User:Egm6321.f09.team4/HW5

 See further comments in google doc for Team 4. Egm6321.f09 19:38, 26 November 2009 (UTC)

= Homework #5: Team 4 =

Problem #1:
(Lecture 29-3 and 29-4)


 * Given :


 * The steady-state heat equation, assuming no heat source and homogenous isotropic material:


 * $$ \begin{align}

\Delta \Psi = \frac{1}{h_{1} h_{2} h_{3}}\ \sum^{3}_{i=1} \frac{\partial}{\partial \xi_{i}}\left[ \frac{h_{1} h_{2} h_{3}}{\left(h_{i}\right)^{2}}\ \frac{\partial \Psi}{\partial \xi_{i}} \right] \end{align} $$


 * Assume spherical coordinates:


 * $$ \begin{align}

\xi_{1} &= r \\ \xi_{2} &= \phi \\ \xi_{3} &= \theta \end{align} $$


 * $$ \begin{align}

h_{1}\left(\xi_{1},\xi_{2},\xi_{3}\right) &= 1 \\ h_{2}\left(\xi_{1},\xi_{2},\xi_{3}\right) &= r\text{ cos }\theta \\ h_{3}\left(\xi_{1},\xi_{2},\xi_{3}\right) &= r \end{align} $$


 * Find :


 * Plug the spherical equations into the steady-state heat equation. Find each term $$i=1,2,3$$.  Term for $$i=1$$ already given in Lecture 24.


 * Solution :


 * (a) Find term for $$i=1$$:


 * First, find $$h_{1} h_{2} h_{3}$$:


 * $$ \begin{align}

h_{1} h_{2} h_{3} &= \left(1\right) \left(r \text{ cos } \theta\right) \left( r \right) \\ &= r^{2}\text{ cos } \theta \end{align} $$


 * Then, plug into steady-state heat equation:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{1}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{1} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{1}} \right] &= \frac{1}{r^{2}\text{ cos } \theta}\ \frac{\partial}{\partial r}\left[ \frac{r^{2}\text{ cos } \theta}{1^{2}} \frac{\partial \Psi}{\partial r}\right] \\

&= \frac{1}{r^{2}\text{ cos } \theta}\ \frac{\partial}{\partial r}\left[ r^{2}\text{ cos } \theta\ \frac{\partial \Psi}{\partial r}\right] \\

&= \frac{1}{r^{2}} \frac{\partial}{\partial r}\left[ r^2\frac{\partial \Psi}{\partial r}\right] \end{align} $$


 * (b) Find term for $$i=2$$:


 * Using a similar procedure as above, we find that:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{2}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{2} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{2}} \right] &=\frac{1}{r^2\text{ cos }\theta }\frac{d}{d\phi }\left [ \frac{r^2 \text{ cos }\theta }{r^2 \text{ cos}^2 \theta }\cdot \frac{d\Psi }{d\phi} \right ] \\

&=\frac{1}{r^2\text{ cos }\theta }\frac{d}{d\phi }\left [ \frac{1}{\text{ cos }\theta} \cdot \frac{d\Psi }{d\phi} \right ] \\

&=\frac{1}{r^2\text{ cos}^2\theta }\frac{d}{d\phi }\left [ \frac{d\Psi }{d\phi} \right ] \end{align} $$


 * (c) Find term for $$i=3$$:


 * As above, we find that:


 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{3}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{3} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{3}} \right] &= \frac{1}{r^2 \text{ cos }\theta} \frac{\partial}{\partial \theta}\left[ \frac{r^2 \text{ cos }\theta}{ r^{2}}\ \frac{ \partial \Psi }{\partial \theta} \right] \\

&=\frac{1}{r^2 \text{ cos }\theta} \frac{\partial}{\partial \theta}\left[ \text{ cos }\theta \cdot \frac{ \partial \Psi }{\partial \theta} \right] \end{align} $$

Contributors: Egm6321.f09.team4.palubin.d 16:28, 2 November 2009 (UTC) Egm6321.f09.Team4.meyerat 00:07, 3 November 2009 (UTC)

Problem #2:
(Lecture 30-2)


 * Given :


 * The L2-ODE-VC:


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


 * Find :


 * Show that:


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


 * is a solution to the ODE where:


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


 * Solution :


 * Solving for the derivatives of R, we get:


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


 * and


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


 * Noting the form of the ODE, we see that:


 * $$r^2 R'' + 2rR'=(r^2 R')'$$


 * Substituting in yields:


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


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


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


 * $$\lambda (\lambda+1)-k=0$$


 * Rearranging this yields our goal:


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


 * Therefore $$R(r)=r^{\lambda}$$ is a solution to the ODE.

Contributors:

Egm6321.f09.team4.palubin.d 16:41, 2 November 2009 (UTC)