This question is the copy from mat.stackexchange.com here. 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?


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.

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  • 1
    $\begingroup$ $-d^2/dx^2+x^2$ is the harmonic oscillator Schrodinger operator, and $0$ is in the resolvent set of this operator. This gives right away that $-z''+x^2z=-f/A$ has a solution in $L^2(\mathbb R)$. Now that's not yet exactly what you want ($b$ is not exactly $x^2$, and $L^2$ is not the same as a derivative bounded), but it seems quite likely that one can get there with more work. $\endgroup$ – Christian Remling Sep 23 '14 at 17:13
  • $\begingroup$ Thank you very much @Frits Veerman and ChristianRemling. I got much knowledge from both of you... $\endgroup$ – Jlamprong Sep 24 '14 at 5:17

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