User:Egm6321.f10.team4/HW6

= Problem 1. The Legendre Equation when n=2 = From Meeting 33, p 33-1.

Given
The general form of the Legendre Equation:

Definitions of the Legendre Polynomials:


 * where $$\displaystyle m = \frac{n}{2}$$.

When n=2 and r=1, the first homogeneous solution, P2(x), is given below as:

Find
Show that the second homogeneous solution is:

using the Method of Variation of Parameters.

Solution
When n=2, the given Legendre Equation is as

Firstly, writing the equation in standard format as

On the assumption that we have one solution, say $$\displaystyle y(x) = u_1(x)$$. We will find the other solution by differentiating $$\displaystyle y(x) = U(x)\cdot {u}_{1}(x)$$ gives

Plugging the above deriviates into the (Eq. 1-6) gives

Since $$ \displaystyle {u}_{1}''+{a}_{1}{u}_{1}'+{a}_{0}{u}_{1} $$ is solution of (Eq. 1-6), the term is zero. We will use $$\displaystyle Z:=U'$$ to reduce the order of Eq. 1-8, via the missing-dependent-variable method,

After dividing the equation in terms of Z by the highest coefficient of Z, now we can find integrating factor h(x),

By definition, $$\displaystyle Z=U'=\frac{k_2}{h(x)}$$, we can go back to U(x) by integrating Z

Since we have U(x) now, we can solve y(x)

By plugging h(x) and u1(x) into above equation, we can get the desired solution Q2(x)

Egm6321.f10.team4.Yoon 00:00, 14 Nov 2010 (UTC) - Primary Author Egm6321.f10.team4.ejm 19:47, 17 November 2010 (UTC) Editor

= Problem 2. Non-homog. L2_ODE_VC using Var. of Paramerters =

From Meeting 33, p 33-1.

Given
Solve the Non-homog L2_ODE_VC with the condition $$\displaystyle f(x)=0 $$. K.p.28, Pb 1.1 ab '''Part. a'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (x-1)y''-xy'+y=0 $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle xy''+2y'+xy=0 $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

General trial solution is given on Lecture note p.32-1 as,
 * {| style="width:100%" border="0" align="left"


 * style="width:10%; padding:10px; border:2px solid #ff0000" |
 * style="width:10%; padding:10px; border:2px solid #ff0000" |

$$ \displaystyle y={x}^{c} {e}^{rx} $$ $$
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

Find
Find solutions of the equations.

'''Part. a'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (x-1)y''-xy'+y=0 $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle xy''+2y'+xy=0 $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

Solution
Taking 1st and 2nd derivatives of the trial solution is

Solution for Part. a
Plugging the derivatives of trial solutions into the equation given (eq 2.1) yields,

Taking the common factor ahead of the equation, Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x term should be survived. Thus, the values for c and r by (Eq 2-7) are,

Substituting the values (Eq.2-1) into the trial solution, the homogeneous solutions are determined as

Solution for Part. b
Plugging the derivatives(from (eq.2-4) to (eq.2-6)) of trial solutions into the equation given (eq 2.2) yields,

Expanding the equation,

Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x terms should be survived. Thus, the values for c and r combinations by (Eq.2-11) are,

Substituting the values (Eq.2-2) into the trial solution, the homogeneous solutions are determined as

Egm6321.f10.team4.Yoon 00:00, 14 Nov 2010 (UTC) - Primary Author

= Problem 3: Direct method = From Meeting 33, p 33-1.

Given
The non-homogeneous Legendre equation with n=1:
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left( 1-x^{2} \right)y''+\left( -2x \right)y'+2y=\frac{1}{\left( 1-x^{2} \right)}

$$ $$
 * $$\displaystyle (Eq. 3-1)
 * }
 * }

Find
The complete solution $$y(x)$$ using the direct method

Solution
First modify Eq (3-1) to look as follows
 * {| style="width:100%" border="0"

\begin{align} y''+\underbrace{\frac{-2x}{(1-{x}^{2})}}_{=:a_1(x)}y'+\underbrace{\frac{2}{1-{x}^{2}}}_{=:a_0(x)}y=\frac{1}{\left( 1-x^{2} \right)^{2}} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-2)
 * }
 * }

We are given the first and second homogeneous solutions [Mtg. 30 p. 2]:


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

$$\displaystyle u_1(x)=x $$ $$
 * $$\displaystyle (Eq. 3-3)
 * }
 * }


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

$$\displaystyle u_2(x)=\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1 $$ $$
 * $$\displaystyle (Eq. 3-4)
 * }
 * }

The next step is to find $$h(x)$$ using the definition below
 * {| style="width:100%" border="0" align="left"

$$h(x)=u^2_{1}(x) \cdot exp \left[{\int a_{1}(x)dx}\right]$$ $$
 * $$\displaystyle (Eq. 3-5)
 * }
 * }

Replace $$\displaystyle a_1(x)$$ from Eq. (3-2) and $$\displaystyle u_1(x)$$ from Eq. (3-3) into Eq. (3-5) to get the following

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

\begin{align} h(x) &=x^{2} \cdot exp\left[ \int{\frac{-2x}{1-x^{2}}dx}\right]\\ &=x^{2} \cdot (x^{2}-1)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-6)
 * }
 * }

The final step will be to compute the particular solution which is defined in the equation below
 * {| style="width:100%" border="0" align="left"

$$y_{p}(x)=u_1(x)\int{ \frac{1}{h(x)}\left[ \int{ h(x)f(x)dx} \right]}dx$$ $$
 * $$\displaystyle (Eq. 3-7)
 * }
 * }

from Eq. (3-2)
 * {| style="width:100%" border="0" align="left"

$$f(x)=\frac{1}{\left( 1-x^{2} \right)^{2}}$$
 * }
 * }

Substituting Eq.(3-3), Eq. (3-6) and $$f(x)$$ into Eq. (3-7), and using the help of Wolfram Alpha for solving the integrals:
 * {| style="width:100%" border="0"

\begin{align} y_{p}(x)&=x\int{\frac{1}{x^{2}(x^{2}-1)}\left[\int{x^2(x^2-1)\frac{1}{(1-x^{2})^{2}}dx}\right]}dx\\ &=-x\int{\frac{1}{x^2(x^2-1)}\underbrace{\left[\int{\frac{x^2}{1-x^2}}dx\right]}_{\frac{1}{2}\left[log\left(\frac{1+x}{1-x}\right)-2x\right]}}dx\\ &=\frac{-x}{2}\underbrace{\int{\frac{log \left(\frac{1+x}{1-x}\right)-2x}{x^2(x^2-1)}}dx}_{-\left[\frac{1}{2}log\left(\frac{1+x}{1-x} \right ) \right ]^2+\frac{1}{x}log\left(\frac{1+x}{1-x} \right )+log\left(\frac{1-x^2}{x^2} \right )}\\ &=\frac{x}{8}log^2\left(\frac{1+x}{1-x} \right )+\frac{1}{2}log\left(\frac{1-x}{1+x} \right )+\frac{x}{2}log\left(\frac{x^2}{1-x^2} \right )\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-8)
 * }
 * }

The complete solution to Eq. (3-1) will be a superposition of the two homogeneous solutions and the particular solution


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

\begin{align} y(x)&=k_1u_{1}(x)+k_2u_{2}(x)+y_{p}(x) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-9)
 * }
 * }

where $$\displaystyle k_1 $$ and $$\displaystyle k_2 $$ are constants. Our final solution is thus:
 * {| style="width:100%" border="0"

$$\begin{align} y(x)=k_1\underbrace{x}_{u_1(x)}+k_2\underbrace{\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1}_{u_2(x)}+\underbrace{\frac{x}{8}log^2\left(\frac{1+x}{1-x} \right )+\frac{1}{2}log\left(\frac{1-x}{1+x} \right )+\frac{x}{2}log\left(\frac{x^2}{1-x^2} \right)}_{y_p(x)} \end{align}$$
 * style="width:40%; padding:10px; border:2px solid #8888aa" |
 * style="width:40%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

Egm6321.f10.team4.petralanda.n 14:20, 16 November 2010 (UTC) Egm6321.f10.team4.ejm 17:26, 17 November 2010 (UTC) Editor

= Problem 4 = From Meeting 34, p 34-1,3

Find
4.1

Show

4.2

The Laplace equation

Solution
4.1 The infinitesimal segment $$\displaystyle ds $$ is defined as:

where

so

Substituting equations 4-3,4,5 into equation 4-1 and reducing,

or

4.2

The Laplace equation is defined as

where

thus

for $$\displaystyle i=1$$

for $$\displaystyle i=2$$

for $$\displaystyle i=3$$

Summing from $$\displaystyle i=1:3$$

Egm6321.f10.team4.Auerbach 05:20, 17 November 2010 (UTC) - Author

= Problem 5 = From Meeting 35, p 35-3,4

Find
5.1

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 $$\displaystyle \left \{ d\xi_k \right \}$$

5.2

Find $$\displaystyle ds^2=dx_idx_i=\sum_{i=1}^{3}(dx_i)^2$$. Identify $$\displaystyle \left \{ dh_i \right \}$$ in terms of $$\displaystyle \left \{ d\xi_j \right \}$$

5.3

Find $$\displaystyle \Delta \psi $$ in cylindrical coordinates

5.4

Find the Bessel equation

Solution
5.1 The infinitesimal segment $$\displaystyle ds $$ is defined as:

where

so

5.2

Substituting equations 5-3,4,5 into equation 5-1 and reducing,

or

so

5.3

The Laplace equation is defined as

where

thus

5.4

Lets start off by assuming

and

so,

divide through by $$\displaystyle ABC $$

lets assume the last term to be

multiplying through by $$\displaystyle \xi_1^2 $$

lets assume the second term to be

multiplying through by $$\displaystyle A $$

The first term can be further reduced

If we now assume

and

We can achieve the Bessel equation

Egm6321.f10.team4.Auerbach 18:20, 17 November 2010 (UTC) - Author

= Problem 6 The Laplace Equation in Cylindrical Coordinates =

From [http://upload.wikimedia.org/wikiversity/en/f/fa/2010_11_05_07_46_02.djvu Mtg. 35]

Given
Converting from rectangular to spherical coordinates:

$$\displaystyle x_1 = x = r \sin \bar\theta \cos \phi $$

$$\displaystyle x_2 = y = r \sin \bar\theta \sin \phi $$

$$\displaystyle x_3 = z = r \cos\bar\theta $$

$$\displaystyle \xi_1 = r $$

$$\displaystyle \xi_2 = \bar\theta $$

$$\displaystyle \xi_3 = \phi $$

where $$\displaystyle \bar\theta $$ is equal to $$\displaystyle \pi / 2 - \theta $$ by the math/physics convention, as opposed to the astronomic convention

Find
the Laplacian, $$\displaystyle \Delta \Psi $$, in terms of spherical coordinates.

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

$$\displaystyle dx_1 = \sin \xi_2 \cos \xi_3 d \xi_1 + \xi_1 \cos \xi_2 \cos \xi_3 d \xi_2 - \xi_1 \sin \xi_2 \sin \xi_3 d \xi_3 $$

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

$$\displaystyle dx_2 = \sin\xi_2 \sin\xi_3 d\xi_1 + \xi_1 \cos\xi_2 \sin\xi_3 d\xi_2 + \xi_1 \sin\xi_2 \cos\xi_3 d\xi_3 $$

$$\displaystyle dx_3 = \frac {\partial x_3}{\partial xi_1} d \xi_1 + \frac {\partial x_3}{\partial xi_2} d \xi_2 + \frac {\partial x_3}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_3 = \cos\xi_2 d\xi_1 - \xi_1 \sin\xi_2 d\xi_2 + 0 $$

$$\displaystyle ds^2 = \sum dx_i^2 $$

$$\displaystyle $$

Using the trigonometric identity

$$\displaystyle \cos^2\alpha + \sin^2\alpha = 1 $$

and regrouping the terms in Eq. 6-1 yields

$$\displaystyle ds^2 = d\xi_1^2 + \xi_1^2 d\xi_2^2 + \xi_1^2 \sin^2 \xi_2 d\xi_3^2 $$

where $$\displaystyle h_i^2 $$ are the coefficients of each term in the above expression.

$$\displaystyle h_1 = 1 $$

$$\displaystyle h_2 = \xi_1 = r $$

$$\displaystyle h_3 = \xi_1 \sin\xi_2 = r \sin\bar\theta $$

For any set of orthogonal curvilinear coordinates, the Laplacian is given by

$$\displaystyle \Delta \Psi = \frac {1}{h_1h_2h_3} \sum \frac {\partial}{\partial \xi_i} (\frac {h_1h_2h_3}{h_i^2} \frac {\partial \Psi}{\partial \xi_i}) $$

for $$\displaystyle i = 1 $$

$$\displaystyle = \frac {1}{\xi_1} ( \frac {\partial}{\partial \xi_1} \frac {\xi_1}{h_1^2} \frac {\partial \Psi}{\partial \xi_1} ) $$

$$\displaystyle = \frac {1}{r^2 \sin\bar\theta} \frac {\partial}{\partial r} (r^2 \sin\bar\theta \frac {\partial \Psi}{\partial r} $$

for $$\displaystyle i = 2 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_2} \frac {\xi_1}{h_2^2} \frac {\partial \Psi}{\partial \xi_2}) $$

$$\displaystyle = \frac {1}{r^2 \sin\bar\theta} \frac{\partial}{\partial \bar\theta} (\frac {r^2 \sin \bar\theta}{r^2} \frac {\partial \Psi}{\partial \bar\theta}) $$

for $$\displaystyle i = 3 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_3} \frac {\xi_1}{h_3^2} \frac {\partial \Psi}{\partial \xi_3}) $$

$$\displaystyle = \frac {1}{r^2 \sin\bar\theta} \frac{\partial}{\partial phi} (\frac {r^2 \sin\bar\theta}{r^2 \sin^2 \bar\theta} \frac {\partial \Psi}{\partial \phi}) $$

Therefore, the expression for the Laplacian in terms of spherical coordinates becomes

128.227.51.176 17:51, 12 November 2010 (UTC)Egm6321.f10.team4.osentowski - Author Egm6321.f10.team4.ejm 20:09, 17 November 2010 (UTC) Editor

= Problem 7 - The Laplace Equation in Elliptical Coordinates=

From [http://upload.wikimedia.org/wikiversity/en/8/83/2010_11_09_15_00_14.djvu Mtg. 36]

Given
Converting from rectangular to elliptical coordinates:

$$\displaystyle x_1 = x = a * cosh \mu cos \nu$$

$$\displaystyle x_2 = y = a * sinh \mu sin \nu$$

$$\displaystyle \xi_1 = \mu $$

$$\displaystyle \xi_2 = \nu $$

Find
the Laplacian, $$\displaystyle \Delta \Psi $$, in terms of elliptical coordinates.

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

$$\displaystyle dx_1 = a \cdot sinh(\mu) \cdot cos(\nu) d\mu - a \cdot cosh(\mu) \cdot sin(\nu) d\nu $$

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

$$\displaystyle dx_2 = a * cosh \mu sin \nu d \mu + a * sinh \mu cos \nu d \nu $$

$$\displaystyle ds^2 = \sum dx_i^2 $$

Regrouping terms in Eq. 7-1 gives

$$\displaystyle = a^2 (sinh^2 \mu cos^2 \nu + cosh^2 \mu sin^2 \nu) d \mu^2 + a^2 (sinh^2 \mu cos^2 \nu + cosh^2 \mu sin^2 \nu) d \nu^2 $$

Using the trigonometric identity

$$\displaystyle cosh^2 \alpha = 1 + sinh^2 \alpha $$

and simplifying the expression gives

$$\displaystyle ds^2 = (a^2 sinh^2 \mu + a^2 sin^2 \nu) d \mu^2 + (a^2 sinh^2 \mu + a^2 sin^2 \nu) d \nu^2 $$

$$\displaystyle h^2 $$ is the coefficient of each term.

$$\displaystyle h_1^2 = h_2^2 = a^2 (sinh^2 \mu + sin^2 \nu) $$

The Laplacian can therefore be expressed as

128.227.51.176 17:51, 12 November 2010 (UTC)Egm6321.f10.team4.osentowski - Author

= Problem 8 - General Form Verification of Legendre Polynomials =

From Meeting 36, p 36-2 & 36-3

Given
With the following set of polynomials


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

\begin{align} P\left ( x \right ) = \sum_{i = 0}^{\left [ n/2 \right ]} (-1)^{i} \frac{(2n-2i)!(x^{n-2i})}{(2^{n})(i!)(n-i)!(n-2i)!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8-1)
 * }
 * }


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

$$\displaystyle P\left ( x \right ) = \sum_{i = 0}^{\left [ n/2 \right ]} \frac{(1)(3)...(2n-2i-1)}{(2^{i})(i!)(n-2i)!}(-1)^{i}x^{n-2i} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-2)
 * }
 * }


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

$$\displaystyle P_{0}\left ( x \right ) = 1 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-3)
 * }
 * }


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

$$\displaystyle P_{1}\left ( x \right ) = x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-4)
 * }
 * }


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

$$\displaystyle P_{2}\left ( x \right ) = \frac{1}{2}(3x^{2}-1) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-5)
 * }
 * }


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

$$\displaystyle P_{3}\left ( x \right ) = \frac{1}{2}(5x^{3}-3x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-6)
 * }
 * }


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

$$\displaystyle P_{4}\left ( x \right ) = \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 8-7)
 * }
 * }

Note: [n/2] = integer part of n/2 e.g. n = 5, n/2 = 2.5, [n/2] = 2

Find
Verify that the Legendre Polynomial Eqns. 8-3, 8-4, 8-5, 8-6, 8-7 can be written as Eqn. 8-1 and Eqn 8-2

Solution
Since Eqn 8-1 is equivalent to Eqn 8-2, we will only use Eqn 8-1 to demonstrate the solution

Using Eqn 8-1

At n = 0:

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

$$\displaystyle P_{0}\left ( x \right )= 1\cdot \frac{1\cdot 1}{1\cdot1\cdot1} $$


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

$$\displaystyle P_{0}\left ( x \right )= 1 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 1:

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

$$\displaystyle P_{1}\left ( x \right )= 1\cdot \frac{2!\cdot x}{2!\cdot1!} $$


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

$$\displaystyle P_{1}\left ( x \right )= x $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 2:

$$\displaystyle P_{2}\left ( x \right )= \sum_{i = 0}^{\left [ 2/2 \right ]} (-1)^{i} \frac{(2(2)-2i)!(x^{(2)-2i})}{(2^{(2)})(i!)((2)-i)!((2)-2i)!}= \sum_{i = 0}^{\left [ 1 \right ]} (-1)^{i} \frac{(4-2i)!(x^{2-2i})}{(4)(i!)(2-i)!(2-2i)!}= $$

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

$$\displaystyle P_{2}\left ( x \right )= \frac{4!\cdot x^2}{4\cdot 2!\cdot2!} -1 \frac{2!}{4\cdot 1!\cdot 1!\cdot 1!}= \frac{24\cdot x^2}{16} - \frac{1}{2} $$


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

$$\displaystyle P_{2}\left ( x \right )= \frac{1}{2}(3x^{2}-1) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 3:

$$\displaystyle P_{3}\left ( x \right )= \sum_{i = 0}^{\left [ 3/2 \right ]} (-1)^{i} \frac{(2(3)-2i)!(x^{(3)-2i})}{(2^{(3)})(i!)((3)-i)!((3)-2i)!}= \sum_{i = 0}^{\left [ 1.5 \right ]} (-1)^{i} \frac{(6-2i)!(x^{3-2i})}{(8)(i!)(3-i)!(3-2i)!} $$

$$\displaystyle P_{3}\left ( x \right )=(-1)^{0} \frac{(6-2(0))!(x^{3-2(0))})}{(8)((0))!)(3-(0))!(3-2(0)))!} + (-1)^{1} \frac{(6-2(1))!(x^{3-2(1))})}{(8)((1))!(3-(1))!(3-2(1)))!} $$

$$\displaystyle P_{3}\left ( x \right )= \frac{6!\cdot x^3}{8\cdot 3!\cdot3!} -\frac{4!}{8\cdot 2!\cdot 1!}= \frac{720\cdot x^2}{8\cdot 6\cdot 6} -\frac{24}{16} $$


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

$$\displaystyle P_{3}\left ( x \right )= \frac{1}{2}(5x^{3}-3x) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 4:

$$\displaystyle P_{4}\left ( x \right )= \sum_{i = 0}^{\left [ 4/2 \right ]} (-1)^{i} \frac{(2(4)-2i)!(x^{(4)-2i})}{(2^{(4)})(i!)((4)-i)!((4)-2i)!}= \sum_{i = 0}^{\left [ 2 \right ]} (-1)^{i} \frac{(8-2i)!(x^{4-2i})}{(16)(i!)(4-i)!(4-2i)!} $$

$$\displaystyle P_{4}\left ( x \right )=(-1)^{0} \frac{(8-2(0))!(x^{4-2(0))})}{(16)((0))!)(4-(0))!(4-2(0)))!} + (-1)^{1} \frac{(8-2(1))!(x^{4-2(1))})}{(16)((1))!(4-(1))!(4-2(1)))!}+(-1)^{2} \frac{(8-2(2))!(x^{4-2(2))})}{(16)((2))!(4-(2))!(4-2(2)))!} $$

$$\displaystyle P_{4}\left ( x \right )= \frac{8!\cdot x^4}{16\cdot 4!\cdot4!} -\frac{6!\cdot x^2}{16\cdot 3!\cdot 2!}+\frac{4!}{16\cdot 2!\cdot 2!}= \frac{1680\cdot x^4}{16\cdot 24} -\frac{120\cdot x^2}{16\cdot 2!}+\frac{24}{64} $$


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

$$\displaystyle P_{4}\left ( x \right )= \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} $$
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Egm6321.f10.team4.Corella.RL 12:54, 17 November 2010 (UTC) - Primary Author

= Problem 9 - Solution Verification of Legendre Equation =

From Meeting 36, p 36-3 Meeting 5, p 5-4 Meeting 24, p 24-1

Given
Equations 8.3-8.7 from Problem 8 and

Legendre Equation


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$$\displaystyle (1-x^{2})\cdot y^{''}-2x\cdot y^{'}+n(n+1)\cdot y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-1)
 * }
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Find
Verify that the Legendre Polynomial Eqns. 8-3, 8-4, 8-5, 8-6, 8-7 are solutions of Eqn. 9-1

Solution
Let $$\displaystyle y_{n} = P_n\left ( x \right )$$, where $$\displaystyle y_{n}^{'} = P_{n}^{'}\left ( x \right ) $$ and $$\displaystyle y_{n}^{} = P_{n}^{}\left ( x \right ) $$

Use $$\displaystyle (1-x^{2})\cdot y_{n}^{''}-2x\cdot y_{n}^{'}+n(n+1)\cdot y_{n} $$

At n = 0

$$\displaystyle y_{0}=P_{0}\left ( x \right )= 1 $$,

$$\displaystyle y_{0}^{'} = 0 $$,

$$\displaystyle y_{0}^{''} = 0 $$

$$\displaystyle (1-x^{2})\cdot (0)-2x\cdot (0)+n(n+1)\cdot (1) = (1-x^{2})\cdot (0)-2x\cdot (0)+(0)((0)+1)\cdot (1) = 0 $$

Verified that Eq. 8-3 is a solution of Eq. 9-1

At n = 1

$$\displaystyle y_{1}=P_{1}\left ( x \right )= x $$,

$$\displaystyle y_{1}^{'} = 1 $$,

$$\displaystyle y_{1}^{''} = 0 $$

$$\displaystyle (1-x^{2})\cdot (0)-2x\cdot (1)+n(n+1)\cdot (x) = (1-x^{2})\cdot (0)-2x\cdot (1)+(1)((1)+1)\cdot (x) = x $$

Verified that Eq. 8-4 is not a solution of Eq. 9-1

At n = 2

$$\displaystyle y_{2}=P_{2}\left ( x \right )= \frac{1}{2}(3x^{2}-1) $$,

$$\displaystyle y_{2}^{'} = 3x $$,

$$\displaystyle y_{2}^{''} = 3 $$

$$\displaystyle (1-x^{2})\cdot (3)-2x\cdot (3x)+n(n+1)\cdot (\frac{1}{2}(3x^{2}-1)) = (1-x^{2})\cdot (3)-2x\cdot (3x)+(2)((2)+1)\cdot (\frac{1}{2}(3x^{2}-1)) = $$

$$\displaystyle 3-3x^{2}-6x^{2}+6\cdot (\frac{1}{2}(3x^{2}-1)) = 2 $$

Verified that Eq. 8-5 is not a solution of Eq. 9-1

At n = 3

$$\displaystyle y_{3}=P_{3}\left ( x \right )= \frac{1}{2}(5x^{3}-3x) $$,

$$\displaystyle y_{3}^{'} = \frac{15\cdot x^{2}}{2}-\frac{3}{2} $$,

$$\displaystyle y_{3}^{''} = 15x $$

$$\displaystyle (1-x^{2})\cdot (15x)-2x\cdot (\frac{15\cdot x^{2}}{2}-\frac{3}{2})+n(n+1)\cdot\frac{1}{2}(5x^{3}-3x)= $$ $$\displaystyle 18x-30x^{3}+(3)((3)+1)\cdot\frac{1}{2}(5x^{3}-3x) $$

$$\displaystyle 18x-30x^{3}+12\cdot\frac{1}{2}(5x^{3}-3x)= 18x-30x^{3}+30x^{3}-18x= 0 $$

Verified that Eq. 8-6 is a solution of Eq. 9-1

At n = 4

$$\displaystyle y_{4}=P_{4}\left ( x \right )= \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} $$,

$$\displaystyle y_{4}^{'} = \frac{35x^{3}}{2}-\frac{15x}{2} $$,

$$\displaystyle y_{4}^{''} = \frac{105x^{2}}{2}-\frac{15}{2} $$

$$\displaystyle (1-x^{2})\cdot (\frac{105x^{2}}{2}-\frac{15}{2})-2x\cdot (\frac{35x^{3}}{2}-\frac{15x}{2})+n(n+1)\cdot(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}) $$

$$\displaystyle -\frac{245x^{4}}{2}+90x^{2}-\frac{15}{2}+(4)((4)+1)\cdot(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}) $$

$$\displaystyle -\frac{245x^{4}}{2}+90x^{2}-\frac{15}{2}+\frac{175x^{4}}{2}-75x^{2}+\frac{15}{2}=-\frac{70x^{4}}{2}+15x^{2} $$

Verified that Eq. 8-7 is not a solution of Eq. 9-1

Egm6321.f10.team4.Corella.RL 21:29, 15 November 2010 (UTC) - Primary Author