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Complex Analysis: an Open Source Textbook

Russell W. Howell, John H. Mathews

Section 10.10 Image of a Fluid Flow

We have already examined several two-dimensional fluid flows and have shown that the image of a flow under a conformal transformation is a flow. The conformal mapping \(w=f(z)=u(x,y)+iv(x,y)\text{,}\) which we obtained by using the Schwarz-Christoffel formula, allows us to find the streamlines for flows in domains in the \(w\) plane that are bounded by straight-line segments.
The first technique involves finding the image of a fluid flowing horizontally from left to right across the upper half-plane \(\mathrm{Im} (z) >0\text{.}\) The image of the streamline \(-\infty \lt t \lt \infty ,\, y = c\) is a streamline given by the parametric equations
\begin{equation*} u=u(t,c) \text{ and } u=v(t,c), \text{ for } -\infty \lt t \lt \infty \end{equation*}
and is oriented in the positive sense (counterclockwise). The streamline \(u=u(t,0), \, v=(t,0)\) is considered to be a boundary wall for a containing vessel for the fluid flow.

Example 10.10.1.

Consider the conformal mapping
\begin{equation*} w = f(z) = \frac{1}{\pi}(z^2-1)^\frac{1}{2} + \frac{1}{\pi}\mathrm{Log}\left[z+(z^2-1)^\frac{1}{2}\right]\text{,} \end{equation*}
which we obtained by using the Schwarz-Christoffel formula. It maps the upper half-plane \(\mathrm{Im}(z) >0\) onto the domain in the \(w\) plane that lies above the boundary curve consisting of the rays \(u \le 0, \; v=1\) and \(u \ge 0, \; v=0\) and the segment \(u=0, \; -1 \le v \le 0\text{.}\) Furthermore, the image of horizontal streamlines in the \(z\) plane are curves in the \(w\) plane given by the parametric equation
\begin{align*} w \amp = f(t+ic) =\frac{1}{\pi}(t^2-c^2-1+i2ct)^\frac{1}{2}\\ \amp + \frac{1}{\pi}\mathrm{Log}\left[ t+ic+(t^2-c^2-1+i2ct)^\frac{1}{2}\right]\text{,} \end{align*}
for \(-\infty \lt t \lt \infty\text{.}\) The new flow is that of a step in the bed of a deep stream and is illustrated in Figure 10.10.2(a). The function \(w=f(z)\) is also defined for values of \(z\) in the lower half-plane, and the images of horizontal streamlines that lie above or below the \(x\) axis are mapped onto streamlines that flow past a long rectangular obstacle, which is illustrated in Figure 10.10.2(b).
Figure 10.10.2. Streamline mappings

Exercises Exercises

For Exercises 1-4, use the Schwarz-Christoffel formula to find a conformal mapping \(w=f(z)\) that will map the flow in the upper half-plane \(\mathrm{Im}(z) >0\) onto the flows indicated.

1.

Use Figure 10.10.3 to find the flow over the vertical segment from 0 to \(i\text{.}\)
Figure 10.10.3. For Exercise 10.10.1
Solution.
\(f\,'(z) = A(z+1)^{-\frac{1}{2}}z(z-1)^{-\frac{1}{2}}=A\frac{z}{(z^2-1)^\frac{1}{2}}\text{.}\) Integrate with the boundary conditions \(f(-1)=0\) and \(f(0)=i\) to get \(w=f(z)=(z^2-1)^\frac{1}{2}\text{.}\)

3.

Use Figure 10.10.5 to find the flow around
(a)
one inclined segment in the upper half-plane.
Solution.
\(w=f(z) = (z-1)^{\alpha}\left(1+\frac{\alpha z}{1-\alpha}\right)\!^{1-\alpha}\text{.}\)
(b)
two inclined segments forming a “V” in the plane.
Figure 10.10.5. For Exercise 10.10.3

5.

For flow around an infinitely long rectangular barrier with a pointed “nose,” find
(a)
the flow up an inclined step, as shown in Figure 10.10.7(a).
Solution.
\(\displaystyle w=f(z)=-1+\int_{-1}^z\frac{(\xi -1)^\frac{1}{4}}{\xi^\frac{1}{4}}\,d\xi\text{.}\) \(w=f(z)=\) \(i+\frac{1}{\pi}\left[4(z-1)^\frac{1}{4}z^\frac{3}{4} - 2\mathrm{Arctan}(1-\frac{1}{z})^\frac{1}{4} +\mathrm{Log}(1-(1-\frac{1}{z})^\frac{1}{4})-\mathrm{Log}(1+(1-\frac{1}{z})^\frac{1}{4})\right]\text{.}\)
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