User:Egm6321.f09.Team2/HW5

 See further comments in google doc for Team 2. Egm6321.f09 19:38, 26 November 2009 (UTC)

Homework Assignment #5 - due Wednesday, 11/4, 21:00 UTC

Problem 1
p.29-4 Complete equations (3) & (4) for i=2 and i=3 respectively.

Problem Statement: From the Laplace equation $$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Bigg[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Bigg]$$ where $$h_1(\xi)=1$$ $$h_2(\xi)=rcos\theta=\xi_1cos\xi_2$$ $$h_3(\xi)=r=\xi_1$$ $$h_1h_2h_3=r^2cos\theta$$

Solve for when $$i=2, i=3$$

When $$i=2$$ $$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Bigg[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Bigg]$$ becomes

$$\mathcal {4}\psi_{i=2}={1\over r^2cos\theta}{\partial \over \partial \phi}\Bigg[{r^2cos\theta\over (rcos\theta)^2}{\partial \psi \over \partial \phi}\Bigg]$$

$$\mathcal {4}\psi_{i=2}={1\over r^2cos\theta}{\partial \over \partial \phi}\Bigg[{r^2cos\theta\over (rcos\theta)^2}{\partial \psi \over \partial \phi}\Bigg]={1\over r^2cos\theta}{\partial \over \partial \phi}\Bigg[{r^2cos\theta\over r^2cos^2\theta}{\partial \psi \over \partial \phi}\Bigg]$$

$$\mathcal {4}\psi_{i=2}={1\over r^2cos\theta}{r^2cos\theta\over r^2cos^2\theta}{\partial \over \partial \phi}\Bigg[{\partial \psi \over \partial \phi}\Bigg]$$

 $$\mathcal {4}\psi_{i=2}={1\over r^2cos^2\theta}\Bigg[{\partial^2 \psi \over \partial \phi^2}\Bigg]$$

When $$i=3$$ $$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Bigg[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Bigg]$$ becomes

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[{r^2cos\theta\over (r)^2}{\partial \psi \over \partial \theta}\Bigg]$$

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[{r^2cos\theta\over r^2}{\partial \psi \over \partial \phi}\Bigg]$$

 $$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[cos\theta{\partial \phi\over \partial\theta}\Bigg]$$

Problem 2
p.30-2 Solve eq.(1) using the trial solution method where $$R(r)=r^\lambda$$

Problem Statement:

Solve $${d\over dr}\Bigg(r^2{dR\over dr}\Bigg)=kR$$ by trial solution: $$R(r)=r^\lambda$$

where,

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

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

$$r^2{dR\over dr}=r^2R'=r^2(\lambda r^{\lambda-1})=\lambda r^{\lambda-1+2}$$

$${d\over dr}\Bigg(r^2{dR\over dr}\Bigg)={d\over dr}\Big(\lambda r^{\lambda-1+2}\Big)=kR$$

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

$$\lambda (\lambda+1)r^\lambda=kr^\lambda$$

Divide by $$r^\lambda$$



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

Contributing Team Members
Joe Gaddone 02:03, 2 November 2009 (UTC) Kumanchik 15:46, 4 November 2009 (UTC) Egm6321.f09.Team2.sungsik 21:25, 4 November 2009 (UTC) Egm6321.f09.Team 2.walker