User:Egm6321.f09.team5.GV/HW6

Problem Statement
Pg. 31-1 Circular Cylinder Coordinates

$$x=x_1=rcos(\theta)=\xi_1cos(\xi_2)$$

$$y=x_2=rsin(\theta)=\xi_1sin(\xi_2)$$

$$z=x_3=\xi_3$$

1.) Find {$$dx_i$$}={$$dx_1$$,$$dx_2$$,$$dx_3$$} in terms of {$$\xi_i$$}={$$\xi_1$$,$$\xi_2$$,$$\xi_3$$} and {$$d\xi_k$$}

2.) $$ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2$$ and identify {$$h_i$$} in therms of {$$\xi_j$$}

3.) Find Laplacian in these curvilinear coordinates

Problem Solution
1.)

Take the derivative of $$x_1, x_2, and x_3$$ with respect to $$\xi_i$$

$$dx_1=\frac{\partial x_1}{\partial \xi_1}d\xi_1+\frac{\partial x_1}{\partial \xi_2}d\xi_2+\frac{\partial x_1}{\partial \xi_3}d\xi_3$$

$$dx_1=cos(\xi_2)d\xi_1-\xi_1sin(\xi_2)d\xi_2+0$$

$$dx_1=cos(\xi_2)d\xi_1-\xi_1sin(\xi_2)d\xi_2$$

$$dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3$$

$$dx_2=sin(\xi_2)d\xi_1-\xi_1cos(\xi_2)d\xi_2+0$$

$$dx_2=sin(\xi_2)d\xi_1-\xi_1cos(\xi_2)d\xi_2$$

$$dx_1=\frac{\partial x_3}{\partial \xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3$$

$$dx_1=0d\xi_1-0d\xi_2+d\xi_3$$

$$dx_1=d\xi_3$$

2.)

Using the given for $$ds^2$$:

$$ds^2=cos^2(\xi_2)(d\xi_1)^2-2\xi_1cos(\xi_2)sin(\xi_2)d\xi_1d\xi_2+(\xi_1)^2sin^2(\xi_2)(d\xi_2)^2+sin^2(\xi_2)(d\xi_1)^2+2\xi_1cos(\xi_2)sin(\xi_2)d\xi_1d\xi_2+(d\xi_3)^2$$

$$ds^2=cos^2(\xi_2)(d\xi_1)^2+(\xi_1)^2sin^2(\xi_2)(d\xi_2)^2+sin^2(\xi_2)(d\xi_1)^2+(d\xi_3)^2 $$

$$ds^2=(d\xi_1)^2+(\xi_1)^2(d\xi_2)^2+(d\xi_3)^2$$

The {$$h_i$$} are the square root of the coefficients on each term so: $$h_1=1, h_2=\xi_1, h_3=1$$

3.)

A general formula for finding the Laplacian of $$\psi$$ in any orthogonal coordinate system is: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1}[\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{1}{\xi_1}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\xi_1\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+(\frac{1}{\xi_1^2}\frac{\partial^2 \psi}{\partial \xi_2^2})+(\frac{\partial^2 \psi}{\partial \xi_3^2})$$

Problem Statement
Pg. 31-1: Spherical Coordinates. Repeat above steps. Use Astronomic Convention.

$$x=x_1=rcos(\theta)cos(\phi)=\xi_1cos(\xi_3)cos(\xi_2)$$

$$y=x_2=rcos(\theta)sin(\phi)=\xi_1cos(\xi_3)sin(\xi_2)$$

$$z=x_3=rsin(\theta)=\xi_1sin(\xi_3)$$

1.) Find {$$dx_i$$}={$$dx_1$$,$$dx_2$$,$$dx_3$$} in terms of {$$\xi_i$$}={$$\xi_1$$,$$\xi_2$$,$$\xi_3$$} and {$$d\xi_k$$}

2.) $$ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2$$ and identify {$$h_i$$} in therms of {$$\xi_j$$}

3.) Find Laplacian in these curvilinear coordinates

Problem Solution
1.)

Take the derivative of $$x_1, x_2, and x_3$$ with respect to $$\xi_i$$

$$dx_1=\frac{\partial x_1}{\partial \xi_1}d\xi_1+\frac{\partial x_1}{\partial \xi_2}d\xi_2+\frac{\partial x_1}{\partial \xi_3}d\xi_3$$

$$dx_1=cos(\xi_3)cos(\xi_2)d\xi_1-\xi_1cos(\xi_3)sin(\xi_2)d\xi_2-\xi_1sin(\xi_3)cos(\xi_2)d\xi_3$$

$$dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3$$

$$dx_2=cos(\xi_3)sin(\xi_2)d\xi_1+\xi_1cos(\xi_3)cos(\xi_2)d\xi_2-\xi_1sin(\xi_3)sin(\xi_2)d\xi_3$$

$$dx_1=\frac{\partial x_3}{\partial \xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3$$

$$dx_1=sin(\xi_3)d\xi_1+0+\xi_1cos(\xi_3)d\xi_3$$

$$dx_1=sin(\xi_3)d\xi_1+\xi_1cos(\xi_3)d\xi_3$$

2.)

Using the given for $$ds^2$$:

$$ds^2=[cos(\xi_3)cos(\xi_2)d\xi_1-\xi_1cos(\xi_3)sin(\xi_2)d\xi_2-\xi_1sin(\xi_3)cos(\xi_2)d\xi_3]^2+[cos(\xi_3)sin(\xi_2)d\xi_1+\xi_1cos(\xi_3)cos(\xi_2)d\xi_2-\xi_1sin(\xi_3)sin(\xi_2)d\xi_3]^2+[sin(\xi_3)d\xi_1+\xi_1cos(\xi_3)d\xi_3]^2$$

$$ds^2=(d\xi_1)^2+(\xi_1)^2cos^2(\xi_3)(d\xi_2)^2+\xi_1^2(d\xi_3)^2$$

The {$$h_i$$} are the square roots of the coefficients on each term so: $$h_1=1, h_2=\xi_1cos(\xi_3), h_3=\xi_1$$

3.)

A general formula for finding the Laplacian of $$\psi$$ in any orthogonal coordinate system is: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1^2cos(\xi_3)}[\frac{\partial}{\partial \xi_1}(\xi_1^2cos(\xi_3)\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{1}{cos(\xi_3)}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\xi_1cos(\xi_3)\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2cos^2(\xi_3)}(\frac{\partial^2 \psi}{\partial \xi_2^2})+\frac{1}{\xi_1^2cos(\xi_3)}\frac{\partial \psi}{\partial \xi_3}(cos(\xi_3)(\frac{\partial \psi}{\partial \xi_3}))$$

Problem Statement
Pg. 31-2 Find Laplacian in the Math/Physics Convention (r,phi,theta_Bar)

$$\theta_{bar}=\frac{\pi}{2}-\theta$$

Problem Solution
$$x=x_1=rsin(\theta_{bar})cos(\phi)=\xi_1sin(\xi_{3bar})cos(\xi_2)$$

$$y=x_2=rsin(\theta_{bar})sin(\phi)=\xi_1sin(\xi_{3bar})sin(\xi_2)$$

$$z=x_3=rcos(\theta_{bar})=\xi_1cos(\xi_{3bar})$$

$$\xi_{3bar}=\frac{\pi}{2}-\xi_3$$

If this is true, then:

$$sin(\xi_3) = cos(\xi_3bar)$$

$$cos(\xi_3) = sin(\xi_3bar)$$

$$\frac{\partial}{\partial\xi_3}=-\frac{\partial}{\partial\xi_3bar}$$

Plugging this into our spherical Laplacian equation derived in Problem #2 (Part 3):

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2sin^2(\xi_{3bar})}(\frac{\partial^2 \psi}{\partial \xi_2^2})-\frac{1}{\xi_1^2sin(\xi_{3bar})}\frac{\partial \psi}{\partial \xi_{3bar}}(sin(\xi_{3bar})(-\frac{\partial \psi}{\partial \xi_{3bar}})$$

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2sin^2(\xi_{3bar})}(\frac{\partial^2 \psi}{\partial \xi_2^2})+\frac{1}{\xi_1^2sin(\xi_{3bar})}\frac{\partial \psi}{\partial \xi_{3bar}}(sin(\xi_{3bar})(\frac{\partial \psi}{\partial \xi_{3bar}})$$

Problem Statement
Pg. 31-2. Solve equation 1 to yield 2 and 3.

Problem Statement
Pg. 31-2.3 Show Equation 6 equals to equation 7

Problem Statement
Pg. 31-3. Show the steps to verify that 1-5, Legendre Polynomials when n=0...4. can be written as 6 or 7

$$ P_{0}(x) = 1 $$  Equation 1 

$$ P_{1}(x) = x $$  Equation 2 

$$ P_{2}(x) = \frac{1}{2}(3x^{2}-1) $$  Equation 3 

$$ P_{3}(x) = \frac{1}{2}(5x^{3}-3x) $$  Equation 4 

$$ P_{4}(x) = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$  Equation 5 

$$ P_{n}(x) = \sum_{0}^{[n/2]}(-1)^{i}\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-i)!(n-2i)!} $$  Equation 6 

$$P_n(x) = \sum_{i=0}^{[n/2]}\frac{1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$$  Equation 7 

Problem Solution
Using Equation 6, when n=0:

$$ P_{0}(x) = \sum_{0}^{0}(-1)^{i}\frac{(-2i)!x^{-2i}}{i!(-i)!(-2i)!} $$

$$ P_{0}(x) = (-1)^{0}\frac{(0)!x^{0}}{0!(0)!(0)!} = 1 $$

Using Equation 6, when n=1:

$$ P_{1}(x) = \sum_{0}^{0}(-1)^{i}\frac{(2-2i)!x^{1-2i}}{2^{1}i!(1-i)!(1-2i)!} $$

$$ P_{1}(x) = (-1)^{0}\frac{(2)!x^{1}}{2^{1}0!(1)!(1)!} = x $$

Using Equation 6, when n=2:

$$ P_{2}(x) = \sum_{0}^{1}(-1)^{i}\frac{(4-2i)!x^{2-2i}}{2^{2}i!(2-i)!(2-2i)!} $$

$$ P_{2}(x) = (-1)^{0}\frac{(4)!x^{2}}{2^{2}0!(2)!(2)!} + (-1)^{1}\frac{(4-2)!x^{2-2}}{2^{2}1!(2-1)!(2-2)!} = \frac{1}{2}(3x^{2}-1)$$

Using Equation 6, when n=3:

$$ P_{3}(x) = \sum_{0}^{1}(-1)^{i}\frac{(6-2i)!x^{3-2i}}{2^{3}i!(3-i)!(3-2i)!} $$

$$ P_{3}(x) = (-1)^{0}\frac{(6)!x^{3}}{2^{3}0!(3)!(3)!} + (-1)^{1}\frac{(6-2)!x^{3-2}}{2^{3}1!(3-1)!(3-2)!} = \frac{1}{2}(5x^{3}-3x)$$

Using Equation 6, when n=4:

$$ P_{4}(x) = \sum_{0}^{2}(-1)^{i}\frac{(8-2i)!x^{4-2i}}{2^{4}i!(4-i)!(4-2i)!} $$

$$ P_{4}(x) = (-1)^{0}\frac{(8)!x^{4}}{2^{4}0!(4)!(4)!} + (-1)^{1}\frac{(8-2)!x^{4-2}}{2^{4}1!(4-1)!(4-2)!} + (-1)^{2}\frac{(8-4)!x^{0}}{2^{4}2!(4-2)!(0)!} = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$

Problem Statement
Pg. 32-1: Verify that p0-p4 eq 1-5 are solutions of the Legendre Equation.

$$ P_{0}(x) = 1 $$  Equation 1 

$$ P_{1}(x) = x $$  Equation 2 

$$ P_{2}(x) = \frac{1}{2}(3x^{2}-1) $$  Equation 3 

$$ P_{3}(x) = \frac{1}{2}(5x^{3}-3x) $$  Equation 4 

$$ P_{4}(x) = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$  Equation 5 

$$ F=(1-x^{2})y^{''} - 2xy^{'} + n(n+1)y = 0 \qquad $$   Lagendre Equation 

Problem Solution
When n=0:

$$P_{0}(x) = 1 = y $$

$$y^{'}=0 \qquad \qquad y^{''} = 0 $$

Using the solution with the Legendre Equation:

$$ F = (1-x^{2}y^{''} - 2xy^{'} + 0 = 0 $$

$$ F = (1-x^{2}(0) - 2x(0) + 0 = 0 $$

$$ 0 = 0 $$

The Legendre Polynomial is indeed a solution to the Legendre Equation.

When n=1:

$$P_{1}(x) = x = y $$

$$y^{'}= 1 \qquad \qquad y^{''} = 0 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 2y = 0 $$

$$ F=(1-x^{2})(0) - 2x(1) + 2(x) = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=1 is a solution of the Legendre Equation

When n=2:

$$ P_{2}(x)=\frac{1}{2}(3x^{2}-1) = y $$

$$y^{'}= 3x \qquad \qquad y^{''} = 3 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 6y = 0 $$

$$ F=(1-x^{2})(3) - 2x(3x) + 6(1.5x^{2} - .5) = 0 $$

$$ F= 3- 3x^{2} - 6x^{2} + 9x^{2} - 3 = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=2 is a solution of the Legendre Equation

When n=3:

$$ P_{3}(x)=\frac{1}{2}(5x^{3}-3x) = y $$

$$y^{'}= 7.5x^{2} - 1.5 \qquad \qquad y^{''} = 15x $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 12y = 0 $$

$$ F=(1-x^{2})(15x) - 2x(7.5x^{2} - 1.5) + 12(2.5x^{3} - 1.5x) = 0 $$

$$ F= 15x - 15x^{3} + 3x + 30x^{3} - 18x = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=3 is a solution of the Legendre Equation

When n=4:

$$ P_{4}(x)= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8}= y $$

$$y^{'}= 17.5x^{3} - 7.5x \qquad \qquad y^{''} = 52.5x^{2}-7.5 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 20y = 0 $$

$$ F=(1-x^{2})(52.5x^{2}-7.5) - 2x(17.5x^{3} - 7.5x) + 20(\frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8}) = 0 $$

$$ F= (87.5x^{4} - 52.5x^{4} -35x^{4}) + (52.5x^{2} + 7.5x^{2} + 15x^{2} - 75x^{2}) + (-7.5 + 7.5) = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=4 is a solution of the Legendre Equation

Problem Statement
Pg. 32-1 Obtain the separate equations for laplace equations of circular cylinder coordinates and Identify the Bessel Diff. Equation

Problem Statement
pg. 33-3: Regarding even and odd equations

Problem Statement
pg. 33-3: Show the functions are even or odd

Problem Statement
Pg. 33-4.

Problem Statement
Pg. 34-1: HW added from pg. 22-2

Problem Statement
Pg. 34-2

Problem Statement
Pg. 35-3: Verify table for Gauss Legendre Quadrature in wikipedia.....

Problem Statement
Pg. 36-4, Show that the statement is true.