Consider second-order ordinary differential equations of the form


I'm interested in general criteria on the function $a(t)$, which guarantee respectively rule out the existence of a positive solution $u(t)>0$, when $a(t)$ is defined on an interval $]b,\infty[$. I believe that under suitable regularity conditions the long-term behaviour of $a(t)$ is crucial and some kind of transition occurs.

As an example, for $a(t)=c/t^2$ with a fixed constant $c \in ]0,\infty[$, WolframAlpha gives a general complex solution for $c \neq 2$ and I guess that a positive solution $u(t)$ exists for $c>2$ but not for $c<2$. Even if this is not correct, this is the kind of phenomena I'm interested in for $a(t)$ as general as possible. Any ideas?

I've taken a look at some research papers on positive solutions of differential equations and related problems, but unfortunately I didn't find anything which fits exactly. Any literature suggestions?


In the case $a(t) = c/t^2$, the general solution is $$u(t) = \frac{2 t^2}{c-2} + a t^{(1+\sqrt{1+4c})/2} + b t^{(1-\sqrt{1+4c})/2}$$ If $c > 2$, the particular solution $2 t^2/(c-2) > 0$. If $c < 2$, that particular solution is negative, and grows faster than the solutions of the homogeneous equation, so there are no positive solutions on $(0,\infty)$.

In the case $a(t) = 2/t^2$, the general solution is

$$ u(t) = \frac{2t^2}{9} - \frac{2 t^2 \ln(t)}{3} + a t^2 + b t^{-1}$$ and the particular solution is negative for $t > 1$ and grows faster than the solutions of the homogeneous equation, so again there are no positive solutions on $(0,\infty)$.

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    $\begingroup$ Thanks for your answer. I'm hoping to generalise this for more general $a(t)$, but it isn't clear to me how this can be done. $\endgroup$ – student19 Jul 5 '18 at 8:36

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