# Why does 3D source potential flow only satisfy Laplace's equation in spherical coordinates and not cylindrical coordinates?

3D source potential flow is given by: $\phi = \frac{-\lambda}{4\pi r}$.

Laplace's equation in cylindrical coordinates is:

$\frac{1}{r}\frac{\partial }{\partial r}\left(r \frac{\partial \phi}{\partial r}\right) + f(\theta,z) = 0$

where $f(\theta,z) = 0$ since $\phi$ is independent of $\theta$ and $z$.

$\frac{1}{r}\frac{\partial }{\partial r}(r \frac{\partial \phi}{\partial r}) = \frac{1}{r}\frac{\partial }{\partial r} \left(\frac{\lambda}{4\pi r}\right) = \left(\frac{-\lambda}{4\pi r^3}\right) \neq 0$

Thus, the 3D potential flow is not a solution to Laplace's equation in cylindrical coordinates.

Now, Laplace's equation in spherical coordinates is:

$\frac{1}{r^2sin(\theta)}\left[\frac{\partial }{\partial r}\left(r^2sin(\theta)\frac{\partial \phi}{\partial r}\right) + f(\theta,\Phi)\right] = 0$

Where $f(\theta,\Phi) = 0$ since $\phi$ is independent of $\theta$ and $\Phi$.

$\frac{1}{r^2sin(\theta)}\frac{\partial }{\partial r} (r^2sin(\theta)\frac{\partial \phi}{\partial r}) = \frac{1}{r^2sin(\theta)}\frac{\partial }{\partial r} (r^2sin(\theta)\frac{\lambda}{4\pi r^2})=\frac{1}{r^2sin(\theta)}\frac{\partial }{\partial r}(f(\theta))=0$

Therefore, source potential flow is a solution to Laplace's equation in spherical coordinates. As I understand it, a solution to Laplace's equation should be independent of coordinate system, so what is going on here?

All equations for this question have come from John D. Anderson Jr., Fundamentals of Aerodynamics Fifth Edition.

• Despite the notation, the $r$ is not the same in the two different coordinate systems. One is distance from a point (the origin) the other is a distance from a line (the $z$ axis). So you haven't described the same potential for cylindrical coordinates. – Dan Apr 11 '18 at 23:37

$\phi = \Lambda ~ln~r$.
$u_r = \frac{\lambda}{r}$ .