One important procedure while formulating a
mathematical model of a physical situation is scaling.
Scaling deals with choosing new, dimensionless,
variables of special type, and reformulating the problem
expressed in these variables. These dimensionless scaling
variables are generated by the procedure of fitting the
graph of a function with a linear manifold - a straight line
(in 2D), or a plane in 3D, or a hyper-plane in higher
dimensions.
Of key importance here is the selection of criteria what is
a good fit, and what is a bad fit. These criteria depend on
the concrete context of the concrete problem to be
solved.
Let us consider here in detail the simplest case of fitting
with a straight line in 2D.
 Given is a function
y  f ( x),x  [a, b].
 Consider the linear change of variables
x  Ax  B, y  Cy  D,
where
1
1
[ B]  [ D]  1, A]  C ]  
[ x]
[ y]
so that the new variables are dimensionless:
[ x]  [ y ]  1.
(1)
 After the change of variables, we have
y  f ( x),x  [a, b],
(2)
where
 xB
f ( x)  Cf 
  D,
 A 
(3)
a  Aa  B,b  Ab  B.
Examples of a quality-of-fitting criterion:
 Normalizing the range of the argument and/or the
range of functional values: let M, m be the largest
and smallest values of f on [a,b], respectively,
and M ,m be the corresponding largest and
smallest value of f on [a, b] . Then:
A and B are chosen so that a  0,b  1 :
A
1
a
,B  
;
ba
ba
C and D are chosen so that m  0,M  1 :
C
1
m
,D 
M m
M m .
If the function in (1) is the solution of an
initial/boundary-value problem for and ODE, then
we have to compute also the derivatives
dy d 2 y
d y d2 y
 2 , etc., in terms of
 2 , etc. The
dx dx
dx dx
respective results are (see the text on change of
variable for functions of one real argument, given in
the material for Wednesday, Week 1):
xB
y C C  x  B 
C
x
, y 
 f

1

,

A
D
D  A 
D
 xB
dy
C
dx 1

f '

,

 ,

d x DA  A 
dx A
dy
dy d x C  x  B 

 f '

dx
dx
D  A 
dx
 xB
C
d 2x

f '' 
, 2  0,
2
2
DA
dx
dx
 A 
d2y
 dy 
d  dx 


 dy  d x  dx 
d 2 y d  dy  d  d x 
 dx  





 
dx
dx 2 dx  dx  dx  dx 


dx
 dx 
d 2 y dx dy d 2 x
 2.

. 2
 xB
C
dx dx dx
d
x

 f '' 
,
3
D  A 
 dx 


dx
and so on. When the values of M and m are known
(or estimates for them), such a normalization of the
function (and its derivatives up to the order of the
ODE in consideration) contributes to achieving
better-conditioned computations.
Exercise: Compute A, B, C, D
for the case of symmetrized normalization: when the
range for
x and y is not [0,1], but [-1,1].
 Let x=t be the time. Typically, in this case a=0
(the process starts at the initial moment 0).
Usually, such a problem has a characteristic time
tc, which is the smallest interval of time in which
a notable change can be observed in the physical
quantity described by the functional dependence
f. We then make the change of variables
to achieve a dimensionless time .
Remark: We have different tc for different problems. If
we study the motion of glaciers, tc can be of order years
while tc can be of order microseconds if we for instance
study nuclear reactions in atoms.
Remark: The normalizing scaling considered above
works best for monotone (increasing or decreasing)
functions. However, most functions are only piecewise
monotone (e.g., f(x)=sin x). In such cases it may be
advisable to first divide the range of x into intervals
where the function f is monotone (for this you will have
to find all local maxima and minima, because these will
be the ends of the respective intervals of monotonicity)
and then do scaling in the same way as above, separately
for every interval of monotonicity.)
Remark: Note that some functions are extremely
inhomogeneous. For example, if you try to follow the
above recommendations for scaling by intervals of
monotonicity for the function f(x)=sin (1/x) for x  (0,1] ,
you will have to consider infinitely many subintervals of
(0,1].
 Scaling in initial/boundary problems for
ODEs/PDEs. The idea is the same as earlier, but
this time:
o scaling of the range of the argument x is
used for normalization (scaling) of the size
of the domain of the problem;
o scaling the range of the functional value y is
used to normalize the size of the coefficients
of the ODEs/PDEs and/or the
initial/boundary conditions
Remark: When scaling initial/boundary-value problems
for differential equations, you have to scale not only
functions but also their derivatives (partial derivatives for
PDEs). This means that you should know how to make a
change of variables for functions of one and more
variables, and for the derivatives of these functions. This
is not very difficult, as you have only to use the chain
rule. Moreover, the chain rule has to be used here only
for a linear change of variable, which means that the
resulting scaling formulas are relatively very simple. For
functions of one argument (and ODEs, respectively),
study the text on the change of variable for functions of
one real argument. This text can be found in the material
for Wednesday Week 1.
Exercise: Using the mentioned material for change of
variable for functions of one argument, check the
correctness of the formulas for
dy d 2 y dx d 2 x dy d 2 y
 2 ,  2  , 2 ,
dx dx
dx dx
dx dx
etc., given above.
Example: (One dimensional heat conduction, see
material for Thursday Week 1.) Consider a tall rod
placed along the x-axis. The rod consists of a material
with heat diffusivity k. We let y=u=u(x,t) denote the
temperature in the point x at time t. The rod has the
temperature 0 degrees at time t=0. We then keep the
ends x=0 and x=l at the temperature T0 and examine how
the temperature spreads in the rod. If the rod is long and
slender, we can consider the problem one dimensional.
The temperature distribution is given by the one
dimensional heat equation
We have the inital value
and the boundary conditions
As characteristic length lc we choose the length l of the
rod and as characteristic temperature uc we choose the
boundary temperature T0. As characteristic time tc we
then choose l2/k, since k has the dimension [k]=L2/T.
Thus we make the the change of variables
The chain rule now gives that
This implies that the heat equation is transformed to
the initial value to
and the boundary conditions to
Altogether, our transformed dimensionless boundary
value problem will be
If we solve the problem we see that the heat will
distribute in the rod as in the figure below.
Remark: If we changed the initial value
to
a natural choice of uc could be the mean value
However, as discussed above, it is very common and
possibly easier to choose