Let us fix $0 \neq \lambda \in \mathbb{R}$. Let us consider the following ODE, on $[0,\infty)$: $$ y^{\prime \prime} (x) + \frac{r e^{-x}}{(1+e^{-x})^2} y(x) = -\lambda^2 y(x).$$ Here $r \ge 1$ is a parameter. Let us consider the solution $e\_r (x)$ which satisfies $e\_r (x) \sim e^{i \lambda x}$ as $x \to +\infty$. How would one approach showing (if this is indeed true) that $$\sup\_{\substack{r \ge 1\\ x \in [0,\infty)}} |e\_r (x)| < \infty.$$ Ideally, I am interested in some wider class of examples, so I am less interested in a "trick" that happens to work in this very particular case, and more in some conceptual approach, but still would like to hear all approaches.


Thank you



Joint boundedness of solutions of a family of Sturm-Liouville ODE
It is bounded. Moreover, $|y(x)|\leq 1$ for $x\geq 0$ for all $r>0$ and $\lambda>0$ (so the estimate is uniform not only in $r$ but in $\lambda$ as well).


This is a special case of the following theorem due to User @Fedja.


Theorem. In the equation $y''+V(x)y=0$, let the potential $V$ be decreasing and
bounded from below by a positive constant. Then every
real solution $y$ has
has an infinite increasing sequence of zeros $x\_n\to+\infty$, and if $$
m\_n=\max\{|f(x)|:x\_n\leq x\leq x\_{n+1}\}$$
then the sequence $m\_n$ is increasing.


Proof. The infinite sequence of zeros is a well-known fact.
Let $x\_n$ be such a zero. Suppose that $f'(x\_n)>0$. We compare $u(t)=y(x\_n+t)$ with $v(t)=-y(x\_n-t)$. Fot $t>0$ they satisfy
the differential equations
$$u''+V(x\_n+t)u=0\quad\mbox{and}\quad v''(t)+V(x\_n-t)v=0.$$
and the same boundary conditions at $t=0$. Since by assumption
$V(x\_n-t)\geq V(x\_n+t)$, Sturm's compasrison theorem gives that
the smallest positive zero $t\_0=x\_n-x\_{n-1}$ of $v$ is at most the smallest positive zero of $u$, and on the interval $(0,t\_0)$ we have $u(t)\geq v(t)$. So $$m\_{n-1}=\max\{v(t):0<t<t\_0\}\leq \{\max|u(t)|: 0<t<t\_0\}\leq m\_n.$$


The result for your problem follows since $e^x/(1+e^x)^2+\lambda^2$ is decreasing for $x>0$, while $\limsup\_{n\to\infty}m\_n=1$ for both real and imaginary parts of your solution, due to your normlization.


