User:Egm6321.f10.team2.oztekin/hw7.2

R*7.2 - Heat Conduction on a Cylinder
From the lecture slide Mtg 40-6

Given
Coordinate equivalents given as in the lecture;


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$$\begin{align}

x=&r(cos(\theta ))=\xi _{1}(cos(\xi _{2}))\\ y=&r(sin(\theta ))=\xi _{1}(sin(\xi _{2}))\\ z=&\xi _{3}

\end{align}$$ (7.2.1)
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Find

 * Find
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$$\begin{align} \begin{matrix} \left \{ dx_{i} \right \}=\left \{ dx_{1},dx_{2},dx_{3} \right \} & \left \{ \xi _{j}\right \}=\left \{ \xi _{1},\xi _{2} ,\xi _{3}\right \} \end{matrix} \end{align}$$ (7.2.2)
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 * Find
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$$\begin{align} ds^{2}=\sum_{i}(dx_{i})^{2}=\sum_{k}(h_{k})^{2}(d\xi _{k})^{2} \end{align}$$ (7.2.3)
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 * Find $$ \Delta u $$ in cylindrical coordinates.


 * Use separation of variable to find the separated equations and compare to the bessel eq. (1)

Solution

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$$\begin{align}

dx_{1}=&\left [ \frac{\partial x_{1}}{\partial \xi _{1}} \right ]d\xi _{1}+\left [ \frac{\partial x_{1}}{\partial \xi _{2}} \right ]d\xi _{2}+\left [ \frac{\partial x_{1}}{\partial \xi _{3}} \right ]d\xi _{3}\\ dx_{2}=&\left [ \frac{\partial x_{2}}{\partial \xi _{1}} \right ]d\xi _{1}+\left [ \frac{\partial x_{2}}{\partial \xi _{2}} \right ]d\xi _{2}+\left [ \frac{\partial x_{2}}{\partial \xi _{3}} \right ]d\xi _{3}\\ dx_{3}=&\left [ \frac{\partial x_{3}}{\partial \xi _{1}} \right ]d\xi _{1}+\left [ \frac{\partial x_{3}}{\partial \xi _{2}} \right ]d\xi _{2}+\left [ \frac{\partial x_{3}}{\partial \xi _{3}} \right ]d\xi _{3}

\end{align}$$ (7.2.4)
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$$\begin{align} dx_{1}=&\frac{\partial \xi _{1}}{\partial \xi _{1}}cos(\xi _{2})d\xi _{1}+\xi _{1}\frac{\partial cos(\xi _{2})}{\partial \xi _{2}}=cos(\xi _{2})d\xi _{1}-\xi _{1}sin(\xi _{2})d\xi _{2}\\ dx_{2}=&\frac{\partial \xi _{1}}{\partial \xi _{1}}sin(\xi _{2})d\xi _{1}+\xi _{1}\frac{\partial sin(\xi _{2})}{\partial \xi _{2}}=sin(\xi _{2})d\xi _{1}+\xi _{1}cos(\xi _{2})d\xi _{2}\\ dx_{3}=&\frac{\partial \xi _{3}}{\partial \xi _{3}}d\xi _{3}=d\xi _{3}\\

\end{align}$$ (7.2.5)
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$$\begin{align} ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} \end{align}$$ (7.2.6)
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$$\begin{align} ds^{2}={\color{Green} (cos\xi _{2})^{2}d\xi _{1}^{2}}+{\color{Blue} \xi _{1}^{2}(sin\xi _{2})^{2}d\xi _{2}^{2}}-{\color{Red} 2\xi _{1}(cos\xi _{2})(sin\xi _{2})d\xi _{1}d\xi _{2}}+{\color{Green} (sin\xi _{2})^{2}d\xi _{1}^{2}}+{\color{Blue} \xi _{1}^{2}(cos\xi _{2})^{2}d\xi _{2}^{2}}+{\color{Red} 2\xi _{1}(cos\xi _{2})(sin\xi _{2})d\xi _{1}d\xi _{2}}+d\xi _{3}^{2} \end{align}$$ (7.2.7)
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$$  \displaystyle ds^{2}=d\xi _{1}^{2}+\xi _{1}^{2}d\xi _{2}^{2}+d\xi _{3}^{2} $$


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$$\begin{align} ds^{2}=\sum_{i}(dx_{i})^{2}=\sum_{k}(h_{k})^{2}(d\xi _{k})^{2}=d\xi _{1}^{2}+\xi _{1}^{2}d\xi _{2}^{2}+d\xi _{3}^{2} \end{align}$$ (7.2.8)
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Equation yields


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$$\begin{align} &h_{1}=1\\ &h_{2}=\xi _{1}\\ &h_{3}=1

\end{align}$$ (7.2.9)
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$$\begin{align} h_{1}h_{2}h_{3}=\xi _{1} \end{align}$$ (7.2.10)
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$$\begin{align} \Delta u =\frac{1}{h_{1}h_{2}h_{3}}\sum_{j=1}^{3}\frac{\partial }{\partial \xi _{j}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{j}^{2}} \frac{\partial u }{\partial \xi _{j}}\right ] \end{align}$$ (7.2.11)
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$$\begin{align} &j=1\\ &\frac{\partial }{\partial \xi _{1}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{1}^{2}} \frac{\partial u }{\partial \xi _{1}}\right ]=\frac{\partial }{\partial \xi _{1}}\left [ \frac{\xi _{1}}{(1)^{2}} \frac{\partial u }{\partial \xi _{1}}\right ] \end{align}$$ (7.2.12)
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$$\begin{align} &j=2\\ &\frac{\partial }{\partial \xi _{2}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{2}^{2}} \frac{\partial u }{\partial \xi _{2}}\right ]=\frac{\partial }{\partial \xi _{2}}\left [ \frac{\xi _{1}}{(\xi _{1})^{2}} \frac{\partial u }{\partial \xi _{2}}\right ] \end{align}$$ (7.2.13)
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$$\begin{align} &j=3\\ &\frac{\partial }{\partial \xi _{3}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{3}^{2}} \frac{\partial u }{\partial \xi _{3}}\right ]=\frac{\partial }{\partial \xi _{3}}\left [ \frac{\xi _{1}}{(1)^{2}} \frac{\partial u }{\partial \xi _{3}}\right ] \end{align}$$ (7.2.14)
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If we substitute in we will get

$$  \displaystyle \Delta u =\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial u }{\partial \xi _{1}} \right )+\frac{1}{\xi _{1}^{2}}\frac{\partial^2 u }{\partial \xi _{2}^2}+\frac{\partial^2 u }{\partial \xi _{3}^2} $$


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$$\begin{align} u=R(\xi _{1})\theta (\xi _{2})Z(\xi _{3}) \end{align}$$ (7.2.15)
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$$\begin{align} \Delta u=\frac{\theta Z}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi_{1} \frac{\partial R}{\partial \xi _{1}} \right )+\frac{RZ}{\xi _{1}^{2}}\frac{\partial^2 \theta }{\partial \xi_{2} ^2}+R\theta \frac{\partial^2 Z}{\partial \xi _{3}^2}=0 \end{align}$$ (7.2.16)
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Dividing through $$ R\theta Z $$


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$$\begin{align} \Delta u=\frac{1}{R\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi_{1} \frac{\partial R}{\partial \xi _{1}} \right )+\frac{1}{\theta \xi _{1}^{2}}\frac{\partial^2 \theta }{\partial \xi_{2} ^2}+\frac{1}{Z} \frac{\partial^2 Z}{\partial \xi _{3}^2}=0 \end{align}$$ (7.2.17)
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$$\begin{align} \frac{1}{R\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi_{1} \frac{\partial R}{\partial \xi _{1}} \right )+\frac{1}{\theta \xi _{1}^{2}}\frac{\partial^2 \theta }{\partial \xi_{2} ^2}=-\frac{1}{Z} \frac{\partial^2 Z}{\partial \xi _{3}^2}=-\lambda ^{2} \end{align}$$ (7.2.18)
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$$\begin{align} \frac{\partial^2 Z}{\partial \xi _{3}^2}+\lambda ^{2}Z=0 \end{align}$$ (7.2.19)
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$$\begin{align} \frac{\xi _{1}}{R}\frac{\partial }{\partial \xi _{1}}\left ( \xi_{1} \frac{\partial R}{\partial \xi _{1}} \right )+\frac{1}{\theta}\frac{\partial^2 \theta }{\partial \xi_{2} ^2}+\xi _{1}^{2}\lambda ^{2}=0 \end{align}$$ (7.2.20)
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$$\begin{align} \frac{\xi _{1}}{R}\frac{\partial }{\partial \xi _{1}}\left ( \xi_{1} \frac{\partial R}{\partial \xi _{1}} \right )+\xi _{1}^{2}\lambda ^{2}=-\frac{1}{\theta}\frac{\partial^2 \theta }{\partial \xi_{2} ^2}=\eta ^{2} \end{align}$$ (7.2.21)
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$$\begin{align} \frac{\partial^2 \theta }{\partial \xi_{2} ^2}+\eta ^{2}\theta=0 \end{align}$$ (7.2.22)
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$$\begin{align} \xi _{1}\frac{\partial }{\partial \xi _{1}}\left ( \xi_{1} \frac{\partial R}{\partial \xi _{1}} \right )+(\xi _{1}^{2}\lambda ^{2}-\eta ^{2})R=0 \end{align}$$ (7.2.23)
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$$\begin{align} \xi _{1}\frac{\partial R}{\partial \xi _{1}}+\xi _{1}^{2}\frac{\partial^2 R}{\partial \xi _{1}^2}+(\xi _{1}^{2}\lambda ^{2}-\eta ^{2})R=0 \end{align}$$ (7.2.24)
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If we change $$ \xi _{1}\lambda =y $$ we will get bessel's equation

$$  \displaystyle y^{2}\frac{\partial^2 R}{\partial y^{2}}+y\frac{\partial R}{\partial y}+(y^{2}-\eta ^{2})R=0 $$


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$$\begin{align} y^{2}=1-t^{2} \end{align}$$ (7.2.25)
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$$\begin{align} \frac{\partial R}{\partial y}=\frac{\partial R}{\partial t}\frac{\partial t}{\partial y} \end{align}$$ (7.2.26)
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$$\begin{align} \frac{\partial R}{\partial y}=\frac{\partial R}{\partial t}\left ( -\frac{y}{\sqrt{1-y^{2}}} \right ) \end{align}$$ (7.2.27)
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$$\begin{align} \frac{\partial^{2} R}{\partial y^{2}}=\frac{\partial }{\partial t}\left ( \frac{\partial R}{\partial t} \left ( -\frac{y}{\sqrt{1-y^{2}}} \right )\right )\frac{\partial t}{\partial y} \end{align}$$ (7.2.28)
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This transformation will lead us to find the bessel equation shown in meeting 27-1.

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