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].
1.1 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
Gm1m2x F12 = , kxk3
and the force on mass 2 due to mass 1 is
Gm1m2x F21 = −F12 = − . kxk3
Since this problem is one-dimensional, we can think of the force, acceleration, and position as scalars (dis- tance 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
2 d x GmBmE m = − , (1) eq:d B dt2 d2
where d = R + x is the distanceeq:d between the center of the ball and the center of the earth (R is the radius of the earth). Simplifying (1) we obtain d2x gR2 = − , (2) eq:x dt2 (R + x)2 2 where we write g = GME/R . 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 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 eq:x If we make this approximation, then (2) becomes
d2x = −g, (4) eq:constcoeff 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 2 the maximal height is xmax = v0/2g. Of course, what we have done here is what is known as an uncontrolledeq:approx approximation! We have no idea, a priorieq:x, how the error we have introduced in the approximation in (3) percolates into errors in the solution 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 ML/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
a b c [xmax] = [m v0g ].
(We will justify this second Ansatz later.) This then becomes
L b L c L = M a = M aLb+cT −b−2c. T T 2
Equating powers, we obtain the system
a = 0, b + c = 1, −b − 2c = 0, or, a = 0, b = 2, c = −1.
Therefore we have v2 x = α 0 . max 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 L M M [R] = L, [v] = , [ρ] = , [µ] = . T L3 LT By the previous logic, we should write a b c d [DF ] = [R v ρ µ ], or ML L b M c M d = La T 2 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 µ d D = αR2−dv2−dρ1−dµd = αR2v2ρ , F Rvρ α where is a scalar. Let us denote µ Π = (5) eq:defofPi Rvρ and we have 2 2 d DF = αR v ρΠ . (6) eq:ad We might want to know the dimensions of Π, so we compute M/LT [Π] = = 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
2 2 d1 DF = R v ρ(α1Π ),
or 2 2 d2 DF = R v ρ(α2Π ), or in fact any linear combination of such terms, namely
n ! 2 2 X dk DF = R v ρ αkΠ . k=1 In general, given any function of Π, this expression is dimensionally correct, so we can write
2 2 DF = R v 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 Rvρc D = αR1+cv1+cρcµ1−c = αRvρ . F µ Thus defining Rvρ Πe = , (7) eq:defofPitilde µ we have c DF = αRvµΠe , and by the same argument of arbitrary powers, we have
DF = Rvµg(Π)e .
Are these two different expressions really different? Note, first of all, that Πe = 1/Π. Moreover, if we set them equal, we obtain ρR2v2f(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]. The nondimensional quantities Π and Πe show up so often in fluid dynamics that they are given names: Πe is known as the Reynolds number, and Π is the P´ecletnumber.
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 D = ρR2v2f(Π), Π = . F 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 Π = Π means that m µ µ = m . Rvρ Rmvmρm
Assuming that we’re using the same fluid, so that ρm = ρ, µm = µ, we then have Rv Rv = Rmvm, or, vm = . Rm
4 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
a1 a2 an [q] = [p1 p2 . . . pn ] (8) eq:mono
Let us assume that the dimensions of each quantity are known, and denote them as
li mi ti l0 m0 t0 [pi] = L M T , [q] = L M T . eq:mono Plugging these into (8) and equating powers, we obtain the three equations
n n n X X X aili = l0, aimi = m0, aiti = t0. (9) eq:lmt i=1 i=1 i=1
We can write these equations efficiently in matrix form as follows. Define the matrix and vectors a1 l l ··· l l 1 2 n a2 0 A = t t ··· t , a = , b = t , 1 2 n . 0 m1 m2 ··· mn . m0 an eq:lmt then (9) becomes Aa = b. (10) eq:A 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 dimen- sionally 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 dimension- ally 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
k X a = a∗ + γ(i)a(i), i=1
5 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) (i) (i) a = (a1 , a2 , . . . , an ),
and similarly for a∗. We first claim that if a(i) ∈ N (A), then
n (i) (i) a Y aj Πi := p := pj j=1 is a dimensionless quantity. To check this, we have
(i) (i) a1 an [P ii] = [p1 ··· pn ] l a(i) m a(i) t a(i) l a(i) m a(i) t a(i) = L 1 1 M 1 1 T 1 1 ··· L n n M n n T n n
Pn l a(i) Pn m a(i) Pn t a(i) = L j=1 j j M j=1 j j T j=1 j j .
However, notice that the three powers which appear in this last expression are the three rows of Aa(i), and (i) 0 0 0 since a is in the kernel of A, these are all zero. Therefore [P ii] = L M T = 1 and is dimensionless. Now we also define ∗ ∗ ∗ a a1 an Q = p = p1 . . . pn . (11) eq:defofQ
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, . . . , γk ∈ R, γ1 γk γ [q] = [QΠ1 ... Πk ] =: [QΠ ]. Of course, this works for any choice of γ, so we can therefore write
∞ X γ q = Q αiΠ 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
n n Y αi Y βi f(p1, . . . , pn) = C1 pi + C2 pi . i=1 i=1 P P By dimensional consistency, we would need to satisfy both i liαi = l0 and i liβi = l0, and similarly for m, t, 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 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 = c2F (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
c2F (θ) = a2F (θ) + b2F (θ),
and assuming that F (θ) 6= 0, we obtain c2 = a2 + b2. 1 Moreover, we can actually use trigonometry to compute F (θ) = 2 sin(θ) cos(θ).
1.7 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) eq:PDE 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
M M M M 3 [u] = , [u ] = , [u ] = , [u ] = . L3 t TL3 xx L5 0 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
a b c d [u] = [x t D u0], or M L2 c M d = LaT b = La−3d+2cT b−cM d. L3 T L3 One solution of the system is to choose b = c = −a/2 and d = 1, giving a 2 −a/2 −a/2 x 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 A = 0 0 0 1 . 0 1 −1 0
7 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 2 quantity will be x /Dt (or a power of it) and we have choseneq:PDE the square root of that. 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 2 F (Π) = β + α e−s /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
Π 1 Z 2 F (Π) = 1 − √ e−s /4 ds, π 0 and therefore we have the general solution √ Z x/ Dt ! 1 −s2/4 u(x, t) = u0 1 − √ e ds . π 0
2 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
d2x −gR2 = . (13) eq:dim dt2 (x + R)2
Let us rescale space and time by the following
t = tcτ, x = xcξ,
where [t] = [tc] and [x] = [xc], so that τ, ξ are nondimensional. By the chain rule, we have
d2 1 d2 2 = 2 2 , dt tc dτ eq:dim so (13) becomes 2 2 xc d ξ −gR −g 2 2 = 2 = 2 , tc dτ (R + xcξ) (1 + xcξ/R)
8 or, rewriting, we have x d2ξ −1 c = . 2 2 xc 2 gtc dτ (1 + R ξ) Moreover, we also have the initial conditions
dξ tc ξ(0) = 0, (0) = v0, dτ xc where v0 is the initial velocity of the projectile. Note that three non-dimensional constants show up in the problem, namely: xc xc tcv0 Π1 = 2 , Π2 = , Π3 = . gtc R xc The first is the ratio of the characteristic acceleration of the problem with respect to the gravitational accel- eration 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
v v2 t = 0 , x = 0 , c g c g and this gives us x v2 Π = c = 0 . 2 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
d2ξ −1 = , ξ(0) = 0, ξ0(0) = 1. dτ 2 (1 + ξ)2
Now, we have chosen wisely, since we know that if we set = 0 in this problem, we have the ODE
ξ00 = −1, ξ(0) = 0, ξ0(0) = 1, which we can solve explicitly as ξ = −τ 2/2 + τ.
9 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
d2ξ −1 dξ = , ξ(0) = 0, (0) = 1, (14) eq:rp dτ 2 (1 + ξ)2 dτ
then how close are ξ and ξ0? 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
0 2 ξ (t) = ξ (t) + ξ1(t) + 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:rpand 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
v 25m/s v2 625m2/s2 t = 0 = ≈ 2.6s, x = 0 = ≈ 63.8m. c g 9.8m/s2 c g 9.8m/s2
This gives us the typical time and length scales for the problem. Moreover, notice that x 62.8m = c = ≈ 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
2 2 R v0 v0 xc = R, tc = , Π1 = ≈ 7 2 2 v0 gR 6.25 × 10 m /s
For human velocities, this is clearly quite small. Again choosing v0 = 25m/s, we have
−6 := Π1 = 9.99 × 10 .
This is again small, but plugging the nondimensional variables into the equation, we obtain
d2ξ 1 dξ = − , ξ(0) = 0, (0) = 1. dτ 2 (1 + ξ)2 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 + γu3, u(x, 0) = u (x). ∂t ∂x2 0
10 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 = xcξ, t = tcτ, u = ucν,
we obtain 2 ∂ν tc ∂ ν 2 3 = D 2 2 + γtcuc ν . ∂t xc ∂ξ Thus we have Dtc 2 Π1 = 2 , Π2 = γtcuc , xc 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.
2 Duc 2 1, γxc
then we can write Π1 = , Π2 = 1 and we have the PDE
∂ν ∂2ν = + ν3, ∂t ∂ξ2
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 2 Duc 2 1, γxc
then we can write Π1 = 1, Π2 = and we have the PDE
∂ν ∂2ν = + ν3, ∂t ∂ξ2
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
Barenblatt.book [1] Grigory Isaakovich Barenblatt. Scaling, self-similarity, and intermediate asymptotics, volume 14 of Cam- bridge Texts in Applied Mathematics. Cambridge University Press, Cambridge, 1996. With a foreword by Ya. B. Zeldovich. Holmes.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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