Lecture 17
Solving the Diffusion equation
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• Past exam papers
• Complete solution from last lecture
• Have a look at homework 2 (due in on 12/12/08)
1 u
u 2
h t
2
Remember Phils Problems and your notes = everything
http://www.hep.shef.ac.uk/Phil/PHY226.htm
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 (4) in regions
containing mass, charge,
sources of heat, etc.
2
The Diffusion equation
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.
2
d
f ( x, t ) 1 f ( x, t )
The 1D diffusion equation has the form
 2
2
dx
h
t
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.
Introduction to PDEs
2
d
x(t )
2
Unstable equilibrium


x(t )
2
dt
2
dx
d
x
mt
mt So
Step 1: Let the trial solution be
 me  mx and 2  m 2e mt  m 2 x
xe
dt
dt
Step 2: The auxiliary is then
m x  x
2
Step 3: General solution for real roots
Step 4: Boundary conditions could
then be applied to find A and B
Thing to notice is that x(t) only
tends towards x=0 in one
direction of t, increasing
exponentially in the other
2
and so roots are
m  
m   is x(t )  Ae t  Be  t
x(t )  Aet
t
Introduction to PDEs
d 2 x(t )
2
Harmonic oscillator



0 x (t )
2
dt
Step 1: Let the trial solution be
Step 2: The auxiliary is then
x
d 2x
mt So dx
mt
 me  mx and 2  m 2e mt  m 2 x
e
dt
dt
m x   x
2
Step 3: General solution for complex
where  = 0 and b =  so
2
and so roots are
m   i
m    ib is x  e t (C sin b t  D cos b t )
x  C sin  t  D cos  t
Step 4: Boundary conditions could then be
applied to find C and D
Thing to notice is that x(t) passes
through the equilibrium position
(x=0) more than once !!!!
x  C sin  t
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

www.falstad.com/mathphysics.html
9. The Fourier series can be used to find the particular 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 
Before we do anything let’s think about hot stuff
Consider 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.
f ( x)  x for 0  x  L 2
f ( x)   x  L for L 2  x  L
x=0
x=L
Is 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 stuff
f ( x)  x for 0  x  L 2
f ( x)   x  L for L 2  x  L
Half-range sine series: f ( x) 
nx
2 d
nx
where
b
sin
b

f
(
x
)
sin
dx

n
n

0
d
d
d
n 1

2 L
nx
2
So bn 
f ( x) sin
dx 

0
L
L
L
L2

0
nx
2
nx
x sin
dx   ( L  x) sin
dx
L
L L. 2
L
L
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….


4L
n
nx
nx
4L
n
f
(
x
)

sin
sin
f
(
x
)

b
sin
b

sin

Find n
So since
then

n
2 2
2
L
d
n 1 n 
2
n 1
n 2 2
This describes the temperature distribution f(x) along the bar in equilibrium
Solving the diffusion equation for relaxation
Consider 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?
Heat flow is governed by the diffusion equation,
(f is the temperature and x is position and t is time)
d 2 f ( x, t ) 1 f ( x, t )
 2
2
dx
h
t
Solving the diffusion equation for relaxation
Heat flow is governed by the diffusion equation,
(f is the temperature and x is position and t is time)
We are looking for solutions of the form
d 2 f ( x, t ) 1 f ( x, t )
 2
2
dx
h
t
f ( x, t )  X ( x)T (t )
Where X(x) is temperature purely as a function of x
and T(t) is temperature purely as a function of time
Step 1: Rewrite using new variables
2
d
X ( x)
1
dT (t )
Substitute f(x,t) back into the Diffusion equation:
T
(
t
)

X
(
x
)
dx 2
h2
dt
Step 2: Rearrange the equation
Separating variables:
1 d 2 X ( x) 1 1 dT (t )
 2
2
X ( x) dx
h T (t ) dt
Solving the diffusion equation for relaxation
1 d 2 X ( x) 1 1 dT (t )
 2
2
X ( x) dx
h T (t ) dt
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.
1 d 2 X ( x)
1 d 2 X ( x)
So
 constant X ( x) (i)
 constant which rearranges to
2
2
X ( x) dx
X ( x) dx
1 1 dT (t )
dT (t )
2

constant


which
rearranges
to

constant
h
T (t )
2
h T (t ) dt
dt
(ii)
Solving the diffusion equation for relaxation
d 2 X ( x)
 constant X ( x)
2
dx
(i)
dT (t )
 constant h 2T (t )
dt
(ii)
Step 4: Decide based on the boundary conditions whether constant is +ve or -ve
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:
d 2 X ( x)
 k 2 X
2
dx
which has general solution
X ( x)  A cos kx  B sin kx
dT (t )
  k 2 h 2T (t )
dt
which has general solution
 k 2 h 2t
T (t )  Ce
(iv)
(iii)
Solving the diffusion equation for relaxation
X ( x)  A cos kx  B sin kx
(iii)
T (t )  Ce
 k 2 h 2t
(iv)
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
Solution to (iii) is
X ( x)  A cos kx  B sin kx
We know that X(0) = 0 so A = 0.
Also since X ( L)  0  B sin kL then kL  n so we can say X ( x)  B sin
1.5
total amplitude
X(x)1
0.5
0
-0.5
x
nx
L
Solving the diffusion equation for relaxation
X n ( x)  Bn sin kn x
Tn (t )  Cn e
Step 6: Write down the special solution for fn (x, t) i.e.
nx
f n ( x, t )  X ( x)T (t )  BnCn sin
e
L
So

t
n
 k n 2 h 2t
f n ( x, t )  X n ( x)Tn (t )
1
 L 
where  n  2 2  

h k
 nh 
nx
f n ( x, t )  X ( x)T (t )  Pn sin
e
L

t
n
where Pn = BnCn
is a special solution of the diffusion equation at one value of n
2
Solving the diffusion equation for relaxation
So
nx
f n ( x, t )  X ( x)T (t )  Pn sin
e
L

t
n
is a special solution of the diffusion equation at one value of n
Step 7: Constructing the general solution for f ( x, t ) 

X
n 1
n
( x)Tn (t )
.
The general solution of our equation
is the sum of all special solutions:
The general solution therefore is
f ( x, t )  
n
nx
f n ( x, t )   Pn sin
e
L
n

t
n
Solving the diffusion equation for relaxation
nx  n
f ( x, t )   Pn sin
e
L
n
t
The general solution is
Step 8: Use Fourier series to find values of Pn
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.
All that remains is calculate the required values of Pn and the harmonics required.
We can do this by relating the Fourier series found at t = 0 to the general solution.
Remember earlier we showed that in equilibrium the temperature profile along the
rod is given by:

4L
n
nx
sin
sin
2 2
2
L
n 1 n 
f ( x,0)  
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 relaxation
nx  n
f ( x, t )   Pn sin
e
L
n
t
The general solution is
Step 8 continued: Use Fourier series to find values of Pn
4L
n
nx
sin
sin
2 2
2
L
n 1 n 

Fourier series at t = 0 is
f ( x,0)  
The general solution at t = 0 is
f ( x,0)   Pn sin
n
By comparing we can see that Pn 
nx
L
4L
n
sin
2 2
n
2
Step 9: Write down full solution of problem
4L
n
nx
f ( x, t )   2 2 sin
sin
e
2
L
n n 

t
n
1
 L 
where  n  2 2  

h k
 nh 
2
Solving the diffusion equation for relaxation
A 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.
4L
n
nx  n
f ( x, t )   2 2 sin
sin
e
2
L
n n 
t
Let’s check that this fulfils all boundary conditions
f ( x,0)  X ( x)T (0)
f ( x, t )  
n
f (0, t )  X (0)T (t )
f ( L, t )  X ( L)T (t )
f ( x, t )  0e
f ( x, t )  
n
4L
n
nx
sin
sin
n 2 2
2
L

This is just the Fourier
series at start
t
n
0
4L
n
sin n  e
sin
2 2
n
2

t
n
0
Solving the diffusion equation for relaxation
A 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.
4L
n
nx
f ( x, t )   2 2 sin
sin
e
2
L
n n 

t
n
1
 L 
where  n  2 2  

h k
 nh 
2
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:
 h t
9 h t
25 h t

4 L 1
x  L2
1
3x  L2
1
5x  L2
f ( x, t )  2  sin
e
 sin
e

sin
e
 .....
 1
L
9
L
25
L

2 2
2 2
2 2
Solving the diffusion equation for relaxation
 h t
9 h t
25 h t

4 L 1 x  L2
1
3x  L2
1
5x  L2
f ( x, t )  2  sin
e
 sin
e
 sin
e
 .....
 1
L
9
L
25
L

2 2
2 2
2 2
Remember how we said earlier that temperature as a function of time was written:
Tn (t )  e

t
n
1
where  n  2 2
h k
The time constant of temperature decay is defined as  n , the time for temperature
to drop to e 1  0.37 of its initial value.
Points of interest
The temperature distribution decays exponentially with time.
1
 L 
n  2 2  

h k
 nh 
2
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.
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.
Our experiment
A good approximation can be obtained by
considering just the first term.
f ( x, t ) 
4L
2
sin
x
L

e
 2 h 2t
L2
At a fixed x, the temperature drop from
t = 0 is:
2 2
f (t )  e

 h t
L2
We know  , L = 0.5m , and h2 = 3×10-5m2s-1 for stainless steel.
So
1 π 2h2
 2 1.2103
1 L
L2
and therefore  1  2 2  844 s So
 h
f (t )  e

t
844
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
Remember that now in the full
equation x is fixed constant
4L
n
nx  n
f ( x, t )   2 2 sin
sin
e
2
L
n n 
t
Our experiment
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
Theory tells us that at a fixed x, the temperature drop from t = 0 should be:
f (t )  e

t
844
since
1 π 2h2
 2 1.2103
1 L
400
350
y = 333.75e-0.0012x
Temperature (degC)
300
Data collected during the
lecture has a decay constant
of 1.2x10-3 as expected!!!!
250
200
150
100
50
0
0
200
400
600
800
Time (seconds)
1000
1200
1400
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.
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.