This question is a variation version from here.
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 $3$-rd derivative is bounded.
What I know so far are:
If the boundedness condition of $3$-rd derivative is ignored, then of course the ODE above has a general solution. The question is, how to choose a solution satisfying the boundedness condition.
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?
An approach (not yet a full solution): using my approximation in the comment write $$\alpha''(x)=\alpha'(x)\left(x\frac{1+3\epsilon+e^{\epsilon x^2}}{1+\epsilon+e^{\epsilon x^2}}\right)+\alpha(x)\left(\frac{1+3\epsilon+e^{\epsilon x^2}}{1+\epsilon+e^{\epsilon x^2}}\right)+x\frac{1+\epsilon}{1+\epsilon+e^{\epsilon x^2}}$$ Since we have the second derivative of $\alpha$ interms of the first two. We can likewise find the third derivative in terms of the first two.
Take one more derivative and substitue for $\alpha''(x)$ where ever it appears. Simplifying, I get
$$\alpha'''(x)=\left(\alpha'(x) (e^{2 x^2 \epsilon} (2 + x^2) + (2 + x^2 + 3 x^2 \epsilon) + e^{x^2 \epsilon} (4 + 2 x^2))+ \alpha(x) x \left(e^{2 x^2 \epsilon }+2 e^{x^2 \epsilon }+1\right) + 2 x^2 \epsilon \left(1-e^{x^2 \epsilon }\right)+\left(x^2+1\right) \left(e^{x^2 \epsilon }+1\right)\right)/\left(e^{x^2 \epsilon }+1\right)^2$$
Now let $x\rightarrow \infty$. The equation simplifies to
$$\alpha'''(x) =\frac{\alpha'(x) x^2 e^{2 x^2 \epsilon}+ \alpha(x) x e^{2 x^2 \epsilon} + x^2 e^{x^2 \epsilon }}{e^{2x^2 \epsilon }} \\ \Rightarrow \alpha'''(x) =\alpha'(x) x^2 + \alpha(x) x + x^2 e^{-x^2 \epsilon } \\ \Rightarrow \alpha'''(x) =\alpha'(x) x^2 + \alpha(x) x $$
It seems that for your $3$rd derivative to be bounded you at least need bounded solutions to the above equation. I haven't worked these out yet (or proved they don't exist).