User:Egm6321.f10.team2.oztekin/HW6

=Problem 6.1 Using variation of parameters method to get $$\displaystyle {{Q}_{2}}(x)$$ from $$\displaystyle {{P}_{2}}(x)$$ =

From the lecture slide Mtg 32- 3

Given

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

$$  \displaystyle P_{2}(x)=\frac{1}{2}(3x^{2}-1) $$     (Eq 1.1) Legendre's equation where n=2
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (1-x^{2})y''-2xy'+6y=0 $$     (Eq 1.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Find

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

$$  \displaystyle Q_{2}(x)=\frac{1}{4}(3x^{2}-1)log(\frac{1+x}{1-x})-\frac{3}{2}x $$     (Eq 1.3) using the variation of parameter method.
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Solution
To begin to solve Q_{2}(x) we must first solve the general form and then plug in.

We start with the equation:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=U(x)u_{1}(x) $$     (Eq 1.4) and take derivatives to get in the Legendre Equation form
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=U*u_{1} $$     (Eq 1.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y'=U'*u_{1}+U*u'_{1} $$     (Eq 1.6)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=U*u_{1}+U*u''_{1}+2U'u'_{1} $$     (Eq 1.7) Plugging Eqns. 1.5-1.7 into the general L2-ODE-VC we get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle U*u_{1}+U*u_{1}+2U'u'_{1}+a_{1}(x)*(U'*u_{1}+U*u'_{1})+a_{0}(x)*(U*u_{1})=0 $$     (Eq 1.8) Combining terms:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle U(u_{1}+a_{1}(x)u'_{1}+a_{0}(x)u_{1})+U'(2u'_{1}(x)a_{1}(x)+a_{1}(x)u_{1}(x))+Uu_{1}(x)=0 $$     (Eq 1.9) Noting that $$ \displaystyle u''_{1}+a_{1}(x)u'_{1}+a_{0}(x)u_{1} $$ is a homogeneous solution to the Legendre solution and therefore =0 we get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle U'(2u'_{1}(x)+a_{1}(x)u_{1}(x))+U''u_{1}(x)=0 $$     (Eq 1.10) To help solve, we set $$ \displaystyle Z=U' $$ so we can use the order reducing method.
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Z(2u'_{1}(x)+a_{1}(x)u_{1}(x))+Z'u_{1}(x)=0 $$     (Eq 1.11) We divide by $$ \displaystyle u_{1}(x) $$ to get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Z\frac{2u'_{1}(x)+a_{1}(x)u_{1}(x)}{u_{1}(x)}+Z'=0 $$     (Eq 1.12) We can now use the Integrating Factor Method
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=exp [\int (\frac{2u'_{1}(x)+a_{1}(x)u_{1}(x)}{u_{1}(x)})dx] $$     (Eq 1.13) Multiplying through by u_{1}(x):
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=exp [\int (a_1(x)+2\frac{u'_1(x)}{u_1(x)})dx] $$     (Eq 1.14) Separating:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=exp [\int a_1(x)dx+\int 2\frac{u'_1(x)}{u_1(x)}dx] $$     (Eq 1.15) We then can solve to get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=u_{1}^{2}(x)exp[\int a_1(x)dx] $$     (Eq 1.16) Now that we have the integrating factor, we need to set up Legendre's Equation (Eq 1.2) so that we can use the integrating factor. This is done below by dividing through by $$ \displaystyle (1-x^2) $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y''-\frac{2x}{1-x^2}y'+\frac{6}{1-x^2}y=0 $$     (Eq 1.17) From this we can see that:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_1(x)=\frac{-2x}{1-x^2} $$     (Eq 1.18) Combining this with the fact that $$ \displaystyle P_2(x)=\frac{1}{2}(3x^2-1) $$ and $$ \displaystyle Q_2(x)=u_1(x)\int \frac{1}{h(x)}dx $$ from above we can plug in out knowns and solve:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Q_2(x)=u_1(x)\int \frac{1}{h(x)}dx $$     (Eq 1.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{1}{u_{1}^{2}(x)exp[\int a_1(x)dx]}dx $$     (Eq 1.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{1}{(\frac{1}{2}(3x^2-1))^{2}exp[\int \frac{-2x}{1-x^2}dx]}dx $$     (Eq 1.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle exp\int \frac{-2x}{1-x^{2}}dx=exp[log(1-x^{2})]=1-x^{2} $$     (Eq 1.22) If we sustitute Eq(1.22) into Eq(1.21) we have
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{4}{(3x^2-1)^{2}(1-x^{2})}dx $$     (Eq 1.23) Let's look at the fraction:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{(3x^{2}-1)^{2}(1-x^{2})}=\frac{Ax+B}{(3x^{2}-1)}+\frac{Cx+D}{(3x^{2}-1)^{2}}+\frac{E}{(1+x)}+\frac{F}{(1-x)} $$     (Eq 1.24) Having compansated denominators we have;
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (Ax+B)(3x^{2}-1)(1-x^{2})+(Cx+D)(1-x^{2})+E(1-x)(3x^{2}-1)^{2}+F(3x^{2}-1)^{2}(1+x)=1 $$     (Eq 1.25) After unpacking the equation we will get five equations for x's orders. Then we can see the coeefficients of the terms as:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle A=0,B=\frac{3}{4},C=0,D=\frac{3}{2},E=\frac{1}{8},F=\frac{1}{8} $$     (Eq 1.26)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{(3x^{2}-1)^{2}(1-x^{2})}=\frac{3}{4}\frac{1}{(3x^2-1)}+\frac{3}{2}\frac{1}{(3x^{2}-1)}+\frac{1}{8}\frac{1}{(x+1)}-\frac{1}{8}\frac{1}{(x-1)} $$     (Eq 1.27)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \int \frac{1}{(3x^{2}-1)(1-x^{2})}=\frac{1}{8}\left [ \frac{-6x}{(3x^{2}-1)} -log(1-x)+log(1+x)\right ] $$     (Eq 1.28) If we sustitute Eq(1.28) into Eq(1.23) we get the second Q
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Q_{2}(x)=\frac{1}{4}(3x^{2}-1)log(\frac{1+x}{1-x})-\frac{3}{2}x $$     (Eq 1.29)
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

= Problem 6.2 Solving Non-homogeneous L2-ODE-VC with different excitation functions =

From the lecture slide From Mtg 32-3 and Mtg 33-1

Given

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

$$ \displaystyle

(x-1)y''-xy'+y=f(x)$$

(Eq 2.1)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

xy''+2y'+xy=f(x)$$

(Eq 2.2)
 * <p style="text-align:right">
 * }

Find
Solve the equations above for $$\displaystyle f(x)=0$$ and $$\displaystyle f(x)=\sin x$$; so


 * a)	Eq 2.1 with $$\displaystyle f(x)=0$$


 * b)	Eq 2.1 with $$\displaystyle f(x)=\sin x$$


 * c)	Eq 2.2 with $$\displaystyle f(x)=0$$


 * d)	Eq 2.2 with $$\displaystyle f(x)=\sin x$$

Solution
First of all applying the generic form of $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$ and then taking the first and the second derivatives will result in the following:


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

$$ \displaystyle

\begin{align} & y'=c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}} \\ & y''=c(c-1){{x}^{c-2}}{{e}^{rx}}+c{{x}^{c-1}}r{{e}^{rx}}+c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}rr{{e}^{rx}} \\ & y''=c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}} \\ \end{align}$$

(Eqs 2.3)
 * <p style="text-align:right">
 * }

The procedure above will be the same for both part(a) and part(c) parts of the problem.

Part a
Now that we found the derivatives, the solutions can be plugged into the Eq 2.1 that is set $$\displaystyle f(x)=0$$:


 * $$\displaystyle (x-1)y''-xy'+y=0$$


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

$$ \displaystyle

(x-1)*\underbrace{[c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}}]}_{y''}-x*\underbrace{[c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}}]}_{y'}+\underbrace{[{{x}^{c}}{{e}^{rx}}]}_{y}=0$$

(Eq 2.4)
 * <p style="text-align:right">
 * }

Since every term in the equation has $$\displaystyle {{e}^{rx}}$$, it can be taken out of the parenthesis to be eliminated:


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

$$ \displaystyle

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

(Eq 2.5)
 * <p style="text-align:right">
 * }

To simplify the equation we will go through the following procedure:


 * $$\displaystyle \begin{align}

& (x-1)[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]-x[c{{x}^{c-1}}+{{x}^{c}}r]+[{{x}^{c}}]=0 \\ & ({{c}^{2}}-c){{x}^{c-1}}+2c{{x}^{c}}r+{{x}^{c+1}}{{r}^{2}}-({{c}^{2}}-c){{x}^{c-2}}-2c{{x}^{c-1}}r-{{x}^{c}}{{r}^{2}}-c{{x}^{c}}-{{x}^{c+1}}r+{{x}^{c}}=0 \\ & {{x}^{c+1}}({{r}^{2}}-r)+{{x}^{c}}(2cr-{{r}^{2}}-c+1)+{{x}^{c-1}}({{c}^{2}}-c-2cr)+{{x}^{c-2}}(c-{{c}^{2}})=0 \\ \end{align}$$

When $$\displaystyle c=1\Rightarrow r=0$$ or $$\displaystyle c=0\Rightarrow r=1$$, which will yield us two homogenous solutions plugging into $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$.


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

$$ \displaystyle

\begin{align} & {{y}_{H1}}={{x}^{0}}{{e}^{1x}}={{e}^{x}} \\ & {{y}_{H2}}={{x}^{1}}{{e}^{0x}}=x \\ \end{align}$$

(Eqs 2.6)
 * <p style="text-align:right">
 * }

Then,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

y={{C}_{1}}{{e}^{x}}+{{C}_{2}}x$$

(Eq 2.7)
 * <p style="text-align:right">
 * }

Part b
When we set $$\displaystyle f(x)=\sin x$$ for Eq 2.1, we get:


 * $$\displaystyle (x-1)y''-xy'+y=\sin x$$

We will rewrite the equation to put in the proper form and find the integrating factor:


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

$$ \displaystyle

y''-\frac{x}{(x-1)}y'+\frac{1}{(x-1)}y=\frac{\sin x}{(x-1)}$$

(Eq 2.8) Plugging in all the values into the integrating factor formula, we get:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

h(x)=\underbrace_{y_{H1}^{2}}{{e}^{-\int{\frac{x}{(x-1)}dx}}}={{e}^{2x}}{{e}^{-\ln (x-1)-x}}={{e}^{x-\ln (x-1)}}={{e}^{x}}{{e}^{\ln \frac{1}{(x-1)}}}=\frac{(x-1)}$$

(Eq 2.9) Now to find the particular solution we will apply the following formula:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

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

(Eq 2.10) Finally combining homogenous solution with the particular solution we get:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

y={{C}_{1}}{{e}^{x}}+{{C}_{2}}x+{{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}dx} \right]dx$$

(Eq 2.11)
 * <p style="text-align:right">
 * }

Part c

 * $$\displaystyle xy''+2y'+xy=0$$


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

$$ \displaystyle

x*\underbrace{[c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}}]}_{y''}+2*\underbrace{[c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}}]}_{y'}+x*\underbrace{[{{x}^{c}}{{e}^{rx}}]}_{y}=0$$

(Eq 2.12) Just like the way it was done in part a, the $$\displaystyle {{e}^{rx}}$$ term will be taken out and the equation will be simplified:
 * <p style="text-align:right">
 * }
 * $$\begin{align}

& x*[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]+2*[c{{x}^{c-1}}+{{x}^{c}}r]+x*[{{x}^{c}}]=0 \\ & ({{c}^{2}}-c){{x}^{c-1}}+2cr{{x}^{c}}+{{r}^{2}}{{x}^{c+1}}+2c{{x}^{c-1}}+2{{x}^{c}}r+{{x}^{c+1}}=0 \\ & {{x}^{c+1}}({{r}^{2}}+1)+{{x}^{c}}(2cr+2r)+{{x}^{c-1}}({{c}^{2}}+c)=0 \\ \end{align}$$

By solving the above equation we get $$\displaystyle c=-1\Rightarrow r=\pm i$$

Plugging these values into $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$, will give us the following equations:


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

$$ \displaystyle

\begin{align} & {{y}_{H1}}={{x}^{-1}}{{e}^{ix}}=\frac{x} \\ & {{y}_{H2}}={{x}^{-1}}{{e}^{-ix}}=\frac{1}{x{{e}^{ix}}} \\ \end{align}$$

(Eqs 2.13)
 * <p style="text-align:right">
 * }

Then,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}$$

(Eq 2.14)
 * <p style="text-align:right">
 * }

Part d
At last but not least Eq 2.2 will be solved with $$\displaystyle f(x)=\sin x$$.


 * $$\displaystyle xy''+2y'+xy=\sin x$$

Rearranging the equation to put in the proper form and find the integrating factor:


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

$$ \displaystyle

y''+\frac{2}{x}y'+y=\frac{\sin x}{x}$$

(Eq 2.15) Plugging in all the values into the integrating factor formula, we get:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

h(x)=\frac{{e}^{\int{\frac{2}{x}dx}}}=\frac{{e}^{\ln {{x}^{2}}}}=\frac{{x}^{2}}={{e}^{2ix}}$$

(Eq 2.16) Now to find the particular solution we will apply the following formula:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

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

(Eq 2.17)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{y}_{P}}(x)=\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.18) Finally combining homogenous solution with the particular solution we get:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}+\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.19)
 * <p style="text-align:right">
 * }

= Problem 6.3 Solving Non-homogenous Legendre Equation using direct menthod =

From the lecture slide Mtg 33-1

Given

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

Find
Solve equation using direct method.

Solution
The two homogeneous were given in class lecture 30-2 as the following:


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

$$ \displaystyle

\begin{align} & {{u}_{1}}(x)=x \\ & {{u}_{2}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \\ \end{align}$$

(Eqs 3.1) $$\displaystyle {{y}_{H}}$$ can be expressed as
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle

{{y}_{H}}={{C}_{1}}x+{{C}_{2}}\left[ \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right]$$

(Eq 3.2)
 * <p style="text-align:right">
 * }

Using the example from the King's book on page 34 we can use the variation of parameters formula that is


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

$$ \displaystyle

y(x)=\int_ – ^{x}{f(s)\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{W(s)} \right\}}ds$$

(Eq 3.3)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{y}_{P}}=\int_ – ^{x}{\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{(1-{{s}^{2}})[{{u}_{1}}(s)u_{2}^{'}(s)-u_{1}^{'}(s)u_{2}^ – (s)} \right\}}ds$$

(Eq 3.4)
 * <p style="text-align:right">
 * }


 * $$\displaystyle W=\frac{1-{{s}^{2}}}$$


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

$$ \displaystyle

\begin{align} {{y}_{P}}(x) & ={{u}_{2}}(x)\int{{{u}_{1}}(x)dx-}{{u}_{1}}(x)\int{{{u}_{2}}(x)dx} \\ & =\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)\int{xdx-}x\int{\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)dx} \\ & =\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)\frac{2}-\left( \frac{{{x}^{3}}-x}{4}\log \left( \frac{1+x}{1-x} \right)-\frac{2} \right) \\ & =\frac{x}{4}\log \left( \frac{1+x}{1-x} \right) \\ \end{align}$$

(Eq 3.5) So we can express the general solution as the following:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

y={{C}_{1}}x+{{C}_{2}}\left[ \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right]+\frac{x}{4}\log \left( \frac{1+x}{1-x} \right)$$

(Eq 3.6)
 * <p style="text-align:right">
 * }

=Problem 6.4: Laplacian in Spherical Coordinate System=

From the lecture slide Mtg 34-1

Given
First we have to indentify our cartesian coorinates into spherical coordinates. Then we can aidentify line segment in spherical coordinate system.Our new coordinates ;
 * {| style="width:100%" border="0"

$$  \displaystyle x_{1}=rcos\theta cos\varphi =\xi _{1}cos\xi _{2}cos\xi _{3} $$     (Eq 4.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle x_{2}=rcos\theta sin\varphi $$     (Eq 4.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle x_{3}=rsin\theta $$     (Eq 4.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle ds^{2}=\vec{ds}.\vec{ds}=(dx_{i}dx_{j})(e_{i}e_{j}) $$     (Eq 4.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \delta _{ij}=\left\{\begin{matrix} 1 &for &i=j \\ 0 &for & i\neq j \end{matrix}\right. $$     (Eq 4.5) In astronomical convention angles are defined as
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & -\frac{\pi }{2}\leq \theta \leq \frac{\pi }{2}\\ & 0\leq \varphi \leq 2\pi \\ \end{align} $$     (Eq 4.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Obtain the steps given in meeting 34

Solution

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

$$  \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} $$     (Eq 4.7) Finding total derivative of first term
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{1}=\frac{\partial x_{1}}{\partial r}dr+\frac{\partial x_{1}}{\partial \theta }d\theta +\frac{\partial x_{1}}{\partial \varphi }d\varphi $$     (Eq 4.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{1}}{\partial r}=cos\theta cos\varphi $$     (Eq 4.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial x_{1}}{\partial \theta }=-rsin\theta cos\theta $$     (Eq 4.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial x_{1}}{\partial \varphi }=-rcos\theta sin\varphi $$     (Eq 4.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{1} =cos\theta cos\varphi dr-rsin\theta cos\varphi d\theta -rcos\theta sin\varphi d\varphi $$     (Eq 4.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{1}^{2}=cos^{2}\theta cos^{2}\varphi dr^{2}+r^{2}sin^{2}\theta cos^{2}\varphi d\theta ^{2}+r^{2}cos^{2}\theta sin^{2}\varphi d\varphi ^{2}+2r^{2}sin\varphi cos\varphi sin\theta cos\theta d\theta d\varphi -2rcos\theta cos^{2}\varphi sin\theta drd\theta -2rcos^{2}\theta cos\varphi sin\varphi drd\varphi $$     (Eq 4.13) Finding total dervative of second term
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{2}=\frac{\partial x_{2}}{\partial r}dr+\frac{\partial x_{2}}{\partial \theta }d\theta +\frac{\partial x_{2}}{\partial \varphi }d\varphi $$     (Eq 4.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{2}}{\partial r}=cos\theta sin\varphi $$     (Eq4.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{2}}{\partial \theta }=-rsin\theta sin\varphi $$     (Eq 4.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{2}}{\partial \varphi }=rcos\theta cos\varphi $$     (Eq 4.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{2}=cos\theta sin\varphi dr-rsin\theta sin\varphi d\theta +rcos\theta cos\varphi d\varphi $$     (Eq 4.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{2}^{2}=cos^{2}\theta sin^{2}\varphi dr^{2}+r^{2}sin^{2}\theta sin^{2}\varphi d\theta^{2} +r^{2}cos^{2}\theta cos^{2}\varphi d\varphi ^{2}+2rcos^{2}\theta sin\varphi cos\varphi drd\varphi -2r^{2}sin\theta sin\varphi cos\theta cos\varphi d\theta d\varphi -2rcos\theta sin\theta sin^{2}\varphi drd\theta $$     (Eq 4.19) Finding total derivative of third term
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{3}=\frac{\partial x_{3}}{\partial r}dr+\frac{\partial x_{3}}{\partial \theta }d\theta +\frac{\partial x_{3}}{\partial \varphi }d\varphi $$     (Eq 4.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{3}}{\partial r}=sin\theta $$     (Eq 4.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{3}}{\partial \theta }=rcos\theta $$     (Eq 4.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \frac{\partial x_{3}}{\partial \varphi \theta }=0 $$     (Eq 4.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{3}=sin\theta dr+rcos\theta d\theta $$     (Eq 4.24)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{3}^{2}=sin^{2}\theta dr^{2}+r^{2}cos^{2}\theta d\theta ^{2}+2rsin\theta cos\theta drd\theta $$     (Eq 4.25)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Collecting all of the differentials Eq 4.13-Eq4.19-Eq4.25 and substitute into ds yields ;


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

$$  \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} $$     (Eq 4.26)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \begin{align} &ds^{2}=cos^{2}\theta cos^{2}\varphi dr^{2}+r^{2}sin^{2}\theta cos^{2}\varphi d\theta ^{2}+r^{2}cos^{2}\theta sin^{2}\varphi d\varphi ^{2}+cos^{2}\theta sin^{2}\varphi dr^{2}+r^{2}sin^{2}\theta sin^{2}\varphi d\theta^{2} +r^{2}cos^{2}\theta cos^{2}\varphi d\varphi ^{2}\\ &+sin^{2}\theta dr^{2}+r^{2}cos^{2}\theta d\theta ^{2}+2r^{2}sin\varphi cos\varphi sin\theta cos\theta d\theta d\varphi -2rcos\theta cos^{2}\varphi sin\theta drd\theta -2rcos^{2}\theta cos\varphi sin\varphi drd\varphi\\ &+2rcos^{2}\theta sin\varphi cos\varphi drd\varphi -2r^{2}sin\theta sin\varphi cos\theta cos\varphi d\theta d\varphi -2rcos\theta sin\theta sin^{2}\varphi drd\theta +2rsin\theta cos\theta drd\theta\\ \end{align} $$     (Eq 4.27)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can collect$$ \displaystyle dr,d\theta,d\varphi$$ together. And we can see that some terms cancel each other and for other terms we hate to manupilate the equation little more.


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

$$  \displaystyle \begin{align} &ds^{2}=cos^{2}\theta (cos^{2}\varphi +sin^{2}\varphi )dr^{2}+sin^{2}\theta dr^{2}+r^{2}sin^{2}\theta (cos^{2}\varphi +sin^{2}\varphi )d\theta ^{2}+r^{2}cos^{2}\theta d\theta ^{2}+r^{2}cos^{2}\theta (sin^{2}\varphi +cos^{2}\varphi )d\varphi ^{2}\\ &+2rsin\theta cos\theta drd\theta -2rcos\theta sin\theta (cos^{2}\varphi +sin^{2}\varphi )drd\theta \\ \end{align} $$     (Eq 4.28) We can obtain the equation given in the lecture 34-1
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle ds^{2}=dr^{2}+r^{2}d\theta ^{2}+r^{2}cos^{2}\theta d\phi ^{2} $$     (Eq 4.29)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Delta \psi =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial \psi }{\partial \xi _{i}}\right ] $$     (Eq 4.30)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle h_{1}.h_{2}.h_{3}=r^{2}cos\theta $$     (Eq 4.31)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle i=1,\xi _{1}=r\Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial r}\left [ \frac{r^{2}cos\theta }{1}\frac{\partial \psi }{\partial r} \right ] $$     (Eq 4.32)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since $$\displaystyle cos\theta $$is constant for r we can take it out from the derivative.
 * {| style="width:100%" border="0"

$$  \displaystyle i=1,\xi _{1}=r\Rightarrow \frac{1}{r^{2} }\frac{\partial }{\partial r}\left [ \frac{r^{2} }{1}\frac{\partial \psi }{\partial r} \right ] $$     (Eq 4.33)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle i=2,\xi _{2}=\theta \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos\theta }{r^{2}}\frac{\partial \psi }{\partial \theta } \right ] $$     (Eq 4.34)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle i=2,\xi _{2}=\theta \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left [ \frac{cos\theta }{1}\frac{\partial \psi }{\partial \theta } \right ] $$     (Eq 4.35)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle i=3,\xi _{3}=\varphi \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \varphi }\left [ \frac{r^{2}cos\theta }{r^{2}cos^{2}\theta }\frac{\partial \varphi }{\partial\varphi } \right ] $$     (Eq 4.36)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since $$\displaystyle cos\theta $$ is constant for \varphi we can take it out from the derivative.


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

$$  \displaystyle i=3,\xi _{3}=\varphi \Rightarrow \frac{1}{r^{2}cos^{2}\theta }\frac{\partial }{\partial \varphi }\left [ \frac{\partial^{2}\psi }{\partial\varphi ^{2} } \right ] $$     (Eq 4.37)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then we can collect all terms into the $$\displaystyle \Delta \psi $$ equation:


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


 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

$$  \displaystyle \Delta \psi =\frac{1}{r^{2}}(r^{2}\frac{\partial \psi }{\partial r})+\frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }(cos\theta \frac{\partial \psi }{\partial \theta })+\frac{1}{r^{2}cos^{2}\theta }\frac{\partial ^2\psi }{\partial \varphi ^2} $$     (Eq 4.38)
 * <p style="text-align:right">
 * }

=Problem 6.5: Laplacian in Cylindrical Coordinate System=

From the lecture slide Mtg 35-3 and Mtg 36-3

Given
Three dimensional cylindrical coordinates were given in the meeting 35-3 as follows:
 * {| style="width:100%" border="0"

$$  \displaystyle x_{1}=rcos\theta =\xi _{1}cos\xi _{2} $$     (Eq 5.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle x_{2}=rsin\theta =\xi _{1}sin\xi _{2} $$     (Eq 5.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle x_{3}=z=\xi _{3} $$     (Eq 5.3) Now,after defining our new curvilinear coordinates,we can define our cartesian coordinates in terms of our new independent variables.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & x_{1}=x_{1}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ &x_{2}=x_{2}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ &x_{3}=x_{3}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ \end{align}$$ (Eq 5.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Repeat steps for cylindrical coordinates and conclude with Bessel's Eq.

Finding dxi's

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

$$  \displaystyle dx_{1}=\frac{\partial x_{1}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{1}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{1}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle\begin{align} &\frac{\partial x_{1}}{\partial \xi _{1}}=cos\xi _{2}\\ &\frac{\partial x_{1}}{\partial \xi _{2}}=-\xi _{1}sin\xi _{2}\\ &\frac{\partial x_{1}}{\partial \xi _{3}}=0\\ \end{align}$$ (Eq 5.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{1}=cos\xi _{2}d\xi _{1}-\xi _{1}sin\xi _{2}d\xi _{2} $$     (Eq 5.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{2}=\frac{\partial x_{2}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{2}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{2}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle\begin{align} &\frac{\partial x_{2}}{\partial \xi _{1}}=sin\xi _{2}\\ &\frac{\partial x_{2}}{\partial \xi _{2}}=\xi _{1}cos\xi _{2}\\ &\frac{\partial x_{2}}{\partial \xi _{3}}=0\\ \end{align}$$ (Eq 5.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{2}=sin\xi _{2}d\xi _{1}+\xi _{1}cos\xi _{2}d\xi _{2} $$     (Eq 5.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle dx_{3}=\frac{\partial x_{3}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{3}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{3}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} &\frac{\partial x_{3}}{\partial \xi _{1}}=0\\ &\frac{\partial x_{3}}{\partial \xi _{2}}=0\\ &\frac{\partial x_{3}}{\partial \xi _{3}}=1\\ \end{align}$$ (Eq 5.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Finding ds^2

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

$$  \displaystyle dx_{1}^{2}=cos^{2}\xi _{2}d\xi _{1}^{2}+\xi _{1}^{2}sin^{2}\xi _{2}d\xi _{2}^{2}-2\xi _{1}cos\xi _{2}sin\xi _{2}d\xi _{1}d\xi _{2} $$     (Eq 5.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{2}^{2}=sin^{2}\xi _{2}d\xi _{1}^{2}+\xi _{1}^{2}cos^{2}\xi _{2}d\xi _{2}^{2}+2\xi _{1}cos\xi _{2}sin\xi _{2}d\xi _{1}d\xi _{2} $$     (Eq 5.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle dx_{3}^{2}=d\xi _{3}^{2} $$     (Eq 5.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle ds^{2}=d\xi _{1}^{2}+\xi _{1}^{2}(sin^{2}\xi _{2}+cos^{2}\xi _{2})d\xi _{2}^{2}+d\xi _{3}^{2}=d\xi _{1}^{2}+\xi _{1}^{2}d\xi _{2}^{2}+d\xi _{3}^{2} $$     (Eq 5.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & h_{1}^{2}=1^{2}\Rightarrow h_{1}=1\\ & h_{2}^{2}=\xi _{1}^{2}\Rightarrow h_{2}=\xi _{1}\\ & h_{3}^{2}=1^{2}\Rightarrow h_{3}=1\\ \end{align}$$ (Eq 5.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h_{1}h_{2}h_{3}=\xi _{1} $$     (Eq 5.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Delta \psi =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial \psi }{\partial \xi _{i}}\right ] $$     (Eq 5.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle i=1\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{1}} \right ) $$     (Eq 5.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle i=2\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{2}}\left ( \frac{\xi _{1}}{\xi _{1}^{2}}\frac{\partial \psi }{\partial \xi _{2}} \right )\Rightarrow \frac{1}{\xi _{1}^{2}}\frac{\partial ^2\psi }{\partial \xi _{2}} $$     (Eq 5.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle i=3\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{3}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{3}} \right )\Rightarrow \frac{\partial^2 \psi }{\partial \xi _{3}^2} $$     (Eq 5.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Delta \psi =\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{1}} \right )+\frac{1}{\xi _{1}^{2}}\frac{\partial^2 \psi }{\partial \xi _{2}^2}+\frac{\partial^2 \psi }{\partial \xi _{3}^2}=0 $$     (Eq 5.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Obtaining Bessel's Eq
To obtain bessel equation we have to use separation of variables.We can identify $$\displaystyle \psi $$ as multiplacation of three functions depending their own variables.
 * {| style="width:100%" border="0"

$$  \displaystyle \psi (\xi _{1},\xi _{2},\xi _{3})=X(\xi _{1})Y(\xi _{2})Z(\xi _{3}) $$     (Eq 5.24)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{Y(\xi _{2})Z(\xi _{3})}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X(\xi _{1})}{\partial \xi _{1}} \right )+\frac{X(\xi _{1})Z(\xi _{3})}{\xi _{1}^{2}}\frac{\partial^2Y(\xi _{2})}{\partial^2 \xi _{2}}+X(\xi _{1})Y(\xi _{2})\frac{\partial^2 Z(\xi _{3})}{\partial \xi _{3}^2}=0 $$     (Eq 5.25) Now, dividing through $$ \displaystyle XYZ$$ we can obtain more useful form.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{X\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+\frac{1}{Y\xi _{1}^{2}}\frac{\partial^2 Y}{\partial \xi _{2}^2}+\frac{1}{Z}\frac{\partial^2 Z}{\partial \xi _{3}^2}=0 $$     (Eq 5.26) Assuming that
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial^2 Z}{\partial \xi _{3}^2}=\alpha ^{2}Z $$     (Eq 5.27) After that our equation forms as
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{1}{X\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+\frac{1}{Y\xi _{1}^{2}}\frac{\partial^2 Y}{\partial \xi _{2}^2}+\alpha ^{2}=0 $$     (Eq 5.28) We can see that after our asssumption $$\displaystyle\triangledown ^{2}\psi $$ is no longer three dimensional instead two dimensional.The Eq(5.28) is more closely to Helmhotz's equation.Multiply whole eq by $$\displaystyle \xi _{1}^{2}$$.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

In addition let's assume
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial^2 Y}{\partial \xi _{2}^2}=-n^{2}Y $$     (Eq 5.29) Which has solutions $$\displaystyle Y=exp(+,-in\xi _{2})$$ where n is an integer. It is now closely related to Bessel's equation.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \xi _{1}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.30)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \xi _{1}^{2}\frac{\partial^2 X}{\partial \xi_{1} ^2}+\xi _{1}\frac{\partial X}{\partial \xi _{1}}+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.31) Say that
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle x=\alpha \xi _{1} $$     (Eq 5.32)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial X}{\partial \xi _{1}}=\frac{\partial X}{\partial x}\frac{\partial x}{\partial \xi _{1}}=\alpha \frac{\partial X}{\partial x} $$ (Eq 5.33)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial }{\partial \xi _{1}}\left ( \alpha \frac{\partial X}{\partial x} \right )=\alpha ^{2}\frac{\partial X}{\partial x} $$ (Eq 5.34) Substituting Eq(5.33) and Eq (5.34) in to Eq (Eq 5.32) We observe
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \xi _{1}^{2}\alpha ^{2}\frac{\partial^2 X}{\partial x^2}+\xi _{1}\alpha \frac{\partial X}{\partial x}+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.35)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"


 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

$$  \displaystyle x^{2}\frac{\partial^2 X}{\partial x^2}+x \frac{\partial X}{\partial x}+(x^{2}-n^{2})X=0 $$     (Eq 5.36)
 * <p style="text-align:right">
 * }

= Problem 6.6 Finding Laplacian in spherical coordinates using Math/Physics convention =

From the lecture slide Mtg 35-4

Given
The Laplacian in spherical coordinates using astronomy convention is ,


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

$$ \displaystyle

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

(Eq 6.1)
 * <p style="text-align:right">
 * }

where $$\displaystyle \theta \in [-\pi /2,+\pi /2]$$

Find
Laplacian in spherical coordinates using Math/Physics convention: $$\displaystyle r,\bar{\theta },\varphi $$ where,


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

$$ \displaystyle

\bar{\theta }=\frac{\pi }{2}-\theta $$

(Eq 6.2)
 * <p style="text-align:right">
 * }

Solution
Sustitute Eq 6.2 into Eq 6.1 we get,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\begin{align} \Delta \Psi & =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\cos }^{2}}(\pi /2-\bar{\theta })}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\cos (\pi /2-\bar{\theta })}\frac{\partial }{\partial (\pi /2-\bar{\theta })}\left( \cos (\pi /2-\bar{\theta })\frac{\partial \Psi }{\partial (\pi /2-\bar{\theta })} \right) \\ & =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\sin }^{2}}\bar{\theta }}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\sin \bar{\theta }}\frac{\partial }{\partial \bar{\theta }}\left( \sin \bar{\theta }\frac{\partial \Psi }{\partial \bar{\theta }} \right) \\ \end{align}$$

(Eq6.3)
 * <p style="text-align:right">
 * }

If we don't know Laplacian in astronomy convention at first
Then the transmation between Cartesian coodinates and spherical coordinates(Math/Physics convention) is,


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

$$ \displaystyle

\begin{align} & x=r\sin \bar{\theta }\cos \varphi \\ & y=r\cos \bar{\theta }\sin \varphi \\ & z=r\cos \bar{\theta } \\ \end{align}$$

(Eqs 6.4)
 * <p style="text-align:right">
 * }

Hence,


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

$$ \displaystyle

\begin{align} & dx=\sin \bar{\theta }\cos \varphi dr+r\cos \bar{\theta }\cos \varphi d\bar{\theta }-r\sin \bar{\theta }\sin \varphi d\varphi \\ & dy=\sin \bar{\theta }\sin \varphi dr+r\cos \bar{\theta }\sin \varphi d\bar{\theta }+r\sin \bar{\theta }\cos \varphi d\varphi \\ & dz=\cos \bar{\theta }dr-r\sin \bar{\theta }d\bar{\theta } \\ \end{align}$$

(Eqs 6.5)
 * <p style="text-align:right">
 * }

Consider the magnitude of an infinitesimal segment $$\displaystyle ds$$,


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

$$ \displaystyle

d{{s}^{2}}=d{{r}^{2}}+{{r}^{2}}d{^{2}}+{{r}^{2}}{{\sin }^{2}}\bar{\theta }d{{\varphi }^{2}}$$

(Eq 6.6)
 * <p style="text-align:right">
 * }

Then we have scale factors,


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

$$ \displaystyle

\begin{align} & {{h}_{r}}=1 \\ & {{h}_}=r \\ & {{h}_{\varphi }}=r\sin \bar{\theta } \\ \end{align}$$

(Eqs 6.7)
 * <p style="text-align:right">
 * }

From lecture slide 34-2 we know the Laplacian for curvilinear coordinates is:


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

$$ \displaystyle

\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\xi }_{i}}}\left[ \frac{h_{i}^{2}}\frac{\partial \Phi }{\partial {{\xi }_{i}}} \right]}$$

(Eq 6.8)
 * <p style="text-align:right">
 * }

Substitute Eqs 6.7 into Eq 6.8 we have,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\sin }^{2}}\bar{\theta }}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\sin \bar{\theta }}\frac{\partial }{\partial \bar{\theta }}\left( \sin \bar{\theta }\frac{\partial \Psi }{\partial \bar{\theta }} \right)$$

(Eq 6.9)
 * <p style="text-align:right">
 * }

= Problem 6.7 Verify expression for $$\displaystyle \Delta \Psi $$ in elliptic coordinates from wikipedia article =

From the lecture slide Mtg 36-1

Given
The most common two-dimension definition of elliptic coordinates $$\displaystyle (\mu, \nu)$$ is



x = a \ \cosh \mu \ \cos \nu $$



y = a \ \sinh \mu \ \sin \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi].$$

Find
The Laplacian $$\displaystyle \Delta \Psi $$ expressed in elliptic coordinates.

Solution
From the information given we have:


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

$$ \displaystyle \begin{align} & dx=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ & dy=a\sinh \mu \cos \nu d\mu +a\cosh \mu \sin \nu dv \\ \end{align}$$

(Eqs 7.1)
 * <p style="text-align:right">
 * }

Consider magnitude of an infinitesimal segment $$\displaystyle ds$$. Before that we should aware that since ellipic coordinates system is among ORTHOGONAL curvilinear coordinates which never have off-diagonal terms in their metric tensor, meaning terms like $$\displaystyle dxdy$$ will never show up in the magnitude of $$\displaystyle ds$$ thus no need to be calculated.


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

$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}} \\ & =({{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu +a{{\cosh }^{2}}\mu {{\sin }^{2}}\nu )(d{{\mu }^{2}}+d{{v}^{2}}) \\ & =\left[ ({{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu )+(a{{\cosh }^{2}}\mu {{\sin }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu ) \right](d{{\mu }^{2}}+d{{v}^{2}}) \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}} \\ \end{align}$$

(Eq 7.2)
 * <p style="text-align:right">
 * }

We can derive scale factors from Eq 7.2,


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

$$ \displaystyle

{{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}$$

(Eq 7.3)
 * <p style="text-align:right">
 * }

From lecture slide 34-2 we know the Laplacian for curvilinear coordinates is:


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

$$ \displaystyle

\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\xi }_{i}}}\left[ \frac{h_{i}^{2}}\frac{\partial \Psi }{\partial {{\xi }_{i}}} \right]}$$

(Eq 7.4)
 * <p style="text-align:right">
 * }

and in two-dimensional case:


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

$$ \displaystyle

\Delta \Psi =\frac{1}\left[ \frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial u}+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial \nu }+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}} \right]$$

(Eq 7.5)
 * <p style="text-align:right">
 * }

from which we can see that the sign before scale factors won't matter since the they will be cancelled.

Then we derive $$\displaystyle \Delta \Psi $$ in two dimensional elliptic coordinates by substituting Eq7.3 into Eq7.5,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}{{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}\left( \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}} \right)$$

(Eq 7.6)
 * <p style="text-align:right">
 * }

Compare Eq 7.6 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

The Laplacian in three dimensinal elliptic coordinate systems
The wikipedia article required for verification also provided some outer links to some 3-D elliptic coordinate systems, hence one may want to verify Laplacians in different 3-D elliptic coordinate systems.

Verification of Laplacian in Elliptic cylindrical coordinates
The most common definition of elliptic cylindrical coordinates $$\displaystyle (\mu, \nu, z)$$ is ,



x = a \ \cosh \mu \ \cos \nu $$



y = a \ \sinh \mu \ \sin \nu $$



z = z \!$$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi)$$.


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

$$ \displaystyle

\begin{align} & dx=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ & dy=a\sinh \mu \cos \nu d\mu +a\cosh \mu \sin \nu dv \\ & dz=dz \\ \end{align}$$

(Eq 7.7)
 * <p style="text-align:right">
 * }

Then the corresponding magnitude of $$\displaystyle ds$$ is


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

$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}} \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}}+d{{z}^{2}} \\ \end{align}$$

(Eq 7.8)
 * <p style="text-align:right">
 * }

Then we have scale factors,


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

$$ \displaystyle

\begin{align} & {{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )} \\ & {{h}_{z}}=1 \\ \end{align}$$

(Eq 7.9)
 * <p style="text-align:right">
 * }

From Eq 7.4 we have


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

$$ \displaystyle

\Delta \Psi =\frac{1}\left[ \frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial u}+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial \nu }+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial z}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right]$$

(Eq 7.10)
 * <p style="text-align:right">
 * }

Substitute Eq 7.9 into Eq 7.10 we have,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}{{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}\left( \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right)$$

(Eq 7.11)
 * <p style="text-align:right">
 * }

Compare Eq 7.11 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

Verification of Laplacian in Prolate spheroidal coordinates
The most common definition of prolate spheroidal coordinates $$\displaystyle (\mu, \nu, \phi)$$ is



x = a \ \sinh \mu \ \sin \nu \ \cos \phi $$



y = a \ \sinh \mu \ \sin \nu \ \sin \phi $$



z = a \ \cosh \mu \ \cos \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, \pi]$$. The azimuthal angle $$\displaystyle \phi$$ belongs to the interval $$\displaystyle [0, 2\pi)$$.

Then we have,


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

$$ \displaystyle

\begin{align} & dx=-a\sin \phi \sinh \mu \cos \nu d\phi +a\cos \phi \cosh \mu \sin \nu d\mu +a\cos \phi \sinh \mu \cos \nu dv \\ & dy=a\cos \phi \sinh \mu \sin \nu d\phi +a\sin \phi \cosh \mu \sin \nu d\mu +a\sin \phi \sinh \mu \cos \nu dv \\ & dz=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ \end{align}$$

(Eq 7.12)
 * <p style="text-align:right">
 * }

Then the corresponding magnitude of $$\displaystyle ds$$ is


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

$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}} \\ & =({{a}^{2}}{{\cosh }^{2}}\mu {{\sin }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu )(d{{\mu }^{2}}+d{{v}^{2}})+{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu d{{\phi }^{2}} \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}}+{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu d{{\phi }^{2}} \\ \end{align}$$

(Eq 7.13)
 * <p style="text-align:right">
 * }

Then we have scale factors,


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

$$ \displaystyle

\begin{align} & {{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )} \\ & {{h}_{\phi }}=a\sinh \mu \sin \nu \\ \end{align}$$

(Eq 7.14)
 * <p style="text-align:right">
 * }

Substitute Eq 7.14 into Eq 7.10 we have,


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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}{{{a}^{3}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )\sinh \mu \sin \nu }\left( \begin{align} & a\cosh \mu \sin \nu \frac{\partial \Psi }{\partial \mu }+a\sinh \mu \cos \nu \frac{\partial \Psi }{\partial \nu }+a\sinh \mu \sin \nu \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}} \\  & +a\sinh \mu \sin \nu \frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+\frac{a({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}{\sinh \mu \sin \nu }\frac{{{\partial }^{2}}\Psi }{\partial {{\phi }^{2}}} \\ \end{align} \right)$$

(Eq 7.15)
 * <p style="text-align:right">
 * }

Compare Eq 7.15 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

= Problem 6.8 Verifying different expressions of Legendre Polynominals =

From the lecture slide Mtg 36-2

Given

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

$$ \displaystyle

{{P}_{0}}(x)=1$$

(Eq 8.1)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{P}_{1}}(x)=x$$

(Eq 8.2)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

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

(Eq 8.3)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x)$$

(Eq 8.4)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

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

(Eq 8.5)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{{{(-1)}^{i}}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-i)!(n-2i)!}}$$

(Eq 8.6)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{\frac{1.3.....(2n-2i-1)}{{{2}^{i}}i!(n-2i)!}{{(-1)}^{i}}}{{x}^{n-2i}}$$

(Eq 8.7)
 * <p style="text-align:right">
 * }


 * $$\displaystyle [n/2]\equiv $$ integer part of n/2

Find
Verify that Eq8.1 through Eq8.5 can be written as Eq8.6 or Eq8.7

Solution
If we expand Eq8.1 through Eq8.5 by using Eq8.6 and Eq8.7 for each “n” according to 0,1,2,3,4, we obtain:

For n=0, and $$\displaystyle [n/2]=0\equiv 0$$
into Eq8.6,


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

$$ \displaystyle

\begin{align} {{P}_{0}}(x) & =\sum\limits_{i=0}^{[0/2]}{{{(-1)}^{i}}\frac{(2\times 0-2i)!{{x}^{0-(2i)}}}{{{2}^{0}}i!(0-i)!(0-2i)!}}={{(-1)}^{0}}\frac{(0-2\times 0)!{{x}^{0-(2\times 0)}}}{1\times 0!(0)!(0-2\times 0)!}=1\times \frac{1\times 1}{1\times 1\times 1\times 1} \\ & =1 \\ \end{align}$$

(Eq 8.8)
 * <p style="text-align:right">
 * }

into Eq8.7,


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

$$ \displaystyle

\begin{align} {{P}_{0}}(x) & =\sum\limits_{i=0}^{[0/2]}{\frac{1.3....(2\times 0-2i-1)}{{{2}^{i}}i!(0-2i)!}{{(-1)}^{i}}}{{x}^{0-(2i)}}=\frac{1....(0-2\times 0-1)}{{{2}^{0}}0!(0-2\times 0)!}{{(-1)}^{0}}{{x}^{0-(2\times 0)}}=\frac{1}{1\times 1\times 1}\times 1\times 1 \\ & =1 \\ \end{align}$$

(Eq 8.9)
 * <p style="text-align:right">
 * }

As seen above, Eq8.8, and Eq8.9 are equal to Eq8.1 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=0.

For n=1, and $$\displaystyle [n/2]=0.5\equiv 0$$
into Eq8.6,


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

$$ \displaystyle

\begin{align} {{P}_{1}}(x) & =\sum\limits_{i=0}^{[1/2]}{{{(-1)}^{i}}\frac{(2\times 1-2i)!{{x}^{1-(2i)}}}{{{2}^{1}}i!(1-i)!(1-2i)!}}={{(-1)}^{0}}\frac{(2-2\times 0)!{{x}^{1-(2\times 0)}}}{2\times 0!(1-0)!(1-2\times 0)!}=1\times \frac{2x}{2\times 1\times 1\times 1} \\ & =x \\ \end{align}$$

(Eq 8.10)
 * <p style="text-align:right">
 * }

into Eq8.7,


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

$$ \displaystyle

\begin{align} {{P}_{1}}(x) & =\sum\limits_{i=0}^{[1/2]}{\frac{1.3...(2\times 1-2i-1)}{{{2}^{i}}i!(1-2i)!}{{(-1)}^{i}}}{{x}^{1-(2i)}}=\frac{1}{{{2}^{0}}0!(1-2\times 0)!}{{(-1)}^{0}}{{x}^{1-(2\times 0)}}=\frac{1}{1\times 1\times 1}x \\ & =x \\ \end{align}$$

(Eq 8.11)
 * <p style="text-align:right">
 * }

As seen above, Eq8.10 and Eq8.11 are equal to Eq8.2 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=1.

For n=2, and $$\displaystyle [n/2]=1\equiv 1$$
into Eq8.6,


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

$$ \displaystyle

\begin{align} {{P}_{2}}(x) & =\sum\limits_{i=0}^{[2/2]}{{{(-1)}^{i}}\frac{(2\times 2-2i)!{{x}^{2-(2i)}}}{{{2}^{2}}i!(2-i)!(2-2i)!}} \\ & =\left( {{(-1)}^{0}}\frac{(4-2\times 0)!{{x}^{2-(2\times 0)}}}{4\times 0!(2-0)!(2-2\times 0)!} \right)+\left( {{(-1)}^{1}}\frac{(4-2\times 1)!{{x}^{2-(2\times 1)}}}{4\times 1!(2-1)!(2-2\times 1)!} \right) \\ & =\left( 1\times \frac{4!{{x}^{2}}}{4\times 0!2!2!} \right)+\left( (-1)\frac{2!{{x}^{0}}}{4\times 1!1!0!} \right) \\ & =\left( \frac{24{{x}^{2}}}{4\times 1\times 2\times 2} \right)-\left( \frac{2\times 1}{4\times 1\times 1\times 1} \right) \\ & =\frac{24{{x}^{2}}}{16}-\frac{2}{4} \\ & =\frac{1}{2}(3{{x}^{2}}-1) \\ \end{align}$$

(Eq 8.12)
 * <p style="text-align:right">
 * }

into Eq8.7,


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

$$ \displaystyle

\begin{align} {{P}_{2}}(x) & =\sum\limits_{i=0}^{[2/2]}{\frac{1.3.....(2\times 2-2i-1)}{{{2}^{i}}i!(2-2i)!}{{(-1)}^{i}}}{{x}^{2-(2i)}} \\ & =\frac{1....(4-2\times 0-1)}{{{2}^{0}}.0!(2-2\times 0)!}{{(-1)}^{0}}{{x}^{2-(2\times 0)}}+\frac{1....(4-2\times 1-1)}{{{2}^{1}}.1!(2-2\times 1)!}{{(-1)}^{1}}{{x}^{2-(2\times 1)}} \\ & =\frac{1\times 3}{1\times 1\times 2}{{x}^{2}}-\frac{1}{2\times 1\times 1} \\ & =\frac{1}{2}(3{{x}^{2}}-1) \\ \end{align}$$

(Eq 8.13)
 * <p style="text-align:right">
 * }

As seen above, Eq8.12 and Eq8.13 are equal to Eq8.3 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=2.

For n=3, and $$\displaystyle [n/2]=1.5\equiv 1$$
into Eq8.6,


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

$$ \displaystyle

\begin{align} {{P}_{3}}(x) & =\sum\limits_{i=0}^{[3/2]}{{{(-1)}^{i}}\frac{(2\times 3-2i)!{{x}^{3-(2i)}}}{{{2}^{3}}i!(3-i)!(3-2i)!}} \\ & =\left( {{(-1)}^{0}}\frac{(6-2\times 0)!{{x}^{3-(2\times 0)}}}{8\times 0!(3-0)!(3-2\times 0)!} \right)+\left( {{(-1)}^{1}}\frac{(6-2\times 1)!{{x}^{3-(2\times 1)}}}{8\times 1!(3-1)!(3-2\times 1)!} \right) \\ & =\left( 1\times \frac{6!{{x}^{3}}}{8\times 0!3!3!} \right)+\left( (-1)\frac{4!{{x}^{1}}}{8\times 1!2!1!} \right)=\left( \frac{720{{x}^{3}}}{8\times 1\times 6\times 6} \right)-\left( \frac{24x}{8\times 1\times 2\times 1} \right) \\ & =\frac{720{{x}^{3}}}{288}-\frac{24x}{16} \\ & =\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

(Eq 8.14)
 * <p style="text-align:right">
 * }

into Eq8.7,


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

$$ \displaystyle

\begin{align} {{P}_{3}}(x) & =\sum\limits_{i=0}^{[3/2]}{\frac{1.3.....(2\times 3-2i-1)}{{{2}^{i}}i!(3-2i)!}{{(-1)}^{i}}}{{x}^{3-(2i)}} \\ & =\frac{1.3...(6-2\times 0-1)}{{{2}^{0}}.0!(3-2\times 0)!}{{(-1)}^{0}}{{x}^{3-(2\times 0)}}+\frac{1....(6-2\times 1-1)}{{{2}^{1}}.1!(3-2\times 1)!}{{(-1)}^{1}}{{x}^{3-(2\times 1)}} \\ & =\frac{1\times 3\times 5}{1\times 1\times 6}{{x}^{3}}-\frac{1\times 3}{2\times 1\times 1}x \\ & =\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

(Eq 8.15)
 * <p style="text-align:right">
 * }

As seen above, Eq8.14 and Eq8.15 are equal to Eq8.4 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=3.

For n=4, and $$\displaystyle [n/2]=2\equiv 2$$
into Eq8.6,


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

$$ \displaystyle

\begin{align} {{P}_{4}}(x) & =\sum\limits_{i=0}^{[4/2]}{{{(-1)}^{i}}\frac{(2\times 4-2i)!{{x}^{4-(2i)}}}{{{2}^{4}}i!(4-i)!(4-2i)!}} \\ & ={{(-1)}^{0}}\frac{(8-2\times 0)!{{x}^{4-(2\times 0)}}}{16\times 0!(4-0)!(4-2\times 0)!}+{{(-1)}^{1}}\frac{(8-2\times 1)!{{x}^{4-(2\times 1)}}}{16\times 1!(4-1)!(4-2\times 1)!}+{{(-1)}^{2}}\frac{(8-2\times 2)!{{x}^{4-(2\times 2)}}}{16\times 2!(4-2)!(4-2\times 2)!} \\ & =\left( 1\times \frac{8!{{x}^{4}}}{16\times 0!\times 4!\times 4!} \right)-\left( 1\times \frac{6!{{x}^{2}}}{16\times 1!\times 3!\times 2!} \right)+\left( 1\times \frac{4!{{x}^{0}}}{16\times 2!\times 2!\times 0!} \right) \\ & =\left( \frac{40320{{x}^{4}}}{16\times 1\times 24\times 24} \right)-\left( \frac{720{{x}^{2}}}{16\times 1\times 6\times 2} \right)+\left( \frac{24}{16\times 2\times 2\times 1} \right) \\ & =\frac{40320{{x}^{4}}}{9216}-\frac{720{{x}^{2}}}{192}+\frac{24}{64} \\ & =\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ \end{align}$$

(Eq 8.16)
 * <p style="text-align:right">
 * }

into Eq8.7,


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

$$ \displaystyle

\begin{align} {{P}_{4}}(x) & =\sum\limits_{i=0}^{[4/2]}{\frac{1.3.....(2\times 4-2i-1)}{{{2}^{i}}i!(4-2i)!}{{(-1)}^{i}}}{{x}^{4-(2i)}} \\ & =\frac{1.3....(8-2\times 0-1)}{{{2}^{0}}.0!(4-2\times 0)!}{{(-1)}^{0}}{{x}^{4-(2\times 0)}}+\frac{1.3....(8-2\times 1-1)}{{{2}^{1}}.1!(4-2\times 1)!}{{(-1)}^{1}}{{x}^{4-(2\times 1)}}+\frac{1....(8-2\times 2-1)}{{{2}^{2}}.2!(4-2\times 2)!}{{(-1)}^{2}}{{x}^{4-(2\times 2)}} \\ & =\frac{1\times 3\times 5\times 7}{1\times 1\times 24}{{x}^{4}}-\frac{1\times 3\times 5}{2\times 1\times 2}{{x}^{2}}+\frac{1\times 3}{4\times 4\times 1}{{x}^{0}} \\ & =\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ \end{align}$$

(Eq 8.17)
 * <p style="text-align:right">
 * }

As seen above, Eq8.16 and Eq8.17 are equal to Eq8.5 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=4.

As a result, Eq8.8 through Eq8.17 for n=0,1,2,3,4 are equal to Eq8.1 through Eq8.5. Therefore, it is verified that Eq8.1 through Eq8.5 can be written as Eq8.6 or Eq8.7.

= Problem 6.9: Verification that the Legendre Polynomials are Solutions to the Legendre Equation =

From the lecture slide Mtg 36-3

Given
The following equations are Legendre Polynomials, Pn(x), where n=0,1,2,3,4:


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

$$ \displaystyle {{P}_{0}}\left( x \right)=1 $$ (Eq 9.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle {{P}_{1}}\left( x \right)=x $$ (Eq 9.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle {{P}_{2}}\left( x \right)=\frac{1}{2}\left(3x^{2}-1\right) $$ (Eq 9.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle {{P}_{3}}\left( x \right)=\frac{1}{2}\left(5x^{3}-3x\right) $$ (Eq 9.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle {{P}_{4}}\left( x \right)=\frac{1}{8}\left(35x^{4}-30x^{2}+3\right) $$    (Eq 9.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Verify that Equations 9.1, 9.2, 9.3, 9.4, and 9.5 are all solutions of the following Legendre Equation:
 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x^2 \right)y'' -2x y' + n\left(n+1\right)y=0 $$ (Eq 9.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
For each equation the following calculations are made:
 * {| style="width:100%" border="0"

$$ \displaystyle y={{P}_{n}}\left( x \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle y'=\frac{d}{dx}[{{P}_{n}}\left( x \right)] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle y''=\frac{d}{dx}[\frac{d}{dx}[{{P}_{n}}\left( x \right)]] $$ Beginning with P0(x) {where n=0}, y=1 and y' and y" are both 0. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \cancel{\left(1-x^2 \right)[0]}^0 \cancel{-2x [0]}^0 + \cancel{(0)\left((0)+1\right)[1]}^0=0 $$ (Eq 9.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For P1(x) {where n=1}, y=x, y'=1 and y"=0. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * {| style="width:100%" border="0"

$$ \displaystyle \cancel{\left(1-x^2 \right)[0]}^0 -2x [1]+(1)\left((1)+1\right)[x]=0 $$ (Eq 9.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \cancel{(-2+2)x}^0=0 $$ (Eq 9.9) For P2(x) {where n=2}, y=1/2(3x2-1), y'=3x and y"=3. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x^2 \right)[3] -2x [3x]+(2)\left((2)+1\right)[\frac{1}{2}\left(3x^2-1\right)]=0 $$ (Eq 9.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \left(3-3x^2\right)-(6x^2)+(9x^2-3)=0 $$ (Eq 9.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \cancel{(-3-6+9)x^2}^0+\cancel{(3-3)}^0=0 $$ (Eq 9.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For P3(x) {where n=3}, y=1/2(5x3-3x), y'=1/2(15x2-3) and y"=15x. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x^2 \right)[15x] -2x \frac{1}{2}\left(15x^2-3\right)+(3)\left((3)+1\right)[\frac{1}{2}\left(5x^3-3x\right)]=0 $$ (Eq 9.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \left(15x-15x^3\right)-(15x^3-3x)+(30x^3-18x)=0 $$ (Eq 9.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \cancel{(-15-15+30)x^3}^0+\cancel{(15+3-18)x}^0=0 $$ (Eq 9.15) For P4(x) {where n=4}, y=1/8(35x4-30x2+3), y'=1/8(140x3-60x) and y"=1/8(420x2-60). Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x^2 \right)[\frac{1}{8}\left(420x^2-60\right)] -2x \frac{1}{8}\left(140x^3-60x\right)+(4)\left((4)+1\right)[\frac{1}{8}\left(35x^4-30x^2+3\right)]=0 $$ (Eq 9.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \left(-\frac{420}{8}x^4+60x^2-\frac{60}{8})\right)+(-35x^4+15x^2)+(\frac{175}{2}x^4-75x+\frac{15}{2})=0 $$ (Eq 9.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \cancel{(-\frac{420}{8}-35+\frac{175}{2})x^4}^0+\cancel{(60+15-75)x^2}^0+\cancel{(-\frac{60}{8}+\frac{15}{2})}^0=0 $$ (Eq 9.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

=Contributions Of Team Members=


 * Problem 1:    Solved by: Dube

Author: Dube

Proofread by: Dube,Oztekin


 * Problem 2:    Solved by: Abudaram

Author: Abudaram, Zou

Proofread by: Ismail,Zou,Abudaram


 * Problem 3:    Solved by: Abudaram

Author: Abudaram, Zou

Proofread by: Ismail,Zou,Abudaram


 * Problem 4:    Solved by: Oztekin

Author: Oztekin

Proofread by: Zou,Oztekin


 * Problem 5:    Solved by: Oztekin

Author: Oztekin

Proofread by: Zou,Oztekin


 * Problem 6:    Solved by: Zou

Author: Zou

Proofread by: Ismail,Zou,Abudaram


 * Problem 7:    Solved by: Zou

Author: Zou

Proofread by: Ismail,Zou,Abudaram


 * Problem 8:    Solved by: Sahin

Author: Sahin, Zou

Proofread by: Ismail,Zou,Abudaram


 * Problem 9:    Solved by: Patterson

Author: Patterson

Proofread by: Oztekin,Patterson

= References =