One dimensional system a Hamiltonian system? I have the following equation of motion:
$$
\dot x = \beta x y 
$$
with $y=1-x$. I would like to see if it is Hamiltonian or not. 
Due to it being one dimensional, I think it should be locally Hamiltonian. I do not know how to define a momentum however. With this, one could check the fundamental Poisson brackets as a test to see if it is endowed with the Poisson algebra. Without this definition to hand, how can I proceed? 
 A: Another approach different from Qmechanic's one. Observe that your differential equation implies that $$\frac{1}{2}\dot{x}^2 -\frac{\beta^2}{2}  x^2 (1-x)^2 =E\:,$$ 
with $E=0$. This is a conservation equation for the second order system
$$\ddot{x} = -\frac{d}{dx} U(x)$$
where 
$$U(x) = -\frac{\beta^2}{2} x^2 (1-x)^2\:.$$
This is a 2D Hamiltoniam system whose Hamiltonian is
$$H(x,p)= \frac{p^2}{2} -\frac{\beta^2}{2} x^2 (1-x)^2\:.$$
The solutions of the initial ODE are exactly those which solve the Hamilton equations of this new system (one is $p= \dot{x}$) such that the
$$p(0):= \beta x(0) (1-x(0))\:.$$
This requirement also fixes the signs (I assume $\beta>0$) by continuity of the solutions. Actually a bit more precise analysis is necessary if $x(0)=0$ or $x(0)=1$.
Addendum.  Actually there is one more even easier possibility. Simply define
$$H(x,p) := p\beta x(1-x)\:.$$
The Hamiltonian equation for $x$ is just your initial equation, which admits a unique solution  when you fix an initial condition $x(0)$ and it does not depend on the variable $p$ and on the remaining equation.
A: *

*A 1D phase space cannot have a regular Poisson structure in any point because of skewsymmetry. (Regular phase spaces are always even-dimensional.)


*However, there is a lot of freedom to embed OP's 1D system into a 2D phase space.
Example: Define fundamental Poisson bracket $$\{x,y\} := \beta x(1-x)y, \qquad y~\neq ~0, \tag{1}$$
and Hamiltonian
$$ H~:=~\ln|y|. \tag{2}$$
Then the first Hamilton's equation is OP's sought-for EOM:
$$\dot{x}~\approx~\{x,H\} ~=~ \{x,y\}\frac{\partial H}{\partial y}~=~ \beta x(1-x).\tag{3}$$
