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

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File:HW5 P1 B Gauss.png

= Summary=

title: File:HW5 P1 B Gauss.png


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User:Egm6341.s11.team4.BM/HW5

=Problem 5.1=

title: User:Egm6341.s11.team4.BM/HW5


text:
User:Yongtao-Li/HW5.4

=Problem 5.4 Using LIBF with equidistant nodes to solve G1DM1.0/D1=

GIVEN
One dimensional Model 1.0/Data set 1.0 on page 2 of Mtg 9:

where $${a_2}\left( x \right){\text{ =  2 + 3x}}$$,$$f\left( {x,t} \right) = 5x $$

1-D LIBF on page 3 of Mtg 26:

FIND
A) Explain how LIBF are used as CBS

B) Plot all LIBF used

C) Use matlab to integrate and solve the weak form

D) Plot $$ u_m^h\ vs\ u$$ and $$ u_m^h\left(0.5 \right)-u\left(0.5 \right)\ vs\ m$$

SOLUTION
A) Take m=4 for example to explain how LIBF works as CBS.

When m=4, we can write the basis function according to Eq. 5.4.4 (N was denoted as basis function from now on):

According to Eq.5.4.2, we should have CBS when x=1, which means that the last basis function should not equal zero when x=1.

At the same time, when x=1, we have the following values for LIBF based on its own properties:

Since we have the weight function as $$ w\left( x \right) = {c_i}N\left( x \right) $$ and from Eq. 5.4.9 we can choose arbitrary value for c from i=1~3 and make c equal zero for the last term, which absolutely satisfy CBS for which have non-zero value at the essential boundary for the last basis function and equal zero for all the others.

B)-D)Integrate Eq. 5.4.1 by parts over the whole domain, we can have

$$ \left[ {w(x)\left( {(2+3x)\frac} \right)} \right]_{}^{x = 1} - \left[ {w(x)\left( {(2+3x)\frac} \right)} \right]_{x = 0}^{} - \int\limits_0^1 {\frac\left( {(2+3x)\frac} \right)} dx + \int\limits_0^1 {w(x).(2+3x)} dx = 0 $$

Since LIBF satisfies CBS, we have w=0 when x=1, and then the above equation can give us the weak form:

If we use LIBF both as trial solution and weight function, the following matlab codes can solve the weak form for m nodes

The above three images show that whatever the value m equals, LIBF show its property which can be used as CBS, since its values vanish all other nodes except just one node where the natural boundary condition happens.

It is not easy to find out the difference from the approximation solution and exact solution, but the gaps do exist since the error is non-zero.

We can find out when the solution convergence and what is the error at x=0.5. The error is so small which indicates that LIBF, as a global basis function, has advantages solving weak form and obviously can satisfy CBS wherever essential boundary happens.

title: User:Yongtao-Li/HW5.4


title: File:Nm1.s11.mtg32.djvu


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User:Parakoopa72

Hello.

title: User:Parakoopa72


text:
Limits To Growth/gallery of footprint calculators

A gallery of Footprint Calculators
Many ecological footprint calculators are available on-line. The assumptions they make and the limits they use differ widely. Some are based primarily on food production, others on energy production, and most consider atmospheric CO2 equilibrium the limiting factor. Here is a list of on-line footprint calculators.

 * The Personal Carbon Calculator is provided by Small World Consulting.
 * The Global Footprint Network (footprintnetwork.org) aims to accelerate the use of the Ecological Footprint — a resource accounting tool that measures how much nature we have, how much we use, and who uses what.
 * The World Wildlife Fund provides a calculator oriented toward living in the UK.
 * WattsOn (www.wattzon.com) calculates your energy consumption in watts (per always) and compares it to a global capacity of 2255 watts per person (15 terawatts / 6.65 billion people). Using an approximate conversion factor of 10 Kilowatt-hour / day ≈ 1 Tonne CO2 / year, the 2255 watts corresponds to approximatly 5.4 Tonnes CO2 per year. [figure out an equivalence in gha]
 * ClimateCare (www.jpmorganclimatecare.com) is owned by JPMorgan Chase & Co. and sells carbon offset credits.
 * The CarbonNeutral Company (www.carbonneutral.com) measures carbon footprints, sells carbon offsets, and consults with companies that want to reduce the carbon emissions.
 * Earth Day Network (www.earthday.org) was founded on the premise that all people, regardless of race, gender, income, or geography, have a moral right to a healthy, sustainable environment. Their footprint calculator expresses the results in terms of the number of equivalent earths it would require to sustain everyone on the planet living like you do.
 * SafeClimate.net (www.safeclimate.net) is a project of the World Resources Institute. They work putting environmental issues on the international agenda.
 * BP (www.bp.com) is a petroleum energy company.
 * The TravelMatters Individual Calculator (www.travelmatters.org) measures how much greenhouse gas you generate as a result of your daily transportation activity.
 * The ClimateCrisis site (www.climatecrisis.net) describes how their calculations are performed.
 * Conservation International (www.conservation.org) is working to help societies adopt a more sustainable approach to development—one that considers and values nature at every turn.
 * Carbon Footprint Ltd (www.carbonfootprint.com) provides Carbon Management services
 * The United States Environmental Protection Agency (www.epa.gov ) website includes a Household Emissions Calculator. The calculations and references used are conveniently summarized.
 * EarthLab.com (www.earthlab.com) provides a practical guide to green living through community interaction, exclusive environmental news, advice from experts and key analysis.
 * Trees, Water & People (www.treeswaterpeople.org) is nonprofit organization that was founded in 1998 by Stuart Conway and Richard Fox, and is staffed by conservationists helping communities to protect, conserve, and manage the natural resources upon which their long-term well-being depends.
 * Co2nsensus (www.co2nsensus.com) is a climate tech company, we have been developing software for businesses and individuals to fight climate change. We also deliver consultancy services for top fashion brands, chemical industries, and many others. We have developed a carbon offset calculator for flights, cars, shipping, events and homes as well.


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User:Egm6341.s11.team4.hylon/hw5

=Problem 5.2 Run Linear State Space Model without Noise and with Gaussian Random Noise and Cauchy Random Noise respectively=

Problem Statement
Refer to lecture notes [[media:nm1.s11.mtg25.djvu|25-2]] and [[media:nm1.s11.mtg25.djvu|26-1]] for detailed description

Given
Linear State Space Model without Random Noise (LSSM):
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{x}_{k+1}}=F{{x}_{k}}

$$ Linear State Space Model without Random Noise (LSSMRN):
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{x}_{k+1}}=F{{x}_{k}}+G{{w}_{k+1}}.

$$ For this specific case here, we choose
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{F}_{n\times n}}={{I}_{n\times n}}+\Delta {{A}_{n\times n}},n=2

$$ And $$I=\left( \begin{matrix}  1 & 0  \\   0 & 1  \\ \end{matrix} \right)$$, $$A=\left( \begin{matrix}   -0.2 & 1  \\   -1 & -0.2  \\ \end{matrix} \right)$$, $$\Delta =0.02$$, $${{x}_{0}}=\left( \begin{matrix}   3  \\   -2  \\ \end{matrix} \right)$$
 * }

Objective
1). Run LSSM and plot $${{x}_{j}},j=0,1,2,\cdots $$ in space $$({{x}^{1}},{{x}^{2}})$$. 2). Find equilibrium point as
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\underset{k\to \infty }{\mathop{\lim }}\,{{x}_{k+1}}=\underset{k\to \infty }{\mathop{\lim }}\,{{F}^{k+1}}{{x}_{0}}=_ – ^ – :_ – ^ – \hat{x}

$$ Plot $$\hat{x}$$ as big red dot and $${{x}_{0}}$$ as big blue dot in the same plot for $${{x}_{j}},j=1,2,\cdots $$ small dots. 3). Let $$G=\left( \begin{matrix} 1 \\   1  \\ \end{matrix} \right)\alpha $$, thus $${{w}_{k+1}}=({{w}_{k+1}})$$. Use Matlab randn to generate $$({{w}_{j}},j=0,1,2,\cdots )$$. Plot $${{x}_{j}},j=0,1,2,\cdots $$for $$\alpha =0.5,1,2$$. 4). Same as 3). but with Cauchy random noise. Hint: Find a Matlab command to generate $${{\theta }_{j}},j=0,1,2,\cdots $$in single slit diffraction.
 * }.

2).Find equilibrium point and run LSSM

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\hat{x}=\underset{k\to \infty }{\mathop{\lim }}\,{{F}^{k+1}}{{x}_{0}}=\underset{k\to \infty }{\mathop{\lim }}\,{{\left( \begin{matrix}  0.996 & 0.02  \\   -0.02 & 0.996  \\ \end{matrix} \right)}^{k+1}}\left( \begin{matrix}   3  \\   -2  \\ \end{matrix} \right)=\left( \begin{matrix}   0  \\   0  \\ \end{matrix} \right)

$$
 * }

4). Run LSSM with Cauchy Random Noise

=Problem 5.8 Proof identities=

Given
Refer to Lecture notes [[media:nm1.s11.mtg30.djvu|30-2]] and [[media:nm1.s11.mtg30.djvu|30-3]] for detailed description.

Objective
Show that

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

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

E_{n}^{T}=\frac{h}{2}\sum\limits_{k=0}^{n-1}{[\int\limits_{-1}^{+1}{{{g}_{k}}(t)dt}-\{{{g}_{k}}(-1)+{{g}_{k}}(+1)\}]}

$$
 * }

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

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} & \int\limits_{-1}^{+1}{(-t)g_{k}^{(-1)}(t)dt}=\int\limits_{-1}^{+1}{{{g}_{k}}(t)dt}-[{{g}_{k}}(-1)+{{g}_{k}}(+1)] \end{align}

$$
 * }

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

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

g_{k}^{(i)}(t)={{(\frac{h}{2})}^{i}}{{f}^{(i)}}(x(t))

$$
 * }

1).Function transformation

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

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} & E_{n}^{T}=I-{{I}_{n}}=\int\limits_{a}^{b}{f(x)dx}-{{T}_{0}}(n) \\ & =\sum\limits_{k=0}^{n-1}{[\int\limits_^{f(x)dx}-\frac{h}{2}\{f({{x}_{k}})+f({{x}_{k+1}})\}]} \\ \end{align}

$$ Where $${{x}_{k}}:=a+kh,_ – ^ – h=(b-a)/n$$. Now, we transfer $$\int\limits_^{f(x)dx}$$ to $$\int\limits_{-1}^{+1}{f(x(t))dt}$$.
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} & x(t)=\frac{{{x}_{k+1}}-{{x}_{k}}}{2}t+\frac{{{x}_{k+1}}+{{x}_{k}}}{2} \\ & =\frac{h}{2}t+\frac{{{x}_{k+1}}+{{x}_{k}}}{2} \\ \end{align}

$$ From above transformation relationship, it’s obviously that
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left\{ \begin{matrix} x(-1)={{x}_{k}} \\ x(0)=\frac{{{x}_{k+1}}+{{x}_{k}}}{2} \\ x(+1)={{x}_{k+1}} \\ dx=\frac{h}{2}dt \\ \end{matrix} \right.

$$ Hence we have:
 * }.
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\begin{align} & E_{n}^{T}=\sum\limits_{k=0}^{n-1}{[\int\limits_{-1}^{+1}{f(x(t))\frac{h}{2}dt}-\frac{h}{2}\{f(x(-1))+f(x(+1))\}]} \\ & =\frac{h}{2}\sum\limits_{k=0}^{n-1}{[\int\limits_{-1}^{+1}{f(x(t))dt}-\{f(x(-1))+f(x(+1))\}]} \\ \end{align}

$$ If we define $${{g}_{k}}(t)=f(x(t))_ – ^ – such_ – ^ – that_ – ^ – x\in [{{x}_{k}},{{x}_{k+1}}]$$, then we have
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

E_{n}^{T}=\frac{h}{2}\sum\limits_{k=0}^{n-1}{[\int\limits_{-1}^{+1}{{{g}_{k}}(t)dt}-\{{{g}_{k}}(-1)+{{g}_{k}}(+1)\}]}

$$
 * }

2).Integrate by part
Integrating by parts, we have
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\begin{align} & \int\limits_{-1}^{+1}{(-t)g_{k}^{(-1)}(t)dt}=[(-t){{g}_{k}}(t)]_{x=-1}^{x=1}-\int\limits_{-1}^{+1}{{{g}_{k}}(t)d(-t)} \\ & =\int\limits_{-1}^{+1}{{{g}_{k}}(t)dt}-[{{g}_{k}}(-1)+{{g}_{k}}(+1)] \\ \end{align}

$$
 * }

3).Successively differentiation
Since we have already defined that $${{g}_{k}}(t)=f(x(t))_ – ^ – such_ – ^ – that_ – ^ – x\in [{{x}_{k}},{{x}_{k+1}}]$$ and from transformation relationship we have $$dx=\frac{h}{2}dt$$, successively differentiate function $${{g}_{k}}(t)$$ with respect to $$t$$, we can get
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

g_{k}^{(i)}(t)={{f}^{(i)}}(x(t))\cdot {{x}^{(i)}}(t)={{(\frac{h}{2})}^{i}}{{f}^{(i)}}(x(t))

$$
 * }

title: User:Egm6341.s11.team4.hylon/hw5


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User:Egm6341.s11.team4.fields/HW5

HW5.6

title: User:Egm6341.s11.team4.fields/HW5


text:
User:Egm6341.s11.team4.fields/HW5.6

=Problem 5.6: Computation of In using CTk(n)=

Refer to lecture slide [[media:nm1.s11.mtg30.djvu|30-1]] for the problem statement.

Given

 * {| style="width:70%" border="0" align="center"

(6.1)
 * $$\displaystyle I:=\int_{a}^{b}\underbrace{\frac{e^x-1}{x}}_{f(x)}dx$$
 *
 * }
 * }

For

 * {| style="width:70%" border="0"

 * $$\displaystyle a = -1$$

 * }
 * }

and

 * {| style="width:70%" border="0"

 * $$\displaystyle b = 1$$

 * }
 * }

Corrected Trapezoidal Rule:

 * {| style="width:70%" border="0" align="center"

(6.2)
 * $$\displaystyle CT_k(n) = CT_{k-1}(n) + a^0_kh^{2k} + O(h^{2(k+1)})$$
 *
 * }
 * }

Where

 * {| style="width:70%" border="0" align="center"

(6.3)
 * $$\displaystyle a_i^0 = d_i \left[ f^{(2i-1)}(b) - f^{(2i-1)}(a) \right]$$
 *
 * }
 * }

and

 * {| style="width:70%" border="0"

 * $$\displaystyle h = \frac{b-a}{n}$$

 * }
 * }

Composite Trapezoidal Rule:

 * {| style="width:70%" border="0" align="center"

(6.4)
 * $$\displaystyle CT_{0}(n) = T_0(n) = \frac{b-a}{n} \left[ \frac{1}{2}f_0 + f_1 + \ldots + f_{n-1} + \frac{1}{2}f_n \right]$$
 *
 * }
 * }

Where

 * {| style="width:70%" border="0"

 * $$\displaystyle d_i = -B_{2i}/(2i)!$$

 * }
 * }

and

 * {| style="width:70%" border="0"

 * $$\displaystyle B_{i} = Bernoulli Number$$

 * }
 * }

Objective
a.) Compute In using CTk(n), for k = 1 and n = 2, 4, 8, 16, ... until error is of order 10-6

b.) Compute In using CTk(n), for k = 2 and n = 2, 4, 8, 16, ... until error is of order 10-6

c.) Compute In using CTk(n), for k = 3 and n = 2, 4, 8, 16, ... until error is of order 10-6

Solution
Using Wolfram Alpha we find that the exact value of the integral to 12 significant digits is:

I = 2.11450175075

a.) Compute In using CTk(n) for k = 1
Using equations 6.2, 6.3 and 6.4 we compute CT1(n)

 * {| style="width:70%" border="0" align="center"

(6.5)
 * $$\displaystyle CT_1(n) = T_0(n) + a^0_1h^{2} + O(h^{4})$$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle a^0_1 = d_1 \left[ f^{(1)}(1) - f^{(1)}(-1) \right] = \frac{-1}{12}\left[1 - 0.264241117657 \right] =    -0.0613132401952$$

 * }
 * }

The MatLab code below was used for different values of n to generate the following table:

function [CTk1,Enk1] = CorTrapk1(a,b,n) %Corrected Trapezoidal Rule for k = 1 %  f(x) = (e^x - 1)/x

h = (b - a)/n; ni = n+1; I = 2.11450175075; a1 = -.061313240195;

x = linspace(a,b,ni);

f = (exp(x) - 1)./x;

f((n/2)+1) = 1;

i = 1;

To = 0;

Toi = zeros(1,ni);

while i <= ni   if i == 1 || i == ni        Toi(i) = 0.5*f(i); else Toi(i) = f(i); end To = To + (Toi(i)); i = i+1; end

To = h*To; CTk1 = To + a1*h^2 Enk1 = abs(I - CTk1)

end

b.) Compute In using CTk(n) for k = 2
Using equations 6.2, 6.3 and 6.5 we compute CT2(n)

 * {| style="width:70%" border="0" align="center"

(6.6)
 * $$\displaystyle CT_2(n) = CT_{1}(n) + a^0_2h^{4} + O(h^{6})$$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle a^0_2 = d_2 \left[ f^{(3)}(1) - f^{(3)}(-1) \right] = \frac{1}{720}\left[0.563436343081 - 0.113928941256 \right] = 0.000624315835 $$

 * }
 * }

The MatLab code below was used for different values of n to generate the following table:

function [CTk2,Enk2] = CorTrapk2(a,b,n) %Corrected Trapezoidal Rule for k = 2 %  f(x) = (e^x - 1)/x

h = (b - a)/n; ni = n+1; I = 2.11450175075; a2 = -.000624315835;

CTk2 = CorTrapk1(a,b,n) + a2*h^4

Enk2 = I - CTk2

end

c.) Compute In using CTk(n) for k = 3
Using equations 6.2, 6.3 and 6.6 we compute CT3(n)

 * {| style="width:70%" border="0" align="center"

(6.7)
 * $$\displaystyle CT_3(n) = CT_{2}(n) + a^0_3h^{6} + O(h^{8})$$
 *
 * }
 * }

 * {| style="width:70%" border="0" align="center"

 * $$\displaystyle a^0_3 = d_3 \left[ f^{(3)}(1) - f^{(3)}(-1) \right] = \frac{-1}{30240}\left[0.395599547802 - 0.071302178109 \right] = -0.000010724119 $$

 * }
 * }

The MatLab code below was used for different values of n to generate the following table:

function [CTk3,Enk3] = CorTrapk3(a,b,n) %Corrected Trapezoidal Rule for k = 3 %  f(x) = (e^x - 1)/x

h = (b - a)/n; ni = n+1; I = 2.11450175075; a3 = -0.000010724119;

CTk3 = CorTrapk2(a,b,n) + a3*h^6

Enk3 = I - CTk3 end

This problem was solved by Erle Fields

title: User:Egm6341.s11.team4.fields/HW5.6


text:
Talk:Electrical Engineering Fundamentals/Basics of Electricity

very confusing paragraph.
"This seems really simple until later on when it is realized that current usually flows in a positive manner from the positive to negative. But as described, the electrons are traveling toward the positive end to balance it out. This is still true and I am not contradicting myself, as we define "current" as going the opposite way from the flow of electrons simply due to convention. The flow of electrons is negative because it is a negatively charged particle moving toward a positively charged area. So what is flowing in the opposite direction? The holes are "flowing" in a positive direction, and if it was possible to chart the change in charge of single atoms one would see that as the negative flows toward the positive, so the positive flows toward the negative."

I think the problem is that 'current' is being used in two ways here, to describe the flow of electrons, and to describe the direction in which the "holes" for electrons to fill are moving. I'm guessing that is what is being described, but as I said, I can't really understand the passage and its meaning. 173.112.128.197 19:58, 20 March 2011 (UTC)

Explanation

Current is the flow of electric charge, it doesnt bother itself with whether that is positive or negative charge. By convention, current direction is described by the positive flow. Because in most everyday conductors, the flow of charge is facilitated by the flow of electrons, which are negatively charged, this convention may be confusing.

However, in many systems its possible to have a flow of strictly positive charges, and its also possible to have one where both negative and positive charges together constitute current. An example is when a person is electrocuted, positve ions (K+ Ca+ Na+) flow towards the negative voltage and negative ions (Cl-) flow towards the positive voltage.

I hope Ive helped. I'll read the article to see if I can clear anything there.DM-Lingit (discuss • contribs) 21:34, 21 November 2015 (UTC) Okay I replaced the misconception where I found it, and also changed "flow of current" to what is correct (and probably what is meant.). For future editors, current is the flow of charge. Flow of current is the flow of the flow of charge, a second order derivative probably meant to be introduced later (around the time of caps and inductors).DM-Lingit (discuss • contribs) 21:34, 21 November 2015 (UTC)

Very helpful!
I found this page to be extremely helpful. Thanks to those who helped create this page. I have an accredited degree in computer science but the fundamentals of electrical engineering are still weak points, and this page helped. Adallace (discuss • contribs) 00:22, 15 August 2017 (UTC)


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text:
Limits To Growth/conversions

Constants, Conversions, and Equivalence Factors
Calculating limits to growth requires many conversions between equivalent, or near equivalent quantities of land area, energy, emissions, and other quantities. This page assembles various quantities useful in these conversions. These have been gathered from a variety of sources (cited and listed at the bottom of this page.) Where reliable sources provide a range of values, this is often presented, and the preferred value, used for calculations in the course, is indicated. This course favors the International System of Units (SI).

Land Area
The following area equivalences are helpful in converting various area measures. These can be easily obtained from Wolfram|Alpha.

Power and Energy
Energy is the ability to do work. Rolling a boulder up a hill represents a particular amount of energy. Power is the rate energy is applied. It takes more power to roll the boulder faster than slower, but the energy is the same if the boulder is moved the same distance, regardless of the speed. The equivalency between various units of power and energy are shown in this table. These can be readily verified at WolframAlpha or Google.

Food and Nutrition
The United Nations Food and Agricultural Organization (FAO) uses the Minimum Dietary Energy Requirement, measured in kcal/person/day, to estimate the prevalence of undernourishment. Dietary energy requirements differ by gender and age, and for different levels of physical activity. Accordingly, minimum dietary energy requirements, the amount of energy needed for light activity and minimum acceptable weight for attained-height, vary by country, and from year to year depending on the gender and age structure of the population.

A listing of MDER by country is available as a spreadsheet from the FAO Food Security Statistics Division.

These have the following statistical characteristics for the years 2004-2006: Because the MDER is expressed as a single aggregate number of kcal / person / day it does not directly account for variations in the distribution of food to individuals, nor the various requirements for fat, protein, and specific nutrients. Therefore the FAO follows a carefully developed statistical Methodology For The Measurement Of Food Deprivation when estimating the number of undernourished people.

Water for Food
Growing food requires significant quantities of fresh water. The amount of water required to grow each food type—"drops per crop"—varies significantly. The following table shows the average amount of water needed to produce 1 kilogram of food (2004 figures) along with the calorie equivalent of each.

If a person requires 2,000 calories / day to meet their minimum nutritional requirements, then the food they consume each day requires 840 liters of water to grow if they eat only wheat and 13,160 liters if they eat only beef.

Carbon Emissions
Burning fossil fuels emits carbon into the atmosphere in the form of carbon dioxide (CO2). This contributes to the overall concentration of Carbon dioxide in Earth's atmosphere. One gram of carbon is contained in 3.67 (44/12) grams of CO2.

Gasoline
Gasoline emits CO2 when it is burned, often as automobile fuel. The amount of carbon dioxide emitted per gallon of motor gasoline burned is 8.92×10-3 metric tons. This is equivalent to 8.92 Kg / Gallon. See: http://www.epa.gov/cleanenergy/energy-resources/refs.html

An automobile getting 20 miles per gallon (mpg) emits 8.93/20 = .4465 Kg CO2 / mile

In 2007, the weighted average combined fuel economy of cars and light trucks combined was 20.4 miles per gallon for the United States. The average vehicle miles traveled in 2007 was 11,720 miles per year.

Natural Gas
Natural Gas emits 0.005 metric tons CO2 / therm when burned. See: http://www.epa.gov/cleanenergy/energy-resources/refs.html

This is equivalent to: Note that emissions of unburned natural gas released to the atmosphere is 21 times higher.
 * 5 Kg / therm and
 * 170 g / Kwh

Fuel Oil
Homes are often heated by burning heating fuel oil. According to the EPA Fuel oil emits 429.61 kg CO2 / barrel which is equivalent to 10.23 kg CO2 / Gallon or 2.70 kg CO2 / liter.

Crude Oil
Crude oil is the basis for petroleum products. When consumed, typically by burning, it emits 0.43 metric tons CO2 / barrel. See: http://www.epa.gov/cleanenergy/energy-resources/refs.html

This is equivalent to 2.7 kg CO2 / liter

Electric power
Electric power is generated in many ways, but primarily by burning fossil fuels. The quantity of CO2 emitted as electricity is generated is listed below. The amount depends on the fuel used and the efficiency of the generating equipment. Because different regions include various types of electric power generation systems, the emissions factors vary by region.

 * 6.91 x 10-4 metric tons CO2 / kWh as an overall average for the US. See: http://www.epa.gov/cleanenergy/energy-resources/refs.html
 * Equivalent to: .691 kg CO2 / kWh
 * Equivalent to: 1.52 pounds CO2 /  kWh
 * Equivalent to: 1,520 pounds CO2 / MWh
 * From Coal fired power plants: .96 kg CO2 / kWh See: http://www.wolframalpha.com/input/?i=carbon+footprint
 * From Natural Gas fired power plants: .6 kg CO2 / kWh See: http://www.wolframalpha.com/input/?i=carbon+footprint
 * From petroleum fired power plants: .9 kg CO2 / kWh See: http://www.wolframalpha.com/input/?i=carbon+footprint

These factors vary by location.
 * See, for example: http://www.epa.gov/cleanenergy/documents/egridzips/eGRID2010V1_0_year07_SummaryTables.pdf
 * and: http://www.inference.phy.cam.ac.uk/withouthotair/cI/page_335.shtml
 * and: http://tonto.eia.doe.gov/FTPROOT/environment/e-supdoc-u.pdf

The book Our Choice, provides these somewhat different figures for the carbon footprints of electricity sources in Grams of CO2 per Kilowatt-Hour:

Various Petroleum Products
From: http://www.defra.gov.uk/environment/business/reporting/pdf/101006-guidelines-ghg-conversion-factors.xls

Gross Calorific Value or higher heating value (HHV) is the Calorific Value under laboratory conditions. Net Calorific Value or lower heating value (LHV) is the useful calorific value in typical real world conditions (e.g. boiler plant). The difference is essentially the latent heat of the water vapor produced (which can be recovered in laboratory conditions).

Converting Kwh / Day to Tonne CO2 / Year
Many ecological footprint-oriented calculations are expressed in Tonnes CO2 / year. How can we convert from Kilowatt-hour / Day to this measure?

First consider Electric power generation, which is high in emissions per Kilowatt-hour.

0.691 kg CO2 / Kilowatt-hour × 365 days / year = 252.215 kg CO2 / Year = 0.252 Tonne CO2 / year

Then consider Natural Gas heating, which is relatively low in emissions per Kilowatt-hour.

0.17 kg CO2 / Kilowatt-hour × 365 days / year	= 62.05 kg CO2 / Year	= 0.062 Tonne CO2 / year

Blending these two calculations to represent a mix of usage gives the very approximate equivalence of 10 Kilowatt-hour / day ≈ 1 Tonne CO2 / year. Equivalently, 400 watts ≈ 1 Tonne CO2 / year.

Carbon Capture
Carbon can be removed from the atmosphere and stored by various biological systems. These are characterized below:

Forests and Crop Lands
The average carbon density of U.S. forests in 2008 was estimated by the EPA at 73 metric tons per hectare, or 29.55 metric tons per acre. This is equivalent to 108.35 metric tons CO2 per acre, or 267.73 metric tons CO2 per hectare.

The carbon content of cropland is calculated by the EPA to be 5.0 metric tons of carbon per hectare, or 2.02 metric tons per acre. This is equivalent to 7.40 metric tons CO2 per acre or 18.28 metric tons CO2 per hectare.

Reforesting by converting cropland to forest increases carbon capture capacity by (29.55 – 2.02) 27.53 metric tons per acre. This is 100.94 Metric Tons CO2 / acre, or 249.42 metric tons CO2 per hectare.

Deforestation, converting forest to cropland decreases carbon capture by (29.55 – 2.02) 27.53 metric tons per acre. This is 100.94 Metric Tons CO2 / acre, or 249.42 metric tons CO2 per hectare.

Deforestation, converting forest to unplanted land decreases carbon capture by 29.55 metric tons per acre. This is 108.35 Metric Tons CO2 / acre, or 249.42 metric tons CO2 per hectare.

Resources

 * The US Government Environmental Protection Agency, Clean Energy Calculations and References
 * Sustainable Energy - without the hot air Quick reference
 * The quick-reference list of conversion factors used by the Bioenergy Feedstock Development Programs at Oak Ridge National Laboratory.
 * U.S. energy Information Administration, Energy Kids Energy Calculators.
 * NIST Guide to the SI, Factors for units listed by kind of quantity or field of science

title: Limits To Growth/conversions


title: File:HW5Part2CompTrapCode.jpg


text: File:Lennard jones potential force.png

title: File:Lennard jones potential force.png


title: File:Nm1.s11.team4.HW5.2.fig7.jpg


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text:
Microsoft Windows/Active Directory

History
High Schools used to use Apple/Mac(s). Then they upgraded to Windows. Now they have Windows XP Professional, Embeded, 7, 98, Server 2008/2003 and Linux. Linux is for networking classes. Windows Embedded is for Cafe Machines. Windows Server 2003 Enterprise is at the main downtown center. Windows 7 Enterprise is what is being installed on new computers.

Active Directory in Schools
Most of each school has active directory on the server. It is connected with an underground trunkline that goes as a Wide Area Network (WAN). Each student and staff member have a Roaming User Profile (RUP) which is stored on the server. Each user can logon a client computer running windows xp professional.

Group Policy Object Editor is where the administrative IT staff secure the computers. They can restrict the start menu, C drive, CMD, more. They force classic start menus on RUP(s). There are different policies for each user on the server.

Single Sign-on
Most schools use a single sign-on tool to manage settings and documents. This is for clients and servers. Administrators manage the passwords and the single sign-on. It provides access to Internal Drives, Special start menus that are there, but not apart of (documents and settings) folder.

My Computer

Local C drive has NTFS File System.

Domain Forests
Students have one. Staff have one. Administrator have one (Built-in accounts).

Local User Profile

Local User Profile (LUP) is a locally stored user profile that was initially stored on a server as a roaming user profile. It has saved profile settings. Everytime a student or staff login to client with roaming user profile their account creates a LUP on the local client.

Quota(s)

Quota(s) are the limits that students and staff get. Some schools allow 20MB disk storage on the network drive. My Documents is where the files are stored. Most school restrict the C drive. My Documents's target is stored on the server under the user account.

Group Policy in Active Directory

Each user is also called a "object". Each object on a server domain can have a group policy set for a group of objects. Group policy is not configured locally most of the time.

Domain Server

The server is the main point of the system. The domain is where the objects are and the Forests connect. Administrative Software Most schools have software for maintance purposes. Bootwiz, SQL2000, Aclient, showname and more.

Rapideploy

Imaging bootworks/RAPIDEPLOY. Version: 6.1.5813.0. Author: Altiris RapiDelpoy. It staff create images of the hard drives and store them on a external HDD. Then they can go to the client and extract the image the HDD. Some images are split into two images.

Rapid Install Package (RIP) Rip(s) are supposed to run Exe(s) they can be created. RIP Conflict editor is an administrative software.

title: Microsoft Windows/Active Directory


title: File:HW5Part2CompSimpCode.jpg


text:
User:EML5526.S11.Team1.HS/HW5

=5.2=

Given
Solve the differential equation $$ \frac{d}{dx} \left[ \left(2+3x \right)  \frac{du}{dx} \right]+5x = 0 $$ in the interval (0,1) with the boundary conditions

$$ u\left(1 \right)= 4 ; \frac{du}{dx} \left(0 \right)= -6 $$

The weak form was obtained in the latter section:

As we do not know $$ \frac{du}{dx} \left(1\right) $$ then we chose $$ w \left(1 \right) = 0 $$ ; moreover $$ u^h \left(1 \right)= 4 $$

1) Polynomial Basis Function
Polynomial basis functions To follows the constrain braking solution, we chose:

Clearly $$ b_1=1 $$, and as we need that $$ b_i \left(1 \right) = 0, i = 1,2,... $$;we chose $$ \beta=1 $$. Then we have $$ U^h \left(x\right)= a_0+a_1 \left(x-1 \right)+a_2 \left(x-1\right)^2+... $$

From the essential boundary condition we have: $$ U^h \left(1 \right)= 4= a_0 $$

Therefore:

In a similar way we have

With the condition: $$ w \left(1 \right)= 0= c_0 $$

Therefore:

The derivative of U and w are:

Replacing in the weak form we have:

Canceling $$ c_i $$ and Organizing we obtain:

Performing the integration we obtain:

From the latter equation the coefficients $$ a_j $$ can be obtained.

The analytical general solution is:

With the boundary conditions we obtain the constants:

 Matlab Code:

2) Fourier Basis Function
Now the differential equation is solved using Fourier basis function and the boundary conditions:

To follow the constrain braking solution we chose:

For Cosine Clearly $$ b_1=1 $$, and as we need that $$ b_j \left(1 \right) = 0,j=0,1,2,... ~$$ ; we chose $$ \beta = 1  $$.

For Sine Clearly $$ b_1=1 $$, and as we need that $$ b_k \left(1 \right) = 0,k=1,2,... ~$$ ; we chose $$ \beta = 1  $$.

Then we have $$ U^h \left(x \right)= a_0 cos \left(x-1 \right)-1 + a_1 cos2 \left( x -1 \right)-1 ... $$

$$ U^h \left(x \right)= a_1 sin \left(x-1 \right) + a_2 sin2 \left( x -1 \right) ... $$

From the essential boundary condition we have: $$ U^h \left(1 \right)= 4=a_0 $$

Therefore:

By constructing a Stiffness & Force Matrices as follows.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

further 3rd equation is satisfied for remaining combinations of i & j.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

We developed a MATLAB code to generate the Force and the stiffness Matrix and We get the following values.

We can now use the relationship which is, $$\displaystyle Kd=F$$ to solve for $$\displaystyle d$$.

Thus the trial solution is:

 Matlab Code:

3) Exponential Basis Function
Now the differential equation is solved using exponential basis function and the boundary conditions:

To follow the constrain braking solution we chose:

Clearly $$ b_1=1 $$, and as we need that $$ b_i \left(1 \right) = 0,i=1,2,... ~$$ ; we chose $$ \beta =1 $$.

Then we have $$ U^h \left(x \right)= a_0+a_1 \left(e^{\left(x-1 \right)}-1\right)+a_2 \left(e^{2\left(x-1\right)}-1\right)+... $$

From the essential boundary condition we have: $$ U^h \left(1 \right)= 4=a_0 $$

Therefore:

In similar way we have:

With the condition: $$ w(1) = 0 = c_1 $$

Therefore:

The derivative of U and w are:

Replacing in the weak form we have:

Canceling $$ c_i $$ and organizing we obtain:

Performing the integration we obtain:

The following figure shows the exact and numerical solutions. Although good approximation was obtained at $$ n = 3 $$, the convergence for an error of order $$10^{-6} $$ was only achieved with $$ n=29 $$ and to $$n $$ greater than $$ 8 $$, a bad-conditioned (close to singular) stiffness matrix was obtained. Figure 2.4 shows the error.

This results shows that polynomial is better basis function because the convergence is obtained faster; moreover for other problems, the integration may not be possible to perform analytically as in this case, making necessary that the program perform the integration numerically which will increase the time of numerical calculations and decrease the accuracy.

 Matlab Code:

title: User:EML5526.S11.Team1.HS/HW5


text:
User:Pumpkin~enwikiversity

Hi There,

I'm a male Australian who is interested in just about anything.

Currently I'm in underground mining so I'm interested in that.

title: User:Pumpkin~enwikiversity


text:
Basic underground mining

So, you dig a tunnel into the ground and hopefully get something worthwhile out of the ground.

Can someone beef this up please?

Lots of things to learn

An underground is necessarily (by definition) under the ground. So the first thing is how do you get under the ground? The first obvious thing is you dig a hole. If the ore body is close to the surface then the mining method used is "open pit mining". Many of the isses about mining are common to open pit and underground mining, but I started with the latter.

Entry Portal or Pithead

If your starting place is a vertical face then the entry point is called a "Portal". However if you dig a hole vertically you need some sort of "winding gear" or "crane" to allow for access. Portal entry has advantages as it allows mobile equipment to move itself into the mine.

In either case you need a tunnel.

Three broad types of tunnel: Vertical (Rise) Horizontal (Drive) and Decline

All of these tunnels may be raw or lined. For safety's sake a lined tunnel is a much better idea. This means that the sides of the tunnel are coated and bolted or some sort of artificial liner (pipe or casing) is inserted into the raw tunnel to give it additional strength and to stop the native material from bursting into the tunnel.

Hanging wall Footwall

Stopes Drive Decline

Stress

Autocompression

Boggers and Bogging (Mucking?) Underground Trucks and Hauling Integrated Tool Carriers (ITs or ITCs) Raise Bore Boxhole

Drilling Jumbo

Blasting Charging Shotfiring Detonators

Shotcreting Meshing Bolting Cable Bolting

Safety and Security Lighting Personal Protective Equipment (PPE)

Tagboards

Danger Signs

Slabs Blocks Boulders Sumps

Mono Pumps

Jumbos

Fans Ventilation Vent Bags Risers

...

The list is endless

title: Basic underground mining


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text:
Paideia High School/Ave verum corpus

Ave verum corpus is a Paideia Unit Plan. These guidelines address teachers for the purpose of guiding instruction. See Paideia Learning Plan for the student's point of view.

Column One
Column One teaching and learning should make up about 10% to 15% of the total scheduled instructional time. It is didactic in nature and uses teacher lectures, text books or other didactic instructional materials, and questioning appropriate to this mode of education. Teaching in this mode encompasses of three facets: Exoridium, Interpretation, and Erudition.

Exordium
The Exordium is the teacher's didactic introduction to the work that is the subject of a Paideia Unit Plan. This introduction consists of both an oral and physical (or electronic) presentation of the work. For longer works, the teacher may limit the oral presentation to key parts of the work. The teacher should read texts live distinctly, accurately, and intelligently. Other works should be orally presented similarly as appropriate to the type of work. In addition, the teacher should provide high quality audio and video recordings of works if possible. Students should have a consumable print copy of the work both electronically and in hard-copy if possible.

Oral Presentation

 *
 *

Written Presentation

 *
 *
 *

Interpretation
Textual Interpretation refers to the teacher's didactic analysis of a written work in terms of the four major questions a demanding reader should ask of a text. Using the term work instead of book, these questions are: (1) What is the text about as a whole? (2) What is being said in detail, and how? (3) Is the text true, in whole or part? and (4) What of it? The Exordium begins to answer the first question because it introduces the whole text both orally and in writing (although the oral presentation may be limited in the case of longer works). It is in the Interpretation stage of Column One instruction that the teacher didactically begins to thoroughly unlock the second question. The third and fourth questions are relevant to Interpretation, but question two receives most of the teacher's attention. It is only when the student begins Column Two activities that a fuller grasp of questions three and four begins to mature in the student's mind. Consistent with the purpose of Column One instruction, the teacher is simply introducing elements of proper interpretation for the student to build on during Column Two and Column Three learning.

These questions can be appropriately modified for analysis of a work in science, mathematics, or the fine arts. Analysis of these types of works proceeds in a manner analagous to that of a text. Inasmuch as imaginative literature can be considered a work of fine art, the analogies drawn in Adler's and Van Doren's How to Read a Book can serve as a guide to other types of works too.

Interpretation of  or
In textual interpretation, the teacher didactically presents the author's key terms in the stage of Interpretation of Words and Terms. In addition, the teacher anticipates unfamiliar general vocabulary. These words and terms are given didactically, either orally or in writing, or both, but the task of interpreting is a Column Two skill of learning that the teacher must coach. Consequently, the purpose of this Column One stage is merely to point out key terms and potentially unfamiliar terms. The student must learn how to interpret. Again, this planning can be adapted to other kinds of works.

Interpretation of Key Terms
The key terms are those few words or phrases the author uses in unique, special, or important ways. . In the Column Two stage, students are coached both to find these key terms and to "unlock" them on their own. This skill is essential to analytical reading. In this Column One stage, however, the teacher points out a list of such terms. Students should understand that this listing is not necessarily exhaustive. In addition, the teacher provides students with a handful of these terms worked out in detail for consideration; it should consist of about three to five terms.
 *
 *
 *

General Vocabulary
The teacher may anticipate general unfamiliar vocabulary and point out the more difficult words either orally or in writing or both. However, defining words using context, a dictionary, or a knowlegable friend or adult (including the teacher) is a Column Two skill that must be coached. Teachers must help students build good habits of knowing the meanings of words. Dictionaries, in both electronic and book form must be available, and students must be taught how to use them. They must also be taught how to ask for definitions--a perfectly acceptible life skill commonly used by demanding readers!

 *
 *
 *

Note: If the teacher points out words in anticipation of their potential difficulty, this should be done in context by giving citations or electronically highlighting the words.

Interpretation of Sentences
One point of didactic interpretation at the level of interpreting sentences is to use grammar to get at the author's meaning. As with interpreting words and terms, this level of interpretation is also a Column Two skill that must be coached. Consequently, at the Column One didactic level, a teacher should choose a handful of the most difficult sentences in the text for demonstration. The teacher will always unlock the grammar of a few important and more difficult sentences for students independent of whether these sentences are key premeses to an argument. As appropriate to the text, a teacher should also consider demonstrating the grammar of sentences that work together as propositions in the author's most important arguments.

Another important point is to demonstrate the meter and prosody in both poetry and prose texts. This aim is in great danger of being completely overlooked or forgotten in a world where oral reading is not nearly as common as it once was. Nevertheless, great speechs often succeed in part because the author understands prosody. Lincoln's Gettysburg Address and King's I Have a Dream serve as striking examples. The teacher should select sentences or versus to demonstrate both meter and rhythm.

 * :
 * : <etc.>
 * <Third Sentence>: <explanation of meter and rhythm, consideration of overall prosody>
 * <Fourth Sentence>: <etc.>

Interpretation of Passages
At the level of passages, the full trivium (grammar, logic, and rhetoric) come into play. First, a teacher should choose one or two of the author's most important arguments for a demonstration of how to use logic as a key to interpreting a text. Next, the teacher should select several passages to demonstrate how they conform (or not) to rhetorical, poetical, and sylistic rules. These rules, of course, must be didactically taught as prerequisites to interpretation of texts.

 * <Argument>: <demonstration of both the grammar of the sentences comprising premeses and of the concluding proposition along with the logic of the argument followed by an explanation of how the arguments help to unravel the author's meaning>
 * <Passage Illustrating Rhetorical Rules>: <demonstration of the rhetorical rules and their value in the success of the author's purpose>
 * <Passage Illustrating Poetical Rules>: <demonstration of the poetical rules and their value in the success of the author's purpose>
 * <Passage Illustration Stylistic Rules>: <demonstration of the stylistic rules and their value in the success of the author's purpose>

Erudition
Erudition refers to all manner of background information assumed by or necessary to understanding or fully appreciating the text.

<Note that these categories are presented alphabetically for ease of organization>

Anthropological References

 * <information relevant to anthropological issues and discoveries--include citations--delete if not used>

Antiquarian References

 * <information about references to things no longer commonly understood because of their situation in ages past--include citations--delete if not used>

Archaeological References

 * <information about details related to archaeological discoveries--include citations--delete if not used>

Biographical References

 * <information about the author--always include>
 * <information about other textual references to important people--include citations--delete if not used>

Cultural References

 * <information and links to cultural references--include citations--delete if not used>

Ethical References

 * <information and links to ethical references--include citations--delete if not used>

Geographical References

 * <links to relevant maps--delete if not used>
 * <information relevant to geographical references--include citations--delete if not used>

Historical References

 * <information and links to historical references--include citations--delete if not used>

Mathematical References

 * <information and links to mathematical references--include citations--delete if not used>

Political References

 * <information and links to political references--include citations--delete if not used>

Religious References

 * <information and links to religious references--include citations--delete if not used>

Scientific References

 * <information and links to scientific references--include citations--delete if not used>

Column Two
Column Two teaching consists of coaching. This mode of instruction aims at helping students to form habitual skills in the language, scientific, and fine arts. Thus, a teacher must correct students as they practice listening, speaking, reading, writing, observing with the senses unaided, observing with the aid of scientific apparatus, measuring, estimating, calculating, and exercising dexterity in the musical and visual arts. Each of these arts in turn rely upon the aquisition of fine and gross motor, imagination, and memory skills. These rules for developing Paideia Unit Plans address teachers. For their counterpart written for students, see Paideia Learning Plan.

Column Two learning comprises 65% to 75% of scheduled learning time. It's chief charactaristic is student activity. Students must be practicing some skill or skills while the teacher corrects him or her. While athletic coaching is an obvious example of this type of instruction, debate coaching, directing a drama, art instruction, and piano lessons also represent coaching. In order to coach well, a teacher must have a repertoire of activities carefully designed to exercised desired skills. Additionally, the teacher must have a clear idea of how to correct the skills as students practice them to ensure their habitual formation.

Mnuemonic Skills
<consider this category for every unit>

Listening Skills

 * <links to additional quality recordings--include citations--delete if not used>
 * <information relevant to listening for grammatical, logical, rhetorical, poetical, and stylistic elements of the work--include citations--delete if not used>

Speaking Skills

 * <imitation of quality oral readings--should be a part of every unit>
 * <recitations from memory--should be some part of every unit>
 * <oral presentations of written work--should be a part of most units>
 * <class discussions--should be a part of every unit>
 * <recitation of the rules of reading from Adler's and Van Doren's How to Read a Book or similar rules>

Reading Skills

 * <read hard-copy with a pencil in hand--always!>
 * <marking up a text both with a pencil and electronically>
 * <inspectional reading notes as taught in Adler's and Van Doren's How to Read a Book>
 * <analytical reading notes as taught in the same work>

Writing Skills

 * <Choose phrases from text and express them in different ways>
 * <Reconstruct a previously disarranged passage from the text>
 * <Compose verses or lines in imitation of the author>
 * <Change a passage or poem of one kind into another kind>
 * <Imitate a passage>
 * <Write a composition imitating the author>
 * <Translate sentences or passages into Latin or another language>

title: Paideia High School/Ave verum corpus


title: File:Fe1.s11.team1.HW5.fig2.3.jpg


title: File:Fe1.s11.team1.HW5.fig2.4.jpg


text:
File:HW521a.tif

= Summary=

title: File:HW521a.tif


text:
Test and treat

HIV: Increased HAART Coverage Associated With 50 Percent Drop in New Infections

'start HAART as soon as found to be infected' - SF Endorses New Policy for Treatment of H.I.V.

Start HAART Immediately Upon Diagnosis - San Francisco To Unveil New Public Health Policy for HIV/AIDS Treatment

Test and Treat Model: Feasibility for Implementation in Africa

Truvada cuts HIV risk significantly in S.F. study

San Francisco Endorses New HIV Strategy: Begin HAART Therapy Immediately

title: Test and treat


text:
User:Egm6341.s11.team5.reiss/HW5

= Problem 5.4- Mass, spring, damper analysis = From [[media:Nm1.s11.mtg29.djvu|Mtg 29-1]]

Given
h=.02

Find
1) Determine the equation of motion of the system in terms of d, c, k, m and, u 2)$$\mathbf{x} = \left\{ {\begin{array}{*{20}{c}} d\\ {\dot d} \end{array}} \right\} = \left\{ {\begin{array}{*{20}{c}} \\

\end{array}} \right\}$$, Find F and G 3)Find $$c_{cr}$$ in terms of k and m such that the system is critically damped 4)Let k=1, m=0.5, $$\mathbf{x_{0}}={\left\lfloor {\begin{array}{*{20}{c}} {0.8}&{ - 0.4} \end{array}} \right\rfloor ^T}$$ a)For u=0, plot $$\mathbf{x_{k}}$$ for $$c = \frac{1}{2}{c_{cr}},{c_{cr}},\frac{3}{2}{c_{cr}}$$ b)For u=0.5 gaussian noise and $$c=\frac{3}{2}{c_{cr}}$$, plot $$\mathbf{x_{k}}$$ c)For u=0.5 cauchy noise and $$c=\frac{3}{2}{c_{cr}}$$, plot $$\mathbf{x_{k}}$$

5.4.1: Equation of Motion
The force of the spring is given by

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

$$ $$
 * $$F_s=-kx
 * $$F_s=-kx
 * $$\displaystyle (4-1)
 * }
 * }

The force of the damper is given by

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

$$ $$
 * $$F_d=-c\dot x
 * $$F_d=-c\dot x
 * $$\displaystyle (4-2)
 * }
 * }

The force of inertia is given by

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

$$ $$ The external force on the system is $$F_e=u$$ The total force on the system, the force balance, is given by
 * $$F_I=m\ddot x
 * $$F_I=m\ddot x
 * $$\displaystyle (4-3)
 * }
 * }

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

$$ $$ Substituting,
 * $$F_I=F_c+F_s+F_e
 * $$F_I=F_c+F_s+F_e
 * $$\displaystyle (4-4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$m\ddot x=-c\dot x-kx+u
 * $$m\ddot x=-c\dot x-kx+u
 * $$\displaystyle (4-5)
 * }
 * }

Rearranging we see that the equation of motion is given by
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\ddot x=-\frac{c}{m}\dot x-\frac{k}{m}x+\frac{1}{m}u
 * $$\ddot x=-\frac{c}{m}\dot x-\frac{k}{m}x+\frac{1}{m}u
 * $$\displaystyle (4-6)
 * }
 * }

5.4.2: Find F and G
We can now designate states such that

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

{x^1} = d\\ {x^2} = \dot d \end{array} $$ $$ Then
 * $$\begin{array}{l}
 * $$\begin{array}{l}
 * $$\displaystyle (4-7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\dot x^1\\ {\dot x^2} \end{array}} \right ) = \left ( {\begin{array}{*{20}{c}} \\
 * $$\dot \mathbf{x}=\left ( {\begin{array}{*{20}{c}}
 * $$\dot \mathbf{x}=\left ( {\begin{array}{*{20}{c}}

\end{array}} \right ) $$ $$ Putting this into matrix form
 * $$\displaystyle (4-8)
 * }
 * }

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

0&1\\ { - \frac{k}{m}}&{ - \frac{c}{m}} \end{array}} \right) \left( {\begin{array}{*{20}{c}} \\
 * $$\dot x = \left( {\begin{array}{*{20}{c}}
 * $$\dot x = \left( {\begin{array}{*{20}{c}}

\end{array}} \right) + \left( {\begin{array}{*{20}{c}} 0\\ {\frac{1}{m}} \end{array}} \right)u $$ $$
 * $$\displaystyle (4-9)
 * }
 * }

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

$$ $$ Where
 * $$\dot \mathbf{x}= \mathbf{A} \mathbf{x} + \mathbf{B}u
 * $$\dot \mathbf{x}= \mathbf{A} \mathbf{x} + \mathbf{B}u
 * $$\displaystyle (4-10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

0&1\\ { - \frac{k}{m}}&{ - \frac{c}{m}} \end{array}} \right) $$
 * $$\mathbf{A}=\left( {\begin{array}{*{20}{c}}
 * $$\mathbf{A}=\left( {\begin{array}{*{20}{c}}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

0\\ {\frac{1}{m}} \end{array}} \right) $$
 * $$\mathbf{B}=\left( {\begin{array}{*{20}{c}}
 * $$\mathbf{B}=\left( {\begin{array}{*{20}{c}}
 * }
 * }

According to equations (1)-(4) of slide 29-7 equation 4-10 can be written as
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$ \mathbf{x_{k+1}}= \underbrace{\left[ \mathbf{I} + h \mathbf{A} \right]}_{\color{blue}\mathbf{F}} \mathbf{x_{k}} + \underbrace{h \mathbf{B}}_{\color{blue}\mathbf{G}}u
 * $$ \mathbf{x_{k+1}}= \underbrace{\left[ \mathbf{I} + h \mathbf{A} \right]}_{\color{blue}\mathbf{F}} \mathbf{x_{k}} + \underbrace{h \mathbf{B}}_{\color{blue}\mathbf{G}}u
 * $$\displaystyle (4-11)
 * }
 * }

So therefore

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

$$ $$
 * $$F = \left[ \mathbf{I} + h \mathbf{A} \right]
 * $$F = \left[ \mathbf{I} + h \mathbf{A} \right]
 * $$\displaystyle (4-12)
 * }
 * }

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

$$ $$ where I is the identity matrix, and A, B, and h are defined above
 * $$G = h \mathbf{B}
 * $$G = h \mathbf{B}
 * $$\displaystyle (4-13)
 * }
 * }

5.4.3: Finding $$c_{cr}$$
Critical damping occurs when the damping ratio is equal to 1. For systems like this one the damping ratio, according to Wikipedia, is given as

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

$$ $$
 * $$\varsigma = \frac{c}
 * $$\varsigma = \frac{c}
 * $$\displaystyle (4-14)
 * }
 * }

Therefore to determine c such that critical damping occurs we simply solve c in terms of m and k, setting the damping ratio equal to one. If we do this we will end up with

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

1 = \frac\\ {c_{cr}} = 2\sqrt {mk} \end{array} $$ $$
 * $$\begin{array}{l}
 * $$\begin{array}{l}
 * $$\displaystyle (4-15)
 * }
 * }

5.4.4: Plotting
Using the given values for the constants we are able to use the definitions of A and B along with equations (4-12) and (4-13) to find the values of the F and G matrices. Substituting we get,

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

F &= \left[ \mathbf{I} + h \mathbf{A} \right]\\ &= \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right) + .02\left( {\begin{array}{*{20}{c}} 0&1\\ { - 2}&{2c} \end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}} 1&.02\\ {-0.04}&{1.04c}\end{array}}\right)\end{align}$$ $$
 * $$\begin{align}
 * $$\begin{align}
 * $$\displaystyle (4-16)
 * }
 * }

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

0\\ {\frac{1}{m}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ {0.04} \end{array}} \right) $$ $$
 * $$G = .02\left( {\begin{array}{*{20}{c}}
 * $$G = .02\left( {\begin{array}{*{20}{c}}
 *
$$\displaystyle (4-17)
 * }
 * }

5.4.4a:$$c=0.5c_{cr}$$
Taking into account $$c = \frac{1}{2}{c_{cr}},{c_{cr}},\frac{3}{2}{c_{cr}}$$ and no input force the equation becomes, respectively,

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

1&.02\\ {-0.04}&{0.7354}\end{array}}\right) \mathbf{x_{k}} $$ $$
 * $$\mathbf{x_{k+1}}=\left( {\begin{array}{*{20}{c}}
 * $$\mathbf{x_{k+1}}=\left( {\begin{array}{*{20}{c}}
 *
$$\displaystyle (4-18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

1&.02\\ {-0.04}&{1.4708}\end{array}}\right) \mathbf{x_{k}} $$ $$
 * $$\mathbf{x_{k+1}}=\left( {\begin{array}{*{20}{c}}
 * $$\mathbf{x_{k+1}}=\left( {\begin{array}{*{20}{c}}
 *
$$\displaystyle (4-19)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

1&.02\\ {-0.04}&{2.206}\end{array}}\right) \mathbf{x_{k}} $$ $$
 * $$\mathbf{x_{k+1}}=\left( {\begin{array}{*{20}{c}}
 * $$\mathbf{x_{k+1}}=\left( {\begin{array}{*{20}{c}}
 *
$$\displaystyle (4-20)
 * }
 * }

The results here are as we would expect. For the under-damped case, $$c=0.5c_{cr}$$ there is overshoot in both the position and the velocity graphs. Overshoot meaning that the value went past the final value and oscillated around the steady state value for a period of time. The critically damped case, $$c=c_{cr}$$ quickly rose to the steady state value and did not overshoot. And the over-damped system, $$c=1.5c_{cr}$$, did not overshoot but rose less quickly than the critically damped case.

5.4.4b: Gaussian noise
Using Gaussian noise as the forcing function for this system and using the over damped value for c the plot of x becomes

Notice that because of the Gaussian noise the system does not come to a steady state value as in the previous example. The system continuously fluctuates, though as time progresses the system fluctuates closely around the steady state values show in figure 1. The Gaussian noise causes instability in the system.

5.4.4c: Cauchy noise
Using Cauchy noise as the forcing function for this system and again using the over damped value for c, the plot of x becomes

Comparing the system's reaction to Cauchy noise and Gaussian noise we can see that the oscillations due to Cauchy noise are much more pronounced than the Gaussian oscillations. This is due to the heavy tails of the Cauchy distribution. The heavy tails cause the system to react to much greater changes in input, which causes the large magnitude spikes seen in figure 3.

Author and proof-reader
[Author] Reiss

[Proof-reader]

title: User:Egm6341.s11.team5.reiss/HW5


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text:
User:Eml5526.s11.team6/hwk5

= Problem 5.1 Solution of the PDE using weak form.=

Strong Form

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

$$ \frac{\partial}{\partial x}\left [ (2+3x)\frac{\partial u}{\partial x} \right ]+5x=0 \qquad \forall  x   \epsilon \left ]0,1   \right [$$ (5.1.1)
 * }
 * }

Boundary Conditions

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

$$ u(0)=4, \qquad -\frac{\partial u}{\partial x}_{(x=1)}=6 $$
 *  (5.1.2)
 * }
 * }

Solve for $$ u(x) $$ and $$ u^h(x) $$
To solve the above problem for $$ u(x) $$ and $$ u^h(x) $$ using appropriate basis functions of the following types until convergence of u^h(0.5) -- 1. Fourier basis function 2. Polynomial basis function 3. Exponential basis function

Introducing the arbitrary weighted function
Weak form can be found as follows, with weighted function

Now, multiply eq. 5.1.1 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in eq. 5.1.4

Using eq. (5.1.2) and eq. (5.1.3), we get

Put eq. (5.1.6) in eq. (5.1.5), we get the weak form as follows

Using the polynomial basis function
For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=0$$ for the constraint breaking solution. Therefore, From eq. 5.1.2 and eq. 5.1.10, we can say that Therefore, eq. 5.1.10 becomes Similarly, the weighted function can be written as Eq. 5.1.3 gives, Therefore eq. 5.1.13 becomes Hence, from eq. 5.2.12 and 5.2.15 Put this in eq. 5.1.7, we get Using eq. 5.1.15 we can find $$ \displaystyle w(1) $$ and put in eq. 5.1.18, we get By integrating and verifying with| Wolfram Alpha, we get The following figure shows comparison of analytical and numerical solution for different values of n. As expected, the numerical solution approaches analytical solution as n increases.

Approximate solution converges to analytical solution at n=4. To minimize the error,higher values of n need to be considered.

Using the exponential basis function
For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=0$$ for the constraint breaking solution. Therefore, From eq. 5.1.2 and eq. 5.1.25, we can say that Therefore, eq. 5.1.25 becomes Similarly, the weighted function can be written as Eq. 5.1.3 gives, Therefore eq. 5.1.28 becomes Hence, from eq. 5.2.27 and 5.2.30 Put this in eq. 5.1.7, we get Using eq. 5.1.30 we can find $$ \displaystyle w(1) $$ and put in eq. 5.1.33, we get By integrating and verifying with| Wolfram Alpha, we get

Following Matlab code is used to find coefficients a_j by changing values of i and j in the above equation After getting the coefficients of numerical solution, we compare numerical solution with analytical solution for different values of n.

Then graph of error that is difference between analytical solution and numerical solution is plotted at x=0.5

As expected, with increase in number of n,numerical solution approaches analytical solution.

From the above graph, error is almost zero for n=6 and above. But to get the error in the order of less than $$10^{-6}$$,we solve it further for higher values of n.

Using the Fourier basis function
The trial solution then becomes the following for n=1 to convergence:

The essential boundary condition,$$u(x=0)=4$$ must be satisfied, and the basis functions must be selected so $$ w(n=0) = 0 $$. Solving the above basis function for u(x=1)=4 yielded the following: $$ {d}_{1}=4, i=1,2,\cdots $$, $$ j=1,2,\cdots $$

The remaining displacement coefficients must be solved for $$d_{j}, \;\;j=1,2,\cdots $$, by constructing the following stiffness and force matrices using the switching Fourier basis function.

Matlab was used to generate the following stiffness and force matrices shown below.

Since Kd=F, d can be solved for using Matlab to perform the matrix algebra. The resulting d matrix is shown below.

The trial solution of the governing equation then becomes the following.

Now we plot for Exact solution u(x) and Approximate solution uh(x) and also Convergence of error vs. number of degrees of freedom (n)

As seen from the Figure 5.1.1 below, it is evident that the approximate solution uh(x) very closely imitates the exact solution u(x) Also it can be seen from Figure 5.1.2. that the error converges to almost zero value when more than 8 basis functions are used

=Problem 5.2 - Solutions using weak form and appropriate families of basis functions=

Strong Form
The strong form as referred from General one dimensional Model 1.0/Data set 1b on page 2 of Mtg 9:-

Boundary Conditions
The Essential Boundary condition as applied on the boundary $$\Gamma_g = [1]$$, is -- $$ u(1)=4 $$, The Natural Boundary condition as applied on the boundary $$\Gamma_h = [0]$$, is -- $$ -\frac{\partial u(x=0)}{\partial x}=6 $$

Solve for $$ u(x) $$ and $$ u^h(x) $$
To solve the above problem for $$ u(x) $$ and $$ u^h(x) $$ using appropriate basis functions of the following types until convergence of $$u^h(0.5)$$ to the order of $$10^{-6}$$ -- 1. Fourier basis function 2. Polynomial basis function 3. Exponential basis function

Solution
Using the technique of integration by parts, we have derived the weak form as follows --

We make following choice for the arbitrary weighted function for our convenience -

Now, multiply Eq. 5.2.2 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in Eq. 5.2.3

Using Eq. (5.2.3) and Eq. (Eq. 5.2.4), we get

From Eq. (5.2.4) in Eq. (5.2.5), we get the weak form as follows -

Now, using the technique from Mtg 18 and Mtg 19 we can write the discretized weak form as -

We choose the basis functions such that $$ w(n=0) = 1 $$, to eliminate the unknown flux at this boundary. Using technique from page -1 Mtg 22, we can write - $$ {{d}_{0}}=(\frac {g}{b_0}) = (\frac {4}{1})= 4, $$ In order to plot the exact solution u(x), solving the above ordinary differential $$ Eq. 5.2.1 $$, we get the exact solution of governing equation as follows -

Now we select legitimate families of basis functions as follows -

Family of Fourier basis function
Now we can write the Approximated trial solution $$ u^h (x) $$ as follows -

Here we have -  [ j = 0,1,2,3,4...]

Now we construct the stiffness matrix as follows:

Note- Here 1st equation is valid when 'i' and 'j' both are even, 2nd expression is valid when 'i' and 'j' both are is odd, 3rd expression is valid for remaining combination of values of 'i' and 'j'

Similarly, the force matrix can be shown as below-

Note- Here 1st expression is valid when 'i' and 'j' both are even, 2nd expression is valid when 'i' and 'j' both are odd

The stiffness matrix and Force matrix are used to solve for the remaining coefficients $${{d}_{j}},_ – ^ – j=1,2,\cdots $$

Also now we need to have convergence of the solutions to the order of 10^(-6), hence we use a Matlab code demonstrated below to build the above matrices which are obtained then as follows-

By the equation that$$Kd=F$$, we can solve the remaining of $$d$$ as

Hence, the trial or approximated solution of the governing equation from $$ Eq. 5.2.1 $$ is-

Now we plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$ and also Convergence of error vs. number of degrees of freedom (n)

As seen from the Figure 5.2.1 below, it is evident that the approximate solution $$ u^h(x) $$ very closely imitates the exact solution $$ u(x) $$ Also it can be seen from Figure 5.2.2. that the error very rapidly converges to almost zero value when 5 or 6 basis functions are used

Introducing the arbitrary weighted function
Weak form can be found as follows, with weighted function

Now, multiply eq. 5.2.1 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in eq. 5.2.14

Using eq. (5.2.2) and eq. (5.2.13), we get

Put eq. (5.2.16) in eq. (5.2.15), we get the weak form as follows

Using the polynomial basis function

For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=1$$ for the constraint breaking solution. Therefore, From eq. 5.2.2 and eq. 5.2.20, we can say that Therefore, eq. 5.2.20 becomes Similarly, the weighted function can be written as Eq. 5.2.13 gives, Therefore eq. 5.2.23 becomes Hence, from eq. 5.2.22 and 5.2.25 Put this in eq. 5.2.17, we get Using eq. 5.2.25 we can find $$ \displaystyle w(0) $$ and put in eq. 5.2.28, we get By integrating and verifying with| Wolfram Alpha, we get Following Matlab code is used to find coefficients a_j by changing values of i and j in the above equation The figure shows comparison of analytical and numerical solution for polynomial function. At n=5,numerical solution approaches analytical solution.Error decreases below $$ 10^{-6}$$ for n=10

Exponential basis function
Using the exponential basis function

For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=1$$ for the constraint breaking solution. Therefore, From eq. 5.2.13 and eq. 5.2.35, we can say that Therefore, eq. 5.2.35 becomes Similarly, the weighted function can be written as Eq. 5.1.3 gives, Therefore eq. 5.2.38 becomes Hence, from eq. 5.2.37 and 5.2.40 Put this in eq. 5.2.17, we get Using eq. 5.2.40 we can find $$ \displaystyle w(1) $$ and put in eq. 5.1.18, we get By integrating and verifying with| Wolfram Alpha1 ,| Wolfram Alpha2,| Wolfram Alpha3 we get

Following Matlab code is used to find coefficients a_j by changing values of i and j in the above equation After getting the coefficients of numerical solution, we compare numerical solution with analytical solution for different values of n.

Then graph of error that is difference between analytical solution and numerical solution is plotted at x=0.5

As expected, with increase in number of n,numerical solution approaches analytical solution.

From the above graph, error is almost zero for n=6 and above. But to get the error in the order of less than $$10^{-6}$$,we solve it further for higher values of n.

=Problem 5.3 Using Lagrange Interpolation Basis Functions to Approximate U(x)=

Given
Refer to lecture slide [[media:Fe1.s11.mtg29.djvu|29-5]] for problem assignment. Given the following Strong Form of a PDE:

Find
1) Solve (Eq. 5.3.1) using the weak form with Lagrange Interpolation Basis Functions (LIBF), equidistant nodes, and m = 4, 6, 8… until convergence of $$ u^{h} \left (0.5 \right ) $$ to $$ O \left ( 10^{-6} \right ) $$.

2) Explain how LIBF are used as Constraint Breaking Solutions.

3) Plot all LIBF used.

4) Plot $$ u \left ( x \right ) $$, $$ u(x)^{h} $$, and convergence (error vs. m)

Solving the PDE using the Exact Solution
First we must obtain the exact solution of (Eq. 5.3.1) in order to determine the error of our approximated function. To determine the exact solution, we take the integral of (Eq. 5.3.1.a) to obtain the following equation

Next, we rearrange (Eq. 5.3.2) and apply the natural boundary condition (Eq. 5.3.1.b) in order to solve for the constant A_1.

Therefore,

Substitute $$ A_1 $$ back into (Eq. 5.3.2), rearrange to isolate the term $$ \frac{du}{dx} $$ and then integrate in order to determine $$ u_{} $$.

Now we substitute the essential boundary condition (Eq. 5.3.1.c) into (Eq. 5.3.5) to solve for the constant $$ A_2 $$.

Solving for $$ A_2 $$ yields

Finally, we substitute (Eq. 5.3.7) into (Eq. 5.3.5) to find the exact solution for $$ \displaystyle u $$.

Determining the Weak Form
Now that we have our exact solution, we can begin to derive our approximated function. The approximated function is found by first obtaining the weak form of (Eq. 5.3.1).

First we multiply the governing partial differential equation (Eq. 5.3.1.a) and the natural boundary condition (Eq. 5.3.1.b) by an arbitrary weight function and integrate over the x domain.

Additionally, the weight function must disappear at the point where the essential boundary condition is applied, therefore,

Using integration by parts on the first term in (Eq. 5.3.9) yields

Substitute (Eq. 5.3.11) into (Eq. 5.3.9) and apply (Eq. 5.3.10) to obtain

Finally, we substitute in the natural boundary condition (Eq. 5.3.1.b) into (Eq. 5.3.12) and rearrange the terms to obtain the weak form of the PDE given in (Eq. 5.3.1)

Approximating U and W using LIBF
Our solution $$ u \left ( x \right ) $$ and our weight functions can be approximated by the following equations:

where $$ c_{i} $$ and $$ d_{j} $$ are constants with respect to x and $$ b_{i} $$ and $$ b_{j} $$ are the LIBF. To approximate our solution u, we divide our x domain into m subdivisions. Additionally, we will use m LIBF to approximate our solution. These LIBF will act as a shape function so that each LIBF will equal 1 at one specific point x, and will equal 0 at all other points, i.e.,

Therefore, for m=4, we take the first 4 Lagrange Interpolation Basis Functions, or more specifically,

Additionally, the weight function must disappear at the essential boundary condition, therefore,

Also, the essential boundary condition (Eq. 5.3.1.c) can be substituted into (Eq. 5.3.14.a) to obtain

Next, we expand (Eq. 5.3.14) into 4 separate terms, i.e.,

Substitute the LIBF into (Eq. 5.3.22) and then take the derivative with respect to x to obtain the following:

Finally, we substitute (Eq. 5.3.20) through (Eq. 5.3.23) into (Eq. 5.3.13). The result can be separated into separate terms multiplied by $$ c_{i} $$.

Equation $$\displaystyle (Eq. 5.3.24)$$ must hold for any arbitrary weight function $$w(x)$$, so long as it satisfies the homogeneous essential boundary condition. It then also follows that it must also hold for any arbitrary $$c_2$$, $$c_3$$, and $$c_4$$, leading to the following algebraic equation:

Pre-multiplying both sides of $$\displaystyle (Eq. 5.3.25)$$ by the inverse of the coefficient matrix on the left side yields:

(Eq. 5.3.28) is combined with (Eq. 5.3.21) and the LIBF to create our approximation for the solution to the PDE, $$ u \left ( x \right ) $$, i.e.,

The same process can be repeated for m=6 and m=8. Results of these further iterations are shown in the sections below.

Explain how LIBF are used as Constraint Breaking Solutions
A Constraint Breaking Solution uses basis functions that abide by the following property,

$$ b_{1}(\beta)\neq 0 $$ and $$ b_{i}(\beta)=0 \quad\text{for}\quad \; i=2,3,...,n $$

where $$ \beta $$ is the location where the essential boundary condition is dictated ( $$ \beta = 0 $$ in our instance).

Upon closer analysis of LIBF, refer to (Eq. 5.3.16-19) for example, it becomes clear that these basis functions abide by this rule. For instance, for basis function $$ b_{1}$$ taken at the point where the essential boundary condition is defined, $$ b_{1} = L_{1,4} \left ( x=0 \right ) = 1 $$. Furthermore, all other LIBF $$ \left ( L_{2,4}, L_{3,4}, \quad\text{and}\quad L_{4,4} \right ) $$ are equal to zero at $$ x=0 $$ because $$ x_{1}=0 $$.

Therefore, substituting physical values of x into our LIBF illustrates that LIBF can be used as Constraint Breaking Solutions.

Plot all LIBF used

Plot u(x), u(x)^{h}, and convergence (error vs. m)

Notice that the approximations and the exact solution are very similar. The differences in the equations are too small for this figure to display.

The graph of the error vs m helps us understand the convergence of our approximated functions as m increases in value. The more exact values of error are:

m=4: 0.0158

m=6: -5.7975e-004

m=8: 2.1571e-005

Thus, we have convergence of our solution on the order of 10^-5 using m=8.

Matlab Code
=Problem 5.4 Using LIBF with equidistant nodes to solve G1DM1.0/D1=

GIVEN
One dimensional Model 1.0/Data set 1.0 on page 2 of Mtg 9:

where $${a_2}\left( x \right){\text{ =  2 + 3x}}$$,$$f\left( {x,t} \right) = 5x $$

1-D LIBF on page 3 of Mtg 26:

FIND
A) Explain how LIBF are used as CBS

B) Plot all LIBF used

C) Use matlab to integrate and solve the weak form

D) Plot $$ u_m^h\ vs\ u$$ and $$ u_m^h\left(0.5 \right)-u\left(0.5 \right)\ vs\ m$$

SOLUTION
A) Take m=4 for example to explain how LIBF works as CBS.

When m=4, we can write the basis function according to Eq. 5.4.4 (N was denoted as basis function from now on):

According to Eq.5.4.2, we should have CBS when x=1, which means that the last basis function should not equal zero when x=1.

At the same time, when x=1, we have the following values for LIBF based on its own properties:

Since we have the weight function as $$ w\left( x \right) = {c_i}N\left( x \right) $$ and from Eq. 5.4.9 we can choose arbitrary value for c from i=1~3 and make c equal zero for the last term, which absolutely satisfy CBS for which have non-zero value at the essential boundary for the last basis function and equal zero for all the others.

B)-D)Integrate Eq. 5.4.1 by parts over the whole domain, we can have

$$ \left[ {w(x)\left( {(2+3x)\frac} \right)} \right]_{}^{x = 1} - \left[ {w(x)\left( {(2+3x)\frac} \right)} \right]_{x = 0}^{} - \int\limits_0^1 {\frac\left( {(2+3x)\frac} \right)} dx + \int\limits_0^1 {w(x).(2+3x)} dx = 0 $$

Since LIBF satisfies CBS, we have w=0 when x=1, and then the above equation can give us the weak form:

If we use LIBF both as trial solution and weight function, the following matlab codes can solve the weak form for m nodes

The above three images show that whatever the value m equals, LIBF show its property which can be used as CBS, since its values vanish all other nodes except just one node where the natural boundary condition happens.

It is not easy to find out the difference from the approximation solution and exact solution, but the gaps do exist since the error is non-zero.

We can find out when the solution convergence and what is the error at x=0.5. The error is so small which indicates that LIBF, as a global basis function, has advantages solving weak form and obviously can satisfy CBS wherever essential boundary happens.

=Problem 5.5 Calculix Guide (Continued)=

Given
Refer to lecture slide [[media:Fe1.s11.mtg29.djvu|29-7]] for problem assignment. The Calculix Guide was originally started in Homework Problem 4.7.

Find
1) For the disk problem in Homework Problem 4.7, extract the node information (node numbers and coordinates) and the element information (element numbers and element nodes).

2) Generate 3 meshes of the same disk with triangular elements (increase the number of elements).

3) Install ccx, run examples, and write a report for "dummies" (explain commands, screenshots, etc...).

1.Extracting Node and Element Information from Calculix
For this tutorial, we will continue with the source code for a disk given in Homework 4.7.

First, we pre-process this code using the keystroke command F10. The disk is then graphically generated.

Now, we can obtain node information by performing the following tasks:

1. Plot the nodes and their global node number labels by entering the command "plot na all" and pressing enter. This is shown in Figure 5.5.1.

2. Next, we change our cursor into a box that can capture data within it. This is done by defining a set using the command "qadd". Type "qadd" and then press enter. Notice that your cursor changes to a miniature box. This is shown in Figure 5.5.2.

3. Now, we redefine the shape of our data capturing box to a more convenient size by typing "r". This will define the upper left corner of our new data capturing box. Move your cursor to where you would like the lower right corner of your data capturing box and type "r" again. Notice that your data capturing box has changed its size to fit the corners defined by your "r" commands. This is shown in Figure 5.5.3.

4. Finally, to capture the requested node information (the node number and its coordinates) simply mouse over a node of interest. Once the node is within your data capturing box, simply press "n". This command will output the node information to your output screen. This is shown in Figure 5.5.4. Once you are done capturing the node information, press "q" to quit using the data capturing box.

We can obtain element information (element number and element nodes) by performing the following tasks:

1. Plot the elements of our disk by typing in the command "plot ea all" and pressing the enter key. Notice that the disk looks significantly different than when the nodes were plotted. This view highlights the elements used in our model. See Figure 5.5.5.

2. Next you set up your data capturing box. This method is the same method that was utilized when finding the node information (outlined in steps 2 and 3 of the node information section).

3. Finally, to capture the requested element information simply mouse over an element of interest. Once the element is within your data capturing box, simply press "e". This command will output the element information to your output screen. This is shown in Figure 5.5.6. Once you are done capturing the element information, press "q" to quit using the data capturing box.

2.Generate three meshes of same disc using triangular elements
Meshes are redefined using commands:

'elty3'

'elty6'

and also by redefining the points

3.Installation procedure and report for dummies
To install CalculiX for Windows, start by downloading the executable file available at http://www.bconverged.com/download.php.

Before installing CalculiX, we recommend that you create a folder that you will want to use as your workspace for CalculiX files.

This executable file installs both-CGX and CCX on your computer.

The procedure to install can be found at Team 6-Homework 4.7 []

Introduction

The basic procedures and commands can be found at CCX-help menu as shown below:

You can also refer to ccx manual below for further details:

 * 1) CCX Manual []

=Problem 5.6 Quadratic Lagrangian Element Basis Functions=

Find
Plot three figures similar to those of the linear Lagrangian element basis functions..

Solution
The generic equation for the quadratic Lagrangian element basis functions (QLEBF) is presented below in Equation 5.6.1;

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

$$ L_{j,n}=\prod_{k=1}^{n}\frac{(x-x_{k})}{(x_{i}-x_{k})} $$
 * $$\displaystyle (Eq.5.6.1) $$
 * }
 * }

Equation 5.6.1 can be expanded to the following equation to fit each of our cases as shown in Equations 5.6.2-4 below with x1=1, x2=2, and x3=3.

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

$$ L_{1,3}=\frac{(x-x_{2})(x-x_{3})}{(x_{1}-x_{2})(x_{1}-x_{3})} $$
 * $$\displaystyle (Eq.5.6.2) $$
 * }
 * }

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

$$ L_{2,3}=\frac{(x-x_{3})(x-x_{1})}{(x_{2}-x_{3})(x_{2}-x_{1})} $$
 * $$\displaystyle (Eq.5.6.3) $$
 * }
 * }

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

$$ L_{3,3}=\frac{(x-x_{1})(x-x_{2})}{(x_{3}-x_{1})(x_{3}-x_{2})} $$
 * $$\displaystyle (Eq.5.6.4) $$
 * }
 * }

Since the quadratic Lagrangian element basis functions have three nodes, three QLEBFs were used to generate the following plots 5.6.1-4 in MATLAB. The code used to generate these plots is presented below.

Comments
Each of the plotted quadratic Lagrangian element basis functions has three nodes as opposed to the two nodes of the linear basis functions due to the n=3 condition. It is observed from the plots that the sum of the basis functions at each node must be equal to 1 regardless of the number of nodes. This also holds true for the linear Lagrangian element basis functions as well.

=Problem 5.7:Solving a strong form with Linear Lagrangian Element Basis Function (LLEBF)=

Given
The given ordinary differential equation is:

Boundary conditions
The Essential Boundary condition is -- $$ u\left(0\right) = 4$$ The Natural Boundary condition is -- $$\frac{du}{dx}\left(1\right) = -6 $$

Descritization
Here we want to use Uniform Descritization such that Element nodes are equidistant with number of elements as $$ nel = 4, 6, 8, ..$$

LLEBF as Constraint Breaking Solution (CBS)
Consider a basis function $$b_{i}$$ such that we have, $$b_{0}\left(\beta\right) = 1$$, and $$b_{i}\left(\beta\right) = 0; i = 1, 2, ...$$. In such a case as per the discussion in Mtg 20, $$b_{i}$$ becomes a Constraint Breaking Solution (CBS)

LLEBFs (N1 and N2 below) for a 2 node linear element can be used to represent the unknown variable u(x)in a given domain which is written as -

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

$$u\left ( x \right )=u_{1}\frac{x-x_{2}}{x_{1} - x_{2}}+u_{2}\frac{x-x_{1}}{x_{2} - x_{1}} =u_{1}N_{1}+u_{2}N_{2} $$
 * $$\displaystyle (5.7.2) $$
 * }
 * }

Here $$u_{1}, u_{2}$$ are the nodal values of the unknown u(x) at the nodes 1 and 2 respectively.

The Lagrange Polynomials follow the property of Partition of Unity i.e. they have the value of 1 at the respective node and 0 at the others nodes. It means the basis function $$ b_{i} $$ carries value as 1 at the i th node and has a zero value at any other nodes in the domain.

Plot all LLEBF used for $$ nel =4 $$
The straight line plots (common for all elements used with $$ nel =4 $$) irrespective of the elemental sizes for the 2 LLEBFs given in Eq. 5.7.2, following the Partition of Unity model are shown in the Figure 5.7.1. Note- In the following Figure 5.7.1., N[i] and N[i+1] are the nodal basis functions and i's denote the nodes

Solve the system using (nel = 4) and increase the number of elements or $$ nel =4,6,8..$$ until the error at x = 0.5 converges to the order of 10^-6
The weighted residual form with weighting function $$ w (x) $$ is given as:

We develop the corresponding weak form using the technique of integration by parts, with $$ w(0) = 0$$ owing to the fact that the flux is unknown at $$ x = 0 $$. The expression for the weak form is as below -

To use the element stiffness matrices to the global stiffness matrix $$ K $$,

The stiffness matrix is obtained as below -

The force matrix is obtained as below -

We assemble the system using a Matlab code given below-

Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$ along x
It can be observed from the Figure 5.7.2 (Using 4 nodes) and Figure 5.7.3 (Using 6 nodes) that the exact solution is nearly linear and can be very well approximated by the LLEBFs considered with least 4 number of nodes and obviously also when the number of nodes is increased to 6. This is as the approximated solution $$ u^h (x) $$ very closely imitates the curve for the exact solution $$ u (x)$$.

Convergence - Plot of error vs. number of nodes (n)
At x = 0.5, to plot the error ( $$ u (x)$$ - $$ u^h (x) $$ ), we use Eq. 5.7.2 with the interpolation functions.

Even with (n = 4), sufficiently low number of elements, the system gives considerable convergence to an order of to 10^-3. We need more elements for it to further converge to the order of of 10^-6. Figure 5.7.4 depicts the error as a function of the number of elements.

=Problem 5.8: Solving Data Set G1DM1.0/D1 Using Linear Lagrangian Element Basis Functions and Uniform Discretization=

Boundary Conditions
The Essential Boundary condition as applied on the boundary $$\Gamma_g = [1]$$, is

The Natural Boundary condition as applied on the boundary $$\Gamma_h = [0]$$, is

Descritization
Here we want to use Uniform Descritization such that Element nodes are equidistant with number of elements as $$ nel = 4, 6, 8, ..$$

Find
1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costraint Breaking Solution. 2. Plot all LLEBF used for $$ nel =4 $$. 3. Use matlab to numerically integrate and find the solution for $$n_{el}$$ = 2,4,6,8... until the error evaluated at x=0.5 converges to the order of $$10^{-6}$$ 4. Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$. 5. Convergence - Plot of error vs. number of nodes (n).

Linear Lagrangian Element Basis Functions as a Constraint Breaking Solution
Linear Lagrangian element basis functions assume the value of the Kronecker Delta at nodes, i.e,

For this case, where $$\displaystyle \Gamma_g = 1$$, define the terminating node, $$\displaystyle x_n = 1$$ so that all except the last of the LLEBF will be equal to zero when evaluated at this point while simultaneously the final basis function, $$ b_n$$ will be equal to unity, i.e.

This breaks the constraints of the weighting coefficients $$\displaystyle \{c_i,\;\;i = 0,2,...,n-1\}$$ and as such any arbitrary set may be chosen, as simply selecting $$\displaystyle c_n = 0$$ will satisfy the homogeneous essential boundary condition.

Solution of the System
In the element sense the basis functions are composed of linear Lagrange equations of the form

Now creating the element shape function matrix for this linear 2-node element:

Now, substituting $$\displaystyle l^e = x_2^e-x_1^e\;$$ into $$ \displaystyle (Eq.5.8.8)$$

The element shape function derivative matrix for these elements is

Now constructing the element stiffness matrix

The element force matrix is

Finally the global stiffness and force matrices are assembled via equations (5.13) and (5.14) (F&B p. 96)

where $$\displaystyle L^e$$ is the element dependent scatter operator.

Plot for Exact solution $$ u (x)$$ and Approximate solution $$ u^h (x) $$ along x
It can be observed from the Figure 5.8.2 (Using 4 nodes) and Figure 5.8.3 (Using 6 nodes) that the exact solution is nearly linear and can be very well approximated by the LLEBFs considered with least 4 number of nodes and obviously also when the number of nodes is increased to 6. This is as the approximated solution $$ u^h (x) $$ very closely imitates the curve for the exact solution $$ u (x)$$.

Convergence - Plot of error vs. number of nodes (n)
Even with (n = 4), sufficiently low number of elements, the system gives considerable convergence to an order of to 10^-3. We need more elements for it to further converge to the order of of 10^-6. Figure 5.7.4 depicts the error as a function of the number of elements.

=Contributing Members=

Eml5526.s11.team6.joglekar 16:38, 23 March 2011 (UTC)

eml5526.s11.team6.lee 23:45, 23 March 2011 (UTC)

Eml5526.s11.team6.tupsakhare 01:30, 24 March 2011 (UTC)

Eml5526.s11.team6.kurth 02:02, 24 March 2011 (UTC)

Eml5526.s11.team6.deshpande 02:12, 24 March 2011 (UTC)

Eml5526.s11.team6.vork 02:37, 24 March 2011 (UTC)

Eml5526.s11.team6.gravois 03:14, 24 March 2011 (UTC)

title: User:Eml5526.s11.team6/hwk5


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text:
Talk:Complex socio-ecological systems/Resilience-based management

Debriefing Resilience 2011 Conference

Facilitators: Flavia and Xanic


text:
Textual Criticism of the Bible

Textual Criticism of the Bible is the study of the Bible with the intention of detecting errors in the text and determining what the original text may have been.

It proceeds by analysing the oldest manuscripts, published versions and translations to produce lists of variant readings. Critics then try to decide which is the best reading. Where no satisfactory reading can be found, critics may make conjectural emendations. Sometimes, these conjectures have later been found in previously unknown manuscripts, such as the Dead Sea Scrolls.

Causes of Textual Corruption
Before the advent of printing, manuscripts had to be copied by hand so were quite rare and expensive. They rarely included the whole Bible, and were often just single books of the Bible.

Textual corruption usually occurs by accident, but sometimes a copyist would deliberately alter what he considered a corrupt or heretical reading. Since such alterations produce sensible readings, they are hard to spot.

Rules for Deciding on the Best Reading
There are no infallible rules for deciding which of several competing rules are the best, and at the end of the day good critical judgment is needed. However, the following rules are often useful.

Alternatives to Textual Criticism
It may be that it is not necessary to amend the text of the Bible. It did not seem to make sense, hence seemed to need amendment, because we did not understand it. However, advances in other areas may permit us to understand the text better.

Archaeology
Archaeology may help us to understand the historical and social background of a passage, hence reveal its meaning.

Comparative Philology
Examination of languages similar to those of the Bible may suggest new meanings of words. Many English words, such as roe, can have several completely different meanings (a kind of deer, fish eggs), which would make a passage seem corrupt if we only knew one meaning and the other was intended. The same may be true of Hebrew and Greek.

Example: Judges 18:7 seems to say "there was no-one to put them to shame in anything" or "there was no-one to insult them in anything". This assumes that the word kulum means "to humiliate" or "to insult". However, it may be equivalent to the Arabic kallama, "to speak", allowing the more intelligible translation "there was no-one uttering a word".

title: Textual Criticism of the Bible