In design controller for a first order system such as: $$\dot{x}=-ax+bu$$ they assume that value of $a>0$. I need to know is it possibile to neglect this assume an design the controller even if $a=0$ as same as if $a>0$ ?
In other words can I design LQR for this system with $a=0$?
On
LQR only requires from the system that you are trying to control that the pair $(a,b)$ is stabilizable (assuming you choose appropriate weights $Q$ and $R$). For this scalar case this means that $b\neq0$ (when $a\leq0$). So when $b\neq0$ then $a$ could be anything, also negative.
It can be noted that LQR does assume that you have perfect knowledge of the system, so know the exact values of $a$ and $b$. If you do not know $a$ exactly then for $a<0$ the closed loop system might be unstable.
If $a = 0$ and $b \neq 0$ you can take
$$ u = -\frac{k}{b}x \tag{1} $$
with $k > 0$, so your dynamics $\dot{x} = b \, x$ will be
$$ \dot{x} = -k \, x $$
Because you can choose $k$ freely (only needs to be positive), you can realize any stable linear 1st order dynamic with $(1)$. So yes, you can use LQR in that case.