Models on Conservation Laws

Joaquim MC. Correia ∗and Jayrold P. Arcede

September 5, 2015

Abstract

This talk is all about Conservation Laws models. At the beginning, we conceptualized a toy modelwhere we build a mathematical equation and find its solution.

1 Introduction

We will begin first to talk about matter. What is matter? In highschool physics, matter is defined tobe anything that occupies space. All objects take up space. Your computer is taking up space on thedesk. You are taking up space on the chair. Objects have mass. Mass is how much there is of an object.Mass is related to how much something weighs. But note that mass and weight are two different things.The unit for mass is a gram. For example, a nickel has the mass of about one gram. Objects that takeup space and have mass are called matter1. We also note that the matter of which we, human being,is made of remains the same, whether on Earth or on the moon. Weight is not a good measure of howmuch matter one has. Hence mass, independent of weight, is used to describe the amount of matter.

Now most variable like matter in our case were looked as continuous variable. Here, we will treatmatter to be composed of “points” (which can thought of atoms, molecules, etc) that don’t have mass,otherwise, the aggregation (or sum) of masses would be infinite. Imagine a chair, which has an infinitemass!(??). This is absurd, and that not make sense. But its better to treat matter as continuous variable,seeing in macroscopic view, so that later, we can talk about density.

Now, the idea of looking matter as a continuous variable originate from Hemholtz. In [McCormmach],we quote the following in verbatim:

In the introductory lectures, Helmholtz discussed in general terms the two ways physicisthave of treating bodies. Depending on the problem at hand, they regard bodies either asaggregates of material points or as volume elements filled with matter....In the first set oflectures..., Helmholtz treated the dynamics of discrete material points, or “mass points”. Themass point is a useful concept for solving problems in which the real form and extension of abody can be neglected.

For the description of the motion of a mass point, which is one of the “first and mostimportant tasks of theoretical physics,” the position of the point must be defined by continuousand differentiable functions of time; otherwise, the point could be in two places at the sametime, disappearing here and instantly reappearing there. Such a discontinuity would violatethe identity of the mass point, which derives from a “fundamental law of experience of allnatural phenomena”:matter cannot be created or destroyed....

In the second set of dynamical lectures, Helmholtz elucidated the central concept of con-tinuously distributed masses, comparing it with the concept of discrete mass points. Thedifference between the two is evident from the calculation of density, or the ratio of mass tovolume: within the picture of continuous masses, we can imagine a close volume of diminishing

∗From his Evora Lecture Series1Matter is discontinuous and broken. It is composed of tiny discrete particles called atoms (looking at microscopic level).

These atoms are held together by strong attractive forces called bonds–this is what gives matter its appearance of continuity(from Gideon Ifianayi, Professor of Chemistry).

1

smallness in which the density approaches a limiting value at a given position; by contrast,within the picture of discrete masses, we can imagine a sufficiently small closed volume con-taining only a few mass points, so that in this case it makes no sense to speak of a limitingvalue of the density.

Although the concept of continuously distributed masses corresponds to our direct senseimpressions of sight and touch, Helmholtz cautioned that we cannot conclude that matter isactually continuous. Of the final division of matter we know nothing; all we can do is formhypotheses about it to explain the properties of bodies, mechanical, thermal, and chemical.We generally work with the hypothesis that the final division is into atoms and moleculargroupings of atoms, the main properties of which correspond to the picture of discrete masspoints interacting through characteristic central forces. But when we treat phenomena bymathematically dividing bodies into volume elements that are large compared with molecularseparations, as we do, for example in hydrodynamics and elasticity theory, we then work withinthe picture of continuously distributed masses rather than atomistically structured ones2. Evenhere, Helmholtz noted, there are problems such as dispersion of light in which the simplepicture of continuously distributed masses is insufficient and the molecular hypothesis mustbe invoked.

After introducing the fundamental concepts for treating the dynamics of continuouslydistributed masses, Helmholtz described how we apply mathematics to them. We introducepartial differential equations, the mathematical language of dynamics of systems with morethan one independent variable. We imagine cell walls distributed throughout a continuousbody, dividing it into masses that retain their identity as the body moves as a whole. Withoutmisunderstanding, we may speak of “mass point” in the present case as we can in that of thedynamics of discrete masses. Only the meaning is different here: the “mass point” now has nodefinite mass; it is a shorthand expression for the corner point of a volume element filled withmass. Exactly as in molecular picture, in the picture of continuous masses the positions of themass points can be differentiated with respect to time; since however, these mass points forma continuum, as they do not in the molecular picture, their velocities can be differentiatednot only with respect to time, to construct their acceleration, but also with respect to thethree spatial coordinates. It is this mathematical distinction—the presence here of spatialcoordinates as variables with which the velocities can be differentiated–that is the “essentialcharacteristic of continuously distributed masses in contrast to systems of discrete mass pointsin which time is the only primitive variable”.

2 Setting

Now, consider [x1, x2] ⊆ R, which can be thought of a (homogeneous) medium, maybe some metal rod,etc. with boundary points at x1 and x2. Then we characterized continuous mass in [x1, x2] by somedensity ρ(x, t), where t can be thought of as time and x as spatial variable. Note that x can be a vector.The density ρ(x, t) is a quantity of interest and usually is unknown. In some applications, ρ(x, t) may bethe temperature of a rod, the pressure of a fluid, the concentration of a chemical or a group of cells, orthe density of cars in a traffic setting.

Suppose we interpret ρ(x, t) as mass points in [x1, x2], then the total mass inside [x1, x2] for sometime t can be expressed mathematically as ∫ x2

x1

ρ(x, t)dx. (1)

Now, since ρ(x, t) is a function of time, then two things can happen to our medium: there will be someinstance which temperature rises which make the medium expands, or other times, make the mediumcontracts, that is there will be some amount of density that goes in or go out in the domain [x1, x2]. Thisflow of density created in expanding or contracting of medium is called flux, denoted by q(x, t). This fluxis observable quantity, which is a function of space and time. Flux represents the amount of density that

2emphasis mine.

2

either goes in, q(x1, t), or comes out, −q(x2, t), of the domain [x1, x2]. To set the orientation of the flowof flux we will agree that to the right of x-axis is positive, otherwise, flux would be negative.

Thus, we can express mathematically, the variation in time of the total density in [x1, x2] or thechange with respect to t of ρ(x, t) or the change of mass in time of Equation (1) as the following:

d

dt

∫ x2

x1

ρ(x, t)dx = q(x1, t)− q(x2, t). (2)

Now, we can interpret, Equation (2) as when temperature rises, then the medium will expand. In thiscase, some quantity ρ(x, t) goes out in the boundary. In x1, q(x1, t) will be negative and in x2, q(x2, t) ispositive. Therefore, equation (2) will be negative. This means that there are more ρ(x, t) quantity thatgoes out in the system, otherwise, (2) is positive.

According to (2), the quantity ρ is neither created nor destroyed: the total amount of ρ containedinside any given interval [x1, x2] can change only due to the flow of ρ(x, t) (flux) across the two boundarypoints x1 and x2. That is why, we call ρ(x, t) a conserved quantity. See the diagram below.

Figure 1: Some insulated medium

Thus, we could easily observe that flux, q(x, t) is a function of density ρ, space x and time t, and canbe written in the following useful notation:

q(x, t) := Φ(ρ, x, t). (3)

2.1 Source or sink

Another factor for the change in ρ(x, t) is due to the source or sink, denoted by δ(x, t). This is thequantity of ρ(x, t) introduced in a medium. This quantity is observable. Suppose in our medium, somecreation of mass happens inside [x1, x2]. Then we could sum up the mass creation inside our domain tobe ∫ x2

x1

λ(x, t)dx, (4)

incorporating (4) to (2), we have

d

dt

∫ x2

x1

ρ(x, t)dx = q(x1, t)− q(x2, t) +

∫ x2

x1

λ(x, t)dx. (5)

To give an emphasis, imagine a closed room of two doors where some crevices in the door or window.We can treat the doors as would be our endpoint x1 and/or x2 and some crevices as source or sink.In this case, our density is the amount of heat enters and leaves the room. To monitor the amount ofheat, we suppose that there is some air-conditioning device installed of this room. Then amount of heatentered or escaped from the room will be measured by the air-con device thru its sensor which regulatesthe temperature inside the room.

3

3 On some technicalities

To proceed mathematically, as we want to get a partial differential equation out of equation (5), we seekto answer the following question.

Question 1. What assumptions in the Henstock sense does the equality holds:

d

dt

∫ x2

x1

ρ(x, t)dx

??︷︸︸︷=

∫ x2

x1

∂

∂tρ(x, t)dx. (6)

This is called Leibniz rule in Calculus.Supposing Leibniz rule is true in this some setting, we can express the left hand side of (5) in the

following manner: ∫ x2

x1

∂

∂tρ(x, t)dx = q(x1, t)− q(x2, t) +

∫ x2

x1

λ(x, t)dx. (7)

By some Fundamental Theorem of Calculus and using the notation in Equation (3), we can write theinflow-outflow of q(x, t) in the domain as

q(x1, t)− q(x2, t) =

∫ x1

x2

∂

∂x(Φ(ρ, x, t)) dx = −

∫ x2

x1

∂

∂x(Φ(ρ, x, t)) dx. (8)

So equation (7) becomes∫ x2

x1

∂

∂tρ(x, t)dx = −

∫ x2

x1

∂

∂x(Φ(ρ, x, t)) dx+

∫ x2

x1

λ(x, t)dx. (9)

In the subsequent, it would be useful to have some shorthand notation:Lets denote

∂

∂tρ(x, t) as ρt(x, t);

and∂

∂x(Φ(ρ, x, t)) as Φx(ρ, x, t).

By these notations and some algebra, (9) can be express as∫ x2

x1

(ρt(x, t) + Φx(ρ, x, t)− λ(x, t)) dx = 0 for all x1, x2 ∈ R (10)

By highlighting that the integral in (10) is zero for all x1, x2 ∈ R, by hiding the parameters ρ, x, and t,we can have the following expression∫ x2

x1

(ρt +Φx − λ) dx = 0 for all x1, x2 ∈ R. (11)

Furthermore, we can write (11) as ∫R(ρt +Φx − λ)χ[x1,x2](x)dx = 0. (12)

where

χ[x1,x2](x) =

{1 if x ∈ [x1, x2]

0 if x /∈ [x1, x2].

Now, Equation (12) can only mean one thing, that is, the integrand should be zero. Therefore,

ρt +Φx − λ = 0, (13)

or in terms of the flux function q, we have the following expression

4

ρt + qx − λ = 0. (14)

Equations (14) or (13) is the partial differential equation (pde) that we are interested about. We wantto find the solution ρ that satisfies this pde.

Consider the following example.

Example 3.1. Suppose in (14), q(x, t) = ρ2(x,t)2 with no source or sink, i.e, λ = 0. Then

∂

∂xq(x, t) =

∂

∂x

ρ2(x, t)

2= ρ · ρx.

So (14) becomesρt + ρ · ρx = 0. (15)

What should be the density ρ, which is the solution of equation (15)? The conservation law in (15) iscalled the inviscid Burgers’ equation.

We want to remark that sometimes we get the derivative of ρ with respect to x, ρx, and not the usualderivative of ρ with respect to t, i.e., ρt. In this case, we could interpret ρx as a transfer of density withrespect to x. Furthermore, if we take another derivative of ρx, which is ρxx, then we have diffusion, sincediffusion equations involve a second order derivative with respect to space.

To make it clear, consider the flux Φ (ρx, ρ, x, t) which now a function of ρx in addition to ρ, x, andt. By differentiating q with respect to x, we have

∂

∂xΦ(ρx, ρ, x, t) = Φx (ρx, ρ, x, t) = ρxx ·D for some coefficient D (16)

Then equation (14) can be written as

ρt +Φx − λ = 0 ⇐⇒ ρt +Dρxx = λ. (17)

If λ = 0, then (17) can be written in the following manner:

∂ρ

∂t= −D∂2ρ

∂x2(18)

Equation (18) is called diffusion equation for some particle density ρ and D > 0 is called a diffusioncoefficient.

3.1 On some general form of flux and source/sink

Now, recall the form of flux q(x, t) = Φ(ρ(x, t), x, t) and suppose the source or sink λ(x, t) = Ψ(ρ, x, t),that is, both are function of density ρ(x, t), which was not considered previously. Then equation (14) canbe written as

∂tρ+ ∂xΦ(ρ(x, t), x, t) = λ⇐⇒ ∂

∂tρ+

chain rule︷ ︸︸ ︷(∂

∂ρΦ · ∂

∂xρ+

∂

∂xΦ · 1

)= λ

⇐⇒ ∂tρ+ (Φρρx +Φx) = λ

⇐⇒ ρt +Φρρx = λ− Φx

(19)

which can also be written asρt + c(ρ, x, t)ρx(x, t) = g(ρ, x, t) (20)

wherec(ρ, x, t) = Φρ(ρ, x, t)

andg(ρ, x, t) = λ(ρ, x, t)− Φx(ρ, x, t). (21)

5

Now, we have some caveat with regards to the above equations (19) and (14). And we feel stronglythat this should be taken with extra care. We have to be vigilant in the following manner: if λ = 0,equation (14) becomes

ρt + qx = 0 a homogeneous pde. (22)

However, equation (20) is not a homogeneous pde for when λ = 0, g = −Φx in (21).Hence,

ρt + cρx = g ̸= 0. (23)

Thus, an extra care should be taken when considering such forms of pde.To put more emphasis, common mistake would be like in the following. Consider the following equation

ρt + qx = λ. (24)

Then equation (24) is a generalization of

ρt +Φρρx = 0,

which is false! For one to not be able to commit this blunder, one should go back to the modelization ofthese equations to get things clear.

4 Differentiation under Integral sign

Before we go further, let’s address the Question 1 that was posed previously.The question is: On what assumptions does the the following equation holds, in the sense of Henstock:

d

dt

∫ x2

x1

ρ(x, t)dx =

∫ x2

x1

∂

∂tρ(x, t)dx. (25)

The rule: the t-derivative of the integral of ρ(x, t) is the integral of the t-derivative of ρ(x, t), is calleddifferentiation under the integral sign. This is what equation (25) is all about.

To answer the question we will see how it works in Riemann sense, in Lebesgue and finally on Henstockintegration. In this manner, we will have a good grasp and foundation of the theory. To accomplish this,we shall gives examples and counterexamples along the way.

To give a brief background, the method of differentiation under the integral sign is due to Leibniz. Itconcerns integrals depending on a parameter, such as∫ 1

0

x2etxdx. (26)

Here t is the extra parameter. (Since x is the variable of integration, x is not a parameter.) In light of(25), we might write such an integral as ∫ b

a

ρ(x, t)dx, (27)

where ρ(x, t) is a function of two variables like ρ(x, t) = x2etx. The integral in (26), is called a gammafunction, which is a generalization of a factorial, it can be solved using integration by parts but usingdifferentiation under integral sign is also quite elegant, we refer the reader to the excellent exposition ofKeith Conrad [Conrad], where our first example is due to him. So we begin.

Example 4.1. Let ρ(x, t) = (2x+ t3)2. Then∫ 1

0

ρ(x, t)dx =

∫ 1

0

(2x+ t3)2dx.

6

An anti-derivative of (2x+ t3)2 with respect to x is 16 (2x+ t3)3, so∫ 1

0

(2x+ t3)2dx =1

6(2x+ t3)3

∣∣∣∣x=1

x=0

=(2 + t3)3 − t9

6

=4

3+ 2t3 + t6.

(28)

This answer is a function of t, which makes sense since the integrand depends on t. We integrate over xand are left with something that depends only on t, not x.

An integral like∫ b

aρ(x, t)dx is a function of t, so we can ask about its t-derivative, assuming that ρ(x, t)

is “nicely behaved”. In this example, the integral is nicely behaved in the since that (28) is a continuousfunction, so no question that we can get the t-derivative. Now, we will compute the t-derivative of (28):

d

dt

∫ 1

0

(2x+ t3)2dx =d

dt

4

3+ 2t3 + t6

=6t2 + 6t5.

(29)

Now, to check if (25) is true, we also compute the t-derivative of the integrand and then get the integralwrt x and we have

∫ 1

0

∂

∂t(2x+ t3)2dx =

∫ 1

0

2(2x+ t3)(3t2)dx

=6t2 + 6t5,

(30)

which agrees to equation (29).If we are used to thinking mostly about functions with one variable, not two, keep in mind that (25)

involves integrals and derivatives with respect to separate variables: integration with respect to x anddifferentiation with respect to t.

We have seen in the example above where differentiation under the integral sign can be carried outflawlessly, but we have not actually stated conditions under which (25) is valid. Something does needto be checked. In [Talvila2], an incorrect use of differentiation under the integral sign due to Cauchy isdiscussed, where a divergent integral is evaluated as a finite expression. Here are two other exampleswhere differentiation under the integral sign does not work. We begin with the following counterexample.

Example 4.2. It is pointed out in [Goel, Example 6] that the formula∫ ∞

0

sinx

xdx =

π

2,

leads to an erroneous instance of differentiation under the integral sign. To see this, rewrite the formulaas ∫ ∞

0

sin(ty)

ydy =

π

2(31)

for any t > 0, by the change of variables x = ty. Then differentiation under the integral sign implies∫ ∞

0

cos(ty)dy = 0,

which doesn’t make sense.

The next example shows that even if both sides of (25) make sense, they need not be equal.

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Example 4.3. For any real numbers x and t, let

ρ(x, t) =

{xt3/(x2 + t2)2, if x ̸= 0 or t ̸= 0,

0, if x = 0 and t = 0.(32)

Let

F (t) =

∫ 1

0

ρ(x, t)dx.

For instance, F (0) =

∫ 1

0

ρ(x, 0)dx =

∫ 1

0

0dx = 0. When t ̸= 0,

F (t) =

∫ 1

0

xt3/(x2 + t2)2dx

=

∫ 1+t2

x2

t3

2u2dx (where u = x2 + t2)

=t

2(1 + t2).

(33)

This formula also works at t = 0, so F (t) = t/(2(1 + t2)) for all t. Therefore, F (t) is differentiableand

F ′(t) =1− t2

2(1 + t2)2

for all t. In particular, F ′(0) = 12 .

Now we compute ∂∂tρ(x, t) and then

∫ 1

0

∂

∂tρ(x, t)dx. Since ρ(0, t) = 0 for all t, ρ(0, t) is differentiable

in t and ∂∂tρ(0, t) = 0. For x ̸= 0, ρ(x, t) is differentiable in t and

∂

∂tρ(x, t) =

xt2(3x2 − t2)

(x2 + t2)3.(34)

Combining both cases (x = 0 and x ̸= 0),

∂

∂tρ(x, t) =

{xt2(3x2−t2)(x2+t2)3 , if x ̸= 0

0, if x = 0(35)

In particular, ∂∂t

∣∣∣t=0

ρ(x, t) = 0. Therefore at t = 0 the left side of equation (25) is F ′(0) = 1/2 and

the right side is

∫ 1

0

∂

∂t

∣∣∣t=0

ρ(x, t)dx = 0. The two sides are unequal!

The problem in this example is that ∂∂tρ(x, t) is not a continuous function of (x, t). Indeed, the

denominator in the formula in (35) is (x2 + t2)3, which has a problem near (0, 0). Specifically, while thisderivative vanishes at (0, 0), it we let (x, t) → (0, 0) along the line x = t, then on this line ∂

∂tρ(x, t) hasthe value 1/(4x), which does not tend to 0 as (x, t) → (0, 0).

We posed the following theorem.

Theorem 4.1. The equationd

dt

∫ b

a

ρ(x, t)dx =

∫ b

a

∂

∂tρ(x, t)dx. (36)

is valid at t = t0, in the sense that both sides exist and are equal, provided the following two conditionshold:

• ρ(x, t) and ∂∂tρ(x, t) are continuous functions of two variables when x is in the range of integration

and t is in some interval around t0,

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• there are upper bounds |ρ(x, t)| ≤ A(x) and | ∂∂tρ(x, t)| ≤ B(x), both being independent of t, such

that∫ b

aA(x)dx and

∫ b

aB(x)dx exist

To understand it better, we will quote verbatim Chapter XIII section 3 of Serge Lang UndergraduateAnalysis book.

4.1 Integration under integral sign: Riemann Sense

Theorem 4.2. Let f be a continuous function of two variables (t, x) defined for t ≥ a and x in somecompact set of numbers S. Assume that the integral∫ ∞

a

f(t, x)dt = limB→∞

∫ B

a

f(t, x)dt

converges uniformly3 for x ∈ S. Let

g(x) =

∫ ∞

a

f(t, x)dt.

Then g is continuous.

Proof. For given x ∈ S we have

g(x+ h)− g(x) =

∫ ∞

a

f(t, x+ h)dt−∫ ∞

a

f(t, x+ h)dt by definition

=

∫ ∞

a

f(t, x+ h)− f(t, x+ h)dt since integrals is linear on continuous functions.

(37)

Given ϵ, select B such that for all y ∈ S we have∣∣∣∣∫ ∞

B

f(t, y) dt

∣∣∣∣ < ϵ

Then by triangle inequality

|g(x+ h)− g(x)| ≤

∣∣∣∣∣∫ B

a

f(t, x+ h)− f(t, x)

∣∣∣∣∣+∣∣∣∣∫ ∞

B

f(t, x+ h)

∣∣∣∣+ ∣∣∣∣∫ ∞

B

f(t, x)

∣∣∣∣ . (38)

We know that f is uniformly continuous on the compact set [a,B] × S. Hence there exists δ such thatwhenever |h| < δ we have

|f(t, x+ h)− f(t, x)| ≤ ϵ/B.

The first integral on the right is then estimated by Bϵ/B = ϵ. The other two are estimated each by ϵ, sowe have a 3ϵ-proof for the theorem.

We shall now prove a special case of the theorem concerning differentiation under the integral signwhich is sufficient for many applications, in particular those of the next chapter. It may be called theabsolutely convergent case.

Theorem 4.3. Let f be a function of two variables (t, x) defined for t ≥ a and x in some intervalJ = [c, d], c < d. Assume that D2f exists, and that both f and D2f are continuous. Assume that thereare functions φ(t) and ψ(t) which are ≥ 0, such that |f(t, x)| ≤ φ(t) and

|D2f(t, x)| ≤ ψ(t),

3A sequence of functions {fn(x)} with domain D converges uniformly to a function f(x) if given any ϵ > 0 there is apositive integer N such that |fn(x)− f(x)| ≤ ϵ for all x ∈ D whenever n ≥ N . (Please note that the above inequality musthold for all x in the domain, and that the integer N depends only on ϵ.

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for all t, x and such that the integrals∫ ∞

a

φ(t) dt and

∫ ∞

a

ψ(t) dt

converge. Let

g(x) =

∫ ∞

a

f(t, x) dt.

Then g is differentiable, and

Dg(x) =

∫ ∞

a

D2f(t, x) dt.

Proof. We have∣∣∣∣g(x+ h)− g(x)

h−∫ ∞

a

D2f(t, x)dt

∣∣∣∣ ≤ ∫ ∞

a

∣∣∣∣f(t, x+ h)− f(t, x)

h−D2f(t, x)

∣∣∣∣ dt. (39)

Butf(t, x+ h)− f(t, x)

h−D2f(t, x) = D2f(t, ct,h)−D2f(t, x).

Select B so large that ∫ ∞

B

ψ(t) dt < ϵ.

Then we estimate our expression by ∫ ∞

a

=

∫ B

a

+

∫ ∞

B

.

Since D2f is uniformly continuous on [a,B]× [c, d], we can find δ such that whenever |h| < δ,

|D2f(t, ct,h)−D2f(t, x)| <ϵ

B.

The integral between a and B is then bounded by ϵ. The integral between B and ∞ is bounded by 2ϵbecause ∣∣∣∣f(t, x+ h)− f(t, x)

h−D2f(t, x)

∣∣∣∣ ≤ 2ψ(t).

This proves our theorem.

Above theorem is the elementary calculus version, that is, in Riemann sense. We will give anotherversion in Lebesgue sense. Our source is [Swartz].

4.2 Integration under integral sign: Lebesgue Sense

First and foremost, we want to say something why people move out from Riemann to a powerful enoughintegral like Lebegue. While the Riemann integral enjoys many desirable properties, it also has severalshortcomings. One of these shortcomings concerns the fact that a general form of the FundamentalTheorem of Calculus does not hold for Riemann integrable functions. Another serious drawback is thelack of ”good” convergence theorems for the Riemann integral. A convergence theorem for an integralconcerns a sequence of integrable functions {fk}∞k=1 which converge in some sense, such a pointwise, to alimit function f and involves sufficient conditions for interchanging the limit and the integral, that is toguarantee limk

∫fk =

∫limk fk. In modern integration theories, the standard convergence theorems are

the Monotone Convergence Theorem, in which the functions converge monotonically, and the BoundedConvergence Theorem, in which the functions are uniformly bounded. These convergence theorems arevalid in the case of Lebesgue, which is also true for Henstock-Kurzweil integral.

Here, we will list as a recall of what we already know the convergence theorems like, monotone,bounded and dominated in the sense of Lebesgue.

We denote the collection of measurable subsets of Rn by Mn.We state the monotone convergence theorem.

10

Theorem 4.4 (Monotone Convergence Theorem). Let E ∈ Mn and {fk}∞k=1 be an increasing sequenceof nonnegative, measurable functions defined on E. Set f(x) = limk→∞ fk(x). Then,

limk→∞

∫E

fk =

∫E

f

For Lebesgue integrable functions, we can get an improvement of the Monotone Convergence Theorembelow.

Corollary 4.1. Let E ∈ Mn and {fk}∞k=1 be an increasing sequence of nonnegative, Lebesgue inte-grable functions defined on E. Set f(x) = limk→∞ fk(x). Then, f is Lebesgue integrable if, and only if,supk

∫Efk <∞. In this case, f is finite a.e..

The Monotone Convergence Theorem is a very useful tool in analysis. However, in many situations, themonotonicity condition is not satisfied by a convergent sequence and other conditions, which guarantee theexchange of the limit and the integral are desirable. We next consider Lebesgue’s Dominated ConvergenceTheorem. This result replaces the monotonicity condition of the Monotone Convergence Theorem by therequirement that the convergent sequence of functions be bounded by a Lebesgue integrable function. Asa corollary of the Dominated Convergence Theorem, we will get the Bounded Convergence Theorem.

Theorem 4.5 (Dominated Convergence Theorem). Let {fk}∞k=1 be a sequence of measurable functionsdefined on a measurable set E. Suppose that {fk}∞k=1 converges to f pointwise almost everywhere andthere is a Lebesgue integrable function g such that |fk(x)| ≤ |g(x)| for all k and almost every x ∈ E.Then, f is Lebesgue integrable and

limk→∞

∫E

fk =

∫E

f

Moreover,

limk→∞

∫E

|fk − f | = 0

If the measure of E is finite, then constant functions are Lebesgue integrable over E. From theDominated Convergence Theorem we get the Bounded Convergence Theorem.

Theorem 4.6 (Bounded Convergence Theorem). Let {fk}∞k=1 be a sequence of measurable functions ona set E of finite measure. Suppose there is a number M so that |fk(x)| ≤M for all k and for almost allx ∈ E. If f(x) = limk→∞ fk(x) almost everywhere, then

limk→∞

∫E

fk =

∫E

f

Now, we the help of DCT, we will apply it to continuity of integrals and “differentiation under integralsign”.

We start with some settings.Let S ⊆ Rn, T ⊆ Rm and f : S × T → R. If f(s, ·) : T → R is integrable for every s ∈ S and

F : S → R is defined by F (s) =∫Tf(s, t)dt, we say that the integral F (s) depends on the parameter s.

We consider how the properties of f are inherited by F.We first consider the continuity of integrals depending on parameters.

Theorem 4.7. Assume f(s, ·) is integrable for every s ∈ S and f(·, t) is continuous at s0 ∈ S for everyt ∈ T. If there exists an integrable function g : T → R such that |f(s, t)| ≤ g(t) for all s ∈ S, t ∈ T , thenF (s) =

∫Tf(s, t)dt is continuous at s0.

Proof. Let {sk}∞k=1 be a sequence from S converging to s0. By the continuity assumption, f(sk, t) →f(s0, t). Since |f(sk, ·)| ≤ g(·), the Dominated Convergence Theorem implies that

F (sk) =

∫T

f(sk, t)dt→ F (s0) =

∫T

f(s0, t)dt

and F is continuous at s0, as we wished to show.

11

Our second application applies to “differentiating under the integral sign”.

Theorem 4.8 (Leibniz’ Rule). Let S = [a, b] and let I be a closed interval. Suppose that

1. ∂f∂s = D1f exist for all s ∈ S and t ∈ I;

2. each function f(s, ·) is integrable over I and there is an integrable function g : I → R such that∣∣∣∣∂f(s, t)∂s

∣∣∣∣ = |D1f(s, t)| ≤ g(t),

for all s ∈ S and t ∈ I.

Then, F (s) =∫If(s, t) dt is differentiable on S and

F ′(s) =

∫I

∂f(s, t)

∂sdt =

∫I

D1f(s, t)dt (40)

Proof. Fix s0 ∈ S and let {sk} be a sequence in S converging to s0 with sk ̸= s0. For t ∈ I, define {hk}by

hk(t) =f(sk, t)− f(s0, t)

sk − s0

and note that hk is integrable over I. By the Mean Value Theorem, for each pair (k, t) there is a z(k,t)between sk and s0 so that

f(sk, t)− f(s0, t)

sk − s0= D1f(z(k,t), t)

which implies that

|hk(t)| =∣∣∣∣f(sk, t)− f(s0, t)

sk − s0

∣∣∣∣ ≤ g(t).

The Dominated Convergence Theorem implies that

F ′(s0) = limk→∞

F (sk)− F (s0)

sk − s0

= limk→∞

∫I

f(sk, t)− f(s0, t)

sk − s0dt

=

∫I

D1f(s0, t)dt

as we wished to show.

4.3 Integration under integral sign: Henstock Sense

In this section, we will learn how Henstock differ with Lebesgue in terms of convergence theorems andits applications to differentiation under integral sign. In this case we follow the exposition by Swartz[Charles].

It can be noted that the principal reason that the Lebesgue integral is favored over the Riemannintegral is the fact that convergence theorems of the form lim

∫Ifk =

∫I(lim fk) hold for the Lebesgue

integral under very general conditions. The major convergence theorems of this type are the MonotoneConvergence Theorem (MCT) and the Dominated Convergence Theorem (DCT). Here we found thatthese same major convergence theorems hold for the gauge or the HK integral, showing that the HKintegral enjoys the same advantages over the Riemann integral as the Lebesgue integral.

We begin by introducing the concept of uniform integrability. This is at the center of the convergencetheorems for the HK integral. Lets do some notation: let I be a closed interval (bounded or unbounded)in R∗4 and fk, f : I → R for k ∈ N. Further, let I be the family of all closed subintervals of I.

4This means that R is extended by adding the two points at infinity, ±∞

12

Definition 4.1. {fk} is uniformly integrable over I if each fk is integrable over I and if for every ϵ > 0there exists a gauge5 γ on I such that ∣∣∣∣S(fk,D)−

∫I

fk

∣∣∣∣ ≤ ϵ

for every k ∈ N whenever D << γ.6

The point of this Definition is, of course, that the same gauge works uniformly for all k. For uniformlyintegrable sequences of integrable functions we have the following convergence theorem.

Theorem 4.9. Let {fk} be uniformly integrable over I and assume that fk → f pointwise. Then f isintegrable over I and lim

∫Ifk =

∫If(=

∫I(lim fk)

).

Now we straight to Monotone Convergence Theorem.

Theorem 4.10 (Monotone Convergence Theorem: MCT). Let fk : I → R be integrable over R andsuppose that fk(t) ↑ f(t) ∈ R for every t ∈ R. If supk

∫Ifk <∞, then

1. {fk} is uniformly integrable over I,

2. f is integrable over I and

3. lim∫Ifk =

∫If(=

∫I(lim fk)

).

The Monotone Convergence Theorem gives a very useful and powerful sufficient condition guaranteeing“passage to the limit under the integral sign”, i.e., lim

∫Ifk =

∫I(lim fk), but the monotone convergence

requirement is often not satisfied.Thus, another convergence result called the Dominated ConvergenceTheorem, which relaxes this requirement.

Theorem 4.11 (Dominated Convergence Theorem: DCT). Let fk, f, g : I → R and assume fk, g areintegrable over I with |fi − fj | ≤ g for all i, j. If lim fk = f pointwise on I, then

1. {fk} is uniformly integrable over I,

2. f is integrable over I and

3. lim∫Ifk =

∫If(=

∫I(lim fk)

).

Remark 4.1. The ”usual dominating hypothesis” in the DCT for the Lebesgue integral is that thereexists an integrable function g such that |fj | ≤ g for all j. Since |fi − fj | ≤ |fi| + |fj | , this hypothesisclearly implies the one in Theorem 4.11 [see also Exercise 4.1]. On the other hand, the hypothesis|fi − fj | ≤ g allows the functions fj to be conditionally integrable (just take any conditionally integrablefunction h and consider the sequence {fj + h}) and is more appropriate for a conditional integral likethe HK integral. Exercise 4.2 gives another equivalent way of phrasing the dominating hypothesis inTheorem 4.11.

Exercise 4.1. Let fk : I → R be integrable over I and fk → f pointwise. Suppose there exists anintegrable function g such that |fk| ≤ g for all k. Show that in this case the conclusion of the DCT canbe improved to read

∫I|fk − f | → 0.

Exercise 4.2. Let fk : I → R be integrable over I. Show that there exists an integrable functiong : I → R satisfying |fk − fj | ≤ g for all k, j if and only if there exist integrable functions g and hsatisfying g ≤ fk ≤ h for all k.

5Any function γ defined on I such that γ(t) is an open interval containing t for each t ∈ I is called a gauge on I6D = {(ti, Ii)} is δ-fine, i.e., ti ∈ Ii ⊂ γ(ti).

13

We now give some examples illustrating the use of the convergence theorems. For this we derive someresults pertaining to integrals which depend upon parameters. Let I be a closed subinterval of R∗ and Sa metric space (substitute R or Rn for S if necessary). If f : S × I → R, we write

f(s, ·) for the function t→ f(s, t) when s ∈ S

andf(·, t) for the function s→ f(s, t) when t ∈ T

If f(s, ·) is integrable over I for each s ∈ S, then F (s) =∫If(s, t) dt defines a function F : S → R.

Functions which depend upon parameters in this way arise in many areas of mathematics, and itis often important to ask if properties of the function f(·, t) are inherited by the function F. We firstconsider the property of continuity.

Theorem 4.12. If f(·, t) is continuous at s0 ∈ S for each t ∈ I and f(s, ·) is integrable over I foreach s ∈ S and there is an integrable function g : I → R such that |f(s, t)| ≤ g(t) for all t ∈ I, thenF (s) =

∫If(s, t)dt is continuous at s0.

Proof. Let sk → s0 in S. Then f(sk, t) → f(s0, t) and |f(sk, t)| ≤ g(t) for every t ∈ I. The DCT impliesthat F (sk) → F (s0) so F is continuous at s0.

For our next application of DCT, we require a version of Leibniz’ Rule for “differentiating under theintegral sign.” Note that when we say integrable, it means that it is HK integrable. There is no differencein the statement between the Leibniz Rule in Lebegue and Henstock. The only things that matter isthat, in Henstock, FTC holds in full generality, and no improper integral anymore.

Theorem 4.13 (Leibniz’ Rule). Let S = [a, b] and let I be a closed interval. Suppose that

1. ∂f∂s = D1f exist for all s ∈ S and t ∈ I;

2. each function f(s, ·) is integrable over I and there is an integrable function g : I → R such that∣∣∣∣∂f(s, t)∂s

∣∣∣∣ = |D1f(s, t)| ≤ g(t),

for all s ∈ S and t ∈ I.

Then, F (s) =∫If(s, t) dt is differentiable on S and

F ′(s) =

∫I

∂f(s, t)

∂sdt =

∫I

D1f(s, t)dt (41)

Proof. Fix s0 ∈ S and let {sk} be a sequence in S converging to s0 with sk ̸= s0. For t ∈ I,

limk

f(sk, t)− f(s0, t)

sk − s0= D1f(s0, t) =

∂f

∂s(s0, t)

and the function {hk} by

hk(t) =f(sk, t)− f(s0, t)

sk − s0

is integrable over I. By the Mean Value Theorem, for each pair (k, t) there is a z(k,t) lying between skand s0 so that

f(sk, t)− f(s0, t)

sk − s0= D1f(z(k,t), t)

which implies that

|hk(t)| =∣∣∣∣f(sk, t)− f(s0, t)

sk − s0

∣∣∣∣ ≤ g(t).

14

The Dominated Convergence Theorem implies that

F ′(s0) = limk→∞

F (sk)− F (s0)

sk − s0

= limk→∞

∫I

f(sk, t)− f(s0, t)

sk − s0dt

=

∫I

D1f(s0, t)dt,

that is, F ′(s0) =∫ID1f(s0, t)dt as desired.

If I is a bounded interval and D1f is continuous over S × I, then D1f is bounded and the function gin Theorem 4.13 can be taken to be a constant. We use this observation in the next example.

Example 4.4. We show A =∫∞0e−t2dt =

√π2 . For this we introduce

F (s) =

∫ 1

0

e−x(1+t2)

(1 + t2)dt.

Note that F (0) = arctan 1 = π/4, and if x > 0

0 ≤ F (x) ≤∫ 1

0

e−x 1

(1 + t2)dt = e−xπ

4.

soF (∞) = lim

x→∞F (x) = 0.

Fix x. We have, by Leibniz’ Rule,

F ′(x) = −e−x

∫ 1

0

e−xt2dt = −(e−x

√x

)∫ √x

0

eu2

du = −(e−x

√x

)g(√x),

where g(z) =∫ z

0e−u2

du. Integrating F ′ from 0 to ∞ gives

−π4= −

∫ ∞

0

(e−x

√x

)g(√x)dx

= −2

∫ ∞

0

ez2

g(z)dz

= −2

∫ ∞

0

g′(z)g(z)dz

= −g(∞)2

= −A2.

(42)

Hence, A =√π2 .

5 Justification of “Conservation law” term

In this section, we shall go back to the topic in conservation laws. Specifically, we want to justify theterm “conservation law”. So we recall the following conservation law:

d

dt

∫ x2

x1

ρ(x, t)dx = q(x1, t)− q(x2, t) +

∫ x2

x1

λ(x, t)dx. (43)

which is the variation in time of some macroscopic quantity given by ρ(x, t), and which has the followingpartial differential equation in divergence form:

ρt + qx = λ (44)

15

where ρ, q and λ are regular enough.Suppose we are observing a physical phenomena where a matter is situated on a bounded set where

no flux coming in and out of the this set. For simplicity, we also suppose that there is no sink or source,that is, λ = 0. See figure below.

Figure 2: Matter on a bounded set

Thus, for x→ ∞, q(x, t) → 0 similary, x→ −∞, q(x, t) → 0, for all time t ≥ 0. This means that

d

dt

∫ ∞

−∞ρ(x, t)dx = q(∞, t)︸ ︷︷ ︸

0

− q(−∞, t)︸ ︷︷ ︸0

+

0︷ ︸︸ ︷∫ ∞

−∞λ(x, t)dx

Therefore,d

dt

∫ ∞

−∞ρ(x, t)dx = 0

This means that ∫Rρ(x, t)dx = C

for some constant C, for all time t ≥ 0. This means that mass is constant for all time, that is, mass isconserve.

5.1 Some notes regarding the conservation law equation in divergence form

We remark that equation (44) a pde in divergence form is a 1-equation in 1-variable if q = Φ(ρ, x, t).Now, if q = Φ̃(ρ) where space varible x and time variable t are not explicit, it would still imply andunderstood as q = Φ̃(ρ) = Φ̃(ρ, x, t) since ρ is a function of x and t.

Moreover, ifq = Φ(ρx, ρ, x, t)

orq = Φ(T, ρx, ρ, x, t)

where T is a fixed but arbitrary temperature, then we could write equation (44) as

ρt + (Φ)x = λ. (45)

Now, suppose that q = Φ(ρ, x, t). Then by chain rule, we can write

(Φ(ρ, x, t))x =∂Φ

∂ρ

∂ρ

∂x+∂Φ

∂x· 1 + 0 = Φρρx +Φx.

Thus, we can write equation (44) asρt +Φρρx +Φx = λ (46)

16

equivalently

ρt + c(ρ, x, t)ρx = g(ρ, x, t) (47)

wherec(ρ, x, t) = Φρ(ρ, x, t)

which can be thought of signal speed of flow and

g(ρ, x, t) = λ(ρ, x, t)− Φx(ρ, x, t). (48)

Now, this time we note that flow of speed

v(x, t) =q(x, t)

ρ(x, t)

is not equal or similar to the signal speed of flow.So, if in equation (47) g ≡ 0, then

ρt + c(ρ, x, t)ρx = 0. (49)

Also, since g ≡ 0 we can haveλ(ρ, x, t) = Φx(ρ, x, t).

What does this means physically (source/sink= change in flux wrt x)?Suppose in (48), there is no source/sink. Then 0 ≡ g ≡ −Φx(ρ, x, t). So we can write (47) as

ρt + cρx = 0 (50)

wherec = Φρ(ρ, x, t),

note that c is dependent on spatial variable x. The presence of x in equation (50) has some relevance.This implies that equation (50) can be of the form

ρt + h(ρ, x, t)ρx = 0 for some h(ρ, x, t) regular enough (51)

which is a quasi-linear pde or

ρt + h(x, t)ρx = 0 for some h(x, t) regular enough (52)

which is a linear pde. These pde’s are not conservation laws (Am I getting this right?). We cannot provethat

d

dt

∫ ∞

−∞ρ(x, t)dx = 0.

5.2 Dispersive-Diffusive Conservation law

Here we go back to the previous subsection where we assume that

q = Ψ(ρx, ρ, x, t)

and as a consequence, we could write equation (44) as

ρt + (Ψ)x = λ. (53)

By chain rule

(Ψ)x = (Ψ(ρx, ρ, x, t))x = Ψρρxx +Ψρρx +Ψx

1︷︸︸︷xx .

Then

ρt + cρx = g (54)

17

wherec = Ψρ and g = λ−Ψρρxx −Ψx

or equivalently

ρt + cρx = g − aρxx (55)

wherec = Ψρ, g = λ−Ψx, and a = Ψρ.

Equation (55) is a parabolic conservation law.Suppose

q = Ψ(ρx, ρ, x, t) := ϕ(ρ, x, t)− ϵρx.

Thenqx = [ϕ(ρ, x, t)]x − [ϵρx]x = ϕρρx + ϕx − ϵρxx.

Then equation (44) becomes

ρt + [ϕρρx + ϕx − ϵρxx] = λ. (56)

that is,ρt + ϕρρx = λ− ϕx + ϵρxx. (57)

which we can write asρt + cρx = g + ϵρxx. (58)

wherec = ϕρ, and g = λ− ϕx.

If g ∼ 0, then equation (58) is a nonlinear diffusion equation. Its a conservation law but not hyperbolic.Moreover, if we have the following pde:

ρt + cρx = g + ϵρxx + δρxxx. (59)

Then equation (59) is a dissipative-dispersive equation, where

1. second term “cρx” is transport

2. “ϵρxx” is a dissipation term also called diffusive term

3. “δρxxx” dispersive term or oscillation.

6 Traffic flow

In this section, we shall begin with an intuition on the units of the quantities we are studying.To understand the unit of quantities we are studying, we will give an explanation below. See table 1

for summary and intuition.

1. Mass as the integral of density on space:To understand, consider a rod whose density ρ is equal to 3, and we have to integrate over interval[3, 19]. Hence we have ∫ 19

3

3dx = 3x∣∣139 = 3(19− 3) = 3× length = mass.

Note that length of a rod can be regarded as a 1-dimensional volume7. Hence, on the unit, we aremultiplying the unit of the density, mass · V ol−1 by another unit, V ol. Thus, whats left out is theunit of mass. So that means that when we integrate over the spatial variable say x, in this case, thelength or volume, we are “adding” 1 to the power of V ol−1, just like the usual when we integratea function. In this case, (V ol)−1+1 = (V ol)0 = 1.

7Similary, area is a 2-dimensional volume and so the 3-dimensional volume is the “Volume” that we know.

18

Table 1: Summary of Quantities and thier units

Quantity Name Symbol Unit Sample Unit

density ρ mass · V ol−1 kg ·m−1

mass∫ρ dx mass kg

density change wrt space x ρx mass · V ol−2 kg ·m−2

density change wrt time t ρt mass · V ol−1 · time−1 kg ·m−1hr−1

Variation in time of the mass ∂∂t

∫ρdx mass · time−1 kg · hr−1

Flux q mass · time−1 kg · hr−1

??∫λ dx mass · time−1 kg · hr−1

Source/sink λ mass · V ol−1 · time−1 kg · hr−1

2. Change of density in space, ρx:

Note that by definition,

ρx(a, t) = lim∆x→0

ρ(a+∆x, t)− ρ(a, t)

∆x=mass · V ol−1

V ol,

since the numerator is subtraction of densities, so it has a density unit, while the numerator is alength, hence it has a unit of “volume” by definition. Therefore, the unit of ρx is mass · V ol−2.

3. To get the unit of flux q and integral of source or sink∫λ dx, we can observe that

∂

∂t

∫ρdx = [q]

x2

x1+

∫λ dx.

This means that whatever the unit of ∂∂t

∫ρdx must also be the unit of the flux and the integral of

source or sink. The unit is mass · time−1.

Therefore, it will now be easy to see that the unit for source or sink λ is mass · time−1V ol−1.

4. Suppose q(x, t) = ϕ(ρ, x, t). Then the unit for ϕρ is

mass · time−1 ·(mass · V ol−1

)−1= V ol · time−1.

Hence, ϕρ as the variation of flux in density can therefore be regarded as speed or velocity.

Also consider the flow speed:

v(t, x) =q(x, t)

ρ(x, t)−→ mass · time−1

mass · V ol−1= V ol/time

which is speed.

6.1 Settings: Assumptions

In our traffic flow example, we shall made the following assumptions.

• For simplicity, we suppose a one-lane road. This will serve our x-axis with the positive orientationgoing to the right direction.

• We assume that on this one-lane road, there were no on/off ramps, i.e., λ = 0. It means that onecar cannot go in or out of the lane.

• We assume that X has a unit distance (dist) kilometer, km. Recall that distance or length issynonymous with volume.

19

• We assume T has a time unit of hour, hr.

• For traffic density ρ, a unit is: n cars/km.

Now, consider the flux as a function of density ρ(X,T ) which is interpreted as how many numbers ofcars in a certain distance, say, kilometers (#cars/km), flux which is also a function of X, this can beinterpreted as curves, ups or downs of the road. Flux is also a function of T , this can be interpreted astime. So when you drive your car, it does matter when you drive during the night or day or your speedcould be limited because of the topography of the road (curvy, zigzag, etc) or even in traffic, that is thedensity ( no. of cars in a distance).

Thus, in reality flux can expressed as

q(X,T ) = Φ(ρ(X,T ), X, T ).

However, for simplicity, we forget about the intricacy of our flux system and we consider that we maybe driving on a straight road during a sunny day free from traffic whatsoever. Here, we can expressedflux by an empirical function:

q(X,T ) = Φ(ρ) = Rρ (ρ− ρmax) ,

where R > 0. So the flux function is not explicit of X and T .Now, flux could also be of the following form, which is an reformulation of the above equation:

Ψ(ρX , ρ) = Φ(ρ)− kρX = Rρ (ρ− ρmax)− kρX ,

where k > 0.We can combine these two formulation of flux into one function and we have

q(X,T ) =

{Φ(ρ) = Rρ (ρ− ρmax) R > 0

Ψ(ρX , ρ) = Φ(ρ)− kρX k > 0(60)

which can be viewed graphically roughly as a parabola.As to the question on how and why we arrive such formulation(?) we will defer for it for the moment

and pick it up later in our discussion.

Figure 3: Traffic flux

In light with Figure 3: when ρ = 0, it means there’s no car, hence no flux, i.e., q = 0. Also, ifρ = ρmax, that is, too much cars in a given distance, then there’s traffic that cars don’t move, hence noflux, i.e., q = 0.

6.2 Into a dimensionless PDE

Now, our aim in this section is to transform the conservation law in divergence form below into a dimen-sionless pde:

ρT + qX = 0. (61)

20

By plugging Ψ(ρX , ρ) into (61), we have

ρT + (Rρ (ρ− ρmax)− kρX)X = 0. (62)

To understand whats going on in (62), we will inspect the unit of each quantities involved in (62). Notethat the unit of ρT is n cars · dist−1 · time−1. Thus, the second term should be the same unit as ρT inorder to compatibly add them.

Now, to know the unit of R, bearing in mind that we are multiplying density to a density andafterwards applying derivative wrt X, hence R unit would make sense if

n cars−1 · dist2 · time−1.

To know the unit of k, remember that ρX has unit n cars · dist−2, so after applying derivative wrt X,the unit becomes n cars · dist−3. Thus, in order for the unit of the third term kρXX to be compatiblewith the unit of ρT , the unit of k should be

dist2 · time−1.

6.2.1 Characteristic scale

One important procedure while formulating a mathematical model of a physical situation is scaling.Roughly speaking, scaling deals with choosing new, usually dimensionless variables and reformulate theproblem expressed in these variables. This is what we were going to do with (62)

So, we begin to normalize variables x, t and u by introducing a characteristic length8 denoted by L0.We have the following

x =X

L0

t =T

L0/ (Rρmax)

u =ρmax

2 − ρ̃ρmax

2

(63)

where ρ̃(x, t) = ρ(X,T ), X and T are given by scaling above, i.e., X = xL0 and T = t (L0/ (Rρmax)).Therefore, u = u(x, t) that is, u is a function now of x and t.

It would be useful to know that given X = xL0 we have

Xx = L0 · xx = L0 · 1 = L0. (64)

and given T = t(

L0

Rρmax

)we obtain

Tt =

(L0

Rρmax

)· tt =

L0

Rρmaxsince tt = 1 (65)

Also, givenρ̃(x, t) = ρ(X,T ) (66)

we have

ρ̃t(x, t) =∂ρ

∂T

∂T

∂t= ρTTt = ρT ·

(L0

Rρmax

)(67)

8There is no unique definition for characteristic length. One can define the length scale as the square root of the surfacearea, as the third root of the volume, as the volume-to-surface ratio, etc. It all depends what length scale you want identify.For our problem (traffic flow on a straight road) the characteristic length is the length of a road (from X = 0 to X = L)and we denote by L0. Is this right, Joaquim?

21

and

ρ̃x(x, t) =∂ρ

∂X

∂X

∂x= ρXXx = ρX · L0. (68)

So to expressed (61) or (62) into dimensionless equation, we will solve what is ut, ux, and uxx becausethese quantities involves ρT , ρX , and ρXX as we will see.

Take note that u can be written as

u = 1− ρ̃ρmax

2

. (69)

Then by chain rule and equation (65), the variation in time of u is

ut = − 1

ρmax/2· ρT · Tt = − 2

ρmax

L0

RρmaxρT = − 2L0

R (ρ2max)ρT . (70)

The variation in space of u

ux =−2

ρmaxρXXx =

−2

ρmaxρXL0 =

−2L0

ρmaxρX . (71)

The variation in space of ux (or the diffusion)

uxx =

[−2L0

ρmaxρX(X,T )

]x

=−2L0

ρmax[ρX(X,T )]x

=−2L0

ρmax(ρX)X Xx · xx

=−2L0

ρmaxρXXL0 · 1

(72)

Hence,

uxx =−2L2

0

ρmaxρXX . (73)

Now, we can write

ρT + (Φ(ρ)− kρX)X = 0 where Φ(ρ) = Rρ (ρ− ρmax)

intoρT +Φ(ρ)X − kρXX = 0, (74)

and by using our dimensionless quantity (70) to (73), we can transform (74) into

Rρ2max

−2L0ut + Φ̃x

1

L0+kρmax

2L20

uxx = 0 (75)

where

Φ̃(x, t) = Φ(X(x), T (t)) (76)

from whence,Φ̃x = ΦXXx = ΦXL0

which means that

ΦX =Φ̃x

L0,

(we substitute it in the second term of (74)).

22

Now, we know that

Φ̃(x, t) = Φ(X(x), T (t)) = Rρ(X(x), T (t)) (ρ(X(x), T (t))− ρmax) ,

and by our assumption in (66), we can write it as

Φ̃(x, t) = Rρ̃(x, t) (ρ̃(x, t)− ρmax) . (77)

We are halfway already to solving Φ̃x in (75), by substituting

ρ̃ =ρmax

2(1− u)

solved in (69) to be plug into (77).Therefore,

Φ̃(x, t) = Rρmax

2(1− u)

(ρmax

2(1− u)− ρmax

)= R

ρmax

2(1− u)

(−ρmax

2(1 + u)

)= −Rρ

2max

4(1− u)(1 + u)

(78)

Therefore,

Φ̃(x, t) = −Rρ2max

4(1− u2), (79)

and by taking the variation in space, it follows that

Φ̃x(x, t) =−Rρ2max

2uux. (80)

So, we can now write (75) into

Rρ2max

−2L0ut −

Rρ2max

2uux · 1

L0+kρmax

2L20

uxx = 0. (81)

Multiplying (81) by −2L0, we have

Rρ2maxut +Rρ2maxuux − kρmax

L0uxx = 0, (82)

and equivalently

Rρ2maxut +Rρ2maxuux =kρmax

L0uxx. (83)

Therefore, our dimensionless equation

ut + uux =k

L0Rρmaxuxx, (84)

which finally we can write as

ut +

(u2

2

)x

= εuxx, (85)

where ε = kL0Rρmax

is known as the Burgers’ equation.

If k = 0, then (86) becomes

ut +

(u2

2

)x

= 0 (86)

which we know as the inviscid Burgers’ equation.

23

6.2.2 Remarks on inviscid Burgers’ and the Burgers’ Equation

References

[Flanders] Harley Flanders. “Differentiation under the Integral Sign”. American Mathematical Monthly,vol. 80 (June-July 1973), p. 615-627.

[Folland] Gerald B. Folland. Real Analysis: Modern Techniques and Their Applications, second ed.Wiley-Interscience, 1999.

[Conrad] K. Conrad. Differentiating under integral sign, http://www.math.uconn.edu/∼kconrad/blurbs/analysis/diffunderint.pdf

[Goel] S. K. Goel and A. J. Zajta. Parametric Integration Techniques. Math. Mag. 62 (1989),318322.

[Charles] Swartz, Charles. Introduction to gauge integrals. Singapore: World Scientific, 2001.

[Swartz] Swartz et al. Theories of integration: The integrals of Riemann, Lebesgue, Henstock-Kurzweiland Mcshane. Second edition, World Scientific.

[McCormmach] Jungnickel, C., & McCormmach, R. (1990). Intellectual Mastery of Nature. TheoreticalPhysics from Ohm to Einstein, Volume 2: The Now Mighty Theoretical Physics, 1870 to1925 (Vol. 2). University of Chicago Press.

[Talvila2] E. Talvila. Some Divergent Trigonometric Integrals, Amer. Math. Monthly 108 (2001), 432–436.

[Talvila] Erik Talvila. “Necessary and Sufficient Conditions for Differentiating Under the IntegralSignhttp://www.math.ualberta.ca/∼etalvila/papers/difffinal.pdf”. American MathematicalMonthly, vol. 108 (June-July 2001), p. 544-548.

The author of this entry has also written an exposition, “Differentiation under the Integral Sign usingWeak Derivativeshttp://gold-saucer.afraid.org/math/diff-int/diff-int.pdf”, containing a proof of Theorem4 along with detailed computational examples.

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