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Mathematics LibreTexts

7.6: More Examples with Pretty Pictures

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    50842
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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 \(\PageIndex{1}\)).

    010 - (7.7.1).svg
    Figure \(\PageIndex{1}\): Linear vortex. (CC BY-NC; Ümit Kaya)

    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 \(\PageIndex{2}\)).

    011 - (7.7.2).svg
    Figure \(\PageIndex{2}\): Two sources. (CC BY-NC; Ümit Kaya)

    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.

    012 - (7.7.3).svg
    Figure \(\PageIndex{3}\): A source in uniform flow. (CC BY-NC; Ümit Kaya)

    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 \(\PageIndex{3}\)).

    013 - (7.7.4).svg
    Figure \(\PageIndex{4}\): A source plus a sink. (CC BY-NC; Ümit Kaya)

    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.