Theorem 4.2.1 The streamfunction is constant along streamlines.
The streamfunction, \(\psi(x, y, t)\text{,}\) is constant along streamlines of the flow (i.e. curves that are everywhere tangent to \(\bu\) at a fixed time \(t\)).
Section 4.2 The streamfunction
Our next task is to introduce the concept of the streamfunction. Remember that in 2D, the irrotational flow led to the equation (4.1.2) and this led to the existence of the potential function. If we begin with incompressibility, however, we have (4.1.4), which can be written as
\begin{equation}
\big(\nabla \times [-v, u, 0]\big)\cdot\bk = \pd{u}{x} + \pd{v}{y} = 0.\tag{4.2.1}
\end{equation}
We can therefore deduce the existence of an analogous function, the streamfunction, \(\psi(x, y, t)\text{,}\) satisfying
\begin{equation}
u = \pd{\psi}{y} \quad \textrm{and} \quad v = -\pd{\psi}{x}.\tag{4.2.2}
\end{equation}
Alternatively and more conveniently, we can write
\begin{equation}
\bu = \nabla \times (\psi \bk).\tag{4.2.3}
\end{equation}
To establish the existence of the streamfunction, we follow a similar proof as in Theorem 4.1.1 but now with the definition that
\begin{equation}
\psi(x, y, t) = \psi_0(t) + \int_0^{\bx} (u \de{y} - v \de{x}),\tag{4.2.4}
\end{equation}
where again \(\psi_0(t)\) is an arbitrary function of \(t\text{.}\) The proof is otherwise identical, relying on establishing the independence of path of the integral using Stokes’ theorem.
Why all this work? The streamfunction has an intuitive interpretation via the following result.
Proof.
The proof follows simply by the fact that
\begin{equation*}
\bu \cdot \nabla \psi = \left(\pd{\psi}{y}, -\pd{\psi}{x}\right) \cdot \left(\pd{\psi}{x}, \pd{\psi}{y}\right) = 0,
\end{equation*}
where the first equality uses (4.2.2).
Since \(\nabla\psi\) is normal to the level sets of \(\psi\text{,}\) and \(\bu\) is orthogonal to \(\nabla\psi\text{,}\) the velocity \(\bu\) is tangent to the level sets. These level sets are therefore streamlines.
A graphical depiction of the property in Theorem 4.2.1 is shown below.
The streamfunction is thus constant on streamlines. Consider two streamlines. The following theorem relates the flux between the streamlines to the values of \(\psi\) on them.
Theorem 4.2.2 Flux between streamlines.
Consider two streamlines \(\psi = \psi_1\) and \(\psi = \psi_2\text{.}\) We assume that the streamlines pass through the points A and B respectively. Consider a contour \(C\) connecting A and B. The flux (net flow of fluid) through \(C\) and hence between the streamlines is
\begin{equation*}
|\textrm{flux}| = |\psi_2 - \psi_1|.
\end{equation*}
We have not specified the sign of the flux as it depends on the direction chosen through \(C\text{.}\)
Proof.
By definition, the flux is given by the integral
\begin{equation*}
\int_C \bu \cdot \bn \, \de{s},
\end{equation*}
where \(C\) is any smooth path joining the two streamlines with unit normal \(\bn\text{,}\) as shown in Figure 4.7.
Note that given the curve, and in consideration of a small arclength element \(\de{s}\text{,}\) the tangent and normal are given by
\begin{equation*}
\bt = \frac{[\de{x}, \de{y}]}{\sqrt{\de{x}^2 + \de{y}^2}} = \frac{[\de{x}, \de{y}]}{\de{s}}, \qquad
\bn = \frac{[\de{y}, -\de{x}]}{\de{s}}.
\end{equation*}
Therefore we can write \(\bn \de{s} = [\de{y}, -\de{x}]\text{.}\) Then the flux is re-written as the following:
\begin{align*}
\textrm{flux} \amp= \int_C \left[\pd{\psi}{y}, -\pd{\psi}{x}\right] \cdot [\de{y}, -\de{x}], \\
\amp= \int_C \left(\pd{\psi}{x} \, \de{x} + \pd{\psi}{y} \, \de{y} \right), \\
\amp= \bigl[ \psi \bigr]_C, \\
\amp= \psi(B) - \psi(A), \\
\amp= \psi_2 - \psi_1.
\end{align*}
In the third line above, \([\psi]_C\) is the change of \(\psi\) along the contour. Note that there is a somewhat arbitrary choice of direction for the contour \(C\text{,}\) as related to the selection of the normal direction \(\bn\text{,}\) and the positivity or negativity of the flux. To be safe, we have taken the absolute value in the statement of the theorem.
An example of the use of the above theorem is given in Exercise 4.7.2.
Example 4.2.3 Flux from the streamfunction.
Let the velocity of a two-dimensional flow be given by the formula
\begin{equation*}
\bu = [3ax^2 - 3ay^2, \, -6axy]
\end{equation*}
where \(a\) is a positive constant. By finding the streamfunction, \(\psi\text{,}\) calculate the volume flux across a curve connecting points \(A = (0, 0)\) and \(B = (1, 1)\) using Theorem 4.2.2.
Solution.
We can check, using the integration of velocity components that
\begin{equation*}
\psi = 3ax^2 y - a y^3 + C.
\end{equation*}
Hence the flux is
\begin{equation*}
Q = \psi(B) - \psi(A) = 2a.
\end{equation*}
We can verify this flux should be positive if the normal is oriented to the right of the curve from A to B.
Finally, note that the velocity potential was governed by Laplace’s equation, \(\nabla^2 \phi = 0\text{.}\) The streamfunction is also governed by the same Laplace’s equation.
Theorem 4.2.4 Streamfunction satisfies Laplace’s equation.
If the flow is also irrotational, then, like the velocity potential, the streamfunction satisfies Laplace’s equation:
\begin{equation}
\pd{^2 \psi}{x^2} + \pd{^2 \psi}{y^2} = \nabla^2 \psi = 0.\tag{4.2.5}
\end{equation}
For some flows, such as the line source in Example 4.1.4, it is easier to work in polar coordinates. Since we know that \(\bu = \nabla \times (\psi \bk)\) by (4.2.3), we can use the conversion of the curl to polar coordinates in (G.1.31) to give
\begin{equation}
u_r = \frac{1}{r} \pd{\psi}{\theta} \quad \textrm{and} \quad u_{\theta} = -\pd{\psi}{r},\tag{4.2.6}
\end{equation}
which allows us to relate the radial and angular velocities to the streamfunction.
Subsection 4.2.1 Elementary flows
Let us return to each of the examples in Section 4.1 and reconsider their corresponding streamfunctions.
Example 4.2.5 Uniform flow.
Find the streamfunction \(\psi\) for the uniform flow of Example 4.1.2, and sketch its streamlines.
Solution.
Integrating \(u = \psi_y = U\cos\alpha\) and \(v = -\psi_x = U\sin\alpha\) gives, up to an arbitrary constant,
\begin{equation*}
\psi = Uy\cos\alpha - Ux\sin\alpha.
\end{equation*}
Example 4.2.6 Stagnation point flow.
Find the streamfunction \(\psi\) for the stagnation point flow of Example 4.1.3, and sketch its streamlines.
Solution.
Integrating \(u = \psi_y = x\) and \(v = -\psi_x = -y\) gives, up to an arbitrary constant, \(\psi = xy\text{.}\) Curves of constant \(\psi\) are the hyperbolae shown in Figure 4.4.
Example 4.2.7 Line source.
Find the streamfunction \(\psi\) for the line source of Example 4.1.4, and sketch its streamlines.
Solution.
From (G.1.30), the velocity is entirely radial,
\begin{equation}
\bu = \pd{\phi}{r}\be_r = \frac{Q}{2\pi r} \be_r.\tag{4.2.7}
\end{equation}
Using (4.2.6), we integrate \(\frac{1}{r} \pd{\psi}{\theta} = \frac{Q}{2\pi r}\) (with \(\pd{\psi}{r} = -u_\theta = 0\)) to give
\begin{equation*}
\psi = \frac{Q}{2\pi} \theta.
\end{equation*}
The lines of constant \(\psi\) are the radial lines of constant \(\theta\text{,}\) as in Figure 4.5.
Note that \(\psi\) is a multi-valued function in Example 4.2.7, gaining a jump of \(Q\) every time the origin is encircled. This is a warning that the standard proof, analogous to Theorem 4.1.1, leading to the existence of a unique streamfunction via (4.2.4), would not apply, because the flow domain (the plane with the origin removed) is not simply connected. However, despite this, the streamfunction provides well-defined predictions of streamlines on the cut plane with e.g. \(\theta \in [0, 2\pi)\text{.}\)
Example 4.2.8 Line source in a uniform flow.
Find the streamfunction \(\psi\) for the line source in a uniform flow of Example 4.1.5, and the equation of its streamlines.
Solution.
By linearity, we add the streamfunctions of the uniform flow (with \(\alpha = 0\)) and the line source:
\begin{equation*}
\psi = Uy + \frac{Q}{2\pi}\theta.
\end{equation*}
The streamlines are given by
\begin{equation*}
Uy + \frac{Q}{2\pi} \theta = \frac{Q}{2\pi}C,
\end{equation*}
having chosen the constant on the right-hand side for convenience. Then using \(y = r\sin\theta\text{,}\) we have
\begin{equation*}
r = \left(\frac{Q}{2\pi U}\right) \frac{C - \theta}{\sin\theta}.
\end{equation*}
Our previous examples were reliant on considering the flows generated by the velocity potentials studied previously. However, we can also consider "fundamental solutions" of the equation \(\nabla^2 \psi = 0\) in their own right. Recall that in deriving the velocity potential of the line source in Example 4.1.4, we considered a radially symmetric problem, where \(\phi = \phi(r)\) is only dependent on the distance from the origin. The same argument shows that a radially symmetric streamfunction \(\psi = \psi(r)\) is also a solution. This leads us to the following example.
Example 4.2.9 Line vortex.
The radially symmetric streamfunction
\begin{equation}
\psi = -\frac{\Gamma}{2\pi}\log r,\tag{4.2.8}
\end{equation}
defined up to a constant, corresponds to a line vortex. Find its velocity \(\bu\text{,}\) and compute the circulation around a circle of radius \(a\) centred at the origin.
Solution.
\begin{equation}
\bu = -\pd{\psi}{r} \be_{\theta} = \frac{\Gamma}{2\pi r} \be_{\theta},\tag{4.2.9}
\end{equation}
so the fluid orbits the origin, with speed \(|\Gamma|/(2\pi r)\) increasing as \(r \to 0\text{.}\)
The circulation around a contour \(C\) is
\begin{equation*}
\oint_C \bu \cdot \de{\bx} = \oint_C (u \, \de{x} + v \, \de{y}),
\end{equation*}
i.e. the sum of the velocity components tangential to \(C\text{.}\) By Stokes’ theorem it equals the flux of vorticity through any surface bounded by \(C\text{.}\) Since the flow is irrotational away from the origin, all of this vorticity is concentrated at the origin. On the circle \(r = a\text{,}\) \(\de{\bx} = a \be_{\theta} \de{\theta}\text{,}\) so
\begin{equation}
\oint_C \bu \cdot \de{\bx} = \int_0^{2\pi} \frac{\Gamma}{2\pi a} \be_{\theta} \cdot \be_{\theta} a \de\theta = \Gamma.\tag{4.2.10}
\end{equation}
The quantity \(\Gamma\) is called the vortex strength, analogous to the source strength \(Q\) in Example 4.1.4. Notice that \(\Gamma \gt 0\) corresponds to flow in the anticlockwise sense, and \(\Gamma \lt 0\) to flow in the clockwise sense.
In Exercise 4.7.1, you will practise the computation of fluxes using line integrals.
