# Condition for a 2nd order differential eqn to have bounded solutions?

1. Aug 25, 2008

### arroy_0205

Suppose I have a 2nd order differential equation
$$a_1y''(x)+a_2y'(x)+a_3y(x)+a_4=0$$
and two conditions y(0), y'(0). Then is there any theorem which gives us the condition under which the solution y(x) will be bounded? Note that x-range is entire real line.
This is a general version of the problem I am facing. In my actual problem I have an equation like
$$Q^{\dagger}Qy(x)=m^2y(x)$$
where dagger denotes hermitian conjugate of the operator. In this case the m^2=0 corresponds to the lowest energy and in principle I should not get any bounded solution with m^2 < 0. However I am also not getting any bounded numerical solution (using mathematica) with m^2 > 0. May be I am using wrong initial conditions. That is why I am asking this question. Can anyone suggest something?
By the way my Q and Q^dagger are
$$Q=-\frac{d}{dy}+p'(y); Q^{\dagger}=\frac{d}{dy}+p'(y)$$
where p(y) is a known function.

2. Aug 25, 2008

### Eidos

Would the Bilateral Laplace transform not help us here?

(edit) Obviously if we restrict our domain to $$\mathbb{R}^{+}$$ then the single sided Laplace transform can tell us when the function is bounded with x. I've found a link on something called Nachbin's Theorem http://en.wikipedia.org/wiki/Generalized_Borel_transform which might be of some use. It seems to generalise the notion of a function being bounded but not necessarily of exponential order. Im assuming here that we can take an integral transform of the differential equation and establish its 'boundedness' that way. What the kernal for the integral transform would be in this particular case though, I have no idea. I hope that this is on the right track...

Last edited: Aug 25, 2008