Friday, January 15, 2021

Looking At Power Series Solutions To Differential Equations

 It's generally agreed that the optimal technique for finding the solution of a general linear differential equation - with variable coefficients - is based on the use of power series.   For starters, recall when we looked at Fourier series (Nov. 24 post)  we began by looking at the Taylor series, e.g.

f(x) = f(x o ) + f ’(x o) × (x – x o) + f ” (x o) (x – x o)2 / 2!  +

 f ”’ (x o) (x – x o)3 / 2!  +   f n (x o) (x– x o)n / n!  +  ….

And note here that a function f(x) may be represented by such a series, provided all derivatives exist at x = x o .  From here we can further assert a function f(x) is analytic at x = x  if  f(x) can be expanded in a power series valid about some point.

To fix ideas consider the general form for the linear differential equation:

a o (x ) +  D n x y  + a 1 (x ) +  D n-1 x y  + …+   a n (x ) y    =   f(x)

In which i (x ), (i = 0,.....n) are polynomials.   Then:

The point x = x  is called an ordinary point of the equation:  o (x o ≠  0

Any point x = x  for which    o (x 1 ) =  is called a singular point of the differential equation.

The point x = x  is called a regular singular point of the general linear DE if the equation can be written in the form:

[(x - x ) n  n  +  (x - x )   b1 (x) D n-1  + 

(x - x  ) n- 2  b1 (x) n-2  +.  .. .....(x - x  ) n- 1  n- 1  (x) D +  n (x) ] y = 0

Where  i (x),  (i =  1,.......n) are analytic at   x = x 1

A key to knowing whether a power series type solution is applicable is to first get the differential equation in the form:

y "  + P (x) y'  + Q (x) y =  j((x)

Where: 

 P(x) = 1 (x)/  2 (x)

Q(x) =   0 (x)/  2 (x)

j((x)  =   g(x) / 2 (x)

Then one can determine if x = 0 is an ordinary point, fulfilling that the power series method is applicable. Consider for example, the DE:

y"  - xy'  + 2y = 0

Here, P(x) = -x  and Q(x) = 2, therefor because they are analytic every value of x,   in particular x = 0,  is an ordinary point.  Again, the power series method is only applicable when x  = 0 is an ordinary point.

Example 2:  Determine whether x = 0 is an ordinary point of the differential equation:

2 x2 y" + 7 x(x + 1) y' - 3 y = 0

Here: P(x) =  7(x + 1)/ 2x

Q(x) = -3/2 x2

Here, neither is analytic at x = 0 (both denominators = 0) so x = 0 is not an ordinary point but a singular point.   At this juncture we note that the use of power series to express a function which is a solution of a given differential equation, is based on the fact there exists a series which contains the necessary arbitrary constants and which converges inside a circle with x = x   and extends out to the singular point nearest to x = x  .

Example 3: Find a power series solution to:

(x - 1) y"'  +  y"   + (x - 1) y' + y = 0

If:  y = y" = 0,  y ' = 1   at x = 0

Solution:

We proceed by first solving for y"' and differentiating:

y"' =   - (x - 1)- 1 y" -  y'  -  (x - 1)- 1  y

y iv   =  - (x - 1)- 1  y"'  + [(x - 1)- 2  - 1] y"(x - 1)- 1  y' + 

(x - 1) - 2  y

v   =  - (x - 1)- 1  iv + 2[(x - 1)- 2  - 1] y"' 

- 2[(x - 1)- 3  + (x - 1)- 1 ] y" + 2(x- 1)- 2 y' - 2( x -1 )- 2 y

At x = 0 we have:

y"'(0) = -1,  iv (0) =  0,  yv (0) =  1

The resulting series is then:

y(x)  =  0 + x + 0 -  1/3! (x2) + 0 + 1/5! (x3) +  ....

y(x) =  x -  1/3! (x2) + 0 + 1/5! (x3)  -   ....

y(x) =  sin x  

Problems:

1.  Give the singular points for each of the following:

(a) (x - 3) y" + (x + 1)y = 0

(b) (x2 + 1) y"'  + y"   -  x2 y  =  0

2. Determine whether x = 0 is an ordinary point of the differential equation:

x2 y " + 2 y'  +  xy = 0 

3. Find the power series solution for the differential equation:

xy" +  x3 y   - 3xy = 0   

That satisfies: y = 0 and y' = 2  at x = 1


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