Mathematical Modeling Lecture Notes - Jeff S. McGough

Math 451 - Mathematical ModelingLecture Notes

Jeff S. McGough

Department of Mathematics and Computer Science,South Dakota School of Mines and Technology

501 E. St. Joseph st.Email: [email protected]

2

Chapter 1

Introduction

1.1 What is Mathematical Modeling?

In this course, a model will be a mathematical description of some object or phenomenon of interest. Thegoal is the gain a description that leads to predictions or a deeper understanding of the object of study.Most students who have finished courses in calculus, physics or other sciences, have seen many modelsdeveloped and used. Some in some ways, this subject, this course is not new. Much of the mathematicsthat is taught is done with the idea of being applied in some manner. The idea that a model will be builtand mathematical analysis will be applied. This is done to answer some question.

Often this subject is described as applied mathematics, however modeling is probably better thought ofas a component of applied mathematics. This being distinguished from pure mathematics. It is commonto hear fields of mathematics, like algebra, topology or number theory being described as “pure” and dif-ferential equations, numerical analysis and probability as “applied”. This is false and misleading. Numbertheory has application in cryptology, algebra in a variety of computing related problems, and topology inphysics (to name just a few application areas). Mathematics tends to arise based on need, so in general, ithas an application. There is clearly differences in the level of abstraction which may play into the terms,but abstraction of the point of mathematics and is not owned by any particular subject area.

Nevertheless the terms persist. Why? They indeed describe two different pursuits. The differencebetween the two has nothing to do the subject and everything to do with approach. Applied mathematiciansthink differently from “pure” mathematicians. This is hardly surprising since it is true that mathematiciansthink differently from physicists both of whom think differently from computer scientists. It is this approachor framework that essentially defines the scientist and provides the intellectual tools required in a particularsubject area.

Assume that you have just completed some mathematical analysis. Where are the results headed?Have you increased your understanding of mathematics or of the object under study? Typically appliedmathematics will close the loop by having the result of the analysis related back to the problem at hand.Modeling then is the first step in the process.

This subject is part science and part art. It is learned by doing, not by watching. It is a very human

3

4 CHAPTER 1. INTRODUCTION

endevour of translating our experience into a formal langage. We have seen many routine tasks becomeautomated by machinery and it is no different for mathematics. Algebra and calculus can be some bysoftware efficiently and quickly. However the process of modeling is still the exclusive domain of organicbeings.

Modeling: Developing a mathematical description which captures the essential elements without un-necessary details. Modeling is a level of focus on the problem at hand. One must capture the essence ofthe problem. The core component of modeling is translating the problem into a formal language.

One common pitfall is to argue about the reality of the model. There is no such thing as absolute reality.“Human conciousness is built to mask details, abstract and generalize.” [JRR Tolkien] For example, yousit in a normal classroom and close your eyes. More than likely you will not remember how many pieces ofchalk or erasers are at the chalk board. You will remember the shape of the room or that the room containsdesks and chairs. Once again, we are capturing the essential details of the room or in the modeling case,the problem or object of interest.

What constitutes a good model?

• Does it reflect the important items under study? Can you see how those items influence the model?

• Does it allow for improved understanding? A good model teaches you in the process of developingthe model. It will give you an increased understanding of the problem. A good model will alwaysgive us something more than just an accurate simulation.

• Does it make useful predictions? Does it say something that you did not know while building themodel?

• Does it allow for validation? It is easy to write a computer program that generates output whichagrees with a phenomenon (a phenomenological model). It is also possible to write conceptual modelswhich don’t allow validation or predictions. Neither is what we will deem as a good model.

Because we are translating observation into mathematics, we will have requirements due to the natureof the language of mathematics. For example, it is normally required that the set of equations be complete;in that all the assumptions and relations have been included to correspond with the object of study. Thisis similar to the notion of completeness in linear algebra. It is also required that the equations for examplebe internally consistent. We are not asking or requiring that the model be a ”good” model (meaninghaving external consistancy or the model interacting with its environment), but just that it has an internalstructure which admits solutions of some form.

Keep in mind that the use or the purpose of the model is a valid aspect of the model. When one asksfor a cup of water, do you really measure an exact cup? A model then is a mathematical description ofour view of some phenomenon or object at some scale.

Here are the Guidlines for building a model suggested by Michael Mersterton-Gibbons (the ABC’s ofModeling):

A. Make Assumptions - A model is a collection of assumptions. To begin, make more assumptionsthan what is given to you. In the simplest case model, it has the most assumptions that are possible for

1.2. A SAMPLE OF THE PROCESS 5

the problem. As one desires to make the problem more complex, assumptions then are taken away fromthe model. Poor agreement with observation indicates the model is lacking, not that the mathematics iswrong (we will assume the the math is done correctly).

B. Borrow Work - ”Don’t reinvent the wheel”. Newton phrased it as standing on the shoulders ofgiants. You don’t gain by doing all alone, we get measured by the results. Use the work by the experts,since you cannot be an expert in all areas. Always cite sources and acknowledge help.

C. Criticize - Be very critical of your work and that of others. This establishes and preserves integrity.Don’t just take the word of an ”expert”. Everyone can make mistakes.

Modeling is a process, not a goal. [And an iterative process as well.] Mark Meerschaert[8] has outlineda five step process or method to approach modeling:

1. Ask the question.

2. Select the modeling approach.

3. Formulate the model.

4. Solve the model.

5. Answer the question.

1.2 A sample of the process

Example 1 How long will it take for your goggles to fall from your lap when your are on the ski lift tothe ground 10 meters below?

Model construction: What do you want to know? What do you know? What can you assume? Startnaming parts for identification. We begin with listing the variables:

• Let t = time

• Let y = position (or height of goggles in meters)

• Let v = velocity (in meters/seconds)

• Let a = acceleration down to earth (in meters/seconds2)

You might ask why we did not list “g = acceleration due to gravity at sea level = -9.806”. This is not avariable for the problem, as the item will remain constant. It is still important and you may also find ituseful to list out constants as well especially if you plan to use labels for them.

What is the initial configuration? We take for initial data that v(0) = 0, y(0) = 10. The basicassumptions we will use are:


1. Acceleration is constant.

2. No other forces acting on goggles, i.e. neglect air resistance.

The next step is to establish the “first principle” or underlying relation between significant variables in theproblem. In this case from Calculus we know that

a =d2y

dt2=dv

dt.

Since we assumed that the acceleration is constant we have that −9.806 = dv/dt. Integration provides−9.806t = v(t) − v(0). The initial data on velocity, v(0) = 0, gives us −9.806t = v(t) = dy/dt. We mayintegrate the velocity formula to recover the position formula, y(t) = y(0) − 4.903t2. The initial data onposition provides y = 10− 4.9t2. Setting y = 0 and solving for t gives us t =

√

10/4.9 for time of flight.

Is this a good model? The model fails to account for two physical properties.

1. We know that air resistance is not zero.

2. We know that acceleration changes with altitude.

Should we include both in the model? No. Why? For the new model we will include a force due to airresistance.

Example 2 How long will it take for your goggles to fall from your lap when your are on the ski lift tothe ground 10 meters below assuming air resistance?

The new model isa = −9.806 + c|v|α

with previous initial conditions and where c and α are measured in the field. We will often find that cwill be fairly small in magnitude and α will be in the range of [0, 4]. We now have an ordinary differentialequation in v.

For the example above

a =d2y

dt2= −9.806 + cvα

thendv

dt= −9.806 + c|v|α.

We will use separation of variable to solve:

dv

(−9.806 + c|v|α)= dt

∫dv

(−9.806 + c|v|α)=

∫

dt = t+ k

We are stuck for 2 reasons:

1.2. A SAMPLE OF THE PROCESS 7

1. There is no antiderivative for the left hand side.

2. Even if you could find a numerical integral for left hand side, it would be impossible to solve for v interms of t.

If α = 1, we can simplify c|v|α = −cv

−∫

dv

(9.806 + cv)=

∫

dt = t+ k

and then work out the integral:

−1

cln(| cv + 9.806 |) = t+ k.

Thuscv + 9.806 = ke−ct

so

v(t) =1

c(ke−ct − 9.806).

Apply initial conditions to find k:

v(0) =1

c(k − 9.806) = 0,

thenk = 9.806.

We now have our velocity term:

v(t) =9.806

c(e−ct − 1).

We have found something new. If we take the limit as t → ∞, the velocity goes to a constant. This isknown as terminal velocity. It is fairly simple to compute the terminal velocity.1

To find the position term, simply integrate and apply initial conditions to obtain:

y(t) = −9.806

c(1

ce−ct + t) + (10 +

9.806

c2)

The time of flight is found by setting y = 0 and solving for t again (use uppercase to distinguish fromgeneral t).:

−9.806

c(1

ce−cT + T ) + (10 +

9.806

c2) = 0.

However, this is much more complicated to solve than the original. Using Maple 9, it does not arriveat a solution in elementary functions. Note that c = 0 causes everything to cancel out. Using Taylor’sexpansion for ecT , we get:

−9.806

c

(1

c

[

1− cT +1

2c2T 2 − 1

6c3T 3 +

1

24c4T 4 − ...

]

+ T

)

+ (10 +9.806

c2) = 0

1Recall that dv/dt = −9.806+ c|v|α. Set dv

dt= 0 then v = ( 9.806

c)

1

α . This makes sense for α > 0 and we obtain the terminalvelocity. We can easily answer the question of how terminal velocity depends on c, which is not easy with numerics.


9.806

c2

[1

2c2T 2 − 1

6c3T 3 +

1

24c4T 4 − ...

]

= 10

T 2 − 1

3cT 3 +

1

12c2T 4 − ... = 10

4.903.

By examining our new model, we notice that we have a perturbation of the original model. Note thatthe basic approximation (of the old model a = −9.806) is contained within our new model in the first term.The rest of the terms are correction terms that account for the air resistance on the goggles. Our newmodel gives us a more accurate description of a free falling goggles; but is a modification of the simplierproblem. This is known as a perturbation result. It is also useful to note that we can recover the originalmodel by taking the limit as c→ 0.

Data has error. One question that arises often is how the error will affect the result. There are severalplaces error can enter in modeling. One is that the parameters and data have errors. Another is thatapproximation used to obtain the model is inaccurate. A third way is that some of the equations in modelcannot be solved exactly.

Example 3 If we vary the parameter c, how does the solution T vary?

Because scales may be dramatically different, we look at relative changes, not absolute changes. Bythis, we mean, if c changes by ∆c we call this the absolute change. The percentage change, ∆c/c is calledthe relative change. For our application the variation we are interested in the relative change

∆t/t

∆c/c→ c

t

dt

dcas c→ 0.

This latter term we call the sensitivity2

S(t, c) =c

t

dt

dc.

To gain the sensitivity, we must find the derivative implicitly.

2tdt

dc− 1

3

(

t3 + c(3t2)dt

dc

)

+1

12

(

2ct4 + c2(4t3)dt

dc

)

+ ... = 0

(

2t− ct2 +c2t3

3

)dt

dc− t3

3+

1

6ct4 + ... = 0.

At c = 0 we note thatdt

dc=t2

6=

10

6(4.903)≈ 1

3.

The sensitivity (using an absolute change in c) can be expressed as

S(t, 0) =1

t

dt

dc

∣∣∣∣c=0

=1

6

√

10

(4.903)

One can conclude that the change from the original problem will not be dramatic, especially when smallvalues of c are concerned.

2This makes sense when c, t 6= 0. When one of the variables is zero, the absolute change may be used.

1.3. MODELING COMPONENTS 9

1.3 Modeling components

Important aspects to the modeling process

1. The process of encapsulation into a mathematical model.

2. The qualitative descriptions of behavior (not numerical descriptions).

3. Verification.

The process of building a mathematical model is a process of learning, focusing, and refining. Theprocess may be sufficient alone. In otherwords, just the model building aspect may provide the insight andanswers needed. Mathematicians often are used to ask the ”right” questions and assist the engineers toa model. Once done the problem may cease to be a problem. It maybe that the elementary aspects of amodel can provide very deep results.

General classes of Models:

• Deterministic - Models that do not involve random variables. Courses like calculus, differentialequations and linear algebra all introduce these types of models.

• Stochastic - Models that do involve random variables. Courses like probability, statistics and stochas-tic processes will introduce these models.

Formulation of a model requires solid principles to build on. Models often begin with basic conservationproperties and then add on observed laws.

1. Conservation Laws - Conservation Laws include conservation of mass, energy, and momentum. Theseare your basic laws and fundamental driving principles in the modeling process. These are built onfirst principles.

2. Constitutive Relations - Constitutive relations are the observed or measured relations and facts.They are the formulas relating observed quantities. Some example of these are the ideal gas las(PV = nRT ), first order reaction (rate = k[A]), and Fourier’s Law (Flux = −k∆u).

Most models will apply at least one conservation law and many constitutive relations.

Rutherford Aris’s “Maxims for Mathematical Modeling”[1]

1. Cast the problem in as elegant form as possible.

2. Choose a sympathetic notation, but don’t become too attached to it.

3. Make variables dimensionless, since this is the only way in which their magnitudes take on generalsignificance. However, do not lose sight of the quantities which may have to be varied later on in theproblem nor forget the physical origin of each part.


4. Use apriori bounds of physical or mathematical origin to keep all variables of the same order ofmagnitude, letting the dimensionless parameters show the relative size of the several terms.

5. Think geometrically. See when you can reduce the number of variables (even at the expense of firsttreating an over simplified problem), but keep in mind the needs of the general case.

6. Use rough and ready methods, but don’t carry them beyond the point of usefulness. (E.g. Isoclinesin the phase plane.)

7. Find critical points and show how the system behaves near them or what the asymptotic behaviouris at long or short times.

8. Check limiting cases and see how they tie in with simplier problems that can be solved explicitly.

9. Use crude approximations, e.g. 1-point collocation. Trade on the analogies they suggest, but remem-ber their limitations.

10. Rearrange the problem. Don’t get fixed ideas on what are the knowns and the unknowns. Be preparedto work with implicit solutions.

11. Neglect small terms, but distinguish between regular and singular pertubations.

12. Use partial insights and despise them not. (E.g. Descartes rule of signs.)

13. These maxims will self-destruct. Make your own!

Notation is important. It is easy to diminish the importance of notation. Everyone has heard “it is onlynotation”. However, notation sets the view, the mindset. It can bias the modeler and modeling process.It can act as a framework to assist, allowing insight and dramatic progress; and equally can hinder orcompletely prevent progress. Selecting good notation is as important as the basic principles of the modelitself. Be aware that notation can be very useful at one point of the process and an hinderance later; sobe able to change during the process.

It is wise to stay with convention when possible, and always use the cleanest notation available. Typi-cally this involves vector, or tensor versions that present the over-all concept without masking ones viewwith unnecessary detail.

1.4 Units

An often overlooked but essential aspect of modeling is fundamental physical quantities involved. Com-monly expressed as units, they play an important role in the modeling process. Attention to units mayoften be the first stage in model rejection or validation. It is prudent to check the units throughout themodeling process, because a velocity given in kg2/s with kg as mass and s as seconds should raise aneyebrow.

Although units can and do provide meaningful checks, models littered with parameters give rise toadministrative or algebraic errors. One way that models are simplified and analysed is to remove the units;

1.4. UNITS 11

a process known as non-dimensionalization. This is nothing new, an early example is using the ratio of thearc of a circle divided by the radius of the circle to measure the resulting angle. This is known as a radian.

Simple units example:

Assume you have the modeldx

dt= ax, where x measure in meters and t is measured in seconds. It is

clear that a must have units 1/sec. We have a modeling error if a is not.

Remove units:

Givendx

dt= ax, define u = x/l, s = t/τ , b = aτ with l in meters, τ in seconds.

d(lu)

d(τs)= a(lu)

du

ds= bu

which is unit free.

More details can be found on this by looking up the Buchingham π theorem.


Chapter 2

Optimization

Many problems in science, engineering and businesses are optimization problems. It is so important thatentire courses, even fields, are devoted to the study, such as, Linear Programming and Operations Research.The following will present a series of problems which ask to find the maxima or minima for some object ofstudy. We start with an elementary example.

Example 1: Assume a particular computer manufacturer makes $120 profit for a particular single unit.Market research has shown that for each $20 rebate or discount off the price, sales increases by 10%.

1. What rebate will maximize profit?

2. What is the sensitivity of the best rebate to the 10% value?

3. How do things change if 10% - 8%?

4. What circumstances can cause profit to decrease?

We must organize our work. Build a habit of structure.

Label and Variable

• P = Total Profit

• Ps = Single unit’s profit

• S = Profit amount when R = 0. (Fixed at $120)

• R = Rabate amount

• n = number of units sold when R = 0 (No rebate, fixed sales amount)

• m = number of units sold when R 6= 0 (Rebate)

• a = rebate motivates sales increase (10%) as decimal form (0.10)

13

14 CHAPTER 2. OPTIMIZATION

• Q =P

n(Scaling P ; average profit)

Model

1. GivenP = Ps ∗mPs = S −R = 120 −Rm = n+

aR

20n

Where the relation,aR

20, implied is linear by $20.

P = (120 −R)(n+aR

20n) = (120 −R)(1 +

aR

20n)

Always check the units to make sure that they make sense.

CallP

n= Q. (Relative Profit)

Q = (120 −R)(1 +aR

20)

dQ

dR= (−1)(1 +

aR

20) + (120 −R)(

a

20) = 60a − 1− aR

10= 0

R =a− 10

a= 60− 10

a(Assume linear at 10% where a = 0.1)

R = 60− 100 = −40

Answer: Raise the price, the rebate does not help profits. Set the price so that the profit per unit =120 + 40 = $160. More market research is needed.

2. What is the S(R, a)? (What is the Sensitivity?)

Q = (120 −R)(1 +aR

20)

dQ

dR= 6a− 1− aR

10(Set = 0)

Repeat if necessary.

Let R = 60− 10

a

What isdR

da= ? What is

dR

da=

10

a2?

S(R, a) ≡ a

R

dR

da=a

R

10

a2=

10

aR

15

For a = 0.1 and R = 20. S =10

0.1(20)= 5

For a = 0.1 and R = −40. S =10

0.1(−40)= −5

2

Interpret these values. Take the magnitude for S and anything < 0 is going to be non-sensitive.

3. If a = 10% − 8% rebate?

Because of the sensitivity expect a linear change. Rebate must be scaled (larger). The rebate needs tobe adjusted by 2%.

4. R moves away from optimal (obviously). How can one improve the model?

Move to a more involved model. Select a business/economic model instead.

Example 2: Build two types of LCD panels: A 19” model and a 21” model. How should we price andproduce these units?

Initial cost

• 19” panel: MSRP: $339 per unit Cost: $195 per unit

• 21” panel: MSRP: $399 per unit Cost: $225 per unit

NOTE: We have no control over vendors, other units will compete against ours.

• Setup for manufacture is a fixed cost: $400,000

Estimates

1. Average selling price seems to drop $0.01 for each additional unit sold.

2. 19” LCD drops $0.003 for each 21” LCD sold.

3. 21” LCD drops $0.004 for each 19” LCD sold.

Total of 10,000 units can be made per year.

1. What should the production levels be set at to maximize profit? Solve without constraints.

Labels and Variable

• n = number of 19” LCDs sold per year

• m = number of 21” LCDs sold per year

• p = selling price of 19” LCD

• q = selling price of 21” LCD


• c = cost of manufacture per year

• R = Revenue per year

• P = Total profit per year

Assumptions (Constitutive Relations)

• p = 339 − 0.01n − 0.003m

• q = 399 − 0.004n − 0.01m

• R = pn+ qm

• C = 400, 000 + 195n + 225m

• P = R− C

Skip constraints due to manufacturer for now, will address them in the next question.

P = pn+ qm− (400, 000 + 195n + 225m)

= (339 − 0.01n + 0.00m)n + (399 − 0.004n − 0.01m)m− (400, 000 + 195n + 225m)

Compute the partials and set them equal to zero: (∂P

∂n= 0 and

∂P

∂m= 0)

∂P

∂n= 144− 0.02n − 0.007m = 0

∂P

∂m= 174 − 0.007n − 0.02m = 0

Result: n ≈ 4, 735 units, and m ≈ 7, 043 units

Substitute n and m into P and get the optimal profit ($554,000)

If no constraints, we are done. But wait! Is this value a max or min? There is a problem with theconstraints (n + m ≤ 10, 000). How sensitive is this result to the measured data? For example: Howsensitive is this to the price elasticity for the 19” LCDs?

Model

Target (isolate) a $0.01 drop in a 19” LCD.

Replace number (0.01) with a variable, lets call it a. So now the profit equation becomes:

P = (339− an+ 0.003m)n + (399 − 0.004n − 0.01m)m− (400, 000 + 195n + 225m) where a = $0.01

Solve∂P

∂n= 0 and

∂P

∂m= 0

17

Which becomes∂P

∂n= 144 − 2an− 0.007m = 0 and

∂P

∂m= 174 − 0.007n − 0.02m = 0

Find n(a) and m(a), the variation between n and a and the variation between m and a. You my useMaple.

Compute∂P

∂nwhere n = 4,735, m = 7,043 and a = 0.01,

Sensitivity of n to a:

S(n, a) =a

n

∂n

∂a≈ −1

S(m.a) =a

m

∂m

∂a≈ 0.3

Also the∂P

∂a= −n2, where

S(P, a) =a

P− n2 ≈ −0.4

Mid to low sensitivity, so the result is OK if measurement errors are made. (1-to-1 sort of response).m is not very sensitive to the change in a. n does very with a, but a very little amount.

2. What about the manufacturing constraints?

A little review of Lagrange Multipliers

Take f(x, y) = −(x2 + y2) + 10. Assume we want to maximize f(x, y) subject to

g(x, y) = x+ y − 2 = 0

⇒ ∇f = λ∇g, (in terms of Normal Vectors).

⇒ ∇f = 〈−2x,−2y〉 and ∇g = 〈1, 1〉

⇒ −2x = λ, −2y = λ and x+ y = 2,

where

x = −T2, y = −λ

2⇒ − λ

2− λ

2= 2

Thusx = 1, y = 1 and λ = −2

End of review and back to the problem at hand.

Model

Assume that a total of 10,000 units (both 19” LCDs and 21” LCDs) can be made per year (n + m).Also assume that only 5,000 19” units and only 8,000 21” units can be made per year.

g(n +m) = n+m− 10, 000


Note that the constraint is g(n,m) ≥ 0. To study max f subject to g ≥ 0, we must also study max f wheng < 0 and max f when g = 0.

Constraints

• n ≥ 0

• m ≥ 0

• n+m ≤ 10, 000

• n ≤ 5, 000

• m ≤ 8, 000

Figure 2.1: Constraints.

Right now the Critical Point (n = 4735 and m = 7043) is outside of the acceptable region for a max.(That is (n+m) > 10, 000, which does not meet the constraint).

There are 3 equations and 3 unknown variables:

1) 144− 0.02n − 0.007m = λ

2) 174− 0.007n − 0.02m = λ

3) n+m = 10, 000

Wheren = 3, 846 units and m = 6, 154 units

What about the other constraints? Sensitivity? Go back and cover it prior to constraints, where:

144 − 2an− 0.007m = λ

174 − 0.007n − 0.02m = λ

n+m = 10, 000

Solve using Maple. In Maple do the following commands:

19

1. P :=

2. diff

3. solve(equ11, ...vars)

4. assign (S)

Thus...

n =50, 000

1, 000a + 3

m = 10, 000 − n

λ =650

1, 000a + 3− 26

For g = 0

⇒ dn

da= −dn

daand

dA

da= − 50million√

1, 000a + 3

Sensitivity:

S(n, a) =n

a

da

dn= −0.77

S(m,a) =a

a

dm

da= 0.48

For Optimization:

Maximize f(x), where x = (x1, x2, x3, ..., xn).

∇f = 0 ⇒ δf

δx1= 0,

δf

δx2= 0, ... ,

δf

δxn= 0

Maximize f(x), where g = 0.

Form: L = f − λg ⇒ ∇L = 0 and g = 0


Chapter 3

Dynamic models, an introduction

A central concept provided by calculus is that of the rate of change, dydx . This gives us the rate of change

of y with respect to x. There are many physical interpretations - velocity, slope, etc. Also, d2ydx2 , which is

the rate of change of velocity. Likewise, there are many interpretations - acceleration, convexity, etc.

Physics provides an enormous collection of physical objects or behaviors with corresponding mathe-matical models. An elementary tool seen in physics is that of Force balance: If an object is at rest, thenthe forces acting on that object must sum to zero (balance).

Example 3: Using Newton’s Second Law: F = ma

Let’s examine simple harmonic motion of a mass-spring system. Let x(t) be the displacement of themass, m, over time, t; Fs = Force due to the spring; Ff = Force due to friction; Fm = Force acting on themass. Then Fm + Fs + Ff = 0 or Fm = −Fs − Ff . This is our conservation law.

Next let’s model the spring, using Hooke’s Law (Constitutive relation):

Fs = k(x− x0)

This is a linear (law) relation and accurate for some range of deviation. Clearly, Fs = g(x) would providea more general model and maybe more accurate. We will assume that x0 = 0.

Then let’s model the friction (Constitutive relation):

Take a linear model

Ff = cv

A better model may include nonlinear aspects, say, Ff = c1v + c2v2 + c3v

3. But we will stick with thelinear model for purposes yet to come.

Last we model the force on the block (mass), using Newton’s 2nd Law (Constitutive relation):

F = ma

21

22 CHAPTER 3. DYNAMIC MODELS, AN INTRODUCTION

Thus we have:ma = kx− cv

Recall from Calculus: v = dxdt and a = d2x

dt2

Then we obtain

md2x

dt2+ c

dx

dt+ kx = 0 (1)

Associated with this are initial conditions: x(0) = x0 and dxdt = v0

Recall this is a basic ordinary differential equation. To solve this, take the simple equation

dx

dt= ax

By separation of variables, we getx(t) = keat

Notice that this simple equation is a subset of the spring-mass system equation (1) in that m = 0,c = −1, and k = a. In general, we cannot expect the same solution behaviour when we set a coefficientto zero that removes the highest derivative. But we use this for motivation, not answers. Since the twoproblems may be related, let’s not re-do the work. The leverage work is already done. So let’s guess thatthis solution is a component of the more general problem.

Assume x(t) = pert and substitute into (1) to get the characteristic equation:

mr2 + cr + k = 0

Then, using the quadratic equation we obtain two solutions: p1er1t and p2e

r2t if r1 and r2 are real anddistinct.

Then by linearity (linear superposition): x(t) = p1er1t + p2e

r2t

We will study linearity so that we can obtain basic models, then we can change the model to handlemore complex cases.

Problem 3 cases (r1 and r2 need not be real solutions):

1. r1 and r2 real valued and distinct:

x(t) = p1er1t + p2e

r2t

2. r1 and r2 reapeated and real:x(t) = p1e

r1t + p2ter2t

3. r1 and r2 complex conjugates (where r1 = α+ iβ and r2 = α− iβ):

x(t) = eαt(p1 cosβt+ p2 sin βt)

3.1. HIGHER ORDER EQUATIONS 23

3.1 Higher Order Equations

Next we will take a look at higher order equations, our goal being to solve an equation of the form

andnx

dtn+ an−1

dn−1x

dtn−1+ · · ·+ a1

dx

dt+ a0x = 0 (1)

In the mass-spring harmonic oscillator system we considered when exploring Newton’s Second Law,recall that we generated a second order equation with three different types of solutions. In that situationwe were dealing with a constant coefficient, linear, ordinary differential equation. Why mention this? Well,we will see throughout this course that labels are important because they often tell us the assumptionsneeded in order to solve a particular system. For example, if an equation is not constant coefficient or notlinear, then we know right away that an analytical solution doesn’t exist for that system.

Coming back to the general equation given above (equation (1)), what if we use the fact that we canwrite dx

dt = ax as x = pert, declare that as our base or final solution, and then plug that result into equation(1)?

You can see that dxdt = rpert, and likewise d2x

dt2= r2pert. If we continue substituting in this manner, we

will obtain the characteristic equation for a higher order problem. Since pert is common to each term, wecan remove it from the final equation through division to obtain

anrn + an−1r

n−1 + · · · + a1r + a0 = 0 (2)

The above equation should look familiar; it is the same as the second order case, just a higher orderversion.

Can we solve such higher order equations analytically? Is there an n-degree version of the quadraticformula? The answer is no. While formulas exist for quadratic, cubic, and quartic equations, it can beshown that no root finder exists for polynomials of 5th degree or higher. So how can we handle suchpolynomials?

In modeling, we can try numerical methods where analytical methods fail. Such techniques mightinclude approximating roots, ”guessing”, and ad hoc methods. If we are fortunate enough that all the ak

are real in equation (2), for example, then we know that the corresponding roots must be real or occur inconjugate pairs – and this dramatically simplifies numerical analysis of the problem.

This brings us back to the three cases we encountered previously when we structure a real model interms of a polynomial equation. We will obtain n roots: r1, r2, · · · rn. Some roots may be distinct, somemay be repeated, and some may be complex conjugate pairs. If we take the non-complex, non-repeatedcollection of roots of the polynomial, we have

p1er1t, p2e

r2t, · · · pnernt

Using these roots, the general solution to equation (1) would be

x(t) = p1er1t + p2e

r2t + · · ·+ pnernt


Finding the roots is certainly not a trivial process, and may in fact be the hardest step involved in generatinga general solution. If you obtain n linearly independent solutions/roots, note that the general solution isthe sum of the n solutions. For the case where you have a single pair of complex conjugate roots, say r1and r2, and the remaining roots are real and distinct, then we can construct a general solution as follows:

r1 = α+ iβ

r2 = α− iβso

x(t) = eαt(p1 cos βt+ p2 sinβt)︸︷︷︸

r1+r2

+p3er3t + · · · + pne

rnt

And finally, if the solution has repeated roots, you must square the t-term.

3.2 Examples/ODE Background

Example 1 Consider the following problem:

d4y

dt4− y = 0, with initial conditions y(0) = 1, y′(0) = −1, y′′(0) = 1, y′′′(0) = −1

This is a fourth degree polynomial and is therefore both theoretically and practically solvable. We canconvert it directly to a characteristic equation of the form r4 − 1 = 0 and then use difference of squares toobtain the roots, which in this case are r = 1,−1, i,−i. From these roots we get the following solutions:

p1et, p2e

−t, p3e0t cos(1t), p4e

0t sin(1t)︸︷︷︸

p3 cos t,p4 sin t

So the general solution isy(t) = p1e

t + p2e−t + p3 cos t+ p4 sin t

The next step is to adapt this general solution to the problem at hand by plugging in the given constraints:

y(0) = p1 + p2 + p3 = 1y′(0) = p1 − p2 + p4 = −1y′′(0) = p1 + p2 − p3 = 1y′′′(0) = p1 − p2 − p4 = −1

This type of problem is known as an initial value problem, where all constraints are given for a singlepoint in time. Initial value problems are well-posed for constant coefficient, linear, ordinary differentialequations such as this one. What you end up with is a linear system to solve. You can pose the problemin matrix terms and solve using Gauss-Jordan elimination. The setup would be

1 1 1 01 −1 0 11 1 −1 01 −1 0 −1

p1

p2

p3

p4

=

1−1

1−1

3.2. EXAMPLES/ODE BACKGROUND 25

To solve the matrix system we just created, we could use LaPlace transforms or a simple tool such asGauss-Jordan elimination. In this case, let’s use Gauss-Jordan. Recall from linear algebra that the goalof Gauss-Jordan elimination is to produce zeros in all elements below the diagonal of the matrix and thensolve the resulting row equations simultaneously. The initial Gauss-Jordan setup is

1 1 1 01 −1 0 11 1 −1 01 −1 0 −1

∣∣∣∣∣∣∣∣

1−1

1−1

To produce zeros in the first element of rows 2, 3, and 4, we might proceed as follows: r2 = r1 − r2;r3 = r1 − r3; r4 = r1 − r4. This produces the intermediate result

1 1 1 00 −2 −1 10 0 −2 00 −2 −1 −1

∣∣∣∣∣∣∣∣

1−2

0−2

We could next reduce row 4: r4 = r4 − r2. This operation gives

1 1 1 00 −2 −1 10 0 −2 00 0 0 −2

∣∣∣∣∣∣∣∣

1−2

00

Now we have zeros below the diagonal and can start solving for p1, p2, p3, and p4.We notice immediately that rows 3 and 4 give us quick solutions for p3 and p4:

−2p3 = 0⇒ p3 = 0− 2p4 = 0⇒ p4 = 0

Armed with this information, we look at row 2 to solve for p2:

−2p2 − p3 + p4 = −2⇒ −2p2 = −2⇒ p2 = 1

Finally, we can solve for p1 using the row 1 equation:

p1 + 1 + 0 + 0 = 1⇒ p1 = 0

So Gauss-Jordan elimination gives us

p1 = 0, p2 = 1, p3 = 0, p4 = 0⇒

0100

This tells us that the general solution to the initial condition problem d4ydt4 − y = 0 is

y(t) = e−t


Example 2d2y

dt2+ y = 0, with initial conditions y(0) = 0, y(1) = 0

Is this a constant coefficient, linear, ordinary differential equation with initial conditions? No. This is aboundary value problem which produces the general solution

y(t) = p1 cos t+ p2 sin t

and gives the single solution y = 0 when the boundary conditions are applied.Example 3

d2y

dt2− y2 = 0, with initial conditions y(0) = 0, y′(0) = 0

This problem is nonlinear, so the ert method does not apply – we don’t get a characteristic polynomial.Although this problem can’t be solved analytically, we can crack it using a numerical method such asRunge-Cutta. Another approach is to use the chain rule on the original equation and then solve byseparating variables:

d2y

dt2=

d

dt(v) =

dv

dy

dy

dt︸︷︷︸

chain rule

=dv

dyv ⇒ v

dv

dy− y2 = 0

Example 4

ydy

dx+ y = 0

Here we have an equation that is not linear and does not have constant coefficients.

Example 5

2d2x

dt2+ t

dx

dt− x = 0

While this equation is linear, it doesn’t have constant coefficients.

3.3 Difference Equations/Discrete Models

We will start our look at difference equations with a Markov process example. While the Markov processis probability-based, here we will apply it to a deterministic problem.

Application (Model) Assume there are two populations of a bird species: one group on the mainland,another group on a large, nearby island. You note that, due to migration, 10% of the birds that live onthe mainland will fly to the island each year. You also determine that 20% of the island population willfly to the mainland at the same time as the mainland-to-island migration occurs. This is a two-componentexample of a much more complicated tracking problem. Such problems are considered discrete becausethe sampling is based on a fixed time interval, like the U.S. census, and the objects involved (birds for ourexample and people for the census) are discrete. We will now set up the system and see how to approachsuch a problem.

Let xn be the number of birds on the mainland at year n. Let yn be the number of birds on the islandat year n. x0 and y0 will represent the starting populations of the two bird groups (initial conditions). We

3.4. LINEAR ALGEBRA REVIEW 27

will assume that the populations are stable – that is, birth and death do not occur so there are no sourceor sink terms. Here are the equations for year one:

x1 = 0.9x0 + 0.2y0

y1 = 0.1x0 + 0.8y0

Note that the entries are positive because they are population states. Also note that the new populationstates are based on the previous population states. We can express the population in a given year in termsof a vector:

~vn =

(xn

yn

)

, ~v0 =

(x0

y0

)

and can express the problem in terms of a matrix-vector multiplication:

~v1 = A~vn, where A =

(0.9 0.20.1 0.8

)

Using this matrix notation, we can express the population for any year as ~vn+1 = A~vn. Now unroll:

A~vn = A(A~vn−1︸︷︷︸

~vn

)

Unroll again:= A(A(A~vn−2)))= A3~vn−2

= A4~vn−3 · · ·More generally, A~vn = An+1~v0. This is an unrolled recursion relation, which is helpful since it effectivelyprovides a solution to the problem. Now we need a formula for

Ak =

(0.9 0.20.1 0.8

)k

We could have a computer calculate a matrix to a power, but that doesn’t help us if we are trying todetermine whether there is a stable population ratio – in other words, what happens to the population ofthe mainland and island birds in the limit?

limk→∞

Ak~v0 = ?

Is there extinction of the species on the mainland or on the island? Before answering this question, areview of linear algebra is in order since linear algebra provides some useful tools to solve difference anddifferential equations.

3.4 Linear Algebra Review

Our review begins with a look at some linear algebra notation. A vector can be declared as

~x =xi

i=1,n=

x1...xn

; x ∈ ℜn


A matrix A (and we will typically deal with square matrices) is defined by

A =ai,j

i,j=1,n=

a11 · · · ain...

. . ....

an1 · · · ann

; A ∈ ℜnxn

Multiplying a matrix by a vector gives a vector:

A~x =

−→a11

. . .

ann

↓ x1...xn

=

a11x1 + a12x2 + · · ·+ a1nxn...an1x1 + an2x2 + · · ·+ annxn

Multiplying two matrices together produces another matrix; this operation is treated as dot products ofrows against columns. In other words, element (1,1) in a product matrix is the dot product of the row 1elements of the first matrix times the corresponding 1st column elements of the second matrix:

AB = C ⇒

←→row 1

(l col 1

)=

(element in row 1, col 1

)

Example: You can rewrite a system of equations as a matrix system and solve using Gauss-Jordanelimination, as we have seen previously.

(2x + y = 4−x + 4y = 7

)

⇔(

2 1−1 4

)(xy

)

=

(47

)

Solving this system gives x = 1, y = 2.

3.4.1 Matrix Theory Elements

In this course, eigenvalues and eigenvectors will be our most important tools from the world of linearalgebra. With that in mind, let us consider an nxn matrix A. We will denote the determinant of A bydetA or |A| and define it as

|A| =∑

i

aijaij

We will say that ~x ∈ ℜn (x is a vector) and λ are an eigenvector, eigenvalue pair if, for x 6= 0,

A~x = λ~x︸︷︷︸

1 equation, 2 unknowns

What does this mean? It is a transformation of the vector; the vector is rotated in space and stretched orshrunk. If the eigenvectors are only scaled and do not undergo a direction change, then they are consideredinvariant and we can say that ”x is a fixed direction under A.” From this information we obtain

A~x− λ~x = 0; 0 = A~x− λ~x = (A− λI)~x


Where I is the identity matrix. This tells us that

(A− λI)~x = ~0, ~x⇒ nullspace of A− λI, A− λI singular

We know from linear algebra that

det(A − λI) = 0

If we expand det(A− λI) as a polynomial in λ, we obtain the characteristic polynomial. This is the samecharacteristic polynomial you may remember from your ODE course, just a different notation:

cnλn + cn−1λ

n−1 + · · · c0 = 0

If we solve this polynomial, we will get a set of n roots λ1, λ2, · · · λn. To find the eigenvector that correspondsto λk we compute (A− λkI) and solve (A− λkI)~x = ~0. Any ~x that satisfies this equation is an acceptableeigenvector; since it is a singular system, there will not be a unique solution. We can now write oureigenvalue, eigenvector pairs: (λ1, ~x1), (λ2, ~x2), · · · , (λn, ~xn).

Facts:

1) If A is triangular (all zeros above or below the diagonal), then the eigenvalues are the diagonalelements of A.

Example:

1 4 50 2 60 0 3

, λ1 = 1, λ2 = 2, λ3 = 3

2) If ~x is an eigenvector of A, it is an eigenvector for Ak.

A~x = λ~x ; multiply by A,A2~x = λA~x = λ2~x

More generally,

Ak~x = λk~x, eigenvalue = λk

Theorem/Fact: Assume that λ1, λ2, · · · λn are distinct eigenvalues. The corresponding eigenvectorsare linearly independent (they point in different directions). This characteristic is necessary when weconstruct a matrix out of eigenvectors and want the matrix to be invertible. In fact, this is the normalcase with eigenvectors.

Armed with this information, we can decompose a matrix into eigenvalues and eigenvectors. Let

A =

a11 · · · a1n...

. . ....

an1 · · · ann

Theorem: Suppose A has linearly independent eigenvectors ~x1, ~x2, · · · ~xn. From these eigenvectors wecan form the matrix S, where

S =[~x1 | ~x2 | · · · ~xn

]∈ ℜnxn n x n matrix)


We also know that S−1 exists since S only has 0 in the null space. We can also state (knowing that theeigenvectors are linearly independent) that S is invertible, det[S] 6= 0, etc. We will now form a new matrixΛ, where

S−1AS = Λ =

λ1 0. . .

0 λn

Λ contains the eigenvalues corresponding to the eigenvectors and preserves the relative order of the pairs.This is known as the diagonalization of A, which is useful in solving systems because diagonal matrices arethe easiest to deal with. Producing Λ from A is known as a similarity transformation.

What is happening here? We are producing a coordinate transformation in both the domain andrange so that in the new coordinate system we will have a decoupling. This ensures that things in the xdirection do not affect things in the y direction. Recall that a good coordinate system is orthogonal, soby decomposing a matrix into its eigenvalues and eigenvectors, we go from a messy system to a simple,orthogonal coordinate system.

What is AS? Can we infer what A−1 is without having to compute the inverse? Argument

AS = A[~x1 | ~x2 | · · · ~xn

]

=[A~x1 | A~x2 | · · · A~xn

]

=[λ1~x1 | λ2~x2 | · · · λn~xn

]

=[~x1 | ~x2 | · · · ~xn

]

λ1

. . .

λn

= SΛ

Now multiply through by S−1 to get

S−1AS = Λ

Using this decomposition, we can solve for A:

A = S︸︷︷︸

vectors

Λ︸︷︷︸

values

S−1︸︷︷︸

vectors

We can use this Jordan form to solve a variety of problems, as we will see throughout this course.

Example: Find the eigenvalues, eigenvectors, and diagonalization of A, where

A =

(5 21 4

)

First compute det(A − λI) = 0, then solve for each λk.

(5− λ 21 4− λ

)

= 0 , then expand to (5− λ)(4− λ)− 2 = 020 − 9λ+ λ2 − 2 = 0, λ2 − 9λ+ 18 = 0


Solving the last equation results in eigenvalues of 6 and 3. Now we plug these eigenvalues in and find thematching eigenvectors – in other words, the directions for which (A − λI) is degenerate. For λ = 3, wemust find a ~v such that (

5− 3 21 4− 3

)(~v1~v2

)

==

(00

)

In some cases we may need to perform Gaussian elimination, but simpler matrices such as this one can besolved by inspection. Here, we get

(1−1

)

For λ = 6, we have(

5− 6 21 4− 6

)(~v1~v2

)

==

(00

)

⇒(

21

)

So now we know that S, the matrix of eigenvectors, is

(2 11 −1

)

To find S−1, multiply S by the identity matrix and perform row operations until S takes on the charac-teristics of the identity matrix:

(2 11 −1

)(1 00 1

)

row ops until

(2 11 −1

)

goes to

(1 00 1

)

The result is that

S−1 =

(13

13

13 −2

3

)

You can verify your work by ensuring that SS−1 = I. So we can now express the matrix A as

[2 11 −1

]

︸︷︷︸

S

[6 00 3

]

︸︷︷︸

Λ

[13

13

13 −2

3

]

︸︷︷︸

S−1

What we have done to matrix A is similar to factoring a polynomial – we decomposed the system into itssimple, most basic components.

At this point we can return to the Markov process example of bird migration between the mainland

and the island. Recall that we needed a formula for Ak, where Ak =

k︷︸︸︷

A ·A · A · A. Using the decompositionwe just performed, we have

Ak = (SΛS−1)(SΛS−1) · · · (SΛS−1)= SΛΛ · · ·ΛS−1

= SΛkS−1

We now have powers of a diagonal matrix (Λ), which is easy to compute:

Λk =

λk1 0

. . .

0 λkn


and

limk→∞

Ak = limk→∞

S

λk1 0

. . .

0 λkn

S−1 = S

limk→∞ λk1 0

. . .

0 limk→∞ λkn

So what happens to Ak is determined only by the eigenvalues: for λk > 1, it will run to ∞; for λk < 1,it will run to 0, and for λk = 1, it will run to 1. To summarize: 1) If all λi are such that |λi| < 1, thenlimk→∞Ak = 0 (converges). 2) If one of λi is |λi| > 1, then limk→∞Ak diverges. We can use these resultsto analyze the behavior of the Markov process for the bird population migration model.

A =

[0.9 0.20.1 0.8

]

⇒ det(A− λI) = 0 givesλ2 − 1.7λ+ 0.7 = 0

This produces the eigenvalues

λ1 = 1︸︷︷︸

boundary case

, λ2 = 0.7

The decomposition is:

A =

[1 11 −2

]

︸︷︷︸

S

[1 00 0.7

]

︸︷︷︸

Λ

[23

13

13 −1

3

]

︸︷︷︸

S−1

So for the bird population model we can now state

(xk (mainland birds)yk (island birds)

)

= Ak

(x0

y0

)

=

(1 11 −2

)(1k 00 0.7k

)(23

13

13 −1

3

)(x0

y0

)

This formula allows us to determine the population at any given time. We can also take the limit to seewhat the long-term population balance will be:

(x0 + y0)

(2313

)

+ (x0 − 2y0)(0.7)k

(13−1

3

)

So

limk→∞

=

(x∞y∞

)

= (x0 + y0)

(2313

)

+ 0

Since x0 + y0 = the total initial populations, we can say that over time 23 of the population will end up on

the mainland and 13 will be on the island.

Summarizing our look at eigenvalues and eigenvectors, we know that decomposing/diagonalizing amatrix A results in:

1)A = SΛS−1

2)Ak = SΛkS−1

Extending this to polynomials of matrices, we can see that if a function is applied to a matrix A, thatfunction is applied only to Λ when A is decomposed:

p(x) = 3x2 + 2x+ 1⇒ p(A) = 3A2 + 2A+ I = S

[p(λ1) 00 p(λ2)

]

S−1


We formally define this property as follows:

f(A) = S

f(λ1) 0. . .

0 f(λn)

S

−1

For example, to find the square root of a matrix, decompose the matrix into A = SΛS−1 and then applythe square root to the diagonal elements of Λ, which of course are the associated eigenvalues.

Example: For this example, we will use the Fibonacci Sequence. Let Fn be a sequence of numbers suchthat Fn is the sum of the two previous values:

Fn+2 = Fn+1 + Fn

This is not in matrix form, instead try:

µk =

(Fk+1

Fk

)

⇒ µk+1 =

(Fk+2

Fk+1

)

then we see:Fk+2 = Fk+1 + Fk

Fk+1 = Fk+1

Thus: (Fk+2

Fk+1

)

=

(1 11 0

)(Fk+1

Fk

)

⇒ µk+1 =

(1 11 0

)

µk ⇒ µk+1 = Aµk

For A:

λ1 =1 +√

5

2λ2 =

1−√

5

2

Solving via the eignenvector/eigenvalue method:

Fk =1√5

[

(1+√

52 )k − (1−

√5

2 )k]

with F0 = 0, F1 = 1.

This is to illustrate going from a second order difference to a lower value equation. An alternate wayto solve this system other than eignenvalues is to use the guess method.

Guess Method:


The guessing method is formulized by the system method. For example, guess Fn = Crn and plug itinto the Fibonacci sequence.

Crn+2 = Crn+1 + Crn

Crn is common so we can cancel it out: r2 = r + 1. So we may solve: r2 − r − 1 = 0, using the quadraticequation:

r =1 +√

5

2r =

1−√

5

2

This method is more general, so it can be a faster way to the solution.

Fn = C1(1 +√

5

2)n + C2(

1−√

5

2)k

Use initial conditions:

F0 = 0 0 = C1 +C2

F1 = 1 1 = C1(1+

√5

2 )n + C2(1−

√5

2 )k

When you solve the two equations, you should get the same answer you get with the eigenvector method.

Example 1:

In this example we will convert a second order differential equation into a 2-D system, start with theequation y′′ + ay′ + by = 0.

Let u =

(yy′

)

, then u′ =

(y′

y′′

)

=

(y′

−ay′ − by

)

. We have no information about y′ so we

will leave it alone. However, we do have information about y′′. Writing y′′ in matrix form, we get:

y′ =

(0 1−b −a

)(yy′

)

We know dudx = Au. Therefore, A =

(0 1−b −a

)

. Now, just apply the eigenvector method to get a

solution.


Solve: dudt = Au, where u =

u1

u2...u3

and A is an n × n matrix.

define: eAt = I +At+ 12A

2t2+ ... is the Taylor Series of ex applied to A.

ddt(e

At) = ddt [I +At+ 1

2A2t2 + ...]

= 0 +A+A2t+ 12A

3t2 + ...= A[I +At+ 1

2A2t2 + ....]

= Aet

thus, eAt solves dudt = Au.

Example 2:

To solve dxdt = −2x + y and dydt = x − 2y, first convert the equations to a system and then compute

eAt.

Putting the equations into a system, you get u =

(xy

)

so dudt =

(−2 11 −2

)

u. Therefore, A =(−2 11 −2

)

.

Now, we need to compute eAt. We’ve said that

f(A) = S

f(λ1) 0. . .

0 f(λn)

S−1

So

eAt = S

eλ1t 0. . .

0 eλnt

S

−1u0


Define S−1u0 = C, where C is the constant vector

c1...cn

. Then u(t) = S

c1eλ1t 0

. . .

0 cneλnt

S =(v1 | v2 | · · · vn

), so

u(t) =(v1 | v2 | · · · vn

)

c1eλ1t 0

. . .

0 cneλnt

⇒ c1eλ1tv1 + c2e

λ2tv2 + · · ·+ cneλntvn

Thus, we have gained the Eigenvector Method.

Example 3:

Solve dudt = Au. Where u =

(10

)

, and A =

(−2 11 −2

)

.

We find the eignenvalues are -1 and -3, and the eigenvectors are

(11

)

and

(1−1

)

, respectively. The

formula directly tells us that

u(t) = c1e−t

(11

)

+ c2e−3t

(1−1

)

,

plugging in t = 0 we get that

(10

)

= c1

(11

)

+ c2

(1−1

)

. When we solve the linear system for c1 and

c2 we find c1 = c2 = 12 .

3.5 Exercises

1. We have the free-fall model a = −9.8 + cvk, where we studied the case k = 1.

(a) What should the sign of c be for k = 1 and k = 2?

(b) Work out the solution in the case k = 2.

(c) Does this present a reasonable answer?

3.5. EXERCISES 37

2. Find the differential equation model for the free fall problem with an alternate set of assumptions.Take the y axis to be positive upward. Let the radius of the planet be R, set the origin (y = 0) atthe surface. Assume that there is no air resistance, but that the force due to gravitiy does change asa function of altitude. Let lower case m stand for the object mass, upper case M stand for the massof the planet and G the gravitational constant.

(a) Find the differential equation for y(t) including only the object data and the gravitational

acceleration. Hint: g =GM

R2

(b) Convert to a differential equation for v(y) (use the ”trick”:dv

dt=dv

dy

dy

dt= v

dv

dy)

(c) Find the escape velocity (smallest initial velocity so that the object does not return) . Hint: vmust stay positive.

3. Solve the initial value problem: y(iv) − y′′ = 0, y(0) = 1, y′(0) = y′′(0) = y′′′(0) = 0.

4. An annual plant is one that lives for a single season. These plants survive by spreading seeds inthe fall which then germinate in the following spring. Some seeds fail to germinate that spring, butdo the following spring. Let pn represent the number of plants for a particular species, a representthe average number of seeds per plant that survive the first winter and germinate, and b representthe average number of seeds per plant that germinate a year following. This can be modeled by thedifference equation: pn+2 = apn+1 + bpn.

(a) What is the criteria for survival of the plant (in terms of a and b)? [ Note that a and b arepositive ]

(b) Is the plant helped or hurt by this two season approach? [ We might assume that whatevergives the seed the ability to survive an extra winter might hinder it from germinating the firstspring, and so a and b are inversely related.]

5. Let A be a 3X3 matrix, with rows: row1 = [2,−1, 0], row2 = [−1, 2,−1], row3 = [0,−1, 2].

(a) Find the diagonalization of A.

(b) Use the previous result to solvedx

dt= Ax, x(0) = (1, 0, 0)t.

(c) Find sqrt(A)


Chapter 4

Finite Dimensional Vector Spaces

Let S be a set of objects. Let x, y, z ∈ S and α, β ∈ R.

Define some operations:

1. x+ y = y + x

2. x+ (y + z) = (x+ y) + z

3. 0 ∈ S, 0 + x = x

4. −x ∈ S, −x+ x = 0

5. α(βx) = (αβ)x

6. (α + β)x = αx + βx

7. α(x+ y) = αx + αy

8. 1x = x, 0x = 0

If the following conditions hold, then we say S is a vector space:

1. x, y ∈ S then x+ y ∈ S

2. α ∈ R, x ∈ S, then αx ∈ S

Examples of vector spaces

1. R2 ⇒ plane. We need to check that we have closure in addition and scalar multiplication in R

2. Ifwe do, then R

2 is a vector space.

Addition formula:

(v1v2

)

+

(u1

u2

)

=

(v1 + u1

v2 + u2

)

39

40 CHAPTER 4. FINITE DIMENSIONAL VECTOR SPACES

Scalar multiplication formula: α

(v1v2

)

=

(αv1αv2

)

Since the conditions hold, R2 is a vector space.

2. Rn ⇒ n-dimensional space.

Let v =

v1v2...

vn

and let u =

u1

u2...

un

.

We know that v + u =

v1 + u1

v2 + u2...

v3 + u3

and αv =

αv1αv2

...αv3

for α ∈ R .

Therefore we can conclude that Rn is a vector space.

3. Polynomials of degree n

P (x) = anxn + an−1x

n−1 + ...+ a1x+ a0

Q(x) = bnxn + bn−1x

n−1 + ...+ b1x+ b0

P (x) +Q(x) and αP (x) are degree n polynomials and thus the set is closed.

4. n × m Matrices

Addition and multiplication are defined and the set is closed.

5. Continuous functions on [a, b].

Let f(x) and g(x) be continuous.

f(x) + g(x) ⇒ continuous αf(x) ⇒ continuous

Linear CombinationFor a given set of vectors x1, x2, ..., xn ∈ S and constants α1, α2, ..., αn ∈ R, we define the following as alinear combination:

α1x1 + α2x2 + ...+ αnxn

Linearly IndependentIf α1x1 + α2x2 + ... + αnxn = 0 implies that αi = 0 ∀i, then we say the collection xi for i = 1, ... , n islinearly independent.

Otherwise we say the set is linearly dependent.

Example. Let

(10

)

,

(11

)

be vectors.

41

α1

(10

)

+ α2

(11

)

= 0

⇒(

1 10 1

)

×(α1

α2

)

=

(00

)

Det

(1 10 1

)

6= 0

Therefore

(α1

α2

)

=

(1 10 1

)

−1 ×(

00

)

=

(00

)

⇒ α1 = α2 = 0. This implies linear independence.

Example. Let

(10

)

,

(11

)

,

(01

)

be vectors.

α1

(10

)

+ α2

(11

)

+ α3

(01

)

= 0

(1 1 00 1 1

)

α1

α2

α3

=

(00

)

⇒ (α1, α2, α3) = (1, -1, 1). This does not equal 0 which implies linear independence.

Spanning SetT is a spanning set if every x ∈ S can be written as a linear combination of elements in T.

Example. Let x =

(10

)

, y =

(01

)

⇒ spans R2.

Any element in R2 ⇒

(v1v2

)

(v1v2

)

= α1

(10

)

+ α2

(01

)

. Solve for α1 and α2.

BasisA basis is a linearly independent spanning set.

The last example is a basis. An example of a basis for R2 is

(10

)

,

(01

)

.

Let’s verify that

(10

)

,

(11

)

is a basis.

(b1b2

)

= α1

(10

)

+ α2

(11

)

=

(1 10 1

)(α1

α2

)


⇒(α1

α2

)

=

(1 10 1

)

−1

(b1b2

)

We arrive at a unique solution, so

(10

)

,

(11

)

is a basis.

Note: Every basis of a fixed vector space has the same number of elements. The number of elementsin a basis is the dimension of the space.

Example In R2, any two non-parallel vectores will work as a basis.

Example The set 1, x, x2 is a basis for the quadratic polynomials.

Question: Is the following a basis for R3?

101

,

010

,

110

,

011

Answer: These four vectors are not a basis for R3 because it has four vectors instead of three. However

the first three vectors in this set do form a basis for R3.

Note: Vectors have direction and magnitude.

Inner ProductIf x, y, z ∈ S and 〈x, y〉 is a bilinear operator mapping S → R or C then we say 〈x, y〉 is an inner productif the following holds:

1. 〈x, y〉 = 〈y, x〉

2. 〈αx, y〉 = α 〈x, y〉

3. 〈x+ y, z〉 = 〈x, z〉 + 〈y, z〉

4. 〈x, x〉> 0 if x 6= 0= 0 iff x = 0

A vector space with an inner product is called an inner product space.

Examples

1. For Rn, let 〈x, y〉 =

n∑

k=1

xkyk.

Then for R2: 〈x, y〉 = x1y1 + x2y2

2. Continuous functions on [0, 1]: C[0, 1]

For f, g ∈ C[0, 1]⇒ 〈f, g〉 =

∫ 1

0f(x)g(x)dx

Note that properties 1 through 4 hold.

43

Functions are Vectors:

Take a function on [0, 1].

Sample at x = 0 and x = 1. ⇒(f0

f1

)

Sample at x = 0, 0.1, 0.2, 0.3, ..., 0.9, 1 ⇒

f0

f0.1

f0.2...f1

re-index ⇒

f1

f2

f3...

f10

Vector =

(x1

xn

)

A vector is a function over a discrete (integer) domain. A function is a vector with a continuousindex.

3. Differentiable functions on [0, 1].

〈f, g〉 =

∫ 1

0f(x)g(x) + f ′(x)g′(x)dx

Note that properties 1 through 4 hold.

NormA norm is a function from S → R

T , with notation ‖ ‖, for which

1. ‖ x ‖> 0 if x 6= 0

2. ‖ x ‖= 0 if x = 0

3. ‖ αx ‖ = | α |‖ x ‖

4. ‖ x+ y ‖ ≤ ‖ x ‖ + ‖ y ‖

Examples

1. Rn: ‖ x ‖2 = [

n∑

i=1

| xi |2]1

2

Note that this is the distance formula. The norm here gives the traditional vector length.

2. Rn: ‖ x ‖p = [

n∑

i=1

| xi |p]1

2 for p > 1

3. ‖ f(x) ‖= [

∫ 1

0[f(x)]2dx]

1

2 for f ∈ C[0, 1]


An inner product will induce a norm. ⇒‖ x ‖= (〈x, x〉) 1

2

For example 3, 〈f, g〉 =

∫ 1

0f(x)g(x)dx will induce the norm.

Example. Let f, g be n times differentiable complex functions on [a, b].

〈f, g〉 =

∫ b

a

n∑

j=0

f (j)(x)g(j)(x)dx

This induces the norm ‖ f ‖= [

∫ b

a

n∑

j=0

| f (j)(x) |2 dx] 1

2

Theorem: Cauchy-Schwartz

For an inner product space: | 〈x, y〉 |2 ≤ ‖ x ‖2‖ y ‖2

Proof: 0 ≤ ‖ x− αy ‖2

0 ≤ 〈x− αy, x− αy〉

0 ≤ 〈x, x〉+ 〈x,−αy〉+ 〈−αy, x〉+ 〈−αy,−αy〉

0 ≤ 〈x, x〉 − α 〈x, y〉 − α 〈y, x〉+ α2 〈y, y〉

Take α to be the projection of x onto y. ⇒ α =〈x, y〉‖ y ‖2 and plug this into the equation.

0 ≤ ‖ x ‖2 −2〈x, y〉‖ y ‖2 Re〈x, y〉+ 〈x, y〉

2

‖ y ‖4 ‖ y ‖2

0 ≤ ‖ x ‖2‖ y ‖2 −2 〈x, y〉2 + 〈x, y〉2

Thus we have | 〈x, y〉 |2 ≤ ‖ x ‖2‖ y ‖2 Q.E.D.

This is used to prove that the triangle inequality holds.

Note: In R3 recall that cos θ =

x · y‖ x ‖‖ y ‖

Define: cos θ =〈x, y〉‖ x ‖‖ y ‖

Examples

1. In R2 take x =

(10

)

, y =

(01

)

cos θ =〈x, y〉‖ x ‖‖ y ‖ =

1 · 0 + 0 · 11 · 1 = 0

45

cos θ = 0⇒ θ = 90

2. If 〈x, y〉 = 0, then we say x and y are orthogonal. The notation is x ⊥ y.Let f(x) = sin(x), g(x) = cos(x), S = C[0, 2π].

〈f, g〉 =

∫ 2π

0sin(x) cos(x)dx = 0

⇒ sin(x) ⊥ cos(x) on [0, 2π]

3. Take f(x) = 1 and g(x) = x over the same space.

〈f, g〉 =

∫ 2π

01 · xdx =

1

2x2 |2π

0 = 2π2

‖ f ‖2= 〈f, f〉 =∫ 2π

012dx = 2π

‖ g ‖2= 〈g, g〉 =

∫ 2π

0x2dx =

8π

3

cos θ =〈f, g〉‖ f ‖‖ g ‖ =

2√

3

2⇒ θ = 30

We found the angle between the two functions 1 and x on [0, 2π]

cos(θ) =〈1, x〉‖ 1 ‖‖ x ‖ =

2√

3

2⇒ θ = 30

There is a clear relation between θ 6= 0 and linear independence. However orthogonal is better thanlinear independence.

Theorem: An orthogonal set of vectors is linearly independent

[note - recall an orthogonal set of vectors means that every distinct element of the set is orthogonal to theother elements i.e. θi ⊥ θj or 〈θi, θj〉 = 0]

Define: inner product equals zero to be the same as orthogonal

〈x, y〉 = 0⇔ x ⊥ y

To have a basis for a space you need to have linear independence, but we don’t use linear independence,we us orthogonality

Let θi be an orthogonal spanning set, then θi will be a basis for the vector space. An orthogonalspanning set is a basis for solving linear systems.

Let f ∈ S where S is a vector space. We can represent f by


f =

n∑

j=1

αj · θj

which is a linear combination of θj

What are the weights? αj

Solve the linear system for αj

Multiply by θk and take an inner product.

〈f, θk〉 =

n∑

j=1

αj 〈θj, θk〉

θj orthogonal ⇒ 〈θj, θk〉 = 0 if k 6= j ‖ θk ‖2 if k = j

⇒ 〈f, θk〉 = αk ‖ θk ‖2

αk =〈f, θk〉‖ θk ‖2

basis weight

This formula is the projection or shadow formula from Calculus III, our axis is two orthogonal vectors.

Problem: What if you only have a linearly independent set and not an orthogonal set?

Process: Gram-Schmidt construct an orthogonal set from a linearly independent set

Given a linearly independent set xi where i = 1, ..., n

θ1 = x1

θ2 = x2 −〈x2, θ1〉‖ θ1 ‖2

θ1

θ3 = x3 −〈x3, θ1〉‖ θ1 ‖2

θ1 −〈x3, θ2〉‖ θ2 ‖2

θ2

θn = xn −n−1∑

j=1

〈xn, θj〉‖ θj ‖2

θj

47

This is an unnormalized Gram-Schmidt process. It starts with any linearly independent set of vectorsand finds a basis.

Example. Given 1, x, x2, x3, ..., xn , produce an orthogonal set using

〈f, g〉 =

∫ 1

0f(x)g(x)dx

θ1 = 1

θ2 = x− 〈x, 1〉‖ 1 ‖2 1

〈x, 1〉 =

∫ 1

0xdx =

1

2x2

∣∣∣∣

1

0

=1

2

‖ 1 ‖2= 〈1, 1〉 =

∫ 1

01dx = 1

θ2 = x− 1

2

Check : 〈θ1, θ2〉 = 0 ?

〈θ1, θ2〉 =

∫ 1

0(1)(x − 1

2)dx =

1

2x2 − 1

2x)

∣∣∣∣

1

0

= 0

θ3 = x2 −⟨x2, θ1

⟩

‖ θ1 ‖2θ1 −

⟨x2, θ2

⟩

‖ θ2 ‖2θ2

⟨x2, 1

⟩=

∫ 1

0x2dx =

1

3x3

∣∣∣∣

1

0

=1

3⟨

x2, x− 1

2

⟩

=

∫ 1

0x2(x− 1

2)dx

=

∫

x3 − 1

2x2dx =

1

4x4 − 1

6x3

∣∣∣∣

1

0

=1

12

θ3 = x2 − 1

3(1) −

12(x− 1

2)

‖ x− 12 ‖2

← This is an orthogonal polynomial.


Define basis: Spanning Linearly independent set, with no redundancy.

There are many choices for a basis, but the best choice is often an orthogonal basis.

Suppose we want to solve Ax = y for y

Change of basis: x→ x′

Relation: x = Cx′ Where C is a matrix. This is a change of coordinates.

x = Cx′, y = Cy′

Convert Ax = y into the new basis:

Ax = y ⇒ ACx′ = Cy′ ⇒ C−1ACx′ = y′

The ideal sysem is where C−1AC = Λ, where Λ is a diagonal matrix.

C−1AC is known as a similarity transform

AC = CΛ⇐

Note: Λx′ = y′

λ1 0. . .

0 λn

x1...xn

=

y′n...y′n

this is easy to solve

By picking the correct similarity transform we can simplify the problem. How do we select this represen-tative system?

AC = Cλ

Take a column of C:

[θ1 | θ2 | · · · | θn]

A[θ1 | θ2 | · · · | θn] = [θ1 | θ2 | · · · | θn]

λ1 0. . .

0 λn

Aθk = λkθk

↓

49

Av = λv

(matrix · vector = λ · vector)This is an eigen value problem.

To solve:

det(A− λI) = 0, solve for λ

Theorem: Let A be an nxn matrix.

1. If A has linearly independent eigenvectors then there is a change of basis in C⋉ so that in the basis

A becomes diagonal.

2. If A is real and has linearly independent eigenvectors then the change of basis is real.

3. If C is the matrix of eigenvectors thenC−1AC = Λ

How do we get this?

LetC = [θ1 | θ2 | · · · | θn]

AC = [Aθ1 | Aθ2 | · · · | Aθn] = [λ1θ1 | λ2θ2 | · · ·λnθn] = [θ1 | · · · θn]

λ1 0. . .

0 λn

= CΛ

Eigenvalues give you a change of basis to diagonalize a change of basis.

Example Computing Eigenvalues

A =

(2 11 2

)

det(A− λI) = det

((2− λ) 1

1 (2− λ)

)

= (2 − λ)2 − 1 = 0

⇒ (2− λ)2 = 1

⇒ (2− λ) = −1, 1


⇒ λ = 1, 3

Forλ = 1

(A− λI) =

((2− 1) 1

1 (2− 1)

)

need: (A− λI)v = 0

⇒(

1 11 1

)(v1v2

)

=

(00

)

⇒ ~v =

(1−1

)

Forλ = 3

(A− λI) =

((2− 3) 1

1 (2− 3)

)

need: (A− λI)v = 0

⇒(−1 11 −1

)(v1v2

)

=

(00

)

⇒ ~v =

(11

)

λ = 1, ~v =

(1−1

)

λ = 3, ~v =

(11

)

What is this theorem trying to show you?

C =

(1 1−1 1

)

AC =

(2 11 2

)(1 1−1 1

)

=

(1 3−1 3

)

=

(1 3−1 3

)

=

(1 1−1 1

)(1 00 3

)

= CΛ

x = cx′

51

y = cy′

Solve (2 11 2

)(x1

x2

)

=

(10

)

Find

(10

)

in the new basis.

(1

0

)

=

(1 1

−1 1

)(12

12

)

We can solve transformed problems:

(1 0

0 3

)(x′1

x′2

)

=

(12

12

)

⇒(x′1

x′2

)

=

(12

16

)

(1 1

−1 1

)(12

16

)

Untransform it to get back to the original and to check that you did it correctly.

Cx′ = x

(1 1

−1 1

)(12

16

)

=

(23

−13

)

= x

Differential Equation is the right hand side.

u′′ + 2u′ + 3u = t

L(u′′ + 2u′ + 3u) = L(t)

Solve the simple problem. Find L−10 ⇒ answer.

Theorem If A has n distinct eigenvalues then it has n linearly independent eigenvectors.

Proof If we have one vector that is linearly independent. By induction: Assume we have k− 1 linearlyindependent eigenvectors. Show that the kth item is linearly independent.

α1x1 + α2x2 + . . .+ αk−1xk−1 + αkxk = 0

Multiply by A


α1A1x1 + α2A2x2 + . . .+ αk−1Ak−1xk−1 + αkAkxk = 0

1) α1λ1x1 + α2λ2x2 + . . . + αk−1λk−1xk−1 + αkλkxk = 0

2) α1λkx1 + α2λkx2 + . . .+ αk−1λkxk−1 + αkλkxk = 0

When we multiply by λk Take the difference of 1− 2

α1(λ1 − λk)x1 + α2(λ2 − λk)x2 + . . .+ αk−1(λk−1 − λk)xk−1 + αk(λk − λk)xk = 0

⇒ α1, α2, . . . , αk−1 = 0

Therefore

αk = 0⇒ α1, . . . , αk = 0→ x1

are linearly independent.

Definition For any matrix A, the adjoint of A is A∗ and is defined by 〈Ax, y〉 = 〈x,A∗y〉

Definition A is said to be self adjoint if A = A∗

Note: A∗ = AT

When A is real: self adjoint ⇔ symmetric.

Spectral Theorem If A is self adjoint, then

1. 〈Ax,X〉 is real for all x

2. All eigenvalues are A are real

3. Eigenvetors of distinct eigenvalues are orthogonal

4. The eigenvectors form an n-dimensional basis

5. The matrix can be diagonalized

Proof:

1. If A = A∗

〈Ax, x〉 = 〈x,A∗x〉 = 〈x,Ax〉 = 〈 ¯Ax, x〉⇒ 〈Ax, x〉 is real

53

2. If Ax = λx〈Ax, x〉 = 〈λx, x〉 = λ〈x, x〉〈Ax, x〉 is real by property 1

〈x, x〉 real

⇒ λ must be real

3. If Ax = x and Ay = µy and λ 6= µλ〈x, y〉 = 〈λx, y〉

= 〈Ax, y〉 = 〈x,A∗y〉 = 〈x,Ay〉 = 〈x, µy〉 = µ〈x, y〉⇒ (λ− µ)〈x, y〉 = 0⇒ 〈x, y〉 = 0⇒ x⊥y

We will take properties 4 and 5 as given from linear algebra.

We can place the eigenvectors into a matrix : Q

Q = [x1 | x2 | . . . | xn]

If A is self adjoint and has disjoint eigenvalues, then Q is an orthogonal matrix. Q is also square.

x1

x2...vn

[x1 | x2 | . . . | xn]

=

‖ x1 ‖2 0 . . . 00 ‖ x2 ‖2 . . . 0. . . . . . . . . . . .0 . . . 0 ‖ xn ‖2

Diagonal matrix

Scale the eigenvectors by their norm. Then Q is called orthogonal. And

QTQ = I

⇒ QT = Q−1

Diagonalization

AQ = A(x1 | x2 | . . . | xn

)=(λ1x1 | λ2x2 | . . . | λnxn

)

= [x1 | x2 | . . . | xn]

λ1 0 . . . 00 λ2 . . . 0

. . . . . . . . . . . .0 . . . 0 λn

= QΛ


⇒ AQ = QΛ

Q−1AQ = Λ

⇓

Q∗AQ = Λ

Q∗ = QT

When values are real.

”The Big Picture” We want to solve Ax = y

What is 〈Ax, y〉 ?

For A self adjoint matrix, 〈Ax, x〉 is a quadratic form for which 〈Ax, x〉 = a constant is an ellipsoid.

Two possibilities:

x21 + x2

2 = c

x21 − x2

2 = c

Courant Matrix Principle: For any real symmetric matrix A

λk = minc

(max(‖ x ‖= 1)(cx = 0)〈Ax, x〉

)

where c is any (k − 1) x n matrix

Fredholm Theorem:

1. The solution of Ax = b is unique if and only if the solution of Ax = 0 is x = 0 where x is the kernal.

a) Assume that for some x 6= 0 implies that Ax = 0. Also, take Ay0 = b build y1 = y0 + αx.

Note: Ay1 = A(y0 + αx) = Ay0 +Aαx = Ay0 = b. Therefore, Ay1 = b and is not unique.

b) Assuming Ax = b is not uniquely solvable,

y1 and y2 are solutions. Let x = y1 − y2 where x is not zero. Ax = A(y1 − y2) = Ay1 − Ay2 whereAyi = b then Ax = b− b = 0 implies Ax = 0 for x 6= 0.

Null vectors use uniqueness where non-uniqueness implies there is a null vector.

Uniqueness must have a null space.

2. The equation Ax = b has a solution if and only if 〈b, v〉 = 0 for every v such that A∗ = 0. If v is suchthat A∗v = 0, then 〈v, b〉 = 〈v,Ax〉 = 〈A∗v, x〉 = 0

55

We now have two results at our fingertips...

The Spectral Theorem which has to do with self adjoint properties, and the Fredholm Theorem whichhas to do with solvability.

Example 1 A =

(1 11 1

)

and the null space: v =

(1−1

)

and Av =

(00

)

Since the Null Vector exists, NOT unique.

Ax =

(33

)

(1 11 1

) (X1

X2

)

=

(33

)

Is there more than one solution?

What about x =

(12

)

as a solution? (This is the particular solution, Xp)

From a Differential Equations Class, we know the General Solution can be written in the form,

Xg = Xp + CXH =

(12

)

+ c

(1−1

)

Example 2: An Adjoint Problem

A∗ =

(1 11 1

)

and v =

(1−1

)

where A∗v = 0 and each element of the null space can be in any

linear combination, for example,

(2−2

)

or

(n−n

)

would work as long as A∗v = 0 holds.

Ax = b(

1 11 1

) (X1

X2

)

=

(b1b2

)

As long as b · v = 0, we can solve when 〈b, v〉 = 0

b1 − b2 = 0, in particular, b1 = b2


Chapter 5

Function Spaces

A metric d in a vector space is a measure of distance between two elements in a vector space.

1. d(x, y) = d(y, x)

Where the distance between x and y = the distance between y and x

2. d(x, y) ≥ 0 and d(x, y) = 0 if x = y

Where the distance is nonnegative

3. d(x, y) ≤ d(x, z) + d(z, y)

Where the shortest distance between two points is a straight line.

For Rnwe know the distance, we need to know for the vector space what distance means.

Definition : A sequence Xn in S is said to have a limit, x, in S if for any ε ≥ 0, there is an integer, N

such that for all n ≥ N, d(Xn,X) ≤ ε.

• Pick a small ε and find N large enough to be in that ε

Definition : A sequence is called a Cauchy Sequence if for any ε ≥ 0, there is an integer Nsuch that for every n,m ≥ N , d(Xn,Xm) ≤ ε.

A Convergent Sequence => a Cauchy Sequence

d(Xn,Xm) ≤ d(Xn,X) + (X,Xm)

A Cauchy Sequence 6=> a Convergent Sequence

An Example Let S = (0, 1) be a real line segment.Xn = 1/n which is Cauchy as n→∞, Xn → 0 /∈ (0, 1)

57

58 CHAPTER 5. FUNCTION SPACES

Lack of Completeness is the problem why these two definitions, a Sequence and a Cauchy Sequence, lookas if they were equal.

Recall Norm

1. ||x|| ≥ 0 if ||x|| = 0, then x = 0

2. ||αx|| = |α|||x||

3. ||x+ y|| ≤ ||x|| + ||y||

A norm will induce a metric ... d(x, y) = ||x− y|| Recall this is the measure of distance.

Examples:

1. Sequences: Xn n = 1, ∞. We have a vector therefore we have a norm.

Norm: ||X|| =( ∞∑

n=1

‖Xn|p)1/p

where d = ||x− y||

2. For continuous Functions on [a, b] : c[a, b]

||f || = max|f(x)| is a sup norm which generates the norm and distance function where d = ||f − g||

3. For c[a, b]

||f || =(∫ b

a‖f |p

)(1/p)

where Lp is the norm and d = ||f − g||

1, 2 and 3 all generate a distance, d.

Definition : A normed linear space is complete if for every Cauchy Sequence in S,is convergent to an element in S.

Example: Take the rational numbers on [0, 2]. The Norm, |x|.

Is this a complete set? No! We can produce a sequence of rational numbers which limits us to anirrational number.

Take√

2, remember√

2 ≈ 1.41423... which is nonrepeating and no periodic sequence.

x = .d1d2d3...dnd1d2d3...dn... if repeating, then its rational.

10nx = d1d2d3...dnd1d2d3...dn...

10nx− x = d1d2d3...dn

(10n − 1)x = d1d2d3...dn

x = d1d2d3...dn

10n−1

59

So for our example, where√

2 ≈ 1.41423

X0 = 1 X3 = 14141000

X1 = 1410 X4 = 14142

10000

X2 = 141100 Xn = something

10n so Xn →√

2 as n → ∞

Example: Continuous function’s on [0, 1]

fn =

0 0 ≤ t < 12 − 1

n12 + n

2 (t− 12 ) 1

2 − 1n ≤ t ≤ 1

2 + 1n

1 12 + 1

n < t ≤ 1

limn→∞

= g(t) where you must use L’Hopitals rule on1

2+n

2(t− 1

2) to solve. Also, g(t) is not continuous.

The PROBLEM:You can’t do approximations if the approximation is outside of the set.

Example : The Cantor Set Starting with the set, [0, 1] [0 ( ) ]1can be broken up into the following sets by extracting components...

[0 ( ) ]1/3 [2/3 ( ) ]1 where 13 = 1

3 ∗ 1

[0 ( ) ]1/9 [2/9 ( ) ]3/9[2/3 ( ) ]7/9 [8/9 ( ) ]1 where 29 = 1

3 ∗ (23 )

The next step would be4

27=

1

3∗ (

2

3)2 so on and so forth.

Remove∞∑

n=0

1

3∗ (

2

3)2 =

1

3

∞∑

n=0

(2

3)2 which we notice to be the Geometric Series.

So, we get1

3∗(

1

1− 23

)

= 1 from [0, 1]

This tells us that the Cantor Middle Thirds Set is a nonempty, closed set.

Continueing with the example, Define the Function

φ(x) =

1 if x ∈ Cantor Set1 if x /∈ Cantor Set

Where φ(x) is known as the Salt and Pepper Function.

∫

φ(x)d(x) does not exist. φ(x) is not continuous.

The problem is that we are mixing limits and functions and can fall our of the space we are working in.


• We can choose a partition to produce any number between 0 and 1.

• The limit of that partion gives us a Riemann integral of any number from 0 to 1.

• We would get 2 Left Hand Side values from the Cantor Set and 2 Right Hand Values, rememberingthat area = length *1, we can get any value desired.

• This also tells us that the Riemann integral is not up to the task.

If the Riemann integral is not up to the task, then what is?

Definition: A set of measure zero:If for every ε < 0, the set can be covered by a collection of open intervalswhose total length is less that ε.

Example: Finite Sets

Let our total length be ε =ε

3+ε

3+ε

3Remember that all finite sets have a measure of 0.

Example: Countable Sets one - to - one with integers

Countable sets have a measure of zero, WHY?

Proof Take Xn to be elements of the set.

Cover X1 by an interval of length ε2

Cover X2 by an interval of lengthε

4

Cover X3 by an interval of lengthε

8

Cover Xn by an interval of lengthε

2n

The Total Length =

∞∑

n=1

ε1

2n= ε

So, if the Riemann integral does not work, then what is the fix?

The Fix: The Lebesgue Integral:Since the Riemann was not up for the task, we will show the Lebesgue integral is. First we must define

what a Lebesgue ingeral is:

Lebesgue Integration

1. If the Riemann integral,

∫

Rf , exists, then the Lebesgue integral

∫

Lf also exists and they are equiv-

alent.

∫

Rf =

∫

Lf

61

2. Linear Properties Hold

For Example:,

∫

L(f + g) =

∫

Lf +

∫

Lg and

∫

L(cf) = c

∫

Lf

3.

∫

set→Af =

∫

set→Bf if A and B differ by a set of measure zero.

4. Lebesgue Dominated Convergence Theorem

If fn → f converges pointwie and |fn| < g then limn→∞

∫

fn =

∫

limn→∞

fn

5. L2[a, b] is a complete space. Start with the continuous functions on [a, b] : c[a, b]

Norm: ||f || =[∫

[a,b]|f |2

]

This norm uses the Lebesgue integral.

It is built on the Lebesgue Dominated Convergence Theorem and bounded by the Cauchy Sequence.Take all Cauchy Sequences in c[a, b]. Form the union of c[a, b] and the limits of the Cauchy Sequence.Call this Space L2[a, b]

Definition Complete inner product space: Hilbert Space L2

Complete normed space: Banach Space Lp

The Completion of c[a, b] using the norm is

||f || =[∫

[a,b](|f |2 + |f ′|2)

]1/2

gives us the Sobolev Space H’

Approximation in Hilbert SpaceModeling is all about approximation, the right approximation.Say we want to approximate some function using our favorite set of functions, which could includeanything from Electrical Engineering, Orbital Mechanics, Thermal Dynamics and so on.

By approximation we mean,

f(x) ∼n∑

i=1

αiφi

where φi is our basis and we want α to minimize our error.

We want ||f(x)−n∑

i=1

αiφi|| to be very small, less than ε given that ε > 0

Given ||f(x)−n∑

i=1

αiφi||2 how would you expand this out?

Note:

• The Norm is generated from the inner product


• ||f(x)||2 =< f(x), f(x) >

By plugging in our f(x) into this equation, we get,

||f(x)−n∑

i=1

αiφi||2

=

n∑

i=1

(αi− < f(x), φi >)2 + ||f(x)||2 −n∑

i=1

< f(x), φi >2

=< f(x)−n∑

i=1

αiφi, f(x)−n∑

i=1

αiφi >

Assuming R,

||f(x)||2 − 2

n∑

n=1

αi < f(x), φi > +

n∑

i=1

n∑

j=1

αiαj < φi, φj >= ERROR = E

We want to minimize the error, f is constant, φ is the basis, and we can choose α to be what we want

dE

dαi= 0

This is easier if φi are an orthogonal set.

We see < φi, φi >=

0 if i 6= j1 if i = j

We want to minimize the error in our inner product space, αi =< c, φi > called the Fourier Coefficient

Select αi =< f(x), φi > where f(x) = α1φ1 + α2φ2

The BEST L2 approximation is:

f(x) ∼n∑

i=1

< f(x), φi > φi

Since L2 is complete, we may take limits...

∞∑

i=1

< f(x), φi > φi = g ∈ L2

Is g(x) = f(x)?

Example

Let φn = sin(nx) which is orthogonal. [0, 2π]

Set < f(x), φn >=

∫ 2π

0f(x) sin(nx)dx. Take f(x) = cos(x)

63

Then, we get

∫ 2π

0cos(x) sin(nx)dx = 0

Then, cos(x, φn) = 0 always.

Here, we can see that the approximation does not work out. Why is this true?

• A function in L2 can differ by a set of measure 0 and still be the same funciton.

• Even though there are an infinitly many number of functions, they do not span the space.

• Completeness to Spanning

– Set Completeness

– Spanning Completeness

We now would like to find the L1 approximation of f(x) using linear terms on the domain [0, 1]

f(x) ∼ α+ βx

We would like to minimize f(x) using the L1 norm

min(α,β)||f(x)− (α+ βx)||

min(α,β)

∫ 1

0|f(x)− (α+ βx)|dx

How do we minimize? We take the derivatives

d

dα

∫ 1

0|f(x)− (α+ βx)|dx

d

dβ

∫ 1

0|f(x)− (α+ βx)|dx

Notion of Completeness

Definition: an orthonormal set is complete if

∞∑

i=1

〈f(x), φi〉φi = f(x) for every f in the Hilbert Space

This is not the set (Cauchy Sequence) complete.

Theorem: A set φi∞i=1 is complete if any of the following equivalent statements hold:

1. f(x) =

∞∑

i=1

〈f(x), φi〉φi

2. for any ε > 0 there exists an N such that for n > N


||f(x)−∞∑

i=1

〈f(x), φi〉φi|| < ε

3. ||f(x)||2 =

∞∑

i=1

〈f(x), φi〉2 (Porseval’s Equality)

4. If 〈f(x), φi〉 = 0 for all i, then f ≡ 0

5. There does not exist ψ such that φi ∪ ψ is orthogonal

Weierstrass Approximation Theorem: For any continuous function, f(x) ∈ C[a, b], and any ε > 0there exists a polynomial p(x) so that

max(a<x<b)

|f(x)− p(x)| < ε

This means we can get arbitrary close to f(x) with a polynomial. This is similar to the Taylor Series. Theproblem with the Taylor series is that it requires differentiability.

To converge uniformly on an interval we will look at the max norm/uniform norm of the function.

What is nice about the max norm is that it allows us to do point-wise approximations. This means we canget every point as close as we need.

Example: of an L2 approximations using a non-polynomial function. That is we will use trigonometricfunctions on the interval [0, 2π]

Let φn =

sinnπ√

π,cosnπ√

π

∞

n=0

One can show that φn is an orthonormal set

∫ 2π

0φnφmdx =

1 n = m0 n 6= m

We can produce a Fourier Expansion

f ∼∞∑

n=0

〈f(x), φn〉φn

∞∑

n=0

〈f(x), φn〉φn︸︷︷︸

cos type

+

∞∑

n=0

〈f(x), φn〉φn︸︷︷︸

sin type

f ∼∞∑

n=0

⟨

f(x),cosnx√

π

⟩cosnx√

π+

∞∑

n=0

⟨

f(x),sinnx√

π

⟩sinnx√

π

We then solve these equations by separating the cos and sin terms

65

⟨

f,cosnx√

π

⟩

=

∫ 2π

0f(x)

cosnx√π

dx

⟨

f,sinnx√

π

⟩

=

∫ 2π

0f(x)

sinnx√πdx

Solving further for n ≥ 1

an ≡ 〈f(x), φn〉 ·1√π

=1√π

∫ 2π

0f(x)

cosnx√π

dx

=1

π

∫ 2π

0f(x) cosnxdx

bn ≡ 〈f(x), φn〉 ·1√π

=1√π

∫ 2π

0f(x)

sinnx√πdx

=1

π

∫ 2π

0f(x) sinnxdx

Note that for n = 0

a0 =1

2π

∫ 2π

0f(x)dx

The Fourier Series approximation of f(x) is

f(x) ∼∞∑

n=0

an cosnx+ bn sinnx

Fourier Series is the projection of the function onto the elements. The summation of the series shouldapproximately equal f(x) in L2. That is

∑

n=0

〈f(x), φn〉φn → f(x)

When and Where does the summation converge?

Theorem: If f(x) is piecewise C1[0, 2π] then the Fourier Series of f converges to

1

2[f(x+) + f(x−)] ∀x ∈ (0, 2π)

This implies:


1. At points of continuity, the series converges point-wise

2. At points of discontinuity, the series converges to the average of the left and right hand limits

3. The endpoints converge at the average of the endpoints.

Different norms imply different styles of convergence.

Example: Let f(x) =

0 0 ≤ x < π1 π ≤ x < 2π

a0 =1

2π

∫ 2π

0f(x)dx =

1

2π

(∫ π

00dx+

∫ 2π

π1dx

)

=1

2

an =1

π

∫ 2π

0f(x) cosnxdx =

1

π

∫ 2π

πcosnxdx =

1

nπsinnx|2π

π = 0

bn =1

π

∫ 2π

0f(x) sinnxdx =

1

π

∫ 2π

πsinnxdx =

−1

nπcosnx|2π

π

=1

nπ((−1)n − 1) =

0 n even−2nπ n odd

So the Fourier Expansion of f(x) is

f(x) ∼ a0 +

∞∑

n=0

an cosnx+ bn sinnx =1

2+

∞∑

n=0

−2

(2n + 1)πsin (2n + 1)x

Fourier Theorem says this series converges point-wise to points of continuity. We know that it convergeson the intervals (0, π) and (π, 2π)

Chapter 6

Integral Equations

Model: Let u(t) be the population at time t. From t0 to t1, ie. ∆t, we know:

1. The function u(t) can produce Au net new individuals

2. Migration implies f(t)

The equation for u(t) can be expressed as

u(t) =

N∑

n=1

Au(tn)∆t+ f(t)∆t

Expressed as an integral, u(t) becomes

u(t) =

∫ t

0Au(τ)dτ + F (t)

Another Model: Let z(x) represent the deflection of a beam under load and described as:

z(x) =

∫ 1

0G(x, y)p(y)dy

where

p(y) = load distribution

G(x, y) = describes material/load

If you rotate the beam, p(y) becomes p(y) = ω2µ(y)z(y); where ω represents the angular velocity andµ represents the mass density.

67

68 CHAPTER 6. INTEGRAL EQUATIONS

Substituting ω and µ, z(x) becomes

z(x) = ω2

∫ 1

0G(x, y)µ(y)z(y)dy

Example: Let’s look at an example where integral equations come into play

Let u solvedu

dt= sinu+ t with the initial condition u(0) = 1

Integratingdu

dtfrom 0 to t gives the equation

∫ t

0

du

dt= u(t)− u(0) = u(t)− 1

∫ t

0

du

dt=

∫ t

0sinu(τ)dτ +

1

2t2

Setting the above equations equal to each other, and solving for u(t) yields

u(t) =1

2t2 + 1 +

∫ t

0sinu(τ)dτ

Conversions: Given the boundary value problem u′′ = f(x) with u(0) = u(1) = 0 and if you let k equalthe piecewise function

k(x, y) =

y(x− 1) 0 ≤ y < x ≤ 1x(y − 1) 0 ≤ x < y ≤ 1

Then the solution to u(x) is an integral equation, which takes the form u(x) =

∫ 1

0k(x, y)f(y)dy. There

are four different types of integral equations that we will look at.

Types of Integral Equations

Fredholm Integral Equation (1st Kind)

u(t) =

∫ b

ak(t, τ)f(τ)dτ

Note: if k(t, τ) = 0 for t < τ this is the same as the Volterra Integral Equation

Fredholm Integral Equation of the 2nd Kind

u(t) =

∫ b

ak(t, τ)f(τ)dτ + f(t)

Volterra Integral Equation

69

if k(t, τ) = 0 when t < τ then

u(t) =

∫ t

ak(t, τ)f(τ)dτ

An important thing to notice, is that all of these integral equations are linear.

Example: We will be looking for a curve that seems linear, but is off-set by the area under it’s curve.

Lets look at the equation below, withdu

dx= 1

u(t) = x+1

2

∫ 1

0u(y)dy

If you let

∫ 1

0u(y)dy = k where k is some constant and then substitute k into the above equation the

result is

u(x) = x+1

2k

Plugging u(x) into the integral equation∫ 1

0(y +

1

2k)dy = k

Integrate and then solve for k gives you k = 1

Substituting this back into u(x) gives the solution for u(x) which is

u(x) = x+1

2

Example: Let

u(x) =5

6+x

3

∫ 1

0u(y)dy

Using the same method as the previous example, let k =

∫ 1

0u(y)dy, and substituting this in gives

u(x) =5

6+x

3k

Plugging this in and solving for k yields

k =

∫ 1

0

(5

6+yk

3

)

dy

=5

6+k

65k

6=

5

6k = 1


Substituting k back in, we get

u(x) =5

6+x

3

Example: This is an integral equation.

u(t) =

∫ b

ak(t, τ)u(τ)dτ + f(t)

Definition : L : H → H is a Bounded Linear Operator if

1) L is Linear – L(αf + βg) = αLf + βLg

2) L is bounded – There is a constant k > 0 such that

‖Lf‖ ≤ k‖f‖

where ‖ ∗ ‖ is the induced norm.

Definition: Norm on an operator

‖L‖ = sup‖Lu‖‖u‖

with u 6= 0

Example:

Lf = f ⇒ ‖Lf‖ = ‖f‖ ⇒ k = 1

Thus, Lf is a bounded linear operator.

Example: Let’s show that Lf is a bounded linear operator, where Lf is

Lf =

∫ 1

0f(x)dx

To prove look at

‖Lf‖2 =

∫ 1

0(Lf)2dt =

∫ 1

0

(∫ 1

0f(x)dx

)2

dt

=

(∫ 1

0f(x)dx

)2 ∫ 1

0dt =

(∫ 1

0f(x)dx

)2

71

=

(∫ 1

0f(x) ∗ 1dx

)2

≤([∫ 1

0f(x)2dx

]1/2 [∫ 1

012dx

]1/2)2

〈f(x), 1〉 ≤ ‖f‖‖1‖⇒ ‖Lf‖2 ≤ ‖f‖2

⇒ ‖Lf‖ ≤ ‖f‖Thus, Lf is a bounded linear operator.

Example: Let’s show that Lf is not a bounded linear operator, where Lf is

Lf =df

dxon [0, 2π]

a) Lf is linearL(αf + βg) = αLf + βLg

b) Is Lf a bounded operator? No. Take f(x) = sin(nx) on [0, 2π]. To prove look at

‖Lf‖2 =

∫ 2π

0(Lf)2dx =

∫ 2π

0(n cos(nx))2dx

= n2

∫ 2π

0cos2(nx)dx =

n2

2

∫ 2π

0(1 + cos(2nx))dx

=n2

2(2π) = n2π

Now look at

‖f‖2 =

∫ 2π

0sin2 nxdx =

1

2

∫ 2π

0(1− cos(2nx))dx

1

2(2π) = π

Is there a k such that ‖Lf‖ ≤ k‖f‖ ?n√π ≤ k√π

Is n ≤ k ∀ n ∈ N? Nope. Thus, it is not a bounded operator.

Definition: We have a Hilbert Schmidt Kernal if

∫ b

a

∫ b

ak2(x, y)dydx <∞

for Lu =∫ ba k(x, y)u(y)dy


Definition: The Adjoint of L is L∗

⇒ 〈Lu, v〉 = 〈u,L∗v〉for all u, v ∈ H, where 〈∗, ∗〉 is an inner product in H. But the Adjoint might not exist.

Example: Let

Lu =

∫ b

ak(x, y)u(y)dy

Finding the adjoint we get

〈Lu, v〉 =

∫ b

a

(∫ b

aku(y)dy

)

v(x)dx

=

∫ b

a

∫ b

ak(x, y)u(y)v(x)dydx =

∫ b

a

∫ b

ak(y, x)u(x)v(y)dxdy

=

∫ b

a

(∫ b

ak(y, x)v(y)dy

)

u(x)dx = 〈u,L∗v〉

Thus we have

L∗v =

∫ b

ak(y, x)v(y)dy

Example: Find the adjoint operator for

Lf =df

dxwith f(0) = f(1) = 0

Then

〈Lf, g〉 =∫ 1

0Lfgdx =

∫ 1

0

df

dxgdx

Using integration by parts we get

= f(x)g(x)|10 −∫ 1

0fdy

dxdx

= −⟨

f,dy

dx

⟩

= 〈f, L∗g〉

Thus the adjoint operator is

L∗g = −dydx

Theorem: If L is a bounded linear operator then

1) L∗ exists

2) L∗ is a bounded linear operator

73

Definition: A set S ∈ H is compact if any sequence xn chosen from S has a convergent subsequence.

Definition: An operator L is compact if it transforms bounded sets into compact sets.

Fredholm Theorem: If L is a compact linear operator then

1) The solution of Lu = f is unique if and only if Lu = 0 has only u = 0 as a solution.

2) The equation Lu = f has a solution if and only if 〈f, v〉 = 0 for every v ∈ N(L∗).

Example: Continuous dependence:

Solve Ax = b for(

1 11 1.00001

) (x1

x2

)

=

(12

)

This matrix represents the equations

x1 + x2 = 1

x1 + 1.00001x2 = 2

Now a small change in the number 1.00001 results in a large change in the solution to this system ofequations. Thus, the solution to this model is very sensitive to small changes in the data, so our model isuseless.

Definition: A problem or model is said to be well-posed if the problem has a unique solution whichdepends continuously on the problem data.

Spectral Theorem: Let K be a compact linear operator and assume that Ku = λu.1. The multiplicity of λ is finite.2. The adjoint of K exists.3. When K = K∗ (self adjoint K), the eigenvalues are real.4. When K = K∗, the eigenfunctions corresponding to distinct eigenvalues are orthogonal.5. Let λn be a collection of distinct eigenvalues, either

i. limn→∞ λn = 0 orii. only a finite number ar non-zero.

6. Self adjoint, compact operators (non-trivial) will have at least one eigenvalue-eigenfunction pair.7. Compact self adjoint non-degenerate operators have an infinite number of orthogonal eigenfunctions.8. If K is a compact self adjoint linear operator then the eigenfunctions are complete over

the range of K.Note that points 7 and 8 imply that a Fourier Series will exist and work.


Example: Consider Lu = −∫ 1

0k(x, y)u(y) dy where k(x, y) =

y(x− 1) 0 < y < x < 1x(y − 1) 0 < x < y < 1

.

⇒ Lu = λu

⇒ −∫ 1

0k(x, y)u(y) dy = λu(x) An eigenvalue problem!

Claim: L is self adjoint. (Assuming compact.)Find the eigenvalues and eigenfunctions.

∫ 1

0k(x, y)u(y) dy = −

∫ x

0k(x, y)u(y) dy −

∫ 1

xk(x, y)u(y) dy

= −∫ x

0y(x− 1)u(y) dy −

∫ 1

xx(y − 1)u(y) dy

= −(x− 1)

∫ x

0y u(y) dy − x

∫ 1

x(y − 1)u(y) dy

= λu(x)

Notice that by the Fundamental Theorem of Calculus that∫ x

0u(y) dy is the inverse of differentiation (with respect to x).

⇒ λdu

dx= −

∫ x

0y u(y) dy + (1− x)[x ·u(x)] −

∫ 1

x(y − 1)u(y) dy + x[(x− 1) ·u(x)]

Differentiating again with respect to x.

⇒ λd2u

dx2= −[x·u(x)] + (x− 1)u(x) = −u(x)

⇒ 0 =d2u

dx2+

1

λu(x) Define µ2 = 1

λ .

0 =d2u

dx2+ µ2 u(x)

We know the solution of this ordinary differential equation is

u(x) = A cos(µx) +B sin(µx).

With u(0) = 0 and u(1) = 1 we obtain the following eigenfunctions for u.

⇒ φn(x) = sin(nπx)

λn =1

n2π2

Chapter 7

Green’s Functions

Many problems in modeling deal with differential operators, but the theory is stated in terms of integraloperators. How can we connect these? We expect the inverse of a differential operator to be an integraloperator.

Formal ArgumentIf L is a differential operator,

Lu = andnu

dxn+ an−1

dn−1u

dxn−1+ ...+ a1

du

dx+ a0

then we expect

L−1u =

∫ b

ag(x, t)u(t) dt.

If it is an inverse, then u = L(L−1u).

u = L

(∫ b

ag(x, t)u(t) dt

)

=

∫ b

aLgu(t) dt = u(x)

⇒∫ b

a[Lg]u(t) dt = u(x) x ∈ [a, b]

⇒ Lg = δ(x, t)

Dirac Delta Function: Theory of DistributionsNormally the Dirac Delta Function, δ(x), is used to describe an impulse at x = 0.We know the following statements about the Dirac Delta Function:

δ(x) = 0 for x 6= 0

∫ b

aδ(x) dx = 1 if x ∈ (a, b).

75

76 CHAPTER 7. GREEN’S FUNCTIONS

Consider the following fucntions:Absolute value

|x| =−x x < 0x x > 0

One sided absolute value

f(x) =

0 x < 0x x > 0

After differentiating once

H(x) = f ′(x) =

0 x < 01 x > 0

After differentiating again

δ(x) = H ′(x) = f ′′(x) =

0 x < 00 x > 0

⇒∫ b

afunction ⇒ 1

⇒∫ b

aδ(x) dx = 1 0 ∈ (a, b)

Just as Ax = b has an inverse x = A−1b, so we want Lu = f to have an inverse u = L−1f .

If L is an nth order differential operator and Lg = δ(x− t) [= δ(x, t)]then

1. The nth derivative of g looks like δ(x − t).2. The (n− 1)th derivative of g looks like H(x− t).3. The (n− k)th derivative of g looks continuous.

Recall:

Lu = anu(n) + an−1u

(n−1) + ...+ a0u

Since Lg = 0 if x 6= t:

1 =

∫ ∞

−∞Lg dx =

∫ t−ǫ

−∞Lg dx+

∫ t+ǫ

t−ǫLg dx+

∫ ∞

t+ǫLg dx

Note that

∫ t−ǫ

−∞Lg dx and

∫ ∞

t+ǫLg dx are both zero. So this gets us

1 =

∫ t+ǫ

t−ǫLg dx =

∫ t+ǫ

t−ǫ[an

dng

dxn+ an−1

dn−1g

dxn−1+ ...+ a0 g] dx

1 = andn−1g

dxn−1

∣∣∣∣

t+ǫ

t−ǫ

+ an−1dn−2g

dxn−2

∣∣∣∣

t+ǫ

t−ǫ

+ ...+ a0

∫ t+ǫ

t−ǫg dx

77

Notice that an−1dn−2g

dxn−2

∣∣∣∣

t+ǫ

t−ǫ

+ ...+ a0

∫ t+ǫ

t−ǫg dx = 0,an−1

dn−2g

dxn−2

∣∣∣∣

t+ǫ

t−ǫ

is continuous,

and also that φ(x)|ǫ−ǫ = φ(ǫ)− φ(−ǫ) = φ+(0)− φ−(0) = 0.

⇒ andn−1g

dxn−1

∣∣∣∣

t+ǫ

t−ǫ

= 1 as ǫ→ 0

Summary: Method to find the Green’s functionGreen’s function is the kernel for the integral inverse operator.

1. Lg(x, t) = 0 for x 6= t.

2.dk

dxk(g(x, t)) is continuous (with k = 0, 1, 2, ..., n − 2).

3.dn−1

dxn−1g(x, t)

∣∣∣∣

t+

t−=

1

an. (Note that this is the jump condition.)

4. g(x, t) is a piecewise defined function.

Next we will consider examples of second order differential equations which are important because themost frequent types of problems seen in Math Modeling occur in this form.

Example: Find the Green’s function ford2u

dx2= f , u(0) = u(1) = 0.

Lu = f ⇒ u = L−1f

Lu = anu(n) + an−1u

(n−1) + ...+ a0u

Using step 1:

Lg = 0 if x 6= t

d2

dx2g = 0 if x 6= t

⇒ g(x, t) = Ax+B

Using step 4:

⇒ g(x, t) =

Ax+B 0 < x < t < 1Cx+D 0 < t < x < 1

Using step 2:Since g is continuous at x = t, g(0) = 0, and g(1) = 0 we get

⇒ g(x, t) =

Ax 0 < x < t < 1C(x− 1) 0 < t < x < 1.

Also, since x = t ⇒ At = C(t− 1) ⇒ C =At

t− 1.

⇒ g(x, t) =

Ax 0 < x < t < 1Att−1 (x− 1) 0 < t < x < 1

where t is a number


Finally, using step 3:

dg

dx=

A x < tAtt−1 t < x

Since1

an=

1

a2=

1

1= 1:

dg

dx

∣∣∣∣

t+

t−=

At

t− 1−A = 1

A

(t

t− 1− 1

)

= 1

A =11

t−1

= t− 1

⇒ g(x, t) =

(t− 1)x 0 < x < t < 1t(x− 1) 0 < t < x < 1

Therefore, for u′′ = f , u(0) = u(1) = 0 we obtain

u =

∫ 1

0g(x, t)f(t) dt (where g(x, t) is given above).

Example: Consider Lu = u′′ + λu where u ∈ L2(−∞,∞).

[∫ ∞

−∞u · u dx < m ⇒

∫ ∞

0f(x) dx < m

]

(Notice u→ 0 as x→ ±∞ and that λ is complex, not real and positive).Using step 1:

Lg = 0 if x 6= t

g′′ + λg = 0 ⇒ m2 + λ = 0 ⇒ m = ±i√λ

⇒ emx ⇒ e±i√

λx

Recall Euler’s Formula: eiθ = cos(θ) + i sin(θ).Adding step 4:

⇒ g(x, t) =

Aei√

λ(x−t) −∞ < t < x <∞Be−i

√λ(x−t) −∞ < x < t <∞

Using steps 2 and 3:

g′(x, t) =

Ai√λei

√λ(x−t) t < x

−Bi√λe−i

√λ(x−t) x < t

g|t+t− = Aei√

λ(t−t) −Bei√

λ(t−t) = 0 ⇒ A = B

Substituting back in we get

g′|t+t− = Ai√λ− (−Ai

√λ) = 1 ⇒ A =

1

2i√λ.

79

⇒ g(x, t) =

1

2i√

λei√

λ(x−t) −∞ < t < x <∞1

2i√

λe−i

√λ(x−t) −∞ < x < t <∞

⇒ g(x, t) =1

2i√λei√

λ|x−t|

Say you want to solve Lu = f ⇒ u = L−1f , then u is as follows:

u(x) =

∫ ∞

−∞g(x, t) f(t) dt =

∫ ∞

−∞

1

2i√λei√

λ|x−t| f(t) dt.

Example: Convert u′′ + λu = u3 where u ∈ L2(−∞,∞) to an integral form.

⇒ u(x) =

∫ ∞

−∞

1

2i√λei√

λ|x−t| u3(t) dt.

⇒ Lu = u3 ⇔ u = L−1(u3)

Adjoint for differential operator:The adjoint of L is L∗: 〈Lu, v〉 = 〈u,L∗v〉.

Example: Consider Lu =1

ω(x)

d

dx

(

ρ(x)du

dx

)

+ q(x)u(x) with α1 u(a) + β1 u′(a) = 0 and

α2 u(b) + β2 u′(b) = 0. This is called a Sturm-Liouville operator (with separated boundary conditions).

(Partial differential equation courses study Lu = f .)Find L∗.

[Deduce the space (functions on [a, b]) and the inner product

(∫ b

au(x) v(x)ω(x) dx = 〈u, v〉

)

.]

〈Lu, v〉 =

∫ b

a(Lu) v ω(x) dx

=

∫ b

a

[1

ω

d

dx

(

ρdu

dx

)

+ q u

]

v ω dx

=

∫ b

a

1

ω

d

dx

(

ρdu

dx

)

v ω dx+

∫ b

aq u v ω dx

Now, integration by parts.

= ρdu

dxv

∣∣∣∣

b

a

−∫ b

aρdu

dx

dv

dxdx+

∫ b

aq u v ω dx

Integration by parts again.

= ρdu

dxv

∣∣∣∣

b

a

− ρ udvdx

dx+

∫ b

aud

dx

(

ρdv

dx

)

dx+

∫ b

aq u v ω dx

Notice that these last two intergrals combine...


= ρdu

dxv

∣∣∣∣

b

a

− ρ udvdx

dx+

∫ b

au

[1

ω

d

dx

(

ρdv

dx

)

+ q v

]

ω dx

... and that the last integral has the inverse we are looking for!

= ρ(b)u′(b) v(b) − ρ(a)u′(a) v(a) − ρ(b) v′(v)u(b) + ρ(a) v′(a)u(a) +

∫ b

au (L∗v)ω dx

Using algebra on the given boundary conditions, this simplifies to just the integral,

=

∫ b

au (L∗v)ω dx

α1 v(a) + β1 v′(a) = 0

also giving us α2 v(b) + β2 v′(b) = 0 which are the boundary conditions for v.

Therefore, the Adjoint for Lu, with boundary conditions, is

⇒ L∗v =1

ω(x)

d

dx

(

ρdv

dx

)

+ q(x) v(x)

α1 v(a) + β1 v′(a) = 0

α2 v(b) + β2 v′(b) = 0

Definition:If L = L∗, D(L) = D(L∗) (their domains are the same), and their boundary conditions are the same, thenwe say L is self adjoint.

Definition:For a differential operator L, the pair φ, λ is called an eigenfunction, eigenvalue pair if Lφ = λφ (φ 6= 0).

Theorem:The eigenfunctions of a self adjoint invertible second order differential operator form a completre set onL2[a, b].

1. Convert to an integral equation: Green’s function.2. Symmetric Green’s function ⇒ Hilbert-Schmidt operator.3. Have a compact self adjoint integral operator.4. Spectral Theorem: Eigenfucntions are complete.

Eigenfunction expansionsFor to solving Lu = f , the general approach is to use eigenfunction expansions.

L: the eigenfunctions & eigenvalues ⇒ φn, λn

u =∞∑

n=1

αn φn

81

f =

∞∑

n=1

βn φn

This is a generalized Fourier Series ⇒ βn = 〈f, φn〉. The problem Lu = f transforms into

L

( ∞∑

n=1

αn φn

)

=

∞∑

n=1

βn φn

. ∞∑

n=1

αn λn φn =∞∑

n=1

βn φn

⇒ αn λn = βn

⇒ αn =βn

λn

Formula: Lu = f for any linear operator

1. f =

∞∑

n=1

βn φn : βn = 〈f, φn〉

(Changing coordinate system.)

2. αn =βn

λn(Transformation into new coordinates.)

3. u =

∞∑

n=1

(βn

λn

)

(Transformation back.)

Lu = f −→ u = L−1f

↓ ↑

Tnu = 〈u, φn〉 T−1 ⇒∞∑

n=1

αn φn

Tnf = 〈f, φn〉↓ ↑

αn λn = βn −→ αn =βn

λn

For Green’s Function:

〈f, g〉 =

∫ b

af(x) g(x) dx, solve Lg(x, t) = δ(x − t).

Consider first the Dirac Delta portion:

δ(x− t) =∞∑

n=1

βn φn(x)

⇒ βn = 〈δ(x− t), φn(x)〉


=

∫ b

aδ(x− t)φn(x) dx = φn(t)

(from the properies of the δ function)

⇒ δ(x− t) =

∞∑

n−1

φn(t)φn(x)

Next consider g:

g =

∞∑

n=1

αn φn

Lg =∞∑

n−1

αn λn φn

Now since Lg = δ:⇒ αn λn = φn(t)

⇒ g(x, t) =∞∑

n−1

φn(t)φn(x)

λn

Therefore, Lu = f ⇒ u = L−1f

L−1v =

∫ b

ag(x, t) v(t) dt, where g(x, t) is given above.

Example: Consider u′′ = 12 sin(7πx), with u(0) = u(1) = 0. This is a problem of the form Lu = f .

Want: u =

∞∑

n=1

αn φn. What are the φn?

Solve Lu = λu ⇒ u′′ − λu = 0.This is an ordinary differential equation with the solution of

u(x) = A cos(√−λx) +B sin(

√−λx), where λ is negative.

Applying the boundary conditions:

u(0) = A cos(0) +B sin(0) = 0

⇒ A = 0

⇒ u(x) = B sin(√−λx)

u(1) = B sin(√−λ) = 0

Either B = 0 or sin(√−λ) = 0.

B = 0 is a trivial solution so...

⇒ sin(√−λ) = 0 when

√−λ = nπ for n ∈ N.

⇒ λ = −n2π2 (Note that λ is negative)

⇒ φn(x) = sin(nπx)

83

Plugging in f :

12 sin(7πx) =∞∑

n=1

βn sin(nπx)

Next, u =

∞∑

n=1

αn sin(nπ x):

⇒ β7 = 12

⇒ βn = 0 for n 6= 7

α7 =β7

λ7where λn = −n2π2

⇒ u(x) =β7

λ7sin(7πx)

⇒ u(x) =12

−49πsin(7πx)


Chapter 8

Eigenfunction Expansions

Solving an operator problem of the form Lu = f may be accomplished with eigenfunction expansions, alsoknown as a Fourier series. The utility of this method is dependent on the ease of finding eigenfunctions.This chapter illustrates the method of eigenfunction expansion solutions and then concludes with someexamples.

For some eigenfunction of an operator, namely functions that solve the problem Lφn = λnφn, aneigenfunction expansion is then the sum

u(x) =

n∑

i=1

ciφi(x)

where the summation is conducted over each eigenfunction. If the operator L is self-adjoint the processof finding the coefficients of the expansion is reduced considerably.

If the eigenfunctions of an operator L are mututally orthogonal with respect to some inner product,then

〈φi, φj〉 = ‖φn‖2 δij .

Proceeding formally, consider the form

Lu− f = 0.

We assert in this formal calculation that an eigenfunction expansion exists, that the operator L acts onthis expansion term-by-term, and the operations of summation and the inner product are interchangable.

Consider the result of taking the inner product of the above form with the eigenfunction φm.

〈Lu− f, φm〉 = 0

85

86 CHAPTER 8. EIGENFUNCTION EXPANSIONS

⟨

L

(n∑

i=1

ciφi(x)

)

− f, φm

⟩

=

⟨(n∑

i=1

ciλiφi(x)

)

− f, φm

⟩

⟨(n∑

i=1

ciλiφi(x)

)

, φm

⟩

− 〈f, φm〉 = 0

The result of this calculation, then, is that

〈f, φm〉 =∑

ciλi 〈φi, φj〉 .

For L self adjoint, then〈f, φm〉 = ciλi ‖φi‖2

and the coefficents have been obtained. Then the potential eigenfunction solution for the problem is of theform

Lu = f → u(x) =∑ 〈f, φm〉

λi ‖φi‖2φi(x)

Using orthogonal polynomials as the orthogonal function of the expansion are good ways to approximatesolutions.

Consider what might result when λi = 0 for some i. For such a situation, the operator has a nullspace.The existance of a zero eigenvalue means that there exists a φi such that Lφi = 0. This operator is notinvertible, and an exact eigenfunction may not be found. The approximate solution is obtained by ignoringthe offending eigenvalue

u(x) =∞∑

j=1,j 6=i

1

λj〈f, φj〉φj

The method for finding an eigenfunction expansion solution to an operator problem is then accomplishedvia these steps

1. Find φn, λn such that Lφn = λnφn.

2. Normalize φn so that ‖φn‖2 = 1.

3. Consider the Fourier series expansion of f

4.

u(x) =∞∑

i=1

1

λi〈f, φi〉φi

Now follows examples to illustrate the method.

Example: Find eigenfunctions for Lu = u′′ with u(0) = u(1) = 0.

87

φ′′ = λφ with φ(0) = φ(1) = 0

has trigonometric solutions for µ 6= 0, namely c1 sin(µx) + c2 cos(µx), where λ = −µ2. Note, that itnot necessary for µ to be real, positive, or otherwise specfied. µ may be found by refering to the boundaryconditions.

For µ = 0, the solution is the the line αx+β which satisfies the boundary conditions only if α = β = 0.To satisfy the Dirichlet boundary conditions, it must be the case that

c1 sin(0) + c2 cos(0) = 0

c1 sin(µ) + c2 cos(µ) = 0

Namely c2 = 0 and µ = nπ.

Then our eigenfunctions become

φn = sin(nπx).

Note that

‖φn(x)‖2 =

∫ 1

0sin(nπx)2dx

=

∫ 1

0

1

2(1− cos(2nπx)) dx

=1

2.

Normalizing the eigenfunctions then results that

φn =√

2 sin(nπx).

Example: Consider the problem u′′ = f where u(0) = u(1) = 0.

The solution to this problem is considered to be of the form

u(x) =∞∑

n=1

cnφn

where the cn are given as

cn =〈f, φn〉λn

.

Example: Consider the above problem, Lu = f , when f = sin(7πx).


The solution, then, is

u(x) =∞∑

n=1

cn√

2 sin(nπx)

where the cn are

cn =

⟨sin(7πx),

√2 sin(nπx)

⟩

−n2π2.

Note that

〈sin(7πx), sin(nπx)〉 =

12 : n = 70 : n otherwise

So, then, the eigenfunction expansion reduces greatly as many terms of the series vanish.

The final solution is u(x) = 1λ7

⟨sin(7πx),

√2 sin(nπx)

⟩√2 sin(nπx) or

u(x) =1

−72π2sin(7πx)

Example: Solve u′′ + u = f(x) such that u(0) = 0 and u′(π) = 0.

1. Solve the eigenvalue problem.

φ′′ = λφ

Note that this has trigonometric solutions for nonzero eigenvalues. To satisfy the boundary condi-tions, then

c1 sin(0) + c2 cos(0) = 0

c1µ cos(µπ)− c2µ sin(µπ) = 0

The boundary conditions require c2 = 0 and that µπ =(n+ 1

2

)π.

The unnormalized eigenfunctions are

φn(x) = sin(

(

n+1

2

)

x)

with eigenvalue

λn = −(

n+1

2

)2

89

2. Normalize the eigenfunctions.

‖φn‖2 =

∫ π

0sin2((n+

1

2)x)dx

=π

2

So, our normalized eigenfunctions are then

φn(x) =

√

2

πsin(

(

n+1

2

)

x)

3. Recast problem in terms of Fourier series

u′′ + u = f(x)∑

aiλiφi(x) +∑

aiφi(x) =∑

〈f, φi(x)〉φi(x)

Note then that it must be the case that anλn + an = 〈f, φn〉.

4. The final solution is then

u(x) =

∞∑

i=1

〈f, φn〉λn + 1

φn(x)

where

φn =

√

2

πsin

((

n+1

2

)

x

)

λn = −(

n+1

2

)2


Chapter 9

Differential Operators

We have spent several chapters on the manipulation of functions. This chapter will focus on the applicationsof our functions and what they truly mean. The following is a list of our current operators and theirapplications.

• du

dx,∂u

∂x: The rate of change with respect to x, slope, or flux

• d2u

dx2,∂2u

∂x2: Acceleration, measure of curvature of u

• ∇f : Gradient, direction of greatest change

• Duf = ∇f · ~u : Directional derivative, flow in some direction

• ∂u

∂n= Dnu = ∇u · ~n : Normal derivative, flux or flow normal to the boundary, common in physical

problems

• We often see ∇u =

⟨∂u

∂x,∂u

∂y,∂u

∂z

⟩

= Uxi+ Uy j + Uz k

• In ℜn: ∇n =

⟨∂u

∂x1,∂u

∂x2. . .

∂u

∂xn

⟩

Note: Outside of 3-space functions, some properties in Calculus

II are not valid. You can extend the gradient out to any space without losing functionality.

• ∇n = grad(u) = vector

• ∇ · F = div (~F ) =∂

∂x1F1 +

∂

∂x2F2 . . . +

∂

∂xnFn

• F = 〈F1, F2, F3〉In many applications, the expansion or contraction of a vector field at a point is basically describinga flow. We can combine the gradients of a function because the gradient is simply a scalar value andthe divergence of ∇u is defined.

91

92 CHAPTER 9. DIFFERENTIAL OPERATORS

• dir(grad(u)) = ∇ · ∇u = ∇2u = ∆u: commonly seen as the Laplacian

• ∆u =n∑

i=1

∂2u

∂x2i

=∂2u

∂x+∂2u

∂y: This contains curvature information and diffusion. This cannot be

applied for 3-space, it is only valid in 2-space. This is an important application for several areas ofstudy, such as electro-magnetic theory and heat transport.

We expect our models will have some basic structure that looks familiar to us. Let’s look at an examplewith heat flow.

Heat Flow in an Insulated Rod

Figure 9.1: Insulated Rod

This example will deal with heat flow in an insulated rod. Heat energy will only move along the x-axis.Also, this rod will have a uniform density. We will model temperature as a function of time and position,T (x, t).

Temperature is measuring energy. We can relate this to heat energy, u.

u = CpρT , where Cp = constant specific heat, ρ = density, and T = temperature at (x,t)

If we look at a single segment of the rod, we determine the heat of the entire rod, E(t):

E(t) = u ·∆x = CpρT∆x

dE(t) = CpρTdx

E(t) =

∫ L

0CpρT (x, t)dx

So, how does temperature change?

∂

∂tE(t) =

∫ L

0Cpρ

∂T

∂xdx

Notice that the heat energy can change in two ways.

93

1. source or sink of energy

2. a flux or diffusion

If we look at diffusion, we must examine the transport of energy across the boundary. Heat flow will benormal to the cross-section. Also, according to Fourier’s Law, heat will flow downhill. Therefore, heat flux(φ), is proportional to the temperature gradient.

φ = −k∂T∂x

with k being some constant. Notice the minus sign in front of k. This is because of the negative flux value,or the downhill effect.

φ(a, t) = energy increase, flow in from the left

−φ(b, t) = flow in on the right

We know this holds true for any interval (a, b). We can generalize this effect to help solve our problemover the interval (0, L). We know the change in energy will be:

dE

dt= φ(0, t) − φ(L, t) +

∫ L

0Qdx

where Q is some accumulated source term

dE

dt= −

∫ L

0

∂φ

∂xdx+

∫ L

0Q(x, t)dx

d

dt[

∫ L

0Cpρ

∂T

∂tdx] = −

∫ L

0

∂φ

∂xdx+

∫ L

0Q(x, t)dx

∫ L

0[Cpρ

∂T

∂t+∂φ

∂x−Q]dx = 0

Now, notice the fact

∫ b

af(x)dx = 0 for any a and b. The only way for f(x) to have an integrand of 0

over any endpoints is for f(x) to be identically zero. So, we then know the following:

Cpρ∂T

∂t+∂φ

∂x−Q = 0

∂T

∂t= − 1

Cpρ

∂φ

∂x+ q

q =Q

Cpρ


This is the differential form of the Conservation of Energy. We can also add Fourier’s Law (φ = −k∂T∂x

)

to our current model to obtain:

∂T

∂t=

1

Cpρ

∂

∂x(k∂T

∂x) + q(x, t)

Now, k may not be a constant that is free from x. k may depend on the location. All Sturm-Liouvilleproblems come from the above form. If we assume k does not depend on x, we can pull out the constant.

a =k

Cpρ,∂T

∂t= a

∂2T

∂x2+ q(x, t)

Now, here is where we must apply initial and boundary conitions. T (0) must have a value. T (x, 0) = f(x)(initial conditions) and T (0, t) = T (L, t) = 0 (boundary conditions)

How do we solve the above situation? In time, the first order equation is your typical initial value prob-lem. The second order equation generates our appropriate transforms. First, determine the appropriateoperator to transform into a partial differential equation. For our example, we need to look at the secondorder equation:

∂2T

∂x2, T (0, t) = T (L, t) = 0

Lu = u′′, u(0) = u(L) = 0

The eigenfunctions are:

∞∑

n=1

Cn(t)sin(nπx

L). Time must be incorporated into this expansion (the Cn(t)

term). This is not normalized.

[

∫ L

0sin2(

nπx

L)dx]

1

2 =

√

L

2

So, we know the following holds true:

T (x, t) =

∞∑

n=1

Cn(t)

√

2

Lsin(

nπx

L)

For our example, let’s take a look at when q = 0. Then:

∂T

∂t= a

∂2T

∂x2, T (x, 0) = f(x), T (0, t) = T (L, t) = 0

We can plug in the series to form the following equation for both values of T (x, t):

95

∞∑

n=1

Cn(t)

√

2

Lsin(

nπx

L) = a

∞∑

n=1

Cn(t)−n2π2

L2

√

2

Lsin(

nπx

L)

Solving, we get ˙Cn(t) = −a−n2π2

L2Cn(t). We can use seperation of variables to get the solution:

Cn(t) = bne−a(−n2π2

L2)t

bn being a constant

Substituting back into our original T (x, t) equation...

T (x, t) =

√

2

L

∞∑

n=1

bne−a(−n2π2

L2)tsin(

nπx

L)

The above equation solves our PDE function. Now, let’s solve this for our initial conditions.

T (x, 0) =

√

2

L

∞∑

n=1

bnsin(nπx

L) = f(x)

Here, bn is the projection of of onto the eigenfunction,⟨

f,√

2Lsin(nπx

L )⟩

=√

2L

∫ L

0f(x)(sin(

nπx

L)dx

T (x, t) =2

L

∞∑

n=1

[

∫ L

0f(x)(sin(

nπx

L)dx]e

−a(−n2π2

L2)tsin(

nπx

L)

Differential Form of the Conservation Law in 3 Dimensions

We now want to study some quanitity with a verbal or understood Conservation Law. The change inthis quantity can occur by transport across the boundary or by internal sources or sinks.

Example: Heat Energy

Heat may flow across the boundary or heat may be produced internally by some process. Variables:

• Ω: Domain in ℜ3

• Γ: Boundary of Ω

• e(t): Internal heat density


Figure 9.2: Object with a boundary in multiple dimensions

• E(t): Total heat energy for Ω

• ρ: Density

• c: Heat Capacity

• u(x, y, z, t): Temperature at any point

• ~n: normal vector to the boundary, assume ||~n|| = 1

• ~m: vector orthogonal to ~n

Using the properties we have previously discussed, we know the following equations hold true:

e(t) = cpu(x, y, z, t)

E(t) =

∫ ∫

Ω

∫

cpu(x, y, z, t)dV

de

dt= Transport + Internal Sources or Sinks

q(x, y, z, t) = some function for sources or sinks

Total sources and sinks = Q(t) =∫ ∫

Ω

∫q(x, y, z, t)dV

Transport: Flow in/out of Ω

If we look at a section of our boundary, we know that the only flow into the domain is caused bya component parallel to the normal vector and by a component orthogonal to the boundary tangent(orthogonal to ~m).

Flux = Flux|| + Flux⊥

97

Flux = φ(x, y, z, t) = 〈φ,~n〉~n+ 〈φ, ~m〉 ~m

〈φ,~n〉 = ~φ · ~n

Flow across the boundary: ~φ · ~n

Total Flux out of the domain :∮∮~φ · ~ndS

So, the following holds true:

dE

dt= −

∮

Γ

∮

~φ · ~ndS +

∫ ∫

Ω

∫

q(x, y, z, t)dV

Notice the minus sign for the surface integral. This is to reverse the flux. Now, flux will be coming intothe domain. The above equation is often seen as the Conservation Law. We also know that the changeof energy over time is as follows:

dE

dt=

d

dt

∫ ∫

Ω

∫

cpu(x, y, z, t)dV =

∫ ∫

Ω

∫

cp∂n

∂tdV

The above equation is a bit too complicated for us to solve at this point. If we can make someassumptions, this problem becomes much easier to solve. Let’s assume the density, ρ, and c, c, do notchange with time. This will allow us to pull them out of the integrals as constants. By doing this, wecannot solve other equations (such as aerodynamic problems, gases, or thermodynamics).

Let’s combine both of these ideas into one idea.

∫ ∫

Ω

∫

cp∂u

∂tdV = −

∮

Γ

∮

~φ · ~ndS +

∫ ∫

Ω

∫

q(x, y, z, t)dV

We can eliminate the surface integral from this equation using the Divergence Theorem. The followingis then true:

∮

Γ

∮

~φ · ~ndS =

∫ ∫

Ω

∫

div(~φ)dV

∫ ∫

Ω

∫

cp∂u

∂tdV = −

∫ ∫

Ω

∫

div(~φ)dV +

∫ ∫

Ω

∫

q(x, y, z, t)dV

By setting the equation equal to 0, the following holds true:

∫ ∫

Ω

∫

[cp∂u

∂t+ div(~φ)− q]dV = 0

Since our domain is arbitrary, we know that:


cp∂u

∂t+ div(~φ)− q = 0

The above is an example of the Conservation Law of Energy which is usually stated in the followingform:

cp∂u

∂t= −div(~φ) + q

Now, using Fourier′s Law just as before we can show:

~φ = −k(x, y, z)∇u

Again, note the negative sign indicating a downhill flow in the direction of the temperature gradient.

Combining these two concepts, we get the following equation:

cp∂u

∂t= div(K(x, y, z)∇u) + q = ∇ · (k∇u) + q

This above is the heat equation that describes linear heat transport in the nth direction.

Chapter 10

PDE Models

10.1 Introduction to Diffusion, Convection and Conservation Laws

Partial Diffferential Equations is a very large subject with a long history. It finds itself in two worlds. Oneis in applications in many areas of the sciences and engineering (this century has witnessed applicationsin the social sciences as well). The other is in a wealth of applications within pure mathematics. It is thisdual nature that makes the study of PDE’s a core part of a mathematical education.

10.1.1 Background: Calc III

In this section we will be examining functions of several variables. We will be doing to look at problemsin 2D and 3D; with up to four independent variables and one dependent variables.

u(x, y, t)andf(x, y, z),

We will graph z = f(x, y), which is a Surface (2D object living in 3D) and w = g(x, y, z), whichis a Hypersurface (3D object living in 4D). Next set z = f(x, y) = c which is a Level Curve and setw = g(x, y, z) = c which is a level Surface; both are Level Sets.

Terminology:

1. Order: The order of an equation is given by the highest order derivative found in the equation.

2. Linear: An equation is said to be linear if the dependent variable and all of it’s derivatives appearas linear terms (with exponent 1).

3. Homogeneous: If u = 0 solves the PDE then it is said to be homogeneous.

Main Features of this section are the following:

99

100 CHAPTER 10. PDE MODELS

Gradient We will let f(x,y,z) equal a scalar function, then grad(f):

∇f =∂f

∂xi+

∂f

∂yj +

∂f

∂zk =< fx, fy, fz > .

∇ = ”del” operator. Gradient gives the steepest increase or decrease. If ∇f = 0, the surface is flat.

Divergence We will let ~F (x, y, z) be a vector function, then div~F :

div~F =∂F1

∂x+∂F2

∂y+∂F3

∂z= ∇ ~F

Divergence determines if vector field is spreading out or contracting.

Example of divergence and gradient: Let u(x,y,z)

We will find the divergence of the gradient by the following:

divgrad(u) = ∇ ∇u = ∇2u = ∆u = div∂u

∂xi+

∂u

∂yj +

∂u

∂zk

= divF1 i+ F2j + F3k =∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= Laplacian Formula

Diffusion How do we model the diffusion of heat energy in a metal object?

Define Domain insert sphere with domain indicated (To understand heat flow, we must break downinto small cubes.) Start with a piece of Ω⇒ D

Step 1) Use the conservation of energy law. Step 2) ⇒ Apply Physical law of heat flow ⇒ Heat model.

Heat in Metal

• e(x,y,z,t) - internal energy at a point. (Heat in elemental object, energy density)

• u(x,y,z,t) - temperature

• c - heat capacity

• ρ - density

e = cρu

This equation determines how much energy is in the total volume. Total internal heat energy

=

∫∫∫

De(x, y, z, t)dV

10.1. INTRODUCTION TO DIFFUSION, CONVECTION AND CONSERVATION LAWS 101

Change is total energy is determined by transport across the boundary plus source and sink terms. Pointsource terms are constants such as q(x,y,z,t). Total internal energy force is equal to total source, which isequal to ∫∫∫

Dq(x, y, z, t)dv

Transport is the flow across the boundary, which is measured by the dot product of point energy flowterm and the normal vector.

We will let ~φ(x, y, z, t) be the point energy flow and the flow across the boundary B will equal.

~φ · ~n = normal

The total energy flow across the boundary is determined by this surface integral.

∫∫

B

~φ · ~ndS

The above equations will be used to determine the conservation of energy.

Conservation of energy

d/dt

∫∫∫

Dedν =

∫∫∫

DqdV −

∫∫

B

~φ · ~ndS

Examine: We want to convert

∮∮

B

~φ · ~ndS to triple integral because Surface Integral are very hard to

compute. We will use the Divergence Theorem in 3-dimesions to do this.

Divergence Thm (3D)

example

∫ b

af(x)dx = F (b)− F (a) =

∑

B

F (x) |B ·n∫

Interior =

∫

Boundary of the antiderivative

∫∫∫

DdivFdV =

∫∫

B

~F · ndζ F is a vector function.

Converting heat transfer across the boundary:

∫∫

B

~φ · ~ndζ =

∫∫∫

Ddiv~φdν

⇒ d/dt

∫∫∫

DedV =

∫∫∫

DqdV −

∫∫∫

Ddiv~φdV

⇒∫∫∫

D

∂e

∂tdν = −

∫∫∫

Ddiv~φdV +

∫∫∫

DqdV


⇒∫∫∫

D[∂e

∂t+ div~φ − q]dV = 0

True for any D.

⇒ ∂e

∂t= −div~φ+ q

Conservation of Energy

∂e

∂t= −div~φ+ q

Convert to single equation and single unknown

Temp and Heat Energye = cρu

• c - heat capacity

• ρ - density

• e - energy

• u - temperature

∂

∂t(cρu) = −div~φ+ q

Realtion between temp and Flux (u and ~φ)

”heat flows downhill”

~φ = −k∇u

Model

∂

∂t(cρu) = −div~φ+ q

Assume that c, ρ, k are constants

⇒ ∂u

∂t= a∇2u+ f

a = k/cρ and f = q/cρ


Heat Equation - Parabolic Partial Differential Equation

Example 1 Metal Rod insert metal rod

also will have the metal rod at some initial temp u(x, 0) = h(x)

insert metal rod

insert graph

Assume No Source term (f = 0)

Model∂u

∂t= a

∂2u

∂x2

u(x, o) = h(x)u(o, t) = T0andu(L, t) = TL

Also note that the heat equation can be written as follows:

∂u

∂t= a∇2u+ f Note: this equation works for numerious dimension

∇2 is known as the Laplacian of u

Observe for numerous dimension the equation will change by inserting the different values for theLaplacian of u:

∇2u =∂2u

∂x2is the 1 Dimensional Representation

∇2u =∂2u

∂x2+∂2u

∂y2is the 2 Dimensional Representation

∇2u =∂2u

∂x2+∂2u

∂y2+∂2u

∂z2is the 3 Dimensional Representation

So the 1 dimensional equation (i.e. the model of the metal rod is) with initial and boundary conditionsgiven above is:

∂u

∂t= a

∂2u

∂x2+ f(x, t)

Now think about a two dimensional heat model. In this case we might be looking at a metal plate.The face of the plate is insulated and heat is applied to the boundaries. We will say that the plate is LxL


and has vertices at (0,0); (L,0) ; (0,L); (L,L). Along the bottom and top edges of the plate, heat appliedis given as h1(x) and h2(x). Along the left and right edges of the plate, heat applies is given as g1(y) andg2(y). Finally we will assume that the plate has an initial temperature distribution of w(x, y).

Equation for a 2-D heat model:

∂u

∂t= a(

∂2u

∂x2+∂2u

∂y2) + f(x, y, t); Equation (1)

u(x, y, 0) = w(x, y) ;this comes from given initial conditions.

Boundary Conditions:

u(x, 0, t) = h1(x) ; u(0, y, t) = g1(y)

u(x,L, t) = h2(x) ; u(L, y, t) = g2(y)

The equation (1) above could be translated to English as follows:

the change in temperature at a point = diffusion + source

Note that f(x, y, t) is the forcing term. Also observe that this equation can also be used to describechemical systems. (For the 2-D case think of a square culture dish!)

One question arises in one’s mind: How do we study these?

For ODE’s we would study the phase plane. The first step is to find the rest points of the ODE by

settingd

dt= 0.

The same is true for PDE’s! We wish to study the steady states which are given when the timederivative is zero.

du

dt= 0

So, for the 2-D model set equation (1) = 0 ; however, we can’t analyize the 2-D case. We will first lookat the 1-D model.

1-D Model Example:

∂u

∂t=∂2u

∂x2; where a = 1, f = 0, and 0 ≤ x ≤ 1

u(x, 0) = x(1− x)u(0, t) = 100

u(1, t) = 0

So, what is the steady state model? To analyize the steady state ignore the initial conditions and holdtime constant (which means no partial derivatives!). Now we get the following equation, and we wish tosolve the boundary value problem...


d2us

dx2= 0 with us(0) = 100 and us(1) = 0

This equation implies that we wish to find an object with zero curvature since the second derivative of uis zero.

By integrating the second order ODE we get:

dus

dx= A

then by integrating again we get:

us = Ax+B; which is a line (zero curvature!!!)

Now use the boundary values to solve us for A and B.

us(0) = 100 = B and us(1) = 0 = A+B = A+ 100; thus A = −100

So the steady state solution for u is:

us(x) = −100x+ 100 = 100(1 − x)

0

20

40

60

80

100

0 0.2 0.4 0.6 0.8 1

100*(1-x)

We will break now from this problem and look at some types of boundary conditions:


1. u(0, t) = a , u(1, t) = b. Note here that the temperature at the endpoints are fixed. This is knownas a Dirichlet Boundary Condition.

2.∂u

∂x(0, t) = a ,

∂u

∂x(1, t) = b. In this condition flow is perscribed and this is known as a Neumann

Boundary Condition. (If a = 0 for the Neumann condition, then this is representive of an insulatedcondition in which no flow across.)

3. u(0, t) + λ∂u

∂x(0, t) = a. This is a mixed condition, known as the Robin Boundary Condition. Tem-

perature and flow can be adjusted on either side. Newton’s laws allow us to combine these elements.

Note that for different boundary conditions that we are given, different solutions are generated.

Let’s look at an example of the mixed case. Here we are given an insulated metal rod with the 0 end

of the rod given by∂u

∂x= 0. The other end of the rod is heated at a fixed temperature of 10 (u = 10).

From 0 ≤ x ≤ 1. Since the rod is insulated, it implies that∂u

∂t= 0. Thus...

∂2us

∂x2= 0 with

∂u

∂x(0) = 0 , us(1) = 10

Following previous work...

us(x) = Ax+B ⇒ dus

dx= A = 0 and us(1) = B = 10⇒ us(x) = 10

Note that Models for Heat Diffusion ⇔ Models for chemical diffusion

i.e. Heat flow is very similar to Chemical diffusion.

An in depth look at components of the heat/chemical diffusion equation:

• ∂u

∂t=change in heat/chemical concentration at point x as a function of time.

• ∂2u

∂x2= diffusion of heat/chemical at x.

• ∂u

∂x= flow or transport term.

A look at 2-D chemical diffusion:

u = [chem]⇒ concentration of chemical.

(∂2u

∂x2+∂2u

∂y2)is the 2-D diffusion term at a point.


∂u

∂t= a(

∂2u

∂x2+∂2u

∂y2) + f(x, y, t); 2-D Equation for chemical process

u(x, y, 0) = g(x, y); Initial concentration.

Note that there is no movement across the boundaries, since we will assume that the system is containedin some sort of glass beaker in which diffusion will not occur beyond the boundaries. Analogy→ insulation: heat :: glass wall : chemical.

Model: String Under Tension

Let u(x, t) solve the wave equation on [0, L]. If we observe a segment of the string with length x, thenthe mass of the segment, m, can be defined by m ∼= ρ

√

x2 +y2, where ρ is the density of the string.

The forces acting on the string are

• Tension forces ~T1 and ~T2 (restoring and springlike forces), and

• Forces related to mass of string segment.

Simplifying assumptions include

1. Small deflections, implying u(x, t) is small,

2. Uniform density and sting,

3. No horizontal motion, and

4. No gravity or air resistance.

Force related to mass can be defined such that

~Fm = m~a = (ρx)∂2u

∂t2.

Force related to tension can be defined such that

~FT = ~T2 sin(θ2)− ~T1 sin(θ1)

From assumption (1), the angle of deflection, θ, is small; thus, sin(θ) ∼= tan(θ), but tan(θ) is approxi-mately the slope of u(x, t). This relation means that sin(θ) ∼= ∂u

∂x . Then,

~FT = ~T2ux(x+x, t)− ~T1ux(x, t).

But, T1 = T2 by assumption (3), so set T = T1. Then,

~FT = ~T (ux(x+x, t)− ux(x, t)) .


The total force is

ρx∂2u

∂t2= ~T (ux(x+x, t)− ux(x, t)) + ExternalForces+AirFriction,

where external forces include other vibrations, gravity, etc. Digression: if air friction would be considered,the force would be proportional to the surface area which is proportional to the length of the segment. Thus,air friction could be represented at being proportional to the velocity of the sting by kxut. Assumingthat the external forces and the forces being attributed to air friction are negligible by assumption (4) andsolving for the time term,

∂2u

∂t2=~T

ρ

(ux(x+x, t)− ux(x, t)

x

)

.

Letting x become really small,

limx→∞

(ux(x+x, t)− ux(x, t)

x

)

= uxx =∂2u

∂x2

Define~Tρ = c2, which implies

∂2u

∂t2= c2

∂2u

∂x2.

This equation is known as the Wave Equation (undamped).

The boundary conditions for this equation are often clamped at the end points such that u(0, t) =u(L, t) = 0. Names for common boundary conditions are

• Zero Boundary Data: Dirichlet’s Boundary Conditions and

• Zero Derivative Boundary Data: Neumann’s Boundary Conditions.

Most stringed instrument has Dirichlet’s boundary conditions, while many wind instruments haveboundary conditions that are a mixture of Dirichlet’s and Neumann’s boundary conditions such thatu(o, t) = ux(L, t) = 0.

Now, let us attempt to solve the linear undamped wave equation, where

∂2u

∂t2= c2

∂2u

∂x2

is the PDE,u(0, t) = u(L, t) = 0

are the boundary conditions, andu(x, 0) = f(x)ut(x, 0) = g(x)

are the initial conditions. This set of equations is a complete model that we hope is also well-posed. Then,for any set of initial conditions, how can we solve this model? Review

Ψ =∞∑

n=1

cnφn


such that

cn = 〈Ψ, φn〉 =

∫ b

aΨΦn(x) dx.

Thus,

u(x, t) =

∞∑

n=1

√

2

Lan(t) sin

(nπx

L

)

.

Analyzing the total energy of a string being modeled with this wave equation can provide significantinsight into the conservation of energy. If we represent total energy as E, kinetic energy as KE, andpotential energy as PE. Then,

E = KE + PE.

The kinetic energy at a point is

KE|pt =m~v2

2,

where ~v is the velocity. The potential energy is

PE =ks

2,

where k is essentially a spring constant and s is the local change in arc length. To find the total kineticenergy of the string, we need to relate the mass of the string to the length of the string such that m = ρx.Then, if we notice that ~v = ∂u

∂t ,

KE =N∑

i=1

ρx2

(∂u

∂t

)2

,

and if k = ~Tx and if s is the slope of the string displacement such that s = ∂u∂x , after replacing the

change with the partial change.

PE =

N∑

i=1

~Tx2

(∂u

∂x

)2

.

Taking the limit as N , the number of partitions of the string, goes to infinity,

KE = limx→∞

(N∑

i=1

ρx2

(∂u

∂t

)2)

=

∫ L

0

ρ

2

(∂u

∂t

)2

dx,

and

PE = limx→∞

(N∑

i=1

~Tx2

(∂u

∂t

)2)

=

∫ L

0

~T

2

(∂u

∂x

)2

dx.

Then,

E =1

2

∫ L

0

(

ρ

(∂u

∂t

)2

+ ~T

(∂u

∂x

)2)

dx.

If we take ~T = ρ = c = 1, then from the initial problem statement utt = uxx, and

E =1

2

∫ L

0

(u2

t + u2x

)dx.


Analyzing the change in total energy of the system with respect to time,

dE

dt=

1

2

d

dt

∫ L

0

(u2

t + u2x

)dx =

1

2

∫ L

0

d

dt

(u2

t + u2x

)dx =

1

2

∫ L

0(ututt + uxuxt) dx.

Simplifying with integration by parts,

dE

dt=

1

2

(∫ L

0ututtdx+ uxut|L0 −

∫ L

0uxxutdx

)

,

which simplifies to

dE

dt=

1

2

(∫ L

0ut (utt − uxx) dx

)

= 0,

after implementing initial conditions and substituting the problem statement. Thus, from the model,

dE

dt= 0⇒ E(t) = E(t = 0).

If utt + kut = uxx with the incorporation of an air resistance term, then utt − uxx = −kut, and

dE

dt=

1

2

(∫ L

0ut (−kut) dx

)

< 0,

implies E goes to zero as t goes to infinity.

How to solve the PDE’s:

PDE:∂u

∂t= a

∂2u

∂x2such that 0 ≤ x ≤ 1 and with t ≥ 0.

Initial Conditions: u(x, 0) = g(x).

Boundary Conditions: u(1, t) = 0 and u(0, t) = 0.

We want the forcing function equal to zero and we want to work for boundary conditions equal to zero.

Question: Solvedy

dx= xy. We would use separation of variables for this problem, and we will reuse

this idea for our other problem.

Assume that u is composed of (F (x)G(t)).

u(x, t) = (F (x)G(t)); plug this into PDE.

⇒ ∂

∂t(F (x)G(t)) = a

∂2

∂x2(F (x)G(t))

⇒ F (x)dG

dt= a

d2F

dx2G(t)

⇒ 1

aG(t)

dG

dt=d2F

dx2

1

F (x)= λ


Thus functions of t have been separated from functions of x. Which means that space is independent oftime.

⇒ (d2F

dx2)(

1

F (x)) = λ , a constant.

and (dG

dt)(

1

aG) = λ , a constant.

The PDE has now been converted into two ODE’s:

1.d2F

dx2− λF = 0 with boundary data: u(0, t) = 0 ⇒ F (0)G(t) = 0 ⇒ F (0) = 0 and u(1, t) = 0 ⇒

F (1) = 0. So, F (0) = F (1) = 0.

2.dG

dt− λaG = 0

We will solve equation 1 first and then use the solutions to solve equation 2.

To solve equation 1, we must assume three possible cases. If λ > 0, λ = 0 , or λ < 0.

Equation 1:d2F

dx2+ (−λ)F = 0

For λ > 0, F”− λF = 0⇒ r2 − λ = 0⇒ two real roots r = ±√λ

⇒ F (x) = c1e√

λx + c2e−√

λx

Apply the boundary condition: F (0) = c1 + c2 = 0, F (1) = c1e√

λ + c2e−√

λ = 0. Now solve the twoequations with two unknowns.

(

1 1

e√

λ e−√

λ

)(c1c2

)

=

(00

)

(c1c2

)

=

(

1 1

e√

λ e−√

λ

)−1(00

)

=

(00

)

So if λ > 0 then F (x) ≡ 0 and then u ≡ 0. This produces a contradiction solution since u(x, 0) = g(x).

Therefore λ > 0 is not a possibility!

When λ = 0⇒ d2F

dx2= 0⇒ F (x) = Ax+B.

F (0) = F (1) = 0 ⇒ A = B = 0 ⇒ F (x) ≡ 0 and u ≡ 0. This also is not a possibility because of thereasons stated above.

Thus the constant must be negative!


If λ < 0 take λ = −µ2 (Force negative).

d2F

dx2+ µ2F = 0⇒ r2 + µ2 = 0, sor = ±

√

−µ2 = ±µi

So,F (x) = c1 cosµx+ c2 sinµx ; with F (0) = F (1) = 0

F (0) = c1 = 0⇒ F (x) = c2 sinµx

F (1) = c2 sinµ = 0⇒ sinµ = 0

So, µ = nπ where n is an integer. Thus λ = −n2π2.

So now we have a family of equations such that:

Fn(x) = cn sinnπx ; however the initial conditions will restrict the possibilities.

Now we will look at equation 2:

dG(t)

dt+ an2π2G(t) = 0 ; now use separation of variables.

⇒ G(t) = c′ne−an2π2t

Now we can construct all of the solutions to the PDE!

un(x, t) = cne−an2π2t(sinnπx)

u(x, t) =

∞∑

n=1

cne−an2π2t(sinnπx) ; is the general solution to the unforced heat equation.

Now we will work with the initial conditions to solve for constants cn.

⇒ u(x, 0) =

∞∑

n=1

cn sinnπx = g(x); from initial conditions.

Now we multiply the equation by sinmπx and we get:

∞∑

n=1

cn sinmπx sinnπx = g(x) sinmπx

Now integrate from x = 0 to x = 1.

∫ 1

0

∞∑

n=1

cn sinmπx sinnπxdx =

∫ 1

0g(x) sinmπxdx

⇒∞∑

n=1

cn

∫ 1

0sinmπx sinnπx dx =

∫ 1

0g(x) sinmπxdx

when n 6= m:∫ 1

0sinnπx sinmπxdx = 0


So, when n = m:

cm

∫ 1

0sin2mπx dx =

∫ 1

0g(x) sinmπxdx

By using some basic trigonometry we get:

cm2

=

∫ 1

0g(x) sinmπxdx

So our final solution is:

u(x, t) =

∞∑

n=1

cne−an2π2t sinnπx

cn = 2

∫ 1

0g(x) sin nπx dx

these combined form solutions to our PDE!

Diffusion Model

PDE:∂u

∂t= a

∂2u


Initial Conditions: f(x, t) = 0.


How do we treat non zero boundary conditions?

Example

∂u

∂t= 2

∂2u


Initial Conditions: u(x, o) = x3 + x+ 1.


For separation of variables to work we require zero BCs. Subtract off steady state solution with thesame boundary conditions. Find the steady state soln.

Solve

∂2us

∂x2= 0

u(0) = 1

u(1) = 3

us = 1 + 2x

w = u− us = u− (2x+ 1)


Find diffusion problem that w solves. Solve in terms of u: u = w+us = w+(2x+1) plug in to originalequation:

∂

∂t(w + (2x+ 1)) = 2

∂2

∂x2(w + (2x+ 1))

∂w

∂t+ 0 = 2

∂2w

∂x2+ 0

∂w

∂t= 2

∂2w

∂x2

w(0, t) + u(0, t) − (2(0) + 1) = 1− 1 = 0

w(1, t) + u(1, t) − (2(1) + 1) = 3− 3 = 0

Example

∂w

∂t= 2

∂2w

∂x2

w(x, 0) = x3 − x

w(0, t) = 0

w(1, t) = 0

a=2

w(x, t) =∞∑

n=1

cne−2n2pi2t sin(nπx)

cn = 2

∫ 1

0(x3 − x) sin(nπx)dx

u(x, t) =∞∑

n=1

cne−2n2pi2t sin(nπx) + 2x+ 1

Uh =∑∞

n=1 cne−2n2pi2t sin(nπx) Up = 2x+ 1

Example

∂u

∂t=∂2u

∂x2+ x

0 ≤ x ≤ 1 and with t ≥ 0.

w(x, 0) = 0


w(0, t) = 0

w(1, t) = 0

Use part of the homogeneous equation.

Homogeneous:∑∞

n=1 cne−n2pi2t sin(nπx)

plug in a solution of the form:

u(x, t) =∞∑

n=1

bn(t) sin(nπx)

x =∞∑

n=1

2[(−1)n+1

nπsin(nπx)]

∂2u

∂2x=

∞∑

n=1

n2π2bn(t) sin(nπx)

∂u

∂t=

∞∑

n=1

∂bn∂t

sin(nπx)

∞∑

n=1

∂bn∂t

sin(nπx) = −∞∑

n=1

n2π2bn sin(nπx) + 2[(−1)n+1

nπsin(nπx)]

∞∑

n=1

[∂bn∂t

+ n2π2bn − 2(−1)n+1

nπ] sin(nπx)

∂bn∂t

+ n2π2bn = 2(−1)(n+ 1)

nπ

Guess a solution! Guess bn = c

(n2π2)c =2(−1)n+1

nπ

c =2(−1)n+1

n3π3

bn = Homogeneous + Particular


bn = cne−n2π2t +

2(−1)n+1

n3π3

Soln:

u(x, t) =

∞∑

n=1

[cne−n2π2t + 2

(−1)n+1

n3π3] sin(nπx)

How do we satisfy the Initial Conditions?

setting t=0

u(x, 0) = [cne0 + 2

(−1)n+1

n3π3] sin(nπx) = 0

So,

cn = 2(−1)n

n3π3

u(x, t) =

∞∑

n=1

2(−1)n

n3π3(e−n2π2t − 1) sin(nπx)

Model: A Vibrating String

Domain:0 ≤ x ≤ 1 and with t ≥ 0

u(x, t)=deflection of the string T (x, t)=tension in the string ρ(x)=the density

TH is the horizontal component of the force

TH(x+ δx, t) = TH

Total force in the verticle direction:

FV = T (x+ δx, t) sin(theta)(x + ∆x, t)− T (x, t) sin(theta)(x, t)

F = ma = m∂2u

∂t2

m ≈ ρ(x)∆x

F = ρ(x)∆x∂2u

∂t2

Force Balance:

ρ(x)∆x∂2u

∂t2= T (x+ ∆x, t) sin(θ)(x+ ∆x, t)− T (x, t) sin(θ)(x, t)


This is more accurate when δx goes to 0.

ρ(x)∆x∂2u

∂2t=

1

∆x[FV ]

ρ(x)∂2u

∂t2= lim

∆x→0

T (x+ ∆x, t) sin(θ)(x+ ∆x, t)− T (x, t) sin(θ)(x, t)

∆x

=∂

∂x(T (x, t) sin(θ)(x+ 1))

=∂u

∂x= tan(θ) =

sin(θ)

cos(θ)

for small θ we end up with sin(θ)

for small deflection T (x, t) = T0- some constant.

T (x, t) sin(θ)(x, t) ≈ T0∂u

∂x

ρ(x)∂2u

∂t2=

∂

∂x(T0

∂u

∂x)

assume a constant density ρ(x) = p0,T0

p0= c2

∂2u

∂t2=T0

p0

∂2u

∂x2

∂2u

∂t2= c2

∂2u

∂x2

One-D Wave Equation:

PDE:∂2u

∂t2= c2

∂2u

∂x20 ≤ x ≤ 1 and with t ≥ 0

u(x, 0) = f(x)

ut(x, 0) = g(x)

u(L, t) = 0

u(0, t) = 0

Energy = E(t) =1

2

∫ L

0(U2

t )dx+c2

2

∫ 1

0(U2

t )dx

Now show that the energy is constant.


∂E

∂t= 0

∂E

∂t=

∫ L

0

∂u

∂t

∂2u

∂t2dx+ c2

∫ L

0

∂u

∂x

∂2u

∂t∂xdx

use integration by parts to solve above equation:

= c2∂u

∂t

∂u

∂x

∣∣∣∣

L

0

− c2∫ L

0

∂2u

∂x2

∂u

∂tdx

∂E

∂t= c2

∫ L

0

∂u

∂x

∂2u

∂x2dx+ c2

∂u

∂x

∂u

∂t

∣∣∣∣

L

0

− c2∫ L

0

∂u

∂x

∂2u

∂x2dx

∂E

∂t= c2∂u∂x

∂u

∂t

for values 0 and 1

u(0, t) = 0,∂

∂t[u(0, t) = 0],

∂

∂t[u(L, t) = 0]

Wave Propagation:

PDE:∂2u

∂t2= c2

∂2u

∂x20 ≤ x ≤ 1 and with t ≥ 0

u(x, 0) = f(x)

ut(x, 0) = g(x)

u(L, t) = 0

u(0, t) = 0

Use separation of variables

u(x, t) = A(t)B(t)

A”B = c2AB”

A”

Ac2=B”

B= λ

A”− c2Aλ = 0

B”−Bλ = 0


r2 − λ = 0

r2 = λ

r =√λ

3 Cases:

For λ = 0 we get a zero solution and B is defined as zero, so this is not a possibility.

For λ > 0 we get 2 positive real root and a zero solution, so λ must be negative.

For λ < 0 we get two solution: cos(√−λx) and sin(

√−λx)

and let λ = −µ so we have cos(µx) and sin(µx)

B(x) = c1 cos(µx) + c2 sin(µx)

after appling BCs B(0) = c1 = 0 and c2 can’t = 0 so sin(µL) must! Which makes µ = πnL

So our solution is B(x) = sin(nπxL )

A”

c2A= λ = −µ2 = −n

2π2

L2

A” +c2n2π2

L2A = 0

Soln: cos(cnπt

L), sin(

cnπt

L)

A(t) = αn cos(cnπt

L) + βn sin(

cnπt

L)

un(x, t) = [αn cos(cnπt

L) + βn sin(

cnπt

L)] sin(

nπx

L)

Solution:

u(x, t) =

∞∑

n=1

un =

∞∑

n=1

[αn cos(cnπt

L) + βn sin(

cnπt

L)] sin(

nπx

L)

αn =2

n

∫ L

0f(x) sin(

nπx

L)dx

βn =2

cnπ

∫ L

0g(x) sin(

nπx

L)dx


Let I be an interval [a,b]. The boundary of I is then a,b. The normal vector to I at x = a, is n = −1i,while the normal vector to I at x = b, is n = 1i.

By the fundamental theorem of calculus,

∫ b

af ′(x)dx = f(b)− f(a)

= f(x)n(x)|x=b + f(x)n(x)|x=a

⇒∑

i=1,2

f(x)n(x)|xi

=∑

x=a,b

f(x)n(x)

By reworking the fundamental theorem of calculus, we find

∫

Idiv(f)dx =

∑

boundary(I)

f(x)n(x)

Now, we will investigate our equation

u(x, t) =∞∑

n=1

[αn coscnπ

Lt+ βn sin

cnπ

Lt] sin

nπx

L

The x profiles are sine waves, below are a few graphs of

sinnπx

L

The pattern seen above continues as n→∞. Take

αn =

1 n = 10 otherwise

, βn = 0

Therefore,

u(x, t) = coscπ

Lt sin

π

Lx

These graphs were produced for n=1. The above graph is known as the standing wave fundamental, with

Amplitude : coscπ

Lt

Wave : sinπx

L

Thus, we find

Period =2L

c, Frequency =

c

2L


where

c2 =tension

density=T0

ρ

thus,

c =

√

T0

ρ

We can interpret the results by stating that as we increase the tension, we increase the frequency. Similarly,when we increase the mass, we decrease the frequency.

Now, we produce the graph for the case when n=2. In this case, we take

αn =

0 n = 11 n = 20 n > 2

, βn = 0

where

u(x, t) = cos2πc

Lt sin

2π

Lx

The frequency is higher for the n=2 case,

Frequency =c

L

Now, let us consider a harpsicord. We have a string of a given length, and we ”pluck the string,” pullingup the string to a certain tension and then letting go.We can model this using the following equations andconditions:

∂2u

∂t2= c2

∂2u

∂x2

We have initial conditions

u(x, 0) =

x 0 ≤ x < 1

21− x 1

2 ≤ x < 1

= f(x)

∂u

∂t(x, 0) = 0

where u(x,0) is the profile of how you pluck the string. The boundary conditions are given by

u(0, t) = u(1, t) = 0

We can also generate a model for a piano. A particular example is given below:

∂2u

∂t2= c2

∂2u

∂x2

u(x, 0) = 0

∂u

∂t(x, 0) =

−x 0 ≤ x < 1

2−(1− x) 1

2 ≤ x ≤ 1

= g(x)


For the harpsicord, βn = 0 ∀n and αn is given by f(x).

For the piano, αn = 0 and βn are given by g(x). Therefore we find

βn = − L

nπcαn

For the fundamental,∣∣∣∣∣

βn

αn

∣∣∣∣∣=

L

nπc

For large n, we find that the amplitude of the overtones for the harpsicord are much smaller than theovertones for the piano. Even though strings are identical and ”plucks” are identical, the overtones aredifferent for the piano when compared to the harpsicord.

10.2 First Order Equations

We begin with functions of two independent variables x and y and the dependent variable u = u(x, y). Apartial differential equation (PDE) is an equation of the form

F

(

u,∂u

∂x,∂u

∂y,∂2u

∂x2,∂2u

∂y∂x,∂2u

∂y2, ...

)

= 0 (10.1)

First Order Equations: In class we derived the Transport PDE. This is the one dimensional waveequation:

ρt − c(ρ)ρx = 0.

This motivates the study of

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u).

The linear version of this equation is

a(x, y)ux + b(x, y)uy = c(x, y)u.

Gradient Method: To get started, we solve a simple problem, the constant coefficient PDE aux+buy = 0.It is noted that this equation is the gradient of u dotted with the constant vector (a, b): ∇u · (ai + bj) = 0.Thus the function must be constant in the direction (a, b). This provides us with the solution u = f(bx−ay).The function f is arbitrary.

Examples:

1. Solve ux − uy = 0, with u(0, y) = ey. From the formula above we see that u(x, y) = f(−x − y).Using the initial value we obtain f(−y) = ey. Thus we have that f(∗) = e−∗. The solution is thenu(x, y) = ex+y.

10.2. FIRST ORDER EQUATIONS 123

2. Solve ux+uy = 0, with u(x, x) = x3. The solution to the pde is u(x, y) = f(x−y). Using u(x, x) = x3,we get f(0) = x3. One notes a problem. We cannot solve this equation. The initial values or thecontraint information was prescribed along a characteristic line.

General Solution Method: The idea for the solution method is obtained from the following relationfrom calculus,

du =∂u

∂xdx+

∂u

∂ydy

(a form of the chain rule). We start with Quasilinear First Order Equations:

a(x, y)ux + b(x, y)uy = c(x, y, u)

and an initial condition, u(ξ(s), η(s)) = φ(s).

This latter expression is a parametric form for the initial condition. Recall that we are looking for asurface u(x, y). The “initial condition” is prescribed by choosing a line (in the x-y plane) and prescribingdata on that line. We represent the line in the x-y plane in parametric form by x = ξ(s) and y = η(s).The variable s is the independent parameter. Examples:

1. u(x, 0) = sin(x). This can be parametrically represented as x = s, y = 0 and then u = sin(s). Thusξ(s) = s, η(s) = 0 and φ(s) = sin(s).

2. u(√x, 2x) = 1/x. We have x = s, y = 2s2 and then u = 1/s2. Note that there is more than one way

to parameterize a line.

y

x

(x(t),y(t))

φ

u

y

u(x(t),y(t))

(x(t),y(t))

x

One can see this initial line as a whole continuum of initial points from which to start a collection ofdifferential equations. Each point on the initial curve has a curve which flows out and lines on the initialsurface. This gives one a way to view the PDE as a system of ODE’s.


φ

u

y

x

Along the curve which arises from the initial curve, the problem reduces to an ordinary differentialequation. Thus we may use the chain rule relation above to simplify the PDE:

du

dt=∂u

∂x

dx

dt+∂u

∂y

dy

dt.

This is a directional derivative. It is a derivative in the direction(dx

dt,dy

dt

)

= (x, y).

This direction is the tangent vector to the curve x = x(t), y = y(t). This curve is known as a characteristiccurve. Along this curve the PDE acts like an ODE. To see this we set

dx

dt= a(x, y),

dy

dt= b(x, y)

thendu

dt= a(x, y)ux + b(x, y)uy = c(x, y, u)

or simplifying and only indicating the t and u dependence

du

dt= c(t)u.

Note that this ODE is valid for every initial point along the starting curve x = ξ(s) and y = η(s). Theserelations must then hold along all lines passing across the initial curve:

dx

dt= a(x, y),

dy

dt= b(x, y),

du

dt= c(x, y, u),

x(0, s) = ξ(s), y(0, s) = η(s), u(0, s) = φ(s).

(10.2)

This can be placed into a solution algorithm for the solution to the problem

a(x, y)ux + b(x, y)uy = c(x, y, u)

u(line) = φ(10.3)


1. Convert the PDE into a system of ODE’s:

a(x, y)ux + b(x, y)uy = c(x, y, u) ⇒

dx

dt= a(x, y),

dy

dt= b(x, y),

du

dt= c(x, y, u).

2. Obtain a parametric representation for the I.C.

u(line) = φ ⇒ x(0, s) = ξ(s), y(0, s) = η(s), u(0, s) = φ(s).

3. Solve for x(t) and y(t) fromdx

dt= a(x, y),

dy

dt= b(x, y),

and then apply the parameterized initial conditions x(0, s) = ξ(s), y(0, s) = η(s). x(t) and y(t) giveyou the characteristics. Note that x and y are functions of two variables: x(t, s), y(t, s).

4. Solve for t and s in terms of x and y: t = t(x, y), s = s(x, y). This is purely an algebraic step, butcan be rather tricky.

5. Solve the ODE:du

dt= c(x(t, s), y(t, s), u).

Apply the final I.C. u(0, s) = φ(s). You then obtain u(t, s) the solution in the t-s coordinate system.

6. Undo the change of variables or remove the t-s variables via u(t(x, y), s(x, y)) = u(x, y)

Examples:

1. ux − 2uy = 0, u(x, 0) = x2.

(a) The ODE system isdx

dt= 1,

dy

dt= −2,

du

dt= 0.

(b)x(0, s) = s, y(0, s) = 0, u(0, s) = s2.

(c) Integrating the equations: x(t) = t+c1, y(t) = −2t+c2. The initial conditions: x(0) = 0+c1 = sand y(0) = 0 + c2 = 0. Thus c1 = s and c2 = 0. We have then x(t, s) = t+ s, y(t, s) = −2t.


(d) Solving for t and s in terms of x and y: t = −y/2 and s = x+ y/2.

(e) Solving the last ODE: u(t) = c3. Using the I.C. we have u(t, s) = s2.

(f) The solution is then u(x, y) = (x+ y/2)2.

2. ux + yuy = 0, u(0, y) = y2.


dt= 1,

dy

dt= y,

du

dt= 0.

(b)

x(0, s) = 0, y(0, s) = s, u(0, s) = s2.

(c) Integrating the equations: x = t+ c1 and y = c2et, using the IC’s, x = t and y = set.

(d) Solving for t and s in terms of x and y: t = x and s = ye−x.

(e) Solving the last ODE: u(t) = c3, thus u(t, s) = s2.

(f) The solution is then u(x, y) = y2e−2x.

3. yux − xuy = x2 + y2, u(x, 0) = ex.


dt= y,

dy

dt= −x, du

dt= x2 + y2.

(b)

x(0, s) = s, y(0, s) = 0, u(0, s) = es.

(c) Integrating the equations: x = c1 cos(t), y = c2 sin(t). Using the IC’s: x = s cos(t), y = s sin(t).

(d) Solving for t and s in terms of x and y: t = tan−1(y/x) and s =√

x2 + y2.

(e) Solving the last ODE:du

dt= x2 + y2 = s2,

u = s2t+ c3. At t = 0, u = es. So c3 = es.

(f) u(x, y) = (x2 + y2) tan−1(y/x) + e√

x2+y2

.

4. yux + xuy = u−1, u(0, y) = y − y2.


dt= y,

dy

dt= x,

du

dt= u−1.

(b)

x(0, s) = 0, y(0, s) = s, u(0, s) = s− s2 = s(1− s).

(c) The equations could be integrated using the Laplace transform or the eigenvector method. Inthis case, one can work them out directly. Note that

d2x

dt2=dy

dt= x


and

x(0, s) = s,d

dtx(0, s) = y(0, s) = s2.

Solving we have x(t, s) = c1et +c2e

−t. At t = 0, we get c1 +c2 = 0 and c1−c2 = s. Inverting the2 × 2 system, c1 = s/2 and c2 = −s/2, then x(t, s) = set/2 − se−t/2, y(t, s) = set/2 + se−t/2.Note you may have seen the hyperbolic trig functions, using these, x(t, s) = s sinh(t) andy(t, s) = s cosh(t).

(d) Solving for t and s in terms of x and y: t = tanh−1(x/y) and s =√

y2 − x2.

(e) Solving the last ODE:du

dt= u−1,

u =√

2t+ c3 and then u(t, s) =√

2t+ s2(1− s)2.

(f) u(x, y) =

√

2 tanh−1(x/y) + (y2 − x2)(1 −√

y2 − x2)2

5. (x− y − x(x2 + y2))ux + (x+ y − y(x2 + y2))uy = u, u(x, 0) = x.

(a) The ODE system is

dx

dt= x− y − x(x2 + y2),

dy

dt= x+ y − y(x2 + y2),

du

dt= u.

(b)x(0, s) = s, y(0, s) = 0, u(0, s) = s.

(c) The method breaks down here because we don’t know how to solve the first two differentialequations.

Nonlinear First Order Equations: these are equations of the form

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u)

which differs from the quasilinear case in that the coefficients a and b can now depend on u. Again we alsohave an initial condition, u(ξ(s), η(s)) = φ(s).

The solution method here differs from the quasilinear case in one important manner. All three ordinarydifferential equations that arise in the process are now coupled. One cannot solve for x and y first, thenplug into the u equation. This reduces the number of steps in the solution algorithm, but increases thedifficulty of each step. As before, we write out the solution algorithm:


a(x, y, u)ux + b(x, y, u)uy = c(x, y, u) ⇒

dx

dt= a(x, y, u),

dy

dt= b(x, y, u),

du

dt= c(x, y, u).


2. Obtain a parametric representation for the I.C.

u(line) = φ ⇒ x(0, s) = ξ(s), y(0, s) = η(s), u(0, s) = φ(s).

3. Solve the ODE’s in step 1, and then apply the initial conditions. You then obtain u(t, s) the solutionin the t-s coordinate system. Again x(t) and y(t) give you the characteristics. Note that x and y arefunctions of two variables: x(t, s), y(t, s).

4. Solve for t and s in terms of x and y: t = t(x, y), s = s(x, y).

5. Undo the change of variables or remove the t-s variables via u(t(x, y), s(x, y)) = u(x, y)

You are not assured of success at each step. Solving the ODE’s or inverting the change of variablesmay not have nice closed form solutions.

Examples:


uux + xuy = y ⇒

dx

dt= u,

dy

dt= x,

du

dt= y.


√

1 + (u)2ux + xyuuy =√

x2 + y2 ⇒

dx

dt=√

1 + (u)2,

dy

dt= xyu,

du

dt=√

x2 + y2.

10.3 Exercises

1. Let u represent the temperature of a bar of length L with insulated ends and no internal sources

(a) Show that if the heat conduction coefficient k(x) is not constant, then we obtain the problemut = (a(x)ux)x, 0 < x < L, u(x, 0) = f(x)ux(0, t) = ux(L, t) = 0. Assume c and p are constant.

(b) Show that the total thermal energy is constant.

2. Let u solve ut = (a(u)ux)x0 < x < L, u(x, 0) = f(x)u(0, t) = T1, u(L, t) = T2

(a) Find the steady state solution when a(u) = u.

10.3. EXERCISES 129

(b) Show that the a(u) = u problem is not linear and the method of seperation of variables fails.

3. Find the general solution for ut = auxx, 0 < x < 1, u(x, 0) = f(x)ux(0, t) = ux(1, t) = 0

4. Let u solve the wave equation on the real line: utt = c2uxx −∞ < x <∞, u(x, 0) = e−x2

ut(x, 0) = 0

(a) Find u (Hint: assume a solution of the form u(x,t) = f(x-ct)+g(x+ct). Plug this into the IC’sand solve the two equations for f & g)

(b) What do the two solution components represent (what are they)? Graph them for t = 0,1,5.

5. A more realistic model of a clamped vibrating string (on [0,L]) involves a damping term.

(a) Find the model if the damping is proportional to string velocity.

(b) What happens to the total kinetic energy E =∫ L0 (ux)2dx.


Chapter 11

Reaction-Diffusion Equation Models

Example: Diffusive limited growth model (Assuming a single population P(x,t).) We will look at thisparticular model with two aspects:

1. Population follows a logistic growth law when constrained in space. This reduces to the logisticgrowth model seen before.

2. Movement through space follows a diffusion law.

We now will build a model. Conservation law: Change in population at a point (x,t) can occur withpopulation growth and decay and population movement. We’d like to translate this statement into math-ematics.

• Change in population:∂P

∂t

• Diffusion of the population: a∂2P

∂x2

• Logistic growth/decay: kP (1− P

L)

We have captured the essential behavior for this model, and by combining the diffusion and growth/decayterms we form the following equation:

∂P

∂t= a

∂2P

∂x2+ kP (1 − P

L)

The nonlinear partial differential equation with a = k = L = 1 above is known as Fischer’s Equation.

Example:(Diffusive Lotka-Volterra model)

Assume that you have two populations u(x,t) and v(x,t) which are, respectively, prey and predator.Local dynamics is that of unlimited predator prey interactions. Both species move via diffusion.

131

132 CHAPTER 11. REACTION-DIFFUSION EQUATION MODELS

∂u

∂t=change in the prey at (x,t)

∂v

∂t=change in the predator at (x,t)

d1∂2u

∂x2= diffusion of the prey

d2∂2v

∂x2= diffusion of the predator, where d1 and d2 are the rates of diffusion

Model(1-dimensional space):∂u

∂t= d1

∂2u

∂x2+ au− buv

∂v

∂t= d2

∂2v

∂x2− cv + duv

We define (au) as the growth term for the prey and (−buv) as the interaction of prey with predator.Also, (−cv) represents the decline of the predator due to no food source, while (duv) is the interaction ofpredator with prey.

Model(2-dimensional space):

Diffusion will change in this setting, but we will continue to use the previously defined diffusion pa-rameters d1 and d2.

∂u

∂t= d1(

∂2u

∂x2+∂2u

∂y2) + au− buv

∂v

∂t= d2(

∂2v

∂x2+∂2v

∂y2)− cv + duv

Due to the interaction of predator and prey, this makes the system nonlinear, which we will approachusing the following steps.

11.1 Nonlinear system approach

1. Find restpoints and nullclines. Change = 0⇒ steady states.

2. Find vector fields: System direction behavior.

3. Stability analysis about the steady states.

4. Generate phase portrait or ”flow picture.”

Recall Fischer’s equation∂P

∂t=∂2P

∂x2+ P (1− P )

11.1. NONLINEAR SYSTEM APPROACH 133

We examine 2 types of boundary conditions:

limx→±∞ P (x, t) = 0. From this statement, we know that the population trails off the further we travelfrom the point we are studying(infinite coastline). The other boundary conditions we will study is a finitecoastline:

P (−L, t) = P (L, t) = 0

1. Get steady states, dependent upon boundary data. Set∂P

∂t= 0

∂2P

∂x2+ P (1− P ) = 0

In this case we have zero boundary conditions and P (x, 0) = Ψ(x).

We’d like to understand what the steady states are.

Direct solve:d2P

dx2+ P (1− P ) = 0

This form suggests that we use the chain rule. Let u =dP

dx

Then,du

dx=d2P

dx2, but we have

du

dx=du

dP

dP

dx= u

du

dP

⇒ udu

dP+ P (1− P ) = 0

This form suggests that we can solve by separation of variables.

⇒ u du = P (P − 1) dP = P 2 − P dP

⇒∫

2u du =

∫

(2P 2 − 2P ) dP

⇒ u2 =2

3P 3 − P 2 + c

Let u =dP

dx. We now find that

dP

dx= ±

√

2

3P 3 − P 2 + c

This suggests that we perform separation of variables once more. From which, we find

∫dP

√23P

3 − P 2 + c=

∫

dx = x+ k


The above equation is a type of an elliptic integral. This does not have an anti-derivative in a finitecombination of elementary functions. We need an analytic representation of the above integral, sincewe can not solve with our conditions.

Phase plane approach - convert to a system for analysis

(a) Restpoints/Nullclines

(b) Vector fields

(c) Stability

(d) Orbits

LetdP

dx= u with

du

dx= P (P − 1)

Then, using the ordered pair (P, u) we find that our rest points are (0, 0) and (1, 0)

2. Vector fields

Notice in our image that we only have vertical motion on the P-axis, while there exists only horizontalmotion when P = 0,1.

3. Stability(

∂∂P (u) ∂

∂u(u)∂

∂P (P 2 − P ) ∂∂u(P 2 − P )

)

=

(0 1

2P − 1 0

)

At the point (0,0) we have the matrix

A =

(0 1−1 0

)

We need to compute the eigenvalues of A in order to understand the stability at the point (0, 0).Thus, we set det(A− λI) = 0 From this, we find

∣∣∣∣

−λ 1−1 −λ

∣∣∣∣= λ2 + 1 = 0

The roots to this equation are ±i. Therefore, this is a center node with circular behavior around(0, 0).

Now, let us examine the stability of the point (1, 0).

A =

(0 11 0

)

Similar to what we did above, we need to compute the eigenvalues of A in order to understand thestability at the point (1, 0). Thus, we set det(A − λI) = 0 From this, we find

∣∣∣∣

−λ 11 −λ

∣∣∣∣= λ2 − 1 = 0

The roots to this equation are ±1. Therefore, the point (1, 0) is a saddle node and is unstable. Anoutward direction corresponds to λ = 1 and an inward direction corresponds to λ = −1.


For λ = 1 we find that

A− λI =

(−λ 11 −λ

)

=

(−1 11 −1

)

We now need to find an eigenvector x,y to make the following statement true.

(−1 11 −1

)(xy

)

=

(00

)

An eigenvector that makes this statement true is x = 1 and y = 1.

4. Orbits

The relation between u and P has already been previously computed,

u2 =2

3P 3 − P 2 + c

Thus, orbits are given by the above equation. What we’ve done previously is form a function H(P, u),

where H(P, u) = u2 + P 2 − 2

3P 3 = c.

Orbits are circular about the origin for small values of P. There exists symmetry about u = 0. Since

this is a population model, we only are concentrating on values of P ≥ 0. At the point (1, 0), c =1

3,

where u =

√

2

3P 3 − P 2 +

1

3. As P → 0, u→

√

1

3.

We now would like to investigate if2

3P 3 − P 2 +

1

3> 0 on 0 ≤ P ≤ 1?

Thus, we could equivalently evaluate P 2(2

3P − 1) +

1

3> 0.

d

dP(2

3P 3 − P 2 +

1

3) = P 2 − P

= P (1− P )

This term is never negative between (0, 1), which suggests there is a curve between u =

√

1

3and

(0,

√

1

3) and (1, 0).

Want P (x) where P (−L) = P (L) = 0. Therefore, we need to find a solution such that at

(x = −LP = 0

)

and

(x = LP = 0

)

Will the time 2L for traversing this distance work with the phase portrait curve?

As we get closer and closer to (1, 0) the speed decreases.


Question: What is the time to travel along the small circular orbit? In order to answer this question, wecan use the linear solution,

dP

dx= u

du

dx= P (P − 1) = P 2 − P

From this, we generate the linear system below, which can be solved:

dP

dx= u

du

dx= −P

Suppose we guess P = v0 sinx. Then, P (0) = 0 and v(0) = v0.

dP

dx= v0 cosx = u

du

dx= −v0 sinx = −P

How long in time(T) does it take to travel fromπ

2to−π2

where the distance traveled is π?

T = π is the quickest transit time possible. Therefore, in order to complete our orbit and solve theproblem we must have 2L ≥ π.

Summary: If L ≥ π

2then,

d2P

dx2+ P (1− P ) = 0

with

P (−L) = P (L) = 0

has a solution, where the time for transit has (π,∞) as output.

1. Examine the steady state:

One solution is: p(x,t) = 0

Also showed:

Recall: As P goes to 1, time goes from π to a maximum time of∞. Suppose L=6 thus π<6. At T=πtrajectory falls short and at T=8 trajectory too large. Therefore you must have L>π. Bifucationplot: This bifurcation has domain dependance.

2. Look at vector fields: Energy:

E(t) =

∫ L

0P 2dx

takedE

dt= 2

∫ L

0PPtdx

= 2

∫ L

0P (Pxx + P (1− P ))dx = 2

∫ L

0(PPxx + P − P 2)dx


Solving using ”integration by parts”

= 2

∫ L

0(−P 2

x − P 2 + P )dx

Fact(with 0 B.C.’s):∫ L

0P 2dx ≤ k

∫ L

0P 2

xdx

Now multiply by -1 to both sides of the inequality

⇒ −∫ L

0P 2

xdx ≤ −k∫ L

0P 2dx

Therefore

= 2

∫ L

0(−P 2

x − P 2 + P )dx ≤ 2

∫ L

0(−kP 2 − P 2 + P )dx ≤ −2(k + 1)E + 2

∫ L

0Pdx

Which impliesdE

dt≤ −2(k + 1)E + 2

∫ L

0Pdx

This says that E is decreasing to some bottom energy level, Emin. ie.:E → Emin.

3. Stability of Steady States: To see if the zero solution is stable we need linearization. Linearize aboutrest point p=0(taylor series coeff.):

d

dp[p(1− p)]|p=0 ⇒ 1− 2p|p=o = 1 = c1

Expand p(1-p) into its taylor series:

p(1− p) = c0 + c1p+ c2p2 + ...... = 0 + 1p − p2 + 0 + 0.....

Using only the linear term then the pde after linearization is:

pt = pxx + p, (withB.C.′s) : p(0, t) = p(l, t) = 0

This works for any initial condition because just finding linearization. Next we will make a guess fora solution to p(x,t):

p(x, t) =

∞∑

n=1

an(t)sin(nπx

L)

Then plugging guess into original pde:

⇒ dan

dt= −(

nπ

L)2an + an = (1− (

nπ

L)2)an

Which implies:

an(t) = αn exp(1− (nπ

L)2)t

Thus the eigenvalues are:

1− (nπ

L)2


For stability:

1− (nπ

L)2 < 0⇒ 1 < (

nπ

L)2 ⇒ 1 <

nπ

L⇒ L < nπ

For n=1 ⇒ L < π. Thus L < π is stable. If L > π at least 1 positive eigenvalue. Therefore L > π isunstable.

Summary: If L < π then p=0 is the only steady state solution which is also stable. Expect p(x,t) →0 as t → ∞. If L > π then p(x,t) → ps(x) as t → ∞.

11.2 Gray-Scott Patterns

Since 1950’s, fundamental studies in reaction kinetics have focused on nonisothermal systems, i.e. wherethermal feedback is a critical element. This is predominant in combustion science and chemical reactorscience. In 1968, Selkov described a particular autocatalytic model of glycolysis. The version of thismodel due to Gray and Scott is investigated here. Gray-Scott wanted to provide the same foundation forisothermal autocatalytic systems, i.e. chemical feedback.

We begin with isothermal autocatalytic systems in the CSTR (continuously flowing, well-stirred, “tank”reactor). “C” Open systems often are better models for nature. “S” Well-stirred assumption to simplifythe mathematics (dropped later).

There are several well known examples of autocatalysis as isothermal feedback.

1. Belousov-Zhabotinskii reaction

2. Halide-based oscillators

3. Arsenite-iodate reaction

4. Enzyme systems

Simplified forms of these have given rise to two standard models: Brusselator, Oregonator. Mathemat-ically these are still complicated. We investigate a model with the overall stoichiometry: A→ B, using thereaction rate law: ka (a=[A]). When the same change is catalyzed by a species Y: A + Y → B + Y, thereaction rate is kay.

Autocatalysis occurs when catalyst is the product. Two frequently encountered cases:(1) Quadratic autocatalysis: A + B→ 2B, [with rate = kab].(2) Cubic autocatalysis: A + 2B→ 3B, [with rate = kab2].

Gray-Scott focus on (2), irreversible cubic autocatalysis. This may be modeled in the following manner.We examine the system:

A + 2B→ 3B, [rate = k1ab2].

B→ C, [rate = k2b].

11.2. GRAY-SCOTT PATTERNS 139

The mass balance in open systems for above:

da

dt′= −k1ab

2 + kf (a0 − a)

db

dt′= k1ab

2 − k2b+ kf (b0 − b)(11.1)

Normally the problem is recast in dimensionless quantities (taking b0 = 0): u - dimensionless reactantconcentration, v - dimensionless catalyst concentration, t - dimensionless time, F - dimensionless “flow”parameter (1 / (residence time)), k - dimensionless catalyst lifetime.

Gray-Scott arrived upon their CSTR Model:

du

dt= −uv2 + F (1− u)

dv

dt= uv2 − (F + k)v

(11.2)

Because nature does not always “well-stir” things. The unstirred version is also of considerable interest:CFUR (continuously feed, unstirred, reactor) built by Harry Swinney et al provide the non-stirred versionof CSTR.

Gray-Scott CFUR Model:

ut = d1∆u− uv2 + F (1− u), x ∈ Ω,vt = d2∆v + uv2 − (F + k)v, x ∈ Ω,

(11.3)

with boundary data: ∂u/∂ν = ∂v/∂ν = 0. Boundary conditions generated by a CFUR are of a no-fluxtype known as zero Neumann conditions. [However, in the some of the numerical simulations often onewants to remove boundary effects and study behaviour generated from the equations, and we then useperiodic data on a square domain: Ω = [0, L] × [0, L]]

Recently in a paper by Pearson [9], numerical studies of the Gray-Scott model produced some interestingpatterns which were proposed as similar to those found by Swinney [7]. Pearson used a simple Euler schemeto integrate the parabolic equations. The goal of this report is to determine what process generates thecomputed patterns. Additional background on the Gray-Scott equations may be found in [4, 5, 3, 6, 11, 12]and on similar models in [10]. The patterns seen are thought to be an example of the Turing process. Theassociate steady state equations are

d1∆u− uv2 + F (1− u) = 0, x ∈ Ω,d2∆v + uv2 − (F + k)v = 0, x ∈ Ω,

(11.4)

with the same BC’s.

11.2.1 Uniform steady state solutions

The steady states of the local dynamics (setting d1 and d2 to zero) give us the uniform steady statesolutions, USS. The point (1, 0) is a USS and will be labeled p1. If F ≥ 4(F + k)2 (or the interior of the


parabolic region defined by F = 12

[

(1/4− 2k) ±√

1/16 − k]

) then there are two further rest points which

are on the intersection of the nullclines (F + k)v + Fu = F and uv = F + k (see Figure 11.1). Solving

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Figure 11.1: Nullclines.

these two equations gives

u1,2 =1

2

[

1±√

1− 4(F + k)2

F

]

, v1,2 =F

2(F + k)

[

1∓√

1− 4(F + k)2

F

]

.

We will call p2 = (u1, v1) = (u+, v−) (lower right) and p3 = (u2, v2) (upper left). This line intersects (1, 0)and (0, F/(F + k)). Later we will make use of the fact that v2 = F is equivalent to F = 4(F + k)2. Thisis the set of points for which there are exactly two restpoints for the vector field. Note that v2 ≤ F (1∓ δ)where 0 < δ < 1. This bounds v in restpoint p2: v

2 ≤ F .

11.2.2 Non-uniform steady state solutions - the elliptic problem

To prove the existence of nonuniform solutions, a bifurcation approach is used (a similar approach to [2]].There are at most three steady states. The solution (1, 0) is always stable and if it exists, the steady statep2 is unstable for the local dynamics. We focus on the third uniform solution, p3. This is denoted (us, vs)in the following analysis.

The system is shifted about the steady state, let u = u− us and v = v − vs and so we have

d1∆u− uv2 − 2 uvvs − uvs2 − usv

2 − 2usvvs − Fu = 0, x ∈ Ω,d2∆v + uv2 + 2 uvvs + uvs

2 + usv2 + 2usvvs − (F + k)v = 0, x ∈ Ω,

(11.5)

with ∇u = ∇v = 0.

Theorem 4 For n = 2, u, v are C2 solutions.

11.2. GRAY-SCOTT PATTERNS 141

We note then that the solutions to the elliptic problem are then bounded with respect to the parametersF, k. Solution continua remain bounded and must branch from and reconnect to the uniform solutions orpersist over the entire range of the parameters. More details are provided by the numerics.

11.2.3 Dynamics

Local dynamics

Setting diffusion to zero, one obtains

ut = −uv2 + F (1− u),vt = uv2 − (F + k)v.

The intersection of the four half spaces

G1(u, v) = u > 0G2(u, v) = v > 0G3(u, v) = −(u− 1) > 0G4(u, v) = −(u+ v − γ) > 0

defines an bounded invariant region for γ > 1 + F/k.

The point (1, 0) is always stable. The other two rest point’s stability are given by the eigenvalues ofthe following

A =

(−(F + v2) −2uv

v2 −(F + k) + 2uv

)

=

(−(F + v2) −2(F + k)

v2 (F + k)

)

.

The determinant is given by det(A) = (v2−F )(F+k) and the trace Tr(A) = −(v2−k) so the eigenvaluesare simply λ = Tr(A)/2 ±

√

Tr(A)2 − 4 det(A)/2. The point where there is a unique interior rest pointwhich gives det(A) = 0 is (F, k) = (1/16, 1/16) giving (u∗, v∗) = (1/2, 1/4). The rest points split and travelup/down the line (F + k)v + Fu = F as one enters the region. The restpoint p2 is a saddle point. Thisfollows simply from recalling that v2 ≤ F which forces the determinant to be positive and expression underthe square root to be large than Tr(A)2.

Restpoint p3 may have stable or unstable (maybe spiral) node structure but is not a saddle point.A Hopf bifurcation occurs if one traverses the correct line in the F, k plane. For this, the trace of thelinearization must be zero, v =

√k, and the determinant must be non-negative, v ≥

√F . Using the first

steady state equation, one obtains

u =F

k + F.

This value may be plugged into the equation (F + k)v + Fu = F giving

(F + k)2 = F√k.

We may compare this line (the Hopf line) to the lower branch of the steady state bifurcation line. The Hopfline lies above and intersects the steady state bifurcation line at (0, 0) and the turning point (1/16, 1/16).


Stability

For the parabolic system, the associated eigenvalue problem is

d1∆α− (v2s + F + λ)α− 2usvsβ = 0

d2∆β + v2sα+ (2usvs − F − k − λ)β = 0.

For the uniform solution (1, 0), this reduces to

d1∆α− (F + λ)α = 0d2∆β − (F + k + λ)β = 0,

We do a basic stability analysis using a Fourier method. Define φn to be the eigenfunctions on Ω, andµn ≥ 0 the eigenvalues for

∆φn = −µnφn.

For (1, 0), we note that the eigenvalues are

λ1 = −F − d1µn < 0, λ2 = −F − k − d2µn < 0

for F ≥ 0. Thus (1, 0) is a stable uniform solution. For the remaining uniform solutions (if they exist), wetry

(θ1, θ2)φn.

Let

M =

[−(d1µn + v2

s + F ) −2(F + k)v2s −d2µn + F + k

]

and then the eigenvalues, λ, may be determined by

det(M − λI) = 0.

IfTr(M) = −µn(d1 + d2)− v2

s + k

anddet(M) = (d1µn + v2

s + F )(d2µn − F − k) + 2(F + k)v2s

= d1d2µ2n + (d2(v

2s + F )− d1(F + k))µn + (F + k)(v2

s − F )

then the solutions are

λ =Tr(M)±

√

Tr(M)2 − 4det(M)

2

It is clear from the structure of M , that as µn gets large, the eigenvalues become real and nega-tive (diagonally dominant matrix). The term D = Tr(M)2 − 4det(M) is strictly increasing in µn (fromdD/dµn = 2(d1 − d2)

2µ+ constant). Thus the magnitude of the imaginary component is strictly decreas-ing. One may conclude that if at µ = 0 the eigenvalues are real, then they are always real. Dynamicalinstabilities occur at the lowest modes when the eigenvalues are complex valued at µ = 0.

Static bifurcation can (and does as one sees later) occur at higher modes. Graphs of the real part of λas a function of µ can be helpful. The choice of which occurs first between static and dynamic bifurcationsis based on the parameters F and k.

11.3. NUMERICS 143

The boundary of the region in parameter space where

(F, k) : supµ≥0ℜ(λ(F, k, µ)) = 0

defines a stability region which generalizes region defined by the Hopf bifurcation. This curve is not thesame as the Hopf curve. It does intersect the origin, but not the turning point of the static bifurcationcurve. One may verify this by plugging into the matrix M and computing the eigenvalues (the first caseis λ = 0 and for the second case λ ≈ .021 for µ ≈ 1550.).

Then the max of λ maybe computed directly. Solving the equations det(M(µ)) = 0 and [det(M(µ))]µ =0, we obtain from the second equation µ = [d1(F + k) − d2(v

2 + F )]/2d1d2. Using the first equation onearrives at

[d1(F + k)− d2(v2 + F )]2 − 4d1d2(F + k)(v2 − F ) = 0.

This equation clearly has no solution when v2 < F . First, it is informative to determine where thiscurve intersects the static bifurcation curve, F − 4(F + k)2 = 0. Since v2 = F , (d1 − 2d2)F + k = 0,which for the Gray-Scott model is the line k = 0. One would also like to determine intersections withthe Hopf curve, F

√k − (F + k)2 = 0. This is done numerically for d1 = 2(10−5) and d2 = 10−5 giving

f = 0.0471, k = 0.06056. Figure 11.3 provides the stability information for d1 = 2(10−5) and d2 = 10−5.

Figure 11.4 graphs a family of curves for fixed F and increasing k. The curves are uniformly increasingin k. This gives one typical bifurcation scenerio; one that the uniform mode becomes unstable first. The xaxis is really a plot of the eigenvalues for the continuous operator (on the present scale, they appear densein the interval). Figure 11.5 again graphs a family of curves for fixed F and increasing k. For this example,note that the uniform mode is not the first mode to become unstable.

For the following, we take Ω to be a rectangle with l1, l2 to be the length scales in the x, y directions.The eigenvalues for the Laplacian on the square (with zero Neumann data) are

µkm = π

√

k2

l21+m2

l22, k = 0, 1, 2, . . . ,m = 0, 1, 2, . . .

Note that for fixed µ, the collection of k and m lie on an ellipse with major and minor axes defined by(l1µ/π)2 and (l2µ/π)2. For the case which Pearson studies, l1 = l2 = l, and so the curve is a circle. Becauseit is symmetric about the m = k line, any eigenvalue µkm for which k 6= m has µkm = µmk, in other words,it occurs in pairs. Thus if m = k does not produce the µkm, the multiplicity of µkm is even. To geteigenvalues of odd multipicity, we need then that k = m produces an eigenvalue. Restricting (relabelingas well) to this set in the Pearson example, we have µm = mπ

√2/l.

This provides the analytical basis for bifurcation of static non-uniform solutions and Hopf bifucationsof non-static solutions. Due to the high mode numbers, direct analysis is not tractable and a numericalapproach is necessary.

11.3 Numerics

For the one dimensional problem, we have

µk = −4N2 sin2

(πk

N

)


and

(φk)j =

sin

(2πkj

N

)

, cos

(2πkj

N

)

, , k = 1, . . . ,N.

Since

µk = −4π2k2

[

sin(

πkN

)

πkN

]2

we note that the smaller eigenvalues are good approximations of the continuous operator where l1,2 = 1.The eigenvalues for the two dimensional operator follows:

µkm = −4N2

[

sin2

(πk

N

)

+ sin2(πm

N

)]

.

Since the domain is very simple and the eigenfunctions for the Laplacian are easily computed, seriesexpansions are straightforward to attempt. Expand the solution out in terms of the eigenfunctions. Letφn solve

∆φn + µnφn = 0

for the domain and boundary conditions in question. We try an expansion

u =∑

anφn, v =∑

bnφn.

For the moment, assume that we may define and compute the following coefficients, cn,

uv2 =(∑

anφn

)(∑

bnφn

)2=(∑

cnφn

)

.

If so, then an algebraic system arises,

dan

dt= −d1µnan − cn(an, bn)− Fan + Fδ(n),

dbndt

= −d2µnbn + cn(an, bn)− (F + k)bn.

(11.6)

2D Parabolic solves

The initial code was a simple Euler scheme. A pattern similar but not identical to the reported patternswas found for a couple of test points. The next step was to really check the results with an implicit scheme(now that the function statement was assured of correctness). The implicit method was a full systemCrank-Nicolson scheme.

First tests were run for time steps of 0.1 for 200, 000 steps. This was to see if the initial behaviourlooked similar. It was not identical.

The speed of the full system solve is not great. An ADI method was adapted to this problem. TheADI code ran about 3.5 times faster (see Table 11.1) with a fixed stepsize of h = 1/10.

Choosing the parameters F, k in the range that Pearson found patterns, between the Hopf curve and thestatic bifurcation curve (the cresent region), more analysis maybe done. Obtaining the exact parameter

11.3. NUMERICS 145

Table 11.1: Comparison of run times for the Full system method and the ADI method.Method nx = 100 nx = 150 nx = 200 nx = 250 nx = 300

Full 516.4 1333.8 2334.5 3594.3 5181.6ADI 151.3 341.9 632.7 962.4 1483.6

Ratio 3.41 3.90 3.69 3.73 3.49

values from Pearson produced other similar patterns to what he obtained. The nonuniform solutionsobtained by Pearson seem to be verified by an implicit method. Thus the patterns seem to be genuine.The steady state solution computed by the adi routine is then passed to an elliptic solver as an initialguess. The output can then be trusted to solve the discrete problem to a residual less than 10−5. Theworm-like patterns persist agreeing with the analytic bifurcation results. The structure of the patterns isvery complex. For the case F = .04 and k = .06 (Figure 11.6), integrating the standard initial conditionto obtain a steady state solution and passing this to the elliptic solver gives a good test case. How thesymmetry breaking occurs and the resultant nonisotropic phenonmenon is still open.


0

0.05

0.1

0.15

0.2

0.25

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

F

k

Static BifurcationStable NodesStable Spirals

Hopf Bifurcation

Figure 11.2: The stability of the upper-left restpoint. The only place there are non stable solutions is inthe thin band near the horizontal axis.

11.3. NUMERICS 147

0

0.05

0.1

0.15

0.2

0.25

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

F

k

Static BifurcationTuring Bifurcation

Hopf Bifurcation

Figure 11.3: Location of the unstable modes. Along the dotted edge the restpoint becomes unstable onthis boundary.


-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

0 1000 2000 3000 4000 5000 6000

ℜ(λ)

µ

Dispersion curves

Figure 11.4: Family of instability curves for F = .038 where k = .057 is the lowest curve with incrementsof ∆k = .0003).

11.3. NUMERICS 149

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0 1000 2000 3000 4000 5000 6000

ℜ(λ)

µ

Dispersion curves

Figure 11.5: Family of instability curves for F = .063 and k = .06149 (increments of .00025).


Figure 11.6: Solution u at t = 20000 for F = .04 and k = .06.

Chapter 12

Appendix

12.1 Solutions

12.1.1 Introduction

1. In the notes, we have the free-fall model a = −9.8 + cv2, where we studies the case k = 1.

1. What should the of c be for k = 1 and k = 2? Using the standard coordinate system, v < 0.Acceleration = gravitational force (down) + air resistance force (up): a = −9.8 + cvk. For the latterterm when k = 1: cv > 0, c must be negative. For the latter term to be positive when k = 2: cv2 > 0,c must be positive.

2. Work out the solution in the case k = 2. We havedv

dt= −g + cv2. Using separation of vari-

ables:dv

−g + cv2= dt, and then integrating: t + k = −

arctanh( cv√gc)√

gcNext, we can solve for v:

v = − tanh(t√gc+ k

√gc)√gc

cand then integrate to get y(t): y =

1

2

ln(tanh(t√gc+ k

√gc)− 1)

c+

12 ln(tanh(t

√gc+ k

√gc) + 1)

c+ k. Apply initial condtions v(0) = 0 and so k = 0, y(0) = y0, we

obtain:

y =1

2

ln(tanh(t√gc)− 1)

c+

1

2

ln(tanh(t√gc) + 1)

c+ y0

3. Does this present a reasonable answer? Yes. The solution limits at a terminal velocity and givessmooth decreasing acceleration up to this.

2. Find the differential equation model for the free fall problem with an alternate set of assumptions. Takethe y axis to be positive upward. Let the radius of the planet be R, set the origin (y = 0) at the surface.Assume that there is no air resistance, but that the force due to gravitiy does change as a function ofaltitude. Let lower case m stand for the object mass, upper case M stand for the mass of the planet andG the gravitational constant.

151

152 CHAPTER 12. APPENDIX

1. Find the differential equation for y(t) including only the object data and the gravitational accelera-

tion. Hint: g =GM

R2y′′ = − GmM

(R+ y)2= − GR2

(R+ y)2.

2. Convert to a differential equation for v(y) (use the ”trick”:dv

dt=dv

dy

dy

dt= v

dv

dy), v

dv

dy= − GR2

(R+ y)2.

3. Find the escape velocity (smallest initial velocity so that the object does not return). Hint: v muststay positive. Solve the ODE (separation of variables) and then solve for v:

v =v0R+ v0y −

√2√

GR(2R2 + 3Ry + y2)

R+ y.

Take the limit of the right as y goes to infinity, we have v = v0 −√

2√GR and so for v = 0, we

obtain: v0 = (2GR)1

2 .

3. Solve the initial value problem: y(iv) − y′′ = 0, y(0) = 1y′(0) = y′′(0) = y′′′(0) = 0. The characteristicequation is r2(r2 − 1) = 0. Then we have y(t) = c1 + c2t + c3e

t + c4e−t. Next we compute the three

derivatives and set t = 0 in the four equations. This gives y(0) = c1 + c3 + c4 = 1, y′(0) = c2 + c3 − c4 =0, y′′(0) = c3 + c4 = 0, y′′′(0) = c3 − c4 = 0. Add the latter two equations and you obtain 2 c3 = 0. Thisgives c3 = 0 and from the last equation we see that c4 = 0 . Plugging these two values into c2 + c3− c4 = 0,we note that c2 = 0 and so we finally obtain that c1 = 1 from the first equation. Thus the solution is y(t)= 1.

4. An annual plant is one that lives for a single season. These plants survive by spreading seedsin the fall which then germinate in the following spring. Some seeds fail to germinate that spring, butdo the following spring. Let pn represent the number of plants for a particular species, a represent theaverage number of seeds per plant that survive the first winter and germinate, and b represent the averagenumber of seeds per plant that germinate a year following. This can be modeled by the difference equation:pn+2 = apn+1 + bpn.

1. What is the criteria for survival of the plant (in terms of a and b)? [ Note that a and b are positive

] Plug in pn = Crn: r2 = ar + b. Solving this for r: r = (1

2a +

1

2

√

a2 + 4b,1

2a − 1

2

√

a2 + 4b). For

survival we need at least one root that is greater than 1 (in magnitude). Since a and b are positive,there is no possibility of complex roots. Setting the first root equal to 1, we obtain a + b = 1. Aslong as a + b >= 1 the plant will survive.

2. Is the plant helped or hurt by this two season approach? [ We might assume that whatever givesthe seed the ability to survive an extra winter might hinder it from germinating the first spring, andso a and b are inversely related.] The second root is smaller than the first. So the growth rate islower with b > 0. In a consistently good environment, the plant is better off using a single seasonapproach. As the environment becomes less consistent, it is clear that the insurance of a two seasonapproach is critical.

5. Let A be a 3X3 matrix, with rows: row1 = [2,−1, 0], row2 = [−1, 2,−1], row3 = [0,−1, 2].

12.1. SOLUTIONS 153

1. Find the diagonalization of A.

A =

12

14

14

0 −14

√2 1

4

√2

−12

14

14

2 0 0

0 2 +√

2 0

0 0 2−√

2

1 0 −1

1 −√

2 1

1√

2 1

2. Use the previous result to solvedx

dt= Ax, x(0) = (1, 0, 0)t. We can compute the solution x(t)

=

12

14

14

0 −14

√2 1

4

√2

−12

14

14

e2t 0 0

0 e((2+√

2)t) 0

0 0 e((2−√

2)t)

1 0 −1

1 −√

2 1

1√

2 1

100

3. Find sqrt(A)√A =

12

14

14

0 −14

√2 1

4

√2

−12

14

14

√2 0 0

0√

2 +√

2 0

0 0√

2−√

2

1 0 −1

1 −√

2 1

1√

2 1

12.1.2 Background

1. Find the phase diagram, direction fields and some sample solution plots (from the direction fields -not by analytic solution methods ) for dy

dt = siny. What is the limit of y(t) as t− >∞ when y(0) = 5.picture Nullclines y = nπ where n is an integer. picture If y(0) = 5 then y(∞) = π since π < 5 < 2π

2. Show the solutions to x′′ + xex2

= 0 are periodic. Let v =dx

dt=>

d2x

dt2= v

dv

dxPlugging this

into the original equation, we get vdv

dx+ xex

2

= 0 => vdv

dx= −xex2

. Integrating both sides we get∫

vdv =

∫

−xex2

dx. Substituting u = x2 we get

∫

vdv =v2

2= −1

2

∫

eudu = −1

2eu+k = −1

2ex

2

+k.

From here we get v2 + ex2

= c. By assuming v and x are independant, we can graph this equation.By choosing a constant c, we can take a slice of the graph and get a closed curve (neither value canrun off to ∞). Since it is a closed curve, we know the solutions are limited and periodic.

3. Two bacteria live on a common food source. In an ideal environment, bacteria A can reproduceby cell division every 2 hours, while bacteria B can reproduce by cell division every 3 hours. Thefood supply is constant and supports a total of 109 bacteria. Each bacteria consumes the sameamount. It is also noted that the relative growth rate decreaces linearly with total population.Normal environmental forces cause 15% population loss per hour in bacteria A and 10% in B. Findthe pair of differential equations which model the populations as a function of time. We can get a

general model ofdA

dt= aA(1− A+B

L)− cA, dB

dt= bB(1− A+B

L)−dB. where a and b are doubling

coeffients. A = A0eat => 2A0 = A0e

2a => 2 = e2a => a = ln(2)2 . Similarly, b = ln(2)

3 . This gives usdA

dt=ln(2)

2A(1 − (A+B)

109)− 0.15A,

dB

dt=ln(2)

3B(1− A+B

109)− 0.1B.

4. A variation of the predator-prey model assumes that the predator will be satiated when food is veryabundant. The following model uses non-biological constants to simplify the numerical computations,

but keeps the essential relations:dx

dt=

2

3x(1 − x

4) − xy

(1 + x),dy

dt=

1

4y(1 − y

x). Find the complete


phase diagram [restpoints & stability, nullclines, direction fields and sample orbits]. First find thenullclines. dx

dt = 0 => x = 0 or y = −16x

2+ 12x+ 2

3 ,dydt = 0 => y = 0 or y = x. This gives us restpoints

of (0,0),(4,0), and (1,1). Now calculate the partial derivatives of the first two equations. dudx =

−x3+3x−3y+23(x+1)2 , du

dy = − xx+1 ,

dvdx = y2

4x2 ,dvdy = x−2y

4x . Stability: near(0,0) saddle and unstable. near (4,0) -

syddle and unstable. near(1,1) -dx

dt=du(1, 1)

dx(x−1)+

du(1, 1)

dy(y−1) =

1

12(x−1)− 1

2(y−1),

dy

dt=

dv(1, 1)

dx(x− 1)+

dv(1, 1)

dy(y− 1) =

1

4(x− 1)− 1

4(y− 1). So

d

dt

(x− 1y − 1

)

=

(112 −1

214 −1

4

)(x− 1y − 1

)

.

From here we get eigenvalues of λ = − 112 +

√14

12 iandλ = − 112 −

√14

12 i. Therefore it is a stable spiralat (1,1).

5. Verify that the origin is an equilibrium point and determine the type of rest point (stability):dx

dt=

x − y + z2,dy

dt= y + z − x2,

dz

dt= z − x + y2. dx(0,0,0)

dt = 0 − 0 + 02 = 0, dy(0,0,0)dt = 0 + 0 − 02 =

0, dz(0,0,0)dt = 0 − 0 + 02 = 0 therefore (0,0,0) is a restpoint. By looking at only the linear terms, we

getd

dt

xyz

=

1 −1 00 1 1−1 0 1

xyz

. λ = 2,1

2+

√3

2i,

1

2−√

3

2i. Since all λ are positive in the

real portion, the node is unstable.

12.1.3 ODE Models

1. Show that dxdt = −y + xf(x2 + y2), dy

dt = x + yf(x2 + y2) is equivalent to drdθ = rf(r2). Use this

to convertdx

dt= −y +

x(1− (x2 + y2))√

(x2 + y2),dy

dt= x +

y(1− (x2 + y2))√

(x2 + y2)to polar form. Determine the

stability of all constant and periodic solutions. r2 = x2+y2 Taking the derivative of both sides we get:r dr

dt = xdxdt +y dy

dt . Substituting in our original equations: r drdt = −xy+x2f(x2+y2)+xy+y2f(x2+y2) =

(x2 + y2)f(x2 + y2) = r2f(r2) This can be simplified to drdt = rf(r2).

dθ

dt=

1

1 + y2

x2

dydt x− dx

dt y

x2=

1

x2 + y2(dy

dtx − dx

dty) =

1

x2 + y2∗ (x2 + xyf(x2 + y2) + y2 − xyf(x2 + y2)) =

x2 + y2

x2 + y2= 1. Now

divide the two differential equations to getdrdtdθdt

=dr

dθ=rf(r2)

1= rf(r2). to solve the example, let

f(x2 + y2) = 1−(x2+y2)√x2+y2

This gives usdx

dt= −y + xf(x2 + y2),

dy

dt= x + yf(x2 + y2). This can be

converted to drdθ = rf(r2) = r 1−r2

r = 1− r2. Since r must be > 0, r = 1 is the only periodic solution

and it is stable since when r > 1, drdθ < 0 and when r < 1, dr

dθ > 0

2. Use the Bendixson-Dulac criterion to show that no limit cycles can exist in the two species com-

petition model:dx

dt=

r1x(k1 − x− ay)k1

,dy

dt=

r2y(k2 − bx− y)k2

, for x,y >= 0. Hint: Take B =

1xy . All constants are positive. Let B = 1

xy . From this we getδ

δx(BF ) =

δ

δx

r1(k1 − x− ayk1y

=

− r1k1y

,δ

δy(BG) =

δ

δy

r2(k2 − bx− y)k2x

= − r2k2x

since all of the constants are positive, − r1

k1y < 0 and

12.1. SOLUTIONS 155

− r2

k2x < 0 therefore − r1

k1y + − r2

k2x < 0 for all x,y > 0 and by the Bendixson-Duloc Criterion no limitcycle can exist.

3. A variation on the Van der pol model isdx

dt= −y+ x− x3

3,dy

dt= x+ y− y5

5. Show that there exists

a periodic solution. Hint: use the Poincare-Bendixson theorem. Find the nullclines. dxdt = 0 => y =

x − x3

3 ,dydt = 0 => x = y5

5 − y. By converting to polar coordinates, we can look at the direction

of the flow with respect to the norm of a surrounding circle. We know 2rdr

dt= 2x

dx

dt+ 2y

dy

dt=

2x(−y + x − x3

3) + 2y(x + y − y5

5= 2x2 − 2

3x4 + 2y2 − 2

5y6 = 2r2 − (

2

3x4 +

2

5y6) knowing that

r2 = x2 + y2 we can see that as r grows, the second term becomes dominant and the sum goes below0 implying that the flow is always going into the circle. Then by excluding the origin, we have aregion in which all boundry points head into the region. Therefore a limit cycle exists according tothe Poincare-Bendixson theorem.

4. Provide an example of a system of two ODE’s whose phase plane has the three properties: a stableperiodic solution, an unstable periodic solution and a periodic solution which is stable from the insideand unstable from the outside. Hint: try working in polar coordinates. In order for a root to be stableon one side and unstable on the other, it must be a repeated root. Also since it is periodic solutionsthat we are looking for it is best to work in polar coordinates. The solution could be something of

the formdr

dt= (r − 1)(2 − r)(r − 3)(r − 2),

dθ

dt= 1

5. Determine the bifurcation structure (bifurcation points and diagram) for

dx

dt= −y + x(µ− (

√

(x2 + y2)− 1)2),dy

dt= x+ (µ− (

√

x2 + y2 − 1)2).

By problem 1 we can convert this to polar form, giving usdr

dt= r ∗ (µ− (r− 1)2),

dθ

dt= 1. This gives

solutions of r = 1 +√µ (stable) r = 1−√µ (unstable) r = 0(stable µ < 1, unstable µ > 1).

12.1.4 PDE Models


Bibliography

[1] R. Aris. Mathematical Modeling Techniques. Dover, New York, 1994.

[2] K. J. Brown and F. A. Davidson. Global bifurcation in the brusselator system. Nonlinear Analysis,

24(12):1713–1725, 1995.

[3] Irving R. Epstein. Complex dynamical behavior in “simple” chemical systems. J. Phys. Chem.,

88:187–198, 1984.

[4] P. Gray and S. K. Scott. Autocatalytic reactions in the isothermal continuous stirred tank reactor:

isolas and other forms of multistability. Chem. Eng. Sci., 38(1):29–43, 1983.

[5] P. Gray and S. K. Scott. Autocatalytic reactions in the isothermal continuous stirred tank reactor:

oscillations and instabilities in the system a + 2b → 3b; b → c. Chem. Eng. Sci., 39(6):1087–1097,

1984.

[6] P. Gray and S. K. Scott. Sustained oscillations and other exotic patterns of behavior in isothermal

reactions. J. Phys. Chem., 89:22–32, 1985.

[7] K. J. Lee, W. D. McCormick, Q. Ouyang, and H. Swinney. Pattern formation by interacting chemical

fronts. Science, 261:192–194, 1993.

[8] M. Meerschaert. Mathematical Modeling. Academic Press, San Diego, 1999.

[9] J. E. Pearson. Complex patterns in a simple system. Science, 261:189–192, 1993.

[10] J. E. Pearson and W. Horsthemke. Turing instabilities with nearly equal diffusion coefficients. J.

Chem. Phys., 90(3):1588–1599, 1989.

[11] S. K. Scott. Isolas, mushrooms and oscillations in isothermal, autocatalytic reaction-diffusion equa-

tions. Chem. Eng. Sci., 42(2):307–315, 1987.

[12] J. A. Vastano, J. E. Pearson, W. Horsthemke, and H. Swinney. Turing patterns in an open reactor.

J. Chem. Phys., 88(10):6175–6181, 1988.

157

Date post:	11-Feb-2016
Category:	Documents
Upload:	seyhan-oezen
View:	40 times
Download:	1 times

Mathematical Modeling Lecture Notes - Jeff S. McGough

Documents