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Systems of Differential Equations

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(1)

Systems of Differential Equations

Systems of differential equations have important applications as they arise quite frequently as models of many engineering

problems.

Solution by elimination

Successive elimination of unknown functions and its derivatives until it reduces a single higher order differential equation

containing one function and its derivatives

Solve the above single equation and then the other unknown functions are found in turn

(2)

. 1 1 1( ) (1) . 2 2 2( ) (2)

1 2 . , , , 1 2 1 2 tan .

1: Re (1)

General Method of Elimination to the Systems of the form

x a x b y f t y a x b y f t

where f and f are given functions a a b b are cons ts

SOLUTION

Step arrange eq

  

  

1 (. ) (3)

1 1

1

.. . . .

(4) 1 1 1

.. . .

( ) 1 (5)

1 1 2 2 2

2: (1)

3: (2) (4)

y x a x f b

x a x b y f

x a x b a x b y f f

for y

Step Differentiate eq

Step Substitute eq in eq

  

  

    

(3)

General Method of Elimination...continued

.. . . .

( ) 1 (6)

1 1 2 2 1 1 1 2

.. .

( ) ( ) (7)

1 2 1 2 2 1

4: (3) (5)

5: (7)

6: sin

x a x b a x b x a x f b f f x a b x a b a b x r

Step Substitute eq in eq

Step Solve eq for x Step Solve for y u g

      

    

(3) eq

(4)

Example 1:

. .

. (1) . (2)

1: Re (1)

. (3)

2: (1)

.. . (4)

3: (3

x y y x

x y y x

Step arrange eq for y y x

Step Differentiate eq x y

Step Substitute eq

Solution

Solve the following systems of differential equations

) (4) .. (5)

in eq x x

(5)

Example 1: Continued…

4: (5)

.. 0

( 2 1) 0 1

1 2

5: (2) .

1 2

1 2

Step Solve eq for x x x

D x

D t t

x c e c e

Step Solve eq for y

t t

y c e c e

x x

y c e c e

 

 

  

 

  

  

(6)

3 4

4 3

. 3 4 (1) . 4 3 (2)

1: Re (1) 4 . 3 (3)

2: (1)

.. 3 . 4 (4) .

Solve the following systems of differential equations

Dx x y

Dy x y

Solution

x x y

y x y

Step arrange eq for y y x x

Step Differentiate eq

x x y

Ste

   

 

 

 

 

3: (2) (4) .. 3 16 12 (5) .

p Substitute eq in eq x   x xy

Example 2

(7)

4: (2) (4)

. .

.. 3 16 3 9 (5) 5: (5) .. 25 0

( 2 25) 0 5

5 5

1 2

6: (3) . 3

4 5 5

2 1 8 2 4

Step Substitute eq in eq

x x x x x

Step Solve eq for x

x x

D x

D

t t

x c e c e

Step Solve eq for x y x x

t t

c e c e y

   

 

 

 

 

 

 

(8)

NUMERICAL

METHODS

(9)

Analytical Methods:

Analytical solution provides excellent insight into the behavior of systems.

Analytical solution can be derived for only a limited class of problems

linear models and those that have simple geometry and dimensionality

Analytical solutions are of limited practical value because most real life problems are non-linear and complex shapes and processes

Today, computers and numerical methods provide an alternative for

such complicated calculations

(10)

Numerical Methods:

Numerical methods are techniques by which mathematical problems are formulated so that they can be solved with arithmetic operations.

They invariably involve large number of tedious arithmetic calculations.

With the development of fast, efficient digital computers, engineering problem solving by numerical methods has increased drastically in recent years.

For effective use of numerical methods understanding

the concept of error is very important.

(11)

Error Definitions

The numerical result is an approximate value of the exact result.

Numerical errors arise from the use of approximations to represent exact mathematical operations and quantities.

The relationship between the true value and the actual value (approximation) can be defined as

 

 

( )

100 true value actual value approximation error

is also called the

error relative error

true value

error percentage relative error r true value

absolute error

 

 

(12)

Notation Represent (true value) Approximate (actu

al value) Error (true valu e – actual valu e)

1/7 0.142 857 142857 142857 … 0.142 857 0.000 000 142857 14

2857 …

ln 2 0.693 147 180 559 945 309 41... 0.693 147 0.000 000 180 559 94 5 309 41…

log10 2 0.301 029 995 663 981 195 21... 0.3010 0.000 029 995 663 981 195 21...

∛ 2 1.259 921 049 894 873 164 76... 1.25992 0.000 001 049 894 873 164 76...

√ 2 1.414 213 562 373 095 048 80... 1.41421 0.000 003 562 373 095 048 80…

e 2.718 281 828 459 045 235 36... 2.718 281 828 459 045 0.000 000 000 000 00 0 235 36...

(13)

Types of errors

1. Experimental errors

errors of data arising from measurements 2. Truncation errors

errors due to chopping off the remaining digits e.g.,

Consider the real numbers 5.6341432543653654 32.438191288

6.3444444444444

To truncate these numbers to 4 decimal digits, we only consider the 4 digits to the right of the decimal point.

The result would be:

5.6341 32.4381 6.3444

(14)

3. Round-off error (rounding error)

errors due to rounding off during computation Rule for rounding off a number to k decimals

1. Decide which is the last digit to keep.

2. Increase it by 1 if the next digit is 6 or more (this is called rounding up) 3. Leave it the same if the next digit is 4 or less (this is called rounding down) 4. Increase it by 1 if the next digit is a 5 followed by one or more non-zero

digits.

5. Round up or down to the nearest even digit if the next digit is a five followed (if followed at all) only by zeroes. That is, increase the rounded digit if it is currently odd; leave it if it is already even.

Example: 3.046 rounded to hundredths is 3.05 (because the next digit [6] is 5 or more)

(15)

Examples:

3.016 is rounded to hundredth is 3.02 (because the next digit (6) is 6 or more

3.013 rounded to hundredths is 3.01 (because the next digit (3) is 4 or less

3.015 rounded to hundredths is 3.02

(because the next digit is 5, and the hundredth digit (1) is odd)

3.045 rounded to hundredths is 3.04 (because the next digit (4) is even)

3.04501 rounded to hundredths is 3.05

(because the next digit is 5, but it is followed by non-zero digits)

(16)

• In decimal notation, every real number is represented by a finite or infin ite sequence of decimal digits

• For machine computation the number must be replaced by a number of f initely many digits

In digital computers the numbers are stored in two ways

1. Fixed point system: The numbers are represented with a fixed number o f decimal places

e.g., 16.78, 1.678, 0.1678, 1.000

2. Floating point system: The numbers are represented with a fixed number of significant digits

3 5 2

. ., 0.1213 10 , 0.1789 10 , -0.5000 10

e g    

(17)

Rules For Significant Digits

 Digits from 1-9 are always significant

 Zeros between two other significant digits are always significant

 One or more additional zeros to the right of both the decimal place and another significant digit are

significant

 Zeros used solely for spacing the decimal point (placeholders) are not significant.

(18)

EXAMPLES

VALUE # OF SIG. DIG. COMMENT

453 3 All non-zero digits are always significant 5057 4 Zeros between 2 sig. dig. are significant 5.00 3 Additional zeros to the right of decimal and a sig. dig. are significant.

0.007 1 Placeholders are not significant

(19)

Multiplying and Dividing

RULE: When multiplying or dividing, your answer may only show as many significant digits as the multiplied or divided measurement showing the least number of significant digits.

Example:

25.47 x 4.50 x 75.75 = 8682.08625 25.47 4 significant digits.

4.50 3 significant digits.

75.75 4 significant digits.

answer can only show 3 significant digits because that is the least number of significant digits in the original problem.

round to the place in order to show only 3 significant digits.

Final answer becomes 8680

(20)

Adding and Subtracting

RULE: When adding or subtracting your answer can only show as many decimal places as the measurement having the fewest number of decimal places.

Example:

13.76 + 24.83 +2.1 = 40.69

2.1 shows the least number of decimal places.

round our answer, 20.69, to one decimal place.

Final answer is 40.7

(21)

Class Exercise

Identify the number of significant digits

1. 0.0005 2. 201.0 3. 6.02 X 105 4. 0.00530 5. 13450

Perform the following calculations and round according to the rule

1. 2.25 + 6

2. 18.640 + 670.445 3. 640 - 627.03 4. 12.09 - 6.7 5. 3.14 x 5.6 6. 0.059 x 6.95 7. 0.003/106 8. 8.5/0.356

(22)

What is a computer?

A computer is an electronic device that executes the instructions in a program.

A computer has four functions:

a. accepts data Input

b. processes data Processing c. produces output Output d. stores results Storage

Hardware: the physical parts of the computer.

Software: the programs (instructions) that tell the computer what to do and designed for end users

(23)

1. Display 2. Motherboard 3. CPU (Microprocessor) 4. Primary storage (RAM) 5. Expansion cards 6. Power supply 7. Optical disc drive 8. Secondary storage (HD) 8. Keyboard

10. Mouse

(24)
(25)

Input: for feeding and transfers information into the memory

Memory: For storing instructions and data

Control unit: Controls input information,

execution of arithmetic operations and output information

Arithmetic unit: Performs arithmetic operations (addition, subtraction, multiplication, division)

Output: For transferring information from memory onto the output sheet.

(26)

CONTROL UNIT

MEMORY

ARITHMERIC UNIT

INPUT OUTPUT

Components of a computer and main functions

(27)

Problem Solving Using Computer

1. Define the problem in full

2. Describe it using a program-oriented language such as Fortran (FORmula TRANslation)

C C++

and feed into the computer

3. Use a compiler which performs an automatic translation of the program into a machine language programs and execute it

(28)

Examples of Real life Problems:

Mathematical models in engineering and science mostly occurs in the form differential equations

1. Ordinary Differential Equations (ODE) Systems with one independent variable e.g.

Consider the reaction A B taking place in a batch reactor.

Then the rate of formation of B is given by

2. Partial Differential Equations (ODE) 3. e.g.

Systems with two or more independent variables e.g. Heat conduction in a slab

dC B kC dtA

2 T T

 

(29)

Finite Difference Methods

•Real life problems results in large sets of simultaneous differential equations, do not have analytical solutions but require the application of numerical techniques

•Finite difference method is the standard method used for the numerical solution of ODE and PDE.

•Finite difference enables to take a differential equation and integrate it

numerically by calculating the values of the function at discrete (finite) number of points

•If a set of finite values is available, such as experimental data, these may be differentiated, or integrated, using the calculus of finite differences.

Numerical differentiation is inherently less accurate

than numerical integration

(30)

Basic Concepts of Finite Difference

In differential calculus, the derivative of a function f(x) is given as

0 0

0 0

0 0

0 0

0 0

0

( ) ( )

( ) '( ) (1)

(1)

( ) ( )

'( )

( ), in t int

[ , ],

x x x

h

f x f x

df x f x lt

dx x x

let h x x

then eq may be approximated by f x h f x

f x lt

h

A function f x which is continuous and differentiable he erval x x can be re

 

 

2 3

0 0 0 0

0 0 0

( )

0 0

( ) ''( ) ( ) '''( )

( ) ( ) ( ) '( )

2! 3!

( ) ( )

... ( )

!

( ) sin

n n

n

n

presented by a Taylor series

x x f x x x f x

f x f x x x f x

x x f x

n R x

where R x is the remainder or truncation error and is represented u g the notation

 

 

1

2 3

0 0 0 0

0 0 0

( )

0 0 1

1

( )

( ) ''( ) ( ) '''( )

( ) ( ) ( ) '( )

2! 3!

( ) ( )

... ( )

! . .

n

n n

n

n

O h

x x f x x x f x

f x f x x x f x

x x f x

n O h

i e error of order h

 

 

(31)

The calculus of finite differences is used in conjunction with a series of discrete values, which can be either

experimental data, such as

3 2 1 1 2 3

or discrete values

y iy iy iy y i iy iy i

of a continuous function ( )

( -3 ) ( -2 ) ( - ) ( ) ( ) ( 2 ) ( 3 ) or by values of a function ( )

( -3 ) ( -2 ) ( - ) ( ) ( ) ( 2 ) ( 3 ) y x

y x h y x h y x h y x y x h y x h y x h f x

f x h f x h f x h f x f x h f x h f x h All these operato

  

  

, lg

rs satisfy distributive commutative and

associative laws of a ebra

(32)

0

2 3

' 0 '' 0 '''

0 0 0 0 0

( ), in t int [ , ]

exp :

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2! 3!

1.

A function f x which is continuous and differentiable he erval x x can be ressed by Taylor seris as

x x x x

f x f x x x f x f x f x

Taylor Series

Series Used

 

0

0

2 3

0

2 3

1 1

1

( )

... ( ) ( )

!

( ) .

1 ... ...

2! 3! ! !

ln(1 ) ... ( 1) ... ( 1)

2 3

2.

n n

n

n

n n

x

n n

n n

x x f x R x

n

where R x is called the remainder

x x x x

e x

n n

x x x x

x x

n n

Maclaurin Series

 

      

     

(33)

- - int

- - - -

-

D differential operator I egral operator

E shift operator

forward difference operator backward difference operator

central difference operator average operat

Linear Symbolic Operators

or

(34)

...

1. :

( ) ( ) '( )

2. :

( ) ( )

int

( )

Symbolic operators continued

Differential operator Dy x dy x y x

dx

Integral operator Iy x x h y x dx

x

the egral operator is equivalent to

Applying symbolic functions to the function y x

 

  

1

the inverse of the differential operator

I D

(35)

...

( ) ( )

2 ( ) ( 2 )

( ) ( )

1 ( ) ( )

( ) ( )

:

3. :

Symbolic operators continued

Ey x y x h E y x y x h E y x n y x nh

E y x y x h E n y x y x nh

The inverse shift operator

Shift operator

 

 

 

  

  

(36)

( )int

2 3

( ) ( ) '( ) ''( ) '''( ) ...

1! 2! 3!

sin

( )

( )

By Expanding the function y x h o a Taylor series about x

h h h

y x h y x y x y x y x

u g the differential

Expression of shift operator E

in terms of differential operator D

     

,

2 2 3 3

( ) ( ) ( ) ( ) ( ) ...

1! 2! 3!

Re

2 2 3 3

( ) 1 ... ( )

1! 2! 3!

operator D to indicate the derivatives of y

h h h

y x h y x Dy x D y x D y x arranging

h h h

y x h

D D D

y x

     

     

(37)

(Continued...)

( )

( )

. . ( ) ( )

( ) ( )

( ) ( )

1

Expression of shift operator E in terms of differential operator D

i e y x h e hD y x Ey x e hD y x

E e hD

y x h e hD y x E e hD

Similarly

 

  

  

(38)

3 2 1 1 2 3

( 3 ) ( 2 ) ( ) ( ) ( ) ( 2 ) ( 3 )

4.

Consider the set of values

y i y i y i y i y i y i y i or the equivalent set

y x h y x h y x h y x y x h y x h y x h The first backward

Backward Finite Differences

     

     

1

( )

( ) ( ) ( )

i y yi i

or

difference of y at i or x is defined as y

y x y x y x h

 

   

(39)

 

   

2 1 1

1 1 2

2 1 2

2

...

sec ( )

i i i i i i

i i i i

i i i i

Continued

y y

y y y y

y y y

o

Backward Finite Differences

The ond backward difference of y at i or x is defined as

y y y y

y

  

  

  

     

2

0

( 1) !

( )! !

( ) ( ) 2 ( ) ( 2 ) general formula of the nth-order

exp

n m

n i i m

m

r

n y

n m m

y x y x y x h y x h

The backward

difference can be ressed as

y

  

     

(40)

( ) ( ) ( )

( )

in t ( )

Re

y x y x y x h

Expression of backward finite difference operator erms of differential operator D

lation between backward finite difference operator and differential operator

   

 

2

(1) ( ) ( ) (2)

(2) (1) ( ) (1 ) ( )

. . (1 )

2 3 3

...

but

y x h e hDy x Substituting in

y x e hD y x i e e hD

h D h D hD

  

   

   

    

(41)

Re

...

,

2 1 2 1 2 2

3 1 3 1 3 3 2 3

.. .

1

lation between backward finite difference operator and differential operator Continued

Similarly

hD hD hD

e e e

hD hD hD hD

e e e e

n e hD n

  

     

   

      

   

(42)

exp sec

2 2 2 3 3 7 4 4 ...

12

3 5

3 3 3 4 4 5 5 ...

2 4

Expansion of the onential terms

and rearrangement yields the following equations for the ond and third

backward difference operators

h D h D h D

h D h D h D

    

    

(43)

(1 ) (1) 1 (2) ln(1 ) (3) ln(1 )

( ) n t

( )

e hD then

e hD hD

Expression of differential operator D i erms of backward finite difference operator

   

  

  

  

 

2 3 4

2 3 4 5

... (4)

2 3 4

.(3) .(4)

... (5)

2 3 4 5

Combining eq with eq hD

 

  

   

   

(44)

( )in t ( )...

2 2 2 3 4 5

3 3

11 5 ... (6)

12 6

Expression of differential operator D erms of backward finite difference operator Continued

h D

h D

Similarly for higher order differential operator

      

n

3 4 5

2 3 4 5

3 7 ... (7)

2 4

.. .

... (8)

2 3 4 5

n n

h D

     

   

    

(45)

Backward difference operators Differential operators 2 2 3 3 ... 2 3 4 ...

2 6 2 3 4

7 11

2 2 2 3 3 4 4 ... 2 2 2 3 4 ...

12 12

3 3 3 3 4 4 2

h D h D

hD hD

h D h D h D h D

h D h D

  

       

         

   5 5 5 ... 3 3 3 3 4 7 5 ...

4 2 4

2 3 4

(1 ) ...

2 3 4

h D h D

n e hD n h D n n

n

      

  

        

(46)

3 2 1 1 2 3

( 3 ) ( 2 ) ( ) ( ) ( ) ( 2 ) ( 3 )

5.

Consider the set of values

y i y i y i y i y i y i y i or the equivalent set

y x h y x h y x h y x y x h y x h y x h The first forward d

Forward Finite Differences

     

     

1

( )

( ) ( ) ( )

i yi yi

or

ifference of y at i or x is defined as y

y x y x h y x

   

(47)

 

   

2 1 1

2 1 1

2 2 1

2

...

=

2

sec ( )

( ) ( 2 ) 2 ( ) ( )

i i i i i i

i i i i

i i i i

Forward Finite Differences Continued

y y

y y y y

y y y

or

The ond forward difference of y at i or x is defined as

y y y y

y

y x y x h y x h y x

  

 

  

     

     

(48)

m n y i m m

n n n

m i m n y

y i y i

y i y i

y i y i

y i y i

y i y i

y i

as ressed be

can difference

finite forward

order nth

the of formula general

The

derived

similarly Higher order forward difference s are

 

  

 

 

 

 

 

 

 

! )!

( !

0 1 ) (

4 1 6 2

4 3 4 4

3 1 3 2

3 3

exp

(49)

( ) ( ) ( ) (1

( )

in t ( )

Re

y x y x h y x

Expression of forward finite difference operator erms of differential operator D

lation between forward finite difference operator and differential operator

   

2

) ( ) ( ) (2)

(2) (1) ( ) ( 1) ( )

. . 1

2 3 3

...

2! 3!

but

y x h ehDy x Substituting in

y x ehD y x i e ehD

h D h D hD

 

  

  

    

(50)

Re

...

,

2 1 2 2 2 1

3 13 3 3 2 3 1

.. .

1

lation between forward finite difference operator and differential operator Continued

Similarly

hD hD hD

e e e

hD hD hD hD

e e e e

n ehD n

     

      

  

(51)

2 3 4 5

1 (1)

ln ln(1 ) (2)

ln(1 )

2 3 4 5

( ) n t

( )

ehD then

ehD hD but

Expression of differential operator D i erms of forward finite difference operator

 

  

   

     

 

 

2 3 4 5

... (3) .(2) .(3)

... (4)

2 3 4 5

Combining eq with eq hD

   

    

(52)

( )in t ( )...

2 2 2 3 4 5

3 3

11 5 ... (5)

12 6

Expression of differential operator D erms of forward finite difference operator Continued

h D

h D

Similarly for higher order differential operator

       

n

3 4 5

2 3 4 5

3 7 ... (6)

2 4

.. .

... (7)

2 3 4 5

n n

h D

     

   

     

(53)

Forward difference operators Differential operators 2 2 3 3 ... 2 3 4 ...

2 6 2 3 4

7 11

2 2 2 3 3 4 4 ... 2 2 2 3 4 ...

12 12

3 5

3 3 3 4 4

2

h D h D

hD hD

h D h D h D h D

h D h D

   

  

      

      

   5 5 ... 3 3 3 3 4 7 5 ...

4 2 4

2 3 4

( 1) ...

2 3 4

h D h D

n e hD n h D n n

n

  

   

  

      

(54)

Form a finite difference table for the following function x = 0.6 0.7 0.8 0.9 1.0

f(x) = 5.9072 6.0092 6.35552 6.9992 8.0000

참조

관련 문서

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