block 1a - lecture 1 soori
TRANSCRIPT
Syllabus…Will be on VISION
• Block 1 - Diff Eqns – Constant Coefficients
• Block 2 – 1st Order Diff Eqns
• Block 3 – Partial differentiation, max and min
• Block 4 – Taylor Series and linear approxn
• Block 5 – Double integrals
Assessment
Assessment Weighting Notes
Exam 80% 2 sectionsSection A – Methods (55%) Section B – Applications (45%) (3 from 5)
AssessmentsExam
20% Opened up for 1 weekMultiple attempts allowed, Best mark taken
To get a D+ you must achieve 25% in the exam!
Books
• Engineering Maths Stroud 6th Edition– Block 1,2,3 and Laplace transforms
• Modern Engineering Mathematics - 3rd Edn
– Glyn James– My favourite but hard examples/ exercises– Covers nearly all of the two modules
• Maths for Engineers - Croft and Davison
Differential equations - Examples
• Chemical kinetics (A,B & C are chemical concentrations)
• LCR Circuits
• Dragster slowing down
CBA BATkdt
dA
01
2
2
iCdt
diR
dt
idL
2kvdt
dv
Properties of Logs and Exps - Revision
yxyx lnlnln
xnxn lnln
xee xx lnln
yxyx eee
y
xyx lnlnln
Dependent & independent variable
• y=x2+3xx-independent variable and y-dependent variable
(since the values of y depends on what we chose for x)
• dx/dt=x+t2
t-independent variablex-dependent variableDependent variable is always the variable being
differentiated.
What is D.E?• D.E is an equation involving an unknown
function y of one or more independent variables x,t and derivatives
• dy/dx=x2
y-unknown function (dependent variable)
x-independent variable
Terminologies• Order of a D.E is the order of the highest
derivative appearing in the equation.
• Degree of D.E is the degree of the highest ordered derivatives.
• Linear: A D.E is said to be linear if the dependent variable and its derivatives occur to the 1st degree only and if there are no products involving the dependent variable or its derivatives.
Homogeneous and Non-homogeneous
• This can be applied to linear equations only.
• If the D.E is arranged such that all the terms containing the dependent variable is on the L.H.S of the equality sign and those terms involving independent variable and constants are on the R.H.S, then in homogeneous equation R.H.S=0
Differential Equations -Terminology
03 tydt
dy
First order
linearhomogeneous D.E.
Variable Coefficient
Printed notes – Pg 3
Terminology
FTdt
ds
dt
sdm
2
2
2
nonlinear
Second order
nonhomogeneous D.E.
More terminology
0 ycdt
dyb
First orderlinear
homogeneous D.E.
y(0) = 5Initial value
problem
1st order D.Es with constant coefficients
0 cydt
dyb
ybcdt
dy)/(
Ktbcy
dtbcy
dy
)/(ln
)/(
Good integrators are good solvers of D.E.s
Let us consider 1st order homogeneous D.E with constant coefficients
Applications
• Bacterial growth in a culture
• Carbon dating (How old fossils are?)
• Spread of diseases
• Population dynamics – no predators/ lots of food
Swine flu infections in the UK 2009
Date Day Confirmed Cases
29 April 0 8
2 May 3 15
7 May 8 34
10 May 11 55
28 May 29 203
3 June 35 381
10 June 42 750
15 June 47 1,320
23 June 55 2,905
24 June 56 3,254
29 June 61 5,937
1 July 63 6,929
Swine Flu Model
kQdt
dQ
80 Q
108.0k
0
1000
2000
3000
4000
5000
6000
7000
8000
0 20 40 60 80
Days
Co
nfi
rme
d C
as
es
Data
Model
A model for the swine flu epidemic
Carbon Dating Problem
Important Note on formation of D.E
• A FUNCTION WITH one ARBITRARY CONSTANT GIVES 1ST ORDER D.E
• An nth order D.E is derived from a function having n-arbitrary constants.
Second Order D.E• Let a,b, and c are constants and f(t) is a
function of t
tfycdt
dyb
dt
yda
2
2
Second order differential equations
Lets make life easier…consider
•2nd order•linear •constant coefficient•Homogeneous because f(t)=0
02
2
ycdt
dyb
dt
yda
General solution
tyBtyA 21
Lets make life even easier
Consider
Solution
02
2
ycdt
dyb
dt
yda
0 ycdt
dyb
tAety
b
c
2nd Order D.Es – Theory 1
Suppose is a solution to
Then
02
2
ycdt
dyb
dt
yda
tAety
02
2
ycdt
dyb
dt
yda
teAdt
dy
teAdt
yd 22
2
2nd Order D.Es – Theory 2
Substituting for
Then
02
2
ycdt
dyb
dt
yda
ydt
dy2
2
dt
yd
02 ttt AeceAbeaA
02 tAecba
2nd Order D.Es – Theory 3
Provided A is nonzero.
So solve the D.E means solve
Good old Euclid!
02
2
ycdt
dyb
dt
yda
02 tAecba
0tAe
02 cba
This is called the auxiliary equation
2nd Order D.Es – Theory 4
•Two real roots, - and +
•One Repeated root, •Complex conjugate roots - , +=
02
2
ycdt
dyb
dt
yda
02 cba
p qi
Three possibilities
a
acbb
2
42
General Solution – Two real roots, - and +
02
2
ycdt
dyb
dt
yda tt BeAety
Two real roots
tsnsAandBareco
BeAety tt
tan
General Solution – Repeated root - 10
2
2
ycdt
dyb
dt
yda
a
b
2
02 cba
02 cba
042 acb
If is a repeated root then,
and
and
General Solution – Repeated root - 10
2
2
ycdt
dyb
dt
yda
02 cba
tBtety
• Differentiate twice
• Sub into the D.E.
1 tBedt
dy t
222
2
tBedt
yd t
0122 ttt BtectBebtaBe
General Solution – Repeated root - 30
2
2
ycdt
dyb
dt
yda
02 cba
• Tidy up
• As is a repeated root,
• We have the result is a solution
0122 ttt BtectBebtaBe
022 tBebatcba
a
b
2
tBtety
General Solution 02
2
ycdt
dyb
dt
yda
02 cba
• Two real roots
• One repeated root
• Two complex conjugate roots - , +=
tt BeAety
teBtAty
p qi
2nd Order D.Es – General Solution
•Two roots, - and +
•Repeated root, •Complex conjugate roots - , +=
02
2
ycdt
dyb
dt
yda
02 cba
p qi
Three possibilities
a
acbb
2
42 tAety
General Solution – Complex roots - 10
2
2
ycdt
dyb
dt
yda
Using Euler’s formula from complex numbers
tqiptqip BeAety
qittqipt BeAeety
qtiqtBqtiqtAety pt sincossincos
qtiBAqtBAety pt sincos
02 cba
- , += p qi
General Solution – Complex roots - 20
2
2
ycdt
dyb
dt
yda
02 cba
Let
qtiBAqtBAety pt sincos
BAC
iBAD
qtDqtCety pt sincos
Summary - General Solution
• Two real roots
• One repeated root
• Two complex conjugate roots - , +=
02
2
ycdt
dyb
dt
yda
02 cba
tt BeAety
teBtAty
p qi
qtDqtCety pt sincos
Small amplitudeOscillations of a pendulum
mm
sinmgF
T
mg
l
02
2
l
g
dt
d
sin
2
2
mgdt
dml
sin
Newton’s 2nd law (F=ma)
2
2
dt
dla
Solution
z =0
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
tBtAt sincos
l
g2
A=1, B=0
=6
Small amplitudeoscillations of a pendulum
T
mg
l
Better model including air resistance
02
2
l
g
dt
d
m
k
dt
d
Standard representation
02 22
2
zdt
d
dt
d
Graphical Solution including air resistance
T
mg
l
z =0.1
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
z =1
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
Graphical Solution Critical damping z=1
T
mg
l
z =2
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
Graphical Solution Over damping 1<z
T
mg
l
Small amplitudeoscillations of a pendulum
T
mg
l
02 22
2
zdt
d
dt
d
l
g
g
l
m
k
2
1z
- natural frequency
z - damping ratio
Why these parameters are interesting will be considered in L8.
Damped systems Analytical solutions
T
mg
l
tBtAet t 22 1sin1cos zz z
BtAet t z
ttBeAet
11 22 zzzz
1<z
1=z
0 <z
Summary
• Oscillating systems– 2nd order linear homogeneous D.Es
– Natural frequency
– Damping ratio
– Standard form
• Forced oscillations and other interesting applications
We need to be able to solve tfycdt
dyb
dt
yda
2
2
02 22
2
ydt
dy
dt
yd z
Forced Vibration
tfxdt
dx
dt
xd 2
2
2
2 z
2nd Order Linear Nonhomogeneous D.E.
tf
• Solution of
– 2nd Order
– Linear
– Constant coefficient
– Non Homogeneous D.Es
tfycdt
dyb
dt
yda
2
2
• Step 1 - Solve homogeneous problem (Complementary function)–
• Step 2 – Find a particular solution to nonhomogeneous D.E. –
• The general solution
Solution StrategyTwo Step Process
ty pi
tBytAytycf 21
tfycdt
dyb
dt
yda
2
2
tfycdt
dyb
dt
yda
2
2
02
2
ycdt
dyb
dt
yda
tytyty picf
• ycf(t) – The complementary function– We can solve these already
• ypi(t) – A particular integral for
Terminology
tfycdt
dyb
dt
yda
2
2
tfycdt
dyb
dt
yda
2
2
02
2
ycdt
dyb
dt
yda
tytyty picf
Worked Example
22
2
2 tydt
dy
dt
yd
Find the general solution
If y(0)=0 and what is y(t)?0)0( dt
dy
Right hand side f(t)
Trial function Undetermined Coefficients
Table of possible trial functions
Polynomials of order n
011
1 ctctctc nn
nn
011
1 dtdtdtd nn
nn
nddd ,, 10
te tAe A
tort cossin tBtA cossin
BA,teorte t cossin tBetAe t cossin
BA,
What's the method?
tytBytAyty pi 21
•Trial function for particular integral comes from f(t)
•Coefficients come from matching terms in D.E.
tfycdt
dyb
dt
yda
2
2
Method of undetermined coefficients
What about more than one type of RHS?...no problem
texdt
dx
dt
xd t 22 22
2
For (worked) example
Find a particular integral for this D.E
One last worked example
Find a particular integral for
texdt
dx
dt
xd 22
2
2
Note to lecturer …This example is degenerate!
Summary
• We can solve
• Well almost, provided we can crack the problem of degeneracy
tfycdt
dyb
dt
yda
2
2
2nd Order Nonhomogeneous DEsTypes of question
• Given the DE find the general solution
• Given the DE find a particular integral
• Given an initial value problem find the solution
Recap
• Undamped Oscillators (Simple Pendulum)– Undamped
– Damped
– Forced
Oscillator models
T
mg
l
02 22
2
zdt
d
dt
d
- natural frequency
z - damping ratio
Undamped pendulum
T
mg
l
tBtAt sincos
02
2
l
g
dt
dl
g2 0z
Standard representation
02 22
2
zdt
d
dt
d
A=1, B=0
=6
z=0
Solution
z =0
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
tBtAt sincos
l
g2
Small amplitudeoscillations of a pendulum
T
mg
l
Better model including air resistance
02
2
l
g
dt
d
m
k
dt
d
Standard representation
02 22
2
zdt
d
dt
d
Under damped Pendulum ( , =6)
z =0.1
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
10 z
Critically damped Pendulum( , =6)1z
z =1
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
Over damped Pendulum( , =6)1z
z =2
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4
Time
ResonanceUndamped but forced vibration
F cos t
T
mg
l
tm
F
dt
d cos2
2
2
l
g
Resonance – SolutionF cos t
T
mg
l
tm
FtBtAt
cossincos
22
ttm
FtBtAt sin
2sincos
Natural frequency = forcing frequency
No resonance F/m=5,
-1
-0.5
0
0.5
1
1.5
0 5 10 15 20
Time
F/m=5,
-1.5
-1
-0.5
0
0.5
1
1.5
0 5 10 15 20
Time
Resonance
F/m=5,
-10-8-6-4-20246810
0 5 10 15 20
Time
Undamped oscillations
• Notice that the amplitude of the oscillations increases with time…… until something breaks!
ApplicationBroken shock absorber pg 44
x(t)
B=0 Ns/mk=600 Ns/m
m=150 kg
V
tV
m
Fgx
m
k
dt
xd
5
1cos
2
2
What speed does resonance occur?