User:Egm6321.f09.team5/HW5

 See further comments in google doc for Team 5. Egm6321.f09 19:38, 26 November 2009 (UTC)

Problem Statement
The Laplace differential operation on orthogonal curvilinear coordinates. is given as follows:

$$ \Delta \Psi = \frac {1}{h_{1}h_{2}h_{3}} \sum_{i=1}^{3} \frac{\delta}{\delta z_{1}} \left[ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\delta\Psi}{\delta z_{i}} \right] $$ Equation 1

From the definition of spherical coordinates the following is identified:

$$ h_{1}(z)= 1 \qquad \qquad h_{2}(z) = r \cos(\theta) \qquad \qquad h_{3}(z) = r $$

$$ dz_{1} = dr \qquad \quad \qquad dz_{2} = d\phi \qquad \qquad \qquad dz_{3} = d\theta $$

Solve for when $$ i=2 $$.

Solution
Using the values for h and z when i=2 in Equation 1, the following is obtained:

$$ \frac {1}{r^{2}\cos(\theta)} \frac{\delta}{\delta\phi} \left[ \frac{r^{2}\cos(\theta)}{r^{2}\cos^{2}(\theta)} \frac{\delta\Psi}{\delta\phi} \right] $$

$$ \frac {1}{r^{2}\cos(\theta)} \frac{\delta}{\delta\phi} \left[ \frac{1}{\cos(\theta)} \frac{\delta\Psi}{\delta\phi} \right] $$

Simplyfing yields:

$$ \frac {1}{r^{2}\cos^{2}(\theta)} \frac{\delta^{2}\Psi}{\delta\phi^{2}} $$

Problem Statement
The Laplace differential operation on orthogonal curvilinear coordinates. is given as follows:

$$ \Delta \Psi = \frac {1}{h_{1}h_{2}h_{3}} \sum_{i=1}^{3} \frac{\delta}{\delta z_{1}} \left[ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\delta\Psi}{\delta z_{i}} \right] $$ Equation 1

From the definition of spherical coordinates the following is identified:

$$ h_{1}(z)= 1 \qquad \qquad h_{2}(z) = r \cos(\theta) \qquad \qquad h_{3}(z) = r $$

$$ dz_{1} = dr \qquad \quad \qquad dz_{2} = d\phi \qquad \qquad \qquad dz_{3} = d\theta $$

Solve for when $$ i=3 $$.

Solution
Using the values for h and z when i=3 in Equation 1, the following is obtained:

$$ \frac {1}{r^{2}\cos(\theta)} \frac{\delta}{\delta\theta} \left[ \frac{r^{2}\cos(\theta)}{r^{2}} \frac{\delta\Psi}{\delta\theta} \right] $$

By Simplification, yields:

$$ \frac {1}{r^{2}\cos(\theta)} \frac{\delta}{\delta\theta} \left[ \cos(\theta) \frac{\delta\Psi}{\delta\theta} \right] $$

Problem Statement
From Euler's Equation as stated in Pg. 15-4 of the lecture notes the following is stated:

$$ r^{2}R^{''} + 2rR^{'} - KR = 0 $$ Equation 1

Using Trial Solution:

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

Show that the following is obtained:

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

Solution
The First step is to identify the differential of the trial solutions:

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

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

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

Replacing these values into Equation 1, the following is obtained:

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

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

$$ r^{\lambda} \left[ \lambda (\lambda - 1) + 2\lambda - K \right] = 0 $$

From here the values inside of the brackets are identified to be equal to 0. The following can be written,

$$ \lambda^{2} - \lambda + 2\lambda - k = 0 $$

$$ \lambda^{2} + \lambda = k $$

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

Contributing Authors
--Egm6321.f09.team5.GV 16:53, 3 November 2009 (UTC)

--Egm6321.f09.team5.risher 13:54, 4 November 2009 (UTC)

68.109.14.142 20:13, 6 November 2009 (UTC)