Regular Singular Points

When we worked out series solutions to differential equations in previous discussions, we always assumed that x0 was an ordinary point, that is p and q converged to their Taylor Series expansions.  What happens if the point is a singular point?

 

Example

Attempt to find the solution to 

        xy'' + y  =  0        y(0)  =  1,        y'(0)  =  2

 

Solution

As before, we let 

      

Take derivatives

       

Now we plug these into the original differential equation

       

Let

        u  =  n - 1        n  =  u + 1

And change back to n to get 

       

Plugging in 0 into the second summand and combining, we get

        

Now comes the twist.  From the initial conditions, we get

        a0  =  1        and        a1  =  2

However, equating coefficients, we get

        a0  =  0

                             -an 
         an+1  =                                   
                          n(n + 1)

We get a contradiction, since a0 cannot be both 0 and 1.  We can conclude that there is no McLaurin series solution to the differential equation.  This does not mean that there is no solution at all, just not a series solution.

What is going on here is that 0 is a singular point for this differential equation.  When we have a singular point, all bets are off as to what type of solution we can have.  Some have no analytic solutions, others have one analytic solution and one non-analytic solution, and still others have two analytic solutions.  There is a classification of singular points that are particularly tame.  We describe this class below.

Definition of Regular Singular Points

Let

          P(x)y'' + Q(x)y' + R(x)y  =  0

be a linear second order differential equation.  Then x0 is called a regular singular point if

         

and

         

are both finite.

A singular point that is not a regular singular point is called an irregular singular point.  Another way of defining singular points is to say that 

        (x - x0)p(x)         and         (x - x0)2 

both have a removable discontinuity at x0.

 

Example

Find all regular and irregular singular points for the differential equation

        (x sin x)y'' + (cos x)y' + ex y  =  0

 

Solution

The zeros of x sin x are

        x  =  kp            for k an integer.

These are the singular points for the equation.  Now we test for regularity.  First test at x  =  0.

       

This limit is undefined hence the singularity at x  =  0 is irregular.  Note that once one of the two limits is not finite, the point is automatically irregular regardless of the other limit.  

Now we test the rest of the points

       

The first factor has a finite limit.  For the second factor, we can use L'Hopital's rule to get

        lim as x goes to k pi of x- k pi divided by sin x = lim as x goes to k pi of 1 over cos x = plus or minus 1

We are half way to being regular.  Now find

       

Since both limits are finite, x =  kp for k nonzero are regular singular points.

We may ask what all the fuss is about?  For differential equations, just about all the points will be nonsingular.  Why can't we just ignore the few points that are singular.  The answer is that the singular point is where the interesting part occurs.  Ignoring the singular point would be like deciding to ignore black holes in the study astronomy because most of space is not a black hole.         

 

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