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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.17)
 * }
 * }

Let


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

\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.19)
 * }
 * }

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


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

\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.22)
 * }
 * }

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


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

\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eqs. 1.28)
 * }
 * }


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

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


 * $$\displaystyle \left( a \right) \left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=f\left( x \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)$$
 * }
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)$$
 * }
 * }


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


 * $$\displaystyle \left( b \right) x{{y}^{''}}+2{{y}^{'}}+xy=f\left( x \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)$$
 * }
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)$$
 * }
 * }

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"


 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)$$
 * }
 * }
 * }


 * Consider a trial solution,


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


 * $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)$$
 * }
 * 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"


 * $$\Rightarrow {{x}^{c+1}}\left[ {{r}^{2}}-r \right]+{{x}^{c}}\left[ 2cr-{{r}^{2}}-c+1 \right]+{{x}^{c-1}}\left[ {{c}^{2}}-c-2cr \right]-{{x}^{c-2}}\left[ {{c}^{2}}-c \right]=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)$$
 * }
 * }
 * }


 * 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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.6)$$
 * }
 * }

.

(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"


 * $$\displaystyle x{{y}^{''}}+2{{y}^{'}}+xy=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.7)$$
 * }
 * }
 * }


 * 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"


 * $$\Rightarrow {{x}^{c+1}}\left( {{r}^{2}}+1 \right)+{{x}^{c}}\left( 2cr+2r \right)+{{x}^{c-1}}\left( {{c}^{2}}+c \right)=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.8)$$
 * }
 * }
 * }


 * 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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.9)$$
 * }
 * }

(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"


 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=\sin (x)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.10)$$
 * }
 * }
 * }


 * Above equation can be rewritten as:


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


 * $$\underbrace{1}_\cdot {{y}^{''}}\underbrace{-\frac{x}{\left( x-1 \right)}}_{{y}^{'}}+\underbrace{\frac{1}{\left( x-1 \right)}}_y=\underbrace{\frac{\sin (x)}{\left( x-1 \right)}}_{F\left( x \right)}$$
 * <p style="text-align: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"


 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.12)$$
 * }
 * }
 * }


 * From 1.1, one homogenous solution is,


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

$${{u}_{1}}\left( x \right)={{y}_}={{e}^{x}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.13)$$
 * }
 * }


 * $$\displaystyle (Eq. 2.12)$$ 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"

$$h\left( x \right)={{e}^{x-\log \left( x-1 \right)}}$$
 * <p style="text-align: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"

$${{y}_{P}}={{e}^{x}}\int\left( \int{\frac{\sin (x)}dx} \right)dx$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.15)$$
 * }
 * 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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.16)$$
 * }
 * }

(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"

$$x{{y}^{''}}+2{{y}^{'}}+xy=\sin \left( x \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.17)$$
 * }
 * }


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


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

$$\underbrace{1}_\cdot {{y}^{''}}+\underbrace{\frac{2}{x}}_{{y}^{'}}+\underbrace{y}_=\underbrace{\frac{\sin (x)}{x}}_{F\left( x \right)}$$
 * <p style="text-align: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"

$${{u}_{1}}\left( x \right)={{y}_}=\frac{x}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.19)$$
 * }
 * }


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


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


 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 * <p style="text-align: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"

$$h\left( x \right)={{e}^{2ix}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.21)$$
 * }
 * }


 * 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"



$${{y}_{P}}=\frac{8x}\left( i-2\left( 1+{{e}^{2ix}} \right)x \right)$$
 * <p style="text-align: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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.23)$$
 * }
 * }

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 $${y}_{H}=x $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.1) $$
 * }
 * }

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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.2) $$
 * }
 * }


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

$$ \frac{dy}{dx}=U \frac{d {u}_{1}}{dx}+\frac{d {U}}{dx}{u}_{1} $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.3) $$
 * }
 * }


 * {| 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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.4) $$
 * }
 * }
 * {| 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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.5) $$
 * }
 * }
 * {| 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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.6) $$
 * }
 * }


 * {| 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:
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.7) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ {Z}^{\prime}+ \left[ \frac{2 {{u}_{1}}^{\prime}}{{u}_{1}}- \frac{2x}{ \left(1- {x}^{2} \right)} \right]Z= \frac{1}{ {u}_{1} {\left(1- {x}^{2} \right)}^{2}} $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.8) $$
 * }
 * }

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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.9) $$
 * }
 * }


 * {| 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"

$$h\left( x \right)={{x}^{2}}\left( 1-{{x}^{2}} \right)$$
 * <p style="text-align: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
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.11) $$
 * }
 * }
 * {| 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"

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}-\frac{1}{2}\log \left( 1-{{x}^{2}} \right) \right]$$
 * <p style="text-align: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"

$$U\left( x \right)={{k}_{1}}+\int{\left( \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}-\frac{1}{2}\frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.13) $$
 * }
 * }

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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.14) $$
 * }
 * }

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"

$$x={{x}_{1}}=r\cos \left( \theta \right)\cos \left( \varphi  \right)$$
 * <p style="text-align:right;">$$\,(Eq. 4.1) $$
 * style= |
 * }


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

$$y={{x}_{2}}=r\cos \left( \theta \right)\sin \left( \varphi  \right)$$
 * <p style="text-align:right;">$$\,(Eq. 4.2) $$
 * style= |
 * }


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

$$z={{x}_{3}}=r\sin \left( \theta \right)$$
 * <p style="text-align: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"

$$d{{s}^{2}}=\sum\limits_{i=1}^{3}$$
 * <p style="text-align:right;">$$\,(Eq. 4.4) $$
 * 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"



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


 * <p style="text-align:right;">$$\,(Eq. 4.5) $$
 * style= |
 * }


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



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


 * <p style="text-align:right;">$$\,(Eq. 4.6) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 4.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

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


 * <p style="text-align:right;">$$\,(Eq. 4.8) $$
 * style= |
 * }


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

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


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

$$d{{x}_{3}}^{2}={{\sin }^{2}}\left( \theta \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\theta }^{2}}+2r\cos \left( \theta  \right)\sin \left( \theta  \right)drd\theta $$
 * <p style="text-align:right;">$$\,(Eq. 4.10) $$
 * 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"



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


 * <p style="text-align:right;">$$\,(Eq. 4.11) $$
 * style= |
 * }


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

$$drd\varphi \to \left[ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)+2r{{\cos }^{2}}\left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right) \right]=0$$
 * <p style="text-align:right;">$$\,(Eq. 4.12) $$
 * style= |
 * }


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



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


 * <p style="text-align:right;">$$\,(Eq. 4.13) $$
 * style= |
 * }


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

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

$$d{{\theta }^{2}}\to \left[ {{r}^{2}}\left( 1-{{\sin }^{2}}\left( \varphi \right) \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}\left( 1-{{\sin }^{2}}\left( \theta  \right) \right) \right]d{{\theta }^{2}}={{r}^{2}}d{{\theta }^{2}}$$
 * <p style="text-align:right;">$$\,(Eq. 4.15) $$
 * style= |
 * }


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

$$d{{\varphi }^{2}}\to \left[ {{r}^{2}}{{\cos }^{2}}\left( \theta \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right) \right]d{{\varphi }^{2}}={{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\varphi }^{2}}$$
 * <p style="text-align:right;">$$\,(Eq. 4.16) $$
 * 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" |
 * <p style="text-align:right;">$$\,(Eq. 4.17) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 4.18) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 4.19) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 4.20) $$
 * 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
 * <p style="text-align:right;">$$\,(Eq. 4.21) $$
 * 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:
 * <p style="text-align:right;">$$\,(Eq. 4.22) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & i=1\to \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right] \\ & i=2\to \frac{\partial }{\partial {{\zeta }_{2}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}}\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]=\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right] \\ & i=3\to \frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right]=\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \\ \end{align}$$
 * <p style="text-align:right;">$$\,(Eq. 4.23) $$
 * 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" |
 * <p style="text-align:right;">$$\,(Eq. 4.24) $$
 * 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" |
 * <p style="text-align:right;">$$\,(Eq. 4.25) $$
 * 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
 * <p style="text-align:right;">$$\displaystyle (Eq 5.1)
 * style= |
 * }


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

{x}_{2}=y= { \xi}_{1} \sin { \xi}_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq 5.2)
 * style= |
 * }


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

{x}_{3}=z= { \xi}_{3} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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} $$
 * <p style="text-align:right;">
 * style= |
 * }


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

$$\displaystyle {dx}_{1} =  \cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\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} $$
 * <p style="text-align:right;">
 * style= |
 * }


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

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

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


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

$$\displaystyle {dx}_{3} =  {d\xi}_{3} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\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} $$
 * <p style="text-align:right;">
 * 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} $$ $$
 * <p style="text-align:right;">$$\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} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 5.9)
 * style= |
 * }


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

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

Now by inspection we see that


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

$$\displaystyle {h}_{1}= 1 ; {h}_{2} =  {\xi}_{1} ; {h}_{3}= 1 $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\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) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 5.12)
 * style= |
 * }

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


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



$$\Delta \Psi =\frac{1}\frac{\partial }{\partial {{\xi }_{1}}}\left( {{\xi }_{1}}\frac{\partial \Psi }{\partial {{\xi }_{1}}} \right)+\frac{1}{{{\xi }_{1}}^{2}}\frac{{{\partial }^{2}}\Psi }{\partial \xi _{2}^{2}}+\frac{{{\partial }^{2}}\Psi }{\partial \xi _{3}^{2}}$$ $$
 * <p style="text-align:right;">$$\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" |

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

$$
 * <p style="text-align:right;">$$\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)$$
 * <p style="text-align:right;">$$\displaystyle (Eq 5.15)
 * <p style="text-align:right;">$$\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$$
 * <p style="text-align:right;">$$\displaystyle (Eq 5.16)
 * <p style="text-align:right;">$$\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$$
 * <p style="text-align:right;">$$\displaystyle (Eq 5.17)
 * <p style="text-align:right;">$$\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"

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


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

\begin{align} \bar{\theta} = \frac {\pi}{2} - \theta\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

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

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

$$ \bar{\theta} = \frac {\pi}{2} -\theta $$
 * <p style="text-align:right;">$$\displaystyle (Eqt. 6.1) $$
 * }
 * }


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

$$ ds^{2} = dr^{2} + r^{2}\cdot cos^{2}\theta \cdot d\phi^{2} + r^{2} \cdot d\theta^{2} $$
 * <p style="text-align:right;">$$\displaystyle (Eqt. 6.2) $$
 * }
 * }

Put Eqt 6.1 into Eqt 6.2


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

$$ d \bar{\theta}= d(\frac {\pi}{2}) - d\theta = 0 - d\theta = - d\theta $$
 * <p style="text-align:right;">$$\displaystyle (Eqt. 6.3) $$
 * }
 * }


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

$$ cos \theta = sin \bar{\theta} $$
 * <p style="text-align:right;">$$\displaystyle (Eqt. 6.4) $$
 * }
 * }

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

Thus, we find h_1, h_2 & h_3


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

\begin{align} h_{1} = 1 \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


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

\begin{align} h_{2} = r \cdot sin(\bar{\theta}) \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


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

\begin{align} h_{3} = r \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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"

$$\displaystyle \left( 1-{{x}^{2}} \right){{y}^{''}}-2x{{y}^{'}}+n\left( n+1 \right)y=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.1)$$
 * }
 * }


 * Legendre polynomials:


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

$$\displaystyle{{P}_{0}}(x)=1$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.2)$$
 * }
 * }


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

$$\displaystyle{{P}_{1}}(x)=x$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.3)$$
 * }
 * }


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

$${{P}_{2}}(x)=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.4)$$
 * }
 * }


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

$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.5)$$
 * }
 * }


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


 * $${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.6)$$
 * }
 * }
 * }

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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.7)$$
 * }
 * }

(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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.8)$$
 * }
 * }

(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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.9)$$
 * }
 * }

(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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.10)$$
 * }
 * }

(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" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9.11)$$
 * }
 * }


 * 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)