User:Egm6321.f11.team3/Hwk7

R*7.1 - Finding ds and Laplace Operator Equivalent in Spherical Coordinates
From the lecture slide Mtg 39-1

Given
Infinitesimal length in Cartesian coordinates given as in Mtg 38-5 ;


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$$\begin{align} ds=dx_{j}.e_{j} \end{align}$$ (7.1.1)
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$$\begin{align} ds^{2}=ds.ds=\left ( dx_{j}.e_{j} \right )\left ( dx_{i}e_{i} \right )=dx_{i}.dx_{j}(e_{i}e_{j}) \end{align}$$ (7.1.2)
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$$\begin{align} e_{i}e_{j}=\delta _{ij} \end{align}$$ (7.1.3)
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Delta function defined as ;


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$$\begin{align} \delta _{ij}=\left \{ \begin{matrix} 1 & for & i=j\\ 0& for  & i\neq j \end{matrix} \right. \end{align}$$ (7.1.4)
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$$\begin{align} ds^{2}=dx_{i}.dx_{i}=\sum_{i=1}^{3}\left ( dx_{i} \right )^{2} \end{align}$$ (7.1.5)
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$$\begin{align} &x_{1}=x=rcos(\theta )cos(\phi )=\xi _{1}cos(\xi _{2})cos(\xi _{3}) \\ &x_{2}=y=rcos(\theta )sin(\phi )=\xi _{1}cos(\xi _{2})sin(\xi _{3}) \\ &x_{3}=z=rsin(\theta )=\xi _{1}sin(\xi _{2}) \\

\end{align}$$ (7.1.6)
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$$\begin{align}

\Delta u =\left [ \frac{1}{h_{1}h_{2}h_{3}} \right ]\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial u }{\partial \xi _{i}}\right ]

\end{align}$$ (7.1.7)
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$$\begin{align} \begin{matrix} \hat{\xi }=(\hat{\xi _{1}},\hat{\xi _{2}},\hat{\xi _{3}}) \end{matrix} \end{align}$$ (7.1.8)
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$$\begin{align} \begin{matrix} h_{1}h_{2}h_{3}=r^{2}cos(\theta ) \end{matrix} \end{align}$$ (7.1.9)
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Find

 * Show that infinitesimal length 'ds' can be written as (5) in meeting 39-1 in spherical coordinates.
 * Derive Laplace operator in spherical coordinates.

Solution

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$$\begin{align} dx_{1}=\left [ \frac{\partial x_{1}}{\partial r} \right ]dr+\left [ \frac{\partial x_{1}}{\partial \theta } \right ]d\theta +\left [ \frac{\partial x_{1}}{\partial \phi } \right ]d\phi

\end{align}$$ (7.1.10)
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$$\begin{align} dx_{2}=\left [ \frac{\partial x_{2}}{\partial r} \right ]dr+\left [ \frac{\partial x_{2}}{\partial \theta } \right ]d\theta +\left [ \frac{\partial x_{2}}{\partial \phi } \right ]d\phi \end{align}$$ (7.1.11)
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$$\begin{align} dx_{3}=\left [ \frac{\partial x_{3}}{\partial r} \right ]dr+\left [ \frac{\partial x_{3}}{\partial \theta } \right ]d\theta +\left [ \frac{\partial x_{3}}{\partial \phi } \right ]d\phi \end{align}$$ (7.1.12)
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$$\begin{align} dx_{1}=\frac{\partial r}{\partial r}cos(\theta )cos(\phi )dr+r\frac{\partial cos(\theta )}{\partial \theta }cos(\phi )d\theta +rcos(\theta )\frac{\partial cos(\phi )}{\partial \phi }d\phi

\end{align}$$ (7.1.13)
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$$\begin{align} dx_{2}=\frac{\partial r}{\partial r}cos(\theta )sin(\phi )dr+r\frac{\partial cos(\theta )}{\partial \theta }sin(\phi )d\theta +rcos(\theta )\frac{\partial sin(\phi )}{\partial \phi }d\phi \end{align}$$ (7.1.14)
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$$\begin{align} dx_{3}=\frac{\partial r}{\partial r}sin\theta dr+r\frac{\partial sin(\theta )}{\partial \theta }d\theta \end{align}$$ (7.1.15)
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$$\begin{align} dx_{1}^{2}=&(cos\theta)^{2} (cos\phi)^{2} dr^{2}+r^{2}(sin\theta)^{2} (cos\phi )^{2}d\theta ^{2}+r^{2}(cos\theta)^{2} (sin\phi)^{2} d\phi ^{2}+2r^{2}sin\phi cos\phi (sin\theta) (cos\theta) d\theta d\phi \\ &-2r(cos\theta) (cos\phi)^{2} (sin\theta) drd\theta -2r(cos\theta) (cos\phi) (sin\phi) drd\phi

\end{align}$$ (7.1.16)
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$$\begin{align} dx_{2}^{2}=&(cos\theta)^{2} (sin\phi)^{2} dr^{2}+r^{2}(sin\theta)^{2} (sin\phi )^{2}d\theta^{2} +r^{2}(cos\theta)^{2} (cos\phi)^{2} d\phi ^{2}+2r(cos\theta )^{2}(sin\phi) (cos\phi) drd\phi \\ &-2r^{2}(sin\theta) (sin\phi) (cos\theta) (cos\phi )d\theta d\phi -2r(cos\theta) (sin\theta) sin^{2}\phi drd\theta \end{align}$$ (7.1.17)
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$$\begin{align} dx_{3}^{2}=(sin\theta)^{2} dr^{2}+r^{2}(cos\theta )^{2}d\theta ^{2}+2r(sin\theta ) (cos\theta) drd\theta \end{align}$$ (7.1.18)
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$$\begin{align} ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} \end{align}$$ (7.1.19)
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$$\begin{align} ds^{2}=&{\color{Green} (cos\theta)^{2} (cos\phi)^{2} dr^{2}}+{\color{Blue} r^{2}(sin\theta)^{2} (cos\phi)^{2} d\theta ^{2}}+{\color{Cyan} r^{2}(cos\theta)^{2} (sin\phi)^{2} d\phi ^{2}}+{\color{Green} (cos\theta)^{2} (sin\phi)^{2} dr^{2}}+{\color{Blue} r^{2}(sin\theta)^{2} (sin\phi)^{2} d\theta^{2}}\\ & +{\color{Cyan} r^{2}(cos\theta)^{2} (cos\phi)^{2} d\phi ^{2}}+(sin\theta)^{2} dr^{2}+r^{2}(cos\theta)^{2} d\theta ^{2}+{\color{Red} 2r^{2}sin\phi cos\phi sin\theta cos\theta d\theta d\phi} -2rcos\theta (cos\phi)^{2} sin\theta drd\theta \\ & -{\color{Orange} 2r(cos\theta)^{2} cos\phi sin\phi drd\phi}+{\color{Orange} 2r(cos\theta)^{2} sin\phi cos\phi drd\phi }-{\color{Red} 2r^{2}sin\theta sin\phi cos\theta cos\phi d\theta d\phi} -2rcos\theta sin\theta sin^{2}\phi drd\theta \\ &+2rsin\theta cos\theta drd\theta \end{align}$$ (7.1.20)
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If we group things together we will get ;

$$  \displaystyle ds^{2}=1.dr^{2}+r^{2}d\theta ^{2}+r^{2}(cos\theta)^{2} d\phi ^{2} $$


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$$\begin{align} &i=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 r}\left [ \frac{r^{2}cos(\theta )}{(1)^{2}} \frac{\partial u }{\partial r}\right ]

\end{align}$$

(7.1.21)
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$$\begin{align} &i=2\\ &\frac{\partial }{\partial \xi _{2}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{2}^{2}} \frac{\partial u }{\partial \xi _{2}}\right ]=\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}} \frac{\partial u }{\partial \theta }\right ] \end{align}$$ (7.1.22)
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$$\begin{align} &i=3\\ &\frac{\partial }{\partial \xi _{3}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{3}^{2}} \frac{\partial u }{\partial \xi _{3}}\right ]=\frac{\partial }{\partial \phi }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}(cos\theta )^{2}} \frac{\partial u }{\partial \phi }\right ] \end{align}$$ (7.1.23)
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If we substitute in


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$$\begin{align} \Delta u =\frac{1}{r^{2}cos(\theta )}\left [ \frac{\partial }{\partial r}\left [ \frac{r^{2}cos(\theta )}{(1)^{2}} \frac{\partial u }{\partial r}\right ]+\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}} \frac{\partial u }{\partial \theta }\right ]+\frac{\partial }{\partial \phi }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}(cos\theta )^{2}} \frac{\partial u }{\partial \phi }\right ] \right ] \end{align}$$ (7.1.24)
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$$  \displaystyle \Delta u =\frac{1}{r^{2}}\left ( r^{2}\frac{\partial u }{\partial r} \right )+\frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left ( cos\theta \frac{\partial u }{\partial \theta } \right )+\frac{1}{r^{2}(cos\theta)^{2} }\frac{\partial ^2 u }{\partial \phi ^2} $$

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.

Find
Find $$\displaystyle \Delta u$$ in spherical coordinates using the math/physics convention.

Solution
According to lecture note 38-6, we can get the following equation Put equation (7.3.13) into equation (7.3.1).

Find
Verify the Laplacian in elliptic coordinates given by

Solution
According to Wikipedia

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu\in [0,2\pi]$$

According to hyperbolic trigonometric identity Put equation (7.4.10) into (7.4.9). Put equation (7.4.12) into (7.4.1).

Given
Parabolic_coordinates

Find
Verify the Laplacian in parabolic corrdinates given by:

Solution
Slove by ourselves

Plunge into equation 7.5.2:

Find
Verify that 7.6.1 thru 7.6.4 are homogeneous solutions of Legendre equation 7.6.6. Show that 7.6.5 can also be written as 7.6.7. Verify that 7.6.1 through 7.6.4 can be obtained from 7.6.5 or 7.6.7.

Solution
We solved this problem on our own. To verify 7.6.1 through 7.6.4 are homogeneous solutions of 7.6.6, from King we first find the first and second derivatives of each equation. Now we plug these solutions into 7.6.6 in place of the y terms for the respective values of n. $$\displaystyle L_2(P_0)=(1-x^2)0-2x(0)+0(0+1)1=0$$ $$\displaystyle L_2(P_1)=(1-x^2)0-2x(1)+1(1+1)x=-2x+2x=0$$ $$\displaystyle L_2(P_2)=(1-x^2)3-2x(3x)+2(2+1)(\frac{1}{2}(3x^2-1))=3-3x^2-6x^2+9x^2-3=0$$ $$\displaystyle L_2(P_3)=(1-x^2)15x-2x(\frac{1}{2}(15x^2-3))+3(3+1)(\frac{1}{2}(5x^3-3x))=15x-15x^3-15x^3+3x+30x^3-18x=0$$ $$\displaystyle L_2(P_4)=(1-x^2)(\frac{105}{2}x^2-\frac{15}{2})-2x(\frac{35}{2}x^3-\frac{15}{2}x)+4(4+1)(\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8})=\frac{105}{2}x^2-\frac{15}{2}-\frac{105}{2}x^4+\frac{15}{2}x^2-35x^4+15x^2+\frac{105}{2}x^4-75x^2+\frac{15}{2}=0$$ Since all terms cancel to zero, the solutions are valid. The equations 7.6.1 through 7.6.4 can be attained from 7.6.5 as shown below,

Find
Find the separated equation for the Laplace equation

Solution
Slove by ourselves

Divided by u:

Find
Show that for n=0: $$\mu \to \pm 1 \Rightarrow \left|Q_0(\mu) \right| \to+ \infty$$

Plot the following Legendre polynomials and functions, $$\displaystyle {P_0,P_1,P_2,P_3} $$ $$\displaystyle {Q_0,Q_1,Q_2,Q_3} $$ Observe the limits of $$P_n(\mu)$$ and $$Q_n(\mu)$$ as $$\mu \to \pm 1$$

Solution
We solved this problem on our own and reviewed old homework afterwards for comparison. We find the first four Legendre polynominials and functions from the Mathworld Wolfram site. They are, $$ \displaystyle P_0(x)=1 $$ $$ \displaystyle P_1(x)=x $$ $$ \displaystyle P_2(x)=\frac{1}{2}(3x^2-1) $$ $$ \displaystyle P_3(x)=\frac{1}{2}(5x^3-3x) $$ $$ \displaystyle Q_0(x)=\frac{1}{2} ln{\frac{1+x}{1-x}} $$ $$ \displaystyle Q_1(x)=\frac{x}{2} ln{\frac{1+x}{1-x}}-1 $$ $$ \displaystyle Q_2(x)=\frac{3x^2-1}{4} ln{\frac{1+x}{1-x}}-\frac{3x}{2} $$ $$ \displaystyle Q_3(x)=\frac{5x^3-3x}{4} ln{\frac{1+x}{1-x}}-\frac{5x^2}{2}+\frac{2}{3} $$ Observe that when x goes to 1, (1-x) is getting smaller. Thus, $$ \displaystyle \lim_{x\rightarrow 1}Q_0(x)=\frac{1}{2} ln{\frac{1+x}{1-x}}=\infty $$ The same situation happens when x goes -1. We now plot each function using the following Matlab script and see that each one becomes asymptotic at +/-1.

clear all; x=-1:.01:1; P0=1; P1=x; P2=.5.*(3.*x.^2-1); P3=.5.*(5.*x.^3-3*x); Q0=.5*log((1+x)./(1-x)); Q1=(x/2).*log((1+x)./(1-x))-1; Q2=(3*x.^2-1)/4.*log((1+x)./(1-x))-(3*x)/2; Q3=(5*x.^3-3*x)/4.*log((1+x)./(1-x))-(5*x.^2)/2+(2/3); figure(1) plot(x,P0,'-',x,P1,'^',x,P2,'*',x,P3,'+') legend('P0','P1','P2','P3') figure(2) plot(x,Q0,'-',x,Q1,'^',x,Q2,'*',x,Q3,'+') legend('Q0','Q1','Q2','Q3')

Given
Consider the non-orthonormal basis $$\{\mathbf b_1,\mathbf b_2,\mathbf b_3\}$$ expressed in the orthonormal basis $$\{\mathbf e_1,\mathbf e_2,\mathbf e_3\}$$ as follows

$$ \mathbf b_i = A_{ij} \mathbf e_i $$

where: $$\displaystyle \mathbf A = [A_{ij}] = \left[ \begin{matrix} 5&2&3\\ 4&5&6\\ 7&8&5 \end{matrix} \right] $$

$$ \mathbf v = -2\mathbf e_1 + 4\mathbf e_2 - 5\mathbf e_3 $$

Find
$${\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that

$$ \mathbf{v}=v_i\mathbf b_i$$

Solution
Solved on our own.

From the given data,

$$ \mathbf b_i = \mathbf A_{ij} \mathbf e_i $$

We need to find $$ \mathbf v_i $$ such that $$ \mathbf{v}=v_i\mathbf b_i$$

So, $$ \mathbf v = \mathbf v_i \mathbf A_{ij} \mathbf e_i $$

We also know that $$ \mathbf v = -2\mathbf e_1 + 4\mathbf e_2 - 5\mathbf e_3 $$

So, $$ \left[ \begin{matrix} -2\\ 4\\ -5 \end{matrix} \right] = \left[ \begin{matrix} v_1\\ v_2\\ v_3 \end{matrix} \right] \mathbf A^T $$

$$\mathbf det(A) = 80 $$

Hence, $$ \left[ \begin{matrix} v_1\\ v_2\\ v_3 \end{matrix} \right] = \mathbf A^{-T} \left[ \begin{matrix} -2\\ 4\\ -5 \end{matrix} \right] $$

and $$\displaystyle \mathbf A^{-T} = \frac{1}{80}\left[ \begin{matrix} 23&-22&3\\-14&-4&26\\3&18&-17 \end{matrix} \right] $$

Therefore, $$ \mathbf v_i = \left[ \begin{matrix} -1.8625\\ -1.475\\ 1.8875 \end{matrix} \right] $$

= References =

= Contributing Members =

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