# The Heat Equation and Diffusion (PowerPoint)

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```					The Heat Equation
and Diffusion

PHYS220 2004
by Lesa Moore
DEPARTMENT OF PHYSICS

Macquarie University 2004   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.

Macquarie University 2004   2
Initial Temperature
Distribution
-3   -2   -1     0         1        2      3   x

   Consider diffusion in 1D (let a thin
copper wire represent a one-
dimensional 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.
Macquarie University 2004           3
The Partial Differential
Equation
   The heat equation is a partial differential
equation (PDE):
u 2
 u
k 2
t   x
   k is the diffusion coefficient.
   Assume the initial distribution is a spike at
x=0 and is zero elsewhere.

Macquarie University 2004      4
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           x0            x
 2u                 (u / x)(t , x  x)  (u / x)(t , x)
   :                 (t , x)  lim
x 2           x 0                    x
Macquarie University 2004                           5
The PDE in full
u    2u
   :                                k 2
t   x

u (t  t , x)  u (t , x)           (u / x)(t , x  x)  (u / x)(t , x) 
   : lim                                 k  lim                                           
t 0            t                   x0                    x                    

       u (t , x  x)  u (t , x) u (t , x)  u (t , x  x) 
                                                            
u (t  t , x)  u (t , x)                         x                         x
   :t 0
lim                               k  lim                                                         
t                  
x 0                           x                          
                                                             
Macquarie University 2004                               6
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            t                  
x 0                           x                          
                                                             

Macquarie University 2004                              7
Simplifying …

1                                      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            t                  
x 0                           x                          
                                                             

Macquarie University 2004                              8
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)

Macquarie University 2004                          9
Modelling in Excel
   Columns are x-values.
   Rows are t-values.
Y           Z           AA            AB           AC
1           -2          -1              0              1        2
2                u(t,x-1)     u(t,x)        u(t,x+1)
3                            u(t+1,x)
4

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.
Macquarie University 2004                      10
   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            Y            Z           AA             AB        AC          AD
1               -3           -2           -1          0            1           2           3
2         0            0            0           1           0            0           0
3         0            0           0.1         0.8         0.1           0           0
4         0           0.01         0.16        0.66       0.16          0.01         0
5       0.001        0.024        0.195        0.56       0.195        0.024       0.001
Macquarie University 2004                                11
   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.

Macquarie University 2004   12
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.
Macquarie University 2004     13
Snapshots
Spread of Heat in 1D: t = 0                                                                       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
0.6                                                                                               0.6
0.5                                                                                               0.5
0.4                                                                                               0.4
0.3                                                                                               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     10   15   20
Space (x) units                                                                                   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

0.6                                                                                               0.6
0.5                                                                                               0.5
0.4                                                                                               0.4
0.3                                                                                               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     10   15   20
Space (x) units                                                                                    Space (x) units

Macquarie University 2004                                                                               14

1
0.9
0.8
t=1
0.7
Measure of heat

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

Macquarie University 2004                    15
   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.

Macquarie University 2004   16
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.
Macquarie University 2004        17
Normalisation of the Gaussian
   Formula for Gaussian with m = 0.
 x 2 / 2 2
e
f ( x) 
 2
   Use a trick for the integral:
                                

 f ( x)dx   f ( x)  f ( y)dxdy
                               

Macquarie University 2004    18
 The integral becomes


 e  x / 2                                        
2    2

              dx  1
  2                            e                  e
 x / 2
2       2
 y 2 / 2 2
dxdy
     2
                                                    

 
1
 e
( x 2  y 2 ) / 2 2
                                               dxdy
 2         

Macquarie University 2004                                   19
                   2       

 But using x 2  y 2  r 2 and              dxdy   d  rdr
                  0        0

                            2        
1
 f ( x)dx   2            d  re
 r 2 / 2 2
dr
                             0        0


1
                2  re          r 2 / 2 2
dr
 2                 0

Macquarie University 2004                                20
 Cancelling

                                      
1
 f ( x)dx   2               2  re        r 2 / 2 2
dr
                                        0


1
 re
 r 2 / 2 2
                                  dr
        0

Macquarie University 2004                       21
 Then use the substitution:

v  r / 2
2         2

dv  (r /  )dr
2

dr  ( 2 / r )dv

Macquarie University 2004   22
 And finally:


e x 2 / 2 2                     
 2 
                dx  1
  2                             r
re v     dv
                                    0           


e
v
                 dv
0

 e       v 
0 
1
Macquarie University 2004            23
   The integral proves that the Gaussian is
normalised to unity – the area under the
curve is one.
 x 2 / 2 2
e
f ( x) 
u    u  2
 2
k 2
t   x
   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.
Macquarie University 2004                           24
What is  ?
   From the Random Walk, we learned that
 √t.
   Try a guess:     2kt
   The Gaussian becomes:
 x / 4 kt
2
e
f ( x, t ) 
4kt
Macquarie University 2004        25
Derivatives of the Gaussian
   Space derivatives:

f      x
     f
x     2kt
 f
2
1   x2    
          1 f
x 2
2kt  2kt 

   The time derivative is left as an exercise …
Macquarie University 2004      26
The Gaussian satisfies the
Heat Equation
   It can be shown that
the heat equation
u    u
2
k 2
is satisfied by our                  t   x
guess.
   The distribution integrates to unity
(conservation of energy).
   The spread of heat is given by  of the
Gaussian (normal) distribution.

Macquarie University 2004        27
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”.

Macquarie University 2004   28
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.

Macquarie University 2004   29
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.

Macquarie University 2004     30
Acknowledgements
   This presentation was based on lecture
material for PHYS220 presented by
Prof. Barry Sanders, 2000-2003.
   Folland, Fourier Analysis and its
Applications, 1992.

Macquarie University 2004   31

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