Definition 4.5.1 Conformal map.
A conformal map is a mapping \(\zeta = g(z)\) with \(g\) analytic and \(g'(z) \neq 0\) in a region \(R\text{.}\)
Section 4.5 Conformal mapping
The essential idea of conformal mapping is as follows. Suppose that we are given a two-dimensional potential flow problem in a region, \(R \subseteq \mathbb{C}\text{,}\) with impermeable boundary \(\partial R\text{.}\) There may be singularities in \(R\) corresponding to sinks, sources, vortices, etc. We then seek a conformal mapping from the \(z\)-plane to the \(\zeta\)-plane via
\begin{equation*}
\zeta = g(z),
\end{equation*}
so that the region \(R\) is mapped to a new region \(\hat{R} \subseteq \mathbb{C}\text{,}\) as shown in Figure 4.12.
The hope is that within the \(\zeta\)-plane, the fluid region is sufficiently simple that a complex potential, say \(F(\zeta)\text{,}\) can be found. This is aided by the fact that sources, sinks and vortices are preserved by the map (Proposition 4.5.2). Typically, we wish for \(\hat{R}\) to be e.g. the upper half-plane or the unit disc, and \(\partial\hat{R}\) the real axis or circumference of the unit disc, respectively. Once found, the complex potential in the \(z\)-plane is obtained by composing \(F\) with the map, i.e.
\begin{equation*}
f(z) = F(g(z)) = \phi(x, y) + \im \psi(x, y).
\end{equation*}
This simple idea turns out to yield many insights into potential flows in two dimensions.
Subsection 4.5.1 The conformal mapping method
The method relies on the following properties of conformal maps.
The derivative \(g'\) may vanish or be infinite at corners of the boundary (e.g. \(z^{\pi/\alpha}\) at \(z = 0\) in Example 4.5.3), which is how the map straightens corners.
Proposition 4.5.2 Properties of conformal maps.
-
Conformal maps preserve angles.
-
If the boundary \(\partial{\hat{R}}\) is a streamline in the \(\zeta\)-plane, then the corresponding boundary \(\partial R\) is a streamline in the \(z\)-plane (and vice versa).
-
A source of strength \(Q\) (or vortex of strength \(\Gamma\)) at \(\zeta = g(c) \in \hat{R}\) in the \(\zeta\)-plane corresponds to a source (or vortex) of the same strength at \(z = c \in R\) in the \(z\)-plane (and vice versa).
Proof (non-examinable).
(1) Near a point \(z_0\text{,}\) \(g(z) - g(z_0) \approx g'(z_0)(z - z_0)\text{,}\) so small line elements are all rotated by \(\arg g'(z_0)\) and scaled by \(|g'(z_0)|\text{;}\) angles between them are unchanged. (2) If \(\Im F\) is constant on \(\partial\hat{R}\text{,}\) then \(\Im f = \Im F(g(z))\) is constant on \(\partial R\text{.}\) (3) Near \(c\text{,}\) \(g(z) - g(c) = g'(c)(z - c)\big(1 + O(z - c)\big)\text{,}\) so \(\log(g(z) - g(c)) = \log(z - c)\) plus a function analytic near \(c\text{.}\)
Subsection 4.5.2 Source in a wedge
Example 4.5.3 Source in a wedge.
Fluid is contained in a wedge with walls at \(\theta = 0\) and \(\theta = \alpha > 0\text{,}\) with the fluid in \(0 < \theta < \alpha\text{.}\) Find the complex potential of a source of strength \(Q\) at the point \(z = c\) in the wedge.
Solution.
Consider the map
\begin{equation}
\zeta = g(z) = z^{\pi/\alpha}.\tag{4.5.1}
\end{equation}
With \(z = r\e^{\im\theta}\text{,}\) we have \(\zeta = r^{\pi/\alpha}\e^{\im\pi\theta/\alpha}\text{,}\) so \(0 < \theta < \alpha\) is mapped to \(0 < \arg\zeta < \pi\text{,}\) the upper half-\(\zeta\)-plane. The ray \(\theta = 0\) is mapped to the positive real axis and the ray \(\theta = \alpha\) to the negative real axis, as shown in Figure 4.13.
In the \(\zeta\)-plane, we must solve for flow in the upper half-plane with a source of strength \(Q\) at \(\zeta_c = c^{\pi/\alpha}\text{,}\) and an impermeable boundary on the real \(\zeta\)-axis. By the method of images (Section 4.4), we place sources at both \(\zeta_c\) and its complex conjugate, \(\overline{\zeta_c}\text{:}\)
\begin{equation}
F(\zeta) = \frac{Q}{2\pi} \log(\zeta - \zeta_c) + \frac{Q}{2\pi} \log(\zeta - \overline{\zeta_c}).\tag{4.5.2}
\end{equation}
Composing with the map, the complex potential in the \(z\)-plane is
\begin{equation}
f(z) = F(g(z)) = \frac{Q}{2\pi} \log(z^{\pi/\alpha} - c^{\pi/\alpha}) + \frac{Q}{2\pi} \log(z^{\pi/\alpha} - \overline{c^{\pi/\alpha}}).\tag{4.5.3}
\end{equation}
By construction (Proposition 4.5.2), the walls of the wedge are streamlines.
Subsection 4.5.3 Standard conformal maps
The exponential map is used to map a channel to a half-plane. Consider a channel of width \(h\) in the region \(0 < \Im z < h\) in the \(z\)-plane. Then
\begin{equation}
\zeta = g(z) = \e^{\pi z/h},\tag{4.5.4}
\end{equation}
maps this channel to the upper half-\(\zeta\)-plane. The correspondence of critical points and points at infinity in the pre-image and the image is shown in Figure 4.14. It is good to see the map as essentially ’unfolding’ the infinite channel, sending points AD to the origin, while sending B to negative infinity and C to positive infinity. The conformal map will preserve the orientation of the boundary, so as we traverse along ABCD, the fluid region is always on the right.
Trigonometric maps are used to map semi-infinite channels into a half-plane. For example, the map
\begin{equation}
\zeta = g(z) = \sin\left(\frac{\pi z}{2a}\right)\tag{4.5.5}
\end{equation}
sends the semi-infinite channel of width \(2a\) in Figure 4.15 to the upper half-plane. The two corners at \(z = \pm a\) are mapped to \(\zeta = \pm 1\text{,}\) respectively.
In Exercise 4.7.8, you will verify this.
Example 4.5.4 Vortex flow in a semi-infinite channel.
Consider the channel shown in the left of Figure 4.16. Insert a vortex of strength \(\Gamma\) at the point \(z = d \in \mathbb{R}^+\text{.}\) Verify that an appropriate conformal map is given by
\begin{equation}
\zeta = g(z) = \sinh\left(\frac{\pi z}{2a}\right),\tag{4.5.6}
\end{equation}
and find where the map sends the relevant critical points of the pre-image. Using the conformal map, find the complex velocity of the flow.
Solution.
The map sends \(z = 0\) and \(z = \pm\im a\) to \(\zeta = 0\) and \(\zeta = \pm\im\text{.}\) On \(x = 0\text{,}\) \(\zeta = \im\sin(\pi y/2a)\) covers the segment between \(\pm\im\text{,}\) and on \(y = \pm a\text{,}\) \(\zeta = \pm\im\cosh(\pi x/2a)\) covers the rest of the imaginary axis. The channel is therefore mapped to the right half-plane, and the vortex to \(\zeta_d = \sinh(\pi d/2a) \gt 0\text{.}\)
By the method of images (Example 4.4.3), with an image vortex of strength \(-\Gamma\) at \(-\zeta_d\text{,}\)
\begin{equation*}
F(\zeta) = -\frac{\im\Gamma}{2\pi}\left[\log(\zeta - \zeta_d) - \log(\zeta + \zeta_d)\right].
\end{equation*}
The complex velocity is then
\begin{equation*}
u - \im v = f'(z) = F'(\zeta)\, g'(z) = -\frac{\im\Gamma}{2\pi}\,\frac{2\zeta_d}{\zeta^2 - \zeta_d^2}\cdot\frac{\pi}{2a}\cosh\left(\frac{\pi z}{2a}\right).
\end{equation*}
Using \(\sinh^2 A - \sinh^2 B = \sinh(A - B)\sinh(A + B)\text{,}\) this simplifies to
\begin{equation*}
u - \im v = -\frac{\im\Gamma \sinh(\pi d/2a) \cosh(\pi z/2a)}{2a \sinh\big(\pi(z - d)/2a\big) \sinh\big(\pi(z + d)/2a\big)}.
\end{equation*}
