text:
User:Egm6322.s09.mafia/HW4

 My second wave of comments is color coded in green. Egm6322.s09 22:27, 2 April 2009 (UTC)

 See my comments below.

After you made a correction for a section with a comment box, you want to put a comment in that same comment box on what you did. Egm6322.s09 13:49, 15 March 2009 (UTC)

=The Classification of Second Order Partial Differential Equations=

Classifications
In the general form of PDE:

$$au_{xx}+2bu_{xy}+cu_{yy}+du_{x}+eu_{y}+fu+g=0$$

it can be classified by the value of $$ac-b^2$$.

$$ac-b^2=det \underline A$$

$$ac-b^2<0 \Rightarrow$$ PDE is hyperbolic.

$$ac-b^2=0\Rightarrow$$ PDE is parabolic.

$$ac-b^2>0\Rightarrow$$ PDE is elliptic.

For example:

In the equation:

$$u_{xx}+u_{yy}=0 $$...(1)

$$\Rightarrow a=1,b=0,c=1$$

$$\Rightarrow ac-b^2=1>0$$

$$\Rightarrow$$ This PDE is elliptic

Another way to find the type of equation (1):

$$\underline \bar A:=\underline J \underline A \underline J^T =\begin{bmatrix} A     & B      \\ B & C \end{bmatrix} =\begin{bmatrix} \bar a     & \bar b      \\ \bar b & \bar c \end{bmatrix} \to$$ preferred form

$$det \underline \bar A=\bar a \bar c-\bar b^2$$

$$\bar a=1; \bar b=0; \bar c=\frac {1}{r^2} (r\ne0)$$

$$\Rightarrow$$The equation (1) is an elliptic PDE.

Relationship Between Classifications and Transformations
Observations from the verification of PDE classifications:

1. Diffusion operator remains elliptic in polar coordinates.

Question:How about a different transformation of coordinate? Would classification remain the same?

2. Does classification make sense if it changes under transformation of coordinate?

The answer is no, because physics (e.g. distribution of temperature as a result of solution of heat equation) must remain the same regardless of how heat equation was solved (under different coordinate system).

Therefore, classification better remains the same under different coordinate system for it to make sense.

Egm6322.s09.three.liu 16:35, 24 April 2009 (UTC)

=The Laplace Equation= Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

The Laplace equation is the heat conduction equation with constant thermal conductivity and no heat generation.

The Laplace equation, symbolically:

$$div(grad \ u)=\triangledown \cdot(\triangledown u)=\triangledown^{2}u$$

Taking the Laplace equation in polar coordinates gives the following.

$$div(grad \ u)=0=u_{rr}+\frac {1} {r} u_{r}+ \frac {1} {r^{2}} u_{\theta \theta}$$

Features of an Axisymmetric Problem
Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

For an axis-symmetric problem, the theta terms drop out.

$$u_{\theta}=u_{\theta\theta}=\cdots =0$$

The equation then reduces to,

$$u_{rr}+\frac {1} {r} u_{r}=0$$

which is an ordinary differential equation (ODE).

In order to solve this ODE one should note that it may be rearranged.

$$u_{rr}+\frac {1} {r} u_{r}=\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr})=0$$

Separating variables and integrating gives the solution:

$$u(r)=A_{0}ln \ r+B_{0}$$

where Ao and Bo are constants.

It should be noted that if the domain $$\omega$$ includes the origin (where r=0) then Ao must be zero for a finite solution.

Separation of Variables
Egm6322.s09.Three.ge 17:58, 24 April 2009 (UTC)

If the problem is not axis-symmetric, the Laplace equation may be solved using separation of variables.

Multiplying the Laplace equation by r2 gives:

$$\begin{matrix} r^{2} \cdot\left \{u_{rr}+\frac {1} {r} u_{r}+ \frac {1} {r^{2}} u_{\theta \theta}=0 \right \}\\

\\ =r^{2}(u_{rr}+\frac {1} {r} u_{r}) +u_{\theta\theta} \end{matrix}$$ ... (a)

Observing that the equation has a portion that depends on r only and a part that depends on theta only, one may assume a solution of the form:

$$u(r,\theta)=F(r)\cdot G(\theta)$$

Thus the solution is a product of 2 functions: one which depends only on r [$$F(r)$$] and the other which only depends on theta [$$G(\theta)$$].

Plugging in this solution into (a) produces:

$$r^{2}G(\theta)\left [\frac{d^{2}F(r)}{dr^{2}}+ \frac{1}{r}\frac{dF(r)}{dr} \right ]+F(r)\frac{d^{2}G}{d\theta^{2}}=0$$

Dividing by $$F(r)G(\theta)$$ and rearranging gives:

$$\frac{1}{F(r)}\left (r^{2}\frac{d^{2}F(r)}{dr^{2}}+ r\frac{dF(r)}{dr} \right )= \frac{-1}{G(\theta)}\frac{d^{2}G}{d\theta^{2}}=n^{2}$$

 Move the figure to the left (instead of displaying it on the right) so to avoid blocking the hide/show link to open up the collapsible box. It is not possible to open up this collapsible box. Egm6322.s09 22:38, 2 April 2009 (UTC)

 Figures moved to left. Egm6322.s09.bit.la 22:38, 2 April 2009 (UTC) And results in the following solution for $$n\neq 0$$.

$$\begin{matrix} F(r)=Ar^{n}+\frac{B}{r^{n}}\\ \\ G(\theta)=C cos(n\theta)+D sin(n\theta) \end{matrix}$$

If n=0:

$$\begin{matrix} F(r)=A_{0}ln \ r +B_{0}\\ \\ G(\theta)=C_{0}\theta+D_{0} \end{matrix}$$

In general, we want the solution to be periodic, such that:

$$ \begin{matrix} k=interger=1,2,3,\cdots\\ u(r,\theta+k2\pi)=u(r,\theta) \end{matrix}$$

The general form of the Laplace equation in polar coordinates takes the form that follows.

$$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$$

Another Axisymmetric Problem
Homework:Derive LP p.14 (1.2.13) --EGM6322.S09.TIAN 17:20, 24 April 2009 (UTC)

$$\begin{matrix} \overline{a}=a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2\\ \\ \overline{b}=a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y \\ \\ \overline{c}=a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2 \end{matrix}$$

Substituting into the equation yields:

$$\overline{ac}-\overline{b}^2 =(a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2)(a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2)$$ - $$[a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y ]^2 $$

Multiplying out the equations and rearranging terms gives:

=$$a^2\phi_x^2\psi_y^2 + 2ab\phi_x^2\psi_x\psi_y  + 2bc\phi_y^2\psi_x\psi_y  +  2bc\phi_x\phi_y\psi_y^2  +   c^2\phi_y^2\psi_y^2  +  2ab\phi_x\phi_y\psi_y^2  +  ac\phi_y^2\psi_y^2  +    4b^2\phi_x\phi_y\psi_x\psi_y  +    ac\phi_x^2\psi_y^2$$

-

$$a^2\phi_x^2\psi_x^2 +  2ab\phi_x^2\psi_x\psi_y  +    2bc\phi_y^2\psi_x\psi_y  +  2bc\phi_x\phi_y\psi_y^2  +  c^2\phi_y^2\psi_y^2  +  b^2\phi_x^2\psi_y^2  +  b^2\phi_y^2\psi_x^2  +  2b^2\phi_x\phi_y\psi_x\psi_y   +    2ab\phi_x\phi_y\psi_x^2  +  2ac\phi_x\phi_y\psi_x\psi_y $$

One can see clearly that the first five terms cancel. This leaves the following equation:

$$\begin{matrix}2ab\phi_x\phi_y\psi_y^2 +  ac\phi_y^2\psi_y^2  +    4b^2\phi_x\phi_y\psi_x\psi_y  +    ac\phi_x^2\psi_y^2\\ -\\ b^2\phi_x^2\psi_y^2 +  b^2\phi_y^2\psi_x^2  +  2b^2\phi_x\phi_y\psi_x\psi_y   +    2ab\phi_x\phi_y\psi_x^2  +  2ac\phi_x\phi_y\psi_x\psi_y \end{matrix}$$

Which, after some manipulation, results in:

$$(ac-b^2)(\phi_x \psi_y - \phi_y \psi_x)^2$$

 What is "LP"? It is not clear how you arrived at the result, which is also wrong. Need more explicit explanation of the derivation, i.e., provide intermediate steps. I also mentioned to use our notation, not the notation by "LP". Egm6322.s09 15:08, 15 March 2009 (UTC)

 My comment above had not been addressed; I did go over the above comment in class. Please take action. Egm6322.s09 22:38, 2 April 2009 (UTC) Comment was addressed, but content was deleted. It has been re-posted, now with the correct result. Egm6322.s09.Three.ge 21:02, 6 April 2009 (UTC)

$$\mathbf{a}\mathbf{c}-\mathbf{b}^2=\left (ac-b^2 \right )\left (\Phi _{x} \psi_{y}-\Phi _{y} \psi_{x}\right )^2$$ page 14.2

Application:

Example:

$$T^{*}(\Theta )=T_0\left (1+cos^2\Theta \right )$$ where T0 is a constant

$$T^{*}(\Theta =0)=2T_0=T^{*}\left (\Theta =2\pi \right )$$

$$T^{*}(\Theta =\frac{\pi}{2} )=T_0=T^{*}(\Theta =\frac{3\pi}{2})$$

$$T^{*}(\Theta)=\frac{3T_0}{2}+\frac{T_0}{2}cos2\Theta$$

Homework:Expand $$\; T^* $$ in terms of $$cos \theta$$ --EGM6322.S09.TIAN 17:23, 24 April 2009 (UTC)

$$T^* (\theta)= \frac {3T_o}{2} + \frac {T_o}{2} cos{2 \theta}$$

Expand it:

$$T^* (\theta)= \frac {3T_0}{2} + \frac {T_0}{2} (2 cos^2 \theta -1)$$ $$=T_0+T_0 cos^2 \theta$$

Principle of Superposition
$$soln=solnT^{*}= \frac{3T_0}{2}=T_{1}^{*}(\Theta )$$

$$+solnT^{*}=\frac{T_0}{2}cos2\Theta=T_{2}^{*}(\Theta )$$

Homework: Prove Law of Superposition is valid

--Egm6322.s09.xyz 16:29, 4 April 2009 (UTC)

The governing PDE is given as $$div\left( gradT\right) = 0$$. This PDE is linear and therefore the solution can be expressed using the Principle of Superposition:

i.e the solution = solution for $$T^{*}\left(\theta\right) = \frac{3T_o}{2} $$ = constant $$+$$ solution for $$T^{*}\left(\theta\right) = \frac{T_o}{2}cos2\theta$$

The proof of the linearity of the $$grad(\cdot)$$ and $$div(\cdot)$$ operators was presented by Team Mafia in R2. For completeness, the relevant portions of the proof are presented again below:

note: $$grad(\cdot)$$ is linear $$grad(u) = \frac {\partial u}{\partial x_i} e_i$$ and $$\frac {\partial u}{\partial x_i} (\cdot)$$ is linear $$\therefore$$ $$grad \left( \alpha u + \beta v \right) = \alpha grad(u) + \beta grad(v)$$

note: $$div(\cdot)$$ is linear because it is another differential operator. Let $$\bar{a}, \bar{b}: \Omega$$ $$\mathbb{R}^3$$ and $$\alpha, \beta \in \mathbb{R}$$ $$\therefore$$ $$div \left( \alpha \bar{a} + \beta \bar{b} \right) = \frac {\partial }{\partial x_i} \left( \alpha a_i + \beta b_i \right) = \alpha \frac {\partial a_i}{\partial x_i} + \beta \frac {\partial b_i}{\partial x_i}$$

The proof of linearity of each operator within the PDE yields the conclusion that the entire PDE is also linear. The Principle of Superposition is applicable to linear PDEs. Applying the Principle of Superposition, the original PDE can be split into two separate parts such that the temperature is

$$T(r,\theta) = T_1(r,\theta)+T_2(r,\theta)$$

The solutions for $$T_1$$ and $$T_2$$ constitutes two separate problems that satisfy the following:

$$div(grad T_1) = 0$$ such that $$T_1(r=a, \theta) = T_1^{*}(\theta)$$

and

$$div(grad T_2) = 0$$ such that $$T_2(r=a, \theta) = T_2^{*}(\theta)$$

$$T(r,\Theta)=T_{1}\left (r,\Theta \right )+T_{2}\left (r,\Theta \right )$$

$$T^{*}\left (\Theta \right )=T_{1}^{*}\left (\Theta \right )+T_{2}^{*}\left (\Theta \right )$$

Problem P:

PDE:

$$\operatorname{Div}( grad T )=0$$

General Solution: Equation 2 p. 20-4.

$$T\left (r=a,\Theta \right )=T^{*}(\Theta )$$

Superposition: $$P=P_{1}+ P_{2}$$

Prob P1: $$\operatorname{Div}( grad T_1 )=0$$

such that:

$$T_{1}(r=a,\Theta)=T_{1}^{*}(\Theta)$$

Prob P2: $$\operatorname{Div}( grad T_2 )=0$$

such that:

$$T_{2}(r=a,\Theta)=T_{2}^{*}(\Theta)$$

Homework: Verification of T1 solution

--Egm6322.s09.xyz 16:33, 4 April 2009 (UTC)

$$T_1$$ represents the solution to the first portion of the temperature profile. It simply states that the temperature is constant. The general form of the solution to the heat equation is given as (see ):

$$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$$

In order for this general form to converge to $$T_1(r,\theta) = \frac{3T_o}{2}$$, all the coefficients need to be equal to zero. i.e. $$A_o = A_n = B_n = C_n = D_n = 0$$

$$T_{1}(r,\Theta)=\frac{3T_0}{2}$$

For the Problem $$T_{2}:T_{2}(a,\Theta)=T_{2}^{*}(\Theta)=\frac{T_0}{2}cos2\Theta$$

$$T_{2}(r,\Theta)=\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right \}$$

Homework: Verification of T2 solution --Egm6322.s09.xyz 17:23, 24 April 2009 (UTC)

For problem P2: $$T_2\left( a, \theta \right) = T_2^{*}\left( \theta \right) = \frac{T_o}{2}cos\left(2\theta\right)$$

$$\blacktriangleright A_o = 0$$ because this is an axisymmetric problem with the origin ( r = 0 ) within the domain. (see Axisymmetric Problems)

$$\blacktriangleright C_n = D_n = 0$$. By inspection of the general form (see above) for the solution to these types of problems, the solution needs to converge to be a function of $$cos\left( 2\theta \right)$$ for $$\theta = \frac{\pi}{2},\frac{3\pi}{2}$$. For these boundary conditions $$sin\left(n \frac{\pi}{2}\right) \not= 0$$ and $$ sin\left(n \frac{3\pi}{2}\right) \not= 0$$, therefore the coefficients of these terms must be equal to zero.

The expression for $$T_2$$ is then $$T_2 \left( r, \theta \right)= \sum_{n=1}^\infty r^n \left[ A_n cos\left( n\theta\right) + B_n sin\left( n\theta\right) \right] $$

Homework: Verification of A & B coefficients for boundary condition at r = a --Egm6322.s09.xyz 16:33, 4 April 2009 (UTC)

At the boundary (r = a), the expression for $$T_2$$ is re-written as:

$$T_2\left(r=a, \theta \right)= \sum_{n=1}^\infty a^n \left[ A_n cos\left( n\theta\right) + B_n sin\left( n\theta\right) \right] $$

such that $$ T_2\left(r=a, \theta \right)= T_2^{*}\left( \theta \right) = \frac{T_o}{2}cos2\theta$$

The resulting temperature profile is a function of $$cos\left( 2\theta\right)$$ only. Upon inspection, the $$sin\left( n\theta\right)$$ term should be forced to go to zero. Otherwise the equality would not be satisfied. Therefore, the coefficient $$B_n = 0$$ for all $$n=1,2,...,\infty$$

Substitution of r = a into the equal for $$T_2$$, yields:

$$a^n \left[ A_n cos \left(n\theta \right)\right] = \frac{T_o}{2}cos2\theta$$

$$\blacktriangleright$$For the case where$$n=2$$: $$a^2 A_2 cos2\theta = \frac{T_o}{2}cos2\theta$$ $$\therefore A_2 = \frac{T_o}{2a^2}$$

$$\blacktriangleright$$For the case where$$n\not=2$$: The equality DOES NOT hold for all values where $$n\not=2$$. Taking the case where $$n=1$$ as an illustrative example, the resulting expression would be:

$$aA_1 cos\theta \not= \frac{T_o}{2}cos2\theta$$

This illustrates that for all $$n\not=2, cosn\theta \not= cos2\theta$$ $$\therefore A_{n\not=2}=0$$

Final Solution:

$$T(r,\Theta)=T_0\left [\frac{3}{2}+\frac{1}{2}\left (\frac{r}{a^2} \right )cos2\Theta \right ]$$

In general, for arbitrary function $$T^{*}\left (\Theta \right )$$ but periodic

i.e. $$T^{*}\left (\Theta +\mathit{K}2\pi \right )=T^{*}(\Theta)$$

for all $$\Theta$$ and any K = constant

this is not periodic (not acceptable)

Homework: Verification of the General Solution

The general solution of the Laplace equation:

$$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$$

Where the partial differential equation governing the problem is:

$$ \nabla ^2T=\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \Theta^2}=0$$

Applying the boundary conditions explained in the previous homework

$$T_{1}(r,\Theta)=\frac{3T_0}{2};T_{2}(r,\Theta)=\frac{T_0}{2}cos2\Theta;A_0=0;C_n=0;D_n=0$$

The resulting solution is:

$$T(r,\Theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right\}$$

$$T(r,\Theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_ncosn\Theta +B_nsinn\Theta \right\}$$

where: rn=an

$$T(a,\Theta)=C_{0}+\sum_{n=1}^{\infty }a^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$$

Fourier Coefficients
Egm6322.s09.Three.nav 13:39, 24 April 2009 (UTC)

How do we derive the fourier coeffcients C0, An and Bn?

An orthogonal basis: {1, cosm$$\theta$$, sinm$$\theta$$} is used.

As defined above, two functions f($$\theta$$), g($$\theta$$) are orthogonal if $$\int_{0}^{2\pi}f(\theta).g(\theta) d\theta= 0$$

$$\blacktriangleright$$Eg. Consider f($$\theta$$)= cos ($$m\theta$$) and g($$\theta$$)= cos($$n\theta$$)

$$ \int_{0}^{2\pi} cos(m\theta).cos(n\theta)d\theta= \int_{0}^{2\pi} \frac{1}{2}\left (cos \left ((m+n)\theta \right)+cos \left((m-n)\theta \right) \right)d\theta $$

$$ = \left [ \left (\frac{1}{2(m+n)} (sin \left ((m+n)\theta \right) \right)+ \left (\frac{1}{2(m-n)} sin \left((m-n)\theta \right) \right) \right]_{0}^{2\pi}$$

$$ = \begin{Bmatrix} 0, if\ m\ \neq n\\ 2\pi, if\ m\ = n

\end{Bmatrix}$$

For more details, here is a link that explains the math in greater detail. Similarly using '1' as the basis function and simplifying, we get $$\int_{0}^{2\pi}cos\theta\left \{1 \right \} d\theta \ or\ \int_{0}^{2\pi}sin\theta \left \{1 \right \} d\theta = 0 $$

Using this knowledge, we proceed to determine the fourier coefficients C0, An and Bn, using the orthogonal basis {1, cosm$$\theta$$, sinm$$\theta$$}

Consider the equation $$T(r,\theta)=C_{0}+\sum_{n=1}^{\infty }r^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$$-(1)

Say we have the Boundary condition T(r=a, $$\theta$$)= ($$ T^{*}\left(\theta \right)$$)

Then, $$T^{*}\left(\theta \right) = C_{0}+\sum_{n=1}^{\infty }a^{n} \left \{A_{n}cosn\Theta +B_{n}sinn\Theta \right\}$$(2)

$$\blacktriangleright$$At n=0, $$T^{*}\left(\theta \right) = C_{0}$$

Multiplying through by {1} and integrating over [0,2$$\pi$$]

$$\Rightarrow \int_{0}^{2\pi}T^{*}\left(\theta \right) d\theta = \int_{0}^{2\pi} C_{0} d\theta$$

$$\Rightarrow \int_{0}^{2\pi}T^{*}\left(\theta \right) d\theta = C_{0} \left (\theta \right)_{0}^{2\pi}$$

$$\Rightarrow C_{0}= \frac {1}{2\pi} \int_{0}^{2\pi}T^{*}\left(\theta \right) d\theta $$

$$\blacktriangleright$$Solving for An in Equation (2), multiply through by cos(m$$\theta$$) and integrate over [0,2$$\pi$$].

Evaluating each term in the resultant equation,

Term on the LHS= \int_{0}^{2\pi}T^{*}\left(\theta \right)cosm\theta d\theta

First term on RHS= $$\int_{0}^{2\pi}C_{0}cosm\theta d\theta $$

$$= C_{0} \int_{0}^{2\pi} \left \{1 \right \} cos m\theta d\theta $$

= 0, by definition of orthogonality

Second term on RHS= $$ \int_{0}^{2\pi} \sum_{n=1}^{\infty}a^{n}A_{n}cosn\theta cosm\theta d\theta $$

$$ = \sum_{n=1}^{\infty} \int_{0}^{2\pi}a^{n}A_{n}cosn\theta cosm\theta d\theta $$

$$ = \begin{Bmatrix} 0, if\ m\ \neq n\\ a^{n}\times A_{n} \times 2\pi, if\ m\ = n

\end{Bmatrix}$$ (by definition of orthogonality)

$$\Rightarrow \int_{0}^{2\pi} \sum_{n=1}^{\infty}a^{n}A_{n}cosn\theta cosm\theta d\theta = a^{n}\times A_{n} \times 2\pi $$

Hence it is seen that though it was assumed m $$\epsilon$$ $$ \Re$$, simplification of the second term in (2) determines that m=n.

Third term on RHS= $$ \int_{0}^{2\pi} \sum_{n=1}^{\infty}a^{n}A_{n}sinn\theta cosm\theta d\theta $$

= 0, by definition of orthogonality

$$\therefore $$ (2) multiplied through by cos($$m\theta$$) or cos($$n\theta$$) and integrated over [0,2$$\pi$$] reduces to

$$\int_{0}^{2\pi}T^{*}\left(\theta \right)cosn\theta d\theta = a^{n}\times A_{n} \times 2\pi$$

$$\Rightarrow A_{n}= \frac {1}{a^{n}\times 2\pi} \int_{0}^{2\pi}T^{*}\left(\theta \right)cosn\theta d\theta $$

We solve similarly for Bn in (2). Multipling through with sin($$m\theta$$) and integrating over [0,2$$\pi$$], it is seen that the first and second terms = 0, by definition. Third term reduces to $$b^{n}\times B_{n} \times 2\pi$$. Then the modified Equation(2) becomes

$$\int_{0}^{2\pi}T^{*}\left(\theta \right)sin n\theta d\theta = a^{n}\times B_{n} \times 2\pi$$

$$\Rightarrow B_{n}= \frac {1}{b^{n}\times 2\pi} \int_{0}^{2\pi}T^{*}\left(\theta \right)sin n\theta d\theta $$

Find the Fourier coefficients C0,An,Bn

$$C_{0}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )d\theta $$

$$A_{n}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )cosn\Theta d\theta $$

$$B_{n}=\frac{1}{2\pi }\int_{\Theta =0}^{2\pi }T^{*}(\theta )sinn\Theta d\theta $$

Due to orthogonality of Fourier basis function $$\left \{1,cosu\Theta,sinu\Theta \right \}$$

 But you were asked to derive the above Fourier coefficients (and therefore the orthogonality property of the Fourier basis functions). Egm6322.s09 20:11, 15 March 2009 (UTC)

Edits to this section were made my Navya on March 27th. Either due to the work of a vandal, or a failure of wikiversity, they were removed.

They are now being placed in by Andrew Lapetina --Egm6322.s09.lapetina 20:56, 27 March 2009 (UTC)

Joseph Fourier 1768-1830

Jean Baptiste Joseph Fourier was a French mathematician and physicist best known for first devising Fourier Series and their applications to heat flow problems. He was born in Auxerre and orphaned at the age of 9. He became a chair at the École Polytechnique at the height of his career. Fourier died in Paris. More information is available here

Homework:Proof of Non Linearity

To show that $$\kappa (u)grad(u)+f(x,y)=0$$ is non linear

Let,

$$L\left( \right):$$ be an operator,such that $$L\left(\ u\right)$$ is linear with respect to $$u$$ if,

$$ L\left(\alpha u+\beta v\right)=\alpha L\left(\ u\right)+\beta L\left(\ v\right)$$

Therefore, in the present problem ,assuming 2D, we have,

$$L\left( \right)=\kappa \left(\overline i\frac{\partial\left( \right) }{\partial x}+\overline j\frac{\partial\left( \right) }{\partial y}\right)+f(x,y)$$

$$L\left(\alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left(\overline i\frac{\partial\left(\alpha u+\beta v \right) }{\partial x}+\overline j\frac{\partial\left(\alpha u+\beta v \right) }{\partial y}\right)+f(x,y)$$

$$\Rightarrow L\left( \alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left(\overline i\alpha\frac{\partial\left(\ u\right) }{\partial x}+\overline i\beta\frac{\partial\left(\ v\right) }{\partial x}+\overline j\alpha\frac{\partial\left(\ u\right) }{\partial y}+\overline j\beta\frac{\partial\left(\ v\right) }{\partial y}\right)+f(x,y)$$

$$\Rightarrow L\left(\alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left[\alpha\left\{\overline i\frac{\partial\left(\ u \right) }{\partial x}+\overline j\frac{\partial\left(\ u \right) }{\partial y} \right\}+\beta\left\{\overline i\frac{\partial\left(\ v\right) }{\partial x}+\overline j\frac{\partial\left(\ v\right) }{\partial y} \right\}\right]+f(x,y)$$

\therefore,we can see that

$$\Rightarrow L\left(\alpha u+\beta v\right)\neq\alpha L\left(\ u\right)+\beta L\left(\ v\right)$$

$$ \therefore \kappa (u)grad (u)+f(x,y)$$ is non linear

Egm6322.s09.bit.gk 20:41, 24 April 2009 (UTC)

Homework:Problem 5.11

Egm6322.s09.bit.gk 20:56, 24 April 2009 (UTC)]]

We can see from the free body diagram ,equating the horizontal forces ,we have

$$ -T_0cos(\alpha)dy+T_0cos(\alpha+d\alpha)dy-T_0cos(\beta)dx+T_0cos(\beta+d\beta)dx=0 $$

and equating the vertical forces, we have,

$$ -T_0sin(\alpha)dy+T_0sin(\alpha+d\alpha)dy-T_0sin(\beta)dx+T_0sin(\beta+d\beta)dx=P(x,y)dxdy $$

where $$P(x,y)$$ is the transverse load acting on the membrane

$$\Rightarrow T_0\left[\left (sin(\alpha+d\alpha) -sin(\alpha)\right )dy+\left (sin(\beta+d\beta)-sin(\beta)\right )dx\right]+P(x,y)dxdy=0$$

When $$\theta$$ is small ,we can assume, $$ sin(\theta)=Tan(\theta)$$

$$\therefore$$ the above equation becomes,

$$T_0\left[\left (tan(\alpha+d\alpha)-tan(\alpha)\right )dy+\left (tan(\beta+d\beta)-tan(\beta)\right )dx\right]+P(x,y)dxdy=0 $$

Let this equation be $$(1)$$

but,$$tan(\alpha)$$ is the slope the membrane with respect to the $$x$$ and $$y$$ axes

where the displacement is given as $$w=w(x,y)$$

$$\therefore$$

$$tan(\alpha)=\left[\frac{\partial w}{\partial x}\right]_{x,y}$$ and $$ tan(\alpha+d\alpha)=\left[\frac{\partial w}{\partial x}\right]_{x+dx,y}$$

$$tan(\beta)=\left[\frac{\partial w}{\partial y}\right]_{x,y}$$ and $$ tan(\beta+d\beta)=\left[\frac{\partial w}{\partial y}\right]_{x,y+dy}$$

Substituting the slopes in $$(1)$$,we have ,

$$T_0\left[\left(\left[\frac{\partial w}{\partial x}\right]_{x+dx,y}-\left[\frac{\partial w}{\partial x}\right]_{x,y}\right)dy+\left(\left[\frac{\partial w}{\partial y}\right]_{x,y+dy}-\left[\frac{\partial w}{\partial y}\right]_{x,y}\right)dx\right]+P(x,y)dxdy=0 $$

Let this be equation $$(2)$$

We have Taylor series expansion as ,

$$f(x+dx,y)=f(x,y)+\frac{\partial f}{\partial x}dx+...$$

substituting taylor series expansion in equation $$(2)$$,we have

$$T_0\left(\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}\right)+P(x,y)=0$$

$$ \Rightarrow T_0\triangledown^{2}w(x,y)+P(x,y)=0 $$

Egm6322.s09.bit.gk 20:39, 24 April 2009 (UTC)

Orthogonal Functions --Egm6322.s09.xyz 16:34, 4 April 2009 (UTC)

The following is the definition of Orthogonal Functions as presented on the Wolfram MathWorld website :

"Two functions $$f(x)$$ and $$g(x)$$ are orthogonal over the inverval $$a\le x\le b$$ with weighting function $$w(x)$$ if

$$\left \langle f(x)|g(x)\right \rangle \equiv \int_{a}^{b} f(x)g(x)w(x)\,dx = 0$$

If, in addition,

$$\int_{a}^{b} f(x)^{2}w(x)\,dx = 1$$

and

$$\int_{a}^{b} g(x)^{2}w(x)\,dx = 1$$

the functions $$f(x)$$ and $$g(x)$$ are said to be orthonormal"

Additional information can be found at Wiki Orthogonality and Dictionary.com

A Third Example of the Laplace Equation
Egm6322.s09.bit.sahin 16:31, 24 April 2009 (UTC)

A domain which is a quadrant of annulus is subjected a boundary conditions that

temperature at $$r=b$$ is $$T\left (r=b,\theta \right )=T_{b}cos4\theta $$

temperature at $$r=a$$ is $$T\left (r=a,\theta \right )=T_{a}cos4\theta$$

where $$T_{a}$$ and $$T_{b}$$ are given constants. Also, the boundaries at $$\theta=0$$ and $$\theta =\pi/2$$ are insulated which means no heat flow at these boundaries. According to the Fourier's law:

$$\underline{q}=\underline{\kappa } \cdot gradT$$

here $$\underline{q}$$ denotes heat flux tensor. Relevant to the insulated conditions,

$$\underline{q}=0\Rightarrow gradT=0\Leftrightarrow \frac{\partial T}{\partial \theta }=0$$ on $$\theta=0,\theta =\pi/2 $$.

Homework: Verification of Insulation

The boundaries are kept insulated which means that there is no heat flow at $$\theta= 0$$ and $$\theta= \pi/2$$. So,

$$grad T=0$$

and

$$grad T= \frac{\partial T}{\partial r}\mathbf{e_{r}}+\frac{1}{r}\frac{\partial T}{\partial \theta}\mathbf{e_{\theta}}=0$$

Since $$\frac{\partial T}{\partial r}=0$$ at insulated surfaces we have

$$\frac{1}{r}\frac{\partial T}{\partial \theta}\mathbf{e_{\theta}}=0$$

So, we obtain that

$$\frac{\partial T}{\partial \theta}=0$$ at $$\theta= 0$$ and $$\theta= \pi/2$$

 How so? In general, express $$\displaystyle {\rm grad} \, T$$ in polar coordinates then deduce $$\displaystyle \partial T / \partial \theta = 0$$. Such approach is important when the insulated boundaries do not coincide with the $$\displaystyle (x,y)$$ Egm6322.s09 20:11, 15 March 2009 (UTC)

Necessary changes were made Egm6322.s09.bit.sahin 16:16, 10 April 2009 (UTC)

The general solution of the Laplace Eq. is

$$T\left (r,\theta \right )=A_{0}lnr+\sum_{n=1}^{\infty }r^{n}\left (A_{n}cosn\theta +B_{n}sinn\theta  \right )+  \sum_{n=1}^{\infty }\frac{1}{r^{n}}\left (C_{n}cosn\theta +D_{n}sinn\theta  \right )+C_{0}$$

Eliminating Terms Based on BC
We can eliminate the following terms that do not satisfy the boundary conditions:

1) $$A_{0}lnr+C_{0}$$, indipendent of $$\theta$$

2) $$B_{n}sinn\theta$$, $$D_{n}sinn\theta$$, cannot satisfy the $$\frac{\partial T}{\partial \theta }\left (r,\theta =0 \right )=0$$

To show the second one, let's differentiate the general solution with respect to $$\theta$$

$$\frac{\partial T}{\partial \theta } =\sum_{n=1}^{\infty }r^{n}n\left (-A_{n}sinn\theta +B_{n}cosn\theta \right )+\sum_{n=1}^{\infty }\frac{1}{r^{n}}n\left (-C_{n}sinn\theta +D_{n}cosn\theta  \right )$$

Since $$sinn\theta=0$$ and $$cosn\theta=1$$ at $$\theta=0$$, $$B_{n}$$ and $$C_{n}$$ must be zero to satify the condition that $$\frac{\partial T}{\partial \theta }=0$$.

Thus the solution has the following form

$$T\left (r,\theta \right )=\sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{C_{n}}{r_{n}} \right )cosn\theta$$

Using the boundary conditions, we have

$$T_{a}cos4\theta =\sum_{n=1}^{\infty }\left (A_{n}a^{n}+\frac{C_{n}}{a_{n}} \right )cosn\theta$$

$$T_{b}cos4\theta =\sum_{n=1}^{\infty }\left (A_{n}b^{n}+\frac{C_{n}}{b_{n}} \right )cosn\theta$$

Since the only term in the boundary condition is the term with $$cos4\theta$$, boundary conditions can only be satisfied for n=4, all other $$A_{n} $$ and $$C_{n} $$ must be zero. Then we have,

$$\begin{bmatrix} a^{8} & 1 \\ b^{8} & 1 \end{bmatrix}\begin{Bmatrix} A_{4}\\C_{4} \end{Bmatrix}=\begin{Bmatrix} a^{4}T_{a}\\b^{4}T_{b}

\end{Bmatrix} $$

Eventually the solution for the temperature distribution is

$$T\left (r,\theta \right )=\left \{\frac{a^{4}b^{4}T_{b}}{\left (b^{8}-a^{8} \right )}\left [\frac{r^{4}}{a^{4}}-\frac{a^{4}}{r^{4}} \right ]-\frac{a^{4}b^{4}T_{a}}{b^{8}-a^{8}}\left [\frac{r^{4}}{b^{4}}-\frac{b^{4}}{r^{4}} \right ] \right \}cos4\theta$$

The figure below shows the plot of the solution for a given data:

$$a=1$$, $$b=2$$, $$T_{a}=5$$, $$T_{b}=20$$

MATLAB Code for Plots --EGM6322.S09.TIAN 17:33, 24 April 2009 (UTC)

--EGM6322.S09.TIAN 17:33, 24 April 2009 (UTC)

 You were asked to plot in both polar coordinates and in cartesian coordinates; the figure shown is a plot in cartesian coordinates; I moved this figure to the left for a better presentation. Also provide the matlab codes used to create these plots. Egm6322.s09 15:08, 15 March 2009 (UTC)

 My comment above had not been addressed in this updated version. Please take action. Egm6322.s09 22:38, 2 April 2009 (UTC)

 Now there it is. EGM6322.S09.TIAN 21:39, 9 April 2009 (UTC)

=The Power Law= --Egm6322.s09.lapetina 02:03, 17 April 2009 (UTC)

The Power Law is a very common relationship in the universe. It is defined as:

$$y=b x^a $$.

The Power Law is observed in classical physics, biology, economics, and many other natural and social sciences. The exponent $$a$$ dominates the nature of the equation. In electrostatics and gravitation, $$a=2$$, while in Stefan-Boltzmann equations, $$a=4$$. More can be found on the Power Law here.

The inverse of the exponent function is the logarithm.

Application of the Power Law
The thermal conductivity of solids is summarized in the following graph.

Plotting here is on a log-log scale. As a result, the slopes appear linear, rather than exponential.

The exponential varying thermal conductivity of solids is very important for solving Fourier's Law:

$$q=- \kappa \; grad \; T$$

where $$q$$ is the heat flux.

From this equation, we can find the units of $$\kappa$$ in the following fashion:

$$q \equiv \left [ \frac{Power}{Unit \; Area} \right ]=\left [ \frac{W}{m^2} \right ]$$ while $$grad \; T= \frac {dT}{dx} = \left [ \frac {K}{m} \right ]$$

Therefore: $$\left [ \kappa \right ]= \left [ \frac {\frac {W}{m^2}}{\frac{K}{m}}\right ] = \frac {W}{mK}$$

If we consider the thermal conductivity $$\kappa$$ as $$\kappa \left ( T \right )$$ where $$T$$ is temperature, we can find the heat flux at any given temperature using the power law by the following equation:

$$\kappa \left ( T \right )=b T^a $$.

This equation can be solved over a given domain if $$a$$ and $$b$$ are known.

Using data from the aforementioned graph, we see that for diamond, $$T \in \left [ 1K, 10.7 K \right ]$$ :

$$a=\frac {log \kappa_2 - log \kappa_1}{log T_2 - log T_1}=\frac {log (1000)-log (0.4)}{log (10.7)-log (1)} \cong 3.39 $$

while $$b \cong 0.4 \frac {W}{mK}$$

so $$\kappa (T) = (0.4) T^{3.39} \frac {W}{mK}$$.

Homework: Determining a and b for a Different Interval --Egm6322.s09.lapetina 02:04, 17 April 2009 (UTC)

For $$T \in \left [ 100 K, 1000 K \right ]$$, we want to find $$\kappa \left ( T \right ) $$. We can estimate $$a$$ for diamond using the slope of graphite parallel to layers:

$$a=\frac {log \kappa_2 - log \kappa_1}{log T_2 - log T_1}=\frac {log (3)-log (80)}{log (1000)-log (100)} \cong -1.43 $$.

Extrapolating $$b$$ backwards shows its value is $$10^7$$.

Therefore, for all $$T \in \left [ 100 K, 1000 K \right ]$$,

$$\kappa (T) = (10^7) T^{-1.43} \frac {W}{mK}$$

The Wave Equation and String Vibration
--Egm6322.s09.lapetina 02:04, 17 April 2009 (UTC)

The Wave Equation can be studied by examining the physics of string vibration in one spatial dimension. An excellent book on this topic is The Theory of Sound by Lord Rayleigh.

 I never mentioned the book by Rossing et al. I mentioned the classic The Theory of Sound by Lord Rayleigh; see the Lecture plan. Egm6322.s09 15:08, 15 March 2009 (UTC)

Correction made. --Egm6322.s09.lapetina 20:57, 27 March 2009 (UTC)

In the accompanying free-body diagram for Case 1, forces in the $$x$$ direction are :

$$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+dx,t \right ) cos \; \theta ( x+dx,t )$$

$$\theta$$ is very small here, therefore $$cos \theta \cong 1-\frac{\theta^2}{2}$$.

Inserting this into the $$x$$ equation results in:

$$\sum F_x = -\tau \left ( x,t \right ) (1-\frac{{\theta(x,t)}^2}{2})+ \tau \left ( x+dx,t \right ) (1-\frac{{\theta (x+dx,t)}^2}{2})$$.

We can neglect second order terms here, since they are very small and nearly cancel, leaving:

$$\tau (x+dx,t)=\tau (x,t)=\tau$$, suggesting $$\tau$$ is constant throughout the string.

In the $$y$$ direction:

$$\sum F_y = \tau sin \; \theta ( x,t )+ \tau sin \; \theta ( x+dx,t )+ P (x,t) dx -m(dx) \ddot w$$

where $$\ddot w={w}_{tt}$$.

Where $$\theta$$ is small, we can simplify the first two terms as:

$$\theta (x+dx,t) -\theta (x,t) \cong \frac{d\theta}{dx} dx= {(w_x)}_{x} dx= {w}_{xx}dx$$

multiplied by the constant $$\tau$$.

Therefore, for an infinitely small length of string $$dx$$,

$$\tau {w}_{xx} +P=m {w}_{tt}$$

Homework: Other Cases for String Shape

--Egm6322.s09.lapetina 02:04, 17 April 2009 (UTC)

Case 4: Vertical Reflection

This is the most trivial of the other three cases. In the $$x$$ direction, this changes only the sign on $$\theta$$. Because this value is squared, we are left with:

$$\tau {w}_{xx} +P=m {w}_{tt}$$

Case 2: Modified Geometry

For cases two and three, we have a slightly more complicated situation, as both ends of the string point down and up, respectively. This negates our ability to simplify the equation in the same way for the string as shown. Looking first at the accompanying figure, we can our case on the left side.

However, recall the string is infinitesimally small. As seen in the accompanying image, we can simply examine a fraction of the infinitesimally small string $$dx_1$$, and encounter the same geometry. In the image the sum of forces in the $$x$$ direction on the left side of the image can be expressed as:

$$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+dx,t \right ) cos \; \theta ( x+dx,t )$$

while the right side (showing a fraction of the infinitesimally small string) can be expressed as:

$$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+{dx}_{1},t \right ) cos \; \theta ( x+{dx}_{1},t )$$

This can be simplified using the assumption:

$$cos \theta \cong 1-\frac{\theta^2}{2}$$,

leaving:

$$\sum F_x = -\tau \left ( x,t \right ) (1-\frac{{\theta(x,t)}^2}{2})+ \tau \left ( x+{dx}_{1},t \right ) (1-\frac{{\theta (x+{dx}_{1},t)}^2}{2})$$.

Neglecting second order terms here, since they are very small and nearly cancel, leaves:

$$\tau (x+{dx}_{1},t)=\tau (x,t)=\tau$$, suggesting $$\tau$$ is constant throughout the string, which by definition continues to $$x=x+dx$$.

This means that even for geometries where the ends of the strings point in the same direction, $$\tau (x,t) = \tau $$.

In the $$y$$ direction, we start with:

$$\sum F_y = \tau sin \; \theta ( x,t )+ \tau sin \; \theta ( x+dx,t )+ P (x,t) dx -m(dx) \ddot w$$.

However, for a fraction of the string $$dx_1$$, (i.e., the modified version shown in the right side of the image), this equation changes to:

$$\sum F_y = \tau sin \; \theta ( x,t )- \tau sin \; \theta ( x+{dx}_1,t )+ P (x,t) dx -m(dx) \ddot w$$.

Now, where $$\theta$$ is small, we can again simplify the first two terms as:

$$\theta (x+dx,t) -\theta (x,t) \cong \frac{d\theta}{dx} dx= {(w_x)}_{x} dx= {w}_{xx}dx$$

multiplied by the constant $$\tau$$.

This leaves us with the same equation as for Cases One and Four:

$$\tau {w}_{xx} +P=m {w}_{tt}$$

Case Three: Vertical Reflection of Case Two

Our signs are merely switched again, as they were between Cases One and Four.

Essentially, no matter what the geometry of the string (so long as it is simple and does not cross, and is always differentiable), the equation of motion is:

$$\tau {w}_{xx} +P=m {w}_{tt}$$.

This is because the original derivation is for a nearly linear string, and the can always be viewed as linear, as:

$$ \delta x \rightarrow 0$$.

An alternative means of solving this problem is to view $$\theta$$ and $$\tau$$ as algebraic quantities rather than physical entities. The accompanying image shows the only free body diagram needed.

Here, we see that:

$$\sum F_x=\tau (x,t)cos \theta (x,t)+\tau (x+dx,t)cos \theta (x+dx,t)=0$$   (1) $$\sum F_y=\tau (x,t) sin \theta (x,t)+\tau (x+dx,t) sin \theta (x+dx,t) +p(x) dx - mdx {w}_{tt} =0$$   (2)

For equation 1, we can assume $$\theta$$ is small, making $$cos (\theta) \cong 1$$, leading to:

$$\tau (x+dx) =-\tau (x)={\tau}_{const} \forall x$$.

Therefore: $$\tau (x+dx)=\tau$$ $$\tau (x)=-\tau$$

This changes equation 2 to:

$$\sum F_y=-\tau sin \theta (x,t)+\tau sin \theta (x+dx,t) +p(x) dx - mdx {w}_{tt} =0$$

where, because $$\theta$$ is small:

$$\sum F_y=0= \tau \frac{\partial \theta}{\partial x} dx+p(x)dx-mdx{w}_{tt} +$$ higher order terms.

This leaves us with:

$$\tau {w}_{xx}+P=m {w}_{tt}$$

 Sign problem in $$\displaystyle \sum F_y$$; inconsistent with Taylor series expansion. The key for the derivation to be independent of any figure of free-body diagram is to treat the tension function $$\displaystyle \tau(x)$$ and the slope angle $$\displaystyle \theta(x)$$ as algebraic quantities so that $$\displaystyle \sum F_x = \tau (x) \cos(x) + \tau(x+dx) \cos (x+dx) = 0$$ and $$\displaystyle \sum F_y = \tau (x) \sin(x) + \tau(x+dx) \sin (x+dx) + p \, dx - m \, dx \, w_{tt} = 0$$. Note the plus signs in both cosine terms in $$\displaystyle \sum F_x$$, and the plus signs in both sine terms in $$\displaystyle \sum F_y$$. This systematic approach is unlike what you used to see in undergraduate courses such as statics, dynamics, etc. See further comments in class. Egm6322.s09 15:08, 15 March 2009 (UTC)

Explanation of sign switch demonstrated in figure is articulated in the text of the original problem.

Alternative solution provided. Image addition will come in near future.

--Egm6322.s09.lapetina 14:33, 16 March 2009 (UTC)

Homework: Derive the Equations of Motion for a Stretched Membrane in Cartesian Coordinates

--Egm6322.s09.lapetina 02:05, 17 April 2009 (UTC)

In two dimensions, the vertical position $$u$$ of a membrane of density $$\rho$$ with constant thickness $$h$$ (and corresponding constant strength and flexibility) and loading $$p(x,y)$$ can be described as:

$$u (x,y,t)$$, where $$x$$ and $$y$$ are Cartesian positions, and $$t$$ is time. With no loading, the membrane rests in the $$x-y$$ plane

The stress in the membrane, $$\sigma$$ is constant throughout. For all time $$t$$ and for all $$x$$ and $$y$$, the membrane surface is differentiable by $$x$$ and $$y$$, and small angle approximations can be made.

As shown in the image, we will examine a small square portion of width $$\Delta x$$ and height $$\Delta y$$.

For this small portion, There are five forces acting on it, the tensions, $$\left \{ T_1,...,T_4 \right \}$$, and the load $$P(x,y)$$. The tensions all act normal to the edge of the small portion, and at an angle to the $$x-y$$ plane, $$\left \{ {\theta}_{1},...{\theta}_{4} \right \}$$, respectively.

The tensions $$\left \{ T_1,...,T_4 \right \}$$are applied at coordinates, $$\left \{ (x+0.5 \Delta y,y_1), (x_1+\Delta x,y_1+0.5 \Delta y), (x_1+0.5 \Delta x,y_1+\Delta y),(x_1,y_1+0.5 \Delta y) \right \}$$ respectively.

Newton's Second Law in the vertical direction is then:

$$\sum F_z=T_1 sin \theta_1+T_2 sin \theta_2+T_3 sin \theta_3+T_4 sin \theta_4=\rho h \Delta x \Delta y \frac{{\partial}^2 u}{\partial t^2}+\Delta x \Delta y p(x,y)$$

Using the small angle approximation,

$$sin {\theta}_n \cong tan {\theta}_n$$

which is approximately equal to the derivative of the membrane in the direction of the tension force at each location.

Therefore:

$$sin \theta_1 \cong -u_y (x_1 +0.5 \Delta x, y_1) $$ $$sin \theta_2 \cong u_x (x_1+\Delta x,y_1+ 0.5 \Delta y)$$

$$sin \theta_3 \cong u_y (x_1 + 0.5 \Delta x, y_1+\Delta y)$$

$$sin \theta_4 \cong u_x (x_1, y+0.5 \Delta y )$$.

The magnitudes of $$\left \{ T_1,...,T_4 \right \}$$ are equivalent to the constant stress multiplied by the area over which it is applied, such that:

$$T_1=\Delta x h \sigma$$ $$T_2=\Delta y h \sigma$$ $$T_3=\Delta x h \sigma$$ $$T_4=\Delta y h \sigma$$

Let us define:

$$x_1+0.5 \Delta x= x_2$$ and $$y_1+0.5 \Delta y= y_2$$.

Substituting these values into Newton's Law and dividing by $$\Delta x$$ and $$\Delta y$$ leaves us with :

$$\rho \frac{{\partial}^2 u}{\partial t^2}+ P(x,y)= \sigma \left [ \frac {u_x (x_1+\Delta x, y_2 )- u_x (x_1,y_2)}{\Delta x} +\frac{u_y (x_2, y_1+\Delta y) - u_y (x_2, y_1)}{\Delta y} \right ] $$.

Taking the limits as $$\Delta x$$ and $$\Delta y$$ approach zero, we are left with a Poisson Equation:

$$m u_{tt} + p(x,y)= \sigma(u_{xx}+u_{yy})$$

where $$m$$ is the mass per unit area.

This is equivalent to:

$$\sigma ( div (grad \; u)) +p(x,y) = m w_{tt}$$.

For steady state, and a massless membrane, the equation becomes:

$$\sigma {\nabla}^2 u(x,y) + p(x,y)=0$$

Applications
$$\tau (div (grad \omega))+ p(x,y) = m W_tt$$

Here, $$div (grad \omega)= \omega_{xx}+\omega_{yy}$$, $$m$$ is mass/unit area.

Wave Equation In 1-D Space (actually a 2-D (x,t) Problem)
--EGM6322.S09.TIAN 17:28, 24 April 2009 (UTC)

$$\tau \omega_{xx} - m \omega_{tt} +p =0$$

$$A= \begin{bmatrix} a & b \\ b & c \end{bmatrix} $$, $$detA=ac-b^2$$

In this case, $$a= \tau >0, b=0, c=-m<0 $$

$$detA=- \tau m <0 \Rightarrow hyperbolic$$

Unsteady Heat Equation
--EGM6322.S09.TIAN 17:24, 24 April 2009 (UTC)

1-D space
$$\frac{d}{dx} (\kappa \frac{du}{dx}) + f = C \frac{du}{dt}$$

Here, $$\kappa$$ is heat conductivity, $$f$$ is heat source, $$C$$ is heat capacity.

We assume $$\kappa$$ is constant.

$$\kappa u_{xx} - C u_t +f =0$$

$$a= \kappa, b=0,  c=0 \Rightarrow  detA=0\Rightarrow parabolic$$

2-D space
$$\frac{d}{dx} (\kappa \frac{du}{dx}) + f = C \frac{du}{dt}$$

We assume $$\kappa$$ is constant here.

$$\kappa (u_{xx} + u_{yy}) - C u_t +f =0$$

General to 3 independent variables (x,y,z)

$$\big\lfloor \partial_x \; \partial_y \;\partial_z  \big\rceil \begin{bmatrix} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{Bmatrix} \partial_x u \\ \partial_y u \\ \partial_z u \end{Bmatrix} $$

Here,$$ \begin{bmatrix} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33} \end{bmatrix} = [A_{ij}]_{3 \times 3}$$

In 2-D case, $$A_{ij}=0 \; \forall \; i,j \; except \; A_{11} = A_{22}= \kappa >0 \Rightarrow detA=0 \Rightarrow parabolic$$

Solution of unsteady heat equation without heat source, which means $$(f=0)$$ in polar coordinate.
Separation of Variables.

$$\frac {1}{r} \frac {\partial} {\partial r} (r \frac {\partial u} {\partial r})$$ $$+ \frac {1}{r^2} \frac {\partial^2 u}{\partial^2 \theta}$$ $$= \frac {\partial u}{\partial t}(1)$$

Here, $$\frac {1}{r} \frac {\partial} {\partial r} (r \frac {\partial u} {\partial r})$$ $$+ \frac {1}{r^2} \frac {\partial^2 u}{\partial^2 \theta}$$ $$=div (grad u)$$

$$u(r,\theta,t)=R(r)\Theta (\theta) T(t)\;(2)$$

Plug $$(2)$$ into $$(1)$$:

$$\frac {1}{rR} \frac {\partial} {\partial r} (r \frac {dR} {dr})$$ $$+ \frac {1}{r^2 \Theta} \frac {d^2 \Theta}{d \theta^2 \theta}$$ $$- \frac {1}{T} \frac {dT}{dt}=0$$

 The solution for problem 5.12 in Selvadurai (2000) is missing. Egm6322.s09 15:15, 15 March 2009 (UTC)

START to Homework:problem 5.12 in Selvadurai (2000)

An unloaded weightless membrane roof over an annular enclosure. The outer circular boundary is r=b, and the inner circular boundary, r=a, is subject to the following displacement:

$$\Delta_{0} + \Delta_{1}sin \theta$$

where $$\Delta_{0}$$ and $$\Delta_{1}$$ are constants.

i) Formulate the Boundary value problem:

The boundary conditions are:
 * 1) $$w(r,\theta)=w(r,\theta+2\pi)$$
 * 2) $$w(b,\theta)=0$$
 * 3) $$w(a,\theta)=\Delta_{0} + \Delta_{1}sin \theta$$
 * 4) $$\frac{\partial w}{\partial r}_{r=b} =0$$

ii) Develop an expression for the membrane, given that:

$$w=A \ ln \ r +B\theta ln \ r +C\theta +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)$$

where $$A, B, C, D, A_{n}, B_{n}, C_{n}, and \ D_{n}$$ are constants.

Solution: Using boundary condition 1, the equation becomes:

$$\begin{matrix} A \ ln \ r +B\theta ln \ r +C\theta +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)\\

=\\ A \ ln \ r +B(\theta+2\pi) ln \ r +C(\theta+2\pi) +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n(\theta+2\pi) +D_{n}cos \ n(\theta+2\pi)) \end{matrix}$$

Clearly,

$$(C_{n}sin \ n\theta +D_{n}cos \ n\theta)=(C_{n}sin \ n(\theta+2\pi) +D_{n}cos \ n(\theta+2\pi))$$

and,

$$\begin{matrix} A \ ln \ r +B\theta ln \ r +C\theta +D =A \ ln \ r +B(\theta+2\pi) ln \ r +C(\theta+2\pi) +D\\ \\ \therefore B=C=0 \end{matrix}$$

Thus the equation simplifies to:

$$w(r,\theta)=A \ ln \ r +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)$$

Using the 3rd boundary condition:

$$w(a,\theta)=\Delta_{0} + \Delta_{1}sin \theta$$

One can clearly see that since the coefficient of the sine term is 1, all n terms that are not equal to one must be zero.

$$\begin{matrix} sin(n\theta)=sin \ \theta\\ \therefore\\ n=1 \end{matrix}$$

And,

$$\begin{matrix} n\neq 1\\ A_{n}=B_{n}=C_{n}=D_{n}=0 \end{matrix}$$

It can also be seen that since there is no cosine term in the boundary equation,

$$D_{1}=0$$

Thus the equation reduces to:

$$w(r,\theta)=A \ ln \ r +D + \left (\overline{A}r+\frac{\overline{B}}{r} \right )(sin \ \theta)$$

Where,

$$\begin{matrix} \overline{A}=A_{1}C_{1}\\ \overline{B}=B_{1}C_{1} \end{matrix}$$

Reapplying boundary condition three gives:

$$\Delta_{0} + \Delta_{1}sin \theta=A \ ln \ a +D + \left (\overline{A}a+\frac{\overline{B}}{a} \right )(sin \ \theta)$$

From this equation it is clearly evident that,

$$\begin{matrix} \Delta_{0}=A \ ln \ a +D\\ \Delta_{1}=\overline{A}a+\frac{\overline{B}}{a} \end{matrix}$$ Egm6322.s09.bit.gk 20:59, 24 April 2009 (UTC) These equations and the last two boundary conditions:

$$\begin{matrix} w(b,\theta)=0 \\ \frac{\partial w}{\partial r}_{r=b} =0 \end{matrix}$$

Give 4 equation to solve for 4 unknown coefficients.

The resulting system of equations may be expressed as a matrices:

$$\begin{bmatrix} sin\theta & \frac{sin\theta}{b^2} & \frac{1}{b}& 0 \\ bsin\theta & \frac{sin\theta}{b}&  \textrm{ln}\ b& \\ a & \frac{1}{a} & 0 & 0 \\ 0& 0&  \textrm{ln}\ a& 0 \end{bmatrix} \begin{Bmatrix} \overline{A}\\ \overline{B}\\ A\\ D \end{Bmatrix}= \begin{Bmatrix} 0\\ 0\\ \Delta_{1}\\ \Delta_{0} \end{Bmatrix}$$

Using the matrix to solve for the constants one finds that:

$$\begin{matrix} \overline{A}=\frac{b}{(b^2-a^2)} \frac{\Delta_{0}}{sin a \ ln \ a}\\ \\ \overline{B}=\frac{-a^2b \Delta_{0} }{(b^2-a^2)sin\theta \ ln \ a}\\ \\ A=\frac{\Delta_{0}}{ln \ a}\\ \\ D=-(1+ln \ b)\frac{\Delta_{0}}{ln \ a} \end{matrix}$$

iii) Calculate the resultant force and moment necessary to maintain the inner rigid disk shaped region in the displaced position.

To find the net force necessary to maintain the inner disk in its position, one must assume the membrane has an effective spring constant of k. The force is then k multiplied by the displacement w. To get the net force we integrate around the edge of the disk.

$$\begin{matrix}F_{net}=k\int_{0}^{2\pi} \Delta_{0}+\Delta_{1}sin\theta \ d\theta\\ \textrm{Note:}\int_{0}^{2\pi}sin\theta \ d\theta=0\\ F_{net}=2\pi k \Delta_{0} \end{matrix}$$

To find the moment necessary to keep the disk in its tilted position requires a bit more consideration. The moment results from an unbalanced force applied at some distance y. It should be noted that since the $$\Delta_{0}$$ term is the same all along the boundary, it does not contribute to the moment necessary to maintain the disk's position.

$$\Delta M_{x}=y \Delta F$$

Where the value of F is already given as:

$$\begin{matrix}\Delta F=ksin\theta \Delta L\\ \Delta L=b \Delta\theta \end{matrix} $$

To get the total moment the sum of the discrete $$\Delta M_{x}$$ should be taken. By increasing the number of $$\Delta M_{x}$$ and simultaneously decreasing their magnitude, the summation becomes on integral.

Thus,

$$\begin{matrix} M_{x}=\int dM_{x}= \int yk\Delta_{1}sin\theta \ bd\theta  \\ \textrm{Note:} \quad y=b \ sin\theta \\ M_{x}= \int b \ sin\theta \ k\Delta_{1}sin\theta \ bd\theta \\ = b^2k\Delta_{1}\int sin^2\theta \ d\theta \end{matrix}$$

This integral must be taken over the entire circle. Thus the moment required is:

$$\begin{matrix} M_{x}= b^2k\Delta_{1}\int_{0}^{2\pi} sin^2\theta \ d\theta \\ =b^2k\Delta_{1}\pi \end{matrix}$$

 The statement of the boundary value problem was incomplete (1st question). Do complete this problem in an updated version of report R4. Egm6322.s09 22:38, 2 April 2009 (UTC) The problem has been completed. Egm6322.s09.Three.ge 19:09, 10 April 2009 (UTC)

=References=

=Signatures= --Egm6322.s09.xyz 03:49, 6 March 2009 (UTC)

Egm6322.s09.Three.ge 19:34, 5 March 2009 (UTC)

--Egm6322.s09.three.liu 19:55, 5 March 2009 (UTC)

--EGM6322.S09.TIAN 21:05, 5 March 2009 (UTC)

--Egm6322.s09.lapetina 22:43, 5 March 2009 (UTC)

Egm6322.s09.Three.nav 16:05, 6 March 2009 (UTC)Egm6322.s09.Three.nav

Egm6322.s09.bit.sahin 18:04, 6 March 2009 (UTC) Egm6322.s09.bit.la 18:08, 6 March 2009 (UTC)

Egm6322.s09.bit.gk 19:20, 6 March 2009 (UTC)

title: User:Egm6322.s09.mafia/HW4


text:
User:Cojan

Welcome ! We have already  resources !

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

title: User:Cojan


text:
Medical experiments at Auschwitz

Most deaths at Auschwitz were caused by medicalized killing. Lead by Professor and Doctor Carl Clauberg, Nazi doctors at Auschwitz performed experiments on helpless prisoners. SS doctors tested the effectiveness of X-rays as a way to sterilize women by administering large doses of radiation. Dr.Clauberg injected chemicals into women's uteruses in an effort to glue them shut.

Dr. Carl Clauberg was given the camp to use humans and animals for experimentation in sterilization. He struggled to come up with a simple and quick way to sterilize many women in a short period of time. In 1943 he developed a drug for injection into the cervix and was able to inject it into a woman during a gynecological exam without them even knowing it was done. This injection would obstruct the fallopian tubes by sending a caustic substance through the cervix. Woman experienced complications frequently, including peritonitis and hemorrhages from the reproductive tract resulting in high fever and sepsis. Often, multiple organ failure and death would follow. Others were killed deliberately in order to perform autopsies.

According to Laurence Rees, Auschwitz; A New History,1 the most infamous doctor at Auschwitz was Josef Mengele, who was also known as the “Angel of Death”. Particularly interested in "research" on identical twins, Mengele performed cruel experiments on them such as inducing diseases in one twin and then killing the other after the first one died in order to perform comparative “autopsies”. Pairs of twins and persons with inherited anomalies were put at the disposal of Dr. Mengele and subjected to specialists’ medical examinations. At the end of all his experiments he would kill them with a lethal injection of phenol to the heart. In the next phase, autopsies would be performed and a comparison of their internal organs would be done. Mengele also experimented with dwarves, injecting them and other prisoners with gangrene in order to study the effects. Experiments like these took place on behalf of German pharmaceutical companies or medical institutes. These doctors and others performed these experiments for their own interests and in order to advance their own careers.

title: Medical experiments at Auschwitz


text:
User:Kylnso/documentary-dfvp212-skate

DOCUMENTARY PRODUCTION - SKATEBARN WEST == 10 minute segment of a documentary.

ITEMS TO TURN IN:

- 10 resource Articles on subject (magazine, newspaper, internet) - 1 Documentary Project Worksheet - Treatment
 * Research Packet
 * Two-Column Script
 * Production Schedule
 * VTR Log and EDL
 * Edited using Non-Linear System
 * Master on Mini-DV, DVCPRO or BetaSP
 * Crew Assignments
 * Equipment List

DUE DATES
WEEK 6 (FEB 16) - 1st draft of script (individual & group) WEEK 7 (FEB 23) - Treatment (individual & group) WEEK 8 (MAR 2)- Production Book: prod sched, crew list, equipment list, final script - VTR LOGS (individual & group) WEEK 9 (MAR 9)- EDL'S (individual & group) WEEK 11 (MAR 23)- individual project & group project

CREW ASSIGNMENTS
•	DREW (DIRECTOR) - Research current happenings / situations (costs) - SCRIPT – 1st draft - SCRIPT – final draft - SHOOT – interviews / park - SHOOT – b-roll - Treatment - Choose music - EDIT

•	KAYLEN (WRITER) - Research different skate ramps / park itself (size) - SCRIPT – 1st draft - SCRIPT – 2nd draft - SHOOT – interview / park - SHOOT – b-roll - Production booklet

•	KYLE (WRITER) - Research events going on there past & future stuff - SHOOT – interviews / park - SHOOT – b-roll - Treatment

•	NATHAN (EDITOR) - Research history on the skate part itself - EDIT – 1st cut - EDIT – 2nd cut - EDIT – Final cut

EQUIPMENT LIST
- Panasonic DVX 100 / Tripod - Sony HVR-Z1U / Tripod - HVR Battery - CGP Battery - Power Strip - Extension Cord - Lowel VIP Kit - Gloves - Reflectors - Gels - A/C Power - Fishpole - Sennheiser Shotgun Mic - XLR Cable - Tote Bag

VTR LOG
http://imaginarydigital.com/downloads/forms/VTR-Log-Sheet.doc http://www.nku.edu/~strobelc/archive/vtrlog.doc

SCRIPT
...

TREATMENT
This documentary will be focused on Skate Barn West, a 45,000 sqare foot indoor skatepark located in Renton. It houses eleven different, unique ramps including the Northwest’s only indoor vert ramp. It is a place where skateboarders, rollerbladers, and BMXers alike can go to escape the weather and have a good time doing what they love. It is a very safe environment where parents feel comfortable leaving their children. In recent years however, Skate Barn has run into financial difficulties which threaten to shut them down. The lease on the building expires May 2009, at which point it appears Skate Barn may have to close its doors forever. Our goal to document the effect that Skate Barn has had on the local communty and the positive impact it has made on all those who are able to take advantage of the space and increase their skills. Also, we wish to expose what has caused these financial problems and attempt to reveal possible solutions.

OUTLINE
- what is it? - why is it important?
 * INTRO

- How long its been there? - Dedicated people/employees? - # of kids from start to now - interview manager and employees and what they know
 * HISTORY

- who goes there? - diff age ranges? - how good they get? - how they feel about the place?
 * PARK

- Events - sponsorship - concession stands - safety - enviroment - equpiment
 * WHAT IS THERE TO OFFER

- what if it went away?
 * TAG / CONCLUSION

PRE-PRODUCTION
- NEW PLAN / NEW IDEA - Create outline - Create questions for manager, kids, parents - 1st draft script - Research - Review script - 2nd draft script - Review script - Finalize script
 * MAR 1
 * MAR 2
 * MAR 4
 * MAR 7

PRODUCTION
- Shoot interview with kids/parents - Shoot skaters (B-ROLL) - Shoot more interviews - Shoot inside (B-ROLL) - Shoot interview with employees/kids - Shoot outside/inside (B-ROLL) - Shoot interview with little kids - Shoot inside (B-ROLL) - Shoot interviews - Pam, #6 skater in world (Sky), Dan with old man vert night - Shoot inside (B-ROLL) - Shoot interviews - manager & co-owner - Shoot inside (B-ROLL)
 * FEB 25
 * MAR 3
 * MAR 10
 * MAR 16
 * MAR 18
 * MAR 20

POST PRODUCTION
- Capture footage - Music - Edit 1st rough cut - Review Edit - Capture footage - Edit 2nd rough cut - Capture footage - Capture footage - Edit 3rd rough cut - Review edit - Edit Final cut - Final cut - Turn in project
 * MAR 12
 * MAR 14 (sat)
 * MAR 16
 * MAR 17
 * MAR 19
 * MAR 20
 * MAR 21
 * MAR 23

title: User:Kylnso/documentary-dfvp212-skate


text:
Biology (A-Level)/Molecules, Cells and Systems

This Course will teach some of the fundamental concepts of cell biology topics which include "DNA,Cell Terminology and structures,Virus's,evolution, Meiosis and Mitosis,And the overall physiology of how the cell works and how we can associate this knowledge with everyday situations". There isn't any complex terminology and this course can easily be taken by anyone with a curiosity into cell biology.

The cell
All life is made up of tiny units called cells.

The two most broadest classification of cells are
 * Prokaryote- without a nucleus(ex. bacteria)
 * Eukaryote- with a nucleus

Cells contain various machinery inside of them called organelles which perform functions for the cell. These organelles are
 * 1) nucleus- which contains the genetic material of the cell[also the site of RNA and DNA activities like replication
 * 2) endoplasmic reticulum- which contains the rough ER(contains ribosomes) and the smooth ER(synthesizes lipids)
 * 3) vacuole- contains nutrients of the cell
 * 5) lysosomes- digestive enzymes
 * 6) vesicles- small body's in the cell
 * 7) Golgi Apparatus- Packager of the cell
 * 8) centrosomes- contain centrioles which aid in mitosis[cell division]
 * 9) cytoskeleteon- organelle that maintains cell shape
 * 10) ribosomes-The ribosomes are a large complex of RNA and protein molecules, This is where proteins are produced

Cell Division
When a cell divides it must go through a process called mitosis. Mitosis is a process that describes the various steps that occur in a cell before and after initial division

The cell cycle as its most commonly called consist of the

 * G1- cell growth begins
 * S- Protein synthesis and chromosomal replication occurs
 * G2- cell now has duplicate copies of chromosomes
 * M Phase- centrioles lock onto each end of the chromosomes and then pull them apart to make two cells

Terms (chromosome - super coiled DNA wounded together with histones)

All of the above is moderated by two molecules called "CDK-Cyclin"

Replication, Transcription, and Translation of DNA+RNA
Transcription

DNA or deoxyribeonucleicacid is the code for life, it contains the instructions for the production of proteins which do various operations in the cell. DNA by nature is a polymer(macromolecule) and consist of four nucleotide bases(adenine,guanine,cytosine,and thymine).It also contains sugar and phosphate groups as a backbone.

There is a general rule in molecule bonds in DNA

A & G are purines C & T are pyrimidines

A--->T(and U if its RNA) & C--->G

DNA is basically the raw code. It is unusable in making proteins until it is coded into something else called "mRNA" mRNA is created when the DNA double helix is unwinded by helicase(an enzyme). afterward stranding RNA nucleotides(AUCG, not T) link with one of the helix's strands, Once mRNA is coded for translation can begin.

if we had a strand of DNA that was   [ACGACGACAGACGTTTTCGAGACAGAC] The complementary RNA strand would be [UGCUGCUGUCUGCAAAAGCUCUGUCUG]

Translation

Each three nucleotides counts as a codon or protein coder(ex. are methionine, and lysine)

with our newly coded strand of mRNA we can being translation and produce a fully functional "polypeptide chain" or protein chain. The proteins are located on the ribosomes called tRNA or transport RNA which transports the proteins so they can link up with their complementary mRNA codons

If we had a codon

 * GCG||CCG|

the complementary tRNA would be protein1{CGC} + protein2{GGC} = polypeptide chain(ex. hemoglobin)

Replication

When a cell divides, it undergoes a process called mitosis and in the "S" phase of this cycle, DNA replicates, for replication and ultimately mitosis to take place (DNA must replicate). There are certain steps to this process. The first step is that DNA must be unwound by an enzyme called helicase. Afterwards, proteins link with the strands to make sure they don't bond back together. DNA polymerase comes into action next. It supplies complementary nucleotides from DNA 5' to 3' end. The leading strand replicates properly, but the lagging strand does not. It is formed in pieces called Okazaki fragments.

ATP
ATP(adenosinetriphosphate) is an molecule that powers your cells--- not finished


text:
User:Biologyprof12

Hello, welcome to my page.I love lecturing on biological topics and editing pages about them

title: User:Biologyprof12


text:
User talk:Biteright

Zoo Dentistry- Exotic Animal Dentistry for Lions, Tigers, and other Exotic Animals. Most experienced (over 30 Years) is Dr. Ed Shagam, Mount Holly, NJ Web bracez.com One of the pioneers of Exotic animal Dentistry in the United States. Root Canals, Extractions, Elephant work, seals, all large cats, hippos, Giraffes.

title: User talk:Biteright


text:
Social entrepreneurship/Evaluation Framework/Triple Bottom Line Reporting

Aim: develop a course on this topic and educate learners about the Triple Bottom Line.


text:
User:R.S.(AUA)

R.S.(AUA) a.k.a. Ramsin Shamouni, 2nd year Medical Student, at American University of Antigua, College of Medicine.

 * W ikiversity School of Medicine, See.....
 * School:Medicine

--Ramsin Shamouni 01:22, 8 March 2009 (UTC)

title: User:R.S.(AUA)


text: File:Paper 2.png

title: File:Paper 2.png


text:
User talk:Catarina

Welcome to my page here at Wikiversity!
My name is Catarina Schmidt, I am a PhD-candidate in Pedagogy at Örebro University in Sweden. My special interest is Literacy.

This page covers issues considering communications- and identity-questions connected to language development and the ability to read and write. Literacy is a wide concept that in many ways includes language and identity, it focuses the ability to read and write as a condition for communication and learning. Communication on a level that enables the individual to take part in society is a necessary ability in the world of today. In the report of an Expert Meeting of literacy assessment, UNESCO published a working definition of literacy which highlihgts the importance of context and use; "Literacy is the ability to identify, understand, interpret, create, communicate, compute and use printed and written materials associated with varying contexts. Literacy involves a continuum of learning to enable an individual to achieve his or her goals, to develop his or her knowledge and pontential, and to participate fully in the wider society." (UNESCO 2005:21, see also http://unesdoc.unesco.org/images/0016/001631/163170e.pdf (page 18).

The concept of literacy opens up for a wider definition - being able to read and write is not only a technical skill. It is much more and includes also the individual as a participant. Each one of us has a history of our own where we have read and written in meaningful contexts. The neccessity of participation, meaningfulness and a functional purpose for my own sake is crucial. Literacy is a neverending story for each one of us. We are always on our way.

Each individual exists of many identities and we all visit, create and participate within many literacies. Identities are always changeable, never the same. I here ad a picture taken from the summer of 2008 (copyright Catarina Schmidt). I name this picture "Floating identies". If you would like to share some thoughts with me or maybe give me some comments on my texts I invite you to do so at the bottom of this page under the headline "Reflections and thoughts".

Below I have added may final paper (The Complexity of Identity) in this course (KOIIA2). The paper discusses the conditions for identity; how it is created and how it is developed. I argue that there is a strong connection between confidence in the written language and identity development. We live in a society where the ability in the written language is immensely important. Through the written language we name ourselves and tell our life stories. We need respect and we ought to respect. Bhabha (1994) argue the neccessity for a "third space of enunciation" which will help us to escape from troublesome binaries like "we and them" and Feinberg (1998) claims the need of a robust recognition in order to give children tools to express themselves in ways that ring true.

The Complexity of Identity

Comments on the paper The Complexity of Identity

Reading logs 1-5
Below I have also added my former reading logs. The reading logs are simply my summaries from the course literature. Sometimes I have added some own reflections, but not always. Above all I have appreciated my reading logs, when I in the end of the couse started to work on my final paper. One can se the reading log as a first attempt of reading and understanding a text, the next time I meet these texts I am sure I will se other things. I have truly appreciated the texts below. The intersectional perspective follows them all, but so do also a strong commitment for an equal and just world.

Reading log 1:

 * The Locaction of Culture by Homi K. Bhabha
 * Language, culture and identity by Philip Riley
 * Generous Betrayal by Unni Wikan

Reading log Bhabha, Wikan, Riley

Reading log 2:

 * Common schools uncommon identities by Walter Feinberg
 * Who does this language belong to? by Avital Feuer
 * Talking voices. Repetition, dialouge and imagery in conversational discourse by Deborah Tannen

Reading log Feinberg, Feuer, Tannen

Reading log 3:

 * Att platsa i en skola för alla by Eva Hjörne and Roger Säljö
 * Crip Theory. Cultural Signs of Queerness and Disability by Robert McRuer
 * Why are all the black kids sitting together in the cafeteria? And Other Conversations About Race by Beverly Daniel Tatum

Reading log McRuer, Tatum, Hjörne & Säljö

Reading log 4:

 * Maktens (o)lika förklädnader. Kön, klass & etnicitet i det postkoloniala Sverige by Paulina de los Reyes, Irene Molina & Diana Mulinari
 * Organisering och intersektionalitet by Ulla Eriksson-Zetterquist & Alexander Styhre
 * Etnicitet. Perspektiv på samhället by Abby Petersson & Mikael Hjerm
 * Identity trouble by Carmen Rosa Caldas-Coulthard & Rick Iedema

de los Reyes, Molin & Mulinari, Eriksson-Zetterquist & Styhre, Peterson & Hjerm, Caldas-Coulthard & Iedema

Reading log 5:

 * Handbook of Multilingualism and Multilingual Communication by Peter Auer & Li Wei
 * Talk in two languages by Joseph Gafaranga
 * Literacy and power by Terese MacCarty

Auer & Wei, Gafaranga & MacCarty

Reflections and thoughts
Några tankar från Ingela

Jag är lite osäker på om de här tankarna ska ligga direkt på din förstasida här, eller om vi ska skapa länkar, men du får gärna flytta mina kommentarer till annan plats om du vill. Tankarna nedan rör först ditt avslutande paper; The complexity of Identity men jag har också lagt till några små kommentarer till dina läsloggar.

The complexity of Identity Ditt paper var mycket intressant att läsa, Catarina! Identitet är ett spännande och intressant område och du har gått in för att reda ut begreppen mer på djupet. Redan i din inledning, på engelska, insåg man vad ditt paper skulle handla om, och du gör precis det du skriver att din avsikt är. För mig kan texten, så här efter sommaren, läsas som en uppfräschning av de kunskaper kurslitteraturen och seminarierna gett, samtidigt som det är intressant att se vad du fastnat för hos de olika forskare du nämner i texten, något som kompletterar min egen läsning och uppfattning. Att identitet är något komplext framgår tydligt i din text.

Hej Ingela!

Det var här jag hade tänkt mig kommentarerna. Jag känner, som du, att identitet är ett mycket komplext begrepp. Antagligen är jag bara i början på detta. Roligt att det gav dig något att läsa mina texter!! Catarina

Läslogg 4 Det är svårt att kommentera något som är en läslogg tycker jag, eftersom man där tar upp sådant man själv fastnar för i texten. Hursomhelst var det intressant att läsa den här loggen som koncentrerar sig på flera texter ur kurslitteraturen. Du lyfter fram texternas centrala tankar kring intersektionalitet, etnicitet-ras, svenskar-invandrare, män-kvinnor och identitet och som läsare får man veta mycket om de texter du skriver om. Har man läst texterna själv så är det intressant att se vad du tar med, och det blir en fördjupning av ens egna läsning, och för den som inte redan läst texterna kan din logg fungera som en introduktion ifall man vill läsa mer om ämnena själv sedan.

Reflektioner och frågor från Jenny R Jag kommer här främst att kommentera din text The Complexity of Identity eftersom innehållet i läsloggarna till viss del redan diskuterats vid seminarierna i kursen.

Centrala begrepp i din text är identitet, jaget, subjekt, intersubjektivitet och identitetsskapande och jag har funderat en del kring vad du lägger in i begreppen. Eftersom jag tror att användingen av dessa begrepp hänger samman med vår förståelse (och ibland en implicit sådan) av identitet vill jag vrida och vända dem. Du skriver inledningsvis att begreppet identitet fokuserar på personen, jaget och subjektet. Jag tolkar detta som att du menar att en identitet är något som finns inom individen att det finns ett jag vilket du ser som synonymt med subjekt inom varje individ. Men vad är då skillnaden mellan identitet och subjekt? På frågan om hur identiteter skapas skriver du att varje individ behöver samspelet med andra för att utveckla identitet och att således att bli till som subjekt kan förstås som en dubbel process. Vidare skriver du senare att "det är mötet mellan subjekt, intersubjektivitet, som krävs för identitetsskpande". Om jag förstår dig rätt så menar du här att subjektet existerar för mötet med andra subjekt, men att identiteter inte gör det - detta kan då således svara på min fråga om vad som skiljer subjekt från identitet. Jag har nyligen läst Moira von Wrights bok Vad eller vem? (2000) som just behandlar frågan om intersubjektivitet utifrån G H Meads teori. Hon skiljer mellan ett punktuellt och ett relationellt perspektiv på subjektivitet. Ur ett punktuellt perpsektiv ses subjektivitet som en individuell produkt av omstädigheter eller egenskaper, isolerade från sitt omedelbara sammanhang, en form av lagbundenhet (2000:32) medan ett relationellt perspektiv ser ser existensen av subjektiviteten som beroende av andra människor i tid och rum och således kontextuellt. Om jag tolkar Wright rätt så finns subjektiviteten inte utan intersubjektiviteten. Intersubjektiviteten handlar således inte om ett möte mellan två oberoende subjekt utan dessa skapas och existerar just genom intersubjektiviteten. Således skapas människan som individ i kommunikationen mellan människor, vilket är ett annorlunda perspektiv än då man ser kommunikationen som ett rum för identitetsutveckling. Angående begreppet intersubjektiviet avser von Wright "koordineringen av subjektiviteterna, dock så att intersubjektiviteten som fenomen inte kan reduceras till enskilda subjektiviteter. Intersubjektivitet omfattar ett samförstånd i en samtidighet, dvs. att deltagarna i en interaktionssituation möts i kommunkation och derlar en verklighet som kommer till dem ur detta möte" (2000:75). Jag har inte här möjlighet att vidare gå in på von Wrights diskussion kring dessa frågor i relation till Meads teori, men förhoppninsvis kan läsningen av hennes bok ge dig en del reflektioner. Jag kan även se likheter med dialogismen såsom den utvecklas från Michael Bakhtin, då han menar att jaget/självet är dialogiskt och en form av relation till den andre. Människan kan endast bli medveten om sig själv gneom andra människor. Som du tar upp i din text är språk cntralt för vår identitet och du skriver att interkation kan ses som basen för identitetskapande. Frågan är dock om språket är centralt för att uttrycka våra identiteter eller om identiteter endast existerar genom/i språket? En sista fundering när det gäller begreppen handlar om identitet och social kategori - vad skiljer begreppen åt och hur förhåller de sig till varandra?

Hej Jenny! Du har satt igång tankar hos mig och det känns som om jag kommer att ha nytta av det. Jag tror att jag som Mead tänker att människan skapas i mötet mellan människor. Det tål att tänka på vilka ord och uttryck man använder och hur mycket man sedan ska använda dem. Det här behöver jag snurra några varv på... Tack för dina tankar! Catarina

My fellow students and course leader at KOIIA 2:
Karin Allard User: Karina

Sara Frödén User:Sara_F

Jenny Rosen User:JennyRosen

Marie Nordmark User:Nordmacca

Ingela Holmström User:Inhol24

Annaliinas sida User:Liinanna

Oliver St John User:Oliverst

Course leader Sangeeta Bagga-Gupta User:Sangeeta_BG

title: User talk:Catarina


text:
Reading log Bhabha, Wikan, Riley

9 March Glad to see your log Catarina! I look forward to the resource that you will create on intersectionality. 130.243.100.240 09:45, 9 March 2009 (UTC)Sangeeta

Läslogg KOIIA 2, vt -09 Catarina Schmidt

Unni Wikan, Generous Betrayal – Politics of Culture in the New Europe

Unni Wikan was born in 1944 and is a professor of social anthropology at the University of Oslo (Wikipedia, 2009-03-01). Wikan leads in her book ”Generous Betrayal” a discussion about culture, cultural rights, human rights, identity and democracy. In the centre of the discussion are young women whose parents have immigrated to Norway or Sweden. Cultural rights are put against human rights when those young women are forced to marry, are denied human democratic rights or in the worst scenario are victims of violence and/or murder. Within anthropology the concept of culture has always being used but not always questioned. Different cultures have been described and taken for granted without being mirrored towards laws and conventions about democracy and human rights signed by societies and nations all over the world. At the same time we are, according to Wikan, building a system of society with traditions and customs containing oppression which stands in open conflict with fundamental human rights. Wikan means therefore that the word culture can be a trap where children and weaker members of a group tend to be losers. The word culture tends to be a label standing for the exotic and strange – which in the name of culture is not being questioned. What those young women in Norway and Sweden, which have been exposed with honor of violence, have in common is that they belong to a patriotic culture associated with traditions which can result in violence and oppression in order to defense the honor of the family. Within this culture there is no room for individualism – variations in aspects of gender, age, ethnicity, social class and education are being denied or reduced. There are simply not many alternatives in being a woman or a man. Since identity is created by gender, age, ethnicity, social class and education the conclusion is a denial of the identity itself. The concept of culture becomes a trap which denies any variations of identity when we instead need the opposite;”a society worth living in must make room for people to craft identities” (p. 74). This sentence from Wikan opens in my opinion up for a meeting between human beings and if the individuals can meet can perhaps in the long run cultures can meet. Because of this Wikan claims that the respect for the individual”may require that”culture” be demoted or resisted” (p. 77). Wikan go as far that she speaks of the concept of culture as”a new concept of race” (p. 79) meaning that it maintains a politic which denies human rights. The author means that since culture is only a concept it has no own autonomic or material existence. Wikan means that culture and power walk hand in hand and that the authority always is resting by those who have the power:

”Some people have the right – or seize the right – to define what is to count and for what, and the result, the authoritative”truth” is often called culture.” (p. 87).

By exaggerating external and internal differences between different cultures and also to prevent differences within a certain culture a false image of a homogeny culture appears, something that will affect those members of the group that from different reasons want to go other ways than the head line. Wikan argues that the concept of culture”covers up the complexity of human existence, the fact that we are both children of”our culture” and unique individuals” (p. 88). All together I understand that Wikan’s recapitulation about the problems regarding the culture concept can be summarized like this:

 * Culture is a concept – not a thing
 * There is no objective definition
 * Culture has no agency – only humans have the power to act
 * Culture has no power – only what people give it
 * Cultural things change
 * Culture and power go hand in hand – some people seize the right to tell what is the “truth”
 * Culture is usually taken to mean difference, not difference and similarity
 * There are internal differences but culture tends to play down them which gives a false picture of homogeneity

(after Wikan, p. 88)

The question ringing in my head after having read Wikan’s book is why are we not doing anything? For me the answer is that we very often are blind mice, but with the help of Wikan we can be at least near sighted.

Homi K. Bhaba, The Location of Culture

Homi K. Bhabha, an Indian theorist of post-colonialism, was born 1949 and is according to Wikipedia (2009-03-01) currently teaching at Harvard University. From what I understand Bhabha in his book “Location of Culture” consequently punctures thoughts that support ideas about any better or worse of cultures. There is, according to the author a great need to redefine the concept of national cultures:

The very concept of homogenous national cultures, the Consensual or contiguous transmission of historical traditions, or ‘organic’ ethnic communities – as the ground of cultural comparativism – are in a profound process of redefinition. The hideous extremity of Serbian nationalism proves that the very idea of a pure, ‘ethnically cleansed’ national identity can only be achieved through the death, literal and figurative, of the complex interweavings of history, and the culturally contingent borderlines of modern nationhood. (p. 7)

His vision includes instead an articulated, clear hybrid of cultures and it is this space that he calls “the third space of enunciation” (p. 53). Within this space Bhabha means that there is a possibility to create a conception of culture that is not based on exotic descriptions which in a one-sided and uninterruptedly way put the others aside as the worse. We can, as I understand Bhabha, achieve this by finding the words where we can talk about ourselves and the others. To ascribe some cultures a kind of intrinsic inherited pureness which is not questionable is, according to Bhabha, untenable. Generations of authors have described meetings between cultures in the old colonies. After having read this book I understand why Bhabha includes so many literary quotations since those give descriptions that cover everything from integration to vast gulfs and oppressing violence between cultures. The Western compulsion to colonize has, of course, had a huge impact on our view of foreign cultures and through colonial and postcolonial authors we can see the possibility to change and, again, the vision of the third space. The different contributions from post colonialism expose both the colonists and the former colonies and have an ambition to explain today’s problems in society as a consequence of the western European colonization. The author’s contribution, as I understand it, is about the cultural and social effects within the post colonialism and by emphasizing the colonists and the colonized dependence from each other he argues that all cultural systems should be constructed in “the third space of enunciation” (p. 53). This, Bhabha means, will help us to escape from “we” and “them” where we are using and giving each other troublesome binaries. Binary oppositions are fatal since they unthinkingly dominate the second, again there is a better and a worse culture and examples might be civilized/savage, rational/irrational, enlightened/ignorant and eventually west/east and white/black. The binaries need according to Bhabha to be destabilized and he argues that culture must be understood to interact in a much more complex manner than those traditional binary oppositions can allow.

To take this perspective would mean that we see ‘racism’ not simply as a hangover from archaic conceptions of the aristocracy, but as part of the historical traditions of civic and liberal humanism that create ideological matrices of national aspiration, together with the concept of ‘a people’ and its imagined community. (p. 359)

The idea of race is present through Bhabas book and as I understand it is something we have to deal with and revise in our time. Bhabas uses, as always, the literature and one example is Fanon from “The fact of blackness”:

‘Dirty nigger!’ Or, simply, Look a Negro’ (p.338)

From this quotation Bhabha draws conclusions and the senses of not only being the individual ‘nigger’ but also of belonging to the marginalized group of black people –but also about developing black identity in human with ideological matrices. And shortly afterwards he again quotes Fanon with the sentence:

You come too late much too late, there will always be a world – a whole white world between you and us. (p. 339)

Where is culture located and how do we see upon culture through the lenses of race? After having read this book I am more aware of the complexity than before but still very confused. Bhabas book is not an easily read book. The language is intense and compact with many adjectives and many, many quotations from the world’s literature describing colonialism, colonists and the aftermaths from it. The author requires a great deal from its reader and I wonder if he could have done it a little easier and hopefully a bit clearer. His theory is also a controversial one and full of demands. It is about you and me, how we look upon ourselves and others. What he wants is this:

What is crucial to such a vision of the future is the belief that we must not merely change the narratives of our histories, but transform our sense of what it means to live, to be, in other times and different spaces, both human and historical. (p. 367)

Philip Riley, Language, Culture and Identity

Philip Riley is emeritus professor of Ethnolinguistics at Nancy University in France. A central theme in Rileys book is communicative practices where language plays the most important role and he claims himself that the thrust of the book is “to study the linguistic and communicative aspects of the social knowledge system” (p. 8). From the very start Riley emphasizes the complexity of language, referring to Herder:

…since language is at one and the same time the tool, the content and the form of human thought, and every act of knowledge is only possible through the medium of language. This interdependence means that the modes of thought and cultures of people can only be studied and analysed in and through their language. (p.9)

Rileys book is a cross-disciplinary contribution showing how social knowledge systems and language work to shape and perform individual and collective identities. In almost every chapter he does a historical review and he cites prominent academics all over the book. In his first chapter he lays the foundation and understanding of the concept of identity and explains that it is socially constructed but also a part of the surrounding society. Riley claims that “the reflection on our interactive experiences – facilitated and canalized by language – enables to become who we are” (p. 16). Society itself, language and interaction are, according to Riley, the very sources of individual identity. In chapter two Riley describes the social knowledge system and the linking and primary mechanism is, again, language. Another crucial ingredient is intersubjectivity which can be explained as a state of shared meaning between the participants and seen by Riley as necessary for the development of identity:

The ability to establish intersubjectivity, to enter into social and meaningful contact with another, is a necessary condition for the formation of identity. (p. 33)

In this way communication and culture are dependent from one another. Riley argues, as Wikan, that the term culture is polysemic, but in a more pragmatic and humoristic way:

Individuals acquire and construct their personal cultural repertoires on the basis of the interactional opportunities available to them. A country-dweller and a town-dweller, a doctor’s child and a labourer’s child, a Muslim and a Catholic will have access to participation in different ranges of interaction……In other words, acquiring culture is not like serving oneself from a soup tureen, where every bowlful has identical contents, it is more like lunching from a vast smorgasbord. (p. 39)

Culture continues to exist through language and Riley seems to mean that culture in itself is knowledge. This knowledge is distributed on different levels; the family level, shared by all in the neighbourhood and then shared by everyone on another level. Riley argues that each kind of knowledge corresponds to an aspect of social identities like family and communities like professions, religious groups or sport clubs. This is for me very close to the concept of literacy – the ability to read and write is always connected with social activities and domains which include relationships and a history in a chronological order. Through all these activities is the competence of communications used and Riley points out how different competences of language are needed, with in my opinion, an outstanding metaphor:

…linguistic competence means being a grammarian, communicative competence means being a speaker and sociocultural competence means being a member. Rather, the three can be imagined like Russian dolls, nesting inside another, so that sociocultural competence includes communicative competence which in turn, includes linguistic competence. (p. 53)

So, the knowledge of culture is communicated in different communicative practices and for this are the competences of language cited above needed. According to Riley communicative practices do not only communicate messages, they also communicate social identity which “is made up of a configuration of memberships and each membership is knowledge-and-language based” (p. 113). And, again, the conclusion is, that if we want to study identity we have to study language. Identity is formed by subject-positions materialized of power relationships and Riley gives a lot of example of this beginning with the gendarme with his uniform and revolver shouting at an innocent boy who in that second becomes a criminal. Communicative practices can also be gendered when we say things like “big boys don’t cry” (p. 139) and “nice girls don’t use words like that” (p. 139).

Riley also dwells over the concept of the stranger, quoting Simmel and Schütz and with the help of the later Riley points out different kinds of mobility addressing membership and belonging. One example is the candidate who wishes to become a member of an exclusive club and another is the new recruit joining the army. In order to form social identity recognition is needed – the stranger has to be recognized by others. Riley refers to Hegel and says:

It is otherness which makes interaction both possible and Necessary. I can only exist as an individual if other individuals exist and if we know it. A human community is a community of recognition: it can only exist to the extent that the individuals concerned recognize the existence of one another. (p.176)

The Me is, according to Riley, seen as the individual’s own perception of his or her social identity while the memberships of social groups form a bundle with roles and positions which the Self occupies. Eventually public perceptions of the Me and the Self form ethos. For me Riley has tried to answer the question about where identity is and how it is created. The answer is a of course complex but also as simple as his last words “just as history tells us who we are, identity is made of the stories we tell ourselves” (p. 244).

title: Reading log Bhabha, Wikan, Riley