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=R*7.1 Infinitesimal Length in Spherical Coordinates= Question from 39-1: Given: A infinitesimal length in spherical coordinates. Find: Show that the infinitesimal length ds in (2) p. 38-6 can be written in spherical coordinates as follows. $$ds^2 = 1 \cdot dr^2 + r^2 d\theta^2 + r^2(\cos\theta)^2\, d \varphi^2$$ $$\displaystyle (Equation\;7.1.1)$$

From lecture notes [[media:pea1.f11.mtg39.djvu|Mtg 39]]

Solution
=

The key is to start with equation (1) on 38-5

 $$\mathbf{ds} = dx_i \mathbf{e}_i \Rightarrow \ ds^2= \mathbf{ds} \cdot \mathbf{ds} = (dx_i \mathbf{e}_i) \cdot (dx_j \mathbf{e}_j) = dx_i\, dx_j(\mathbf{e}_i \cdot \mathbf{e}_j)$$ $$\displaystyle (Equation\;7.1.2)$$

Next, note the following expressions.

$$\mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij}$$ $$\delta_{ij} := \begin{cases}1, & \text{for} i = j \\ 0, & \text{for} i \ne j \end{cases}$$

The form of the infinitesimal length ds is expressed in Cartesian coordinates. $$ds^2 = dx_i dx_i = \sum_{i=1}^3(dx_i)^2$$ This can be rewritten in the following form, which is the key for derivation steps, which are rigorous with trigonometric identities. $$ds^2 = dx_1^2 + dx_2^2 + dx_3^2$$ $$\displaystyle (Equation\;7.1.3)$$ The spherical coordinates can be utilized. Simply identify the appropriate identities. Recall that we are using the "astronomical" coordinate system. $$x = r\cos {\theta} \cos {\phi} = \xi_1 \cos{\xi_2}\cos{\xi_3}$$ $$y = r\cos {\theta} \sin {\phi} = \xi_1 \cos{\xi_2} \sin{\xi_3}$$ $$z = r\sin {\theta} = \xi_1 \sin{\xi_2}$$ Now, make the trigonometric substitutions for each term in 7.1.3.

$$ds^2 = (dr\cos{\theta} \cos{\phi} - (r\sin{\theta} \cos{\phi} d\theta + r\cos{\theta} \sin{\phi} d\phi))^2 +(dr\cos{\theta} \sin{\phi} - (r\sin{\theta} \sin{\phi} d\theta - r\cos{\theta} \cos{\phi} d\phi))^2 + (dr\sin{\theta} + r\cos{\theta} d\theta)^2$$ In such long math equations, simple variable substitutions allow easier management of squared terms (FOIL). $$\alpha = (r\sin{\theta} \cos{\phi} d\theta + r\cos{\theta} \sin{\phi} d\phi)$$ $$\beta = (r\sin{\theta} \sin{\phi} d\theta - r\cos{\theta} \cos{\phi} d\phi)$$ The substitutions then give the shortened form, which allows derivation of the squared terms. $$ds^2 = (dr\cos{\theta} \cos{\phi} - \alpha)^2 +(dr\cos{\theta} \sin{\phi} - \beta)^2 + (dr\sin{\theta} + r\cos{\theta} d\theta)^2$$ Simply apply algebraic FOIL method to the equation above and get the terms below. <p style="font-size:125%" align="center">$$ds^2 = (dr\cos{\theta} cos{\phi})^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 + (dr\cos{\theta} \sin{\phi})^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2$$ The algebraic manipulation is long, but proper addition and subtraction eliminate unnecessary terms. The key is to appply trigonometric identities and grouping with the identity of $$\sin^2{x} + \cos^2{x} = 1$$. <p style="font-size:125%" align="center">$$ds^2 = (dr^2\cos^2{\theta}(cos^2{\phi} + \sin^2{\phi}) - 2\alpha dr \cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + dr^2 \sin{\theta}^2 + 2rdr \sin{\theta} \cos{\theta}d\theta + r^2 \cos^2{\theta} d{\theta}^2$$ <p style="font-size:125%" align="center">$$ds^2 = dr^2\cos^2{\theta} + dr^2\sin^2{theta} - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2$$ <p style="font-size:125%" align="center">$$ds^2 = dr^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2 $$ <p style="text-align:right;">$$\displaystyle (Equation\;7.1.4)$$ The next set of steps involves solutions utilizing our substitutions of $$\alpha$$ and $$\beta$$. Recall the "inner" and "outer" terms from the FOIL multiplications. Each of those terms must now be expanded and simplified.

<p style="font-size:125%" align="center">$$A = - 2\alpha dr\cos{\theta} \cos{\phi}$$ <p style="font-size:125%" align="center">$$A = - 2rdr(\sin{\theta}\cos{\theta}\cos^2{\phi}d\theta + \cos^2{\theta} \cos{\phi} \sin{\phi}d{\phi})$$

<p style="font-size:125%" align="center">$$B = - 2\beta dr\cos{\theta} \sin{\phi}$$ <p style="font-size:125%" align="center">$$B = - 2rdr(\sin{\theta} \cos{\theta} \sin^2{\phi} d\phi - \cos^2{\theta} \cos{\phi} \sin{\phi} d\phi)$$ <p style="font-size:125%" align="center">$$A + B = -2rdr \sin{\theta}\cos{\theta}(\cos{\phi}^2 + \sin^2{\phi})d\theta = -2rdr \sin{\theta} \cos{\theta}d\theta$$

Recall equation 7.1.4. The terms A and B can be put back into the equation to get the simplification. <p style="font-size:125%" align="center">$$ds^2 = dr^2 + A + \alpha^2 + B + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2 $$ <p style="font-size:125%" align="center">$$ds^2 = dr^2 - 2rdr \sin{\theta} \cos{\theta} d\theta + \alpha^2 + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2 $$ Next, the terms for $${\alpha}^2 + {\beta}^2$$ must be expanded. <p style="font-size:125%" align="center">$${\alpha}^2 = r^2 \sin^2{\theta} \cos^2{\phi} d{\theta}^2 + 2r^2 \sin{\theta} \cos{\theta} \cos{\phi} \sin{\phi} d\theta d\phi r^2 \cos^2{\theta} \sin^2{\phi} d{\phi}^2$$ <p style="font-size:125%" align="center">$${\beta}^2 = r^2 \sin^2{\theta} \sin^2{\phi} d{\theta}^2 - 2r^2 \sin{\theta} \cos{\theta} \cos{\phi} \sin{\phi} d\theta d\phi r^2 \cos^2{\theta} \cos^2{\phi} d{\phi}^2$$ <p style="font-size:125%" align="center">$${\alpha}^2 + {\beta}^2 = r^2\sin^2{\theta} d{\theta}^2 (\cos^2{\phi} + \sin^2{\phi}) + r^2 \cos^2{\theta}(\sin^2{\phi} + \cos^2{\phi})d{\phi}^2$$ <p style="font-size:125%" align="center">$${\alpha}^2 + {\beta}^2 = r^2\sin^2{\theta} d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2$$ Moving along, all inner and outer terms from the FOIL calculation are added and simplified below.

<p style="font-size:125%" align="center">$$ds^2 = dr^2 - rdr \sin{\theta} \cos{\theta} d\theta + r^2 \cos^2{\theta} d{\theta}^2 + r^2 \sin^2{\theta} d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2 + 2rdr \sin{\theta} \cos{\phi} d\theta - 2rdr \sin{\theta} \cos{\theta} d\theta$$ Finally, after addition and subtracting terms above, the solution is derived in the desired form to solve the problem. <p style="font-size:125%" align="center">$$ds^2 = dr^2 + r^2 d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2$$ Each coordinate line has a tangent vector with a magnitude of h. <p style="font-size:125%" align="center">$$dr^2 = 1 * dr^2 = {h_1}^2{\xi_1}^2$$ <p style="font-size:125%" align="center">$${h_1}^2 = 1$$ <p style="font-size:125%" align="center">$${h_1} = 1$$ Similarly, the other h parameters can be found for each variable. <p style="font-size:125%" align="center">$$r^2 d{\theta}^2 = {h_2}^2{\xi_2}^2$$ <p style="font-size:125%" align="center">$${h_2}^2 = r^2$$ <p style="font-size:125%" align="center">$${h_2} = r$$ <p style="font-size:125%" align="center">$$r^2 \cos^2{\theta} d{\phi}^2 = {h_3}^2{\xi_3}^2$$ <p style="font-size:125%" align="center">$${h_3}^2 = r^2 \cos^2{\theta}$$ <p style="font-size:125%" align="center">$${h_3} = r\cos{\theta}$$

=R*7.2- Laplacian in Cylindrical Coordinate System=

Given
$$\begin{align} & {{x}_{1}}=x=r\cos \phi ={{\xi }_{1}}\cos {{\xi }_{2}} \\ & {{x}_{2}}=y=r\sin \phi ={{\xi }_{1}}\sin {{\xi }_{2}} \\ & {{x}_{3}}=z={{\xi }_{3}} \\ \end{align}$$

Find
$$\begin{align} & \text{(1) Express }\left\{ d{{x}_{i}} \right\} \text{ in terms of } \left\{ d{{\xi }_{i}} \right\} \text{ and }\left\{ {{\xi }_{i}} \right\} \\ & \text{(2) Find } d{{s}^{2}}=\sum\limits_{i=1}^{3}{dx_{i}^{2}}=\sum\limits_{i=1}^{3}{h_{i}^{2}\cdot d\xi _{i}^{2}} \text{ and express } \left\{ {{h}_{i}} \right\}\text{ in terms of } \left\{ {{\xi }_{i}} \right\} \\ & \text{(3) Find Laplacian } \Delta u \\ & \text{(4) Obtain Bessel Equation by solving }\Delta u =0 \\ \end{align}$$

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

Part 2

$$\displaystyle ds^2=(\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2)^2 + (\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2)^2 + (d\xi_3)^2=(d\xi_1)^2+\xi_1^2 (d \xi_2)^2+(d\xi_3)^2 = (h_1)^2(d\xi_1)^2 + (h_2)^2(d\xi_2)^2 + (h_3)^2(d\xi_3)^2 $$ $$\displaystyle ds^2=dr^2+r^2d \theta^2 + dz^2 $$ 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, $$\displaystyle \Delta u = \frac{1}{\xi_1} \left [\frac{\partial}{\partial \xi_1} \left (\frac{\xi_1}{1^2} \frac{\partial u }{\partial \xi_1}\right ) +\frac{\partial}{\partial \xi_2} \left (\frac{\xi_1}{\xi_1^2}\frac{\partial u }{\partial \xi_2}\right )+\frac{\partial}{\partial \xi_3} \left ( \frac{\xi_1}{1^2}\frac{\partial u }{\partial \xi_3} \right ) \right ] =\frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2} $$

part 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*7.3- Laplacian in Spherical Coordinate System with Math/Phy Convention=

Given
The Laplacian Equation is given in general form - $$\Delta u$$.

Find
The expression of Laplacian in spherical coordinates for math/physics convention. From lecture notes [[media:pea1.f11.mtg41.djvu|Mtg 41]]

Solution
The approach is to derive the Laplacian in spherical coordinates with the astronomical convention. <p style="font-size:125%" align="center">$$(\xi_1, \xi_2, \xi_3) = (r, \theta, \phi)$$ Then we will convert the final form to the math/physics equation using the standard transformation identity. From meeting 39, the general equation for the Laplacian in curvilinear coordinates is given. <p style="font-size:125%" align="center">$$\Delta{u} = \frac{1}{h_1 h_2 h_3} \sum^3_{i=1} \frac{\partial}{\partial \xi_i} \left[\frac{h_1 h_2 h_3}{(h_i)^2} \frac{\partial u}{\partial \xi_i} \right]$$ <p style="text-align:right;">$$\displaystyle (Equation\;7.3.1)$$

The spherical coordinates from R*7.1 can be utilized. Recall the appropriate identities using the "astronomical" coordinate system. <p style="font-size:125%" align="center">$$x = r\cos {\theta} \cos {\phi} = \xi_1 \cos{\xi_2}\cos{\xi_3}$$ <p style="font-size:125%" align="center">$$y = r\cos {\theta} \sin {\phi} = \xi_1 \cos{\xi_2} \sin{\xi_3}$$ <p style="font-size:125%" align="center">$$z = r\sin {\theta} = \xi_1 \sin{\xi_2}$$ We derived the solution in R*7.1 that can be reapplied to R*7.3. <p style="font-size:125%" align="center">$$ds^2 = dr^2 + r^2 d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2$$ Each coordinate line has a tangent vector with a magnitude of h. <p style="font-size:125%" align="center">$$dr^2 = 1 * dr^2 = {h_1}^2{\xi_1}^2$$ <p style="font-size:125%" align="center">$${h_1}^2 = 1$$ <p style="font-size:125%" align="center">$${h_1} = 1$$ <p style="font-size:125%" align="center">$$r^2 d{\theta}^2 = {h_2}^2{\xi_2}^2$$ <p style="font-size:125%" align="center">$${h_2}^2 = r^2$$ <p style="font-size:125%" align="center">$${h_2} = r$$ <p style="font-size:125%" align="center">$$r^2 \cos^2{\theta} d{\phi}^2 = {h_3}^2{\xi_3}^2$$ <p style="font-size:125%" align="center">$${h_3}^2 = r^2 \cos^2{\theta}$$ <p style="font-size:125%" align="center">$${h_3} = r\cos{\theta}$$ Based on the identities derived in R*7.1, the following expression can be utilized for the Laplacian equation. <p style="font-size:125%" align="center">$${h_1}*{h_2}*{h_3} = 1 * r * r\cos{\theta}$$ <p style="font-size:125%" align="center">$${h_1}{h_2}{h_3} = r^2\cos{\theta}$$ The Laplacian has three terms for the summation in sigma notation i= 1, 2, 3. Each must be derived separately and summed based on the identities derived from R*7.1. The total is multiplied by $$({h_1}{h_2}{h_3})^-1$$. Before the latter is applied, let us derive each term for i = 1, 2, and 3 below. <p style="font-size:125%" align="center">$${h_1} = 1$$ <p style="font-size:125%" align="center">$$d\xi_1 = dr$$ <p style="font-size:125%" align="center">$$\partial {\xi_1} = \partial r$$ <p style="font-size:125%" align="center">$$\frac{\partial}{\partial \xi_1} \left[\frac{h_1 h_2 h_3}{(h_1)^2} \frac{\partial u}{\partial \xi_1} \right] = \frac{\partial}{\partial r} \left[\frac{r^2 \cos \theta}{(1)^2} \frac{\partial u}{\partial r} \right]$$ Next, we derive the 2nd term. <p style="font-size:125%" align="center">$${h_2} = r$$ <p style="font-size:125%" align="center">$$d\xi_2 = d\theta$$ <p style="font-size:125%" align="center">$$\partial {\xi_2} = \partial {\theta}$$ <p style="font-size:125%" align="center">$$\frac{\partial u}{\partial \xi_2} \left[\frac{h_1 h_2 h_3}{(h_2)^2} \frac{\partial u}{\partial \xi_2} \right] = \frac{\partial}{\partial \theta} \left[\frac{r^2 \cos \theta}{(r)^2} \frac{\partial u}{\partial \theta} \right]$$

Moving along, we derive the 3rd term. <p style="font-size:125%" align="center">$${h_3} = r\cos{\theta}$$ <p style="font-size:125%" align="center">$$d\xi_3 = d\phi$$ <p style="font-size:125%" align="center">$$\partial {\xi_3} = \partial {\phi}$$ <p style="font-size:125%" align="center">$$\frac{\partial u}{\partial \xi_3} \left[\frac{h_1 h_2 h_3}{(h_3)^2} \frac{\partial u}{\partial \xi_3} \right] = \frac{\partial}{\partial \phi} \left[\frac{r^2 \cos \theta}{(r \cos{\theta})^2} \frac{\partial u}{\partial \phi} \right]$$ Finally, sum all three terms and multiply each by $$({h_1}{h_2}{h_3})^{-1}$$. Equation 7.3.1 is repeated again for clarity. <p style="font-size:125%" align="center">$$\Delta{u} = \frac{1}{h_1 h_2 h_3} \sum^3_{i=1} \frac{\partial}{\partial \xi_i} \left[\frac{h_1 h_2 h_3}{(h_i)^2} \frac{\partial u}{\partial \xi_i} \right]$$ Simply sum and multiply appropriate terms for the Laplacian. <p style="font-size:125%" align="center">$$\Delta{u} = \frac{1}{r^2 \cos{\theta}}[\frac{\partial}{\partial r} \left[\frac{r^2 \cos{\theta}}{(1)^2} \frac{\partial u}{\partial r} \right] + \frac{\partial}{\partial \theta} \left[\frac{r^2 \cos \theta}{(r)^2} \frac{\partial u}{\partial \theta} \right] + \frac{\partial}{\partial \phi} \left[\frac{r^2 \cos \theta}{(r \cos{\theta})^2} \frac{\partial u}{\partial \phi} \right]]$$ From analysis, the equation can be factored with the following terms: $$r$$ and $$\cos{\theta}$$. <p style="font-size:125%" align="center">$$\Delta u = \frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2 \cos^2\theta} \frac{\partial^2 u}{\partial \phi^2} + \frac{1}{r^2 \cos \theta} \frac{\partial}{\partial \theta} \left(\cos \theta \frac{\partial u}{\partial \theta} \right)$$ Now, this equation must be converted to the math/physics convention. The following identities can be utilized. <p style="font-size:125%" align="center">$$(\xi_1, \xi_2, \xi_3) = (r, \bar \theta, \phi)$$ <p style="font-size:125%" align="center">$$\bar \theta := \frac{\pi}{2} - \theta$$ In order to solve the Laplacian for the math/physics convention, the following two identities are utilized. <p style="font-size:125%" align="center">$$\partial \theta = \partial {\bar \theta}$$ <p style="font-size:125%" align="center">$$\cos{\theta} = \cos{\frac{\pi}{2} - {\bar \theta}} = \cos{\frac{\pi}{2}} \cos{\bar \theta} + \sin{\frac{\pi}{2}} \sin{\bar \theta} = \sin{\bar \theta}$$ Therefore, simply make the appropriate substitutions of $$\partial \theta = \partial \bar \theta$$ and $$\cos{\theta} = \sin{\bar \theta}$$. <p style="font-size:125%" align="center">$$\Delta u = \frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2 \sin^2{\bar \theta}} \frac{\partial^2 u}{\partial \phi^2} + \frac{1}{r^2 \sin {\bar \theta}} \frac{\partial}{\partial {\bar \theta}} \left(\sin{\bar \theta} \frac{\partial u}{\partial {\bar \theta}} \right)$$

= R*7.4 Laplacian In Elliptic Coordinates =

Given
See wikipedia reference of ellipitic coordinates.

Laplacian operator:


 * $$\Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=3}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right]$$

== Find ==

Verify the Laplacian in elliptic coordinates:


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

Solution
Using the reference listed above, we find that

$$\displaystyle \xi_1=\mu$$

$$\displaystyle\xi_2=\nu$$

$$\displaystyle h_1=h_2=a\sqrt{\sinh^2\mu+\sin^2\nu}$$

The reference does not give a derivation for each magnitude h of the tangential vector to the coordinate line. However, team 1 provides a derivation here based on class principles. The key is to define each variable with its 2nd and 3rd derivatives for the infinitesimal summation of ds. <p style="font-size:125%" align="center">$$x_1 = x = a\cosh{\mu} \cos{\nu}$$ <p style="font-size:125%" align="center">$$dx_1 = a(\sinh{\mu} \cos{\nu}d\mu - \cosh{\mu} \sin{\nu}d\nu$$ <p style="font-size:125%" align="center">$${dx_1}^2 = a^2 \sinh^2{\mu} \cos^2{\nu}du^2 - 2a^2 \sinh{\mu} \cos{\nu} \cosh{\mu} \sin{\nu} d\mu d\nu + a^2 \sinh^2{\mu} ]cos^2{\nu}d{\nu}^2$$ <p style="font-size:125%" align="center">$$x_2 = x = a\sinh{\mu} \sin{\nu}$$ <p style="font-size:125%" align="center">$$dx_2 = a(\cosh{\mu} \sin{\nu}d\mu + \sinh{\mu} \cos{\nu}d\nu$$ <p style="font-size:125%" align="center">$${dx_2}^2 = a^2 \cosh^2{\mu} \sin^2{\nu} du^2 + 2a^2 \sinh{\mu} \cos{\nu} \cosh{\mu} \sin{\nu} d\mu d\nu + a^2 \sinh^2{\mu} ]cos^2{\nu}d{\nu}^2$$ Sum all terms to get ds squared. <p style="font-size:125%" align="center">$$ds^2 = {dx_1}^2 + {dx_2}^2$$ <p style="font-size:125%" align="center">$$ds^2 = a^2 \sinh^2{\mu} \cos^2{\nu} du^2 - 2a^2 \sinh{\mu} \cos{\nu} \cosh{\mu} \sin{\nu} d\mu d\nu + a^2 \sinh^2{\mu}]cos^2{\nu}d{\nu}^2 + a^2 \cosh^2{\mu} \sin^2{\nu} du^2 + 2a^2 \sinh{\mu} \cos{\nu} \cosh{\mu} \sin{\nu} d\mu d\nu + a^2 \sinh^2{\mu} ]cos{\nu}^2d{\nu}^2$$ Add all terms to simplify - note +/- signs. Then group by similar variables. <p style="font-size:125%" align="center">$$ds^2 = a^2[( \sinh^2{\mu} \cos^2{\nu} + \cosh^2{\mu} \sin^2{\nu})d{\mu}^2 + (\cosh^2{\mu} \sin^2{\nu} + \sinh^2{\mu} \cos^2{\nu})d{\nu}^2]$$ Substitute the following hyperbolic identity: $$\cosh^2{\mu} = 1 + \sinh^2{\mu}$$. The simplification progresses below. <p style="font-size:125%" align="center">$$ds^2 = a^2[( \sinh^2{\mu} \cos^2{\nu} + \sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu})d{\mu}^2 + (\sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu} + \sinh^2{\mu} \cos^2{\nu})d{\nu}^2]$$ Next, the standard trigonometric subsitution can be applied: $$\cos^2{x} = 1 - \sin^2{x}$$. This leads to the next simplification. <p style="font-size:125%" align="center">$$ds^2 = a^2[( \sinh^2{\mu} ( 1 - \sin^2{\nu}) + \sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu})d{\mu}^2 + (\sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu} + \sinh^2{\mu} (1 - \sin^2{\nu}))d{\nu}^2]$$ <p style="font-size:125%" align="center">$$ds^2 = a^2[( \sinh^2{\mu} ( 1 - \sin^2{\nu}) + \sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu})d{\mu}^2 + (\sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu} + \sinh^2{\mu} (1 - \sin^2{\nu}))d{\nu}^2]$$ <p style="font-size:125%" align="center">$$ds^2 = a^2[( \sinh^2{\mu} ( 1 - \sin^2{\nu}) + \sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu})d{\mu}^2 + (\sin^2{\nu} + \sinh^2{\mu} \sin^2{\nu} + \sinh^2{\mu} (1 - \sin^2{\nu}))d{\nu}^2]$$ <p style="font-size:125%" align="center">$$ds^2 = a^2[( \sinh^2{\mu} + \sin^2{\nu})d{\mu}^2 + (\sin^2{\nu} + \sinh^2{\mu})d{\nu}^2]$$ <p style="font-size:125%" align="center">$$ds^2 = {h_1}^2{\xi_1}^ + {h_2}^2{\xi_2}^2 = a^2[( \sinh^2{\mu} + \sin^2{\nu})d{\mu}^2 + (\sin^2{\nu} + \sinh^2{\mu})d{\nu}^2]$$ Therefore, the h parameters are derived below. <p style="font-size:125" align="center">$${h_1}^2 = {h_2}^2 = a^2(\sinh^2{\mu} + \sin^2{\nu})$$ <p style="font-size:125" align="center">$$h_1 = h_2 = a\sqrt{(\sinh^2{\mu} + \sin^2{\nu})}$$

Therefore, using the definition of Laplacian operator in the above section,

$$\nabla^2u=\Delta u=\frac{1}{h_1h_2}\left[\frac{\partial}{\partial\xi_1}\left(\frac{h_1h_2}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right)+\frac{\partial}{\partial\xi_2}\left(\frac{h_1h_2}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right)\right]$$

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

Therefore we arrive at our solution.

= R7.5 Laplacian In Parabolic Coordinates =

Given
See wikipdia reference of parabolic coordinates.

Laplacian operator:


 * $$\Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=3}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right]$$

== Find ==

Verify the Laplacian in parabolic coordinates:


 * $$\nabla^2u=\frac{1}{\mu^2+\nu^2}\left(\frac{\partial^2u}{\partial\mu^2}+\frac{\partial^2u}{\partial\nu^2}\right)$$

Solution
Using the reference listed above, we find that

$$\displaystyle \xi_1=\sigma$$

$$\displaystyle\xi_2=\tau$$

$$\displaystyle h_1=h_2=\sqrt{\sigma^2+\tau^2}$$

The reference did not give a derivation for each value of h. However, our team 1 will give a brief derivation here. The form of the infinitesimal length ds is expressed in parabolic coordinates. <p style="text-align:center;">$$ds^2 = dx_i dx_i = \sum_{i=1}^3(dx_i)^2$$ This can be rewritten in the following form, which is the key for derivation steps with parabolic coordinates in two dimensions. <p style="font-size:125%" align="center">$$ds^2 = dx_1^2 + dx_2^2$$ <p style="text-align:right;">$$\displaystyle (Equation\;7.5.1)$$ The key is to identify the proper partial derivatives for 7.5.1. There are two variables for parabolic coordinates, so ds must include partial deriviates for each: $$\sigma$$ and $$\tau$$. <p style="font-size:125%" align="center">$$ x_1 = \sigma*\tau$$ <p style="font-size:125%" align="center">$$ dx_1 = d\sigma*\tau + d\tau*\sigma$$ <p style="font-size:125%" align="center">$$ x_2 = {\frac{1}{2}}({\tau}^2 - {\sigma}^2)$$ <p style="font-size:125%" align="center">$$ dx_2 = \tau*d\tau - \sigma*d\sigma$$ Now, square each term and add below. <p style="font-size:125%" align="center">$$ {dx_1}^2 + {dx_2}^2 = {d\sigma*\tau}^2 + 2\tau \sigma d\sigma d\tau + {\sigma*d\tau}^2 + {\tau*d\tau}^2 - 2\tau \sigma d\sigma d\tau + {\sigma*d\sigma}^2$$ By rearranging the terms above, the values of each h can be determined. <p style="font-size:125%" align="center">$$ {dx_1}^2 + {dx_2}^2 = ({\tau}^2 + {\sigma}^2){d\sigma}^2 + ({\tau}^2 + {\sigma}^2){d\tau}^2 = {h_1}^2{d\sigma}^2 + {h_2}^2{d\tau}^2$$ This gives the identity we need to derive each value of h. <p style="font-size:125%" align="center">$$ {h_1}^2 = {h_2}^2 = ({\tau}^2 + {\sigma}^2)$$ <p style="font-size:125%" align="center">$$ h_1 = h_2 = \sqrt{{\tau}^2 + {\sigma}^2}$$

I will use the following definitions to match the syntax given in the problem.

$$\displaystyle \mu:=\sigma$$

$$\displaystyle \nu:=\tau$$

Therefore, using the definition of Laplacian operator in the above section,

$$\nabla^2u=\Delta u=\frac{1}{h_1h_2}\left[\frac{\partial}{\partial\xi_1}\left(\frac{h_1h_2}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right)+\frac{\partial}{\partial\xi_2}\left(\frac{h_1h_2}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right)\right]$$

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

Therefore we arrive at our solution.

=R*7.6 - Verify Legendre Polynomials 0 through 4 =

Find
Verify that the first five Legendre Polynomials

= R*7.7 Find the separated equation for the Laplace Equation in Parabolic coordinates= == Given == The Laplacian in parabolic coordinates given by
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \nabla^2 u=\frac{1}{\mu^2+\nu^2}\left(\frac{\partial^2 u}{\partial \mu^2}+\frac{\partial^2 u}{\partial \nu^2}\right)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;7.7.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;7.7.1)
 * }
 * }

Find
Find the separated equation for the Laplace equation in Pararbolic coordinates.

Solution
Separation of variables: $$\displaystyle u(\mu,\nu)=U(\mu)V(\nu)$$ Substituting it into Equation (7.7.1), $$\displaystyle \Delta u=0=\frac{1}{\mu^2+\nu^2}\left(V \frac{\partial^2 U}{\partial \mu^2}+ U \frac{\partial^2 V}{\partial \nu^2}\right)$$ Cancel $$\displaystyle \frac{1}{\mu^2+\nu^2} $$ and $$\displaystyle UV$$, $$\displaystyle \frac{1}{U}\,\frac{\partial^2 U}{\partial \mu^2}+ \frac{1}{V}\,\frac{\partial^2 V}{\partial \nu^2}=0$$ Rearranging it, we can obtain that $$\displaystyle \frac{1}{U}\,\frac{\partial^2 U}{\partial \mu^2} = - \frac{1}{V}\,\frac{\partial^2 V}{\partial \nu^2}=k(const.)$$ Therefore, the separated equations are:

= R*7.8 - Plot Legendre Functions for $$n=0 \text{ to } 3$$ =

Given
From [[media:pea1.f11.mtg42.djvu|Mtg 42]] Pg 42-2

Legendre polynomials

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ & {{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1) \\ & {{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

Legendre functions

$$\begin{align} & {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{Q}_{1}}(x)=\frac{1}{2}x\log \left( \frac{1+x}{1-x} \right)-1 \\ & {{Q}_{2}}(x)=\frac{1}{4}(3{{x}^{2}}-1)\log \left( \frac{1+x}{1-x} \right)-\frac{3}{2}x \\ & {{Q}_{3}}(x)=\frac{1}{4}(5{{x}^{3}}-3x)\log \left( \frac{1+x}{1-x} \right)-\frac{5}{2}{{x}^{2}}+\frac{2}{3} \\ \end{align}$$

Find
The plot of the Legendre polynomials (figure 1) and functions (figure 2) together:

Show that for $$ n = 0 $$: $$\scriptstyle \mu \to \pm 1 \Rightarrow \left|Q_0 (\mu) \right| \to + \infty$$

Observe the limits of $$P_n(\mu)$$ and $$Q_n(\mu)$$ as $$\scriptstyle \mu \rightarrow \pm 1$$.

Solution
See the requested solutions plotted below.



What has been shown

Clearly with $$ n = 0 $$ for $$\scriptstyle \mu \to \pm 1 $$ the value of $$ \left|Q_0 (\mu) \right|$$ goes to $$ + \infty$$ as can be seen in the plot for $$Q_0$$ above

The limits of $$P_n(\mu)$$ and $$Q_n(\mu)$$ as $$\scriptstyle \mu \rightarrow \pm 1$$ are as follows:

$$P_0(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+1$$

$$P_0(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$+1$$

$$P_1(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+1$$

$$P_1(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$-1$$

$$P_2(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+1$$

$$P_2(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$+1$$

$$P_3(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+1$$

$$P_3(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$-1$$

$$Q_0(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+ \infty$$

$$Q_0(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$- \infty$$

$$Q_1(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+ \infty$$

$$Q_1(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$+ \infty$$

$$Q_2(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+ \infty$$

$$Q_2(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$- \infty$$

$$Q_3(\mu)$$ as $$\scriptstyle \mu \rightarrow + 1$$ = $$+ \infty$$

$$Q_3(\mu)$$ as $$\scriptstyle \mu \rightarrow - 1$$ = $$+ \infty$$

So the limits depend on whether $$ n $$ is odd or even

= R7.9 Express Vector As Linear Combination of Basis Vectors =

Given
Non-orthonormal basis:


 * $$\displaystyle{\mathbf b_1,\mathbf b_2,\mathbf b_3}$$

Orthonormal basis:


 * $$\displaystyle{\mathbf e_1,\mathbf e_2,\mathbf e_3}$$

Expressing the non-orthonormal in the orthonormal basis:


 * $$\displaystyle\mathbf b_i=A_{ij}\mathbf e_j$$


 * where


 * $$\mathbf A=[A_{ij}]=\left[\begin{matrix}5&2&3\\4&5&6\\7&8&5\end{matrix}\right]$$

Vector $$\displaystyle\mathbf v$$ in $$\displaystyle\mathbb{R}^3$$:


 * $$\displaystyle-2\mathbf e_1+4\mathbf e_2-5\mathbf e_3$$

Definition of Gram Matrix:


 * $$\boldsymbol\Gamma=[\Gamma_{ij}]=[(\mathbf b_i\cdot\mathbf b_j)]\in\mathbb R^{n\times n}$$

Finding components of $$\displaystyle\mathbf v$$


 * $$\displaystyle{v_j}=\mathbf\Gamma^{-1}\{\mathbf b_i\mathbf v\}$$

== Find ==


 * $$\displaystyle{\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that $$\displaystyle\mathbf=v_i\mathbf b_i$$

Solution
To begin, find each entry of the Gram Matrix:

$$\Gamma_{11}=\mathbf b_1\cdot\mathbf b_1=25+4+9=38$$

$$\Gamma_{12}=\mathbf b_1\cdot\mathbf b_2=20+10+18=48$$

$$\Gamma_{13}=\mathbf b_1\cdot\mathbf b_3=35+16+25=76$$

$$\Gamma_{21}=\mathbf b_2\cdot\mathbf b_1=48$$

$$\Gamma_{22}=\mathbf b_2\cdot\mathbf b_2=16+25+36=77$$

$$\Gamma_{23}=\mathbf b_2\cdot\mathbf b_3=28+40+30=98$$

$$\Gamma_{31}=\mathbf b_3\cdot\mathbf b_1=76$$

$$\Gamma_{32}=\mathbf b_3\cdot\mathbf b_2=98$$

$$\Gamma_{33}=\mathbf b_2\cdot\mathbf b_3=49+64+25=128$$

Thus

$$\boldsymbol\Gamma=\left[\begin{matrix}38&48&76\\48&77&98\\76&98&128\end{matrix}\right]$$

And

$$\boldsymbol\Gamma^{-1}=\left[\begin{matrix}-0.017&-0.086&0.076\\-0.086&0.060&0.005\\0.076&0.005&0.041\end{matrix}\right]$$

Also we can see that

$$\displaystyle\mathbf b_1\cdot\mathbf v=-10+8-15=-17$$

$$\displaystyle\mathbf b_2\cdot\mathbf v=-8+20-30=-18$$

$$\displaystyle\mathbf b_3\cdot\mathbf v=-14+32-25=-7$$

So then

$$\displaystyle\mathbf b_i\cdot\mathbf v=\left[\begin{matrix}-17\\-18\\-7\end{matrix}\right]$$

Therefore by matrix multiplication

$$\displaystyle\Gamma^{-1}\{\mathbf b_i\cdot\mathbf v\}=\left[\begin{matrix}1.305\\0.347\\-1.095\end{matrix}\right]$$

This implies our solution is

= References for Report 7 = Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 38) [[media:pea1.f11.mtg38.djvu|Mtg 38]] 10 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 39) [[media:pea1.f11.mtg39.djvu|Mtg 39]] 15 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 40) [[media:pea1.f11.mtg40.djvu|Mtg 40]] 17 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 41) [[media:pea1.f11.mtg41.djvu|Mtg 41]] 17 Nov 2011. http://en.wikipedia.org/wiki/Elliptic_coordinates http://en.wikipedia.org/wiki/Parabolic_coordinates http://en.wikipedia.org/wiki/Spherical_coordinates http://math.arizona.edu/~goriely/M322/PDE_Handout_1x2.pdf Stevens, David E, and Masoud Olia. "Fundamentals of Engineering Exam, 2nd Ed." Hauppauge, NY: Barron's Educational Series, Inc., 2008, p. 27. Print.

= Team Work Distribution =

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=Book Derivation of Quadratic Shape Functions (p. 82-83)= The book derives the quadratic shape functions with an inverse manipulation of the matrix of polynomials. $$ \mathbf{d^e} = \begin{bmatrix} \theta_1^e \\ \theta_2^e \\ \theta_3^e \end{bmatrix} $$ $$ \mathbf{M^e} = \begin{bmatrix} 1 & x_1^e & x_1^{e^2} \\ 1 & x_2^e & x_2^{e^2} \\ 1 & x_3^e & x_3^{e^2} \end{bmatrix} $$ $$ \mathbf{\alpha^e} = \begin{bmatrix} \alpha_0^e \\ \alpha_1^e \\ \alpha_2^e \end{bmatrix} $$ $$\theta^e = \mathbf{p}(\mathbf{M^e}^{-1} = \mathbf{N^e}\mathbf{d^e} = \sum_{I=1}^{n_{en}} N_I^e(x)\theta_I^e$$ The derivation of the quadratic shape function leads to this initial solution. $$\mathbf{N^e} = \frac{2} \begin{bmatrix} (x-x_2^e)(x-x_3^e) -2(x-x_1^e)(x-x_3^e) (x-x_1^e)(x-x_2^e) \end{bmatrix} $$ In order for the 1st shape function to vanish at nodes 2 and 3 but have a value of 1 at node 1, then the following derivation applies. $$N_1^e(x) = \frac{(x-x_2^e)(x-x_3^e)}{(x_1^e-x_2^e)(x_1^e-x_3^e)}$$ Similarly, the 2nd and 3rd shape functions follow similar derivations. $$N_2^e(x) = \frac{(x-x_1^e)(x-x_3^e)}{(x_2^e-x_1^e)(x_2^e-x_3^e)}$$ $$N_3^e(x) = \frac{(x-x_1^e)(x-x_2^e)}{(x_3^e-x_1^e)(x_3^e-x_2^e)}$$

=EML 5526 In-Class Derivation - Lecture 8=