Chapter 2 - Series Solutions of Linear Ordinary Differential Equations Flashcards

1
Q

What is an elementary function?

A

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

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2
Q

How are power series related to differential equations?

A
  • 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
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3
Q

List the three well known types of elementary equation

A

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

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4
Q

Algebraic Functions

A

-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)

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5
Q

Combining to Form Elementary Functions

A
  • 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
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6
Q

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

A

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
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7
Q

Radius of Convergence

A

-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

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8
Q

Term by Term

A
  • 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

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9
Q

Ratio Test

A
-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
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10
Q

Power Series

e^x

A

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

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11
Q

Power Series

sinx

A

sinx = x - x^3/3! + x^5/5! + …

-for all x

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12
Q

Power Series

1/(1-x)

A

1/1-x = Σ x^n = 1 + x + x² + x^3 +…

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

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13
Q

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

A
  • 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
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14
Q

Analytic Solutions of ODEs

Steps

A

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 = Σan
x^n
-to obtain the general solution

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15
Q

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

A

-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

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16
Q

Ordinary Point

Definition

A

-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

17
Q

Singular Point

Definition

A

-any point that is not ordinary is singular

18
Q

Legendre’s Equation

A
-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
19
Q

Analyticity Theorem

A
-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|
20
Q

Regular Singular Point

Definition

A
-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
21
Q

Irregular Singular Point

Definition

A

-any singular point that is not regular is irregular

22
Q

What is the Frobenius Method used for?

A

-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

23
Q

The Frobenius Method for Regular Singular Points

Steps

A

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)

24
Q

The Frobenius Method - Regular Singular Points

A
  • 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
25
Q

The Frobenius Method - General Recurrence Relation

A

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

26
Q

The Frobenius Method - Theorem

A

-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 = c
y1lnx + x^r Σbnx^n , for some equations c will turn out to be 0, in that case a second Frobenius solution exists

27
Q

The Frobenius Method - Irregular Singular Points

A
  • 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
28
Q

Bessel’s Equation

Description

A

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
29
Q

Bessel’s Equation

Solutions

A

-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

30
Q

cos(x)

Power Series

A

cosx = 1 - x²/2 + x^4/4! - …

31
Q

Frobenius Method

Indicial Equation with Double Roots

A
  • 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