Ch5-Sec(5.4): Euler Equations; Regular Singular Points

Recall that the point x0 is an ordinary point of the equation

if p(x) = Q(x)/P(x) and q(x)= R(x)/P(x) are analytic at at x0. Otherwise x0 is a singular point.

Thus, if P, Q and R are polynomials having no common factors, then the singular points of the differential equation are the points for which P(x) = 0.

0)()()(2

2

yxRdx

dyxQ

dx

ydxP

Euler Equations

A relatively simple differential equation that has a regular singular point is the Euler equation,

where , are constants.

Note that x0 = 0 is a regular singular point.

The solution of the Euler equation is typical of the solutions of all differential equations with regular singular points, and hence we examine Euler equations before discussing the more general problem.

0][ 2 yyxyxyL

Solutions of the Form y = xr

In any interval not containing the origin, the general solution of the Euler equation has the form

Suppose x is in (0, ), and assume a solution of the form y = xr. Then

Substituting these into the differential equation, we obtain

or

or

0][ 2 yyxyxyL

)()()( 2211 xycxycxy

21 )1(,, rrr xrryxryxy

0)1(][ rrrr xxrxrrxL

0)1(][ rrrxxL rr

0)1(][ 2 rrxxL rr

Quadratic Equation

● Thus, after substituting y = xr into our differential equation, we arrive at

and hence

Let F(r) be defined by

We now examine the different cases for the roots r1, r2.

0,0)1(2 xrrxr

2

4)1()1( 2 r

))(()1()( 212 rrrrrrrF

Real, Distinct Roots

If F(r) has real roots r1 r2, then

are solutions to the Euler equation. Note that

Thus y1 and y2 are linearly independent, and the general solution to our differential equation is

21 )(,)( 21rr xxyxxy

.0 allfor 0112

11

12

12

1121

21

21

2121

21

21

xxrr

xrxr

xrxr

xx

yy

yyW

rr

rrrr

rr

rr

0,)( 2121 xxcxcxy rr

Example 1

Consider the equation

Substituting y = xr into this equation, we obtain

and

Thus r1 = 1/2, r2 = -1, and our general solution is

0,032 2 xyyxyx

21 )1(,, rrr xrryxryxy

0112

012

013)1(2

03)1(2

2

rrx

rrx

rrrx

xrxxrr

r

r

r

rrr

0,)( 12

2/11 xxcxcxy

2 4 6 8 1 0x

1

2

3

4y x 12/1)( xxxy

Equal Roots

If F(r) has equal roots r1 = r2, then we have one solution

We could use reduction of order to get a second solution; instead, we will consider an alternative method.

Since F(r) has a double root r1, F(r) = (r - r1)2, and F'(r1) = 0.

This suggests differentiating L[xr] with respect to r and then setting r equal to r1, as follows:

1)(1rxxy

0,ln)(

2ln]ln[

][

)1(][

12

12

1

21

21

2

xxxxy

xrrrrxxxxL

rrxr

xLr

rrxrrxxL

r

rrr

rr

rrr

Equal Roots

Thus in the case of equal roots r1 = r2, we have two solutions

Now

Thus y1 and y2 are linearly independent, and the general solution to our differential equation is

xxxyxxy rr ln)(,)( 1121

.0 allfor 0

ln1ln

1ln

ln

12

1211

12

111

121

21

1

11

11

11

xx

xxrxrx

xrxxr

xxx

yy

yyW

r

rr

rr

rr

0,lnln)( 1112121 xxxccxxcxcxy rrr

Example 2Consider the equation

Then

and

Thus r1 = r2 = -2, our general solution is

0,0452 xyyxyx

21 )1(,, rrr xrryxryxy

02

044

045)1(

045)1(

2

2

rx

rrx

rrrx

xrxxrr

r

r

r

rrr

0,ln)( 221 xxxccxy

1 2 3 4 5x

2

1

1

2yx 2ln1)( xxxy

Complex Roots

Suppose F(r) has complex roots r1 = + i, r2 = - i, with 0. Then

Thus xr is defined for complex r, and it can be shown that the general solution to the differential equation has the form

However, these solutions are complex-valued. It can be shown that the following functions are solutions as well:

0,lnsinlncoslnln

lnlnlnlnln

xxixxee

eeeeexxix

xixxixrxr r

0,)( 21 xxcxcxy ii

xxxyxxxy lnsin)(,lncos)( 21

Complex Roots

The following functions are solutions to our equation:

Using the Wronskian, it can be shown that y1 and y2 are linearly independent,

and thus the general solution to our differential equation can be written as

xxxyxxxy lnsin)(,lncos)( 21

0,lnsinlncos)( 21 xxxcxxcxy

0for0lnsin,lncos 12 xxxxxxW

Example 3

Consider the equation

Then

and

Thus r1 = -i, r2 = i, and our general solution is

0,02 xyyxyx

21 )1(,, rrr xrryxryxy

01

0)1(

0)1(

2

rx

rrrx

xrxxrr

r

r

rrr

0,lnsinlncos

lnsinlncos)(

21

02

01

xxcxc

xxcxxcxy

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

3

2

1

1

2

3yx

2 4 6 8 10 12 14x

3

2

1

1

2

3yx

150

lnsinlncos)(

x

xxxy

5.00

lnsinlncos)(

x

xxxy

Solution Behavior

Recall that the solution to the Euler equation

depends on the roots:

where r1 = + i, r2 = - i.

The qualitative behavior of these solutions near the singular point x = 0 depends on the nature of r1 and r2.

Also, we obtain similar forms of solution when t < 0. Overall results are summarized on the next slide.

0][ 2 yyxyxyL

,lnsinlncos)(:complex ,

ln)(:

)(:

2121

2121

2121

1

21

xxcxxcxyrr

xxccxyrr

xcxcxyrrr

rr

General Solution of the Euler Equation

The general solution to the Euler equation

in any interval not containing the origin is determined by the roots r1 and r2 of the equation

according to the following cases:

where r1 = + i, r2 = - i.

02 yyxyx

))(()1()( 212 rrrrrrrF

,lnsinlncos)(:complex ,

ln)(:

)(:

2121

2121

2121

1

21

xxcxxcxyrr

xxccxyrr

xcxcxyrrr

rr

Shifted Equations

The solutions to the Euler equation

are similar to the ones given in previous slide:

where r1 = + i, r2 = - i.

002

0 yyxxyxx

,lnsinlncos)(

:complex ,

ln)(:

)(:

02001

21

002121

020121

1

21

xxxcxxxxcxy

rr

xxxxccxyrr

xxcxxcxyrrr

rr

Solution Behavior and Singular PointsIf we attempt to use the methods of the preceding two sections to solve the differential equation in a neighborhood of a singular point x0, we will find that these methods fail.

This is because the solution may not be analytic at x0, and hence will not have a Taylor series expansion about x0. Instead, we must use a more general series expansion.A differential equation may only have a few singular points, but solution behavior near these singular points is important.For example, solutions often become unbounded or experience rapid changes in magnitude near a singular point.Also, geometric singularities in a physical problem, such as corners or sharp edges, may lead to singular points in the corresponding differential equation.

Numerical Methods and Singular PointsAs an alternative to analytical methods, we could consider using numerical methods, which are discussed in Chapter 8.

However, numerical methods are not well suited for the study of solutions near singular points.

When a numerical method is used, it helps to combine with it the analytical methods of this chapter in order to examine the behavior of solutions near singular points.

Solution Behavior Near Singular PointsThus without more information about Q/P and R/P in the neighborhood of a singular point x0, it may be impossible to describe solution behavior near x0.

It may be that there are two linearly independent solutions that remain bounded as x x0; or there may be only one, with the other becoming unbounded as x x0; or they may both become unbounded as x x0.

If a solution becomes unbounded, then we may want to know if y in the same manner as (x - x0)-1 or |x - x0|-½, or in some other manner.

Classifying Singular PointsOur goal is to extend the method already developed for solving

near an ordinary point so that it applies to the neighborhood of a singular point x0.

To do so, we restrict ourselves to cases in which singularities in Q/P and R/P at x0 are not too severe, that is, to what might be called “weak singularities.”

It turns out that the appropriate conditions to distinguish weak singularities are

0)()()( yxRyxQyxP

finite. is )(

)(lim finite is

)(

)(lim 2

0000 xP

xRxxand

xP

xQxx

xxxx

Regular Singular PointsConsider the differential equation

If P and Q are polynomials, then a regular singular point x0 is singular point for which

For more general functions than polynomials, x0 is a regular singular point if it is a singular point with

Any other singular point x0 is an irregular singular point.

finite. is )(

)(lim finite is

)(

)(lim 2

0000 xP

xRxxand

xP

xQxx

xxxx

.at analytic are )(

)(

)(

)(0

200 xx

xP

xRxxand

xP

xQxx

0)()()( yxRyxQyxP

Example 4: Legendre EquationConsider the Legendre equation

The point x = 1 is a regular singular point, since both of the following limits are finite:

Similarly, it can be shown that x = -1 is a regular singular point.

0121 2 yyxyx

0

1

11lim

1

11lim

)(

)(lim

,11

2lim

1

21lim

)(

)(lim

12

2

1

20

1210

0

0

xx

xx

xP

xRxx

x

x

x

xx

xP

xQxx

xxxx

xxxx

Example 5Consider the equation

The point x = 0 is a regular singular point:

The point x = 2, however, is an irregular singular point, since the following limit does not exist:

02322 2 yxyxyxx

022

lim22

2lim

)(

)(lim

,022

3lim

22

3lim

)(

)(lim

022

0

20

20200

0

0

x

x

xx

xx

xP

xRxx

x

x

xx

xx

xP

xQxx

xxxx

xxxx

22

3lim

22

32lim

)(

)(lim

2220

0

xx

x

xx

xx

xP

xQxx

xxxx

Example 6: Nonpolynomial Coefficients (1 of 2)

Consider the equation

Note that x = /2 is the only singular point.

We will demonstrate that x = /2 is a regular singular point by showing that the following functions are analytic at /2:

0sincos2/ 2 yxyxyx

xx

xx

x

x

x

xx sin

2/

sin2/,

2/

cos

2/

cos2/ 2

2

2

Example 6: Regular Singular Point (2 of 2)

Using methods of calculus, we can show that the Taylor series of cos x about /2 is

Thus

which converges for all x, and hence is analytic at /2. Similarly, sin x is analytic at /2, with Taylor series

Thus /2 is a regular singular point of the differential equation.

0

121

2/!)12(

)1(cos

n

nn

xn

x

,2/!)12(

)1(1

2/

cos

1

21

n

nn

xnx

x

0

22/!)2(

)1(sin

n

nn

xn

x

Exercise 47: Bessel Equation

Consider the Bessel equation of order

The point x = 0 is a regular singular point, since both of the following limits are finite:

0222 yxyxyx

22

222

0

20

200

lim )(

)(lim

,1lim)(

)(lim

0

0

x

xx

xP

xRxx

x

xx

xP

xQxx

xxx

xxx