Ideal Fluid Flow #
Suppose we have an irrotational fluid:
\[\curl \vec V = 0\]If \( \vec V \) is curl-free, then there exists a potential \( \phi \) such that
\[\vec V = \grad \phi\]In 2D, for example, this means
\[\vec V = U \vu{x} + V \vu{y} \\ \pdv{\phi}{x} = U \qquad \pdv{\phi}{y} = V\]Physically, an inviscid flow is the result of zero viscosity. Generally this is a good approximation of fluid flow as long as we are far away from any boundaries.
If the fluid is also incompressible:
\[\div \vec V = 0 \qquad \rightarrow \qquad \pdv{U}{x} + \pdv{V}{y} = 0\]Then this implies the existence of a stream function \( \psi \) such that
\[\pdv{\psi}{y} = U \qquad \pdv{\psi}{x} = - V\]because we can substitute into the divergence-free condition:
\[\pdv{}{x} \pdv{\psi}{y} - \pdv{}{y} \pdv{\psi}{x} = 0\]The divergence of a curl is always zero:
\[\div (\curl \vec V) = 0 \qquad \rightarrow \qquad \nabla ^2 \phi = 0\]Both \( \phi \) and \( \psi \) are harmonic functions, so the function
\[F(z) = \phi + i \psi\]is analytic.
Example: uniform flow
\[F(z) = U_0 z\]\[U = U_0 = \pdv{\phi}{x} \quad \rightarrow \quad \phi = U_0 x + C\] \[V = 0 = \pdv{\phi}{y} \qquad \rightarrow \qquad \phi \text{ cannot depend on y}\] \[U = U_0 = \pdv{\psi}{y} \qquad \rightarrow \qquad \phi = U_0 y + C\] \[V = 0 = - \pdv{\psi}{x} \qquad \rightarrow \qquad \psi \text{ cannot depend on x}\]
Example: Uniform flow at angle \( \alpha \)
We can just rotate everything by \( \alpha \) and
\[F(z) = U_0 z e^{i \alpha}\]Example: Source flow
Given a point source at the origin (think a line source perpendicular to the z-plane), mass conservation implies
\[Q = 2 \pi r \rho = \text{ const. }\]Radial flow velocity \( V_0 \) is
\[V_0 = \frac{Q / \rho}{2 \pi r}\]We can define the complex velocity \( W(z) \)
\[W(z) \equiv \pdv{F(z)}{z} = U - i V\]where we have used the fact that \( F(z) \) is analytic. Then in this example,
\[W(z) = V_0 (\cos \theta - i \sin \theta) \\ = \frac{Q / \rho}{2 \pi r} e^{- i \theta} \\ = \frac{Q / \rho}{2 \pi} \frac{1}{z}\]\[F(z) = \int _0 ^z W(z) \dd z = \frac{Q / \rho}{2 \pi} \ln z + C\] \[\psi = \text{Im} F = \frac{Q / \rho}{2 \pi} \theta\]
Again, the streamlines are rays pointing out from the origin.
As a side note, we see that when we have to integrate \( 1/z \), we introduce the nowhere-analytic \( \ln z \) and therefore we introduce a branch cut and have to restrict the argument. Why is this?
Example: Vortex flow
\[U = - V_0 \sin \theta\] \[V = V_0 \cos \theta\]
We define the conserved quantity “circulation” \( \Gamma \) \[\Gamma = \oint \vec V \cdot \dd \vec l = 2 \pi r V_0\]
Conservation of angular momentum tells us that
\[U = - \frac{\Gamma}{2 \pi r} \sin \theta\] \[V = \frac{\Gamma}{2 \pi r } \cos \theta\] \[W = - \frac{i \Gamma}{2 \pi r} (\cos \theta - i \sin \theta) = - \frac{i \Gamma}{2 \pi r} \frac{1}{z}\] \[F(z) = \int ^z W(z) \dd z = - \frac{i \Gamma}{2 \pi } \ln z + C\]
This is very similar to the source flow case. That extra factor of \( i \) makes all the difference. Radial source flow has a purely real coefficient for \( \ln z \), and purely rotational vortex flow has an imaginary coefficient for \( \ln z \). Cool.