Solving the KG equation in Rindler coordinates Let us consider a massive Klein-Gordon scalar field $\phi$ satisfying $$(\Box+m^2)\phi=0$$
I want to solve this in Rindler coordinates, choosing $\phi$ to be positive frequency with respect to the generator of boosts $\partial_\eta$. The idea is to find a set of basis modes to quantize the field in the Rindler frame.
Now, we have the $\Box$ operator given by
$$\Box\phi=e^{-2a\xi}(-\partial_\eta^2+\partial_\xi^2)\phi$$
Thus the equation becomes
$$e^{-2a\xi}(-\partial_\eta^2\phi+\partial_\xi^2\phi)=-m^2\phi$$
Imposing that $\partial_\eta \phi=-i\omega \phi$ we have $\phi(\eta,\xi)=A(\xi)e^{-i\omega \eta}$ and hence after some manipulation we get the equation for $A$
$$A''(\xi)+\omega^2A(\xi)+m^2 e^{2a\xi}A(\xi)=0.$$
I now set $u(\xi)=\frac{m}{a}e^{a\xi}$. Perfoming the change of variables and dividing through by $a^2$ gives
$$u^2 \dfrac{d^2A}{du^2}+u\dfrac{dA}{du}+\frac{\omega^2}{a^2}A+u^2A=0.$$
Defining $\alpha = -i\omega/a$ we then have
$$u^2 \dfrac{d^2A}{du^2}+u\dfrac{dA}{du}+(u^2-\alpha^2)A=0.$$
This is Bessel's equation and its solution is
$$A(u)=c_1 J_\alpha (u)+c_2 Y_\alpha(u).$$
Using $u = u(\xi)$ gives $A(\xi)$ and solves the problem. 
The issue is, usually boundary conditions allow us to eliminate $c_1$ or $c_2$. If we want a solution regular at $u=0$ we would have $c_2 =0$ for instance.
But here I'm failing to see this. For instance, $u=0$ corresponds to $\xi\to -\infty$. I don't see why impose regularity there.
How can we proceed? We must leave both $c_1,c_2$, or is there some further manipulation we can do?
Edit: Rindler coordinates are defined by
$$t=\dfrac{1}{a}e^{a\xi}\sinh(a\eta),\quad x=\dfrac{1}{a}e^{a\xi}\cosh(a\eta)$$
 A: I'll do it in 2d as the extra dimensions are trivial.
If $x$, $t$ are Minkowski coordintes and we we set
$$
t= e^{\xi} \sinh \tau\nonumber\\
x= e^{\xi} \cosh \tau\nonumber
$$
then $-\infty<\xi< \infty$, $-\infty<\tau < \infty$ covers the Rindler wedge. 
In these coordinates the  metric $ds^2= dx^2-dt^2$ becomes
$$
ds^2= e^{2\xi}(d\xi^2-d\tau^2).
$$
The usual Rindler coordinate  sets $r= e^{\xi}$ so that
$$
ds^2 = dr^2-r^2 d\tau^2.
$$
For the KG equation the $\xi$, $\tau$ coordinates are nicer as
$$
(-\nabla^2+m^2)\psi(x)=0
$$
becomes (after multiplication by $e^{2\xi}$)
$$
\left(-\frac{\partial^2}{\partial \xi^2}+\frac{\partial^2}{\partial \tau^2}+m^2 e^{2\xi}\right)\psi(\xi,\tau)=0.
$$
The real eigenfunctions of the equation
$$
\left(-\frac{d^2}{d\xi^2}+m^2 e^{2\xi}\right) \psi_\nu(\xi)=\nu^2 \psi_\nu(\xi)
$$
are
$$
\psi_\nu(\xi)=\left( \frac{2\nu \sinh  \nu \pi}{\pi^2}\right)^{1/2} K_{i\nu}(m e^{\xi}), \quad 0<\nu<\infty
$$
where $K_{i\nu}(x)$ is the Bessel K function of purely imaginary order.
The eigenfunctions  have been normalized to  obey
$$
\int_{-\infty}^{\infty} \psi_\mu(\xi)\psi_\nu(\xi)\, d\xi = \delta(\nu-\mu),
$$ 
and are the ingredients of the Kontorovich–Lebedev transform (There is Wikipedia article on this) which makes it clear that this set of solutions  is both orthogonal and complete.
Thus the KG equation has solutions
$$
\psi_\nu(\xi,\tau) =e^{-i\nu \tau}\psi_\nu(\xi).
$$ 
There is no need to worry about boundary conditions as, in  the $\xi$, $\tau$ coordinates, we are in the Weyl limit-point case. 
Here are a few more details taken from my notes:
Consider the Bessel function
$$
{\rm K}_{i\nu}(x)= \int_0^\infty e^{-x\cosh u}\cos \nu u\, du\nonumber\\
=\frac 12 \left(\frac{x}{2}\right)^{i\nu} \int_0^\infty \exp\left(-t-\frac{x^2}{4t}\right) t^{-i\nu -1}\,dt.\nonumber
$$
of purely imaginary order. The function  ${\rm K}_{i\nu}(x)$ is real for $x\in (0,\infty)$, and ${\rm K}_{i\nu}(x)={\rm K}_{-i\nu}(x)$.
For small $x$
$$
{\rm K}_{i\nu}(x)\sim \frac{i\pi}{2\sinh \pi \nu}\left\{ \frac{(x/2)^{i\nu}}{\Gamma(1+i\nu)}- \frac{(x/2)^{-i\nu}}{\Gamma(1-i\nu)}\right\}\nonumber\\
= \sqrt\frac{\pi}{\nu \sinh \pi \nu} \left\{ e^{i\alpha} (x/2)^{i\nu} + e^{-i\alpha} (x/2)^{-i\nu}\right\}\nonumber
$$
for some real $\alpha$. We have used, at the last step,
$$
\Gamma(1+i\nu)\Gamma(1-i\nu)= \frac{\pi \nu}{\sinh \pi \nu}.
$$
These functions  therefore satisfy the orthogonality property
$$
\frac{1}{\pi^2}\int_0^\infty \frac{dx}{x} {\rm K}_{i\mu}(mx){\rm K}_{i\nu}(mx)= \frac{\delta(\mu-\nu)}{2\nu \sinh \nu \pi},
$$
and a completeness condition
$$
\frac{1}{\pi^2} \int_0^\infty 2\nu\sinh \nu\pi \,{\rm K}_{i\nu}(x){\rm K}_{i\nu}(x')\,d\nu= x \delta(x-x').
$$
and provide the  Kontorovich-Lebedev  transform pair
$$
\tilde f(\nu)\equiv K[f](\nu)= \int_0^\infty {\rm K}_{i\nu}(x) f(x)\,dx,\nonumber\\
f(x)= \frac{1}{\pi^2 x} \int_0^\infty 2\nu \sinh \nu\pi\, {\rm K}_{i\nu}(x) \tilde f(\nu)\,d\nu.\nonumber
$$
There is also a Mehler-type convergent 
$$
\frac{1}{\pi^2} \int_0^\infty 2\nu\sinh \nu(\pi-\epsilon)  {\rm K}_{i\nu}(x){\rm K}_{i\nu}(y)\,d\nu= \frac{xy}{\pi}  \sin\epsilon  \frac{{\rm K}_1(\sqrt{x^2+y^2-2xy \cos\epsilon})}{\sqrt{x^2+y^2-2xy \cos\epsilon}}.
$$
