### Author Topic: Thanksgiving bonus 2  (Read 2055 times)

#### Victor Ivrii

• Elder Member
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##### Thanksgiving bonus 2
« on: October 06, 2018, 05:06:19 AM »
Read section 2.1.1* and solve the problem (10 karma points). As a sample see solutions to odd numbered problems

Problem 4. 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.
$$x^2-y^2+2ixy.$$

#### Junya Zhang

• Full Member
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• Karma: 29
##### Re: Thanksgiving bonus 2
« Reply #1 on: October 06, 2018, 11:58:38 AM »
Let $f(x,y) = u+iv = x^2-y^2 + 2ixy$.
Then
$$\bar{f}(z,y) = x^2 - y ^2 -2ixy$$ $$u(x,y)=x^2-y^2$$ $$v(x,y)=2xy$$
The given function represents a locally sourceless and irrotational flow since $\bar{f}$ is analytic on $\mathbb{C}$.
However, $f$ is neither globally sourceless nor globally irrotational.
$$\frac{\partial{v}}{\partial{x}} = 2y$$ $$\frac{\partial{u}}{\partial{y}} = -2y$$ $$\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}} = 4y$$
This shows that $f$ is not globally irrotational.
$$\frac{\partial{u}}{\partial{x}} = 2x$$ $$\frac{\partial{v}}{\partial{y}} = 2x$$ $$\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 4x$$
This shows that $f$ is not globally sourceless.
« Last Edit: October 06, 2018, 12:25:43 PM by Junya Zhang »