Ch 5.4: Euler Equations;Regular Singular Points

Recall that for equation

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

Example 1: Bessel and Legendre EquationsBessel Equation of order :

The point x = 0 is a singular point, since P(x) = x2 is zero there. All other points are ordinary points.

Legendre Equation:

The points x = 1 are singular points, since P(x) = 1- x2 is zero there. All other points are ordinary points.

0222 yxyxyx

0121 2 yyxyx

Euler Equations

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

where , are constants.

Note that x0 = 0 is a singular point.

The solution of the Euler equation is typical of the solutions of all differential equations with 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

)()()( 2211 xycxycxy

21 )1(,, rrr xrryxryxy

0)1(][ rrrr xxrxrrxL

0)1(][ 2 rrxxL rr

0)1(][ rrrxxL rr

0][ 2 yyxyxyL

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

))(()1()( 212 rrrrrrrF

2

4)1()1( 2 r

Real, Distinct Roots

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

are solutions to the Euler equation. Note that

Thus y1 and y2 form fundamental solutions, 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 = -1/3, r2 = 1, and our general solution is

21 )1(,, rrr xrryxryxy

0113

0123

01)1(3

0)1(3

2

rrx

rrx

rrrx

xrxxrr

r

r

r

rrr

0,)( 23/1

1 xxcxcxy

0,03 2 xyyxyx

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 form fundamental solutions, 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 2

Consider the equation

Then

and

Thus r1 = r2 = -3, our general solution is

21 )1(,, rrr xrryxryxy

0,0972 xyyxyx

03

096

097)1(

097)1(

2

2

rx

rrx

rrrx

xrxxrr

r

r

r

rrr

0,ln)( 321 xxxccxy

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 form fundamental solutions, and thus the general solution to our differential equation can be written as

xxxyxxxy lnsin)(,lncos)( 21

0,lnsinlncos)( 21 xxxcxxcxy

Example 3

Consider the equation

Then

and

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

0,042 xyyxyx

21 )1(,, rrr xrryxryxy

04

04)1(

04)1(

2

rx

rrrx

xrxxrr

r

r

rrr

0,ln2sinln2cos

ln2sinln2cos)(

21

02

01

xxcxc

xxcxxcxy

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. Discuss.

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

,lnsinlncos)(:complex ,

ln)(:

)(:

2121

2121

2121

1

21

xxcxxcxyrr

xxccxyrr

xcxcxyrrr

rr

0][ 2 yyxyxyL

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.

,lnsinlncos)(:complex ,

ln)(:

)(:

2121

2121

2121

1

21

xxcxxcxyrr

xxccxyrr

xcxcxyrrr

rr

02 yyxyx

))(()1()( 212 rrrrrrrF

Shifted Equations

The solutions to the Euler equation

are similar to the ones given in Theorem 5.5.1:

where r1 = + i, r2 = - i.

,lnsinlncos)(

:complex ,

ln)(:

)(:

02001

21

002121

020121

1

21

xxxcxxxxcxy

rr

xxxxccxyrr

xxcxxcxyrrr

rr

002

0 yyxxyxx

Example 5: Initial Value Problem (1 of 4)

Consider the initial value problem

Then

and

Using the quadratic formula on r2 + 2r + 5, we obtain

1)1(,1)1(,01062 2 yyyyxyx

21 )1(,, rrr xrryxryxy

01042

0106)1(2

0106)1(2

2

rrx

rrrx

xrxxrr

r

r

rrr

ir 212

2042

Example 5: General Solution (2 of 4)

Thus = -1, = 2, and the general solution of our initial value problem is

where the last equality follows from the requirement that the domain of the solution include the initial point x = 1.

To see this, recall that our initial value problem is

,ln2sinln2cos

ln2sinln2cos)(1

21

1

1

2

1

1

xxcxxc

xxcxxcxy

1)1(,1)1(,01062 2 yyyyxyx

Example 5: Initial Conditions (3 of 4)

Our general solution is

Recall our initial value problem:

Using the initial conditions and calculus, we obtain

Thus our solution to the initial value problem is

xxcxxcxy ln2sinln2cos)( 12

11

1,112

121

21

1

cccc

c

xxxxxy ln2sinln2cos)( 11

1)1(,1)1(,01062 2 yyyyxyx

Example 5: Graph of Solution (4 of 4)

Graphed below is the solution

of our initial value problem

Note that as x approaches the singular point x = 0, the solution oscillates and becomes unbounded.

xxxxxy ln2sinln2cos)( 11

1)1(,1)1(,01062 2 yyyyxyx

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

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.

Example 1Consider the following equation

which has a singular point at x = 0. It can be shown by direct substitution that the following functions are linearly independent solutions, for x 0:

Thus, in any interval not containing the origin, the general solution is y(x) = c1x2 + c2 x

-1.

Note that y = c1 x2 is bounded and analytic at the origin, even though Theorem 5.3.1 is not applicable.

However, y = c2 x -1 does not have a Taylor series expansion

about x = 0, and the methods of Section 5.2 would fail here.

12

21 )(,)( xxyxxy

,022 yyx

Example 2Consider the following equation

which has a singular point at x = 0.

It can be shown the two functions below are linearly independent solutions and are analytic at x = 0:

Hence the general solution is

If arbitrary initial conditions were specified at x = 0, then it would be impossible to determine both c1 and c2.

0222 yyxyx

221 )(,)( xxyxxy

221)( xcxcxy

Example 3Consider the following equation

which has a singular point at x = 0.

It can be shown that the following functions are linearly independent solutions, neither of which are analytic at x = 0:

Thus, in any interval not containing the origin, the general solution is y(x) = c1x

-1 + c2 x -3.

It follows that every solution is unbounded near the origin.

32

11 )(,)( xxyxxy

,0352 yxyyx

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 Points

Consider the differential equation

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

Any other singular point x0 is an irregular singular point, which will not be discussed in this course.

finite. is )(

)(lim finite is

)(

)(lim 2

0000 xP

xRxxand

xP

xQxx

xxxx

0)()()( yxRyxQyxP

Example 4: Bessel EquationConsider 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

Example 5: 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 6Consider 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