this has general solution
X ( x)  A cos kx  B sin kx
and in the same way
T (t )  C cos kct  D sin kct
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
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.)
X ( x)  A cos kx  B sin kx
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.
n
So k can only take certain values k n 
where n is an integer
L
So we have X n ( x)  Bn sin
nx
L
for n = 1, 2, 3, ….
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
Step 6: Solve for the boundary conditions for T(t)
T (t )  C cos kct  D sin kct
From previous page k n 
n
L
By standard trigonometric manipulation we can rewrite this as
 nct

Tn (t )  E cosk n ct  n   E cos
 n 
 L

 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
Step 7: Write down the special solution for Y(x,t)
Hence we have special solutions:
Yn ( x, t )  X n ( x)Tn (t )  Bn sin k n x cosnt  n   Bn sin
nx
 nct

cos
 n 
L
 L

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)
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
Step 8: Constructing the general solution for Y(x,t)
We have special solutions:
Yn ( x, t )  Bn sin
nx
 nct

cos
 n 
L
 L

Bearing in mind the superposition principle, the general
solution of our equation is the sum of all special solutions:

y ( x, t )   Bn sin
n 1
nx
 nct

cos
 n 
L
 L

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)
y ( x, t )  B1 sin
x
3x
 ct

 3ct

cos
 1   B3 sin
cos
 3 
L
L
 L

 L

NB. The Fourier series is a further example of the superposition principle.
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
Step 8 continued: Constructing the general solution for Y(x,t)
Since y ( x, t ) 
nx
 nct

B
sin
cos




n
n
L
 L

n 1

At t = 0 the string is at rest, i.e.
y
 0 , if we differentiate we find
t t 0

B nc
y
nx  nct

  n
sin
sin 
 n   0
t n 1
L
L
 L

 nct

 n   0 for all n and x, and this is only true if
For this to be true sin 
 L


So the general solution becomes
y ( x, t )   Bn sin
n 1
nx
nct
cos
L
L
n  0
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
Step 9: Use Fourier analysis to find values of Bn
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.
Half-range sine series

y ( x)   bn sin
n 1
nx
L
where
2 L
nx
bn   y ( x) sin
dx
0
L
L
It can be shown (see Phils
Problems 5.10) that this
shape can be represented by
y ( x) 
y ( x) 
8d
x

L
sin
2

2 xd
L
y ( x)  
2 xd
 2d
L
8d
3x
8d
5x
sin

sin
 ...
2
2
9
L 25
L
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
Step 9 continued: Use Fourier series to find values of Bn
Since Fourier series at t = 0 is
y ( x) 

and the general solution is
8d
x

L
sin
2
y ( x, t )   Bn sin
n 1

8d
3x
8d
5x
sin

sin
 ...
2
2
9
L 25
L
nx
nct
cos
L
L
then at t = 0 the
general solution is y ( x,0)   Bn sin nx
L
n 1

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 …..
y ( x,0) 
8d  x 1
3x 1
5x 1
7x

sin

sin

sin

sin


 for
L 9
L 25
L 49
L
 2 
0 xL
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
Step 10: Finding the full solution for all times
The solution at t = 0 is
y ( x,0) 
8d  x 1
3x 1
5x 1
7x

sin

sin

sin

sin



L 9
L 25
L 49
L
 2 
But we also know that the general solution at all times is

y ( x, t )   Bn sin
n 1
nx
nct
cos
L
L
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:y ( x, t ) 
8d  x
ct 1 3x
3ct 1
5x
5ct 1
7x
7ct

sin
cos

sin
cos

sin
cos

sin
cos



L
L 9
L
L
25
L
L
49
L
L
 2 
The One-Dimensional Wave Equation
Find the solution to the wave equation to predict the
displacement of a guitar string of length L at any time t
 2 y ( x , t ) 1  2 y ( x, t )
 2
x 2
c
t 2
What does this all mean ????
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.
y ( x, t ) 
8d  x
ct 1 3x
3ct 1
5x
5ct 1
7x
7ct

sin
cos

sin
cos

sin
cos

sin
cos



L
L 9
L
L
25
L
L
49
L
L
 2 
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.
SUMMARY of the procedure used to solve PDEs
 2 y ( x , t ) 1  2 y ( x, t )
1. We have an equation
 2
x 2
c
t 2
with supplied boundary conditions
2. We look for a solution of the form y ( x, t )  X ( x)T (t )
3. We find that the variables ‘separate’
1 d 2 X ( x)
1 d 2T (t )
 2
N
X ( x) dx 2
c T (t ) dt 2
4. We use the boundary conditions to deduce the polarity of N.
e.g. N  k 2
5. We use the boundary conditions further to find allowed values of k and hence X(x).
X ( x)  A cos kx  B sin kx so X n ( x)  Bn sin nx
L
6. We find the corresponding solution of the equation for T(t).
 nct

T
(
t
)

E
cos



n
T (t )  C cos kct  D sin kct n
L


7. We hence write down the special solutions. Yn ( x, t )  Bn sin nx cos nct  n 
L
 L

8. By the principle of superposition, the general solution is the sum of all special

nx
 nct

solutions..
y ( x, t )  B sin
cos
 

n 1
n
L
 L
n

9. The Fourier series can be used to find the full solution at all times.
y ( x, t ) 
8d  x
ct 1 3x
3ct 1
5x
5ct 1
7x
7ct

sin
cos

sin
cos

sin
cos

sin
cos



L
L 9
L
L
25
L
L
49
L
L
 2 
3D 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,
1
F (k ) 
(2 ) 3 / 2

f (r )e ik.r dV
if
all space
and
1,  a  x  a,  b  y  b,  c  z  c
f ( x, y , z )  
otherwise
0
k  k x i  k y j  k z k and r  xi  yj  zk
The integral is just the product of three 1D integrals, and is thus easily evaluated:
Just integrating over x gives
Mistake in notes
a
I  e
a
ikx x
a

 e 
  e ikx a   eikx a
dx  
  
ik x
 ik x  a 
ikx x
   e
 
 
ikx a
 e ikx a 

ik x

ik b
 ik b
a
b
c
1
1  eikx a  e ikx a  e y  e y  eikz c  e ikz c 
ik y y
ik x x
ik z z




F (k x , k y , k z ) 
e dx  e dy  e dz 
32 
32 


(2 )  a
(2 ) 
ik x
ik y
ik z


b
c

This is therefore a product of three sinc functions, i.e.
8 sin( k x a) sin( k y b) sin( k z c)
8abc
F (k x , k y , k z ) 

sinc (k x a ) sinc (k y b) sinc (k z c)
(2 ) 3 2
kx
ky
kz
(2 ) 3 2
Polar 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:
x  r sin cos  ,
y  r sin sin  ,
r  x2  y 2  z 2
z  r cos
.
2. 3D Integrals in Spherical Polars
The volume element is dV  r sin  drd d (given on data sheet).
2
To cover over all space, we take 0  r  ,
0     , 0    2 .
Example 1 Show that a sphere of radius R has volume 4R3/3.
2

R
So
R
3
4R 3
2
2
2
 r 
V   dV   r sin  dr d d   d  sin  d  r dr   0  cos  0   
3
sphere
0
0
0
 3 0
Polar Coordinate Systems
3. 2 in Spherical Polars: Spherical Solutions
1  2 
1
 

As given on the data sheet,   2
r

sin




r r  r  r 2 sin   

2
1
2

 2 2
2
 r sin  
Example 3 Find spherically symmetric solutions of Laplace’s Equation 2V(r) = 0.
(Spherically symmetric’ means that V is a function of r but not of  or .)
Therefore we can say  2V (r )  1 d  r 2 d V (r )   0
2
r dr 
dr

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.
Polar Coordinate Systems
 2V (r ) 
1 d  2 d

r
V
(
r
)

0
2
r dr  dr

Multiplying both sides by r2 gives
Integrating both sides gives r 2
This rearranges to
d
A
. V (r ) 
dr
r2
d  2 d

V (r )   0
r
dr  dr

d
V ( r )  A where A is a constant.
dr
and so ….
Integrating we get the general solution: V ( r )  
A
B
r
We’ve just done Q3(i) of the homework backwards!!! (see earlier note in red)
Extra tips for the exam
When we write
y (t )  A cos t
and say
We mean that y = 5 when t = 0
When we say prove that
y (0)  5
y (0)  A cos  (0)  A  5
y (t )  A cos t
is a solution of
d2y
2



y
2
dt
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
xe
mt
You also need to know the different forms of the complementary solutions