# Another question dealing with Frobenius method

1. Feb 8, 2005

### 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.

2. Feb 8, 2005

### Hurkyl

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

3. Feb 8, 2005

### 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.

4. Feb 8, 2005

### Hurkyl

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

5. Feb 8, 2005

### 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.

6. Feb 8, 2005

### Hurkyl

Staff Emeritus
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.

7. Feb 8, 2005

### meteorologist1

Yes I agree with you now. Thanks.