**A velocity field that is a gradient of a scalar potential field is irrotational**

It is a direct result from vector calculus that the curl of the gradient of a scalar is always zero, $\nabla \times \nabla \phi = 0$:

$\nabla \times \nabla \phi = \nabla \times (\frac{\partial \phi}{\partial x}\hat{x} + \frac{\partial \phi}{\partial y}\hat{y} + \frac{\partial \phi}{\partial z}\hat{z})$

$\nabla \times \nabla \phi = (\frac{\partial^{2} \phi}{\partial z \partial y} - \frac{\partial^{2} \phi}{\partial y \partial z})\hat{x} - (\frac{\partial^{2} \phi}{\partial z \partial x} - \frac{\partial^{2} \phi}{\partial x \partial z})\hat{y} + (\frac{\partial^{2} \phi}{\partial y \partial x} - \frac{\partial^{2} \phi}{\partial x \partial y})\hat{z} = 0$

**Incompressible flow is divergence free**

By definition, small fluid elements in an incompressible flow do not change in density. Therefore the *material derivative* of density is zero in an incompressible flow: $\frac{D \rho}{Dt} = 0$.

The differential statement of conservation of mass is:

$\frac{D \rho}{Dt} + \nabla \cdot \rho \vec{v} = 0$

The first term is zero for an incompressible flow, which leads to:

$\nabla \cdot \rho \vec{v} = 0$

$\rho(\nabla \cdot \vec{v}) = 0$

$\nabla \cdot \vec{v} = 0$

A divergence free velocity field is an equivalent statement of conservation of mass for an incompressible flow.

**Bernoulli's equation is valid in the full fluid domain in potential flow**

The Euler equations are a statement of conservation of momentum in an inviscid flow:

$\rho \frac{\partial \vec{v}}{\partial t} + \rho(\vec{v} \cdot \nabla)\vec{v} + \nabla p + \rho g \nabla z = 0$

The second term can be rewritten through a vector identity:

$(\vec{v} \cdot \nabla)\vec{v} = -\vec{v} \times \omega + \nabla (\frac{| \vec{v} |^{2} }{2})$

Vorticity is zero in irrotational flow, so Euler's equation simplifies to:

$\rho \frac{\partial \nabla \phi}{\partial t} + \rho \nabla(\frac{| \vec{v} | ^{2}}{2}) + \nabla p + \rho g \nabla z$

$\nabla(\rho \frac{\partial \phi}{\partial t} + \rho(\frac{| \vec{v} |^{2}}{2}) + p + \rho gz) = 0$

$\rho \frac{\partial \phi}{\partial t} + \rho(\frac{| \vec{v} |^{2}}{2}) + p + \rho gz = constant$

This is a statement of the unsteady version of Bernoulli's equation, which is valid not just along streamlines, but between points anywhere in the fluid domain.

**A dipole flow element is a combination of a point source and sink**

Combine the potentials of a source and sink, both strength $m$ and located on the x-axis separated by distance $2a$:

$\phi = \frac{m}{2\pi}[ln(\sqrt{(x - a)^{2} + y^{2}}) - ln(\sqrt{(x + a)^{2} + y^{2}})]$

Let the dipole strength, $\mu$, be equal to a combination of the sink/source strengths and the distance: $\mu = 2ma$. Take the limit as the source and sink get closer together and $\mu$ remains constant:

$\phi = \textrm{lim}_{a\rightarrow 0} \frac{\mu}{4\pi a}[ln(\sqrt{(x - a)^{2} + y^{2}}) - ln(\sqrt{(x + a)^{2} + y^{2}})]$

$\phi = \frac{\mu}{2\pi} \frac{x}{x^{2} + y^{2}}$

This same methodology can be used to prove a more general case of a dipole at any angle.