User:Egm6321.f11.team2/HW7

=R*7.1 - Infinitesimal Length in Spherical Coordinates =

Find
Show that ds2 in spherical coordinates is as given below:

$$\displaystyle ds^2 = dr^2 + r^2d\theta^2 + r^2\cos^2\theta\, d\phi^2$$

Solution
This problem was solved without referring to past solutions.

Differentiating 7.1.1 yields

$$\displaystyle dx_1 = \cos\theta\cos\phi\,dr - r\sin\theta\cos\phi\,d\theta - r\cos\theta\sin\phi\,d\phi$$

Squaring this expression gives

Similarly for 7.1.2:

$$\displaystyle dx_2 = \cos\theta\sin\phi\,dr - r\sin\theta\sin\phi\,d\theta + r\cos\theta\cos\phi\,d\phi$$

And again for 7.1.3:

$$\displaystyle dx_3 = \sin\theta\,dr + r\cos\theta\,d\theta$$

Finally, substituting 7.1.5 &mdash; 7.1.7 into 7.1.4 and making much use of the trigonometric identity $$\displaystyle \cos^2 a + \sin^2 a = 1$$, we find

$$\displaystyle \begin{align} ds^2 = &\sum_{i=1}^3 dx_i^{\,2} \\ = & \,( \cos^2\theta \cos^2\phi + \cos^2\theta \sin^2\phi + \sin^2\theta ) dr^2 \\ &+ ( r^2 \sin^2\theta \cos^2\phi + r^2 \sin^2\theta \sin^2\phi + r^2 \cos^2\theta) d\theta \\ &+ ( r^2 \cos^2\theta \sin^2\phi + r^2 \cos^2\theta \cos^2\phi) d\phi^2 \\ &- 2r\cos\theta\sin\theta \cancel{( \cos^2\theta + \sin^2\phi - 1 )} dr\,d\theta \\ \\ = &\,\left[\cos^2\theta ( \cos^2\phi + \sin^2\phi ) + \sin^2\theta \right] dr^2 \\ & + r^2 \left[ \sin^2\theta (\cos^2\phi + \sin^2\phi) + \cos^2\theta \right] d\theta^2 \\ & + r^2\cos^2\theta (\sin^2\phi + \cos^2\phi) d\phi^2 \\ \\ = &\, dr^2 + r^2 d\theta^2 + r^2\cos^2\theta\,d\phi^2 \end{align} $$

=R*7.2 - Heat Conduction on a cylinder =

Given
$$\displaystyle x_1 = r \cos \theta = \xi_1 cos \xi_2 $$ $$\displaystyle x_2 = r \sin \theta = \xi_1 sin \xi_2 $$ $$\displaystyle x_3 = z = \xi_3 $$

Find
1. Find $$\displaystyle \left \{ dx_i \right \}=\left \{dx_1,dx_2,dx_3\right \} $$ in terms of $$\displaystyle \left \{ d\xi_j \right \}=\left \{d\xi_1,d\xi_2,d\xi_3\right \} $$ and $$\displaystyle \left \{ d\xi_k \right \}$$ 2. Find $$\displaystyle ds^2=\sum_{i=1}(dx_i)^2= \sum_{k}(h_k)^2(d\xi_k)^2$$. Identify $$\displaystyle \left \{ dh_i \right \}$$ in terms of $$\displaystyle \left \{ d\xi_j \right \}$$ 3. Find $$\displaystyle \Delta u $$ in cylindrical coordinates

4. Use seperation variable and compare to the Bessel equation.

Solution
This problem was solved without referring to past solutions.

part 1

$$\displaystyle ds^2=\sum_{k=1}^3(h_k)^2(d\xi_k)^2= \sum_{i=1}^{3}(dx_i)^2 $$ $$\displaystyle dx_i=\frac{\partial x_i}{\partial \xi_1} d\xi_1+\frac{\partial x_i}{\partial \xi_2} d\xi_2+\frac{\partial x_i}{\partial \xi_3} d\xi_3 $$ $$\displaystyle dx_1 =\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2+(0) d\xi_3 $$ $$\displaystyle dx_2 =\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2+(0) d\xi_3 $$ $$\displaystyle dx_3 =(0) d\xi_1+(0) d\xi_2+(1) d\xi_3 $$

Part 2

$$\displaystyle ds^2=(\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2)^2 + (\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2)^2 + (d\xi_3)^2=(d\xi_1)^2+\xi_1^2 (d \xi_2)^2+(d\xi_3)^2 = (h_1)^2(d\xi_1)^2 + (h_2)^2(d\xi_2)^2 + (h_3)^2(d\xi_3)^2 $$ $$\displaystyle ds^2=dr^2+r^2d \theta^2 + dz^2 $$ Thus, $$\displaystyle h_1(\xi)=1 $$ $$\displaystyle h_2(\xi)=r=\xi_1 $$ $$\displaystyle h_3(\xi)=1 $$

Part 3 $$\displaystyle \Delta u = \frac{1}{h_1h_2h_3}\sum_{i=1}^{3} \frac{\partial}{\partial \xi_i}\left [ \frac{h_1h_2h_3}{h_i^2}\frac{\partial u }{\partial \xi_i} \right] $$ Where $$\displaystyle h_1h_2h_3=\xi_1=r $$ So, $$\displaystyle \Delta u = \frac{1}{\xi_1} \left [\frac{\partial}{\partial \xi_1} \left (\frac{\xi_1}{1^2} \frac{\partial u }{\partial \xi_1}\right ) +\frac{\partial}{\partial \xi_2} \left (\frac{\xi_1}{\xi_1^2}\frac{\partial u }{\partial \xi_2}\right )+\frac{\partial}{\partial \xi_3} \left ( \frac{\xi_1}{1^2}\frac{\partial u }{\partial \xi_3} \right ) \right ] =\frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2} $$

part 4

$$\displaystyle \Delta u = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2}=0 $$ $$\displaystyle u = X(\xi_1)Y(\xi_2)Z(\xi_3) $$ $$\displaystyle \frac{Y Z}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{X Z }{\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+X Y \frac{\partial^2 Z }{\partial \xi_3^2}=0 $$ divide through by  $$\displaystyle XYZ $$ $$\displaystyle \frac{1}{X \xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+\frac{1}{Z} \frac{\partial^2 Z }{\partial \xi_3^2}=0 $$ Assuming that $$\displaystyle \frac{1}{Z} \frac{\partial^2 Z }{\partial \xi_3^2}=A $$ $$\displaystyle \frac{1}{X \xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+A=0 $$ multiplying through by $$\displaystyle \xi_1^2 $$ $$\displaystyle \frac{\xi_1}{X } \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y}\frac{\partial^2 Y }{\partial \xi_2^2}+A \xi_1^2 =0 $$

assumeing for second term $$\displaystyle \frac{1}{Y}\frac{\partial^2 Y }{\partial \xi_2^2}=-B $$ $$\displaystyle \frac{\xi_1}{X } \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) -B+A \xi_1^2 =0 $$ multiplying through by $$\displaystyle X $$ $$\displaystyle \xi_1\frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +(A \xi_1^2-B)X =0 $$ The Original equation is following that $$\displaystyle \xi_1^2 \frac{\partial^2 X }{\partial \xi_1^2}+\xi_1 \frac{\partial X }{\partial \xi_1}+(A \xi_1^2-B)X =0 $$

Lets define $$\displaystyle \xi_1 =x, \,\,\,\,\, X =y $$ Thus, the above equations is following that $$\displaystyle x^2 y''+xy'+(A x^2-B)y =0 $$

=R*7.3 -Laplacian operator in spherical coordinate using Math/Physics convention =

Given
Spherical coordinate using Math/Physics convention:

$$\displaystyle \begin{cases} x_{1}=r\cos\overline{\theta}\cos\varphi=\xi_{1}\cos\xi_{2}\cos\xi_{3}\\ x_{2}=r\cos\overline{\theta}\sin\varphi=\xi_{1}\cos\xi_{2}\sin\xi_{3}\\ x_{3}=r\sin\overline{\theta}=\xi_{1}\sin\xi_{2}\end{cases}$$

(7.3.1) Where $$\displaystyle \overline{\theta}=\frac{\pi}{2}-\theta $$

Find
Find $$\displaystyle \triangle u$$ in this coordinate.

Solution
From 7.3.1, we can obtain

$$\displaystyle \begin{cases} x_{1}=r\sin\theta\cos\varphi=\xi_{1}\cos\xi_{2}\cos\xi_{3}\\ x_{2}=r\sin\theta\sin\varphi=\xi_{1}\cos\xi_{2}\sin\xi_{3}\\ x_{3}=r\cos\theta=\xi_{1}\sin\xi_{2}\end{cases} $$ (7.3.2) Thus $$\displaystyle \begin{cases} dx_{1}=\mathrm{cos}\phi\, r\,\mathrm{cos}\theta\, d\theta+\mathrm{cos}\phi\,\mathrm{sin}\theta\, dr-\mathrm{sin}\phi\, r\,\mathrm{sin}\theta\, d\phi\\ dx_{2}=\mathrm{sin}\phi\, r\,\mathrm{cos}\theta\, d\theta+\mathrm{sin}\phi\,\mathrm{sin}\theta\, dr+\mathrm{cos}\phi\, r\,\mathrm{sin}\theta\, d\phi\\ dx_{3}=\mathrm{cos}\theta\, dr-r\,\mathrm{sin}\theta\, d\theta\end{cases}$$ (7.3.3) Since $$ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}$$, then $$\displaystyle ds^{2}={dr}^{2}+{r}^{2}\,{d\theta}^{2}+{r}^{2}\,{\mathrm{sin}^2\theta}\,{d\phi}^{2}$$ (7.3.4) So $$\displaystyle h_{1}=1,\; h_{2}=r,\; h_{3}=r\sin\theta$$ (7.3.5) The Laplacian operator is as follows: $$\displaystyle \Delta u = \frac{1}{h_1h_2h_3}\sum_{i=1}^{3} \frac{\partial}{\partial \xi_i}\left [ \frac{h_1h_2h_3}{h_i^2}\frac{\partial u }{\partial \xi_i} \right ] $$ (7.3.6) hence $$\displaystyle \Delta u=\frac{1}{r^{2}\sin\theta}\left(\frac{\partial}{\partial r}\left(\frac{r^{2}\sin\theta}{1}\frac{\partial u}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\frac{r^{2}\sin\theta}{r^{2}}\frac{\partial u}{\partial\theta}\right)+\frac{\partial}{\partial\phi}\left(\frac{r^{2}\sin\theta}{r^{2}\sin^{2}\theta}\frac{\partial u}{\partial\phi}\right)\right)$$ (7.3.7)

$$\displaystyle \Delta u=\frac{1}{r^{2}\sin\theta}\left(\frac{\partial}{\partial r}\left(r^{2}\sin\theta\frac{\partial u}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial u}{\partial\theta}\right)+\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\theta}\frac{\partial u}{\partial\phi}\right)\right)$$ (7.3.8)

=R*7.4 - Laplacian in elliptic coordinates =

Given
The most common definition of elliptic coordinates $$\displaystyle (\mu, \nu)$$ is

$$ x = a \ \cosh \mu \ \cos \nu $$

$$ y = a \ \sinh \mu \ \sin \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi].$$

Find
Verify the Laplacian in elliptic coordinates given by

Solution
Elliptical coordinates are

First, we assume

And then, we differentiate (7.4.2) and (7.4.3) with respect to each direction (i.e. x1 and x2).

And then,

Since $$\displaystyle ds^2 = \sum_{i} (dx_i)^2= \sum_{k} {(h_k)^2 (d \xi _k)^2} $$,

And the Laplacian equals

Thus,

=R 7.5 - Laplacian in parabolic coordinates. =

Given
Two dimensional parabolic coordinates $$ (\mu,\nu) $$ are defined by the equations:

Find
Verify the Laplacian in parabolic coordiantes as:

Solution
Differentiating first equation: Differentiating second equation: First equation becomes: Second equation becomes: Now: From Equation 7.5.6: Comparing with equation 7.5.6 we analayze that: Laplace operator for two dimensional system in general curvilinear coordinates is given as: Putting the value of $$ h_1, h_2 $$ in equation 7.5.8 we get

=R*7.6 - =

Given
The Legendre differential equation: $$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = 0$$(7.6.1) The 1st homogeneous solutions of the Legendre differential equation for n = 0,1,2,3,4. $$\displaystyle P_0(x) = 1$$(7.6.2) $$\displaystyle P_1(x) = x$$(7.6.3) $$\displaystyle P_2(x) = \frac{1}{2} (3x^2 - 1)$$(7.6.4) $$\displaystyle P_3(x) = \frac{1}{2} (5x^3 - 3x)$$(7.6.5) $$\displaystyle P_4(x) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$(7.6.6) Which can be generalized as $$\displaystyle P_n(x) = \sum^{[n/2]}_{i=0} (-1)^i \frac{(2n-2i)! x^{n-2i}}{2^n i! (n-i)! (n-2i)!}$$(7.6.7) Where [n\2] = integer part of $$\frac{n}{2}$$

Find
(A) Verify that equations (7.6.2)-(7.6.6) are homogeneous solutions of the Legendre differential equation. (B) Show that equation (7.6.7) can be rewritten as
 * $$\displaystyle P_n(x) = \sum^{[n/2]}_{i=0} \frac{1 \cdot 3 \cdots (2n-2i-1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$$(7.6.8)

(C) Verify that equations (7.6.2)-(7.6.6) can be obtained from equations (7.6.7) and (7.6.8).

Solution
This problem was solved without referring to past solutions.

Solution Part A
To show that equations (7.6.2)-(7.6.6) are homogeneous solutions of their respective Legendre equations, we must first show these equations. For n = 0,
 * $$\displaystyle (1-x^2) y'' - 2 x y' = 0$$

For n = 1,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 2y = 0$$

For n = 2,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 6y = 0$$

For n = 3,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 12y = 0$$

For n = 4,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 20y = 0$$

For n=0, $$\displaystyle P_0(x) = 1$$ $$\displaystyle P_0'(x) = 0$$ $$\displaystyle P_0''(x) = 0$$ Substituting $$\displaystyle P_0(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) 0 - 2 x 0 = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=1, $$\displaystyle P_1(x) = x$$ $$\displaystyle P_1'(x) = 1$$ $$\displaystyle P_1''(x) = 0$$ Substituting $$\displaystyle P_1(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) 0 - 2 x 1 + 2x = 0$$
 * $$\displaystyle - 2x + 2x = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=2, $$\displaystyle P_2(x) = \frac{1}{2} (3x^2 - 1)$$ $$\displaystyle P_2'(x) = 3x$$ $$\displaystyle P_2''(x) = 3$$ Substituting $$\displaystyle P_2(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) 3 - 2 x (3x) + 6(\frac{1}{2} (3x^2 - 1)) = 0$$
 * $$\displaystyle 3 - 3x^2 - 6x^2 + 9x^2 - 3 = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=3, $$\displaystyle P_3(x) = \frac{1}{2} (5x^3 - 3x)$$ $$\displaystyle P_3'(x) = \frac{1}{2} (15x^2 - 3)$$ $$\displaystyle P_3''(x) = 15x$$ Substituting $$\displaystyle P_3(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) (15x) - 2 x (\frac{1}{2} (15x^2 - 3)) + 12(\frac{1}{2} (5x^3 - 3x)) = 0$$
 * $$\displaystyle 15x - 15x^3 -  15x^3 + 3x + 30x^3 - 18x = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=4, $$\displaystyle P_4(x) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$ $$\displaystyle P_4'(x) = \frac{35}{2}x^3 - \frac{15}{2}x$$ $$\displaystyle P_4''(x) = \frac{105}{2}x^2 - \frac{15}{2}$$ Substituting $$\displaystyle P_4(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) (\frac{105}{2}x^2 - \frac{15}{2}) - 2 x (\frac{35}{2}x^3 - \frac{15}{2}x) + 20(\frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}) = 0$$


 * $$\displaystyle \frac{105}{2}x^2 - \frac{15}{2} - \frac{105}{2}x^4 + \frac{15}{2}x^2 - \frac{70}{2}x^4 + \frac{30}{2}x^2 + \frac{175}{2}x^4 - \frac{150}{2} x^2 + \frac{15}{2} = 0$$

Which proves it is a homogeneous equation of the Legendre equation.

Solution Part B
Looking at equations (7.6.7) and (7.6.8) we can see that the terms that must be equal for the equations to be equal is
 * $$\displaystyle \frac{(2n-2i)!}{2^n (n-i)!}$$(7.6.9)

and
 * $$\displaystyle \frac{1 \cdot 3 \cdots (2n-2i-1)}{2^i}$$(7.6.9)

These can be shown to be equal by experimentation. For n = 1,2,3,4 and i = 0,1,2 we see that equation (7.6.9) becomes For n = 1 i=0 :$$\displaystyle \frac{(2-0)!}{2^1 (1-0)!} = 1$$ For n = 2 i=0 :$$\displaystyle \frac{(4-0)!}{2^2 (2-0)!} = 3$$ i=1 :$$\displaystyle \frac{(4-2)!}{2^2 (2-1)!} = \frac{1}{2}$$ For n = 3 i=0 :$$\displaystyle \frac{(6-0)!}{2^3 (3-0)!} = 15$$ i=1 :$$\displaystyle \frac{(6-2)!}{2^3 (3-1)!} = \frac{3}{2}$$ For n = 4 i=0 :$$\displaystyle \frac{(8-0)!}{2^4 (4-0)!} = 105$$ i=1 :$$\displaystyle \frac{(8-2)!}{2^4 (4-1)!} = \frac{15}{2}$$ i=2 :$$\displaystyle \frac{(8-4)!}{2^4 (4-2)!} = \frac{3}{4}$$ We see that equation (7.6.10) becomes For n = 1 i=0 :$$\displaystyle \frac{1}{2^0} = 1$$ For n = 2 i=0 :$$\displaystyle \frac{1 \cdot 3}{2^0} = 3$$ i=1 :$$\displaystyle \frac{1}{2^1} = \frac{1}{2}$$ For n = 3 i=0 :$$\displaystyle \frac{1 \cdot 3 \cdot 5}{2^0} = 15$$ i=1 :$$\displaystyle \frac{1 \cdot 3}{2^1} = \frac{3}{2}$$ For n = 4 i=0 :$$\displaystyle \frac{1 \cdot 3 \cdot 5 \cdot 7}{2^0} = 105$$ i=1 :$$\displaystyle \frac{1 \cdot 3 \cdot 5}{2^1} = \frac{15}{2}$$ i=2 :$$\displaystyle \frac{1 \cdot 3}{2^2} = \frac{3}{4}$$ This shows that equations (7.6.9) and (7.6.10) are equivalent, which proves equation (7.6.8) is equivalent to equation (7.6.7).

Solution Part C
Equations (7.6.2)-(7.6.6) can be obtained from equation (7.6.7) and (7.6.8). For n = 0, $$\displaystyle P_0(x) = \sum^{0}_{i=0} (-1)^i \frac{(2(0)-2i)! x^{(0)-2i}}{2^{(0)} i! ((0)-i)! ((0)-2i)!}$$ $$\displaystyle P_0(x) = (-1)^0 \frac{(2(0)-2(0))! x^{(0)-2(0)}}{2^{(0)} (0)! ((0)-(0))! ((0)-2(0))!}$$ Since (0)! = 1, $$\displaystyle P_0(x) = 1 \frac{1 (1)}{1 (1) (1) (1)} = 1$$ For n = 1, $$\displaystyle P_1(x) = \sum^{0}_{i=0} (-1)^i \frac{(2(1)-2i)! x^{(1)-2i}}{2^{(1)} i! ((1)-i)! ((1)-2i)!}$$ $$\displaystyle P_1(x) = (-1)^0 \frac{(2-2(0))! x^{(1)-2(0)}}{2 (0)! ((1)-(0))! ((1)-2(0))!}$$ $$\displaystyle P_1(x) = (1) \frac{2 x}{2 (1) (1) (1)} = x$$ For n = 2, $$\displaystyle P_2(x) = \sum^{1}_{i=0} (-1)^i \frac{(2(2)-2i)! x^{(2)-2i}}{2^{(2)} i! ((2)-i)! ((2)-2i)!}$$ $$\displaystyle P_2(x) = (-1)^0 \frac{(2(2)-2(0))! x^{(2)-2(0)}}{2^{(2)} (0)! ((2)-(0))! ((2)-2(0))!} + (-1)^1 \frac{(2(2)-2(1))! x^{(2)-2(1)}}{2^{(2)} (1)! ((2)-(1))! ((2)-2(1))!}$$ $$\displaystyle P_2(x) = 1 \frac{(4)! x^2}{4 (1) (2) (2)} - 1 \frac{2 x^0}{4}$$ $$\displaystyle P_2(x) = \frac{3 x^2}{2} - \frac{1}{2}$$ $$\displaystyle P_2(x) = \frac{1}{2} (3 x^2 - 1)$$ For n = 3, $$\displaystyle P_3(x) = \sum^{1}_{i=0} (-1)^i \frac{(2(3)-2i)! x^{(3)-2i}}{2^{(3)} i! ((3)-i)! ((3)-2i)!}$$ $$\displaystyle P_3(x) = (-1)^0 \frac{(2(3)-2(0))! x^{(3)-2(0)}}{2^{(3)} (0)! ((3)-(0))! ((3)-2(0))!} + (-1)^1 \frac{(2(3)-2(1))! x^{(3)-2(1)}}{2^{(3)} (1)! ((3)-(1))! ((3)-2(1))!}$$ $$\displaystyle P_3(x) = \frac{6! x^3}{8 (3!) (3!)} - \frac{6! x}{8 (2!)}$$ $$\displaystyle P_3(x) = \frac{5 x^3}{2} - \frac{3 x}{2}$$ $$\displaystyle P_3(x) = \frac{1}{2} (5 x^3 - 3x)$$ For n = 4, $$\displaystyle P_4(x) = \sum^{2}_{i=0} (-1)^i \frac{(2(4)-2i)! x^{(4)-2i}}{2^{(4)} i! ((4)-i)! ((4)-2i)!}$$ $$\displaystyle P_4(x) = \frac{8! x^4}{16 (4!) (4!)} - \frac{6! x^2}{16 (3!) (2!)} + \frac{4! x^0}{16 (2!) (2!)}$$ $$\displaystyle P_4(x) = \frac{35 x^4}{8} - \frac{15 x^2}{4} + \frac{3}{8}$$

=R*7.7 - Separating Laplace's Equation in Elliptical and Parabolic 2D Coordinates =

Given
Laplace's Equation in Elliptical Coordinates is

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

Laplace's Equation in Parabolic Coordinates is

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

Find
Separate the Laplace equations in both elliptical and parabolic coordinates.

Solution
This problem was solved without referring to past solutions. After the problem was solved, it was checked against F10 Team6's solution for the elliptic case.

We begin with elliptical coordinates. We'll try a solution of the form

$$\displaystyle u(\mu,\nu) = \Mu(\mu)\Nu(\nu)$$

Plugging this into Laplace's equation yields

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

Multiplying both sides of the equation by $$\displaystyle a^2\left(\sinh^2\mu + \sin^2\nu\right)$$ and dividing by $$\displaystyle \Mu(\mu)\Nu(\nu)$$ yields

$$\displaystyle \frac{1}{\Mu}\frac{\partial^2\Mu}{\partial\mu^2} = - \frac{1}{\Nu}\frac{\partial^2\Nu}{\partial\nu^2}$$

The only way this equation can hold is if both sides are equal to a constant k. Thus, the separated equations are

$$\displaystyle \frac{\partial^2\Mu}{\partial\mu^2} = k\Mu$$

and

$$\displaystyle \frac{\partial^2\Nu}{\partial\nu^2} = -k\Nu$$

For parabolic coordinates, the procedure is much the same. We use the same form for u, but this time we multiply both sides of the equation by $$\displaystyle \left(\mu^2 + \nu^2\right)$$ before dividing by $$\displaystyle \Mu(\mu)\Nu(\nu)$$:

$$\displaystyle \begin{align} 0 &= \frac{1}{\mu^2 + \nu^2}\left( \Nu \frac{\partial^2 \Mu}{\partial \mu^2} + \Mu \frac{\partial^2 \Nu}{\partial \nu^2}\right) \\ \frac{1}{\Mu}\frac{\partial^2 \Mu}{\partial \mu^2} &= -\frac{1}{\Nu}\frac{\partial^2 \Nu}{\partial \nu^2} \end{align} $$

As before, both sides must be equal to a constant k, so the separated equations are again

$$\displaystyle \frac{\partial^2\Mu}{\partial\mu^2} = k\Mu$$

and

$$\displaystyle \frac{\partial^2\Nu}{\partial\nu^2} = -k\Nu$$

=R*7.8 - Plot legendre polynomial function and observe the limit of P and Q=

Given
$$\displaystyle y=A \cdot P_n(x)+B \cdot{Q_n(x)} $$

Legendre Polynomials are following that $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ where $$\displaystyle m = \frac{n}{2}$$. $$\displaystyle {P}_{0}(x) = 1 \,\,\,\, {P}_{1}(x) = x \,\,\,\,{P}_{2}(x) = \frac{1}{2}(3x^2-1)\,\,\,\,{P}_{3}(x) = \frac{1}{2}(5x^3-3x,) $$ $$\displaystyle {Q}_{0}(x) = \frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right) \,\,\,\,{Q}_{1}(x) = \frac{1}{2}x \mathrm{log}\left(\frac{1+x}{1-x}\right)-1 \,\,\,\, $$ $$\displaystyle {Q}_{2}(x) = \frac{1}{4}(3x^2-1)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x \,\,\,\,$$ $$\displaystyle {Q}_{3}(x) = \frac{1}{4}(5x^3-3x)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{5}{2}x^2+\frac{2}{3} $$

Find
Show that for n=0; $$\displaystyle \mu \to \pm1 => Q_0(\mu) \to oo$$ Plot $$\displaystyle \begin{align} &\left[P_0, P_1, P_2, P_3\right]\\ &\left[Q_0, Q_1, Q_2, Q_3\right]\\ \end{align} $$ Oberve the limit of $$\displaystyle P_n(\mu) \,\,\, and\,\,\, Q_n(\mu) \,\,\,as\,\,\, \mu \to \pm 1 $$

Solution
This problem was solved by referring to past solutions. Egm6321.f10.team4/HW7.

From the given equations, legendre polynomial equation is following that This is the legendre polynomial graph. The above graph is a function of Q When n=oo, $$\displaystyle P_n(\mu) \,\,\, and\,\,\, Q_n(\mu) \,\,\,as\,\,\, \mu \to \pm 1 $$ It shows that a higher order number is shown the Runge's phenomenon at the edge of graph.

=R7.9 - Finding $$ v_i$$ =

Given
Consider the non-orthonormal basis $$ \{\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}\}$$, expressed in the orthonomal basis $$ \{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\}$$ as folllows: $$\displaystyle \mathbf{b}_{i}=A_{ij}\mathbf{e}_{j}$$ (7.9.2) here $$\displaystyle \mathbf{A}=\left[A_{ij}\right]=\left[\begin{array}{ccc} 5 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 5\end{array}\right]$$ and $$\displaystyle \mathbf{v}=-2\mathbf{e}_{1}+4\mathbf{e}_{2}-5\mathbf{e}_{3}$$ (7.9.3)

Find
$$\displaystyle{\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that $$\displaystyle \mathbf{v}=v_i\mathbf b_i$$

Solution
From 7.9.2 and 7.9.3, we have $$\displaystyle \mathbf{v}=v_{i}A_{ij}\mathbf{e}_{j}$$ (7.9.4) $$\displaystyle \mathbf{\mathbf{A}}^{T}\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.9.5) $$\displaystyle \left[A_{ij}\right]^{T}\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.9.6) Since $$\displaystyle det(\mathbf{A})=-80 \neq 0$$

we have $$\displaystyle \left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[A_{ij}\right]^{-T}\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]=\left[\begin{array}{ccc} \frac{23}{80} & -\frac{11}{40} & \frac{3}{80}\\ -\frac{7}{40} & -\frac{1}{20} & \frac{13}{40}\\ \frac{3}{80} & \frac{9}{40} & -\frac{17}{80}\end{array}\right]\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.9.7) thus $$\displaystyle \left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -\frac{149}{80}\\ -\frac{59}{40}\\ \frac{151}{80}\end{array}\right]$$ (7.9.8)

=Contributing Members= Egm6321.f11.team2.Xia I solved R*7.3 and R7.9, proofread R*7.2 and R*7.5 Egm6321.f11.team2.johri.y 03:08, 7 December 2011 (UTC) I solved R 7.5 and proofread R*7.1 and R*7.4 Egm6321.f11.team2.epps 03:38, 7 December 2011 (UTC) I solved R*7.1 and R*7.7.

Egm6321.f11.team2.kim 15:33, 7 December 2011 (UTC) I solved R*7.2 and R*7.8, proofread R*7.1, R*7.3, R7*.4 and R7.9 Egm6321.f11.team2.rho17:47, 7 December 2011 (UTC) I solved R*7.4, proofread R*7.5, R*7.7, and R7.8. Egm6321.f11.team2.wise 20:06, 7 December 2011 (UTC) I solved R*7.6, and proofread R7.9