ODE lecture 1

ENGR90024 COMPUTATIONAL FLUID DYNAMICS

Lecture O01

Ordinary Differential Equations (ODEs):!

Introduction & Taylor series!

(Pages 1-4 of the Printed Lecture Notes)!

!

Mathematical equations that govern fluid flow

OpenFoam

Supercomputer

x PREFACE

(a) (b)

(c) (d)

(e) (f)

Figure 1: Some example applications in science and engineering which involve thesolution of ordinary and partial di↵erential equations.

x PREFACE

(a) (b)

(c) (d)

(e) (f)


x PREFACE

(a) (b)

(c) (d)

(e) (f)


Lecture & Workshop timetable3 MODULES!•ORDINARY DIFFERENTIAL EQUATIONS (ODE)!•PARTIAL DIFFERENTIAL EQUATIONS (PDE)!•OPENFOAM(HTTP://WWW.OPENFOAM.ORG)

Lecture & Workshop timetable

Week Starting Tue-morning Tues-afternoon Wed-Workshop Thu

1 Mar 3 Lecture on ODEs Prof Andrew Ooi

Lecture on ODEs Prof Andrew Ooi

Workshop on Linux/C++/ODEs



Lecture on ODEs Prof Andrew Ooi Workshop on ODEs Lecture on OpenFOAM

Dr Stephen Moore


Lecture on ODEs Prof Andrew Ooi Workshop on OpenFOAM Lecture on OpenFOAM

Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore

6 Apr 7 Lecture on PDEs A/Prof Malcolm Davidson

Lecture on PDEs A/Prof Malcolm Davidson

Consultation for Assignment 1



Lecture on PDEs A/Prof Malcolm Davidson Workshop on OpenFOAM Lecture on PDEs

A/Prof Malcolm Davidson

Apr 21 Break

Break

Break

Break


Lecture on PDEs A/Prof Malcolm Davidson Workshop on PDEs Lecture on PDEs


9 May 5 Lecture on PDEs A/Prof Malcolm Davidson




Lecture on PDEs A/Prof Malcolm Davidson Workshop on PDEs Lecture on OpenFOAM

Dr Stephen Moore

11 May 19 Lecture on OpenFOAM Dr Stephen Moore

Lecture on OpenFOAM Dr Stephen Moore Workshop on OpenFOAM Lecture on OpenFOAM

Dr Stephen Moore



Dr Stephen Moore

Assignment 1 due!18th April

Assignment 2 due!30th May









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore








Apr 21 Break

Break

Break

Break









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Doug McDonell-309!11am-12pm









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore








Apr 21 Break

Break

Break

Break









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Alan Gilbert Theatre 2!5:15pm-6:15pm









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore








Apr 21 Break

Break

Break

Break









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Doug McDonell-503!2:15pm-3:15pm









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore








Apr 21 Break

Break

Break

Break









Dr Stephen Moore



Dr Stephen Moore



Dr Stephen Moore



Old Engineering - EDS 1!12:00pm-2:00pm

Assessment

•10% Workshops (weekly)-1% satisfactory completion of every workshop-11 workshops in total so can afford to miss 1

•10% Assignment 1 (Due 18th April)-ODE & OpenFOAM

•20% Assignment 2 (Due 30th May)-PDE & OpenFOAM

•60% end of semester exam -ODE, PDE & OpenFOAM-Exam is a Hurdle. Need to pass the exam to pass this subject!!!!

Student comments“Open foam should be introduced in the workshops much earlier than they were”

• Last year, OpenFOAM workshops were not introduced until week 4• We will be introducing OpenFOAM workshop in week 3. There could also be

! OpenFOAM material in workshops in weeks 1 and 2.

“Assignments should have been released earlier”• Assignment 1 (due week 7) will be released week 1 or 2 of the semester• We will release assignment 2 in week 8 (last year was released in week 10).

Ordinary Differential EquationsIn the most generic form, ordinary differential equations (ODE) can bewritten as

g

✓dN�

dtN, · · · , d

3�

dt3,d2�

dt2,d�

dt,�, t

◆= 0

where N is the order of the ODE.

a(t)d3�

dt3+ b(t)

d2�

dt2+ c(t)

d�

dt+ d(t)�+ e(t) = 0

then the ODE is linear.

(O01.1)

(O01.2)

If every term in the ODE is linear, then the ODE is linear. If even one!term of the ODE is nonlinear, then the ODE is nonlinear.

If g is a linear function, e.g.

If g is a nonlinear function, then the ODE is nonlinear.

Example O01.1: For the ODEs shown below, state the order of the ODE and if the ODE is linear/nonlinear. a, b and c are constants.!

(a) t2d2�

dt2+ at

d�

dt+ b� = c

(b)d2�

dt2� a

�1� �2

� d�dt

+ � = 0

(a) t2d2�

dt2+ at

d�

dt+ b� = c

Let’s look at equation (a).

Linear Linear Linear

Linear

Since all terms in the above equations are linear, the ODE is a linear ODE

Let’s now look at (b)

(b)d2�

dt2� a

�1� �2

� d�dt

+ � = 0

d2�

dt2� a

d�

dt+ �2 d�

dt+ � = 0

LinearLinear

Linear

Nonlinear

Let’s now look at (b)

(b)d2�

dt2� a

�1� �2

� d�dt

+ � = 0

d2�

dt2� a

d�

dt+ �2 d�

dt+ � = 0

LinearLinear

Linear

Nonlinear

Since the third term in the above equation is nonlinear, the ODE is a nonlinear ODE!

End of Example O01.1

In many engineering problems, Eq. (O01.1) can be written as

Eq. (O01.3) is solved usually solved in the domain

with the initial condition

(O01.3)d�

dt= f(t,�)

tmin

t tmax

�(tmin) = �min

Initial Value Problem

d�

dt= f(t,�)

tmin

t tmax

�(tmin) = �min


Mathematical MethodsExact (true) solution

Approximate solution

Numerical/Computer Methods

Compare

Example O01.2: !Solve the following ordinary differential equation (ODE)!!!!!with ɸ(t=0)=0. Use MATLAB to plot the solution for 0 < t < 8.

d�

dt= 1� � (O01.4)

d�

dt= f(t,�)

tmin

t tmax

�(tmin) = �min


Mathematical MethodsExact (true) solution



Compare

d�

dt= 1� �

d�

dt+ � = 1

Use integrating factor, multiply both sides by et

etd�

dt+ et� = et

d

dt

�et�

�= et

Integrating gives

1

et� = et +K

Where K is a constant. We are given that ɸ=0 when t=0. Substituting!into the above equation gives

1⇥ 0 = 1 +K

K = �1Thus the solution to Eq. (O01.4) is

� = 1� e�t

e0 ⇥ 0 = e0 +K

function MPO01p2() !!t=0:0.1:8; phi=1-exp(-t); !plot(t,phi); !xlabel('t'); ylabel('\phi(t)');

t=[0.0 0.1 0.2 0.3 0.4........8.0]

phi=[0.0 0.0952 0.1813 0.2592 ........ 1.0]

1-e-0.0 1-e-0.2 1-e-8.0

1-e-0.1 1-e-0.3

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

φ(t)

d�

dt= f(t,�)

tmin

t tmax

�(tmin) = �min


Mathematical MethodsExact solution



Compare

� = 1� e�t

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

φ(t)

d�

dt= f(t,�)

tmin

t tmax

�(tmin) = �min

Initial Value Problem Approximate solution


Taylor series

TheTaylor series expansion of a function ɸ(t) about the pointtl can be written as

tl+1=tl+Δt.

�(tl+1) = �(tl) +�t

1!

d�

dt

��tl+

�t2

2!

d2�

dt2

��tl+

�t3

3!

d3�

dt3

��tl+ . . .

(O01.5)�(tl+1) =1X

n=0

(tl+1 � tl)n

n!dn�

dtn

��tl

(O01.6)

Equation (O01.5) can be expanded to look like

Example O01.3: Find the Taylor series for the function!!!!Plot the first few terms of this function and show that you can get closer to the original function if you use the more terms in the series.

�(t) = sin(t)

�(tl+1) = �(tl) +�t

1!

d�

dt

��tl+

�t2

2!

d2�

dt2

��tl+

�t3

3!

d3�

dt3

��tl+ . . .

If tl=0, then tl+1=Δt=t

�(t) = �(0) +t

1!

d�

dt

��0

+t2

2!

d2�

dt2

��0

+t3

3!

d3�

dt3

��0

+ . . .

You are given that ɸ(t)=sin(t), so

ɸ(0)=sin(0)=0!dɸ/dt(0)=cos(0)=1.0!

d2ɸ/dt2(0)=-sin(0)=0.0!d3ɸ/dt3(0)=-cos(0)=-1.0!d4ɸ/dt4(0)=sin(0)=0.0!d5ɸ/dt5(0)=cos(0)=1.0

From Eq. (O01.6), the Taylor’s series can be written as

Substituting into the Taylor series Eq. (O01.6) gives!

�(t) = sin(t) = t� (1/6)t3

+(1/120)t5

�(1/5040)t7

+ . . .

�(t) = t

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

tφ

�(t) = t�(1/6)t3

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

t−(1/6) t3φ

�(t) = t�(1/6)t3+(1/120)t5

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

t−(1/6) t3+(1/120) t5φ

�(t) = t�(1/6)t3+(1/120)t5�(1/5040)t7

−4 −3 −2 −1 0 1 2 3 4−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

t−(1/6) t3+(1/120) t5−(1/5040) t7φ

Sometimes it is convenient to group all higher order terms of theTaylorsseries into a single term. For example,

�(tl+1) = �(tl) +�t

1!

d�

dt(tl) +

�t2

2!

d2�

dt2(tl) +

�t3

3!

d3�

dt3(tl) +

�t4

4!

d4�

dt4(tl) + . . .

can be written as

or

�(tl+1) = �(tl) +�t

1!

d�

dt(⇠2)

where tl < ξ1 < tl+1 , tl < ξ2 < tl+1 and ξ1 ≠ ξ2

(O01.7)

(O01.8)

�(tl+1) = �(tl) +�t

1!

d�

dt(tl) +

�t2

2!

d2�

dt2(⇠1)

�(tl+1) = �(tl) +�t

1!

d�

dt(tl) +

�t2

2!

d2�

dt2(tl) +

�t3

3!

d3�

dt3(tl) +

�t4

4!

d4�

dt4(tl) + . . .

�(tl+1) = �(tl) +�t

1!

d�

dt(tl)+

�t2

2!

d2�

dt2(⇠)

where ξ is somewhere between tl and tl+1 i.e. tl < ξ < tl+1

Exercise O01.4: For the function!!!!!!

�(t) = sin(t)show the validity of Eqs. (O01.7) & (O01.8) by finding !finding values of ξ1 and ξ2 for tl=0.0 and Δt =π/2.

Equation (O01.8), repeated here

tl+1 = tl +�t

tl ⇠2 tl+1

�(tl+1) = �(tl) +�td�

dt(⇠2)

tl+1 = tl +�t

tl ⇠2 tl+1

�(tl+1) = �(tl) +�td�

dt(⇠2)

For this example

tl = 0.0�t = ⇡/2

So

�(t) = sin(t)

tl+1 = 0 + ⇡/2 = ⇡/2

Hence, equation (O01.8), says that

�(tl+1) = sin(tl+1) = sin(⇡/2) = 1

=

�(tl) = sin(tl) = sin(0) = 0

+1 0 ⇡/2

d�

dt(⇠2) = cos (⇠2)

cos (⇠2)

1 = 0 + ⇡/2 cos (⇠2)

0 ⇠2 ⇡/2

Illustrate graphically

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

φ

�t

sin(⇡/2)d�

dt(⇠2 = 0)

sin(0) ⇡/2

sin(0) +�td�

dt(⇠2 = 0)

� = sin(t)

sin(0) +�td�

dt(⇠2 = 0) > sin(⇡/2)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

φ

sin(⇡/2)

sin(0)

� = sin(t)d�

dt(⇠2 = ⇡/2)

�tsin(0) +�t

d�

dt(⇠2 = ⇡/2)

sin(0) +�td�

dt(⇠2 = ⇡/2) < sin(⇡/2)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

φ

� = sin(t)

sin(⇡/2)

sin(0)

d�

dt(⇠2 = 1.2)

�t

sin(0) +�td�

dt(⇠2 = 1.2)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

φ

� = sin(t)

sin(⇡/2)

sin(0)

d�

dt(⇠2 = 1.2)

�t

sin(0) +�td�

dt(⇠2 = 1.2)

sin(0) +�td�

dt(⇠2 = 1.2) 6= sin(⇡/2)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

φ

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

φ

⇠2 = 0

⇠2 = ⇡/2

sin(0) +�td�

dt(⇠2 = 0)

sin(0) +�td�

dt(⇠2 = ⇡/2)

sin(⇡/2)

sin(⇡/2)

From the top graph, it is shown that for ξ2=0, the equation

evaluates to a value greater than sin(p/2). From the bottom graph, it is shown that for ξ2= p/2, the equation

evaluates to a value less than sin(p/2). Hence, in order for

sin(0) +�td�

dt(⇠2)

sin(0) +�td�

dt(⇠2)

sin(0) +�td�

dt(⇠2) = sin(⇡/2)

0<ξ2<p/2.

sin(⇡/2) = sin(0) +

⇡

2

cos(⇠2)

cos(⇠2) =2

⇡

⇠2 = 0.8807

Finding the value of ξ2 is easy

�t �t

�t�t

⇠2 = 0.0 ⇠2 = 0.5

⇠2 = 0.7 ⇠2 = 0.8807

ODE lecture 1

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