Block 1A - Lecture 1 Soori

transcript

EPS Mathematics 3

Delivered by: Soori Prashant Kumar

p.k.soori@hw.ac.uk

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!

• 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

Properties of Logs and Exps - Revision

yxyx lnlnln

xnxn lnln

xee xx lnln

yxyx eee

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

First order

linearhomogeneous D.E.

Variable Coefficient

Printed notes – Pg 3

Terminology

nonlinear

Second order

nonhomogeneous D.E.

More terminology

0 ycdt

First orderlinear

homogeneous D.E.

y(0) = 5Initial value

problem

1st order D.Es with constant coefficients

0 cydt

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

108.0k

0 20 40 60 80

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

Second order differential equations

Lets make life easier…consider

•2nd order•linear •constant coefficient•Homogeneous because f(t)=0

General solution

tyBtyA 21

Lets make life even easier

Consider

Solution

0 ycdt

2nd Order D.Es – Theory 1

Suppose is a solution to

Substituting for

02 ttt AeceAbeaA

02 tAecba

Provided A is nonzero.

So solve the D.E means solve

Good old Euclid!

02 tAecba

02 cba

This is called the auxiliary equation

•Two real roots, - and +

•One Repeated root, •Complex conjugate roots - , +=

02 cba

Three possibilities

General Solution – Two real roots, - and +

yda tt BeAety

Two real roots

tsnsAandBareco

BeAety tt

General Solution – Repeated root - 10

02 cba

042 acb

If is a repeated root then,

02 cba

tBtety

• Differentiate twice

• Sub into the D.E.

1 tBedt

0122 ttt BtectBebtaBe

02 cba

• Tidy up

• As is a repeated root,

• We have the result is a solution

0122 ttt BtectBebtaBe

022 tBebatcba

tBtety

General Solution 02

02 cba

• Two real roots

• One repeated root

• Two complex conjugate roots - , +=

tt BeAety

teBtAty

2nd Order D.Es – General Solution

•Two roots, - and +

•Repeated root, •Complex conjugate roots - , +=

02 cba

Three possibilities

42 tAety

General Solution – Complex roots - 10

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

02 cba

qtiBAqtBAety pt sincos

qtDqtCety pt sincos

Summary - General Solution

• Two real roots

• One repeated root

• Two complex conjugate roots - , +=

02 cba

tt BeAety

teBtAty

qtDqtCety pt sincos

Small amplitudeOscillations of a pendulum

sinmgF

Newton’s 2nd law (F=ma)

Solution

0 1 2 3 4

tBtAt sincos

A=1, B=0

Small amplitudeoscillations of a pendulum

Better model including air resistance

Standard representation

Graphical Solution including air resistance

z =0.1

0 1 2 3 4

Graphical Solution Critical damping z=1

0 1 2 3 4

Graphical Solution Over damping 1<z

- natural frequency

z - damping ratio

Why these parameters are interesting will be considered in L8.

Damped systems Analytical solutions

tBtAet t 22 1sin1cos zz z

BtAet t z

ttBeAet

11 22 zzzz

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

Forced Vibration

2nd Order Linear Nonhomogeneous D.E.

• Solution of

– 2nd Order

– Linear

– Constant coefficient

– Non Homogeneous D.Es

tfycdt

• Step 1 - Solve homogeneous problem (Complementary function)–

• Step 2 – Find a particular solution to nonhomogeneous D.E. –

• The general solution

Solution StrategyTwo Step Process

tBytAytycf 21

tfycdt

tytyty picf

• ycf(t) – The complementary function– We can solve these already

• ypi(t) – A particular integral for

Terminology

tfycdt

tytyty picf

Worked Example

2 tydt

Find the general solution

If y(0)=0 and what is y(t)?0)0( dt

Right hand side f(t)

Trial function Undetermined Coefficients

Table of possible trial functions

Polynomials of order n

1 ctctctc nn

1 dtdtdtd nn

nddd ,, 10

te tAe A

tort cossin tBtA cossin

BA,teorte t cossin tBetAe t cossin

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

Method of undetermined coefficients

What about more than one type of RHS?...no problem

xd t 22 22

For (worked) example

Find a particular integral for this D.E

One last worked example

Find a particular integral for

Note to lecturer …This example is degenerate!

Summary

• We can solve

• Well almost, provided we can crack the problem of degeneracy

tfycdt

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

• Undamped Oscillators (Simple Pendulum)– Undamped

– Damped

– Forced

Oscillator models

- natural frequency

z - damping ratio

Undamped pendulum

tBtAt sincos

A=1, B=0

Solution

0 1 2 3 4

tBtAt sincos

Better model including air resistance

Under damped Pendulum ( , =6)

z =0.1

0 1 2 3 4

Critically damped Pendulum( , =6)1z

0 1 2 3 4

Over damped Pendulum( , =6)1z

0 1 2 3 4

ResonanceUndamped but forced vibration

F cos t

d cos2

Resonance – SolutionF cos t

FtBtAt

cossincos

FtBtAt sin

2sincos

Natural frequency = forcing frequency

No resonance F/m=5,

0 5 10 15 20

F/m=5,

0 5 10 15 20

Resonance

F/m=5,

-10-8-6-4-20246810

0 5 10 15 20

Undamped oscillations

• Notice that the amplitude of the oscillations increases with time…… until something breaks!

ApplicationBroken shock absorber pg 44

B=0 Ns/mk=600 Ns/m

m=150 kg

What speed does resonance occur?

Block 1A - Lecture 1 Soori

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