User:Egm6321.f10.team5/hw6

=Problem 1 - Showing Using Variation of Parameters = From (meeting 32 page 3)

Given
$$ P_2(x) = \frac 1 2 (3x^2-1) \,$$ $$\displaystyle (Eq. 1.1) $$

$$ Q_2(x) = \frac 1 4 (3x^2-1) log(\frac {1+x} {1-x}) - \frac 3 2 x \,$$ $$\displaystyle (Eq. 1.2) $$

Find
Assuming Eq. 1.1 is the first homogeneous solution to the Legendre function show that Eq. 1.2 is a second homogeneous solution using the variation of parameters method.

Solution
First we must identify the Equation that Eq. 1.1 and 1.2 are solutions to. The Legendre function is written as

$$ 0 = (1-x^2)y'' - 2xy' + n(n+1)y \,$$ $$\displaystyle (Eq. 1.3) $$

For a second order equation $$n=2$$ so the Legendre function becomes

$$ 0 = (1-x^2)y'' - 2xy' + 6y \,$$ $$\displaystyle (Eq. 1.4) $$

To obtain the second homogeneous solution the general form of a homogeneous L2-ODE-VC will be needed,

$$ 0 = a_2y'' + a_1 y' + a_0 y \,$$ $$\displaystyle (Eq. 1.5) $$

So we use the general solution form,

$$ y(x)=U(x) u_1(x) \,$$ $$\displaystyle (Eq. 1.6) $$

Where $$U(x)$$ is unknown, $$u_1(x) = P_2(x)$$ and $$u_2(x)$$ is the second homogeneous solution in this case. So we now solve for $$u_2(x)$$ in a general form,

$$ u_2(x)=U(x)u_1(x) \,$$ $$\displaystyle (Eq. 1.7) $$

$$ u_2'=U'u_1+Uu_1' \,$$ $$\displaystyle (Eq. 1.8) $$

$$ u_2=Uu_1+2U'u_1'+Uu_1'' \,$$ $$\displaystyle (Eq. 1.9) $$

Plug Eqs. 1.7, 1.8, 1.9 into Eq. 1.5 and simplify,

$$ a_2(Uu_1+2U'u_1'+Uu_1)+a_1(U'u_1+Uu_1')+a_0Uu_1=0 \,$$ $$\displaystyle (Eq. 1.10) $$

$$ U(a_2u_1+a_1u_1'+a_0u_1)+U'(2a_2u_1'+a_1u_1)+Ua_2u_1=0 \,$$ $$\displaystyle (Eq. 1.11) $$

This can be simplified further noting that $$U$$ is multiplied by the original general form of the L2-ODE-VC, which equals $$0$$. So that completely eliminates one term in Eg. 1.11. We then get the resulting equation,

$$ U'(2a_2u_1'+a_1u_1)+U''a_2u_1=0 \,$$ $$\displaystyle (Eq. 1.12) $$

Then we force $$a_2 = 1$$, use the identity $$Z=U'$$ to reduce the order of Eq. 1.12, and divide through by $$u_1$$ to get,

$$ Z(\frac{2u_1'} {u_1} + a_1)+Z'=0 \,$$ $$\displaystyle (Eq. 1.13) $$

This now makes the equation of first order. We can then use the integrating factor method to make the equation exact and thus easier to solve for $$U(x)$$. According to Eq. (1) on p.10-3 of the class notes the integrating factor for a linear first order ODE is of the form,

$$ h(x) = exp[\int a_0(s) ds] \,$$ $$\displaystyle (Eq. 1.14) $$

In this case $$a_0 = (\frac{2u_1'} {u_1} + a_1)$$ so Eq. 1.14 becomes,

$$ h(x) = exp[\int (\frac{2u_1'} {u_1} + a_1) dx] \,$$ $$\displaystyle (Eq. 1.15) $$

Simplifying Eq. 1.15 results in,

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$ $$\displaystyle (Eq. 1.16) $$

Then multiply Eq. 1.13 through by Eq. 1.16 and simplify we find that

$$ hZ' + h'Z = (hZ)' = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.17) $$

Solving for $$Z$$ results in

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.18) $$

Invoking the identity $$Z = U'$$ to solve for $$ U $$.

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.19) $$

Then plug Eq. 1.19 back into Eq. 1.6 to get the expression,

$$ y(x) = u_1 [\int \frac {k_2} {h} dx + k_1] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.20) $$

Then multiply through by the $$u_1$$ term,

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.21) $$

Knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.22) $$

and $$ k_1 = 1 $$

Now we make the Legendre function, Eq. 1.4, fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to sole for $$h(x)$$

$$ 0 = y'' - \frac {2x} {(1-x^2)} y' + \frac {6} {(1-x^2)}y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.23) $$

Therefore,

$$ a_1 = \frac {-2x} {(1-x^2)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.24) $$

$$ a_0 = \frac {6} {(1-x^2)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.25) $$

Now plug Eq. 1.22 and Eq. 1.1 back into Eq. 1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = (\frac 3 2 x^2 - \frac 1 2)^2 (x^2 -1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.26) $$

Then plug Eq. 1.26 and Eq. 1.1 into Eq. 1.22 to get the expression for the 2nd homogeneous solution. We will also assume $$k_2=-1$$ for reasons that will become clear later. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = [\frac 1 2 (3 x^2 - 1)] (\int \frac{-1} {(\frac 3 2 x^2 - \frac 1 2)^2 (x^2 - 1)}) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.27) $$

This simplifies to

$$ u_2(x) = \frac 1 2 (3 x^2 - 1) \frac 1 2 [\frac{-6x} {3x^2 - 1} - log (1-x) + log (x+1)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.28) $$

This equation then simplifies to the form of Eq. 1.2

$$ u_2(x) = Q_2(x) = \frac 1 4 (3x^2-1) log(\frac {1+x} {1-x}) - \frac 3 2 x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.29) $$

Author and proof-reader
 [Author] Rob Carroll

[Proof-reader]

=Problem 2 - Solve Nonhomogeneous L2_ODE_VC= From (meeting 32 page 3)

Given
From p.28 problem 1.1 of King we are given

Part 1 A) $$ (x-1)y'' - xy'+y = f(x) = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.1) $$

with solution

$$ u_1(x) = e^x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.2) $$

and

B) $$ xy'' + 2y'+ xy = f(x) = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.1) $$

with solution

$$ u_1(x) = x^{-1}sin(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.2) $$

Part 2

Instead of homogeneous functions, make $$f(x) = sin(x)$$ for Eqs. 2.1.1 and 2.2.1 and find the general solutions.

Find
Show that $$u_1$$ in parts 1 and 2 are homogeneous solutions to their respective ODE's. Then use variation of parameters to find $$u_2$$ for Eq. 2.1.1 and Eq. 2.2.1

Part 1
Part A

First we must show that Eq. 2.1.2 is a solution to Eq. 2.1.1. To do this we simply plug Eq. 2.1.2 in for $$y$$ in Eq. 2.1.1. To do this we first must find the 1st and 2nd derivatives of Eq. 2.1.2

$$ u_1' = e^x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.3) $$

$$ u_1'' = e^x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.4) $$

Then plug Eq. 2.1.2, 2.1.3 and 2.1.4 into Eq. 2.1.1

$$ (x-1)e^x - x e^x + e^x = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.5) $$

It is then clear that the RHS reduces to zero, which matches the LHS, so we have confirmed that Eq. 2.1.2 is a solution of Eq. 2.1.1.

Next, we wil find the 2nd homogeneous solution to Eq. 2.1.1 using variation of parameters. to do this we use the general form of the homogeneous L2-ODE-VC,

$$ a_2y'' + a_1y' + a_0y = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.6) $$

To solve the general form we use the general form of the solution,

$$ y(x)=U(x) u_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.7) $$

Where $$U(x)$$ is unknown, $$u_1(x)$$ is Eq. 2.1.2 and $$u_2(x)$$ is the second homogeneous solution in this case. So we now solve Eq. 2.1.6 for $$y(x)$$ in a general form. To do this we must have the 1st and 2nd derivatives of $$y(x)$$,

$$ y =U(x)u_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.7) $$

$$ y'=U'u_1+Uu_1' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.8) $$

$$ y=Uu_1+2U'u_1'+Uu_1'' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.9) $$

Plug Eqs. 2.1.7, 2.1.8, 2.1.9 into Eq. 2.1.6 and simplify,

$$ a_2(Uu_1+2U'u_1'+Uu_1)+a_1(U'u_1+Uu_1')+a_0Uu_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.10) $$

$$ U(a_2u_1+a_1u_1'+a_0u_1)+U'(2a_2u_1'+a_1u_1)+Ua_2u_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.11) $$

This can be simplified further noting that $$U$$ is multiplied by the original general form of the L2-ODE-VC, which equals $$0$$. So that completely eliminates one term in Eg. 2.1.11. We then get the resulting equation,

$$ U'(2a_2u_1'+a_1u_1)+U''a_2u_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.12) $$

Then we force $$a_2 = 1$$, use the identity $$Z=U'$$ to reduce the order of Eq. 2.1.12, and divide through by $$u_1$$ to get,

$$ Z(\frac{2u_1'} {u_1} + a_1)+Z'=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.13) $$

This now makes the equation of first order. We can then use the integrating factor method to make the equation exact and thus easier to solve for $$U(x)$$. According to Eq. (1) on p.10-3 of the class notes the integrating factor for a linear first order ODE is of the form,

$$ h(x) = exp[\int a_0(s) ds] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.14) $$

In this case $$a_0 = (\frac{2u_1'} {u_1} + a_1)$$ so Eq. 2.1.14 becomes,

$$ h(x) = exp[\int (\frac{2u_1'} {u_1} + a_1) dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.15) $$

Simplifying Eq. 2.1.15 results in,

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.16) $$

Then multiply Eq. 2.1.13 through by Eq. 2.1.16 and simplify we find that

$$ hZ' + h'Z = (hZ)' = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.17) $$

Solving for $$Z$$ results in

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.18) $$

Invoking the identity $$Z = U'$$ to solve for $$ U $$.

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.19) $$

Then plug Eq. 2.1.19 back into Eq. 2.1.7 to get the expression,

$$ y(x) = u_1 [\int \frac {k_2} {h} dx + k_1] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.20) $$

Then multiply through by the $$u_1$$ term,

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.21) $$

Knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.22) $$

and $$ k_1 = 1 $$

Now we make the Eq. 2.1.1 fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to solve for $$h(x)$$

$$ 0 = y'' - \frac {x} {(x-1)} y' + \frac {1} {(x-1)}y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.23) $$

Therefore,

$$ a_1 = \frac {-x} {(x-1)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.24) $$

$$ a_0 = \frac {1} {(x-1)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.25) $$

Now plug Eq. 2.1.22 and Eq. 2.1.2 back into Eq. 2.1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = (\frac {e^x} {x-1} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.26) $$

Then plug Eq. 2.1.26 and Eq. 2.1.2 into Eq. 2.1.22 to get the expression for the 2nd homogeneous solution. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = e^x (k_2 \int \frac {(x-1)} {e^x}) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.27) $$

This simplifies to

$$ u_2(x) = e^x \frac {-k_2x} {e^x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.28) $$

After the negative sign is absorbed into the constant $$k_2$$, the 2nd homogeneous solution simplifies to the form

$$ u_2(x) = k_2 x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.29) $$

Part B

First we must show that Eq. 2.2.2 is a solution to Eq. 2.2.1. To do this we simply plug Eq. 2.2.2 in for $$y$$ in Eq. 2.2.1. To do this we first must find the 1st and 2nd derivatives of Eq. 2.2.2

$$ u_1' = -x^{-2} sin(x) + x^{-1} cos(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.3) $$

$$ u_1'' = 2x^{-3} sin(x) - 2x^{-2} cos(x) - x^{-1} sin(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.4) $$

Then plug Eq. 2.2.2, 2.2.3 and 2.2.4 into Eq. 2.2.1

$$ [2x^{-2} sin(x) - 2x^{-1} cos(x) - sin(x)] + [2x^{-1} cos(x) - 2x^{-2} sin(x)] + sin(x) = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.5) $$

It is then clear that the RHS reduces to zero, which matches the LHS, so we have confirmed that Eq. 2.2.2 is a solution of Eq. 2.2.1.

Next, we wil find the 2nd homogeneous solution to Eq. 2.2.1 using variation of parameters. to do this we use the same techniques to find the general solutions to $$h(x)$$, $$Z$$ , and $$U$$ as we did in Part 1. So we can use Eqs. 2.1.16, 2.1.18 , 2.1.19 , and 2.1.21 shown below.

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.16) $$

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.18) $$

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.19) $$

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.21) $$

Again, knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.22) $$

and $$ k_1 = 1 $$

Now we make the Eq. 2.2.1 fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to solve for $$h(x)$$

$$ 0 = y'' - \frac {2} {x} y' + y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.6) $$

Therefore,

$$ a_1 = \frac {2} {x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.7) $$

$$ a_0 = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.8) $$

Now plug Eq. 2.2.7 and Eq. 2.2.2 into Eq. 2.1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = sin^2(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.9) $$

Then plug Eq. 2.2.9 and Eq. 2.2.2 into Eq. 2.1.22 to get the expression for the 2nd homogeneous solution. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = [\frac {sin(x)} {x}] [k_2 \int \frac {1} {sin^2(x)}] dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.10) $$

This simplifies to

$$ u_2(x) = [\frac {sin(x)} {x}] [-k_2cot(x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.11) $$

The 2nd homogeneous solution then simplifies to the form

$$ u_2(x) = k_2 [\frac {-cos(x)} {x}] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.12) $$

Part 2
Part A

To find the general solution of

$$ sin(x) = (x-1)y'' - xy' + y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.30) $$

we assume the general solution is of the form

$$ y(x) = A(x) u_1 + B(x) u_2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.31) $$

Since we already know $$u_1$$ and $$u_2$$ from part 1 A we need to find $$A(x)$$ and $$B(x)$$. To do this we utilize the equations,

$$ A(x) = - \int \frac 1 W u_2(x) f(x) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.32) $$

and

$$ B(x) = \int \frac 1 W u_1(x) f(x) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.33) $$

where W is the Wronskian. The wronskian is the determinant of a simple 2x2 matrix in this case. It will simplify to,

$$ W =u_1 u_2' - u_2 u_1' = e^x -xe^x = e^x (1-x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.34) $$

Now we plug Eq. 2.1.34 and 2.1.29 into Eq. 2.1.32, where the constant will be neglected in this case for $$u_2$$ and $$f(x) = sin(x)$$ so we get,

$$ A(x) = - \int \frac {x sin(x)} {e^x (1-x)} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.35) $$

For simplicity we will leave Eq. 2.1.35 in its integral form. Now we calculate $$B(x)$$. Plug Eq. 2.1.2 and Eq. 2.1.34 into Eq. 2.1.33 to get

$$ B(x) = \int \frac {sin(x)} {(1-x)} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.36) $$

Again, for simplicity we will leave Eq. 2.1.36 in its integral form. So we then plug Eqs. 2.1.35, 2.1.36 into Eq. 2.1.31 to get our general solution for Part A,

$$ y(x) = -e^x \int \frac {x sin(x)} {e^x (1-x)} dx + x \int \frac {sin(x)} {(1-x)} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.37) $$

Part B We will now use the same procedure to find the general solution to the nonhomogeneous equation,

$$ sin(x) = xy'' + 2y' + xy \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.13) $$

Our Wronskian is now,

$$ W = u_1 u_2' - u_2 u_1' = (x^{-1} sin(x)) (x^{-2}cos(x) + x^{-1}sin(x)) - (-x^{-1}cos(x)) (-x^{-2}sin(x) + x^{-1}cos(x)) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.14) $$

which simplifies all the way down to,

$$ W = x^{-2} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.15) $$

So the coefficients of our general solution $$A(x)$$ and $$B(x)$$ can be abtained by pluging Eq. 2.2.15, 2.2.12, and 2.2.2 into Eq. 2.1.32 and 2.1.33 as they are needed,

$$ A(x) = - \int x^2 \frac {-cos(x)} {x} sin(x) dx = \frac 1 8 [sin(2x)-2xcos(2x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.16) $$

$$ B(x) = \int x^2 \frac {sin(x)} {x} sin(x) dx = \frac {x^2} {4} - \frac 1 4 x sin(2x)- \frac 1 8 cos(2x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.17) $$

Now we plug Eq. 2.2.17, 2.2.16, 2.2.2 and 2.2.12 into Eq. 2.1.31 to obtain the general solution for Eq. 2.2.13.

$$ y(x) = \frac {sin(x)} {8x} [sin(2x)-2xcos(2x)] - \frac {cos(x)} {x} [\frac {x^2} {4} - \frac 1 4 x sin(2x)- \frac 1 8 cos(2x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.18) $$

Author and proof-reader
[Author] Rob Carroll

[Proof-reader]

=Problem 3 - Solve Nonhomogeneous Legendre Equation= From (meeting 33 page 1)

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

Find
Solve non-homogeneous Legendre equation where n=1 using direct method.

Solution
The solution of this equation is to be given in the following form. We already determined $$u_1(x)$$ and $$u_2(x)$$ in the lecture. Now derive the particular solution. Deriving $$h(x)$$ first. Then we move to the particular solution. The general solution is then

Author and proof-reader
[Author] S Jo

[Proof-reader]

=Problem 4 - Solve Axisymmetric Problem= From (meeting 34 page 2)

Given
Spherical Coordinate System

Find
Derive that $$ds^2 = (h_1)^2 (d \xi_1)^2 + (h_2)^2 (d \xi_2)^2 + (h_3)^2 (d \xi_3)^2 = (dr)^2 + r^2 (d\theta)^2 + r^2 \cos^2 \theta (d\phi)^2$$

Derive $$\Delta \psi_2$$

Derive $$\Delta \psi_3$$

Derive $$\Delta \psi$$

Solution
First transform cartesian coordinate to spherical coordinate. We simplify the above equation and get Thus the following relations are determined $$\Delta \psi$$ is defined as For $$\Delta \psi_2$$, we have For $$\Delta \psi_3$$, we have With the $$\Delta \psi_1$$ already derived in the class, we get

Author and proof-reader
[Author] S Jo

[Proof-reader]

=Problem 5 - Circular Cylindrical Coordinate Problem= From (meeting 35 page 3)

Given

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

\begin{array}{*{20}l} x_1 & = & r \cos \theta & = & \xi_1 \cos \xi_2 \\ x_2 & = & r \sin \theta & = & \xi_1 \sin \xi_2 \\ x_3 & = & z & = & \xi_3 & \end{array} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (5.1)
 * }
 * }

Find
1. Find $$\displaystyle \{{\rm d} x_i\} = \{{\rm d} x_1, {\rm d} x_2, {\rm d} x_3\} $$ in terms of $$\displaystyle \{\xi_j\} = \{\xi_1, \xi_2, \xi_3\} $$ and $$\displaystyle \{{\rm d} \xi_k\} $$

2. Find $$\displaystyle {\rm d} s^2 = \sum_i ({\rm d} x_i)^2 = \sum_k (h_k)^2 ({\rm d} \xi_k)^2 $$. Identify $$\displaystyle \{h_i\}$$ in terms of $$\displaystyle \{ \xi_j \}$$

3. Find $$\displaystyle \Delta \psi $$ in cylin. coord.

Part 1. Finding { $$\displaystyle dx_i$$ }
[Background knowledge] :  Cylindrical Coordinates


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

$$\displaystyle \begin{array}{*{20}l} {\rm d}x_1 & = & \cos \xi_2 \cdot {\rm d}\xi_1 - \xi_1 \cdot \sin \xi_2 \cdot {\rm d} \xi_2 \\
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |

{\rm d}x_2 & = & \sin \xi_2 \cdot {\rm d}\xi_1 + \xi_1 \cdot \cos \xi_2 \cdot {\rm d} \xi_2 \\

{\rm d}x_3 & = & {\rm d} \xi_3 \end{array} $$
 * }
 * }
 * }


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


 * }
 * }
 * }

===Part 2. Finding $$\displaystyle ds^2$$, Identify { $$\displaystyle h_i$$ }=== 1. Finding $$\displaystyle ds^2$$


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

ds^2 = (dx_1)^2 + (dx_2)^2 + (dx_3)^2 $$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle = \underbrace{\cos^2 \cdot \xi_2 \cdot (d \xi_1)^2 - 2 \cdot \xi_1 \cdot \sin \xi_2 \cdot \cos \xi_2 \cdot d \xi_1 \cdot d \xi_2 + \xi_1^2 \cdot \sin^2 \xi_2 \cdot (d \xi_2)^2}_{(dx_1)^2} $$

$$\displaystyle + \underbrace{\sin^2 \xi_2 \cdot (d \xi_1)^2 + 2 \cdot \xi_1 \cdot \sin \xi_2 \cdot \cos \xi_2 \cdot d \xi_1 \cdot d \xi_2 + \xi_1^2 \cdot \cos^2 \xi_2 \cdot (d \xi_2)^2}_{(dx_2)^2} $$

$$\displaystyle + \underbrace{(d \xi_3)^2}_{(dx_3)^2} $$
 * }
 * }
 * }

Therefore,


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

$$\displaystyle ds^2 = (d \xi_1)^2 + \xi_1^2 \cdot (d \xi_2)^2 + (d \xi_3)^2 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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


 * }
 * }
 * }

2. Identify { $$\displaystyle h_i$$ }


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

ds^2 = (h_1)^2(d \xi_1)^2 + (h_2)^2(d \xi_2)^2 + (h_3)^2(d \xi_3)^2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }

According to the lecture [http://upload.wikimedia.org/wikiversity/en/a/ab/2010_11_04_14_51_31.djvu Mtg. 34]

{$$\displaystyle h_i $$} are magnitude of tangent vetors to curv. coord.

So, we just select positive part of them.

Therefore,


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

$$\displaystyle h_1 = 1, \quad h_2 = \xi_1, \quad h_3 = 1 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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


 * }
 * }
 * }

Part 3. Finding $$\displaystyle \Delta \psi$$
$$\displaystyle \Delta \psi = \frac{1}{h_1h_2h_3}\sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left [ \frac{h_1h_2h_3}{(h_i)^2} \frac{\partial \psi}{\partial \xi_i} \right ] $$

$$\displaystyle \Delta \psi = \frac{1}{\xi_1} \left [ \frac{\partial}{\partial \xi_1} \left [ \frac{\xi_1}{(1)^2} \frac{\partial \psi}{\partial \xi_1} \right ]+ \frac{\partial}{\partial \xi_2} \left [ \frac{\xi_1}{(\xi_1)^2} \frac{\partial \psi}{\partial \xi_2} \right ]+ \frac{\partial}{\partial \xi_3} \left [ \frac{\xi_1}{(1)^2} \frac{\partial \psi}{\partial \xi_3} \right ] \right ] $$

$$\displaystyle \Delta \psi =  \frac{\partial}{\partial \xi_1} \frac{\partial \psi}{\partial \xi_1} + \frac{\partial}{\partial \xi_2} \left [ \frac{1}{(\xi_1)^2} \frac{\partial \psi}{\partial \xi_2} \right ]+ \frac{\partial}{\partial \xi_3} \frac{\partial \psi}{\partial \xi_3} $$

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

$$\displaystyle \Delta \psi =  \frac{\partial^2 \psi}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2}+ \frac{\partial^2 \psi}{\partial z^2} $$

Author and proof-reader
[Author] Oh, Sang Min

[Proof-reader]

=Problem 6 - Spherical Coordinate Problem= From (meeting 35 page 4)

Given
We are given the following relations,
 * $$\displaystyle x_1 = r \sin \bar{\theta} \cos \phi =  \xi_1 \sin \xi_2 \cos \xi_3 $$
 * $$\displaystyle x_2 = r \sin \bar{\theta} \sin \phi =  \xi_1 \sin \xi_2 \sin \xi_3 $$
 * $$\displaystyle x_3 = r \cos \bar{\theta} =  \xi_1 \cos \xi_2$$

Noting that $$\displaystyle \xi_1=r$$, $$\displaystyle \xi_2=\bar{\theta}$$, and $$\displaystyle \xi_3=\phi$$

Find
We are to derive the laplacian, $$\Delta\Psi$$, in spherical coordinates using the math/physics convention in which,
 * $$\bar\theta=\frac{\pi}{2}-\theta$$

Solution
We first calculate the total derivatives, $$dx_i$$, in spherical coordinates using the given relations. We have,
 * $$\displaystyle dx_1=\frac{\partial x_1}{\partial \xi_1}d \xi_1 + \frac {\partial x_1}{\partial \xi_2} d \xi_2 + \frac{\partial x_1}{\partial\xi_3}d\xi_3=sin\xi_2 cos \xi_3 d\xi_1 + \xi_1 cos \xi_2 cos \xi_3 d \xi_2 - \xi_1 sin \xi_2 sin \xi_3 d \xi_3 $$
 * $$\displaystyle dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3=dx_2=sin\xi_2 sin\xi_3 d\xi_1 + \xi_1 cos\xi_2 sin\xi_3 d\xi_2 + \xi_1 sin\xi_2 cos\xi_3 d\xi_3 $$
 * $$\displaystyle dx_3=\frac{\partial x_3}{\partial\xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3=dx_3=cos\xi_2 d\xi_1 - \xi_1 sin\xi_2 d\xi_2 + 0 $$

Next, $$(ds)^2$$ is found. We have,
 * $$\displaystyle (ds)^2=\sum (dx_i)^2 $$

By substituting the given relations and using the first Pythagorean identity, $$(ds)^2$$ simplifies to,
 * $$\displaystyle ds^2=d\xi_1^2+\xi_1^2d\xi_2^2+\xi_1^2sin^2\xi_2d\xi_3^2$$

We note that $$h_i^2$$ is the coefficient of $$\xi^2_i$$. Listing the coefficients,
 * $$\displaystyle h_1=1$$
 * $$\displaystyle h_2=\xi_1$$
 * $$\displaystyle h_3=\xi_1 sin\xi_2$$

The Laplacian for orthogonal curvilinear coordinates can be expressed as
 * $$\displaystyle \Delta\Psi=\frac{1}{h_1h_2h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left(\frac {h_1h_2h_3}{(h_i)^2}\cdot\frac{\partial\Psi}{\partial\xi_i}\right)$$

We substitute the values for $$h_i$$ found above to yield
 * $$\displaystyle \Delta\Psi=\frac{1}{\xi^2_1sin\xi_2}\left[\frac{\partial}{\partial\xi_1}\left(\xi^2_1sin\xi_2\frac{\partial\Psi}{\partial \xi_1}\right)

+\frac{\partial}{\partial\xi_2}\left(sin\xi_2\frac{\partial\Psi}{\partial \xi_2}\right) +\frac{\partial}{\partial\xi_3}\left(\frac{1}{sin\xi_2}\frac{\partial\Psi}{\partial \xi_3}\right)\right]$$ We use the given relations to yield the Laplacian in spherical coordinates,
 * $$\Delta\Psi =\frac {1}{r^2 sin \bar\theta} \left[ \frac {\partial}{\partial r} \left(r^2 sin \bar\theta \frac {\partial\Psi}{\partial r} \right)

+ \frac {\partial}{\partial \bar\theta} \left(sin \bar\theta \frac {\partial \Psi}{\partial \bar\theta} \right) + \frac {\partial}{\partial \phi} \left(\frac {1}{sin \bar\theta} \frac {\partial \Psi}{\partial \phi}\right) \right] $$

Author and proof-reader
[Author]
 * Raul Riveros

[Proof-reader]

=Problem 7 - Elliptic Coordinate Problem= From (meeting 36 page 1)

Given
Elliptic coordinates as described in the Wikipedia article.

Find
Verify expression for $$ \Delta \psi \! $$ in elliptic coordinates by performing the following:

1. Find $$ \big\{ dx_i \big\} = \big\{ dx_1, dx_2 \big\} \! $$ in terms of $$ \big\{ \xi_j \big\} = \big\{ \xi_1, \xi_2 \big\} \! $$ and $$ \big\{ d\xi_k \big\} \! $$

2. Find $$ ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2 \! $$. Identify $$ \big\{ h_i \big\} \! $$ in terms of $$ \big\{ \xi_j \big\} \! $$.

3. Find $$ \Delta \psi \! $$ in elliptic coordinates.

Part 1
From the Wikipedia article, the elliptic coordinates are defined as shown in (7.1) and (7.2).

Part 3
For orthogonal curvilinear coordinates (2D), the Laplacian operator may be defined as

Since $$ h_1 $$ and $$ h_2 $$ are equal, (7.10) becomes

Author and proof-reader
[Author] Michael Steinberg

[Proof-reader]

=Problem 8 - Verification genernal form of Legendre poly.= From (meeting 36 page 2)

Given

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

P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-1)
 * }
 * }


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

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-2)
 * }
 * }


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

P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-3)
 * }
 * }


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

P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-4)
 * }
 * }


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

P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-5)
 * }
 * }

Find
Verify that (8-1) - (8-5) can be written as


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

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i (2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-6)
 * }
 * }

or


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

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} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (8-7)
 * }
 * }

$$\displaystyle [\frac{n}{2}]$$ = interger part of  $$ \displaystyle \ \frac{n}{2} $$

Solution
Background knowledge :  Factorial , Floor and ceiling functions

We tried to discuss about background knowledges and how to verify clearly and efficiently

In this time, we verified using both equations such that $$\displaystyle (8-6) $$ and $$\displaystyle (8-7) $$

Part 1. Verification using $$\displaystyle (8-6)$$
1. In case of $$\displaystyle n = 0 $$


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

P_0(x) = \sum_{i=0}^{[0/2]} \frac{(-1)^i (2 (0) - 2 i)! x^{0 - 2i}}{2^0 i! (0-i)! (0-2i)!} \ = \sum_{i=0}^{0} \frac{(-1)^i ( - 2 i)! x^{- 2i}}{i! (-i)! (-2i)!} \ = \frac{(-1)^0 \cdot 0! \cdot x^0}{0! \cdot 0! \cdot 0!}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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




 * }
 * }
 * }


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

$$\displaystyle P_0(x) = 1 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

2. In case of $$\displaystyle n = 1 $$


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

P_1(x) = \sum_{i=0}^{[1/2]} \frac{(-1)^i (2 (1) - 2 i)! x^{1 - 2i}}{2^1 i! (1-i)! (1-2i)!}= \sum_{i=0}^{0} \frac{(-1)^i \ (2 - 2 i)! \ x^{1 - 2i}}{2 \ i! \ (1-i)! \ (1-2i)!}= \frac{(-1)^0 \cdot 2! \cdot x^1}{2 \cdot 0! \cdot 1! \cdot 1!} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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




 * }
 * }
 * }


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

$$\displaystyle P_1(x) = x $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

3. In case of $$\displaystyle n = 2 $$


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

P_2(x) = \sum_{i=0}^{[2/2]} \frac{(-1)^i (2 (2) - 2 i)! x^{2 - 2i}}{2^2 i! (2-i)! (2-2i)!}= \sum_{i=0}^{1} \frac{(-1)^i \ (4 - 2 i)! \ x^{2 - 2i}}{4 \ i! \ (2-i)! \ (2-2i)!}= \underbrace{\frac{(-1)^0 \cdot 4! \cdot x^2}{4 \cdot 0! \cdot 2! \cdot 2!}}_{\frac{3}{2}x^2} \ + \ \underbrace{\frac{(-1)^1 \cdot 2! \cdot x^0}{4 \cdot 1! \cdot 1! \cdot 1!}}_{-\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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

$$\displaystyle P_2(x) = \frac{1}{2}(3x^2-1) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

4. In case of $$\displaystyle n = 3 $$
 * {| style="width:100%" border="0" align="left"

P_3(x) = \sum_{i=0}^{[3/2]} \frac{(-1)^i (2 (3) - 2 i)! x^{3 - 2i}}{2^3 i! (3-i)! (3-2i)!}= \sum_{i=0}^{1} \frac{(-1)^i (6 - 2 i)! x^{3 - 2i}}{8 \ i! \ (3-i)! \ (3-2i)!} = \underbrace{\frac{(-1)^0 \cdot 6! \cdot x^{3}}{8 \cdot 0! \cdot 3! \cdot 3!}}_{\frac{5}{2}x^3} + \underbrace{\frac{(-1)^1 \cdot 4! \cdot x^1}{8 \cdot 1! \cdot 2! \cdot 1!}}_{-\frac{3}{2}x}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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

$$\displaystyle P_3(x) = \frac{1}{2}(5x^3-3x) $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

5. In case of $$\displaystyle n = 4 $$
 * {| style="width:100%" border="0" align="left"

P_4(x) = \sum_{i=0}^{[4/2]} \frac{(-1)^i (2 (4) - 2 i)! x^{4 - 2i}}{2^4 i! (4-i)! (4-2i)!}= \sum_{i=0}^{2} \frac{(-1)^i \ (8 - 2 i)! \ x^{4 - 2i}}{16 \ i! \ (4-i)! \ (4-2i)!}$$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle = \underbrace{\frac{(-1)^0 \cdot 8! \cdot x^4}{16 \cdot 0! \cdot 4! \cdot 4!}}_{\frac{35}{8}x^4} \ + \ \underbrace{\frac{(-1)^1 \cdot 6! \cdot x^{2}}{16 \cdot 1! \cdot 3! \cdot 2!}}_{- \frac{15}{4}x^2} \ + \ \underbrace{\frac{(-1)^2 \cdot 4! \cdot x^0}{16 \cdot 2! \cdot 2! \cdot 0!}}_{\frac{3}{8}}$$
 * }
 * }
 * }


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

$$\displaystyle P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

Part 2. Verification using $$\displaystyle (8-7)$$
1. In case of $$\displaystyle n = 0 $$
 * {| style="width:100%" border="0" align="left"

P_0(x) = \sum_{i=0}^{[0/2]} \frac{1}{2^i \ i! \ (-2i)!} (-1)^i \ x^{-2i} = \sum_{i=0}^{0} \frac{1}{2^0 \cdot 0! \cdot 0!} (-1)^0 \cdot x^0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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




 * }
 * }
 * }


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

$$\displaystyle P_0(x) = 1 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

2. In case of $$\displaystyle n = 1 $$
 * {| style="width:100%" border="0" align="left"

P_1(x) = \sum_{i=0}^{[1/2]} \frac{1}{2^i \ i! \ (1-2i)!} (-1)^i \ x^{1-2i} = \sum_{i=0}^{0} \frac{1}{2^0 \cdot 0! \cdot 1!} (-1)^0 \cdot x^1 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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




 * }
 * }
 * }


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

$$\displaystyle P_1(x) = x $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

3. In case of $$\displaystyle n = 2 $$
 * {| style="width:100%" border="0" align="left"

P_2(x) = \sum_{i=0}^{[2/2]} \frac{1 \cdot 3 \cdot \cdot \cdot 4-2i-1}{2^i \ i! \ (2-2i)!} (-1)^i \ x^{2-2i} = \underbrace{\frac{1 \cdot 3}{2^0 \cdot 0! \cdot 2!} (-1)^0 \cdot x^2}_{\frac{3}{2}x^2} + \underbrace{\frac{1}{2^1 \cdot 1! \cdot 0!} (-1)^1 \cdot x^0}_{- \frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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

$$\displaystyle P_2(x) = \frac{1}{2}(3x^2-1) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

4. In case of $$\displaystyle n = 3 $$
 * {| style="width:100%" border="0" align="left"

P_3(x) = \sum_{i=0}^{[3/2]} \frac{1 \cdot 3 \cdot \cdot \cdot 6-2i-1}{2^i \ i! \ (3-2i)!} (-1)^i \ x^{3-2i} = \underbrace{\frac{1 \cdot 3 \cdot 5}{2^0 \cdot 0! \cdot 3!} (-1)^0 \cdot x^3}_{\frac{5}{2}x^3} + \underbrace{\frac{1 \cdot 3}{2^1 \cdot 1! \cdot 1!} (-1)^1 \cdot x^1}_{- \frac{3}{2}x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * }


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

$$\displaystyle P_3(x) = \frac{1}{2}(5x^3-3x) $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

5. In case of $$\displaystyle n = 4 $$
 * {| style="width:100%" border="0" align="left"

P_4(x) = \sum_{i=0}^{[4/2]} \frac{1 \cdot 3 \cdot \cdot \cdot 8-2i-1}{2^i \ i! \ (4-2i)!} (-1)^i \ x^{4-2i}$$
 * $$\displaystyle
 * $$\displaystyle

$$\displaystyle \ = \underbrace{\frac{1 \cdot 3 \cdot 5 \cdot 7}{2^0 \cdot 0! \cdot 4!} (-1)^0 \cdot x^4}_{\frac{35}{8}x^4} + \underbrace{\frac{1 \cdot 3 \cdot 5}{2^1 \cdot 1! \cdot 2!} (-1)^1 \cdot x^2}_{- \frac{15}{4}x^2} + \underbrace{\frac{1 \cdot 3}{2^2 \cdot 2! \cdot 0!} (-1)^2 \cdot x^0}_{\frac{3}{8}} $$
 * }
 * }
 * }


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




 * }
 * }
 * }


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

$$\displaystyle P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }


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




 * }
 * }
 * }

Author and proof-reader
[Author] [http://en.wikiversity.org/wiki/User_talk:Egm6321.f10.team5.oh/hw6#Problem_8_-_Verification_genernal_form_of_Legendre_poly. Oh, Sang Min]

[Proof-reader] Mike Faraone

=Problem 9 - Verification solution of Legendre eq.= From (meeting 36 page 3)

Given

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

P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (9-1)
 * }
 * }


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

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (9-2)
 * }
 * }


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

P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (9-3)
 * }
 * }


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

P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (9-4)
 * }
 * }


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

P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (9-5)
 * }
 * }

Find
Verify that $$\displaystyle (9-1) \ - \ (9-5)$$ are solution of Legendre equation
 * {| style="width:100%" border="0" align="left"

(1-x^2) y'' - 2 x y' + n(n+1) y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (9-6)
 * }
 * }

Solution
1. In case of $$\displaystyle n=0, \ (9-1)$$


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

$$\displaystyle y = 1, \ y' = 0, \ y'' = 0$$
 * }
 * }
 * }


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

$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \cdot 0 - 2x \cdot 0 + 0 \cdot (0+1) \cdot 1 = 0 $$
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2. In case of $$\displaystyle n=1, \ (9-2)$$


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$$\displaystyle y = x, \ y' = 1, \ y'' = 0$$
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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \cdot 0 - 2x \cdot 1 + 1 \cdot (1+1) \cdot x = 0 $$
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3. In case of $$\displaystyle n=2, \ (9-3)$$


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$$\displaystyle y = \frac{1}{2} (3 x^2 - 1), \ y' = 3x, \ y'' = 3$$
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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = \underbrace{(1-x^2) \cdot 3 - 2x \cdot 3x + 2 \cdot (2+1) \cdot \frac{1}{2} (3 x^2 - 1)}_{(-3 - 6 + 9)x^2 + (3 -3)} = 0 $$
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4. In case of $$\displaystyle n=3, \ (9-4)$$


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$$\displaystyle y = \frac{1}{2} (5 x^3 - 3x), \ y' = \frac{3}{2} (5 x^2 - 1), \ y'' = 15x$$
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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = \underbrace{(1-x^2) \cdot 15 x - 2 x \cdot \frac{3}{2} (5 x^2 - 1)+ 3 \cdot (3+1) \cdot \frac{1}{2} (5 x^3 - 3x)}_{15 x - 15 x^3 - 15x^3 + 3x + 30 x^3 - 18x} = 0 $$
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5. In case of $$\displaystyle n=4, \ (9-5)$$


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$$\displaystyle y = \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8}, \ y' = \frac{35}{2} x^3 - \frac{15}{2} x, \ y'' = \frac{105}{2} x^2 - \frac{15}{2}$$
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$$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = (1-x^2) \left(\frac{105}{2} x^2 - \frac{15}{2} \right) - 2 x \left(\frac{35}{2} x^3 - \frac{15}{2} x\right) + 4(4+1) \left(\frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8}\right)$$
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$$\displaystyle \ = \frac{105}{2} x^2 - \frac{15}{2} - \frac{105}{2} x^4 + \frac{15}{2} x^2 - 35 x^4 + 15 x^2 +  \left(\frac{175}{2} x^4 - 75 x^2 + \frac{15}{2}\right)$$

$$\displaystyle \ = (- \frac{105}{2} x^4 + \frac{175}{2} x^4 - \frac{70}{2} x^4) + (\frac{105}{2} x^2 + \frac{15}{2} x^2  + \frac{30}{2}x^2 - \frac{150}{2} x^2) + (\frac{15}{2}- \frac{15}{2}) = 0$$
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Therefore, $$\displaystyle (9-1) - (9-5) $$ are clearly solution of Legendre equation such that $$\displaystyle (9-6) $$
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Author and proof-reader
[Author] [http://en.wikiversity.org/wiki/User_talk:Egm6321.f10.team5.oh/hw6#Problem_9_-_Verification_solution_of_Legendre_eq. Oh, Sang Min]

[Proof-reader] Mike Faraone

=References=

=Contributing members=
 * Oh, Sang Min Author of problem 5, 8 and 9
 * Raul Riveros authored Problem 6
 * S Jo Authored Problem 3 and Problem 4
 * Mike Faraone Proof Read 7 & 8. Solved Part 3 of Problem 5