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Section 4.1 The velocity potential

In this chapter, we focus on 2D flows where the velocity vector is given by
\begin{equation*} \bu(x, y, t) = u(x, y, t)\bi + v(x, y, t)\bj = [u, v]. \end{equation*}
With the velocity given as above, the vorticity is then
\begin{equation} \bomega = \nabla \times \bu = \left(\pd{v}{x} - \pd{u}{y}\right)\bk.\tag{4.1.1} \end{equation}
If we assume that the flow is irrotational according to Definition 3.5.3, then \(\nabla \times \bu = 0\) and
\begin{equation} \pd{v}{x} - \pd{u}{y} = 0.\tag{4.1.2} \end{equation}
Further, we know that if the flow is irrotational, then there exists a velocity potential, \(\phi\text{,}\) such that \(\bu = \nabla \phi\text{.}\) Thus the velocities are expressed as
\begin{equation} u = \pd{\phi}{x} \quad \textrm{and} \quad v = \pd{\phi}{y}.\tag{4.1.3} \end{equation}
The above result about irrotational flows is a standard result in Vector Calculus, but we will re-state the result here for reference, and provide a review of its proof.

Proof (non-examinable).
Define
\begin{equation*} \phi(\bx, t) \equiv \phi_0(t) + \int_C \bu \cdot \de\bx, \end{equation*}
where \(C\) is any contour connecting an arbitrary origin point to the point \(\bx\) (changing the origin point will change the "constant" of integration \(\phi_0(t)\)).
We can verify, using the definition of differentiation, and the fundamental theorem of calculus, applied along each of the three coordinate directions, that \(\nabla \phi = \bu\) as desired.
The key is to prove that the above definition is independent of the choice of contour \(C\text{.}\) To this end, consider two contours, \(C_1\) and \(C_2\text{,}\) both with the same origin point, \(O\text{,}\) and end point \(P\text{.}\) Then the contour \(C_1 - C_2\) is a closed contour beginning and ending at \(O\text{.}\)
Figure 4.2. Proof of the uniqueness of the potential
By Stokes’ theorem,
\begin{equation*} \oint_{C_1 - C_2} \bu \cdot \de\bx = \iint_S (\nabla \times \bu) \cdot \bn \, \de{S}, \end{equation*}
where \(S\) is any surface with bounding curve \(C_1 - C_2\) (such a surface lying in \(D\) exists because \(D\) is simply connected), and with unit normal \(\bn\) positively oriented with the bounding curve. However, the right-hand side is zero by irrotationality, and therefore
\begin{equation*} \int_{C_1} \bu \cdot \de\bx = \int_{C_2} \bu \cdot \de\bx, \end{equation*}
and our choice of curve in the definition of the potential is irrelevant.
The converse direction of the theorem follows directly from the fact that "curl grad equals zero", i.e. \(\nabla \times \nabla \phi = 0\text{.}\)
Let us return to discussing the setting of potential flow.
In addition to being irrotational, we further assume that the flow is incompressible. Therefore from Theorem 3.4.1,
\begin{equation} \nabla \cdot \bu = \pd{u}{x} + \pd{v}{y} = \pd{^2\phi}{x^2} + \pd{^2\phi}{y^2} = 0.\tag{4.1.4} \end{equation}
This is the crucial result, which is that in potential flows, we need only solve the Laplace equation:
\begin{equation} \nabla^2 \phi = 0,\tag{4.1.5} \end{equation}
within the flow region. This is effectively a single linear equation for the single unknown \(\phi\text{.}\) However, for different problems, the boundary conditions can render even this "simple" problem difficult.
Once the velocity potential \(\phi\) has been found, the velocities in the flow can be recovered from the relationship (4.1.3). The pressure in the flow also follows from Bernoulli’s equation. For the situation of a steady potential flow, following Theorem 3.5.8, it is
\begin{equation} \frac{p}{\rho} + \frac{1}{2} |\nabla \phi|^2 + \chi = \textrm{constant}.\tag{4.1.6} \end{equation}

Subsection 4.1.1 Elementary flows

The next three examples will introduce you to the elementary flows consisting of uniform flow, stagnation point flow, and line source/sink flows. You will also investigate the notion of a source "strength".

Example 4.1.2 Uniform flow.

Find the velocity potential \(\phi\) and the velocity \(\bu\) of a uniform flow of speed \(U\) directed at angle \(\alpha\) to the \(x\)-axis.
Solution.
The velocity is constant, \(\bu = U[\cos\alpha, \sin\alpha]\text{.}\) Integrating \(\pd{\phi}{x} = U\cos\alpha\) and \(\pd{\phi}{y} = U\sin\alpha\) gives, up to a constant, the linear function
\begin{equation*} \phi(x, y) = Ux \cos\alpha + Uy \sin\alpha, \end{equation*}
which clearly satisfies \(\nabla^2\phi = 0\text{.}\) The streamlines are shown below.
Figure 4.3. Streamlines (or velocity field) of uniform flow with \(U = 1\) and \(\alpha = \pi/4\text{.}\)

Example 4.1.3 Stagnation point flow.

Show that \(\phi = \frac{1}{2} (x^2 - y^2)\) is the velocity potential of an incompressible flow, and find the velocity \(\bu\text{.}\)
Solution.
Since \(\pd{^2\phi}{x^2} + \pd{^2\phi}{y^2} = 1 - 1 = 0\text{,}\) Laplace’s equation is satisfied. The velocity is
\begin{equation*} \bu = \nabla\phi = [x, -y], \end{equation*}
which vanishes at the origin. This is stagnation point flow, shown below.
Figure 4.4. Streamlines (or velocity field) of stagnation point flow.

Example 4.1.4 Line source.

A line source ejects fluid from the origin equally in all directions. Find a radially symmetric potential \(\phi = \phi(r)\) satisfying Laplace’s equation, and the corresponding velocity \(\bu\text{.}\) By computing the flux of fluid across a circle centred at the origin, relate the constant in \(\phi\) to the rate \(Q\) at which fluid is produced.
Solution.
In plane polar coordinates,
\begin{equation*} \nabla^2 \phi = \frac{1}{r} \pd{}{r}\left(r \pd{\phi}{r}\right) + \frac{1}{r^2} \pd{^2 \phi}{\theta^2} = 0. \end{equation*}
With \(\phi = \phi(r)\text{,}\) direct integration gives
\begin{equation*} \phi = \frac{Q}{2\pi} \log r, \end{equation*}
where we have set an additive constant to zero, and written the leading constant as \(Q/(2\pi)\) in anticipation of the result below. The velocity is
\begin{equation*} \bu = \nabla \phi = \pd{\phi}{r} \be_r + \frac{1}{r}\pd{\phi}{\theta} \be_{\theta} = \frac{Q}{2\pi r} \be_r = \frac{Q}{2\pi r^2} [x, y], \end{equation*}
using \(\be_r = [\cos\theta, \sin\theta]\text{.}\) The flow is directed radially outwards. It is called a "line" source because in \((x, y, z)\) the source runs parallel to the \(z\)-axis.
Figure 4.5. Streamlines (or velocity field) of line source flow.
The flux (the flow per unit time) of fluid crossing a closed contour \(C\) with unit normal \(\bn\) is \(\int_C \bu \cdot \bn \, \de{s}\text{.}\) Take \(C\) to be the circle \(r = a\text{,}\) on which \(\bn = \be_r\) and the arc-length element is \(\de{s} = a \, \de\theta\text{.}\) Then
\begin{equation*} \int_C \bu \cdot \bn \, \de{s} = \int_0^{2\pi} \frac{Q}{2\pi a} \, (a \, \de\theta) = Q. \end{equation*}
Therefore, \(Q\text{,}\) the strength of the source, is the rate at which fluid is produced. If \(Q \lt 0\text{,}\) we refer to the flow as a line sink.
Crucially, the governing equation is Laplace’s equation, which is linear, so any sum of elementary flows is also an admissible flow.

Example 4.1.5 Line source in a uniform flow.

Find the velocity potential \(\phi\) and the velocity \(\bu\) of a line source of strength \(Q\) at the origin placed in a uniform flow of speed \(U\) in the \(x\)-direction.
Solution.
Adding the potentials of Example 4.1.2 (with \(\alpha = 0\)) and Example 4.1.4,
\begin{equation*} \phi = Ux + \frac{Q}{2\pi} \log r = Ux + \frac{Q}{2\pi} \log \sqrt{x^2 + y^2}, \end{equation*}
and so
\begin{equation*} \bu = [U, 0] + \frac{Q}{2\pi(x^2 + y^2)} [x, y]. \end{equation*}
The streamlines are shown below. Where do you think the stagnation point lies in this flow?
Figure 4.6. Streamlines (or velocity field) of a line source in a uniform flow with \(U = 1\) and \(Q = 1\text{.}\)
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