Another question dealing with Frobenius method

[meteorologist1]

This is another question I have trouble proving:

Suppose the coefficients of the equation: w'' + p(z)w' + q(z)w = 0 are analytic and single-valued in a punctured neighborhood of the origin. Suppose it is known that the function w(z) = f(z) ln z is a solution, where f is analytic and single-valued in a punctured neighborhood. Deduce that f is also a solution.

Thanks for your help.

 

[Hurkyl]

Can't you just make that substitution into the DE?

 

[meteorologist1]

When I make the substitution it becomes very messy and it is not easy to see that f is a solution. There are terms without the natural log which I can't see how they would vanish.

 

[Hurkyl]

Actually, it's clearly not true. Take p(z) = q(z) = 0. Try the solution w(z) = 1.

 

[meteorologist1]

Hmmm, it's probably because p and q are required to have poles at z = 0. p(z) = q(z) = 0 is analytic at z = 0.

 

[Hurkyl]

The nice thing about being everywhere analytic is that you're also analytic on a punctured neighborhood of the origin. Whether they're analytic at the origin or not is irrelevant.

If you're still not convinced, try making your own differential equation whose coefficients are singular at the origin. Mine had 1/z as a solution.

 

[meteorologist1]

Yes I agree with you now. Thanks.