Lecture 21 Lecture 21 Revision session Revision session http://www.hep.shef.ac.uk/Phil/PHY226.htm Remember Phils Problems and your notes = everything Remember I’m available for questions all through Christmas
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
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
22ndnd order homogeneous ODE order homogeneous ODE
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
22ndnd order homogeneous ODE order homogeneous ODE
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
22ndnd order homogeneous ODE order homogeneous ODE
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!!!!!
)( 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.
22ndnd order homogeneous ODE order homogeneous ODE
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!!!!!
(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
22ndnd order homogeneous ODE order homogeneous ODE
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
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
Complexity, Symmetry and the Cosine TransformComplexity, Symmetry and the Cosine Transform
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
Complexity, Symmetry and the Cosine TransformComplexity, Symmetry and the Cosine Transform
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
The One-Dimensional Wave Equation
(i) (ii)
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
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t
The One-Dimensional Wave Equation
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
Polar Coordinate SystemsPolar Coordinate Systems
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.
Polar Coordinate SystemsPolar Coordinate Systems
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