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# 7.6: More Examples with Pretty Pictures

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## Example $$\PageIndex{1}$$: Linear Vortex

Analyze the flow with complex potential function

$\Phi (z) = i \log (z). \nonumber$

Solution

Multiplying by $$i$$ switches the real and imaginary parts of $$\log (z)$$ (with a sign change). We have

$\Phi = -\theta + i \log (r). \nonumber$

The stream lines are the curves $$\log (r)$$ = constant, i.e. circles with center at $$z = 0$$. The flow is clockwise because the potential $$\phi = -\theta$$ increases in the clockwise direction (Figure $$\PageIndex{1}$$).

This flow is called a linear vortex. We can find $$F$$ using $$\Phi '$$.

$\Phi ' = \dfrac{i}{z} = \dfrac{y}{r^2} + i \dfrac{x}{r^2} = \phi_x - i\phi_y. \nonumber$

So

$F = (\phi_x, \phi_y) = (y/r^2, -x/r^2). \nonumber$

(By now this should be a familiar vector field.) There are no stagnation points, but there is a singularity at the origin.

## Example $$\PageIndex{2}$$: Double Source

Analyze the flow with complex potential function

$\Phi (z) = \log (z - 1) + \log (z + 1). \nonumber$

Solution

This is a flow with linear sources at $$\pm 1$$ with the level curves of $$\psi = \text{Im} (\Phi)$$ (Figure $$\PageIndex{2}$$).

We can analyze this flow further as follows.

• Near each source the flow looks like a linear source.
• On the $$y$$-axis the flow is along the axis. That is, the $$y$$-axis is a streamline. It’s worth seeing three different ways of arriving at this conclusion.

Reason 1: By symmetry of vector fields associated with each linear source, the $$x$$ components cancel and the combined field points along the $$y$$-axis.

Reason 2: We can write

$\Phi (z) = \log (z - 1) + \log (z + 1) = \log ((z - 1)(z + 1)) = \log (z^2 - 1). \nonumber$

So

$\Phi '(z) = \dfrac{2z}{z^2 - 1}. \nonumber$

On the imaginary axis

$\Phi ' (iy) = \dfrac{2iy}{-y^2 - 1}. \nonumber$

Thus,

$F = (0, \dfrac{2y}{y^2 + 1}) \nonumber$

which is along the axis.

Reason 3: On the imaginary axis $$\Phi (iy) = \log (-y^2 - 1)$$. Since this has constant imaginary part, the axis is a streamline.

Because of the branch cut for $$\log (z)$$ we should probably be a little more careful here. First note that the vector field $$F$$ comes from $$\Phi ' = 2z/(z^2 - 1)$$, which doesn’t have a branch cut. So we shouldn’t really have a problem. Now, as $$z$$ approaches the $$y$$-axis from one side or the other, the argument of $$\log (z^2 - 1)$$ approaches either $$\pi$$ or $$-\pi$$. That is, as such limits, the imaginary part is constant. So the streamline on the $$y$$-axis is the limit case of streamlines near the axis.

Since $$\Phi '(z) = 0$$ when $$z = 0$$, the origin is a stagnation point. This is where the fields from the two sources exactly cancel each other.

## Example $$\PageIndex{3}$$: A Source in Uniform Flow

Consider the flow with complex potential

$\Phi (z) = z + \dfrac{Q}{2\pi} \log (z). \nonumber$

This is a combination of uniform flow to the right and a source at the origin (Figure $$\PageIndex{2}$$). It shows that the flow looks like a source near the origin. Farther away from the origin the flow stops being radial and is pushed to the right by the uniform flow.

Since the components of $$\Phi '$$ and $$F$$ are the same except for signs, we can understand the flow by considering

$\Phi '(z) = 1 + \dfrac{Q}{2\pi z}.$

Near $$z = 0$$ the singularity of $$1/z$$ is most important and

$\Phi ' \approx \dfrac{Q}{2\pi z}.$

So, the vector field looks a linear source. Far away from the origin the $$1/z$$ term is small and $$\Phi ' (z) \approx 1$$, so the field looks like uniform flow.

Setting $$\Phi '(z) = 0$$ we find one stagnation point

$z = -\dfrac{Q}{2\pi}.$

It is the point on the $$x$$-axis where the flow from the source exactly balances that from the uniform flow. For bigger values of $$Q$$ the source pushes fluid farther out before being overwhelmed by the uniform flow. That is why $$Q$$ is called the source strength.

## Example $$\PageIndex{4}$$: Source + Sink

Consider the flow with complex potential

$\Phi (z) = \log (z - 2) - \log (z + 2). \nonumber$

This is a combination of source $$(\log (z - 2))$$ at $$z = 2$$ and a sink $$(- \log (z + 2))$$ at $$z = -2$$ (Figure $$\PageIndex{3}$$).

7.6: More Examples with Pretty Pictures is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.