How to choose the limits of integration to find the velocity field of a circular pipe from the Laplace operator?

Question

I have a cylindrical pipe where the pressure gradient($G$), the kinematic viscosity ($\nu$), and density of liquid ($\rho$) are constants. The no-slip boundary condition may be taken at the wall of the pipe, so $u(R) = 0$. The velocity $u(r)$ satisfies the Navier-Stokes equation:

$$\nabla^2 u = \frac{1}{r}\frac{d}{dr}(r\frac{du}{dr}) = -\frac{G}{\nu \rho}$$

I now know that I should multiply by r and do a first integration from 0 to r.

$$\int d(r\frac{du}{dr}) = r(\frac{du}{dr}) = -\int_{0}^{r} r\frac{G}{\nu \rho} dr = -\frac{r^2}{2}\frac{G}{\nu \rho} $$

Next, I divide by r and integrate from r to R.

$$\int du = u(R) - u(r) = -\int_{r}^{R}\frac{r}{2}\frac{G}{\nu \rho} dr = -\frac{G}{4 \nu \rho}(R^2-r^2)$$

Applying the no-slip boundary condition.

$$u(r) = \frac{G}{4\nu\rho}(R^2-r^2) = \frac{GR^2}{4\nu\rho}\left (1-\left (\frac{r}{R}\right)^2 \right).$$

Question: How does one relate the physical dimensions of the pipe to the Laplace equation in $r$? How does one know to use $0$ to $r$ for the first integration and then $r$ to $R$ for the second integration?

Answer 1

I think that the first integral you should evaluate the term r du/dr between the limits as the the integral of the right hand side, then you Will get only the term rdu/dr @ r=r because rdu/dr @ r=0 is zero (that is the maximum point of the velocity profile). Maybe that could give a better interpretation. When you evaluate the last integral you use the limits on both sides (as it should be) and apply the boundary condition.

In the first integral you are applying a "differential boundary condition", so you need to evaluate at some point in the domain r and the boundary condition (r=0). For the second integral is the same.

It is only my opinion and I am open to any better interpretation that someone else can provide.

Regards