 |
|
|
|
11/29/11 - Power-Series solutions to ODEs with non-constant coefficients
|
|
|
|
 | Power-Series Solutions to ODEs
|
|
|
|
|
 | ODEs with non-constant coefficients often can be solved with a power series.
|
|
|
|
|
| A function f is analytic at x0 provided f can be represented by an infinite power series about x0 and the series converges for all x in some interval containing x0:
|
|
|
|
|
| If y'' + P(x)y' + Q(x)y = 0 and the functions P(x) and Q(x) are both analytic at the point x0, then x0 is called an ordinary point for the equation. A point which is not an ordinary point is called a singular point.
Given the ODE: (1-x2)y'' - 2xy' + n(n+1)y = 0, we first put the ODE into standard form
We then get:
P(x) and Q(x) are both analytic everywhere except at x = ±1. That means that x = ±1 are singular points. Every other value of x is an ordinary point. The ODE can be solved by a power series for values of x which are ordinary points.
|
|
|
|
|
| 
All values of x are ordinary points, so a power-series solution exists:
Substitute series into the original differential equation:
Remove terms from each summation until remaining terms have the same power of x:
Re-index the summations to get a single summation:
Set coefficients for each power of x to zero to get recurrence relation among coefficients:
Calculate the first few terms of the series (note that all the coefficients with odd indices are zero):
First few terms of the series:
The blue line in the graph below is the first few terms of the power-series solution:
|
|
|
|
|
| 
Put the equation in standard form to determine the interval of existence:
Note the singularity (point where the derivative is undefined) at x = 2. Since the initial condition is given at x = 0, the interval of existence of the solution is [0,2) and a power-series solution exists over that interval:
Substitute into original equation:
Distribute (2-x) through first summation:
Pull out constant terms. Remaining terms in summations have x1 power:
Re-index summations:
Set coefficients to zero, get recurrence relation:
Solution:
The blue line solution exists over the interval [0, 2).
|
|
|
|
|
| 
All values of x are ordinary points, so a power-series solution exists:
Substitute into original equation:
Distribute:
Pull out constant terms, remaining terms have x1 power:
Re-index:
Set coefficients to zero, get recurrence relation:
Calculate first few terms:
Write first few terms of series:
Note that the solution is a linear combination (depending on the initial values y(0) and y'(0)) of two independent series.
|
|
|
|
|
| Convergence of the series:
|
|
|
|
|
| Ratio Test: For each series to converge, we require that the absolute value of the ratio of two successive terms be less than 1:
|
|
|
|
|
| 
which says that as you add more terms you increase the radius of convergence. For n = 8 terms, the radius of convergence is about 3. For n = 24 terms, the radius of convergence is about 5. For an infinite number of terms, the radius of convergence is ∞.
|
|
|
|
|
| Read Section 11.2, pp. 543-554.
|
|
|
|
|
| Do exercises p. 554 (5,13,17,27).
|
|
|
|
|
 |
|
;){kind=small}
;){kind=small}
;){kind=small}
;){kind=small}
;){kind=small}
;){kind=small}
;){kind=small}
