User:Egm6321.f09.team3/HW5

 See further comments in google doc for Team 3. Egm6321.f09 19:38, 26 November 2009 (UTC)

= Problem 1: Solving for $$ \Delta \Psi $$ =

Given
The differential equation:

$$ \Delta \Psi = \frac{1}{h_1 \cdot h_2 \cdot h_3} \sum_{i=1}^3 \frac {\partial}{\partial \xi_i} \left[ \frac{h_1 \cdot h_2 \cdot h_3}{(h_i)^2} \cdot \frac {\partial \psi}{\partial \xi_i}\right] $$

where

$$ (h_1,h_2,h_3) = (1, r \cdot \cos \theta, r) $$

$$ (\xi_1, \xi_2, \xi_3) = (r, \theta, \varphi) $$

Find
$$ \Delta \Psi_i $$ for i = (1, 2, 3)

Solution
for i = 1:

$$ \Delta \Psi_1 = \frac{1}{r^2 \cdot \cos \theta} \cdot \frac {\partial}{\partial r} \left[ \frac{r^2 \cdot \cos \theta}{(1)^2} \cdot \frac {\partial \psi}{\partial r}\right] $$

for i = 2:

$$ \Delta \Psi_2 = \frac{1}{r^2 \cdot \cos \theta} \cdot \frac {\partial}{\partial \theta} \left[ \frac{r^2 \cdot \cos \theta}{(r \cdot \cos \theta)^2} \cdot \frac {\partial \psi}{\partial \theta}\right] $$

for i = 3:

$$ \Delta \Psi_3 = \frac{1}{r^2 \cdot \cos \theta} \cdot \frac {\partial}{\partial \varphi} \left[ \frac{r^2 \cdot \cos \theta}{(r)^2} \cdot \frac {\partial \psi}{\partial \varphi}\right] $$

Combined:

$$ \Delta \Psi = \frac{1}{r^2 \cdot \cos \theta} \left[\frac {\partial}{\partial r} \left( r^2 \cdot \cos \theta \cdot \frac {\partial \psi}{\partial r}\right)\right] + \frac {\partial}{\partial \theta} \left[ \frac{1}{\cos \theta} \cdot \frac {\partial \psi}{\partial \theta}\right]  +  \frac {\partial}{\partial \varphi} \left[ \cos \theta \cdot \frac {\partial \psi}{\partial \varphi}\right] $$

Simplified:


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$$ \Delta \Psi = \frac{1}{r^2} \frac{\partial}{\partial r}\left[r^2 \frac{\partial \psi}{\partial r}\right] + \frac{1}{r^2 \cdot \cos \theta} \left[\frac{\partial }{\partial \theta}\left(\frac {1}{\cos \theta} \frac {\partial \psi}{\partial \theta}\right)\right] + \frac{1}{r^2} \frac{\partial ^2 \psi}{\partial \varphi^2} $$
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= Problem 2: Prove $$ \lambda (\lambda +1) = k $$ =

Given
The differential equation:

$$ r^2R'' + 2rR' - kR = 0 $$

where

$$ R(r) = r^{\lambda} $$

Find
Prove $$ \lambda (\lambda +1) = k $$

Solution
First, take the derivatives of R:

$$R' = \lambda r^{\lambda - 1}$$

$$R'' = \lambda (\lambda - 1) r^{\lambda - 2}$$

Substitute in:

$$ r^2 \cdot \lambda (\lambda - 1) r^{\lambda - 2} + 2r \cdot \lambda r^{\lambda - 1} - k \cdot r^{\lambda} = 0 $$

Simplify:

$$ \lambda (\lambda - 1) r^{\lambda} + 2 \lambda r^{\lambda} - k r^{\lambda} = 0 $$

Divide by $$ r^{\lambda} $$:

$$ \lambda^2 - \lambda + 2\lambda = k $$

Simplify:


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$$ \lambda(\lambda + 1) = k $$
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, as expected!

= Contributing Team Members =

Angela Diggs Egm6321.f09.team3.diggs 02:05, 4 November 2009 (UTC)

Catherine Snow EGM6321.f09.Team3.cgs 15:22, 4 November 2009 (UTC)

Barrett Mooney Egm6321.f09.team3.mooney 13:10, 3 November 2009 (UTC)

Francisco Kuljevan --Egm6321.f09.team3.Kuljevan.Francisco 04:56, 4 November 2009 (UTC)