### Author Topic: Thanksgiving bonus 4  (Read 2839 times)

#### Victor Ivrii ##### Thanksgiving bonus 4
« on: October 06, 2018, 05:08:29 AM »
Read section 2.1.1* and solve the problem (10 karma points). As a sample see solutions to odd numbered problems

Problem 8. 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.
$$\cosh(x)\cos(y)+i\sinh(x)\sin(y).$$

#### Tianfangtong Zhang

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•   • Posts: 16
• Karma: 15 ##### Re: Thanksgiving bonus 4
« Reply #1 on: October 06, 2018, 10:32:20 PM »
$cosh(x)cos(y)+isinh(x)sin(y)$
Then
$u=cosh(x)cos(y)$
$v=sinh(x)sin(y)$
$\frac{du}{dx} = sinh(x)cos(y)$
$\frac{du}{dy}= -cosh(x)sin(y)$
$\frac{dv}{dx} = cosh(x)sin(y)$
$\frac{dv}{dy} = sinh(x)cos(y)$
since $\frac{du}{dx} \neq - \frac{dv}{dy}$
$\frac{du}{dy} \neq \frac{dv}{dx}$
Thus, the flow is not globally sourceless and is not irrotational.

#### Tianfangtong Zhang

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•   • Posts: 16
• Karma: 15 ##### Re: Thanksgiving bonus 4
« Reply #2 on: October 06, 2018, 10:58:53 PM »
sorry, there is a typo. The solution should be the given function doesn't represent a locally sourceless and an irrotational flow

#### Victor Ivrii ##### Re: Thanksgiving bonus 4
« Reply #3 on: October 07, 2018, 12:49:10 AM »
First, correct your typesetting, replacing cos by \cos and so on

Also not \frac{du}{dx} but \frac{\partial u}{\partial x} and so on (on Quiz/Test/Exam in any math discipline starting from Calculus II it will cost you.

On Page 92 (not in problems) there are 2 errors in formulae. Find them and type a correct version.

#### Min Gyu Woo

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• Karma: 12 ##### Re: Thanksgiving bonus 4
« Reply #4 on: October 07, 2018, 02:36:47 PM »
Let $f(u,v) = u + iv$ where $u = \cosh{x}\cos{y}$ and $v=\sinh{x}\sin{y}$

Note that

\begin{equation*}

\begin{aligned}

\frac{\partial u}{\partial x} &= \sinh{x}\cos{y}\\
\frac{\partial v}{\partial y} &= \sinh{x}\cos{y}\\
\frac{\partial u}{\partial y} &= -\cosh{x}\sin{y}\\
\frac{\partial v}{\partial x} &= \cosh{x}\sin{y}

\end{aligned}

\end{equation*}

Since

\begin{equation*}

\begin{aligned}

\frac{\partial u}{\partial x} &=  \frac{\partial v}{\partial y} \\

\frac{\partial u}{\partial y} &\neq -\frac{\partial v}{\partial x}

\end{aligned}

\end{equation*}

We know that $f$ is not analytic on $D$ by the contrapositve of Cauchy-Riemann Theorem.

So, $f$ is not sourceless or irrotational on $D$.

The error you are talking about is that the textbook used $dx$ instead of $\partial x$

Thus,  the proper equations should be

\begin{equation*}

\begin{aligned}

\int_{\gamma} u dx+v dy &= \iint_{\Omega} \left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right]dxdy=0 \\

\int_{\gamma} u dy - v dx &= \iint_{\Omega} \left[\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right]dxdy=2 \text{ area}(\Omega)

\end{aligned}

\end{equation*}