User:Egm6321.f09.team5/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: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

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

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

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.

Equation 1:

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

Equation 2:

$$\lambda_1=n$$

Equation 3:

$$\lambda_2=-(n+1)$$

Problem Solution
To solve, distribute the lambda in equation 1:

$$\lambda^2+\lambda=n(n+1)$$

This can be rearranged to show that it is a quadratic equation:

$$\lambda^2+\lambda-n(n+1)=0$$

This can be solved with the quadratic equation:

$$\lambda_{1,2}=\frac{-1\pm \sqrt{1^2-4(1)(-n(n+1))}}{2(1)}$$

$$\lambda_1=\frac{-1+\sqrt{1+4n(n+1)}}{2}=\frac{-1+\sqrt{(2n+1)^2}}{2}$$

$$\lambda_1=n$$

$$\lambda_2=\frac{-1-\sqrt{1+4n(n+1)}}{2}=\frac{-1-\sqrt{(2n+1)^2}}{2}$$

$$\lambda_2=-(n+1)$$

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

Equation 6:

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

Equation 7:

$$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^ix^{n-2i}\frac{1*3*...(2n-2i-1)}{2^ii!(n-2i)!}$$

Problem Solution
Solution

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
Utilizing Equation 5 from lecture 31 page 3, we will calculate the Legendre polynomial for n = 0:4. We will then compare our calculated value of the Legendre polynomial with those provided in class.

for n=0

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

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

for n=1

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

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

$$P_{1}(x)=2!x/2$$

$$P_{1}(x)=x$$ for n=2

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

$$P_{2}(x)=\frac{24*x^2}{16}-\frac{2}{4}$$ $$P_{2}(x)=0.5(3x^2-1)$$ for n=3

$$P_{3}(x)=\frac{(2*3)!x^3}{2^3*6*6} - \frac {(2(2)-2)!x}{2^3*2}$$

$$P_{3}(x)=\frac{720x^3}{288}-\frac{24x}{16}$$ $$P_{3}(x)= \frac{1}{2}(5x^3-3x)$$

for n=4

$$P_4(x)=\frac{8!x^4}{2^4*4!*4!}-\frac{(8-2)!x^2}{2^4*3!}+\frac{(8-4)!}{2^4*2*(4-2)!}$$

$$P_4(x)=\frac{40320x^4}{9216}-\frac{360x^2}{96}+\frac{24}{64}$$ $$P_4(x)=\frac{35x^4}{8}-\frac{15x^2}{4}+\frac{3}{8}$$

These values obtained utilizing Equation 6 are identical to the Legendre polynomials given in class.

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

$$x^2y''+xy'+(x^2-\nu^2)y=0$$

Problem Solution
Laplace equation:

$$\nabla^2\psi=0$$

Having derived this in Problem #1:

$$\nabla^2\psi=\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+(\frac{1}{\xi_1^2}\frac{\partial^2 \psi}{\partial \xi_2^2})+(\frac{\partial^2 \psi}{\partial \xi_3^2})=0$$

Making an assumption:

$$\psi(\xi_1,\xi_2,\xi_3)=X_1(\xi_1)X_2(\xi_2)X_3(\xi_3)$$

Plugging this into the Laplace equation:

$$\nabla^2\psi=X_2X_3\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+X_1X_3(\frac{1}{\xi_1^2}\frac{\partial^2 X_2}{\partial \xi_2^2})+X_1X_2(\frac{\partial^2 X_3}{\partial \xi_3^2})=0$$

To simplify, divide the equation by:

$$X_1X_2X_3$$

Which yields:

$$\nabla^2\psi=\frac{1}{X_1}\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+\frac{1}{X_2}\frac{1}{\xi_1^2}(\frac{\partial^2 X_2}{\partial \xi_2^2})+\frac{1}{X_3}(\frac{\partial^2 X_3}{\partial \xi_3^2})=0$$

Assuming that:

$$\frac{1}{X_3}(\frac{\partial^2 X_3}{\partial \xi_3^2})=constant$$

Then a separation constant can be used for this term:

$$\frac{1}{X_3}(\frac{\partial^2 X_3}{\partial \xi_3^2})=\alpha^2$$

$$\frac{1}{X_1}\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+\frac{1}{X_2}\frac{1}{\xi_1^2}(\frac{\partial^2 X_2}{\partial \xi_2^2})+\alpha^2=0$$

Multiplying by $$\xi_1^2$$:

$$\frac{\xi_1}{X_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+\frac{1}{X_2}(\frac{\partial^2 X_2}{\partial \xi_2^2})+\xi_1^2\alpha^2=0$$

Now, assuming that:

$$\frac{1}{X_2}(\frac{\partial^2 X_2}{\partial \xi_2^2})=constant$$

Then a separation constant can be used for this term as well:

$$\frac{1}{X_2}(\frac{\partial^2 X_2}{\partial \xi_2^2})=-\beta^2$$

$$\frac{\xi_1}{X_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})-\beta^2+\xi_1^2\alpha^2=0$$

Multiplying by $$X_1$$:

$$\xi_1\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+X_1(\xi_1^2\alpha^2-\beta^2)=0$$

Using the chain rule:

$$\xi_1(\frac{\partial X_1}{\partial \xi_1})+\xi_1^2(\frac{\partial^2 X_1}{\partial \xi_1^2})+X_1(\xi_1^2\alpha^2-\beta^2)=0$$

Rearranging:

$$\xi_1^2(\frac{\partial^2 X_1}{\partial \xi_1^2})+\xi_1(\frac{\partial X_1}{\partial \xi_1})+(\xi_1^2\alpha^2-\beta^2)X_1=0$$

Divide by $$\xi_1^2$$:

$$\frac{\partial^2 X_1}{\partial \xi_1^2}+\frac{1}{\xi_1}(\frac{\partial X_1}{\partial \xi_1})+(\alpha^2-\frac{\beta^2}{\xi_1^2})X_1=0$$

Have to transform variables, so assume:

$$X_1=y(x)$$

$$\frac{\partial X_1}{\partial \xi_1}=\frac{\partial y(\alpha\xi_1)}{\partial \xi_1}=\frac{\partial y(\alpha\xi_1)}{\partial \alpha\xi_1}\frac{\partial \alpha\xi_1}{\partial \xi_1}=\alpha\frac{\partial y(\alpha\xi_1)}{\partial \alpha\xi_1}$$

$$\frac{\partial^2 X_1}{\partial \xi_1^2}=\alpha^2\frac{\partial^2 y(\alpha\xi_1)}{\partial (\alpha\xi_1)^2}$$

$$\xi_1^2(\alpha^2\frac{\partial^2 y(\alpha\xi_1)}{\partial (\alpha\xi_1)^2})+\xi_1(\alpha\frac{\partial y(\alpha\xi_1)}{\partial \alpha\xi_1})+(\xi_1^2\alpha^2-\beta^2)y(\alpha\xi_1)=0$$

If:

$$x=\alpha\xi_1$$

$$\nu=\beta$$

Then:

$$x^2\frac{\partial^2 y(x)}{\partial x^2}+x\frac{\partial y(x)}{\partial x}+(x^2-\nu^2)y(x)=0$$

Which is Bessel's differential equation:

$$x^2y''+xy'+(x^2-\nu^2)y=0$$

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

$$ f = \sum{i}{}g_{i} $$

Show that if {gi} is odd then f is odd. Show that if {gi} is even then f is even.

Problem Solution
An even function is defined as follows:

$$ f(-x) = f(x) $$

An odd function is defined as follows:

$$ f(-x) = -f(x) $$

The properties for even and odd functions state the following: Properties of Even and Odd Functions

The sum of even functions is even 

The sum of odd functions is odd

By this rule if g{i} is odd then f will also be add. Also if g{i} is even then f will also be even.

Problem Statement
pg. 33-3

Given:

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

Show $$ P_{2k}(x)\qquad \qquad $$     is even k=0,1,2...

Show $$ P_{2k+1}(x) \qquad \qquad $$    is odd

Problem Solution
$$ P_{2k}(x) = \sum_{0}^{k}(-1)^{i}\frac{(4k-2i)! \color{blue}{(x)^{2k-2i}}}{2^{2k}i!(2k-i)!(2k-2i)!} $$

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

Since the terms highlighted in blue are raised to an even power, this makes them equivalent. This equivalence makes the Polynomial is considered to be even.

$$ P_{2k+1}(x) = \sum_{0}^{[\frac{2k+1}{2}]}(-1)^{i}\frac{(4k+2-2i)! \color{red}{(x)^{2k+1-2i}}}{2^{2k+1}i!(2k+1-i)!(2k+1-2i)!} $$

$$ P_{2k+1}(-x) = \sum_{0}^{[\frac{2k+1}{2}]}(-1)^{i}\frac{(4k+2-2i)! \color{red}{(-x)^{2k+1-2i}}}{2^{2k+1}i!(2k+1-i)!(2k+1-2i)!} $$

Since the terms highlighted in red will always be different since they will be raised to an odd power, this makes them not equivalent. Because of this the polynomial is considered to be odd.

Problem Statement
Pg. 33-4.

$$q(x)=\sum_{i=0}^4c_ix^i$$

$$c_0=3$$, $$c_1=10$$, $$c_2=15$$, $$c_3=-1$$, $$c_4=5$$

1.) find {$$a_i$$} such that $$q=\sum_{i=0}^4a_iP_i$$

2.) plot $$q=\sum_{i=0}^4c_ix^i=\sum_{i=0}^4a_iP_i$$

Problem Solution
1.)

$$P_i(x)=Ax_i$$


 * Where $$P_i(x)$$ represents Legendre polynomials

$$\left[\begin{matrix} P_0(x) \\ P_1(x) \\ P_2(x) \\ P_3(x) \\ P_4(x) \end{matrix}\right]= \left[\begin{matrix} 1 \\ x \\ \frac{3x^2}{2}-\frac{1}{2} \\ \frac{5x^3}{2}-\frac{3x}{2} \\ \frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} \end{matrix}\right]= \left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \\ \end{matrix}\right]\left[\begin{matrix} 1 \\ x \\ x^2 \\ x^3 \\ x^4 \end{matrix}\right]$$

We know that:

$$q(x)=a_iP(x)=c_ix^i$$

If:

$$P_i(x)=Ax^i$$

Then:

$$a_iAx_i=c_ix_i$$

Simplifying:

$$a_i=A^{-1}c_i$$

$$c_i=\left[\begin{matrix} 3 \\ 10 \\ 15 \\ -1 \\ 5 \\ \end{matrix}\right]$$

$$A=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \\ \end{matrix}\right]$$

$$A^{-1}=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 \\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} \\ \end{matrix}\right]$$

$$a_i=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 \\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} \\ \end{matrix}\right]\left[\begin{matrix} 3 \\ 10 \\ 15 \\ -1 \\ 5 \\ \end{matrix}\right]$$

$$a_i=\left[\begin{matrix} 3 \\ 10 \\ 11 \\ 5.6 \\ 10.314 \\ \end{matrix}\right]$$

Problem Statement
Pg. 34-1: HW added from pg. 22-2 King et al. Pg 28. Problem 1.1b

$$ xy^{''} + 2y^{'} + xy = 0 $$          Equation 0

1) Check for exactness, If not exact can it be made exact using integration factor.

2) Use the following trial solutions to find the homogeneous solution

3) Use the undetermined factor method to find the second homogeneous solution, knowing a homogeneous solution

1)
The First step is to check for exactness using the following conditions:

Condition 1: The ODE must be of the form:

$$F=f(x,y,p)y^{''}+g(x,y,p)$$            Equation 1

Where $$ P=y^{'} $$

Condition 2: The following Conditions must be met:

$$    \ f_{xx} + 2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}  \qquad $$      Equation 2

$$    \ f_{xp} + pf_{yp} + 2f_{y} = g_{pp}  \qquad $$         Equation 3

$$ f(x)=x $$

$$ g(x)= 2p + xy $$

$$ f_{x}= 1 \qquad \qquad  g_{y}= x $$

$$ f_{xx}= 0  \qquad \qquad g_{yp}= 0 $$

$$ f_{xy}= 0  \qquad \qquad  g_{x}= y  $$

$$ f_{y}= 0        \qquad \qquad \qquad  g_{xp}= 0 $$

$$ f_{yy}= 0    \qquad \qquad  g_{p}= 2 $$

$$ f_{yp}= g_{pp}= f_{xp}=0 $$

Equation 2 yields:

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

$$ 0 \neq -x $$

This shows that the ODE is not exact.

The next step is to check whether it can be made exact by using the integration factor method:

The integration factor to be used will be:

$$ h(x,y) = x^{m}y^{n} $$

multiplying the integration factor by the ODE yields the following:

$$ \left [ x^{m}y^{n} \right ] \left[ xy^{''} + 2y^{'} + xy \right ] = 0 $$

$$ x^{m+1}y^{n}y^{''} + 2x^{m}y^{n}y{'} + x^{m+1}y^{n+1} = 0 $$

The next step is to check once again for exactness:

$$ f = x^{m+1}y^{n} $$ $$ g= 2x^{m}y^{n}p + x^{m+1}y^{n+1} $$

$$ f_{x}= y^{n}(m+1)x^{m} \qquad \qquad  g_{y}= 2x^{m}ny^{n-1}p + x^{m+1}(m+1)y^{n} $$

$$ f_{xx}= y^{n}(m+1)mx^{m-1}  \qquad \qquad g_{yp}= 2x^{m}ny^{n-1} $$

$$ f_{xy}= ny^{n-1}(m+1)x^{m}  \qquad \qquad  g_{x}= 2y^{n}pmx^{m-1} + y^{n+1}(m+1)x^{m} $$

$$ f_{y}= x^{m+1}ny^{n-1}        \qquad \qquad \qquad  g_{xp}= 2y^{n}mx^{m-1} $$

$$ f_{yy}= x^{m+1}n(n-1)y^{n-2}    \qquad \qquad  g_{p}= 2x^{m}y^{n} $$

$$ f_{yp}= g_{pp}= f_{xp}=0 $$

Using Equation 3 in order to check for exactness yields the following:

$$ 0 + p(0) + 2(x^{m+1}y^{n-1}n \neq 0 $$

Therefore the new ODE is also concluded not to be exact.

2)
The next step is to find a homogeneous solution by using different trial solutions as follows:

Trial Solution 1

$$ y(x)=\exp^{rx} $$

$$ y^{'}(x) = r\exp^{rx} $$

$$ y^{''}(x) = r^{2}\exp^{rx} $$

using these results back in Equation 0 the following is obtained:

$$ x(r^{2}\exp^{rx}) + 2r\exp^{rx} + x\exp^{rx} = 0 $$

$$ \exp^{rx} (r^{2}(x) + 2r + x) = 0 $$

$$ r^{2}x + 2r + x = 0 $$

This trial solution is abandoned as it does not yield a root for the homogeneous solution

Trial Solution 2

$$ y(x)=x\exp^{rx} $$

$$ y^{'}(x) = (xr\exp^{rx} + x^{'}\exp^{rx}) $$

$$ y^{}(x) = r^{2}x\exp^{rx} + rx^{'}\exp^{rx} + x^{'}r\exp^{rx} + x^{}\exp^{rx} $$

Using these results in Equation 0 the following is obtained:

$$ x^{2}r^{2} + 2rxx^{'} + xx^{''} + 2xr + 2x^{'} = 0 $$

This trial solution is abandoned as it will not yield a root for the homogeneous solution

Trial Solution 3

$$ y(x)=x^{-1}\exp^{rx} $$

$$ y^{'}(x) = (rx^{-1}\exp^{rx} - x^{-2}\exp^{rx}) $$

$$ y^{''}(x) = \exp^{rx}(-2rx^{-2} + r^{2}x^{-1} + 2x^{-3}) $$

Using these results in Equation 0 the following is obtained:

$$\exp^{rx} \left [ r^{2} + 1 \right ] = 0 $$

From the previous equation the following result is found:

$$ r = \sqrt[]{-1} = i $$

$$ y(x) = x^{-1}\exp^{ix} $$

Using the de Moivre relationships the following form is obtained for the homogeneous solution:

$$ y(x)= x^{-1}(cosx + isinx) $$

3)
Given a homogeneous solution the full solution can be found using the undetermined factor method. The full solution is assumed to be of the form:

$$ y(x) = U(x)u_{1}(x) $$

Where $$ u_{1}(x) $$ is the known homogeneous solution.

$$ u_{1}(x)= \frac{sin(x)}{x} $$

Equation 0 is organized as follows:

$$ y^{''} + \frac{2}{x}y^{'} + y = 0 $$

where:

$$ a_{0}=1 \qquad \qquad a_{1} = 2x^{-1} $$

Because of the undetermined factor method the following can be written:

$$ 0 = U^{'} \left [ x^{-2}\sin(x) + 2x^{-1}\cos(x) \right ] + U^{''}\frac{\sin(x)}{x} $$  Equation 3

A variable is defined as follows:

$$ z=U^{'} $$

Equation 3 can be rewritten as follows:

$$ z\left [ x^{-2}\sin(x) + 2\cos(x)x^{-1} \right] + z^{'} \frac{\sin(x)}{x} $$

Solving the previous L1.ODE.VC using integration by parts and then solving for $$ U $$ the following is identified:

$$ u_{2}(x) = u_{1} \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt $$

$$ \int{}^{t} a_{1}(s)ds = 2ln(x) $$

$$ \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt = \int{}^{x}\frac{t^{2}}{\sin(t)^{2}}t^{-2} = -cot(x)  $$

$$ u_{2}(x) = \frac{\sin(x)}{x}\frac{-\cos(x)}{\sin(x)} = -\frac{\cos(x)}{x}  $$

This result confirms the results found in part 2 for the full homogeneous solution.

Problem Statement
Pg. 34-2

Problem Statement
Pg. 35-3: Verify table for Gauss Legendre Quadrature in wikipedia, for {xj} and {wj}, j= 1,...,n and n=1,...,5

Analytical Method to determine weights:

$$ \frac{-2}{(n+1)(P^{'}_{n}(x_{j})P_{n+1}(x_{j})} \qquad \qquad $$   Equation 1

After this Evaluate them numerically with the Abramowitz & Stegun Method.

Numerical Method to determine weights:

$$w_i = \frac{2}{\left( 1-x_i^2 \right) (P'_n(x_i))^2} \,\! \qquad \qquad $$  Equation 2

Problem Solution
When n=1

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

The value for x is the root of the polynomial:

$$ x_{i} = 0 $$

The weight is calculated using Equation 1.

$$ \frac{-2}{(2)(1)(1)(\frac{-1}{2})} = 2 $$

A matlab routine can be setup that will first find the roots of the Legendre Polynomial and then using Equation 1 to find the weights. The code for n=5 is as follows:

n=5 >> p=[7.875 0 -8.75 0 1.875 0]; >> roots(p) ans = 0    -0.9062     -0.5385     0.9062     0.5385     >> weight(0) ans = 0.0813    >> weight(-.9062) ans = 0.0233    >> weight(.9062) ans = 0.0233    >> weight(.5385) ans = -0.0865    >> weight(-.5385) ans = -0.0865    function w= weightnum(x) pp=39.375*(x^4)-26.25*(x^2)+1.875; pp1= 14.4375*(x^6) - 19.6875*(x^4) + 6.5625*(x^2)-2.1875; %n=5 w= -2/((6)*(pp)*(pp1));

The Following Results are Obtained:

The Results are the same as those presented in Wikipedia.

The Next step is to setup a Matlab routine that will calculate the weight evaluating it numerically with Equation 2.

n=2 >> p=[1.5 0 -.5]; >> roots(p) ans = 0.5774    -0.5774     >> weightnum(-.5774) ans = 0.9999    >> weightnum(.5774) ans = 0.9999    function w= weightnum(x) pp=39.375*x^4-26.25*x^2+1.875; %n=5 w= 2/((1-x^2)*(pp)^2);

Either method can be considered accurate to calculate the weights of the Polynomials.

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

Show that:

$$Q_0(x)=\frac{1}{2}log(\frac{1+x}{1-x})=tanh^{-1}(x)$$

Problem Solution
For the equations to be equal:

$$tanh(Q_0)=x$$

$$tanh(Q_0)=\frac{e^{2Q_0}-1}{e^{2Q_0}+1}$$

$$tanh(Q_0)=\frac{e^{2(\frac{1}{2}log(\frac{1+x}{1-x}))}-1}{e^{2(\frac{1}{2}log(\frac{1+x}{1-x}))}+1}$$

$$tanh(Q_0)=\frac{e^{log(\frac{1+x}{1-x}))}-1}{e^{log(\frac{1+x}{1-x}))}+1}$$

$$tanh(Q_0)=\frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1}$$

$$tanh(Q_0)=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}$$

$$tanh(Q_0)=\frac{2x}{2}$$

$$tanh(Q_0)=x$$

Which shows that:

$$Q_0(x)=\frac{1}{2}log(\frac{1+x}{1-x})=tanh^{-1}(x)$$

Contributing Authors
--Egm6321.f09.team5.GV 03:03, 18 November 2009 (UTC) --Egm6321.f09.team5.bear 14:42, 18 November 2009 (UTC) --Egm6321.f09.team5.risher 19:08, 18 November 2009 (UTC) --Egm6321.f09.team5.Jason16:13, 18 Noverber 2009 (UTC)