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Lecture 17Lecture 17
Solving the Diffusion equationSolving the Diffusion equation
http://www.hep.shef.ac.uk/Phil/PHY226.htmRemember Phils Problems and your notes = everything
Only 5 lectures left
• Come to see me before the end of term• I’ve put more sample questions and answers in Phils Problems• Past exam papers• Complete solution from last lecture• Have a look at homework 2 (due in on 12/12/08)
t
u
hu
2
2 1
Poisson’s equation
t
uiVuu
m
22
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 (4) 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
The Diffusion equationThe Diffusion equation
f(x, t) is the quantity that diffuses. It usually describes a chemical or heat diffusing through a region where h2 is the diffusion constant typically 1×10-4 m 2s -1 for metals.
In classical physics, almost all time dependent phenomena may be described by the wave equation or the diffusion equation. Smaller than the micrometer scale, diffusion is often the dominant phenomenon.
The 1D diffusion equation has the form t
txf
hdx
txfd
),(1),(
22
2
Step 4: Boundary conditions could then be applied to find A and B
Step 2: The auxiliary is then and so roots are
Step 1: Let the trial solution be So andmtx e mxmedt
dx mt xmemdt
xd mt 222
2
Unstable equilibrium
xxm 22 m
Step 3: General solution for real roots is m
)()( 2
2
2
txdt
txd
tt BeAetx )(
Introduction to PDEsIntroduction to PDEs
Thing to notice is that x(t) only tends towards x=0 in one direction of t, increasing exponentially in the other
tAetx )(
t
Step 1: Let the trial solution be So and
Introduction to PDEsIntroduction to PDEs
Harmonic oscillator
mtx e mxmedt
dx mt xmemdt
xd mt 222
2
Step 2: The auxiliary is then and so roots are xxm 22 im
Step 3: General solution for complex is
where = 0 and = so
)cossin( tDtCex t
tDtCx cossin
im
Thing to notice is that x(t) passes through the equilibrium position (x=0) more than once !!!!
Step 4: Boundary conditions could then be applied to find C and D
)()( 2
02
2
txdt
txd
tCx sin
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 particular 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
3sin9
1cossin
8),(
2
www.falstad.com/mathphysics.html
Before we do anything let’s think about hot stuffBefore we do anything let’s think about hot stuffConsider a metal bar heated along it’s length. The ends are placed in ice water and so are held at 0ºC and the middle section is heated by a gas burner.
After a while we reach an equilibrium or steady state temperature distribution along the rod which, let’s say is given by f(x), the temperature distribution plot below.
x = 0 x = Lforxxf )( 20 Lx
Lxxf )( LxL 2forIs there a way of describing the shape of the temperature distribution in terms of an infinite series of sine terms ???????
Yes, it’s called the Half range sine series !!!!
Before we do anything let’s think about hot stuffBefore we do anything let’s think about hot stuff
forxxf )( 20 Lx
Lxxf )( LxL 2for
1
sin)(n
n d
xnbxf
d
n dxd
xnxf
db
0sin)(
2 Half-range sine series: where
. L
L
LL
n dxL
xnxL
Ldx
L
xnx
Ldx
L
xnxf
Lb
2
2
00
sin)(2
sin2
sin)(2
So
Full solution is given in the notes, but we don’t want to waste time doing number crunching so let’s go straight to the answer….
2sin
422
n
n
Lbn
1
sin)(n
n d
xnbxf
L
xnn
n
Lxf
n
sin2
sin4
)(1
22
Find So since then
This describes the temperature distribution f(x) along the bar in equilibrium
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxationConsider a metal bar heated along it’s length. The ends are placed in ice water and so are held at 0ºC and the middle section is heated by a gas burner.
After a while we reach an equilibrium or steady state temperature distribution along the rod which, let’s say is given by f(x), the temperature distribution plot below.
At t = 0 we switch off the burner and allow the rod to cool.
What is the function f(x,t) that describes the temperature of the rod at any point along its length at any time as it cools?
t
txf
hdx
txfd
),(1),(
22
2Heat flow is governed by the diffusion equation,
(f is the temperature and x is position and t is time)
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
t
txf
hdx
txfd
),(1),(
22
2Heat flow is governed by the diffusion equation,
(f is the temperature and x is position and t is time)
Step 1: Rewrite using new variables
Substitute f(x,t) back into the Diffusion equation:
Separating variables:
Step 2: Rearrange the equation
We are looking for solutions of the form )()(),( tTxXtxf
Where X(x) is temperature purely as a function of x
and T(t) is temperature purely as a function of time
dt
tdTxX
htT
dx
xXd )()(
1)(
)(22
2
dt
tdT
tThdx
xXd
xX
)(
)(
11)(
)(
122
2
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
Step 3: Equate to a constant
Now we have separated the variables. The above equation can only be true for all x, t if both sides are equal to a constant.
So which rearranges to (i)
which rearranges to (ii)
dt
tdT
tThdx
xXd
xX
)(
)(
11)(
)(
122
2
constant)(
)(
12
2
dx
xXd
xX )(constant
)(
)(
12
2
xXdx
xXd
xX
constant)(
)(
112
dt
tdT
tTh )(constant
)( 2 tThdt
tdT
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
Step 4: Decide based on the boundary conditions whether constant is +ve or -ve
(i) )(constant)(
2
2
xXdx
xXd
We are told in the boundary conditions that both ends of the rod are held at 0°C at all times.
We therefore choose a negative constant, –k2, to give LHO type solutions that will allow X(x) to be zero at more than one value of x, and rearrange to get two ODEs:
Xkdx
xXd 22
2 )( kxBkxAxX sincos)(
)()( 22 tThk
dt
tdT thkCetT
22
)(
which has general solution
which has general solution
(ii) )(constant)( 2 tTh
dt
tdT
(iii)
(iv)
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
kxBkxAxX sincos)( thkCetT22
)(
Also since then
Solution to (iii) is kxBkxAxX sincos)(
Step 5: Solve for the boundary conditions for X(x)
We know that X(0) = X(L) = 0 meaning that the temperature is zero at the ends
nkL L
xnBxX
sin)( so we can say kLBLX sin0)(
We know that X(0) = 0 so A = 0.
(iii) (iv)
X(x)
x
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
where
Step 6: Write down the special solution for fn (x, t) i.e.
where Pn = BnCn
n
t
nnn eL
xnCBtTxXtxf
sin)()(),(2
22
1
hn
L
khn
xkBxX nnn sin)( thknn
neCtT22
)(
)()(),( tTxXtxf nnn
n
t
nn eL
xnPtTxXtxf
sin)()(),(So
is a special solution of the diffusion equation at one value of n
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
1
)()(),(n
nn tTxXtxf
n
t
nn eL
xnPtTxXtxf
sin)()(),(So
is a special solution of the diffusion equation at one value of n
The general solution therefore is
n
t
nn
nne
L
xnPtxftxf sin),(),(
.
Step 7: Constructing the general solution for
The general solution of our equation is the sum of all special solutions:
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
The general solution is
n
t
nne
L
xnPtxf sin),(
All that remains is calculate the required values of Pn and the harmonics required.
Step 8: Use Fourier series to find values of Pn
We can do this by relating the Fourier series found at t = 0 to the general solution.
The general solution is the solution to the diffusion equation for all values of n summed from 1 to infinity. For our specific example we only want our solution to contain those harmonics required to fulfil the boundary conditions.
Remember earlier we showed that in equilibrium the temperature profile along the rod is given by:
L
xnn
n
Lxf
n
sin2
sin4
)0,(1
22
By comparing this with the general solution at the top of the page when t = 0 we can assign values to Pn .
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
The general solution is
n
t
nne
L
xnPtxf sin),(
Step 8 continued: Use Fourier series to find values of Pn
L
xnn
n
Lxf
n
sin2
sin4
)0,(1
22
By comparing we can see that
n
n L
xnPxf
sin)0,(The general solution at t = 0 is
Fourier series at t = 0 is
2sin
422
n
n
LPn
Step 9: Write down full solution of problem
n
t
neL
xnn
n
Ltxf
sin
2sin
4),(
22where
2
22
1
hn
L
khn
Let’s check that this fulfils all boundary conditions
This is just the Fourier series at start
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxationA metal bar is heated along it’s length. The ends are placed in ice water and so are held at 0ºC and the middle section is heated by a gas burner. At t=0 heating stops.
)0()()0,( TxXxf
)()0(),0( tTXtf
)()(),( tTLXtLf
n
t
neL
xnn
n
Ltxf
sin
2sin
4),(
22
n L
xnn
n
Ltxf
sin2
sin4
),(22
00),(
n
t
etxf
0sin2
sin4
),(22
n
t
nenn
n
Ltxf
Now we can stick in appropriate values of h and find how the temperature profile changes over time. It can be shown that f(x, t) with increasing time looks like this:
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxationA metal bar is heated along it’s length. The ends are placed in ice water and so are held at 0ºC and the middle section is heated by a gas burner. At t=0 heating stops.
n
t
neL
xnn
n
Ltxf
sin
2sin
4),(
22where
2
22
1
hn
L
khn
.....
5sin
25
13sin9
1sin1
14),(
2
22
2
22
2
22 259
2L
th
L
th
L
th
eL
xe
L
xe
L
xLtxf
Solving the diffusion equation for relaxationSolving the diffusion equation for relaxation
Points of interest
The temperature distribution decays exponentially with time.
2
22
1
hn
L
khn
The time constant of the decay is proportional to k -2, and therefore also L2. So the longest wavelengths (such as the fundamental) last longest.
.....
5sin
25
13sin9
1sin1
14),(
2
22
2
22
2
22 259
2L
th
L
th
L
th
eL
xe
L
xe
L
xLtxf
To know exactly how the temperature profile changes with time then we need all the terms. But usually a very good approximation can be obtained by considering just the first term.
n
t
n etT
)(
Remember how we said earlier that temperature as a function of time was written:
22
1
khn
The time constant of temperature decay is defined as , the time for temperature to drop to of its initial value.
n37.01 e
where
Our experimentOur experiment
2
22
sin4
),(2
L
th
eL
xLtxf
A good approximation can be obtained by considering just the first term.
At a fixed x, the temperature drop from t = 0 is:
2
22
)( L
th
etf
We know , L = 0.5m , and h2 = 3×10-5m2s-1 for stainless steel.
32
22
1
102.11
L
hπ
sh
L844
22
2
1
So and therefore So 844)(t
etf
We must record temperature at a fixed location on the rod at regular intervals, then plotting it as a function of time and fitting an exponential to find the decay time constant
n
t
neL
xnn
n
Ltxf
sin
2sin
4),(
22
Remember that now in the full equation x is fixed constant
Our experimentOur experiment
Theory tells us that at a fixed x, the temperature drop from t = 0 should be:
32
22
1
102.11
L
hπ
844)(t
etf
We recorded temperature at a fixed location on the rod at regular intervals, then plotted it as a function of time and fitted an exponential to find the decay time constant
since
y = 333.75e-0.0012x
0
50
100
150
200
250
300
350
400
0 200 400 600 800 1000 1200 1400Time (seconds)
Tem
per
atu
re (
deg
C)
Data collected during the lecture has a decay constant of 1.2x10-3 as expected!!!!
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.
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; explain in general terms what spherical harmonics are.
