User:Egm6321.f10.team03.sigillo/Hwk6

=Problem 4: Infinitesimal segment ds in spherical coordinates and spherical Laplace equation"=

Given
The spherical coordinates in the cartesian coordinate system are given as
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$$x={{x}_{1}}=r\cos \left( \theta \right)\cos \left( \varphi  \right)$$
 * $$\,(Eq. 4.1) $$
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$$y={{x}_{2}}=r\cos \left( \theta \right)\sin \left( \varphi  \right)$$
 * $$\,(Eq. 4.2) $$
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$$z={{x}_{3}}=r\sin \left( \theta \right)$$
 * $$\,(Eq. 4.3) $$
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Find
The infinitesimal segment ds and the Laplace equation for spherical coordinates have to be found.

Solution
Segment ds: The infinitesimal segment ds in the three dimensional space can be expressed as
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$$d{{s}^{2}}=\sum\limits_{i=1}^{3}$$ Thus the first step is to derive expressions for $$d{{x}_{1}}$$,$$d{{x}_{2}}$$ and $$d{{x}_{3}}$$
 * $$\,(Eq. 4.4) $$
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$$d{{x}_{1}}=\frac{\partial \left( r\cos \left( \theta \right)\cos \left( \varphi  \right) \right)}{\partial r}dr+\frac{\partial \left( r\cos \left( \theta  \right)\cos \left( \varphi  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\cos \left( \theta  \right)\cos \left( \varphi  \right) \right)}{\partial \varphi }d\varphi =\underbrace{\cos \left( \theta  \right)\cos \left( \varphi  \right)dr}_{a}-\underbrace{r\cos \left( \varphi  \right)\sin \left( \theta  \right)d\theta }_{b}-\underbrace{r\cos \left( \theta  \right)\sin \left( \varphi  \right)d\varphi }_{c}$$


 * $$\,(Eq. 4.5) $$
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$$d{{x}_{2}}=\frac{\partial \left( r\cos \left( \theta \right)\sin \left( \varphi  \right) \right)}{\partial r}dr+\frac{\partial \left( r\cos \left( \theta  \right)\sin \left( \varphi  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\cos \left( \theta  \right)\sin \left( \varphi  \right) \right)}{\partial \varphi }d\varphi =\underbrace{\cos \left( \theta  \right)\sin \left( \varphi  \right)dr}_{d}-\underbrace{r\sin \left( \varphi  \right)\sin \left( \theta  \right)d\theta }_{e}+\underbrace{r\cos \left( \theta  \right)\cos \left( \varphi  \right)d\varphi }_{f}$$


 * $$\,(Eq. 4.6) $$
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$$d{{x}_{3}}=\frac{\partial \left( r\sin \left( \theta \right) \right)}{\partial r}dr+\frac{\partial \left( r\sin \left( \theta  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\sin \left( \theta  \right) \right)}{\partial \varphi }d\varphi =\sin \left( \theta  \right)dr+r\cos \left( \theta  \right)d\theta $$

The squared values for Eq. 4.5, 4.6 and 4.7 are
 * $$\,(Eq. 4.7) $$
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$$\begin{align} & d{{x}_{1}}^{2}={{\left( a-b-c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2bc-2ac-2ab \right) \\ & \quad \ \ \ ={{\cos }^{2}}\left( \theta \right){{\cos }^{2}}\left( \varphi  \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)d{{\varphi }^{2}} \\ & \quad \quad \ \ +2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)d\theta d\varphi  \\ & \quad \quad \ \ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)drd\varphi  \\ & \quad \quad \ \ -2r\cos \left( \theta \right){{\cos }^{2}}\left( \varphi  \right)\sin \left( \theta  \right)drd\theta  \\ \end{align}$$


 * $$\,(Eq. 4.8) $$
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$$\begin{align} & d{{x}_{2}}^{2}={{\left( d-e+f \right)}^{2}}=\left( {{d}^{2}}+{{e}^{2}}+{{f}^{2}}+2df-2de-2ef \right) \\ & \quad \quad ={{\cos }^{2}}\left( \theta \right){{\sin }^{2}}\left( \varphi  \right)d{{r}^{2}}+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right)d{{\varphi }^{2}} \\ & \quad \quad \ \ +2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)drd\varphi  \\ & \quad \quad \ \ -2r\cos \left( \theta \right)\sin \left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)drd\theta  \\ & \quad \quad \ \ -2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)d\theta d\varphi  \\ \end{align}$$
 * $$\,(Eq. 4.9) $$
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$$d{{x}_{3}}^{2}={{\sin }^{2}}\left( \theta \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\theta }^{2}}+2r\cos \left( \theta  \right)\sin \left( \theta  \right)drd\theta $$
 * $$\,(Eq. 4.10) $$
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Adding the Eqs. 4.8, 4.9 and 4.10 together will be performed in the following manner: regroup the common terms, factorize and add
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$$d\theta d\varphi \to \left[ 2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)-2{{r}^{2}}\cos \left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right) \right]d\theta d\varphi =0$$


 * $$\,(Eq. 4.11) $$
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$$drd\varphi \to \left[ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)+2r{{\cos }^{2}}\left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right) \right]=0$$
 * $$\,(Eq. 4.12) $$
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$$\begin{align} & drd\theta \to \left[ -2r\cos \left( \theta \right){{\cos }^{2}}\left( \varphi  \right)\sin \left( \theta  \right)-2r\cos \left( \theta  \right)\sin \left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)+2r\cos \left( \theta  \right)\sin \left( \theta  \right) \right]drd\theta  \\ & \quad \quad \quad \ =2r\cos \left( \theta \right)\sin \left( \theta  \right)\left[ -{{\cos }^{2}}\left( \theta  \right)-{{\sin }^{2}}\left( \theta  \right)+1 \right]drd\theta =0 \\ \end{align}$$


 * $$\,(Eq. 4.13) $$
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$$d{{r}^{2}}\to \left[ {{\cos }^{2}}\left( \theta \right){{\cos }^{2}}\left( \varphi  \right)+{{\cos }^{2}}\left( \theta  \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+\left( 1-{{\cos }^{2}}\left( \theta  \right) \right) \right]d{{r}^{2}}=1d{{r}^{2}}$$
 * $$\,(Eq. 4.14) $$
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$$d{{\theta }^{2}}\to \left[ {{r}^{2}}\left( 1-{{\sin }^{2}}\left( \varphi \right) \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}\left( 1-{{\sin }^{2}}\left( \theta  \right) \right) \right]d{{\theta }^{2}}={{r}^{2}}d{{\theta }^{2}}$$
 * $$\,(Eq. 4.15) $$
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$$d{{\varphi }^{2}}\to \left[ {{r}^{2}}{{\cos }^{2}}\left( \theta \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right) \right]d{{\varphi }^{2}}={{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\varphi }^{2}}$$
 * $$\,(Eq. 4.16) $$
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As it can be seen only Eq. 4.14, 4.15 and 4.16 are non-zero. Thus the infinitesimal segment ds is just the summation of these three expressions:
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$$d{{s}^{2}}=\sum\limits_{i=1}^{3}=1d{{r}^{2}}+{{r}^{2}}d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta \right)d{{\varphi }^{2}}$$
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 * <p style="text-align:right;">$$\,(Eq. 4.17) $$
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Spherical Laplace equation:

The Laplace equation can be written in general as
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$$\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\zeta }_{i}}}}\left[ \frac\frac{\partial \Psi }{\partial {{\zeta }_{i}}} \right]$$ To get the spehrical Laplace equation we re-define the following variables
 * <p style="text-align:right;">$$\,(Eq. 4.18) $$
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$$\begin{align} & r={{\zeta }_{1}} \\ & \theta ={{\zeta }_{2}} \\ & \varphi ={{\zeta }_{3}} \\ \end{align}$$ Equation 4.17 can now be written as
 * <p style="text-align:right;">$$\,(Eq. 4.19) $$
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$$d{{s}^{2}}=1{{\left( d{{\zeta }_{1}} \right)}^{2}}+{{\zeta }_{1}}^{2}{{\left( d{{\zeta }_{2}} \right)}^{2}}+{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right){{\left( d{{\zeta }_{3}} \right)}^{2}}={{\left( {{h}_{1}}\left( \zeta \right) \right)}^{2}}{{\left( d{{\zeta }_{1}} \right)}^{2}}+{{\left( {{h}_{2}}\left( \zeta  \right) \right)}^{2}}{{\left( d{{\zeta }_{2}} \right)}^{2}}+{{\left( {{h}_{3}}\left( \zeta  \right) \right)}^{2}}{{\left( d{{\zeta }_{3}} \right)}^{2}}$$ where the coefficients $${{h}_{i}}\left( \zeta \right)$$ are
 * <p style="text-align:right;">$$\,(Eq. 4.20) $$
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$$\begin{align} & {{h}_{1}}\left( \zeta \right)=1 \\ & {{h}_{2}}\left( \zeta \right)={{\zeta }_{1}} \\ & {{h}_{3}}\left( \zeta \right)={{\zeta }_{1}}\cos \left( {{\zeta }_{2}} \right) \\ \end{align}$$ Now Eq. 4.18 can be re-written as
 * <p style="text-align:right;">$$\,(Eq. 4.21) $$
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$$\Delta \Psi =\frac{1}{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\zeta }_{i}}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{i}}} \right]$$ for the different cases where the index variable i has values of 1,2 and three, the terms of the summation are:
 * <p style="text-align:right;">$$\,(Eq. 4.22) $$
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$$\begin{align} & i=1\to \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right] \\ & i=2\to \frac{\partial }{\partial {{\zeta }_{2}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}}\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]=\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right] \\ & i=3\to \frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right]=\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \\ \end{align}$$
 * <p style="text-align:right;">$$\,(Eq. 4.23) $$
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So that the Laplace equation can be finally written as being:
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$$\Delta \Psi =\frac{1}{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\left\{ \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right]+\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]+\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \right\}$$
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 * <p style="text-align:right;">$$\,(Eq. 4.24) $$
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Additoinal links and useful informations on spherical coordinates and Laplace equations
The following page delivers a short description on how to convert cartesian coordinates into spherical or cylindrical coordinates in Matlab with an already existing function:

Matlab-spherical-transformation

This Wolfram page gives a more in depth insight into spherical cooridnates that is worth to take a look at. Note, that the angle $$\theta $$ was selected to be 90° minus the $$\theta $$ used in the problem above.

Wolfram-SphericalCoordinates

In this page the Wolfram-team shows how to solve the Laplace equation with spherical coordinates using the method of separation of vaiables, like discussed in class:

Wolfram-Laplaces Equation SphericalCoordinates

This paper from the University of Texas-State shows also the solution of the Laplace equation, described in a very elegant way:

Texas-State-University:spherical_laplace.pdf

Contributing Members
Solved and posted by Egm6321.f10.team03.sigillo 19:21, 12 November 2010 (UTC)

test
=Problem 7: Laplace equation in elliptic coordinates=

Given
The article for elliptic coordinates discussed in class is given with it's Laplace equation. The cartesian coordinates are expressed as

where a is the distance from the origin to the foci of the ellips and

Find
The expression for the Laplace equation in the mentioned article needs to be verified using the steps learned for cylindrical and spherical coordinates

Solution
Like in probelm 4, we find an expression for the infinitesimal segment ds in the elliptic coordinate system though:

Therefore we find first the expressions for $$d{{x}_{1}}$$ and $$d{{x}_{2}}$$, that are:

Because of the axis symmetry we can exclude the dependency on a and cancel all terms related to $$da$$. The squared term of Eq. 7.5 and 7.6 are:

To find the expression for the segment ds we regroup, factorize and add the common terms:

Thus the segment ds is just a summation over the $$d{{\mu }^{2}}$$ and $$d{{\nu }^{2}}$$ terms:

Now, the infinitesimal segment ds can be expressed in general as where

We recall that the Laplace equation can be written as:

And in our specific case we can re-write the equation as:

Which is the same expression for the Laplace equation as in the article discussed in class.

Contributing Members
Solved and posted by Egm6321.f10.team03.sigillo 21:32, 13 November 2010 (UTC)