Is a bounded solution of this PDE a constant? I have a problem: 

Suppose a bounded function $u(x, y) \in C^{2}(\mathbb{R}^{2})$ is a solution of $$u_{xx}-2u_{xy}+\beta u_{yy}+2\beta u_{x}-\beta^{2} u_{y}=0.\tag{1}$$ 
  Is it true that $u\equiv\operatorname{const}$ when 
a) $\beta>5$, 
b) $\beta<-5$?

I reduce equation to canonical form and have no other idea. Do you have one?
I get the following canonical form in case a):
$(\beta-1)u_{rr}+(\beta-1)u_{tt}+\beta(2-\beta)u_{r}-\sqrt{\beta-1}u_{t}=0$ in coordinates $r=x+y,t=-\sqrt{\beta-1}x$;
in case b):
$-2\sqrt{1-\beta}u_{rt}+\beta(\sqrt{1-\beta}-1)u_{r}+\beta(\sqrt{1-\beta}+1)u_{t}=0$ in coordinates $r=y+(1-\sqrt{1-\beta})x,t=y+(1+\sqrt{1-\beta})x$.
 A: Here is a sketched answer:
Consider a linear change of coordinates $(x,y)\to (z,w)$:
$$\begin{align} x~=~& 2z+w, \cr 
y~=~&-\beta z +\alpha w, \cr  
\alpha~:=&~\frac{\beta}{\beta-2} , \cr 
\beta~\neq~& 2.\end{align}\tag{A}$$
Then the chain rule yields
$$\begin{align} \partial_z 
~=~&  2\partial_x - \beta \partial_y, \cr 
\partial_w ~=~&  \partial_x + \alpha \partial_y, \end{align}\tag{B}$$
and conversely,
$$\begin{align} (2\alpha+\beta) \partial_x ~=~&  \alpha\partial_z + \beta \partial_w, \cr  (2\alpha+\beta) \partial_y ~=~&   2\partial_w - \partial_z. \end{align}\tag{C}$$
Then OP's PDE (1) becomes "diagonal"
$$\begin{align} 0~=~&u_{xx}-2u_{xy}+\beta u_{yy} +2 \beta u_x -\beta^2 u_y\cr
~=~&u_{xx}-u_{yz}+ \beta u_z\cr
~=~&\frac{(\beta -2)^2}{\beta^2}u_{ww} +\frac{\beta -1}{\beta^2}u_{zz} + \beta u_z.  \end{align}\tag{D}$$

*

*In the elliptic case $\beta>1 \wedge \beta\neq 2$ a Liouville-type argument prohibits non-constant bounded solutions.


*In the hyperbolic case $\beta<1\wedge \beta\neq 0$, the dissipative $\beta u_z$-term prevents non-constant bounded solutions. A solution always increases exponentially at one of the asymptotic regions $z=\pm\infty$.


*In the case $\beta=0$, the full solution to OP's PDE (1) is of the form
$$u(x,y)=f(2x+y)+g(y),\tag{E}$$
which clearly has non-constant bounded solutions.
