Maths Material From Bits

-1-

Numerical Differentiation & Integration

Study Material for MATH ZC232

ENGINEERING MATHEMATICS II

1. Introduction :

The aim of this chapter is to enable the reader to understand numerical

integration. Numerical integration is needed to evaluate the integral of a

function given in data form as

x x0 x1 x2 … xn

Value of f=f(x) f0 f1 f2 … fn

or 1

0 2

x

dxx1

sinxe, which can not be evaluated by other methods.

Basis of numerical integration of b

adxxf is to replace the function xf

by a polynomial nn

2210 xa...xaxaaxp , which satisfies the

conditions ni ...2,1,0,xpxf ii . This polynomial is known as

interpolating polynomial.

-2-

2. Interpolation :

A polynomial xp is said to interpolate f(x) at 1n,...xx,x n10 distinct

points if ii xfxp

for i = 0,1,2…n. It is shown that for given such

data, there exists unique polymenial of degree n. The value of ixf is

denoted by if .

Suppose the function is given in data form as

x x0 x1 x2 x3 … xn

f(x) f0 f1 f2 f3 … fn

Let xp = nn

2210 xa....xaxaa

be such a polynomial. To satisfy

the above data, we must have iinin

2i2i10 fxpxa....xaxaa

i = 0,1,2…n. (n+1) conditions. These are (n+1) linear equations in

n10 ...aa,a , (n+1) unknowns. These equations can be solved. The solution

of these equations exists and it is unique for n10 ...xx,x distinct values.

Solving these equations enhance too much work, therefore there are some

other techniques to get the same polynomial. Depending on the techniques

used, forms of the interpolating polynomial are different and known

-3-

accordingly as Langrangian, Newton’s devided difference, Newtons

forward difference form and so on.

First existence and unequencess are shown with the help of Lagrangian

polynomial.

2.1 Lagrangian Inlerpolating polynomial

Let n10 ...xx,x be (n+1) distinct points. Define (n+1) polynomials

ni1ii1iioi

n1i1i0i xx...xxxx...xx

xx...xxxx...xxxl i = 0,1,2…n.

Each of these polynomial is of degree n. These polynomials are called

Lagrangian polynomials for the points n10 ...xx,x .

Consider the polynomial ,xl...fxlf...xlfxlfxp nnii1100

which is sum of polynomials of degree n multiplied by some constant.

Note ijfor

ijfor

1

0xl ji

Now inniiii11i00i xl...fxlf...xlfxlfxp

= 0 + 0 + if + 0.

= if

-4-

Which is true for i = 0,1,2…n. Also xp is polynomial of degree n .

Hence xp is interpolating polynomial of xf . Let us denote xp by

xpn . This shows existence of interpolating polynomial.

Now to prove uniqueness, let xpn and xq n interpolate xf at

n10 ...xx,x , (n+1) distinct points. Therefore iin fxp , iin fxq ,

i = 0,1,2…n.

Consider xqxpxh nn . xh is polynomial of degree n

because

xpn and xq n are so. Also 0ffxqxpxh iiinini

i = 0,1,2,3…n. This shows that xh vanishes at (n+1) distinct points. A

polynomial of degree n has (n+1) zeros only if

xh 0

xpn

xq n .

Hence interpolating polynomial is unique. The interpolating polynomial

n

0iiin xlfxp is called Lagrangian form of interpolation.

Error in interpolation, E(x) = f(x) - pn(x)

= (x – x0) (x – x1)…(x – xn) !1n

?f 1n

.

-5-

for some suitable n10 ...xx, xcontaining ba,? .

2.2 Newton’s Forward Interpolating Polynomial:

For deriving Newton-Cote’s formula of integration to be discussed later,

Newton’s forward form of interpolating polynomial is required. This

Newton’s form is suitable when abscissas (points) n10 x,...,x,x are

equidistant.

Let n210 x,...,xx,x be ( n+1) equidistant points. Further let

hx-x i1i for i = 0,1,2…(n-1).

Forward differences are defined as :

Zero order ii0 ff? ,

First order i1ii1 fff?

Second order i1ii2 ?f?ff?

i1i2i ff2f

nth order difference i1-n

in f?f? .

in f? = i

n2-in21in1in f1...fncfncf and so on.

-6-

Forward differences can be computed in the table form conveniently as

below.

x f(x) ?f

f? 2

0x 0f

0?f

1x 1f 02f?

1?f

01-n f?

2x 2f 12f? 0

n f?

(2.1)

11-n f?

22f?

1-nx 1-nf

1-n?f

nx nf

To get the values of a particular column, values of preceeding column are

subtracted as 010 ff?f , 0102 ?f?ff? , 0

21

20

3 f?f?f?

and

so on. Newton’s forward form of interpolating polynomial is

shxpxp 0nn

-7-

= 0

n0

200 f?

1.2.3...n

1ns...1ss...f?

2

1ssfsf .

= n

0i0

if?i

s (2.2)

Where h

xxs 0 .

Error term in this interpolating polynomial form = 1)n1n fh1n

s

(2.3)

Example 2.1 Generate the forward difference table and find interpolating

polynomial for the data :

x 0 .2 .4 .6 .8

f .12 .46 .74 .9 1.2

Hence interpolate the value of f(0.1).

Forward difference table

-8-

x f ?f

f? 2 f? 3 f? 4

0x 0 .12

0.34?f0

1x .2 .46 -0.06

0.28?f1 -0.06

2x .4 .76 -0.12 0.32

0.16?f2

.26

3x .6 .90 0.14

0.30?f3

4x .8 1.2

The values at the top row are of 0x , 0f , 0?f , 02f? , 0

3f? , 04f? . We get a

polynomial of degree 4 for the data given at 5 points with h = 0.2.

For interpolation at 0.1, x = 0.1, so 2

1

2.

01.00

h

xxs .

Therefore, bionomial coefficients are

8

1

21

12

1

2

1

2

s,

2

11

s1,

0

s

16

1

1.2.3 2

1

2

1

1.2.3

2s1ss3

s 23

-9-

128

5

1.2.3.42

1

2

1

1.2.3.4

3s2s1ss4

s 25

23

Using the polynomial

04

03

02

0004 f?4

sf?

3

sf?

2

sfsfshxp

0.32128

50.06

16

10.26-

8

10.34

2

10.12 .

28125.00.1p4

Hence f(0.1) 0.28125.

Example 2.2 prepare the forward difference table for the data

x -1 0 1 2 3

f 10 2 0 10 62

Using Newton’s forward interpolating polynomial, find approximate

value of f(-0.5).

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Forward difference table

x f ?f

f? 2 f? 3

-1 10

-8

0 2 6

-2 6

1 0 12 24

10 30

2 10 42

52

3 62

h = 1, x = -0.5 , x0 = -1

2

1

1

10.5

h

x-xs 0

As in the above example

128

5

4

s,

16

1

3

s,

8

1

2

s,

2

1

1

s1,

0

s.

04

03

02

00044 f?4

sf?

3

sf?

2

sf

1fshxp

spx

24 128

5-6

16

16

8

18-

2

110

2

1p4 .

16

75

16

1561296

16

15

8

3

4

3410

= 4.8125.

-11-

8125.45.0p4 ,

Therefore f(-0.5) 4.8125.

Exercise 2.3. Prepare the table of forward differences for the data

x -4 -2 0 2 4

f(x) 174 4 2 24 310

Using Newtonls forward interpolating polynomial, find the approximate

value of f(-3).

Exercise 2.4. Prepare the table of forward differences for the data

x 0.2 0.3 0.4 0.5

F(x) 0.848 0.817 0.824 0.875

and hence using Newton’s forward interpolating polynomial approximate

f(0.25).

3. Numerical Differentiation:

Numerical differentiation is required when functions is given in data form

as

-12-

x x0 x1 x2 x3 … xn

f(x) f0 f1 f2 f3 … fn

Simple formulas of first order and second order derivatives are also

required for solving partial differential equation later in the course. Some

of formulas can be derived easily. Taylorl’s series of a real function f(x) of

single variable x, is f(x+h) = f (x) +h f (x) +!2

2h fh (x)+…

Where f(x) and its all derivatives are continuous in the nbd of x.

Now suppose derivative is required at xi, when values of x are given with

spacing h.

...xf2

hxfhxfhxf i

2

iii

(A)

h

xfhxfxf ii

i by neglecting higher order terms of h.

h

fff i1i

i

(3.1)

similarly ...xf2

hxfhxfhxf i

2

iii

(B)

h

hxfxfxf ii

i , by neglecting higher order terms.

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h

fff 1-ii

i

(3.2)

On adding (3.1) and (3.2), we get

2h

hxfhxfxf ii

i

2h

fff 1-i1i

i

(3.3)

Adding (A) and (B), we get

...xfhxf2hxfhxf i2

iii

2iii

i h

hxfx2fhxfxf

21ii1i

i h

f2fff

(3.4)

by neglecting 3h and higher order terms.

If we want to use more data values to find the derivative of a function at a

point, then we can use

h

1

dx

ds sh.,x x ,xp _~xf 0n

0n

02

00n f?n

s...f?

2

sf

1p

sfx (D)

ds

dp

h

1

dx

ds

ds

dp

dx

df nn (3.5)

-14-

similarly we can find higher order derivatives by differentiating xpn in

(D) further.

Example 3.1. write first derivative of f(x) at x=0.1,0.2,0.3 where f(x) is

given by

x 0.1 0.2 0.3 0.4 0.5 0.6

f(x) 0.425 0.475 0.400 0.450 0.525 0.675

Using (3.1)

.05.0.1

0.050

0.1

.42500.475

h

0.1f0.2f1f

.75.00.1

0.075-

0.1

.47500.400

h

0.2f0.3f.2f

5..00.1

0.05

0.1

.40000.450

h

0.3f0.4f.3f

Example 3.2. Find the 2nd derivative xf

at 0.3,0.4,0.5 for the function

given in example 3.1 above.

Using (3.4)

12.5.01

0.125

0.01h

0.2f.3f 20.4f0.3f

2

2.50.01

0.25

0.01h

0.3f.4f 20.5f0.4f

2

-15-

5.70.01

0.075

01.0

4.0.5f 20.6f0.5f

f.

4. Numerical Integration

Methods of numerical integration can be classified in two types. First type

known as Newton cote’s formulas are derived when abscissas are already

fixed for interpolating polynomial used which are discussed in sections

4,5. Second type of formulas is known as Gaussian Quadrature discussed

in section 6 and for deriving these formulas abscissas are not preassigned

for the polynomial, but are chosen to increase accuracy by using lower

degree polynomial, which is generally nearly half of degree as of that used

in Newton Cotes.

Newton Cote’s Formulas

To evaluate the integral b

axf ,dx xf is replaced by an interpolating

polynomial, which interpolates f(x) at some points n10 ...xxx spread over

the interval ba, . In many cases ax 0 and bx n .

The general principle of numerical integration of b

a dx,xf

b

a

b

a n dxxp -~dx xf is

-16-

b

a nni100 dxxlf...xlfxlf

nn1100 Af...AfAf . from (2.1)

Where n

0ii xl are Lagrangian polynomials, for the points .,...xx,x n10

and b

a ii 0,1,2...n.i dx,xlA

sA1i are called weights.

Error in integration E(I) b

adx,xE

Where E(x) = f(x) – pn(x), error in interpolation.

4.1 Basic Trapezoidal Rule

Suppose b

adxxf is to be evaluated.

f(x) is replaced by a polynomial of degree one, which interpolates f(x) at a

and b. Taking ax 0 , bx1 , h = b-a, the interpolating polynomial of

degree 1 is 001 fsfxp .

Therefore b

a

x

x dx xf dxxf

1

0

1

0

1

0

x

x

x

x 001 dxfsf dxxp ~

-17-

Since 0sshxx xhds,dx sh,xx 000

1sshxhxshxxxx 00011 .

1

0 0000 ?f2

1fhhdsff I s

10010 ff2

hff

2

1fh

Writing ab for h, we have Trapezoidal Rule

b

abfaf

2

a-b -~dxxf (4.1)

This is known as basic Trapezoidal Rule.

Error in the above integration rule is obtained by integrating (2.3), ie

b

a

1n1nT 1n with dx,afh1n

sIE , and some ba,a

b

a

1

0

32 dsaf1ss

2

hdxafh

1.2

1ss

dx = h ds

1

0

23

dsss?f2

h for suitable ba,?

Using mean-value theorem of integrals,

0,1on continous xf and 0,1in sign changenot does ss 2 .

?f12

hIE

3T

-18-

?f12

a-b 3

. (4.2)

This is Error in Basic Trapezoidal rule.

4.2 Composite Trapezoidal Rule

In the above Trapezoidal rule (4.1), if the interval [a,b] is relatively large,

then the error in (4.2) will also be large. Therefore, interval [a,b] is

subdivided into n equivalent subintervals separated by b,...xx,xa n10

hn

abwith , 0,1,2...n.i ih,xx 0i

a b

0x 1x 2x 3x 4x nx .

For Composite Trapezoidal Rule :

b

a

x

x

x

x

x

x

x

x dxxf dxt...xf dxxf dx xf xf

n

0

1

0

2

1

n

1n

applying Trapezoidal rule on each of above sub intervals

n1n2110

x

xff

2

h...ff

2

hff

2

h -~dxxf

n

0

.

n1n210 f2f...2f2ff2

h

(4.3)

and result in (4.3) is denoted by fICT .

-19-

This is known as composite Trapezoidal rule.

Simularly adding the result of (4.2) over such n subintervals.

Error in composite Trapezordal rule

n321

3

?f...?f?f?f12

h

?fn12

h 3

for some suitable [a,b] under the assumption of

continuity of xf on [a,b]. Since hn = b – a.

Error ?f12

hab 2

. (4.4)

Example 4.1 Use composite Trapezoidal rule to evaluate the integral of

the function f(x) as given by

0x 1x 2x 3x 4x 5x

x 0.1 0.2 0.3 0.4 0.5 0.6

f(x) 0.425 0.475 0.400 0.450 0.575 0.675

0f 1f 2f 3f 4f 5f

Using (4.3)

543210CT ff...fff2f

2

hfI , h = 0.1

-20-

.6750.5750.4500.40000.4752.4252

0.1

= 0.2632.

Example 4.2 Evaluate the integrel 2

1 2

2x

dxx1

e using composite

Trapezoidal rule with 6 functional values ie h = 0.2.

For this problem 2 x1.8, x1.6, x1.4, x1.2, x1,x 543210

and 2

2x

x1

exf .

Now applying composite Trapezoid Rule (4.3).

2

1

4

2

3.6

2

3.2

2

2.8

2

2.42

2

2x

5

e

1.81

e

1.61

e

1.41

e

1.21

e2

2

e

2

0.2dx

x1

eI

10.91968.63166.89115.55564.517623.69450.1I

= 6.58059.

Exercise 4.3 Find the area below the curve of y = f(x), above x – axis and

between x = - 1 and x = 3, using compositive Trapezoidal rule when f(x) is

given by

-21-

x -1 -0.5 0 0.5 1 1.5 2 2.5 3

f(x) 7 5 3.5 4 5.5 6 6.5 5 4.5

Exercise 4.4 Use composite Trapezoidal rule to evaluate the integral

2

0

-x

dxx1

e2

, with h = .25

Exercise 4.5 Use composite Trapezoidal rule with values of intgrand at 7

equidistant points to evaluate the integral

0dx

x

sin x 1

0

0sin Note .

Exercise 4.6 2

1

2x 0.05850dxe true value

By trapezoidal = 0.1 [e-2+2 (e-2.4+e-2.8+e-3.2+e-3.6)+e-4]

= 0.05928.

Exercise 4.7 2

1

2x 6045.23dxe true value

By trapezoidal = 0.1 [e2+2 (e2.4+e2.8+e3.2+e3.6)+e4]

= 23.6045.

-22-

5. Simpson’s 3

1 Rule

Trapezoidal rule has been derived by replacing the integrand by a

polynomial of degree 1. Now Simpson’s 3

1 rule is derived using

interpolating polynomial of degree 2. Similarly Simpson’s 8

3 rule can be

derived by using polynomial of degree 3.

5.1 Basic Simpson’s 3

1 rule

Consider b

adxxf ;

a 2

ba b

Interval [a,b] is divided into 2 equal subintervals with 2

abh

and let

b x,2

ba xa,x 210 .

2

0

2

0

x

x 2

b

a

x

xdxxp~dxxf~dxxf

Simpson’s 3

1 rule, denoted by

2

0

x

x 2S dxxpfI

2

0

x

x 02

00 dxf?1.2

1ssfsf from (2.2)

-23-

Where sh

x-x 0 , dx = hds, 20 x xand 0sxx

2s2h xshx 00

Therefore

2

0 02

00S dsh f?

1.2

1ssfsffI

02

00 f?3

1f22fh

012010 f2ff3

12f2f2fh

210 f4ff3

h

bf2

ba4faf

6

a-b (5.1)

This is known as basic Simpson’s 3

1 rule.

Error in basic Simpson’s 3

1 rule requires some more results.

Therefore, this is stated only.

Error in Simpson’s 3

1 rule.

-24-

?f90

hIE IV

5S for some suitable ba,? ,

2

abh

(5.2)

5.2 Composite Simpson’s 3

1 rule

Basic Simpson’s 3

1 rule can be used to evaluate the integral but it gives

more error because values of the function are used only at 3 points,

therefore like compositive Trapezoidal rule, composite Simpson’s 3

1 rule

is derived.

Let [a,b] be divided into 2n equal subintervals ie h,2n

ab

and let

,ihx xa,x oi0 i = 0,1,2…(2n-1), bx 2n .

4

2

2x

2-2n

2

0

2x

0

x

x

x

x

x

x

b

a

x

xdxxf...dxxfdxxfdxxfdxxf .

Applying Simpson’s 3

1 rule for each integral, we get

2n12n2-2n432210CS f4ff...f4fff4ff

3

hfI

2n12n2-2n43210 f4f2f...2f4f2f4ff3

h. (5.3)

This is known as composite Simpson’s 3

1 rule.

-25-

Adding error terms (5.2) for each pair of subintervals.

Error in composite Simpson’s 3

1 rule ?f

180

abh IV4

for some

suitable ba,

abnh2

(5.4)

Example 5.1 Using composite Simpson’s 3

1 rule, evaluate the integral for

the data

x -1 -0.5 0 0.5 1 1.5 2 2.5 3

f(x) 7 5 3.5 4 5.5 6 6.5 5 4.5

For this problem h = 0.5.

Value of integral

(4.5)54(6.5)264(5.5)244(3.5)25473

0.5

3

5.1225.420132411167207

3

5.0

= 40.83.

-26-

Example 5.2 Evaluate the integral 2

0 2

2x

dxx1

e, using composite

Simpson’s 3

1 rule with function values at 9 points ( h = .25).

2 x1.75, x1.5, x1.25, x1, x.75, x.5, x.25, x0,x 876543210

2

0 8765432102

2x

ff 4f 2f 4f 2f 4f 2f 4f3

hdx

x1

e.

2.174608.51

ef 1.55,

.251

ef 1,

01

ef

2

1

22

.5

1

0

0

.75414441.251

ef .694528,3

11

ef ,86828.2

75.1

ef

2

2.5

5

25

42

1.5

3

.91965.1041

ef 1.813209,1

1.751

ef ,1801652.6

5.11

ef

4

82

3..5

72

3

6

After substituting these values

2

0 2

2x

.10-~9.99673x1

e.

Using higher order interpolating polynomials many such other formulas

are obtained.

Exercise 5.3 Find area for data given in exercise 4.3 using composite

Simpson’s rule.

-27-

Exercise 5.4 Evaluate integral 2

0

x

dxx1

e2

, with h = 0.25 using

composite Simpson’s rule.

Exercise 5.5 Evaluate the integral 1

0 2

x

dxx1

sinxe, using composite

Simpson’s rule. Take h = 0.1.

6. Gaussian Quadratures.

Now, some results are discussed, when abscissas are not preassigned but

are to be chosen so that integral rule becomes exact for a polynomial of a

degree as high as possible.

To evaluate the integral b

adxxf where x w,xg xwxf

is

weight function which is non-negative on [a,b].

We consider

b

a

22nnn1100

b

a?agxgA...xgAxgAdxxgxwdxxf

10 A ,A … nA are known as weights.

-28-

In the above result 1010 A ,A ... x,x are unknown and are to be found so

that integration rule.

b

a nn00 xgA...xgAdxxgxw (6.1)

becomes exact for polynomials of degree upto 12n .

In general we take 12n2 x... , x x,1,xg and the (2n+1) equations.

b

a

inn

i11

io0

i xA...xAxAdxxwx .

i = 0,1,2 … (2n+1) , are solved.

The rules of type (6.1) are known as Gaussian Quadratures.

Because it is difficult to solve the above equations for large n

3,

properties of orthogonal polynomials are used to get n10 x... x,x in the

rule (6.1).

To get (n+1) point rule, abscissas ie n10 ,...xx,x are found to be roots of

th1n 1n,xp

degree polynomial from the corresponding set of

orthogonal polynomials on [a,b] with weight function w(x). The order of

-29-

derivative of g(x), is taken 22n

twice the degree of 1nP

in error

terms.

6.1 Two point Gauss-Legendre Quadrature

Let 1

1-

IV1100 ?f a xf Axf Adxxf .

Here weight function 1xw . Forth order of derivative of f in error term

is taken equal to the number of unknowns 1010 A ,A ... x,x . However if

value of 0a , then next order of derivative is taken in the error term.

On substituting , x, x x,1,xf 32 in above we get 4 equations.

10 A A 2

1100 xA xA 0

211

200 xA xA

3

2

311

300 xA xA 0

solution of these equations give.

1 A A ,3

1 x,

3

1x 1010

So 2 point Gauss-Legendre Quadrative is

-30-

3

1f

3

1f-~dx xf

1

1-

(6.1)

To find a in error term ?afIE IV .

Write 4xxf in

dxxf1

1-

?f a3

1f

3

1f IV

4! a9

1

9

1

5

2

135

1a

Therefore

?f135

1IE IV for some suitable 1 1, ?

(6.2)

Sumilarly or using Legendre polynomial of degree 3 i.e

x2

3 x

2

5xp 3

3 , which gives three abscissas.

53 x0, x,

5

3x 210 .

Three point Gauss-Legendre qudrature is

1

1 5

3f

9

50f

9

8

5

3f

9

5dxxf (6.3)

with error term 15750

?fIE

VI

. (6.4)

-31-

6.2 Two Point Gauss-Chebyshev rule

Let 1

1

IV11002

?g axg Axg A dxx1

xg

On substitution we get

32 x, x x,1,xg

10 AAp

1100 xAxA0

211

200 xAxA

2

p

311

300 xAxAO

On solving above equations, we get

2

pAA ,

2

1 x,

2

1x 1010

1

1 2 2

1g

22

1g

2

pdx

x1

xg

(6.5)

To get in the error term, substituting 4x in

-32-

?g a2

1g

22

1g

2

pdx

x1

xg IV1

1 2

We get error term

4!

?g

2

pIE

IV

3

(6.6)

for some 1 1, ? .

Simularly three point Gauss-Chebyshev rule is obtained. Or using

Chebyshev polynomial of degree 3 ie 3x4xxp 33 , which gives three

abscissas 2

3 x0, x,

2

3x 210

Three point Gauss-Chebyshev rule is

1

1 2 2

3g

3

p0g

3

p

2

3g

3

p-~dx

x1

xg (6.7)

with error term 6!

?f

2

pIE

VI

5 (6.8)

for some 1 1, ? .

Example 6.1 Using 3 point Gauss-Legendre quadralure evaluate the

entegral.

1

1 2dx

x1

sin xx .

-33-

Using formula (6.3).

1

1 2 5

3f

9

50f

9

8

5

3-f

9

5 dx

x1

sin xx

Here2x1

sin xx xf

1.599999

.5417744

9

50

9

8

1.599999

.5417744

9

5 ~I

.338609 338609.9

5

= 0.37623.

Example 6.2 Using 3-point Gauss-Chebyshev quadrature, evaluate the

integral.

1

1 2

x

dxx-1

ex1.

Three point Gauss-Qudrature formula is

1

1 2

x

2

3 0g

2

3-g

3

p dx

x-1

ex1g

Here .7574258 2

3g ,ex1 xg x

3.24338 2

3g , 1 0g

-34-

On substituting these values.

1

1 2

x

5.23683 dx x-1

ex1.


1

x dxe2

.

Using

(a) Two point Gauss-Legendre Quadrature.

(b) Three point Gauss-Legendre Quadrature.


1 2

x

dxx-1

e2

.

Using

(a) Two point Gauss-Chebyshev Quadrature.

(b) Three point Gauss-Chebyshev Quadrature.

Note : Gauss-Legendre Quadratures can be used to evaluate g

adxxf by

changing to 1

1dxuF by substitution

2

baabux .

9/22/2010

1

MATH ZC232Engineering Mathematics II

Lecture 1

Dr. Deepmala Agarwal

4/8/2010 MATH ZC232 1

Basics of Laplace TransformDefinition

Let f be a function defined for . Then the integral

denoted by is said to be the Laplace Transform of f provided the integral converges.

Examples• 1

• 2. L(eat) = Exp[-(s – a)t] dt = 1/[s – a) for s a.

0t0

dttfe st

tfL

s

stdteL

sest 11

00

0s

4/8/2010 MATH ZC232 2

• Linearity property: L[f(t) + g(t)] = L[f(t) + L[g(t)]• If we differentiate first example wrt s, on both sides we get

• Again differentiating, we get

• Continuing in this way we get

• Therefore, L(tn) = n ! / s(n+1)

02

0

11s

dtteorsds

ddtedsd stst

03

2 2s

dtte st

10

!n

nst

sndtte

4/8/2010 MATH ZC232 3

Important Formulae

2222

2222

2222

1

)sin()cos(

)cosh()sinh(

)cos()sin(

1...3,2,1,!

basbbteL

basasbteL

kssktL

kskktL

kssktL

kskktL

aseLn

sntL

tata

tan

n

4/8/2010 MATH ZC232 4

Piecewise Continuity• A function f is said to be piecewise continuous

on [0, ), if , in any interval 0 a t b, there are atmost a finite number of points tk , k = 1 to n, at which f has finite discontinuities and continuous on each open interval (tk – 1 , tk).

• Exponential OrderA function f is said to be of exponential order c if there exists a constant c, such that for some M > 0 and T > 0 we have |f(t) M Exp (ct) for all t > T.

4/8/2010 MATH ZC232 5

4/8/2010 MATH ZC232 6

LECTURE 1

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4/8/2010 MATH ZC232 14

Inverse Laplace Transform (ILT)• We need function f(t) given its LT F(s) such that

f(t) is called Inverse Laplace transform of F(s).• In general approach to find inverse LT, we use previous knowledge of LTs of some

important forms of functions.

• To begin with, we use partial fractions. As done in example 1 (p.197)

Using standard results (p. 196)

• Note: Some results are listed on p. 196. Memorize them for use in problems.

sFtfL

430

1

26

25

15

16

421962

ssssssss

ttt eeesss

ssL 422

1

301

625

516

42196

4/8/2010 MATH ZC232 15

Some Formulae for Inverse Transform

221

221

221

221

221

221

11

1

)sin()cos(

)cosh()sinh(

)cos()sin(

1...3,2,1,!

basbLbte

basasLbte

kssLkt

kskLkt

kssLkt

kskLkt

asLen

snLt

tata

tan

n

4/8/2010 MATH ZC232 16

ILT Example

• Find ILT of

on taking ILT, we get

93 2sss

93

61

361

93 22 ss

ssss

963

96361

22 sss

s

ttess

sL t 3sin613cos

61

61

933

21

4/8/2010 MATH ZC232 17

Solution of Initial Value Problems (IVP)

• We need Theorem 4.4 mentioned below.• If are continuous on and are of exponential order

and is piece-wise continuous on then

• Example Solve the IVP using LT

1,,, nffff ,0nf ,0

000 121 nnnnn ffsfssFstfL

196 yyy 00y 10y

196 LyLyLyLs

sYyssYysysYs 1906002

ss

ssYss 111962

22 3331

sC

sB

sA

ssssY

2334

391

91

ssssY tt eety 33

34

91

91On taking ILT

4/8/2010 MATH ZC232 18

LECTURE 1

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4

Transform of Derivatives

)0()0()0()()(

)0()0()()(

)0()()(

'''23'''

'2''

'

ffsfssFstfL

ffssFstfL

fsFstfL

4/8/2010 MATH ZC232 19

Example Problem on IVP

• Using LT, solve the IVP

Sol. On taking LT on both sides

teydt

yd 642

2

10y 40y

164

syLyL 1

64002

ssYysysYs

16442

sssYs

164

41

2 ss

ssY

1644

41 2

2 ssss

ssY

12225

14103

2

2

sssss

sssssY

12

21

125

ssssssYOn taking ILT

tt eety 22

4/8/2010 MATH ZC232 20

4/8/2010 MATH ZC232 21

4/8/2010 MATH ZC232 22

4/8/2010 MATH ZC232 23

4/8/2010 MATH ZC232 24

LECTURE 1

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4/8/2010 MATH ZC232 30

LECTURE 1

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LECTURE 1

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4/8/2010 MATH ZC232 38

First Translation Theorem (4.6, p. 204)

• Consider

Replacing s by s-a on both sides gives

Note: If you know LT of f(t), then you can write LT ofby replacing s by s-a in F(s).

• If and a is any real number then

• Consider the following example

sFtfL asFtfeL at

1!

nn

sntL 1

5

5!

nnt

snteL 6

53

3!5

steL t

sFdttfe st

0

asFdttfee atst

0

tfeat

4/8/2010 MATH ZC232 39

Examples

1. We know that L[t7] = 7! / s8

Therefore, we get L[e- 5t t7] = 7! / (s + 5)8

We also get L[e101t t7] = 7! / (s – 101)8

2. We know that L[sin 4t] = 4 / (s2 +16).Therefore we get L[e- 3t sin 4t] = 4 / [(s +3)2 +16).

4/8/2010 MATH ZC232 40

Unit Step Function or Heaviside Function

The unit step function or Heaviside function U(t-a) is defined to be

atatatU 0,0

,1)(

4/8/2010 MATH ZC232 41

Second Translation Theorem

If F(s) = L{f(t)} and a > 0, then

L{f(t-a) U(t-a)} = e-as F(s)

4/8/2010 MATH ZC232 42

LECTURE 1

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8

)2(421

41 ts ee

sL U(t-2)

)2

(3cos9

2/2

1 tes

sL s U )2

(t

LECTURE 1

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1

6/8/2010 MATH ZC232 1


Lecture 2


6/8/2010 MATH ZC232 1

6/8/2010 MATH ZC232 2

First Translation Theorem (4.6, p. 204)• Consider

Replacing s by s-a on both sides gives

Note: If you know LT of f(t), then you can write LT ofby replacing s by s-a in F(s).

• If and a is any real number then

• Consider the following example

sFtfL asFtfeL at

1!

nn

sntL 1

5

5!

nnt

snteL 6

53

3!5

steL t

sFdttfe st

0

asFdttfee atst

0

tfeat

6/8/2010 MATH ZC232 2

6/8/2010 MATH ZC232 3

Examples

1. We know that L[t7] = 7! / s8

Therefore, we get L[e- 5t t7] = 7! / (s + 5)8

We also get L[e101t t7] = 7! / (s – 101)8

2. We know that L[sin 4t] = 4 / (s2 +16).Therefore we get L[e- 3t sin 4t] = 4 / [(s +3)2 +16).

6/8/2010 MATH ZC232 3

6/8/2010 MATH ZC232 4

Unit Step Function or Heaviside Function

The unit step function or Heaviside function U(t-a) is defined to be

atatatU 0,0

,1)(

6/8/2010 MATH ZC232 4

6/8/2010 MATH ZC232 5

Second Translation Theorem


L{f(t-a) U(t-a)} = e-as F(s)Hence,

U(t-a)

6/8/2010 MATH ZC232 5

)()}({1 atfsFeL as

6/8/2010 MATH ZC232 6

)2(421

41 ts ee

sL U(t-2)

)2

(3cos9

2/2

1 tes

sL s U )2

(t

LECTURE 2

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LECTURE 2

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6/8/2010 MATH ZC232 15

Derivatives of Laplace Transform (4.8, p. 215)

• If is known, then differentiating both sides wrt s gives

Differentiating n times

Note: You can use this result to find LT of if LT of f(t) i.e. F(s) is known.

• Example: Given

Similarly differentiating once more

sFdttfetfL st

0

sFdttfte st

0

sFdttfet nstn

0

sFdttfte nnnst 10

tft n

2555sin 2s

tL2222 25

10251015sin

ss

ssttL

32

2

32

22

42

2222

2532510

2542510

2522251025105sin

ss

sss

sssssttL

6/8/2010 MATH ZC232 15

6/8/2010 MATH ZC232 16

Combined Examples

• Example 1 Find LT of

Sol. From results on p. 193,

Using translation theorem 4.6 (p. 204)

Now using theorem 4.8 (p. 215)

tte t 3sin2

933sin 2s

tL

1343

9233sin 22

2

sssteL t

134313sin 2

2

ssdsdtteL t

22 134126

sss

contd…6/8/2010 MATH ZC232 16

6/8/2010 MATH ZC232 17

Convolution Theorem (4.9)• Convolution Theorem

If f(t) and g(t) are piece-wise continuous on and of exponential order, then

where F(s) and G(s) are LT of f(t) and g(t) respectively.

• Writing in inverse form

• Example Use convolution theorem to find

where

therefore,

,0

duutgufLgfLt

0

sGsF

duutgufsGsFLt

0

1

22

1

11

sL

tftfss

L1

11

122

1

11sinsin 2s

tLttf

tt

duutuduutguftgtf00

sinsin

tttduttut

cossin21cos2cos

21

06/8/2010 MATH ZC232 17

6/8/2010 MATH ZC232 186/8/2010 MATH ZC232 18

LECTURE 2

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6/8/2010 MATH ZC232 226/8/2010 MATH ZC232 22

6/8/2010 MATH ZC232 23

The Dirac Delta Function

• The function characterized by the following two properties is called the Dirac Delta Function

i.

ii.

)( 0tt

0

0

,,00)( tt

tttt

x

dttt0

0 1)(

6/8/2010 MATH ZC232 24

THEOREM 4.11

• The laplace transform of the Dirac delta function for >00t

0)}({ 0stettL

LECTURE 2

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System of Linear Differential Equation

• When the initial conditions are specified, the Laplace transform reduces a system of Linear differential equations with constant coefficients to a set of simultaneous algebraic equations in the transformed functions.

6/8/2010 MATH ZC232 32

Example

• Use the Laplace transform to solve

Subject to

0440410

2"21

21"1

xxxxxx

1)0(,0)0(,1)0(,0)0( '22

'11 xxxx

6/8/2010 MATH ZC232 33

Solution

• Taking the Laplace transform of both equations, we get

whereand

from the initial conditions we get

0)(4)0()0()()(4

0)(4)(10)0()0()(

2'222

21

21'111

2

sXxsxsXssXsXsXxsxsXs

)}({)( 11 txLsX )}({)( 22 txLsX

6/8/2010 MATH ZC232 34

Solving these equations for and using partial fractions we get

1)()4()(4

1)(4)()10(

22

1

212

sXssXsXsXs

)(1 sX

125/6

25/1)( 221 ss

sX

taking Laplace inverse, we get

tttx 32sin532sin

102)(1

(a)

(b)

6/8/2010 MATH ZC232 35

By substituting the value of in (a), we get)(1 sX

125/3

25/2)( 222 ss

sX

hence

tttx 32sin10

32sin52)(2

6/8/2010 MATH ZC232 36

Exercise 4.6

Q.9 Use the Laplace transform to solve

Subject tot

dtyd

dtxd

tdt

yddt

xd

42

2

2

2

22

2

2

2

0)0(',0)0(,0)0(',8)0( yyxx

LECTURE 2

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6/8/2010 MATH ZC232 37

Solution

• Taking the Laplace transform of both equations, we get

whereand

from the initial conditions we get

222

322

4)0(')0()()0(')0()(

2)0(')0()()0(')0()(

sysysYsxsxsXs

sysysYsxsxsXs

)}({)( txLsX )}({)( tyLsY

6/8/2010 MATH ZC232 38

Solving these equations for , we get

sssYsX

sssYsX

84)()(

82)()(

4

5

)(sX

ssssX 821)( 45

taking Laplace inverse, we get

831

241)( 34 tttx

(a)

(b)

6/8/2010 MATH ZC232 39

By substituting the value of in (a), we get)(sX

4521)(ss

sY

hence

34

31

241)( ttty

LECTURE 2

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Lecture 3


9/8/2010 MATH ZC232 1

Power Series

A series of the form

is known as a power series centered at .

)2(...0

2210

n

nn axcaxcaxccxy

a

Example (1):

0

32

!...

!3!21

n

nx

nxxxxe

is a power series about the origin.9/8/2010 MATH ZC232 2

Convergence

A power series of the form (2) is said to converge at aspecified value of x if the limit of the partial sum

nN

nn

NaxcLim

0

exists. The limiting value is called the sum of the powerseries. It is clear that the power series converges atsince

ax

02

210 ... caacaacc

9/8/2010 MATH ZC232 3

Interval of Convergence

Every power series has an interval of convergenceabout . Within this interval, the power seriesconverges and diverges outside the interval.

ax

Example (2):The power series about

................1 32

0

xxxxn

n

Converges for:Diverges for: 1x

0x

),1()1,(x)1,1(x 1xor

or9/8/2010 MATH ZC232 4

Radius of convergence

For every power series, there exist a positive real numberR , called the radius of convergence with the propertythat the series (2) converges for and divergesfor .

RaxRax

Note: If then every x satisfies If then no x satisfies0R

R Rax

Rax

In the example (2) the radius of convergence is .1R

9/8/2010 MATH ZC232 5

Test for convergenceThe ratio test is often used to determined the convergenceof power series. Suppose

Lc

caxaxcaxc

n

n

nn

n

nn

nLimLim 1

11

.0 ncn

If - The series convergesIf - The series divergesIf - The test fails

1L1L1L

Now, the formula for radius of convergence is

1n

n

n cc

R Lim9/8/2010 MATH ZC232 6

LECTURE 3

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2

Power series solution about ordinary point

)1(02

2

yxQdxdyxP

dxyd

The behavior of the solution of the equation

Near a point x0 depends on the behavior of the itscoefficient functions and near this point. A pointx0 is said to be an ordinary point of the differentialequation (1) if both and are well-behaved/analyticat x0 . So the functions P(x) and Q(x) can be representedby power series. The point that is not an ordinary point isknown as singular point.

xP xQ

xP xQ

9/8/2010 MATH ZC232 7

Example (3): Consider the equationHere and . These functions

are analytic at all the points.

Example (4): Consider the equation

Here

0" yy0xP 1xQ

06'2''12 yxyyx

;1

22x

xxP1

62x

xQ

Clearly, or are the only singularpoints of the equation. All other values of x areordinary points.

012x 1x

9/8/2010 MATH ZC232 8

Theorem 5.1 Existence of power seriessolutions

If is an ordinary point of the differentialequation (1), then there exist two linearly independentsolutions in the form of a power series centered at .A series solution converges at least on some intervaldefined by , where R is the distancefrom to the nearest singular point.

0xx

0x

0xRxx 0

9/8/2010 MATH ZC232 9

Working rules to solve the differentialequation by series solution aboutordinary points (say x = 0):

Assume its solution to be of the form

Calculate and substitute the value ofin the given equation.Equate to zero the coefficient of the various powersof x and determine in terms ofSubstituting the values of in theassumed solution, we get the desired series solutionhaving as its arbitrary constant.

......2210

nn xcxcxccxy

'',' yy '',', yyy

,...,, 432 ccc ., 10 cc,...,, 432 ccc

10 ,cc9/8/2010 MATH ZC232 10

9/8/2010 MATH ZC232 11

xk

9/8/2010 MATH ZC232 12

LECTURE 3

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(k+1)

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LECTURE 3

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9/8/2010 MATH ZC232 19

Example (5): Solve Legendre’s equation

Sol: It is clear that the coefficient functions

01'2"1 2 yppxyyx

;1

22xxxP

211

xppxQ

are analytic at the origin. The origin is therefore anordinary point. Let us assume a solution of the form

n

nn xcxy

0

1

1'n

nn xncy

On differentiating, we calculate

p is a real number

9/8/2010 MATH ZC232 20

Putting these values into the given differentialequation, we get

2

2

1'' n

nn xcnny

0121101

12

2

2

n

nn

n

nn

n

nn xcppxncxxcnnx

After shifting the summation indices, we get

01

211

210

2

11

2

2

2

n

nn

n

nn

n

nn

xcxccpp

xnccxxcnnx

9/8/2010 MATH ZC232 21

0111

2211

210

21

2

2

2

n

nn

n

nn

n

nn

n

nn

xcppxcppcpp

xncxcxcnnxcnn

0111

221162

210

21

2

2

432

n

nn

n

nn

n

nn

n

nn

xcppxcppcpp

xncxcxcnnxcnnxccor

0111

2211262

210

21

22232

n

nn

n

nn

n

nn

n

nn

xcppxcppcpp

xncxcxcnnxcnnxcc

or

9/8/2010 MATH ZC232 22

0121

62121

22

3120

n

nnn xcnpnpcnn

xccppccpp

After combining the terms of same order

Equating the coefficients of various powers of x onboth the sides, we have

021 20 ccpp

0621 31 ccpp

0121 2 nn cnpnpcnn

02 !21 cppc

13 !321 cppc

nnnnpnp

n cc 211

2

.........4,3,2n9/8/2010 MATH ZC232 23

When n takes the values 2,3, 4….

024 !4312

)4)(3(32 cppppcppc

135 !54231

)5)(4(43 cppppcppc

046 !653142

)6)(5(54 cppppppcppc

157 !7642531

)7)(6(65 cppppppcppc

9/8/2010 MATH ZC232 24

LECTURE 3

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5

By inserting these coefficients into the assumed solution, we obtained

.....!7

642531!5

4231!3

21

........!6

53142!4

312!2

11

7

53

1

6

42

0

xpppppp

xppppxppxc

xpppppp

xppppxpp

cy

9/8/2010 MATH ZC232 25

Solutions about singular points

Singular points

Regular Singular Points Irregular Singular Points

A singular point x0 is said tobe a regular singular point ofthe DE. (1) if

xPxxxp 0

xQxxxq 20

are both analytic at x0.

A singular point that isnot regular is known asirregular point of thedifferential equation.

9/8/2010 MATH ZC232 26

Example: Classify the singular points of

Sol: Writing the equation in the standard form

05'23''422 yyxyx

04

5'423'' 2222

yx

yx

xy

Here;

223

2xxxP 22 4

5x

xQ

Clearly, and are the singular point of thegiven equation.

2x 2x

9/8/2010 MATH ZC232 27

Case I: Check whether is regular or not.2x

2232

xxPxxp 2

2

252

xxQxxqand

Both and are analytic at . Hence is a regular singular point of the given differential equation.

xp xq 2x2x

Case II: Check whether is regular or not.2x

2232

xxxPxxp 2

2

252

xxQxxqand

Since is not analytic at . Hence is airregular singular point of the given differential equation.

2xxp 2x

9/8/2010 MATH ZC232 28

Frobenius TheoremIf is a regular singular point of the differentialequation (1), then there exists at least one solution of theform

0xx

00

000

n

rnn

n

nn

r xxcxxcxxy

where the number r is a constant to be determined. Theseries will converge at least on some interval .Rxx 00

Note (1): For the sake of simplicity, we shall alwaysassume in solving diff. eq. that the regular singular pointis x = 0.Note (2): If x = 0 is a irregular singular point than it is notpossible to find any solution of the form .

0n

rnn xcy9/8/2010 MATH ZC232 29

Example: SolveSol: Writing the given equation into standard form

0'1''2 yyxxy

Here

021'

21'' y

xy

xxy

;2

1xxxP x

xQ21

;2

1 xxxPxp 22 xxQxxq

Both and are analytic at x = 0 . So, x = 0 is a regular singular point of the given equation.By Frobenius theorem, substituting into the given equation.

xp xq

0n

rnn xcy

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LECTURE 3

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6

011200

1

0

2

n

rnn

n

rnn

n

rnn xcxcrnxxcrnrnx

012000

1

0

1

n

rnn

n

rnn

n

rnn

n

rnn xcxcrnxcrnxcrnrn

0112200

1

n

rnn

n

rnn xcrnxcrnrn

or

or

0112200

1

n

nn

n

nn

r xcrnxcrnrnxor

or 011221201

110

n

nn

n

nn

r xcrnxcrnrnxcrrx

1nk nk9/8/2010 MATH ZC232 31

011221120

11

0k

nkk

r xcrkcrkrkxcrrx

)(012 Arr

)(;011221 1 Bcrkcrkrk kk

.......3,2,1,0k

Equation (A) is called the indicial equation of theproblem. The indicial roots are

;21

1r 02r

9/8/2010 MATH ZC232 32

Case I: For equation (B) gives21

1r

01211121

21

1 kk ckckk

or ;121 k

cc kk 2

3k.......3,2,1,0k

;1.20

1cc

!2.22.2 201

2ccc

;!3.23.2 3

023

ccc!4.24.2 4

034

ccc

Putting the values of k , we get

9/8/2010 MATH ZC232 33

Similarly the nth term is given by

!.21 0

ncc n

n

n

Thus for the indicial root we obtained

0

2/1

1

2/11 !2

)1(!2

)1(1)(n

nn

n

n

nn

n

xn

xn

xxy

,21

1r

This solution is valid for all x since the series convergesabsolutely for all x.9/8/2010 MATH ZC232 34

Case II: For equation (B) gives02r

.......3,2,1,0k121 k

cc kk

Putting the values of k , we get

;1

01

cc 3.1301

2ccc

;5.3.15

023

ccc7.5.3.17

034

ccc

9/8/2010 MATH ZC232 35

Similarly the nth term is given by

12....7.5.3.11 0

ncc

n

n

Thus for the indicial root we obtained,02r

,12....7.5.3.1

11)(1

2n

n

n

xn

xy

The general solution is

xDyxCyy 21

|| x

,0x

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1


Lecture 4


11/8/2010 MATH ZC232 1


Singular points



xPxxxp 0

xQxxxq 20



11/8/2010 MATH ZC232 2


0xx

00

000

n

rnn

n

nn

r xxcxxcxxy



0n


Qs. 1 Solve

03 ''' yyyx (1)

X= 0 is a regular singular point

11/8/2010 MATH ZC232 4

Substitute0n

rnn xcy

in (1) and simplifying that, we get

00

1''' )233)((3n

rnn

n

rnn xcxcrnrnyyyx

SOLUTION

11/8/2010 MATH ZC232 5

Taking xr common and rearranging rest of the terms, we get

01

10

''' )133)(1()23(3k

kkk

r xccrkrkxcrrxyyyx

= 0 [from (1)]

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LECTURE 4

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2

Now equate the coefficients of powers of x, we get

0)23( 0crr (2)

..,.........2,1,0,0)133)(1( 1 kccrkrk kk (3)

In equation (2), nothing will be gained if c0 = 0, therefore

0)23( rr (4)

Equation (4) is called the indicial equation of the givenproblem and the two values of ‘r’ are called as indicial roots.These values shall play important role in finding out thesolution of the given problem.

11/8/2010 MATH ZC232 7

These two roots may be:(i) distinct and not differing by an integer(ii) equal, or (iii) distinct and differing by an integer, like in our problem

Therefore, from (4) r1= 0, r2 = 2/3.

From equation (3), we get

)133)(1(1 rkrkcc k

k (5)

11/8/2010 MATH ZC232 8

Substituting the values of r1 and r2 (5), we getr1= 0,

)13)(1(1 kkc

c kk k = 0, 1, 2,……… (6)

r2 = 2/3,

)1)(53(1 kkc

c kk k = 0, 1, 2,……… (7)

11/8/2010 MATH ZC232 9

,1.5

01

cc

,8.5!.22.8

012

ccc

,11.8.5!.33.11

023

ccc !

,14.11.8.5!.44.14

034

ccc.

,)23........(14.11.8.5!.

0

nnccn

Substituting k = 0, 1, 2…in (6), we get

11/8/2010 MATH ZC232 10

Substituting k = 0, 1, 2…in (7), we get

,1.10

1cc

,4.1!.24.2

012

ccc

,7.4.1!.37.3

023

ccc

,10.7.4.1!.410.4

034

ccc.

,)23........(7.4.1!

1 0

nncc

n

n

11/8/2010 MATH ZC232 11

Therefore, there will be two solutions corresponding to thetwo values of r, i.e. r1 = 0 and r2 = 2/3, which are as follows(after omitting c0 from each term)

0

01 )23....(7.4.1.!

11n

nxnn

xy

1

3/22 )23....(11.8.5.!

11n

nxnn

xy

11/8/2010 MATH ZC232 12

LECTURE 4

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3

Therefore, the complete solution of the given problem is

y = c1 y1 + c2 y2

1

3/22

0

01 )23....(11.8.5.!

11)23....(7.4.1.!

11n

n

n

n xnn

xcxnn

xcy

11/8/2010 MATH ZC232 13

Case 1.

If r1 and r2 are distinct and do not differ by an integer,there exists two independent solution y1(x) and y2(x),(like in the given example) of the form

0n

rnn xcy

11/8/2010 MATH ZC232 14

Case 2

If r1 - r2 = N, where N is a positive integer,then there exists two independent solution y1(x) and y2(x) of the form

0, 00

11 cxcxy

n

rnn

0,ln 00

122 bxbxxyCxy

n

rnn

where C is a constant that could be zero.

11/8/2010 MATH ZC232 15

Case 3

If r1 = r2, then there exists two linearly independent solutiony1(x) and y2(x) of the form

0, 00

11 cxcxy

n

rnn

0,ln 00

122 bxbxxyxy

n

rnn

The solution y2 (x) in this case can also be obtained using y1(x).

11/8/2010 MATH ZC232 16

y=0

92)(,)( 2 xQxxxxp

11/8/2010 MATH ZC232 17

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LECTURE 4

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4

2rxrkx

or

11/8/2010 MATH ZC232 19

xr

ng xr

11/8/2010 MATH ZC232 20

11/8/2010 MATH ZC232 21

11/8/2010 MATH ZC232 22

11/8/2010 MATH ZC232 23

C1

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LECTURE 4

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5

23

11389

389 2

3

CC

11/8/2010 MATH ZC232 25

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1

MATH ZC 232ENGINEERING MATHEMATICS II

LECTURE 5

Dr. Saroj Kumar Sahani

9/22/2010 MATH ZC232 1

Numerical Solutionof

Differential Equation

9/22/2010 2MATH ZC232

3

Euler’s Method

Step size, h

x

y

x0,y0

True value

y1,Predictedvalue

00,, yyyxfdxdy

SlopeRunRise

01

01xxyy

00 , yxf

010001 , xxyxfyy

hyxfy 000 ,Figure 1. Graphical interpretation of the first step of Euler’s method

9/22/2010 MATH ZC232

4

Euler’s Method

Step size

h

True Value

Predicted value

x

y

Figure 2. General graphical interpretation of Euler’s method

nnnn yxhfyy ,1

ii xxh 1

Divide the region of interest [a,b]up into discrete values of x=nh,n=0,1,2,….N spaced at intervalh=(b-a)/N. Use the forwarddifference approximation for thedifferential coefficient:

xn Xn+1

Yn

Yn+1

hyy

yyxf nnnnn

1',

9/22/2010 MATH ZC232

5

How to write Ordinary Differential Equation

Example

50,3.12 yeydxdy x is rewritten as

50,23.1 yyedxdy x

In this case yeyxf x 23.1,

How does one write a first order differential equation in the form of

yxfdxdy ,

9/22/2010 MATH ZC232

6Contd.

5.820122 23 xxxdxdy

We will solve the ODE-IVP from x=0 to x=4.0 in steps of h=0.5.

01 xwheny

The exact solution is given by the expression: 15.81045.0 234 xxxxy

We have to use the formula:

here

nnnn yxhfyy ,1

txh 5.0

5.85.005.05.01,005.0

yyfyy

21875.35.0y

Similarly, other values are calculated as given in the table next:

Euler Method – Example 1

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7

x 0 0.5 1 1.5 2 2.5 3 3.5 4

ytrue 1 3.218 3 2.218 2.718 2 4 4.718 3

yEuler 1 5.25 5.87 5.125 4.5 4.75 5.87 7.12 7

It will be noted that Euler’s method although simple to use,introduces gives erroneous results when function is highly non-linear. It is hence used for those equations where power ofdependent terms is not greater than 2.


9/22/2010 MATH ZC232

8


Use the Euler method to solve

The algorithm for the Euler method is

1ydxdy 0)0(y

nnn fhyy 1

For our example this gives:

)1(1 nnn yhyy9/22/2010 MATH ZC232

9


n xn yn fn= - yn+1 yn+1= yn+hfn

0 0 0.000 1.000 0.100

1 0.1 0.100 0.900 0.190

2 0.2 0.190 0.810 0.271

3 0.3 0.271 0.729 0.344

4 0.4 0.344 0.656 0.410

5 0.5 0.410 0.590 0.469

6 0.6 0.469 0.531 0.522

7 0.7 0.522 0.478 0.570

8 0.8 0.570 0.430 0.613

9 0.9 0.613 0.387 0.6519/22/2010 MATH ZC232

10


xey 1Analytic solution

0

0.2

0.4

0.6

0.8

0

0.25 0.5

0.75

1

1.25

Exact

Numerical

y

x

9/22/2010 MATH ZC232

11

Truncation Errors

There are

• Local truncation errors - error from application at a singlestep

• Propagated truncation errors - previous errors carriedforward

The sum is “global truncation error”.

9/22/2010 MATH ZC232

12

Errors in Euler’s Method

It can be seen that Euler’s method has large errors. This can beillustrated using Taylor series.

...!3

1!2

1 31

,3

32

1,

2

2

1,

1 nnyx

nnyx

nnyx

nn xxdx

ydxxdx

ydxxdxdyyy

nnnnnn

...),(''!3

1),('!2

1),( 31

211 nnnnnnnnnnnn xxyxfxxyxfyxfyy

As you can see the first two terms of the Taylor series

nnnn yxhfyy ,1

The true error in the approximation is given by

...!3,

!2, 32 hyxfhyxfE nnnn

t

are the Euler’s method.

2hEt

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3

13

Local Truncation Error

....!3

1!2

1))(,()()( 33

32

2

2

hdx

ydhdx

ydxyxhfxyhxynn

nnnn

• This is only the Euler algorithm for the first step when we know f(xn,y(xn)).

• This gives the local truncation error.

• Local truncation error for Euler algorithm is second order O(h2).

• The global truncation error in Euler method is O(h).

Euler Algorithm Truncation Error

9/22/2010 MATH ZC232

14

The Local truncation error in the approximation of yn+1 is given by

2

!2)(" hcyEt 1nn xcx

The upper bound on the absolute value of error is

!2

2hM )("max1

xyMnn xxx

where

9/22/2010 MATH ZC232

15

Euler method corresponds to keeping the first two terms inthe Taylor series.

Accuracy is low (O(h2)).

Error propagation is a problem.

Step size is important.

Problems with the Euler method

9/22/2010 MATH ZC232

16

Improved / Modified Euler’s Method

• Improved Euler’s method employs Euler’s method describedpreviously as a prediction step (called predictor formula) to start with.This gives first approximation of yn+1 denoted by

• The corrected value of yn+1 is obtained by using following formulacommonly called corrector formula:

• Note that predictor-corrector sequence must be followed. Calculationalways starts with predictor formula, then corrector formula is used.The corrected value is used in the next prediction.

1ny

2,, 11

1nnnn

nnyxfyxfyy

9/22/2010 MATH ZC232

17

Modified Euler Method

Predictor step (Euler method):

Two stage predictor-corrector method:

Corrector step :

nnnn yxhfyy ,*1

2,, *

111

nnnnnn

yxfyxfhyy

Figure 2. General graphical interpretation of Modified Euler’s method

9/22/2010 MATH ZC232

18

Example (Modified Euler Method)

Use the Modified Euler method to solve

1ydxdy 0)0(y

),(*1 nnnn yxfhyy


)1(*1 nnn yhyy

Predictor step (Euler method):

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4

19



Corrector step :

115.0 *11 nnnn yyhyy

2,, *

111

nnnnnn

yxfyxfhyy

9/22/2010 MATH ZC232

20


n

0 0 0.000 1.000 0.100 0.900 0.095

1 0.1 0.095 0.905 0.186 0.815 0.181

2 0.2 0.181 0.819 0.263 0.737 0.259

3 0.3 0.259 0.741 0.333 0.667 0.329

4 0.4 0.329 0.671 0.396 0.604 0.393

nx ny nf *1ny *

1nf 1ny

9/22/2010 MATH ZC232

21


xey 1Analytic solution

0

0.2

0.4

0.6

0.8

0

0.25 0.5

0.75

1

1.25

Mod. Euler

Exact

y

x

9/22/2010 MATH ZC232

22

RUNGE-KUTTA SECOND ORDER (RK-2) METHOD

9/22/2010 MATH ZC232

23

Runge-kutta methods propagate a solution over aninterval by combining information from severalEuler style steps, and then using the informationobtained to match a Taylor series expansion up tosome order.

Runge-Kutta methods are more symmetric than Euler’s method.

Runge-Kutta methods use a trial step at the midpoint of an interval to cancel out lower-order error terms.

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Example: Consider the IVP

020,2 yyxyxy

Use Euler’s method with h=0.3 to compute the approximate value of y (0.6)

Solution : Euler’s Method

Here 2, nnnnn xyxyxf

1 ,n n n ny y h f x f

9/22/2010 MATH ZC232

34

4.120203.02

23.02,

000

0001

xyxyxfhyy

453.023.04.13.03.04.1

23.04.123.04.1

,

110

111

1112

xyhxxyx

yxfhyy

953.06.02022 yhxyxyy

9/22/2010 MATH ZC232

35

RUNGE-KUTTA SECOND ORDER (RK-2) METHOD

9/22/2010 MATH ZC232

36

Runge-kutta methods propagate a solution over aninterval by combining information from severalEuler style steps, and then using the informationobtained to match a Taylor series expansion up tosome order.

Runge-Kutta methods are more symmetric than Euler’s method.

Runge-Kutta methods use a trial step at the midpoint of an interval to cancel out lower-order error terms.

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37

Runge-Kutta 2nd Order Method

Runge Kutta 2nd order method is given by

211 bkakyy nn

wherenn yxhfk ,1

1112 , kqyhpxhfk nn

For0)0(),,( yyyxf

dxdy

9/22/2010 MATH ZC232

38

Find out the constants a,b,p1,q1 such that

,1ba ,2/11bp 2/11bq

Three equations in four unknowns

Then the method reduces to the improved Euler method.

111 qp,2/1baFor

9/22/2010 MATH ZC232

39

Runge-Kutta Method of order 2

211

12

1

21

,,

KKyY

KyhxfhKyxfhK

nn

nn

nn

9/22/2010 MATH ZC232

40

Consider the IVP

10,32 yyxy

Using R-K method of only 2, approximate y (0.2) with h = 0.1

R-K Method of order 2

211

12

1

21

,,

KKyY

KyhxfhKyxfhK

nn

nn

nn

9/22/2010 MATH ZC232

41

We have

nnnn

n

yxyxfyxyxf32,

32,

For n = 0, we have

3.013021.0

32, 00001 yxhyxfhK

41.01.41.09.32.01.0

3.0131.00232

,

100

1002

hKyhxh

kyhxfhK

21001 211.0 kkyyhxyy 11 0.3 0.41 1.355

29/22/2010 MATH ZC232

42

For n = 1, we have

4265.0355.131.21.032

,

11

111

yxhyxfhK

726.14265.0355.131.01.021.0

32,

111

1112

kyhxhKyhxfhK

45125.2

726.14265.021355.1

212.0 2112 KKyyy

9/22/2010 MATH ZC232

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43

RUNGE-KUTTA FOURTH ORDER (RK-4) METHOD

9/22/2010 MATH ZC232

44

Runge – Kutta Method of order 4

1

12

23

4 3

,

,2 2

,2 2,

n n

n n

n n

n n

K h f x y

khK h f x y

khK h f x y

K h f x h y K

1 1 2 3 41 2 2 , 0,1, 2,6n nY y K K K K n

9/22/2010 MATH ZC232

45

• Fourth Order Runge-Kutta Method is a popular numerical method for solving IVP of the type just described. It is 4th order accurate method.

• The solution yn+1 is approximated using the formula:

• The parameters k are calculated using the formulae:

Basics

43211 2261 kkkkyy nn

34

23

12

1

,2,2

2,2

,

kyhxhfk

kyhxhfk

kyhxhfk

yxhfk

nn

nn

nn

nn

9/22/2010 MATH ZC232

46

Ex : Consider the IVP

32

22

yyxy

Do one iteration of RKM of order 4 with h = 0.1

Solution : Here

1.0,3,2,

00

22

hyxyxyxf

9/22/2010 MATH ZC232

47

For n=0, we have

3.1941.0

,20

20

001

yxhyxfhK

22

0

2

0

1002

22

2,

2

Kyhxh

KyhxhfK

9/22/2010 MATH ZC232

48

23

2 2

,2 2

0.1 1.75250.1 2 32 2

1.9226

n nkhK h f x y

864.29226.131.021.0

,

22

230

20

3004

Kyhxh

KyhxfhK

11 0 1 2 3 46 2 2 4.919y y K K K K

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49

Example 1

• Perform four steps of each of 4th-order Runge Kutta onthe IVP

to estimate y(1).

'( ) ( ) 1(0) 1

y x y x x xy

1,1,4/3,2/1,4/1,0,4/1 043210 yxxxxxh

9/22/2010 MATH ZC232

50

547623.0,

761963.02,2

765625.02,2

1,

34

2003

1002

001

kyhxfk

kyhxfk

kyhxfk

yxfk

nn

138168.0,

338213.02,2

347604.02,2

547946.0,

3114

2113

1112

111

kyhxfk

kyhxfk

kyhxfk

yxfk

808217.0

226 432101 kkkkhyy

722477.0

226 432112 kkkkhyy

Step 2:

Step 1:

9/22/2010 MATH ZC232

51

307173.0,

081682.02,2

065707.02,2

138761.0,

3224

2223

1222

221

kyhxfk

kyhxfk

kyhxfk

yxfk

741777.0

226 432123 kkkkhyy

888036.0,

585038.02,2

55756.02,2

306333.0,

3334

2333

1332

331

kyhxfk

kyhxfk

kyhxfk

yxfk

88676.0

226 432134 kkkkhyy

Step 3:

Step 4:

9/22/2010 MATH ZC232

52

We will solve the following ODE-IVP from x=1 to x=1.5 in steps of 0.1.xyy 2

Using the previous formulae @ x=0.1

2715.0,

2343.02,2

231.02,2

2.0,

34

23

12

1

kyhxhfk

kyhxhfk

kyhxhfk

yxhfk

nn

nn

nn

nn Using the previous formula @ x=1.1

23367.12261

432111.1 kkkkyy

x 1 1.1 1.2 1.3 1.4 1.5

y 1 1.2337 1.5527 1.9937 2.6116 3.4902

True 1 1.2337 1.5527 1.9937 2.6116 3.4902

Note that no error has been introduced upto 4 decimal accuracy.

Example 2

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LECTURE 5

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LECTURE 5

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LECTURE 5


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LECTURE 5(cont)

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10

We will solve the following ODE-IVP from x=1 to x=1.5 in steps of 0.1.xyy 2

Using the previous formulae @ x=0.1

2715.0,

2343.02,2

231.02,2

2.0,

34

23

12

1

kyhxhfk

kyhxhfk

kyhxhfk

yxhfk

nn

nn

nn

nn Using the previous formula @ x=1.1

23367.12261

432111.1 kkkkyy

x 1 1.1 1.2 1.3 1.4 1.5

y 1 1.2337 1.5527 1.9937 2.6116 3.4902

True 1 1.2337 1.5527 1.9937 2.6116 3.4902

Note that no error has been introduced upto 4 decimal accuracy.

Example 2

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LECTURE 5(cont)

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9/22/2010 18MATH ZC232

LECTURE 5(cont)

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LECTURE 5(cont)

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LECTURE 5(cont)

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LECTURE 5(cont)

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9/22/2010 MATH ZC232 37

System of Differential Equations First, let us consider first order differential equations

1001r2111 yxy,,...yy,yx,f

dxdy

2002r2122 yxy,,...yy,yx,f

dxdy

r00rr21rr yxy,,...yy,yx,f

dxdy

for large system of equations y1 1, y2,…yn are dependent variables and second

suffixes are for the values of independent variable x at x j, i.e.

yij denotes the value of yi at x j

9/22/2010 38MATH ZC232

All the methods used for solving a single equation with one dependent variable

and one independent variable can be extended for the system of differential

equations (6.61).

For convenience, the modified algorithms are shown for a pair of equations.

10012111 yxy,y,yx,f

dxdy

20022122 yxy,y,yx,f

dxdy

9/22/2010 39MATH ZC232

Runge-Kutta Method for System of EquationsAlgorithm of forth order Runge-Kutta for nth step

2n1nn11 y,y,xhfk

2n1nn22 y,y,xhfk

÷2

ky,2ky,

2hxhfl 2

2n1

1nn11

÷2

ky,2ky,

2hxhfl 2

2n1

1nn22

÷2ly,

2ly,

2hxhfm 2

2n1

1nn11

9/22/2010 40MATH ZC232

÷2l

y,2l

y,2h

xhfm 22n

11nn22

22n11nn11 my,myh,xhfp

22n11nn22 my,myh,xhfp

1111n1,1n1, p2m2lk61yy

2222n2,1n2, p2m2lk61yy

9/22/2010 41MATH ZC232

HIGHER ORDER DIFFERENTIAL EQUATIONS

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LECTURE 5(cont)

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Higher order differential equations:

If the differential equation is of higher order then that can be written as a

equivalent set of first order differential equations:

Consider the equation

yx,hyyx,yyx,dxdyyx,

dxdyyx,

dxyd 2

432

2

12

2

and 00 xy,xy are given, functions areh and ,,, 4321 of x and y.

denote dxdy

dxyd,y

dxdy 1

2

2

1

9/22/2010 43MATH ZC232

denoting y by y1 and dx

dydx

yddx

yd ,ydxdy

dxdy 2

21

2

2

2

21 .

Therefore above equations can be written as

0121121 xy ,y,yx,fy

dxdy

2141322

2211

2 yyyyyx,hdxdy

002212 xyxy , y,yx,f .

The above second order equation is written equivalent to a pair of first order differential equation. Similarly a nth order differential equation can be written equivalent to set of n equations each of first order

9/22/2010 44MATH ZC232

Example 3: Use Runge-kutta method of order 4 to find approximate value of

y(1.1) and 1.1y with spacing h = 0.1

11y and 1y(1) 2x,ydxdy

ydx

yd2

2

equivalent system of equations

zy,x,fzdxdy

1

2xyyzdxdz zy,x,fyyz2x 2

1z(1)z 1y(1)y 00 .

9/22/2010 45MATH ZC232

Applying Runge-kutta method of order four.

n = 0

0.11 0.1hzz,y,xhfk 000011

0112 0.1yzy2xhz,y,xhfk 000000022

9/22/2010 46MATH ZC232

Example 2: Write the second order differential equation

x32

22

2x xesinxxy

dxdyx

dxyde

with initial conditions 00 yxy and 00 yxy

equivalent to a pair of first order equations.

Let zy,x,fzdxdy

1

x222x xesin xxyzxdxdze

yzxex sin xxedxdz 22x2x = f2(x, y, z)

0000 yxz ,yxy .

9/22/2010 47MATH ZC232

Applying Runge-kutta method of order four.

n = 0

0.11 0.1hzz,y,xhfk 000011

0112 0.1yzy2xhz,y,xhfk 000000022

9/22/2010 48MATH ZC232

LECTURE 5(cont)

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2k

z,2k

y,2hxhfl 2

01

0011 = 0.1

2k

z,2k

y,2hxhfl 2

01

0022

1.051 1.052

2.120.1

01.051.052.10.1

9/22/2010 49MATH ZC232

9/22/2010 50MATH ZC232

0.11 0.10zh2l

z,2l

y,2hxhfm 0

20

10011

2lz,

2ly,

2hxhfm 2

01

0022

2l

z2l

z2l

yh2x0.1 20

20

100

01.051.052.10.1

0.1mzhmz,myh,xhfp 202010011

2010022 mz,myh,xhfp

1020100 mymz myhx2h

01.11.12.2h

111101 p2m2lk61yy~1.1y

1.10.10.20.20.1611

1p2m2lk61zz~1.1z 222201

9/22/2010 51MATH ZC232

Example : Use Runge -Kutta method of order 4 to find approximate value of

y(1.1) and 1.1y with spacing h = 0.1

11yand1y(1)2x,ydxdyy

dxyd2

2

equivalent system of equations

zy,x,fzdxdy

1

2xyyzdxdz zy,x,fyyz2x 2

1z(1)z1y(1)y 00

9/22/2010 52MATH ZC232

Applying Runge-Kutta method of order four.

n = 0

0.110.1hzz,y,xhfk 000011

01120.1yzy2xhz,y,xhfk 000000022

9/22/2010 53MATH ZC232

÷2

kz,

2k

y,2hxhfl 2

01

0011 = 0.1

÷2

kz,

2k

y,2hxhfl 2

01

0022

÷ 1.0511.052

2.120.1

01.051.052.10.1

9/22/2010 54MATH ZC232

LECTURE 5(cont)

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10

0.110.10zh2l

z,2l

y,2h

xhfm 02

01

0011 ÷

÷2l

z,2l

y,2hxhfm 2

01

0022

÷÷÷2l

z2l

z2l

yh2x0.1 20

20

100

01.051.052.10.1

0.1mzhmz,myh,xhfp 202010011

9/22/2010 55MATH ZC232

2010022 mz,myh,xhfp

1020100 mymzmyhx2h

01.11.12.2h

111101 p2m2lk61

yy~1.1y

1.10.10.20.20.1611

1p2m2lk61zz~1.1z 222201

1.1,(1.1y 1(1.1y

9/22/2010 56MATH ZC232

MULTI-STEP METHODS

9/22/2010 57MATH ZC232

Multi-step Methods :

•

To compute the value of y at xn+1, if the values of y, at more than one proceeding tabular points are used, then method is known as multistep method.

Suppose given values or computed values of y at xn, xn-1,…xn-m(m +1) are used to compute the value of y at xn+1, then the method is multistep.

9/22/2010 58MATH ZC232

Multistep Methods (cont.):

Let x0, x1,…xN be equispaced tabular points with spacingh.

Suppose we are to solve the differential equation withsome values of y available at preceding tabular points.Integrating as

1n x

pn x1ny

pny y)dxf(x,dy

9/22/2010 59MATH ZC232

CONTINUED

m

0iin

iim f

is

1xp

F(x) at xn, xn-1,…xn-m. Therefore above integral takes the form

(leaving the error term of interpolation) since x = xn + sh

dsfis

1hyym

0iin

ii1

p pn1n

9/22/2010 60MATH ZC232

Since y = y(x) is a function of x, so f(x,y(x)) = F(x) is replaced by interpolating polynomial

LECTURE 5(cont)

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11

Adams-bashforth formulas

This is known as fourth order Adams bashforth.

(A1)

Error in this formula is

3n2n1nnn1n 9f37f59f55f24hyy

1n3-n(IV)5 xfor x fh

720251

9/22/2010 61MATH ZC232

Adams-Moulton formulas

• To derive these formulas f(x,y) = F(x) is replaced bypolynomial pm+1 of degree (m+1),

(B1)with the error term

2n1nn1n1nn1n f5f19fy,x9f24hyy

5 (IV)n 2 n 1

19 h f , for x x720

9/22/2010 62MATH ZC232

Predictor-Corrector Pairs with Modifiers

• Consider the pair of multistep formulas

1n1nnnn(0)

1n y,xf 59y,x55f24hyy

3n3n2n2n y,x9fy,x37f

(0) 5 (IV)n-3 n 1

251E h f , x x720

with error term

9/22/2010 63MATH ZC232

2n2nn1nnn(0)

1n1nn(c)

1n y,xfy,x5fy,x19fy,x9f24hyy

1n2-n(IV)5c x x,f h

72019E

In (A1) to keep suffices non-negative minimum n = 3 so taking n = 3,

0011223330

4 y,x9fy,x37fy,x59fy,x55f24hyy

we note that to begin the computation 04y , we should have starting values 00 y,x ,

11 y,x , 22 y,x , 33 y,x . These values should be given with the equation or should be computed using single step methods i.e. Euler’s, Taylor’s or Runge-kutta methods. Then computed value y4 is denoted by 0

4y , because, this is to be used in (B1). Now to compute y4 value of from (B1),

9/22/2010 64MATH ZC232

we need 11 y,x , 22 y,x , 33 y,x , and 044 y,x , 0

4y is computed with help of

(A1), now computed y4 from (B1) is denoted by 14y , which is supposed to be

improved value over 04y . Looking at this sequence of computation of 0

4y and

14y , 0

4y is called predicted value obtained from a formula (A1), called predictor,

because 1ny does not appear on right side of (A1), 14y is called corrected value

obtained from formula (B1), called corrector formula because already 1ny

appears on right side of (B1).

9/22/2010 65MATH ZC232

Example 1. Compute the value of y at x = 1.4 as a solution of

using predictor corrector pair with x0 = 1, y0 = 0.649,

x1 = 1.1, y1 = 0.731 x2 = 1.2, y2 = 0.854,

x3 = 1.3, y3 = 1.028

0.1 h spacing where1,xxydxdy 2

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LECTURE 5(cont)

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1xxyyx,f 2

0.649110.6491xyxf 20000

1.0141.10.7311.11xyxf 221111

1.46481xyxf 22222 , 2.02641xyxf 2

3333

9/22/2010 67MATH ZC232

on substituting in modified = 2.72998

123*43

14 f5f19f9f

24h

yy

1.0141.464852.0264192.729989240.11.028

172469.1y 14 ,

24 10242263.01.2642741.172469

72019D

24 100.2422631.2645061.4yxy

= 1.26208

similarly computation may continue to compute y(1.5), y(1.6) and so on.

9/22/2010 68MATH ZC232

Multi-Step Methods

• To compute the value of y at xn+1, if the values of y at more than one preceding tabular points are used, then method is known as multi-step method.

• Suppose given values or computed values of y at xn, xn-1,…xn-m+1 are used to compute the value of y at xn+1, then the method is multi-step.

Contd.9/22/2010 69MATH ZC232

Multi-Step Methods

Let x0, x1,…xN be equi-spaced tabular points with spacing h.Suppose we are to solve the differential equation with some values of y

available at preceding tabular pointsintegrating as

Since y = y(x) is a function of x,so f(x,y(x)) = F(x) is replaced by interpolating polynomial given next

1n x

pn x1ny

pny y)dxf(x,dy

Contd.

yxfdxdy ,

9/22/2010 70MATH ZC232

m

0iin

iim f

is

1xp

F(x) at xn, xn-1,…xn-m.

Therefore above integral takes the form

, (leaving the error term of interpolation) since x = xn + sh

, x = xn + sh,

where interpolates

dsfis

1hyym

0iin

ii1

p pn1n

9/22/2010 71MATH ZC232

Adams-Bashforth Formula

We have

This is known fourth order Adams-Bashforth Formula.Error in this formula is given by



720251

9/22/2010 72MATH ZC232

LECTURE 5(cont)

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13

Adams-Moulton Formula • To derive this formula f(x,y) = F(x) is replaced by polynomial pm+1 of

degree (m+1),

error term is2n1nn1n1nn1n f5f19fy,x9f

24hyy

1n2n(IV)5 xfor x ,fh

72019

9/22/2010 73MATH ZC232

Predictor-Corrector Pairs with Modifiers• Consider the pair of multi-step formulas

1n1nnnn(0)

1n y,xf 59y,x55f24hyy 3n3n2n2n y,x9fy,x37f

1n3-n(IV)5p x x,f h

720251E

with error term

9/22/2010 74MATH ZC232

2n2nn1nnn(0)

1n1nn(c)


1n2-n(IV)5c x x,f h

72019E

9/22/2010 75MATH ZC232

1In (A ) to keep suffices nonnegative minimum n = 3 so taking n = 3,

0011223330





9/22/2010 76MATH ZC232





4y and

14y , 0





9/22/2010 77MATH ZC232


using predictor-corrector pair with x0 = 1, y0 = 0.649

x1 = 1.1, y1 = 0.731 x2 = 1.2, y2 = 0.854 x3 = 1.3, y3 = 1.028


9/22/2010 78MATH ZC232

LECTURE 5(cont)

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1xxyyx,f 2

0.649110.6491xyxf 20000

1.0141.10.7311.11xyxf 221111

1.46481xyxf 22222 , 2.02641xyxf 2

3333

9/22/2010 79MATH ZC232


123

*

43

1

4f5f19f9f

24

hyy

1.0141.464852.0264192.72998924

0.11.028 ´´´

172469.1y 1

4

9/22/2010 80MATH ZC232

24 10242263.01.2642741.172469

72019D

24 100.2422631.2645061.4yxy

= 1.26208


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LECTURE 5(cont)

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LECTURE 7


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LECTURE 7

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9/22/2010 MATH ZC232 9

9/22/2010 MATH ZC232 10

MULTI-STEP METHODS

9/22/2010 11MATH ZC232

Multi-step Methods :

•

To compute the value of y at xn+1, if the values of y, at more than one proceeding tabular points are used, then method is known as multistep method.

Suppose given values or computed values of y at xn, xn-1,…xn-m(m +1) are used to compute the value of y at xn+1, then the method is multistep.

9/22/2010 12MATH ZC232

LECTURE 7

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3

Multistep Methods (cont.):

Let x0, x1,…xN be equispaced tabular points with spacingh.

Suppose we are to solve the differential equation withsome values of y available at preceding tabular points.Integratingas

1n x

pn x1ny

pny y)dxf(x,dy

9/22/2010 13MATH ZC232

CONTINUED

m

0iin

iim f

is

1xp

F(x) at xn, xn-1,…xn-m. Therefore above integral takes the form

(leaving the error term of interpolation) since x = xn + sh

dsfis

1hyym

0iin

ii1

p pn1n

9/22/2010 14MATH ZC232

Since y = y(x) is a function of x, so f(x,y(x)) = F(x) is replaced by interpolating polynomial

Adams-bashforth formulas

This is known as fourth order Adams bashforth.

(A1)

Error in this formula is



720251

9/22/2010 15MATH ZC232

Adams-Moulton formulas • To derive these formulas f(x,y) = F(x) is replaced by

polynomial pm+1 of degree (m+1),

(B1)

with the error term

2n1nn1n1nn1n f5f19fy,x9f24hyy

5 (IV)n 2 n 1

19 h f , for x x720

9/22/2010 16MATH ZC232

Predictor-Corrector Pairs with Modifiers

• Consider the pair of multistep formulas

1n1nnnn(0)

1n y,xf 59y,x55f24hyy

3n3n2n2n y,x9fy,x37f

(0) 5 (IV)n-3 n 1

251E h f , x x720

with error term

9/22/2010 17MATH ZC232

2n2nn1nnn(0)

1n1nn(c)


1n2-n(IV)5c x x,f h

72019E

9/22/2010 MATH ZC232 18

LECTURE 7

Page 3 of 16

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4

In (A1) to keep suffices non-negative minimum n = 3 so taking n = 3,

0011223330





9/22/2010 19MATH ZC232





4y and

14y , 0





9/22/2010 20MATH ZC232


using predictor corrector pair with x0 = 1, y0 = 0.649,

x1 = 1.1, y1 = 0.731 x2 = 1.2, y2 = 0.854,

x3 = 1.3, y3 = 1.028


9/22/2010 21MATH ZC232

1xxyyx,f 2

0.649110.6491xyxf 20000

1.0141.10.7311.11xyxf 221111

1.46481xyxf 22222 , 2.02641xyxf 2

3333

9/22/2010 22MATH ZC232


123*43

14 f5f19f9f

24hyy

1.0141.464852.0264192.729989240.11.028

172469.1y 14 ,

24 10242263.01.2642741.172469

72019D

24 100.2422631.2645061.4yxy

= 1.26208


9/22/2010 23MATH ZC232

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LECTURE 7

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LECTURE 7

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22/09/2010 MATHZC 232 31

Solve the differential Equation by four step Adam Bashforth Predictor and three order Adam Moulton Corrector Pair.

22/09/2010 MATHZC 232 32

22/09/2010 MATHZC 232 33

22/09/2010 MATHZC 232 34

22/09/2010 MATHZC 232 35

Fourier Series

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LECTURE 7

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For all

For all

LECTURE 7

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LECTURE 7

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9

Partial Differential Equations (PDEs)

• Objectives of study– What are PDEs?– Solution of linear PDEs in two independent variables– Classification of linear 2nd order PDEs– Separation Of Variables method for product solution– Initial & boundary conditions & their role in solution– Use of boundary conditions to get unique solution– Special PDEs in general practice

LECTURE 7

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What are PDEs?• Definition:

– A linear 2nd order partial differential equation is expressed in the general form:

– A, B, C…… are functions of x and y– x and y are called “independent variables”– u is called “dependent variable”– The PDE is linear because power of any term is 1.– The PDE is 2nd order because order of highest derivative is 2.– The DE is “partial” because more than one independent variables are present– G can be zero (homogeneous) or non-zero (non-homogeneous).

• Solution of PDE:– It is a function of two independent variables x and y that satisfies the equation

in some region of xy-plane.– Particular solution is more useful. It applies to a specific situation and is elegant.

General solution applies to all situations but it is difficult to obtain.

GFuyuE

xuD

yuC

yxuB

xuA 2

22

2

2

yxu ,

….Classification Next

Classification & Some PDEs• General form for linear 2nd order PDE is

• Coefficients of second order derivatives (A, B, C) are considered.• Classification as:

Hyperbolic whenParabolic whenElliptic when

• Examples of some PDEs– One dimensional wave equation (Hyperbolic)

– Heat equation (Parabolic)

– Laplace equation (Elliptic)

02

22

2

2

FuyuE

xuD

yuC

yxuB

xuA

042 ACB042 ACB042 ACB

tT

xTk 2

2

2

2

2

22

tu

xua

02

2

2

2

yT

xT

…..SOV Method Next

Separation Of Variables Method (SOV)• Finds solution in terms of product of functions, called product solution.• We will take up an example problem (p. 688) to find product solution of

• Note that we have not mentioned any boundary conditions because we want a generally applicable particular solution of the problem.Sol.:

STEP I: Assume product solution formassume that product solution is of the form:eq. (2) will satisfy the original PDE.

STEP II: Obtain separated variable form

yu

xu 42

2

yYxXyxu ,

YXx

xXyYyYxXxxx

u2

2

2

2

and YXyyYxXyYxX

yyu 4444

YXYX 4

YY

XX4

……?

(1)

(3)

(2)

SOV Method Continued…….• An Argument: On observing eq. (3), we find that:

– LHS is independent of y– RHS is independent of x– LHS=RHS

• This implies that:– Both sides are independent of x and y– Thus each side is a constant

• Denote this constant by 2 OR 2. (Separation Constant)

• Three possibilities exist: 2 >0 OR 2 <0 OR 2 =0.

STEP III: Examine the three possibilities & find general solution

……?

SOV Method Continued…….• Case I: 2 >0

Eq. (4) & (5) are ODEs whose solutions are found by usual methods.Solution of (4) is:Solution of (5) is:

c1, c2 and c3 are arbitrary constants that are evaluated by use of boundary conditions.Using eq. (2)

Note: We need three boundary conditions to evaluate three arbitrary constants.

……?

04

4

2

2

XX

YY

XX

02YYand

(4)

(5)

xcxcX 2sinh2cosh 21yecY

2

3

yecxcxcyxu2

321 2sinh2cosh, required

SOV Method Continued…….• Case II: 2 <0




……?

04

4

2

2

XX

YY

XX

02YYand

(6)

(7)

xcxcX 2sin2cos 54yecY

2

6

yecxcxcyxu2

654 2sin2cos, required

LECTURE 7

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SOV Method Concluded

….Initial & Boundary Conditions Next

Case III:Case III: 22 =0=0

Eq. (Eq. (88) & () & (99) are) are ODEsODEs whose solutions are found by usual methods.whose solutions are found by usual methods.Solution of (Solution of (88) is:) is:Solution of (Solution of (99) is: ) is:

cc77, , cc88 and and cc99 are are arbitrary constantsarbitrary constants that are evaluated by use of that are evaluated by use of boundary conditionsboundary conditions..Using eq. (Using eq. (22))

Note:Note: We need three boundary conditions to evaluate three arbitrary constants.We need three boundary conditions to evaluate three arbitrary constants.

Method could work because we could separate variables in the form of eq. (3)Method could work because we could separate variables in the form of eq. (3)

0

04

X

YY

XX

0Yand

(8)

(9)

87 cxcX9cY

987, ccxcyxu required

Common Boundary Conditions & PDE Solutions• Boundary conditions & initial conditions allow unique solutions.• Boundary conditions define physical reality at the extreme ends of body.• Source of information to connect solution with environment.• Initial conditions are conditions specified at initial time.• Boundary conditions are specified at physical boundaries (length extremes).• No. of boundary conditions equal to order of highest order derivative.• Three types of boundary conditions:

– Dirichlet type: to specify value of dependent variable u

– Neumann type: to specify no change in dependent variable normal to boundary

– Robin type: to specify non-zero change in dependent variable normal to boundary

0, utLu

0Lxx

u

mLx

utLuhxu ,

…..Use of BCs Next

Using Boundary Conditions to Get Unique Solution

• We saw how SOV Method gave generally applicable particular solutions.

• The constants (c1, c2, etc.) are evaluated using the boundary conditions of the type stated previously.

• Unknown variable at a boundary is assigned some value and values of independent and unknown variable values are inserted in particular solutions.

• A set of algebraic equations is obtained, which can be solved to obtain values or expressions for the arbitrary constants.

• Note: PDE solution by SOV Method also involves solution of Ordinary Differential Equations (section 2.3 & 3.3).

…..Special PDEs Next

Heat Equation

• Describes heat conduction in a rod in terms of T(x,t).• k is called thermal conductivity (k>0)• See p. 692-693 for assumptions & derivation.• Solution:

– Two independent variables x and t and T is dependent variable– Needs one initial condition (IC)– Needs two boundary conditions (BC)– IC: T(x,0)=f(x) for 0<x<L to describe initial distribution of temperature.– BC 1: T(0,t)=T0

– BC 2: T(L,t)=TL

– Method of choice: Separation of Variables

tT

xTk 2

2

0 Lx x+ x

c/s area A

t>0

Lx0 0t

…Wave Equation Next

One Dimensional Wave Equation

• Describes transverse vibrations/waves (perpendicular to x-axis) and u(x,t) is the displacement.

• See p. 693-694 for assumptions & derivation.• Solution:

– Two independent variables x and t and u is dependent variable– Needs two ICs– Needs two BCs– IC 1:– IC 2:– BC 1:– BC 2– Method of choice: Separation of Variables

2

2

2

22

tu

xua Lx0 0t

xfxu 0,

xgtu

t 0

0,0 tu0,tLu

t>0

….Laplace Equation Next

Laplace Equation

• Describes steady state distribution of temperature.• Solution:

– x and y are independent variables and T is dependent variable

– Needs four BCs

– BC 1:

– BC 2:

– BC 3:

– BC 4:


02

2

2

2

yT

xT

ax0 by0

00xx

T

0axx

T by0

ax000,xT

xfbxT ,

x

y

a

b

LECTURE 7

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Please Note

• Read classification method for PDEs on p. 690.

• Read Section 2.3 & 3.3 to revise methods of solution of linear ODEs.

• See Separation of Variables (SOV) method on p. 688-689 once again. Solve some problems from Exercise 13.1, p. 691.

• Understand the significance of ICs and BCs in obtaining particular solution.

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For all

For all

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LECTURE 8

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GFuyuE

xuD

yuC

yxuB

xuA 2

22

2

2

yxu ,








02

22

2

2

FuyuE

xuD

yuC

yxuB

xuA


tT

xTk 2

2

2

2

2

22

tu

xua

02

2

2

2

yT

xT


LECTURE 8

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yu

xu 42

2

yYxXyxu ,

YXx

xXyYyYxXxxx

u2

2

2

2

and YXyyYxXyYxX

yyu 4444

YXYX 4

YY

XX4

……?

(1)

(3)

(2)







……?





……?

04

4

2

2

XX

YY

XX

02YYand

(4)

(5)


2

3

yecxcxcyxu2






……?

04

4

2

2

XX

YY

XX

02YYand

(6)

(7)


2

6

yecxcxcyxu2









0

04

X

YY

XX

0Yand

(8)

(9)

87 cxcX9cY






0, utLu

0Lxx

u

mLx

utLuhxu ,

…..Use of BCs Next

LECTURE 8

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…..Special PDEs Next

Heat Equation



– BC 2: T(L,t)=TL


tT

xTk 2

2

0 Lx x+ x

c/s area A

t>0

Lx0 0t

…Wave Equation Next





2

2

2

22

tu

xua Lx0 0t

xfxu 0,

xgtu

t 0

0,0 tu0,tLu

t>0

….Laplace Equation Next

Laplace Equation



– Needs four BCs

– BC 1:

– BC 2:

– BC 3:

– BC 4:


02

2

2

2

yT

xT

ax0 by0

00xx

T

0axx

T by0

ax000,xT

xfbxT ,

x

y

a

b

Please Note





LECTURE 8

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LECTURE 10Tutorial 2


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Repeat the previous example with Modefied Euler’s Algorithm

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Question

10 Feb 2009 MATH ZC232, Eng Maths II 19

2

' ,1' 5 3 45 ( , , )2

x

y z then

z z y e g x y z

0 0 0

1 0

1 0 0 0

2 0 1

2 0 0 1 0 1

0, 2, 1, 0.10.1

( , , ) 0.1 (0,2,1) 2.80( / 2) 0.1 2.4 0.24( / 2, / 2, / 2)

0.1 (0.005,2.05,2.4) 3.394

x y z hk hzl h g x y z gk h z ll h g x h y k z l

g


3 0 2

3 0 0 2 0 2

4 0 3

4 0 0 3 0 3

( / 2) 0.1 2.69 0.27( / 2, / 2, / 2)

0.1 (0.05,2.12,2.697) 3.4790.1 4.4789 0.4478

( , , ) 0.1 (0.1,2.269,4.478)4.208

k h z ll h g x h y k z l

gk h z ll h g x h y k z l g


1 0 1 2 3 4

1 0 1 2 3 4

1 2( )6

12 0.1 2(0.24 0.2697) 0.4478 2.261261 2( )6

11 2.80 2(3.394 3.479) 4.208 4.4596

y y k k k k

z z l l l l

(0.1) 2.2612,(0.1) 4.459

yy

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by

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Solve the system by Runge-Kutta Method of order 4.

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LECTURE 11

Rajesh Bhatt

9/22/2010 MATH ZC232 1



9/22/2010 MATH ZC232 2







GFuyuE

xuD

yuC

yxuB

xuA 2

22

2

2

yxu ,


9/22/2010 MATH ZC232 3







02

22

2

2

FuyuE

xuD

yuC

yxuB

xuA


tT

xTk 2

2

2

2

2

22

tu

xua

02

2

2

2

yT

xT


9/22/2010 MATH ZC232 4





yu

xu 42

2

yYxXyxu ,

YXx

xXyYyYxXxxx

u2

2

2

2

and YXyyYxXyYxX

yyu 4444

YXYX 4

YY

XX4

……?

(1)

(3)

(2)

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……?9/22/2010 MATH ZC232 6

LECTURE 11

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……?

04

4

2

2

XX

YY

XX

02YYand

(4)

(5)


2

3

yecxcxcyxu2


9/22/2010 MATH ZC232 7





……?

04

4

2

2

XX

YY

XX

02YYand

(6)

(7)


2

6

yecxcxcyxu2


9/22/2010 MATH ZC232 8








0

04

X

YY

XX

0Yand

(8)

(9)

87 cxcX9cY


9/22/2010 MATH ZC232 9





0, utLu

0Lxx

u

mLx

utLuhxu ,

…..Use of BCs Next9/22/2010 MATH ZC232 10







…..Special PDEs Next9/22/2010 MATH ZC232 11

Heat Equation



– BC 2: T(L,t)=TL


tT

xTk 2

2

0 Lx x+ x

c/s area A

t>0

Lx0 0t

…Wave Equation Next9/22/2010 MATH ZC232 12

LECTURE 11

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2

2

2

22

tu

xua Lx0 0t

xfxu 0,

xgtu

t 0

0,0 tu0, tLu

t>0

….Laplace Equation Next9/22/2010 MATH ZC232 13

Laplace Equation



– Needs four BCs

– BC 1:

– BC 2:

– BC 3:

– BC 4:


02

2

2

2

yT

xT

ax0 by0

00xx

T

0axx

T by0

ax000,xT

xfbxT ,

x

y

a

b

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Please Note





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MATH ZC232

Lecture 13By

Rajesh Bhatt

03 Sep 2010 1MATH ZC232

03 Sep 2010 2MATH ZC232

Complex Numbers

• A complex number is any number of the form z=x+iy where x and y are real numbers and i is the imaginary unit.

• Modulus or absolute value of z is denoted by

2 2z x y

03 Sep 2010

Polar Form

• A nonzero complex number z =x+iy can be written as

• Where r is the modulus of z.• is the argument of z is written as

( cos ) ( sin )z r i r

a rg z

03 Sep 2010

03 Sep 2010 5MATH ZC232

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03 Sep 20107MATH ZC232

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Neighborhood Point

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MATH ZC232

Lecture 14By

Rajesh Bhatt

1MATH ZC2326/09/2010

2MATH ZC2326/09/2010

3MATH ZC2326/09/2010

4MATH ZC2326/09/2010

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7MATH ZC2326/09/2010

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LECTURE 14

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LECTURE 14

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LECTURE 14

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LECTURE 14

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LECTURE 15

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LECTURE 15

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LECTURE 15

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LECTURE 15

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LECTURE 15

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LECTURE 15

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MATH ZC232

Lecture 16By

Rajesh Bhatt

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LECTURE 16

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LECTURE 17Rajesh Bhatt

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Taylor’s Theorem Let f be analytic within a domain D and let 0z be a point in D. Then f has the series representation

00

0

8!

kk

k

f zf z z z

k Valid for the largest circle C with center at 0z and radius R that lies entirely within D.

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This series is called the Taylor series for f centered at 0z . A Taylor series with center ,00z

0

0, 7

!

kk

k

ff z z

k is referred to as a Maclaurin series.

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Maclaurin series of some functions are given below

0

2

!...

!2!11

k

kz

kzzze

(12)

0

1253

!121...

!5!3sin

k

kk

kzzzzz

(13)

0

242

!21...

!4!21cos

k

kk

kzzzz

(14)

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.....!4!2

1cosh42 zz

z

.....!5!3

sinh53 zz

zz

......11

1 32 zzzz (15)

1

122 ......321

11

k

kkzzzz

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Theorem 19.10 Laurent Series Let f be analytic within the annular domain Ddefined by .0 Rzzr Then f has the series representation

k

kk zzazf 0 (3)

LECTURE 17

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Valid for .0 Rzzr The coefficients ak are given by

c kk kdszssf

ia ......,,2,1,0,

21

10 (4)

Where C is a simple closed curve that lies entirely within D and has 0z in its interior.

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Example 1

.....!7!5!3

11sin 42

23

zzzz

zzf (2) This series converges for all z except z=0, that is the series converges for |z| > 0.

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Example 2

Expand 11

zzzf in a Laurent series valid for

(a) ,10 z (b) ,1 z (c) ,110 z (d) 11 z .

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(a) zzzf

111

,

.....11 32 zzzz

zf

......11 2zzz

zf0<|z|<1

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(b) z

zzf

11

112

.....11111

322 zzzzzf

.......111432 zzz

or

.1for ly equivalentor 11 zz

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(c) 1111

zzzf

111

11

zz

......11111

1 32 zzzz 1|1|0 z

.....1111

1 2zzz .

110 z

LECTURE 17

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(d) 1

11

11

111

11

12

zzzz

zf

......1

11

11

111

1322 zzzz

......

11

11

11

11

5432 zzzz

11for converges series final the,111 zz .

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Example 3 Expand 11

zzzf in a Laurent series

valid for 221 z .

.1

1121 zfzf

zzzf

2211

1 zzzf

221

121

z

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.......2

22

22

2121

3

3

2

2 zzz

.......2

22

22

221

4

3

3

2

2

zzz

This series converges for .22or122 zz

Furthermore,

211

12

121

11

12

zzzz

zf

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........2

12

12

112

132 zzzz

........

21

21

21

21

432 zzzz Converges for 21or12

1 zz . Substituting these two results in (13) then gives

......2

22

22

221

21

21

21

21...... 4

3

3

2

2234

zzzzzzz

zf

This representation is valid for .221 z

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Example 4

Expand zezf3

in a Laurent series valid for z0 .

From (12) of Section 19.2 we know that for all finite z,

......!3!2

132 zzze z

(14)

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By replacing z in (14) by ,0,3 zz we obtain

the Laurent series

.......!33

!2331 3

3

2

23

zzze z

This series is valid for z0 .

LECTURE 17

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If a complex function f fails to be analytic at a point 0zz , then this point is said to be a singularity or a singular point of the function. For example, the complex numbers iziz 2and2 are singularities of the function 4/ 2zzzf

because f is discontinuous at each of these points.

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0zz is said to be an isolated singularity of the function f if there exists some deleted neighborhood, or punctured open disk,

00 of0 zRzz throughout which f is analytic. Suppose z = z0 is an isolated singularity of a function f. Hence

k k k

kk

kk

kk zzazzazzazf

1 0000 … (1)

is the Laurent series valid for Rzz 00

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1 0kk

k

zza

in the Laurent series (1) is called principal part of the series.

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An isolated singular point 0zz of a complex function f is given a classification depending on whether the principal part (2) of its Laurent expansion (1) contains zero, a finite number, or an infinite number of terms. Principal part is zero, that is, all the coefficients ka in (2) are zero, then 0zz is called a removable singularity.

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Principal part contains a finite number of nonzero terms then 0zz a pole. If, in this case, the last nonzero coefficient in (2) is a 1na n , then we say that 0zz is a pole of order n. If 0zz is a pole of order 1, then principal part (2) contains exactly one term with coefficient 1a . A pole of order 1 is commonly called a simple pole. Principal part (2) contains infinitely many nonzero terms, then 0zz is called an essential singularity.

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Removable Discontinuity Proceeding as we did in (2) of section 19.3, we see from

.......!5!3

1sin 42 zzz

z (2)

That is 0z is a removable singularity of the

function zzzf sin

.

LECTURE 17

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Principal part

........,!5!3

1sin 3

2

zzzz

z

In example 3 of section 19.3 we showed that the Laurent expansion of

210for valid31/1 2 zzzzf is

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Principal part

.......16

181

141

121

2

zzz

zf

Since 02a , we conclude that 1z is a pole of order 2.

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Pole of Order n If the function f and g are analytic at 0zz and fhas a zero of order n at 0zz and 00zg , then the function zfzgzF / has a pole of order n at 0zz .

The coefficient 01

1 of zza in the Laurent series called the residue of the function f at the isolated singularity 0z . We shall use the notation

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01 , Res zzfa

denote the residue of f at 0z . If f has a simple pole at 0zz , then

zfzzzzfzz

00 lim0

, Res . (1)

If f has a pole of order n at 0zz , then

zfzzdzd

nzzf n

n

n

zz01

1

0 lim0

!11, Res . (2)

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Suppose a function f can be written as a quotient zhzgzf / , where g and h are analytic at

0zz . If 00zg and if the function h has a zero of order 1 at 0z then f has simple pole at 0zzand

0

00 '

, Reszhzgzzf

.

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The polynomial 14z can be factored as 4321 zzzzzzzz , where 4321 and,, zzzz

are the four distinct roots of the equation 014z. It follows from 19.11 that the function

11

4zzf

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Has four simple poles. Now from (10) of Section 17.2 we have

4/74

4/53

4/32

4/1 ,,, iiii ezezezez .

iez

zzfs i

241

241

41

41,Re 4/3

31

1

iez

zzfs i

241

241

41

41,Re 4/9

32

2

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iez

zzfs i

241

241

41

41,Re 4/15

33

3

iez

zzfs i

241

241

41

41,Re 4/21

34

4

.

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Cauchy’s Residue Theorem Let D be a simply connected domain and C a simple closed contour lying entirely within D. if a function f is analytic on and within C, except at a finite number of singular points nzzz ,......,, 21

within C, then

.,Res21

n

kkc

zzfidzzf (5)

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Evaluation by the Residue Theorem Example 1

Evaluate cdz

zz 3112 , where

The contour C is the rectangle defined by a

,1,1,4,0 yyxx and (b) the contour C is the circle 2z .

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(a) Since both poles 3z and1z lie within the square, we have from the residue theorem

.3,Re1,Re231

12 zfszfsidz

zzc .

.041

412

3112 idz

zzc from example 2 of Th. 19.13

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(b) Only the pole z =1 lies within the circle 2z , we

ii

zfsidzzzc

2412

1,Re231

12

.

LECTURE 17

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Example 2

Evaluate ,462

2cdz

zz

where the contour C is the circle 2 iz The integrand has simple poles at -2i and 2i. Nowsince only 2i lies within the contour C, it follows from (5) that

.2,Re2462

2 izfsidzz

zc

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izizzizizfs

iz 22622lim2,Re

2

ii

ii

223

446

ci

iiidz

zz 23

2232

462

2

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Example 3

c

z

dzzz

e34 5 , where the contour C is the circle .2z

Since 55 334 zzzz we see that the integrand has a pole of order 3 at 0z . Since only 0zlies within the given contour, we have

c

z

zfsidzzz

e 0,Re25 34

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5lim

!212 3

32

2

0 zzez

dzdi

z

z

.12517

5178lim 3

2

0i

zezzi

z

z

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Example 4

cdzz ,tan where the contour C is the circle .2z

Integrand tan zzz cos/sin has simple poles at the points where 0cos z . ,....2,1,0,2/12 nnz

since only 2/and2/ are within the circle 2z

, we have

2,Re

2,Re2tan zfszfsidzz

c .

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From (4) with ,sin'and,cos,sin zzhzzhzzg

12/sin

2/sin2

,Re zfs and

12/sin

2/sin2

,Re zfs

Therefore .4112tan iidzzc

LECTURE 17

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Example 5

,/3 dzec

z where the contour C is the circle 1z .

We have seen, 0z is an essential singularity of the integrand zezf /3 The Laurent series of f at 0z gives 30,Re zfs

. Hence from (5) we have

izfsidzec

z 60,Re2/3.

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Integrals of the Forum dF sin,cos20 . Convert

into a complex integral where the contour C is the unit circle centered at the origin. Contour can be parameterized by 20,sincos ieiz .

diedz i ,

2cos

ii ee

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iee ii

2sin ,

11

21sin,

21cos, zz

izz

izdzd .

c izdzzz

izzF 11

21,

21

, Where C is 1z .

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Example 1: Evaluate

2

0 2cos21 d

.

cdz

zzz

22 1414

'14 2

12

022 zzzz

zzz

zzf

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Where 32and32 10 zz . Since only 1z is inside the unit circle C, we have

czzfsidz

zzz

122,Re2

14 .

20

211

11

limlim,Rezz

zdzdzfzz

dzdzzfs

zzzz

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361lim 3

0

0

1 zzzz

zz

ci

izzfsi

idz

zzz

361.2.4,Re2.4

1441

122

2

0 2 334

cos21 d

.

LECTURE 17

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Art 19.5 Q.3 Find Residue using Laurent series

21

32

2

64

264

zz

zz

zzzzf

......22

132 2

2zzz

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20...23

233...

22 2

2

zzz

zz Residue = -3 II Method

0,00,00 zqp is a simple pole residue =

326

00

qp

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Q.5 Laurent series of .......2

221

2

2

2

22 z

ze z

z = 0 is an essential singularity. Residue at

z = 0 is coefficient of .01z

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Q.4 3,Re3

1sin3 2 zfsz

zzf

.......!7!5!3

753 zzzzzSin

53 3!51

3!31

31

31sin

zzzz

32

3!51

3!313

31sin2

zzz

zz

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This is the Laurent series whose principal part contains infine no. of terms z = -3 is an essential singular point. (Res. [f(z), -3)

31

zofcoeff in Laurent series. !3

1

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Q.11 2,1,321

345 2

zzzfzg

zzzzzzF

z = -3 are zeros of f(z) and g(z) is not zero at these points: These are simple poles of F(z). Residue at

62

123234511

2

11 zzzzLtzfzLtz

zz

Residue at

LECTURE 17

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313134522

2

22 zzzzLtzfzLtz

zz Residue at

2134533

2

33 zzzzLtzFzLtz

zz 302

60

iidzzzz

zz

zC

10303162321

345

4:

2

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Q.13 0.cos

32z

zzzzf

is a zero of order of the denominator and z = 0 is a not a zero of the numerator. Hence z = 0 is a pole of order 2. Similarly z is a pole of order 3.

Residue at zfzdzdLtz

z

2

0 !110

34

030

sincos3cosz

zzzLtz

zdzdLt

zz 43

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Residue at

zfzdzdLtz

z

32

2

!21

4

2

22

2 cos2sin21cos

21

zzzzz

dzdLt

zz

dzdLt

zx

3

cos2sin21

zzzz

dzdLt

z

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6

23 cos2sin23sin2cossin21

zzzzzzzzzLt

z

4

cos2sin3sin2cossin21

zzzzzzzzzLt

z

4

2 621

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Q.17 C zzdz

221 Singular points are z = 1, z = -2, z = 1 is a simple pole and z= -2 is a pole of order 2. Both singular points are outside of the circle .

21z The function

is analytic inside and on C

The value of the integral is zero by Cauchy Goursat Theorem.

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(b) 23: zC . Then z = 1is inside the contour and

z = -2 is outside zfzLtzfsz

11,Re1

91

21

21 zLtz By Cauchy residue theorem

C

izz

dz9

221 2

LECTURE 17

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(c) .3: zc Both singular points are inside the contour.

Resdue at zfz

dzdLtz

z

2

22

!112

91

11

11

222 zLt

zdzdLt

zz

091

912

21 2C

izz

dz

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Evaluate by residue theorem.

19. C

z dzez 21

3

(a) 5z (b) 2iz (c) 13z z=0 is an essential singular point.

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.......!31

!2111 642

31

3 2

zzzzez z

.......!31

!21

33

zzzz

is the Laurent series.

Residue = .21

!21

. In (a) and (b) z = 0 lies inside the contour. In both

iidzezC

z

2122

13

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(c) z = 0 lies outside 13z

021

3

C

zez by C.G. Th.

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21. 33:

1342 izCzz

dz

C The singular point are given by

25216401342 zorzz

ii 322

64

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Distance bet 3i and -2 + 3i = Distance bet (0,3) and (-2, 3)

304 2

i32 is inside the contour

Distance bet -2-3i and 3i

LECTURE 17

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i323334 2 outside of the

circle 33iz

Residue at izizizLti

iz 32323232

32

362

61

iiI

i

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Q.26 C

zz

dz 2,14 is the contour.

3,2,1,011 42

41

4 kezorzik

iieeezii

,1,,1,,1 23

12

1,011 22 izzz are the singular

points.

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All are poles of order 1 as denominator of these points = 0 put 0zq and 0zp at these points. All the singular points are inside the contour.

Residue = 341zzq

zp

Residue at 411

Residue at ii

41

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Residue at iii

41

41

3 and residue at 411

Sum of the residues = 0

0I

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Q.25 C

z

zCdzzze .2:,

12 Both singular points are inside the circle both are simple poles

22

zz ez

zezqzpresidue

Residue at 21 eisz

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Residue at 21

1eisz

22

1

1

2

eeidzzze

C

z

i1cosh2

LECTURE 17

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Q.29 C

zdz,cot C is the rectangle defined by

xx ,21

1,1 yy

zzzf

sincos

3,2,1,0,,0sin nnzz

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Three singular points z = 1,2,3 are inside the contour. All these are simple poles since

0,0,0 zqzqzp at these points residue 1

coscos

zz

zqzp

Remain same for all poles

c

iizdz 61112cot

LECTURE 17

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LECTURE 19

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LECTURE 19

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LECTURE 19

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LECTURE 19

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Q.11 2,1,321

345 2

zzzfzg

zzzzzzF

z = -3 are zeros of f(z) and g(z) is not zero at these points: These are simple poles of F(z). Residue at

62

123234511

2

11 zzzzLtzfzLtz

zz

Residue at

9/22/2010

313134522

2

22 zzzzLtzfzLtz

zz Residue at

2134533

2

33 zzzzLtzFzLtz

zz 302

60

iidzzzz

zz

zC

10303162321

345

4:

2

9/22/2010

Q.13 0.cos

32z

zzzzf

is a zero of order of the denominator and z = 0 is a not a zero of the numerator. Hence z = 0 is a pole of order 2. Similarly z is a pole of order 3.

Residue at zfzdzdLtz

z

2

0 !110

34

030

sincos3cosz

zzzLtz

zdzdLt

zz 43

9/22/2010

Residue at

zfzdzdLtz

z

32

2

!21

4

2

22

2 cos2sin21cos

21

zzzzz

dzdLt

zz

dzdLt

zx

3

cos2sin21

zzzz

dzdLt

z

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6

23 cos2sin23sin2cossin21

zzzzzzzzzLt

z

4

cos2sin3sin2cossin21

zzzzzzzzzLt

z

4

2 621

LECTURE 20

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2

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Q.17 C zzdz

221 Singular points are z = 1, z = -2, z = 1 is a simple pole and z= -2 is a pole of order 2. Both singular points are outside of the circle .

21z The function

is analytic inside and on C

The value of the integral is zero by Cauchy Goursat Theorem.

9/22/2010

(b) 23: zC . Then z = 1is inside the contour and

z = -2 is outside zfzLtzfsz

11,Re1

91

21

21 zLtz By Cauchy residue theorem

C

izz

dz9

221 2

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(c) .3: zc Both singular points are inside the contour.

Resdue at zfz

dzdLtz

z

2

22

!112

91

11

11

222 zLt

zdzdLt

zz

091

912

21 2C

izz

dz

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Evaluate by residue theorem.

19. C

z dzez 21

3

(a) 5z (b) 2iz (c) 13z z=0 is an essential singular point.

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.......!31

!2111 642

31

3 2

zzzzez z

.......!31

!21

33

zzzz

is the Laurent series.

Residue = .21

!21

. In (a) and (b) z = 0 lies inside the contour. In both

iidzezC

z

2122

13

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(c) z = 0 lies outside 13z

021

3

C

zez by C.G. Th.

LECTURE 20

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3

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21. 33:

1342 izCzz

dz

C The singular point are given by

25216401342 zorzz

ii 322

64

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Distance bet 3i and -2 + 3i = Distance bet (0,3) and (-2, 3)

304 2

i32 is inside the contour

Distance bet -2-3i and 3i

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i323334 2 outside of the

circle 33iz

Residue at izizizLti

iz 32323232

32

362

61

iiI

i

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Q.26 C

zz

dz 2,14 is the contour.

3,2,1,011 42

41

4 kezorzik

iieeezii

,1,,1,,1 23

12

1,011 22 izzz are the singular

points.

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All are poles of order 1 as denominator of these points = 0 put 0zq and 0zp at these points. All the singular points are inside the contour.

Residue = 341zzq

zp

Residue at 411

Residue at ii

41

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Residue at iii

41

41

3 and residue at 411

Sum of the residues = 0

0I

LECTURE 20

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4

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Q.25 C

z

zCdzzze .2:,

12 Both singular points are inside the circle both are simple poles

22

zz ez

zezqzpresidue

Residue at 21 eisz

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Residue at 21

1eisz

22

1

1

2

eeidzzze

C

z

i1cosh2

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Q.29 C

zdz,cot C is the rectangle defined by

xx ,21

1,1 yy

zzzf

sincos

3,2,1,0,,0sin nnzz

9/22/2010

Three singular points z = 1,2,3 are inside the contour. All these are simple poles since

0,0,0 zqzqzp at these points residue 1

coscos

zz

zqzp

Remain same for all poles

c

iizdz 61112cot

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LECTURE 20

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1

MATH ZC232

Engineering Mathematics II

Tutorial 1


13/8/2010 MATH ZC232 1

Important Formulae

2222

2222

2222

1

)sin()cos(

)cosh()sinh(

)cos()sin(

1...3,2,1,!

basbbteL

basasbteL

kssktL

kskktL

kssktL

kskktL

aseLn

sntL

tata

tan

n

13/8/2010 MATH ZC232 2

Inverse Transform

221

221

221

221

221

221

11

1

)sin()cos(

)cosh()sinh(

)cos()sin(

1...3,2,1,!

basbLbte

basasLbte

kssLkt

kskLkt

kssLkt

kskLkt

asLeLn

snLt

tata

tan

n

13/8/2010 MATH ZC232 3

13/8/201013/8/2010 MATH ZC232MATH ZC232 4

000 121 nnnnn ffsfssFstfL

duutgufLgfLt

0sGsF

duutgufsGsFLt

0

1

sFdttfte nnnst 10

asFtfeL at


L{f(t-a) U(t-a)} = e-as F(s)

0)}({ 0stettL

Transform of Derivatives

)0()0()0()()(

)0()0()()(

)0()()(

'''23'''

'2''

'

ffsfssFstfL

ffssFstfL

fsFstfL

13/8/2010 MATH ZC232 6

TUTORIAL 1

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2

Find

(a) Since

So by shifting theorem

And finally using multiplication theorem

13/8/2010 MATH ZC232 7

2[cos ]1

sts

L

22[ co 2

1 (2s ]

)te t s

sL

2

22 2

2[ cos ] 2 3 41 (2 ) 5 4

t s s ss

dte tds s s

L

Question 1}cos{ 2 tteL t

(b) We have

By Formula

13/8/2010 MATH ZC232 8

12

12

-2

62 9

6 33 2 9

2e sin 3t

s

s

t

L

L

13/8/2010 MATH ZC232 9

2 2( ) 0 (0) 4 ( ) 4 /s y x y sy y x sL L2

2

4( )4 5s y xs

s+ L

22 2

44

(4

1)y xss s

s+

L+

3

2

2

2

5( 44

)y xs

s ss

L+

By Partial Fraction

Taking Laplace transform on both sides, we get

Question 2

Solve y”(x) + 4y(x) = 4t, y(0)=1, y’(0)=5

13/8/2010 MATH ZC232 10

Finally Taking Laplace inverse Transform we get

( ) cos 2 2sin 2y x x x x

Example : Find

Since

Taking Laplace transform we get

2 1 cos2 3 1 1sin 3 1 1 cos 6 22 2

tt t

2 2

3

2 1sin 3 1 1 cos

18 3 si

6 22

si 23

n 1 n6

L

s ss s

t L L t

13/8/2010 11

2sin 3 1L t

MATH ZC232

Here, we have to find L-1 of , therefore

decomposing into partial fraction we get

Taking Laplace inverse transform we have

22

1

1

s

s

222

1 1 12( 1) 4( 1) 4( 1)

1

1 s s ss

s

1 1 1

2 21

21 1 1 1 1

( 1)1 1

2 4 ( 1) 4 ( 1)1 ssL L L L

s s s

13/8/2010 12

Example : Find 122

1

1

sLs

MATH ZC232

TUTORIAL 1

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3

1 1 1 122 2

1 1 1 1 1( 1) 4 ( 1) 4 ( 1)

1 1 12 4

121

4

1

t t t

sLs s

L L Ls s

te e e

13/8/2010 13MATH ZC232

Equation is

Taking Laplace transform on both sides we get

3 2 1 2 , 0 0 0y y y t t y y

3 2 1 2L y L y L y L t L t

2 2(0) '(0) 3 (0) 2 s ss L y sy y sL y y L y e e

13/8/2010 14

Example : Find solution ofusing Laplace Transform.

3 2 1 2 ,y y y t t0 0 0y y

MATH ZC232

2 2 2

2 13 2 2

s s s s s se e e e e eL ys ss s

2 2 2 1H 2 1 H 1t t ty t e e e t e t

2 23 2 s sL y s s e e

Applying Initial conditions, we get

Finally Taking Laplace inverse transform we get

13/8/2010 15MATH ZC232

Power Series

A series of the form

is known as a power series centered at .

)2(...0

2210

n

nn axcaxcaxccxy

a

Example (1):

0

32

!...

!3!21

n

nx

nxxxxe

is a power series about the origin.13/8/2010 MATH ZC232 16

Power series solution about ordinary point

)1(02

2

yxQdxdyxP

dxyd

The behavior of the solution of the equation

Near a point x0 depends on the behavior of the itscoefficient functions and near this point. A pointx0 is said to be an ordinary point of the differentialequation (1) if both and are well-behaved/analyticat x0 . So the functions P(x) and Q(x) can be representedby power series. The point that is not an ordinary point isknown as singular point.

xP xQ

xP xQ

13/8/2010 MATH ZC232 17

Working rules to solve the differentialequation by series solution aboutordinary points (say x = 0):

Assume its solution to be of the form

Calculate and substitute the value ofin the given equation.Equate to zero the coefficient of the various powersof x and determine in terms ofSubstituting the values of in theassumed solution, we get the desired series solutionhaving as its arbitrary constant.

......2210

nn xcxcxccxy

'',' yy '',', yyy

,...,, 432 ccc ., 10 cc,...,, 432 ccc

10 ,cc13/8/2010 MATH ZC232 18

TUTORIAL 1

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4


Singular points



xPxxxp 0

xQxxxq 20



13/8/2010 MATH ZC232 19


0xx

00

000

n

rnn

n

nn

r xxcxxcxxy



0n


Case 1.

If r1 and r2 are distinct and do not differ by an integer,there exists two independent solution y1(x) and y2(x),(like in the given example) of the form

0n

rnn xcy

13/8/2010 MATH ZC232 21

Case 2

If r1 - r2 = N, where N is a positive integer,then there exists two independent solution y1(x) and y2(x) of the form

0, 00

11 cxcxy

n

rnn

0,ln 00

122 bxbxxyCxy

n

rnn

where C is a constant that could be zero.

13/8/2010 MATH ZC232 22

Case 3

If r1 = r2, then there exists two linearly independent solutiony1(x) and y2(x) of the form

0, 00

11 cxcxy

n

rnn

0,ln 00

122 bxbxxyxy

n

rnn

The solution y2 (x) in this case can also be obtained using y1(x).

13/8/2010 MATH ZC232 23

13/8/2010 24

Here

Hence, x=0 is a regular singular point

0 2

22 2 2

0 2

1

1 1

xp x x x P x xx

xq x x x Q x x xx

0 0lim 1, lim 1x x

p x q x

Example

MATH ZC232

TUTORIAL 1

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5

Assume the solution of the formAnd putting in original equation we get,

0

n cn

n

y a x

2 2 1

0 0

2

0

1

( 1) 0

n c n cn n

n n

n cn

n

x a n c n c x x a n c x

x a x

13/8/2010 25MATH ZC232

2 1 0cHence, indicial equation is therefore, which has two equal root , hence the differential equation has only one Frobenious Series solution, Equating the coefficients of (n+k)th term we get

1c

13/8/2010 26MATH ZC232

Hence Frobenious Series solution of given equation is

13/8/2010 27MATH ZC232

13/8/2010 28

So putting c=1 in general recursion relation we get

MATH ZC232

13/8/2010 29

Here

Hence, x=0 is a regular singular point

0 2

2

2 2 20 2

1

1144

xp x x x P x xx

xq x x x Q x x x

x

0 0

1lim 1, lim4x x

p x q x

Example

MATH ZC232

13/8/2010 30MATH ZC232

TUTORIAL 1

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6

13/8/2010 31

Shifting the summation index, we get

MATH ZC232

13/8/2010 32

Hence, indicial equation is therefore

Equating the coefficient of to zero, we get n cx

MATH ZC232

13/8/2010 33

Similarly

Putting n=2, 3, 4, 5, ….

MATH ZC232

13/8/2010 34

Hence Frobenious Series solution of given equation is

Putting all these coefficients in assumed series solution, we get

MATH ZC232

13/8/2010 35

21 1( ) , ( )2 2

xP x x x Q x

0

m nn

ny x a x

Example: Solve

Here , so is a

regular singular point of the given differential equation. So let the

Solution of differential equation be of the form

0x

Differentiating above equation we have

1

0

2

0

' ( )

'' ( )( 1)

n mn

n

n mn

n

y a m n x

y a m n n m x

2

2 2

1 2 1 02 2

d y x dy ydx x dx x

MATH ZC232

13/8/2010 36

00

11 0

22 1

1 1. : ( 1) 02 21 1. : ( 1) ( 1) 02 2

1 1. : ( 2)( 1) ( 2) ( 1) 02 2

Coeff of x a m m m

Coeff of x a m m m a m

Coeff of x a m m m a m

1 2112

m and m

Putting the values of derivatives in given equation and then equating to zero the various powers of x, we have

So indicial equation becomes which gives1 1( 1) 02 2

m m m

MATH ZC232

TUTORIAL 1

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7

13/8/2010 37

01 0

12 1 0

21 1 52.1 .22 22 2 41 1 7 353.2 .32 2

aa a

aa a a

1 1,m

Since , , there exists two independent Frobenoius Series

solution of given equation which can be obtained by substituting values of m

in above coefficients.

1 23 Z2

m m

For

MATH ZC232

13/8/2010 38

0

1 0

1

2 1 0

12

1 1 1 1 1.2 2 2 2 2

11 12

3 1 1 3 1 2 2.2 2 2 2 2

aa a

aa a a

21 ,2

mFor

MATH ZC232

13/8/2010 39

21

1/2 22

2 415 35

112

y c x x x

c x x x

Let , then the two solutions are therefore, 0 1a

21

1/2 22

2 415 35

112

y x x x

y x x x

Hence complete or general solution of given differential equation is

MATH ZC232

TUTORIAL 1

Page 7 of 7

Maths Material From Bits

Documents

x x x x

f x p x

f x pis

also0 f f x q x p x

f x pn n

x p x fi i

i1f f f

f x p fori