Part I (One doesn't take into account viscosity)
One way to generate needed pressure difference will be to use some kind of pump as shown on the first picture:
$$\Delta P = P_{\text{pump}} - P_{0}$$
where $P_{0}$ stands for atmospheric pressure. Note however that in this setup you will run out of fluid after time $\frac{L}{V}$ (unless you modify it)
You can also consider another setup:
$$\Delta P = mgh $$
Here fluid itself will generate pressure difference. However this it may be considered to be constant only within short amount of time. After some time pressure difference will decrease.
Finally you can of course place your pipe into stream of fluid. In this case you will just have to fix it so it doesn't move. In this case this pipe will be somewhat secondary in the sense that it will have nothing to do with generation of pressure difference. It may be produced in external liquid by any of aforementioned methods.
Part II (Laminar flow of viscous incompressible fluid at constant temperature)
Flow produced the same way as in previous case enters pipe while having uniform velocity profile. We imply so called no-slip condition which states that speed of fluid layer near the pipe surface equals zero. Due to presence of viscosity($\approx$ friction between the molecules of fluid) aforementioned layer starts to slow down the one nearest to it and so on. This forces uniform velocity profile to change itself as shown in the picture:
After some time velocity profile reaches certain(parabolic) form and doesn't change afterwards(fully developed velocity profile).
Now let us analyze in details a ring-shaped differential volume element of radius r, thickness $dr$ and length $dx$ oriented coaxially with the pipe, as shown in the picture:
Since speed of every such volume is constant we state that sum of forces acting on it equals zero. From one hand there are $\tau_{r+dr}$ and $\tau_{r}$ - shear stresses due to interaction of the nearest layers of the fluid.
Since $\tau$ has radial dependence there appears disbalance in forces acting on a given volume. This generates the same(up to sign) pressure difference in x direction.
$$2 \pi r dr P_x - 2 \pi r dr P_{x+dx} +2 \pi r dr \tau_r - 2 \pi r dr \tau_{r+dr} = 0$$
Once we take limit $dr,dx \to 0$ we obtain:
$$r \frac{dP}{dx}+ \frac{d(r\tau)}{dr} = 0$$
If one takes $\tau = \mu(du/dr)$ where $\mu$ is some constant one can also show that $\frac{dP}{dx} = \text{const}$
PS: Aforementioned explanation is heavily based on the following article. See 8-3 and 8-4.
Hope it helps!
