One-Dimensional Diffusion on the Real Line:

Theory and Experiment

Kyle Claassen

Bethel College

11 May 2010

Contents

1 Introduction 31.1 Transport Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Modeling Advection-Diffusion: Partial Differential Equations . . . . . . . . . 31.3 Notation of Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Derivation of the Advection-Diffusion Equation 52.1 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Diffusive and Advective Mass Flux . . . . . . . . . . . . . . . . . . . . . . . 6

3 Diffusion on the Real Line 73.1 Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Invariance Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 Uniqueness Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Solution of the Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . 13

4 Advection-Diffusion on the Real Line 16

5 Inhomogeneous Advection-Diffusion Initial Value Problem 18

6 Experiment 236.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6.2.1 Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246.2.2 Initial Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.2.3 Launching the Experiment . . . . . . . . . . . . . . . . . . . . . . . . 26

6.3 Data Analysis and Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.3.1 Image Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.3.2 “Pure” Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.3.3 Diffusion with Advection . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

7 Future Considerations 387.1 Further Research Topics and Applications . . . . . . . . . . . . . . . . . . . 387.2 Modeling “Double Diffusion” . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

A The Error Function 40

2

1 Introduction

1.1 Transport Phenomena

Phenomena involving the transport of mass are readily observed in our everyday environ-

ment. The motion of wind, the propagation of a pollutant in the ocean, and the slow

dispersion of an odor from a spilled bottle of cologne or perfume are all examples of mass

moving from one location to another. Two mass transport phenomena are of particular

interest: diffusion and advection.

According to [5], the transport mechanism of diffusion “has two primary properties: it is

random in nature, and transport is from regions of high concentration to low concentration,

with an equilibrium state of uniform concentration.” At the macroscopic level, diffusion often

manifests itself as an observable spectrum of intensity, or a concentration gradient (see Figure

1). Certainly the odor of the spilled cologne bottle varies continuously in intensity as one

moves closer to the source. Dyes in liquids behave similarly, forming a visible concentration

gradient. Further, [5] defines advection as the “[non-random] transport associated with the

Figure 1: Formation of a Concentration Gradient

[bulk] flow of a fluid.” Water flowing in a river is a good example of an advective process, as

the bulk motion of the water “pushes” matter along with it.

1.2 Modeling Advection-Diffusion: Partial Differential Equations

The combination of diffusion and advection is important in many applications of fluid dy-

namics, and thus a mathematical study of the advection-diffusion equation is in order. The

advection-diffusion equation will be derived in the generality of three dimensions, but as a

3

more tractable introduction, only solutions in one dimension will be subsequently explored.

The one-dimensional advection-diffusion equation,

∂u

∂t= k

∂2u

∂x2− v(t)u

is a linear, second-order partial differential equation relating the concentration of mass u

in terms of two variables: x (the single spatial dimension) and t (time). As with ordinary

differential equations (ODEs), initial value problems often appear in applications, so the

solution of the advection-diffusion initial value problem

∂u

∂t= k

∂2u

∂x2− v(t)u (x ∈ R, t ∈ R+)

u(x, 0) = φ(x)

will be examined and applied as a model to experimental data.

1.3 Notation of Partial Derivatives

As a notational convenience, it is common to write partial differential equations using sub-

scripts to denote differentiation with respect to the subscripted variable. For example,

∂

∂xu(x, t) = ux(x, t).

Though perhaps a slight abuse of notation, if u = u(x, t), then a subscripted x may also

denote differentiation with respect to the first argument, while a subscripted t denotes differ-

entiation with respect to the second argument. So, assuming constant a, b, one could write

the following:

∂2

∂x2u(ax, bt) = a2uxx(ax, bt)

∂

∂tu(ax, bt) = but(ax, bt).

4

Using subscripts, the advection diffusion equation may be expressed slightly more com-

pactly as

ut = kuxx − v(t)u.

2 Derivation of the Advection-Diffusion Equation

2.1 Continuity Equation

Let u = u(x, y, z, t) denote the concentration (mass per unit volume) of a substance, and

let S be an arbitrary surface surrounding it. Suppose particles of the substance move with

velocity ~v. The total mass of the substance contained in S is given by

M =

˚S

u dV,

and the time-rate of change of this mass is thus

dM

dt=

d

dt

˚S

u dV =

˚S

ut dV. (1)

One may also compute dMdt

via the surface integral

dM

dt= −‹∂S

~F · n dS,

where ~F is the flux of mass across the boundary ∂S of S, i.e. the amount of mass that

flows through the boundary per unit area per unit time, and n is the outward-oriented unit

normal to S. By the divergence theorem,

‹∂S

~F · n dS =

˚S

∇ · ~F dV,

5

hence

dM

dt= −˚

S

∇ · ~F dV. (2)

Equating (1) and (2) gives

˚S

ut dV = −˚

S

∇ · ~F dV.

Since S is an arbitrary surface, the integrands must be equal, yielding the continuity equation,

ut = −∇ · ~F . (3)

2.2 Diffusive and Advective Mass Flux

The flux ~F in (3) is determined by two phenomena: diffusion and advection. According to

Fick’s first law, the flux due to diffusion is given by the linear function ~Fd = −k∇u, where

k is the diffusion coefficient. While this “law” is phenomenological, it makes physical sense.

Indeed, it is well-known that −∇u gives the direction of fastest decrease of the function u.

In words, then, Fick’s first law states that mass moves from regions of high concentration

to regions of low concentration, preferring the direction in which concentration is reduced

as quickly as possible. The flux due to advection is most easily derived from dimensional

analysis:

Flux =mass

area · time=

mass

volume· length

time.

Thus it is reasonable to adopt ~Fa = u~v as the advective mass flux. Combining these fluxes,

we have

~F = ~Fd + ~Fa = −k∇u+ u~v. (4)

6

Substituting (4) into the continuity equation (3) gives

ut = −∇ · (−k∇u+ u~v)

= ∇ · (k∇u)−∇ · (u~v) .

Assuming k and ~v are independent of position, we achieve the three-dimensional advection-

diffusion equation,

ut = k∇2u− ~v · ∇u. (5)

In one dimension, this reduces to

ut = kuxx − vux. (6)

For our purposes, k will be constant and v = v(t) will be a function of time alone.

3 Diffusion on the Real Line

3.1 Initial Value Problem

Ignoring the effects of advection, we take v(t) ≡ 0 in (6) to obtain the one-dimensional

diffusion equation,

ut = kuxx. (7)

If this differential equation is coupled with the initial condition u(x, 0) = φ(x) and its domain

taken to be the real line, then we have the following initial value problem:

ut = kuxx (x ∈ R, t ∈ R+)

u(x, 0) = φ(x).

(8)

7

We will proceed to examine the properties of solutions of this initial value problem, eventually

deriving a systematic method of solution.

3.2 Invariance Properties

As given in [6], in order to find a solution to (8) we will first investigate five invariance

properties of the diffusion equation (7).

Theorem 3.2.1. The following properties are true of the diffusion equation (7):

1. The translation u(x− y, t) of any solution u(x, t) is another solution for any fixed y.

2. Any derivative of a solution is again a solution.

3. A linear combination of solutions is again a solution.

4. If S(x, t) is a solution, then so is the convolution

u(x, t) =

ˆ ∞−∞

S(x− y, t)g(y) dy

for any function g(y) such that this improper integral converges.

5. If u(x, t) is a solution, so is the dilated function u(√ax, at) for any a > 0.

Proof. (1) Suppose u is a solution of the diffusion equation. Then

∂

∂tu(x− y, t)− k ∂

2

∂x2u(x− y, t)

= ut(x− y, t)− kuxx(x− y, t)

= 0.

(2) Suppose u is a solution of the diffusion equation. Take the derivative with respect to

8

x of both sides:

utx = kuxxx

(ux)t = k(ux)xx by equality of mixed partials.

Thus ux is a solution. By the same method, it is easy to see that the time derivative is also

a solution, as well as any derivatives of higher order.

(3) Suppose that u and v are solutions. Consider the linear combination αu+ βv:

(αu+ βv)t − k(αu+ βv)xx

= αut + βut − kαuxx − kβvxx

= α(ut − kuxx) + β(vt − kvxx)

= α · 0 + β · 0 since u and v solve the diffusion equation

= 0.

Thus αu+ βv is a solution.

(4) If u(x, t) =´∞−∞ S(x − y, t)g(y) dy for any function such that this improper integral

converges, we have

ut =

ˆ ∞−∞

St(x− y, t)g(y) dy by Leibniz’ rule

uxx =

ˆ ∞−∞

Sxx(x− y, t)g(y) dy

9

Hence

ut − kuxx =

ˆ ∞−∞

[St(x− y, t)− kSxx(x− y, t)] g(y) dy

=

ˆ ∞−∞

0 · g(y) dy since S(x− y, t) is a solution by (1)

= 0.

Thus u solves the diffusion equation.

(5) Let u(x, t) be a solution of the diffusion equation, and a ∈ R+. By the chain rule,

∂

∂tu(√ax, at) = aut(

√ax, at)

∂2

∂x2u(√ax, at) =

∂

∂x

[√aux(√ax, at)

]=√a√auxx(

√ax, at)

= auxx(√ax, at).

Thus

∂

∂tu(√ax, at)− k ∂

2

∂x2u(√ax, at) = a

(ut(√ax, at)− kuxx(

√ax, at)

)= a · 0 since u is a solution

= 0.

Therefore u(√ax, at) is also a solution. �

3.3 Uniqueness Considerations

Even though we have not derived a general way to find a solution of the initial value problem

(8), we can investigate the uniqueness of such a solution. In doing so, we will first heuristically

demonstrate the following lemma:

10

Lemma 3.3.1. u(x, t) ≡ 0 is the unique solution to the initial value problem

ut = kuxx

u(x, 0) = 0.

(9)

Proof. Rewrite the diffusion equation as kuxx−ut = 0 and integrate over an arbitrary simple

region D, bound by a piecewise-smooth, simple closed curve ∂D in the xt-plane:

¨D

(kuxx − ut) dA = 0 =

˛∂D

u dx+ kux dt by Green’s Theorem.

Since D is an arbitrary region, ∂D is an arbitrary closed path, and we gather that the vector

field

~F (x, t) = (u(x, t), kux(x, t))

is conservative. Apply the divergence theorem to ~F :

¨D

∇ · ~F dA =

˛∂D

~F · d~s¨D

∇ · ~F dA = 0 since ~F is conservative

¨D

(ux + kuxt) dA = 0

ux + kuxt = 0 since D is an arbitrary region

(u+ kut)x = 0

ut +1

ku = f(t) for arbitrary f : R→ R. (10)

Note that (10) is an ODE in t with initial condition u(0) = 0 and that it may be solved with

11

the integrating factor et/k:

d

dt

(et/ku

)= et/kf(t)

et/ku =

ˆ t

0

es/kf(s) ds+ C(x) where C(x) is an arbitrary function of x.

Utilize the initial condition u(0) = 0. Then C(x) = 0 for all x, and we have

u(x, t) = e−t/kˆ t

0

es/kf(s) ds. (11)

Note that (11) is independent of x. Thus u(x, t) = g(t) for arbitrary g : R → R, and the

vector field simplifies to ~F (x, t) = (u(x, t), 0). Moreover, since ~F is conservative, we also

know that its curl vanishes, i.e.

~0 = ∇× ~F =

∣∣∣∣∣∣∣∣∣∣x t z

∂/∂x ∂/∂t ∂/∂z

u(x, t) 0 0

∣∣∣∣∣∣∣∣∣∣= −ut(x, t) z.

It follows that ut(x, t) ≡ 0, hence u(x, t) = h(x) for arbitrary h : R → R. We have shown

that g(t) = u(x, t) = h(x), so it must be that u(x, t) is constant for all x, t. Finally, the

initial condition gives the result u(x, t) = 0 for all x, t. �

At this time it must be mentioned that the proof of Lemma 3.3.1, while convincing, is

not completely correct. A (pathological) counterexample can be found in [1], where it is

asserted that

u(x, t) =∞∑n=0

f (n)(t)x2n

(2n)!

where

f(t) =

exp(−1/t2) if t > 0

0 if t = 0

12

is a solution to the initial value problem

ut = uxx (x ∈ R, t ∈ R+)

u(x, 0) = 0.

Clearly this solution is not identically zero, so the initial value problem of Lemma 3.3.1 is

actually not unique. However, it is shown in [1] that this initial value problem is in fact

unique for solutions u(x, t) satisfying the growth condition |u(x, t)| ≤ C1 exp(C2 x2) for all

|x| sufficiently large and C1, C2 being positive constants to be chosen as necessary. We will

only consider such solutions, as they are sufficient in applications.

Making appropriate assumptions about the growth of solutions, we are now prepared to

prove the uniqueness of the initial value problem (8).

Theorem 3.3.2. The solution u(x, t) to the diffusion initial value problem (8) is unique.

Proof. Suppose u(x, t) and v(x, t) are solutions of (8). By linearity, w(x, t) = u(x, t)−v(x, t)

solves the initial value problem with initial condition w(x, 0) = φ(x) − φ(x) = 0. By

Lemma 3.3.1, w(x, t) ≡ 0 is the unique solution to this special initial value problem, hence

u(x, t) = v(x, t). Therefore, the solution to (8) is unique. �

3.4 Solution of the Initial Value Problem

Motivated by the invariance properties of section 3.2, we will “guess” a particular solution of

the diffusion initial value problem (8) (see [6]). Then, an application of invariance property

(4) of Theorem 3.2.1 provides a way to “construct” solutions for any initial condition φ(x).

Noting invariance property (5) of Theorem 3.2.1, one might guess that a solution to the

diffusion equation (7) has the form

Q(x, t) = g

(x√4kt

)(12)

13

since this function is unchanged by the transformation Q(x, t)→ Q(√ax, at). That is,

Q(√ax, at) = g

( √ax√

4kat

)= g

(x√4kt

)= Q(x, t).

We will call such functions dilation-invariant. Moreover, consider the dilation-invariant

special initial condition given by

Q(x, 0) =

0 if x < 0

1 if x > 0.(13)

Now, let p = x/(√

4kt) so that Q(x, t) = g(p). Via the chain rule, we have

Qt =dg

dp

∂p

∂t=

(x√4k

)(− 1

2t3/2

)g′(p) =

−x2t√

4ktg′(p)

Qx =dg

dp

∂p

∂x=

1√4kt

g′(p)

Qxx =dQx

dp

∂p

∂x=

1

4ktg′′(p).

It follows that

0 = Qt − kQxx

=−x

2t√

4ktg′(p)− k

(1

4ktg′′(p)

)= − 1

4t

[2

(x√4kt

)g′(p) + g′′(p)

]= − 1

4t[2pg′(p) + g′′(p)] .

Since t > 0, it must be that 2pg′(p) + g′′(p) = 0. Solve this second-order ODE by letting

14

h(p) = g′(p) and using the integrating factor ep2. Then

d

dp

[ep

2

h(p)]

= 0

ep2

h(p) = C1

g′(p) = h(p) = C1e−p2 .

Therefore,

g(p) = C1

ˆ p

0

e−r2

dr + C2

Q(x, t) = C1

ˆ x/√4kt

0

e−r2

dr + C2.

Clearly this solution is only valid for t > 0. Taking the limit as t → 0+, we can utilize the

initial condition (13) ofQ(x, t) to find C1 and C2. For x > 0, Q(x, 0) = 1, and x/√

4kt→ +∞

as t→ 0+. Thus

1 = C1

ˆ ∞0

e−r2

dr + C2 = C1

√π

2+ C2.

Moreover, for x < 0, Q(x, 0) = 0, and x/√

4kt→ −∞ as t→ 0+. So,

0 = C1

ˆ −∞0

e−r2

dr + C2 = −C1

√π

2+ C2.

Solving simultaneously for C1, C2 in

C1

√π2

+ C2 = 1

−C1

√π2

+ C2 = 0

yields C1 = 1/√π and C2 = 1/2. Therefore, the particular solution is

Q(x, t) =1

2+

1√π

ˆ x/√4kt

0

e−r2

dr. (14)

15

Since Q(x, t) is a solution, so is Qx(x, t) by invariance property (2). Define

S(x, t) = Qx(x, t) =1√

4πkte−x

2/(4kt) (15)

and consider the solution (by invariance properties (1) and (4)) given by

u(x, t) =

ˆ ∞−∞

S(x− y, t)φ(y) dy (t > 0) (16)

for any function φ(x) such that the integral converges. We will demonstrate that (16) indeed

solves the initial value problem (8). At this point, we need only show that limt→0+

u(x, t) = φ(x).

Suppose that φ(x) is a bounded and continuous function on R (except allowing a jump

discontinuity at x = 0) with lim|x|→∞

φ(x) = 0 in such a way that the integral converges. Note

that S(x, t) is a Gaussian with unit area and variance 2kt. As stated in [4], “the limit of the

Gaussian behaves like the Dirac delta function” δ (x); that is, as t → 0+, S(x, t) → δ (x).

Thus we take

limt→0+

ˆ ∞−∞

S(x− y, t)φ(y) dy =

ˆ ∞−∞

δ (x− y)φ(y) dy

= φ(x) by properties of the Dirac delta function.

Therefore (16) is a solution of the diffusion initial value problem (8).

4 Advection-Diffusion on the Real Line

Having solved the diffusion initial value problem (8), let us consider the advection-diffusion

initial value problem on the real line:

ut = kuxx − v(t)ux (x ∈ R, t ∈ R+)

u(x, 0) = φ(x)

(17)

16

To solve this equation, we will employ the following change of variables:

ξ(x, t) = x− α(t)

η(x, t) = t

(18)

where α(t) =´ t0v(p) dp. By the chain rule,

ut =∂u

∂η

∂η

∂t+∂u

∂ξ

∂ξ

∂t

= uη − v(t)uξ

= uη − v(η)uξ (19)

ux =∂u

∂ξ

∂ξ

∂x+∂u

∂η

∂η

∂x

= uξ

uxx =∂uξ∂ξ

∂ξ

∂x+∂uξ∂η

∂η

∂x

= uξξ.

Substituting these results into the differential equation of (17), we have

uη − v(η)uξ = kuξξ − v(η)uξ (20)

uη = kuξξ.

This is the ordinary diffusion equation! Moreover, the initial condition is invariant through

this change of variables since, when t = 0, ξ(x, 0) = x− α(0) = x. Therefore the solution is

given by (16):

u(ξ, η) =

ˆ ∞−∞

S(ξ − y, η)φ(y) dy.

17

Restoring the original variables x and t yields

u(x, t) =

ˆ ∞−∞

S(x− y − α(t), t)φ(y) dy. (21)

Some insight into the impetus for the change of variables (18) is warranted. Exercise

2.4.18 of [6] indicates that a change of variables ξ = x − vt is appropriate to solve the

advection-diffusion equation with constant velocity v. Solving with constant velocity gives

the outline for the proof of the more general case v = v(t) given above. One can see

in (20) that it is key for the v(η)uξ terms to cancel in order to achieve a reduction to the

ordinary diffusion equation. Thus, α(t) must be an antiderivative of v(t) so that (19) contains

the desired term. Further, for the initial condition to be invariant through the change of

variables it is necessary that α(0) = 0. It follows that α(t) =´ t0v(p) dp is the appropriate

antiderivative.

Moreover, (21) provides insight into the nature of advection-diffusion. Notice that

if u(x, t) is a solution of the advection-diffusion initial value problem (17) and w(x, t)

is a solution of the ordinary diffusion initial value problem (8), then (21) reveals that

u(x, t) = w(x− α(t), t). This appeals to intuition, as α(t) physically represents the dis-

placement of a mass due to advection. That is, one can conceive of advection-diffusion as

ordinary diffusion occurring in a frame of reference moving in the positive x-direction at

velocity v(t).

5 Inhomogeneous Advection-Diffusion Initial Value Problem

Some will find the previous derivation of the solution to the ordinary diffusion initial value

problem (8) to be somewhat unsatisfying, as (12) required a clever guess based on keen

observations (i.e. the invariance properties of section 3.2). Indeed, “guessing” and changing

variables are valid techniques which are often used. However, integral transforms can also

prove to be powerful tools in solving partial differential equations, especially inhomogeneous

18

ones. With this in mind, we now turn our attention to the more general inhomogeneous

advection-diffusion initial value problem on x ∈ R, t ∈ R+ given by

ut(x, t) = kuxx(x, t)− v(t)ux(x, t) + f(x, t)

u(x, 0) = φ(x).

(22)

In this equation, the source term f(x, t) accounts for an insertion or extraction of mass

from the system as it evolves with time. Specifically, f(x, t) represents the time-rate of

change of concentration due to external factors, such as a source or a sink. One can derive

the differential equation of (22) by applying the same procedure adopted in deriving the

advection-diffusion equation and introducing an additional flux ~Fs due to the source. That

is, the total flux through a three-dimensional surface is given by ~F = ~Fd + ~Fa + ~Fs, and

substituting this flux into the continuity equation (3) yields

ut = k∇2u− ~v · ∇u−∇ · ~Fs. (23)

Via dimensional analysis, one finds that ∇· ~Fs represents a change in concentration per unit

time; moreover, a flux is positive by convention if mass leaves a region through a surface

surrounding it, so an input of mass into the system (i.e. increase in concentration) may be

modeled by a negative flux through the surface. Therefore, we take f(x, y, z, t) = −∇· ~Fs as

the desired source term, and (23) becomes (after simplifying to one dimension)

ut(x, t) = kuxx(x, t)− v(t)ux(x, t) + f(x, t). (24)

As an introduction to solving the diffusion (heat) initial value problem (8) via Fourier

Transforms, [3] is a quite readable text. Here, we will use such techniques to solve (22),

taking the Fourier Transform of the differential equation and letting U(ω, t) = F [u(x, t)]

19

and F (ω, t) = F [f(x, t)]:

F [ut(x, t)] = kF [uxx(x, t)]− v(t)F [ux(x, t)] + F [f(x, t)]

dU(ω, t)

dt= −kω2U(ω, t)− iωv(t)U(ω, t) + F (ω, t).

(Here i =√−1.) In conjunction, transform the initial condition, letting Φ(ω) = F [φ(x)]:

U(ω, 0) = F [φ(x)] = Φ (ω) .

For now, fix ω and (for simplicity) write U = U(t), F = F (t), and Φ(ω) = Φ so that

dU(t)

dt+[kω2 + iωv(t)

]U(t) = F (t),

F [φ(x)] = Φ.

This is an ODE in t which can be solved via the integrating factor ekω2t+iωα(t), where

α(t) =´ t0v(p) dp. It follows that

d

dt

[ekω

2t+iωα(t) U(t)]

= ekω2t+iωα(t)F (t)

ekω2t+iωα(t) U(t) =

ˆ t

0

ekω2s+iωα(s)F (s) ds+ C.

At t = 0, we have U(0) = C, hence C = Φ. So,

U(t) = e−(kω2t+iωα(t))[ˆ t

0

ekω2s+iωα(s)F (s) ds+ Φ

]=

ˆ t

0

e−kω2(t−s)e−iω[α(t)−α(s)]F (s) ds+ e−kω

2te−iωα(t)Φ.

20

Reintroduce ω and apply the Inverse Fourier Transform:

u(x, t) = F−1 [U(ω, t)] = F−1[ˆ t

0

e−kω2(t−s)e−iω[α(t)−α(s)]F (ω, s) ds

]+ F−1

[e−kω

2te−iωα(t)Φ(ω, t)].

By an application of Fubini’s Theorem, commute the Inverse Fourier Transform with the

integral to obtain

u(x, t) =

ˆ t

0

F−1[e−kω

2(t−s)e−iω[α(t)−α(s)]F (ω, s)]ds

+ F−1[e−kω

2te−iωα(t)Φ(ω, t)].

(25)

Let A(ω, t, s) = e−kω2(t−s)e−iω[α(t)−α(s)] and B(ω, t) = e−kω

2te−iωα(t). Apply the Fourier Con-

volution Theorem in (25):

u(x, t) =

ˆ t

0

ˆ ∞−∞F−1 [A] (x− y, t, s)f(y, s) dy ds

+

ˆ ∞−∞F−1 [B] (x− y, t) φ(y) dy

(26)

Note that A and B are both of the form e−ω2ae−iωb. Using transform tables, we find that

F−1[e−ω

2a]

=1

2√πa

e−x2/(4a) (a > 0)

F−1[e−iωb

]= δ (x− b)

where δ(x) is the Dirac delta function. Thus, applying the Fourier Convolution Theorem to

21

F−1 [A] and F−1 [B] , we have

F−1 [A] =

ˆ ∞−∞

1

2√πk(t− s)

exp

[− (x− z)2

4k(t− s)

]δ (z − [α(t)− α(s)]) dz

=1√

4πk(t− s)exp

[−(x− [α(t)− α(s)])2

4k(t− s)

]

F−1 [B] =

ˆ ∞−∞

1

2√πkt

exp

[−(x− z)2

4kt

]δ (z − α(t)) dz

=1√

4πktexp

[−(x− α(t))2

4kt

].

Recall S(x, t) from (15). Then

F−1 [A] = S(x− [α(t)− α(s)], t− s)

F−1 [B] = S(x− α(t), t).

Substituting these results into (26) gives the final solution to (22):

u(x, t) =

ˆ t

0

ˆ ∞−∞

S(x− y − [α(t)− α(s)], t− s)f(y, s) dy ds

+

ˆ ∞−∞

S(x− y − α(t), t) φ(y) dy.

(27)

It can be shown that (27) indeed satisfies the initial value problem (22), given of course

that these integrals exist. Moreover, notice that the second term of (27) is the solution to

the homogeneous advection-diffusion equation (21). This agrees with ODE intuition that

the solution of the inhomogeneous equation will be the superposition of the homogeneous

solution with a “particular” solution.

22

6 Experiment

6.1 Preliminaries

Here we will experimentally investigate a diffusion process and explore the modeling capa-

bilities afforded by the diffusion and advection-diffusion initial value problems, (8) and (17)

respectively. Of course, these models assume the substances involved obey Fick’s law, an

assumption that is certainly not warranted in all cases. Nevertheless, in devising a method

to quantify a diffusion process, we are already aware of many of the parameters that we

should expect to quantify through experimental data or through inference from this data.

Such parameters include

• the mass concentration u as a function of position x and time t

• the initial concentration distribution u(x, 0) = φ(x)

• the diffusion coefficient k

• the displacement due to advection α(t).

Naturally, obtaining the mass concentration u as a function of position and time is the

most important challenge, as all other items will rely on this data set. For this experiment,

digital image analysis is used to extract a “concentration profile” from successive images

of the diffusion process, each taken at known times. The specifics of this analysis will be

discussed shortly.

It is important that the diffusion process occurs in an approximately one-dimensional

domain if (8) and (17) are to be applicable. In designing the experiment, a 10 mL test tube

with an inner diameter of 1.3 cm was determined to be sufficiently one-dimensional. While

such a test tube still allows some freedom in three dimensions, its diameter is such that the

effects of capillary action will be insignificant, an important feature of the setup.

Moreover, if the diffusion coefficient k and displacement function α(t) are to be deter-

mined, the initial condition φ(x) of the model needs to be “nice.” That is, we must be able

23

to solve (8) in a closed form and use properties of the solution to deduce k and α(t) from the

position-time-concentration data. An initial condition in which there is a distinct interface

between the solvent and the constant-concentration solute is indeed nice enough, and we

will attempt to achieve it. To do so, a viscous solvent is necessary, as a non-viscous one

would permit mixing too easily and prevent an adequate description of the initial condition.

Due to low cost and ease of availability, off-the-toy-shelf “bubble solution” was chosen as

the “viscous” solvent, while a solution of water and blue food coloring was chosen as the

solute. It is not readily known whether the diffusion of water in “bubble solution” behaves

according to Fick’s law, but using the experimental data we will also attempt to conclude

whether or not the nature of diffusion in these substances is Fickian.

6.2 Procedure

6.2.1 Equipment

The following equipment was used to implement the data collection:

• Tripod-mounted SBIGTM Charge Coupled Device (CCD)

• Computer with CCDOPSTM image collection software

• 10 mL test tube of 1.3 cm inner diameter with cork stopper

• Stand with test-tube clamp

• 100 mL graduated cylinder

• Small-diameter glass funnel

• Eye dropper

• Imperial R© Miracle Bubble solution

• Blue food coloring (FD&C Blue #1)

24

6.2.2 Initial Setup

Prepare a dyed water solution by filling the graduated cylinder with 100 mL of water, then

squeezing 14 drops of blue dye into it. Mix the solution thoroughly. This quantity of dyed

water will serve many experiments, as only a few milliliters will be used for each.

In CCDOPS, establish a link to the CCD and configure its internal cooling system to

maintain 10◦ C. Click “Camera Setup” and use the following settings:

• Temperature regulation: active

• Setpoint: 10.00

This will help mitigate the effect of changes in ambient temperature on image contrast. Also

configure the “Auto Grab” feature to collect images at regular intervals. Click “Grab” and

use the following settings:

• Exposure time: 0.050 (seconds)

• Dark frame: none

• Image size: full

• Exposure delay: 0

• Special processing: auto grab.

After confirming these settings, the Auto Grab setup dialog will appear. Use the following

settings:

• Directory: AutoGrab (browse to locate the desired target directory)

• Type: TIFF

• Number of exposures: 100

• Exposure interval 1800 (seconds)

• Per-image notification: none.

The camera will be collecting images at regular 30 minute intervals. Press escape to cancel

this collection.

25

Position the stand and test tube clamp so that the test tube is oriented vertically, and

adjust the position and height of the tripod so that the test tube is large in the field of view.

Use the CCDOPS “Focus” mode to aid this positioning. If necessary, adjust the camera lens

f-stop so that the image contrast is viable. It might be necessary to use a white poster board

(or something similar) as a background to ensure that the lighting is uniform from the top

to the bottom of the image, i.e. that the background is not significantly brighter or darker

anywhere.

6.2.3 Launching the Experiment

In CCDOPS, enable “Focus” mode, as this will become helpful with last-second configura-

tions. Fill a 10 mL test tube approximately half full of bubble solution. If small bubbles

form at the surface, use an eye dropper or pipette to remove them so that the surface is as

“smooth” as possible. Wait a few minutes for residual bubbles to rise and exit the solution.

Secure the test tube in the test tube clamp, oriented vertically. Using an eye dropper and

funnel, very slowly drop the dyed water solution into the test tube. For best results, hold the

funnel so that its tip touches the test tube wall. The trickle of dyed water should begin to

fill the remainder of the test tube, sitting atop the bubble solution without mixing. Gently

swirl the tip of the funnel around the circumference of the test tube while filling so that the

trickle of dyed water is as evenly distributed as possible (see Figure 2). When full, place a

stopper cork in the tube in order to prevent evaporation. At this point, readjust the f-stop

as necessary for viable contrast, and ensure that the background lighting is as uniform as

possible from the top to the bottom of the image. Moreover, be sure that the test tube is

perfectly vertical, not crooked in any way. Finally, click “Grab” in CCDOPS and quickly

verify the settings. Once image collection has begun, 100 images will be taken at 30 minute

intervals. Return in about two days to gather the images for analysis.

26

Figure 2: Initial Test Tube Configuration

6.3 Data Analysis and Modeling

6.3.1 Image Analysis

From the digital images it is possible to quantify concentration by taking “dark” pixels to

mean high concentration and “light” pixels to represent low concentration. Instead of using

standard units for concentration, e.g. “grams per milliliter,” it is much more convenient to

use relative concentrations, with the dyed water being “100% concentrated” and the bubble

solution being “0% concentrated.” Using the Diffusion Image Analysis software [2], a plot of

the relative concentration vs. position, or a concentration profile, could be determined for

each image.

The Diffusion Image Analysis software “Input Directory” should be set to the path where

the experiment images reside, and a path should be given to an “Output Directory” where

the concentration profiles should be stored (see Figure 4). An “Image Interval” of 0.5 (hours)

is used, and three “Smoothing Passes” are sufficient to reduce image “noise.” Set the “Image

Search Pattern” to “AutoGrab*.TIF” so that only images with this file name convention are

considered as input data. A concentration profile is computed by assigning a rectangular

region to the test tube, with the “Rect. X” and “Rect. Y” fields corresponding to the pixel

27

coordinates of the upper-left corner of the rectangle and “Rect. Length” and “Rect. Width”

fields corresponding to the dimensions of this rectangle (see Figure 3). Initially the “Origin”

field is set to zero. In these experiments, diffusion occurs in a downward direction, hence the

downward-pointing radio button for “Direction” should be checked. Clicking the “Generate

Profile” button copies the concentration profile data for the very first image in the data set

(presumed to be taken at time t = 0) to the Windows clipboard.

Figure 3: Initial Profile Rectangle

Using the Vernier LoggerProTM data analysis program, this initial concentration profile

was plotted and is shown in Figure 4. Taking the mean of the high and low concentration

regions provides a calibration for concentration, i.e. what numeric values correspond to

“100% concentration” and “0% concentration.” These values should be entered into the

“Concentration Zero” and “Concentration One” fields of the “Calibration” group. The

resulting concentration profiles will have concentrations ranging in value from zero to one

based on the darkness of the pixels, averaged across each row of the previously defined

28

rectangle. Moreover, the x-value of the inflection point from this initial concentration profile

should be entered in the “Origin” field of the “Initial Profile” group so that the initial

interface between the dyed water and bubble solution is assigned the position x = 0. The

“Pixels/Unit Length” field should be set to a value of one, as calculations were found to be

much more reliable (less rounding error) when working on this scale. Measured values can

be easily transformed into “real units” after the more difficult calculations have been done.

Figure 4: Logger Pro - Initial Profile/Calibration

Once the calibration information has been entered, clicking the “Generate Profiles” but-

ton outputs the concentration profile for each image in the data set, storing it as a text

file containing two columns: x (position) and u (concentration). The “Compute [Auxil-

iary] Data” button calculates the diffusion coefficient and displacement of the concentration

29

profiles, which will be used in section 6.3.3. Lastly, the information obtained from the “Com-

pute [Auxiliary] Data” button can be entered into the “Additional Gnuplot Code” field of

the “PDE Data” group, and clicking the ”Output Images” button generates plots of the con-

centration profiles superposed with the predicted concentration profile as derived in section

6.3.3. See Figure 5 for an example of a full configuration.

Figure 5: Diffusion Image Analysis - Full Configuration

30

6.3.2 “Pure” Diffusion

Applying the ordinary diffusion initial value problem (8) as a model, we will solve the

following:

ut = kuxx

u(x, 0) = φ(x) =

1 if x < 0

0 if x > 0.

(28)

Applying (16), we have

u(x, t) =

ˆ ∞−∞

S(x− y, t)φ(y) dy

=

ˆ 0

−∞

1√4πkt

e−(x−y)2/(4kt) dy.

Substitute r = (x− y)/√

4kt, dr = −dy/√

4kt.

u(x, t) = − 1√π

ˆ x/√4kt

∞e−r

2

dr

=1√π

ˆ ∞x/√4kt

e−r2

dr. (29)

Since´e−r

2dr cannot be further evaluated in terms of elementary functions, we will proceed

to use the error function, defined by

erf (x) =2√π

ˆ x

0

e−r2

dr.

31

See Appendix A for further discussion of this special function. Expressing (29) in terms of

the error function, we have

u(x, t) =1

2

[2√π

ˆ ∞0

e−r2

dr − 2√π

ˆ x/√4kt

0

e−r2

dr

]

=1

2

[2√π·√π

2− erf

(x√4kt

)]=

1

2

[1− erf

(x√4kt

)]. (30)

Figure 6 demonstrates the behavior of this solution with time.

Figure 6: Behavior of (30) with Time

Note that the inflection point of his solution remains fixed at x = 0 for all times. Using the

Diffusion Image Analysis program, a plot of the concentration versus position was extracted

for each image, a few of which are shown in Figure 7. Clearly the inflection points of these

profiles do not remain fixed at x = 0, though the shape of the graph is consistent with

that of (30). The fact that the theoretical curve appears to be a translation of the actual

32

Figure 7: Concentration Profiles at Various Times

concentration profile indicates the possible presence of advection. At this point we will

pursue advection-diffusion as a model instead of ordinary diffusion.

6.3.3 Diffusion with Advection

Here we will assume the advection-diffusion initial value problem (17) as a model, with the

velocity v(t) yet unknown. Equation (21) tells us that the solution of the advection-diffusion

initial value problem is merely a translation of the ordinary diffusion initial value problem.

So, in the case of advection-diffusion, (30) becomes

u(x, t) =1

2

[1− erf

(x− α(t)√

4kt

)](31)

where α(t) =´ x0v(p) dp.

Using this model, we can extract from the concentration profile data the displacement

due to advection, α(t), as well as the diffusion coefficient k. Indeed, the Diffusion Image

33

Analysis software (via the “Compute [Auxiliary] Data” button) computes the discrete points

of α(t) from each image by locating the position in the profile at which the concentration

is closest to 0.5, as it is easy to see in (31) that u(α(t), t) = 1/2. Having located α(t), it is

possible to solve for k in (31):

k =1

4t

[x− α(t)

erf−1 (1− 2u(x, t))

]2

for any x in the concentration profile corresponding to time t. For the sake of simplicity, the

Diffusion Image Analysis software uses the concentration at x = 0 to obtain

k =1

4t

[α(t)

erf−1 (1− 2u(0, t))

]2. (32)

The output of the “Compute [Auxiliary] Data” button is shown in Figure 8. First note

Figure 8: α, k vs. t.

34

that the values of k computed using (32) are very close to the values of k resulting from a

curve fit. This gives confidence that the method is sound. It is a problem, however, that

the value of k is not constant throughout the entire process. This means that the diffusion

does not obey Fick’s law, the core phenomenological assumption underlying the models.

Indeed, many diffusion processes are non-Fickian, and such mixing problems are often quite

complex. Despite these difficulties, we will continue to develop a model which can quantify

the concentration.

Notice the interesting shape of α(t) in Figure 8. The graph is strictly increasing with

time, and as t becomes large enough, the graph is essentially linear; i.e. the advection process

approaches a terminal velocity. One might expect a terminal velocity, as a linear drag model

is often applicable in viscous fluids. As linearity becomes dominant in the graph of α(t),

the plots of k appear to become more and more constant. Indeed, over the last 20 hours of

Figure 8, the value of k changes by only 25%. This perhaps indicates that the process obeys

Fick’s law after an initial “mixing period,” and we will follow this line of reasoning. While

the physics behind the behavior of α(t) is unclear, it is not difficult to empirically determine

a simple function that fits the α vs. t data.

To determine an appropriate function α(t), begin with the description

α(t) = At+Bf(t) (33)

where f(t) is an unknown function satisfying the following properties:

1. f(0) = 0, since it is necessary that α(0) = 0

2. f ′(t) > 0 for all t > 0

3. limt→∞

f(t) = 1, i.e. α(t) is linear for sufficiently large t

4. limt→∞

f ′(t) = 0.

Properties (2) and (3) follow from the observation that α(t) is a strictly increasing function

and that it is desirable for f(t) to asymptotically approach unity. Property (4) is a conse-

35

quence of property (3), but it is helpful to state it separately. Superposing properties (3) and

(4), we obtain limt→∞

[f ′(t) + Cf(t)] = C for real numbers C. In combination with property

(1), this suggests that we solve the initial value problem given by

f ′(t) + Cf(t) = C

f(0) = 0.

The solution f(t) = 1−e−Ct is easily found to a satisfactory function for C > 0. Substituting

into (33), we have

α(t) = At+B(1− e−Ct

)(34)

where parameters A,B, and C are to be determined by a curve-fit to the α vs. time data.

Such a fit is demonstrated in Figure 8. Over three experiments, the following values of

k,A,B, and C were calculated:

Figure 9: Advection-Diffusion Model Data

Steady-State k (cm2/s)Date Computed Curve-Fit A (cm/s) B (cm) C (s−1)

27 February 2010 3.83×10−7 4.19×10−7 3.17×10−6 0.215 7.78×10−5

9 March 2010 3.68×10−7 3.60×10−7 2.66×10−6 0.268 3.67×10−5

15 March 2010 3.83×10−7 6.06×10−7 2.82×10−6 0.255 7.00×10−5

Mean: 3.78×10−7 4.62×10−7 2.88×10−6 0.246 6.15×10−5

Standard Deviation: 8.42×10−9 1.29×10−7 2.58×10−7 0.0278 2.18×10−5

The (computed) diffusion coefficient k is extremely consistent across the experiments,

and the values of A,B, and C are also somewhat consistent. Using these parameters, one

could reasonably predict the distribution of colored water in the bubble solution during the

tested time interval of approximately two days (∼2×105 s). See Figure 10 for a “goodness

of fit” comparison of the model and the data.

36

Figure 10: Advection-Diffusion Model with Data

(a) t = 5 hours (b) t = 20 hours

(c) t = 30 hours (d) t = 40 hours

6.4 Conclusions

The experimental data revealed that the binary mixture of dyed water in “bubble solution”

does not initially behave according to Fick’s law. However, the advection displacement func-

tion α(t) eventually exhibits linear behavior, during which the diffusion coefficient becomes

constant. Hence Fickian diffusion might become an appropriate model after enough time

has passed. Moreover, we find that the model

u(x, t) =1

2

[1− erf

(x− α(t)√

4kt

)]

where k = 3.78 × 10−7cm2/s, α(t) = (2.88 × 10−6)t + 0.246(

1− e−(6.15×10−5)t)

reasonably

approximates the diffusion process for t ∼ 105 s across multiple experiments (see Figure 10).

Indeed, for particularly large times, the predicted concentration profile is almost indistin-

37

guishable from the data.

7 Future Considerations

7.1 Further Research Topics and Applications

This project opens many possibilities for future mathematical research. It would be inter-

esting to investigate boundary value problems in which a finite interval of the real line is

taken as the spatial domain, i.e. modeling a finite container. Of course, models of diffusion

in higher dimensions, while more complex, would likely prove more useful in our three-

dimensional world. Also, it would be interesting to explore diffusion as a stochastic process,

invoking results from probability theory and statistics as opposed to the pure differential

equations approach taken here. There are numerous applications of advection-diffusion in

environmental science (see [5]), and it would be satisfying to apply the advection-diffusion

model to a real-world problem.

7.2 Modeling “Double Diffusion”

Certainly the biggest question arising from this research is that of the physics governing the

diffusion of two different materials into each other; indeed the experiment conducted here

involved dyed water diffusing into bubble solution, and vice versa. The behavior of this

diffusion was modeled as advection-diffusion because a curve could be made to fit the data

quite well, but there is unfortunately little physical motivation for advection as a mode of

mass transport in this system. The “double diffusion” due to the properties of two different

materials (e.g. water, bubble solution) is a much more likely cause of the observed advection-

like behavior. A potential model of this situation deserves some attention.

Suppose we account for the concentration of the two materials separately. Let u(x, t) and

v(x, t) denote the concentrations of species A and B, respectively. We will consider the mass

flux of species A to obey Fick’s law, but with an additional term accounting for the effects

38

of species B. That is, the mass flux of species A will have a term that is proportional to

the negative gradient of the concentration of species A (Fick’s law), and another term that

is proportional to the negative gradient of the concentration of species B. The idea behind

such a model is that as species B diffuses from high to low concentration, its interaction

with species A will induce more flux of species A. The mass flux of species B can be modeled

similarly, and we obtain

FA = −k1ux − k2vx

FB = −k3vx − k4ux

where k1, k2, k3, and k4 are proportionality constants analogous to diffusion coefficients. Sub-

stituting these fluxes into the continuity equation, we achieve a coupled system of partial

differential equations. Moreover, if species A is initially at 100% concentration for x < 0 and

species B is at 100% concentration for x > 0, we finally achieve the following initial value

problem:

ut = k1uxx + k2vxx (x ∈ R, t ∈ R+)

vt = k3vxx + k4uxx

u(x, 0) =

1 if x < 0

0 if x > 0

v(x, 0) =

0 if x < 0

1 if x > 0.

(35)

More investigation would be necessary to determine if (35) is well-posed and has sufficient

initial data to be solved. Thus, a study of systems of partial differential equations would be

an appropriate avenue of future research.

39

A The Error Function

It is sometimes convenient to express solutions of the diffusion equation in terms of the error

function, defined by

erf (x) =2√π

ˆ x

0

e−r2

dr.

For example, in terms of the error function (14) becomes

Q(x, t) =1

2

[1 + erf

(x√4kt

)].

In addition to providing a more compact form, the error function also affords greater ease

of computation since many mathematics/scientific software programs (e.g. gnuplot) contain

a construct for the error function while relatively few can perform integration.

In becoming familiar with this special function, it is easy to verify the following useful

properties:

1. erf (x) is strictly increasing on R (and thus has an inverse)

2. erf (x) is an odd function

3. erf (∞) = 1

4. erf (−∞) = −1

5. ddx

erf (x) = 2√πe−x

2

6. erf (x) changes inflection at x = 0.

40

References

[1] John Rozier Cannon, The one-dimensional heat equation, Addison-Wesley Publishing

Company, Menlo Park, CA, 1984.

[2] Kyle M. Claassen, Senior seminar custom diffusion image analysis software,

http://www.bethelks.edu/academics/undergrad research/lookup project.php?project id=116,

2010.

[3] Stanley J. Farlow, Partial differential equations for scientists and engineers, Dover, New

York, 1993.

[4] Sadri Hassani, Mathematical physics: A modern introduction to its foundations, Springer-

Verlag, New York, 1999.

[5] Scott A. Socolofsky and Gehard H. Jirka, Special topics in mixing and transport processes

in the environment, https://ceprofs.civil.tamu.edu/ssocolofsky/ocenx89/book.html,

2005.

[6] Walter A. Strauss, Partial differential equations: An introduction, John Wiley & Sons,

Inc., Hoboken, NJ, 2008.

41