This question is the copy from mat.stackexchange.com [here](https://math.stackexchange.com/questions/934856/existence-of-a-bounded-function-satisfying-a-second-order-differential-equation). I requestioned here due to the very limited responses there.


Let $\phi:\mathbb{R}\mapsto\mathbb{R}$ be the standard normal density, $$\phi(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}{2}}, \forall x\in\mathbb{R}.$$
Given $0<\sigma\le 1$. I am working with the differential equation 
$$\alpha''(u) [\phi(u) + \frac 1\sigma \phi(u/\sigma)] - \alpha'(u) [u\phi(u) + \frac 1{\sigma^3} u\phi(u/\sigma)] - \alpha(u) [\phi(u) + \frac 1{\sigma^3} \phi(u/\sigma)] = \frac 1{\sigma^5} u\phi(u/\sigma).$$


I am looking for a solution of the above equation such that its $n-$th derivative is bounded for some positive integer $n$.


What I know so far is:


1. If the boundedness condition of $n-$th derivative is ignored, then of course the ODE above has general solution. The question is, how to choose a solution satisfying the boundedness condition.
2. For $\sigma=1$, I could take $\alpha(u)=-\frac14u$ that satisfies all conditions. I got this function because in this case the general solution for the above ODE is $\alpha(u)= -\frac 14 u + (a + b \Phi(u))/\phi(u)$. So we can choose $a=b=0$. However, for $\sigma<1$, it makes my problem much harder. Any suggestion?



Existence of bounded $n-$th derivative of the solution of differential equation
The Liouville transformation works as follows. Take the differential operator
\begin{equation}
L = \frac{d^2}{d x^2} + a\_1(x) \frac{d}{d x} + a\_2(x),
\end{equation}
where $a\_1 \in C\_1$ and $a\_2 \in C\_0$. Then, defining
\begin{equation}
A(x) = \text{exp}\left(-\frac{1}{2} \int^x a\_1(\xi)d\xi\right)
\end{equation}
and
\begin{equation}
b(x) = -\frac{1}{2} \frac{d a\_1}{d x} - \frac{1}{4} a\_1^2 + a\_2,
\end{equation}
we can transform the equation $L\,y(x) = f(x)$ into
\begin{equation}
A(x) \left[\frac{d^2}{d x^2} + b(x)\right] z(x) = f(x),
\end{equation}
where $y(x) = A(x) z(x)$.


Your equation is of the form $L y = f$ when you divide by the function multiplying $\alpha''$. You can then calculate 
\begin{equation}
A(u) = (2\pi)^{-\frac{1}{4}}\left(\sigma\phi(u) + \phi(u/\sigma)\right)^{-\frac{1}{2}}.
\end{equation}
Calculating the function $b(u)$ is more involved, but should pose no problem (it's a little long to post here). Dividing by $A(x)$ yields
\begin{equation}
\left[\frac{d^2}{d x^2} + b(x)\right] z(x) = \frac{f(x)}{A(x)} \quad(\*).
\end{equation}


So far, so good: $b(x)$ and $f(x)/A(x)$ are both smooth functions. However, the problem is that $b(x)$ is unbounded: it behaves as $-x^2$ for $|x|\gg1$. Note that $f(x)/A(x)$ decays superexponentially as $x \to \pm \infty$, so that's no problem.


Roughly speaking, that means that in the `far field' $|x|\gg1$, the solution $z(x)$ will behave like the solutions to the homogeneous equation
\begin{equation}
\frac{d^2 z}{d x^2} - x^2 z = 0,
\end{equation}
which in turn behave as a linear combination of $e^{\pm x^2}$. If their behaviour would *not* be superexponential, then we would have an exponential dichotomy. This would allow us to choose the two linearly independent solutions such that one decays as $x \to -\infty$ and the other one decays as $x \to +\infty$. Using the method of variation of parameters (or using Green's function), one could then formally express the solution to the inhomogeneous problem $(\*)$ in terms of those functions, and investigate whether it would be possible for this inhomogeneous solution to be bounded.


However, for superexponential growth/decay, I'm not sure whether this works. You could always try to mimic the approach sketched above, but I'm not sure which problems you could encounter. I think that the expertise of @ChristianRemling will be very useful here, and I hope he is able to be of further assistance.


