Consider a neutral conductor with $n$ free carriers that are uniformly distributed inside the conductor when there is no electric field, $E_z = 0$. From the continuity equation, I can write down the charge drift-diffusion equation as
$ \frac{\partial n}{\partial t} = D \frac{\partial^2 n}{\partial z^2} - \mu E_z \frac{\partial n}{\partial z} $
where $D$ allows for diffusion, $\mu$ is the mobility, $E_z$ is an electric field applied, and only the the $z$-direction is focused on for simplicity. As I already said, initially $E_z = 0$ so the $n$ is uniform throughout the conductor.
Now the electric field is turned on. What happens? Well, intuitively one one would surmise that the free charges move against the electric field (assuming electrons here) which will leave the conductor positively charged on one end and negatively charged on the other. However, if I solve the continuity equation, I get $n$ to be uniform still -- nothing happens. You can see why this is, mathematically, from the above equation. The gradient of $n$ is initially 0 which sets $\partial n/\partial t = 0$ and nothing interesting comes of it. Compare this to the situation where you start with a charge gradient - i.e. a delta impulse at $z = 0$ - where the drift of the charge impulse can see in the time evolution.
Why is the math not matching my expectations?