mathsoc.unsw.edu.au · Part I: Linear ODEs Dynamical Systems Table of Contents 1 Part I: Linear ODEs Linear Diﬀerential Operators Wronskian and Linear Independence Methods Power

Part I: Linear ODEs Dynamical Systems

UNSW Mathematics Society presents

MATH2121/2221 Revision Seminar

(Higher) Theory & Applications ofDifferential Equations

Seminar I / II

Written by Abdellah Islam, Presented by Gerald HuangT2, 2020


Table of Contents

1 Part I: Linear ODEsLinear Differential OperatorsWronskian and Linear IndependenceMethodsPower SeriesBessel EquationLagrange Equation (2221 only)

2 Dynamical SystemsLipschitz (2221 only)Linear Dynamical SystemsMatricesEquilibrium PointsFirst Integrals (2221 only)

2 / 87MATH2121/2221 Revision SeminarWritten by Abdellah Islam, Presented by Gerald Huang


Part I: Linear ODEs



Differential Operators

Definition 1: Linear differential operatorsDefine the linear differential operator L of order m to be

Lu(x) =m∑

j=0aj(x) · Dju(x)

= amDmu(x) + am−1Dm−1u(x) + · · ·+ a0u,

where Dju = d juduj and D0u = u.

Definition 2: Singular ODEsAn ODE is said to be singular with respect to [a, b] if the leadingcoefficient vanishes for any x ∈ [a, b].



Differential Operators

Definition 1: Linear differential operatorsDefine the linear differential operator L of order m to be

Lu(x) =m∑

j=0aj(x) · Dju(x)

= amDmu(x) + am−1Dm−1u(x) + · · ·+ a0u,

where Dju = d juduj and D0u = u.

Definition 2: Singular ODEsAn ODE is said to be singular with respect to [a, b] if the leadingcoefficient vanishes for any x ∈ [a, b].



Homogeneous and inhomogeneous ODEsDefinition 3: Homogeneous ODEsAn ODE is said to be homogeneous if the right hand side is 0.That is, we have a differential equation of the form

Lu = 0.

Example: u′′ + u′ + u = 0.

Definition 4: Inhomogeneous ODEsAn ODE is said to be inhomogeneous if the right hand side is not0. Then we can write the differential equation as

Lu = f .

Example: u′′ + u′ + u = cos(x).




Lu = 0.

Example: u′′ + u′ + u = 0.


Lu = f .





Lu = 0.

Example: u′′ + u′ + u = 0.


Lu = f .





Lu = 0.

Example: u′′ + u′ + u = 0.


Lu = f .




Initial-valued problems (IVP)

Definition 5: Initial-valued problemsConsider an m-th order differential equation

Lu = f , on [a, b] (1)

along with the values

u(a) = v0, u′(a) = v1, ..., u(m−1)(a) = vm−1. (2)

The problem (1) with (2) is called an initial-valued problem.

Example: u′ + u = x , u(0) = 0.

Solution: u(x) = x − 1 + e−x .



Initial-valued problems (IVP)

Definition 5: Initial-valued problemsConsider an m-th order differential equation

Lu = f , on [a, b] (1)

along with the values

u(a) = v0, u′(a) = v1, ..., u(m−1)(a) = vm−1. (2)

The problem (1) with (2) is called an initial-valued problem.

Example: u′ + u = x , u(0) = 0.

Solution: u(x) = x − 1 + e−x .



Unique solutionsTheorem 1: Unique solution for IVPIf f is continuous on [a, b] and the ODE Lu = f is non-singular on[a, b], then the IVP (1) and (2) has a unique solution.

Theorem 2: Homogeneous equation solutionIf L is a linear m-th order differential operator and non-singular on[a, b], then the set of all solutions to the homogenous equationLu = 0 on [a, b] forms a vector space of dimension m.

What does this mean?: The solution space has a basis ofdimension m, with elements u1, ..., um. And so every solution to thehomogeneous equation can be written as a linear combination ofthis basis:

u(x) = c1u1(x) + ...+ cmum(x), ∀x ∈ [a, b].This is called the general solution.















Unique solutions

We can do the same for an inhomogeneous equation Lu = f byfixing a particular solution uP . Then for any solution u,L(u − up) = f − f = 0 and so we can write u − up as a linearcombination of the homogeneous equation basis:

u(x)− up(x) = c1u1(x) + ...+ cmum(x), ∀x ∈ [a, b].

Rearranging, we have the general solution for an inhomogeneousdifferential equation:

u(x) = uP(x) + c1u1(x) + ...+ cmum(x), ∀x ∈ [a, b].



Solving an inhomogeneous DE

Example 1Solve the second order ODE

2u′′ + u′ − u = 10 sin(x).

Characteristic equation 2t2 + t − 1 = 0. Solutions t = −1, 12 so a

basis for the homogeneous solution space is {e−x , ex/2}. Particularsolution guess uP(x) = A cos(x) + B sin(x). Substituting givesA = −1 and B = −3, so uP(x) = − cos(x)− 3 sin(x). We canwrite the general solution as

u(x) = c1e−x + c2ex/2 − cos(x)− 3 sin(x).





2u′′ + u′ − u = 10 sin(x).

Characteristic equation 2t2 + t − 1 = 0.

Solutions t = −1, 12 so a







2u′′ + u′ − u = 10 sin(x).


basis for the homogeneous solution space is {e−x , ex/2}.

Particularsolution guess uP(x) = A cos(x) + B sin(x). Substituting givesA = −1 and B = −3, so uP(x) = − cos(x)− 3 sin(x). We canwrite the general solution as






2u′′ + u′ − u = 10 sin(x).


basis for the homogeneous solution space is {e−x , ex/2}. Particularsolution guess uP(x) = A cos(x) + B sin(x).

Substituting givesA = −1 and B = −3, so uP(x) = − cos(x)− 3 sin(x). We canwrite the general solution as






2u′′ + u′ − u = 10 sin(x).


basis for the homogeneous solution space is {e−x , ex/2}. Particularsolution guess uP(x) = A cos(x) + B sin(x). Substituting givesA = −1 and B = −3, so uP(x) = − cos(x)− 3 sin(x).

We canwrite the general solution as






2u′′ + u′ − u = 10 sin(x).






Linear Independence

Definition 6: Linear independenceLet u1(x), ..., um(x) be functions on some interval I ⊂ R. If thereexists non-zero constants a1, ..., am such that

a1u1(x) + ...+ amum(x) = 0, ∀x ∈ I,

then we say that u1, ..., um are linearly dependent.

If the above equation only holds true for all constants zero, thenwe say that u1, ..., um are linearly independent.



Linear Independence

Definition 6: Linear independenceLet u1(x), ..., um(x) be functions on some interval I ⊂ R. If thereexists non-zero constants a1, ..., am such that

a1u1(x) + ...+ amum(x) = 0, ∀x ∈ I,

then we say that u1, ..., um are linearly dependent.If the above equation only holds true for all constants zero, thenwe say that u1, ..., um are linearly independent.



Wronskian

Definition 7: WronskianThe Wronskian of the functions u1, ..., um is the m ×mdeterminant

W (x) = W (x ; u1, ..., um) = det(Di−1uj).

For example, a 3× 3 Wronskian is

W (x) = W (x ; u1, ..., um) = det

u1 u2 u3u′1 u′2 u′3u′′1 u′′2 u′′3

.



Wronskian

Definition 7: WronskianThe Wronskian of the functions u1, ..., um is the m ×mdeterminant

W (x) = W (x ; u1, ..., um) = det(Di−1uj).

For example, a 3× 3 Wronskian is

W (x) = W (x ; u1, ..., um) = det

u1 u2 u3u′1 u′2 u′3u′′1 u′′2 u′′3

.



Lemmas

Lemma 1If u1, ..., um are linearly dependent over an interval I ⊂ R thenW (x ; u1, ..., um) = 0 for all x ∈ I.

Lemma 2If u1, ..., um are solutions to the ODE

am(x)u(m)(x) + am−1(x)u(m−1)(x) + ...+ a0(x)u(x) = 0

on an interval I ⊂ R, then the Wronskian satisfies

am(x)W ′(x) + am−1(x)W (x) = 0 ∀x ∈ I.



Lemmas

Lemma 1If u1, ..., um are linearly dependent over an interval I ⊂ R thenW (x ; u1, ..., um) = 0 for all x ∈ I.

Lemma 2If u1, ..., um are solutions to the ODE

am(x)u(m)(x) + am−1(x)u(m−1)(x) + ...+ a0(x)u(x) = 0

on an interval I ⊂ R, then the Wronskian satisfies

am(x)W ′(x) + am−1(x)W (x) = 0 ∀x ∈ I.



Wronskian

Example 2: MATH2221 2014 T2 2.iii).bGiven the functions u1, u2, prove that if they are solutions to asecond-order, homogeneous linear differential equation

a2(x)u′′ + a1(x)u′ + a0(x)u = 0,

then the Wronskian W satisfies

a2(x)W ′ + a1(x)W = 0.

W = u1u′2 − u′1u2, W ′ = u1u′′2 − u′′1 u2.

a2W ′ + a1W = a2(u1u′′2 − u′′1 u2) + a1(u1u′2 − u′1u2)= u1(a2u′′2 + a1u′2)− u2(a2u′′1 + a1u′1).



Wronskian


a2(x)u′′ + a1(x)u′ + a0(x)u = 0,


a2(x)W ′ + a1(x)W = 0.

W = u1u′2 − u′1u2, W ′ = u1u′′2 − u′′1 u2.




Wronskian


a2(x)u′′ + a1(x)u′ + a0(x)u = 0,


a2(x)W ′ + a1(x)W = 0.

W = u1u′2 − u′1u2, W ′ = u1u′′2 − u′′1 u2.




Wronskian

Add and subtract a0u1u2:

a2W ′ + a1W = u1(a2u′′2 + a1u′2 + a0u2)− u2(a2u′′1 + a1u′1 + a0u1).

Since u1 and u2 are solutions to the ODE, the RHS is 0.



Wronskian

Add and subtract a0u1u2:

a2W ′ + a1W = u1(a2u′′2 + a1u′2 + a0u2)− u2(a2u′′1 + a1u′1 + a0u1).

Since u1 and u2 are solutions to the ODE, the RHS is 0.



Linear Independence of solutions

Theorem 3: Linear independenceLet u1, ..., um be solutions to the non-singular, linear, homogeneousm-th order ODE Lu = 0 on the interval [a, b]. Then either

W (x) = 0 and the m solutions are linearly dependent,

or

W (x) 6= 0 and the m solutions are linearly independent.



Polynomial solution guess

Theorem 4Let L = p(D) be a linear differential operator of order m withconstant coefficients. Assume that p(0) 6= 0. Then for any integerr ≥ 0, there exists a unique polynomial uP of degree r such thatLuP = x r .

This means that if our linear ODE has a polynomial on the RHS,we should guess a polynomial of the same degree for our particularsolution.



Polynomial solution guess

Theorem 4Let L = p(D) be a linear differential operator of order m withconstant coefficients. Assume that p(0) 6= 0. Then for any integerr ≥ 0, there exists a unique polynomial uP of degree r such thatLuP = x r .

This means that if our linear ODE has a polynomial on the RHS,we should guess a polynomial of the same degree for our particularsolution.



Exponential solution guess

Theorem 5Let L = p(D) and µ ∈ C. If p(µ) 6= 0, then the function

uP(x) = eµx

p(µ)

satisfies LuP = eµx .

This means that we should guess a multiple of eµx when the RHSof a linear ODE is eµx if it is not already a solution of thehomeogeneous solution.



Exponential solution guess

Theorem 5Let L = p(D) and µ ∈ C. If p(µ) 6= 0, then the function

uP(x) = eµx

p(µ)

satisfies LuP = eµx .

This means that we should guess a multiple of eµx when the RHSof a linear ODE is eµx if it is not already a solution of thehomeogeneous solution.



Polynomial + exponential

Theorem 6Let L = p(D) and assume µ ∈ C. If p(µ) 6= 0 then for any integersr ≥ 0, there exists a unique polynomial v of degree r such that

uP(x) = v(x)eµx

satisfies Lup = x r eµx .

So if the RHS of the inhomogeneous linear ODE is a polynomialtimes an exponential, and the exponential isnt in the homogeneoussolution, then the guess for the particular solution should be apolynomial times exponential.



Polynomial + exponential

Theorem 6Let L = p(D) and assume µ ∈ C. If p(µ) 6= 0 then for any integersr ≥ 0, there exists a unique polynomial v of degree r such that

uP(x) = v(x)eµx

satisfies Lup = x r eµx .

So if the RHS of the inhomogeneous linear ODE is a polynomialtimes an exponential, and the exponential isnt in the homogeneoussolution, then the guess for the particular solution should be apolynomial times exponential.



General solutions

Example 3: MATH2221 2014 T2 2.ii)

Let p(z) = (z − 1)(z + 2)2(z2 + 1) and D = ddx .

Write down the general solution uH of the 5-th order, linearhomogeneous ODE p(D)u = 0.

Write down the form of a particular solution uP to theinhomogeneous ODE

p(D)u = e−2x + x2 + cos(x).

The zeros of p(z) are 1, −2 (multiplicity 2), i and −i . So thehomogeneous ODE has general solution

uH(x) = c1ex + c2e−2x + c3xe−2x + c4 cos(x) + c5 sin(x).



General solutions

Now consider the inhomogeneous ODE. The solution form willneed to contain a x2e−2x term (since all lower powers are in thehomogeneous solution), as well a second order polynomial, as wellas x cos(x) and x sin(x) terms (since cos(x) and sin(x) are in thehomogeneous solution).

Putting these together,

uP(x) = a1x2e−2x + (a2x2 + a3x + a4) + x(a5 cos(x) + a6 sin(x)).



General solutions

Now consider the inhomogeneous ODE. The solution form willneed to contain a x2e−2x term (since all lower powers are in thehomogeneous solution), as well a second order polynomial, as wellas x cos(x) and x sin(x) terms (since cos(x) and sin(x) are in thehomogeneous solution). Putting these together,

uP(x) = a1x2e−2x + (a2x2 + a3x + a4) + x(a5 cos(x) + a6 sin(x)).



Reduction of order

Theorem 7: Reduction of orderIf we know a solution u1(x) 6= 0 to the ODE

u′′ + p(x)u′ + q(x)u = 0

then we can find a second solution

u2(x) = u1(x)∫ 1

u1(x)2 exp (∫

p(x)dx)dx .



Reduction of orderExample 4Find the general solution to

x2y ′′ + 2xy ′ − 2y = 0

given that y1(x) = x is a solution.

We need to rewrite the ODE in the form from Theorem 3:

y ′′ + 2x y ′ − 2

x2 y = 0.

Then exp (∫

p(x)dx) = exp(∫ 2

x dx)

= exp (2 ln(x)) = x2.Substituting into the reduction of order formula,

y2(x) = x∫ 1

x4 dx = − 13x2 .




x2y ′′ + 2xy ′ − 2y = 0



y ′′ + 2x y ′ − 2

x2 y = 0.

Then exp (∫

p(x)dx) = exp(∫ 2

x dx)

= exp (2 ln(x)) = x2.

Substituting into the reduction of order formula,

y2(x) = x∫ 1

x4 dx = − 13x2 .




x2y ′′ + 2xy ′ − 2y = 0



y ′′ + 2x y ′ − 2

x2 y = 0.

Then exp (∫

p(x)dx) = exp(∫ 2

x dx)

= exp (2 ln(x)) = x2.Substituting into the reduction of order formula,

y2(x) = x∫ 1

x4 dx = − 13x2 .



Annihilator method (2221 only)

This method gives us a way to find particular solutions to aninhomogeneous differential equation. Start with an ODE

Lu = f (x).

We find a differential operator A(D) which annihilates f (x),meaning A(D)f (x) = 0. For example D3 annihilates x2. We applythis differential operator to both sides of our ODE:

A(D)Lu = 0.

A solution to this homogeneous differential equation will be aparticular solution to the original differential equation.





Lu = f (x).

We find a differential operator A(D) which annihilates f (x),meaning A(D)f (x) = 0. For example D3 annihilates x2.

We applythis differential operator to both sides of our ODE:

A(D)Lu = 0.






Lu = f (x).


A(D)Lu = 0.






Lu = f (x).


A(D)Lu = 0.




Annihilator method (2221 only)Example 5Find a particular solution to the ODE

y ′′ − 2y ′ + y = ex + sin(x).

The LHS differential operator is L(D) = (D − 1)2.

The function ex

is annihilated by (D − 1). The function sin(x) is annihilated by(D2 + 1). So the annihilator is A(D) = (D − 1)(D2 + 1). Applythe annihilator to both sides:

A(D)L(D)y(x) = (D − 1)3(D2 + 1)y(x) = 0.

Solutions to the characteristic equation here is 1 (multiplicity 3), iand −i . The particular solution is of the form

yP(x) = c1ex + c2xex + c3x2ex + c4 sin(x) + c5 cos(x).




y ′′ − 2y ′ + y = ex + sin(x).

The LHS differential operator is L(D) = (D − 1)2. The function ex

is annihilated by (D − 1).

The function sin(x) is annihilated by(D2 + 1). So the annihilator is A(D) = (D − 1)(D2 + 1). Applythe annihilator to both sides:

A(D)L(D)y(x) = (D − 1)3(D2 + 1)y(x) = 0.






y ′′ − 2y ′ + y = ex + sin(x).


is annihilated by (D − 1). The function sin(x) is annihilated by(D2 + 1).

So the annihilator is A(D) = (D − 1)(D2 + 1). Applythe annihilator to both sides:

A(D)L(D)y(x) = (D − 1)3(D2 + 1)y(x) = 0.






y ′′ − 2y ′ + y = ex + sin(x).


is annihilated by (D − 1). The function sin(x) is annihilated by(D2 + 1). So the annihilator is A(D) = (D − 1)(D2 + 1).

Applythe annihilator to both sides:

A(D)L(D)y(x) = (D − 1)3(D2 + 1)y(x) = 0.






y ′′ − 2y ′ + y = ex + sin(x).



A(D)L(D)y(x) = (D − 1)3(D2 + 1)y(x) = 0.

Solutions to the characteristic equation here is 1 (multiplicity 3), iand −i .

The particular solution is of the form





y ′′ − 2y ′ + y = ex + sin(x).



A(D)L(D)y(x) = (D − 1)3(D2 + 1)y(x) = 0.






The first two terms in this particular solution are contained in thehomogeneous solution of the original ODE, so we discard them:

yP(x) = c3x2ex + c4 sin(x) + c5 cos(x).

Substitute into the original ODE, coefficients are c3 = 12 , c4 = 0

and c5 = 12 . So our particular solution is

yP(x) = 12x2ex + 1

2 cos(x).







and c5 = 12 .

So our particular solution is

yP(x) = 12x2ex + 1

2 cos(x).







and c5 = 12 . So our particular solution is

yP(x) = 12x2ex + 1

2 cos(x).



Variation of parametersIf we have a linear, 2nd order, inhomogeneous ODE with leadingcoefficient 1,

Lu = u′′ + p(x)u′ + q(x)u = f (x).

Let u1, u2 be a basis for the homogeneous solution space, and letW (x) be the Wronskian W (x ; u1, u2).Then a particular solution tothe inhomogeneous equation is

u(x) = v1(x)u1(x) + v2(x)u2(x),

where

v ′1(x) = −u2(x)f (x)W (x) , and v ′2(x) = u1(x)f (x)

W (x) .




Lu = u′′ + p(x)u′ + q(x)u = f (x).

Let u1, u2 be a basis for the homogeneous solution space, and letW (x) be the Wronskian W (x ; u1, u2).

Then a particular solution tothe inhomogeneous equation is

u(x) = v1(x)u1(x) + v2(x)u2(x),

where


W (x) .




Lu = u′′ + p(x)u′ + q(x)u = f (x).

Let u1, u2 be a basis for the homogeneous solution space, and letW (x) be the Wronskian W (x ; u1, u2).Then a particular solution tothe inhomogeneous equation is

u(x) = v1(x)u1(x) + v2(x)u2(x),

where


W (x) .



Variation of parametersExample 6: MATH2121 2018 T2 1.i)Use the variation of parameters method to solve

y ′′ − 2y ′ + y = ex cos(x).

Homogeneous solution has characteristic equation t2 − 2t + 1 = 0so u1(x) = ex and u2(x) = xex .

The Wronskian is

W (x) = det(

ex xex

ex (x + 1)ex

)= e2x .

Thenv ′1(x) = −xexex cos(x)

e2x = −x cos(x),

andv ′2(x) = exex cos(x)

e2x = cos(x).




y ′′ − 2y ′ + y = ex cos(x).

Homogeneous solution has characteristic equation t2 − 2t + 1 = 0so u1(x) = ex and u2(x) = xex .The Wronskian is

W (x) = det(

ex xex

ex (x + 1)ex

)= e2x .


e2x = −x cos(x),


e2x = cos(x).




y ′′ − 2y ′ + y = ex cos(x).

Homogeneous solution has characteristic equation t2 − 2t + 1 = 0so u1(x) = ex and u2(x) = xex .The Wronskian is

W (x) = det(

ex xex

ex (x + 1)ex

)= e2x .


e2x = −x cos(x),


e2x = cos(x).



Variation of parameters

Integrating both v1 and v2 we have

v1(x) = −x sin(x)− cos(x), and v2(x) = sin(x).

Then a particular solution is

u(x) = (−x sin(x)− cos(x)) ex + (sin(x)) xex

= −ex cos(x).



Variation of parameters

Integrating both v1 and v2 we have

v1(x) = −x sin(x)− cos(x), and v2(x) = sin(x).

Then a particular solution is

u(x) = (−x sin(x)− cos(x)) ex + (sin(x)) xex

= −ex cos(x).



Power seriesconsider a general second-order, linear, homogeneous ODE

Lu = a2(x)u′′ + a1(x)u′ + a0(x)u = 0,

or equivalently

Lu = u′′ + p(x)u′ + q(x)u = 0

with p(x) = a1(x)a2(x) and q(x) = a0(x)

a2(x) .

Assume that a0, a1, a2 areanalytic at 0 (ie. they have local convergent power series), anda2(0) 6= 0. Then p and q are analytic at 0, so we can find powerseries expansions of both:

p(z) =∞∑

k=0pkzk , q(z) =

∞∑k=0

qkzk for |z | < ρ,

where ρ > 0.




Lu = a2(x)u′′ + a1(x)u′ + a0(x)u = 0,

or equivalently

Lu = u′′ + p(x)u′ + q(x)u = 0


a2(x) . Assume that a0, a1, a2 areanalytic at 0 (ie. they have local convergent power series), anda2(0) 6= 0. Then p and q are analytic at 0,

so we can find powerseries expansions of both:

p(z) =∞∑

k=0pkzk , q(z) =

∞∑k=0

qkzk for |z | < ρ,

where ρ > 0.




Lu = a2(x)u′′ + a1(x)u′ + a0(x)u = 0,

or equivalently

Lu = u′′ + p(x)u′ + q(x)u = 0


a2(x) . Assume that a0, a1, a2 areanalytic at 0 (ie. they have local convergent power series), anda2(0) 6= 0. Then p and q are analytic at 0, so we can find powerseries expansions of both:

p(z) =∞∑

k=0pkzk , q(z) =

∞∑k=0

qkzk for |z | < ρ,

where ρ > 0.29 / 87

MATH2121/2221 Revision SeminarWritten by Abdellah Islam, Presented by Gerald Huang


Power series

Theorem 8If coefficients p(z) and q(z) are analytic for |z | < ρ, then theformal power series for the solution u(z) constructed in theprevious slide, is also analytic for |z | < ρ.

This means that if we find where p(z) and q(z) are both analytic,the power series solution u(z) is also analytic in this region.



Power series

Theorem 8If coefficients p(z) and q(z) are analytic for |z | < ρ, then theformal power series for the solution u(z) constructed in theprevious slide, is also analytic for |z | < ρ.

This means that if we find where p(z) and q(z) are both analytic,the power series solution u(z) is also analytic in this region.



Power seriesExample 7: MATH2121 2018 T2 1.iii)We aim to construct a series solution to the ODE about theordinary point x0 = 0:

(1− x2)y ′′ − 2xy ′ + 20y = 0, y(0) = 1, y ′(0) = 0,

of the formy(x) =

∞∑n=0

Anxn.

Give the recurrence relation for the coefficients An.

Explain from the recurrence relation that one of the series willterminate yielding a polynomial solution, and the other does not.

Write down the polynomial solution.



Power seriesNote that

y ′(x) =∞∑

n=1nAnxn−1, y ′′(x) =

∞∑n=2

n(n − 1)Anxn−2.

Then we substitute into the ODE:Ly = y ′′ + (−x2y ′′ − 2xy ′ + 20y)

=∞∑

n=2n(n − 1)Anxn−2 +

∞∑n=2−n(n − 1)Anxn

−∞∑

n=12nAnxn +

∞∑n=0

20Anxn

=∞∑

n=2n(n − 1)Anxn−2 +

∞∑n=0−n(n − 1)Anxn

−∞∑

n=02nAnxn +

∞∑n=0

20Anxn.




y ′(x) =∞∑

n=1nAnxn−1, y ′′(x) =

∞∑n=2

n(n − 1)Anxn−2.

Then we substitute into the ODE:Ly = y ′′ + (−x2y ′′ − 2xy ′ + 20y)

=∞∑

n=2n(n − 1)Anxn−2 +

∞∑n=2−n(n − 1)Anxn

−∞∑

n=12nAnxn +

∞∑n=0

20Anxn

=∞∑

n=2n(n − 1)Anxn−2 +

∞∑n=0−n(n − 1)Anxn

−∞∑

n=02nAnxn +

∞∑n=0

20Anxn.



Power series

We combine the last three sums as follows.

Ly =∞∑

n=2n(n − 1)Anxn−2 +

∞∑n=0

(−n2 − n + 20)Anxn.

In the first sum, we change n to n + 2.

Ly =∞∑

n=0(n + 1)(n + 2)An+2xn +

∞∑n=0

(−n2 − n + 20)Anxn.

We can finally combine both sums.

Ly =∞∑

n=0[(n + 1)(n + 2)An+2 − (n + 5)(n − 4)An] xn.



Power series

Since Ly = 0, we need (n + 1)(n + 2)An+2− (n + 5)(n− 4)An = 0.Rearranging,

An+2 = (n + 5)(n − 4)(n + 1)(n + 2)An.

We have A0 = 1 and A1 = 0, so all odd terms are zero and theeven terms terminate Since A6 = 0. Hence the polynomial solutionis

y(x) = 1− 10x2 + 353 x4.



Power series


An+2 = (n + 5)(n − 4)(n + 1)(n + 2)An.

We have A0 = 1 and A1 = 0, so all odd terms are zero and theeven terms terminate Since A6 = 0.

Hence the polynomial solutionis

y(x) = 1− 10x2 + 353 x4.



Power series


An+2 = (n + 5)(n − 4)(n + 1)(n + 2)An.

We have A0 = 1 and A1 = 0, so all odd terms are zero and theeven terms terminate Since A6 = 0. Hence the polynomial solutionis

y(x) = 1− 10x2 + 353 x4.



Power series

Example 8: MATH2221 2015 T2 1.iii)Consider the ODE

(1 + z2)u′′ − zu′ − 3u = 0.

Find the recurrence relation satisfied by the coefficients Ak in anypower series solution:

u =∞∑

k=0Akzk .

Show that A5 = A7 = A9 = ... = 0.

Hence find the solution for which u(0) = 0, u′(0) = 6.




u′(z) =∞∑

k=1kAkzk−1, u′′(z) =

∞∑k=2

k(k − 1)Akzk−2.

Substitute into the ODE:Lu = u′′ + (z2u′′ − zu′ − 3u)

=∞∑

k=2k(k − 1)Akzk−2 +

∞∑k=2

k(k − 1)Akzk

−∞∑

k=1kAkzk −

∞∑k=0

Akzk

=∞∑

k=2k(k − 1)Akzk−2 +

∞∑k=0

k(k − 1)Akzk

−∞∑

k=0kAkzk −

∞∑k=0

Akzk .




u′(z) =∞∑

k=1kAkzk−1, u′′(z) =

∞∑k=2

k(k − 1)Akzk−2.

Substitute into the ODE:Lu = u′′ + (z2u′′ − zu′ − 3u)

=∞∑

k=2k(k − 1)Akzk−2 +

∞∑k=2

k(k − 1)Akzk

−∞∑

k=1kAkzk −

∞∑k=0

Akzk

=∞∑

k=2k(k − 1)Akzk−2 +

∞∑k=0

k(k − 1)Akzk

−∞∑

k=0kAkzk −

∞∑k=0

Akzk .



Power seriesCombine the last three sums.

Lu =∞∑

k=2k(k − 1)Akzk−2 +

∞∑k=0

(k2 − 2k − 3)Akzk .

In the first sum, change k to k + 2.

Lu =∞∑

k=0(k + 1)(k + 2)Ak+2zk +

∞∑k=0

(k2 − 2k − 3)Akzk .

Finally, we can combine the sums.

Lu =∞∑

k=0(k + 1) [(k + 2)Ak+2 + (k − 3)Ak ] zk .



Power series

Since Lu = 0, we need (k + 2)Ak+2 + (k − 3)Ak = 0. Rearranging,

Ak+2 = −k − 3k + 2Ak .

Since A5 = −3−33+2A3 = 0, then all odd terms past A5 are zero. Also

A1 = 6 and A3 = 4.The even terms start at A0 = 0 so all even terms past this arezero. Hence

u(z) = 6z + 4z3.



Power series


Ak+2 = −k − 3k + 2Ak .


A1 = 6 and A3 = 4.

The even terms start at A0 = 0 so all even terms past this arezero. Hence

u(z) = 6z + 4z3.



Power series


Ak+2 = −k − 3k + 2Ak .


A1 = 6 and A3 = 4.The even terms start at A0 = 0 so all even terms past this arezero. Hence

u(z) = 6z + 4z3.



Singular/Cauchy-Euler ODEs

For singular ODEs, we only need to check the case when theleading coefficient vanishes at the origin.

A second-order Cauchy-Euler ODE has the form

Lu = ax2u′′ + bxu′ + cu = f (x),

where a, b, c are constants with a 6= 0. This is singular at x = 0.Applying this differential operator L to x r ,

Lx r = [ar(r − 1) + br + c] x r ,

we can see that x r is a solution to the homogeneous equationLu = 0 iff

ar(r − 1) + br + c = 0.




For singular ODEs, we only need to check the case when theleading coefficient vanishes at the origin.A second-order Cauchy-Euler ODE has the form

Lu = ax2u′′ + bxu′ + cu = f (x),

where a, b, c are constants with a 6= 0. This is singular at x = 0.

Applying this differential operator L to x r ,

Lx r = [ar(r − 1) + br + c] x r ,


ar(r − 1) + br + c = 0.




For singular ODEs, we only need to check the case when theleading coefficient vanishes at the origin.A second-order Cauchy-Euler ODE has the form

Lu = ax2u′′ + bxu′ + cu = f (x),

where a, b, c are constants with a 6= 0. This is singular at x = 0.Applying this differential operator L to x r ,

Lx r = [ar(r − 1) + br + c] x r ,


ar(r − 1) + br + c = 0.




Lemma 3Suppose there are distinct solutions r1, r2 to the equationar(r − 1) + br + c = 0. That is, r1 6= r2. Then the general solutionof the homogeneous equation Lu = 0 is

u(x) = C1x r1 + C2x r2 , x > 0.

Lemma 4Suppose there is one solution r1 to ar(r − 1) + br + c = 0. Thenthe general solution to the homogeneous equation Lu = 0 is

C1x r1 + C2x r1 log(x), x > 0.




Lemma 3Suppose there are distinct solutions r1, r2 to the equationar(r − 1) + br + c = 0. That is, r1 6= r2. Then the general solutionof the homogeneous equation Lu = 0 is

u(x) = C1x r1 + C2x r2 , x > 0.

Lemma 4Suppose there is one solution r1 to ar(r − 1) + br + c = 0. Thenthe general solution to the homogeneous equation Lu = 0 is

C1x r1 + C2x r1 log(x), x > 0.



Cauchy-Euler ODEs

For a particular solution to the inhomogeneous Cauchy-Eulerequation

ax2u′′ + bxu′ + cu = x r ,

we can use the particular solution guess u(x) = αx r .



Cauchy-Euler ODEsExample 9: MATH2121 2016 T2 2.i)Find the general solution of the Cauchy-Euler ODE

2x2y ′′ + 7xy ′ + 3y = 13x1/4, x > 0.

We want to solve the equation ar(r − 1) + br + c = 0, wherea = 2, b = 7 and c = 3. That is, 2r2 + 5r + 3 = 0. Solutions arer1 = −1 and r2 = −3/2. Using Lemma 3, we have generalhomogeneous solution

yH(x) = C1x−1 + C2x−3/2.

A particular solution can be found by applying variation ofparameters, giving us

yP(x) = −8x1/4.

So y(x) = C1x−1 + C2x−3/2 − 8x1/4.




2x2y ′′ + 7xy ′ + 3y = 13x1/4, x > 0.

We want to solve the equation ar(r − 1) + br + c = 0, wherea = 2, b = 7 and c = 3. That is, 2r2 + 5r + 3 = 0. Solutions arer1 = −1 and r2 = −3/2.

Using Lemma 3, we have generalhomogeneous solution

yH(x) = C1x−1 + C2x−3/2.


yP(x) = −8x1/4.

So y(x) = C1x−1 + C2x−3/2 − 8x1/4.




2x2y ′′ + 7xy ′ + 3y = 13x1/4, x > 0.


yH(x) = C1x−1 + C2x−3/2.


yP(x) = −8x1/4.

So y(x) = C1x−1 + C2x−3/2 − 8x1/4.




2x2y ′′ + 7xy ′ + 3y = 13x1/4, x > 0.


yH(x) = C1x−1 + C2x−3/2.


yP(x) = −8x1/4.

So y(x) = C1x−1 + C2x−3/2 − 8x1/4.




2x2y ′′ + 7xy ′ + 3y = 13x1/4, x > 0.


yH(x) = C1x−1 + C2x−3/2.


yP(x) = −8x1/4.

So y(x) = C1x−1 + C2x−3/2 − 8x1/4.



Frobenious normal formA frequent form of ODE that appears in many applications can bewritten in Frobenious normal form:

z2u′′ + zP(z)u′ + Q(z)u = 0,where P(z) and Q(z) are analytic at 0. Let P0 = P(0) andQ0 = Q(0), and define a series F as

F (z ; r) = z r∞∑

k=0Ak(r)zk .

Consider the equation r(r − 1) + P0r + Q0 = 0 with solutions r1and r2.Lemma 5If r1 6= r2, then f (z ; r1) is a solution to the Frobenious normal formODE. If r1 − r2 is not a whole number, then a second linearlyindependent solution is F (z ; r2).



Bessel function

The Bessel equation with parameter ν is:

z2u′′ + zu′ + (z2 − ν2)u = 0.

This ODE is in Frobenious normal form, with indicial polynomial:

I(r) = (r + ν)(r − ν),

and we seek a series solution:

u(z) =∞∑

k=0Akzk+r .

We assume Re(ν) ≥ 0, so r1 = ν and r2 = −ν.



Bessel functionWith the normalisation:

A0 = 12νΓ(1 + ν)

the series solution is called the Bessel function of order ν and isdenoted:

Jν(z) = (z/2)νΓ(1 + ν)

[1− (z/2)ν

1 + ν+ (z/2)4

2!(1 + ν)(2 + ν) − ...].

And from the functional equation Γ(1 + z) = zΓ(z):

Jν(z) =∞∑

k=0

(−1)k(z/2)2k+ν

k!Γ(k + 1 + ν)



Bessel function

If ν is not an integer, then a second linearly independent, solutionis:

J−ν(z) =∞∑

k=0

(−1)k(z/2)2k−ν

k!Γ(k + 1− ν) .

For an integer ν = n ∈ Z, since Γ(n + 1) = n!, we have:

Jn(z) =∞∑

k=0

(−1)k(z/2)2k+n

k!(k + n)! .

Also, since 1Γ(z) = 0 for z = 0,−1,−2, ..., we find that Jn and J−n

are linearly independent; in fact:

J−n(z) = (−1)nJn(z).



Bessel function

Example 10: MATH2221 2015 T2 2.ii)1 Use term-by-term differentiation to prove that for ν ∈ R and x > 0:

ddx (xνJν(x)) = xνJν−1(x).

2 Hence evaluate the definite integral:

I =1∫

0

x 72 J 1

2(x)dx .



Bessel function

Note thatxνJν(x) =

∞∑k=0

xν (−1)k(x/2)2k+ν

k!Γ(k + 1 + ν) .

Moving the xν into the fraction,

xνJν(x) =∞∑

k=0

(−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν) .

Take one term in this sum and differentiate with respect to x :

ddx

((−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν)

)= (−1)k(2k + 2ν)x2k+2ν−1

22k+νk!Γ(k + 1 + ν) .



Bessel functionSince Γ(k + 1 + ν) = (k + ν)Γ(k + 1 + ν), then

ddx

((−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν)

)= (−1)k(2k + 2ν)x2k+2ν−1

22k+νk!(k + ν)Γ(k + ν) .

However there is a factor of 2(k + ν) in both the numerator anddenominator. So

ddx

((−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν)

)= (−1)k(x)2k+2ν−1

22k+ν−1k!Γ(k + ν) .

Finally, factor out a factor xν :

ddx

((−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν)

)= xν (−1)k(x/2)2k+ν−1

k!Γ(k + ν) .



Bessel functionSince the derivative is linear,

ddx (xνJν(x)) =

∞∑k=0

ddx

((−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν)

).

Substituting in the derivative we found,ddx (xνJν(x)) = xν

∞∑k=0

(−1)k(x/2)2k+ν−1

k!Γ(k + ν) .

The RHS here is just xνJν−1(x), so we are done.

To find the

integral1∫0

x 72 J 1

2(x)dx , we separate and use integration by parts:

1∫0

x2 · x32 J 1

2(x)dx

where u′ = x 32 J 1

2(x) and v = x2.



Bessel functionSince the derivative is linear,

ddx (xνJν(x)) =

∞∑k=0

ddx

((−1)k(x)2k+2ν

22k+νk!Γ(k + 1 + ν)

).

Substituting in the derivative we found,ddx (xνJν(x)) = xν

∞∑k=0

(−1)k(x/2)2k+ν−1

k!Γ(k + ν) .

The RHS here is just xνJν−1(x), so we are done. To find the

integral1∫0

x 72 J 1

2(x)dx , we separate and use integration by parts:

1∫0

x2 · x32 J 1

2(x)dx

where u′ = x 32 J 1

2(x) and v = x2.



Bessel functionApplying integration by parts:

I = [x2.x32 J 3

2(x)]10 −

1∫0

2x · x32 J 3

2(x)dx

= J 32(1)− 2

1∫0

x52 J 3

2(x)dx .

However by our previous result, ddx (xνJν(x)) = xνJν−1(x). So the

antiderivative of x5/2J3/2 is x5/2J5/2, hence

= J 32(1)− 2[x

52 J 5

2(x)]10

= J 32(1)− 2J 5

2(1).



Legendre equation (2221 only)The Legendre equation with paramater ν is:

(1− z2)u′′ − 2zu′ + ν(ν + 1)u = 0.

This ODE is not singular at z = 0, so the solution has an ordinaryTaylor series expansion:

u =∞∑

k=0Akzk .

The Ak must satisfy:

(k + 1)(k + 2)Ak+2 − [k(k + 1)− ν(ν + 1)]Ak = 0.

The recurrence relation is:

Ak+2 = (k − ν)(k + ν + 1)(k + 1)(k + 2) Ak for k ≥ 0.



Legendre equation (2221 only)We have:

u(z) = A0u0(z) + A1u1(z)where:

u0(z) = 1− ν(ν + 1)2! z2 + (ν − 2)ν(ν + 1)(ν + 3)

4! z−...

and:

u1(z) = z− (ν − 1)(ν − 2)3! z3+ (ν − 3)(ν − 1)(ν + 2)(ν + 4)

5! z5−...

Suppose now that ν = n is a non-negative integer. If n is even,then the series for u0(z) terminates, whereas if n is odd, then theseries for u1(z) terminates. The terminating solution is then calledthe Legendre polynomial of degree n and is denoted by Pn(z)with the normalisation:

Pn(1) = 1.



Dynamical Systems



Dynamical Systems

State variables are natural variables which depending on a singleindependent variable. A dynamical system is a natural processdescribed by these state variables. The state of a system at a giventime is described by the values of the state variables at thatinstant.

Note that any nth order ODE can be writen as a system of firstorder ODEs (not vice versa):

dnydtn = g

(y , dy

dt , ...,dn−1ydtn−1

)dxdt = f (x1, x2, ..., xn)



Non-autonomous ODEs

Definition 8A system of ODEs of the form:

dxdt = F(x)

is said to be autonomous.

Definition 9In a non-autonomous system, F may depend explicitly on t:

dxdt = F(x, t).



Lipschitz (2221 only)Definition 10The number L ∈ R is a Lipschitz constant for a functionf : [a, b]→ R if

|f (x)− f (y)| ≤ L |x − y | ∀x , y ∈ [a, b].

We say that the function f is Lipschitz if a Lipschitz constant for fexists.

Theorem 9If f is Lipschitz, then f is uniformly continuous.

Lemma 6If f : I → R is differentiable and f ′ is continuous on I, then f isLipschitz.




|f (x)− f (y)| ≤ L |x − y | ∀x , y ∈ [a, b].







|f (x)− f (y)| ≤ L |x − y | ∀x , y ∈ [a, b].






Lipschitz Vector Field (2221 only)We extend the definition of Lipschitz to vector fields.

Definition 11A vector field F : S ⊆ Rm → Rn is Lipschitz on S if

||F(x)− F(y)|| ≤ L ||x− y|| ∀ x, y ∈ S.

Here,

||x|| =

m∑j=1

x2j

1/2

denotes the Euclidean norm of the vector x ∈ Rm.

We say that F(x, t) is Lipschitz in x if, for all t:

||F(x, t)− F(y, t)|| ≤ L ||x− y|| , ∀x, y ∈ Rm.



Lipschitz Vector Field (2221 only)We extend the definition of Lipschitz to vector fields.

Definition 11A vector field F : S ⊆ Rm → Rn is Lipschitz on S if

||F(x)− F(y)|| ≤ L ||x− y|| ∀ x, y ∈ S.

Here,

||x|| =

m∑j=1

x2j

1/2

denotes the Euclidean norm of the vector x ∈ Rm.

We say that F(x, t) is Lipschitz in x if, for all t:

||F(x, t)− F(y, t)|| ≤ L ||x− y|| , ∀x, y ∈ Rm.



Existence and Uniqueness Theorem (2221only)

We want to find solutions to a non-autonomous system. Thefollowing theorem guarantees a unique solution for anon-autonomous system, under certain conditions.

Theorem 10The initial value problem defined by:

dxdt = F(x, t), x(t0) = x0

has a unique solution x(t) over a time interval |t − t0| < α ifF(x, t) is continuous and Lipschitz.

In fact, x(t) is continuous and differentiable.



Notes

The existence of solutions follow from continuity in x and t.

The uniqueness of solutions follow from the Lipschitz condition in x .

The theorem is a local existence theorem. It provides for theexistence of solutions over a finite time interval.



Lipschitz (2221 only)

Example 11

f (x) = 2√

x , x ∈ R

Where is f Lipschitz?

Since f ′(x) = 1√x is continuous on x > 0, then f is Lipschitz for

x > 0 (Lemma 6). At 0, f is not Lipschitz since if it was,

|f (x)− f (y)| = 2∣∣√x −√y

∣∣≤ L |x − y |

2L ≤

∣∣√x +√y∣∣

which cannot hold true for x and y both 0.




Example 11

f (x) = 2√

x , x ∈ R



x > 0 (Lemma 6).

At 0, f is not Lipschitz since if it was,

|f (x)− f (y)| = 2∣∣√x −√y

∣∣≤ L |x − y |

2L ≤

∣∣√x +√y∣∣





Example 11

f (x) = 2√

x , x ∈ R



x > 0 (Lemma 6). At 0, f is not Lipschitz since if it was,

|f (x)− f (y)| = 2∣∣√x −√y

∣∣≤ L |x − y |

2L ≤

∣∣√x +√y∣∣




Linear systems of ODEs

Definition 12We say that the n × n, first-order system of ODEs:

dxdt = F(x, t)

is linear if the RHS has the form:

F(x, t) = A(t)x + b(t)

for some n × n matrix-valued function A(t) = [ai ,j(t)] and avector-valued function b(t) = [bi (t)].

The linear first-order system is autonomous when A and b areconstant.



Global Existence and Uniqueness

We have a stronger existence result in the linear case:

Theorem 11If A(t) and b(t) are continuous for 0 ≤ t ≤ T , then the linearinitial-value problem

dxdt = A(t)x + b(t) with x(0) = x0,

has a unique solution x(t) for 0 ≤ t ≤ T .



A special case

It is much easier to work with the special case when A(t) = A is aconstant n × n matrix and b(t) = 0:

dxdt = Ax.

The general solution to this system is

x(t) =n∑

i=1cieλi tvi ,

where λi is the i-th eigenvalue with corresponding eigenvector viand c1, ..., cn are constants.



A special case

It is much easier to work with the special case when A(t) = A is aconstant n × n matrix and b(t) = 0:

dxdt = Ax.

The general solution to this system is

x(t) =n∑

i=1cieλi tvi ,

where λi is the i-th eigenvalue with corresponding eigenvector viand c1, ..., cn are constants.



Initial-valued systemRecall that for an n × n complex matrix A, we define theexponential of a matrix as

eA =∞∑

k=0

Ak

k! = I + A + 12!A2 + 1

3!A3 + · · · .

Consider the initial-valued system

dxdt = Ax, x(0) = x0.

Then we can write the solution as a matrix exponential,

x(t) = etAx0.

The issue is calculating etA, which requires finding Ak for k ∈ N.To do this efficiently, we look to diagonalisation.




eA =∞∑

k=0

Ak

k! = I + A + 12!A2 + 1

3!A3 + · · · .


dxdt = Ax, x(0) = x0.


x(t) = etAx0.





eA =∞∑

k=0

Ak

k! = I + A + 12!A2 + 1

3!A3 + · · · .


dxdt = Ax, x(0) = x0.


x(t) = etAx0.




Diagonalising a matrixDefinition 13An n × n complex matrix A is diagonalisable if there exists anon-singular matrix n× n matrix M such that M−1AM is diagonal.

Theorem 12An n × n complex matrix A is diagonalisable if and only if thereexists a basis {v1, ..., vn} for Cn, where v1, ..., vn are eigenvectorsof A. In fact the columns of M are the eigenvectors of A,M = (v1| · · · |vn), and M−1AM = Λ where:

Λ =

λ1. . .

λn

for eigenvalues λi corresponding to eigenvector vi .



Diagonalising a matrixDefinition 13An n × n complex matrix A is diagonalisable if there exists anon-singular matrix n× n matrix M such that M−1AM is diagonal.

Theorem 12An n × n complex matrix A is diagonalisable if and only if thereexists a basis {v1, ..., vn} for Cn, where v1, ..., vn are eigenvectorsof A. In fact the columns of M are the eigenvectors of A,M = (v1| · · · |vn), and M−1AM = Λ where:

Λ =

λ1. . .

λn

for eigenvalues λi corresponding to eigenvector vi .



Matrix Powers

In general since M−1AM = Λ, we can efficiently calculate Ak :

Ak =k times︷︸︸︷

MΛM−1 ·MΛM−1 · · ·MΛM−1 = MΛkM−1.

This is better because

Λk =

λk1

. . .λn

.In fact, due to the taylor series of the exponential function, we cansimplify the solution to the initial-valued system further.



Matrix Powers

In general since M−1AM = Λ, we can efficiently calculate Ak :

Ak =k times︷︸︸︷

MΛM−1 ·MΛM−1 · · ·MΛM−1 = MΛkM−1.

This is better because

Λk =

λk1

. . .λn

.In fact, due to the taylor series of the exponential function, we cansimplify the solution to the initial-valued system further.



Exponential of a diagonalisable matrix

Theorem 13If A = MΛM−1 is diagonalisable, then:

eA = MeΛM−1 and eΛ =

eλ1

. . .eλn

.

We can now find the exponential of tA:

etA = MetΛM−1, and etΛ =

etλ1

. . .etλn

.



Exponential of a diagonalisable matrix

Theorem 13If A = MΛM−1 is diagonalisable, then:

eA = MeΛM−1 and eΛ =

eλ1

. . .eλn

.We can now find the exponential of tA:

etA = MetΛM−1, and etΛ =

etλ1

. . .etλn

.



ExampleExample 12: MATH2121 2016 T2 2.iii)For an n × n matrix A.

1 State the definition of eA.

2 Show that if Av = λv, then eAv = eλv.

The definition of the matrix exponential is

eA =∞∑

k=0

Ak

k! = I + A + A2

2! + A3

3! + · · · .

If Av = λv then

eAv =( ∞∑

k=0

1k!Ak

)v

=∞∑

k=0

1k!(Akv

).



ExampleExample 12: MATH2121 2016 T2 2.iii)For an n × n matrix A.

1 State the definition of eA.

2 Show that if Av = λv, then eAv = eλv.

The definition of the matrix exponential is

eA =∞∑

k=0

Ak

k! = I + A + A2

2! + A3

3! + · · · .

If Av = λv then

eAv =( ∞∑

k=0

1k!Ak

)v

=∞∑

k=0

1k!(Akv

).



Example

But we can find Akv:

Akv = Ak−1λv = · · · = λkv.

Hence

eAv =∞∑

k=0

1k!λ

kv

=( ∞∑

k=0

λk

k!

)v.

This is the exponential function Taylor series at the point λ, so

eAv = eλv.



Example


Akv = Ak−1λv = · · · = λkv.

Hence

eAv =∞∑

k=0

1k!λ

kv

=( ∞∑

k=0

λk

k!

)v.


eAv = eλv.



Example


Akv = Ak−1λv = · · · = λkv.

Hence

eAv =∞∑

k=0

1k!λ

kv

=( ∞∑

k=0

λk

k!

)v.


eAv = eλv.



Equilibrium points

Definition 14We say that a ∈ Rn is an equilibrium point for the dynamicalsystem dx

dt = F(x) ifF(a) = 0.

Suppose a is an equilibrium point for the system dxdt = F(x).


dxdt = F(x), x(0) = a.

Then the solution is the constant function x(t) = a.



Equilibrium points

Definition 14We say that a ∈ Rn is an equilibrium point for the dynamicalsystem dx

dt = F(x) ifF(a) = 0.

Suppose a is an equilibrium point for the system dxdt = F(x).


dxdt = F(x), x(0) = a.

Then the solution is the constant function x(t) = a.



Stable Equilibrium

Definition 15An equilibrium point a is stable if for every ε > 0, there existsδ > 0 such that whenever ||x0 − a|| < δ, the solution of

dxdt = F(x), x(0) = x0

satisfies||x(t)− a|| < ε ∀ t > 0.

Intuitively: if a solution starts close enough to the stableequilibrium point, then they will remain close to the stableequilibrium point.



Stable Equilibrium

Definition 15An equilibrium point a is stable if for every ε > 0, there existsδ > 0 such that whenever ||x0 − a|| < δ, the solution of

dxdt = F(x), x(0) = x0

satisfies||x(t)− a|| < ε ∀ t > 0.

Intuitively: if a solution starts close enough to the stableequilibrium point, then they will remain close to the stableequilibrium point.



Asymptotic StabilityThis is a stronger form of stability, on a particular subset of Rn.

Definition 16Let N be an open subset of Rn that contains an equilibrium pointa. We say that a is asymptotically stable in N if a is stable, andwhenever x0 ∈ N the solution of

dxdt = F(x) x(0) = x0

satisfiesx(t)→ a as t →∞.

We call N a domain of attraction for a.

Intuitively: not only do the solutions stay close to the stableequilibrium point, but they also approach the equilibrium point ast goes to infinity.



Asymptotic StabilityThis is a stronger form of stability, on a particular subset of Rn.

Definition 16Let N be an open subset of Rn that contains an equilibrium pointa. We say that a is asymptotically stable in N if a is stable, andwhenever x0 ∈ N the solution of

dxdt = F(x) x(0) = x0

satisfiesx(t)→ a as t →∞.

We call N a domain of attraction for a.

Intuitively: not only do the solutions stay close to the stableequilibrium point, but they also approach the equilibrium point ast goes to infinity.



Linear constant case

Consider the linear system

dxdt = Ax + b, x(0) = x0,

where det(A) 6= 0.

Then the unique equilibrium point of thissystem is

a = −A−1b,

and the solution to the system is

x(t) = a + etA (x0 − a) .




Consider the linear system

dxdt = Ax + b, x(0) = x0,

where det(A) 6= 0. Then the unique equilibrium point of thissystem is

a = −A−1b,

and the solution to the system is

x(t) = a + etA (x0 − a) .




Theorem 14Consider the previous linear constant coefficient system. Letλ1, ..., λn be the eigenvalues of A. The equilibrium pointa = −A−1b is:

1 Stable if and only if Re(λj) ≤ 0 for all j .2 asymptotically stable if and only if Re(λj) < 0 for all j .

In the second case, the domain of attraction is the whole of Rn.



Classification of 2D Linear Systems



Improper Node

Here λ1 6= λ2 ∈ R. Which eigenline orbits tend to depends on theeigenvalues.



Deficient Node

Here λ1 = λ2 ∈ R, and the eigenspace is deficient (out of thescope of the course).



Star Node

Here λ1 = λ2 6= 0 and all nonzero vectors are eigenvectors.Therefore, all orbits are either being attracted or repelled by theequilibrium points.



Saddle Point

Here, λ1 < 0 < λ2. This means λ1 is attracting, whilst λ2 isrepelling.



Centre

Here λ1 = iβ = λ2, where β ∈ R. If eigenvalues are purelyimaginary, orbits are given by ellipses around the eigen-plane aseit = cos(t) + i sin(t).



SpiralHere λ1 = α + iβ = λ̄2. If eigenvalues are in conjugate pairs, realparts of the orbit will be defined bye−λt = e−Re(()λ)t(c1 cos(t) + c2 sin(t)), which will spiral inward oroutward, clockwise or anticlockwise depending on values ofRe (()λ) and Im (()λ).



Equilibrium pointsExample 13: MATH2121 2018 T2 2.iii)Solve for x(t) and y(t) and determine the type and stability of theequilibrium point of the following system of differential equations:

dxdt = x + y ;

dydt = 2x .

To find the equilibrium point, we solve x + y = 0 and 2x = 0simultaneously. The only solution is the point (0, 0). The systemcan be written as

ddt

(xy

)=(

1 12 0

)(xy

).



Equilibrium pointsExample 13: MATH2121 2018 T2 2.iii)Solve for x(t) and y(t) and determine the type and stability of theequilibrium point of the following system of differential equations:

dxdt = x + y ;

dydt = 2x .

To find the equilibrium point, we solve x + y = 0 and 2x = 0simultaneously. The only solution is the point (0, 0). The systemcan be written as

ddt

(xy

)=(

1 12 0

)(xy

).



Equilibrium points

The eigenvalues of(

1 12 0

)are λ1 = −1 and λ2 = 2.

Since λ1 < 0 < λ2, the equilibrium point (0, 0) is a saddle point.Since Re(λ2) > 0 then (0, 0) is an unstable equilibrium point.Hence (0, 0) is an unstable saddle point.



Equilibrium points

The eigenvalues of(

1 12 0

)are λ1 = −1 and λ2 = 2.

Since λ1 < 0 < λ2, the equilibrium point (0, 0) is a saddle point.

Since Re(λ2) > 0 then (0, 0) is an unstable equilibrium point.Hence (0, 0) is an unstable saddle point.



Equilibrium points

The eigenvalues of(

1 12 0

)are λ1 = −1 and λ2 = 2.

Since λ1 < 0 < λ2, the equilibrium point (0, 0) is a saddle point.Since Re(λ2) > 0 then (0, 0) is an unstable equilibrium point.Hence (0, 0) is an unstable saddle point.



First integrals (2221 only)

Definition 17A function G : Rn → R is a first integral for a system of ODEs

dxdt = F(x)

if G (x(t)) is constant for every solution x(t).

Geometrically: G is a first integral iff

∇G(x) ⊥ F(x) for all x.




Definition 17A function G : Rn → R is a first integral for a system of ODEs

dxdt = F(x)

if G (x(t)) is constant for every solution x(t).

Geometrically: G is a first integral iff

∇G(x) ⊥ F(x) for all x.




Example 14Consider the system of ODEs

dx1dt = x1x2,

dx2dt = −x2

1 .

Prove that G(x) = x21 + x2

2 is a first integral.

Set the function F(x) as the RHS of the system:

F(x) =(

x1x2−x2

1

).



First integrals (2221 only)We want to find the gradient of G :

∇G(x) =(

2x12x2

).

So

∇G(x) · F(x) =(

2x12x2

)·(

x1x2−x2

1

)= 2x2

1 x2 − 2x2x21 = 0.

This means that all solutions to the system of ODEs must bemapped to a constant under G . That is, if (x1, x2) is a solutionthen x2

1 + x22 = C for some C . In other words, all the solutions to

the system of ODEs lie on some circle around the origin. Whatthis tells us is that if x1 = r cos(t) then x2 must be r sin(t) forwhatever value of r > 0.


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mathsoc.unsw.edu.au · Part I: Linear ODEs Dynamical Systems Table of Contents 1 Part I: Linear ODEs Linear Diﬀerential Operators Wronskian and Linear Independence Methods Power