User:Egm6321.f11.team3/Hwk7

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R*7.1 - Finding ds and Laplace Operator Equivalent in Spherical Coordinates[edit]

From the lecture slide Mtg 39-1 [1]

Given[edit]

Infinitesimal length in Cartesian coordinates given as in Mtg 38-5 [2];

\begin{align}
ds=dx_{j}.e_{j}
\end{align}

(7.1.1)

\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)


\begin{align}
e_{i}e_{j}=\delta _{ij}
\end{align}

(7.1.3)

Delta function defined as ;

\begin{align}
\delta _{ij}=\left \{ \begin{matrix}
1 &  for & i=j\\ 
0& for   & i\neq j
\end{matrix} \right.
\end{align}

(7.1.4)

\begin{align}
ds^{2}=dx_{i}.dx_{i}=\sum_{i=1}^{3}\left ( dx_{i} \right )^{2}
\end{align}

(7.1.5)

\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)

\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)

\begin{align}
\begin{matrix}
\hat{\xi }=(\hat{\xi _{1}},\hat{\xi _{2}},\hat{\xi _{3}})
\end{matrix}
\end{align}

(7.1.8)

\begin{align}
\begin{matrix}
h_{1}h_{2}h_{3}=r^{2}cos(\theta )
\end{matrix}
\end{align}

(7.1.9)


Find[edit]

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


  • Derive Laplace operator in spherical coordinates.


Solution[edit]

\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)

\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)

\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)



\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)

\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)

\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)

\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)

\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)

\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)

\begin{align}
ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}
\end{align}

(7.1.19)

\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)

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}


\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)

\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)

\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)

If we substitute in

\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)



   \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[edit]

From the lecture slide Mtg 40-6 [3]

Given[edit]

Coordinate equivalents given as in the lecture;

\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)


Find[edit]

  • Find

\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)

  • Find

\begin{align}
ds^{2}=\sum_{i}(dx_{i})^{2}=\sum_{k}(h_{k})^{2}(d\xi _{k})^{2}
\end{align}

(7.2.3)

  • Find  \Delta u in cylindrical coordinates.
  • Use separation of variable to find the separated equations and compare to the bessel eq. (1) [4]

Solution[edit]

\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)

\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)

\begin{align}
ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}
\end{align}

(7.2.6)


\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)



   \displaystyle
   ds^{2}=d\xi _{1}^{2}+\xi _{1}^{2}d\xi _{2}^{2}+d\xi _{3}^{2}


\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)

Equation yields

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

\end{align}

(7.2.9)


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

(7.2.10)

\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)

\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)

\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)

\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)

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}



\begin{align}
u=R(\xi _{1})\theta (\xi _{2})Z(\xi _{3})
\end{align}

(7.2.15)

\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)

Dividing through  R\theta Z

\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)

\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)

\begin{align}
\frac{\partial^2 Z}{\partial \xi _{3}^2}+\lambda ^{2}Z=0
\end{align}

(7.2.19)

\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)

\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)

\begin{align}
\frac{\partial^2 \theta }{\partial \xi_{2} ^2}+\eta ^{2}\theta=0
\end{align}

(7.2.22)

\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)

\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)

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


\begin{align}
y^{2}=1-t^{2}
\end{align}

(7.2.25)

\begin{align}
\frac{\partial R}{\partial y}=\frac{\partial R}{\partial t}\frac{\partial t}{\partial y}
\end{align}

(7.2.26)

\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)

\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)

This transformation will lead us to find the bessel equation shown in meeting 27-1.

R*7.3 Find \displaystyle \Delta u in spherical coordinates[edit]


Given[edit]


\displaystyle 
\Delta u =\frac{1}{h_{1}h_{2}h_{3}}\sum^{3}_{i=1}\frac{\partial}{\partial \xi_{i}}[\frac{h_{1}h_{2}h_{3}}{(h_{i}^{2})}\frac{\partial u}{\partial \xi_{i}}]

(7.3.1)



Find[edit]


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

Solution[edit]


According to lecture note 38-6, we can get the following equation

\displaystyle 
(\xi_{1},\xi_{2},\xi_{3})=(r,\bar\theta,\varphi)

(7.3.2)


\displaystyle 
x_{1}=x=r\sin\bar \theta \cos\theta=\xi_{1}\sin\xi_{2}\cos\xi_{3}

(7.3.2)


\displaystyle 
x_{2}=y=r\sin\bar \theta \sin\theta=\xi_{1}\sin\xi_{2}\sin\xi_{3}

(7.3.3)


\displaystyle 
x_{3}=z=r\cos\bar\theta=\xi_{1}\cos\xi_{2}

(7.3.4)


\displaystyle 
dx_{1}=\frac{\xi_{1}\sin\xi_{2}\cos\xi_{3}}{d\xi_{1}}\frac{d\xi_{1}}{dx_{1}}+\frac{\xi_{1}\sin\xi_{2}\cos\xi_{3}}{d\xi_{2}}\frac{d\xi_{2}}{dx_{1}}+\frac{\xi_{1}\sin\xi_{2}\cos\xi_{3}}{d\xi_{3}}\frac{d\xi_{3}}{dx_{1}}

(7.3.5)


\displaystyle 
=\sin\xi_{2}\cos\xi_{3}d\xi_{1}+\xi_{1}\cos\xi_{2}\cos\xi_{3}d\xi_{2}-\xi_{1}\sin\xi_{2}\sin\xi_{3}d\xi_{3}


\displaystyle 
dx_{2}=\frac{\xi_{1}\sin\xi_{2}\sin\xi_{3}}{d\xi_{1}}\frac{d\xi_{1}}{dx_{2}}+\frac{\xi_{1}\sin\xi_{2}\sin\xi_{3}}{d\xi_{2}}\frac{d\xi_{2}}{dx_{2}}+\frac{\xi_{1}\sin\xi_{2}\sin\xi_{3}}{d\xi_{3}}\frac{d\xi_{3}}{dx_{2}}

(7.3.6)


\displaystyle 
=\sin\xi_{2}\sin\xi_{3}d\xi_{1}+\xi_{1}\cos\xi_{2}\sin\xi_{3}d\xi_{2}+\xi_{1}\sin\xi_{2}\cos\xi_{3}d\xi_{3}


\displaystyle 
dx_{3}=\frac{\xi_{1}\cos\xi_{2}}{d\xi_{1}}\frac{d\xi_{1}}{dx_{2}}+\frac{\xi_{1}\cos\xi_{2}}{d\xi_{2}}\frac{d\xi_{2}}{dx_{2}}+\frac{\xi_{1}\cos\xi_{2}}{d\xi_{3}}\frac{d\xi_{3}}{dx_{2}}

(7.3.7)


\displaystyle 
=\cos\xi_{2}d\xi_{1}-\xi_{1}\sin\xi_{2}d\xi_{2}

(7.3.8)


\displaystyle 
(dx_{1})^{2}=(\sin\xi_{2}\cos\xi_{3}d\xi_{1})^{2}+(\xi_{1}\cos\xi_{2}\cos\xi_{3}d\xi_{2})^{2}+(\xi_{1}\sin\xi_{2}\sin\xi_{3}d\xi_{3})^{2}

(7.3.9)


\displaystyle 
+2\xi_{1}\sin\xi_{2}\cos\xi_{2}\cos^{2}\xi_{3}d\xi_{1}d\xi_{2}-2\xi_{1}^{2}\cos\xi_{2}\sin\xi_{2}\cos\xi_{3}\sin\xi_{3}d\xi_{2}d\xi_{3}-2\xi_{1}\sin^{2}\xi_{2}\cos\xi_{3}\sin\xi_{3}d\xi_{1}d\xi_{3}


\displaystyle 
(dx_{2})^{2}=(\sin\xi_{2}\sin\xi_{3}d\xi_{1})^{2}+(\xi_{1}\cos\xi_{2}\sin\xi_{3}d\xi_{2})^{2}+(\xi_{1}\sin\xi_{2}\cos\xi_{3}d\xi_{3})^{2}+

(7.3.10)


\displaystyle 
+2\xi_{1}\cos\xi_{2}\sin\xi_{2}\sin^{2}\xi_{3}d\xi_{1}d\xi_{2}+2\xi_{1}^{2}\sin\xi_{2}\cos\xi_{2}\sin\xi_{3}\cos\xi_{3}d\xi_{2}d\xi_{3}+2\xi_{1}\sin^{2}\xi_{2}\sin\xi_{3}\cos\xi_{3}d\xi_{1}d\xi_{2}


\displaystyle 
(dx_{3})^{2}=(\cos\xi_{2}d\xi_{1})^{2}+(\xi_{1}\sin\xi_{2}d\xi_{2})^{2}-2\xi_{1}\sin\xi_{2}\cos\xi_{2}d\xi_{1}d\xi_{2}

(7.3.11)


\displaystyle 
(ds)^{2}=(dx_{1})^{2}+(dx_{2})^{2}+(dx_{3})^{2}=(d\xi_{1})^{2}+(\xi_{1}\cos\xi_{2}\cos\xi_{3}d\xi_{2})^{2}+(\xi_{1}\sin\xi_{2}\sin\xi_{3}d\xi_{3})^{2}

(7.3.12)


\displaystyle 
+(\xi_{1}\cos\xi_{2}\sin\xi_{3}d\xi_{2})^{2}+(\xi_{1}\sin\xi_{2}\cos\xi_{3}d\xi_{3})^{2}+(\xi_{1}\sin\xi_{2}d\xi_{2})^{2}


\displaystyle 
=(d\xi_{1})^{2}+(\xi_{1}d\xi_{2})^{2}+(\xi_{1}\sin\xi_{2}d\xi_{3})^{2}


\displaystyle 
h_{1}=1,\ h_{2}=\xi_{1},\ h_{3}=\xi_{1}\sin\xi_{2}

(7.3.13)


Put equation (7.3.13) into equation (7.3.1).

\displaystyle 
\Delta u=\frac{1}{\xi_{1}^{2}\sin\xi_{2}}(\frac{\partial}{\partial \xi_{1}}[\xi_{1}^{2}\sin\xi_{2}\frac{\partial u}{\partial \xi_{1}}]+\frac{\partial}{\partial \xi_{2}}[\sin\xi_{2}\frac{\partial u}{\partial \xi_{2}}]+\frac{\partial}{\partial \xi_{3}}[\frac{1}{\sin\xi_{2}}\frac{\partial u}{\partial \xi_{3}}])

(7.3.14)


R*7.4 Laplacian in elliptic coordinates[edit]


Given[edit]


\displaystyle 
\Delta u =\frac{1}{h_{1}h_{2}}\sum^{2}_{i=1}\frac{\partial}{\partial \xi_{i}}[\frac{h_{1}h_{2}}{(h_{i}^{2})}\frac{\partial u}{\partial \xi_{i}}]

(7.4.1)



Find[edit]


Verify the Laplacian in elliptic coordinates given by

\displaystyle 
\Delta u=\frac{1}{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}(\frac{\partial^{2}u}{\partial \mu^{2}}+\frac{\partial^{2}u}{\partial \nu^{2}})

(7.4.2)



Solution[edit]


According to Wikipedia [5]

\displaystyle 
x=a\cosh\mu\cos\nu,\ \ y=a\sinh\mu\sin\nu

(7.4.3)


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

\displaystyle 
x_{1}=a\cosh\mu\cos\nu=a\cosh\xi_{1}\cos\xi_{2},\ \ x_{2}=a\sinh\mu\sin\nu=a\sinh\xi_{1}\sin\xi_{2},\ \ (\mu,\nu)=(\xi_{1},\xi_{2})

(7.4.4)


\displaystyle 
dx_{1}=\frac{a\cosh\xi_{1}\cos\xi_{2}}{d\xi_{1}}\frac{d\xi_{1}}{dx_{1}}+\frac{a\cosh\xi_{1}\cos\xi_{2}}{d\xi_{2}}\frac{d\xi_{2}}{dx_{1}}

(7.4.5)


\displaystyle 
=a\sinh\xi_{1}\cos\xi_{2}d\xi_{1}-a\cosh\xi_{1}\sin\xi_{2}d\xi_{2}


\displaystyle 
dx_{1}^{2}=(a\sinh\xi_{1}\cos\xi_{2}d\xi_{1})^{2}+(a\cosh\xi_{1}\sin\xi_{2}d\xi_{2})^{2}-2a^{2}\cosh\xi_{1}\sinh\xi_{1}\cos\xi_{2}\sin\xi_{2}d\xi_{1}d\xi_{2}

(7.4.6)


\displaystyle 
dx_{2}=\frac{a\sinh\xi_{1}\sin\xi_{2}}{d\xi_{1}}\frac{d\xi_{1}}{dx_{1}}+\frac{a\sinh\xi_{1}\sin\xi_{2}}{d\xi_{2}}\frac{d\xi_{2}}{dx_{1}}

(7.4.7)


\displaystyle 
=a\cosh\xi_{1}\sin\xi_{2}d\xi_{1}+a\sinh\xi_{1}\cos\xi_{2}d\xi_{2}


\displaystyle 
dx_{2}^{2}=(a\cosh\xi_{1}\sin\xi_{2}d\xi_{1})^{2}+(a\sinh\xi_{1}\cos\xi_{2}d\xi_{2})^{2}+2a^{2}\cosh\xi_{1}\sinh\xi_{1}\cos\xi_{2}\sin\xi_{2}d\xi_{1}d\xi_{2}

(7.4.8)


\displaystyle 
ds^{2}=dx_{1}^{2}+dx_{2}^{2}=a^{2}[\sinh^{2}\xi_{1}\cos^{2}\xi_{2}+\cosh^{2}\xi_{1}\sin^{2}\xi_{2}]d\xi_{1}^{2}+a^{2}[\sinh^{2}\xi_{1}\cos^{2}\xi_{2}+\cosh^{2}\xi_{1}\sin^{2}\xi_{2}]d\xi_{2}^{2}

(7.4.9)


According to hyperbolic trigonometric identity

\displaystyle 
\sinh^{2}\xi_{1}(1-\sin^{2}\xi_{2})+\cosh^{2}\xi_{1}\sin^{2}xi_{2}=\sinh^{2}\xi_{1}+\sin^{2}\xi_{2}(\cosh^{2}\xi_{1}-\sinh^{2}\xi_{1})=\sinh^{2}\xi_{1}+\sin^{2}\xi_{2}

(7.4.10)


Put equation (7.4.10) into (7.4.9).

\displaystyle 
ds^{2}=a^{2}(\sinh^{2}\xi_{1}+\sin^{2}\xi_{2})d\xi_{1}^{2}+a^{2}(\sinh^{2}\xi_{1}+\sin^{2}\xi_{2})d\xi_{2}^{2}

(7.4.11)


\displaystyle 
h_{1}=h_{2}=a(\sinh^{2}\xi_{1}+\sin^{2}\xi_{2})^{\frac{1}{2}}

(7.4.12)


Put equation (7.4.12) into (7.4.1).

\displaystyle 
\Delta u=\frac{1}{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}((\frac{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}\frac{\partial^{2}u}{\partial \mu^{2}})+ (\frac{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}\frac{\partial^{2}u}{\partial \nu^{2}}))

(7.4.13)


\displaystyle 
\Delta u=\frac{1}{a^{2}(\sinh^{2}\mu+\sin^{2}\nu)}(\frac{\partial^{2}u}{\partial \mu^{2}}+\frac{\partial^{2}u}{\partial \nu^{2}})


R*7.5-Laplacian in parabolic coordinates[edit]


Given[edit]


\displaystyle 
x=\mu \nu \,, \, y=\frac{1}{2}(\nu ^{2}-\mu ^{2})

(7.5.1)

Parabolic_coordinates

\displaystyle 
\bigtriangleup u=\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \varepsilon_{i} }[\frac{h_{1}h_{2}h_{3}}{(h_{i})^{2}}\frac{\partial u}{\partial \varepsilon _{i}}]

(7.5.2)


Find[edit]


Verify the Laplacian in parabolic corrdinates given by:

\displaystyle 
\bigtriangleup u=\frac{1}{\mu ^{2}+\nu ^{2}}(\frac{\partial^2 u}{\partial \mu^2}+\frac{\partial^2 u}{\partial \nu^2})

(7.5.3)


Solution[edit]


Slove by ourselves

\displaystyle 
dx=d(\mu \nu )=d\mu\nu  +\mu d\nu

(7.5.4)

\displaystyle 
dy=d(\frac{1}{2}(\nu ^{2}-\mu ^{2}))=\nu d\mu -\mu d\nu

(7.5.5)

\displaystyle 
ds^{2}=dx^{2}+dy^{2}=d\mu ^{2}\nu ^{2}+\mu ^{2}d\nu ^{2}+2d\mu d\nu \mu \nu +\nu ^{2}d\nu^{2} -2\mu \nu d\mu d\nu +\mu ^{2}d\mu ^{2}

(7.5.6)

\displaystyle 
\Rightarrow ds^{2}=d\mu ^{2}(\nu ^{2}+\mu ^{2})+d\nu ^{2}(\nu ^{2}+\mu ^{2}),so,\, h_{\nu }=h_{\mu }=\sqrt{\nu ^{2}+\mu ^{2}}

(7.5.7)

Plunge into equation 7.5.2:

\displaystyle 
\bigtriangleup u=\frac{1}{h_{1}h_{2}}[\frac{\partial }{\partial \mu }(\frac{h_{1}h_{2}}{h_{1}^{2}}\frac{\partial u}{\partial \mu })+\frac{\partial }{\partial \nu }(\frac{h_{1}h_{2}}{h_{2}^{2}}\frac{\partial u}{\partial \nu })]

(7.5.8)

\displaystyle 
\Rightarrow \bigtriangleup u=\frac{1}{\mu ^{2}+\nu ^{2}}(\frac{\partial^2 u}{\partial \mu^2}+\frac{\partial^2 u}{\partial \nu^2})

(7.5.9)

R*7.6 Verify Homogeneous Solution of Legendre Equation[edit]


Given[edit]


\displaystyle 
P_0(x)=1\,,P_1(x)=x

(7.6.1)


\displaystyle 
P_2(x)=\frac{1}{2}(3x^2-1)

(7.6.2)


\displaystyle 
P_3(x)=\frac{1}{2}(5x^3-3x)

(7.6.3)


\displaystyle 
P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}

(7.6.4)


\displaystyle 
P_n(x)=\sum^{n/2}_{i=0}(-1)^i\frac{(2n-2i)!\,x^{n-2i}}{2^ni!(n-i)!(n-2i)!}

(7.6.5)


\displaystyle 
L_2(y)=(1-x^2)y''-2xy'+n(n+1)y=0

(7.6.6)


\displaystyle 
P_n(x)=\sum^{n/2}_{i=0}\frac{1\cdot 3\cdot\cdot\cdot(2n-2i-1)}{2^ii!(n-2i)!}(-1)^ix^{n-2i}

(7.6.7)


Find[edit]


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[edit]

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.

\displaystyle 
P_0(x)=1\,\,,P_0(x)'=0\,\,,P_0(x)''=0


\displaystyle 
P_1(x)=x\,\,,P_1(x)'=1\,\,,P_1(x)''=0


\displaystyle 
P_2(x)=\frac{1}{2}(3x^2-1)\,\,,P_2(x)'=3x\,\,,P_2(x)''=3


\displaystyle 
P_3(x)=\frac{1}{2}(5x^3-3x)\,\,,P_3(x)'=\frac{1}{2}(15x^2-3)\,\,,P_3(x)''=15x


\displaystyle 
P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}\,\,,P_4(x)'=\frac{35}{2}x^3-\frac{15}{2}x\,\,,P_4(x)''=\frac{105}{2}x^2-\frac{15}{2}


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,

\displaystyle 
P_0(x)=\sum^{n/2}_{i=0}(-1)^0\frac{(2(0)-2(0))!\,x^{(0)-2(0)}}{2^{0}(0)!(0-0)!(0-2(0))!}=(1)\frac{(1)\,x^{0}}{(1)(1)(1)(1)}=1



\displaystyle 
P_1(x)=\sum^{n/2}_{i=0}(-1)^0\frac{(2(1)-2(0))!\,x^{(1)-2(0)}}{2^{1}(0)!(1-0)!(1-2(0))!}=(1)\frac{(2)\,x}{(2)(1)(1)(1)}=x



\displaystyle 
P_2(x)=\sum^{n/2}_{i=0}(-1)^0\frac{(2(2)-2(0))!\,x^{(2)-2(0)}}{2^{2}(0)!(2-0)!(2-2(0))!} + (-1)^1\frac{(2(2)-2(1))!\,x^{(2)-2(1)}}{2^{2}(1)!(2-1)!(2-2(1))!}=(1)\frac{(24)\,x^2}{(4)(1)(2)(2)}-\frac{(2)}{(4)(1)(1)(1)}=\frac{1}{2}(3x^2-1)



\displaystyle 
P_3(x)=\sum^{n/2}_{i=0}(-1)^0\frac{(2(3)-2(0))!\,x^{(3)-2(0)}}{2^{3}(0)!(3-0)!(3-2(0))!} + (-1)^1\frac{(2(3)-2(1))!\,x^{(3)-2(1)}}{2^{3}(1)!(3-1)!(3-2(1))!}=(1)\frac{(720)\,x^3}{(8)(1)(6)(6)}-\frac{(24)x}{(8)(1)(2)(1)}=\frac{1}{2}(5x^3-3x)



\displaystyle 
P_4(x)=\sum^{n/2}_{i=0}(-1)^0\frac{(2(4)-2(0))!\,x^{(4)-2(0)}}{2^{4}(0)!(4-0)!(4-2(0))!} + (-1)^1\frac{(2(4)-2(1))!\,x^{(4)-2(1)}}{2^{4}(1)!(4-1)!(4-2(1))!} + (-1)^2\frac{(2(4)-2(2))!\,x^{(4)-2(2)}}{2^{4}(2)!(4-2)!(4-2(2))!} = \frac{(40320)\,x^4}{(16)(1)(24)(24)}-\frac{(720)x^2}{(16)(1)(6)(2)} + \frac{(24)x}{(16)(2)(2)(1)} = \frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}



R*7.7-Obtain the separated equations for the Laplace equation in Parabolic coordinates[edit]


Given[edit]


\displaystyle 
\bigtriangleup u=\frac{1}{\mu ^{2}+\nu ^{2}}(\frac{\partial^2 u}{\partial \mu^2}+\frac{\partial^2 u}{\partial \nu^2})

(7.7.1)


Find[edit]


Find the separated equation for the Laplace equation

Solution[edit]


Slove by ourselves

\displaystyle 
Let:u(\mu ,\nu )=A(\mu )B(\nu )

(7.7.2)

\displaystyle 
\Rightarrow \frac{B(\nu )}{\mu ^{2}+\nu ^{2}}(\frac{\partial^2 A(\mu )}{\partial \mu ^2})+\frac{A(\mu )}{\mu ^{2}+\nu ^{2}}(\frac{\partial^2 B(\nu )}{\partial \nu ^2})

(7.7.3)

Divided by u:

\displaystyle 
\Rightarrow \frac{1}{A(\mu )(\mu ^{2}+\nu ^{2})}(\frac{\partial^2 A(\mu )}{\partial \mu ^2})+\frac{1}{B(\nu )(\mu ^{2}+\nu ^{2})}(\frac{\partial^2 B(\nu )}{\partial \nu ^2})=0

(7.7.4)

\displaystyle 
Assume:\frac{1}{A(\mu ^{2}+\nu ^{2})}(\ddot{A})=-\frac{1}{B(\mu ^{2}+\nu ^{2})}(\ddot{B})=K

(7.7.5)

\displaystyle 
\Rightarrow \ddot{A}=KA,\, -\ddot{B}=KB

(7.7.6)

R*7.8 Heat Conduction on a Sphere[edit]


Find[edit]


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[edit]

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')



Legendre Polynomials 1.jpg

Legendre Functions 1.jpg

R7.9: To Find vi[edit]


Given[edit]


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[edit]


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

 \mathbf{v}=v_i\mathbf b_i

Solution[edit]


 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[edit]

  1. Lecture slide Mtg 39-1 [1]
  2. Lecture slide Mtg 38-5 [2]
  3. Lecture slide Mtg 40-6 [3]
  4. Lecture slide Mtg 27-1 [4]
  5. Elliptic coordinates [5]

Contributing Members[edit]

Problem Assignments
Problem Number Lecture(Mtg) Assigned To
R*7.1 Mtg 39-1 Ismail
R*7.2 Mtg 40-5 Ismail
R*7.3 Mtg 40-6 YungSheng Chang
R*7.4 Mtg 41-1 YungSheng Chang
R*7.5 Mtg 41-2 Chao Yang
R*7.6 Mtg 41-5 Robert Anderson
R*7.7 Mtg 41-6 Chao Yang
R*7.8 Mtg 42-2 Robert Anderson
R7.9 Mtg 42-6 Prashant Gopichandran
Table of Contributions
Name Problems Solved Problems Checked Signature
Robert Anderson R*7.6,R*7.8 Robert Anderson 06:15, 29 November 2011 (UTC)
YungSheng Chang R*7.3,R*7.4 R*7.8 YungSheng Chang 22:08, 21 November 2011 (UTC)


Chao Yang R*7.5,R*7.7 R*7.6 Chao Yang 00:08, 28 November 2011 (UTC)


Prashant Gopichandran R7.9 Prashant Gopichandran 18:41, 7 December 2011 (UTC)
Ismail R*7.1, R*7.2 R*7.5,R*7.7 Ismail 23:39, 6 December 2011 (UTC)