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TRANSCRIPT
-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.
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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
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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)
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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
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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.
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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.
Exercise 6.3 Evaluate the integral 1
1
x dxe2
.
Using
(a) Two point Gauss-Legendre Quadrature.
(b) Three point Gauss-Legendre Quadrature.
Exercise 6.4 Evaluate the integral 1
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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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
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4/8/2010 MATH ZC232 23
4/8/2010 MATH ZC232 24
LECTURE 1
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LECTURE 1
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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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6/8/2010 MATH ZC232 1
MATH ZC232Engineering Mathematics II
Lecture 2
Dr. Deepmala Agarwal
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
If F(s) = L{f(t)} and a > 0, then
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 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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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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MATH ZC232Engineering Mathematics II
Lecture 3
Dr. Deepmala Agarwal
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
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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
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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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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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MATH ZC232Engineering Mathematics II
Lecture 4
Dr. Deepmala Agarwal
11/8/2010 MATH ZC232 1
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.
11/8/2010 MATH ZC232 2
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 xcy11/8/2010 MATH ZC232 3
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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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
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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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2rxrkx
or
11/8/2010 MATH ZC232 19
xr
ng xr
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C1
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5
23
11389
389 2
3
CC
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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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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.
Euler Method – Example 1
9/22/2010 MATH ZC232
8
Euler Method – Example 2
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
Euler Method – Example 2
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
Euler Method – Example 2
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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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
For our example this gives:
)1(*1 nnn yhyy
Predictor step (Euler method):
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19
Example (Modified Euler Method)
For our example this gives:
Corrector step :
115.0 *11 nnnn yyhyy
2,, *
111
nnnnnn
yxfyxfhyy
9/22/2010 MATH ZC232
20
Example (Modified Euler Method)
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
Example (Modified Euler Method)
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
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43
RUNGE-KUTTA FOURTH ORDER (RK-4) METHOD
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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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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
9/22/2010 MATH ZC232
9/22/2010 53MATH ZC232
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LECTURE 5
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LECTURE 5
Page 10 of 11
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9/22/2010 66MATH ZC232
LECTURE 5
Page 11 of 11
9/22/2010
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
9/22/2010 MATH ZC232 3
9/22/2010 MATH ZC232 4
9/22/2010 MATH ZC232 5
9/22/2010 MATH ZC232 6
LECTURE 5(cont)
Page 1 of 14
9/22/2010
2
9/22/2010 MATH ZC232 7
9/22/2010 MATH ZC232 8
9/22/2010 MATH ZC232 9
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
9/22/2010 MATH ZC232
9/22/2010 11MATH ZC232
9/22/2010 12MATH ZC232
LECTURE 5(cont)
Page 2 of 14
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9/22/2010 17MATH ZC232
9/22/2010 18MATH ZC232
LECTURE 5(cont)
Page 3 of 14
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4
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9/22/2010 22MATH ZC232
9/22/2010 23MATH ZC232
9/22/2010 24MATH ZC232
LECTURE 5(cont)
Page 4 of 14
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5
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9/22/2010 MATH ZC232 28
9/22/2010 MATH ZC232 29
9/22/2010 MATH ZC232 30
LECTURE 5(cont)
Page 5 of 14
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6
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9/22/2010 MATH ZC232 34
9/22/2010 MATH ZC232 35
9/22/2010 MATH ZC232 36
LECTURE 5(cont)
Page 6 of 14
9/22/2010
7
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
9/22/2010 42MATH ZC232
LECTURE 5(cont)
Page 7 of 14
9/22/2010
8
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)
Page 8 of 14
9/22/2010
9
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)
Page 9 of 14
9/22/2010
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)
Page 10 of 14
9/22/2010
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
9/22/2010 66MATH ZC232
LECTURE 5(cont)
Page 11 of 14
9/22/2010
12
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
3n2n1nnn1n 9f37f59f55f24hyy
1n3-n(IV)5 xfor x fh
720251
9/22/2010 72MATH ZC232
LECTURE 5(cont)
Page 12 of 14
9/22/2010
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)
1n y,xfy,x5fy,x19fy,x9f24hyy
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
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 76MATH 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 77MATH 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
9/22/2010 78MATH ZC232
LECTURE 5(cont)
Page 13 of 14
9/22/2010
14
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
on substituting in modified = 2.72998
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
similarly computation may continue to compute y(1.5), y(1.6) and so on.
9/22/2010 81MATH ZC232
LECTURE 5(cont)
Page 14 of 14
9/22/2010
1
MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 7
Dr. Saroj Kumar Sahani
9/22/2010 MATH ZC232 1
Numerical Solutionof
Differential Equation
9/22/2010 2MATH ZC232
9/22/2010 MATH ZC232 3
9/22/2010 MATH ZC232 4
9/22/2010 MATH ZC232 5
9/22/2010 MATH ZC232 6
LECTURE 7
Page 1 of 16
9/22/2010
2
9/22/2010 MATH ZC232 7
9/22/2010 MATH ZC232 8
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
Page 2 of 16
9/22/2010
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
3n2n1nnn1n 9f37f59f55f24hyy
1n3-n(IV)5 xfor x fh
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)
1n y,xfy,x5fy,x19fy,x9f24hyy
1n2-n(IV)5c x x,f h
72019E
9/22/2010 MATH ZC232 18
LECTURE 7
Page 3 of 16
9/22/2010
4
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 19MATH 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 20MATH 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
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
on substituting in modified = 2.72998
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
similarly computation may continue to compute y(1.5), y(1.6) and so on.
9/22/2010 23MATH ZC232
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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.
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Fourier Series
9/22/2010 MATH ZC232 36
LECTURE 7
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For all
For all
LECTURE 7
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LECTURE 7
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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
Eq. (6) & (7) are ODEs whose solutions are found by usual methods.Solution of (6) is:Solution of (7) is:
c4, c5 and c6 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
(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:
– Method of choice: Separation of Variables
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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MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 8
Dr. Saroj Kumar Sahani
LECTURE 8
Page 1 of 21
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For all
For all
LECTURE 8
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LECTURE 8
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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
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
LECTURE 8
Page 6 of 21
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7
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
Eq. (6) & (7) are ODEs whose solutions are found by usual methods.Solution of (6) is:Solution of (7) is:
c4, c5 and c6 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
(6)
(7)
xcxcX 2sin2cos 54yecY
2
6
yecxcxcyxu2
654 2sin2cos, required
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
LECTURE 8
Page 7 of 21
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8
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:
– Method of choice: Separation of Variables
02
2
2
2
yT
xT
ax0 by0
00xx
T
0axx
T by0
ax000,xT
xfbxT ,
x
y
a
b
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.
LECTURE 8
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MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 10Tutorial 2
Dr. Saroj Kumar Sahani
9/22/2010 MATH ZC232 1
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Repeat the previous example with Modefied Euler’s Algorithm
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LECTURE 10
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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
10 Feb 2009 MATH ZC232, Eng Maths II 20
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
10 Feb 2009 MATH ZC232, Eng Maths II 21
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 10
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1
MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 11
Rajesh Bhatt
9/22/2010 MATH ZC232 1
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
9/22/2010 MATH ZC232 2
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
9/22/2010 MATH ZC232 3
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
9/22/2010 MATH ZC232 4
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)
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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
……?9/22/2010 MATH ZC232 6
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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
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SOV Method Continued…….• Case II: 2 <0
Eq. (6) & (7) are ODEs whose solutions are found by usual methods.Solution of (6) is:Solution of (7) is:
c4, c5 and c6 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
(6)
(7)
xcxcX 2sin2cos 54yecY
2
6
yecxcxcyxu2
654 2sin2cos, required
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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
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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 Next9/22/2010 MATH ZC232 10
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 Next9/22/2010 MATH ZC232 11
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 Next9/22/2010 MATH ZC232 12
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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 Next9/22/2010 MATH ZC232 13
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:
– Method of choice: Separation of Variables
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
• 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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MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 12
Rajesh Bhatt
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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
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MATH ZC232
Lecture 14By
Rajesh Bhatt
1MATH ZC2326/09/2010
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MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 15
Rajesh Bhatt
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Lecture 16By
Rajesh Bhatt
1MATH ZC23210/09/2010
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MATH ZC 232ENGINEERING MATHEMATICS II
LECTURE 17Rajesh Bhatt
9/22/2010
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
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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
9/22/2010
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
LECTURE 17
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6
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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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7
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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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8
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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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9
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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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10
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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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11
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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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12
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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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13
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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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1
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MATH ZC232Engineering Mathematics II
Rajesh BhattLecture 18
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LECTURE 18
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2
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LECTURE 18
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LECTURE 18
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LECTURE 18
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MATH ZC232Engineering Mathematics II
Rajesh BhattLecture 19
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LECTURE 19
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2
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LECTURE 19
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3
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MATH ZC232Engineering Mathematics II
Rajesh BhattLecture 20
9/22/2010
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
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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
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
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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
LECTURE 20
Page 1 of 6
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.
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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
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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
Page 2 of 6
3
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21. 33:
1342 izCzz
dz
C The singular point are given by
25216401342 zorzz
ii 322
64
9/22/2010
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
9/22/2010
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
Page 3 of 6
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
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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
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1
MATH ZC232
Engineering Mathematics II
Tutorial 1
Dr. Deepmala Agarwal
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
If F(s) = L{f(t)} and a > 0, then
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
Page 1 of 7
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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
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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
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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
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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
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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
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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
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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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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.
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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 xcy13/8/2010 MATH ZC232 20
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
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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.
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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).
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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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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
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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
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Hence Frobenious Series solution of given equation is
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So putting c=1 in general recursion relation we get
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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
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Shifting the summation index, we get
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Hence, indicial equation is therefore
Equating the coefficient of to zero, we get n cx
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Similarly
Putting n=2, 3, 4, 5, ….
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Hence Frobenious Series solution of given equation is
Putting all these coefficients in assumed series solution, we get
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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
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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
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TUTORIAL 1
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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
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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
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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
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