NATIONAL OPEN UNIVERSITY OF NIGERIA SCHOOL OF SCIENCE AND TECHNOLOGY COURSE CODE: MTH 302- COURSE TITLE: ELEMENTARY DIFFERENTIAL EQUATIONS II
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NATIONAL OPEN UNIVERSITY OF NIGERIA
SCHOOL OF SCIENCE AND TECHNOLOGY
COURSE CODE: MTH 302-
COURSE TITLE: ELEMENTARY DIFFERENTIAL EQUATIONS II
MTH 302- ELEMENTARY DIFFERENTIAL EQUATIONS II Course Team: Dr. O.J. Adeniran (Writer) – UNI. of Agric. Abeokuta Dr. Bankole Abiola (Editor) – NOUN Dr. Bankole Abiola (Programme Leader) – NOUN
Dr S.O. Ajibola (Course Coordinator) – NOUN
NATIONAL OPEN UNIVERSITY OF NIGERIA
CONTENT Module 1: Series Solution of Ordinary Differential Equation UNIT 1: Series Solution of Differential Equation UNIT2: Euler Equation UNIT3: Indicial Equation with Difference of Roots- Positive Integer and Logarithmic case UNIT 4: Boundary Value Problems UNIT 5: Sturm and Liouville Problem ODULE 1: SERIES SOLUTION OF ORDINARY DIFFERENTIAL EQUATION UNIT 1: SERIES SOLUTION OF DIFFERENTIAL EQUATIONS 1.0 Introduction 2.0 Objectives 3.0 Main Content 3.1. Series Solution of Differential Equation 3.2 Method of finding radius of Convergence 3.3 Ordinary points and Singular points of the Differential Equation 3.3.1 Solution near Ordinary Point 4.0 Conclusion 5.0 Summary 6.0 Tutor Marked Assignment 7.0 References/ Further Readings 1.0. Introduction: A large class of ordinary differential equations possesses solution expressible, over a certain interval, in terms of power series. In this unit we are going to investigate methods of obtaining such solutions. 2.0. Objectives: At the end of this unit you should be able to - determine radius of convergence of series - apply series solution method to solving differential equation - determine ordinary point, and singular points of the differential equation
3.0 MAIN CONTENT 3.1. Series Solution of Ordinary Differential Equation An expression of the form
...)(...)(...)(0
0010 +−=+−+−+ ∑∞
=
nyn
n
nn xxAxxAxxAA (1)
is called the power series. To determine for what values of x the series (1) converges we use ratio test
01
0
101 limlim )(
)(xxL
T
T
xxA
xxA
n
n
nn
n
nn
n
−==−−= +
∞→
++
∞→ρ
Where n
n
n A
AL 1lim +
∞→= (2)
The series is convergent when1<ρ , divergent when 1>ρ . The test
fails if 1=ρ . L
1=ρ is called the radius of convergence
The series converges when
RL
xx =<− 10 (radius of convergence)
diverges when
RL
xx =>− 10
(i) If L is zero, the series converges for all Values of x (ii) If L is infinite, the series converges only at the point 0xx = (iii) If L is finite, then the series converges, when
RL
xx =<− 10 (radius of convergence) and diverges if
Lxx
10 >−
If nyn
n
xxa )(0
−∑∞
=
and nyn
n
xxb )(0
−∑∞
=
Converge to )(xf and )(xg respectively, for 10 ρ<− xx (radius of
convergence) 01 >ρ , then the following are true for 10 ρ<− xx .
(i)Two series can be added and subtracted term wise, and
nynn
n
xxbaxgxf ))(()()(0
−+=± ∑∞
=
(ii) The series can be multiplied and n
nn
nn
n
nyn
n
xxCxxbxxaxgxf )()()()()( 00
000
−=−−= ∑∑∑∞
=
∞
=
∞
=
Where 022110 ... babababaC nnnnn ++++= −−
If 0)( 0 ≠xg , the series
nyn
n
xxdxg
xf)(
)(
)(
0
−=∑∞
=
although formula for nd is complicated if nyn
n
xxaxf )()(0
−=∑∞
=
,then )(xf is
continuous has derivates of all orders for 10 ρ<− xx . and f ′ , f ′′ , f ′′′ … can be computed
by differentiating the series. Thus
!
)( 0
n
xfa
n
n = or ny
n
n
xxn
xfxf )(
!
)()( 0
0
−=∑∞
=
……. (3)
(3) is called the Taylor series for function f at 0xx = A function f that has Taylor series expansion about 0xx =
ny
n
n
xxLn
xfxf )(
)()( 0
0
−=∑∞
=
With a radius of convergence 0>ρ is said to be analytic at 0xx = . The polynomial is analytic at every point, thus sums, differences, products, quotients (excepts at the zeroes of the denominator) of polynomials are analytic at every point.
(i) Determine the radius of convergence of the power series
(i) nn
n
x20∑
∞
=
(ii) 2
0
)12()(
n
xxf
n
n
+=∑∞
=
(iii) 2
1
2
21lim == +
∞→n
n
n
ρ
(iv)2
1
1lim ==
+∞→ n
n
n A
Aρ (v) 2
111
2
1
2
)1(22
01
2
2 limlim =
+=+=→
+∞→ n
n
n nn
n
n
ρ
3.2. Determining the Radius of Convergence If we obtain the Taylor series of a function )(xf about a point 0x , then the radius of convergence of the series is equal to the distance of the point 0x from the nearest singularity.
Remark about a change in the index of summation
(a) 22
0
22
02
++
∞
=
++
∞
=
∞
=∑∑∑ == k
kk
nn
n
nn
n
xaxaxa
(b) nn
n
nn
n
xannxann 20
2
2
)1)(2()1( +
∞
=
−∞
=++=− ∑∑
( c) nn
n
nn
n
xaxa 22
2
0−
∞
=
+∞
=∑∑ =
(d) kpnmkn
n
pnmn
kn
xaxa ++++
∞
=
++
∞
=∑∑ =
0
3.3 Ordinary Points and Singular Points of the Differential Equations
We consider the differential equation
0)()()(2
2
=++ yxRdx
dyxQ
dx
ydxP (4)
(we assume that )(xP , )(xQ and )(xR are polynomials)
(a) if 0)( 0 ≠xP , then 0x is an ordinary point of the equation (1), or
)(
)(
xR
xQP = ,
)(
)(
xP
xRQ =
y , P , Q are analytic at the point 0xx = , then 0x is the ordinary point of the equation. (b) If the functions )(xP , )(xQ and )(xR are polynomials having no
common factors, the singular points of equation (1) are the points for which
0)( =xP (5)
(c) If )(
)()( 0lim
0xP
xQxx
xx
−→
is finite
and )(
)()( 2
0lim0
xP
xQxx
xx
−→
is finite
Then the point 0xx = is called the REGULAR SINGULAR POINT of equation (3)
(d) Any singular point of equation (3) that is not regular singular point is called an irregular singular point.
3.3.1 Solution Near An Ordinary Point Let us consider the equation
0)()()( =+′+′′ yxRyxQyxP (6) Where )(xP , )(xQ and )(xR are polynomials. 0x is the ordinary point of the equation (6). Assuming that )(xy φ= is a solution of (6) and )(xφ has a Taylor Series
nyn
n
xxaxy )()(0
−== ∑∞
=
φ (7)
Now we know that
!
)(
m
xa
m
m
φ= (8)
We can write (1) 0)()( =+′+′′ yxqyxPy
where )(
)(
xR
xQP = ,
)(
)(
xP
xRq =
∴ qyyPy −′−=′′ (9) or yqyqyPypy ′−′−′′−′′′−=′′′ (10) (It is natural to assume that )(xyy = , )(xyy ′=′ at 0xx = and 1)0( ay = ,
2)0( ay =′ , we can easily calculate the coefficient na a, provided that we could compute infinitely many derivatives of p and q existing at 0x . Thus p and q must have some condition for line calculation of na . It has been proved that.
)(
)(
xR
xQP = ,
)(
)(
xP
xRq = are analytic at 0x ., then the general solution of (6)
is )()()( 211000
xyaxyaxxay nn
n
+=−=∑∞
=
Where na and 1a are arbitrary 1y and 2y are linearly independent series solutions which are analytic at 0x .
We shall illustrate the method by examples. Example 1. Solve the equation
04 =+′′ yy
near the ordinary point 0=x Solution: we assume the solution as
nn
n
xay ∑∞
=
=0
(1)
2
0
)1( −∞
=
−=′′ ∑ nn
n
xanny (2)
Substituting these values in the equation yields
04)1(0
2
0
=+− ∑∑∞
=
−∞
=
nn
n
nn
n
xaxann (3)
or 04)1( 22
2
2
0
=+− −−
∞
=
−∞
=∑∑ n
nn
nn
n
xaxann (4)
or [ ] 04)1( 22
20
=+− −−
∞
=
∞
=∑∑ n
nn
nn
xaann (5)
Because the first two terms of the first sum in (4) are zero. We now use the fact that for a power series to vanish identically over any interval, each coefficient in the series must be zero Recurrence relation :
04)1( 2 =+− −nn aann or )1(
4 2
−−= −
nn
aa n
n , 2≥n
Now we calculate in coefficients
1.2
4 02
aa
−= , 2.3
4 13
aa
−= , 3.4
4 24
aa
−= , 4.5
4 35
aa
−=
)12(2
4 222 −
−= −
kk
aa k
k , )2)(12(
4 1212 kk
aa k
k +−= −
+
From above we have
22202...2 ....!2
4)1(. −
−= k
kk
kn aaak
aaa
02 !2
4)1(a
ka
kk
k
−= , 112 )!12(
4)1(a
ka
kk
k +−=+
Hence we can write in solution
[ ] [ ]
xSinaxCosa
xk
kaxk
a
xak
xaxak
a
xaxaxaaxay
kk
k
kk
k
kkk
k
kkk
k
kk
k
kk
k
nn
n
22
12
)2()!12(
)1(2
2
1)2(
!2
)1(1
)!12(
4)1(
!2
4)1(
10
12
01
2
10
121
01
20
10
1212
11
22
10
0
+=
+−++−+=
+−++−+=
+++==
+∞
=
∞
=
+∞
=
∞
=
++
∞
=
∞
=
∞
=
∑∑
∑∑
∑∑∑
Example 2: Solve the equation
046)1( 2 =−′−′′− yyxyx near the ordinary point 0=x Solution: we assume the solution
nn
n
xay ∑∞
=
=0
(1)
The only singular points of the equation in the finite plane are 1=x and 1−=x . Hence we show that the solution is valid in 1<x with 0a and 1a
arbitrary coefficients
046)1()1(000
2
0
=−−−−− ∑∑∑∑∞
=
∞
=
∞
=
−∞
=
nn
n
nn
n
nn
n
nn
n
xaxnaxannxann
or nn
n
nn
n
xannxann )45()1( 2
0
2
0
++−− ∑∑∞
=
−∞
=
or nn
n
nn
n
xannxann ))4)(1()1(0
2
0
++−− ∑∑∞
=
−∞
=
Let us shift the index the second series.
0))2)(1()1( 22
2
2
0
=+−−− −−
∞
=
−∞
=∑∑ n
nn
nn
n
xannxann (2)
In equation (2), the coefficient of each power of x must the zero.
2
2−
+= nn a
n
na , for 2≥n (3)
(3) is called recurrence relation. A recurrence relation is a special kind of difference equation.
,...6,4,2=n and ,...7,5,3=n
02 2
4aa = , 13 3
5aa =
24 4
6aa = 35 5
7aa =
. .
. .
222 2
12−
+= kk ak
ka 1212 12
32−+ +
+= kk ak
ka
1≥k 02 )1( aka k += Similarly, 1≥k
112 3
32a
ka k
+=+
Hence the solution nn
n
xay ∑∞
=
=0
[ ] [ ]
22
31
20
12
11
2
10
1212
11
22
10
)1(3
)3(
)1(
3
32)1(1
x
xxa
x
a
xk
xaxka
xaxaxaay
k
k
k
k
kk
k
kk
k
−−+
−=
+++++=
+++=
+∞
=
∞
=
++
∞
=
∞
=
∑∑
∑∑
Example 3. Solve the equation
0)1(4)1( 2 =−−′−+′′ yxyxy about the ordinary point 1=x Solution: we assume the solution
nn
n
xay )1(0
−=∑∞
=
( 1)
We first translate the axes, putting
ux =−1 , dx
dy
dx
du
du
dy =.
dx
dy
du
dy =1. The equation becomes
0422
2
=−+ uydu
dyu
du
yd
Then we assume the solution n
nn
uay ∑∞
==
0
04)1( 1
0
1
0
2
0
=−+− +∞
=
+∞
=
−∞
=∑∑∑ n
nn
nn
n
nn
n
uaunauann
Collecting the terms
0)4()1( 1
0
2
0
=−+− +∞
=
−∞
=∑∑ n
nn
nn
n
uanuann
Shifting the index from n to 3−n in the second series
0)7()1( 213
3
2
0
=−+− −−
∞
=
−∞
=∑∑ n
nn
nn
n
uanuann
Therefore 0a and 1a are arbitrary and for remainder, we have 02 2 =a
3≥n
3)1(
7−−
−= nn ann
na
10a arbitrary 11a arbitrary 010 =a
03 2.3
4aa
−= 14 3.4
3aa
−= 04.5
225 =−= aa
36 5.6
1aa
−= 0)6(7
047 == aa 0() 58 == aa
69 8.9
2aa
−= 09.10
3710 =−= aa 01 =a
- - - - - -
333 )13(3
73−−
−= kk akk
ka 013 =+ka , 2≥k 022 =+ka , 1≥k
1≥k : [ ][ ][ ] 63 )13...(8..5.2)3....(9.6.3
)73...(2)...1)(4()1(a
kk
ka
k
k −−−−−=
[ [ ][ ][ ] )
4
1(
)13...(8..5.2)3....(9.6.3
)73...(2)...1)(4()1(1 4
13
10 uuau
kk
kay k
k
k
++−
−−−−+= ∑∞
=
Now substitute 1−= xu
[ [ ][ ][ ] )
4
1(
)13...(8..5.2)3....(9.6.3
)73...(2)...1)(4()1(1 4
13
10 uuau
kk
kay k
k
k
++−
−−−−+= ∑∞
=
4.0 Conclusion: In this unit we have attempted the series solution method to Ordinary Differential Equations. In the subsequent unit we are going to discuss more about this method in greater details. You are supposed to master this unit properly to be well equipped for the next unit. 5.0 Summary: Recall that in this unit we discuss power series and radius of convergence for the series. We also applied the series to solve differential equations. We derived the singular and ordinary points for each of the series solutions. Study this unit properly before going to the next unit.
6.0. Tutor Marked Assignments 1. Determine a lower bound for the radius of convergence of series solution about each given point 0x for each of the following differential equations. (i) 04)32( 2 =+′+′′−− yyxyxx , 40 =x , 40 −=x and 00 =x (ii) 04)1( 3 =+′+′′+ yyxyx , 00 =x , and 20 =x 2. Determine whether each of the points 0,1− and 1 is an ordinary point, or regular singular point or irregular singular point for the following differential equation, (i) 032)1(2 224 =+′+′′− yxyxyxx (ii) 0)1(2)3( 2 =−+′−′′+ yxyxyx 7.0 REFERENCES/FURTHER READINGS EARL. A. CODDINGTON: An Introduction to Ordinary Differential Equations. Prentice-Hall of India FRANCIS B. HILDEBRAND: Advanced Calculus for Applications, Prentice-Hall, New Jersey EINAR HILLE: Lectures on Ordinary Differential Equations, Addison – Wesley Publishing Company, London.
UNIT 2: EULER EQUATION. 1.0 Introduction 2.0 Objectives 3.0 Main Content 3.1. Euler Equation 3.2 Series solution near a regular point 3.3 Indicial equation with equal roots 3.4 Indicial equation with difference of roots, a positive integer and Non- Logarithmic case. 4.0 Conclusion 5.0 Summary 6.0 Tutor Marked Assignment 7.0 References/ Further Readings 1.0 Introduction: In this unit we deal with a class of differential equation normally refer to as Euler Equation. This type of equation usually possesses solutions that are classified as regular singular points of the differential equations. Series solution of this class of equation must be attempted with different approach. We shall see this in our treatment of this system of equation in this unit. 2.0 Objective: At the end of this unit you should be able to - differentiate Euler equations from others. - use series solution approach to solve these categories of equations - solve problems relating to Euler equation 3.0. MAIN CONTENT 3.1. Euler Equation
0)(2
22 =++= y
dx
dyx
dx
ydxyL βα (1)
is known as Euler equation. It is easy to see that 0=x is a regular singular point of (1) In any interval not including the origin, (1) has a general solution of the form.
)()( 2211 xycxycy += ,
1y and 2y are linear, independent solution. Here we assume that (1) has a solution of the form
2xy =
)()()()( 222222 rFxxxxxxxL =+′+′′= βα Where
βα ++−= rrrrF )1()( (2) If r r is a root of the equation
2
)1()1( 2
1
βαα ur
−−+−−= (3)
2
)1()1( 2
2
βαα ur
−−−−−= (4)
))(()( 21 rrrrrF −−=∴ Case I 04)1( 2 >−− βα , then the roots are real and unequal and ),( 21 rr xxW is non-vanishing for )21 rr ≠ and 0>x . Thus the general solution is
2121
rr xcxcy += 0>x
case ll 04)1( 2 =−− βα , then 2
)1(21
−−== αrr
and we have only one solution 1)(1
rxxy = of the differential equation. We can obtain the second solution by the method of reduction. We consider a different approach to obtain the solution.
)()( rFxxL rr = If 1rr = , then
0)()( 111 == rFxxL rr
Now 21)()( rrrF −= , if we differentiate
)(rF i.e )(2)( 1rrrF −=′∴ and then set 1rr = , if given 0)( =′ rF ,it suggest that
)]([)( rFxr
xLr
rr
∂∂=
∂∂
rrr xrrrrxFxxxL )()(log)log( 1−+= We set 1rr = , thus
0)log( 1 =xxL r xxy r log1
2 =∴ 0>x is the second solution of (1) Thus the general solution is
1)log( 212rxxccy += , 0>x
Case III 04)1( 2 <−− βα ), in this case, the root are complex, say µλ ir +=1 , µλ ir +=
Thus the general solution is µλµλ iri xxcxcy 1)( 212
−+ +=
][ 21
µµλ ii xcxcx −+=
][log
2log
1
xixi ececxµµλ −+=
]logsin()logcos([ 21 xcxcx µµλ += It is always possible to obtain a real valued solution of Euler equation (1) in the interval, by making the following changes
ξd
dx
d)1(−= ,
2
2
2
2
ξd
d
dx
d =
in the equation, we have
,022
22 =+= u
d
du
d
ud βξ
ξξ
ξ 2ξ 02 >ξ
It is obtained as above. Since xx for 0>x
ξ=− x for 0>x
It follows that we read one, to replace for x by x in the above solution to
obtain real valued solution valid in any interval not containing the origin To solve the Euler equation (1)
02 =+′+′′ yyxyx βα in any interval not containing the origin substitute rxy = and compute the root 1r and 2r of the equation
0)1()( 2 =+−+= βα rrrF If the roots are real and unequal
If the roots are real and equal
1)log( 21
rxxxccy +=
If the roots are complex
xcxcxy logsin(logcos(( 21 µµλ +=
For an Euler equation of the form
21
21
rrxcxcy +=
0)()( 02
0 =+′−+′′− yyxxyxx βα Change the independent variable by
rxoxt )−= or suppose the solution
rxoxy )( −= Note: The situation for a general second order differential equation with a regular singular point is similar to that for an Euler equation. 2. Another method of obtaining the solution of Euler Equation
02 =+′+′′ yyxyx βα Solution: We make the change of variable zex = or xz log= and 0>x .
dz
dy
xdx
dz
dz
dy
dx
dy 1==
dz
dy
xdx
dz
dz
yd
xdx
yd =−=2
2
2
2 11
Substituting there value in the equation
0)1(2
2
=+−+ ydx
dz
dx
yd βα
This is an equation with constants coefficients The auxiliary equation is
0)1(2 =+−+ yrr βα (i) If 1r and 2r are real and unequal. 212
2121
1rrzrzr xcxcececy +=+=
(ii) If the roots are equal i.e zrezccy )( 21 +=∴ . 1)log( 21rxcc α+=
(iii) If the roots are complex )sinhcos( 21 zczcey z += µλ
)logsin()logcos(( 21 xcxcx µµλ ++= 3.2 Series solution near a regular singular point Consider the equation
0)()()( =+′+′′ yxRyxQyxP (1)
Assume that 0=x is a regular singular point of (1) means that )(
)()(
xP
xxQxxP = =
and )(
)()(
22
xP
xRxxqx = have finite limits as 0→x and are analytics at 0xx = for
some interval about the origin
(i) can be written 0)]([)]([ 22 =+′+′′ yxqxyxxpxyx
But nn
n
xPxxp ∑∞
=
=0
)(
nn
n
xqxqx ∑∞
=
=0
2 )(
0]..................[......].....[ 1012 =+++++′++++′′ yxqxqqyxpxppxyx n
xno (2)
If all the coefficient use zeros, except 0P and nq , then (2) reduces to Euler equation, which was discussed previously. If some of the nP and nq . ≥n will not be zero. However the essential character of the solution remains the same. It is natural to seek the solution of the form of “Euler Solution” the power series.
nn
n
r xaxy ∑∞
=
=0
(3)
As part of our problem we have to determine
(1) The values of r for which equation (1) has a solution of the form (3)
(2) The recurrence relation for the na
(3) The radius of convergence of the series nn
n
xa∑∞
=0
We shall illustrate the method by example Example I: Find the series solution of the equation
02)1(2 =−′++′′ yyxyx (1) Solution: 0=x is the regular singular point of the equation. We assume line solution
rnn
n
xay +∞
=∑=
0
(2)
Direct substitution of y in (2) given
02)()()1)((200
1
00
=−++++−++ +∞
=
+∞
=
−+∞
=
+∞
=∑∑∑∑ rn
nn
znn
n
rnn
n
rnn
n
xaxarnxarnxarnrn
Now we shift the index of the second series in (3). We get
0)2()122)((00
=−++−++ +∞
=
−+∞
=∑∑ rn
nn
lrnn
n
xarnxarnrn (3)
Once more we reason that the total coefficient of each power of x in the left member of (4) must vanish. The second summation does start the contribution, until 1=n . Hence the equation determinants c and an are given by
0=n 0)12( 0 =− arr , but 00 ≠a (5) 0)12( =−∴ rr (6)
(6) is called the indicial equation.
2
11 =∴r , 02 =r 1≥n
0)3()122)( 1 =−++−++ −nn arvarnrn Recurrence relation
1)122)((
3−−++
−+−=∴ nn arnrn
rna (7)
(i) Take 2
1=r
3.2
)3( 01
aa
−=
5.4
)1( 12
aa
−=
7.6
)1( 23
aa
−=
…………
12(.2
)52( 1
+−
−= −
nn
ana n
n
)]12.....(7.5.3)][2........(6.4.2[
)]52)...(1)(1])(3[()1( 0
+−−−−
=nn
ana
n
n
Omitting the constant 0a , we may write the particular solution as
)12)(12)(32(12
3)1( 2
1
11 +−−
−++
∞
=∑
nnnn
xxy
n
nn
n
yz (8)
Next task is to find the solution corresponding to the root 02 =c . The recurrence relation becomes1≥n .
0)3()12( 1 =−+− −nn bnbnn
0r 0)12(
)3(1 =
−−= −nn bnn
nb
1.1
)2( 01
bb
−=
3.2
)1( 12
bb
−=
5.3
)1( 22
bb
−=
)12(
)0( 1
−= −
nn
bb n
n
0=∴ nb if )12(
)3( 1
−−
−= −
nn
bnb n
n
0=∴ nb if 3≥n
01 bb =∴ , 012 6
1
6
1bbb ==∴
The solution is
]3
121[ 2
02 xxby ++=
The general solution is 21 ByAyy +=
Note: The roots of indicial equation are unequal and do not differ by integer 3.3Indicial equation with equal roots
Example 2. Solve the equation
0)21(32 =−+′+′′ yxyxyx (1) Solution: 0=x is regular singular point of (i) We assume the solution
rnn
n
xay +∞
=∑=
0
(2)
Substituting this value in (2), we have
02)(3)1)(( 1
0000
=−+++−++ ++∞
=
+∞
=
+∞
=
+∞
=∑∑∑∑ rn
nn
rnn
n
rn
n
rnn
n
xaxaxrnxarnrn
Shifting the index
02]1)(2)[( 11
2
0
=−++++ +−
∞
=
+∞
=∑∑ rn
nn
rnn
n
xaxarnrn (3)
The indicial equation 0122 =++ rr
1−=∴r The recurrence relation is
21
)1(
2
++= −
rn
aa n
n (4)
,1≥n In which
,1≥n2
0
)]12)........(32)(2[(
2
++++=
nr
aa
n
n (5)
,)(),(1
2 rnn
n
xraxrxy +∞
=∑+=∴ (6)
in which
,1≥n2)]1)........(32)(2[(
2)(
++++=
nrrra
n
n (7)
Let as write yxyxyxyL )21(3)( 2 −+′+′′=
The y of equation (6) has been so determined that for that y the Eight member of (8) reduced to a single term the0=n . Thus
yxyxyxyL )21(3)( 2 −+′+′′= (8) A solution of the original differential equation is a function y for which
0)( =yL . Now taking lr −= makes 0)]1,([ =−xyL Now differentiate each member of (9) with respect to
])1[()],([ 22 xrr
rxydr
+∂∂=∂
xxrxrrxydr
l log)1()1(2)],([ 222 +++=∂ (10)
From (9) and 10, it can be seen early that the two solution of the equation 0)( =yL are
)],([1 rxyy = )1,([ −= xyy (11)
and 12 )],([ −=∂∂= rrxyr
y (12)
rnn
n
r xraxrxy +∞
=∑+= )(),(
1
xxaxraxxrxydr
rnn
n
rnn
n
log)2()(log)],(11
2 +∞
=
+∞
=∑∑ +′+=∂
rnn
n
xraxrxy +∞
=
′+= ∑ )(log),(1
rnn
n
xaxy −∞
=
−+∴= ∑ )1(log1
1
rnn
n
xaxy −∞
=
− −′+∴= ∑ )1(1
11
( )2)]12)...(3)2[(
2)(
++++ nrrra
n
n
)]2log(......)2[log(22log)(log rnrra nn +++++=
]1
1........
3
1
2
1)[(2)(
+++
++
+=′′
nrrrrara nn
1−=r , we obtain
2)1(
2)1(
na
n
n =−
]1
.......3
1
2
11[
)1(
22)1(
2 nna
n
n +++−=−′∴
We write
nH n
1.......1
2
1 +=
The solutions are
2
1
1
11 )!(
2
n
xxy
nn
n
−∞
=
− ∑+=
2
11
112 )!(
2log
n
xHxxy
nn
n
n
−+∞
=∑−=
The general solution, valid for all finite 0≠x is 21 ByAyy += .
3.3: Indicial Equation with difference of Roots a Positive Integer, non Logarithmic case Solve line equation
02)4( =+′+−′′ yyxyx (1)
Solution: We assume the solution
rnn
n
xay +∞
=∑=
0
(2)
11
1
1
0
)3()6)(()( −+−
∞
=
−+∞
=
−+−++=∴ ∑∑ rnn
n
rnn
n
xarnxarnrnyL (3)
The Indicial equation is
0)5( =−cc 02 =∴c , 51 =c
505 =−=∴ s We reason that we hope for two power series solutions, one starting with an
0x and term, one with an 5x term. If we use the longer root 5=c , then the 0x term would never enter. Thus we use the smaller roots 0=c , then the trial solution of the form.
nn
n
xay ∑∞
=
=0
(4)
has a chance of picking up both solutions because the 5=n sn = term does contain 5x
0)()3()5()( 11
1
1
0
=−−−=∴ −−
∞
=
−∞
=∑∑ yLxanxannyL n
nn
nn
n
0)( =yL
1)3()5( −−=− nn anann (5)
1=n 024 01 =+− aa oaa2
11 =∴
2=n 06 12 =+− aa oaa12
12 =
3=n 006 13 =+− aa 03 =a
4=n 04 34 =+− aa 04 =a
5=n 02.0 45 =+ aa 04 =a
6≥n
1.6
3 66
aa =
2.7
4 67
aa =
---------------
)5(
))(3( 1
−−
= −
nn
ana n
n
1)5](......8.7.6[
)3....(5.4.3 5
−−
=nn
anan
1)5()1)(2(
5.4.3 5
−−−=
nnnn
aan
Therefore, with 0a and 5a arbitrary, the general solution may be written
)12
1
2
11( 2
0 xxay ++=
)2_(1(1)5(
60[
6
55 −−−
+∑∞
= nnnn
xxa
n
n
)2
11( 2
0 xxa ++=
])3)(4)(5(1
60[
5
0
55 +++
+=+∞
=∑
nnnn
xxa
n
n
Example 2 Solve the equation 03)34( =+′++′′ yyxyx (1) Solution 0=x is the regular singular point of (1). We assume the solution
rnn
n
xay +∞
=∑=
0
(2)
1
0
)1)(()( −+∞
=
−++=∴ ∑ rnn
n
xarnrnyL
rnnn
n
rnn
n
xararnxa +∞
=
−+∞
=
−+++ ∑∑ )1)((340
1
0
rnn
n
xa +∞
=∑+
0
3
1
0
)3)(()( −+∞
=
+++∑ rnnn
n
xrarnayL
rnn
n
xrna +∞
=
+++∑ )1(30
(3)
0=n , we get the indicial equation 0)3( =+rr
0=r , 3−
1
0
)3)(()( −+∞
=
+++∑ rnn
n
xrarnayL
11
1
)(3 −+−
∞
=
++∑ rnn
n
xrna
Using the smaller root 31 −=r
4
0
)3()( −∞
=
−∑ nn
n
nnayL
Recurrence relation is
1)3(3)3( −−−=− nn annna
1=n oaa )2)(3()1)(2(1 −−=−
2=n 12 )1)(3()2)(1( aa −−=−
3=n 23 )0)(3()3)(0( aa −=
4>n 1
3−
−= nn an
a
34 4
3aa
−=
45 5
3aa
−=
……………….
1
3−
−= nn an
a
36)3(
anl
an
−=∴
12
21
30( −−− += xaxaxtay
1
)1(6)
2
93(
33
33
12
230 n
xaxaxxa
nn
n
−−∞
=
−− −+++− ∑
This is the required solution 4.0. Conclusion We have looked at various problems involving Euler equations in this unit their various form of indicial equations. In the next unit we shall consider indicial equation of positive integer and logarithmic case.
5.0 Summary You will recall in this unit that a general form of Euler equation was given. We also consider various form of Euler equations. You are required to master this unit very well before proceeding to other units. 6.0. Tutor Marked Assignment: (1). Solve the equation
0)21(32 =−+′+′′ yxyxyx
(2).Solve the equation 02)4( =+′+−′′ yyxyx
7.0 REFERENCES/FURTHER READINGS 1. EARL. A. CODDINGTON: An Introduction to Ordinary Differential Equations. Prentice-Hall of India 2. FRANCIS B. HILDEBRAND: Advanced Calculus for Applications, Prentice-Hall, New Jersey 3. EINAR HILLE: Lectures on Ordinary Differential Equations, Addison – Wesley Publishing Company, London.
UNIT3: INDICIAL EQUATION WITH DIFFERENCE OF ROOTS A POSITIVE INTEGER, LOGARITHMIC CASE 1.0 Introduction 2.0 Objectives 3.0 Main Content 3.1. Indicial Equation with difference of roots, positive integer and logarithmic case 3.2 Fourier Series 3.3 Orthogonality of a set of series and cosines 4.0 Conclusion 5.0 Summary 6.0 Tutor Marked Assignment 7.0 References/ Further Readings 1.0. Introduction In unit 2 we have considered indicial equations where logarithm case is not considered. We shall undertake to consider the positive and logarithm cases in this unit. 2.0 Objectives At the end of this unit you should be able to - to solve differential equation whose indicial equation has a positive integer - to solve differential equation whose indicial equation has roots with logarithmic case. 3.0. MAIN CONTENT
3.1. Indicial Equation with Difference of roots a positive integer, logarithmic case.
We illustrate this method by an example Solve the equation
0)31()1(2 =+−′−+′′ yxyxxyx
Solution: We assume the solution
rnn
n
rnn
n
xarnxarnrnyL +−
∞
=
+∞
=
++−−+++∴ ∑∑ 110
)2()1)(19)(
The indicial equation is 0)1)(1( =−+ rr (Putting )0=n (1)
1,1−=r 1≥n , the recurrence relation
)]1).....(1()[2(
)2( 0
−+++++
=∴nrrrr
anran (2)
)]12).....(1()[2(
)2( 0
10 −+++
++=∴
+∞
=∑
nrrr
xanrxay
rn
n
r
It follows that
ro xarryL )1)(1)(( −+∴
For 1=r , only one solution can be obtained. Note: for 1−=r , since there is no power series with skilling 1−x , we suspect the presence 1−=r . Choose 10 += ra , we have
]12).....(12(2)[2(
)2(1()1(),(
1 −++++++++=
+∞
=∑
nr
xnrrxrrxy
rn
n
r
We can obtain two solutions with respect to For which
cxrrrxyL )1()1(),([ 2 −+=∴ We use the same argument as that of equal roots Putting
)1,(1 −= xyy
12 )],([ −=∂∂= rrx
r
yy
21
)2(
)4(
)2(
)3)(1()1(),( +
+
+++
+++++== r
r
r
xr
r
rr
xrrrrxyy
)]12).....(2)[(2(
)2(
3 −++++++
+∞
=∑
nrr
xnr rn
n
Differentiate with respect to r .
)2(
)3)(1(log),(
1
+++++=
∂∂ +
rr
xrrxxxry
r
rr
{ } {2
11
)2(
)4(
2
1
2
1
3
1
1
1 22
+−
++++−
+−
++
+
+
rurrr
xr
rrr
)12)........(3)(2)[(2(1
1.....
3
1
2
1(
1
2
1
22
1)22(
3 −++++−+
++
++
−−+
−++
+++
+∞
=∑
nrrrnrrrrrn
xn rn
n
Putting 1−=e , get
)]2...(2.1)[1(
)1(3.0.0
1
31 −−
++−+=−∞
−=∑
n
xnxxxy
n
n
o
)]2...(2.1)[1(
}2
1.....
2
11(11
1
1{)1(
}113
1{32log
1
3
0112 −−
−+++−+−
++
++−−−+=
−∞
=∑ n
nnxn
xxxxyy
n
n
)!2(
)1(3
1
31 −
++−=−∞
=∑
n
xnxy
n
n
)]2(
])1(1[2log
12
3
112 −
+−−−−+=
−−
∞
=
− ∑n
xHnxxxyy
nn
n
Problem: Find line general series solution of the D.E
0642
2
=++ ydx
dy
dx
ydx
and show that it can be expressed in line form
Solution: 0=x is a regular singular point of D.E. Assuming the solution Substituting in the D.E
rnn
n
xay +∞
=∑=
0
Changing the index
0)122)((2 1
1
1
0
=++++ −+∞
=
−+∞
=∑∑ rn
nn
rnn
n
xaxarnrn .
The indicial equation is
0)12(2 =+rr , 0=r , 2
1
The recurrence relation is
)]2...(2.1)[1(
)1(3.0.0
1
31 −−
++−+=−∞
−=∑
n
xnxxxy
n
n
o
1122)(2(2
)1(−+++
= nn arnn
a , 1≥n
(i) 0=r , then
1)12(2(
)1(−+
−= nn ann
a
,1=n 01 )3)(2(
)1(aa
−=
12 )5)(4(
)1(aa
−=
Thus
01 )3)()!12(
)1(a
na
+−=
Hence the solution is
)!12(
)1(
001 +
−= ∑∞
= n
xay
h
n
xSinx
a
n
x
x
ann
n
0
12
0
0
)!12(
)2
1()1(
=+
−=
+∞
=∑
(ii) 2
1−=r , then
1)2)(12(
)1(−−
−= nn ann
a
1=n , )2)(1(
)1( 01
aa
−=
)4)(3(
)1( 12
aa
−=
------------------------------
!2
)1( 0
n
aa
n
n
−=
xx
a
n
x
x
ay
n
n
cos)]2(
)1( 0
0
02 =
−−= ∑
∞
=
Hence the general solution is
xy
1= ( )xBSinxACos +
3.2. Fourier series 1. Orthogonality: A set of function is ),.....}(),....(),({ 0 xfxfxf n said to be
an orthogonal set with respect to the weight function )(xw over the interval bxa ≤≤ if
0)()()( =∫ dxxfxfxw mn
b
a for nm ≠
0≠ for nm = Orthogonality is a property widely encountered in certain branches of mathematics. Much use is made of the representation of functions in series of the form
)(0
xfc nnn∑
∞
=
In which the nc are numerical coefficients and )}({ xf n is an orthogonal set is
3.3 Orthogonality of a set of series and cosines:
We shall consider the set of function
),(c
xnSin
π ...........3,2,1=n
),(c
xnCos
π ...........3,2,1,0=n
or
),(c
xSin
π ),
2(
c
xSin
π ),
3(
c
xSin
π…………… )(
c
xnSin
π…………….
1, ),(c
xnCos
π ),
2(
c
xCos
π ),
3(
c
xCos
π…………… )(
c
xnCos
π……………..
is orthogonal with respect to the weight function 1)( =xw over the interval cxc ≤≤−
i.e.
0=∫− dxc
xkCos
c
xnSin
c
c
ππ, where nk ≠ .
: Before we prove the result, we give some definition to shorten the proof.
(a) Even function: A function )(xgy = is said to be even y )()( xgxg =−
For all x .
(b) Odd function: A function )(xhy = is odd if )()( xhxhy −−== For all x.
Example: Sinx is an odd function Cosx is an even function
(c) Most function are neither even or odd example x. (is in one function 0)( =xf
(d) If )(xg is an even function then as well as even)
dxxgdxxgcc
c)(2)(
0∫∫ =−
Consider the integral.
01 == ∫−
dxc
xkCos
c
xnSinI
c
c
ππ for all k and n .
Follows at once from the facts that the integrand is an odd function of x. It does not dependent upon one fact and k and n are integers
,2 dxc
xkSin
c
xnSinI
ππ∫= nk ≠
Take
βπ
πβ dc
dxc
x == ,
βββπ
π
λπdknCosknCos
c])()(
2+−−∫− βββ
ππ
λπdSinkSinn
c∫−
=
[ ] ππ
ββπ −+
+−−
−=kn
knSin
kn
knSinc )()(
2
Since kn − and kn + are ve+ integers. 0=
Finally we consider the integral Where 0=n , ..,.........1 0≠n ,
[ ] ππ
ββπ −+
+−−−=
kn
knSin
kn
knSinc )()(
2
0=
dxc
nxSindx
c
xnSinI
c
c
c
c
224 ∫∫ −−
== π
let f(x) the continuous and differentiable at every point in an interval cxc ≤≤− except for a most finite number of points and at more points, let
)(xf and )(xf ′ have right and left-hand limits Note:
The notation )0( +cf is used to denote line right-hand limit of cx → as )(xf from alone, i.e.
)0( +cf and )(lim xfcn +→
=
Similarly
=− )0( cf )(lim xfcn =→
=
Denotes, the limit of f(x) as approaches c. Since Fourier series for )(xf may not converge to the value )(xf every where. It is customary to replace the equals sign in equation (8) by the symbol ~which may be read “has for its Fourier Series” we write
c
xnb
c
xnaaxf nn
n
ππsincos(
2
1~)(
10 ++∑
∞
=
,
Where na and nb are given by (a) and (10) Example: Construct the Fourier series, over the interval ,02 ≤≤− x for the function defined by
2)( =xf , ,02 ≤≤− x 2=x , 20 << x
Solution: Now )(xf ~ )2
sin2
cos(2
1
10
xnb
xnaa nn
n
ππ ++∑∞
=
In which
;2
cos)(2
1 2
2dx
xnxfan
π∫−= 2,......1,0=n
and
;2
sin)(2
1 2
2dx
xnxfbn
π∫−= 2,......1,=n
dxxn
xdxxn
an 2cos
2
1
2cos
2
1 2
0
0
2
ππ∫∫ +=∴
− (a)
If 0≠n , then
] 0220
2 ]2
cos)2
(2
sin2
[2
1
2[sin
2 xnxnx
xnan
ππ
ππ
ππ
++=∴ −
22
)cos1(2
ππ
n
n−−=
For 0=n , from (a), we get
0a =3
2
1=∴ nb
Thus we write
]2
sin1
2cos
)1(1[2
2
3~)(
221
xn
n
xn
nxf
n
n
ππ
ππ
+−−− ∑∞
=
.
Example 2 Obtain the Fourier series over the interval π− toπ for the function 2x
Solution: We know
]sincos[2
1~
10
2 nxbaax nnx
nn
++∑∞
=
for ππ <<− x , where
;cos1 2 nxdxxan ∫−=
π
ππ ,.......2,1,=n
;sin1 2 nxdxxbn ∫−=
π
ππ ,.......2,1,=n
2x is an even function, nxsin is an odd function, thus nxx sin2 is an odd function Hence.
0=nb . for every n .
032
2
]sin2cos2sin
[2 ππ
ππ n
nx
n
xx
n
nxxan −+= ∫− for 0≠n
From which
22
)1(4]
cos2[
2
nn
nan
−== πππ
, ,........2,1=n
2
)1(4]
cos2[
2
nnan
πππ
−== ,........2,1=n
for 0=n
32
3
2[
2 232
00
ππππ
π=== ∫ dxxa
Therefore, in the interval; ππ <<− x
21
22 cos)1(
43
~n
nxx
n
n
−=+ ∑∞
=
π
Indeed, because of condition of line function involved, we write
,cos)1(
43
~2
1
22
n
nxx
n
n
−=+ ∑∞
=
πfor ππ ≤≤− x .
2. Fourier Sine Series: Sometimes it is desirable to expand the function
)(xf in a series involving Sine function only. In order to get a Sine series for )(xf we introduced a function )(xg defined as follows
)()( xfxg = cx <<0 )( xf −−= 0<<− xc
Thus )(xg is an odd function over the interval 0<<− xc . Hence
0cos(2
1~)(
10 =+∑
∞
=
dxc
xnaaxg n
n
π, ,......1,0=n
It follows that
0cos)(1
0
=∫− dxc
xnxg
ca
c
cn
π ,......1,0=n
(Note integrand is an odd function) and that
dxc
xnxg
cb
c
cn
πsin)(
1∫−=
c
xnxf
c
c
c
πsin)(
2∫−=
Thus
c
xnbxf n
n
πsin~)(
1∑
∞
=
+ cxo <<
)(xf Where dxc
xnxf
cb
c
cn
πsin)(
2
0∫−= ........,.........2,1=n
Example: Expand 2)( xxf == = in a Fourier Sine Series over the interval
10 << x Solution: At once we write, for 10 << x 0
xnbx nn
πsin~1
2 ∑∞
=
In which
xdxnxbc
cn πsin2∫−=
3221
0 )(
cos2
)(
sin2cos2
ππ
ππ
ππ
n
xn
n
xnx
n
xnx ++= ∫
3333
cos2cos2cos[2
ππ
ππ
ππ
n
xn
n
x
n
n ++−=
Hence the Fourier Sine Series, over 10 << x for is 2x
]2)1(2cos
[23333 πππ
πnnn
n n−+−=
3. Fourier Cosine Series: In order to expand the function )(xf in a series
involving cosine function only, such series is called Fourier Cosine Series. We define
)()( xfxh = cx <<0 )( xf −= 0<<− xc It follows that )(xh is an even function of x .
c
xnb
c
xnaaxh nn
n
ππsincos(
2
1~)(
10 ++∑
∞
=
dxc
xnxf
cdx
c
xnxh
ca
cc
cn
ππcos)(
2cos)(
10∫∫ =
−
cx <<0
But 0sin)( == ∫− dxc
xnxh
c
ib
c
cn
π
Thus we have
dxc
xnxfaxf
c πcos)((
2
1~)(
00 ∫+
in which
dxc
xnxf
ca
c
cn
πcos)(
2∫−=
Example: Solution: At once we have
c
xnaaxf n
n
πcos
2
1~)(
10 ∑
∞
=
+ in which
in which
dxc
xn
ca
c
n
πcos
20∫
c
c
xn
n
c
c
xnx
n
c
c 02 ]cos)(sin
.[
2 ππ
ππ
+=
c
c
xnn
n
c
c 02 ]cos)
.[(
2 πππ
−=
)cos1(2
22π
πn
n−−= , 0≠n
The coefficient 0a is readily obtained
cc
cxdx
ca
c
cn ==∫− 2
22 2
Thus the Fourier Cosine Series over the interval cx <<0 the function
xxf =)( is
c
xn
n
uccxf
k
ππ
cos)1(1
2
1~)(
21
2
−−− ∑∞
=
4.0. Conclusion: You have learnt about indicial equations where the roots are positive and logarithmic. You have also learnt about Fourier series and odd functions. 5.0 Summary: You are required to study materials in this unit very well before proceeding to the next units. 6.0 Tutor Marked Assignments: (1). Find the general series solution of the D.E
0642
2
=++ ydx
dy
dx
ydx
(2).Construct the Fourier series, over the interval ,02 ≤≤− x for the function defined by
2)( =xf , ,02 ≤≤− x 2=x , 20 << x
7.0 REFERENCES/FURTHER READINGS 1. EARL. A. CODDINGTON: An Introduction to Ordinary Differential Equations. Prentice-Hall of India 2. FRANCIS B. HILDEBRAND: Advanced Calculus for Applications, Prentice-Hall, New Jersey 3. EINAR HILLE: Lectures on Ordinary Differential Equations, Addison – Wesley Publishing Company, London.
UNIT 4: BOUNDARY VALUE PROBLEMS 1.0 Introduction 2.0 Objectives 3.0 Main Content 3.1. Boundary Value Problems 3.2 Eigen Values and Eigen Functions 4.0 Conclusion 5.0 Summary 6.0 Tutor Marked Assignment 7.0 References/ Further Readings 1.0 Introduction In this unit, we will discuss some of the properties of boundary valve problems for linear second order equation. This class of differential equations is very useful for practical applications. We shall devote some time in studying them in this unit. 2.0 Objectives: At the end of this unit you should be able to
- classify second order differential equations into homogeneous and non-homogeneous.
- differentiate between eigen values and eigen functions - solve related eigen value problems
3.0 MAIN CONTENT 3.1 Boundary Value Problems The linear differential equation )()()()( xgyxRyxQyxP =+′+′′ (1) was classified homogeneous if, 0)( =xg , and non-homogeneous otherwise. Similarly, a linear boundary condition
cyaya =′′+ )0()0( 21 (2) A boundary value problem is homogeneous if both its differential equation and in boundary conditions are homogeneous. If not then it is non- homogeneous.
A typical linear homogeneous second order boundary value problem is of the form.
0)()()( =+′+′′ yxRyxQyxP (3) 10 << x ,
0)0()0( 21 =′′+ yaya (4)
0)0()( 21 =′+ ybIyb (5)
Most of the problems, we will discuss are of the form given by (3) to (5).
3.2. Eigen Values and Eigen Functions Consider the differential equation 0),(),( =+′+′′ yxqyxpy λλ 10 << x (1) The boundary conditions
0)0()0( 21 =′+ yaya (2) 0)0()( 21 =′+ ybIyb (3)
Where λ is arbitrary parameter. Clearly the solution of (1) depends on x and λ and can be written as
),(),( 211 λλ xcxycy += , (4) Where 1y and 1y are a fundamental solution of (1). Substituting for 1y in the boundary condition (2) and (3), yield.
0)],([)],(),([ 2122111 =′+′+ λλλ xyacoyaoyac (5)
0)],([)],(),([ 2222111 =′+′+ λλλ iybciybiybc (6) A set of two linear homogeneous algebraic equations for the constant. Such a set has solutions (other than 021 == cc if and only if the determinant of coefficients )(λD vanishes i.e.
)],(),([ 2111 λλ oyaoyac ′+ ),(),( 2221 λλ oyaoya ′+
=)(λD )],(),( 211 λλ iybiyb ′+ )],(),( 2221 λλ iybiyb ′+ =0 Values satisfying this determinant equation are the eigenvalues of the boundary- value problems (1), (2) and (3) Corresponding to each Eigen value is at least one non- trivial solution. An Eigen function Note: We will consider problems namely only real eigen value Example I consider the equation
0=+′′ yy λ (1) ),0(y 0)1( =y (2) Solution: 0=+′′ yy λ , the solution is xcxcy sincos 11 += , (3) By the boundary conditions
01 =c s
0sin2 =λc
02 ≠c , otherwise 0=y is the solution πλλ n=⇒= 0sin , .......,.........2,1=n
or 22 4, ππλ nn= (4) (4) gives the eigen values of (1). If we consider 1=λ
xcxcy sincos 21 +=∴
10 c= by (2) 00sin 22 =⇒==∴ cIcy
0 = c1 by ( 2)
0=∴ y Hence 1=λ is the eigen -function The eigen function are
xncyn πsin2=∴ ……………. (5)
2.,.........2,1 cn = ………. c1 is an arbitrary constant
Example 2: Find the real eigen values eigen -function of the boundary value problem
0=+′′ yy λ 0)0( =y 0)( =′ ly
.The solution is 0=+′′ yy λ
xcy λcos1=∴ (1) 0)0( =y , given
01 =c . Also xcy λλ coscos2=′
But 0)( ==′ ly , yields ⇒== 0cos2 lc λλ
0cos2 == lc λ
l
nl
2
)12( πλ − , ........,.........2,1=n
(2) gives eigen value
]2
)12sin[
l
xnyn
π−= , ........,.........2,1=n (2)
(3) gives eigen functions Examples 0=+′′ yy λ 0)0( ==′y 0)1( ==′y Solution: The solution is xccxcy λλ sincos 21 +=′ (1) xxcxcy λλ cossin 21 +−=′ (2)
021 =+ µcc
µcos
1 µ
µcos
=0
Thus the eigen value are given by the equation µµ tan= . (3)
xccy n µµµ sin22 +−= If nµ is the root of (3), then eigen function is
xy nnnn µµµ cossin −= (4)
If 0=λ , then the solution is
22 ccy +−= y
1cy =′
021 =+∴ cc
Hence the solution is
)1(1 −= xcy
thus 0=λ is also an eigen value µµ tan=
,49.4~1λ 2
)12(~
πλ +nN
Example 5. 0=+′′ yy λ 0=y , 0)0()1( =′+ yy
Solution xcxcy λλλλ cossincos 21 ++=
02 =c ,
λλ cot=∴
The eigen values are given by equation (3). The eigen function are xy nn λ= Where the root of is nλ is the root of the equation
λλ cot=
x=λ , xy cot=
πλ )1( −= nn ..,.........3,2≥n 22)1( πλ −= nn for large n
Example 6. Consider the problem
0=+′′ yy λ )0(=y , 0)0( =′y
Show that if mφ ,and nφ are eigen function corresponding to the eigen value
mλ and nλ Respectively, then
0)()(0
=∫ dxxxm
lφφ
Provided that nm λλ ≠ . Solution: 0=+′′ mmm φλφ 0=+′′ mmn φλφ 0=+′′ nmn φλφ
0=+′′ nmmnm φφλφφ (1)
0=+′′ nnnmn φφλφφ (2)
0)()(00
=+′′′′ ∫∫ dxdxxx nm
l
mnm
lφφλφφ (3)
000
=+′′′′ ∫∫ dxdx mn
l
nmn
lφφλφφ (4)
0)()()()(000
=+′−′′′ ∫∫∫ dxdxxxxx nm
l
mnm
l
nm
lφφλφφφφ
dxxxdxxxxxoll nm
l
omnm
l
nmnmnm )()(])()()()([)0()()()(0
φφλφφφφφφφφφ ′′′+′′′−′′−′−′′−′′′ ∫∫
or By boundary value conditions
0)()()()(00
=+′′ ∫∫ dxxxdxxx nm
l
mnm
lφφλφφ (5)
Subtract (4) from (5), we have
0)()((0
=− ∫ dxxx nm
l
mn φφλλ
If mn λλ ≠ , then
0)()(0
=′′∫ dxxx nm
lφφ
Example 14. Hyperbolic function
2cos
xxeex
−
, 2
sinhxx ee −−
)sinh((cosh) xdx
d =
)(cosh)(sin xhxdx
d =
(a) Solution of the problem is
04 =− λr , Take 4µλ =
044 =− µr The solution is
xcxcxcxcy µµµµµ sinhcossincos 22
321 +++= (1) The boundary condition
031 =+ cc
031 =− cc
01 =⇒ c and 03 =c
0sinhsin0sinsin 4242 =+−=+=∴ lclclccy µµµµ
0sinhsin0sinsin 4242 =+−=+=∴ lclclccy µµµµ
0sinhsin0sinsin 4242 =+−=+=∴ lclclccy µµµµ
0sinhsin0sinsin 4242 =+−=+=∴ lclclccy µµµµ 0sinh0sin ==∴ ll µµ ....,.........2,1=n
πµ nl =∴sin
xl
nyn
πsin=∴ ....,.........2,1=n
Is the eigen-function 4.0 Conclusion We have been able to study some eigen-value problems in this unit. This unit must be mastered properly before moving to the next unit. 5.0 Summary Recall that the linear differential equation )()()()( xgyxRyxQyxP =+′+′′ (1) was classified homogeneous if, 0)( =xg , and non-homogeneous otherwise. Similarly, a linear boundary condition
cyaya =′′+ )0()0( 21 (2) A boundary value problem is homogeneous if both its differential equation and in boundary conditions are homogeneous. If not then it is non- homogeneous. We also classified some equations into eigen value problem depending upon whether the determinant of the eigen value of the problem is zero or not. Read carefully and re work all exercises and problems in this unit for better understanding. 6.0 Tutor Marked Assignment: 1. Consider the problem
0=+′′ yy λ )0(=y , 0)0( =′y
Show that if mφ ,and nφ are eigen function corresponding to the eigen value
mλ and nλ Respectively, then
0)()(0
=∫ dxxxm
lφφ
Provided that nm λλ ≠ . 2. Find the real eigen- values and eigen -function of the boundary value problem
0=+′′ yy λ 0)0( =y 0)( =′ ly
7.0 REFERENCES/FURTHER READINGS 1. EARL. A. CODDINGTON: An Introduction to Ordinary Differential
Equations. Prentice-Hall of India 2. FRANCIS B. HILDEBRAND: Advanced Calculus for Applications,
Prentice-Hall, New Jersey 3. EINAR HILLE: Lectures on Ordinary Differential Equations,
Addison – Wesley Publishing Company, London.
Module 2: Sturm Liouville Boundary Value Problems and Special Functions UNIT1: Sturm and Liouville Problem 1.0 Introduction 2.0 Objectives 3.0 Main Content 3.1. Sturm and Lioville Problems 4.0 Conclusion 5.0 Summary 6.0 Tutor Marked Assignment 7.0 References/ Further Readings 1.0. Introduction We solved some partial differential equations by the method of separation of variables. In the last step we expanded a certain function in a Fourier series, i.e. as the sum of an infinite series of sine and cosine functions. It is of fundamental importance that the eigen functions of a more general class of boundary values problems can be used as a basis for series expansions, which have properties similar to Fourier Series. Such eigen- functions series are useful in extending the method of separation of values to a larger class of problems in partial differential equation. The class of boundary value problem we will discuss is associated with the names of Sturm and Liouville. 2.0 Objectives: After studying this unit you should be able - to solve partial differential equation using Sturm and Liouville methods - solve correctly the associated Tutor Marked Assignments 3.0 MAIN CONTENT 3.1. Sturm and Liouville Problem We introduce the operator
yxqyxpyL )(])([][ +′−= (1)
yxryL )(][ λ= (2)
0)()(])([ =+−′′ yxryxqyxP λ (3) on the interval 10 << x , together with the boundary condition
0)0()0( 21 =′+ yaya (4) 0)1()( 21 =′+ ybIyb (5)
We shall assume that qp, and r are continuous functions in the interval ].1,0[
0)(,0)( >> xrxP for all x in 10 ≤≤ x .
(i) Lagrange’s identity: let u and v be functions having continuous second derivatives on the interval 10 ≤≤ x . Then
dxuULUUL ])[][(1
0−∫
)]()()()()[( xuxuxuxuxp ′−′−= (6) Solution 1:
dxuxquxpudxUUL })()([{][1
0
1
0′−∫∫
dxuquupuxuxpuypu })({)()()(1
0+′−+′+′−= ∫
)()()()()[(])[][(1
0xuxuxuxupxdxUULuuL ′−′−=−∴ ∫
This is known as Lagrange’s identity if u and u satisfy (5) and (4) R.H.S = )]1()1()1()1()[1( uuuup ′−′− )]0()0()0()0()[0( uuuup ′−′+
)]1()1()1()1()[1(2
1
2
1 uub
buu
b
bp +−−
)]0()0()0()0()[0(2
1
2
1 uua
auu
a
ap +−+
0= Thus we have
0][][{1
0=−∫ dxuuLuu
(ii) Show that all the eigen value of the Sturm-Liouville problem
yxryL )()( λ= A With boundary conditions
0)0()0( 21 =′+ yaya B 0)1()( 21 =′+ ybIyb
are real. Proof: let us suppose there exists a complex eigen value iv+= µλ will 0≠v and corresponding to this value is the eigen function )()()( xIVxUxQ += Where at least one of them is not identically zero. Now Q satisfies the differential equation rQQL λ=)[ RQQL π=][
or Qu = and Qu =
)()]()({1
0
1
0λλ ′−=− ∫∫ dxQQLQQL
0)()()( =dxxQxQxr
or 0)()()[(2 221
0=+∫ dxxVxVxrr ………. (1)
Since 0)( >xr for all x in 10 ≤≤ x (1) 0=⇒ v This contradicts the original hypothesis. Hence the eigen value of Sturm-Liouville problem are real. (iii) If 1Q and 2Q are eigenvalues of the Sturm-Liouville problem (A)
and (B), corresponding to eigen valves 1λ and 2λ , respectively , and 21 λλ + , then
( ) 0)()( 21
1
0=∫ dxxQxQxr
)([ xr is called the weight function and it is an orthogonal property of
eigenfunction] Proof: - 111][ rQQL λ
222 ][ rQQL λ
If we let = 1QU = and 2QU = then
dxUULUuL ]}[][{1
0−∫
0)()()( 21
1
021 =− ∫ dxxQxQxrλλ
Hence the result (iv) Let us now consider a more general boundary value problem for
the differential equation ][][ yMyL λ= , 10 << x
Where L and M are linear homogeneous differential operations of orders n and n respectively.
yxpyxpyxpyxpyL nnnn )()(........)()(][ 1
)1(1
)(0 +′+++= −
−
yxryxyxryxryM mmmm )()(2........)()(][ 1
)1(1
)(0 +′+++= −
− Where mn > . In addition to the differential equation a set of n linear homogeneous boundary conditions at 0=x , 1=x is also prescribed. If the relations
0][][(1
0=−∫ dxuuLuuL
0][][(1
0=−∫ dxuuMuuM
are line for every pair of functions u and u , which are −n lines continuously differentiable on l,0 and which satisfy un given boundary conditions, then the given boundary valve problem is said to be self adjoint. Problem I. Show that the Sturm-Liouville problems
yxqyxPyL )(])([)( +′− yxryM )()( λ=
(i) dxuuMuUM ][][[1
0−∫
dxuxruuxrU ])()([1
0λλ −∫
0=
For every pair of u, u
0][][[1
0=−∫ dxuuLuUL
as shown previously. Hence it is self-adjoint Problem
(a) 02 =+′+′′ yyy 0=y , 0)( == Iy
Solution yyyyL 2)( +′+′′=
(i) dxuuuuuuuU )2()2([1
0+′+′′−+′+′′∫
,21
0udxu′−= ∫ are true for every pair of function u and u, which are n-
times continuously differentiable on ],[ lo which satisfy un given boundary value problem is said to be self-adjoint. Solution: Sturm-Liouville problems
yxqyxpyL )(])([)( +′−= yxryM )()( λ=
(i) dxuuMuUM )]()([1
0−∫
dxxruxrU )]()([1
0λλ −∫
0= as shown previously. Hence it is not self-adjoint. Problem
02 =+′+′′ yyy 0=y , 0)( == Iy \ Solution yyyyL 2)( +′+′′= 0=y , 0)( == Iy
dxuuuuuuuU )]2()2([1
0+′+′′−+′+′′∫
0)]()([1
0=−∫ dxuuLuUL
as shown previous. Hence it self-adjoint Problem (a) Problem
02 =+′+′′ yyy 0=y , 0)( == Iy \ Solution yyyyL 2)( +′+′′= 0=y , 0)( == Iy
dxuuuuuuuU )]2()2([1
0+′+′′−+′+′′∫
021
0=′−= ∫ udxu , u′ and u′ are continuous in the interval 10 ≤≤ x . Hence it is
not zero. Thus it is not self-adjoint. Problem(b)
02)( 2 =+′+′′+ yyxyxx 0=′y , 0)1(2)1( =′+= yy
02)1()( 2 =+′′+′′+= yyxUxyL , 0)0( ==′y 0)1(2)1( =′+= yy ,
yyxyxyL +′+′′+= 2)1()( 2
0)( =yM
0)]2()1[([ 21
0=+′+′′′+∫ dxuuuuxU
It is Sturm-Liouville problem. (c) ,yyy λ=+′′ 0)1()0(0)( =′−′−= yyy 0)1()0(0 =′−′−=′ yyy Solution
yyxyxyL +′+′′+= 2)1()( 2 yyM =)(
(i)
dxUMuM )()((1
0µµ −∫
0)(1
0=−∫ dxuuuu
(ii) ])()(1
0dxUuu µ ′′−′′∫
dxuuuu )(1
0′′−′′= ∫
uuuu ′−′= )1()1()1()1([ uuuu ′−′= )]0()1()0()1([ uuuu ′−− )0()()0()1([ uIuuu −=
The right side is not zero. Hence it is not self-adjoint. Problem 3 Consider the differential equation
02 =++′′ yyy λ \
With boundary conditions
0)()0( =−= Iyy , 0)1()0( =′−=′ yy
(a) Show that the problem is self-adjoint even though it is not a Sturm-Liouville problem.
(b) Find all eigenvalues and eigenfunctions of the given problem
Solution: yyL ′′=)( yyM −=)(
(i) 0])()([1
0=−−∫ dxuuuUM
0)]()([1
0=−−−∫ dxuuuU
(ii) dxuuuuu ]([1
0′′−′′∫
dxuuuudxuuuu ′′+′−′′−′= ∫∫1
0
1
0()(
)]()()()1[( IuIuIu ′−′= )]0()0()0()0[( uuu ′−′− )]0()0()0()0[( uuu ′−′− 0)]0()0()0()0[( =′−′− uuu Hence it is self-adjoint The solution of the equation is
xcxcy λsincos 21 += Applying the boundary conditions, we have
0cos1(sinsin 21 =−+ λλλλ cc
0sin1(cos 21 =+− λλ cc
Thus λλ sin λλ cos1( − 1cos −λ λsin =0
Or
0=λ or 2)2( πλ −= n , ..,.........2,1=n 00 =λ 1)(0 =xϕ
⇒=− 0cos1( λλ
2)2( πλ −= nn ,2cos)( xnxQn π=∴ ,2sin)( xnxQn π=∴
,2cos1 xny π= ,2sin2 xny π=
,2cos1 xny π= ,2cos22 xnxny ππ=
xnnxnn
xnxnyyW
ππππππ2cos2sin2
2sin2cos( )21 −
=
xnnxnxn ππππ 22 sin22cos2 + 02 ≠πn x←0
Between 10 ≤≤ x Thus the eigen functions are lining independent. Problem 5:
Consider the Sturm-Liouville problems yxryxqyxp )()(])([ λ=+′− a1
0)1()1(,,0)0()0( 221 =′+=′+ ybybyaya
Where p, q and z continuous function in the interval 10 ≤≤ x . (a) show that if λ is an eigen-value and ϕ a corresponding eigen
function, then
)0()0()1()1()( 2
2
12
2
1221
0
21
0φφφφλ p
a
ap
b
bdxqpdxeQ −++′= ∫∫
Provided that 02 ≠a and 02 ≠b How this result be modified if 02 =a or 02 =b
(b) Show that if 0)( ≥xq and if 2
1
b
b. and
1
2
a
a− are non-negative, then the
eigen-value λ Is non negative (c) Show that the eigen-value λ is strictly 10 ≤≤ x under q(x) = 0 for
each x in 0,x,1 Solution
22 )(])([)( QrqQQxpQxr +′′−=λ Thus
dxQQxpqQdxQxr )))(()( 21
0
21
0′−= ∫∫λ
Integrating by parts, we have
dxQpQpQdxqQ )]([ 21
0
21
0′+′− ∫∫
From boundary condition, we have obtain the result
)1()1(2
1 Qb
bQ ′=′
)0()0(2
1 Qa
aQ ′−=′
Putting these values on the right side and we obtain the result if or 02 =a or 02 =b , then the first boundary condition reduces to
0=y 0)1( =⇒ Q or 0)( =⇒ oQ
The result reduces to
)()1(( 2
2
121
0
21
0IQp
b
bdxQpqQdxrQ ′+= ∫∫λ
or
)()0(( 22
2
1221
0
21
0IQQ
a
adxQpqQdxrQ ′+= ∫∫λ
(b) In a Sturm Lioville problem, we always assume that
,0)( >xp ,0)( >xr
By given condition 0)( >xr for all x in 10 ≤≤ x 02 >Q for all 10 ≤≤ x . Now we impose condition, so that right side of the equation in (a) is ve+
The second and third term are ve+ y2
1
b
b and
2
1
a
a− are non-negatives
Now
dxqQ21
0∫ is ve+ in order that
(c) If 0)( =xq for all x 10 ≤≤ x then λ is strictly. 4.0 Conclusion We have studied the Sturm Lioville problem in this unit . You are to master this unit properly so that you will be able to solve the problems that follow.
5.0 Summary Recall that Sturm Lioville problems are usually problems associated with eigen values problems of partial differential equations which we have dealt with in this unit. In our subsequent course in mathematics in this Programme, we will have cause to deal with it again particularly when will shall study Partial Differential Equation. 6.0 Tutor Marked Assignment
Consider the problem
0)1(2 =++′−′′ uyy λ 0)1(,0)0( == yY
1) Show that this problem is not self-adjoint 2) Show that all eigenvalues are real 3) Show that the eigenfunctions are not orthogonal. (with respect to the weight function arising from the coefficients of …….. in the differential equation.
7.0 REFERENCES/FURTHER READINGS 1. EARL. A. CODDINGTON: An Introduction to Ordinary Differential
Equations. Prentice-Hall of India 2. FRANCIS B. HILDEBRAND: Advanced Calculus for Applications,
Prentice-Hall, New Jersey 3. EINAR HILLE: Lectures on Ordinary Differential Equations,
Addison – Wesley Publishing Company, London.
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