Chapter 2 - Series Solutions of Linear Ordinary Differential Equations

Question 1

What is an elementary function?

Answer 1

-an elementary function is one that can be expressed as a power series expansion

Question 2

How are power series related to differential equations?

Answer 2

• we can calculate power series solutions of differential equations which have no elementary functions as solutions
• they are also an alternative method to solving differential equations which do have a known function as their solution

Question 3

List the three well known types of elementary equation

Answer 3

1) algebraic functions
2) the exponential function e^x
3) trigonometric functions

Question 4

Algebraic Functions

Answer 4

-solutions y(x) of any equation of the form Pn(x)y^n + … + P1(x)y + Po(x) = 0
-polynomials which correspond to n=1, P1=-1
=> y = Po(x)
-rational functions which correspond to n=1 and general P1(x)
=> y = - Po(x) / P1(x)
-roots of rational functions e.g. n=2, P1(x)=0
=> y = √[-Po(x)/P2(x)

Question 5

Combining to Form Elementary Functions

Answer 5

• algebraic combinations of other types,
  e. g. tan(x) = sin(x)/cos(x)
• function of a function
  e. g. sin(x*e^x)
• inverse functions
  e. g. log(x) or tan^(-1)(x)
• -but only a finite number of such operations

Question 6

Taylor Expansion of a function f(x) about a point xo

Answer 6

f(x) = Σ f(‘n)(xo)/n! * (x-xo)^m

• where f(‘n)(xo) is the nth derivative of f evaluated at xo
• and the sum is taken from 0 to infinity

Question 7

Radius of Convergence

Answer 7

-the Taylor series about a point xo is only valid within a radius of convergence of that point i.e. valid for:
|x-xo| < rc (rc>0)
rc = lim |an / an+1|
-where the limit is taken as n tends to infinity and the n and n+1 are subscripts
-if rc tends to infinity then we say that the series converges for all x

Question 8

Term by Term

Answer 8

• the Taylor series can be differentiated term by term

- in fact it is infinitely differentiable and each of the resulting power series has the SAME radius of convergence rc

Question 9

Ratio Test

Answer 9

-consider the power series:
Σ an * x^n
-where the sum is taken between 0 and infinity
-for x=0 it evidently has the sum ao 
-for x≠0 we use the ration test:
lim |an+1*x^(n+1) / an*x^n| 
= lim |an+1 / an | * |x| = L
-where the limit is taken as n tends to infinity and the n and n+1 following a are subscripts
-if |L|<1 the series converges
-if |L|>1 the series diverges
-if |L|=1 it is unknown

Question 10

Power Series

e^x

Answer 10

e^x = Σ x^n/n!
-where the sum is taken from n=0 to n=infinity
e^x = 1 + x + x²/2 + x^3/3 + …
-for all x

Question 11

Power Series

sinx

Answer 11

sinx = x - x^3/3! + x^5/5! + …

-for all x

Question 12

Power Series

1/(1-x)

Answer 12

1/1-x = Σ x^n = 1 + x + x² + x^3 +…

-where the sum is taken from n=0 to infinity

Question 13

Where can an elementary function be represented as a power series?

Answer 13

• given an elementary function we can represent it as a power series wherever the function is analytic
• e.g. for √x , the function is not analytic at xo=0 so has no Taylor series there, but at xo=1 a Taylor series can be defined

Question 14

Analytic Solutions of ODEs

Steps

Answer 14

y’’ + p(x)y’ + q(x)y = 0
-start by supposing that the solution y takes the form:
y = Σanx^n = ao + a1x + a2x² +…
-where the sum is taken between 0 and infinity
-differentiate to get an expression for y’
-differentiate again for an expression for y’’ (if necessary)
-sub into the original differential equation
-collect terms by power of x e.g. a1-ao + (2a2-a1)x + (3a3-a2)x² + …. + ((n+1)an+1-an)x^n+…
-the only way for a series to be identically zero is for all coefficients to be equal to zero
-setting each coefficient equal to zero write the other coefficients in terms of constants ao, a1, a2
-sub these expression into the original
y = Σanx^n
-to obtain the general solution

Question 15

When can an analytic solution for the following equation be found?
y’’ + p(x)y’ + q(x)y = 0

Answer 15

-in order to build analytic solutions to equations of this form we must demand that functions p and q are not only continuous but also analytic
-most examples will be derived from:
S(x)y’’ + P(x)y’ + Q(x)y = 0
-where S, P and Q are polynomials with no common factor
-so p(x) = P(x) / S(x)
-and q(x) = Q(x) / S(x)
-p and q are analytic if they can be written as a power series

Question 16

Ordinary Point

Definition

Answer 16

-suppose the equation:
y’’ + p(x)y’ + q(x)y = 0
-has coefficients which are analytic at the point x=xo:
p(x) = po + p1(x-xo) + p2(x-xo)² + …
q(x) = qo + q1(x-xo) + q2(x-xo)² + …
-for |x-xo|0
-then xo is said to be an ordinary point of the ODE

Question 17

Singular Point

Definition

Answer 17

-any point that is not ordinary is singular

Question 18

Legendre’s Equation

Answer 18

-the equation:
(1-x²)y'' - 2xy' + λy = 0
-has singular points at xo=±1
-the point xo=0 is an ordinary point
-the radius of convergence o the series representation of:
p(x) = -2x/(1-x²) , q(x) = λ/(1-x²)
-about xo=0 is 1 since:
|±1 - 0| = 1

Question 19

Analyticity Theorem

Answer 19

-let xo be an ordinary point of :
y'' + p(x)y' + q(x)y = 0
-with p(x) and q(x) analytic for 
|x-xo|0
-then there exists two linearly independent solutions of the form :
y(x) = Σ an(x-xo)^n
-also for |x-xo|

Question 20

Regular Singular Point

Definition

Answer 20

-start with:
S(x) y'' + P(x) y' + Q(x) y = 0
-the point S(xo)=0 is a singular point
-if:
lim (x-xo)*P(x)/S(x) is finite
AND lim (x-xo)²*Q(x)/S(x) is finite where both limits are taken as x->xo
-THEN xo is a regular singular point

Question 21

Irregular Singular Point

Definition

Answer 21

-any singular point that is not regular is irregular

Question 22

What is the Frobenius Method used for?

Answer 22

-the analyticity theorem tells us that ordinary points have analytic solutions with power series expansions
-when you have a singularity the Frobenius method allows you to find a solution any way, of the form:
y=x^r (ao + a1*x + … + anx^n + …)
-where you can calculate r

Question 23

The Frobenius Method for Regular Singular Points

Steps

Answer 23

1) start by considering expansions of the form:
y = x^r (ao + a1x + … + anx^n + …) = x^r Σanx^n
where the sum is taken between n=0 and n=∞
2) differentiate to obtain expressions for y’ and y’’
-sub into the differential equation:
Cx² y’’ + xp(x)y’ + q(x)*y = 0
3) to find the indicial equation, find the lowest possible power of x remembering that n can take values between 0 and ∞, then sub this value of n into the coefficient of this lowest power of x and equate to zero
4) since ao≠0, find the values of r that are solutions to the indicial equation, then consider each value of r corresponds to a solution to the equation, consider each separately
5) given a value of r, start with x^0, find the values of n required for this power and sub into to the coefficients, equate to zero to find a1 in terms of ao
6) for a given value of r, find the value of n required for x^1 and sub into the coefficients and equate to zero, solve for a2 in terms of
7) continue for the desired number of coefficients for each value of r (i.e. for each solution)

Question 24

The Frobenius Method - Regular Singular Points

Answer 24

• for equations with regular singular points, the Frobenius method although not always necessary will always work, and sometimes will be necessary so it is easiest just to use it
• in the case where it wasn’t necessary you get r=0 so x^r=1 and you end up with a normal power series solution

Question 25

The Frobenius Method - General Recurrence Relation

Answer 25

R.H.S. = F(n+r)*
L.H.S. = known stuff
-where F(r) is the indicial equation

Question 26

The Frobenius Method - Theorem

Answer 26

-given a second order differential equationof the form:
Cx² y’’ + xp(x)y’ + q(x)y = 0
-the indicial equation F(r)=0 generates solutions r1≥r2
–if r1>r2 and r1-r2 is not a positive integer then there are two Frobenius solutions
–if r1=r2 then there is one Frobenius solution and one wih logs
–if r1-r2 is a positive integer then:
y2 = cy1lnx + x^r Σbnx^n , for some equations c will turn out to be 0, in that case a second Frobenius solution exists

Question 27

The Frobenius Method - Irregular Singular Points

Answer 27

• the Frobenius method can be applied at irregular singular points
• there are several outcomes, there may be a contradiction e.g. ao=0 so no solution is found, it may be that the radius of convergence turns out to be 0 so the solution diverges for all x
• in some cases a Frobenius (or even an analytic) solution can be found HOWEVER there is no theorem that guarantees this in the case of an IRregular singular point

Question 28

Bessel’s Equation

Description

Answer 28

x² y’’ + x y’ + (x²- ν²)y = 0

• there is a regular singularity at x=0
• the indicial equation is F(r) = r² - ν² = 0

Question 29

Bessel’s Equation

Solutions

Answer 29

-from the indicial equation we have;
r1 = +ν , r2 = -ν
-if r1-r2=2ν is not an integer then we are guaranteed two Frobenius solutions
-if ν is an integer we obtain one independent Frobenius solution and the second solution requires a logarithmic term: y2 = y1logx + Σbnx^n
-if ν is a half integer then r1-r2 is an integer AND it can be shown that we have two independent Frobenius solutions which are elementary functions

Question 30

cos(x)

Power Series

Answer 30

cosx = 1 - x²/2 + x^4/4! - …

Question 31

Frobenius Method

Indicial Equation with Double Roots

Answer 31

• find the first solution, y1, using the frobenius method
• find the Wronskian of the differential equation using Abel’s formula
• use reduction of order to find the second solution y2
• while calculating y2 you can let ao=1 and then for the general formula at the end just create a linear superposition of y1 and y1 so you end up with two arbitrary constants in the general formula as expected