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Applied Mathematics
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I

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Mathematical modelling

2 Dimensional analysis and Scaling
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1 Dimensions and units
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2 Laws and unit free laws
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3 Pi theorem
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2 Example 2: Heat transfer problem
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4 Scaling
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3 Perturbation methods
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1 Regular perturbations
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1
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3
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2 Big O and little o notation
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2 Singular perturbations
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3 Boundary layers
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4 The WKB approximation
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1 Normed linear spaces
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1 Higher derivatives
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2 Several functions
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More problems
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1 Equilibria and stability
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2 Continuous dynamical systems
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2 Stationary orbits and stability
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6 Introduction to partial differential
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equations
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7 Sturm-Liouville problems
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9 Integral equations
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1 Volterra equations
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2 Fredholm equations
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Some words
These lecture notes are based mainly on the book Applied Mathematics: Third edition
written by J
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www
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se/~larserik/
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I strongly recommend to go the sources for
a better and further exposition on the selected topics
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I follow the idea that
mathematics is learnt through exercises!

v

vi

SOME WORDS

Chapter 1
What is Applied Mathematics
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Applied mathematics deals with all the stages for solving these problems,
namely:
1
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2
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3
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In case that they disagree
qualitatively, go back and reformulate the problem
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1
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In order to get it, first, we need to propose a model
(mathematical model) that describes it
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This is
done by solving it
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If they agree, we will say that the model describes the phenomenon
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Usually, this rethinking process means that, while we constructed
the first model, we discarded some things in order of getting a simple model
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We will
learn how to deal with several of the steps involved in this process
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1

2

CHAPTER 1
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Figure 1
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Part I
Mathematical modelling

3

5
Dimensional analysis and scaling methods deal with the first stage in applied mathematics: finding a mathematical model
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6

Chapter 2
Dimensional analysis and Scaling
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2
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In dimensional analysis we should distinguish between two different but related concepts:
dimension and unit
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Example 1
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Example 2
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A set of fundamental dimensions (units) is a set of dimensions (units) from which
every dimension (unit) can be generated
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See Table 2
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Observation 3
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For example, it could happen that in
economics we use dimensions like: population, wealth, happiness
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For example, in the SI system there are plenty of them: velocity, acceleration, frequency,
energy, force
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2 for other examples
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DIMENSIONAL ANALYSIS AND SCALING
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1: SI fundamental dimensions
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Derived Unit
Dimension
Herz
Frequence
Radian
Angle
Newton
Force
Pascal
Pressure
Joule
Energy, work, heat
Watt
Power
Coulomb
Electric charge, quantity of electricity
Volt
Electrical potential difference
Farad
Electrical capacitance
Ohm
Electrical resistance
Lux
Illuminance

Equivalence to fundamental dimensions
1/T
1
M L/T 2
M/(LT 2 )
M L2 /T 2
M L2 /T 3
AT
2
M L /(T 3 A)
A2 T 4 /(M L2 )
M L2 /(T 2 A)
C/(M 2 )

Table 2
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Dimensionless
dimensions are expressed as 1
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2
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2
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Let’s define what is a law
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, qn in
m < n fundamental dimensions L1 ,
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, qn ) = 0
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With these, we can create the dimension matrix
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
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Given a system that depends on the position (q), velocity
(p) and mass (m), the law of energy preservation in its most classical setting says that the
sum of the kinetical and potential energy is constant
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2

Thus, in this example, the function f depends on three variables, p, q, m and it is
f (p, q, m) = m

p2
+ V (q) − C
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The force F needed to extend (or compress) a spring by some
distance L is proportional to this distance
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Hence, the function f in this case is
f (F, L) = F − kL
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This observation
should be keeped in mind
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DIMENSIONAL ANALYSIS AND SCALING
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Supose that there is an atomic explosion
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A shockwave is then
propagated from it
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Then, we have
f (r, t, ρ, E) = 0
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Once we talk about relations between dimensions/units/quantities, equations
appear
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If we
know the law, then we know exactly their relation, but just knowing that there is a law tells
us that there is some relation
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More concretely,
given a law that depends on n quantities q1 ,
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, Lm ,
f (q1 ,
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That is,
L
f (ˆ
q1 ,
...
An example of a unit free law is
1
f (x, g, t) = x − gt2 = 0,
2

(2
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ˆ = λ1 L, Tˆ = λ2 T then, since g has units in L/T 2 , we get that
If L
f (ˆ
x, gˆ, tˆ) = 0
if and only if Equation (2
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2
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Theorem 8
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, qn ) = 0
be a unit free physical law that relates the dimensioned quantities q1 ,
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Let L1 ,
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2
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PI THEOREM
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Then there are n − r independent
dimensionless quantities
π1 ,
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, qn
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Moreover, the physical law above is equivalent with an equation
F (π1 ,
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I will not prove this theorem
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What it is important in this theorem is the information that we can get from it
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Suppose that a given a unit free law can be reduced to a just one dimensionless
variable π1
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Now, since we can suppose that, generally, zeros of functions of one
variable are discrete, then π1 can only attain discrete values
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This means that we get a relation between the variables qi of the form
α

q1 1,1 · · · qnαn,1 = C
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Example 10
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Then, this law is equivalent to one with
F (π1 , π2 ) = 0
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This function g can be deteremined by means of experiments
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DIMENSIONAL ANALYSIS AND SCALING
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Let f : Rn → R be a smooth function
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Furthermore, if z = (z1 ,
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, zi−1 , zi+1 ,
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2
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1

Example 1: Atomic bomb
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that depends on the radius r, the air density ρ, the time t and the energy E
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The dimension matrix is


1 0 −2 0
0 1 2 −3
...
Hence, Pi Theorem asserts that there is just 1 dimensionless quantity π1 that can be
formed from these 4 quantities
...

With a little bit of algebra we get that
π1 =

r5 ρ
t2 E

so, we deduce from Principle 11 that
r5 ρ
= C,
t2 E
where C is (an unknown) constant
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The former are vague versions of the latter
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4
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2
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2

13

Example 2: Heat transfer problem
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We want to determine the temperature u as a function of r and t
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As in the previous example we have that the dimensions are [t] = T , [r] = L, [u] = K,
[e] = E, [c] = EK −1 L−3 and [k] = L2 /T
...

Doing some algebra we get
r
π1 = √
kt
and
3
uc
π2 = (kt) 2
...

e
kt
So, the temperature u behaves like


e
r
− 32
u = (kt) g √

...
4

Scaling
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Scaling is about
scaling the variables in their correct magnitude
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For example, it is not the same measuring
time when we study galaxies or when we study atomic reactions
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Every problem has its own scales (for each of the dimensions)
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Once the characteristic scale is identified, a new dimensionless variable is formed by
dividing the former with the latter
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14

CHAPTER 2
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characteristic scale could be something around tc = 106 years, and the dimensionless time
will be
t
t¯ =
...
This process is called non-dimensionalization
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15

Chapter 3
Perturbation methods
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Example 12
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g
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In this setting, if we consider the problem
of 3 masses with one of them much smaller with respect the other two (e
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Sun-Earthsatellite system) then we have a perturbed system
...

Example 13
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For ε = 0 the equation has solutions x(0) = ±1
...
It
is natural to expect that for values of ε small enough, solution to the equation will be close
to x(0) = 1
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Question 14
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, y n) , ε) = 0,
satisfying that for ε = 0 a solution to it is known
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Perturbation methods do not only deal with ODEs, but also with PDEs,
integral equations
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17

18

CHAPTER 3
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3
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The basic idea behind regular perturbations is the one behind Example 13: We do not need
to perform any change in the equation and the Taylor expansion works fine
...
Consider the initial value problem


mv 0 = −av + bv 2
,
v(0) = V0

with b  a
...
1)

bV0
 1
...
1) when ε = 0 is

where ε =

y0 (t) = e−t
...
1) are of the form
y(t) = y0 (t) + εy1 (t) + ε2 y2 (t) + · · ·
and substituting it into Equation (3
...
o
...
= −y0 (t)−εy1 (t)−ε2 y2 (t)+ε y0 (t) + εy1 (t) + ε2 y2 (t) +h
...
t
...
o
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= −y0 (t) − εy1 (t) − ε2 y2 (t) + εy0 (t)2 + ε2 2y0 (t)y1 (t) + +h
...
t
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3
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REGULAR PERTURBATIONS
...

Swedish
mathematician
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3
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1

19

Henri Poincar´e
...
18541912
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The Poincar´e-Lindstedt method is used for uniformly approximate periodic solutions in
perturbed systems, when the period of the perturbed periodic solution changes with respect
ε
...
Consider the Duffing equation
x00 + x + εx3 = 0
with initial conditions
x(0) = 1, x0 (0) = 0
...

If we perform the ansatz that the perturbed solution is of the form
x(t) = x0 (t) + εx1 (t) + h
...
t
...
2)

then we get that
3
1
(cos(3t) − cos(t)) − t sin(t)
...
Since the Taylor
expansion in Equation (3
...
It does not approximate any periodic solution!
The way we overcome this is by also letting power series in the time variable
...
o
...

Notice that ω0 = 1
...


x(0)
˙
=0

20

CHAPTER 3
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Then, expanding in power series
x(τ ) = x0 (τ ) + εx1 (τ ) + h
...
t
...

Their solutions are
x0 (τ ) = cos(τ ),


1
3
x1 (τ ) =
(cos(3τ ) − cos(τ )) + ω1 −
τ sin(τ )
...

8

Exercise 18
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Exercise 19
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Use the Poincar´e-Lindstedt method for computing the periodic solutions with that initial
conditions
...
1
...


We write f (s) = O(g(s)), as s → A if there exists a constant C > 0 such that for all s ≈ A
|f (s)| < C|g(s)|
...

We write f (s) = o(g(s)), as s → A if
f (s)
= 0
...

Exercise 20
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x2 = o(x) as x → 0
...
2
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21

2
...

3
...

4
...

This notation is very useful because it help us to specify how, among other things, the
approximate solutions computed using perturbation methods approximate the true solutions
...
In Example 17 it could be proved that the true solution x(τ ) and the approximate solution x0 (τ ) + εx1 (τ ) satisfy
x(τ ) − (x0 (τ ) + εx1 (τ )) = O(ε2 )
...
2

Singular perturbations
...
The small parameter multiplies the highest derivative in an ODE
...

2
...

For example,
εx4 + x + 1 = 0
...
The problem occurs in infinite domains
...
When singular points occur in the domain of interest
...
When the equations have multiple scales
...

y 0 = g(x, y)
Let’s give an explicit example
...
Consider the algebraic equation
εx5 + x − 1 = 0
...
3)

Notice that for ε = 0 Equation (3
...
Hence, there is a family of solutions given by the Taylor expansion
x(ε) = 1 + εx1 + ε2 x2 + h
...
t
...
PERTURBATION METHODS
...
What is wrong with Equation (3
...

Let’s try to find a way of computing them
...
Applying this change of variables to Equation (3
...

(3
...
4) is 1 and we obtain
1

y 5 + y − ε 4 = 0
...
3)
...

Now, if we want a Taylor-like expansion of these four roots, we proceed as in the regular
case for the y variable and obtain, for each of them, an expansion of the form
1

1

y(ε) = y0 + ε 4 y1 + (ε 4 )2 y2 + h
...
t
...
4)
...
Give a second order approximation of the following algebraic equations:
1
...

2
...

3
...

For the case of how to solve ODEs with singular perturbations, see Sections 3
...
4
...
3

Boundary layers
...


3
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BOUNDARY LAYERS
...
Consider the boundary value problem
 00
 εy + (1 + ε)y 0 + y = 0,
y(0) = 0,

y(1) = 1
...
5)

Notice that this example has an explicit solution,
x

y(x) =

e−x − e− ε
1

e−1 − e− ε

,

but we will only use it for checking if our approximation method succeeds
...
1
for a visual representation of it
...


3

ε=0
...
01
ε=0
...
5

2

1
...
5

0
0

0
...
4

0
...
8

1

Figure 3
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The boundary value problem (3
...

This is overcome by approximating the solution to (3
...

Outer layer:

24

CHAPTER 3
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The outer layer is the one away x = 0
...

Inner layer:
We scale the boundary problem (3
...

y¨ +
2
f (ε)
f (ε)

(3
...

2
f (ε) f (ε) f (ε)
ε
We will choose f (ε) so that the leading coefficient f (ε)
2 has the same order as another and
the other two are small in comparison
...

x
Then, an approximate solution of Equation (3
...


Now, the problem is finding the value of constant a in the inner approximation so the
solutions match
...
For example, when x = O( ε)
...


ε→0

ε→0

With this, we obtain that a = e
...
Perform the same analysis as in Example 24 in the initial value problem
 0
εy + y = e−x ,
y(0) = 0
...
4

The WKB approximation
...


(3
...
8)
(3
...
4
...


25

Example 26
...

Let’s consider Equation (3
...
The method consists on doing the ansatz that the solution
is of the form
u(x)
y(x) = e ε
...
Finally, using a regular perturbation of f
f = f0 + εf1 + h
...
t
...

4q(x)
Hence, we obtain an approximation of the form
p
q 0 (x)
f (x) = ± q(x) − ε
+ O(ε2 )
...

q(x)

Exercise 27
...
εy 00 + xy = 0, 0 < ε  1
...
y 00 + λ cos(x)y = 0, 1  λ
...
PERTURBATION METHODS
...

The calculus of variations deals with the study of minimizing/maximizing functionals
...
Examples are: minimize the length of a curve,
maximize the area given a fixed length, minimize the area
...
1

Variational problems
...

A global minimum will be a local minimum for all neighbourhoods
...

If the function f is differentiable, a necessary condition of being a local minimum is that
∇f (x0 ) = 0
...

Exercise 28
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Question 29
...

Definition 30
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A functional is a
map
J : X → R
...
CALCULUS OF VARIATIONS
...


• C 0 (U, Rm ), where U ⊂ Rn , is the space of continuous maps
f : U → Rm

with norm
kf k0 := sup kf (x)k
...


k=0

• A(U, C), where U ⊂ C, is the space of analytic maps
f :U →C
with norm
kf kU := sup |f (x)|
...

Let’s see some examples of functionals
...


• Let x0 ∈ R, then J : C 0 (R, R) → R with
J(f ) := f (x0 )

is a functional
...

• Let a, b ∈ R, then J : C 2 (R, R) → R with
Z b
J(f ) :=
f (x)(f 00 (x)2 + cos(f (x)))dx
a

is a functional
...
2
...


29

• (Arclength) Let a, b ∈ R, then J : C 1 (R, R) → R with
Z bp
J(f ) :=
1 + f 0 (x)2 dx
a

is a functional
...
(Brachistochrone problem) Let p = (0, b) and q = (a, 0) be two points lying
on a vertical plane under the force of the gravity g (vertical)
...

Prove that the time that a bead takes to travel across the wire, starting at p and finishing
at q, is
Z a p
1 + y 0 (x)
p
dx
...
This function L is called the Lagrangian
...
2
4
...
1

Necessary conditions for extrema
...


A normed linear space V is a vector space equipped with a norm
...
kyk = 0 if and only if y = 0
...
kyk ≥ 0 for all y ∈ V
...
kayk = |a|kyk for all a ∈ C and all y ∈ V
...
ky1 + y2 k ≤ ky1 k + ky2 k for all y1 , y2 ∈ V
...
Prove that the vector spaces in Notation 31 are normed vector spaces with the
norms specified there
...
2
...


Given a functional J : A ⊂ X → R, its directional derivative with direction v at the
point y0 is (if it exists)
d
δJ(y0 , v) := J(y0 + εv)|ε=0
...
CALCULUS OF VARIATIONS
...

Exercise 35
...


1

Z

y 2 dx, y0 = cos(x), v = sin(x)
...

Z
J(y) =

1

y 02 dx, y0 = cos(x), v = sin(x)
...


1

Z

cos(y)dx, y0 = x, v = x2
...

Theorem 36
...
If y0 ∈ A is a minimum (maximum) of J, then
δJ(y0 , v) = 0
for all v where the directional derivative exists
...
Consider the functional J : C 0 ([2, 4], R) → R,
Z 4
J(y) =
y(x)2 dx
...


4
...


The simplest problem in calculus of variations is to consider the functional
Z b
J(y) =
L(x, y, y 0 )dx

(4
...
The function
L should satisfy that it is twice differentiable in [a, b] × R2
...


4
...
THE SIMPLEST PROBLEM
...
Prove that in the simplest problem
Z b
δJ(y0 , v) =
∂y L(x, y, y 0 )v + ∂y0 L(x, y, y 0 )v 0 dx
...
2)
a

for all v with v(a) = v(b) = 0
...
2) is equivalent to

Z b
d
0
0
∂y L(x, y0 , y0 ) − ∂y0 L(x, y0 , y0 ) vdx = 0
...

Theorem 39
...
1) and defined for functions y ∈ C 2 [a, b]
with the extra condition y(a) = A, y(b) = B, a necessary condition for y0 being a minimum
(maximum) is that
d
∂y L(x, y0 , y00 ) − ∂y0 L(x, y0 , y00 ) = 0
...
3)
dx
Observation 40
...
If
Z b
f (x)g(x)dx = 0
a

for all twice differentiable functions g with g(a) = g(b) = 0, then f is identically the zero
function
...
3) is called Euler-Lagrange equation
...

Exercise 41
...

Also, if ∂x L = 0, then
L − y 0 ∂y0 L = C,
with C being a constant
...


32

CHAPTER 4
...


Exercise 42
...

Z

1

((y 0 )2 + 3y + 2x)dx,

J(y) =
0

with y(0) = 0, y(1) = 1
...


1

Z

p
1 + (y 0 )2 dx,

J(y) =
0

with y(0) = a, y(1) = b
...

Z
J(y) =
0

a

p
1 + (y 0 )2
p
dx,
2g(b − y)

with y(0) = b, y(a) = 0
...
4

Generalizations
...


4
...
1

Higher derivatives
...
For example, the second-order problem
Z b
L(x, y, y 0 , y 00 )dx,
a

where y ∈ C 4 [a, b] satisfying the boundary conditions y(a) = A1 , y(b) = B1 , y 0 (a) = A2 , y 0 (b) =
B2
...

dx
dx
More generally, in the case of having the Lagrange function L depending on the derivatives of y up to the n-th order, then y ∈ C 2n [a, b], y should satisfy boundary conditions up
the (n − 1)-th derivatives, and the Euler-Lagrange equations are
d
d2
dn
∂y0 L + 2 ∂y00 L + · · · + (−1)n n ∂yn−1) L = 0
...
Find the extremals of the functional
Z 2p
1 + (y 00 )2 dx,
∂y L −

0
0

0

with y(0) = 0, y (0) = 1, y(2) = 1, y (2) = 1
...
5
...


4
...
2

33

Several functions
...
For example, if two are involved, we
get the functional
Z b
L(x, y1 , y10 , y2 , y20 )dx,
J(y) =
a

with boundary conditions y1 (a) = A1 , y2 (a) = A2 , y1 (b) = B1 , y2 (b) = B2
...

dx 2

4
...
3

Natural boundary conditions
...
For example, consider the functional
Z

b

L(x, y, y 0 )dx,

a

with boundary conditions y(a) = A and y(b) free
...
5

d
∂y0 L = 0,
dx
∂y0 L(b, y(b), y 0 (b)) = 0
...


Exercise 44
...

Exercise 45
...

Exercise 46
...

Z
J(y) =

1

(y 2 + y 02 − 2y sin(x))dx,

0

where y(0) = 1 and y(1) = 2
...

Z
J(y) =
1

where y(1) = 1 and y(2) = 0
...
CALCULUS OF VARIATIONS
...


2

Z

(y 2 + y 02 + 2yex )dx,

J(y) =
0

where y(0) = 0 and y(2) = 1
...
Find the Euler-Lagrange equation of the functional
Z b
p
f (x) 1 + y 02 dx,
a

and solve it for y(a) = A, y(b) = B
...
Find an extremal for
Z

2

J(y) =
1

p
1 + y 02
dx,
x

where y(1) = 0 and y(2) = 1
...
Show that the area of a surface given by the graph of a function z = f (x, y)
defined on a domain D is given by the double integral
ZZ q
1 + (∂x f )2 + (∂y f )2 dxdy
...

Prove that the surface given by z = arctan yx satisfies Equation (4
...

Could you give an idea of the proof of Equation (4
...
Find the extremals of
1
...

2
...



1 2 02 1 2
x y − y dx,
2
8

(4
...

A dynamical system is a rule for time evolution on a state space
...

Example 51
...
Then an
example of a dynamical system on this phase space is
xn+1 = xn − 1
...

Example 52
...

Another example of a dynamical system on this phase space is
x˙ = −1
...

As shown in the previous two examples, there are different types of dynamical systems
...

In this notes we will concentrate in two types of dynamical systems
...
When the time evolution is continuous, we will
consider dynamical systems described by ODEs
x˙ = F (x, t),
where x ∈ X
...
Questions that are usually asked are: do all particles converge
to a point? Do all particles converge to a set? Are there stationary particles? How is the
evolution of the particles near a stationary one?
35

36

5
...
DYNAMICAL SYSTEMS
...


As said before, a discrete dynamical system is given by the evolution of a system of the form
xn+1 = f (xn , n)
...
Consider the dynamical system that models the evolution of your savings in
your bank account
...
Describe the evolution of all initial states x0 under this
system
...
Find a closed formula of the forward iterations of the (linear) dynamical
system
xn+1 = Axn
in terms of the eigenvalues and eigenvectors of the matrix A
...

Apply this to the system with


2 1
A=

...

Usually this is not the case
...
1
...


Definition 55
...
That is,
f (x∗ ) = x∗
...
Find all fixed points of the system in Exercise 53
...
Find all fixed points of the system
xn+1 = x2n + a
...
Given a discrete dynamical system xn+1 = f (xn ), a periodic orbit of
period k is a point x∗ such that its forward evolution is k-periodic
...

Exercise 59
...


5
...
DISCRETE DYNAMICAL SYSTEMS
...
Find all fixed points and period 2 orbits of the system

xn+1 = 1 − ax2n + byn

...

Once they are computed, the next question that one should ask is: how is the behaviour
of nearby orbits? Do they converge to the fixed point? Are they repelled? In order of
answering these questions, we should introduce linear stability
...
Given a dynamical system xn+1 = f (xn ), we will say that a fixed (or periodic) point x∗ is asymptotically stable if all points y in a neighbourhood of it converge
satisfy
lim d(f n (y), f n (x∗ )) = 0
...

Exercise 62
...
Similarly,
it is asymptotically stable if |1 + r| > 1
...

Theorem 63
...

If the spectrum of the lineariztion contains an eigenvalue with modulus larger than 1,
then it is asymptotically unstable
...

The procedure in Theorem 63 is sometimes called the study of the linear behaviour
...

Exercise 64
...

Exercise 65
...
4 and b = 0
...


38

CHAPTER 5
...


5
...


As we saw, ODEs define continuous dynamical systems
...

Some ODEs are easy to solve analytically, while others do not have known analytic
solution
...

Exercise 66
...
Write down the solution of it in terms of the eigenvalues/eigenvectors
of it
...

For the ODEs that no known explicit solution is known, other methods for studying
them are needed
...

For the sake of simplicity, from now on we will only consider autonomous ODEs: ∂t F (x, t) =
0
...
2
...


An ODE
x˙ = F (x)
defines a vector field
...
Vector
fields are very useful for the understanding of the dynamics
...
Have you ever seen the weather
forecast? The wind field is a vector field! A dust particle will flow under the wind field
following the vector directions
...
Of course, this last equation is just an approximation of
how solutions of the vector fields behave
...

Exercise 67
...

x˙ = 1
...

x˙ = x
...
2
...

3
...



5
...
Consider the following vector fields in polar coordinates
...




2
...


5
...
2

r˙ = 1
θ˙ = 1

r˙ = r3 − r
θ˙ = 1


r˙ = 0
θ˙ = r

Stationary orbits and stability
...
This is equivalent
to x˙ = 0
...

As in the discrete case, the stability of a stationary orbit is dictated by its linearization
around it
...
Consider an ODE
x˙ = F (x),
with a stationary orbit x0
...
Similarly, if part of the spectrum is on the right hand side, it is asymptotically
unstable
...
DYNAMICAL SYSTEMS
...
Find the stationary orbits of the following ODEs and study their stability:
1
...

2
...


5
...
3



x˙ = y
y˙ = cos(x) + 21 y



x˙ = x2 + y 2 − 1
y˙ = x − y

Periodic orbits
...
Finding periodic orbits is not an easy task and requires, in general,
advance techniques out of the scope of this course
...

Exercise 71
...
Polar coordinates:


2
...
3



r˙ = r3 − r
θ˙ = 1

x˙ = x3 − x + y 2 x − y
y˙ = y 3 − y + x2 y + x

Chaotic systems
...
This means that if x0 ∈ U then, for all y0 close to x0
d(xn , yn )
diverge
...

Observation 72
...
It involves two other conditions: that the system has a dense set of periodic
orbits and that it is topologically mixing: all neighborhoods in U mix under the action of the
system
...
3
...


41

Exercise 73
...
That is, prove that it has a dense set of periodic orbits and its sensitive to initial
conditions
...
Convince yourself that the discrete dynamical system defined on the circle
[0, 1]/Z
xn+1 = f (xn ) (mod 1),
where
f (x) = 2x (mod 1)
is chaotic
...

Where could we find this dynamical system?

42

CHAPTER 5
...


Chapter 6
Introduction to partial differential
equations
...
They model
chemical reactions, physical laws, reaction-diffusion systems in biology and chemistry, gas
dynamics, fluid dynamics
...

The goal of this chapter is to give an introduction to this topic
...
1

Some examples
...
(Heat equation
...

The operator ∆, the laplacian, is defined as ∆u = nk=0 uxk xk
...
(Wave equation
...

Example 77
...
)
The reaction-diffusion equation
ut = ∆u + f (u)
describes how the concentration of one or more substances distributed in space changes under
the influence of two processes: diffussion which causes the substances to spread out, and
reactions which causes the substances to transform themselves
...
INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS
...
2

Basic concepts

We will concentrate our efforts in second-order PDEs
...

If the PDE has a variable that represents time, then the PDE studies the evolution
of the solutions with respect time
...
(These are just conventions)
...
Just as the general solution of an
ODE depends on some constants, the solution of a PDE depends on some arbitrary function
...

Exercise 78
...

ut = cos(x)
...

utx = x
...

uxx = y
...
Solve the following PDEs:
1
...

2
...

3
...

Use the change of coordinates z = x − ct, t = t
...
Find all solutions of the heat equation ut = kuxx of the form u(x, t) = U (z),
with z = √xkt
...
We will study some examples
where analytic solutions are known
...
Nevertheless, we
will see that some PDEs can be classified into three categories: elliptic (those that govern
equilibrium phenomena), hyperbolic (those that govern wave propagation) and parabolic
(those that govern diffusion processes)
...
3
...

Example 81
...


• Hyperbolic:
utt = uxx
...


6
...


PDEs can be of two types: linear and nonlinear
...

A linear PDE (just with two variables x, t) is of the form:
L(∂x , ∂t , x, t)u(x, t) = f (x, t),
where f is a known function and u is the unknown
...

Exercise 82
...

uxx + cos(x)ux = 0
...

uxx + u2x = 0
...
L(u + w) = Lu + Lw
...
L(cu) = cLu, for c ∈ R
...
This property
is that if u1 and u2 are solutions of the PDE
Lu = 0,
then any linear combination c1 u1 + c2 u2 , with ci R, is also a solution of it
...
1)

46

CHAPTER 6
...


is of the form
up + c1 u1 + · · · + cn un ,
where up is a particular solution of Equation (6
...

Another type of superposition is that, if u(x, t; α) are all solutions of the equation Lu = 0,
then
Z ∞
g(α)u(x, t; α)dα
−∞

is a solution also for every function g
...
Prove that
−(x−α)2
1
e 4kt , t > 0
u(x, t; α) = √
4πkt

is a solution of the heat equation
ut = kuxx
...


6
...


Laplace’s equation is the prototype of elliptic equations
...

Its solutions model equilibrium problems because, for example, these are the time independent solutions of the heat and the wave equations
...

• (Dirichlet condition
...

• (Neuman condition
...

dn |∂Ω

6
...
EVOLUTION PROBLEMS
...

Theorem 84
...

¯ then it is unique
...

Exercise 85
...

Write it down in polar coordinates
...
Consider the Laplace’s equation in R3 :
uxx + uyy + uzz = 0
...

Exercise 87
...
That is, u(x, y, z) = ϕ(x2 + y 2 + z 2 )
...
5

Evolution problems
...

An initial value problem for it is defined once we fix u(x, 0 = u0 (x)
...

Exercise 88
...
Let’s discretize it by considering just N values of u, uk ,
equidistant along the [0, l] interval
...


48

CHAPTER 6
...


Observation 89
...
This is the same as saying
the they are globally attracting solutions
...
Consider the two dimensional heat equation on the unit square Ω = [0, 1]2
with boundary condition u|∂Ω = 0
...

Exercise 91
...

Prove that the zeros of F are steady-states
...
Consider the reaction-diffusion equation
ut = ux x + u(1 − u),
with x ∈ [− π2 , π2 ] and Dirichlet conditions u(± π2 , t) = 3
...

Show that u(x, t) = 1+cos(x)

6
...


In this section we will see how to solve linear PDEs using eigenfunction expansions
...
2)

where B(u) = 0 are the boundary conditions (Dirichlet or Neumann)
...
2) is, if possible, finding a basis of eigenfunctions vk , with associated eigenvalues λk ,
satisfying the eigenvalue problem

Lvk = λk vk , vk ∈ Ω

...
2) is
X fk
vk (x)
...

λk

6
...
EIGENFUNCTION EXPANSIONS
...
Consider the PDE
uxx + uyy = f
on Ω = [0, π]2 with Dirichlet condition u|∂Ω = 0
...

Exercise 94
...

Exercise 95
...


49

50

CHAPTER 6
...


Chapter 7
Sturm-Liouville problems
...
1)

defined on an interval [a, b] with boundary conditions

a1 y(a) + a2 y 0 (a) = 0
b1 y(b) + b2 y 0 (b) = 0
defined a Sturm-Liouville problem, (SLP)
...

Notice that solutions of Equation (7
...
A value λ for which Equation (7
...

Regular SLP problems satisfy that they have infinite eigenvalues with associated eigenvectors of finite multiplicity
...
Moreover,
all eigenvalues are real
...
Find all eigenvalues and eigenvectors of the regular SLP problems:
1
...

2
...

3
...


51

52

CHAPTER 7
...


Chapter 8
Theory of transforms
...

The applied transformations are usually of integral type: Laplace transform and Fourier
transform
...
As we will see, the theory behind
these two transforms is parallel to each other
...
For example, Fourier and Laplace transforms are
useful for replacing derivatives by algebraic equations
...
Let’s illustrate the idea behind transform methods
...
But, if
the system

 2X + 3Y + Z
X +Y

3X + 5Y − Z

we perform the logarithm transform, we get
= log(8)
= log(7) ,
= 0

where X = log(X), Y = log(Y ), Z = log(Z)
...
Once we
have its solutions, we pull them back with the help of the inverse of the logarithm transform,
the exponential
...
1

Laplace transform
...


54

CHAPTER 8
...


We usually denote by capital letters the transformed function: L(y) = Y
...
Compute the Laplace transforms of the following functions:
1
...

2
...

3
...

4
...

The inverse of the Laplace transfrom is defined as
Z a+i∞
1
−1
Y (s)est dt,
L (Y )(t) =
2πi a−i∞
where the integration path is a vertical line on the complex plane from bottom to top and
a is chosen in a way that all singularities of the function Y lie on the left side of the vertical
line with real part a
...

Theorem 99
...


k=0

Theorem 100
...

0

Then, the Laplace transform satisfies that
L(y1 ∗ y2 )(s) = Y1 (s)Y2 (s)
...

Exercise 101
...

Exercise 102
...
That is, L(af +bg) =
aL(f ) + bL(g), where a and b are constants
...
1, where some of the most common Laplace transforms appear
...
With the help of Laplace transforms, solve the following problems:

8
...
LAPLACE TRANSFORM
...
1: Laplace transforms
...


56

CHAPTER 8
...

1
...

2
...

3
...

u(x, 0) = 0, x > 0,

u(0, t) = 0, t > 0
...
The PDE

ut − uxx = 0, x > 0, t > 0,



u(x, 0) = 0, x > 0,
u(0, t) = 1, t > 0,



u(x, t) is bounded
...
The PDE


 ut − kuxx = 0, x > 0, t > 0,
u(x, 0) = 0, x > 0,

u(0, t) = g(t), t > 0
...


8
...


The Fourier transform is defined as
Z

∞

F(y)(s) =

y(t)eist dt
...

The inverse of the Fourier transform is
Z ∞
1
−1
yˆ(t)e−ist dt
...
The definition of the Fourier transform presented in these notes differs
from others given in other textbooks
...

The Fourier transform satisfies similar properties to the ones satisfied by the Laplace
transform
...
2
...


57

Exercise 105
...
That is, F(af +bg) =
aF(f ) + bF(g), where a and b are constants
...
Compute the Fourier transform of n-th derivative y (n) and of the convolution
y1 ∗ y2
...
2 there are some Fourier transforms
...
2: Fourier transforms
...


Exercise 107
...
The ODE
u00 − u = f (x), x ∈ R
...
The PDE (Heat equation)
ut − uxx = 0, x ∈ R, t > 0, u(x, 0) = f (x)
...
The PDE (Wave equation)
utt − c2 uxx = 0, x ∈ R, t > 0, u(x, 0) = f (x), ut (x, 0) = g(x)
...
The PDE (Laplace equation on the half-plane)
uxx + uyy = 0, x ∈ R, y > 0, u(x, 0) = f (x)
...
The PDE (Laplace equation on a strip)
uxx + uyy = 0, x ∈ R, b > y > 0, u(x, 0) = f (x), u(x, b) = g(x)
...
The PDE (Advection-diffusion equation)
ut − cux − uxx = 0, x ∈ R, t > 0, u(x, 0) = f (x)
...
The PDE (Non-homogeneous heat equation)
ut = uxx + F (x, t), x ∈ R, t > 0, u(x, 0) = 0
...
3

CHAPTER 8
...


Other transforms
...

For example, when dealing with periodic functions y : R → R, y(x) = y(x + 1), there is
the (periodic) Fourier transform
Z 1
y(t)e−2πist dt
...

Example 108
...
For example, the heat equation on the circle
...

Another example is when dealing with sequences
...

k=0

This transform is useful in finite differences equations
...

An integral equation is an equation where the unknown is a function and integrals are
involved
...

Z

1

f (x)dx = f (2)
...

Z

x

f (t)dt = f (0) + f (x)
...

For example, given an ODE
x(t)
˙
= f (x(t), t),
(9
...


x(t) = x(0) +

(9
...
Prove that a solution of Equation (9
...
1)
...
The former is of the form
Z x
k(x, y)u(y)dy − λu(x) = f (x), a ≤ x ≤ b
a

while the latter is
Z

b

k(x, y)u(y)dy − λu(x) = f (x), a ≤ x ≤ b
...
The function k is
usually called the kernel
...
INTEGRAL EQUATIONS
...
Notice that both problems look similar
...
As we will see, this small detail changes dramatically the way each problem is
addressed
...
Notice that both equations can be written
in the form
(K − λId)u = f,
(9
...
Hence, the equations will have a solution u if
the function f is on the range of the linear operator K − λId
...
if it is invertible:
u = (K − λId)−1 f
...
If the operator K − λId fails to be invertible, it is still possible that for
some (but not all) f Equation (9
...

To study the invertibility of K − λId it is important to understand for which λs the
eigenvalue equation
Ku = λu
is satisfied
...

The following exercise shows why studying the spectrum of a linear operator A is useful
for solving linear systems
...
Consider the real symmetric n × n matrix A
...
Use the fact that there exists
an orthogonal basis of eigenvectors, and that the eigenvalues are all real
...
1

Volterra equations
...


(9
...
Let’s see some
of these
...
Suppose that the kernel k does not depend on the first variable x (k(x, t) =
g(t))
...
4) satisfies the ODE
u0 (x) =

1
(g(x)u(x) − f 0 (x))
...
1
...


61

Exercise 116
...


x

Z

u(t)dt = u(x) + x
...


Z

x

tu(t)dt = 2u(x) + cos(x)
...
Suppose that the kernel k in Equation (9
...

Prove that the solution of Equation (9
...

(Hint: Remember that the Laplace transform of the convolution is the product of the
Laplace transforms
...
Solve the Volterra equations
1
...


u(x) +
0

2
...


u(x) =
0

In general, Volterra equations are solved by means of the Picard’s method
...

λ

The solution is of the form
u=

∞
X

ˆ n f,
K

(9
...

ˆ This series is called the Neuwhere K
mann series
...

Theorem 119
...
5)
...
Since solution (9
...
It can be proven that
n

k
n
ˆ n f | ≤ max |f | (b − a) max | λ |
...
INTEGRAL EQUATIONS
...

max |f |
n!
n=N +1

(9
...
Prove that an upper bound of (9
...


(9
...
Find approximate solutions to the following Volterra equations using Neumann series:
1
...


u(x) + λ
0

2
...


λu(x) +
0

9
...


As said before, Fredholm equations are of the form
Z

b

k(x, y)u(y)dy − λu(x) = f (x), a ≤ x ≤ b
...
8)

a

In the case of Volterra equations we saw that all the linear equations have a solution,
given by the Neumann series
...

However, as we will see, there are cases that we can treat
...
2
...


A Fredholm equation with degenerate kernel is one that its kernel k(x, y) can be expressed
in the form
n
X
αi (x)βi (x)
...
2
...


63

In this special case, the solution to the Fredholm equation (9
...
Notice that it is equivalent to
Z b
n
X
αi (x)
βi (y)u(y)dy − λu(x) = f (x)
...
9)
a

i=0

Let’s denote by (f, g) the integrals
Z

b

f (y)g(y)dy
...
9) by βj (x) and integrating with respect x we obtain the n
linear equations of the form
n
X
(αi , βj )(βi , u) − λ(βj , u) = (βj , f )
...
10)

where A is the matrix with (i, j) entry (αi , βj ), and w and f are vectors with entries (βi , u)
and (βj , fj )
...
10) has a solution w, then a solution to the Fredholm equation
with degenerate kernel will be
!
n
X
1
u(x) =
−f (x) +
αi (x)wi
...
Notice that the linear system (9
...

It is easily proven in this case the following theorem, sometimes called the Fredholm
alternative
...
Consider the Fredholm equation (9
...
Then, if λ is
not an eigenvalue of the matrix A, the problem has a unique solution
...

Exercise 125
...

0

Exercise 126
...

0

64

CHAPTER 9
...


9
...
2

Symmetric kernels
...
With these kind of kernels,
the eigenvalue problem
Ku = λu
satisfies that if an eigenvalue exists, it is real, and all the eigenvectors corresponding to
distinct eigenvalues are orthogonal
...
Nevertheless, we can give some conditions for their existence
...
If the Fredholm equation satisfies that its kernel is symmetric, continuous
and non-degenerate, then the eigenvalue problem
Ku = λu
has infinite eigenvalues λi , each with finite multiplicity, such that then can be ordered
0 < · · · < |λ2 | < |λ1 |
with limn→∞ = 0
...


k=1

The coefficients ak are equal to

Rb
a

f (x)φk (x)dx
...
See Exercise
114 for an analogue solution
...
Find the eigenvalues and eigenvectors of the operator
Z 1
Ku(x) =
(1 − |x − y|)u(y)dy
...
3

Perturbation methods
...
These methods
could be very helpful for solving nonlinear integral equations or, more generally, integrodifferential equations
...


9
...
PERTURBATION METHODS
...
Find approximate solutions of the following equations by means of perturbation series around ε = 0:
1
...


0

2
...


0

3
...


66

CHAPTER 9
...


Appendices

67

Appendix A
Solving some ODEs
...
1

First order linear ODEs
...


(A
...
1) by a function f (x), obtaining
f (x)y 0 + f (x)p(x)y = f (x)q(x)
...


(A
...
The solution to Equation (A
...


Thus, we get that
(f (x)y)0 = f (x)p(x),
so

A
...


Second order linear ODEs
...

First, we find solutions to the homogeneous equation
y 00 + p(x)y 0 + q(x)y = 0
...
3)

70

APPENDIX A
...


These are of the form
yH (x) = Ay1 (x) + By2 (x)
...
3) is found by finding a particular solution of the
form
yP (x) = A(x)y1 (x) + B(x)y2 (x),
with the extra condition
A0 (x)y1 (x) + B 0 (x)y2 (x) = 0

---