User:EGM6321.F10.TEAM1.HW6

Find
Show that Eq1.3 is a valid solution to 1.1 based on Eq1.2, using the variation of parameters method.

Solution
First let's put the Legendre equation into the standarm form.

$${y}''-\frac{2x}{1-x^{2}}{y}'+\frac{2}{1-x^{2}}y=0$$ We are given the first homogenous solution of this differential equation so we know we can use variation of parameters to solve for the second solution.

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

$$\begin{align} & y=Uu_{1} \\ & {y}'={U}'u_{1}+U{u}'_{1} \\ & {y}={U}u_{1}+2{U}'{u}'_{1}+U{u}''_{1} \\ & {U}u_{1}+2{U}'{u}'_{1}+U{u}_{1}+\frac{2}{x}({U}'u_{1}+U{u}'_{1})+Uu_{1}=0 \\ & U({u}_{1}+\frac{2}{x}{u}'_{1}+u_{1})+{U}'(2{u}'_{1}+\frac{2}{x}u_{1})+{U}u_{1}=0 \\ \end{align}$$ We know that the given solution when plugged into the differential equation will give us zero as seen in the following. Therefore, we can use a reduction of order method and integrate through to solve for the second homogenous solution.

$$\begin{align} & {u}''_{1}+\frac{2}{x}{u}'_{1}+u_{1}=0 \\ & \text{Let }z(x)={U}' \\ & z(2{u}'_{1}-\frac{2x}{1-x^{2}}u_{1})+{z}'u_{1}=0 \\ & 2\int{\frac{{u}'_{1}}{u_{1}}dx}-2\int{\frac{x}{1-x^{2}}dx}+\int{\frac{z}dx}=0 \\ & \ln (u^{2}_{1})+\ln (x^{2}-1)+\ln (z)=C_{1} \\ & z(x)=C_{1}\frac{1}{u^{2}_{1}(x^{2}-1)} \\ & {U}'=z(x) \\ & U=C_{2}+C_{1}\int{\frac{dx}{u^{2}_{1}(x^{2}-1)}} \\ & U=C_{2}+C_{1}\int{\frac{4}{(3x^{2}-1)^{2}(x^{2}-1)}}dx \\ & \text{Used Wolfram Alpha for this integration}\text{.} \\ & \int{\frac{4}{(3x^{2}-1)^{2}(x^{2}-1)}}dx=\frac{1}{2}(\frac{6x}{3x^{2}-1}+\ln \frac{1-x}{1+x}) \\ & U=C_{2}+C_{1}[\frac{1}{2}(\frac{6x}{3x^{2}-1}+\ln \frac{1-x}{1+x})] \\ & y=Uu_{1}=UP_{2} \\ & y=\frac{C_{2}}{2}(3x^{2}-1)+C_{1}[\frac{3}{2}x+\frac{1}{4}(3x^{2}-1)\ln \frac{1-x}{1+x}] \\ & y=\frac{C_{2}}{2}(3x^{2}-1)-C_{1}[\frac{1}{4}(3x^{2}-1)\ln \frac{1+x}{1-x}-\frac{3}{2}x] \\ \end{align}$$

Author Paul Lang

Find
Solve Eq2.1 and Eq2.2. reference King p28, problem 1.1ab

Solution
First let's solve equation 2.1. First let's put the differential equation into the standard form. $${y}''-\frac{x}{x-1}{y}'+\frac{1}{x-1}y=\frac{f(x)}{x-1}=F(x)$$ We assume a trail solution.

$$\ u_{1}=\exp (x)$$

$$\begin{align} & y=Uu_{1} \\ & {y}'={U}'u_{1}+U{u}'_{1} \\ & {y}={U}u_{1}+2{U}'{u}'_{1}+U{u}''_{1} \\ \end{align}$$ Now we plug these 3 equations above back into the original differential equation and rearrange. $$\begin{align} & {U}u_{1}+2{U}'{u}'_{1}+U{u}_{1}-\frac{x}{x-1}({U}'u_{1}+U{u}'_{1})+\frac{1}{x-1}Uu_{1}=F(x) \\ & U({u}_{1}-\frac{x}{x-1}{u}'_{1}+\frac{1}{x-1}u_{1})+{U}'(2{u}'_{1}-\frac{x}{x-1}u_{1})+{U}u_{1}=F(x) \\ \end{align}$$ Because we assumed that our trial solution is indeed a solution of the differential equation it must satisfy the homogenous case.

$$\begin{align} & {u}''_{1}-\frac{x}{x-1}{u}'_{1}+\frac{1}{x-1}u_{1}=0 \\ & \text{Let }z(x)={U}' \\ & z(2{u}'_{1}-\frac{x}{x-1}u_{1})+{z}'u_{1}=F(x) \\ \end{align}$$

$$\begin{align} & \text{Let }a_{0}=(2\frac{{u}'_{1}}{u_{1}}-\frac{x}{x-1}) \\ & \text{Let }A(x)=\frac{F(x)}{u_{1}}=\frac{f(x)}{u_{1}(x-1)} \\ & {z}'+a_{0}z=A(x) \\ \end{align}$$ We solve by using an integrating factor. $$\begin{align} & h(x)=\exp \{\int{a_{0}dx\}} \\ & \int{a_{0}dx}=\int{2\frac{{u}'_{1}}{u_{1}}dx-\int{\frac{x}{x-1}dx}} \\ & h(x)=\exp \{\ln (u{}_{1}^{2})-x-\ln (x-1)\} \\ & h(x)=\exp \{\ln [\exp (2x)]-x-\ln (x-1)\} \\ & \therefore h(x)=\frac{\exp (x)}{x-1} \\ \end{align}$$ Therefore we know the following. $$\begin{align} & z=\frac{x-1}{\exp (x)}(k_{2}+\int{\frac{\exp (x)}{x-1}Adx)} \\ & {U}'=z \\ & U=k_{1}+\int{k_{2}\frac{x-1}{\exp (x)}}dx+\int{\frac{x-1}{\exp (x)}\int{\{\frac{\exp (x)}{x-1}A(x)dx\}dx}} \\ & y=Uu_{1} \\ & \therefore y=k_{1}\exp (x)-k_{2}\frac{\exp (x)}{\exp (x)}x-\frac{\exp (x)}{\exp (x)}x\int{\frac{\exp (x)}{x-1}A(x)dx} \\ \end{align}$$

Now let's solve the differential equation 2.2. Again we start off by putting it into the standard form.

$${y}''+\frac{2}{x}{y}'+y=\frac{f(x)}{x}=F(x)$$ Assume a trial solution and due to foresight of the problem take the derivative of the assumed solution.

$$\begin{align} & u_{1}=\frac{\sin (x)}{x} \\ & {u}'_{1}=\frac{x\cos (x)-\sin (x)}{x^{2}} \\ \end{align}$$

The rest of the solution is completed in the same manor as the above solution.

$$\begin{align} & y=Uu_{1} \\ & {y}'={U}'u_{1}+U{u}'_{1} \\ & {y}={U}u_{1}+2{U}'{u}'_{1}+U{u}''_{1} \\ & {U}u_{1}+2{U}'{u}'_{1}+U{u}_{1}+\frac{2}{x}({U}'u_{1}+U{u}'_{1})+Uu_{1}=F(x) \\ & U({u}_{1}+\frac{2}{x}{u}'_{1}+u_{1})+{U}'(2{u}'_{1}+\frac{2}{x}u_{1})+{U}u_{1}=F(x) \\ & {u}''_{1}+\frac{2}{x}{u}'_{1}+u_{1}=0 \\ & \text{Let }z(x)={U}' \\ & z(2{u}'_{1}+\frac{2}{x}u_{1})+{z}'u_{1}=F(x) \\ & \text{Let }a_{0}=(2\frac{{u}'_{1}}{u_{1}}+\frac{2}{x}) \\ & \text{Let }A(x)=\frac{F(x)}{u_{1}}=\frac{f(x)}{u_{1}x} \\ & {z}'+a_{0}z=A(x) \\ & h(x)=\exp \{\int{a_{0}dx\}} \\ & \int{a_{0}dx}=\int{2\frac{{u}'_{1}}{u_{1}}dx+\int{\frac{2}{x}dx}} \\ & h(x)=\exp \{2\int{\frac{x\cos (x)-\sin (x)}{x^{2}}}\frac{x}{\sin (x)}dx+2\ln (x)\} \\ & h(x)=\exp \{2\int{\frac{\cos (x)}{\sin (x)}dx-2\int{\frac{dx}{x}+\ln (x^{2})}}\} \\ & h(x)=\exp \{2\ln [\sin (x)]\} \\ & \therefore h(x)=\sin ^{2}(x) \\ & z=\frac{1}{\sin ^{2}(x)}\{k_{2}+\int{\sin ^{2}(x)A(x)dx\}} \\ & {U}'=z \\ & U=k_{1}+\int{k_{2}\frac{1}{\sin ^{2}(x)}}dx+\int{\frac{1}{\sin ^{2}(x)}\int{\{\sin ^{2}(x)A(x)dx\}dx}} \\ & \int{\frac{1}{\sin ^{2}(x)}dx=\int{\csc ^{2}(x)dx}}=-\cot (x) \\ & y=Uu_{1} \\ & y=k_{1}\frac{\sin (x)}{x}-k_{2}\frac{\sin (x)}{x}\frac{\cos (x)}{\sin (x)}-\frac{\sin (x)}{x}\frac{\cos (x)}{\sin (x)}\int{\sin ^{2}(x)A(x)dx} \\ & \therefore y=k_{1}\frac{\sin (x)}{x}-k_{2}\frac{\cos (x)}{x}-\frac{\cos (x)}{x}\int{\sin ^{2}(x)A(x)dx} \\ \end{align}$$

Author Paul Lang

Find
Solve Eq3.1 using the direct method. reference King p34 example; also reference Mtgs 29 and 30; see (2) p29-3

Solution
The complementary function is: Knowing: Interchange parameters to obtain $$ P_1(x)=u_1(x) \ $$ and $$ Q_1(x)=u_2(x) \ $$. Next, given the variation of parameters formula: where Obtain particular solution by rewriting Eq3.4 as: Solve for $$ W_0 \ $$ as $$ s \Rightarrow \ 0 \ $$ Given: and Substituting Eq3.7 and Eq3.8 into Eq3.5 results in: As $$ s \Rightarrow \ 0 \ $$ Eq3.9 becomes: Substituting and solving results in: Reducing results in:

Author EGM6321.F10.TEAM1.WILKS 22:47, 13 November 2010 (UTC) EGM6321,F10.TEAM1.WILKS

Find
Solve for and

Solution
Author EGM6321.f10.team1.Kim.TH

Find
1) Solve for in terms of and

2) Find: and identify $$ \left \{h_i \right\} $$ in terms of $$ \left \{\xi _j \right\} $$

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

4) Do separation of variables to recover Bessel's function:

Solution
1) find $$ dx_i\ $$

find $$ dx_1 \ $$

find $$ dx_2 \ $$

find $$ dx_3 \ $$ combine $$ dx_i \ $$

identify $$ ds^2 \ $$

find $$ dx_1^2 \ $$

find $$ dx_2 \ $$

find $$ dx_3 \ $$

combine $$ dx_i^2 \ $$

identify $$ h_i \ $$ by 5.7 and 5.29

find $$ \Delta\ \Psi\ $$ for orthogonal coordinates,

solve summation piecewise for $$ xi_1 \ $$

for $$ xi_2 \ $$

for $$ xi_3 \ $$

recombine

4) Bessel's function is given as:

assuming the solution $$ \phi=0 $$

now, taking the laplacian of this:

substitute in 5.46

divide by $$X_1X_2X_3$$

define a variable $$ \alpha $$. Since this is the only term dependent on $$ \xi_3 $$ it must be constant.

substitute in 5.50 and multiply by $$ \xi_1^2 $$

Now, the second term is the only one dependent on $$ \xi_2 $$ so that it can be redefined as a constant term

define variables $$ x=\xi_1\alpha and X_1=y(x)$$

Author Egm6321.f10.team1.allison 00:19, 10 November 2010 (UTC)

Find
Find $$ \Delta\ \Psi\ $$ in spherical coordinates using the mathematical/physics notation as shown in Eq6.1.

Solution
spherical coordinates in math/science notation can be described as:

find $$ dx_i \ $$

find $$ ds^2 \ $$

which can be simplified to:

identify h: from this the h's are determined to be:

find $$ \Delta\ \Psi\ $$ for orthogonal coordinates,

Author Egm6321.f10.team1.allison 23:42, 10 November 2010 (UTC)

Find
1) Verify expression for $$ \Delta\ \Psi\ $$ in elliptical coordinates from wikipedia article, located at: http://en.wikipedia.org/wiki/Elliptic_coordinates. Use the same steps as in problems 6.5 and 6.6. 2) Obtain the separate equations by separation of variables.

Solution
Elliptical coordinates are defined as:

find $$ x_i's $$

find $$ \Delta\ \Psi\ $$ for orthogonal coordinates,

2) find the seperate equations

divide by $$X_1X_2 $$

for this to be true both terms must be constants.

by the same process,

7.25 and 7.26 could then be solved for $$X_1 $$ and $$ X_2 $$ could then be substituted back to solve $$\Psi $$ Author Egm6321.f10.team1.allison 16:57, 17 November 2010 (UTC)

Given
Where $$\ [n/2] $$ = the integer part of $$\ n/2 $$

Find
Verify that Eq8.3 thru Eq8.7 can be written in the form of Eq8.1 or Eq8.2

Solution
Because equations 8.1 and 8.2 are equivalent, it is only necessary to prove that equations 8.3 through 8.7 may be written in the form of equation 8.1.

1) Verify that equation 8.3 may be written in the form of equation 8.1.

Equation 8.3 is given as:

By observing the subscript in equation 8.3, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.10 yields:

After performing the necessary calculations, equation 8.11 becomes:

This expression of equation 8.1 is equivalent to equation 8.3.

Therefore, equation 8.3 may be written as equation 8.1

2) Verify that equation 8.4 may be written in the form of equation 8.1.

Equation 8.4 is given as:

By observing the subscript in equation 8.4, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.15 yields:

After performing the necessary calculations, equation 8.16 becomes:

This expression of equation 8.1 is equivalent to equation 8.4.

Therefore, equation 8.4 may be written as equation 8.1

3) Verify that equation 8.5 may be written in the form of equation 8.1.

Equation 8.5 is given as:

By observing the subscript in equation 8.5, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.20 yields:

After performing the necessary calculations, equation 8.21 becomes:

This expression of equation 8.1 is equivalent to equation 8.5.

Therefore, equation 8.5 may be written as equation 8.1

4) Verify that equation 8.6 may be written in the form of equation 8.1.

Equation 8.6 is given as:

By observing the subscript in equation 8.6, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.25 yields:

After performing the necessary calculations, equation 8.26 becomes:

This expression of equation 8.1 is equivalent to equation 8.6.

Therefore, equation 8.6 may be written as equation 8.1

5) Verify that equation 8.7 may be written in the form of equation 8.1.

Equation 8.7 is given as:

By observing the subscript in equation 8.7, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.30 yields:

After performing the necessary calculations, equation 8.31 becomes:

This expression of equation 8.1 is equivalent to equation 8.7.

Therefore, equation 8.7 may be written as equation 8.1

It has thus been verified that equations 8.3 through 8.7 may be written as such.

Therefore, equations 8.3 through 8.7 may be written in the form of equation 8.1 or 8.2

Author EGM6321.f10.team1.russo 02:40, 12 November 2010 (UTC)

Find
Verify that Eq9.3 thru Eq9.7 are solutions of Legendre Equations as given in Eq9.1 or Eq9.2

Solution
Since equations 9.2 and 9.1 are identical, it is only necessary to prove that equations 9.3 through 9.7 are solutions to equation 9.1. This will be accomplished by taking the first and second derivatives of each potential solution equation, substituting the results into equation 9.1, and then verifying that there exists an integer $$\ n $$ for which the Legendre differential equation is satisfied:

1) Verify that equation 9.3 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.3 yields:

and taking the second derivative of equation 9.3 yields:

Substituting $$\ P_0(x)$$, $$\ P_0'(x)$$, and $$\ P_0''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=0 $$, equation 9.11 becomes:

And when $$\ n=-1 $$ equation 9.11 becomes:

when $$\ n=0 $$ or when $$\ n=-1 $$

2) Verify that equation 9.4 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.4 yields:

and taking the second derivative of equation 9.4 yields:

Substituting $$\ P_1(x)$$, $$\ P_1'(x)$$, and $$\ P_1''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Dividing both sides of equation 9.18 by $$\ x $$ and rearranging the remaining terms yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=1 $$, equation 9.19 becomes:

And when $$\ n=-2 $$ equation 9.19 becomes:

when $$\ n=1 $$ or when $$\ n=-2 $$

3) Verify that equation 9.5 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.5 yields:

and taking the second derivative of equation 9.5 yields:

Substituting $$\ P_2(x)$$, $$\ P_2'(x)$$, and $$\ P_2''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Rearranging equation 9.26 yields:

Dividing both sides of equation 9.27 by $$\ \frac{1}{2}(3x^2 - 1) $$ yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=2 $$, equation 9.28 becomes:

And when $$\ n=-3 $$ equation 9.28 becomes:

when $$\ n=2 $$ or when $$\ n=-3 $$

4) Verify that equation 9.6 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.6 yields:

and taking the second derivative of equation 9.6 yields:

Substituting $$\ P_3(x)$$, $$\ P_3'(x)$$, and $$\ P_3''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Rearranging equation 9.35 yields:

Dividing both sides of equation 9.36 by $$\ \frac{1}{2}(5x^3 - 3x) $$ yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=3 $$, equation 9.37 becomes:

And when $$\ n=-4 $$ equation 9.37 becomes:

when $$\ n=3 $$ or when $$\ n=-4 $$

4) Verify that equation 9.7 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.7 yields:

and taking the second derivative of equation 9.7 yields:

Substituting $$\ P_4(x)$$, $$\ P_4'(x)$$, and $$\ P_4''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Rearranging equation 9.44 yields:

Dividing both sides of equation 9.45 by $$\ \left(\frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}\right) $$ yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=4 $$, equation 9.46 becomes:

And when $$\ n=-5 $$ equation 9.46 becomes:

when $$\ n=4 $$ or when $$\ n=-5 $$

It has thus been verified that equations 9.3 through 9.7 are solutions to equation 9.1.

Therefore, equations 9.3 through 9.7 are valid solutions to Legendre's differential equation as shown in equation 9.1

Author EGM6321.f10.team1.russo 23:08, 11 November 2010 (UTC)

Signatures
Solved problem 1 -- EGM6321.F10.TEAM1.Lang.P 70.185.113.180 11:47, 17 November 2010 (UTC)

Solved problem 2 -- EGM6321.F10.TEAM1.Lang.P 70.185.113.180 11:47, 17 November 2010 (UTC)

Solved problem 3 -- EGM6321.F10.TEAM1.WILKS 22:47, 13 November 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Solved problem 4 -- EGM6321.f10.team1.Kim.TH

Solved problem 5 -- Egm6321.f10.team1.allison 16:58, 17 November 2010 (UTC)

Solved problem 6 -- Egm6321.f10.team1.allison 16:58, 17 November 2010 (UTC)

Solved problem 7 -- Egm6321.f10.team1.allison 16:58, 17 November 2010 (UTC)

Solved problem 8 -- EGM6321.f10.team1.russo 02:55, 12 November 2010 (UTC)

Solved problem 9 -- EGM6321.f10.team1.russo 23:09, 11 November 2010 (UTC)