User:Egm6321.f11.team2.kim/PR7

=R*7.2 - Heat Conduction on a cylinder =

Given
$$\displaystyle x_1 = r \cos \theta = \xi_1 cos \xi_2 $$ $$\displaystyle x_2 = r \sin \theta = \xi_1 sin \xi_2 $$ $$\displaystyle x_3 = z = \xi_3 $$

Find
1. Find $$\displaystyle \left \{ dx_i \right \}=\left \{dx_1,dx_2,dx_3\right \} $$ in terms of $$\displaystyle \left \{ d\xi_j \right \}=\left \{d\xi_1,d\xi_2,d\xi_3\right \} $$ and $$\displaystyle \left \{ d\xi_k \right \}$$ 2. Find $$\displaystyle ds^2=\sum_{i=1}(dx_i)^2= \sum_{k}(h_k)^2(d\xi_k)^2$$. Identify $$\displaystyle \left \{ dh_i \right \}$$ in terms of $$\displaystyle \left \{ d\xi_j \right \}$$ 3. Find $$\displaystyle \Delta u $$ in cylindrical coordinates

4. Use seperation variable and compare to the Bessel equation.

Solution
part 1

$$\displaystyle ds^2=\sum_{k=1}^3(h_k)^2(d\xi_k)^2= \sum_{i=1}^{3}(dx_i)^2 $$ $$\displaystyle dx_i=\frac{\partial x_i}{\partial \xi_1} d\xi_1+\frac{\partial x_i}{\partial \xi_2} d\xi_2+\frac{\partial x_i}{\partial \xi_3} d\xi_3 $$ $$\displaystyle dx_1 =\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2+(0) d\xi_3 $$ $$\displaystyle dx_2 =\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2+(0) d\xi_3 $$ $$\displaystyle dx_3 =(0) d\xi_1+(0) d\xi_2+(1) d\xi_3 $$

Part2

$$\displaystyle ds^2=(\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2)^2 + (\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2)^2 + (d\xi_3)^2=(d\xi_1)^2+\xi_1^2 (d \xi_2)^2+(d\xi_3)^2 = (h_1)^2(d\xi_1)^2 + (h_2)^2(d\xi_2)^2 + (h_3)^2(d\xi_3)^2 $$ $$\displaystyle ds^2=dr^2+r^2d \theta^2 + dz^2 $$ Thus, $$\displaystyle h_1(\xi)=1 $$ $$\displaystyle h_2(\xi)=r=\xi_1 $$ $$\displaystyle h_3(\xi)=1 $$

Part 3 $$\displaystyle \Delta u = \frac{1}{h_1h_2h_3}\sum_{i=1}^{3} \frac{\partial}{\partial \xi_i}\left [ \frac{h_1h_2h_3}{h_i^2}\frac{\partial u }{\partial \xi_i} \right] $$ Where $$\displaystyle h_1h_2h_3=\xi_1=r $$ So, $$\displaystyle \Delta u = \frac{1}{\xi_1} \left [\frac{\partial}{\partial \xi_1} \left (\frac{\xi_1}{1^2} \frac{\partial u }{\partial \xi_1}\right ) +\frac{\partial}{\partial \xi_2} \left (\frac{\xi_1}{\xi_1^2}\frac{\partial u }{\partial \xi_2}\right )+\frac{\partial}{\partial \xi_3} \left ( \frac{\xi_1}{1^2}\frac{\partial u }{\partial \xi_3} \right ) \right ] =\frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2} $$

5.4

$$\displaystyle \Delta u = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2}=0 $$ $$\displaystyle u = X(\xi_1)Y(\xi_2)Z(\xi_3) $$ $$\displaystyle \frac{Y Z}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{X Z }{\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+X Y \frac{\partial^2 Z }{\partial \xi_3^2}=0 $$ divide through by  $$\displaystyle XYZ $$ $$\displaystyle \frac{1}{X \xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+\frac{1}{Z} \frac{\partial^2 Z }{\partial \xi_3^2}=0 $$ Assuming that $$\displaystyle \frac{1}{Z} \frac{\partial^2 Z }{\partial \xi_3^2}=A $$ $$\displaystyle \frac{1}{X \xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+A=0 $$ multiplying through by $$\displaystyle \xi_1^2 $$ $$\displaystyle \frac{\xi_1}{X } \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y}\frac{\partial^2 Y }{\partial \xi_2^2}+A \xi_1^2 =0 $$

assumeing for second term $$\displaystyle \frac{1}{Y}\frac{\partial^2 Y }{\partial \xi_2^2}=-B $$ $$\displaystyle \frac{\xi_1}{X } \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) -B+A \xi_1^2 =0 $$ multiplying through by $$\displaystyle X $$ $$\displaystyle \xi_1\frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +(A \xi_1^2-B)X =0 $$ The Original equation is following that $$\displaystyle \xi_1^2 \frac{\partial^2 X }{\partial \xi_1^2}+\xi_1 \frac{\partial X }{\partial \xi_1}+(A \xi_1^2-B)X =0 $$

Lets define $$\displaystyle \xi_1 =x, \,\,\,\,\, X =y $$ Thus, the above equations is following that $$\displaystyle x^2 y''+xy'+(A x^2-B)y =0 $$ =R*7.8 - =

Given
$$\displaystyle y=A \cdot P_n(x)+B \cdot{Q_n(x)} $$

Legendre Polynomials are following that $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ where $$\displaystyle m = \frac{n}{2}$$. $$\displaystyle {P}_{0}(x) = 1 \,\,\,\, {P}_{1}(x) = x \,\,\,\,{P}_{2}(x) = \frac{1}{2}(3x^2-1)\,\,\,\,{P}_{3}(x) = \frac{1}{2}(5x^3-3x,) $$ $$\displaystyle {Q}_{0}(x) = \frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right) \,\,\,\,{Q}_{1}(x) = \frac{1}{2}x \mathrm{log}\left(\frac{1+x}{1-x}\right)-1 \,\,\,\, $$ $$\displaystyle {Q}_{2}(x) = \frac{1}{4}(3x^2-1)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x \,\,\,\,$$ $$\displaystyle {Q}_{3}(x) = \frac{1}{4}(5x^3-3x)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{5}{2}x^2+\frac{2}{3} $$

Find
Show that for n=0; $$\displaystyle \mu \to \pm1 => Q_0(\mu) \to oo$$ Plot $$\displaystyle \begin{align} &\left[P_0, P_1, P_2, P_3\right]\\ &\left[Q_0, Q_1, Q_2, Q_3\right]\\ \end{align} $$ Oberve the limit of $$\displaystyle P_n(\mu) \,\,\, and\,\,\, Q_n(\mu) \,\,\,as\,\,\, \mu \to \pm 1 $$

Solution
From the given equations, legendre polynomial equation is following that This is the legendre polynomial graph. The above graph is a function of Q $$\displaystyle P_n(\mu) \,\,\, and\,\,\, Q_n(\mu) \,\,\,as\,\,\, \mu \to \pm 1 $$ It shows that a higher order number is shown the Runge's phenomenon at the edge of graph.