Download a review sheet for test #4

ODE
Test #4 Review
Page 1 of 14
Section 5.1: Review of Power Series

1. Definition of convergence of a power series: A power series
a x  x 
n 0
m
n
n
0
is said to
converge at a point x if lim  an  x  x0  exists for that x.
m 
n
n 0

2. Definition of absolute convergence of a power series: A power series
a x  x 
n 0

said to converge absolutely at a point x if
 a x  x 
n 0
n
0
n
n
0
n
is
converges.
a. Absolute convergence implies convergence…
3. Ratio test: If, for a fixed value of x, lim
n 
an 1  x  x0 
an  x  x0 
n 1
n
 x  x0 lim
n 
an 1
 x  x0 L , then
an
the power series converges absolutely at that value of x if x  x0 L  1, and diverges if
x  x0 L  1. If x  x0 L  1, then the test is inconclusive.

4. If
a x  x 
n 0
n
n
0
converges at x = x1, it converges absolutely for x  x0  x1  x0 , and if
it diverges at x = x1, it diverges for x  x0  x1  x0 .
5. Radius and interval of convergence: The radius of convergence  is a nonnegative

number such that
a x  x 
n 0
n
0
n
converges absolutely for x  x0   and diverges for
x  x0   .
a. Series that converge only when x = x0 are said to have  = 0.
b. Series that converge for all x are said to have  = .
c. If  > 0, then the interval of convergence of the series is x  x0   .
ODE
Test #4 Review

Given that

Page 2 of 14
 an  x  x0  and  bn  x  x0  converge for x  x0   …
n
n 0
n
n 0

6. Sum of series:


 an  x  x0    bn  x  x0     an  bn  x  x0 
n
n 0
n
n 0
n
n 0
7. Product and Quotient of series:



n
n
a
x

x


0
 n
   bn  x  x0  
 n 0
  n 0

2
3
2
3
  a0  a1  x  x0   a2  x  x0   a3  x  x0   ... b0  b1  x  x0   b2  x  x0   b3  x  x0   ...




 a0 b0  b1  x  x0   b2  x  x0   b3  x  x0   ...
2
3


 a1  x  x0  b0  b1  x  x0   b2  x  x0   b3  x  x0   ...
2

a  x  x  b  b  x  x   b  x  x 
3


 b  x  x   ...
 a2  x  x0  b0  b1  x  x0   b2  x  x0   b3  x  x0   ...
2
3
3
0
0
1
0
2
0
2
2
3
3
3
0
...
 a0b0   a0b1  a1b0  x  x0    a0b2  a1b1  a2b0  x  x0    a0b3  a1b2  a2b1  a3b0  x  x0   ...
2

 n

n
    ak bn  k   x  x0 
n 0  k 0

To do a quotient, can write as a multiplication and equate terms:

a x  x 
n 0

n
b  x  x 
n 0
n
n
0
n




n
n
n
n
  d n  x  x0     d n  x  x0     bn  x  x0     an  x  x0 
n 0
 n 0
  n 0
 n 0
0
OR can do long division…
8. Derivatives of a series:

d 
n
n 1
a
x

x

nan  x  x0 




n
0


dx  n0
 n1
2


d 
n
n2
a
x

x

n  n  1 an  x  x0 



0
2  n

dx  n0
 n2
9. Taylor series:
f  n   x0 
n
 x  x0  is the Taylor series for a function f(x) about the point x = x0.

n!
n 0

3
ODE
Test #4 Review
Page 3 of 14
10. Equality of series: every corresponding term is equal…
11. Analytic functions: have a convergent Taylor series with non-zero radius of convergence
about some point x = x0.
12. Shift of index of Summation…
Section 5.2: Series Solutions Near an Ordinary Point, Part I
d2y
dy
Will consider homogeneous equations of the form P  x  2  Q  x   R  x  y  0 where the
dx
dx
polynomial coefficients are polynomials, like:
The Bessel equation: x 2 y  xy   x 2   2  y  0
The Legendre equation: 1  x 2  y  2 xy     1 y  0
Ordinary point: a point x0 such that P(x0)  0.  y  p  x  y  q  x  y  0
Singular point: a point x0 such that P(x0) = 0.  p(x) or q(x) becomes unbounded (See sections
5.4 – 5.7)
To solve a differential equation near an ordinary point using a power series technique:


Assume a power series form of the solution y  x    an x n which converges in the
n 0
interval x  x0  





Plug the power series into the equation and equate like terms
Obtain a recurrence relation for the coefficients of higher-order terms in terms of earlier
coefficients.
If possible, try to find the general term, which is an explicit function for the coefficients
in terms of the index of summation.
Write the final answer in terms of those coefficients.
Check the radius of convergence.
ODE
Test #4 Review
Page 4 of 14
Section 5.3: Series Solutions Near an Ordinary Point, Part II
To justify the statement that a solution of P  x  y  Q  x  y  R  x  y  0  y  p  x  y  q  x  y

m
can be written as y    x    an  x  x0  , we must be able to compute m!am      x0  for
n
n 0
any order derivative m based only on the information given by the IVP.
Note:
So it seems that p and q need to at least be infinitely differentiable at x0, but in addition they also
need to be analytic at x0. In other words, they must have Taylor series expansions that converge
in some interval about x0.
Theorem 5.3.1 (due to Fuchs)
If x0 is an ordinary point of the differential equation P  x  y  Q  x  y  R  x  y  0 (i.e.,
pQ
P
and q  R are analytic at x0), then its general solution is
P

y   an  x  x0   a0 y1  x   a1 y2  x  ,
n
n 0
where a0 and a1 are arbitrary, and y1 and y2 are two power series solutions that are analytic at x0.
The solutions y1 and y2 form a fundamental set of solutions. Also, the radius of convergence of
y1 and y2 is at least as large as the minimum of the radii of convergence of p and q.
Note:
From the theory of complex-valued rational expressions, it turns out that radius of convergence
of a power series of a rational expression about a point x0 is the distance from x0 to the nearest
zero of the (complex-valued) denominator.
ODE
Test #4 Review
Page 5 of 14
Section 5.4: Euler Equations; Regular Singular Points
P  x  y  Q  x  y  R  x  y  0
Singular points are x = x0 such that P(x0) = 0.
Euler equations have singular points at x = 0:
x 2 y   xy   y  0
To solve Euler equations, make an assumption for the solution of y = xr.
When the roots are real and distinct, then the two fundamental solutions are obtained.
When the roots are repeated, multiply the first fundamental solution by ln(x) to obtain the second
fundamental solution.
When the roots are complex, r    i  , the fundamental solutions are
y1  x cos   ln  x   and y2  x sin   ln  x  
In summary, the solutions of x 2 y   xy   y  0 are:
 r1
r2
c1 x  c2 x , given r1 and r2 are distinct real roots of r  r  1   r    0

y  x    c1  c2 ln  x   x r1 , given r1 is a real double root of r  r  1   r    0


 c1 cos   ln  x    c2 sin   ln  x   x , given   i  are complex roots of r  r  1   r    0


For negative x values, one can make the substitution    x , and show that, for any x value, the
solutions are:
r
r

c1 x  c2 x
r

y  x    c1  c2 ln  x   x


 c1 cos   ln  x    c2 sin   ln  x   x

1
2
1


We do not have a general theory of how to handle any possible singularity of
P  x  y  Q  x  y  R  x  y  0 . So, for the next few sections, we will restrict the discussion to
the power series solution of this type of equation near regular singular points.
A regular singular point at x = x0 is a singular point with the additional restrictions that
Q  x
2 R  x
lim  x  x0 
is finite , and lim  x  x0 
is finite .
x  x0
x  x0
P  x
P  x
ODE
Test #4 Review
Page 6 of 14
Section 5.5: Series Solutions Near a Regular Singular Point, Part I
Big Idea: According to Frobenius, it is valid to assume a series solution of the form
y   x  x0 
r

a x  x 
n 0
n
n
0
to a second-order linear differential equation near a regular singular
point x = x0.
Recall:
If x = x0 is a regular singular point of the second order linear equation
P  x  y  Q  x  y  R  x  y  0 ,
then lim x
x  x0
Q  x
R  x
 lim xp  x   finite and lim x 2
 lim x 2 q  x   finite .
x  x0
P  x  x  x0
P  x  x  x0


This means xp  x    pn  x  x0  and x 2 q  x    qn  x  x0  are convergent for some radius
n
n 0
n
n 0
about x0.
Thus, we can write the original equation as: x 2 y   xp  x   xy   x 2 q  x   y  0
If all the coefficients pn and qn are zero except for p0 and q0, then the equation reduces to
x 2 y  p0 xy  q0 y  0 , which is an Euler equation. In fact, this is called the corresponding
Euler equation when all the coefficients pn and qn are not zero, and the roots of this
corresponding Euler equation play a role in the solution called “the exponents at the singularity.”
Since the equation we are trying to solve looks like an equation with “Euler coefficients” times
power series, we will look for a solution that is of the form of an “Euler solution” times a power
series:
y   x  x0 
r


a x  x   a x  x 
n 0
n
n
0
n 0
n
r n
0
As part of the solution, we must determine:
 The values of r that make this a valid solution.
 The recurrence relation for the coefficients an.
 The radius of convergence.
The theory behind a solution of this form is due to Frobenius.
ODE
Test #4 Review
Page 7 of 14
To solve a linear second order equation near a regular singular point using the method of
Frobenius:
 Identify singular points and verify they are regular.


Assume a solution of y   an  x  x0 
r n
and its derivatives:
n 0

y   an  r  n  x  x0 
n 0
r  n 1

, y   an  r  n  r  n  1 x  x0 
r  n2
n 0

Substitute the assumed solution and its derivatives into the given equation.
o You may have to re-write coefficients in terms of  x  x0  …

Shift indices so that all series solutions have  x  x0 

“Spend” any terms needed so that all series start at the same index value.
o This should result in a term that looks like the characteristic equation of the
corresponding Euler equation.
o This is called the indicial equation.
o The roots of the indicial equation are called the exponents of the singularity.
o They are the same as the roots of the corresponding Euler equation.
r n
Set the coefficient of  x  x0  to zero to get the recurrence relation.



r n
in the general term.
Use the recurrence relation with each exponent to get the general term for each of the two
solutions for each exponent.
o If the exponents of the singularity are equal or differ by an integer, then it is only
valid to get the series solution for the larger root. (What to do for the second
solution will be covered in 5.6).
Compute the radius of convergence for each solution.
ODE
Test #4 Review
Page 8 of 14
5.6: Series Solutions Near a Regular Singular Point, Part II
To solve a linear second order equation near a regular singular point using the method of
Frobenius REGARDLESS OF THE EXPONENTS OF THE INDICIAL EQUATION:
 Identify singular points and verify they are regular.
 Find the exponents of the singularity by solving the corresponding Euler equation
Q  x
R  x
x 2 y  p0 xy  q0 y  0 , where p0  lim x
and q0  lim x 2
, or using the indicial
x 0
x 0
P  x
P  x
equation.


Assume a solution of y  x r1  an  x  x0  .
n
n 0


Substitute the assumed solution and its derivatives into the given equation.
r n
Shift indices so that all series solutions have  x  x0  in the general term.


“Spend” any terms needed so that all series start at the same index value.
o This should result in a term with a factor that looks like the characteristic equation
of the corresponding Euler equation; this is the indicial equation.
r n
Set the coefficients of  x  x0  to zero to get the recurrence relation.

Get the second solution using the theorem below…
ODE
Test #4 Review
Page 9 of 14
Theorem 5.6.1: General Solution of a Second Order Equation with Real Exponents of the
Singularity near a Regular Singular Point
Consider the differential equation x 2 y  x  xp  x   y   x 2 q  x   y  0 , where x = 0 is a regular
singular point. Then xp  x  and x 2 q  x  are analytic at x = 0 with convergent power series


expansions xp  x    pn x n and x 2 q  x    qn x n for x   where   0 is the minimum of
n 0
n 0
the radii of convergence for xp  x  and x q  x  . Let r1 and r2 be the roots of the indicial equation
2
F  r   r  r 1  p0r  q0  0 , with r1  r2 if r1 and r2 are real. Then in either the interval
   x  0 or 0  x   , there exists a solution of the form


r 
y1  x   x 1 1   an  r1  x n 
 n 1

where the an  r1  are given by the recurrence relation
n 1
F  r  n  an   ak  r  k  pn  k  qn  k   0
k 0
with a0 = 1 and r = r1. There are three cases for the second solution:

If r1  r2 is not zero or a positive integer, then in either the interval    x  0 or 0  x   ,
there exists a second solution of the form


r 
y2  x   x 2 1   an  r2  x n  .
 n 1

The an  r2  are determined by the same recurrence relation as the an  r1  , with with a0 = 1 and
r = r2. These power series solutions converge for at least x   .


If r1 = r2 then the second solution is of the form


r 
y2  x   y1  x  ln x  x 1  bn  r1  x n  .
 n 1

If r1  r2 = N, a positive integer, then the second solution is of the form


r 
y2  x   ay1  x  ln x  x 2 1   cn  r2  x n  .
 n1

The coefficients an , bn  r1  , cn  r2  and the constant a can be determined by substituting the form
of the series solution for y2 into the original differential equation. The constant a may turn out to
be zero. Each of these series converge at least for x   and defines an analytic function near x
= 0.
In all three cases, the two solutions form a fundamental set of solutions.
ODE
Test #4 Review
Page 10 of 14
Section 5.7: Bessel’s Equation
Bessel’s Equation: x 2 y  xy   x 2  2  y  0





x = 0 is a regular singular point
The roots of the indicial equation are .
The value of  is called the “order” of the equation.
The first solution for a given value of  is called the “Bessel function of the first kind of
order ,” and is denoted by J(x).
The second solution for a given value of  is called the “Bessel function of the second
kind of order ,” and is denoted by Y(x).
Bessel Equation of Order Zero (i.e.,  = 0): x 2 y  xy  x 2 y  0
 The roots of the indicial equation are r1 = r2 = 0.
m
m



1 x 2 m 
1 x 2 m


 y1  x   a0 1   2 m
 J 0  x   1   2m
2 
2
m 1 2
 m!
 m1 2  m ! 

J0(x)  1 as x  0.


 2 2

J0  x  
 cos  x   as x   .
4
x 

1


y2  x   J 0  x  ln  x   
m 1
o
1 1
Hm  1  
2 3
 1
2
2m

m 1
Hm
 m!
2
x2m
m
1
1
  ; i.e., a partial sum of the harmonic series up to m.
m k 1 k
ODE
Test #4 Review
Page 11 of 14

The traditional Bessel function of the second kind of order zero is a linear combination of
y2  x  and J 0  x  :

Y0  x  

 y  x      ln  2   J 0  x  
 2
m 1

1 H m 2 m 

2
Y0  x      ln  2   J 0  x    2 m
x 
2
 
m 1 2
 m!

o  is the “Euler-Mascheroni” constant:   lim  H n  ln  n    0.577 215 665
2
n
2

Y0  x  


 2 2

Y0  x   
 sin  x   as x   .
4
x 


ln  x  as x  0 .
1
ODE
Test #4 Review
Page 12 of 14
1

Bessel Equation of Order One-Half (i.e.,  = ½ ): x 2 y  xy   x 2   y  0
4

 The roots of the indicial equation are r1 = + ½ , r2 = - ½ .
m
m


1 
1 x 2 m 
1 x 2 m1


1
1
2
2
 y1  x   x 1  
 x 2 sin  x 
x 
m  0  2m  1 !
 m1  2m  1! 
1



 2 2
By convention, J 1  x   
 sin  x  , x > 0.
2
x 
n

   1n x 2 n
1 x 2 n1 
cos  x 
sin  x 

y2  x   x  a0 
 a1 
 a1 1
  a0
1
n 1  2n  1 ! 
 n 1  2n !
x 2
x 2

The Bessel function of the second kind of order one-half is:
1
2
1

 2 2
J 1  x  
 cos  x  , x > 0.
2
x 
ODE
Test #4 Review
Page 13 of 14
Bessel Equation of Order One (i.e.,  = 1): x 2 y  xy   x 2  1 y  0

The roots of the indicial equation are r1 = 1, r2 = -1.
2m

 1  x 
x   1 x
J1  x    2 m

 
2 m 0 2  m  1 !m ! m 0  m  1!m !  2 
series can be written as powers of (x/2)…
m






(Note: J1(x) = ½ y1(x) so that the
3 
 2 2

J1  x   
 cos  x   as x   .
4 
x 

m

1  H m  H m1  2 m 

1
y2  x    J1  x  ln  x   1  
x , x > 0
x  m1 22 m m ! m  1!

The traditional Bessel function of the second kind of order one is:
2
Y1  x     y2  x      ln  2   J1  x  

2
Y1  x    as x  0 .
x
1

2 m 1
J1(x)  0 as x  0.
1

m
3 
 2 2

Y1  x   
 sin  x   as x   .
4 
x 

ODE
Test #4 Review
Page 14 of 14
Bessel Equations of Higher Positive Integer Order (i.e.,  = positive integer,  > 1):
x 2 y  xy   x 2  2  y  0

The roots of the indicial equation are r1 = , r2 = -.

 1  x 2 m
J  x   
 
m 0  m   !m !  2 

J(x)  0 as x  0.


 2  1   as x   .
 2  2
J  x   

 cos  x 
4
x 


m

1



Weisstein, Eric W. "Bessel Function of the First Kind." From MathWorld--A Wolfram
Web Resource.
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.htmlhttp://mathworld.wolfr
am.com/BesselFunctionoftheFirstKind.html
J  x  cos    J   x 
Y  x   
sin  
Y  x   
  1!  2 
  as x  0 .
x


 2  1   as x   .
 2  2
Y  x   

 sin  x 
4
x 


Weisstein, Eric W. "Bessel Function of the First Kind." From MathWorld--A Wolfram
Web Resource.
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.htmlhttp://mathworld.wolfr
am.com/BesselFunctionoftheSecondKind.html
1

