text:
User:Egm6321.f10.team03/Hwk6

= Problem 1-Second Homogeneous Solution of Legendre Equation for n=1 =

Given
Given the Legendre Equation where $$\displaystyle n=1$$,

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

\left(1-x^2\right)y^{\prime\prime}-2xy^\prime+2y=0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.1)
 * }
 * }

and the 1st homogeneous solution $$\displaystyle P_2$$
 * {| style="width:100%" border="0"

P_2\left(x\right) = \frac{1}{2}\left(3x^2-1\right). $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

Find
Show that the second homogeneous solution $$\displaystyle Q_2 $$ is
 * {| style="width:100%" border="0"

Q_2\left(x\right) = \frac{1}{4}\left(3x^2-1\right)log\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3)
 * }
 * }

by using variation of parameters method.

Solution
We will use variation of parameters to find the 2nd homogeneous solution, $$\displaystyle Q_2\left(x\right) $$, to the Legendre Equation, given the 1st homogeneous solution, $$\displaystyle P_2\left(x\right) $$.

Variation of Parameters, Homogeneous Case $$\displaystyle \left( f\left(x\right)= 0 \right) $$
The solution to the Legendre Equations is of the form

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

y\left(x\right)=U\left(x\right)u_1\left(x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.4)
 * }
 * }

The above form of the solution contains both homogeneous solutions, $$\displaystyle P_2$$ and $$\displaystyle Q_2$$ and, for cases where $$\displaystyle f\left(x\right)\neq0$$, the particular solution $$\displaystyle y_p$$.

Let

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

u_1\left(x\right)=P_2\left(x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.5)
 * }
 * }

We will need the first and second derivatives of $$\displaystyle y\left(x\right) $$. For clarity, I will omit the arguments.

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

y = U u_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.6)
 * }
 * }

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

y^\prime = U^\prime u_1 + U u_1^\prime $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.7)
 * }
 * }

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

y^{\prime\prime} = U^{\prime\prime} u_1 +2U^\prime u_1^\prime + U u_1^{\prime\prime} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.8)
 * }
 * }

Compare

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

a_2 y^{\prime\prime}+a_1 y^\prime+a_o y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.9)
 * }
 * }

to the Legendre Equation where $$\displaystyle n=1$$,

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

\left(1-x^2\right)y^{\prime\prime}-2xy^\prime+2y=0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.10)
 * }
 * }

In order for $$\displaystyle a_2$$ to equal 1, we will divide the Legendre Equation by $$\displaystyle \left(1-x^2\right) $$. This gives

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

y^{\prime\prime}-\frac{2xy^\prime}{\left(1-x^2\right)}+\frac{2y}{\left(1-x^2\right)}=0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.11)
 * }
 * }

By setting Eq 1.9 and Eq 1.11 equal to each other,

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

a_2 y^{\prime\prime}+a_1 y^\prime+a_o y = \underbrace{1}_{a_2} y^{\prime\prime}+\underbrace{\frac{-2x}{\left(1-x^2\right)}}_{a_1}y^\prime + \underbrace{\frac{2}{\left(1-x^2\right)}}_{a_0}y, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.12)
 * }
 * }

we see that

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

\begin{align} a_2\left(x\right) &= 1,\\ a_1\left(x\right) &= \frac{-2x}{\left(1-x^2\right)},\\ a_0\left(x\right) &= \frac{2}{\left(1-x^2\right)}.\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eqs. 1.13)
 * }
 * }

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

\begin{align} \left[y^ = U u_1\right]\times a_0 \\ \left[y^\prime = U^\prime u_1 + U u_1^\prime\right]\times a_1 \\ \left[y^{\prime\prime} = U^{\prime\prime} u_1 +2U^\prime u_1^\prime + U u_1^{\prime\prime}\right]\times a_2 \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eqs. 1.14)
 * }
 * }

Summing the left and right hand side of Eqs. 1.14 gives

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

a_2 y^{\prime\prime}+a_1 y^\prime+a_o y = U\cancelto{0}{\left[u_1^{\prime\prime} + a_1 u_1^\prime + a_0 u_1 \right]} + U^\prime \left[ a_1 u_1 + 2 u_1^\prime \right] + U^{\prime\prime} \left[u_1\right]=0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.15)
 * }
 * }

The term $$\displaystyle\left[u_1^{\prime\prime} + a_1 u_1^\prime + a_0 u_1 \right]$$ goes to zero because $$\displaystyle u_1$$ is a homogeneous solution to the Legendre Equation. Therefore, Eq. 1.15 reduces to

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

U^\prime a_1 u_1 + 2 U^\prime u_1^\prime + U^{\prime\prime} u_1 = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.16)
 * }
 * }

If we let $$\displaystyle Z:=U^\prime $$, we can use the order reduction method to help us solve for $$\displaystyle Q_2$$. We can rewrite Eq. 1.16 as

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

Z a_1 u_1 + 2 Z u_1^\prime + Z^\prime u_1 = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.17)
 * }
 * }

Regrouping we get

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

Z \underbrace{ \left( \frac {a_1 u_1 + 2 u_1^\prime } {u_1}\right)}_{\tilde{a}_0} + Z^\prime \underbrace{\left( \frac{u_1}{u_1} \right)}_{\tilde{a}_1=1} = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.17)
 * }
 * }

Let

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\begin{align} &\tilde{a}_1\left(x\right) = 1,\\ &\tilde{a}_0\left(x\right) = \frac {a_1 u_1 + 2 u_1^\prime } {u_1}.\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.18)
 * }
 * }

Because Eq. 1.17 is in the form of L1_ODE_CC, we can write the integrating factor in the form of

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

h\left(x\right) = \exp{\left[\int \tilde{a}_0 \left(x\right)\,dx\right]} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.19)
 * }
 * }

Substituting in for $$\displaystyle \tilde{a}_0 \left(s\right) $$ gives

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\begin{align} h\left(x\right) &= \exp{\left[\int \tilde{a}_0 \left(x\right)\,dx\right]}\\ &= \exp{\left[\int \frac {a_1\left(x\right) u_1\left(x\right) + 2 u_1^\prime\left(x\right) } {u_1\left(x\right)}\,dx\right]}\\ &= \exp{\left[\int a_1\left(x\right)\,dx + \int\frac{2u_1^\prime\left(x\right)} {u_1\left(x\right)}\,dx\right]}\\ &= \exp{\left[\int a_1\left(x\right) \,dx + 2\log u_1\left(x\right)\right]}\\ h\left(x\right) &= u^2_1\left(x\right)\exp{\left[\int a_1\left(x\right) \,dx\right]}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.20)
 * }
 * }

Now that we have $$\displaystyle h\left(x\right) $$, we can solve for $$\displaystyle Z\left(x\right) $$.

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

Z\left(x\right) = \frac{1}{h\left(x\right)} \left[k_2 + \int h\left(x\right) f\left(x\right) \,dx\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.21)
 * }
 * }

Because $$\displaystyle f\left(x\right) = 0 $$, Eq. 1.21 becomes

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

Z\left(x\right) = \frac{k_2}{h\left(x\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.22)
 * }
 * }

We can now back out $$\displaystyle U\left(x\right) $$ from $$\displaystyle Z\left(x\right) $$, because $$\displaystyle Z:=U^\prime $$.

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\begin{align} &U\left(x\right) = \int Z\left(x\right)\,dx \\ &U\left(x\right) = k_1 + \int Z\left(x\right)\,dx \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.23)
 * }
 * }

Now that we have an expression for $$\displaystyle U\left(x\right) $$, we can solve for $$\displaystyle y\left(x\right) $$.

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

\begin{align} &y\left(x\right) = U\left(x\right)u_1\left(x\right)\\ &y\left(x\right) = \left(k_1 + \int Z\left(x\right)\,dx \right)u_1\left(x\right)\\ &y\left(x\right) = k_1 \underbrace{u_1\left(x\right)}_{P_2\left(x\right)} + k_2 \underbrace{u_1\left(x\right) \int \frac{1}{h\left(x\right)}\,dx}_{Q_2\left(x\right)}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.24)
 * }
 * }

Recall that $$\displaystyle P_2\left(x\right) $$ and $$\displaystyle Q_2\left(x\right) $$ are the 1st and 2nd homogeneous solutions, respectively.

Our Particular Case
Recall that

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

P_2\left(x\right) = \frac{1}{2}\left(3x^2-1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.25)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

h\left(x\right) = u^2_1\left(x\right)\exp{\left[\int a_1\left(x\right) \,dx\right]} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.26)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

a_1\left(x\right) = \frac{-2x}{\left(1-x^2\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.27)
 * }
 * }

Using Eqs. 1.24, we can now solve for $$\displaystyle Q_2\left(x\right) $$.
 * {| style="width:100%" border="0"

\begin{align} Q_2\left(x\right) &= u_1\left(x\right) \int \frac{1}{h\left(x\right)}\,dx\\ &= \frac{1}{2}\left(3x^2-1\right) \int \frac{1}{u^2_1\left(x\right)\exp{\left[\int a_1\left(x\right) \,dx\right]}}\,dx\\ &= \frac{1}{2}\left(3x^2-1\right) \int \frac{1}{\left(\frac{1}{2}\left(3x^2-1\right)\right)^2\exp{\left[\int \frac{-2x}{\left(1-x^2\right)} \,dx\right]}}\,dx\\ &= 2\left(3x^2-1\right) \int \frac{1}{\left(3x^2-1\right)^2\exp{\left[\log{\left(x^2-1\right)}\right]}}\,dx\\ &= 2\left(3x^2-1\right) \int \frac{1}{\left(3x^2-1\right)^2\left(x^2-1\right)}\,dx\\ &= 2\left(3x^2-1\right) \left[-\left( \frac{1}{8}\right)\left( \frac{6x}{3x^2-1} + \log{\left(1-x\right)} - \log{\left(1+x\right)} \right) \right]\\ &= \frac{1}{4}\left(3x^2-1\right) \left( \frac{-6x}{3x^2-1} - \log{\left( \frac{1-x}{1+x} \right)} \right)\\ &= \frac{1}{4}\left( -6x + \left(3x^2-1\right) \log{\left(\frac{1+x}{1-x}\right)}\right)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eqs. 1.28)
 * }
 * }

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

$$\displaystyle Q_2\left(x\right) = \frac{1}{4}\left(3x^2-1\right) \log{\left(\frac{1+x}{1-x}\right)}-\frac{3}{2}x $$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq. 1.29)
 * }
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team3.franklin 04:18, 15 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 14:32, 15 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 22:41, 15 November 2010 (UTC)
 * Proofread by James Roark 13:19, 16 November 2010 (UTC)

= Problem 2 - Solution of L2-ODE-VC=

Given

 * Given two $$\displaystyle L2-ODE-VC $$,

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

 * $$\displaystyle \left( a \right) \left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=f\left( x \right)$$
 *
$$\displaystyle (Eq. 2.1)$$
 * }
 *
$$\displaystyle (Eq. 2.1)$$
 * }
 * }

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

 * $$\displaystyle \left( b \right) x{{y}^{''}}+2{{y}^{'}}+xy=f\left( x \right)$$
 *
$$\displaystyle (Eq. 2.2)$$
 * }
 *
$$\displaystyle (Eq. 2.2)$$
 * }
 * }

Find

 * (1) Solve $$\displaystyle Eq.2.1$$ and $$\displaystyle Eq.2.2$$ for the case: $$\displaystyle f\left( x \right)=0$$

 * (2) Solve $$\displaystyle Eq.2.1$$ and $$\displaystyle Eq.2.2$$ for the case: $$\displaystyle f\left( x \right)=\sin x$$

(1.1) Solve $$\displaystyle Eq.2.1$$ for $$\displaystyle f\left( x \right)=0$$ (Homogeneous Case)

 * For the case of $$\displaystyle f\left( x \right)=0$$, $$\displaystyle Eq.2.1$$ can be written as:

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

 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=0$$
 *
$$\displaystyle (Eq. 2.3)$$
 * }
 * }
 * }

 * Consider a trial solution,

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

 * $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$
 *
$$\displaystyle (Eq. 2.4)$$
 * }
 * where $$\displaystyle c$$ and $$\displaystyle r$$ are constants to be determined.
 * }
 * where $$\displaystyle c$$ and $$\displaystyle r$$ are constants to be determined.

 * Taking derivative,

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

 * $$\displaystyle{{y}^{'}}=c{{x}^{c-1}}{{e}^{rx}}+r{{x}^{c}}{{e}^{rx}}$$
 * }
 * }
 * }

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

 * $$\displaystyle{{y}^{''}}=c\left( c-1 \right){{x}^{c-2}}{{e}^{rx}}+2cr{{x}^{c-1}}{{e}^{rx}}+{{r}^{2}}{{x}^{c}}{{e}^{rx}}$$
 * }
 * }
 * }

 * Now substitute $$\displaystyle y$$ and whose derivatives in $$\displaystyle (Eq. 2.3)$$,implies:

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

 * $$\left( x-1 \right){{e}^{rx}}\left[ c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c}} \right]-x{{e}^{rx}}\left[ c{{x}^{c-1}}+r{{x}^{c}} \right]+{{x}^{c}}{{e}^{rx}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow c\left( c-1 \right){{x}^{c-1}}-c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c}}-2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c+1}}-{{r}^{2}}{{x}^{c}}-c{{x}^{c}}-r{{x}^{c+1}}+{{x}^{c}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{x}^{c+1}}\left[ {{r}^{2}}-r \right]+{{x}^{c}}\left[ 2cr-{{r}^{2}}-c+1 \right]+{{x}^{c-1}}\left[ {{c}^{2}}-c-2cr \right]-{{x}^{c-2}}\left[ {{c}^{2}}-c \right]=0$$
 *
$$\displaystyle (Eq. 2.5)$$
 * }
 * }
 * }

 * Solving $$\displaystyle (Eq. 2.5)$$ using Wolfram Alpha give constants $$\displaystyle c$$ and $$\displaystyle r$$

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

\begin{array}{*{20}l} i) & c=0 & r=1\\  ii) & c=1 & r=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Now homogeneous solutions are obtained by substituting $$\displaystyle c$$ and $$\displaystyle r$$ in $$\displaystyle (Eq. 2.4)$$

 * First Homogeneous solution is

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

 * $${{y}_}={{x}^{0}}{{e}^{1x}}={{e}^{x}}$$
 * }
 * }
 * }

 * and second homogeneous solution is

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

 * $${{y}_}={{x}^{1}}{{e}^{0x}}=x$$
 * }
 * }
 * }

 * Therefore solution of $$\displaystyle (Eq. 2.3)$$ is the linear combination of two homogeneous solution

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}={{C}_{1}}\cdot {{e}^{x}}+{{C}_{2}}\cdot x$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.6)$$
 * }
 * }

.

(1.2) Solve $$\displaystyle Eq.2.2$$ for $$\displaystyle f\left( x \right)=0$$ (homogeneous)

 * For $$\displaystyle f\left( x \right)=0$$, $$\displaystyle Eq.2.2$$ reduces to:

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

 * $$\displaystyle x{{y}^{''}}+2{{y}^{'}}+xy=0$$
 *
$$\displaystyle (Eq. 2.7)$$
 * }
 * }
 * }

 * Considering a trial solution for y $$\displaystyle (Eq. 2.4)$$, and substituting y and its derivatives in $$\displaystyle (Eq. 2.7)$$,yields:

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

 * $$x{{e}^{rx}}\left[ c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c}} \right]+2{{e}^{rx}}\left[ c{{x}^{c-1}}+r{{x}^{c}} \right]+{{x}^{c+1}}{{e}^{rx}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow c\left( c-1 \right){{x}^{c-1}}+2cr{{x}^{c}}+{{r}^{2}}{{x}^{c+1}}+2c{{x}^{c-1}}+2r{{x}^{c}}+{{x}^{c+1}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{x}^{c+1}}\left( {{r}^{2}}+1 \right)+{{x}^{c}}\left( 2cr+2r \right)+{{x}^{c-1}}\left( {{c}^{2}}+c \right)=0$$
 *
$$\displaystyle (Eq. 2.8)$$
 * }
 * }
 * }

 * By Solving $$\displaystyle (Eq. 2.8)$$ using Wolfram Alpha yields constants $$\displaystyle c$$ and $$\displaystyle r$$

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

\begin{array}{*{20}l} i) & c=-1 & r=i\\  ii) & c=-1 & r=-i\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * So homogenous solutions are obtained by plugging $$\displaystyle c$$ and $$\displaystyle r$$ in $$\displaystyle (Eq. 2.4)$$

 * First Homogenous solution is

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

 * $${{y}_}={{x}^{-1}}{{e}^{ix}}=\frac{x}$$
 * }
 * }
 * }

 * and second homogenous solution is

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

 * $${{y}_}={{x}^{-1}}{{e}^{-ix}}=\frac{x}$$
 * }
 * }
 * }

 * Hence solution of $$\displaystyle (Eq. 2.7)$$ is the linear combination of two homogenous solutions

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}={{C}_{1}}\cdot \frac{x}+{{C}_{2}}\cdot \frac{x}$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.9)$$
 * }
 * }

(2.1) Solve $$\displaystyle Eq.2.1$$ for $$\displaystyle f\left( x \right)=\sin x$$

 * For the case of $$\displaystyle f\left( x \right)=\sin x$$, $$\displaystyle Eq.2.1$$ can be written as:

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

 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=\sin (x)$$
 *
$$\displaystyle (Eq. 2.10)$$
 * }
 * }
 * }

 * Above equation can be rewritten as:

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

 * $$\underbrace{1}_\cdot {{y}^{''}}\underbrace{-\frac{x}{\left( x-1 \right)}}_{{y}^{'}}+\underbrace{\frac{1}{\left( x-1 \right)}}_y=\underbrace{\frac{\sin (x)}{\left( x-1 \right)}}_{F\left( x \right)}$$
 *
$$\displaystyle (Eq. 2.11)$$
 * }
 * }
 * }

 * Method of variation of parameters could be used to obtain the complete solution of $$\displaystyle (Eq. 2.11)$$.

 * Integrating factor for the above equation [[media:2010_10_26_15_07_13.djvu|Page3, Mtg 29 (b)]]is,

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

 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 *
$$\displaystyle (Eq. 2.12)$$
 * }
 * }
 * }

 * From 1.1, one homogenous solution is,

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

$${{u}_{1}}\left( x \right)={{y}_}={{e}^{x}}$$
 *
$$\displaystyle (Eq. 2.13)$$
 * }
 * }

 * $$\displaystyle (Eq. 2.12)$$ becomes:

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

$$h\left( x \right)={{e}^{2x}}\cdot \exp \left[ \int{\left( -\frac{x}{\left( x-1 \right)} \right)dx} \right]$$
 * }
 * }

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

$$h\left( x \right)={{e}^{x-\log \left( x-1 \right)}}$$
 *
$$\displaystyle (Eq. 2.14)$$
 * }
 * }

 * Particular solution is given by[[media:2010_10_28_14_00_19.djvu|page1, Mtg 30 (c)]],

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

 * $${{y}_{P}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}}\left( \int{h\left( x \right)\frac{F\left( x \right)}{{{u}_{1}}(x)}dx} \right)dx$$
 * }
 * }
 * }

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

$${{y}_{P}}={{e}^{x}}\int\left( \int{{{e}^{-\log \left( x-1 \right)}}\cdot \frac{\sin (x)}{\left( x-1 \right)}dx} \right)dx$$
 * }
 * }

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

$${{y}_{P}}={{e}^{x}}\int\left( \int{\frac{\sin (x)}dx} \right)dx$$
 *
$$\displaystyle (Eq. 2.15)$$
 * }
 * Now complete solution is obtained by combining $$\displaystyle (Eq. 2.15)$$ and $$\displaystyle (Eq. 2.6)$$,
 * Now complete solution is obtained by combining $$\displaystyle (Eq. 2.15)$$ and $$\displaystyle (Eq. 2.6)$$,

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}+{{y}_{P}}$$
 * }
 * }

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

$$y={{C}_{1}}\cdot {{e}^{x}}+{{C}_{2}}\cdot x+{{e}^{x}}\int\left( \int{\frac{\sin (x)}dx} \right)dx$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.16)$$
 * }
 * }

(2.2) Solve $$\displaystyle Eq.2.2$$ for $$\displaystyle f\left( x \right)=\sin x$$

 * For $$\displaystyle f\left( x \right)=\sin x$$, $$\displaystyle Eq.2.2$$ becomes:

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

$$x{{y}^{''}}+2{{y}^{'}}+xy=\sin \left( x \right)$$
 *
$$\displaystyle (Eq. 2.17)$$
 * }
 * }

 * $$\displaystyle Eq. 2.17$$can be rewritten as:

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

$$\underbrace{1}_\cdot {{y}^{''}}+\underbrace{\frac{2}{x}}_{{y}^{'}}+\underbrace{y}_=\underbrace{\frac{\sin (x)}{x}}_{F\left( x \right)}$$
 *
$$\displaystyle (Eq. 2.18)$$
 * }
 * }

 * As in the previous case, method of variation of parameters could be used to obtain the complete solution of $$\displaystyle (Eq. 2.17)$$.

 * From 1.2, one homogenous solution is,

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

$${{u}_{1}}\left( x \right)={{y}_}=\frac{x}$$
 *
$$\displaystyle (Eq. 2.19)$$
 * }
 * }

 * Integrating factor for $$\displaystyle Eq. 2.18$$ is,

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

 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 *
$$\displaystyle (Eq. 2.20)$$
 * }
 * }
 * }

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

$$h\left( x \right)=\frac\cdot \exp \left[ \int{\left( \frac{2}{x} \right)dx} \right]$$
 * }
 * }

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

$$h\left( x \right)={{e}^{2ix}}$$
 *
$$\displaystyle (Eq. 2.21)$$
 * }
 * }

 * Particular solution is given by,

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

$${{y}_{P}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}}\left( \int{h\left( x \right)\frac{F\left( x \right)}{{{u}_{1}}(x)}dx} \right)dx$$
 * }
 * }

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

$${{y}_{P}}=\frac{x}\int\left( \int{{{e}^{2ix}}\frac{\sin (x)}{\left( \frac{x} \right)x}dx} \right)dx$$
 * }
 * }

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

$${{y}_{P}}=\frac{x}\int\left( \int{{{e}^{ix}}\sin (x)dx} \right)dx$$
 * }
 * }

 * Simplifying using Wolfram Alpha,

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

$${{y}_{P}}=\frac{x}\int\left( \frac{ix}{2}-\frac{4} \right)dx$$
 * }
 * }

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

$${{y}_{P}}=\frac{x}\cdot \frac{8}\left( i-2\left( 1+{{e}^{2ix}} \right)x \right)$$
 * }
 * }

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

$${{y}_{P}}=\frac{8x}\left( i-2\left( 1+{{e}^{2ix}} \right)x \right)$$
 *
$$\displaystyle (Eq. 2.22)$$
 * }
 * }

 * Therefore complete solution is the combination of $$\displaystyle (Eq. 2.22)$$ and $$\displaystyle (Eq. 2.9)$$,

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}+{{y}_{P}}$$
 * }
 * }

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

$$y={{C}_{1}}\cdot \frac{x}+{{C}_{2}}\cdot \frac{x}+\frac{8x}\left( i-2\left( 1+{{e}^{2ix}} \right)x \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.23)$$
 * }
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team3.Sudheesh 01:09, 15 November 2010 (UTC)
 * Proofread by

=Problem 3: Solution to Non-homogeneous Legendre Equation=

Given
Given a Non-homogeneous Legendre Equation with $$n=1$$ in the form

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

$$ \left(1- {x}^{2} \right) \frac{ {d}^{2}y}{ d{x}^{2}}-2x \frac{dy}{dx}+2y= \frac{1}{1- {x}^{2}} $$ And a homogenous solution given in meeting  of $${y}_{H}=x $$
 *
$$\displaystyle (Eq. 3.1) $$
 * }
 * }

Find
Find the solution to equation 3.1 using the direct method.

Solution
A solution can be assumed to be

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

$$ y \left(x \right)=U {u}_{1} \left(x \right) $$ When $$U$$ is unknown and $$ {u}_{1} \left(x \right) $$ is the known homogeneous solution. The derivatives of equation 3.2 are given by
 *
$$\displaystyle (Eq. 3.2) $$
 * }
 * }

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

$$ \frac{dy}{dx}=U \frac{d {u}_{1}}{dx}+\frac{d {U}}{dx}{u}_{1} $$
 *
$$\displaystyle (Eq. 3.3) $$
 * }
 * }

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

$$\frac{ {d}^{2}y}{ d{x}^{2}}=U\frac{ {d}^{2} {u}_{1}}{ d{x}^{2}}+2 \frac{dU}{dx}\frac{d {u}_{1}}{dx}+\frac{ {d}^{2}U}{ d{x}^{2}}\frac{d {u}_{1}}{dx} $$ Plugging equations 3.2, 3.3, and 3.4 into equation 3.1, we find
 *
$$\displaystyle (Eq. 3.4) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$\left(1- {x}^{2} \right) \left[ {U}^{\prime \prime } {u}_{1}+2 {U}^{\prime } { {u}_{1}}^{\prime }+U {{u}_{1}}^{\prime \prime } \right]-2x \left[U {{u}_{1}}^{\prime }+ {U}^{\prime } {u}_{1} \right] +2 \left[U {u}_{1} \right]= \frac{1}{1- {x}^{2}} $$ Noting that substitution of homogeneous solution in the given equation yields zero, we can cancel those terms out, yielding
 *
$$\displaystyle (Eq. 3.5) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$  {u}_{1} \left(1- {x}^{2} \right) {U}^{\prime \prime}+ \left[2 \left(1- {x}^{2}  \right) { {u}_{1}}^{\prime}-2x {u}_{1}\right] {U}^{\prime}= \frac{1}{1- {x}^{2}} $$ Use the Reduction of Order method 0. Let $$ Z= {U}^{\prime} $$, and equation 3.6 becomes
 *
$$\displaystyle (Eq. 3.6) $$
 * }
 * }

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

$${u}_{1} \left(1- {x}^{2} \right) {Z}^{\prime}+ \left[2 \left(1- {x}^{2} \right) { {u}_{1}}^{\prime}-2x {u}_{1}\right] {Z}= \frac{1}{1- {x}^{2}}   $$ Rearranging:
 *
$$\displaystyle (Eq. 3.7) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ {Z}^{\prime}+ \left[ \frac{2 {{u}_{1}}^{\prime}}{{u}_{1}}- \frac{2x}{ \left(1- {x}^{2} \right)} \right]Z= \frac{1}{ {u}_{1} {\left(1- {x}^{2} \right)}^{2}} $$
 *
$$\displaystyle (Eq. 3.8) $$
 * }
 * }

The integrating factor is
 * {| style="width:100%" border="0"

$$h\left( x \right)=exp\left[ \int{\left( \frac{2{{u}_{1}}^{\prime }}-\frac{2x}{\left( 1-{{x}^{2}} \right)} \right)}dx \right]$$ Substituting in our homogeneous solution and integrating gives
 *
$$\displaystyle (Eq. 3.9) $$
 * }
 * }

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

$$h\left( x \right)=exp\left[ log\left( 1-{{x}^{2}} \right)+2log\left( x \right) \right]$$
 * }
 * }

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

$$h\left( x \right)={{x}^{2}}\left( 1-{{x}^{2}} \right)$$
 *
$$\displaystyle (Eq. 3.10) $$
 * }
 * }

The solution to equation 3.8 can be written as

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

$$Z\left( x \right)=\frac{1}{h\left( x \right)}\left[ {{k}_{2}}+\int{\left[ h\left( x \right)\frac{f\left( x \right)}{{{u}_{1}}(x)} \right]}dx \right]$$ Substituting in
 *
$$\displaystyle (Eq. 3.11) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}+\int{\left[ \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}{x{{\left( 1-{{x}^{2}} \right)}^{2}}} \right]}dx \right]$$
 * }
 * }

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

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}+\int{\left[ \frac{x}{\left( 1-{{x}^{2}} \right)} \right]}dx \right]$$
 * }
 * }

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

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}-\frac{1}{2}\log \left( 1-{{x}^{2}} \right) \right]$$
 *
$$\displaystyle (Eq. 3.12) $$
 * }
 * }

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

$$U\left( x \right)={{k}_{1}}+\int{Z}\left( x \right)dx$$
 * }
 * }

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

$$U\left( x \right)={{k}_{1}}+\int{\left( \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}-\frac{1}{2}\frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 *
$$\displaystyle (Eq. 3.13) $$
 * }
 * }

Therefore,

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

$$y\left( x \right)=U\cdot {{u}_{1}}\left( x \right)$$
 * }
 * }

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

$$y\left( x \right)={{k}_{1}}x+{{k}_{2}}x\int{\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}dx}-\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * }
 * }

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

$$y\left( x \right)={{k}_{1}}x+{{k}_{2}}x\cdot \frac{1}{2}\left( -\frac{2}{x}-\log (1-x)+\log 1+x \right)-\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * }
 * }

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

$$y\left( x \right)={{k}_{1}}\underbrace{x}_+{{k}_{2}}\left( \underbrace{\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1}_ \right)-\underbrace{\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx}_$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 3.14) $$
 * }
 * }

Contributing Members

 * Solved and posted by James Roark 22:16, 15 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 02:11, 16 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 18:30, 17 November 2010 (UTC)

=Problem 4: Infinitesimal segment ds in spherical coordinates and spherical Laplace equation"=

Given
The spherical coordinates in the cartesian coordinate system are given as
 * {| style="width:100%" border="0"

$$x={{x}_{1}}=r\cos \left( \theta \right)\cos \left( \varphi  \right)$$
 *
$$\,(Eq. 4.1) $$
 * style= |
 * }

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

$$y={{x}_{2}}=r\cos \left( \theta \right)\sin \left( \varphi  \right)$$
 *
$$\,(Eq. 4.2) $$
 * style= |
 * }

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

$$z={{x}_{3}}=r\sin \left( \theta \right)$$
 *
$$\,(Eq. 4.3) $$
 * style= |
 * }

Find
The infinitesimal segment ds and the Laplace equation for spherical coordinates have to be found.

Solution
Segment ds: The infinitesimal segment ds in the three dimensional space can be expressed as
 * {| style="width:100%" border="0"

$$d{{s}^{2}}=\sum\limits_{i=1}^{3}$$
 *
$$\,(Eq. 4.4) $$
 * style= |
 * }

The general expression for $$d{{x}_{i}}$$ is

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

$$d{{x}_{i}}=\frac{\partial {{x}_{i}}}{\partial r}dr+\frac{\partial {{x}_{i}}}{\partial \theta }d\theta +\frac{\partial {{x}_{i}}}{\partial \varphi }d\varphi $$
 * style= |
 * }

Thus the expressions for $$d{{x}_{1}}$$,$$d{{x}_{2}}$$ and $$d{{x}_{3}}$$
 * {| style="width:100%" border="0"

$$d{{x}_{1}}=\frac{\partial \left( r\cos \left( \theta \right)\cos \left( \varphi  \right) \right)}{\partial r}dr+\frac{\partial \left( r\cos \left( \theta  \right)\cos \left( \varphi  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\cos \left( \theta  \right)\cos \left( \varphi  \right) \right)}{\partial \varphi }d\varphi =\underbrace{\cos \left( \theta  \right)\cos \left( \varphi  \right)dr}_{a}-\underbrace{r\cos \left( \varphi  \right)\sin \left( \theta  \right)d\theta }_{b}-\underbrace{r\cos \left( \theta  \right)\sin \left( \varphi  \right)d\varphi }_{c}$$

 *
$$\,(Eq. 4.5) $$
 * style= |
 * }

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

$$d{{x}_{2}}=\frac{\partial \left( r\cos \left( \theta \right)\sin \left( \varphi  \right) \right)}{\partial r}dr+\frac{\partial \left( r\cos \left( \theta  \right)\sin \left( \varphi  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\cos \left( \theta  \right)\sin \left( \varphi  \right) \right)}{\partial \varphi }d\varphi =\underbrace{\cos \left( \theta  \right)\sin \left( \varphi  \right)dr}_{d}-\underbrace{r\sin \left( \varphi  \right)\sin \left( \theta  \right)d\theta }_{e}+\underbrace{r\cos \left( \theta  \right)\cos \left( \varphi  \right)d\varphi }_{f}$$

 *
$$\,(Eq. 4.6) $$
 * style= |
 * }

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

$$d{{x}_{3}}=\frac{\partial \left( r\sin \left( \theta \right) \right)}{\partial r}dr+\frac{\partial \left( r\sin \left( \theta  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\sin \left( \theta  \right) \right)}{\partial \varphi }d\varphi =\sin \left( \theta  \right)dr+r\cos \left( \theta  \right)d\theta $$

The squared values for Eq. 4.5, 4.6 and 4.7 are
 *
$$\,(Eq. 4.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & d{{x}_{1}}^{2}={{\left( a-b-c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2bc-2ac-2ab \right) \\ & \quad \ \ \ ={{\cos }^{2}}\left( \theta \right){{\cos }^{2}}\left( \varphi  \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)d{{\varphi }^{2}} \\ & \quad \quad \ \ +2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)d\theta d\varphi  \\ & \quad \quad \ \ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)drd\varphi  \\ & \quad \quad \ \ -2r\cos \left( \theta \right){{\cos }^{2}}\left( \varphi  \right)\sin \left( \theta  \right)drd\theta  \\ \end{align}$$

 *
$$\,(Eq. 4.8) $$
 * style= |
 * }

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

$$\begin{align} & d{{x}_{2}}^{2}={{\left( d-e+f \right)}^{2}}=\left( {{d}^{2}}+{{e}^{2}}+{{f}^{2}}+2df-2de-2ef \right) \\ & \quad \quad ={{\cos }^{2}}\left( \theta \right){{\sin }^{2}}\left( \varphi  \right)d{{r}^{2}}+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right)d{{\varphi }^{2}} \\ & \quad \quad \ \ +2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)drd\varphi  \\ & \quad \quad \ \ -2r\cos \left( \theta \right)\sin \left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)drd\theta  \\ & \quad \quad \ \ -2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)d\theta d\varphi  \\ \end{align}$$
 *
$$\,(Eq. 4.9) $$
 * style= |
 * }

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

$$d{{x}_{3}}^{2}={{\sin }^{2}}\left( \theta \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\theta }^{2}}+2r\cos \left( \theta  \right)\sin \left( \theta  \right)drd\theta $$
 *
$$\,(Eq. 4.10) $$
 * style= |
 * }

Adding the Eqs. 4.8, 4.9 and 4.10 together will be performed in the following manner: regroup the common terms, factorize and add
 * {| style="width:100%" border="0"

$$d\theta d\varphi \to \left[ 2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)-2{{r}^{2}}\cos \left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right) \right]d\theta d\varphi =0$$

 *
$$\,(Eq. 4.11) $$
 * style= |
 * }

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

$$drd\varphi \to \left[ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)+2r{{\cos }^{2}}\left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right) \right]=0$$
 *
$$\,(Eq. 4.12) $$
 * style= |
 * }

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

$$\begin{align} & drd\theta \to \left[ -2r\cos \left( \theta \right){{\cos }^{2}}\left( \varphi  \right)\sin \left( \theta  \right)-2r\cos \left( \theta  \right)\sin \left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)+2r\cos \left( \theta  \right)\sin \left( \theta  \right) \right]drd\theta  \\ & \quad \quad \quad \ =2r\cos \left( \theta \right)\sin \left( \theta  \right)\left[ -{{\cos }^{2}}\left( \theta  \right)-{{\sin }^{2}}\left( \theta  \right)+1 \right]drd\theta =0 \\ \end{align}$$

 *
$$\,(Eq. 4.13) $$
 * style= |
 * }

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

$$d{{r}^{2}}\to \left[ {{\cos }^{2}}\left( \theta \right){{\cos }^{2}}\left( \varphi  \right)+{{\cos }^{2}}\left( \theta  \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+\left( 1-{{\cos }^{2}}\left( \theta  \right) \right) \right]d{{r}^{2}}=1d{{r}^{2}}$$
 *
$$\,(Eq. 4.14) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$d{{\theta }^{2}}\to \left[ {{r}^{2}}\left( 1-{{\sin }^{2}}\left( \varphi \right) \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}\left( 1-{{\sin }^{2}}\left( \theta  \right) \right) \right]d{{\theta }^{2}}={{r}^{2}}d{{\theta }^{2}}$$
 *
$$\,(Eq. 4.15) $$
 * style= |
 * }

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

$$d{{\varphi }^{2}}\to \left[ {{r}^{2}}{{\cos }^{2}}\left( \theta \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right) \right]d{{\varphi }^{2}}={{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\varphi }^{2}}$$
 *
$$\,(Eq. 4.16) $$
 * style= |
 * }

As it can be seen only Eq. 4.14, 4.15 and 4.16 are non-zero. Thus the infinitesimal segment ds is just the summation of these three expressions:
 * {| style="width:100%" border="0"

$$d{{s}^{2}}=\sum\limits_{i=1}^{3}=1d{{r}^{2}}+{{r}^{2}}d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta \right)d{{\varphi }^{2}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 4.17) $$
 * style= |
 * }

Spherical Laplace equation:

The Laplace equation can be written in general as
 * {| style="width:100%" border="0"

$$\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\zeta }_{i}}}}\left[ \frac\frac{\partial \Psi }{\partial {{\zeta }_{i}}} \right]$$ To get the spehrical Laplace equation we re-define the following variables
 *
$$\,(Eq. 4.18) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & r={{\zeta }_{1}} \\ & \theta ={{\zeta }_{2}} \\ & \varphi ={{\zeta }_{3}} \\ \end{align}$$ Equation 4.17 can now be written as
 *
$$\,(Eq. 4.19) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$d{{s}^{2}}=1{{\left( d{{\zeta }_{1}} \right)}^{2}}+{{\zeta }_{1}}^{2}{{\left( d{{\zeta }_{2}} \right)}^{2}}+{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right){{\left( d{{\zeta }_{3}} \right)}^{2}}={{\left( {{h}_{1}}\left( \zeta \right) \right)}^{2}}{{\left( d{{\zeta }_{1}} \right)}^{2}}+{{\left( {{h}_{2}}\left( \zeta  \right) \right)}^{2}}{{\left( d{{\zeta }_{2}} \right)}^{2}}+{{\left( {{h}_{3}}\left( \zeta  \right) \right)}^{2}}{{\left( d{{\zeta }_{3}} \right)}^{2}}$$ where the coefficients $${{h}_{i}}\left( \zeta \right)$$ are
 *
$$\,(Eq. 4.20) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{h}_{1}}\left( \zeta \right)=1 \\ & {{h}_{2}}\left( \zeta \right)={{\zeta }_{1}} \\ & {{h}_{3}}\left( \zeta \right)={{\zeta }_{1}}\cos \left( {{\zeta }_{2}} \right) \\ \end{align}$$ Now Eq. 4.18 can be re-written as
 *
$$\,(Eq. 4.21) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Delta \Psi =\frac{1}{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\zeta }_{i}}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{i}}} \right]$$ for the different cases where the index variable i has values of 1,2 and three, the terms of the summation are:
 *
$$\,(Eq. 4.22) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & i=1\to \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right] \\ & i=2\to \frac{\partial }{\partial {{\zeta }_{2}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}}\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]=\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right] \\ & i=3\to \frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right]=\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \\ \end{align}$$
 *
$$\,(Eq. 4.23) $$
 * style= |
 * }

So that the Laplace equation can be finally written as being:
 * {| style="width:100%" border="0"

$$\Delta \Psi =\frac{1}{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\left\{ \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right]+\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]+\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \right\}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 4.24) $$
 * style= |
 * }

Or explicit in spherical coordinates the Laplace equation would then be, after replacing $${{\zeta }_{i}}$$ with the respective spherical coordinate:
 * {| style="width:100%" border="0"

$$\Delta \Psi =\frac{1}\frac{\partial }{\partial r}\left[ {{r}^{2}}\frac{\partial \Psi }{\partial r} \right]+\frac{1}{{{r}^{2}}\cos \left( \theta \right)}\frac{\partial }{\partial \theta }\left[ \cos \left( \theta  \right)\frac{\partial \Psi }{\partial \theta } \right]+\frac{1}{{{r}^{2}}{{\cos }^{2}}\left( \theta  \right)}\left[ \frac{{{\partial }^{2}}\Psi }{\partial {{\varphi }^{2}}} \right]$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 4.25) $$
 * style= |
 * }

Additoinal links and useful informations on spherical coordinates and Laplace equations
The following page delivers a short description on how to convert cartesian coordinates into spherical or cylindrical coordinates in Matlab with an already existing function:

Matlab-spherical-transformation

This Wolfram page gives a more in depth insight into spherical cooridnates that is worth to take a look at. Note, that the angle $$\theta $$ was selected to be 90° minus the $$\theta $$ used in the problem above.

Wolfram-SphericalCoordinates

In this page the Wolfram-team shows how to solve the Laplace equation with spherical coordinates using the method of separation of vaiables, like discussed in class:

Wolfram-Laplaces Equation SphericalCoordinates

This paper from the University of Texas-State shows also the solution of the Laplace equation, described in a very elegant way:

Texas-State-University:spherical_laplace.pdf

Contributing Members

 * Solved and posted by Egm6321.f10.team03.sigillo 19:21, 12 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 23:25, 15 November 2010 (UTC)

=Problem 5: Derive the Laplacian in Cylindrical Coordinates=

Given
From the lecture notes p 35-3
 * {| style="width:100%" border="0"

{x}_{1}=x= { \xi}_{1} \cos { \xi}_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq 5.1)
 * style= |
 * }

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

{x}_{2}=y= { \xi}_{1} \sin { \xi}_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq 5.2)
 * style= |
 * }

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

{x}_{3}=z= { \xi}_{3} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq 5.3)
 * style= |
 * }

Find
$$ 1) $$ Find $$ \left\{  {dx}_{i}\right\}=  \left\{  {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$ in terms of $$  \left\{  {\xi}_{j}\right\}=  \left\{  {\xi}_{1}, {\xi}_{2}, {\xi}_{3}\right\} $$ and $$ \left\{  {d\xi}_{k}\right\} $$

$$ 2) $$ Find $${ds}^{2} = \sum_{i} { \left(d {x}_{i} \right)}^{2}=  \sum_{i}  {\left(  {h}_{i}\right)}^{2}  { \left( {d\xi}_{i} \right)}^{a}$$ Identify $$ \left\{  {h}_{i}\right\} $$ in terms of $$ \left\{  {\xi}_{i}\right\} $$

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

$$ 4) $$ Do seperation of variable to recover Bessel's equation

$$ \left\{ {dx}_{i}\right\} $$
To solve we find the total derivatives of $$ \left\{ {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$

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

{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} $$ $$ Substituting $$\displaystyle (Eq 5.1) $$ into $$\displaystyle (Eq 5.4) $$ we get
 *
$$\displaystyle (Eq 5.4)
 * style= |
 * }

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

{dx}_{1} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \cos {\xi}_{2}\right){d\xi}_{3} $$
 *

 * style= |
 * }

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

$$\displaystyle {dx}_{1} =  \cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq 5.5)
 * style= |
 * }

Similarly,

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

{dx}_{2} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \sin {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \sin {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \sin {\xi}_{2}\right){d\xi}_{3} $$
 *

 * style= |
 * }

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

$$\displaystyle {dx}_{2} =  \sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq 5.6)
 * style= |
 * }

Since there is no projection on $$ {x}_{3}$$ from $$\displaystyle (Eq 5.4) $$

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

$$\displaystyle {dx}_{3} =  {d\xi}_{3} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq 5.7)
 * style= |
 * }

$$ {\left(ds\right)}^{2} $$
To solve we use the relationship

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

{ds}^{2} = { \left(d {x}_{1} \right)}^{2}+ { \left(d {x}_{2} \right)}^{2}+ { \left(d {x}_{3} \right)}^{2} $$
 *

 * style= |
 * }

And substituting equations 5.5 through 5.7 we obtain the following

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

{ds}^{2} = { \left(\cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} \right)}^{2}+ { \left(\sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} \right)}^{2}+ { \left(d {\xi}_{3} \right)}^{2} $$ $$
 *
$$\displaystyle (Eq 5.8)
 * style= |
 * }

Expanding equation 5.8 and combining like terms gives us

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

{ds}^{2} = \cancelto{1}{\left( {\cos {\xi}_{2}}^{2}+ { \sin {\xi}_{2}}^{2} \right)} {\left( {d\xi}_{1} \right)}^{2}+ \cancelto{{{\xi}_{1}}^{2}}{\left({{\xi}_{1}}^{2}{\cos {\xi}_{2}}^{2}+ {  {{\xi}_{1}}^{2}\sin {\xi}_{2}}^{2} \right)}  {\left( {d\xi}_{2} \right)}^{2}+  {\left( {d\xi}_{3} \right)}^{2} $$ $$
 *
$$\displaystyle (Eq 5.9)
 * style= |
 * }

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

$$ {ds}^{2}={\left( {d\xi}_{1} \right)}^{2}+ {{\xi}_{1}}^{2}{\left( {d\xi}_{2} \right)}^{2}+ {\left( {d\xi}_{3} \right)}^{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq 5.10)
 * style= |
 * }

Now by inspection we see that

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

$$\displaystyle {h}_{1}= 1 ; {h}_{2} =  {\xi}_{1} ; {h}_{3}= 1 $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq 5.11)
 * style= |
 * }

$$\Delta \Psi $$
To obtain the Laplacian in cylindrical coordinates we use the following equation

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

\Delta \Psi = \frac{1}{ {h}_{1}{h}_{2} {h}_{3}} \sum_{i=1}^{3}  \frac{\partial }{\partial  {\xi}_{i}} \left(  \frac{{h}_{1}{h}_{2} {h}_{3}}{ {\left( {h}_{i} \right)}^{2}} \frac{\partial  \Psi}{\partial {\xi}_{i}}\right) $$ $$
 *
$$\displaystyle (Eq 5.12)
 * style= |
 * }

Now we plug in our values for $$ {h}_{i}$$ and expand the summation

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

$$\Delta \Psi =\frac{1}\frac{\partial }{\partial {{\xi }_{1}}}\left( {{\xi }_{1}}\frac{\partial \Psi }{\partial {{\xi }_{1}}} \right)+\frac{1}{{{\xi }_{1}}^{2}}\frac{{{\partial }^{2}}\Psi }{\partial \xi _{2}^{2}}+\frac{{{\partial }^{2}}\Psi }{\partial \xi _{3}^{2}}$$ $$
 *
$$\displaystyle (Eq 5.13)
 * style= |
 * }

Finally we replace the $$ \xi s $$ with the cylindrical coordinate convention to obtain

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

 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |

$$\Delta \Psi =\frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial \Psi }{\partial r} \right)+\frac{1}\frac{{{\partial }^{2}}\Psi }{\partial {{\theta }^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}}$$

$$
 *
$$\displaystyle (Eq 5.14)
 * style= |
 * }

Do seperation of variable to recover Bessel's equation
Assume,

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

$$
 * $$\psi (r,\phi ,z)=R(r)\cdot \Theta (\phi )\cdot Z(z)$$
 *
$$\displaystyle (Eq 5.15)
 *
$$\displaystyle (Eq 5.15)
 * style= |
 * }

Substitute above equation in $$\displaystyle Eq 5.14 $$ and equate to zero:

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

 * $$\Delta \psi =\Theta \cdot Z\cdot \frac{1}{r}\frac{d}{dr}\left( r\frac{dR}{dr} \right)+R\cdot Z\cdot \frac{1}\frac{{{d}^{2}}\Theta }{d{{\phi }^{2}}}+R\cdot \Theta \cdot \frac{{{d}^{2}}Z}{d{{z}^{2}}}=0$$
 * style= |
 * }
 * }

Divide the above equation by $$\displaystyle R\cdot \Theta \cdot Z$$

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

 * $$\frac{1}{R\cdot r}\frac{d}{dr}\left( r\frac{dR}{dr} \right)+\frac{1}{\Theta \cdot {{r}^{2}}}\frac{{{d}^{2}}\Theta }{d{{\phi }^{2}}}+\frac{1}{Z}\frac{{{d}^{2}}Z}{d{{z}^{2}}}=0$$
 * style= |
 * }
 * }

Now consider

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

 * $$\frac{1}{Z}\frac{{{d}^{2}}Z}{d{{z}^{2}}}=q$$
 * style= |
 * }
 * }

Implies:

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

 * $$\frac{1}{R\cdot r}\frac{d}{dr}\left( r\frac{dR}{dr} \right)+\frac{1}{\Theta \cdot {{r}^{2}}}\frac{{{d}^{2}}\Theta }{d{{\phi }^{2}}}+q=0$$
 * style= |
 * }
 * }

Multiply the above equation by $$\displaystyle {{r}^{2}}$$, yields:

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

$$
 * $$\frac{r}{R}\frac{d}{dr}\left( r\frac{dR}{dr} \right)+\frac{1}{\Theta }\frac{{{d}^{2}}\Theta }{d{{\phi }^{2}}}+{{r}^{2}}{{q}^{2}}=0$$
 *
$$\displaystyle (Eq 5.16)
 *
$$\displaystyle (Eq 5.16)
 * style= |
 * }

Now assume,

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

 * $$\frac{1}{\Theta }\frac{{{d}^{2}}\Theta }{d{{\phi }^{2}}}=-{{\nu }^{2}}$$
 * style= |
 * }
 * }

Therfore $$\displaystyle Eq 5.16 $$ becomes,

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

$$
 * $${{r}^{2}}\frac{{{d}^{2}}R}{d{{r}^{2}}}+r\frac{dR}{dr}+\left( {{(rq)}^{2}}-{{\nu }^{2}} \right)R=0$$
 *
$$\displaystyle (Eq 5.17)
 *
$$\displaystyle (Eq 5.17)
 * style= |
 * }

Now assume $$\displaystyle \text{x=rq}$$

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

 * $$\frac{dR}{dr}=\frac{dR}{dx}\frac{dx}{dr}=\frac{dR}{dx}q$$
 * style= |
 * }
 * }

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

 * $$\frac{{{d}^{2}}R}{d{{r}^{2}}}={{q}^{2}}\frac{{{d}^{2}}R}{d{{x}^{2}}}$$
 * style= |
 * }
 * }

Substitute above two equations in $$\displaystyle Eq 5.17 $$

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

 * $${{r}^{2}}{{q}^{2}}\frac{{{d}^{2}}R}{d{{x}^{2}}}+rq\frac{dR}{dx}+\left( {{x}^{2}}-{{\nu }^{2}} \right)R=0$$
 * style= |
 * }
 * }

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

$${{x}^{2}}\frac{{{d}^{2}}R}{d{{x}^{2}}}+x\frac{dR}{dx}+({{x}^{2}}-{{\nu }^{2}})R=0$$ $$ the above equation is the Bessel Equation
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq 5.18)
 * style= |
 * }

Contributing Members
Solved and Posted by--Egm6321.f10.team3.cook 02:45, 16 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 02:58, 16 November 2010 (UTC)
 * Solved and posted part4 and proofread by Egm6321.f10.team3.Sudheesh 20:19, 17 November 2010 (UTC)

=Problem 6 : Spherical coordinates using Math/Phys. convention=

Given
the equation below has been determined by Eqt 4.17

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

\begin{align} ds^{2} &= \sum_{i=1}^{3} ({dx_{i}})^{2} = \sum_{k=1}^{3} (h_{k})^{2} \cdot (\xi)^{2}\\ &= dr^{2} + r^{2}\cdot cos^{2}(\phi) \cdot d \phi^{2} + r^{2} \cdot d\theta^{2}\\ &= (h_{1})^{2} \cdot dr^{2} + (h_{2})^{2} \cdot d \phi^{2} + (h_{3})^{2} \cdot d\theta^{2}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

\begin{align} \bar{\theta} = \frac {\pi}{2} - \theta\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Find
Find $$ \psi $$ in spherical coordinate using Math/Phys. convention

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

$$ \bar{\theta} = \frac {\pi}{2} -\theta $$
 *
$$\displaystyle (Eqt. 6.1) $$
 * }
 * }

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

$$ ds^{2} = dr^{2} + r^{2}\cdot cos^{2}\theta \cdot d\phi^{2} + r^{2} \cdot d\theta^{2} $$
 *
$$\displaystyle (Eqt. 6.2) $$
 * }
 * }

Put Eqt 6.1 into Eqt 6.2

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

$$ d \bar{\theta}= d(\frac {\pi}{2}) - d\theta = 0 - d\theta = - d\theta $$
 *
$$\displaystyle (Eqt. 6.3) $$
 * }
 * }

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

$$ cos \theta = sin \bar{\theta} $$
 *
$$\displaystyle (Eqt. 6.4) $$
 * }
 * }

by using Eqt 6.3 & 6.4, we obtain new coordinates & h_1, h_2, h_3

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

\begin{align} ds^{2} &= dr^{2} + r^{2}\cdot sin^{2}\bar{\theta} \cdot d\phi^{2} + r^{2} \cdot d\theta^{2}\\ &= {h_{1}}^2 \cdot dr^{2} + {h_{2}}^2 \cdot d\phi^{2} + {h_{3}}^2 \cdot d\theta^{2} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Thus, we find h_1, h_2 & h_3

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

\begin{align} h_{1} = 1 \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

\begin{align} h_{2} = r \cdot sin(\bar{\theta}) \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

\begin{align} h_{3} = r \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

The laplacian for any function $$ \delta \Psi $$ can be determined by

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

\begin{align} \Delta \psi = \frac {1}{h_{1} \cdot h_{2} \cdot h_{3}} \cdot \sum_{i=1}^{3} \frac {\partial}{\partial \xi_i} [\frac {h_{1} \cdot h_{2} \cdot h_{3}}{(h_{i})^{2}} \cdot \frac {\partial \psi}{\partial \xi_{i}}] \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

for i=1

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

\begin{align} \Delta \psi_{1} &= \frac {1}{h_{1} \cdot h_{2} \cdot h_{3}} \cdot \frac {\partial}{\partial \xi_1} [\frac {h_{1} \cdot h_{2} \cdot h_{3}}{(h_{1})^{2}} \cdot \frac {\partial \psi}{\partial \xi_{1}}]\\ &= \frac {1}{1 \cdot r \cdot sin \bar{\theta} \cdot r} \frac {\partial}{\partial r} [\frac {1 \cdot r \cdot sin \bar{\theta} \cdot r}{(1)^{2}} \cdot \frac {\partial \psi}{\partial r}]\\ &= \frac {1}{r^{2} \cdot sin \bar{\phi}} \frac {\partial}{\partial r} [r^{2} \cdot sin \bar{\theta} \frac {\partial \psi}{\partial r}]\\ &= \frac {1}{r^{2}} \frac {\partial}{\partial r} [r^{2} \frac{\partial \psi}{\partial r}]\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

for i=2

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

\begin{align} \Delta \psi_{2} &= \frac {1}{h_{1} \cdot h_{2} \cdot h_{3}} \cdot \frac {\partial}{\partial \xi_2} [\frac {h_{1} \cdot h_{2} \cdot h_{3}}{(h_{2})^{2}} \cdot \frac {\partial \psi}{\partial \xi_{2}}]\\ &= \frac {1}{1 \cdot r \cdot sin \bar{\theta} \cdot r} \frac {\partial}{\partial \phi} [\frac {1 \cdot r \cdot sin \bar{\theta} \cdot r}{(r \cdot sin \bar{\theta})^{2}} \cdot \frac {\partial \psi}{\partial \phi}]\\ &= \frac {1}{r^{2} \cdot sin \bar{\theta}} \frac {\partial}{\partial \phi} [\frac {1}{sin \bar{\theta}} \frac {\partial \psi}{\partial \phi}]\\ &= \frac {1}{r^{2} \cdot sin^{2} \bar{\theta}} \frac {\partial}{\partial \phi} [\frac {\partial \psi}{\partial \phi}]\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

for i=3

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

\begin{align} \Delta \psi_{3} &= \frac {1}{h_{1} \cdot h_{2} \cdot h_{3}} \cdot \frac {\partial}{\partial \xi_3} [\frac {h_{1} \cdot h_{2} \cdot h_{3}}{(h_{3})^{2}} \cdot \frac {\partial \psi}{\partial \xi_{3}}]\\ &= \frac {1}{1 \cdot r \cdot sin \bar{\theta} \cdot r} \frac {\partial}{\partial \bar {\theta}} [\frac {1 \cdot r \cdot sin \bar{\theta} \cdot r}{(r)^{2}} \cdot \frac {\partial \psi}{\partial \bar {\theta}}]\\ &= \frac {1}{r^{2} \cdot sin \bar{\theta}} \frac {\partial}{\partial \bar {\theta}} [sin \bar{\theta} \cdot \frac {\partial \psi}{\partial \bar {\theta}}]\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

As a result, we determine $$ \Delta \psi $$

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

\begin{align} \Delta \psi = \frac {1}{r^{2}} \frac {\partial}{\partial r} [r^{2} \frac{\partial \psi}{\partial r}] + \frac {1}{r^{2} \cdot sin^{2} \bar{\theta}} \frac {\partial}{\partial \phi} [\frac {\partial \psi}{\partial \phi}] + \frac {1}{r^{2} \cdot sin \bar{\theta}} \frac {\partial}{\partial \bar {\theta}} [sin \bar{\theta} \cdot \frac {\partial \psi}{\partial \bar {\theta}}] \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team03.Hong SJ 19:02, 15 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 23:53, 15 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 02:59, 16 November 2010 (UTC)
 * Proofed and edited by James Roark 13:56, 16 November 2010 (UTC)

=Problem 7: Laplace equation in elliptic coordinates=

Given
Elliptic coordinate system

 * [[Image:Sudheesh ellipse.jpg|center|650px]]

The article for elliptic coordinates discussed in class is given with it's Laplace equation. The cartesian coordinates are expressed as

where a is the distance from the origin to the foci of the ellips and

Find
The expression for the Laplace equation in the mentioned article needs to be verified using the steps learned for cylindrical and spherical coordinates

Solution
Like in probelm 4, we find an expression for the infinitesimal segment ds in the elliptic coordinate system though:

Therefore we find first the expressions for $$d{{x}_{1}}$$ and $$d{{x}_{2}}$$, that are:

Because of the axis symmetry we can exclude the dependency on a and cancel all terms related to $$da$$. The squared term of Eq. 7.5 and 7.6 are:

To find the expression for the segment ds we regroup, factorize and add the common terms:

Thus the segment ds is just a summation over the $$d{{\mu }^{2}}$$ and $$d{{\nu }^{2}}$$ terms:

Now, the infinitesimal segment ds can be expressed in general as where

We recall that the Laplace equation can be written as:

And in our specific case we can re-write the equation as:

Which is the same expression for the Laplace equation as in the article discussed in class.

Contributing Members

 * Solved and posted by Egm6321.f10.team03.sigillo 21:32, 13 November 2010 (UTC)
 * Drew and uploaded figure and proofread solution byEgm6321.f10.team3.Sudheesh 02:52, 17 November 2010 (UTC)

=Problem 8: Legendre Equation=

Given
$$Eqt(7) = \sum_{i=0}^{n/2} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} $$

$$Eqt(8) = \sum_{i=0}^{n/2} (-1)^{i} \frac {1 \times 3 \times 5 \times \cdot \cdot \cdot \times (2n-2i-1) \cdot x^{n-2i}} {2^{i}\cdot i! \cdot (n-2i)!} $$

Find
Verify that Eqt(7) is identical to Eqt(8)

Solution
Assume that Eqt(7) = Eqt(8)

$$Eqt(7) = \sum_{i=0}^{n/2} \cancel {(-1)^{i}} \cdot \frac {(2n-2i)! \cdot \cancel {x^{n-2i}}} {2^{n}\cdot \cancel {i!} \cdot (n-i)! \cdot \cancel {(n-2i)!}} $$

$$Eqt(8) = \sum_{i=0}^{n/2} \cancel {(-1)^{i}} \cdot \frac {1 \times 3 \times 5 \times \cdot \cdot \cdot \times (2n-2i-1) \cdot \cancel {x^{n-2i}}} {2^{i}\cdot \cancel {i!} \cdot \cancel {(n-2i)!}} $$

Thus, we can verify the equation below

$$ \sum_{i=0}^{n/2} \frac {(2n-2i)!} {2^{n} \cdot (n-i)!} =  \sum_{i=0}^{n/2} \frac {1 \times 3 \times 5 \times \cdot \cdot \cdot \times (2n-2i-1)} {2^{i}} $$

Let's focus on the left hand side

$$ [LHS] = \sum_{i=0}^{n/2} \frac {(2n-2i)!} {2^{n} \cdot (n-i)!} = \sum_{i=0}^{n/2} \frac {1 \times 2 \times 3 \times 4 \times 5 \cdot \cdot \cdot \times (2n-2i-1) \cdot (2n-2i)} {2^{n-i} \cdot 2^{i} \cdot (n-i)!} $$

replace constant & factoral by pi to easily arrange the equation

$$ 2^{n-i} = \prod\limits_{k = 1}^{n - i} 2 $$

$$ (n-i)! = \prod\limits_{k = 1}^{n - i} k $$

Therefore, [LHS]

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

\begin{align} [LHS] &= \sum_{i=0}^{n/2} \frac {1 \times 2 \times 3 \times 4 \times 5 \cdot \cdot \cdot \times (2n-2i-1) \cdot (2n-2i)} {2^{i} \cdot \prod\limits_{k = 1}^{n - i} 2 \cdot \prod\limits_{k = 1}^{n - i} k}\\ &= \sum_{i=0}^{n/2} \frac {1 \times 2 \times 3 \times 4 \times 5 \cdot \cdot \cdot \times (2n-2i-1) \cdot (2n-2i)} {2^{i} \cdot \prod\limits_{k = 1}^{n - i} 2k } \\ &= \sum_{i=0}^{n/2} \frac {1 \times 2 \times 3 \times 4 \times 5 \cdot \cdot \cdot \times (2n-2i-1) \cdot (2n-2i)} {2^{i} \times 2 \times 4 \times 6 \cdot \cdot \cdot \times 2(n-i) }\\ &= \sum_{i=0}^{n/2} \frac {1 \times \cancel {2} \times 3 \times \cancel {4} \times 5 \cdot \cdot \cdot \times (2n-2i-1) \cdot \cancel {(2n-2i)}} {2^{i} \times \cancel {2} \times \cancel {4} \times \cancel {6} \cdot \cdot \cdot \times \cancel {2(n-i)} }\\ &= \sum_{i=0}^{n/2} \frac {1 \times 3 \times  5 \cdot \cdot \cdot \times (2n-2i-1)} {2^{i}}  \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

As a result, Eqt(7) is identical to Eqt(8).

Given

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

\begin{align} &Eqt(2) : {P}_{0}(x) =  1\\ &Eqt(3) : {P}_{1}(x) =  x\\ &Eqt(4) : {P}_{2}(x) =  \frac {1}{2}(3x^{2}-1)\\ &Eqt(5) : {P}_{3}(x) =  \frac {1}{2}(5x^{3}-3x)\\ &Eqt(6) : {P}_{4}(x) =  \frac {35}{8}x^{4} - \frac {15}{4}x^{2} + \frac {3}{8}\\ &Eqt(7) : {P}_{n}(x) =  \sum_{i=0}^{n/2} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Find
Verify that Eqt(2)~(6) can be writthen as Eqt(7)

Solution

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

\begin{align} Eqt(2) : {P}_{0}(x) &= \sum_{i=0}^{0} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &= 1 \cdot \frac {1 \cdot 1} {1\cdot 1 \cdot 1 \cdot 1}\\ &= 1\\ Eqt(3) : {P}_{1}(x) &= \sum_{i=0}^{1/2} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &=(-1)^{0} \cdot \frac {2! \cdot x^{1}} {2^{1}\cdot 0! \cdot 1! \cdot 1!} = 1 \cdot \frac {\cancel {2} \cdot x} {\cancel {2} \cdot 1 \cdot 1 \cdot 1} \\ &= x\\ Eqt(4) : {P}_{2}(x) &= \sum_{i=0}^{2/2} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &=(-1)^{0} \cdot \frac {(4)! \cdot x^{2}} {2^{2}\cdot 0! \cdot (2)! \cdot (2)!} + (-1)^{1} \cdot \frac {(4-2)! \cdot x^{2-2}} {2^{2}\cdot 1! \cdot (2-1)! \cdot (2-2)!}\\ &= 1 \cdot \frac {4 \cdot 3 \cdot 3 \cdot x^{2}}{4 \cdot 1 \cdot 2 \cdot 2} + (-1) \cdot \frac {2 \cdot 1}{4 \cdot 1 \cdot 1 \cdot 1} = \frac {3}{2}x^{2}-\frac {1}{2})\\ &= \frac {1}{2}(3x^{2}-1)\\ Eqt(5) : {P}_{3}(x) &= \sum_{i=0}^{3/2} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} \\ &=(-1)^{0} \cdot \frac {6! \cdot x^{3}} {2^{3}\cdot 0! \cdot 3! \cdot 3!} + (-1)^{1} \cdot \frac {4! \cdot x^{1}} {2^{3}\cdot 1! \cdot 2! \cdot 1!} \\ &= 1 \cdot \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^{3}} {8 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 2} + (-1) \cdot \frac {4 \cdot 3 \cdot 2 \cdot x^{1}} {8\cdot 1 \cdot 2 \cdot 1}  = \frac {5}{2}x^{3} - \frac {3}{2} x \\ &= \frac {1}{2}(5x^{3}-3x)\\ Eqt(6) : {P}_{4}(x) &=  \sum_{i=0}^{4/2} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &= (-1)^{0} \cdot \frac {8! \cdot x^4} {2^{4}\cdot 0! \cdot 4! \cdot 4!} + (-1)^{1} \cdot \frac {6! \cdot x^2} {2^{4}\cdot 1! \cdot 3! \cdot 2!} + (-1)^{2} \cdot \frac {4! \cdot x^0} {2^{4}\cdot 2! \cdot 2! \cdot 0!} \\ &= 1 \cdot \frac {8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^4} {16 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2} + (-1) \cdot \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^2} {16 \cdot 1 \cdot 3 \cdot 2 \cdot 2} + 1 \cdot \frac {4 \cdot 3 \cdot 2 \cdot 1} {16 \cdot 2 \cdot 2 \cdot 1} \\ &= \frac {35}{8}x^{4} - \frac {15}{4}x^{2} + \frac {3}{8}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team03.Hong SJ 22:37, 14 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 02:10, 16 November 2010 (UTC)

= Problem 9 - Verification of First Homogenous Solutions of Legendre Equation=

Given

 * Legendre Equation,

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

$$\displaystyle \left( 1-{{x}^{2}} \right){{y}^{''}}-2x{{y}^{'}}+n\left( n+1 \right)y=0$$
 *
$$\displaystyle (Eq. 9.1)$$
 * }
 * }

 * Legendre polynomials:

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

$$\displaystyle{{P}_{0}}(x)=1$$
 *
$$\displaystyle (Eq. 9.2)$$
 * }
 * }

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

$$\displaystyle{{P}_{1}}(x)=x$$
 *
$$\displaystyle (Eq. 9.3)$$
 * }
 * }

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

$${{P}_{2}}(x)=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$
 *
$$\displaystyle (Eq. 9.4)$$
 * }
 * }

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

$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$
 *
$$\displaystyle (Eq. 9.5)$$
 * }
 * }

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

 * $${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$
 *
$$\displaystyle (Eq. 9.6)$$
 * }
 * }
 * }

Find

 * Verify that $$\displaystyle Eqs. 9.2-9.6 $$ are solutions of Legendre equation.

(1) n=0,$$\displaystyle {{P}_{0}}(x)=1$$

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

\begin{array}{*{20}l} y=1 & {{y}^{'}}=0 & {{y}^{''}}=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 0-2x\times 0+0\times 1=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.7)$$
 * }
 * }

(2) n=1,$$\displaystyle{{P}_{1}}(x)=x$$

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

\begin{array}{*{20}l} y=x & {{y}^{'}}=1 & {{y}^{''}}=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 0-2x\times 1+2\times x=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.8)$$
 * }
 * }

(3) n=2,$${{P}_{2}}(x)=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$

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

\begin{array}{*{20}l} y=\frac{1}{2}\left( 3{{x}^{2}}-1 \right) & {{y}^{'}}=3x & {{y}^{''}}=3\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 3-2x\times 3x+6\times \frac{1}{2}\left( 3{{x}^{2}}-1 \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.9)$$
 * }
 * }

(4) n=3,$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$

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

\begin{array}{*{20}l} y=\frac{1}{2}\left( 5{{x}^{3}}-3x \right) & {{y}^{'}}=\frac{1}{2}\left( 15{{x}^{2}}-3 \right) & {{y}^{''}}=15x\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 15x-2x\times \frac{1}{2}\left( 15{{x}^{2}}-3 \right)+12\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.10)$$
 * }
 * }

(5) n=4,$${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$

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

\begin{array}{*{20}l} 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}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

 * $$\left( 1-{{x}^{2}} \right)\times 15x-2x\times \frac{1}{2}\left( 15{{x}^{2}}-3 \right)+12\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$
 * }
 * }
 * }

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

$$=\frac{105}{2}{{x}^{2}}-\frac{105}{2}{{x}^{4}}-\frac{15}{2}+\frac{15}{2}{{x}^{2}}-\frac{70}{2}{{x}^{4}}+\frac{30}{2}{{x}^{2}}+\frac{175}{2}{{x}^{4}}-\frac{150}{2}{{x}^{2}}+\frac{15}{2}$$
 * }
 * }

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

$$=\frac{2}\left( 175-105-70 \right)+\frac{2}\left( 105+15+30-150 \right)+\frac{15}{2}-\frac{15}{2}=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.11)$$
 * }
 * }

 * Hence $$\displaystyle Eqs. 9.2-9.6 $$ are solutions of Legendre equation.

Contributing Members

 * Solved by Egm6321.f10.team3.Schulze 04:04, 15 November 2010 (UTC)

 * Posted and rechecked solution by Egm6321.f10.team3.Sudheesh 04:04, 15 November 2010 (UTC)

title: User:Egm6321.f10.team03/Hwk6


text: File:2010 11 10 14 40 12.djvu

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:2010 11 10 14 40 12.djvu


text:
Biodiversity

Biodiversity is the total of all the genetic variation contained within the biosphere (living world).

Biodiversity may be considered locally or globally. Local biodiversity can cover any amount of space. Examples of local biodiversity would be the number of species found in a local park, in a city, or on a continent. Global biodiversity covers the entire world.

Biodiversity is measured in four main ways:

 * taxonomic diversity (usually measured at the species diversity level)
 * ecological diversity often viewed from the perspective of ecosystem diversity
 * morphological diversity which stems from genetic diversity and molecular diversity
 * functional diversity which is a measure of the number of functionally disparate species within a population (e.g. different feeding mechanism, different motility, predator vs prey, etc.)

(see /Introduction/ for references)

Learning Tasks

 * (Distribution) Look on the distribution of biodiversity on the global map. Biodiversity is not evenly distributed on global map. Could you identify factors that can serve as indicators for the diversity in specific regions and identify causes for loss of biodiversity on the globe. Start from the consideration, that diversity of all living things (biota) depends on temperature, precipitation, altitude, soils, geography and the presence of other species living in symbiosis or due to dependencies in a food chain. The study of the spatial distribution of organisms, species and ecosystems, is the science of biogeography..
 * What are human factors that have an impact on the distribution of biodiversity?
 * Assume you get environmental conditions as information for a specific area on the globe and/or the geolocation of the area and/or the populations density and/or pollution. What are the most important factor estimate biodiversity in specific region?
 * Consider climate change scenarios and estimate the impact on biodiversity due to changes in the environmental conditions. Search for scientific evidence in peer-reviewed journals about the link between biodiversity and climate change.

Comments Distribution of Biodiversity
Diversity consistently measures higher in the tropics and in other localized regions such as the Cape Floristic Region and lower in polar regions generally. Rain forests that have had wet climates for a long time, such as Yasuní National Park in Ecuador, have particularly high biodiversity.

Terrestrial biodiversity is thought to be up to 25 times greater than ocean biodiversity. Forests harbour most of Earth's terrestrial biodiversity. The conservation of the world's biodiversity is thus utterly dependent on the way in which we interact with and use the world's forests. A new method used in 2011, put the total number of species on Earth at 8.7 million, of which 2.1 million were estimated to live in the ocean. However, this estimate seems to under-represent the diversity of microorganisms. Forests provide habitats for 80 percent of amphibian species, 75 percent of bird species and 68 percent of mammal species. About 60 percent of all vascular plants are found in tropical forests. Mangroves provide breeding grounds and nurseries for numerous species of fish and shellfish and help trap sediments that might otherwise adversely affect seagrass beds and coral reefs, which are habitats for many more marine species.

Learning Resources

 * /Introduction/
 * Biodiversity loss

Page Information
This page was based on the following wikipedia-source page:
 * Biodiversity https://en.wikipedia.org/wiki/Biodiversity
 * Datum: 3/31/2022
 * Wikipedia2Wikiversity-Converter: https://niebert.github.com/Wikipedia2Wikiversity

title: Biodiversity


text:
User:Mago5

I am new to this entire process, although I have been a Wikipedia reader for years! I am intrigued by two possibilities, but have not yet begun to explore either. I first am attracted to the possibility that I might be able to take online courses via Wikiversity, but have not yet been able to discern how to do so. I thought by creating this account, I might be able to further that process. Secondly, the idea that one can "deposit" ones knowledge here for world view, consumption and preservation is highly exciting. I doubt I have any specific knowledge that isn't already noted here; nonetheless, the concept is exciting to think about. Well, this is all I have to say for the moment, but I will return! Mago5

title: User:Mago5


text:
User talk:Mago5

 Hello Mago5, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Mago5


text:
Agape

Agape Law seems in many ways to be an oxymoron. In the writings of Paul the Apostle, the word 'agape' is used as a contrast to the word 'law'. One could of course argue that the word 'law' in Paul refers to the law of Judaism and that the word 'agape' then refers to the law of the early Christian church. However, as soon as the expressions of Agape, viz. to be kind, compassionate, caring and so forth, become compulsory or legally binding, the essential Pauline notion of Agape is diminished. Agape is an attitude towards others which is not obligatory. One chooses that attitude. Paul does imply that Christians "must have this attitude which is in Christ" but I do not think we need to make much of the 'must'.

title: Agape


text:
Ideas in Geometry/Instructive examples/Monty Hall Lesson

Monty Hall Problem

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car; behind the other two, goats. You pick on door, say No. 1. The host, who knows what lies behind each of the doors, opens another door, say No. 3, which has a goat. He then asks you, " Would like like to pick door No.2?" Is it in your best interest to switch or stay with the same door.

How Should we go about solving this problem?

- Let's first think about the possible door outcomes:

Right? So we choose door one (*). The host opens a door with a goat (o). So, based of of this, the probability must be 1/2 of the time or 50%. Because of this, it doesn't matter if you stay or switch.

WRONG

But why not? There are only two choices? - The critical thing to know is that the host does not have a choice in the door he opens. He must always open a door that has a goat behind it after the contestant has made their choice. - As long as the 3 choices are equally likely, there can never be a probability of 1/2 since the host's choice does not have an effect on the outcome.

'''So what is the best thing to do? Should we stay or should we switch?'''

- Let's start by thinking of all the possible outcomes behind the doors:

- To figure out the outcomes of staying or switching, let's assume that you always choose door one. The results would be as follows:

- Our outcome when we stay yields a car one out of three times or 1/3 of the time. If we were to switch doors, this would yield a car two out of three times or 2/3 or the time.


text:
User:Egm6321.f10.team6.lee/HW6

=Problem 1 Legendre's Equation =

Problem Statement
Legendre's Equation is, for n=0,1,2,...

when n=2,

Required
If we were given the 1st homogeneous solution,

Show that the 2nd homogeneous solution is

Solution
For the general homogeneous L2_ODE_VC,

let's put the solution like eqn(1.6) to find out the 2nd homogeneous solution.

Then,

Plug in to the eqn(1.5),

So, we can get L2_ODE_VC with missing dep. var. U. To reduce order, let

then,

integrating factor for this ODE would be,

multiply eqn(1.11) to eqn(1.10) on both side.

Therefore, from eqn(1.6),

So we can figure out what the 2nd homogeneous solution is if we have the 1st homogeneous solution.

Let's rewrite the given problem eqn(1.2).

From the problem statement, we have the 1st homogeneous solution, eqn(1.2).

Let's find the integrating factor first. from eqn(1.11),

Hence,

This is the same to the eqn(1.4).

Required
Solve the L2_ODE_VC when$$\begin{align} f(x)=0 \end{align}$$.

problem a
Using general trial solution,

hence,

Plug in eqn(2.4) to eqn(2.1). Then,

Then,

we can rewrite the eqn(2.5) in terms of x.

we can determine set of constants c and r from the eqn(2.6).

i.e.

problem b
using the same procedure to the problem a, let

hence,

Plug in eqn(2.4) to eqn(2.2). Then,

Then,

we can rewrite the eqn(2.9) in terms of x.

we can determine set of constants c and r from the eqn(2.10).

i.e.

title: User:Egm6321.f10.team6.lee/HW6


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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:35, 11 November 2010 (UTC)

title: User talk:Clubguppy


text:
User talk:Vanished user 565478

 Hello Osteveliam, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:35, 11 November 2010 (UTC)

title: User talk:Vanished user 565478


text:
User talk:Wangen

 Hello Wangen, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:35, 11 November 2010 (UTC)

Thx, Thank you, Danke, merci --Wangen 15:54, 13 November 2010 (UTC)

title: User talk:Wangen


text:
User talk:ViralStorm~enwikiversity

 Hello ViralStorm, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:36, 11 November 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called ViralStorm. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name ViralStorm~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:46, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:24, 19 April 2015 (UTC)

title: User talk:ViralStorm~enwikiversity


text:
User talk:Slapshot136

 Hello Slapshot136, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:36, 11 November 2010 (UTC)

title: User talk:Slapshot136


text:
User talk:Himacharya

 Hello Himacharya, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:36, 11 November 2010 (UTC)

title: User talk:Himacharya


text:
User talk:Ani~enwikiversity

 Hello Ani, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:36, 11 November 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Ani. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Ani~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:28, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:03, 19 April 2015 (UTC)

title: User talk:Ani~enwikiversity


text:
User talk:Haytham abulela

 Hello Haytham abulela, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:36, 11 November 2010 (UTC)

title: User talk:Haytham abulela


text:
User talk:Avellar

 Hello Avellar, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. -- –SJ + > 07:36, 11 November 2010 (UTC)

title: User talk:Avellar