Bessel- Legendre de 1(Beamer)

7/28/2019 Bessel- Legendre de 1(Beamer)

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OutlinePower series

ExamplesExerciseAnswers

References

Special Functions: Bessel functions and Legendre

Polynomials

March 01 , 2013

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Outline of Topics

1 Power seriesReal Analytic FunctionsSingular points

The Indicial equationComplete solutionMethod of reduction of order

2 Examples

3 Exercise

4 Answers

5 References

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References

Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order

Introduction

A power series is an infinite series of the form

m=0

am(x x0)m = a0 + a1(x x0) + a2(x x0)

2 + ...,

where the variable x, center x0, and the co-efficients a0, a1, ... arereal.

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The power series method is the standard basic method for solvinglinear differential equations with variable co-efficients. It givessolutions in the form of power series.

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A real function p(x) is called analytic at a point x = x0 if it can berepresented by a power series in powers of (x x0) with radius ofconvergence R > 0.

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Existence of power series solution

Theorem 2.1

If p(x), q(x) and r(x) are analytic at x = x0, then every solution of

y

+ p(x)y

+ q(x)y = r(x)

is analytic at x = x0 and can thus be represented by a power series

in powers of (x x0) with radius of convergence R > 0.

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Ordinary and singular points

y

+ p(x)y

+ q(x)y = 0 (1)A point x0

is said to be a regular(ordinary) point of the D.E. (1) if bothp(x) and q(x) are analytic at x = x0.

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Ordinary and singular points

y

+ p(x)y

+ q(x)y = 0 (1)A point x0

is said to be a regular(ordinary) point of the D.E. (1) if bothp(x) and q(x) are analytic at x = x0.

is said to be a singular point of the D.E. (1) if either p(x) orq(x) or both are not analytic at x = x0.

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Classification of singular points:

regular(removable) singular points.

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irregular singular points.

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Power seriesExamples

ExerciseAnswers

References





A singular point x0 is called a regular singular point if both(x x0)p(x) and (x x0)

2q(x) are analytic.

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A singular point x0 is called a regular singular point if both(x x0)p(x) and (x x0)

2q(x) are analytic.On the other hand, if either (x x0)p(x) or (x x0)

2q(x) or both

are not analytic at x0, then x = x0 is called an irregular singularpoint.

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ExerciseAnswers

References


Examples:

Any real number is a regular point of

y

+ (x2 + 1)y

+ (x3 + 2x2 + 3x)y = 0.

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Examples:


y

+ (x2 + 1)y

+ (x3 + 2x2 + 3x)y = 0.

1 are the only singular points of

(1 x2)y

+ 2y

+ (x3 3)y = 0.

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Examples:


y

+ (x2 + 1)y

+ (x3 + 2x2 + 3x)y = 0.


(1 x2)y

+ 2y

+ (x3 3)y = 0.

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ExerciseAnswers

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Examples:


y

+ (x2 + 1)y

+ (x3 + 2x2 + 3x)y = 0.


(1 x2)y

+ 2y

+ (x3 3)y = 0.

Both 1 are regular singular points.

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ExerciseAnswers

References


Examples:

The only singular points of

x3(x 1)y

+ 2(x 1)y

+ 5xy = 0

are

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Examples:


x3(x 1)y

+ 2(x 1)y

+ 5xy = 0

are 0 and 1.

1 is an

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Examples:


x3(x 1)y

+ 2(x 1)y

+ 5xy = 0

are 0 and 1.

1 is an regular singular point.

0 is an

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References

ySingular pointsComplete solutionMethod of reduction of order

Examples:


x3(x 1)y

+ 2(x 1)y

+ 5xy = 0

are 0 and 1.

1 is an regular singular point.

0 is an irregular singular point.

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References

ySingular pointsComplete solutionMethod of reduction of order

Exercise: Classify the singular points of

1

x(x 1)y

+ y

+ (x2 1)y = 0.

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References

Singular pointsComplete solutionMethod of reduction of order


1

x(x 1)y

+ y

+ (x2 1)y = 0.

2 Legendres equation with n=1:

(1 x2)y

2xy

+ 2y = 0.

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References



1

x(x 1)y

+ y

+ (x2 1)y = 0.

2 Legendres equation with n=1:

(1 x2)y

2xy

+ 2y = 0.

3

(x2 4)y

(2x 8)y = 0.

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E lReal Analytic FunctionsSi l i


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The Form of the solution:

If x0 is a regular singular point of (1), then (1) admits at least one

nontrivial series solution of the form

y1(x) = (x x0)r

m=0

am(x x0)m

,

where r is a constant.

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E lReal Analytic FunctionsSi l i t


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In fact r is a root of the quadratic equation obtained by equatingto zero the co-efficient of least power of x in the D.E. (1).

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The quadratic equation is called the indicial equation associatedwith the differential equation(1).

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The quadratic equation is called the indicial equation associatedwith the differential equation(1).

Remark: If x0 is an ordinary point , then r=0

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A second linearly independent solution y2(x) may be obtainedusing the method of reduction of order.

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Working rule:

Given a solution y1(x), let y2(x) be of the form

y2(x) = y1(x)v(x).

Substituting in

y

+ p(x)y

+ q(x)y = 0,

we get

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pExerciseAnswers

References

g pComplete solutionMethod of reduction of order

Working rule:


y2(x) = y1(x)v(x).

Substituting in

y

+ p(x)y

+ q(x)y = 0,

we get

u

+ u

2

y

1

y1+ p(x)

= 0, where u = v

,

-a first order differential equation(reduced order).

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pExerciseAnswers

References

g pComplete solutionMethod of reduction of order

Working rule:


y2(x) = y1(x)v(x).

Substituting in

y

+ p(x)y

+ q(x)y = 0,

we get

u

+ u

2

y

1

y1+ p(x)

= 0, where u = v

,

-a first order differential equation(reduced order).Separating the variables and integrating, we get

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ExerciseAnswers

References

Complete solutionMethod of reduction of order

Working rule:


y2(x) = y1(x)v(x).

Substituting in

y

+ p(x)y

+ q(x)y = 0,

we get

u

+ u

2

y

1

y1+ p(x)

= 0, where u = v

,


v(x) =

1y21

e

p(x)dxdx.

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Real Analytic FunctionsSingular pointsC l l i
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Working rule:


y2(x) = y1(x)v(x).

Substituting in

y

+ p(x)y

+ q(x)y = 0,

we get

u

+ u

2

y

1

y1+ p(x)

= 0, where u = v

,


v(x) =

1y21

e

p(x)dxdx.

Thus

y2(x) = y1(x)

1

y21

e

p(x)dx

dx.

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ExerciseAnswers

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Example1

y1(x) = cosx is a solution of

y

+ y = 0.

y2(x) = cosx v(x) v(x) =

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Real Analytic FunctionsSingular pointsComplete solution
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Example1


y

+ y = 0.


1cos2(x)

e

0dxdx = tanx

y2(x) = sinx.

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Example1


y

+ y = 0.


1cos2(x)

e

0dxdx = tanx

y2(x) = sinx.

Thus the complete solution is

c1cosx + c2sinx, where c1 and c2

are

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Example1


y

+ y = 0.


1cos2(x)

e

0dxdx = tanx

y2(x) = sinx.

Thus the complete solution is

c1cosx + c2sinx, where c1 and c2

are arbitrary constants.

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Example 2

y1(t) = 1 + t is a solution of

y

+2(t + 1)

t(t + 2)y

2

t(t + 2)y = 0.

Now y2(t) = (1 + t)v(t)

v(t) =

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Example 2

y1(t) = 1 + t is a solution of

y

+2(t + 1)

t(t + 2)y

2

t(t + 2)y = 0.

Now y2(t) = (1 + t)v(t)

v(t) =

1

(1 + t)2e

(

dt

t+

dt

t + 2)dt

=

1

(1 + t)21

t(t + 2)dt

=1

2ln

t

t + 2

+

1

t + 1.

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AnswersReferences

pMethod of reduction of order

Complete solution:

The general solution of the differential equation

y

+ p(x)y

+ q(x)y = 0

isy(x) = c1y1(x) + c2y2(x), where c1 and c2


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Exercise


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AnswersReferences

Example1:

Obtain the general solution of

y

+ y = 0

in powers of (x-0).

Solution:

Note that x=0 is a

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AnswersReferences

Example1:


y

+ y = 0

in powers of (x-0).

Solution:

Note that x=0 is a regular point .

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AnswersReferences

Example1:


y

+ y = 0

in powers of (x-0).

Solution:

Note that x=0 is a regular point . the differential equation admits a series solution of the form

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AnswersReferences

Example1:


y

+ y = 0

in powers of (x-0).

Solution:

Note that x=0 is a regular point . the differential equation admits a series solution of the form say

y(x) =

m=0

am(x 0)m+r

.

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Substituting and simplifying,

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Substituting and simplifying,we get r=0 as a root of the indicial equation .

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AnswersReferences

Anda2m+1 =

a2m1

(2m + 1)(2m)

,

a2m+2 = a2m

(2m + 2)(2m + 1)

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Therefore

y(x) = a0

1

x2

2! +

x4

4! ...

+ a1

x

x3

3!

x5

5!

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Therefore

y(x) = a0

1

x2

2! +

x4

4! ...

+ a1

x

x3

3!

x5

5!

i.e., y(x) = a0cosx + a1sinx, where a0 and a1 are arbitrary.

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Example2:


x2y

+ (x2 3x)y

+ 3y = 0

in powers of (x-0).

Solution:

Note that x=0 is a

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Example2:


x2y

+ (x2 3x)y

+ 3y = 0

in powers of (x-0).

Solution:

Note that x=0 is a regular singular point .

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Example2:


x2y

+ (x2 3x)y

+ 3y = 0

in powers of (x-0).

Solution:

Note that x=0 is a regular singular point . the differential equation admits a series solution of the form

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Example2:


x2y

+ (x2 3x)y

+ 3y = 0

in powers of (x-0).

Solution:

Note that x=0 is a regular singular point . the differential equation admits a series solution of the form say

y(x) =

m=1

am(x 0)m+r

.

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Substituting and simplifying, we get

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(r(r 1) 3r + 3)a0xr

+

m=1

{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+

The indicial equation is

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(r(r 1) 3r + 3)a0xr

+

m=1

{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+


r2 4r + 3 = 0, ( as a0 = 0)

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(r(r 1) 3r + 3)a0xr

+

m=1

{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+


r2 4r + 3 = 0, ( as a0 = 0)

The roots are

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(r(r 1) 3r + 3)a0xr

+

m=1

{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+


r2 4r + 3 = 0, ( as a0 = 0)

The roots are r=3 and 1.

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Substituting r=3 and equating the co-efficients of xm+r on eithersides, we get the following recurrence relation:

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am =(m + 2)am1

(m + 3)(m 1) + 3

, m 1.

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am =(m + 2)am1

(m + 3)(m 1) + 3

, m 1.

Thus

a1 = a0 , a2 =

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am =(m + 2)am1

(m

+ 3)(m 1) + 3

, m 1.

Thus

a1 = a0 , a2 =1

2a0 , a3 =

a03

, a4 =a0

4, ....

y1(x) = a0x3

ex

.

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To find y2(x), apply the method of reduction of order with a0 = 1.

Lety2 = y1v.

Then

v(x) =

ex

x3dx.

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Lety2 = y1v.

Then

v(x) =

ex

x3dx.

i.e., v =1

2x2

1

x+

1

2lnx +

1

6x +

1

48x2 + ....

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Lety2 = y1v.

Then

v(x) =

ex

x3dx.

i.e., v =1

2x2

1

x+

1

2lnx +

1

6x +

1

48x2 + ....

y2(x) = (1

2x

1

2x2 +

3

4x3 ...) +

1

2x3exlnx.

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Lety2 = y1v.

Then

v(x) =

ex

x3dx.

i.e., v =1

2x2

1

x+

1

2lnx +

1

6x +

1

48x2 + ....

y2(x) = (1

2x

1

2x2 +

3

4x3 ...) +

1

2x3exlnx.

Thusy(x) = c1y1(x) + c2y2(x), where c1 and c2


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Have some Patience?

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Have some Patience?

yes...1 Obtain the general solution of

x(x + 1)y

(2x + 1)y = 0

in powers of (x-0).2 Obtain the general solution of

(1 x2)y

2xy

+ 2y = 0

in powers of (x-0).3 Obtain a series solution of

(8x2)y

+ 10xy

(1 + x)y = 0

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Still Have Patience?

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Still Have Patience?

Good...4 Obtain a series solution of

(4x2)y

+ 2x(2 x)y

(1 + 3x)y = 0

about 0.5 Obtain the general solution of

xy

+ 3y

+ 4x3y = 0

in powers of (x-0).

6 Obtain the general solution of

(x 1)2y

+ (x 1)y

4y = 0

about x=1. (Hint: put x-1=t)

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S ill ??
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Still ??

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Still ??

Very Good...

7 Obtain the general solution of

(1 + x)x2y

(1 + 2x)xy

+ (1 + 2x)y = 0

about x=0.

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Answers:

Try yourself.

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A
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Answers:

Try yourself.

1

y(x) = a1x + a2x2

.

2

y(x) = a1x + a0

1 x2

1

3x4

1

5x6 + ...

.

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Power seriesExamplesExerciseAnswers

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Answers:

3 r=1/4,-1/2;

y1(x) =

m=0

amxm+

1

4 ,

where am =a0

2mm!7.11.15....(4m 1)(4m + 3),

m = 1, 2, ... a0 is arbitrary.

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Answers:

4 r=1/2,-1/2;

y1(x) =

m=1

a0

2m1(m 1)!x

m1

2 , where a0 is a arbitrary constant.

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Answers:

5 r=0,-2;y1(x) = a0x

2 sin(x2),

v(x) =cot(x2)

2; y2(x) =

x2cos(x2)

2.

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Answers:

6 r=2,-2;

y1(t) = a0t2

,

v(t) =t4

4; y2(x) =

t2

4

y(x) = c1(x 1)2 + c2(x 1)

2.

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Answers:

7 r=1,1; y1(x) = a0x,

v(x) = x + ln x; y2(x) = x(x + ln x).

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For more details , you can click on Series Solutions of DifferentialEquations.

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References
http://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://find/


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Erwin Kreyzig., Advanced Engineering Mathematics.

B.S. Grewal, Higher Engineering Mathematics.

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Thank You

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Bessel- Legendre de 1(Beamer)

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