- #1
arroy_0205
- 129
- 0
Suppose I have a 2nd order differential equation
[tex]
a_1y''(x)+a_2y'(x)+a_3y(x)+a_4=0
[/tex]
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
[tex]
Q^{\dagger}Qy(x)=m^2y(x)
[/tex]
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
[tex]
Q=-\frac{d}{dy}+p'(y); Q^{\dagger}=\frac{d}{dy}+p'(y)
[/tex]
where p(y) is a known function.
[tex]
a_1y''(x)+a_2y'(x)+a_3y(x)+a_4=0
[/tex]
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
[tex]
Q^{\dagger}Qy(x)=m^2y(x)
[/tex]
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
[tex]
Q=-\frac{d}{dy}+p'(y); Q^{\dagger}=\frac{d}{dy}+p'(y)
[/tex]
where p(y) is a known function.