Differential Equations coursework

DIFFERENTIAL EQUATIONS

GIANG B KHUONG

Contents

1. What is Ordinary Differential Equations (ODE)? 2

2. Basic/Logistic Model and Its Modification 3

3. What is a solution? 5

4. Initial Value Problems (IVP) 8

5. Fundamental Theorem of Ordinary Differential Equations (FTODE) 10

6. Slope Field 11

7. Numerical Method: Euler’s Method 14

8. Linearization of Systems of Differential Equations 16

9. Superposition: Characteristic property of linear equations 19

10. Eigenvalues, eigenvectors, and eigen solutions 22

11. Nullclines 36

12. Oscillations 39

13. Conserved and Monotone Quantities 41

14. Analytic methods 43

1

2 GIANG B KHUONG

1. What is Ordinary Differential Equations (ODE)?

Ordinary Differential Equations (or ODEs) are equations with derivatives in them, and

where unknowns are functions.

Example:

Q = population of orchids in garden, in thousands

t = time, in weeks

dQ

dt= 0.8Q(1− Q

12)− 0.12Q

To build differential equations:

• make assumptions on which the model will be based

• describe variables and parameters to be used

• translate into math symbols

Example:

The rate of change of the orchid population is precisely 1500/week

P = orchid population, in thousands

t = time, in weeks

dP

dt= 7

The unknown is the function P (t).

DIFFERENTIAL EQUATIONS 3

2. Basic/Logistic Model and Its Modification

(1) Basic Models

Basic models of population growth are based on the assumptions that the rate

of growth of the population is proportional to the size of the population.

dP

dt= rP

where r is the constant rate of growth, and P is the population.

Example:

The rate of change of fish in river A annually is exactly 15%.

F = fish population in river A, in thousands

t = time, in years

dF

dt= 0.15F

(2) Logistic Models

Logistic models are based on the assumptions that the rate of growth of the

population is proportional to the size of the population but that increase may be

limited, e.g. growth rate of fish depends on the availability of volume of water.

dP

dt= rP (1− P

K)

where r is the constant rate of growth, and K is the availability of habitat.

Example:

The growth rate of aphids in kale patch is 7% per week. The kale patch can

sustain a maximum of 1.5 million aphids.

A = aphid population, in thousands

4 GIANG B KHUONG

t = time, in weeks

dA

dt= 0.07A(1− A

1500)

(3) Modifications of Logistic Models

(a) Harvesting: −f(P )

Example:

The growth rate of aphids in kale patch is 7% per week. The kale patch

can sustain a maximum of 1.5 million aphids.

Also, the aphids are harvested at a constant rate of 10% per week.

A = aphid population, in thousands

t = time, in weeks

dA

dt= 0.07A(1− A

1500)− 0.1A

(b) Hatchery: +f(P )

Example:

The growth rate of fish in river A monthly is 2%. The river can sustain

up to 15,000 fish.

Environmentalists are trying to replenish the fish stock in that river by

releasing 5,000 fish in the river every month.

F = fish population in river A, in thousands

t = time, in years

dF

dt= 0.2F (1− F

15) + 5


3. What is a solution?

As mentioned before, solutions to ODEs are functions, instead of numbers like in alge-

braic equations.

A function is a solution to an ODE if the equation is true when we plug it in.

Example:

Proof that y(t) = 42e0.5t is a solution to

ODE1 =dy

dt= 0.5y

Proof:

LHS =d

dt[y(t)] =

d

dt(42e0.5t) = 42(0.5)e0.5t = 21e0.5t

RHS = 0.5y = (0.5)(42e0.5t) = 21e0.5t

LHS = RHS

Hence, y(t) = 42e0.5t is a solution to ODE1.

General Solutions

Example:

There is a collection of functions in the form y(t) = Ce0.5t to ODE1, where C is a

constant.

Proof:

LHS =d

dt[y(t)] =

d

dt(Ce0.5t) = (0.5)(Ce0.5t)

6 GIANG B KHUONG

RHS = 0.5y = (0.5)(Ce0.5t)

LHS = RHS

Hence, y(t) = Ce0.5t is a solution to ODE1, where C is any constant.

Graph:

Figure 1. C = −2,−1, 0, 1, 2

Observation:

Any point on the plane lies one exactly one solution curve.

General features of general solutions:

• lots of solutions

• the solutions fill the plane


.

Equilibrium Solutions

Equilibrium solutions are constant functions. The interpretation of such functions is

steady-state, i.e. no dynamics.

(1) Equilibrium solutions of Basic models

Example:

Consider the following basic growth model

dP

dt= 0.7P

For equilibrium solutions, P (t) = constant, hence dPdt = 0

0 = 0.7P

P (t) = 0

P (t) = 0 is the equilibrium solution for this basic growth model.

(2) Equilibrium solutions of Logistic models

Example:

Consider this logistic growth model

dP

dt= 0.7P (1− P

9)


0 = 0.7P (1− P

9)

8 GIANG B KHUONG

either P = 0 or 1− P9 = 0

P (t) = 0 andP (t) = 9 are the equilibrium solutions for this logistic model.

(3) Equilibrium solutions of Logistic models with modifications

Example:

Consider this modified logistic growth model

dP

dt= 0.7P (1− P

9) + 2


0 = 0.7P (1− P

9) + 2

0 = 0.7P − 0.07

9P 2 + 2

0 = P 2 − 15P − (15

0.07)2

either P = −15 or P = 30

P (t) = −15 andP (t) = 30 are the equilibrium solutions for this modified logistic

model.

4. Initial Value Problems (IVP)

For equations of the formdy

dt= f(t, y)

it seems like there are many solutions; in fact, the solutions fill the plane, i.e. each point

(tA, yA) lies on a solution.

So, given an equation, does there exist a solution? If so, is this solution unique?


An Initial Value Problem (IVP) put additional requirement so that to satisfy an IVP,

the solution graph has to pass through a point on the plane. This makes IVP have one

unique solution.

Example:

Find y(t) such that

dydt = f(t, y)

y(to) = yo

y(t) will not only be solution to the ODE, but it will also satisfy the additional requirement

to pass through (to, yo). Note: that to can be any arbitrary starting point t.

Example:

Consider the following IVP:

dydt = cos (t)

y(0) = 7

Solve:

First look at ODE dydt = cos (t)

y(t) = sin (t) + C

is a solution for any C.

Figure 2. C = −5, 0, 7

10 GIANG B KHUONG

y(0) = 7

7 = sin (t) + C

C = 7

Figure 3. C = 7

.

5. Fundamental Theorem of Ordinary Differential Equations (FTODE)

Consider the following IVP:

dydt = f(t, y)

y(to) = yo

If f(t, y) and dfdt (t, y) are continuous at y(to) = yo then there exists a unique solution

nearby.


Example:

dydt = t3 cos (y)

y(1) = π

Here f(t, y) = t3 cos (y)

• is this continuous near (t, y) = (1, π)?

• look at dfdt (t, y) = t3(− sin (y))

is this continuous near (1, π)?

Both questions have answers ”Yes”, which means that there exists a function y(t) such

that:

• y(t) = π

• y is defined for t nearby t = 1

• dydt = t3 cos (y)

• there is only one such function

Important consequences:

We cannot have two solutions meet.

6. Slope Field

The ODE tells us the slope of the function. This gives us a sense of what the growth

looks like.

We can use that information to plot a slope field.

Example:

(1) Basic Growth Model

dP

dt= rP

12 GIANG B KHUONG

(a) r > 0

Figure 4. r > 0

P = 0, slope = 0

P = 1, slope = r

P = 2, slope = 2r

These slopes are independent of t. At each point on the t-p plane, we have a

slope.

The slope arrows are tangent to any solution curve, i.e. solutions follow the

slope values.

Thus slope field tells us what graphs of solutions look like.


(b) r < 0

Figure 5. r < 0

Initial conditions nearby to P = 0 have solutions which approach P = 0

Terminology:

• If nearby solutions approach an equilibrium, we say that the equilibrium is stable.

• If nearby solutions diverge from an equilibrium, we say that the equilibrium is unstable.

14 GIANG B KHUONG

(2) Logistic Model

dP

dt= rP (1− P

K)

r > 0,K > 0

Figure 6. Equilibrium at P = 0 and P = K

P = 0, slope = 0

P = K, slope = 0

P = 2K, slope = r(2K)(1− 2KK ) = −2Kr

P (t) = 0 is an unstable equilibrium because nearby solutions diverge from P = 0.

P (t) = K is an stable equilibrium because nearby solutions converge towards

P = K.

.

7. Numerical Method: Euler’s Method

Euler’s Method approximate numerically the change of y as t changes a ∆t, as oppose

to the slope field, which offers a graphical approximation.


Consider the following IVP: dydt = f(y)

y(0) = yo

to = 0 yo = yo

t1 = ∆t y1 = yo + f(yo)∆t

t2 = 2∆t y2 = y1 + f(y1)2∆t

t3 = 3∆t y3 = y2 + f(y2)3∆t

. .

. .

. .

tk = k∆t yk = yk−1 + f(yk−1)k∆t

.

Example:

ODE2 = dy

dt = 0.3y

y(0) = 1

Choose ∆t = 12 = 0.5

We get a sequence of points:

(to, yo) to = 0

(t1, y1) t1 = to + ∆t = 0 + 0.5 = 0.5

(t2, y2) t2 = t1 + 2∆t = 0 + 2(0.5) = 1

(t3, y3) t3 = t2 + 3∆t = 0 + 3(0.5) = 1.5

.

.

.

16 GIANG B KHUONG

8. Linearization of Systems of Differential Equations

(1) Why should we linearize a system of ODEs?

Linear equations are continuous, hence it is easy to solve IVPs to these equations.

Example:

ODE = dy

dt = 3y

y(0) = A

General solution: y(t) = Ce3t

y(0) = Ce3(0) = C

Hence, C = A

The solution to this IVP is y(t) = Ae3t

Linearization is trying to approximate non-linear equations at a certain point.

Usually we choose the equilibrium point to get an approximated linear equation for

a new variable which represents the deviation from the equilibrium.

(2) How to linearize systems of ODEs?

If we have a non-linear function f(yk), it can be locally approximated by a

straight line.


g(y) = f(yk) + f ′(yk)(y − yk)

We say that g(y) approximates f(y) near yk

f(y) = f(yk) + f ′(yk)(y − yk) + ...

We usually call this the Taylor series.

The linearization of f at yk is linear part of the Taylor series f(y) ≈ f(yk) +

f ′(yk)(y − yk)

By Linearization of Systems of ODEs, we can learn ’essentially everything’ we

want to know near any equilibrium yk by looking at the linearization.

dxdt = f(x, y)

dydt = g(x, y)

To approximate f(x, y) near some point (xA, yA):

f(x, y) ≈ f(xA, yA) +df

dx(xA, yA)(x− xA) +

df

dy(xA, yA)(y − yA)

Example:

Linearize the following system at (2, 6)

dxdt = 0.08x(1− x

8 )− 0.01xy

dydt = 0.04y − 0.02xy

• f(2, 6) = 0.16(34)− 0.12

dfdx = 0.08(1− x

8 )+0.08x(18)−0.01y → dfdx(2, 6) = 0.06−0.02−0.06 = 0.02

dfdy = −0.01x → df

dy (2, 6) = −0.02

f(x, y) ≈ 0 + (−0.02)(x− 2) + (−0.02)(y − 6)

• g(2, 6) = 0.04(6)− 0.02(2)(6)

dgdx = −0.02y → dg

dx(2, 6) = −0.12

18 GIANG B KHUONG

dgdy = 0.04− 0.02x → dg

dy (2, 6) = 0.04− 0.02(2) = 0

g(x, y) ≈ 0 + (−0.12)(x− 2) + 0(y − 6)

∴

dxdt ≈ −0.02(x− 2)− 0.02(y − 6)

dydt ≈ 0.12(x− 2)

(3) How to analyze properties of systems of ODEs via linearization?

The system of ODEs

dxdt = ax+ by

dydt = cx+ dy

can be written in matrix form as follow:

d

dt

xy

=

ax+ by

cx+ dy

=

ax+ by

cx+ dy

=

a b

c d

xy

Let

xy

= ~y we have:

d

dt~y =

a b

c d

~yTo check whether a function is a solution to a linearized system of ODEs or not,

just plug it in.

Example:

d

dt~y =

2 3

7 6

~yShow that ~y1(t) = e−t

−1

1

and ~y1(t) = e9t

3

7

are solutions.


(a) Examine ~y1(t) = e−t

−1

1

LHS = d

dt [~y1(t)] = −e−t−1

1

= e−t

1

−1

RHS =

2 3

7 6

e−t−1

1

= e−t

1

−1

LHS = RHS

∴ ~y1(t) = e−t

−1

1

is a solution.

(b) Examine ~y1(t) = e9t

3

7

LHS = d

dt [~y2(t)] = 9e9t

3

7

= e9t

27

63

RHS =

2 3

7 6

e9t3

7

= e9t

27

63

LHS = RHS

∴ ~y2(t) = e9t

3

7

is a solution.

9. Superposition: Characteristic property of linear equations

(1) Superposition Theorem:

If y1 and y2 are solutions to a linear equation, then αy1 + βy2 is also a solution

to the linear equation for all number α, β.

The quantity αy1 + βy2 is called a linear combination (of y1 and y2)

20 GIANG B KHUONG

Example:

Assume that w1 and w2 solve:

dw

dt= rw

Examine αw1 + βw2

d

dt[αw1 + βw2] = α

dw1

dt+ β

dw2

dt= α(rw1) + β(rw2) = r(αw1 + βw2)

This works because multiplication and derivation share some properties.

(2) Application of the Superposition theorem:

We can use the Superposition theorem to generate a general solution form from

a few solutions.

Example 1:

We know from example in Section 8, number (3) that ~y1(t) = e−t

−1

1

and

~y1(t) = e9t

3

7

are solutions for the system:

d

dt~y =

2 3

7 6

~yBy the Superposition theorem,

Y (t) = α~y1(t) + β ~y2(t)

i.e. Y (t) = αe−t

−1

1

+ βe9t

3

7


Example 2:

Solve the following IVP:

ddt~y =

2 3

7 6

~yy(0) =

1

9

We know that

Y (t) = αe−t

−1

1

+ βe9t

3

7

is a solution for all α, β.

→ Find α, β to match initial condition.

We want1

9

= αe−(0)

−1

1

+ βe9(0)

3

7

1

9

= α

−1

1

+ β

3

7

1

9

=

−α+ 3β

α+ 7β

We then need to solve:

1 = −α+ 3β

9 = α+ 7β

Adding the two equations give: 10 = 10β

β = 1

1 = −α+ 3(1)

22 GIANG B KHUONG

−2 = −α

α = 2

∴ Y (t) = 2e−t

−1

1

+ e9t

3

7

solves the IVP.

10. Eigenvalues, eigenvectors, and eigen solutions

(1) Transforming the linearized problems into eigenvalue problems

From last chapter, we had a linearized, first order ODE:

d

dt~y =

a b

c d

~yWe look for solutions of the form:

~y(t) = eλt

xy

where λ, x, y are real numbers.

Assuming that ~y(t) solves the ODE, we have:

λeλt

xy

=

a b

c d

eλtxy

Since eλt 6= 0, we can divide by eλt. We then want λ

xy

such that:

a b

c d

xy

= λ

xy

We now have an eigenvalue problem. Find constant λ and vector ~V such that:

a b

c d

~V = λ~V


λ is called the eigenvalue.

~V is called the eigenvector.

Given the matrix, we can find the eigenvalue λ and the associated eigenvector

~V . Then we can can form the solution eλt~V .

(2) How to solve eigenvalue problem

Take the example from the previous section:

a b

c d

xy

= λ

xy

a b

c d

xy

− λxy

=

0

0

Observe that the matrix

λ 0

0 λ

has the same effect as the scalar quantity λ.

∴

a b

c d

xy

−λ 0

0 λ

xy

=

0

0

a b

c d

−λ 0

0 λ

xy

=

0

0

In general, we have:

a− λ b

c d− λ

xy

=

0

0

We do not want x = y = 0, hence we are looking for non-trivial solutions for

this system. For non-trivial solutions to exist, the determinant

∣∣∣∣∣∣a− λ b

c d− λ

∣∣∣∣∣∣ has

to be equal to zero.

24 GIANG B KHUONG

i.e.

(a− λ)(d− λ)− bc = 0

We solve the quadratic equation for λ1 and λ2. To find the associated eigenvector,

plug the found λ back into the eigenvalue problem.

a b

c d

xy

= λ1

xy

ax+ by = λ1x

cx+ dy = λ1y

Then choose a pair of x, y that satisfies the pair of equations. That is the

associated eigenvector ~V1 for λ1. Do the same for λ2

We, then, have two ”eigen” solutions eλ1t ~V1 and eλ2t ~V2.

By the principle of superposition, we have the general solution:

~Y (t) = αeλ1t ~V1 = βeλ2t ~V2

Example: Find the general solution for the following ODE.

d

dt~y =

2 1

1 2

~y1. Find eigenvalues λ1, λ22 1

1 2

xy

= λ

xy

2 1

1 2

xy

−λ 0

0 λ

xy

=

0

0

DIFFERENTIAL EQUATIONS 252− λ 1

1 2− λ

xy

=

0

0

∣∣∣∣∣∣2− λ 1

1 2− λ

∣∣∣∣∣∣ = 0

∴ (2− λ)(2− λ)− (1)(1) = 0

(2− λ)2 − 12 = 0

(2− λ− 1)(2− λ+ 1) = 0

λ1 = 1 and λ2 = 3

2. Find the associated eigenvectors ~V1, ~V2

λ = 3 2 1

1 2

xy

= 3

xy

2x+ y = 3x

x+ 2y = 3yy = x

x = y

∴ We choose the eigenvector ~V1 =

1

1

.

λ = 1 2 1

1 2

xy

= 1

xy

2x+ y = x

x+ 2y = y

26 GIANG B KHUONGy = −x

x = −y

∴ We choose the eigenvector ~V1 =

1

−1

.

We, then, have two ”eigen” solutions e3t

1

1

and et

1

−1

.

By the principle of superposition, we have the general solution:

~Y (t) = αe3t

1

1

+ βet

1

−1

Figure 7. Phase Plane Plot

For IVPs, solve for specific α, β.


(3) Real eigenvalues

There are three cases for real eigenvalues λ:

• λ1 is positive and λ2 is negative

• both λ1 and λ2 are positive

• both λ1 and λ2 are negative

Case 1: One λ is positive and one λ is negative

d

dt~y =

3 8

3 5

~yλ1 = 9 and λ2 = −1

Associated eigenvectors are ~V1 =

4

3

and ~V2 =

−2

1

The general solution is ~Y (t) = αe9t

4

3

+ βe−t

−2

1

Figure 8. λ1 > 0, λ2 < 0

28 GIANG B KHUONG

The two straight lines are the eigenvectors. In this case, we have one going in

(towards (0, 0)), corresponding to λ < 0 and one going out (away from (0, 0)),

corresponding to λ > 0.

This configuration is called a saddle.

Case 2: Both λ1 and λ2 are positive

d

dt~y =

5 0

17 3

~yλ1 = 5 and λ2 = 3


2

17

and ~V2 =

0

1

The general solution is ~Y (t) = αe5t

2

17

+ βe3t

0

1

Figure 9. λ1 > 0, λ2 > 0


Here, we have both eigenvector going out (away from (0, 0)), corresponding to

both λ > 0.

This configuration is called a source.

Case 3: Both λ1 and λ2 are negative

d

dt~y =

−6 −2

4 0

~yλ1 = −4 and λ2 = −2


1

−1

and ~V2 =

−1

2

The general solution is ~Y (t) = αe−4t

1

−1

+ βe3t

−1

2

Figure 10. λ1 < 0, λ2 < 0

30 GIANG B KHUONG

In this case, we have both eigenvector going in (towards (0, 0)), corresponding

to both λ < 0.

This configuration is called a sink.

(4) Complex/unreal eigenvalues

(a) Tools to understand complex solutions

(i) What is a complex number?

A complex number is a number that can be put in the form a + bi,

where a and b are real numbers and i is called the imaginary unit, where

i2 = −1.

Example:

λ2 + 4 = 0

λ2 = −4

λ = ±2i

If we include i in our world, all quadratic can be solved:

aλ2 + bλ+ c = 0

λ =−b±√b2 − 4ac

2a

If b2 − 4ac ≥ 0, then we have our usual real solutions.

If b2 − 4ac ≤ 0, then we have complex solutions.

All usual algebra works for complex numbers.


(ii) Euler’s Formula

We can express eit as:

eit = cos(t) + i sin(t)

This is from Taylor’s series expansion.

eit = 1 +it

1!+it2

2!+it3

3!+it4

4!...

= 1 + it− t2

2!+it3

3!+t4

4!+it5

5!...

= (1− t2

2!+t4

4!− ...) + i(t+

t3

3!+t5

5!...)

= cos t+ i sin t

(b) Solve for complex eigenvalues

d

dt~y =

1 1

−5 3

~y(1− λ)(3− λ) + 5 = 0

λ1 = 2 + 2i and λ2 = 2− 2i

λ1 = 2 + 2i 1 1

−5 3

xy

= (2 + 2i)

xy

x+ y = (2 + 2i)x

−5x+ 3y = (2 + 2i)y

y = (1 + 2i)x

∴ ~V1 =

1

1 + 2i

32 GIANG B KHUONG

The first eigen solution is:

~Y1(t) = e(2+2i)t

1

1 + 2i

= e2te2it

1

1 + 2i

= e2t(cos(2t) + i sin(2t))

1

1

+ i

0

2

=

e2t cos(2t)

1

1

− e2t sin(2t)

0

2

+ i

e2t cos(2t)

0

2

+ e2t sin(2t)

1

1

λ1 = 2− 2i 1 1

−5 3

xy

= (2− 2i)

xy

x+ y = (2− 2i)x

−5x+ 3y = (2− 2i)y

y = (1− 2i)x

∴ ~V2 =

1

1− 2i


The second eigen solution is:

~Y2(t) = e(2−2i)t

1

1− 2i

= e2te−2it

1

1− 2i

= e2t(cos(−2t) + i sin(−2t))

1

1

− i0

2

=

e2t cos(2t)

1

1

− e2t sin(2t)

0

2

− ie2t cos(2t)

0

2

+ e2t sin(2t)

1

1

Observation:

• eigenvalues come in pairs λ = a± bi

• eigenvectors come in pairs ~V = ~Vreal + ~Vimaginary

• eigen solutions come in pairs

General solution:

~Y (t) = α ~Y1(t) + β ~Y2(t)

= (α+ β)

e2t cos(2t)

1

1

− e2t sin(2t)

0

2

+ i(α− β)

e2t cos(2t)

0

2

+ e2t sin(2t)

1

1

Set α = β = 12 , we have REAL ~Y (t) =

e2t cos(2t)

1

1

− e2t sin(2t)

0

2

Set α = 1

2i , β = −12i , we have REAL ~Y (t) =

e2t cos(2t)

0

2

+ e2t sin(2t)

1

1

34 GIANG B KHUONG

We can then generate another general solution:

~Y (t) = A

e2t cos(2t)

1

1

− e2t sin(2t)

0

2

+B

e2t cos(2t)

0

2

+ e2t sin(2t)

1

1

⇒ If eigenvalues λ are complex, then both the real and imaginary part of ~Y1(t)

or ~Y2(t) are REAL solutions which can be used to form REAL general solution.

(5) Different types of complex eigenvalues

There are three cases for complex eigenvalues λ = a± bi:

• Real part a of λ > 0

• Real part a of λ < 0

• Real part a of λ = 0

Case 1: Real part a of λ > 0

Use the example from (4), we have this following graph:

Figure 11. a of λ = a+ bi > 0


This configuration is called spiral source since all solutions are moving away

from the center (0, 0).

Case 2: Real part a of λ < 0

d

dt~y =

−1 −1

−1 0

~yλ = −1

2± i√

3

2


This configuration is called spiral sink since all solutions are moving towards

the center (0, 0).

36 GIANG B KHUONG

Case 3: Real part a of λ = 0

d

dt~y =

1 2

−5 −1

~yλ = ±3i


11. Nullclines

Definition:

Nullclines offer a way to analyze ODEs behavior in between equilibrium points. This

gives a more general behavior of the ODEs than qualitative analysis like linearization and

eigenstuffs analysis.

For any system dxdt = f(x, y)

dydt = g(x, y)


the x-nullcline is the set of points (x, y) where f(x, y) is zero, i.e. the level curve where

f(x, y) is zero. Similarly, the y-nullcline is the set of points where g(x, y) is zero.

Along the x-nullcline,the x-component of the vector field is zero, hence, the vector field

is vertical, i.e. either pointing straight up or straight down. Similarly, the vector field along

the y-nulcline will be pointing either right or left, since its y-component is zero.

The intersections of these nullclines are the equilibria. because both f(x, y) and g(x, y)

are zero at equilibria.

Example:

Consider the following system:

dxdt = 2x(1− x

2 )− xy)

dydt = −y + xy

Using the normal linearization and eigenstuffs analysis yield the following phase plane:

Figure 14. Assemble phase plane around equilibria

38 GIANG B KHUONG

The nullclines phase plane of this system of ODEs is the skeleton of the behavior, as

shown below:

Figure 15. Nullclines phase plane

Assemble both phase plane results in the following picture:

Figure 16. Final phase plane


12. Oscillations

According to Newton’s second law of motion, we have the equation:

F = ma = md2y

dt2

where m =mass (constant) and y = position as a function of time.

Hooke’s law also offered another force model for describe mass on springs, where:

Together, we have:

F = md2y

dt2= −ky

Rearranging the equation, we have:

d2y

dt2+k

my = 0

This is called the Simple Harmonic Oscillator (SHO).

The general form of SHO is the second order linear ODE:

md2y

dt2+ b

dy

dt+ ky = f(t)

where md2ydt2

is the factor from Newton’s law, bdydt is the damping factor, ky is the oscillating

factor, and f(t) is the external force factor; or the forcing/driving factor.

When f(t) = 0, the system is called homogeneous; otherwise, it is called inhomoge-

neous.

To analyze the second order SHO, we need to convert the second order equation to

a system of first order ODEs. Then, we can use our normal qualitative methods like

linearization and eigenstuffs analysis.

To convert this second order SHO to a first order system of ODEs, we need to introduce

a new variable v = dydt . This make d2y

dt2= dv

dt .

40 GIANG B KHUONG

We obtain the system:

dydt = v

mdvdt + bv + ky = f(t)

For homogeneous system, we guess the solution to be in the form of yh(t) = eλt. We,

then pluck this guess into the equations to check and figure out λ. We work out the general

homogeneous solution using normal method of eigenvalues λ and eigenvectors.

For inhomogeneous system, the solution is made up of 2 parts: the homogeneous and

the particular solution, i.e. y(t) = yh(t) + yp(t). This works because of the Superposition

theorem discussed in chapter 9.

We find the homogeneous solution by setting the equation equals to zero, and use the

same method as solving homogeneous equation. For example, we have inhomogeneous

equation:

md2y

dt2+ b

dy

dt+ ky = f(t)

We find the homogeneous part of the solution by setting

md2y

dt2+ b

dy

dt+ ky = 0

and using the guessing method as discussed above.

For the particular part of the solution, we guess the solution to be in the form of the

forcing factor, i.e. in the form of whatever f(t) is.

For example, if f(t) = 7t, then the particular solution is in the form yp(t) = A+ Bt; if

f(t) = e−10t, then yp(t) = Aert; if f(t) = cos 3t, then yp(t) = A cos 3t+B sin 3t, etc.


13. Conserved and Monotone Quantities

Conserved Quantities

A quantity is conserved along a solution curve in the phase plane if ddt(quantity) = 0

An example of conserved quantity is is energy, as studied in Physics, which consists of

kinetic energy, which measures how large velocity is, and potentl energy, which measures

the energy inherent in the position of the system.

E =1

2mv2 + V (y)

where 12mv

2 is kinetic energy and V (y) is potential energy.

d

dt[E] =

d

dt

(1

2mv2 + V (y)

)= mv

dv

dt+ V ′(y)

dy

dt(dy

dt= v and

dy

dt= − k

my)

= v

(m−kmy + V ′(y)

)= v

(−ky + V ′(y)

)The quantity E is conserved if d

dt [E] = 0, which happens when V (y) = ky. We then

choose V (y) = 12ky

2 so that E is conserved.

Thus, we have the conserved quantity

E =1

2mv2 +

1

2ky2

However, there are many different conserved quantity, depends on the system of SHO.

We can draw the energy plane and phase plane of the SHO with conserved quantity as

following:

42 GIANG B KHUONG

Figure 17. Energy plane: V (y)

Figure 18. Phase plane y − v

The energy and phase plane above can be describing a mass attached to a spring, sliding

on ice, i.e. no friction/damping: At whatever initial displacement y0 from equilibrium the

mass was set, it will oscillate between y0 and −y0 forever, because there is no damping and

no forcing acting on the system (the reason why the phase plane is a circle.)

Analyzing the energy diagram’s behavior:

We have the initial condition: at t = 0, (y0, v0)

E0 =1

2y0

+ 1

2v0

2


E is conserved, hence the solutions stay on whatever initial E = E0.

If V0 is greater than 0, then initially y increases until solution runs into the curve given

by V (y). If the system stop at this point then that point is an equilibrium.

If this point is not an equilibrium then dydt is negative, and y begins to decreases until it

hits the other side of the solution.

Look at Figure 17. Our solutions oscillates at height E0 between potential walls. If t0

was large then it will oscillate at a higher altitude decreases.

If the solution/quantity oscillates but either decreases or increases, then it is called a

monotone quantity.

14. Analytic methods

Separation of variables:

We may work out the general solution by re-arranging equation to separate the variables,

and integrate both sides.

Example:

dy

dt= et cos (y)

1

eydy

dt= cos (y)

e−ydy

dt= cos (y)∫ T

te−y

dy

dtdt =

∫ T

tcos (y)dx (y0 = y(t0))

−e−t∣∣∣y(T )y(t0)

= sin (t)∣∣∣Tt0

e−y(T ) = − sin (T ) + sin (t0) + e−y0

y(T ) = − ln(− sin (T ) + sin (t0) + e−y0

)

44 GIANG B KHUONG

We can set sin (t0) + e−y0 = C, since this is just some constant that are determined by

initial condition (t0, y0).

Our final solution is

y(t) = − ln [C − sin (t)]

Usage on Logistic Model:

dPdt = rP

(1− P

K

)P (0) = P0

We now see that dPdt = rP

(1− P

K

)is separable.

dP

dt= rP

(1− P

K

)1

P(1− P

K

) dPdt

= r

∫ T

0

1

P(1− P

K

) dPdtdt =

∫ T

0rdt

∫ PT

P0

1

P(1− P

K

) dPdt

= rT

We now need to find an anti-derivative for 1P(1− P

K )by partial fractions, as we have known

from Calc II.

We have now


1

P(1− P

K

) =1

P+

1K

1− PK

=1

P+

1

K − P

Thus: ∫ PT

P0

1

P+

1

K − PdP = rT

ln∣∣∣ P

K − P

∣∣∣∣∣∣∣∣P (T )

P0

= rT

ln∣∣∣ P (T )

K − P (T )

∣∣∣− ln∣∣∣ P0

K − P0

∣∣∣ = rT

ln∣∣∣ P (T )K−P (T )

P0K−P0

∣∣∣ = rT

P (T )K−P (T )

P0K−P0

= erT

P (T )

K − P (T )=

(P0

K − P0

)erT

P (T ) =K(

P0K−P0

)erT

1 +(

P0K−P0

)erT

P (T ) =K(

K−P0P0

)e−rT + 1

With the method of separation of variables, we can actually work out the solution to

the system of ODEs directly.

Differential Equations coursework

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