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User:Egm6322.s09.Three.nav

PROJECT # 2
Project

The word linear comes from the latin word ‘linearis’ which means created by lines.

'''Proof to show that the order of double differentiation is independent of the variables i.e uxy= uyx Assume that there exist uniform number of intervals of length 'h' in both 'x' and 'y' directions'''

Then

ux= $$\lim_{h\to 0}\frac {u(x+h,y)-u(x,y)}{h}$$

uy= $$\lim_{h\to 0}\frac {u(x,y+h)-u(x,y)}{h}$$

uxy= $$ \lim_{h\to 0}\frac {\frac{\partial u(x+h,y) }{\partial y}-\frac{\partial u(x,y)}{\partial y}}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{\left [\frac{u(x+h,y+h)-u(x+h,y)}{h} \right ]-\left [\frac{u(x,y+h)-u(x,y)}{h} \right ]}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{u(x+h,y+h)-u(x+h,y)-u(x,y+h)+u(x,y)}{h^{2}}$$---(1) uyx= $$ \lim_{h\to 0}\frac {\frac{\partial u(x,y+h) }{\partial x}-\frac{\partial u(x,y)}{\partial x}}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{\left [\frac{u(x+h,y+h)-u(x,y+h)}{h} \right ]-\left [\frac{u(x+h,y)-u(x,y)}{h} \right ]}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{u(x+h,y+h)-u(x+h,y)-u(x,y+h)+u(x,y)}{h^{2}}$$---(2) Hence As is evident from (1) and (2), the order of double differentiation is indeed independent of the variables.

PROJECT#3
To do: Transform the Laplace Equation into Polar coordinates and compare with published results

The Laplace's Equation is often encountered in electrostatics, heat and mass transfer theory and in other areas of mechanics and physics. Simply worded, the Laplace's Equation states that div(grad u)=0.

It has the following mathematical form:

In Cartesian coordinates (x,y)$$: \ \ u_{xx}+u_{yy}= 0 $$

In Polar coordinates (r,$$\theta$$)$$: \ \ \ \ \ u_{rr}+\frac{u_{r}}{r}+\frac{u_{\theta\theta} }{r^{2}}=0 $$

where x= rcos($$\theta$$) and y= rsin($$\theta$$)

$$ Recognize\ that\ u_{xx}+u_{yy}= \begin{bmatrix} \partial _{x}\ \partial_{y} \end{bmatrix} \begin{bmatrix} \partial _{x} \\ \partial_{y}\end{bmatrix} \left \{u \right \} =\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}$$

where $$ J= \begin{bmatrix} \frac {\partial r}{\partial x} \ \frac {\partial r}{\partial y}\\ \frac {\partial \theta }{\partial x} \ \frac{\partial \theta}{\partial y} \end{bmatrix}$$

Denote cos($$\theta$$)=C and sin($$\theta$$)=S

$$ \Rightarrow J= \begin{bmatrix} \ C \ S \\\frac{-S}{r} \ \frac{C}{r} \end{bmatrix}$$

Differentiating by parts

$$\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}+\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$---(A)

where the terms in the angled brackets are kept fixed throughout differentiation.

Evaluating the RHS of equation (A) term by term '''First term= $$\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \} $$(1)

$$ JJ_{T}=\begin{bmatrix} \ C \ S \\ \frac{-S}{r} \ \frac{C}{r} \end{bmatrix}\begin{bmatrix} \ C \ -S/r \\ S \ \ C/r \end{bmatrix}= \begin{bmatrix} 1 \  0 \\  0 \ \frac{1}{r^{2}} \end{bmatrix}$$

$$\Rightarrow \ (1)= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle \begin{bmatrix} 1 \ \ \ 0 \\ \ 0 \ \frac{1}{r^{2}} \end{bmatrix} \right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$

$$\Rightarrow\ (1)=\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\begin{bmatrix} \partial_{r} \\ \frac{\partial_{\theta}}{r^2}\end{bmatrix}\left \{u \right \}$$

$$\Rightarrow\ (1)= \frac {\partial^2{u}}{\partial{r^2}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial{\theta^2}} $$ ''' Second term= $$\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}'''$$-(2)

By definition of the Jacobian, note that the term in the angled brackets is the same as $$\begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}$$

$$\Rightarrow\ (2)= \left \langle \begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}J^{T} \right \rangle \begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left\{u\right \}$$

Evaluating the term in angled brackets for equation(2)

$$\left \langle \begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}J^{T} \right \rangle= \begin{bmatrix} \partial_{x}C+\partial_{y}S \ -\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r} \end{bmatrix}$$

$$= \begin{bmatrix} \frac{1}{r} \ 0 \end{bmatrix}$$

$$\Rightarrow\ (2)= \begin{bmatrix} \frac{1}{r} \ 0 \end{bmatrix}\begin{bmatrix} \partial_{r} \\ \partial_{\theta} \end{bmatrix} \left \{u \right \} $$

$$\Rightarrow\ (2)= \frac{1}{r} \partial_{r}u$$

Finally putting terms (1) and (2) together into equation (A)

$$u_{xx}+u_{yy}= \begin{bmatrix} \partial _{x}\ \partial_{y} \end{bmatrix} \begin{bmatrix} \partial _{x} \\ \partial_{y}\end{bmatrix} \left \{u \right \} =\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}$$

$$\Rightarrow\ (A)= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}+\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$

$$\Rightarrow\ u_{xx}+u_{yy}= \frac {\partial^2{u}}{\partial{r^2}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial{\theta^2}}+\frac{1}{r}\frac {\partial{u}}{\partial{r}}$$

Hence we have proved, using transformation of coordinates, that:

$$ u_{xx}+u_{yy} =\ u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta\theta}$$

Comparing with published results, we see that our derived result matches perfectly.

Rotation of Coordinates
What is the form of the Jacobian when implementing a rotation of coordinates?

Consider one coordinate system, x-y (cartesian) and another coordinate system, $$\bar{x}-\bar{y}$$, obtained by rotating x-y clockwise by angle $$\theta$$ (as shown in the figure)

Assuming $$\widehat{\underline{i}}$$ and $$\widehat{\underline{j}}$$ are the unit vectors in the new coordinate system, $$\hat{i}$$ and $$\hat{j}$$ are the unit vectors in the old coordinate system

Then

$$ \widehat{\underline{i}}= icos\left(\theta \right)+ jsin\left(\theta \right)$$

$$ \widehat{\underline{j}}= -isin\left(\theta \right)+ jcos\left(\theta \right)$$

Comparing the above vector equations with those given in the example of Affine Mapping  considered previously,

$$\begin{matrix} \widehat{\underline{i}}=m \widehat{i}+n \widehat{j}\\ \widehat{\underline{j}}=p \widehat{i}+q \widehat{j} \end{matrix}$$

We see that m= cos$$\theta$$, n= sin$$\theta$$, p= -sin$$\theta$$, q= cos$$\theta$$

Hence the Jacobian ( J ) is given by

$$\underline{J}= \underline{E}= \begin{bmatrix} \frac{q}{\left (mq-np \right )}&\frac{-p}{\left (mq-np \right )} \\ \frac{m}{\left (mq-np \right )}&\frac{-n}{\left (mq-np \right )} \end{bmatrix}$$

$$\because$$mq-np= $$ \left(cos\theta \times cos\theta \right)$$ - $$\left(sin\theta \times -sin\theta \right)$$ = $$ cos^{2}\theta+ sin^{2}\theta$$ = 1

$$ \therefore \underline{J}= \underline{E}= \frac {1} {mq-np}\begin{bmatrix} \ \ q \ \ -p \\ -n \ \ \ m \end{bmatrix} = \frac {1} {1}\begin{bmatrix} cos\theta \ \ \ sin\theta \\ -sin\theta \ \ cos\theta \end{bmatrix} = \begin{bmatrix} cos\theta \ \ \ sin\theta \\ -sin\theta \ \ cos\theta \end{bmatrix}$$

PROJECT#5
Particularizing the general solution:

Case (i): Domain $$ \mho $$ is a full circle i.e. $$\theta \ \epsilon \ [0,2\pi]$$

$$ \Rightarrow$$ Solution has to be periodic w.r.t $$\theta$$

$$ \Rightarrow u(r,\theta+2k\pi, t) = u(r,\theta,t) $$, for any non-negative integer 'k'.

This implies that solution for $$\theta$$($$\theta$$), which consists of sines and cosines (periodic functions) is only possible only for integers n= 1,2,...

$$ \Rightarrow \sqrt{\rho}= n $$

$$ \Rightarrow \rho= n^{2} $$

Also replace the continuous integral $$ \int d\rho$$ by a discrete sum $$\sum_{0}^{n}$$ in Eqn.(2)

Now the equation becomes

$$ u(r,\theta,t)= \int_{0}^{\infty} d\lambda \left [\sum_{0}^{n} exp(-\lambda t)\left (\bar{B}(\rho, \lambda)sin(\sqrt{\rho}\theta) + \bar{C}(\rho,\lambda)cos(\sqrt{\rho}\theta) \right)\left (D(\rho, \lambda)J_{\sqrt{\rho}}(\sqrt{\lambda}r)+(E(\rho, \lambda)Y_{\sqrt{\rho}}(\sqrt{\lambda}r \right) \right] $$

An important point to remember here is that the above solution, obtained as a result of assuming periodicity, is applicable to the annulus domain.

Q: What if the domain is a solid disk, instead of an annulus?

In the plots of $$ J_{n}(z)$$ and $$Y_{n}(z)$$ Vs z, it was seen that $$Y_{n}(z) \to -\infty \ as\ z\ \to \ z^{+} $$

For the solution to remain meaningful, this cannot be allowed to happen. Hence we force $$E(\rho, \lambda)$$ = 0.

Hence the equation reduces to

$$ u(r,\theta,t)= \int_{0}^{\infty} d\lambda \left [\sum_{0}^{n} exp(-\lambda t)\left (\bar{\bar{B}}(\rho, \lambda)sin(\sqrt{\rho}\theta) + \bar{\bar{C}}(\rho,\lambda)cos(\sqrt{\rho}\theta) \right)J_{\sqrt{\rho}}(\sqrt{\lambda}r)\right] $$

where $$\bar{\bar{B}}= \bar{B}(\rho,\lambda)D(\rho,\lambda)$$ and

$$\bar{\bar{C}}= \bar{C}(\rho,\lambda)D(\rho,\lambda)$$

Remembering that $$\sqrt{\rho} = n$$, the above equation can be rewritten as

$$ u(r,\theta,t)= \int_{0}^{\infty} d\lambda \left [\sum_{0}^{n} exp(-\lambda t)\left (\bar{\bar{B}}(\rho, \lambda)sin(n\theta) + \bar{\bar{C}}(\rho,\lambda)cos(n\theta) \right)J_{n}(\sqrt{\lambda}r)\right] $$

DERIVATION OF FOURIER COEFFICIENTS
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

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

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

PROJECT#6
The Leibniz Integral Rule, named after Gottfried Leibniz accomplishes differentiation of an integral. It tells us that

$$\frac{\mathrm{d} }{\mathrm{d} x}\left(\int^{B(x)}_{A(x)}F(x,\xi)d\xi \right)= \int^{B(x)}_{A(x)}\frac{\partial F(x,\xi)}{\partial x}d\xi+\left(F(x, \xi= B(x))\frac{\mathrm{d} B}{\mathrm{d} x} \right)- \left(F(x, \xi= A(x))\frac{\mathrm{d} A}{\mathrm{d} x} \right)$$ --(1)

This can be derived from the fundamental theorem of calculus as shown in this Wikipedia link. Leibniz_integral_rule

The Leibniz Theorem is applied as a tool in deriving most of the conservation laws in Fluid Mechanics. Consider a control volume that is neither fixed nor moving, but is simply bounded by surfaces a(t) and b(t). These surfaces are moving with a velocity different from that of the local fluid velocity. Then the differential of the integral, as given by the Leibniz theorem, is written as

$$\frac{\mathrm{d} }{\mathrm{d} t}\left(\int^{b(t)}_{a(t)}F(x,t)dx \right)= \int^{B(x)}_{A(x)}\frac{\partial F(x,t)}{\partial t}dx+\left(F(x= b(t),t)\frac{\mathrm{d} B}{\mathrm{d} t} \right)- \left(F(x= a(t),t))\frac{\mathrm{d} A}{\mathrm{d} t} \right)$$-(2)

where a(t), b(t) form the boundaries of the control volume considered and are moving at a speed dA/dt and dB/dt respectively.

Multiplying through by dS (the surface area of a differential control volume) and generalizing further, we obtain the following form:

$$\frac{\mathrm{d} }{\mathrm{d} t}\left(\int^{b(t)}_{a(t)}F(x,t)dV \right)= \int^{b(t)}_{a(t)}\frac{\partial F(x,t)}{\partial t}dV+\int_{S(t)}\mathbf{dS.u_{s}}F$$--(3)

The surface integral over area S(t) in (3) includes the contributions from both the boundaries, thus excluding the need for the previous form of the equation (2).

For a control volume that is fixed, i.e. u= 0, (3) becomes

$$\frac{\mathrm{d} }{\mathrm{d} t}\left(\int^{b(t)}_{a(t)}F(x,t)dV \right)= \int^{b(t)}_{a(t)}\frac{\partial F(x,t)}{\partial t}dV$$--(4)

For a material volume i.e. a control volume in which the boundaries are moving at a velocity equal to the fluid velocity, u, (3) becomes

$$\frac{\mathrm{D} }{\mathrm{D} t}\left(\int^{b(t)}_{a(t)}F(x,t)dV \right)= \int^{b(t)}_{a(t)}\frac{\partial F(x,t)}{\partial t}dV+\int_{S(t)}\mathbf{dS.u_{s}}F$$--(5)

Eqn.(5) is sometimes referred to as the Reynold's Transport Theorem'. We use the term D/Dt to convey that we are defining the property of a material control volume. This theorem is often used in formulating the basic conservation laws in fluid mechanics.

Applying Gauss' theorem, which states that $$\oint_{S}\mathbf{X.dS}= \int_{V}(\nabla.X)dV$$, (5) can be rewritten as

$$\frac{\mathrm{D} }{\mathrm{D} t}\left(\int^{b(t)}_{a(t)}F(x,t)dV \right)= \int^{b(t)}_{a(t)}\frac{\partial F(x,t)}{\partial t}dV+\int_{V(t)}\nabla.(u_{s}F)dV$$(6)

Consider F as a physical property

F= $$\rho$$ gives us the equation for Conservation of Mass. For a steady flow system, the mass of the system i.e.$$\rho$$dV remains constant with time. Hence from(6)

$$0= \int^{b(t)}_{a(t)}\frac{\partial \rho}{\partial t}dV+\int_{V(t)}\nabla.(\rho u_{s})dV$$

(or)

$$\int_{V(t)}\left (\frac{\partial \rho}{\partial t}+\nabla.(\rho u) \right) dV = 0 $$---(7)

Similarly F= linear momentum= $$\rho$$v gives us the equation for Conservation of Momentum.

$$\int_{V(t)}\left (\frac{\partial \rho v}{\partial t}+\nabla.(\rho vu) \right) dV = 0 $$

Hence we see the relevance of the Leibniz Integral rule in the context of Fluid Mechanics and the fundamental role it plays in deriving the basic laws that the complex science of Fluid Mechanics is built on.

PROJECT#7
Problem:  Given the general,modified form of a PDE: $$\lambda_{1}\bar{x}^2+\lambda_{2}\bar{y}^2+d\bar{x}+e\bar{y}+f=0$$--(1)

Derive the canonical form representing a parabola from (1).

The canonical form representing a parabola is given by the following equation

$$\bar{\xi}^{2}- \bar{\eta} = 0$$ --(2)

where $$\bar{\xi}$$ and $$\bar{\eta}$$ represent coordinates in a different coordinate system.

So how do we derive (2) from (1)? We apply rotation once and translation of coordinates twice to $$\bar{x}$$ and $$\bar{y}$$.

Step(1)- First Translation

Define $$\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}= \begin{bmatrix}\bar{x}+r \\ \bar{y}\end{bmatrix}$$

Substituting $$\bar{\bar{x}}$$ and $$\bar{\bar{y}}$$ in (1), we get

$$\lambda_{1}(\bar{\bar{x}}-r)^2+\lambda_{2}\bar{\bar{y}}^2+d(\bar{\bar{x}}-r)+e\bar{\bar{y}}+f=0$$

$$\Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+\lambda_{2}\bar{\bar{y}}^2+\bar{\bar{x}}(d-2\lambda_{1}r)+e\bar{\bar{y}}+\lambda_{1}r^{2}+f=0$$--(3)

For equation (3) to resemble (2), we need the terms: $$\bar{\bar{y}}^2, \bar{\bar{x}} \ and \ \lambda_{1}r^{2}+f$$ to vanish.

Note that for a parabola, the equation of conics: $$Ax^2+Bxy+Cy^2+\bar{d}x+\bar{y}+\bar{f}=0$$, $$B^2-4AC=0$$ i.e only one eigenvalue exists. Thus assume $$\lambda_{2}$$= 0.

Also in order for the $$\bar{\bar{x}}$$ term to vanish, we need its coefficient to be equal to 0 $$\Rightarrow d- 2\lambda_{1}r=0$$

$$\Rightarrow r= \frac{d}{2\lambda_{1}r}$$

Substituting $$\lambda_{2}$$ and r in (3), the equation reduces to $$\lambda_{1}\bar{\bar{x}}^{2}+e\bar{\bar{y}}+\frac{d}{4\lambda_{1}^{2}}+f=0$$-(4)

Denote g= $$\frac{d}{4\lambda_{1}^{2}}+f$$. Thus (4) reduces to

$$\Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+e\bar{\bar{y}}+g=0$$-(5)

Step(2)- Rotation

Eqn.(4) can be rewritten as $$\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor\begin{Bmatrix} \lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix}\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}+\left \lfloor 0 \ e \right \rfloor\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}+g=0$$ (6)

Now apply rotation of coordinates by defining a new coordinate system $$(\xi, \eta)$$ where $$\begin{bmatrix} \xi \\ \eta \end{bmatrix}= \mathbf{J_{\beta}}\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}$$

where $$\mathbf{J_{\beta}}= \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix}$$

Due to $$\mathbf{J_{\beta}}$$ being an orthogonal matrix, $$\mathbf{J_{\beta}}^{-1}= \mathbf{J_{\beta}}^{T}= \mathbf{J_{\beta}}$$

$$\Rightarrow \begin{bmatrix}\bar{\bar{x}}\\\bar{\bar{y}} \end{bmatrix}= \mathbf{J_{\beta}}^{-1}\begin{bmatrix}\xi\\\eta \end{bmatrix}= \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}$$

Substituting this rotation of coordinates in (6), we get

$$\begin{bmatrix}\xi\ \eta \end{bmatrix} \mathbf{J_{\beta}} \begin{Bmatrix}\lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix} \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}+\left \lfloor 0 \ e \right \rfloor \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}+g=0$$

$$\Rightarrow \begin{bmatrix}\xi\ \eta \end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{Bmatrix}\lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix} \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{bmatrix}\xi\\\eta \end{bmatrix}+\left \lfloor 0 \ e \right \rfloor \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{bmatrix}\xi\\\eta \end{bmatrix}+g=0$$

Simplifying the above equation further, we obtain

$$\xi^2- \eta+g=0$$ ---(7)

Step(3)- Second Translation

Now define $$\begin{bmatrix}\bar{\xi}\\ \bar{\eta}\end{bmatrix}= \begin{bmatrix}\xi \\ \eta-g \end{bmatrix}$$

This is a valid translation because $$\lambda_{1}$$ which is a function of the coefficients A,C (as defined in the equation of conics) is a constant value. Hence g which is a function of $$\lambda_{1}$$ and f is also aconstant. Thus all we are doing in the second translation is translate the $$\bar{\bar{y}}$$ by a constant. Hence making this substitution in (7) gives us the final canonical form (2)

$$\bar{\xi}^{2}- \bar{\eta} = 0$$

Lecture#38
The General form for the Equation of Conics is given below:

$$\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2+f-\frac{d^2}{4\lambda_{1}}-\frac{e^2}{4\lambda_{2}}=0$$ ---(1)

Define $$ g= -(f-\frac{d^2}{4\lambda_{1}}-\frac{e^2}{4\lambda_{2}})$$. Then (1) reduces to the form

$$\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2= g$$--(2)

Here $$\lambda_{1}$$ and $$\lambda_{2}$$ denote the eigenvalues obtained by solving the eigenvalue problem for the original equation of conics: $$Ax^2+Bxy+Cy^2+\bar{d}x+\bar{e}y+\bar{f}=0$$

Equation (2) can be further reduced to equations representing the canonical forms for an ellipse, hyperbola and parabola.

Case (i):Ellipse: $$\lambda_{1}$$ and $$\lambda_{2}$$ >0 $$\Rightarrow$$ g>0

The canonical form representing the ellipse then reduces to: $$\xi^2+\eta^2=1$$ --(3)

Case (ii): Hyperbola: $$\lambda_{1}$$ and $$\lambda_{2}$$ <0 $$\Rightarrow g\neq 0 $$

The canonical form representing the hyperbola takes two different forms depending on the sign of 'g'.

Case ii(a): g>0 $$\to \xi^2-\eta^2=+1$$ ---(4a) (This is also true for $$\lambda_{1} < 0$$,$$\lambda_{2} > 0$$ and g>0)

Case ii(b): g<0 $$\to \xi^2-\eta^2=-1$$ ---(4b) (This is also true for $$\lambda_{1} < 0$$,$$\lambda_{2} > 0$$ and g<0) Case ii(c) $$\bar{\xi}\bar{\eta}=\pm 1$$--(4c)

This is a different representation of the canonical form for hyperbolas. It is derived from either of the equations (4a) or (4b)

Note that $$[\xi,\eta]$$ represent a different coordinate system using which (2) has been transformed into the various canonical forms shown. The derivation of forms(3),(4a) and (4b) are shown in separate sections. The derivation of form (4c) from (4a)or (4b) is shown below.

Define a new coordinate system $$[\bar{\xi}, \bar{\eta}]$$ in terms of the old coordinate system $$[\xi, \eta]$$ such that $$ \begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}= \begin{Bmatrix}\xi-\eta \\ \xi+\eta\end{Bmatrix}$$

$$ \Rightarrow \begin{Bmatrix}\bar{\xi}\\ \bar{\eta}\end{Bmatrix}= \begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}\begin{Bmatrix}\xi-\eta \\ \xi+\eta\end{Bmatrix}$$

$$\Rightarrow \begin{Bmatrix}\xi \\ \eta \end{Bmatrix}= \begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}^{-1}\begin{Bmatrix} \bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \frac {1}{2}\begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}$$

Eqn.(4a) or (4b) can be re-written in the form: $$\begin{Bmatrix}\xi \ \eta \end{Bmatrix}\begin{bmatrix}1 \ \ \ \ 0 \\ 0 \ -1 \end{bmatrix}\begin{Bmatrix}\xi \\ \eta \end{Bmatrix}= \pm 1$$--(5)

Rewriting (5) in the new coordinate system $$[\bar{\xi}, \bar{\eta}]$$, we get:

$$ \begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\frac{1}{2}\begin{bmatrix} \ \ 1 \  -1 \\ -1 \ -1 \end{bmatrix}\begin{bmatrix}1 \ \ \ \  0 \\ 0 \ -1 \end{bmatrix}\frac{1}{2}\begin{bmatrix} \ \ 1 \   -1 \\ -1 \ -1 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$ \Rightarrow \frac{1}{4} \begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\begin{bmatrix} \ \ 1 \  -1 \\ -1 \ -1 \end{bmatrix}\begin{bmatrix}1 \ \ \ \  0 \\ 0 \ -1 \end{bmatrix}\begin{bmatrix} \ \ 1 \   -1 \\ -1 \ -1 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$\Rightarrow \frac{1}{4}\begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\begin{bmatrix}\ \ \ 0 \ -2 \\ -2 \ \ \ \ 0 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$\Rightarrow \bar{\xi}\bar{\eta}=\pm 1$$ -(4c)

Thus we see that by a rotation of coordinates, form (4a) or (4b), which is the standard canonical form representign a hyperbola reduces to form (4c). This simplified form is used mostly in algebraic calculations and even in high-school geometry lessons.

Term Project: Understanding the Plasma-Wall Problem
I am a doctoral candidate enrolled in the Mechanical & Aerospace Engineering Department at the University of Florida, Gainesville. I work in the Computational Plasma Dynamics Laboratory & Testing Facility under the supervision of Dr.Subrata Roy.

On earth, we know and have learnt to live with only three states of matter. However Sir William Crookes identified Plasma, the fourth state in 1879. It is by far the most common form of matter. It is present in the stars and in the tenuous space between them. It is present, though invisible, in almost 99.9% of the visible universe. Ideally plasma consists of neutrals, electrons and ions. However commercial plasma is ionized gas. In order to separate electrons and ions from atoms, energy is needed and this energy can be in any form, thermal, electric etc. Most commonly, a gas is passed through a pair of electrodes and an electric field. This electric field provides the energy needed to ionize the gas into atoms and ions. Plasma has helped create a better understanding of the world we live in. It provides us the prospect of abundant energy. It provides alternatives to surface cleaning, waste removal, in the medical industry and many more.

In my research, we focus on Dielectric Barrier Discharges(DBDs) & their applications. Commerically produced plasma is classified into two types- Volume Plasma (wherein the plasma is produced in the discharge space between two electrodes) & Surface Plasma (wherein both electrodes are embedded on the same surface). DBDs are a kind of surface plasma. They are generated between two electrodes with a dielectric barrier in between them. The electrodes are separated by a gap of a few millimetres. Typically DBDs are operated at voltages of 1-100 kV and frequencies of 50Hz-1MHz. When electric field is sufficiently high to cause breakdown of the discharge gas, a large number of micro-discharges can be observed emanating from the electrodes.

Current Research Focus- Plasma Sterilization Plasma Sterilization is a process in which low-temperature plasma is utilized to kill micro-organisms. Our research focus is on using Dielectric Barrier Discharges for this purpose. Plasma is produced usually by passing a gas through an electric field and ionizing it. Dielectric Barrier Discharges are a special case, in which two electrodes are embedded into the top and bottom of a dielectric surface (an insulator). An electric field is produced between these two electrodes. When a gas is passed through this electric field, it is ionized, producing ions, electrons, neutrals and UV photons. All of these species collide with the micro-organism, reacting with its essential elements and forming harmless compounds. The micro-organism, when devoid of its essential elements becomes harmless and is thus inactivated. More information about preliminary experiments on this subject is given here.

Abstract
One of the oldest problems that has sparked a lot of research in plasma physics is the plasma-wall problem. The sheath is a region of electric field that exists between the plasma and the boundary into which it runs. The basic problem of plasma running into a wall needs to be understood in order to better visualize some of the other aspects of plasma physics. In this paper, we attempt to understand the processes happening in the sheath region. There exist a set of governing equations, akin to the governing equations of fluid dynamics that outline the physical processes running in the plasma. These equations are discretized using a simple, first order finite difference scheme and then solved using the Newton-Raphson Method, as described in a previous paper[3]. The boundary conditions applied however are those pertaining to the sheath region existing between plasma and a solid wall. The objective in conducting this analysis is to investigate the accuracy of the numerical scheme used. If the numerical scheme is accurate, the electron density at the wall should go to zero. Further a probability density function is constructed using electron density data generated from the solver code.

Introduction
To completely describe the state of plasma, we would need to write down all the particle locations and velocities and describe the electromagnetic field in the plasma. But considering the zillions of particles, this would be an impossible task! In order to make our life easier, we construct models which simplify the processes taking place in plasma.

Models are of two types:
 * Fluid Model- This describes plasmas in terms of smoothed quantities like density and averaged velocity around each position. A general description is the two-fluid picture, where ions and electrons are described separately. Fluid models usually model the particles using a Maxwell-Boltzmann Distribution. However they are usually unable to capture accurately finer structures such as beams or double layers.
 * Kinetic Model- While being more accurate than the fluid model, it is usually more computationally intensive. It describes the particle velocity distribution function at each point in the plasma and therefore does not need to assume a distribution function like the fluid model. The more commonly used Particle-In-Cell (PIC) technique includes kinetic information by following the trajectories of a large number of individual particles

Understanding plasma sheaths is perhaps one of the oldest problems in plasma physics. A plasma sheath is the localized electric field that separates plasma from a material boundary. It confines the more mobile species (electrons) in the plasma and accelerates the less mobile species (ions) out of the plasma and toward the walls. This means that in the typical case, where the electrons are more mobile than the positively charged ions, the electric field in the sheath points towards the boundary.This behaviour can be further influenced by external processes such as variations in the applied electric field or the imposition of an external magnetic field.

Modeling of the behaviour of species in the Sheath Region
Many numerical models have been developed to describe Sheaths. The simplest visualization of this process is the behaviour of electrons and ions in the sheath region. Consider a planar slab of collisionless plasma, modeled in 1D space, using a fluid approach and assuming a Boltzmannian electron distribution. The governing equations for such a model are given below:

Continuity Equation for Ion flux:

$$ \frac{\partial (n_{i}v_{i})}{\partial x}= zn_{e}$$ ---(1)

Continuity Equation for Electron flux:

$$ \frac{\partial (n_{e}v_{e})}{\partial x}= zn_{e}$$ ---(2)

Momentum of ions:

$$ \frac{\partial (Mn_{i}v_{i}^{2})}{\partial x}+en_{i}\frac{\partial(\psi)}{\partial{x}}= 0 $$ ---(3)

Momentum of electrons:

$$ \frac{\partial (mn_{e}v_{e}^{2}+kn_{e}T_{e})}{\partial x}-en_{e}\frac{\partial(\psi)}{\partial{x}}= 0$$(4)

Poisson Equation:

$$ \epsilon_{0}\frac{\partial^2 \psi}{\partial x^2}= -e(n_{i}-n_{e})$$ --(5)

where k is the Boltzmann Constant, m is the electron mass, M is the ion mass, Te is the electron temperature,ve is the electron velocity, viis the ion velocity, $$\psi$$ is the electric potential, z is the frequency of direct ionization, e is the electron charge, $$\epsilon_{0}$$ is the permittivity of free space, niis the ion density and ne is the electron density.

More details on these equations and their non-dimensionalization can be obtained from the paper by Sternberg and Godyak on Asymptotic Matching and the Sheath Edge

For our purpose, we take the normalized versions of Equations (1)-(5) used in the paper by Roy & Gaitonde.

$$\frac{\partial{(y_{i}V_{i}})}{\partial{x}}= y_{e}= exp(-\phi)$$--(5)

$$ V_{i}\frac{\partial{V_{i}}}{\partial{x}}-E+V_{i}\frac {y_{e}}{y_{i}}=0$$---(6)

$$ \frac{\partial{E}}{\partial{x}}- \kappa^{2}(y_{i}-y_{e})= 0$$ --(7)

$$ E= \frac{\partial{\phi}}{\partial{x}}$$(8)

where $$y_{i}$$ is the normalized ion density, $$y_{e}$$ is the normalized electron density, $$V_{i}$$ is the normalized ion velocity, E is the electric field strength and $$\phi$$ is the electric potential.

Solution through Numerical Schemes
A finite difference scheme is applied to discretize the first derivatives in Equations(5)-(7). Equation(8) is just needed to calculate the electric potential $$\phi$$ Finite difference schemes are usually applied for complex, non-linear partial differential equations, wherein the solution space is divided into n number of grid points and the required value at each grid point is calculated. For the single derivative, there are typically three types of schemes: Forward difference, Backward difference and Central Difference. Each of these have a different order of accuracy and are applied depending on how accurate the numerical scheme applied has to be.

Forward Difference$$\Rightarrow \frac{\partial {A}}{\partial x}= \frac {A_{n+1}-A_{n}}{\delta x}$$

Backward Difference$$\Rightarrow \frac{\partial {A}}{\partial x}= \frac {A_{n}-A_{n-1}}{\delta x}$$

Central Difference $$\Rightarrow \frac{\partial {A}}{\partial x}= \frac {A_{n+1}-A_{n-1}}{2\delta x}$$

where An+1 is the value of A(the quantity being calculated) at the n+1th grid point and An-1 is the value of A at the n-1th grid point.

The Central Difference Scheme usually has an order of accuracy higher than the other two schemes but this does not necessarily mean that using this scheme might give us the most accurate solution. The nature of the PDE being solved should also be taken into account. This is where we define terms like 'stability' and 'consistency' of a finite difference scheme. For more information, Computational Fluid Dynamics- John.D.Anderson is a very informative book to read. The basics of fundamental numerical schemes as well as more complex ones are outlined along with illustrative applications to common problems in Fluid Mechanics like the Burger's Equation.
 * Hence once Equations (5)-(7) are discretized using forward difference, they reduce to a set of non-linear equations in which the unknowns are yi, Vi and E. This set of non-linear equations is solved using the Newton-Raphson Method. Once these are known, ye can be calculated using Equation(5).

Newton Raphson Method

Consider a typical problem that gives N functional relations to be zeroed involving variables xi, i=1,2....N.

f(xi)= 0, i=1,2....N

Each of these functions can be expanded into a Taylor Series i.e $$ f(x_{i}+\delta x_{i})= f(x_{i})+ \delta x_{j}\frac {\partial{f_{i}}}{\partial{x_{j}}}+O(\delta x^2)$$

Assuming $$f \to 0$$, we get $$ \Sigma \alpha_{ij}\delta x_{ij}= \beta_{ij} $$ where $$ \beta_{ij}= -f_{i} $$

$$\alpha_{ij} $$ is known as the Jacobian. Here Equations 5-7 are the equations that are discretized and contain three unknowns. So $$\alpha$$ is going to be a 3x3 matrix and $$\beta$$ will be a 3x1 matrix. $$\delta x_{ij}$$ is then solved for iteratively in MATLAB.

 * Now, once the electron density distributions and ion-density distributions are calculated, how do we estimate what happens down the line, say after like 100 iterations? Does the electron density in the Sheath Region remain zero always, as it should? Does the ion-density increase or decrease? This check achieves two purposes: The accuracy of the numerical scheme is verified and the sheath behaviour after a long time is understood. This check is performed with the help of a probability density analysis. A MATLAB code was written to simulate the entire process described above. This code furthe takes the electron-density data generated, divides it into 100 blocks and constructs a probability density function. This is repeated for every iteration so that we know the nature of the pdf after about 100 iterations.

Results

Fig(1) shows the published results from the paper by Roy et.al[3]. In this paper, equations (1-7) to (1-10) are solved using a second-order time accurate, implicit finite element method using a computational domain extending from x=0 to the wall. Fig. 1(a) is a plot of ion density versus position x. Fig. 1(b) is the plot of ion velocity as a function of position, x. Fig. 1(c) is the plot of electric potential as a function of x, while Fig. 1(d) represents the plot of electric field versus x. As expected, since the more mobile species is confined within the plasma and the less mobile species is accelerated out, it is seen accordingly that at xw~0.71, the ion velocity increases. Since electrons are being confined, one would expect electric field at wall to rise (more charge being accumulated). This is seen in Fig. 1(c) and consequently in Fig. 1(d).

In Fig.(2), the plasma parameter distributions obtained from the MATLAB code used are given. In fig. 2(a), as can be seen, ion density prediction matches the trend as predicted in Fig.1(a) i,e ion density decreases in the direction of the wall. This is expected. Fig. 2(b), which shows the plot of ion velocity versus x also seems to match the trend predicted in 1(b). It makes senses too because one would expect ion velocity to increases as sheath approaches. Fig.2(c) makes perfect sense because electron density decreases to zero as wall is approached. The value of x at which ne goes to zero is a little earlier than expected (at x=0.2). Also to be noted is the important fact that ion density at wall does not go to zero while electron density does go to zero, which simply confirms the fact the electrons stay behind in the plasma while ions accelerate towards the wall. Fig. 2(d) is a plot of the potential versus x, which also matches the plot shown in fig.1(c). This tells us that the code is predicting the behaviour of the species in the sheath correctly to some extent. The linear trend present in fig. 2(a), 2(b), 2(d) is of slight concern as we are looking for more of an exponential trend. This could be because of an error in the code or due to slight errors introduced by the numerical scheme used. Note that the results in fig.(1) are the result of a finite element code while the results in fig. (2) are those computed from a finite difference scheme. Issues of accuracy tend to exist.

Fig. 3 shows the pdf analysis for the electron density. Electron density data is taken from the code, divided into 100 blocks and a pdf is constructed. Fig.3(a) shows this pdf as a bar diagram. Further the convergence of this pdf is investigated by calculating moments as high as Ethumb|left|Fig.3:PDF analysis for electron density dataNavya4.jpg There are a couple of important points to note in these two figures. Fig.3(a) shows that there is a 4% probability of electron density going to zero at the wall. Fig. 4(a) shows that the ion density does not go to zero at the wall. Fig. 3(b) and Fig. 4(b) confirm convergence of the pdf for both electron density and ion density data. Note also that the electron density is steady after about 40 iterations while the ion density is steady after about 20 iterations

Conclusion & Future Work
The governing equations for the plasma wall problem were solved using a finite difference scheme to discretize them. The electron and ion density are calculated using a MATLAB code. The computed distributions of ion and electron density exhibit the trend shown in the published results as shown in Fig.(1). The linear trend may be due to accuracy errors in solver code. However the ion and electron density plots are very similar to ion and electron behaviour in sheath region. Also the PDF analysis shown in Fig. (3) and (4) for electron and ion density data respectively shows that there is convergence after 100 iterations, proving that the results predicted by the code are accurate. Also we see that there is a 4% probability of the electron density going to zero. This value might further reduce, provided the accuracy errors in the code are resolved.

Future work involves using more complex, higher order numerical schemes to discretize the governing equations. Also a simplistic case of the plasma-wall phenomenon was examined here. More complex phenomenon can be introduced into this simulation in order to further understand this phenomenon.

title: User:Egm6322.s09.Three.nav


text: User:Davidsoiseth

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

title: User:Davidsoiseth


text:
User:Cis4914.s09/Documentation

 Document your work here.

Cis4914.s09 18:32, 4 February 2009 (UTC)

 * Document in details all steps that you did.
 * Reread the documentation to see whether you had made any implicit assumptions; if yes, document explicitly these implicit assumptions.

Egm6322.s09 15:45, 19 February 2009 (UTC)

= Things to do =
 * test

= Documentation =

Why we chose Ubuntu 8.04 (Hardy Heron)
Ubuntu has an uncommonly steady release cycle. A new release is made available every six months with LTS versions coming out once every four cycles. Regular releases are maintained for 1.5 years. LTS releases are maintained for 3+ years. Because of the stability provided by LTS releases we decided to go with Ubuntu 8.04 (Hardy Heron) since it is the latest LTS version.

Keyboard shortcuts
Switch to command line interface (CLI) Ctrl-Alt-F1 Switch to GUI Ctrl-Alt-F7

Set up It's All Text to edit wiki articles
Documenting intensive command line work and configuration files can be very cumbersome with normal wiki text boxes. The tool 'It's All Text!' will invoke an external editor of your choice instead of having to use the text boxes provided by the wiki. If you have firefox 2 or 3 and use Windows, Linux, or Mac then you are ready to begin installation.

You will need to install gvim in order for this to work under Linux distributions. For Ubuntu: sudo apt-get install vim-gnome You can download and install the 'It's All Text!' plugin here. After you install it Firefox will prompt you to restart. Go ahead and restart Firefox. Now you need to go to a wiki page and click the 'edit' button or hit 'alt-shift-e'. You should see a little 'edit' bubble at the bottom right of the text box area. Right click it and select 'Preferences'. We now need to enter the path to the editor of our choice. The path '/usr/bin/vim' will not work as we need the graphical version. Enter '/usr/bin/gvim' into the empty box and hit the 'close' button to save the changes.

Now when you are on a wiki edit page clicking the 'edit' bubble will bring up a gvim window to edit in. After making changes you can type :w to send the changes back to the wiki. (the wiki text box will flash yellow) Hit 'alt-tab' to change windows back to the wiki and hit 'alt-shift-p' to preview the changes.

More tips and tricks with 'It's All Text!' can be found here.

= Preparing base system in hydra3 =

Register an MSDNAA account
The Microsoft Developers Network Academic Alliance (MSDNAA) product is a licensing program  donated to the College of Engineering by Microsoft which allows all College of Engineering units,  students, faculty, and staff to use various Microsoft software products and online resources  for academic and non-profit research purposes.  Students may obtain an official copy of Microsoft Windows XP through MSDNAA. Send an e-mail to software@eng.ufl.edu asking whether you have an account at on MSDNAA or not and if there is an account then what is the username and password
 Normally UF engineering students have a MSDNAA account created automatically but
 if one does not have an account then he/she can Register here.
 Fill in the form with appropriate data and if you are not registered within
 two business days then email software@eng.ufl.edu with a request for creating a MSDNAA account.

 You want to document whom you contacted to know that "Normally UF engineering students have a MSDNAA account created automatically". Give contact info (name, phone, e-mail address, etc.) Egm6322.s09 14:21, 26 February 2009 (UTC)

Download Windows XP
After your account has been created, you may log into the MSDNAA system here.
 In case you have forgotten your password: (a) Click here and enter your gatorlink email-id as your username. The password will be e-mailed to your gatorlink email
 (b) If step (a) fails, then send an email to software@eng.ufl.edu reporting the problem with your UF-ID
 After you have logged in, click on the "Software" tab on the upper left side
 Either browse through the list of softwares shown or enter the name
 of the specific software you want to download in the blank next to "Search" and click "Go" tab
 Click on the tab "Add to cart".
 Click on the "I Agree" tab
 Next click on the "Download" button
 If asked "Are you sure you want to run this software?" click on "Run"
 Type the desired path where you want to download and click on "Resume"
 After Downloading, the process of Unpacking will automatically start, you need not do anything now
 After unpacking is complete click on "Launch Install"
 The folder will open where the ISO file has been downloaded
 Enter an empty CD in the CD ROM and burn the ISO file on this CD. This CD is called the Live CD

Installing Windows XP
Boot up the computer with the above Windows XP Installation disk.
 Click any button to boot from the CD
 Setup will start loading files
 Press ENTER to setup Windows XP
 Click on F8 to agree to the agreement
 Use UP and DOWN arrows to select an item from the list, select C and Press ENTER.
 To continue setup press C
 Use the UP and DOWN arrows and Select "Format partition using NTFS file system (quick)" and Press ENTER
 Press F to format the drive
 Setup will start formatting
 Setup will then copy all the files
 Setup then initializes Windows XP configuration
 Then the Computer reboots
 When the computer reboots, either remove the live CD or when asked to press any button to start setup
 do not press any button
 Installing windows phase will start which will take around 35 minutes
 In the Regional and Language option click on "Customize" and select English (US)and click Next
 Enter the name as clesm-g and organization as CLESM and click Next
 Enter the product key which has been provided with the software.
 If you have forgotten the product key then log into your MSDNAA account here
 You can see the list of softwares you have downloaded, under the software which you have downloaded on your live CD click on "Download" which will lead you to another page
 Note down the serial number, it is your product key and click Next
 Enter the computer name and administrator password and click Next
 In the Date and time settings, set the time zone to GMT – 5:00 Eastern Time (US and Canada) and enter the date and time and click Next
 For Network Settings, select Typical Settings and click Next
 In the next step, choose the option No to the question "Do you want this computer to be a member of a domain?" and click Next
 Windows XP gets installed and system restarts

To set up the network do the following:
 After you have logged in click on Start>Settings>Control Panel>Network connections
 Right click on the Local Area Connection and click on Properties
 Under the Networking tab, select the item which ends with "TCP/IP" in brackets and click on Properties
 Select the option where you will have to enter the IP address, Subnet mask and Default gateway.
 Enter the following:

 Static IP address: 10.227.237.37
 Subnet mask: 255.255.255.0
 Default gateway: 10.227.237.1

 Then select the option you will have to enter the DNS server addresses:
 Preferred DNS: 128.227.116.4
 Alternative DNS: 128.277.128.24
 Click on OK

Partitioning the drive
Boot up the computer with the Ubuntu CD already in the drive slot.
 You will encounter a menu with 5 different options. Select the 'Install Ubuntu' option.
 In case it leads to warning or does not load for some reason, reboot the system by clicking Ctrl+Alt+Delete and when you
 reach the menu click F4 and choose Safe Graphics mode and then select the 'Install Ubuntu' option
 You will soon see a screen prompting you for the language selection. Make sure 'English' is selected and hit the Forward button.
 Next comes the Time Zone. Select 'New York' and hit the Forward button again.
 Now comes the keyboard layout. Select 'USA' and hit a few keys in the text area to ensure this setting is correct and hit Forward again.

 We are now ready to partition the drive.
 Our goal is the following:
 1 x 8GB ntfs (for Microsoft Windows XP, already done)
 7 x 4GB ext3 (1 for Ubuntu Hardy, the rest for future installs)
 1 x 1GB swap (linux swap space)

 At the first menu select 'Manual' and then hit forward when it asks how you would like to partition the disk.
 The partition table should look like the following:
 /dev/sda
 /dev/sda1 ntfs
 free space

 Click on the 'free space' and hit the 'New partition' button.
 For 'Type for the new partition' select 'Primary'.
 For 'New partition size in MB' write in 4000.
 For 'Location for the new partition' select 'Beginning'.
 For 'Use as' select Ext3.
 You can leave 'Mount point' empty for now.
 Note that when you are back at the main partition screen the partition size will not be exactly 4000 MB. This is expected.
 Repeat the above steps until you have 7 partitions of 4 GB. Be sure that for these you select 'Logical' for 'Type for the new partition'.
 Otherwise, we will be limited to only 4 partitions total.
 There is now about 3.6 GB left. Use 2.6 GB of this for another logical partition.
 Now comes time to create our swap space. There should be 1GB left. Click the 'free space' and select 'New partition'.
 Everything will be the same except for the 'Use as' area. Select 'swap area' and click Ok.
 Now we need to select our '/' or root mount point.
 Click on /dev/sda2 and click 'edit'.
 For 'Mount point' select '/'. This is the partition we will be installing Hardy to.
 You will receive some warnings about mount points for the logical partitions. You can ignore these.

 The partition table should now look like the following:
 /dev/sda
 /dev/sda1 <=primary (our windows partition)
 /dev/sda2 <=primary (where we will install ubuntu)
 /dev/sda5 <=logical (note the jump in number)
 /dev/sda6
 /dev/sda7
 /dev/sda8
 /dev/sda9
 /dev/sda10
 /dev/sda11
 /dev/sda12

 We are now done partitioning. Click Forward to continue.

Installing Hardy
You will now be at a screen requesting your personal information. Fill in the appropriate fields and hit 'Forward'.
 You will now be asked about importing accounts. Do not. Hit 'Forward' to continue.
 The screen you are on now will give you an overview of your installation. Hit 'Install' when you feel comfortable.
 Depending on hardware this process can take 15 minutes to an hour.
 You will eventually be prompted to restart the computer. Do this and take the disk out when prompted.
 When the computer boots back up you'll be taken to a screen with a few boot options. For Ubuntu, select the topmost option and hit the enter key.
 If you make it to a log in screen, success!

Network information for hydra3
IP: 10.227.237.37
 Subnet mask: 255.255.255.0
 Gateway: 10.227.237.1 <--note that this is now 10.xxx instead of 128.xxx
 DNS: 128.227.116.4

 For Galileo:
 the correct settings in the file /etc/network/interfaces are:

auto eth0
 iface eth0 inet static
 address 128.227.237.55 (galileo)
 gateway 128.227.237.1
 network 128.227.237.0
 broadcast 128.227.237.127

The DNS nameservers in the file /etc/resolv.conf are:

search mae.ufl.edu
 nameserver 128.227.128.24
 nameserver 128.227.116.4
 nameserver 128.227.42.51

Resolve internet connection problem in Hydra3
System>Administration>Network
 Click on the unlock tab and enter password.
 Change Gateway address for eth0 to: 10.227.237.1
 Click OK.

Success. An internet connection has been established.
 There appears to have been an issue with the host (hydra3) not being under the same subnet as the gateway. Whoops.

System updates
Many updates have been made since the creation of the Hardy .iso install CD.
 To update the repository information and download/install these updates do the following.
 sudo apt-get update
 sudo apt-get upgrade
 The following will output to the terminal screen.
 Enter 'y' and hit 'enter' to download and install the updates.
 Reading package lists... Done Building dependency tree Reading state information... Done The following packages have been kept back: bind9-host dnsutils libbind9-30 libisccc30 libisccfg30 linux-generic linux-headers-generic linux-image-generic linux-restricted-modules-generic ssl-cert The following packages will be upgraded: acpi-support acpid alacarte anacron app-install-data-commercial apt apt-utils avahi-autoipd avahi-daemon base-files bsdutils compiz-fusion-plugins-main console-setup cpp-4.2 cupsys cupsys-bsd cupsys-client cupsys-common dbus dbus-x11 deskbar-applet eject eog evince evolution evolution-common evolution-data-server evolution-data-server-common evolution-exchange evolution-plugins firefox firefox-3.0 firefox-3.0-gnome-support firefox-gnome-support foo2zjs gcalctool gcc-4.2 gcc-4.2-base gdb gdm gedit gedit-common gksu gnome-about gnome-cards-data gnome-desktop-data gnome-games gnome-games-data gnome-panel gnome-panel-data gnome-power-manager gnome-system-monitor grub gstreamer0.10-tools gtk2-engines gtk2-engines-murrine gtk2-engines-pixbuf gtk2-engines-ubuntulooks gtkhtml3.14 guidance-backends gvfs gvfs-backends gvfs-fuse hal hal-info hpijs hplip hplip-data initramfs-tools initscripts iproute jockey-common jockey-gtk language-pack-en language-pack-en-base language-pack-gnome-en language-pack-gnome-en-base libavahi-client3 libavahi-common-data libavahi-common3 libavahi-compat-libdnssd1 libavahi-core5 libavahi-glib1 libavahi-ui0 libc6 libc6-i686 libcamel1.2-11 libcupsimage2 libcupsys2 libdbus-1-3 libebook1.2-9 libecal1.2-7 libedata-book1.2-2 libedata-cal1.2-6 libedataserver1.2-9 libedataserverui1.2-8 libegroupwise1.2-13 libexchange-storage1.2-3 libexif12 libffi4 libfreetype6 libgadu3 libgcc1 libgdata-google1.2-1 libgdata1.2-1 libgksu2-0 libglib2.0-0 libglibmm-2.4-1c2a libgnome-desktop-2 libgnutls13 libgomp1 libgphoto2-2 libgphoto2-port0 libgstreamer0.10-0 libgtk2.0-0 libgtk2.0-bin libgtk2.0-common libgtkhtml3.14-19 libgtksourceview2.0-0 libgtksourceview2.0-common libgvfscommon0 libgweather-common libgweather1 libhal-storage1 libhal1 libisc32 liblcms1 libldap-2.4-2 liblwres30 libnautilus-extension1 libnm-glib0 libnm-util0 libnspr4-0d libnss3-1d libntfs-3g23 libpanel-applet2-0 libpango1.0-0 libpango1.0-common libpcre3 libperl5.8 libpoppler-glib2 libpoppler2 libpurple0 libsmbclient libsmbios1 libsnmp-base libsnmp15 libsoup2.4-1 libssl0.9.8 libstdc++6 libtiff4 libvorbis0a libvorbisenc2 libvorbisfile3 libwnck-common libwnck22 libxml2 libxml2-utils libxslt1.1 linux-headers-2.6.24-19 linux-headers-2.6.24-19-generic linux-image-2.6.24-19-generic linux-restricted-modules-2.6.24-19-generic linux-restricted-modules-common login logrotate module-init-tools mount myspell-en-gb myspell-en-us myspell-en-za nautilus nautilus-data nautilus-sendto nautilus-share network-manager ntfs-3g ntpdate openoffice.org-base-core openoffice.org-calc openoffice.org-common openoffice.org-core openoffice.org-draw openoffice.org-gnome openoffice.org-gtk openoffice.org-help-en-gb openoffice.org-help-en-us openoffice.org-impress openoffice.org-l10n-common openoffice.org-l10n-en-gb openoffice.org-l10n-en-za openoffice.org-style-human openoffice.org-thesaurus-en-us openoffice.org-writer openssl passwd pciutils perl perl-base perl-modules pidgin pidgin-data poppler-utils procps python-apt python-gobject python-gtkhtml2 python-libxml2 python-uno python2.5 python2.5-minimal rdesktop readahead samba-common smbclient sudo sysv-rc sysvutils thunderbird-locale-en-gb totem totem-common totem-gstreamer totem-mozilla totem-plugins ttf-opensymbol tzdata ubuntu-docs ufw update-manager update-manager-core update-notifier update-notifier-common util-linux util-linux-locales vim-common vim-tiny vinagre xkb-data xserver-xorg-video-amd xserver-xorg-video-ati xserver-xorg-video-geode xserver-xorg-video-intel xsltproc xterm xulrunner-1.9 xulrunner-1.9-gnome-support yelp 247 upgraded, 0 newly installed, 0 to remove and 10 not upgraded. Need to get 270MB of archives. After this operation, 1139kB of additional disk space will be used. Do you want to continue [Y/n]?

Moving /home to it's own partition
The following process was outlined in this guide as well as this guide.

 To take a look at our partition table run the following: sworkman@hydra3:~$ sudo fdisk -l Disk /dev/sda: 30.0 GB, 30020272128 bytes
 255 heads, 63 sectors/track, 3649 cylinders
 Units = cylinders of 16065 * 512 = 8225280 bytes
 Disk identifier: 0xa033a033
 Device Boot     Start         End      Blocks   Id  System
 /dev/sda1  *           1         736     5911888+   7  HPFS/NTFS
 /dev/sda2            737        1222     3903795   83  Linux
 /dev/sda3           1223        3649    19494877+   5  Extended
 /dev/sda5           1223        1708     3903763+  83  Linux
 /dev/sda6           1709        2194     3903763+  83  Linux
 /dev/sda7           2195        2680     3903763+  83  Linux
 /dev/sda8           2681        3166     3903763+  83  Linux
 /dev/sda9           3167        3519     2835441   83  Linux
 /dev/sda10          3520        3649     1044193+  82  Linux swap / Solaris

sda5 through sda9 are all formated to ext3 and ready for use. We'll use sda5.

First we need to create a mount point.
 We'll create it under /mnt. sudo mkdir /mnt/newhome Next, mount the partition: sudo mount -t ext3 /dev/sda5 /mnt/newhome Next, copy files over from old /home:
 A regular cp will not do the job completely, so we'll use a command from the Debian archiving guide. The following command would not work using sudo. You must be logged in as root. cd /home find. -depth -print0 | cpio --null --sparse -pvd /mnt/newhome Unmount the new partition: sudo umount /mnt/newhome Backup the old /home: sudo mv /home /old_home Recreate the /home directory: sudo mkdir /home Mount the new /home: sudo mount /dev/sda5 /home Now we have to tell Ubuntu to mount the new home directory on boot. Add the following line to /etc/fstab: /dev/sda5 /home ext3 nodev,nosuid 0 2 Now it's time to reboot and see if everything goes smoothly.
 There appears to be permission problems with the $HOME/.dmrc file and .ICEauthority file. The following procedure will fix this.
 Boot the computer into Recovery Mode:
 Restart the computer and select the second entry (it will have 'Recovery Mode' next to it) After the boot up process is complete you'll be presented with a menu. Select the 'Drop to root shell prompt' entry. Enter the following commands (note that there is a set of commands for each affected user): chown -R sworkman:sworkman /home/sworkman chmod 644 /home/sworkman/.dmrc chmod 644 /home/sworkman/.ICEauthority chown -R vql:users /home/vql chmod 644 /home/vql/.dmrc chmod 644 /home/vql/.ICEauthority chown -R clesm-g:users /home/clesm-g chmod 644 /home/clesm-g/.dmrc chmod 644 /home/clesm-g/.ICEauthority exit You will now be back at the same menu as before. Check the 'Continue Normal Boot Process' and see if you able to log in without problems. After you have checked to make sure everything is in working order you can remove the old mount point we made under /mnt and the old home directory under /old_home sudo rm -r /mnt/newhome sudo rm -r /old_home Good to go now. Users are in their own partition and permission problems have been resolved.

Setting up root account
Ubuntu no longer sets a root password upon installation (security reasons). The following process outlines the steps needed to enable the root user: stephen@nix:~$ su Password: su: Authentication failure stephen@nix:~$ sudo passwd root [sudo] password for stephen: Enter new UNIX password: Retype new UNIX password: passwd: password updated successfully stephen@nix:~$ su Password: root@nix:/home/stephen# whoami root root@nix:/home/stephen# exit exit stephen@nix:~$

Configuring SSH
The client is installed by default but to be able to log into the machine the server package is needed. Install using the following: sudo apt-get install openssh-server No further configuration is necessary. If you would like to change the default behavior of the openssh server application you can edit the following file: sudo vim /etc/ssh/sshd_config

= Preparing base system in euclid =

OS Installations
The goal here was to have Euclid running in a similar fashion to Hydra3.
 Euclid already had a fresh install of Windows XP so this partition was left alone.
 Using the steps outlined above for the Hydra3 installation, Euclid received a fresh install of Ubuntu Hardy Heron.
 There was one problem with the installation for Euclid. Booting into the install disk resulted in the following error:
 user not known to the underlying authentication module This was easily worked around by doing the following:
 Boot into the install disk like normal, selecting English as your primary language.
 Hit 'F4' to bring up different boot options.
 Select 'Safe Graphics Mode.'
 Now you can select 'Install Ubuntu' and it will proceed as it should.

Network information for euclid
IP: 10.227.237.42
 Subnet Mask: 255.255.255.0
 Gateway: 10.227.237.1
 DNS: 128.227.116.4, 128.227.116.1, 128.227.128.24, 128.227.42.51
 Search: mae.ufl.edu

Euclid customizations
The goal at this point is to have Hydra3 and Euclid running in the same manner. All changes to Hydra3 mentioned above (moving the /home partition, configuring an ssh server, etc) were all done with Euclid. For clarity and space I am not going to repeat these steps again here. Just note that everything done for Hydra3 was also done for Euclid.

= Grid computing =

SciLab

ProActive

ProActive applications

Find out how to access this applications web page from the home page of ProActive.

ProActive
ProActive is an open source software package for parallel, distributed, and multi-core programming. It will be the base of our grid framework for use with Scilab. The official site for ProActive can be found here.

Downloading ProActive
From the home page (linked above), click on the 'Downloads' tab and select 'ProActive Programming'.
 You will be directed to a page requesting some personal information and an email. Fill these fields out and click 'Go to the download page' button.
 You will now have to agree to the GNU General Public License for ProActive. Click the 'I agree' button at the bottom of the page to continue.
 You will finally be at the download page. There are a lot of different packages hosted on this page. The one we want is at the top under the heading "ProActive Programming: Java Parallel Frameworks -- Enterprise-ready Java Grid". There's no reason not to grab the most recent version so go ahead and click on 'ProActive Programming 4.0.2 (zip)'. A download dialog box will open. Save to the directory of your choice.

Installing ProActive
We need to first install the Sun Java Development Kit (JDK) 5.0. Other packages will be marked for installation such as the JRE which we also need.: sudo apt-get install sun-java5-jdk We now need to set the environment variable for the JDK (note BASH syntax!) Add the following to the .bashrc file in your home directory as well: export JAVA_HOME=/usr/lib/jvm/java-1.5.0-sun-1.5.0.16 ProActive will be installed beneath the /etc directory. Lets start by unzipping the the downloaded file. Assuming you saved the .zip to your home directory run the following command: unzip ProActive-4.0.2.zip A new directory will be present on the desktop with the same name (minus the .zip extension of course). We'll now move this under /etc using the following: sudo mv ProActive-4.0.2 /etc/proactive We now need to set the classpath for using ProActive to write applications.
 First cd to the install directory: cd /etc/proactive Next, enter the following (note that this is BASH syntax!): PA_PATH=$(pwd) export CLASSPATH=.:${PA_PATH}/lib/dist/ProActive_examples.jar:\ ${PA_PATH}/lib/dist/ProActive/ProActive.jar:${PA_PATH}/lib/dist/ibis:\ ${PA_PATH}/lib/dist/asm-2.2.1.jar:\ ${PA_PATH}/lib/dist/log4j.jar:${PA_PATH}/lib/dist/xercesImpl.jar:\ ${PA_PATH}/lib/dist/fractal.jar:${PA_PATH}/lib/dist/bouncycastle.jar:\ ${PA_PATH}/lib/dist/javassist.jar We now need to edit the java.policy file to set permissions: sudo vim /etc/java-1.5.0-sun/security/java.policy Now add the following lines: grant codeBase "file:/etc/proactive/*" { permission java.security.AllPermission; }; ProActive supports logging through log4j. It's a nice feature, so we might as well use it. Install log4j by running the following: sudo apt-get install liblog4j1.2-java The option -Dlog4j.configuration=file:pathToFile will specify which logging policy file to use within ProActive. An example of such a file is below. If you choose not to use one, ProActive will use the file proactive-log4j by default. # This file represents the log4j config file for non regression tests. # It is given as property (in the proactive.xml) when starting nonregression tests # The default logging level is INFO # The root logger logs in the test.log file log4j.rootLogger=INFO,CONSOLE # If INFO is enabled Jetty is too verbose at startup log4j.logger.org.mortbay = WARN ########### The following are existing categories in ProActive #log4j.logger.proactive = INFO #log4j.logger.proactive.classloader = DEBUG #log4j.logger.proactive.events = DEBUG #log4j.logger.proactive.runtime = DEBUG #log4j.logger.proactive.body = DEBUG #log4j.logger.proactive.mop = DEBUG #log4j.logger.proactive.groups = DEBUG #log4j.logger.proactive.sync_call = DEBUG #log4j.logger.proactive.deployment = DEBUG #log4j.logger.proactive.deployment.log = DEBUG #log4j.logger.proactive.deployment.process = DEBUG #log4j.logger.proactive.deployment.filetransfer = DEBUG,CONSOLE #log4j.logger.proactive.filetransfer = DEBUG #log4j.logger.proactive.nfe = FATAL #log4j.logger.proactive.gc = DEBUG #log4j.logger.proactive.ft = DEBUG #log4j.logger.proactive.ft.cic = DEBUG #log4j.logger.proactive.ft.pml = DEBUG #log4j.logger.proactive.communication.transport.http = DEBUG #log4j.logger.proactive.communication.rmi = DEBUG #log4j.logger.proactive.communication.ssh = DEBUG #log4j.logger.proactive.communication.transport.http = DEBUG #log4j.logger.proactive.migration = DEBUG #log4j.logger.proactive.communication.requests = DEBUG #log4j.logger.proactive.examples = DEBUG #log4j.logger.proactive.components = DEBUG #log4j.logger.proactive.components.requests = DEBUG #log4j.logger.proactive.components.activity = DEBUG #log4j.logger.proactive.components.bytecodegeneration = DEBUG #log4j.logger.proactive.components.adl = DEBUG #log4j.logger.proactive.components.gui = DEBUG #log4j.logger.proactive.p2p = DEBUG #log4j.logger.proactive.p2p.startservice = DEBUG #log4j.logger.proactive.p2p.vn = DEBUG #log4j.logger.proactive.p2p.vn.service = DEBUG #log4j.logger.proactive.p2p.nodes = DEBUG #log4j.logger.proactive.p2p.acquaintances = DEBUG #log4j.logger.proactive.p2p.first_contact = DEBUG #log4j.logger.proactive.p2p.skeletons = DEBUG #log4j.logger.proactive.p2p.skeletons.worker = DEBUG #log4j.logger.proactive.p2p.skeletons.manager = DEBUG #log4j.logger.proactive.p2p.daemon = DEBUG #log4j.logger.proactive.security = DEBUG #log4j.logger.proactive.security.node = DEBUG #log4j.logger.proactive.security.session = DEBUG #log4j.logger.proactive.security.body = DEBUG #log4j.logger.proactive.security.manager = DEBUG #log4j.logger.proactive.security.request = DEBUG #log4j.logger.proactive.security.runtime = DEBUG #log4j.logger.proactive.security.policy = DEBUG #log4j.logger.proactive.security.policyserver = DEBUG #log4j.logger.proactive.security.crypto = DEBUG #log4j.logger.proactive.security.psm = DEBUG #log4j.logger.proactive.skeletons = DEBUG #log4j.logger.proactive.skeletons.taskpool = DEBUG #log4j.logger.proactive.skeletons.structure = DEBUG #log4j.logger.proactive.skeletons.environment = DEBUG #log4j.logger.proactive.skeletons.application = DEBUG #log4j.logger.proactive.skeletons.diagnosis	 = DEBUG #log4j.logger.proactive.skeletons.system = DEBUG #log4j.logger.proactive.masterworker = DEBUG #log4j.logger.proactive.masterworker.workermanager = DEBUG #log4j.logger.proactive.masterworker.pinger = DEBUG #log4j.logger.proactive.masterworker.repository = DEBUG #log4j.logger.proactive.masterworker.workers = DEBUG #log4j.logger.proactive.configuration = DEBUG #log4j.logger.proactive.remoteobject = DEBUG #log4j.logger.proactive.jmx = DEBUG #log4j.logger.proactive.jmx.mbean = DEBUG #log4j.logger.proactive.jmx.notification = DEBUG ################ Appenders #################### # # Appender output can be configured by using a pattern layout # See: http://logging.apache.org/log4j/1.2/apidocs/org/apache/log4j/PatternLayout.html # #   - %c the category of the logging event #  - %d the date #  - %m he application supplied message #  - %n the platform dependent line separator character or characters #  - %p the priority of the logging event #  - %t the name of the thread that generated the logging event #  - %X{hostname}         the hostname #  - %X{id@hostname}      the VMID and the hostname #  - %X{shortid@hostname} the short VMID and the hostname (a collision can occur between two shortids, you should use id@hostname) #  - %X{runtime}          the ProActive runtime url (does not work very well since a different MDC is associated to each thread) # CONSOLE appender is used by default log4j.appender.CONSOLE=org.apache.log4j.ConsoleAppender log4j.appender.CONSOLE.layout=org.apache.log4j.PatternLayout log4j.appender.CONSOLE.layout.ConversionPattern=%X{shortid@hostname} - [%p %20.20c{2}] %m%n # Appender FILE writes to the file "tests.log". # This file is recreated a file for each run log4j.appender.FILE=org.apache.log4j.FileAppender log4j.appender.FILE.File=tests.log log4j.appender.FILE.Append=false log4j.appender.FILE.layout=org.apache.log4j.PatternLayout log4j.appender.FILE.layout.ConversionPattern=%5p [%t]: %m%n

Testing ProActive
To make sure everything is working properly we should run a few examples. The examples are located under /etc/proactive/examples. A few interesting examples to run are: The Reader/Writer Problem: /etc/proactive/examples/readers/reader.sh The Dining Philosophers Problem: /etc/proactive/examples/philosophers/philosophers.sh The N-Body Simulation (for use later, as a deployment example for multiple machines): /etc/proactive/examples/nbody/nbody.sh

= Scilab = Scilab is an open source platform for numerical computation and is the reason for which the ProActive grid interface was installed. The Scilab homepage can be found here. It is important to note that the current version of ProActive currently only supports up to version 4.1.2 of Scilab. The current stable version of Scilab is 5.1.

Downloading Scilab
For Scilab to function properly we will need to install version 4.1.2 from SOURCE. The download page is here or alternatively you can just download the file directly.

Installing Scilab
The first thing we need to do is unzip the .tar.gz file we downloaded. You may run the following: tar -xvzf scilab-4.1.2-src.tar.gz We'll now have the folder scilab-4.1.2 on the desktop. Let's move (and rename it to something easier) to /etc like we did with the ProActive installation. sudo mv scilab-4.1.2 /etc/scilab Now we can start the real installation. There are a few requirements we need to meet before we can start. Scilab requires a C compiler, a Fortran compiler, tcl, and a few other dependencies. Let's go ahead and install all these by running the following: sudo apt-get install gcc gfortran tcl tk-dev libxaw7-dev Go to the newly created Scilab directory: cd /etc/scilab We need to configure the system by issuing the following command: ./configure --with-gcc --with-gfortran If configure finishes without any errors you may continue. If it does you likely have some dependency issues. DO NOT CONTINUE until configure finishes successfully.

Now it's compile time. This may take about 15-20 minutes to complete. While still in the /etc/scilab directory run the following: make all We now need to install it into our path. Run the following in the /etc/scilab directory (if unsuccessful run 'make distclean' before attempting again): sudo make install Comment out the following line in the scilab.star file. sudo vim /etc/scilab/scilab.star Here's the line to comment out. It's on line 196: //show_startupinfo; We now need to complete the installation of ProActive. Go to the ProActive directory: cd /etc/proactive/compile Now it's time to compile. Run the following: ./build compile Attempting to run a test of Scilab from within ProActive resulted in the following error: sworkman@hydra3:/etc/proactive/examples/scilab$ ./scilab_test.sh .: 10: Can't open ./../../.././scripts/unix/env.sh Well, it turns out this file and scilab_env.sh are in the wrong place (or at minimum referenced incorrectly). Run the following commands to put them in the right place. mv /etc/proactive/examples/scilab/scilab_env.sh /etc/proactive/scripts/unix/scilab/scilab_env.sh mv /etc/proactive/examples/env.sh /etc/proactive/scripts/unix/env.sh We need to make a quick change to /etc/proactive/examples/scilab/scilab.sh as well. vim /etc/proactive/examples/scilab/scilab.sh Make corrections to look like the following: PROACTIVE=/etc/proactive Now we should be able to launch the Scilab Grid ToolBox: cd /etc/proactive/examples/scilab ./scilab.sh Success.
 * 1) PROACTIVE=$workingDir/../../../.

title: User:Cis4914.s09/Documentation


text:
User:Kylnso

STORY BOARD http://www.mcli.dist.maricopa.edu/authoring/studio/guidebook/storyboard.html

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

CASTING CALL http://tpsonline.org/auditions/ http://www.seattleactor.com/news/modules.php?name=ANews

Winter Quarter 2009 (dec-mar)
User:Kylnso/documentary-dfvp212
 * Documentary - strippers

User:Kylnso/documentary-dfvp212-skate
 * DOC - SKATE

User:Stevenarntson
 * Steve Arntson's class

User:Kaylene_so
 * My page for Steve's class

http://jp216.aisites.com/dae/
 * Audio class

Spring Quarter 2009 (mar-jun)
User:Kylnso/studioproduction-dfvp200
 * Studio Production

User:Kylnso/digitalfilmmaking
 * Digtal Filmmaking

User:Kylnso/windbreakRZ
 * Producing & Directing

title: User:Kylnso


text:
User talk:Stevenarntson/group spaces/analytical writing/group 4

 * Hi Group 4,
 * Glad to see some things starting to appear here--I can see a pretty useful framework underlying this already; keep amplifying your research, add links to other places, etc. Actually, the movie we're going to watch in class today, The Future of Food, might be pretty useful to you.
 * Stevenarntson 14:58, 11 February 2009 (UTC)

 * Hi Group 4,
 * Great additions this week. Generally, I'd like to see some more developed introductions to the topics--right now they somewhat begin with a barrage of information, and could spend a little more time setting up the situation with a good, evocative introduction. Secondly, your "conclusion" section should be titled "solutions" instead; I'd like to see this grow into a larger piece--you're doing a good job right now with outlining the problem, and it makes me curious about how we as a culture can get out of it! Lastly, make sure you use a consistent method of citations--either [1],[2], or (author, page#).
 * Best, Stevenarntson


text:
User talk:Kylnso

test test Kylnso 19:42, 4 February 2009 (UTC)

Hello Anonnymous 19:43, 4 February 2009 (UTC)

title: User talk:Kylnso


text:
Category:Checkuser templates

Templates that help make things easier to track, or to investigate, or to report status of inquiries.

title: Category:Checkuser templates


text:
Category:Spam blacklist templates

This category holds templates used for working on the blacklist.

title: Category:Spam blacklist templates


text: Category:Image with comment templates

title: Category:Image with comment templates


text:
User:Robert Holzhauser

Robert Earl Holzhauser born in Denver, Colorado, August 31, 1967.

title: User:Robert Holzhauser


text: Category:Standardised user warning templates


text:
User:Egm6322.s09.Three/Report2

Reviewing from the previous report

Report 1 problem: diffusion operator

$$\mathcal D (u):=div(\kappa \cdot grad\ u)$$

$$=\frac{\partial }{\partial x_{i}}(\kappa_{ij} \frac{\partial u}{\partial x_{j}})$$

Expanding:

$$\sum_{i=1}^{2} \sum_{j=1}^{2}\frac{\partial }{\partial x_{i}}(\kappa_{ij} \frac{\partial u}{\partial x_{j}})$$

$$\frac{\partial }{\partial x_{1}}(\kappa_{11} \frac{\partial u}{\partial x_{1}})+\frac{\partial }{\partial x_{1}}(\kappa_{12} \frac{\partial u}{\partial x_{2}})+\frac{\partial }{\partial x_{2}}(\kappa_{21} \frac{\partial u}{\partial x_{1}})+\frac{\partial }{\partial x_{2}}(\kappa_{22} \frac{\partial u}{\partial x_{2}})$$

Linear if $$\alpha$$ and $$\beta$$ are constants:

$$\mathcal L(\alpha u+\beta v)=\alpha \mathcal L(u)+ \beta \mathcal L(v)$$

$$\mathcal D(\alpha u+\beta v)=\alpha \mathcal D(u)+ \beta \mathcal D(v)$$

$$\forall \ u,v; \ \Omega\rightarrow  \Re, \ \forall \ \alpha, \beta \ \in \Re$$

$$div[\underline{\kappa} \cdot grad(\alpha u+ \beta v)] $$

Gradient is linear.

$$grad(\alpha u+ \beta v)=\alpha grad \ u+\beta grad \ v$$

Matrix multiplication is a linear operation.

$$\underline{A}(\alpha \cdot \underline{x}+\beta \cdot \underline{y})= \alpha \underline{A} \cdot \underline{x}+ \beta \underline{A} \cdot \underline{y}$$

$$\underline{\kappa}grad \cdot(\alpha \underline{u}+\beta\underline{v})= \alpha \underline{\kappa} \cdot grad\ u + \beta \underline{\kappa} \cdot grad\ v$$

Vector fields a, b : $$\Omega \rightarrow R^{3}$$

=Transformation of Coordinates=

Definition of a linear operator
The word linear comes from the latin word ‘linearis’ which means created by lines.

An operator is said to be linear if it has both Additivity and homogeneity.

\forall \ u,v; \ \Omega\rightarrow  \Re\\
 * Additive: $$\begin{matrix}

\mathcal L(u+v)=\mathcal L(u) + \mathcal L(v)

\end{matrix}$$

\forall \alpha \in \Re, \ \forall u: \Omega \rightarrow \Re\\
 * Homogeneous: $$\begin{matrix}

\mathcal L (\alpha u)=\alpha \mathcal L(u)

\end{matrix}$$

Equivalence between $$\mathcal L (\alpha u+ \beta v)=\alpha \mathcal L(u)$$

To prove additivity, set $$alpha=beta=1$$=beta=1. To prove homogeneity set $$\beta=0$$.

Homogeneity of $$\mathcal L(\cdot)$$ select either $$alpha=0$$ or $$u=0$$.

Image of zero under $$\mathcal L$$ is zero if $$\mathcal L$$  is a linear operator.

$$\mathcal L(0)=0$$

Example of Linear map for $$R^n$$ (n-dimensional case)
 R n $$\rightarrow$$ $$\mathbb{R}$$n is a set-to-set map.

If x $$\in$$  R  n; y $$\in$$ $$\mathbb{R}$$n,

then x $$\rightarrow$$ y=Ax is a point-to-point map.

A $$\in$$ $$\mathbb{R}$$n×n, and is a n×m matrix.

D (domain) =  R n

R (range) = $$\mathbb{R}$$n

General Case
When $$\mathbb{R}$$m $$\rightarrow$$ $$\mathbb{R}$$n; m $$\neq$$ n

If x $$\in$$ $$\mathbb{R}$$n; y $$\in$$ $$\mathbb{R}$$n; and y=Ax,

then A is a n×m matrix, A $$\in$$ $$\mathbb{R}$$n×m

Now consider: y=Ax+b;

y is a n×1 matrix; A is a n×m matrix; x is a m×1 matrix; b is a n×1 matrix.

M($$\cdot$$): $$\mathbb{R}$$m $$\rightarrow$$ $$\mathbb{R}$$n

If x $$\in$$ $$\mathbb{R}$$m; y $$\in$$ $$\mathbb{R}$$n;

then x $$\longmapsto$$ y=Ax+b

So clearly,M( 0 )= b $$\neq$$ 0 $$\Rightarrow$$ M($$\cdot$$) is not a linear map, and M($$\cdot$$) is not homogeneous, either.

Example of Rotation

In the case y=Rx+b

$$\begin{Bmatrix} y_1 \\ y_2 \end{Bmatrix}$$ = $$\begin{bmatrix} R_{11} &R_{12}    \\ R_{21} &R_{22} \end{bmatrix}$$$$\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix}$$ + $$\begin{Bmatrix} b_1 \\ b_2 \end{Bmatrix}$$

 Note:  If A $$\Leftrightarrow$$ B, then A is equivalent to B.

If A $$\Rightarrow$$ B, then A is sufficient condition for B;

If A $$\Leftarrow$$ B, then B is sufficient condition for A, or A is necessary condition for B.

[A $$\Rightarrow$$ B]$$\Leftrightarrow$$ [negation of B $$\Rightarrow$$ negation of A]

Transformation of Coordinates and the Jacobian
Assume $$R^m$$ and $$R^n$$ are expressions of vectors (tensors). Divergent maps of vector field (vector valued function) into a scalar function, in other words, domain and range of div(·) are functions spaces.

 Note:  In P9.1, it can't be transformed to coordinates of Eqn(5) by 2nd order linear PDEs (e.g. Eqn(2))

In coordinates:

$$x$$=$$\Phi$$($$\bar{x}$$,$$\bar{y}$$)

$$y$$=$$\psi$$($$\bar{x}$$,$$\bar{y}$$)

in coordinate transformation of Eqn(5) in P9-1

$$u$$(x,y)=$$u$$($$\Phi$$($$\bar{x}$$,$$\bar{y}$$),$$\psi$$($$\bar{x}$$,$$\bar{y}$$))

=$$u$$($$\bar{x}$$,$$\bar{y}$$)           (abuse of notation by using "u")

=$$\bar{u}$$($$\bar{x}$$,$$\bar{y}$$)      (more rigorous notation)

 Example：

Let $$u(x)=ax+b$$

$$ x=x(\bar{x})

=sin\bar{x} =\Phi(\bar{x}$$)

$$u(x)=u(\Phi(\bar{x}))

=asin(\bar{x})+b

=u(\bar{x})

=\bar{u}(\bar{x})$$

$$u(\bar{x})$$is an abuse of notation; $$\bar{u}(\bar{x})$$ is more rigorous.

$$u_x(x,y)=\frac{\partial u}{\partial x}(x,y)=u_x(\Phi(\bar{x},\bar{y}),\Psi(\bar{x},\bar{y})) =\frac{\partial}{\partial x}\bar{u}(\bar{x},\bar{y}) =\frac{\partial \bar{u}}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial x}+\frac{\partial \bar{u}}{\partial \bar{y}}\frac{\partial \bar{y}}{\partial x}$$

 Define:  $$\bar{x}=\bar{x}(x,y)$$=$$\bar{\Phi}(x,y)$$; $$\bar{y}=\bar{y}(x,y)$$=$$\bar{\Psi}(x,y)$$

then $$u_y(x,y)=\frac{\partial}{\partial y}\bar{u}(\bar{x},\bar{y})=\bar{u}_{\bar{x}}\frac{\partial \bar{x}}{\partial y}+\bar{u}_{\bar{y}}\frac{\partial \bar{y}}{\partial y}$$

 Matrix form:

$$\partial_xu=[\frac{\partial \bar{x}}{\partial x},\frac{\partial \bar{y}}{\partial x}]\begin{Bmatrix} \partial_\bar{x} \\ \partial_\bar{y} \end{Bmatrix}(\bar{u})$$

$$ \begin{Bmatrix} \partial_x \\ \partial_y \end{Bmatrix}=\begin{bmatrix} \frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}     \\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y} \end{bmatrix} \begin{Bmatrix} \partial_\bar{x} \\ \partial_\bar{y} \end{Bmatrix}$$

 Jacobian Matrix:  $$\begin{bmatrix} \frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}     \\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y} \end{bmatrix}$$; (Sometimes it is defined as transformation of Jacobian matrix)

Note: An easier and more general expression:

$$(x_1,...,x_n)\longmapsto(\bar{x}_1,...,\bar{x}_n)$$

Indicial notation:

$$\bar{x}_i=\bar{x}_i(x_1,...,x_n)$$

$$\underline{J}$$=$${[\frac{\partial \bar{x}_i}{\partial x_j}]}$$n×n

i: row index; j: column index.

=2nd order PDEs : Alternative presentation, from particular to general =

PDE's linear with respect to 2nd derivative, but still not linear in general
Intro: In order to learn more about PDE's in general, we decided to go from particular examples to general expressions.

For the general 2nd order PDE:

$$div[\bar{\kappa}\cdot grad\, u]+f(x,y,u,u_{x},u_{y})=0$$

Where, $$\bar{\kappa}$$ is the conductivity tensor (2nd order) and:

$$\bar{\kappa}=\bar{\kappa}(x,y)$$

The portion of the equation with the 2nd order derivatives is given by: $$div[\bar{\kappa}\cdot grad\, u]$$

The rest of the equation, $$f(x,y,u,u_{x},u_{y})=0$$, may be a function of x,y, the unknown function u, or its first order derivatives.

e.g.

$$div[\bar{\kappa}(x,y)\cdot grad\, u]+ ax^{2}+ by + \sqrt{u}+ (u_{x})^{4}+2(u_{y})^{2}=0$$

Note that:

$$D_{1}(\cdot)=div[\bar{\kappa}(x,y)\cdot grad\, u]$$

is linear, and,

$$D_{2}(\cdot):=D_{1}+(\cdot)^{\frac{1}{2}}+[(\cdot)_{x}]^{4}+[(\cdot)_{y}]^{2}$$

is nonlinear. Therefore:

$$D_{3}:=D_{2}+ax^{2}+by$$

is also nonlinear.

Show PDE is linear with respect to 2nd order derivative, nonlinear as a whole.

PDE:Particular to General
2nd order PDE's, linear in all orders. i.e. linear with respect to 1st order & zeroeth order.

General form:

$$au_{xx}+2bu_{xy}+cu_{yy}+du_{x}+eu_{y}+fu+g=0...(1)$$

coefficients can be a function of x,y, or other constants in general if they are a function of u then the PDE is said to be quasilinear.

$$ \begin{pmatrix} \partial _{x} & \partial_{y} \end{pmatrix} \begin{pmatrix} a & b\\ c & d \end{pmatrix}\begin{Bmatrix} \partial_{x}u\\ \partial_{y}u \end{Bmatrix}+\begin{pmatrix} d & e \end{pmatrix}\begin{Bmatrix} \partial_{x}u\\ \partial_{y}u \end{Bmatrix}+fu+g=0...(2)$$

let $$au_{xx}+2bu_{xy}+cu_{yy}=\alpha$$

$$\partial_{x}=\frac{\partial}{\partial x}(\cdot)$$ $$\partial_{y}=\frac{\partial}{\partial y}(\cdot) $$

$$\begin{bmatrix} \partial_{x} & \partial_{y} \end{bmatrix}\begin{Bmatrix} au_{x}+bu_{y}\\ bu_{x}+cu_{y} \end{Bmatrix}=\begin{Bmatrix} (au_{x})_{x}+(bu_{y})_{x}\\ (bu_{x})_{y}+(cu_{y})_{y} \end{Bmatrix}$$

Assuming a,b,c are constants, uxy=uyx.

$$[a_{x}u_{x}]+[b_{x}u_{y}]+[b_{y}u_{x}]+[...] $$

Eqn (2)=> Eqn (1) only if a,b,c are constant, but d,e,f,g may be functions of x & y. eqn(2) is more general.

$$(au_{x})_x+(bu_{x})_y+(bu_{y})_x+(cu_{y})_y+du_{x}+eu_{y}+fu+g=0$$ (3)

Showing equation 2 and equation 1 are the same form

$$(au_{x})_x+(bu_{x})_y+(bu_{y})_x+(cu_{y})_y+du_{x}+eu_{y}+fu+g=0$$

$$[au_{xx}+a_{x}u_{x}]+[2bu_{xy}+b_{y}u_{x}+b_{x}u_{y}]+[cu_{yy}+c_{y}u{y}]+du_{x}+eu_{y}+fu+g=0$$

=>

$$au_{xx}+2bu_{xy}+cu_{yy}+[d+a_{x}+b_{y}]u_{x}+[e+b_{x}+c_{y}]u_{y}+fu+g=0$$

Noting that:

$$\overline{d}=[d+a_{x}+b_{y}]$$

$$\overline{e}=[e+b_{x}+c_{y}]$$

the format of equation 1 is recovered.

PDEs linear in all orders (more particular)
Proof to show that the order of double differentiation is independependent of the variables i.e uxy= uyx

Assume that there exist uniform number of intervals of length 'h' in both 'x' and 'y' directions

Then

ux= $$\lim_{h\to 0}\frac {u(x+h,y)-u(x,y)}{h}$$

uy= $$\lim_{h\to 0}\frac {u(x,y+h)-u(x,y)}{h}$$

uxy= $$ \lim_{h\to 0}\frac {\frac{\partial u(x+h,y) }{\partial y}-\frac{\partial u(x,y)}{\partial y}}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{\left [\frac{u(x+h,y+h)-u(x+h,y)}{h} \right ]-\left [\frac{u(x,y+h)-u(x,y)}{h} \right ]}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{u(x+h,y+h)-u(x+h,y)-u(x,y+h)+u(x,y)}{h^{2}}$$---(1) uyx= $$ \lim_{h\to 0}\frac {\frac{\partial u(x,y+h) }{\partial x}-\frac{\partial u(x,y)}{\partial x}}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{\left [\frac{u(x+h,y+h)-u(x,y+h)}{h} \right ]-\left [\frac{u(x+h,y)-u(x,y)}{h} \right ]}{h}$$

$$\Rightarrow \lim_{h\to 0}\frac{u(x+h,y+h)-u(x+h,y)-u(x,y+h)+u(x,y)}{h^{2}}$$---(2)

Hence as is evident from (1) and (2), the order of double differentiation is indeed independent of the variables.

=List of Contributing Members=

Egm6322.s09.Three.ge 17:13, 6 February 2009 (UTC) team coordinator

Egm6322.s09.three.liu 15:01, 6 February 2009 (UTC)

Egm6322.s09.Three.nav 15:38, 6 February 2009 (UTC)

title: User:Egm6322.s09.Three/Report2