Author Topic: Thanksgiving bonus 3  (Read 553 times)

Victor Ivrii

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Thanksgiving bonus 3
« on: October 06, 2018, 05:07:03 AM »
Read section 2.1.1* and solve the problem (10 karma points). As a sample see solutions to odd numbered problems

Problem 6. Decide whether or not the given function represents a locally sourceless and/or irrotational flow. For those that do, decide whether the flow is globally sourceless and/or irrotational. Sketch some of the streamlines.
$$
e^y\cos(x)+ie^y\sin(x).
$$
« Last Edit: October 06, 2018, 05:14:53 PM by Victor Ivrii »

Junya Zhang

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Re: Thanksgiving bonus 3
« Reply #1 on: October 06, 2018, 12:30:41 PM »
Let $f(x,y) = u+iv = e^y\cos(x)+ie^y\sin(x)$.
Then
$$\bar{f}(z,y) = e^y\cos(x)-ie^y\sin(x)$$ $$u(x,y)=e^y\cos(x)$$ $$v(x,y)=e^y\sin(x)$$
The given function represents a locally sourceless and irrotational flow since $\bar{f}$ is analytic on $\mathbb{C}$.

Note that
$$\frac{\partial{v}}{\partial{x}} = e^y\cos(x)$$ $$\frac{\partial{u}}{\partial{y}} = e^y\cos(x)$$ $$\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}} = 0$$
This shows that $f$ is globally irrotational.
$$\frac{\partial{u}}{\partial{x}} = -e^y\sin(x)$$ $$\frac{\partial{v}}{\partial{y}} = e^y\sin(x)$$ $$\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0$$
This shows that $f$ is globally sourceless.
So $f$ is both globally sourceless and globally irrotational.

Victor Ivrii

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Re: Thanksgiving bonus 3
« Reply #2 on: October 06, 2018, 05:15:17 PM »
No double dipping