This is somewhat unrelated to what I normally do in mathematics, which is why it may be obvious to some of you, but I was puzzled by this:


If we look for classical solutions on $[0,1]$ to


$$-y''(x) = \lambda y(x)$$ with initial conditions $y(0)=1, y'(0)=0$ then the solution is 
$y\_{\lambda}(x)=\cos(\sqrt{\lambda}x).$ Now consider $f(\lambda)=y\_{\lambda}(1)$
Now, if we allow $\lambda$ to be complex (which is why I call it $z$ in the following) and always pick the correct branch of the square root, then a plot showed me that we have in fact $ \frac{1}{\left\lvert f(z) \right\rvert} =O\left({\left\lvert \textbf{Im}(z) \right\rvert^{-1}}\right)$ by which I mean that there is a constant $C\_R>0$ such that 


$\frac{{\left\lvert \textbf{Im}(z) \right\rvert}}{\left\lvert f(z) \right\rvert} \le C\_R$ for $z \in \mathbb{C}\backslash \mathbb{R}\cap B(0,R) $ and $B(0,R)$ is an arbitrary ball in the complex numbers.


Numerically, I found some evidence that if $V$ is an even and let's say smooth function with respect to $\frac{1}{2}$ then the same holds for solutions to 
$-y''(x)+V(x)y(x)=\lambda y(x)$ with the same initial conditions.
I am not really sure how to show this theoretically, if it is even true? 


Best 


Kinzlin



Solutions to Schrödinger equation parameter dependence
For the equation $-y''+V(x)y=zy$, and the solution defined by $y(0,z)=1,\; y'(0,z)=1$, if we set $f(z)=y(1,z)$, then
$$f(z)=\cos\sqrt{z}+O(|z|^{-1/2}\exp(|\Im z|),\quad z\to\infty.$$
The only assumption is that $V$ is continuous on $[0,1]$.
See, for example, Levitan, Sargsjan, Introduction to spectral theory, AMS 1975,
Chapter I, Lemma 2.2. 


