# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Thanksgiving bonus => Topic started by: Victor Ivrii on October 06, 2018, 05:07:03 AM

Title: Thanksgiving bonus 3
Post by: Victor Ivrii 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.
$$e^y\cos(x)+ie^y\sin(x).$$
Title: Re: Thanksgiving bonus 3
Post by: Junya Zhang 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.
Title: Re: Thanksgiving bonus 3
Post by: Victor Ivrii on October 06, 2018, 05:15:17 PM
No double dipping