User:Egm6321.f11.team4/HW7

= Problem R*7.1 - Infinitesimal length in spherical coordinate = From [[media:Pea1.f11.mtg39.djvu|Mtg 39-1]]

Find
Show that the infinitesimal length $$\displaystyle ds $$ in the equation (1.1) can be written in spherical coordinate as follows: Derive the Laplace operator in spherical coordinates.

Solution
Compute $$\displaystyle dx_1, dx_2, dx_3 $$ first,

Then, compute the squares of (1.3), (1.4), and (1.5), Substitute the equations (1.6), (1.7) and (1.8) into the equation (1.1), then, express the equation in terms of $$\displaystyle dr, d\theta$$ and $$\displaystyle d\phi$$,

The Laplace equation in the spherical coordinates is defined by: where,

Substitute the equations (1.11), (1.12) and (1.13) into (1.10),

Therefore $$\displaystyle u $$ can be computed by: Therefore, the Laplace equation for the spherical coordinates is:

Author
Contributed by Shin

= Problem 7-2  Heat conduction on a cylinder = From [[media:Pea1.f11.mtg40.djvu|Mtg 40-5]]

Find
1) Solve for in terms of and

2) Find: Identify $$ \left \{h_i \right\} $$ in terms of $$ \left \{\xi _i \right\} $$

3) Find $$ \Delta u\ $$ in cylindrical coordinates.

4) Use separation of variable to find the separated equations and compare to the Bessel equation.

Solution
1)

find $$ dx_1 \ $$

find $$ dx_2 \ $$

find $$ dx_3 \ $$ Therefore, we can get 2)

Therefore, we can get with 3)


 * {| style="width:100%" border="0"

$$  \displaystyle \Delta u =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{(h_{i})^{2}} \frac{\partial u }{\partial \xi _{i}}\right ] $$      (2.25) For $$\displaystyle i=1$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{h_{1}h_{2}h_{3}}\frac{\partial }{\partial \xi _{1}}\left [ \frac{h_{1}h_{2}h_{3}}{(h_{1})^{2}} \frac{\partial u }{\partial \xi _{1}}\right ]=\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial u }{\partial \xi _{1}} \right ) $$      (2.26) For $$\displaystyle i=2$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{h_{1}h_{2}h_{3}}\frac{\partial }{\partial \xi _{2}}\left [ \frac{h_{1}h_{2}h_{3}}{(h_{2})^{2}} \frac{\partial u }{\partial \xi _{2}}\right ]=\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{2}}\left ( \frac{\xi _{1}}{\xi _{1}^{2}}\frac{\partial u }{\partial \xi _{2}} \right )= \frac{1}{\xi _{1}^{2}}\frac{\partial ^2u }{\partial \xi _{2}} $$      (2.27) For $$\displaystyle i=3$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{h_{1}h_{2}h_{3}}\frac{\partial }{\partial \xi _{3}}\left [ \frac{h_{1}h_{2}h_{3}}{(h_{3})^{2}} \frac{\partial u }{\partial \xi _{3}}\right ]=\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{3}}\left ( \xi _{1}\frac{\partial u }{\partial \xi _{3}} \right )= \frac{\partial^2 u }{\partial \xi _{3}^2} $$      (2.28) Therefore, we can get
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Delta u =\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial u }{\partial \xi _{1}} \right )+\frac{1}{\xi _{1}^{2}}\frac{\partial^2 u }{\partial \xi _{2}^2}+\frac{\partial^2 u }{\partial \xi _{3}^2}=0 $$      (2.29) 4）
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Bessel Equation:
 * {| style="width:100%" border="0"

$$  \displaystyle (1-x^2)y''-2xy'+(x^2-\nu^2)y=0, \ \nu \in \mathbb R $$ (2.30) Assume that $$\displaystyle u=R(r)\Theta(\theta)Z(z)$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Taking the Laplacian of this:

So we have

divide by $$\displaystyle R\Theta Z$$

Let $$\begin{align} \alpha^2:=-\frac{1}{Z}\left(\frac{\partial^2Z}{\partial z^2}\right) \end{align}$$

Then, (2.23) can be written as

Now, define $$\displaystyle \nu^2:=\frac{1}{\Theta}\left(\frac{\partial^2 \Theta}{\partial\theta^2}\right)$$

We have

We define $$\displaystyle x:=r\alpha, y:=R $$

Author
Contributed by Kexin Ren

=''' Problem 7-3. Find $$\Delta u$$ in spherical coordinates using the math/physics convention '''= From [[media:Pea1.f11.mtg40.djvu|Mtg 40-6]]

Find
Find $$ \Delta u $$ in spherical coordinates using the math/physics convention

Solution
From (3.2)

Put (3.3) into (3.1)

Laplace operator in general curvilnear cood is

i=1

i=2

i=3

From (3.6),(3.7) and (3.8),

Author
Chung

= Problem R*7.4 Laplacian in Elliptic Coordinates = From [[media:Pea1.f11.mtg41.djvu|Mtg 41-1]]

Given
The Laplacian operator for general coordinates is given by $$\nabla^{2} u = \Delta u = \frac{1}{h_1h_2h_3} \sum^3_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2h_3}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The elliptic coordinates are given by $$\displaystyle \xi_1 = \mu $$ $$\displaystyle \xi_2 = \nu $$. The scale factors for the elliptic coordinates $$\displaystyle (\mu, \nu)$$ are given by $$\displaystyle h_{\mu} = h_{\nu} = a \sqrt{\sinh^2{\mu} + \sin^2{\nu}}$$.

Find
Verify the Laplacian in elliptic coordinates is given by $$\displaystyle \nabla^{2} u = \Delta u= \frac{1}{a^{2} \left( \sinh^{2} \mu + \sin^{2}\nu \right)} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

Solution
Since this is 2-D problem, the Laplacian operator for general coordinates is reduces to $$\Delta u = \frac{1}{h_1h_2} \sum^2_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The above equation expands to $$\Delta u = \frac{1}{h_1 h_2} \left[ \frac{\partial}{\partial\xi_1} \left( \frac{h_1 h_2}{(h_1)^2} \frac{\partial u}{\partial\xi_1} \right) + \frac{\partial}{\partial\xi_2} \left( \frac{h_1h_2}{(h_2)^2} \frac{\partial u}{\partial\xi_2} \right) \right] $$. Since $$\displaystyle h_{\mu} = h_{\nu} $$, the following terms cancel out: $$\displaystyle \cancel{\frac{h_1 h_2}{(h_1)^2}} $$ and $$\displaystyle \cancel{\frac{h_1 h_2}{(h_2)^2}} $$. With $$\displaystyle \xi_1 = \mu $$, $$\displaystyle \xi_2 = \nu $$ and $$\displaystyle h_{\mu} = h_{\nu} = a \sqrt{\sinh^2{2\mu} - \sin^2{2\nu}}$$, the Laplacian reduces to $$\displaystyle \Delta u= \frac{1}{a^{2} \left( \sinh^{2} \mu + \sin^{2}\nu \right)} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

Author
Egm6341.f11.team4.allen 03:21, 6 December 2011 (UTC)

= Problem R7.5 Laplacian in Parabolic Coordinates = From [[media:Pea1.f11.mtg41.djvu|Mtg 41-2]]

Given
The Laplacian operator for general coordinates is given by $$\nabla^{2} u = \Delta u = \frac{1}{h_1h_2h_3} \sum^3_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2h_3}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The parabolic coordinates are given by $$\displaystyle \xi_1 = \mu $$ $$\displaystyle \xi_2 = \nu $$. The scale factors for the parabolic coordinates $$\displaystyle (\mu, \nu)$$ are given by $$\displaystyle h_{\mu} = h_{\nu} = \sqrt{\mu^2 + \nu^2}$$.

Find
Verify the Laplacian in elliptic coordinates is given by $$\displaystyle \nabla^{2} u = \Delta u= \frac{1}{\mu^2 + \nu^2} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

Solution
Since this is 2-D problem, the Laplacian operator for general coordinates is reduces to $$\Delta u = \frac{1}{h_1h_2} \sum^2_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The above equation expands to $$\Delta u = \frac{1}{h_1 h_2} \left[ \frac{\partial}{\partial\xi_1} \left( \frac{h_1 h_2}{(h_1)^2} \frac{\partial u}{\partial\xi_1} \right) + \frac{\partial}{\partial\xi_2} \left( \frac{h_1h_2}{(h_2)^2} \frac{\partial u}{\partial\xi_2} \right) \right] $$. Since $$\displaystyle h_{\mu} = h_{\nu} $$, the following terms cancel out: $$\displaystyle \cancel{\frac{h_1 h_2}{(h_1)^2}} $$ and $$\displaystyle \cancel{\frac{h_1 h_2}{(h_2)^2}} $$. With $$\displaystyle \xi_1 = \mu $$, $$\displaystyle \xi_2 = \nu $$ and $$\displaystyle h_{\mu} = h_{\nu} = \sqrt{\mu^2 + \nu^2} $$, the Laplacian reduces to $$\displaystyle \Delta u= \frac{1}{\mu^2 + \nu^2} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

Author
Egm6341.f11.team4.allen 03:21, 6 December 2011 (UTC)

=''' Problem 7-6. Verify the homogeneous solutions of the Legendre eq. '''= From [[media:Pea1.f11.mtg41.djvu|Mtg 41-5]]

Find
Verify the homogeneous solutions of the Legendre eq.

Author
Chung

= Problem 7-7 Separated Equations for Laplace equation in elliptic and parabolic coordinates = From [[media:Pea1.f11.mtg41.djvu|Mtg 41-6]]

Given
Laplace equation in elliptic and parabolic coordinates,

$$\displaystyle \nabla^{2} u = \Delta u= \frac{1}{a^{2} \left( \sinh^{2} \mu + \sin^{2}\nu \right)} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) = 0$$

$$\displaystyle \nabla^{2} u = \Delta u= \frac{1}{\mu^2 + \nu^2} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) = 0$$

Find
The separated equations for above equations in Eq. (7.1) and (7.2)

Solution
Consider Eq. (7.1) first, using separation of variable, we write

$$\displaystyle u(\mu, \nu) = A(\mu)B(\nu) $$

Therefore both second derivatives in Eq. (7.1) can be written as,

$$\displaystyle \frac{\partial^{2} u}{\partial \mu^{2}} = B(\nu) \frac{d^{2} A(\mu)}{d \mu^{2}} $$

$$\displaystyle \frac{\partial^{2} u}{\partial \mu^{2}} = A(\mu) \frac{d^{2} B(\nu)}{d \nu^{2}} $$

We also note that the term, $$\displaystyle \frac {1} {a^{2} (\sinh^{2} \mu + \sinh^{2} \nu)} $$ is always positive and is never zero, hence we can eliminate this from Eq. (7.1) to get separated equations,

$$\displaystyle \frac{d^{2} A}{d \mu^{2}}. \frac{1}{A(\mu)} = \frac{d^{2} B(\nu)}{d \nu^{2}}.\frac{1}{B(\nu)} = k = constant $$

Since, common factor in Eq. (7.2) is also always positive and hence can be eliminated, and we are left with same equation as we got for Eq. (7.1), and again performing separation of variables we will get the same separated equations as above.

Author
Ankush

= Problem 7-8 Plot Legendre Functions = From [[media:Pea1.f11.mtg42.djvu|Mtg 42-2]]

Given
Legendre polynomials (functions /first homogeneous solutions)


 * $$\begin{align}

& {{P}_{0}}(\mu)=1 \\ & {{P}_{1}}(\mu)=\mu \\ & {{P}_{2}}(\mu)=\frac{1}{2}(3{{\mu}^{2}}-1) \\ & {{P}_{3}}(\mu)=\frac{1}{2}(5{{\mu}^{3}}-3\mu) \\ \end{align}$$

Legendre functions (second homogeneous solutions)


 * $$\begin{align}

& {{Q}_{0}}(\mu)=\frac{1}{2}\log \left( \frac{1+\mu}{1-\mu} \right) \\ & {{Q}_{1}}(\mu)=\frac{1}{2}\mu\log \left( \frac{1+\mu}{1-\mu} \right)-1 \\ & {{Q}_{2}}(\mu)=\frac{1}{4}(3{{\mu}^{2}}-1)\log \left( \frac{1+\mu}{1-\mu} \right)-\frac{3}{2}\mu \\ & {{Q}_{3}}(\mu)=\frac{1}{4}(5{{\mu}^{3}}-3\mu)\log \left( \frac{1+\mu}{1-\mu} \right)-\frac{5}{2}{{\mu}^{2}}+\frac{2}{3} \\ \end{align}$$

Find
Plot the following figures

$$\displaystyle Fig.1: \left\{P_0,P_1,P_2,P_3\right\}$$

$$\displaystyle Fig.2: \left\{Q_0,Q_1,Q_2,Q_3\right\}$$

Observe the limits of $$\displaystyle P_n(\mu)$$ and $$\displaystyle Q_n(\mu)$$ as $$\displaystyle \mu \to \pm 1$$

Solution
The following is a matlab code.
 * {| style="width:100%" border="0" align="left"

The following is the plot of $$\displaystyle \left\{P_0,P_1,P_2,P_3\right\}$$. The following is a matlab code.
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

The following is the plot of $$\displaystyle \left\{Q_0,Q_1,Q_2,Q_3\right\}$$. So we can find $$\displaystyle \left|Q_n(\mu)\right|\to +\infty$$ as $$\displaystyle \mu \to \pm 1$$
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

Author
Contributed by Kexin Ren

=Problem 7.9 - express in the orthonomal basis= From [[media:Pea1.f11.mtg42.djvu|Mtg 42-6]]

Given
The non-orthonomal basis: $$\displaystyle \mathbf b = \{ \mathbf b_1, \mathbf b_2, \mathbf b_3 \} $$ expressed in the orthonomal basis: $$\displaystyle \mathbf e = \{ \mathbf e_1, \mathbf e_2, \mathbf e_3 \} $$ as follows:

where: $$\displaystyle \mathbf A = [A_{ij}] = \left[ \begin{matrix} 5&2&3\\ 4&5&6\\ 7&8&5 \end{matrix} \right] $$

Find
$$\displaystyle \{ \mathbf v_i \} $$ such that:

Solution
where: $$\displaystyle \mathbf c = \left[ \begin{matrix} -2\\ 4\\ -5 \end{matrix} \right] $$.

Since $$\displaystyle det \mathbf A = 80 \ne 0$$, we can get from (9.1):

where: $$\displaystyle \mathbf A^{-T} = \frac{1}{80}\left[ \begin{matrix} 23&-22&3\\-14&-4&26\\3&18&-17 \end{matrix} \right] $$.

So:

So:

Author
YuChen

=Contributing Members=