The Heat Equation
and Diffusion
PHYS220 2004
by Lesa Moore
DEPARTMENT OF PHYSICS
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1
Diffusion of Heat



The diffusion of heat through a material
such as solid metal is governed by the
heat equation.
We will not try to derive this equation.
We will compare results from the heat
equation with our studies of the random
walk.
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2
Initial Temperature
Distribution
-3


-2
-1
0
1
2
3
x
Consider diffusion in 1D (let a thin
copper wire represent a onedimensional lattice).
Let u(t,x) be the heat at point x at time
t, with x and t integers, u(t=0,x=0)=1
and u(t=0,x)=0 if x is not zero.
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The Partial Differential
Equation

The heat equation is a partial differential
equation (PDE):
2
u
 u
k 2
t
x


k is the diffusion coefficient.
Assume the initial distribution is a spike at
x=0 and is zero elsewhere.
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Partial Derivatives


For functions of more than one variable,
the partial derivative is the rate of change
with respect to one variable with the other
variable(s) fixed.
:u (t , x)  lim u (t  t , x)  u(t , x)
t


:
:
t 0
t
u
u (t , x  x)  u (t , x)
(t , x)  lim
x 0
x
x
 2u
(u / x)(t , x  x)  (u / x)(t , x)
(t , x)  lim
2
x 0
x
x
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The PDE in full
u
 2u
k 2
t
x

:

: lim

u (t , x  x)  u (t , x) u (t , x)  u (t , x  x) 




u (t  t , x)  u (t , x)

x

x
 k  lim
:lim

t 0
x 0
t

x




u (t  t , x)  u (t , x)
(u / x)(t , x  x)  (u / x)(t , x) 

 k  lim

t 0
t
x
 x0
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Converting to a
Difference Equation


Don’t take the limits as intervals approach
zero.
Take finite time steps (t=1) and finite
positions steps (x=1).
u (t , x  x)  u (t , x) u (t , x)  u (t , x  x) 




u (t  t , x)  u (t , x)

x

x
lim
 k  lim

t 0
x 0
t

x




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Simplifying …
1
1
u (t , x  x)  u (t , x) u (t , x)  u (t , x  x) 




u (t  t , x)  u (t , x)

x

x
lim
 k  lim

t 0
x 0
t

x




1
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Rearranging …

Want all t+1 terms on l.h.s. and
everything else on r.h.s.
u(t  1, x)  u(t, x)  k u(t , x  1)  u(t , x)  u(t, x)  u(t, x 1)
u(t  1, x)  k u(t , x  1)  2u(t, x)  u(t , x 1)  u(t, x)
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Modelling in Excel


Columns are x-values.
Rows are t-values.
Y
1
2
3
4
Z
-2
-1
u(t,x-1)
AA
AB
0
u(t,x)
u(t+1,x)
AC
1
2
u(t,x+1)
u(t  1, x)  k u(t , x  1)  2u(t, x)  u(t , x 1)  u(t, x)

The difference equation relates each cell to
three cells in the row above.
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The Excel Spreadsheet


The first row (t=0) is all zeros except for the
initial spike: u(t=0,x=0) = 1.
The same formula is entered in every cell
from row 2 down:


A1 holds the value of k (k = 0.1)
AA3=$A$1*(Z2-2*AA2+AB2)+AA2
X
1
2
3
4
5
Y
-3
0
0
0
0.001
Z
-2
0
0
0.01
0.024
AA
-1
0
0.1
0.16
0.195
AB
0
1
0.8
0.66
0.56
AC
1
0
0.1
0.16
0.195
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AD
2
0
0
0.01
0.024
3
0
0
0
0.001
11
Filling the Spreadsheet

In Excel, it is easiest to insert the
formula in the top left cell of the range,
select the range and use Ctrl+R, Ctrl+D
to fill the range:

-20 ≤ x ≤ 20; 0 ≤ t ≤ 60.
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Boundary Conditions


What happens at the boundaries?
Setting columns at x=±21 equal to zero
stops the spatial evolution of the model –
is this a problem?


Provided that values in neighbouring
columns (x=±20) are still small at the end of
the simulation, the choice of boundary
conditions is not so important.
u=0 is equivalent to an absorbing boundary.
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Snapshots
Spread of Heat in 1D: t = 11
1
1
0.9
0.9
0.8
0.8
0.7
0.7
Measure of heat
Measure of heat
Spread of Heat in 1D: t = 0
0.6
0.5
0.4
0.3
0.6
0.5
0.4
0.3
0.2
0.2
0.1
0.1
0
0
-20
-15
-10
-5
0
5
10
15
20
-20
-15
-10
Space (x) units
0
5
10
15
20
10
15
20
Space (x) units
Spread of Heat in 1D: t = 3
Spread of Heat in 1D: t = 51
1
1
0.9
0.9
0.8
0.8
0.7
0.7
Measure of heat
Measure of heat
-5
0.6
0.5
0.4
0.3
0.6
0.5
0.4
0.3
0.2
0.2
0.1
0.1
0
0
-20
-15
-10
-5
0
5
10
15
20
-20
-15
-10
-5
0
5
Space (x) units
Space (x) units
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Plotting the Heat Spread
Spread of Heat in 1D
1
0.9
Measure of heat
0.8
t=1
0.7
t=11
0.6
t=21
0.5
t=31
0.4
t=51
0.3
t=81
0.2
0.1
0
-20
-15
-10
-5
0
5
10
15
20
Space (x) units
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Spreadsheet Results

Conservation of heat can be
demonstrated by adding the values in a
row (a row is a time step).



Values in a row should add to 1.
Checking the sum in a row is good test
of numerical accuracy.
Heat diffusion looks like a Gaussian
distribution.
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The Distribution


The simulation satisfies conservation of
energy (total heat along a row = 1).
Does the Gaussian distribution satisfy this
condition too (area under curve = 1)?



The initial spike can be thought of as a very
sharp, very narrow Gaussian.
For t>0, need to integrate the Gaussian.
“Normalised” if integral yields unity.
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Normalisation of the Gaussian

Formula for Gaussian with m = 0.
f ( x) 

e
 x 2 / 2 2
 2
Use a trick for the integral:






 f ( x)dx   f ( x)  f ( y)dxdy
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 The integral becomes
e

  2


 x 2 / 2 2

1
dx 

 2

1

 2

e
 x / 2
2

 
 e

2
e
 y 2 / 2 2
dxdy

 ( x 2  y 2 ) / 2 2
dxdy
  
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1
 f ( x)dx   2
1

 2
2


0
0
  dxdy   d  rdr
 But using x 2  y 2  r 2 and

 
2

 d  re
0
 r 2 / 2 2
dr
0

2  re
 r 2 / 2 2
dr
0
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 Cancelling

1
 f ( x)dx   2

1


 re

2  re
r 2 / 2 2
dr
0
 r 2 / 2 2
dr
0
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 Then use the substitution:
v  r / 2
2
2
dv  (r /  )dr
2
dr  ( / r )dv
2
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 And finally:
e

  2


 x 2 / 2 2

dx  1






re
0  r

v
2

dv

dv
e



0
 e
v

v 
0
1
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
The integral proves that the Gaussian is
normalised to unity – the area under the
curve is one.
u
 u
k 2
t
x
2


f ( x) 
e
 x 2 / 2 2
 2
But the heat equation is a function of x and
t, and uses a constant k.
k and t must be included in the  term of
the Gaussian if we are to say our model
satisfies this distribution.
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What is  ?

From the Random Walk, we learned that
 √t.


Try a guess:
  2kt
The Gaussian becomes:
f ( x, t ) 
e
 x / 4 kt
2
4kt
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Derivatives of the Gaussian

Space derivatives:
f
x

f
x
2kt
2
2


 f
1 x

 1 f

2
x
2kt  2kt 

The time derivative is left as an exercise …
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The Gaussian satisfies the
Heat Equation



It can be shown that
the heat equation
is satisfied by our
guess.
u
 u
k 2
t
x
2
The distribution integrates to unity
(conservation of energy).
The spread of heat is given by  of the
Gaussian (normal) distribution.
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Diffusion and the
Random Walk



The initial temperature spike grows into
a Gaussian distribution according to the
1D heat equation.
The width  grows in proportion to the
square root of elapsed time.
Heat and diffusion can be understood in
terms of the “random walk”.
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Other Conditions



The initial condition may not be a spike,
but could be some initial distribution:
u(x,0)=g(x).
The boundary conditions may not be
absorbing, but could be continuous.
The thermal diffusivity constant k may
not be constant, but may vary with x or t.
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Summary




The heat equation is a PDE.
By separating space and time variables, we see
that a Gaussian that spreads as √t is a solution.
We can model the differential equation as a
difference equation in Excel and see the same
effect.
The spread of heat is a physical example of a
random walk.
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Acknowledgements


This presentation was based on lecture
material for PHYS220 presented by Prof.
Barry Sanders, 2000-2003.
Additional Reference:

Folland, Fourier Analysis and its
Applications, 1992.
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