Parabolic velocity profile in laminar pipe flow

Consider steady laminar flow in a horizontal circular pipe of radius $R$. Take a coaxial cylinder of fluid of radius $r$ and length $L$.

        <-- tau (on surface)
   p1 ->  ____________  <- p2
         |   cylinder |  radius r, length L
   ------|------------|-------- axis
         |____________|

Force balance (no acceleration): pressure force on the ends = shear force on the curved surface.

$$(p_1 - p_2)\pi r^2 = \tau\,(2\pi r L) \quad\Rightarrow\quad \tau = \frac{(p_1 - p_2)}{L}\frac{r}{2} = -\frac{dp}{dx}\frac{r}{2}$$

So the shear stress varies linearly from zero at the axis to $\tau_0$ at the wall.

Newton's law of viscosity. With $y = R - r$, $\tau = \mu\dfrac{du}{dy} = -\mu\dfrac{du}{dr}$:

$$-\mu\frac{du}{dr} = -\frac{dp}{dx}\frac{r}{2} \quad\Rightarrow\quad \frac{du}{dr} = \frac{1}{2\mu}\frac{dp}{dx}\,r$$

Integrate:

$$u = \frac{1}{4\mu}\frac{dp}{dx}\,r^2 + C$$

Boundary condition (no slip): $u = 0$ at $r = R$, so $C = -\dfrac{1}{4\mu}\dfrac{dp}{dx}R^2$.

$$\boxed{u = -\frac{1}{4\mu}\frac{dp}{dx}\left(R^2 - r^2\right)}$$

This is the equation of a parabola in $r$ ($u$ falls as $r^2$), with its maximum at the axis:

$$u_{max} = -\frac{1}{4\mu}\frac{dp}{dx}R^2, \qquad \frac{u}{u_{max}} = 1 - \frac{r^2}{R^2}$$

The mean velocity is $V = u_{max}/2$, so the velocity distribution is parabolic. (For an inclined pipe, replace $p$ by the piezometric pressure $p + \gamma z$.)