Topic 1: Nondimensionalization, Scaling, and Units
Course Notes, Math 558
Spring 2012
Barenblatt.book
Holmes.new.book
This section is a selection of material related to include Chapters 0–3 of [1], Chapters 1–3 of [2].
1
Dimensional Analysis
1.1
Example 1. Throw a ball
Consider the following example: We throw a ball directly upwards from the surface of the Earth. What is
its maximum height?
We need two physical laws to describe this system:
1. Newton’s Second Law. The acceleration of a body is directly proportional to its force and inversely
proportional to its mass; moreover, the acceleration is parallel to the form. This is typically said as
“Force equals mass times acceleration” or F = ma.
2. Newton’s law of universal gravitation. Every body in the universe attracts every other body with
a gravitational force, which force is proportional to each of the masses of the objects and inversely
proportional to the square of the distance between the objects. In equations, if x is the vector from
mass 1 to mass 2, then the force on mass 1 due to mass 2 is
F12 =
Gm1 m2 x
3
kxk
,
and the force on mass 2 due to mass 1 is
F21 = −F12 = −
Gm1 m2 x
3
kxk
.
Since this problem is one-dimensional, we can think of the force, acceleration, and position as scalars (distance from the ground). So if we define mB as the mass of the ball, mE as the mass of the earth, then we
have the differential equation for x(t), the height above the ground, as
mB
d2 x
GmB mE
=−
,
2
dt
d2
(1)
where d = R + x is the distance
between the center of the ball and the center of the earth (R is the radius of
eq:d
the earth). Simplifying (1) we obtain
d2 x
gR2
=
−
,
(2)
dt2
(R + x)2
where we write g = GME /R2 . Clearly this problem, being second-order, requires two integration constants,
but these can be specified by x(0) and x0 (0), the initial position and velocity. Let us denote the maximal
height by xmax .
1
eq:d
eq:x
This equation can actually be solved exactly, but it is tedious and requires special functions (inverse
sinh, for example) and in general is a big mess. Let us delay this solution for now.
Idea 1. The first idea we might have is to say that x(t) R for all t, and thus
x(t) + R ≈ R.
(3)
eq:approx
(4)
eq:constcoe
eq:x
If we make this approximation, then (2) becomes
d2 x
= −g,
dt2
which we can easily solve as
g
x(t) = − t2 + x0 (0)t + x(0).
2
0
Writing x(0) = 0 and x (0) = v0 , it is easy to see that the time of maximal height is t∗ = v0 /g, and therefore
the maximal height is xmax = v02 /2g.
Of course, what we have done here is what is known as an uncontrolled
approximation! We have no idea,
eq:approx
3)
percolates
into errors in the solution
a priori,
how
the
error
we
have
introduced
in
the
approximation
in
(
eq:x
of (2). Studying the effects of this approximation is exactly what we will study in Topic 2 of this course in a
few weeks.
eq:constcoeff
Idea 2. Let us imagine that we didn’t know how to solve (4), and look at the units of the problem.
There are three physical dimensions that appear in this problem, length, time, and mass; we denote these
as L, T , M . Clearly, the units of velocity are L/T , acceleration is L/T 2 , force is M L/T 2 , etc. (If we didn’t
know these, we could deduce the first two from the definitions and the third from Newton’s Second Law!)
Now, let us make the Ansatz that the maximal height depends only on g, mB , and v0 , so that
xmax = f (g, mB , v0 ).
If this is true, then these quantities must have the same units, i.e.
[xmax ] = [f (g, mB , v0 )],
and let us make the further Ansatz that this function can be written as a monomial, so that
[xmax ] = [ma v0b g c ].
(We will justify this second Ansatz later.) This then becomes
L = Ma
b c
L
L
= M a Lb+c T −b−2c .
T
T2
Equating powers, we obtain the system
a = 0, b + c = 1, −b − 2c = 0, or, a = 0, b = 2, c = −1.
Therefore we have
xmax = α
v02
.
g
This is consistent with the exact answer derived above, although is a weaker statement (here we only know
that there is a constant out front and not that this constant is α = 1/2). However, notice that we had to insert
much less information into the problem to obtain this solution and did not have to know how to solve an
ODE.
2
1.2
Example 2. Drag on a sphere
Imagine a sphere moving through a fluid, we want to compute the force due to drag on the sphere. We
postulate that it should depend on the dimensional quantities R, v, ρ, µ: R is the radius of the sphere, v is
the velocity of the sphere, ρ is the density of the fluid, and µ is the (dynamic) viscosity of the fluid. The
units of these quantities are
[R] = L,
[v] =
L
,
T
[ρ] =
M
,
L3
[µ] =
M
.
LT
By the previous logic, we should write
[DF ] = [Ra v b ρc µd ],
or
ML
= La
T2
b c d
M
L
M
T
L3
LT
Expanding these out and equating powers, we obtain the system
a + b − 3c + d = 1,
c + d = 1,
b + d = 2.
There are only three equations, but four unknowns! So there will not be a unique solution and there is (at
least) one free variable.
For now, let us make the choice of d as the free variable, and then we obtain
a = 2 − d,
b = 2 − d,
c = 1 − d,
giving us
DF = αR
2−d 2−d 1−d d
v
ρ
2 2
µ = αR v ρ
where α is a scalar. Let us denote
Π=
µ
Rvρ
d
,
µ
Rvρ
(5)
eq:defofPi
(6)
eq:ad
and we have
DF = αR2 v 2 ρΠd .
We might want to know the dimensions of Π, so we compute
[Π] =
M/LT
= 1.
L(L/T )(M/L3 )
eq:ad
We say that Π is dimensionless. Now, since Π is dimensionless, (6) is dimensionally correct no matter what
the choice of α, or d. Therefore we could write
DF = R2 v 2 ρ(α1 Πd1 ),
or
DF = R2 v 2 ρ(α2 Πd2 ),
or in fact any linear combination of such terms, namely
n
X
2 2
DF = R v ρ
!
αk Π
dk
.
k=1
In general, given any function of Π, this expression is dimensionally correct, so we can write
DF = R2 v 2 ρf (Π)
3
for some unknown function f . Since Π is nondimensional, f (Π) is nondimensional for any function f , and
therefore we cannot proscribe the solution further.
Now, notice that we made a choice of d as a free parameter in the derivation above. Would anything
have changed had we made another choice there? For example, let’s now say that c is free. Solving, we
would obtain
a = 1 + c, b = 1 + c, d = 1 − c,
giving
DF = αR
1+c 1+c c 1−c
v
ρ µ
= αRvρ
Rvρ
µ
c
.
Thus defining
e = Rvρ ,
Π
µ
(7)
we have
e c,
DF = αRvµΠ
and by the same argument of arbitrary powers, we have
e
DF = Rvµg(Π).
e = 1/Π. Moreover, if we set
Are these two different expressions really different? Note, first of all, that Π
them equal, we obtain
ρR2 v 2 f (Π) = Rvµg(1/Π),
Rvρ
f (Π) = g(1/Π),
µ
f (Π) = Πg(1/Π).
Thus there is a functional relationship between f and g; if we know one, we know the other. So these
expressions are equivalent even though they do not seem so at first glance.
Now, one might ask how one determines theHolmes.new.book
unknown function f (or g), and this is something that
should be done by experiment. See Figure 1.3 of [2].
e show up so often in fluid dynamics that they are given names:
The nondimensional quantities Π and Π
e is known as the Reynolds number, and Π is the Péclet number.
Π
1.3
Using dimensional analysis for scale models
Let us imagine that we have specific values for the quantities R, v, ρ, µ in mind, but we want to know the
drag a sphere would experience without building the object itself and making a measurement. Can we
figure out how to build a scale model of the sphere, and then embed it in another physical experiment to
get the measurement.
We know that
µ
DF = ρR2 v 2 f (Π), Π =
.
Rvρ
As long as we know f (Π), then we are done. So let us build a model, with model parameters Rm , vm , ρm , µm
so that Πm = Π. Then f (Πm ) = f (Π), and we can measure the former to get the value for the latter. The
restriction that Πm = Π means that
µ
µm
=
.
Rvρ
R m vm ρ m
Assuming that we’re using the same fluid, so that ρm = ρ, µm = µ, we then have
Rv = Rm vm , or, vm =
4
Rv
.
Rm
eq:defofPit
As an example, let us say that we wanted to measure the drag force on a sphere of radius 1000m at a given
velocity v. This is much to large to build, but we could, for example, build a sphere with radius Rm = 2m
and then choose velocity vm = 500v.
1.4
Buckingham Pi Theorem
The questions we have from the example above are clear. Will such a procedure always work? Will we
have choices? When we are given a choice during the procedure, will this affect matters significantly? The
answer to this is given in a theorem that we prove below.
We will state and prove the theorem in the case that the only dimensions available to us are mass, length,
and time, for concreteness. There could be other dimensional quantities (e.g. charge) but it will be easy to
see at the end how to modify the statements when there are other dimensions.
Let us consider a physical quantity q which depends on the n physical quantities p1 , p2 , . . . , pn . We have
the relationship
q = f (p1 , . . . , pn ),
and let us assume a monomial dependence of the units as
[q] = [pa1 1 pa2 2 . . . pann ]
(8)
eq:mono
(9)
eq:lmt
Let us assume that the dimensions of each quantity are known, and denote them as
[pi ] = Lli M mi T ti ,
[q] = Ll0 M m0 T t0 .
eq:mono
Plugging these into (8) and equating powers, we obtain the three equations
n
X
ai li = l0 ,
i=1
n
X
ai mi = m0 ,
i=1
n
X
ai ti = t0 .
i=1
We can write these equations efficiently in matrix form as follows. Define the matrix and vectors


a1




l1
l2 · · · ln
l0
 a2 


t2 · · · tn  , a =  .  , b =  t0  ,
A =  t1
 .. 
m1 m2 · · · mn
m0
an
eq:lmt
then (9) becomes
(10)
Aa = b.
eq:A
So, the question remains: does (10) have a solution? If so, is it unique?
eq:A
Definition 1. We say that p1 , . . . , pn are dimensionally complete if (10) has a solution for every q, and dimensionally incomplete if it does not. Equivalently, we say that p1 , . . . , pN are dimensionally complete if the matrix A
has rank 3.
We are now in position to state the theorem:
Theorem 1. Assume that q = f (p1 , . . . , pn ) is a dimensionally homogeneous relation and p1 , . . . , pn are dimensionally complete. Then there exists a function F such that q = QF (Π1 , . . . , Πk ), where Πi are dimensionless products
of the pi , [Q] = [q], and k is the dimension of the kernel of A.
eq:A
Proof. We know that all solutions to (10) can be written in the form
a = a∗ +
k
X
i=1
5
γ (i) a(i) ,
eq:A
eq:A
where a∗ is a particular solution to (10), γ (i) ∈ R, and a(i) all lie in the kernel of A. We will write
(i)
(i)
a(i) = (a1 , a2 , . . . , a(i)
n ),
and similarly for a∗ . We first claim that if a(i) ∈ N (A), then
(i)
Πi := pa
:=
n
Y
(i)
a
pj j
j=1
is a dimensionless quantity. To check this, we have
(i)
a
a(i)
[P ii ] = [p1 1 · · · pnn ]
(i)
(i)
(i)
(i)
(i)
(i)
= Ll1 a1 M m1 a1 T t1 a1 · · · Lln an M mn an T tn an
(i)
j=1 lj aj
Pn
=L
M
(i)
Pn
j=1
mj aj
T
(i)
j=1 tj aj
Pn
.
However, notice that the three powers which appear in this last expression are the three rows of Aa(i) , and
since a(i) is in the kernel of A, these are all zero. Therefore [P ii ] = L0 M 0 T 0 = 1 and is dimensionless.
Now we also define
∗
a∗
a∗
Q = pa = p1 1 . . . pnn .
(11)
We can check that [Q] = [q] in a manner similar to checking that [P ii ] = 1 above. From this it follows that
for any gamma1 , . . . , γ k ∈ R,
1
k
[q] = [QΠγ1 . . . Πγk ] =: [QΠγ ].
Of course, this works for any choice of γ, so we can therefore write
q=Q
∞
X
α i Πγ i
i=1
for some choice of αi (some of which may be zero) and thus we have the general nonlinear function q =
QF (Π1 , . . . , Πk ).
Moreover, given a choice of basis of the kernel of A (i.e. choosing a(i) ), this specifies the nondimensional
quantities Πi . If we were to make a different choice of basis here, this would give different Π’s.
Remark 1. We assumed in the theorem that the law q = f (p1 , . . . , pn ) is homogeneous. Is this a valid assumption?
For example, consider a relationship of the form
f (p1 , . . . , pn ) = C1
n
Y
i
pα
i + C2
i=1
n
Y
pβi i .
i=1
P
By dimensional consistency, we would need to satisfy both i li αi = l0 and
which would mean that we must satisfy both
Aα = Aβ = a
and therefore we can combine the two terms.
1.5
Example 3. Computing the yield of a nuclear device
See course lecture.
6
P
i li βi
= l0 , and similarly for m, t,
eq:defofQ
1.6
Example 4. Pythagoras’ Theorem
Consider a right-triangle ABC where B is the right angle. Define c as the length of the hypotenuse AC and
θ as the angle between AB and AC. It is not hard to see that specifying c and θ completely determines the
triangle, and therefore the area of the triangle is given by a function f (c, θ).
Notice that θ is nondimensional, [c] = L and [area] = L2 , therefore
area = c2 F (θ).
Now drop a perpendicular from vertex B to the hypotenuse, breaking the original triangle into two
smaller ones. These are both right triangles with one interior angle equal to θ; the first has hypotenuse a,
the second hypotenuse b. Therefore we have
c2 F (θ) = a2 F (θ) + b2 F (θ),
and assuming that F (θ) 6= 0, we obtain
c2 = a2 + b2 .
Moreover, we can actually use trigonometry to compute F (θ) =
1.7
1
2
sin(θ) cos(θ).
Example 5. Diffusion equation
If we consider the density of (e.g.) a chemical in solution, where we denote said density at x and t by u(x, t),
then the diffusion equation is given by
ut = Duxx ,
(12)
where D is the diffusion coefficient.
Let us imagine that we post this problem on the domain x ∈ (0, ∞) and t > 0. Moreover, we assume that
u(x, 0) = 0 for all x (zero concentration at time zero) and we inject the chemical at x = 0 so that u(0, t) = u0 .
We also append the boundary condition u(∞, t) = 0. Computing dimensions, we have
[u] =
M
,
L3
[ut ] =
M
,
T L3
[uxx ] =
M
,
L5
[u0 ] =
M3
.
L
To make the PDE dimensionally consistent, we must have [D] = L2 /T . If we assume that the concentration
u(x, t) is a function of x, t, D, and u0 , we obtain
[u] = [xa tb Dc ud0 ],
or
M
= La T b
L3
L2
T
c M
L3
d
= La−3d+2c T b−c M d .
One solution of the system is to choose b = c = −a/2 and d = 1, giving
a
x
2 −a/2 −a/2
u = αx t
D
u0 = αu0 √
.
Dt
√
Thus we have the nondimensional Π = x/ Dt and we have u = αF (Π) for some function F .
Note, of course, that if we consider the matrix A as in the theorem, we have


1 0 2 −3
1 .
A= 0 0 0
0 1 −1 0
7
eq:PDE
It is easy to see that this matrix has a one-dimensional nullspace (it clearly has column rank 3) and we
can check that a nullvector is (2, −1, −1, 0)t . Therefore we have no real choice here, our nondimensional
quantity will be x2 /Dt (or a power of it) and we have chosen
the square root of that.
eq:PDE
Plugging the expression u(x, t) = αF (Π) into the PDE (12), we obtain
Π
F 00 (Π) = − F 0 (Π).
2
We also have the boundary conditions
F (0) = 1, F (∞) = 0.
The general solution of this system is
Π
Z
e−s
F (Π) = β + α
2
/4
ds.
0
It is not hard to see (exercise below!) that
∞
Z
2
e−s
/4
ds =
√
π,
0
and therefore plugging in the initial conditions we obtain that
1
F (Π) = 1 − √
π
Π
Z
e−s
2
/4
ds,
0
and therefore we have the general solution
u(x, t) = u0
2
1
1− √
π
Z
√
x/ Dt
!
e
−s2 /4
ds .
0
Scaling and nondimensionalization
We will revisit a couple of the earlier problems and nondimensionalize them (i.e. remove the dimensions
from the problems to obtain purely mathematical problems).
2.1
Projectile problem revisited
Consider the projectile problem above. Recall that we have the ODE given by
d2 x
−gR2
=
.
2
dt
(x + R)2
Let us rescale space and time by the following
t = tc τ,
x = xc ξ,
where [t] = [tc ] and [x] = [xc ], so that τ, ξ are nondimensional. By the chain rule, we have
d2
1 d2
=
,
dt2
t2c dτ 2
eq:dim
so (13) becomes
xc d2 ξ
−gR2
−g
=
=
,
2
2
2
tc dτ
(R + xc ξ)
(1 + xc ξ/R)2
8
(13)
eq:dim
or, rewriting, we have
−1
xc d2 ξ
=
.
gt2c dτ 2
(1 + xRc ξ)2
Moreover, we also have the initial conditions
dξ
tc
(0) = v0 ,
dτ
xc
ξ(0) = 0,
where v0 is the initial velocity of the projectile. Note that three non-dimensional constants show up in the
problem, namely:
xc
t c v0
xc
.
Π1 = 2 , Π2 = , Π3 =
gtc
R
xc
The first is the ratio of the characteristic acceleration of the problem with respect to the gravitational acceleration of the earth; the second is the characteristic size of the problem compared to the earth’s radius, and
the third is the characteristic velocity with respect to the initial velocity of the problem.
Now, notice that we have two degrees of freedom to choose (xc , tc ) and three quantities which we can
change. Of course it is unclear what to do here since there are so many choices, so we have rules of thumb
on how to proceed here.
Rules of Thumb on nondimensionalization:
1. (always) Make as many nondimensional constants equal to one as possible.
2. (usually) Make the constants that appear in the initial or boundary conditions equal to one.
3. (usually) If there is a nondimensional constant that, if we were to set it equal to zero, would simplify
the problem significantly, allow it to remain free and then see when we can make it small.
Using the guidance above, we definitely want to choose Π3 = 1. Moreover, we saw above that if the Π2
term disappeared, the problem becomes very simple, so Π2 should remain free, and we then set Π1 = 1.
This means that we choose the characteristic scales of the problem as
tc =
v0
,
g
xc =
v02
,
g
and this gives us
Π2 =
xc
v2
= 0.
R
gR
Now, we expect this to be small if the projectile is our throwing a ball. We definitely don’t expect the
characteristic height of this problem to be significant when compared to R. Similarly, notice that the other
expression is the ratio of the kinetic energy of the ball to the potential energy of the ball at time zero. Since
we don’t expect to be able to throw a ball into orbit, we expect this ratio to be small as well. Since Π2 is
small, we denote it by , and we obtain the rescaled, nondimensional ODE
d2 ξ
−1
=
,
dτ 2
(1 + ξ)2
ξ(0) = 0,
ξ 0 (0) = 1.
Now, we have chosen wisely, since we know that if we set = 0 in this problem, we have the ODE
ξ 00 = −1,
ξ(0) = 0,
which we can solve explicitly as ξ = −τ 2 /2 + τ .
9
ξ 0 (0) = 1,
So the question one can (and should!) ask at this point is how much the addition of an in the problem
changes things. More specifically, if we consider the problem
−1
d2 ξ =
,
2
dτ
(1 + ξ)2
ξ (0) = 0,
dξ (0) = 1,
dτ
(14)
then how close are ξ and ξ 0 ? This is a useful question, since we know the latter exactly. Of course, we need
to be careful about what we mean by close here. Now, for example, imagine that we know how to write
ξ (t) = ξ 0 (t) + ξ1 (t) + 2 ξ2 (t) + . . .
and we can guaranteed that ξ1 (t) is bounded over some time interval. Then we have a good approximation
to the solution ξeq:rp
and we can make it better and better as → 0. If we can do this, then we call the
perturbation in (14) a regular perturbation; we will study these extensively in the next section of the course.
(We will see for this problem that we can do so.)
Just to get a handle on numbers here, let us assume that the initial velocity was 25m/s (this is about 55
mph so pretty fast for a human!). We then have
tc =
25m/s
v0
=
≈ 2.6s,
g
9.8m/s2
xc =
v02
625m2 /s2
=
≈ 63.8m.
g
9.8m/s2
This gives us the typical time and length scales for the problem. Moreover, notice that
=
62.8m
xc
=
≈ 9.85 × 10−6 .
R
6378.1km
Since is so small, our approximation will probably work quite well (we verify this later). However, solving
the approximate problem is easy: the time of maximum height occurs at τ = 1 (or t = tc = 2.6s) and
therefore the maximum height is ξ = 1/2 (or xmax = xc /2 = 31.9m).
2.2
A different scaling for projectile problem
What if, on the other hand, we had chosen Π2 = Π3 = 1? Then we would have
xc = R,
tc =
R
,
v0
Π1 =
v02
v02
≈
gR
6.25 × 107 m2 /s2
For human velocities, this is clearly quite small. Again choosing v0 = 25m/s, we have
:= Π1 = 9.99 × 10−6 .
This is again small, but plugging the nondimensional variables into the equation, we obtain
d2 ξ
1
=−
,
2
dτ
(1 + ξ)2
ξ(0) = 0,
dξ
(0) = 1.
dτ
When we take the limit as → 0, we obtain an equation where the derivatives disappear, and in fact is not a
differential equation at all. This is actually a singular perturbation; we will deal with these kinds of problems
as well, but after regular perturbations.
2.3
Diffusion equation, revisited
Recall the diffusion equation we considered above. We can even add a nonlinear term to it, as follows:
∂u
∂2u
= D 2 + γu3 ,
∂t
∂x
10
u(x, 0) = u0 (x).
eq:rp
This is an example of a reaction-diffusion equation; the polynomial term is a (local) reaction of the substance
whose concentration we are tracking. Rescaling with
x = xc ξ,
we obtain
t = tc τ,
u = uc ν,
tc ∂ 2 ν
∂ν
= D 2 2 + γtc u2c ν 3 .
∂t
xc ∂ξ
Thus we have
Π1 =
Dtc
,
x2c
Π2 = γtc u2c ,
as our nondimensional quantities. This lets us know what parameters we would choose to have (relative)
small diffusion or (relative) large diffusion. If Π1 Π2 , i.e.
Du2c
1,
γx2c
then we can write Π1 = , Π2 = 1 and we have the PDE
∂ν
∂2ν
= 2 + ν3,
∂t
∂ξ
or the small diffusion scaling regime. This lets us know how we would make this small, e.g. say the constants
D, γ are fixed, we could either take uc small (small concentrations) or xc large (long lengthscales) to get the
small diffusion regime.
Similarly, if Π1 Π2 , or
Du2c
1,
γx2c
then we can write Π1 = 1, Π2 = and we have the PDE
∂ν
∂2ν
=
+ ν 3 ,
∂t
∂ξ 2
or the small reaction scaling regime. This can be obtained by looking at large concentrations, or really small
lengthscales.
We will study both of these asymptotic problems below as well; we will see that the small reaction
regime is basically a regular perturbation, but the small diffusion regime is a singular perturbation. As we
saw in the previous case, when we have a small parameter multiplying a derivative term, setting the small
parameter to zero makes that term disappear, and this changes the form of the equation significantly.
References
nblatt.book
[1] Grigory Isaakovich Barenblatt. Scaling, self-similarity, and intermediate asymptotics, volume 14 of Cambridge Texts in Applied Mathematics. Cambridge University Press, Cambridge, 1996. With a foreword by
Ya. B. Zeldovich.
es.new.book
[2] Mark H. Holmes. Introduction to the foundations of applied mathematics, volume 56 of Texts in Applied
Mathematics. Springer, New York, 2009.
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