Lecture 1 Review Basic Concept

8/8/2019 Lecture 1 Review Basic Concept

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APPLIED MATHEMATICS ANDMODELING FOR CHEMICAL

ENGINEERS

Lecture 1: ODE REVIEW


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FORMULATION OF PHYSICOCHEMICAL

PROBLEMS

Step 1.Fuzzy Logic Drawing a Picture

Step 2: Bringing together of all applicable

physical and chemical information, conservationlaws, and rate expressions

Step 3. Setting down of finite or differential

volume elements, followed by writing the

conservation laws.

Step 4. Apply Mathematical Solution Method

2


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Real-World Problem

PhysicochemicalProcesses

Mathematical Model

Solution

Interpretation of Results

Revision of

Model

(if necessary)

Differential equations arisein many engineering

problems as mathematical

models of various physical

systems.

3


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1.1 REVIEW OF MATHEMATICAL BASIC CONCEPT

Functions and Equations

Let f(x) name ofa function

x=anumber or some other entity

Representation of a function analytic expression of afunction to find the value of a given function

Differential Equationsa relationbetweena functionand its derivatives of various orders; equations

containing derivatives or differentials of one ormore variables

4


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Definition of Terms

Derivative

The derivative of a function represents an infinitesimal change in the

function with respect to one of its variables;

Measurement of how a function changes when the values of its inputs

change.

How much a quantity is changing at some given point.

Orderthe order of the highest-order derivative that appears in the equation

Degree - the power to which the highest-order derivative is raised, in a

differential equation. A linear differential equation has degree 1.

Dependent variable variables that denote values of a function; derivative of

the variable occurs

Independent variable may take on any value in the domain of the functions

which the dependent variable stands for; equation involves one or more

derivatives with respect to a particular independent variable5


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1.2 CLASSIFICATION OF DIFFERENTIAL EQUATION: (ACCORDING TO

NUMBER OF INDEPENDENT VARIABLE AND KIND OF DERIVATIVES

THEY INVOLVE)

1. Ordinary Differential Eqution,ODE contains one or severalderivatives of an unknownfunction (dependent variable)with respect to a single

independent variable

2. Partial Differential Equation,PDE containat least one

partialderivative of somedependent variable; involves anunknown function of two ormore independent variablesand its partialderivatives;

" 2 ' cos

dependent variable

independent variable

y y y x

y

x

!

!

!

2 2

2 2

dependent variable

, , independent variable

u u u

tx y

u

x y t

x x x !

xx x

!

!

6


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1.3 CLASSIFICATIONOF ODE(ACCORDING TO THE WAY IN WHICH A DEPENDENT VARIABLEAND ITS

INDICATED DERIVATIVE APPEAR)

a. Linear ODEs

Linear ina set of one or more of its dependentvariables if and only if each term of the equation

which contains a variable of the set or any oftheir derivatives is of the first degree in thosevariables and their derivatives

There are no multiplications among dependentvariables and their derivatives, that is, allcoefficients are functions of independent variables.

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1 '0 1 1

0

...

( ) 0 t r ugh ut s m i t r l

n nn na x y a x y a x y a x y f x

a x

!

{

" 4 ' 2 cos y xy y x !

8

Example:

Linear differential equation: A differential equation in

which the highest-order derivative is not raised to a

power, but is simply multipliedby a constant. For instance

22

2

22

d y dyx

dxdy

!

General form of the First Order DIFEQ


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2

2si

4 2 cos

u vu v t

tx

y yy y x

x x !xx

dd d !

9

b. Nonlinear ODEs not linear in some dependent

variable; not linear in the set of all of its dependentvariables

Examples:


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EXERCISES: IDENTIFY THE EQUATIONS

WHETHERODE ORPDE, LINEAROR

NONLINEAR

10

''' '' '

'

2

2

2

1. 6 11 6

2. 0

3.

4.

5.

x y y y y e

d xy

xydx

u uu

tx

uu

x y x y

x y d y x y dx

!

!

x x!

xx

x x x !

x x x x

!


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1.4 CONCEPT OF SOLUTION

Solution of a given first-order equation on

some open interval a


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TYPES OF SOLUTION

Implicit solution:

Explicit Solution:

General Solution: anonempty set of solutionsspecifiedby an expression which contains atleast one parameter; function involving anarbitrary constant, c

Particular Solution: each individual solution of adifferential equation; the solution when wechoose a specific value of c

, 0, an implicit functionH x y !

y h x!

12


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1.5 FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS

Linear in y and y

Occurs frequently inmodels

Solvable

Involve only the first derivative of the

unknown function,y, andmay containyand

given functions of x.

14

, , 0 o

,

F x y y

y f x y

d!

d!

2 3

2

1. cos

si 12. 4 6 cos

1

y x

x y x y y y

x x

d!

d d !

Example:


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15


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LINEAR ODE

Standard Linear form:

16

xqyxpynotxqyxpy

!

!

'

'

:t i


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Problem 1.General Solution Variable Separable

12 2 2variabl s

1001 0.01 10

Separating tranf rmdy dydx dx

y yp ! p !

3

1 .11

General

Solution

cy x c c

p ! !

18

21 0.01y yd!

Note: Introduce the constant of integration immediately when the

integration is performed.

2

si s100

Integrating

bothp

101

10 a

rctan

10

y

x c

!

0r t 0.

0

simplif

divide

p !


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Problem 2: Exponential Growth or Decay

ywhen , , ln ln =y

y y y y y

y

d dd d " ! !

yh n , lny

y ydd

"

19

y kyd!

1 Se arable dykdxyp !

3 ln , ByCalculus

yy

y

ddp !


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4

General kx

Solution

y c

y c

c

y ce

c e

c e

y

"

!

p !

p !

p !

p |

%

%

2ln

Integrate y kx cp ! %

20

3,

Takin kx c kx c

Ex p nnetialsy e e e ! !% %


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INITIAL VALUE PROBLEM

Differential equation with together with an initial

condition:

given values: x0andy0

Initial condition: 0 0y x y!

21

0 0, , y f x y y x yd! !


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PROBLEM 3: INITIAL VALUE PROBLEM

22

22 1 , 0 1 6xe y x yd! !


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1.7A REDUCTION TO SEPARABLE FORM

Case 1. Differential Equation of the form:

Case 2: Transformations

23

yy g

x

d!

v ay bx k!


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Problem 4: Case 1-Initial Value problems

24

4 23 cos , y 1 0,y y

xy y x ux x

d! ! !


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PROBLEM 5: CASE 2

25

1 2 4

1 2

Hi t: Us 2

y xy

y xy x Y

d!

!


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1.7B MODELING: SEPARABLE EQUATIONS

Problem 6: Newtons Law of Cooling

A thermometer, reading 5C, is brought into a

room

whose tem

pera

ture is 22C.

On

em

in

utelater the thermometer reading is 12C. How long

does it take until the reading is practically 22C,

say 21.9C?

26


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Problem 7: Radiocarbondating.

What shouldbe the 6C14 content (in percent ofy0) of a fossilized tree that is claimed to be

3000 years old?

27


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1.8 EXACT FIRST-ORDER DIFFERENTIAL

EQUATIONS:

INTEGRATING FACTORS

Differential Form:

: ,Function u x y!

28

, ,M x y dx N x y dy !

u u

du dx dyx y

x x! x x

1 ca be wr e Eqdu p !

(1)

By i tegration

Ge eral Solution:

p ,u x y c!

(2)

(3)


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Comparing Eq. (1) and (2):

2 2

,M N

y y x x x y

x x x x! !

x x x x x x

M N

y x

x x!

x x

29

,

u u

M Nx y

x x

! !x x (4)

Test for Exactness:

Note: M and N have continuous first partialderivative:

(5)

From Eq(4):

u Mdx k y! u Ndy l y!

Constants of int gration:

;

uk

y

ul

x

x!

x

x!

x


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PROBLEM 1: AN EXACT EQUATION

30

22 0xydx x dy !


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PROBLEM 2: TEST FOR EXACTNESS - IVP

31

2sin 2 sinh cos 2 cosh 0, 0 1x yd x x ydy y ! !


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1.9 REDUCTION TO EXACT FORM.

INTEGRATINGFACTORS, F

Givennonexact equation:

0FPdx FQdy !

y y x x F P FP F Q FQ !

32

, , 0P x y d x Q x y dy !

Multiply Eq.(6) by F = F(x,y):

(6)

Exact Equation: (7)

Exactn

ess Cond

ition

: FP F Qy xx x

!x x (8)

By the product rule: (9)


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Golden Rule: Look for Fthat depend only onone

variable.

y x FP F Q FQd!

33

, 0,

substitute in q.(9)x

Let F F x F y

dFF F

dx

! !

d! !

Dividing by FQ and rearranging:

1 1dF P Q F dx Q y x

x x! x x

(10

)

(11)


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Theorem 1: Integrating factor F(x)If (6) is such that the right side of (11), R, depends only on x, then (6)has an integrating

factor F =F(x), which is obtained by integrating (11) and taking exponential on bothsides.

34

expF x R x dx! (12)

:If F F y! 1 1dF Q

F d y P x y

!

x x

(13)

Theorem 2: Integrating factor F(y)If (6) is such that the right side of (13 ), depends only on y, then (6)has an integrating

factor F =F(y), which is obtained from (13):R%

expF y R y y! %


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PROBLEM 4: FIND AN INTEGRATING

FACTOR

35

2cosh cos sinh sin x ydx x ydy!


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Alternative standard form for homogeneous,

first-order differential equation:

37

dy ygdx x !

The theoryofthesubstitution ,or /

or

orSubstitution

y ux u y x

y ux

dy duu x dy udx xdu

dx

dx

duu x g u xdu g u u dx

dx

! !

!

! !

p ! ! -

l s i e ticll ,

0, t e e ati is se ara le at t e tset

dy yI f u u

dx x

u

! !

|

du dx

u u x!


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1.11 LINEAR FIRST-ORDER DIFFERENTIAL

EQUATION

cannot contain products, powers, or othernonlinear combination ofyory

Divide the equationby F(x) and rename coefficients:

38

dy

F x G x y H xdx

!

dy

p x y r xdx

!Linear in the unknown functionyand itsderivativey, whereaspandrmay be any givenfunction of x.

If r(x) is zero for allxin the interval, the equation is homogeneous.

0y p x yd !


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Separating variables:

( hen 0)

: 0 and 0 rivial solution

cc e y

Note c y x

! s "

! | p

39

( ) , ldy p x dx y p x dx cy! !

p x dx y x ce!

Taking exponentials onboth sides


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dy p x y r xdx

!

40

Nonhomogeneous equation

General Solution:

h h

y x e e rdx c

h p x dx

! -

!


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2ND ORDER LINEAR HOMOG ODE

SLinear in y, y, y

Solution method: find y1 and y2 independent

solutions

Independent solutions: y2 is not a constant

multiple of y1 nor is y1 is not a constant

multiple of y2

(y1=0, y2 nonzero)

41

nonzeroyyIf

cyy

cyy

yy

21

21

22

,0!

{

{

ddd

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Lecture 1 Review Basic Concept