# User:Egm6321.f11.team3/Hwk7

## R*7.1 - Finding ds and Laplace Operator Equivalent in Spherical Coordinates

From the lecture slide Mtg 39-1 [1]

### Given

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

 {\displaystyle {\begin{aligned}ds=dx_{j}.e_{j}\end{aligned}}} (7.1.1)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.2)

 {\displaystyle {\begin{aligned}e_{i}e_{j}=\delta _{ij}\end{aligned}}} (7.1.3)

Delta function defined as ;

 {\displaystyle {\begin{aligned}\delta _{ij}=\left\{{\begin{matrix}1&for&i=j\\0&for&i\neq j\end{matrix}}\right.\end{aligned}}} (7.1.4)
 {\displaystyle {\begin{aligned}ds^{2}=dx_{i}.dx_{i}=\sum _{i=1}^{3}\left(dx_{i}\right)^{2}\end{aligned}}} (7.1.5)
 {\displaystyle {\begin{aligned}&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{aligned}}} (7.1.6)
 {\displaystyle {\begin{aligned}\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{aligned}}} (7.1.7)
 {\displaystyle {\begin{aligned}{\begin{matrix}{\hat {\xi }}=({\hat {\xi _{1}}},{\hat {\xi _{2}}},{\hat {\xi _{3}}})\end{matrix}}\end{aligned}}} (7.1.8)
 {\displaystyle {\begin{aligned}{\begin{matrix}h_{1}h_{2}h_{3}=r^{2}cos(\theta )\end{matrix}}\end{aligned}}} (7.1.9)

### 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

 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.10)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.11)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.12)

 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.13)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.14)
 {\displaystyle {\begin{aligned}dx_{3}={\frac {\partial r}{\partial r}}sin\theta dr+r{\frac {\partial sin(\theta )}{\partial \theta }}d\theta \end{aligned}}} (7.1.15)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.16)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.17)
 {\displaystyle {\begin{aligned}dx_{3}^{2}=(sin\theta )^{2}dr^{2}+r^{2}(cos\theta )^{2}d\theta ^{2}+2r(sin\theta )(cos\theta )drd\theta \end{aligned}}} (7.1.18)
 {\displaystyle {\begin{aligned}ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}\end{aligned}}} (7.1.19)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.1.20)

If we group things together we will get ;

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


 {\displaystyle {\begin{aligned}&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{aligned}}} (7.1.21)
 {\displaystyle {\begin{aligned}&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{aligned}}} (7.1.22)
 {\displaystyle {\begin{aligned}&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{aligned}}} (7.1.23)

If we substitute in

 {\displaystyle {\begin{aligned}\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{aligned}}} (7.1.24)

${\displaystyle \displaystyle \Delta u={\frac {1}{r^{2}}}\left(r^{2}{\frac {\partial u}{\partial r}}\right)+{\frac {1}{r^{2}cos\theta }}{\frac {\partial }{\partial \theta }}\left(cos\theta {\frac {\partial u}{\partial \theta }}\right)+{\frac {1}{r^{2}(cos\theta )^{2}}}{\frac {\partial ^{2}u}{\partial \phi ^{2}}}}$


## R*7.2 - Heat Conduction on a Cylinder

From the lecture slide Mtg 40-6 [3]

### Given

Coordinate equivalents given as in the lecture;

 {\displaystyle {\begin{aligned}x=&r(cos(\theta ))=\xi _{1}(cos(\xi _{2}))\\y=&r(sin(\theta ))=\xi _{1}(sin(\xi _{2}))\\z=&\xi _{3}\end{aligned}}} (7.2.1)

### Find

• Find
 {\displaystyle {\begin{aligned}{\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{aligned}}} (7.2.2)
• Find
 {\displaystyle {\begin{aligned}ds^{2}=\sum _{i}(dx_{i})^{2}=\sum _{k}(h_{k})^{2}(d\xi _{k})^{2}\end{aligned}}} (7.2.3)
• Find ${\displaystyle \Delta u}$ in cylindrical coordinates.
• Use separation of variable to find the separated equations and compare to the bessel eq. (1) [4]

### Solution

 {\displaystyle {\begin{aligned}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{aligned}}} (7.2.4)
 {\displaystyle {\begin{aligned}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{aligned}}} (7.2.5)
 {\displaystyle {\begin{aligned}ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}\end{aligned}}} (7.2.6)

 {\displaystyle {\begin{aligned}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{aligned}}} (7.2.7)

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


 {\displaystyle {\begin{aligned}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{aligned}}} (7.2.8)

Equation yields

 {\displaystyle {\begin{aligned}&h_{1}=1\\&h_{2}=\xi _{1}\\&h_{3}=1\end{aligned}}} (7.2.9)

 {\displaystyle {\begin{aligned}h_{1}h_{2}h_{3}=\xi _{1}\end{aligned}}} (7.2.10)
 {\displaystyle {\begin{aligned}\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{aligned}}} (7.2.11)
 {\displaystyle {\begin{aligned}&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{aligned}}} (7.2.12)
 {\displaystyle {\begin{aligned}&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{aligned}}} (7.2.13)
 {\displaystyle {\begin{aligned}&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{aligned}}} (7.2.14)

If we substitute in we will get

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


 {\displaystyle {\begin{aligned}u=R(\xi _{1})\theta (\xi _{2})Z(\xi _{3})\end{aligned}}} (7.2.15)
 {\displaystyle {\begin{aligned}\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{aligned}}} (7.2.16)

Dividing through ${\displaystyle R\theta Z}$

 {\displaystyle {\begin{aligned}\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{aligned}}} (7.2.17)
 {\displaystyle {\begin{aligned}{\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{aligned}}} (7.2.18)
 {\displaystyle {\begin{aligned}{\frac {\partial ^{2}Z}{\partial \xi _{3}^{2}}}+\lambda ^{2}Z=0\end{aligned}}} (7.2.19)
 {\displaystyle {\begin{aligned}{\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{aligned}}} (7.2.20)
 {\displaystyle {\begin{aligned}{\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{aligned}}} (7.2.21)
 {\displaystyle {\begin{aligned}{\frac {\partial ^{2}\theta }{\partial \xi _{2}^{2}}}+\eta ^{2}\theta =0\end{aligned}}} (7.2.22)
 {\displaystyle {\begin{aligned}\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{aligned}}} (7.2.23)
 {\displaystyle {\begin{aligned}\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{aligned}}} (7.2.24)

If we change ${\displaystyle \xi _{1}\lambda =y}$ we will get bessel's equation

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


 {\displaystyle {\begin{aligned}y^{2}=1-t^{2}\end{aligned}}} (7.2.25)
 {\displaystyle {\begin{aligned}{\frac {\partial R}{\partial y}}={\frac {\partial R}{\partial t}}{\frac {\partial t}{\partial y}}\end{aligned}}} (7.2.26)
 {\displaystyle {\begin{aligned}{\frac {\partial R}{\partial y}}={\frac {\partial R}{\partial t}}\left(-{\frac {y}{\sqrt {1-y^{2}}}}\right)\end{aligned}}} (7.2.27)
 {\displaystyle {\begin{aligned}{\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{aligned}}} (7.2.28)

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

## R*7.3 Find ${\displaystyle \displaystyle \Delta u}$ in spherical coordinates

### Given

 ${\displaystyle \displaystyle \Delta u={\frac {1}{h_{1}h_{2}h_{3}}}\sum _{i=1}^{3}{\frac {\partial }{\partial \xi _{i}}}[{\frac {h_{1}h_{2}h_{3}}{(h_{i}^{2})}}{\frac {\partial u}{\partial \xi _{i}}}]}$ (7.3.1)

### Find

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

### Solution

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

 ${\displaystyle \displaystyle (\xi _{1},\xi _{2},\xi _{3})=(r,{\bar {\theta }},\varphi )}$ (7.3.2)

 ${\displaystyle \displaystyle x_{1}=x=r\sin {\bar {\theta }}\cos \theta =\xi _{1}\sin \xi _{2}\cos \xi _{3}}$ (7.3.2)

 ${\displaystyle \displaystyle x_{2}=y=r\sin {\bar {\theta }}\sin \theta =\xi _{1}\sin \xi _{2}\sin \xi _{3}}$ (7.3.3)

 ${\displaystyle \displaystyle x_{3}=z=r\cos {\bar {\theta }}=\xi _{1}\cos \xi _{2}}$ (7.3.4)

 ${\displaystyle \displaystyle dx_{1}={\frac {\xi _{1}\sin \xi _{2}\cos \xi _{3}}{d\xi _{1}}}{\frac {d\xi _{1}}{dx_{1}}}+{\frac {\xi _{1}\sin \xi _{2}\cos \xi _{3}}{d\xi _{2}}}{\frac {d\xi _{2}}{dx_{1}}}+{\frac {\xi _{1}\sin \xi _{2}\cos \xi _{3}}{d\xi _{3}}}{\frac {d\xi _{3}}{dx_{1}}}}$ (7.3.5)

 ${\displaystyle \displaystyle =\sin \xi _{2}\cos \xi _{3}d\xi _{1}+\xi _{1}\cos \xi _{2}\cos \xi _{3}d\xi _{2}-\xi _{1}\sin \xi _{2}\sin \xi _{3}d\xi _{3}}$

 ${\displaystyle \displaystyle dx_{2}={\frac {\xi _{1}\sin \xi _{2}\sin \xi _{3}}{d\xi _{1}}}{\frac {d\xi _{1}}{dx_{2}}}+{\frac {\xi _{1}\sin \xi _{2}\sin \xi _{3}}{d\xi _{2}}}{\frac {d\xi _{2}}{dx_{2}}}+{\frac {\xi _{1}\sin \xi _{2}\sin \xi _{3}}{d\xi _{3}}}{\frac {d\xi _{3}}{dx_{2}}}}$ (7.3.6)

 ${\displaystyle \displaystyle =\sin \xi _{2}\sin \xi _{3}d\xi _{1}+\xi _{1}\cos \xi _{2}\sin \xi _{3}d\xi _{2}+\xi _{1}\sin \xi _{2}\cos \xi _{3}d\xi _{3}}$

 ${\displaystyle \displaystyle dx_{3}={\frac {\xi _{1}\cos \xi _{2}}{d\xi _{1}}}{\frac {d\xi _{1}}{dx_{2}}}+{\frac {\xi _{1}\cos \xi _{2}}{d\xi _{2}}}{\frac {d\xi _{2}}{dx_{2}}}+{\frac {\xi _{1}\cos \xi _{2}}{d\xi _{3}}}{\frac {d\xi _{3}}{dx_{2}}}}$ (7.3.7)

 ${\displaystyle \displaystyle =\cos \xi _{2}d\xi _{1}-\xi _{1}\sin \xi _{2}d\xi _{2}}$ (7.3.8)

 ${\displaystyle \displaystyle (dx_{1})^{2}=(\sin \xi _{2}\cos \xi _{3}d\xi _{1})^{2}+(\xi _{1}\cos \xi _{2}\cos \xi _{3}d\xi _{2})^{2}+(\xi _{1}\sin \xi _{2}\sin \xi _{3}d\xi _{3})^{2}}$ (7.3.9)

 ${\displaystyle \displaystyle +2\xi _{1}\sin \xi _{2}\cos \xi _{2}\cos ^{2}\xi _{3}d\xi _{1}d\xi _{2}-2\xi _{1}^{2}\cos \xi _{2}\sin \xi _{2}\cos \xi _{3}\sin \xi _{3}d\xi _{2}d\xi _{3}-2\xi _{1}\sin ^{2}\xi _{2}\cos \xi _{3}\sin \xi _{3}d\xi _{1}d\xi _{3}}$

 ${\displaystyle \displaystyle (dx_{2})^{2}=(\sin \xi _{2}\sin \xi _{3}d\xi _{1})^{2}+(\xi _{1}\cos \xi _{2}\sin \xi _{3}d\xi _{2})^{2}+(\xi _{1}\sin \xi _{2}\cos \xi _{3}d\xi _{3})^{2}+}$ (7.3.10)

 ${\displaystyle \displaystyle +2\xi _{1}\cos \xi _{2}\sin \xi _{2}\sin ^{2}\xi _{3}d\xi _{1}d\xi _{2}+2\xi _{1}^{2}\sin \xi _{2}\cos \xi _{2}\sin \xi _{3}\cos \xi _{3}d\xi _{2}d\xi _{3}+2\xi _{1}\sin ^{2}\xi _{2}\sin \xi _{3}\cos \xi _{3}d\xi _{1}d\xi _{2}}$

 ${\displaystyle \displaystyle (dx_{3})^{2}=(\cos \xi _{2}d\xi _{1})^{2}+(\xi _{1}\sin \xi _{2}d\xi _{2})^{2}-2\xi _{1}\sin \xi _{2}\cos \xi _{2}d\xi _{1}d\xi _{2}}$ (7.3.11)

 ${\displaystyle \displaystyle (ds)^{2}=(dx_{1})^{2}+(dx_{2})^{2}+(dx_{3})^{2}=(d\xi _{1})^{2}+(\xi _{1}\cos \xi _{2}\cos \xi _{3}d\xi _{2})^{2}+(\xi _{1}\sin \xi _{2}\sin \xi _{3}d\xi _{3})^{2}}$ (7.3.12)

 ${\displaystyle \displaystyle +(\xi _{1}\cos \xi _{2}\sin \xi _{3}d\xi _{2})^{2}+(\xi _{1}\sin \xi _{2}\cos \xi _{3}d\xi _{3})^{2}+(\xi _{1}\sin \xi _{2}d\xi _{2})^{2}}$

 ${\displaystyle \displaystyle =(d\xi _{1})^{2}+(\xi _{1}d\xi _{2})^{2}+(\xi _{1}\sin \xi _{2}d\xi _{3})^{2}}$

 ${\displaystyle \displaystyle h_{1}=1,\ h_{2}=\xi _{1},\ h_{3}=\xi _{1}\sin \xi _{2}}$ (7.3.13)

Put equation (7.3.13) into equation (7.3.1).

 ${\displaystyle \displaystyle \Delta u={\frac {1}{\xi _{1}^{2}\sin \xi _{2}}}({\frac {\partial }{\partial \xi _{1}}}[\xi _{1}^{2}\sin \xi _{2}{\frac {\partial u}{\partial \xi _{1}}}]+{\frac {\partial }{\partial \xi _{2}}}[\sin \xi _{2}{\frac {\partial u}{\partial \xi _{2}}}]+{\frac {\partial }{\partial \xi _{3}}}[{\frac {1}{\sin \xi _{2}}}{\frac {\partial u}{\partial \xi _{3}}}])}$ (7.3.14)

## R*7.4 Laplacian in elliptic coordinates

### Given

 ${\displaystyle \displaystyle \Delta u={\frac {1}{h_{1}h_{2}}}\sum _{i=1}^{2}{\frac {\partial }{\partial \xi _{i}}}[{\frac {h_{1}h_{2}}{(h_{i}^{2})}}{\frac {\partial u}{\partial \xi _{i}}}]}$ (7.4.1)

### Find

Verify the Laplacian in elliptic coordinates given by

 ${\displaystyle \displaystyle \Delta u={\frac {1}{a^{2}(\sinh ^{2}\mu +\sin ^{2}\nu )}}({\frac {\partial ^{2}u}{\partial \mu ^{2}}}+{\frac {\partial ^{2}u}{\partial \nu ^{2}}})}$ (7.4.2)

### Solution

According to Wikipedia [5]

 ${\displaystyle \displaystyle x=a\cosh \mu \cos \nu ,\ \ y=a\sinh \mu \sin \nu }$ (7.4.3)

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

 ${\displaystyle \displaystyle x_{1}=a\cosh \mu \cos \nu =a\cosh \xi _{1}\cos \xi _{2},\ \ x_{2}=a\sinh \mu \sin \nu =a\sinh \xi _{1}\sin \xi _{2},\ \ (\mu ,\nu )=(\xi _{1},\xi _{2})}$ (7.4.4)

 ${\displaystyle \displaystyle dx_{1}={\frac {a\cosh \xi _{1}\cos \xi _{2}}{d\xi _{1}}}{\frac {d\xi _{1}}{dx_{1}}}+{\frac {a\cosh \xi _{1}\cos \xi _{2}}{d\xi _{2}}}{\frac {d\xi _{2}}{dx_{1}}}}$ (7.4.5)

 ${\displaystyle \displaystyle =a\sinh \xi _{1}\cos \xi _{2}d\xi _{1}-a\cosh \xi _{1}\sin \xi _{2}d\xi _{2}}$

 ${\displaystyle \displaystyle dx_{1}^{2}=(a\sinh \xi _{1}\cos \xi _{2}d\xi _{1})^{2}+(a\cosh \xi _{1}\sin \xi _{2}d\xi _{2})^{2}-2a^{2}\cosh \xi _{1}\sinh \xi _{1}\cos \xi _{2}\sin \xi _{2}d\xi _{1}d\xi _{2}}$ (7.4.6)

 ${\displaystyle \displaystyle dx_{2}={\frac {a\sinh \xi _{1}\sin \xi _{2}}{d\xi _{1}}}{\frac {d\xi _{1}}{dx_{1}}}+{\frac {a\sinh \xi _{1}\sin \xi _{2}}{d\xi _{2}}}{\frac {d\xi _{2}}{dx_{1}}}}$ (7.4.7)

 ${\displaystyle \displaystyle =a\cosh \xi _{1}\sin \xi _{2}d\xi _{1}+a\sinh \xi _{1}\cos \xi _{2}d\xi _{2}}$

 ${\displaystyle \displaystyle dx_{2}^{2}=(a\cosh \xi _{1}\sin \xi _{2}d\xi _{1})^{2}+(a\sinh \xi _{1}\cos \xi _{2}d\xi _{2})^{2}+2a^{2}\cosh \xi _{1}\sinh \xi _{1}\cos \xi _{2}\sin \xi _{2}d\xi _{1}d\xi _{2}}$ (7.4.8)

 ${\displaystyle \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}=a^{2}[\sinh ^{2}\xi _{1}\cos ^{2}\xi _{2}+\cosh ^{2}\xi _{1}\sin ^{2}\xi _{2}]d\xi _{1}^{2}+a^{2}[\sinh ^{2}\xi _{1}\cos ^{2}\xi _{2}+\cosh ^{2}\xi _{1}\sin ^{2}\xi _{2}]d\xi _{2}^{2}}$ (7.4.9)

According to hyperbolic trigonometric identity

 ${\displaystyle \displaystyle \sinh ^{2}\xi _{1}(1-\sin ^{2}\xi _{2})+\cosh ^{2}\xi _{1}\sin ^{2}xi_{2}=\sinh ^{2}\xi _{1}+\sin ^{2}\xi _{2}(\cosh ^{2}\xi _{1}-\sinh ^{2}\xi _{1})=\sinh ^{2}\xi _{1}+\sin ^{2}\xi _{2}}$ (7.4.10)

Put equation (7.4.10) into (7.4.9).

 ${\displaystyle \displaystyle ds^{2}=a^{2}(\sinh ^{2}\xi _{1}+\sin ^{2}\xi _{2})d\xi _{1}^{2}+a^{2}(\sinh ^{2}\xi _{1}+\sin ^{2}\xi _{2})d\xi _{2}^{2}}$ (7.4.11)

 ${\displaystyle \displaystyle h_{1}=h_{2}=a(\sinh ^{2}\xi _{1}+\sin ^{2}\xi _{2})^{\frac {1}{2}}}$ (7.4.12)

Put equation (7.4.12) into (7.4.1).

 ${\displaystyle \displaystyle \Delta u={\frac {1}{a^{2}(\sinh ^{2}\mu +\sin ^{2}\nu )}}(({\frac {a^{2}(\sinh ^{2}\mu +\sin ^{2}\nu )}{a^{2}(\sinh ^{2}\mu +\sin ^{2}\nu )}}{\frac {\partial ^{2}u}{\partial \mu ^{2}}})+({\frac {a^{2}(\sinh ^{2}\mu +\sin ^{2}\nu )}{a^{2}(\sinh ^{2}\mu +\sin ^{2}\nu )}}{\frac {\partial ^{2}u}{\partial \nu ^{2}}}))}$ (7.4.13)

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

## R*7.5-Laplacian in parabolic coordinates

### Given

 ${\displaystyle \displaystyle x=\mu \nu \,,\,y={\frac {1}{2}}(\nu ^{2}-\mu ^{2})}$ (7.5.1)
 ${\displaystyle \displaystyle \bigtriangleup u={\frac {1}{h_{1}h_{2}h_{3}}}\sum _{i=1}^{3}{\frac {\partial }{\partial \varepsilon _{i}}}[{\frac {h_{1}h_{2}h_{3}}{(h_{i})^{2}}}{\frac {\partial u}{\partial \varepsilon _{i}}}]}$ (7.5.2)

### Find

Verify the Laplacian in parabolic corrdinates given by:

 ${\displaystyle \displaystyle \bigtriangleup u={\frac {1}{\mu ^{2}+\nu ^{2}}}({\frac {\partial ^{2}u}{\partial \mu ^{2}}}+{\frac {\partial ^{2}u}{\partial \nu ^{2}}})}$ (7.5.3)

### Solution

Slove by ourselves

 ${\displaystyle \displaystyle dx=d(\mu \nu )=d\mu \nu +\mu d\nu }$ (7.5.4)
 ${\displaystyle \displaystyle dy=d({\frac {1}{2}}(\nu ^{2}-\mu ^{2}))=\nu d\mu -\mu d\nu }$ (7.5.5)
 ${\displaystyle \displaystyle ds^{2}=dx^{2}+dy^{2}=d\mu ^{2}\nu ^{2}+\mu ^{2}d\nu ^{2}+2d\mu d\nu \mu \nu +\nu ^{2}d\nu ^{2}-2\mu \nu d\mu d\nu +\mu ^{2}d\mu ^{2}}$ (7.5.6)
 ${\displaystyle \displaystyle \Rightarrow ds^{2}=d\mu ^{2}(\nu ^{2}+\mu ^{2})+d\nu ^{2}(\nu ^{2}+\mu ^{2}),so,\,h_{\nu }=h_{\mu }={\sqrt {\nu ^{2}+\mu ^{2}}}}$ (7.5.7)

Plunge into equation 7.5.2:

 ${\displaystyle \displaystyle \bigtriangleup u={\frac {1}{h_{1}h_{2}}}[{\frac {\partial }{\partial \mu }}({\frac {h_{1}h_{2}}{h_{1}^{2}}}{\frac {\partial u}{\partial \mu }})+{\frac {\partial }{\partial \nu }}({\frac {h_{1}h_{2}}{h_{2}^{2}}}{\frac {\partial u}{\partial \nu }})]}$ (7.5.8)
 ${\displaystyle \displaystyle \Rightarrow \bigtriangleup u={\frac {1}{\mu ^{2}+\nu ^{2}}}({\frac {\partial ^{2}u}{\partial \mu ^{2}}}+{\frac {\partial ^{2}u}{\partial \nu ^{2}}})}$ (7.5.9)

## R*7.6 Verify Homogeneous Solution of Legendre Equation

### Given

 ${\displaystyle \displaystyle P_{0}(x)=1\,,P_{1}(x)=x}$ (7.6.1)

 ${\displaystyle \displaystyle P_{2}(x)={\frac {1}{2}}(3x^{2}-1)}$ (7.6.2)

 ${\displaystyle \displaystyle P_{3}(x)={\frac {1}{2}}(5x^{3}-3x)}$ (7.6.3)

 ${\displaystyle \displaystyle P_{4}(x)={\frac {35}{8}}x^{4}-{\frac {15}{4}}x^{2}+{\frac {3}{8}}}$ (7.6.4)

 ${\displaystyle \displaystyle P_{n}(x)=\sum _{i=0}^{n/2}(-1)^{i}{\frac {(2n-2i)!\,x^{n-2i}}{2^{n}i!(n-i)!(n-2i)!}}}$ (7.6.5)

 ${\displaystyle \displaystyle L_{2}(y)=(1-x^{2})y''-2xy'+n(n+1)y=0}$ (7.6.6)

 ${\displaystyle \displaystyle P_{n}(x)=\sum _{i=0}^{n/2}{\frac {1\cdot 3\cdot \cdot \cdot (2n-2i-1)}{2^{i}i!(n-2i)!}}(-1)^{i}x^{n-2i}}$ (7.6.7)

### 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.

 ${\displaystyle \displaystyle P_{0}(x)=1\,\,,P_{0}(x)'=0\,\,,P_{0}(x)''=0}$

 ${\displaystyle \displaystyle P_{1}(x)=x\,\,,P_{1}(x)'=1\,\,,P_{1}(x)''=0}$

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

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

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

Now we plug these solutions into 7.6.6 in place of the y terms for the respective values of n.

${\displaystyle \displaystyle L_{2}(P_{0})=(1-x^{2})0-2x(0)+0(0+1)1=0}$

${\displaystyle \displaystyle L_{2}(P_{1})=(1-x^{2})0-2x(1)+1(1+1)x=-2x+2x=0}$

${\displaystyle \displaystyle L_{2}(P_{2})=(1-x^{2})3-2x(3x)+2(2+1)({\frac {1}{2}}(3x^{2}-1))=3-3x^{2}-6x^{2}+9x^{2}-3=0}$

${\displaystyle \displaystyle L_{2}(P_{3})=(1-x^{2})15x-2x({\frac {1}{2}}(15x^{2}-3))+3(3+1)({\frac {1}{2}}(5x^{3}-3x))=15x-15x^{3}-15x^{3}+3x+30x^{3}-18x=0}$

${\displaystyle \displaystyle L_{2}(P_{4})=(1-x^{2})({\frac {105}{2}}x^{2}-{\frac {15}{2}})-2x({\frac {35}{2}}x^{3}-{\frac {15}{2}}x)+4(4+1)({\frac {35}{8}}x^{4}-{\frac {15}{4}}x^{2}+{\frac {3}{8}})={\frac {105}{2}}x^{2}-{\frac {15}{2}}-{\frac {105}{2}}x^{4}+{\frac {15}{2}}x^{2}-35x^{4}+15x^{2}+{\frac {105}{2}}x^{4}-75x^{2}+{\frac {15}{2}}=0}$

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,

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

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

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

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

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

## R*7.7-Obtain the separated equations for the Laplace equation in Parabolic coordinates

### Given

 ${\displaystyle \displaystyle \bigtriangleup u={\frac {1}{\mu ^{2}+\nu ^{2}}}({\frac {\partial ^{2}u}{\partial \mu ^{2}}}+{\frac {\partial ^{2}u}{\partial \nu ^{2}}})}$ (7.7.1)

### Find

Find the separated equation for the Laplace equation

### Solution

Slove by ourselves

 ${\displaystyle \displaystyle Let:u(\mu ,\nu )=A(\mu )B(\nu )}$ (7.7.2)
 ${\displaystyle \displaystyle \Rightarrow {\frac {B(\nu )}{\mu ^{2}+\nu ^{2}}}({\frac {\partial ^{2}A(\mu )}{\partial \mu ^{2}}})+{\frac {A(\mu )}{\mu ^{2}+\nu ^{2}}}({\frac {\partial ^{2}B(\nu )}{\partial \nu ^{2}}})}$ (7.7.3)

Divided by u:

 ${\displaystyle \displaystyle \Rightarrow {\frac {1}{A(\mu )(\mu ^{2}+\nu ^{2})}}({\frac {\partial ^{2}A(\mu )}{\partial \mu ^{2}}})+{\frac {1}{B(\nu )(\mu ^{2}+\nu ^{2})}}({\frac {\partial ^{2}B(\nu )}{\partial \nu ^{2}}})=0}$ (7.7.4)
 ${\displaystyle \displaystyle Assume:{\frac {1}{A(\mu ^{2}+\nu ^{2})}}({\ddot {A}})=-{\frac {1}{B(\mu ^{2}+\nu ^{2})}}({\ddot {B}})=K}$ (7.7.5)
 ${\displaystyle \displaystyle \Rightarrow {\ddot {A}}=KA,\,-{\ddot {B}}=KB}$ (7.7.6)

## R*7.8 Heat Conduction on a Sphere

### Find

Show that for n=0:

${\displaystyle \mu \to \pm 1\Rightarrow \left|Q_{0}(\mu )\right|\to +\infty }$

Plot the following Legendre polynomials and functions,

${\displaystyle \displaystyle {P_{0},P_{1},P_{2},P_{3}}}$

${\displaystyle \displaystyle {Q_{0},Q_{1},Q_{2},Q_{3}}}$

Observe the limits of ${\displaystyle P_{n}(\mu )}$ and ${\displaystyle Q_{n}(\mu )}$ as ${\displaystyle \mu \to \pm 1}$

### 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,

${\displaystyle \displaystyle P_{0}(x)=1}$

${\displaystyle \displaystyle P_{1}(x)=x}$

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

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

${\displaystyle \displaystyle Q_{0}(x)={\frac {1}{2}}ln{\frac {1+x}{1-x}}}$

${\displaystyle \displaystyle Q_{1}(x)={\frac {x}{2}}ln{\frac {1+x}{1-x}}-1}$

${\displaystyle \displaystyle Q_{2}(x)={\frac {3x^{2}-1}{4}}ln{\frac {1+x}{1-x}}-{\frac {3x}{2}}}$

${\displaystyle \displaystyle Q_{3}(x)={\frac {5x^{3}-3x}{4}}ln{\frac {1+x}{1-x}}-{\frac {5x^{2}}{2}}+{\frac {2}{3}}}$

Observe that when x goes to 1, (1-x) is getting smaller. Thus,

${\displaystyle \displaystyle \lim _{x\rightarrow 1}Q_{0}(x)={\frac {1}{2}}ln{\frac {1+x}{1-x}}=\infty }$

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')


## R7.9: To Find vi

### Given

Consider the non-orthonormal basis ${\displaystyle \{\mathbf {b} _{1},\mathbf {b} _{2},\mathbf {b} _{3}\}}$ expressed in the orthonormal basis ${\displaystyle \{\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\}}$ as follows

${\displaystyle \mathbf {b} _{i}=A_{ij}\mathbf {e} _{i}}$

where: ${\displaystyle \displaystyle \mathbf {A} =[A_{ij}]=\left[{\begin{matrix}5&2&3\\4&5&6\\7&8&5\end{matrix}}\right]}$
${\displaystyle \mathbf {v} =-2\mathbf {e} _{1}+4\mathbf {e} _{2}-5\mathbf {e} _{3}}$

### Find

${\displaystyle {\mathbf {v} _{i}}\in \mathbb {R} ^{3\times 1}}$ such that

${\displaystyle \mathbf {v} =v_{i}\mathbf {b} _{i}}$

### Solution

 Solved on our own.


From the given data,

${\displaystyle \mathbf {b} _{i}=\mathbf {A} _{ij}\mathbf {e} _{i}}$

We need to find ${\displaystyle \mathbf {v} _{i}}$ such that ${\displaystyle \mathbf {v} =v_{i}\mathbf {b} _{i}}$

So, ${\displaystyle \mathbf {v} =\mathbf {v} _{i}\mathbf {A} _{ij}\mathbf {e} _{i}}$

We also know that ${\displaystyle \mathbf {v} =-2\mathbf {e} _{1}+4\mathbf {e} _{2}-5\mathbf {e} _{3}}$

So, ${\displaystyle \left[{\begin{matrix}-2\\4\\-5\end{matrix}}\right]=\left[{\begin{matrix}v_{1}\\v_{2}\\v_{3}\end{matrix}}\right]\mathbf {A} ^{T}}$

${\displaystyle \mathbf {d} et(A)=80}$

Hence, ${\displaystyle \left[{\begin{matrix}v_{1}\\v_{2}\\v_{3}\end{matrix}}\right]=\mathbf {A} ^{-T}\left[{\begin{matrix}-2\\4\\-5\end{matrix}}\right]}$

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

Therefore, ${\displaystyle \mathbf {v} _{i}=\left[{\begin{matrix}-1.8625\\-1.475\\1.8875\end{matrix}}\right]}$

# References

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

 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)